Consider the PDE problem for the wave eq. $$\begin{align} u_{tt}-u_{xx}+2u & = 0, & 0<x<1, t \in \mathbb{R} \\ u_x(0,t) = 0,u_x(1,t) & = 0, & t>0 \\ u(x,0) = 0, u_t(x,0) & =\cos(2 \pi x), & 0<x<1 \end{align}$$ Define the energy solution $u(x,t)$ by $$ E(t) = \frac{1}{2} \int_0^1 ((u_t(x,t))^2+(u_x(x,t))^2+2(u(x,t))^2)dx$$ Show that the energy is conserved, i.e. that $E(t)$ is independent of $t$.

I know I need to determine $E'(t)$. However I need some basic knowledge of how to compute $\frac{d}{dt} E(t)$.

Is this correct ? \begin{align} \frac{d}{dt}E(t) & = \int_0^1 u_tu_{tt} + u_x u_{xt}+2uu_t dx \\ & = \int_0^1 u_t u_{tt} +2uu_t-u_tu_{xx} dx \\ & = \int_0^1 u_t(u_{tt}+2u-u_{xx})dx \\ & = \int_0^1 u_t \cdot 0 dx \\ & = 0 \end{align} where I have used that (integrating by parts) we have \begin{align} \int_0^1 u_xu_{xt} dx & = \left[ u_x(1,t)u_t(1,t)-u_x(0,t)u_t(0,t) \right] - \int_0^1 u_t u_{xx} dx \\ & = \left[ 0 \cdot u_t(1,t)-0 \cdot u_t(0,t) \right] - \int_0^1 u_t u_{xx} dx \end{align}


I consider this to be homework and will therefore give a hint. The proof of Evans, chapter 2.4.3 can be adapted easily. When taking the time derivative, note that the integral is independent of $t$, hence you can pull in the derivative. Then apply the chain rule on all terms. The only difference to Evans is the term $$2u(x,t)^2.$$ You will easily find the derivative using the chain rule and checking on how it worked with $u_x$.

Edit: Your approach is correct. In a written exam, I would include the argumentation why you computed the energy and what does this mean for the solution. In fact, since $E'(t)$ for all $t>0$, you know that the energy is constant, namely $$E(t)=E(0)=\int_0^1 \cos^2(2\pi x)dx=\frac12\,,$$ and therefore conserved.

Note that it immediately follows the uniqueness of the solution, and some smoothness properties in terms of Sobolev norms. But I guess that's not part of your exam.

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    $\begingroup$ It isn't homework. Just preparing for an exam. However I have tried something. Is it correct (see question above). $\endgroup$ – ANYN11 Dec 13 '15 at 22:11

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