A Poincaré–Bendixson theorem for hybrid dynamical systems on directed graphs

Abstract

The purpose of this work is to obtain restrictions on the asymptotic structure of two-dimensional hybrid dynamical systems. Previous results have been achieved by the authors concerning hybrid dynamical systems with a single impact surface and a single state space. Here, this work is extended to hybrid dynamical systems defined on a directed graph; each vertex corresponds to a state space and each directed edge corresponds to an impact.

Appendices

Appendix

A Technical lemmas

Before we state the following lemmas, we need to define the discrete $\omega$-limit set. Given a discrete dynamical system, $T:S\rightarrow S$, the discrete $\omega _d$-limit set is defined as

$$\begin{aligned} \omega _d(x) = \left\{ y\in S : \exists N_n\rightarrow \infty ~s.t.~ \lim _{n\rightarrow \infty } \, T^{N_n}(x)=y\right\} . \end{aligned}$$

Lemma 1

(See, Lemma 4.2 in [4]) Let $P:[a,b]\rightarrow [a,b]$ be $C^1$ and injective. Then for all $x\in [a,b]$, $\omega _d(x)$ is either a single point or two points. In either case, all trajectories approach a periodic orbit.

Proof

By being injective, the map P is automatically monotone. Let $I:=[a,b]$ and define the fixed-point set $F:=\{x\in I : P(x)=x\}$ and consider $x\in I$. If $x\in F$ we are done, so assume $x\not \in F$.

First, assume that P is (not necessarily strictly) increasing. Then, since $P(x)>x$ implies $P^2(x)>P(x)$ and $P(x)<x$ implies $P^2(x)<P(x)$, it follows (since $x\not \in F$) that $\{P^n(x)\}$ is a monotone and bounded sequence and therefore converges. Hence, $\omega _d(x)$ is a singleton.

To complete the proof, consider the case where P is (not necessarily strictly) decreasing. Since $\omega _d(x;P)=\omega _d(x;P^2)\cup \omega _d(P(x);P^2)$, the previous paragraph shows that $\omega _d(x)$ is either two points or a singleton. $\square$

Lemma 2

(See, Lemma 6.1 in [4]) Let $S\subset {\mathbb {R}}$ be a finite set and $P:S\rightarrow S$. Then, for all $x\in S$, there exists $N\ne M$ large enough such that $P^N(x)=P^M(x)$ where

$$\begin{aligned} P^N(x) = \underbrace{P\circ P\circ \ldots \circ P}_{N\mathrm {~times}}(x). \end{aligned}$$

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In particular, $\omega _d(x)$ is a periodic orbit.

Proof

This follows immediately from the Pigeonhole principle. We have a function $f_x:{\mathbb {Z}}^+\rightarrow S$, $m\mapsto P^m(x)$, which cannot be injective. Therefore, there exists N and M such that $f_x(N)=f_x(M)$. $\square$

Lemma 3

The set $S_{(1,2)}^\infty$ is either a point or an interval.

Proof

We start by defining the sets $S_{(1,2)}^m$. Let P be the return map, $P:U\subset S_{(1,2)} \rightarrow S_{(1,2)}$. Then, $S_{(1,2)}^1 := \{x\in S_{(1,2)} : P(x)\in S_{(1,2)}\}$, that is, points in $S_{(1,2)}$ that return to $S_{(1,2)}$ at least once. Similarly, $S_{(1,2)}^m$ are points of $S_{(1,2)}$ that return to $S_{(1,2)}$ at least m times. This allows us to express $S_{(1,2)}^\infty$ as

$$\begin{aligned} S_{(1,2)}^\infty = \bigcap _{m=1}^\infty \,S_{(1,2)}^m. \end{aligned}$$

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If we can show that each $S_{(1,2)}^m$ is an interval, then the desired result follows due to nesting. The base case is satisfied by (Q.4) via $S_{(1,2)}^0 := S_{(1,2)}$. We will continue by induction. For each m, we can iterate by

$$\begin{aligned} S_{(1,2)}^{m+1} = S_{(1,2)}^m \cap P^{-1}\left( S_{(1,2)}^m \right) . \end{aligned}$$

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Therefore, if $S_{(1,2)}^m$ and $P^{-1}\left( S_{(1,2)}^m \right)$ are both intervals, so is $S_{(1,2)}^{m+1}$. All that is left to prove is $P^{-1}\left( S_{(1,2)}^m \right)$ is an interval. Before we can study the structure of $P^{-1}\left( S_{(1,2)}^m \right)$, we first define the family of functions

$$\begin{aligned} \mu _i : {\mathscr {D}}_i\subset {\mathscr {X}}_i\rightarrow S_{(i,i+1)}. \end{aligned}$$

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where ${\mathscr {D}}_i = \{ x\in {\mathscr {X}}_i : \exists t>0 \text { with } \varphi _t^i(x)\in S_{(i,i+1)}\}$ and $\mu _i(x) = \varphi _t^i(x)\in S_{(i,i+1)}$. These functions can be thought of as “projecting” onto the sets $S_{(i,i+1)}$. What remains to show is that $\varDelta _n(S_{(n,1)})\cap {\mathscr {D}}_1$ is an interval. If we can show this, we are done because $\varDelta _{(n,1)}^{-1}\left( \varDelta _{(n,1)}(S_{(n,1)})\cap {\mathscr {D}}_1\right)$ is also an interval, and we can march backward around the cycle and end up on with a subinterval of $S_{(1,2)}$.

Since $S_{(n,1)}$ is diffeomorphic to an interval, so is the set $\varDelta _{(n,1)}(S_{(n,1)})$. Let the map $h:\varDelta _{(n,1)}(S_{(n,1)})\rightarrow [a,b]$ be a diffeomorphism. Define the points ${\tilde{a}}$ and ${\tilde{b}}$ as

$$\begin{aligned} \begin{aligned} {\tilde{b}}&= \max \{ x\in [a,b] : o^+_c(h^{-1}(x))\cap S_{(1,2)} \ne \emptyset \}, \\ {\tilde{a}}&= \min \{ x\in [a,b] : o^+_c(h^{-1}(x))\cap S_{(1,2)} \ne \emptyset \}. \end{aligned} \end{aligned}$$

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We claim that $\varDelta _{(n,1)}(S_{(n,1)})\cap {\mathscr {D}}_1 = h^{-1}[{\tilde{a}},{\tilde{b}}]$. By the choice of ${\tilde{a}}$ and ${\tilde{b}}$, we know that $\varDelta _{(n,1)}(S_{(n,1)})\cap {\mathscr {D}}_1\subset h^{-1}[{\tilde{a}},{\tilde{b}}]$. To show the other direction, we consider the region bounded by the four curves: $h^{-1}[{\tilde{a}},{\tilde{b}}]$, ${\tilde{o}}^+_c(h^{-1}({\tilde{a}}))$, ${\tilde{o}}^+_c(h^{-1}({\tilde{b}}))$, and $S_{(1,2)}$ (where ${\tilde{o}}^+(x)$ is the forward orbit of x until it impacts $S_{(1,2)}$). By assumption (Q.5) and uniqueness of solutions, any curve starting on $h^{-1}[{\tilde{a}},{\tilde{b}}]$ must intersect $S_{(1,2)}$ before leaving the region described above (see, Lemma 10 in [4]). Therefore, $h^{-1}[{\tilde{a}},{\tilde{b}}]\subset \varDelta _n(S_n)\cap {\mathscr {D}}_1$. $\square$

Lemma 4

Let $({\mathscr {N}},{\mathscr {E}},{\mathscr {X}},S,\varDelta ,f)$ be a GHDS where $({\mathscr {N}},{\mathscr {E}})$ is a cycle. Let $x_0\in S_{(i-1,i)}{\setminus } \text {fix}(f_{i-1})$ be such that there exists a time, $T_0>0$, where $\varphi ^i_{T_0}(\varDelta _{(i-1,i)}(x_0))\in S_{(i,i+1)}$ ($\varphi _t^i$ is the continuous flow corresponding to ${\dot{x}}=f_i(x)$ in vertex i). Additionally, assume that the flow intersects the surface, $S_{(i,i+1)}$, transversely. Then there exists an $\varepsilon > 0$ and a $C^1$ function $\tau :{\mathscr {B}}_{\varepsilon }(x_0)\cap S_{(i-1,i)} \rightarrow {\mathbb {R}}^+$ such that for all $y\in {\mathscr {B}}_{\varepsilon }(x_0)\cap S_{(i-1,i)}$, $\varphi _{\tau (y)}^i(\varDelta _{(i-1,i)}(y))\in S_{(i,i+1)}$.

Proof

Define the function

$$\begin{aligned} F_i:(0,+\infty )\times S_{(i-1,i)} \rightarrow {\mathscr {X}}_i, \end{aligned}$$

by $F_i(t,x) = H_{(i,i+1)}(\varphi _t^i(\varDelta _{(i-1,i)}(x)))$. It follows from Theorem 1 in Section 2.5 in [12] that $F_i\in C^1({\mathbb {R}}^+\times S_{(i-1,i)})$. This allows the use of the implicit function theorem. By the assumptions of the lemma, we know that $F_i(T_0,x_0)=0$. Differentiating $F_i$ with respect to time yields

$$\begin{aligned} \frac{\partial F_i}{\partial t}(T_0,x_0) = \left. \frac{\partial H_{(i,i+1)}}{\partial y}\right| _{y=\varphi _{T_0}^i(\varDelta _{(i-1,i)}(x_0))\in S_{(i,i+1)}} \cdot f_i\left( \varphi _{T_0}^i(\varDelta _{(i,i+1)}(x_0))\right) \ne 0.\nonumber \\ \end{aligned}$$

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The first factor is nonzero because of assumption (G.4) and the second is nonzero because we are away from fixed points of the continuous flow. Their inner product is nonzero because of the transversality condition. This allows the use of the implicit function theorem (see, Theorem 9.28 in [14]) to show that there exists a neighborhood of $x_0$ and a $C^1$ function $\tau$ with the desired properties. $\square$

Note that this implies that the maps $P_i:U_i\subset S_{(i-1,i)}\rightarrow S_{(i,i+1)}$ are all $C^1$. Because the composition of $C^1$ maps is still $C^1$, the map $P:= P_1\circ \ldots \circ P_n : U\subset S_{(1,2)} \rightarrow S_{(1,2)}$ is $C^1$.