# Consistency of the Duhem Model with Hysteresis

Authors: Mohammad Fuad Mohammad Naser; Fayçal Ikhouane
Journal: Mathematical Problems in Engineering
 (2013)

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## Abstract

The Duhem model, widely used in structural, electrical, and mechanical engineering, gives an analytical
description of a smooth hysteretic behavior. In practice, the Duhem model is mostly used within the following black-box approach: given a set of experimental input-output data, how to tune the model so that its output matches the experimental data. It may happen that a Duhem model presents a good
match with the experimental real data for a specific input but does not necessarily keep significant physical properties which are inherent to the real data, independent of the exciting input. This paper presents a characterization of different classes of Duhem models in terms of their consistency with the hysteresis behavior.


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## Body

## 1. Introduction

Hysteresis is a nonlinear behavior encountered in a wide variety of processes including biology, optics, electronics, ferroelectricity, magnetism, mechanics, structures, among other areas. The detailed modeling of hysteresis systems using the laws of Physics is an arduous task, and the obtained models are often too complex to be used in applications. For this reason, alternative models of these complex systems have been proposed [1–5]. These models do not come, in general, from the detailed analysis of the physical behavior of the systems with hysteresis. Instead, they combine some physical understanding of the system along with some kind of black-box modeling.One of the popular models for hysteresis is the Duhem model proposed in [6]. The generalized form of the Duhem model consists of an ordinary differential equation of the form x˙=f(x,u)g(u˙), where u is the input and x is the state or the output [7]. Other special forms of the model have been used, like the form x˙=f1(x,u)max{u˙,0}+f2(x,u)min{u˙,0} [8] or the semilinear form x˙=(Ax+Bu)g(u˙) [9]. Other important special cases of the Duhem model are the LuGre model of friction [10], the Dahl model of friction [11], and the Bouc-Wen model of hysteresis [12, 13]. The Duhem model has been used to represent friction [7], electromagnetic behavior [14, 15], or hysteresis in magnetorheological dampers [16].In the current literature, the Duhem model is mostly used within the following black-box approach: given a set of experimental input-output data, how to adjust the Duhem model so that the output of the model matches the experimental data? The use of system identification techniques is one practical way to perform this task. Once an identification method has been applied to tune the Duhem model, the resulting model is considered as a “good” approximation of the true hysteresis when the error between the experimental data and the output of the model is small enough. Then, this model is used to study the behavior of the true hysteresis under different excitations. By doing this, it is important to consider the following remark. It may happen that a Duhem model presents a good match with the experimental real data for a specific input but does not necessarily keep significant physical properties which are inherent to the real data, independent of the exciting input. In the current literature, this issue has been considered in [17, 18] regarding the passivity/dissipativity of Duhem model.In this paper, we investigate the conditions under which the Duhem model is consistent with the hysteresis behavior. The concept of consistency is formalized in [19] where a general class of hysteresis operators is considered. The class of operators that are considered in [19] are the causal ones, with the additional condition that a constant input leads to a constant output. For these classes of systems, consistency has been defined formally. This property is useful in system modeling and identification as it limits the search for the system's parameters to those regions where consistency holds. From the results of [19], it can be concluded that to check consistency one has to consider the sequence of inputs uγ(t)=u(t/γ),t≥0,  γ>0 and the corresponding sequence of outputs xγ with x˙γ=f(xγ,uγ)g(u˙γ). For the Duhem model to represent a hysteresis system, it is necessary that the sequence of functions t→xγ(γt) converges uniformly when γ→∞. In this paper, we seek necessary conditions and sufficient ones for this uniform convergence to hold.This paper is organized as follows. Section2 presents the needed mathematical background. The problem statement is formalized in Section 3. A classification of functions g that is used throughout the paper, is introduced in Section 4. Sections 5 and 6 present necessary conditions and sufficient ones for the Duhem model to be consistent with the hysteresis behavior. Conclusions are given in Section 7.


## 2. Background Results

This section summarizes the results obtained in [19].A functionν:ℝ→ℝ is said to be increasing (resp., decreasing) if t1<t2⇒ν(t1)<ν(t2) (resp. ν(t1)>ν(t2)), and it is said to be nondecreasing (resp. nonincreasing) if t1<t2⇒ν(t1)≤ν(t2) (resp., ν(t1)≥ν(t2)).The Lebesgue measure onℝ is denoted μ. A subset of ℝ is said to be measurable when it is Lebesgue measurable. Consider a function p:I⊂ℝ+=[0,∞)→ℝm for some interval I; the function p is said to be measurable when p is (M,B)-measurable where B is the class of Borel sets of ℝm and M is the class of measurable sets of ℝ+. For a measurable function p:I⊂ℝ+→ℝm, ∥p∥∞,I denotes the essential supremum of the function |p| on I where |·| is the Euclidean norm on ℝm. When I=ℝ+, and it is denoted simply ∥p∥∞.Consider the Sobolev spaceW1,∞(ℝ+,ℝn) of absolutely continuous functions u:ℝ+→ℝn, where n is a positive integer. For this class of functions, the derivative u˙ is defined almost every where with ∥u∥∞<∞ and ∥u˙∥∞<∞. Endowed with the norm ∥u∥1,∞=max(∥u∥∞,∥u˙∥∞), W1,∞(ℝ+,ℝn) is a Banach space [20].Foru∈W1,∞(ℝ+,ℝn), let ρu:ℝ+→ℝ+ be the total variation of u on [0,t]; that is, ρu(t)=∫0t‍|u˙(τ)|dτ∈ℝ+. The function ρu is well defined as u˙∈Lloc1(ℝ+,ℝn). (Lloc1(ℝ+,ℝn) is the space of locally integrable functions ℝ+→ℝn). It is nondecreasing and absolutely continuous. Denote ρu,max=limt→∞ρu(t). If ρu,max=ρu(t) for some t∈ℝ+, let Iu=[0,ρu,max] (in this case ρu,max is necessarily finite). On the other hand, if ρu,max>ρu(t) for all t∈ℝ+, let Iu=[0,ρu,max) (in this case ρu,max may be finite or infinite).Lemma 1.
Letu∈W1,∞(ℝ+,ℝn) be nonconstant. Then, there exists a unique function ψu∈W1,∞(Iu,ℝn) that satisfies ψu○ρu=u.Consider the linear time scale changesγ(t)=t/γ, for any γ>0 and t≥0. Let Ξ be a set of initial conditions. Let ℋ be an operator that maps the input function u∈W1,∞(ℝ+,ℝn) and initial condition ξ0∈Ξ to an output in L∞(ℝ+,ℝm), where m is a positive integer. That is ℋ:W1,∞(ℝ+,ℝn)×Ξ→L∞(ℝ+,ℝm). We consider causal operators such that   for all (u1,ξ0),(u2,ξ0)∈W1,∞(ℝ+,ℝn)×Ξ, if u1=u2  in  [0,τ], then ℋ(u1,ξ0)=ℋ(u2,ξ0) in [0,τ], [1, page 60].Let(u,ξ0)∈W1,∞(ℝ+,ℝn)×Ξ and let y=ℋ(u,ξ0)∈L∞(ℝ+,ℝm). In the rest of this work, only causal operators are considered. Additionally, we consider that the following holds.Assumption 2.
Let(u,ξ0)∈W1,∞(ℝ+,ℝn)×Ξ and y=ℋ(u,ξ0)∈L∞(ℝ+,ℝm); if ∃θ∈ℝ+ such that u is constant on [θ,∞); then y is constant on [θ,∞).Lemma 3.
There exists a unique functionφu∈L∞(Iu,ℝm) that satisfies φu○ρu=y. Moreover, one has ∥φu∥∞,Iu≤∥y∥∞. If y is continuous on ℝ+, then φu is continuous on Iu and one has ∥φu∥∞,Iu=∥y∥∞.Definition 4.
Letu∈W1,∞(ℝ+,ℝn) and initial condition ξ0∈Ξ be given. The operator ℋ is said to be consistent with respect to input u and initial condition x0 if and only if the sequence of functions {φu○sγ}γ>0 converges in L∞(Iu,ℝm) as γ→∞.It is shown in [19] that for hysteresis process, the sequence of functions {φu○sγ}γ>0 converges in L∞(Iu,ℝm) as γ→∞. This fact shows that consistency is a mathematical property that any model of hysteresis should satisfy.


## 3. Problem Statement

The generalized Duhem model is defined for almost allt≥0 by [7]
(1)x˙(t)=f(x(t),u(t))g(u˙(t)),(2)x(0)=x0,
where x0 and state x(t) take values in ℝm for some positive integer m, input u∈W1,∞(ℝ+,ℝ), function f:ℝm×ℝ→ℝm×r is continuous, where r and m are positive integers. Finally, g:ℝ→ℝr is continuous and satisfies g(0)=0. Observe that if u is constant; then x(t)=x0,  for all  t≥0. For this reason, we consider only nonconstant inputs u in this paper.Sinceg is continuous and u˙∈L∞(ℝ+,ℝ), we have g○u˙∈L∞(ℝ,ℝr). The differential equation (1) satisfies Carathéodory conditions, thus, for each initial state x0∈ℝm, (1) has an absolutely continuous solution that is defined on an interval of the form [0,T),T>0 [21, page 4].Consider the time scale changesγ(t)=t/γ,γ>0,t≥0. When the input u○sγ is used instead of u, the system (1)-(2) becomes
(3)x˙γ(t)=f(xγ(t),u○sγ(t))g(1γu˙○sγ(t)),(4)xγ(0)=x0,
where xγ is the maximal solution of (3). When γ=1, system (3)-(4) reduces to (1)-(2). For any γ>0, define σγ:ℝ+→ℝm as σγ=xγ○s1/γ. System (3)-(4) can be rewritten as
(5)σγ(t)=x0+γ∫0tf(σγ(τ),u(τ))g(1γu˙(τ))dτ,
for all γ>0 and for almost all t∈[0,ωγ), where [0,ωγ) is the maximal interval of existence of the solution σγ.Observe that Lemma3 implies that for any γ>0 there exists a unique function xu○sγ∈L∞(Iu,ℝm) such that xu○sγ○ρu○sγ=xγ (when γ=1, we get xu○ρu=x). The latter equality is equivalent to xu○sγ○ρu=σγ. According to Definition 4, the system (1)-(2) is consistent with respect to (u,x0) if and only if the sequence of functions xu○sγ converges in L∞(Iu,ℝm).Proposition 5.
The system (1)-(2) is consistent with respect to (u,x0) in the sense of Definition 4 if and only if the sequence of function σγ converges in L∞(ℝ+,ℝm) as γ→∞.Proof.
To prove the if part, define the causal operatorℋ:W1,∞(ℝ+,ℝn)×ℝm→L∞(ℝ+,ℝm) that maps (u,x0) to x, where x is given in (1)-(2). Assume that there exists σ*∈L∞(ℝ+,ℝm) such that limγ→∞∥σγ-σ*∥∞=0. We know from (5), that σγ is a sequence of continuous functions. Thus, the function σ* is continuous as a uniform limit of continuous function. Lemma 3 implies that there exists a unique continuous function xu*∈L∞(Iu,ℝm) such that xu*○ρu=σ*. Let ϱ∈Iu. Since ρu is continuous, there exists some t≥0 such that ϱ=ρu(t). We get from the relation σγ=xu○sγ○ρu that for all γ>0: |xu○sγ(ϱ)-xu*(ϱ)|=|xu○sγ○ρu(t)-xu*○ρu(t)|≤∥σγ-σ*∥∞. This implies that ∥xu○sγ-xu*∥∞,Iu≤∥σγ-σ*∥∞ so that limγ→∞∥xu○sγ-xu*∥∞,Iu=0, which means that the system (1)-(2) is consistent with respect to (u,x0).
To prove the only if part, assume thatlimγ→∞∥xu○sγ-xu*∥∞,Iu=0, then the relation xu○sγ○ρu=σγ implies that for almost all t≥0(6)|σγ(t)-xu*○ρu(t)|=|xu○sγ○ρu(t)-xu*○ρu(t)|≤∥xu○sγ-xu*∥∞,Iu.
Thus, we have ∥σγ-xu*○ρu∥∞≤∥xu○sγ-xu*∥∞,Iu so that limγ→∞∥σγ-xu*○ρu∥∞=0.Proposition5 implies that the consistency of the system (1)-(2) can be investigated by studying the uniform convergence of the sequence of functions σγ instead of xu○sγ. Thus, we know from Section 2 that the system (1)-(2) is a hysteresis only if σγ converges uniformly as γ→∞.Problem.In this paper, our objective is to derive necessary conditions and sufficient ones for the uniform convergence of the sequence of functions σγ as γ→∞.


## 4. Classification of Functiong

This section introduces a classification for the functiong that is used throughout the paper.Definition 6.
LetG∈C0([t1,t2],ℝ)(C0([t1,t2],ℝ)={p:[t1,t2]→ℝsuch that p is continuous on [t1,t2]}). The right and left local fractional derivatives of G at t3∈(t1,t2) with respect to order λ>0 are defined respectively as follows [22]:
(7)d+λG(t3)=Γ(1+λ)limκ→t3+G(κ)-G(t3)(κ-t3)λ∈ℝ,d-λG(t3)=Γ(1+λ)limκ→t3-G(t3)-G(κ)(t3-κ)λ∈ℝ,
where Γ is the gamma function.The local fractional derivative of a vector-valued function is the vector of local fractional derivatives of its components.Definition 7.
The functiong∈C0(ℝ,ℝr) is said to be of class λ>0  if  g(0)=0 and the quantities d+λg(0)  and  d-λg(0) exist, are finite, and at least one of them is nonzero.Proposition 8.
The function
g
∈
C
0
(
ℝ
,
ℝ
r
)
is of class
λ
if and only if
(8)
∀
ϑ
∈
ℝ
,
o
n
e
h
a
s
lim
γ
→
∞
γ
λ
g
(
ϑ
γ
)
=
g
*
(
ϑ
)
,
where g*∈C0(ℝ,ℝr) is defined as
(9)g*(ϑ)={ϑλlimκ→0+g(κ)κλϑ≥0,(-ϑ)λlimκ→0-g(κ)(-κ)λϑ<0.Proof.
Immediate using of the change of variablesκ=ϑ/γ.Proposition 9.
If the functiong∈C0(ℝ,ℝr) is of class λ; then limγ→∞∥γλg(u˙/γ)-g*(u˙)∥∞=0, where g*∈C0(ℝ,ℝr) is defined in (9).Proof.
The result is trivial whenu is constant. Assume that u is nonconstant. Given ε>0. Since g∈C0(ℝ,ℝr) is of class λ, there exists some cε; that depend solely on ε, such that
(10)|g(ϑ)ϑλ-d+λg(0)Γ(1+λ)|<ε∥u˙∥∞λ,whenever0<ϑ<cε,|g(ϑ)(-ϑ)λ-d-λg(0)Γ(1+λ)|<ε∥u˙∥∞λ,whenever-cε<ϑ<0.
The boundedness of u˙ implies that there exists a positive constant γ* such that |u˙(t)|/γ<cε,  for  all  γ>γ*. Thus, we have for all γ>γ*,
(11)|γλg(u˙/γ)(u˙)λ-d+λg(0)Γ(1+λ)|<ε∥u˙∥∞λ,wheneveru˙>0,|γλg(u˙/γ)(-u˙)λ-d-λg(0)Γ(1+λ)|<ε∥u˙∥∞λ,wheneveru˙<0.
Thus, we get from (9) that
(12)|γλg(u˙γ)-g*(u˙)|≤|u˙|λmax{|γλg(u˙)|u˙|λ-d+λg(0)Γ(1+λ)|,|γλg(u˙)|u˙|λ-d-λg(0)Γ(1+λ)|}<ε,∀γ>γ*,
which completes the proof.Proposition 10.
If the functiong∈C0(ℝ,ℝr) is of class λ1 for some λ1>0; then it cannot be of any class λ2>0 different than λ1.Proof.
Assume that the functiong∈C0(ℝ,ℝr) is of class λ1 and λ2 with λ1<λ2. Then,
(13)d+λ1g(0)Γ(1+λ1)=limκ→0+g(κ)κλ1=limκ→0+κλ2-λ1g(κ)κλ2=0×d+λ2g(0)Γ(1+λ2)=0,d-λ1g(0)Γ(1+λ1)=limκ→0-g(κ)(-κ)λ1=limκ→0-|κ|λ2-λ1g(κ)|κ|λ2=0×d-λ2g(0)Γ(1+λ2)=0,
which contradicts the fact that g is of class λ1.Proposition 11.
If the functiong∈C0(ℝ,ℝr) is of class λ>0; then there exists g0∈C0(ℝ,ℝ+), such that
(14)|γλg(ϑγ)|≤g0(ϑ),∀γ>1,∀ϑ∈ℝ.Proof.
see AppendixA.


## 5. Necessary Conditions

The objective of this section is to derive necessary conditions for the uniform convergence of the sequence of functionsσγ as γ→∞.The standard way to ensure that the system (1)-(2) admits a unique solution is to prove that the right-hand side of (1)-(2) is Lipschitz with respect to x. A function ν:D⊆ℝm×ℝ+→ℝm is Lipschitz with respect to x if there exists a summable function l:ℝ+→ℝ+ such that |ν(α1,t)-ν(α2,t)|≤l(t)|α1-α2|, for almost all t≥0 and for all α1,α2∈ℝm that satisfy (t,α1),(t,α2)∈D [21].Lemma 12.
Assume that the system (1)-(2) has a unique global solution for each input u∈W1,∞(ℝ+,ℝ) and initial condition x0∈ℝm. Assume that the function g is of class λ>0. Suppose that there exists a continuous function Q:ℝ+×ℝ+×ℝ+→ℝ+ such that
(15)|x(t)|≤Q(|x0|,∥u∥∞,∥u˙∥∞),∀t≥0,
for each initial state x0∈ℝm and each input u∈W1,∞(ℝ+,ℝ). Assume that the system (1)-(2) is consistent with respect to (u,x0); that is, there exists qu∈L∞(ℝ+,ℝm) such that limγ→∞∥σγ-qu∥∞=0 (see Proposition 5); thenIfλ=1, one has
(i)
qu∈W1,∞(ℝ+,ℝm).(ii)
one has for allt≥0 that
(16)qu(t)=x0+∫0tf(qu(τ),u(τ))g*(u˙(τ))dτ,whereg* is given in (9).Ifλ∈(0,1), one has
(i)
qu∈C0(ℝ+,ℝm)∩L∞(ℝ+,ℝm).(ii)
qu(0)=x0.(iii)
one has for almost allt≥0 that
(17)f(qu(t),u(t))g*(u˙(t))=0,whereg* is defined in (9).Ifλ>1, one has qu(t)=x0, for all t≥0.Proof.
By (15), the fact that u∈W1,∞(ℝ+,ℝn), the continuity of the function Q, and the relation ∥u∥∞=∥u○sγ∥∞, for all γ>0, there exists some a>0 independent of γ, and γ0>0 such that
(18)|xγ(t)|≤Q(|x0|,∥u∥∞,∥u˙∥∞γ)≤a,∀t≥0,∀γ>γ0,
where xγ is given in (3)-(4). Thus,
(19)∥xγ∥∞≤a,∀γ>γ0.
On the other hand, we conclude from Lemma 3 that xu○sγ∈C0(Iu,ℝm)∩L∞(Iu,ℝm) and ∥xu○sγ∥∞,Iu=∥xγ∥∞, for all γ>0. Hence, the continuity of xu○sγ and (19) imply that
(20)|σγ(t)|=|xu○sγ○ρu(t)|≤∥xu○sγ∥∞,Iu=∥xγ∥∞≤a,∀t≥0,∀γ>γ0.
Thus, the continuity of f and g, the boundedness of u˙, and Proposition 11, imply that there exists a constant b>0 independent of γ such that γλ|f(σγ(τ),u(τ))g(u˙(τ)/γ)|≤b, for almost all τ≥0, for all γ>1. Thus, we can apply the Dominated Lebesgue Theorem [23] to get
(21)limγ→∞γλ∫0tf(σγ(τ),u(τ))g(u˙(τ)γ)dτ=∫0tf(xu*○ρu(τ),u(τ))g*(u˙(τ))dτ,∀t≥0.
On the other hand, since qu is continuous as a uniform limit of continuous sequence of functions, we have qu∈C0(ℝ+,ℝm)∩L∞(ℝ+,ℝm) and qu(0)=x0 (note that σγ(0)=x0, for all γ>0).
Whenλ=1, we obtain from (21) and (5) that qu(t)=x0+∫0tf(qu(τ),u(τ))g*(u˙(τ))dτ, for all t≥0. Thus, the continuity of the functions f and g* along with the boundedness of the functions qu, u, and u˙ implies that the function q˙u is bounded. Therefore, qu∈W1,∞(ℝ+,ℝm) and (16) is satisfied.
Whenλ∈(0,1), we get from inequality (20) that ∥σγ-x0∥∞/γ1-λ→0 as γ→∞. Moreover, (5) can be written for all t≥0 as
(22)σγ(t)-x0γ1-λ=γλ∫0tf(σγ(τ),u(τ))g(u˙(τ)γ)dτ.
The fact that limγ→0∥σγ-x0∥∞/γ1-λ=0, along with (21) and (22) implies that
(23)∫0tf(xu*○ρu(τ),u(τ))g*(u˙(τ))dτ=0,∀t≥0,
which proves (17).
Finally, whenλ>1, (5) implies for all t≥0 that
(24)σγ(t)-x0=γ1-λγλ∫0tf(σγ(τ),u(τ))g(u˙(τ)γ)dτ,
and thus, we get from (21) that limγ→∞σγ(t)=x0, for all t≥0. Therefore, the uniqueness of limits and the continuity of qu imply that qu(t)=x0, for all t≥0.Remark 13.
Observe that forλ>1, the fact that qu(t)=x0,  for  all  t≥0, means that system  (1)-(2)  does not represent a hysteresis behavior [24].Remark 14.
For the caseλ∈(0,1), (17) and the fact that qu(0)=x0 imply that f(x0,u(0))g(u˙(0))=0, whenever u˙(0) exists.Example 15.
Consider the Following LuGre model [10]:
(25)x˙=u˙-θ|u˙|ζ(u˙)x=[1x][u˙-θ|u˙|ζ(u˙)]=f(x)g(u˙),(26)x(0)=x0,
where parameter θ is the stiffness, x∈ℝ is the average deflection of the bristles and is the output of the system, x0∈ℝ is the initial condition, u∈W1,∞(ℝ+,ℝ) is the relative displacement, and is the input of the system. The function ζ:ℝ→ℝ is defined as
(27)ζ(α)=FC+(FS-FC)e-|α/vs|,∀α∈ℝ,
where FC>0 is the Coulomb friction force, FS>0 is the stiction force, and vs>0 is the Stribeck velocity.
System (25)-(26) has a unique global solution [21, page 5]. On the other hand, the sequence of function σγ is given by (see (5))
(28)σ˙γ(t)=u˙(t)-θ|u˙(t)|ζ(u˙(t)/γ)σγ(t),for  almost  allt≥0.
The following facts are proved in Example 29.(i)
There existγ1,E>0 such that |xγ(t)|≤E,  for all  γ>γ1, where xγ is the output when we use input u○sγ instead of u (see system (3)-(4)).(ii)
∥σγ-qu∥∞→0 as γ→∞, where the function qu∈W1,∞(ℝ+,ℝm) is defined for all t≥0 as
(29)qu(t)=x0+∫0t(u˙(τ)-θ|u˙(τ)|FSqu(τ))dτ.Thus, all conditions of Lemma 12 are satisfied.
Now, we have to find the value ofλ and the function g*. We have
(30)limκ→0+g(κ)κ=[1-θFS],limκ→0-g(κ)(-κ)=[-1-θFS].
Thus, the function g∈C0(ℝ+,ℝr) in (25) is of class λ=1 (see Definition 7) and the function g*∈C0(ℝ+,ℝr) in (9) is defined as
(31)g*(ϑ)=[ϑ-θFS|ϑ|],∀ϑ∈ℝ.
Therefore, by applying Lemma 12, it follows that the system (29) satisfies (16).Simulations. Take θ=105N/m, vs=0.001m/s, FS=1.5 N, FC=1.0 N, x(0)=0N, and u(t)=10-4sin(t) m, for all t≥0 (values taken from [7]). Figure 1(a) shows that the graphs {(σγ(t),u(t)),t≥0} converge to the hysteresis loop {(qu(t),u(t)),t≥0} as γ→∞. This is the main characteristic of a hysteresis system. Also, observe that {(σγ(t),u(t)),t≥0} are different for different values of γ. This is what is called “rate-dependent” property of the model (25)-(26). Figure 1(b) presents the graph of (σγ-qu)(t) versus t; we observe that σγ-qu converges uniformly to the zero function as γ→∞ which means that σγ converges uniformly to qu when γ→∞. The graph of qu(t) versus t is presented in Figure 1(c).(a)σγ(t) versus u(t) for different values of γ. (b) (σγ-qu)(t) versus t for different values of γ. (c) qu(t) versus t. The plots are for the LuGre model (25)-(26), where the solid lines are for qu.
(a)
(b)
(c)


## 6. Sufficient Conditions

This section presents sufficient conditions for the uniform convergence of the sequence of functionsσγ as γ→∞ (and hence for consistency of the system (1)-(2) with respect to (u,x0)). The main results of this section are given in Lemmas 20, 23,  and 27.


### 6.1. Classλ∈(0,1) Functions

In this subsection, sufficient conditions for the uniform convergence ofσγ as γ→∞, are derived when the function g is of class λ∈(0,1).Definition 16 (see [25]).
A continuous functionβ:ℝ+→ℝ+ is said to belong to class 𝒦∞ if it is increasing,   satisfies β(0)=0,  and limt→∞β(t)=∞.The following lemma generalizes Theorem4.18 in [25, page 172]. Indeed, in [25], continuous differentiability is needed, while in Lemma 17, we only need absolute continuity. Also, in [25], the inequality on the derivative of the Lyapunov function is needed everywhere, while in Lemma 17 it is needed only almost everywhere.Lemma 17.
Consider a functionz:[0,ω)⊆ℝ+→ℝ+, where ω>0 is finite or infinite. Assume the following.(1)
The functionz is absolutely continuous on each compact interval of [0,ω).(2)
There existz1≥0 and z2>0 such that z1<z2, z(0)<z2 and
(32)z˙(t)≤0foralmostallt∈[0,ω)thatsatisfyz1<z(t)<z2.Then, z(t)≤max(z(0),z1),for all  t∈[0,ω).Proof.
see AppendixB.Example 18.
We want to study the stability of the following system(33)x˙(t)=-x3(t)+u(t),x(0)=x0,
where x0 and state x take values in ℝ, and input u∈W1,∞(ℝ+,ℝ). System (33) has an absolutely continuous solution that is defined on an interval of the form [0,ω) [21, page 4].
Letz:[0,ω)→ℝ+ be such that z(t)=x2(t),  for  all  t∈[0,ω). The function z is absolutely continuous on each compact subset of [0,ω) because x is absolutely continuous. Thus, condition (1) in Lemma 17 is satisfied.
We have for almost allt∈[0,ω) that
(34)z˙(t)=2x(t)·x˙(t)=2x(t)(-x3(t)+u(t))≤-2z2(t)+2∥u∥∞z(t).
Thus,
(35)z˙(t)≤0foralmostallt∈[0,ω)thatsatisfy∥u∥∞2/3<z(t).
Therefore, condition (2) in Lemma 17 is satisfied with z1=∥u∥∞2/3 and z2 can be any positive real number such that z2>max(z(0),z1)=max(x02,∥u∥∞2/3). Thus, we deduce from Lemma 17 that z(t)≤max(z(0),∥u∥∞2/3)=max(x02,∥u∥∞2/3),for allt∈[0,ω), and hence |x(t)|≤max(|x0|,∥u∥∞3), for all  t∈[0,ω).Corollary 19.
Consider a functionz:[0,ω)⊆ℝ+→ℝ+, where ω>0 may be infinite. Assume the following.(1)
The functionz is absolutely continuous on each compact subset of [0,ω).(2)
There exist a class𝒦∞  function β:ℝ+→ℝ+ and constants z1≥0, z2>0, and z3≥0 such that max{β-1(z3),z1,z(0)}<z2, and
(36)z˙(t)≤-β(z(t))+z3foralmostallt∈[0,ω)thatsatisfyz1<z(t)<z2.Then, z(t)≤max{z(0),z1,β-1(z3)},for all  t∈[0,ω).Proof.
We have from (36) that
(37)z˙(t)≤0for    almost    allt∈[0,ω)that  satisfymax{β-1(z3),z1}<z(t)<z2,
and hence the result follows directly from Lemma 17.Although the latter corollary follows immediately from Lemma17, it is useful in many situations [25].Lemma 20.
Suppose that the system (1)-(2) has a unique solution and that the function g is of class λ∈(0,1). Assume that there exists qu∈W1,∞(ℝ+,ℝm) such that for almost all t≥0(38)qu(0)=x0,f(qu(t),u(t))g*(u˙(t))=0.
For all γ>0, define  yγ:ℝ+→ℝm as
(39)yγ(t)=σγ(t)-qu(t)=xγ(γt)-qu(t),
for all t∈[0,ωγ), where [0,ωγ) is the maximal interval of existence of solution σγ in (5). Suppose that we can find a continuously differentiable function V:ℝm→ℝ+ such that(1)
there exists a functionδ1:ℝ+→ℝ+ that satisfies
(40)δ1(γ)⟶0asγ⟶∞,(2)
there exist constantsγ*,δ2>0, continuous functions R1,R2:ℝ+→ℝ+ and 𝒦∞ class functions β1,β2,β3:ℝ+→ℝ+, satisfying
(41)β1(|α|)≤V(α)≤β2(|α|),∀α∈ℝm,(42)dV(α)dα|α=yγ(t)·f(yγ(t)+qu(t),u(t))g(u˙(t)γ)≤-1γλβ3(|yγ(t)|)+1γR1(|yγ(t)|)foralmostallt∈[0,ωγ),∀γ>γ*thatsatisfyδ1(γ)<|yγ(t)|<δ2,(43)|dV(α)dα|≤R2(|α|),∀α∈ℝm.
Then,(i)
there existE,γ*>0 such that for all γ>γ*:  ωγ=+∞ and ∥xγ∥∞≤E, where xγ is given in (3)-(4).(ii)
limγ→∞∥σγ-qu∥∞=0.Proof.
From (5) and (39), we get for all γ>0 and almost all t∈[0,ωγ) that yγ(0)=0 and
(44)y˙γ(t)=γf(yγ(t)+qu(t),u(t))g(u˙(t)γ)-q˙u(t).
For any γ>0, define Vγ:(0,ωγ]→ℝ+ as Vγ(t)=V(yγ(t)),for all t∈(0,ωγ]. Note that function Vγ is absolutely continuous on each compact subset of [0,ωγ) as a composition of a continuously differentiable function V and an absolutely continuous function yγ. The derivative of V along with trajectories (44) is given for almost all t∈[0,ωγ) and all γ>0 by
(45)V˙γ(t)=dV(α)dα|α=yγ(t)·y˙γ(t)=dV(α)dα|α=yγ(t)·[γf(yγ(t)+qu(t),u(t))-q˙u(t)].
By (40), there exists some γ1>γ*, such that β2○δ1(γ)<β1(δ2), for all γ>γ1. Let Ωγ=(β2○δ1(γ),β1(δ2)). By (41), we have for any γ>γ1, for almost all t∈[0,ωγ) that
(46)Vγ(t)∈Ωγ⟹δ1(γ)<|yγ(t)|<δ2.
Thus, we deduce from (42), (43), (45), and (46) that
(47)V˙γ(t)≤-γ1-λβ3(|yγ(t)|)+R1(|yγ(t)|)+∥q˙u∥∞R2(|yγ(t)|),foralmostallt∈[0,ωγ),∀γ>γ1thatsatisfyVγ(t)∈Ωγ.
Therefore, (46) and the continuity of the functions R1  and  R2 imply that there exists a constant a>0 that does not depend on γ, such that
(48)V˙γ(t)≤-γ1-λβ3(|yγ(t)|)+a,for  almost  allt∈[0,ωγ),∀γ>γ1thatsatisfyVγ(t)∈Ωγ.
Thus, we deduce from (41) that
(49)V˙γ(t)≤-βγ(Vγ(t))+a,foralmostallt∈[0,ωγ),∀γ>γ1thatsatisfyVγ(t)∈Ωγ,
where βγ:ℝ+→ℝ+ is defined as βγ=γ1-λβ3○β2-1.
On the other hand, sinceλ∈(0,1), there exists γ2>γ1 such that βγ-1(a)=β2○β3-1(aγλ-1)<β1(δ2),for  allγ>γ2. Hence, Corollary 19 and the fact that Vγ(0)=0,forallγ>0, imply that Vγ(t)≤max{βγ-1(a),β2○δ1(γ)}=max{β2○β3-1(aγλ-1),β2○δ1(γ)},forallt∈[0,ωγ)and  for  allγ>γ2. Therefore, (41) implies for allγ>γ2 and for allt∈[0,ωγ) that
(50)|yγ(t)|≤max{β1-1○β2○β3-1(aγλ-1),β1-1○β2○δ3(γ)}.
Thus, ωγ=+∞,for allγ>γ2. Furthermore, (40), (50), and the fact that λ∈(0,1) imply that ∥yγ∥∞=∥σγ-qu∥∞→0 as γ→∞. This proves the consistency with respect to (u,x0) because of Proposition 5.
Moreover, by (50), there exist some E>0,γ*>γ2, such that
(51)∥σγ∥∞≤E,∀γ>γ*.
On the other hand, let γ>γ*. Since xγ is continuous, Lemma 3 ensures that ∥xu○sγ∥∞,Iu=∥xγ∥∞. Let ϱ∈Iu. Due to the continuity of ρu, there exists some t≥0 such that ϱ=ρu(t) and thus (51) and the continuity of σγ lead to |xu○sγ(ϱ)|=|xu○sγ○ρu(t)|=|σγ(t)|≤∥σγ∥∞≤E,  for  allγ>γ*. Therefore, ∥xu○sγ∥∞=∥xγ∥∞≤E,for  allγ>γ*, which completes the proof.Remark 21.
Forλ∈(0,1),  if the function qu∈C0(ℝ+,ℝm)∩L∞(ℝ+,ℝm) in Lemma 20, such that qu=R(u) for some R:ℝ→ℝm, then the graphs {(σγ(t),u(t)),t≥0} converge to the curve R as γ→∞. Hence, (1)-(2) is not a hysteresis because the hysteresis loop cannot be a function [24]. This fact is illustrated in Example 22.Example 22.
Consider the following semilinear Duhem model:(52)x˙=(Ax+Bu)g(u˙),x(0)=x0,
where A is a Hurwitz m×m matrix (i.e., every eigenvalue of A has a negative real part), vector B and state x take values in ℝm. The right-hand side of (52) is Lipschitz and thus the system has a unique solution [21]. Take an input u∈W1,∞(ℝ+,ℝ) such that A-1Bu(0)=-x0 and that |u˙(t)|≥M for almost all t∈ℝ and for some M>0. Assume that the function g:ℝ→ℝ+ is of class λ∈(0,1) and that d+λg(0),d-λg(0)>0. Thus, there exists L>0 such that g*(ϑ)≥L|ϑ|λ, for all ϑ∈ℝ, where the function g* is defined in (9). On the other hand, Proposition 9 states that limγ→∞∥γλg(u˙/γ)-g*(u˙)∥∞=0. This means that there exists γ1>0 such that we get for almost all t≥0, and all γ>γ1 that
(53)γλg(u˙(t)γ)>g*(u˙(t))-LMλ2.
Thus, the facts that |u˙|≥M and g*(·)≥L|·|λ imply that
(54)g(u˙(t)γ)>LMλ2γλ,for  almost  allt≥0,∀γ>γ1.
The function qu∈W1,∞(ℝ+,ℝ) which is defined as qu=-A-1Bu satisfies (38) because A-1Bu(0)=-x0.
SinceA is Hurwitz, there exists a m×m positive-definite matrix P such that [25, page 136]
(55)PA+ATP=-I,
where I is the identity matrix. Consider the continuously differentiable quadratic Lyapunov function candidate V:ℝm→ℝ such that V(α)=αTPα, for all α∈ℝm. Since P is symmetric, we have for all α∈ℝm that
(56)λmin(P)|α|2≤V(α)=αTPα≤λmax(P)|α|2,
where λmax(P) and λmin(P) are, respectively, the maximum and minimum eigenvalues of the matrix P. This shows that (41) is satisfied with β1(η)=λmin(P)η2 and β2(η)=λmax(P)η2 for all η>0. Since P is symmetric, we have the following matrices derivation:
(57)dV(α)dα=(P+PT)α=2Pα,∀α∈ℝm.
Thus, we get
(58)|dV(α)dα|=2|Pα|≤2|P||α|,∀α∈ℝm,
where |P| is the induced 2-norm for the matrix P and hence (43) is satisfied with R2(η)=2|P|η,  for all  η≥0. From (57), we have  for  all  α∈ℝm that
(59)dV(α)dα·Aα=2Pα·Aα=αT(PA+ATP)α=-|α|2.
Therefore, (54) implies that for almost all t∈ℝ+ and γ>0 that
(60)dV(α)dα|α=yγ(t)·[A(yγ(t)+qu(t))+Bu(t)]g(u˙(t)γ)=-g(u˙(t)γ)|yγ(t)|2≤-LMλ2γλ|yγ(t)|2,
where yγ is defined in (39). Thus, (42) is satisfied with β3(η)=(LMλ/2)η2, for all η≥0 and R1(η)=0, for all η≥0.
Letδ1 be the zero function. Then (40) is verified. Take δ2,γ* arbitrary in ℝ+ (say δ2=1, γ*=1). Hence, all conditions of Lemma 20 are satisfied. Thus, it follows from Lemma 20 that there exist some E,γ*>0 such that for all γ>γ*, the solution of (52) is global with |xγ(t)|≤E, for all t≥0. Moreover, the operator which maps (u,x0) to x is consistent. In particular, we have ∥σγ-qu∥∞=∥σγ+A-1Bu∥∞→0 as γ→∞.
As a conclusion, the graphs{(σγ(t),u(t)),t≥0} converge to the graph of the linear function R:ℝ→ℝm, which is defined as R(α)=-A-1Bα, for all α∈ℝ. This means that for λ∈(0,1), the model (52) does not represent a hysteresis (see Remark 21).Simulations. Take m=1, B=1.0, A=-1.0, and x0=0. Let g(ϑ)=|ϑ|, for all ϑ∈ℝ, then d+λg(0)=d-λg(0)=Γ(3/2)=Γ(1/2)/2=π/2>0. Let u∈W1,∞(ℝ+,ℝ) be the function of period 2 such that u(t)=t, for all t∈[0,1], and u(t)=2-t, for all t∈[1,2]. Then, we have |u˙(t)|=1, for  almost  all t≥0. We also have qu=-A-1Bu=u. Figure 2(a) shows that the graph {(σγ(t),u(t)),t≥0} collapses into the identity function when γ→∞. This happens because of the fact that qu=u and Remark 21. Figure 2(b) shows that the sequence of functions σγ converges uniformly to qu=u as γ→∞.(a)σγ(t) versus u(t) for different values of γ. (b) σγ(t) versus t for different values of γ. In each plot, the function qu is the solid line.
(a)
σ
γ
(
t
) versus u(t) for system  (52)
(b)
σ
γ
(
t
) versus t for system  (52)


### 6.2. Classλ=1 Functions

In this subsection, we consider classλ=1 functions. (A function g∈C0(ℝ,ℝr) is of class λ=1 if g(0)=0 and the limits limκ→0+g(κ)/|κ| and limκ→0-g(κ)/|κ| exist, are finite, and at least one of them is nonzero (see Section 5)). The main results of this subsection are given in Lemmas 23 and 27.Lemma 23.
Assume the following.(1)
The system (1)-(2) has a unique global solution.(2)
For the functiong in system (1)-(2), there exist a1,a2∈ℝr such that
(61)g(ϑ)={a1ϑϑ≥0,a2ϑϑ<0.Then, the sequence of functions σγ of (5) is independent of γ and the operator which maps (u,x0) to x is consistent.Proof.
By condition(2), the right-hand side of (5) is independent of γ. Thus, the solution σγ of (5) is independent of γ. Since xu○sγ○ρu=σγ, the function xu○sγ is also independent of γ (this is the so-called “rate-independent hysteresis”) and hence consistency holds.Example 24.
Consider Bouc's hysteresis model [13] as follows:
(62)x˙=-c|u˙|x+Φ′(u)u˙,
where c>0, Φ∈C1(ℝ,ℝ), input u∈W1,∞(ℝ+,ℝ), and Φ′(u)=dΦ(α)/dα|α=u.
The right-hand side of (62) is Lipschitz with respect to x. Thus, the system has a unique solution. Furthermore, we have
(63)x˙≤-|Φ′(u)u˙|+Φ′(u)u˙≤0,whenx≥1c|Φ′(u)|,x˙≥|Φ′(u)u˙|+Φ′(u)u˙≥0,whenx≤-1c|Φ′(u)|.
Thus, |x|≤max{|Φ′(u)|/c,|x(0)|}, for each input u∈W1,∞(ℝ+,ℝ) and each initial state x(0)∈ℝ. Since u is bounded and Φ′(u) is continuous, the solution x of (62) is bounded and hence global. Hence, condition (1) in Lemma 23 is satisfied. Equation (62) can be written as
(64)x˙=f(x,u)g(u˙)=[-cx+Φ′(u)cx+Φ′(u)][max{0,u˙}min{0,u˙}].
Clearly, the function g is of class λ=1 and satisfies condition (2) in Lemma 23. This fact implies that the operator which maps (u,x(0)) to x is consistent and σγ is independent of γ.Simulations. Let c=1, Φ(α)=α3/3, for  all  α∈ℝ,  x(0)=0, and input u(t)=sin(t),forallt≥0. The function σγ is independent of γ and is plotted in Figure 3(a). Furthermore, Figure 3(b) shows a rate-independent hysteresis behavior; that is graphs {(σγ(t),u(t)),t≥0} are the same for different values of γ.(a)σγ(t) versus t. (b) σγ(t) versus u(t) for Boucs hysteresis model (62).
(a)
σ
γ
(
t
) versus t
(b)
σ
γ
(
t
) versus u(t)Proposition 25.
Letu∈W1,∞(ℝ+,ℝ) be nonconstant. There exists a unique function vu∈L∞(Iu,ℝ) that is defined by vu○ρu=u˙. Moreover, ∥vu∥∞,Iu≤∥u˙∥∞. Assume that u˙ is nonzero on a set A⊆ℝ that satisfies μ(ρu(ℝ∖A))=0. Then, vu is nonzero almost everywhere.Proof.
Consider the left-derivative operatorΔ- defined on W1,∞(ℝ+,ℝ) by [Δ-(u)](t)=limτ↑t((u(τ)-u(t))/(τ-t)). The operator Δ- is causal as [Δ-(u)](t) depends only on values of u(τ) for τ≤t, and we have Δ-(u)=u˙ almost every where as u∈W1,∞(ℝ+,ℝ) so that Δ-(u)∈L∞(ℝ+,ℝ). The operator Δ- satisfies Assumption 2. The first part of Proposition 25 follows immediately from Lemma 3. Now, let B={ϱ∈Iu/vu(ϱ)=0}, then B⊆ρu(ℝ∖A) which implies that μ(B)=0.Remark 26.
Observe that ifu˙  is nonzero almost everywhere, then μ(ℝ∖A)=0 so that by [26] we have μ(ρu(ℝ∖A))=0 as ρu is absolutely continuous. An example in which u˙ does not need to be nonzero almost everywhere, is when u is constant on some interval, or on a finite number of intervals, or an infinite number of intervals such that this infinite number has measure zero (e.g., countable).Lemma 27.
Letu∈W1,∞(ℝ+,ℝ) be such that u˙ is nonzero on a set A⊆ℝ that satisfies μ(ρu(ℝ∖A))=0. Consider the semilinear Duhem model with m=1 and λ=1  as follows:
(65)x˙=(Ax+Bu+C)g(u˙),x(0)=x0,
where A=[a1a2⋯ar]≠0, B, and C are 1×r row vectors, state x∈ℝ and function g∈C0(ℝ,ℝr) and of class λ=1, and nonconstant input u∈W1,∞(ℝ+,ℝ). Denote
(66)limκ→0+g(κ)|κ|=G+*=[G+,1*G+,2*⋮G+,r*],limκ→0-g(κ)|κ|=G-*=[G-,1*G-,2*⋮G-,r*].
For any i∈{1,2,…,r}, assume that
(67)G+,i*,G-,i*≥0,wheneverai<0,G+,i*,G-,i*≤0,wheneverai>0.
Suppose that there exists some i0∈{1,2,…,r}, such that
(68)ai0≠0,|G+,i0*|+|G-,i0*|>0.
Then,(i)
there existE,γ1>0 such that ∥xγ∥∞≤E,  for all  γ>γ1  (xγ is the output of the system (65) when we use the input u○sγ instead of the input u (see system (3)-(4))).(ii)
There exists a functionqu∈W1,∞(ℝ+,ℝ) such that ∥σγ-qu∥1,∞→0, as γ→∞ where ∥·∥1,∞ is the norm of the Banach space W1,∞(ℝ+,ℝm) (and hence the consistency of the system (65) is guaranteed with respect to input u and initial condition x0).(iii)
qu(0)=x0. Furthermore, for almost all t≥0, one has
(69)q˙u(t)=(Aqu(t)+Bu(t)+C)g*(u˙(t)),where the functiong*∈C0(ℝ,ℝ) is defined as in (9), that is;
(70)g*(ϑ)={ϑG+*ϑ≥0,-ϑG-*ϑ<0.Proof.
The semilinear Duhem model has a unique global solution due to the Lipschitz property of the right-hand-side [21, page 5].
For anyi∈{1,2,…,r} and any κ∈ℝ∖{0}, let Gi(κ) be the ith component of the function g(κ)/|κ|. From (66), there exists some constant ζ>0, such that for all  i∈{1,2,…,r}(71)G+,i*-|G+,i*|2<Gi(κ)<G+,i*+|G+,i*|2,wheneverκ∈(0,ζ),G-,i*-|G-,i*|2<Gi(κ)<G-,i*+|G-,i*|2,wheneverκ∈(-ζ,0).
Let i∈{1,2,…,r}. We get from (67) and (71) that
(72)aiGi(κ)≤-max{|ai||G+,i*|2,|ai||G-,i*|2}=-Mi, wheneverκ∈(-ζ,ζ)∖{0},ai≠0.
For the case ai=0, take Mi=0. Let M=∑i=1r‍Mi. From (67) and (68), we get M>0. Thus, we get from (72) that
(73)Ag(κ)|κ|=∑i=1r‍aiGi(κ)≤-M<0,∀κ∈(-ζ,ζ)∖{0}.
By Proposition 25, a function vu∈L∞(Iu,ℝ) can be defined almost every where as vu○ρu=u˙ with vu≠0 almost every where The boundedness of vu implies that there exists γ0>0 such that vu(ϱ)/γ∈(-ζ,ζ) for almost all ϱ∈Iu and all γ>γ0. Thus, we deduce from (73) that for all γ>γ0,
(74)γAg(vu(ϱ)/γ)|vu(ϱ)|≤-M,for  almost  allϱ∈Iu.
Let H:ℝ→ℝr be a function such that
(75)H(κ)={G+*κ>0,G-*κ<0.
For any γ>0, define χγ:Iu→ℝr as
(76)χγ(ϱ)={γg(vu(ϱ)/γ)|vu(ϱ)|-H(vu(ϱ))vu(ϱ)≠0,0otherwise.
Since vu∈L∞(Iu,ℝ), relations (66) imply that
(77)limγ→∞∥χγ∥∞=0.
This result can be easily checked using the same techniques used in the proof of Proposition 9. Now, consider the system
(78)h˙=(Ah+Bψu+C)(H○vu),(79)h(0)=x0,
where state h∈ℝ. The differential equation (78) verifies Carathéodory conditions with Lipschitz property with respect to h. Thus, system (78)-(79) has a unique and absolutely continuous local solution [21, page 4]. Consider the Lyapunov function W=h2. We deduce from (78)-(79) that W(0)=x02 and that
(80)W˙=2A(H○vu)W+2(Bψu+C)(H○vu)W.
We get from (67) and (68) that AH(vu(ϱ))≤-∑i=1r‍max{|ai||G+,i*|,|ai||G-,i*|}=-2M for almost all ϱ∈Iu. Thus, the boundedness of the function ψu (see Lemma 1) along with (80) imply that W˙≤-4MW+D1W, for some D1>0. This leads to W˙≤0, whenever W≥(D1/4M)2. Therefore, Lemma 17 and the fact that W(0)=x02 imply that W≤max{x02,(D1/4M)2} which means that the solution of the system (78)-(79) is bounded and hence is global (i.e., is defined on Iu). On the other hand, the relation σγ=xu○sγ○ρu implies that σ˙γ=|u˙|x˙u○sγ○ρu. Thus, we obtain from systems (65) and (5) and the relations vu○ρu=u˙, and ψu○ρu=u, that xu○sγ(0)=x0 and that
(81)x˙u○sγ(ϱ)=γ(Axu○sγ(ϱ)+Bψu(ϱ)+C)g(vu(ϱ)/γ)|vu(ϱ)|
for all γ>0, for almost all ϱ∈[0,τγ)⊆Iu, where [0,τγ) is the maximal interval of existence [21, page 4]. For any γ>0, let yγ:Iu→ℝ be defined as yγ=xu○sγ-h. Since xu○sγ(0)=h(0)=x0, the system (81) can be written for all γ>0, for almost all ϱ∈[0,τγ) as
(82)y˙γ(ϱ)=γAg(vu(ϱ)/γ)|vu(ϱ)|yγ(ϱ)+(Ah(ϱ)+Bψu(ϱ)+C)χγ(ϱ),(83)yγ(0)=0.
For any γ>0, consider the Lyapunov function Vγ:[0,τγ)→ℝ+ with Vγ(ϱ)=yγ2(ϱ),  for all  ϱ∈[0,τγ). By (82) and the boundedness of both ψu and the solution of (78)-(79), there exists some D2>0 such that for almost all ϱ∈[0,τγ) and all γ>γ0,
(84)V˙γ(ϱ)≤2γAg(vu(ϱ)/γ)|vu(ϱ)|Vγ(ϱ)+D2∥χγ∥∞Vγ(ϱ).
Thus, we get from (74) that
(85)V˙γ(ϱ)≤0,for    almost    allϱ∈[0,τγ),∀γ>γ0,that    satisfyVγ(ϱ)>(D2∥χγ∥∞2M)2.
Therefore, Lemma 17 and the fact that Vγ(0)=0 imply that Vγ(ϱ)≤(D2∥χγ∥∞/2M)2 for all γ>γ0 and almost all ϱ∈[0,τγ), and hence we obtain, for almost all ϱ∈[0,τγ) that
(86)|yγ(ϱ)|=|xu○sγ(ϱ)-h(ϱ)|≤D22M∥χγ∥∞,∀γ>γ0,
which implies that [0,τγ)=Iu,  for  all  γ>γ0 and (77) implies that
(87)limγ→∞∥xu○sγ-h∥∞,Iu=0.
On the other hand, the continuity of xγ implies that ∥xu○sγ∥∞,Iu=∥xγ∥∞,  for all  γ>γ0 (see Lemma 3). Thus, there exists some E>0 and γ1>γ0 with
(88)∥xu○sγ∥∞=∥xγ∥∞≤E,∀γ>γ1.
Moreover, we get from (78) and (81) for all γ>γ1 that
(89)x˙u○sγ-h˙=(Axu○sγ+Bψu+C)χγ+A(xu○sγ-h)(H○vu).
Thus, by the boundedness of functions u˙ and ψu, and the relation (88), there exist positive constants D3 and D4 independent of γ, such that
(90)|x˙u○sγ-h˙|≤D3∥χγ∥∞+D4∥xu○sγ-h∥∞,∀γ>γ1,
which means that xu○sγ converges to h in W1,∞(Iu,ℝ) as γ→∞ because of (77) and (87). Define qu∈C0(ℝ+,ℝ) as qu=h○ρu. Since for all q˙u=|u˙|h˙○ρu, relations (78)-(79) imply for all t≥0 that
(91)qu(t)=x0+∫0t(Aqu(τ)+Bu(τ)+C)g*(u˙(τ))dτ.
Moreover, using the relation σγ=xu○sγ○ρu, it can be easily verified that limγ→∞∥σγ-qu∥1,∞=0.Remark 28.
In Lemma27, the sequence of functions σγ converges in W1,∞(ℝ+,ℝ) as γ→∞. This result is stronger than the one obtained in Lemma 20, where the convergence is only in L∞(ℝ+,ℝ). An application to this stronger convergence is given in the following example.Example 29.
The LuGre model is described by [10] as follows:
(92)x˙=u˙-θ|u˙|ζ(u˙)x=[1x][u˙-θ|u˙|ζ(u˙)]=f(x)g(u˙),x(0)=x0,(93)F=θx+c1x˙+c2u˙,
where parameters θ, c1, and c2>0 are, respectively, the stiffness, damping, and viscous friction coefficients, x∈ℝ is the average deflection of the bristles, x0∈ℝ is the initial state, u∈W1,∞(ℝ+,ℝ) is the relative displacement and is the input of the system, and F is the friction force and is the output of the system. The function ζ:ℝ→ℝ is defined as
(94)ζ(α)=FC+(FS-FC)e-|α/vs|,∀α∈ℝ,
where FC>0 is the Coulomb friction force, FS>0 is the stiction force, and vs>0 is the Stribeck velocity.
The LuGre model can be written in the form of the system (65) with A=[a1a2]=[01], B=[00], and C=[10]. We have
(95)limκ→0+g(κ)|κ|=G+*=[G+,1*G+,2*]=[1-θFS],limκ→0-g(κ)|κ|=G-*=[G-,1*G-,2*]=[-1-θFS].
Clearly, conditions (67) and (68) are satisfied. Thus, Lemma 27 implies that ∥σγ-qu∥1,∞→0, as γ→∞, where the functions σγ:ℝ+→ℝ and qu∈W1,∞(ℝ+,ℝ) are defined for all t≥0 as
(96)σγ(t)=x0+∫0t(u˙(τ)-θ|u˙(τ)|ζ(u˙(τ)/γ)qu(τ))dτ,(97)qu(t)=x0+∫0t(u˙(τ)-θ|u˙(τ)|FSqu(τ))dτ.
Also, there exist some E, γ1>0 such that for all γ>γ1, the solution of (92) is global with ∥xγ∥∞≤E.Now, the following analysis is not a part of Lemma27, but it follows straightforwardly from it.LetFγ be the output of the system when we use the input u○sγ instead of u. We obtain from (93) for almost all t≥0 that
(98)Fγ(t)=θxγ(t)+c1x˙γ(t)+c2γu˙○sγ(t),∀γ>0,
which leads to
(99)σγ,F(t)=θσγ(t)+c1σ˙γ(t)γ+c2γu˙(t),∀γ>0,
where σγ,F:ℝ+→ℝm is defined as σγ,F(t)=Fγ(γt), for all γ>0 and for all  t≥0.Since∥σγ-qu∥1,∞→0, as γ→∞, we have limγ→∞∥σγ-qu∥∞=0 and limγ→∞∥σ˙γ-q˙u∥∞=0. Thus, we get from the boundedness of u˙ and (99) that
(100)limγ→∞∥σγ,F(t)-θqu∥∞=0,
which means that the operator which maps input u and initial state x0 to output F is consistent.The conclusion of the analysis is that the hysteresis loop of the LuGre model is{(θq˙u(t),u(t)),t≥0}, where qu is given in (97). Observe that this conclusion has been obtained due to the convergence of σγ in W1,∞(ℝ+,ℝ) (see Remark 28).Simulations. Take θ=4 N/m, vs=0.001m/s, FS=3N, FC=1N, c1=1N·s/m,  c2=1N·s/m,  and x(0)=0 N. Let u∈W1,∞(ℝ+,ℝ) be a function of period 2 that is measured in meters such that u(t)=t, for all t∈[0,1] s, and u(t)=2-t, for all t∈[1,2]s. Figure 4(a) shows the uniform convergence of σγ,F to θqu as γ→∞. Figure 4(b) shows that the graphs {(σγ(t),u(t)),t≥0} converge to the hysteresis loop {(θqu(t),u(t)),t≥0}. Figures 4(c) and 4(d) show that the sequence of functions σγ and σ˙γ converges uniformly to qu and q˙u as γ→∞. Figures 4(e) and 4(f) give the graphs of the functions qu and q˙u.(a)σγ,F(t) versus t. (b) σγ,F(t) versus u(t). (c) σγ(t) versus t. (d) σ˙γ(t) versus t. (e) qu(t) versus t. (f) q˙u(t) versus t. The plots are for system (92) of Example 29. In each plot, qu and q˙u are the solid black lines.
(a)
(b)
(c)
(d)
(e)
(f)


## 6.1. Classλ∈(0,1) Functions

In this subsection, sufficient conditions for the uniform convergence ofσγ as γ→∞, are derived when the function g is of class λ∈(0,1).Definition 16 (see [25]).
A continuous functionβ:ℝ+→ℝ+ is said to belong to class 𝒦∞ if it is increasing,   satisfies β(0)=0,  and limt→∞β(t)=∞.The following lemma generalizes Theorem4.18 in [25, page 172]. Indeed, in [25], continuous differentiability is needed, while in Lemma 17, we only need absolute continuity. Also, in [25], the inequality on the derivative of the Lyapunov function is needed everywhere, while in Lemma 17 it is needed only almost everywhere.Lemma 17.
Consider a functionz:[0,ω)⊆ℝ+→ℝ+, where ω>0 is finite or infinite. Assume the following.(1)
The functionz is absolutely continuous on each compact interval of [0,ω).(2)
There existz1≥0 and z2>0 such that z1<z2, z(0)<z2 and
(32)z˙(t)≤0foralmostallt∈[0,ω)thatsatisfyz1<z(t)<z2.Then, z(t)≤max(z(0),z1),for all  t∈[0,ω).Proof.
see AppendixB.Example 18.
We want to study the stability of the following system(33)x˙(t)=-x3(t)+u(t),x(0)=x0,
where x0 and state x take values in ℝ, and input u∈W1,∞(ℝ+,ℝ). System (33) has an absolutely continuous solution that is defined on an interval of the form [0,ω) [21, page 4].
Letz:[0,ω)→ℝ+ be such that z(t)=x2(t),  for  all  t∈[0,ω). The function z is absolutely continuous on each compact subset of [0,ω) because x is absolutely continuous. Thus, condition (1) in Lemma 17 is satisfied.
We have for almost allt∈[0,ω) that
(34)z˙(t)=2x(t)·x˙(t)=2x(t)(-x3(t)+u(t))≤-2z2(t)+2∥u∥∞z(t).
Thus,
(35)z˙(t)≤0foralmostallt∈[0,ω)thatsatisfy∥u∥∞2/3<z(t).
Therefore, condition (2) in Lemma 17 is satisfied with z1=∥u∥∞2/3 and z2 can be any positive real number such that z2>max(z(0),z1)=max(x02,∥u∥∞2/3). Thus, we deduce from Lemma 17 that z(t)≤max(z(0),∥u∥∞2/3)=max(x02,∥u∥∞2/3),for allt∈[0,ω), and hence |x(t)|≤max(|x0|,∥u∥∞3), for all  t∈[0,ω).Corollary 19.
Consider a functionz:[0,ω)⊆ℝ+→ℝ+, where ω>0 may be infinite. Assume the following.(1)
The functionz is absolutely continuous on each compact subset of [0,ω).(2)
There exist a class𝒦∞  function β:ℝ+→ℝ+ and constants z1≥0, z2>0, and z3≥0 such that max{β-1(z3),z1,z(0)}<z2, and
(36)z˙(t)≤-β(z(t))+z3foralmostallt∈[0,ω)thatsatisfyz1<z(t)<z2.Then, z(t)≤max{z(0),z1,β-1(z3)},for all  t∈[0,ω).Proof.
We have from (36) that
(37)z˙(t)≤0for    almost    allt∈[0,ω)that  satisfymax{β-1(z3),z1}<z(t)<z2,
and hence the result follows directly from Lemma 17.Although the latter corollary follows immediately from Lemma17, it is useful in many situations [25].Lemma 20.
Suppose that the system (1)-(2) has a unique solution and that the function g is of class λ∈(0,1). Assume that there exists qu∈W1,∞(ℝ+,ℝm) such that for almost all t≥0(38)qu(0)=x0,f(qu(t),u(t))g*(u˙(t))=0.
For all γ>0, define  yγ:ℝ+→ℝm as
(39)yγ(t)=σγ(t)-qu(t)=xγ(γt)-qu(t),
for all t∈[0,ωγ), where [0,ωγ) is the maximal interval of existence of solution σγ in (5). Suppose that we can find a continuously differentiable function V:ℝm→ℝ+ such that(1)
there exists a functionδ1:ℝ+→ℝ+ that satisfies
(40)δ1(γ)⟶0asγ⟶∞,(2)
there exist constantsγ*,δ2>0, continuous functions R1,R2:ℝ+→ℝ+ and 𝒦∞ class functions β1,β2,β3:ℝ+→ℝ+, satisfying
(41)β1(|α|)≤V(α)≤β2(|α|),∀α∈ℝm,(42)dV(α)dα|α=yγ(t)·f(yγ(t)+qu(t),u(t))g(u˙(t)γ)≤-1γλβ3(|yγ(t)|)+1γR1(|yγ(t)|)foralmostallt∈[0,ωγ),∀γ>γ*thatsatisfyδ1(γ)<|yγ(t)|<δ2,(43)|dV(α)dα|≤R2(|α|),∀α∈ℝm.
Then,(i)
there existE,γ*>0 such that for all γ>γ*:  ωγ=+∞ and ∥xγ∥∞≤E, where xγ is given in (3)-(4).(ii)
limγ→∞∥σγ-qu∥∞=0.Proof.
From (5) and (39), we get for all γ>0 and almost all t∈[0,ωγ) that yγ(0)=0 and
(44)y˙γ(t)=γf(yγ(t)+qu(t),u(t))g(u˙(t)γ)-q˙u(t).
For any γ>0, define Vγ:(0,ωγ]→ℝ+ as Vγ(t)=V(yγ(t)),for all t∈(0,ωγ]. Note that function Vγ is absolutely continuous on each compact subset of [0,ωγ) as a composition of a continuously differentiable function V and an absolutely continuous function yγ. The derivative of V along with trajectories (44) is given for almost all t∈[0,ωγ) and all γ>0 by
(45)V˙γ(t)=dV(α)dα|α=yγ(t)·y˙γ(t)=dV(α)dα|α=yγ(t)·[γf(yγ(t)+qu(t),u(t))-q˙u(t)].
By (40), there exists some γ1>γ*, such that β2○δ1(γ)<β1(δ2), for all γ>γ1. Let Ωγ=(β2○δ1(γ),β1(δ2)). By (41), we have for any γ>γ1, for almost all t∈[0,ωγ) that
(46)Vγ(t)∈Ωγ⟹δ1(γ)<|yγ(t)|<δ2.
Thus, we deduce from (42), (43), (45), and (46) that
(47)V˙γ(t)≤-γ1-λβ3(|yγ(t)|)+R1(|yγ(t)|)+∥q˙u∥∞R2(|yγ(t)|),foralmostallt∈[0,ωγ),∀γ>γ1thatsatisfyVγ(t)∈Ωγ.
Therefore, (46) and the continuity of the functions R1  and  R2 imply that there exists a constant a>0 that does not depend on γ, such that
(48)V˙γ(t)≤-γ1-λβ3(|yγ(t)|)+a,for  almost  allt∈[0,ωγ),∀γ>γ1thatsatisfyVγ(t)∈Ωγ.
Thus, we deduce from (41) that
(49)V˙γ(t)≤-βγ(Vγ(t))+a,foralmostallt∈[0,ωγ),∀γ>γ1thatsatisfyVγ(t)∈Ωγ,
where βγ:ℝ+→ℝ+ is defined as βγ=γ1-λβ3○β2-1.
On the other hand, sinceλ∈(0,1), there exists γ2>γ1 such that βγ-1(a)=β2○β3-1(aγλ-1)<β1(δ2),for  allγ>γ2. Hence, Corollary 19 and the fact that Vγ(0)=0,forallγ>0, imply that Vγ(t)≤max{βγ-1(a),β2○δ1(γ)}=max{β2○β3-1(aγλ-1),β2○δ1(γ)},forallt∈[0,ωγ)and  for  allγ>γ2. Therefore, (41) implies for allγ>γ2 and for allt∈[0,ωγ) that
(50)|yγ(t)|≤max{β1-1○β2○β3-1(aγλ-1),β1-1○β2○δ3(γ)}.
Thus, ωγ=+∞,for allγ>γ2. Furthermore, (40), (50), and the fact that λ∈(0,1) imply that ∥yγ∥∞=∥σγ-qu∥∞→0 as γ→∞. This proves the consistency with respect to (u,x0) because of Proposition 5.
Moreover, by (50), there exist some E>0,γ*>γ2, such that
(51)∥σγ∥∞≤E,∀γ>γ*.
On the other hand, let γ>γ*. Since xγ is continuous, Lemma 3 ensures that ∥xu○sγ∥∞,Iu=∥xγ∥∞. Let ϱ∈Iu. Due to the continuity of ρu, there exists some t≥0 such that ϱ=ρu(t) and thus (51) and the continuity of σγ lead to |xu○sγ(ϱ)|=|xu○sγ○ρu(t)|=|σγ(t)|≤∥σγ∥∞≤E,  for  allγ>γ*. Therefore, ∥xu○sγ∥∞=∥xγ∥∞≤E,for  allγ>γ*, which completes the proof.Remark 21.
Forλ∈(0,1),  if the function qu∈C0(ℝ+,ℝm)∩L∞(ℝ+,ℝm) in Lemma 20, such that qu=R(u) for some R:ℝ→ℝm, then the graphs {(σγ(t),u(t)),t≥0} converge to the curve R as γ→∞. Hence, (1)-(2) is not a hysteresis because the hysteresis loop cannot be a function [24]. This fact is illustrated in Example 22.Example 22.
Consider the following semilinear Duhem model:(52)x˙=(Ax+Bu)g(u˙),x(0)=x0,
where A is a Hurwitz m×m matrix (i.e., every eigenvalue of A has a negative real part), vector B and state x take values in ℝm. The right-hand side of (52) is Lipschitz and thus the system has a unique solution [21]. Take an input u∈W1,∞(ℝ+,ℝ) such that A-1Bu(0)=-x0 and that |u˙(t)|≥M for almost all t∈ℝ and for some M>0. Assume that the function g:ℝ→ℝ+ is of class λ∈(0,1) and that d+λg(0),d-λg(0)>0. Thus, there exists L>0 such that g*(ϑ)≥L|ϑ|λ, for all ϑ∈ℝ, where the function g* is defined in (9). On the other hand, Proposition 9 states that limγ→∞∥γλg(u˙/γ)-g*(u˙)∥∞=0. This means that there exists γ1>0 such that we get for almost all t≥0, and all γ>γ1 that
(53)γλg(u˙(t)γ)>g*(u˙(t))-LMλ2.
Thus, the facts that |u˙|≥M and g*(·)≥L|·|λ imply that
(54)g(u˙(t)γ)>LMλ2γλ,for  almost  allt≥0,∀γ>γ1.
The function qu∈W1,∞(ℝ+,ℝ) which is defined as qu=-A-1Bu satisfies (38) because A-1Bu(0)=-x0.
SinceA is Hurwitz, there exists a m×m positive-definite matrix P such that [25, page 136]
(55)PA+ATP=-I,
where I is the identity matrix. Consider the continuously differentiable quadratic Lyapunov function candidate V:ℝm→ℝ such that V(α)=αTPα, for all α∈ℝm. Since P is symmetric, we have for all α∈ℝm that
(56)λmin(P)|α|2≤V(α)=αTPα≤λmax(P)|α|2,
where λmax(P) and λmin(P) are, respectively, the maximum and minimum eigenvalues of the matrix P. This shows that (41) is satisfied with β1(η)=λmin(P)η2 and β2(η)=λmax(P)η2 for all η>0. Since P is symmetric, we have the following matrices derivation:
(57)dV(α)dα=(P+PT)α=2Pα,∀α∈ℝm.
Thus, we get
(58)|dV(α)dα|=2|Pα|≤2|P||α|,∀α∈ℝm,
where |P| is the induced 2-norm for the matrix P and hence (43) is satisfied with R2(η)=2|P|η,  for all  η≥0. From (57), we have  for  all  α∈ℝm that
(59)dV(α)dα·Aα=2Pα·Aα=αT(PA+ATP)α=-|α|2.
Therefore, (54) implies that for almost all t∈ℝ+ and γ>0 that
(60)dV(α)dα|α=yγ(t)·[A(yγ(t)+qu(t))+Bu(t)]g(u˙(t)γ)=-g(u˙(t)γ)|yγ(t)|2≤-LMλ2γλ|yγ(t)|2,
where yγ is defined in (39). Thus, (42) is satisfied with β3(η)=(LMλ/2)η2, for all η≥0 and R1(η)=0, for all η≥0.
Letδ1 be the zero function. Then (40) is verified. Take δ2,γ* arbitrary in ℝ+ (say δ2=1, γ*=1). Hence, all conditions of Lemma 20 are satisfied. Thus, it follows from Lemma 20 that there exist some E,γ*>0 such that for all γ>γ*, the solution of (52) is global with |xγ(t)|≤E, for all t≥0. Moreover, the operator which maps (u,x0) to x is consistent. In particular, we have ∥σγ-qu∥∞=∥σγ+A-1Bu∥∞→0 as γ→∞.
As a conclusion, the graphs{(σγ(t),u(t)),t≥0} converge to the graph of the linear function R:ℝ→ℝm, which is defined as R(α)=-A-1Bα, for all α∈ℝ. This means that for λ∈(0,1), the model (52) does not represent a hysteresis (see Remark 21).Simulations. Take m=1, B=1.0, A=-1.0, and x0=0. Let g(ϑ)=|ϑ|, for all ϑ∈ℝ, then d+λg(0)=d-λg(0)=Γ(3/2)=Γ(1/2)/2=π/2>0. Let u∈W1,∞(ℝ+,ℝ) be the function of period 2 such that u(t)=t, for all t∈[0,1], and u(t)=2-t, for all t∈[1,2]. Then, we have |u˙(t)|=1, for  almost  all t≥0. We also have qu=-A-1Bu=u. Figure 2(a) shows that the graph {(σγ(t),u(t)),t≥0} collapses into the identity function when γ→∞. This happens because of the fact that qu=u and Remark 21. Figure 2(b) shows that the sequence of functions σγ converges uniformly to qu=u as γ→∞.(a)σγ(t) versus u(t) for different values of γ. (b) σγ(t) versus t for different values of γ. In each plot, the function qu is the solid line.
(a)
σ
γ
(
t
) versus u(t) for system  (52)
(b)
σ
γ
(
t
) versus t for system  (52)


## 6.2. Classλ=1 Functions

In this subsection, we consider classλ=1 functions. (A function g∈C0(ℝ,ℝr) is of class λ=1 if g(0)=0 and the limits limκ→0+g(κ)/|κ| and limκ→0-g(κ)/|κ| exist, are finite, and at least one of them is nonzero (see Section 5)). The main results of this subsection are given in Lemmas 23 and 27.Lemma 23.
Assume the following.(1)
The system (1)-(2) has a unique global solution.(2)
For the functiong in system (1)-(2), there exist a1,a2∈ℝr such that
(61)g(ϑ)={a1ϑϑ≥0,a2ϑϑ<0.Then, the sequence of functions σγ of (5) is independent of γ and the operator which maps (u,x0) to x is consistent.Proof.
By condition(2), the right-hand side of (5) is independent of γ. Thus, the solution σγ of (5) is independent of γ. Since xu○sγ○ρu=σγ, the function xu○sγ is also independent of γ (this is the so-called “rate-independent hysteresis”) and hence consistency holds.Example 24.
Consider Bouc's hysteresis model [13] as follows:
(62)x˙=-c|u˙|x+Φ′(u)u˙,
where c>0, Φ∈C1(ℝ,ℝ), input u∈W1,∞(ℝ+,ℝ), and Φ′(u)=dΦ(α)/dα|α=u.
The right-hand side of (62) is Lipschitz with respect to x. Thus, the system has a unique solution. Furthermore, we have
(63)x˙≤-|Φ′(u)u˙|+Φ′(u)u˙≤0,whenx≥1c|Φ′(u)|,x˙≥|Φ′(u)u˙|+Φ′(u)u˙≥0,whenx≤-1c|Φ′(u)|.
Thus, |x|≤max{|Φ′(u)|/c,|x(0)|}, for each input u∈W1,∞(ℝ+,ℝ) and each initial state x(0)∈ℝ. Since u is bounded and Φ′(u) is continuous, the solution x of (62) is bounded and hence global. Hence, condition (1) in Lemma 23 is satisfied. Equation (62) can be written as
(64)x˙=f(x,u)g(u˙)=[-cx+Φ′(u)cx+Φ′(u)][max{0,u˙}min{0,u˙}].
Clearly, the function g is of class λ=1 and satisfies condition (2) in Lemma 23. This fact implies that the operator which maps (u,x(0)) to x is consistent and σγ is independent of γ.Simulations. Let c=1, Φ(α)=α3/3, for  all  α∈ℝ,  x(0)=0, and input u(t)=sin(t),forallt≥0. The function σγ is independent of γ and is plotted in Figure 3(a). Furthermore, Figure 3(b) shows a rate-independent hysteresis behavior; that is graphs {(σγ(t),u(t)),t≥0} are the same for different values of γ.(a)σγ(t) versus t. (b) σγ(t) versus u(t) for Boucs hysteresis model (62).
(a)
σ
γ
(
t
) versus t
(b)
σ
γ
(
t
) versus u(t)Proposition 25.
Letu∈W1,∞(ℝ+,ℝ) be nonconstant. There exists a unique function vu∈L∞(Iu,ℝ) that is defined by vu○ρu=u˙. Moreover, ∥vu∥∞,Iu≤∥u˙∥∞. Assume that u˙ is nonzero on a set A⊆ℝ that satisfies μ(ρu(ℝ∖A))=0. Then, vu is nonzero almost everywhere.Proof.
Consider the left-derivative operatorΔ- defined on W1,∞(ℝ+,ℝ) by [Δ-(u)](t)=limτ↑t((u(τ)-u(t))/(τ-t)). The operator Δ- is causal as [Δ-(u)](t) depends only on values of u(τ) for τ≤t, and we have Δ-(u)=u˙ almost every where as u∈W1,∞(ℝ+,ℝ) so that Δ-(u)∈L∞(ℝ+,ℝ). The operator Δ- satisfies Assumption 2. The first part of Proposition 25 follows immediately from Lemma 3. Now, let B={ϱ∈Iu/vu(ϱ)=0}, then B⊆ρu(ℝ∖A) which implies that μ(B)=0.Remark 26.
Observe that ifu˙  is nonzero almost everywhere, then μ(ℝ∖A)=0 so that by [26] we have μ(ρu(ℝ∖A))=0 as ρu is absolutely continuous. An example in which u˙ does not need to be nonzero almost everywhere, is when u is constant on some interval, or on a finite number of intervals, or an infinite number of intervals such that this infinite number has measure zero (e.g., countable).Lemma 27.
Letu∈W1,∞(ℝ+,ℝ) be such that u˙ is nonzero on a set A⊆ℝ that satisfies μ(ρu(ℝ∖A))=0. Consider the semilinear Duhem model with m=1 and λ=1  as follows:
(65)x˙=(Ax+Bu+C)g(u˙),x(0)=x0,
where A=[a1a2⋯ar]≠0, B, and C are 1×r row vectors, state x∈ℝ and function g∈C0(ℝ,ℝr) and of class λ=1, and nonconstant input u∈W1,∞(ℝ+,ℝ). Denote
(66)limκ→0+g(κ)|κ|=G+*=[G+,1*G+,2*⋮G+,r*],limκ→0-g(κ)|κ|=G-*=[G-,1*G-,2*⋮G-,r*].
For any i∈{1,2,…,r}, assume that
(67)G+,i*,G-,i*≥0,wheneverai<0,G+,i*,G-,i*≤0,wheneverai>0.
Suppose that there exists some i0∈{1,2,…,r}, such that
(68)ai0≠0,|G+,i0*|+|G-,i0*|>0.
Then,(i)
there existE,γ1>0 such that ∥xγ∥∞≤E,  for all  γ>γ1  (xγ is the output of the system (65) when we use the input u○sγ instead of the input u (see system (3)-(4))).(ii)
There exists a functionqu∈W1,∞(ℝ+,ℝ) such that ∥σγ-qu∥1,∞→0, as γ→∞ where ∥·∥1,∞ is the norm of the Banach space W1,∞(ℝ+,ℝm) (and hence the consistency of the system (65) is guaranteed with respect to input u and initial condition x0).(iii)
qu(0)=x0. Furthermore, for almost all t≥0, one has
(69)q˙u(t)=(Aqu(t)+Bu(t)+C)g*(u˙(t)),where the functiong*∈C0(ℝ,ℝ) is defined as in (9), that is;
(70)g*(ϑ)={ϑG+*ϑ≥0,-ϑG-*ϑ<0.Proof.
The semilinear Duhem model has a unique global solution due to the Lipschitz property of the right-hand-side [21, page 5].
For anyi∈{1,2,…,r} and any κ∈ℝ∖{0}, let Gi(κ) be the ith component of the function g(κ)/|κ|. From (66), there exists some constant ζ>0, such that for all  i∈{1,2,…,r}(71)G+,i*-|G+,i*|2<Gi(κ)<G+,i*+|G+,i*|2,wheneverκ∈(0,ζ),G-,i*-|G-,i*|2<Gi(κ)<G-,i*+|G-,i*|2,wheneverκ∈(-ζ,0).
Let i∈{1,2,…,r}. We get from (67) and (71) that
(72)aiGi(κ)≤-max{|ai||G+,i*|2,|ai||G-,i*|2}=-Mi, wheneverκ∈(-ζ,ζ)∖{0},ai≠0.
For the case ai=0, take Mi=0. Let M=∑i=1r‍Mi. From (67) and (68), we get M>0. Thus, we get from (72) that
(73)Ag(κ)|κ|=∑i=1r‍aiGi(κ)≤-M<0,∀κ∈(-ζ,ζ)∖{0}.
By Proposition 25, a function vu∈L∞(Iu,ℝ) can be defined almost every where as vu○ρu=u˙ with vu≠0 almost every where The boundedness of vu implies that there exists γ0>0 such that vu(ϱ)/γ∈(-ζ,ζ) for almost all ϱ∈Iu and all γ>γ0. Thus, we deduce from (73) that for all γ>γ0,
(74)γAg(vu(ϱ)/γ)|vu(ϱ)|≤-M,for  almost  allϱ∈Iu.
Let H:ℝ→ℝr be a function such that
(75)H(κ)={G+*κ>0,G-*κ<0.
For any γ>0, define χγ:Iu→ℝr as
(76)χγ(ϱ)={γg(vu(ϱ)/γ)|vu(ϱ)|-H(vu(ϱ))vu(ϱ)≠0,0otherwise.
Since vu∈L∞(Iu,ℝ), relations (66) imply that
(77)limγ→∞∥χγ∥∞=0.
This result can be easily checked using the same techniques used in the proof of Proposition 9. Now, consider the system
(78)h˙=(Ah+Bψu+C)(H○vu),(79)h(0)=x0,
where state h∈ℝ. The differential equation (78) verifies Carathéodory conditions with Lipschitz property with respect to h. Thus, system (78)-(79) has a unique and absolutely continuous local solution [21, page 4]. Consider the Lyapunov function W=h2. We deduce from (78)-(79) that W(0)=x02 and that
(80)W˙=2A(H○vu)W+2(Bψu+C)(H○vu)W.
We get from (67) and (68) that AH(vu(ϱ))≤-∑i=1r‍max{|ai||G+,i*|,|ai||G-,i*|}=-2M for almost all ϱ∈Iu. Thus, the boundedness of the function ψu (see Lemma 1) along with (80) imply that W˙≤-4MW+D1W, for some D1>0. This leads to W˙≤0, whenever W≥(D1/4M)2. Therefore, Lemma 17 and the fact that W(0)=x02 imply that W≤max{x02,(D1/4M)2} which means that the solution of the system (78)-(79) is bounded and hence is global (i.e., is defined on Iu). On the other hand, the relation σγ=xu○sγ○ρu implies that σ˙γ=|u˙|x˙u○sγ○ρu. Thus, we obtain from systems (65) and (5) and the relations vu○ρu=u˙, and ψu○ρu=u, that xu○sγ(0)=x0 and that
(81)x˙u○sγ(ϱ)=γ(Axu○sγ(ϱ)+Bψu(ϱ)+C)g(vu(ϱ)/γ)|vu(ϱ)|
for all γ>0, for almost all ϱ∈[0,τγ)⊆Iu, where [0,τγ) is the maximal interval of existence [21, page 4]. For any γ>0, let yγ:Iu→ℝ be defined as yγ=xu○sγ-h. Since xu○sγ(0)=h(0)=x0, the system (81) can be written for all γ>0, for almost all ϱ∈[0,τγ) as
(82)y˙γ(ϱ)=γAg(vu(ϱ)/γ)|vu(ϱ)|yγ(ϱ)+(Ah(ϱ)+Bψu(ϱ)+C)χγ(ϱ),(83)yγ(0)=0.
For any γ>0, consider the Lyapunov function Vγ:[0,τγ)→ℝ+ with Vγ(ϱ)=yγ2(ϱ),  for all  ϱ∈[0,τγ). By (82) and the boundedness of both ψu and the solution of (78)-(79), there exists some D2>0 such that for almost all ϱ∈[0,τγ) and all γ>γ0,
(84)V˙γ(ϱ)≤2γAg(vu(ϱ)/γ)|vu(ϱ)|Vγ(ϱ)+D2∥χγ∥∞Vγ(ϱ).
Thus, we get from (74) that
(85)V˙γ(ϱ)≤0,for    almost    allϱ∈[0,τγ),∀γ>γ0,that    satisfyVγ(ϱ)>(D2∥χγ∥∞2M)2.
Therefore, Lemma 17 and the fact that Vγ(0)=0 imply that Vγ(ϱ)≤(D2∥χγ∥∞/2M)2 for all γ>γ0 and almost all ϱ∈[0,τγ), and hence we obtain, for almost all ϱ∈[0,τγ) that
(86)|yγ(ϱ)|=|xu○sγ(ϱ)-h(ϱ)|≤D22M∥χγ∥∞,∀γ>γ0,
which implies that [0,τγ)=Iu,  for  all  γ>γ0 and (77) implies that
(87)limγ→∞∥xu○sγ-h∥∞,Iu=0.
On the other hand, the continuity of xγ implies that ∥xu○sγ∥∞,Iu=∥xγ∥∞,  for all  γ>γ0 (see Lemma 3). Thus, there exists some E>0 and γ1>γ0 with
(88)∥xu○sγ∥∞=∥xγ∥∞≤E,∀γ>γ1.
Moreover, we get from (78) and (81) for all γ>γ1 that
(89)x˙u○sγ-h˙=(Axu○sγ+Bψu+C)χγ+A(xu○sγ-h)(H○vu).
Thus, by the boundedness of functions u˙ and ψu, and the relation (88), there exist positive constants D3 and D4 independent of γ, such that
(90)|x˙u○sγ-h˙|≤D3∥χγ∥∞+D4∥xu○sγ-h∥∞,∀γ>γ1,
which means that xu○sγ converges to h in W1,∞(Iu,ℝ) as γ→∞ because of (77) and (87). Define qu∈C0(ℝ+,ℝ) as qu=h○ρu. Since for all q˙u=|u˙|h˙○ρu, relations (78)-(79) imply for all t≥0 that
(91)qu(t)=x0+∫0t(Aqu(τ)+Bu(τ)+C)g*(u˙(τ))dτ.
Moreover, using the relation σγ=xu○sγ○ρu, it can be easily verified that limγ→∞∥σγ-qu∥1,∞=0.Remark 28.
In Lemma27, the sequence of functions σγ converges in W1,∞(ℝ+,ℝ) as γ→∞. This result is stronger than the one obtained in Lemma 20, where the convergence is only in L∞(ℝ+,ℝ). An application to this stronger convergence is given in the following example.Example 29.
The LuGre model is described by [10] as follows:
(92)x˙=u˙-θ|u˙|ζ(u˙)x=[1x][u˙-θ|u˙|ζ(u˙)]=f(x)g(u˙),x(0)=x0,(93)F=θx+c1x˙+c2u˙,
where parameters θ, c1, and c2>0 are, respectively, the stiffness, damping, and viscous friction coefficients, x∈ℝ is the average deflection of the bristles, x0∈ℝ is the initial state, u∈W1,∞(ℝ+,ℝ) is the relative displacement and is the input of the system, and F is the friction force and is the output of the system. The function ζ:ℝ→ℝ is defined as
(94)ζ(α)=FC+(FS-FC)e-|α/vs|,∀α∈ℝ,
where FC>0 is the Coulomb friction force, FS>0 is the stiction force, and vs>0 is the Stribeck velocity.
The LuGre model can be written in the form of the system (65) with A=[a1a2]=[01], B=[00], and C=[10]. We have
(95)limκ→0+g(κ)|κ|=G+*=[G+,1*G+,2*]=[1-θFS],limκ→0-g(κ)|κ|=G-*=[G-,1*G-,2*]=[-1-θFS].
Clearly, conditions (67) and (68) are satisfied. Thus, Lemma 27 implies that ∥σγ-qu∥1,∞→0, as γ→∞, where the functions σγ:ℝ+→ℝ and qu∈W1,∞(ℝ+,ℝ) are defined for all t≥0 as
(96)σγ(t)=x0+∫0t(u˙(τ)-θ|u˙(τ)|ζ(u˙(τ)/γ)qu(τ))dτ,(97)qu(t)=x0+∫0t(u˙(τ)-θ|u˙(τ)|FSqu(τ))dτ.
Also, there exist some E, γ1>0 such that for all γ>γ1, the solution of (92) is global with ∥xγ∥∞≤E.Now, the following analysis is not a part of Lemma27, but it follows straightforwardly from it.LetFγ be the output of the system when we use the input u○sγ instead of u. We obtain from (93) for almost all t≥0 that
(98)Fγ(t)=θxγ(t)+c1x˙γ(t)+c2γu˙○sγ(t),∀γ>0,
which leads to
(99)σγ,F(t)=θσγ(t)+c1σ˙γ(t)γ+c2γu˙(t),∀γ>0,
where σγ,F:ℝ+→ℝm is defined as σγ,F(t)=Fγ(γt), for all γ>0 and for all  t≥0.Since∥σγ-qu∥1,∞→0, as γ→∞, we have limγ→∞∥σγ-qu∥∞=0 and limγ→∞∥σ˙γ-q˙u∥∞=0. Thus, we get from the boundedness of u˙ and (99) that
(100)limγ→∞∥σγ,F(t)-θqu∥∞=0,
which means that the operator which maps input u and initial state x0 to output F is consistent.The conclusion of the analysis is that the hysteresis loop of the LuGre model is{(θq˙u(t),u(t)),t≥0}, where qu is given in (97). Observe that this conclusion has been obtained due to the convergence of σγ in W1,∞(ℝ+,ℝ) (see Remark 28).Simulations. Take θ=4 N/m, vs=0.001m/s, FS=3N, FC=1N, c1=1N·s/m,  c2=1N·s/m,  and x(0)=0 N. Let u∈W1,∞(ℝ+,ℝ) be a function of period 2 that is measured in meters such that u(t)=t, for all t∈[0,1] s, and u(t)=2-t, for all t∈[1,2]s. Figure 4(a) shows the uniform convergence of σγ,F to θqu as γ→∞. Figure 4(b) shows that the graphs {(σγ(t),u(t)),t≥0} converge to the hysteresis loop {(θqu(t),u(t)),t≥0}. Figures 4(c) and 4(d) show that the sequence of functions σγ and σ˙γ converges uniformly to qu and q˙u as γ→∞. Figures 4(e) and 4(f) give the graphs of the functions qu and q˙u.(a)σγ,F(t) versus t. (b) σγ,F(t) versus u(t). (c) σγ(t) versus t. (d) σ˙γ(t) versus t. (e) qu(t) versus t. (f) q˙u(t) versus t. The plots are for system (92) of Example 29. In each plot, qu and q˙u are the solid black lines.
(a)
(b)
(c)
(d)
(e)
(f)


## 7. Conclusion

This paper presented a classification of the possible Duhem models in terms of their consistency with the hysteresis behavior. Three classes of models have been considered in relation with the range of a parameterλ. For λ>1, it has been shown that the corresponding generalized Duhem model does not represent a hysteresis behavior. For 0<λ<1, it has been shown that the semilinear Duhem model is not compatible with a hysteresis behavior. In all other cases, necessary conditions and sufficient ones have been derived to insure the consistency of the Duhem model with the hysteresis property. Numerical simulations have been carried out to illustrate the obtained results.



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