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In $u_t=u(1-u-\beta v)+u_{xx}$, $v_t=-\delta uv+v_{xx}$, the line $v=1-u$ is invariant when $\beta =1-\delta$. On it $u$ satisfies Fisher's equation with growth coefficient $\delta$. The reduction also requires compatible initial and boundary data: parameter tuning alone does not reduce arbitrary two-field solutions. Positive concentrations and positive $\beta$ in this normalization require $0<\delta <1$.

This criterion concerns travelling-front solutions of Fisher's equation.

For diffusion coefficient $D$ and leading-edge growth rate $r>0$, a Fisher-KPP-type pulled front has minimum speed $2\sqrt{Dr}$.

Fisher's equation is a reaction-diffusion equation combining diffusion with logistic growth. For positive $D,r,K$, one-dimensional fronts joining $K$ to zero have minimum speed $2\sqrt{Dr}$. The phase-plane construction of a Fisher travelling front and the linearized leading edge explain this threshold.

The linear spreading speed is the propagation speed predicted by the linearized dynamics near an unstable state. Forinsert an exponentially decreasing leading-edge mode $u=e^{-\lambda (x-ct)}$, $\lambda >0$. This givesThe minimum occurs at $\lambda =\sqrt{r/D}$ and equals $c_{\ast }=2\sqrt{Dr}$. Equivalently, the spatial equation $D{\lambda }^2-c\lambda +r=0$ has a repeated positive root at that speed.

For Fisher's equation, this is the Fisher-KPP minimum wave speed. It is a leading-edge calculation, not a universal speed theorem for every reaction-diffusion equation. A pulled travelling front is selected by that linear leading edge, while a sufficiently strong nonlinear interior can select a larger speed. Initial-data decay and nonlinear assumptions determine whether an observed invasion actually travels at the linear prediction.

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 $D=0$ areFor $0<\delta <2$, choose $k=\sqrt{1-\delta /2}$, $\beta v_0=2-\delta$, and setIndeed $U^{\prime }=kU(1-U)$ and $V^{\prime }/V=-2kU$, 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.

The travelling-wave reduction of a reaction-diffusion system of Fisher's equation isMultiplying by $U^{\prime }$ and integrating gives $c{\int }_{\mathbb{R}}(U^{\prime }{)}^{2}dz=1/6$, so the decreasing invasion front has $c>0$. The linearization at $U=0$ has spatial eigenvaluesFor $0<c<2$ they are complex: approaching zero spirals in the $(U,U^{\prime })$ plane and changes the sign of $U$, which is inadmissible for a density. Thus $c\ge 2$ is necessary.

It is also sufficient. Write $V=U^{\prime }$ and choose $\ell =(c-\sqrt{c^2-4})/2>0$. The triangleis forward invariant for $U^{\prime }=V$, $V^{\prime }=-cV-U(1-U)$. On $V=0$, $V^{\prime }\le 0$. On $V=-\ell U$,The boundary $U=1$ points left, and the boundary at $U=0$ consists only of the origin. The unstable branch from the saddle $(1,0)$ enters this triangle, where $U$ decreases and remains nonnegative. Its limiting state must have $V=0$ and $U(1-U)=0$; since it has left $U=1$, it tends to $(0,0)$. Therefore a monotone positive front exists for every $c\ge 2$, including the double-root endpoint:This is the Fisher-KPP minimum wave speed, obtained with an explicit invariant-region existence argument.