Discrete-time map for an impulsive Goodwin oscillator with a distributed delay

Abstract

A system of impulsive integro-differential equations representing a hybrid counterpart of the renown Goodwin oscillator is considered. The continuous part of the system possesses a cascade structure and contains a distributed bounded delay. The impulses impacting the continuous part are modulated in amplitude and frequency by the continuous output thus implementing an impulsive feedback. This kind of mathematical models appears in mathematical biology and computational medicine. Applying a version of the linear chain trick, it is demonstrated that a discrete-time (Poincaré) map can be constructed to capture the main dynamical properties of the system in hand.

Proof of Theorem 1

The following lemma plays a key role in the proof. Let $\widetilde{K}(t)$ be a matrix valued function of a scalar argument t:

$$\begin{aligned} \widetilde{K}(t) = \int _t^h K(s) \mathrm {e}^{A_0(t-s)}\mathrm{d}s. \end{aligned}$$

Lemma 2

Let $t_n<t<t_{n+1}$ where $n\ge 2$. Then any solution x(t) of (18), (19) satisfies the ordinary differential equation

$$\begin{aligned} x' =Dx-(D-A_0)\eta (t), \end{aligned}$$

(35)

where the vector-valued function $\eta (t)$ is defined as follows:

Case (i) $T_n>h$, $T_{n-1}>h$. Then

$$\begin{aligned} \eta (t)={\left\{ \begin{array}{ll} \lambda _n \mathrm {e}^{A_0(t-t_n)}B, &{}t_n<t\le t_n+\tau , \\ \lambda _n \widetilde{K}(t-t_n)\,B_0, &{} t_n+\tau <t\le t_n+h, \\ 0, &{} t_n+h<t< t_{n+1}. \end{array}\right. } \end{aligned}$$

Case (ii) $T_n<h$, $T_{n-1}>h$. Then

$$\begin{aligned} \eta (t)={\left\{ \begin{array}{ll} \lambda _n \mathrm {e}^{A_0(t-t_n)}B, &{}t_n<t\le t_n+\tau , \\ \lambda _n \widetilde{K}(t-t_n)\,B_0, &{} t_n+\tau <t< t_{n+1}. \end{array}\right. } \end{aligned}$$

Case (iii) $T_n<h$, $T_{n-1}<h$. Then

$$\begin{aligned} \eta (t)={\left\{ \begin{array}{ll} \lambda _n \mathrm {e}^{A_0(t-t_n)}B + \lambda _{n-1} \widetilde{K}(t-t_{n-1})\,B_0, &{}t_n<t\le t_{n-1}+h, \\ \lambda _n \mathrm {e}^{A_0(t-t_n)}B, &{}t_{n-1}+h<t\le t_n+\tau , \\ \lambda _n \widetilde{K}(t-t_n)\,B_0, &{} t_n+\tau <t< t_{n+1}. \end{array}\right. } \end{aligned}$$

Case (iv) $T_n>h$, $T_{n-1}<h$. Then

$$\begin{aligned} \eta (t)={\left\{ \begin{array}{ll} \lambda _n \mathrm {e}^{A_0(t-t_n)}B + \lambda _{n-1} \widetilde{K}(t-t_{n-1})\,B_0, &{}t_n<t\le t_{n-1}+h, \\ \lambda _n \mathrm {e}^{A_0(t-t_n)}B, &{}t_{n-1}+h<t\le t_n+\tau , \\ \lambda _n \widetilde{K}(t-t_n)\,B_0, &{} t_n+\tau <t\le t_n+h, \\ 0, &{} t_n+h<t< t_{n+1}. \end{array}\right. } \end{aligned}$$

Proof of Lemma 2

From (13), one gets

$$\begin{aligned} A_1=(D-A_0) {\mathfrak {K}}(A_0)^{-1}. \end{aligned}$$

Now (6) can be rewritten as

$$\begin{aligned} x' =Dx+(D-A_0)\left( {\mathfrak {K}}(A_0)^{-1} x_d - x\right) . \end{aligned}$$

(36)

Without loss of generality (cf. Lemma 1), let the continuous part of (18) be readily given in the form of (14). Let also the matrix B be partitioned as

$$\begin{aligned} B = \begin{bmatrix} B_1 \\ B_2 \end{bmatrix}, \end{aligned}$$

where the dimensions of the vectors $B_1$, $B_2$ correspond to those of u, v, respectively. The matrix D can be represented as (16) and

$$\begin{aligned} D-A_0 = \begin{bmatrix} 0&0\\ \bar{W}{\mathfrak {K}}(U)&0 \end{bmatrix}. \end{aligned}$$

(37)

It follows from (14) that

$$\begin{aligned} {\mathfrak {K}}(A_0) = \begin{bmatrix} {\mathfrak {K}}(U)&0\\ *&* \end{bmatrix}, \qquad {\mathfrak {K}}(A_0)^{-1}x_d-x = \begin{bmatrix} {\mathfrak {K}}(U)^{-1}u_d-u \\ * \end{bmatrix}, \end{aligned}$$

where asterisks stand for any matrix blocks of appropriate size. Hence (36) and (37) yield

$$\begin{aligned} x' =Dx-(D-A_0)\eta (t), \end{aligned}$$

(38)

where

$$\begin{aligned} \eta (t)=\begin{bmatrix} u(t)-{\mathfrak {K}}(U)^{-1} u_d(t) \\ * \end{bmatrix}, \end{aligned}$$

$u_d(t)$ is defined by (15) and $*$ is any column of a suitable size. Equation (38) holds for any time interval where the function x(t) is smooth, i.e., in any interval that does not contain the points $t_0,t_1,\ldots$.

Define the time intervals

$$\begin{aligned} L_k=\{t\,: \, t_k<t<t_{k+1}\},\quad k=0,1,\ldots . \end{aligned}$$

For brevity, denote $u_k^- = u(t_k^-)$, $u_k^+ = u(t_k^+)$. One has $u_k^+= u_k^- +\lambda _k B_1$ and

$$\begin{aligned} u(t) ={\left\{ \begin{array}{ll} \mathrm {e}^{U(t-t_k)}u_k^+, &{}t\in L_k, \\ \mathrm {e}^{U(t-t_k)}u_k^-, &{}t\in L_{k-1}. \end{array}\right. } \end{aligned}$$

(39)

Suppose $t\in L_n$, $n\ge 2$. To prove Lemma 2, it suffices to show that $\eta (t)$ can be chosen as

$$\begin{aligned} \eta (t) = {\left\{ \begin{array}{ll} 0 &{}\text{ if }\quad t-h\in L_n,\; t-\tau \in L_n; \\ \lambda _n \mathrm {e}^{A_0(t-t_n)}B &{}\text{ if }\quad t-h\in L_{n-1},\; t-\tau \in L_{n-1}; \\ \lambda _{n-1} \widetilde{K}(t-t_{n-1})\,B_0 &{}\text{ if }\quad t-h\in L_{n-1},\; t-\tau \in L_n; \\ \lambda _n \mathrm {e}^{A_0(t-t_n)}B+ \lambda _{n-1} \widetilde{K}(t-t_{n-1})\,B_0 &{}\text{ if }\quad t-h\in L_{n-2},\; t-\tau \in L_{n-1}. \end{array}\right. } \end{aligned}$$

(a) Let $t-h\in L_n$, $t-\tau \in L_n$. Since $t\in L_n$, (39) implies $u(t) = \mathrm {e}^{U(t-t_n)} u_n^+$. Moreover, since $t-s\in L_n$ for $\tau \le s\le h$ and from (39), one obtains

$$\begin{aligned} u_d(t)=\int _\tau ^h K(s)\,\mathrm {e}^{U(t-t_n-s)}\,\mathrm{d}s\, u_n^+ = {\mathfrak {K}}(U) u(t). \end{aligned}$$

Hence x(t) is a solution of (8) and $\eta (t)\equiv 0$ is justified.

(b) Let $t-h\in L_{n-1}$, $t-\tau \in L_{n-1}$. Hence $t-s\in L_{n-1}$ for $\tau \le s\le h$ and making use (39) yields

$$\begin{aligned} u_d(t) = \int _\tau ^h K(s)\,\mathrm {e}^{U(t-s-t_n)}\,\mathrm{d}s\, u_n^- = {\mathfrak {K}}(U) \mathrm {e}^{U(t-t_n)} u_n^-. \end{aligned}$$

Since $u_n^- = u_n^+ -\lambda _n B_1$, one gets

$$\begin{aligned} \mathrm {e}^{U(t-t_n)} u_n^- = u(t) -\lambda _n \mathrm {e}^{U(t-t_n)} B_1. \end{aligned}$$

(40)

Thus

$$\begin{aligned} u(t)-{\mathfrak {K}}(U)^{-1} u_d(t) = \lambda _n \mathrm {e}^{U(t-t_n)} B_1. \end{aligned}$$

Since

$$\begin{aligned} \mathrm {e}^{A_0t}B=\begin{bmatrix} \mathrm {e}^{Ut}&0\\ *&* \end{bmatrix} \begin{bmatrix} B_1\\ B_2 \end{bmatrix}= \begin{bmatrix} \mathrm {e}^{Ut}B_1 \\ * \end{bmatrix}, \end{aligned}$$

where asterisks stand for some blocks, $\eta (t)= \lambda _n \mathrm {e}^{A_0(t-t_n)} B$ can be taken in (38).

(c) Let $t-h\in L_{n-1}$, $t-\tau \in L_n$. Then $t-h<t_n<t-\tau$, so $t-s\in L_n$ if $\tau <s<t-t_n$ and $t-s\in L_{n-1}$ if $t-t_n<s<h$. Hence

$$\begin{aligned} u(t-s)={\left\{ \begin{array}{ll} \mathrm {e}^{U(t-t_n-s)} u_n^+ &{}\text{ if }\quad \tau < s < t -t_n. \\ \mathrm {e}^{U(t-t_n-s)} u_n^- &{}\text{ if }\quad t-t_n< s <h. \end{array}\right. } \end{aligned}$$

Then

$$\begin{aligned} u_d(t) = \int _{\tau }^{t-t_n} K(s)\,\mathrm {e}^{U(t-t_n-s)}\,\mathrm{d}s\, u_n^+ +\int _{t-t_n}^h K(s)\,\mathrm {e}^{U(t-t_n-s)}\,\mathrm{d}s\, u_n^- . \end{aligned}$$

Using (40), rewrite the previous line as

$$\begin{aligned} u_d(t)= {\mathfrak {K}}(U)u(t) -\lambda _n \int _{t-t_n}^h K(s)\,\mathrm {e}^{U(t-t_n-s)}\,\mathrm{d}s\,B_1. \end{aligned}$$

leading to

$$\begin{aligned} u(t)- {\mathfrak {K}}(U)^{-1}u_d(t) = \lambda _n {\mathfrak {K}}(U)^{-1} \int _{t-t_n}^h K(s)\mathrm {e}^{U(t-t_n-s)} \mathrm{d}s B_1. \end{aligned}$$

Thus $\eta (t)$ in (38) can be chosen as

$$\begin{aligned} \eta (t)=\lambda _n {\mathfrak {K}}(A_0)^{-1} \int _{t-t_n}^h K(s)\mathrm {e}^{A_0(t-t_n-s)} \mathrm{d}s B= \lambda _n \widetilde{K}(t-t_n) B_0. \end{aligned}$$

(d) Let $t-h\in L_{n-2}$, $t-\tau \in L_{n-1}$. Hence $t-h<t_{n-1}<t-\tau$, so $t-s\in L_{n-1}$ if $\tau <s<t-t_{n-1}$ and $t-s\in L_{n-2}$ if $t-t_{n-1}<s<h$. Hence

$$\begin{aligned} u_d(t) = \int _{\tau }^{t-t_{n-1}} K(s)\mathrm {e}^{U(t-t_n-s)}\, \mathrm{d}s\, u_n^- + \int _{t-t_{n-1}}^h K(s)\mathrm {e}^{U(t-t_{n-1}-s)}\, \mathrm{d}s\, u_{n-1}^-. \end{aligned}$$

Since

$$\begin{aligned} u^+_{n-1}= u^-_{n-1} +\lambda _{n-1} B_1, \quad u^-_n = \mathrm {e}^{UT_{n-1}} u^+_{n-1}, \end{aligned}$$

it can be concluded that

$$\begin{aligned} u^+_{n-1}= \mathrm {e}^{-UT_{n-1}} u^-_n - \lambda _{n-1} B_1 \end{aligned}$$

and hence

$$\begin{aligned} u_d(t) = {\mathfrak {K}}(U) \mathrm {e}^{U(t-t_n)} u_n^- - \lambda _{n-1} \int _{t-t_{n-1}}^h K(s)\mathrm {e}^{U(t-t_{n-1}-s)}\, \mathrm{d}s\, B_1. \end{aligned}$$

It is deduced from (40) that

$$\begin{aligned} u(t)- {\mathfrak {K}}(U)^{-1}u_d(t)= \lambda _n \mathrm {e}^{U(t-t_n)} B_1 +\lambda _{n-1} {\mathfrak {K}}(U)^{-1} \int _{t-t_{n-1}}^h K(s)\mathrm {e}^{U(t-s-t_{n-1})}\, \mathrm{d}s\, B_1. \end{aligned}$$

Thus

$$\begin{aligned} \eta (t)=\lambda _n \mathrm {e}^{A_0(t-t_n)} B + \lambda _{n-1} \widetilde{K}(t-t_{n-1})\, B_0. \end{aligned}$$

can be taken in (38),

The proof of Lemma 2 is complete. The corollary below follows immediately. $\square$

Corollary 1

The function $\eta (t)$ in Lemma 2 is continuous in $(t_n,t_{n+1})$ and piecewise smooth. For $n\ge 2$ in its intervals of smoothness $\eta (t)$ satisfies the differential equation

$$\begin{aligned} \eta '(t)=A_0\eta (t)-H(t), \end{aligned}$$

where H(t) is defined as follows

Case (i) $T_n>h$, $T_{n-1}>h$. Then

$$\begin{aligned} H(t) ={\left\{ \begin{array}{ll} 0, &{}t_n<t<t_n+\tau , \\ \lambda _n K(t-t_n) B_0, &{}t_n+\tau <t<t_n+h, \\ 0, &{} t_n+h<t<t_{n+1}. \end{array}\right. } \end{aligned}$$

Case (ii) $T_n<h$, $T_{n-1}>h$. Then

$$\begin{aligned} H(t) ={\left\{ \begin{array}{ll} 0, &{}t_n<t<t_n+\tau , \\ \lambda _n K(t-t_n) B_0, &{}t_n+\tau <t<t_{n+1}. \end{array}\right. } \end{aligned}$$

Case (iii) $T_n<h$, $T_{n-1}<h$. Then

$$\begin{aligned} H(t) ={\left\{ \begin{array}{ll} \lambda _{n-1} K(t-t_{n-1}) B_0, &{}t_n<t<t_{n-1}+h, \\ 0, &{}t_{n-1}+h<t<t_n+\tau , \\ \lambda _n K(t-t_n) B_0, &{}t_n+\tau <t<t_{n+1}. \end{array}\right. } \end{aligned}$$

Case (iv) $T_n>h$, $T_{n-1}<h$. Then

$$\begin{aligned} H(t) ={\left\{ \begin{array}{ll} \lambda _{n-1} K(t-t_{n-1}) B_0, &{}t_n+\tau <t<t_{n-1}+h, \\ 0, &{}t_{n-1}+h<t<t_n+\tau , \\ \lambda _n K(t-t_n) B_0, &{}t_n+\tau <t<t_n+h, \\ 0, &{} t_n+h<t<t_{n+1}. \end{array}\right. } \end{aligned}$$

Now all the prerequisites for the proof of Theorem 2 are in place.

For $t_n<t< t_{n+1}$, $n\ge 2$, introduce the function $z(t)=x(t)-\eta (t)$, where x(t) is a solution of (18), (19). From (35) and Corollary 1, on those subintervals of $(t_n,t_{n+1})$, where $\eta (t)$ is smooth, it applies that

$$\begin{aligned} z'(t)=Dz(t)+H(t). \end{aligned}$$

(41)

Since $x(t^-_{n+1})=\bar{x}_{n+1}$, $x(t^+_n)=\bar{x}_n+\lambda _n B$, from (41) it follows

$$\begin{aligned} \bar{x}_{n+1} = \eta (t^-_{n+1})+ \mathrm {e}^{D T_n} (\bar{x}_n +\lambda _n B - \eta (t^+_n)) +\int _{t_n}^{t_{n+1}} \mathrm {e}^{D(t_{n+1}-s)} H(s)\, \mathrm{d}s. \end{aligned}$$

(42)

Consider four previously defined cases (i)–(iv).

Case (i) From Lemma 2, the following boundary conditions are obtained

$$\begin{aligned} \eta (t_n^+) = \lambda _n B, \quad \eta (t_{n+1}^-)=0. \end{aligned}$$

Since

$$\begin{aligned} \int _{t_n+\tau }^{t_n+h} K(s-t_n)\mathrm {e}^{D(t_{n+1}-s)} \, \mathrm{d}s= \int _{\tau }^h K(s) \mathrm {e}^{D(T_n-s)} \, \mathrm{d}s = \mathrm {e}^{DT_n} {\mathfrak {K}}(D), \end{aligned}$$

(43)

(42) takes the form

$$\begin{aligned} \bar{x}_{n+1} = \mathrm {e}^{D T_n} \bar{x}_n +\lambda _n\mathrm {e}^{DT_n}{\mathfrak {K}}(D) B_0=Q_n. \end{aligned}$$

This implies the claim of Theorem 2 for case (i).

Case (ii) The boundary conditions are

$$\begin{aligned} \eta (t_n^+) = \lambda _n B, \quad \eta (t_{n+1}^-)= \lambda _n \widetilde{K}(T_n)\, B_0. \end{aligned}$$

Then (42) takes the form

$$\begin{aligned} \bar{x}_{n+1} = \mathrm {e}^{D T_n} \bar{x}_n + \lambda _n \int _{T_n}^h K(s) \mathrm {e}^{A_0(T_n-s)} \mathrm{d}s\, B_0 +\lambda _n \int _{\tau }^{T_n} K(s) \mathrm {e}^{D(T_n-s)} \mathrm{d}s\, B_0. \end{aligned}$$

Since

$$\begin{aligned} \int _{\tau }^{T_n} K(s) \mathrm {e}^{D(T_n-s)} \mathrm{d}s= & {} \int _{\tau }^h K(s) \mathrm {e}^{D(T_n-s)} \mathrm{d}s - \int _{T_n}^h K(s) \mathrm {e}^{D(T_n-s)} \mathrm{d}s \\= & {} \mathrm {e}^{DT_n} {\mathfrak {K}}(D)-\int _{T_n}^h K(s) \mathrm {e}^{D(T_n-s)} \mathrm{d}s, \end{aligned}$$

one finally arrives at

$$\begin{aligned} \bar{x}_{n+1} = \mathrm {e}^{D T_n} \bar{x}_n + \lambda _n\mathrm {e}^{D T_n}{\mathfrak {K}}(D)B_0- \lambda _n \int _{T_n}^h K(s) G(T_n-s)\, \mathrm{d}s, \end{aligned}$$

where G(t) is defined by (22). Since $T_n={\varPhi }(C\bar{x}_n)$ and $\lambda _n=F(C\bar{x}_n)$,

$$\begin{aligned} \bar{x}_{n+1}=Q_n - \lambda _n R(T_n). \end{aligned}$$

Case (iii) The boundary conditions are

$$\begin{aligned} \eta (t_n^+) = \lambda _n B+ \lambda _{n-1} \widetilde{K}(T_{n-1})\, B_0, \quad \eta (t_{n+1}^-)= \lambda _n \widetilde{K}(T_n)\, B_0. \end{aligned}$$

Since

$$\begin{aligned} \int _{t_n}^{t_{n-1}+h} K(s-t_{n-1})\mathrm {e}^{D(t_{n+1}-s)}\,\mathrm{d}s= \mathrm {e}^{DT_n} \int _{T_{n-1}}^h K(s)\mathrm {e}^{D(T_{n-1}-s)}\,\mathrm{d}s \end{aligned}$$

(44)

and

$$\begin{aligned} \int _{t_n+\tau }^{t_{n+1}} K(s-t_n)\mathrm {e}^{D(t_{n+1}-s)}\,\mathrm{d}s= & {} \mathrm {e}^{DT_n} \int _{\tau }^{T_n} K(s) \mathrm {e}^{-Ds} \mathrm{d}s \\= & {} \mathrm {e}^{DT_n} {\mathfrak {K}}(D) - \int _{T_n}^h K(s) \mathrm {e}^{D(T_n-s)} \mathrm{d}s, \end{aligned}$$

equality (42) takes the form

$$\begin{aligned} \bar{x}_{n+1}= & {} \mathrm {e}^{D T_n} \bar{x}_n + \lambda _n \mathrm {e}^{DT_n} {\mathfrak {K}}(D) B_0 \\&-\lambda _n \int _{T_n}^h K(s)G(T_n-s)\,\mathrm{d}s + \lambda _{n-1}\mathrm {e}^{DT_n} \int _{T_{n-1}}^h K(s)G(T_{n-1}-s)\,\mathrm{d}s. \end{aligned}$$

This implies

$$\begin{aligned} \bar{x}_{n+1}=Q_n-\lambda _nR(T_n)+ \lambda _{n-1}\mathrm {e}^{DT_n}R(T_{n-1}). \end{aligned}$$

Case (iv) The boundary conditions are

$$\begin{aligned} \eta (t_n^+) = \lambda _n B+ \lambda _{n-1} \widetilde{K}(T_{n-1})\, B_0, \quad \eta (t_{n+1}^-)= 0. \end{aligned}$$

Using (43), (44) results in

$$\begin{aligned} \bar{x}_{n+1}= & {} \mathrm {e}^{D T_n} \bar{x}_n + \lambda _n \mathrm {e}^{DT_n} {\mathfrak {K}}(D) B_0 \\&+\lambda _{n-1}\mathrm {e}^{DT_n} \int _{T_{n-1}}^h K(s)G(T_{n-1}-s)\,\mathrm{d}s. \end{aligned}$$

Hence

$$\begin{aligned} \bar{x}_{n+1}=Q_n+ \lambda _{n-1}\mathrm {e}^{DT_n}R(T_{n-1}). \end{aligned}$$

The proof of Theorem 2 is complete.