$\require{cancel}$
The "energy method" of showing the uniqueness of solution for the initial boundary value problem of the wave equation $$u_{tt}=c^2u_{xx}$$ (in a bounded domain) seems to require that the solution has continuous second partial derivatives.
On the other hand, if the initial displacement $~f(x)~$ is not smooth, say like saw tooth wave, then the sharp corners persist in the travelling wave "solution"; the standard explanation being that it is a solution in the "weak sense."
My question is, what can we say about the uniqueness of such "weak solution" ?
Could there be more than one weak (and preferably continuous) solution?
Edit: The energy method is the following: Suppose that $u(x,t)$ is a solution of the initial boundary value problem \begin{align} u_{tt}&=c^2u_{xx},&0&<x<L,&t&>0\\ u(x,0)&=0,&0&\leq x\leq L\\ u_t(x,0)&=0,&0&\leq x\leq L\\ u(0,t)&=u(L,t)=0,&t&\geq0 \end{align} Define $$E(t)=\frac{1}{2}\int_0^L(c^2u_x^2+u_t^2)\,dx.$$ Then \begin{align} \frac{dE}{dt}&=\int_0^L(c^2u_xu_{xt}+u_tu_{tt})\,dx\tag{1}\label{from}\\ &=\int_0^Lc^2u_xu_{tx}\,dx+\int_0^Lu_tu_{tt}\,dx\tag{2}\label{swt}\\ &=[c^2u_xu_t]_0^L-\int_0^Lc^2u_{xx}u_{t}\,dx+\int_0^Lu_tu_{tt}\,dx\tag{3}\label{ftc}\\ &=c^2(u_x(L,t)\cancel{u_t(L,t)}-u_x(0,t)\cancel{u_t(0,t)})+\int_0^Lu_t(u_{tt}-c^2u_{xx})\,dx=0. \end{align} Therefore $E(t)$ is constant. Now, $u(x,0)=0$ implies that $u_x(x,0)=0$, and $u_t(x,0)=0$ by assumption, so $$E(t)=E(0)=\int_0^L(c^2u_x(x,0)^2+u_t(x,0)^2)\,dx=0.$$ Hence, $u_x=0$ and $u_t=0$, which implies that $u(x,t)$ is constant. Since $u(x,0)=0$ by assumption, $u(x,t)=0$.
Note that going from (\ref{from}) to (\ref{swt}), the order of differentiation was changed $\partial_t\partial_x\to\partial_x\partial_t$, which requires continuous second partial derivatives. Also, going from (\ref{swt}) to (\ref{ftc}) (fundamental theorem of calculus) would be problematic if $u_x$ and $u_t$ are not continuous in $x$.