Consider the wave equation $u_{xx}-u_{tt}=0$ on the unit interval $x\in[0,1]$. Take mixed boundary conditions ($\alpha_{1,2}^2+\beta_{1,2}^2 \neq 0$) \begin{align*} \alpha_1 u(0,t) + \beta_1u_x(0,t) &= 0,\\ \alpha_2 u(1,t) + \beta_2u_x(1,t) &= 0. \end{align*} For classical solutions this equation has a conserved quantity $$ E=\int_0^1 \!\big(u_x(x,t)^2 + u_t(x,t)^2\big)\,\mathrm{d} x + \frac{\alpha_2\beta_2}{\alpha_2^2 + \beta_2^2}\big(u(1,t)^2+u_x(1,t)^2\big)- \frac{\alpha_1\beta_1}{\alpha_1^2 + \beta_1^2}\big(u(0,t)^2+u_x(0,t)^2\big). $$ For (say) Dirichlet boundary conditions the last two terms vanish so it is natural to define a Hilbert space $\mathcal{H}=H_0^1([0,1])\oplus L^2([0,1])$ (here $H_0^1$ is the Sobolev space with first derivatives in $L^2$ and satisfying Dirichlet conditions) with inner product $$ \langle(u,w),(u',w')\rangle = \int_0^1 \!\big(u_x\overline{u'_x} + w\overline{w'}\big)\,\mathrm{d} x. $$ The (unbounded) operator $L[(u,w)] = (w,u_{xx})$ encodes the wave equation via $$ \frac{\mathrm{d}}{\mathrm{d}t}(u(t),w(t))=(w(t),u_{xx}(t)) = L[(u(t),w(t))]. $$ $iL$ is essentially self-adjoint (indeed, the basis vectors $(\sin(n\pi x),\sin(m\pi x))$ are easily seen to be analytic for $iL$) so we can use it to generate a strongy continuous unitary evolution map $U(t)$. This then gives a notion of a weak solution.
For general boundary conditions, however, $E$ is not positive definite, so this procedure breaks down when defining the Hilbert space. Is there some way to circumvent this issue? With certain boundary conditions it is possible to get exponentially growing solutions so I suppose there is some kind of instability at play here, but it still seems like such a simple problem that a general theory should be available.
