Reduction of a kinetic model for ${\mathrm{Na}}^{+}$ channel activation, and fast and slow inactivation within a neural or cardiac membrane

A 15-state kinetic model for ${\mathrm{Na}}^{+}$ channel gating that describes the coupling between three activation sensors, a two-stage fast inactivation process, and slow inactivated states may be reduced to equations for a 6-state system by application of the method of multiple scales. By expressing the occupation probabilities for closed states and the open state in terms of activation and fast inactivation variables, and assuming that activation has a faster relaxation than inactivation and that the activation sensors are mutually independent, the kinetic equations may be further reduced to rate equations for activation, and coupled fast and slow inactivation that describe spike frequency adaptation, a repetitive bursting oscillation in the neural membrane, and a cardiac action potential with a plateau oscillation. The fast inactivation rate function is, in general, dependent on the activation variable $m\left(t\right)$ but may be approximated by a voltage-dependent function, and the rate function for entry into the slow inactivated state is dependent on the fast inactivation variable.

Figure 1

State diagram for ${\mathrm{Na}}^{+}$ channel gating where horizontal transitions represent the activation of three voltage sensors (DI, DII, and DIII) that open the pore, and vertical transitions represent the two-stage fast inactivation process of the DIV voltage sensor and the inactivation motif.

Figure 2

The state diagram for ${\mathrm{Na}}^{+}$ channel gating in Fig. 1 may be reduced to an eight-state model when ${\beta }_{ik}\gg {\delta }_{ik}$, ${\gamma }_{ik}\gg {\alpha }_{ik}$, and ${\gamma }_{ik}+{\beta }_{ik}$ is greater than the activation and deactivation rate functions, for each $k$, where the derived rate functions ${\rho }_k$ and ${\sigma }_k$ are defined in Eqs. (26) and (27).

Figure 3

The state diagram for ${\mathrm{Na}}^{+}$ channel gating in Fig. 2 may be reduced to a seven-state model when ${\alpha }_{I1}\gg {\rho }_1$ and ${\sigma }_1\gg {\beta }_{I1}$, where the derived rate functions ${\widehat{\rho }}_1$ and ${\widehat{\sigma }}_1$ are defined in Eqs. (30) and (31).

Figure 4

During the action potential solution of Eqs. (18, 19, 20, 21, 22, 23, 24, 25) for $C_1$, $C_2$, $C_3$, $O,$ and $I_1$–$I_4$ (solid line), Eqs. (18), (22), and (23) for $C_1$, $I_1$, and $I_2$ may be approximated by Eqs. (28), (29), and (32) (dotted line), when ${\alpha }_{I1}\gg {\rho }_1$ and ${\sigma }_1\gg {\beta }_{I1}$, and $n$ and $V$ are determined by Eqs. (16) and (17). The rate functions are ${\alpha }_m=0.1(V+35)/\{1-exp[-(V+35)/10]\}$, ${\beta }_m=4exp[-(V+60)/18]$, ${\alpha }_{C1}=3{\alpha }_m$, ${\beta }_{C1}={\beta }_m$, ${\alpha }_{C2}=2{\alpha }_m$, ${\beta }_{C2}=2{\beta }_m$, ${\alpha }_O={\alpha }_m$, ${\beta }_O=3{\beta }_m$, ${\alpha }_{I1}={\alpha }_{C1}$, ${\beta }_{I1}=0.016{\beta }_{C1}$, ${\alpha }_{I2}=2{\alpha }_{C2}$, ${\beta }_{I2}=2{\beta }_{C2}$, ${\alpha }_{I3}=2{\alpha }_O$, ${\beta }_{I3}=2{\beta }_O$, ${\alpha }_{ik}=1$, ${\gamma }_{ik}=22.2$, ${\beta }_{ik}=exp[-V/10]$, ${\delta }_{i1}=2.5$, ${\delta }_{i2}={\delta }_{i3}={\delta }_{i4}=0.04$, ${\rho }_k={\alpha }_{ik}/(1+{\beta }_{ik}/{\gamma }_{ik})$ for $k=1$ to 4, ${\sigma }_1={\delta }_{i1}/(1+{\gamma }_{i1}/{\beta }_{i1})$, ${\sigma }_2={\sigma }_3={\sigma }_4=0.016{\sigma }_1$, ${\alpha }_n=0.01(V+50)/(1-exp[-(V+50)/10])$, ${\beta }_n=0.125exp[-(V+60)/80]$ (${\mathrm{ms}}^{-1}$), and ${\overline{g}}_{\text{Na}}=120\mathrm{mS}/{\mathrm{cm}}^2$, ${\overline{g}}_{\text{K}}=36\mathrm{mS}/{\mathrm{cm}}^2$, ${\overline{g}}_L=0.3\mathrm{mS}/{\mathrm{cm}}^2$, $V_{\text{Na}}=55\mathrm{mV}$, $V_{\text{K}}=-75\mathrm{mV}$, $V_L=-60\mathrm{mV}$, $C=1\mu \mathrm{F}/{\mathrm{cm}}^2$, and $i_e=1\mu \mathrm{A}/{\mathrm{cm}}^2$.

Figure 5

State diagram for ${\mathrm{Na}}^{+}$ channel gating in Fig. 3 may be reduced to a five state model when the transition rates between fast inactivated states are larger than inactivation and recovery rates.

Figure 6

The solution of a ${\mathrm{Na}}^{+}$ channel eight-state kinetic model, Eqs. (18, 19, 20, 21, 22, 23, 24, 25) for $C_1$, $C_2$, $C_3$, $O,$ and $I_1$–$I_4$ (solid line) may be approximated by the solution of a five state model, Eqs. (33, 34, 35, 36, 37) (dotted line), where $I_1$–$I_4$ are calculated from Eqs. (32) and (42, 43, 44), and $n$ and $V$ are determined by Eqs. (16) and (17). The rate functions are ${\alpha }_m=0.1(V+35)/\{1-exp[-(V+35)/10]\}$, ${\beta }_m=4exp[-(V+60)/18]$, ${\alpha }_{C1}=3{\alpha }_m$, ${\beta }_{C1}={\beta }_m$, ${\alpha }_{C2}=2{\alpha }_m$, ${\beta }_{C2}=2{\beta }_m$, ${\alpha }_O={\alpha }_m$, ${\beta }_O=3{\beta }_m$, ${\alpha }_{I1}={\alpha }_{C1}$, ${\beta }_{I1}=0.016{\beta }_{C1}$, ${\alpha }_{I2}=2{\alpha }_{C2}$, ${\beta }_{I2}=2{\beta }_{C2}$, ${\alpha }_{I3}=2{\alpha }_O$, ${\beta }_{I3}=2{\beta }_O$, ${\alpha }_{ik}=1$, ${\gamma }_{ik}=22.2$, ${\beta }_{ik}=exp[-V/10]$, ${\delta }_{i1}=2.5$, ${\delta }_{i2}={\delta }_{i3}={\delta }_{i4}=0.04$, ${\rho }_k={\alpha }_{ik}/(1+{\beta }_{ik}/{\gamma }_{ik})$ for $k=1$ to 4, ${\sigma }_1={\delta }_{i1}/(1+{\gamma }_{i1}/{\beta }_{i1})$, ${\sigma }_2={\sigma }_3={\sigma }_4=0.016{\sigma }_1$, ${\alpha }_n=0.01(V+50)/\{1-exp[-(V+50)/10]\}$, ${\beta }_n=0.125exp[-(V+60)/80]$ (${\mathrm{ms}}^{-1}$), and ${\overline{g}}_{\text{Na}}=120\mathrm{mS}/{\mathrm{cm}}^2$, ${\overline{g}}_{\text{K}}=36\mathrm{mS}/{\mathrm{cm}}^2$, ${\overline{g}}_L=0.3\mathrm{mS}/{\mathrm{cm}}^2$, $V_{\text{Na}}=55\mathrm{mV}$, $V_{\text{K}}=-75\mathrm{mV}$, $V_L=-60\mathrm{mV}$, $C=1\mu \mathrm{F}/{\mathrm{cm}}^2$, and $i_e=1\mu \mathrm{A}/{\mathrm{cm}}^2$.

Figure 7

If the ${\mathrm{Na}}^{+}$ channel inactivation and recovery rates are an order of magnitude smaller than the activation and deactivation rates, then the occupation probabilities $C_1=m_{1}h$, $C_2=m_{2}h$, $C_3=m_{3}h$, and $O=m_{O}h$ calculated from the solution of Eqs. (45, 46, 47, 48, 49) for $m_1$, $m_2$, $m_3$, $m_O$, and $h$ (solid line) may be approximated by the open and closed state probabilities obtained from the solution of Eqs. (49) and (52, 53, 54, 55) (dotted line), where $n$ and $V$ are determined by Eqs. (16) and (17). The rate functions are ${\alpha }_m=0.1(V+35)/\{1-exp[-(V+35)/10]\}$, ${\beta }_m=4exp[-(V+60)/18]$, ${\alpha }_{C1}=3{\alpha }_m$, ${\beta }_{C1}={\beta }_m$, ${\alpha }_{C2}=2{\alpha }_m$, ${\beta }_{C2}=2{\beta }_m$, ${\alpha }_O={\alpha }_m$, ${\beta }_O=3{\beta }_m$, ${\alpha }_{I1}={\alpha }_{C1}$, ${\beta }_{I1}=0.016{\beta }_{C1}$, ${\alpha }_{I2}=2{\alpha }_{C2}$, ${\beta }_{I2}=2{\beta }_{C2}$, ${\alpha }_{I3}=2{\alpha }_O$, ${\beta }_{I3}=2{\beta }_O$, ${\alpha }_{ik}=1$, ${\gamma }_{ik}=22.2$, ${\beta }_{ik}=exp[-V/10]$, ${\rho }_k={\alpha }_{ik}{\gamma }_{ik}/({\beta }_{ik}+{\gamma }_{ik})$ for $k=1$ to 4, ${\delta }_{i1}=2.5$, ${\delta }_{i2}={\delta }_{i3}={\delta }_{i4}=0.04$, ${\sigma }_1={\delta }_{i1}/(1+{\gamma }_{i1}/{\beta }_{i1})$, ${\sigma }_2={\sigma }_3={\sigma }_4=0.016{\sigma }_1$, ${\alpha }_n=0.01(V+50)/\{1-exp[-(V+50)/10]\}$, ${\beta }_n=0.125exp[-(V+60)/80]$ (${\mathrm{ms}}^{-1}$), and ${\overline{g}}_{\text{Na}}=120\mathrm{mS}/{\mathrm{cm}}^2$, ${\overline{g}}_{\text{K}}=36\mathrm{mS}/{\mathrm{cm}}^2$, ${\overline{g}}_L=0.3\mathrm{mS}/{\mathrm{cm}}^2$, $V_{\text{Na}}=55\mathrm{mV}$, $V_{\text{K}}=-75\mathrm{mV}$, $V_L=-60\mathrm{mV}$, $C=1\mu \mathrm{F}/{\mathrm{cm}}^2$, and $i_e=1\mu \mathrm{A}/{\mathrm{cm}}^2$.

Figure 8

The voltage dependence of the ${\mathrm{Na}}^{+}$ channel HH inactivation rate function ${\alpha }_h+{\beta }_h$ (dotted line), where ${\alpha }_h=0.07exp[-(V+60)/20]$ and ${\beta }_h=1/\{1+exp[-(V+30)/10]\}$ may be approximated by the expressions in Eqs. (58) and (60) where the rate functions are defined as ${\alpha }_m=0.1(V+35)/(1-exp[-(V+35)/10])$, ${\beta }_m=4exp[-(V+60)/18]$, ${\alpha }_{C1}=3{\alpha }_m$, ${\beta }_{C1}={\beta }_m$, ${\alpha }_{C2}=2{\alpha }_m$, ${\beta }_{C2}=2{\beta }_m$, ${\alpha }_O={\alpha }_m$, ${\beta }_O=3{\beta }_m$, ${\alpha }_{I1}={\alpha }_{C1}$, ${\beta }_{I1}=0.016{\beta }_{C1}$, ${\alpha }_{I2}=2{\alpha }_{C2}$, ${\beta }_{I2}=2{\beta }_{C2}$, ${\alpha }_{I3}=2{\alpha }_O$, ${\beta }_{I3}=2{\beta }_O$, ${\alpha }_{ik}=1$, ${\gamma }_{ik}=22.2$, ${\beta }_{ik}=exp[-V/10]$, ${\delta }_{i1}=2.5$, ${\delta }_{i2}={\delta }_{i3}={\delta }_{i4}=0.04$, ${\rho }_k={\alpha }_{ik}/(1+{\beta }_{ik}/{\gamma }_{ik})$ for $k=1$ to 4, ${\sigma }_1={\delta }_{i1}/(1+{\gamma }_{i1}/{\beta }_{i1})$, ${\sigma }_2={\sigma }_3={\sigma }_4=0.016{\sigma }_1$ (${\mathrm{ms}}^{-1}$).

Figure 9

The solution of a ${\mathrm{Na}}^{+}$ channel 12-state kinetic model, Eqs. (4)–(15) (solid line) may be approximated by $C_1={(1-m)}^{3}h$, $C_2=3m{(1-m)}^{2}h$, $C_3=3m^2(1-m)h$, $O=m^{3}h$, $I=I_2+I_3+I_4=1-h$ (dotted line), where $m$ and $h$ satisfy Eqs. (56) and (59), and $n$ and $V$ are determined by Eqs. (16) and (17). The conditions for the reduction are that (1) the two-stage inactivation process satisfies ${\beta }_{ik}\gg {\delta }_{ik}$ and ${\gamma }_{ik}\gg {\alpha }_{ik}$, for each $k$ (see Fig. 1), (2) ${\alpha }_{I1}\gg {\rho }_1$ and ${\sigma }_1\gg {\beta }_{I1}$ (see Fig. 2), and (3) the transition rates between fast inactivated states are an order of magnitude larger than inactivation and recovery rates (see Fig. 3). The rate functions are ${\alpha }_m=0.1(V+35)/\{1-exp[-(V+35)/10]\}$, ${\beta }_m=4exp[-(V+60)/18]$, ${\alpha }_{C1}=3{\alpha }_m$, ${\beta }_{C1}={\beta }_m$, ${\alpha }_{C2}=2{\alpha }_m$, ${\beta }_{C2}=2{\beta }_m$, ${\alpha }_O={\alpha }_m$, ${\beta }_O=3{\beta }_m$, ${\alpha }_{I1}={\alpha }_{C1}$, ${\beta }_{I1}=0.016{\beta }_{C1}$, ${\alpha }_{I2}=2{\alpha }_{C2}$, ${\beta }_{I2}=2{\beta }_{C2}$, ${\alpha }_{I3}=2{\alpha }_O$, ${\beta }_{I3}=2{\beta }_O$, ${\alpha }_{ik}=1$, ${\gamma }_{ik}=22.2$, ${\beta }_{ik}=exp[-V/10]$, ${\delta }_{i1}=2.5$, ${\delta }_{i2}={\delta }_{i3}={\delta }_{i4}=0.04$, ${\rho }_k={\alpha }_{ik}/(1+{\beta }_{ik}/{\gamma }_{ik})$ for $k=1$ to 4, ${\sigma }_1={\delta }_{i1}/(1+{\gamma }_{i1}/{\beta }_{i1})$, ${\sigma }_2={\sigma }_3={\sigma }_4=0.016{\sigma }_1$, ${\alpha }_n=0.01(V+50)/\{1-exp[-(V+50)/10]\}$, ${\beta }_n=0.125exp[-(V+60)/80]$ (${\mathrm{ms}}^{-1}$), and ${\overline{g}}_{\text{Na}}=120\mathrm{mS}/{\mathrm{cm}}^2$, ${\overline{g}}_{\text{K}}=36\mathrm{mS}/{\mathrm{cm}}^2$, ${\overline{g}}_L=0.3\mathrm{mS}/{\mathrm{cm}}^2$, $V_{\text{Na}}=55\mathrm{mV}$, $V_{\text{K}}=-75\mathrm{mV}$, $V_L=-60\mathrm{mV}$, $C=1\mu \mathrm{F}/{\mathrm{cm}}^2$, and $i_e=1\mu \mathrm{A}/{\mathrm{cm}}^2$.

Figure 10

The solution of a ${\mathrm{Na}}^{+}$ channel 12-state kinetic model, Eqs. (4)–(15) (solid line) may be approximated by $C_1={(1-m)}^{3}h$, $C_2=3m{(1-m)}^{2}h$, $C_3=3m^2(1-m)h$, $O=m^{3}h$, $I=I_2+I_3+I_4=1-h$ (dotted line), where $m$ and $h$ satisfy Eqs. (56) and (59), and $n$ and $V$ are determined by Eqs. (16) and (17). The rate functions are ${\alpha }_m=7.45exp[0.5V/25]$, ${\beta }_m=0.8exp[-0.9V/25]$, ${\alpha }_{C1}=3{\alpha }_m$, ${\beta }_{C1}={\beta }_m$, ${\alpha }_{C2}=2{\alpha }_m$, ${\beta }_{C2}=2{\beta }_m$, ${\alpha }_O={\alpha }_m$, ${\beta }_O=3{\beta }_m$, ${\alpha }_{I1}={\alpha }_{C1}$, ${\beta }_{I1}=0.01{\beta }_{C1}$, ${\alpha }_{I2}=2{\alpha }_{C2}$, ${\beta }_{I2}=0.2{\beta }_{C2}$, ${\alpha }_{I3}=2{\alpha }_O$, ${\beta }_{I3}=0.2{\beta }_O$, ${\beta }_{i1}=2000exp[-2.4V/25]$, ${\beta }_{i2}=200exp[-2.4V/25]$, ${\beta }_{i3}=20exp[-2.4V/25]$, ${\beta }_{i4}=2exp[-2.4V/25]$, ${\delta }_{i1}=1$, ${\delta }_{i2}={\delta }_{i3}={\delta }_{i4}=0.1$, ${\alpha }_{ik}=2.1$, ${\gamma }_{ik}=25$, ${\rho }_k={\alpha }_{ik}/(1+{\beta }_{ik}/{\gamma }_{ik})$, ${\sigma }_k={\delta }_{ik}/(1+{\gamma }_{ik}/{\beta }_{ik})$, for $k=1$ to 4, ${\alpha }_n=0.01(V+50)/\{1-exp[-(V+50)/10]\}$, ${\beta }_n=0.125exp[-(V+60)/80]$ (${\mathrm{ms}}^{-1}$), and ${\overline{g}}_{\text{Na}}=20\mathrm{mS}/{\mathrm{cm}}^2$, ${\overline{g}}_{\text{K}}=10\mathrm{mS}/{\mathrm{cm}}^2$, ${\overline{g}}_L=1\mathrm{mS}/{\mathrm{cm}}^2$, $V_{\text{Na}}=40\mathrm{mV}$, $V_{\text{K}}=-90\mathrm{mV}$, $V_L=-80\mathrm{mV}$, $C=1\mu \mathrm{F}/{\mathrm{cm}}^2$, and $i_e=1\mu \mathrm{A}/{\mathrm{cm}}^2$.

Figure 11

State diagram for ${\mathrm{Na}}^{+}$ channel gating where horizontal transitions represent the activation of three voltage sensors (in the domains DI, DII, and DIII) that open the pore, and vertical transitions represent the two-stage fast inactivation process to states $I_1\left(t\right)$ to $I_4\left(t\right)$, and slow inactivation to states $S_2\left(t\right)$ to $S_4\left(t\right)$ (in domain DIV).

Figure 12

State diagram for ${\mathrm{Na}}^{+}$ channel gating in Fig. 11 may be reduced to an eleven state model when ${\beta }_{ik}\gg {\delta }_{ik}$, ${\gamma }_{ik}\gg {\alpha }_{ik}$, and ${\gamma }_{ik}+{\beta }_{ik}$ is greater than the activation and deactivation rate functions, for each $k$, where the derived rate functions are ${\rho }_k$ and ${\sigma }_k$ defined in Eqs. (26) and (27).

Figure 13

The 11-state system for ${\mathrm{Na}}^{+}$ channel gating in Fig. 12 may be reduced to a 10-state system when ${\alpha }_{I1}\gg {\rho }_1$ and ${\sigma }_1\gg {\beta }_{I1}$, where the derived rate functions ${\widehat{\rho }}_1$ and ${\widehat{\sigma }}_1$ are defined in Eqs. (30) and (31).

Figure 14

The ten state system for ${\mathrm{Na}}^{+}$ channel gating in Fig. 13 may be reduced to a six-state system when the transition rates between fast inactivated states $I_2\left(t\right)$ to $I_4\left(t\right)$, and between slow inactivated states $S_2\left(t\right)$ to $S_4\left(t\right)$ are larger than inactivation and recovery rates.

Figure 15

The solution of a ${\mathrm{Na}}^{+}$ channel 15-state kinetic model, Eqs. (61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75) (solid line) may be approximated by $C_1={(1-m)}^3{h}_{f}s$, $C_2=3m{(1-m)}^2{h}_{f}s$, $C_3=3m^2(1-m)h_{f}s$, $O=m^3{h}_{f}s$, $I=I_2+I_3+I_4=(1-h_f)s$, and $S=S_2+S_3+S_4=1-s$ (dotted line), where $m$, $h_f$, and $s$ satisfy Eqs. (56), (105), and (106), and $n$ and $V$ are determined by Eqs. (16) and (76). The conditions for the reduction are that (1) the two-stage inactivation process satisfies ${\beta }_{ik}\gg {\delta }_{ik}$ and ${\gamma }_{ik}\gg {\alpha }_{ik}$, for each $k$ (see Fig. 11), (2) ${\alpha }_{I1}\gg {\rho }_1$ and ${\sigma }_1\gg {\beta }_{I1}$ (see Fig. 12), and (3) the transition rates between fast inactivated states $I_2$–$I_4$, and between slow inactivated states $S_2$–$S_4$ are at least an order of magnitude larger than inactivation and recovery rates (see Fig. 13). The decrease in the slow inactivation probability $s$ limits the number of spikes (spike frequency adaptation), and the stationary state of the system is stable when the recovery rate $\nu$ for slow inactivation is sufficiently small. The rate functions are ${\alpha }_m=0.1(V+43.9)/(1-exp[-(V+43.9)/10])$, ${\beta }_m=0.11exp[-V/19.1]$, ${\alpha }_{C1}=3{\alpha }_m$, ${\beta }_{C1}={\beta }_m$, ${\alpha }_{C2}=2{\alpha }_m$, ${\beta }_{C2}=2{\beta }_m$, ${\alpha }_O={\alpha }_m$, ${\beta }_O=3{\beta }_m$, ${\alpha }_{I1}={\alpha }_{C1}$, ${\beta }_{I1}=0.0135{\beta }_{C1}$, ${\alpha }_{I2}={\alpha }_{C2}$, ${\beta }_{I2}={\beta }_{C2}$, ${\alpha }_{I3}={\alpha }_O$, ${\beta }_{I3}={\beta }_O$, ${\alpha }_{ik}=0.9$, ${\gamma }_{ik}=25$, ${\beta }_{ik}=2exp[-V/10]$, ${\delta }_{i1}=2.5$, ${\delta }_{i2}={\delta }_{i3}={\delta }_{i4}=0.0135{\delta }_{i1}$, ${\rho }_k={\alpha }_{ik}/(1+{\beta }_{ik}/{\gamma }_{ik})$, ${\sigma }_k={\delta }_{ik}/(1+{\gamma }_{ik}/{\beta }_{ik})$, for $k=1$ to 4, $\mu =0.047/\{1+exp[-(V+17)/10\left]\right\}$, $\nu =0.00001exp(-V/25){\alpha }_n=0.007(V+58.9)/\{1-exp[-(V+58.9)/10]\}$, ${\beta }_n=0.038exp(-V/80)$ (${\mathrm{ms}}^{-1}$), and ${\overline{g}}_{\text{Na}}=12\mathrm{mS}/{\mathrm{cm}}^2$, ${\overline{g}}_{\text{K}}=3\mathrm{mS}/{\mathrm{cm}}^2$, ${\overline{g}}_L=0.03\mathrm{mS}/{\mathrm{cm}}^2$, $V_{\text{Na}}=50\mathrm{mV}$, $V_{\text{K}}=-77\mathrm{mV}$, $V_L=-54.4\mathrm{mV}$, $j=4$, $C=1\mu \mathrm{F}/{\mathrm{cm}}^2$, and $i_e=1\mu \mathrm{A}/{\mathrm{cm}}^2$.

Figure 16

The solution of a ${\mathrm{Na}}^{+}$ channel 15-state kinetic model, Eqs. (61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75) (solid line) may be approximated by $C_1={(1-m)}^3{h}_{f}s$, $C_2=3m{(1-m)}^2{h}_{f}s$, $C_3=3m^2(1-m)h_{f}s$, $O=m^3{h}_{f}s$, $I=I_2+I_3+I_4=(1-h_f)s$ and $S=S_2+S_3+S_4=1-s$ (dotted line), where $m$, $h_f$, and $s$ satisfy Eqs. (56), (105), and (106), and $n$ and $V$ are determined by Eqs. (16) and (76). The decrease in the slow inactivation probability $s$ terminates the burst of spikes, and as the slow variable relaxes during the subthreshold oscillation, the stationary state of the subsystem loses its stability when the recovery rate $\nu$ for slow inactivation is sufficiently large, and the bursting oscillation resumes. The rate functions are ${\alpha }_m=0.1(V+43.9)/\{1-exp[-(V+43.9)/10]\}$, ${\beta }_m=0.11exp[-V/19.1]$, ${\alpha }_{C1}=3{\alpha }_m$, ${\beta }_{C1}={\beta }_m$, ${\alpha }_{C2}=2{\alpha }_m$, ${\beta }_{C2}=2{\beta }_m$, ${\alpha }_O={\alpha }_m$, ${\beta }_O=3{\beta }_m$, ${\alpha }_{I1}={\alpha }_{C1}$, ${\beta }_{I1}=0.0135{\beta }_{C1}$, ${\alpha }_{I2}={\alpha }_{C2}$, ${\beta }_{I2}={\beta }_{C2}$, ${\alpha }_{I3}={\alpha }_O$, ${\beta }_{I3}={\beta }_O$, ${\alpha }_{ik}=0.9$, ${\gamma }_{ik}=25$, ${\beta }_{ik}=2exp[-V/10]$, ${\delta }_{i1}=5.5$, ${\delta }_{i2}={\delta }_{i3}={\delta }_{i4}=0.0135{\delta }_{i1}$, ${\rho }_k={\alpha }_{ik}/(1+{\beta }_{ik}/{\gamma }_{ik})$, ${\sigma }_k={\delta }_{ik}/(1+{\gamma }_{ik}/{\beta }_{ik})$, for $k=1$ to 4, $\mu =0.141/\{1+exp[-(V+17)/10\left]\right\}$, $\nu =0.0001exp(-V/25)$, ${\alpha }_n=0.007(V+58.9)/\{1-exp[-(V+58.9)/10]\}$, ${\beta }_n=0.038exp[-V/80]$, (${\mathrm{ms}}^{-1}$), and ${\overline{g}}_{\text{Na}}=12\mathrm{mS}/{\mathrm{cm}}^2$, ${\overline{g}}_{\text{K}}=3\mathrm{mS}/{\mathrm{cm}}^2$, ${\overline{g}}_L=0.03\mathrm{mS}/{\mathrm{cm}}^2$, $V_{\text{Na}}=50\mathrm{mV}$, $V_{\text{K}}=-77\mathrm{mV}$, $V_L=-54.4\mathrm{mV}$, $j=4$, $C=1\mu \mathrm{F}/{\mathrm{cm}}^2$, and $i_e=1\mu \mathrm{A}/{\mathrm{cm}}^2$.

Figure 17

The solution of a ${\mathrm{Na}}^{+}$ channel 15-state kinetic model, Eqs. (61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75) (solid line) may be approximated by $C_1={(1-m)}^3{h}_{f}s$, $C_2=3m{(1-m)}^2{h}_{f}s$, $C_3=3m^2(1-m)h_{f}s$, $O=m^3{h}_{f}s$, $I=I_2+I_3+I_4=(1-h_f)s$, and $S=S_2+S_3+S_4=1-s$ (dotted line), where $m$, $h_f$, and $s$ satisfy Eqs. (56), (105), and (106), $n$ and $V$ are determined by Eqs. (16) and (76), and the rate of recovery from inactivation ${\sigma }_1$ is sufficiently small to generate a cardiac plateau. The rate functions are ${\alpha }_m=0.1(V+34.3)/\{1-exp[-(V+34.3)/15]\}$, ${\beta }_m=4exp[-(V+59.3)/25]$, ${\alpha }_{C1}=3{\alpha }_m$, ${\beta }_{C1}={\beta }_m$, ${\alpha }_{C2}=2{\alpha }_m$, ${\beta }_{C2}=2{\beta }_m$, ${\alpha }_O={\alpha }_m$, ${\beta }_O=3{\beta }_m$, ${\alpha }_{I1}={\alpha }_{C1}$, ${\beta }_{I1}=0.0135{\beta }_{C1}$, ${\alpha }_{I2}=10{\alpha }_{C2}$, ${\beta }_{I2}={\beta }_{C2}$, ${\alpha }_{I3}={\alpha }_O$, ${\beta }_{I3}={\beta }_O$, ${\alpha }_{ik}=0.012$, ${\gamma }_{ik}=25$, ${\beta }_{ik}=79exp(-2.3V/25)$, ${\delta }_{i1}=0.1$, ${\delta }_{i2}=0.0135{\delta }_{i1}$, ${\delta }_{i3}={\delta }_{i4}=0.00135{\delta }_{i1}{\rho }_k={\alpha }_{ik}/(1+{\beta }_{ik}/{\gamma }_{ik})$, ${\sigma }_k={\delta }_{ik}/(1+{\gamma }_{ik}/{\beta }_{ik})$, for $k=1$ to 4 (${\mathrm{ms}}^{-1}$), $\mu =0.11exp(0.1V/25)$ (${\mathrm{s}}^{-1}$), $\nu =0.025exp(-1.95V/25)$ (${\mathrm{s}}^{-1}$), ${\alpha }_n=0.015(V+25)/\{1-exp[-(V+25)/10]\}$, ${\beta }_n=0.5exp[-(V+65)/80]$ (${\mathrm{s}}^{-1}$), and ${\overline{g}}_{\text{Na}}=36\mathrm{mS}/{\mathrm{cm}}^2$, ${\overline{g}}_{\text{K}}=3\mathrm{mS}/{\mathrm{cm}}^2$, ${\overline{g}}_L=2\mathrm{mS}/{\mathrm{cm}}^2$, $V_{\text{Na}}=55\mathrm{mV}$, $V_{\text{K}}=-80\mathrm{mV}$, $V_L=-58.5\mathrm{mV}$, $j=1$, $C=12\mu \mathrm{F}/{\mathrm{cm}}^2$, and $i_e=27\mu \mathrm{A}/{\mathrm{cm}}^2$.

Figure 18

The solution of a ${\mathrm{Na}}^{+}$ channel 15-state kinetic model, Eqs. (61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75) (solid line) may be approximated by $C_1={(1-m)}^3{h}_{f}s$, $C_2=3m{(1-m)}^2{h}_{f}s$, $C_3=3m^2(1-m)h_{f}s$, $O=m^3{h}_{f}s$, $I=I_2+I_3+I_4=(1-h_f)s$, and $S=S_2+S_3+S_4=1-s$ (dotted line), where $m$, $h_f$, and $s$ satisfy Eqs. (56), (105), and (106), $n$ and $V$ are determined by Eqs. (16) and (76), and the rate of recovery from inactivation ${\sigma }_1$ is increased to generate a cardiac action potential with a plateau oscillation. The rate functions are ${\alpha }_m=0.1(V+34.3)/(1-exp[-(V+34.3)/15])$, ${\beta }_m=4exp[-(V+59.3)/25]$, ${\alpha }_{C1}=3{\alpha }_m$, ${\beta }_{C1}={\beta }_m$, ${\alpha }_{C2}=2{\alpha }_m$, ${\beta }_{C2}=2{\beta }_m$, ${\alpha }_O={\alpha }_m$, ${\beta }_O=3{\beta }_m$, ${\alpha }_{I1}={\alpha }_{C1}$, ${\beta }_{I1}=0.0135{\beta }_{C1}$, ${\alpha }_{I2}=10{\alpha }_{C2}$, ${\beta }_{I2}={\beta }_{C2}$, ${\alpha }_{I3}={\alpha }_O$, ${\beta }_{I3}={\beta }_O$, ${\alpha }_{ik}=0.012$, ${\gamma }_{ik}=25$, ${\beta }_{ik}=79exp(-2.3V/25)$, ${\delta }_{i1}=0.12$, ${\delta }_{i2}=0.0135{\delta }_{i1}$, ${\delta }_{i3}={\delta }_{i4}=0.00135{\delta }_{i1}{\rho }_k={\alpha }_{ik}/(1+{\beta }_{ik}/{\gamma }_{ik})$, ${\sigma }_k={\delta }_{ik}/(1+{\gamma }_{ik}/{\beta }_{ik})$, for $k=1$ to 4 (${\mathrm{ms}}^{-1}$), $\mu =0.11exp[0.1V/25]$ (${\mathrm{s}}^{-1}$), $\nu =0.025exp[-1.95V/25]$ (${\mathrm{s}}^{-1}$), ${\alpha }_n=0.015(V+25)/\{1-exp[-(V+25)/10]\}$, ${\beta }_n=0.5exp[-(V+65)/80]$ (${\mathrm{s}}^{-1}$), and ${\overline{g}}_{\text{Na}}=36\mathrm{mS}/{\mathrm{cm}}^2$, ${\overline{g}}_{\text{K}}=3\mathrm{mS}/{\mathrm{cm}}^2$, ${\overline{g}}_L=2\mathrm{mS}/{\mathrm{cm}}^2$, $V_{\text{Na}}=55\mathrm{mV}$, $V_{\text{K}}=-80\mathrm{mV}$, $V_L=-58.5\mathrm{mV}$, $j=1,C=12\mu \mathrm{F}/{\mathrm{cm}}^2$, and $i_e=27\mu \mathrm{A}/{\mathrm{cm}}^2$.