Proof in multi-variable functions I am facing this question and can't get my head around where to start. I have tried a couple of things.
let $r=\sqrt{x^2+y^2+z^2}$ and $u=f(r)$, and $f_{xx} + f_{yy}+f_{zz}=0$. Then prove that there exists $a$,$b$ $\in \mathbb{R}$ where $f(r)=\frac{a}{r}+b$. I know the rules say that you should write some stuff about the things you have tried, but I can't wrap my head around where to start.
 A: The question is a bit unclear, but you are essentially asking for solutions of Laplace's equation, $\nabla^2 U=0$, in spherical coordinates. And yes, it so happens that $U(r,\theta,\phi)=\frac{a}{r}+b ~$ is a particular (but not general) solution.
This comes from the fact that the scalar Laplacian is
$$\nabla^2 U(r,\theta,\phi)=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial U}{\partial r}\right)+\frac{1}{r^2\sin^2(\phi)}\frac{\partial^2 U}{\partial \theta^2}+\frac{1}{r^2\sin(\phi)}\frac{\partial}{\partial \phi}\left(\sin(\phi)\frac{\partial U}{\partial \phi}\right).$$
Assuming no $\theta$ or $\phi$ dependence this can be reduced to
$$\nabla^2 U(r)=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial U}{\partial r}\right)=0.$$
We notice that a constant function $U(r)=b$ works. Multiplying both sides by $r^2$,
$$\frac{\partial}{\partial r}\left(r^2\frac{\partial U}{\partial r}\right)=0$$
$$r^2\frac{\partial U}{\partial r}=A$$
$$\frac{\partial U}{\partial r}=\frac{A}{r^2}\implies U(r)=\frac{-A}{r}=\frac{a}{r}.$$
We have found two linearly independent solutions, and since this ODE is of second order, these must be the only two solutions. Thus the "general" solution is a linear combination:
$$U(r)=\frac{a}{r}+b.$$
