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?


1 Answer 1


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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