 Solution

Let $c_{\uparrow }(w)$ denote the increasing-front speed from the preceding solution. Reversing that heteroclinic reverses its speed, so a decreasing back has speed $c_{\downarrow }(w)=-c_{\uparrow }(w)$. A rigidly travelling pulse has one speed throughout, requiringThis is the correct use of the integral formula: compare the oppositely oriented transitions, including their signs, rather than equating two increasing-front speeds.

For completeness, write $f(v)-w=-(v-r_1)(v-r_2)(v-r_3)$, $r_1<r_2<r_3$. The explicit increasing frozen-recovery cubic front isSubstitution, using $v_X=(v-r_1)(r_3-v)/\sqrt{2}$, verifies the equation. Since $r_1+r_2+r_3=1+a$, speed matching givesFor $0<a<1/2$ this selects the appropriate positive back level within the cubic's three-root range. The high branch connects $B$ to $C$ slowly, and the low branch supplies recovery after $D$.

The stated positive-speed, rising-front orientation imposes $a<1/2$ at leading order. The printed range $0<a<1$ by itself does not guarantee that pulse construction: at $a=1/2$ the leading front is stationary, while for $a>1/2$ the calculated increasing-front speed is negative. Existence and stability of a full finite-$ϵ$ pulse also require the slow connections; speed matching is its leading necessary construction, rather than a proof for every parameter in the printed range.