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Let $u$ be a solution to $$\begin{cases}\Box u =0,\; \; (t,x)\in \mathbb{R}_+ \times \mathbb{R}^2\\(u,u_t)\restriction_{t=0} = (f,g),\end{cases}$$ where $f,g$ are smooth functions with compact support. I'd like to show that $|u(x,t)|\leq C t^{-\frac 12}$.

To do so, I try to use the Poisson formula given the solution as $$u(t,x) = \underbrace{\int_{B(x,t)} \frac{g(y)}{\sqrt{t+|y-x|} \sqrt{t-|y-x|}} \text{d}y }_{I}+\underbrace{ \partial_t \int_{B(x,t)} \frac{f(y)}{\sqrt{t+|y-x|} \sqrt{t-|y-x|}} \text{d}y}_{II}.$$

I split integral $I$ in $I_a$ where in $I_a$, I integrate in the region $t-|y-x|\geq 1$. However in the region $1\geq t-|y-x|$ I don't know how to conlude. To be more precise, I would like to have the following estimate

$$\int_{B(x,t)\cap \{1\geq t-|y-x|\}} \frac{g(y)}{\sqrt{t+|y-x|} \sqrt{t-|y-x|}} \text{d}y \leq \frac{C}{\sqrt{t}},$$ where $C$ is independent from $x$ and $t$.

Has someone a hint for that?

Peter
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1 Answers1

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Since $g$ is bounded and has bounded support, we have $g\le M\chi_{B(0,R)}$ for some $M,R>0$. So, the integral (I) can be estimated as follows, letting $r=|y-x|$: $$ I\le M\int_0^t \frac{\operatorname{Area}( S(x,r)\cap B(0,R))}{\sqrt{(t-r)(t+r) }} \,dr \le \frac{M}{\sqrt{t}}\int_{\max(0,|x|-R)}^{\min(t,|x|+R)} \frac{ 4\pi R^2 }{\sqrt{t -r }} \,dr $$ Here $S(x,r)$ is a sphere; its intersection with $B(0,R)$ is empty outside of the stated interval for $r$. Also, a piece of sphere contained within $B(0,R)$ cannot have area more than $4\pi R^2$, see Surface area of a convex set less than that of its enclosing sphere?

The latter integral is taken over a subinterval of length $2R$ contained in $[0,t]$. Since the function $1/\sqrt{t-r}$ is increasing, the largest this integral can be is $$ \int_{t-2R}^t \frac{ 4\pi R^2 }{\sqrt{t -r }} \,dr = 8\pi R^{2}(2R)^{1/2} $$

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