Oscillatory Behaviour on a Non-autonomous Hybrid SIR-Model

Abstract

We study the impact of some abstract agent intervention on the disease spread modelled by a SIR-model with linear growth infectivity. The intervention is meant to decrease the infectivity, which are activated by a threshold on the number of infected individuals. The coupled model is represented as a nonlinear non-autonomous hybrid system. Stability and reduction results are obtained using the notions of non-autonomous attractors, Bohl exponents, and dichotomy spectrum. Numerical examples are given where the number of infected individuals can oscillate around a equilibrium point or be a succession of bump functions, which are validated with a tool based on the notion of $\delta$-complete decision procedures for solving satisfiability modulo theories problems over the real numbers and bounded $\delta$-reachability. These findings seem to show that hybrid SIR-models are more flexible than standard models and generate a vast set of solution profiles. It also raises questions regarding the possibility of the agent intervention been somehow responsible for the shape and intensity of future outbreaks.

Appendix

Proof

(Lemma 1) Let $h=y-x$ for $x,y\in \varSigma _1$. Recall that the mean value inequality, for a vector value function $F:\mathbb {R}\times \varSigma _1\rightarrow \mathbb {R}^3$, says that when the Jacobian matrix of F at $w=x+\tau h$, i.e. $J_F(w)$, is uniformly bounded by some constant $L>0$ for any $\tau \in$ [0,1] and $t\in \mathbb {R}$, then

$$ | F(t,x+h)-F(t,x)|\le L\,|h|.$$

Hence, the function F in Eq. (2.3) is locally Lipschitz continuous in the second variable since

$$\begin{aligned} J_F(w(t))= \left( \begin{array}{ccc} -\alpha -\beta _\xi (t) w_2(t) &{} \zeta -\beta _\xi (t) w_1(t) &{} 0\\ \beta _\xi (t) w_2(t) &{} \beta _\xi (t) w_1(t)-(\zeta +\alpha +\gamma ) &{} 0\\ 0 &{} \gamma &{} -\alpha \end{array}\right) \end{aligned}$$

is uniformly bounded, because $\beta _\xi (t)\in [\beta _*,\beta ^*]$ and w(t) is bounded for any $\tau \in$ [0,1], by definition. So, the Picard-Lindelöf theorem ensures the existence and uniqueness of solution in each node of the hybrid system. Its not difficult to see that the solution is globally defined. Further, the hybrid system Fig. 2 is deterministic and has only one jump condition in each node, so we conclude the proof.

Proof

(Lemma 4) First note that

$$\begin{aligned} I^*(t)^{-1}&=\int _{-\infty }^t \beta _\xi (r)\varphi (t,r)^{-1}\,dr =\hat{\varphi }(t)^{-1}\lim _{a\rightarrow +\infty }\int _{-a}^t \beta _\xi (r)\hat{\varphi }(r)dr\\&=\hat{\varphi }(t)^{-1}\int _{0}^t \beta _\xi (r)\hat{\varphi }(r)dr +\hat{\varphi }(t)^{-1}\lim _{a\rightarrow +\infty }\int _{-a}^0 \beta _\xi (r)\hat{\varphi }(r)dr \end{aligned}$$

with $t\in [0,\mathcal {T}]$. We consider two cases: (a) $\xi =0$; and (b) $\xi \ge 0$.

(a) For $\xi =0$, i.e. when $\beta _\xi (t)\equiv \beta _0$ is constant and $\hat{\varphi }(r)= e^{\xi _0r}$, we have

$$\begin{aligned} I^*(t)^{-1}&=\beta _0e^{-\xi _0t}\int _{0}^t e^{\xi _0r}dr+\beta _0e^{-\xi _0t}\lim _{a\rightarrow +\infty }\int _{-a}^0 e^{\xi _0r}dr\\&=\frac{\beta _0}{\xi _0}\left( 1-\lim _{a\rightarrow +\infty }e^{-\xi _0 a}\right) . \end{aligned}$$

From which, we obtain: (i) for $\xi _0\le 0$, then $(S^*(t),I^*(t))=(1,0)$; and (b) for $\xi _0>0$, we recover the expected values

$$\begin{aligned} I^*(t)=(\beta _0-\alpha -\zeta )\beta _0^{-1}\quad \hbox { and }\quad S^*(t)=(\alpha +\zeta )\beta _0^{-1}. \end{aligned}$$

(18)

(b) In general, for $\xi \ne 0$ and $t\in \overline{\mathcal {T}}_{t_0}$, using (12) and integration by parts, we have

$$\begin{aligned} I^*(t)^{-1}\hat{\varphi }(t)&=\int _{0}^t (\beta _0+\xi r)e^{\xi _0r+\frac{1}{2}\xi r^2}dr+\beta _0\lim _{a\rightarrow +\infty } \int _{-a}^0 e^{\xi _0r}dr\\&=\int _{0}^t (\beta _0+\xi r)e^{\xi _0r+\frac{1}{2}\xi r^2}dr+\lim _{a\rightarrow +\infty }\frac{\beta _0}{\xi _0}\left( 1-e^{-\xi _0 a}\right) \\&=\hat{\varphi }(t)-1-(\alpha +\zeta )G(t)+\lim _{a\rightarrow +\infty }\frac{\beta _0}{\xi _0}\left( 1-e^{-\xi _0 a}\right) . \end{aligned}$$

Therefore, we obtain: (i) for $\xi _0\le 0$, then $(S^*(t),I^*(t))=(1,0)$; and (ii) for $\xi _0>0$, we get

$$\begin{aligned} I^*(t)=\frac{\beta _0-\alpha -\zeta }{\xi _0+\beta _0\hat{\varphi }(t)^{-1}-\xi _0[1+(\alpha +\zeta )G(t)]\hat{\varphi }(t)^{-1}}\quad \hbox { and}\quad S^*(t)=1-I^*(t). \end{aligned}$$

(19)

In particular, when $\xi \rightarrow 0$, $G(t)\rightarrow (1-\hat{\varphi }(t))\xi _0^{-1}$ so we recover the values (18).

To prove the stability result, assume $\xi _0>\max \{0,-\xi \mathcal {T}\}$. This means that $\beta _*>\alpha +\zeta$. For any solution I(t), $D=I(t)-I^*(t)$ satisfy

$$ D'(t)=(\beta _\xi (t)-\alpha -\zeta )D(t)\ge (\beta _*-\alpha -\zeta )D(t) \quad \Rightarrow \,\,\, D'(t)^2\le D(0)^2\, e^{-2|\beta _*-\alpha -\zeta |t}, $$

so $D(t)\rightarrow 0$ and $I(t)-I^*(t)\rightarrow 0$ as $t\rightarrow +\infty$. Because $|S(t)-S^*(t)|=|1-I(t)-(1-I^*(t))|=|I^*(t)-I(t)|$, we have the desired conclusion.

Proof

(Lemma 5) Recall $\beta _\xi (t)\in [\beta _*,\beta ^*]$ for $t\in \mathbb {R}$. We consider three cases: (A) $\beta ^*<\alpha +\zeta$, (B) $\beta ^*=\alpha +\zeta$, and (C) $\beta _*>\alpha +\zeta$. (A) Using the second equation of (10) and $0\le S\le 1$, we obtain $(I^2)' = 2I\,I' =2\left( \beta _\xi S-\alpha -\zeta \right) I^2\le 2\left( \beta ^*-\alpha -\zeta \right) I^2$ which implies that $I(t)^2\le I(0)^2 e^{-2|\beta ^*-\alpha -\zeta |t}$ as $t\rightarrow +\infty ,$ meaning that $E_0$ is (asymptotically) stable. (B) Consider Eq. (11), so for all $I>0$, we have $I'= (\beta _\xi -\alpha -\zeta )I-\beta _\xi I^2 = -\beta _\xi I^2< -\frac{\beta _\xi }{\beta _*} I^2<0$ and $0\le I(t) \le \lim _{t\rightarrow +\infty }\frac{\beta _*I(0)}{\beta _*+\beta _\xi I(0)t}.$ Thus, $S(t)-S^*(t)=S(t)-1=-I(t)\rightarrow 0$ as $t\rightarrow +\infty$. Hence, $E_0$ is (asymptotically) stable. (C) Suppose $I(t)\le \epsilon \in$ [0,1], so $S(t)\ge 1-\epsilon$. Then

$$S' \le \left[ \alpha +\zeta -\beta _\xi (1-\epsilon )\right] I \quad \hbox { and }\quad I' \ge \left[ \beta _*(1-\epsilon )-\alpha -\zeta \right] I,$$

so I is strictly increasing if $0<I(t)\le \epsilon$ and $\epsilon < 1-(\alpha +\zeta )\beta _*^{-1}$. In particular, for any solution if $(S(t),I(t))\rightarrow (1,0)$ then we have that I(t) is strictly increasing (i.e. a contradiction), so $E_0$ is unstable.

Proof

(Lemma 6) Recall that $\beta _0>\max \{0,-\xi \mathcal {T}\}$, $\beta _*=\min \{\beta _0,\beta _0+\xi \mathcal {T}\}$ and $\beta ^*=\max \{\beta _0,\beta _0+\xi \mathcal {T}\}$. So, by direct computation, we get

$$\max \{\beta _0,\beta _0+\xi \mathcal {T}\}\le \alpha +\zeta \,\Leftrightarrow \, \max \{0,\xi \mathcal {T}\}\le -\xi _0 \,\Leftrightarrow \, \xi _0 \le \min \{0,-\xi \mathcal {T}\},$$

$$\min \left\{ \beta _0,\beta _0+\xi \mathcal {T}\right\}>\alpha +\zeta \,\Leftrightarrow \, \min \left\{ 0,\xi \mathcal {T}\right\}>-\xi _0 \,\Leftrightarrow \, \xi _0>\max \left\{ 0,-\xi \mathcal {T}\right\} ,$$

for which Lemma 5 implies that $E_0$ is globally asymptotically stable w.r.t. $\varSigma _0$, when $\xi _0 \le \min \{0,-\xi \mathcal {T}\}$, and unstable when $\xi _0>\max \left\{ 0,-\xi \mathcal {T}\right\}$. The statements in this lemma are then a direct consequence of $\mathcal {T}\ge 0$ and $\beta _-<0<\beta _+$.

Proof

(Lemma 8) First, suppose the system (10) is defined on $t\in J\subseteq \mathbb {R}$. By applying Lemma 7, we conclude that the linear part of (10) has the dichotomy spectrum $-(\alpha +\zeta )+[\underline{\beta }_{J}(\beta _\xi ),\overline{\beta }_{J}(\beta _\xi )]$. Hence, by propositions 4.9 and 4.10 in [19], Lemma 5 is still valid when we replace globally asymptotically stable by uniformly asymptotically stable and $\beta _*,\beta ^*$ by $\underline{\beta }_{J}(\beta _\xi ),\overline{\beta }_{J}(\beta _\xi )$, respectively. This tell us that it is expect to occur a bifurcation of (10) when $\underline{\beta }_{J}(\beta _\xi )=\overline{\beta }_{J}(\beta _\xi )=\alpha +\zeta$.

We have $\underline{\beta }_{\overline{\mathcal {T}}_{0}}(\beta _\xi )=\overline{\beta }_{\overline{\mathcal {T}}_{0}}(\beta _\xi )=\beta _0+\frac{1}{2}\xi \mathcal {T}$. In fact, note that $\beta _\xi$ is a integrable bounded function in $\overline{\mathcal {T}}_{0}$. For $s,t \in \overline{\mathcal {T}}_{0}$ and $w\in \mathbb {R}$, let

$$\begin{aligned}&F(s,t,w) = \frac{1}{t-s}\int _s^t \beta _\xi (r)-w\,dr = \beta _0+\frac{1}{2}\xi (t+s) -w,\\&\sup _{s\le t, (s,t)\in [0,\mathcal {T}]^2} F(s,t,w) =\sup _{t\le s, (s,t)\in [0,\mathcal {T}]^2} F(s,t,w)= \beta _0+\frac{1}{2}\xi \mathcal {T} -w. \end{aligned}$$

Hence, $\underline{\beta }_{\overline{\mathcal {T}}_{0}}(\beta _\xi )=\overline{\beta }_{\overline{\mathcal {T}}_{0}}(\beta _\xi )=\beta _0+\frac{1}{2}\xi \mathcal {T}$. So, replacing the obtained values in $\overline{\beta }_{\overline{\mathcal {T}}_{0}}(\beta _\xi )\le \alpha +\zeta$, $\underline{\beta }_{\overline{\mathcal {T}}_{0}}(\beta _\xi )>\alpha +\zeta$ and recalling that $\xi _0=\beta _0-\alpha -\zeta$, we simplify to

$$\beta _0+\frac{1}{2}\xi \mathcal {T}\le \alpha +\zeta \,\Leftrightarrow \, \xi _0 \le -\frac{1}{2}\xi \mathcal {T} \,\hbox { and }\,\beta _0+\frac{1}{2}\xi \mathcal {T}>\alpha +\zeta \,\Leftrightarrow \, \xi _0>-\frac{1}{2}\xi \mathcal {T},$$

which confirms the above bifurcation point of (10).

By Lemma 7, we have that $\underline{\beta }_{\mathbb {R}}(\beta _\xi )=\min \{\beta _0,\beta _0+\xi \mathcal {T}\}$ and $\overline{\beta }_{\mathbb {R}}(\beta _\xi )=\max \{\beta _0,\beta _0+\xi \mathcal {T}\}$. In the same way, for $t\in \mathbb {R}$, we have

$$\max \{\beta _0,\beta _0+\xi \mathcal {T}\}\le \alpha +\zeta \,\Leftrightarrow \, \xi _0 \le \min \{0,-\xi \mathcal {T}\},$$

$$\min \{\beta _0,\beta _0+\xi \mathcal {T}\}>\alpha +\zeta \,\Leftrightarrow \, \xi _0>\max \{0,-\xi \mathcal {T}\},$$

The statement in the lemma is a consequence of $\mathcal {T}\ge 0$ and $\beta _-<0<\beta _+$.

Proof

(Lemma 9) Let $v_1=\psi _I(0)$ and $v_2=\psi _R(0)$. The functions $\psi _I, \psi _R$ are continuous so there exist $T>0$ such that their signs are preserved in [0, T], so from (14) they prescribe the monotonicity of S(t), I(t), R(t) in $t\in [0,T]$. Now, it is enough to explicitly construct the map $(v_1,v_2)\mapsto (S(0), I(0), R(0))$ as

$$ S(0) = \frac{v_1+q}{\beta _0},\,\, I(0) = -\frac{\alpha (-\beta _0+v_1+q)}{\beta _0 (\alpha +\gamma -v_2)},$$

$$ R(0) = -\frac{\gamma ^2+(v_1-\beta _0-v_2+\alpha +\zeta )\gamma +v_2(\beta _0- v_1-\alpha -\zeta )}{\beta _0(\alpha +\gamma -v_2)}, $$

where $q=\alpha +\zeta +\gamma$, $\beta _0=\beta _\xi (0)$ and $v_1\in (-q, -q+\beta _0)$, $v_2\in (\gamma ,\gamma +\alpha )$ (which ensure $0<S(0),I(0),R(0)<1$).

Proof

(Lemma 11) Consider the (reduced) system (17). The dichotomy spectrum (of the linear part) is

$$\varSigma _A=[-\alpha ,-\alpha ]\cup [\underline{\beta }_{\overline{\mathcal {T}}}(\beta _\xi )-\alpha -\zeta -\gamma ,\overline{\beta }_{\overline{\mathcal {T}}}(\beta _\xi )-\alpha -\zeta -\gamma ].$$

Recall $\underline{\beta }_{\overline{\mathcal {T}}_{0}}(\beta _\xi )=\overline{\beta }_{\overline{\mathcal {T}}_{0}}(\beta _\xi )=\beta _0+\frac{1}{2}\xi \mathcal {T}$, First, suppose $\xi _0>-\alpha -\frac{1}{2}\xi \mathcal {T}$ so we have the (ordered) dichotomy spectrum

$$\varSigma _A=[-\alpha ,-\alpha ]\cup \left[ \xi _0+\frac{1}{2}\xi \mathcal {T},\xi _0+\frac{1}{2}\xi \mathcal {T}\right] $$

and $a_1b_2 = -\alpha \left( \xi _0+\frac{1}{2}\xi \mathcal {T}\right) >0$, meaning it is a system of type-I. Since $a_i=b_i$, we also have that, for $m\in \mathbb {N}^2$, to be of type-II is the same as

$$\begin{aligned}&0\not \in \left\{ -\alpha m_1 + \left( \xi _0+\frac{1}{2}\xi \mathcal {T}\right) m_2+\alpha , -\alpha m_1 + \left( \xi _0+\frac{1}{2}\xi \mathcal {T}\right) m_2-\xi _0-\frac{1}{2}\xi \mathcal {T}\right\} , \end{aligned}$$

Which is true, since the inclusion $a m_1 +b m_2 \in \{a,b\}$ with $a=\alpha$, $b=\left| \xi _0+\frac{1}{2}\xi \mathcal {T}\right|$, do not have integer solutions. Hence, it is of type-II. If we suppose $\xi _0<-\alpha -\frac{1}{2}\xi \mathcal {T}$ so we have the (ordered) dichotomy spectrum

$$\varSigma _A=\left[ \xi _0+\frac{1}{2}\xi \mathcal {T},\xi _0+\frac{1}{2}\xi \mathcal {T}\right] \cup [-\alpha ,-\alpha ],$$

the conclusions are the same.

Then system (17) is locally analytically equivalent to its linear part $w'=A(t)w$, by applying Lemma 10. From (16), there exists a matrix Q(t), with determinant $\gamma ^{-1}\beta _\xi ^{-1}(\beta _\xi -\zeta -\gamma )(\zeta +\gamma )$, such that $w=Q(t) x$, so $x'=Q^{-1}(t)A(t)Q(t)x$ and then applying the Jordan canonical form transformation $x=Bv$, i.e.

$$ x'= \left( \begin{array}{ccc} 0 &{} 0 &{} 0\\ 0 &{} \xi _0+\xi t &{} 0\\ 0 &{} -\gamma (\zeta +\gamma )^{-1}(\alpha +\xi _0+\xi t) &{} -\alpha \end{array}\right) x \quad \hbox { and }\quad B = \left( \begin{array}{ccc} 1 &{} 0 &{} 0\\ 0 &{} 0 &{} 1\\ 0 &{} \frac{\gamma }{\zeta +\gamma } &{} -\frac{\gamma }{\zeta +\gamma } \end{array}\right) , $$

gives the expected result.