Page 374 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 22: Solution of Laplace Differential Equation

Notes
2
2
1 d X 1 d Y 2
2 = 2 .
X dx Y dY
Here we have taken the negative constant because it suits the boundary conditions.
Therefore the corresponding differential equations are
2
2
d X 2 d X 2
2 X = 0 and 2 Y 0
dx dx
whose general solutions are
X = A cos x B sin x
y y
and Y = Ce De
Hence

y
x
V ( , ) = XY ( cos x B sin x )(Ce y De y ) ...(vii)
A
using boundary condition ( ), we get C = 0
Otherwise V as y and hence
x
y
V ( , ) = ( cosA x B sin x )e y . (we have put D = 1)
and using boundary condition (iii), we have
sin a = 0

n
or = (n 1, 2, 3,...)
a
Thus for each value of n, we have
n n y /a
x
V n ( , ) = B n sin xe (n 1,2,3,....) ...(viii)
y
a
and therefore for different values of n, the solution may be taken as

y
x
x
y
V ( , ) = V n ( , )
n 1
n n y /a
or ( , )V x y = B sin xe ...(ix)
n
a
n 1
Using boundary condition (iv), we have
n
V ( ,0) = B n sin x f ( )
x
x
a
n 1
which gives
a
2 n
B n = f ( ) sin xdx ...(x)
x
a a
0

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