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Consider the wave equation $$ u_{tt}-\Delta u=0, \ x\in\mathbb{R}^n, \ t>0, \ u(x,0)=\phi(x), \ u_t(x,0)=\psi(x), $$ with compactly supported $\phi$ and $\psi$.

I want to show that for $n=3$ there is a constant $C$ such that

$$ |u(x,t)|<\frac{C}{1+t} \ x\in\mathbb{R}^3, \ t>0. $$

I believe that I want to use Kirchoff's formula which is given by, $$ u(x,t)=\int\limits_{\partial B(x,t)} [t\psi(y)+\phi(y)+(y-x)\centerdot\nabla\phi(y)] \, dS_y/A_t $$ where $A_t$ is the surface area of the ball $B(x,t)$.

I really have no idea where to start on this problem. I thought that I could use the fact that $\phi$ and $\psi$ are bounded on their support. But I can't seem to get the desired bound.

Any suggestions?

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1 Answer 1

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Since $\psi$ and $\phi$ have compact support, we know there exists an $R>0$ such that their support lies in a ball of radius $R$. And the surface area of $\partial B(x,t) \cap B(0,R)$ is bounded by $c R^2$ for some constant $c>0$. Hence $$ \int\limits_{\partial B(x,t)} |\phi(y)| \, dS_y \le \min\{c R^2, A_t \} \|\phi\|_\infty , $$ $$ \int\limits_{\partial B(x,t)} t |\psi(y)| \, dS_y \le t \min\{c R^2, A_t \} \|\psi\|_\infty , $$ $$ \int\limits_{\partial B(x,t)} |(y-x)\cdot\nabla\phi(y)| \, dS_y \le 2t \min\{c R^2, A_t \} \|\nabla\phi\|_\infty . $$ So $$ |u(x,t)| \le \min\{c R^2, A_t \} (\|\phi\|_\infty + t \|\psi\|_\infty + 2t \|\nabla\phi\|_\infty) / A_t ,$$ and since $A_t = 4\pi t^2$, the result follows by considering the cases $t \gg 1$ and $t \ll 1$ separately.

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