# $u$ is a spherical wave $\iff$ $f$ and $g$ are radial

Consider the Cauchy problem for the wave equation : $$\begin{cases} u_{tt}= \Delta u \\u(x,0)=f(x) \\ u_t(x,0)= g(x)\end{cases}$$ with $$t>0$$ and $$x\in\mathbb{R^d}$$

This is an exercise from Stein's Fourier Analysis an Introduction chapter 6 :

A spherical wave is a solution $$u(x; t)$$ of the Cauchy problem for the wave equation in $$\mathbb{R^d}$$, which as a function of $$x$$ is radial. Prove that $$u$$ is a spherical wave if and only if the initial data $$f; g \in \mathcal S$$ are both radial, where $$\mathcal S$$ denotes the Schwartz space.

I have already done the $$\Leftarrow$$ direction , my doubt is on the $$\Rightarrow$$ :

My thought: if $$u$$ is a solution then $$u(x,0) = f(x)$$ and because $$u(x,0)$$ depends only on $$x$$ and $$u$$ is radial as a function of $$x$$ (by the definition of spherical wave above) then $$f$$ is radial by definition since $$h(x)$$ is radial if $$h(x)= h_0(|x|)$$ for some $$h_0$$ . In an analogous way we obtain the result for $$g$$.

Is this ok ? Or is there a better way to prove it ?

• I'd look at the contrapositive (if $f$ is not radial....) and use the continuity at $t$ goes to $0$. – Paul Nov 26 '18 at 21:11
• The hard part of the "complete solution" is to find the Green function is radial so radial initial conditions is equivalent to radial solution. – reuns Nov 27 '18 at 4:12