I recently encountered this problem in PDE class involving a concept I have never met.
Consider the mixed problem for wave equation
\begin{align} u_{tt} &= u_{xx} & \text{in } (0,\infty) \times \Omega \\ u(t,0) &= u(t,1) = 0 & &\\ u(0,x) &= u_{0}(x) & & \\ u_{t}(0,x) &= v_{0}(x) & & \end{align}
Prove that
$$\frac 1 2 \|u_{t}(t) \|_{L^2}^{2} + \frac 1 2 \|u_{x}(t) \|_{L^2}^{2} = \frac 1 2 \|v_{0}\|_{L^2}^{2} + \frac 1 2 \|u_{0}\|_{L^2}^{2}$$
for all $t \ge 0$.
My approach
I know that the ODE governing the evolution of each component $\frac{d^2u_{n}}{ dt^2}=-λ_{n}u_{n}$ has the energy conservation property $\frac{d}{dt} (\frac 12 v_{n}(t)^2+ \frac{λ_{n}}{2}u_{n}(t)^2)=0$, where $v_{n}(t):=\frac{du_{n}}{dt}$
But how do I use this to prove the statement? All help appreciated, thanks.
From the boundary conditons I suppose your $\Omega$ is the interval $[0,1]\,.$ Apart from bad typesetting I also think that the correct form of the conservation equation should use the squares of the $L^2$-norms: $$ \frac{1}{2}||u_t||^2_{L^2}+\frac{1}{2}||u_x||^2_{L^2}=\text{const.} $$ Proof. The sum of kinetic and potential energy is \begin{align} E(t)=\frac{1}{2}\int_0^1u^2_t(t,x)+u^2_x(t,x)\,dx\,. \end{align} Therefore, \begin{align} \frac{d}{dt}E(t)&=\int_0^1u_{tt}(t,x)\,u_t(t,x)+u_{tx}(t,x)\,u_x(t,x)\,dx\\ &=\int_0^1u_{xx}(t,x)\,u_t(t,x)+u_{xt}(t,x)\,u_x(t,x)\,dx\,. \end{align} Integrating the second term by parts gives \begin{align} \frac{d}{dt}E(t) &=\underbrace{\int_0^1u_{xx}(t,x)\,u_t(t,x)-u_t(t,x)\,u_{xx}(t,x)\,dx}_{0}\\ &\,\,\tag{1} +\Big[u_t(t,x)\,u_x(t,x)\Big]_{x=0}^{x=1}\,. \end{align} From the boundary condition $u(t,0)=u(t,1)=0$ it follows that $u_t(t,0)=0$ and $u_t(t,1)=0\,.$ Therefore the boundary term in (1) also vanishes.