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Let $M = \mathbb{R}^3 \setminus B_1$ where $B_1$ is the unit ball (equip $M$ with the euclidean metric for simplicity, but it will be replaced by an arbitrary asymptotically flat metric). Let $N: L^2(\partial M) \to L^2(\partial M)$ be the Dirichlet to Neumann map defined by $N(f) = \frac{\partial u}{\partial r}$ where $u$ is the unique harmonic extension of $f$ on $M$ going to $0$ at infinity.

Consider the operator $H: H^1(\partial M) \to L^2(\partial M)$ defined by $H(f) = |\nabla f|^2 - N(f)^2 $ where $\nabla$ is the covariant derivative with respect to the round sphere $\partial M$.

What do we know about the solvability of the PDE: $$H(f) = g$$ where $g \in L^2 (\partial M)$? (or any other function space that makes the question easy to answer). How do I approach this problem? Also $H$ should be a pseudo-differential operator. How do I go about finding its order?

What are the different methods/tools for showing an operator like $H$ is surjective?

What if $M$ is equipped with an arbitrary asymptotically flat metric? Does that affect the solvability? It is known that the first and 0^th order of $N$ only depend on the metric and second fundamental form on $\partial M$; that might suggest that that the solvability is independent of the metric on $M$ but only depends on intrinsic and extrensic geometry on $\partial M$.

Any suggestions or references are appreciated.

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  • $\begingroup$ This seems to be stated in a peculiar way. Isn't H(f) just the square of the tangential gradient (which would make the harmonic extension irrelevant)? $\endgroup$
    – Michael Renardy
    Commented Oct 31, 2020 at 20:54
  • $\begingroup$ $\nabla$ is the covariant derivative with respect to $\partial M$. I edited the question to make this clear. So it doesn't cancel with $N(f)^2$. $\endgroup$
    – Laithy
    Commented Oct 31, 2020 at 21:18
  • $\begingroup$ What is the covariant derivative with respect to the round sphere? The tangential gradient or something else? $\endgroup$
    – Andrew
    Commented Nov 1, 2020 at 15:26
  • $\begingroup$ Yes the tangential gradient (angular derivatives). Also, notice that $f$ is a function on the sphere. Is there a better term for it? Let me know and I will edit the question. In the case of the round sphere, $\nabla f = \frac{\partial f}{\partial \theta} \frac{\partial}{\partial \theta} + \frac{1}{\sin \theta}\frac{\partial f}{\partial \phi} \frac{\partial}{\partial \phi}$ $\endgroup$
    – Laithy
    Commented Nov 1, 2020 at 20:10
  • $\begingroup$ Have you tried the model case: $M$ is a halfspace and $\partial M$ is a plane? It's easy to write out $H(f)$ but the equation $H(f) = g$ doesn't look promising even for this. $\endgroup$
    – Andrew
    Commented Nov 5, 2020 at 12:09

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