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For the decreasing front connecting $1$ to $0$, the balanced stationary travelling-wave reduction of a reaction-diffusion system has first integral $D{U}_x^2/2-U^2(1-U{)}^2/4=0$. Choosing the negative derivative gives $U_x=-U(1-U)/\sqrt{2D}$ and hence $U=[1+e^{(x-q)/\sqrt{2D}}{]}^{-1}$. This logistic profile remains an exact travelling wave when the reaction parameter differs from $1/2$: using $z=(x-ct-q_0)/\sqrt{D}$, its derivative is $U_z=-U(1-U)/\sqrt{2}$, and the wave equation is satisfied for $c=\sqrt{D/2}(1-2r)$. Thus the speed changes with the potential bias while the shape, apart from translation, does not.

There is a consistency issue in the printed model. If $w\equiv w_0$ is imposed while retaining its evolution equation, then $0=\beta v-\gamma w_0$ forces $v\equiv \gamma w_0/\beta$. Thus the full two-equation system has no nonconstant travelling wave with exactly constant $w$. Its constant solution exists only when $I=w_0-f(\gamma w_0/\beta )$. The intended wave calculation must treat the recovery variable as externally frozen, leaving the scalar reaction-diffusion equationFor this frozen-recovery reduction, letThe cubic polynomial $f$ has a local minimum at $a_{-}$ and a local maximum at $a_{+}$. Hence the conditiongives three distinct real roots $r_1<r_2<r_3$ of $g$, withThe outer roots are stable for the scalar spatially uniform dynamics and the middle root is unstable. Seek a travelling wave $v(x,t)=V(z)$, where $z=x-ct$. The travelling-wave reduction of a reaction-diffusion system givesLet $\delta =r_3-r_1$ and chooseIt has $V(-\infty )=r_3$, $V(+\infty )=r_1$ andSubstitution leavesConsequently the frozen-recovery cubic front isReflection $x\mapsto -x$ reverses the front and its speed. A nonzero speed requires $r_2\ne (r_1+r_3)/2$; equality gives a stationary front and is the Maxwell balance for a scalar reaction-diffusion front. Since the roots sum to $1+\alpha$, this stationary value is $I=w_0-f((1+\alpha )/3)$. For the simple admissible choice $I=w_0$, the roots are $0,\alpha ,1$ andThis particular front moves when $\alpha \ne 1/2$. At $\alpha =1/2$, other currents inside the displayed open interval give moving fronts.

Write the reaction term as $f(u;r)=u(1-u)(u-r)$ and assume $D>0$. At $r=1/2$, multiply the stationary travelling-wave reduction of a reaction-diffusion system $D{U}^{\prime \prime }+f(U;1/2)=0$ by $U^{\prime }$ and integrate once. Since $U^{\prime }=0$ at both asymptotic phases and ${\int }_0^{U}f(s;1/2)ds=-U^2(1-U{)}^2/4$, the integration constant is zero:The boundary orientation selects $U^{\prime }=-U(1-U)/\sqrt{2D}$. Separating variables gives the logistic front of a bistable cubic equation, with arbitrary center $q_0$:The arbitrary translation is expected from translation invariance.

To develop the requested perturbation theory, let $r=1/2+\epsilon a$ with signed $a=\pm 1$, introduce slow time $T=\epsilon t$, and put $z=(x-q(t))/\sqrt{D}$. Seekwhere $U_0(z)=(1+e^{z/\sqrt{2}}{)}^{-1}$. The explicit $T$ derivative of $\epsilon U_1$ is second order. Since ${\partial }_{r}f=-u(1-u)$, the first-order equation isDifferentiation of $U_0^{\prime \prime }+f(U_0;1/2)=0$ gives $L{U}_0^{\prime }=0$. The self-adjoint differential operator $L$ has this decaying translation mode, and $U_1\to 0$ at both ends because the phases remain exactly $1$ and $0$. Multiplying the first-order equation by $U_0^{\prime }$ and integrating by parts makes its left side zero. Therefore the translation-mode solvability for a bistable front requiresBoth integrals can be evaluated without a special-function formula. The decreasing profile takes $U$ from $1$ to $0$ and has $U_0^{\prime }=-U(1-U)/\sqrt{2}$, soandThe resulting speed isThis derives the velocity by the Fredholm solvability condition, rather than substituting a guessed speed.

With this value, the forcing actually cancels pointwise: $-c_1{U}_0^{\prime }+a{U}_0(1-U_0)=0$. Thus $L{U}_1=0$. The decaying kernel is only the translation direction: an independent solution is $U_0^{\prime }{\int }^z(U_0^{\prime }{)}^{-2}d\zeta$, which grows at at least one infinity. Fixing the center by $U_1(0)=0$ consequently gives $\boxed{U_1=0}$. In this particular bistable cubic reaction-diffusion equation the absence of shape deformation holds to all orders. Indeed $U_0^{\prime \prime }+f(U_0;r)=-(r-1/2)U_0(1-U_0)$, and the exact moving-frame equation has left side $-(\dot{q}/\sqrt{D})U_0^{\prime }$. Thereforeis the exact front for every $0<r<1$. For $r<1/2$, $c>0$ and the $u=1$ phase advances to the right; for $r>1/2$ the $u=0$ phase advances to the left. The small-bias speed is consistent with the prescribed $\dot{q}/\sqrt{D}=O(\epsilon )$ scaling.

For a nonconstant invasion travelling wave, the relevant end states are $w(-\infty )=k$ and $w(+\infty )=0$. These fronts do not exist at every nonnegative speed: the phase plane calculation gives the threshold $2\sqrt{k}$. The constant solutions $w\equiv 0$ and $w\equiv k$ are, of course, nonnegative travelling waves for every speed, including zero.

Put $z=x-ct$ and $p=w^{\prime }$. The travelling-wave reduction of a reaction-diffusion system givesThe equilibrium points in the phase plane are $(k,0)$ and $(0,0)$. At $(k,0)$ the characteristic equation is $r^2+cr-k=0$, so there is one positive and one negative eigenvalue. Thus it is a saddle equilibrium. Its descending unstable manifold has $w<k$, $p<0$ and locallyAt $(0,0)$ the eigenvalues areFor $c>2\sqrt{k}$ it is a stable node; for $0<c<2\sqrt{k}$ it is a stable spiral. The repeated-root case $c=2\sqrt{k}$ must be included in the existence argument rather than inferred from distinct roots.

Suppose $c\ge 2\sqrt{k}$. Choose $a>0$ with $a(c-a)\ge k$; for example $a=(c-\sqrt{c^2-4k})/2$. Consider the closed triangular regionOn the horizontal side $p=0$, the vector field has $p^{\prime }=-w(k-w)\le 0$, pointing into $\mathcal{T}$. On the vertical side $w=k$, it has $w^{\prime }=p\le 0$, again pointing inward. For the sloping side use $h=p+aw$, so that on $h=0$,Thus trajectories cannot cross into $h<0$. The remaining boundary point $(0,0)$ is an equilibrium point and cannot be crossed in finite time by a nonconstant trajectory, by uniqueness theorem for ordinary differential equations. This proves forward invariance of $\mathcal{T}$. The argument works unchanged at $c=2\sqrt{k}$ with $a=\sqrt{k}$.

The descending unstable manifold of $(k,0)$ initially enters $\mathcal{T}$. Its forward trajectory remains in this compact region, so it exists for all positive $z$. Within the interior, $p<0$: a first contact with $p=0$ at $0<w<k$ would have $p^{\prime }<0$, contrary to arrival from below. Hence $w$ decreases strictly and has a limit $\ell \in [0,k)$. To identify the limit, its omega-limit set lies on the line $w=\ell$ and is invariant. Any trajectory remaining on this line has $p=w^{\prime }=0$; its second equation then forces $\ell (k-\ell )=0$. Since $\ell <k$, it follows that $\ell =0$, and the only possible omega-limit set is $(0,0)$. Backward in $z$, the unstable manifold tends to $(k,0)$ by its construction. We have therefore obtained a positive decreasing heteroclinic orbit and hence a nonnegative travelling wave for every $c\ge 2\sqrt{k}$.

For completeness, the threshold is necessary for such a front. When $0<c<2\sqrt{k}$, put $\nu =\sqrt{k-c^2/4}$, $X=w$, and $Y=(p+cw/2)/\nu$. Near the zero equilibrium point the system becomesFor a nonzero trajectory its polar angle satisfiesA trajectory approaching zero therefore makes infinitely many rotations and must cross $X=w=0$. It cannot be a nonnegative front. At $c=0$, the first integralis conserved. It has values $k^3/6$ and $0$ at the two desired end states, so no heteroclinic orbit connects them. ConsequentlyThis is the phase-plane construction of a Fisher travelling front, including the critical speed. The statement for all $c\ge 0$ must distinguish these nonconstant fronts from the two constant travelling waves.

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.