Let $k$ be a non-zero real number. Consider the problem

$$ \nabla^2 \phi = 0 \ \ \ \mbox{for} \ \ \ 1 \leq r \leq 2, \ \ \ \ \alpha\phi + \frac{\partial\phi}{\partial r} = k\cos\theta \ \ \ \mbox{on} \ \ r = 1, \ \ \alpha\phi + \frac{\partial\phi}{\partial r} = 0 \ \ \ \mbox{on} \ \ r = 2 $$

What can you say about existence and uniqueness of the solution to the problem in terms of $\alpha$?

For uniqueness, I am aware e.g. that for $\alpha > 0$ it holds (well-known for Robin conditions) and that for $\alpha = 0$ it holds up to additive constants.

However, now with the Fourier series approach I got a unique solution (respecting the general solution to Laplace's equation) for $\alpha \neq 0, \frac{5\pm \sqrt{97}}{12}, \frac{1}{2\log 2}$, and existence for $\alpha \neq \frac{5\pm \sqrt{97}}{12}$.

My solutions for $\alpha = \frac{1}{2\log 2}$ are $c(2\log 2 - \log r) + (A_1r + B_1/r)\cos\theta$ where $c$ is arbitrary and $A_1, B_1$ satisfy $k=( \alpha + 1)A_1 + (\alpha - 1)B_1$ and $(2\alpha + 1)A_1 + (\frac{\alpha}{2} - \frac{1}{4})B_1 = 0$. How does this happen, as $\frac{1}{2\log 2} > 0$?

Any help appreciated!

  • $\begingroup$ How do you know the solution you got using the Fourier series approach is unique? and why exactly is the fact that $1/(2log2)>0$ bothers you? $\endgroup$ – MOMO Oct 25 '18 at 21:30
  • $\begingroup$ I just started from the general solution to Laplace in Polar Coordinates (which involves $A + B\log r + \sum(A_nr^n+B_nr^{-n})(C_n\cos(n\theta) + D_n\sin(n\theta))$ and substituted in the boundary conditions. Also, 1/(2log2) > 0 bothers me, since as said above, for positive $\alpha$ there must be unique solution (Robin conditions). $\endgroup$ – DesmondMiles Oct 27 '18 at 0:51
  • $\begingroup$ This general solution is not necessarily unique so you did not prove uniqueness for $\alpha\not = 1/(2log2)$. But even if you did, it still doesn't contradict the fact that there is unique solution for $\alpha = 1/(2log2)$, so there is no problem. $\endgroup$ – MOMO Oct 27 '18 at 20:46
  • $\begingroup$ No, I did as follows: 1)I have found two solutions for 1/(2log2) using the general one (it doesn't matter how I found them - it is just that both are solutions) 2) Uniqueness for $\alpha > 0$ (Robin conditions) follows by applying the Divergence theorem in a suitable way. $\endgroup$ – DesmondMiles Oct 28 '18 at 0:56
  • $\begingroup$ This is the first time you mentioned you have found two solutions. You already know it is not possible, so one of your solutions must be false. If you will describe how you found those solution maybe someone can point your mistake. $\endgroup$ – MOMO Nov 3 '18 at 9:56

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