Given $\Sigma_\varepsilon = \left\lbrace (x,y)\in\mathbb{R}^2:\left\lvert\sqrt{x^2+y^2}-1\right\rvert\le \varepsilon \right\rbrace$ for some small $\varepsilon>0$.

I want to determine analytically all the functions $u$ and constants $\lambda$ which satisfy the following conditions:

  1. $-\Delta u=\lambda u \;$ for $\,(x,y) \in \Sigma_\varepsilon$


  1. $ \;\vec{n} \cdot \nabla u=0\;$ for $\,(x,y) \in \partial\Sigma_\varepsilon$ (the boundary of $\Sigma_\varepsilon$)

where $\Delta$ is the standard Laplacian, $ \vec{n}$ is the normal vector (normal to the boundary of $\Sigma_\varepsilon$), and $\nabla$ is the gradient.

I found $u=a\sin\left(n\theta\right)$, $u=b\cos \left(n\theta\right)$ and $u=a\sin\left(n\theta\right)+b\cos \left(n\theta\right)$ for $n \in \mathbb{Z}$ (which are constant along normal direction) works.

Are there any other solutions? In particular, are there any solution which are not constant along normal direction?


Separation of variables works for this equation. The region you're dealing with is an annulus where $1-\epsilon \le r \le 1+\epsilon$, and the normal derivative condition for a solution $u(r,\theta)$ translates to $u_{r}(r,\theta)=0$ for $r=1\pm \epsilon$. The Laplacian in cylindrical coordinates is $$ \nabla^2 u = \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2u}{\partial\theta^2}. $$ Assuming solutions of the form $u(r,\theta)=R(r)\Theta(\theta)$ in $\nabla^2u=-\lambda u$ leads to $$ \frac{1}{r}(rR'(r))'\Theta(\theta)+\frac{R(r)}{r^2}\Theta''(\theta)=-\lambda R(r)\Theta(\theta) \\ R''(r)\Theta(\theta)+\frac{1}{r}R'(r)\Theta(\theta)+\frac{R(r)}{r^2}\Theta''(\theta)=-\lambda R(r)\theta(\theta) $$ Dividing by $R(r)\Theta(\theta)$ allows a separation of variables $$ \frac{R''(r)}{R(r)}+\frac{1}{r}\frac{R'(r)}{R(r)}+\frac{1}{r^2}\frac{\Theta''(\theta)}{\Theta(\theta)}=-\lambda \\ r^2\frac{R''(r)}{R(r)}+r\frac{R'(r)}{R(r)}+\lambda r^2 = -\frac{\Theta''(\theta)}{\Theta(\theta)} \\ r^2\frac{R''(r)}{R(r)}+r\frac{R'(r)}{R(r)}+\lambda r^2 = \mu,\;\;\; \mu=-\frac{\Theta''(\theta)}{\Theta(\theta)} \\ R'(1-\epsilon)=0=R'(1+\epsilon), \;\; \Theta(0)=\Theta(2\pi),\Theta'(0)=\Theta'(2\pi) $$ Periodicity in $\theta$ gives $\mu=n^2$ for some $n=0,\pm 1,\pm 2,\cdots$, and solutions $\Theta_n(\theta)=e^{in\theta}$. The radial equation is $$ r^2R''(r)+rR'(r)+(\lambda r^2-n^2)R(r) = 0 \\ R'(1-\epsilon) = 0 = R'(1+\epsilon). $$ This is Bessel's ODE, which has independent solutions $J_n(\sqrt{\lambda}r), K_n(\sqrt{\lambda}r)$. The values of $\lambda_{n,k}$ are determined by the endpoint conditions, and determine the proper linear combination of these Bessel functions. This is a Sturm-Liouville eigenfunction equation. Every fixed $n=0,1,2,3,\cdots$ determines a separate set of $\lambda$ that will work. There will be solutions that are not constant in $\theta$; any solutions for $n\ne 0$ and $\lambda\ne 0$ will be non-constant in $\theta$.


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