Consider the problem
$$u_{tt}=u_{xx},\quad 0<x<1,\quad t>0$$
$$u|_{x=0}=u|_{x=1}=0$$
$$u|_{t=0} = 0,\quad u_{t}|_{t=0}=x(1-x)$$
Compute $$\int_{0}^{\frac{1}{2}}u^2_t+u^2_x dx$$
i know that $$E(t)=E(0)=\int_{0}^{1}u^2_t+u^2_x dx=\int_{0}^{1}x^2(1-x)^2 dx=\frac{1}{30}$$ via energy conservation and i guess the result of the unknown integral is half of it (1/60), but how to prove it?
Note that the problem is invariant under $x \mapsto 1 - x$, and that the configuration is symmetric w.r.t. to $x = \frac12$. Then, we can use $$\int_0^1\text dx = \int_0^{1/2}\text dx + \int_{1/2}^1\text dx$$ and perform the change of variable $\xi = 1 -x$ in one of the integrals: $$\int_0^1\text dx = \int_0^{1/2}\text dx - \int_{1/2}^{0}\text d\xi = 2 \int_0^{1/2}\text dx .$$ Using conservation of energy over $x$ belonging to $[0,1]$, we find $\int_0^{1/2} u_t^2 + u_x^2\,\text d x = 1/60$.