*3.1. Equilibrium Points*

The model described by the system of differential Equation (6) has two stationary states, the first one corresponds to the disease-free equilibrium and the second to the endemic equilibrium, which we will denote *P*<sup>0</sup> and *P*<sup>∗</sup> respectively. To determine both states we must calculate the critical points of the system (6) by setting *dIE*(*t*) *dt* <sup>=</sup> *dI*(*t*) *dt* <sup>=</sup> *dV*(*t*) *dt* <sup>=</sup> *dT*(*t*) *dt* <sup>=</sup> 0. Thus, we have

$$\begin{aligned} 0 &= \beta TV - \beta TV e^{-\delta\_{I\_E}\Lambda} - \delta\_{I\_E}I\_E\\ 0 &= \beta TV e^{-\delta\_{I\_E}\Lambda} - \delta\_I I\\ 0 &= N\delta\_I I - CV - \beta TV\\ 0 &= \Lambda - \beta TV - \mu\_0 T. \end{aligned} \tag{11}$$

The disease-free equilibrium point of a model are solutions of steady state in the absence of infection. For this case, we must consider *IE* = 0, *I* = 0, *V* = 0, and *T* > 0, in the system (11). Then *<sup>P</sup>*<sup>0</sup> will be of the form *<sup>P</sup>*<sup>0</sup> = (0, 0, 0, *<sup>T</sup>*0), where *<sup>T</sup>*<sup>0</sup> <sup>=</sup> <sup>Λ</sup> *μ*0 . Therefore,

$$P^0 = \left(0, 0, 0, \frac{\Lambda}{\mu\_0}\right).$$

On the other hand, we can determine the basic reproductive number using the next generation matrix methodology. With the terms of infection and viral production in the mathematical model (11), matrices F and V are given by

$$\mathbb{F} = \begin{pmatrix} 0 & 0 & \beta T^0 \left(1 - e^{-\delta\_{I\_E} \Lambda} \right) \\ 0 & 0 & \beta T^0 e^{-\delta\_{I\_E} \Lambda} \\ 0 & 0 & 0 \end{pmatrix}, \ \mathbb{V} = \begin{pmatrix} \delta\_{I\_E} & 0 & 0 \\ 0 & \delta\_I & 0 \\ 0 & N \delta\_I & \mathbb{C} + \beta T^0 \end{pmatrix}.$$

where *<sup>T</sup>*<sup>0</sup> <sup>=</sup> <sup>Λ</sup> *μ*0 , which it is the number of target cells before infection. Thus, the basic reproductive number R0, is calculated as the spectral radius of the matrix given by

$$\begin{split} \mathbb{F}\mathbb{V}^{-1} &= \begin{pmatrix} 0 & 0 & \beta^{T}\mathbb{I}^{0}\left(1 - e^{-\delta\_{\bar{l}\_{E}}\Lambda}\right) \\ 0 & 0 & \beta^{T}\mathbb{I}^{0}e^{-\delta\_{\bar{l}\_{E}}\Lambda} \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} \frac{1}{\delta\_{\bar{l}\_{E}}} & 0 & 0 \\ 0 & -\delta\_{\bar{l}} & 0 \\ 0 & -\frac{N}{\mathbb{C} + \beta^{T}\mathbb{I}^{0}} & \frac{1}{\mathbb{C} + \beta^{T}\mathbb{I}^{0}} \end{pmatrix} \\ &= \begin{pmatrix} 0 & -\frac{N\beta^{T}\mathbb{I}^{0}\left(1 - e^{-\delta\_{\bar{l}\_{E}}\Lambda}\right)}{\beta^{T}\mathbb{I}^{0} + \mathbb{C}} & \frac{\beta^{T}\mathbb{I}^{0}\left(1 - e^{-\delta\_{\bar{l}\_{E}}\Lambda}\right)}{\beta^{T}\mathbb{I}^{0} + \mathbb{C}} \\ 0 & -\frac{N\beta^{T}\mathbb{I}^{0}e^{-\delta\_{\bar{l}\_{E}}\Lambda}}{\beta^{T}\mathbb{I}^{0} + \mathbb{C}} \end{pmatrix} . \end{split}$$

Therefore, the basic reproductive number R<sup>0</sup> is given by

$$\mathcal{R}\_0 = \frac{N\beta\Lambda\sigma^{-\delta\_{I\_E}\Lambda}}{C\mu\_0 + \beta\Lambda}.\tag{12}$$

Now, the endemic equilibrium point of a model is its steady-state solutions in the presence of infection or disease, for which it must be considered *IE* > 0, *I* > 0, *V* > 0 and *T* > 0 in the system (11.) Then *P*∗ will be of the form *P*∗ = (*I*∗ *<sup>E</sup>*, *I*∗, *V*∗, *T*∗). In this case, from the system (11) the following equalities are obtained

$$I\_E^\* = \frac{\beta T^\* V^\* \left(1 - e^{-\delta\_{I\_E} \Lambda} \right)}{\delta\_{I\_E}}. \tag{13}$$

$$I^\* = \frac{\beta T^\* V^\* e^{-\delta\_{l\_E}\Lambda}}{\delta\_I}.\tag{14}$$

$$I^\* = \frac{V^\*}{N\delta\_I} (\mathbb{C} + \beta T^\*). \tag{15}$$

$$T^\* = \frac{\Lambda}{\beta V^\* + \mu\_0}.\tag{16}$$

Replacing (16) in (14) and (15), is obtains

$$I^\* = \frac{\beta \Lambda V^\* e^{-\delta\_{\tilde{E}} \Lambda}}{\delta\_I (\beta V^\* + \mu\_0)} = \frac{V^\*}{N \delta\_I} \left( \mathbb{C} + \frac{\beta \Lambda}{\beta V^\* + \mu\_0} \right). \tag{17}$$

Then

$$V^\* = \frac{N\beta\Lambda e^{-\delta\_{l\_E}\Lambda} - \mathbb{C}\mu\_0 - \beta\Lambda}{\mathbb{C}\beta} = \frac{(\mathbb{C}\mu\_0 + \beta\Lambda)\frac{N\beta\Lambda e^{-\delta\_{l\_E}\Lambda}}{\mathbb{C}\mu\_0 + \beta\Lambda} - 1}{\mathbb{C}\beta}.\tag{18}$$

Thus, if R<sup>0</sup> > 1, then *Nβ*Λ*e* <sup>−</sup>*δIE*<sup>Δ</sup> <sup>−</sup> *<sup>C</sup>μ*<sup>0</sup> <sup>−</sup> *<sup>β</sup>*<sup>Λ</sup> <sup>&</sup>gt; 0. Hence, we can write *<sup>V</sup>*<sup>∗</sup> as

$$V^\* = \frac{(\mathbb{C}\mu\_0 + \beta\Lambda)(\mathcal{R}\_0 - 1)}{\mathbb{C}\beta}. \tag{19}$$

Next, we replace (18) in (17) to get

$$I^\* = \frac{N\beta\Lambda e^{-\delta\_{l\_E}\Lambda} - \mathbb{C}\mu\_0 - \beta\Lambda}{\delta\_I \pounds^{\delta\_{l\_E}\Lambda} (Ne^{-\delta\_{l\_E}\Lambda} - 1)} = \frac{(\mathbb{C}\mu\_0 + \beta\Lambda)(\mathcal{R}\_0 - 1)}{\delta\_I \pounds^{\delta\_{l\_E}\Lambda} (Ne^{-\delta\_{l\_E}\Lambda} - 1)}.\tag{20}$$

Now, substituting (18) in (16) one gets

$$T^\* = \frac{\Lambda \mathbb{C}}{(\mathbb{C}\mu\_0 + \beta \Lambda)(\mathcal{R}\_0 - 1) + \mathbb{C}\mu\_0}. \tag{21}$$

Finally, we replace (18) and (21) in (13) to obtain

$$I\_E^\* = \frac{\left(N\notin\Lambda e^{-\delta\_{l\_E}\Lambda} - \mathbb{C}\mu\_0 - \beta\Lambda\right)\left(e^{\delta\_{l\_E}\Lambda} - 1\right)}{\delta\_I\beta e^{\delta\_{l\_E}\Lambda}\left(Ne^{-\delta\_{l\_E}\Lambda} - 1\right)} = \frac{\left(\mathbb{C}\mu\_0 + \beta\Lambda\right)\left(\mathcal{R}\_0 - 1\right)\left(e^{\delta\_{l\_E}\Lambda} - 1\right)}{\delta\_I\beta e^{\delta\_{l\_E}\Lambda}\left(Ne^{-\delta\_{l\_E}\Lambda} - 1\right)}\tag{22}$$

Note that *I*∗ *<sup>E</sup>* <sup>&</sup>gt; 0, *<sup>I</sup>*<sup>∗</sup> <sup>&</sup>gt; 0, *<sup>V</sup>*<sup>∗</sup> <sup>&</sup>gt; 0 and *<sup>T</sup>*<sup>∗</sup> <sup>&</sup>gt; 0 if only if <sup>R</sup><sup>0</sup> <sup>=</sup> *<sup>N</sup>β*Λ*<sup>e</sup>* −*δIE*<sup>Δ</sup> *<sup>C</sup>μ*<sup>0</sup> <sup>+</sup> *<sup>β</sup>*<sup>Λ</sup> <sup>&</sup>gt; 1. Thus, *Ne*−*δIE*<sup>Δ</sup> > *Cμ*<sup>0</sup> + *β*Λ *<sup>β</sup>*<sup>Λ</sup> <sup>&</sup>gt; 1.
