**Corollary 1.**


**Proof.** (i) The real eigenvalue

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

is positive because R0(*τ*0) > 1. Then, the disease-free equilibrium (9) is unstable for *τ* = *τ*0. (ii) Suppose that there exits *τ*<sup>1</sup> ≥ 0 such that R0(*τ*1) < 1 and the disease-free equilibrium (9) is locally asymptotically stable for *τ* = *τ*1. As *τ* → R0(*τ*) is decreasing, we have R0(*τ*) < 1 for all *τ* ≥ *τ*1. Then, the real eigenvalue *λ*<sup>0</sup> is negative for all *τ* ≥ *τ*1. The other eigenvalues given by

$$\bar{\Delta}(\mathfrak{r}, \lambda) := P(\mathfrak{r}, \lambda) - Q(\mathfrak{r}, \lambda)e^{-\lambda \cdot \mathfrak{r}} = 0\_{\prec}$$

with

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

which cannot cross the imaginary axis when we increase *τ* starting from *τ*1. The diseasefree equilibrium (9), thus, keeps the same stability as for *τ* = *τ*1, i.e., it remains locally asymptotically stable for all *τ* ≤ *τ*1.

(iii) Let us consider the case *τ* = 0. Then, Equation (11) becomes

$$(\lambda + \mu)(1 - \theta) = 0.$$

This means that *λ* = −*μ* < 0. Then, the disease-free equilibrium (9) is locally asymptotically stable for *τ* = 0. We use the previous property, (ii), to conclude that the disease-free equilibrium (9) is locally asymptotically stable for all *τ* ≥ 0.

One can prove this corollary directly by proving that Equation (11) has no root with a nonnegative real part, and therefore, the stability of the endemic equilibrium is determined by the sign of R<sup>0</sup> − 1. Indeed, let us suppose by contradiction the existence of such a root *λ* = *α* + *iω*, with *α* ≥ 0. Then, we obtain

$$\frac{|P(\alpha + i\omega)|}{|Q(\alpha + i\omega)|} = e^{-a\tau} \le 1.$$

This gives the following inequality:

$$[P(\alpha) - Q(\alpha)][P(\alpha) + Q(\alpha)] + \left(1 - e^{-2\mu\tau}\theta^2\right)\omega^2 \le 0,$$

with

$$P(a) - Q(a) = (1 - e^{-\mu \tau} \theta)(a + \mu) + (1 - e^{-\mu \tau}) K f'(\Phi^{-1}(\sigma/\mu)) > 0.$$

This leads to a contradiction.

Now, we focus on the endemic equilibrium (10):

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

**Proposition 3.** *Suppose that* R0(0) > 1*. Then, for all τ* ≥ 0 *such that* R0(*τ*) > 1*, the endemic steady-state is locally asymptotically stable.*

**Proof.** We suppose that R0(*τ*) > 1, and we consider the endemic equilibrium. The characteristic equation becomes

$$\Delta(\tau,\lambda) := \lambda^2 + A\lambda + \mu\beta I^\* - e^{-\mu\tau}e^{-\lambda\tau} \left[\theta\lambda^2 + (\theta A + Kf'(S^\*))\lambda + \theta\mu\beta I^\*\right] = 0,\tag{13}$$

with *A* := *μ* + *βI*∗ > 0. We follow the same steps as in the proof of the stability of the disease-free equilibrium. First, we consider the case *τ* = 0. Then, the endemic steadystate becomes 

$$(S^\*, I^\*, \mu^\*) = \left(\frac{\mu}{\beta} \; , \frac{\sigma}{\mu} - \frac{\mu}{\beta} \; , \frac{Kf(\mu/\beta)}{1-\theta}\right).$$

with R0(0) = *βσ*/*μ*2, and the characteristic equation is reduced to

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

We obtain two eigenvalues *λ*<sup>1</sup> = −*μ* < 0 and *λ*<sup>2</sup> = −*μ* − *β*(*I*<sup>∗</sup> − *S*∗) = −*μ*(R0(0) − 1) < 0. Hence, the endemic steady-state is locally asymptotically stable for *τ* = 0. We then look for pure imaginary roots *λ* = ±*iω*, *ω* > 0. Then, splitting the real and imaginary parts in (13), we have the following system:

$$\begin{cases} \omega \Gamma \cos(\omega \tau) e^{-\mu \tau} + \rho\_\omega \sin(\omega \tau) e^{-\mu \tau} &= \omega (A + \psi f'(\mu/\beta)), \\\rho\_\omega \cos(\omega \tau) e^{-\mu \tau} - \omega \Gamma \sin(\omega \tau) e^{-\mu \tau} &= \omega^2 - \mu \beta I^\*, \end{cases}$$

with *ρω* := (*ω*<sup>2</sup> <sup>−</sup> *μβI*∗)*θe*−*μτ* and <sup>Γ</sup> = (*θ<sup>A</sup>* <sup>+</sup> *<sup>ψ</sup><sup>f</sup>* (*μ*/*β*))*e*−*μτ*. Using the identity cos2(*ωτ*) + sin2(*ωτ*) = 1, we obtain the polynomial equation in *ω*:

$$\begin{array}{rcl} [1 - (\theta \varepsilon^{-\mu \tau})^2] \omega^4 + [2\mu \beta I^\*(\theta \varepsilon^{-\mu \tau})^2 - 2\mu \beta I^\* + (A + \psi f'(\mu/\beta))^2 - \Gamma^2] \omega^2 \\ + (\mu \beta I^\*)^2 (1 - (\theta \varepsilon^{-\mu \tau})^2) &=& 0. \end{array}$$

Let us set

$$N := \frac{1}{1 - (\theta e^{-\mu \tau})^2} [2\mu \beta I^\* (\theta e^{-\mu \tau})^2 - 2\mu \beta I^\* + (A + \psi f'(\mu/\beta))^2 - \Gamma^2] \quad \text{and} \quad X := \omega^2 \cdot \Gamma$$

Then, we obtain the following polynomial:

$$X^2 + NX + (\mu \beta I^\*)^2 = 0.\tag{14}$$

First, if *N* > 0, then the Routh–Hurwitz criterion implies that all roots of (14) have a negative real part. This leads to a contradiction.

Suppose now that *N* ≤ 0. We compute the discriminant of Equation (14) to obtain

$$
\Delta = N^2 - 4(\mu \beta I^\*)^2 = (N + 2\mu \beta I^\*)(N - 2\mu \beta I^\*).
$$

As we supposed that *N* ≤ 0, we have that *N* − *μβI*<sup>∗</sup> < 0. On the other hand, we have that

$$(1 - (\theta e^{-\mu \tau})^2)(N + 2\mu \beta I^\*) = (A + Kf'(\mu/\beta))^2 - \Gamma^2.$$

But, as *θ* < 1 and *e*−*μτ* < 1, we obtain

$$
\Gamma^2 = [(\theta A + Kf'(\mu/\beta))e^{-\mu \tau}]^2 < (A + Kf'(\mu/\beta))^2.
$$

We conclude that Δ < 0. Therefore, there is no positive real root of (14) and *ω* > 0 cannot exist. Furthermore, as for the disease-free equilibrium, if Δ has infinitely many different zeros, *<sup>λ</sup><sup>n</sup>* <sup>∈</sup> <sup>C</sup>, *<sup>n</sup>* <sup>∈</sup> <sup>N</sup>, then

$$\lim\_{n \to +\infty} |\lambda\_n| = +\infty.$$

The sequence *<sup>λ</sup>n*, *<sup>n</sup>* <sup>∈</sup> <sup>N</sup>, satisfies

$$e^{\lambda\_n \tau} \left( \frac{P(\lambda\_n)}{\lambda\_n^2} \right) - \frac{Q(\lambda\_n)}{\lambda\_n^2} = 0,$$

with

$$P(\lambda) := \lambda^2 + A\lambda + \mu \beta I^\* \quad \text{and} \quad Q(\lambda) := e^{-\mu \tau} \left[ \theta \lambda^2 + (\theta A + K f'(S^\*)) \lambda + \theta \mu \beta I^\* \right].$$

We have

$$\lim\_{n \to \infty} \left| \frac{P(\lambda\_n)}{\lambda\_n^2} - 1 \right| = 0 \quad \text{and} \quad \lim\_{n \to \infty} \left| \frac{Q(\lambda\_n)}{\lambda\_n^2} - \theta e^{-\mu \tau} \right| = 0.$$

Then, as for the case of the disease-free steady-state, the closed forms of the roots *λ<sup>n</sup>* are

$$
\bar{\lambda}\_p = \frac{1}{\pi} \ln(\theta) - \mu + \frac{2p\pi}{\pi} i, \quad p \in \mathbb{Z}.
$$

Thus, all roots such that |*λn*| → +∞ have a negative real part for *n* large enough. We conclude that all roots of (13) have a negative real part, and then, the endemic equilibrium is locally asymptotically stable.

**Remark 1.** *As for the proof of Corollary 1, we can also prove directly by contradiction that there is no root λ* = *α* + *iω of* (13) *with α* ≥ 0*.*
