Page 388 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 23: Wave and Diffusion Equations by Separation of Variable

u(x, 0) = f(x) ...(iii) Notes
u
and = g(x) ...(iv)
t t 0
It is obvious from the equation (i), that u is a function of x and t. Therefore we suppose that the
solution of equation is of the form by
u(x, t) = X(x)T(t)
or u(x, t) = XT(say) ...(v)
where X is a function of x only and T is that of t only.
Substituting this solution in (i), we have

2
2
1 d X 1 1 d T
. .
X dx 2 c 2 T dt 2
Now L.H.S. is a function of the independent variable x, while R.H.S. is a function of independent
variable t. Therefore both sides cannot be equal unless both reduce to a constant value. Hence
2
2
1 d X 1 1 d T 0 or or 2
2
. .
X dx 2 c 2 T dt 2
Therefore in the three cases, we have
2
2
d X d T
0, 0,
dx 2 dt 2
2
2
d X 2 X 0, d X 2 2 0,
c T
dx 2 dt 2
2
2
d X 2 X 0, d X 2 2 0
c T
dx 2 dt 2
The general solutions in the above three cases are
(a) X = Ax + B, T = Ct + D
(b) X = Ae x Be x , T = Ce ct De ct
(c) X = A cos x B sin x , T = cos ct D sin ct
Using boundary conditions and the solution (a), we have
u(0, t) = X(0) T(t) = 0

and u(L, t) = X(l) T(t) = 0
which gives either T(t) = 0 or X(0) = X(L)= 0
But T(t) 0 otherwise we get
u(x, t) = 0
Therefore X(0) = X(L) = 0
Using this in solution (a), we have
X(0) = B = 0
and X(L) = AL + B = 0
Giving A = B = 0. Hence X(x) = 0 and therefore u (x, t) = 0 which is absurd. This proves that (a)
cannot be solution of the wave equation (i).

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