The solution of a non-stationary problem of diffraction at an impedance wedge using tabulated functions

THE SOLUTION OF A NON-STATIONARY PROBLEM OF DIFFRACTION AT AN IMPEDANCE WEDGE USING TABULATED FUNCTIONS* ht. P.

SAKHAROVA and A.F.

FILIPPOV

Moscow 12 iuay 1966)

(Received

THE solution of the stationary problem of the diffraction of a plane wave at a wedge with impedance boundary conditions has been obtained in [l, 21. It is expressed by a Sommerfeld integral in terms of a special function Yp(z), for which tables exist [31. The solution of a similar non-stationary problem in [41 is expressed in terms of integrals with respect to rather complicated complex functions. The solution of the same problem is expressed below in terms of the tabulated function Y@(z). The asymptotics of the function Y@(z) are investigated as Im Z-XI and the asymptotics of the solution of the non-stationary diffraction problem as r + 0 (in the vicinity of the edge of the wedge) and as t -+m. Sect ions 1 - 4 were written by A.F. Filippov and Sections 5 and 6 by M.P. Sakharova. In a following paper an investigation will be carried out into the solution in the vicinity of the wave front, including the neighbourhood of the boundary of the shadow, and the uniqueness of the solution will be proved.

1. Formulation of the problem We are

seeking

the solution

of the equation

Utt = in a region

in the form of an angle un =Y+W

*

Zh.

c2 (uxx + uyy)

vychisl.

Mat.

(cp= mat.

Fiz.

@), 7.

191 < @ with boundary conditions Un =

Y-Ut

568

- 579.

3,

128

((p = 196’7.

-@)

(1.2)

The

solution

of

prob

a non-stationary

Zen

of

129

diffraction

(u, is the derivative along the inward normal to the boundary), polar coordinates r, 9,

(the upper or lower signs must be taken together). tion must represent the plane wave &=

f(t+

$N+

ysin c

fl

)

=f

(

If

or in

t < 0 the solu-

t+ i-cos(cpC

p,)

(1.4)

(incident wave) or the sum of the incident wave and waves reflected from the sides of the angle, if such waves exist for t < 0. Here c > 0, and 0 and S LT~ given real, and yk complex constants,

f(-r)

is a given

function

equal to zero

if T < 0.

For there to be no extraneous source at the vertex of the angle at least one of the following two conditions must be satisfied: (a) u, ut are bounded, ru,(t, approaches any point of the axis

r, 9) - 0, when the point r = 0;

(t,

r,

9)

(b) the function u is equal to amV /at”‘, where v is some function which satisfies equation (1.1) and the conditions (1.3) and (a).

2.

The reflected wave

In Sections 2 - 4 only the case 20 > ‘II, ‘II - @ < p c @ is considered (then there is only one reflected wave, (see figure). The results obtained are extended to the case of any p and 0 in the following paper. When the wave (1.4)

Ui =hf

is reflected

from the line

q~ = 0 a wave

(t++s(9-2@+B)),

(2.1)

coefficient, dependis formed, where kl = k+(@ - p) is the reflection ing on the angle @ - p between the face cp = Q, and the direction 9 = p, from which the wave comes, and on the form of the boundary conditions. Substituting the sum u. + u1 in the boundary condition (1.3) with 9 = 0, k_ from the face cp = -@ we find k+(O - PI. The refl ec ti on coefficient

M.P.

130

Sakhatova

and

A.F.

Filippov

is expressed in a similar way

sin@---w-

k+((-))=

sin 8 + cy+ a sine -CYsin 8 + cy-

qq=

(2.2)

*

In the case where IT- QI< p < @ reflection occurs only from the face 9 = @ (see figure). Diffraction of the wave occurs if t = 0, when the

incident wave is incident on the vertex of the angle, and is propagated in all directions with speed c, occupying the circle r < ct. Outside the circle there are only incident end reflected waves, so that if r&ct the solution u is equal to 0 in the shadow zone--,@
U=

3.

@0 i u, + u1

if if

Representation of the solution by means of an analytic function

We shall first of all seek a solution wave has a step form, i.e. in (1.4) no -

f0

(t

(2.3)

~--~<<<2~---~-~, 2@--_--n<<<@.

+

:eOS{q

-

/3)),

fO(T)

=

0

if

for the case where the incident

7 < 0,

fo(7) =r: i if

?T10. (3.1)

In this case the solution

for t < 0 is the same as the function

(3.1)

The

solution

of

a non-stationary

prob

Zen of

diffraction

131

and does

not change when t, rr and cp are changed to At, AF, mid cp, Where any number. Since the equation and bo~dary conditions do not change here either, the solution fox 2 > 0 also must not change (in view of the identity theorem). The solution with t > 0 deneods only on r/t and ‘p. Transferring in (1.1) to the coordinates t, r, and (9, assuming ct 3 r cash Q and assuming that the function u depends only on g, and n, we obtain from (1.1) and (1.3)

h

>

0

is

ff 7t 1 0, which corresponds by the formulae (2.3). Taking obtain

to F (1.41,

=

ct.

function rd is expressed and (3.1) into account, we

the

(2.1)

Up to the present time the method of solving the problem has been the same as in Cd, In t41 the solution is also sought in the form u(cp, n) = Re Cr(q~+ in). In our case, where the numbers yk in (1.2), and then also the solution, are complex we shall seek the solution in the form

We can put a,ny complex harmonic function form (3.4) equal to a constant d = d, + iii,

n(cp. r)) = u1 + iu, if n = 0,

in the

Since uk - dk, k = 1, 2 are harmonic functions, equal to zero if Q = 0, analytic functions r, exist such that uk(cp, qf - tlk = Re li’&.(p t in), In view of the principle of symmetr: fik is extended analytically into the region q < 0 so that Re ilk is an tilleven and Im Lib an even function Of ?J. Then uk(g, + in) - ML(q) - in) = 2(Uk - dk). Passing to the functions u = “1 t iu,, !i = :J, + it!Jzs we obtain (3.4). In particular, when the function (A is real, Ii, 3 0 and (3.4) becomes u=Re Ei+~i. The function u(cp, O), by virtue of (3.31, has jumps equal to 1 and at the points ‘pl = S - N, ‘p2 = 2Q, - i-, - w. The function z~((P, f 0 1, m = 1, 2, where k,

132

M.P.

Sakharova

and

A.F.

Filippov

G&T, d =‘/2~(cpm-~---irl)--~(cpm-~++irl)l (the branch of the logarithm is selected tion

has a jump 4~

--n
in accordance with the condiwhen 9 = q,,,. Therefore the

when q = 0. 191<0 u - izi I ax- ikzzln function keeps a constant value. Consequently i

U-

has no jumps and

ikt

-zi --z2=

al

xc

(3.5)

$(u'(rps$)--(P--irl))+U(--cD,O),

(3.6)

where the function II+ is regular (has no singularities) if [PI< 0. Now from (3.6) and (3.5) it follows that the function u can be expressed by the formula (3.41, where

icmln(qpm-cp- iq),

U(~+W=U'(q+W+ii

(3.7) d= u(-a&O),

m=i ci

=Id/n,

C2 =

kl / ax, and the function II* is regular. Although U(q + iq)

is discontinuous and analytic

if q = 0, the derivative

in the strip

U’(~cp+ iq) is cant inuous

1~1 < @, except for the points rp+ iq =(pm,

where U’ has simple poles with residues ic,, Substituting two equalities

expression

(3.4)

(* sin8 - cy*)U’(f@ They are valid

if

in (3.2)

m = 1, 2.

and varying iq by 6. we obtain

+ 0) = (* sin43+ cyk)U'(fQ,- 6). (3.8)

0 = iq, 0 < q < 00, and since they do not vary when

6 is changed to - 0, they are also valid if and (2.2) we have

‘00 > b > oo-

From (3.6) (3.9)

/c+(8) U’(u) + e) = U’(@ -q,

k_(eju~(-@-08) = uJ(-a + e).

The function CJ’(q t iq) is defined in the strip IqI
The

extended

solution

of

to the strips

The function virtue of (3.9)

a

non-stationary

101 <

(2k

+ 1)1),

k = 2,

U’ so obtained has poles also at the points

(-1)~(p*n--2n@), The leading

problem

parts

133

diffraction

... .

3.

at the points

r&=0,1,2 I...)

of II’ at the poles

of

q1 and q2 and by

n=--1,-2

are found from (3.7)

,....

(3.10)

and (3.9).

Thus

Moreover, since, in view of the equality ct = r cash q we have - f-up = tut = +, coth q, for condition (1.6) to be satisfied for t > 0 it is sufficient to require that

IU’(‘p+iq)I
as q+-*+oo, const > 0.

a= Thus we must find

a function

tJ’(q

+ iq)

satisfying

l’pl<@;

(3.12) the equations

(3.8),

having no more than two poles q1 and p2 in the strip 191 40, having poles (3.11) and satisfying condition (3.12) (we consider for the present that ‘pl and q2 do not lie on the faces 9 = f 0 and that y+ are not purely imaginary; then .when 9 = f 0 there will he no poles).

4. The solution of the problem We show that we can put U’(z)

=

&z+n)-s(-)1,

where s (z) is the function obtained and defined by the formulae

‘uw

s(z) = ---(T(z), V(B)

in [21 with

(T(z)=

(4.1) cpo=

p, sin Brt =

2g cos2Q3 sin 2&t? - sin 2Ef3’

t

cy*

3c = g’(4.2)

(4.3)

XY@

(z-w---JI--+e_)w(

2--4+-~-e-),

M.P.

134

YcJ(z)= ii it n=i

the function In fact

m=l

Sakharova

A.F.

Filippov

[I- (2@(2n_l);n(m_-1, ))z](-i)m+‘; (4*4) 2

Ya is tabulated by virtue

and

[31.

of equation

(5)

of

[21

On changing once from 0 to 9 + TI, and again from 8 to 0 - TT, subtracting one equality from the other and passing from s to I/’ in accordance with (4. l), we find that CJ’ satisfies equation (3.8). By virtue of formula (4) of [21 the function (4.1) satisfies the condition (3.11). According to (4.3) IRe z 1 < 2@+33n/2, a~+ Re fL.

Since

and (4.4), Y’o(z) has no poles in the strip and Y(z) has none in the strip ]Rezl ( dj+

e+ =

arc sin ey*,

in the case where Re y+ > 0 the

functions s (z t a) have the same poles o(2 f JT), i.e. =m+(-l)“p+2m@, Only two of these

fall

in the strip

1Re zI &

n=0,1,2,.,.,-I,-2 in the strip

IRe z[ 9

CD as

,....

(4.6)

@.

Condition (3. 12) is satisfied by virtue of (6.3) obtained Thus the function L” satisfies all the required conditions.

below.

We write the expression with r
22(Cr, cp)=

u(hct,cp)+ W(Gr,f-P),

w =

1 ?y$(z) 2

&,

chtj

=t.

cp-irl i4.7)

In the case where TT- UJ< p < $ the function u(c, ct, 9) is determined from (2.3), i.e. it is equal to the sum of the incident and reflected waves, replaced by zeros in the shadow zones. This sum has a constant value in each of the three sectors when r
The

solution

of

a

non-stationary

problem

of

135

diffraction

We shall denote by S(cp) the sum of the residues of the function s(z) at the poles, falling in the strip 9 - IT < Re L < ‘p f TT. We show that

u(t,

ct,cp)=

(4.8)

S(q).

According to formula (4) of [21, the residue of the function equal to 1 if s = 8. The residues at other poles (see (4.G)) mined by means of equations (5) of [21, written in the form

s(z) is are deter(4.9)

s(z+(D)=

--k+(z)s(--z+(D),

s(-z-a)=-k-(z)s(z-O),

where k, and .k_ are the same as in (2.2). Poles of the function s (2)

a

2D--3

Residues

1

k+ (=‘-3)

Suppose,

as in Sect ion 2, that

S(o)

=

We obtain

-2G--p

p-40

k_ (Q---P)

20 >

a~, a~-

k,(@-3)

0 <

p <

k-(3@+)

0.

0

if

--<\(<<---,

1

if

fJ--rc
1 +k+(iD-fi)

if

.., . ..

Then

(4.10)

On the other hand for the problem of diffraction of the wave (3.1) when r = ct, by virtue of (3.1) and (2.1) we have ‘lo = 1, u1 = k, = k+(O - p). Now from (2.3) and (4.10) it follows that S(q) E u(t, ct, 9). Since 28is((p) is equal to the integral of the function S(Z) along a closed rectangular contour with vertices p - TT- in, 9 t TI - iq, 9 f IT + iq, q~ - ?r + iq, from (4.7), (4.8) and (4. l), we have

(4.11) rl =

In (4+

1/ ( +)‘-

10 .

Fromula (4.11) is the solution of the problem of the diffraction of the wave (3.1) when r < ct; the function s(z) is expressed by the formulae (4.2) - (4.4). The solution

u* of the problem of Section

1 for

any incident

wave of

M.P.

136

Sakharova

and

A.F.

Filippov

the form (1.4)) where f(v) = 0 if T < 0, is expressed in terms of the solution u(t, r, 9) which is found

In fact u satisfies equation (1.1) and the boundary conditions, and so does u* also; if t < 0 the function u represents the incident wave (3.1) (or the sum of the incident and reflected waves), and by virtue of (4.12) u* is identical with the right-hand side of (1.4) (or the sum of (1.4) and the reflected wave (2.1)). Substituting (4.7) in (4.12), integrating by parts and making the change of variable r = t - c-k cash u, we find that the solution IL*of the problem of Section 1 is equal to the sum of the incident and refleeted waves if ct r equal to the same sum plus the diffracted wave W*

where f is the same as in (1.4),

and U’(z)

is defined in (4.1)

- (4.4).

The solution of problems on the diffraction of a cylindrical wave at a wedge with boundary conditions (1.2) and of multiple diffraction at vertices of polygons in c51 are expressed in terms of the fun& ion :s(g, p) . Putting

m(q, b) = i)lJC/ xu’(cp),

in formula (4.6)

of [51, where

II’ is defined by the formulae (4.1) - (4.4) of this paper, we obtain a representation of the solutions of these problems also in terms of the tabulated function YQ.

5. The asymptotic expansion of the function 'Y@(z) as Im 2 -*f a, The results of Section 5 are used to verify that conditions (3.12) are satisfied and to investigate the solution of the diffraction problem as r/t + 0. In [61 the following representation of the function Y@(z) with [Re tl < 2@ is found, which is convenient for obtaining the asymptotic formulae

The

solution

We introduce We have

of

a non-stationary

the notat ion

E=

problem

n /4@;

if

of

0 <

137

diffraction

@ <

51, then ‘4s

$ < 2.

(5.2)

We find tie asymptotic consider the function

series In (‘I+

for the integral T as Im z + + Q. We shall sh2Ev/cos2&) in the strip (Re z 1 < 29,

in which it is regular for all v, 0 &J < ;o. We introduce a small quantity E = cos-a$ as Im z -+ f co. Taking a = c-1 Arsh 1, we obtain

The integral

?’ can be represented

I’=

(--I)“+’&nK,(g,

;

integral

( “h;;

&(zJ=

dv.

sinh2
ev = (ly + 1 + 13 l/E, k=o

grlv chvdy

45.5)

1 -= F- (---iIk (,‘yTl sh 2g.v ch v ,rJwy + 1

-

j'y +

1

in a series

in negative

+ Ii)-_(WS.

powers of y we obtain

(5.6)

k=Un=o

x

(5.4)

0

T” we make the change of variables

1 -= chv

Expanding

form of an expansion

n

?I=1

In the

in tie

(I++y-;.;2+

_&+$_...)-(2k+f)Ez

_=

y-*-m

i

-j

k~Om=il

y-k/'"&&)*

Iu.P.

138

Equating

Clto (Q =

(-

Sakharova

the coefficients

of

A.F.

and

identical

Filippov

powers of y we find

1) k 2--W+‘E,

Substituting

The expression

(5.6)

in the

integral

under the integral

I” we obtain

in (5.9)

is put in the form

where .%’is an integer such that q - l,i:V < q; in the case where Y = 0 the sum from p = 1 to p = N is replaced by zero. Then

f,

~(-l)p-iep-’+(-l)NEN

=

p=i If q is an integer ln((1 + E)/E). If difference of two the second from 0 duced to the Euler of E. We obtain

4-P

m..yN-p+_

s ll+w

(5.10)

N = q - 1 and the integral in (5.10) is equal to is not an integer it is put in the form of the integrals: the first between the limits 0 to CD, and to 1. By the change (1 + .ey)-l = x the first is rebeta-function and the second is expanded in a series

q

neq-N--i

sinn(q-IV) Substituting this formula into ln(1 + E) in a series we find

(5.10)

and (5.9)

and expanding

The

of

solution

a non-stationary

prob

Zem of

diffract

is not an integer)

ion

139

->

I km (P+l)

(from the

last

Finally

sum the term with ,v = q is excluded).

we obtain

I

=

for the integral

5,__-(--I)“-’ TL==i

Jkm

n

Ckm(c) from (5.6),

in (5.12) we confine ourselves (5.7) we obtain: when c # n f

I = Moe +

2

(-

n

f

H,

Ikm from (5.11)

and

to terms of order below E* In E, = 0, 1, 2, . . .

$4, n

1) k 2-W+WE ne~k+‘/z)/l

(5.13)

O
when e =

(5.12)

k=Om=O

where f(,($) is taken from (5.4), q from (5.9). If using

(5.3)

4-*/2)/E

and R is an integer

(5.14) Mi =4si&(e)+$

Ckm(E) !* ___ __ k---O?b--Om + ((k _1_‘/2j - 1. b

The asterisk in the summation sign means that the term whose denominator is zero is excluded from the sum. We calculate the coefficient y!e first of variable sinh*
for $ -C ‘h. Making the change K,(S) (see (5.4) for r~ = l),

140

/d.P. Sa.hharova

and A.F.

Filzppov

On the other hand, on multiplying the left and right-hand equality (5.6) by y and integrating we obtain co

c

ydy

sh2@vchv

l

0

Then the coefficient

(see

C7,

p.

of

Gwn(~) k=dJm=O

m+(k+i/2)/%-

3

%O with 0 < $ 4 M is equal to 00

jj‘&)=

sides

s

---_

YdY

ogsh2.$vchv

ODshs,$vdv fo chv

(5.15)

3181).

In the case where %
Then when % < 5 < 3/2 the coefficient

The second

integral

converges

uniformly

‘(0 is equal to

if 6 <

I Im c I < 6 (6 > 0 is as small as we please). analytic function of fj in the given region. the difference between the two integrals

Re E <

3/~ -

6,

Consequently it is an If ‘4 < 5 < $4 it is equal to

Joining the first of these with the first inteqral in (5.17) and using formula (5.15) we find that the sum of the two integrals in (5.17) is equal to

7’he

solution

T!~isis also true snd the function ‘5 < 5 < 3/2 also

o_fa non-stationary problem of diffraction

if !$ d 5 < 3/z by virtue of the integrals (5,18} being analytic. in the above region,

141

In (5. I!“), Then when

This is also true in the case where g > 31’2, g’+ rz -f‘/2, but for the prod of this it is necessary to transpose to the left-hand side alf. the terms af the series (5.6) with m = 0, k C 5 - ?L The ~~~~n~~~ afguments are siqiizr.

In the case where Ej = ‘4 the coefficient ;I, in (5.14), according to (5.4) aEd (5.16) is equal to the slim 4f the two integrals in (5.17) plus 3. Since the integral converges uniformly near 5 = ?;l, this sum is (5. I@, i.e. is eqwl to equal to the limit as E -* % of the expression In 2 - ~/it:. Consequently ,

fn

the

ease where

c=

n + 5s we similarly

Mf-_---

rc 4 +_(I’)” 2n 4T ,-

obtain

k-w

In 2 + j--&y-~ . -I-

M.P. Sakharova

142

v = 1, p = min(4E,

and

A.F.

Filippow

2c f 1); M is an integer,

25

-

2 < ,Jy < 25

-

1;

if

c = 1/2we have C(z) =

v=p=2;

D(z) = 1/3x,

T&/n,

if c = 0 + $$, I, 2 1 is an integer we have C(z) )1 = 2< t I, D(t) of the same form as in (5.20). Formula

was obtained

(5.19)

for

=

F(--l)%I

S-C, v = 1,

[Re t 1 < 2@. Using the functional

eqirei; ion

y’,(z+2~)=Y,(z(formula (12) all Re z.

of

[21)

20)

of the problem

diffraction

(z),

-&

it is proved that relation

6. The asymptotics

s

ctg $ + “1

We write the asymptotic representation defined by formulae (4.2) - (4.4).

(5.19)

solution of as r/t -+ 0

is valid

for

the

as Im z -+ f co for We have

the function

the upper signs being taken as Im z -+ a, the lower as Im z -\I- a~. Substituting (5.19) in (4.3) and (4.2) we obtain

as Im z -+ + co, where

B=

4EA4cos 2@ Y(B)

2sinE(n+8+-0-)sin~(0~--_-)

cos Tcg

’ -

if E#1n+ 2 sin 2&3

$12?

c,

=

(

z[sin

Hence from (4.1)

258+ - sin ZT$L]

we have for

all

c > 0

if

5=

n + ‘12.

The

solution

of

a non-stationatn

problem

of

143

diffraction

IIKiZ-+&CQ,

U’(z) = 3?i~BLe+2iEZ + 0 (Pe*+z),

(6.3)

where

The estimate it

(3.12)

is easy to obtain

d?$/dF

as

r/t

+

follows

from (6.3).

an asymptotic

Using formulae

representtltion

for

(3.4)

and (6.3),

au/ci\cp, au./&,

0.

For the solution u(t, F, cp) of the problem of the diffraction of the wave (3.1) we can also write an asymptotic representation for small values of-” r/t = c/Gosh q. Substituting in (4.11) the asumptotic representation (6.2) of the function s(z) we obtain for all c

Trans

Eated

by

H.F. Cleaves

REFERENCES 1.

~iALYUZHI~ETS,G.D. Some generalizations of the method of reflections in the theory of the dfffract ion of sinusoidal waves. Diss . Dot . inst. Akad. Nauk SSSR, 1950. fiz.-matem. n., Moscow, Fiz.

2.

~ALYUZ~IN~TS, G.D. Perturbation. reflection and radiation of surface waves on a wedge with given impedance faces. Dokl. Akad. Nauk SSS.?, ,121, 3, 436 - 439. 1958.

3.

ZAVADSKII, V.Yu. and SAKSARUVA,M.P. Tables of special functions y@(z) (Tablitsy spetsial’noi funktsii y@(z)). Gtchet Akusticheskogo instituta (Report of the Akoustic Institute), Moscow. 1961.

4.

PAPADOPOULOS, V.M. Pulse diffraction by an iml?erfeCtlY wedge, J. Aust. Math. Sot. 2, 1, 97 - 106, 1961.

5.

FILIPPOV, A.F. Exact expressions with a circular front, Prikl.

6.

for nat.

repeatedly mekh.

28,

reflecting

diffracted 6.

1083

-

waves 1091. 1964.

MALYUHINET,G.D. Das Sommerfeldische Integral und die Lasung von Reugungsaufgabe~ in Winkelgebieten. Ann. Phys. 6, l-2, 107 - 112‘1960.

.il.P.

144

7.

GRADSHT~I~, I.S. anti products Fizmatgiz,

Sakharova

and

and RYXHIK, I.M. (Tablitsy

n.foscow,

1962.

int3gralov,

A.F.

Tables

Filippov

of integrals,

summ, rgadov

sums,

series

i proizbedenii),