Let $x \in \mathbb{R}^3$, $t \in [1,\infty)$ and $ u(x,t)$ be a solution of the PDE
$$\partial_{t}^{2}u - \Delta u = 0 \\ u(x,0) = 0 \\ \partial_{t} u(x,0) = v(x)$$
where $v$ and $\partial_{x_i}v$ are both integrable on $\mathbb{R}^3$ for all $1 \leq i \leq 3$.
Show that there exists $C > 0$ such that $|u(x,t)| \leq \frac{C}{t}$
I'm not too sure how to approach this. Before this I just learned about solving some PDEs with Fourier Transforms, but conditions usually involved the initial functions being in Schwartz Space. Any help is appreciated
$\DeclareMathOperator{\p}{\partial} $Recall Kirchoff's formula for the solution to the wave equation with initial conditions $u(x,0)=g$ and $\p_tu(x,0)=v(x)$: $$u(x,t)=\frac{1}{4\pi t^2}\int_{\p B(x,t)}g(y) + \nabla g(y)\cdot(y-x) + tv(y)~d\sigma(y).$$ In your case $g=0$ and so the representation formula simplifies significantly. Now notice that for $y\in\p B(x,t)$ that the outward pointing unit normal is $\nu=\frac{y-x}{t}.$ It then follows that \begin{align*} u(x,t) & = \frac{1}{4\pi t^2}\int_{\p B(x,t)}tv(y)\frac{y-x}{t}\cdot\nu ~d\sigma(y) \\ & = \frac{1}{4\pi t^2}\int_{B(x,t)}\operatorname{div}\left(v(y)(y-x)\right) ~d y \\ & = \frac{1}{4\pi t^2}\int_{B(x,t)}3v(y) + \nabla v(y)\cdot(y-x)~d y. \end{align*} Note that we needed to use the divergence theorem to avoid working with a surface integral.
Now for $t \geq 1$ we have \begin{align} |u(x,t)|&\leq\frac{1}{4\pi t^2}\left(3\|v\|_{L^1}+\int_{B(x,t)}\|\nabla v(y)\|\|y-x\|~dy\right)\\ & \leq\frac{1}{4\pi t^2}\left(3\|v\|_{L^1}+t\|\|\nabla v\|\|_{L^1}~dy\right)\\ & \leq \frac{3\|v\|_{L^1}+\|\|\nabla v\|\|_{L^1}}{4\pi t}. \end{align} So by taking $C=\frac{3\|v\|_{L^1}+\|\|\nabla v\|\|_{L^1}}{4\pi}$ you have the desired result. In the above $\|\|\nabla v\|\|_{L^1}$ just means the $L^1$-norm of $\|\nabla v\|$.