I'm currently working on a project involving Fischer's equations:
$$u_t=u(1-u) + u_{xx}.$$
Admitting a travelling wave solution $u(x,t)=q(x-ct)$ and letting $p=q'$, this becomes a system of ordinary differential equations:
$$ q'=p, \\ p'=−q(1−q)−cp.\\ $$
We find fixed points in $(0,0)$ and $(1,0)$ (saddle). The type of fixed point in $(0,0)$ is dependent on $c$, where for $c=0$ we find a center, for $0<c<2$ a stable spiral, for $-2<c<0$ an unstable spiral, and for $c>2$ and $c<-2$ a stable and unstable equilibrium respectively.
If we require for a physically relevant solution that $q(x-ct)>0$ for all $x-ct$ and that the solution is bounded, then we cannot admit any solution $-2<c<2$ since any bounded orbit will assume negative values while approaching or coming from the equilibrium at $(0,0)$.
I've plotted phase portraits for $c=3$ and $c=-3$ with nullclines (black dashed), stable (green) and unstable (red) manifolds:
Here we can clearly see that solutions for $c<-2$ are unbounded, whereas values for $c>2$ left of the right stable manifold are bounded.
Now my question is whether this would imply that only right traveling (positive $c$) wave solutions are allowed, since a left traveling wave would imply a negative $c$, which fails our requirements for physically meaningful solutions. Does Fischer's equation allow for left-traveling waves?