**Property 1.**


**Proof.** (i) First, we seek for pure imaginary roots *λ* = ±*iω*, *ω* > 0, of (11). Splitting in (11) the real and the imaginary part, we obtain

$$\begin{cases} \gamma \cos(\omega \tau) e^{-\mu \tau} + \theta \omega \sin(\omega \tau) e^{-\mu \tau} &=& \mu + K f'(\Phi^{-1}(\sigma/\mu)),\\ -\theta \omega \cos(\omega \tau) e^{-\mu \tau} + \gamma \sin(\omega \tau) e^{-\mu \tau} &=& -\omega,\end{cases}$$

where *γ* := *θμ* + *K f* (Φ−1(*σ*/*μ*)). Solving this system, we find that

$$\begin{cases} \cos(\omega \tau) &=& \frac{1}{(\theta^2 \omega^2 + \gamma^2)\epsilon^{-\mu \tau}} \Big[ \theta \omega^2 + \gamma (\mu + K f'(\Phi^{-1}(\sigma/\mu))) \Big], \\\sin(\omega \tau) &=& \frac{1}{(\theta^2 \omega^2 + \gamma^2)\epsilon^{-\mu \tau}} \Big[ -\omega \gamma + \theta \omega (\mu + K f'(\Phi^{-1}(\sigma/\mu))) \Big]. \end{cases}$$

Using the trigonometric identity cos2(*ωτ*) + sin2(*ωτ*) = 1, we obtain that *ω* satisfies

$$(\theta^2 \omega^2 + \gamma^2)(e^{-\mu \tau})^2 = \omega^2 + (\mu + \mathcal{K}f'(\Phi^{-1}(\sigma/\mu)))^2.$$

Then, we obtain

$$\omega^2 = \frac{(\gamma e^{-\mu \tau} - (\mu + K f'(\Phi^{-1}(\sigma/\mu))))((\mu + K f'(\Phi^{-1}(\sigma/\mu))) + \gamma e^{-\mu \tau})}{1 - \theta^2 e^{-2\mu \tau}}.\tag{12}$$

Notice that 1 <sup>−</sup> *<sup>θ</sup>*2*e*−2*μτ* <sup>&</sup>gt; 0 and

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

Thus, the right-hand side of (12) is negative, which is absurd. We conclude that the Equation (11) has no purely imaginary root.

(ii) As the function Δ¯ is entire, the set of its roots cannot have accumulation points. Furthermore, Δ¯ has at most countably many zeros. 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

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

It is clear that

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

Then, there is a sequence (*λ <sup>n</sup>*)*n*∈<sup>N</sup> of roots of the equation

$$e^{\lambda \tau} - \theta e^{-\mu \tau} = 0$$

such that

$$\lim\_{n \to \infty} (\lambda\_n - \lambda'\_n) = 0.$$

This means that the closed-forms of the roots *λ<sup>n</sup>* are the complex numbers

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

Since *<sup>θ</sup>* <sup>∈</sup> (0, 1), we have that <sup>R</sup>*e*(*λ*¯ *<sup>p</sup>*) <sup>&</sup>lt; 0. Thus, all roots such that <sup>|</sup>*λn*| → <sup>+</sup><sup>∞</sup> have a negative real part for *n* large enough.

We state the stability of the disease-free equilibrium (9):
