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For the travelling wave coordinate $\xi =x+st$, the chain rule givesFast-slow matching for a calcium pulse joins two fast transitions, $B\to C$ and $D\to A$, joined by slow portions $C\to D$ and $A\to B$. Write the nullcline as $V=g(U)$. The resting point has $U_B=\mu$, $V_B=g(\mu )$ in the leading matched description.

Put $\xi =\epsilon X$. Multiplying the first equation by $\epsilon$ gives $U_{XX}=s(U_X+\gamma V_X)+\epsilon (U-\mu )$. At leading order, integrate and use $U_X\to 0$, $(U,V)\to (U_B,V_B)$ at the start of the front. Together with the second equation this givesThe fast equilibria are intersections of $V=g(U)$ and $V=V_B-(U-U_B)/\gamma$. As the slope $-1/\gamma$ changes, there may be one or three simple intersections, with two distinct intersections at a tangency. At a simple point of the nullcline, $f_U=-f_V{g}^{\prime }$, and the sign convention above/below it gives $f_V<0$. The Jacobian matrix of the fast system hasFor the usual three-intersection configuration, the two outer points have $1+\gamma g^{\prime }>0$, hence negative determinant and one positive and one negative eigenvalue: they are saddle points. The middle point has positive determinant; it is linearly stable if $s^2<-f_V$ and unstable if $s^2>-f_V$. Zero trace is a possible oscillatory stability threshold, with a nonlinear classification requiring further information about $f$.

To select the front speed, follow the appropriate branch of the unstable manifold of $B$ and vary $s$ until it meets the stable manifold of $C$. This is a shooting method for a heteroclinic orbit. Follow the high-$U$ slow branch from $C$ to $D$; the falling front obeys the same fast equations with the integration constant anchored at $D$. Choose $D$ so that its descending saddle connection reaches a low-$U$ point $A$ at that same speed. The remaining slow branch returns toward the resting point, completing the matched spike. The construction requires the indicated connections; a nullcline sketch alone does not specify their numerical speed or prove their existence for every possible kinetic function $f$.

On the slow portions, $f(U,V)=0$ at leading order, so $V=g(U)$ and $V^{\prime }=g^{\prime }(U)U^{\prime }$. Neglecting the small diffusion term yieldsOn the outer rising branches the coefficient on the left is positive. Recovery from $C$ toward $D$ requires decreasing $U$, hence $\mu <U_D$; the low branch from $A$ back toward rest requires increasing $U$, hence $U_A<\mu$. Therefore $\boxed{U_A<\mu <U_D}$. These inequalities distinguish the recovery branches from a trajectory that would move away from the required matching points.