Topological entropy of switched linear systems: general matrices and matrices with commutation relations

Abstract

This paper studies a notion of topological entropy for switched systems, formulated in terms of the minimal number of trajectories needed to approximate all trajectories with a finite precision. For general switched linear systems, we prove that the topological entropy is independent of the set of initial states. We construct an upper bound for the topological entropy in terms of an average of the measures of system matrices of individual modes, weighted by their corresponding active times, and a lower bound in terms of an active-time-weighted average of their traces. For switched linear systems with scalar-valued state and those with pairwise commuting matrices, we establish formulae for the topological entropy in terms of active-time-weighted averages of the eigenvalues of system matrices of individual modes. For the more general case with simultaneously triangularizable matrices, we construct upper bounds for the topological entropy that only depend on the eigenvalues, their order in a simultaneous triangularization, and the active times. In each case above, we also establish upper bounds that are more conservative but require less information on the system matrices or on the switching, with their relations illustrated by numerical examples. Stability conditions inspired by the upper bounds for the topological entropy are presented as well.

Appendices

A Proof of Lemma 1

As a direct consequence of the definition (12), the maximal weighted average $\bar{a}$ satisfies

$$\begin{aligned} \bar{a}(T) \ge \max \Bigg \{ \dfrac{1}{T} \sum _{p \in \mathcal P} a_p \tau _p(T),\, 0 \Bigg \} = \max \Bigg \{ \sum _{p \in \mathcal P} a_p \rho _p(T),\, 0 \Bigg \} \qquad \forall T > 0, \end{aligned}$$

and thus $\limsup _{T \rightarrow \infty }\, \bar{a}(T) \ge \max \{\hat{a},\, 0\}$.

It remains to prove that the reverse inequality holds as well. The definition (11) of the asymptotic weighted average $\hat{a}$ implies that for each $\delta > 0$, there is a large enough $T_\delta ' \ge 0$ such that $\sum _{p \in \mathcal P} a_p \rho _p(t) < \hat{a} + \delta$ for all $t > T_\delta '$. For a $T > T_\delta '$, let

$$\begin{aligned} t^*(T) := \mathrm {arg\,max}_{t \in [0, T]} \sum _{p \in \mathcal P} a_p \tau _p(t). \end{aligned}$$

Then, $\sum _{p \in \mathcal P} a_p \tau _p(t^*(T)) \ge 0$. If $t^*(T) \in (T_\delta ', T]$, then

$$\begin{aligned} \begin{aligned} \bar{a}(T)&= \dfrac{1}{T} \sum _{p \in \mathcal P} a_p \tau _p(t^*(T)) \le \dfrac{1}{t^*(T)} \sum _{p \in \mathcal P} a_p \tau _p(t^*(T)) \\&= \sum _{p \in \mathcal P} a_p \rho _p(t^*(T)) < \hat{a} + \delta . \end{aligned} \end{aligned}$$

Otherwise $t^*(T) \in [0, T_\delta ']$, and thus

$$\begin{aligned} \bar{a}(T) = \dfrac{1}{T} \sum _{p \in \mathcal P} a_p \tau _p(t^*(T)) \le \dfrac{a_m t^*(T)}{T} \le \dfrac{\max \{a_m, 0\}\, T_\delta '}{T} \end{aligned}$$

with the constant $a_m := \max _{p \in \mathcal P} a_p$. Combining the two cases above yields

$$\begin{aligned} \bar{a}(T) \le \max \{\hat{a} + \delta ,\, \max \{a_m, 0\}\, T_\delta '/T\} \qquad \forall T > T_\delta '. \end{aligned}$$

Hence

$$\begin{aligned} \bar{a}(T) \le \max \{\hat{a},\, 0\} + \delta \qquad \forall T > T_\delta := \max \{T_\delta ',\, \max \{a_m, 0\}\, T_\delta '/\delta \}. \end{aligned}$$

As $\delta > 0$ is arbitrary, we have $\limsup _{T \rightarrow \infty }\, \bar{a}(T) \le \max \{\hat{a},\, 0\}$.

B Proof of Lemma 2

• 1. As $R(\hat{x}) \subset B_{A_\sigma }(\hat{x}, \varepsilon , T)$ for all $\hat{x} \in G(\theta )$, we have

  $$\begin{aligned} K = \bigcup _{\hat{x} \in G(\theta )} R(\hat{x}) \subset \bigcup _{\hat{x} \in G(\theta )} B_{A_\sigma }(\hat{x}, \varepsilon , T). \end{aligned}$$

  Then, (4) implies that the grid $G(\theta )$ is $(T, \varepsilon )$-spanning, and thus

  $$\begin{aligned} \log S(A_\sigma , \varepsilon , T, K) \le \log |G(\theta )| = \sum _{i=1}^{n} \log (2 \lfloor 1/\theta _i \rfloor + 1) \le \sum _{i=1}^{n} \log (2/\theta _i + 1). \end{aligned}$$

  Consequently, the definition of entropy (5) implies

  $$\begin{aligned} h(A_\sigma )\le & {} \lim _{\varepsilon \searrow 0} \limsup _{T \rightarrow \infty } \sum _{i=1}^{n} \dfrac{\log (2/\theta _i + 1)}{T} \\= & {} \lim _{\varepsilon \searrow 0} \limsup _{T \rightarrow \infty } \sum _{i=1}^{n} \dfrac{\log (1/\theta _i)}{T}+ \lim _{\varepsilon \searrow 0} \limsup _{T \rightarrow \infty } \sum _{i=1}^{n} \dfrac{\log (2 + \theta _i)}{T}, \end{aligned}$$

  where the last term equals 0 if (17) holds.

• 2. For all distinct points $\hat{x}, \hat{x}' \in G(\theta )$, as $\hat{x}' \notin R(\hat{x})$ and $B_{A_\sigma }(\hat{x}, \varepsilon , T) \subset R(\hat{x})$, we have $\hat{x}' \notin B_{A_\sigma }(\hat{x}, \varepsilon , T)$. Then, (6) implies that the grid $G(\theta )$ is $(T, \varepsilon )$-separated, and thus

  $$\begin{aligned}&\log N(A_\sigma , \varepsilon , T, K) \ge \log |G(\theta )| \\&= \sum _{i=1}^{n} \log (2 \lfloor 1/\theta _i \rfloor + 1) > \sum _{i=1}^{n} \log (\max \{2/\theta _i - 1,\, 1\}). \end{aligned}$$

  Consequently, the definition of entropy (5) implies

  $$\begin{aligned} \begin{aligned} h(A_\sigma )&\ge \lim _{\varepsilon \searrow 0} \limsup _{T \rightarrow \infty } \sum _{i=1}^{n} \dfrac{\log (\max \{2/\theta _i - 1,\, 1\})}{T} \\&= \lim _{\varepsilon \searrow 0} \limsup _{T \rightarrow \infty } \sum _{i=1}^{n} \dfrac{\log (1/\theta _i)}{T} + \lim _{\varepsilon \searrow 0} \limsup _{T \rightarrow \infty } \sum _{i=1}^{n} \dfrac{\log (\max \{2 - \theta _i,\, \theta _i\})}{T}, \end{aligned} \end{aligned}$$

  where the last term equals 0 if (17) holds.

C Proof of Corollary 2

First, the definition (11) of $\hat{a}$ and the subadditivity of limit suprema imply

$$\begin{aligned} \hat{a} \le \sum _{p \in \mathcal P} \limsup _{t \rightarrow \infty }\, a_p \rho _p(t) \le \sum _{p \in \mathcal P} \max \{a_p,\, 0\} \limsup _{t \rightarrow \infty } \rho _p(t) = \sum _{p \in \mathcal P} h(a_p) \hat{\rho }_p; \end{aligned}$$

which, combined with (28), implies the upper bound (29). For the case where the limits $\lim _{t \rightarrow \infty } \rho _p(t)$ exist and $a_p \ge 0$ for all $p \in \mathcal P$, the inequalities in the derivation above becomes equalities due to the additivity of limits and $\max \{a_p,\, 0\} = a_p$. Second, the definition (11) of $\hat{a}$ implies

$$\begin{aligned} \hat{a} \le \limsup _{t \rightarrow \infty } \bigg ( \max _{p \in \mathcal P}\, a_p \bigg ) \sum _{p \in \mathcal P} \rho _p(t) = \max _{p \in \mathcal P}\, a_p \le \max _{p \in \mathcal P}\, h(a_p), \end{aligned}$$

which, combined with (28), implies the upper bound (30).

D Proof of Lemma 6

First, we establish the upper bound in (36). For each $p \in \mathcal P$, as $N_p$ is nilpotent, there is a positive integer $k_p$ such that $N_p^{k_p} = 0$. Let $k_s := \sum _{p \in \mathcal P} k_p$, which is finite as the index set $\mathcal P$ is finite. Define the weighted average matrix over [0, t] by

$$\begin{aligned} N(t) := \sum _{p \in \mathcal P} N_p \rho _p(t) \in \mathbb C^{n \times n}. \end{aligned}$$

For all $t \ge 0$, as $\{N_p: p \in \mathcal P\}$ is a commuting family, we have $(N(t))^{k_s} = 0$. Also, $\Vert N(t)\Vert \le \eta _M := \max _{p \in \mathcal P} \Vert N_p\Vert$. Hence for all $v \in \mathbb C^n$, we have

$$\begin{aligned} \big \Vert e^{N(t)\, t} v \big \Vert= & {} \Bigg \Vert \left( \sum _{k=0}^{k_s-1} \dfrac{(N(t))^k t^k}{k!} \right) v \Bigg \Vert \le \left( \sum _{k=0}^{k_s-1} \dfrac{\eta _M^k t^k}{k!} \right) \Vert v\Vert \\\le & {} c_\delta \left( \sum _{k=0}^{k_s-1} \dfrac{\delta ^k t^k}{k!} \right) \Vert v\Vert \le c_\delta e^{\delta t} \Vert v\Vert \qquad \forall t \ge 0 \end{aligned}$$

with $c_\delta := \max \{(\eta _M/\delta )^{k_s - 1},\, 1\} > 0$.

Second, we establish the lower bound in (36). As $\Vert {-N(t)}\Vert = \Vert N(t)\Vert \le \eta _M$ for all $t \ge 0$, the proof above also implies that for all $v \in \mathbb C^n$, we have

$$\begin{aligned} \Vert v\Vert = \big \Vert e^{-N(t)\, t} e^{N(t)\, t} v \big \Vert \le c_\delta e^{\delta t} \big \Vert e^{N(t)\, t} v \big \Vert , \end{aligned}$$

that is, $\Vert e^{N(t)\, t} v\Vert \ge c_\delta ^{-1} e^{-\delta t} \Vert v\Vert$ for all $t \ge 0$.

E Computation of $h(D_{\sigma _2})$ using (34) in Example 3

Recall from footnote 2 that $\sigma = 1$ on $[t_{2k}, t_{2k+1})$ and $\sigma = 2$ on $[t_{2k+1}, t_{2k+2})$, where $t_0 = 0$, $t_1 = 1$, and $t_k = 9^{k-1} + 9^{k-2}$ for all $k \ge 2$. Hence

(68)

and thus

$$\begin{aligned} \begin{aligned} a_1^1 \tau _1(t) + a_2^1 \tau _2(t) = 3 \tau _2(t) - \tau _1(t)&= {\left\{ \begin{array}{ll} 3.6 t_{2k} - t, &{}t \in [t_{2k}, t_{2k+1}), \\ 3 t - 3.6 t_{2k+1}, &{}t \in [t_{2k+1}, t_{2k+2}) \end{array}\right. } \\ a_1^2 \tau _1(t) + a_2^2 \tau _2(t) = 2 \tau _1(t) - \tau _2(t)&= {\left\{ \begin{array}{ll} 2 t - 2.7 t_{2k}, &{} t \in [t_{2k}, t_{2k+1}), \\ 2.7 t_{2k+1} - t, &{}t \in [t_{2k+1}, t_{2k+2}). \end{array}\right. } \end{aligned} \end{aligned}$$

Then, $\bar{a}_1$ and $\bar{a}_2$ in (34) satisfy

$$\begin{aligned} \begin{aligned} \bar{a}_1(T)&= \dfrac{1}{T} \max _{t \in [0, T]} a_1^1 \tau _1(t) + a_2^1 \tau _2(t) \\&= {\left\{ \begin{array}{ll} 2.6 t_{2k}/T, &{}T \in [t_{2k}, t_{2k+1} + 8 t_{2k}/3), \\ 3 - 3.6 t_{2k+1}/T, &{}T \in [t_{2k+1} + 8 t_{2k}/3, t_{2k+2}) \end{array}\right. } \\ \bar{a}_2(T)&= \dfrac{1}{T} \max _{t \in [0, T]} a_1^2 \tau _1(t) + a_2^2 \tau _2(t)\\&= {\left\{ \begin{array}{ll} 1.7 t_{2k+1}/T, &{}T \in [t_{2k+1}, t_{2k+2} + 4 t_{2k+1}), \\ 2 - 2.7 t_{2k+2}/T, &{}T \in [t_{2k+2} + 4 t_{2k+1}, t_{2k+3}). \end{array}\right. } \end{aligned} \end{aligned}$$

Hence

$$\begin{aligned} \bar{a}_1(T) + \bar{a}_2(T) = {\left\{ \begin{array}{ll} 17.9 t_{2k}/T, &{}T \in [t_{2k+1}, t_{2k+1} + 8 t_{2k}/3), \\ 3 - 1.9 t_{2k+1}/T, &{}T \in [t_{2k+1} + 8 t_{2k}/3, t_{2k+2}), \\ 25.1 t_{2k+1}/T, &{}T \in [t_{2k+2}, t_{2k+2} + 4 t_{2k+1}), \\ 2 - 0.1 t_{2k+2}/T, &{}T \in [t_{2k+2} + 4 t_{2k+1}, t_{2k+3}). \end{array}\right. } \end{aligned}$$

Therefore,

$$\begin{aligned} h(D_{\sigma _2}) = \limsup _{T \rightarrow \infty }\, \bar{a}_1(T) + \bar{a}_2(T) = \max \{1.99,\, 2.79\} = 2.79. \end{aligned}$$

F Proof of Lemma 7

We regard (43) as a family of scalar differential equations (recall that here $\xi _\sigma ^k$ denotes the k-th scalar component of $\xi _\sigma$):

$$\begin{aligned} \begin{aligned} \dot{\xi }_\sigma ^1&= a_\sigma ^1 \xi _\sigma ^1 + b_\sigma ^{1, 2} \xi _\sigma ^2 + \cdots + b_\sigma ^{1, n} \xi _\sigma ^n, \\ \dot{\xi }_\sigma ^2&= a_\sigma ^2 \xi _\sigma ^2 + b_\sigma ^{2, 3} \xi _\sigma ^3 + \cdots + b_\sigma ^{2, n} \xi _\sigma ^n, \\&\,\,\, \vdots \\ \dot{\xi }_\sigma ^{n-1}&= a_\sigma ^{n-1} \xi _\sigma ^{n-1} + b_\sigma ^{n-1, n} \xi _\sigma ^n, \\ \dot{\xi }_\sigma ^n&= a_\sigma ^n \xi _\sigma ^n, \end{aligned} \end{aligned}$$

and prove Lemma 7 by mathematical induction. For brevity, let

$$\begin{aligned} \psi _{i,j}(t) := b^{i, j}_{\sigma (t)} e^{\eta _j(t) - \eta _i(t)}, \qquad i, j \in \left\{ 1, \ldots , n\right\} . \end{aligned}$$

Then, $\varPsi$ in (50) can be written as

$$\begin{aligned} \varPsi \left( t, \mathcal C_{k,l,i}\right) = \sum _{(c_0, \ldots , c_i) \in \mathcal C_{k,l,i}} \int _0^t \int _0^{s_1} \cdots \int _0^{s_{i-1}} \prod _{j=1}^{i} \left( \psi _{c_{j-1}, c_j}(s_j) \hbox {d}s_j \right) . \end{aligned}$$

(69)

1.1 F.1 The basis of induction

For the n-th scalar differential equation $\dot{\xi }_\sigma ^n = a_\sigma ^n \xi _\sigma ^n$, the state-transition function is defined by

$$\begin{aligned} \phi _n(t, s) := e^{\eta _n(t) - \eta _n(s)}, \qquad t \ge s \ge 0. \end{aligned}$$

Hence the n-th scalar component of $\xi _\sigma (x, t)$ satisfies $\xi _\sigma ^n(x, t) = e^{\eta _n(t)} x_n$, that is, (50) holds for $k = n$.

1.2 F.2 The inductive step

For an arbitrary $m \in \{1, \ldots , n-1\}$, suppose that $\xi _\sigma ^k(x, t)$ satisfy (50) for all $k \in \{m+1, \ldots , n\}$. For the m-th differential equation

$$\begin{aligned} \dot{\xi }_\sigma ^m = a_\sigma ^m \xi _\sigma ^m + \sum _{k=m+1}^{n} b_\sigma ^{m, k} \xi _\sigma ^k, \end{aligned}$$

the state-transition function is defined by

$$\begin{aligned} \phi _m(t, s) := e^{\eta _m(t) - \eta _m(s)}, \qquad t \ge s \ge 0. \end{aligned}$$

By variation of constants, the m-th scalar component of $\xi _\sigma (x, t)$ satisfies

$$\begin{aligned} \begin{aligned} \xi _\sigma ^m(x, t)&= e^{\eta _m(t)} \left( x_m + \sum _{k=m+1}^{n} \int _0^t e^{-\eta _m(s_1)} b_{\sigma (s_1)}^{m, k} \xi _\sigma ^k\left( x, s_1\right) \hbox {d}s_1 \right) \\&= e^{\eta _m(t)} \left( x_m + \sum _{k=m+1}^{n} \int _0^t \psi _{m,k}(s_1) \left( x_k + \sum _{l=k+1}^{n} \sum _{i=1}^{l-k} x_l \varPsi \left( s_1, \mathcal C_{k,l,i}\right) \right) \hbox {d}s_1 \right) \\&= e^{\eta _m(t)} \left( x_m + \sum _{k=m+1}^{n} x_k \int _0^t \psi _{m,k}(s_1) \hbox {d}s_1\right. \\&\left. \quad + \sum _{k=m+1}^{n} \sum _{l=k+1}^{n} \sum _{i=1}^{l-k} x_l \int _0^t \psi _{m,k}(s_1) \varPsi (s_1, \mathcal C_{k,l,i}) \hbox {d}s_1 \right) . \end{aligned} \end{aligned}$$

Based on the definition (52) of $\mathcal C_{k,l,i}$ and the formula (69) of $\varPsi$, we have

$$\begin{aligned} \int _0^t \psi _{m,k}(s_1) \hbox {d}s_1 = \varPsi (t, \mathcal C_{m,k,1}) \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\int _0^t \psi _{m,k}(s_1) \varPsi \left( s_1, \mathcal C_{k,l,i}\right) \hbox {d}s_1\\&= \sum _{\left( c_1, \ldots , c_{i+1}\right) \in \mathcal C_{k,l,i}} \int _0^t \int _0^{s_1} \cdots \int _0^{s_i} \psi _{m,k}(s_1) \prod _{j=2}^{i+1} \left( \psi _{c_{j-1},c_j}(s_j) \hbox {d}s_j \right) \hbox {d}s_1 \\&= \sum _{\left( c_0, \ldots , c_{i+1}\right) \in \{m\} \times \mathcal C_{k,l,i}} \int _0^t \int _0^{s_1} \cdots \int _0^{s_i} \prod _{j=1}^{i+1} \left( \psi _{c_{j-1},c_{j}}(s_j) \hbox {d}s_j \right) \\&= \varPsi (t, \{m\} \times \mathcal C_{k,l,i}). \end{aligned} \end{aligned}$$

Changing the order of summation, we obtain

$$\begin{aligned} \sum _{k=m+1}^{n} \sum _{l=k+1}^{n} \sum _{i=1}^{l-k} x_l \varPsi (t, \{m\} \times \mathcal C_{k,l,i})&= \sum _{l=m+2}^{n} \sum _{k=m+1}^{l-1} \sum _{i=1}^{l-k} x_l \varPsi \left( t, \{m\} \times \mathcal C_{k,l,i}\right) \\&= \sum _{l=m+2}^{n} \sum _{i'=2}^{l-m} \sum _{k=m+1}^{l-i'+1} x_l \varPsi \left( t, \{m\} \times \mathcal C_{k,l,i'-1}\right) , \end{aligned}$$

where in the last step we also let $i' = i + 1$. Next, we prove that the family of sets $\{\{m\} \times \mathcal C_{k,l,i'-1}: k = m+1, \ldots , l-i'+1\}$ forms a partition of $\mathcal C_{m,l,i'}$.

• For all $(c_1, \ldots , c_{i'}) \in \mathcal C_{k_1,l,i'-1}$ and $(c'_1, \ldots , c'_{i'}) \in \mathcal C_{k_2,l,i-1}$ with $k_1 \ne k_2$, as $c_1 = k_1 \ne k_2 = c'_1$, we have $(c_1, \ldots , c_{i'}) \ne (c'_1, \ldots , c'_{i'})$. Hence the sets in $\{\{m\} \times \mathcal C_{k,l,i'-1}: k = m+1, \ldots , l-i'+1\}$ are pairwise disjoint.

• For all $(c_1, \ldots , c_{i'}) \in \mathcal C_{k,l,i'-1}$, as $c_1 = k \ge m+1$ and $c_{i'} = l$, we have $(m, c_1, \ldots , c_{i'}) \in \mathcal C_{m,l,i'}$. Hence $\bigcup _{k=m+1}^{l-i'-1} \{m\} \times \mathcal C_{k,l,i'-1} \subset \mathcal C_{m,l,i'}$.

• For all $(c_0, \ldots , c_i') \in \mathcal C_{m,l,i'}$, as $c_1 \ge c_0 + 1 = m + 1$ and $c_1 \le c_{i'} - (i' - 1) = l - i' + 1$, we have $k := c_1$ satisfies $m + 1 \le k \le l - i' + 1$ and $(c_0, \ldots , c_{i'}) \in \{m\} \times \mathcal C_{k,l,i'-1}$. Hence $\mathcal C_{m,l,i'} \subset \bigcup _{k=m+1}^{l-i'-1} \{m\} \times \mathcal C_{k,l,i'-1}$.

Therefore,

$$\begin{aligned} \sum _{l=m+2}^{n} \sum _{i'=2}^{l-m} \sum _{k=m+1}^{l-i'+1} x_l \varPsi \left( t, \{m\} \times \mathcal C_{k,l,i'-1}\right) = \sum _{l=m+2}^{n} \sum _{i'=2}^{l-m} x_l \varPsi \left( t, \mathcal C_{m,l,i'}\right) . \end{aligned}$$

Combining the results above, we obtain

$$\begin{aligned} \xi _\sigma ^m(x, t) = e^{\eta _m(t)} \left( x_m + \sum _{l=m+1}^{n} \sum _{i=1}^{l-m} x_l \varPsi \left( t, \mathcal C_{m,l,i}\right) \right) , \end{aligned}$$

that is, (50) holds for $k = m$. Therefore, mathematical induction implies that (50) holds for all $k \in \{1, \ldots , n\}$.

G Proof of Lemma 8

For every $k \in \{1, \ldots , n\}$, following the formula (50) and the triangle inequality, the k-th scalar component of $\xi _\sigma (x, t)$ satisfies

$$\begin{aligned} |\xi _\sigma ^k(x, t)| \le e^{Re(\eta _k(t))} |x_k| + \sum _{l=k+1}^{n} \left( \sum _{i=1}^{l-k} e^{Re(\eta _k(t))} |\varPsi (t, \mathcal C_{k,l,i})| \right) |x_l|. \end{aligned}$$

First, following the definitions (46) of $\bar{d}_i$ and (51) of $\eta _i$, we have

$$\begin{aligned} Re\left( \eta _i(t)\right) \le Re\left( \eta _{i-1}(t)\right) + \bar{d}_i(t)\, t \qquad \forall t \ge 0, \forall i \in \left\{ 2, \ldots , n\right\} . \end{aligned}$$

Hence

$$\begin{aligned} \begin{aligned} |\varPsi (t, \mathcal C_{k,l,i})|&\le \sum _{(c_0, \ldots , c_i) \in \mathcal C_{k,l,i}} b_M^i \int _0^{t} \int _0^{s_1} \cdots \int _0^{s_{i-1}} \prod _{j=1}^{i} \left( e^{Re(\eta _{c_j}(s_j) - \eta _{c_{j-1}}(s_j))} \hbox {d}s_j \right) \\&\le \sum _{\left( c_0, \ldots , c_i\right) \in \mathcal C_{k,l,i}} b_M^i t^i \prod _{j=1}^{i} \left( \max _{s_j \in [0, T]} e^{Re\left( \eta _{c_j}(s_j) - \eta _{c_{j-1}}(s_j)\right) } \right) \\&= \sum _{\left( c_0, \ldots , c_i\right) \in \mathcal C_{k,l,i}} b_M^i t^i e^{\sum _{j=1}^{i} \bar{d}_{c_j}(t)\, t} \\&\le \sum _{\left( c_0, \ldots , c_i\right) \in \mathcal C_{k,l,i}} b_M^i t^i e^{\sum _{j=k+1}^{l} \bar{d}_j(t)\, t}, \end{aligned} \end{aligned}$$

where the last inequality follows partially from the definition (52) of the sets $\mathcal C_{k,l,i}$. As $b_M^i t^i$ and $\sum _{j=k+1}^{l} \bar{d}_j(t)\, t$ are independent of the choice of $(c_0, \ldots , c_i) \in \mathcal C_{k,l,i}$ and the latter is also independent of the choice of $i \in \{1, \ldots , l-k\}$, and the set $\mathcal C_{k,l,i}$ can be characterized by the combinations of $i-1$ increasing integers from $k+1$ to $l-1$, we have

$$\begin{aligned} \sum _{i=1}^{l-k} e^{Re(\eta _k(t))} |\varPsi (t, \mathcal C_{k,l,i})|\le & {} \left( \sum _{i=1}^{l-k} |\mathcal C_{k,l,i}| b_M^i t^i \right) e^{Re(\eta _k(t)) + \sum _{j=k+1}^{l} \bar{d}_j(t)\, t} \\= & {} \left( \sum _{i=1}^{l-k} \left( {\begin{array}{c}l-k-1\\ i-1\end{array}}\right) b_M^i t^i \right) e^{Re(\eta _k(t)) + \sum _{j=k+1}^{l} \bar{d}_j(t)\, t}\\\le & {} (b_M t + 1)^{l - k} e^{Re\left( \eta _k(t)\right) + \sum _{j=k+1}^{l} \bar{d}_j(t)\, t}, \end{aligned}$$

where the last inequality follows partially from the binomial formula. Hence

$$\begin{aligned} \begin{aligned} |\xi _\sigma ^k(x, t)|&\le e^{Re(\eta _k(t))} |x_k| + \sum _{l=k+1}^{n} \Big ( (b_M t + 1)^{l - k} e^{Re(\eta _k(t)) + \sum _{j=k+1}^{l} \bar{d}_j(t)\, t} |x_l| \Big ) \\&\le e^{Re(\eta _k(t))} \sum _{l=k}^{n} \Big ( (b_M t + 1)^{l - k} e^{\sum _{j=k+1}^{l} \bar{d}_j(t)\, t} |x_l| \Big ). \end{aligned} \end{aligned}$$

Note that the upper bound for $|\xi _\sigma ^k(x, t)|$ above is decreasing in k. Indeed, the upper bound for $|\xi _\sigma ^{k-1}(x, t)|$ satisfies

$$\begin{aligned} \begin{aligned}&e^{Re(\eta _{k-1}(t))} \sum _{l=k-1}^{n} \Big ( (b_M t + 1)^{l-k+1} e^{\sum _{j=k}^{l} \bar{d}_j(t)\, t} |x_l| \Big ) \\&\quad \ge e^{Re(\eta _{k-1}(t)) + \bar{d}_k(t)\, t} \sum _{l=k}^{n} \Big ( (b_M t + 1)^{l-k} e^{\sum _{j=k+1}^{l} \bar{d}_j(t)\, t} |x_l| \Big ) \\&\quad \ge e^{Re(\eta _k(t))} \sum _{l=k}^{n} \Big ( (b_M t + 1)^{l-k} e^{\sum _{j=k+1}^{l} \bar{d}_j(t)\, t} |x_l| \Big ). \end{aligned} \end{aligned}$$

Hence we obtain (53) by taking the upper bound for $|\xi _\sigma ^1(x, t)|$ (recall that we take $\Vert \cdot \Vert$ to be the $\infty$-norm; see Remark 1).

Second, recall $c_0 = k$ and $c_i = l$, and let $s_0 := t$ and

$$\begin{aligned} a_m^i := \max _{p \in \mathcal P} Re(a_p^i), i \in \{1, \ldots , n\}. \end{aligned}$$

Following the definition (51) of $\eta _i$, we have

$$\begin{aligned} Re(\eta _i(t) - \eta _i(\tau )) = \sum _{p \in \mathcal P} Re(a_p^i) (\tau _p(t) - \tau _p(\tau )) \le a_m^i (t - \tau ) \qquad \forall t \ge \tau \ge 0, \forall i \in \{1, \ldots , n\}. \end{aligned}$$

Hence

$$\begin{aligned} \begin{aligned} e^{Re(\eta _k(t))} |\varPsi (t, \mathcal C_{k,l,i})|&\le \sum _{(c_0, \ldots , c_i) \in \mathcal C_{k,l,i}} b_M^i \int _0^{s_0} \cdots \int _0^{s_{i-1}} e^{Re(\eta _{c_i}(s_i))} \prod _{j=1}^{i}\\&\quad \left( e^{Re(\eta _{c_{j-1}}(s_{j-1}) - \eta _{c_{j-1}}(s_j))} \hbox {d}s_j \right) \\&\le \sum _{(c_0, \ldots , c_i) \in \mathcal C_{k,l,i}} b_M^i \int _0^{s_0} \cdots \int _0^{s_{i-1}} e^{a_m^{c_i} s_i} \prod _{j=1}^{i} \Big ( e^{a_m^{c_{j-1}}(s_{j-1} - s_j)} \hbox {d}s_j \Big ) \\&\le \sum _{(c_0, \ldots , c_i) \in \mathcal C_{k,l,i}} b_M^i e^{\max _{0 \le j \le i} a_m^{c_j} t} \Bigg ( \int _0^{s_0} \cdots \int _0^{s_{i-1}} \prod _{j=1}^{i} \hbox {d}s_j \Bigg ) \\&\le \sum _{(c_0, \ldots , c_i) \in \mathcal C_{k,l,i}} b_M^i t^i e^{\max _{k \le j \le l} a_m^j t}, \end{aligned} \end{aligned}$$

where the last inequality follows partially from the definition (52) of the sets $\mathcal C_{k,l,i}$. As $b_M^i t^i$ and $\max _{k \le j \le l} a_m^j t$ are independent of the choice of $(c_0, \ldots , c_i) \in \mathcal C_{k,l,i}$ and the latter is also independent of the choice of $i \in \{1, \ldots , l-k\}$, and the set $\mathcal C_{k,l,i}$ can be characterized by the combinations of $i-1$ increasing integers from $k+1$ to $l-1$, we have

$$\begin{aligned} \begin{aligned} \sum _{i=1}^{l-k} e^{Re(\eta _k(t))} |\varPsi (t, \mathcal C_{k,l,i})|&\le \left( \sum _{i=1}^{l-k} |\mathcal C_{k,l,i}| b_M^i t^i \right) e^{\max _{k \le j \le l} a_m^j t} \\&= \left( \sum _{i=1}^{l-k} \left( {\begin{array}{c}l-k-1\\ i-1\end{array}}\right) b_M^i t^i \right) e^{\max _{k \le j \le l} a_m^j t}\\&\le \left( b_M t + 1\right) ^{l - k} e^{\max _{k \le j \le l} a_m^j t}, \end{aligned} \end{aligned}$$

where the last inequality follows partially from the binomial formula. Hence

$$\begin{aligned} |\xi _\sigma ^k(x, t)| \le e^{Re(\eta _k(t))} |x_k| + \sum _{l=k+1}^{n} \Big ( (b_M t + 1)^{l - k} e^{\max _{k \le j \le l} a_m^j t} |x_l| \Big ). \end{aligned}$$

Note that the upper bound for $|\xi _\sigma ^k(x, t)|$ above is decreasing in k. Indeed, the upper bound for $|\xi _\sigma ^{k-1}(x, t)|$ satisfies

$$\begin{aligned} \begin{aligned}&e^{Re(\eta _{k-1}(t))} |x_{k-1}| + \sum _{l=k}^{n} \Big ( (b_M t + 1)^{l-k+1} e^{\max _{k-1 \le j \le l} a_m^j t} |x_l| \Big ) \\&\quad \ge e^{a_m^k t} |x_k| + \sum _{l=k+1}^{n} \Big ( (b_M t + 1)^{l-k} e^{\max _{k \le j \le l} a_m^j t} |x_l| \Big ) \\&\quad \ge e^{Re(\eta _k(t))} |x_k| + \sum _{l=k+1}^{n} \Big ( (b_M t + 1)^{l-k} e^{\max _{k \le j \le l} a_m^j t} |x_l| \Big ). \end{aligned} \end{aligned}$$

Hence we obtain (54) by taking the upper bound for $|\xi _\sigma ^1(x, t)|$ (recall that we take $\Vert \cdot \Vert$ to be the $\infty$-norm; see Remark 1).

H Computation of an upper bound for $h(U_{\sigma _2})$ using (44) in Example 4

Following (68) in “Appendix E”, we have

$$\begin{aligned} \begin{aligned} a_1^1 \tau _1(t) + a_2^1 \tau _2(t)&= 3 \tau _2(t) - \tau _1(t) \\&= {\left\{ \begin{array}{ll} 3.6 t_{2k} - t, &{}t \in [t_{2k}, t_{2k+1}), \\ 3 t - 3.6 t_{2k+1}, &{}t \in [t_{2k+1}, t_{2k+2}) \end{array}\right. } \\ \left( a_1^2 - a_1^1\right) \, \tau _1(t) + \left( a_2^2 - a_2^1\right) \, \tau _2(t)&= 3 \tau _1(t) - 4 \tau _2(t) \\&= {\left\{ \begin{array}{ll} 3 t - 6.3 t_{2k}, &{}t \in [t_{2k}, t_{2k+1}), \\ 6.3 t_{2k+1} - 4 t, &{}t \in [t_{2k+1}, t_{2k+2}). \end{array}\right. } \end{aligned} \end{aligned}$$

Then, $\bar{a}_1$ and $\bar{d}_2$ in (44) satisfy

$$\begin{aligned} \begin{aligned} \bar{a}_1(T)&= \dfrac{1}{T} \max _{t \in [0, T]} a_1^1 \tau _1(t) + a_2^1 \tau _2(t) \\&= {\left\{ \begin{array}{ll} 2.6 t_{2k}/T &{}T \in [t_{2k}, t_{2k+1} + 8 t_{2k}/3); \\ 3 - 3.6 t_{2k+1}/T, &{}T \in [t_{2k+1} + 8 t_{2k}/3, t_{2k+2}) \end{array}\right. } \\ \bar{d}_2(T)&= \dfrac{1}{T} \max _{t \in [0, T]} (a_1^2 - a_1^1)\, \tau _1(t) + \left( a_2^2 - a_2^1\right) \, \tau _2(t)\\&= {\left\{ \begin{array}{ll} 2.3 t_{2k+1}/T, &{}T \in [t_{2k+1}, 2 t_{2k+2} + 5 t_{2k+1}/3); \\ 3 - 6.3 t_{2k+2}/T, &{}T \in [2 t_{2k+2} + 5 t_{2k+1}/3, t_{2k+3}). \end{array}\right. } \end{aligned} \end{aligned}$$

Hence

$$\begin{aligned} 2 \bar{a}_1(T) + \bar{d}_2(T) = {\left\{ \begin{array}{ll} 25.9 t_{2k}/T, &{}T \in [t_{2k+1}, t_{2k+1} + 8 t_{2k}/3); \\ 6 - 4.9 t_{2k+1}/T, &{}T \in [t_{2k+1} + 8 t_{2k}/3, t_{2k+2}); \\ 49.1 t_{2k+1}/T, &{}T \in [t_{2k+2}, 2 t_{2k+2} + 5 t_{2k+1}/3); \\ 3 - 1.1 t_{2k+2}/T, &{}T \in [t_{2k+2}+ 5 t_{2k+1}/3, t_{2k+3}). \end{array}\right. } \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} h(U_{\sigma _2})&= \limsup _{T \rightarrow \infty }\, 2 \bar{a}_1(T) + \bar{d}_2(T) = \max \{2.88,\, 5.46\} = 5.46. \end{aligned} \end{aligned}$$