# Seperation of Variables (double check please

1. May 5, 2006

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

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

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