Polytope Lyapunov Functions for Stable and for Stabilizable LSS

Abstract

We present a new approach for constructing polytope Lyapunov functions for continuous-time linear switching systems (LSS). This allows us to decide the stability of LSS and to compute the Lyapunov exponent with a good precision in relatively high dimensions. The same technique is also extended for stabilizability of positive systems by evaluating a polytope concave Lyapunov function (“antinorm”) in the cone. The method is based on a suitable discretization of the underlying continuous system and provides both a lower and an upper bound for the Lyapunov exponent. The absolute error in the Lyapunov exponent computation is estimated from above and proved to be linear in the dwell time. The practical efficiency of the new method is demonstrated in several examples and in the list of numerical experiments with randomly generated matrices of dimensions up to 10 (for general linear systems) and up to 100 (for positive systems). The development of the method is based on several theoretical results proved in the paper: the existence of monotone invariant norms and antinorms for positively irreducible systems, the equivalence of all contractive norms for stable systems and the linear convergence theorem.

Appendix

Proof of Theorem 3

It is split into four steps. First we construct a special positively homogeneous monotone convex functional $\varphi$ and prove (step 1) that it is actually a norm. Then, in step 2, we establish the extremality property of $\varphi$, i.e., that it is non-increasing on each trajectory. Thus, we have an extremal norm. In steps 3 and 4, we show the existence of a generalized trajectory on the unit sphere starting at an arbitrary point. This is done by considering a special convex optimal control problem and applying the Banach–Alaoglu compactness theorem.

In view of (2), it suffices to consider the case $\sigma ({\mathcal {A}}) = 0$. We take an arbitrary norm monotone with respect to K, for example, $\Vert x\Vert = (b,x)$, where $b \in \mathrm{int}\, K^*$, where $K^*$ is a dual cone (15). For $t \ge 0$ and $z \in K$, denote $l(z, t) \, = \, \sup \, \{\Vert x(t)\Vert \, , \ A \in {\mathcal {U}}\, [0,t]\, , \, x(0) = z\}$. For every fixed t, the function $l(\cdot , t)$ is a seminorm on K, i.e., it is positively homogeneous and convex, as a supremum of homogeneous convex functions. Moreover, Corollary 4 implies that l is non-decreasing in z, i.e., if $z_1 \ge _K z_2$, then $l(z_1, t) \ge l(z_2, t)$ for each t. The function $\varphi (x) = \sup \limits _{t \in {\mathbb {R}}_+} l(x, t)$ is, therefore, also a monotone seminorm on K as the supremum of monotone seminorms. Moreover, $\varphi (x) \ge l(x,0) = \Vert x\Vert$; hence, $\varphi$ is positive.

Step 1. Let us show that $\varphi (x) < \infty$ for all x, i.e., $\varphi$ is a norm on K. Denote by ${\mathcal {L}}$ the set of points $x \in K$ such that $\varphi (x) < \infty$. Since $\varphi$ is convex, homogeneous and monotone on K, it follows that either ${\mathcal {L}}= K$ or ${\mathcal {L}}$ is a face of K. Writing $\tilde{{\mathcal {L}}}$ for the linear span of ${\mathcal {L}}$, we are going to show that $A \tilde{{\mathcal {L}}} \subset \tilde{{\mathcal {L}}}$ for each $A \in {\mathcal {A}}$. If this is not the case, then $A_0z \notin \tilde{{\mathcal {L}}}$ for some $z \in {\mathcal {L}}$ and $A_0 \in {\mathcal {A}}$, hence $(I + tA_0)z \notin \tilde{{\mathcal {L}}}$ for all $t \in (0, \tau ]$, and therefore $e^{\, t\, A_0} z \notin \tilde{{\mathcal {L}}}$ for all $t \in (0, \tau ]$, where $\tau > 0$ is small enough. Hence, for the control function $A(t) \equiv A_0, \, \ t \in [0, \tau ]$ and for $x(0) = z$ we have $x(\tau ) \notin \tilde{{\mathcal {L}}}$, and consequently, $\varphi (z) \ge \varphi (x(\tau )) = +\infty$, which is a contradiction. Thus, unless ${\mathcal {L}}= K$, the set $\tilde{{\mathcal {L}}}$ is a face plane invariant for all operators from ${\mathcal {A}}$. From the K-irreducibility, it follows that either ${\mathcal {L}}= K$ (in which case the proof is completed) or ${\mathcal {L}}= \{0\}$. It remains to show that the latter is impossible.

If ${\mathcal {L}}= \{0\}$, then $\varphi (z) = +\infty$ for all $z \in K \setminus \{0\}$. Let $K_1 = \{z \in K, \, \Vert z\Vert = 1\}$. For every natural n, denote by ${\mathcal {H}}_{\, n}$ the set of points $z \in K_1$, for which there exists a number $\tau = \tau (z) \le n$ and a trajectory starting at z such that $\Vert x(\tau )\Vert > 2$. Since ${\mathcal {H}}_{\, n}$ is open and $\cup _{n =1}^{\infty }{\mathcal {H}}_{\, n} \, = \, K_1$, from the compactness we conclude that $\cup _{n = 1}^{N}{\mathcal {U}}_{\, n} \, = \, K_1$ for some natural N. Thus, $\tau (z) \le N$ for all $z \in K_1$. Whence, starting from an arbitrary point $x_0 \in K_1$ one can consequently build a trajectory x(t) and sequences $\{x_n\}, \{t_n\}$ an $\{\tau _n\}$ such that $t_0 = 0, t_n = \sum _{k=0}^{n-1}\tau (x_k)$, $x_k = x(t_k)$. For this trajectory, $\Vert x(t_n)\Vert \, > \, 2^n$ and $t_n \le nN$, hence $\Vert x(t_n)\Vert \, > \, e^{t_n\ln 2 / N}$. Therefore, $\sigma ({\mathcal {A}}) \ge \frac{\ln 2}{N} >0$, which contradicts the assumption. Thus, the case ${\mathcal {L}}= \{0\}$ is impossible; hence, $\varphi$ is a norm.

Step 2. By definition, for any trajectory x(t) the function $\varphi (x(t))$ is non-increasing in t. Indeed, suppose $t_1 < t_2$; then $\varphi (x(t_1))$ is the supremum of $\Vert y(t)\Vert \, , \, t \in [t_1, +\infty )$ over all possible trajectories $y(\cdot )$ on the half-line $[t_1, +\infty )$ with the initial condition $y(t_1) = x(t_1)$. This set of trajectories includes $x(\cdot )$. Hence this supremum is not smaller than $\varphi (t_2)$, which is the supremum over a narrower set of trajectories $y(\cdot )$ on the half-line $[t_2, +\infty )$ with the initial condition $y(t_2) = x(t_2)$.

Step 3. Thus, we have found a norm $\Vert x\Vert = \varphi (x)$ which is non-increasing in t on every trajectory x(t). In this norm, the function l(z, t) is non-increasing in t for each $z \in K$. Hence, the limit $F(z) = \lim \limits _{t \rightarrow +\infty }l(z, t)$ exists for every $z \in K$. Let us show that F is a norm we are looking for. First of all, this is a monotone seminorm on K as a limit of monotone seminorms. Second, F(x(t)) is non-increasing in t on every trajectory x(t). Finally, by definition of l(z, t), we have

$$\begin{aligned} \sup _{A(\cdot ) \in {\mathcal {U}}\, [0,\tau ]\, , \, x_0 \, = \, z }\ F(x(\tau )) \quad = \quad F(z)\ , \qquad t > 0\, . \end{aligned}$$

(27)

For every $\tau > 0$, we denote by ${\mathcal {Q}}_{\, \tau }$ the set of control functions $A(\cdot ) \in {\mathcal {U}}\, [0, \tau ]$, for which the supremum in the left-hand side of (27) is attained. To show that ${\mathcal {Q}}_{\tau }$ is non-empty, we consider, for an arbitrary $a \in K^*$, the following optimal control problem:

$$\begin{aligned} \left\{ \begin{array}{l} \bigl (a , x(\tau ) \bigr )\ \rightarrow \ \max \\ \dot{x} \, = \, A\, x\\ x(0)\, = \, z\\ A(t) \in \mathrm{co}\, ({\mathcal {A}})\, , \ t \in [0,\tau ] \end{array}\right. \end{aligned}$$

(28)

Since this problem is linear in the control function $A(\cdot )$, the set $\mathrm{co}\, ({\mathcal {A}})$ is convex and compact, and the objective function $\bigl (a , x(\tau ) \bigr )$ is linear, it possesses the optimal solution $(\bar{A}, \bar{x}) \in L_1[0, \tau ] \times W^1_1 [0, \tau ]$ (see, e.g., [18, 29]). Now we take a maximizing sequence $\{x_i(\cdot )\}_{i = 1}^{\infty }$, for which $F(x_i(\tau )) \rightarrow F(z)$ in (27) as $i \rightarrow \infty$. By the compactness, without loss of generality it can be assumed that the sequence $x_i(z)$ converges to some point $y \in K$ as $i \rightarrow \infty$. Taking $a \in \partial \, F(y)$ (the subdifferential of F at the point y), and solving problem (28) for that a, we obtain $\bigl (a , \bar{x}(\tau ) \bigr ) = F(z)$, and hence $F(\bar{x}(\tau )) = F(z)$. Thus, ${\mathcal {Q}}_{\, \tau }\, \ne \, \emptyset$ for each $\tau > 0$. Note that this set is compact in the weak-* topology of the space $L_1[0, \tau ]$ due to Banach–Alaoglu theorem. Furthermore, the family $\{{\mathcal {Q}}_{\, \tau }\}_{\tau > 0}$ is embedded: ${\mathcal {Q}}_{\, \tau _2} \subset {\mathcal {Q}}_{\, \tau _1}$ if $\tau _2 > \tau _1$. Indeed, if $\bar{x} \in {\mathcal {Q}}_{\, \tau _2}$, then $F(\bar{x}(\tau _2)) = F(z)$, and hence $F(\bar{x}(t))$ equals identically to F(z) on the segment $[0, \tau _2]$. Therefore, it equals identically to F(z) on a smaller segment $[0, \tau _1]$, and so $\bar{x} \in {\mathcal {Q}}_{\, \tau _1}$. Since an embedded system of non-empty compact sets has a non-empty intersection, it follows that there exists a control function $\bar{A}$, whose trajectory $\bar{x}$ with $\bar{x} (0) = z$ possesses the property $F(\bar{x} (t)) \equiv F(\bar{x} (0))\, , \ t \in [0, +\infty )$.

Step 4. Thus, we have proved that the seminorm F is invariant: It is non-increasing in t on any trajectory x(t), and for every starting point, there is a trajectory, on which F is identically constant. It remains to show that F is a norm, i.e., F(z) is finite and positive for every $z \in K\setminus \{0\}$. Since F(z) is defined as a limit of a non-increasing function as $t \rightarrow +\infty$, we have $F(z) < \infty$. The positivity is proved by contradiction. Let ${\mathcal {M}}= \{z \in K, \ F(z) = 0\}$. Since F is a seminorm, it follows that ${\mathcal {M}}$ is either entire K or a face of K. If this is a face of K, then as in Step 1 we conclude that its linear span is a common invariant face plane for ${\mathcal {A}}$, which by the irreducibility implies ${\mathcal {M}}= \{0\}$, and the proof is completed. If ${\mathcal {M}}= K$, then $l(z, t) \rightarrow 0$ as $t \rightarrow \infty$ for every $z \in K$. Take an arbitrary $x_0 \in \mathrm{int}\, K$. There exists a constant $c > 0$ such that for every $x \in K$ inequality $\Vert x\Vert \le c$ implies $x \, \le _K \, \frac{1}{2}\, x_0$. Let n be such that $l(x_0, n) < c$. Hence, $x(n) \, \le _K \, \frac{1}{2}\, x(0)$ for every trajectory x(t) with $x(0) = x_0$. Applying now Corollary 4 and iterating k times, we get $x(kn) \, \le _K \, 2^{-k}\, x_0$. Since the norm is monotone, it follows that $\Vert x(kn)\Vert \, \le \, 2^{-k}\Vert x_0\Vert \, , \, k \in {\mathbb {N}}$. On the other hand, since the norm is non-increasing in t on every trajectory, we see that $\Vert x(t)\Vert \, \le \, 2^{- \bigl [\frac{t}{n} \bigr ]}\Vert x_0\Vert$, where the brackets denote the integer part. Since for every $y_0 \in K$ there is a constant C such that $y_0 \le _K Cx_0$, it follows that for every trajectory y(t) one has $\Vert y(t)\Vert \, \le \, C 2^{- \bigl [\frac{t}{n} \bigr ]}\Vert y_0\Vert$. Therefore, $\sigma ({\mathcal {A}}) \le - \frac{\ln 2}{n} < 0$. The contradiction concludes the proof. $\square$

Proof of Proposition 6

Replacing the family ${\mathcal {A}}$ by ${\mathcal {A}}+ hI$, where $h>0$ is large enough, it may be assumed that $A \ge _K I$ for all $A \in {\mathcal {A}}$. In this case the family ${\mathcal {B}}\, = \, \{B = A - I \ | \ A \in {\mathcal {A}}\}$ consists of K-irreducible operators that leave K invariant. Take arbitrary vectors $x \in K \, , \, x^*\in K^* \, , \ \Vert x\Vert = \Vert x^*\Vert = 1$. For every $B \in {\mathcal {B}}$, we denote $p_B(x^*, x)\, = \, \max \limits _{n=0, \ldots , d-1}(x^*,B^nx)$. If $p_B(x^*, x) = 0$, then x is contained in a face of K invariant with respect to B. This contradicts irreducibility of B. Thus, the function p is strictly positive. Hence, by the compactness, there is $a > 0$ such that $p_B(x^*, x) \ge a$ for all $B \in {\mathcal {B}}$ and all $x \in K , x^*\in K^* \, , \, \Vert x\Vert = \Vert x^*\Vert = 1$. Assume for the moment that the family ${\mathcal {A}}$ is finite: ${\mathcal {A}}= \{A_1, \ldots , A_m\}$ and $A_k = I +B_k$. Every product $P = \prod _{k=1}^{md}A_k$ of md operators contains at least d equal terms, say, $A_i$. Then $P = \prod _{k=1}^{md}(I+B_k) \ge \sum _{n=0}^{d-1}B_i^n$, and therefore, $(x^*, Px) \, \ge \, p_{B_i}(x^*, x)\, \ge \, a$. In case of general compact set ${\mathcal {A}}$, we take its $\varepsilon$-net ${\mathcal {A}}_{\varepsilon } = \{A_i\}_{i=1}^{m(\varepsilon )}$, and to every $A \in {\mathcal {A}}$, we associate the closets element from ${\mathcal {A}}_{\, \varepsilon }$ (if there are several closest elements, we take any of them). There is a function $c(\varepsilon )$ such that $c(\varepsilon ) \rightarrow 0, \varepsilon \rightarrow 0$, and for every product $\varPi$ of length at most d of operators from ${\mathcal {A}}$, we have $\Vert \varPi - \varPi '\Vert \le c(\varepsilon )$, where $\varPi '$ is the corresponding product of operators from ${\mathcal {A}}_{\varepsilon }$. Hence, for every product P of length $md(\varepsilon )$, we have $(x^*, Px) \, \ge \, a - c(\varepsilon )$. Taking $\varepsilon$ small enough, so that $c(\varepsilon ) < a/2$, we see that $(x^*, Px) \, \ge \, a/2$, for every product P of length $d m(\varepsilon )$ of operators from ${\mathcal {A}}$. Since this holds for all $x^* \in K^*\, , \, \Vert x^*\Vert = 1$, it follows that $Px \in \mathrm{int}(K)$ for every $x \in K\, , \, \Vert x\Vert = 1$. Consequently, the cone PK is embedded in K. Let $K'$ be the convex hull of all cones PK taken over all products $P \in {\mathcal {A}}^{\, d m(\varepsilon )}$. The cone $K'$ is embedded in K and $AK' \subset K'\, , \ A \in {\mathcal {A}}$; hence, A is $K'$-Metzler, which completes the proof. $\square$

Proof of Theorem 5

In view of (2), it suffices to consider the case $\check{\sigma }({\mathcal {A}}) = 0$. Take a positive antinorm on K, for example, $g(x) = (b,x)$, where $b \in \mathrm{int}\, K^*$. For $t \ge 0$ and $z \in K$, denote $r(z,t)=\inf \,\{g(x(t)),\,A \in {\mathcal {U}}\, [0,t],\,x(0)=z\}$.

Step 1. For every fixed t, the function $r(\cdot , t)$ is an antinorm on K. Corollary 4 implies that r is non-decreasing in z. The function $\psi (x) = \inf \limits _{t \in {\mathbb {R}}_+} r(x, t)$ is, therefore, also an antinorm on K as the infimum of antinorms. This antinorm is non-decreasing in t for every trajectory x(t) (the proof is the same as in Theorem 3). It remains to show that $\psi$ is not identically zero. For an arbitrary $x_0 \in \mathrm{int}\, K$, there is a constant $c > 0$ such that inequality $g(x) \le c$ implies $x \, \le _K \, \frac{1}{2}\, x_0$. If $\psi (x_0) = 0$, then there is $n > 0$ such that $r(x_0, n) < c$. Hence, there is a control function $\bar{A} (\cdot )$ on the segment [0, n] and the corresponding trajectory $\bar{x}$ with $\bar{x}(0) = x_0$ and $\bar{x}(n) \, \le _K \, \frac{1}{2}\, \bar{x}(0)$. If now $A(\cdot )$ is the periodic extension of the control function $\bar{A} (\cdot )$ to ${\mathbb {R}}_+$ with period n, then the corresponding trajectory $x(\cdot )$ satisfies $x(kn) \, \le _K \, 2^{-k}\, \bar{x}(0)$. Since the antinorm g is monotone on K (Lemma 8), we have $g(x(kn)) \le 2^{-k} g(x_0)\, , \, k \in {\mathbb {N}}$. Moreover, g is positive, and hence it is equivalent to any norm on K. Therefore, $\Vert x(kn)\Vert \le C\, 2^{-k}\, , \, k \in {\mathbb {N}}$, and hence $\check{\sigma }\le - \frac{\ln 2}{n}$, which contradicts the assumption. Thus, $\psi$ is an extremal antinorm, which concludes the proof of the first part.

Step 2. Now let us show that if all operators of ${\mathcal {A}}$ are K-irreducible, then there is a positive invariant antinorm. Take the extremal antinorm $\psi$ on K constructed in the previous step and consider the function $f(z) = \lim \limits _{t \rightarrow +\infty }r(z, t)$, where we now denote $r(z, t) \, = \, \inf \, \{\psi (x(t))\, , \ A \in {\mathcal {U}}\, [0, t]\, , \, x(0) = z\}$. Since $\psi (x(t))\,$ is non-decreasing in t, so is $\, r(z, t)$. Hence, this limit exists for every $z \in K$, maybe it becomes $+\infty$. If $f(x) = +\infty$ for some $x \in K$, then f is equal to $+\infty$ in the whole interior of K, since for every $z \in \mathrm{int}(K)$ there is a constant c such that $cz >_K x$.

Now we invoke Corollary 6: There exists a cone $K'$ embedded in K such that every trajectory x(t) starting in $K'$ remains in $K'$. We see that if $f(x) = +\infty$ for some $x \in K$, then $f(x_0) = +\infty$ for every point $x_0 \in K'\setminus \{0\}$. Take an arbitrary $x_0 \in K'\setminus \{0\}$. As we saw, $f(x) = +\infty$ for some $x \in K$, then $f(x_0) = +\infty$. Since $\psi$ is positive and continuous on $K'$, there exists a constant $C > 0$ such that for every $x \in K'$ inequality $\psi (x) \ge C$ implies $x \, \ge _K \, 2\, x_0$. If $f(x_0) = +\infty$, then there is $q > 0$ such that $r(x_0, q) > C$. Hence, $x(q) \, \ge _K \, 2\, x(0)$ for every trajectory x(t) with $x(0) = x_0$. Applying Corollary 4 and iterating k times, we get $x(kq) \, \ge _K \, 2^{\, k}\, x_0$. Since $\psi$ is monotone, it follows that $\psi (x(kq)) \, \ge \, 2^{\, k}\psi (x_0)$. On the other hand, $\psi$ is non-decreasing in t on every trajectory, consequently $\psi (x(t)) \, \ge \, 2^{\bigl [\frac{t}{q} \bigr ]}\psi (x_0)$. Since $\psi$ is equivalent to any norm on $K'$, it follows that $\check{\sigma }({\mathcal {A}}) \ge \frac{\ln 2}{n} > 0$. The contradiction shows that $f(x)< +\infty$ for every $x \in K$, i.e., f is an antinorm on K.

Consider now the optimization problem (28), where the maximum is replaced by minimum. It always has a solution $(\bar{x}, \bar{A})\, \in \, W_1^1\times L_1[0, \tau ]$ for which $f(\bar{x} (\tau )) = F(z)$. Therefore, the set $P_{\, \tau }$ of control functions $A \in {\mathcal {U}}\, [0, \tau ]$ for which f(x(t)) is equal identically to f(z) on the segment $[0, \tau ]$ is non-empty. Since the sets $\{P_{\, \tau }\}_{\, \tau \in {\mathbb {R}}_+}$ form an embedded system of non-empty compact sets, they have a common point $\bar{A} (\cdot ) \in {\mathcal {U}}\, [0, +\infty )$, for which $f(\bar{x}(t))=f(z), \,t\in [0, +\infty )$. Whence, f is an invariant antinorm. $\square$