Page 364 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 22: Solution of Laplace Differential Equation

and Notes

1 0
C = 0 d ...(8)
0 2 2
0
substituting into (6), we obtain
n
2C r sinn
r
n
( , ) = 0 0 .... for odd ...(9)
R n 2
n 1
Self Assessment
r
1. Find the potential u ( , ) in the exterior of a unit sphere satisfying the relation
2 u 1
r sin u 0
r r sin
under the conditions

u (1,0) = cos2

and lim ( , )u r = 0
r
22.2.1 Solution of Laplace s Equation in Spherical Polar Co-ordinates

The Laplace equation in spherical polar co-ordinates is given by

2 V V 1 V 1 2 u
r 2 2r sin = 0 ...(1)
r 2 r sin sin 2 2

we apply here a separation of variable s method and write the solution of (1) in the form
r
r
V ( , , ) = R ( ) ( ) ( ) ...(2)
where R is a function of r only, that of and that of only. Substituting in (1) we get

2
2
r d R 2r dR 1 d sin d sin 2 = 1 d 2 ...(3)
R dr 2 R dr sin 2 d d d 2
Since both sides are functions of different independent variables hence each side should be equal
2
to some constant. Let this constant be . Then equation (3) gives
d 2 = 0 ...(4)
d 2

2
2
r d R 2r dR 1 d d 2
and 2 = sin 2 ...(5)
R dr R R sinq d d sin
Again in (5) both sides are functions of different variables and hence both will be equal to a
constant say (n n 1). This gives us from (5)

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