Page 402 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 23: Wave and Diffusion Equations by Separation of Variable

If we use the boundary conditions (ii) and (iii) we observe that (i) and (ii) do not constitute the Notes
solution as they give A = B = 0 i.e. X = 0 and hence V (x, t) = 0, which is absurd.
Using boundary conditions (ii) and (iii) the solution (iii) gives.

X 0 A 0 and X L 0 B sin L 0.

Now B 0 otherwise X 0 and hence V x ,t 0.
Therefore

sin L 0
or L n

n
or ,n 1,2,3,.....
L
Hence for each value of n.

n n 2 2 2 /L 2
c t
V , x t B sin xe
n n
L
are solution of (i) satisfying the given boundary condition. Therefore for each value of n, we
take the solution as

V n , x t V n , x t
n 1

n 2 2 2 2
or V n , x t B n sin . x e n c t /L ...(vi)
n 1 L
Using initial condition, we have

n
V x ,0 B n sin x f x
n 1 L
which gives

L
2 n
B n F ( )sin . x dx ...(vii)
x
L L
0
Thus (vi) with coefficient (vii) is the solution of one dimensional heat equation in (i).
Case II: Let L be the length of a uniform wire whose end x = 0 is kept at 0 temperature and other
end x = L is kept at constant temperature t and we have to obtain the temperature function of the
0
wire as t increases, the initial temperature being t .
1
Hence boundary conditions are
V 0,t 0 ...(viii)

V , L t t for all t ...(ix)
0
and initial condition is

V x ,0 t i ...(x)

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