Page 371 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes Therefore we have
2
V
2 = 0 ...(ii)
So Laplace equation expressed in spherical polar co-ordinates reduces to
2 2
V 2 V 1 V 1 V
r 2 r r r 2 2 r 2 tan r = 0 ...(iii)
The general solution of this equation can be written in the form

V = A r n B n n 1 P n (cos ) ...(iv)
n
n 0 r
The potential satisfies the boundary conditions

V = f ( ) when r 0 and Lt V 0 ...(v)
r
Potential in the Region outside the spherical surface

According to the second boundary condition of equation ( ), the potential may not be zero at
r = . Therefore in the region outside the spherical surface no positive powers of r are admissible
in the solution of Laplace s equation. Thus in the general solution we should have A = 0 and so
n
B
V = n n 1 n (cos ) for r > a ...(vi)
P
r
n 0
The coefficients B are to be determined. This can be done by making use of the first boundary
n
of equation ( ). Hence from (vi) we get
B
V = F ( ) f (cos ) n n 1 P n (cos ) ...(vii)
a
n 0
Let cos = u then
B
V = f ( ) n n 1 n ( ) ...(viii)
u
u
P
a
n 0
To obtain the value of the general coefficient B , we multiply both sides of equation (viii) with
n
P (u) and integrate with respect to u in between the limit 1 to +1 we obtain
n
1 1
B n 2
u
u
u
f ( )P n ( )du = n 1 [P n ( ) ]du
a
1 1
All other integrals vanish because of the orthogonal property of P (u).
n
1
1 2B n
u
u
f ( )P n ( )du = n 1
a (2n 1)
1
(2n 1) n 1
or B n = .a f ( ) P n (cos )sin d ...(ix)
2
0
This gives us the value of the coefficient B . Hence the potential outside the spherical surface is
n
given by equation (viii) with B given by equation (ix).
n
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