I am studying the spherically symmetric wave equation on the Schwarzschild background in Eddington-Finkelstein coordinates $(v,r)$. I want to numerically integrate the $v$-time-evolution of such a wave. However, I realised that I seem to misunderstand something, because the following problem arises:

For integration of the wave equation, I apply a Finite-Difference-Scheme to the spatial derivatives. In order for the simulation to be numerically stable, the grid sizes $(\Delta v,\Delta r)$ must satisfy the Courant-condition

$$\frac{\Delta r}{\Delta v} > \frac{u}{C}$$

where $C < 1$ is the Courant-factor and $u$ is the speed of the wave. Eddington Finkelstein coordinates are constructed such that radially ingoing null geodesics satisfy $v = \text{const}$. If this is the case, then -- when integrating the differential equation -- in order to obtain the solution for the next time step $(v_0 + \Delta v)$ at some point $r = r_0$, each spatial point $r > r_0$ must be considered, since they all may influence the solution at $(v_0 + \Delta v, r_0)$. For a usual number of spatial gridpoints, this would require an asymmetric FDA-Scheme of an order much too high to be practical.

What is the flaw in my thinking here?

  • $\begingroup$ Your inequality sign in the Courant condition is the wrong way around. $\endgroup$
    – TimRias
    Oct 17, 2022 at 5:48
  • $\begingroup$ Ah, yes of course. Thanks for pointing this out. $\endgroup$
    – Octavius
    Oct 17, 2022 at 10:30
  • $\begingroup$ I don't see anything wrong with your thinking; I think you're correctly finding that you can't use explicit time stepping here. Go back to your wave equation and write it in terms of your coordinates. I'd bet you'll find that it separates into "time" equations (involving a derivative wrt $v$) and purely spatial equations (involving just derivatives wrt $r$ and/or angles). The latter are called "hypersurface equations". With luck, you may be able to separate the equations into one group that can be evolved between hypersurfaces, and another that have to be solved again on each hypersurface. $\endgroup$
    – Mike
    Oct 17, 2022 at 14:29


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