100 FUNDAMENTAL SOLUTIONS

This can be transformed to a simpler form by taking spherical polar

coordinates and integrating twice to give

VTj(x)

= A-& Ρ^(η)ά8

(4.58)

8πΓ

^|

=

ι

The integral is now taken round a unit circle which has centre at the

source point (the origin here) and lying in the plane perpendicular to

the line joining

3c

to the origin at a distance r.

For a general source point ξ (4.58) becomes

Ufj

(3c,

t)

=

-,,1

-?, <j>

FT,

l

(f) ds (4.59)

For an isotropic material (4.59) may be evaluated to give the Kelvin

solution:

„

sft

a__>

(iz*.^-· .<*-«<*-«

4πμ|χ-|1 \4(l-v) " 4(1

—

v)

|3c —^|

\χ-ξ\

In the general case (4.59) must be evaluated numerically.

4.6 NUMERICAL SOLUTIONS (ANISOTROPIC BODIES)

We have seen that the determination of the fundamental solution

involves taking transforms of both sides of our original equation with

respect to the eigenfunctions φ

η

, and later inverting these trans-

forms.

These processes can only be performed in closed form for a very

limited number of special cases. When we have a governing equation

with different coefficients such as with anisotropic media, a numerical

approach for the evaluation of our solution becomes essential.

Suppose for simplicity we are working in rectangular Cartesian

coordinates, then the operator if may be represented as a polynomial

combination of the derivatives d/dx

i9

i = 1, . . ., n. We can write

ΞΡ

(

αι

έ

,α

°'

where a

0

, a

u

a

2

, . . . are constants independent of x. Then the

^

p

i

fl

'^

fl

»·^'-

'

(461)

FUNDAMENTAL SOLUTIONS 101

equation for the fundamental solution will be

<?(ι**(χ,ξ)) = δ(χ-ξ) (4.62)

Transforming this equation a sufficient number of times we obtain an

equation which for constant a

0

, a

u

a

2

is

π v./2 P(ßoAa\Ki, -a

2

KiKj, . . .) ü* (κ, ξ) = 1 (4.63)

(2π)

where n is the number of coordinate directions, hence

VKti-rT-.

Ϊ

(464)

P(a

0

,

ia

x

K

i9

-a

2

KiK

j9

. . .)

then the fundamental solution is given by the inverse transform:

-*

<*»- ί^ ί".

,^ί*-·.;"„■..,

d

«

«

4

·

65

>

In the case where a

0

, a

i9

. . . are functions of space the transfor-

mations must be done numerically, i.e. the integrations involved in the

transformations are performed numerically. Equation (4.65) will then

consist of tables of values of Μ*(Κ,£) for various values of the

parameter κ and the source coordinate ξ. The inverse transform

equation (4.65) then becomes the summation of

the

integrand for the

points where ü* has been calculated.

These numerical integrations will involve a truncation of

the

range

of integration of the integrals which in turn means that an approxi-

mation has been made to the Dirac delta function on the right-hand

side. Figures 4.4(a) and (b) show the representation of the δ function

for various integration ranges.

Example 4.7

Considering the equation of Example 4.6, i.e.

dx

we found the fundamental solution defined by

2

+ K

2

U = 0 (a)

^ +

κ

2

ι**

= δ(χ-ξ) (b)

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