 Solution

Put $X=z/ϵ$. The travelling equations become $c{v}_X=f(v)-w+v_{XX}$ and $c{w}_X=ϵ(\gamma v-w)$. Hence, for nonzero $O(1)$ speed and $ϵ\gamma \ll 1$, $w$ is constant to leading order across the front.

Writing $p=v_X$ gives the planar system $v_X=p$, $p_X=cp-f(v)+w$. At the outer cubic roots $A=(V_{-},0)$ and $B=(V_{+},0)$, $f^{\prime }(V_{\pm })<0$. The Jacobian determinant is $f^{\prime }(V_{\pm })<0$, so both are saddle points. The increasing front is a heteroclinic orbit from the unstable manifold of $A$ to the stable manifold of $B$. Varying $c$ supplies the shooting parameter required to match these manifolds. For this cubic one can also construct it explicitly, as in the next solution.

Multiply the second-order equation by $v_X$ and integrate across the front. Because $v_X\to 0$ at both ends, the term $\int v_{XX}{v}_{X}dX=[v_X^2/2{]}_{-\infty }^{\infty }$ vanishes. The remaining identity isThe denominator is positive for a nonconstant front, so its direction is determined by the signed area between the cubic and the horizontal recovery line: Maxwell balance for a scalar reaction-diffusion front gives zero speed when that area vanishes.

The derivative in the denominator must use the inner coordinate $X$. If one instead uses the source's original $z$ and denotes $J_z=\int v_z^{2}dz$, then $J_z={ϵ}^{-1}\int v_X^{2}dX$ and the identity is $c=\int (f-w)dv/(ϵJ_z)$. Thus the printed $v^{\prime }(z)$ is valid only if $z$ has silently been reused for the rescaled inner coordinate; with the original coordinate it misses a factor $ϵ$.