Consider the following conditions:
- $u_{tt}=c^2 u_{xx} \quad\quad\quad\quad\quad\space \space \space x\in (0,L), t>0$
- $u(0,x)=u_0(x) \quad\quad\quad\quad x\in [0,L]$
- $u_t(0,x)=u_1(x) \quad\quad\quad\space \space \space x\in [0,L]$
- $u(t,0)=u(t,L)=0 \quad\space\space \space \space t\in [0,T]$
The Energy Function: $E(t)=\int_{0} ^{L} c^2(u_x)^2 dx \space + \space \int_{0} ^{L} (u_t)^2 dx $
Show the energy conservation: For each twice continuously partially differentiable solution $u$: $[0,T] \times [0,L] \rightarrow \mathbb{R} $ of the conditions above for the wave equation the following relation applies: $E(t)=E(0), \space$ for $\space t\geq 0$
After that show that there is at most one twice continuously partially differentiable solution for the above given conditions.
My attempt:
$E'(t)=\int_{0} ^{L} 2 u_t u_{tt}+ 2 c^2 u_x u_{xt} dx $
Integrating the second part gives:
$2c^2 \int_{0} ^{L} u_x u_{xt} dx = 2 c^2 (u_x u_t-\int_{0} ^{L} u_t u_{xx} dx)$
$u_x u_t $ vanishes because of $u_t (t,0)=u_t (t,L)=0$
Now $E'(t)=\int_{0} ^{L} 2 u_t u_{tt}-2 c^2 u_t u_{xx} dx\quad $ with $\quad u_{tt}-c^2 u_{xx}=0 $
$E'(t)=0$, I get to this point and now I don't have an idea of how to proceed
$$E'(t) =0\implies E(t) =C$$ for some constant $C$. In particular $E(0) =C$ so $E(t) =E(0) $ for every value of $t$. To prove uniqueness suppose you have two solutions $u_1$ and $u_2$, and consider $u=u_1-u_2$. The function $u$ solves the homogeneus equation with boundary conditions all equal to zero, which implies that the energy is zero. It implies that $$0=E(t)=\int_0^L \left(c^2(u_x)^2+(u_t)^2\right)dx\implies c^2(u_x)^2+(u_t)^2=0\implies c^2(u_x)^2=0=(u_t)^2\implies u(t,x)=0.$$ The reasoning is as follow: we have an integral of a positive function over a positive domain, which implies that the integrand is zero, but the sum of two real squares is zero if and only if both are zero. It implies that $u(x,t)$ is constant and takes the same value as its boundary, which is zero.