3.2 Laplace PDE outside disk

Problem Solve \(u_{xx}+u_{yy}=0\) outside disk \(x^{2}+y^{2}>1\) with boundary condition \(xy^{2}\) when \(x^{2}+y^{2}=a\). Where \(a=1\) in this problem.

The first step is to convert the boundary condition to polar coordinates. Since \(x=r\cos \theta ,y=r\sin \theta \), then at the boundary \(u\left ( r,\theta \right ) =r\cos \theta \left ( r\sin \theta \right ) ^{2}\). But \(r=1\) (the radius). Hence at the boundary, \(u\left ( 1,\theta \right ) =f\left ( \theta \right ) \) where \begin{align*} f\left ( \theta \right ) & =\cos \theta \sin ^{2}\theta \\ & =\cos \theta \left ( 1-\cos ^{2}\theta \right ) \\ & =\cos \theta -\cos ^{3}\theta \end{align*}

But \(\cos ^{3}\theta =\frac{3}{4}\cos \theta +\frac{1}{4}\cos 3\theta \). Therefore the above becomes\begin{align} f\left ( \theta \right ) & =\cos \theta -\left ( \frac{3}{4}\cos \theta +\frac{1}{4}\cos 3\theta \right ) \nonumber \\ & =\frac{1}{4}\cos \theta -\frac{1}{4}\cos 3\theta \tag{1} \end{align}

The above is also seen as the Fourier series of \(f\left ( \theta \right ) \). The PDE in polar coordinates is\[ u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta }=0 \] The solution is known to be\begin{equation} u\left ( r,\theta \right ) =\frac{c_{0}}{2}+\sum _{n=1}^{\infty }r^{-n}\left ( c_{n}\cos \left ( n\theta \right ) +k_{n}\sin \left ( n\theta \right ) \right ) \tag{2} \end{equation} Since the above solution is the same as \(f\left ( \theta \right ) \) when \(r=1\), then equating (2) when \(r=1\) to (1) gives\[ \frac{1}{4}\cos \theta -\frac{1}{4}\cos 3\theta =\frac{c_{0}}{2}+\sum _{n=1}^{\infty }\left ( c_{n}\cos \left ( n\theta \right ) +k_{n}\sin \left ( n\theta \right ) \right ) \] By comparing terms on both sides, this shows by inspection that\begin{align*} c_{0} & =0\\ c_{1} & =\frac{1}{4}\\ c_{3} & =\frac{-1}{4} \end{align*}

And all other \(c_{n},k_{n}\) are zero. Using the above result back in (2) gives the solution as \begin{equation} u\left ( r,\theta \right ) =\frac{r^{-1}}{4}\cos \theta -\frac{r^{-3}}{4}\cos 3\theta \tag{3} \end{equation} This is 3D plot of the solution


This is a contour plot