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Singular and fundamental solutions to potential and elasticity problems for a riemann surface
COMPUTER METHODS NORTH-HOLLAND
IN APPLIED
MECHANICS
AND ENGINEERING
83 (1990) 211-230
SINGULAR AND FUNDAMENTAL SOLUTIONS TO POTENTIAL AND ELASTICITY PROBLEMS FOR A RIEMANN SURFACE Uhich HEISE institut fiir
Technische Mechanik,
Technische Hochschule,
D-5100 Aachen,
Fed. Rep. Germany
Received 20 February 1989 Revised manuscript received 23 August 1989
The solutions can be used as kernels of integral equations by the aid of which it will be possible to solve certain engineering problems advantageously. Fields of application of such integral equations and a survey of the paper are given in Section 1.
1. Introduction We consider a logarithmic Riemann surface, i.e., an infinitely extended helical surface with an infinite number of sheets (see e.g. [1,2]). The origin of the co-ordinates is a branch point of infinite order, The surface can be pictured as a spiral staircase with an infinite number of turnings, an infinite exterior radius and a vanishing interior radius. To illustrate this we may express the height z of a point of this staircase in cylinder co-ordinates by z = E *9. (The central angle cp is not limited to the interval 0-2~. t: is small since we deal with plane problems.) In this paper singular and fundamental solutions are given to the potential equation, to the bipotential equation and to Navier’s differential equation (equilib~um equation) for the plane elasticity problem. These solutions (not only the fundamental solutions) can be used as kernels of integral equations by the aid of which it should be possible to solve certain problems advantageously. In the following we outline some promising fields of application. Integral equations of the considered type will enable us to solve problems where the domains have overlapping parts (see Fig. l), e.g. certain helical springs. Other integral equations are definitely unsuited for such purposes. A body with a slit (see Fig. 2) can be easily treated, since points at the opposite sides of the slit facing each other have different co-ordinates. In conventional integral equations such points are identical and measures have to be taken to overcome the resulting difficulties [3,4]. A technically eminently important particular case of a slit is a crack (see Fig. 3). The proposed integral equations should be suited for the treatment of edge crack problems. However, for this purpose it is necessary first to investigate the singularity of the given solutions at the origin of the co-ordinates or to treat the crack tip separately. A standard integral equation method for the treatment of crack problems makes use of solutions for an infinite medium with a built-in crack [5,6]. The solutions for the Riemann surface have the advantage of being more simple. ~4S-78~/90/$03.50
0
1990 - Elsevier Science Publishers
B.Y. (North-Holland)
212
U. Heise, Singular and fundamental
Fig. 1. Domain with an overlapping part. The ‘pitch of the helical surface’ is drawn on an exaggerated scale.
solutions to problems for a Riemann surface
Fig. 2. Domain with a silt
The solutions under consideration are based on polar co-ordinates. Hence, the integral equations can possibly be approached analytically or annular sectors for circular sectors (see Fig. 4). Analytical solutions of integral equations are necessary for the testing of numerical methods. (In the case of the circular sector the singularity of the kernels at the origin of the co-ordinates might be troublesome.) Successful treatment of circular sectors perhaps opens up possibilities for the more easy solution of problems dealing with domains with corners. In this paper we consider plane problems. The solutions for the Riemann surface are based on polar co-ordinates. Possibly the subject can be generalized to three dimensions. One has to formulate the problems in cylinder co-ordinates without the restriction that the central angle cp covers only the interval 0-27~. Then one has to find a suitable transformation of the ‘Riemann space’, which penetrates itself, onto the full space and to construct singular solutions of the equation. The generalization is intellectually tempting since the human brain cannot envisage
Fig. 3. Domain with a crack.
Fig. 4. Annular
and circular sector.
U. Heise,
Singular and fundamental
solutions to problems for a Riemann surface
213
the 3-D solutions should have the same a space which penetrates itself. Nevertheless, advantages as those for plane problems: Bodies with a slit can be easily treated, and probably cracks also, if the tip line is straight. Helical elastic springs with arbitrary cross section, the neighbouring turnings of which are adjacent to each other, will not lead to difficulties, at least not in principal. In Section 2 singular solutions to the potential and the bi-potential problem are established. We start from the differential equations formulated in polar co-ordinates and use a transformation mapping the Riemann surface onto an infinite plane. Singular solutions of the differential equations in the transformed space are determined. Then these solutions are retransformed. Section 3 deals with the plane elastostatic problem. In Section 3.1 we transform Navier’s equation. The author did not succeed in finding directly a solution to the transformed equation by the aid of a standard method (see [7, S]), but tried to derive it from a solution to the bipotential equation. As a guiding rule for this purpose Section 3.2 shows how the ordinary Kelvin solution to Navier’s equation can be obtained from the ordinary r’ In rsolution to the bi-potential equation. Then, using the results of Section 3.2, however, not carrying out each step by strict analogy, a singular solution is established in Section 3.3 which satisfies the transformed Navier equation. This solution is retransformed in Section 3.4. In Section 4 it is shown that the solution to Navier’s equation represents a fundamental solution. The most economical possibility of exploiting numerically the formulae developed in this paper would be to modify the algorithm of an available computer program package. One has to exchange the kernel function in the program (e.g., in the case of the plane elasticity problem Kelvin’s solution) by the corresponding solution for the Riemann surface, or even better to provide in the program a branch with the solution for the Riemann surface parallel to a branch with the classical solution.
2. Solutions surface
to the potential
and to the bipotential
problem for a logarithmic
Riemann
The subject of this section is to establish solutions to the equations AF(x, y) = 6(x - i, y - y)
(2.1)
AAF(x, y) = 6(x - X, y - y)
(2.2)
and
for a Riemann 2.1.
surface.
Tranformation
of the differential operators
We start with the Laplace operator
for polar co-ordinates
r, cp
(2.3)
and introduce
the transformation:
U. Heise, Singular and fundamental
214
p=lnr,
p=lnF,
4’90,
r=ep,
q=(P,
f=e”,
9’47
(2.4), (2.5), (2.6), (2.7) (2.8),(2.9),(2.10),(2.11)
CP’G,
4”=q-g=cp-cp,
p^=p-p=1n;, r = ed = r ’
solutions to problems for a Riemann surface
(2.12), (2.13) (2.14), (2.15)
cp-cp’q”, 2
/k2=j?+q2=
(
1,;
)
+(cp-+)2,
(2.16) = sinh p” ,
r
r
= = cash p” + sinh p” ,
; =
r
(2.17), (2.18)
cash p” - sinh p” .
(2.19), (2.20)
With a,F=‘a
r
F,
P
afrF= L (a',, - d,)F r2
(2.21), (2.22)
7
t$+,F = d:4F,
(2.23)
AF = -$ (dtp + a’,,)F =
e-2P(at, + a’,,)F
(2.24)
field point
Fig. 5. Cartesian
and polar co-ordinates
in the untransformed field point
Fig. 6. Co-ordinates
in the transformed
space.
space.
U. He&e, Singular and fundamental
solutions to problems for a Riemann surface
21.5
we find AF=ePzP
(2.23,
A* = a2, + a;, , 1
b*F,
(2.27)
AAF = ePzp A” eT2’ A* F , A*(A* - 4~3,+ 4)F.
AAF=e-4P 2.2. ~ru~sfor~utio~
(2.26)
(2.28)
of Dirffc’s 6 -function
The a-function of the right-hand sides of (2.1) and (2.2) has to be transformed in the same way as the differential operators on the left-hand sides. The following gives a brief outline of the procedure. The &function can be defined as 1 B(x, x, y, 9, n) = nlt2 i 0 S(x -
for(x-X)2+(y-Y)2
x, y - jq = iii@B(x, x, y, y, n)
WP -6
q - S) = Fi
(2.29)
for (x - x)2 + ( y - 7)’ > n2 ,
B(P,
,
(2.30) (2.31)
P, q, 64.
Setting x = ep cos q ,
y=ePsinq,
(2.32), (2.33)
xi=e”cosq,
y=e”sinq,
(2.34), (2.35)
we obtain C(P, PT 4, 4,4
= W(P,
qh WP, 41,
1 = i 7rn
fore2P+e2~-2ep+pcoS(q-~)~~2,
2
e 2P +
e2F _
zep+P
q),
j%%
for e2p + e2P - 2ep+P cos
cos (q
9),4
(2.36)
10 Using
Y(P,
-
4)
= e2F[e2fp-pt = e’@[(
p
-
+ p)”
(q
- 4) > n2 .
1 - 2epep cos + (q
-
4)2
+
(q
-
**-3 )
i)]
(2.37)
we have for yt2<< 1, C(p, P, 4, 4, n) =
WP,
1 2
= / ?T?z 0
I
Pt 97 47 4
for e2P[( p
-
ji)”
+ (q
-
tj)*]
<
n*
,
(2.38) fore2P[(p-j)2+(q-q)2]>n2.
U. Heise, Singular and fundamental
216
solutions to problems for a Riemann surface
With the definition e-2P
E(p, p, 4, $3 n) = lTn2 i 0
for(p-P)2+(q-~)2<122,
(2.39)
for(p-fi)*+(q-q)*>12’,
we find at last 6(x -
x, y -
y) = ljFo B(x,
x, y, y,n) = FyoC(p, p, 4,
4,4
= liio D( p, p, q, 434 = liio Jqp, $747434 = e-*’
liio B( p, p, q, 4, n) = e-*’ 6 *(p - p, q - 4)
(2.40) (2.41) (2.42) 2.3. Transformation of the differential equations From (2.1), (2.25) and (2.41) follows (2.43)
A*F=a*(p-p,q-q), and from (2.12), (2.13), (2.26) and (2.42): 11
(2.44), (2.45)
with I”=a;,=a:,.l
Using (2.2), (2.28) and (2.41) we find e-2P A*(A* _ 4 ap+4)F=6*(p-p,q-q),
(2.46)
epp e -’ A*(A* - 4a, + 4)F = 6 *(p - p, q - 4) ,
(2.47)
and setting (2.48)
-1 we have epp A*(A* - 4a, + 4)G ep = S *( p - j, (A* -2a,
q - 4)
+ l)(A* + 2a, + l)G = S*(p -p,
j (&2a,+i)(i+2a,-+l)G=@,@).)
,
q - 4))
(2.49) (2.50) (2.51)
U. Heise,
A possible method
Singular and fundamental
solutions to problems for a Riemann surface
of solving (2.51) would be to superimpose
(i - 2~3, + l)(i
+ 2~3, + l)G = 0
217
suitable singular solutions of
for k # 0.
(2.52)
2.4. Solutions to the equations The solution to (2.44) is well-known: F=Zl;;lnk.
(2.53)
Hence, the fundamental
solution to the potential
equation for the Riemann
surface reads as (2.54)
Setting G = G, = e*;‘@f ,
(2.55)
(2.52) changes into (2.56) i.e., functions of the form f* sin @, f. sin 4. In k , Differentiation
cos
cosGwith if = 0 are solutions, e.g.,
q”.ln k.
(2.57)
of (2.57) with respect to G or p” yields further solutions:
cos 4”In k + $ sin 4” , 7p” sin 4” , k By linear combination 4” sin . q k
_Z
2
sin 4 In k -I -$ cos @ ,
p” 1 cos 4” .
(2.58)
k
we find that
s” cos s”
7
k
(2.59)
are also solutions. Setting G=
(;,=,*~h,
(2.60)
(2.52) changes into ektP (i + 2 & 49(i
+ 2)h = 0 ,
(2.61)
U. H&e,
218
Singular and fundamental
solutions to problems for a Riemann surface
i.e., functions of the form h *sinh p”, h * cash p” with (i + 2)h = 0 are solutions, sinh p”. N,(fik)
cash p” - &,(A/%)
,
where N,, is a Bessel function. solutions:
Differentiation
(2.62)
,
of (2.62) with respect to 6 and p” yields further
d,-(sinh p”- NJV%))
,
+(cosh p” - N,(V%))
,
a,-(sinh p”* N,(tik))
,
a,-(cash F - N,(V%))
,
i(sinh p” * N,(V%)) By linear combination
i(cosh p” - No@‘%))
,
e.g.,
(2.63)
.
we find that cash p”. a,&(~‘%)
are also solutions. We are particularly linearly independent:
interested
G, = cos 4”In k,
(2.64)
in those solutions which are symmetrical
G, = 3
sin 4”,
in p” and in 4” and
G, =ap (; sinh p”- N,(fik))
,
(2.65) G, = ; cash p”. N,(fik)
Since N,(V?k)
,
G, = ii (5
cash p”. N,,(fik))
.
diverges as (~/IT ) In k one could believe at first sight that cos 4 In k
G=;coshp”.N,((x’%)-&
(2.66)
satisfies (2.51), i.e., represents a fundamental solution to the bipotential equation for the Riemann surface. However, closer inspection shows that this is not the case. Retransformation of (2.66) yields, with (2.48), F = &(I+’ + ?)N,(fik)
- g
cos (cp - Cp)In k2 .
I
(2.67) I
Starting from (2.67) we will establish in Section 3 a fundamental solution to the equation for the elasticity problem. For this purpose we will preliminarily use in place of N,(fik) a function N(fik2) which fulfills (2.70): N(fik2)
= 2N@k)
,
(2.68)
2(i + 2)N,,(v’??k) = (a,,= ; ak+ 2) N(fik2) =22/2N’+8k2N”+2diiN’+2N=0,
(2.69)
U. He&e, Singular and fundamental
*N”(Z)
Another
+
interesting
N’(z) + g
4
solutions to problems for a Riemann surface
219
NC.4= 0.
(2.70)
solution of (2.52) is
(2.71) Retransformation
of (2.71) yields together _
F=--&
s(p-+)sin(q-+)-XT
with (2.48):
v2 r2 - r2
In ; * N~(t/zk)
, -w
i
I
(2.72) This solution to the bipotential equation for the Riemann fundamental solution, which, however, has to be proved.
surface possibly
represents
a
3. Fundamental solution of the plane elasticity problem equation (Navier’s equilibrium equation) for a logarithmic Riemann surface 3. f . ~r~~~o~~~~~u~
of ~avier~s equation
We consider Navier’s equation for the plane elastostatic problem (plane stress) with a single force as the load:
u, and u, are Cartesian components of the displacement vector, G is the shear modulus and Y = 1 Im is Poisson’s ratio. We replace the Cartesian vector components by radial and tangential components ~1,and Us (see Fig. 7) and transform the differential operators on the left-hand sides of (3.1) and (3.2) into polar co-ordinates:
(3.3)
(34
U. Heise, Singular and fundamental solutions to problems for a Riemann surface
220
Fig. 7. Displacement vector.
With the aid of p=lnr,
a;, = A(a’PP -a,) g
9=4p,
T r
1. a2 g
ov
J_a2
=
g
i
44 ’
a,=
1 ap, r
-la =$a,, r* p r
(3.5), (3.6), (3.7), (3.8) (3.9), (3.10), (3.11)
1 -.l=L.l,
the differential e
(3.12)
r*
2
-2p
operators 5
(3.3), (3.4) are transformed
into
+ ai, - 21~~ + [a’,, - 3a$,)
{m(l2G
-+ I-+24,“, + [a’,, + a,@,))
2
(3.13) e -2P
5
+
k4a’,,
+
3 a,14 + [a’,, + 2 a2,,-
WJ
m;,- a,14 + r-a”,,+ llu,>>
(3.14)
and Dirac’s delta function (see (2.41)) into (3.15)
6(x - X, y - y) = e -‘“s*(p-p,q-q). By the final transformation p”=p-p.
(3.16), (3.17)
4”=4-4,
we obtain for Navier’s equation
--&
{m([2a',s
=-s"(p",&
+a&
-2]24,+
[a& -3e+,)+
(--a&u,+[a&+a,]u,>} (3.18)
U. Heise, Singular and fundamental
---&
{m([a;,
+ 3a+,
+ [a&
solutions to problems jh
+ 2a&
- 11~~ + ([a’,,
a Riemann surface
- a+,
+ r-a’,
+
221
llu,)>
= -&p, 4”).
(3.19)
It is remarkable that the main parts (i.e., the derivatives of second order) of the differential operators in (3.18), (3.19) and the differential operators in (3.1), (3.2) are analogous. 3.2. Method to derive the fundamental
solution to Navier’s equation from a solution to the
bipotential equation
To obtain the fundamental solution to Navier’s equation (3.18), (3.19) for the Riemann surface by a standard method is, to say the least, complicated and the author did not succeed in it. It is simpler to derive the solution directly from the solution (2.67) to the bipotential equation for the Riemann surface. However, since the author has been guided by intuition the method is not ‘systematic’ and to understand it detailed knowledge of singular solutions (dislocations, dipoles) of the elastic problem is necessary [9,10]. We show in this section how the well-known fundamental solution (3.20) (Kelvin’s solution) to the ‘classical’ Navier equation (3. l), (3.2) can be constructed with the aid of the fundamental solution pz In pl(87~) to the bipotential theory. In order to find essential parts of the solution for the Riemann surface we must only proceed analogously in the next section. In this section we partially apply the subscript notation for tensorial quantities (the subscripts cover the numbers 1 and 2) and use some formulae given in [9]. The radius vectors are explained by Fig. 5. The fundamental solution to Navier’s equation (3.1), (3.2) (Kelvin’s solution) is x”.x”.
-( 3m - l)Sij In p + (m + 1) + P =- 4i
&
A
{ -(3m
+ + 6, I
I
- l)Sij In p - (m + 1) ei’x~~xm
+ (m + z)C)
(3.20)
The constant term in (3.20) is unimportant, yields for m = 1:
eij is the permutation
tensor. In particular,
(3.20)
(3.21)
222
U. Heise, Singular and fundamental solutions to problems for a Riemann surface
In reverse, the general solution (3.20) can easily be obtained from the particular case (3.21). One has only to multiply the first term by (3m - 1)/(2m) and the second term by (m + l)/ (2m) Airy’s stress function (xS). of the elastic field caused by a wedge dislocation is, apart from a constant factor, identical with the bipotential solution [9]: (x8),
ZE
-g
.
%
;
(3.22)
p*ln p.
Airy’s stress function (XC). j describing the elastic field caused by an edge dislocation (the edge dislocation vector can be interpreted as a dipole of wedge dislocations) is obtained from (3.22) by differentiation with respect to the source point variable Xl [9]:
;
G m+l
(3.23)
-~y+P+;)[-‘,-].
e,,a, is the gradient vector rotated by 90”. By differentiation with respect to the field point variable x, (gradient rotated by 90”) we define the stress function vector (TC)~,~ which, too, describes the elastic field of an edge dislocation [9]: CTc)i.j
=
eik ak(Xc)‘j
’ - eik dkejm a,( X8)
(3.24)
.
(3.24) differs from (3.21) by a constant factor only. By comparison m=l 1 m (uR)i,j = - -2G2 m+l
eik akej,
we find, using (3.22),
am(xs)w
(3.25)
3.3. Ccmstruction of a solution to Navier’s equation for the Riemann surface The sought-for solution is a tensor which can be interpreted physically as the displacement field in an elastic Riemann surface caused by a single force. We relate the tensorial quantities
U. Heise,
Singular and fundamental
solutions to problems for a Riemann surface
223
to polar co-ordinates (not to Cartesian co-ordinates). u,, -uV, R,, R, are the radial and tangential components of the displacement vector at the field point and of the force at the source point. The four tensor components of the solution are denoted by (uR),,, (uR)+, (uR>,,, (“R)$ F or example (uR),, represents the radial component of the displacement vector caused by a unit force in tangential direction (see Fig. 8). In order to construct a solution for an elastic Riemann surface consisting of a fictitious material with m = 1 we adapt formula (3.25) to polar co-ordinates:
(3.26) (3.27)
(3.28)
where F(r, f, cp, Cp)is, at least preliminarily, the Riemann surface. Given k* = (In ;)’ + (cp N(V%‘)
the solution (2.67) to the bipotential
equation for
Cp)’,
(3.29)
)
(3.30)
= 2N,(V%)
(2.67) reads -
F = &(r2 + f2)N(fik2)
& rrcos (40 - Cp)In
k2 .
(3.31)
We obtain from (3.28) and (3.31):
u
Fig. 8. Components
= ‘p luR lcpi’
of the displacement
vector
at the field point caused by a single force acting at the source point.
17. Heise, Singular and fundamental
224
solutions to problems for a Riemann surface
m=l
w% =J--
(2 (;
32G
+ f) [-P2N’(lm’)
- 4(47 - ~)W(vw)])
1 -(-$[-l+~(~-(p)2]cos((p-~)-~(p-~)sin(ip-~) + 8nG - In m=l
k2 cos(q
&
W),i =
- +)
(3.32)
,
(4(9 - Cp)[ +fi
; N’
(fik2)
+
2 (; + F) In b
ZV”(fik”)])
1 -(-~[-(,-,)-+,,ln;lcos((P-(P) +%TG + $ In ; sin (p - (p) - In
k2
(3.33)
sin (q - (p)> ,
m=l
(UR)~i = L 32G (4(~p-@)[+fi;N’(fik~)+2(;+ 1 +8nG -
(--$
[(v
- (P)-
$
f)ln
(P - 6)ln
;]
+$ln;sin(rp-+)+lnk’sin(q-(o)
>
cos(9
;N”(fik2)])
- (P)
(3.34)
,
m=l CUR>@
=
L 32G (2 (; + f) [ -fiN’(fik2) -4 (; - f) fl
-
In ; N’
4 (In ;)’
N”(fik2)]
(fik2))
+&(-~[-1+~(ln~)2]cos(rp-~)-lnk2cos(~-(P)). With the aid of (2.4)-(2.20),
eqs. (3.32)-(3.35)
+ g
cash
p”N’(fik’)
[ -/
+ g
sinh
p”N’(fik2)
[ 8
+ m
1
In
k2
change into
IT]
_&I
-
&
+ &
I.
-cos 4” -sin 4” sin 4 -cos 4”
(3.35)
$
$
cos 4” [ -$
‘71
sin 4” [ -;“I
{] (3.36)
U. Heise, Singular and fundamental solutions to problems for a Riemann surface
225
with vanishing right-hand sides instead of the &functions, the call (3*18), (3.19), homogeneous equations. By multiplication of these equations by m - 1, setting m = 1 and insertion of (3.36) we verify that (3.36) is a solution for the special case m = 1. It is important that the special properties of the function N are not used for the verification, i.e., N need not be a Bessel function but can be an arbitrary function of the argument V’2k’. It can also easily be shown that the last summand of (3.36) fulfills the homogeneous equations (3.18), (3.19) for m = -1. Hence, multiplication of this summand by an arbitrary factor (3.20) suggests the choice of (3m - 1) l(2m) as factor) and multiplication of the remaining six summands of (3.36) by the factor (m + 1)/(2m), which vanishes for m = - 1, yields a solution of the homogeneous equations (3.18), (3.19) for the particular case m = - 1. This solution holds good also for m = 1 since (3 0 1 - 1)/(2 + 1) = (1 + 1)/(2 *1). The considerations of Section 3.2 could lead us to expect that the solution obtained is valid for arbitrary values of m. Unfortunately, however, this is not true. We may superimpose on the solution just proposed which is valid for m = 1 and m = - 1, the ((m + 1)/(2m) - (3 m - 1) /(2m))-fold of any solution valid for m = -1. It can easily be seen that
wlz
fulfills the homogeneous
equations
kZN”(fik2)
+ -$
k2N”‘(tik2)
+ v%V”(t/Zk2)
(3.18), (3.19) for m = -1 if
N’(v’?k’)
= 0,
(3.38)
= 0 ,
(3.39)
i.e.,
4kzN~‘Ir(~kz)
Equation
= 0 .
+ 6~N’~‘(~k2)
(3.40)
(3.38) can be written as zN”+ N'=O
(3.41)
and has a solution (we do not need the function N but only its derivatives):
N&L IT z’ N”=_2
I Tr z2 ’
N’(t/Zk’)
N~(~k*)
= :
= -;
(3.42), (3.43)
,
--&
&
.
(3.44), (3.45)
It is important that (3.41) differs from the differential equation (2.70) for the Bessel function N occurring in the solution of the bipotential equation. From (3.36) and (3.37) we obtain in the manner just described a solution valid for arbitrary m:
226
U. He&e, Singular and fundamental
=
& -$(cash
solutions to problems for a Riemann surface
p” - cos 4”)
1 ‘(coshpcos+l + ~ITG k2
1
+ 16-G
In k2
- 1 -1 + &rG k2
-cos $ c sin @
;yjc
-sin 4” .P3m-1 -cos 4”I m -q”cosq” 4”sin 4”
p” sinh p” 4”cash @ -q”cosh p” psinh p”
m+l
---
(3.46)
m
Inserting into (3.46)
(3.49)
1 r2 - J2 =ZrF’
(3.50)
we obtain the solution in the final form given in Section 3.4. 3.4. The fundamental solution We give a solution to Navier’s equation for the Riemann shown that it is a fundamental solution): (uR),, (r@,i =--
(uR),, (“R)V$
surface (in Section 4.2 it will be
1 (9 - 13;)”
-
-2cos(cp
- 40)
-(q--5;)ln;
U. Heise,
Singular and fundamental
solutions to problems for a Riemann surface
[ -2E;P) - Cp)
+2sin(q
1 In + 16lTG
k2
-cos
1 +Kzk2
1
lr$;’
(qJ -
1
-sin(cp-6) 3~2-1 m -cos (cp - (P)
$q
sin (cp - (p)
r* _
227
72
L!
In ;
r7
$C
r2 + r2
-~(q-‘p)
(cp - 4)
qln;
1
2(1- m) m ’
sin (9 - Cp) -cos (q - (p) + 2(V - cp) cos (9 - (p) sin (cp - (p) [
(3.51)
I1 k* =
(In ;)’ + (cp -
cp)’,
(3.52)
-QJ<(p,‘p<+co.
4. Proof that the singular solution is a fundamental
solution
4.1. Auxiliary formulae
In order to consider under which circumstances the limits of certain integrals extended over a circular area F with radius p vanish, we introduce polar co-ordinates: p”=kcos$,
4”= k sin $ ,
dF=kdkd+.
(4.1), (4.2), (4.3)
We find -(I
-y
p4
p”l@’
k”‘y-P’ldk=O
cos”~sinY~d~~lim
dF=
kP
In k dF =
cosa $sinY cC,d+.lim
k*““‘lnkdk=O
ifa+y+l*O. We denote by Pkl ( p”, @) sums of terms p”ll@’with From (4.4) and (4.5) follows that lim p-0
lim p-0
P&, I (F) I (F)
4”) kP
dF=O
P&(p”,q”)lnkdF=O
(4.5) K, a,
y
non-negative
integers and (Y+ y 2
K.
ifK--+laO, ifK+l>O.
(4.7)
228
U. He&e, Singular and fundamental
solutions to problems for a Riemann surface
Using the abbreviation
Qho (p”, s”) = ijlpD (p”, 4”) =
I& (p”,4”)
(4.8)
kP
or Qho ( p”,@) = P”o1(p”, 4”) in k ,
(4‘9)
we write (4.6) and (4.7) as QLo(p”,q”)dF=O Differentiation
ifn+120.
with respect to p” or 4, respectively,
(4.10) decreases the index by one:
rct1
K+l
PO1
‘_
(Phl.lnk)‘=&i*lnk+$!,
k&+2 ’
(4.11), (4.12) (4.13), (4.14)
Finally we have
j%
(Q~o(~,~))‘~‘d~=O
f (F)
ifn-ma-l,
(4.15)
where II is the degree of the factor of the logarithm or n is the difference of the degrees of numerator and denominator. Obviously
(Pk(p”, i#‘-’
dF = 0 s
(4.16)
4.2. Proof for the solution to Navier’s equation for the elasticity problem Inserting the expansions -2
-4
-3
sinhj?=@+‘T++..,
coshp=l+pT+pG+...,
cos~=l-~+~-*-.’
cash ;--cos into (3.46), we obtain
-2 9
(4.17), (4.18)
-3 g4
2
4”= k-i2
sink=+_%_...,
p”” - g4 + I . . 24
(4.19,
(4.20) (4.21)
U. Heise,
Singular and fundamental
solution
to broiled
for a
p”” -6 (f4+... _P”%+i3+__* 2
+Pbl(p”,
q”)Ink+
61
G, k2
3
= S,G
R~e~n~
229
surface
i24”-+if3+ . . .
..li i;4- 4”4 +* II 2
2(1-
m)
m
6
4”) + up”>
s”),
(4.22)
(4.23)
We have split up the solution into two parts, the first one of which, S,(J?, G), is formally identical with the classical fundamental solution (3.20) to Navier’s equation (3-l), (3.2). Next, we decompose the operator of Navier’s equation (3.18), (3.19) for the Riemann surface into two summands 0, and 0,:
G 0, & m_l
-2m [ 3ma,--a,
-3m a,- + a< -m+l ’ I
(4.26)
U. Heise, Singular and fundamental
230
where 0, is the operator O,S,(p”, 4”) = -6
solutions to problems for a Riemann surface
of the classical Navier equation
(3.1), (3.2), i.e., (4.27)
4”).
In order to prove that our solution is a fundamental
solution we have to show that
(0, + O,)(S, + S,) = -s”(P, 4”) *
(4.28)
(O,+O,)(S,+S,)=O
(4.29)
Since fork#O,
it is only necessary for the following integral extended over a circle F at the origin, the radius p of which tends to zero, to be equal to unity:
;i
I (F)
(0, + O,)(S, + S,) dF = -1 ,
(4.30)
i.e., due to (4.27), that the integral J= lim (O,S, + O,S, + O,S,) dF I (F) P--*‘J
(4.31)
vanishes. 0, is a differential operator of the second order and 0, an operator of the first order. From (4.23), (4.24) and (4.31) together with (4.14), (4.15) it can be concluded that J=limp_o (F) [(Quo)” + (Quo)’ + (Quo)‘] dF i = lim p-0
I (F)
[Quo + Quo + Quo] dF = 0.
(4.32)
References W.F. Osgood, Lehrbuch der Funktionentheorie, Vol. I (Chelsea Publishing, Bronx, NY, 1928) Section 8. B.A. Fuchs and B.V. Shabat, Functions of a complex variable and some of their applications, Vol. 1 (Pergamon, Oxford, 1964). T.A. Cruse, Two dimensional BIE fracture mechanics analysis, Appl. Math. Modelling 2 (1978) 287-293. U. Heise, Spectra of integral operators for ellipse and hyperbola-shaped boundaries, J. Elasticity (1990) to appear. M.D. Snyder and T.A. Cruse, Boundary integral equation analysis of cracked anisotropic plates, Internat. J. Fracture 11 (1975) 315-328. H. Mews, Calculation of stress intensity factors for various crack problems with the boundary element method, in: C.A. Brebbia, W.L. Wendland and G. Kuhn, eds., Boundary Elements IX, Vol. 2 (Springer, Berlin, 1987) 259-278. Faltung von Distributionen. Teil 1: Zur Berechnung von Fundamentallosungen, Teil [71 N. Ortner, Regularisierte 2: Eine Tabelle von Fundamentallosungen, Z. Angew. Math. Phys. 31 (1980) 133-173. of fundamental solutions, in: C.A. Brebbia, ed., Topics in Boundary Element 181 N. Ortner, Construction Research, to appear. of the singularity method for the formulation of plane elastostatic boundary value [91 U. Heise, Application problems as integral equations, Acta Mech. 31 (1978) 33-69. of integral equations of the Rizzo type and of Kupradze’s functional IlO1 U. Heise, Systematic compilation equations for boundary value problems of plane elastostatics, J. Elasticity 10 (1980) 23-56.
IN APPLIED
MECHANICS
AND ENGINEERING
83 (1990) 211-230
SINGULAR AND FUNDAMENTAL SOLUTIONS TO POTENTIAL AND ELASTICITY PROBLEMS FOR A RIEMANN SURFACE Uhich HEISE institut fiir
Technische Mechanik,
Technische Hochschule,
D-5100 Aachen,
Fed. Rep. Germany
Received 20 February 1989 Revised manuscript received 23 August 1989
The solutions can be used as kernels of integral equations by the aid of which it will be possible to solve certain engineering problems advantageously. Fields of application of such integral equations and a survey of the paper are given in Section 1.
1. Introduction We consider a logarithmic Riemann surface, i.e., an infinitely extended helical surface with an infinite number of sheets (see e.g. [1,2]). The origin of the co-ordinates is a branch point of infinite order, The surface can be pictured as a spiral staircase with an infinite number of turnings, an infinite exterior radius and a vanishing interior radius. To illustrate this we may express the height z of a point of this staircase in cylinder co-ordinates by z = E *9. (The central angle cp is not limited to the interval 0-2~. t: is small since we deal with plane problems.) In this paper singular and fundamental solutions are given to the potential equation, to the bipotential equation and to Navier’s differential equation (equilib~um equation) for the plane elasticity problem. These solutions (not only the fundamental solutions) can be used as kernels of integral equations by the aid of which it should be possible to solve certain problems advantageously. In the following we outline some promising fields of application. Integral equations of the considered type will enable us to solve problems where the domains have overlapping parts (see Fig. l), e.g. certain helical springs. Other integral equations are definitely unsuited for such purposes. A body with a slit (see Fig. 2) can be easily treated, since points at the opposite sides of the slit facing each other have different co-ordinates. In conventional integral equations such points are identical and measures have to be taken to overcome the resulting difficulties [3,4]. A technically eminently important particular case of a slit is a crack (see Fig. 3). The proposed integral equations should be suited for the treatment of edge crack problems. However, for this purpose it is necessary first to investigate the singularity of the given solutions at the origin of the co-ordinates or to treat the crack tip separately. A standard integral equation method for the treatment of crack problems makes use of solutions for an infinite medium with a built-in crack [5,6]. The solutions for the Riemann surface have the advantage of being more simple. ~4S-78~/90/$03.50
0
1990 - Elsevier Science Publishers
B.Y. (North-Holland)
212
U. Heise, Singular and fundamental
Fig. 1. Domain with an overlapping part. The ‘pitch of the helical surface’ is drawn on an exaggerated scale.
solutions to problems for a Riemann surface
Fig. 2. Domain with a silt
The solutions under consideration are based on polar co-ordinates. Hence, the integral equations can possibly be approached analytically or annular sectors for circular sectors (see Fig. 4). Analytical solutions of integral equations are necessary for the testing of numerical methods. (In the case of the circular sector the singularity of the kernels at the origin of the co-ordinates might be troublesome.) Successful treatment of circular sectors perhaps opens up possibilities for the more easy solution of problems dealing with domains with corners. In this paper we consider plane problems. The solutions for the Riemann surface are based on polar co-ordinates. Possibly the subject can be generalized to three dimensions. One has to formulate the problems in cylinder co-ordinates without the restriction that the central angle cp covers only the interval 0-27~. Then one has to find a suitable transformation of the ‘Riemann space’, which penetrates itself, onto the full space and to construct singular solutions of the equation. The generalization is intellectually tempting since the human brain cannot envisage
Fig. 3. Domain with a crack.
Fig. 4. Annular
and circular sector.
U. Heise,
Singular and fundamental
solutions to problems for a Riemann surface
213
the 3-D solutions should have the same a space which penetrates itself. Nevertheless, advantages as those for plane problems: Bodies with a slit can be easily treated, and probably cracks also, if the tip line is straight. Helical elastic springs with arbitrary cross section, the neighbouring turnings of which are adjacent to each other, will not lead to difficulties, at least not in principal. In Section 2 singular solutions to the potential and the bi-potential problem are established. We start from the differential equations formulated in polar co-ordinates and use a transformation mapping the Riemann surface onto an infinite plane. Singular solutions of the differential equations in the transformed space are determined. Then these solutions are retransformed. Section 3 deals with the plane elastostatic problem. In Section 3.1 we transform Navier’s equation. The author did not succeed in finding directly a solution to the transformed equation by the aid of a standard method (see [7, S]), but tried to derive it from a solution to the bipotential equation. As a guiding rule for this purpose Section 3.2 shows how the ordinary Kelvin solution to Navier’s equation can be obtained from the ordinary r’ In rsolution to the bi-potential equation. Then, using the results of Section 3.2, however, not carrying out each step by strict analogy, a singular solution is established in Section 3.3 which satisfies the transformed Navier equation. This solution is retransformed in Section 3.4. In Section 4 it is shown that the solution to Navier’s equation represents a fundamental solution. The most economical possibility of exploiting numerically the formulae developed in this paper would be to modify the algorithm of an available computer program package. One has to exchange the kernel function in the program (e.g., in the case of the plane elasticity problem Kelvin’s solution) by the corresponding solution for the Riemann surface, or even better to provide in the program a branch with the solution for the Riemann surface parallel to a branch with the classical solution.
2. Solutions surface
to the potential
and to the bipotential
problem for a logarithmic
Riemann
The subject of this section is to establish solutions to the equations AF(x, y) = 6(x - i, y - y)
(2.1)
AAF(x, y) = 6(x - X, y - y)
(2.2)
and
for a Riemann 2.1.
surface.
Tranformation
of the differential operators
We start with the Laplace operator
for polar co-ordinates
r, cp
(2.3)
and introduce
the transformation:
U. Heise, Singular and fundamental
214
p=lnr,
p=lnF,
4’90,
r=ep,
q=(P,
f=e”,
9’47
(2.4), (2.5), (2.6), (2.7) (2.8),(2.9),(2.10),(2.11)
CP’G,
4”=q-g=cp-cp,
p^=p-p=1n;, r = ed = r ’
solutions to problems for a Riemann surface
(2.12), (2.13) (2.14), (2.15)
cp-cp’q”, 2
/k2=j?+q2=
(
1,;
)
+(cp-+)2,
(2.16) = sinh p” ,
r
r
= = cash p” + sinh p” ,
; =
r
(2.17), (2.18)
cash p” - sinh p” .
(2.19), (2.20)
With a,F=‘a
r
F,
P
afrF= L (a',, - d,)F r2
(2.21), (2.22)
7
t$+,F = d:4F,
(2.23)
AF = -$ (dtp + a’,,)F =
e-2P(at, + a’,,)F
(2.24)
field point
Fig. 5. Cartesian
and polar co-ordinates
in the untransformed field point
Fig. 6. Co-ordinates
in the transformed
space.
space.
U. He&e, Singular and fundamental
solutions to problems for a Riemann surface
21.5
we find AF=ePzP
(2.23,
A* = a2, + a;, , 1
b*F,
(2.27)
AAF = ePzp A” eT2’ A* F , A*(A* - 4~3,+ 4)F.
AAF=e-4P 2.2. ~ru~sfor~utio~
(2.26)
(2.28)
of Dirffc’s 6 -function
The a-function of the right-hand sides of (2.1) and (2.2) has to be transformed in the same way as the differential operators on the left-hand sides. The following gives a brief outline of the procedure. The &function can be defined as 1 B(x, x, y, 9, n) = nlt2 i 0 S(x -
for(x-X)2+(y-Y)2
x, y - jq = iii@B(x, x, y, y, n)
WP -6
q - S) = Fi
(2.29)
for (x - x)2 + ( y - 7)’ > n2 ,
B(P,
,
(2.30) (2.31)
P, q, 64.
Setting x = ep cos q ,
y=ePsinq,
(2.32), (2.33)
xi=e”cosq,
y=e”sinq,
(2.34), (2.35)
we obtain C(P, PT 4, 4,4
= W(P,
qh WP, 41,
1 = i 7rn
fore2P+e2~-2ep+pcoS(q-~)~~2,
2
e 2P +
e2F _
zep+P
q),
j%%
for e2p + e2P - 2ep+P cos
cos (q
9),4
(2.36)
10 Using
Y(P,
-
4)
= e2F[e2fp-pt = e’@[(
p
-
+ p)”
(q
- 4) > n2 .
1 - 2epep cos + (q
-
4)2
+
(q
-
**-3 )
i)]
(2.37)
we have for yt2<< 1, C(p, P, 4, 4, n) =
WP,
1 2
= / ?T?z 0
I
Pt 97 47 4
for e2P[( p
-
ji)”
+ (q
-
tj)*]
<
n*
,
(2.38) fore2P[(p-j)2+(q-q)2]>n2.
U. Heise, Singular and fundamental
216
solutions to problems for a Riemann surface
With the definition e-2P
E(p, p, 4, $3 n) = lTn2 i 0
for(p-P)2+(q-~)2<122,
(2.39)
for(p-fi)*+(q-q)*>12’,
we find at last 6(x -
x, y -
y) = ljFo B(x,
x, y, y,n) = FyoC(p, p, 4,
4,4
= liio D( p, p, q, 434 = liio Jqp, $747434 = e-*’
liio B( p, p, q, 4, n) = e-*’ 6 *(p - p, q - 4)
(2.40) (2.41) (2.42) 2.3. Transformation of the differential equations From (2.1), (2.25) and (2.41) follows (2.43)
A*F=a*(p-p,q-q), and from (2.12), (2.13), (2.26) and (2.42): 11
(2.44), (2.45)
with I”=a;,=a:,.l
Using (2.2), (2.28) and (2.41) we find e-2P A*(A* _ 4 ap+4)F=6*(p-p,q-q),
(2.46)
epp e -’ A*(A* - 4a, + 4)F = 6 *(p - p, q - 4) ,
(2.47)
and setting (2.48)
-1 we have epp A*(A* - 4a, + 4)G ep = S *( p - j, (A* -2a,
q - 4)
+ l)(A* + 2a, + l)G = S*(p -p,
j (&2a,+i)(i+2a,-+l)G=@,@).)
,
q - 4))
(2.49) (2.50) (2.51)
U. Heise,
A possible method
Singular and fundamental
solutions to problems for a Riemann surface
of solving (2.51) would be to superimpose
(i - 2~3, + l)(i
+ 2~3, + l)G = 0
217
suitable singular solutions of
for k # 0.
(2.52)
2.4. Solutions to the equations The solution to (2.44) is well-known: F=Zl;;lnk.
(2.53)
Hence, the fundamental
solution to the potential
equation for the Riemann
surface reads as (2.54)
Setting G = G, = e*;‘@f ,
(2.55)
(2.52) changes into (2.56) i.e., functions of the form f* sin @, f. sin 4. In k , Differentiation
cos
cosGwith if = 0 are solutions, e.g.,
q”.ln k.
(2.57)
of (2.57) with respect to G or p” yields further solutions:
cos 4”In k + $ sin 4” , 7p” sin 4” , k By linear combination 4” sin . q k
_Z
2
sin 4 In k -I -$ cos @ ,
p” 1 cos 4” .
(2.58)
k
we find that
s” cos s”
7
k
(2.59)
are also solutions. Setting G=
(;,=,*~h,
(2.60)
(2.52) changes into ektP (i + 2 & 49(i
+ 2)h = 0 ,
(2.61)
U. H&e,
218
Singular and fundamental
solutions to problems for a Riemann surface
i.e., functions of the form h *sinh p”, h * cash p” with (i + 2)h = 0 are solutions, sinh p”. N,(fik)
cash p” - &,(A/%)
,
where N,, is a Bessel function. solutions:
Differentiation
(2.62)
,
of (2.62) with respect to 6 and p” yields further
d,-(sinh p”- NJV%))
,
+(cosh p” - N,(V%))
,
a,-(sinh p”* N,(tik))
,
a,-(cash F - N,(V%))
,
i(sinh p” * N,(V%)) By linear combination
i(cosh p” - No@‘%))
,
e.g.,
(2.63)
.
we find that cash p”. a,&(~‘%)
are also solutions. We are particularly linearly independent:
interested
G, = cos 4”In k,
(2.64)
in those solutions which are symmetrical
G, = 3
sin 4”,
in p” and in 4” and
G, =ap (; sinh p”- N,(fik))
,
(2.65) G, = ; cash p”. N,(fik)
Since N,(V?k)
,
G, = ii (5
cash p”. N,,(fik))
.
diverges as (~/IT ) In k one could believe at first sight that cos 4 In k
G=;coshp”.N,((x’%)-&
(2.66)
satisfies (2.51), i.e., represents a fundamental solution to the bipotential equation for the Riemann surface. However, closer inspection shows that this is not the case. Retransformation of (2.66) yields, with (2.48), F = &(I+’ + ?)N,(fik)
- g
cos (cp - Cp)In k2 .
I
(2.67) I
Starting from (2.67) we will establish in Section 3 a fundamental solution to the equation for the elasticity problem. For this purpose we will preliminarily use in place of N,(fik) a function N(fik2) which fulfills (2.70): N(fik2)
= 2N@k)
,
(2.68)
2(i + 2)N,,(v’??k) = (a,,= ; ak+ 2) N(fik2) =22/2N’+8k2N”+2diiN’+2N=0,
(2.69)
U. He&e, Singular and fundamental
*N”(Z)
Another
+
interesting
N’(z) + g
4
solutions to problems for a Riemann surface
219
NC.4= 0.
(2.70)
solution of (2.52) is
(2.71) Retransformation
of (2.71) yields together _
F=--&
s(p-+)sin(q-+)-XT
with (2.48):
v2 r2 - r2
In ; * N~(t/zk)
, -w
i
I
(2.72) This solution to the bipotential equation for the Riemann fundamental solution, which, however, has to be proved.
surface possibly
represents
a
3. Fundamental solution of the plane elasticity problem equation (Navier’s equilibrium equation) for a logarithmic Riemann surface 3. f . ~r~~~o~~~~~u~
of ~avier~s equation
We consider Navier’s equation for the plane elastostatic problem (plane stress) with a single force as the load:
u, and u, are Cartesian components of the displacement vector, G is the shear modulus and Y = 1 Im is Poisson’s ratio. We replace the Cartesian vector components by radial and tangential components ~1,and Us (see Fig. 7) and transform the differential operators on the left-hand sides of (3.1) and (3.2) into polar co-ordinates:
(3.3)
(34
U. Heise, Singular and fundamental solutions to problems for a Riemann surface
220
Fig. 7. Displacement vector.
With the aid of p=lnr,
a;, = A(a’PP -a,) g
9=4p,
T r
1. a2 g
ov
J_a2
=
g
i
44 ’
a,=
1 ap, r
-la =$a,, r* p r
(3.5), (3.6), (3.7), (3.8) (3.9), (3.10), (3.11)
1 -.l=L.l,
the differential e
(3.12)
r*
2
-2p
operators 5
(3.3), (3.4) are transformed
into
+ ai, - 21~~ + [a’,, - 3a$,)
{m(l2G
-+ I-+24,“, + [a’,, + a,@,))
2
(3.13) e -2P
5
+
k4a’,,
+
3 a,14 + [a’,, + 2 a2,,-
WJ
m;,- a,14 + r-a”,,+ llu,>>
(3.14)
and Dirac’s delta function (see (2.41)) into (3.15)
6(x - X, y - y) = e -‘“s*(p-p,q-q). By the final transformation p”=p-p.
(3.16), (3.17)
4”=4-4,
we obtain for Navier’s equation
--&
{m([2a',s
=-s"(p",&
+a&
-2]24,+
[a& -3e+,)+
(--a&u,+[a&+a,]u,>} (3.18)
U. Heise, Singular and fundamental
---&
{m([a;,
+ 3a+,
+ [a&
solutions to problems jh
+ 2a&
- 11~~ + ([a’,,
a Riemann surface
- a+,
+ r-a’,
+
221
llu,)>
= -&p, 4”).
(3.19)
It is remarkable that the main parts (i.e., the derivatives of second order) of the differential operators in (3.18), (3.19) and the differential operators in (3.1), (3.2) are analogous. 3.2. Method to derive the fundamental
solution to Navier’s equation from a solution to the
bipotential equation
To obtain the fundamental solution to Navier’s equation (3.18), (3.19) for the Riemann surface by a standard method is, to say the least, complicated and the author did not succeed in it. It is simpler to derive the solution directly from the solution (2.67) to the bipotential equation for the Riemann surface. However, since the author has been guided by intuition the method is not ‘systematic’ and to understand it detailed knowledge of singular solutions (dislocations, dipoles) of the elastic problem is necessary [9,10]. We show in this section how the well-known fundamental solution (3.20) (Kelvin’s solution) to the ‘classical’ Navier equation (3. l), (3.2) can be constructed with the aid of the fundamental solution pz In pl(87~) to the bipotential theory. In order to find essential parts of the solution for the Riemann surface we must only proceed analogously in the next section. In this section we partially apply the subscript notation for tensorial quantities (the subscripts cover the numbers 1 and 2) and use some formulae given in [9]. The radius vectors are explained by Fig. 5. The fundamental solution to Navier’s equation (3.1), (3.2) (Kelvin’s solution) is x”.x”.
-( 3m - l)Sij In p + (m + 1) + P =- 4i
&
A
{ -(3m
+ + 6, I
I
- l)Sij In p - (m + 1) ei’x~~xm
+ (m + z)C)
(3.20)
The constant term in (3.20) is unimportant, yields for m = 1:
eij is the permutation
tensor. In particular,
(3.20)
(3.21)
222
U. Heise, Singular and fundamental solutions to problems for a Riemann surface
In reverse, the general solution (3.20) can easily be obtained from the particular case (3.21). One has only to multiply the first term by (3m - 1)/(2m) and the second term by (m + l)/ (2m) Airy’s stress function (xS). of the elastic field caused by a wedge dislocation is, apart from a constant factor, identical with the bipotential solution [9]: (x8),
ZE
-g
.
%
;
(3.22)
p*ln p.
Airy’s stress function (XC). j describing the elastic field caused by an edge dislocation (the edge dislocation vector can be interpreted as a dipole of wedge dislocations) is obtained from (3.22) by differentiation with respect to the source point variable Xl [9]:
;
G m+l
(3.23)
-~y+P+;)[-‘,-].
e,,a, is the gradient vector rotated by 90”. By differentiation with respect to the field point variable x, (gradient rotated by 90”) we define the stress function vector (TC)~,~ which, too, describes the elastic field of an edge dislocation [9]: CTc)i.j
=
eik ak(Xc)‘j
’ - eik dkejm a,( X8)
(3.24)
.
(3.24) differs from (3.21) by a constant factor only. By comparison m=l 1 m (uR)i,j = - -2G2 m+l
eik akej,
we find, using (3.22),
am(xs)w
(3.25)
3.3. Ccmstruction of a solution to Navier’s equation for the Riemann surface The sought-for solution is a tensor which can be interpreted physically as the displacement field in an elastic Riemann surface caused by a single force. We relate the tensorial quantities
U. Heise,
Singular and fundamental
solutions to problems for a Riemann surface
223
to polar co-ordinates (not to Cartesian co-ordinates). u,, -uV, R,, R, are the radial and tangential components of the displacement vector at the field point and of the force at the source point. The four tensor components of the solution are denoted by (uR),,, (uR)+, (uR>,,, (“R)$ F or example (uR),, represents the radial component of the displacement vector caused by a unit force in tangential direction (see Fig. 8). In order to construct a solution for an elastic Riemann surface consisting of a fictitious material with m = 1 we adapt formula (3.25) to polar co-ordinates:
(3.26) (3.27)
(3.28)
where F(r, f, cp, Cp)is, at least preliminarily, the Riemann surface. Given k* = (In ;)’ + (cp N(V%‘)
the solution (2.67) to the bipotential
equation for
Cp)’,
(3.29)
)
(3.30)
= 2N,(V%)
(2.67) reads -
F = &(r2 + f2)N(fik2)
& rrcos (40 - Cp)In
k2 .
(3.31)
We obtain from (3.28) and (3.31):
u
Fig. 8. Components
= ‘p luR lcpi’
of the displacement
vector
at the field point caused by a single force acting at the source point.
17. Heise, Singular and fundamental
224
solutions to problems for a Riemann surface
m=l
w% =J--
(2 (;
32G
+ f) [-P2N’(lm’)
- 4(47 - ~)W(vw)])
1 -(-$[-l+~(~-(p)2]cos((p-~)-~(p-~)sin(ip-~) + 8nG - In m=l
k2 cos(q
&
W),i =
- +)
(3.32)
,
(4(9 - Cp)[ +fi
; N’
(fik2)
+
2 (; + F) In b
ZV”(fik”)])
1 -(-~[-(,-,)-+,,ln;lcos((P-(P) +%TG + $ In ; sin (p - (p) - In
k2
(3.33)
sin (q - (p)> ,
m=l
(UR)~i = L 32G (4(~p-@)[+fi;N’(fik~)+2(;+ 1 +8nG -
(--$
[(v
- (P)-
$
f)ln
(P - 6)ln
;]
+$ln;sin(rp-+)+lnk’sin(q-(o)
>
cos(9
;N”(fik2)])
- (P)
(3.34)
,
m=l CUR>@
=
L 32G (2 (; + f) [ -fiN’(fik2) -4 (; - f) fl
-
In ; N’
4 (In ;)’
N”(fik2)]
(fik2))
+&(-~[-1+~(ln~)2]cos(rp-~)-lnk2cos(~-(P)). With the aid of (2.4)-(2.20),
eqs. (3.32)-(3.35)
+ g
cash
p”N’(fik’)
[ -/
+ g
sinh
p”N’(fik2)
[ 8
+ m
1
In
k2
change into
IT]
_&I
-
&
+ &
I.
-cos 4” -sin 4” sin 4 -cos 4”
(3.35)
$
$
cos 4” [ -$
‘71
sin 4” [ -;“I
{] (3.36)
U. Heise, Singular and fundamental solutions to problems for a Riemann surface
225
with vanishing right-hand sides instead of the &functions, the call (3*18), (3.19), homogeneous equations. By multiplication of these equations by m - 1, setting m = 1 and insertion of (3.36) we verify that (3.36) is a solution for the special case m = 1. It is important that the special properties of the function N are not used for the verification, i.e., N need not be a Bessel function but can be an arbitrary function of the argument V’2k’. It can also easily be shown that the last summand of (3.36) fulfills the homogeneous equations (3.18), (3.19) for m = -1. Hence, multiplication of this summand by an arbitrary factor (3.20) suggests the choice of (3m - 1) l(2m) as factor) and multiplication of the remaining six summands of (3.36) by the factor (m + 1)/(2m), which vanishes for m = - 1, yields a solution of the homogeneous equations (3.18), (3.19) for the particular case m = - 1. This solution holds good also for m = 1 since (3 0 1 - 1)/(2 + 1) = (1 + 1)/(2 *1). The considerations of Section 3.2 could lead us to expect that the solution obtained is valid for arbitrary values of m. Unfortunately, however, this is not true. We may superimpose on the solution just proposed which is valid for m = 1 and m = - 1, the ((m + 1)/(2m) - (3 m - 1) /(2m))-fold of any solution valid for m = -1. It can easily be seen that
wlz
fulfills the homogeneous
equations
kZN”(fik2)
+ -$
k2N”‘(tik2)
+ v%V”(t/Zk2)
(3.18), (3.19) for m = -1 if
N’(v’?k’)
= 0,
(3.38)
= 0 ,
(3.39)
i.e.,
4kzN~‘Ir(~kz)
Equation
= 0 .
+ 6~N’~‘(~k2)
(3.40)
(3.38) can be written as zN”+ N'=O
(3.41)
and has a solution (we do not need the function N but only its derivatives):
N&L IT z’ N”=_2
I Tr z2 ’
N’(t/Zk’)
N~(~k*)
= :
= -;
(3.42), (3.43)
,
--&
&
.
(3.44), (3.45)
It is important that (3.41) differs from the differential equation (2.70) for the Bessel function N occurring in the solution of the bipotential equation. From (3.36) and (3.37) we obtain in the manner just described a solution valid for arbitrary m:
226
U. He&e, Singular and fundamental
=
& -$(cash
solutions to problems for a Riemann surface
p” - cos 4”)
1 ‘(coshpcos+l + ~ITG k2
1
+ 16-G
In k2
- 1 -1 + &rG k2
-cos $ c sin @
;yjc
-sin 4” .P3m-1 -cos 4”I m -q”cosq” 4”sin 4”
p” sinh p” 4”cash @ -q”cosh p” psinh p”
m+l
---
(3.46)
m
Inserting into (3.46)
(3.49)
1 r2 - J2 =ZrF’
(3.50)
we obtain the solution in the final form given in Section 3.4. 3.4. The fundamental solution We give a solution to Navier’s equation for the Riemann shown that it is a fundamental solution): (uR),, (r@,i =--
(uR),, (“R)V$
surface (in Section 4.2 it will be
1 (9 - 13;)”
-
-2cos(cp
- 40)
-(q--5;)ln;
U. Heise,
Singular and fundamental
solutions to problems for a Riemann surface
[ -2E;P) - Cp)
+2sin(q
1 In + 16lTG
k2
-cos
1 +Kzk2
1
lr$;’
(qJ -
1
-sin(cp-6) 3~2-1 m -cos (cp - (P)
$q
sin (cp - (p)
r* _
227
72
L!
In ;
r7
$C
r2 + r2
-~(q-‘p)
(cp - 4)
qln;
1
2(1- m) m ’
sin (9 - Cp) -cos (q - (p) + 2(V - cp) cos (9 - (p) sin (cp - (p) [
(3.51)
I1 k* =
(In ;)’ + (cp -
cp)’,
(3.52)
-QJ<(p,‘p<+co.
4. Proof that the singular solution is a fundamental
solution
4.1. Auxiliary formulae
In order to consider under which circumstances the limits of certain integrals extended over a circular area F with radius p vanish, we introduce polar co-ordinates: p”=kcos$,
4”= k sin $ ,
dF=kdkd+.
(4.1), (4.2), (4.3)
We find -(I
-y
p4
p”l@’
k”‘y-P’ldk=O
cos”~sinY~d~~lim
dF=
kP
In k dF =
cosa $sinY cC,d+.lim
k*““‘lnkdk=O
ifa+y+l*O. We denote by Pkl ( p”, @) sums of terms p”ll@’with From (4.4) and (4.5) follows that lim p-0
lim p-0
P&, I (F) I (F)
4”) kP
dF=O
P&(p”,q”)lnkdF=O
(4.5) K, a,
y
non-negative
integers and (Y+ y 2
K.
ifK--+laO, ifK+l>O.
(4.7)
228
U. He&e, Singular and fundamental
solutions to problems for a Riemann surface
Using the abbreviation
Qho (p”, s”) = ijlpD (p”, 4”) =
I& (p”,4”)
(4.8)
kP
or Qho ( p”,@) = P”o1(p”, 4”) in k ,
(4‘9)
we write (4.6) and (4.7) as QLo(p”,q”)dF=O Differentiation
ifn+120.
with respect to p” or 4, respectively,
(4.10) decreases the index by one:
rct1
K+l
PO1
‘_
(Phl.lnk)‘=&i*lnk+$!,
k&+2 ’
(4.11), (4.12) (4.13), (4.14)
Finally we have
j%
(Q~o(~,~))‘~‘d~=O
f (F)
ifn-ma-l,
(4.15)
where II is the degree of the factor of the logarithm or n is the difference of the degrees of numerator and denominator. Obviously
(Pk(p”, i#‘-’
dF = 0 s
(4.16)
4.2. Proof for the solution to Navier’s equation for the elasticity problem Inserting the expansions -2
-4
-3
sinhj?=@+‘T++..,
coshp=l+pT+pG+...,
cos~=l-~+~-*-.’
cash ;--cos into (3.46), we obtain
-2 9
(4.17), (4.18)
-3 g4
2
4”= k-i2
sink=+_%_...,
p”” - g4 + I . . 24
(4.19,
(4.20) (4.21)
U. Heise,
Singular and fundamental
solution
to broiled
for a
p”” -6 (f4+... _P”%+i3+__* 2
+Pbl(p”,
q”)Ink+
61
G, k2
3
= S,G
R~e~n~
229
surface
i24”-+if3+ . . .
..li i;4- 4”4 +* II 2
2(1-
m)
m
6
4”) + up”>
s”),
(4.22)
(4.23)
We have split up the solution into two parts, the first one of which, S,(J?, G), is formally identical with the classical fundamental solution (3.20) to Navier’s equation (3-l), (3.2). Next, we decompose the operator of Navier’s equation (3.18), (3.19) for the Riemann surface into two summands 0, and 0,:
G 0, & m_l
-2m [ 3ma,--a,
-3m a,- + a< -m+l ’ I
(4.26)
U. Heise, Singular and fundamental
230
where 0, is the operator O,S,(p”, 4”) = -6
solutions to problems for a Riemann surface
of the classical Navier equation
(3.1), (3.2), i.e., (4.27)
4”).
In order to prove that our solution is a fundamental
solution we have to show that
(0, + O,)(S, + S,) = -s”(P, 4”) *
(4.28)
(O,+O,)(S,+S,)=O
(4.29)
Since fork#O,
it is only necessary for the following integral extended over a circle F at the origin, the radius p of which tends to zero, to be equal to unity:
;i
I (F)
(0, + O,)(S, + S,) dF = -1 ,
(4.30)
i.e., due to (4.27), that the integral J= lim (O,S, + O,S, + O,S,) dF I (F) P--*‘J
(4.31)
vanishes. 0, is a differential operator of the second order and 0, an operator of the first order. From (4.23), (4.24) and (4.31) together with (4.14), (4.15) it can be concluded that J=limp_o (F) [(Quo)” + (Quo)’ + (Quo)‘] dF i = lim p-0
I (F)
[Quo + Quo + Quo] dF = 0.
(4.32)
References W.F. Osgood, Lehrbuch der Funktionentheorie, Vol. I (Chelsea Publishing, Bronx, NY, 1928) Section 8. B.A. Fuchs and B.V. Shabat, Functions of a complex variable and some of their applications, Vol. 1 (Pergamon, Oxford, 1964). T.A. Cruse, Two dimensional BIE fracture mechanics analysis, Appl. Math. Modelling 2 (1978) 287-293. U. Heise, Spectra of integral operators for ellipse and hyperbola-shaped boundaries, J. Elasticity (1990) to appear. M.D. Snyder and T.A. Cruse, Boundary integral equation analysis of cracked anisotropic plates, Internat. J. Fracture 11 (1975) 315-328. H. Mews, Calculation of stress intensity factors for various crack problems with the boundary element method, in: C.A. Brebbia, W.L. Wendland and G. Kuhn, eds., Boundary Elements IX, Vol. 2 (Springer, Berlin, 1987) 259-278. Faltung von Distributionen. Teil 1: Zur Berechnung von Fundamentallosungen, Teil [71 N. Ortner, Regularisierte 2: Eine Tabelle von Fundamentallosungen, Z. Angew. Math. Phys. 31 (1980) 133-173. of fundamental solutions, in: C.A. Brebbia, ed., Topics in Boundary Element 181 N. Ortner, Construction Research, to appear. of the singularity method for the formulation of plane elastostatic boundary value [91 U. Heise, Application problems as integral equations, Acta Mech. 31 (1978) 33-69. of integral equations of the Rizzo type and of Kupradze’s functional IlO1 U. Heise, Systematic compilation equations for boundary value problems of plane elastostatics, J. Elasticity 10 (1980) 23-56.