K$ is as follows:

\[eq:godunovwithKT\] $$\begin{aligned}
\left( 1 - \frac{k}{4} A_{x} \right) u^{\star} & = \left( 1 + \frac{k}{4} A_{x} 
\right) u(t) \label{eq:godunovwithKTA} \\
u_{E} &= u^{\star} + k A_{y} u^{\star} + h (1-\frac{u^{\star}}{\mathcal{K}(t)})
u^{\star} \label{eq:godunovwithKTB} \\
\left( 1 - \frac{k}{2} A_{y} - \frac{h}{2} (1-\frac{u_{E}}{\mathcal{K}(t+h)}) 
\right) u^{\star \star} & =
\left( 1 + \frac{k}{2} A_{y} + \frac{h}{2} (1-\frac{u^{\star}}{\mathcal{K}(t)}) 
\right) u^{\star} \label{eq:godunovwithKTC} \\
\left( 1 - \frac{k}{4} A_{x} \right) u(t+h) & = 
\left( 1 + \frac{k}{4} A_{x} \right) u^{\star \star},
\label{eq:godunovwithKTD}\end{aligned}$$

where we have again suppressed the $\mathbf{x}$ dependence of $\mathcal{K}(\mathbf{x},t)$ for simplicity of notation.

Fisher/KPP on geographical maps {#subsec:geographicalmaps}
-------------------------------

In order to solve Fisher/KPP on an irregular domain such as a geographical map, it is sufficient to break down the map into $x$- and $y$-direction segments, imposing a $u=0$ boundary condition at their endpoints; the solver can act on each segment independently, alternating the direction [@leveque2007; @ritchmyer1967] of integration as discussed above. This approach lends itself also to efficient parallelization. Appendix \[sec:code\] and Figure \[fig:mapsegmentation\] illustrate in detail our scheme with a sample Matlab code.

Note that holes in $\mathcal K$ maps can represent the partial closing of land bridges without the necessity of re-segmenting land portions, as in 