*3.2. Steady-States and Basic Reproduction Number*

In this subsection, we compute the steady-states of the system and study their local asymptotic stability. Let us consider a steady-state (*S*∗, *I*∗, *u*∗) of (4). We then have

$$\begin{cases} 0 &= \sigma - \beta I^\* S^\* - \mu S^\* - Kf(S^\*) + (1 - \theta)e^{-\mu \tau} u^\*, \\ 0 &= \begin{pmatrix} \beta S^\* - \mu \end{pmatrix} I^\*, \\ u^\* &= Kf(S^\*) + \theta e^{-\mu \tau} u^\*. \end{cases} \tag{7}$$

The third equation of the system (7) leads to

$$
u^\* = \frac{Kf(S^\*)}{1 - \theta e^{-\mu \tau}}.
$$

We remark that, as *<sup>θ</sup>* <sup>∈</sup> (0, 1), we always have 1 <sup>−</sup> *<sup>θ</sup>e*−*μτ* <sup>&</sup>gt; 0.

Assume that *I*∗ = 0. Then, the first equation of (7) implies that *S*∗ satisfies the equation:

$$\Phi(S^\*) := S^\* + \frac{K(1 - e^{-\mu \tau})}{\mu(1 - \theta e^{-\mu \tau})} f(S^\*) = \frac{\sigma}{\mu}. \tag{8}$$

Since Φ is increasing, tends to infinity, and satisfies Φ(0) = 0, we obtain the existence of a unique *S*<sup>∗</sup> such that Φ(*S*∗) = *σ*/*μ*. We can therefore write *S*<sup>∗</sup> = Φ−1(*σ*/*μ*). Thus, we obtain the disease-free steady-state:

$$(S^\*, I^\*, \mu^\*) = \left(\Phi^{-1}(\sigma/\mu) \; , \; 0 \; , \; \frac{Kf(\Phi^{-1}(\sigma/\mu))}{1 - \theta e^{-\tau \mu}}\right) . \tag{9}$$

Now, suppose that *S*∗ = *μ*/*β*. We immediately obtain that

$$
\mu^\* = \frac{Kf(\mu/\beta)}{1 - \theta e^{-\mu \tau}}.
$$

We inject these two last expressions into the first equation of (7) to obtain

$$I^\* = \frac{\sigma}{\mu} - \Phi\left(\frac{\mu}{\beta}\right),$$

where the function Φ is given by the expression (8). Then, *I*∗ > 0 exists if and only if

$$
\Phi^{-1}\left(\frac{\sigma}{\mu}\right) > \frac{\mu}{\beta}
$$

.

.

Let us define the basic reproduction number R<sup>0</sup> as a threshold for the existence of the endemic steady-state:

$$\mathcal{R}\_0 := \frac{\beta}{\mu} \Phi^{-1} \left( \frac{\sigma}{\mu} \right).$$

Hence, we obtain the existence of a unique endemic steady-state:

$$(S^\*, I^\*, \mu^\*) = \left(\frac{\mu}{\beta}, \frac{\sigma}{\mu} - \Phi\left(\frac{\mu}{\beta}\right), \frac{Kf(\mu/\beta)}{1 - \theta e^{-\mu \tau}}\right) \tag{10}$$

if and only if

$$\mathcal{R}\_0 > 1.$$

We will need to know the variation of R<sup>0</sup> as a function of *τ* ∈ [0, +∞). We have the following lemma.

**Lemma 1.** *The basic reproduction number τ* ∈ [0, +∞) → R0(*τ*) *is a decreasing function such that* 

$$\mathcal{R}\_0(0) = \frac{\beta \sigma}{\mu^2} \quad \text{and} \quad \mathcal{R}\_0(\infty) = \frac{\beta}{\mu} \Phi\_{\infty}^{-1} \left(\frac{\sigma}{\mu}\right) \sigma$$

*with*

$$\Phi\_{\infty}(\mathcal{S}) := \mathcal{S} + \frac{\mathcal{K}}{\mu} f(\mathcal{S}).$$

**Proof.** Consider the dependence of the function Φ on *τ* by putting

$$\Phi(S,\tau) := S + \frac{K(1 - e^{-\mu\tau})}{\mu(1 - \theta e^{-\mu\tau})} f(S).$$

We want to study the variation of *<sup>τ</sup>* → R0(*τ*) = *<sup>β</sup> μ* Φ−<sup>1</sup> *σ μ*, *τ* . Then, we set *S*(*τ*) := Φ−1(*σ*/*μ*, *τ*). In fact, we have Φ(*S*(*τ*), *τ*) = *σ*/*μ*. By differentiating, we deduce that

$$\frac{\partial \Phi}{\partial S}(S(\tau), \tau) \frac{d}{d\tau} S(\tau) + \frac{\partial \Phi}{\partial \tau}(S(\tau), \tau) = 0.$$

Thus, we obtain

$$\frac{d}{d\tau}\mathcal{R}\_0(\tau) = \frac{\beta}{\mu} \frac{d}{d\tau}S(\tau) = -\frac{\beta}{\mu} \frac{\frac{\partial \Phi}{\partial \tau}(S(\tau), \tau)}{\frac{\partial \Phi}{\partial S}(S(\tau), \tau)} < 0.$$

Then, the function *τ* → R0(*τ*) is decreasing on [0, +∞). Furthermore, by taking *τ* = 0 and *τ* → +∞ in the equality:

$$\Phi(S,\tau) := S(\tau) + \frac{K(1 - e^{-\mu\tau})}{\mu(1 - \theta e^{-\mu\tau})} f(S(\tau)) = \frac{\sigma}{\mu'}$$

we obtain that

$$\mathcal{R}\_0(0) = \frac{\beta \sigma}{\mu^2} \quad \text{and} \quad \mathcal{R}\_0(\infty) = \frac{\beta}{\mu} \Phi\_{\infty}^{-1} \left(\frac{\sigma}{\mu}\right).$$

,

with

$$\Phi\_{\infty}(\mathcal{S}) := \mathcal{S} + \frac{K}{\mu}f(\mathcal{S}).$$

We conclude from Lemma 1 that R0(∞) < R0(*τ*) < R0(0), for all *τ* ∈ (0, +∞), and we immediately obtain the following result.

**Proposition 2.** *Suppose that τ* ≥ 0*. Then, we have three cases:*


We linearize System (4) around any equilibrium (*S*∗, *I*∗, *u*∗), and we obtain

$$\begin{cases} \begin{array}{rcl} S'(t) &=& -\beta I^\*S(t) - \beta S^\*I(t) - \mu S(t) - Kf'(S^\*)S(t) + (1-\theta)e^{-\mu\tau}u(t-\tau), \\ I'(t) &=& \beta I^\*S(t) + \beta S^\*I(t) - \mu I(t), \\ u(t) &=& Kf'(S^\*)S(t) + \theta e^{-\mu\tau}u(t-\tau). \end{array} \end{cases}$$

We look for solutions of the form *S*(*t*) = *e*−*λ<sup>t</sup> S*0, *I*(*t*) = *e*−*<sup>λ</sup>tI*<sup>0</sup> and *u*(*t*) = *e*−*λ<sup>t</sup> u*0, *t* > 0, and *<sup>λ</sup>* <sup>∈</sup> <sup>C</sup>. Then, we obtain the following characteristic equation

$$
\Delta(\lambda, \tau) := \begin{vmatrix}
\lambda + \beta I^\* + \mu + Kf'(S^\*) & \beta S^\* & -(1 - \theta)e^{-\mu \tau}e^{-\lambda \tau} \\
\end{vmatrix} = 0.
$$

We develop the determinant and obtain

$$\begin{array}{rcl} \Delta(\boldsymbol{\tau},\boldsymbol{\lambda}) &=& (\lambda-\beta\boldsymbol{S}^{\*}+\mu)\left[ (\lambda+\mu)(1-\theta e^{-\mu\boldsymbol{\tau}}e^{-\lambda\boldsymbol{\tau}})+Kf'(\boldsymbol{S}^{\*})(1-e^{-\mu\boldsymbol{\tau}}e^{-\lambda\boldsymbol{\tau}})\right] \\ &+(\lambda+\mu)\beta I^{\*}(1-\theta e^{-\mu\boldsymbol{\tau}}e^{-\lambda\boldsymbol{\tau}}),\\ &=& 0. \end{array}$$

First, consider the disease-free equilibrium (9). Then, the characteristic equation becomes

$$\left[\lambda - \mu(\mathcal{R}\_0 - 1)\right] \left[ (\lambda + \mu)(1 - \theta e^{-\mu \tau} e^{-\lambda \tau}) + \mathcal{K}f'(\Phi^{-1}(\sigma/\mu))(1 - e^{-\mu \tau} e^{-\lambda \tau}) \right] = 0.$$

We have an immediate real eigenvalue

$$
\lambda\_0 = \mu(\mathcal{R}\_0 - 1).
$$

The other eigenvalues are solutions of the transcendental equation:

$$
\bar{\Delta}(\lambda) := P(\lambda) - Q(\lambda)e^{-\lambda \tau} = 0,\tag{11}
$$

with

$$P(\lambda) := \lambda + \mu + K f'(\Phi^{-1}(\sigma/\mu)) \quad \text{and} \quad Q(\lambda) := e^{-\mu \tau} [\theta \lambda + \theta \mu + K f'(\Phi^{-1}(\sigma/\mu))].$$

We have the following properties:
