This is the equation given, $$\begin{array}{l} u_{tt}=a^{2}\left(u_{x x}+u_{y y}\right), \\ \left\{\begin{array}{l} \left.u\right|_{t=0}=\varphi(x, y), \\ \left.u_{t}\right|_{t=0}=\psi(x, y) . \end{array}\right. \end{array}$$ Its solution $u(x,y,t)$ can be given by Poisson's formula,
$$ \begin{align*} u(x,y,t) = {} & \frac{1}{2\pi a}\biggl[\frac{\partial}{\partial t} \iint_{\Sigma_{at}^M} \frac{\varphi(\xi,\eta)d\xi d\eta}{\sqrt{a^2t^2-(\xi-x)^2-(\eta-y)^2}} \\ & + \iint_{\Sigma_{at}^M}\frac{\psi(\xi,\eta)d\xi d\eta} {\sqrt{a^2t^2-(\xi-x)^2-(\eta-y)^2}}\biggr]. \end{align*} $$
If the initial data $\varphi$ and $\psi$ has compact support (for any fixed point M = ($x_0$, $y_0$) $∈$ $\mathbb{R} ^2$ , there exists some $ρ > 0$ such that both $φ$ and $ψ$ vanish outside $Σ^M _ρ$ and are bounded in $Σ^M _ρ$).
For any given $(x_0,y_0)$, how to prove
$$u(x_0,y_0,t)=O(t^{-\frac{1}{2}}),t\to\infty.$$
The estimation I can get is $O(t^{-1}),$
\begin{align*} u(x_0,y_0,t) = {} & \frac{1}{2\pi a}\biggl[\frac{\partial}{\partial t} \iint_{\Sigma_{at}^M} \frac{\varphi(\xi,\eta)d\xi d\eta}{\sqrt{a^2t^2-(\xi-x_0)^2-(\eta-y_0)^2}} \\ & + \iint_{\Sigma_{at}^M}\frac{\psi(\xi,\eta)d\xi d\eta} {\sqrt{a^2t^2-(\xi-x_0)^2-(\eta-y_0)^2}}\biggr] \\ ={} & \frac{1}{2\pi a}\bigg[\frac{\partial}{\partial t}\int_0^{at}\int_0^{2\pi} \frac{\varphi(x_0+r\cos\theta,y_0+r\sin\theta)}{\sqrt{a^2t^2-r^2}}r d\theta d r \\ & +\int_0^{at}\int_0^{2\pi}\frac{\psi(x_0+r\cos\theta,y_0+r\sin\theta)} {\sqrt{a^2t^2-r^2}}rd\theta d r\bigg] \\ ={} & \frac{1}{2\pi a}\bigg[\frac{\partial}{\partial t} \int_0^{\rho}\int_0^{2\pi}\frac{\varphi(x_0+r\cos\theta,y_0+r\sin\theta)} {\sqrt{a^2t^2-r^2}}rd\theta d r \\ & + \int_0^{\rho}\int_0^{2\pi}\frac{\psi(x_0+r\cos\theta,y_0+r\sin\theta)}{\sqrt{a^2t^2-r^2}}rd\theta d r\bigg] \\ ={} & \frac{1}{2\pi a}\bigg[\int_0^{\rho}\int_0^{2\pi}\frac{\partial}{\partial t} \frac{\varphi(x_0+r\cos\theta,y_0+r\sin\theta)}{\sqrt{a^2t^2-r^2}}rd\theta d r \\ & + \int_0^{\rho}\int_0^{2\pi}\frac{\psi(x_0+r\cos\theta,y_0+r\sin\theta)} {\sqrt{a^2t^2-r^2}}rd\theta d r\bigg]. \end{align*} Thus when $t\to +\infty$, \begin{split} |u(x_0,y_0,t)|&\leq \frac{1}{2\pi a}\left[2\pi C\int_0^{\rho}-a^2t(a^2t^2-r^2)^{-\frac{3}{2}}r d r+2\pi C\int_0^{\rho}\frac{r}{\sqrt{a^2t^2-r^2}}dr\right]\\ &=\frac{1}{2\pi a}\left[2\pi C\left(a-\frac{a^2t}{\sqrt{a^2t^2-\rho^2}}\right)+2\pi C(at-\sqrt{a^2t^2-\rho^2})\right] \\ & = O(t^{-1}). \end{split}
