Page 368 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 22: Solution of Laplace Differential Equation

For solution which exist for r = 0, then B = 0. Notes
The complete solution is given by summing over m or

n
m
µ
V = A r n A e im B e im P n m ( ) ...(19)
m 0,1,2,...
n 0,1,2,...
x
Since P n m ( ) involves mth derivative of P n ( ) which is polynomial of degree n, so for m > n
x
P n m ( ) = 0 ...(20)
m
for m > n. Defining S, the surface Harmonic by

µ
S = A e im B e im P m ( ) ...(21)
n
n
If S is independent of , then
n
dS
n
d = 0
So S has only m = 0 value hence
n
µ
S = P n ( ). In the case V becomes
n
V = (A r n B r n 1 )P ( ) For m = cos ...(22)
µ
n
n
Example 1: Gravitational Potential Due to Uniform Circular Ring
1 of a reference Cartesian
,
z
Let us consider a particle of mass m situated at a point (x y 1 , )
1
coordinate system, then the gravitational potential due to this mass at the point with coordinate
y
z
x
( , , ) is given by
mass m
V = ...(i)
distance {(x x 1 ) 2 (y y 1 ) 2 (z z 1 ) }
2
We know that potential V, satisfies Laplace equation
2
V = 0 ...(ii)
in matter free space.
Now, we have to calculate the gravitational potential at any point due to a uniform circular ring
of small cross-section, lying in the x y plane and with its centre situated at the point O, (Figure
22.3).
Obviously, the gravitational potential is symmetric about the z-axis and so it should be
independent of the angle . The potential V, therefore may be written with following form:

B
V = A r n n n 1 P n (cos ) ...(iii)
n
n 0 r

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