Consider the problem $$\begin{cases} u_{tt}=u_{xx}-2\alpha u_x+\alpha^2u,&(x,t)\in\mathbb{R}^2\\ u(x,0)=f(x),\ u_t(x,0)=g(x),&x\in\mathbb{R}, \end{cases}$$ where $\alpha\geq 0$ is constant, $f\in C^2(\mathbb{R})$ and $g\in C^1(\mathbb{R})$.
Assuming the existence of solution, how do I prove it is unique? I would prefer to use energy functional, that is, if I had two solutions $u_1$ and $u_2$, then
their difference, $u:=u_1-u_2$, would be a solution to the problem with $f=g=0$. Now, I want to define a function $E=E(t)$ of the form
$$E(t)=\int^{\beta(t)}_{\alpha(t)}F(x,t,u,u_t,u_x)dx$$ with the following properties:
$(i)$ $F(x,t,u,u_t,u_x)\geq 0$, and if $F(x,t,u,u_t,u_x)=0$, then $u=0$;
$(ii)$ $E'(t)\leq 0$ for every $t$;
$(iii)$ $\alpha(0)=\beta(0)$, $\alpha(t)\leq\beta(t)$ for every $t$.
Usually, property $(ii)$ will follow from the PDE and deriving under the integral sign, using Leibniz's rule, etc... After I ahd found such $E$, I would get $F=0$, so $u=0$ by $(i)$, so $u_1=u_2$.
I'm trying to do the following: Let's search for $F$ of the form $F=u_t^2+ u_x^2-\alpha^2u^2+G$ for some $G$. Then
$$E'(t)=\dfrac{d}{dt}E(t)=\left(\int_{\alpha(t)}^{\beta(t)}2u_tu_{tt}+2u_xu_{xt}-2\alpha^2uu_tdx\right)+H=\left(\int_{\alpha(t)}^{\beta(t)}2u_t(u_{tt}-u_{xx}-\alpha^2u)dx\right)+H_1$$ for some $H$ and $H_1$. Now, the problem is that I can't find something "good" to put in $F$ so that, when deriving with respect to $x$ and using product rule for integrals will give me $2u_tu_x$.
