Inhomogeneous Wave Equation Energy Method Let $u$ be a solution to the following equation:
$$
u_{tt}=u_{xx}−u^3
$$
Assume that $u(x,0) =u_t(x,0) = 0$ for all $x\in[a, b]$.  Prove that $u(x, t) = 0$ if $a+t < x < b−t$.
I initially thought of using Duhamel's principle on this but realized that the cube in the integrand would make things a little tricky. I was also thinking of using an energy functional and showing that it was $0$ everywhere and nonincreasing. What would the functional be, and what would/should the domain of integration be?
 A: Note that, thanks to the Leibniz integral rule,
\begin{align}
&\frac{\rm d}{{\rm d}t}\int_{a+t}^{b-t}\left(\frac{1}{2}u_t^2+\frac{1}{2}u_x^2+\frac{1}{4}u^4\right){\rm d}x\\
&=\int_{a+t}^{b-t}\left(u_tu_{tt}+u_xu_{xt}+u^3u_t\right){\rm d}x-\frac{1}{2}\left(u_t^2+u_x^2+\frac{1}{2}u^4\right)\Bigg|_{x=b-t}-\frac{1}{2}\left(u_t^2+u_x^2+\frac{1}{2}u^4\right)\Bigg|_{x=a+t}.
\end{align}
Besides, the first term in above yields, following the usual integration by parts,
\begin{align}
&\int_{a+t}^{b-t}\left(u_tu_{tt}+u_xu_{xt}+u^3u_t\right){\rm d}x\\
&=\int_{a+t}^{b-t}\left(u_tu_{tt}+u^3u_t\right){\rm d}x+\int_{a+t}^{b-t}u_xu_{xt}{\rm d}x\\
&=\int_{a+t}^{b-t}\left(u_tu_{tt}+u^3u_t\right){\rm d}x+\int_{a+t}^{b-t}u_x{\rm d}u_t\\
&=\int_{a+t}^{b-t}\left(u_tu_{tt}+u^3u_t\right){\rm d}x+u_xu_t|_{a+t}^{b-t}-\int_{a+t}^{b-t}u_tu_{xx}{\rm d}x\\
&=\int_{a+t}^{b-t}\left(u_{tt}-u_{xx}+u^3\right)u_t{\rm d}x+u_xu_t|_{a+t}^{b-t}\\
&=u_xu_t|_{a+t}^{b-t},
\end{align}
where the last step is a gift from the governing equation.
Thanks to these results,
\begin{align}
&\frac{\rm d}{{\rm d}t}\int_{a+t}^{b-t}\left(\frac{1}{2}u_t^2+\frac{1}{2}u_x^2+\frac{1}{4}u^4\right){\rm d}x\\
&=u_xu_t|_{a+t}^{b-t}-\frac{1}{2}\left(u_t^2+u_x^2+\frac{1}{2}u^4\right)\Bigg|_{x=b-t}-\frac{1}{2}\left(u_t^2+u_x^2+\frac{1}{2}u^4\right)\Bigg|_{x=a+t}\\
&=-\frac{1}{2}\left(u_t^2-2u_tu_x+u_x^2+\frac{1}{2}u^4\right)\Bigg|_{x=b-t}-\frac{1}{2}\left(u_t^2+2u_tu_x+u_x^2+\frac{1}{2}u^4\right)\Bigg|_{x=a+t}\\
&=-\frac{1}{2}\left[\left(u_t-u_x\right)^2+\frac{1}{2}u^4\right]\Bigg|_{x=b-t}-\frac{1}{2}\left[\left(u_t+u_x\right)^2+\frac{1}{2}u^4\right]\Bigg|_{x=a+t}\\
&\le 0.
\end{align}
This immediately leads to
$$
\int_{a+t}^{b-t}\left(\frac{1}{2}u_t^2+\frac{1}{2}u_x^2+\frac{1}{4}u^4\right){\rm d}x\le\int_a^b\left(\frac{1}{2}u_t^2+\frac{1}{2}u_x^2+\frac{1}{4}u^4\right)\Bigg|_{t=0}{\rm d}x=0,
$$
where the last step is a gift from the initial conditions. Therefore,
$$
u=0
$$
holds for all $x\in\left(a+t,b-t\right)$ when this interval is non-degenerated.
