Showing a solution of a PDE is bounded. 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  
 A: $\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 $\|\nabla v\|$.
