Page 412 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 23: Wave and Diffusion Equations by Separation of Variable

The general solution is obtained by summing over all values of i.e. Notes

2 2
k th
u A e i J k r ...(xiv)
0
i
i
i 1
where the arbitrary constants A must be obtained from the initial conditions i.e. at t = 0, u = f (r).
i
Putting t = 0 in (xiv), we have
f r A J k r
i 0 i ...(xv)
i 1
Here A are now obtained as
i
a
2
A i 2 r f r J k r dr , i 1,2,.... ...(xvi)
i
0
2
a J k a 0
i
1
Example 1: Determine the solution of one dimensional heat equation under the following
boundary and initial conditions:
V ( ) V= ( ,L t ) 0 t= > 0
0,t
and V x ,0 x 0 x L where L is the length of the bar.
Solution: Proceeding as before for Case I; we have

n x n 2 2 2 /L 2
a t
V , x t B n sin .e
n 1 L
L
2 n
where B x .sin xdx
n
L L
0
Integrating by parts, we get
2 L
B n cosn
n
2L 1 n x n 2 2 2 /L 2
a t
Therefore V x ,t cosn sin .e
n L
n 1
Example 2: A rectangular plate bounded by the lines x = 0, y = 0, x = a, y = b has an initial
distribution of temperature given by.
x y
y
V x , ,0 A sin sin
a b
The edges are kept at zero temperature and the plane faces are impervious to heat. Find V at any
point and at a time.
Solution: We have the heat equation as

2V 2V 1 2V
x 2 y 2 c 2 t

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