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Let $r,\theta$ be the usual polar coordinates in $\mathbb{R^2}$, let $\Omega$ be the unit disc $r<1$ and recall that the Laplacian is given by $$\nabla^2u=\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}.$$ Use separation of variables to solve, \begin{align} -\nabla^2u&=\lambda u \ \ \ \ \text{in}\ \ \Omega \\ u&=0\ \ \ \ \ \ \ \text{in}\ \ \partial\Omega. \end{align}

We seek $u=X(x)Y(y)$ for \begin{align} -\nabla^2u-\lambda u&=0 \\ -\left(\partial_{xx}u+\partial_{yy}u\right)-\lambda u&=0 \\ -\frac{X''}{X}-\lambda&=\frac{Y''}{Y}=\mu \ \ \ \ \text{($\mu$ is a separation constent)} \end{align} We then consider three cases ($\mu=0, \ \mu<0, \ \mu>0$) for each equation. But what are the boundary conditions for $X$ and $Y$ if $u=0$?

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  • $\begingroup$ Why not instead look for a solution of the form $u(r, \theta) = R(r) \Theta (\theta)$? $\endgroup$ – Kenny Wong Oct 28 '18 at 9:18
  • $\begingroup$ Yes, I should have probably done this. Though this does not really change the nature of my question. In that case, what would the boundary conditions for $r$ and $\theta$ be if $u=0$. $\endgroup$ – M B Oct 28 '18 at 9:50
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    $\begingroup$ Well, in order for $u = 0$ to hold on $\partial \Omega$, you would need $R(r = 1) = 0$. You would also need $R(r = 0)$ to be non-infinite. And you would need $\Theta(\theta = 0) = \Theta(\theta = 2\pi)$, so that the solution is consistent with itself. $\endgroup$ – Kenny Wong Oct 28 '18 at 9:58
  • $\begingroup$ How did you determine this? I don't quite understand how this was derived $\endgroup$ – M B Oct 28 '18 at 10:07
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    $\begingroup$ math.okstate.edu/people/binegar/4263/4263-l15.pdf $\endgroup$ – user3417 Oct 28 '18 at 19:08

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