Harmonic function properties on $\mathbb R^3$

Question

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.

Answer 1

This problem has an important background in geomagnetism. When planning the MAGSAT satellite mission (1979/80) to determine the spherical harmonic coefficients of the Earth's magnetic field from space, Backus (JGR, 1970) showed that a measurement of the total field intensity $||\nabla V||$ on a spherical shell is in general not sufficient to uniquely determine $V$ (not regarding trivial non-uniqueness due to gauge and sign). He did this by explicitly constructing some counterexample by means of similar arguments as used in the question and comments here. As a consequence, MAGSAT became the first mission carrying a vector magnetometer instead of the much simpler absolute field sensors. In relation to the problem as posed here, Backus (Quart. Journ. Mech. and Applied Math., 21, 195-221 , 1968) proved the following theorem:

THEOREM 5: Suppose $\phi$ and $\phi'$ are harmonic outside some open bounded set $W$ in $\mathbf{R}^n$ and vanish at infinity. Suppose that $| \nabla \phi| = | \nabla \phi'|$ outside some sphere which contains $W$. If $n \geq 3$ then one of the two functions $\phi -\phi'$ and $\phi +\phi'$ vanishes identically outside $W$.

Answer 2

I don't have a solution for the general question, but consider a special case where we pick $\phi$ to be a simple dipole, $$ \phi = {\cos{\theta} \over r^2} = \phi_{10} {Y_{10} \over r^2} $$ where $\phi_{10} = {1 \over 2} \sqrt{{3 \over \pi}}$ is a constant which isn't really important. This choice of $\phi$ is harmonic, and so the summation expression I gave above will reduce to the following: $$ \sum_{nm} \psi_{nm} \left({1 \over r} \right)^{n+5} \left[ 2(n+1) \cos{\theta} Y_{nm} - \sin{\theta} \partial_{\theta} Y_{nm} \right] = 0 $$ Applying the recurrence relation for the spherical harmonics: \begin{align} x Y_{nm} &= \alpha_{nm} Y_{n-1,m} + \alpha_{n+1,m} Y_{n+1,m} \\ (1-x^2)\partial_x Y_{nm} &= (n+1)\alpha_{nm} Y_{n-1,m} - n \alpha_{n+1,m}Y_{n+1,m} \end{align} where $x = \cos{\theta}$ and $$ \alpha_{nm} = \sqrt{ \frac{(n+m)(n-m)}{(2n+1)(2n-1)} } $$ Putting these recurrence relations into the above sum, and then relabeling indices to isolate all the $Y_{nm}$ terms, we get $$ \sum_{nm} Y_{nm} \left( 3(n+2)\alpha_{n+1,m}\psi_{n+1,m} + (n+1)\alpha_{nm} \psi_{n-1,m} \right) = 0 $$ To get this I had to assume $\psi_{|m|-1,m} = 0$. Since the $Y_{nm}$ are complete, this must mean that $$ 3(n+2)\alpha_{n+1,m}\psi_{n+1,m} + (n+1)\alpha_{nm} \psi_{n-1,m} = 0 $$ or, $$ \psi_{n+1,m} = -\frac{(n+1)}{3(n+2)} \frac{\alpha_{nm}}{\alpha_{n+1,m}} \psi_{n-1,m} $$ So this indicates I am free to choose any $\psi_{|m|,m}$ I want, and then the relation will determine $\psi_{|m|+2k,m}$ for all positive integers $k$.

This means that there are in fact nontrivial solutions $\psi$ which satisfy $\nabla \phi \cdot \nabla \psi = 0$ for my specific choice of $\phi$ (dipole field). In fact, there are infinitely many choices of $\psi$ which will satisfy that relation. At this point I don't have any intuition as to what these $\psi$ solutions look like geometrically, compared to my choice of $\phi$. Maybe it would be worth plotting some surface maps to see if there is some interesting pattern which would motivate a solution to the general problem.