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Evaluating the wave equation $u_{tt} = c^2 u_{xx}$ on $x \in (0,L)$ and $t \in (0,\infty)$ my lecture notes say:

Let us assume that the solution is of the form $$u(x,t) = X(x)T(t)$$ with two functions $X:(0,L) \to \mathbb{R}$ and $T:(0,\infty) \to \mathbb{R}$.

My question is where did they get this from? I understand how the solution to the wave equation can be split into $u(x,t) = f(x+ct) + g(x-ct)$ with functions $f,g:\mathbb{R} \to \mathbb{R}$ but I can't see where they got $X(x)T(t)$ (my printed lecture notes don't contain any intermediate steps or many proofs and I missed a fair few lecture due to illness). Also my lecture notes then say:

Then we obtain $$X''(x) - \lambda X(x) = 0, \quad T''(t) - c^2 \lambda T(t) = 0.$$

And I can't see how he got this either? I can get the result $$X(x)T''(t) = c^2X''(x)T(t)$$ when I differentiate $u(x,t) = X(x)T(t)$ with respect to $t$ and $x$ twice and then substitute back into the wave equation but I can't see where $\lambda$ comes from? It just appears with no explanation? Any help would be appreciated!

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1 Answer 1

up vote 7 down vote accepted

First, the separation of variables (a common technique) is an assumption that does not hold true for all of the solutions. Instead, it is only there to find a certain kind of simple solution (namely, the separable ones), and from there we can build more solutions by linearity (every linear combination of solutions will also be a solution, by linearity). Indeed, these linear combinations cannot generally themselves be expressed in the separated form, like how $a+b$ cannot be expressed as $f(a)g(b)$.

Second, the idea is separate into two sides:

$$\frac{T''(t)}{T(t)}=c^2\frac{X''(x)}{X(x)}=?$$

Notice that the LHS is a function of $t$ and does not depend on $x$, while likewise the middle is a function purely of $x$ and does not depend on $t$. Thus both sides do no depend on either $t$ or $x$; it is a fixed constant. Call the constant $c^2\lambda$ and put it where the $?$ is. After that, turn it into two separate equations and clear the denominators again to obtain $T''=c^2\lambda T$ and $X''=\lambda X$.

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Just to add a bit to your explanation, there is a really cool fact that $L^2(\mathbb{R}^{n+m}) \cong L^2(\mathbb{R}^n) \otimes L^2(\mathbb{R}^m)$, it blew my mind when I learned about it :) –  Alexei Averchenko May 7 '12 at 16:30
    
Thanks for your help! –  user26069 May 9 '12 at 11:51

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