 Solution

For $z=ct-x$ the chain rule gives ${\partial }_t=cd/dz$, ${\partial }_x=-d/dz$, ${\partial }_{xx}=d^2/d{z}^2$. ThusAs $z$ increases from the material ahead of the wave towards that behind it, the singular pulse construction is $A\to B\to C\to D\to A$. The front $AB$ raises $v$ at approximately fixed $w=0$ from the low root to the high root. On the plateau $BC$, $f(v)\simeq w$ on the right attracting branch while $w$ rises slowly. The back $CD$ decreases $v$ at approximately fixed $w=w_0$ to the left attracting branch. The recovery tail then lowers $w$ back to zero on that branch, returning to $A=(0,0)$. Both sharp transitions have thickness $O(ϵ)$ in $z$; the other segments have the slower recovery scale.

The figure labels all four points and the front/back orientations. This frozen-recovery construction needs $ϵ\gamma \ll 1$ as well as $ϵ\ll 1$ if $\gamma$ is large, so that the recovery variable remains nearly fixed in an inner transition. It also needs an admissible positive front speed for the chosen orientation, addressed below.