# Radial solution in Wave equation

Let $u$ be a solution of the initial wave equation in $\mathbb{R}^n$: $$u_{tt}-\Delta u = 0 \mbox{ in } \mathbb{R}^n\times\mathbb{R}_+ \quad ; \quad u(x,0) = g(x) \quad ; \quad u_t(x,0) = h(x).$$ Assume $g$ and $h$ are smooth. Then show that $u$ is radial if and only if $g$ and $h$ are radial.

Note: Solutions are given in Evens' book.

1) For $n\geq 2$ even, we have $$u(x,t)=\frac{1}{\gamma_n}\left[ \partial_t \left( \frac{1}{t}\partial_t \right)^\frac{n-2}{n} \left( t^n \int_{B_t(x)}^{avg} \frac{g}{(t^2-|y-x|^2)^\frac{1}{2}}dy \right) + \left( \frac{1}{t}\partial_t \right)^\frac{n-2}{n} \left( t^n \int_{B_t(x)}^{avg} \frac{h}{(t^2-|y-x|^2)^\frac{1}{2}}dy \right)\right],$$ where $\gamma_n=2\cdot4\cdots n$.

2) For $n\geq 3$ odd, we have $$u(x,t)=\frac{1}{\gamma_n}\left[ \partial_t \left( \frac{1}{t}\partial_t \right)^\frac{n-3}{n} \left( t^{n-2} \int_{\partial B_t(x)}^{avg} g dS_y \right) + \left( \frac{1}{t}\partial_t \right)^\frac{n-3}{n} \left( t^{n-2} \int_{\partial B_t(x)}^{avg} h dS_y \right)\right],$$ where $\gamma_n=1\cdot3\cdots n-2$.

Can we just look at the formulae and say "since $g$ and $h$ are radial, $u$ is radial. We're done in 1 direction"? How about the other direction?

Thank you.

For instance, since all derivatives are with respect to $t$ in the representation formulas, it's enough to prove that the integrals are radial functions of $x$. I'll only treat one, the others being similar. Let's look at the integral for $g$ in the case of even dimension, call it $G$, so that $$G(x,t)= \int_{B_t(x)} \frac{g(y)}{(t^2 + |y-x|^2)^{1/2}} dy = \int_{B_t(0)} \frac{g(x+y)}{(t^2 + |y|^2)^{1/2}}dy.$$ Take $A\in O(n)$ an orthogonal matrix so that we want to prove $G(x,t)=G(Ax,t)$ for every $x$ and $t$. But notice that $g(Ax+y)=g(A(x+A^{-1}y))=g(x+A^{-1}y)$ where we used that $g$ is radial for the last equality. Moreover, again by orthogonality of $A$ we have $|Az|^2=|z|^2$ for every $z\in \mathbb{R}^n$. Therefore, using all this and a change of variables we arrive at $$G(Ax,t)= \int_{B_t(0)} \frac{g(Ax+y)}{(t^2 + |y|^2)^{1/2}}dy= \int_{B_t(0)} \frac{g(x+A^{-1}y)}{(t^2 + |y|^2)^{1/2}}dy= \int_{B_t(0)} \frac{g(x+y)}{(t^2 + |Ay|^2)^{1/2}}dy= G(x,t).$$ We conclude $G$ is radial.
For the other direction, just notice that if $u(x,t)$ is radial in $x$ for every $t$, in particular it's radial for $t=0$, but $u(x,0)=g(x)$ so that $g$ is radial. As before, we also have that $u_t$ is radial whenever $u$ is so that $h$ is also radial.
Edit: A more elementary approach: Recall that the solution of the intitial value problem is unique. Then, assume $g,\ h$ are radial and $u$ is the solution of this problem. Then for any orthogonal transformation $A$ we set $u_A=u(t,Ax)$, and notice that $g_A=g$ and $h_A=h$ so that by invariance of $\Delta$ under orthogonal transformations we get that $u_A$ is a solution to the same problem as $u$ and by uniqueness $u_A=u$ and therefore $u$ is radial.
• @dh87: I just added an easier approach to prove that $u$ is radial. Hope it helps. Commented Apr 7, 2015 at 1:55