Page 361 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes 22.2 Circular Harmonics

Laplace equation in cylindrical co-ordinates is given by

1 V 1 2 V 2 V
r = 0 ...(1)
r r r r 2 2 z 2
Assume that V is independent of co-ordinates z, we then have

1 V 1 2 V
r = 0 ...(2)
r r r 2 2
We now attempt to find a solution of this equation of the form.
F
r
V = F 1 ( ) ( ) ...(3)
2
Substituting this in (2), we have
2
r
F 1 ( ) d r dF 2 F 2 ( ) d F 1 ( )
r dr dr r 2 d 2 = 0 ...(4)
2
,
Multiplying by r and dividing by F F we have
1
2
2
2
1 2 d F 2 dF 2 1 d F 2
r 2 r = 1 n ...(5)
F 1 dr dr F d 2
1
Since L.H.S. is a function of r and the R.H.S. is a function of so each one of them is a constant.
We thus have the two solutions.
2
d F
2
1 n F ...(6)
d 2 1 = 0
and
2
2 d F 1 dF 2 2
r r n F 2 = 0 ...(7)
dr dr
The solutions are separable. The solution of (6) is given by
F 1 = A cosn B sin n ...(8)

Also it is easily verified that the solution of (7) is

F 2 = Cr n Dr n , if n 0 ...(9)
If n = 0, we have the solution

F 2 = C 0 log r D 0 ...(10)
Where A, B, C and D are arbitrary constants. The solution of Laplace equation in cylindrical co-
ordinates when V is independent of the co-ordinate z are called circular harmonics. The circular
harmonics are then

V = (A B )(C logr D ) degree zero
0 0 0 0
V = (A cosn B sin )(C r n D r n ) degree n ...(11)
n n n n

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