I am trying to prove a finite speed of propagation result for weak solutions of hyperbolic equations, but getting stuck.
For simplicty, consider the simplest hyperbolic equation $$u_t = u_x +f; \hskip5pt u(0,x)=u_0(x)$$ for $u,f \colon [0,T]\times \mathbb{R} \to \mathbb{R}^d$, $f\in L^2 ([0,T]\times \mathbb{R})$.
If $u\in \mathcal{C}^1([0,T]\times \mathbb{R})$ is a classical solution, then for a localized energy estimate, set $r(t)=r_0-t$ giving \begin{align} \frac{d}{dt}\left(\int_{B_{r(t)}(x_0)}|u|^2dx\right)&=2\int_{B_{r(t)}(x_0)}(u_t\cdot u)dx+r'(t)\int_{\partial B_{r(t)}(x_0)} |u|^2d \sigma(x) \\ &=2\int_{B_{r(t)}(x_0)}((u_x+f)\cdot u)dx +r'(t)\int_{\partial B_{r(t)}(x_0)} |u|^2d \sigma(x) \\ &=2\int_{B_{r(t)}(x_0)}(f\cdot u)dx +(1+r'(t))\int_{\partial B_{r(t)}(x_0)} |u|^2d \sigma(x) \\ &\leq \int_{B_{r(t)}(x_0)}(|f|^2+|u|^2)dx \end{align}which integrated gives $$\int_{B_{r(t)}(x_0)}|u(t,x)|^2dx \leq e^t\left( \int_{B_{r_0}(x_0)} |u_0(x)|^2 dx + \int_0^t\int_{B_{r({t})}(x_0)}|f(t,x)|^2dxdt \right) $$ and thus is follows that we have finite speed of propagation:
(FSP) --- If $\mathrm{supp}(u_0)\subset \{|x-x_1|\leq r_1\}$ and $\mathrm{supp}(f)\subset \{|x-x_1|\leq r_1+t \} $, then $\mathrm{supp}(u)\subset \{|x-x_1|\leq r_1+t \} $ ---
Now, if $u \in L^2([0,T]\times \mathbb{R}^n)$ only solves the equation only in the weak sense, that is $$ \int_0^T \int_{\mathbb{R}^d} (u\cdot(\phi_t-\phi_x)+f\cdot \phi)(t,x) dxdt + \int_{\mathbb{R}^d}u_0(x)\cdot\phi(0,x) dx =0 $$ holds for any $\phi\in \mathcal{C}^\infty_c((-\infty,T]\times \mathbb{R}^d)$, then we cannot justify the forgoing compitations, so can we still obtain the result (FSP)?
I first tried using a Galerkin method, approximating the weak solution by classical solutions to a finite dimensional problem (by means of taking $L^2$ projections of $f$), but this runs into difficulties as we have to cut off these projections so that the approximates satisfy themselves FSP, and then have difficulties proving convergence. What's worse is that this method in the end only gives weak $L^2$ convergence to the solution, which doesn't give the almost everywhere convergence required. This makes me think that an approximation argument is not the way to go, but there may be a way to justify the computations formally for weak solutions to give the local energy estimate, but I am unsure how to proceed here...
I know I'm late but you can use approximations.
Let $S(\mathbb R^d)$ be the space of Schwartz functions.
Let $(u_{0,n}(x))_{n=1}^{\infty} \subset S(\mathbb R^d)$ be a sequence of Schwartz functions converging to $u_0(x)$ in $L^2(\mathbb R^d)$.
Let $(f_n(t,x))_{n=1}^{\infty} \subset C^{\infty}([-T,T]; L^2(\mathbb R^d))$ be a sequence of smooth functions converging to $f(t,x)$ in $L^2([-T,T] \times \mathbb R^d)$ and such that $f_n(t,x)$ is Schwartz for all fixed $t$.
For the homogeneous equation, you can write $u(t,x) = (U(t)u_0)(x)$, where $U(t)$ is a $C_0$ semi-group on $L^2(\mathbb R^d)$ given by convolution with the kernel $K(t,x) = \mathcal{F}_{\xi}^{-1}(e^{i\xi t})$.
By Duhamel's formula, for the non-homogeneous equation, the solution is
$$u(t,x) = U(t)u_0 + \int_{0}^t U(t-s)f(s)ds$$
If $u_n(t,x)$ solves $v_t = v_x + f_n$, $v_0 = u_{0,n}$, then
$$u_n(t,x) = U(t)u_{0,n} + \int_{0}^t U(t-s)f_n(s)ds$$
converges to $u(t,x)$ in $C^0([-T,T]; L^2(\mathbb R^d))$.
Moreover, $u_n(t,x) \in C^1([-T,T];S(\mathbb R^d))$ and for such solutions, your localized energy inequality holds.
Then, take the limit $n \to +\infty$ in the energy inequality for $u_n(t,x)$.