* Where there is no curvature, there is no change from one step to the next.
* So, in the above image $f_{i-1}^{n+1} = f_{i-1}^n$, and $f_{i+1}^{n+1} = f_{i+1}^n$.
$$f_{i-1}^{n+1} = f_{i-1}^n + d(f_{i-2}^n - 2f_{i-1}^n + f_i^n) = f_{i-1}^n + d(1\cdot 0 - 2\cdot 1 + 1\cdot 2) = f_{i-1}^n + d(0) = f_{i-1}^n.$$
where $k_{\eta}$ is the wavenumber (inverse wavelength), and $c_{\eta}$ is a complex coefficient.
to get $$\frac{f_i^{n+1}}{f_i^n} = 1+2d(\cos(k_{\eta}\Delta x)-1) = G,\,\,\mbox{say}.$$
* The upper bound in this equation (right-most inequality) is always satisfied.
* For the lower bound, rearranging, we have
$$ -2\le2d(\cos(k_{\eta}\Delta x)-1),$$
or
$$d\le\frac{1}{1-\cos(k_{\eta}\Delta x)}.$$