Suppose that $|c(x)| < C$ for all $x \in \mathbb{R},$ and that $\phi$ and $\psi$ are smooth functions. Show that there exists at most one solution to $$ u_{tt} - (c(x))^2 u_{xx} = -u_t$$ where $u(x,0) = \phi$ and $u_t(x,0) = \psi.$
The only way that comes to mind is energy methods, letting $w$ be the difference of two solutions to the PDE, and then letting $$ E(t) := \frac{1}{2} \int w_t^2 + c(x)^2 w_x^2 \; dx.$$ Noting that we have $E(0) = 0,$ due to everything being linear, we can differentiate with respect to time and integrating by parts. After doing so and using the PDE, one gets, $$ E'(t) = \int -w_t^2 - 2w_tw_xc(x)c'(x) \; dx \leq 2||c'(x)||_\infty E(t),$$ where it has been assumed that either $w_x$ and/or $w_t$ vanish as $|x| \to \infty,$ so no surface term appears when integrating by parts. The last inequality comes from the fact that $ab \leq \frac{1}{2} (a^2 + b^2)$ and $\int -w_t^2$ is nonpositive. From here, its easy to see that $E(t) \leq 0$ for all time.
The problem with this solution is, we aren't given any information about $c'(x)$ or the behavior of $w_x$ and $w_t$ as $|x| \to \infty.$ Also, no where did I use the fact that $|c|$ is bounded. Does anyone see how to get around this, perhaps by using an entirely different method to show uniqueness? Thanks in advance for any suggestions/proofs!
I think you just have the "wrong" energy here. I'm assuming that $c(x)$ is also bounded from below away from zero: that is $0 < \gamma^2 \le c(x)^2 < C^2$. Otherwise, the operator $- c(x)^2 \Delta$ is no longer elliptic which would raise another problem. However, if we do have this bound from below, we can define the energy $$E(t) = \frac 1 2 \int \left(\frac{w_t^2}{c(x)^2} + w_x^2\right) dx.$$ The advantage of defining it this way is that you never pass a derivative onto $c(x)$: $$E'(t) = \int \left(\frac{w_tw_{tt}}{c(x)^2}+ w_xw_{xt}\right)dx = \int\left(\frac{w_{tt}}{c(x)^2} - w_{xx} \right)w_t dx = - \int \frac{w_t^2}{c(x)^2} dx \le 0.$$ As for the behaviour of $w_x$ and $w_t$ as $\lvert x \rvert \to \infty$: for the original equation, you would typically assume that $\phi, \psi$ have compact support. Then using the finite wave speed, any solution will have compact support at any time $t > 0$ [note the support could be growing, but at any finite time, it will be compact]. This will assure that $E$ is actually well-defined (i.e. finite) since $w$ would have compact support for each $t > 0$. These assumptions are often swept under the rug since by even defining $E(t)$ in that way, you have sort of implicitly assumed some nice properties for $w_x, w_t$.