= Solution
\b[Yes.] An immobile $Y$ <concentration> can still change locally through reaction while diffusing $X$ transports activation. Vanishing <diffusion> of one component does not prohibit a propagating reaction front.
In the invariant special case $v=0$, the first equation is already <Fisher's equation> with growth coefficient one, so it has travelling fronts regardless of $D$. A <one-diffuser reaction front> also gives a coupled example with nonzero $v$. For $z=\xi-c\tau$, the equations with $D=0$ are
$$
U''+cU'+U(1-U-\beta V)=0,\qquad cV'=\delta UV.
$$
For $0<\delta<2$, choose $k=\sqrt{1-\delta/2}$, $\beta v_0=2-\delta$, and set
$$
\boxed{U(z)=\frac1{1+e^{-kz}},\quad V(z)=v_0[1-U(z)]^2,\quad c=-\frac\delta{2k}.}
$$
Indeed $U'=kU(1-U)$ and $V'/V=-2kU$, giving the second equation. In the first, division by $U(1-U)$ leaves $k^2(1-2U)+ck+1-\beta v_0(1-U)$, whose constant and $U$ coefficients both vanish with these choices. Ahead of the left-moving wave $U=0,V=v_0$; behind it $U=1,V=0$. This explicitly demonstrates a travelling front with substrate depletion but no $Y$ <diffusion>, rather than relying solely on the uncoupled special case.
Back to article page
