### 3.1 Laplace PDE inside disk

Solve $$u_{xx}+u_{yy}=0$$ on disk $$x^{2}+y^{2}<1$$ with boundary condition $$xy^{2}$$ when $$x^{2}+y^{2}=a$$. Where $$a=1$$ in this problem. Express solution in $$x,y$$

Solution

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} \fbox{$u\left ( r,\theta \right ) =\frac{r}{4}\cos \theta -\frac{r^3}{4}\cos 3\theta$} \tag{3} \end{equation} This solution is now converted to $$xy$$ using the formula\begin{align*} r^{n}\cos n\theta & =\sum _{\substack{k=0\\even}}^{n}\begin{pmatrix} n\\ k \end{pmatrix} x^{n-k}\left ( -1\right ) ^{\frac{k}{2}}y^{k}\\ & =\sum _{\substack{k=0\\even}}^{n}\frac{n!}{k!\left ( n-k\right ) !}x^{n-k}\left ( -1\right ) ^{\frac{k}{2}}y^{k} \end{align*}

For $$n=1$$ the above gives\begin{align} r\cos \theta & =\frac{1!}{0!\left ( 1-0\right ) !}x^{1-0}\left ( -1\right ) ^{0}y^{0}\nonumber \\ & =x \tag{4} \end{align}

And for $$n=3$$\begin{align} r^{3}\cos 3\theta & =\frac{3!}{0!\left ( 3-0\right ) !}x^{3-0}\left ( -1\right ) ^{0}y^{0}+\frac{3!}{2!\left ( 3-2\right ) !}x^{3-2}\left ( -1\right ) ^{1}y^{2}\nonumber \\ & =x^{3}-3xy^{2} \tag{5} \end{align}

Using (4,5) in (3) gives the solution in $$x,y$$\begin{equation} \fbox{$u\left ( x,y\right ) =\frac{1}{4}x-\frac{1}{4}\left ( x^3-3xy^2\right )$} \tag{6} \end{equation} This is now veriﬁed that is satisﬁes the PDE $$u_{xx}+u_{yy}=0$$.\begin{align*} \frac{\partial u}{\partial x} & =\frac{1}{4}-\frac{1}{4}\left ( 3x^{2}-3y^{2}\right ) \\ \frac{\partial ^{2}u}{\partial x^{2}} & =-\frac{6}{4}x \end{align*}

And\begin{align*} \frac{\partial u}{\partial y} & =\frac{6}{4}xy\\ \frac{\partial ^{2}u}{\partial y^{2}} & =\frac{6}{4}x \end{align*}

Therefore $$\frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{\partial y^{2}}=0$$.

Now the boundary conditions $$u\left ( x,y\right ) =xy^{2}$$ are also veriﬁed. This condition applies when $$x^{2}+y^{2}=1$$ or $$y^{2}=1-x^{2}$$. Substituting this into (6) gives$u\left ( x,y\right ) _{@D}=\frac{1}{4}x-\frac{1}{4}\left ( x^{3}-3x\overset{y2}{\overbrace{\left ( 1-x^{2}\right ) }}\right )$ Simplifying gives\begin{align*} u\left ( x,y\right ) _{@D} & =\frac{1}{4}x-\frac{1}{4}\left ( x^{3}-\left ( 3x-3x^{3}\right ) \right ) \\ & =\frac{1}{4}x-\frac{1}{4}x^{3}+\frac{1}{4}\left ( 3x-3x^{3}\right ) \\ & =\frac{1}{4}x-\frac{1}{4}x^{3}+\frac{3}{4}x-\frac{3}{4}x^{3}\\ & =x-x^{3}\\ & =x\left ( 1-x^{2}\right ) \\ & =xy^{2} \end{align*}

Veriﬁed. This is 3D plot of the solution This is a contour plot 