102 FUNDAMENTAL SOLUTIONS

+ 3

12

(b)

Figure 4.4 Dirac delta function (ap-

proximate)

:

(a) integration range = 7,

integration increment =0.1; (b) integ-

ration range = 40, integration

increment = 0.2

(c)

to be given by

Β

*

(χ

·

ο=

έΓ-ί?^)

βχρ[ίκ

'

ίί

"

χ)]ακ

'

which can be evaluated by contour integration to give

u*(x, ζ) = --sin (ΚΓ)

2K

where r is the positive distance between x and ξ. In Figures 4.5(a) and

(b) the integral (c) has been represented as a truncated integral which

is evaluated numerically. This demonstrates the numerical method of

finding u* which assumes that we know the form of the integrand of

equation (c).

4.7 THE METHOD OF IMAGES

So far we have a method of finding the fundamental solution in empty

space. In some circumstances our problem region may be confined in

FUNDAMENTAL SOLUTIONS 103

Figure 4.5 Approximate fundamental solution for the one-dimensional Helmholtz

equation, κ = 1.1: (a) integration range = 5, integration increment = 0.2;

(b) integration range = 20, integration increment = 0.2

some regular way and it may be more convenient to find a

fundamental solution specific to a region.

The simplest case is that of

a

semi-infinite space such as may occur

in a foundation or fluids problem. The surface of the soil or fluid may

make it more convenient to work with a semi-infinite space funda-

mental solution. We choose this solution to satisfy the boundary

condition on the interface identically; in this way we shall not need to

put elements on the surface when using the boundary integral

method.

To obtain these solutions it is necessary to consider their physical

interpretation. Consider for the moment a two-dimensional space

and take coordinates as in Figure 4.6. The interface is along the x

l

axis

and the normal to this is measured in the positive x

2

direction. We

shall require that du/δη = 0 on this interface.

The fundamental solution is the field due to a point source in

infinite space and so the field in the presence of our boundary x

2

= 0

will represent our half space fundamental solution. Consider a source

strength m at a point (ξ

ί9

—a).

104 FUNDAMENTAL SOLUTIONS

*2

Image strength m'

j

n

Interface

du

dn

=

0

Observation point

"* Source strength m

(ί,-α)

Figure 4.6 Derivation of

the

fundamental solution in a half

plane

The field

mw*

from this source will be reflected somehow at x

2

= 0,

depending on the boundary condition applied there. In order to

represent this reflection we place an image source of strength m'

behind the interface. The strength of

m!

required will depend on the

kind of interface to be represented. We may then forget the existence

of the surface (x

2

= 0) and adjust

m'

in order to satisfy the boundary

condition there. The observed field at any point (x

l9

x

2

) will then be

the superposition of the

field

from both sources (as we have assumed a

linear operator for our governing differential equation).

This field will be given by

w(x

1

) = mw*(x

1

,x

2

; ξ

ΐ9

-a) + m'u*(x

u

x

2

'^

u

a) (4.66)

where

u*(x

l9

x

2

; ξ

ΐ9

ζ

2

) is the empty space fundamental solution for a

source at (ξ

ΐ9

ξ

2

) observed at (x

1?

x

2

). Then applying the boundary

condition

du du

^- = τ— = 0 οηχ

2

= 0

en cx

2

(4.67)

we obtain

m

du*(

Xl

,0^

u

-a)

+ m

,^*(*i,0;^a)

= Q (4 6g)

dx

7

dx

0

which for the two-dimensional Laplace equation becomes

m —

[In

Jx\ + (x

2

-a)

2

] +

™'

—

[In

Jx\ + (x

2

+ 0)

2

]

= 0 on x

2

= 0,

V

x

x

(4.69)

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