Page 359 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes
If we use cylindrical co-ordinates ( , , )r z given by
x = r cos
y = r sin ...(3)

z = z
Then 2 V in this co-ordinate system is given by

1 V 1 2 V 2 V
2 = r ...(4)
V r r r r 2 2 z 2
So Laplace differential equation in cylindrical co-ordinates is given by

1 V 1 2 V 2 V
r = 0
r r r r 2 z 2
2 V 1 V 1 V 2 V
or, 2 2 2 2 = 0 ...(5)
r r r r z

Here V is a function of r, and z. Let us suppose the solution of (5) as
V = R ( ) ( ) ( ) ...(6)
Z
r
r
r
Where ( )R r is a function of r, is a function of and Z is a function of z only. This method is
known as method of separation of variable. Substituting in (6) and dividing by R Z, we have
2
2
1 d R 1 dR 1 d 2 1 d Z
= ...(7)
2
R dr 2 Rr dr r 2 d 2 Z dz 2
Now the right hand side is only a function of z whereas L.H.S. is function of r and , so each side
must be constant i.e.
2
2
1 d R 1 dR 1 d 2 1 d z 2
R 2 dr Rr dr r 2 d 2 = z dz 2 ...(8)
2
Where is a negative constant. This gives us
2
1 d R 1 dR 1 d 2 2
R dr 2 rR dr r 2 d 2 = ...(9)

and
2
d Z 2 Z = 0
dz 2 ...(10)
The equation (9) can be rewritten as

2
r d R r dR 2 2 1 d 2
r
R dr 2 R dr = d 2 ...(11)
Keeping in view the same argument, we have from (11)

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