Consider the Cauchy problem for wave equation in space $\mathbb R^3$
$$\begin{cases}\partial_t^2u-\Delta u=0\\
(u,\partial_t u)|_{t=0}=(0,u_1)\end{cases}$$
for $u_1\in C^1(\mathbb R^3)$. This problem has an integral representation for the solution namely, Kirchoff's formula:
$$u(t,x)=\frac{1}{4\pi t}\int_{S(x,t)} u_1(y)d\sigma$$
where $S(x,t)=\partial B(x,t)=\{y\in \mathbb R^3| |x-y|=t\}$ and $\sigma$ is a surface measure on the sphere. I am trying to prove that there exists a $C>0$, such that for all $t\in\mathbb R_{\ge 0}$ and $u_1\in W^{1,1}(\mathbb R^3)$,
$$\Vert u(t,\cdot)\Vert_{L^\infty}\le C\bigg(\frac{\Vert \nabla u_1\Vert _{L^1}}{|t|}+\frac{\Vert u_1\Vert _{L^1}}{t^2}\bigg). $$
My attempt: For $\hat y:=\frac{y}{|y|}$, using divergence theorem,
$$ \begin{aligned}
\int_{S(x,t)} u_1(y)d\sigma&=\int_{S(0,t)}u_1(x+y)\,(\hat y\cdot n )\,d\sigma\\
&= \int_{B(0,t)} \hat y\cdot\nabla u_1(x+y)\,dy+ \int_{B(0,t)}(\nabla \cdot \hat y)\,u_1(x+y)\,dy
\end{aligned}$$
so
$$\bigg|\int_{S(x,t)} u_1(y)d\sigma\,\bigg|\le\int_{B(0,t)}|\nabla u_1(x+y)|\,dy+
\bigg|\int_{B(0,t)}\frac{2u_1(x+y)}{|y|}dy\,\bigg|$$
we can bound the first term by $\Vert \nabla u_1\Vert _{L^1}$ so it suffices to prove
$$\bigg|\int_{B(0,t)}\frac{u_1(x+y)}{|y|}dy\,\bigg|\le C\bigg(\Vert \nabla u_1\Vert _{L^1}+\frac{\Vert u_1\Vert _{L^1}}{|t|}\bigg)$$
for all $x\in \mathbb R^3$ i.e.
$$ \Big\Vert \bigg(\frac{\mathbf 1_{B(0,t)}}{|\cdot|}\bigg)\star u_1\Big\Vert_{L^\infty}\le C\bigg(\Vert \nabla u_1\Vert _{L^1}+\frac{\Vert u_1\Vert _{L^1}}{|t|}\bigg)$$
and this is where I got stuck. I tried using Young's inequality for convolutions and Sobolev embedding but none of it seem to give the bound with the factor $\frac{1}{|t|}$ I want.
Thank you in advance for any help.
Best Answer
By the divergence theorem, \begin{align*}t \int_{S(x,t)} u(y)\, d \sigma(y) &= \int_{B(x,t)} \mathrm{div}((y-x)u)\, dy \\ &=3 \int_{B(x,t)} u (y)\, d y+ \int_{B(x,t)} \nabla u(y) \cdot (y-x) \, d y \\ &\leqslant 3 \| u \|_{L^1}+ t \| \nabla u \|_{L^1} . \end{align*} Hence, $$u(t,x) \leqslant \frac3{4 \pi t^2} \| u \|_{L^1}+ \frac1{4\pi t} \| \nabla u \|_{L^1} $$ as required.