Multiscale analysis for a vector-borne epidemic model

Abstract

Traditional studies about disease dynamics have focused on global stability issues, due to their epidemiological importance. We study a classical SIR-SI model for arboviruses in two different directions: we begin by describing an alternative proof of previously known global stability results by using only a Lyapunov approach. In the sequel, we take a different view and we argue that vectors and hosts can have very distinctive intrinsic time-scales, and that such distinctiveness extends to the disease dynamics. Under these hypothesis, we show that two asymptotic regimes naturally appear: the fast host dynamics and the fast vector dynamics. The former regime yields, at leading order, a SIR model for the hosts, but with a rational incidence rate. In this case, the vector disappears from the model, and the dynamics is similar to a directly contagious disease. The latter yields a SI model for the vectors, with the hosts disappearing from the model. Numerical results show the performance of the approximation, and a rigorous proof validates the reduced models.

Appendix A

Proof of Theorem 3

The proof of Theorem 3 is divided in several lemmas.

We write system (4) in a more concise form as

$$\begin{aligned} \dot{\mathbf{W}}&= \fancyscript{F}(\mathbf{W},Z),\\ \epsilon \dot{Z}&= \fancyscript{G}(\mathbf{W},Z). \end{aligned}$$

with $\mathbf{W}(t)=(X(t),Y(t))^t$ and $\fancyscript{F}$ and $\fancyscript{G}$ being the appropriate entries of the right hand side of (4).

We write

$$\begin{aligned} \mathbf{W}=\mathbf{W}^0+\epsilon {\hat{\mathbf{W}}}^0+\epsilon \mathbf{Q}\quad \text{ and }\quad Z=Z^0+\hat{Z}^0+\epsilon \bar{Z}, \end{aligned}$$

with

$$\begin{aligned} {\hat{\mathbf{W}}}^0=\frac{X_0}{\bar{\sigma }Y_0+\bar{\mu }_v}\left( Z_0-\frac{\bar{\sigma }Y_0}{\bar{\sigma }Y_0+\bar{\mu }_v}\right) \mathrm{e}^{-(\bar{\sigma }Y_0 +\bar{\mu }_v)t/\epsilon }(1,-1)^t. \end{aligned}$$

Notice that since ${\hat{\mathbf{W}}}^0$ is bounded, we need only to prove that $(\mathbf{Q},\bar{Z})$ exist, are bounded, and are unique. In this case, we then take $C=\Vert (\mathbf{Q},\bar{Z})\Vert _\infty$.

First, we observe that

$$\begin{aligned} \dot{\mathbf{W}^0}\!+\!\epsilon \dot{{\hat{\mathbf{W}}}^0}\!=\!\fancyscript{F}(\mathbf{W}^0,Z^0\!+\!\hat{Z}^0)\!+\!K(t)(1,\!-\!1)^t \quad \text{ and }\quad \dot{\hat{Z}^0}\!=\!\fancyscript{G}(\mathbf{W}^0,Z^0\!+\!\hat{Z}^0)\!+\!L(t), \end{aligned}$$

where

$$\begin{aligned} K(t)=\delta \hat{Z}^0(t/\epsilon )(X^0(t)-X_0) \quad \text{ and }\quad L(t)=\bar{\sigma }\hat{Z}^0(t/\epsilon )(Y^0(t)-Y_0). \end{aligned}$$

In particular, because of the fast decay of $\hat{Z}^0$, and because $L(0)=K(0)=0$, it follows that there exists a constant $C>0$, such that

$$\begin{aligned} \int \limits _0^\infty K(t)\,\mathrm{d}t,\quad \int \limits _0^\infty L(t)\,\mathrm{d}t\le C\epsilon ^2. \end{aligned}$$

Since $\fancyscript{F}$ and $\fancyscript{G}$ are quadratic, we write:

$$\begin{aligned} \fancyscript{F}(\mathbf{W},Z)&= \fancyscript{F}(\mathbf{W}^0,Z^0+\hat{Z}^0)+\epsilon D_{\mathbf{W}}\fancyscript{F}(\mathbf{W}^0,Z^0+\hat{Z}^0)(\mathbf{Q}+{\hat{\mathbf{W}}}^0)\\&+ \epsilon D_{Z}\fancyscript{F}(\mathbf{W}^0,Z^0+\hat{Z}^0)\bar{Z}+\epsilon ^2\delta {\bar{X}}\bar{Z}\begin{pmatrix} -1\\ 1 \end{pmatrix}; \\ \fancyscript{G}(\mathbf{W},Z)&= \fancyscript{G}(\mathbf{W}^0,Z^0+\hat{Z}^0)+\epsilon D_{\mathbf{W}}\fancyscript{G}(\mathbf{W}^0,Z^0+\hat{Z}^0)(\mathbf{Q}+{\hat{\mathbf{W}}}^0)\\&+ \epsilon D_{Z}\fancyscript{G}(\mathbf{W}^0,Z^0+\hat{Z}^0)\bar{Z}+\epsilon ^2\bar{\sigma }\bar{Y}\bar{Z}. \end{aligned}$$

where $\mathbf{Q}=({\bar{X}},\bar{Y})^t$.

Let $T(t,s)$ be the fundamental solution to the linearised system

$$\begin{aligned} \begin{pmatrix} \dot{\mathbf{Q}}\\ \epsilon \dot{\bar{Z}} \end{pmatrix} =\begin{pmatrix} D_{\mathbf{W}}\fancyscript{F}(\mathbf{W}^0,Z^0+\hat{Z}^0)&{}D_{Z}\fancyscript{F}(\mathbf{W}^0,Z^0+\hat{Z}^0)\\ D_{\mathbf{W}}\fancyscript{G}(\mathbf{W}^0,Z^0+\hat{Z}^0)&{}D_{Z}\fancyscript{G}(\mathbf{W}^0,Z^0+\hat{Z}^0) \end{pmatrix} \begin{pmatrix} \mathbf{Q}\\ \bar{Z}\end{pmatrix} \end{aligned}$$

(12)

Then direct integration yields

Lemma 1

The functions $(\mathbf{Q},\bar{Z})$ satisfy the following integral equation:

$$\begin{aligned} \left( \begin{array}{l} \mathbf{Q}\\ \bar{Z}\end{array} \right)&= \!T(t,0) \left( \begin{array}{l} Q_0\\ \bar{Z}_0 \end{array} \right) + \int \limits _0^tT(t,s) \left( \begin{array}{ll} \epsilon \delta {\bar{X}}\bar{Z}&{} \left( \begin{array}{c} 1\\ -1 \end{array}\right) \\ &{} \bar{\sigma }\bar{Y}\bar{Z}\\ \end{array}\right) \,\mathrm{d}s \nonumber \\&+\int \limits _0^tT(t,s) \left( \begin{array}{l} D_W{\fancyscript{F}} \cdot {\hat{\mathbf{W}}}^0(s)\\ D_W{\fancyscript{G}} \cdot {\hat{\mathbf{W}}}^0(s)-\dot{Z}_0(s) \end{array}\right) \, \mathrm{d}s \nonumber \\&- \frac{1}{\epsilon ^2}\int \limits _0^tT(t,s) \left( \begin{array}{ll} \epsilon K(s) &{} \left( \begin{array}{l} 1\\ -1 \end{array}\right) \\ &{} L(s)\\ \end{array}\right) \,\mathrm{d}s. \nonumber \\ \end{aligned}$$

(13)

Moreover, the last term is bounded uniformly in $\epsilon$.

We also have the following large time behaviour result for the linearised system (12):

Lemma 2

Let $(\mathbf{Q}(t),\bar{Z}(t))$ be a solution to (12). Then

$$\begin{aligned} \lim _{t\rightarrow \infty }(\mathbf{Q}(t),\bar{Z}(t))=\mathbf 0 . \end{aligned}$$

In particular, the solutions to (12) are bounded uniformly in time for any given $\epsilon$. Moreover, they are also uniformly bounded in $\epsilon \le 1$, for all $t\ge 0$.

Proof

For notation convenience, let us write (12) as

$$\begin{aligned} \begin{pmatrix} \dot{\mathbf{Q}}\\ \epsilon \dot{\bar{Z}} \end{pmatrix} = A(\mathbf{W}^0,Z^0+\hat{Z}^0) \begin{pmatrix} \mathbf{Q}\\ \bar{Z}\end{pmatrix}. \end{aligned}$$

Fix $\epsilon >0$ and $(X_0,Y_0,Z_0)$. From Theorem 2, we know that

$$\begin{aligned} \lim _{t\rightarrow \infty } \left( \mathbf{W}^0(t),Z^0(t)+\hat{Z}^0(t/\epsilon )\right) =(\mathbf{W}^*,Z^*), \end{aligned}$$

where $(\mathbf{W}^*,Z^*)$ is the globally asymptotic stable equilibrium given by Theorem 1. Therefore, there exists $T>0$, such that $t>T$ implies that $A$ is negative-definite. Since $\hat{Z}^0\rightarrow 0$, as $\epsilon \rightarrow 0$. We can choose $T$ such this holds for all $0<\epsilon \le 1$.

Because of the continuity of the solution with respect to the initial conditions, we have that $T$ is a continuous function of the initial conditions. Since these lie on a compact set, we can pick $T$ such that $A$ is negative definite for all $0<\epsilon \le 1$ and for all $(X_0,Y_0,Z_0)$, such that $X_0,Y_0\ge 0, X_0+Y_0\le 1$ and $0<Z_0\le 1$. But then, for any such $(X_0,Y_0,Z_0)$ and $0<\epsilon \le 1$, we have that

$$\begin{aligned} \lim _{t\rightarrow \infty }T(t,T)=0. \end{aligned}$$

Therefore, for any initial condition $(Q_0,\bar{Z}_0)^t$, we have

$$\begin{aligned} \lim _{t\rightarrow \infty }T(t,0)\begin{pmatrix}Q_0\\ \bar{Z}_0\end{pmatrix}= \lim _{t\rightarrow \infty }T(t,T)T(T,0)\begin{pmatrix}Q_0\\ \bar{Z}_0\end{pmatrix}=0. \end{aligned}$$

We can now conclude the proof as follows: $\square$

Proof

(Proof of Theorem 3)

First, we observe that the nonlinear term in (13) is locally Lipschitz, hence a standard fixed point yields existence and uniqueness for $0\le t <t_0$ , for some, possibly small, $t_0$.

From the decaying of ${\hat{\mathbf{W}}}^0$ and from Lemma 1, we conclude that the two last terms of (13) are uniformly bounded in time and $\epsilon$.

Moreover, Lemma 2 implies that the first two terms on the right hand side of (13) are also uniformly bounded in time and in $\epsilon$, for sufficiently small $\epsilon$. Thus, we conclude that the same also holds true for $(Q,R)^t$. Therefore, the solution to (13) is globally defined in time, and it is bounded uniformly in $\epsilon$, if the latter is sufficiently small. $\square$