Suppose $u$ is a solution of the two-dimensional wave equation $$u_{tt}-c^2 \Delta u=f(x)$$ with initial values $u(0),u_t(0)$ that have support on the disc $x_1^2+x_2^2 \le 1$. Up to what time can you be sure that $u=0$ at the point $(x_1,x_2)=(2,3)$?
My attempt:
I think it depends on the point farthest from $(2,3)$ in the disc, which is $(-\frac{-2 \sqrt {13}}{13},-\frac{-3 \sqrt {13}}{13})$. Then by the range of influence of wave equation, the time should be $t=\frac{14+2\sqrt {13}}{c}$, where $c$ is constant of two dimensional wave equation $u_{tt}-c^2 \Delta u=f(x)$, $14+2\sqrt {13}$ is the distance of $(2,3)$ and $(-\frac{-2 \sqrt {13}}{13},-\frac{-3 \sqrt {13}}{13})$.
Is that right? Can anyone help? Thanks so much!:)
The initial values at the point on the unit circle closest to $(2,3)$ are the ones that first influence the function value at this point.
Due to Evans Book, Chapter 2.4.3, Theorem 6, we know that if $u$ is a solution to the wave equation (with speed $c$ normed to 1 for the moment) and $u(0)\equiv u_t(0)\equiv 0$ on the ball $B(x_0,t_0)$, then $u\equiv 0$ within the cone $$C=\{(x,t)\,|\,0\leq t\leq t_0,|x-x_0|\leq t_0-t\}.$$ This easily extends to the case of constant speed $c$ different to $1$ by simply scaling, so our condition is $u(0)\equiv u_t(0)\equiv 0$ within $B(x_0,ct_0)$.
What's therefore left to do is to compute $\operatorname{dist}(x_0,B(0,1))$, with $x_0=(2,3)$: Since you suppose that the support of $u(0),u_t(0)$ is within the unit disc $B(0,1)$, the searched-for $t_0$ is given by $t_0=\operatorname{dist}(x_0,B(0,1))/c$.
But this distance is just the Euclidean norm minus the radius of the unit circle, that is $$||x_0||=\sqrt{2^2+3^2}-1=\sqrt{13}-1.$$ This finally leads to $$t_0=\frac{\sqrt{13}-1}{c}.$$
Remark: With your approach, you were on the way to answer a different question, that is: How long do I have to wait while measuring at $x_0$ until I can be sure that my initial data $u(0),u_t(0)$ vanish. You have however a sign error for the farthest point, which is in fact $$x_\text{far}=\left(-\frac{2}{\sqrt{13}},-\frac{3}{\sqrt{13}}\right).$$ I can't follow at all the second calculation, but the time (I call it $t_1$) should be $$t_1=\frac{\sqrt{13}+1}{c},$$ since the distance to $t_1$ is exactly $2$ larger than the distance to $t_0$.
I sketched the situation in Wolfram alpha.