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Unit 22: Solution of Laplace Differential Equation

In most applications of circular harmonics, V is usually single-valued function of . So if we Notes
change by 2 , we reach the conclusion

r
r
V ( , 2 ) = V ( , ) ...(12)
It is necessary that n take integer values. So a general single valued solution of Laplace equation
is obtained by summing over n i.e.

V = a 0 logr (a n cosn b n sinn )r n 1 n (q n cosn p n sinn ) c 0 ...(13)
n 1 n 1 r

c
where a 0 , a n , b n , q and p and 0 are constants.
n
n
Example: Find the steady state temperature in the region inside a cylinder, the two
halves of the cylinder are ‘thermally insulated from each other, and the upper half of it is kept at
,
.
temperature 1 while the lower half is kept at temperature 2 It is assumed that cylinder is so
long in the z-direction that the temperate is independent of z.
Solution: To solve this problem, let ( , , , )r z t be the temperature that satisfies heat equation

= 2 ...(1)
t

In the steady state is independent of t so that we have to solve Laplace equation
2
= 0 ...(2)
in the region inside the cylinder and satisfy the boundary conditions

= 1 at r R 0 ...(3)
= 2 at r R 2
we do this by taking the general solution independent of z as, we have

= a 0 log r r n (a n cosn B n sinn ) c 0 r n 1 (q n cosn f n sinn ) ...(4)
n 1 n 1
and use the boundary conditions (3). We first see that the temperature must be finite

Figure 22.1

r

x

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