Sources: this is an old advanced calculus exam question, which I think is asking for harmonic functions.
The problem statement is:
Suppose $F(r)$ is a smooth function of $r$ for $r>0$ . Define $\Phi(x) = F(|x|)$ with $x = (x_1, ...,x_d) \in \mathbb{R}^d$ and $|x| = (\large \sum_i^dx_i^2)^{\frac{1}{2}}$.For $d=2$ and $d=3$, find all possible functions $F(r)$ so that $$\sum_{i=1}^d (\frac {\partial}{\partial x_i})^2\Phi(x) = 0.$$
Any hints or comments are welcome.
So, it appears that this problem statement is just a tricky way of asking to find all functions $\Phi$ such that
$$\Phi_{x_1x_1} + \Phi_{x_2x_2} = 0 $$
and
$$\Phi_{x_1x_1} + \Phi_{x_2x_2} + \Phi_{x_3x_3} = 0$$
I.e., we are looking for all functions that are harmonic. But what is weird about this question is that it doesn't really specify a domain in which we are to find these harmonic functions. Unless the domain is simply understood to be for all $| x| \ne 0$.
Thanks,
It's easy enough to carry out the calculation for all $d$. $$ \frac{\partial^2 }{\partial x_j^2}F(r)=\frac{\partial}{\partial x_j}\left(F'(r)\frac{x_j}{r}\right)=F''(r)\frac{x_j^2}{r^2}+F'(r)\frac{1}{r}-F'(r)\frac{x_j^2}{r^3} $$ Adding these together gives \begin{align} \nabla^2 F(r) &= F''(r)+F'(r)\frac{d}{r}-F'(r)\frac{1}{r} \\ &= F''(r)+F'(r)\frac{d-1}{r}. \end{align} In order to have a harmonic function $F(r)$ for $d \ge 2$, it is necessary and sufficient that, for $d > 1$, $$ rF''(r)+(d-1)F'(r)=0 \\ (r^{d-1}F'(r))'=0,\\ F'(r)=\frac{C}{r^{d-1}}. $$ For $d=2$, the solutions are $$ F(r)=A\ln r + B $$ For $d\ge 3$, the solutions are $$ F(r)= A\frac{1}{r^{d-2}}+B. $$