Dynamics of a Piecewise-Linear Morris–Lecar Model: Bifurcations and Spike Adding

Abstract

Multiple-timescale systems often display intricate dynamics, yet of great mathematical interest and well suited to model real-world phenomena such as bursting oscillations. In the present work, we construct a piecewise-linear version of the Morris–Lecar neuron model, denoted PWL-ML, and we thoroughly analyse its bifurcation structure with respect to three main parameters. Then, focusing on the homoclinic connection present in our PWL-ML, we study the slow passage through this connection when augmenting the original system with a slow dynamics for one of the parameters, thereby establishing a simplified framework for this slow-passage phenomenon. Our results show that our model exhibits equivalent behaviours to its smooth counterpart. In particular, we identify canard solutions that are part of spike-adding transitions. Focusing on the one-spike and on the two-spike scenarios, we prove their existence in a more straightforward manner than in the smooth context. In doing so, we present several techniques that are specific to the piecewise-linear framework and with the potential to offer new tools for proving the existence of dynamical objects in a wider context.

Appendices

Local Flow

In this appendix, we present some local properties of the flow of system (2) when the parameters are contained in the region ${\mathcal {P}}$, see (4).

System (2) can be written in terms of its associated linear system

$$\begin{aligned} \mathbf {{\dot{x}}} = M_{i,j}{\textbf{x}}+b_{i,j}, \end{aligned}$$

where $M_{i,j}$ and $b_{i,j}$ are the matrix and the vector components associated to the linear region $R_{i,j}$. The local flow of (2) in the region $R_{i,j}$ is determined by the trace, the determinant and the discriminant of $M_{i,j}$: $\textrm{tr}_{i,j}$, $\det _{i,j}$, and $\Delta _{i,j}$, respectively. Let us denote ${\textbf{p}}_{i,j}=-M_{i,j}^{-1}b_{i,j}$. If ${\textbf{p}}_{i,j}\in R_{i,j}$, this point is an equilibrium point of the system; otherwise, it is a virtual equilibrium (Di Bernardo et al. 2008).

Depending on the region, the matrices, traces, determinants and equilibrium points are as follows. Matrix

$$\begin{aligned} M_{1,1} = \begin{pmatrix} -1/\varepsilon &{} \quad -1/\varepsilon \\ 0 &{}\quad -1 \end{pmatrix}, \end{aligned}$$

with $\det _{1,1}=1/\varepsilon >0$, $\textrm{tr}_{1,1}=-1/\varepsilon -1<0$ and $\Delta _{1,1}=(1/\varepsilon -1)^2\ge 0$. When point ${\textbf{p}}_{1,1}=(I,0)$ is an equilibrium point it is an attractor node. Matrix

$$\begin{aligned} M_{2,1} = \begin{pmatrix} \delta /\varepsilon &{}\quad -1/\varepsilon \\ 0 &{} \quad -1 \end{pmatrix}, \end{aligned}$$

with $\det _{2,1}=-\delta /\varepsilon >0$, $\textrm{tr}_{2,1}=\delta /\varepsilon -1<0$ and $\Delta _{2,1}=(\delta /\varepsilon +1)^2\ge 0$. Again, when point ${\textbf{p}}_{2,1}=(\frac{\lambda +I}{-\delta },0)$ is an equilibrium point it is an attractor node. Matrix

$$\begin{aligned} M_{3,1} = \begin{pmatrix} k/\varepsilon &{}\quad -1/\varepsilon \\ 0 &{}\quad -1 \end{pmatrix}, \end{aligned}$$

with $\det _{3,1}=-k/\varepsilon <0$. Since the determinant is negative, when the point ${\textbf{p}}_{3,1}=(-\frac{\beta +I}{k},0)$ is an equilibrium, it is a saddle. The eigenvectors associated to the eigenvalues $\lambda _1=k/\varepsilon$, $\lambda _2=-1$ are $v_1=(1,0)$, $v_2=(k+\varepsilon ,1)$, respectively. These directions over the equilibrium point define the unstable manifold $W^u({\textbf{p}}_{3,1})$ and the stable manifold $W^s({\textbf{p}}_{3,1})$, respectively. Matrix

$$\begin{aligned} M_{3,2} = \begin{pmatrix} k/\varepsilon &{}\quad -1/\varepsilon \\ l &{}\quad -1 \end{pmatrix}. \end{aligned}$$

with $\det _{3,2}=(l-k)/\varepsilon >0$, $\textrm{tr}_{3,2}=k/\varepsilon -1$ and $\Delta _{3,2}=\frac{1}{\varepsilon ^2}\Big ((k+\varepsilon )^2-4l\varepsilon \Big )$. When the point ${\textbf{p}}_{3,2}=\frac{1}{k-l}(n-\beta -I,kn-Il-\beta l)$ is an equilibrium point it has either a focus or a node type dynamics. In particular, for sufficiently large l the point ${\textbf{p}}_{3,2}$ is a repelling focus with eigenvalues $\rho \pm \theta \,\textrm{i}$, where $\rho =(k-\varepsilon )/(2\varepsilon )$ and $\theta =\sqrt{4l\varepsilon -(k+\varepsilon )^2}/(2\varepsilon )$. In such a case, the local expression of the orbit of system (2) with initial condition ${\textbf{q}}_0=(x_0,y_0)^T \in R_{3,2}$ is given by:

$$\begin{aligned} \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = P^{-1}e^{Jt}P({\textbf{q}}_0+A^{-1}{\textbf{a}}_0)-A^{-1}{\textbf{a}}_0, \end{aligned}$$

where

$$\begin{aligned} P^{-1} = \begin{pmatrix} \frac{\rho +1}{l} &{}\quad \frac{\theta }{l} \\ 1 &{}\quad 0 \end{pmatrix}, \quad e^{Jt} = e^{\rho t}\begin{pmatrix} \cos (\theta t) &{}\quad \sin (\theta t) \\ -\sin (\theta t) &{}\quad \cos (\theta t) \end{pmatrix}, \\ A = \begin{pmatrix} k/\varepsilon &{}\quad -1/\varepsilon \\ l &{}\quad -1 \end{pmatrix}, \quad {\textbf{a}}_0 = \left( \frac{1}{\varepsilon }(\beta +I), n\right) ^T. \end{aligned}$$

We finally consider matrix

$$\begin{aligned} M_{4,3} = \begin{pmatrix} -1/\varepsilon &{}\quad -1/\varepsilon \\ 0 &{}\quad -1 \end{pmatrix}, \end{aligned}$$

which satisfies that $M_{4,3}=M_{1,1}$. Hence, when ${\textbf{p}}_{4,3}=(\gamma -1+I,1)$ is an equilibrium point then it is an attractor node. The expressions of the orbits in the half-space $\{x>1\}$ are given by

$$\begin{aligned} \begin{aligned}&x(t) = C_1e^{-t/\varepsilon }-\frac{C_2}{1-\varepsilon }e^{-t} - 1 + \gamma + I, \\&y(t) = C_2e^{-t}+1, \end{aligned} \end{aligned}$$

(A1)

where $C_1,C_2\in {\mathbb {R}}$.

Let us consider the point ${\textbf{p}}=(1,-1+\gamma +I)^T$ on the separation line $\{x=1\}$ of the region $R_{4,3}$. Denoting by $\dot{{\textbf{p}}}$ the vector field of system (2) evaluated at ${\textbf{p}}$ it follows that $\dot{{\textbf{p}}}=M_{4,3}{\textbf{p}}+b_{4,3}=(0,2-\gamma -I)^T$. Hence, provided $2-\gamma -I\ne 0$, the set $\{{\textbf{p}},\dot{{\textbf{p}}}\}$ is a base of ${\mathbb {R}}^2$ called a Krylov base, and the orbits of system (2) can be expressed in terms of this base as $\Gamma (t)=u(t){\textbf{p}}+v(t)\dot{{\textbf{p}}}$. From Theorem 5 of Llibre et al. (2002), as long as $\Gamma (t)\subset R_{4,3}$, function

$$\begin{aligned} H(u,v)=(u+\lambda _1v)^{\lambda _2}(u+\lambda _2v)^{-\lambda _1}, \end{aligned}$$

(A2)

is constant along $\Gamma (t)$, where $\lambda _1,\lambda _2$ are the eigenvalues of matrix $M_{4,3}$. Therefore, the point transformation defined by the flow in region $R_{4,3}$ on the line $\{x=1\}$ can be expressed as $\pi (z)$, where the point ${\textbf{p}}+z\dot{{\textbf{p}}}$ maps onto the point ${\textbf{p}}+\pi (z)\dot{{\textbf{p}}}$, through the implicit relation $H(1,z)=H(1,\pi (z))$.

For a sufficiently smooth function h(z), and to simplify the notation, let us consider the following abuse of notation $H(z)=H(1,h(z))$. It follows that

$$\begin{aligned} H''(z)&= D (\lambda _1 - \lambda _2) (\lambda _1 h(z) + 1)^{\lambda _2 - 2} (\lambda _2 h(z) + 1)^{-\lambda _1 - 2} \\&\quad \Big (h'(z)^2 (D (\lambda _1 - \lambda _2 + 1) h(z)^2 - 1) \\&\quad - h(z) (\lambda _1 h(z) + 1) (\lambda _2 h(z) + 1) h''(z) \Big ), \end{aligned}$$

$$\begin{aligned} H'''(z)&= D (\lambda _1 h(z) + 1)^{\lambda _2 - 3} (\lambda _2 h(z) + 1)^{-\lambda _1 - 3} \\&\quad \Big ( (\lambda _2 - \lambda _1) h(z) h^{(3)}(z) (\lambda _1 h(z) + 1)^2 (\lambda _2 h(z) + 1)^2 - (\lambda _1 - \lambda _2) h'(z)^3 \\&\quad (D (\lambda _1 - \lambda _2 + 2) h(z) (D (\lambda _1 - \lambda _2 + 1) h(z)^2 - 3) - 2 (T)) \\&\quad + 3 (\lambda _1 - \lambda _2) (\lambda _1 h(z) + 1) (\lambda _2 h(z) + 1) h'(z) h''(z) \\&\quad (D(\lambda _1 - \lambda _2 + 1) h(z)^2 - 1) \Big ), \end{aligned}$$

where $T=\lambda _1+\lambda _2$ and $D=\lambda _1\lambda _2$. Hence, by equalling $H''(0)=H''(\pi (0))$ and $H'''(0)=H'''(\pi (0))$, we get the first terms of the time series expansion of $\pi (z)$, where

$$\begin{aligned} \pi (z)= -z + \frac{2T}{3}z^2 + O(z^3), \end{aligned}$$

which describes the flow in $R_{4,3}$ in a neighbourhood of $z=0$, corresponding with the coordinate in the Krylov base of the contact point $(1,k+\beta +I)$.

Finally, by direct integration, the solution of system (7) in the half-space $\{x>1\}$ is given by

$$\begin{aligned} \begin{aligned}&x_{4,3}(t) = C_1e^{-t/\varepsilon }-\frac{C_2}{1-\varepsilon }e^{-t}+\sigma t - 1 + \gamma -\sigma \varepsilon + I_0 \\&y_{4,3}(t) = C_2e^{-t}+1 \\&I(t) = \sigma t + I_0 \end{aligned} \end{aligned}$$

(A3)

We notice that $I(t) = \sigma t + I_0$ is the same expression for any region. On the other hand, orbits in region $R_{3,2}$ for the ${\textbf{x}}(t)=(x_{3,2}(t),y_{3,2})^t$ variables are given by

$$\begin{aligned} {\textbf{x}}(t) = P^{-1}e^{Jt}P{\textbf{x}}_0+{\textbf{x}}_p, \end{aligned}$$

(A4)

where P, $e^{Jt}$ are the ones given previously, ${\textbf{x}}_0\in {\mathbb {R}}^2$ and ${\textbf{x}}_p$ is a particular solution; for instance, in our case we have taken the solution ${\textbf{x}}_p=(a_{p}t+b_{p},(ka_p+\sigma )t+kb_p-\varepsilon a_p+\beta +I_0)$, with $a_p=\sigma /(l-k)$ and $b_p=\frac{(k-\varepsilon )\sigma /(l-k)+\sigma +\beta -n+I_0}{l-k}$.

2-Spikes Maximal Canards

In this appendix, we provide all the partial derivatives of the equations in system (34) with respect to the variables $\{\sigma ,y_1,y_2,y_3,z_1,z_2,z_3,\tau _1,{\tilde{\tau }}_2,\tau _3\}$ and evaluated at the point $(0,{\hat{y}}_1(\varepsilon ),0,{\hat{y}}_1(\varepsilon ),I(\varepsilon ),I(\varepsilon ),I(\varepsilon ),\tau (\varepsilon ),0,\tau (\varepsilon ))$.

We first compute the partial derivative with respect to $y_1$. Since this variable only appears in the 2nd, 4th and 5th equations, it follows that $\frac{\textrm{d}\phi _i}{\textrm{d}y_1}=0$ for $i\ne 2,4,5$. Moreover,

$$\begin{aligned} \frac{\textrm{d}\phi _2}{\textrm{d}y_1}=-1, \quad \lim _{{\tilde{\tau }}_2\rightarrow 0}\frac{\textrm{d}\phi _4}{\textrm{d}y_1}=\frac{-1}{k+\varepsilon },\quad \lim _{{\tilde{\tau }}_2\rightarrow 0}\frac{\textrm{d}\phi _5}{\textrm{d}y_1}=0. \end{aligned}$$

Variable $y_2$ only appears in the 5th, 7th and 8th equations, so

$$\begin{aligned} \frac{\textrm{d}\phi _5}{\textrm{d}y_2}=-1,\quad \frac{\textrm{d}\phi _7}{\textrm{d}y_2}=\frac{e^{ \frac{\tau (\varepsilon )}{\varepsilon } }-e^{-\tau (\varepsilon )}}{1-\varepsilon }, \quad \frac{\textrm{d}\phi _8}{\textrm{d}y_2}=e^{-\tau (\varepsilon )}. \end{aligned}$$

Similarly, variable $y_3$ only appears in the 8th and 9th equations, so

$$\begin{aligned} \frac{\textrm{d}\phi _8}{\textrm{d}y_3}=-1, \quad \frac{\textrm{d}\phi _9}{\textrm{d}y_3}=\frac{k}{k+\varepsilon }. \end{aligned}$$

Variable $z_1$ only appears in 1st and 3rd equations, thus

$$\begin{aligned} \frac{\textrm{d}\phi _1}{\textrm{d}z_1}=1-e^{- \frac{\tau (\varepsilon )}{\varepsilon } }, \quad \frac{\textrm{d}\phi _3}{\textrm{d}z_1}=1. \end{aligned}$$

Variable $z_2$ appears in the 3rd, 4th and 5th equations, thus

$$\begin{aligned} \frac{\textrm{d}\phi _3}{\textrm{d}z_2}=-1, \quad \lim _{{\tilde{\tau }}_2\rightarrow 0}\frac{\textrm{d}\phi _4}{\textrm{d}z_2}=\frac{1}{k}, \quad \lim _{{\tilde{\tau }}_2\rightarrow 0}\frac{\textrm{d}\phi _5}{\textrm{d}z_2}=0. \end{aligned}$$

Variable $z_3$ appears in the 6th, 7th and 9th equations, and so

$$\begin{aligned} \lim _{{\tilde{\tau }}_2\rightarrow 0}\frac{\textrm{d}\phi _6}{\textrm{d}z_3}=0, \quad \frac{\textrm{d}\phi _7}{\textrm{d}z_3}=1-e^{-\frac{\tau (\varepsilon )}{\varepsilon } }, \quad \frac{\textrm{d}\phi _9}{\textrm{d}z_3}=1. \end{aligned}$$

Variable $\sigma$ appears in the 1st, 3rd, 4th, 6th, 7th and 9th equations, and its derivatives are

$$\begin{aligned}{} & {} \frac{\textrm{d}\phi _1}{\textrm{d}\sigma }=\frac{\textrm{d}\phi _7}{\textrm{d}\sigma }=\varepsilon (e^{- \frac{\tau (\varepsilon )}{\varepsilon } }-1)+\tau (\varepsilon ), \quad \frac{\textrm{d}\phi _3}{\textrm{d}\sigma }=\frac{\textrm{d}\phi _9}{\textrm{d}\sigma }=\tau (\varepsilon ), \\{} & {} \lim _{{\tilde{\tau }}_2\rightarrow 0}\frac{\textrm{d}\phi _4}{\textrm{d}\sigma }=\frac{\varepsilon }{k^2}, \quad \lim _{{\tilde{\tau }}_2\rightarrow 0}\frac{\textrm{d}\phi _6}{\textrm{d}\sigma }=1. \end{aligned}$$

Variable $\tau _1$ appears in the 1st, 2nd, 3rd equations, and so

$$\begin{aligned} \frac{\textrm{d}\phi _1}{\textrm{d}\tau _1}=\frac{k+\beta +I(\varepsilon )}{\varepsilon }, \quad \frac{\textrm{d}\phi _2}{\textrm{d}\tau _1}=e^{-\tau (\varepsilon )}, \quad \frac{\textrm{d}\phi _3}{\textrm{d}\tau _1}=0. \end{aligned}$$

Variable ${\tilde{\tau }}_2$ appears in the 4th, 5th and 6th equations, and its derivatives are

$$\begin{aligned} \lim _{{\tilde{\tau }}_2\rightarrow 0}\frac{\textrm{d}\phi _4}{d{\tilde{\tau }}_2}=0,\quad \lim _{{\tilde{\tau }}_2\rightarrow 0}\frac{\textrm{d}\phi _5}{d{\tilde{\tau }}_2}=0,\quad \lim _{{\tilde{\tau }}_2\rightarrow 0}\frac{\textrm{d}\phi _6}{d{\tilde{\tau }}_2}=z_2-z_3. \end{aligned}$$

Variable $\tau _3$ only appears in the 7th, 8th and 9th equations, so

$$\begin{aligned} \frac{\textrm{d}\phi _7}{\textrm{d}\tau _3}=\frac{k+\beta +I(\varepsilon )-{\hat{y}}_1(\varepsilon )}{\varepsilon },\quad \frac{\textrm{d}\phi _8}{\textrm{d}\tau _3}=(1-{\hat{y}}_1(\varepsilon ))e^{-\tau (\varepsilon )},\quad \frac{\textrm{d}\phi _9}{\textrm{d}\tau _3}=0. \end{aligned}$$

Homoclinic Connection

In this appendix, we obtain system (32) from system (29), and we show that both systems are equivalent for $a<1$. Additionally, we compute the Jacobian matrix of system (32) at the solution ${\textbf{v}}_0=(0,\tau (\varepsilon _0),0,0,{\hat{y}}_1(\varepsilon ),I(\varepsilon _0),1,\varepsilon _0)$. Finally, we show that this Jacobian has a minor of maximal range.

In order to do that, let us consider the change of variables (31). From this, we obtain the following expression for the first equation in system (29):

$$\begin{aligned} {\tilde{\phi }}_1 = \frac{1-a}{1-(1-a)k}\left[ (ak+\beta +I) e^{\frac{\rho }{\theta }\sqrt{1-a}{\tilde{\tau }}_1}\Big (\xi _{1,1} \frac{\sin (\sqrt{1-a}{\tilde{\tau }}_1)}{\sqrt{1-a}}\right. \\ \left. -\cos (\sqrt{1-a}{\tilde{\tau }}_1)\Big ) - 1 +(k+\beta +I) \right] . \end{aligned}$$

We notice that functions

$$\begin{aligned} \frac{\sin (\sqrt{1-a}{\tilde{\tau }}_1)}{\sqrt{1-a}},\quad \cos (\sqrt{1-a}{\tilde{\tau }}_1), \quad e^{\frac{\rho }{\theta }\sqrt{1-a}{\tilde{\tau }}_1}, \end{aligned}$$

are well-defined analytical functions at $a=1$, and

$$\begin{aligned} \frac{\sin (\sqrt{1-a}{\tilde{\tau }}_1)}{\sqrt{1-a}}&= {\tilde{\tau }}_1+\frac{{{{\tilde{\tau }}_1}^{3}} \left( a-1\right) }{6} + O((1-a)^2), \\ \cos (\sqrt{1-a}{\tilde{\tau }}_1)&= 1+{\tilde{\tau }}_1^2(a-1)/2 + O((1-a)^2), \\ e^{\frac{\rho }{\theta }\sqrt{1-a}{\tilde{\tau }}_1}&= 1+\frac{\left( \varepsilon - k\right) }{2\sqrt{\varepsilon }} {\tilde{\tau }}_1 \left( a-1\right) + O((1-a)^2). \end{aligned}$$

Hence, we may consider ${\tilde{\phi }}_1=(1-a){\hat{\phi }}_1$, where ${\hat{\phi }}_1={\tilde{\tau }}_1(k+\beta +I)/\sqrt{\varepsilon }-1+O(1-a)$. This gives us the theoretical value for the variable ${\tilde{\tau }}_1=\sqrt{\varepsilon }/(k+\beta +I)$.

The 2nd equation in system (29) is given by:

$$\begin{aligned} {\tilde{\phi }}_2&= \frac{1-a}{1-(1-a)k}\left[ (ak+\beta +I) \Big (\xi _{2,1} \frac{\sin (\sqrt{1-a}{\tilde{\tau }}_1)}{\sqrt{1-a}} e^{\frac{\rho }{\theta }\sqrt{1-a}{\tilde{\tau }}_1} \right. \\&\quad \left. +(1-e^{\frac{\rho }{\theta }\sqrt{1-a}{\tilde{\tau }}_1})\frac{\cos (\sqrt{1-a}{\tilde{\tau }}_1)}{1-a}\Big ) - (1-(1-a)k)g_1 \right] \end{aligned}$$

Again, since the expression $(1-e^{\frac{\rho }{\theta }\sqrt{1-a}{\tilde{\tau }}_1})\frac{\cos (\sqrt{1-a}{\tilde{\tau }}_1)}{1-a}$ is well defined and analytical at $a=1$, we can consider ${\tilde{\phi }}_2=(1-a){\hat{\phi }}_2$ where ${\hat{\phi }}_2=\frac{{\tilde{\tau }}_1^2(k+\beta +I)}{2}-g_1+O(1-a)$. This gives us the theoretical value for variable $g_1=\frac{\varepsilon }{2(k+\beta +I)}$.

The 3rd and 4th equations in system (29) become

$$\begin{aligned} {\hat{\phi }}_3&= (2-\gamma -I)(e^{- \frac{\tau _2}{\varepsilon } }-1)+\frac{1-(1-a)g_1}{1-\varepsilon } (e^{-\tau _2}-e^{- \frac{\tau _2}{\varepsilon } }) = 0 \\ {\hat{\phi }}_4&= -(1-(1-a)g_1)e^{-\tau _2}+1 - {\hat{y}}_1(\varepsilon ) - (1-a)g_2 = 0. \end{aligned}$$

The 5th equation is then given by:

$$\begin{aligned} {\tilde{\phi }}_5 = \frac{1-a}{1-(1-a)k}\left[ (ak+\beta +I) e^{\frac{ \rho }{\theta } \sqrt{1-a}{\tilde{\tau }}_3}\Big (\xi _{5,1} \frac{\sin (\sqrt{1-a}{\tilde{\tau }}_3)}{\sqrt{1-a}}\right. \\ \left. -\cos (\sqrt{1-a}{\tilde{\tau }}_3)\Big ) - 1 +(k+\beta +I) \right] , \end{aligned}$$

which can be simplified again as ${\tilde{\phi }}_5=(1-a){\hat{\phi }}_5$ where ${\hat{\phi }}_5=-(k+\beta +I)\sqrt{\varepsilon }{\tilde{\tau }}_3/k-1+O(1-a)$. This gives us the value ${\tilde{\tau }}_3=-k/(\sqrt{\varepsilon }(k+\beta +I))$. Finally, ${\tilde{\phi }}_6$ is given by

$$\begin{aligned} {\tilde{\phi }}_6&= \frac{1-a}{1-(1-a)k}\left[ (ak+\beta +I) e^{\frac{\rho }{\theta }\sqrt{1-a}{\tilde{\tau }}_3}\Big ( \xi _{6,1} \frac{\sin (\sqrt{1-a}{\tilde{\tau }}_3)}{\sqrt{1-a}} \right. \\&\quad + \xi _{6,2} \cos (\sqrt{1-a}{\tilde{\tau }}_3)\Big ) + \frac{1}{1-a}((ak+\beta +I) \\&\quad - (1-(1-a)k)({\hat{y}}_1(\varepsilon )+(1-a)g_2)) \Big ]. \end{aligned}$$

This expression can be written as ${\tilde{\phi }}_6=(1-a){\hat{\phi }}_6$ where ${\hat{\phi }}_6=-(k+\beta +I){\tilde{\tau }}_3\sqrt{\varepsilon }(\sqrt{\varepsilon }{\tilde{\tau }}_3+2(k+\varepsilon ))/(2k)-g_2+O(1-a)$. This provides the value $g_2=(2(k+\varepsilon )-k/(k+\beta +I))/2$.

We conclude that system (32) is equivalent to system (29) when $a<1$. Moreover, system (32) is defined at $a=1$ and

$$\begin{aligned} \tilde{{\textbf{v}}}_0=\left( \frac{\sqrt{\varepsilon }}{k+\beta +I}, \tau , -\frac{k/\sqrt{\varepsilon }}{k+\beta +I}, \frac{\varepsilon /2}{k+\beta +I}, k+\varepsilon -\frac{k/2}{k+\beta +I}, I,1,\varepsilon \right) , \end{aligned}$$

is a solution, where $\tau =\tau (\varepsilon )$ and $I=I(\varepsilon )$ are given in Lemma 3.

Let us now compute the Jacobian matrix of system (32) at $\tilde{{\textbf{v}}}_0$ and show that there exists a minor with maximal range. In particular, the partial derivatives of equation ${\hat{\phi }}_i$ with $i=1,2,\ldots ,6$ with respect to the variables $({\tilde{\tau }}_1,\tau _2,{\tilde{\tau }}_3,g_1,g_2,I)$ produce the minor

$$\begin{aligned} A= \begin{pmatrix} \frac{k+\beta +I}{\sqrt{\varepsilon }} &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad I_{1} \\ {\tilde{\tau }}_{1,2} &{} \quad 0 &{} \quad 0 &{} \quad -1 &{} \quad 0 &{} \quad I_{2} \\ 0 &{} \quad \frac{k+\beta +I-1+e^{-\tau }}{\varepsilon } &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad 1-e^{-\frac{\tau }{\varepsilon }} \\ 0 &{} \quad e^{-\tau } &{} \quad 0 &{} \quad 0 &{} \quad 0 &{} \quad -\frac{k+\varepsilon }{k} \\ 0 &{} \quad 0 &{} \quad \frac{(k+\beta +I)\sqrt{\varepsilon }}{k} &{} \quad 0 &{} \quad 0 &{} \quad I_{5} \\ 0 &{} \quad 0 &{} \quad {\tilde{\tau }}_{3,6} &{} \quad 0 &{} \quad -1 &{} \quad I_{6} \\ \end{pmatrix} \end{aligned}$$

where not all the entries has been explicitly computed. In fact, ${\tilde{\tau }}_{i,j}$ stands for $\frac{\partial {\hat{\phi }}_j}{\partial {\tilde{\tau }}_i}$ and $I_j$ stands for $\frac{\partial {\hat{\phi }}_j}{\partial I}$. Therefore,

$$\begin{aligned} \det (A)= -\frac{(k+\beta +I)^2}{k} \begin{vmatrix} \dfrac{k+\beta +I-1+e^{-\tau }}{\varepsilon }&1-e^{- \frac{\tau }{\varepsilon } } \\ e^{-\tau }&-\dfrac{k+\varepsilon }{k} \end{vmatrix} \end{aligned}$$

(C5)

where $(k+\beta +I)\ne 0$ and the second term is precisely the determinant calculated in Lemma 3, which is nonzero.