A Survey of the Hysteretic Duhem Model

Abstract

The Duhem model is a simulacrum of a complex and hazy reality: hysteresis. Introduced by Pierre Duhem to provide a mathematical representation of thermodynamical irreversibility, it is used to describe hysteresis in other areas of science and engineering. Our aim is to survey the relationship between the Duhem model as a mathematical representation, and hysteresis as the object of that representation.

Appendices

Appendix

1.1 On the Existence and Uniqueness of Solutions of Differential Equations

In this section we present some existence and uniqueness theorems for the solutions of ordinary differential equations. To this end, let $\mathcal {D}$ be a domain, that is an open connected subset of $\mathbb {R} \times \mathbb {R}^{n}$ where $n>0$ is an integer. Let $(t_0,x_0) \in \mathcal {D}$ and let $a,b \in \; ]0,\infty [.$ Define the parallelepiped $Q_{a,b}$ by

$$\begin{aligned} Q_{a,b}=\left\{ (t,w) \in \mathbb {R} \times \mathbb {R}^{n} \mid |t-t_0| \le a,|w-x_0| \le b\right\} . \end{aligned}$$

(127)

We say that the map $F:\mathcal {D} \rightarrow \mathbb {R}^{n}$ satisfies the Carathéodory conditions on the domain $\mathcal {D}$ if Conditions (i)–(iii) hold on any parallelepiped $Q_{a,b} \subset \mathcal {D}$ [61, p. 68].

• (i) The function F is defined and continuous in w for almost all t;

• (ii) the function F is measurable in t for each fixed w;

• (iii) for each $Q_{a,b} \subset \mathcal {D}$ there exists a measurable function $m_{Q_{a,b}} \in L^1\big ([t_0-a,t_0+a],\mathbb {R}\big )$ such that

  $$\begin{aligned}&|F(t,w)| \le m_{Q_{a,b}}(t), \; \forall w \in \mathbb {R}^{n} \text { and for almost all } \nonumber \\&t \in [t_0-a,t_0+a] \text { satisfying } (t,w) \in Q_{a,b}. \end{aligned}$$

  (128)

Now, consider the differential equation

$$\begin{aligned} \dot{x}(t)&= F\big (t,x(t)\big ), \end{aligned}$$

(129)

$$\begin{aligned} x(t_0)&= x_0, \end{aligned}$$

(130)

where $F:\mathcal {D} \rightarrow \mathbb {R}^{n}$ satisfies the Carathéodory conditions on the domain $\mathcal {D} \subset \mathbb {R} \times \mathbb {R}^{n}$ and $(t_0,x_0) \in \mathcal {D}$.

Theorem 10

[61, p. 68] The differential equation (129)–(130) has a solution on some nonempty open interval $I \ni t_0,$ in the sense that there exists an absolutely continuous function $x:I\rightarrow \mathbb {R}^{n}$ such that the following properties (i)–(iii) are satisfied.

• (i) The initial condition (130) holds;

• (ii) $\forall t \in I$ we have $\big (t,x(t)\big ) \in \mathcal {D}$;

• (iii) and the differential equation (129) is satisfied almost everywhere in I.

A lower bound on the size of the interval I is obtained by solving the inequality

$$\begin{aligned} \int _{t_0-c}^{t_0+c} m_{Q_{a,b}}(t) \, \text {d}t \le b, \end{aligned}$$

(131)

where $a,b \in \; ]0,\infty [$ are chosen so that $(t_0,x_0) \in Q_{a,b} \subset \mathcal {D}$. Observe that the function $c \rightarrow \int _{t_0-c}^{t_0+c} m_{Q_{a,b}}(t) \, \text {d}t$ is continuous and is zero at $c=0$. This implies that there exists at least a $0<c\le a$ such that (131) holds. Then we have $]t_0-c,t_0+c[ \,\subset I$ [61, p. 69].

Theorem 11

[61, p. 70 and p. 80] Assume that $F:\mathcal {D}\rightarrow \mathbb {R}^{n}$ satisfies the Carathéodory conditions on the domain $\mathcal {D}$. Let x be a solution of the differential equation (129)–(130) defined on some interval I. Then x may be extended as a solution of (129)–(130) to a maximal interval of existence $]\omega _-,\omega _+[$ and $\big (t,x(t)\big ) \rightarrow \partial \mathcal {D}$ as $t \rightarrow \omega _{\pm }$, where $\partial \mathcal {D}$ is the boundary of $\mathcal {D}$.

Theorem 12

[29, p. 5] Assume that $F:\mathcal {D}\rightarrow \mathbb {R}^{n}$ satisfies the Carathéodory conditions on the domain $\mathcal {D}$. Assume that there exists a function $l \in L^1\big (J , \mathbb {R}_+ \big )$ for every finite interval $J \subset \mathbb {R}$ which satisfies the following. For almost all $t \in \mathbb {R}$ and $\forall w_1,w_2 \in \mathbb {R}^{n}$ such that $(t,w_1),(t,w_2) \in \mathcal {D}$ we have

$$\begin{aligned} \big |F(t,w_1)-F(t,w_2)\big | \le l(t)|w_1-w_2|. \end{aligned}$$

(132)

Then in the domain $\mathcal {D}$ there exists at most one solution to the differential equation (129)–(130).

The local Lipschitz condition (132) can be relaxed as follows [29, p. 5].

$$\begin{aligned} \big (F(t,w_1)-F(t,w_2)\big )\cdot&\, (w_1-w_2) \le l_{1}(t)|w_1-w_2|^2, \nonumber \\&\text { for almost all } t \ge t_0, \end{aligned}$$

(133)

$$\begin{aligned} \big (F(t,w_1)-F(t,w_2)\big )\cdot&\,(w_1-w_2) \ge -l_{2}(t)|w_1-w_2|^2, \nonumber \\&\text { for almost all } t \le t_0, \end{aligned}$$

(134)

where the product is understood as the scalar product if $F(t,w_1),F(t,w_2),w_1,w_2$ are vectors; the functions $l_{1},l_{2} \in L^1\big (J , \mathbb {R}_+ \big )$ for every finite interval $J \subset \mathbb {R}$, and $w_1,w_2 \in \mathbb {R}^{n}$ are such that $(t,w_1),(t,w_2) \in \mathcal {D}$.

Finally we provide a result we could not find in the literature, and which is useful to the present paper.

Lemma 12

Suppose that the application $F:\mathbb {R} \times \mathbb {R} \rightarrow \mathbb {R}$ satisfies the Carathéodory conditions on the domain $\mathbb {R}^2$. Assume that there exists $k \in [0,\infty [$ such that

$$\begin{aligned} \big (F(t,w_1)-F(t,w_2)\big )\cdot (w_1-w_2) \le k|w_1-w_2|^2, \nonumber \\ \text {for almost all } t \ge t_0, \forall w_1,w_2 \in \mathbb {R}. \end{aligned}$$

(135)

Then the differential equation (129)–(130) has exactly one solution defined on $[t_0,\infty [$.

Proof

From Theorems 10, 11, and 12 it follows that there exists a unique solution x to the differential equation (129)–(130) defined on a maximal interval of existence $[t_0,\omega _+[$ where $\omega _+ \in \; ]t_0,\infty ]$. Assume that $\omega _+ < \infty$, and let $w \in \mathbb {R}$ be fixed. It comes from Theorem 11 that $\exists \, t_w \in \;]t_0,\omega _+[$ such that $\forall t \in [t_w,\omega _+[$ we have $|x(t)|> |w|$. Consider the case $\forall t \in [t_w,\omega _+[,\,x(t)> |w| \ge w$ (a similar proof holds for the case $\forall t \in [t_w,\omega _+[,\,x(t) <- |w|$). Then Inequality (135) leads to

$$\begin{aligned} F(t,x(t)) \le F(t,w)+k\big (x(t)-w\big ), \text { for almost all } t \in [t_w,\omega _+[. \end{aligned}$$

(136)

Integrating both sides of (136) on the time interval $[t_w,t]$ it follows that

$$\begin{aligned} |x(t)|&=x(t)=x(t_w)+\int _{t_w}^t F(s,x(s))\,\text {d}s \nonumber \\&\le C+k \int _{t_w}^t |x(s)|\,\text {d}s, \forall t \in [t_w,\omega _+[, \\C&= x(t_w)+\int _{t_w}^{\omega _+}|F(s,w)|\,\text {d}s +k|w|(\omega _+-t_w) <\infty . \nonumber \end{aligned}$$

(137)

Using Gronwall’s lemma [32, p. 24] it comes from Inequality (137) that

$$\begin{aligned} |x(t)| \le Ce^{t-t_w} \le Ce^{\omega _+ -t_w}, \forall t\in [t_w,\omega _+[. \end{aligned}$$

(138)

Inequality (138) contradicts the fact that $|x(t)| \rightarrow \infty$ as $t \rightarrow \omega _+$.

Proof of Lemma 13

Lemma 13

Let $u \in W^{1,\infty }(\mathbb {R}_+,\mathbb {R})$ be non constant. There exists a unique function $v_{u}\in L^{\infty }\left( I_{u},\mathbb {R}\right)$ that is defined by $v_{u}\circ \rho _{u}=\dot{u}$. Moreover, $\Vert v_u\Vert _{I_u} \le \Vert \dot{u}\Vert$ and $v_{u}$ is nonzero almost everywhere on $I_u$.

Proof

The operator $\Delta _-$ defined in Section 11.2 is causal and satisfies Assumption 3. Using Lemma 3 it follows that $v_{u}\in L^{\infty }\left( I_{u},\mathbb {R}\right)$ and $\Vert v_u\Vert _{I_u} \le \Vert \dot{u}\Vert$. Now, define the following sets:

$$\begin{aligned} A&=\{\varrho \in I_u \mid v_u(\varrho )=0\}, \\ B&=\{t \in \mathbb {R}_+ \mid \dot{u}(t) =0\}, \\ B_1&=\{t \in \mathbb {R}_+ \mid \dot{\rho }_u(t) \text{ is } \text{ not } \text{ defined } \text{ at } t\}, \\ B_2&=\{t \in \mathbb {R}_+ \mid \dot{u}(t) \text{ is } \text{ defined, } \dot{\rho }_u(t) \text { is defined, and} \\&\quad \quad |\dot{u}(t)| \ne \dot{\rho }_u(t)\}, \\ C&=\{t \in \mathbb {R}_+ \mid \dot{\rho }_u(t) =0\}. \end{aligned}$$

Since $\rho _u$ is absolutely continuous on $\mathbb {R}_+$, we get from [45, Corollary 3.41] that $\mu (B_1)=0$. Since $\dot{u}\in L^{\infty }\left( \mathbb {R}_+,\mathbb {R}\right)$ we get from [45, Lemma 3.31] that $\dot{\rho }_u = |\dot{u}|$ almost everywhere on $\mathbb {R}_+$, which implies that $\mu (B_2)=0$. Also, from [45, Corollary 3.14] it follows that $\mu \big (\rho _u(C)\big )=0.$ Since $\rho _u$ is absolutely continuous on $\mathbb {R}_+$, and since $\mu (B_1)=\mu (B_2)=0$ it follows from [45, Corollary 3.41] that $\mu \big (\rho _u(B_1)\big )=\mu \big (\rho _u(B_2)\big )=0$. Now, observe that $B \subset C \cup B_1 \cup B_2$, thus $\rho _u(B) \subset \rho _u(C) \cup \rho _u(B_1) \cup \rho _u(B_2)$ which implies that $\mu \big (\rho _u(B)\big )=0$. Since $A=\rho _u(B)$ it follows that $\mu (A)=0$.

Proof of Theorem 8

We get from Equation (68) that $\exists \delta _1>0$ such that $\forall w \in (0,\delta _1)$ we have $\left| \bar{g}_1(w)-1\right| <\frac{1}{2}$, and $\exists \delta _2>0$ such that $\forall w \in ( -\delta _2,0)$ we have $\left| \bar{g}_2(w)+1\right| <\frac{1}{2}$. Define

$$\begin{aligned} \gamma _0 = \frac{\Vert \dot{u}\Vert }{\min (\delta _1,\delta _2)}. \end{aligned}$$

(139)

Observe that $0<\gamma _0<\infty$ since $u \in \Lambda _{u_{\min },u_{\max },\alpha _1,T}$. Let $\gamma \in \; ]\gamma _0,\infty [$ be fixed, and define $x_\gamma = \mathcal {H}_s(u \circ s_\gamma ,x_{0})$. From Equations (63) and (64) we get

$$\begin{aligned} {\begin{matrix} x_\gamma (t)={}&{}x_0+\int _0^t g_1\big (\dot{u}_\gamma (\tau )\big ) \big ( A_1 x_\gamma (\tau )+ B_1u_\gamma (\tau )+ E_1 \big ) \\ &{} +g_2\big (\dot{u}_\gamma (\tau )\big ) \big ( A_2 x_\gamma (\tau )+ B_2 u_\gamma (\tau )+ E_2 \big ) \,\text {d}\tau , \forall t \in \mathbb {R}_+ \end{matrix}} \end{aligned}$$

(140)

where $u_\gamma = u \circ s_\gamma$. Consider the change of variable $\tau ^\prime = \frac{\tau }{\gamma }$, then

$$\begin{aligned} {\begin{matrix} x_\gamma (t)={}&{}x_0+\gamma \int _0^{\frac{t}{\gamma }} g_1\left( \frac{\dot{u}(\tau ^\prime )}{\gamma }\right) \big [ A_1 x_\gamma (\gamma \tau ^\prime )+ B_1 u(\tau ^\prime ) + E_1 \big ] \\ &{} +g_2\left( \frac{\dot{u}(\tau ^\prime )}{\gamma }\right) \times \big [ A_2 x_\gamma (\gamma \tau ^\prime )+ B_2 u(\tau ^\prime )+ E_2 \big ] \, \text {d}\tau ^\prime , \\ {} &{} \forall t \in \mathbb {R}_+. \end{matrix}} \end{aligned}$$

(141)

Define $\sigma =\frac{t}{\gamma }$ and $z:\mathbb {R}_+ \rightarrow \mathbb {R}$ by $z(\sigma )=x_\gamma (\gamma \sigma ), \forall \sigma \in \mathbb {R}_+$; then

$$\begin{aligned} {\begin{matrix} z(\sigma )={}&{}x_0+\gamma \int _0^{\sigma } g_1\left( \frac{\dot{u}(\tau ^\prime )}{\gamma }\right) \big (A_1 z(\tau ^\prime )+ B_1 u(\tau ^\prime )+ E_1 \big ) \\ &{} +g_2\left( \frac{\dot{u}(\tau ^\prime )}{\gamma }\right) \big (A_2 z(\tau ^\prime )+ B_2 u(\tau ^\prime )+ E_2 \big ) \, \text {d}\tau ^\prime , \\ &{} \forall \sigma \in \mathbb {R}_+. \end{matrix}} \end{aligned}$$

(142)

For any $m \in \mathbb {N}$ define $z_{m}:[0,T] \rightarrow \mathbb {R}$ by

$$\begin{aligned} z_{m}(\sigma )=z(\sigma +mT),\forall \sigma \in [0,T]. \end{aligned}$$

(143)

The objective of the following analysis is to show that the sequence $\{z_m\}_{m \in \mathbb {N}}$ converges in the Banach space $C^0\left( [0,T],\mathbb {R}\right)$ endowed with the norm $\Vert \cdot \Vert _{[0,T]}$. To this end, we prove that $\{z_m\}_{m \in \mathbb {N}}$ is a Cauchy sequence. For any $m_1,m_2 \in \mathbb {N}$ define

$$\begin{aligned} z_{m_1,m_2}=z_{m_1}-z_{m_2}. \end{aligned}$$

(144)

Owing to the T–periodicity of both u and $\dot{u}$ it follows from Equations (142)–(144) that

$$\begin{aligned} \dot{z}_{m_1,m_2}(\sigma )=&\;\gamma \left( A_1 g_1\left( \frac{\dot{u}(\sigma )}{\gamma }\right) +A_2 g_2\left( \frac{\dot{u}(\sigma )}{\gamma }\right) \right) z_{m_1,m_2}(\sigma ),\nonumber \\&\forall \sigma \in \;]0,\alpha _1[\;\cup \;]\alpha _1,T[. \end{aligned}$$

(145)

Let $\sigma \in (0,\alpha _1)$ then $\dot{u}(\sigma ) \ge 0$ since $u \in \Lambda _{u_{\min },u_{\max },\alpha _1,T}$. We study two cases: $\dot{u}(\sigma )> 0$ and $\dot{u}(\sigma ) = 0$.

Case $\dot{u}(\sigma )> 0$. Since $0<\frac{\dot{u}(\sigma )}{\gamma }< \frac{\Vert \dot{u}\Vert }{\gamma _0}\le \delta _1$ it follows that $\left| \bar{g}_1\left( \frac{\dot{u}(\sigma )}{\gamma }\right) -1\right| <\frac{1}{2}$ which, using Equation (66), leads to

$$\begin{aligned} \frac{3A_1}{2}\dot{u}(\sigma ) \le \gamma A_1 g_1\left( \frac{\dot{u}(\sigma )}{\gamma }\right) \le \frac{A_1}{2}\dot{u}(\sigma ). \end{aligned}$$

(146)

Case $\dot{u}(\sigma ) = 0$. In this case, Inequality (146) holds by definition of the function $g_1$. That is we have

$$\begin{aligned} \frac{3A_1}{2}\dot{u}(\sigma ) \le \gamma A_1 g_1\left( \frac{\dot{u}(\sigma )}{\gamma }\right) \le \frac{A_1}{2}\dot{u}(\sigma ), \forall \sigma \in \; ]0,\alpha _1[. \end{aligned}$$

(147)

Similarly, it can be shown that

$$\begin{aligned} \frac{3A_2}{2}\dot{u}(\sigma ) \le \gamma A_2 g_2\left( \frac{\dot{u}(\sigma )}{\gamma }\right) \le \frac{A_2}{2}\dot{u}(\sigma ), \forall \sigma \in \; ]\alpha _1,T[. \end{aligned}$$

(148)

Now, define the function $V:[0,T] \rightarrow \mathbb {R}$ by

$$\begin{aligned} V(\sigma )=\frac{1}{2}z^2_{m_1,m_2}(\sigma ), \forall \sigma \in [0,T]. \end{aligned}$$

(149)

Then, V is continuous on [0, T] and is $C^1$ on $]0,\alpha _1[\;\cup \; ]\alpha _1,T[$. From Eq. (145) we obtain

$$\begin{aligned} \dot{V}(\sigma )=&\;2\gamma \left( A_1 g_1\left( \frac{\dot{u}(\sigma )}{\gamma }\right) +A_2 g_2\left( \frac{\dot{u}(\sigma )}{\gamma }\right) \right) V(\sigma ),\nonumber \\&\forall \sigma \in \;]0,\alpha _1[ \;\cup \;]\alpha _1,T[. \end{aligned}$$

(150)

Combining Eqs. (150), (147) and (148) it follows that

$$\begin{aligned} \dot{V}(\sigma )&\le A_1\dot{u}(\sigma )V(\sigma ), \forall \sigma \in \;]0,\alpha _1[,\end{aligned}$$

(151)

$$\begin{aligned} \dot{V}(\sigma )&\le A_2\dot{u}(\sigma )V(\sigma ), \forall \sigma \in \;]\alpha _1,T[. \end{aligned}$$

(152)

Define the continuous function $W:[0,\alpha _1] \rightarrow \mathbb {R}$ as being the solution of the following differential equation

$$\begin{aligned} \dot{W}(\sigma )&= A_1\dot{u}(\sigma )W(\sigma ), \forall \sigma \in \; ]0,\alpha _1[,\end{aligned}$$

(153)

$$\begin{aligned} W(0)&=V(0). \end{aligned}$$

(154)

Integrating (153)–(154) gives

$$\begin{aligned} W(\sigma )=V(0)\exp \left( \frac{A_1}{\gamma }\big (u(\sigma )-u_{\min } \big )\right) , \forall \sigma \in [0,\alpha _1]. \end{aligned}$$

(155)

Using the Comparison Lemma [42, p. 102] it comes from Eqs. (151), (153), (154), and (155) that

$$\begin{aligned} V(\alpha _1) \le W(\alpha _1) = V(0)\exp \big (A_1\left( u_{\max }-u_{\min } \right) \big ). \end{aligned}$$

(156)

Using a similar argument on the interval $[\alpha _1,T]$ it follows that

$$\begin{aligned} V(T) \le W(\alpha _1)\exp \big (A_2\left( u_{\min }-u_{\max } \right) \big ). \end{aligned}$$

(157)

As a conclusion, we have proved that

$$\begin{aligned} V(T)&\le r V(0), \end{aligned}$$

(158)

$$\begin{aligned} 0<r&=\exp \big ((A_1-A_2)(u_{\max }-u_{\min })\big ) <1, \end{aligned}$$

(159)

$$\begin{aligned} \Vert V\Vert _{[0,T]}&\le V(0). \end{aligned}$$

(160)

Note that (160) is due to the inequality $\dot{V}(\sigma ) \le 0$, $\forall \sigma \in \;]0,\alpha _1[\;\cup \;]\alpha _1,T[$ b‘ecause of Inequalities (151)–(152).

Combining Eqs. (158), (149), (144), and (143) we get

$$\begin{aligned} \Big [ z\big ( (m_1+1)T\big )-z\big ( (m_2+1)T\big )\Big ]^2&\;\le r \Big [ z\big ( m_1T\big ) -z\big ( m_2T\big )\Big ]^2,\nonumber \\&\forall m_1,m_2 \in \mathbb {N}. \end{aligned}$$

(161)

An argument by induction shows that from (161) we get

$$\begin{aligned} {\begin{matrix} V(0)=&{}\;\frac{1}{2}\Big [ z\big ( m_1T\big )-z\big ( m_2T\big )\Big ]^2 \\ &{}\le \frac{1}{2}r^{\min (m_1,m_2)} \Big [ z\left( 0\right) -z\big (| m_2-m_1|T\big )\Big ]^2 \\ &{} \le 2 r^{\min (m_1,m_2)} \Vert z\Vert ^2, \forall m_1,m_2 \in \mathbb {N}. \end{matrix}} \end{aligned}$$

(162)

Observe that, owing to Theorem 5, we have $\Vert z\Vert <\infty$. Hence, from Eqs. (162), (160), (149), (144), and (159) it comes that $\{z_m\}_{m \in \mathbb {N}}$ is a Cauchy sequence. Therefore there exists $z_{\infty } \in C^0\left( [0,T],\mathbb {R}\right)$ such that $\lim _{m \rightarrow \infty } \Vert z_m - z_{\infty }\Vert _{[0,T]}=0$. Thus we get $\lim _{m \rightarrow \infty }\left| z_m(0)-z_{\infty }(0)\right| =0$ and

$\lim _{m \rightarrow \infty } \left| z_m(T)-z_{\infty }(T)\right| =0$. Note that $z_m(0)=z(mT)$ and $z_m(T)=z\big ( (m+1)T \big )$ by (143). Take $m_1=m$ and $m_2=m+1$ in Inequality (162). Then we get $\lim _{m \rightarrow \infty } \left| z(mT)-z\big ( (m+1)T \big )\right| =0$. All these facts show that we have

$$\begin{aligned} z_{\infty }(0)=z_{\infty }(T). \end{aligned}$$

(163)

Combining Eqs. (142) and (143) it comes that

$$\begin{aligned} z_m(\sigma )&\;=z_m(0)+\gamma \int _0^{\sigma } g_1\left( \frac{\dot{u}(\tau )}{\gamma }\right) \big [A_1 z_m(\tau )+B_1 u(\tau ) + E_1 \big ] \nonumber \\&\quad +g_2\left( \frac{\dot{u}(\tau )}{\gamma }\right) \big (A_2 z_m(\tau )+B_2 u(\tau )+ E_2 \big ) \, \text {d}\tau ,\nonumber \\&\forall \sigma \in [0,T], \forall m \in \mathbb {N}. \end{aligned}$$

(164)

Note that $\Vert z_m\Vert \le \Vert z\Vert < \infty$. Also, $\left| \frac{\dot{u}(\tau )}{\gamma } \right| \le \frac{\Vert \dot{u}\Vert }{\gamma _0}$ so that, by the continuity of the functions $g_1$ and $g_2$ we have $\left| g_1\left( \frac{\dot{u}(\tau )}{\gamma }\right) \right| \le k_1$ and $\left| g_2\left( \frac{\dot{u}(\tau )}{\gamma }\right) \right| \le k_2$, where $k_1, k_2 \in \mathbb {R}_+$ are independent of $\tau$ and m. This means that the term under the integral in Eq. (164) is bounded by a constant independent of $\tau$ and m. Using the Lebesgue Dominated Convergence Theorem it follows from (164) that

$$\begin{aligned} z_\infty (\sigma )&=z_\infty (0)+\gamma \!\int _0^{\sigma }\!\! g_1\left( \frac{\dot{u}(\tau )}{\gamma }\right) \big [ A_1 z_\infty (\tau )+B_1 u(\tau )+E_1 \big ] \nonumber \\ {}&\quad +g_2\left( \frac{\dot{u}(\tau )}{\gamma }\right) \big [ A_2 z_\infty (\tau )+B_2 u(\tau )+E_2 \big ] \,\text {d}\tau ,\nonumber \\&\forall \sigma \in [0,T]. \end{aligned}$$

(165)

Define $\bar{z}_\gamma :\mathbb {R}_+ \rightarrow \mathbb {R}$ by

$$\begin{aligned} \bar{z}_\gamma (\sigma +mT)=z_\infty (\sigma ),\forall \sigma \in [0,T],\forall m \in \mathbb {N}. \end{aligned}$$

(166)

Then it comes from Eqs. (166), (165) and (163) that $\bar{z}_\gamma$ is T–periodic and

$$\begin{aligned} \bar{z}_\gamma (\sigma )&=\bar{z}_\gamma (0)+\gamma \int _0^{\sigma } g_1\left( \frac{\dot{u}(\tau )}{\gamma }\right) \big [ A_1 \bar{z}_\gamma (\tau )+ B_1 u(\tau )+E_1 \big ] \nonumber \\&\quad +g_2\left( \frac{\dot{u}(\tau )}{\gamma }\right) \big [ A_2 \bar{z}_\gamma (\tau )+ B_2 u(\tau )+E_2 \big ]\, \text {d}\tau ,\nonumber \\&\forall \sigma \in \mathbb {R}_+. \end{aligned}$$

(167)

As a conclusion, we have proved that there exists

$$\begin{aligned} x_{0,\gamma }=\bar{z}_\gamma (0) \end{aligned}$$

(168)

such that

$$\begin{aligned} \mathcal {H}_s(u \circ s_\gamma ,x_{0,\gamma })=\bar{z}_\gamma \circ s_\gamma \end{aligned}$$

(169)

is $T \gamma$–periodic.

To prove the uniqueness of $x_{0,\gamma }$ we use an argument similar to the one used for the proof of the existence. Take $\gamma> \gamma _0$ and suppose that there exists $x^\prime _{0,\gamma }$ such that $\mathcal {H}_s(u \circ s_\gamma ,x^\prime _{0,\gamma })$ is $T \gamma$–periodic. Define $\bar{z}_\gamma ^\prime : \mathbb {R}_+ \rightarrow \mathbb {R}$ by $\bar{z}_\gamma ^\prime =\mathcal {H}_s(u \circ s_\gamma ,x^\prime _{0,\gamma }) \circ s_{\frac{1}{\gamma }}$. Then, $\bar{z}_\gamma ^\prime (0)=x^\prime _{0,\gamma }$ and $\bar{z}_\gamma ^\prime$ satisfies Eq. (167) with $\bar{z}_\gamma$ replaced by $\bar{z}_\gamma ^\prime$. Considering the difference $\varepsilon =\bar{z}_\gamma -\bar{z}_\gamma ^\prime$ it follows that $\varepsilon$ satisfies Eq. (145) with $z_{m_1,m_2}$ replaced by $\varepsilon$. A function V can be defined as in Eq. (149) with $z_{m_1,m_2}$ replaced by $\varepsilon$ which leads to Inequality (158). Since $V(0)=V(T)$ owing to the T–periodicity of V, it follows that $V(0)=0$ as V is nonnegative. Thus $x^\prime _{0,\gamma }=x_{0,\gamma }$.

Proof of Theorem 9

Let $\gamma \in \;]\gamma _0,\infty [$ where $\gamma _0$ is given by Eq. (139). From Eq. (147) it follows that

$$\begin{aligned} \frac{3A_1}{2}\big (u_{\max }-u_{\min } \big ) \le \int _0^{\tau }\frac{3A_1}{2}\dot{u}(t)\text {d}t \le \int _0^{\tau }\gamma A_1 g_1\left( \frac{\dot{u}(t)}{\gamma }\right) \text {d}t, \end{aligned}$$

(170)

and

$$\begin{aligned} \left| \gamma A_1 g_1\left( \frac{\dot{u}(\tau )}{\gamma }\right) \right| \le \frac{3|A_1|}{2}\Vert \dot{u}\Vert , \forall \tau \in \;]0,\alpha _1[. \end{aligned}$$

(171)

Also, From Eq. (148) it follows that

$$\begin{aligned} \frac{3A_2}{2}\big (u_{\min }-u_{\max } \big ) \le \int _{\alpha _1}^{\tau }\frac{3A_2}{2}\dot{u}(t)\text {d}t \le \int _{\alpha _1}^{\tau }\gamma A_2 g_2\left( \frac{\dot{u}(t)}{\gamma }\right) \text {d}t, \end{aligned}$$

(172)

and

$$\begin{aligned} \left| \gamma A_2 g_2\left( \frac{\dot{u}(\tau )}{\gamma }\right) \right| \le \frac{3A_2}{2}\Vert \dot{u}\Vert , \forall \tau \in \; ]\alpha _1,T[. \end{aligned}$$

(173)

Equations (170)–(173) show that we can apply the Lebesgue Dominated Convergence Theorem in (103) so that we get

$$\begin{aligned} \lim _{\gamma \rightarrow \infty } \bar{z}_\gamma (0)= \bar{z}(0)=\theta . \end{aligned}$$

(174)

Observe that using the same theorem we can show that $\forall \sigma \in [0,T]$ we have $\lim _{\gamma \rightarrow \infty } |\bar{z}_\gamma (\sigma )-\bar{z}(\sigma )|=0$. However, this simple convergence does not imply Theorem 9; we need to prove the uniform convergence of $\bar{z}_\gamma$ to $\bar{z}$ on the interval [0, T]. This is the aim of the following analysis.

Inequalities (170)–(173) along with Eqs. (99), (101) and (102) lead to

$$\begin{aligned} \Vert \bar{z}_\gamma \Vert _{[0,T]} \le c_1, \forall \gamma \in \; ]\gamma _0,\infty [ \end{aligned}$$

(175)

where $c_1 \in \mathbb {R}_+$ is independent of $\gamma$.

On the other hand, it can be checked that Eqs. (93), (94), (90), (104), (105) lead to

$$\begin{aligned} \dot{\bar{z}}(\sigma )&= \dot{u}(\sigma ) \left( A_1 \bar{z}(\sigma )+ B_1 u(\sigma )+ E_1 \right) , \; \forall \sigma \in \; ]0,\alpha _1[, \end{aligned}$$

(176)

$$\begin{aligned} \dot{\bar{z}}(\sigma )&= \dot{u}(\sigma ) \left( A_2 \bar{z}(\sigma )+ B_2 u(\sigma )+ E_2 \right) , \; \forall \sigma \in \;]\alpha _1,T[. \end{aligned}$$

(177)

Define the function $V:[0,T] \rightarrow \mathbb {R}$ by the relation

$$\begin{aligned} V_\gamma (\sigma )=\frac{1}{2}\big (\bar{z}(\sigma )- \bar{z}_\gamma (\sigma )\big )^2, \forall \sigma \in [0,T]. \end{aligned}$$

(178)

Take $\sigma \in \;]0,\alpha _1[$, then it comes from Eqs. (167) and (176) that

$$\begin{aligned} \dot{V}_\gamma (\sigma )=&\,\big (\bar{z}(\sigma )- \bar{z}_\gamma (\sigma )\big )\Big [\dot{u}(\sigma )\big (A_1 \bar{z}(\sigma )+ B_1 u(\sigma )+ E_1 \big ) \nonumber \\&-\gamma g_1\left( \frac{\dot{u}(\sigma )}{\gamma }\right) \big (A_1 \bar{z}_\gamma (\sigma ) + B_1 u(\sigma )+ E_1 \big )\Big ]\nonumber \\ =&\,\big (\bar{z}(\sigma )- \bar{z}_\gamma (\sigma )\big )\big ( B_1u(\sigma )+E_1\big ) \left( \dot{u}(\sigma )-\gamma g_1\left( \frac{\dot{u}(\sigma )}{\gamma }\right) \right) \nonumber \\ {}&+A_1\big (\bar{z}(\sigma )- \bar{z}_\gamma (\sigma )\big ) \left( \dot{u}(\sigma )-\gamma g_1\left( \frac{\dot{u}(\sigma )}{\gamma }\right) \right) \bar{z}(\sigma ) \\&+2A_1\gamma g_1\left( \frac{\dot{u}(\sigma )}{\gamma }\right) V_\gamma (\sigma ),\forall \sigma \in \;]0,\alpha _1[, \forall \gamma> \gamma _0. \nonumber \end{aligned}$$

(179)

Let $\varepsilon>0$. From Eqs. (66) and (68) it follows that $\exists \delta _\varepsilon>0$ such that $\forall w \in \;]0,\delta _\varepsilon [$ we have $\left| \bar{g}_1(w)-1\right| <\frac{\varepsilon }{\Vert \dot{u}\Vert }$. Thus, $\exists \gamma _\varepsilon =\min \left( \gamma _0,\frac{\Vert \dot{u}\Vert }{\delta _\varepsilon } \right)$ such that $\forall \gamma> \gamma _\varepsilon$ we have

$$\begin{aligned} \left| \gamma g_1\left( \frac{\dot{u}(\sigma )}{\gamma }\right) -\dot{u}(\sigma )\right| \le \varepsilon , \forall \sigma \in \;]0,\alpha _1[. \end{aligned}$$

(180)

Combining Eqs. (178)–(180) along with Inequalities (175) and (147) it comes that

$$\begin{aligned} \dot{V}_\gamma (\sigma ) \le A_1 \dot{u}(\sigma ) V_\gamma (\sigma )+c_2 \varepsilon \sqrt{V_\gamma (\sigma )}, \forall \sigma \in \;]0,\alpha _1[, \forall \gamma> \gamma _\varepsilon . \end{aligned}$$

(181)

where $c_2 \in \mathbb {R}_+$ is independent of $\gamma$. Define the continuous function $W:[0,\alpha _1] \rightarrow \mathbb {R}_+$ as the solution of the following differential equation

$$\begin{aligned} \dot{W}(\sigma )&= A_1\dot{u}(\sigma )W(\sigma )+c_2\varepsilon \sqrt{W(\sigma )}, \forall \sigma \in \;]0,\alpha _1[, \end{aligned}$$

(182)

$$\begin{aligned} W(0)&=V(0). \end{aligned}$$

(183)

Integrating (182)–(183) gives

$$\begin{aligned} W(\sigma )&=e^{A_1u(\sigma )}\Big ( \sqrt{V(0)}e^{-\frac{A_1}{2}u_{\min }} +\frac{c_2}{2} \varepsilon \int _0^{\sigma } e^{-\frac{A_1}{2}u(\tau )} \text {d}\tau \Big )^2,\nonumber \\&\forall \sigma \in [0,\alpha _1],\nonumber \\&\le \; e^{A_1u_{\min }}\Big ( \sqrt{V(0)}e^{-\frac{A_1}{2}u_{\min }} +\frac{c_2}{2} \varepsilon \int _0^{\alpha _1} e^{-\frac{A_1}{2}u(\tau )} \text {d}\tau \Big )^2,\nonumber \\&\forall \sigma \in [0,\alpha _1]. \end{aligned}$$

(184)

Using the Comparison Lemma [42, p. 102] it follows from (181)–(184) that

$$\begin{aligned} V_\gamma (\sigma ) \le&\;e^{A_1u_{\min }}\Big ( \sqrt{V(0)}e^{-\frac{A_1}{2}u_{\min }} +\frac{c_2}{2} \varepsilon \int _0^{\alpha _1} e^{-\frac{A_1}{2}u(\tau )} \text {d}\tau \Big )^2,\nonumber \\&\forall \sigma \in [0,\alpha _1],\forall \gamma> \gamma _\varepsilon . \end{aligned}$$

(185)

Equations (185), (174) and (178) show that

$\lim _{\gamma \rightarrow \infty }\Vert V_\gamma \Vert _{[0,\alpha _1]}=0$. A similar argument on the interval $[\alpha _1,T]$ shows that $\lim _{\gamma \rightarrow \infty }\Vert V_\gamma \Vert _{[0,T]}=0$. The uniform convergence of $\bar{z}_\gamma$ (restricted to the interval [0, T]) to $\bar{z}$ has thus been demonstrated, which completes the proof.

Proof of Lemma 8

(i) $\Rightarrow$ (ii). From Eq. (80) and $C\ne 0$ it comes that $\forall \varrho _1,\varrho _2 \in [0,\rho _u(T)]$ we have $\varphi _u^\circ (\varrho _1)=\varphi _u^\circ (\varrho _2) \Leftrightarrow x_u^\circ (\varrho _1) = x_u^\circ (\varrho _2)$. Condition (i) implies that $\forall \nu \in [u_{\min },u_{\max }]$ we have $\xi _1(\nu )=\xi _2(\nu )$. Therefore $\forall \nu \in \;]u_{\min },u_{\max }[$ we have $\dot{\xi }_1(\nu )=\dot{\xi }_2(\nu ).$ Thus we get from (91)–(92) that

$$\begin{aligned} \xi _1(\nu )=\xi _2(\nu )=\frac{B_1-B_2}{A_2-A_1}\nu +\frac{E_1-E_2}{A_2-A_1}, \forall \nu \in \;]u_{\min },u_{\max }[. \end{aligned}$$

(186)

Consider the functions $f_1,f_2,f_3,\mathbf {0}:\;]u_{\min },u_{\max }[\; \rightarrow \mathbb {R}$ defined by $\forall \nu \in \;]u_{\min },u_{\max }[, f_1(\nu )=1,$ $f_2(\nu )=\nu,$ $f_3(\nu )= e^{A_1(\nu -u_{\min })},$ and $\mathbf {0}(\nu )=0.$ Then Eq. (186) along with (93)–(94) lead to

$$\begin{aligned} {\begin{matrix} &{}\left( \frac{E_1-E_2}{A_2-A_1}+\frac{E_1}{A_1}+\frac{B_1}{A_1^2}\right) f_1+\left( \frac{B_1-B_2}{A_2-A_1}+\frac{B_1}{A_1}\right) f_2\\ &{} -\left( \frac{B_1}{A_1}u_{\min } +\frac{E_1}{A_1}+\frac{B_1}{A_1^2}+\theta \right) f_3=\mathbf {0}, \end{matrix}} \end{aligned}$$

(187)

$$\begin{aligned} {\begin{matrix} &{} \left( \frac{E_1-E_2}{A_2-A_1}+\frac{E_2}{A_2}+\frac{B_2}{A_2^2}\right) f_1+\left( \frac{B_1-B_2}{A_2-A_1}+\frac{B_2}{A_2}\right) f_2 \\ &{}- \left( \frac{B_2}{A_2}u_{\min } +\frac{E_2}{A_2}+\frac{B_2}{A_2^2}+\theta \right) f_3=\mathbf {0}. \end{matrix}} \end{aligned}$$

(188)

Consider the vector space of functions $\{p:\;]u_{\min },u_{\max }[\; \rightarrow \mathbb {R}\}$ with its usual binary operations of vector addition and scalar multiplication. Then the functions $f_1,f_2,f_3$ are linearly independent vectors so that, owing to Eqs. (187)–(188), we must have

$$\begin{aligned} \frac{E_1-E_2}{A_2-A_1}+\frac{E_1}{A_1}+\frac{B_1}{A_1^2}&=0, \end{aligned}$$

(189)

$$\begin{aligned} \frac{B_1-B_2}{A_2-A_1}+\frac{B_1}{A_1}&=0,\end{aligned}$$

(190)

$$\begin{aligned} \frac{B_1}{A_1}u_{\min } +\frac{E_1}{A_1}+\frac{B_1}{A_1^2}+\theta&=0,\end{aligned}$$

(191)

$$\begin{aligned} \frac{E_1-E_2}{A_2-A_1}+\frac{E_2}{A_2}+\frac{B_2}{A_2^2}&=0,\end{aligned}$$

(192)

$$\begin{aligned} \frac{B_1-B_2}{A_2-A_1}+\frac{B_2}{A_2}&=0, \end{aligned}$$

(193)

$$\begin{aligned} \frac{B_2}{A_2}u_{\min } +\frac{E_2}{A_2}+\frac{B_2}{A_2^2}+\theta&=0. \end{aligned}$$

(194)

Simple calculations show that Eqs. (189)–(194) lead to (96)–(97).

(ii) $\Rightarrow$ (i). It can be checked that Eqs. (96)–(97) lead to (189)–(194) so that the opertor $\mathcal {H}_o$ has a trivial hysteresis loop with respect to all $(u,x_0) \in \Lambda _{u_{\min },u_{\max },\alpha _1,T} \times \mathbb {R}$.

Proof of Lemma 10

Using Eq. (40) the functions $F_1,F_2: \mathbb {R}^2 \rightarrow \mathbb {R}$ are given by

$$\begin{aligned} F_1(x_1,v) =&\frac{A_1-A_2}{2}x_1 + \frac{B_1-B_2}{2}v+\frac{E_1-E_2}{2}, \end{aligned}$$

(195)

$$\begin{aligned} F_2(x_1,v) =&\frac{A_1+A_2}{2}x_1 + \frac{B_1+B_2}{2}v+\frac{E_1+E_2}{2}. \end{aligned}$$

(196)

Then Assumption 7 holds since $A_1 \ne A_2$. The anhysteresis function is

$$\begin{aligned} f_{\text {an}}(v)=-\frac{B_1}{A_1}v+\frac{E_2-E_1}{A_1-A_2}, \forall v \in \mathbb {R} \end{aligned}$$

(197)

where (114) has been used. For every pair $(x_0,u_0) \in \mathbb {R}^2$, let $\omega _{\Phi ,1}(\cdot ,x_0,u_0):[u_0,\infty ) \rightarrow \mathbb {R}$ be the solution z of $z(\sigma )-x_0=\int _{u_0}^{\sigma } A_1 z(\tau ) + B_1 \tau + E_1\,\text {d}\tau$, for all $\sigma \in [u_0,\infty [$ and let $\omega _{\Phi ,2}(\cdot ,x_0,u_0):\;]-\infty ,u_0] \rightarrow \mathbb {R}$ be the solution z of $z(\sigma )-x_0=\int _{u_0}^{\sigma } A_2 z(\tau )+B_2\tau +E_2\,\text {d}\tau$, for all $\sigma \in \; ]-\infty ,u_0]$. Then

$$\begin{aligned} \omega _{\Phi ,1}(\sigma ,x_0,u_0) =&\, \frac{A_1B_1u_0+A_1E_1+B_1}{A_1^2}e^{(\sigma -u_0)A_1}\nonumber \\&-\frac{A_1B_1\sigma +A_1E_1+B_1}{A_1^2}\nonumber \\&\,+e^{(\sigma -u_0)A_1}x_0,\forall \sigma \in [u_0,\infty [, \end{aligned}$$

(198)

$$\begin{aligned} \omega _{\Phi ,2}(\sigma ,x_0,u_0) =&\, \frac{A_2B_2u_0+A_2E_2+B_2}{A_2^2}e^{(\sigma -u_0)A_2}\nonumber \\&-\frac{A_2B_2\sigma +A_2E_2+B_2}{A_2^2} \nonumber \\&\,+e^{(\sigma -u_0)A_2}x_0,\forall \sigma \in \;]-\infty ,u_0]. \end{aligned}$$

(199)

Equations (198)–(199) are valid since $A_1\ne 0$ and $A_2 \ne 0$. Define the function $\omega _{\Phi }(\cdot ,x_0,u_0)$ by Eq. (41). Then, the intersecting function $\Omega$ should satisfy

$$\begin{aligned} \omega _{\Phi }\big (\Omega (x_1,v) ,x_1,v\big )=f_{\text {an}}\big (\Omega (x_1,v)\big ),\forall (x_1,v) \in \mathbb {R}^2. \end{aligned}$$

(200)

Define

$$\begin{aligned} M_1=&\;\Big (B_1 A_1^{-1}\big (A_2^{-1}-A_1^{-1} \big )-E_1 A_1^{-1}+E_2A_2^{-1}\Big ) \frac{A_2}{A_1-A_2}, \end{aligned}$$

(201)

$$\begin{aligned} M_2=&\;\Big (B_1 A_1^{-1}\big (A_2^{-1}-A_1^{-1} \big )-E_1 A_1^{-1}+E_2A_2^{-1}\Big ) \frac{A_1}{A_1-A_2}. \end{aligned}$$

(202)

Note that $M_1>0$ and $M_2<0$ owing to (114)–(116). Combining (197)–(200) and (114)–(116) it follows from the definition of function $\Omega$ (in Sect. 8.3) that

$$\begin{aligned} \Omega (x_1,v) = {\left\{ \begin{array}{ll}&{}v-\frac{1}{A_1} \log \left( 1+\frac{x_1 - f_{\text {an}}(v)}{M_1} \right) \text { if } \quad x_1 \ge f_{\text {an}}(v),\\ &{}v-\frac{1}{A_2} \log \left( 1-\frac{f_{\text {an}}(v)-x_1}{M_2} \right) \text { if } \quad x_1 \le f_{\text {an}}(v), \end{array}\right. } \end{aligned}$$

(203)

where $\log$ sets for the natural logarithm. The function $\varsigma$ in Eq. (42) can be determined explicitly as

$$\begin{aligned} \varsigma (x_1,v) =&\; x_1v-\frac{1}{A_1}x_1+\frac{E_1}{A_1}v+\frac{B_1}{2A_1}v^2 -M_1 \Omega (x_1,v) \nonumber \\&+\frac{E_2-E_1}{A_1(A_1-A_2)} \quad \text { if } \quad x_1 \ge f_{\text {an}}(v), \end{aligned}$$

(204)

$$\begin{aligned} \varsigma (x_1,v) =&\; x_1v-\frac{1}{A_2}x_1+\frac{E_2}{A_2}v+\frac{B_1}{2A_1}v^2 -M_2 \Omega (x_1,v) \nonumber \\&+\frac{E_2-E_1}{A_2(A_1-A_2)} \quad \text { if } \quad x_1 \le f_{\text {an}}(v). \end{aligned}$$

(205)

It can be checked that

$$\begin{aligned} \varsigma (x_1,v) =-\frac{B_1}{2A_1}v^2 \quad \text {if} \quad x_1=f_{\text {an}}(v). \end{aligned}$$

(206)

The fact that Inequality (39) holds for any input $u \in AC(\mathbb {R}_+,\mathbb {R})$ and any initial condition $x_0 \in \mathbb {R}$ follows from Theorem 3. However, $\varsigma$ is not nonnegative: it can be checked that for any fixed $x_1$ we have $\lim _{v \rightarrow \pm \infty } \varsigma (x_1,v)= -\infty$.

The aim of the following analysis is to show that $\forall (x_1,v) \in \mathbb {R} \times \left[ \frac{1}{A_1},\frac{1}{A_2} \right]$ we have $\varsigma (x_1,v) \ge 0$. To this end, observe that, from (114) and (206), we have

$$\begin{aligned} \varsigma (x_1,v) \ge 0 \quad \text {whenever} \quad x_1=f_{\text {an}}(v). \end{aligned}$$

(207)

Now, fix $v \in \left[ \frac{1}{A_1},\frac{1}{A_2} \right]$. From (203)–(204) and (114)–(116) it follows that

$$\begin{aligned} \lim _{x_1 \rightarrow \infty } \varsigma (x_1,v) = \infty .\end{aligned}$$

(208)

Suppose that there exists $x_2 \in \;]f_{\text {an}}(v),\infty [$ such that $\varsigma (x_2,v)<0$. Then, from (207)–(208) it follows that $\varsigma (\cdot ,v)$ should have a minimum at $x_3 \in \; ]f_{\text {an}}(v),\infty [$ such that $\varsigma (x_3,v)<0$. A necessary condition for this to happen is $\frac{\partial \varsigma }{\partial x_1}(x_3,v)=0$. It can be checked from Eq. (204) that this last equality cannot hold. A similar argument can be used for Equation (205).

Proof of Lemma 11

Observe that, for Theorem 5 to hold, it is needed that $A_1$ and $-A_2$ are both stable. Since $n=1$, this condition translates into $A_1<0$ and $A_2>0$ so that the results of Theorems 5, 6, and 7 apply.

The proof is done in two steps. In Step 1 we consider a specific T–periodic input $u \in W^{1,\infty }(\mathbb {R}_+,\mathbb {R})$ and an arbitrary initial condition $x_0$. Using Theorem 7 it follows that the function $\varphi _u^\circ$ that characterizes the hysteresis loop satisfies the differential state equation (79) and the output equation (80). The aim of Step 1 is to find the initial state $x_u^\circ (0)$ since the latter may be different from $x_0$. In Step 2 we use the knowledge of $x_u^\circ (0)$ to prove that, if Assumption 8 holds, then the relations (251)–(252) hold.

Step 1. Let $\alpha \in \;]0,1[;$ define $\varrho _1=1$, $\varrho _2=2-\alpha$, $\varrho _3=3-2\alpha$, $\varrho _4=4-2\alpha$. Note that $0<\varrho _1<\varrho _2<\varrho _3<\varrho _4$. We consider the $\varrho _4$-periodic input $u : \mathbb {R}_+ \rightarrow \mathbb {R}$ defined on the interval $[0,\varrho _4]$ by

$$\begin{aligned} u(\varrho )&=\varrho , \forall \varrho \in [0,\varrho _1], \end{aligned}$$

(209)

$$\begin{aligned} u(\varrho )&=2-\varrho , \forall \varrho \in [\varrho _1,\varrho _2], \end{aligned}$$

(210)

$$\begin{aligned} u(\varrho )&=2\alpha -2+\varrho , \forall \varrho \in [\varrho _2,\varrho _3], \end{aligned}$$

(211)

$$\begin{aligned} u(\varrho )&=4-2\alpha -\varrho , \forall \varrho \in [\varrho _3,\varrho _4]. \end{aligned}$$

(212)

Observe that $u(0)=0,$ $u(\varrho _1)=1,$ $u(\varrho _2)=\alpha,$ $u(\varrho _3)=1,$ $u(\varrho _4)=0,$ and that $u \in W^{1,\infty }(\mathbb {R}_+,\mathbb {R}).$ Observe also that $|\dot{u}(\varrho )|=1$ for almost all $\varrho \in \mathbb {R}_+$ so that $\rho _u$ is the identity function which gives $\psi _u=u.$ Let $x_0 \in \mathbb {R}$ and consider the scalar semilinear Duhem model with input u and initial condition $x_0$ (Eqs. (63)–(65)). Since all conditions of Theorem 7 hold, we get from Equality (80) that

$$\begin{aligned} \varphi _u^\circ (\varrho )=C x_u^\circ (\varrho ) + D u(\varrho ), \forall \varrho \in [0,\varrho _4], \end{aligned}$$

(213)

where $x_u^\circ$ satisfies the differential equation (79). To find the initial condition $x_u^\circ (0)$ we compute $x_u^\circ (\varrho _k), k=1,\ldots ,4$ as a function of $x_u^\circ (0)$ and we use the fact that, by Theorem 7, we have $x_u^\circ (0)=x_u^\circ (\varrho _4).$ We start by computing $x_u^\circ (\varrho _1)$ as a function of $x_u^\circ (0).$ In the interval $[0,\varrho _1],$ the differential equation (79) becomes

$$\begin{aligned} \frac{\text{ d }x_u^\circ }{\text{ d }\varrho }(\varrho ) = A_1x_u^\circ (\varrho )+B_1u(\varrho )+E_1, \forall \varrho \in \;]0,\varrho _1[. \end{aligned}$$

(214)

Equation (214) can be solved explicitly and it gives

$$\begin{aligned} x_u^\circ (\varrho _1) =e^{\varrho _1 A_1 }x_u^\circ (0)+e^{\varrho _1 A_1 } \int _{0}^{\varrho _1} e^{-\tau A_1} \left( B_1u(\tau )+E_1\right) \, \text{ d }\tau . \end{aligned}$$

(215)

Taking into account Eq. (209) it follows that

$$\begin{aligned} x_u^\circ (1)&=e^{ A_1 }x_u^\circ (0)+ \beta _{11}, \end{aligned}$$

(216)

$$\begin{aligned} \beta _{11}&=A_1 ^{-1} \Big [\left( -1 -A_1^{-1} +A_1^{-1}e^{ A_1} \right) B_1 + \left( -1 + e^{ A_1} \right) E_1\Big ]. \end{aligned}$$

(217)

In the interval $[\varrho _1,\varrho _2]$, the differential equation (79) becomes

$$\begin{aligned} \frac{\text{ d }x_u^\circ }{\text{ d }\varrho }(\varrho ) = -A_2x_u^\circ (\varrho )-u(\varrho )B_2-E_2, \forall \varrho \in \;]\varrho _1,\varrho _2[. \end{aligned}$$

(218)

Equation (218) can be solved explicitly and it gives

$$\begin{aligned} x_u^\circ (\varrho _2) =&\;e^{-(\varrho _2-\varrho _1) A_2 }x_u^\circ (\varrho _1)\nonumber \\&-e^{-\varrho _2 A_2 } \int _{\varrho _1}^{\varrho _2} e^{\tau A_2} \left( u(\tau )B_2+E_2\right) \, \text{ d }\tau . \end{aligned}$$

(219)

Taking into account Eq. (210) it follows that

$$\begin{aligned} x_u^\circ (\varrho _2)&=e^{(\alpha -1) A_2 }x_u^\circ (1) + \beta _{21}e^{A_2 \alpha }+\beta _{22} \alpha +\beta _{23}, \end{aligned}$$

(220)

$$\begin{aligned} \beta _{21}&=A_2 ^{-1} e^{-A_2}\left( B_2(1+A_2^{-1})+E_2\right) , \end{aligned}$$

(221)

$$\begin{aligned} \beta _{22}&=-A_2^{-1}B_2, \end{aligned}$$

(222)

$$\begin{aligned} \beta _{23}&=-A_2^{-1}\left( A_2^{-1}B_2+E_2 \right) . \end{aligned}$$

(223)

In the interval $[\varrho _2,\varrho _3]$, the differential equation (79) becomes

$$\begin{aligned} \frac{\text{ d }x_u^\circ }{\text{ d }\varrho }(\varrho ) = A_1x_u^\circ (\varrho )+B_1u(\varrho )+E_1, \forall \varrho \in \;]\varrho _2,\varrho _3[. \end{aligned}$$

(224)

Equation (224) can be solved explicitly and it gives

$$\begin{aligned} {\begin{matrix} x_u^\circ (\varrho _3) = &{}\;e^{(\varrho _3-\varrho _2) A_1 }x_u^\circ (\varrho _2)\\ &{}+e^{\varrho _3 A_1 } \int _{\varrho _2}^{\varrho _3} e^{-\tau A_1} \left( B_1u(\tau )+E_1\right) \, \text{ d }\tau . \end{matrix}} \end{aligned}$$

(225)

Taking into account Eq. (211) it follows that

$$\begin{aligned} x_u^\circ (\varrho _3)&=e^{(1-\alpha ) A_1 }x_u^\circ (\varrho _2)+\beta _{31}\alpha e^{-A_1\alpha } +\beta _{32} e^{-A_1\alpha }+ \beta _{33}, \end{aligned}$$

(226)

$$\begin{aligned} \beta _{31}&=A_1 ^{-1} B_1e^{A_1}, \end{aligned}$$

(227)

$$\begin{aligned} \beta _{32}&= A_1 ^{-1} e^{A_1} \left( A_1^{-1}B_1+E_1\right) , \end{aligned}$$

(228)

$$\begin{aligned} \beta _{33}&=A_1^{-1}\left( -(1+A_1^{-1})B_1-E_1 \right) . \end{aligned}$$

(229)

In the interval $[\varrho _3,\varrho _4]$, the differential equation (79) becomes

$$\begin{aligned} \frac{\text{ d }x_u^\circ }{\text{ d }\varrho }(\varrho ) = -A_2x_u^\circ (\varrho )-u(\varrho )B_2-E_2, \forall \varrho \in \;]\varrho _3,\varrho _4[. \end{aligned}$$

(230)

Eq. (230) can be solved explicitly and it gives

$$\begin{aligned} x_u^\circ (\varrho _4) =&\;e^{-(\varrho _4-\varrho _3) A_2 }x_u^\circ (\varrho _3)\nonumber \\ {}&-e^{-\varrho _4 A_2 } \int _{\varrho _3}^{\varrho _4} e^{\tau A_2} \left( u(\tau )B_2+E_2\right) \, \text{ d }\tau . \end{aligned}$$

(231)

Taking into account Eq. (212) it follows that

$$\begin{aligned} x_u^\circ (\varrho _4)&=e^{- A_2 }x_u^\circ (\varrho _3)+\beta _{41}, \end{aligned}$$

(232)

$$\begin{aligned} \beta _{41}&= -A_2^{-1}\Big [B_2\left( -e^{-A_2}-A_2^{-1}e^{-A_2}+A_2^{-1} \right) \nonumber \\&\quad +E_2\left( 1-e^{-A_2} \right) \Big ]. \end{aligned}$$

(233)

Now we use the relation $x_u^\circ (0)=x_u^\circ (\varrho _4)$ to find $x_u^\circ (0)$ using Eqs. (216)–(217), (220)–(223), (226)–(229) and (232)–(233). We get

$$\begin{aligned} x_u^\circ (0)&=\frac{\beta _{51} + \beta _{52}e^{(A_2-A_1)\alpha }+ \beta _{53} \alpha e^{-A_1\alpha }+\beta _{54}e^{-A_1\alpha } }{1+\beta _{55}e^{(A_2-A_1)\alpha }}, \end{aligned}$$

(234)

$$\begin{aligned} \beta _{51}&=\beta _{33}e^{-A_2}+\beta _{41}, \end{aligned}$$

(235)

$$\begin{aligned} \beta _{52}&=\beta _{21}e^{A_1-A_2}+\beta _{11}e^{A_1-2A_2}, \end{aligned}$$

(236)

$$\begin{aligned} \beta _{53}&=\beta _{22}e^{A_1-A_2} + e^{-A_2}\beta _{31}, \end{aligned}$$

(237)

$$\begin{aligned} \beta _{54}&=\beta _{23}e^{A_1-A_2}+e^{-A_2}\beta _{32},\end{aligned}$$

(238)

$$\begin{aligned} \beta _{55}&=-e^{2(A_1-A_2)}. \end{aligned}$$

(239)

Note that, since $0<\alpha <1$, $A_1<0$ and $A_2>0$ it follows that $0<e^{(2-\alpha )(-A_2+A_1)}<1$ so that the denominator in Equ. (234) is nonzero.

Step 2. By Assumption 8 it follows that $\varphi _u^\circ (\varrho _1)=\varphi _u^\circ (\varrho _3)$. This means that $x_u^\circ (1)=x_u^\circ (\varrho _3)$ because $C \ne 0$. Since $x_u^\circ (0)$ has been computed explicitly, $x_u^\circ (1)$ and $x_u^\circ (\varrho _3)$ are available explicitly using Eqs. (216)–(217) and (226)–(229) respectively. We get

$$\begin{aligned} {\begin{matrix} x_u^\circ (1)=&{}\;\frac{e^{A_1}}{1+\beta _{55}e^{(A_2-A_1)\alpha }} \cdot \Big (\beta _{51} + \beta _{52}e^{(A_2-A_1)\alpha } \\ &{}+ \beta _{53} \alpha e^{-A_1\alpha }+\beta _{54}e^{-A_1\alpha } \Big )+\beta _{11}, \end{matrix}} \end{aligned}$$

(240)

$$\begin{aligned} {\begin{matrix} x_u^\circ (\varrho _3)={}&{}\;\frac{e^{2A_1-A_2}}{1+\beta _{55}e^{(A_2-A_1)\alpha }}\cdot \Big (\beta _{51} e^{(A_2-A_1) \alpha } \\ &{}+\beta _{52}e^{2(A_2-A_1)\alpha }+ \beta _{53} \alpha e^{(A_2-2A_1)\alpha }\\ {} &{}+\beta _{54}e^{(A_2-2A_1)\alpha } \Big ) + \beta _{61} e^{(A_2-A_1) \alpha }\\ {} &{}+ \beta _{62}\alpha e^{-A_1 \alpha } + \beta _{63}e^{-A_1 \alpha } + \beta _{33}, \end{matrix}} \end{aligned}$$

(241)

where

$$\begin{aligned} \beta _{61}&= \left( \beta _{21} +\beta _{11}e^{-A_2}\right) e^{A_1},\end{aligned}$$

(242)

$$\begin{aligned} \beta _{62}&=\beta _{22}e^{A_1}+\beta _{31},\end{aligned}$$

(243)

$$\begin{aligned} \beta _{63}&=\beta _{23}e^{A_1}+\beta _{32}. \end{aligned}$$

(244)

Our aim in the following analysis is to find the conditions under which we have $x_u^\circ (1)=x_u^\circ (\varrho _3)$ for all inputs u that satisfy the relations (209)–(212). This means finding the conditions under which we have $x_u^\circ (1)=x_u^\circ (\varrho _3)$ for all $\alpha \in \;]0,1[$. In the equality $x_u^\circ (1)=x_u^\circ (\varrho _3)$ we multiply both terms with $1+\beta _{55}e^{(A_2-A_1)\alpha }$ so that we get from Equalities (240)–(244) that

$$\begin{aligned} \beta _{71}+\beta _{72}e^{(A_2-A_1)\alpha }+&\;\beta _{73}\alpha e^{-A_1\alpha }+\beta _{74}e^{-A_1\alpha }=0, \nonumber \\&\forall \alpha \in \; ]0,1[, \end{aligned}$$

(245)

where

$$\begin{aligned} \beta _{71}&=e^{A_1}\beta _{51}+\beta _{11}-\beta _{33}, \end{aligned}$$

(246)

$$\begin{aligned} \beta _{72}&=e^{A_1}\beta _{52}+\beta _{11}\beta _{55}-e^{2A_1-A_2}\beta _{51}-\beta _{61}-\beta _{55}\beta _{33}, \end{aligned}$$

(247)

$$\begin{aligned} \beta _{73}&=e^{A_1}\beta _{53}-\beta _{62},\end{aligned}$$

(248)

$$\begin{aligned} \beta _{74}&=e^{A_1}\beta _{54}-\beta _{63}. \end{aligned}$$

(249)

Consider the functions $f_1,f_2,f_3,f_4,\mathbf {0}:\;]0,1[\; \rightarrow \mathbb {R}$ defined by $\forall \alpha \in \;]0,1[, f_1(\alpha )=1$, $f_2(\alpha )=e^{(A_2-A_1)\alpha }$, $f_3(\alpha )=\alpha e^{-A_1\alpha }$, $f_4(\alpha )=e^{-A_1\alpha }$, and $\mathbf {0}(\alpha )=0$. Then Eq. (245) can be written as

$$\begin{aligned} \beta _{71}\cdot f_1+\beta _{72}\cdot f_2+\beta _{73}\cdot f_3+\beta _{74}\cdot f_4=\mathbf {0}. \end{aligned}$$

(250)

Consider the vector space of functions $\{p:\;]0,1[\; \rightarrow \mathbb {R}\}$ with its usual binary operations of vector addition and scalar multiplication. Then the functions $f_1,f_2,f_3,f_4$ are linearly independent vectors so that, owing to Eq. (250), we have $\beta _{71}=\beta _{72}=\beta _{73}=\beta _{74}=0$ since $\beta _{ij}$ is independent of $\alpha$ for all possible i and j.

We start by solving Equation $\beta _{73}=0$. Combining Eqs. (248), (237), (243), (222), and (227) it comes that

$$\begin{aligned} A_2^{-1} B_2 = A_1^{-1}B_1. \end{aligned}$$

(251)

Now we solve Equation $\beta _{71}=0$. Combining Eqs. (251), (246), (235), (229), (233), and (217) it follows that

$$\begin{aligned} B_1 A_1^{-1}\left( A_2^{-1}-A_1^{-1} \right) -E_1 A_1^{-1}+E_2A_2^{-1}=0. \end{aligned}$$

(252)

It can be checked that Equalities (251)–(252) imply that $\beta _{71}=\beta _{72}=\beta _{73}=\beta _{74}=0$.

Lemma 11 follows from Lemma 8.