Links between topology of the transition graph and limit cycles in a two-dimensional piecewise affine biological model

Abstract

A class of piecewise affine differential (PWA) models, initially proposed by Glass and Kauffman (in J Theor Biol 39:103–129, 1973), has been widely used for the modelling and the analysis of biological switch-like systems, such as genetic or neural networks. Its mathematical tractability facilitates the qualitative analysis of dynamical behaviors, in particular periodic phenomena which are of prime importance in biology. Notably, a discrete qualitative description of the dynamics, called the transition graph, can be directly associated to this class of PWA systems. Here we present a study of periodic behaviours (i.e. limit cycles) in a class of two-dimensional piecewise affine biological models. Using concavity and continuity properties of Poincaré maps, we derive structural principles linking the topology of the transition graph to the existence, number and stability of limit cycles. These results notably extend previous works on the investigation of structural principles to the case of unequal and regulated decay rates for the 2-dimensional case. Some numerical examples corresponding to minimal models of biological oscillators are treated to illustrate the use of these structural principles.

Appendices

Appendix A: Proofs

Lemma 1.

Proof

In regular domains, the evolution of the system is described by continuous affine differential equations. Therefore, for a given an initial condition, the solution of system (2) in each regulatory domain is unique.

Moreover, the switching segments are all transparent. Thus, when a trajectory reaches a switching segment, it can be continued into its contiguous regular domain.

Therefore, a solution of the system which does not cross a switching point is unique, which ends the proof. $\square$

Lemma 2.

Proof

The proof of this lemma can be straightforwardly derived from the definition of a 2-cyclic attractor, which is the union of two transition cycles sharing at least a transition in common and which does not contain vertices from which the system can escape the 2-cyclic attractor. $\square$

Lemma 3.

Proof

First, two different trajectories cannot intersect in a point which is not a switching point according to Lemma 1. Then assume that they intersect on a switching point. The two trajectories would then pass through two different domains before reaching the switching point. This implies that $C^2$ would contain more than 1 branching vertex which is forbidden by Lemma 2. $\square$

Lemma 4.

Proof

Let $l$ be the length of the entering switching segment of $f_k$. Changing the direction of the entering axis is equivalent to transform $z \rightarrow l-z$. Moreover:

$$\begin{aligned} \dfrac{d\left[ f_k(l-z)\right] }{dz}=-\dfrac{df_k}{dz}(l-z) \hbox { and } \dfrac{d^2\left[ f_k(l-z)\right] }{dz^2}=\dfrac{d^2f_k}{dz^2}(l-z) \end{aligned}$$

Thus changing the direction of the entering axis changes the monotonicity but does not change the concavity of $f_k$.

Changing the direction of the escaping axis is equivalent to transform $f_k(z) \rightarrow l-f_k(z)$. Moreover:

$$\begin{aligned} \dfrac{d\left[ l - f_k(z) \right] }{dz}=-\dfrac{df_k}{dz}(z) {\hbox { and }} \dfrac{d^2 \left[ l-f_k(z) \right] }{dz^2}=-\dfrac{d^2f_k}{dz^2}(z) \end{aligned}$$

Therefore, changing the direction of the escaping axis changes the monotonicity and concavity of $f_k$ which ends the proof. $\square$

Proposition 1.

Proof

let $f_k^1,\,f_k^2,\,f_k^3$ and $f_k^4$ be the elementary maps of the four parallel motifs (listed in Fig. 4): $D^{{i(j-1)}} \rightarrow D^{ij} \rightarrow D^{i(j+1)},\,D^{i(j+1)} \rightarrow D^{ij} \rightarrow D^{i(j-1)},\,D^{(i+1)j} \rightarrow D^{ij} \rightarrow D^{(i-1)j}$ and $D^{(i-1)j} \rightarrow D^{ij} \rightarrow D^{(i+1)j}$ respectively, with the entering and escaping axes oriented along axis [0,$x$), for $f_k^1$ and $f_k^2$, and [0,$y$), for $f_k^3$ and $f_k^4$.

The analytical expression of $f_k^1$ and $f_k^2$ are derived by replacing in Eq. 5:

• $(x(0),y(0)) = (z + \theta ^{i-1}_x, \theta ^{i-1}_y)$ and $(x(t), y(t)) = (f_k(z) + \theta ^{i-1}_x, \theta ^{i}_y)$ for $f_k^1$;

• $(x(0),y(0)) = (z + \theta ^{i-1}_x, \theta ^{i}_y)$ and $(x(t), y(t)) = (f_k(z) + \theta ^{i-1}_x, \theta ^{i-1}_y)$ for $f_k^2$:

  $$\begin{aligned} f_k^1(z)&= \left( z + \theta ^{i-1}_x - \phi _x^{ij} \right) \cdot \left( \frac{ \theta ^{i}_y - \phi _y^{ij}}{\theta ^{i-1}_y - \phi _y^{ij}} \right) ^{\frac{d^{ij}_x}{d^{ij}_y}} + \phi _x^{ij} - \theta ^{i-1}_x {\hbox { with }} \phi _y^{ij} > \theta ^i_y .\\ f_k^2(z)&= \left( z + \theta ^{i-1}_x - \phi _x^{ij} \right) \cdot \left( \frac{ \theta ^{i-1}_y - \phi _y^{ij}}{\theta ^{i}_y - \phi _y^{ij}} \right) ^{\frac{d^{ij}_x}{d^{ij}_y}} + \phi _x^{ij} - \theta ^{i-1}_x {\hbox { with }} \phi _y^{ij} < \theta ^{i-1}_y . \end{aligned}$$

Let $a = \theta ^{i-1}_x - \phi _x^{ij} ,\, b = \frac{ \theta ^{i}_y - \phi _y^{ij}}{\theta ^{i-1}_y - \phi _y^{ij}}$ and $c = \frac{d^{ij}_x}{d^{ij}_y}$.

Then we have:

$$\begin{aligned} f_k^1(z)&= b^c \cdot \left( z + a \right) - a\\ f_k^2(z)&= (1/b)^c \cdot \left( z + a \right) - a \end{aligned}$$

Therefore, $f_k^1$ and $f_k^2$ are affine functions. As $b > 0 ,\, f_k^1$ and $f_k^2$ are increasing affine functions.

The analytical expression of $f_k^3$ and $f_k^4$ are derived by exchanging $x$ and $y$ in the expression of $f_k^1$ and $f_k^2$ respectively. Thus, $f_k^3$ and $f_k^4$ are also increasing affine functions.

Finally, according to Lemma 4, changing the orientation of either the entering or escaping axis changes the monotonicity of an elementary map, which ends the proof. $\square$

Proposition 2.

Proof

Let $f_k^1, f_k^2, f_k^3, f_k^4, f_k^5, f_k^6, f_k^7, f_k^8$ be the elementary maps of the 8 perpendicular motifs (listed in Fig. 5): $D^{(i+1)j} \rightarrow D^{ij} \rightarrow D^{i(j+1)},\,D^{(i-1)j} \rightarrow D^{ij} \rightarrow D^{i(j-1)},\,D^{i(j+1)} \rightarrow D^{ij} \rightarrow D^{(i-1)j},\,D^{i(j-1)} \rightarrow D^{ij} \rightarrow D^{(i+1)j},\,D^{i(j+1)} \rightarrow D^{ij} \rightarrow D^{(i+1)j},\,D^{i(j-1)} \rightarrow D^{ij} \rightarrow D^{(i-1)j},\,D^{(i+1)j} \rightarrow D^{ij} \rightarrow D^{i(j-1)}$ and $D^{(i-1)j} \rightarrow D^{ij} \rightarrow D^{i(j+1)}$ respectively, and $S_k^i$ the switching point at the intersection of the entering and escaping segment of $f_k^i$. We assume that the origin of the entering and escaping axes of $f_k^i$ is $S_k^i$.

The analytical expression of $f_k^1,\,f_k^2,\,f_k^3$ and $f_k^4$ are derived by replacing in Eq. 5:

• $(x(0),y(0)) = (\theta ^{i}_x, \theta ^{i}_y - z)$ and $(x(t), y(t)) = (\theta ^{i}_x - f_k(z), \theta ^{i}_y)$ for $f_k^1$;

• $(x(0),y(0)) = (\theta ^{i-1}_x, z + \theta ^{i-1}_y)$ and $(x(t), y(t)) = (f_k(z) + \theta ^{i-1}_x, \theta ^{i-1}_y)$ for $f_k^2$;

• $(x(0),y(0)) = (\theta ^{i-1}_x + z, \theta ^{i}_y)$ and $(x(t), y(t)) = (\theta ^{i-1}_x, \theta ^{i}_y - f_k(z))$ for $f_k^3$;

• $(x(0),y(0)) = (\theta ^{i}_x - z, \theta ^{i-1}_y)$ and $(x(t), y(t)) = (\theta ^{i}_x, \theta ^{i-1}_y + f_k(z))$ for $f_k^4$:

  $$\begin{aligned} f_k^1(z)&= - \left( \frac{\theta ^{i}_y - \phi _y^{ij}}{\theta ^{i}_y - \phi _y^{ij} - z} \right) ^{\frac{d^{ij}_x}{d^{ij}_y}} \cdot \left( \theta ^{i}_x - \phi _x^{ij} \right) \\&+\,\, \theta ^{i}_x - \phi _x^{ij} {\hbox { with }} \phi _x^{ij} < \theta ^{i}_x {\hbox { and }} \phi _y^{ij} > \theta ^{i}_y \\ f_k^2(z)&= - \left( \frac{\theta ^{i-1}_y - \phi _y^{ij}}{\theta ^{i-1}_y - \phi _y^{ij} + z} \right) ^{\frac{d^{ij}_x}{d^{ij}_y}} \cdot \left( \phi _x^{ij} - \theta ^{i-1}_x \right) \\&+\,\, \phi _x^{ij} - \theta ^{i-1}_x \quad {\hbox {with}}\quad \phi _x^{ij} > \theta ^{i-1}_x {\hbox { and }} \phi _y^{ij} < \theta ^{i-1}_y \\ f_k^3(z)&= - \left( \frac{\theta ^{i-1}_x - \phi _x^{ij}}{\theta ^{i-1}_x - \phi _x^{ij} + z} \right) ^{\frac{d^{ij}_y}{d^{ij}_x}} \cdot \left( \theta ^{i}_y - \phi _y^{ij} \right) \\&+\, \theta ^{i}_y - \phi _y^{ij} {\hbox { with }} \phi _y^{ij} < \theta ^{i}_y {\hbox { and }} \phi _x^{ij} < \theta ^{i-1}_x \\ f_k^4(z)&= - \left( \frac{\theta ^{i}_x - \phi _x^{ij}}{\theta ^{i}_x - \phi _x^{ij} - z} \right) ^{\frac{d^{ij}_y}{d^{ij}_x}} \cdot \left( \phi _y^{ij} - \theta ^{i-1}_y \right) \\&+\,\, \phi _y^{ij} - \theta ^{i-1}_y {\hbox { with }} \phi _y^{ij} > \theta ^{i-1}_y {\hbox { and }} \phi _x^{ij} > \theta ^{i}_x \end{aligned}$$

$f_k^i(z)$ for $i=\left\{ 1,2,3,4\right\}$ are thus of the form:

$$\begin{aligned} - \left( \frac{b}{b+z} \right) ^{c}a+a \end{aligned}$$

with $\left( a,b,c\right) > 0$. The first and second derivative of this function are respectively positive and negative. Therefore, $f_k^1,\,f_k^2,\,f_k^3$ and $f_k^4$ are increasing and strictly concave.

The analytical expression of $f_k^5, f_k^6, f_k^7, f_k^8$ are derived by exchanging $x$ and $y$ in the expression of $f_k^1, f_k^2, f_k^3, f_k^4$ respectively. Thus, $f_k^5, f_k^6, f_k^7, f_k^8$ are also increasing and concave.

Finally, according to Lemma 4, changing the orientation of the entering axis or the escaping axis of an elementary map changes its monotonicity, while concavity only changes with orientation of the escaping axis, which ends the proof. $\square$

Theorem 1.

Proof

Let the entering and escaping switching segments of all the motifs composing $C$ be oriented towards the outside of $C$. Let $f_k$ be the elementary maps of $C$.

In this case, the entering and escaping axis of the parallel motifs composing $C$ have the same orientation. Therefore, according to Proposition 1, the elementary maps of the parallel motifs of $C$ are increasing functions.

The entering and escaping axis of the perpendicular motifs composing $C$ both point either towards or away from the switching point at the intersection of the entering and escaping segment of these motifs. Therefore, according to Proposition 2, the elementary maps of the perpendicular motifs of $C$ are increasing functions.

The first return map $f$ of $C$ is thus an increasing and continuous function as the composite of increasing and continuous functions.

Moreover, if $f(x)$ is increasing, $-f(-x)$ is also increasing. Therefore, changing the orientation of the axis where $f$ is calculated does not change the monotonicity of $f$, which ends the proof. $\square$

Theorem 2.

Proof

By taking $d^{ij}_x = d^{ij}_y$ in the analytical expression of the elementary maps determined in the proofs of Propositions 1 and 2, it is straightforward to deduce that the elementary maps of C are linear fractional functions (i.e. homographic functions). The first return map of C is thus homographic as composite of homographic functions, and has, therefore, a constant and strict concavity. $\square$

Theorem 3.

Proof

Let the entering and escaping axes of the elementary maps of $C$ be oriented towards the outside of $C$. By definition of a transition cycle with no turn change, the perpendicular motifs composing $C$ are either all clockwise or all counterclockwise.

Thus, the origins of the entering and escaping axes of the perpendicular motifs are the switching point at the intersection of the entering and escaping segments of these motifs. Then, according to Proposition 2, the elementary maps of the perpendicular motifs of $C$ are strictly concave and increasing functions.

Moreover, the elementary maps of parallel motifs of a transition cycle are increasing affine functions (Proposition 1).

From the expression of the 2nd derivative of the composite of two functions: $\frac{d^2(f \circ g)}{dx^2}(x)= \left( \frac{df}{dx} \circ g\right) (x) \cdot \frac{d^2g}{dx^2}(x) + \left( \frac{d^2f}{dx^2} \circ g\right) (x) \cdot \left( \frac{dg}{dx}(x)\right) ^2$, we deduce that the composite of two increasing concave (resp. an increasing concave, and an increasing and strictly concave) functions is concave (resp. strictly concave) (and obviously increasing).

Therefore, the first return map $f$ of $C$ is strictly concave. Finally, it is straightforward to see that if $f(x)$ is strictly concave, $-f(-x)$ is strictly convex. Thus changing the direction of the axis of $f$ changes the concavity of $f$, which ends the proof. $\square$

Theorem 4.

Proof

${\widehat{f}}$ is continuous in each interval $[ 0, \alpha [$ and $] \alpha , l ]$ (Theorem 1). To study if ${\widehat{f}}$ can be continuously extended at $x = \alpha$, one approach is to analyze the uniqueness of the solution of initial condition $x = \alpha$ until the first return in $D_C$.

$D_S$ is the unique branching domain of $C^2$ (Lemma 2). Thus the trajectory will not cross a switching point before it enters $D_S$. The trajectory is therefore unique before it reaches the switching point S (Lemma 1).

Let $t_1$ be the time when the trajectory reaches S (Fig. 11). For $t=t_1$, the PWA differential equations are not defined. In order to define the solution on S, we use the Filippov approach (Gouzé and Sari 2002) as we will see later in the proof. We now make use of the assumption on the property of $SG_{S \rightarrow T}$.

Part (a). Assume $SG_{S \rightarrow T}$ is composed of 4 vertices (trajectories merge within 4 domains).

Let the regular domains composing $SG_{S \rightarrow T}$ be labeled $D^1,\,D^2,\,D^3$ and $D^4$ (as in Fig. 11 left). We can rewrite the vector field $f^i(x,y)=(f^i_x(x),f^i_y(y))$ in $D^i$ as follows:

$$\begin{aligned} f^i(X)=\varGamma ^i(\phi ^{i}-X) \end{aligned}$$

with

$$\begin{aligned} X=(x,y), \phi ^{i}=(\phi ^{i}_x,\phi ^{i}_y) \hbox { and } \varGamma ^i=[ \begin{array}{cc} d^{i}_{x} &{}\quad 0 \\ 0 &{}\quad d^{i}_{y} \end{array} ] \end{aligned}$$

• (1) Without loss of generality, we can consider the case where the separatrix is in $D^1$ with:

  $$\begin{aligned} \phi ^{1}_{x}-\theta _x > 0 {\hbox { and }} \phi ^{1}_{y}-\theta _y < 0 . \end{aligned}$$

• (2) To avoid stable sliding modes (excluded by Assumption 2), we have to exclude the following four cases:

  • (2.a) $\phi ^{3}_{x}-\theta _x < 0 {\hbox { from }} (1)$

  • (2.b) $\phi ^{2}_{y}-\theta _y > 0 {\hbox { from }} (1)$

  • (2.c) $\phi ^{4}_{y}-\theta _y >0 {\hbox { and }} \phi ^{3}_{y}-\theta _y < 0$

  • (2.d) $\phi ^{2}_{x}-\theta _x >0 {\hbox { and }} \phi ^{4}_{x}-\theta _x < 0$

• (3) We also need to satisfy the assumption that trajectories merge within $D^2,\,D^3$ or $D^4$, that is:

  • either (3.a) $\phi ^{3}_{y}-\theta _y < 0 {\hbox { and }} \phi ^{4}_{x}-\theta _x < 0\quad ({\hbox { merging in }} D^2)$

  • or (3.b) $\phi ^{2}_{x}-\theta _x > 0 {\hbox { and }} \phi ^{4}_{y}-\theta _y > 0\quad ({\hbox { merging in }} D^3)$

  • or (3.c) $\phi ^{3}_{y}-\theta _y < 0 {\hbox { and }} \phi ^{2}_{x}-\theta _x > 0\quad ({\hbox { merging in }} D^4)$

  [trajectories cannot merge in more than 1 domain due to (2)].

• (4) Following Fillipov approach, the vector field on $S=(\theta _x,\theta _y)$ is a vector in the convex hull of the adjacent vector fields, computed at $S$. That is:

  $$\begin{aligned} f(S)&\in \overline{co} \left\{ f^1, f^2, f^3, f^4 \right\} \\&= \left\{ a_1\varGamma ^1(\phi ^{1}-S) + a_2\varGamma ^2(\phi ^{2}-S) + a_3\varGamma ^3(\phi ^{3}-S) \right. \\&\left. +\,\, a_4\varGamma ^4(\phi ^{4}-S): a_i > 0, \ \sum _{i=1}^4 a_i=1 \right\} \end{aligned}$$

Consider the case (3.a). From (1), (2.b) and (2.c), we have: $\phi ^{i}_{y}-\theta _y < 0$ for all $i$. Hence the vertical component of $f(S)$ is negative as well so a trajectory starting at S may evolve towards $D^2$ or $D^4$. Moreover, from (2.d) we have: $\phi ^{2}_{x}-\theta _x < 0$. Therefore the horizontal component on $D^2$ and $D^4$ is oriented towards $D^2$. It follows that there exists only one absolutely continuous trajectory from S (evolving towards $D^2$) that satisfies the differential inclusion in $S$ and the piecewise affine system. The case (3.b) is similar, but on the domains $D^3$ and $D^4$ [$\phi ^{i}_{x}-\theta _x > 0$ for all $i$ using (1), (2.a) and (2.d), and $\phi ^{3}_{y}-\theta _y > 0$ from (2.c)].

For the case (3.c), we have from (1) an (2): $\phi ^{i}_{y}-\theta _y < 0$ and $\phi ^{i}_{x}-\theta _x > 0$ for all $i$. Thus, the vertical and horizontal component of $f(S)$ are negative and there exists only one absolutely continuous trajectory from S (evolving towards $D^4$) that satisfies the differential inclusion in $S$ and the piecewise affine system.

Therefore, if $SG_{S \rightarrow T}$ is composed of 4 vertices, the trajectory can be extended continuously in S ($t=t_1$). As $D_S$ is the unique branching domain of $C^2$, the trajectory will not cross a switching point until its first return in $D_C$ (at $t=t_2$). The solution of initial condition $x=\alpha$ is thus unique for $t_1 < t \le t_2$ (Lemma 1). This solution is therefore unique for $0 \le t \le t_2$. ${\widehat{f}}$ can then be extended by continuity at $x = \alpha$.

Part (b). Assume $SG_{S \rightarrow T}$ is composed of more than 4 vertices (trajectories merge within more than 4 domains) (see Fig. 11 right).

• (1) Let’s again assume without loss of generality that the separatrix is in $D^1$ ($D^1$ is the unique branching domain in $C^2$), that is:

  $$\begin{aligned} \phi ^{1}_{x}-\theta _x > 0 {\hbox { and }} \phi ^{1}_{y}-\theta _y < 0 . \end{aligned}$$

• (2) To avoid sliding modes (excluded by Assumption 2), one needs:

  $$\begin{aligned} \phi ^{3}_{x}-\theta _x > 0 {\hbox { and }} \phi ^{2}_{y}-\theta _y < 0 {\hbox { from }} (1) \end{aligned}$$

• (3) For a merging of the two trajectories within more than 4 domains, either $D^2$ or $D^3$ cannot communicate with $D^4$. This means that:

  • (3.a) either $\phi ^{4}_{x}-\theta _x > 0$ and $\phi ^{2}_{x}-\theta _x < 0$

  • (3.b) or $\phi ^{4}_{y}-\theta _y < 0$ and $\phi ^{3}_{y}-\theta _y > 0$

• (4) All together the information on the focal points yields:

  • (4.a) $\phi ^{1}_{x}-\theta _x > 0 , \phi ^{3}_{x}-\theta _x > 0 ,\, \phi ^{4}_{x}-\theta _x > 0$ and $\phi ^{2}_{x}-\theta _x < 0$

  • (4.b) or $\phi ^{1}_{y}-\theta _y < 0 ,\, \phi ^{2}_{y}-\theta _y < 0 ,\, \phi ^{4}_{y}-\theta _y < 0$ and $\phi ^{3}_{y}-\theta _y > 0$

Given the convex hull at $S$, in each case there are at least two absolutely continuous solutions to the system that satisfy the differential inclusion in $S$ and the piecewise affine differential system. In case (4.a), a solution from S evolves to $D^2$; another solution from S to $D^3$ or $D^4$ (depending on the vertical coordinates). In case (4.b), a solution from S evolves to $D^3$; another solution from S to $D^2$ or $D^4$ (depending on the horizontal coordinates).

Therefore, if $SG_{S \rightarrow T}$ is composed of more than 4 vertices, there are at least two distinct trajectories emerging from S. According to Lemma 3, two different trajectories cannot intersect. Each will thus cross $D_C$ in two distinct points after a first return. ${\widehat{f}}$ then cannot be extended by continuity at $x = \alpha$, which ends the proof.$\square$

Theorem 5.

Proof

First, we deduce from Theorems 2 and 3 that $f$ has a constant and strict concavity, and from Theorem 1 that $f$ is continuous.

Let $g(x)=f(x)-x$. Thus $g$ is continuous, and $\frac{dg}{dx}$ is strictly monotone and has at most one zero. Therefore, by applying the intermediate value theorem on each interval where $g$ is strictly monotone (one interval if $\frac{dg}{dx}$ has no zeros, two if $\frac{dg}{dx}$ has one zero), we deduce that $g$ has at most 2 zeros i.e. $f$ has at most two fixed points. Then, there are at most two limit cycles in $D^C$.

Moreover, if $f$ admits two fixed points at $x=x_1$ and $x=x_2,\,\frac{dg}{dx}(x_1) \cdot \frac{dg}{dx}(x_2)<0$. In addition, $\frac{df}{dx}>0$ (Theorem 1). Thus one of the two limit cycles is stable and the other one is unstable (Strogatz 2001, p.281), which ends the proof. $\square$

Lemma 5.

Proof

Let all the axes of the elementary maps $f_k$ composing $f$ be oriented towards the outside of C. We have: $f=f_n \circ f_{n-1}\circ \ldots f_1$. Let $I_k$ be the interval of definition of $f_k$ and $l_k$ the length of the entering switching segment where $f_k$ is calculated. Let $g_k = f_{k} \circ \ldots \circ f_1$ for $k \in \left\{ 1,2,\ldots , n \right\}$.

• (1) To prove statement (a), assume that $C$ has no inside branching transition. This means that each $I_j$ is of form:

  $$\begin{aligned} I_j = [ 0 , \alpha _j ] (\hbox { for some } \alpha _j \ge 0 ) \hbox { or } I_j = \emptyset \end{aligned}$$

• (2) Assume that $I \ne \emptyset$, that is, $\exists a\ge 0: a \in I$. This statement is equivalent to:

  $$\begin{aligned} \forall j \in \left\{ 1,2, \ldots n-1 \right\} , g_j(a) \in I_{j+1} \end{aligned}$$

  We also have that $I_j \ne \emptyset$ for all $j$ (otherwise $I = \emptyset$).

• (3) To get a contradiction, assume that $0 \notin I$. Then:

  $$\begin{aligned} \exists p: \forall i \in \left\{ 1,\ldots , p-2 \right\} , g_i(0) \in I_{i+1} {\hbox { and }} g_{p-1}(0) \notin I_p \end{aligned}$$

Now since $I_p = [ 0 , \alpha _p ]$ (for some $\alpha _p \ge 0$), we have that:

$$\begin{aligned} g_{p-1}(a) \leqslant \alpha _p {\hbox { from (2), and }} g_{p-1}(0) > \alpha _p {\hbox { from (3) }}. \end{aligned}$$

Since $g_{p-1}$ is increasing (because all axes are oriented outside of $C$, see Propositions 1 and 2), we have:

$$\begin{aligned} \alpha _p < g_{p-1}(0) \leqslant g_{p-1}(a) \leqslant \alpha _p \end{aligned}$$

which is a contradiction ($\alpha _p < \alpha _p$). Therefore $0 \in I$ which ends the proof of statement (a). Statement (b) can be straightforwardly proved using the same reasoning. $\square$

Theorem 6.

Proof

$f$ is continuous according to Theorem 1. Assume that the axis where $f$ is calculated is oriented to the outside of $C$, and that $C$ is a transition cycle with no turn change. Then $f$ is strictly concave (Theorem 3). Assume moreover that $C$ contains no inside branching transition. Then we deduce from Lemma 5a that $I$ contains 0 and $f(0) \geqslant 0$.

Let $g(x)=f(x)-x$. $g$ is continuous, strictly concave and $g(0) \geqslant 0$. Therefore, by applying the intermediate value theorem on each interval where $g$ is strictly monotone, we show that $g$ has either no zero, a single zero, or two zeros one of which being $x=0$. Therefore $D^C$ contains either no limit cycle or a single limit cycle.

Moreover if $f$ admits a non-zero fixed point for $x=x_0,\,\frac{dg}{dx}(x_0)<0$. Therefore, it corresponds to a stable limit cycle, which ends the proof. Note that the case $f(0)=0$ corresponds to an equilibrium point of the system but also belongs to a switching domain. By Assumption 1, it cannot be a focal point, so it is a Filippov-type equilibrium. This case can only happen when $C$ consists of 4 consecutive perpendicular motifs with the same orientation (see transition cycle $C_1$ in Fig. 7 for an example). $\square$

Theorem 7.

Proof

Assume that the axis where $f$ is calculated is oriented to the outside of $C$, and that $C$ contains no outside branching transition. Then we deduce from Lemma 5b that $I$ contains $l$ and $f(l)\le l$. From Assumption 1, we further deduce that $f(l) < l$.

Assume the system admits an unstable limit cycle in $D^C$, that is $\exists x_0$: $f(x_0) = x_0$ and $\vert \frac{df}{dx}(x_0) \vert > 1$ (Strogatz 2001, p.281). Since $f$ is an increasing function, $\frac{df}{dx}(x_0)>1$. Therefore, $\exists x_1>x_0$: $f(x_1)>x_1$.

Assume either $C$ has no turn change or the decay rates are equal. Then, $f$ has a constant and strict concavity (Theorems 2 and 3). Moreover, $f$ is a continuous and increasing function (Theorem 1).

Let $g(x)=f(x)-x$. We have $g(l)<0,\,g(x_1)>0$ and $g$ is continuous, and has a constant and strict concavity. Therefore, by applying the intermediate value theorem, we show that $f$ admits a single fixed point $x=x_2$. Moreover, $\frac{dg}{dx}(x_2)<0$. Thus, $x=x_2$ corresponds to a stable limit cycle (Strogatz 2001, p.281) and the system admits a single stable limit cycle in $D^C$, which ends the proof. $\square$

Theorem 8.

Proof

Let ${\widehat{f}}$ be the 2-cycle first return map of $C^2$ calculated on the oriented segment $D_{C}, {\widehat{f}}_1$ and ${\widehat{f}}_2$ the first return maps of $C_1$ and $C_2$ respectively, and $I_1 =[ 0,a [$ and $I_2=] a,l]$ the intervals of definition of ${\widehat{f}}_1$ and ${\widehat{f}}_2$.

Let ${\widehat{g}}(x)={\widehat{f}}(x)-x$. Assume either $C_1$ and $C_2$ have no turn change, or the decay rates are equal. Thus, according to Theorems 2 and 3, ${\widehat{g}}$ has a constant and strict concavity in each interval $I_1$ and $I_2$. Assume also that $SG_{S \rightarrow T}$ is composed of 4 vertices. Thus ${\widehat{f}}$ and therefore ${\widehat{g}}$ is continuous (Theorem 4).

Assume moreover the system admits two stable limit cycles in $D^{C^2}$, that is there exists $x_1 <a$ and $x_2>a$ such that ${\widehat{g}}(x_1)={\widehat{g}}(x_2)=0$.

By applying the intermediate value theorem, we show that ${\widehat{g}}$ admits a unique fixed point $x_3 \in ]x_1,x_2 [$ which corresponds to an unstable limit cycle separating the basins of attraction of the two stable limit cycles, which ends the proof. $\square$

Appendix B: First return maps for the applications considered

See Figs. 12, 15

Fig. 15

First return maps for the discontinuous case. Positive fixed points of the maps (e.g. positive intersections between the maps $f$ (in blue) and line of equation $f(z)=z$ (in red)) correspond to limit cycles: a one large stable limit cycle; b one small stable limit cycle; c one large stable limit cycle, one unstable limit cycle and one stable equilibrium point at $z=0$; d one small and one large stable limit cycle, and one unstable limit cycle; e one small and one large stable limit cycles; f one stable equilibrium point; g one large stable limit cycle and one stable equilibrium point at $z=0$. For cases a–f, the parameter values are indicated in the caption of Fig. 14. For (g), the parameter values are the same as the other cases except: $\theta _{x} = 2,\,k_{4x} = 0$ (color figure online)

Fig. 16

Simulations of the continuous differential equations translated from the discontinuous example studied in Sect. 5.2. a Numerical simulations of the continuous model in the phase plane. The initial conditions of the trajectories (shown in blue) which converge to a stable point and a limit cycle are $(25.7,90)$ and $(25.7,100)$ respectively. The initial conditions of the trajectories simulated in reverse-time (shown in red) are: $(26.5,64.475),\,(26.5, 64.465),\,(27.1,62.4),\,(27.1,62.41)$. The trajectories simulated in reverse-time give an approximation of the separatrix curve which delimits the basins of the two attractors. Two additional unstable points appear: one at $(27.35,55)$ and the other at $(26.8,63.5)$. b Numerical simulation of the first return map from and to the half line of equation: $\left\{ x=25.7, y \ge 0 \right\}$ (black dashed line in Fig. a). The fixed point $x_1=97.8$ of the first return map corresponds to the stable limit cycle towards which the trajectory starting from $(25.7,100)$ converges (Figure (a)). The parameter values are: $n=10$ , $\theta _{x} = 25,\,\theta _{y}^1 = 50,\,\theta _{y}^2 = 70,\,\theta _{y}^3 = 90,\,k_{1x} = 5,\,k_{2x} = 50,\,k_{3x} = 55,\,k_{4x} = 30,\,k_{1y} = 190,\,d_x=1$ and $d_y=1$ (color figure online)

Appendix C: Translation to the continuous differential framework

The class of PWA biological model corresponding to the discontinuous case (Sect. 5.2) is translated into a continuous ordinary differential model by replacing step functions $s^+(x,\theta )$ and $s^-(x,\theta )$ by Hill function of the form $\frac{x^{n}}{x^{n} + \theta ^{n}}$ and $\frac{\theta ^{n}}{x^{n} + \theta ^{n}}$, respectively. The continuous differential system obtained is:

$$\begin{aligned} \left\{ \begin{array}{l} \dfrac{dx}{dt} = [ k_{1x} + k_{2x} \cdot \frac{y^{n}}{y^{n} + (\theta _{y}^1)^n} ] \cdot \frac{(\theta _{y}^2)^{n}}{y^{n} + (\theta _{y}^2)^{n}} \cdot \frac{x^{n}}{x^{n} + \theta _{x}^{n}} \\ \quad +\,\, \frac{y^{n}}{y^{n} + (\theta _{y}^3)^{n}} \cdot [ k_{3x} + k_{4x} \cdot \frac{x^{n}}{x^{n} + \theta _{x}^{n}} ] - d_x \cdot x \\ \dfrac{dy}{dt} = k_{1y} \cdot \frac{\theta _{x}^{n}}{x^{n} + \theta _{x}^n} - d_y \cdot y \end{array}\right. \end{aligned}$$

(8)

Simulations of this model for parameter setting corresponding to the situation where the discontinuous case admits a large stable limit cycle and an equilibrium point (case (g), Fig. 15) and the computation of a first return map are presented in Fig. 16. Interestingly, the simulations show an abrupt change of the derivative of the first return map (at $x=91.9$, Fig. 16b) which corresponds to the discontinuity observed in the corresponding PWA model (Fig. 15).