II. Analytical Solution Methods

(2.1.1)

Ordinary differential equations

II. A N A L Y T I C A L

SOLUTION

631

METHODS

2.1 ORDINARY DIFFERENTIAL EQUATIONS

2.1.1. Direct integration Ordinary differential equations in groundwater flow only occur in steady onedimensional flow, in steady radial-symmetric flow and in more-dimensional and also in non-steady flow if the concerning partial differential equations have been reduced to ordinary equations by means of one or more integral transformations. They are always linear and almost always of the second order and may be homogeneous or non-homogeneous (see Section 2.1.2). For direct integration it is necessary that the differential equation is homogeneous or eventually non-homogeneous, but then with a constant value for the term that makes the equation non-homogeneous. Example 1. One-dimensional steady flow in a leaky aquifer with vertical infiltration. Differential equation: d2(,o dx 2

q9 q -t)~2 K D

- - O,

~2 __ K D c .

General solution" x

x

q)(x) -- Aler + B l e - r + q c

or

(x) + B 2 c o s h (x)~ + q c .

qg(x)-A2sinh ~

The first form is suitable for a semi-infinite field and the second for a finite field. This solution has been obtained by first solving the corresponding homogeneous differential equation d299

q9

dx 2

~2

=0,

and next adding the particular solution qX 2 ~o --

to it.

KD

= qc

632

Analytical solution methods

(2.1.1)

In the case of a semi-infinite field with zero head at x - 0 (see Fig. 1), the boundary values become"

/------]-99 7///////////////////////////~

99(0)- 0

and

d99 1 x = A1 - - e x dx )~

Fig. 1. Vertical infiltration in a leaky aquifer.

d99 (cx)) -- 0, d-~

--

1 x B1 e-r" ~ '

~dx( o o ) -- 0 gives A1 -- 0 while 99(0) -- 0 gives B1 -- - q c . becomes (see 123.54, Part A):

The solution thus

~o - qc(1 - e - ~ ) .

E x a m p l e 2. Radial-symmetric steady flow in a leaky aquifer. Differential equation: d299 1 d99 J dr 2 r dr

99 ---0, I. 2

)2 - K D c .

This is a homogeneous equation and is known as the modified Bessel differential equation of zero order (see Section 3.3.2) with the general solution r

r

99(r) -- AIo(-£) + B K o ( - ~ ) , in which I0 and K0 are modified Bessel functions of zero order and of the first and second kind, respectively.

iih~,I __

Radial flow towards a circular basin the water level of which sinks by an amount h gives a boundary value for the drawdown at infinity of zero and at the border of h. As 10(cx)) = oo, then A = 0 while 99(R) - h - B K0 (~). The solution thus becomes:

~///////////////~99 c

I I

i

KD

4

R

X0( ) x0(r)

Fig. 2. Radial flow towards a circular basin.

~o(r) - h ~ R "

(See Fig. 2 and 223.23, Part A.) E x a m p l e 3. The ordinary differential equations obtained from partial differential equations by means of integral transformations are almost always of the form: d2cpt dx 2

N99t + M -- 0

or

d299t

d-7 q

1 d99t

r dr

N~t + M -- 0,

(2.1.2)

Ordinary differential equations

633

where 99t = transformed head or drawdown and N and M are functions independent of opt or x or r with N > 0. The general solutions then become respectively (see Examples 1 and 2): M (t9t - A e x "/W + Be - x "/~ - 4 - m N

and

(/9 t -

M AIo(r x / N ) + B Ko(r v/-N) 4- - N

2.1.2. Variation of parameters We consider an ordinary differential equation of the second order, which is nonhomogeneous and linear: d2y dx 2

dy

-F f ( x ) - - : - + g ( x ) y -- h ( x ) ,

(1)

assuming that f , g and h are continuous on an open interval I. We shall obtain a particular solution of (1) by using the method of variation of parameters as follows. The corresponding homogeneous equation d2y dx 2

dy

(2)

-F f ( x ) - : - + g ( x ) y -- O,

has a general solution yh (x) on I, which is of the form yh(X) "- ClYl(X) Jr- c2Y2(X)

with Cl and c2 as constants. The method consists in replacing ¢1 and c2 by functions of x: A(x) and B(x) to be determined such that the resulting function yp(X) = A ( x ) y l ( x ) q- B ( x ) y 2 ( x )

(3)

is a particular solution of (1). By differentiating (3) we obtain dyp dA dyl dB dy2 dx = dx y l -+- A--~x + ~-x Y2 + B dx

We assume that we can determine A and B such that dA dB dx Y 1 + -dTx Y2

(4)

o. d Vp

This reduces the expression for ~ dyp = A d y l + B dy2 dx dx dx

to the form:

(5)

634

Analytical solution methods

(2.1.2)

By differentiating this expression, we obtain

d2yp

dA dyl d2yl dB dy2 d2y___~2 dx2 = d--~-d--x-4- a d - ~ -t dx dx l- B dx2"

(6)

By substituting equations (3), (5) and (6) into equation (1) and collecting terms containing A and terms containing B, we readily obtain d2yl

dyl

(d2y2

dy2

A(-d----xZ 4- f--~---xx 4-gyl) 4-B\--d-----xZ 4- f---~-xx -tI

gy2)

dA dyl dB dy2 ~ =h. dx dx dx dx

Since yl and Y2 are solutions of the homogeneous equation (2), this reduces to dA dyl dB dy2 I =h. dx dx dx dx Equation (4) is" dA

dB

dx Y 1 + -d-SY2

o.

This is a system of two linear algebraic equations for the unknown functions dB The solution is obtained by Cramer's rule" and 7~-" A

dy2

Yl

Y2

dyl

dy2

dx

dx

--

Yl-~x-

d__~A dx

dyl --

W,

Y2 d x

the Wronskian of Yl and Y2.

AA--

0

Y2

h(x)

dr2

/

-- -y2h(x),

AB--

Yl dyl -a-Z

0 -- ylh(x). h(x)

We find

dA dx

y2h(x)

- - - ~ W

and

dB dx

ylh(x)

= ~ . W

(7)

Clearly, W :/: 0, because the functions yl and Y2 constitute a fundamental system. By integrating equation (7) we have f y2h(x)

a(x) -- - ] J

W

dx

and

B(x) -- f ylh(x) ~7 dx.

(8)

These integrals exist because h(x) is continuous. By substituting these expressions for A and B into (3), we obtain the desired particular solution of equation (1)" W dx. yp(X) -- --Yl f y2h W dx 4- y2 f ylh

(9)

Ordinary differential equations

(2.1.2)

635

The general solution of equation (1) then becomes" Y--Yh +Yp, which can readily be verified by substituting this function in (1). If the constants of integration in equation (9) are left arbitrary, then equation (9) represents the general solution of equation (1). Then the result can be written as: y -- A ( X ) y l ( x ) + B(x)y2(x) with dA dx

y2h(x) W

'

~

dB dx

ylh(x) W

( 1O)

and dy2 dyl W - - yl-~x- -- Y2 dx

Example 4. In an infinite field the groundwater head initially is a given function of x (see Fig. 3), for a confined aquifer with transmissivity K D and storage coefficient S. Owing to the gravity, the heads will diminish in a fixed manner, whereas the groundwater will move horizontally in a lateral direction until finally all heads will be equal.

KD, S x

Fig. 3. Given initial function in a confined aquifer. The main objective of this initial value problem is the determination of the depletion function, that is the function that shows the variation of the heads with time everywhere in the field. If the initial function is chosen arbitrarily, the boundary value problem can be written down as: 02(t9 __ f12 0(/9 with f12 = S Ox 2 at KD ' q)(x, O) -- f (x),

qg(-oo, t) - 99(00, t) - 0.

Laplace transformation with respect to t gives (see Section 2.2.2-1): d2q3 dx 2

~(-oo,

s) -

C,(oo, s) -

O.

636

(2.1.2)

Analytical solution methods

The differential equation is an ordinary one and non-homogeneous:

d2q3 dx 2

fl 2 S ~) - - - - fl 2 f ( x ) "

The solutions of the homogeneous differential equation e -~x'F. The Wronskian becomes"

dq)2

w - ,~1 ~d-x , ~ 2 ~

dq31 -d--~-

t

In this case h(x) - - f i 2 f ( x ) , dA

dx

=

fl2 f (x)e-l~x'fi

-2fi,~

are

q)l

-

e ~x~ and

q~2 --

e~ X4-i ( - fl ~/7) e - ~X~ - e - ~X'/-i fl V~et~ X~ = - 2 fl ~/7.

so, according to equation (10)" and

dB

dx

=

fl2 f (x)ePX'fi

-2fi~

and A --

2~/7

f ( x ) e -[~x'/-i dx ÷ Cl,

B -- 2~/-s

f (x)e t~x~ dx ÷

C2.

The general solution is q5 -- A(x)q31 ÷ B(x)q52 -- A ( x ) e z x ~ + B ( x ) e -zx'/i. As qS(cx), s) - 0, it follows that A (co) -- 0, and from ~5(-oo, s) - 0 it follows that B(-cx)) must be zero. Suitable expressions for A ( x ) and B ( x ) then become: fl f f e~X°~ B (x) -- 2~/7 ~ f (x0) dx0

A (x) -- 2~/_~

f (xo)e -~x°'F dx0,

fl qS(x, s) -- 2V/7

e~ f (x0)e -~(x° -x),/-i dx0 ÷ ~ -fl~

and f (xo)e -~(x-x°)4~ dxo,

or 99(x, s) -- 2~-s

_

~

~

f+~

f (xo)e -[~lx-x°l'/~ dx0.

This is the solution of the transformed boundary value problem. Inverse Laplace transform leads to the desired solution (see Section 2.2.2-3b, equation (41)) q0(x t ) - - f l f+oo ' 2v'~7 oo f (x0) exp (see 1 12.01 of Part A).

{ fi2(X_X0)2} -

4t

dx0

Ordinary differential equations

(2.1.3)

637

2.1.3. Use of matrix functions for solving problems in multi-layer systems A groundwater flow system will be referred to here as a multi-layer system or multiple-aquifer system, if it consists of more than one aquifer, separated by one or more semi-permeable layers and if it is assumed that in the aquifers only horizontal flow takes place and in the semi-permeable layers only vertical flow. Then the differential equations for each aquifer also contain state variables (heads or drawdowns) of adjacent aquifers. In that case, a simultaneous treatment of the set of differential equations, one for each aquifer, is necessary, which is a condition for applying the solution method with use of matrix functions.

~EQ

Consider the geohydrological scheme of Fig. 4, where a n-layer system consists of y//# ~ / / / / / / / / / / / / / / / / / / / / Cl n aquifers of transmissivity T1 = K1D1, i T2 -- K2 D2 . . . . , Tn = Kn Dn and n + '.~ Q1 qgl T1 = KI D1 t 1 semi-permeable layers with resistances m ~///////////////////// C2 Cl, c2 . . . . , Cn+l in days. It is supposed r that above the cl-layer and below the cn+li > I layer a fixed water level is maintained, I ~///////////////////// which is chosen to be the reference level. Further, the assumptions are made that on~////////,///////////// Ci+l ly horizontal flow takes place in the aquifers and only vertical flow in the semipermeable layers. The amounts of water passing the semi-permeable layers depend I ~//////////////////// cn on the difference in head on both sides of i]",,,~Qn ~Pn Tn = KnDn the layers and on the value of the resistance c, and are assumed to be distributed ~//////////////////// Cn+l equally over the thickness of the aquifers, ~n+a - - 0 thus occurring as separate terms in the Fig. 4. n-layer system with wells. differential equations for the aquifers (see Section 1.4.2-6, equation (37)). These terms, for instance, for aquifer 1 become: ~oo=0

991 - - q90

K1DlCl

991 - - q92 -k-

K1Dlc2

=

(all

@ a12)~Ol - - a12992,

a s qg0 -

0,

and for aquifer i"

qgi -- qgi- 1 + qgi -- qgi+1 = KiDici KiDici+l 1

1

--aii~Oi-1 Jr- (aii -Jr-ai(i+l))qgi -- ai(i+l)qgi+l,

if we put ~ - - aii, Tici+l = ai(i+l), etc. As an example of flow in a multiple-aquifer system, consider groundwater flow in the system, caused by withdrawal of water with a fully penetrating line well

638

Analytical solution methods

(2.1.3)

with screens in all aquifers and discharges Q l, Q2, . . . , Q,,, as given in Fig. 4. The (ordinary) differential equations for this system for steady flow are:

d2~l

1 d~l

dr 2

r dr

d2~2

1 d~2

dr 2

r dr

= (all -4- a12)~l -- a12992, -- -a22~Ol + (a22 -+- a23)q92 -- a23993, (11)

d2~oi

1 dqpi

dr 2

r dr

d2tpn

-- --aii99i-1 + (aii + ai(i+l))qgi -- ai(i+l)qgi+l,

1 dqgn

dr 2

r dr

-- --ann~On-1 -4- (ann + an(n+l))~On,

and the boundary values: qgi(c~) -- 0,

lim r~0

( dqgi] r

--

dr ]

Oi

= -qi

2zr~

(i - 1 2 . . . . n). ' '

Using matrix notation, this system of n simultaneous differential equations can be written in c o m p a c t form: Vr2tp -- Atp,

(d)

¢p(c¢) - 0,

lim r ~ r-+0

- --q'

(12)

in which V2 =

d2

!

dr 2

1 d

,

r dr

--a12 a22 -I- a23

f a l l -+- a12 --a22

0

0

--a23 .

A

m

0

°°°

--aii

aii -4- ai(i+l)

--ai(i+l)

°

o

.

,

q--

O

qn

ann -f- an(n+l) j

0 0

ql q2

~o2 ~o--

--ann

°°°

•

o

o

The system matrix A is a square n x n matrix; tp, q and 0 are column vectors, each with n elements; Atp is also a column vector with n elements.

Ordinary differential equations

(2.1.3)

639

To solve the vector equation vZcp - Acp, we consider the corresponding single differential equation d2q9 dr 2

1 dq) _I-

-- pq),

r dr

with the general solution:

q) -- aKo(r~/-fi) + blo(rv/-p)

(a and b constants).

Now suppose ¢p -- c{aKo(rv/-fi) + blo(r~/-fi)} -- cq) is a solution of equation (12), with Cl c2

Cn

is a column vector with n constants. As V 2~o -- cV rq) 2 - cpq9 , and also length:

(all + al2)Cl

-

a12c2

2

Acqg, we find: Ac - pc, or written at

V r ~o -

-- pCl

-- a22cl -at- (a22 -+ a23)c2 -- a23c3 -- pc2

(13) -- annCn-1

n t-

(ann

+ an(n+l))Cn

-- pCn.

The vector c = 0 is a solution of this set of equations for any value of p. A value of p for which equation (13) has a solution c -¢ 0 is called an eigenvalue of the matrix A. The corresponding solutions c are called eigenvectors of A. To determine the eigenvalues and eigenvectors, we write equation (13) as (all -

+ a12 -

a22cl

p)cl

-

a12c2

-k- ( a 2 2 q-- a 2 3 -

z

p)c2

-

a23c3

0

-- 0

(14)

-- annCn-1 + (ann -k- an(n+l) -- p)Cn

-- O.

This h o m o g e n e o u s set of linear equations has a non-trivial solution if the determinant of the coefficients is zero" all

+

a12 -

ma22

p

--a12

a22 + a23 - p

• • •

0

--a23

D ( p ) -0 -- 0,

--ann

a n n n t- a n ( n + 1) - - P

(~5)

(2.1.3)

640

Analytical solution m e t h o d s

D(p) -- d e t ( A - pl), where I is the n-rowed unit matrix •

0 I

1

• ° •

0

...

0

Ooo

"

m °

0

...

•

1

By developing D(p) we obtain a polynomial of n-th degree in p. So the eigenvalues of the square matrix A are the roots of the equation (15). In general, there are n solutions p l , p2, . . . , p,,. Once the eigenvalues have been calculated, the corresponding eigenvectors can be determined from the system (14). We have thus found that a solution for ~oi of Vr2q~ = A~p is: q~i -- ei {ai

Ko(r~/-~) 4- bi I o ( r x / ~ ) },

where ei is the eigenvector c i of the matrix A, corresponding to the eigenvalue Pi. There are n solutions, so n

~0 -- E ei { ai K o (r ~/~t ) 4- bi lo (r ~PT) }. i=1

With ~o(c~) - 0 ,

all constants bi become zero and so

n

q~ - - E ei ai K o (r x / ~ l ), i=l

or written at length •

991 -- ellal K o ( r ~ - { ) + e21a2Ko(r~/~) 4 - . . . 4- enlanKo(r~/~n), q92 -- el2al Ko(r ~ff{) 4- e22a2Ko(r ~ff2) 4 - " " 4- en2an Ko(r ~/~n), (16)

gOn - elnalKo(r~ff~) + e2na2Ko(r~/~) + ' "

+ ennanKo(r~n).

In matrix notation equation (16) can be written as: (17)

tp - - E F a ,

where the matrix E is composed of the eigenvectors ei as follows" ell

e21

•••

enl

el2

e22

...

en2

E - (elez-..en) •

eln

• •o

e2n

•••

enn

Ordinary differential equations

(2.1.3)

641

and is called the eigenmatrix of A,

0

K0 ( ; x / ~ )

0

Ko(r~) F

--function diagonal matrbc,

m

0

Ko (r x/~-~n)

and al a

m

al

- column vector of constants.

an

That equation (17) is identical to equation (16) can easily be proved by applying the rules for matrix multiplication. The constants vector a can be solved with the help of the boundary conditions for the well-screens"

lim(rd)

m

_q,

r--+0

as follows: Differentiation with respect to r of equation (16) gives" d r d---~cp - - -

e i a i r ~-fi-[ K 1 ( r ~fi-[t ) , i=1

,im(rd.)

r-+O

f/

-- -- E

eiai

--

-Ea

-- -q,

i=1

from which a - E - l q , where E -1 - inverse eigenmatrix of A. The end solution of the problem then becomes with equation (17)" ~o(r) -- E F ( r ) E - l q ,

(18)

which means that, in matrix notation, the solution consists of a function diagonal matrix, premultiplied by the eigenmatrix of the system matrix, post-multiplied by the inverse eigenmatrix, delivering a new matrix and this matrix multiplied by the column vector of the discharges. The diagonal matrix F contains the same function as will be found in solving the problem for a single leaky aquifer, but now with the eigenvalues of the system matrix as arguments. The question suggests itself whether this is a general rule

(2.1.3)

642

Analytical solution methods

and not only a result of this particular problem. Let us consider an arbitrary square matrix of the n-th order, and its eigenmatrix C:

M

mll

m12

...

mln

m21

m22

.. •

m2n

~

o

,

,

C-(Cl

•

C2 . . .

Cn),

°o

mnl

mn2

•••

mnn

with C i eigenvectors of M. Then M C - (Me1 Me2 . . . Men). As ci is an eigenvector of M corresponding to the eigenvalue ,ki, it is clear that M C - (~1Cl ,k2c2 . . . )~,c,), from which M C - CD, with D the diagonal matrix of the eigenvalues of M:

O

)~1

0

...

0

0

)~2

...

0

Oo•

•

...

,kn

•

0

0

•

Postmultiplying both sides with C -1 we obtain (19)

M - CDC-1.

So an arbitrary square matrix can be written as the diagonal matrix of its eigenvalues, premultiplied by its eigenmatrix and postmultiplied by its inverse eigenmatrix. For the matrix A of the geohydrological system, this means that A - E D E -1. The matrix A 2 then can be written as: A 2 --

E D E - 1 E D E -1

_ EDZE -1

with 0 2 the diagonal matrix of ,k2. This result can easily be extended to yield: A m = EDmE -1

and to the power series f ( A ) f (A) - E

~ Em=l O/mAm;

°lmEDmE-1 - E

m=l

Otm D m

E -1

m=l

Hence: (20)

f ( A ) - E F E -~ with 0

o

.

•

f (,k2) F

~

0 0

o

0

0

o

o

o

s(L)

Ordinary differential equations

(2.1.3)

643

We have thus obtained the important result that an arbitrary function of the arbitrary square matrix A can be written as the product of three matrices, provided that the function f ( A ) can be developed in a power series. The product consists of a diagonal matrix, containing the functions f()~l), f().2), . . . , f ()~n) in which

A has been replaced by its eigenvalues, premultiplied and postmultiplied by the eigenmatrix and inverse eigenmatrix of A, respectively. Now we may write the solution (18) as: ¢p(r) - K o ( r x / A ) q

with q i -

Qi 2rc----~i"

(21)

This result generalizes the well-known formula for steady flow towards a fully penetrating well in a leaky aquifer: Q ~o(r) -- ~ K o ( r ~ / - a l l ) 2 r c K D

1 with a l l -- ~ . K D c

(22)

(See 215.14, Part A.) Many properties of matrix functions are similar to the properties of analogue scalar functions and this enables us to solve systems of simultaneous differential equations in a way quite similar to solving a single differential equation, as may be shown by the solution with use of the matrix functions of the foregoing problem.

Example 5. dZcP ~ = Acp, dr 2 r dr

¢p(oo) -- 0,

(d.)

lim r r-+O ~ r

--q"

General solution: ¢p -- Ko(r~/A)a + Io(r~/A)b with a and b as column vectors containing constants. As ¢p(e~) = 0, it follows that b = 0, dcp = - x / A K 1 (r x/A)a, dr

lim r x/AK, (r v/-A) - I, r-+O

SO

lim

= -a = -q,

r--+0

and therefore ¢p(r) - K0(r~/A)q

(see 720.03, Part A).

Although the use of matrix functions in multiple-aquifer systems is very convenient and enables us to solve rather complicated problems in an easy way, leading in most cases to simple analytical formulas, the analytical evaluation of the ob-

644

Analytical solution methods

(2.1.3)

tained matrix functions will become very complicated if more than two aquifers are involved. However, nowadays effective methods are available to evaluate matrix functions numerically by means of computer programs. For a two-layer system it is possible to derive analytical expressions for an arbitrary function of the matrix that describes that system. Consider the function f ( A ) , where

al )

--a22

a22 -t- a23

(see A in equation (12) with n - 2). According to equation (20) we have to develop

f(A)--E(f(pl) 0

0 )E-1 f(P2) '

and we start with the determination of the eigenvalues pl and p2. tion (15) we obtain

D(p) --

all -t- a12 -- p --a22

--a12 a22 + a23 -- p

From equa-

-- 0,

which means that pl and p2 are the roots of the quadratic equation" p2 _ (all -+- a12 ~- a22 + a23)P -+- alia22 + alia23 -+- a12a23 -- 0. For p -

Pl the system (14) becomes" (all -t- a12 -- pl)Cl -- a12c2 -- 0, --a22cl + (a22 + a23 -- pl)c2 -- 0,

from which Cl

a12

c2

all + a12 -- Pl

and an eigenvector el belonging to Pl becomes: el-(

a12

).

a l l -+- a12 -- Pl

In a similar manner we find: e2--(

a12 ). all + al2 -- P2

The eigenmatrix E of A then becomes: E--(

012 all + a12 -- Pl

a12 ) all -k- a12 -- P2 "

(23)

Ordinarydifferentialequations E

Det

= al2(all

+ a12 -

E-l--

P2) - a l 2 ( a l l

-- P 2 )

As f(A) - E (f(Pl) \ 0

--(all

Pl) -- al2(Pl

-

P2), and thus

2_al2)

+ a12 -- P l )

0)

f(P2)

645

+ a12 -

ta,l+al2_

1 al2(Pl

(2.1.3)

a12

"

E-

we find, after some mathematical manipulations, taking into account that from equation (23) it follows that P l + P2 = a l l + a12 + a22 + a23

and

P i p 2 - - a l i a 2 2 + a l i a 2 3 + a12a23,

that f(A) = Pl

-- P2

(24) X ( (all +al2--P2).f(Pl)--(all +al2--Pl)f(p2)

\

--a22 {f ( P l ) - f (P2)}

--a12 {f(Pl ) - f (P2) } (a22 +a23 -- P2)f (Pl)-- (a22 +a23 -- P l ) f (P2)

/

E x a m p l e 6. Example 4 for 2 layers • 9 ( r ) - Ko(rv/A)q, if q - (q~) (pumping in the upper aquifer): ~p(r) =

ql ((all+ P l -- P2

a12-

p2)Ko(r~-7)-

(all-Jr-a12--a22 { Ko (r ~P-7) -- Ko (r ~ / ~ ) }

pl)Ko(r~-~)) "

E x a m p l e 7.

d ///////////////////'//////////////////,//,/////~ C1

T1 ~//////////////////////,/~

Another example of a two-layer system is a fully penetrating canal or river with a level different from the reference level of the system (for instance, the polder level; see Fig. 5). d2

C2

dx 2 ~o -- Ag,

r2

9(0) - h,

~o(~) =o,

Nergg Fig. 5. Fully penetrating canal in a 2layer system.

q~-

, q92

A--

(all +a12 a12) a23--0, h --a22

a22

'

"

646

Analytical solution methods

(2.1.3)

The solution is" ~o(x) - e-X4Ah

(see 710.12, Part A)

which, according to equation (24) can be written:

--

(qgl) ~02

-

h ((all--p2)e-X4~-(--(all_Pl)e-X~) Pl -- P2 --P2 e-x4Ti + Pl e - x 4 ~

with Pl and p2 the roots of the quadratic equation" p2 _ (all -k- a12 -k- a22)P -+- alla22 -- O. R e m a r k 1. In the foregoing examples the matrix ~/A occurs, and although ~/A cannot be developed in a series, it is still allowable to write ~/A - EA D(v/-~/)EA 1, because ~/~/~-

EAD(~/)EA1EAD(v/-~/)EA 1 -- EAD(v/-~/)D(v/-~/)EA 1 = EAD(Xi)EA 1 -- A,

so indeed ~ is a matrix, that multiplied by itself yields A. ~ is not unique, as can be shown, for example, for a (2 x 2) matrix:

0

,

0 0 -

0

-4

0) S

l

0

because all 4 solutions, if multiplied by themselves, yield A. The square root of a (3 x 3) matrix has even eight solutions, four of them only differing in sign from the other four. In general, the square root of a n x n matrix has 2n solutions. It can be shown that the system matrix A of the multiple-aquifer system has real and positive eigenvalues, so that there is always a solution of ~/A with real and positive eigenvalues too, the only solution that satisfies the boundary conditions. R e m a r k 2. Matrix multiplication is commutative if the factors involved are functions

of the same matrix f (A)g(A) = g(A) f (A), because f ( A ) g ( A ) - EAD{ f (~,i) }EA1EAD{g(~,i) }EA 1

: EAD{f(Xi)g(Xi)}EA 1 = EAD{g(Xi)f(Li)}EA 1 -- g(A) f (A).

(2.2.1)

Partial differential equations

647

2.2. PARTIAL DIFFERENTIAL EQUATIONS

2.2.1. Separation of variables If two or more independent variables occur in the differential equation to solve, for instance, x and t, r and t or r, z and t, we have to do with partial differential equations, and unlike ordinary differential equations, a general solution with a number of constants does not exist. In many cases, however, it is possible to reduce the partial differential equation to two ordinary ones, by applying the m e t h o d o f separating variables.

Consider, for instance, the partial differential equations for one-dimensional and for radial symmetric non-steady flow:

02(/9

and

02cP = f12 O~p

1 0(/9

f12 0(/9

r Or

Ot

We assume that solutions of these equations are such that the independent variables occur separately in them: qg(x, t) = X ( x ) T ( t ) and ~0(r, t) = R ( r ) T ( t ) where X, R and T are functions of x, r and t only. Differentiation of ~0(x, t) and qg(r, t) twice with respect to x and r, respectively, and once with respect to t, yields: d2X

dT

and

- - fl 2 X ~

T dx2

dt

T

d2R

T dR

dT

-~. . . . ~r 2 r dr

fl ZR ~ dt

Deviding these expressions by T X and TR, respectively, we obtain 1

d2X

X dx 2

=

f12 dT

1 d2R

and

T dt

R dr 2

1

t

dR

=

rR dr

f12 d T T dt

The expressions on the left side of these equations depend only on x or r and on the right side only on t. Hence, both sides of the equations must be equal to a constant, for instance, p. Thus,

1 d 2X X dx2

=

/~2 dT

= p

T dt

1 d 2R

and

R dr2

~

1

dR

rR dr

=

f12 dT T dt

=p,

from which three ordinary differential equations are obtained: d2X dx 2

p X - O,

d2R

1 dR

dT

p

dt

f12 T

-- 0

and (1)

dr 2

t

r dr

pR

-0.

In these equations p is still arbitrary. The solution of these equations depends on the initial and boundary conditions of the problem, which will be made clear by the following examples.

648

(2.2.1)

Analytical solution methods

E x a m p l e 1. In an infinite confined aquifer, the initial head of the groundwater is a given function of x: ~o(x, 0) = f ( x ) . This is the same problem as in Example 4 of Section 2.1.2; the objective is to determine the depletion function. The boundary value problem can be written as:

02q 9 __ f12 0(19 OX2 Ot '

qP(X, O ) - f (x),

~0(-cxz, t) = 99(cx~, t) = 0 .

We assume ~p(x, t) = X ( x ) T ( t ) and, according to equation (1), we find d2X dx 2

pX-O

and

If we take p positive, say p

dT dt

p

j~2 T

-- 0.

the general solution for X becomes

- - 0/2,

o~2

X (x) = c le ux + c2e -~x,

and for T:

T (t) -- c3e~'Tt

and for ~p(x, t) = X ( x ) T ( t ) : (p(x, t ) -

~t

(Ae ~x + Be-~X)e~

.

Because increasing values of x and t lead to increasing exponential functions and thus to increasing values of the head, this solution has no physical meaning. Therefore we choose p negative, say p = _0/2, and then we find a solution: ~o(x, t) -- {A cos(o/x) + B sin(o/x)} exp

- ~-~t .

Since A and B are arbitrary, we may consider these quantities as functions of 0/ and write A = A(0/) and B = B(0/). Since the differential equation is linear and homogeneous, the superposition principle is valid (Section 2.3.1) and the function ~o(x, t) -

f0

{a(0/) cos(o/x) + B(0/) sin(o/x)} exp

(

)

- ~2 t do/

is then a solution, provided this integral exists and can be differentiated twice with respect to x and once with respect to t. According to the initial condition ~o(x, 0) = f ( x ) , it follows that oo

f ( x ) --

L

{A(0/) cos(o/x) + B(0/) sin(o/x)} do/.

This is a Fourier integral representation for f (x) (see Section 2.2.3-1, equation (60)), whence A(0/) -- --

f ( k ) cos(0/)~) dk

and

i f +~ f(~.) sin(oeX) d~,

B(0/) -- --

7Z"

cx3

(2.2.1)

Partial differential equations

649

SO

~o(x, t) -- --

?

f()~) { cos(o/x)cos(o/)~) + sin(o/x) sin(o/~.) } d)~

Jr

cx~ o/2

x exp ( - fi-:ft) do/

= --1 fo~f_~-,c f (~.) cos(o/x - o/)~) exp ( - o/2 t ) d)~ do/. Jr

~

Assuming that we may invert the order of integration, we obtain:

i f +°°f()~) fo °~ cos(o/x -

qg(x, t) -- -Jr

o/)~) exp ( -

o 2)

do/d),..

cxD

The inner integral can be evaluated by the use of the formula f0 ° e -a~2 cos(2bo/) do/-- ~1 ~

e b,,2

(a > 0),

to obtain by differentiating both sides of equation (21) of Section 3.1.1 with respect to x, giving

fl2(X --)v) 2

/~ [ + ~

_

qg(x, t)

2 ~ - - [ J_oo

f ( L ) exp

--

4t

} d)~ (see 112.01 of Part A).

Example 2. In a circular island with radius R the initial head is a constant amount h higher than the surrounding surface water level. There is a resistance w against outflowing groundwater at the boundary r -- R. The aquifer is assumed to be confined. The objective is to determine the depletion curve as a function of space r and time t: ~o = ~0(r, t) = head.

R Fig. 1. Depletion function for a circular island with entrance resistance. 2----~-~ Oq -t-

Or 2 09o

~(o,

Or

1 099 _- -/ 3 2 099 r ar at' t) = o ,

099 --(R,t) Or

=

~o(r, t) Kw

99(r, 0) = h.

650

Analytical solution methods

(2.2.1)

Application of separation of variables gives, according to (1): d 2R dr2

1 dR

t

r dr

pR = 0

with ~p(r, t) = R(r)T(t). solution qg(r, t ) -

and

dT

p

dt

f12

T =0

If we choose p positive, say p

=

C~2, we find a general

{AKo(oer) + BIo(o~r) } exp (~2 t).

This solution has no physical meaning because increasing values of time lead to increasing heads. Therefore, we take p negative, say p -- -or 2, finding as a general solution:

01"2t)" qg(r, t) -- {AJo(otr) + BYo(otr)} exp ( - ~ , 0~o 0---7-= { _ AotJ1 (oer)

-

B o l Y 1 (fir)}

exp

--

~--~t .

~Or( 0 , t) - - 0 and as Y1 (0) - - o o B must be zero, so °12t)

~p(r, t) -- AJo(c~r) exp ( - ~-g .

__ 0~o (R t) -- -AOtJl (cZR) exp ( Or

'

-

-~

°t2t) ,

t)---KwA Jo(c~R)exp(-~-~

from which

¢zJl (ot R ) = ~ Jo (oeR ), Kw ocn from which ot - W with ol,, being a root of OtnJl(Oln) - 8Jo(Otn), with e -There are infinitely many roots oln (n - 0, 1 ' 2, " " .) and thus ~o(r, t) may be Kw" represented by

oo

~o(r, t) -- Z

(olnr) (ot2t) enJo\---R- exp f12R2 ,

n=0

according to the superposition principle (see Section 2.3.1), provided that, according to the initial value qg(r, 0) = h, the constant h may be represented by

ot2r~ c~lr'] h -- A1Jo (--~--) -J- A 2Jo (--R- j

+-.-

if

oe. J1 (Otn) -- 8 Jo (Otn).

In Section 2.2.4-3 it will be shown that under certain conditions an arbitrary function can be represented by a so-called Fourier-Bessel series:

Partial differential equations

( nx)

C¢3

f (x) -- ~

(2.2.1)

cnJo ,,--~-/

651

with

n--O

2

2 o~n cn = a2 (or2 + 82)jg(O/n )

fo

( OlnX

x f (x) 3"o

', a

) dx,

with o6, being the roots of OgnJl(Ogn) -- 8Jo(ogn). (See equations (123) and (124) of Section 2.2.4.) oo ffnr ~ So, if h - ~--~n=oAn Jo ( ~ , then 2

06,

2

An = R2 (ol2 + e2) j2 (Oen)

°lnr dr. fo R hrJo( ---~-)

Otnr. The integral in this expression becomes, with u - --y-

h fo R rJo (O~nr) ~ dr-

hR2[uJl(u)]Un hR2foUn °e2 u Jo ( u ) d u - oe---Tn o hR 2

J1 (Otn) -

h R 2 e 3'o(O~n)

Oln

Thus, the final solution becomes" oo e J0(~nr---R--) 99(r, t) -- 2h S (oe2 + e2)J0(oln) exp

2t ) f12R2

R ,

e-- K w

n--O

(see 236.05, Part A) with otn being the roots of OlnJ 1 (Ogn) -- 6 J o ( o l n ) . In general, the method of separation of variables, applied to practical problems, leads to Fourier and Fourier-Bessel integrals and series, as has been shown in the foregoing examples. The method, therefore, forms an inportant part of the development of the theory of Fourier and Hankel transformations as will be shown in the Sections 2.2.3 and 2.2.4. As soon as knowledge about these transformation methods has been acquired application of them to practical problems leads to much easier and quicker solutions than the method of separation of variables.

2.2.2. Laplace transformation

1. Fundamental properties The integral transformation techniques are based on the fact that, by means of a skillful substitution in integral form, one of the dimensions of the differential equation for the groundwater flow can be eliminated, and by continuing this procedure the partial differential equation can be reduced to an ordinary differential equation or even to a common algebraic equation, which usually will not yield any difficulties in their solution with help of the likewise transformed boundary values. The solutions obtained in this way must undergo a reverse or some reverse transformations in order to obtain the desired solution of the problem.

(2.2.2-1)

652

Analytical solution methods

The Laplace transformation generally is used to eliminate the time variable; so it is most frequently applied to non-steady problems. There are many textbooks on operational methods in which the Laplace transformation is thoroughly treated; here we will restrict ourselves to developing the Laplace transformation method briefly, giving a number of theorems and examples, necessary to understand the way in which most of the unsteady problems in this book have been solved. Let F(t) be a given function which is defined for all positive values of t. We multiply F(t) by e -st and integrate with respect to t from zero to infinity. Then, if the resulting integral exists, it is a function of s, say F(s) and is independent of t" F ( s ) --

f0

L{F}.

e-StF(t) d t -

(2)

This operation on the function F(t) is called the Laplace transformation of F(t), and the new function F(s) is called the Laplace transform of F(t), which will usually be indicated by L{F} or by means of a bar: F(s) or shortly F. Furthermore, the original function F(t) is called the inverse transform or inverse of F(s) and will be denoted by L -1 {F}, that is

F(t) -- L -1 {F}. In geohydrological problems we generally wish to find a solution for the head or drawdown qg. Besides a function of t, if the problem is unsteady, q9 may be also a function of the space variables x, y, z, r; during the Laplace transformation with respect to t, however, they will be considered as constants. In the following we will assume that q9 is a function of x and t and ~3 or L {~0} a function of x and s" q~ -- ~o(x, t),

q3 -- qg(x, s)

L(cp(x, t)} --

e -st~p(x, t) dt - q3(x, s)

so

and

L -l{q3(x, s)} - ~o(x, t).

T h e o r e m 1 (Linearity theorem). The Laplace transformation is a linear operation,

that is, for any functions f (t) and g(t) whose Laplace transforms exist and any constants a and b, we have L { a f ( t ) + bg(t)} - a L { f } + bL{g} - a f + b~, for, by definition, L { a f ( t ) + bg(t)} -

-

-

e-St{af(t) + bg(t)} dt a

/o

e - ' t f (t) dt + b

=af(s)+b~(s).

fo

e -st g(t) dt

(3)

(2.2.2-1 )

Partial differential equations

Theorem

65 3

2 (Transforms of derivatives). I f 99 = 99(x, t), then L ~

--

--

fo fo

e - s t ~ dt 8t

ji

e -st d 9 9 - [99e-St]0 -

= -99(x, O) + s

99 de -st

99e-st dt.

So,

{~}

L -~

- s~(x, s) - e(x, o),

(4)

or in words: the Laplace transform o f the first derivative o f a function is the Laplace transform o f the f u n c t i o n multiplied by s minus the initial value o f the function (if t is the time variable).

By applying (4) to the second-order derivative ~~)t2 , we obtain:

1" 0299 L|

099

099 099

= s[sL{99} - 99(x, 0)] - -~-(x, 0),

or L { 02(/9

-5-ff } - s

2@

099

(x, s) - s~o(x, o) - -57(x, o),

and, in general: 0n99 } __ S n n-1 99(X, O) -- S n - 2 0 9 9 ( X , O) . . . . L --~ ~)(x, s) - s Ot

- 0~-~99(x O) Otn-1

'

(5)

•

Transform of the derivative with respect to other variables becomes:

fo ~ -- ~O f ~ Ox

fo ~ e-St 99 dt - ~O~ (x, Ox

s)

654

(2.2.2-1)

Analytical solution methods

and also

L[ 02¢p~} -- aX 202 ~°°

e-'e~o dt -

a2q9 OX2 "

(6)

Theorem 3 (Transforms of integrals). L

/fot

f(r)dr

J

--f(s).

S

(7)

dg

Take g(t) -- fo f (r) dr, then 77 - f (t) and according to (4): L { f ( t ) } -- L { -dT} dg - s L { g ( t ) } - g(O). Clearly g(O) - 0 and therefore L {g(t) } -- { L {f (t) }. Thus also

ll

s, J - fo

qg(x, r) dr.

(8)

Theorem 4 (First shifting theorem). If L{qg(x, t)} = ~(x, s), then L { e atqg(x, t)} -- ~(x, s - a),

(9)

because L { e at

~(x, l)}

e at~o(x, t)e -st dt

--

fo ° qg(x, t)e -(s-a)t dt -- ~(x, s - a);

--

that is, the substitution o f s - a f o r s in the transform corresponds to the multiplication o f the original function by e at (shifting along the s-axis).

Theorem 5 (Second shifting theorem). If q3(x, s) = L{qg(x, t)}, then multiplying q5 by e -tos, where to is some positive constant, we have e-t°S~(x, s) --

•

/o

e-S(r+t°)qg(x, r) dr.

Substituting t = r + to, we obtain: e-t°S~(x, s) --

f?o

e -stqg(x, t -- to) dt.

Partial differential equations

(2.2.2-1)

655

f(t)

Ua(t)

/ a

f ( t - a)

f(t-

a)Ua(t )

I

a

I

i

"-"---'----"

/

[ i

t

a

t

Fig. 2. Shifting along the t-axis and unit step function. We need to write this integral as an integral from 0 to oe. For this purpose we may replace ~p(x, t - to) by the function, which is zero on the interval 0 ~< t < to, and is equal to qg(x, t - to), when t > to. The function

qg(x, t - to)Uto(t) --

0

for t < to,

qg(x

,t-t0)

fort >to

has the desired property, where Uto (t) is the unit step function, defined as follows"

b/to (t)

-

l0 /1

whent to.

Therefore

e-tos(9(x, s) -- L{(p(x, t - to)Uto(t)},

(10)

that is, the multiplication of the transform of a function by e -tos yields the Laplace transform of the original function, shifted along the t-axis over an amount to and multiplied by the unit step function Uto(t). An illustrative example of a function f ( t - a ) U a ( t ) , where f ( t ) is an arbitrary function of (t), which is also defined for t < 0, is shown in Fig. 2. Theorem

6

(Convolution theorem). If the convolution of two functions F(t) and

G(t) is denoted by F(t) • G(t) and is defined by F(t) • G(t) -

f0 t F ( r ) G ( t -

r)dr,

(11)

656

(2.2.2-1)

Analytical solution methods

the Laplace transform o f this function o f t is the product of the transforms o f the functions: (12)

L { F ( t ) • G(t) } -- -ff(s)-G(s), which can be made plausible as follows (no proof)" -ff (s)-G(s) -- -if(s)

f000

G ( r ) e -st dr --

f000

G ( r ) F ( s ) e -st dr.

From Theorem 5 we know that F ( s ) e -st -- L { F ( t - r)ur(t)}. It will be convenient to define F ( t ) to be zero when t < 0; then F ( t - r) will be zero for t < r and the unit step-function ur(t) can be omitted. Thus,

f0

-ff(s)G(s) --

G(r)

f0

e -st F ( t - r) dt dr.

Here the boundaries of the first integral refer to r, whereas the boundaries of the second integral relate to t. We assume that the order of integration can be interchanged, finding: -F(s)--G(s) --

f0 f0 e -st

G ( r ) F ( t - r) dr dt

in which the boundaries of the second integral relate to r. However, F ( t - r) is zero for r > t, so the boundary oo may be replaced by t and we have

-ff (s)-G(s) -- L

{/ot

G ( r ) F ( t - r) dr

}

--L{F,G}.

It can easily be shown that fo G ( r ) F ( t - r) dr - fo F ( t - r ) G ( r ) dr, which means that the convolution operation is commutative. T h e o r e m 7 (Derivatives of transforms). ~ ( x , s) -Og) 5s -

02( Os 29

f0 ° e_St

_

e-Stq)(x, t) dt,

(-t~(~,

fo c~ e-Stt2 q)(x,

t) dt - L { - t ~ ( x , t~ },

t) dt -- L {t2qo(x, t)},

(13)

(2.2.2-2)

Partial differential equations

657

so that, in general:

Osn

= L{(-t)"~o(x, t)}.

(14)

Theorem 8 (Integral of a transform). Consider the transform F ( p ) (Laplace transform with letter p instead of s)

fr__F ( p )

dp -

fr fOCXaF(t)e -pt dt dp

foC~OF(t) fr

-

e -pt dp dt

-- f o ° ° f ( t ) ( - ~ ) ( e - r t - e - ' ~ t ) d t . Letting r approach infinity, we have: F ( p ) dp -

provided that limt_+0 ( - ~ )

t

e

dt - L

t

'

(15)

exists (not proved here).

2. Some examples of Laplace transforms The transform of a constant c may be derived directly from the definition of the Laplace transform (2)"

L{c}--c

fo

e-stdt---

e -st o o _ _

]

S

0

(16)

S

1 and t h u s L { 1 } - - 7" Also, by direct derivation"

L {t/~} --

fo °

Substitute x - s t , L,

t/~e-st dt.

then

,!1 -

S

Sk+l

f cx~Xke-x dx

1

= (by definition) --z-rzF(k + 1) S,~-v,

(gamma function, see Section 3.4.1).

So,

L{t k} -

F ( k + l)

sk+l

(k > - 1 )

(17)

658

Analytical solution methods

(2.2.2-2)

For k - m (m integer and positive) m!

L {t m }

--

sm+l

(18)

,

2 L{t 2} -- ~ ,

1 -~,

L{t}-

1

L{1} -- - , s

k-

~,

(19)

s~/s- = 2s4rJ -'

1

L {eat } - -

{1}

L

fo

e ate -st

1

F(½) - ~

--

,/7

dt

fo

--

•

(20)

e -(s-a)t

[e_(S_a)t] ~

with a a

dt

constant.

s-a

Thus, 1

L {e at } --

(21)

s--a 1 _ - a t } and with Theorem 1 and equation (21): L{sinh(at)} - L{~1 e a t - ~c

1

1

1

1

L{sinh(at)}--~s_a-2s+a

a

-- s 2 - a 2"

(22)

Also 1

L{cosh(at)}

--

1

2s - a

1 -i-

1

.2 s. +. a.

s S2

-

(23) a2.

Substituting a - ib in equations (22), (23) we find: b L{sin(bt)} -- s2 + b2

L{cos(bt)} -

F (t) d2F dt 2

s 2 s +

- - c o s 2 (at),

and

(24)

(25)

b 2"

dF dt

-- - 2 a sin(at) cos(at),

= 2a2{ s i n 2 ( a t ) _ cos2(at)} -- 2a2{1 - 2 c o s 2 ( a t ) } - 2a2{1 - 2F(t)}.

(2.2.2-2)

Partial differential equations

659

According to Theorem 2" L{ d 2 F -~

and as F ( O ) -

dF

-- s F ( 0 )

] -- s 2 F ( s )

(o),

1 and -d-/ dF (0) -- 0 we find:

s2F(s)

- s -- 2 a 2 { 1 s

2if(s) }.

Solving for F ( s ) yields s 2 ~- 2 a 2 L{ cos2(at)}

-S ( S 2 nt-

(26)

4a2) "

L{ sin 2(at)} -- L{1 - cos 2(at)} - L{1} - L{ cos 2(at)} 1

s 2 + 2a 2

2a 2

S

S (S 2 + 4 a 2)

s (s 2 -1-- 4 a 2)

(27)

L{t sin(cot)}. If we put F ( t ) --sin(cot), then according to Theorem 7

dF(s) L { t F ( t ) } -- - ~ ; ds L{sin(cot)}-

F(s)-

CO S 2 _.]._CO2

(see equation (24))

and thus L {t sin(cot) } --

Also with F ( t ) L

t

m

F ( p ) --

2cos (28)

(S 2 _Jr_0)2)2"

sin(cot)"

i 1 --

/~(p) dp

according to Theorem 8;

09

p2 -k- o92

(see equation (24))

and thus p2 -k- 092 --

arctan co

.,.

=

2

arctan

- arctan

;

SO

L

t

-- arctan

.

(29)

660

(2.2.2-3)

Analytical solution methods

The Laplace transformation of an impulse function (see Section 1.5.1-2, Fig. 4 and equation (3)) can be derived from the second shifting theorem (Theorem 5 of Section 2.2.2-1). For instance, a discharge impuls I ( Q ) = Q i a ( t - to)

is defined as an instantaneous abstraction of a finite quantity of water Pi [L 3] from a groundwater body with a discharge Qi [L 3T-l] during a very short time dt. The delta function means that 1 6(t - to) -

I(Q)

for t -- to, f o r t -/=to.

0

(Kronecker delta)

We write Pi - Qi A t and thus Qi with Pi a fixed quantity (see Fig. 3) and

At

ai

I(Q)--

Qiuto(t)-

Qiuto+zxt(t),

where u is the unit step function (see Theorem 5). Then according to Theorem 5, equation (10) we have

to to+At Fig. 3. Discharge impulse.

Qie-StO _ Qi e-s(to+At) = Pie_St ° 1 - e -Ats L{I(Q)}

=

s

s

Ats

and thus lim L { I ( Q ) }

= Pie -st°.

At--+0

(30)

3. I n v e r s e t r a n s f o r m s

By means of Laplace transformation, applied to a differential equation for nonsteady groundwater flow, with boundary values and an initial value, a new differential equation is obtained, in which the time variable has been eliminated, as follows from Theorem 2, applied to the non-steady one-dimensional differential equation, 02~0 __ f12 0(/9 OX 2

Ot

which gives, after Laplace transformation: d2q5 dx 2

t~2{s,~- ~(x, o))=o,

Partial differential equations

(2.2.2-3)

661

in which 93 - q3(x, s) is a function independent of time. The solution of this ordinary differential equation, called the subsidiary equation, together with the also transformed boundary values for x, is a function of the Laplace parameter s. In order to obtain the desired solution of the problem, the transformed solution q3(x, s) which has been found must undergo a reverse Laplace transformation (see, for instance, Example 4 of Section 2.1.2).

a. Application of the inversion theorem There are a number of ways to find the inverse transform of a given function of s. In the first place there are explicit formulas for the inverse transform L -1 {F(s)} as is the case for the transform L{F(t)}. The most useful of these formulas involves an integral in the complex plane, where s is a complex variable, which is stated in

Theorem 9 (Laplace inversion theorem).

f y+iot

F(t)-

L-I{F(s)}-

1 lim eSt-ff(s) ds. 27ri o~-+oo,,×_io,

(31)

The integration takes place in the complex plane, with respect to s - x + i y, along the straight line x - ?, from - o o to +oo; 9/ is so large that all singularities of F(s) lie to the left of the line x - 9/ (see Fig. 5). There are conditions on F(s) or F(t) for the validity of equation (31), but neither these nor the proof of the theorem will be considered here, which implies that it should be verified subsequently whether the solution that was found by means of this theorem satisfies the differential equation and the initial and boundary conditions of the problem. Dependent on the character of the complex function F(s), there are two standard methods of procedure to put the line integral of equation (31) in real form. Two examples of geohydrological problems, the solutions of which have been obtained using each of these procedures, follow below.

Example 3. N~

b

a

Fig. 4. Strip with open water storage.

One-dimensional finite flow in a strip with width 2b. Zero flux at x = 0 and open water storage at x - b for a confined aquifer (Fig. 4). Sudden drawdown h of the surface water level.

Analytical solution methods

(2.2.2-3)

662

Differential equation for one-dimensional non-steady flow" Oq2~0 ._ f12 a~O

ax 2

at '

where 9 - 9 ( x , t) - drawdown and f12 = KD" s The initial condition is: ~p(x t ) = {0 [ ' [ h

f o r 0 <~x < b , forx >b.

Atx-0, ~Ox( 0 ' t ) - 0 . The condition for open water storage at x - b becomes (see Section 1.5.1-3c2°):

K D ~--~-~

O9 (b t)

Ox (b, t) - - a 0-7

'

"

Laplace transformation with respect to t applied to the differential equation with initial and boundary conditions, yields (see Theorem 2 of Section 2.2.2-1): d2~

fl2Sq3 --" 0,

dx 2 dq5 x- (-bd'

d~

~(0

S) -- 0,

dx a s) = -~{sdp(b,KD

' s) - h},

with q3 = q3(x, s). The general solution of this ordinary differential equation is q3 -- A cosh(flx ~/~) + B sinh(flx ~/s). As ~dx( 0 ' s) = 0, it follows that B - - 0 dq5 (b, s) = Aft sinh(fib~7) = dx

a

K D { As cosh(flb ~/7) -- h },

and thus as

A fl ~ sinh(flb ~/7) + K--D cosh(flbx/~)

ah = KD

The transformed solution then becomes: h ~(x,s)

=

cosh(flx ~Cs)

-

S flKa D" sinh(flb,f~) ~

+ cosh(/3 b Vts)

Application of the inversion theorem gives, according to equation (31): 1

lira

f y+iot

e st~(x, s) ds.

(32)

Partial differential equations

(2.2.2-3)

663

The integrand eStq)(x, s) is a single-valued function of s, because all roots of s vanish, as can easily be shown by using the series for the hyperbolic functions: Z2

coshz=

Z4

1+~.

Z3

+~.v + " "

and

sinhz-z+~.v

Z5

+~+'''"

The points for which the integrand becomes infinite are s - 0 and the zeros of the denominator: f l K D sinh(flb4ff)

a

+ cosh(flb,v/7) - O,

47

fib,v'7 c o t h ( f l b ~ ) --

-

~2 K Db a

or

bS

As coth z -- i cot(iz), these zeros are the roots c~,, (n - 1, 2 , . . . ) of bS withO--~,

OenCOtOen----O

O/n

a

- - i f l b ~ d T , so that ~

/~b and Sn --

C

+Or

R

B

o,2

iol n

=

V

So 1

-- Ol

A

f12 b2 "

Thus, so = 0 and s~ are single poles of the integrand, lying on the origin and on the negative x-axis. Now consider the line integral as a part of a closed curve which should not pass through any of the poles, such as, for instance, the circle A B C in Fig. 5. It follows from Cauchy's integral theorem that the integral along the closed curve A B C A is equal to 2zri times the sum of the residues of the integrand at its poles within the contour. It can be shown that as the radius of the large circle tends to infinity, the integral over the arc C B A tends to zero. Thus, at the limit, the line integral of equation (32) is equal to 2zri times the sum of the residues of its integrand at the poles s - 0 and

Fig. 5. Contour for a single-valued function. Sn --- -

The residue at s - 0 Res [e st q3(x s--O

°'2 ( n - - 1 2,

f12 b2

,

becomes" h

h

-

fl K D ~ a

b

.qt_ 1

0 + 1

..

.)

.

664

Analytical solution methods

(2.2.2-3)

f (s) with If we write e st q3(x, s) as heSt sg(s)

f (s) -- cosh(flxv~)

and

g(s) --

fl K D sinh(flb~/7) a

47

+ cosh (13b x/~-),

then the residue at s - Sn becomes:

Res s=s,,

he m f (s) ] S~sg(S)

,,'=s,,

ot2

n With Sn =-~2"/,2 and ~ - - t ~ - - - f iol f we find

f(sn) _ c o s h l f l x (

}_tiff) iOtn __COS( b°lnX~/,

fl K D 1 sinh (flb ~/7) + fl2 K Db cosh (/3 b V/'s-) a 2s ~/7 2as

d d--Tg ( s ) --

fib + ~ - ~ sinh (,6 b,~/J-),

d cls

s.-:-g(s.)

-

-'

1 fl2KDb flKD sinh(-io6,) + ~ cosh(-o6,) otn 2a a - 2 i~--T

io.

-~)

sinh(-iOtn)

0 0 1 sinc~n+ cosoen-sinotn 2a~n 22 c~" " fin

Using these results and remembering that sin o~, =

cos oln, we finally have from

equation (32)

oo c ° s ( ~ n-X ~be x p] (

go(x, t) -- 1 +-----~+ 2hO Z

n=l

(0 +

l) 2 -+-Ot 2)

with otn as the roots of ol,~cotun - - 0 - - ~

~°t2t )

COSO~n

(n-

1, 2 . . . . ).

(33)

Partial differential equations

(2.2.2-3)

665 The graphical construction of c,,, is shown in Fig. 6. If t tends to infinity, the series in equation (33) vanish and a steady state is reached with

Y y=t

/

zr/2

Ot1 / z r

3zr/2 i

O~2

2zr h qg(x, oo) - hs = 1 + 0 '

;

which means that the amount of groundwater released from the aquifer equals the amount of water responsible for the rise h - hs of the open water Fig. 6. Roots of otn cot ogn a level in the storage basin: bsh,, - a(h-hs). Solution (33) has been used for finding the depletion function for an initial head h, given in 138.02 of Part A by putting the head ~o'(x, t) - h-qo(x, t) and by modifying --

hS

0 (O~,2 + l) 2 --t- l))COSOI n

Example

sin c~,,

to

.,,(1 +

for practical reasons.

L9

4.

i~~ __h[~///////////////////~., q9 ~ c

R Fig. 7. Leaky aquifer around a circular basin.

Radial flow in a leaky aquifer towards a circular basin with radius R in which the surface water level is "suddenly" lowered by an amount h. Above the semipermeable layer, a constant level is maintained, which serves as a reference level for the drawdown qo(x, t). The differential equation for non-steady radial

flow in leaky aquifers becomes (see Section 1.4.2-6): 02@

1 8~0

Or2

t r Or

99 ~2 =

f12 0(/9 Ot

The initial condition is ~p(r, O) --

! 0 / h

f o r r > R, for r - - R.

with ~2 _

K Dc

and f12 =

KD

666

Analytical solution methods

(2.2.2-3)

The boundary conditions are cp(R, t) -- h and ~p(cx~, t) - 0. Laplace transformation with respect to t gives:

d293 t-

dr 2

r dr

fl2s Jr-

~-~

qg-O,

~(R

s)-

'

-

93(oo, s ) - - 0

s'

with the solution (see Example 2 of Section 2.1.1):

h Ko(flr~/s + ~2-~) ~ ( r , s ) --

(34)

-

s

+

We consider the function -if(r, s)

-

Ko(flr~/~) KO (flRvC~)" The inverse transform of this function

yields, according to equation (31),

f y+iot e st Ko(flr~/C#) 1 lim ds. F(r, t) -- L -1 {F(r, s)} -- 2yri ~--,~ ay-iot K o ( fl R v/-# )

F +or

Now consider this line integral as a part of a closed curve which does not pass through the singular points of the integrand. These singular points are the zeros of Ko(~R~/-#) and the point s = 0. The integrand is doublevalued, owing to the occurrence of ~/J, except at the point s = 0, which therefore is a so-called branch point.

1

jJ¢

We choose a contour such that the value of the contour integral, according to the theorem of Cauchy, becomes zero, which means that the

A -o~

singularities must lie outside this contour.

The contour A B C D E F

of

Fig. 8 satisfies this condition, as it is known from the theory of Bessel Fig. 8. Contour around a branch point.

functions that no zeros of Ko(fl R ~/'#) are enclosed by this contour. In the limiting case, c~ and Pl tend to infin-

ity and P0 approaches zero. Thus we may write, according to Cauchy's theorem:

Partial differential equations

2rci F(r, t) -- lim +

(2.2.2-3)

F

lim Pl -+oo p0-+0

+

lim

f f (r' s)eSt ds -

667

Pl-+cxzlimfa B

-ff (r' s)eSt ds

££-

P0 --+0

lim

£-

E

Pl -+oo

F

P0 -+0 pl -+oo

F(r, s)e st ds

F(r, s)e st ds + lim

C

F(r, s)e st ds +

D

F ( r , s)e st ds.

It can be shown that the integrals along the parts A B and E F of the large circle become zero if pl tends to infinity. Along the small circle s - poe i° and ds -

i poe i° dO, ~

- ~p-oe½i° and thus f

lim /

if(r, s)e st dt

PO--+0 J C D

= lim P0 -+0

yr

F

exp () . t p 0 i°. e

n"

Ko(flr~-~e½ i°) i io) ipo ei° dO - O, Ko(flR~e7

Ko(ax) =1. x-+O Ko(bx)

as lim

Along B C is s -- pe-~ri= - p

and ds - - d p ; ~ / 7 -

~I-pe -1iTr -

-i~/-~. We have

from equations (79), (74) of Section 3.3.3" Jr

Ko(flr ~/7) -- K o ( - i f l r v/--fi) -- i -~ { Jo(flr x/-fi) + i Yo(flr v/-fi) }, and so F ( r , s)e st ds - - f ~ e -pt Jo(firv/-fi) + iYo(flr~/-fi)

lim

p,~oo

po-+O

c

Jo(~Rq/-P) + iYo(flRv/-fi) •

Along D E we find, with s - pe '~r - - p , ds - - d p ,

dp.

(35)

1.

v/}- - ~/-~e7 'Jr - i~'-~ and

from equation (84) of Section 3.3.3:

Xo(flr

lim ;o -+°

pl -+oo

~/7)

--

Ko(ifir ~/-p)

D -F (r, E

-

s ) e 't d s - - -

-i

Jr { Jo (fir q/-fi ) - i Yo ( fi r ,g/-fi) }, -~

fo °°

e -p'

Jo(flr ~f-d) - i Yo(flr~Ffi) dp.

Jo (fl R V/-~) - i Yo (fl R ~/~)

(36)

668

Analytical solution methods

(2.2.2-3)

Thus, only the line integrals along the negative real axis contribute to the end value of the line integral along AF. Together they yield ((35) + (36)):

BC+DE -if(r, s)e st ds = 2i fo ~ J°(flRv/-fi)Y°(~r~/-fi)- Jo(~r~-fi)Yo(~R~/~) pt Jg (fl R ~/-fi) + r~ (fl R ~-fi) e- dp , as can easily be verified. Substituting u -/~Rff-~, we find for the inverse transform of F(r, s)"

F(r, t ) -

rcflaR2

fo

-

J~(u) + Yg(u)

exp

(

u2t flZ R2 ) u du

or shortly

F(r, t) -- 7rf12R2

h(u, r) exp

as the inverse transform of F(r, s)

--

f12R2 u du, Ko(flr~ )

Ko(flR~/~ ) .

(37)

Then, according to the first shifting

1. theorem (Theorem 4, equation (9) of Section 2.2.2-1), we find, with 77 - ~21)2 = c-g" g -1 {F(r, s + r/)} --

F(r, t)e -~t

and from equation (34), applying Theorem 3: u2c

rp(r, t) -- rrfl2R2

)

flZR2

h(u, r) exp

r/r u du dr.

As exp

/~2 R 2

r/'r dr --

1 - exp

u2

f12R2

f12 R 2

r/t

,

7]

we find

¢p(r, t) -- ~Jr

{

R------Th(u, r) 1 - exp u2+~

(

/~2R 2

r/t)} du.

As

u2t Ko(~R~/s + r/)

YrflZR 2

h(u, r) exp

f12 R2

r/t)u du

(38)

Partial differential equations

(2.2.2-3)

669

is, by definition of the Laplace transform:

Ko(fir~/s+rl) Ko(flR~/s + rl)

2

-- 7rfl2R 2

2 7rf12R2

foO°fo °°

h(u, r) exp

fo ~ h(u, r)

(

u2t

f12R2

) rlt - st u du dt

u du

u2 . fl2 R--"7-- nt- 7] -]- S

Letting s approach zero, we have:

Ko(C£) _ _2 fo ~ ~ h (uu ,R2 r)du.

Ko(-~)

rc

bt 2 -Jr- --~

The right side of this equation equals the first term of equation (38). Thus, the end solution of our problem may be written as"

~o(r, t) -- h K ° ( - ~ ) 2 h f o ° ° U Ko(R)

n"

R------yh(u,r) exp (

U2 -]- 7

u2t ]~2R2

~/t) du

(39)

with

h(u r) ,

-

Jo(u)YO(R u) -- Yo(u)Jo(R u) Jg(u) + y2(u)

1 and

/7-

1

fl2)v 2 ---" c S "

The first term of equation (39) represents the steady state solution (see 223.22 and 223.23 of Part A).

b. Solutions obtained without use of the Inversion Theorem Though all Laplace transformed functions F(s) can be inversely transformed to the original function F(t) by means of the inversion theorem, in general, it is a laborious undertaking, which requires a considerable knowledge of complex analysis, as is shown in the foregoing Examples 3 and 4. Therefore, methods have been developed to obtain reverse transforms in an easier way, generally starting from already known transforms, collected in a table of transforms and derived from these by means of the various theorems, discussed in Section 2.2.2-1. For instance, the function F ( s ) - e -h'/;: which, together with related functions, occurs frequently in one-dimensional problems in semi-infinite aquifers, can be transformed back to the original function by constructing a linear differential equation for F(s) and next by determining the inverse differential equation for F(t) and solving for F(t), as follows" T -- e_k,/;:,

dF _ ds

__k -kvq, 2~/s e

from which dZF 2 dF ds 7 -~ s ds

k 2 -s F -- 0.

d2ff ds 2

k__ -kvG _jr_k2 e-kv~ 4s x/s e 4s

670

(2.2.2-3)

Analytical solution methods

The inverse Laplace transform yields (see Theorems 6 and 7): 4t2F - 2

fot

r F ( r ) dr - k 2

fot

F ( r ) dr - O.

Differentiation with respect to t gives" d F - + 8t F - 2t F - k2 F -- 0 4t2--~

4t 2

or

dF dt

+ (6t - k 2) F -- 0,

from which k2 (c--constant).

F(t)--ct-3/2exp(---~)

ds

2~/~e

- L{-tF(t)}

-- L

- c t -1/2 exp

~

(Theorem 7)

or k2 L-1l!e -k~} --c't-1/2exp(---~) 47 1

1

c!

and f o r k - 0 "

(see equation (20))

from which !

c --

1

and

c

k --

2v~

Thus we have found: L-l{e-k47}-

k 2t~

k2 exp ( - - ~--~)

(k

>

O)

(40)

and 1

1 e-k47

}

exp(-~--~)

(also k - 0).

(41)

Example 5. One-dimensional groundwater flow in a semi-infinite leaky aquifer towards open water without entrance resistance. The drawdown of the surface water level is an arbitrary f u n c t i o n of the time F ( t ) , with F(0) = 0; ~0(x, t) = drawdown. Fig. 9. Leaky aquifer with open boundary.

(2.2.2-3)

Partial differential equations

0299

99t9 __ f12 099

OX 2

~2

~ 2 __ K D c ,

-~'

e ( x , o) - o,

671

~2 __

KD'

99(oo, t) - 0 .

99(0, t) -- F ( t ) ,

Laplace transformation with respect to t gives"

d2q3

( 2S

dx 2

fl

1) ÷ -2~ 0 -- O,

~ ( 0 , s) -- F ( s ) ,

~ ( o o , s) -- 0

m with the solution ~ ( x , s) - F ( s ) e x p ( - f l x ~ / s

+ rl), in which 7/--

fl21).2 - -

1

7s"

If we write C,(s) -- e x p ( - f l x ~/s + rl), then according to the convolution theorem (Theorem 6, equations (11) and (12)):

99(x, t) -- L -1 {q3(x, s)} -- F ( t ) • G ( t ) -- f0 t F ( t - r ) G ( r ) dr. F ( t ) is arbitrary, but G ( t ) can be obtained from equation (40), using Theorem 4:

~x

G ( t ) --

2t ~ - 7

/

f12X2

k

4t

exp /

rlt),

and thus

99(x, t) --

X/o'

2~/~-

Substituting 0/2 __ --

F ( t - r ) r -3/2 exp

f12x2 4r ' from which r

--

t

4r

fl4ot2 2x2

,

2

)

Or dr.

and r -3/2 dr -

- 2 d ( r - 1/2) _

- 2 d ( ~-7) 2~ -- _ 4 doe, while the boundaries 0 and t of the integral are replaced by ~x cx~ and 7-~' respectively, 99(x, t) can be written: I

99(x

,

t) -- ~ _

F t

40/2

exp

-- 0/2

X2 4~.20/2 doe.

(42)

2,,/7 (See Part A, 123.31.) If a s u d d e n d r a w d o w n h takes place which is kept constant thereafter, then -- 0 for t -- 0 and - h for t > 0 while F ( s ) - h__. s Equation (42) then becomes" F(t)

99(x,t) -- ~

exp 2,/7

-

x2) 4)~20/2 do/.

672

Analytical solution methods

(2.2.2-3)

This integral can be evaluated by making use of the solution of the following indefinite integral:

f exp(-a2x2 - bx-4)dx =

# 4a

{ e2ab erf(ax

+ -b) + x

e-2aherf( a x b-- - ) } + x

constant,

which can easily be verified by differentiating the right side of the equation. We find:

~o(x, t) -- ~h{ e~x erfc (fl~

+

~)

x

(fix

+ e-~ erfc 2~/7

~)}

(43)

or qg(x, t) -- hP(~, ~-~) - polder function (see Section 3.1.2 and Part A, 123.32) 1 We have also found the inverse Laplace transformation" with r / - fl21~2 = ~"~.

~ ,lies ~~ / -,

(~ ,~)

(44)

If we consider a gradual drawdown of the surface water level, for instance, according to a linear function F(t) = at we have, from equation (42):

2a qg(x, , ) -

~

F

{t

x Iv 247

2at ff~

-~_

f12X2 4ot2 )exp (_ot2

exp ( - ot2

X2 4,k2ot2)dot

x 2

4~2ot2)dot

247

afl2x2 fz~Xt exp ( _

ote

X2 )dot 4)~2ot2 ~'"

The first integral equals at P (~, ~ ) and the second integral can be evaluated, by substituting c o - - ~ , 1 to lafl2x)~Pconj(~,~-i) (conjugate polder function). Thus: ~o(x, t ) - - a t P

1

,

(x~

+ =a~2xXPconj

2

4~

)

(Part A, 123.34).

(45)

For this problem we could have started directly from the transformed solution qS(x, s) with F(s) -- ~ (equation (18) with k - 1) finding the inverse transform:

~ {~

/ t'(~

~ ,con~(~

,~)

~46,

(2.2.2-3)

Partial differential equations Example

673

6.

Continuous abstraction of groundwater from a leaky infinite aquifer by means of a fully penetrating line source. The discharge is an arbitrary function o f t. ~0(r, t) = drawdown. See Fig. 10.

c

Fig. 10. Line well in a leaky aquifer.

0299

1 0(t9

(t9

Or 2 ~ r Or

2 0(t9

)v 2-

)v2 = fl --~'

qg(r, O) -- O,

99(00, t) -- O,

K Dc, f12 ._ Oqo)

lim r r -+O -~r

---

KD' Q(t) 2 rc K D

Laplace transformation with respect to t gives" d293

ldq3

dr 2 ~

(

1) fl2s --~

r dr

~a(oo, s) - 0 ,

q9 - 0,

m

lim r

r -+ O

---

-~r

2 7r K D

with solution: ~(r, s ) with 1 " 1 -

Q(s) ~Ko(flr~/s 27r K D

1 fl2~2

1 c"-S"

_ --

(47)

+ 71)

The inverse transform may be obtained if the inverse

transform of Ko(k,,/7) is known. In Section 3.3.2 the inverse Laplace transform L -1 {(~/7)~Kv (a~/}-)} -a~ 1 e x p ( - @ ) has been derived in a similar way as has (2t)v+ been done for L -1 {ekv~s} (equation 40). So 1

k2

5 oxp(- at),

(48)

Now, applying the Theorems 4 and 6 to q3(r, s) we have: ~p(r, t) --

1

47rKD

fot

Q(t-

r)exp

(

-

f12r2

4r

- ~r

)dr -r- ,

and with c o - r/r" ~o(r, t) - 4 r r K D

(see Part A, 215.12).

Q(t - - ) e x p r/

- co

4~.2co

09

(49)

674

Analytical solution methods

(2.2.2-3)

If a sudden abstraction Qo takes place, which is kept constant thereafter, then Q(t)-Ofort-OandQ0 f o r t > 0, while ( ~ ( s ) - a0.s Equation (49) then becomes:

QOforlt exp ( - 09

qg(r, t) -- 4zrKD

r2 ) ~de° , 4X2co co

(50)

or

Qo W r/t qg(r, t) - 4zr K-------D ' (well-function of Hantush, Section 3.2.2 and Part A, 215.13). We have also found the invers transform: _

1

L_l{ 1 Ko(k~/s + ~)} -- ~ W (r/t, k~/-~).

(51)

S

If a sudden abstraction Q0 is followed by an exponentially increasing function of the time: Q(t) Q0e ~2t from which ~)(s) - Q0 qS(r,s) then becomes"

Qo Ko(flr~/s + rl) and q3(r, s) -- 2re K D q3(r, s

+ 0/2) - -

s

--

0l2

Oo Ko(flrv/s + ot2 + ~) -- L {e -~2t qg(r, t)}

2rcKD

and thus Qo eo~2t ~ W { ( o l

~o(r, t) -- 4rrKD

2.2.3. Fourier

2 -°t-

rl)t, flrx/ot 2 -at- r/} (Part A, 215.16).

(52)

transformations

1. Fourier series and integrals Practically every periodic function occurring in geohydrological problems can be represented by a trigonometric series, also called a Fourier series, for instance, an arbitrary function f(x) with period 21"

f (x) -- E

anCOS --'~-/ + bn sin \---i-- /

n=0 (x)

~?/X

oo

22"X

- - a o + E a n c o s ( - - - ~ - ) + ~ bn sin (--~-) n=l

= ao 4- a l cos --~

n=l

q- a2 cos \

1

4- a3 cos \ - - 7 / q - . . .

rex (2rex (3rex +blsin(-T-)+b2sin\ l )+bgsin\---~)+....

(53)

Partial differential equations

(2.2.3- l)

675

The determination of the coefficients an and bn is done by integrating both sides of (53) with respect to x between the boundaries -1 and +l. We first determine a0 by integrating both sides from -1 to +l"

f+'

f+'

l f(x) dx-ao

l dx +

f+' {an (.,~x l

~-"

COS

(~]]

1 ) +bnsin\--T- /

dx

n=l

The first term on the right equals 2a0/, whereas all the other integrals are zero, as can be readily seen by performing the integrations. Hence we find for a0

,f+'t

ao -- -~

f (x) dx.

(54)

Next we multiply (53) by cos(~-~), where m is any fixed positive integer, and then integrate from - l to +l, finding:

t

f (x) cos

1

dx

---/-) + b,~sin (.~] \---~-/}] f7' [a0+ /a cos (.~

COS

myrx

1 ) dx.

n=l

The first term on the right side f ~ l a0 c o s ( m- rTc x - ) d x - 0

t

bn sin

1 ]

cos

1

again and also

dx - 0

because sin(nJrx -7-) is an odd function, cos(m~x ~ dx an even function and their product l / an odd function, yielding zero if integrated between boundaries which are at the same distance from the origin at both sides of the origin. Further:

(nyrx )cos( mrcx )dx cos,,,-5I

f+l

{~x 1an f+'l 1 f_+, --[--(m + n) } dx + -~ l

2an

cos

cos --~(m - n) dx.

The first integral on the right side again equals zero, whereas the second integral is zero if m, -y: n and equals anl if m - n. Thus we find for an"

if_ +l n~x~ l f (x) cos ( - - - ~ / d x .

an -- -[

(55)

676

Analytical solution methods

(2.2.3-1)

By applying a similar procedure for the determination of bn in (53) (multiplying by sin(mJrx - 7 - J~ and integrating from - l to -t-l) we find:

bn

--

1 f+l

7

l

nzrx

(56)

f (x) sin (----[--) dx.

If f ( x ) is an even function, we have

1 fol f (x) dx,

(t/WX)

a n - 72 fol f (x) cos --7-- dx

ao -- 7

and

bn - O,

so that f ( x ) can be represented by the cosine series:

0(3

nTrx,~.

f (x) -- ao + ~-~an COS(--7-- /

(57)

n---1

If f (x) is an odd function, we find a sine series

oo

nyrx

sin (-7-)

(58)

n=l with

bn --

72f0'

f (x) sin \ - - ~ / d x .

If a function f (x) is given only over the interval between x = 0 and x = l, both the cosine series and the sine series will represent this function in the interval 0 < x < l. Outside this interval the cosine series (57) will represent the even periodic extension, or continuation of f ( x ) , having the period 2l, and the sine series (58) will represent the odd periodic continuation of f ( x ) with the period 21 (see Fig. 11). The series (57) and (58) with coefficients are called half-range expansions of the given function f (x). They will have important applications in connection with the solution of partial differential equations, especially if periodic functions are involved, but also in finite regions, where parts of non-periodic functions may be considered as parts of periodic functions. Since, of course many practical problems do not involve periodic functions or finite regions, it is desirable to generalize the method of Fourier series to include non-periodic functions as well. in general, if we have a periodic function fT (x) of period T and let T approach infinity, then the resulting function f ( x ) is no longer periodic. The Fourier series of f T (x ) is

f r (x ) -- ao + Z

n--1

an COS

(2 nx) T

+ b,, sin

(2 nx)j T

'

(2.2.3-1)

Partial differential equations

677

f(x)

i

jl

given function f(x) it, X

even continuation of f(x) I

J

f

-2l

I I i i

-l

2l I

f(x)

odd continuation of f(x) I

/

J

Y

-l

-21

21 J

Fig. 11.

Even

and odd

continuation

of a periodic

X

i i

x

f

function.

in which

if,

ao -- -~

T

T

r f r (x) dx 2 T

b~ -- T

f r (x) cos

a~ - ~

'

r fr (x) sin

T

dx

T

2

and (59)

dx.

2

If we use the short notation

Otn =

2rrn T ' then

27r(n + 1) O / n + 1 __ Of n m

AIy

2zrn

m

2jr --

T

T

T

2

and

--

T

Ac~ =

Jr

and thus,

if= T

fT (X) -- ~

oo

T

cos(olnX) Aol

r fT (X) dx + -2

ofT (X) COS(OtnX) dx

Jr

n=l

2

T

+ --1 ~~ sin(olnx) Aot f ~ 22"

n=l

fT (X) sin(olnX) dx. T 2

678

(2.2.3-1)

Analytical solution methods

We now let T approach infinity; then T1 approaches zero and so also the first term 2rr of f T ( x ) . Furthermore, Aot -- T approaches zero and it seems plausible that the infinite series in (59) becomes an integral from 0 to c~, which represents f ( x ) , namely,

lf0~ {A(ot) cos(xot) 4- B(ot) sin(xot) } dot,

f ( x ) -- --

(60)

Jr

f ( x ) cos(otx) dx

A(ot) --

and

B(ot) --

f (x) sin(otx) dx.

¢x:)

This is a representation of f (x) by a so-called Fourier integral. It is clear that this approach merely suggests the representation (60), but by no means establishes it; in fact, there are some conditions for the validity of (60), of which the most important is that the integral

F

~ If(x)l dx

exists.

(x)

The simplifications for even and odd functions are quite similar to those in the case of a Fourier series representation of f ( x ) . If f ( x ) is an even function, then B(ot) = 0 in (60) and

If °~A (ot) cos(xot) dot

f (x) -- Jr

with A(ot) -- 2

f

ox:)

f (x) cos(otx) dx.

(61)

If f ( x ) is an odd function, then A(ot) = 0, and B(ot) sin(xot) dot

f (x) -- -Jr

with Oo

B(ot) -- 2

L

f (x) sin(otx) dx.

(62)

2. Finite Fourier transformations a. Finite Fourier sine transformation

By this transformation the operation is understood in which the unknown function , l ) and the product is integrated with respect to x from 0 q)(x) is multiplied by sin ~mrx

Partial differential equations

(2.2.3-2)

679

to l, thus obtaining a new function, denoted as Sn{{p(x)} or as 93(n), which is independent of x" q~(n) -- fo l {p(x) sin (nrcX 1 ) dx

(n-l,2,

..).

(63)

For instance, the finite sine transform of a constant c is: g3(n) -- c

fo l sin (I12TX)dx _ nTr Cl {l l /

--C0S(/'/~)};

so

Sn{c} --

Sn {x } -

cl nTl"

{1 - (-1)n}.

f0' x sin

--~~

nTr

l

(64)

'fo'

/dx-

XCOS

riTZ"

l

o

+ ~ nzr

+ nZ:rr2

l n~

x d cos cos

sin

-T-]

dx

l / o

Thus" 12

Sn{x} -- ~ ( - 1 )

"+l.

riTZ"

(65)

The main property from which the finite Fourier sine transformation derives its value for solving certain partial differential equations, is, that it reduces the differentialquotient of the second order, ~Ox2 ' to the transformed function itself, as can be shown as follows"

Snl 0293

ff~x2} -

102(t9

-~-Tx2s i n (

I /

sin(T) d(~° ~xx )"

dx

Partial integrating gives:

£n{O2q 9

099

~x2}-[-~-xSin( =

1

117~X~ l

IITg fol

cos

d(p,

l

/]o

l /

l

as the first term on the right side becomes zero. yields"

F}-

,

°c°S,,Tj o (T) -

(IITgX 0~9 dx

cos\

/ )0x

Once more partial integration

~osin\

1 )dx.

680

(2.2.3-2)

Analytical solution methods

The second integral equals the transformed function q3(n). Hence:

Sn { 0299

-

n~

+ -7- {99(0)- (- 1)nq:,(1)}.

-

(66)

Thus, for finite regions along a space variable x, y or z, the finite Fourier sine transformation deserves consideration if the heads (or drawdowns) on both sides of the interval are given. According to the Fourier analysis developed in the preceeding section, the function qg(x) can be represented in the interval from 0 to l by (see equation (58)): oo

~p(x)--~bnsin(

l

/'

n--1

2 l mrx 2 2 in which bn -- 7 fo qg(x)sin(--T-)dx - 7Sn{qg} - 793. Thus, the inverse transformation, turns out to be very simple:

S/1 { ~ ( ~ ) } I ~ (X) l

2 ~ 7 Z n--1

nrcx

~(~)sin( - U )

(67)

Example 7.

.

.

.

.

One-dimensional finite flow through a confined aquifer in a strip with at one side a constant surface water level and at the other side a sudden fall of the surface water level, which is thereafter kept constant, qo(x, t) = drawdown. See Fig. 12.

i;

.

Fig. 12. Strip with open boundaries. 02q 9

Ox 2 -- fl

209 O'---t

--0, ~o(0 t) '

with f12 __

S

K D ' qg(x, O) - O, 9(b t) - J O ' [ h

fort-O, fort>O.

A finite Fourier sine transformation with respect to x gives, according to (66)"

-

nTr)2

-~

~

nrch

~---(-1

)n

--

fl2dq3

d-t-'

q3(n,O)-O.

Partial differential equations

(2.2.3-2)

681

The transformed solution becomes"

~(n' t) - hb (-1)n exP

- ( n2~t }) - ~

-n---~hb(-1)n"

(68)

The inverse transform of -,,--Y hh (_l)n - ~TY(-1 hb2 )n+l becomes ~ as is given by (65). So with (67) we have:

hx qg(x, t)

-

b

2h ~-~ ( - 1 ) n+l 7r n=l

sin

n

(nTrx](n27r2t) f12b2 . b / exp

(69)

The first term on the right side represents the steady state as would be expected (see 134.02 and 134.03 of Part A).

b. Finite Fourier cosine transformation This transformation consists of multiplying the unknown function qg(x) by c o s ( ~ -~-) and integrating the product with respect to x between the boundaries 0 and l, yielding a new function, denoted by C,,{qg(x)} or ~(n), which is independent of x" q3(n) --

~o(x) cos --7-- dx

(n - 1,2 . . . . ).

(70)

Examples:

(nrcx ) cl n yrs x t c o s \ I d x - nzr sin \ 1

[

C.{c}-ct

)1' .

0'

SO

C.{c} - o .

(71)

Cn{x} - £ l

(nrcx )

1

xcos\

1

dx-

nzr

,[ x sin ,2[

l

/ 0

nzr

nzr

///221.2 COS \ T

'

fo

xdsin sin

\

---7--

I

dx

0

12

.2~2 { cos(.~)

- 1};

SO

C.{x} -

12

(72)

682

Analytical solution methods

(2.2.3-2)

The main property of the finite Fourier cosine transformation may be derived as follows:

(/'/7I'X ~

fo I cos (F/TFX 0(/9 \-'T-)d(-~x )

cos \ - - i - - t dx -

--

I / o + -7-

~x cos

sin \ ~

] dq9

Oqg(1)cos(nrr) _ Oq) nzr [ (nzrx]] l O--~~xx (0) + -l- ~osin \ - 7 - - 7 o 12

~pcos ---7-- dx;

SO Cn{ 02(/9

/'/27/'2 12 q3 +

_~x2 } _

(_l)n ~O~o O~o(0). x ( 1 ) - ~x

(73)

Like the sine transformation, also the finite Fourier cosine transformation reduces the differential quotient of the second order to the transformed function itself, together with boundary values at both sides of the interval. As these boundary values concern the fluxes, it is obvious that the finite Fourier cosine transformation should be applied to finite regions, if the derivatives are given at both sides of the interval. The function ~p(x) can be represented over the interval from 0 to l by (see equation (57))

(n x)

oo cp(x) -- ao + Z

an cos --7--

n=l

with

1/o1

ao -- 7

and

qg(x)dx

an -- -[

q)(x) cos - - ~

dx,

or

2 2~ an -- 7 C n { 9 } - 7q)(n)

and

1 a 0 - 793(0),

in which q~(0) represents the value of q3 for n - 0. Thus, the inverse transformation becomes:

1 c; 1

-

-

7

2 ~ (o) + 7

(nrrx) cos

-7-

•

(74)

n---1

Example 8. Non-steady one-dimensional flow in a confinied aquifer, caused by constant groundwater abstraction by means of fully penetrating well-screens at mutual distances 2b; q)(x, t) - drawdown (see Fig. 13).

(2.2.3-2)

Partial differential equations

683

=~q

=*q

I

KD, S

X

b

iI

b

2b

Fig. 13. Abstraction from well-screens.

Only the interval between x -- 0 and x = b has to be considered and qg(x) there will be represented by the cosine series (cf. Fig. 10), because of the even periodic extension. The boundary value problem can be written down as: 02q 9 __ f12 0(t9

f12 __

57'

S XD '

Oq)

a--f-~ Ox (o ' t) - o ,

q)(x, O) - O,

q

Ox ( b ' t )

2KD

Finite Fourier cosine transformation with respect to x gives, according to (73): -

/'/7r) 2 q __ f12 dq~ -b--- q~+(-1)"2K----D -d-t-'

~3(n,0)-0

with the transformed solution: O(n t) -- (-- 1)n b2q ' 2n27r 2KD

1 -- exp

t f12b2

(75) •

The inverse transform needs the value q3(0, t) (see equation (74)). Letting n 2 - m approach zero in (75) we find, after applying once the rule of l'H6pital: (o(0, t) -

qt 2bS'

cp(x, t) -

2bS

so that ÷

E Yr2 K D ,,=l

n2

cos

---if--/

1 - exp

flZb2 t

(76)

(see 135.02 of Part A). e. Finite Fourier resistance transformations In the previous examples we have seen that one-dimensional problems in finite regions, concerning partial differential equations, can be simplified considerably by means of finite Fourier transformations, the choice between sine and cosine

(2.2.3-2)

684

Analytical solution methods

transformation depending on the character of the boundary conditions, as the sine transformation is related to given heads and the cosine transformation to given fluxes at both sides of the finite region. The question arises whether for more complicated boundary conditions and for various combinations thereof at both sides of the interval, generalized Fourier transforms are also available, leading to simplifications in a similar way. Indeed, a general theory has been developed in which special series as solutions of a generalized differential equation, satisfy conditions of the form 099 c199 q- c2-~x - 0

at x - a

and

Oq9

c3q9-Jr-c4 Ox = 0

at x -- b

for the interval a ~< x ~< b. Here C1, C2, C3 and C4 are given as constants, at least one in each condition being different from zero (Sturm-Liouville problem). This general theory will not be discussed here; only some problems that may be solved by means of Fourier transformations will be described. As to the boundary conditions, we will restrict ourselves to the interval 0 ~< x ~ b with either 99(0) -- 0 or ~0x (0) - 0 and 399

CliO(b) + c2-2--(b) -- O. Ox

In practical problems the latter condition occurs if there is a certain resistance against outflow of groundwater from an aquifer into open water, or inversely inflow of surface water into an aquifer (see Section 1.5.1-3c1°). If on the other side of the finite region the head is given we may apply the so-called finite Fourier sine "resistance" transformation as can be shown by the following example.

Example 9.

•

In a finite confined aquifer the initial head is an arbitrary function of x. There is a constant zero head at x - 0 and an entrance resistance w at x - b. See Fig. 14. The boundary value problem for the head ~p(x, t) becomes: 0299 __ f12 0(/9

f12 __

-dT'

S xo

b Fig. 14. Strip with entrance resistance.

q)(x, O) - f (x),

99(0, t) - O,

aqg(b, t) 3x

qg(b, t) Kw

Separation of variables (see Section 2.2.1) yields a general solution ~o(x, t) -- {A c o s ( p x ) + B sin(px)} exp ( -

PZt)

Y

'

Partial differential equations

(2.2.3-2)

685

From ~o(0, t) - - 0 we may conclude that A - - 0 .

ox(b't)--Bpc°s(pb)exp

-~--~t

p2t)

- - ~ es ixnp((p- b~)- ~ K w

or

pbcos(pb) -

Kw

sin(pb),

if we assume B -fi 0. Thus, p must satisfy, with p - g, c~ cot or - - e O/X

~0(x, t ) -

B sin ( T - )

b ,and xw

ot2t exp (

flZb2 )"

As there are infinitely many solutions of c~ - Oln (n -- 1, 2 , . . . ) , assuming or0 -- 0, ~o(x, t) may be represented by

99(x, t) - Z

Bn sin

exp

flZb2

(77)

,

n=l

provided that, according to the initial condition ~o(x, 0) - f ( x ) , an arbitrary function can be represented by the series" OO

f (x) -- Z

Bn sin (°lnx--ff-)

(78)

n=l

b ( n - - 1 ' 2 , . . )" " with o~n being the roots of otn cot otn - - e Kw To find Bn we multiply both sides of equation (78) by sin( --F-) ~mx and integrate with respect to x from 0 to b; O/m being also a root of ot cot ot - - e (m - 1, 2 . . . . ).

fo b f

OlmX OlnX~ . sin (---b-- / dx f0 hsin \---if--)

(OlmX)dx -- Zn=ln n b

(x) sin \

For m ¢ n we have

fo b s i n

(\ Olmx b ) s i n ( - - -°l'nX ff--)dx

'f0 [ /"

-- ~

cos

J /"

~(oen - otto) - cos

1 b - ~ sin(oln - Olm) 2 otn - C~m 1 b ( sin otn cos O/m - 2 c~,~ - o/m 1 b ( sin ~xn cos Otm m

2 O/n - -

O/m

1

~(Otn + O~m)

b

2 C~n + elm sin

dx

sin(or. + elm)

O~m COS O/n)

sin Olm cos c~n)

(79)

686

Analytical solution methods

(2.2.3-2)

1

b

2

+

an

( sin an

c o s a m --[-- sin a m COS a n )

am

b 0/2 __ 0l 2

(a m sin or, COS a

m -- a n

sin

a m COS a n ) .

N o w an cos an - - e sin an and a m COS a m - - --8 sin a m . This substituted in the last expression between brackets yields zero. So we have found that

fb if an and

sin

(amX) n --if-- sin (~,X)dx_Oform~ b / are roots of a cot a -- - e . If m -- n we find:

a m

fobsin 2 (\ anX ) dx -- 2 b

1 -- cos

1 (

--~b

_1 2b

1

(1-

1 an

dx - - b - - - sin(2an) 2 4ct,

b

sinancosan

an cot ~n a 2 + a 2 cotZan

)

-

1 ( 2

b 1

)- (

1

cota.

)

an 1 + cot2an

_1b 1 + 2 a 2 + e2

)

Thus, the right-hand side of (79) is reduced to the single term, belonging to m - n, whence we find for B,"

Bn

=-

2

1

- ~ b 1 + ot2+e2

~obf ( x ) s i n

(anX]dx. b /

This value for Bn, substituted in equation (77), gives the solution of the problem. N o w we define the finite Fourier sine resistance transformation of a function, that function multiplied by sin(Z-~), .where a , (n - 1, 2 , . . . ) are the roots of a c o t a - e , and integrated with respect to x from 0 to b, thus yielding a function, which is independent of x

Snr t . .,[~(xI,~ -- ~(n) --

fb ~(x)

sin

(anY)/dx

(80)

\ b

Kw" b

with an cot an - - e It is then shown that this transformation reduces the differential quotient of the second order to the transformed function itself and the boundary value at x - O, using the boundary condition at x - b, concerning the entrance resistance.

f0h 2 --

-~x sin

( nX o-

f0h OnX -~

cos \ b

d9

Partial differential equations

--

(2.2.3-2)

[ ~(19 (OlnX) -~x sin

687

Oln

b /

(OgnX)]b __ Ogn 2fob

g~ocos --if-

099

O/n

o

Otn

~-

(OgnX)

qgsin --ff-/dx

Ot2

= sin Oen-0-Tx( b ) - --b-v(b)COSOen + -b-V(0)- ~-~q~(n). As ~(b)ax - - Kw ' we have, with ten cos oe,~+ ~ sinoe,, - 0 , 2 C~n ,.,

Snr { ~x2

O~n

- -~o(,,)

+ -b--~o(o).

(81)

According to the definition (80) for this transformation, we have found

2

Bn = -b

f(n)

e_

1 ÷ ~+e2

and, referring to equation (78), we may represent an arbitrary function

2~

_f(n~

f ( x ) - ~ n=o 1 + ~2n27_e2

f(x) by

sin (---~-/,

(82)

where f (x) sin (~nX --U)

f ( n ) - fo

dx

and otn are the roots of Otn cot o6~ = - e . Equation (82) represents the

inversefinite

Fourier sine resistance transformation. We can now solve the problem of Example 9 very quickly: the transformed differential equation with transformed initial condition becomes: /~2 dq~

2

b

-d7 + ~ o ( n )

- 0,

~(,,, 0) -

b )dxo fo f (x0) sin (~nX°

with the solution q3(n, t) -- exp

fleZ;b2

f (x0) sin ( 06, b x0 )dxo,

and the inverse transform, according to equation (82), is: OQ

2 ~(x, t) - g ~

n--0

Oln X

sin(s ) exp ( 1+ ot2+8 2

2t fobf (xo)sin (°lnX° b )dxo

%

f12b2)

with Oen being the roots of c~cot c~ = - e - - K - V h (see 137.44 of Part A).

(83)

Analytical solution methods

(2.2.3-2)

688

If the flux at x -- 0 is given, application of the finite Fourier cosine resistance transformation is suitable. This transformation arises if a function of x is multiplied by cos (~"x --y-), and integrated with respect to x from 0 to b, thus yielding a function which is independent of x

Cnr{99(x)} -- ~o(n) -- fo b 99(x) cos (ku nbx ) dx

(84)

b In a similar way as where ot~ (n - 0, 1, 2 , . . . ) are the roots of o~ tan o~ - e - K--w" has been done for the sine resistance transformation, it can easily be shown that

Cnr { 0299

°12

099

~7x2} - - ~ ( , , )

- ~(0),

(85)

and that the inverse finite Fourier cosine resistance transformation is given by

2 oo

q3(n)

C~r 1{q3(n)} -- 99(x) -- ~ ~~o- 1 + =2+e 2 c o s (

with Otn being the roots of ot tan o~ - e -

b /

(86)

b

K--7"

Example 10. Axial-symmetric groundwater flow in a leaky aquifer, towards a partially penetrating well with discharge Q. The drawdown is a function of r and z in the steady state: 99 = 99(r, z). The boundary value problem is given by (Fig. 15):

Z

:,a

99=0

/////.,¢ ~ / / / / / / / / / / / / / / / / / / / , 4

Ir-|

C

S,K D

b

0299 ~---1 099 -Jr- 0299 --0 -Or -z-

O--~( r, O) -- 0 Oz

Fig. 15. Partially penetrating well in a leaky aquifer.

r- 0 -Yr-r)

m

'

'

O--~( r, D) -- - ~99(r, D) Kc Oz

for 0 <~ z ~< a and for b ~< z ~< D,

0 lim ( r 099

r-fir

m

Q

2rcK(b - a)

for a ~< z ~< b, while 99(o0, z) = 0.

We apply the cosine resistance transformation with respect to z over the interval 0 4 z ~< D, as for z - 0 ~ - 0. According to equation (85) the transformed az

differential equation becomes

1 d~

d2q3 t dr 2 r dr

2

oln O2q3 - O.

Partial differential equations

(2.2.3-2)

689

The transform of the discontinuous boundary condition along the z-axis (r can be written as:

lim

r

r--+0

57

cos

--D--/

dz - -

_-_

Q

COS

27cK(b - a)

Q

_D{

27r K (b - a) Otn

O)

( OtnZ dz otnb

t~na

sin ( - - - ~ - ) - s i n ( - - D - ) }

D = - Qn ~ • O/n

The boundary condition qg(oo, z) - 0 becomes q3(cx~, n) - O. The solution of this ordinary differential equation is

D (Otnr) (o(r, n) -- Q n Oln - - Ko --D--

(87)

and the inverse transform according to equation (86):

go(r,

z) --

Q

~

with c~,, (n -- 0, 1, 2 , . . . ) Part A).

1 sin

m = ot,~

7 r K ( b - a)

( - ~ ) -sin (~"]--b---, ~ cos 1 + ~2+e2

being the roots of o~tano~ -

e -

OfnZ ~

ofnr

-6D

(cf.

532.06 of

3. Infinite Fourier transformations a. Infinite Fourier sine transformation The operation in which an unknown function ~p(x) is multiplied by sin(ctx) and the product is integrated with respect to x from zero to infinity, is called the infinite Fourier sine transformation. This transformation yields a new function which depends on ot and is independent of x

-

- f0

~o(x) sin(otx) dx.

For instance, S{e -~x } - f o e-ex sin(olx) dx -

(89)

k2+ot2 • (Laplace integral, see equa-

tion (24) of Section 2.2.2-2.) An infinite Fourier transformation only has significance if f0~ I~p(x)[ dx exists. For instance, the transform of a constant has no sense. As

690

Analytical solution methods

(2.2.3-3)

sin(ox d( t

fo

---

~ x sin(otx) o - ot

cos(otx) dcp

--

~ 99 sin(otx) - otq9cos(otx) 0 -

l

q9 sin(otx) dx

__ __ot2~ _~_otqg(0), if both q9 and ~Ox are zero at infinity, S [ 02q9

} _ _~2~ + ,~o(o),

(90)

provided that 99(o0) - ~ ( o 0 ) - 0. This means that the infinite Fourier sine transformation reduces the differential quotient of the second order to the transformed function itself and the value of the function at x - 0, and thus should be applied to problems in semi-infinite regions with given head (or drawdown) at x - 0, and both zero head (drawdown) and zero flux at x - o0. The Fourier analysis of nonperiodic functions shows that an odd function ~p(x) over the interval from zero to infinity, may be represented by (see equation (62)):

if0

~0(x) -- -Jr

B(ot) sin(xot) dot,

with B(ot) -- 2 f o qg(x) sin(otx) dx - 2~b(ot), from which it follows that the inverse infinite Fourier sine transform can be written as"

~(x~- s-l{~(~}-

2

~(ot) sin(xot) doe.

(91)

E x a m p l e 11.

~o= H

~°=0

c K

lz Fig. 16. Leaky aquifer with different water lev-

els on top.

Two semi-infinite lakes or polders separated by an impermeable screen (small dike) on top of a very thick leaky aquifer. Steady groundwater flow takes place, caused by a difference in level between the two lakes (polders). The flow is symmetrical with respect to the z-axis, which is taken as positive downwards; so we need only to consider the infinite quarter of a plane between the positive parts of the x- and z-axis (see Fig. 16).

(2.2.3-3)

Partial differential equations

691

The boundary value problem becomes: 0299

0299

H (/9(0, Z) - - T '

"} oqg2 - - O,

OX 2

a~o (x 0) _ ~,~9(x' 0) K O--z

'

~o(c~, z) - 0 ,

~a~°(x, co) - 0.

c

Oz

Infinite Fourier sine transformation with respect to x gives (see equation (90)):

d2rp

2~

dz 2

Hoe

c~

+

K c d ~ (ol O) -- ~(cg, O)

__ O,

2

dz

and

d~ (or, oo) -- 0 dz

'

H o with the general solution ~b - A e -'~z + B e '~z -J 2t~,

~~bd_._5_=

- A ot e - ~ z +

B oee ~ z

and

dz from which B -

~d---L(c~,o o ) dz

0,

O;

d~b

K c - 7 - ( o ~ , O) -- - K c o t A az

-- ~(c~, O) - A q-

H

or

H

A-

1

2oe 1 + K cot

Thus we have the transformed solution:

^

H(

qg(c~, z) -- ~ 1

__

1

1-

e -~ +Kcoe

1

Kc ' l+Kcot

N o w ot(l+Kcot) - - ot to equation (91):

~o(x, z) -- -H- J i ° °

)

.

(92)

and thus the inverse transformation becomes, according

-1 sin(xol) dc~ - -H- f o ° ° -1e -z~ sin(xoe) doe

Jr

19l

Jr

+ -H- fo ~ ~ K c Jr

1 + Kcot

og

e -z~ sin (x c~) do~.

The second integral in this expression is the Laplace transform of I13/ sin(xoe) (see equation (29)) and is equal to arctan(z), x which becomes T 7r for z - 0 ; so the first integral equals 2 and together the first and second intgerals become"

-{Jr

2

arctan

-

- - - arctan Jr x

z

"

Thus we find" ~o(x,z)-~arctan Jr

x

Kc

+ Jr

sin(xoe)e -z~ dc~,

(93)

1 +Kcoe

the first term being the solution for a confined aquifer (c - 0). (See 335.13, Part A.)

692

(2.2.3-3)

Analytical solution methods

b. Infinite Fourier cosine transformation By this integral transformation the operation is understood in which an unknown function q)(x) is multiplied by cos(oex) and the product is integrated with respect

to x from zero to infinity, yielding a new function, independent of x:

c{~0(x)} - ~(~) - fo ~ ~o(x) cos(olx) dx.

(94)

For instance, {

1

}

C k2 q- x2

f0C~c°s(°lx)~

--

k 2 -t- x 2 dx

_Jr ~k ~e-2k

The main property of the infinite Fourier cosine transformation is that it reduces the differential quotient of the second order of the differential equation to the transformed function itself and the derivative of the function at x = 0:

C { 32q9 0(/9 ~ x 2 } - -ty2~ - ~X (0),

(95)

provided that ~o(c~) - ~3x( c ~ ) - 0, as can easily be derived in a similar way as has been done for the infinite sine transformation. From equation (95) we may conclude that the infinite cosine transformation should be applied to problems in semi-infinite regions with given flux at x = 0 and both zero head (drawdown) and zero flux at x = co. An even function ~p(x) over the interval from zero to infinity may, according to (61), be represented by

~0(x) - -I f Jr

c~ A (or) cos(xot) dot

with A(a) - 2 f ~ ~o(x)cos(~x)dx - 2q3(ot), from which it immediately follows that the inverse Fourier cosine transform can be written down as:

~(X)- C -1 {(p(Ol)}_ __jfor2c~ ~b(oe) cos(xoe) doe.

(96)

Partial differential equations

(2.2.3-3)

693

E x a m p l e 12 (see Fig. 17).

z"/ / / / _

a/

/~/~ K1DI ~/

2D2/////

=~Q

Cl/////////////////,;'//,

K2D2

K1D1 x

a

Abstraction of groundwater by a fully penetrating well from a leaky aquifer that has a discrete inhomogeneity, caused by different transmissivities of the aquifer and different resistances of the semipermeable layers at both sides of a straight line (for instance, a geological fault), this line not passing through the well. We choose the y-axis along the straight line and the x-axis through the well. The problem then becomes symmetrical with respect to the xaxis. Along the boundary x -- 0 we assume ~01(0, y) = ~02(0, y) = ~0g, a still unknown function of y, where ~01(x, y) represents the drawdown in the aquifer at the left side of the boundary (x < 0) and ~02(x, y) that for x > 0.

,, I I I

I I I

Fig. 17. Well in a leaky aquifer with discrete inhomogeneity along a straight line.

As usual in problems with discrete inhomogeneities, we consider each homogeneous part separately and try to find a solution for each part. These solutions contain common boundary values, which are still unknown and can be solved by applying the laws of continuity, as follows: l°x~
Ox 2

-+-

0 2991

991

Oy2

~2

~o~( - o o , y) - O, ~ol (x, ~ ) - O.

-0

with ~2 _ K1D1c1, 0o1(0, y) - qOg(y),

~(x,

Oy

O) - 0

and

Analytical solution methods

(2.2.3-3)

694

Owing to the boundary condition ~Oy (x ' 0) - 0 we apply the infinite Fourier cosine transformation with respect to y, giving according to equation (95):

d2~l dx 2

(

0l2 -Jr-

~..~)" q)l - 0,

(Pl (-oe, 0/) - 0 and ~bl(0, 0/) -- ~g,

with the solution: (pl(X, 0 / ) - q3gexp (x

(97)

o/2-1- ~-~1).

2 ° x >~ 0: We consider the flow pattern at the right side of the y-axis as the result of the superposition of two simpler flow fields, caused by" (a) (b)

abstraction of Q by the well, but with an open boundary along the y-axis and a drawdown q)g along the y-axis.

Case (a) gives, with )~2

9a

_

Q

2Jr K2 D2

_

K2D2c2"

[ Ko{_£Tx/(x_a)2+y2}-Ko{-.~2x/(x-k-a)2-k-y2 1 1 ] }

and as

f0 c~ Ko(p,/a~ + y2)cos(~y)dy- ~Jr exp(--av/0/2 + p2) , / ~ + P~

(a >/0)

(see equation (217) in Section 3.3.5), we find for the transformed solution ~a"

~(x, c~) -

Q 4K2D2

{ (

1 exp 2 + x7

-Ix -al

-exp(-[x+al~/ot2+~~2)

*)

2 "-I- ~--~2

}.

Case b can be solved in a similar way as q91"

(pb(x,0/) -- ~geXp(--x~/ot2q--~2), and thus ,~ 2 ( x , o,) = ¢,,, + ,~ h .

(98)

(2.2.3-3)

Partial differential equations

695

The continuity at x = 0 is responsible for the equation: d~l K1D1--~x

(0, ot)

-

-

K2 D2 -~"~x' d~2 x(O, ot) ,

and by differentiating the expressions in (97) and (98) and putting x - 0: 1 -- "~Q exp 2 _~_ )'-~1

K1 D1 ~g

- a

2 _1_ ~)~I

- q3gK2 D2

2 --l- )~I'

from which Q

exp(-a?2+~

2) (99)

g)g (ot) - '2 K1D l ? 2 _.l_.__lflq._ K 2D 2v/ ot2 .+__122 The inverse transformation yields, with equation (96): exp ( - a ? 2 + ~ ) q)g(Y)- --Jr Q fo ~

cos(yot) dot,

K1Ol~ot2-~- ~11 -t-K2O2?2nt- ~2

exp (x~ot2 + ~ - a ? 2 + 1~) 991(x, y) - --n" Q fo °°

cos(yot) dot

(x negative)

(100)

and

Q [=o/1, +a~2+ y2}] ~2(x. y~- 2.=2z~2 g,/u-a~2+ y2j -,~o{g,/(x +~/o" exp/ (x+a,(~+~/~ -(see Part A, 370.01).

cos(yot) dot

696

Analytical solution methods

(2.2.3-3)

Remark. If we replace the leaky aquifers in Example 12 (Fig. 17) by confined ones and maintain the different transmissivities, we find a very interesting and unexpected result for the flow pattern in the unpumped aquifer. The flow now is no longer steady, but quasi-steady, which means that the groundwater velocities are uniquely defined everywhere in the field, as distinct from the potential function which contains an arbitrary constant. This can be shown by letting Cl and c2 approach infinity, 1/Xl and 1/~.2 thus becoming zero in the expression for qgl (x, y) in equation (100), which yields the integral

fo ~ e -(a-x)°l o{

cos(yot) dot

(x negative)

which does not converge for the lower boundary. However, if we consider the derivatives of qgl with respect to x and y, we find, with ~1 - ~: - cx~:

0(/91 _-Ox

Q 7r(K1D1 + K2D2)

f0 °

Q

e -~(a-x) cos(yot) dot

a-x

zr(K1 D1 + K2D2) ( a - X) 2 -4- y2 and =

Oy

-o

zr(K1D1 + K2D2) Q

f0

e -~"-x) sin(yot) dot y

zr(K1 D1 + KzD2) ( a - x) 2 q-- y2 (Laplace integrals, see equations (24) and (25)). Both equations, integrated, give:

O 991(x' Y) -- -2re(KID1 + K2D2)

In {(a - x)2 + y2} + c

(x negative).

Because the quasi-steady drawdown, caused by abstraction of a fully penetrating well in an infinite confined aquifer, situated in the origin is determined by"

~o(r)

--

Q c - 2zr K D

= c - ~ l n ( xQ 4Jr K D

lnr--qg(x y ) - - c '

Q 2rr K D

In x/x 2 -q- y2

2 + y2 ),

it follows that the flow pattern at the left side of the y-axis in our problem can be considered as caused by a pumping well at the point (a, 0) in a confined aquifer with a transmissivity that equals the average value of the two given values K1 D1 and K2D2 (see Fig. 18, where K1 D1 < K2D2). So in the left half plane, where the unpumped aquifer lies, the streamlines are straight lines and the equipotential lines are circles with their centre at the well location.

Partial differential equations

(2.2.3-3)

697

Y

T \\

K2D2 "\\

[_

I-

\

o

___

Fig. 18. Well in a confined aquifer with different transmissivities on both sides of a straight line.

e. Infinite Fourier resistance transformation This transformation may be derived in a somewhat artificial way from the finite Fourier sine or cosine resistance transformation (see Section 2.2.3-2c), as follows. Consider the problem of Example 9 (Fig. 14) where the initial value is given by equation (78): oo

f(x)--~B,,sin(~-)

with

n-- 1

B,,

and

c~,, c o t

--

-2 1 ~ fo h f ( x ) sin ( _ ~ ) b 1 + oe2+~z

oe,, - -

--e

--

dx

b K w •

In order to find solutions for a semi-infinite aquifer as drawn in Fig. 19, we have to shift the origin to the point x -- b in Fig. 14, convert the x-direction and let b approach infinity. Then x has to be replaced by b - x and thus sin( --y-) '~'x becomes sin (oe,,

Semi-infinite field with entrance resistance. Fig. 19.

Oln x

cos oe,, sin -\(ot,,X__b_)"

698

(2.2.3-3)

Analytical solution methods

If we write F ( x ) for f ( x ) and ot for ~ , we have b

sin otn Kw sin C~n = - K wot cos otn,

or

otn COS otn --"

and thus s i n ( ~ -~) becomes - c o s o t n { s i n ( o t x ) + Kwot cos(otx)}, from which 2

COS 2 ot n

F(x)--Z~I+

{ sin(otx) + Kwot cos(otx)}A(ot)

7

n=0

ot2+s 2

with A(ot) - f bo F(x){sin(otx) + Kwot cos(otx)} dx. As otn are the points of intersection of the curve yl -- tan or,, and the straight Kw line through the origin y2 - ---y-otn, it can easily be shown that if b approaches infinity, C~n approaches nzr (n = 0, 1, 2 , . . . ) (cf. Fig. 6). We introduce Aot with Aot -- (,+l),rh ,,~rh ---- _,rh and replace the factor ~2 in F(x) by -2Aot 7 - , and evaluate COS 2 otn

1+

(otn2 -'1" t32) COS 2 otn

+

~

t3 2

/32 s i n 2 otn -1-" 132 COS 2 otn

+

2+

+

b2

132

+

+

1

ot 2 K 2 w 2 nL b 2 nt- b K w

K2w2ot2

gto b °

q-- 1 q

Now we have 2 ~ F ( x ) - - - - yt" 2. n=0 A(ot) --

sin(xot) + Kwot cos(xot) . k-~ ,A(ot) Aot KZw2ot2

fo b F(x){sin(otx)

q - 1 --1

with

b

+ gwot cos(otx)} dx

and

ot - -b--. otn

Now letting b approach infinity and putting dot for Aot, it seems plausible that the infinite series for F(x) becomes an integral from zero to infinity, which represents F(x): 2 t "~ sin(xot) + Kwotcos(xot) F ( x ) - -- Jo At(or) K2w2ot 2 dot Jr 1+ A'(ot) -

with

(101)

F(x){sin(otx) + gwot cos(otx)} dx.

Now if we define the infinite Fourier resistance transformation as Fr{qg(x)} -- qb(ot) -

f0 °

qo(x){sin(otx) + Kwot cos(otx)} dx,

(102)

Partial differential equations

(2.2.3-3)

699

the inverse transformation becomes, with equation (101):

Kwol cos(xc~) f0 oK)q3(o~) sin(xu)1 ++ K2to2o/2 dol.

(/9(X) -- Fr-l{(p(of)} __ __2 7/"

(103)

The transform of ~i)x2 becomes: Fr{O2(/9

~x2} - foCX~ 0299 {sin(o~x)+Kwotcos(otx)Idx -

{ sin(o,x) + K,,,o, cos(,,,.,,:)} d

=

{ sin(oex)+ Kw~ cos(otx)} 0 { cos(otx) - Kwot sin(otx) } d~o

- c~ :

in( x) + K

cos(

x)}

-- 0e~0{COS(OtX) -- K w u sin(c~x)} o _--

_~O~°(O)Kwa Ox

+ a~o(O) -

C~2~

ol2q),

provided that ~o(oo)

-

7--(00)

-

ox

o.

With the boundary condition K ~a~o( 0 ) Fr{

Oq2~ 572} - - ~ '

~w , we find (104)

provided that ~o(oo) -

-7(00)

ox

-

-

o.

Example 13. Q 5/////2 /////~/~///x~/////////////////////////>c j.

r

a

l z Fig. 20. Finite line well in a thick leaky aquifer.

Abstraction of groundwater by a finite line well from a very thick leaky aquifer; the discharge Q is uniformly distributed along the well-screen. The drawdown in the steady state is a function of r and z: ~0 = ~0(r, z) (see Fig. 20).

700

(2.2.3-3)

Analytical solution methods

02(t9 ~- --10q) -+- 0299 --0

----g Or - r -~r

Oz 2

[r Oq) ~ _

r-+olimIV -~-r/

0

' 2re Kl

I °

(1) (1) for a - -~l < z < a +-~l , (1) (1) for0~ a + -~l ,

Ocp(r, 0) -- ~p(r,0) 0-7 K-----~'

~o(c~, z) - 0 ,

099 ¢p(r, oo) -- -~-z(r, oo) --0.

We apply the infinite Fourier resistance transformation with respect to z. The differential equation is transformed into d2~ dr 2

}

1 d~b - - O f 2 ~ r dr

"-- O,

while Kcot cos(otz)} dz

Q [a+ll{sin(otz)+ 2rr g l Ja-½l

lim ( r d-~r~) -

r--+O

Q [_cos(~z)+Kcusin(~z)]a+i; 2rr K l ~ --

[ Kcoesin{u.(a+ Q 2rr K l u 1l

1l) } - K c o e sin {ol(a - 11) }

-2

/

cos a(a

Q {Kcot cos(aot)sin ( 2 ) + zr K l ot

Q

-2

1 sin(aot)sin ( ~ ) }

sin (l~_ ] F ( a , ~),

rr K l ot

\21

if we use a short notation for F(z, u) - sin(zot) + K c u cos(zoo). The transformed solution is ~b(r, or) --

Q zr K l u

sin

('2)

F(a,u)Ko(r~),

(105)

and the reverse transform according to equation (103): fo °° sin ( ~ ) F ( a , ~) q)(r, z) -- 7r2Q 2K l -£-(] ~ -~ ~-~2-ot-~i K o ( r ~ ) F ( z , ~ ) d u , with F(z, c~) = sin(zoe) + Kcot cos(zot). (See Part A, 522.05.)

(106)

Partial differential equations

(22.4-1)

701

2.2.4. Hankel transformations

1. Orthogonal functions The Fourier analysis and the closely connected Fourier transformations of the preceding section are based on the fact that for the determination of the Fourier coefficients a,, and bn the property has been used that integrals of the form

f_,

sin

l

)cos( \

1

)dx

and

1

f_'

cos (---7-- / cos (

t

m.. /

)dx

become zero (the second integral for m :/= n). The set of functions cos( n~x ~] (n - 1 ' 2 ' ' " .) thus has the property that if two l distinct functions of this set are multiplied by each other and the product is integrated over the interval - l ~< x ~< l, the result becomes zero. Such a set of functions is called an orthogonal set of functions. A set of functions g! (x), g2(x), g 3 ( x ) , . . , is said to be orthogonal over a finite interval a ~< x ~< b if

f~

b gm (X)gn(X) dx -- 0

for m # n.

For m - n the integral f,h g2 (x) dx will generally not be equal to zero. The positive square root of this last expression is called the norm of gm(X) and is denoted by Ilgmll:

g2 (x) dx - IIgm II.

(107)

Clearly, an orthogonal set gl, g2, . . . on an interval a ~< x ~< b whose functions have norm 1 satisfies the relations

gm (x) gn (x) dx --

0

when m # n, m = l, 2, .. .,

1

whenm=n,n--

1,2 . . . . .

Such a set is called an orthonormal set of functions on the interval a ~< x ~ b. Obviously, from an orthogonal set we may obtain an orthonormal set by dividing each function by its norm. The set of functions gm(x) = sin(mx), namely sinx, sin(2x), sin(3x) . . . . may serve as an example of an orthogonal set on the interval - J r ~< x ~< Jr, because

f_

Usin(mx) sin(nx) dx - 0

for m :/: n.

71"

The norm Ilgmll equals ~

as Ilgmll 2 -- f_~ s i n 2 ( m x ) d x - rr.

702

(2.2.4-1)

Analytical solution methods

Hence, the corresponding orthonormal set consists of the functions sinx

sin(2x)

sin(3x) ,

. . . . .

Because of the fact that the possibility of representing an arbitrary function f ( x ) by a Fourier series is based on the orthogonality of the goniometric functions (see Section 2.2.3-1) on a certain interval, it is tempting to represent given functions in terms of any other orthogonal set gn (x) in the form oo

f (x) -- E

Cngn(X) -- Clgl (x) + c2g2(x) + ' " ,

(108)

n=l and determine the coefficients cn in a similar way as for the Fourier series, as follows: Multiply both sides of equation (108) by gm(X) and integrate over the interval a ~< x ~< b on which the functions are orthogonal:

b

L

c~ Lb f (x ) gm (X ) dx -- E Cn gn (x ) gm (X ) dx .

n=l

If n ~ m, all integrals on the right-hand side become zero except that only the integral for n = m does not, but becomes IIg,,IIz. So the formula for the determination of the constants is:

1yah

c,, = ilgnll2

f(x)gn(x) dx.

(109)

If the series (108) converges and represents f ( x ) , it is called a generalized Fourier series of f ( x ) , and its coefficients are called Fourier constants of that series, represented by equation (109). Some important sets of real functions occurring in applications are not orthogonal but have the property that, for some function p(x),

L

b p(X)gm(X)gn(X) dx -- O,

when m ¢- n.

(11 O)

Such a set is then said to be orthogonal with respect to the weight function p(x) on the interval a ~< x <~ b. The norm of gm is now defined as

I[gm I[ = ~fa b p (x) g2m (x) dx,

(111)

(2.2.4-1)

Partial differential equations

703

and if the norm of each function gm is 1, the set is said to be orthonormal on that interval with respect to p ( x ) . If we set hm -- X/c-figm, then equation (11 l) becomes

f

b h m ( x ) h n ( x ) dx - 0

(m ¢ n),

that is, the functions hm form an orthogonal set in the usual sense. 2. Series a n d integrals o f Bessel f u n c t i o n s

Consider the differential equations (Bessel's equations, see Section 3.3.1): d2y 1 dy dx 2 ~--x d-ffx + k 2 y - 0 '

and

d2y 1 dy dx 2 + x-d-fix- + 12y - 0; it follows that u -- J o ( k x ) is a solution of the first and v = Jo(lx) a solution of the second equation, in which J o ( k x ) and Jo(lx) are Bessel functions of the first kind and order zero (Section 3.3.1); so it is allowed to write d2u 1 du dx 2 q- x-~x- q- k2u - 0

and

d2v 1 dv ~ -k- x~xx- -}- 1 2 v - 0

or

1 d (du) x if- k2u -- 0 x dx ~xx

and

1 d (dr) 12 -~ x + v =0. x dx ~xx

Multiplying the first equation by x v and the second by x u and subtracting the second equation from the first, we find d(xdU d(xdV Vdx ~ x ) - U~x-x d-x-x) + ( k 2 - 12)xuv - O. Integration with respect to x gives

f u xd -- f ud(x-dyx)-f dv

_

dv = x u dx

As

do du f x~

f

d. }- y dx .-~x dv

dv du - x v x dx dx ~

-- f x ~do ~du dx -- f x ~du dr, we have

(k 2 - I 2)

f

x u v dx - x

-dx

(X~x)

v~

.

_

o

704

(2.2.4-2)

Analytical solution methods

Inserting u = Jo(kx) and v = Jo(Ix) in this expression, we get (k 2 _ / 2 ) / xJo(kx)Jo(lx) dx = x { - lJo(kx)J1 (lx) + kJo(lx)J1 (kx) }, or

f xJo(kx)Jo(lx) dx = k 2 x_ l 2 {kJo(lx)J1 (kx) - IJo(kx)J1 (tx) }

(112)

It should be noted that the general formula

/ xJv(kx)J~(lx) dx -

x {kJv(Ix)Jo+l(kx) -IJv(kx)Jo+l(Ix)} k 2 _ 12

(113)

may be derived in a similar manner. In particular it follows from equation (112) that the definite integral over the interval 0 ~< x ~< a becomes

fo

a x Jo(kx)Jo(lx) dx -

a

k 2 _

l2

{kJo(la)Jl(ka) -lJo(ka)Jl(la)}

•

Now set ka = 0/1 and la - 0/2 and assume that 0/1 and o/2 are two roots of the equation J0(c~) = 0; then Jo(la) = 0 and Jo(ka) = 0 and thus,

a / fo°xJo ( aa' X )/j o ( 0/2X)dx

=0.

(114)

From this equation it can be stated that, according to equation (110), the set of functions J0 (~"x ---y-) (n = 0, 1, 2, ...), is orthogonal with respect to the weight function p(x) = x if 0/n (n = 0, 1, 2 . . . . ) are the roots of the equation J0(0/) = 0. 0 When k - l, the result in equation (112) assumes the indeterminate form +. By using l'H6pital's theorem it can be shown that

f x j 2 ( k x ) d x - 2 x e { J ~ ( k x ) + j2(kx)},

(115)

from which f oa x j2 (kx) dx - ~la2jZ(ka) , so that the norm of Jo( ~"x a ,] has now been determined (see equation (111)): a

1]J°( o/n' ) a~x I 1 - ~

Jl(~n).

(116)

Using the results (114) and (116) we may conclude, referring to Section 2.2.4-1, equations (108) and (109), that under certain conditions an arbitrary function f ( x )

Partial differential equations

(2.2.4-2)

705

may be represented by a so-called Fourier-Bessel series of the form

oo

OllX

c0.0 v ) +

n----O Cn ----

'

aZJ2(oln)

c1Jo( Ol2x) a

+...

1o"x f (x ) Jo (o.x) dx, \ a

with (117)

if Ofn (rt "--0, 1, 2 , . . . ) are the roots of Jo(ot~) = 0. The Fourier-Bessel series are powerful tools in treating geohydrological problems involving radial and axial-symmetric flow, especially in finite regions, as they are the basis for the finite Hankel transformations, to be discussed in the next section (Section 2.2.4-3). Since, of course, many practical problems in radial- or axial-symmetric flow involve semi-infinite instead of finite regions, it is desirable to know whether an arbitrary function can also be represented by an infinite integral of one or more Bessel functions. Now, indeed, it can be proved that, under certain conditions, a function f ( r ) can be represented by the so-called Hankel integral

f (r) -- fo ~

otA (ot) J,(rot) dot,

in which

A(ot) --

f0 °

r f (r)J~(otr) dr.

(118)

The proof of this statement will not be given here; instead, the following derivation would make it acceptable. Integrals of the type

f txJu(at)J~(bt) dt have been investigated by Weber and Schafheitlin, and they found that discontinuities occur in these integrals at a = b for certain relations between X, lZ and v. A well known Weber-Schafheitlin integral is (see Section 3.3.5, equation (156)):

{ bu-1

oo f0

Jtz(at)Jtz-l(bt) dt -

a~ 0

for 0 < b < a (/1. > 0) forb>a >0.

From this we have

r ~ fo ° Jv(r~)J~_l(Xot)d~

-

-

JX ~-l I0

for0r >0.

706

(2.2.4-2)

Analytical solution methods

Both sides of this equation, multiplied by )~f(~.) and integrated with respect to )~ between zero and infinity, yields

r ~ f0 cx~Jv(r(~)

f0 cxz)~f()~)Jv-l(~)~) d)~ do~ -- fo r )~f()~)

d)~.

Now, differentiating both sides with respect to r and making use of the general d differentiation formula for the Bessel functions T;{x~Jv(x)} -- x~Jv_l(x) (equation ( 2 2 ) o f Section 3.3.1), we find r~

f0

f0

otJv_l (rc~)

or w i t h / z - - v -

f (r) --

Lf(~) Jv-l(a)~)d~ dc~- r V f ( r )

1

fo

otJ~(rot)

fo

~.f (~.) Ju (c~.) d)~ do~,

which is identical with equation (118).

3. Finite Hankel transformations Considering the representation of a function by a Fourier-Bessel series as given in equation (117), we may understand the definite integral in the expression for cn as an operation performed on that function and may call this operation the finite Hankel transformation. As we want to apply this transformation to the solution of problems in radial flow or axial-symmetric flow, we replace the function f ( x ) in (117) by qg(r). The finite Hankel transformation is then the operation in which the unknown function qg@) is multiplied by the factor r Jo( --R--) ~"r and the product is integrated with respect to r over the interval 0 ~< r ~< R, thus obtaining a new function, denoted as Hn {qg(r)} or as q3(n), which is independent of r" Hn{~o(r)} -- ~(n) -- fo R rqg(r)Jo( °enr R '~ / dr

(119)

with c~n (n = 0, 1, 2 , . . . ) being the roots of J0(ot,) = 0. From equation (117) we can immediately determine the inverse transformation

2~

q3(n) Otnr ~ J0 \--~--), ~ o ( r ) - H n - - ' l ~ ( n ) } - R--2-n=o J~(o~n)

(120)

again, with or,, being the roots of Jo(a,) = O. The main property from which the finite Hankel transformation derives its value

d299 in solving radial and axial-symmetric problems, is, that it reduces the terms ~-7~-+

Partial differential equations

(2.2.4-3)

707

1 ~ which always occur in the differential equations, governing these problems to rdr' the transformed function itself, as can be shown as follows:

~ + r ~ r ] - ~ r~r ~r) --

_ _ _d ( r - ~ ) d r /o Rr J o ( °tnr]l R }rdr

(Otnr dqgin ---[Jo\ R )rdrJ0-

_

r ) d (r~rr) f0R Jo (otn \--R-

fo R dq9 otnr l rd-rrdJ°( R ]

dq9 R Oln fo R Otnr] + ~ rJl (---~-j d~ -[~0(~O~nr]~r ~r]0

--

,~)r - --~

h-Tr+ --Y-~, ~-T qgd r J l \

R

)~]0

"

Now

Otnr d---r ---~-J1

R ]

(otnr)Otn R Jo\ R --~-

o~n2 ( ~ ) R---~Jo

r

(equation (22), Section 3.3.1), whence we have, as J0(c~n) - 0 , d299 1 d99 0q9 c~n Hn -~r 2 + -- O~n J1( otn ) 99( R ) - l i m r ~b(n) r--&r r~O -~r ~ "

{

]

(

)

(121)

Hence, application of the finite Hankel transformation is useful in radial or axialsymmetric problems involving wells with given discharge at r - 0 and with a given head (drawdown) distribution at r = R.

Example 14.

7 Q

i

r KD, S

Fig. 21. Well in a circular island.

Non-steady flow of groundwater through a confined aquifer caused by abstraction of a constant discharge Q by a fully penetrating well at the centre of a circular island. ~p - qg(r, t) drawdown (see Fig. 21).

708

Analytical solution methods

(2.2.4-3)

02(t9 1 0(t9 __ f12099 (p(r, 0) - - 0 ,

S

(p(R,t) --0,

lim r, = r~O ( a ~ r ) 2rc K D

for t > O.

Finite Hankel transformation with respect to r gives 2

Q

Otn

2rc K D

f12

R 2 (~ - -

~b(n, 0) -- 0,

dt '

an ordinary differential equation with initial value, and solution

o R2{1 -- exp

~ ( n , t) -- 2 7r K D ot---~n

t

fl2R2

"

The inverse Hankel transformation immediately yields qg(r,t) --

Q

k

{anr{

Jo~--k-)

1 -exp

(

2t ) }

%

Now

jO(--.-) °r n=0 is the Fourier-Bessel representation, over the interval 0 ~< r ~< R, of the function 1 in(rE ) as is shown in Section 3.3.4, equation (115); so we get

2

(p(r, t ) -

Q In 2rc K------D

/r"--)

k

Q J° ( z - ~ ) e x p 7r K D n=O °tz J~(°tn)

/ o t) flZR2,

where an are the roots of Jo(~n) - 0 (see 241.02, Part A). If t tends to infinity the series vanishes and the well known solution for the steady state is obtained, given by the first term in the expression for (p(r, t). The Hankel transformation technique, though in a somewhat modified way, enables us to solve also problems in radial flow with more complicated boundary values. Like the Fourier transformation, the Hankel transformation can be extended to the so-called finite Hankel resistance transformation, which should be applied in particular to problems involving entrance resistance against outflow of groundwater from an aquifer into open water, or inversely, inflow of surface water into an aquifer (see Section 1.5.1-3c1°). We will derive the necessary equations for this transformation, starting from the following theoretical example.

Partial differential equations

(2.2.4-3)

709

Example 15.

t!

~

In a circular island, with radius R, the initial head in the confined aquifer is an arbitrary function of r. There is no flux at r - 0 and entrance resistance w at r -- R (Fig. 22). The boundary value problem for the head ~p(r, t) can be written as

m

02~0 Or 2

R

10q 9 ~

r-Or

S )

2099( ._ fl

f12 _

-~

-- K D

'

Fig. 22. Circular island with entrance resistance. 0__~

~p(r, O) -- f (r),

099

Or (0, t) -- O,

K~(R, Or

t) --

~o(r, t) w

To solve the differential equation we apply the method of separation of variables (see Section 2.2.1), which yields a general solution

(- y,). O/2

+

From ~Or (0, t) -- 0 we may conclude that B -- O, so

o/2

99(r, t) -- AJo(o/r) exp ( - ~-f

t)

The boundary value at r - R gives:

Kw~(R Or

'

t) -- - K w A o / J l

(o/R) exp

-

-y

t

-- - A J o ( o / R ) exp

-

-y

t

,

from which K wo/J1 (o/R) -- J0 (o/R). Writing o/,, for o/R and e for X--8-y,we have found that o / - W has to satisfy the relation

0tn

~ . J1 ( ~ )

- ~ Jo (~,,),

while o/2 t

o/n r

q)(r't)--AJokR)exp

(

f12R2 ) •

As there are infinitely many solutions of otn (n represented by

0, 1, 2 . . . . ), ~o(r, t) may be

OO q)(r,t)--ZAnJ0 n=0

R lexp f12 R 2

'

(122)

710

Analytical solution methods

(2.2.4-3)

according to the superposition principle (see Section 2.3.1-1), provided that, according to the initial condition ~0(r, 0) - f(r), an arbitrary function can be represented by the series oo

f (r) " ~

(123)

AnJo(~-~- )

n--0 R

with 0/n being the roots of 0/,,J1 (0/,,) - e Jo(0/~), where e - y ¥ . Now,

as

r {aJo(br)Jl(ar) - bJo(ar)Jl(br)} f r Jo(ar) Jo(br) dr -- a 2 -b2 (see equation 112), over the interval 0 <~ r ~< R we have: R

fo

l~ r Jo(ar) Jo(br) dr = a2 _ b2 {aJo(bR)Jl(aR) - bJo(aR)Jl(bR)}.

If we assume that

0/1

--

a R and

0/2

--

b R are roots of the equation

0~n J1 (0/n) - -

e J0 (0/,), we find 0/1

E

aJo(bR)J1 (ag) -- --~ J0(0/2)J1 (0/1) -- ~ J0(0/2) Jo(0/1), and also

bJo(an)J~(bg)-

~Jo(O~)J~(o~2)-~Jo(O~)Jo(a2),

O/2

8

and therefore

fo R r JOk--~--)Jo(~--~) (0/lr dr - O . From this result it follows that, according to equation (114), the set of functions J0(-q-~) is orthogonal with respect to the weight function p(r) - r if 0/n (n 0, 1, 2 , . . . ) are the roots of the equation 0/n J1 (o/n) - e J0(0/n). ( Otmr , Multiplication of both sides of expression (123) by rJo~--~-) in which am is also a root of 0/J1 (0/) - eJ0(a), and integration with respect to r from 0 to R, gives

f0 r f (r)Jo

dr - ~ n--O

A.

f0 rJo('--~-/Jo o.r

dr

(m -- 0, 1, 2 , . . . ) . All terms of the series with m -¢- n become zero and only the term m - n is left, i.e."

An

fo"

r J~

dr.

(2.2.4-3)

Partial differential equations

711

From (115) it follows that fo R r J~ \( -Otnr - ~ ) dr - !R'{g(~n> 2

#(~.>}

+

So, we find 2 1 fo R rf(r)Jo (otnr) dr An = ~R 2 JZ(c~n) + JZ(t~n) \ R

(124)

with C~nthe roots of OlnJ1(Oln) -- EJo(oln). This value of An substituted in (122) gives the solution of the problem. We now consider the definite integral in the expression for An, replacing f (r) by qg(r), as the finite Hankel resistance transformation Olnr ) Hnr{q)(r)} -- ~(n) -- fo R rq)(r)Jo\ R dr

(125)

with o~. being the roots of OtnJ1 (O/n) -- E J0 (O/n).

From equations (122) and (123) we get the inverse transformation" ~o(r)-

2

Hn~l{~(n)}-

~

Otnr

6(n)

~ Z n--O

jo2(O~n)-I-J~(otn)Y0\ R )"

The transform of ~dr 2 + 1r ~ becomes

~r {'d(rd.} r~r Tr)-

foRr'0(,.r~r ".r ld(rd~r) dr

(~.r (rd~ -/o . '°,,) d

.,)r

]o- fo"d~ r--~rdJ°\C"F R /

=[jo(Otnr dq)lR Otnfor \ R )rdr_lo + R r J l ( -T)d~ Otnr = [ jo ( Unr , d q) C~nr ( otnr R --~-)r--~r + --~-q)Jl\ R )]0 Otn~'oRRq) d{ r

J, (-~) }

d~o

= Yo(otn)R-~r(R ) + Oln(tg(R)Yl(Oln) -

lim

(d~)~.2

r-~0 r ~r

-

~

~b(n)

(126)

712

Analytical solution methods

(2.2.4-3)

d~o (R) -- - / ¢~o(R) The boundary condition for the resistance at r - R, namely ?-7 w ' makes the first two terms in the last equation (setting e - K---~)equal to -eJo(0/n)q)(R) + 0~nJ1 (0/n)qg(R), and therefore equal to zero if 0/,, are the roots of 0/,, J1 (0/,,) - eJo(a,,). Thus, we have found:

Hnr{d2q)

ldq9

~r2 + --r_~_r } _ _ r--+0 li m

(rdq 9 ~r

)_

2 0/n ,, -~q)(n).

(127)

The problem of Example 15 can now be solved very quickly. The transformed differential equation becomes: 2 R2 q3 =

d~b dt

~nr]

~(n, O) - f R rf (r)Jo(---ff-/ dr - f (n)

with

the solution of which is

o~2t

e(.,

The reverse transformation then gives, according to equation (126)

2 oo

2t

f(n)

~o( r, t) -- - ~ n ~ ° j g ( 0/n ) + J ~ ( 0/n) J 0 ( - ~ )

exp (

flZ'R2)

with f(n) - fog rf(r)Jo(f-~-)dr and 0/n being the roots of 0/,,Jl(0/n) - ~J0(0/n), where e - ~ (see Part A, problem 236.04).

4. Infinite Hankel transformation The operation in which an unknown function ¢p(r) is multiplied by the factor r Jo(0/r) and the product is integrated with respect to r from zero to infinity, is called the infinte Hankel transformation. This transformation yields a new function, which is independent of r and will be denoted as H{qg(r)} or shortly as ~b(0/)" q3(0/) -- H{go(r)} --

f0 °

(128)

rqg(r)Jo(0/r) dr,

Jo(c~r) in this expression is the Bessel function of the first kind and of zero order. The infinite Hankel transform of r1e-Cr' for instance, becomes lTe-CrI -- fo~ e-Cr Jo(0/r) dr -

40/2 -Jr-c2

(Laplace integral).

It is easy to show that, if the function qg(r) satisfies the condition that both q) and d__e dr vanish for r --+ oo H{d2g° I&P ~ r 2 -+- -d--r-r }r

(r dq9 - - r~olim ~ - r ) - 0/2~(0/).

(129)

Partial differential equations

(2.2.4-4)

713

Hence, application of the infinite Hankel transformation is possible for radial and axial-symmetric problems, for which the horizontal groundwater velocity at r - 0 is given and the head (or drawdown) and velocity at infinity can be assumed to be zero. Now, from equation (118) it follows that, according to equation (128), for v - 0 A(ot) can be replaced by f(ot) and so the inverse infinite Hankel transformation becomes simply" 99(r) -- H-l{~b(c~)}- fo ~ ot~(c~) Jo(rc~) dot,

(130)

Example 16. I I

I I

I I

I I

I

I

Consider vertical infiltration of water with a constant velocity q [L T - l ] , uniformly distributed over a circular area with radius R, into an assumed semi-infinite aquifer (see Fig. 23). This infiltration will cause an axialsymmetric rise of the original piezometric head, which after some time becomes steady and will then be a function of r and z: q9 = 99(r, z) = head.

I

I I I I I

1" r

z

Fig. 23. Verticial infiltration on a circular area. This boundary value problem can be translated mathematically as follows: 0299 1 399 0299 Or 2 i f - -r -~r q- ~3z 2 - - 0 ,

qg(c~,z)--0

oo(r,

-

o,

and also

099 ~(~ Or

'

q

a---V(r, o) Oz

z)-0,

0

099 ~(0 Or

'

z)-0,

for 0 ~ r < R forr>

R.

Infinite Hankel transformation with respect to r of the differential equation becomes d2~ dz 2

o/2~ - 0

(see equation (129))

714

(2.2.4-4)

Analytical solution methods

and transformation of the discontinuous boundary condition yields R

d(~ (ol, O) - fo ~ oqq9 d--z -~-z(r, O)r Jo(ar) dr = - qK [Jo r Jo(otr) dr =

qR

- ~ J1 (orR), K~

whereupon the transformed boundary value problem can be written as d2(p dz 2

c~2~b -- 0

q3(c~, oo) -- 0

'

and

qR

"~d--E~(c~, 0) - - ~ J 1 dz Kot

(otR).

The solution of this ordinary differential equation is ff(~, z) =

q R J1 (orR)e -~z Kot2

Inverse infinite Hankel transformation gives the desired solution q R fo c¢ 1 Jl (ROt)Jo(rot)e_Z~ dot go(r, z) -- --~

(see Part A, 523.02). This solution can be evaluated in infinite series and thus calculated for every value of r and z. Along the coordinate axes the integral function reduces to transcendental functions, for example, along the z-axis: q R fo ~ -g1 Jl(R(z)e -z~ d(z ~ ( o, z) - -~-

K

(~//R 2 +

Z2 -

Z)

(Laplace integral),

and along the r-axis the Weber-Schafheitlin integral (see Section 3.3.5, equation (170)):

q) ( r, O) --

q R f0 c¢ 1 Jl(R~)Jo(rot) dol -- 2 q R E ( r 2 ) --K --d 7r K -~

2q (r 2 -- R2) K (R -

7rKr

2)

-rT

+ 2qr E (R -~

2)

-rT

for 0 ~< r < R, f o r r > R,

in which K (z) and E(z) represent complete elliptic integrals of the first and second kinds, respectively (see Section 3.5).

2.2.5. Conformal transformation 1. Complex analytic functions

If x and y are real variables, then z -- x + i y is said to be a complex variable. Consider two complex variables z = x + iy and w = u + iv, and suppose that

Partial differential equations

(2.2.5-1)

715

a relation is given such that to each value in some region of the complex z-plane there corresponds one or more values of w in a well-defined manner. Then w is said to be a complex function of z, defined in that region, and we write

w--f

(z)

w - w(z).

or simply

If to each z in that region there corresponds only one value of w = f ( z ) , then the function f ( z ) is said to be single-valued. A function which is not single-valued, is called multi-valued. Since w depends on z -- x + iy and w = u + iv, it is clear that, in general, u depends on x and y, and so does v. We may, therefore, write

(131)

w -- f ( z ) -- u(x, y) ÷ iv(x, y),

and this shows that a complex function f ( z ) is equivalent to two real functions u(x, y) and v(x, y). It is known from complex analysis that a number of basic concepts, such as limits, continuity, derivatives, etc., concerning complex functions are quite similar to those in real calculus. So, for instance, a complex function f (z) is said to be differentiable at a point z = z0 if the limit df f'(z0) -- -=--(z0) -- lim f(z0 + Az) -- f(z0) Az-+0 AZ dz

exists.

(132)

A basic statement in complex analysis is the definition of analyticity of a function. A function f ( z ) is called analytic at a point z -- z0 if it is defined, and has a derivative, at every point in some neighbourhood of z0.

B

z+Az

AYt=iA

x

Consider the function w = f ( z ) in equation (131) to be analytic in a domain D of the zplane. Then, by definition, f ( z ) has a derivative according to equation (132), everywhere in D, where Az may approach zero along any path. We may set Az = Ax + i A y . Choosing path A in Fig. 24, we let Ay approach zero first and then Ax -+ 0. After Ay becomes zero, Az - Ax and by equation (132)

Fig. 24. Paths in the z-plane.

f ' ( z ) -- lim

u(x ÷ Ax, y) + i v(x + Ax, y) - {u(x, y) ÷ iv(x, y)} Ax

Ax-+0

=

lim

Ax-+0

u (x + Ax, y) -- u (x, y) Ax

+

i

lim Ax---,0

v(x + Ax, y) - v(x, y) Ax

716

(2.2.5-1 )

Analytical solution methods

Since f'(z) exists, the last two real limits exist. They are the partial derivatives of u and v with respect to x. Hence, f'(z) can be written

dw Ou Ov = +i~. dz Ox Ox

f'(z)-

(133)

Similarly, if we choose path B in Fig. 24, we let Ax approach zero first and then Ay --+ 0. After Ax becomes zero, Az -- i Ay and f'(z)--

lim

Ay--+O

u(x, y + Ay) -- u(x, y) v(x, y + Ay) -- v(x, y) + i lim iAy Ay--+O iA y

that is,

dw

f ' (z) . . . .

dz

Ou

Ov

= -i ~ + -0y 0y

(134)

because 71 -_ - i • The existence of f'(z) thus implies the existence of the four partial derivatives in (133) and (134). By equating the real and the imaginary parts of the right-hand sides of (133) and (134) we obtain

Ou Ov =~ Ox Oy

and

Ou Oy

~=-~.

Ov Ox

(135)

These basic relations between the partial derivatives of the real and imaginary parts of a complex analytic function are called the Cauchy-Riemann equations. These equations are fundamental because they are not only necessary but also sufficient for a function to be analytic, which can be stated by the following T h e o r e m 10. If two real functions u(x, y) and v(x, y) of two real variables x and y

have continuous first partial derivatives that satisfy the Cauchy-Riemann equations in some domain D, then the comnplex function f (z) = w = u(x, y) + iv(x, y) is analytic in D. The proof of this theorem will not be given here. It can also be proved that the derivative of an analytic function f ( z ) is itself analytic, so that u(x, y) and v(x, y) will have continuous partial derivatives of all orders. In particular, the mixed second derivatives of these functions will be equal:

02u OxOy

=

02u OyOx

and

02v OxOy

=

02v OyOx

Differentiating the Cauchy-Riemann equations of equation (135), we thus obtain

02U

021)

02U

02V

OX2

OxOy'

Oy2

OxOy'

02U

OxOy

02V =

Oy2'

OZu and

OxOy

02l) =

OX2"

(2.2.5-1)

Partial differential equations

717

This yield the following important result: Theorem 11. The real part and the imaginary part of a complex function f (z) u(x, y ) + i v ( x , y) that is analytic in a domain D are solutions of Laplace's equation, for two dimensions, 02u 02u V 2 u - - Ox 2 + ~ - 0 Oy2

021)

and

V2v-~+~-0, OX 2

021)

Oy2

in D and have continuous second partial derivatives in D.

A solution of Laplace's equation, having continuous second-order partial derivatives is called a harmonic function. Hence, the real and imaginary parts of an analytic function are harmonic functions. This is one of the main reasons for the great practical importance of complex analysis for solving geohydrological problems, because we have found in Section 1.2.4, equation (37) that the potential function ok(x, y) and the stream function ~ ( x , y) in steady two-dimensional groundwater flow satisfy the Cauchy-Riemann conditions, thus being the real part and imaginary part, successively, of a complex analytic function, that we shall denote by f2: f2 -- q~ + i~,

(136)

f2 is called the complex potential function for the flow. The (real) potential function 4~( - K~p) and the stream function 7r are, according to Theorem 11, harmonic and are thus solutions of Laplace's equation as has been shown already in Section 1.4.2-3, equation (16) (for the three-dimensional case) and Section 1.4.2-5, equation (26). Let us consider two-dimensional horizontal steady groundwater flow in a confined aquifer, which means that the motion of the groundwater is the same in all planes parallel to the x y-plane, the velocity being parallel to that plane. It then suffices to consider the motion of the fluid in the x y-plane. If we put z = x + i y then the xy-plane is assumed to be the complex z-plane and f2 becomes a function of z: or shortly

~2 -- F(z),

,f2 - ,f2(z).

From equations (133) and (134) it follows that

dz

=

~

Ox

+

i ~

Ox

-

-i

Oy

I

Oy

and also with equation (135): d~Q

O@

dz

Ox

.o4,

t ~

Oy

- - - Vx +

ivy,

(137)

718

(2.2.5-1)

Analytical solution methods

which gives the relation between the complex potential function and the D a r c y velocities in the two coordinate directions (cf. equation (32) of Section 1.2.3). If one of the three functions I2, q~ and 7r is known, the other two are determined and, in general, can be found easily.

Example 17. Radial flow of groundwater in a confined aquifer towards a fully penetrating well with a constant discharge Q. At a distance R from the well the drawdown is assumed to be zero. After some time a steady state will be reached with a drawdown s of the original head as a function of r in the domain 0 ~ r ~< R (see Fig. 25).

KD

R Fig. 25. Pumping well in a

circular island.

At a distance r (< R) from the well the velocity vector is directed towards the well and has the magnitude l)r = + K - -

ds dr'

in which the positive sign means that the velocity has the direction of the increasing drawdown, whereas Darcy's law requires the negative sign in the case of variable heads. The law of conservation of mass (volume, if the groundwater is homogeneous) applied to a circular cylinder with radius r and thickness D, gives ds

Q = - 2 r c r Dvr = - 2 r r K D r m dr

(negative sign, because the flow direction is negative and the discharge is a positive number). Q dr ds = - ~ - -

2rr K D r '

from which

For r - R the drawdown s - 0 ' so s q~ = - Ks becomes Q ln(R ) ~ b - 27rD

s - Q

2rrKD

~

Q

2rr K D

ln(~)

"

In r + constant.

Then the potential function

which is negative for r < R.

In Cartesian coordinates we have ~ b - 4-Y-B Q ln( x2+v2 R2 )' as r -- v/X ~ + y2.

(2.2.5-1 )

Partial differential equations

719

To find the stream function, we make use of the Cauchy-Riemann conditions" Od?

Q

Ox

2x

_-- 0gr

from which

~P -

Oy

4reD x 2 4- y2

Qx f dy 27c D J X 2 + y2

or

~P -- ~2zrQD arc t an ( Y ) 4- f (x ' Q Ox

=

v2 2reD 1 + x~

---~

4-f'(x)--

so

f'(x)-O

y

2Jr D x 2 4- y2

+ f ' ( x ) --

o4) Oy

Now

0.__.~¢= Q y 0y 2zr D x 2 + y2'

and

f (x) - c (constant),

from which it follows that Q a r c t a n ( y ) + c. ~P-- 2zrD x If we choose gt -- 0 along the x-axis (y - 0), we have c - 0, and ~p becomes"

~--

Q arctan( y)_ . 2zrD x

Now, from equation (136), we get

2rrD

R

) + i arctan(Y)

'

from which t'2 -- 2zrD as In z - In Izl + i arg z. The expression for 4, determines a set of equipotential lines, which are, in this case a set of circles with the well location as a centre: X 2 4- y2 _

R2

e x p (4zrDq~\ Q )'

and with parameter ¢ (negative or zero). The streamlines are straight lines through the origin with parameter Or.

y -- x tan (2zr D~p

Q ).

720

(2.2.5-2)

Analytical solution methods

2. Conformal mapping A continuous real function y -- f ( x ) of a real variable x can be exhibited graphically by plotting a curve in the Cartesian xy-plane; this curve is called the graph of the function. In the case of a complex function w -- w(z) -- w(x + iy) =

u(x, y) + iv(x, y), the situation is more complicated, because each of the complex variables w and z is represented by the points in the complex plane. This suggests the use of two separate complex planes for the two variables: one the z-plane, in which the point z = x + iy is to be plotted, and the other the w-plane, in which the corresponding point w = u + iv is to be plotted. In this way, the function w = w(z) defines a correspondence between points of these two planes. This correspondence is called a mapping or transformation of points in the z-plane onto points in the w-plane, and we say that w(z) maps its domain of definition in the z-plane onto its range of values in the w-plane. The point w0 = w(zo) corresponding to a point z0 is called the image point. If z moves along some curve C and w(z) is continuous (not a constant), the corresponding point w - w(z) will, in general, travel along a curve C* in the w-plane. This curve is then called the image of the curve C, and the word "image" applies also to regions or other point sets. To investigate the specific properties of a mapping defined by a given analytic function w(z), we may consider the images of the straight lines x - const, and y = const, in the w-plane. Another possibility is the study of the images of the circles ]zl = constant and the straight lines through the origin. Conversely, we may consider the curves defined by u(x, y) -- constant and v(x, y) = constant in the z-plane. These curves are called the level curves of u and v. An important geometrical property of the mappings defined by analytic functions is their conformality. A mapping in the plane is said to be angle-preserving, or conformal, if it preserves angles between oriented curves in magnitude as well as in sense; that is, the images of any two such curves, make the same angle of intersection as the curves, both in magnitude and direction, which leads to the following theorem: T h e o r e m 12. The mapping defined by an analytic function w -- f (z) is conformal,

except at points where the derivative -dZz dw is zero. To prove this theorem we represent a curve C in the complex z-plane in the parametric form

z --z(t)--x(t)+iy(t), where t is a real parameter.

Partial differential equations

(2.2.5-2)

721

The tangent to C at a point z0 - z(to) is defined as the limiting position of the straight line through z0 and another point Zl - z(to + At) as zl approaches z0 along C, that is, as At --+ 0. So, the tangent to C at z0 is represented by dz (to) - - dx dv d-T d"t(t0) + i-d7 (t0), and the angle between this tangent and the positive x-axis dz is arg ~/-(to). Consider now the mapping given by an analytic function w - w ( z ) defined in a domain containing the curve C. Then the image of C under this mapping is a curve C* in the w-plane represented by w = w(t) = w{z(t)} = u(t) + iv(t).

The point zo -- z(to) corresponds to the point w(to) of C* , and -d-F dw (t0) represents the tangent to C* at this point. Now, by the chain rule, dw

dw dz

dt

dz dt

o

do) hence, if dw (Zo) -¢ 0, we see that -d-F(t0) -¢ 0 and C* has a unique tangent at

dw

w(to), making an angle arg-dy(to) with the positive u-axis. Since the argument of

a product of two complex numbers equals the sum of the arguments of the factors, we have dw dw dz arg ~ (to) -- arg -7-(zo) + arg -7-(to). 0Z

at

Thus, under the mapping the tangent to C at z0 is rotated through the angle between the two tangents dw dz dw arg ~ (t0) - arg ~- (t0) - arg ~(zo). dz Since the expression on the right is independent of the choice of C, we see that this angle is independent of C, that is, the transformation w -- f ( z ) rotates the tangents of all the curves through z0 through the same angle arg -d?-z dw (Z0). Hence, two curves through zo whose tangents form a certain angle at z0 are mapped onto curves, the tangents of which form the same angle in both sense and magnitude at the image point w0. If we apply the foregoing considerations to the complex potential function ,(2 (z) in (136) we see that this function maps the relevant domain in the z-plane onto a domain in the S2-plane in a conformal way, that is, at every point in the domain where -d2-z dS? --/: 0, the angle between the images of two curves is the same as the angle between the curves, both in sense and in magnitude. In particular, the curves defined by ~b -- const, and ~ = const, in the S-2-plane, which are perpendicular to each other, are the images of the potential lines and streamlines in the z-plane, which are theretore also perpendicular to each other, as already has been proved in another way (see Section 1.2.4).

722

(2.2.5-3)

Analytical solution methods

The practical importance of conformal mapping for solving two-dimensional steady groundwater flow results from the fact that the total flow region in the physical p la n e (the z-plane) can be mapped onto a flow region in another plane without losing its character, as it can easily be proved that harmonic functions remain harmonic under a change of variables arising from a conformal transformation. Consequently, if it is required to find a solution of Laplace's equation of two independent variables in a given region D with known boundary conditions, it may be possible to find a conformal mapping which transforms D into some simpler region D*, for instance, a half plane or a circular disk. Then we may solve Laplace's equation subject to the transformed boundary conditions in D*. The resulting solution when carried back to D by the inverse transformation will be the solution of the original problem. The following special mappings, the quadratic transformation and the Joukowski transformation, will make this clear. 3. The quadratic transformation

An important transformation, frequently applied in potential flow, is the quadratic mapping 11) ~

(138)

Z 2.

If we switch to polar coordinates, we have Z -- re iO

and

w -- R e i~°

and thus

R e i~° -- r2e 2i0.

Then R - - r 2 and q) - 20 and.we see that circles r - r0 - constant are mapped onto circles R = r 2 -- constant, and rays 0 - 00 - constant onto rays q9 20o - constant. In particular, the positive real axis (0 - 0) is mapped onto the positive real axis in the w-plane, and the positive imaginary axis (0 - 2) in the z-plane is mapped onto the negative real axis in the w-plane. So the first quadrant 0<~0~<~-rr is mapped upon the entire upper half of the w-plane (Fig. 26). ly

v

z-plane

w-plane o

0 = 67.5 ° ~

0=45°

X

Fig. 26. Mapping w - -

z 2.

1 4

1

21 4

4 u

Partial

differential

(2.2.5-3)

equations

As the right angles between circles and rays in the z-plane remain right between their images in the w-plane, we see that the mapping is conformal, dw dw where at points for which -Uz -¢ 0 holds. As -~z - 2z, we see that at the (z -- 0) this derivative becomes zero and the mapping is not conformal: the are doubled there in this case. In rectangular coordinates the transformation w - z 2 becomes u + i 1) - -

x 2 -

y2 -k- 2 i x y

y2

and

723

angles everyorigin angles

and therefore U -- X 2 --

v -- 2 x y .

We see that the level curves of u and v are equilateral hyperbolas with the lines y = i x and the coordinate axes for asymptotes, while these curves are the orthogonal trajectories of each other. Another property of this mapping is that every point w ( u o , vo) is the image of two points z(xo, Yo) and z ( - x o , - y o ) in the z-plane. Finally, we determine the images of the straight lines x -- const, and y - const. The line x -- c has the image U - - C2 _

V -- 2cy.

y2,

Eliminating y from these equations, we find l) 2 - - 4 C 2 ( C 2 - - U ) .

This is a parabola with focus at the origin and opening to the left. Similarly, the image of a line y - c can be represented in the form U2 - - 4 C 2 ( C 2 + U ) ,

a parabola with focus at the origin and opening to the right. E x a m p l e 18. Y,l ~

z-plane

'"

P(zo)

l a/"~O

,

F i g . 27. ary.

W e l l , near a h y p e r b o l i c b o u n d -

Abstraction of groundwater by a fully penetrating well from a confined aquifer in the neighbourhood of a fully penetrating river bend. The shape of this bend can be approximated by an equilateral hyperbola, while the water level is assumed to be constant.

724

(2.2.5-3)

Analytical solution methods

We choose the asymptotes of the hyperbola as the coordinate axes; the equation of the hyperbola then becomes: xy -

~ a 2.

We consider the drawdown ~p(x, y) as a function of x and y caused by abstraction of a discharge Q in P ( x o , yo) and set ~0 - 0 at the hyperbolic boundary. To solve this problem, i.e., to find the flow pattern (streamlines and equipotential lines), we consider the physical xy-plane as the complex z-plane and try to find the complex potential £2(z) which level curves determine the required stream and potential functions of the flow. One level curve is already known: the equipotential line q5 - K99 - 0, which coincide with the hyperbola. Now from the mapping of w - z 2 (see the example in this section), we know that the level curves u and v are equilateral hyperbolas, their images in the w-plane being straight lines. So it is obvious to apply this transformation to our problem. Then the first quadrant of the physical z-plane is mapped upon the upper half w-plane and as v - 2 x y the hyperbolic boundary x y - - gal2 is mapped upon the straight line v - a 2 (Fig. 28) The image of point P ( z o ) , the location of the well, becomes P ( w o ) where wo -- uo + i vo -- x~ - y~ + 2ixoYo.

For further simplification, we apply a second mapping ¢ - w - ia 2, where ~" - rl + i 0, which consists of a t r a n s l a t i o n of the w-plane through a distance a 2 in the direction of the negative v-axis, yielding the g-plane with the straight boundary v - a 2 now becoming the 0-axis (0 - 0) of this plane (Fig. 28). We see that the original problem, to be solved in the z-plane, has been reduced to a much simpler problem in the g-plane by means of a transformation according to the function (z) = z 2 - ia 2.

The solution for the drawdown q)(rl, O) in the g-plane is the well-known solution for a well near a straight boundary with constant level, or also a positive well

1)0

I [

w-plane •

A

P(Wo)

?

I I IA

. . . . . .

a2

g-plane _

P(~o)

o0

?

I I

I t

I

t

I ,

Uo

Fig. 28.

w-plane and ~'-plane.

L

O0

Partial differential equations

(2.2.5-3)

725

(abstraction) at the point ~'o -- r/o +iOo and a negative well (infiltration) of the same strength at the point ~o - r/o - i 0o, that is

m

___.__~Q In { ( r / - / 7 0 ) 2 ÷ ( t~ + 0 0 ) 2 } 4rcKD ( r / - r/o) 2 + (0 - 0o) 2

(see Example 5 of Section 2.3.1-2). It can easily be shown that q~ - Kq9 is the real part of the complex potential function

~" -

~'o

S 2 ( ~ - ) - 2rrD As ~" -- z 2 - ia 2, ~0 - z 2 - ia 2 - x 2 - y2 + i ( 2 x o Y o _ a 2) and thus ~0 xg - z~ - i(2xoYo - a 2) -- z.~)÷ ia 2, we find for the inverse transform

O In ( z2 _z2 _ _z 2ia 2)

(see Part A, 337.04).

~2 (z) - 2reD

The real part of ,f2 is the potential function qS:

Q In { (x2- y 2

q~(x, y) --

4zrD

X2 + y2)2 ÷ 4 ( x y + XOYO- a2) 2 }

(X 2 __ y2 __ X 2 ÷ y 2 ) 2 ÷

4(xy

-

xoYo) 2 -

'

and the imaginary part the stream function ~p:

Er2 xy+xoyoa2,

~p(x, y) --

12xyxoyo, j]

Q arctan - arctan D X 2 __ y 2 _ _ X02 ÷ yo2 X 2 __ y 2 _ _ X02 ÷ yo2

2zr

•

Example 19.

.~. . . .

-_ _ _ - . . . - _ - _ _ - _ - _ - - . . _ . _ . - _ - _ . . _ . _ . . , _ , _ . _ . . . _ . . . . . .

_._._.._

,,.~

x

A

B

R

Fig. 29. Flow to a semi-infinite canal.

C

A

Flow in a confined aquifer towards a semi-infinite fully penetrating straight canal with constant drawdown h. The drawdown at a point C ( R , O ) is kept at zero (reference point). The x-axis is an axis of symmetry of which the negative part is a potential line (4~ = - K h ) and the positive part a streamline, chosen ~p -- 0 (Fig. 29).

(2.2.5-3)

726

Analytical solution methods

The complex potential plane 12 is shown in Fig. 30. Now we have to find the function S2 (z) that maps the z-plane onto the I2-plane. We see that the negative x-axis is mapped onto the line ~b - K h and the point B(0, 0) onto the point B * ( - K h , 0). The image of the posi* * B C tive x-axis becomes the 4~-axis between - K h and +oo of the O-plane. The A* 4' -Kh point c(R, 0) becomes the origin of the S-2-plane. If we translate the gt-axis Fig. 30. O-plane. through a distance K h to the left by means of the transformation w - S2 + K h, we see that the relation between the z-plane and the w-plane is the reverse of the mapping w - z 2 of Example 18, that is

ai

im

Z

aw 2

--

or

z

--

a(I2

+

Kh) 2

in which a is some complex constant. R and (point C*), so a - /~2h2

For z -- R (point C) we have ,f2 - 0

S-2- K h ( £

-1)

is the required solution of the problem in complex form (see Part A, 323.14). If we switch to polar coordinates, we have I2

= ~{

cos ( 0 ) + i sin ( 0 ) } - 1

Kh

and if we introduce the dimensionless variables S2 - Kh' s2 q~ -- K h '

-

ap ~ P K h' --

r -

r R'

x-

x -R

y and

y -

R

,

we find

rcos2(°/

r - ~(1 + cosO)-

~p___2= r s i n 2 ( 0 ) - ~r ( ~ - c o s o )

-

~l(v/X2 + y_2 + x)

~1V/X_ ( ~ + y~ -

and

~).

From this we may easily derive the equation for the set of streamlines with parameter ~, y2 = 4ap2(~2 + x), which are parabolas with focus at the origin and opening

(2.2.5-4)

Partial differential equations

727

to the right and the equation for the set of equipotential lines with parameter ¢ y2 __ 4(¢ + 1)2{(¢ + 1) 2 - x}, which are parabolas with focus at the origin and opening to the left. 4. The Joukowski transformation The mapping defined by R2

w -- z + - -

(139)

is called the Joukowski transformation, which is important for solving special problems in potential flow. Since dw dz

R2

-~-l

(z + R ) ( z -

--

Z2

R)

Z2

the mapping is conformal except at the points z - R and z - - R where d__~_w dz becomes zero. These points correspond to w - 2R and w - - 2 R , respectively. From (139) we find z 2 - w z + R 2 - 0 or 1

(140)

1 v/w2 _ 4R 2

To any value w there correspond two values of z, except to w = 2R and w = - 2 R for which there is only one image in the z-plane (R and - R , respectively). Hence, the points w -- + 2 R are branch points in the w-plane. Consequently, (140) maps the z-plane onto a two-sheeted Riemann surface, the two sheets being connected crosswise from w -- - 2 R to w -- 2R (Fig. 31), and this mapping is one to one. (Here it is assumed that the reader is more or less familiar with the representation of many-valued functions by Riemann surfaces.)

--

w-plane

z-plane b I f /

(

\\

/

/

/ -2R ,a I \ \

x \

/ \

/ J ..._.

_....

Fig. 31. The Joukowski transformation.

\ \ 2R ! _ /a u / / J

728

(2.2.5-4)

Analytical solution m e t h o d s

If we set z -- r e i°, then w -- u + i 1) -- r e i° + ~R2e_iO, from which it follows that

and

u-(r-t-R~Zr)COsO

v-(r-R~2r)

sin0

and also u2

1)2

a--7 + ~

- 1,

a--r+~

R2

where

and

b-

ir - ~ . R21

F

F

The circles r -- constant are thus mapped onto ellipses whose principal axes lie in the u and v axes and have the lengths 2a and 2b, respectively. Since a 2 - b 2 -- 4, independent of r, these ellipses are confocal, with foci at w = - 2 R and w = 2R. The circle r = R maps onto the line segment from w -- - 2 R to w = 2R. The circles r -- ot R and r - _R (o~ --fi 1) are mapped onto the same ellipse in the w-plane, 0t corresponding to the two sheets of the Riemann surface. Hence, the interior of the circle r -- R corresponds to one sheet, and the exterior to the other. Furthermore, from the expressions for u and v we obtain /,/2

1)2

cos 2 0

sin 2 0

= 4R 2 u2

l) 2

4R 2 cos 2 0

4R 2 sin 2 0

or

---

l.

The lines 0 -- constant are thus mapped onto the hyperbolas which are the orthogonal trajectories of those ellipses as they have the same foci as the ellipses (w - - 2 R and w - 2R). The direction of the asymptotes is defined by 2R sin 0 t a n c~ -

4

-

~

=

+tan

0

2R c o s 0 and thus ¢z - 4-0 which means that at infinity in the w-plane the image of a line in the z-plane through the origin has the same direction as that line. This is important in connection with problems involving uniform flow, as a Joukowski transformation does not change the direction of the uniform flow. E x a m p l e 20. A circular impermeable screen (for instance, around a building pit) is placed in a confined aquifer in which a uniform flow takes place. What will be the effect of this circular obstacle upon the flow pattern?

Partial differential equations

(2.2.5-4)

Y

729

We choose the positive x-axis in the flow direction and apply the Joukowski mapping R2

w--zq-~ Z

which maps the circle onto a line segment along the u axis, that is in the direction of the uniform flow, this direction being the same in both planes, as we have seen. So this line segment in the w-plane does not offer any resistance to the flow and the flow is purely uniFig. 32. Circular i m p e r m e a b l e screen in uniform flow. form. If the thickness of the aquifer is D and the strength of the uniform flow q (dimension L2T-1), the drawdown in the w-plane will be q5 - Kq9 -- qu, assuming zero drawdown along the v-axis. Then the complex potential a2(w) will become: a2 - ~Dw. The inverse transformation yields (see Part A, 336.11) x

from which the potential function and the stream function can be derived easily. In polar coordinates" ~b- ~

r +

cos 0

and

---~ r-~

.2)

sin0,

r

and in Cartesian coordinatesqx

q~---~-(l+x2

R 2

+y2)

and

qY

O----~-(1-x2

R 2

+y2).

According to equation (137) the total velocity at a point in the flow field is v -- I da2

q

R2

1'-71

At the point A(0, R) in the z-plane, Z A - - i R and thus VA = --~, which means that the velocity at A is twice the velocity of the uniform flow, independent of the radius R! Example 21. A fully penetrating water basin of elliptical shape in a confined aquifer in which uniform flow takes place. The flow makes an arbitrary angle otrr with one of the principal axes of the ellipse and has the strength q [L2T -1] (see Fig. 33). The equation of the ellipse in the z-plane is x2 y2 a--~, -~-1.

730

(2.2.5-4)

Analytical solution m e t h o d s

Y

q

ya .

.

.

.

x i

'''I

Fig. 33. Elliptical water basin in uniform flow.

Foci F1 and F2 in (c, 0) and ( - c , 0) with c 2 = a 2 - b 2. We know that the Joukowski transformation maps circles of one plane onto ellipses in another plane and inversely. As the problem of a circular basin in uniform flow is easier to solve than that of an elliptical one, we look for a Joukowski transformation that maps the ellipse in the z-plane onto a circle in the w-plane, say with radius P0. Considering the inverse mapping, there is one circle with radius R in the w-plane which is mapped onto the line segment from z = - c to z = c of the z-plane. So c corresponds to 2R (compare Fig. 31 with Fig. 33) and according to (139) we apply the transformation c2

z - w-F - - . 4w

(a) c2

We set w - p e in, then z - x + i y - p e in + -g-floe-in. from which c2

(,o + V)cos. x2

and

y--

( c2) p-~p

sinr/

or

y2

+

2

=1.

This ellipse represents the ellipse from

x2.-+7 ~v2-

c2

c2

a2 -- (Po + ~po)

1 for p - P0 where P0 has to be solved

and

which, if a > b, yields two solutions

(2.2.5-4)

Partial differential equations

1

and

po -- -~ (a + b)

731

1

P0 -- ;-(a - b). z

So, by means of the transformation 1v/z2

1

c2

the ellipse x2

y2

a--~- +

-1

in the z-plane is mapped onto two circles in the w-plane, a large circle with radius P 0 - gl(a + b) and a small circle with radius P0 - 51(a - b)" Now, the mapping according to the positive sign in the expression for w, that is 1

1

1/3 -- ~z -1t- ~v'/z 2 - c 2, transforms the right half of the ellipse (x > 0) into the right half of the large circle, and the left half of the ellipse (x < 0) into the left half of the small circle. For the mapping w - g1 z - ~ l ~ / Z 2 - C2 the inverse holds: the left side of the ellipse maps onto the left side of the large circle and the right side of the ellipse onto the right side of the small circle. So the required transformation of the outer region of the ellipse, the ellipse itself included, into the outer region of the circle with radius Po - ~l(a -+- b) this circle included, needs the mapping according to 1

w -- - z -+- sign x 2

v/z 2 -

c2

"

u

(b) Now we have to solve the problem in the w-plane, that is a circular basin in uniform flow (Fig. 34). The radius of the circle is P 0 - ~l(a + b) while the direction of the uniform flow remains unchanged under the Joukowski transformation, as has been proved before.

Fig. 34. w-plane.

In the first place we apply the transformation //31 ~ w e

-init

which means that the w-plane is rotated clockwise through an angle 7ra~, as IOle irJl -- ,oe i(rl-rr°~)

and thus

r/1 - r / - n'oe.

Analytical solution methods

(2.2.5-4)

732

In the wl-plane the circle remains the same but the uniform flow is parallel to the ul-axis. Secondly, we apply the Joukowski transformation of Example 20, with R - i po, that is

1)2 2Po

W2 -- Wl -- ~ ,

Wl

U2

-2po

which maps the circle onto a line segment along the v2-axis (Fig. 35) as can easily be shown. This line segment with assumed zero potential does not constitute any obstruction to the uniform flow, as it is perpendicular to the direction of the flow. So the solution in the wz-plane is simply q

Fig. 35. w2-plane.

S"2 -- ~ w2,

D

like as in Example 20. The solution in the wl-plane then becomes

_(W l - ~

q S2--D

Wl

and in the w-plane

q(we-iOlrr

P2eiOerr)

(c)

The end solution is equation (c) with w according to equation (b). If we write C2

C2 q- 4 p ~ e 2i~7r )

a"2 -- q e -i~Tr w -+-

D

4w

4w

c2

and write z - w + ~ according to equation (a) we find with equation (b)" q e -ic~rr { Z --

C2 k- (a -Jr b ) 2 e 2i°trr

/

2(z + sign x N/z 2 - c '2) ---- qe-i~ZrD { z - c 2 + (a + b)2e 2i~rr (z - sign x ~ z 2 2c 2

c2)}

Partial differential equations

(2.2.5-4)

7 33

or

S2 -- q I c2e-i~zr -- (a 4-b)2e ic~zr 2c 2 D I

+

c2e -i°err -Jr-(a +

b ) 2 e i°tn"

sign xv/z 2 - C2}.

2c 2

Now c2e -i~Tr + (a +

b ) 2 e i°trr

1

-- ~{ cos(c~zr)- i sin(oen')}

2c 2

a+b

+ 2(a - b) { c°s(crzr) + i sin(otzr/} a a -b

cos(otzr) +

b a -b

i sin(oe:rr),

and in the same way

c2e -i°~zr -- (a + b)2e i°~zr 2c 2

a

a -- b

cos (otzr)

a-b

i sin(c~zr).

The end result therefore is

S2 (z) - D(a q- b) [{a cos(crzr) + ib sin(crzr)} signxv/z 2 - c 2 - {b cos(c~zr) + ia sin(crzr)}z] (see 324.02 of Part A). Some special characteristics of the flow field will be considered in the following.

dX2dz

D(aq- b) {a cos(crJr) + ib sin(otzr)}

signxz C2

~/Z 2 __

-- {b cos(otzr) + ia sin(otrr)}] -- -Vx + ivy with Vx and l)y being the velocities the field where both these velocities are the stagnation points, for which

in the coordinate directions. At the points in are zero, there is no flow at all. These points dg2 - 0 holds, that is -d-rz

z2{a cos(~zr) + ib sin(otzr)} 2 - (z 2 - c2){b cos(crrr) + ia sin(otrr) }2 which can be evaluated to Zs -- + {a sin(c~zr) - ib cos(aTr) }.

734

Analytical solution methods

(2.2.5-4)

As Xs -- ± a sin(otzr) and ys -- Tb cos(otzr), we see that the stagnation points lie on x2

v2

the ellipse ~ + ~ = 1. It will be interesting to know the amount of groundwater that enters and leaves the elliptical basin as a result of the interrupted uniform flow. For that purpose, we determine the values of the stream function ~ ( x , y) along the circumference of the ellipse. The easiest way to do this is to make use of the parameter representation of the ellipse: Ze -- a cos t 4- i b sin t,

where

Xe --

a cos t and Ye - - b sin t.

This expression for Z e , substituted in I2 (z), gives the values of the complex potential function ff2e along the edge of the basin. As Ze2 _ C 2 - a 2 cos 2 t + 2iab sin t cos t b 2 sin 2 t - a 2 + b 2 - b 2 cos 2 t 4- 2iab sin t cos t - a 2 sin 2 t - (b cos t 4- ia sin t) 2, we have v/Ze2 - c 2

4-(b cos t + ia sin t) or

--

~/

(b Z 2 -- C 2 --

4-

Xe

a ) + i~ye

•

The parametric representation of the ellipse applied to the mapping according to (b) yields 2We -- a cos t + ib sin t 4- (b cos t 4- ia sin t) = (a 4- b)(cos t 4- i sin t), from which we may conclude that the positive sign here determines the mapping of the total ellipse in the z-plane onto the total large circle with radius l(a + b ) in the w-plane and the negative sign the mapping onto the small circle with radius L2 ( a - b) " So we have to substitute x/z 2 - c 2 - aXe b + i ag Ye in the expression for

S?(z) q D(a-b)

~e-2e - -

-

[{a cos(otyr)+ ib sin(c~yr)}

(b

a

aXe +i-~ye

)

{b cos(otn')+ ia sin(~yr)}(/e +/Ye)]

or

S'2e--" q SO ~e

--0

(a + b)i {

Xe

a

Ye } sin(oeyr) + --b-cos(olrr)

(as it ought to be) and

a P e = ~ q (a + b)[ -b-Yecos(otyr)

-

- - sin(otyr) }. Xea

-

~be - t - i l ~ e ,

(2.2.5- 5 )

Partial differential equations

735

The values of lpe at the stagnation points $1 and $2 for which zs -- +{a sin(otrr) i b cos(otn') } become lfirel -- -~q(a + b)

and

1/re2 = --~q (a + b),

respectively. Now the streamlines towards and from the points $1 and $2 are the outermost streamlines that reach the basin (see Fig. 33); the total flow through the basin is therefore determined by the difference between the two values of the stream function: Qe - 2 q ( a + b)

which is independent of the direction of the uniform flow! 5. The S c h w a r z - C h r i s t o f f e l transformation

In the previous section the solution of geohydrological problems by means of conformal mapping required a foreknowledge of the mapping properties of certain complex functions, such as the quadratic function and the Joukowski function. A more systematical way of finding a suitable function that maps the physical plane onto the S2-plane is the method, discovered independently by two German mathematicians, Schwarz and Christoffel, which will be discussed in the following. Consider a region in the z-plane bounded by a given polygon with vertices z l, z2, . . . , z,,-1, zn (Fig. 36). Now we try to find a function that maps this region onto the entire upper half of the w-plane; this means that the circumference of the polygon must coincide with the real axis u. The vertices zi (i - 1, 2 . . . . , n) have images ui (i -- 1, 2 , . . . , n)

\\ \\

Z2 c

t/

I t

/

tO

}

-I /'/n

//1 p

Fig. 36. Schwarz-Christoffel transformation.

*

U2

t

I

u3

u4

I ¸

Un_ 1 p*

U

736

(2.2.5-5)

Analytical solution methods

for instance, and one point P (here chosen between Z n - 1 and zn) becomes the point P* at infinity of the u-axis. It is assumed, as it were, that the polygon has been cut at P and stretched along the real axis of the w-plane from minus infinity to plus infinity. Suppose that this transformation can be realized by means of the function z F ( w ) (it turned out that the determination of this function is easier than that of the inverse function w = f ( z ) ) , then F ( w ) should be analytical in the whole region, eventually with the exception of a limited number of isolated points (poles) where dz = F ' (w) has no finite value. a~ The image of w in z is conformal, which means that angles between curves in the w-plane remain the same as the angles between their images in the z-plane, both in sense and in magnitude, except at points where ~d z - 0 . However, as we see, at the vertices zi and ui, the angles between the adjacent sides of the polygon in the z-plane differ from the corresponding angles in the w-plane, the latter all having the magnitude 0 or rr. So the transformation z = F ( w ) is not conformal at dz - oo there. The form of the derivative the vertices, which means that ~dz _ 0 or T~ of the function thus may be: dz

-- F ' ( w ) -- c l ( w

-

ul)al(w

-

u2)a2(w

-

u3) a3"'"

dw X (W--Un_

1) an-1 ( W - - U n ) an .

(141)

dz = 0 for w -- un (analytical, not conformal) and if a,, < 0, then If an > 0, then T~ ddz __w_w= C~ for w -- Un (not analytical, pole). Now we let a point z travel along the sides of the polygon, its image w then following the real u-axis. Suppose that z lies between z l and z2 (Fig. 36) on the line segment ZlZ2 with parameter representation z(t) = t ÷ i(tanOlt + c). The direction of the tangent at a point z to the curve z(t) is determined by a r g ( ~ ) (see Theorem 12, Section 2.2.5-2), which in this case is equal to the direction of the straight line itself: dz = 1 + / t a n 01 dt

arg(d~t ) - arctan(tan 01) - 01.

dz As long as z is situated between Z l and za, the value of arg(7?) remains constant and is equal to the angle between the straight line through Zl and z2 and the positive

x-axis. The image point w of z shows a similar behaviour: as long as w finds itself between Ul and U2 the value of arg( -aS-) dw remains constant and is equal to zero. Now z -

F ( w ) and

dz

dz dw

dt

dw dt

(2.2.5-5)

Partial differential equations

737

and from this a r g ( - ~ - ~ ) - arg(dd-@Zw)+ a r g ( - ~ )

-arg(

dz

-01,

if z lies between z l and Z2. As soon as the travelling point z passes Z2 in order to pursue its way along the line segment z2z3 the value of arg(-~) jumps by an amount or2, whereas arg(-~) dz remains zero, so that arg(T~ ) is increased by an amount or2 as well. From equation (141) we have arg ( ff----~zw) -- arg C1

-3t.

alarg(w

-+- an-1 arg(w

-

-

Ul) @

a2 arg(w

U n _ l ) -Jr- an

-

u 2 ) -~-

a3 arg(w -- u3) + . . -

arg(w - Un).

Now, the argument of the difference of two complex numbers wl and W2 equals the angle between the straight line that connects the points wl and w2 and the positive real axis (see Fig. 37). So if w lies on the real axis between u l and u2 (see Fig. 36) then a r g ( w - u l ) = 0, a r g ( w - u 2 ) = Jr, a r g ( w - u3) = Jr . . . . . a r g ( w - Un-1) = Jr and arg(w - un) = 0. From thiswe find

1)

W2

W 1 -- 1132

0 U F i g . 37. A r g u m e n t o f the d i f f e r e n c e of two complex numbers.

arg

-- arg cl + azzr + a3zr + .-. + an_lzr

if w lies between u l and

U2,

and

arg(ff~Zw) - arg cl + a3rc + a47r 4- ...-4- an_ire if w moves between U2 and u3. Hence, we may conclude that arg( T-~) dz remains constant, if z moves between two vertices of the polygon and decreases by an amount airr as soon as it passes a vertex. Compared with the increase of Ofi we found before, we have Ol i - - --7rai and thus dz

dw

=

Cl(//) -

,~1

Ul)--Y-(w

-

~2 u 2 ) -'5-- • • • ( t o -

Un)

Otn

Jr - - Cl

i~i

(113 -

c~i

ui)--Y

i=1

from which

clf

n

°ti

1--I(w - ui) - ~ d w + i--1

C 2.

(142)

Analytical solution methods

(2.2.5-5)

738

This result is called the transformation function of Schwarz-Christoffel. That arg(dd~zw) also remains constant if w moves from Un-1 to u, via the point P* at infinity, can be shown as follows: If w moves between u,-1 and + o e arg(dd-~zw)-arg Cl, and if w moves between - o e and u, arg (dd--~Zw) - =

arg

cl

arg Cl

+ a l : r r -+- a 2 7 r -k- . . -

tyl

~

o l 2 -1- . . . .

+

a._lzr

O/n-1

-

O/n

- arg c1 - 2zr - arg C l .

The transformation (142) maps the real u-axis onto a polygon P' of which the sides make angles O~i with each other. As long as the real constants ui (i = 1, 2 . . . . , n) and the complex constants Cl and c2 are arbitrary, the polygon P' will not coincide with the given polygon P in the z-plane and will not even be similar (equal form) to P. The constants ui are responsible for the similarity of the two polygons P' and P and it is easy to see that if two polygons with n vertices have the same angles, they become similar if n - 2 connected sides of P' have a common ratio to the corresponding sides of P; this condition is expressed by means of n - 3 equations in the n real unknowns ui. Thus, three of the numbers ui can be chosen arbitrarily, the point at infinity included; the remaining n - 3 unknowns can then be determined in the transformation of the u-axis onto the polygon P' which is similar to the given polygon P. Now the size and position of P' still have to be adjusted to match those of P by introducing the appropriate constants cl and c2, as multiplication of a mapping function by a complex number causes a rotation combined with a dilatation or contraction and addition of a complex number to a mapping function gives a translation. If one point is chosen as the point at infinity, for instance, u,, = cx~, then the factor w - u, vanishes in the Schwarz-Christoffel equation ( w - u, becomes 1) as can be seen as follows: equation (141) could also have been written as dz - - C 3 ( l / ) w /,tl ) a l ( W m

.

b. / 2 ). a 2 .

.

(1

~)""

dw with c3 instead of C l . If u. --> oe, then -~ --+ 0 and (1 - ~)~. --~ 1. Un Un The Schwarz-Christoffel transformation is a powerful method for solving geohydrological problems concerning flow fields of which the boundaries are straight lines, making fixed angles with each other, thus representing the contour of a polygon. An additional advantage of the Schwarz-Christoffel mapping is that also degenerated polygons, such as semi-infinite or infinite strips, are mapped onto the real u-axis. By means of the Schwarz-Christoffel transformation, we may find z as a function of w, but also the I2-plane can be mapped onto the w-plane by means of a Schwarz-Christoffel transformation, finding I2 as a function of w as well. Elimination of w from z = z ( w ) and I2 = I2(w) yields the required solution I2 = S-2(z)

Partial differential equations

(2.2.5-5)

739

or the inverse solution z -- z(Y2). The following examples may give an impression of the valuable features of the Schwarz-Christoffel transformation for solving geohydrological problems. E x a m p l e 22.

~o=0

Two semi-infinite reservoirs separated by an impermeable screen lie on top of an aquifer with transmissivity KD. The reservoirs have different levels, causing a steady flow of groundwater through the aquifer (Fig. 38). The differential equation with boundary values for the drawdown ~0 = ~0(x, z) can be written

rp=H

as:

Fig. 38. Two semi-infinite reservoirs with different levels.

0299

02(t9 -

H -0

OX 2 1"- OZ 2 -o(x, o ) -

o,

~9(O,z) -

'

7'

~o(~,z)

-0

'

Oq°(x ' D ) - O

(positive z-axis chosen downwards; aquifer thickness D). We now introduce the complex physical plane ( - x + i z. I I

Jr

i

- 7 .... _ v _

T

'

O

t I I

Jr

1/,=0 B Z

I I

i/

Fig. 39. Physical plane.

Jr at A is twice an amount 7, so o/A - - Yr.

m

2

t

2

The flow domain is bounded by a semiinfinite strip which may be considered as a degenerated polygon with three vertices A, B and C, one vertex lying at infinity (Fig. 39). The angles between the sides of the polygon at B and C are ~2 ' whereas the jump

740

(2.2.5-5)

Analytical solution methods

¢ =IKH

-(

) "

(ap = 0)

8

[

c

I

T

1

-1

l

+1

¢=0

i g

1)

Fig. 40.

w-plane.

Applying the Schwarz-Christoffel transformation (142) we have

d~

dw

=

A(w

1 b/A)-I(w

-

--

UB)-2(N

1 -- blc)-2

with A being an arbitrary complex constant. We choose UA -- co, UB -- --1 and uc -- + 1 (see Fig. 40) with which we find d~" dw

~ 1 = A(w + 1 ) - 7 ( w - 1)-7

and thus

~" -- A

f

dw ~/W 2-

1

+B

with B also an arbitrary complex constant. ~" - A arccoshw + B

or

~" - (A1 + ia2) ln(w + v/w 2

i) + B1 + iB2.

The point C in the g-plane, for which ~'c - 0, has its image in the w-plane, for which wc - 1, so that 0 - (A1 + i A2) In 1 + B1 + i B2, from which B1 - B2 - 0. Point B" (B -- iD and w8 - - 1 whence iD -- (A1 +iAz)iJr, from which A1 - ¥D and A 2 -

0. The Schwarz-Christoffel transformation thus becomes

~ ' - - - aDr c c o s h w 7r

or

w-cosh

(Tr()D --=- .

Partial d i f f e r e n t i a l e q u a t i o n s

(2.2.5-5)

741

The groundwater flow takes place from the top of the aquifer (horizontal boundary with KH q) = 0) towards the line B C (vertical boundary with drawdown 99 - 1H) with a streamline ~ - 0 along the bottom of the aquifer, whereas at the point C the stream function. theoretically tends to infinity. In the 22-plane, 7~ IC therefore, the flow domain is a semi-infinite strip with C being the point at infinity of Fig. 41. ~-plane. the imaginary axis ~ (see Fig. 41) and A B (~ - 0 ) along the real axis qS. Now we also map the 22-plane onto the w-plane by means of the SchwarzChristoffel transformation A

B

r

d22

1

= A(w

dw

or, as

1

U A ) - ~ ( W -- U B ) - ~ (W -

-

UA - - CX~, UB

d22

-- --1 and

uc --

Uc) -1

+1,

A

dw

(w-

1)~/W + 1

and thus 22--A

f

dw (w-1)~/w+l

+B

or

22 -- (A1 ÷ iA2) In

dw+ 1 - ,/5)

~ww ÷ 1 ÷ ~

+ B1 ÷ iB2.

Point A" WA - - CX~, 22A - - 0 - - (A1 + i A 2 ) In 1 + B1 + and B2 - - 0. Point B" WB - - - - 1 ' 22B - - -5-KH = (A1 ÷ i A 2 ) l n ( - 1 ) which A1 - - 0 and A 2 KH Hence 2~ • 22

2Jr

-i In

from which B1 -- 0

-- (A1 ÷ i A z ) i J r ,

from

~/w + [ +

With w -cosh(--~-) we have ~ the end result becomes: - ~2jr i In

iB2,

+ 1

g-g) ~ - 1 cosh(2-5) + 1

with 22 - ~b + i ~ and

~ -

x + iz

- v /cosh(--~) or

÷ 1

- v / 2cosh2(~-5)

,f2 - ~K H i In coth ( zr~",~ Jr \4DI

(see Part A, 355.1 1).

and thus

742

Analytical solution methods

(2.2.5-5)

Equating the real and imaginary parts of this equation we find, making use of some standard formulas of complex analysis,

4~ (x

'

z) -

{

KH sin(~) - - - - arctan ~x Jr sinh(~-fi)

/

and ~rz / cos(~-$) ~p(x, z) - -

7r

arctanh

cosh(~--~) "

E x a m p l e 23. Horizontal flow between an infinite and a semi-infinite fully penetrating canal which are parallel to each other in a confined aquifer. The canals have different levels. The differential equation with boundary values for the drawdown q) = ~o(x, y) can be written as qg(X, 0) = 0,

02(/9 ] 02q9 = 0, OX2 Oy 2

qg(X, b) = h

0q° (+oo, y) = 0,

qg(-cx~, y) = 0,

for x >~ 0,

q)(x, cxz) -- 0.

Ox

The physical plane is shown in Fig. 42; we see that the infinite canal A B with drawdown q9 = 0 loses water at one side, which flows horizontally through the aquifer and reaches the semi-infinite canal with drawdown h on both sides. The contour of the degenerated polygon becomes A B D A with angles O / a - - 2 7 " f , O/B - - 7"f, c~o -- - z r and the Schwarz-Christoffel transformation gives:

7/"

2

i iI

i I

dp = Kh .

.

.

.

.

.

.

.

.

.

- - - -

I

--

i

.

.

.

.

.

,,

t

ap=0

I

C A

5/"

A .

.

i

4>=0

B

x

~ 2

Fig. 42. Flow between an infinite and a semi-infinite canal.

--

(2.2.5-5)

Partial differential equations

743

dz

dw

A(w

=

-

UA)-2(W

w e take U A - - (:X), U B - - 0

dz dw

w-

- - UD) +1.

-- UB)-I(w

a n d u D - - 1.

1

- - A ~

w

and z - A ( w -

l n w ) + B. Point D: wo = 1, zo = i b - - A + B, from which B -- i b - A .

-e* -c* •

I

E

C

Point C: w c -- real and negative (between A and B on the u-axis), for instance, w c = - c * (c* > 0).

1 I

B

u

D

zc i b-

A

= 0 = A(-c*

-

ln c* -

A , from which A -- L and c* +

l n c * + 1 = 0, B = i b - A so that

Fig. 43. w-plane.

in-) +

= ib-~

~'

b z - --(w - lnw - 1 + in). 7/"

A

E

If we choose the streamline ~ - 0 through point D, coming from a still unknown point E on the infinite canal, the flow domain in the ,f2-plane becomes an infinite strip (Fig. 44). The Schwarz-Christoffel transformation that maps the S2-plane onto the wplane satisfies

A

~=0

D Kh

dS2 ¢

A(tO

dw UA

-- UA)-I(to

-- UB) -1

(X) and UB - - 0 .

--

Hence

B

B

dS2

A

dw

to

Fig. 44. S2-plane.

Point D" w o - - 1, I2o -- K h -- B .

and

I-2 -- A l n w

+ B.

744

(2.2.5-5)

Analytical solution methods

Point E: wE = real and negative (between A and C on the u-axis), for instance, WE = --e* (e* > O) X?E = 0 = A lne* + iArr + K h from which e* = 1 and A=i~;

Kh 7r

hence Kh Y2 -- ~ ( 7 r

+ i In w)

or w - e irr(1 - K-' s?h) -- _e_iTrs? g---hSubstituting this value of w in the expression for z as a function of w, we find the end result z---with

b (n'X2 i rr Kh

z = x + iy

e

_i 7rS2

-~-1

)

(see Part A, 323.01),

and ~2 = ~b + i~p.

2.2.6. Successive transformations

In problems involving several independent variables, a great economy in calculations can be achieved by making integral transforms successively with regard to those variables. Either a Laplace transformation may be used first to remove the timevariable, followed by other integral transforms on the space-variables, or successive integral transforms may be used on the space-variables. This method of successive transformations makes it possible to reduce a partial differential equation to an algebraic equation in the multiply transformed dependent variable, which can be solved directly. The obtained solution will now still have to be subjected to a number of inverse transformations in order to get the desired solution. The several transformations and inverse transformations may be undertaken in different sequences, yielding solutions that may differ considerably from each other, dependent on the way in which the solution has been obtained. However, as the solution of a well-posed boundary value problem is unique, the obtained solutions must be various forms of the same result. In this way, alternative expressions for the same solution are obtained, a fact that may be very useful for further evaluation of the solution. To illustrate this method of successive transformations we consider the following examples.

Partial differential equations

(2.2.6)

745

Example 24. Cf. Example l0 of Section 2.2.3-2 (Fig. 15). Non-steady, threedimensional axial-symmetric flow of groundwater, caused by abstraction of a constant discharge Q from a leaky aquifer by means of a partially penetrating well. The boundary value problem, becomes, with 99 - ~o(r, z, t) - drawdown: 02(t9 1 O q) Or 2 q_ _r -~r q

02(t9

0(t9 07~2 _ fi2 a t '

99(r, z, O) - 0

OCP(r, O, t) - O, Oz

'

099 lim(r ) -- F ( z ) - r-+o -~r

O--~(r, D t) Oz '

{0

-

~o(r, D, t) Kc

for0~
-27rK(b-a)

f o r a ~
q ) ( ~ , z, t) - - 0 .

The solution of this boundary value problem will be shown step by step, while interchanging the steps will yield a number of interesting relations between the obtained results. First step. Laplace transformation with respect to the time-variable t yields, according to Theorem 2 of Section 2.2.2-1 and equation (16): 02@ 10q~ 02@ ~)r 2 _+_ _r -OTr q_ ~ z 2 _ f12 s (# -- O ,

a--~-~(r, 0, s) - 0 Oz

a-~ (r, D s) - _ ~ ( r , D, s) '

(O~)

lira r r~ O

--

-~r

- ( r, z , s ) , (p- -- ~o

Oz

F(Z) s

=

'

Kc

'

forO~
{ 0

Q

1

-- 2 rc K ( b -- a) s

for a <~ z <~ b.

¢(c~, z, s) - 0 . Finite Fourier cosine resistance transformation with respect to the space variable z gives, according to (84) and (85):

S e c o n d step.

d2~ l d~ dr 2 t r dr

o~2 Dn2( ~ _ f i 2 S ~ ) __ O,

~- -- ~p(r, (P °/n' s) '

' i mo( r -dTr d ~ ) - - f 0 D F ( z ) c ° s ( --~-) r-~ °enz dz

= -

-

2 7 r K ( b - a) s

cos

\--D

dz - - ~ S n

SOen

OllZ a with P _ 2Jr/((h-a) Q and Sn - s i n ( ~ -) - sin(--b--), ~(oo, otn, s) D the roots of ot tan oe -- c = x---7"

0, while Oen are

746

Analytical solution methods

(2.2.6)

Third step. Infinite Hankel transformation, finally, with respect to the space variable r yields according to equation (129) with co instead of or" ^

PD& --

2

^

^

CO2~_ ~C~n ~ _

fl2S~ -- 0 "

S Oln

This is an algebraic equation with the solution"

PD&

^

~0(oo, ~ ,

s) =

(a)

~2

"

s~.(co: + ~ + ~:s)

Expression (a) represents the threefold transformed solution of the problem, which requires three inverse transformations in order to obtain the end solution q9 as a function of the variables r, z and t. The result (a) is independent of the order of transformations; so, for instance, if we denote the Laplace, Fourier and Hankel transformations shortly by their first letter, the sequences L FH, L HF, F LH, F H L , H L F and H F L all give the same result (a), as can easily be verified. Now, the inverse transformations will be given in the same sequence LaplaceFourier-Hankel as the transforms.

Fourth step. We apply first the inverse Laplace transformation to equation (a). Setting 2 M _ co2 _l_ °t._.E. n

D2

for a moment, we have:

, --

s ( m + fl2S)

1(, --

m

s

,) m

'

s + -fl

and thus (see equations (19) and (21))" Mt

~(co an t ) ' '

PDSn c~nM ( 1 - e

e2

).

(b)

Fifth step. The inverse finite Fourier cosine resistance transformation applied to expression (b), gives, according to equation (86)

~(co, z, t) = 2P

~ Sn cos (1-e " n=00en M 1 + ~2+62

-

t~2)

(c)

Partial differential equations

(2.2.6)

747

Sixth step. The last step consists of the inverse infinite Hankel transformation of (c), which becomes, according to equation (130):

(p(r, z, t) -

Q

~

rcK(b - a)

f0

D,

(OlnZ]

sin ( - q - ~ ) - s i n ( ° ' ' a l 8

Oen(1 -31- o~2q_t¢2)

=

O)2 --i-- ~2

{, .xp(-

COS --D--]

a'2+~ } f12

(d)

t) Jo(rw)dw,

provided, that it is allowed to interchange the summation and integration. With 06, as the roots of ot tan ot - s - ~D and f12 - ~KSD ' the expression (d) represents the end result. It can readily be verified that also the six possibilities of the inverse transform sequences all lead to the same result (d). So it may be stated that, in general, if we apply successive transformations and inverse transformations to all the independent variables of a differential equation and its boundary values, the end result will be

independent of the order of the transforms and inverse transforms. If we apply successive transformations to all but one independent variables instead of to all of them, an ordinary differential equation will then be left, which generally may be solved for the untransformed variable. Application of the still necessary inverse transformations will give the end result, which may be different from the result found before. For instance, in our example, if we should have omitted the third step and have solved the ordinary differential equation with boundary values for ~(r, otn, s), after the second step, we would have found: (cf. Example 6 in Section 2.2.2-3):

((°2)

.... ~(r, an , s) -- PDSn Ko r

- ~n + fl2s

(e)

SOl n

Inverse Laplace transformation of (e) becomes (see equation (51)):

P D S,~ q3(r, Ogn, t) -- ~ W Or'n

( °tzv

olnr

,',2,--2'

'

(f)

and finally inverse Fourier transformation, according to (86) gives _

qg(r, z ' t)

Q

~

2rcK(b - a) n~0 =

sin ( - ~ ) - sin (~-~-~) otn(1 "_ql_ot2q_82) --7-.--

co. ) (<, o.r)

× COS \ - - 5

W

fl202,

(g)

[)

with o~n being the roots of ot tan c~ - s - ~D and W - well-function (see Part A, 532.05).

748

Analytical solution methods

(2.2.6)

Here we have found an alternative expression for the end solution of the problem. Compared with solution (d), it is obvious that we have also found the equality:

f0 °

002 + 00 p2 { 1 - exp ( -

002/32+P2t)

jJo(r00)do) _l_~Wk_~,pr).( p2t

(h)

If we had first applied the inverse Hankel transformation to equation (a) during the fourth step, we would have found:

¢)(r, Otn, s) -- PDSn fo ~ S Oln

~co 002 + ~

Jo(r00) do).

(i)

_it_ f12 S

The expressions (e) and (i) for ~ must be equal and thus we have found that, in general,

f0 c~z002_+_ 09 p2 Jo(r00) do) - Ko(pr),

(j)

which may also be proved in other ways (equation (214) of Section 3.3.5). The alternative expression (g) for the end solution has the advantage, in this case, compared with expression (d), that it has a simpler composition, as the well-function, though defined as an integral (see equation (50)) can be evaluated numerically somewhat more easily than the integral of equation (d). Thus, it can be stated that, in general, it is worthwhile to investigate alternative solutions for a problem. Remark. The steady state of the problem may be derived from equation (d) with t --+ c~. According to equation (j) it becomes:

q)(r,z)-

Q

.K(b-a

oo

~~0 =

Otn a

sin(~)-sin( E

) (OtnZ --if) cos - 5 )

(Olnr

)(k)

If the well is fully penetrating, we get, with a = 0, b = D and e = O~ntan Otn: .

. . . cos Ko Jr K D = otn + sin Otncos Oln \--D \--D

(1)

D with O~n being the roots of ot tan ot - e - K--S" This formula represents the exact axial-symmetric solution for the abstraction of groundwater with a constant discharge from a leaky aquifer by means of a fully penetrating well. Compared with the "classical" solution of de Glee: Q Ko qg(r, z) = 2zrK------D

(r)

(m)

Partial differential equations

(2.2.6)

749

with )~ = ~ / K D c , in which it is assumed that only horizontal flow of groundwater takes place and that the leakage through the semi-permeable layer is uniformly distributed over the thickness D of the aquifer (cf. Section 1.4.2-6), we see that in formula (1) also vertical flow has been taken into account. Solution (m) is valid only for large values of the leakage factor c, as can be shown as follows. For large values of c, the product K c becomes large with respect to the thickness D D of the aquifer and thus e - 27~ becomes small. Now, if we consider the roots o~,, (n = 0, 1, 2, 3 , . . . ) of c~ tan ol = e as the points of intersection of the two curves yl - t a n c ~

and

y2--,

8 o/

a tangent and a hyperbola successively (see Fig. 45), we see that for small values of e the values of oe all tend to rr, 2rr, 3~ . . . . except the first one which approaches the value x/7. Because of the factor sin otn, all terms of the series in expression (1), for which n > 0 disappear. As or0 is small, sin a~0 may be replaced by o~0, cos oe0 by 1 and cos(~-~-j) also by 1, so that the only term of the series that is left, becomes: r

1

r

So with )~ - ~ / K D c , the result is expression (m).

- i! ,/2 /

I

I

I

i Ol

Fig. 45. Roots of ~ tan ct = e

2

2 2

750

Analytical solution methods

(2.2.6)

Example 25 (see Fig. 12). We consider the problem of Example 7 of Section 2.2.3-2 and apply first the Laplace transformation with respect to t, which leads to: d2q3

fl2sq5= 0,

q3(0 s) = 0,

dx 2

q3(b, s) = h

'

s

(a)

according to Theorem 2 of Section 2.2.2-1 and equation (16). Secondly, finite Fourier sine transformation with respect to x gives, according to equation (66): -

q5 -

(-1

b

0

s

with the solution: ~(n,s) =

(_ 1)n+l nrr h b7

(b)

~s + (,¢)2.

If we write ~ ( n , s ) = ( - - 1 ) n+l

1 )

hb (1 /'/~

s+~-~ n2rr2

s

'

the inverse Laplace transformation becomes (see eqs. (19) and (21))

O(n' t) -- (-1)n+l hb { 1 - eXP

n2rr2t ) } f12b2

,

(c)

which corresponds to equation (68). Inverse Fourier transformation then gives equation (69). If we solve the differential equation with boundary values (a) directly, we find h sinh (fix ~/s-) q~(x, s) -- s sinh(flb4~ ) w

,

(d)

First inverse Fourier transformation, applied to solution (b), also gives an expression for q3(x, s):

~(x, s ~ -

2Zrhbzs k. - , ~2s(-1)n+ln (nrcx] + ()'~'2 sin ~ b J "

(e~

The right sides of both (d) and (e) must be equal, thus, in general, with a = fl~/s: cx)

a2b2 nt- n27r 2 sin n=l

(n x) --if--

sinh a -- ~

sinh(ab)

for - b < x < b.

(f)

Partial differential equations

(2.2.6)

751

The inverse Laplace transformation of (d) can be performed in two different ways. First, we apply the inversion theorem (Theorem 9 of Section 2.2.2-3a, equation (31)). We find: y +iot

99(x, t ) -

1 lim e stqS(x, s) ds. 2zri ~-,c~ ~,×-i~

(g)

The integrand eSt,(x, s) is a single-valued function of s, because all roots vanish, as can easily be shown by using the series representation for the hyperbolic sine function: Z3

Z5

sinhz -- z + ~ + ~ + . . . The points for which the integrand becomes infinite, are s - 0 and the zeros of sinh(flbv/7 ). sinh(flb~/7) - - i s i n ( i / 3 b ~ ) - 0, from which i f l b ~ n - nrc (n - 1 . .2. . .) and so sn - _(,-r)2 ~ . Thus, so and s,, are single poles of the integrand, lying in the origin and on the negative x-axis. Now, as in Example 3 of Section 2.2.2-3, we consider the line integral (g) as a part of a closed curve which should not pass through any of the poles, such as the circle A B C of Fig. 5. The solution then will be the sum of the residues at the poles. The residue at s - 0 becomes Res [e stqS(x s)] - [se stqS(x, s)]s= 0 = s=O

'

hx

-b

and Res [eSt~b(x s)] -- ~ ,,,=s,,

'

sinh (/3x x/~-)

n=, = Z

s ~ sinh(flb~/7) h exp{-(~_~.)2

he s' sinh (fix x/s-) ] s=s.

-- Z

n=,

~,~b

~

t} sinh(n-'~-i !

,,'~ 2"-7-COS h(,,,~) "-7-

n= 1

= h ~c~ - 2 i sin/'~Jrx _ T _ ) e x p { _ ( ~ nJr ) 2t_" } n=l

--inrc cos(nrr)

Thus, the end solution is: 2h ~

hx

o(x, t ) - T +

-

-

Y/"

n--I

(-1) n

- -ns i n ( t-/ ~;X) e x p{-

as has already been found (equation (69)).

//7/" 2

cosh(/3bxF)J.,,=s,

752

(2.2.6)

Analytical solution methods

Secondly, we may write equation (d) as" h

h e~ x ~ - e - ~ x ~ ~(x,s)

-

-

e-~x~

s et3b~ (1 - e-Z/~b4g:)"

s e~b,/-~ - e-~b,/-~

Now, as

e~ x ~ -

1 is the sum of the geometric series O0

1 + y + y2 + . . . .

~

yn,

if lYl < 1,

n=O

we have for positive s" OO

~ ( x , s) - -h{ e_~(b_x),/-~ - e - ~ ( h + ~ }

S

e_2n~bvG

n=0

or h

h oo

OO

Cp(x, S) -- -- Z s

e-Cl"F{(2n+l)b-x}

-- ~ S

n---0

e -~'fd{(2n+l)b+x}

(i)

n--0

The inverse Laplace transformation can be found if the general inverse transform

L,/se1

-k./~

v/

is known. Now, from equation (40) we know that k k2 g-l{e-k4~:}_ 2 t ~ / ~ exp ( - - 4-t).

So, applying Theorem 3 of Section 2.2.2-1, we see that L_l{le_k,fi}_ s

k

2~/~-

f0 t -3/2 k2 ~_~ r exp ( - ~--r-r)dr -

2 K247

e -~2 dot

= erfc(2~/7 ( - complementary error function, see Section 3.1.1). The inverse transform of (i) then becomes" O(3

n--0 - h ~ e ~r f c n--0

(j) [ fl~{(2n+l)b+x}].

(2.2.6)

Partial differential equations

753

Compared with equation (h), this equation can be considered as an alternative solution of the problem. Remark. If we consider the same problem of Example 7, Section 2.2.3-2, but for a leaky aquifer and we assume that only the differential equation changes, while the initial and boundary conditions remain the same, we have: 02(/9

(/9

OX 2

)v 2 = fl

2 0q)

q)(x 0) - - 0 . .

with/32 -

Ot

S

and

)v 2 - - K D c

-- KD

99(0 t ) - - 0 . .

and

'

qg(b t ) -

l0 | h

fort-0, fort>0.

Laplace transformation with respect to t and finite Fourier sine transformation with respect to x gives:

b s

'

with the solution: }(n,s)

-

( - 1)n-+-I nwr h b s (~-)2-ql-/~2S--~- ~2

If we set r/-- fll~2.. = ~1 it is easy to verify that after inverse Laplace transformation, we find

n 2 7 r 2 + flZb27]

-

flab2

+ rl t

,

and after inverse Fourier transformation: 99(x, t) - 27rh Z n ( - 1 ) n + l sin ~.-b---) 1 - exp n=l ; 2 7 2 ~ fl2b2r]

-

n f1262 +/7 t

.

Now we can make use of the equality found before, as given by equation (f) to evaluate the series related to the unit term between the brackets: sinh(~) oo ( _ 1),,+, n sin (--if-) exp g)(x, t) -- h ~ - 2zrh ~ 11271. 2 /~2 sinh (-~) n=l -11- ~2

-

+ f12b2

t rl

"

The first term at the right side of this expression represents the steady state of the groundwater flow, which also follows after solving the differential equation for the steady flow of this problem (without ~a~o ).

(2.2.7)

754

2.2.7. Table

Survey

of integral

Analytical solution m e t h o d s

transformations

1

Transform

Laplace

Main property

Invers transform

L{9(t)} = qS(s)

(Section 2.2..2) = f o qO(t)e-St dt

a by means of operations •-

s~(s)

and from tabels.

~(o)

b by integration in the complex plane.

Fourier

s.{~o(x)} = ~(n)

A. Finite

= fob ~o(x)sin (mrXb *~ dx

1. Sine

,

s,{a->-} = (~)~(~)

s2J{(o(n)}

dx 2

--~ ~ - { ~ ( 0 )

= ~o(x)

2 X-'-,oo

~enX

• -- ~/--.n=l Vat )

sin

lnzrx

k---b--/

'

(-1)n~(b)}

(Section 2.2.3-2a)

2. Cosine (Section

G {~o(x)} = ~ ( n ) [117rX

- fo~ ~o(x) cos ~ ~, ) dx

d29

=

C n l {q~(n) } -- qg(x)

(torT) 2 ¢(n)

+ ( - 1y ~dx (b)

2.2.3-2b)

3. Sine

C.{d-V}

S n r {q)(X)} = q~(n)

(~.x b /~ dx

S

nr

=

(Section

with ogn being the roots of ogcotog

"-

s~rl {q3(n) } = qo(x).

J d~9 I. I d~2 I

Resistance

fo~ ~o(x)sin

+2 EL~ ~(~ cos (.~x_;_~

a_,e dx (0)

~; r"~ (n) + T~"9 9(0)

" a n +e

=

-e

~b

-

2.2.3-2c)

4. Cosine

Cnr {qg(x) } -- q3(n)

c2} {~(n)} = ~0(x)

Cnr { d--~dx 2 }

Resistance

- fo~ ~o(x)cos ( - ~ ) 4 2

(Section

with ogn being the roots of og tan og = e -

"--

~qS(n)

dd-~x(0)

' oln +e

gi~ h

2.2.3-2c)

Fourier

s{~0(x)} = ¢(c~)

B. Infinite

= f o ~p(x) sin(ogx) dx

1. Sine

dx 2

--

= --~2~a(~) + ~0(0)

,, • "-- ~2 f0c~ ~0(=) sln(xog) dog

,

and q) (cx~)

•- ~ (~) = o

(Section 2.2.3-3a)

2. Cosine

c{~o(x)} = ~ ( ~ )

c{~} dx 2

(Section

= fo~ ~o(x) cos(ogx) dx

= -~(~

I

~(o~

c -~1¢(~)1- - ~(x) = ~2 f o ~(~

cos(x,~ a~

• and ~o(oo) -

2.2.3-3b)

~dx (oc,) = 0 3. Resistance

F,.{~o(x)} = ~ ( ~ )

(Section

= f o ~o(~){ sin(~)

2.2.3-3c)

+ K woe cos(ogx) } dx

d2cp

Fr{ d__~ } __ _og2¢(og) and q)(cxz) • -- dd2 x (00) -- 0

,,

FI.--1 {q3(og)} ._ 2 fo

1+~u12~2 ' { sm(xog)

+ K wog cos(xog) } dog

.,

Partial differential equations

(2.2.7)

755

T a b l e 1 (Continued) Transform

Hankel

Hn{~p(r)} -- ~b(n)

A. Finite

= f o r ~ ( r ) J o ( " "Rr ]~ dr

1. Jo

with otn being the roots

(Section

of Jo(ot) -- 0

Main property d2~o 1 d~o

Hn { ~r 2 + r -d7 } •2 ^

-

Invers transform

H n 1 {~b(n)} = ~o(r) ^

,-

:7

=o

+otn J1 (Oen)~o(R) limr-+O (r d~o

aT)

2.2.4-3)

2. Resistance

Hnr {Cp(r) } = ~ ( n )

(Section

.__

2.2.4-3)

with otn being the roots

[ Otnr f o r~o(r)Jo,--R--) dr

H . {dr~2 + 7~~r } "--

4~(n) R l i m r ~ 0 (r dd~r)

Hn-1 {~b(n)} = ~o(r) I_

_2 p o e . 93(n) R2 Z-.,n=l j,~(oen)+j2(oe,,)

.

Jo ( -a',r r)

of o~J 1 (or) -- eJ0(ot) • and e --

R Kw

B. Infinite

H{~o(r)} = q3(ot)

(Section

= fo

2.2.4-4)

rcp(r)Jo(otr) dr

d2~o 1 d~o H{ d--~ + 7?-7}

-

2^ ~ ~o(~)

H - l { ~ ( o t ) } = 99(r)

. - fo~o ~o(ot)Jo(rot) ^ dot d~o

limr-+0 (r aT) and ~p(oo) - ~ (~) = o

.

756

(2.3.1-1)

Analytical solution methods

2.3. SOLUTIONS, DERIVED F R O M KNOWN SOLUTIONS

2.3.1. Superposition 1. Principles Although the most general differential equations for groundwater flow, developed in Section 1.4.2 (equations (3) and (4)), are non-linear because of the terms with (Op~2 (~.¥ ~)p)2 a n d (~p~2 ~ . j , the simplified differential equations that are used to solve ,,0x j , geohydrological problems analytically, such as the equations (9) and (10) of Section 1.4.2-2, are all linear. This means that the state variables, mostly the head or drawdown ~0 and its derivatives with respect to space and time, that appear in the differential equation, are of the first degree. Consider, for instance, the partial differential equation 02q9

O2q9 ax---5 + Oy 2

~

__ /~2 0q9

x--5 -

(1)

a---7'

with )v2 -- K D c and f12 __ ~ which is used for non-steady horizontal flow in a KD' leaky isotropic aquifer. This is a linear differential equation, which is homogeneous moreover (see Section 2.1.2, equation (2)). Now it is easy to show that if ~ol, ~02, . . . , ~0n are all solutions of equation (1), independently of each other, a new function q9 -- Cl~01 at- C2q92 -at - . . . -{- Cn~0n

can be composed which will be a solution of equation (1) as well, if cl, C2, . - ' , are constants. This is a basic property which may be characterized as follows"

Cn

Theorem 1. If a solution of a homogeneous linear differential equation is multiplied by any constant, the resulting function is also a solution. If two or more solutions are added, the resulting function is also a solution of that equation. This theorem is called the superposition principle or linearity principle and is of great practical and theoretical importance, because it enables us to generate more complicated solutions from simpler ones, but it must be kept in mind that it does not hold for non-homogeneous linear equations or non-linear equations. In the previous sections we have met already with some applications of this theorem. In Section 2.2.1, Example 2, we found the solution o/r

(-

0t2t -

-

-

f12R2)

of the differential equation 02([3 0r2t

10q3

r Or --

f12 Oq9 ~at,

with ol being a root of the equation otJl(~) - sJo(~) - 0 (A, R, ~, ~ and s are all constants). As there are infinitely many roots o~ - o~,~ (n - 0 , 1,2, 3 . . . . ) there are

Solutions, derivedfrom known solutions (2.3.1- l)

757

also infinitely many solutions qgn(r, t), which may be taken together in an infinite series, according to the superposition theorem, yielding the new solution:

oo

(Oenr)

qg(r,t)--ZAnJo

--~

(ol2t)

exp

(2)

fl2R 2 '

n---0

provided that this series and also the series that result from differentiating equation (2) twice with respect to r and once with respect to t, all converge. In Example 1 of Section 2.2.1 we have the solution 99(x, t) -- {A(ol) cos(czx) 4- B(cz) sin(olx)} exp ( - -c~2t -~) of the differential equation ~~)x2 - f12~ with ot and A(ot) and B(ot) being arbitrary ~}t ' constants. Then according to the superposition principle, also the integral function ~o(x t) -'

f0

{A(o~) cos(o~x) + B(o~) sin(o~x)} exp

(

- °t2t] do~

(3)

/~2 ,]

is then a solution of the differential equation, provided that this integral exists and can be differentiated twice with respect to x and once with respect to t. So far, we considered superposition of more or less arbitrary solutions of a linear homogeneous differential equation. However, for the application of the superposition principle to solutions of boundary value problems it is not sufficient that the solutions satisfy the differential equation (which must be linear and homogeneous). In that case, also the boundary values play a role. It will be clear that the following theorem must hold: Theorem 2. The superposition principle for solutions of boundary value problems

is valid if the separate problems are governed by a linear homogeneous differential equation and if the resulting solution satisfies that equation and also boundary conditions and eventually the initial condition. Example 1. Q

h

~o=0

[

d

:,1 KD, S

Y Q

,/? -

x

Consider a fully penetrating well with constant discharge Q in a semi-infinite confined aquifer at a distance d from a fully penetrating straight open water body. The original surface water level is equal to the original groundwater level and is supposed to be the reference level for the head 99 (~o = 0). At the time t -- 0, the surface water rises by an amount h and the discharge of the well is started (see Fig. 1). The head ~p is a function of x, y and t: ~0 = 99(x, y, t) and the boundary value problem can be written as: 0299 0299 ._ f12 099 -Ox - 3 -t- Oy2 0---7

Fig. 1. Well near a straight open boundary.

with/32 ---

S K D'

758

Analytical solution m e t h o d s

(2.3.1-1)

q)(x,y 0 ) - - {0 ' h lim r r~ O

forx >0, for x - - O ,

= -~r

Oqg(c~,y,t)=O, q)(O, y , t) -- h,

Ox

for r - xf(x - d ) 2 + y2, 2 rr g D

O--~ ( x , O, t) -- O, Oy

- - ( x , -t- c ~ , t) Oy

O.

This problem includes a combination of one-dimensional flow from an open water boundary and abstraction of water from a well. To solve this problem it is therefore obvious to apply the superposition principle in one way or another, as the differential equation is linear and homogeneous. First, we solve the separate problems: a. Non-steady one-dimensional flow of groundwater from a straight open water boundary through a confined aquifer (see Fig. 2). 02(t9 __ f12 0(/9 ax---g --

fi2

-~'

S -- K D '

Fig.2. One-dimensionalflow from an open water boundary.

qO(X t) -- head, '

{0

for x > O,

h

for x -- O,

qg(x O) -' !a

~o(0, t) -- h,

Ox ( o o , t) -- O.

Laplace transformation with respect to t gives (equations (4), (6) and (16) of Section 2.2.2)" d2q3

fl2sq) -- 0,

q)(0, s) -

dx 2

h

dq5

-,

--(oo,

s

dx

s) - 0

with the solution: qS(x, s) =

h -e

s

The inverse transform of e -t~x47 is, according to equation (40) of Section 2.2.2: /~x

/~2x2

2t~Texp(

4t )

and thus, from equation (8) of Section 2.2.2:

hflx qg(x, t) -- 24%-

.~-3/2 exp

fo'

(

- f12x2) d.c. 4r

Solutions, derived from known solutions

(2.3.1-1)

759

With the substitution 0/2 -- f12x2 4r ' we find Z"-3/2 dr -- --2

dz " - 1 / 2 ~-~- - 2

d( 20/'] -\fix/

40/

_ m

~x

dot

fly giving and the new boundaries of the integral: oo and g-~, e -a~2 d o / - h erfc( fix

q)a(X, t) -- h - - ~

(a)

2,/7

(see Part A, 123.02). b. Non-steady radial flow of groundwater in a confined aquifer towards a fully penetrating well with a constant discharge (see Fig. 3). 02(/9

Fig. 3. Radial flow towards a well.

1 0(t9

Or---T -F r Or --

f12 0(/9 -57'

99(r, t) -- drawdown, lim r r--+0

---

2= K D

f12

S -- K D ,

~p(r, O) - O, ~0(oo, t) --0.

Laplace transformation with respect to t gives (see equations (4), (6) and (16) of Section 2.2.2): d2q3

1 dq3

dr 2

r dr dq3)

lim r

r--+O

~

/~2S q) - - 0 , ----

Q 2rc K Ds

~(oo, s) - 0

with the solution: ~ Ks o ( f l r x / ~ . ~(r, s) -- 2 7 r Q KD

The inverse transform of Ko(flr~/7) is, according to equation (48) of Section 2.2.2:

1 2t exp (

f12r2 4t )

and thus, from equation (8) of Section 2.2.2"

760

Analytical solution methods

(2.3.1-1)

99(r, t)

Q

-

4rc K D

fo t exp ( f 1 2 r 2 )

-4r.

With the substitution u -

f12r2 4r

dT

~.

r

we find or _

du and the new boundaries of u

flZr2 the integral: oo and --W-, giving

qgb(r, t)

--

QfOO 47rKD

2r2

e -u

a7

du ~

U

Q

(f12r2)

4rrKD

4t

-- ~ E 1

(b)

(see Part A, 215.03), the solution for a well located in the origin of the xy-plane. The solution for a well in the point x --- a, y -- 0 becomes, if q) is considered as the head:

-Q

(pc(x'y't)--47rK'---'-~

E1

[ e2 {(x 47

a) 2

+

y2

] }.

(c)

Both solutions (a) and (c) for qg~ and qgc, respectively, are solutions of the differential equation

02(/9

02q9

0(/9

OX----~ -F Oy2 __ f12 at and according to the superposition theorem also qg(x, y, t) = 99a+~0c is a solution of that equation. However, as qga(0, y, t) = h and

{f12

Q E1 (a 2 -+- y qgc(0, y, t) -- - 47rK--------~ ~

2)}

the boundary value for x = 0 is not satisfied, as qg(0, y, t) = q)a(0, y, t ) + qgc(0, y, t) :/: h. So q0(x, y, t) = q9a --i-qgc is not a solution of the boundary value problem.

a

0

.e

Fig. 4. Well near a straight open water boundary.

Obviously, in order to find the right solution we must add to solution (a) not the solution of a well in an infinite field, but the solution for a well near a straight open water boundary, as given in Fig. 4, which in this case will be (q9 is head):

Solutions, derived from known solutions (2.3. l- l)

(#d(X, y,

t)

761

f12

= ------~Q E1 [-~-- { (x -ql-a)2 -l- 22}] 47c K D

Q

4rr K--------~

(d)

E1

4t-

{(x - a )

2

y2 +

}

(well and image well, see sub 2 of this section). The right solution then becomes: ~o(x, y, t) = ~0a(x, t) + 99d(x, y, t), which satisfies both the differential equation and the boundary conditions and initial condition of the posed problem. Remark. A steady state will be reached if t approaches infinity, i.e.: q)(x, y) = h +

Q

ln { (X - a)2 + Y 2 } (x-k-a) 2 + y 2 " 4rr K D

From this example we may learn that if a solution of a boundary value problem is known, a second solution of the differential equation may be superposed upon the first one, but only if it does not affect the initial value and the boundary values belonging to the first solution and inversely if the first solution does not influence the initial value and the boundary values of the second solution. For instance, in this example, solution (Pc did affect the boundary value h of solution ~0a for x - 0, while solution 99d did not, because ~0d(0, y, t) = 0. On the other hand, the solution 99a does not affect the boundary value at the well-screen of solution qgc: the total discharge remains Q, because the total flux entering and leaving the well-screen, caused by the one-dimensional flow described by ~0a, is zero. This last property, arising from the law of continuity, leads to the following theorem: Theorem 3. In a certain groundwater flow field the superposition method is valid for a finite number of all kinds of abstraction means or infiltration means, such as wells, well-screens, drains, infiltration canals and others, provided that the concerning boundary conditions relate to the total fluxes only and, .for instance, not to potentials (heads or drawdowns) or combinations of fluxes and potentials. Thus, for instance, the flow field caused by an arbitrary number of wells or drains or combinations of them with fixed discharges (which may be constant or given functions of the time), can be determined by adding the flow fields caused by the separate wells or drains. Example 2. A simple example of application of this theorem is a well field consisting of n wells with yields Q1, Q 2 , . . . , Q,. The drawdown at a certain point

762

(2.3.1-1)

Analytical solution methods

and at a certain time, caused by the i th well may be expressed as q9i (X,

Qi

y, t) -- 2rr K----~~ (x, y, t)

(i -- 1, 2 . . . . , n)

(4)

in which ~ ( x , y, t) is a dimensionless function, depending on the location of the point, the time passed since the beginning of the discharge, the location of the i th well (xi, yi) and on the characteristics of the aquifer and the well-screen. For a fully penetrating well in a confined aquifer, for instance, j~(x, y, t) becomes: j~(x, y, t) -- ~E1 -~-{(x

-xi)

2 -at-

(y

-

yi) 2

,

with El(z) being the exponential integral (Part A, 215.03). The total drawdown, according to the superposition principle then becomes: 1

n

~p(x, y, t) -- 2rcKD E

Qifi(x , y, t).

(5)

i=1

It is obvious that the superposition principle, applied to well fields, will simplify calculations on groundwater intake areas and on drainage of building pits, considerably. If, instead of the discharges, the drawdowns at the well-screens are kept constant, the principle of superposition according to equation (5) is no longer valid because the drawdown at a well-screen is influenced by the pumping of the other wells. Then, before formula (5) can be used the yields must be computed from the given drawdowns. In Theorem 3 it is stated that application of the superposition method is limited to a finite number of abstraction means, but in most cases the theorem is also valid for an infinite number; however, there are exceptions and one of these will be discussed under sub 2 of this section. Also an infinite number of infinitely small discharges may be superposed, which is shown in the following example.

Example 3.

l /,-

z0l

l Q

P

The drawdown, caused by abstraction of a discharge Q from a line well of length l on top of a semiinfinite aquifer (Fig. 5) can be determined by integrating the drawdown caused by a point well along the length of the well. The drawdown caused by a point well in the origin is equal to Q

Z Fig. 5. Line well in a semiinfinite field.

~p -

4rr Kp

Q =

4rr K ~/r 2 +

(6) Z2

(2.3.1-1)

Solutions, derived from known solutions

and for a well in point P(O ' zo) with strength

dq)-

763

2__QQ. 21

Q dzo 4re K l v l r 2 + (z

go) 2

-

Integration from -1 to +l yields for the total drawdown"

Q f+] Q

99(r' z) -

dzo

, v/r2 _+_(g ZOi2 and so

99 -- 4 r c K l

{ arcsinh( l + z

4rr K------1

r

arcsinh( I - z

)+

)}"

r

(7)

This is the solution for the drawdown if the discharge is uniformly distributed along the well-screen (Part A, 522.04). The drawdown distribution along the well-screen is not uniform in this case. An important application of Theorem 2 is the m e t h o d o f simplification. This method implies that the solution of a problem, for instance, the solution for the drawdown 99 is supposed to be the sum of two other solutions 991 and 992 of which 991 is the already known solution of a simplified problem, related to the problem to be solved, and a still unknown function 992 which must be determined making use of the condition that 991 + 992 has to satisfy the differential equation and the boundary conditions and initial condition.

Example 4.

I

Groundwater flow in a confined aquifer from a straight surface water boundary with entrance resistance, towards a fully penetrating line well with discharge Q. The differential equation with boundary values for the drawdown 99 = 99(x, y) becomes (see Fig. 6):

P X

Q

=~

~o=0

0299 /..1~p

i

i w

a

.+

Ox--7 + ~

- O,

a~

~(O, y)

--(0, Ox

J

'

0299

y) = ~ , Kw

~0(oo, y) = 0,

KD

Fig. 6. Well near a boundary with entrance resistance. ~90_~,( x , O) _ O Oy

lim ( 0r 9 9 ) - r ---~ O

-~F

f o r x ~>0 a n d x 7/= a , Q 2 rr K D

for r2

--

(X - - a )

y ( x , c~ ) - O ,

2 -at-

v2 .

764

Analytical solution methods

(2.3.1-1)

Now, assume that the solution 99(x, y) - 991(x, y ) + 99:(x, y), where 991 is the solution for a well near an open boundary (without entrance resistance) and 992 a still unknown function, dependent on K w . The solution for 991, the simplified solution, is"

Q

ln{ (x -+- a)2 -+- Y2 } (x - a ) 2 + y2

991(x, y) -- 47cKD

(8)

(well and image well, see sub 2 of this section). As both 99 and 991 are solutions of the differential equation 0299 t 0299 = 0, OX 2 Oy 2 992 is also a solution of this equation. The boundary values for (/92 follow from:

992 -- (./9- 991 0992 Ox

099 Ox

and

0992 099 = Ox Ox

0991 Ox

Q [ 2(x + a) / 4re K D (x -+- a) 2 + y2

or 2(x - a ) / (X - - a ) 2 -k- y2

099 Q 4a 0 99---~2 x ( 0 y) - -~x ( 0 ' y) - 4 rc K D a 2 -4- y2 991(0, y) + 992(0, y) Kw

992(0, y) Kw

99(0, y) Kw

Q

a

r c K D a 2 -4- y2

Qa r r K D ( a 2 + y2)

Q a J r K D a: + y : '

992(00, y) - 99((x~, y ) 0(/92 _ 0(/9 Oy -- Oy

and so

991((x~, y) - 0 -

0(t91 _ 099 Oy -- Oy

0 ' O,

___~Q { 2y 47r K D (x -t- a) 2 + y2

2y (x - - a ) 2 --t- y2

!

and so °~--22(x, o) - -099 0-T(x,0)-0-0 Oy lim r r--+O -~r

-lim

r~O

r

Or

forxCa,

r

-Or

---

2rr K D - ( -

for r 2 -- (x - a ) 2 + y2, from which T~)~°2(X ' 0) -- 0 ~o(x, oo) - ~01(x, oo) - 0 - 0 - 0.

Q

2rcKD

also for x

-

)-o

a, 992(x co) -

Solutions, derived from known solutions (2.3.1-1)

765

The boundary value problem for 992(X, y) thus becomes: 02(/92 02992 OX2 ] Oy 2 = O,

992(0, Y)

0992 Ox (0, y) --

992(00, y) -- O,

0992 (X, O) -- 0 Oy '

Kw

Q

a

7 r K D a 2 4- y 2 '

(D2(X, 00) -- 0

"

Owing to the boundary condition ~0V( x ' 0) - 0 we apply the infinite Fourier cosine transformation with respect t o y (see Section 2.2.3-3), giving according to equations (94) and (95): d2 (~2 _ 0/2 (P2 - 0 dx 2 d(~2 dx

~(0,

~) =

~b2(oo, or) - O,

(~2 (0, oe) Kw

Qa

JrKD

f0 ~ cos(oey) a 2+y2 dy-

~2(0, oe)

Q

Kw

2KD

_e-aO~

with the solution: (~2(X, oe) __

Q Kw e-(X+a)~ 2 K D 1 + Kwoe

From equation (96) we find the inverse transform: Q

(#2(X, y) -- r r K D

fo o°

Kw e -(x+")~ cos(yoe) doe 1 + Kwoe

(9)

and finally 99(x, y) - q91(x, y) + 992(X, y) (equations (8) and (9)). Remark. As cos(yoe) - Re[e-iy ~] is, with z - x + iy, [ f o °° Kw e -(z+a)~doe] q)2(x, y) - r r KQD Re 1 + Kwoe

and thus S'22 -- ~2 q - i ~2 --

_

Q [~ reD Jo

Q exp(Z + a

-- - ~

Kw e -(z+a)°t doe 1 + Kwoe

z +a,]

....K t o ) E l ( K 1 1 )

,I

[substitute u - (z+a)(x--!y+oe)] with E1 - exponential integral and a"22the complex potential function (see Section 2.2.5-1, equation (136)). The end solution, written as a complex function thus becomes: Q

z

2 r r D { l n ( z - a+ (Part A, 337.21).

a

)+2exp\

(z+a z+a K w ) E 1( )}

(10)

766

Analytical solution methods

(2.3.1-2)

2. Method of images A very important application of the superposition principle is the so-called method of images for solving geohydrological problems involving straight or circular boundaries with prescribed conditions. Instead of theoretical discussions, we shall show the principles of this method by giving some examples. E x a m p l e 5.

A fully penetrating line well in an aquifer near a fully penetrating straight surface water body (channel, river, lake, etc.) with a constant water level (see Fig. 7). P e The straight open water boundary has been -'hQ x Q~,a a -tchosen along the y-axis; the groundwater a =~ level in the aquifer is initially the same as , ' . ~o=0 the surface water level and assumed to be zero (q9 = 0). Pumping of groundwater with a discharge Q from the line well at P (a, 0) yields a steady flow of water from KD the surface water body through the porous medium towards the well, causing a drawdown qg(x, y) everywhere in the half plane Fig. 7. Well near an open boundary. (x > 0, --c~ < y < +cx~) except at x = 0. If we imagine an infinite aquifer without surface water boundary and another line well at point P ' ( - a , 0) with a recharge (negative discharge) of strength Q equal to the discharge at P (a, 0), then the y-axis (x = 0) is the locus of the points that have equal distances to P and P'. Such a point undergoes influences from both wells that are equal but of opposite sign, resulting, according to the superposition principle, in an unchanged head (q9 = 0). So the problem of a well in a half plane near a straight open boundary can be replaced by a discharge- and a recharge well of equal strength in an infinite field. The solution for the drawdown caused by the discharge well at P (a, 0) in an infinite field is: l

I I I

•" - - . . ~

_

q)=c-

2rc K D

lnr

with r - v/(x

a)2 + y2,

and for the drawdown (negative) caused by the recharge well at P ' ( - a , 0) in an infinite field: (,o' = c' +

Q In r' 2~rK D

with r ' = x//(x + a) 2 + y2.

Solutions, derived from known solutions

(2.3.1-2)

767

Making use of the principle of superposition, we find for the solution of a discharge well and a recharge well of equal strength in an infinite field: Q l n { ( x + a ) 2 + y2} 99 -- 4zrK--------D (x - a ) 2 -k- y2 '

(1 1)

as c + c' = 0 from the condition ~o(0, y) = 0. If we define a discharge well as a positive well, giving a positive d r a w d o w n and a recharge well as a negative well, causing a rise of the groundwater heads (negative drawdown), we have seen that the combination of a positive and a negative well in an infinite field has the same effect as a positive well near a straight open boundary in a semi-infinite field. (Here, and in the sequel we assume all wells and image wells to have the same strength.) On the other hand, two positive wells or two negative wells in an infinite field is the substitute for a well near a straight impervious boundary in a semi-infinite field, as can be deduced from symmetrical arguments and also derived mathematically by differentiating the total drawdown

~o -

Q In [{(x - a) 2 + y2}{(x + a) 2 + y2}] + c 4 7r K D

(12)

(for this case c + c' --/=0): Oq) _ Ox -

Q

{

2(x+a) } a) 2 + y2 -+- (x -nt- a) 2 -+- y2

2(x - a)

47rK-------D ( x -

and putting x -- 0, from which a~0(0, y) -- 0. This means that no flux of water takes place through the boundary along x = 0. In general, all kind of solutions for a single well in an infinite field, such as solutions for steady or non-steady flow in confined or leaky aquifers or multilayer systems, may be used for practising the method of images. An example for non-steady flow has already been given in sub 1 of this section (equation (d); see Fig. 4). E x a m p l e 6. As in Example 5 (Fig. 7), but for a leaky aquifer, the solution becomes:

~o -

2 rc K D

in which r -- v/(x - a) 2 + y2, r' -- v/(x + a) 2 + y2, )~ _ ~/~2Dc -- leakage factor and c -- resistance against vertical flow of the semi-permeable layer.

(2.3.1-2)

768

Analytical solution methods

Example 7. Y

,,~ Q!-_)._

P(xo, Yo) 1

i

Iy°....Xo2 Q(-)

I

Flow towards a discharge well in an infinite quarter of a plane bounded by two straight boundaries, perpendicular to each other; one of them is an open water boundary, the other is impervious.

I

;,,'~Q(+)

Fig. 8. Well near an open and an impervious boundary. The differential equation with boundary values becomes, with q9 -

qg(x, y),

0<~x < c~ and 0 ~< y < c~" 02~0

0299

OX----5+----Oy2 = 0 ,

°--~(x, Oy

o) - o,

~o(0, y) - 0, ~o(x, ~ )

(0~0) Q lim r = r~0 -~r 2zr K D '

qg(c~, y) - 0,

- 0,

with r - ~/(x - x 0 ) 2 -1- (y - Y0)2

It is not difficult to see that this problem of one well in a quarter of a plane may be replaced by a system of four wells in an infinite plane; namely, the original well together with three image wells, the place and character (positive or negative) of the latter depending on the location of the original well with regard to the boundaries and the character of the boundaries (open or closed). In the present case, the image wells are a positive well at (xo,-yo) and two negative wells at (-xo, yo) and ( - x o , - y o ) , respectively. The solution for x >~ 0, y >/0 thus becomes" Q l n [ { ( x + xo)2 + ( Y - Y°)2}{(x + x°)2 + (Y + Y°)2}] q9 - 4rrK------D {(x - xo) 2 + (y - yo)2}{(x - xo) 2 + (Y + Yo)2} "

(14)

Example 8.

Fig. 9. Drain in a confined aquifer.

An infinite drain with discharge q [L2T -1] on top of a confined aquifer with thickness D (Fig. 9). In order to make use of the method of images, we have to realize that two parallel boundaries need an infinite number of image wells or drains, as is shown in Fig. 10, where the original aquifer first has been replaced by an aquifer of thickness 2D and a drain in the middle with discharge 2q.

Solutions, derived from known solutians

(2.3.1-2)

As the boundaries are impermeable, the image drains must be positive (discharge drains) and be located at the points (0, + 2 n D ) with n = 0, 1, 2 . . . . . The solution for the drawdown caused by a drain with discharge 2q in an infinite field is, if the drain is situated in the origin:

2q z = -6D 2q -4D +

2q -2D .,~ 2q

+

769

2D % 2q

q~ -- c -

4D % 2q

= c

6D 2q 8D % 2q Fig. 10. Infinite n u m b e r of positive drains in an infinite field.

q ~ lnr JrK q ln(x 2 + 2JrK

Z2

).

For a two-dimensional steady or quasi-steady problem it is, in general, convenient to work with the complex potential solutions because of the simpler expressions for the latter. Then for one drain in an infinite field we have (cf. Example 17 of Section 2.2.5-1): q £2 - c - - - I n ( .

The physical plane is now considered as the complex plane ( - x + i z and the image drains are located at the points ( - + 2 i n D (n - 1, 2 , . . . ) . Now, according to the superposition principle, the drain together with its image drains should give: ,(2 -- co

q In ( + cl - q ln(( - 2i D) + Jr

Jr

C2 - - q Jr

In(( + 2i D)

+ c3 - q ln(~" - 4 i D ) + c4 - q ln(~" + 4 i D ) + . . Jr Jr OQ

--c

- q In ~ - q ~ 7/"

(X3

ln(( - 2 i n D ) -

Jr

q E

ln(( + 2 i n D )

or

Jr n-- 1

n-- 1

0(3

S2-c

- q In ( - q ~ Jr

In(( 2 + 4n2D2).

Jr n--1

Unfortunately, the series in this expression diverges, as the general term does not approach zero with increasing n; so it was not permitted to apply the superposition theorem in this case.

(2.3.1-2)

770

Analytical solution methods

However, if we apply this theorem to the groundwater velocities, caused by each of the drains, as expressed in the relation (see equation (137) of Section 2.2.5-1): dX?

d~

= - Vx + i vz,

we find: dX?

q(1

d~ =

rc - ~ + ~ - 2 i D

q

1

1

1

1

+ ~'+2iD +~-4iO

+~+4iD

) +""

+

Jr

=

~-2 ÷ 4n2D 2

"

As

coth(Jvx) -

1 2x ~ 1 7//,/ ÷ --7/" k~l= x ÷ 2 k 2 '

we find d£2 _ d~" -

q coth(Zr~') 2D 2--D '

as can easily be verified. The solution for the total complex function, which is also the solution for the posed problem, thus becomes" X2 - c

7l"

In sinh

(15) rr~

(Part A, 356.01) from which follows 4) - R e [ ~ ] - cl - ~ In Isinh(g-5)l if c = Cl ÷ ic2, and as 4) - K~0 we find for the drawdown: ~o(x, z) -- c - 2~r----K

2-D

÷ sin2 2-D

"

(16)

With some slight adaptations this formula is also valid for an "infinite" row of wells of equal strengths Q and at equal distances d from each other in a confined aquifer of thickness D: ~0(x, y) - c - 47r K D

-d-

+ sin2 --d-

"

(17)

These solutions are important for finding solutions for more complicated problems concerning drains and wells between open and closed boundaries, for instance, the following example. * Formula 1.421-4 of Tables of Integrals, Series and Products of Gradshteyn and Ryzhik.

Solutions, derived from known solutions

(2.3.1-2)

771

E x a m p l e 9.

A discharge well in a confined aquifer of thickness D between two parallel open boundaries. We suppose a -¢ g1b which implies the use of a more complicated system of image wells in order to replace the flow field shown in Fig. 11 by an infinite field.

............. Y I ..................................

a I I ............1................................. ................

x

Fig. 11. Well between parallel open boundaries.

a4b+a 4b-a

"~a2b+a 2b - a

This system is shown in Fig. 12 and consists of an infinite number of positive wells situated at the points (0, a -t- 2rib) (n - 0, 1, 2 . . . . ) and an infinite number of negative wells located at the points ( 0 , - a - t - 2 n b ) (n - 1, 2 . . . . ). Both the positive and the negative wells have mutual distances of 2b. If we apply the formula for a row of wells (see Example 8) we find for the positive wells: q)l(X, Y ) - - C l -

)------~Q 4 r Inc [sinh2 K D( -zrx ~

:...... :......:........-...--.-.-...~-.-.-.....,.,.,.

+ sin2 \(7r(Y2bS-a))] -a

x

Q

and for the negative wells" 992(X, y) -- C2 _Jr_~4 r rQK D ln[sinh2 ( -~-) 7rx

~-2b + a

+ s i n 2 ( 7 r ( y]+2 ab) )

-2b -a

m

and together: Q ~-4b + a

-4b -a

Fig. 12. Infinite number of positive and neg-

ative wells.

Q q) -- 4rr K D In

[sinh2(~_~) q_ sin 2 (~r(v+a) ] sinh2(~-~) + sin 2 (Tr(2~-a))

(Part A, 356.12) as cl +

C2

--

(18)

0 because q9 -- 0 for y -- 0.

Analytical solution methods

(2.3.1-2)

772

Remark.

The problem of Example 9 can also be solved by means of a finite Fourier sine transformation with respect to y (Section 2.2.3-2a). To make the boundary value at the location of the drain suitable for this transformation, we consider the discharge Q of the line drain as the limit of a discharge g1 Q at both sides of an infinite screen with length l, if l approaches zero (see Fig. 13).

............. Y.l ................................................ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

liill~aQ .llb ................] ....I............................... x

Fig. 13. Well-screen between two parallel open boundaries.

The boundary value along the y-axis for the well-screen becomes" for 0 ~< y < a and for (a + l) < y ~< b,

0 Oq°(O y) --

Q

Ox '

2KDI

for a < y < (a + 1).

Finite sine transformation gives: dq3(0 , n )

dx

_

Q 2KDI

f a + t sin ( ~ )

dy.

Letting l approach zero, we have" lim dq3 t~0 d--x-x(0, n) -

Q I f~:'+t sin ("~---Y-v)dY ] _- 0 - i n d e f i n i t e . 2KD 1 t--,0 0

Applying the rule of l'H6pital, by differentiating the numerator and the denominator with respect to l, we find lim dq3 t-+o ~x-x(0'

Q [sinlnyr(a+l)]] n) -- 2 K----D b t-+o

Q sin(nzra) 2K D --if-

o

The transformed differential equation becomes: d2q3 dx 2

(~_)2q3 -- 0

and as q3(oo, n) - 0 we find for the transformed solution: nyra Qb 1 (o(x, n) - 2rrK D n sin (---b--) exp (

nTrx]

- ---~-/

and thus"

sin( nrr T )a sin( nrry )exp( n=l

nrrx ). b

(19)

Solutions, derived from known solutions

A s s i n z l s i n z2 -

~1 {COS(Z 1 -

COS(Z 1 -3t- Z2)} , a n d

Z2) -

CXD

E

773

(2.3.1-2)

Cx3

le-,Lx n cos(ny) -- Re Z

n-- 1

_n1 e-nZ.

with z - x + i y,

n-- 1

and OO

E n--1

1-un -- - - l n ( 1 -- u), Ft

we can show after some mathematical exertion, that equation (18) is another form of equation (19)*. E x a m p l e 10.

Groundwater flow between a straight open / \ boundary and a circu/ N / \ / \ lar basin with a conI I stant drawdown h in ', " d / a confined aquifer. X / N / x / -x .. Both boundaries are penetrating the aquifer fully (see Fig. 14). At first sight, this problem of steady groundwater flow should be solved by adding to the solution of flow to a circular basin in an infinite field the solution of flow from a circuFig. 14. Circular basin near a straight open water boundary. lar image basin with rise h of the level at the other side of the straight boundary, according to the method of images. Y

- KL/

h

k=

However, on further consideration, we may notice that the solution of the image basin will affect the values of the drawdown inside the real basin, which are supposed to be constant. So a direct application of the method of images is impossible in this case. On the other hand, if we consider the solution of a well and an image

* Formula 1.462 of Tables of Integrals, Series and Products of Gradshteyn and Ryzhik.

774

(2.3.1-2)

Analytical solution methods

well of Example 5, we see that we can write solution (11) as

4rc K Dq9 = In { (x + a) 2 + y2 (x - a ) 2 + y2 }" Q Setting

A

-

exp(4Zr

K QD~o),

(20)

we have A ( x - a ) 2 -k- Ay 2 -- (x -k-a) 2 -+- y2 or x2(A -

1) + yZ(A - 1) + a2(A - 1) - 2ax(A + 1) - 0, giving" A+I (x - a ~ - - i )

2

+

4a2A

y2 --

(A - 1)2'

the equation for a set of circles with centres on the positive x-axis. The parameter of this set is A and as A is dependent on qg, we see that the set of circles represents the equipotential lines, belonging to the flow pattern of the well and image well at (a, 0) and ( - a , 0), respectively. Now we assume that the circumference of the circular basin of our problem coincides with one of these equipotential circles; the problem is thus then brought back to that of a positive well and negative image well, with still unknown location. But from equation (20) it follows that for (x - d) 2 q- y2 __ R 2, the equation of the circular basin, the equipotential q9 must be equal to the drawdown h: (x+a) 2+R 2- (x-d) 2 = In { (1 - a ) 2 + R 2 _ ( x _ d ) 2 }

4rc K Dh

Q

or

4:rrK Dh = ln{ 21(d + a) + a 2 + R 2 - d 2 } 2x(d - a) + a 2 + R 2 - d 2 " Q As the left side of this equation is a constant, the right side must be independent of x from which it follows that a 2 - d 2 - R 2 and

47r K Dh Q

= In

d + ~/d 2 - R 2 d - ~/d 2 - R 2

or Q

u

2Jr K Dh arccosh(-~)"

The solution for the drawdown qg(x, y) becomes:

I ~o--

h 2 arccosh(-~) h

(Part A, 335.04).

In

{ (x +

~/d 2 - R2) 2 -+- y2

for

(x - ~/d2 - R'~-+- y2 } for

(x - d )

2 + y2 > / R 2,

(x - d )

2 + y2 <~ R 2

(21)

Solutions, derived from known solutions (2.3.1-3)

775

3. Discharge impulses A discharge impulse is an instantaneous abstraction of a finite quantity of water from a groundwater body, for instance, a discharge Qi during an infinitely short time dt from a line well in an aquifer, yielding a total abstraction Pi = Qi dt. A recharge impulse is an instantaneous injection of a finite quantity of water Pi into a groundwater body, that is, a recharge Qi during an infinitely short time dt. The dimension of Pi is [L3]. A discharge impulse causes a non-steady flow field, starting from an initial drawdown distribution ~0 = 0 at all points of the field with the exception of the impulse location where the drawdown becomes infinitely large. In course of time the drawdowns will gradually spread, originating from the impulse centre, reach a maximum and then decrease to zero if the time tends to infinity. The next example will make this clear. Example

11.

Instantaneous abstraction Pi [L 3] of groundwater from a fully penetrating line well in a confined aquifer. The drawdown is a function of r and t: ~0 = ~0(r, t).

KD]S.

r

Fig. 15. Discharge impulse in a line well. The boundary value problem can be written as:

0299

1 0(/9

/~2 0(/9

f12

-

~p(r, 0) = 0

(O~p)

lim r r-+0 -~r

-~' {__

'

KD

for r > 0, -

S -

r -Or

cp(oo, t) = 0,

Qi

fort-0,

27r K D

0

Pi -- Qi dt.

for t > 0,

Laplace transformation with respect to t gives:

d299 dr 2

t

1 d~ r dr

flesq9 "-0,

~)(OO, S) = 0 ,

lim

(d~a) r ---

r-+O --dTr

Pi 2rc K D

(cf. equation (30) for to = 0 of Section 2.2.2-2) with the solution:

Pi Ko (fi r v/7) . ~(r, s) -- 27rKD The inverse Laplace transformation becomes (cf. equation (48) of Section 2.2.2-3):

Pi e x p ( 99(r, t) -- 47r K D------7 (Part A, 215.01).

f12r2 ) 4t

(22)

776

Analytical solution methods

(2.3.1-3)

For t = 0 ~0(r, 0) = 0 for every value of r with the exception of r = 0 where the drawdown ~0(0, 0) becomes infinitely large. A control on the correctness of equation (22) is that at every time t > 0 the total quantity of groundwater released from storage must be equal to Pi. The total quantity of water d V, released from the aquifer between the two cylinders with length D (thickness of the aquifer) and radii r and r + dr, respectively, at the time t, is:

Pi S exp dV - 27rr dr Scp - 2 K Dt

f12r2 ) r dr 4t

and thus

Pi S 2KDt

V=

f12r2

exp (

4t ) r d r - - P i [ e x p (

f0 °°

f12r2 4t ) ] 0 - - P i .

For a fixed distance r0 from the line well, the drawdown as a function of the time, the time-drawdown line, follows from equation (22) by putting r - r0. The curve begins with qg(r0, 0) - 0, reaches a maximum at t - tmax and tends to zero again for t approaching infinity. The maximum can be determined, if we set u -

O~p Ot

=

-

Pi 4zr K Dt 2

e -u

+

~

Pi e

4re K Dt

-u-

u t

=

-

Pi

- - e - "

4rr K Dt 2

(1 -

f12r2 4t

'

from

u),

which becomes zero for u - 1' from which tmax -- f12r2 Sr2 and 99max(r0, tmax) -4 --" 4xo Pi e-1 this result being independent of the transmissivity K D So if we perform ~r2S , a pumping test, by giving a fully penetrating line well in a confined aquifer a discharge impulse Pi m 3 of water and measuring the values, qgmax and tmax of the time-drawdown curve for an observation well at distance r0 from the pumping well, we can directly determine the values for S and KD. A first application of the impulse function(s) in combination with the superposition principle is the solution of initial value problems. An example of an initial value problem has been given in Example 4 of Section 2.1.2, concerning one-dimensional flow. In the following example a two-dimensional initial value problem will be discussed. Example

12.

Z

~o=0

~888~ y

KD, S '

X

Fig. 16. Two-dimensional initial val-

ue problem.

In an infinite field the groundwater head in a confined aquifer with transmissivity K D and storage coefficient S is initially an arbitrary function of x and y (see Fig. 16). Owing to the gravity, the heads will decrease and the groundwater will move horizontally with velocity components in x- and y-direction until all heads are equal. These movements are governed by the depletion function, which is the function that shows the variation of the heads with time everywhere in the field.

Solutions, derivedfrom known solutions

(2.3. I-3)

777

In order to determine this depletion function, we may regard the original head

~(x, y, O) = F(x, y) as being due to infinitely many recharge impulses at t = 0. From the definition of the storage coefficient S (Section 1.3.5-1) it follows that at the vertical through the point (x0, y0) at t = 0 an amount of water is stored equal to F(xo, yo)S dx0 dy0. This small quantity of water, considered as due to a small recharge impulse dPi gives, according to equation (22), a head distribution dqg(x, y, t) -

F (xo, Yo)S dxo dyo

47r K Dt

f12

exp[-- --~{(X --Xo)2 + (y-- yo)2}].

The total head distribution at any time and location, the depletion function, then becomes: ~o(x, y, t) --

f12 f+oof+oo F (xo, yo)

4rrt

oo

x exp [ (see 313.01, Part A).

~o=0

~o=h

~o=0

~o=0

(23)

oo fi2 ~-7 {(x --XO)2--~ - ( y -

20)2}] dxo dyo

A special case of this example is the depletion function, starting from a constant head h in the first quadrant and zero head everywhere else (Fig. 17).

Fig. 17. Constant head in the first quadrant as initial value. Here

~9(x, y, O) -- F(x, y) --

h

0

for x > 0 and y > 0, forx <0or y <0.

Equation (23) then becomes: q)(x, y, t ) -

f12h fOOOexp{

~-

f12

-- - ~ ( X --XO) 2 } dxo

/ '2 x

As

exp

/

- -4-7(y - yo) 2 dyo

= -e -c°2 do). h -.f~ x~ e -°t2 doe - ~ 7/" 2v/t 247

fz ~00 e -a~2 doe --

f O e -°~2 doe + ~0 -c~ e -°t2 doe z

q7 erfz + ,/7 erf(-oo)

= - ~

2

2

-- - ~ ( 1

+ erfz),

778

Analytical solution methods

(2.3.1-3)

we find: ~o(x, y, t) -- ~h{ 1 + erf (fix_ 2~/t) }{ 1 + erf (2@t) }

(24)

(Part A, 313.02).

A second application of the impulse function(s) in combination with the superposition principle is the integration of impulse functions with respect to time, yielding continuous functions such as continuous discharges or recharges at wells or drains. Example 13. As an example we choose the line well of Example 11, now with a discharge Q(t) which is an arbitrary function of time. We regard this discharge as due to an infinite number of successive discharge impulses, starting at t = 0. At the time to, the discharge impulse Q(to)dt0 takes place which gives at the time t (> to) a drawdown according to equation (22):

Q(to)dto e x p { f12r2 } dgo = 4rrKD(t - to) 4(t - t0--------~' and in total:

t ~p(r, t) -- 4rrKD

l 2r j

dto

Q(to) exp - 4(t -----to) t - t o f12r2

which by substituting ) ~ - 4(t-to) can be written as

l f~ ( ~o(r, t) -- 41rKD 2r2 Q t

flere)le-Xd) ~ 4X

~.

(25)

4t

(Part A, 215.02). As the Exponential Integral is defined as: El(Z)

-

f

e~ e-)~

)~ d)~ (see Section 3.2.1),

z

we find for a constant discharge Q0:

Qo E l ( fl 2r2 ) ~p(r, t) - 41r K--------D

4t

(26)

(cf. Part A, 215.03) and for a discharge that increases linearly with the time Q(t) = At (A is a constant):

A f~ ( ~o(r, t) - 4~rKD 2r2 t 4t

/32r2)le-)~ 41. X d~,

Solutions, derived from known solutions (2.3.1-3) and with

779

p2r2

u =

o

4t

~o(r,t)_

At fu°° ( ~ ) _xd)~ 4Jr K D 1e

and as E,(z) is defined as E, (z) -- z'-1 fz ~ --e 1 )~n we find

~p(r, t) =

At{

-)" d~.

f12r2 }

f12r2

4rrK--------~ E l ( 4t ) - E 2 (

(27)

41 )

(Part A, 215.04). A third application of the impulse function(s) in combination with the superposition principle is the possibility to write down directly the solution for the heads caused by vertical infiltration (precipitation or artificial recharge) into an aquifer, which will become clear from the following example. Example

14.

I rP

q(t)

~sxa~$,1,,1,,1, $ , b , l , , l , ~

1,

b

]

b

-b

ii K D ' S

Vertical infiltration over an infinite strip of width 2b into a confined aquifer in which the flow is supposed to be merely horizontal. The infiltration velocity q [LT -1] is an arbitrary function of the time and equally distributed over the area of the strip. The mathematical translation of this onedimensional problem is:

02(t9 099 OX 2 ~ q(t) K D = fl 2 --~,

+b x

~p = ~o(x, t),

f12

-- K D ' ~p(x, O) = O,

Fig. 18. Infinite strip with vertical infiltration.

a---~(o, t) - o

Ox q-

q(t) 0

'

~ o ( i o o t) - o,

'

for-b b.

In order to solve this problem, we may regard the vertical infiltration as an infinite number of successive recharge impulses in line wells everywhere in the infiltration area. For this reason, we consider the flow field as two-dimensional for a moment

(2.3.1-3)

780

Analytical solution methods

and find that at the time to at the point (xo, yo) a recharge impulse takes place equal to d Pi -

q (to)

dxo dyo dto,

which will give at the time t (> to) and at the point (x, y) a rise of the head, according to equation (22)" d~o-

q (to) dxo d yo dto 4re K D ( t - to)

/~2 /] F exp L - 4 ( t - to){(x - x o ) 2 + ( y - yo)2lj.

All impulses together form the vertical infiltration and will give a total head:

~o(x, y, t) --

1 f_+bf_+c~Lt q(t°) 4rcKD

x exp

b

~

f12

F

L- 4 ( t -

t -- to

to){(x- xo)2 + ( y -

/"] yo)2/j dxo dyo dto.

Now

f12 (y _ yo)2 [

4(t - to)

/ dyo

may be evaluated with the substitution f i ( y - Yo)

2~/t - to to

2~/t - to [ + ° ° e -~2 dol = - ~ ~ / t

- to { - erf(-oo) + erf(oo) }

J- ~ 2~/-~-~/t - t o

/3 while

f +bexp { - p2 (x _ xo)2 I dxo = b

4(t -- to)

and g 2 where g l - 2~(x+b) ~

-

2~/t - to [g2 e_~2 dol,

fl(x-b) and thus equals 2,/~-~'

~~'-~ '° [ere/~x 2. +~/,o -ere/~x 2~ -~/] - 'o

J~

781

Solutions, derived from known solutions (2.3.1-3)

So if we set r = t - to the end result becomes: ~p(x, t) - - ~

q(t - r) erf

2~/~-

- erf

2,/7

dr

(28)

(Part A, 124.22). For q(t) -- q as a constant, we have to solve integrals of the type / - f0t e r f ( 2 ~ r ) dr

k (k >/0). (see Section 3.1.1, equation (22)), which equals t{1 - 4i2erfc(~--~)} Due to the condition k >~ 0, we must split up the solution (28) for q(t) = q -- const into a solution for - b ~< x ~< b and one for x < - b and x > b, as follows: 99(xt) -

'

qt

fl(b - I x l )

qt

[2i2erfc{ f i ( ]2~/7 x l - b) }-

- ~ - [ 1 - 2i2erfc{

2~/t

} -2i2

erfc{ ~(b + Ixl) 2x/7 }]

2i2erfc{

fl(Ix] Jr- b) 247 }]

for Ix] < b, for Ixl > b

(Part A, 124.23), with i2erfcz - f ~ i e r f c t d t and ierfct - ftC~erfcudu (Section 3.1.1). So far we have considered discharge- and recharge impulses in line wells that fully penetrated confined aquifers, extending to infinity. But also impulses in point wells, well-screens, cylindrical surfaces, disks and others may be performed and not only in infinite confined aquifers. The solutions for the heads or drawdowns of these impulses may be determined in a similar way as has been done for the impulse in the line well (Example 11 of this section). From these solutions, the solutions of a great many problems can also be written down immediately in the form of definite integrals. 4. Hydrological screens

A very important application of the superposition principle is the creation of socalled hydrological screens; these are fictitious screens through which either no water particles pass or flow takes place only in one (desired) direction. Hydrological screens that are impermeable for flow of groundwater can be obtained by discharging or recharging groundwater from or into aquifers in which a natural, more or less uniform, flow takes place. Example 15. A simple example of a hydrological screen arises if a fully penetrating line well with a constant discharge Q [L3T -1] is placed in an infinite confined aquifer which is subject to a uniform flow with strength q [L2T -1] (Fig. 19). The flow is two-dimensional and quasi-steady and is the superposition of the flow field caused by a single well in an infinite field and the uniform flow.

782

(2.3.1-4)

Analytical solution methods

~=0

q

.q +

a 71=-~-fi S

x

q

q

q

Fig. 19. Well in uniform flow. The solution, written as a complex potential thus becomes: 2zrQD l n z + ~qz

$2 - c

(Part A, 325.51)

(29)

with z - x + i y dX2 dz

Q = Vx - i v v

"

=

1

q

~_ m . 2rr D z D

If both Vx and Vy are zero, then ~dI2 - 0

-

(

Q

ys -

Zs - 2zr q

0, Xs -

Q)

27r q

and we find the stagnation point

.

The potential function q~ - K qg, with ~o = drawdown, becomes Q

~b - c -

4rr---D

l n ( x 2 q_ y 2 ) q_ DX

(a)

and the stream function:

--

Q arctan(Y) + D y ' 2zrD

(b)

if we choose the positive x-axis for ap - 0. The set of streamlines with parameter ~ can be represented by x--ycot\

(2~rq

2rrD

Q y---~lp

)

.

(c)

Solutions, derived from known solutions (2.3.1-4)

783

In polar coordinates the stream function becomes: q ~(r, 0) - -D- r sin 0

QO 2arD ,

(d)

and the set of streamlines with parameter ~p may be represented by:

r=

2rcD~ + QO 2zrqsin0 '

(e)

which means that every value of ~p represents a streamline, for which r is a function of 0. The values of ~ vary from - ~ to +e~; the values of ~ for which r = 0 follow from equations (d) or (e): ~(o,o)

=

QO 2rrD

The streamlines belonging to of the well. As 0 varies from R D ' the end-values included, their way from left to right. flowing towards the well is streamlines ~ - 0 and ~ p from equation (e): QO r

-~

2rrq sin 0 Q(O - 2zr) 2zr q sin 0

these values all pass through the origin, the location 0 to 27r, we see that the streamlines with values 0 to all reach the well, while all other streamlines pursue This means that the total quantity of groundwater separated from the rest of the groundwater by the D" Q The equations for these two streamlines follow

for 0 ~< 0 ~< zr, (f) for Jr ~<0~<2rr

and in Cartesian coordinates from equation (c): x - - y cot(2rCqy)

(30)

y.

As y approaches zero, we have

{

y

x -- lim ~,, v--,0 tan( Q , )

} -- -0 -- indefinite. 0

Applying the rule of l'H6pital, we find x - 2-~q, the stagnation point. The equation (30) is the equation for the hydrological screen, a straight cylinder with length D and with the shape of a hyperbola, symmetrical with respect to the xaxis, going through the stagnation point and approaching asymptotically the straight lines y - + ~q. This fictitious screen may be considered as an impermeable obstacle that separates the groundwater flowing to the well from the original uniform flow.

784

(2.3.1-4)

A n a l y t i c a l solution m e t h o d s

The practical importance of such a screen for p o l l u t e d a r e a s is obvious, because a pollution inside the screen cannot escape to the stream field outside the screen and thus always reaches the abstraction well. Only the phenomenon of transversal dispersion of the contaminants may cause a slight deviation of the solute from the theoretical streamlines; so the hydrological screen has to designed amply around the polluted area in order to isolate that area from the surrounding flow field. This example of an o p e n h y d r o l o g i c a l s c r e e n has the disadvantage that because of the quasi-steady character of the flow field, the drawdowns may become unacceptably large. An improvement therefore will be the addition of a recharge well of the same strength as the discharge well as shown in the following example. Example 16. A discharge well and a recharge well of equal strength Q, fully penetrating a confined aquifer in which a uniform flow of strength q takes place. The mutual distance between the two wells is 21. The wells are situated such that their line of connection is parallel to the direction of the uniform flow and the recharge well lies upstream of the discharge well (Fig. 20). Again applying the principle of superposition, we find for the complex potential:

Q ln(Zz +l)l

I2 - 2zr----D

+ D z'

(31)

with z - x + i y dS2 _ dz

-

Ql

1

Jr D z 2 - 1 2

q + -D -

vx - i vy"

The location of the stagnation points follows from Vx Z2 .

.i2_~. . Q t zrq

-

v,,r -

x 2 - y2 + 2 i x y .

hydrological screen

q

q

2,. s_ I I I I I_

r

i

.---.--------~"~

I

I

_

21

I

i

Fig. 20. Positiveand negative well in uniform flow.

0

dS2 - O" and thus -dT-z

Solutions, derived from known solutions

(2.3.1-4)

785

Equating the real and imaginary parts of both sides, we find for the stagnation points:

X s --

--t-

2

+ ~ rrq

and

Ys - 0.

If we choose gr = 0 for y = 0 and Ix l > 1, the stream function becomes:

Q larct

-- 2zrD

n(Y)-ar

x+l

tan(Y)}+ q x 1 DY'

(a)

which can also be written as: x 2 _

12 _

y 2 _+_ 21y

27rq cot(---~-y,,

2rrD Q ~).

(b)

For ~p -- 0 and ~ -- --QD ' we can derive from equation (b) the equation for the hydrological screen: 2zrq)

x 2 _ l 2 _ y2 + 2 l y cot ---~--y ,

(32)

a straight cylinder with the shape of an ellipse and passing through the stagnation points. As distinct from Example 15, we have to do here with a c l o s e d h y d r o l o g i c a l s c r e e n which has the great advantage that inside such a screen a steady groundwater flow takes place that hardly effects the groundwater heads outside it. Moreover, it is possible to wash out a pollution in the aquifer inside the cylinder with clean water and thus purify the contaminated soil and groundwater to a certain degree. The case of one abstraction well and one injection well may easily be extended to more wells in different places, thus allowing of creating any desirable shape of the hydrological screen, which can be seen in Fig. 21 where five injection wells have been placed opposite one abFig. 21. One discharge well with 5 recharge wells in uniform straction well. But in flow. order to obtain a steady state inside a closed hydrological screen, one must take care that:

786 a. b. c.

(2.3.1-4)

Analytical

solution

methods

the total discharge of the positive wells is equal to the total recharge of the negative wells, the configuration of the wells is symmetrical with respect to a line (chosen as x-axis) parallel to the uniform flow, the discharge well(s) lie(s) downstream of the recharge well(s).

In this way, taking into account that a slight transversal dispersion may occur, any pollution of the underground in uniform flow can be isolated by a suitable location and strength of positive and negative wells, while at the same time the underground is sanitated in situ. If no appreciable natural groundwater flow takes place, it is impossible to create impermeable hydrological screens by means of positive and negative wells. However, it is possible in that case to construct closed fictitious screens that allow water particles to pass only in one direction, namely to the inside, thus representing a water r e p e l l i n g s c r e e n for the groundwater inside that screen, together with an eventual pollution. Example 17 will make this clear.

Example 17. y

A confined aquifer with fully penetrating wells. One well with discharge Q in the centre of a circular row of n recharge wells, regularly distributed along the circumference of the circle, each with recharge Q. Suppose the recharge wells are located at the points Zm (m -- 0, 1, 2 . . . . . n - 1) and the positive x-axis has been chosen through the point z0. The complex potential caused by one recharge well at the point Zm in an infinite field has the value

Q

Zo

x

Fig. 22. Discharge well in the centre of a circular row of n recharge wells.

Q ,.~"-2m --- Cm @

27r D n

ln(z

- - Zm)

and for all recharge wells together, according to the superposition principle: Q 2re D n

The points

Q

m=0

(m -- 0 , 1 , 2 . . . . , n -

R n e i2rrm =

R n.

Thus we have £2 =

In { ( z - z o ) ( z - z l )

2rrm

Re i-7-

0 as

(Zm) n =

2zr D n

. . . (z-zn-l)

}+c.

are regularly distributed along the circumference of the circle with

Zm

radius R; so Zm Z n -- R n =

,,- 1 ~'-~ l n ( z - - Z m ) + C --

Q 2rr D n

ln(z" - R") + c

1). Now Zm are the roots of

(2.3.1-4)

Solutions, derived from known solutions

787

and together with the discharge well:

Q

In (z" - R n

~2 -- 2rr D n

z '~

(33)

)"

Written in polar coordinates, with z - re i°, equation (33) becomes" Q ln{1-(R)ne-i"°} a'2 -- 27rD-----n = ~Q

In [ 1 - ( R ) ~ { c o s ( n 0 ) - i

27r D n

sin(n0)} 1.

A s q ~ - K 9 9 - - 4~rDn 0 lnll -- ( E ) n e- i,,o 12, we find for the drawdown:

{ (R) -

Q In 1 + ~o( r, O) - 4 7r K D--------~

2n

r

-2(R)n

cos(n0) }.

0~o

Q

- 2 n R 2 n r -2n-1 + 2 n R n r n-1 cos(n0)

Or

4rr K D n

1 + ( R ) 2n -- 2(~)" cos(n0)

and thus

Vr(r,O) -

099

K~

Or

=

Q

1

(r, - ( r

cos( 0

2 7 r D r 1 + ( E ) 2 , , - 2(R],~ 7-, cos(n0)

represents the velocity of the groundwater in the direction of the origin. For r -- R this velocity becomes Q v r ( R , O) -

-27cDR

1 - cos(n0) 2 - 2 cos(n0)

47r R D '

provided that cos(n0) -J: 1 or 0 -J: 2mn'. ?l This surprising result means that, with the exception of the locations of the recharge wells, everywhere on the circumference of the recharge circle the velocity of the groundwater is a constant, directed towards the origin, independent of 0 or norK!

So groundwater inside the circular cylinder with radius R and length D cannot leave the cylinder and a pollution may thus be isolated and washed out. This cylinder may be considered as a water repelling hydrological screen for the water inside the screen.

788

(2.3.2)

Analytical solution methods

2.3.2. Product solutions

Besides solutions obtained by adding solutions of simpler problems as has been discussed in the previous section, solutions may also be constructed by multiplying solutions of simpler problems. These so-called product solutions are, in general, solutions of non-steady problems with two or three space variables and are composed of the solutions of related problems in one-dimensional flow. Consider the homogeneous differential equation for three-dimensional non-steady flow of groundwater through homogeneous isotropic ground (equation (10) of Section 1.4.2-2):

0299 -1-0299 nt-- 02 q9- fi2 099 Ox 2 OY 2 OZ 2 --~,

f12 with

Ss --K

(34)

and the three differential equations for one-dimensional non-steady flow in the same ground in the three coordinate directions:

02(/91 ___ f12 0(t91 OX 2

Ot '

02(t92 __ fi2 0(/92 Oy 2

at '

02(/93 ._ f12 0993 OZ 2

at

then, if ~o:(x,t), ~02(y,t) and ~p3(z,t) are successively solutions of these onedimensional differential equations, their product 99(x, y, z, t) -- q91 (x, t)q92(y, t)tp3(z, t) is a solution of the three-dimensional differential equation (34), which can be shown simply by writing:

02(/9 OX 2 - -

02(/91 q92993 OX 2 '

02(/9 Oy2 - -

02(/92 991993 Oy2 '

02(p 02 2

02q03 --- 991992 OZ 2

and

099 0991 0992 0(/93 0"--'7"= (/92(/93'~ -Jr- (/91gO3-'~ -1- (/91992 Ot and substituting these expressions in equation (34). Similar considerations are the basis for product solutions in three-dimensional axial-symmetric flow: if qgl (r, t) is a solution of the differential equation O2qgl ~ 1 O(pl = /~2 0991 Or 2

r Or

Ot

for a radial-symmetric problem, and 992(2, t) a solution of the one-dimensional equation

~2~2 __ f12 ~992 OZ2 bS '

Solutions, derived from known solutions

(2.3.2)

789

then ~o(r, z, t) = 991(r, t)992(Z , t) is a solution of the differential equation 02(/9 1 0(/9 0299 /~2 099 Or---T - t - -r -Sir + ~OZ2 = at for axial-symmetric flow, as can easily be verified. In order to apply this method of product solutions to boundary value problems, we have to investigate the types of initial conditions and boundary conditions that will make such an application possible. It will be clear that in the three-dimensional case, if the initial condition qg(x, y, z, 0) = F(x, y, z) should be equal to the product of the initial conditions of the one-dimensional problems, this is only possible if, theoretically, F ( x , y, z) = f ( x ) g ( y ) h ( z ) . In practice, f (x), g(y) and h(z) will almost always be constants. The boundaries along which the boundary conditions are given should be straight lines (planes) perpendicular to the coordinate directions, for instance, along x = a or y - - 0 , etc., and must have the general form A1 ~---fx + Bl~O - - 0

for x -- xo,

A2-~y -+- B 2 q 0 - 0

for y = Yo,

A3-7- + B3~o - - 0 oz

for z = zo,

(35)

where A,, and Bn (n = l, 2, 3) are constants, either of which may be zero, so that the cases of zero head and zero flux are included, while x0, y0 and z0 may vary from zero to plus or minus infinity. Now we may use the solutions for the related one-dimensional problems with similar boundary conditions as factors for the product solution. If the boundary condition along x = x0 for the one-dimensional case is 0991 A1-2--(x0, t) + Bl~Ol (x0, t) -- 0, Ox

the condition along x = x0 for the three-dimensional problem becomes, with ~p = ~p1992~P3: A1 ~ x (xo, y, z, t) + Bl~O(Xo, y, z, t) -- Al~P2~o3--~-x (x 0, t) -+- B1~02~03~01(xo, t) -- 992993 A1--2---(x0, t) nt- Ol(fll(XO, t)

as it should be.

-- 0

790

(2.3.2)

Analytical solution methods

Similar considerations hold for the other coordinate directions.

Example 18.

~p=h

~o=0

~o=0

x

Two-dimensional flow of groundwater in a confined aquifer towards two straight open boundaries, perpendicular to each other, starting from an initial head h, compared with the zero head along the boundaries which is kept constant (Fig. 23). The boundary value problem can be written as:

02(/9

~5

g----W- hi

0299 _ f12 099

ax---~ + ~y~ rp(x,y,0)-h

~888~

x •

KD, S

qg(O, y, t) -- O,

Fig. 23. Two-dimensional flow

qg(x, O, t) - - 0 ,

to open boundaries.

with f12 _

0-7 forx >0,

S

- KD' y>0,

Oqg(~, Y, t) -- O, Ox

e)a___~(x, , cxz, t ) - O. Oy

The boundary conditions are in agreement with equations (35) as for x0 - 0 A1 = 0, for xo -- cxz B1 -- 0, for Y0 - 0 A2 - 0 and for y0 - e~ B2 -- 0, while also the initial condition may be considered as the product of f ( x ) and g(y) with f ( x ) - ~ and g ( y ) - ~/-h, for instance. The related one-dimensional solution in the x-direction has to satisfy:

02(/91 __ f12 0f/91 Ox 2

~01(0, t ) - 0

8t

and

991(x, 0) - v/-h, O~l.(c~,t)-0, Ox

and in the y-direction:

02(/92 = fl20q 92 Oy 2 0--t-'

~02(y, 0) -- v/h,

0(/92 q)2(0, t) - 0

and

oY (c~, t) - O.

From these conditions it follows that the method of product solutions is applicable. The solution for the head q91(x, t) can be obtained from equation (a) of Example 1 of Section 2.3.1-1 which is the solution for the same one-dimensional problem but with zero initial head and head h along the open boundaries; so

erfc

ere( )

Solutions, derived from known solutions

(2.3.2)

791

In a similar way, we find q92(y t ) -

4'-herf(flY)

'

2,/7"

Thus the solution of the two-dimensional problem becomes" q)(x y t ) , , (Part A, 334.02).

h eft( fix \ \ ) e f t ( f l y ) k2v/7 k2v/7

(36)

Example 19. Y

Two-dimensional flow of groundwater in a confined aquifer towards three straight surface water boundaries with entrance resistances, two of them parallel to each other with entrance resistance c2 and the third one perpendicular to them with entrance resistance cl. The initial head is h compared with the zero head along the boundaries, which is kept constant.

~o=0

dll'll~llIl',',',l',',l',Ill;II',ll',l',

C2

~o=h for t = O ~o=0

Cl

X C2

-b

Fig. 24. T w o - d i m e n s i o n a l flow in a semi-infinite strip.

The differential equation with boundary values and initial value is, for the head ~p -- ~p(x, y, t)" 02q 9 OX 2 1

02q 9 = fi20(t9 fi 2 __ Oy 2 --~' -- K D '

99(x,y,0)-h

forx >0and

K 099(0, y, t) -- 99(0, y' t), OX

Cl

K 099(x, b, t) - q)(x, b, t) Oy c2 '

-b


099(oo, y, t) -- 0, OX

K OqO(x, - b , t) -- q)(x, - b , t) Oy c2

The boundary conditions may be compared with the equations (35). For x - 0 we have A1 - K and B1 -- Cl1 . For x - ec B 1 - 0 . __1 Fory-bA2-KandB2c2" Fory---b A2-K andB21 C2 " As the boundaries are straight planes perpendicular to the coordinate directions, and also the boundary conditions and the initial condition are in agreement with the demands underlying the application of product solutions, we may write the solution as"

q)(x, y, t) -- 991(x, t)q)2(y, t),

(37)

792

(2.3.2)

Analytical solution methods

in which 991(x, t) is the solution of the one-dimensional boundary value problem: 32991

23991

3X 2 = fl

K

3"7'

991(X, O) -- h l,

0991 991 (0, t) (0, t) = ~ , OX Cl

0991 OX

~(oo,

t) = 0 ,

and 992(y, t) is the solution of the one-dimensional boundary value problem: 02q 92 __ f12 0992 Oy 2 oq--7-'

-

K

992(Y, O) -- h2,

0992 (b, t) _ ~,992(b' t) Oy c2

K 0992 . . . .( b, t) - 992(-b, t) Oy c2

In order to determine 991(x, t) we apply the Laplace transformation with respect to t, giving: d2q31

fl2(Sqgl -- hi) -- 0,

dx 2

dq51

K-T-(0, s) -tax

qgl (0, S) Cl

dq~l

~ ( o o , s) =0, dx

with the solution qgl(X, S ) -

hi (1 s

e-t~x'fi

)

1 + Xc~,/-i

"

1 If we define the resistance parameter as 0 -- f12K2c2 [T- 1 ], we find from tables for

Laplace transforms, the inverse transform 991(X, t) -- hi e r f\(2f ~l x/ ] + h 1R ( ~ f l X ~ ,

(38)

(Part A, 126.12), where the resistance function R is defined as R(x

-- e2X+y2erfC(y ) + y

(see Section 3.1.3).

For the solution of 992(y, t) we apply the finite Fourier cosine resistance transformation (see Section 2.2.3-2, the equations (84), (85) and (86)), with q32 -- q~2(n, t)" 2 dq~2 O/n f12 b2 q~2 -dt '

q32(n, 0) -- h2 foot, cos ( °lbn Y /'] d y --

h2b O/n

sin Otn,

Solutions, derived from known solutions

(2.3.3)

793

with the solution:

h2b

(

q32(n, t) -- ~ot,, sin c~,, exp

°t2t ),z flZb2

and the inverse transform: oo

_v) (

sin oe,,

~02(y, t) -- 2h2 ~ (1 -+- ,,e, cos (or,, y n--o Oln ol2q-g2)

f12b2 )

(39)

b with c~,, being the roots of oe tan oe = e = Kc 2• Now, with h lh2 : h (for instance, h l = h2 -- v/h ) the solution of our twodimensional problem is equation (37) in combination with equation (38) and equation (39) (Part A, 354.08). 2.3.3. Periodic flow solutions

Periodic functions are of great practical importance for the solution of geohydrological problems because they not only serve as the basis for the Fourier transformations (Section 2.2.3) but in many cases they also allow a general function to be written as a Fourier series, consisting of sine- and/or cosine functions, thus reducing the solution of a problem to a much simpler one. In many groundwater problems a periodic function of the time is involved in one of the boundary conditions, for instance, seasonal fluctuations of the precipitation or tidal fluctuations of surface water. As we have seen in Section 2.2.3-1, periodic functions may be represented by series of sine functions and cosine functions; so it will be sufficient to investigate the influence of a simple sine function (or cosine function) of the time on the behaviour of the groundwater flow, that under such circumstances is referred to as periodic flow. In the following we will make use of the general trigonometric function

f (t) = A sin(cot + 3)

27/"

with c o - ~ T'

(40)

in which A -- the amplitude, T = the period, and 3 - an arbitrary constant. Theoretically, solutions for periodic flow problems may be obtained from solutions involving arbitrary functions of the time by substituting a sine function of the time. In most cases, this will yield complicated expressions with definite integrals that can hardly or not at all be evaluated to simpler results. However, there are methods of solving these problems in a particular manner, which will be the subject of this section. To show the method, we start with a simple example.

794

Analytical solution methods

(2.3.3)

Example 20.

2h:

One-dimensional groundwater flow in a confined semi-infinite aquifer towards open water, without entrance resistance. There are tidal fluctuations of the open water according to the trigonometric function.

KD, S

x

F ( t ) - h sin(cot + 8)

Tidal fluctuations along a half-plane.

with c o -

Fig. 25.

27/" T'

in which h = amplitude, T -- period and 8 an arbitrary constant (Fig. 25). The drawdown or head qg(x, t) is a function of the distance from the open boundary and of the time and can be solved from"

0299 __ f12 0(/9 with 1~2= S Ox 2

at

99(x, 0) -- 0

KD '

for x > O,

99(0, t) -

F ( t ) -- h sin(cot + 8),

~o(oo, t) - - 0 . We first solve the problem, assuming F ( t ) to be an arbitrary function of time. Laplace transformation gives"

d2q3 dx 2

f12

sq3 - O,

(p(O, s) - F(s),

q3(oo, s) -- O,

with the solution q3(x, s) - F(s)e - ~ * ~ .

(a)

According to the convolution theorem (Theorem 6 of Section 2.2.2-1) and equation (40) of Section 2.2.2-3, the inverse Laplace transformation yields:

~p(x, t) -

flXfot

2~.~

With the substitution ot2

gO(X, t) -- ~2

F(t-

_ -

-

f12x2 47: '

ft~x F ( t 2qri

r)exp

(f12x2) 4r

dT

r~r-

for x > 0.

(b)

we find

fl2X2)4ot 2 e -~2 dot

for x > 0,

the solution for an arbitrary fluctuation of the open water with time.

(c)

Solutions, derived from known solutions

(2.3.3)

795

Secondly, we replace F ( t ) by h sin(cot + 3), giving ~p(x, t) -

--~

sin co t

+ 3 e -~2 dc~.

(d)

f (oe) dot,

(e)

5-Z Now,

/~x

f(oe) doe -247

/0

f(oe) dot --

/0

f12x2

with f(c~) - sin{co(t - 4¥gY) + ~}e -~2. The second integral on the right-hand side of equation (e) is a transient disturbance caused by starting the oscillation of the surface water at time t - 0; it dies away as t increases, leaving the first integral as the solution for large t:

v/~

sin

l(

co t

40/2

+ ~

/

e -a2 d~.

(f)

As sin 0 is the imaginary part of e iO, we have ~o(x, t) -- Im

2h ei(O~t+3)

exp

- - 13l2 - -

icofl2x2 40e-------5--- dc~,

which can be evaluated, with the help of Section 2.2.2-3 if we set there in equation ( 4 3 ) ) 2 1 andt-oo, as -

-

iwfl2

~p(x, t) -- Im[hei(°~t+~)e-t~x'/7-d] or, as ~ -

~~/2(1 + i )

99(x, t) -- Im[hei(~°t+~)-½45/3x~(l+i)]. So, with c o ' - fix/-~, we find

~o(x, t) - he - J x sin(cot - c o ' x + 3),

(41)

which means that the head (drawdown) 99 at every fixed distance x - x0 is represented by a steady oscillation with period T 2__~and amplitude he-"/x0 and with a time lag co'x0 with respect to the surface water. The solution (d) is the non-steady solution of the problem if the oscillation of the surface water starts at the time t - 0; it tends to the "steady" solution (t) or (41) with increasing values of t. This solution is a special steady state solution because it still contains the time variable and the storage coefficient S. So the head ~o keeps -

796

Analytical solution methods

(2.3.3)

changing with time but has become periodical with the same period T as that of the surface water movement, but with an amplitude and a difference in phase that are both functions of the distance only. It can easily be verified that equation (41) satisfies the differential equation 0299 __ ~}x2 - -

f120~0 and the boundary conditions for x - 0 and x - oo. Of course, it no at

longer satisfies the initial condition qg(x, 0) - 0 for x > 0. It would be convenient to find a direct method to solve the steady oscillation represented by equation (41); therefore we make use of this result and seek for a solution that has the general form: q0(x, t) -- f (x) sin {cot + ~ - g (x) },

(g)

in which f ( x ) and g(x) are functions of x only and must be determined such that f ( 0 ) - h, g(0) - 0 and f ( o o ) - 0. Now ~p(x, t) may be considered as the imaginary part of a complex function 4~(x, t) as follows: ~0(x, t) - Im[4~(x, t)] - Im[f(x)e i{rot+a-g(x)}]

or

~o(x, t ) - Im[F(x)e i(°~t+a)] with F(x) - f (x)e -ig(x),

(h)

where F(x) is a complex function of x only. As a2~° ax2 - Im[ a2~1 ax2 j and ~a~0 _ im[a~l at J' we have the differential equation for qS: a2~

= /~2 atp

OX 2

at

which becomes, with equation (h)"

02 Fei(O)t+~) __ fl2icoF(x)ei(°)t+~) Ox 2 02F OX 2

or

ifl2coF - 0 .

~0(0, t) -- h sin(cot + ~), but also ~o(0, t) -- Im[~b(O, t)] -- Im[F(O)e i(°)t+~)] - F(O) sin(cot + ~) from which F ( 0 ) - h. In a similar way we find F(oo) - - 0 . Thus, the differential equation with boundary values for the complex variable F as a function of x, becomes" d2F dx 2

i fi ZcoF - 0 ,

F (0) - h, F (oo) - O,

(2.3.3)

Solutions, derived from known solutions

797

with solution: F ( x ) -- he -flx#7-d = he -°/x(l+i)

(i)

if c o ' - flv/~. Compared with equation (h), we see that and

f (x) - he -°Ix

g ( x ) - +co'x.

These values, substituted in equation (g) give the result (41), obtained before. The function F ( x ) has been solved from a differential equation with boundary values similar to the Laplace transformed function qS(x, s) in equation (a). So the solution (i) can be derived immediately from solution (a) by replacing F(s) and s by h and ico, respectively. Then F ( x ) must be multiplied by e i(c°t+6) to find ~b(x, t), the complex function the imaginary part of which equals the function ~0(x, t) that was looked for. In this way, also more complicated problems concerning periodic flow may be solved by making use of the Laplace transforms of already known solutions for non-periodic flow, as we shall see in the following examples. Example

~

21.

$

Periodic precipitation on a circular island, surrounded by open water with a constant level without entrance resistance. Assumed horizontal flow in a confined aquifer. We first consider the case of a precipitation being an arbitrary function of the time p ( t ) with p(0) = 0 (Fig. 26).

$ $~ p(t]

m

.

.

.

.

"r

= 0c

.

KD, s

I

Fig. 26. Precipitation on a cir-

cular island. We have" 02(/9 1 099 p (t ) 2 0(t9 Or 2 +--¢-~ =/3 ~ r -~r K D Ot

~p(r, O) -- O,

f12

with

-8--;-r(0, t) -- 0

and

S

=

K D '

¢p(R, O) - O.

Laplace transformation with respect to t gives" d2q3 1 dq3 •dr 2 t

/?(s) flZsq) @

r dr

KD

- 0

d (o s)-o,

'

dr

'

s) - 0 ,

with the solution:

1 99(r, s) -- S

} s

Io(~R~/-s)

"

(a)

798

Analytical solution methods

(2.3.3)

The inverse Laplace transformation can be obtained by means of the Laplace Inversion Theorem (Theorem 9 of Section 2.2.2-3), but is rather complicated; it is easier to apply a finite Hankel transformation with respect to r to the differential equation with boundary values of which equation (a) was a solution, making use of the equations (119) and (121) of Section 2.2.4-3:

2

13(s) R 2

~" ~ - ~ 2 s ~ 4 J~ (~.) - 0, R2 K D O/n an algebraic equation with the solution

~(n, s) --

/3(s)

R2 J1 (o/n)

~--7

.

(b)

fl2KD s-+- flZR2 Inverse Hankel transformation yields, according to equation (120): 2 oo s) -

-(s) p

._0s+

Jo(°tnr~

TJ

(c)

O/n Jl (o/n) '

with O/n being the roots of J0(o/) = 0. Inverse Laplace transformation gives the end solution:

T ) f0 ' 2~ J0(c~nr qg(r, t) -- ~ n=0 O/nJ1 (O/n) exp {

2 f12R2( t - r ) } p ( r ) d r ,

(42)

with O/n being the roots of J0(o/) = 0 (Part A, 233.15). Secondly, we consider periodic precipitation and substitute p sin(mr + 5) for p ( r ) with p as a constant, in equation (42) and let t approach infinity in order to find the steady oscillation of the groundwater head which we may expect. The evaluation of the combined series and integral in equation (42) becomes rather complicated; therefore, like in Example 20, we assume a steady state of the form:

qg(r, t) -- f ( r ) sin{cot + ~ - g(r)},

(d)

in which f ( r ) and g(r) are functions of r only and rp(r, t) is considered as the imaginary part of 4) (r, t): ~o(r, t) -- Im[q~(r, t)] -- Im[f(r)ei{a't+a-g(r)l],

or

~o(r, t) = Im[F(r)e i(°°t+a)] with (r) - f (r)e -ig(r). F(r) is a complex function of r only.

(e)

Solutions, derived from known solutions

(2.3.3)

799

The differential equation for 99" 0299

1 Oq9

p sin(cot + 3)

Or 2 4-_r -~r 4-

0q)

= ~2 at

KD

leads to"

[O2~b

10q5

Im 7r2-r 24--r-~-r 4- KPD ei(Oot+a)]- fi2im[0q~__~_] from which d 2F 1 dF dr 2 t r dr while -dT-r dF (0) F(r)-

--

f12

P

icoF 4- ~ = 0, KD

0 and F ( R ) -- 0, with the solution

~S

1-

Io(~n47-£)

I

(f)

This solution is identical to the solution (a) if we replace there fi(s) by p, s by ico and q3(r, s) by F(r), in a similar way to that found in the previous example (equation (i) compared with equation (a)). The solution for the steady oscillations becomes, if we combine the equations (e) and (f)" q)(r, t) -- = I m - - 1 ico

Io(fl r ~ ) ] Io(~ R ,/~) ! e~(~'+~

(43)

(Part A, 238.21). Making use of the Kelvin functions ber(x) and bei(x) as the real and imaginary parts of Io(x~/7) (Section 3.3.3) Io(xv/Tl) -- ber(x) + i bei(x) - Mo(x)e iOo(x) with Mo (x) -- v/ber 2 (x) 4- bei 2 (x)

and

00 (x) - arctan [ ~ ] , b(x) ei

we find" ~o(r, t) -

p sin cot + coS

rr } (44)

p Mo (fl r.v/-~) sin cot + 3 + 0o (fl r x/~) - 0o (fl R VG) - -}- .

~oS Mo(~/~/-£)

R e m a r k . In practice, a periodic precipitation will generally be composed of a constant precipitation and a variation more or less according to a sine function, for

800

Analytical solution methods

(2.3.3)

instance: p(t) -- Po + P sin(cot + ~), in which po and p are constants with po ~> P. In that case, the steady state solution for a constant precipitation on a circular island P0 (R 2 _ r 2) ~p(r)- 4KD

(45)

must be added to equation (44). Solution (45) may be obtained by direct solution of the differential equation for the steady state: 1 dq0

po

dZg° + -~r -t -- 0 dr 2 r KD

with ~o(R) -- 0

and

d~o

-a-~-r(0) - 0,

namely:

1 d [r d~p

d (r dqo por d---~ --~r ) - - K D

po -~,

r-dr

-gr ) - dcp po r2 r~ +Cl, dr 2KD po r2

qo --

4K D

dqo ~ = dr

por 2KD

{ Cl , r

+ cl In r + c2. poR 2

From ~ ( 0 ) -- 0 it follows that cl - 0 and from ~o(R) - 0 it follows that c2 = 4/(D, which leads to equation (45). This result may also be derived from equation (42) by replacing p ( r ) by po and letting t approach infinity. Then, as

Po

2

fot /

f12R2

exp

= poexp ( ~ - ~ ~ 2 ) _ [12R 2po exp --

fot

exp (/32R2] dr

- -----

Ot2

f12 R2

-- /~2R2p0 -012

1-exp(/32R2)

exp

- 1 \ f12 R 2 1

,

p(r, t) -- 2p°R2 J°(~-~) 1 - exp K D n=O or3 J1 (O/n) with an being the roots of Jo(o~) - 0 . 2poR2 ~ KD

Jo (~-~)

n=O ~3 J1 (~n)

f12R2

The first term

,

(g)

Solutions, derived from known soluticns

(2.3.3)

801

in this equation can be evaluated by comparing the two solutions (a) and (c) for 93(r, s) from which it follows that

io(flrv/7 ) }

ls 1 - I0(-flR-~

oo

2~ 0

must be equal to

_~)

Jo ( °tnr ~2

= oln(s + - ~ ) J l ( c ~ ) or, in general, if we set a -- fl V/s-:

oo

jo(~_~_)

1

,,=0 Otn (or2 -4- a 2 R e) J1 (Otn) = -2a- -2-R~

{

Io(ar) }

(h)

1 - Io(aR)

In particular, for a - - 0 we have oo

n--O

go(Unr ] R !

R2

~3nJ1 (~n) --

r2

(46)

8R 2

which can easily be verified by developing the modified Bessel functions in the right-hand side of equation (h) in infinite series according to

lz 2 (¼z2)2 I o ( z ) - 1 -4-~

9- (2!) 2 -4- (3!) 2

•

°

°

°

So equation (g) becomes" P0

(R 2 _ r 2)

~o(r, t) -- 4K----D

2po R2

KD

J0(~RZ) exp

,,=o c~ J1 (c~)

("2')

f12R2 ,

(47)

with c~,~ being the roots of J0(ot) - 0 . The first term of this solution represents the steady state which we found before. From the Examples 20 and 21, we may conclude that the following theorem holds: Theorem 4. If the solution of a periodic flow problem tends to a steady oscil-

lation, it can be derived from the Laplace transform of the solution of a similar problem involving an arbitrary function G(t) of the time instead of the sine function A sin(cot + ~), according to the following rules: 1° Determine the function F, which is a complex function of space only, from the Laplace transformed function ~, which is a function of the same space variable(s) and s, by replacing the transformed arbitrary function G(s) by the amplitude A of the sine function and s by i co. 2 ° Multiply F by e i(~°t+a) to find the complex function qb, which is a function of space and time.

0

0

Analytical solution methods

(2.3.3)

802

Take the imaginary part of ¢ which equals the solution q) (head or drawdown) of the problem. Verify whether q) satisfies the differential equation and the boundary values (not the initial value) in order to be sure that a steady oscillation can be reached, a condition for the validity of this theorem.

Example 22. We take Example 9 of Section 2.3.1-2, a discharge well between two parallel open boundaries (Fig. 11), and suppose that the discharge is an arbitrary function of time: Q - Q(t) with Q ( 0 ) - 0. Then we have:

02(/9 t 02q9 -- fl2Oq ~ ) Oy2

OX 2

with

f12 -

S

at

lim ( r Orp) = - Q(t) r-+0 -~r 27r K D ~0(cx~, y, t) -- 0

,

~o(x, y, 0) - 0

KD

and

'

for r -- V/X2 + ( y - a) 2, qg(x, 0, t) - ~0(x, b, t) - 0.

Finite Fourier sine transformation with respect to y gives (see remark concerning Fig. 13): OX 2

T

(0 --

0(o (0, n t) 0-7 '

-~ ,

~O(X , 11,

O(t) sin (nrra,~ 2KD --7/

--

and

q3(c~, n, t) - 0.

Laplace transformation with respect to t yields:

d2~ dx 2

{(nTr) 2 } '-if- q-flgs ~ = 0 '

dq3(O,n S ) d---x '

Q(s) sin nrra

and

,}(oo, ,,, s ) = o , with the solution

(o(x n s ) '

'

Q(s) sin

--2K-----D

--b----/

M.

with M n -

+

Inverse Fourier transformation gives:

O_.(s) ~

nrra']

qS(x, y, s) - b / ~ D ~ s i n ( - - - b - / s i n \

n=l

(arty b )

e -M~x Me

(a)

which still needs an inverse Laplace transformation in order to get the solution for the drawdown q) as a function of space and time, caused by an arbitrary timedependent discharge. In the context of this section we shall not perform this transformation but make use of equation (a) for the determination of the steady oscillations of the drawdown, caused by a periodic abstraction, for instance:

(2.3.3)

Solutions, derived from known solutions

803

Q(t) = Qo + Q sin(cot + 3) 27r with co -- -7and Q0 ~> Q, which describes a discharge composed of a constant part and a part that varies according to a goniometric function. The solution is the superposition of the solution for a constant abstraction given in Example 9 (equation (18)) and the solution for a sinusoidal abstraction. Now the latter may be written down immediately from equation (a) according to Theorem 4:

q)(x, y, t) =

Q bKD

I

nrra nzry Im e i(wt+3) ~-'~sin --if-- sin - - ~

e-xv/(-cff--)2+ifl 2w ]

( ) ( ) V/(t--~) 2 q- ifl2co

n=l

'

which may be evaluated, if we put V/( ~.)2 n~r nt_ ico - pn + iq, to"

qg(x, y, t) --

(nrca ~ ( n r c y ) e -~xp" Q Z sin \--if--/sin \ - 7 v/P 2 ÷ q2 b f l K D n=l

x sin {c o t - g x q n -

(48)

arctan(q--L) + 3 } .

Pn

Verification, of whether this solution satisfies the differential equation and boundary values may be most easily performed by investigating whether the function cx~

(nrca] F(x, y) = Q Z s i n \ - 7 / s i n b K D n=l

(nrcy] e

-x

v/(--if-)2+i[32co nTr

b ] v/(n~_)2_t_ ig2co

is a solution of 02F OX 2

-t--

02F Oy 2

fl z i coF = O,

with F(x,O) = F ( x , b ) = 0 and F(oo, y) = 0, which can be easily shown. As qg(x, t) - Im[F(x)ei(~°t+a)], it follows that qg(x, t) is a solution of the posed boundary problem. 2.3.4. Solutions in anisotropic soils

The general three-dimensional differential equation for non-steady flow of homogeneous groundwater through homogeneous anisotropic ground in which the main directions of the anisotropy are parallel with the chosen coordinate directions, has been given in Section 1.4.2-2, equation (9):

02(,0

0299

02

099

K x - ~ x2 + K "v ~Oy 2 + Kz ----~-~ Og 2 = S s ~a t

(49)

804

(2.3.4)

Analytical solution methods

If the main directions of the anisotropy (as is assumed here) are horizontal (x and y) and vertical (z), also horizontal and vertical boundary conditions, which are present or will be assumed to be present in almost all geohydrological problems suitable for analytical solutions, can easily be expressed in terms of the different conductivities. For instance, flow through a semi-permeable horizontal layer often serves as a boundary condition for problems in leaky aquifers, and may be described as" Vz - - Kz Oq) = Oz

A~ c

for anisotropic soils, where c = the leakage factor (cf. Section 1.3.3-3). Problems in such anisotropic soils may be solved in a similar manner to that which has been employed in the previous examples for isotropic soils, as the following example may show. Example

_

23.

....

Two-dimensional steady flow through an anisotropic leaky aquifer from a fixed water level above the semi-permeable layer towards a vertical surface water boundary with entrance resistance. The differential equation with boundary values for the drawdown q9 = qg(x, z) becomes:

e - 0

~'////////////////////////,//C

0299

Fig. 27. Anisotropicleaky aquifer with entrance resistance. Oq9

02(/9

K~~x2 + Kz OZ 2 = 0

h -99(0, z)

-

K~ 7x(O,

-

I
z) -

099

w

,

q)(x, D) , c

~o(~, z) -

099 (x,0) Oz

~

_

O,

0. _

Finite Fourier cosine resistance transformation with respect to z (see Section 2.2.32c) gives, with q 3 - q3(x, otn)" d2q3 Kx dx 2

2 Kz °tn -~--~q3 -- 0,

with c~n being the roots of ot tan c~ - e - K o__0__ zc' dO - K x -7-(0, Otn) (IX

hD

(o(0, o~,~) sin(otn) -

-

~

tOOln

,

q3(oe, Otn) - O. W

The solution is, with ~ 2 _ K_...Lz. Kx

~(x, ~ ) -

h D 2 sin (Otn)

u~nx e

Otn( D + I~ Kx wetn )

o

Solutions, derived from known solutions

(2.3.4)

805

Inverse finite Fourier cosine resistance transformation yields the end solution for the anisotropic case" I~C~nX

99(x, z) -- 2 h D

Z

sin(ot~)e

D E

n=O otn(D + lzKxwotn)(1 +

ol.z] cos (--D--/'

(50)

with oe,~ being the roots of ot tan oe - e - Kzc D " The solution for the isotropic case follows immediately from equation (50) by putting Kx -- Kz -- K and t h u s / z - 1" sin(otn)e

OlnX

z~ ~o(x, z) -- 2 h D Z n=o C~n(D + Kwotn)(1 +

E

COS(0/nZ ~--5)

(51)

D" (Part A, 355.15), with or,, being the roots of ot tan ot - e - K---~. Inversely, deriving the anisotropic solution (50) from the isotropic solution (51) is less simple but is of more importance, because most solutions of geohydrological problems concerning two-dimensional or three-dimensional flow through aquifers are available for the isotropic cases. In order to find a general method to perform such derivations, we must compare relations between variables in isotropic flow with corresponding relations in anisotropic flow. For that reason we consider the isotropic case as a model and the anisotropic case as the prototype and try to find the several scale factors underlying the similitude of model and prototype. The general differential equation for the anisotropic case (prototype) is equation (49) and for the isotropic case (model): 0299m

0299m

O2@m

0~m

Km Ox2 +Kin Oy~m + K i n 0~m =Ssm'0~m'

(52)

in which the index m refers to the model. We rewrite the equations (49) and (52) as"

Kx 0299 K y 0299 0299 Ss 0~9 t t = Kz OX 2 Kz Oy2 OZ 2 Kz at

0299m 0299m 0299m

and

(a)

Ssm 099m

Ox-----~ -~- Oy2 + Oz2 = Km Otto

(b)

and set /z 2 -

Kz

K~

and

va = Kz. Kv

(53)

806

(2.3.4)

Analytical solution methods

As the piezometric head 99 is defined as q) - z + ~ , it is obvious to choose the same scale for q) and z and even the scale factor equal to 1 in order to ensure that the definition for the piezometric head remains unchanged. So q)m -- q9

and

Zm -- z.

(c)

Then, as the corresponding terms of the differential equations (a) and (b) should be the same, we find: Xm-/zx

and

Ym

vy

(d)

and

Ss

Ssm Kmtm

=

Kzt

•

(e)

The discharge equations for the prototype are:

Oq)

0q) IdQy[ - Ky- Z- dx dz

[dQx[ - KX-~x dy dz,

oy

and

099 Id Qz I = Kz -~z dx dy, and for the model: 0q)m Id Qxm[ -- Km ~ dym dzm,

060~ IdQvm[ = Km -T''' dxm dzm " 0Yrn

and

0qgm [dQzml - Km~zm dxm dym. If we choose, in general, scale factor 1 for Q, we have

Q x m - Qx,

Q y m - Qy,

Q z m - Qz,

(f)

and according to equations (d): 0p

0q)m

0(,/9

0~0m

099

0q)m

099

Kx ~x d y dz - Km~x m dym dzm - K m - ~ x V d y dz, K y -ff-fy dx d z - Km ~y m-

0q) dxm dzm - Km v~y/z dx dz,

and

099

gz -~z dx d y - Kin-g--- dxm d ym - Km-2--/z dx v d y, oz OZm from which 1

Km -- Iz Kx __ _v Ky -KZ~ v IX tzv

(g)

Solutions, derived from known solutions w/ ~Kz a n d v -

or w i t h # -

(2.3.4)

807

W/ K~--7, Kz . (h)

Km -- v / K x K y .

The value Km = /zv K---zzfrom equation (g) substituted in equation (e) gives" Ssm

Ssm#

Kmtm

P

Ss

Kztm

Kzt

For reasons which will become clear in the following, we choose: SS m - -

(i)

SS

and thus"

0)

tm -- l z v t .

The t r a n s p o r t e q u a t i o n s for the prototype are (see Section 1.2.1)" dx

090

dy

ne d--7 = - g x -~x ,

Oqo

ne d--7 = - K v" Oy '

dz

Oqo

n e - - = - Kz - - , dt Oz

and for the model: dxm nem dtm

dym __

099m Km OXm

dzm

0(/gm

nero dtm

Km OZm

nemdt m

3q)m -K'n 5~m '

in which ne -- effective porosity (Section 1.3.2-2). Comparison of the corresponding equations and substituting the already obtained scale factors according to the equations (c), (d), (g) and (j), yields" lzdx nem~ -lzvdt

lZ v

Kx~

Oq)

#Ox '

vdy v Oqo nem~ -- - - - K v lz v dt lZ r o y dz n e m ~

Kz 09o ~-

lz vd t

lZ V Oz '

from which nero -- he, f r o m w h i c h nero - he,

from which nero -- he,

all three equations yielding the same scale for ne, determining for practical reasons also the same scale factor 1 for the volumetric porosity n (Section 1.3.2-2) nem -- ne

and

nm -- n.

(k)

808

(2.3.4)

Analytical solution methods

From these equations the scale factors for the velocities in the directions of the coordinates also follow: 1)x l)xm = -

v

,

I)ym--

Uy ----:-

/z

and

l)zm--

1)z ~.

/zv

(1)

The scale factor for the volume of a part of the ground follows from the volume d V of a small box with sides dx, dy and dz" dV - dx dy dz and dVm - dxm dym dzm /~ dx v dy dz, namely (m)

Vm - i~ v V.

Also, the scale factors for the water volume Vw and the soil volume Vs, which follow respectively from Vw = n V, Vwm - - nm Vm and Vs = (1 - n) V, Vsm - - ( | - n)m Vm, are equal to/zv, because nm -- n (equation (k)). So the condition V = Vw + Vs from the prototype also holds for the model: Vm - - Vwm -~- Vsm.

Now we see that if we had chosen Ssm :~ Ss instead of Ssm -- Ss (equation (i)), for instance, tm - t ' yielding Ssm - % # v ' we would have found nm - ~ #v instead of t/m - - t/.

Then, though the scale factor for V would have remained unchanged (equation (m)), the scale factors for Vw and Vs would have become different from the scale factor for V and thus Vwm + Vsm ~ Vm, which is not logical. The boundary conditions along planes to which the main directions of the anisotropy are parallel (here horizontal and vertical planes); in particular boundary conditions involving entrance resistances (Section 1.5.1-3cl °) may serve to find scale factors for the latter. For instance, along a horizontal semi-permeable layer with resistance c, the boundary condition for the prototype is: _KzOq) = Aq9 Oz c

and for the model: Oqgm

A(flm

- K m OZm --

As K m - - #-'-~, Kz (tim

Cm

- - (/9

and Zm - z, according to the equations (g) and (c), respec-

tively, we have Kz Oq)

Aq)

ILl) OZ

Cm

from which it follows that Cm- ~vc.

(n)

Solutions, derived from known solutions

(2.3.4)

809

Along a vertical surface water boundary with entrance resistance, the boundary condition is" 099

Acp --

-X~ ~

099 or

w~

-

Kv

Oy

A~p --

Wy

for the prototype and for the model: AC, Om - - K m ~0(Pm X m __ m tOln

and

O~pm

- Km -0Y-ram -

A~om tOm

or according to (g), (c) and (d): # Kx 099

A99

V

tom

#OX

and

v #

O(# roy

Wm

from which tom -- Utox -- #toy.

(0)

From Ss -- (,-~g 1-,, + -~w)pg n (Section 1.3.5, equation (57)), it follows that, because both Ss and n remain the same for the prototype and the model (equations (i) and (k)), also the scale factors for Eg, Ew, /9 and g should be chosen equal to 1. So, Egm-

Ewm-

Eg,

Pm - - P ,

Ew,

(p)

(q)

Ym - - P m g m - - Y.

As M - p V, the scale factor for the mass becomes, according to the equations (m) and (q)"

Mm -- # v M .

(r)

From the definition for the piezometric head 99 1 for the pressure Pm -- P.

z +

, we find the scale factor

(s)

The relations between the piezometric head and the stream function ~ are for axialsymmetric flow for the prototype

Kr ~

--

2rcr

Oz

and for the model 099m | Ogtm (see Section 1 . 2 . 4 ) Km 0r-ran -- 27rrm OZm

810

(2.3.4)

Analytical solution methods

As for axial-symmetric flow with the z-axis as axis of revolution Kr - - K x - - K y we have from equation (h)" K m - Kr and from

V

~

I~

rm -- V//./,2x2 "-l- v 2 y 2 -- lzr, so that ~

KF

099

Olpm

1

---"

lzOr

2rclzr Oz

from which for axial-symmetric flow

~m -- ~ [L3T-1] •

(t)

The relations between the piezometric head and the stream function ~p are for twodimensional flow in vertical planes for the prototype

0q)m = 0ff~m and for the model: Km 0~m OZm from which

0~0 _ 0~p Kx-~x-az

0q)

Kx lzox

01pm

=

Oz

,

as Kv

-

-

Kx and thus v - #.

This means that for two-dimensional flow in vertical planes

(u)

~rm -- % [L2T-1]. In two-dimensional flow in horizontal planes we have

099

0~r

Kx-~x - 5z

0~0m and

Km

OXm

0~m =

Oym

for the prototype and the model respectively, from which ~--'gx Oq9 _

O~m

v

roy

lzOx

So for two-dimensional flow in horizontal planes we find l~rm -- ~

[L2T-1].

(v)

The knowledge of the various scale factors obtained for the variables and parameters in the relation between isotropic and anisotropic flow of groundwater, enables us to transform known solutions for groundwater flow in isotropic soils into desired solutions for analogous flow in anisotropic soils, which can be recorded in the following theorem.

Solutions, derived from known solutions

(2.3.4)

81 1

Theorem 5. An analytical solution for flow of homogeneous groundwater through an anisotropic homogeneous medium of which the main directions of the anisotropy are parallel to the planes subjected to boundary conditions, can be derived directly from the analytical solution for the analogous problem in an isotropic medium, according to the following rules.

1° Keep the state variables ~o (piezometric head or water table), p (pressure) and Q (total discharge) unchanged. 2 ° Multiply the space variables x, y and z and also constant dimensions in the directions of x, y and z by #, v and 1, respectively, in which ~ Kz

~ Kz

(z remains unchanged).

3 ° Multiply the time variable t by #v. 4 ° Replace K by v/KxKy. 5 ° Keep Ss (specific storage coefficient), the effective porosity ne and the porosity n unchanged. 6 ° Multiply the resistance in a horizontal layer c by #v and the resistance w in a vertical plane by v and Wy in a vertical plane by lz. 7 ° Keep the axial-symmetric stream function ~ [L3T -1] and the two-dimensional stream function ~ [LZT -1] in horizontal planes unchanged and multiply the 1 two-dimensional stream function ~ [LZT -1] in vertical planes by 7" 8 ° Multiply the velocities Vx, vy and vz in the coordinate directions [LT -1] by ± p~ Lu and --~, respectively. This theorem, applied to the isotropic solution (51) of Example 23, yields the anisotropic solution (50), which can easily be verified, as in two-dimensional anisotropic flow K. = Ky and thus # = v and #v = #2; then x has to be replaced by ttx, K by Kx ' w by /zw and c by #2c, yielding K x bDt 2 c ~_ KD for E, the other zc parameters (99, h, D, c~,, and z) remaining unchanged. Example 24. For the derivation of the analytical solution of a non-steady threedimensional groundwater flow problem in an anisotropic medium, we choose Example 24 of Section 2.2.6 involving axial-symmetric flow, caused by abstraction of a constant discharge from a leaky aquifer by means of a partially penetrating well. The solution for isotropic conditions is given there by equation (g) and becomes, if we transform qg(r, z, t) into ~0(x, y, z, t) with r - v/X 2 + y2:

Q

O0

sin ( - ~ ) -

--b-J sin (°tna~

qg(x, y, z, t) -- 2rcK(b - a) =

(54) ( ~n z X COS

ot2K t Otnv/Xe + ye } D

812

Analytical solution methods

(2.3.4)

D with O~n being the roots of o~tan o~ - e - K--~." In order to find the solution for the anisotropic case we must, according to the transformation Theorem 5, replace K by v/Kx Ky (rule 4), t b y / x v t (rule 3), x and y b y / x x and vy, respectively (rule 2) and c b y / ~ v c (rule 6), the other parameters remaining unchanged (rules 1, 2 and 5). So

D

g__+

--

v / K x KyiJ, v c

D

v/X2 _+_ y2 ~

V//j2X 2 nt_ v2y2

Kzc '

Oln2Kt

2 Otn~KxKylzVt

Ss D 2

Ss D 2

2

Oln Kzt Ss D 2 '

and thus the anisotropic solution becomes for 3 dimensions"

~(x y, z, t) -'

Q

~~0 sin (z_~k) _ sine(~""ID'

2rcv/KxKy(b - a) =

Otn(1 -+- ~2+e2)

(55)

OlnZ )

{ Oln2Kzt u"v/lz2x2+v2y2] x c o s \ - - D - - w SsD2, D ' D

with Oen being the roots of ot tan ot - e - 2-7zC" For axial-symmetric flow we have Kx - Ky -- Kr and v - / , ; and V///~2x2 --~ v2 y 2 -- l Z v / X 2 -at- y2 -- lzr.

so v/K~ Ky -- Kr

Example 25. q

~,4&$, R

s~

R Kr

Fig. 28. Vertical infiltration from a circular area.

Vertical infiltration from a circular area (see Fig. 28) on top of a confined aquifer of "infinite" thickness. The infiltration rate q [L T -1 ] is uniformly distributed over the circular area with radius R, so that the total recharge Q = 7r R2q. The horizontal conductivity is the same in all directions (Kx = Ky = Kr) and differs from the vertical conductivity. So the flow is axial-symmetric. The solution for the isotropic case (see Part A, 523.02) for the steady state is:

1 J1 (Rot) Jo(rol)e -z~ du, ~o(r, z) -- ~q R fo ~ --£

(56)

~(r, z) - 27rqRr fo ~ -1 J1 (Rot)J1 (rot)e -z~ dot, Ol

(57)

and

Solutions, derived from known solutions

(2.3.4)

813

and for the non-steady state:

qg(r,z,t)- qRKfo ee -oeljl(Roe)J°(roe)Pc°nj( oez2'oeS'] doe'/3 ]

(58)

where J0 and J1 are Bessel functions of the first kind and order zero and one, respectively, and Pconj is the conjugate polder function (Section 3.1.2). The solutions (56), (57) and (58) can easily be derived by means of a Hankel transformation and a Laplace transformation (Sections 2.2.4-4 and 2.2.2-3). The transformation of these solutions to solutions for axial-symmetric anisotropy can be performed by replacing K by Kr (rule 4 of Theorem 5 with Kx -- Kv = Kr), q by ~ (rule 8 for a velocity in vertical direction), R and r by # R and #r, respectively (rule 2, because v = #) and t by /,2t (rule 3), the other parameters remaining constant (qg, z, ~P, Ss, according to the rules 1, 2, 7 and 5, respectively). Then 7,/7 = K ~ s b e c o m e s C K r u Ss 2 t " = CKzt TS~' and the solutions for the analogous problem in axial-symmetric anisotropic ground for the steady state become: 99(r, Z) --

qR f ~

1 - J1 (#Roe) Jo(#roe)e -z~ doe,

(59)

#Kr J0 oe

(60)

~/(r, z) - 2rcqRr fo ~ -1J l ( l z R o e ) J l ( l z r o e ) e -z~ doe,

oe

and for the non-steady state:

~p( r, z, t) -

# KR

oo oel J l ( # R oe) Jo ( # r oe) Pc onj ~

oeV--~s

doe.

(61)

Up till now we have considered anisotropic soils of which the main directions of the anisotropy were horizontal and vertical while the boundary conditions were also given in horizontal and vertical planes. If, however, the boundary planes are not perpendicular to the main directions of the anisotropy, it is no longer possible to derive simple rules for the transformation of solutions of problems on isotropic soils to solutions of problems on anisotropic soils.

2.3.5. Approximate solutions for phreatic flow and for density flow with interface 1. Phreatic f l o w Phreatic flow means groundwater flow in a phreatic aquifer, that is, an aquifer that contains a free water level, the so-called water table or phreatic surface, where the water pressure equals the atmospheric pressure (see Section 1.2.2). Unlike a confined aquifer, a phreatic aquifer has a thickness that depends on the height of the (fluctuating) water table the shape of which is unknown beforehand.

814

(2.3.5-1)

Analytical solution methods

As exact solutions for phreatic flow cannot always be found easily because of the complicated boundary conditions along the phreatic surface (see Section 1.5.2-1), it is worthwhile to look for approximate solutions, at the same time maintaining the character of phreatic flow, that is, taking into account the variable thickness of the aquifer. A considerable simplification of the calculations on phreatic flow can be obtained by assuming only horizontal flow, which, in general, will be allowable if the boundary values along vertical planes do not or hardly not contain vertical flow components. In Fig. 29 some examples are given of flow through phreatic aquifers. In Figs 29a and 29b the vertical boundaries relate to fully penetrating surface water bodies, with or without entrance resistance and to a fully penetrating well-screen, all boundaries giving only horizontal flow. Here the assumption of only horizontal flow in the whole aquifer will be a good approximation of the exact phreatic flow. In Fig. 29c a horizontal drain at the bottom of a phreatic aquifer causes vertical flow along the vertical plane of symmetry through the drain; here the assumption of only horizontal flow is not allowable in the neighbourhood of the drain. This assumption was first suggested by Dupuit for a particular case and later on generalized by Forchheimer. Only horizontal flow means that in a vertical plane all points have the same piezometric head q9 which in phreatic flow coincides with the water table and in the sequel will be denoted by h. Darcy's law, which gives the specific discharges in x=~q _

_

I

to

a

Fig. 29. Examples of phreatic flow.

b

Solutions, derived from known solutions (2.3.5- l)

815

and y-directions, now becomes" Vx

-

Oh - K o x~ ,

Vy

--

-

K

~

Oh , Oy

(a)

while the total discharge over a vertical becomes: Oh qx -- - K h o x

Oh -Khoy

qy

(b)

The continuity equation in this case becomes" Oqx

Oqv Oh + -7-- + Sph-7-7 --0, 0---7oy

(62)

in which the term Sph ,-57 oh describes the amount of water that will be taken into storage if the water table rises or will be released from storage if the water table falls. Here Sph is the phreatic storage coefficient or specific yield (dimension [0]), in contrast with the elastic storage coefficient S which originates from the compressibility of the ground and the groundwater (see Section 1.3.5-!). As the latter is very much smaller than the specific yield: S << Sph, the elastic storage coefficient has been neglected and omitted in equation (62). As O q x --- K h O 2 heax o-x- - = - K (Oh)2~xx and

O2h ( o')h'] 2 Oqv = - K h ~ K . ! o y\ Oy2 Oy

the differential equation (62) appears to be non-linear. However, if we consider the steady state (ah _ 0) and write 1 oq(~Kh 2)

qx -- -

and

Ox

0(1Kh 2) qy -

-

Oy

,

equation (62) becomes"

02(1Kh 2) OX 2

O 2 ( 2 K h 2)

+

Oy 2

= 0

(63)

1 and this is a linear differential equation in ~Kh 2° We set

1

CDF-- -2 K h 2

(64)

and call it the "Forchheimer" potential, which has the dimension of a discharge [L3T-1]. Equation (63) now becomes"

02~F O2q~F OX 2

+

Oy 2

= o.

(65)

816

(2.3.5-1 )

Analytical solution methods

The Forchheimer potential thus is a solution of the two-dimensional Laplace differential equation and therefore the real part of a complex function S2F, with S2F -- ~OF+ i~F

(66)

(see Section 2.2.5-1). The stream function OF is the conjugate function of q~F and has also the dimension of a discharge [L3T-1]. The Cauchy-Riemann conditions are also valid:

a4~F

ag~F

a~F

Ox

Oy '

Oy

a~PF with Ox

OdpF

Oh = Kh~--hvx--qx Ox Ox OC/)F Oh = Kh~--hvy--qy. Oy Oy

(67)

and

(c)

As a consequence of the similitude of the complex Forchheimer function I2F with the complex function I2 for confined aquifers, the solutions according to the Forchheimer approximation for steady two-dimensional flow in phreatic aquifers are identical to the solutions for analogous problems in confined aquifers, apart from some adaptations in relation to the boundary values. Phreatic flow with accretion (see Section 1.5.2-1), for instance, constant precipitation p [LT-1], may also be approximated in this way, the precipitation not acting as a boundary condition with a vertical flow component, but as a quantity, hypothetically generated in the aquifer and uniformly distributed over the height h of the water table, in order to maintain the assumption of horizontal flow only. The precipitation now appears as a separate term in the differential equation, as can be shown as follows.

I ~_

P

I

[

'~

hreatic surface

I I

qx

qx +

I I - -f-fl-~

/

,

/

~dy .....

dx

Fig. 30. Phreatic surface with precipitation.

Considering a vertical column with small cross-section dx dy and average height h in a phreatic aquifer with precipitation p (see Fig. 30) and applying the law of continuity, we see that per unit of time at the left side an amount of water vxh dy -- qx dy enters the column and at the right side an amount of water equal to Oqxx dx) dy leaves the col(q x + "g£' umn, the difference ~~)x dx dy being the total amount of water that leaves the column in

Solutions, derived from known solutions (2.3.5- l)

817

x-direction per unit of time. Similarly, the total amount of water that leaves the iJqr column per unit of time in y-direction is T dy dx. Together they must be equal to the replenishment of the column per unit of time by the precipitation, namely p dx dy. So we have: Oqx Ox

Oqy =p Oy

or with equations (c):

O2~F -~-02q~F - ~ - p - 0. Ox 2

(68)

Oy 2

This differential equation is very similar to the corresponding equation for steady two-dimensional flow in confined aquifers with precipitation:

0299

02(./9

p

---~ Ox + ~Oy2 + K D = 0.

(d)

The solutions for qSF are almost identical to those for q9 and can easily be derived from the latter. For two-dimensional radial-symmetric flow, the continuity equation becomes (see Fig. 31):

P

dQr - 27cr drp. r

Qr+dQr

(e)

With dh Qr - 27rrhvr - - 2 7 c r h K ~ dr

dr

Fig.31. Radial phreatic flow with precipitation.

and with q~F -- ~1 Kh2 , we have Qr - -27cr ~ 1 d ( d~bF] r + -0 r dr --~-r/ P

or

and thus, according to equation (e):

d2~bF l d~bF ~ + +p-0 dr 2 r -~r

'

(69)

the differential equation for two-dimensional radial-symmetric steady flow in a phreatic aquifer, according to the Forchheimer approach, with or without precipitation (p = 0 in the latter case). Also this differential equation has its counterpart for groundwater flow in confined aquifers, namely:

d2~o

1 d~o

p

-dr - T + -r -d-~r + K D = 0.

(f)

All solutions of the linear differential equations (68) and (69) and also solutions of particular cases of these equations like p = 0, or qSF independent on y (onedimensional flow), are almost identical to solutions of corresponding problems in confined aquifers and may be derived, according to the following theorem:

818

Analytical solution methods

(2.3.5-1)

Theorem

6. The approximate solution dpFfor a phreatic groundwater flow problem,

for which one of the differential equations (68) and (69) holds, particular cases included, may be found fromthe solution q9for the analogous problem in a confined aquifer, as follows: 1° Replace in the expression for ~p: q9 - A f ( x , y ) + B or cp -- A f ( x ) + B or q9 -- A f (r) + B the known constants A and B by the unknown constants cl and c2 (also if B - 0). 2 ° Determine the constants c l and c2 by means of the differential equation and

boundary conditions belonging to the phreatic flow problem. Some examples may serve for the application of this theorem. Example

26.

A circular basin with radius R and constant level H is lying in a phreatic aquifer in which uniform flow takes place with a discharge q [L 2T -1] (Fig. 32). In a confined aquifer the solution for the complex

Y ----->, q

h=H

R

potential S2 is:

q (z _ R 2

s2--

phreatic surface

-7)

(Part A, 336.01)

and for the drawdown ~o:

H

qx ~0---~(1-x2+y2 Fig. 32. Circular basin and uniform flow in a phreatic aquifer. According to Theorem 6, the solution ~PF becomes:

R2 ) 1Kh2 ~F -- ClX 1 -- X2 + y2 q-C2 -- ~ • Now h -- H for x 2 + y2 _ R 2, so c2 -- ~1 K H 2 and 0~bF _

qx --

Cl -Jr- Cl R 2

OX

y2 x 2 (X2 _+_y2)2' _

while limx~oo qx - q, so cl - - q and thus

epF---

1KH2

R2)

--qx 1-- X2_{_ y2 (

"

R2 )"

Solutions, derived from known solutions (2.3.5- l)

819

The stream function becomes:

!/fF

-

R2 - q y ( 1 + X2 + y 2 )

and the complex potential function: 1KH2

,.~'F-- ~ z

-q

(

z-~

R2) 27

,

and can easily be verified. Remark. Uniform flow in a phreatic aquifer means that the discharge q is constant everywhere; as q - - K h ~ ,dh it follows that the velocity Vx - - K ~ dh is not constant.

Example 27. Constant rainfall p [LT-1] on a rectangular island with a phreatic aquifer and surrounded by surface water with a constant level (Fig. 33). For the analogous problem with a confined aquifer the solution for the head ~o can be found by means of a finite Fourier sine transformation with respect to y (Part A, 363.05):

======================================================================

t -

~o=H

aye

Fig. 33. Rainfall on a rectangular phreatic aquifer.

4b2p ~o(x, y) -- 7r3E D

Z

1 --

m=l,3 ....

(mna cosh~--~)

~

sin

(m.,) . b

According to Theorem 6, the solution for q~F becomes:

[

q~F(x, y ) - - c, m=,,3 ~ .... 1 -

cosh(m~a )_Ty_

cm yt +c2.

m33 sin \ b

(f)

820

(2.3.5-1)

Analytical solution methods

Now h -- H and thus qSF -- 21 K H 2 for x -- 0 and x -- a and for y -- 0 and

1 H 2. ~p(x, y) is a solution of the differential y -- b, from which we find c2 - 2K equation (d) and q~F(X, y) must satisfy the differential equation (68), from which it follows directly that cl -

1

qSF(X, y) -- -~ K

H2

4b2p

rr3 " So the phreatic solution becomes:

1

+ -~ py(b - y)

4b2 p

oo m = l , 3 ....

1 cosh{ ~--~( a - 2x)} (tarry) m3 cosh(mzra~ sin 2b ] b

as

~ Z

1 ~

sin

(mzry) 7r 3 b - -~y(b-

y),

m = l , 3 ....

which may be verified by applying the finite Fourier sine transformation (Section 2.2.3-2a) with respect to y from 0 to b to the right side of this equation.

E x a m p l e 28. Constant rainfall on a circular island with a phreatic aquifer, surrounded by surface water with a constant level and with entrance resistance along the boundaries (Fig. 34). The solution for the analogous problem with a confined aquifer is (Part A, 236.08):

,,+++++++,+,~+4g~

R Fig. 34.

p (R 2 _ r 2 ) q p R w ~ 0 ( r ) - 4K---~ - 2---D--

Rainfall on a circular phreat-

(g)

ic aquifer with entrance resistance. According to Theorem 6, this solution can be written as" qg(r) - Ar 2 + B, with p A --

4KD

pR 2 and

B --

4KD

pRw q-

2D '

the solution for ~bF(r) thus becoming; ~bF(r) -- Cl r2 -+-C2. If we set at r -- R the (unknown) head h ( R ) - hR, we have

1

qSF(R)- = K h 2" -- Cl R 2 --]'-C2, Z from which

C2-

~1 K h ~

- c l R 2 and thus ~bF(r) -- ~I K h 2

must satisfy the differential equation (69)" d~F _ 2c 1r and

--

c 1(R

dz~bF - dr 2

--

2 -- r 2 ).

2C 1

,

q~F

and with

Solutions, derived from known solutions

821

(2.3.5-2)

equation (69): 2cl + 2c! + p -- 0, it follows that

C 1

--

--~ and thus

p q~F(r) -- ~1 K h 2 + ~-(R 2 - r2).

The boundary value at r - R gives v(R) dh hR - H K d--~-(R) w

(h)

--

h R-H W

or

and

dqSF dh h R (h R - H ) d-T -(R) - KhR--dTr (R) - - - w From equation (h) we have d a - ~ ( R ) - - P R , so that he(hR - H)

w

=

pR

or

2

h 2 - HhR

pRw

2

= 0

'

from which h R -- ~! H + ~1VH2 + 2 p R w " this expression substituted in equation (h) gives the end solution. With w = 0 we find hR = H and qSF(R) = H. 2. Interface flow Interface flow will be understood here as flow of fresh water along a sharp interface that completely separates the fresh-water body from underlying immobile salt water. Like phreatic flow also for interface flow it holds that exact analytical solutions cannot easily be found because of the complicated boundary conditions along the interface (see Section 1.5.2-2). Therefore it is obvious to apply some simplifications of the conditions for interface flow in order to obtain (approximate) solutions in a more easy way. The main assumptions are:

1° 2° 3° 4°

....

Sharp interface between fresh and salt water. No flow of the salt water. Only horizontal flow of the fresh water. Neglection of the dimensions of the region where fresh water flows into the sea.

_71_H__~ ~-----~~~~

P ~ x ~ c e~ ~,,..~xgtex ~.-,-

sea kzvel

•

..

Fig. 35. Phreatic aquifer with interface.

Calculations based on these assumptions were first made by Herzberg and later on independently by BadonGhyben. We may distinguish between calculations for phreatic conditions and for confined conditions. In a phreatic aquifer near the sea with a sharp interface, the pressures at a point P of the interface must be the same for the fresh water and for the salt water (Fig. 35):

822

Analytical solution methods

(2.3.5-2)

ysH = yf(H + h) or h = otH with ot -- Y s - Y f = interface parameter Yf

(70)

with ys and yf being the specific weights of salt and fresh water, respectively (Section 1.3.1.- 1). Darcy's law gives the specific discharges in x- and y-directions for an isotropic medium: Oh Vx -- - K o x

and

Oh Vy -- - K ~ . 3y

The discharges over a vertical become, as a consequence of the assumption of only horizontal flow: qx -- - K ( H

Oh -+-h)-~x,

qy - - K ( H

Oh -Jr-h) Oy

and with equation (70): OH qx = --Kol(1 + ~ ) H Ox

OH -Kol(1 + ~ ) H Oy

qy

We set ~H -- --1 Koe(1 + oe)H2 2

(71)

and call it the "Herzberg potential", which has the dimension of a discharge [L3T-1]. Now we have

qx --

0~H Ox

and

qy--

0~H [L 2T -1 ]. Oy

The continuity equation with precipitation becomes, for two-dimensional flow: Oqx ~ Oqy Oh OH 0---~" - "~y -'1- Sph-~ "-[-"tle--'~- -- P,

and thus the differential equation 02~H 02~H q---[--p OX2 Oy2

_

_

OH

(OtSph+ ne) at

(72)

Solutions, derived f r o m known solutions

(2.3.5-2)

823

with Sph -- specific yield and ne - effective porosity. The term Sph ~;)h describes the amount of water that will be taken into storage if the phreatic level rises, and the #H term ne ,-57- describes the amount of fresh water that replaces the salt water if the interface goes down. For radial-symmetric flow, the general differential equation for the Herzberg potential becomes: 02q~n 1 0q~n Or 2 ~ r Or ~- p

fresh

H

--

( O f S p h -nt-

0H ne) 0--7"

(73)

In a c o n f i n e d a q u i f e r near the sea with a sharp interface and sea level hs, we find for the pressure ps of the salt water at a point P on the interface ps = ys(H + hs) (see Fig. 36), with H being the depth of the interface below the upper edge of the aquifer.

/ ~i:lll.

P

Fig. 36. Confined aquifer with an interface. The salt water level in point A is hs and the corresponding fresh water level there is h f - ys ×f h s. Now if we take the fresh water level hf as reference level for the piezometric head h, we find for the fresh water pressure at P: pf -- yf(H + hf + h). Now again, Ps = Pf, so ys(H + hs) = Ff(H + hf + h), from which, with Fshs = yfhf, we have FsH = Ff(H + h) and thus h=o~H

withot=

ys-yf

(see equation (70))

Vf the same result as for the phreatic aquifer. The discharge over a vertical, however, is different from that for phreatic conditions because of the reduced thickness of that part of the aquifer that contributes to the flow of the fresh water (H instead of H+h): Oh qx -- - K H ~ Ox '

Oh qv -- - K H ~ " Oy

824

(2.3.5-2)

Analytical solution methods

and with h --o~H" Oh qx -- - K ° l H ox

OqbH Ox '

Oh qy -- - K o l H oy

aeH Oy

if we set 1

(74)

C])H-- - Kol H 2. 2

Now the differential equations (72) and (73) are with Sph -- 0 also valid for confined aquifers, provided that equation (74) is used instead of equation (71). These equations are non-linear, but the steady state versions (a/-/ -57 _ 0) are linear differential equations: 02t~H 02t~H Ox----5- 4- OYz + p -- O,

(75)

d2t~n [- -1 d~H 4- p -- 0, ----5dr - r - ~ - r

(76)

with CH -- 1Kct(1 4- ot)H 2 for a phreatic aquifer and Cn - 1KotH2 for a confined aquifer. These equations are very similar to the corresponding equations for confined aquifers without interface (see equations (d) and (f) of Section 2.3.5-1) and therefore all solutions of the equations (75) and (76) and also solutions of particular cases of these equations, like p - 0, or ~n independent of y (one-dimensional flow) are almost identical to solutions of corresponding problems in confined aquifers without interface and may be derived, according to the following theorem: Theorem 7. The approximate solution CH for a groundwater flow problem involving a steady sharp interface between two fluids of different density, for which one of the differential equations (75) or (76) holds, particular cases included, may be found from the solution cp for the analogous problem in a confined aquifer without interface, as follows:

1° Replace in the solution for 99" ~o -- A f (x, y) 4- B

or

cp -- A f (x) 4- B

or

~p -- A f (r) 4- B

the known constants A and B by the unknown constants cl and ¢2 (also if 0

B --0). Determine the constants Cl and c2 by means of the differential equation and boundary conditions belonging to the interface problem.

Solutions, derived from known solutions

Example

(2.3.5-2)

825

29.

A long earthen dam in the sea with precipitation p and discharge Q from a well at the location (x0, y0). The aquifer is assumed to be phreatic and of sufficient thickness to guarantee the development of a complete interface (Fig. 37). The differential equation for the Herzberg potential ~bH is equation (75) and the boundary conditions:

Y0 I I i

x0 B

B

Q

e vel

p ,[,,l,,l,,l,,l,,I,,l,,l,,l,,l,,l,~$~,l,$,l,,[;,l,

sea_ _

!

- _

~H(B, y)

2

-- ~H(--B, y) -- O,

Fig. 37. Earthen dam in the sea with precipitation and discharge from a well.

lim (r o~H r--+0

Q

"--~-r ) --

2rr

for r -- v/(X -- x0) 2 ~- (y -- y0) 2.

The solution for the analogous problem in a confined aquifer without interface for the head 99 is" P

(B 2 - x 2) -

go(x, y) -- 2K---D

Q

I cosh{

} + cos

2B

}

]

4zr K D In cosh{rr{~iv0,} - cos {rr(2-~x0) } '

where q9 is a solution of the differential equation" 02('/9-~ 0299 -t

Ox 2

Oy 2

P

KD

=0

with boundary value for the well" 0qg) Q lim r --r-+0 -~r 2rrKD

for r -- v/(x - xo) 2 + (y - yo) 2.

Comparison with the differential equation (75) and the boundary value for the well in this problem immediately gives the solution for qSH by replacing ~ by p and

826

(2.3.5-2)

A n a l y t i c a l solution m e t h o d s

Q by Q. With qSn -- $K0/(1 1 KD + 0/) H 2, we have the solution for H"

H 2 __

P (B 2 _ 0/(1 + 0/)K

x 2)

c°sh{ zr(v-y°) }'2B o ,n[ cosh { rr(Y-YO) 2:fro/(1 + 0/)K

"-I'-COS { n'(x+x0)2B} ] 2B } -- COS { n'(2;0) }

and h 2 - 0/2 H 2. The smallest value of H will be at the well-screen, for instance, with good approximation at the point (xo, yo + rw) with rw - well radius:

P

(B2

x 2) _

0/(1 + 0/) K

cosh,, ~ j + cos ( ~ )

Q

2~0/(1 + 0/) K In

cosh(nrw ] B _TTj_ 1

this expression determining the relation between the precipitation p and the discharge Q if the admissible upconing of the salt water is given. In practice, H is much larger than h; for instance, water of the North Sea has an average Cl-content of 18000 mg/1 and according to Fig. 2 of Section 1.3.1-1 a density of 1025 kg/m 3. Then, as y = p g we find 0/ --

)/s -- Yf yf

=

Ps -- Pf pf

=

1025 -- 1000 1000

= 0.025

and thus H = 40h. In contrast with phreatic water without interface as has been discussed in the first part of this section, the thickness of the fresh water body, responsible for the flow of the groundwater varies considerably and even becomes zero at the outflow into the sea, while also the outflow opening is reduced to zero. In this context, especially the assumptions 3 ° and 4 ° namely mere horizontal flow and neglect of the outflow opening become questionable, the more the flow approaches the outflow region. For this reason, it is worthwhile to compare a Herzberg approximation with an exact solution and investigate for which regions this approximation is still admissible. This will be done in the following example. Example 30. A long earthen dam in the sea of width 2b considered as a confined very thick aquifer, with constant infiltration p (Fig. 38). First we determine the e x a c t s o l u t i o n of the problem, which must satisfy the two-dimensional Laplace equation 8299 0x 2

t

0299 0z 2

=0.

We choose the x-axis along the upper edge of the aquifer and the positive z-axis downwards, the origin lying in the centre of the dam. If we set ~l - x + i z, then 4~ - K~0 is the real part of a complex function S2 - q~ + i gt which is a function

Solutions, derived from known solutions

(2.3.5-2)

827

¢ 5 ¢ ¢

B . . . . . . .

p$$,1,$,1,$ W,4,,1, 4, 4,,1,1,~4 4,,l,,b,l,,l,,~ .................... ~=o fresh D

(¢//

/

Vz

Fig. 38. Earthen dam in the sea with precipitation and exact interface" hodograph of the flow field. of rl = x 4- i z, where ~, the stream function, is the imaginary part of a'2 (see Section 2.2.5-1). The general solution X2 = £2(r/) cannot always be found as an explicit function of r/and therefore we write F(£2, r / ) = 0 .

(a)

If we consider the real and imaginary parts of equation (a) we may write F(£2, 77) = f (ck, ~r, x, z) 4- i g ( ~ , ~, x, z) = 0

(b)

and we have both f@5, O , x , z ) = O

and

g(qS,~,x,z)=O.

(c)

Now we try to find functions f and g which satisfy the boundary conditions. The boundary conditions are (cf. Section 1.5.2-1, equation (10)): Along the x-axis (z = 0) gr = p x for 0 ~< x ~< b (along D C ) and q5 = 0 for x >~ b (along C B , point B still being unknown). Along the z-axis (x = 0) fr = 0 (along D A ) . Along the interface, which shape is still unknown, 05 = ~ K z and gr = 0. The function px(qb - olKz) 4- Oh(qb, d/, x, z) = 0

(d)

with h being an arbitrary function of qS, gr, x and z (h --/: 0), with ~ - 0 already satisfies the boundary values along D A (x = 0) and along the interface (q~ = olKz). For z = 0 we have ~p = p x and q5 = 0, so that if we substitute z = 0 in equation (d) we should get: ( p x - ~)q5 = 0, for instance, h(d,b, ~, x, z) -- zk(q5, ~, x, z) - q5.

828

(2.3.5-2)

Analytical solution methods

Equation (d) then becomes: p x ( ~ - o~z,:z) + ~ {z~(~, ~ , x, z) - ~} - o

(e)

or

(f)

~c/) + ozpKxz - pxqb - grzk(dp, ~p, x, z) = O.

All boundary values of the problem are satisfied by this equation and if it corresponds with f = 0 or g -- 0 (equations (c)), the real and imaginary parts of a complex function S2(r/), then the latter is the solution of the problem. 1 2 Now 7r~b is the imaginary part of ~S2 , as £2 - q~+igr and if2 2 - - ~bz-~rz-k-2ioS~. Also ~ p K x z is the imaginary part of lc~Kprl 2 and if we set k - - p we have -pxc]) + pzgr as the imaginary part of - i p £ 2 ~ , as - i p I 2 r l = -ip{(dp + i ~ ) ( x + iz)} = -ip{(dpx - ~ z ) + i(grx + qSz)}. The complex solution thus becomes: 1

2

1

S22 + =otKprl 2 2

ip.f-2rl + c

(g)

0

with c as a real constant. The real part of equation (g) becomes: 1

~(q~2

_7~2

1

) + " ~ K p ( x 2 -- Z2) + p(~px + qbZ) + c -- O.

(h)

The head go(x, O) along C D can be determined by substituting z = 0 and 7r = p x in (h): 1 2 - -~p2x2 1 -~4) + ~1 X p x 2 + 1 2

1

2

~4, + ~ p ( p + ~ / ~ ) ~

p 2X 2 + c - o,

or

+c-O.

For x = b go(x, O) and also q~(x, O) = O, from which 1

(i)

c -- - - ~ p ( p -t- o t g ) b 2.

So the end solution F(S-2, r/) -- 0 becomes: 1 $2 2 -~- ~otKptl2 2

ipX2rl

1 -~p(p + otK) b2 - O.

(77)

The real part f(4~, ~, x, z) = 0 becomes 1

1

1

~(4~2 _ ~p2) + _~c~Kp(x 2 _ z 2) + p(grx + dpz) - ~ p ( p + ~ K ) b 2 -- O, (78)

Solutions, derived from known solutions (2.3.5-2)

829

and the imaginary part g(¢, ~p, x, z) = O: ¢ ~ + 0/pKxz - pCx + p O z = O.

(79)

If we substitute ~k -- 0 and ¢ -- 0/Kz in equation (78), we find the equation for the interface: 1 °l 2 K 2 z 2 nt-

-2

1

2

1

0/Kpx2

1

- -0/2Kpz2 + 0/KPz2 - -2p ( p + 0/K)b2 -- 0

or

0/K(0/K + p ) z 2 + 0/Kpx 2 - p(0/K + p)b 2 - 0

and thus P b2

7..,2

p

-- 0/K

x 2 -- H 2

p + oeK

"

(80)

This line intersects the x-axis at B where z = 0, from which

XB - -

For x - 0

+ 0/ K

b,/ V

~ I(

"

(J )

we have the maximum value of H: !

Hmax - - b /

P

V0/K"

(k)

The interface line is an ellipse for which the long axis is represented by (j) and the short axis by (k). The variation of the piezometric head ~Pi along the interface can be derived directly from equation (80), as (/9i - - c ~ H . The piezometric head ~p(x, 0) along C D becomes, with z - 0 and ~p - p x substituted in equation (h): ~2

+

p(p

+ 0/K)x 2 _

p ( p + 0/K)b2 _ 0

and thus, with ¢ = K~p(x, 0): (/92 (X, 0) =

p ( p + 0/K) K2 ( b2 -- X2).

(81)

The maximum head ~o(0, 0) is b (p(O, O) - -7; v / p ( p + 0/K). lk

(1)

Equation (81) is an ellipse with the long axis b and short axis represented by equation (1).

830

(2.3.5-2)

Analytical solution methods

The velocities everywhere in the flow field can be obtained by differentiating (78) and (79) with respect to x and z; for instance, differentiation of (79) with respect to x gives: Og

O~

O~

~4)

o7,

+ 4 -g2x 4- otp K z - pc/) - p X -~x 4- p z .-~x - 0

Ox -

and w i t h v x = - o x -~ --

_

~Oz and V z -

~4)

_ Oz -

( p x - ~/)Vx + ( p z + ~)Vz + p ( o t K z -

~x we find: ~ ) = O,

(m)

( p z + ~)Vx + (7z -- pX)Vz -- p ( o t K x + ap) = O,

(n)

Bg

and in the same way 57 = 0 yields

giving two equations in Vx and Vz from which Vx and Vz are determined as functions of ~b, 7r, x and z and thus as functions of x and z. Differentiation with respect to x and z of the real part, f (q~, gr, x, z) = O, of the solution gives also the results represented by (m) and (n). Especially the velocities along the boundaries of the flow field can easily be obtained from (m) and (n). For instance, along the interface we have gr - 0 and = o t K z , which substituted in these equations gives: pXVx + ( p + o t K ) z v z = 0

and

( p + o t K ) z v x - pXVz - p o t K x = 0 .

Elimination of ( p 4 - o t K ) z from these two equations gives 2

2

vx + v z + o t K v z - - O ,

(o)

which is the boundary condition along the interface, representing the circle in the hodograph representation (see Fig. 38). Along A D the velocities become, with ~ - 0 and x - 0 from equation (n) v~ = 0 and from equation (m): Vz =

p(4, - o~Kz) pz+¢

•

(p)

Point D (z - 0) gives Vz - p and point A (4) - o t K z ) gives Vz - 0 (stagnation point because also Vx = 0). Along D C we have z = 0 and ~ = p x . From equation (m) it follows that Vz = p and from equation (n) p(p + otK)x Vx =

.

(q)

(2.3.5-2)

Solutions, derived from known solutions

831

Point D (x -- 0) gives Vx -- 0 and point C (x - b, q~ - 0) gives Vx - cx~. Along C B ( z - 0, ¢ - 0) the velocities become vx - 0 and p(aKx VZ

+ ~)

---

(r)

--px

Point C (x - b, ~ - p b ) gives Vz - cx~ (cavitation point because also Vx - e~) and point B (gr - - 0 ) gives Vz -- - o t K . Now, applying the Herzberg approximation for a confined aquifer with precipitation, we must solve the differential equation d2q~H dx 2

+p--0,

with q~H - ~1 K o t H 2, dq~H (0) -- 0 dx

and

CH(b) -

0,

with the solution P 2 - x 2), ~H -- -~-(b

from which

H 2 = P._~_(b 2 X2). otK -

(82)

This equation represents an ellipse with long axis b and short axis b,/ff-~. If we compare this result with the exact solution of the interface,V~'"according to equation (80), we see that in both cases the interface is an ellipse, having a short axis of the same length (cf. equation (k)) and a long axis, the length of which differs by an amount b~P

+ c~K

o~K

-b

(cf. equation (j)).

p is, in general, very small compared with c~K. For instance, p - 0.002 m/day, ot - 0 . 0 2 5 and K - 10 m/day; then of K V/P + = 1.004 otK

and the difference in length of the long axis of the two ellipses, which corresponds with the dimension of the outflow opening in the exact solution, becomes very small, in this case 0.004b, which means that if b - 500 m the outflow opening is only 2 m. So in this particular case the Herzberg solution is an extremely good approximation of the exact shape of the interface.

832

(2.3.6-i)

2.3.6.

The

reciprocity

Analytical solution methods

principle

1. Description and mathematical proof 8. The drawdown of the piezometric head at an arbitrary point A in a

Theorem

heterogeneous anisotropic porous medium with homogeneous groundwater, caused by pumping with a discharge Q in another arbitrary point B at the time t after the beginning of the pumping is equal to the drawdown in B as the result of pumping in A with the same discharge Q and after the same time t. This theorem describes the so-called reciprocity principle and is valid for , J , , e%~ I I • "° ", " " I homogeneous groundwater in heterogeneous anisotropic soils of whatever complex composition (Fig. 39). (In the sequel we take it for granted eB ,' I that by "groundwater" is meant "homogeneous groundwater".) In order to prove this principle, we '- S~_ ~. , . , ,s ~, .... I make use of the divergence theorem ' " I • % 8 • • of Gauss, applied to a closed groundwater body with volume V and suri o s ~/ face A in which a vector field for the clay specific discharge or bulk velocity v J • • P has been generated. This means, in l oA • " , general, that at every point of the groundwater body the velocities Vx, Vy and Vz in the directions of the coordinate axis x, y and z, respecFig. 39. Cross section of anisotropic heterogeneous tively, are functions of x, y, z and soil. t. Then, according to the divergence theorem, we have at the time t, per unit of time: I

I

s

,,

S

%

e

I

'~

~

p

fffdivvdV-ffAo, dA, OVx

OVy

(83

OVz

in which div v - -~x + ~ ÷ -~z and v, - v . n is the component of the vector v in the direction of the outer normal of the surface A. If we apply equation (83) to steady flow, we find, according to the continuity principle: d i v v - 0 (Section 1.4.2-3, equation (12)), and thus

f fa Vn dA -- O,

(84)

Solutions, derivedfrom knownsolutions (2.3.6-1)

833

which means that the total amount of groundwater that flows across the surface A equals zero, provided that no sources or sinks exist within the body, enclosed by A. A second application of the divergence theorem leads to Green's formulas for groundwater flow, as follows: Let f and g be scalar functions of x, y and z and let the vector v in equation (83) be related to f and g by v-fgradg-

f Vg.

Then div v - div(fVg)

Of ag O2g Of Og ~-f Oeg + Of Og + f O2g 0-7 0--7+ f ~ ~ Oy Oy ---~ ay ~ ~z 022 = fV2g + V f . Vg and vn--v.n--(fVg).n--f(Vg-n)-According to equation (83) we find

ag f On.

Green'sfirst formula:

Og dA. fff
(85)

By interchanging f and g, we obtain a similar formula. On subtracting this formula from equation (85) we find Green's second formula:

f f f(fV2g-gV2f)dV-f

Og

Of

(86)

If both f and g satisfy Laplace's differential equations g 2 f - 0 and V2g - O, respectively, we have:

Og Of f £ (f-~n-g-~n) dA --0.

(87)

In particular, if the functions 99a(x, y, z) and 99t,(x, y, z) both represent the piezometric head of a groundwater flow and they satisfy the differential equations vZq)a -- 0 and vZqgb --0 respectively, equation (87) becomes

ff (

g)a On

991,~

dA -- 0

This equation serves as the basis of the proof of the reciprocity principle.

(88)

(2.3.6-1)

834

Analytical solution methods

First Theorem 8 will be proved for s t e a d y f l o w . Let the points A and B belong to a closed groundwater Qa V body with volume V, whose boundary is a piecewise smooth orientable surn zI B[] face A (Fig. 40). x/ Assume that pumping from a A with a discharge Qa gives a steady drawdown distribution ~0a and a bulk velocity v, and pumping from B with a discharge Q b causes a steady drawFig. 40. Abstraction of water from A and B. down distribution ~0h and a bulk velocity vb in that groundwater body; ~Pa, ~Ph and the components of Va and vb (Vxa, Vya, Vza, Vxh, vyh and Vzh) are functions of x, y and z. Qb

=~

If the soil is isotropic the velocity vectors are proportional to the negative gradients of the piezometric heads (Section 1.2.3, equation (31)): Va = - K g ~ o a

and

Vb = - K V ~ o h ,

and if besides the soil is homogeneous with K = constant, the functions ~p, and ~Ph will satisfy the differential equations of Laplace. In that case equation (88) is valid. It is important to know if equation (88) also holds for flow through a n i s o t r o p i c h e t e r o g e n e o u s ground. Then the bulk velocity is no longer proportional to the gradient of the piezometric head as its components are defined by

Vx = - Kx -~x ,

vy -

- K y -~y

and

vz - - K z -~z ,

in which K x , Ky and Kz are different functions of x, y and z. Then Green's second formula (equation (86)) is not valid and we need an extension thereof to show that also for anisotropic ground a generalization of formula (88) holds. For that purpose we compose the vector U - - ~OaVb -- qgbVa

and apply Gauss' divergence theorem thereto. Then formula (83) takes the form:

fff

Solutions, derived from known solutions (2.3.6- l)

835

in which Van and vhn are the components of va and vh in the direction of the outer normal of A. Now div(99aVb)

--

(/9 a

div vh + V~o..vh

and

div(99bVa) -- q)h d i v v . + Vq)b'Va, and on subtracting these equations and setting div v,, - 0 and div vh - 0 as the flow is steady, we find div(q)aVb) -- div(~OhVa) = Vq)a "Vb -- VqOb "Va

3~pa

Oq)a

099h

099b

Furthermore, Vx,, - - K x ~ax and vxh - - K x ;-57-x and thus 3q)a a--7- v x h

3q)h 09% 399b t- Kx 3q)h 399a -~-x v x" - - K x -~x a x -~x a x = 0

and also 3q)a 3y Vyb

099h ~Vva

-- O

and

O~Oa 099h ~3z vzh - -~z Vza -- O.

So the volume integral of equation (89) becomes zero and we have found an extension of Green's theorem for anisotropic soils with steady flow:

f

(90)

f a (99,, Vhn -- 99h Van) d A -- O.

A condition for equation (90) is, that no sources or sinks exist within the body enclosed by A. So we have to choose for A the surface indicated by Fig. 41, where A = A' 4- AA 4- AB in which A' is the outer boundary of the groundwater body and A A and A B are the surfaces of the filters used by pumping in A and B. In practice, there is always a surface A' (finite or infinite) where either the head or the flux or a combination thereof is equal to zero: a~

+ fly,,

-

0

along A' in which a and fl are constants, including zero, but not at the same time. Fig. 41. Outer and inner boundaries of a groundwater body.

836

Analyticalsolutionmethods

(2.3.6-1)

Then, along A'

fi qga V b n - - (tgb V a n m

fi

__ ~ Ua n 12bn _4_ ~ Vbn V a n ~ Ol Ol

0

and thus the surfaces A A and A B are left in the double integral of equation (90) which now becomes:

ffaA(~O~,vb~--~OhV~'~)dA+ffa8(qga Vbn

- - qgb l ) a n )

d A = O,

which, of course, is also valid for drawdowns instead of heads. Now we assume that pumping from A causes a drawdown ~0aa at A A and a drawdown ~oha at A B and also that pumping from B gives a drawdown qgbh at A B and a drawdown 99ab at A A, the first index of qOi.j referring to the location of the drawdown and the second to the location of the well. Then we have:

°a,,ffavbdA--°"hffA

A vhndA B

(91)

0. B

f

fA A Dan dA is the velocity, perpendicular to the surface A A and caused by abstraction of water from filter A, integrated over the surface AA and thus is equal to the discharge Qa, while in the same way

f fasvhndA-- Qh. f

fA A l)bn dA is the velocity, perpendicular to the surface A A, integrated over A A, caused by abstraction of water, not from filter A but from filter B and therefore equals zero, because the total flux entering and leaving the well-screen at A as a consequence of the flow field described by ~0h, equals zero. In the same way

f fA8 VandA -- O. So equation (91) is reduced to: - - qgab (,Oab ~

Qa + qgba

qOba Q b

-'- O ,

or if

Qa = Qb (92)

by which the reciprocity principle is proved for steady flow. Secondly, the theorem will be proved for unsteadyflow. ~0,,, ~oh, va, vh and the components of va and vh are now also functions of the time t and the divergence

Solutions, derived from known solutions

(2.3.6- l)

837

of the bulk velocity is not equal to zero as a result of the compressibility of the ground (Section 1.4.2-2, equations (6) and (7))"

Oq),, div v,, - - Ss 0-t

and

Oqgh div vh - - Ss 3---7'

(a)

where Ss is the specific storage (in general a function of x, y and z). The volume integral in the divergence theorem of Gauss applied to the time-dependent vector U --

q)a Vb - - qOb V a

now does not vanish but becomes equal to

The dependence of time can be eliminated by applying the Laplace integral transformation to the functions that depend on t (Section 2.2.2). Instead of the vector u - q)aVb --(pbVa, which depends on x, y, z and t, we now start from the vector V --

q)a Vb - - (#b V a ,

which depends on x, y, z and s. In this case e -stq).(t) dt

qSa --

and

% -

e - ' t Va(t) dt,

etc.

It must be noted that 9 -# fi because the Laplace transform of a product of two functions of time is not equal to the product of the transforms of the separate functions (Section 2.2.2-1, Theorem 6). Applying the divergence theorem to the vector ~, we find:

dA. Now div(q3a~h) = q3a div~h + Vq3,. vh, div(q3h%) = q3h d i v % + Vq3h" %, and on subtracting these equations: div ~ = ~b, div vh - q3h div % + Vq3a • vh - Vq3h • %.

OG , div ~', =

3x

+

Ovv,, OVz,, ( Ov~,, ,~ O-y + = 3z 3x /

= (div Va) - - S s \-=--t,ot/ according to equation (a).

!

(b)

838

Analytical solution methods

(2.3.6-1)

As (~-g) = S~a -qga(0) (Section 2.2.1-1 Theorem 2) and the drawdown g)a at at the time t - 0 is equal to zero, we find div~'a - -Sssq)a and in the same way div~,b -- - & s O b . Then, the first two terms on the right of equation (b) become q3a div g'h - q3b div re,, - - S s s ~ a ~ b + Sssqghqga - O,

Vq)a" Vb -- Vq)b "Va -

--

aqga OX

'~)xb

aqga aqga -. -~- --~Z Uzb -~y Uvb

nt-

-

aq3b _ .

aqSb _

. . . 3x Vxa

OqSh _

. . 3y vya

OZ

l)za .

°af-fi x and if K~ is independent of time:

Uxa -- - K x

aqSa _

(O~-[~a

Vxa -

\ ax

-- - Kx

and also

a¢h f)xb -- -- Kx

ax

and thus O~)a

a-Tf'x

Oq9b _

-T2x " * ' - - K x

a~a a~h t- K xa ah ~ ~a ao = O.

5x

ax

ax ax

A similar derivation may be performed for the coordinate directions y and z, from which it follows that also the second two terms of the right side of equation (b) become equal to zero, and thus div ~' = O, which means that also the triple integral of the divergence theorem equals zero and consequently also the double integral:

f

f A (~af)bn -- ~hf)an) d A -- O.

(93)

We assume that also Qa and Qb are functions of time and their Laplace transforms are ~)a and Qb. Then we may consider qSa and Va as the drawdown and the bulk velocity caused by an abstraction ~)a; the same is valid for the functions with index b. Now mathematical analysis of equation (93) in a similar way as has been done for steady flow, and if also the boundary condition olqo + flVn - 0 has been transformed to otq3 + firOn -- O, will result in: Q a q g a b - Qb~ba

for every value of the parameter s. If Qa is the same function of s as Ob, q~ab will be the same function of s as ~h,,, and inasmuch as the inverse Laplace transform of a function is essentially unique,

Solutions, derived from known solutions (2.3.6- l)

839

we can conclude that q)ah is the same function of t as qgba, if Qa is the same function of t as Qb. The reciprocity principle enables us to derive a drawdown at a certain point A, caused by abstracting water from B, if the drawdown at B as a result of pumping water from A, is known. As such it is the theoretical representative of the category of solutions, derived from known solutions, which are the subject and title of this Section 2.3. The practical applications are, in general, limited to checking existing or knewly found analytical solutions, especially those concerning layered soils and multilayered systems. Example 31. An aquifer, infinitely extended under open water, consists of two horizontal layers with different transmissivities K1 D1 and KzD2 (Fig. 42). From each layer an arbitrary amount of groundwater is abstracted by means of a partially penetrating well. The wells are located in a vertical chosen as the axis of revolution for the axial-symmetric flow. Each layer has its own differential equation with boundary values for the steady drawdowns (/91 (r, z) and 992(r, z). For 0 ~< z ~< D1 we have:

02991 1 0991 02991 -----5Or + -r ~ + 0 Z 2 "-- 0,

~01 -- q91(r, z)

~=0

al i

K1

|

Q1¢= ' '

bl

D1

A

q)g(r) a2

b2 B'

K2

;ol(r, z)

ro

-IA' I I I I I I I

ro

r ~ i

i

'' ==~a2 D2 i

go2(r, Z)

|

Fig. 42. Layered aquifer under open water and with partially penetrating wells.

840

Analytical solution methods

(2.3.6-1)

lim [r

0qgl (r, z ) ] -

Ol {

r--,o -~r

-

0

2:7rK1 (bl - a l )

for al < z < bl, for0~
andbl
(t91 (OO, Z) -- O.

~Pl(r, 0) -- 0, (/91(r, D1) -- ~Og(r) a still unknown function. In the lower layer (D1 ~< z ~< (D1 -q- D2)) we find: 02q92

1 0(/92

-Or - 5 - + -r ~

02992 q)2 -- q92(r, Z),

+ ~oz 2 = 0 ,

Q2 lim r

r--,0 -~r

(r, z)

--

2zr K2 (b2 - a2)

0

for (D1 + a2) < z < (D1 + b2), for D1 <~ z < (D1 + a2) and (D1 -+-b2) < z ~< (D1 + D2),

(/92(00, Z) -- O,

q92(r, D1) -- qgg(r),

O~p2 ~(r,

D1 + D2) -- 0.

Oz

For the solution of this problem, we apply the method of separation of variables (see Section 2.2.1). We assume that ~pl(r,z)= FI(r)GI(z) and ~02(r, z ) = F2(r)G2(z), in which F1 (r) and F2(r) are functions of r only, and G1 (z) and G2(z) are functions of z only. Then we have for i -- 1, 2:

d2Fi Gi dFi d2Gi Gi-=--z-.-t " ~-Fi " =0 dr z r dr dz 2 or

1 d2Fi

1 d2Gi

1 dFi

"F//( dr 2 + r - - d T ) -

Gi dz 2 "

The expression on the left side of this equation depends only on r and on the right side only on z. Hence, both sides must be equal to the same constant, for instance, o~2, from which two ordinary differential equations are obtained: d 2 Fi 1 dFi dr 2 t r dr

ot2 Fi -- 0

and

d2Gi

~ + ot2Gi - O, dz 2 •

with the general solutions:

Fi(r)

-- ¢ilKo(otir)

Gi(z)

- - ci3

-t- c i 2 I o ( o l i r )

sin(otiz) -+- ¢i4

and

COS(0tiZ).

Solutions, derivedfrom known solutions (2.3.6-1) The boundary condition 99i(oo, z) = ci2 = O. With Ai - - C i l C i 3 and Bi =

~.(oo)Gi(z) CilCi4,

841

-- 0 gives

F,(oo)

=

0

and thus

we find:

(a)

qgi(r, Z) -- {Ai sin(o~iz) -q- Bi COS(CeiZ)}Ko(ceir). For z = D1 the functions q)l and 992 become: q)l (r, D1) -- {A1 sin(o
and

992(r, D1) -- {A2 sin(o<2D1) ÷ B2 cos(o~2D1) }Ko(o<2r). Both expressions must be equal to the same function g)g(r) of r, from which it follows that o
G1 = A1 sin(o
The boundary conditions along z = 0, z = D1 and z = D1 -+- D2 become, for G1 and G2:

GI(O) --O,

GI(D1) -- Gz(D1) -- g,

dG2 ~(D1 dz

-nt- D2) - - 0 ,

from which G1 -- g

G2=g

sin(o
and

(b)

cos{or(D1 + D2 -- Z)}

(c)

cos(o/D2)

In these expressions, the constants g and c~ still have to be determined. The continuity condition along the boundary z = D1 becomes: 0(/91 0(/92 X l - ~ Z (D1) -- K 2 - ~ g (D1), dG1 (D1) -- K2

or

dG2 --(D1),

(d)

and thus, by differentiating the equations (b) and (c): K1 cot(o
(e)

842

Analytical solution methods

(2.3.6-1)

This expression is a goniometric equation for oe and so oe has infinitely many solutions C~k (k -- 1, 2, 3 , . . . ) . Hence, the equations (a) become" sin(otkZ) Ko(otkr)

cpl(r, z) -- ~

k--1

and

(f)

g(k) sin(otkD1)

cx)

C_OS{Otk(D1 + D2 - z)! Ko(otkr). cos(otkD2)

~p2(r, z) -- Z g(k) k=l

(g)

The function g(k) is a function of Oek and may be determined with the help of the boundary values at r - 0. We now write the functions G1 and G2 from equations (b) and (c) for a special value of oe, for instance, Olk, as G I (k, z) and Gz(k, z). Then by differentiating equation (f) with respect to r we have: oo

a~o---~l(r, z) -- - Z G1 (k, z)otkK1 (akr), Or k=l and thus oo

lim r r-+O

-

- ~

G1

as lim

,

{xK,(x)}- 1;

x-+O

k=l

also oo

lim ( r. = - ~ G2(k , z) r--+0 --~-r ] k=l The boundary conditions at r - 0 therefore become: c~ al(k,

Z) --

k=l

sin(otkZ)

~

g(k)sin(oekD1)

k=l

(h)

/

Q1

=

oo

oo

for0~
andal
cos{oek(D1 + D2 - Z)}

G2(k, z) - ~ k--1

for al < Z < bl,

2rr K1 (bl - a l ) 0

g(k)

k=l

cos(otkD2)

(i) for (D1 + a) < z < (D1 + b2), =

2rr K2 (b2 - a2) 0

for D1 <~ z < (D1 + a 2 ) and (D1 + b2) < z <~ (D1 + D2).

From the equations (h) and (i) we have to solve g(k), which can be done as follows. Besides the function Gl(k, z), we consider the function Gl(l, z) in which Olk of

Solutions, derived from known solutions (2.3.6-1)

843

G1 (k, z) has been replaced by another root of equation (e), namely oel. The functions a l (k, z) and a l (1, z) satisfy the differential equations: d2G1 (k, z) dz 2 + ot~G1 (k, z) - 0

d2G1 (1 z) + c~2Gl(l, z) - 0. dz 2 '

and

Multiply the first equation by G j(1, z) and the second by G l(k, z), subtract the second from the first and integrate with respect to z between z = 0 and z = D~: _

--

2 f0D1G1 (k, z)G1 (l, z) dz f D1

Gl(l,

d2Gl(k ' dz2

Z)

z)

dz -

fD,

(k, z) } -- fo D1 Gl(l, z) d { dGl-d~ --

Gl(l,z)

Gl(k

'

z)

d2Gl(l z) ' dz dz2

foD,Gl(k,z) d{ dG 1(/, Z) } dz

dGl(k, z) _ G l(k z) dGl(I,z)] D1

dz

dz

[ DI dG1 (k, z) dG1 (1, z) J0

dz

fO D1 dGl(/, z) dGl(k, z) dz. dz dz

dz +

dz

o

The last two terms cancel each other, so that

-

2 f0D1Gl(k,z)Gl(l,z)dz (J)

dGl(k, = g(l) dz

D 1 ) -- g ( k ) d G 1 (1 D )

----~Z- '

1 ,

as G1 (k, 0) -- G1 (l, 0) -- 0 and G1 (1, D1) = g(l) and G1 (k, D1) = g(k). In a similar way, for the second layer we find:

_

fo °1+ 2 G2(k,z)G2(l,z)dz 1

dG2(k,D)+ = - g (I) ---~-z 1

g

(k) dG2(l --dz

(k) D1) '

"

If we now multiply equation (j) by K1 and equation (k) by K2 and add both equations, we find, making use of the continuity equation (d), the important result:

K1fo l Gl(k,z)Gl(1, z) dz + K2 fD ol+D2 G2(k, z)G2(I, z) dz =- 0, 1

provided that k =/: 1.

(1)

844

Analytical solution methods

(2.3.6-1)

Now we multiply both sides of equation (h) with K1GI(I, z) and integrate with respect to z between z = 0 and z --- D1, yielding

K1

foD1G1 (l, z) ~

Q1

Gl(k, z) dz - 2yr(bl - al)

f~bl Gl(l,z)dz. 1

(m)

k=l

Next, we multiply both sides of equation (i) by K2G2(I, z) and integrate with respect to z over the interval D1 <~ z ~< (D1 + D2): +D2 G2(l, z) K2 fo D1

Gz(k, z) dz

f D1 +b2 / G2 (l, z) dz. (n) 27r(b2 - a2) j Dl+a2

Q2

-

k=l

1

Now, addition of the equations (m) and (n) gives an infinite series, for which all the terms with k --/: l vanish, according to equation (1) and only the terms with k - l are left. Hence equation (m) + equation (n) gives: K1

foD1G 2(k, z) dz +

K2

fOO1+D2 G 2 (k, z) dz 1

=

(o)

,

Q1 2yr(bl - a l )

G1 (1 z) dz + 2yr(b2 - a2) d Dl+a 2

1

G2(l, z) dz.

Now, substituting the expressions for G1 and G2 from equations (b) and (c) in equation (o) gives: g2

foD1 sinZ(°lz)

K1

dz + Kzg 2 fD°l+DZ c°s2{~(Ol -t- O2 -- Z)} dz 1 COS2 (0l D2)

sin 2 (ol D1)

Qlg

--

[bl

2yr(bl - al) Jal

Q2g

sin(olz) dz sin(olD1)

f D1Wb2cos{or(D1 +

+ 2yv(b2 - a2) ]dDl+a 2

D 2 - z)}

dz,

cos(otD2)

which may be evaluated to: g2 {

K2 D2 K1 D1 _ K1 cot(olD1) + + K2 tan(or D2) 2 sin2 (olD1) 2or 2 cos2(olD2) 2~

Q1

cos(oral) -- cos(otbl)

-- g 2yr(bl - al) +

}

Q2 2Jr(b2 - a2)

ol sin(czD1) sin(otD2 - ota2) - sin(olD2 -- orb2)/" ot cos(olD2)

The second and fourth terms on the left side cancel each other according to equation (e), and we may solve for g. Substituting the value for g in the equations (f) and (g), we find the final solutions.

Solutions, derived from known solutions

(2.3.6-1)

oo Wl + W2 sin(otkZ) Ko (Otkr ) q91(r, z) -- ~ ~ + U2 sin(~k D1)

845

and

k=l

992(r' g ) -

c~ Wl + W2 COS{Olk(D1 + D2 -- Z)} Ko(oekr), ~ ~11 + U2 cos(c~kD2) k=l

(94)

with

U1W1 =

W2

K1D1

sin 2 (C~kD1)

,

U2-

K2D2

cos 2 (O/kD2)'

Q1 cos(otkal) -- cos(otkbl) xr~(bl - al) sin(~kD1)

and

Q2

sin(c~kD2 -- otka2) -- sin(c~kD2 -- otkb2)

:rrOlk (b2 -- a2)

cos(otkD2)

with O~k being the roots of tan(otD1)tan(otD2) -- L K2 according to equation (e) (Part A, 562.01). We may now verify these solutions by applying the reciprocity principle for the two locations A and B (see Fig. 42). Therefore, we must modify the solutions ~01 and q92 for partially penetrating wells into solutions for point wells. A point well at point A(0, bl) may be considered as the limit of the partially penetrating well with length ll = bl - a l if 11 approaches zero. Then W1 becomes:

W1 "--

Q1 lim [cos{oek(bl- l l ) } - cos(otkbl)-[_ Q1 sin(c~kbl) 7rc~k sin(otkD1) lifO [ ll J Jr sin(oekD1)

So, if we abstract only an amount of water from point A(0, bl), we find with Q1 - Q and Q2 - 0 (so also W2 - 0) from equation (94) the drawdown at the point B(ro, D1 + a2)"

Q w-~°° 7r

k=l

1

sin(~kbl) cos{c~k(D2 -- a2)}

UI + U 2

sin(ak D1) cos(¢k D2)

Ko(akro).

(p)

Now, pumping in B(ro, Ol + a2) with an amount Q will give a drawdown ~P,,b at point A(0, bl), which is the same drawdown at point A'(ro, bl) as a result of pumping in B'(0, D1 + a2) with a discharge Q, because of symmetry. A point well at B'(0, D1 + a2) may be considered as the limit of the partially penetrating well with length 12 -- b 2 - a2 if 12 approaches zero. Then W2 becomes

846

(2.3.6-2)

W2

[sin{otk(D2-a2)}-sin{otl~(D2-a2-12)}]

lim

Q2

Analytical solution methods

YrOCkCOS(OtkD2) /2 --+0

12

Q2cos{~k(D2 --a2)} :It COS(Otk D2)

Pumping an amount of water only from point B'(0, D1 + a2) we find, with Q1 - 0 and Q2 - Q from equation (94), the drawdown in the point A'(ro, bl)" Q ~ qga'b' -- ~

1

COS{Otk(D2 -- a2)} sin(otkbl)

COS(OtkD2)sin(olkD1)

' ~ U1 --I-- U2 k=l

Ko(u~ro).

(q)

Comparing equation (p) with equation (q), we see that qgatb, - - q g b a , and as ~o,,,b, -~Oab, we have found 9ab -- ~Oba, the reciprocity of discharges and drawdowns for the points A and B in Fig. 42. 2. The reciprocity principle in multi-layer systems A multi-layer system has been described in Section 2.1.3 and consists of more than one aquifer, separated by one or more semi-permeable layers, while the flow in the aquifers is assumed to be merely horizontal and the flow through the semi-permeable layers merely vertical (see Fig. 4 of Section 2.1.3). The principle difference with a layered soil, as discussed in the previous section, where for each layer the drawdown has its own (three-dimensional) differential equation with boundary values, is that, in a multi-layer system, the (two-dimensional) differential equation for the drawdown of each separate aquifer also contains terms referring to drawdowns in adjacent aquifers. As the vertical flow in the aquifers is neglected, the pumping takes place by means of wells that fully penetrate the distinct aquifers; then, the reciprocity principle refers to groundwater abstractions from line wells instead of point wells and to drawdowns being the same along a vertical in an aquifer.

Qb :=~

Qa

Cl

i

K1D1

B'

c2 K2D2 c3 r I

A '11

K3D3

I I I

Fig. 43. Three-layer system.

B • c4

For instance, pumping an amount Q. from A in aquifer 1 gives a drawdown qgba in B in aquifer 3, which must be equal to the drawdown qgab i n A as a result of pumping an amount Qb from B, if Q. = Qb, as is shown schematically in the three-layer system of Fig. 43. Suppose that r is the horizontal distance between A and B, then (Pb., the drawdown in B, is a function of r, if the location of the well in A is taken as the origin and qg.b the same function of r if the origin is chosen in the vertical through B.

Solutions, derived f r o m k n o w n solutions

(2.3.6-2)

847

From the supposed homogeneity of the layers, it follows that pumping an amount Qh from A' in aquifer 3, lying in the same vertical as A, will give a drawdown ~0h,,,, in B' in aquifer 1 in the same vertical as B, which is equal to 99,h, the drawdown in A caused by pumping in B. So, q)h',' = ~0,h and as q)c,h = qghc, we have q)h',' = qgh,, which means that pumping in aquifer 1 from a vertical gives a drawdown in aquifer 3 at a horizontal distance r, which is equal to the drawdown at the same distance in aquifer 1, caused by pumping from aquifer 3 in the same vertical. Because r may be chosen arbitrary, the drawdown line as a function of r in aquifer 3, caused by pumping Q1 in aquifer 1, namely q)31(r) is equal to q)13(r), the distance-drawdown line in aquifer 1 as a result of pumping Q3 in aquifer 3 by means of a well in the same vertical as the well in aquifer 1, provided that Q1 = Q3. In order to check the reciprocity principle also for multi-layer systems, we consider the n-layer system of Fig. 4, Section 2.1.3, where we have to show that pumping Qi from aquifer i gives a drawdown q).ji(r) in aquifer j, which is equal to the drawdown qgij(r) in aquifer i, caused by pumping Qj from aquifer j, the wells being located in the same vertical, if Qi -- Qj.

za

Instead of checking the reciprocity 9=0

y/////////~//////////////////////////////////////////~ Cl principle in the general n-layer sys,

1

=~ Q 1

q91

tem by means of the solution

T1 =

U//'////////////////////////////////////////////////~////,~

ili

C2

¢ p ( r ) - Ko(rx/A)q,

I I I I

2

=¢" Q2

Ill

3

~°2

T2 = K2D2

r

I I I I I I

, l:::~ 0 3

which has been derived in Section 2.1.3, and in order to avoid very inq)3

tricate matrix analysis, we will con-

T3 = K3D3

I I

(.U////////J/.)//////////////////////////////////////////~ I

¢4

Fig. 44, and consider groundwater

I I

4

= , ~' Q4

q94

sider a four-layer system as given in

T4 = K4D4

flow, caused by abstraction of wa-

I

ter by means of fully penetrating line

I

c,

~o - 0

wells with discharges Qi (i = 1, 2, 3,4).

Fig. 44. Four-layer system.

The ordinary differential equations for steady flow are, if we put 1 -- aii Tt ci

and

1 Ti ci + l

= ai(i+l),

848

Analytical solution methods

(2.3.6-2)

with i -

1--4 for Ti and i -

d2rpl

1 dq)l

dr 2

r dr

d2q92

+-

r dr

d2q)3

1 dq)3

~-~ dr 2

d2q94

dr 2

Ci"

-- ( a l l -t- al2)fPl -- a12q)2,

1 dq92

dr 2

1-5 for

r dr

= --a22q)l -k- (a22 -t- a23)q92 -- a23993,

(a) -- --a33q)2 -+- (a33 -i- a34)q93 -- a34q)4,

1 dq94 ~ -- -a44993 -at- (a44 -k-a45)q94,

r dr

and the boundary conditions are: lim

(Pi ( 0 0 ) -- O,

( dqoi~ r

r--+0 "~f, /

Qi

-

=

27rTi

-qi

(i -- 1 2 3, 4). ' '

For the solution of this boundary value problem, we apply the infinite Hankel transformation (Section 2.2.4-4). Then we have d2~0i 1 d99i H { --~-r2 + r - ~ - r ] - - l i mr----~0 in which ~)i(0/) -- f~x~

(rd~Oi "] _ --~-r/

r~oi(r)Jo(0/r)dr,

0/2~i (0/),

and use this expression to reduce the differ-

ential equations (a) to four algebraic equations in g3i (oe)" ( a l l + a12 -at- 0/2)(p1 -- a12932 -- ql, -- a22~1 + (a22 + a23 + 0/2)~2 -- a23933 -- q2,

(b)

--a33(P2 nt- (a33 "at- a34 "-i-0/2)(p3 --a34(P4 -- q3, --a44(P3 -t- (a44 q - a 4 5 + 0/2)(p4 -- q4.

The solution of these 4 linear equations for the 4 unknowns ~i (i - 1, 2, 3, 4) can be written in the form: ~ b l - D1 -D-' in which D and

O

~b2-

Di

D2

D

,

~b3-

D3

D

and

D4

~b4- ~ , D

(c)

are determinants of the 4th order:

all + a12 + 0/2 --a22 0 0

--a12 a22 -k- a23 + 0/2 --a33 0

0

0

--a23 a33 -at- a34 -k- 0/2 --a44

0 --a34 a44 nt- a45 --t- 0/2

Solutions, derived from known solutions

(2.3.6-2)

849

and D1, D2, D3 and D4 are derived from D, by replacing the elements of the columns 1, 2, 3 and 4 by the elements of the column matrix

q2

, respectively.

q3 q4

For instance, D3 becomes"

all + a12 nt- 0/2

--a12

ql

0

--a22

a22 nt- a23 nt- 0/2

q2

0

0 0

--a33 0

q3 q4

--a34 a44 -+-a45 -t-0/2

D3--

Now, if we p u m p only in aquifer 1 with a discharge Q, we have

Q ql

--

2:rr T1

,

q2--q3

--q4--0

and D3 becomes"

all + a12 nt- 0/2

--a12

ql

0

--a22

a22 + a23 nt- 0(2

0

0

0 0

--a33 0

0 0

--a34 a44 q- a45 + 0/2

D3--

--a22

a22 + a23 -t-- 0/2

0

0 0

--a33 0

--a34 a44 q- a45 -t- 0/2

=ql

= qla22a33(a44 -+- a45 -+- 0/2),

and thus

D3 q~ ~31 -- - ~ = --~a22a33(a44 -+- a45 -+- 0/2).

If we p u m p only in aquifer 3 with a discharge Q, we have

Q q3 -- 27r T~'

q l - q2 - q4 - 0 ,

(d)

Analytical solution m e t h o d s

(2.3.6-2)

850

and D1 becomes" 0

--a12 a22 + a23 + or2

q3

--a33

--a23 a33 + a34 -t- ot 2

0

0

-a44

0 D1 --

0

0

0

--a12

0

--a23

= q3 a22 + a23 + ot 0

0 --a34 a44 -t- a45 -k- ol 2

--a44

0 a44+a45+ot

= q 3 a 12a23 (a44 -Jr- a45 nt- t3/2),

and thus q313

(e)

D1 q3 -- - - a 1 2 a 2 3 (a44 + a45 -+- 19/2) • D D

-- ~

Comparing solution (d) with solution (e) we see that Q

1

1

O q3a 12a23 -- ~

q 1a22a33 - ~27rT1 T2c2 T3c3

1

1

T1 c2 T2c3

and thus ~31 ~ (P13"

As the inverse Hankel transform is essentially unique, we also have q931 - ~013. In the same way, we may easily find that ~012 - ~021, (/914 - - 9)41, (/923 - - q932' q924 - - q942 and (/934 - - q943• For non-steady flow the differential equations (a) Si ~)~Pi have to be modified by adding the -terms T , ~ (i - 1, 2, 3, 4) to the right sides. i The initial values are assumed to be zero: qgi(r, 0) - 0. Now, if we apply the infinite Hankel transformation with respect to r and the Laplace transformation with respect to t, we find a set of 4 algebraic equations for the 4 unknown functions qSi(o~, s), which is similar to the equations (b), provided that tY2 in each equation is replaced qi ( i - 1 ' 2, 3 ' 4) " by o/2 -+-/~2s, with/32 - ~Si and qi by .-7. The results (d) and (e) for steady flow now become, respectively" ql

~31-

- - ~ a 2 2 a 3 3 ( a 4 4 + a45-t-ol 2 -t- fl2s)

-

q3

a, a 3(a44 +

+

and

+

from which it follows that ~31 - ~13 and thus as the inverse Hankel and Laplace transforms are unique: q931 -q913. Also for other values of n, the reciprocity principle can be demonstrated in a similar manner for a n-layer system, using determinants for solving the differential equations transformed to algebraic equations, the order of the determinants being equal to n.

Solutions, derived from known solutions (2.3.6-2)

851

Example 32. In Fig. 45a a double-layer system is given, consisting of two leaky aquifers, separated and confined by 3 semiprevious layers. Both aquifers will be drained in succession by means of a steadily discharging pumping well that is screened only throughout the 11 aquifer in question. In the same way that has been used I I f I for a 4-layer system, we will deter,, S1, T1 mine the various drawdown distribu'5/'///J/J////////////////////////////////////////////////~ C2 tions, in the non-steady state, by applying Laplace and Hankel transfor$2, T2 mations with respect to t and r successively. First we consider Fig. 45a c where pumping with a discharge Q1 ~o=0 takes place in the upper aquifer, yielding a drawdown line ~oll (r, t) in the Fig. 45a. Double-layer system. Pumping the upupper aquifer and a drawdown line per aquifer. ~021(r, t) in the lower aquifer. The doubly transformed functions ~11 (ol, s) and ~21 (or, s) may then be solved from the two algebraic linear equations (cf. equations (b) for the 4-layer system)" Q1

=>

~r

( a l l -[--a12 -+-og 2 -+- f l 2 S ) ~ l l -

ql

a12~21 -- - - , s

(f)

--a22~11 + (a22 + a23 _+_ o/2 .qt_/~2s)~21 _ 0, DI D2 with the solutions" ~ll -- -D-a n d ~21 "- --D-, in which

a l l -+- a12 -+- Ot2 + O

--a22 -

fl2S

--a12 a22 -+- a23 + Ot2 -+- fl2S

( a l l 9- a12 -+- ot 2 -+- f 1 2 s ) ( a 2 2 + a23 q- ot 2 --[- f l 2 s ) -- a12a22,

or

D

-- ¢s 2

-Jr-(aot 2 -k d)s + Og4 -Jr-bot 2 -+- e

a - - fl 2 -k- fl 2, 2

with

b - - a 11 -+- a 12 + a22 nt- a23 ,

d -- (all +- a l Z ) f l 22 -k (a22 -+- a23)fi~

and

c - - f121 f12, 2

e -- a l l a l 2 -Jr- a12a23 -+- a l l a 2 3 .

(g)

852

(2.3.6-2)

Analytical solution methods

Further, D1--

q_.l. s

--a12

0

a22 -nt- a23 nt- 012 + fl 2s

all + a12 + 0/2 -t-- fl2s

q_l

--a22

0

D2-

_ q__}_l(a22 + a23 -k- 0/2 + / ~ 2 s )

and

S

ql - --a22. s

First we apply the inverse Laplace transformation to ~11 and ~21. Therefore, we write D as D -- cs 2 + (a0/2 -% d ) s + 0/4 + b0/2 jr_ e -- c(s + Xl)(S --I-x2),

in which a0/2 + d xl -t- x2 -

0/4 + b0/2 + e and

c

X lX2 -

(h) •

c

Obviously, X l and x2 are the roots of the quadratic equation c x 2 - (a0/2 -t-

d)x +

0/4 -t- b0/2 -t- e

-

0

(i)

(Xl and x2 are both positive, because all constants a, b, c, d and e are positive). Now ql (a22 nt- a23 -Jr-0/2 %. fl2s) ~11 (0/, S) =

and

CS(S + Xl)(S "-{-X2)

(j)

qla22

,~~ 2 1 ,~ o ~ , s ~ -

CS(S nt- Xl)(S "-i-X2)

The inverse Laplace transform of the function

1 f(s)

_

-- (s -+- X l ) ( S + x2)

1(1

1)

Xl - x2 s --1-x2

s --i- Xl

is 1

F ( t ) -- ~ ( e

Xl while L - l { Isf ( s ) } 1

XlX2

1 +

-x2t - e-Xlt)

(Section 2.2.2-2, equation (21)),

x2

-

1

-

f0 (e-X2r - - e - X l r ) d r ,

~1-~

X2

_Xl t - ~ Xl _x2t) e Xl -- X2 Xl -- X2 e

which can be evaluated to (Section 2.2.2-1, equation (8)).

Consequently, for the inverse Laplace transform of ~11 and ~21 we obtain: @11(0/, l) -- q1(a22 + a23 + 0/2) ( 1 + ~ - -X2 e CXlX2 _%

q lfl2 C(Xl --X2)

Xl - - X 2 (e-Xate-Xlt)

-xlt -- ~ Xl e Xl - - X 2

-x2t)

Solutions, derived from known solutions

(2.3.6-2)

853

and @21 (oe, t) -- qla22

1 -+- ~ X2 e

CXlX2

_Xl t

-- ~ Xel

Xl --X2

-x2t )

.

Xl --X 2

Inverse Hankel transformation gives the final solutionq-2-1L °c oe(a22 nt- a23 -4- 0l 2)

9911 (r, t) --

C

XlX 2

x

l+~--e X1

-C

X2 ~

- x 1t

-

X2

~ X1

Jo(oer) X1

XlX 2

Q2

] doe

-xlt -- e -x2t) doe,

~

X2 e

X 1 ~ X2

_Xl t

-

(95) Xl

---~~e

X 1 ~ X2

_x2t' ~

]

d~.

Secondly, we will derive the drawdown lines 9912(r, t) and (/922(r, t) in the upper and lower aquifer respectively, caused by p u m p i n g with a discharge Q2 in the lower aquifer (see Fig. 45b).

=~

r

X2

~Jo(oer)(e X 1 ~ X2

~o21(r, t) -- qla22 f0 ec oe Jo(roe) ( 1 + C

_x2 t

e

~

~o=0

S1, T1 5"//,,'2Y//////////////////////////////////////////////ZC 2

82, T2

~o=0 Fig. 45b. Double-layer system. Pumping the lower aquifer.

The doubly transformed f u n c t i o n s ~12 a n d ~22 can then be solved from the linear equations( a l l ~L_ a , 2 + 0l 2 J r - / ~ s ) ~ 1 2

-- a12~2 2 -- 0

- - a 2 2 ~ 1 2 -+- (a22 -~- a23 -~- 0/2 _4_ f i 2 s ) ~ 2 2 _

q2. s

and

(k)

The determinant D of this set of equations is the same as the determinant of the equations (f) and is given by equation (h): D - c(s + Xl)(S + X2),

854

(2.3.6-2)

Analytical solution methods

so q2a12

g912(0/, s) --

and

CS(S + X1)(S -at- X2)

O)

q2(all Jr- a12 -+- 0/2 _+. fl2s)

^

q~22(0/, S) --

CS(S "nt- X1)(S -'l- X2)

The inverse Laplace and Hankel transforms will yield, in a similar way to that in the foregoing problem:

rplz(r, t ) -

q2a12 fo ~ ~0/ ¢

XlX2

-xlt - ~ Xl - x 2 t ) du, Jo(r0/) ( 1 + ~ X2 e e Xl X2 Xl -- X2

q922(r , t) -- qZc f0 ~ 0/(all +Xlx2al2nt- 0/2) Jo(r0/) X2

x

-Xlt

1+ ~ e Xl

~

qzf12 fo ~ ~ ¢

X1

-

x2

Xl

0/

_x2 t

~ e

--

X2

] dot

(96)

Jo(r0/) (e -x't - e -x2t) d0/.

Xl -- X2

According to the reciprocity principle, q921 in equation (95) should be equal to (/912 in equation (96) if Q1 = Q2 and indeed the right-hand sides of both equations have the same integral form and differ only with respect to the constants before the integral signs. But if we assume that Q1 = Q2 = Q, we find q2a12 c

Q 1 1 2zr T2 T1c2 c

Q 1 1 2zr T1 T2c2 c

qla22 c

and consequently q912(r, t) - q)21(r, t) if Q1 - Q2 for all values of r and t. The steady state can be obtained by letting t approach to infinity in the solutions (95) and (96), which gives, with C X l X 2 - ot4 nt_ b0/2 + e (equation (h))"

qgll (r) -- ql

f0 o° 0/(a22 -+- a23 -+- 0/2) Jo(r0/) do~, 0/4 + b0/2 + e

(/921(r) -- qla22

~

oo

u Jo(r0/) du, ot4 q._ b0/2 + e

(m)

0/ + e Jo(r0/) du, q)12(r) -- q2a12 ~0 °° 0/4 + b0/2

~22 (r) -- q2

fo ~ 0/(all q- a12 -at- 0/2) Jo(r0/) du. 0/4 -Jr-b0/2 -}- e

Putting 0/4 -~- b0/2 -Jr- e -- (0/2 q_ yl)(0/2 -k- Y2), where Yl + Y2 - b and Yly2 - e and thus yl and y2 are the roots of the quadratic equation y2 _ by + e -- O,

(n)

Solutions, derived from known solutions (2.3.6-2)

855

we find"

1

1

1

0/4 4- b0/2 4- e

(0/2 4- yl)(0/2 4- Y2)

~yl ( - 0 y2 / 2

a22 4- a23 4- 0/2

0/4 4- b0/2 4- e

l (a22 4-a23 -- Yl Y l - Y2 0/2 4- Y l

a l l 4- a12 4- 0/2

1

0/4 4- b0/2 4- e

Yl - Y2

[all 4- a12 -- Yl 0/2 4- Yl

1)

1 4- Yl

0/2 4- Y2 '

a22 4-a23 - Y2) 0/2 4- Y2

and

a l l 4- a12 -- Y2'~. 0/2 4- Y2

)

Substituting these equalities in the equations (m), integrals of the form

4- k2 J0(r0/)do/ f0 °° 0/2 0/ occur, which m a y be replaced by Ko(kr) (Section 3.3-5, equation (214)). Thus, the equations (m) become"

(/911

ql ~{(a22 Y2 -- Yl

4- a23 - y l ) K o ( r ~ ) -- (a22 4- a23 -- y2)Ko(r~iC~)},

q)21

q912 --

(97)

q l a22 { Ko (r ~y-~) - Ko (r ~ / ~ ) }, Y2 -- Yl

q2a12

Y2 - Yl q2 q922 -- ~ { ( a l l y2 -- Yl

{ K0 (r ~y]-) - Ko (r ~y-~) }, 4- a12 -- y l ) K o ( r ~ )

(98)

-- (all 4- a 1 2 - y 2 ) K o ( r ~ / ~ ) } . Solution (97) is the s a m e as that found in Section 2.1.3, E x a m p l e 6, as Yl 4- y2 b -- a l l 4- a l2 4- a22 4- a23 and thus a22 + a23 -- Yl -- - - ( a l l -4- a12 -- Y2) a22 4- a23 -- Y2 -- - - ( a l l + a12 -- Yl).

and