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Let $M = \mathbb{R}^3 \setminus B_1$ where $B_1$ is unit ball. I am trying to solve the following PDE for $f$:

$$\Delta f -\frac{ f }{r^2}+ \frac{ \left. f \right|_{\partial M}}{r^2} = 0, \qquad \text{on} \, M$$ $$f + a \partial_r f = h, \qquad \text{on} \, \partial M$$ $$\lim_{|x|\to \infty} f= 0 $$

where $a>0$ is a constant, $r = |x|$, $h \in L^2(\partial M)$ (or any nice function space you want). Also, $\left. f \right|_{\partial M}$ is the function $f$ restricted to $r=1$; you can see it as $f(\frac{x}{|x|})$.

Note that this PDE has a nonlocal part, namely $\left. f \right|_{\partial M} $, and so it's not really a quasilinear PDE.

Also note that it is easy to see right away that $\sup_{x \in M} f(x) \leq \max\{\sup_{x\in \partial M}f,0\} $. This can be proven by contradiction; if the maximum is in the interior, then $\Delta f<0$ implying $f - \left. f \right|_{\partial M} <0$, which is a contradiction.

How do I prove existence and uniqueness? Can Perron method be applied?

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    $\begingroup$ What is $r$? $\partial M=S^2$? In the first equation summands seem to have different domains. It is not clear also what the third equation means. The limit is a function in at least 2 variables. $\endgroup$
    – markvs
    Commented Jan 31, 2021 at 5:05
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    $\begingroup$ What do you mean by $\left. f \right|_{\partial M}$? Is supposed to be a function of the angular variables at $r=1$, or is it supposed to be a constant, implying that $f$ needs to be constant on $\partial M$? The latter case adds an addition boundary condition. The two boundary conditions might overspecify the system, meaning that the solution might not exist. $\endgroup$
    – Igor Khavkine
    Commented Jan 31, 2021 at 8:52
  • $\begingroup$ @dodd I edited the question. $r=|x|$. Yes, $\partial M = S^2$. The third equation says the function goes to $0$ at infinity. $\endgroup$
    – Laithy
    Commented Jan 31, 2021 at 14:07
  • $\begingroup$ @IgorKhavkine $\left. f \right|_{\partial M}$ is the function restricted to the boundary. So it's the function $f$ at $r=1$. Or you can think of it as $f(\frac{x}{|x|})$. $\endgroup$
    – Laithy
    Commented Jan 31, 2021 at 14:09

1 Answer 1

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One can try the Kelvin transforn and expansion of solutions into series of spherical harmonics in $B_1$. Since the equation is rotationally inveriant it should lead to simple enough equations for the harmonics' coefficients.

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  • $\begingroup$ you get an ODE that looks similar to the PDE: $\lambda''(r) + 2r \lambda'(r) - b\lambda(r) + \lambda(1)$ with boundary conditions $\lambda(1) + a \lambda'(1) = $known. (b is some constant depending on the eignevalue of the harmonics). How do I solve that? It also has a nonlocal component. $\endgroup$
    – Laithy
    Commented Jan 31, 2021 at 23:07
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    $\begingroup$ Replace $\lambda(1)$ by an unspecified parameter $c$ to make the ODE local and then add the additional boundary condition $\lambda(1)=c$. Presumably the ODE you obtain can be solved exactly using Bessel or Hankel functions, either by hand or by using some standard symbolic computing package (Maple, Mathematica, SAGE, etc.). You may end up with some implicit equation for $c$ that then requires further analysis to solve but at least no further differential equations are involved. $\endgroup$
    – Terry Tao
    Commented Jan 31, 2021 at 23:49
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    $\begingroup$ It should read $r^2\lambda''(r)+2r\lambda'(r)-b\lambda(r)+\lambda(1)=0$. This is an Euler equation. No Bessel functions needed. $\endgroup$
    – Michael Renardy
    Commented Feb 1, 2021 at 4:09
  • $\begingroup$ Thank you Andrew, Terry, and Michael. :) I was trying to use Perron method since solutions to the PDE satisfy the maximum principle. But I was stuck. This is easier; it seems I get existence and uniqueness for the PDE. $\endgroup$
    – Laithy
    Commented Feb 1, 2021 at 6:04

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