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INa=GˉNamNa3hNajNa(VmENa)INa,b=GˉNa,b(VmENa)ENa=VTlnnaonaidmNadt=αmmNa(αm+βm)dhNadt=αhhNa(αh+βh)djNadt=αjmNa(αj+βj)αm=3.2ms1exprel((Vm+47.13)/10)βm=0.08ms1exp(Vm/11)αh=0.135ms1exp((Vm+80)/6.8)βh=7.69231+exp((Vm+10.66)/11.1)ms1αj=(127140exp(0.2444Vm)3.474105exp(0.04391Vm))Vm+37.781+exp(0.311(Vm+79.23))ms1βj=0.3exp(2.535107Vm)1+exp(0.1(Vm+32))ms1\begin{align} \mathrm{I_{Na}} &= \bar G_{Na} \cdot m_{Na}^{3} \cdot h_{Na} \cdot j_{Na} (V_m-E_{Na}) \\ \mathrm{I_{Na,b}} &= \bar G_{Na,b} (V_m - E_{Na}) \\ E_{Na} &= V_T \ln \frac{na_o}{na_i} \\ \frac{dm_{Na}}{dt} &= \alpha_{m} - m_{Na}(\alpha_{m} + \beta_{m}) \\ \frac{dh_{Na}}{dt} &= \alpha_{h} - h_{Na}(\alpha_{h} + \beta_{h}) \\ \frac{dj_{Na}}{dt} &= \alpha_{j} - m_{Na}(\alpha_{j} + \beta_{j}) \\ \alpha_{m} &= 3.2 \mathrm{ms}^{-1} exprel(-(V_m + 47.13) / 10) \\ \beta_{m} &= 0.08 \mathrm{ms}^{-1} \exp(-V_m / 11) \\ \alpha_{h} &= 0.135 \text{ms}^{-1} \exp(-(V_m+80)/6.8) \\ \beta_{h} &= \frac{7.6923}{ 1 + \exp(-(V_m+10.66)/11.1) } \mathrm{ms}^{-1} \\ \alpha_{j} &= (-127140 \exp(0.2444 V_m)-3.474 \cdot 10^{-5}\exp(-0.04391 V_m))\frac{V_m + 37.78}{1 + \exp(0.311( V_m + 79.23))} \mathrm{ms}^{-1} \\ \beta_{j} &= 0.3 \frac{\exp(-2.535 \cdot 10^{-7}V_m)}{1 + \exp(-0.1(V_m + 32))} \mathrm{ms}^{-1} \\ \end{align}