First, I have a very limited knowledge of P.D.E.. this question may look easy to others, but I struggled and couldn't solve.
$u(x, t)$ is a continuous function having first and second partial derivatives, and $u$ and its first partial derivatives are periodic in $x$ with period 1. Furthermore, $u_{tt}=u_{xx}$. Then we have the following:
$\frac{1}{2} \int_0^1 (u_t^2(x, t) + u_x^2(x, t))dx$
is always constant, independent of t.
I tried taking partial derivatives of the expression inside the integral and show that they are zero, but this approach was really getting me nowhere.. looks like there's another way of doing it. Any help would be greatly appreciated.
Let me start with $$ K(t)=\frac 12\int_0^1 u_{t}^2dx. $$ We have $$ \frac{dK}{dt}(t)=\int_0^1u_tu_{tt}dx=\text{(due to the equation)}=\int_0^1u_tu_{xx} dx. $$ Now use the integration by parts $$ \int_0^1u_tu_{xx}dx=u_tu_{x}|_0^1-\int_0^1 u_{x}u_{xt}dx=-\frac 12\frac{d}{dt}\int_0^1u_x^2\,dx=:-\frac{dT}{dt}(t). $$ Hence we have that $$ \frac{dK}{dt}(t)=-\frac{dT}{dt}(t)\implies K(t)+T(t)=E, $$ where $E$ is a constant.