# Partial DIfferential Equations

1. Oct 29, 2005

### stunner5000pt

Solve using separation of variables and find particular solution
$$\frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} - u = 0 \ for\ 0 <x<1, t >0$$
$$\frac{\partial u}{\partial t} (x,0) = 0$$
$$0,t) = u(1,t) = 0$$
to assume u(x,t) = X(x) T(t)
then $$\frac{X''}{X} = \frac{T'' - T}{T} = \lambda$$
solving for X(x) yields $$X(x) = C_{1} \cos{\sqrt{\lambda} x} + C_{2} \sin{\sqrt{\lambda} x}$$
and i get C1 = 0 and C2 assumed to be 1 and $\lambda = \sqrt{n \pi}$
now for T(t),
$$T'' - (1 + n^2 pi^2) T = 0$$
now here is where i am stuck...
$$C_{1} \cos{\sqrt{1 + n^2 \pi^2} t}+ C_{2} \sin{\sqrt{1 + n^2 \pi^2} t}$$
i'm not quite sure what to do after this
the answer in my book is $$\cos{\sqrt{n^2 \pi^2 -1}} t} \sin (n \pi x)$$
i got the x part right... but what about the t part

Last edited: Oct 29, 2005
2. Oct 29, 2005

### Physics Monkey

Hi stunner5000pt,

Sorry, I got all confused there, and I read your differential equation backwards. Your problem is simply that $$\lambda < 0$$ in order to obtain sines and cosines and meet the boundary condition for the x part. In reality, you should have $$- \lambda = n^2 \pi^2$$ and not $$\lambda = n^2 \pi^2$$.

Once you have solved the t part, apply the boundary condition you haven't used yet.

Last edited: Oct 29, 2005
3. Oct 29, 2005

### stunner5000pt

the DE is
$$\frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} - u = 0$$

2 negative signs

what is wrong with the equation?

4. Oct 29, 2005

### Physics Monkey

Yeah, my fault, I read it as
$$\frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial t^2} - u = 0,$$
and then realized my mistake as I was thinking about the solution.

5. Oct 29, 2005

### stunner5000pt

uh.. the t part was the part at which i was stuck.

the condition given is $$\frac{\partial u}{\partial t} (x,0) = 0$$
im not quite sure howto relate this to what i have already...

6. Oct 29, 2005

### Physics Monkey

First, as I said, your general solution to the t part in your first post is incorrect, it should be
$$T(t) = C_1 \sin{(\sqrt{n^2 \pi^2 -1} \,t )} + C_2 \cos{(\sqrt{n^2 \pi^2 - 1}\, t)}$$
Now, regarding the boundary condition, what do you mean you don't know how to relate it to what you have? Just take the derivative of $$u(x,t) = X(x)T(t)$$ with respect to t and set it equal to zero for t=0.

7. Oct 29, 2005

### stunner5000pt

why should the generla solution for T be that?

8. Oct 29, 2005

### Physics Monkey

Read my post 2 above, and try to figure it out.

9. Oct 30, 2005

### stunner5000pt

so why should $- \lambda = n^2 \pi^2$
i m not sure about the 'conditions on the x part' to which you are referring. And wouldnt the solution become imaginary if that was the case?

10. Oct 30, 2005

### Physics Monkey

What is the solution to
$$\frac{1}{X}\frac{d^2 X}{dx^2} = \lambda,$$
isn't it
$$X = C_1 e^{\sqrt{\lambda} x} + C_2 e^{-\sqrt{\lambda} x}.$$
you get real exponentials and not real sines and cosines if $$\lambda > 0$$. If you try applying the boundary conditions with real exponentials then you find $$C_1 = C_2 = 0$$. Take it from here.

11. Oct 30, 2005

### stunner5000pt

i get it now lambda mustb e imaginary iof you are going to use Eler's identity and expand E into cos and sin. ANd thus carries onto the T(t) part where C1-0 and C2 =1 and the solution is what appears in the book

thank you for your help! I appreciate it!