7
$\begingroup$

In this question, I suggested that the Couran-Friedrichs-Lewy (CFL) condition for the wave equation in polar coordinates reads

$$C = 2c\frac{\Delta t}{\Delta r \Delta \phi} \leq C_\max \enspace ,$$

where $c$ is the phase speed. I suggested this from an intuitive point of view, and it worked in that example. Nevertheless, This is probably not right, And I could not find an expression for this case.

Question: What is the CFL condition in polar coordinates?

This question was asked before in Math.SE:

$\endgroup$
3
$\begingroup$

I was going to write a comment, but the equation seems to view better in answers..

I assume Von Neumann analysis is the proper approach to derive this equation, but a coordinate transformation from the cartesian CFL condition (I took from wikipedia) is not somehow equivalent? Specifically:

\begin{equation} \Delta t \sum_{i=1}^3 \frac{u_i}{\Delta x_i} = \Delta t \left( \frac{u_r}{\Delta r} + \frac{u_{\phi}}{r \Delta \phi} + \frac{u_z}{\Delta z} \right) < C_{max} \\ \end{equation}

$\endgroup$
  • $\begingroup$ That is definitely a clever option, although that implies that a discretization that includes the origin is really problematic for the selection of $\Delta t$. $\endgroup$ – nicoguaro Mar 5 '17 at 20:58
  • $\begingroup$ Regarding the von Neumann analysis, I thought about that, but how is the expansion? If one thinks to translate the analysis from the 1D case, it might involve a Fourier-Bessel expansion? Also, the concept of wave-number for polar coordinates is not as natural as it is in Cartesian ones. $\endgroup$ – nicoguaro Mar 5 '17 at 21:01
  • 1
    $\begingroup$ I'm not sure how the expansion would pan out, or, for that matter if you'll be able to define a criterion without a transcendental function.. As for your first comment, though, I would argue that, analytically your argument is true but, the CFL condition should be measured at collocated cell centers, so the first radial element is located at $r=0+0.5 \Delta r$, meaning that there's no issue at the origin. Also, this analog is in agreement when $r$ is very large and the elements are nearly rectangular.. $\endgroup$ – Charles Mar 5 '17 at 22:01

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.