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Use $\xi =x-\hat{X}(T)$, $T=\epsilon t$. At order $\epsilon$, the moving-profile expansion givesThe slow time derivative of $\epsilon R$ is of order ${\epsilon }^2$. Multiplication by $A_0^{\prime }$ and integration eliminate $\mathcal{L}R$ by the Fredholm solvability condition. Thus the leading velocity isFor $\nu (x,T)=K\delta (x+cT)$ the Dirac delta samples $\xi =-\hat{X}-cT$. From the first-order profile equation,ConsequentlyThe prefactor $q$ in the printed profile is important for the position shift in $Z$. This is a leading-order velocity in slow time; the physical velocity is $d\hat{X}/dt=\epsilon {\hat{X}}_T+O({\epsilon }^2)$.

The relative position obeys the autonomous moving-defect locking of a cubic-quintic front equation $Z_T=c-K{\mathrm{sech}}^2(\beta Z/2)$. For $K\ne 0$, a finite constant-velocity front must have constant $Z$, and hence must travel with the defect at ${\hat{X}}_T=-c$. Such locked fronts exist exactly whenFor the strict inequality, their positions areLinearizing the relative-position equation gives ${\eta }_T=K\beta {\mathrm{sech}}^2(\beta Z_{\ast }/2)\tanh (\beta Z_{\ast }/2)\eta$. Therefore the root with $K{Z}_{\ast }<0$ is stable and the other root unstable. At $c=K$, they coalesce at $Z=0$ in a saddle-node bifurcation. Since $Z_T=K{\beta }^2{Z}^2/4+O(Z^4)$ there, the double root attracts from only one side and is not two-sided asymptotically stable. Locking within its existence range still requires an initial relative position in the stable root's basin; beyond the unstable root the front escapes instead.

These positional stability conclusions also describe the leading weakly perturbed front. Indeed $v=A_0^{\prime }>0$ has no zeros, and $\mathcal{L}=({\partial }_{\xi }+v^{\prime }/v)({\partial }_{\xi }-v^{\prime }/v)$ is nonpositive in the $L^2$ inner product. The limiting coefficients $f^{\prime }(0)=-3{\alpha }^2/16$ and $f^{\prime }(\sqrt{q})=-3{\alpha }^2/4$ are negative. Thus the unperturbed shape modes are damped, with the neutral translation mode singled out by the solvability condition; the defect determines its stability at first order.

If $c\ne 0$ and the locking inequality fails, $Z_T$ has no finite zero and has the sign of $c$. The separation grows without bound, the localized forcing of the front decays exponentially, and ${\hat{X}}_T\to 0$: the defect passes the front, which approaches a stationary position in laboratory coordinates. The same escaping behavior occurs outside the locking basin when roots exist. If $c=0$ and $K\ne 0$, the stationary defect drives the front away from it with ever decreasing speed; $Z$ grows in magnitude logarithmically rather than tending to a finite locked position. Finally, when $K=0$, the front remains stationary for every $c$; when also $c=0$, every front position is a neutral equilibrium. These degenerate cases must not be included by dividing by $K$.