= 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, requiring
$$
\boxed{c=c_\uparrow(0)=c_\downarrow(w_0),\qquad c_\uparrow(w_0)=-c_\uparrow(0).}
$$
This 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> is
$$
v(X)=r_1+\frac{r_3-r_1}{1+\exp[-(r_3-r_1)X/\sqrt2]},\qquad c_\uparrow(w)=\frac{r_1+r_3-2r_2}{\sqrt2}.
$$
Substitution, using $v_X=(v-r_1)(r_3-v)/\sqrt2$, verifies the equation. Since $r_1+r_2+r_3=1+a$, speed matching gives
$$
c_\uparrow(0)=\frac{1-2a}{\sqrt2},\qquad r_2(w_0)=\frac{2-a}{3},\qquad\boxed{w_0=f\!\left(\frac{2-a}{3}\right)=\frac{2(2-a)(1-2a)(1+a)}{27}.}
$$
For $0<a<1/2$ 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 $D$.
\b[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-$\epsilon$ pulse also require the slow connections; speed matching is its leading necessary construction, rather than a proof for every parameter in the printed range.
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