Uniqueness of solutions to $u_{tt} - c^{2}u_{xxxx} + au_{t} = 0$ The problem I am working on is to show that there is a unique compactly supported solution to the PDE $u_{tt} - c^{2}u_{xxxx} + au_{t} = 0$, $(x, t) \in \mathbb{R} \times [0, \infty)$ with $u(x, 0)= \phi(x)$ and $u_{t}(x, 0) = \psi(x)$ where $\phi, \psi$ are smooth and compactly supported.
Suppose I had 2 solutions $u_{1}$ and $u_{2}$ for the above equation. Then let $v = u_{1} - u_{2}$. Thus $v_{tt} - c^{2}v_{xxxx} + av_{t} = 0$, $v(x, 0) = 0$, $v_{t}(x,0) = 0$. I think if I can define the correct energy, I'll be able to solve the problem.
If I let $E(t) := \frac{1}{2}\int_{\mathbb{R}}u_{t}^{2} - c^{2}u_{xx}^{2}\, dx$, then I can show that $E'(t) \leq 0$ for all $t > 0$. The problem is $E(t)$ is not necessarily $\geq 0$. Is there a way to use another energy function to prove uniquness?
 A: I'm not sure if this will help. But if you use Fourier transform on the system, you obtain a Sturm-Liouville problem, namely:
\begin{equation}
\frac{d}{dt} \left(e^{at}\hat u_t\right) = c^2 e^{at}t^4\hat u\\
\hat u(\xi, 0) = \hat \phi(\xi) \ \ \ \text{and} \ \ \ \hat u_t(\xi, 0) = \hat \psi(\xi)
\end{equation}
I believe the compactness of functions $\psi$ and $\phi$ ensures that it is a regular Sturm-Liouville problem, while their smoothness ensures uniqueness of Fourier Transform. But whether a unique solution exists or not, I'm not sure.
A: My opinion is that there is a sign mistake in the problem statement. Indeed, up to a sign change, this PDE models the dynamic deflection of a damped Euler-Bernoulli beam. To show this, we define the Lagrangian density $\mathcal{L} = K - V$ with $K = \tfrac12 (u_t)^2$ and $V = \tfrac12 c^2 (u_{xx})^2$ representing the kinetic energy and the potential energy. Furthermore, we introduce the Rayleigh dissipation function $D = \tfrac12 a (u_t)^2$. The corresponding quasi-variational problem leads to the following equation of motion with non-conservative forces (pseudo Euler-Lagrange equation) $$
\frac{\partial \mathcal{L}}{\partial u} - \frac{\partial}{\partial t} \frac{\partial \mathcal{L}}{\partial u_t} + \frac{\partial^2}{\partial x^2} \frac{\partial \mathcal{L}}{\partial u_{xx}} = \frac{\partial D}{\partial u_t} ,
$$ i.e., $$
u_{tt} + c^2 u_{xxxx} = -a u_t ,
$$
which is similar to our initial PDE (up to the sign in front of the fourth-order spatial derivative). This remark leads to considering instead the non-negative Hamiltonian energy $$
E(t) = \int_{\Bbb R} (K + V)\, dx \geq 0
$$
such that $$
\begin{aligned}
E'(t) &= \int_{\Bbb R} \left(u_tu_{tt} + c^2u_{xx}u_{xxt}\right) dx \\
&= c^2 \int_{\Bbb R} \left(-u_tu_{xxxx} + u_{xx}u_{xxt}\right) dx - a \int_{\Bbb R} (u_t)^2 dx \\
&= c^2 \int_{\Bbb R} u_{xxx} \left(u_{tx} - u_{xt}\right) dx - a \int_{\Bbb R} (u_t)^2 dx \\
&= - a \int_{\Bbb R} (u_t)^2 dx \leq 0 .
\end{aligned}
$$ where the PDE, integration by parts and the equality of mixed partials were used. With this definition, the energy method can be applied in the usual way (see also this related post).
