Assume $u:\mathbb{R^{3}\times \mathbb{R}\rightarrow \mathbb{R}}$ is smooth and satisfies $u_{tt}=\Delta u, u(x,0)=0, u_{t}(x,0)=h(x)$ for $x \in \mathbb{R^{3}}$. Fix $t>0$ and let $f(x)=u(x,t-|x|), g(x)=u_{t}(x,t-|x|)$. Show $div(|x|^{-1}Df+x|x|^{-3}f+2x|x|^{-2}g)=0$.
I firstly use product rule to write $div(|x|^{-1}Df+x|x|^{-3}f+2x|x|^{-2}g)=|x|^{-1}\Delta f+\nabla |x|^{-1} \cdot f+x|x|^{-3}\nabla f+\nabla (x|x|^{-3}) \cdot f+2x|x|^{-2}\nabla g + \nabla (2x|x|^{-2}) \cdot g$
then I computed $\nabla |x|^{-1}=-x|x|^{-3}, \nabla x|x|^{-2} =-|x|^{-2} $ and $\nabla x|x|^{-3}=-2|x|^{-3} $
though I am not sure if it's right.
But after some computation it doesn't seem to give me a zero divergence.
Please help. Thanks!