Does $\Delta u = f(|x|)$ mean that $u=v(|x|)$ for some $v$? If $u$ solves $\Delta u = f(|x|)$ in $B_1 \subset \mathbb{R}^n$ for some continuous function $f: [0,\infty) \to \mathbb{R}$ does that mean that $u(x) = v(|x|)$ for some function $v$?
I think this is true result about radially symmetric solutions, but I may need need one more condition that $u = 0$ on $\partial B_1$. Anyway, I am having a difficult time showing this result. 
 A: Hint: This might not be true 
Take $$u(x) = \sum_{i=1}^{n}x_i \implies \Delta u  = 0 $$
or 
 $$u(x) =\sum_{i=1}^{n}x_i^4\implies \Delta u  = 12|x|^2$$
 $$u(x) =\sum_{i=1}^{n}(x_i^4 +x_i)\implies \Delta u  = 12|x|^2 $$
A: Assume that the boundary conditions produce a unique solution, or at least unique up to a constant.
The magic of linearity allows you to split $u$ into two components,
$$ u(x) = w(x) + v(\lvert x \rvert), $$
where $\Delta w = 0 $ and $w$ satisfies the boundary conditions, and $\Delta v = f $ and $v$ satisfies homogeneous boundary conditions; $v$ is radial, at least in this situation of overall uniqueness (one can explicitly produce a radial solution with one-dimensional integration, and this is only one). Of course it also follows that $w$ is unique. One can also write down an expression for $w$ explicitly using the appropriate Dirichlet/Green's function integral.
An alternative in low dimensions would be to use spherical harmonic expansions, where one has a radial component and then a load of angular-dependent components.
A: Hint What happens if you take $u : \Bbb R^n \to \Bbb R$ to be any nonradial harmonic function (e.g., the projection onto the first component)?
