# Solving the 2D Poisson equation with variable boundary location

I am trying to find $z(r,\phi)$ from the 2D Poisson equation in polar coordinates: $$\frac{1}{r}\frac{\partial}{\partial r}\left(r \frac{\partial z}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2z}{\partial \phi^2}=C \tag{1}$$ where $C$ is a constant and the following boundary conditions apply: $$z^{(1,0)} (0,\phi)=0 \tag{2}$$ $$z (r_0(\phi),\phi)=0 \tag{3}$$ $$z^{(0,1)} (r,0)=0 \tag{4}$$ $$z^{(0,1)} (r,\pi/2)=0 \tag{5}$$

where $z^{(1,0)}=\partial z/\partial r$ and $z^{(0,1)}=\partial z/\partial \phi$.

The part I cannot wrap my head around is how to work with the dirichlet boundary condition on a variable boundary $r=r_0(\phi)$ in $(3)$.

Can someone guide me through the steps to find $z(r,\phi)$?

If it helps $r_0(\phi)$ is an ellipse, i.e. $r_0(\phi)=\frac{a b}{\sqrt{b^2 \cos^2\phi + a^2 \sin^2\phi}}$.

Following up on the comment by @Dmoreno to use a different coordinate system: $x=a r \cos \phi$ and $y=b r \cos \phi$. Indeed this transform BC$(3)$ into $z(1,0)=0$, which seems convenient. However, transforming the original equation $(1)$ for this coordinate system results in the following nasty equation: $$\left(\frac{\cos^2\phi}{a^2}+\frac{\sin^2\phi}{b^2}\right)\frac{\partial^2 z}{\partial r^2}+\left(\frac{\cos^2\phi}{b^2 r^2}+\frac{\sin^2\phi}{a^2 r^2}\right)\frac{\partial^2z}{\partial \phi^2}+\\ \left(\frac{\cos^2\phi}{b^2 r}+\frac{\sin^2\phi}{a^2 r}\right)\frac{\partial z}{\partial r}+ \frac{2 \cos\phi \sin \phi}{r^2}\left(\frac{1}{a^2}-\frac{1}{b^2}\right)\frac{\partial z}{\partial \phi} + \\ \frac{2 \cos\phi \sin \phi}{r}\left(\frac{1}{b^2}-\frac{1}{a^2}\right)\frac{\partial^2 z}{\partial \phi \partial r} = C\tag{6}$$

Which I don't think I can solve anymore in the framework of the poisson equation. Any other input/suggestions on solving this system?

• I guess you are interested in solving the 2D Poisson equation and you have decided to go with polar coordinates, $x = r \cos \phi$, $y = r \sin \phi$. Two ideas: 1) maybe obtaining an expression for the Laplacian with the change of variables: $x = a r \cos{\phi}$, $y = b r \sin{\phi}$ will do the trick? 2) Note that you cannot, generally speaking, set a boundary condition for $r = 0$ since this is a geometric singularity. Instead of setting $z_r = 0$, wouldn't $\lim_{r\to 0}|z(r)| < \infty$ be the right boundary condition for $z$? – Dmoreno Aug 19 '14 at 17:38
• @Dmoreno I will certainly try the other choice of coordinates, indeed that would turn the boundary condition simply into $r=r_0=1$. I'm not sure I completely understand your second point, I can see that $r=0$ is indeed a singular point in the differential equation, but would the boundary condition be something like $\lim_{r\to0}$ $z_r\to0$ (sorry about the notation, don't quite know how to write that) ? – Michiel Aug 19 '14 at 18:52
• Hi @Michiel. I mean that you cannot force $z$ or its derivative to be any prescribed function since the solution might be singular at $r = 0$. The usual way to deal with this is to set the singular part of the solution (which usually are Bessel functions, logarithms, etc.) to zero as a way to keep $z$ bounded, i.e., $\lim_{r\to 0} |z| < \infty$ (which usually leads to a bounded derivative). Cheers! – Dmoreno Aug 19 '14 at 19:06
• @Dmoreno ok, I think I get it now. Thanks! I actually know roughly what the solution to this problem should look like (because of the physical shape it represents) and I know that there isn't a singularity at $r=0$, but I will work with the mathematically correct formulation you propose! – Michiel Aug 19 '14 at 19:32
• This recent paper looks like it'd provide the relevant Green's function. Unfortunately, it's behind a paywall and so I can't verify that. Another earlier paper by the same author is also behind a paywall but one which I can actually read; I'll see if I can figure out how to adapt the method there for this case. – Semiclassical Aug 26 '14 at 14:46

• Wow, Awesome, thanks! Minor correction, you missed a minus sign in $A_1$ i.e. $A_1=\frac{C}{4}\frac{b^2-a^2}{a^2+b^2}$ – Michiel Aug 28 '14 at 20:15
• Subtracting out the $r^2/4$ part of the solution seems to have worked like magic! This is a very elegant answer to something I thought had a much more complicated solution. What motivates that substitution apriori? – rajb245 Aug 28 '14 at 23:39
• @rajb245 I think you have to decide that you want to eliminate $C$ from the pde :) – Bernhard Aug 29 '14 at 6:17