# Seperation of Variables (double check please

1. May 5, 2006

### FrogPad

Seperation of Variables (double check please :)

I have a final coming up, and I want to make sure I have this method down.

Q: For the second-order wave equation $u_{tt}=u_{xx}$, the substitution of $u=A(x)B(t)$ will give second-order equations for A nd B when the x and t variables are seperated. From $B''/B=A''/A=\omega^2$, find all solutions of the form $u=A(x)B(t)$

Assume: $$u(x,t)=A(x)B(t)$$

$$\frac{A(x)B''(t)}{A(x)B(t)}=\frac{A''(x)B(t)}{A(x)B(t)}$$

$$\frac{B''}{B}=\frac{A''}{A}=-\omega^2$$ is a second order ODE of the form:

$$B''+\omega^2 B = 0$$

Solving yields:
$$B(t)=c_1 \cos \omega t + c_2 \sin \omega t$$ with the assumption that $$\omega > 0$$

and:
$$A(x)=d_1 \cos \omega x + d_2 \sin \omega x$$

therefore:
$$u(x,t) = A(x)B(t)= (c_1 \cos \omega t + c_2 \sin \omega t)(d_1 \cos \omega x + d_2 \sin \omega x)$$

And this is simply all the solutions right? It seems really straightforward, but sometimes when I think it is... it totally isn't. Thanks

Last edited: May 5, 2006
2. May 5, 2006

### HallsofIvy

Staff Emeritus
First, why are you assuming that the constant value is $-\omega^2$? Is there a boundary condition you didn't mention?

No, that is not "all the solutions". Any linear combination of solutions, for different $\omega$ is also a solution. The general sum would involve a sum over all possible values of $\omega$. What the possible values are goes back to that boundary condition I mentioned.

3. May 5, 2006

### FrogPad

The question is verbatim from the text. I used $-\omega^2$ as the assumpition, as I was told to in the question itself.

I'm not exactly sure what you are saying that any linear combination satisfies the equation. I see that any $\omega$ value satisfies u_tt = u_xx. Since there is no boundary condition can't $\omega$ be anything?

4. May 6, 2006

### FrogPad

Ahh.. Well there really isn't anything else in my book about seperation of variables, but I've read up a little more on them. I understand what you mean by the linear combination and the factor. Since the "terms" each will differentiate to 0 they can each have a constant in front and are indpendent of each other. Interesting.

My professor did not go into seperation of variables for more than 25 minutes the whole semester. We only had three problems to do also...

so anyways. Thanks for the comments man, it allowed me to read up on some interesting stuff.

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