Show that a wave equation $\rho u_{tt}=Tu_{xx}$ has solution $u=0$ if $u(x,0)=u_t(x,0)=0$.
Thoughts: This is easy using the general solution to wave equations
$$u(x,t)=\frac12(\phi(x+ct)+\phi(x-ct))+\frac1{2c}\int_{x-ct}^{x+ct}\psi(s)\,ds$$
with $c=\sqrt{T/\rho}$ and $\phi$ and $\psi$ as initial displacement and velocity respectively. Obviously the form above will be $0$. However, I'm trying to show that the only solution is $0$ using conservation laws.
Conservation Laws: We have $E=KE+PE$ (energy is kinetic plus potential) and $dE/dt=0$. In particular,
$$KE=\frac12\int_D\rho u_t^2\,dx$$ $$PE=\frac12\int_D Tu_x^2\,dx$$
Question: I can't think of a good strategy to show that the conservation of energy implies that my PDE has only the trivial solution, even though intuitively it makes a lot of sense.
You know that $E(t)' =0$ where $$ E(t) = \frac{1}{2}\int (\rho u_t^2 + T u_x^2) dx = \frac{1}{2}\int (\rho u_t^2 - T u u_{xx}) dx $$ via integration by parts (assuming there is no wave at infinity). Clearly $$ E(0) = 0 $$ by the initial conditions. Thus we see $$ \frac{1}{2}\int (\rho u_t^2 + T u_x^2) dx = 0 \quad \forall t \in \mathbb{R}_+ $$ Since the integrand is non-negative, we see that $$ u_t = u_x = 0 \quad \forall x,t \in \mathbb{R} \times \mathbb{R}_{+}$$ along with the initial conditions, we conclude $$ u \equiv 0 $$