iISS and ISS dissipation inequalities: preservation and interconnection by scaling

Abstract

In analysis and design of nonlinear dynamical systems, (nonlinear) scaling of Lyapunov functions has been a central idea. This paper proposes a set of tools to make use of such scalings and illustrates their benefits in constructing Lyapunov functions for interconnected nonlinear systems. First, the essence of some scaling techniques used extensively in the literature is reformulated in view of preservation of dissipation inequalities of integral input-to-state stability (iISS) and input-to-state stability (ISS). The iISS small-gain theorem is revisited from this viewpoint. Preservation of ISS dissipation inequalities is shown to not always be necessary, while preserving iISS which is weaker than ISS is convenient. By establishing relationships between the Legendre–Fenchel transform and the reformulated scaling techniques, this paper proposes a way to construct less complicated Lyapunov functions for interconnected systems.

Appendix

1.1 Appendix 1: Proof of Theorem 1

Property (17) implies the existence of $w_L\in (0,\infty )$ and a sequence $\{s_i\}$ of real numbers such that $\lim _{i\rightarrow \infty }s_i=\infty$ and $\lim _{i\rightarrow \infty }\alpha (s_i)<\sigma (w_L)$. By virtue of (8), if $\liminf _{s\rightarrow \infty }\mu ^\prime (s)=\infty$ holds, then

$$\begin{aligned} \limsup _{|x|\rightarrow \infty } \mu ^\prime (V(x))\{-\alpha (V(x))+\sigma (w_L)\}=\infty . \end{aligned}$$

(99)

On the other hand, the assumptions $\hat{\alpha }\in \mathcal {P}$ and $\hat{\sigma }\in \mathcal {K}$ imply

$$\begin{aligned} \limsup _{|x|\rightarrow \infty } -\hat{\alpha }(W(x))+\hat{\sigma }(|w|)<\infty , \quad \forall |w|\in \mathbb {R}_+ . \end{aligned}$$

(100)

The contradiction between (99) and (100) arising from (18) indicates that (19) must hold. If $\lim _{s\rightarrow \infty }\alpha (s)$ exists, property $\lim _{i\rightarrow \infty }\alpha (s_i)<\sigma (w_L)$ holds for any sequence $\{s_i\}$ of real numbers such that $\lim _{i\rightarrow \infty }s_i=\infty$. Hence, the claim (20) follows from (100).

1.2 Appendix 2: Proof of Theorem 2

The decomposition (11) yields

$$\begin{aligned}&\mu ^\prime (V(x))\left[ -\alpha (V(x))+\sigma (|w|)\right] \nonumber \\&\quad =-b\alpha (V(x))+b\sigma (|w|)+ \lambda (V(x))\left[ -\alpha (V(x))+\sigma (|w|)\right] . \end{aligned}$$

(101)

Obviously, in the case of $\lambda (s)\equiv 0$, inequality (15) holds with $\hat{\alpha }=b\alpha \circ \mu ^{-1}\in \mathcal {P}$ and $\hat{\sigma }=b\sigma \in \mathcal {K}$ which are identical with the pair (23a), (23b). The assertions about $\hat{\alpha }\in \mathcal {K}$ and $\hat{\alpha }\in \mathcal {K}_\infty$ are straightforward. Note that $\omega$ is irrelevant in this case. Hence, the rest of the proof assumes $\lambda (s)\not \equiv 0$. Property (12) implies $\lambda (s)>0$ for all $s\in (0,\infty )$. Since $\mu ^\prime$ is non-decreasing, so is $\lambda$.

First, suppose that $\liminf _{l\rightarrow \infty }\alpha (l)>0$. This clearly guarantees the existence of $\tilde{\alpha }\in \mathcal {K}$ satisfying (25). It is also straightforward that there exists a continuous function $\omega : \mathbb {R}_+\rightarrow \mathbb {R}_+$ satisfying (26). Following the idea in [36], we evaluate $\lambda (V(x))\left[ -\alpha (V(x))+\sigma (|w|)\right]$ in (101) in the two cases $\tilde{\alpha }(V(x))\ge (\mathbf {Id}+\omega )\circ \sigma (|w|)$ and $\tilde{\alpha }(V(x))\le (\mathbf {Id}+\omega )\circ \sigma (|w|)$ separately. Due to the non-decreasing property of $\lambda$ and $\mathbf {Id}+\omega \in \mathcal {K}_\infty$, the combination of the evaluation in the two cases yields (15) with (23). Notice that (22) implying $\lim _{s\rightarrow \infty }\lambda (s)< \infty$ for $\lim _{s\rightarrow \infty }\tilde{\alpha }(s)< \infty$ ensures $\lambda \circ \tilde{\alpha }^{\ominus }\circ (\mathbf {Id}+\omega )\circ \sigma (s)$ is well-defined for all $s\in \mathbb {R}_+$. The non-decreasing property of $\lambda$ and $\mathbf {Id}+\omega \in \mathcal {K}_\infty$ yields $\hat{\sigma }\in \mathcal {K}$. It is verified that

$$\begin{aligned} (\mathbf {Id}-(\mathbf {Id}+\omega )^{-1}) \circ (\mathbf {Id}+\omega )=(\mathbf {Id}+\omega )-\mathbf {Id}=\omega . \end{aligned}$$

Due to $\mathbf {Id}+\omega \in \mathcal {K}_\infty$, we have

$$\begin{aligned} (\mathbf {Id}-(\mathbf {Id}+\omega )^{-1}) =\omega \circ (\mathbf {Id}+\omega )^{-1} \end{aligned}$$

(102)

which gives

$$\begin{aligned} \hat{\alpha }= [\omega \circ (\mathbf {Id}+\omega )^{-1}\circ \tilde{\alpha }\circ \mu ^{-1}] [\lambda \circ \mu ^{-1}] +b\alpha \circ \mu ^{-1} . \end{aligned}$$

(103)

From (26) it follows that $\omega \circ (\mathbf {Id}+\omega )^{-1}\circ \tilde{\alpha }(s)>0$ holds for all $s\in (0,\infty )$, and $\omega \circ (\mathbf {Id}+\omega )^{-1}\circ \tilde{\alpha }\in \mathcal {P}$. Thus, we have $\hat{\alpha }\in \mathcal {P}$. Finally, Eq. (103) also implies $\hat{\alpha }\in \mathcal {K}$ (resp. $\hat{\alpha }\in \mathcal {K}_\infty$) if $\omega , \tilde{\alpha }, \alpha \in \mathcal {K}$ (resp. $\omega , \tilde{\alpha }, \alpha \in \mathcal {K}_\infty$).

Next, suppose that $\liminf _{l\rightarrow \infty }\alpha (l)=0$. Then property (24) implies that $\tilde{\alpha }^\ominus (s)=\infty$ for all $s\in \overline{\mathbb {R}}_+$ by virtue of the definition of $\ominus$. Since $L:=\lim _{l\rightarrow \infty }\lambda (l)< \infty$ is ensured by (22), the formula (23b) gives $\hat{\sigma }=(b+L)\sigma \in \mathcal {K}$ which is independent of $\omega$. The choice (24) also implies $\hat{\alpha }\in \mathcal {P}$ for (23a) for each given $\omega$. On the other hand,

$$\begin{aligned} b\sigma (|w|)+\lambda (V(x))\sigma (|w|)\le (b+L)\sigma (|w|) ,\quad \forall x\in \mathbb {R}^N, \ \forall w\in \mathbb {R}^M \end{aligned}$$

(104)

holds. From (24) we also obtain

$$\begin{aligned}&b\alpha (V(x))+\lambda (V(x))\alpha (V(x)) \nonumber \\&\quad \ge b\alpha \circ \mu ^{-1}(W(x)) + [\lambda \circ \mu ^{-1}(W(x))][\tilde{\alpha }\circ \mu ^{-1}(W(x))] \end{aligned}$$

(105)

$$\begin{aligned}&\quad \ge b\alpha \circ \mu ^{-1}(W(x)) \nonumber \\&\qquad +[\lambda \circ \mu ^{-1}(W(x))] [(\mathbf {Id}- (\mathbf {Id}+\omega )^{-1})\circ \tilde{\alpha }\circ \mu ^{-1}(W(x))] \end{aligned}$$

(106)

for all $x\in \mathbb {R}^N$ and $w\in \mathbb {R}^M$ Applying these inequalities to (101), we arrive at not only (23), but also (23) with (27) with ${\varOmega }\rightarrow \infty$.

Finally, suppose that $\lim _{s\rightarrow \infty }\mu ^\prime (s)< \infty$. Defining $L:=\lim _{l\rightarrow \infty }\mu ^\prime (l)< \infty$ again yields (104). Independently, (105) follows from (25). Thus, (101) is bounded from above by $-\hat{\alpha }(W(x))+\hat{\sigma }(|w|)$ defined by

$$\begin{aligned} \hat{\alpha }(s)= [\tilde{\alpha }\circ \mu ^{-1}(s)] [\lambda \circ \mu ^{-1}(s)] +b\alpha \circ \mu ^{-1}(s) , \quad \hat{\sigma }=(b+L)\sigma (s) . \end{aligned}$$

These functions are identical with taking ${\varOmega }\rightarrow \infty$ in (23) with (27) for each $s\in \mathbb {R}_+$.

1.3 Appendix 3: Proof of Theorem 3

In the case of $\lambda (s)\equiv 0$, the claim holds true obviously from $\alpha , \sigma \in \mathcal {K}$ and (10) since (34) gives $\hat{\alpha }=b\alpha \circ \mu ^{-1}\in \mathcal {K}$ and $\hat{\sigma }=b\sigma \in \mathcal {K}$. Therefore, the rest considers the case of $\lambda (s)\not \equiv 0$. First, suppose that (30)–(32) are satisfied with a continuous function $\omega : \mathbb {R}_+\rightarrow \mathbb {R}_+$. Following the proof of Theorem 2 with $\tilde{\alpha }=\alpha$, we obtain $\hat{\sigma }\in \mathcal {K}$ in (34b). Note that property (28) is guaranteed by (32). The function $\hat{\alpha }\in \mathcal {K}$ which is obtained as in (23a) with $\tilde{\alpha }=\alpha$ and satisfies (15) is only of class $\mathcal {P}$. Hence, write (23a) as $\eta + b\alpha \circ \mu ^{-1}$ by defining $\eta$ as in (35). Rewrite $\eta \in \mathcal {P}$ as

$$\begin{aligned} \eta (s)&= \left[ \left( \frac{\alpha \circ \mu ^{-1} (s)}{(\mathbf {Id}+\omega )^{-1}\circ \alpha \circ \mu ^{-1}(s)} -1\right) \right] \nonumber \\&\quad \cdot \left[ (\mathbf {Id}+\omega )^{-1}\circ \alpha \circ \mu ^{-1}(s)\right] . \left[ \lambda \circ \mu ^{-1}(s)\right] . \end{aligned}$$

(107)

Applying (30) and (32) to (107) with the help of $\mu \in \mathcal {K}_\infty$, one arrives at

$$\begin{aligned} \liminf _{s\rightarrow \infty }\eta (s)\ge \lim _{s\rightarrow \infty }[\lambda \circ \alpha ^{\ominus }\circ (\mathbf {Id}+\omega )\circ \sigma ]\sigma . \end{aligned}$$

Since this inequality implies $\liminf _{s\rightarrow \infty }\eta (s)>0$ and we have $\alpha \circ \mu ^{-1}\in \mathcal {K}$ in addition, there always exists a continuous function $k: \mathbb {R}_+\rightarrow \mathbb {R}_+$ such that (36) and (37) are fulfilled. Defining $\hat{\alpha }$ as in (34a) with (36) and (37) ensures $\hat{\alpha }\in \mathcal {K}$ and $\hat{\alpha }(s)\le \eta (s)+ b\alpha \circ \mu ^{-1}(s)$ for all $s\in \mathbb {R}_+$. Thus, the preservation of the iISS dissipation inequality (9) under the scaling $\mu$ is established by $\hat{\alpha }, \hat{\sigma }\in \mathcal {K}$ given in (34). Furthermore, by virtue of $\limsup _{l\rightarrow \infty } k(l)=1$ and $\hat{\alpha }\in \mathcal {K}$, property (33) follows from (10), (30) and (107). This proves the preservation of the ISS dissipation inequality.

Finally, replace the pair of (31) and (32) by (27) with ${\varOmega }\rightarrow \infty$ in the case of $\lim _{s\rightarrow \infty }\mu ^\prime (s)< \infty$. Define $L:=\lim _{l\rightarrow \infty }\lambda (l)< \infty$. Then (34) becomes

$$\begin{aligned} \hat{\alpha }=k [\mu ^\prime \circ \mu ^{-1}][\alpha \circ \mu ^{-1}] , \quad \hat{\sigma }=(b+L)\sigma , \end{aligned}$$

(108)

and clearly satisfies $\hat{\sigma }$, $\hat{\sigma }\in \mathcal {K}$. Using (36) and (37) to modify (23) of Theorem 2 verifies that the functions in (108) achieve the preservation the iISS dissipation inequality (9) under the scaling $\mu$. For (108), by virtue of $\limsup _{l\rightarrow \infty } k(l)=1$ and $\hat{\alpha }\in \mathcal {K}$, property (33) is implied by (10). This establishes the preservation the ISS dissipation inequality.

1.4 Appendix 4: Proof of Proposition 2

The existence of $\tau , \varphi \ge 0$ fulfilling (45) is straightforward from $c>1$. Property (44) with $\tau <c$ implies (39). Due to the property $\alpha _{i}\circ \alpha ^{\ominus }(s)\le s$ for all $s\in \mathbb {R}_+$, property (40) follows if

$$\begin{aligned} \lim _{s\rightarrow \infty }(\tau -1)\left[ \alpha (s)\right] ^\varphi \beta (s) \ge \lim _{s\rightarrow \infty } [\tau \sigma (s)]^\varphi [\beta \circ \alpha ^\ominus \circ \tau \sigma (s)] . \end{aligned}$$

(109)

The non-decreasing property of $\beta$ and (39) guarantee

$$\begin{aligned} \lim _{s\rightarrow \infty }\beta (s) \ge \lim _{s\rightarrow \infty } [\beta \circ \alpha ^\ominus \circ \tau \sigma (s)] . \end{aligned}$$

Thus, if

$$\begin{aligned} \lim _{s\rightarrow \infty }(\tau -1)\left[ \alpha (s)\right] ^\varphi \ge \lim _{s\rightarrow \infty } [\tau \sigma (s)]^\varphi \end{aligned}$$

(110)

is met, property (109) is satisfied. In the case where (45) holds for $\varphi =0$, we can easily verify (110). Therefore, we next assume that (45) holds for some $\varphi >0$. Property (110) is satisfied if we have

$$\begin{aligned} \lim _{s\rightarrow \infty }(\tau -1)^{\frac{1}{\varphi }}\alpha (s) \ge \lim _{s\rightarrow \infty } \tau \sigma (s) . \end{aligned}$$

This property is achieved if

$$\begin{aligned} (\tau -1)^{-\frac{1}{\varphi }}\tau \le c \end{aligned}$$

(111)

since we have (44). Property (111) is secured by (45).

1.5 Appendix 5: Proof of Proposition 4

In the case of $\mu ^\prime (s)\equiv b$, the implications (52) and (22) do not require anything, which proves the claim. Suppose that $\mu ^\prime (s)\not \equiv b$. Since $\mu \in \mathcal {K}_\infty$, property (49) is equivalent to

$$\begin{aligned} (\lambda +b)\alpha -\kappa \circ \lambda \in \mathcal {P}. \end{aligned}$$

(112)

Clearly, this property implies

$$\begin{aligned} (\lambda (s)+b)\alpha (s)-\kappa \circ \lambda (s)\ge 0 , \quad \forall s\in \mathbb {R}_+ . \end{aligned}$$

(113)

Recalling $\mu ^\prime (s)\not \equiv b$, properties $\lim _{s\rightarrow \infty }\lambda (s)>0$ and $\kappa \in \mathcal {K}_\infty$ imply $\lim _{s\rightarrow \infty }\kappa \circ \lambda (s)>0$. Since $\kappa ^\prime$ is of $\mathcal {K}_\infty$, property (113) requires $\liminf _{s\rightarrow \infty }\alpha (s)>0$. Hence, property (52) must hold. Next, suppose $\lim _{s\rightarrow \infty }\mu ^\prime (s)=\infty$ which means $\lim _{s\rightarrow \infty }\lambda (s)=\infty$. Property $\kappa ^\prime \in \mathcal {K}_\infty$ in (113) again implies $\lim _{s\rightarrow \infty }\alpha (s)=\infty$. Therefore, (22) must hold.

1.6 Appendix 6: Proof of Theorem 4

First, we assume that (64) holds. Let $L:=\lim _{l\rightarrow \infty }\lambda (l)\le \infty$. Suppose that $\alpha \circ \lambda ^\ominus$ is piecewise differentiable on the interval [0, L). Let $\kappa ^\prime$ be any class $\mathcal {K}_\infty$ function satisfying

$$\begin{aligned} \frac{1}{\tau }\alpha \circ \lambda ^{\ominus }(s) \le \kappa ^\prime (s) \le \frac{1}{\tau }\alpha \circ \lambda ^{\ominus }(s)+ \frac{1}{\tau } s\,[(\alpha \circ \lambda ^{\ominus })^\prime (s)], \quad \forall s\in [0,L) , \end{aligned}$$

(114)

where the second inequality is evaluated at all differentiable points. Note that $(\alpha \circ \lambda ^\ominus )^\prime (s)\ge 0$ holds for almost all $s\in [0,L)$ due to $\alpha$, $\lambda \in \mathcal {K}$. Therefore, in the case of $\lim _{s\rightarrow \infty }\alpha (s)< \infty$, the existence of a function $\kappa ^\prime \in \mathcal {K}_\infty$ satisfying (114) follows from assumption (22) since $\lim _{s\rightarrow \infty }\lambda (s)< \infty$ and $\alpha , \lambda \in \mathcal {K}$. In the case of $\lim _{s\rightarrow \infty }\alpha (s)= \infty$, the existence is guaranteed by $\alpha \circ \lambda ^{\ominus }\in \mathcal {K}_\infty$ under the assumption of (64).

Let $\kappa$ denote the antiderivative of $\kappa ^\prime$ satisfying $\kappa (0)=0$. Then $\kappa \in \mathcal {K}_\infty$ follows from $\kappa ^\prime \in \mathcal {K}_\infty$. Define the map $\overline{\mathbb {R}}_+\rightarrow \overline{\mathbb {R}}_+$ as

$$\begin{aligned} \overline{\kappa }(s)=\displaystyle \frac{1}{\tau }s\,[\alpha \circ \lambda ^\ominus (s)] , \quad s\in \overline{\mathbb {R}}_+ , \end{aligned}$$

(115)

which satisfies

$$\begin{aligned}&\overline{\kappa }^\prime (s)= \frac{1}{\tau }\alpha \circ \lambda ^{\ominus }(s)+ \frac{1}{\tau } s\,[(\alpha \circ \lambda ^{\ominus })^\prime (s)] \end{aligned}$$

(116)

for almost all $s\in [0,L)$. Thus, from (114) we obtain

$$\begin{aligned} \kappa (s)\le \overline{\kappa }(s) , \quad \forall s\in [0,L) . \end{aligned}$$

(117)

This property together with (115) yields

$$\begin{aligned}&\kappa \circ \lambda (s)\le \overline{\kappa }\circ \lambda (s) =\frac{1}{\tau }\lambda (s)\alpha (s) , \quad \forall s\in \mathbb {R}_+. \end{aligned}$$

(118)

Hence, we have

$$\begin{aligned} \hat{\alpha }_L(s)= [\mu ^\prime \circ \mu ^{-1}(s)][\alpha \circ \mu ^{-1}(s)] -\kappa \circ \lambda \circ \mu ^{-1}(s) \ge \hat{\alpha }_{D,\tau }(s) , \ \forall s\in \mathbb {R}_+ \end{aligned}$$

(119)

and $\hat{\alpha }_{D,\tau }\in \mathcal {K}$ by virtue of $\mu \in \mathcal {K}_\infty$ and $\tau >1$. Therefore, we arrive at (61) with $\hat{\alpha }_L\in \mathcal {K}$.

Next, applying $\lambda \in \mathcal {K}$ to both sides of the first inequality in (114) from the right, one obtains

$$\begin{aligned} \frac{1}{\tau }\alpha (s)\le \kappa ^\prime \circ \lambda (s) , \quad \forall s\in \mathbb {R}_+ . \end{aligned}$$

Applying the non-decreasing function $\alpha ^\ominus (\tau s)$ defined for $s\in [0,\lim _{l\rightarrow \infty }\alpha (l)/\tau )$ again yields

$$\begin{aligned} s\le \kappa ^\prime \circ \lambda \circ \alpha ^\ominus (\tau s) , \quad \forall s\in [0,\lim _{l\rightarrow \infty }\alpha (l)/\tau ) . \end{aligned}$$

Applying $(\kappa ^\prime )^{-1}\in \mathcal {K}_\infty$ to the above from the left, one obtains

$$\begin{aligned} (\kappa ^\prime )^{-1}(s) \le \lambda \circ \alpha ^{\ominus }(\tau s) , \quad \forall s\in [0,\lim _{l\rightarrow \infty }\alpha (l)/\tau ) . \end{aligned}$$

(120)

Here, recalling $\ell \kappa (s)=s(\kappa ^\prime )^{-1}(s)-\kappa \circ (\kappa ^\prime )^{-1}(s)$ and $\kappa , \kappa ^\prime \in \mathcal {K}_\infty$, we have $\ell \kappa (s)\le s(\kappa ^\prime )^{-1}(s)$ for all $s\in \mathbb {R}_+$. Thus,

$$\begin{aligned} \ell \kappa (s)\le s\,[\lambda \circ \alpha ^\ominus (\tau s)] , \quad \forall s\in [0,\lim _{l\rightarrow \infty }\alpha (l)/\tau ) . \end{aligned}$$

Hence, we have

$$\begin{aligned}&\ell \kappa \circ \sigma (s) \le [\lambda \circ \alpha ^\ominus \circ \tau \sigma (s)]\sigma (s) , \quad \forall s\in [0,\lim _{l\rightarrow \infty }\sigma ^\ominus \circ \tau ^{-1}\alpha (l)) . \end{aligned}$$

Since we have the implication

$$\begin{aligned} \lim _{l\rightarrow \infty }\alpha (l)<\lim _{l\rightarrow \infty }\tau \sigma (l) \ \Rightarrow \ \lambda \circ \alpha ^\ominus \circ \tau \sigma (s) =L , \quad \forall s\in [\lim _{l\rightarrow \infty }\sigma ^\ominus \circ \tau ^{-1}\alpha (l),\infty ) , \end{aligned}$$

by virtue of (22), we arrive at

$$\begin{aligned} \min \{ \ell \kappa \circ \sigma (s), L\sigma (s) \} \le [\lambda \circ \alpha ^\ominus \circ \tau \sigma (s)]\sigma (s) , \ \forall s\in \mathbb {R}_+ . \end{aligned}$$

Comparing this with (42) yields

$$\begin{aligned} \hat{\sigma }_L(s)= \min \left\{ \ell \kappa \circ \sigma (s), L\sigma (s)\right\} +b\sigma \le \hat{\sigma }_{D,\tau }(s) , \quad \forall s\in \mathbb {R}_+ \end{aligned}$$

(121)

which implies (62) with $R=\infty$ and (63).

If $\alpha \circ \lambda ^\ominus : [0,L)\rightarrow \mathbb {R}_+$ is not piecewise differentiable, the above arguments hold true by replacing (114) with

$$\begin{aligned} \kappa ^\prime (s)= \frac{1}{\tau }\alpha \circ \lambda ^{\ominus }(s), \quad \forall s\in [0,L) . \end{aligned}$$

(122)

Note that (117) is guaranteed again since $\kappa ^\prime \in \mathcal {K}_\infty$ is chosen, due to (22).

Finally, suppose that (64) does not hold, i.e., assume that $\lim _{s\rightarrow \infty }\lambda (s)< \infty$ and $\lim _{s\rightarrow \infty }\alpha (s)=\infty$ are satisfied. Let $q>0$ be arbitrary. Consider $p\in (0,\infty )$ which has yet to be determined. Let $\tilde{\lambda }\in \mathcal {K}_\infty$ be defined as

$$\begin{aligned} \tilde{\lambda }(s)=\lambda (s)+q\max \{s-p,0\} , \quad \forall s\in \mathbb {R}_+ . \end{aligned}$$

(123)

Obviously, $\tilde{L}:=\lim _{l\rightarrow \infty }\tilde{\lambda }(l)=\infty$, $\tilde{L}>L:=\lim _{l\rightarrow \infty }\lambda (l)$ and in addition,

$$\begin{aligned}&\lambda (s)\le \tilde{\lambda }(s) , \quad \forall s\in \mathbb {R}_+ \end{aligned}$$

(124)

$$\begin{aligned}&\lambda (s)=\tilde{\lambda }(s) , \quad \forall s\in [0,p] . \end{aligned}$$

(125)

Assume that $\alpha \circ \tilde{\lambda }^{-1}: \mathbb {R}_+\rightarrow \mathbb {R}_+$ is piecewise differentiable. Let $\kappa ^\prime$ be any class $\mathcal {K}_\infty$ function satisfying

$$\begin{aligned} \frac{1}{\tau }\alpha \circ \tilde{\lambda }^{-1}(s) \le \kappa ^\prime (s) \le \frac{1}{\tau }\alpha \circ \tilde{\lambda }^{-1}(s)+ \frac{1}{\tau } s\,[(\alpha \circ \tilde{\lambda }^{-1})^\prime (s)], \quad \forall s\in \mathbb {R}_+, \end{aligned}$$

(126)

where the second inequality is evaluated at all differentiable points. Due to $\alpha , \tilde{\lambda }\in \mathcal {K}_\infty$, we have $(\alpha \circ \tilde{\lambda }^{-1})^\prime (s)\ge 0$ for almost all $s\in \mathbb {R}_+$. Thus, the existence of a function $\kappa ^\prime \in \mathcal {K}_\infty$ satisfying (126) is guaranteed by virtue of $\alpha \circ \tilde{\lambda }^{-1}\in \mathcal {K}_\infty$. Let $\kappa$ denote the antiderivative of $\kappa ^\prime$ satisfying $\kappa (0)=0$. Then $\kappa \in \mathcal {K}_\infty$ follows from $\kappa ^\prime \in \mathcal {K}_\infty$. Define $\overline{\kappa }\in \mathcal {K}_\infty$ by

$$\begin{aligned} \overline{\kappa }(s)=\displaystyle \frac{1}{\tau }s\,[\alpha \circ \tilde{\lambda }^{-1}(s)] , \quad s\in \mathbb {R}_+ . \end{aligned}$$

(127)

Since $\overline{\kappa }^\prime (s)= \frac{1}{\tau }\alpha \circ \tilde{\lambda }^{-1}(s)+ \frac{1}{\tau } s\,[(\alpha \circ \tilde{\lambda }^{-1})^\prime (s)]$ holds for almost all $s\in \mathbb {R}_+$, from (126) we obtain

$$\begin{aligned} \kappa (s)\le \overline{\kappa }(s) , \quad \forall s\in \mathbb {R}_+ . \end{aligned}$$

(128)

This property, (127) and (124) give

$$\begin{aligned} \kappa \circ \lambda (s)\le \overline{\kappa }\circ \lambda (s)= \frac{1}{\tau }\lambda (s)\,[\alpha \circ \tilde{\lambda }^{-1}\circ \lambda (s)] \le \frac{1}{\tau }\lambda (s)\alpha (s) , \quad \forall s\in \mathbb {R}_+. \end{aligned}$$

Hence, by virtue of $\mu \in \mathcal {K}_\infty$ and $\tau >1$, we arrive at (61) with $\hat{\alpha }_L\in \mathcal {K}$. On the other hand, taking inverse of both sides of the first inequality in (126) yields

$$\begin{aligned} \tilde{\lambda }\circ \alpha ^{-1}(\tau s) \ge (\kappa ^\prime )^{-1}(s) , \quad \forall s\in \mathbb {R}_+ . \end{aligned}$$

(129)

From $\ell \kappa (s)\le s(\kappa ^\prime )^{-1}(s)$ for all $s\in \mathbb {R}_+$ it follows that

$$\begin{aligned}&\ell \kappa \circ \sigma (s) \le [\tilde{\lambda }\circ \alpha ^{-1}\circ \tau \sigma (s)]\sigma (s) , \quad \forall s\in \mathbb {R}_+ . \end{aligned}$$

Due to (125), we have

$$\begin{aligned}&\ell \kappa \circ \sigma (s) \le [\lambda \circ \alpha ^{-1}\circ \tau \sigma (s)]\sigma (s) , \quad \forall s\in [0,\sigma ^\ominus \circ \tau ^{-1}\alpha (p)) . \end{aligned}$$

Since $\alpha$ is of class $\mathcal {K}_\infty$, for any given $R\in (0,\infty )$, there exists $p\in (0,\infty )$ such that $R=\sigma ^\ominus \circ \tau ^{-1}\alpha (p)$ holds. Therefore,

$$\begin{aligned} \min \{ \ell \kappa \circ \sigma (s), L\sigma (s) \} \le [\lambda \circ \alpha ^\ominus \circ \tau \sigma (s)]\sigma (s) , \ \forall s\in [0,R) . \end{aligned}$$

Using $L=\lim _{l\rightarrow \infty }\lambda (l)$ and (42), we obtain (62) and (63). If $\alpha \circ \tilde{\lambda }^\ominus : \mathbb {R}_+\rightarrow \mathbb {R}_+$ is not piecewise differentiable, the above arguments hold true by replacing (126) with

$$\begin{aligned} \kappa ^\prime (s)= \frac{1}{\tau }\alpha \circ \tilde{\lambda }^{-1}(s), \quad \forall s\in \mathbb {R}_+ . \end{aligned}$$

(130)

This completes the proof.

1.7 Appendix 7: Proof of Proposition 6

Apply Theorem 2 and Remark 3 to each subsystem ${\varSigma }_i$ with $\omega _i(s)=(\tau -1)s$ for $\tau >1$, $\tilde{\alpha }_i=\alpha _i$ and $\lambda _i=\mu _i^\prime$. It can be verified that (52), (25) and (26) are satisfied. Property (77) also guarantees (28). The formulas in (29) with $\tilde{\alpha }_i=\alpha _i\in \mathcal {K}$ yield

$$\begin{aligned}&\hat{\alpha }_i= \left( 1-\frac{1}{\tau }\right) \left[ \alpha _{i}\circ \mu _i^{-1}(s) \right] ^{\varphi +1} \left[ \sigma _{3-i}\circ \mu _i^{-1}(s)\right] ^{\varphi +1} , \end{aligned}$$

(131)

$$\begin{aligned}&\hat{\sigma }_i= \left[ \alpha _{i}\circ \alpha _i^{\ominus }\circ \tau \sigma _i(s) \right] ^\varphi \left[ \sigma _{3-i}\circ \alpha _i^{\ominus }\circ \tau \sigma _i(s)\right] ^{\varphi +1}\sigma _i(s) \end{aligned}$$

(132)

Property $\alpha _{i}\circ \alpha ^{\ominus }(s)\le s$ for all $s\in \mathbb {R}_+$ and property (77) with $1<\tau \le c$ in (79) imply

$$\begin{aligned} \hat{\sigma }_i(s)&\le \tau ^\varphi \left[ \sigma _{3-i}\circ \alpha _i^{\ominus }\circ \tau \sigma _i(s)\right] ^{\varphi +1}[\sigma _i(s)]^{\varphi +1} \nonumber \\&\le \tau ^\varphi \left[ \displaystyle \frac{1}{c} \alpha _{3-i}(s)\right] ^{\varphi +1}[\sigma _i(s)]^{\varphi +1} , \quad s\in \mathbb {R}_+ . \end{aligned}$$

(133)

Thus, we have

$$\begin{aligned} \sum _{i=1}^2\mu _i^\prime (V_i)\{-\alpha _i(V_i)+\sigma _i(V_{3-i})\} \le -\sum _{i=1}^2[\sigma _{3-i}\circ \mu _i^{-1}(W_i)]^{\varphi +1}q_i(W_i), \end{aligned}$$

(134)

where $W_i=\mu _i(V_i)$ and

$$\begin{aligned} q_i(s)= \left( 1-\frac{1}{\tau }\right) [\alpha _i\circ \mu _i^{-1}(s)]^{\psi +1} -\tau ^{\psi } \left[ \frac{1}{c}\alpha _i\circ \mu _i^{-1}(s)\right] ^{\psi +1} . \end{aligned}$$

The existence of $\epsilon >0$ such that

$$\begin{aligned} q_i(s)\ge \epsilon [\alpha _i\circ \mu _i^{-1}(s)]^{\psi +1} , \quad s\in \mathbb {R}_+ \end{aligned}$$

(135)

is guaranteed by (79). From (134), (135), (73) and (74) it follows that W is a Lyapunov function proving GAS of $x=0$ of (72).

1.8 Appendix 8: Proof of Proposition 7

Property (80) together with (78) implies

$$\begin{aligned} \left\{ \displaystyle \lim _{s\rightarrow \infty }\alpha _i(s)=\infty \ \quad \mathrm{or}\quad \ \lim _{s\rightarrow \infty }\mu _i^\prime (s)<\infty \right\} , \ i=1,2. \end{aligned}$$

(136)

In the case of $\lim _{s\rightarrow \infty }\mu _i^\prime (s)<\infty$, $i=1,2$, iISS and ISS of system (72) are easily verified by incorporating

$$\begin{aligned} \mu _i^\prime (V_i)\theta _i(|w_i|) \le \lim _{l\rightarrow \infty }\mu _i^\prime (l)\theta _i(|w_i|) , \quad i=1,2 \end{aligned}$$

(137)

into the proof of Proposition 6. In the remaining case, property (136) allows one to invoke a technique proposed in [8, 14] as indicated by [12, Proposition 12].

1.9 Appendix 9: Proof of Theorem 5

Suppose that $\tau , \varphi \ge 0$ satisfy (45). Then $\tau <c$ implies $({\tau }/{c})^\varphi >({\tau }/{c})^{\varphi +1}$. Hence, property (79) is met. By virtue of (77), Propositions 6 and 7 prove all the claims.

1.10 Appendix 10: Proof of Theorem 6

First, the function $\lambda _{i,\psi }$ defined by (87) for each $i=1,2$ is of class $\mathcal {K}$ for all $\psi >0$ since $\alpha _i, \sigma _{3-i}\in \mathcal {K}$. With the help of (80), property (87) with $\psi >0$ also ensures

$$\begin{aligned} \left\{ \lim _{s\rightarrow \infty }\alpha _i(s)< \infty \ \Leftrightarrow \ \lim _{s\rightarrow \infty }\lambda _{i,\psi }(s)< \infty \right\} , \ i=1,2 \end{aligned}$$

(138)

for all $\psi >0$, which corresponds to (64) as well as (22). Thus, for arbitrary given $\psi >0$, Corollary 4 is applicable to the two pairs $(\alpha _i,\sigma _i)$, $i=1,2$, and the formula (66) with (65) and $k=0$, which is exactly (88), guarantees (61) and (62) with $R=\infty$, provided that $\kappa _{i,\psi }: [0,L_{i,\psi })\rightarrow \mathbb {R}_+$ is continuously differentiable, and that (67), (68) and (69) hold in terms of $\kappa _{i,\psi }$ for each $i\in \{1,2\}$. Recall that the arguments to derive (119) and (121) allow $\kappa _{i,\psi }$ to be defined on only the interval $[0,{L_i,\psi })$ in (88) instead of the entire $\mathbb {R}_+$. To confirm the continuous differentiability of $\kappa _{i,\psi }$ and (67), (68) and (69), using (87) and (88) and continuous differentiability of $\alpha _i$ and $\sigma _{3-i}$, we first obtain

$$\begin{aligned}&\lambda _{i,\psi }^\prime (s) = (\alpha _i(s)\sigma _{3-i}(s))^\psi \left\{ \frac{\psi \alpha _i^\prime (s)\sigma _{3-i}(s)}{\alpha _i(s)}+ (\psi +1)\sigma _{3-i}^\prime (s)\right\} , \ \forall s\in (0,\infty ) \end{aligned}$$

(139)

$$\begin{aligned}&\kappa _{i,\psi }\circ \lambda _{i,\psi }(s) =\frac{1}{\tau }(\alpha _i(s)\sigma _{3-i}(s))^{\psi +1} , \ \forall s\in \mathbb {R}_+ \end{aligned}$$

(140)

$$\begin{aligned}&[\kappa _{i,\psi }\circ \lambda _{i,\psi }]^\prime (s) = \frac{\psi +1}{\tau }(\alpha _i(s)\sigma _{3-i}(s))^\psi \left\{ \alpha _i^\prime (s)\sigma _{3-i}(s)+\alpha _i(s)\sigma _{3-i}^\prime (s) \right\} , \nonumber \\&\qquad \forall s\in \mathbb {R}_+ . \end{aligned}$$

(141)

The chain rule $[\kappa _{i,\psi }\circ \lambda _{i,\psi }]^\prime (s)= \left( \kappa _{i,\psi }^\prime \circ \lambda _{i,\psi }(s)\right) \lambda _{i,\psi }^\prime (s)$ yields

$$\begin{aligned} \kappa _{i,\psi }^\prime \circ \lambda _{i,\psi }(s)&= [\kappa _{i,\psi }\circ \lambda _{i,\psi }]^\prime (s) \frac{1}{\lambda _{i,\psi }^\prime (s)} \nonumber \\&= \frac{1}{\tau }\alpha _i(s) \left( 1+G_{i,\psi }(s)\right) , \end{aligned}$$

(142)

where

$$\begin{aligned} G_{i,\psi }(s)&=\frac{\alpha _i^\prime (s)\sigma _{3-i}(s)}{\psi \alpha _i^\prime (s)\sigma _{3-i}(s) +(\psi +1)\alpha _i(s)\sigma _{3-i}^\prime (s)} , \\&=\frac{1}{\psi +(\psi +1)\dfrac{\alpha _i(s) \sigma _{3-i}^\prime (s)}{\alpha _i^\prime (s)\sigma _{3-i}(s)}} . \end{aligned}$$

Property $\alpha _i, \sigma _{3-i}\in \mathcal {K}$ implies $\alpha _i(s)\ge 0$ and $\sigma _{3-i}(s)\ge 0$ for $s\in \mathbb {R}_+$. Hence, for each $\psi >0$, assumption (86) and $\alpha _i\in \mathcal {K}$ guarantee that $\kappa _{i,\psi }^\prime \circ \lambda _{i,\psi }(s)$ given by (142) exits and is strictly increasing for all $s\in \mathbb {R}_+$. By definition, it also holds that

$$\begin{aligned} \lim _{s\rightarrow \infty }G_{i,\psi }(s)<\infty \end{aligned}$$

(143)

for any $\psi >0$. Therefore, properties $\kappa _{i,\psi }^\prime \circ \lambda _{i,\psi }(0)=0$ and $\lambda _{i,\psi }\in \mathcal {K}$ ensure for all $\psi >0$ that $\kappa _{i,\psi }^\prime (s)$ exists for $s\in [0,L_{i,\psi })$, and

$$\begin{aligned}&\kappa _{i,\psi }^\prime (0)=0 \end{aligned}$$

(144)

$$\begin{aligned}&\kappa _{i,\psi }^\prime (s) \text{ is } \text{ strictly } \text{ increasing } \forall s\in [0,L_{i,\psi }) . \end{aligned}$$

(145)

Recalling that $L_{i,\psi }< \infty$ implies $\lim _{s\rightarrow \infty }\alpha _i(s)< \infty$ due to (138), from (142) and (143) it follows that

$$\begin{aligned} L_{i,\psi }<\infty \ \Rightarrow \lim _{s\rightarrow \,L_{i,\psi }^-}\kappa _{i,\psi }^\prime (s)<\infty . \end{aligned}$$

(146)

Therefore, it is proved that $\kappa _{i,\psi }$ is continuously differentiable and fulfills and (67), (68) and (69) for each $i=1,2$. We can now invoke Corollary 4. By virtue of $\tilde{\alpha }_{i,\psi }=\lambda _{i,\psi }\alpha -\kappa _{i,\psi }\circ \lambda _{i,\psi }$ that can be verified from (118) with $\kappa =\overline{\kappa }$, substituting (119) and (121) into (134) and (135) in the proof of Proposition 6, there exists $\epsilon >0$ satisfying (90). Hence, (i) is proved. Finally, following the arguments used to prove Propositions 6 and 7, the proof of (ii) is completed.