*3.1. Basic Reproduction Number*

To determine the initial exponential growth rate and the basic reproduction number R0, we use the equations for the infected compartments *I*(*t*) considered in the form:

$$\frac{dI(t)}{dt} = \beta \frac{S(t)}{N} \left. I(t) - \beta \frac{S(t-\tau)}{N} \right. \tag{10}$$

Suppose that, at the beginning of the epidemic, *S*(*t*) ≈ *S*<sup>0</sup> and *S*(*t* − *τ*) = *S*0. Then, from (10), we have

$$\frac{dI(t)}{dt} = \beta \frac{\mathcal{S}\_0}{N} \left( I(t) - I(t-\tau) \right).$$

Substituting *I*(*t*) = *I*<sup>0</sup> *eλ<sup>t</sup>* , we have

$$
\lambda = \beta \frac{S\_0}{N} \left( 1 - e^{-\lambda \tau} \right). \tag{11}
$$

Let <sup>G</sup>(*λ*) = *<sup>β</sup> <sup>S</sup>*<sup>0</sup> *N* <sup>1</sup> <sup>−</sup> *<sup>e</sup>*−*λτ* . Clearly, (11) has a solution *λ* = 0 and a non-zero solution with the sign determined by G (0). Denote

$$\mathcal{R}\_0 = \mathcal{G}'(0) = \beta \tau \frac{S\_0}{N}.$$

Let R<sup>0</sup> > 1. This implies, G (0) > 1. We observe that G(*λ*) is an increasing function of *<sup>λ</sup>* and <sup>G</sup>(*λ*) <sup>→</sup> *<sup>β</sup> <sup>S</sup>*<sup>0</sup> *<sup>N</sup>* as *λ* → ∞. This implies that Equation (11) has a positive solution, i.e., there exists *λ*<sup>∗</sup> > 0 such that G(*λ*∗) = *λ*∗. If R<sup>0</sup> < 1, i.e., G (0) < 1, then

$$\mathcal{G}'(\lambda) = \beta \tau \frac{\mathcal{S}\_0}{N} e^{-\lambda \tau} < 1$$

for all *λ* ≥ 0. Therefore, the equation G(*λ*) = *λ* has no positive solution.

Hence, the basic reproduction number is given by the following expression:

$$\mathcal{R}\_0 = \beta r \frac{S\_0}{N}.\tag{12}$$

Letting *λ*ˆ = *λ*/(*β <sup>S</sup>*<sup>0</sup> *<sup>N</sup>* ), we can write Equation (11) in the following form:

$$
\hat{\lambda} = 1 - e^{-\mathcal{R}\_0 \hat{\lambda}}
$$

.

The solution of this equation determines the initial exponential growth rate, which depends on the single parameter R0.
