I'm trying to find the fixed points of the Hodgkin-Huxley model, given by the system of equations:
$$ dV/dt=f_V(V,m,h,n)=\frac{1}C_M[I_{ext}-\bar{g}_{Na}m^3h(V-V_{Na})-\bar{g}_Kn^4(V-V_K)-\bar{g}_l(V-V_l)], $$ $$ dm/dt=f_m(V,m)=α_m(V)(1-m)-β_m(V)m, $$ $$ dh/dt=f_h(V,h)=α_h(V)(1-h)-β_h(V)h, $$ $$ dn/dt=f_n(V,n)=α_n(V)(1-n)-β_n(V)n $$ with,
$$ α_m(V) = \frac{(V+45)/10}{(1-\exp(-(V+45)/10)} $$
$$ β_m(V) = 4\exp(-(V+70)/18) $$
$$ α_h(V) = 0.07\exp(-(V+70)/20) $$
$$ β_h(V) = \frac{1}{1+\exp(-(V+40)/10)} $$
$$ α_n(V) = 0.1\frac{(V+60)/10}{1-\exp(-(V+60)/10)} $$
$$ β_n(V) = 0.125\exp(-(v+70)/80) $$
and $$ C_M = 1, $$ $$ \bar{g}_{Na} = 120 $$ $$ \bar{g}_K = 36, $$ $$ \bar{g}_l = 0.3, $$ $$ V_{Na} = 45, $$ $$ V_K = -82, $$ $$ V_l = -59 $$
I went ahead and set the derivatives to zero, and solved for $V$, $m$, $n$, and, $h$ in MatLab using vpasolve. I got a single fixed point at $(-69.898, 0.593, 0.054, 0.319)$. However, in my research, (Wang, Chen, Fei. Analysis and Control of the Bifurcation of the Hodgkin-Huxley Model, Table 2) I see results yielding three fixed points. Moreover, one of these fixed points should correspond with the stable resting potential of the cell membrane, which in my model is found to be $-66.831$, which does not even correspond with the one fixed point I did find. I would really appreciate any guidance as to where I may have gone wrong, but I apologize in advance for formatting issues or if my question is misguided.
