Energy conservation for the wave equation I recently encountered this problem in PDE class involving a concept I have never met, it states:

$ u_{tt} - u_{xx} = 0 ; \space \space 0 < x < 1 ; \space \space t > 0 $
$ u(0,t)=u(1,t)=0 $
$ u(x,0) = x(1-x) $
$ u_t (x,0) = 0 $
We are to use the law of energy conservation for this problem to determine the sum:
$$ \sum_{k=1}^{\infty} \frac{1}{(2k-1)^4} $$

All I can do here is separate variables to solve the problem via eigenfunctions expansions but I do not believe I have encountered the term Energy Conservation for the wave equation and how it can be used to find this sort of sum, I know about Green's functions and how they can be used to find the sums of inverse eigenvalues, so I need someone to help me with this in telling me what this term means and possibly direct me to a source or demonstration of this sort of solution. All help appreciated thanks
 A: We solve this by computing the Fourier series of $u = x(1-x)$ and then evaluating $\int_0^1 u_x^2 {\rm d}x$ (the energy) in two different ways.
The Fourier series of $u(x,0) = x(1-x)$ on $[0,1]$ is
$$u(x,0) = \frac{8}{\pi^3}\sum_{n=1}^\infty\frac{\sin[\pi (2n-1) x]}{(2n-1)^3}$$
The energy of the system is given as $H = \frac{1}{2}\int_0^1 u_x^2 + u_t^2{\rm d}x$ as derived in the other answer. At $t=0$ we have $u_t = 0$ so the second term does not contribute. 
First by using $u(x,0)=x(1-x)$ we find
$$H = \frac{1}{2}\int_0^1(1-2x)^2{\rm d}x = \frac{1}{6}$$
and secondly by using the Fourier series representation of the solution we also have
$$\begin{align}H &= \frac{32}{\pi^4}\int_0^1 \sum_{n=1}^\infty\sum_{m=1}^\infty \frac{\cos[\pi (2n-1) x]\cos[\pi (2m-1) x]}{(2n-1)^2(2m-1)^2} {\rm d}x \\&= \frac{16}{\pi^4}\sum_{n=1}^\infty \frac{1}{(2n-1)^4}\end{align}$$
Equating the two results gives us
$$\sum_{n=1}^\infty \frac{1}{(2n-1)^4} = \frac{\pi^4}{96}$$
As you can see this problem has little to do with the acctual PDE and energy conservation per se, and is more just an exercise in working with Fourier series.
A: For the first part note that 
$$
\frac{d}{dt}(u_t^2/2)=u_tu_{tt}=u_tu_{xx},
$$
then after an integration by parts and up to boundary terms, you will find the Hamiltonian
$$
H=\frac {1}{2}\int_0^1(u_t^2+u_x^2)dx
$$ 
or the energy, a conserved quantity. For the value of the sum I think you have to expand with respect to a suitable base of [0,1].
