6
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $z$? $\endgroup$ – Dmoreno Aug 19 '14 at 17:38
  • $\begingroup$ @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) ? $\endgroup$ – Michiel Aug 19 '14 at 18:52
  • $\begingroup$ 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! $\endgroup$ – Dmoreno Aug 19 '14 at 19:06
  • 1
    $\begingroup$ @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! $\endgroup$ – Michiel Aug 19 '14 at 19:32
  • 2
    $\begingroup$ 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. $\endgroup$ – Semiclassical Aug 26 '14 at 14:46
3
+200
$\begingroup$

Finding the general solution of the PDE (below) is not the more difficult part of the task :

enter image description here

As usual, the arduous part is to determine, among the infinite number of solutions which one fits with the boundary conditions :

enter image description here

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Wow, Awesome, thanks! Minor correction, you missed a minus sign in $A_1$ i.e. $A_1=\frac{C}{4}\frac{b^2-a^2}{a^2+b^2}$ $\endgroup$ – Michiel Aug 28 '14 at 20:15
  • $\begingroup$ You are right. I made the correction. $\endgroup$ – JJacquelin Aug 28 '14 at 20:51
  • $\begingroup$ 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? $\endgroup$ – rajb245 Aug 28 '14 at 23:39
  • $\begingroup$ @rajb245 I think you have to decide that you want to eliminate $C$ from the pde :) $\endgroup$ – Bernhard Aug 29 '14 at 6:17
  • $\begingroup$ @rajb245 : The motivation is to simplify the original PDE and reduce it to an homogeneous PDE. For that, the usual way is to search for a convenient change of function. Of course, it is not always possible. But, it was possible for this kind of PDE. $\endgroup$ – JJacquelin Aug 29 '14 at 7:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.