Let $\Omega$ be an annulus with inner radius $R_1$ and outer radius $R_2$, and $\beta>0$. Let $u$ solve the equation $$ \begin{cases} -\Delta u=1\quad &\mbox{in $\Omega$}\\ \frac{\partial u}{\partial\nu}+\beta u=0\quad &\mbox{on } \partial \Omega \end{cases} $$ Is $u$ necessarily radial for every $\beta>0$? If so, is it possible to explicitly write down the solution?
1 Answers
Solutions in $C^2(\Omega) \cap C^1(\bar{\Omega})$ are indeed unique (though, of course, this can get swapped for Sobolev spaces or others if you prefer, but the basic idea is similar). If you have two solutions $u_1$ and $u_2$, let $u = u_1 - u_2$, multiply the PDE they satisfy by $u$ and integrate by parts. You'll find that $$ \int_\Omega |\nabla u|^2 + \int_{\partial \Omega} \beta |u|^2 =0 $$ and hence that $0 = u = u_1 - u_2$ in $\Omega$.
Once you have uniqueness in hand, you can just propose a radial ansatz and see if it works. If it does, then that's the solution. A radial solution $u(x) = v(|x|)$ must satisfy $$ v''(r) + \frac{(n-1)}{r} v'(r) = -1 $$ and thus $$ (r^{n-1} v'(r)) ' = - r^{n-1}. $$ Integrate twice to get the general form with two free coefficients, and then plug into the BCs to determine the coefficients. I'll leave this to you to work out.
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1+1 Great answer. As a small comment for the OP, once you know solutions are unique another method to prove radial symmetry is to prove that $v(x) :=u( O x)$ is also a solution to the BVP where $O$ is an arbitrary orthogonal matrix. By uniqueness, this implies that $u$ is invariant under rotations about the origin, so it is radial. – JackT Oct 20 '22 at 12:31
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Yes, good point @JackT. Thanks for the addition! – Glitch Oct 20 '22 at 19:15