PDE: separation of variables problem

1. Nov 18, 2005

eckiller

I am to reduce the following PDE to 2 ODEs and find only the particular solutions:

u_tt - u_xx - u = 0; u_t(x,0) = 0; u(0,t) = u(1,t) = 0

I guess u = X(x)T(t), and plug u_tt, u_xx into PDE and divide by u to get:

T''/T = X''/X + 1 = K

I solve X'' + (1-K)X = 0 first.

From characteristic equation, r = +- sqrt(1-K)

Becausse of boundary conditions, we must have 1-K < 0

So r = +- i sqrt(K-1)

(I think I make an error somewhere around here...)

==> u = c1 cos(sqrt(K-1)x) + c2 sin(sqrt(K-1)x)

From boundary conditions, c1 =0, and sin(sqrt(K-1)) = 0 => sqrt(K-1) = n*pi

=> K-1 = n^2*pi^2
=> K = n^2*pi^2 + 1

Correct so far?

Now for T:

T'' - KT = 0

T'' - (n^2*pi^2 + 1)T = 0

r = +- sqrt(n^2*pi^2+1)

I think my K = n^2*pi^2 + 1 is wrong because it is strictly positive and I don't think an e^t solution will satisfy the initial condition.

The book's answer is cos( sqrt(n^2*pi^2-1)t) * sin(n*pi*x)

2. Nov 18, 2005

saltydog

When you separate variables and equate to a separation constant, usually need to check the boundary conditions against a positive, negative, and zero separation constant. For now, these types of equations just have a negative separation constant. That is:

$$\frac{T^{''}}{T}=\frac{X^{''}}{X}+1=K=-\lambda;\quad \lambda>0$$

So, need to solve:

$$\frac{X^{''}}{X}+1=-\lambda$$

or:

$$X^{''}+X(1+\lambda)=0$$

Solving for the roots:

$$m=\pm \sqrt{-(1+\lambda)}$$

Thus have:

$$X(x)=A_1Cos(\sqrt{1+\lambda}x)+A_2Sin(\sqrt{1+\lambda}x)$$

Substituting the boundary conditions yield:

$$\lambda=(n\pi)^2-1;\quad n>0$$

and:

$$X(x)=A_2Sin(n\pi x)$$

Can you do T now?

Edit: Also, I think you need another constraint to obtain a particular solution, you know, wave equations usually specify the initial velocity as well.

Last edited: Nov 18, 2005