Consider the semilinear wave equation in $[0,t_0] \times \mathbb R^d$ :
$$\square u = \pm |u|^{p-1}u$$
With subcritical/critical power $1 < p \leq \frac{d+2}{d-2}$.
It is easy to show by energy method that any two $C^2$ classical solutions $u(t,x)$ & $v(t,x)$ with same initial conditions on $\{0\} \times B(x_0, t_0)$ will coincide on the backwards light cone $K(x_0,t_0) = \{(t,x) : 0 \leq t \leq t_0, 0 \leq |x-x_0| \leq t_0-t\}$ (c.f. T. Tao, Nonlinear dispersive equations, Proposition 3.3).
I've seen this property being used for solutions in energy space $C^0([0,T], \dot{H}^1 \times L^2)$. I guess one should be able to generalize the above proposition using some approximation argument but I could not find a reference with a proof.
My main problem is that the proof using energy method uses the fact that a $C^2$ classical solution is locally bounded. This does not hold for energy space solutions.
My second problem is that if $u(t,x)$ is an energy space solution, then we can approximate $u(0,x)$, $\partial_t u(0,x)$ and $\square u(t,x)$ by smooth functions $u_{0,n}(x)$, $u_{1,n}(x)$, $g_n(t,x)$, and get an approximate classical solution $u_n(t,x)$ to the inhomogeneous equation
\begin{align*}
\square u_n(t,x) &= g_n(t,x), \quad (t,x) \in [0,t_0] \times \mathbb R^d \\
u_n(0,x) &= u_{0,n}(x), \quad x \in \mathbb R^d \\
\partial_t u_n(0,x) &= u_{1,n}(x), \quad x \in \mathbb R^d
\end{align*}
but I don't know whether we can approximate $u(t,x)$ by classical solutions $u_n(t,x)$ solving
\begin{align*}
\square u_n(t,x) &= \pm |u_n|^{p-1}u_n, \quad (t,x) \in [0,t_0] \times \mathbb R^d \\
u_n(0,x) &= u_{0,n}(x), \quad x \in \mathbb R^d \\
\partial_t u_n(0,x) &= u_{1,n}(x), \quad x \in \mathbb R^d
\end{align*}
Any help or reference is welcomed.
Best Answer
You can approach this dually using that for classical solutions the finite propagation speed holds. This argument is similar in spirit to this answer of mine for low regularity uniqueness for the linear wave equation.
1
First write $w = u-v$. Then you see that $w$ solves a linear wave equation of the form
$$ \Box w = G(u,v) w $$
where the potential $G(u,v) \lesssim |u|^{p-1} + |v|^{p-1}$. Pretend now that $G$ is a smooth function of $(t,x)$.
2
Suppose for convenience that $x_0 = 0$. You want to prove that $w(t,\cdot) |_{B(0,t_0 - t)} = 0$ for $t\in (0,t_0)$. So it is enough to show that $\int w(t,x) f(x) ~dx = 0$ for every $f\in C^\infty_c(B(0,t_0-t))$. We will try to do so using a duality argument.
Fix $T\in (0,t_0)$, and take $f\in C^\infty_c(B(0,t_0-T))$. Solve the wave equation $$ \Box \varpi = G \varpi $$ with initial data $\varpi(0,T) = 0$ and $\partial_t\varpi(0,T) = f$. Under the assumption that $G$ is smooth, we have that $\varpi$ is a smooth function with compact support for all time. In fact, for $t\in (0,T)$ we have that $\mathrm{supp}~ \varpi \subset B(0,t_0 - t)$ using the finite speed of propagation for classical solutions.
3
That $w$ is an energy solution implies that the following identity holds for any test function $\varpi$:
$$ \int_0^T \int_{\mathbb{R}^d} \Box w \varpi - w \Box \varpi = \int_{\mathbb{R}^d} w \partial_t\varpi - \partial_t w \varpi \Big|_{t = 0}^T $$
Using the support property for $\varpi$ as derived above, you have that the right hand integral vanishes at $t = 0$ since $w,\partial_t w$ vanishes on $B(0,t_0)$. The integral at $t = T$ has only one term, and that is $\int w f$.
The left hand side however vanishes: since the two functions solve the same (linear) equation you have $LHS = \int Gw~v - w ~Gv = 0$.
4
The above works assuming that $G$ is smooth. In general, $G$ is not. Replace $G$ by $G_\epsilon$ through mollification. And replace $\varpi$ correspondingly by $\varpi_\epsilon$. Then it suffices to show that
$$ \int_0^T \int_{\mathbb{R}^d} (G - G_\epsilon) w \varpi_\epsilon \to 0 $$
Let me do the critical case for convenience; the argument should be similar for the subcritical cases with some adjustment of the exponents.
Noting that $p-1 = \frac{4}{d-2}$, by Strichartz inequality, $G$ belongs to space-time norm $L^1([0,T]; L^q_x)$ for any $q\in [\frac{d}{2},d]$. And the mollification will converge in that norm. Since $w$ is energy class you have that it is uniformly bounded in $L^\infty_t L^{2d/(d-2)}_x$.
The Strichartz estimates together with Gronwell's inequality can be used to show that if $G_\epsilon \in L^1_t L^d_x$, then
$$ \|\varpi_\epsilon(t)\|_{\dot{H}^1} \lesssim e^{C(T-t) \|G_\epsilon\|_{L^1_t L^d_x}} \|f\|_{L^2} $$
This shows that $\|\varpi_\epsilon(t)\|_{L^\infty_t L^{2d/(d-2)}}$ is uniformly bounded on the time interval $[0,T]$. And hence the fact that $G-G_\epsilon$ converges to zero in $L^1_t L^{d/2}$ gives that $|\int w(T,x) f(x) ~dx | = 0$ after taking the limit.