Canard-induced mixed-mode oscillations and bifurcation analysis in a reduced 3D pyramidal cell model

Abstract

This study entails investigation of mixed-mode oscillations (MMOs) with a conductance-based pyramidal cell (PC) model located in the entorhinal cortex layer V. This six-dimensional neuron model was reduced to three dimensions by analysis of the voltage-dependent timescales to illustrate a regime in which the MMOs are generated. Additionally, the mechanism of generation of the MMOs under antiepileptic drug conditions was illustrated in the 3D PC model. Combined with geometric singular perturbation theory (GSPT), this work shows that there is a range of parameters under which the reduced model explains the emergence of MMOs caused by an underlying canard phenomenon. In particular, we theoretically calculate the number of subthreshold oscillations using the relationship with the eigenvalue ratio of the singular perturbation system at the folded singular node, which is consistent with numerical simulations. Furthermore, a slow–fast dynamics analysis of the 3D PC model is performed, where two slow/one fast and one slow/two fast systems with the layer problem and the reduced problem are considered to explain the trajectory on the critical manifold. General one- and two-parameter bifurcation types are also discussed in this work. The first Lyapunov coefficient of the Hopf bifurcation can decide whether the bifurcation is supercritical or subcritical. Bogdanov–Takens (BT) bifurcation was also analyzed in this study and associated with three bifurcation curves near the BT point. Finally, studies on the GSPT and bifurcation analysis are of great importance for further understanding the complex dynamic behaviors and crucial roles of the signal transmission and information processing pathways of the biological nervous system.

Appendices

Appendix A

First, we rewrite the system for Eq. (4) as

$$\begin{aligned} \left\{ \begin{aligned}&\displaystyle \dot{V}=F_1(V,m_{k_s},n),\\&\dot{m_{k_s}}=F_2(V,m_{k_s}),\\&\dot{n}=F_3(V,n) \end{aligned}\right. \end{aligned}$$

(61)

where

$$\begin{aligned} \left\{ \begin{aligned} \displaystyle F_1&=\frac{1}{C}[I_\mathrm{app}-g_\mathrm{ks}m_{k_s}(V-E_\mathrm{K})\\&\quad -g_\mathrm{Nap}m_\mathrm{Nap_{\infty }}(V-E_\mathrm{Na})-g_Kn^4(V-E_\mathrm{Na})\\&\quad -g_\mathrm{Na}m^3_\mathrm{Na_{\infty }}(V)h_\mathrm{Na_{\infty }}(V)(V-E_\mathrm{Na})\\&\quad -g_\mathrm{L}(V-E_\mathrm{L})],\\ F_2&=\frac{m_{k_s\infty }(V)-m_{k_s}}{\tau _{k_s}(V)},\\ F_3&=\frac{n_{\infty }(V)-n}{\tau _{n}(V)} \end{aligned}\right. \end{aligned}$$

(62)

where $m_{\mathrm{Na}\infty }(V)$, $\alpha _\mathrm{mNa}(V)$, $\beta _\mathrm{mNa}(V)$, $h_{\mathrm{Na}\infty }(V)$, $\alpha _\mathrm{hNa}(V)$, $\beta _\mathrm{hNa}(V)$, $m_{k_{s\infty }}(V)$, $\alpha _{n}(V)$, $\beta _{n}(V)$, $\tau _{n}(V)$ and $n_\infty (V)$ are defined as shown in Table 1.

The Jacobian matrix can be expressed as

$$\begin{aligned} A =\left( \begin{array}{c@{\quad }c@{\quad }c} \displaystyle \frac{\partial F_1}{\partial V} &{} \displaystyle \frac{\partial F_1}{\partial m_{k_s}} &{} \displaystyle \frac{\partial F_1}{\partial n} \\ \displaystyle \frac{\partial F_2}{\partial V} &{} \displaystyle \frac{\partial F_2}{\partial m_{k_s}} &{}\displaystyle \frac{\partial F_2}{\partial n}\\ \displaystyle \frac{\partial F_3}{\partial V} &{} \displaystyle \frac{\partial F_3}{\partial m_{k_s}} &{}\displaystyle \frac{\partial F_3}{\partial n} \end{array} \right) , \end{aligned}$$

where

$$\begin{aligned} \displaystyle \frac{\partial F_1}{\partial V}= & {} \displaystyle G(V),\\ \frac{\partial F_1}{\partial m_\mathrm{ks}}= & {} \displaystyle -5.779162000V-520.1245800,\\ \frac{\partial F_1}{\partial n}= & {} -29.33333334n^3(V-55),\\ \frac{\partial F_2}{\partial V}= & {} 0.001709401709e^{(-0.1538461538V-3.538461537)}\\&/(e^{(-0.1538461538V-3.538461537)}+1)^2,\\ \frac{\partial F_2}{\partial m_\mathrm{ks}}= & {} -1/90,\\ \frac{\partial F_2}{\partial n}= & {} 0,\\ \frac{\partial F_3}{\partial V}= & {} H(V),\\ \frac{\partial F_3}{\partial m_\mathrm{ks}}= & {} 0,\\ \frac{\partial F_3}{\partial n}= & {} (12.50000000(e^{(-0.1V-2.7)}-1))\\&\quad e^{(-(1/80)V-37/80)}/(V+27), \end{aligned}$$

$$\begin{aligned} \frac{\partial F_1}{\partial V}= & {} G(V)\nonumber \\= & {} -5.779162000m_{k_s}-7.333333334n^4\nonumber \\&+0.2426666667e^{-(1/20)V-37/20} \nonumber \\&\quad (V-55)/((e^{(-0.1V-2.3)}-1)\nonumber \\&\quad ((-0.1V-2.3)/(e^{(-0.1V-2.3)}-1) \nonumber \\&+\,4e^{(1/18)V+23/18)}(0.07e{(-(1/20)V-37/20)}\nonumber \\&+\,1/(e^{(-0.1V-0.7)}+1)))\nonumber \\&+\,(0.002333333334(-5.2V-119.6))\nonumber \\&\quad e^{(-(1/20)V-37/20)}(V-55)/((e^{(-0.1V-2.3)}-1)\nonumber \\&\quad ((-0.1V-2.3)/(e^{(-0.1V-2.3)}-1)\nonumber \\&+\,4e^{(1/18)V+23/18)}(0.07e^{(-(1/20)V-37/20)}\nonumber \\&+\,1/(e^{(-0.1V-0.7)}+1)))\nonumber \\&-\,(0.04666666667(-5.2V-119.6))\nonumber \\&\quad e^{(-(1/20)V-37/20)}/((e^{(-0.1V-2.3)-1)}\nonumber \\&\quad ((-0.1V-2.3)/(e^{(-0.1V-2.3)}-1)\nonumber \\&+\,4e^{(1/18)V+23/18)}(0.07e^{(-(1/20)V-37/20)}\nonumber \\&+\,1/(e^{(-0.1V-0.7)}+1)))\nonumber \\&-\,(0.004666666667(-5.2V-119.6))\nonumber \\&\quad e^{(-(1/20)V-37/20)}(V-55)\nonumber \\&\quad e^{(-0.1V-2.3)}/((e^{(-0.1V-2.3)}-1)^2\nonumber \\&\quad ((-0.1V-2.3)/(e^{(-0.1V-2.3)}-1)\nonumber \\&+\,4e^{((1/18)V+23/18))}(0.07e^{(-(1/20)V-37/20)}\nonumber \\&+\,1/(e^{(-0.1V-0.7)}+1)))\nonumber \\&+\,(0.04666666667(-5.2V-119.6))\nonumber \\&\quad e^{(-(1/20)V-37/20)}(V-55)\nonumber \\&\quad (-0.1/(e^{(-0.1V-2.3)}-1)+(0.1(-0.1V-2.3))\nonumber \\&\quad e^{(-0.1V-2.3)}/(e^{(-0.1V-2.3)}-1)^2\nonumber \\&+\,(2/9)e^{((1/18)V+23/18)})/((e^{(-0.1V-2.3)}-1)\nonumber \\&\quad ((-0.1V-2.3)/(e^{(-0.1V-2.3)}-1)\nonumber \\&+\,4e^{((1/18)V+23/18)})^2(0.07e^{(-(1/20)V-37/20)}\nonumber \\&+\,1/(e^{(-0.1V-0.7)}+1)))\nonumber \\&+\,(0.04666666667(-5.2V-119.6))\nonumber \\&\quad e^{(-(1/20)V-37/20)}(V-55)\nonumber \\&\quad (-0.0035e^{(-(1/20)V-37/20)}\nonumber \\&+\,.1e^{(-0.1V-0.7)}/(e^{(-0.1V-0.7)}+1)^2)\nonumber \\&\quad /((e^{(-0.1V-2.3)}-1)((-0.1V-2.3)\nonumber \\&\quad /(e^{(-0.1V-2.3)}-1)\nonumber \\&+\,4e^{(1/18)V+23/18)}(0.07e^{(-(1/20)V-37/20)}\nonumber \\&+\,1/(e^{(-0.1V-0.7)}+1))^2)-0.06666666667\nonumber \\ \end{aligned}$$

(63)

$$\begin{aligned} \displaystyle \frac{\partial F_3}{\partial V}= & {} H(V)\nonumber \\= & {} -\,(12.5(-0.01/((e^{(-0.1V-2.7)}-1)\nonumber \\&\quad (-(0.01(V+27))/(e^{(-0.1V-2.7)}-1)\nonumber \\&+0.125e^{(-(1/80)V-37/80)}))-(0.001(V+27))\nonumber \\&\quad e^{(-0.1V-2.7)}/((e^{(-0.1V-2.7)}-1)^2\nonumber \\&\quad (-(0.01(V+27))/(e^{(-0.1V-2.7)}-1)\nonumber \\&+0.125e^{(-(1/80)V-37/80)}))+(0.01(V+27))\nonumber \\&\quad (-0.01/(e^{(-0.1V-2.7)}-1)-(0.001(V+27))\nonumber \\&\quad e^{(-0.1V-2.7)}/(e^{(-0.1V-2.7)}-1)^2\nonumber \\&-\,0.0015625e^{(-(1/80)V-37/80)})/((e^{(-0.1V-2.7)}-1)\nonumber \\&\quad (-(0.01(V+27))/(e^{(-0.1V-2.7)}-1)\nonumber \\&+\,0.125e^{(-(1/80)V-37/80)})^2)))(e^{(-0.1V-2.7)}-1)\nonumber \\&\quad e^{(-(1/80)V-37/80)}/(V+27)\nonumber \\&+\,(12.5(-(0.01(V+27))/((e^{(-0.1V-2.7)}-1)\nonumber \\&\quad (-(0.01(V+27))/(e^{(-0.1V-2.7)}-1)\nonumber \\&+\,0.125e^{(-(1/80)V-37/80)}))-n))(e^{(-0.1V-2.7)}-1)\nonumber \\&\quad e^{(-(1/80)V-37/80)}/(V+27)^2\nonumber \\&+\,(1.25(-(0.01(V+27))/((e^{(-0.1V-2.7)}-1)\nonumber \\&\quad (-(0.01(V+27))/(e^{(-0.1V-2.7)}-1)\nonumber \\&+\,0.125e^{(-(1/80)V-37/80)}))-n))e^{(-0.1V-2.7)}\nonumber \\&\quad e^{(-(1/80)}V-37/80)/(V+27)\nonumber \\&+\,(0.15625(-(0.01(V+27))/((e^{(-0.1V-2.7)}-1)\nonumber \\&\quad (-(0.01(V+27))/(e^{(-0.1V-2.7)}-1)\nonumber \\&+\,0.125e{(-(1/80)V-37/80)}))-n))\nonumber \\&\quad (e^{(-0.1V-2.7)}-1)e^{(-(1/80)V-37/80)}/(V+27)\nonumber \\ \end{aligned}$$

(64)

We calculate the equilibrium of the system for Eq.(4) at point $H_2$ (when $g_\mathrm{ks} =8.668743$) is $(V_0, {m_{k_s0}},n_0) = (-41.402667, 0.055662, 0.252883)$. The corresponding Jacobian matrix at point $H_2$ is

$$\begin{aligned} A|_{H_2} = \left( \begin{array}{c@{\quad }c@{\quad }c} -0.48999 &{} -280.8518602 &{} 45.72923277\\ 0.00008985310312 &{} -1/90 &{}0\\ 0.002475075649&{} 0&{} -0.1767754695 \end{array} \right) , \end{aligned}$$

which has a pair of conjugate eigenvalues $\lambda$ and $\bar{\lambda }$, where $\lambda =iw$, $w=0.0784$. Therefore, the Hopf bifurcation occurred at $H_2$ point as shown in Fig. 8. Set

$$\begin{aligned} \begin{aligned} q&=\left( \begin{array}{c} 0.9999 \\ 0.0002 - 0.0011i \\ 0.0117 - 0.0052i \end{array} \right) ,\\ p_0&=\left( \begin{array}{c} - 0.0003i\\ 0.9978 \\ -0.0353 - 0.0564i \end{array} \right) , \end{aligned} \end{aligned}$$

which satisfy that $Aq=iwq$, $Ap_0=-iwp_0$, $A^Tp=-iwp$, and $\langle p,q \rangle =1$. Here $\langle p, q \rangle =\bar{p}_1q_1+\bar{p}_2q_2+\bar{p}_3q_3$ is the standard scalar product in $\mathbf{C} ^3$.

By calculation, we can obtain

$$\begin{aligned} p=\left( \begin{array}{c} -1.6-2.8i \\ 9400.7-5396.9i \\ -637.6-340.4i \end{array} \right) , \end{aligned}$$

To compute the first Lyapunov coefficient, we move the equilibrium of the system to the origin of coordinate by making the following transformation

$$\begin{aligned} \left\{ \begin{array}{l} V=\xi _1+V_0,\\ m_{k_s}=\xi _2+{m_{k_s0}},\\ n=\xi _3+n_0. \end{array}\right. \end{aligned}$$

(65)

where $(V_0,{m_{k_s0}},n_0)=(-41.402667,0.055662,0.252883)$.

By this transformation, system for Eq. (23) changes into

$$\begin{aligned} \left\{ \begin{aligned} \displaystyle \frac{\mathrm{d}\xi _1}{\mathrm{d}t}&=\frac{1}{C}[I_\mathrm{app} -g_\mathrm{ks}(\xi _2+{m_{k_s0}}) (\xi _1+V_0-E_\mathrm{K})\\&\quad -g_\mathrm{K}(\xi _3+n_0)^4(\xi _1+V_0-E_\mathrm{Na})\\&\quad -g_\mathrm{Na}m^3_\mathrm{Na_{\infty }}(\xi _1+V_0)h_\mathrm{Na_{\infty }}\\&\quad (\xi _1+V_0)(\xi _1+V_0-E_\mathrm{Na})\\&\quad -g_\mathrm{L}(\xi _1+V_0-E_\mathrm{L})]\\ \displaystyle \frac{\mathrm{d}\xi _2}{\mathrm{d}t}&=\frac{m_{k_s\infty }(\xi _1+V_0)-(\xi _2+{m_{k_s0}})}{\tau _{k_s}(\xi _1+V_0)}\\ \displaystyle \frac{\mathrm{d}\xi _3}{\mathrm{d}t}&=\frac{n_{\infty }(\xi _1+V_0)-(\xi _3+n_0)}{\tau _{n}(\xi _1+V_0)} \end{aligned}\right. \end{aligned}$$

(66)

This system can also be represented as

$$\begin{aligned} \dot{x}=Ax+F(x),~x\in \mathbf{R} ^3, \end{aligned}$$

(67)

where $A=A|_{H_2}$, $F(x)=\displaystyle \frac{1}{2}B(x,x)+\displaystyle \frac{1}{6}C(x,x,x)+O(\Vert x\Vert ^4)$, B(x, y) and C(x, y, z) are symmetric multilinear vector functions which take on the planar vectors $x=(x_1,x_2,x_3)^\mathrm{T}$, $y=(y_1,y_2,y_3)^\mathrm{T}$, $z=(z_1,z_2,z_3)^\mathrm{T}$. In coordinates, we have

$$\begin{aligned} B_i(x,y)= & {} \left. \sum \limits _{j,k=1}^3\frac{\partial ^2F_i(\xi )}{\partial \xi _j \partial \xi _k}\right| _{\xi =0}x_j y_k,i=1,2,3 \end{aligned}$$

(68)

$$\begin{aligned} C_i(x,y,z)= & {} \left. \sum \limits _{j,k,l=1}^3\frac{\partial ^3F_i(\xi )}{\partial \xi _j \partial \xi _k \partial \xi _l}\right| _{\xi =0}x_j y_k z_l,i=1,2,3\nonumber \\ \end{aligned}$$

(69)

where $\xi =(\xi _1,\xi _2,\xi _3)^\mathrm{T}$.

It is easy to calculate

$$\begin{aligned} B_1(x,y)= & {} -5.779162000x_1y_2\\&+(-0.4743733935)x_1y_3\\&+(-5.779162000)x_2y_1\\&+(-29.33333334(\xi _3+0.252883)^3)x_3y_1\\&+(-88.00000002(\xi _3+.252883)^2\\&\quad (\xi _1-96.402667))x_3y_3,\\ B_2(x,y)= & {} 0.00001228465090x_1y_1,\\ B_3(x,y)= & {} (-0.005036448192-0.01845124157\xi _3)x_1y_1\\&+0.2189409839x_1y_3+0.2189409839x_3y_1\\ C_1(x,y,z)= & {} -88.00000002(\xi _3+0.252883)^2x_1y_3z_3\\&-88.00000002(\xi _3+0.252883)^2x_3y_1z_3\\&-88.00000002(\xi _3+0.252883)^2x_3y_3z_1\\&-176(\xi _3+0.252883)(\xi _1-96.402667)x_3y_3z_3,\\ C_2(x,y,z)= & {} 0,\\ C_3(x,y,z)= & {} (0.0004494324415+0.001667115304\xi _3)x_1y_1z_1\\&-0.01845124157(x_1y_1z_3+x_3y_1z_1) \end{aligned}$$

Then take $\xi =(\xi _1,\xi _2,\xi _3)^\mathrm{T}=\mathbf{0}$, there are

$$\begin{aligned} B(x,y)= & {} \left( \begin{array}{c} -5.779162000(x_1y_2+x_2y_1)-0.4743733935(x_1y_3+x_3y_1)+542.5141x_3y_3\\ 0.00001228465090x_1y_1 \\ -0.005036448192x_1y_1+ 0.2189409839x_1y_3+0.2189409839x_3y_1 \end{array} \right) ,\\ C(x,y,z)= & {} \left( \begin{array}{c} -5.6276(x_1y_3z_3+x_3y_1z_3+x_3y_3z_1)+429.06x_3y_3z_3\\ 0\\ 0.0004494324415x_1y_1z_1-0.01845124157(x_1y_1z_3+x_3y_1z_1) \end{array} \right) . \end{aligned}$$

So we can simply calculate

$$\begin{aligned}&C(q,q,\bar{q})\\&\quad =\left( \begin{array}{c} 1.742413351068001\times 10^{-4} - 4.874140221600001\times 10^{-4}i\\ 0\\ 1.762492000503097\times 10^{-5} \end{array} \right) . \end{aligned}$$

$\langle p,C(q,q,\bar{q})\rangle = -0.010151675869331 + 0.007267260943468i$

$$\begin{aligned}&B(q,\bar{q})=\left( \begin{array}{c} 0.075523675405321\\ 1.228219409266651\times 10^{-5}\\ 8.726574863159161\times 10^{-5} \end{array} \right) .\\&A^{-1}B(q,\bar{q})=\left( \begin{array}{c} 1.001243699152279\times 10^{-5}\\ -2.957137884176276\times 10^{-8}\\ 9.082122179169386\times 10^{-8} \end{array} \right) .\\&B(q,A^{-1}B(q,\bar{q}))\\&\quad =\left( \begin{array}{c} 6.797131952840937\times 10^{-7}-1.676413529615720\times 10^{-7}i\\ 1.229869931697938\times 10^{-10}\\ -4.891624430061922\times 10^{-9}-1.139909059203593\times 10^{-8}i \end{array} \right) . \end{aligned}$$

$\langle p,B(q,A^{-1}B(q,\bar{q}))=7.5371686766656\times 10^{-6} + 8.43812282046107\times 10^{-6}i$

$$\begin{aligned}&B(q,q)\\&\quad =\left( \begin{array}{c} 0.046184512877321 - 0.048367240759569i\\ 1.228219409266651\times 10^{-5}\\ 0.000087265748632 - 0.002276758533937i \end{array} \right) .\\&(2iwE-A)^{-1}B(q,q)\\&\quad =\left( \begin{array}{c} -1.249468665174975 + 0.179439435913080i\\ -1.233254792815857 + 0.328687812096022i\\ -0.014661087710834 + 0.002637391934155i \end{array} \right) .\\&B(\bar{q},(2iwE-A)^{-1}B(q,q))\\&\quad =\left( \begin{array}{c} 7.136575872105496 - 1.052954468813263i \\ -1.534775143352734\times 10^{-5} + 2.204130392802322\times 10^{-6}i \\ -0.000322278445668 - 0.001289129315673i \end{array} \right) . \end{aligned}$$

$\langle p,B(q,A^{-1}B(q,\bar{q}))\rangle = -7.982119604896726 +22.317274949636339i$.

The first Lyapunov coefficient is an index to judge the stability of the Hopf equilibrium, which is first applied in the two-dimensional system. However, for high-dimensional systems, we provide another expression which is calculated in the center manifold, as follows [56]:

$$\begin{aligned} l_1(0)= & {} \displaystyle \frac{1}{2w}\mathrm{Re}\{\langle p,C(q,q,\bar{q})\rangle \\&-2\langle p,B(q,A^{-1}B(q,\bar{q}))\rangle \\&+\langle p,B(\bar{q},(2iwE-A)^{-1}B(q,q))\rangle \}\\= & {} -50.971213999383991<0. \end{aligned}$$

Appendix B

First, we rewrite the system for Eq. (23) as

$$\begin{aligned} \frac{\mathrm{d}X}{\mathrm{d}t}= & {} F(X,\mu )\nonumber \\= & {} \left( \begin{array}{c} M_1(X,\mu )\\ M_2(X,\mu )\\ M_3(X,\mu ) \end{array}\right) , \end{aligned}$$

(70)

where $X=(V, m_{k_s}, h)^\mathrm{T}, \mu =(I, g_\mathrm{ks})^\mathrm{T}$, and

$$\begin{aligned} \left\{ \begin{aligned} \displaystyle M_1(X,\mu )&=\frac{1}{C}[I_\mathrm{app}-g_{k_s}m_{k_s}(V-E_\mathrm{K})\\&\quad -g_\mathrm{Nap}m_\mathrm{Nap_{\infty }}(V-E_\mathrm{Na})\\&\quad -g_Kn^4(V-E_\mathrm{Na})\\&\quad -g_\mathrm{Na}m^3_\mathrm{Na_{\infty }}(V)h_\mathrm{Na_{\infty }}(V)(V-E_\mathrm{Na})\\&\quad -g_\mathrm{L}(V-E_\mathrm{L})],\\ M_2(X,\mu )&=\frac{m_{k_s\infty }(V)-m_{k_s}}{\tau _\mathrm{ks}(V)},\\ M_3(X,\mu )&=\frac{n_{\infty }(V)-n}{\tau _{n}(V)} \end{aligned}\right. \nonumber \\ \end{aligned}$$

(71)

where $m_{\mathrm{Na}\infty }(V)$, $\alpha _\mathrm{mNa}(V)$, $\beta _\mathrm{mNa}(V)$, $h_{\mathrm{Na}\infty }(V)$, $\alpha _\mathrm{hNa}(V)$, $\beta _\mathrm{hNa}(V)$, $m_{K_{s\infty }}(V)$, $\alpha _{n}(V)$, $\beta _{n}(V)$, $\tau _{n}(V)$ and $n_\infty (V)$ are illustrated in Table 1.

Afterwards, the Taylor series of $F(X,\mu )$ around $(X_0, \mu _0)$ can be obtained by

$$\begin{aligned} \begin{aligned} F(X, \mu )&=DF(X_0,\mu _0)(X-X_0)+F_{\mu }(X_0,\mu _0)(\mu -\mu _0)\\&\quad + \frac{1}{2} D^2 F(X_0,\mu _0)(X-X_0,X-X_0)\\&\quad +F_{\mu X}(X_0,\mu _0)(\mu -\mu _0, X-X_0)+\cdots . \end{aligned} \end{aligned}$$

Note

$$\begin{aligned}&A_1\triangleq ~DF(X_0, \mu _0)\nonumber \\&\quad =\left. \ \left( \begin{array}{c@{\quad }c@{\quad }c} \frac{\partial M_1}{\partial V} &{} \frac{\partial M_1}{\partial m_\mathrm{ks}} &{}\frac{\partial M_1}{\partial n}\\ \frac{\partial M_2}{\partial V} &{} \frac{\partial M_2}{\partial m_\mathrm{ks}} &{}\frac{\partial M_2}{\partial n}\\ \frac{\partial M_3}{\partial V} &{} \frac{\partial M_3}{\partial m_\mathrm{ks}} &{}\frac{\partial M_3}{\partial n}\\ \end{array} \right) \right| _{(X_0, \mu _0)} \end{aligned}$$

(72)

$$\begin{aligned}&\quad =\left( \begin{array}{c@{\quad }c@{\quad }c} -4.5727088564696&{} -80.30077644 &{} 804.9883276\\ 0.00001784512307 &{} -1/90 &{} 0\\ 0.001155776073 &{} 0 &{} -0.2088980861\\ \end{array} \right) \end{aligned}$$

(73)

$$\begin{aligned}&F_{\mu }(X_0, \mu _0)\nonumber \\&\quad =\left. \ \left( \begin{array}{c@{\quad }c} \frac{\partial M_1}{\partial I} &{} \frac{\partial M_1}{\partial g_\mathrm{ks}} \\ \frac{\partial M_2}{\partial I} &{} \frac{\partial M_2}{\partial g_\mathrm{ks}} \\ \frac{\partial M_3}{\partial I} &{} \frac{\partial M_3}{\partial g_\mathrm{ks}} \\ \end{array} \right) \right| _{(X_0, \mu _0)}\nonumber \\&\quad =\left( \begin{array}{c@{\quad }c} 0.6666666667&{} -63.66524300 \\ 0 &{} 0 \\ 0 &{} 0 \\ \end{array} \right) . \end{aligned}$$

(74)

The matrix $A_1$ has three eigenvalues, specifically, 0, 0, and $-4.776117677231149$. Set $P=(p_1, p_2, P_0)$ be an invertible matrix, which satisfies

$$\begin{aligned} P^{-1}AP=\left( \begin{array}{c@{\quad }c} J_0&{}0\\ 0&{}J_1\end{array} \right) , \end{aligned}$$

where

$$\begin{aligned} J_0=\left( \begin{array}{c@{\quad }c} 0&{}1\\ 0&{}0\end{array} \right) ,~J_1=-4.776117677231149, \end{aligned}$$

$p_1$, $p_2$ are generalized eigenvectors of the matrix $A_1$, which correspond to the double-zero eigenvalue and $P_0$ includes the generalized eigenvectors of the matrix $J_1$. Thus, we conclude

$$\begin{aligned} p_1= & {} ( 1, 0.000843325052647- 0.002154679950670i, \\&0.005754280113072-0.000206016175909i )^\mathrm{T}, \\ p_2= & {} ( 1, -0.074291732336324+ 0.193917461907164i, \\&0.369290755503175+0.361481860671471i )^\mathrm{T},\\ P_0= & {} ( -2.670201010988380\times 10^5,1, \\&67.571842715765911 )^\mathrm{T}. \end{aligned}$$

If we define $P^{-1}=(q_1, q_2, Q_0^\mathrm{T})^\mathrm{T}$, then

$$\begin{aligned} q_1= & {} ( 0.0218378-0.0075046i, -9.9541435\\&+230.1753368i, 86.4424250-33.0619089i )^\mathrm{T}, \\ q_2= & {} ( 0.0002377- 0.0000984i, -1.8560778\\&-1.9317156i,0.9669274 - 0.3601848i )^\mathrm{T},\\ Q_0= & {} ( -0.0000037-0.00000003i, -0.0000442\\&+0.0008548i, 0.0003273-0.0001252i ). \end{aligned}$$

By calculating related expressions in [51], we get

$$\begin{aligned} a= & {} \frac{1}{2}p_1^\mathrm{T}(q_2\cdot D^2F(X_0, \mu _0))p_1\\= & {} -3.642039599379696\times 10^{-5} \\&+ 1.650857475385467\times 10^{-5}i, \\ b= & {} p_1^\mathrm{T}(q_1\cdot D^2F(X_0, \mu _0))p_1\\&+ p_1^\mathrm{T}(q_2\cdot D^2F(X_0, \mu _0))p_2\\= & {} -0.001154465217808 + 0.002708751706266i, \\ S_1= & {} F_{\mu }^\mathrm{T}(X_0, \mu _0)q_2\\= & {} (0.1584924085\times 10^{-3}-0.6558824307\times 10^{-4}i, \\&-0.01513568655+0.006263537149i)^\mathrm{T}, \\ S_2= & {} \displaystyle \bigg [\frac{2a}{b} (p_1^\mathrm{T}(q_1\cdot D^2F(X_0, \mu _0))p_2 \\&+ p_2^\mathrm{T}(q_2\cdot D^2F(X_0, \mu _0))p_2) \\&- \displaystyle p_1^\mathrm{T}(q_2\cdot D^2F(X_0, \mu _0))p_2)\bigg ]\\&\quad F_{\mu }^\mathrm{T}(X_0, \mu _0)q_1 \\&- \frac{2a}{b}\sum \limits _{i=1}^{2}(q_i\cdot (F_{\mu X}(X_0, \mu _0)\\&- ((P_0J_1^{-1}Q_0)F_{\mu }(X_0, \mu _0))^\mathrm{T}\times D^2F(X_0, \mu _0)))p_i \\&+ (q_2 \cdot (F_{\mu X}(X_0, \mu _0)\\&-((P_0J_1^{-1}Q_0)F_{\mu }(X_0, \mu _0))^\mathrm{T} \\&\times D^2F(X_0, \mu _0)))p_1 \\= & {} (904.4677894-558.3855372i, -3517.914512\\&+23378.45005i)^\mathrm{T}. \end{aligned}$$

where

$$\begin{aligned} A1\triangleq & {} \left. \ \left( \begin{array}{c@{\quad }c@{\quad }c} \frac{\partial ^2 M_1}{\partial ^2 V^2} &{} \frac{\partial ^2 M_1}{\partial Vm_\mathrm{ks}} &{}\frac{\partial ^2 M_1}{\partial Vn}\\ \frac{\partial ^2 M_1}{\partial m_\mathrm{ks}V} &{} \frac{\partial ^2 M_2}{\partial m_\mathrm{ks}m_\mathrm{ks}} &{}\frac{\partial ^2 M_1}{\partial m_\mathrm{ks}n}\\ \frac{\partial ^2 M_1}{\partial nV} &{} \frac{\partial ^2 M_1}{\partial nm_\mathrm{ks}} &{}\frac{\partial ^2M_3}{\partial nn}\\ \end{array} \right) \right| _{(X_0, \mu _0)}\nonumber \\= & {} \left( \begin{array}{c@{\quad }c@{\quad }c} 0.02968286213&{} -.8319926667&{} -16.60324570\\ -.8319926667 &{} 0 &{} 0\\ -16.60324570 &{} 0 &{} 2919.445095\\ \end{array} \right) .\end{aligned}$$

(75)

$$\begin{aligned} A2\triangleq & {} \left. \ \left( \begin{array}{c@{\quad }c@{\quad }c} \frac{\partial ^2 M_2}{\partial V^2} &{} \frac{\partial ^2 M_2}{\partial Vm_\mathrm{ks}} &{}\frac{\partial ^2 M_2}{\partial Vn}\\ \frac{\partial ^2 M_2}{\partial m_\mathrm{ks}V} &{} \frac{\partial ^2 M_2}{\partial m_\mathrm{ks}m_\mathrm{ks}} &{}\frac{\partial ^2 M_2}{\partial m_\mathrm{ks}n}\\ \frac{\partial ^2 M_2}{\partial nV} &{} \frac{\partial ^2 M_2}{\partial nm_\mathrm{ks}} &{}\frac{\partial ^2M_3}{\partial nn}\\ \end{array} \right) \right| _{(X_0, \mu _0)}\nonumber \\= & {} \left( \begin{array}{c@{\quad }c@{\quad }c} -0.2687471610\times 10^{-5}&{} 0 &{} 0\\ 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0\\ \end{array} \right) . \end{aligned}$$

(76)

$$\begin{aligned} A3\triangleq & {} \left. \ \left( \begin{array}{c@{\quad }c@{\quad }c} \frac{\partial ^2 M_3}{\partial V^2} &{} \frac{\partial ^2 M_3}{\partial Vm_\mathrm{ks}} &{}\frac{\partial ^2 M_3}{\partial Vn}\\ \frac{\partial ^2 M_3}{\partial m_\mathrm{ks}V} &{} \frac{\partial ^2 M_3}{\partial m_\mathrm{ks}m_\mathrm{ks}} &{}\frac{\partial ^2 M_3}{\partial m_\mathrm{ks}n}\\ \frac{\partial ^2 M_3}{\partial nV} &{} \frac{\partial ^2 M_3}{\partial nm_\mathrm{ks}} &{}\frac{\partial ^2M_3}{\partial nn}\\ \end{array} \right) \right| _{(X_0, \mu _0)}\nonumber \\= & {} \left( \begin{array}{c@{\quad }c@{\quad }c} -0.1340960884\times 10^{-3}&{} 0 &{} 0.008085695255\\ 0 &{} 0 &{} 0\\ 0.008085695255 &{} 0 &{} 0\\ \end{array} \right) .\nonumber \\ \end{aligned}$$

(77)

And there is

$$\begin{aligned} i_j\cdot D^2F(X_0, \mu _0)= & {} i^{(1)}_j\cdot A1+i^{(2)}_j\nonumber \\&\cdot A2+i^{(3)}_j\cdot A3; \end{aligned}$$

(78)

where $i=p,q , \ \ and \ \ j=1,2,3$.

Note

$$\begin{aligned} G1\triangleq & {} \left. \ \left( \begin{array}{c@{\quad }c@{\quad }c} \frac{\partial ^2 M_1}{\partial IV} &{} \frac{\partial ^2 M_1}{\partial Im_\mathrm{ks}} &{} \frac{\partial ^2 M_1}{\partial In}\\ \frac{\partial ^2 M_1}{\partial g_\mathrm{ks}V} &{} \frac{\partial ^2 M_1}{\partial g_\mathrm{ks}m_\mathrm{ks}} &{} \frac{\partial ^2 M_1}{\partial g_\mathrm{ks}n}\\ \end{array} \right) \right| _{(X_0, \mu _0)}\nonumber \\= & {} \left( \begin{array}{c@{\quad }c@{\quad }c} 0&{} 0 &{}0\\ -.6596326667 &{} -64.34413800 &{}0\\ \end{array} \right) . \end{aligned}$$

(79)

$$\begin{aligned} G2\triangleq & {} \left. \ \left( \begin{array}{c@{\quad }c@{\quad }c} \frac{\partial ^2 M_2}{\partial IV} &{} \frac{\partial ^2 M_2}{\partial Im_\mathrm{ks}} &{} \frac{\partial ^2 M_2}{\partial In}\\ \frac{\partial ^2 M_2}{\partial g_\mathrm{ks}V} &{} \frac{\partial ^2 M_2}{\partial g_\mathrm{ks}m_\mathrm{ks}} &{} \frac{\partial ^2 M_2}{\partial g_\mathrm{ks}n}\\ \end{array} \right) \right| _{(X_0, \mu _0)}\nonumber \\= & {} \left( \begin{array}{c@{\quad }c@{\quad }c} 0&{} 0 &{}0\\ 0 &{} 0 &{}0\\ \end{array} \right) . \end{aligned}$$

(80)

$$\begin{aligned} G3\triangleq & {} \left. \ \left( \begin{array}{c@{\quad }c@{\quad }c} \frac{\partial ^2 M_3}{\partial IV} &{} \frac{\partial ^2 M_3}{\partial Im_\mathrm{ks}} &{} \frac{\partial ^2 M_3}{\partial In}\\ \frac{\partial ^2 M_3}{\partial g_\mathrm{ks}V} &{} \frac{\partial ^2 M_3}{\partial g_\mathrm{ks}m_\mathrm{ks}} &{} \frac{\partial ^2 M_3}{\partial g_\mathrm{ks}n}\\ \end{array} \right) \right| _{(X_0, \mu _0)}\nonumber \\= & {} \left( \begin{array}{c@{\quad }c@{\quad }c} 0&{} 0 &{}0\\ 0 &{} 0 &{}0\\ \end{array} \right) . \end{aligned}$$

(81)

In addition, $P_0J_1^{-1}Q_0=$

$$\begin{aligned} \left( \begin{array}{c@{\quad }c@{\quad }c} -0.2048-0.0016i&{} -2.4728+47.7885i&{} 18.3013-6.9978i\\ 7.6681\times 10^{-7}+5.9616\times 10^{-9}i &{} 0.9261\times 10^{-5}-0.1790\times 10^{-3}i &{} -0.6854\times 10^{-4}+0.2621\times 10^{-4}i\\ 0.5181\times 10^{-4}+4.0284\times 10^{-7}i &{} 0.6258\times 10^{-3}-0.01209i &{} -0.0046+0.0018i\\ \end{array} \right) .\nonumber \\ \end{aligned}$$

(82)

because

$$\begin{aligned} i_j\cdot F_{\mu X}(X_0, \mu _0)=i^{(1)}_j\cdot M_1+i^{(2)}_j\cdot M_2+i^{(3)}_j\cdot M_3;\nonumber \\ \end{aligned}$$

(83)

where $i=p,q$, and $j=1,2,3$. And, there is

$$\begin{aligned} Mi=\left( \begin{array}{c@{\quad }c} Gi\\ \mathbf{0} \end{array} \right) .\nonumber \\ \end{aligned}$$

(84)

If we regard $\bar{\lambda }_1$, $\bar{\lambda }_2$ as bifurcation parameters, where $\bar{\lambda }_1=I+124.484261$, $\bar{\lambda }_2=g_\mathrm{ks}-1.247989$, then

$$\begin{aligned} \beta _1= & {} S_1^\mathrm{T}(\mu -\mu _0)\nonumber \\= & {} (0.15849\times 10^{-3}-0.65588\times 10^{-4}i)\bar{\lambda }_1\nonumber \\&+(-0.01514+0.0063i )\bar{\lambda }_2\end{aligned}$$

(85)

$$\begin{aligned} \beta _2= & {} S_2^\mathrm{T}(\mu -\mu _0)\nonumber \\= & {} (904.467789-558.385537i)\bar{\lambda }_1\nonumber \\&+(-3517.9145+23378.45005i)\bar{\lambda }_2. \end{aligned}$$

(86)

According to Theorem 1 in Ref. [51], the system for Eq. (23) at $X = X_0$, $\mu \approx \mu _0$, is locally topologically equivalent to

$$\begin{aligned} \left\{ \begin{aligned} \displaystyle \frac{\mathrm{d}z_1}{\mathrm{d}t}&=z_2,\\ \displaystyle \frac{\mathrm{d}z_2}{\mathrm{d}t}&=\beta _1+\beta _2z_1+az_1^2+bz_1z_2\\&=(0.1584924\times 10^{-3}-0.6558824\times 10^{-4}i)\bar{\lambda }_1\\&\quad +(-0.01513570+0.0062635i)\bar{\lambda }_2\\&\quad +((904.4677894-558.3855372i)\bar{\lambda }_1\\&\quad +(-3517.914512+23378.45005i)\bar{\lambda }_2)z_1\\&\quad +(-3.642039599379696\times 10^{-5}\\&\quad +1.650857475385467\times 10^{-5}i)z_1^2\\&\quad +(-0.001154465217808 + 0.002708751706266i)z_1z_2. \end{aligned}\right. \end{aligned}$$

(87)

In addition, the transformation of variables is made as follows:

$$\begin{aligned} t= & {} \Big |\frac{b}{a}\Big |t_1=\displaystyle 73.636207910921414t_1,\\ z_1= & {} \frac{a}{b^2}\eta _1\\= & {} \displaystyle \frac{(-0.3642039599\times 10^{-4} +0.1650857475385467\times 10^{-4}i)}{(-0.1154465218\times 10^{-2} +0.2708751706266\times 10^{-2}i)^2}\eta _1\\= & {} (1.535670905720935 - 4.348896894741531i)\eta _1,\\ z_2= & {} {\mathrm{sign}}\left( \frac{b}{a}\right) \frac{a^2}{b^3}\eta _2\\= & {} \displaystyle \frac{(-0.3642039599\times 10^{-4}+0.1650857475385467 \times 10^{-4}i)^2}{(-0.1154465218\times 10^{-2}+0.2708751706266\times 10^{-2}i)^3}\eta _2\\= & {} (0.055292389946590 - 0.029422219768626i)\eta _2, \end{aligned}$$

system (56) becomes

$$\begin{aligned} \left\{ \begin{aligned} \displaystyle \frac{\mathrm{d}\eta _1}{\mathrm{d}t_1}&=\eta _2,\\ \displaystyle \frac{\mathrm{d}\eta _2}{\mathrm{d}t_1}&=\bar{\beta }_1+\bar{\beta }_2\eta _1+\eta _1^2+s\eta _1\eta _2, \end{aligned}\right. \end{aligned}$$

(88)

where

$$\begin{aligned} \bar{\beta }_1= & {} \displaystyle \frac{b^4}{a^3}\beta _1\\= & {} (0.16106309-0.12134901i)\bar{\lambda }_1\\&+(-15.38118074 +11.58857174i)\bar{\lambda }_2,\\ \bar{\beta }_2= & {} \displaystyle \frac{b^2}{a^2}\beta _2\\= & {} (-2.594634\times 10^6- 5.146558\times 10^6i)\bar{\lambda }_1\\&+(1.246540\times 10^8 + 2.990890\times 10^7i)\bar{\lambda }_2,\\ s= & {} {\mathrm{sign}}(ab)=1. \end{aligned}$$

Since

$$\begin{aligned}&4\bar{\beta }_1-\bar{\beta }_2^2=0\\&\quad \Longleftrightarrow (1.97549304\times 10^{13}-2.67068651\times 10^{13}i)\bar{\lambda }_1^2\\&\qquad +(3.39007357\times 10^{14}+1.43828345\times 10^{15}i)\bar{\lambda }_1\bar{\lambda }_2\\&\qquad +(-1.46440823\times 10^{16}-7.45652823\times 10^{15}i)\bar{\lambda }_2^2\\&\qquad +(0.644252-0.485396i)\bar{\lambda }_1+(-61.524723\\&\qquad +46.354287i)\bar{\lambda }_2=0,\\&\bar{\beta }_1=0 \\&\quad \Longleftrightarrow \bar{\lambda }_1=(-54.261969203299586\\&\qquad +49.778808881697962i)\bar{\lambda }_2,\\&\bar{\beta }_2<0 \\&\quad \Longleftrightarrow \bar{\lambda }_2<(0.029048533873225 \\&\qquad + 0.034316968354156i)\bar{\lambda }_1,\\&\bar{\beta }_1+\displaystyle \frac{6}{25}\bar{\beta }_2^2=o(\bar{\beta }_2^2) \\&\quad \Longleftrightarrow (-4.74118330\times 10^{12} + 6.40964762\times 10^{12}i)\bar{\lambda }_1^2\\&\qquad +(-8.13617656\times 10^{13}-3.45188027\times 10^{14}i)\bar{\lambda }_1\bar{\lambda }_2\\&\qquad +(1.30146021\times 10^{-17}-6.62682341\times 10^{-18}i)\bar{\lambda }_2^2\\&\qquad +(-0.36420396\times 10^{-4}+0.16508575\times 10^{-4}i)\bar{\lambda }_1\\&\qquad +(-0.11544652\times 10^{-2}+0.27087517\times 10^{-2}i)\bar{\lambda }_2\\&\quad =o(\Vert (\bar{\lambda }_1,\bar{\lambda }_2)\Vert ^2). \end{aligned}$$

Appendix C

First, we rewrite the system for Eq. (23) as

$$\begin{aligned} \frac{\mathrm{d}X}{\mathrm{d}t}= & {} F(Y,\delta )\nonumber \\= & {} \left( \begin{array}{c} M_1(Y,\delta )\\ M_2(Y,\delta )\\ M_3(Y,\delta ) \end{array}\right) , \end{aligned}$$

(89)

where $Y=(V,~m_\mathrm{ks},~h)^\mathrm{T}, ~\delta =(~g_\mathrm{ks},E_\mathrm{Na})^\mathrm{T}$, and

$$\begin{aligned} \left\{ \begin{aligned} \displaystyle M_1(Y,\delta )&=\frac{1}{C}[I_\mathrm{app}-g_{k_s}m_{k_s}(V-E_\mathrm{K})\\&\quad -g_\mathrm{Nap}m_\mathrm{Nap_{\infty }}(V-E_\mathrm{Na})\\&\quad -g_Kn^4(V-E_\mathrm{Na})\\&\quad -g_\mathrm{Na}m^3_\mathrm{Na_{\infty }}(V)h_\mathrm{Na_{\infty }}(V)(V-E_\mathrm{Na})\\&\quad -g_\mathrm{L}(V-E_\mathrm{L})],\\ M_2(Y,\delta )&=\frac{m_{k_s\infty }(V)-m_{k_s}}{\tau _\mathrm{ks}(V)},\\ M_3(Y,\delta )&=\frac{n_{\infty }(V)-n}{\tau _{n}(V)} \end{aligned}\right. \nonumber \\ \end{aligned}$$

(90)

where $m_{\mathrm{Na}\infty }(V)$, $\alpha _\mathrm{mNa}(V)$, $\beta _\mathrm{mNa}(V)$, $h_{\mathrm{Na}\infty }(V)$, $\alpha _\mathrm{hNa}(V)$, $\beta _\mathrm{hNa}(V)$, $m_{K_{s\infty }}(V)$, $\alpha _{n}(V)$, $\beta _{n}(V)$, $\tau _{n}(V)$ and $n_\infty (V)$ are kept consistent with Table 1.

Let us consider the Taylor series of $F(Y,\delta )$ around $(Y_0,\delta _0)$,

$$\begin{aligned} \begin{aligned} F(Y, \delta )&=DF(Y_0,\delta _0)(Y-Y_0)+F_{\delta }(Y_0,\delta _0)(\delta -\delta _0)\\&\quad + \frac{1}{2} D^2 F(Y_0,\delta _0)(Y-Y_0,Y-Y_0)\\&\quad +F_{\delta Y}(Y_0,\delta _0)(\delta -\delta _0, Y-Y_0)+\cdots . \end{aligned} \end{aligned}$$

Note

$$\begin{aligned} A_1\triangleq & {} DF(Y_0, \delta _0)\nonumber \\= & {} \left. \ \left( \begin{array}{c@{\quad }c@{\quad }c} \frac{\partial M_1}{\partial V} &{} \frac{\partial M_1}{\partial m_{k_s}} &{}\frac{\partial M_1}{\partial n}\\ \frac{\partial M_2}{\partial V} &{} \frac{\partial M_2}{\partial m_{k_s}} &{}\frac{\partial M_2}{\partial n}\\ \frac{\partial M_3}{\partial V} &{} \frac{\partial M_3}{\partial m_{k_s}} &{}\frac{\partial M_3}{\partial n}\\ \end{array} \right) \right| _{(Y_0, \delta _0)} \end{aligned}$$

(91)

$$\begin{aligned}= & {} \left( \begin{array}{c@{\quad }c@{\quad }c} a_0&{} -524.0657322 &{} 2375.144340\\ 0.00000473130416 &{} -1/90 &{} 0\\ 0.0006020199582 &{} 0 &{} -0.1517339434\\ \end{array} \right) . \end{aligned}$$

(92)

$$\begin{aligned} a_0= & {} -9.199900247+3.838047651/((1/e^{3.8246904}-1)\nonumber \\&(-3.8246904/(1/e^{3.8246904}-1)+33.48582990))\nonumber \\&-9.280672978/((1/e^{3.8246904}-1)^2\nonumber \\&(-3.8246904/(1/e^{3.8246904}-1)\nonumber \\&+33.48582990)e^{3.8246904})\nonumber \\&+(92.80672978(-.1/(1/e^{3.8246904}-1)\nonumber \\&-0.38246904/((1/e^{3.8246904}-1)^2e^{3.8246904})\nonumber \\&+1.860323883))/((1/e^{3.8246904}-1)\nonumber \\&(-3.8246904/(1/e^{3.8246904}-1)+33.48582990)^2)\nonumber \\ \end{aligned}$$

(93)

By calculation, there is

$$\begin{aligned} A_1\triangleq & {} DF(Y_0, \delta _0)\nonumber \\= & {} \left. \ \left( \begin{array}{c@{\quad }c@{\quad }c} \frac{\partial M_1}{\partial V} &{} \frac{\partial M_1}{\partial m_{k_s}} &{}\frac{\partial M_1}{\partial n}\\ \frac{\partial M_2}{\partial V} &{} \frac{\partial M_2}{\partial m_{k_s}} &{}\frac{\partial M_2}{\partial n}\\ \frac{\partial M_3}{\partial V} &{} \frac{\partial M_3}{\partial m_{k_s}} &{}\frac{\partial M_3}{\partial n}\\ \end{array} \right) \right| _{(Y_0, \delta _0)} \end{aligned}$$

(94)

$$\begin{aligned}= & {} \left( \begin{array}{c@{\quad }c@{\quad }c} -9.443041117(nearly)&{} -524.0657322 &{} 2375.144340\\ 0.000004731304161 &{} -1/90 &{} 0\\ 0.0006020199582 &{} 0 &{} -0.1517339434\\ \end{array} \right) .\nonumber \\ \end{aligned}$$

(95)

Then,

$$\begin{aligned} F_{\mu }(Y_0, \delta _0)= & {} \left. \ \left( \begin{array}{c@{\quad }c} \frac{\partial M_1}{\partial g_\mathrm{ks}} &{} \frac{\partial M_1}{\partial E_\mathrm{Na}} \\ \frac{\partial M_2}{\partial g_\mathrm{ks}} &{} \frac{\partial M_2}{\partial E_\mathrm{Na}} \\ \frac{\partial M_3}{\partial g_\mathrm{ks}} &{} \frac{\partial M_3}{\partial E_\mathrm{Na}} \\ \end{array} \right) \right| _{(Y_0, \delta _0)}\nonumber \\= & {} \left( \begin{array}{c@{\quad }c} -69.96982574&{} 4.188172505048008 \\ 0 &{} 0 \\ 0 &{} 0 \\ \end{array} \right) .\nonumber \\ \end{aligned}$$

(96)

The matrix $A_1$ has three eigenvalues, that is, 0, 0, and $-9.594213398933288$. Let $P=(p_1, p_2, P_0)$ be an invertible matrix, which satisfies

$$\begin{aligned} P^{-1}AP=\left( \begin{array}{c@{\quad }c} J_0&{}0\\ 0&{}J_1\end{array} \right) , \end{aligned}$$

where

$$\begin{aligned} J_0=\left( \begin{array}{c@{\quad }c} 0&{}1\\ 0&{}0\end{array} \right) ,~J_1= -9.594213398933288, \end{aligned}$$

$p_1$, $p_2$ are generalized eigenvectors of the matrix $A_1$ corresponding to the double-zero eigenvalues and $P_0$ indicates the generalized eigenvectors of the matrix $J_1$. Then we get

$$\begin{aligned} p_1= & {} ( 1, 6.859055672332738\times 10^{-4}, \\&\quad 0.004124660623372 )^\mathrm{T}, \\ p_2= & {} ( 1, 0.062156774117970, 0.004124660623372 )^\mathrm{T},\\ P_0= & {} ( -2.025467388383832\times 10^6, 1, \\&-1.568466182380163\times 10^4 )^\mathrm{T}. \end{aligned}$$

If we define $P^{-1}=(q_1, q_2, Q_0^\mathrm{T})^\mathrm{T}$, so

$$\begin{aligned} q_1= & {} ( 2.1635881181719, -0.0238844412902, \\&\quad 0.0000005626868 )^\mathrm{T}, \\ q_2= & {} ( -16.2678683997221, 16.2678683997221, 0 )^\mathrm{T},\\ Q_0= & {} ( -279.3999317518015 , 3.0853980363227, \\&-0.0001364201346 ). \end{aligned}$$

After calculating related expressions in Ref. [51], we get

$$\begin{aligned} a= & {} \frac{1}{2}p_1^\mathrm{T}(q_2\cdot D^2F(Y_0, \delta _0))p_1\\= & {} 7.790675229\times 10^{-8}, \\ b= & {} p_1^\mathrm{T}(q_1\cdot D^2F(Y_0, \delta _0))p_1 \\&+ p_1^\mathrm{T}(q_2\cdot D^2F(Y_0, \delta _0))p_2\\= & {} 0.00004837327504, \\ S_1= & {} F_{\delta }^\mathrm{T}(Y_0, \delta _0)q_2\\= & {} (1138.259917, -68.13263915)^\mathrm{T}, \\ S_2= & {} \displaystyle \bigg [\frac{2a}{b} (p_1^\mathrm{T}(q_1\cdot D^2F(Y_0, \delta _0))p_2\\&+ p_2^\mathrm{T}(q_2\cdot D^2F(Y_0, \delta _0))p_2) \\&- \displaystyle p_1^\mathrm{T}(q_2\cdot D^2F(Y_0, \delta _0))p_2)\bigg ]\\&\quad F_{\delta }^\mathrm{T}(Y_0, \delta _0)q_1 \\&- \frac{2a}{b}\sum \limits _{i=1}^{2}(q_i\cdot (F_{\delta Y}(Y_0, \delta _0)\\&- ((P_0J_1^{-1}Q_0)F_{\delta }(Y_0, \delta _0))^\mathrm{T}\times D^2F(Y_0, \delta _0)))p_i \\&+ (q_2 \cdot (F_{\delta Y}(Y_0, \delta _0)\\&-((P_0J_1^{-1}Q_0)F_{\delta }(Y_0, \delta _0))^\mathrm{T} \\&\times D^2F(Y_0, \delta _0)))p_1 \\= & {} (-9.87502004880888\times 10^{11}, \\&-3.34555762806193\times 10^{11})^\mathrm{T}. \end{aligned}$$

where

$$\begin{aligned} A1\triangleq & {} \left. \ \left( \begin{array}{c@{\quad }c@{\quad }c} \frac{\partial ^2 M_1}{\partial ^2 V^2} &{} \frac{\partial ^2 M_1}{\partial Vm_\mathrm{ks}} &{}\frac{\partial ^2 M_1}{\partial Vn}\\ \frac{\partial ^2 M_1}{\partial m_\mathrm{ks}V} &{} \frac{\partial ^2 M_2}{\partial m_\mathrm{ks}m_\mathrm{ks}} &{}\frac{\partial ^2 M_1}{\partial m_\mathrm{ks}n}\\ \frac{\partial ^2 M_1}{\partial nV} &{} \frac{\partial ^2 M_1}{\partial nm_\mathrm{ks}} &{}\frac{\partial ^2M_3}{\partial nn}\\ \end{array} \right) \right| _{(Y_0, \delta _0)}\nonumber \\= & {} \left( \begin{array}{c@{\quad }c@{\quad }c} 0.02403267096&{} -4.979393334&{} -19.20015775\\ -4.979393334 &{} 0 &{} 0\\ -19.20015775 &{} 0 &{} 8206.603836\\ \end{array} \right) . \end{aligned}$$

(97)

$$\begin{aligned} A2\triangleq & {} \left. \ \left( \begin{array}{c@{\quad }c@{\quad }c} \frac{\partial ^2 M_2}{\partial V^2} &{} \frac{\partial ^2 M_2}{\partial Vm_\mathrm{ks}} &{}\frac{\partial ^2 M_2}{\partial Vn}\\ \frac{\partial ^2 M_2}{\partial m_\mathrm{ks}V} &{} \frac{\partial ^2 M_2}{\partial m_\mathrm{ks}m_\mathrm{ks}} &{}\frac{\partial ^2 M_2}{\partial m_\mathrm{ks}n}\\ \frac{\partial ^2 M_2}{\partial nV} &{} \frac{\partial ^2 M_2}{\partial nm_\mathrm{ks}} &{}\frac{\partial ^2M_3}{\partial nn}\\ \end{array} \right) \right| _{(Y_0, \delta _0)}\nonumber \\= & {} \left( \begin{array}{c@{\quad }c@{\quad }c} -7.238523899\times 10^{-7}&{} 0 &{} 0\\ 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0\\ \end{array} \right) . \end{aligned}$$

(98)

$$\begin{aligned} A3\triangleq & {} \left. \ \left( \begin{array}{c@{\quad }c@{\quad }c} \frac{\partial ^2 M_3}{\partial V^2} &{} \frac{\partial ^2 M_3}{\partial Vm_{k_s}} &{}\frac{\partial ^2 M_3}{\partial Vn}\\ \frac{\partial ^2 M_3}{\partial m_{k_s}V} &{} \frac{\partial ^2 M_3}{\partial m_\mathrm{ks}m_{k_s}} &{}\frac{\partial ^2 M_3}{\partial m_{k_s}n}\\ \frac{\partial ^2 M_3}{\partial nV} &{} \frac{\partial ^2 M_3}{\partial nm_{k_s}} &{}\frac{\partial ^2M_3}{\partial nn}\\ \end{array} \right) \right| _{(Y_0, \delta _0)}\nonumber \\= & {} \left( \begin{array}{c@{\quad }c@{\quad }c} -0.6425180442\times 10^{-4}&{} 0 &{} 0.005262992907\\ 0 &{} 0 &{} 0\\ 0.005262992907 &{} 0 &{} 0\\ \end{array} \right) .\nonumber \\ \end{aligned}$$

(99)

And there is

$$\begin{aligned} i_j\cdot D^2F(Y_0, \delta _0)= & {} i^{(1)}_j\cdot A1+i^{(2)}_j\nonumber \\&\cdot A2+i^{(3)}_j\cdot A3; \end{aligned}$$

(100)

where $i=p,q , \ \ and \ \ j=1,2,3$.

Note

$$\begin{aligned} G1\triangleq & {} \left. \ \left( \begin{array}{c@{\quad }c@{\quad }c} \frac{\partial ^2 M_1}{\partial g_\mathrm{ks}V} &{} \frac{\partial ^2 M_1}{\partial g_\mathrm{ks}m_{k_s}} &{} \frac{\partial ^2 M_1}{\partial g_\mathrm{ks}n}\\ \frac{\partial ^2 M_1}{\partial E_\mathrm{Na}V} &{} \frac{\partial ^2 M_1}{\partial E_\mathrm{Na}m_{k_s}} &{} \frac{\partial ^2 M_1}{\partial E_\mathrm{Na}n}\\ \end{array} \right) \right| _{(Y_0, \delta _0)}\nonumber \\= & {} \left( \begin{array}{c@{\quad }c@{\quad }c} -0.664816&{} -70.16460267 &{}0\\ -0.00179970 &{} 0 &{}19.20015775\\ \end{array} \right) . \end{aligned}$$

(101)

$$\begin{aligned} G2\triangleq & {} \left. \ \left( \begin{array}{c@{\quad }c@{\quad }c} \frac{\partial ^2 M_2}{\partial g_\mathrm{ks}V} &{} \frac{\partial ^2 M_2}{\partial g_\mathrm{ks}m_{k_s}} &{} \frac{\partial ^2 M_2}{\partial g_\mathrm{ks}n}\\ \frac{\partial ^2 M_2}{\partial E_\mathrm{Na}V} &{} \frac{\partial ^2 M_2}{\partial E_\mathrm{Na}m_{k_s}} &{} \frac{\partial ^2 M_2}{\partial E_\mathrm{Na}n}\\ \end{array} \right) \right| _{(Y_0, \delta _0)}\nonumber \\= & {} \left( \begin{array}{c@{\quad }c@{\quad }c} -0.664816&{} 0 &{}0\\ 0 &{} 0 &{}0\\ \end{array} \right) . \end{aligned}$$

(102)

$$\begin{aligned} G3\triangleq & {} \left. \ \left( \begin{array}{c@{\quad }c@{\quad }c} \frac{\partial ^2 M_3}{\partial g_\mathrm{ks}V} &{} \frac{\partial ^2 M_3}{\partial g_\mathrm{ks}m_{k_s}} &{} \frac{\partial ^2 M_3}{\partial g_\mathrm{ks}n}\\ \frac{\partial ^2 M_3}{\partial E_\mathrm{Na}V} &{} \frac{\partial ^2 M_3}{\partial E_\mathrm{Na}m_{k_s}} &{} \frac{\partial ^2 M_3}{\partial E_\mathrm{Na}n}\\ \end{array} \right) \right| _{(Y_0, \delta _0)}\nonumber \\= & {} \left( \begin{array}{c@{\quad }c@{\quad }c} 0&{} 0 &{}0\\ 0 &{} 0 &{}0\\ \end{array} \right) . \end{aligned}$$

(103)

In addition,

$$\begin{aligned}&P_0J_1^{-1}Q_0\nonumber \\&\quad =\left( \begin{array}{c@{\quad }c@{\quad }c} -5.898508054\times 10^7&{} 6.513689910\times 10^5&{} -28.80012381\\ 29.12171329 &{} -0.3215894736 &{} 0.00001421900149\\ -4.567642247\times 10^5 &{} 5044.022140&{} -0.2230202299\\ \end{array} \right) .\nonumber \\ \end{aligned}$$

(104)

because

$$\begin{aligned} i_j\cdot F_{\delta Y}(Y_0, \delta _0)= & {} i^{(1)}_j\cdot M1+i^{(2)}_j\nonumber \\&\cdot M2+i^{(3)}_j\cdot M3; \end{aligned}$$

(105)

where $i=p,q$, and $j=1,2,3$. And, there is

$$\begin{aligned} Mi=\left( \begin{array}{c@{\quad }c} Gi\\ \mathbf{0} \end{array} \right) . \end{aligned}$$

(106)

If we choose $\bar{\lambda }_1$, $\bar{\lambda }_2$ as bifurcation parameters where $\bar{\lambda }_1=g_\mathrm{ks}-7.469090$, $\bar{\lambda }_2=E_\mathrm{Na}-138.951322$, then

$$\begin{aligned} \beta _1= & {} S_1^\mathrm{T}(\delta -\delta _0) \end{aligned}$$

(107)

$$\begin{aligned}= & {} (1138.259917)\bar{\lambda }_1 \nonumber \\&+(-68.13263915)\bar{\lambda }_2, \end{aligned}$$

(108)

$$\begin{aligned} \beta _2= & {} S_2^\mathrm{T}(\delta -\delta _0) \end{aligned}$$

(109)

$$\begin{aligned}= & {} (-9.87502004880888\times 10^{11})\bar{\lambda }_1\nonumber \\&+(-3.34555762806193\times 10^{11})\bar{\lambda }_2. \end{aligned}$$

(110)

Due to Theorem 1 in Ref. [51], the system for Eq. (23) at $Y = Y_0, \delta \approx \delta _0$, is locally topologically equivalent to

$$\begin{aligned} \left\{ \begin{aligned} \displaystyle \frac{\mathrm{d}z_1}{\mathrm{d}t}&=z_2,\\ \displaystyle \frac{\mathrm{d}z_2}{\mathrm{d}t}&=\beta _1+\beta _2z_1+az_1^2+bz_1z_2\\&=(1138.259917)\bar{\lambda }_1+(-68.13263915)\bar{\lambda }_2\\&\quad +((-9.87502004880888\times 10^{11})\bar{\lambda }_1\\&\quad +(-3.34555762806193\times 10^{11})\bar{\lambda }_2)z_1\\&\quad +(7.790675229\times 10^{-8})z_1^2\\&\quad +0.00004837327504z_1z_2. \end{aligned}\right. \nonumber \\ \end{aligned}$$

(111)

By making the following transformation of variables

$$\begin{aligned} t= & {} \Big |\frac{b}{a}\Big |t_1\\= & {} \displaystyle 620.9124834t_1,\\ z_1= & {} \frac{a}{b^2}\eta _1\\= & {} \displaystyle \frac{7.790675229\times 10^{-8}}{(0.00004837327504)^2}\eta _1\\= & {} 33.29385754\eta _1,\\ z_2= & {} {\mathrm{sign}}\left( \frac{b}{a}\right) \frac{a^2}{b^3}\eta _2\\= & {} \displaystyle \frac{(7.790675229\times 10^{-8})^2}{(0.00004837327504)^3}\eta _2\\= & {} 0.05362085388\eta _2, \end{aligned}$$

system (112) becomes

$$\begin{aligned} \left\{ \begin{aligned} \displaystyle \frac{\mathrm{d}\eta _1}{\mathrm{d}t_1}&=\eta _2,\\ \displaystyle \frac{\mathrm{d}\eta _2}{\mathrm{d}t_1}&=\bar{\beta }_1+\bar{\beta }_2\eta _1+\eta _1^2+s\eta _1\eta _2, \end{aligned}\right. \end{aligned}$$

(112)

where

$$\begin{aligned} \bar{\beta }_1= & {} \displaystyle \frac{b^4}{a^3}\beta _1\\= & {} (1.318068887\times 10^7)\bar{\lambda }_1\\&+(-7.889543551\times 10^5)\bar{\lambda }_2,\\ \bar{\beta }_2= & {} \displaystyle \frac{b^2}{a^2}\beta _2\\= & {} (-3.807139312\times 10^{17})\bar{\lambda }_1\\&+(-1.289820568\times 10^{17})\bar{\lambda }_2,\\ s= & {} {\mathrm{sign}}(ab)=1. \end{aligned}$$

Since

$$\begin{aligned}&4\bar{\beta }_1-\bar{\beta }_2^2=0 \\&\quad \Longleftrightarrow (-1.449430974\times 10^{35})\bar{\lambda }_1^2\\&\qquad +(-9.821053180\times 10^{34})\bar{\lambda }_1\bar{\lambda }_2\\&\qquad +(-1.663637098\times 10^{34})\bar{\lambda }_2^2\\&\qquad +(5.272275548\times 10^7)\bar{\lambda }_1\\&\qquad +(-3.155817420\times 10^6)\bar{\lambda }_2=0,\\&\bar{\beta }_1=0 \\&\quad \Longleftrightarrow \bar{\lambda }_1 =0.05985683775\bar{\lambda }_2,\\&\bar{\beta }_2<0 \\&\quad \Longleftrightarrow \bar{\lambda }_2>-2.951681347\bar{\lambda }_1,\\&\bar{\beta }_1+\displaystyle \frac{6}{25}\bar{\beta }_2^2=o(\bar{\beta }_2^2) \\&\quad \Longleftrightarrow (3.478634338\times 10^{34})\bar{\lambda }_1^2\\&\qquad +(2.357052763\times 10^{34})\bar{\lambda }_1\bar{\lambda }_2\\&\qquad +(3.992729035\times 10^{33})\bar{\lambda }_2^2\\&\qquad +(1.318068887 \times 10^7)\bar{\lambda }_1\\&\qquad +(-7.889543551\times 10^5)\bar{\lambda }_2\\&\quad =o(\Vert (\bar{\lambda }_1,\bar{\lambda }_2)\Vert ^2). \end{aligned}$$

Appendix D (tables)

See Tables 1, 2, 3 and 4.

Table 1 Standard parameter values for the PC model

Table 2 Data related to the special bifurcation points with two-parameter (I, $g_\mathrm{ks}$)

Table 3 Data related to the special bifurcation points with two-parameter ($g_\mathrm{ks}$, $E_\mathrm{Na}$)

Table 4 Data related to the special bifurcation points