= Solution
Add a nonconducting inactivated state to the <continuous-time Markov chain>:
$$
C\mathrel{\mathop{\rightleftarrows}^{k_o(V)}_{k_c(V)}}O
\mathrel{\mathop{\rightleftarrows}^{k_i(V)}_{k_r(V)}}I,
\qquad p_C+p_O+p_I=1.
$$
For example, $\dot p_O=k_op_C-(k_c+k_i)p_O+k_rp_I$ and $\dot p_I=k_ip_O-k_rp_I$. <Depolarization> first increases $p_O$, then transfers population into $I$ by <ball-and-chain inactivation>. A small recovery rate during the pulse permits a sustained decline of $I_K=N\gamma p_O(V-E_K)$. <Repolarization> favors recovery, eventually restoring the closed but available population. Several sequential closed states can represent the cooperative movements of four <voltage-sensing domains>; three states are the minimal explanation of rise followed by decay, rather than a quantitative fit to the missing trace.
Loss of outward <potassium> <electric current> weakens <repolarization> and can broaden an <action potential>. Reduced residual <potassium> <electrical conductance> can reduce the subsequent <hyperpolarization> and permit faster repeated firing. Conversely, prolonged <depolarization> can maintain <sodium> <ion-channel inactivation>, delay recovery and lower the maximum firing frequency or cause depolarization block. The sign of the frequency change depends on the other <ion channels> and recovery kinetics; rapid <potassium> <ion-channel inactivation> alone does not fix it.
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