Let $X$ be the set of all harmonic functions external to the unit sphere on $\mathbb R^3$ which vanish at infinity, so if $V \in X$, then $\nabla^2 V(\mathbf{r}) = 0$ on $\mathbb R^3 - S(2)$ and $\lim_{r \rightarrow \infty} V(r) = 0$. Now consider a function $f: X \rightarrow \mathbb R$, defined by $$ f(V)(\mathbf{r}) = || \nabla V(\mathbf{r}) ||^2 $$ For some given $V \in X$, I am looking for all functions $W \in X$ which satisfy $$ f(V) = f(W) $$ Certainly $W = \pm V$ will satisfy the condition. Can anyone find nontrivial solutions for $W$?

**My approach so far:**

The condition on $V$ and $W$ is $$ \nabla V \cdot \nabla V = \nabla W \cdot \nabla W $$ By defining $\phi = V + W$ and $\psi = V - W$, this is equivalent to $$ \nabla \phi \cdot \nabla \psi = 0 $$ I then tried expanding $\nabla \phi$ and $\nabla \psi$ in a basis of vector spherical harmonics and plugging into the above formula. This step makes use of the fact $\nabla^2 \phi = \nabla^2 \psi = 0$ and leads to the following condition on the expansion coefficients: $$ \nabla \phi \cdot \nabla \psi = \sum_{nm,n'm'} \phi_{nm} \psi_{n'm'} \left( \frac{1}{r} \right)^{n+n'+4} \left( (n+1)(n'+1) Y_{nm} Y_{n'm'} + \partial_{\theta} Y_{nm} \partial_{\theta} Y_{n'm'} + \frac{1}{\sin^2{\theta}} \partial_{\phi} Y_{nm} \partial_{\phi} Y_{n'm'} \right) $$ Its not clear to me how to proceed from here, or whether this is even the correct approach to take. I could get rid of the sum over $n',m'$ by integrating both sides over a unit sphere and using the orthogonality relations for the spherical harmonics. Doing this gives: $$ \sum_{nm} (n+1)(2n+1) \phi_{nm} \psi_{nm} = 0 $$ though I'm not sure that yields any additional insight. I would appreciate any ideas.

isconstant, we can assume that it is $1$, in which case $(\phi,\psi)$ is a harmonic morphism from the domain to $\mathbb{R}^2$. Hence its fibers are lines, which your global assumptions do not allow. $\endgroup$ – Robert Bryant Jun 30 '17 at 11:15