二维波动方程柯西问题解的衰减阶估计技术问询
大家好，我最近在研究二维波动方程的柯西问题时，碰到了一个关于解的衰减阶估计的困惑，想跟各位请教一下：
首先，我们的问题基于以下二维波动方程柯西问题：
$$\begin{array}{l}
u_{tt}=a^{2}\left(u_{x x}+u_{y y}\right), \
\left{\begin{array}{l}
\left.u\right|{t=0}=\varphi(x, y), \
\left.u{t}\right|{t=0}=\psi(x, y) .
\end{array}\right.
\end{array}$$
它的解可通过泊松公式给出：
$$
\begin{align*}
u(x,y,t) = {}
& \frac{1}{2\pi a}\biggl[\frac{\partial}{\partial t}
\iint{\Sigma_{at}^M}
\frac{\varphi(\xi,\eta)d\xi d\eta}{\sqrt{a2t2-(\xi-x)2-(\eta-y)2}} \
& + \iint_{\Sigma_{at}^M}\frac{\psi(\xi,\eta)d\xi d\eta}
{\sqrt{a2t2-(\xi-x)2-(\eta-y)2}}\biggr].
\end{align*}
$$
已知初始数据$\varphi$和$\psi$具有紧支集：对任意固定点$M=(x_0,y_0)\in\mathbb{R}2$，存在$\rho>0$，使得$\varphi$和$\psi$在$\SigmaM_\rho$外全为0，且在$\Sigma^M_\rho$内有界。
现在我需要证明，当$t\to\infty$时，$u(x_0,y_0,t)=O(t{-\frac{1}{2}})$，但自己推导出来的结果却是$O(t{-1})$，我的推导过程如下：
$$\begin{align*}
u(x_0,y_0,t) = {}
& \frac{1}{2\pi a}\biggl[\frac{\partial}{\partial t}
\iint_{\Sigma_{at}^M}
\frac{\varphi(\xi,\eta)d\xi d\eta}{\sqrt{a2t2-(\xi-x_0)2-(\eta-y_0)2}} \
& + \iint_{\Sigma_{at}^M}\frac{\psi(\xi,\eta)d\xi d\eta}
{\sqrt{a2t2-(\xi-x_0)2-(\eta-y_0)2}}\biggr] \
={} & \frac{1}{2\pi a}\bigg[\frac{\partial}{\partial t}\int_0{at}\int_0{2\pi}
\frac{\varphi(x_0+r\cos\theta,y_0+r\sin\theta)}{\sqrt{a2t2-r^2}}r d\theta d r \
& +\int_0{at}\int_0{2\pi}\frac{\psi(x_0+r\cos\theta,y_0+r\sin\theta)}
{\sqrt{a2t2-r^2}}rd\theta d r\bigg] \
={} & \frac{1}{2\pi a}\bigg[\frac{\partial}{\partial t}
\int_0{\rho}\int_0{2\pi}\frac{\varphi(x_0+r\cos\theta,y_0+r\sin\theta)}
{\sqrt{a2t2-r^2}}rd\theta d r \
& + \int_0{\rho}\int_0{2\pi}\frac{\psi(x_0+r\cos\theta,y_0+r\sin\theta)}{\sqrt{a2t2-r^2}}rd\theta d r\bigg] \
={} & \frac{1}{2\pi a}\bigg[\int_0{\rho}\int_0{2\pi}\frac{\partial}{\partial t}
\frac{\varphi(x_0+r\cos\theta,y_0+r\sin\theta)}{\sqrt{a2t2-r^2}}rd\theta d r \
& + \int_0{\rho}\int_0{2\pi}\frac{\psi(x_0+r\cos\theta,y_0+r\sin\theta)}{\sqrt{a2t2-r^2}}rd\theta d r\bigg].
\end{align*}$$
当$t\to +\infty$时，我做了如下估计：
$$\begin{split}
|u(x_0,y_0,t)|& \leq \frac{1}{2\pi a}\left[2\pi C\int_0{\rho}-a2t(a2t2-r2){-\frac{3}{2}}r d r+2\pi C\int_0{\rho}\frac{r}{\sqrt{a2t2-r2}}dr\right]\
&=\frac{1}{2\pi a}\left[2\pi C\left(a-\frac{a2t}{\sqrt{a2t2-\rho2}}\right)+2\pi C(at-\sqrt{a2t2-\rho^2})\right] \
& = O(t^{-1}).
\end{split}$$
显然这个结果和需要证明的$O(t^{-\frac{1}{2}})$不符，想请教各位我哪里出错了，或者应该用什么方法才能得到正确的衰减阶估计？
备注：内容来源于stack exchange，提问作者Zydragon
