# Numerical dispersion in FDTD

I was reading the book "Computational Electrodynamics: The FDTD method" by Taflove and Hagness, probably the most cited book when it comes to the FDTD method in Electromagnetics. In the second chapter of the book, the authors investigate the numerical dispersion of an 1-D numerical wave. They show that when the Courant number S is exactly 1, $$S\doteq c_0 \delta t/\delta x = 1$$, then the dispersion relation becomes exact, that means $$k=\hat{k}$$ (where $$k$$ is the continuous-wave wavenumber, and $$\hat{k}$$ is the numerical-wave wavenumber). This is found to be true, regardless of the space and time increments $$\delta x, \delta t$$. Later, they study the general case of a dispersive numerical wave. They show that there is a value for the grid sampling resolution per free-space wavelength $$N_{\lambda}=\lambda_0/\delta x$$, when the transition from $$\hat{k}$$ real to $$\hat{k}$$ complex happens. This value is $$N_{\lambda}|_{transition}=2\pi S/\arccos(1-2S^2)$$. However, the last relationship states that for $$S=1$$, we have a corresponding $$N_{\lambda}|_{transition}=4$$, which means that the space increment $$\delta x$$ must me such that $$\delta x < \lambda_0/4$$, in order to have a real (and thus physical) wavenumber. So, I understand that we cave two contradictory piece of information here. One says that when $$S=1$$ (the so-called "magic step"), the dispersion relation will always be exact, and the other says that even when $$S=1$$, we need a fine enough spatial resolution in space in order to avoid numerical dispersion. What am I missing here?

• It's worth your time moving beyond finite difference to finite volume techniques, if only for the theoretical motivation. – A rural reader Feb 14 at 4:51