Let $u:\mathbb{R}\times (0,+\infty) \to \mathbb{R}$ is a classical solution to the problem \begin{align}u_{tt}-u_{xx}+u&=0~;~x\in \mathbb{R}, t>0 \\u(x,0)&=\varphi(x)=0~;~x\in \mathbb{R}\\u_t(x,0)&=\psi(x)=0~;~x\in \mathbb{R}\end{align} such that for each fixed $t$, $x \mapsto u(x,t)$ is a function with compact support. I need to show that $u\equiv 0$.
Attempt : Note that, if we multiply $u_t$ with the eqution we get \begin{align}u_{tt}u_t-u_{xx}u_t+uu_t=0 \implies \frac{\partial}{\partial t} \left(\frac{1}{2}u_t^2+\frac{1}{2}u_x^2+\frac{1}{2}u^2\right)+\frac{\partial}{\partial x}(-u_tu_x)=0\end{align} Now integrating we get \begin{align}\int_{\mathbb{R}} \frac{\partial}{\partial t} \left(\frac{1}{2}u_t^2+\frac{1}{2}u_x^2+\frac{1}{2}u^2\right)\,\mathrm{d}x+\int_{\mathbb{R}}\frac{\partial}{\partial x}(-u_tu_x)\,\mathrm{d}x=0\end{align} The second term vanished as $x \mapsto u(x,t) \in \mathcal{C}_0$, giving us $$\frac{\partial}{\partial t}E(t)=0 \implies E(t)=\mathrm{constant}$$ Where $$E(t):=\int_{\mathbb{R}}\left(\frac{1}{2}u_t^2+\frac{1}{2}u_x^2+\frac{1}{2}u^2\right)\,\mathrm{d}x$$ Also we have $E(0)=0$, then this would imply $E(t)=0\,; \,\forall\, t>0$. This is possible because $u(x,0)=0$. Also $u(x,t)$ needs to the zero identically for all $t$ to guarantee that $E(t)=0$ for all $t$. Which implies $u\equiv 0$.
Does my solution seems alright? Would appreciate any corrections and hints, thank you!