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No credit card required 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
*
·
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
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
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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