# If $u$ is the solution of wave equation with compactly supported initial data, does $tu(x,t)$ stay bounded?

Let $u$ be a solution of the $2$-dimensional wave equation whose initial data $g$ and $h$ vanish outside $B_1 \subset \mathbb{R}^2$. Show for any fixed $(x,y)$, $t\mapsto t \cdot u(x,y,t)$ remains bounded, as $t\to\infty$. Is this still true for the $3$ dimensional wave equation?

We have that $$\begin{cases} u_{tt} - \Delta u = 0 & \mbox{ in} \, \, \mathbb{R}^2 \times (0,\infty)\\ u(x,0)=g(x) & \mbox{ on} \, \, B_1\times \left\{t=0\right\} \\ u(x,0) = 0 & \mbox{ on} \, \, \mathbb{R}^2 \setminus B_1 \times \left\{t=0\right\} \\ u_t(x,0)=h(x) & \mbox{ on} \, \, B_1 \times \left\{t=0\right\} \\ u_t(x,0) = 0 & \mbox{ on} \, \, \mathbb{R}^2\setminus B_1 \times \left\{t=0\right\} \end{cases}$$ I am unsure how to show that the mapping $(x,y),t \mapsto t\cdot u(x,y,t)$ remains bounded as $t\to\infty$. As far as the second question in 3-dimensional space, my intuition leads me to believe that this will not be true. I know the odd dimensional cases of the wave equation are solved by introducing a dummy variable and solving it in an even dimension (or at least that's how its been done in my lectures).

I suppose you are expected to use Kirchhoff's formulas. In $n=2$ dimensions, the solution is $${\displaystyle u(x,t)={\frac {1}{2}}\left[\partial _{t}\left({\frac {t^{2}}{|B_{t}(x)|}}\int _{B_{t}(x)}{\frac {g(y)}{(t^{2}-|y-x|^{2})^{\frac {1}{2}}}}dy\right)+ \left({\frac {t^{2}}{|B_{t}(x)|}}\int _{B_{t}(x)}{\frac {h(y)}{(t^{2}-|y-x|^{2})^{\frac {1}{2}}}}dy\right)\right]}$$
The fraction $t^2/|B_t(x)|$ simplifies to $1/\pi$, and it should become clear that the second integral is $O(1/t)$ while the first is even $O(1/t^2)$, as the differentiation brings the power down.
In three dimensions, $$u(x, t) = {\frac {1}{3}}\left[\partial _{t}\left({\frac {t}{|\partial B_{t}(x)|}}\int _{\partial B_{t}(x)}gdS\right)+\left({\frac {t}{|\partial B_{t}(x)|}}\int _{\partial B_{t}(x)}hdS\right)\right]$$ Here $\dfrac {t}{|\partial B_{t}(x)|} = \dfrac{1}{4\pi t}$. The surface integrals are both bounded. So yes, it appears $u(x,t) = O(1/t)$ in this case as well.