Stability of a delayed SARS-CoV-2 reactivation model with logistic growth and adaptive immune response

Abstract

This paper develops and analyzes a SARS-CoV-2 dynamics model with logistic growth of healthy epithelial cells, CTL immune and humoral (antibody) immune responses. The model is incorporated with four mixed (distributed/discrete) time delays, delay in the formation of latent infected epithelial cells, delay in the formation of active infected epithelial cells, delay in the activation of latent infected epithelial cells, and maturation delay of new SARS-CoV-2 particles. We establish that the model’s solutions are non-negative and ultimately bounded. We deduce that the model has five steady states and their existence and stability are perfectly determined by four threshold parameters. We study the global stability of the model’s steady states using Lyapunov method. The analytical results are enhanced by numerical simulations. The impact of intracellular time delays on the dynamical behavior of the SARS-CoV-2 is addressed. We noted that increasing the time delay period can suppress the viral replication and control the infection. This could be helpful to create new drugs that extend the delay time period.

1. Introduction

In November 2019, a dangerous type of virus called severe acute respiratory syndrome coronavirus 2 (SARS-CoV-2) appeared, and it infects the human body and may lead to death. SARS-CoV-2 causes Coronavirus disease 2019 (COVID-19). World Health Organization (WHO) reported in the COVID-19 weekly epidemiological update of 16 January 2022 that, over 323 million confirmed cases and over 5.5 million deaths worldwide [1]. Some symptoms can appear on COVID-19 patients such as: fever, cough, fatigue, sputum production, headache, dyspnoea, diarrhea and hemoptysis [2]. The primary way that people become infected with SARS-CoV-2 is when they are exposed to respiratory fluids carrying the infectious virus. To reduce the SARS-CoV-2 transmission, preventive measures must be implemented such as hand washing, using of face masks, physical and social distancing, disinfection of surfaces and getting COVID-19 vaccine. Fortunately, WHO approved the following COVID-19 vaccines: Moderna, Oxford/AstraZeneca, Sinovac, Janssen (Johnson & Johnson), Pfizer/BioNTech, Sinpharm (Beijing), Serum Institute of India, Novavax and Bharat Biotech [3]. Beside vaccination, scientists and researchers are working hard to create new effective drugs for COVID-19 patients.

SARS-CoV-2 is a single-stranded RNA virus, belonging to the Coronaviridae family. Epithelial cells with angiotensin-converting enzyme 2 (ACE2) receptors are attacked by SARS-CoV-2 [4]. These target cells founded at the respiratory tracks including lungs, nasal and trachea/bronchial tissues [5]. The immune response plays an essential role in controlling the disease progression and clearing the SARS-CoV-2 infection. Adaptive immune response is based on Cytotoxic $T$ lymphocytes (CTLs) which kill virus-infected cells and antibodies which neutralize the viruses.

Beside biological and medical research, mathematical modeling of infection diseases were attracted the interest of several researchers. Several epidemiological mathematical models for COVID-19 were proposed to forecast disease severity and assist policy makers for inferring disease-control interventions (see e.g., [6], [7], [8], [9], [10], [11], [12], [13], [14], [15], [16], [17]).

1.1. Mathematical models of within-host SARS-CoV-2 infection

Mathematical models of within-host SARS-CoV-2 dynamics can help researchers to understand the replication cycle of SARS-CoV-2 and the immune system response to the viral infection. Moreover, these models enable the merits of different types of antiviral drug therapies to be assessed in individual COVID-19 patients [18]. Many scientists have been interested in modeling and analyzing SARS-CoV-2 dynamics within the host (see the review paper [19]).

• •
  Target cell-limited model:Hernandez-Vargas and Velasco-Hernandez [20] proposed the following target cell-limited model forSARS-CoV-2 infection:

  $$\left\{\begin{array}{c}\dot{T}\left(t\right)=-\xi V\left(t\right)T\left(t\right),\hfill \\ \dot{I}\left(t\right)=\xi V\left(t\right)T\left(t\right)-d_{I}I\left(t\right),\hfill \\ \dot{V}\left(t\right)=kI\left(t\right)-d_{V}V\left(t\right),\hfill \end{array}\right)$$ (1)

  where $T\left(t\right)$, $I\left(t\right)$ and $V\left(t\right)$ are the concentrations of healthy target cells, active infected cells and SARS-CoV-2 particles, at time $t$, respectively. Parameters $d_I$ and $d_V$ are death rate constants of infected cells and SARS-CoV-2 particles, respectively, $\xi$ is the infection rate constant. Abuin et al. [21] studied the mathematical analysis of model (1). The effect of the antiviral Pharmacodynamic therapy which reduce the production of infectious SARS-CoV-2 particles was studied using control theory.
• •
  Target cell-limited model with latent infected cells:Model (1) was extended in [20] by including two classes of infected cells, latent infected cells and active infected cells as:

  $$\left\{\begin{array}{c}\dot{T}\left(t\right)=-\xi V\left(t\right)T\left(t\right),\hfill \\ \dot{L}\left(t\right)=\xi V\left(t\right)T\left(t\right)-\alpha L\left(t\right),\hfill \\ \dot{I}\left(t\right)=\alpha L\left(t\right)-d_{I}I\left(t\right),\hfill \\ \dot{V}\left(t\right)=kI\left(t\right)-d_{V}V\left(t\right),\hfill \end{array}\right)$$ (2)

  where $L\left(t\right)$ is the concentration of latent infected cells. Parameter $\alpha$ represents the activation rate constant of latent infected cells. Ke et al. [22] developed some mathematical models for within-host dynamics of SARS-CoV-2 and fitted them to real data. They supported a quantitative framework for concluding the influence of vaccines and therapeutics on the infectiousness of COVID-19 patients and for assessing rapid testing strategies. Based on model (2), Pinky and Dobrovolny developed a mathematical model for SARS-CoV-2 and other respiratory viruses coinfection within a host [23]. It was reported that SARS-CoV-2 progression can be suppressed by other viruses when the infections occur at the same time. Gonçalves et al. [24] modified model (2) by including the absorption effect. The last equation of model (2) was modified as: $\dot{V}\left(t\right)=kI\left(t\right)-d_{V}V\left(t\right)-\xi V\left(t\right)T\left(t\right)$. The model was fitted using real data. The results showed that, less drug efficacy is required to reduce the peak viral load when the treatment is starting before the symptom onset. Wang et al. [25] introduced a within-host SARS-CoV-2 dynamics model with two types of target cells (pneumocytes and lymphocytes). They fitted their model, model (1) and model (2) with real data of COVID-19 patients and non-human primates. The results showed that model with two target cells significantly improves the fit. The effect of antiviral drugs or anti-inflammatory treatments combined with interferon on the viral load and recovery time were studied.
• •
  Model with effector cells: The following model describes the interaction between effector cells and SARS-CoV-2 [20]:

  $$\left\{\begin{array}{c}\dot{V}\left(t\right)=bV\left(t\right)\left(1-\frac{V\left(t\right)}{K}\right)-d_{V}V\left(t\right)-cV\left(t\right)E\left(t\right),\hfill \\ \dot{E}\left(t\right)=s+hE\left(t\right)\left(\frac{V^m\left(t\right)}{V^m\left(t\right)+a^m}\right)-d_{E}E\left(t\right),\hfill \end{array}\right)$$ (3)

  where $b$ is viral replication rate, $K$ is the maximum carrying capacity of the viruses, $E$ is the concentration of effector cells and $c$ is killing rate constant of infected cells by effector cells. The parameter $s=d_{E}E\left(0\right)$ represents the effector cell homeostasis, $h$ denotes the proliferation rate of effector cells and $d_E$ is death rate constant of the effector cells, $a$ is a saturation constant, and $m$ denotes the width of the sigmoidal function. Model (3) was used in many works (see e.g. [26], [27]). In [26], differential evolution algorithm was applied to fit model (3) in case of $m=2$ with experimental data. Blanco-Rodriguez et al. [27] elucidated the key parameters that define the course of the COVID-19 deviating from severe to critical case. Hernandez-Vargas and Velasco-Hernandez [20] used Akaike information criterion to compare between different models. The models were fitted with real data from 9 patients with COVID-19. It was shown that model (3) was better fitting than logarithmic decay and exponential growth models, model (1) and model (2).
• •
  Model with constant regeneration of target cells:Following the basic within-host viral dynamics model presented by Nowak and Bangham [28], Li et al. [29] formulated the following within host SARS-CoV-2 model:

  $$\left\{\begin{array}{c}\dot{T}\left(t\right)=d_T(T\left(0\right)-T\left(t\right))-\xi V\left(t\right)T\left(t\right),\hfill \\ \dot{I}\left(t\right)=\xi V\left(t\right)T\left(t\right)-d_{I}I\left(t\right),\hfill \\ \dot{V}\left(t\right)=kI\left(t\right)-d_{V}V\left(t\right),\hfill \end{array}\right)$$ (4)

  where $d_{T}T\left(0\right)$ is the regeneration rate constant of the healthy epithelial cells and $T\left(0\right)$ is the concentration of healthy epithelial cells without virus. The model’s parameters were estimated by using Chest radiograph score data in [29]. Sadria and Layton [30] formulated a within-host SARS-CoV-2 infection model with constant regeneration of target cells to simulate the effect of three drug therapies, Remdesivir, an alternative (hypothetical) therapy and transfusion therapy convalescent plasma. It was suggested that therapies are more effective when they applied early, one or two days post symptoms onset [30]. Danchin et al. [31] extended model (4) by including the effect of antibodies. Ghosh [32] formulated a mathematical model that describes the SARS-CoV-2 dynamics with constant regeneration of target cells. Both innate and adaptive immune responses were included. The model was fitted with real data and the effect of different antiviral drugs was addressed. Du and Yuan [5] proposed a within-host model of SARS-CoV-2 infection with constant regeneration of target cells. They studied the influence of the interaction between adaptive and innate immune responses on the peak of viral load in COVID-19 patients. They showed that, temporarily suppress the adaptive immune response to avoid interfering with the innate immune response, it may allow the innate immunity to get rid of the virus more efficiently.

Stability of analysis of within-host SARS-CoV-2 dynamics models is one of the powerful tools that can provide researchers with better understanding about the dynamics of the virus and how the immune system control and clear the virus. Stability analysis of model (3) was studied by Almocera et al. [33]. It was mentioned that the SARS-CoV-2 may replicate fast enough to overcome the response of the effector cells and cause infection. Hattaf and Yousfi [34] developed a SARS-CoV-2 dynamics model with CTL immune response and cell-to-cell infection. The global stability of the three steady states of the model was studied. Ghanbari [35] extended the model presented in [34] by using fractional derivatives by. Chatterjee and Al Basir [36] studied a SARS-CoV-2 infection model with treatment and CTL immune response. Mondal et al. [37] developed and analyzed a five-dimensional within-host SARS-CoV-2 dynamics model which includes both CTL and antibody immune responses. Elaiw et al. [38] developed and proved the global stability of a SARS-CoV-2/cancer coinfection model with two immune responses: cancer-specific CTL immune response and SARS-CoV-2-specific antibody. Mathematical modeling and analysis of SARS-CoV-2/HIV coinfection dynamics were studied in [39] and [40]. Nath et al. [41] studied the mathematical analysis of model (4). They proved both local and global stability of the two steady states of the model.

Optimal control theory was applied to determine optimal treatment strategies for infected patients with different viruses such as HIV [42], [43], HBV [44], HCV [45]. On the basis of the basic within-host viral dynamics model presented by Nowak and Bangham [28], Chhetri et al. [46] formulated and analyzed within-host SARS-CoV-2 dynamics model under the effect of immunomodulating and antiviral drug therapies. Optimal drug interventions were determined. It was suggested that the combination of immunomodulating and antiviral drug therapies is most effective. In [47], fractional differential equations were used in formulating within-host SARS-CoV-2 model with non-lytic and lytic immune responses. Two types of antiviral drugs were included as control inputs, one for blocking the infection and the other for inhibiting the viral production. Optimal antiviral drugs were determined by solving the fractional optimal control problem.

In most of the above mentioned works, the proliferation of the healthy target cells was not considered. Fatehi et al. [18] and Fadai et al. [48] developed SARS-CoV-2 dynamics models by assuming that the healthy epithelial cells follow logistic growth in the absence of SARS-CoV-2. However, mathematical analysis of these models was not studied.

It was observed experimentally that there exist a time lag between infection of a target cell and the release of new virions [49]. Therefore, several SARS-CoV-2 dynamics models were developed using ordinary differential equations (ODEs) by splitting the infected cells into two compartments, latent infected cells and active (productive) infected cells (see e.g. [18], [20], [22], [23], [25], [30], [50]). Latent infected cells contain viruses but do not produce them until they are activated. These model assumed that one infected, the cell immediately becomes a latent infected cell. Further, these models neglected the time for the latent infected cells to be activated [51]. Furthermore, the maturation time of the new viruses was not considered. To incorporate these time lags we need to formulate the SARS-CoV-2 dynamics using delay differential equations (DDEs). DDEs models can characterize the effect time delay on the dynamical behavior of the virus.

In a very recent work [52], we have studied the global stability of a delayed SARS-CoV-2 infection model with antibody immune response. However, the CTL immune response has not considered. Therefore, our contribution in the present paper is to develop and analyze a within-host SARS-CoV-2 infection model with both CTL and antibody immune responses. The model includes: (i) both latent and active infected epithelial cells, (ii) logistic growth term for the healthy epithelial cells, and (iii) four time delays, the times form the SARS-CoV-2 particles contact the healthy epithelial cells to become latent/active infected cells, the reactivation time of latent infected cells, and the maturation time of new virions. The basic and global properties of the model are studied. To support the theoretical results we performed some numerical simulations. The effect of time delay on the dynamics of SARS-CoV-2 is addressed.

Overall, our proposed model and its analysis can be useful for better understanding the within-host SARS-CoV-2 dynamics under the effect of both CTL and antibody immune responses. In addition, our study can be helpful to develop coinfection dynamics model with more aggressive variants of SARS-CoV-2 like Alpha, Beta, Gamma, Delta, Lambda and Omicron.

2. Model development

This section gives a brief description of the model under consideration. The model takes the form

$$\dot{T}\left(t\right)=\lambda -d_{T}T\left(t\right)+rT\left(t\right)\left(1-\frac{T\left(t\right)}{T_{max}}\right)-\xi T\left(t\right)V\left(t\right),$$ (5)

$$\dot{L}\left(t\right)=\eta \xi {\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }V(t-\psi )T(t-\psi )d\psi -\alpha L\left(t\right)-d_{L}L\left(t\right),$$ (6)

$$\dot{I}\left(t\right)=(1-\eta )\xi {\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }V(t-\psi )T(t-\psi )d\psi +\alpha e^{-n_3{\tau }_3}L(t-{\tau }_3)-d_{I}I\left(t\right)-\omega I\left(t\right)C\left(t\right),$$ (7)

$$\dot{V}\left(t\right)=k{e}^{-n_4{\tau }_4}I(t-{\tau }_4)-d_{V}V\left(t\right)-uV\left(t\right)A\left(t\right),$$ (8)

$$\dot{A}\left(t\right)=qV\left(t\right)A\left(t\right)-d_{A}A\left(t\right),$$ (9)

$$\dot{C}\left(t\right)=\sigma I\left(t\right)C\left(t\right)-d_{C}C\left(t\right),$$ (10)

where $T\left(t\right)$, $L\left(t\right)$, $I\left(t\right)$, $V\left(t\right),A\left(t\right)$ and $C\left(t\right)$ represent the concentrations of healthy epithelial cells, latent infected cells, active infected cells, SARS-CoV-2 particles, antibodies and CTLs at time $t$, respectively. The healthy epithelial cells are regenerated with constant $\lambda$ and are proliferated with logistic growth rate $rT\left(1-\frac{T}{T_{max}}\right)$, where $r$ is the rate of growth and $T_{max}$ is the maximum capacity of healthy epithelial cells in the human body. Healthy epithelial cells are infected by SARS-CoV-2 at rate $\xi VT$. Parameter $\eta \in (0,1)$ is the part of the healthy epithelial cells that enters the latent state, while $\alpha$ is the activation rate constant of latent infected cells. $kI$ is the rate at which active infected cells produces SARS-CoV-2 particles. The infected cells are killed by CTLs at rate $\omega CI$. $uAV$ is the neutralization rate of SARS-CoV-2 due to antibody immunity. The terms $qAV$ and $\sigma CI$ refer to the proliferation rate of antibodies and CTLs, respectively. The parameters $d_T$, $d_L$, $d_I$, $d_V$, $d_A$ and $d_C$ are the death rate constants of healthy epithelial cells, latent infected cells, active infected cells, SARS-CoV-2 particles, antibodies and CTLs, respectively. The factor $f\left(\psi \right)e^{-n_1\psi }$ represents the probability that healthy epithelial cells touched by SARS-CoV-2 particles at time $t-\psi$ survived $\psi$ time units and become latent infected cells at time $t$. The factor $g\left(\psi \right)e^{-n_2\psi }$ is the probability that healthy epithelial cells touched by SARS-CoV-2 particles at time $t-\psi$ survived $\psi$ time units and become active cells infected at time $t$. Here, $\psi$ is random taken from probability distribution functions $f\left(\psi \right)$ and $g\left(\psi \right)$ over the intervals $[0,{\tau }_1]$ and $[0,{\tau }_2]$, respectively. ${\tau }_1$ and ${\tau }_2$ are the upper limits of the delay periods. ${\tau }_3$ is the period of time during which latent infected cells are activated to produce active infected cells. ${\tau }_4$ is the time it takes from the newly released viruses to be mature and then infectious. Factors $e^{-n_3{\tau }_3}$ and $e^{-n_4{\tau }_4}$ represent the survival rates of latent infected cells and viruses during their delay periods $[t-{\tau }_3,t]$ and $[t-{\tau }_4,t]$, respectively. The functions $f\left(\psi \right):\left[0,{\tau }_1\right]\to \left[0,\infty \right)$ and $g\left(\psi \right):\left[0,{\tau }_2\right]\to \left[0,\infty \right)$ are the distribution functions which satisfy the following conditions:

$$\text{(i)}f\left(\psi \right)>0,g\left(\psi \right)>0,$$

$$\text{(ii)}{\int }_0^{{\tau }_1}f\left(\psi \right)d\psi =1,{\int }_0^{{\tau }_2}g\left(\psi \right)d\psi =1,$$

$$\text{(iii)}{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }d\psi <\infty ,{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }d\psi <\infty ,n_1,n_2>0.$$

Let

$$F={\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }d\psi \text{and}G={\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }d\psi .$$

Hence $0<F,G\le 1$.

The initial conditions of system (5)–(10) are:

$$T(\varkappa )={\phi }_1(\varkappa ),L(\varkappa )={\phi }_2(\varkappa ),I(\varkappa )={\phi }_3(\varkappa ),V(\varkappa )={\phi }_4(\varkappa ),A(\varkappa )={\phi }_5(\varkappa ),C(\varkappa )={\phi }_6(\varkappa ),$$$${\phi }_i(\varkappa )\ge 0,\varkappa \in [-\kappa ,0],i=1,2,\dots ,6,$$ (11)

where $\kappa =max\{{\tau }_1,{\tau }_2,{\tau }_3,{\tau }_4\}$, ${\phi }_i\in \mathcal{C}([-\kappa ,0],{\mathbb{R}}_{\ge 0}),i=1,2,\dots ,6$, and $\mathcal{C}$ is the Banach space of continuous functions mapping the interval $[-\kappa ,0]$ to ${\mathbb{R}}_{\ge 0}$ with

$$\Vert {\phi }_i\Vert =\underset{-\kappa \le \varkappa \le 0}{sup}\left|{\phi }_i(\varkappa )\right|\text{for}{\phi }_i\in \mathcal{C}.$$

By the fundamental theory of functional differential equations [53], system (5)–(10) with initial conditions (11) has a unique solution.

Remark 1

When the CTL immune response is not considered, then model (5)–(10) will lead to the model presented in [52]. Further, if we neglect the logistic growth rate term, time delays, CTL immune response and antibody immune response, then system (5)–(10) will lead to system (4), where $\lambda =d_T\text{(}T\left(0\right)\text{)}$.

3. Basic properties

This section proves the basic properties of system (5)–(9) including the non-negativity and boundedness of solutions. Moreover, it lists all possible steady states and their existence conditions.

For the non-negativity and boundedness of solutions for the system (5)–(10), we have the following theorem:

Theorem 1

Let${\left(T\left(t\right),L\left(t\right),I\left(t\right),V\left(t\right),A\left(t\right),C\left(t\right)\right)}^{\prime }$be arbitrary solution of system(5)–(10)with initial conditions (11) . Then, $\left(T\left(t\right),L\left(t\right),I\left(t\right),V\left(t\right),A\left(t\right),C\left(t\right)\right)\prime$ are non-negative on $[0,+\infty )$ and ultimately bounded.

Proof

Let us write system (5)–(10) in the matrix form $\dot{\Theta }\left(t\right)=H\left(\Theta \left(t\right)\right)$, where $\Theta ={(T,L,I,V,A,C)}^{\prime }$, $H=(H_1,H_2,H_3,H_4,$ ${H_5,H_6)}^T$, and

$$H\left(\Theta \left(t\right)\right)=\left(\begin{array}{c}\hfill H_1\left(\Theta \left(t\right)\right)\hfill \\ \hfill H_2\left(\Theta \left(t\right)\right)\hfill \\ \hfill H_3\left(\Theta \left(t\right)\right)\hfill \\ \hfill H_4\left(\Theta \left(t\right)\right)\hfill \\ \hfill H_5\left(\Theta \left(t\right)\right)\hfill \\ \hfill H_6\left(\Theta \left(t\right)\right)\hfill \end{array}\right)=\left(\begin{array}{c}\hfill \lambda -d_{T}T\left(t\right)+rT\left(t\right)\left(1-\frac{T\left(t\right)}{T_{max}}\right)-\xi V\left(t\right)T\left(t\right)\hfill \\ \hfill \eta \xi {\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }V(t-\psi )T(t-\psi )d\psi -\alpha L\left(t\right)-d_{L}L\left(t\right)\hfill \\ \hfill (1-\eta )\xi {\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }V(t-\psi )T(t-\psi )d\psi +\alpha e^{-n_3{\tau }_3}L(t-{\tau }_3)-d_{I}I\left(t\right)-\omega I\left(t\right)C\left(t\right)\hfill \\ \hfill k{e}^{-n_4{\tau }_4}I(t-{\tau }_4)-d_{V}V\left(t\right)-uV\left(t\right)A\left(t\right)\hfill \\ \hfill qV\left(t\right)A\left(t\right)-d_{A}A\left(t\right)\hfill \\ \hfill \sigma I\left(t\right)C\left(t\right)-d_{C}C\left(t\right)\hfill \end{array}\right).$$

We see that the function $H$ satisfies the following condition:

$$H_i\left(\Theta \left(t\right)\right){\mid }_{{\Theta }_i=0,\Theta \left(t\right)\in {\mathbb{R}}_{\ge 0}^6}\ge 0,i=1,2,\dots ,6.$$

Using Lemma 2 in [54], any solution of system (5)–(10) with the initial states (11) is such that $\Theta \left(t\right)\in {\mathbb{R}}_{\ge 0}^6$ for all $t\ge 0$. Hence, ${\mathbb{R}}_{\ge 0}^6$ is positively invariant for the system (5)–(10).

Next, we prove the ultimate boundedness of the solutions. From Eq. (5), we have

$$\dot{T}\left(t\right)=\lambda -d_{T}T\left(t\right)+rT\left(t\right)\left(1-\frac{T\left(t\right)}{T_{max}}\right)-\xi V\left(t\right)T\left(t\right)\le \lambda -d_{T}T\left(t\right)+rT\left(t\right)\left(1-\frac{T\left(t\right)}{T_{max}}\right).$$ (12)

From the inequality (12) and the comparison principle, we obtain ${lim\; sup}_{t\to \infty }T\left(t\right)\le T_0$, where $T_0$ is the positive root of $\lambda -d_{T}T+rT\left(1-\frac{T}{T_{max}}\right)=0$ and is given by:

$$T_0=\frac{T_{max}}{2r}\left[r-d_T+\sqrt{{(r-d_T)}^2+\frac{4r\lambda }{T_{max}}}\right].$$ (13)

Now, we define

$$W_1\left(t\right)={\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }T(t-\psi )d\psi +\frac{1}{\eta }L\left(t\right).$$

Then, we get

$${\dot{W}}_1\left(t\right)={\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\dot{T}(t-\psi )d\psi +\frac{1}{\eta }\dot{L}\left(t\right)={\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\left[\lambda -d_{T}T(t-\psi )+rT(t-\psi )\left(1-\frac{T(t-\psi )}{T_{max}}\right)-\xi V(t-\psi )T(t-\psi )\right]d\psi +\xi {\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }V(t-\psi )T(t-\psi )d\psi -\frac{\alpha }{\eta }L\left(t\right)-\frac{d_L}{\eta }L\left(t\right)={\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\left(-\frac{r}{T_{max}}{T}^2(t-\psi )+rT(t-\psi )+\lambda \right)d\psi -d_T{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }T(t-\psi )d\psi -\frac{\alpha +d_L}{\eta }L\left(t\right).$$

Let us define $\Gamma \left(T\right)=-\frac{r}{T_{max}}{T}^2+rT+\lambda$. Then to find the maximum value of $\Gamma \left(T\right)$ we find

$${\Gamma }^{\prime }\left(T\right)=-\frac{2r}{T_{max}}T+r=0\Rightarrow T=\frac{T_{max}}{2}$$

and

$${\Gamma }^{″}\left(T\right)=-\frac{2r}{T_{max}}<0.$$

Then

$$\Gamma \left(\frac{T_{max}}{2}\right)=-\frac{r}{T_{max}}{\left(\frac{T_{max}}{2}\right)}^2+r\left(\frac{T_{max}}{2}\right)+\lambda =\frac{r{T}_{max}}{4}+\lambda .$$

Let $N_1=\frac{r{T}_{max}+4\lambda }{4}>0$ and $q_1=min\{d_T,\alpha +d_L\}$, then

$${\dot{W}}_1\left(t\right)\le F{N}_1-q_1{W}_1\left(t\right)\le N_1-q_1{W}_1\left(t\right).$$

Therefore, ${lim\; sup}_{t\to \infty }{W}_1\left(t\right)\le \frac{N_1}{q_1}$. Since $T\left(t\right)\ge 0$ and $L\left(t\right)\ge 0$, then ${lim\; sup}_{t\to \infty }L\left(t\right)\le \frac{\eta N_1}{q_1}=p_1$. To prove the ultimate boundedness of $I\left(t\right)$ and $C\left(t\right)$, we define

$$W_2\left(t\right)={\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }T(t-\psi )d\psi +\frac{1}{1-\eta }I\left(t\right)+\frac{\omega }{(1-\eta )\sigma }C\left(t\right).$$

Then, we obtain

$${\dot{W}}_2\left(t\right)={\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\dot{T}(t-\psi )d\psi +\frac{1}{1-\eta }\dot{I}\left(t\right)+\frac{\omega }{(1-\eta )\sigma }\dot{C}\left(t\right)={\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\left[\lambda -d_{T}T(t-\psi )+rT(t-\psi )\left(1-\frac{T(t-\psi )}{T_{max}}\right)-\xi V(t-\psi )T(t-\psi )\right]d\psi +\xi {\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }V(t-\psi )T(t-\psi )d\psi -\frac{d_I}{1-\eta }I\left(t\right)+\frac{\alpha e^{-n_3{\tau }_3}}{1-\eta }L(t-{\tau }_3)-\frac{\omega }{1-\eta }C\left(t\right)I\left(t\right)+\frac{\omega }{1-\eta }I\left(t\right)C\left(t\right)-\frac{\omega d_C}{(1-\eta )\sigma }C\left(t\right)\le {\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\left(-\frac{r}{T_{max}}{T}^2(t-\psi )+rT(t-\psi )+\lambda \right)d\psi +\frac{\alpha e^{-n_3{\tau }_3}}{1-\eta }{p}_1-d_T{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }T(t-\psi )d\psi -\frac{d_I}{1-\eta }I\left(t\right)-\frac{\omega d_C}{(1-\eta )\sigma }C\left(t\right)\le {\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\left(\frac{r{T}_{max}+4\lambda }{4}\right)d\psi +\frac{\alpha e^{-n_3{\tau }_3}}{1-\eta }{p}_1-d_T{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }T(t-\psi )d\psi -\frac{d_I}{1-\eta }I\left(t\right)-\frac{\omega d_C}{(1-\eta )\sigma }C\left(t\right)\le \frac{r{T}_{max}+4\lambda }{4}+\frac{\alpha }{1-\eta }{p}_1-d_T{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }T(t-\psi )d\psi -\frac{d_I}{1-\eta }I\left(t\right)-\frac{\omega d_C}{(1-\eta )\sigma }C\left(t\right).$$

Let $N_2=\frac{r{T}_{max}+4\lambda }{4}+\frac{\alpha }{1-\eta }{p}_1>0$ and $q_2=min\{d_T,d_I,d_C\}$, then

$${\dot{W}}_2\left(t\right)\le N_2-q_2{W}_2\left(t\right).$$

This implies that ${lim\; sup}_{t\to \infty }{W}_2\left(t\right)\le \frac{N_2}{q_2}$. Since $I\left(t\right)\ge 0$ and $C\left(t\right)\ge 0$, then ${lim\; sup}_{t\to \infty }I\left(t\right)\le \frac{(1-\eta )N_2}{q_2}=p_2$ and ${lim\; sup}_{t\to \infty }C\left(t\right)\le \frac{(1-\eta )\sigma N_2}{\omega q_2}=p_3$. To prove the ultimate boundedness of $V\left(t\right)$ and $A\left(t\right)$, we consider

$$W_3\left(t\right)=V\left(t\right)+\frac{u}{q}A\left(t\right).$$

This gives

$${\dot{W}}_3\left(t\right)=k{e}^{-n_4{\tau }_4}I(t-{\tau }_4)-d_{V}V\left(t\right)-uAV\left(t\right)\left(t\right)+uV\left(t\right)A\left(t\right)-\frac{u{d}_A}{q}A\left(t\right)=k{e}^{-n_4{\tau }_4}I(t-{\tau }_4)-d_{V}V\left(t\right)-\frac{u{d}_A}{q}A\left(t\right)\le k{e}^{-n_4{\tau }_4}I(t-{\tau }_4)-q_3[V\left(t\right)+\frac{u}{q}A\left(t\right)]\le k{p}_2-q_3{W}_3\left(t\right),$$

where $q_3=min\{d_V,d_A\}$. Hence, ${lim\; sup}_{t\to \infty }{W}_3\left(t\right)\le \frac{k{p}_2}{q_3}=p_4$. Since $V\left(t\right)\ge 0$ and $A\left(t\right)\ge 0$, then ${lim\; sup}_{t\to \infty }V\left(t\right)\le p_4$, and ${lim\; sup}_{t\to \infty }A\left(t\right)\le \frac{q}{u}{p}_4$. The above analysis proves that $T\left(t\right),L\left(t\right),I\left(t\right),V\left(t\right),A\left(t\right)$ and $C\left(t\right)$ are ultimately bounded.

3.1. Steady states

This subsection computes all steady states of system (5)–(10) and the threshold parameters that guarantee the existence of these steady states. For system (5)–(10) we define the basic reproduction number ${\mathcal{R}}_0$ as [51]:

$${\mathcal{R}}_0=\frac{k\xi e^{-n_4{\tau }_4}{T}_0}{d_I{d}_V}\left[\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }d\psi +(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }d\psi \right]=\frac{k\xi e^{-n_4{\tau }_4}{T}_0}{d_I{d}_V}\left(\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}F+(1-\eta )G\right).$$

For convenience, let $\rho =\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}F+(1-\eta )G$. Then, ${\mathcal{R}}_0$ can be rewritten as:

$${\mathcal{R}}_0=\frac{k\xi e^{-n_4{\tau }_4}{T}_0}{d_I{d}_V}\rho .$$

Let $SS=(T,L,I,V,A,C)$ be any steady state of system (5)–(10) satisfying the following system of equations:

$$0=\lambda -d_{T}T+rT\left(1-\frac{T}{T_{max}}\right)-\xi TV,$$ (14)

$$0=\eta F\xi TV-(\alpha +d_L)L,$$ (15)

$$0=(1-\eta )G\xi TV+\alpha e^{-n_3{\tau }_3}L-d_{I}I-\omega CI,$$ (16)

$$0=k{e}^{-n_4{\tau }_4}I-d_{V}V-uAV,$$ (17)

$$0=qAV-d_{A}A,$$ (18)

$$0=\sigma CI-d_{C}C.$$ (19)

By solving system (14)–(19), we get five steady states:

• (i)Healthy steady state $S{S}_0=(T_0,0,0,0,0,0)$, where $T_0$ is given by Eq. (13).
• (ii)
  Infected steady state with inactive immune response $S{S}_1=(T_1,L_1,I_1,V_1,0,0)$, where

  $$T_1=\frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k\xi \rho }=\frac{T_0}{{\mathcal{R}}_0},$$

  $$L_1=\frac{\eta }{\alpha +d_L}F\xi T_1{V}_1,$$

  $$I_1=\frac{d_V{e}^{n_4{\tau }_4}}{k}{V}_1,$$

  $$V_1=\frac{\lambda k{e}^{-n_4{\tau }_4}\rho }{d_I{d}_V}+\frac{r}{\xi }-\left(\frac{d_T}{\xi }+\frac{r{d}_I{d}_V{e}^{n_4{\tau }_4}}{k{\xi }^2{T}_{max}\rho }\right).$$

  Assume that $d_T-r+\frac{r{T}_1}{T_{max}}>0$, then we get

  $$d_T-r+\frac{r}{T_{max}}\frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k\xi \rho }>0.$$ (20)

  We note that

  $${\mathcal{R}}_0>1⟺\frac{T_{max}}{2r}\left[r-d_T+\sqrt{{(r-d_T)}^2+\frac{4r\lambda }{T_{max}}}\right]>\frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k\xi \rho }⟺\sqrt{{(r-d_T)}^2+\frac{4r\lambda }{T_{max}}}>\frac{2r{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi T_{max}\rho }-(r-d_T).$$

  From inequality (20) we have $\frac{2r{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi T_{max}\rho }-(r-d_T)>0$. Then

  $${\mathcal{R}}_0>1⟺\frac{4r\lambda }{T_{max}}>\frac{4r^2{d}_I{d}_V{e}^{2n_4{\tau }_4}}{k^2{\xi }^2{T}_{max}^2{\rho }^2}-\frac{4r{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi T_{max}\rho }(r-d_T)⟺r\lambda >\frac{r^2{d}_I{d}_V{e}^{2n_4{\tau }_4}}{k^2{\xi }^2{T}_{max}{\rho }^2}-\frac{r^2{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi \rho }+\frac{r{d}_T{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi \rho }⟺\frac{\lambda k{e}^{-n_4{\tau }_4}\rho }{d_I{d}_V}+\frac{r}{\xi }-\left(\frac{d_T}{\xi }+\frac{r{d}_I{d}_V{e}^{n_4{\tau }_4}}{k{\xi }^2{T}_{max}\rho }\right)>0⟺V_1>0.$$

  Thus, $S{S}_1$ exists when ${\mathcal{R}}_0>1$ and $d_T-r+\frac{r{T}_1}{T_{max}}>0$. At this steady state the virus exists while the immune response is inhibited.
• (iii)
  Infected steady state with only active antibody immunity $S{S}_2=(T_2,L_2,I_2,V_2,A_2,0)$, where

  $$T_2=\frac{T_{max}}{2r}\left[r-d_T-\frac{d_A\xi }{q}+\sqrt{{\left(r-d_T-\frac{d_A\xi }{q}\right)}^2+\frac{4r\lambda }{T_{max}}}\right],$$

  $$L_2=\frac{d_A\eta \xi F}{q(\alpha +d_L)}\frac{T_{max}}{2r}\left[r-d_T-\frac{d_A\xi }{q}+\sqrt{{\left(r-d_T-\frac{d_A\xi }{q}\right)}^2+\frac{4r\lambda }{T_{max}}}\right]=\frac{d_A\eta \xi F{T}_2}{q(\alpha +d_L)},$$

  $$I_2=\frac{d_A\xi }{q{d}_I}\frac{T_{max}}{2r}\left[r-d_T-\frac{d_A\xi }{q}+\sqrt{{\left(r-d_T-\frac{d_A\xi }{q}\right)}^2+\frac{4r\lambda }{T_{max}}}\right]\rho =\frac{d_A\xi T_2}{q{d}_I}\rho ,$$

  $$V_2=\frac{d_A}{q},$$

  $$A_2=\frac{d_V}{u}\left[\frac{k\xi e^{-n_4{\tau }_4}}{d_I{d}_V}\frac{T_{max}}{2r}\left(r-d_T-\frac{d_A\xi }{q}+\sqrt{{\left(r-d_T-\frac{d_A\xi }{q}\right)}^2+\frac{4r\lambda }{T_{max}}}\right)\rho -1\right]=\frac{d_V}{u}\left(\frac{k\xi e^{-n_4{\tau }_4}{T}_2}{d_I{d}_V}\rho -1\right).$$

  We note that $S{S}_2$ exists when $\frac{k\xi e^{-n_4{\tau }_4}{T}_2}{d_I{d}_V}\rho >1$. We define the antibody immunity activation number ${\mathcal{R}}_1^A$ as:

  $${\mathcal{R}}_1^A=\frac{k\xi e^{-n_4{\tau }_4}{T}_2}{d_I{d}_V}\rho .$$

  which determines when the antibody immunity is activated. Thus, $A_2=\frac{d_V}{u}\left({\mathcal{R}}_1^A-1\right)$. We note that $A_2>0$ when ${\mathcal{R}}_1^A>1$. Thus, $S{S}_2$ exists when ${\mathcal{R}}_1^A>1$.
• (iv)
  Infected steady state with only active CTL immunity $S{S}_3=(T_3,L_3,I_3,V_3,0,C_3)$, where

  $$T_3=\frac{T_{max}}{2r}\left[r-d_T-\frac{d_{C}k\xi e^{-n_4{\tau }_4}}{d_V\sigma }+\sqrt{{\left(r-d_T-\frac{d_{C}k\xi e^{-n_4{\tau }_4}}{d_V\sigma }\right)}^2+\frac{4r\lambda }{T_{max}}}\right],$$

  $$L_3=\frac{d_C\eta k\xi F{e}^{-n_4{\tau }_4}}{d_V\sigma (\alpha +d_L)}\frac{T_{max}}{2r}\left[r-d_T-\frac{d_{C}k\xi e^{-n_4{\tau }_4}}{d_V\sigma }+\sqrt{{\left(r-d_T-\frac{d_{C}k\xi e^{-n_4{\tau }_4}}{d_V\sigma }\right)}^2+\frac{4r\lambda }{T_{max}}}\right],=\frac{d_C\eta k\sigma \xi F{e}^{-n_4{\tau }_4}{T}_3}{d_V\sigma (\alpha +d_L)},$$

  $$I_3=\frac{d_C}{\sigma },$$

  $$V_3=\frac{d_{C}k{e}^{-n_4{\tau }_4}}{d_V\sigma },$$

  $$C_3=\frac{d_I}{\omega }\left[\frac{k\xi e^{-n_4{\tau }_4}{T}_3}{d_I{d}_V}\left(\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}F+(1-\eta )G\right)-1\right]=\frac{d_I}{\omega }\left(\frac{k\xi e^{-n_4{\tau }_4}{T}_3\rho }{d_I{d}_V}-1\right).$$

  We note that $C_3>0$ when $\frac{k\xi e^{-n_4{\tau }_4}{T}_3\rho }{d_I{d}_V}>1$. Then, we define the CTL immunity activation number ${\mathcal{R}}_1^C$ as:

  $${\mathcal{R}}_1^C=\frac{k\xi e^{-n_4{\tau }_4}{T}_3\rho }{d_I{d}_V}.$$

Therefore, $S{S}_3$ exists when ${\mathcal{R}}_1^C>1$.

• 1.
  Infected steady state with both active antibody and CTL immune responses $S{S}_4=(T_4,L_4,I_4,V_4,A_4,C_4)$, where

  $$T_4=\frac{T_{max}}{2r}\left[r-d_T-\frac{d_A\xi }{q}+\sqrt{{\left(r-d_T-\frac{d_A\xi }{q}\right)}^2+\frac{4r\lambda }{T_{max}}}\right]=T_2,$$

  $$L_4=\frac{d_A\eta \xi F}{q(\alpha +d_L)}\frac{T_{max}}{2r}\left[r-d_T-\frac{d_A\xi }{q}+\sqrt{{\left(r-d_T-\frac{d_A\xi }{q}\right)}^2+\frac{4r\lambda }{T_{max}}}\right]=\frac{d_A\eta \xi F{T}_4}{q(\alpha +d_L)},$$

  $$I_4=\frac{d_C}{\sigma }=I_3,$$

  $$V_4=\frac{d_A}{q}=V_2,$$

  $$A_4=\frac{d_{C}kq{e}^{-n_4{\tau }_4}}{d_A\sigma u}-\frac{d_V}{u}=\frac{d_V}{u}\left[\frac{d_{C}kq{e}^{-n_4{\tau }_4}}{d_V{d}_A\sigma }-1\right],$$

  $$C_4=\frac{\sigma }{d_C\omega }\left[\frac{d_A\xi T_4}{q}\left(\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}F+(1-\eta )G\right)-\frac{d_I{d}_C}{\sigma }\right]=\frac{d_I}{\omega }\left[\frac{d_A\sigma \xi T_4}{d_I{d}_{C}q}\rho -1\right].$$

  We see that $A_4$ and $C_4$ exist when $\frac{d_{C}kq{e}^{-n_4{\tau }_4}}{d_V{d}_A\sigma }>1$ and $\frac{d_A\sigma \xi T_4}{d_I{d}_{C}q}\rho >1$. Now, we define

  $${\mathcal{R}}_2^C=\frac{d_A\sigma \xi T_4}{d_I{d}_{C}q}\rho .$$

  Hence $A_4$ and $C_4$ can be rewritten as:

  $$A_4=\frac{d_V}{u}\left[\frac{{\mathcal{R}}_1^A}{{\mathcal{R}}_2^C}-1\right],C_4=\frac{d_I}{\omega }\left[{\mathcal{R}}_2^C-1\right]\text{.}$$

  Therefore, $S{S}_4$ exists when ${\mathcal{R}}_1^A>{\mathcal{R}}_2^C$ and ${\mathcal{R}}_2^C>1$. Here, ${\mathcal{R}}_2^C$ refers to the competed CTL immunity number.

Lemma 1

System (5) – (10) has four threshold parameters ${\mathcal{R}}_0>0$ , ${\mathcal{R}}_1^A>0$ , ${\mathcal{R}}_1^C>0$ and ${\mathcal{R}}_2^C>0$ , such that:

• (i) if ${\mathcal{R}}_0\le 1$ , then there exists only one steady state $S{S}_0$ ;
• (ii) if ${\mathcal{R}}_1^A\le 1<{\mathcal{R}}_0$ , ${\mathcal{R}}_1^C\le 1$ and $d_T-r+\frac{r{T}_1}{T_{max}}>0$ , then there exist two steady states $S{S}_0$ and $S{S}_1$ ;
• (iii) if ${\mathcal{R}}_1^A>1$ and ${\mathcal{R}}_2^C\le 1$ , then there exist only three steady states $S{S}_0$ , $S{S}_1$ and $S{S}_2$ ;
• (iv) if ${\mathcal{R}}_1^C>1$ and ${\mathcal{R}}_1^A\le {\mathcal{R}}_2^C$ , then there exist only three steady states $S{S}_0$ , $S{S}_1$ and $S{S}_3$ ;
• (v) if ${\mathcal{R}}_1^A>{\mathcal{R}}_2^C>1$ , then there exist five steady states $S{S}_0$ , $S{S}_1$ , $S{S}_2$ , $S{S}_3$ and $S{S}_4$ .

Remark 2

We note that our proposed model has five steady states, while the model presented in [52] (where the CTL immune response is not considered) has three steady states. Therefore, our proposed model can be more suitable to describe the reaction of the immune system against SARS-CoV-2 infection.

4. Global properties

In this section, the global asymptotic stability of the five steady states $S{S}_i$, $i=0,1,2,3,4$ will be established by using Lyapunov approach and applying LaSalle’s invariance principle. Denote $(T,L,I,V,A,C)=\left(T\left(t\right),L\left(t\right),I\left(t\right),V\left(t\right),A\left(t\right),C\left(t\right)\right)$.

Theorem 2

Suppose that The steady state $S{S}_0$ is globally asymptotically stable (GAS) when ${\mathcal{R}}_0\le 1$ .

Proof

Let

$$\mathcal{H}\left(x\right)=x-1-lnx,$$

and define a Lyapunov function ${\mathcal{V}}_0(T,L,I,V,A,C)$ as:

$${\mathcal{V}}_0=\rho T_0\mathcal{H}\left(\frac{T}{T_0}\right)+\frac{\alpha e^{-n_3{\tau }_3}}{\alpha +d_L}L+I+\frac{d_I{e}^{n_4{\tau }_4}}{k}V+\frac{d_{I}u{e}^{n_4{\tau }_4}}{kq}A+\frac{\omega }{\sigma }C+{\mathcal{U}}_0\left(t\right),$$

where and

$${\mathcal{U}}_0\left(t\right)=\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }{\int }_{t-\psi }^t\xi T\left(\varphi \right)V\left(\varphi \right)d\varphi d\psi +(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }{\int }_{t-\psi }^t\xi T\left(\varphi \right)V\left(\varphi \right)d\varphi d\psi +\alpha e^{-n_3{\tau }_3}{\int }_{t-{\tau }_3}^{t}L\left(\varphi \right)d\varphi +d_I{\int }_{t-{\tau }_4}^{t}I\left(\varphi \right)d\varphi .$$

Clearly, ${\mathcal{V}}_0(T,L,I,V,A,C)>0$ for all $T,L,I,V,A,C>0$, and ${\mathcal{V}}_0(T_0,0,0,0,0,0)=0$. The derivative of ${\mathcal{U}}_0\left(t\right)$ is computed as:

$$\frac{d{\mathcal{U}}_0\left(t\right)}{dt}=\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\xi TVd\psi -\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\xi T(t-\psi )V(t-\psi )d\psi +(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\xi TVd\psi -(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\xi T(t-\psi )V(t-\psi )d\psi +\alpha e^{-n_3{\tau }_3}L-\alpha e^{-n_3{\tau }_3}L(t-{\tau }_3)+d_{I}I-d_{I}I(t-{\tau }_4)=\rho \xi TV-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\xi T(t-\psi )V(t-\psi )d\psi -(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\xi T(t-\psi )V(t-\psi )d\psi +\alpha e^{-n_3{\tau }_3}L-\alpha e^{-n_3{\tau }_3}L(t-{\tau }_3)+d_{I}I-d_{I}I(t-{\tau }_4).$$

Hence, $\frac{d{\mathcal{V}}_0\left(t\right)}{dt}$ along the solutions of system (5)–(10) is given by:

$$\frac{d{\mathcal{V}}_0}{dt}=\rho \left(1-\frac{T_0}{T}\right)\dot{T}+\frac{\alpha e^{-n_3{\tau }_3}}{\alpha +d_L}\dot{L}+\dot{I}+\frac{d_I{e}^{n_4{\tau }_4}}{k}\dot{V}+\frac{d_{I}u{e}^{n_4{\tau }_4}}{kq}\dot{A}+\frac{\omega }{\sigma }\dot{C}+\frac{d{\mathcal{U}}_0\left(t\right)}{dt}.$$

By using system (5)–(10), we obtain

$$\frac{d{\mathcal{V}}_0}{dt}=\rho \left(1-\frac{T_0}{T}\right)\left[\lambda -d_{T}T+rT\left(1-\frac{T}{T_{max}}\right)-\xi TV\right]+\frac{\alpha e^{-n_3{\tau }_3}}{\alpha +d_L}\left[\eta {\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\xi T(t-\psi )V(t-\psi )d\psi -(\alpha +d_L)L\right]+(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\xi T(t-\psi )V(t-\psi )d\psi +\alpha e^{-n_3{\tau }_3}L(t-{\tau }_3)-d_{I}I-\omega CI+\frac{d_I{e}^{n_4{\tau }_4}}{k}\left[k{e}^{-n_4{\tau }_4}I(t-{\tau }_4)-d_{V}V-uAV\right]+\frac{d_{I}u{e}^{n_4{\tau }_4}}{kq}\left[qAV-d_{A}A\right]+\frac{\omega }{\sigma }\left[\sigma CI-d_{C}C\right]+\rho \xi TV-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\xi T(t-\psi )V(t-\psi )d\psi -(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\xi T(t-\psi )V(t-\psi )d\psi +\alpha e^{-n_3{\tau }_3}L-\alpha e^{-n_3{\tau }_3}L(t-{\tau }_3)+d_{I}I-d_{I}I(t-{\tau }_4).$$

By using the steady state conditions at $S{S}_0$, we obtain $\lambda =d_T{T}_0-r{T}_0\left(1-\frac{T_0}{T_{max}}\right)$, then

$$\lambda -d_{T}T+rT\left(1-\frac{T}{T_{max}}\right)=(T_0-T)\left(d_T-r+\frac{r{T}_0}{T_{max}}+\frac{rT}{T_{max}}\right).$$

Therefore, we deduce that

$$\frac{d{\mathcal{V}}_0}{dt}=-\rho \left(d_T-r+\frac{r{T}_0}{T_{max}}+\frac{rT}{T_{max}}\right)\frac{{(T-T_0)}^2}{T}+\left(\rho \xi T_0-\frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k}\right)V-\frac{d_I{d}_{A}u{e}^{n_4{\tau }_4}}{kq}A-\frac{d_C\omega }{\sigma }C\le -\rho \left(d_T-r+\frac{r{T}_0}{T_{max}}\right)\frac{{(T-T_0)}^2}{T}+\left(\rho \xi T_0-\frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k}\right)V-\frac{d_I{d}_{A}u{e}^{n_4{\tau }_4}}{kq}A-\frac{d_C\omega }{\sigma }C=-\rho \left(d_T-r+\frac{r{T}_0}{T_{max}}\right)\frac{{(T-T_0)}^2}{T}+\frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k}\left(\frac{\rho \xi T_{0}k{e}^{-n_4{\tau }_4}}{d_I{d}_V}-1\right)V-\frac{d_I{d}_{A}u{e}^{n_4{\tau }_4}}{kq}A-\frac{d_C\omega }{\sigma }C=-\rho \left(d_T-r+\frac{r{T}_0}{T_{max}}\right)\frac{{(T-T_0)}^2}{T}+\frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k}\left({\mathcal{R}}_0-1\right)V-\frac{d_I{d}_{A}u{e}^{n_4{\tau }_4}}{kq}A-\frac{d_C\omega }{\sigma }C.$$

At the steady state, we have $\lambda =d_T{T}_0-r{T}_0\left(1-\frac{T_0}{T_{max}}\right)$ which implies that $d_T-r+\frac{r{T}_0}{T_{max}}>0$. It follows that $\frac{d{\mathcal{V}}_0}{dt}\le 0$ when ${\mathcal{R}}_0\le 1$. Also, $\frac{d{\mathcal{V}}_0}{dt}=0$ when $T=T_0$, $A=0$ and $C=0$ and $\left({\mathcal{R}}_0-1\right)V=0$. The solutions of system (5)–(10) converge to $M_0^{\prime }$ the largest invariant subset of $M_0=\left\{(T,L,I,V,A,C)\right|\frac{d{\mathcal{V}}_0}{dt}=0\}$. For any elements in $M_0^{\prime }$we have $T=T_0$ and $A=C=0$ and

$$\left({\mathcal{R}}_0-1\right)V=0.$$ (21)

Let us consider two cases:

(i) ${\mathcal{R}}_0<1$, then Eq. (21) yields $V=0$. Since $M_0^{\prime }$is invariant then $\dot{V}=0$ and from Eq. (8) we get

$$0=\dot{V}=k{e}^{-n_4{\tau }_4}I⟹I\left(t\right)=0,\text{ for all }t$$ (22)

and hence $\dot{I}=0$. From Eq. (7) we obtain

$$0=\dot{I}=\alpha e^{-n_3{\tau }_3}L⟹L\left(t\right)=0,\text{ for all }t.$$ (23)

It follows that $M_0^{\prime }=\left\{S{S}_0\right\}$.

(ii) ${\mathcal{R}}_0=1$, we have $T=T_0$, then $\dot{T}=0$. From Eq. (5) we obtain

$$0=\dot{T}\left(t\right)=\lambda -d_T{T}_0+r{T}_0\left(1-\frac{T_0}{T_{max}}\right)-\xi T_{0}V\left(t\right)⟹V\left(t\right)=0,\text{ for all }t\text{.}$$

Eqs. (22)–(23) give $L\left(t\right)=I\left(t\right)=0$, for all $t$and hence $M_0^{\prime }=\left\{S{S}_0\right\}$.

By LaSalle’s invariance principle (LIP) [55], we conclude that $S{S}_0$ is GAS when ${\mathcal{R}}_0\le 1$.

Theorem 3

The steady state $S{S}_1$ is GAS when ${\mathcal{R}}_1^A\le 1<{\mathcal{R}}_0$ , ${\mathcal{R}}_1^C\le 1$ and $d_T-r+\frac{r{T}_1}{T_{max}}>0$ .

Proof

Define a function ${\mathcal{V}}_1(T,L,I,V,A,C)$ as:

$${\mathcal{V}}_1=\rho T_1\mathcal{H}\left(\frac{T}{T_1}\right)+\frac{\alpha e^{-n_3{\tau }_3}}{\alpha +d_L}{L}_1\mathcal{H}\left(\frac{L}{L_1}\right)+I_1\mathcal{H}\left(\frac{I}{I_1}\right)+\frac{d_I{e}^{n_4{\tau }_4}}{k}{V}_1\mathcal{H}\left(\frac{V}{V_1}\right)+\frac{d_{I}u{e}^{n_4{\tau }_4}}{kq}A+\frac{\omega }{\sigma }C+{\mathcal{U}}_1\left(t\right),$$

where

$${\mathcal{U}}_1\left(t\right)=\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_1{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }{\int }_{t-\psi }^t\mathcal{H}\left(\frac{T\left(\varphi \right)V\left(\varphi \right)}{T_1{V}_1}\right)d\varphi d\psi +(1-\eta )\xi T_1{V}_1{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }{\int }_{t-\psi }^t\mathcal{H}\left(\frac{T\left(\varphi \right)V\left(\varphi \right)}{T_1{V}_1}\right)d\varphi d\psi +\alpha e^{-n_3{\tau }_3}{L}_1{\int }_{t-{\tau }_3}^t\mathcal{H}\left(\frac{L\left(\varphi \right)}{L_1}\right)d\varphi +d_I{I}_1{\int }_{t-{\tau }_4}^t\mathcal{H}\left(\frac{I\left(\varphi \right)}{I_1}\right)d\varphi .$$

Clearly, ${\mathcal{V}}_1(T,L,I,V,A,C)>0$ for all $T,L,I,V,A,C>0$, and ${\mathcal{V}}_1(T_1,L_1,I_1,V_1,0,0)=0$. Then, $\frac{d{\mathcal{U}}_1\left(t\right)}{dt}$ is given by:

$$\frac{d{\mathcal{U}}_1\left(t\right)}{dt}=\rho \xi TV-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\xi T(t-\psi )V(t-\psi )d\psi -(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\xi T(t-\psi )V(t-\psi )d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_1{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +(1-\eta )\xi T_1{V}_1{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +\alpha e^{-n_3{\tau }_3}\left(L-L(t-{\tau }_3)+L_{1}ln\left(\frac{L(t-{\tau }_3)}{L}\right)\right)+d_I\left(I-I(t-{\tau }_4)+I_{1}ln\left(\frac{I(t-{\tau }_4)}{I}\right)\right).$$

Hence, $\frac{d{\mathcal{V}}_1\left(t\right)}{dt}$ is given by:

$$\frac{d{\mathcal{V}}_1}{dt}=\rho \left(1-\frac{T_1}{T}\right)\dot{T}+\frac{\alpha e^{-n_3{\tau }_3}}{\alpha +d_L}\left(1-\frac{L_1}{L}\right)\dot{L}+\left(1-\frac{I_1}{I}\right)\dot{I}+\frac{d_I{e}^{n_4{\tau }_4}}{k}\left(1-\frac{V_1}{V}\right)\dot{V}+\frac{d_{I}u{e}^{n_4{\tau }_4}}{kq}\dot{A}+\frac{\omega }{\sigma }\dot{C}+\frac{d{\mathcal{U}}_1\left(t\right)}{dt}.$$

By using the derivatives in Eqs. (5)–(10), we get

$$\frac{d{\mathcal{V}}_1}{dt}=\rho \left(1-\frac{T_1}{T}\right)\left[\lambda -d_{T}T+rT\left(1-\frac{T}{T_{max}}\right)-\xi TV\right]+\frac{\alpha e^{-n_3{\tau }_3}}{\alpha +d_L}\left(1-\frac{L_1}{L}\right)\left[\eta {\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\xi T(t-\psi )V(t-\psi )d\psi -(\alpha +d_L)L\right]+\left(1-\frac{I_1}{I}\right)\left[(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\xi T(t-\psi )V(t-\psi )d\psi +\alpha e^{-n_3{\tau }_3}L(t-{\tau }_3)-d_{I}I-\omega CI\right]+\frac{d_I{e}^{n_4{\tau }_4}}{k}\left(1-\frac{V_1}{V}\right)\left[k{e}^{-n_4{\tau }_4}I(t-{\tau }_4)-d_{V}V-uAV\right]+\frac{d_{I}u{e}^{n_4{\tau }_4}}{kq}\left[qAV-d_{A}A\right]+\frac{\omega }{\sigma }\left[\sigma CI-d_{C}C\right]+\rho \xi TV-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\xi T(t-\psi )V(t-\psi )d\psi -(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\xi T(t-\psi )V(t-\psi )d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_1{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +(1-\eta )\xi T_1{V}_1{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +\alpha e^{-n_3{\tau }_3}\left(L-L(t-{\tau }_3)+L_{1}ln\left(\frac{L(t-{\tau }_3)}{L}\right)\right)+d_I\left(I-I(t-{\tau }_4)+I_{1}ln\left(\frac{I(t-{\tau }_4)}{I}\right)\right).$$

Then,

$$\frac{d{\mathcal{V}}_1}{dt}=\rho \left(1-\frac{T_1}{T}\right)\left[\lambda -d_{T}T+rT\left(1-\frac{T}{T_{max}}\right)\right]+\rho \xi T_{1}V-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\frac{L_1\xi T(t-\psi )V(t-\psi )}{L}d\psi +\alpha e^{-n_3{\tau }_3}{L}_1-(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\frac{I_1\xi T(t-\psi )V(t-\psi )}{I}d\psi -\alpha e^{-n_3{\tau }_3}\frac{I_{1}L(t-{\tau }_3)}{I}+d_I{I}_1+\omega I_{1}C-\frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k}V-d_I\frac{V_{1}I(t-{\tau }_4)}{V}+\frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k}{V}_1+\frac{d_{I}u{e}^{n_4{\tau }_4}}{k}{V}_{1}A-\frac{d_I{d}_{A}u{e}^{n_4{\tau }_4}}{kq}A-\frac{d_C\omega }{\sigma }C+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_1{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +(1-\eta )\xi T_1{V}_1{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +\alpha e^{-n_3{\tau }_3}{L}_{1}ln\left(\frac{L(t-{\tau }_3)}{L}\right)+d_I{I}_{1}ln\left(\frac{I(t-{\tau }_4)}{I}\right).$$

By using the steady state conditions at $S{S}_1$:

$$\lambda =d_T{T}_1-r{T}_1\left(1-\frac{T_1}{T_{max}}\right)+\xi T_1{V}_1,$$

$$\alpha e^{-n_3{\tau }_3}{L}_1=\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}F\xi T_1{V}_1,$$

$$d_I{I}_1=\rho \xi T_1{V}_1=\frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k}{V}_1,$$

we get,

$$\lambda -d_{T}T+rT\left(1-\frac{T}{T_{max}}\right)=(T_1-T)\left(d_T-r+\frac{r{T}_1}{T_{max}}+\frac{rT}{T_{max}}\right)+\xi T_1{V}_1.$$

Further, we obtain

$$\frac{d{\mathcal{V}}_1}{dt}\le -\rho \left(d_T-r+\frac{r{T}_1}{T_{max}}\right)\frac{{(T-T_1)}^2}{T}+\rho \xi T_1{V}_1-\rho \xi T_1{V}_1\frac{T_1}{T}+\left(\rho \xi T_1-\frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k}\right)V-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_1{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\frac{L_{1}T(t-\psi )V(t-\psi )}{L{T}_1{V}_1}d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_{1}F-(1-\eta )\xi T_1{V}_1{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\frac{I_{1}T(t-\psi )V(t-\psi )}{I{T}_1{V}_1}d\psi -\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_{1}F\frac{I_{1}L(t-{\tau }_3)}{I{L}_1}+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_{1}F+(1-\eta )\xi T_1{V}_{1}G-\rho \xi T_1{V}_1\frac{V_{1}I(t-{\tau }_4)}{V{I}_1}+\rho \xi T_1{V}_1+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_1{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +(1-\eta )\xi T_1{V}_1{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_{1}Fln\left(\frac{L(t-{\tau }_3)}{L}\right)+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_{1}Fln\left(\frac{I(t-{\tau }_4)}{I}\right)+(1-\eta )\xi T_1{V}_{1}Gln\left(\frac{I(t-{\tau }_4)}{I}\right)+\frac{d_{I}u{e}^{n_4{\tau }_4}}{k}\left(V_1-\frac{d_A}{q}\right)A+\omega \left(I_1-\frac{d_C}{\sigma }\right)C.$$

We have $\rho \xi T_1-\frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k}=0$.  Now, using the following equalities:

$$ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)+ln\left(\frac{L(t-{\tau }_3)}{L}\right)=ln\left(\frac{T_1}{T}\right)+ln\left(\frac{L(t-{\tau }_3)V_1}{L_{1}V}\right)+ln\left(\frac{L_{1}T(t-\psi )V(t-\psi )}{L{T}_1{V}_1}\right),$$

$$ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)+ln\left(\frac{I(t-{\tau }_4)}{I}\right)=ln\left(\frac{T_1}{T}\right)+ln\left(\frac{I(t-{\tau }_4)V_1}{I_{1}V}\right)+ln\left(\frac{I_{1}T(t-\psi )V(t-\psi )}{I{T}_1{V}_1}\right),$$

we get

$$\frac{d{\mathcal{V}}_1}{dt}\le -\rho \left(d_T-r+\frac{r{T}_1}{T_{max}}\right)\frac{{(T-T_1)}^2}{T}+\rho \xi T_1{V}_1-\rho \xi T_1{V}_1\frac{T_1}{T}-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_1{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\frac{L_{1}T(t-\psi )V(t-\psi )}{L{T}_1{V}_1}d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_{1}F-(1-\eta )\xi T_1{V}_1{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\frac{I_{1}T(t-\psi )V(t-\psi )}{I{T}_1{V}_1}d\psi -\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_{1}F\frac{I_{1}L(t-{\tau }_3)}{I{L}_1}+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_{1}F+(1-\eta )\xi T_1{V}_{1}G-\rho \xi T_1{V}_1\frac{V_{1}I(t-{\tau }_4)}{V{I}_1}+\rho \xi T_1{V}_1+\rho \xi T_1{V}_{1}ln\left(\frac{T_1}{T}\right)+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_1{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{L_{1}T(t-\psi )V(t-\psi )}{L{T}_1{V}_1}\right)d\psi +(1-\eta )\xi T_1{V}_1{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{I_{1}T(t-\psi )V(t-\psi )}{I{T}_1{V}_1}\right)d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_{1}Fln\left(\frac{L(t-{\tau }_3)V_1}{L_{1}V}\right)+(1-\eta )\xi T_1{V}_{1}Gln\left(\frac{I(t-{\tau }_4)V_1}{I_{1}V}\right)+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_{1}Fln\left(\frac{I(t-{\tau }_4)}{I}\right)+\frac{d_{I}u{e}^{n_4{\tau }_4}}{k}\left(V_1-\frac{d_A}{q}\right)A+\omega \left(I_1-\frac{d_C}{\sigma }\right)C.$$ (24)

By using the equality:

$$ln\left(\frac{L(t-{\tau }_3)V_1}{L_{1}V}\right)+ln\left(\frac{I(t-{\tau }_4)}{I}\right)=ln\left(\frac{I_{1}L(t-{\tau }_3)}{I{L}_1}\right)+ln\left(\frac{V_{1}I(t-{\tau }_4)}{V{I}_1}\right),$$

and rearranging the $R.H.S$. of (24), we get

$$\frac{d{\mathcal{V}}_1}{dt}\le -\rho \left(d_T-r+\frac{r{T}_1}{T_{max}}\right)\frac{{(T-T_1)}^2}{T}-\rho \xi T_1{V}_1\left[\frac{T_1}{T}-1-ln\left(\frac{T_1}{T}\right)\right]-\rho \xi T_1{V}_1\left[\frac{V_{1}I(t-{\tau }_4)}{V{I}_1}-1-ln\left(\frac{V_{1}I(t-{\tau }_4)}{V{I}_1}\right)\right]-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_1{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\left[\frac{L_{1}T(t-\psi )V(t-\psi )}{L{T}_1{V}_1}-1-ln\left(\frac{L_{1}T(t-\psi )V(t-\psi )}{L{T}_1{V}_1}\right)\right]d\psi -\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_{1}F\left[\frac{I_{1}L(t-{\tau }_3)}{I{L}_1}-1-ln\left(\frac{I_{1}L(t-{\tau }_3)}{I{L}_1}\right)\right]-(1-\eta )\xi T_1{V}_1{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\left[\frac{I_{1}T(t-\psi )V(t-\psi )}{I{T}_1{V}_1}-1-ln\left(\frac{I_{1}T(t-\psi )V(t-\psi )}{I{T}_1{V}_1}\right)\right]d\psi +\frac{d_{I}u{e}^{n_4{\tau }_4}}{k}\left(V_1-\frac{d_A}{q}\right)A+\omega \left(I_1-\frac{d_C}{\sigma }\right)C=-\rho \left(d_T-r+\frac{r{T}_1}{T_{max}}\right)\frac{{(T-T_1)}^2}{T}-\xi T_1{V}_1\mathcal{H}\left(\frac{T_1}{T}\right)-\rho \xi T_1{V}_1\mathcal{H}\left(\frac{V_{1}I(t-{\tau }_4)}{V{I}_1}\right)-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_1{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\mathcal{H}\left(\frac{L_{1}T(t-\psi )V(t-\psi )}{L{T}_1{V}_1}\right)d\psi -\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_1{V}_{1}F\mathcal{H}\left(\frac{I_{1}L(t-{\tau }_3)}{I{L}_1}\right)-(1-\eta )\xi T_1{V}_1{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\mathcal{H}\left(\frac{I_{1}T(t-\psi )V(t-\psi )}{I{T}_1{V}_1}\right)d\psi +\frac{d_{I}u{e}^{n_4{\tau }_4}}{k}\left(V_1-\frac{d_A}{q}\right)A+\omega \left(I_1-\frac{d_C}{\sigma }\right)C.$$

Since $d_T-r+\frac{r{T}_1}{T_{max}}>0$, then we get

$$d_T-r+\frac{r{T}_1}{T_{max}}=d_T-r+\frac{r{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi T_{max}\rho }>0⟹d_T-r+\frac{d_A\xi }{q}+\frac{2r{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi T_{max}\rho }>0⟹\frac{2r{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi T_{max}\rho }-\left(r-d_T-\frac{d_A\xi }{q}\right)>0.$$

We obtain

$${\mathcal{R}}_1^A\le 1⟺\frac{T_{max}}{2r}\left[r-d_T-\frac{d_A\xi }{q}+\sqrt{{\left(r-d_T-\frac{d_A\xi }{q}\right)}^2+\frac{4r\lambda }{T_{max}}}\right]\le \frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k\xi \rho }⟺\sqrt{{\left(r-d_T-\frac{d_A\xi }{q}\right)}^2+\frac{4r\lambda }{T_{max}}}\le \frac{2r{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi T_{max}\rho }-\left(r-d_T-\frac{d_A\xi }{q}\right)⟺\frac{4r\lambda }{T_{max}}\le \frac{4r^2{d}_I^2{d}_V^2{e}^{2n_4{\tau }_4}}{k^2{\xi }^2{T}_{max}^2{\rho }^2}-\frac{4r{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi T_{max}\rho }\left(r-d_T-\frac{d_A\xi }{q}\right)⟺r\lambda \le \frac{r^2{d}_I^2{d}_V^2{e}^{2n_4{\tau }_4}}{k^2{\xi }^2{T}_{max}{\rho }^2}-\frac{r^2{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi \rho }+\frac{r{d}_T{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi \rho }+\frac{r{d}_I{d}_V{d}_A{e}^{n_4{\tau }_4}}{kq\rho }⟺\frac{\lambda k{e}^{-n_4{\tau }_4}\rho }{d_I{d}_V}+\frac{r}{\xi }-\left(\frac{d_T}{\xi }+\frac{r{d}_I{d}_V{e}^{n_4{\tau }_4}}{k{\xi }^2{T}_{max}\rho }\right)\le \frac{d_A}{q}⟺V_1\le \frac{d_A}{q}.$$

Also, since $d_T-r+\frac{r{T}_1}{T_{max}}>0$, then we get

$$d_T-r+\frac{r{T}_1}{T_{max}}=d_T-r+\frac{r{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi T_{max}\rho }>0⟹d_T-r+\frac{d_{C}k\xi e^{-n_4{\tau }_4}}{d_V\sigma }+\frac{2r{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi T_{max}\rho }>0⟹\frac{2r{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi T_{max}\rho }-\left(r-d_T-\frac{d_{C}k\xi e^{-n_4{\tau }_4}}{d_V\sigma }\right)>0.$$

We obtain

$${\mathcal{R}}_1^C\le 1⟺\frac{T_{max}}{2r}\left[r-d_T-\frac{d_{C}k\xi e^{-n_4{\tau }_4}}{d_V\sigma }+\sqrt{{\left(r-d_T-\frac{d_{C}k\xi e^{-n_4{\tau }_4}}{d_V\sigma }\right)}^2+\frac{4r\lambda }{T_{max}}}\right]\le \frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k\xi \rho }⟺\sqrt{{\left(r-d_T-\frac{d_{C}k\xi e^{-n_4{\tau }_4}}{d_V\sigma }\right)}^2+\frac{4r\lambda }{T_{max}}}\le \frac{2r{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi T_{max}\rho }-\left(r-d_T-\frac{d_{C}k\xi e^{-n_4{\tau }_4}}{d_V\sigma }\right)⟺\frac{4r\lambda }{T_{max}}\le \frac{4r^2{d}_I^2{d}_V^2{e}^{2n_4{\tau }_4}}{k^2{\xi }^2{T}_{max}^2{\rho }^2}-\frac{4r{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi T_{max}\rho }\left(r-d_T-\frac{d_{C}k\xi e^{-n_4{\tau }_4}}{d_V\sigma }\right)⟺r\lambda \le \frac{r^2{d}_I^2{d}_V^2{e}^{2n_4{\tau }_4}}{k^2{\xi }^2{T}_{max}{\rho }^2}-\frac{r^2{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi \rho }+\frac{r{d}_T{d}_I{d}_V{e}^{n_4{\tau }_4}}{k\xi \rho }+\frac{r{d}_I{d}_C}{\sigma \rho }⟺\frac{d_V{e}^{n_4{\tau }_4}}{k}\left[\frac{\lambda kr{e}^{-n_4{\tau }_4}}{d_V}+\frac{r^2{d}_I}{\xi \rho }-\left(\frac{r{d}_T{d}_I}{\xi \rho }+\frac{r^2{d}_I^2{d}_V{e}^{n_4{\tau }_4}}{k{\xi }^2{T}_{max}\rho }\right)\right]\le \frac{r{d}_I{d}_C}{\sigma \rho }⟺\frac{d_V{e}^{n_4{\tau }_4}}{k}\left[\frac{\lambda k{e}^{-n_4{\tau }_4}\rho }{d_I{d}_V}+\frac{r}{\xi }-\left(\frac{d_T}{\xi }+\frac{r{d}_I{d}_V{e}^{n_4{\tau }_4}}{k{\xi }^2{T}_{max}\rho }\right)\right]\le \frac{d_C}{\sigma }⟺I_1\le \frac{d_C}{\sigma }.$$

Thus, $\frac{d{\mathcal{V}}_1}{dt}\le 0$ when ${\mathcal{R}}_1^A\le 1$, ${\mathcal{R}}_1^C\le 1$ and $d_T-r+\frac{r{T}_1}{T_{max}}>0$. Also, $\frac{d{\mathcal{V}}_1}{dt}=0$ when $T=T_1$, $L=L_1$, $I=I_1$, $V=V_1$, $A=0$ and $C=0$. Thus, the largest invariant subset of $M_1=\left\{(T,L,I,V,A,C)|\frac{d{\mathcal{V}}_1}{dt}=0\right\}$ is $M_1^{\prime }=\left\{S{S}_1\right\}$. Using LIP we get that $S{S}_1$ is GAS when ${\mathcal{R}}_1^A\le 1<{\mathcal{R}}_0$, ${\mathcal{R}}_1^C\le 1$ and $d_T-r+\frac{r{T}_1}{T_{max}}>0$.

Theorem 4

The steady state $S{S}_2$ is GAS when ${\mathcal{R}}_1^A>1$ , ${\mathcal{R}}_2^C\le 1$ and $d_T-r+\frac{r{T}_2}{T_{max}}>0$ .

Proof

Define a function ${\mathcal{V}}_2(T,L,I,V,A,C)$ as:

$${\mathcal{V}}_2=\rho T_2\mathcal{H}\left(\frac{T}{T_2}\right)+\frac{\alpha e^{-n_3{\tau }_3}}{\alpha +d_L}{L}_2\mathcal{H}\left(\frac{L}{L_2}\right)+I_2\mathcal{H}\left(\frac{I}{I_2}\right)+\frac{d_I{e}^{n_4{\tau }_4}}{k}{V}_2\mathcal{H}\left(\frac{V}{V_2}\right)+\frac{d_{I}u{e}^{n_4{\tau }_4}}{kq}{A}_2\mathcal{H}\left(\frac{A}{A_2}\right)+\frac{\omega }{\sigma }C+{\mathcal{U}}_2\left(t\right),$$

where

$${\mathcal{U}}_2\left(t\right)=\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_2{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }{\int }_{t-\psi }^t\mathcal{H}\left(\frac{T\left(\varphi \right)V\left(\varphi \right)}{T_2{V}_2}\right)d\varphi d\psi +(1-\eta )\xi T_2{V}_2{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }{\int }_{t-\psi }^t\mathcal{H}\left(\frac{T\left(\varphi \right)V\left(\varphi \right)}{T_2{V}_2}\right)d\varphi d\psi +\alpha e^{-n_3{\tau }_3}{L}_2{\int }_{t-{\tau }_3}^t\mathcal{H}\left(\frac{L\left(\varphi \right)}{L_2}\right)d\varphi +d_I{I}_2{\int }_{t-{\tau }_4}^t\mathcal{H}\left(\frac{I\left(\varphi \right)}{I_2}\right)d\varphi .$$

Clearly, ${\mathcal{V}}_2(T,L,I,V,A,C)>0$ for all $T,L,I,V,A,C>0$, and ${\mathcal{V}}_2(T_2,L_2,I_2,V_2,A_2,0)=0$. We have

$$\frac{d{\mathcal{U}}_2\left(t\right)}{dt}=\rho \xi TV-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\xi T(t-\psi )V(t-\psi )d\psi -(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\xi T(t-\psi )V(t-\psi )d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_2{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +(1-\eta )\xi T_2{V}_2{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +\alpha e^{-n_3{\tau }_3}\left(L-L(t-{\tau }_3)+L_{2}ln\left(\frac{L(t-{\tau }_3)}{L}\right)\right)+d_I\left(I-I(t-{\tau }_4)+I_{2}ln\left(\frac{I(t-{\tau }_4)}{I}\right)\right).$$

Hence,

$$\frac{d{\mathcal{V}}_2}{dt}=\rho \left(1-\frac{T_2}{T}\right)\dot{T}+\frac{\alpha e^{-n_3{\tau }_3}}{\alpha +d_L}\left(1-\frac{L_2}{L}\right)\dot{L}+\left(1-\frac{I_2}{I}\right)\dot{I}+\frac{d_I{e}^{n_4{\tau }_4}}{k}\left(1-\frac{V_2}{V}\right)\dot{V}+\frac{d_{I}u{e}^{n_4{\tau }_4}}{kq}\left(1-\frac{A_2}{A}\right)\dot{A}+\frac{\omega }{\sigma }\dot{C}+\frac{d{\mathcal{U}}_2\left(t\right)}{dt}=\rho \left(1-\frac{T_2}{T}\right)\left[\lambda -d_{T}T+rT\left(1-\frac{T}{T_{max}}\right)-\xi TV\right]+\frac{\alpha e^{-n_3{\tau }_3}}{\alpha +d_L}\left(1-\frac{L_2}{L}\right)\left[\eta {\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\xi T(t-\psi )V(t-\psi )d\psi -(\alpha +d_L)L\right]+\left(1-\frac{I_2}{I}\right)\left[(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\xi T(t-\psi )V(t-\psi )d\psi +\alpha e^{-n_3{\tau }_3}L(t-{\tau }_3)-d_{I}I-\omega CI\right]+\frac{d_I{e}^{n_4{\tau }_4}}{k}\left(1-\frac{V_2}{V}\right)\left[k{e}^{-n_4{\tau }_4}I(t-{\tau }_4)-d_{V}V-uAV\right]+\frac{d_{I}u{e}^{n_4{\tau }_4}}{kq}\left(1-\frac{A_2}{A}\right)\left[qAV-d_{A}A\right]+\frac{\omega }{\sigma }\left[\sigma CI-d_{C}C\right]+\rho \xi TV-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\xi T(t-\psi )V(t-\psi )d\psi -(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\xi T(t-\psi )V(t-\psi )d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_2{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +(1-\eta )\xi T_2{V}_2{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +\alpha e^{-n_3{\tau }_3}\left(L-L(t-{\tau }_3)+L_{2}ln\left(\frac{L(t-{\tau }_3)}{L}\right)\right)+d_I\left(I-I(t-{\tau }_4)+I_{2}ln\left(\frac{I(t-{\tau }_4)}{I}\right)\right).$$ (25)

Simplifying Eq. (25), we get

$$\frac{d{\mathcal{V}}_2}{dt}=\rho \left(1-\frac{T_2}{T}\right)\left[\lambda -d_{T}T+rT\left(1-\frac{T}{T_{max}}\right)\right]+\rho \xi T_{2}V-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\frac{L_2\xi T(t-\psi )V(t-\psi )}{L}d\psi +\alpha e^{-n_3{\tau }_3}{L}_2-(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\frac{I_2\xi T(t-\psi )V(t-\psi )}{I}d\psi -\alpha e^{-n_3{\tau }_3}\frac{I_{2}L(t-{\tau }_3)}{I}+d_I{I}_2+\omega I_{2}C-\frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k}V-d_I\frac{V_{2}I(t-{\tau }_4)}{V}+\frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k}{V}_2+\frac{d_{I}u{e}^{n_4{\tau }_4}}{k}{V}_{2}A-\frac{d_I{d}_{A}u{e}^{n_4{\tau }_4}}{kq}A-\frac{d_{I}u{e}^{n_4{\tau }_4}}{k}{A}_{2}V+\frac{d_I{d}_{A}u{e}^{n_4{\tau }_4}}{kq}{A}_2-\frac{d_C\omega }{\sigma }C+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_2{\int }_0^{{\tau }_2}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +(1-\eta )\xi T_2{V}_2{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +\alpha e^{-n_3{\tau }_3}{L}_{2}ln\left(\frac{L(t-{\tau }_3)}{L}\right)+d_I{I}_{2}ln\left(\frac{I(t-{\tau }_4)}{I}\right).$$ (26)

At the steady state $S{S}_2$ we have:

$$\lambda =d_T{T}_2-r{T}_2\left(1-\frac{T_2}{T_{max}}\right)+\xi T_2{V}_2,$$

$$\alpha e^{-n_3{\tau }_3}{L}_2=\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}F\xi T_2{V}_2,$$

$$d_I{I}_2=\rho \xi T_2{V}_2=\frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k}{V}_2+\frac{d_{I}u{e}^{n_4{\tau }_4}}{k}{A}_2{V}_2,$$

$$V_2=\frac{d_A}{q},$$

then we get,

$$\lambda -d_{T}T+rT\left(1-\frac{T}{T_{max}}\right)=(T_2-T)\left(d_T-r+\frac{r{T}_2}{T_{max}}+\frac{rT}{T_{max}}\right)+\xi T_2{V}_2.$$

By using the above conditions, the derivative in (26) is transformed into

$$\frac{d{\mathcal{V}}_2}{dt}\le -\rho \left(d_T-r+\frac{r{T}_2}{T_{max}}\right)\frac{{(T-T_2)}^2}{T}+\rho \xi T_2{V}_2-\rho \xi T_2{V}_2\frac{T_2}{T}+\left(\rho \xi T_2-\frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k}-\frac{d_{I}u{e}^{n_4{\tau }_4}}{k}{A}_2\right)V-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_2{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\frac{L_{2}T(t-\psi )V(t-\psi )}{L{T}_2{V}_2}d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_{2}F-(1-\eta )\xi T_2{V}_2{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\frac{I_{2}T(t-\psi )V(t-\psi )}{I{T}_2{V}_2}d\psi -\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_{2}F\frac{I_{2}L(t-{\tau }_3)}{I{L}_2}+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_{2}F+(1-\eta )\xi T_2{V}_{2}G-\rho \xi T_2{V}_2\frac{V_{2}I(t-{\tau }_4)}{V{I}_2}+\rho \xi T_2{V}_2+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_2{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +(1-\eta )\xi T_2{V}_2{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_{2}Fln\left(\frac{L(t-{\tau }_3)}{L}\right)+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_{2}Fln\left(\frac{I(t-{\tau }_4)}{I}\right)+(1-\eta )\xi T_2{V}_{2}Gln\left(\frac{I(t-{\tau }_4)}{I}\right)+\omega \left(I_2-\frac{d_C}{\sigma }\right)C.$$

The conditions of $S{S}_2$ imply that

$$\rho \xi T_2-\frac{d_I{d}_V{e}^{n_4{\tau }_4}}{k}-\frac{d_{I}u{e}^{n_4{\tau }_4}}{k}{A}_2=0.$$

Now, using the following equalities:

$$ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)+ln\left(\frac{L(t-{\tau }_3)}{L}\right)=ln\left(\frac{T_2}{T}\right)+ln\left(\frac{L(t-{\tau }_3)V_2}{L_{2}V}\right)+ln\left(\frac{L_{2}T(t-\psi )V(t-\psi )}{L{T}_2{V}_2}\right),$$

$$ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)+ln\left(\frac{I(t-{\tau }_4)}{I}\right)=ln\left(\frac{T_2}{T}\right)+ln\left(\frac{I(t-{\tau }_4)V_2}{I_{2}V}\right)+ln\left(\frac{I_{2}T(t-\psi )V(t-\psi )}{I{T}_2{V}_2}\right),$$

we get

$$\frac{d{\mathcal{V}}_2}{dt}\le -\rho \left(d_T-r+\frac{r{T}_2}{T_{max}}\right)\frac{{(T-T_2)}^2}{T}+\rho \xi T_2{V}_2-\rho \xi T_2{V}_2\frac{T_2}{T}-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_2{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\frac{L_{2}T(t-\psi )V(t-\psi )}{L{T}_2{V}_2}d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_{2}F-(1-\eta )\xi T_2{V}_2{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\frac{I_{2}T(t-\psi )V(t-\psi )}{I{T}_2{V}_2}d\psi -\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_{2}F\frac{I_{2}L(t-{\tau }_3)}{I{L}_2}+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_{2}F+(1-\eta )\xi T_2{V}_{2}G-\rho \xi T_2{V}_2\frac{V_{2}I(t-{\tau }_4)}{V{I}_2}+\rho \xi T_2{V}_2+\rho \xi T_2{V}_{2}ln\left(\frac{T_2}{T}\right)+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_2{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{L_{2}T(t-\psi )V(t-\psi )}{L{T}_2{V}_2}\right)d\psi +(1-\eta )\xi T_2{V}_2{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{I_{2}T(t-\psi )V(t-\psi )}{I{T}_2{V}_2}\right)d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_{2}Fln\left(\frac{L(t-{\tau }_3)V_2}{L_{2}V}\right)+(1-\eta )\xi T_2{V}_{2}Gln\left(\frac{I(t-{\tau }_4)V_2}{I_{2}V}\right)+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_{2}Fln\left(\frac{I(t-{\tau }_4)}{I}\right)+\frac{d_C\omega }{\sigma }\left(\frac{d_A\sigma \xi T_4}{d_I{d}_{C}q}\rho -1\right)C.$$

By using the equality:

$$ln\left(\frac{L(t-{\tau }_3)V_2}{L_{2}V}\right)+ln\left(\frac{I(t-{\tau }_4)}{I}\right)=ln\left(\frac{I_{2}L(t-{\tau }_3)}{I{L}_2}\right)+ln\left(\frac{V_{2}I(t-{\tau }_4)}{V{I}_2}\right),$$

and rearranging the $R.H.S$. of $\frac{d{\mathcal{V}}_2}{dt}$, we get

$$\frac{d{\mathcal{V}}_2}{dt}\le -\rho \left(d_T-r+\frac{r{T}_2}{T_{max}}\right)\frac{{(T-T_2)}^2}{T}-\rho \xi T_2{V}_2\left[\frac{T_2}{T}-1-ln\left(\frac{T_2}{T}\right)\right]-\rho \xi T_2{V}_2\left[\frac{V_{2}I(t-{\tau }_4)}{V{I}_2}-1-ln\left(\frac{V_{2}I(t-{\tau }_4)}{V{I}_2}\right)\right]-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_2{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\left[\frac{L_{2}T(t-\psi )V(t-\psi )}{L{T}_2{V}_2}-1-ln\left(\frac{L_{2}T(t-\psi )V(t-\psi )}{L{T}_2{V}_2}\right)\right]d\psi -\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_{2}F\left[\frac{I_{2}L(t-{\tau }_3)}{I{L}_2}-1-ln\left(\frac{I_{2}L(t-{\tau }_3)}{I{L}_2}\right)\right]-(1-\eta )\xi T_2{V}_2{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\left[\frac{I_{2}T(t-\psi )V(t-\psi )}{I{T}_2{V}_2}-1-ln\left(\frac{I_{2}T(t-\psi )V(t-\psi )}{I{T}_2{V}_2}\right)\right]d\psi .+\frac{d_C\omega }{\sigma }\left({\mathcal{R}}_2^C-1\right)C.=-\rho \left(d_T-r+\frac{r{T}_2}{T_{max}}\right)\frac{{(T-T_2)}^2}{T}-\rho \xi T_2{V}_2\mathcal{H}\left(\frac{T_2}{T}\right)-\rho \xi T_2{V}_2\mathcal{H}\left(\frac{V_{2}I(t-{\tau }_4)}{V{I}_2}\right)-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_2{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\mathcal{H}\left(\frac{L_{2}T(t-\psi )V(t-\psi )}{L{T}_2{V}_2}\right)d\psi -\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_2{V}_{2}F\mathcal{H}\left(\frac{I_{2}L(t-{\tau }_3)}{I{L}_2}\right)-(1-\eta )\xi T_2{V}_2{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\mathcal{H}\left(\frac{I_{2}T(t-\psi )V(t-\psi )}{I{T}_2{V}_2}\right)d\psi +\frac{d_C\omega }{\sigma }\left({\mathcal{R}}_2^C-1\right)C.$$

We see that $\frac{d{\mathcal{V}}_2}{dt}\le 0$ when ${\mathcal{R}}_1^A>1$, ${\mathcal{R}}_2^C\le 1$ and $d_T-r+\frac{r{T}_2}{T_{max}}>0$. Also, $\frac{d{\mathcal{V}}_2}{dt}=0$ when, $T=T_2$, $L=L_2$, $I=I_2$, $V=V_2$ and $C=0$. The solutions of system (5)–(10) tend to $M_2^{\prime }$ the largest invariant subset of $M_2=\left\{(T,L,I,V,A,C)\right|\frac{d{\mathcal{V}}_2}{dt}=0\}$. For each element in $M_2^{\prime }$ we have $V=V_2$ then $\dot{V}=0$ and from Eq. (8) we have $0=\dot{V}=k{e}^{-n_4{\tau }_4}{I}_2-d_V{V}_2-uA{V}_2$, which gives $A\left(t\right)=A_2$, for all $t$. It follows that $M_2^{\prime }=\left\{S{S}_2\right\}$. By applying LIP we get that $S{S}_2$ is GAS when ${\mathcal{R}}_1^A>1$, ${\mathcal{R}}_2^C\le 1$ and $d_T-r+\frac{r{T}_2}{T_{max}}>0$.

Theorem 5

The steady state $S{S}_3$ is GAS when ${\mathcal{R}}_1^C>1$ , $\frac{{\mathcal{R}}_1^A}{{\mathcal{R}}_2^C}\le 1$ and $d_T-r+\frac{r{T}_3}{T_{max}}>0$ .

Proof

Define a ${\mathcal{V}}_3(T,L,I,V,A,C)$ as:

$${\mathcal{V}}_3=\rho T_3\mathcal{H}\left(\frac{T}{T_3}\right)+\frac{\alpha e^{-n_3{\tau }_3}}{\alpha +d_L}{L}_3\mathcal{H}\left(\frac{L}{L_3}\right)+I_3\mathcal{H}\left(\frac{I}{I_3}\right)+\frac{(d_I+\omega C_3)e^{n_4{\tau }_4}}{k}{V}_3\mathcal{H}\left(\frac{V}{V_3}\right)+\frac{(d_I+\omega C_3)u{e}^{n_4{\tau }_4}}{kq}A+\frac{\omega }{\sigma }{C}_3\mathcal{H}\left(\frac{C}{C_3}\right)+{\mathcal{U}}_3\left(t\right),$$

where

$${\mathcal{U}}_3\left(t\right)=\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_3{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }{\int }_{t-\psi }^t\mathcal{H}\left(\frac{T\left(\varphi \right)V\left(\varphi \right)}{T_3{V}_3}\right)d\varphi d\psi +(1-\eta )\xi T_3{V}_3{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }{\int }_{t-\psi }^t\mathcal{H}\left(\frac{T\left(\varphi \right)V\left(\varphi \right)}{T_3{V}_3}\right)d\varphi d\psi +\alpha e^{-n_3{\tau }_3}{L}_3{\int }_{t-{\tau }_3}^t\mathcal{H}\left(\frac{L\left(\varphi \right)}{L_3}\right)d\varphi +(d_I+\omega C_3)I_3{\int }_{t-{\tau }_4}^t\mathcal{H}\left(\frac{I\left(\varphi \right)}{I_3}\right)d\varphi .$$

We have ${\mathcal{V}}_3(T,L,I,V,A,C)>0$ for all $T,L,I,V,A,C>0$, and ${\mathcal{V}}_3(T_3,L_3,I_3,V_3,0,C_3)=0$. Then, we have

$$\frac{d{\mathcal{U}}_3\left(t\right)}{dt}=\rho \xi TV-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\xi T(t-\psi )V(t-\psi )d\psi -(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\xi T(t-\psi )V(t-\psi )d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_3{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +(1-\eta )\xi T_3{V}_3{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +\alpha e^{-n_3{\tau }_3}\left(L-L(t-{\tau }_3)+L_{3}ln\left(\frac{L(t-{\tau }_3)}{L}\right)\right)+(d_I+\omega C_3)\left(I-I(t-{\tau }_4)+I_{3}ln\left(\frac{I(t-{\tau }_4)}{I}\right)\right).$$

Hence,

$$\frac{d{\mathcal{V}}_3}{dt}=\rho \left(1-\frac{T_3}{T}\right)\dot{T}+\frac{\alpha e^{-n_3{\tau }_3}}{\alpha +d_L}\left(1-\frac{L_3}{L}\right)\dot{L}+\left(1-\frac{I_3}{I}\right)\dot{I}+\frac{(d_I+\omega C_3)e^{n_4{\tau }_4}}{k}\left(1-\frac{V_3}{V}\right)\dot{V}+\frac{(d_I+\omega C_3)u{e}^{n_4{\tau }_4}}{kq}\dot{A}+\frac{\omega }{\sigma }\left(1-\frac{C_3}{C}\right)\dot{C}+\frac{d{\mathcal{U}}_3\left(t\right)}{dt}=\rho \left(1-\frac{T_3}{T}\right)\left[\lambda -d_{T}T+rT\left(1-\frac{T}{T_{max}}\right)-\xi TV\right]+\frac{\alpha e^{-n_3{\tau }_3}}{\alpha +d_L}\left(1-\frac{L_3}{L}\right)\left[\eta {\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\xi T(t-\psi )V(t-\psi )d\psi -(\alpha +d_L)L\right]+\left(1-\frac{I_3}{I}\right)\left[(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\xi T(t-\psi )V(t-\psi )d\psi +\alpha e^{-n_3{\tau }_3}L(t-{\tau }_3)-d_{I}I-\omega CI\right]+\frac{(d_I+\omega C_3)e^{n_4{\tau }_4}}{k}\left(1-\frac{V_3}{V}\right)\left[k{e}^{-n_4{\tau }_4}I(t-{\tau }_4)-d_{V}V-uAV\right]+\frac{(d_I+\omega C_3)u{e}^{n_4{\tau }_4}}{kq}\left[qAV-d_{A}A\right]+\frac{\omega }{\sigma }\left(1-\frac{C_3}{C}\right)\left[\sigma CI-d_{C}C\right]+\rho \xi TV-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\xi T(t-\psi )V(t-\psi )d\psi -(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\xi T(t-\psi )V(t-\psi )d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_3{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +(1-\eta )\xi T_3{V}_3{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +\alpha e^{-n_3{\tau }_3}\left(L-L(t-{\tau }_3)+L_{3}ln\left(\frac{L(t-{\tau }_3)}{L}\right)\right)+(d_I+\omega C_3)\left(I-I(t-{\tau }_4)+I_{3}ln\left(\frac{I(t-{\tau }_4)}{I}\right)\right).$$ (27)

Simplifying Eq. (27) we get

$$\frac{d{\mathcal{V}}_3}{dt}=\rho \left(1-\frac{T_3}{T}\right)\left[\lambda -d_{T}T+rT\left(1-\frac{T}{T_{max}}\right)\right]+\rho \xi T_{3}V-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\frac{L_3\xi T(t-\psi )V(t-\psi )}{L}d\psi +\alpha e^{-n_3{\tau }_3}{L}_3-(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\frac{I_3\xi T(t-\psi )V(t-\psi )}{I}d\psi -\alpha e^{-n_3{\tau }_3}\frac{I_{3}L(t-{\tau }_3)}{I}+d_I{I}_3+\omega I_{3}C-\frac{(d_I+\omega C_3)d_V{e}^{n_4{\tau }_4}}{k}V-(d_I+\omega C_3)\frac{V_{3}I(t-{\tau }_4)}{V}+\frac{(d_I+\omega C_3)d_V{e}^{n_4{\tau }_4}}{k}{V}_3+\frac{(d_I+\omega C_3)u{e}^{n_4{\tau }_4}}{k}{V}_{3}A-\frac{(d_I+\omega C_3)d_{A}u{e}^{n_4{\tau }_4}}{kq}A-\frac{d_C\omega }{\sigma }C+\frac{d_C\omega }{\sigma }{C}_3+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_3{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +(1-\eta )\xi T_3{V}_3{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +\alpha e^{-n_3{\tau }_3}{L}_{3}ln\left(\frac{L(t-{\tau }_3)}{L}\right)+(d_I+\omega C_3)I_{3}ln\left(\frac{I(t-{\tau }_4)}{I}\right).$$ (28)

The components of $S{S}_3$ satisfy

$$\lambda =d_T{T}_3-r{T}_3\left(1-\frac{T_3}{T_{max}}\right)+\xi T_3{V}_3,$$

$$\alpha e^{-n_3{\tau }_3}{L}_3=\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}F\xi T_3{V}_3,(d_I+\omega C_3)I_3=\rho \xi T_3{V}_3=\frac{(d_I+\omega C_3)d_V{e}^{n_4{\tau }_4}}{k}{V}_3,$$

$$I_3=\frac{d_C}{\sigma }.$$

Then we get,

$$\lambda -d_{T}T+rT\left(1-\frac{T}{T_{max}}\right)=(T_3-T)\left(d_T-r+\frac{r{T}_3}{T_{max}}+\frac{rT}{T_{max}}\right)+\xi T_3{V}_3.$$

By using the above equalities, the derivative in Eq. (28) is transformed into

$$\frac{d{\mathcal{V}}_3}{dt}\le -\rho \left(d_T-r+\frac{r{T}_3}{T_{max}}\right)\frac{{(T-T_3)}^2}{T}+\rho \xi T_3{V}_3-\rho \xi T_3{V}_3\frac{T_3}{T}-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_3{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\frac{L_{3}T(t-\psi )V(t-\psi )}{L{T}_3{V}_3}d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_{3}F-(1-\eta )\xi T_3{V}_3{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\frac{I_{3}T(t-\psi )V(t-\psi )}{I{T}_3{V}_3}d\psi -\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_{3}F\frac{I_{3}L(t-{\tau }_3)}{I{L}_3}+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_{3}F+(1-\eta )\xi T_3{V}_{3}G-\rho \xi T_3{V}_3\frac{V_{3}I(t-{\tau }_4)}{V{I}_3}+\rho \xi T_3{V}_3+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_3{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +(1-\eta )\xi T_3{V}_3{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_{3}Fln\left(\frac{L(t-{\tau }_3)}{L}\right)+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_{3}Fln\left(\frac{I(t-{\tau }_4)}{I}\right)+(1-\eta )\xi T_3{V}_{3}Gln\left(\frac{I(t-{\tau }_4)}{I}\right)+\frac{(d_I+\omega C_3)u{e}^{n_4{\tau }_4}}{k}\left(V_3-\frac{d_A}{q}\right)A.$$

Now, using the following equalities:

$$ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)+ln\left(\frac{L(t-{\tau }_3)}{L}\right)=ln\left(\frac{T_3}{T}\right)+ln\left(\frac{L(t-{\tau }_3)V_3}{L_{3}V}\right)+ln\left(\frac{L_{3}T(t-\psi )V(t-\psi )}{L{T}_3{V}_3}\right),$$

$$ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)+ln\left(\frac{I(t-{\tau }_4)}{I}\right)=ln\left(\frac{T_3}{T}\right)+ln\left(\frac{I(t-{\tau }_4)V_3}{I_{3}V}\right)+ln\left(\frac{I_{3}T(t-\psi )V(t-\psi )}{I{T}_3{V}_3}\right),$$

we get

$$\frac{d{\mathcal{V}}_3}{dt}\le -\rho \left(d_T-r+\frac{r{T}_3}{T_{max}}\right)\frac{{(T-T_3)}^2}{T}+\rho \xi T_3{V}_3-\rho \xi T_3{V}_3\frac{T_3}{T}-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_3{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\frac{L_{3}T(t-\psi )V(t-\psi )}{L{T}_3{V}_3}d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_{3}F-(1-\eta )\xi T_3{V}_3{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\frac{I_{3}T(t-\psi )V(t-\psi )}{I{T}_3{V}_3}d\psi -\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_{3}F\frac{I_{3}L(t-{\tau }_3)}{I{L}_3}+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_{3}F+(1-\eta )\xi T_3{V}_{3}G-\rho \xi T_3{V}_3\frac{V_{3}I(t-{\tau }_4)}{V{I}_3}+\rho \xi T_3{V}_3+\rho \xi T_3{V}_{3}ln\left(\frac{T_3}{T}\right)+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_3{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{L_{3}T(t-\psi )V(t-\psi )}{L{T}_3{V}_3}\right)d\psi +(1-\eta )\xi T_3{V}_3{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{I_{3}T(t-\psi )V(t-\psi )}{I{T}_3{V}_3}\right)d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_{3}Fln\left(\frac{L(t-{\tau }_3)V_3}{L_{3}V}\right)+(1-\eta )\xi T_3{V}_{3}Gln\left(\frac{I(t-{\tau }_4)V_3}{I_{3}V}\right)+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_{3}Fln\left(\frac{I(t-{\tau }_4)}{I}\right)+\frac{(d_I+\omega C_3)d_{A}u{e}^{n_4{\tau }_4}}{kq}\left(\frac{d_{C}kq{e}^{-n_4{\tau }_4}}{d_V{d}_A\sigma }-1\right)A.$$

By using the equality:

$$ln\left(\frac{L(t-{\tau }_3)V_3}{L_{3}V}\right)+ln\left(\frac{I(t-{\tau }_4)}{I}\right)=ln\left(\frac{I_{3}L(t-{\tau }_3)}{I{L}_3}\right)+ln\left(\frac{V_{3}I(t-{\tau }_4)}{V{I}_3}\right),$$

and rearranging the $R.H.S$. of $\frac{d{\mathcal{V}}_3}{dt}$, we get

$$\frac{d{\mathcal{V}}_3}{dt}\le -\rho \left(d_T-r+\frac{r{T}_3}{T_{max}}\right)\frac{{(T-T_3)}^2}{T}-\rho \xi T_3{V}_3\left[\frac{T_3}{T}-1-ln\left(\frac{T_3}{T}\right)\right]-\rho \xi T_3{V}_3\left[\frac{V_{3}I(t-{\tau }_4)}{V{I}_3}-1-ln\left(\frac{V_{3}I(t-{\tau }_4)}{V{I}_3}\right)\right]-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_3{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\left[\frac{L_{3}T(t-\psi )V(t-\psi )}{L{T}_3{V}_3}-1-ln\left(\frac{L_{3}T(t-\psi )V(t-\psi )}{L{T}_3{V}_3}\right)\right]d\psi -\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_{3}F\left[\frac{I_{3}L(t-{\tau }_3)}{I{L}_3}-1-ln\left(\frac{I_{3}L(t-{\tau }_3)}{I{L}_3}\right)\right]-(1-\eta )\xi T_3{V}_3{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\left[\frac{I_{3}T(t-\psi )V(t-\psi )}{I{T}_3{V}_3}-1-ln\left(\frac{I_{3}T(t-\psi )V(t-\psi )}{I{T}_3{V}_3}\right)\right]d\psi +\frac{(d_I+\omega C_3)d_{A}u{e}^{n_4{\tau }_4}}{kq}\left(\frac{{\mathcal{R}}_1^A}{{\mathcal{R}}_2^C}-1\right)A.=-\rho \left(d_T-r+\frac{r{T}_3}{T_{max}}\right)\frac{{(T-T_3)}^2}{T}-\rho \xi T_3{V}_3\mathcal{H}\left(\frac{T_3}{T}\right)-\rho \xi T_3{V}_3\mathcal{H}\left(\frac{V_{3}I(t-{\tau }_4)}{V{I}_3}\right)-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_3{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\mathcal{H}\left(\frac{L_{3}T(t-\psi )V(t-\psi )}{L{T}_3{V}_3}\right)d\psi -\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_3{V}_{3}F\mathcal{H}\left(\frac{I_{3}L(t-{\tau }_3)}{I{L}_3}\right)-(1-\eta )\xi T_3{V}_3{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\mathcal{H}\left(\frac{I_{3}T(t-\psi )V(t-\psi )}{I{T}_3{V}_3}\right)d\psi +\frac{(d_I+\omega C_3)d_{A}u{e}^{n_4{\tau }_4}}{kq}\left(\frac{{\mathcal{R}}_1^A}{{\mathcal{R}}_2^C}-1\right)A.$$

We see that $\frac{d{\mathcal{V}}_3}{dt}\le 0$ when ${\mathcal{R}}_1^C>1$, $\frac{{\mathcal{R}}_1^A}{{\mathcal{R}}_2^C}\le 1$ and $d_T-r+\frac{r{T}_3}{T_{max}}>0$. Also, $\frac{d{\mathcal{V}}_3}{dt}=0$ when, $T=T_3$, $L=L_3$, $I=I_3$, $V=V_3$ and $A=0$. The solutions of system (5)–(10) tend to $M_3^{\prime }$ the largest invariant subset of $M_3=\left\{(T,L,I,V,A,C)\right|\frac{d{\mathcal{V}}_3}{dt}=0\}$. For each element in $M_3^{\prime }$ we have $I=I_3$ then $\dot{I}=0$ and Eq. (7) becomes $0=\dot{I}=(1-\eta )G\xi T_3{V}_3+\alpha e^{-n_3{\tau }_3}{L}_3-d_I{I}_3-\omega I_{3}C$, which gives $C\left(t\right)=C_3$, for all $t$. It follows that $M_3^{\prime }=\left\{S{S}_3\right\}$. LIP implies that $S{S}_3$ is GAS when ${\mathcal{R}}_1^C>1$, $\frac{{\mathcal{R}}_1^A}{{\mathcal{R}}_2^C}\le 1$ and $d_T-r+\frac{r{T}_3}{T_{max}}>0$.

Theorem 6

The steady state $S{S}_4$ is GAS when ${\mathcal{R}}_1^A>{\mathcal{R}}_2^C>1$ and $d_T-r+\frac{r{T}_4}{T_{max}}>0$ .

Proof

Define a function ${\mathcal{V}}_4(T,L,I,V,A,C)$ as:

$${\mathcal{V}}_4=\rho T_4\mathcal{H}\left(\frac{T}{T_4}\right)+\frac{\alpha e^{-n_3{\tau }_3}}{\alpha +d_L}{L}_4\mathcal{H}\left(\frac{L}{L_4}\right)+I_4\mathcal{H}\left(\frac{I}{I_4}\right)+\frac{(d_I+\omega C_4)e^{n_4{\tau }_4}}{k}{V}_4\mathcal{H}\left(\frac{V}{V_4}\right)+\frac{(d_I+\omega C_4)u{e}^{n_4{\tau }_4}}{kq}{A}_4\mathcal{H}\left(\frac{A}{A_4}\right)+\frac{\omega }{\sigma }{C}_4\mathcal{H}\left(\frac{C}{C_4}\right)+{\mathcal{U}}_4\left(t\right),$$

where

$${\mathcal{U}}_4\left(t\right)=\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_4{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }{\int }_{t-\psi }^t\mathcal{H}\left(\frac{T\left(\varphi \right)V\left(\varphi \right)}{T_4{V}_4}\right)d\varphi d\psi +(1-\eta )\xi T_4{V}_4{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }{\int }_{t-\psi }^t\mathcal{H}\left(\frac{T\left(\varphi \right)V\left(\varphi \right)}{T_4{V}_4}\right)d\varphi d\psi +\alpha e^{-n_3{\tau }_3}{L}_4{\int }_{t-{\tau }_3}^t\mathcal{H}\left(\frac{L\left(\varphi \right)}{L_4}\right)d\varphi +(d_I+\omega C_4)I_4{\int }_{t-{\tau }_4}^t\mathcal{H}\left(\frac{I\left(\varphi \right)}{I_4}\right)d\varphi .$$

We have ${\mathcal{V}}_4(T,L,I,V,A,C)>0$ for all $T,L,I,V,A,C>0$, and ${\mathcal{V}}_4(T_4,L_4,I_4,V_4,A_4,C_4)=0$. Then, we have

$$\frac{d{\mathcal{U}}_4\left(t\right)}{dt}=\rho \xi TV-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\xi T(t-\psi )V(t-\psi )d\psi -(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\xi T(t-\psi )V(t-\psi )d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_4{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +(1-\eta )\xi T_4{V}_4{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +\alpha e^{-n_3{\tau }_3}\left(L-L(t-{\tau }_3)+L_{4}ln\left(\frac{L(t-{\tau }_3)}{L}\right)\right)+(d_I+\omega C_4)\left(I-I(t-{\tau }_4)+I_{4}ln\left(\frac{I(t-{\tau }_4)}{I}\right)\right).$$

Hence,

$$\frac{d{\mathcal{V}}_4}{dt}=\rho \left(1-\frac{T_4}{T}\right)\dot{T}+\frac{\alpha e^{-n_3{\tau }_3}}{\alpha +d_L}\left(1-\frac{L_4}{L}\right)\dot{L}+\left(1-\frac{I_4}{I}\right)\dot{I}+\frac{(d_I+\omega C_4)e^{n_4{\tau }_4}}{k}\left(1-\frac{V_4}{V}\right)\dot{V}+\frac{(d_I+\omega C_4)u{e}^{n_4{\tau }_4}}{kq}\left(1-\frac{A_4}{A}\right)\dot{A}+\frac{\omega }{\sigma }\left(1-\frac{C_4}{C}\right)\dot{C}+\frac{d{\mathcal{U}}_4\left(t\right)}{dt}=\rho \left(1-\frac{T_4}{T}\right)\left[\lambda -d_{T}T+rT\left(1-\frac{T}{T_{max}}\right)-\xi TV\right]+\frac{\alpha e^{-n_3{\tau }_3}}{\alpha +d_L}\left(1-\frac{L_4}{L}\right)\left[\eta {\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\xi T(t-\psi )V(t-\psi )d\psi -(\alpha +d_L)L\right]+\left(1-\frac{I_4}{I}\right)\left[(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\xi T(t-\psi )V(t-\psi )d\psi +\alpha e^{-n_3{\tau }_3}L(t-{\tau }_3)-d_{I}I-\omega CI\right]+\frac{(d_I+\omega C_4)e^{n_4{\tau }_4}}{k}\left(1-\frac{V_4}{V}\right)\left[k{e}^{-n_4{\tau }_4}I(t-{\tau }_4)-d_{V}V-uAV\right]+\frac{(d_I+\omega C_4)u{e}^{n_4{\tau }_4}}{kq}\left(1-\frac{A_4}{A}\right)\left[qAV-d_{A}A\right]+\frac{\omega }{\sigma }\left(1-\frac{C_4}{C}\right)\left[\sigma CI-d_{C}C\right]+\rho \xi TV+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_4{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +(1-\eta )\xi T_4{V}_4{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +\alpha e^{-n_3{\tau }_3}\left(L-L(t-{\tau }_3)+L_{4}ln\left(\frac{L(t-{\tau }_3)}{L}\right)\right)+(d_I+\omega C_4)\left(I-I(t-{\tau }_4)+I_{4}ln\left(\frac{I(t-{\tau }_4)}{I}\right)\right).$$ (29)

Eq. (29) can be written as:

$$\frac{d{\mathcal{V}}_4}{dt}=\rho \left(1-\frac{T_4}{T}\right)\left[\lambda -d_{T}T+rT\left(1-\frac{T}{T_{max}}\right)\right]+\rho \xi T_{4}V-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\frac{L_4\xi T(t-\psi )V(t-\psi )}{L}d\psi +\alpha e^{-n_3{\tau }_3}{L}_4-(1-\eta ){\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\frac{I_4\xi T(t-\psi )V(t-\psi )}{I}d\psi -\alpha e^{-n_3{\tau }_3}\frac{I_{4}L(t-{\tau }_3)}{I}+d_I{I}_4+\omega I_{4}C-\frac{(d_I+\omega C_4)d_V{e}^{n_4{\tau }_4}}{k}V-(d_I+\omega C_4)\frac{V_{4}I(t-{\tau }_4)}{V}+\frac{(d_I+\omega C_4)d_V{e}^{n_4{\tau }_4}}{k}{V}_4+\frac{(d_I+\omega C_4)u{e}^{n_4{\tau }_4}}{k}{V}_{4}A-\frac{(d_I+\omega C_4)d_{A}u{e}^{n_4{\tau }_4}}{kq}A-\frac{(d_I+\omega C_4)u{e}^{n_4{\tau }_4}}{k}{A}_{4}V+\frac{(d_I+\omega C_4)d_{A}u{e}^{n_4{\tau }_4}}{kq}{A}_4-\frac{d_C\omega }{\sigma }C+\frac{d_C\omega }{\sigma }{C}_4+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_4{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +(1-\eta )\xi T_4{V}_4{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +\alpha e^{-n_3{\tau }_3}{L}_{4}ln\left(\frac{L(t-{\tau }_3)}{L}\right)+(d_I+\omega C_4)I_{4}ln\left(\frac{I(t-{\tau }_4)}{I}\right).$$ (30)

The steady state $S{S}_4$ satisfies the following:

$$\lambda =d_T{T}_4-r{T}_4\left(1-\frac{T_4}{T_{max}}\right)+\xi T_4{V}_4,$$

$$\alpha e^{-n_3{\tau }_3}{L}_4=\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}F\xi T_4{V}_4,(d_I+\omega C_4)I_4=\rho \xi T_4{V}_4=\frac{(d_I+\omega C_4)d_V{e}^{n_4{\tau }_4}}{k}{V}_4+\frac{(d_I+\omega C_4)u{e}^{n_4{\tau }_4}}{k}{V}_4{A}_4,$$

$$V_4=\frac{d_A}{q},I_4=\frac{d_C}{\sigma },$$

then we get,

$$\lambda -d_{T}T+rT\left(1-\frac{T}{T_{max}}\right)=(T_4-T)\left(d_T-r+\frac{r{T}_4}{T_{max}}+\frac{rT}{T_{max}}\right)+\xi T_4{V}_4.$$

By using the above conditions, the derivative in (30) is transformed into

$$\frac{d{\mathcal{V}}_4}{dt}\le -\rho \left(d_T-r+\frac{r{T}_4}{T_{max}}\right)\frac{{(T-T_4)}^2}{T}+\rho \xi T_4{V}_4-\rho \xi T_4{V}_4\frac{T_4}{T}+\left(\rho \xi T_4-\frac{(d_I+\omega C_4)d_V{e}^{n_4{\tau }_4}}{k}-\frac{(d_I+\omega C_4)u{e}^{n_4{\tau }_4}}{k}{A}_4\right)V-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_4{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\frac{L_{4}T(t-\psi )V(t-\psi )}{L{T}_4{V}_4}d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_{4}F-(1-\eta )\xi T_4{V}_4{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\frac{I_{4}T(t-\psi )V(t-\psi )}{I{T}_4{V}_4}d\psi -\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_{2}F\frac{I_{4}L(t-{\tau }_3)}{I{L}_4}+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_{4}F+(1-\eta )\xi T_4{V}_{2}G-\rho \xi T_4{V}_4\frac{V_{4}I(t-{\tau }_4)}{V{I}_4}+\rho \xi T_4{V}_4+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_4{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +(1-\eta )\xi T_4{V}_4{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_{4}Fln\left(\frac{L(t-{\tau }_3)}{L}\right)+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_{4}Fln\left(\frac{I(t-{\tau }_4)}{I}\right)+(1-\eta )\xi T_4{V}_{4}Gln\left(\frac{I(t-{\tau }_4)}{I}\right).$$

The steady state conditions of $S{S}_4$ imply that

$$\rho \xi T_4-\frac{(d_I+\omega C_4)d_V{e}^{n_4{\tau }_4}}{k}-\frac{(d_I+\omega C_4)u{e}^{n_4{\tau }_4}}{k}{A}_4=0.$$

Now, using the following equalities:

$$ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)+ln\left(\frac{L(t-{\tau }_3)}{L}\right)=ln\left(\frac{T_4}{T}\right)+ln\left(\frac{L(t-{\tau }_3)V_4}{L_{4}V}\right)+ln\left(\frac{L_{4}T(t-\psi )V(t-\psi )}{L{T}_4{V}_4}\right),$$

$$ln\left(\frac{T(t-\psi )V(t-\psi )}{TV}\right)+ln\left(\frac{I(t-{\tau }_4)}{I}\right)=ln\left(\frac{T_4}{T}\right)+ln\left(\frac{I(t-{\tau }_4)V_4}{I_{4}V}\right)+ln\left(\frac{I_{4}T(t-\psi )V(t-\psi )}{I{T}_4{V}_4}\right),$$

we get

$$\frac{d{\mathcal{V}}_4}{dt}\le -\rho \left(d_T-r+\frac{r{T}_4}{T_{max}}\right)\frac{{(T-T_4)}^2}{T}+\rho \xi T_4{V}_4-\rho \xi T_4{V}_4\frac{T_4}{T}-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_4{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\frac{L_{4}T(t-\psi )V(t-\psi )}{L{T}_4{V}_4}d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_{4}F-(1-\eta )\xi T_4{V}_4{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\frac{I_{4}T(t-\psi )V(t-\psi )}{I{T}_4{V}_4}d\psi -\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_{4}F\frac{I_{4}L(t-{\tau }_3)}{I{L}_4}+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_{4}F+(1-\eta )\xi T_4{V}_{4}G-\rho \xi T_4{V}_4\frac{V_{4}I(t-{\tau }_4)}{V{I}_4}+\rho \xi T_4{V}_4+\rho \xi T_4{V}_{4}ln\left(\frac{T_4}{T}\right)+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_4{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }ln\left(\frac{L_{4}T(t-\psi )V(t-\psi )}{L{T}_4{V}_4}\right)d\psi +(1-\eta )\xi T_4{V}_4{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }ln\left(\frac{I_{4}T(t-\psi )V(t-\psi )}{I{T}_4{V}_4}\right)d\psi +\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_{4}Fln\left(\frac{L(t-{\tau }_3)V_4}{L_{4}V}\right)+(1-\eta )\xi T_4{V}_{4}Gln\left(\frac{I(t-{\tau }_4)V_4}{I_{4}V}\right)+\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_{4}Fln\left(\frac{I(t-{\tau }_4)}{I}\right).$$

By using the equality:

$$ln\left(\frac{L(t-{\tau }_3)V_4}{L_{4}V}\right)+ln\left(\frac{I(t-{\tau }_4)}{I}\right)=ln\left(\frac{I_{4}L(t-{\tau }_3)}{I{L}_4}\right)+ln\left(\frac{V_{4}I(t-{\tau }_4)}{V{I}_4}\right),$$

and rearranging the $R.H.S$. of $\frac{d{\mathcal{V}}_4}{dt}$, we get

$$\frac{d{\mathcal{V}}_4}{dt}\le -\rho \left(d_T-r+\frac{r{T}_4}{T_{max}}\right)\frac{{(T-T_4)}^2}{T}-\rho \xi T_4{V}_4\left[\frac{T_4}{T}-1-ln\left(\frac{T_4}{T}\right)\right]-\rho \xi T_4{V}_4\left[\frac{V_{4}I(t-{\tau }_4)}{V{I}_4}-1-ln\left(\frac{V_{4}I(t-{\tau }_4)}{V{I}_4}\right)\right]-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_4{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\left[\frac{L_{4}T(t-\psi )V(t-\psi )}{L{T}_4{V}_4}-1-ln\left(\frac{L_{4}T(t-\psi )V(t-\psi )}{L{T}_4{V}_4}\right)\right]d\psi -\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_{4}F\left[\frac{I_{4}L(t-{\tau }_3)}{I{L}_4}-1-ln\left(\frac{I_{4}L(t-{\tau }_3)}{I{L}_4}\right)\right]-(1-\eta )\xi T_4{V}_4{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\left[\frac{I_{4}T(t-\psi )V(t-\psi )}{I{T}_4{V}_4}-1-ln\left(\frac{I_{4}T(t-\psi )V(t-\psi )}{I{T}_4{V}_4}\right)\right]d\psi .=-\rho \left(d_T-r+\frac{r{T}_4}{T_{max}}\right)\frac{{(T-T_4)}^2}{T}-\rho \xi T_4{V}_4\mathcal{H}\left(\frac{T_4}{T}\right)-\rho \xi T_4{V}_4\mathcal{H}\left(\frac{V_{4}I(t-{\tau }_4)}{V{I}_4}\right)-\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_4{\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }\mathcal{H}\left(\frac{L_{4}T(t-\psi )V(t-\psi )}{L{T}_4{V}_4}\right)d\psi -\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}\xi T_4{V}_{4}F\mathcal{H}\left(\frac{I_{4}L(t-{\tau }_3)}{I{L}_4}\right)-(1-\eta )\xi T_4{V}_4{\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }\mathcal{H}\left(\frac{I_{4}T(t-\psi )V(t-\psi )}{I{T}_4{V}_4}\right)d\psi .$$

We see that $\frac{d{\mathcal{V}}_4}{dt}\le 0$ when ${\mathcal{R}}_1^A>{\mathcal{R}}_2^C>1$ and $d_T-r+\frac{r{T}_4}{T_{max}}>0$. Also, $\frac{d{\mathcal{V}}_4}{dt}=0$ when, $T=T_4$, $L=L_4$, $I=I_4$ and $V=V_4$. The solutions of system (5)–(10) tend to $M_4^{\prime }$ the largest invariant subset of $M_4=\left\{(T,L,I,V,A,C)\right|\frac{d{\mathcal{V}}_4}{dt}=0\}$. For each element in $M_4^{\prime }$ we have $V=V_4$, $I=I_4$ then $\dot{V}=0$, $\dot{I}=0$ and from Eq. (8) we have $0=\dot{V}=k{e}^{-n_4{\tau }_4}{I}_4-d_V{V}_4-uA{V}_4$, which gives $A\left(t\right)=A_4$, for all $t$. Further, from Eq. (7) we have $0=\dot{I}=(1-\eta )G\xi T_4{V}_4+\alpha e^{-n_3{\tau }_3}{L}_4-d_I{I}_4-\omega I_{4}C$, which gives $C\left(t\right)=C_4$, for all $t$. It follows that $M_4^{\prime }=\left\{S{S}_4\right\}$. LIP implies that $S{S}_4$ is GAS when ${\mathcal{R}}_1^A>{\mathcal{R}}_2^C>1$ and $d_T-r+\frac{r{T}_4}{T_{max}}>0$

Based on the above findings, we summarize the existence and global stability conditions for all steady state points in Table 1.

Table 1.

Steady states and their global stability conditions for model (5)–(10).

Steady state	Global stability conditions
$S{S}_0=(T_0,0,0,0,0,0)$	${\mathcal{R}}_0\le 1$
$S{S}_1=(T_1,L_1,I_1,V_1,0,0)$	${\mathcal{R}}_1^A\le 1<{\mathcal{R}}_0$, ${\mathcal{R}}_1^C\le 1$ and $d_T-r+\frac{r{T}_1}{T_{max}}>0$
$S{S}_2=(T_2,L_2,I_2,V_2,A_2,0)$	${\mathcal{R}}_1^A>1$, ${\mathcal{R}}_2^C\le 1$ and $d_T-r+\frac{r{T}_2}{T_{max}}>0$
$S{S}_3=(T_3,L_3,I_3,V_3,0,C_3)$	${\mathcal{R}}_1^C>1$, ${\mathcal{R}}_1^A\le {\mathcal{R}}_2^C$ and $d_T-r+\frac{r{T}_3}{T_{max}}>0$
$S{S}_4=(T_4,L_4,I_4,V_4,A_4,C_4)$	${\mathcal{R}}_1^A>{\mathcal{R}}_2^C>1$ and $d_T-r+\frac{r{T}_4}{T_{max}}>0$

5. Numerical simulations

In this section, we execute numerical simulations to enhance the results of Theorems 2–6. Besides, we study the impact of time delays on the dynamical behavior of the system. Let us take a particular form of the probability distributed functions as:

$$f\left(\psi \right)=\delta (\psi -{\psi }_1),g\left(\psi \right)=\delta (\psi -{\psi }_2),$$

where $\delta (.)$ is the Dirac delta function. When ${\tau }_i\to \infty ,i=1,2$, we have

$${\int }_0^{\infty }f\left(\psi \right)d\psi =1,{\int }_0^{\infty }g\left(\psi \right)d\psi =1.$$

We have

$${\int }_0^{\infty }\delta (\psi -{\psi }_i)e^{-n_i\psi }d\psi =e^{-n_i{\psi }_i},i=1,2.$$

Moreover,

$${\int }_0^{\infty }\delta (\psi -{\psi }_i)e^{-n_i\psi }T(t-\psi )V(t-\psi )d\psi =e^{-n_i{\psi }_i}T(t-{\psi }_i)V(t-{\psi }_i),i=1,2.$$

Hence, model (5)–(10) becomes:

$$\dot{T}\left(t\right)=\lambda -d_{T}T\left(t\right)+rT\left(t\right)\left(1-\frac{T\left(t\right)}{T_{max}}\right)-\xi V\left(t\right)T\left(t\right),$$ (31)

$$\dot{L}\left(t\right)=\eta \xi e^{-n_1{\psi }_1}V(t-{\psi }_1)T(t-{\psi }_1)-\alpha L\left(t\right)-d_{L}L\left(t\right),$$ (32)

$$\dot{I}\left(t\right)=(1-\eta )\xi e^{-n_2{\psi }_2}V(t-{\psi }_2)T(t-{\psi }_2)+\alpha e^{-n_3{\tau }_3}L(t-{\tau }_3)-d_{I}I\left(t\right)-\omega C\left(t\right)I\left(t\right),$$ (33)

$$\dot{V}\left(t\right)=k{e}^{-n_4{\tau }_4}I(t-{\tau }_4)-d_{V}V\left(t\right)-uV\left(t\right)A\left(t\right),$$ (34)

$$\dot{A}\left(t\right)=qV\left(t\right)A\left(t\right)-d_{A}A\left(t\right),$$ (35)

$$\dot{C}\left(t\right)=\sigma C\left(t\right)I\left(t\right)-d_{C}C\left(t\right).$$ (36)

The basic reproduction number of model (31)–(36) is given by:

$${\mathcal{R}}_0=\frac{\xi T_{0}k{e}^{-n_4{\tau }_4}}{d_I{d}_V}\left(\frac{\alpha \eta e^{-n_3{\tau }_3}}{\alpha +d_L}{e}^{-n_1{\psi }_1}+(1-\eta )e^{-n_2{\psi }_2}\right).$$ (37)

To solve system (31)–(36) we use the MATLAB solver dde23. Without loss of generality let us consider for simplicity that ${\psi }_1={\psi }_2={\tau }_3={\tau }_4=\tau$. The values of some parameters of model (31)–(36) are chosen as $\lambda =0.11$, $r=0.01$, $T_{max}=12$, $\eta =0.5$, $\alpha =4.08$, $\omega =0.001$, $k=0.22$, $u=0.05$, $d_T=0.01$, $d_L=10^{-3}$, $d_I=0.13$, $d_V=4.36$, $d_A=0.028$, $d_C=0.06$, $n_1=10^{-3}$, $n_2=0.11$, $n_3=1$, and $n_4=1$. The remaining parameters of the model will be varied. In fact, obtaining real measurements from COVID-19 patients is difficult, therefore we have chosen the values of the model’s parameters just to conduct the numerical simulations. However, if one can get real data, then the parameters of the model can be estimated and the model can be validated. We select three different sets of initial conditions for (31)–(36):

$$\mathbf{Initial}\text{-}\mathbf{1}:\left(T(\varkappa ),L(\varkappa ),I(\varkappa ),V(\varkappa ),A(\varkappa ),C(\varkappa )\right)=(8,0.004,0.2,0.01,14,20),$$

$$\mathbf{Initial}\text{-}\mathbf{2}:\left(T(\varkappa ),L(\varkappa ),I(\varkappa ),V(\varkappa ),A(\varkappa ),C(\varkappa )\right)=(7,0.006,0.4,0.02,15,30),$$

$$\mathbf{Initial}\text{-}\mathbf{3}:\left(T(\varkappa ),L(\varkappa ),I(\varkappa ),V(\varkappa ),A(\varkappa ),C(\varkappa )\right)=(6,0.008,0.6,0.03,16,40),$$

where $\varkappa \in [-\tau ,0]$.

5.1. Stability of steady states

In this subsection we address the stability of the five steady states with $\tau =0.1$, and $\xi ,q$ and $\sigma$ are varied.

Case 1 (Stability of$S{S}_0$): $\xi =0.05$, $q$$=0.5$ and $\sigma$ $=0.09$. Using these data, we compute ${\mathcal{R}}_0=0.1910<1$. According to Theorem 2, $S{S}_0$ is GAS and the SARS-CoV-2 is predicted to be completely removed from the body. From Fig. 1, we can see that the numerical results agree with the results of Theorem 2. We observe that, the concentration of healthy epithelial cells is increased and converged to its normal value $T_0=11.4891$, while the concentrations of latent infected cells, active infected cells, SARS-CoV-2 particles, antibodies and CTLs are extremely decaying and tend to zero.

Fig. 1.

Solutions of system (31)–(36) with three initial conditions when ${\mathcal{R}}_0\le 1$. (a) Healthy epithelial cell; (b) Latent infected cells; (c) Active infected cells; (d) SARS-CoV-2 particles; (e) Antibodies; (f) CTLs.

Case 2 (Stability of$S{S}_1$): $\xi$$=0.5$, $q$$=0.5$ and $\sigma$ $=0.05$. Using these data, we compute ${\mathcal{R}}_1^A=0.5907<1$, ${\mathcal{R}}_1^C=0.6014<1<{\mathcal{R}}_0=1.9102$ and $d_T-r+\frac{r{T}_1}{T_{max}}=0.005>0$. According to Theorem 3, $S{S}_1$ is GAS. From Fig. 2, we see that there is an agreement between the numerical and results of Theorem 3. In addition, the states of the system converge to the steady state $S{S}_1=(6.0147,0.0098,0.5816,0.0266,0,0)$.

Fig. 2.

Solutions of system (31)–(36) with three initial conditions when ${\mathcal{R}}_1^A\le 1<{\mathcal{R}}_0$, ${\mathcal{R}}_1^C\le 1$ and $d_T-r+\frac{r{T}_1}{T_{max}}>0$. (a) Healthy epithelial cell; (b) Latent infected cells; (c) Active infected cells; (d) SARS-CoV-2 particles; (e) Antibodies; (f) CTLs.

Case 3 (Stability of$S{S}_2$): $\xi$$=0.5$, $q$$=1.4$ and $\sigma =0.05$. Using these data, we compute ${\mathcal{R}}_1^A=1.1574>1,{\mathcal{R}}_2^C=0.6014<1$ and $d_T-r+\frac{r{T}_2}{T_{max}}=0.0058>0$. According to Theorem 4, $S{S}_2$ is GAS. The numerical solutions displayed in Fig. 3 is consistent with results of Theorem 4. Further, the states of the system converge to the steady state $S{S}_2=(6.9615,0.0085,0.507,0.02,13.7267,0)$.

Fig. 3.

Solutions of system (31)–(36) with three initial conditions when ${\mathcal{R}}_1^A>1$, ${\mathcal{R}}_2^C\le 1$ and $d_T-r+\frac{r{T}_2}{T_{max}}>0$. (a) Healthy epithelial cell; (b) Latent infected cells; (c) Active infected cells; (d) SARS-CoV-2 particles; (e) Antibodies; (f) CTL cells.

Case 4 (Stability of$S{S}_3$): $\xi$$=0.5$, $q$$=1.4$ and $\sigma$ $=0.16$. Using these data, we compute ${\mathcal{R}}_1^C=1.2384>1,\frac{{\mathcal{R}}_1^A}{{\mathcal{R}}_2^C}=0.8561<1$ and $d_T-r+\frac{r{T}_3}{T_{max}}=0.0062>0$. According to Theorem 5, $S{S}_3$ is GAS and this is shown in Fig. 4. We can see that, the states of the system converge to the steady state $S{S}_3=(7.4490,0.0078,0.375,0.0171,0,30.993)$.

Fig. 4.

Solutions of system (31)–(36) with three initial conditions when ${\mathcal{R}}_1^C>1$, $\frac{{\mathcal{R}}_1^A}{{\mathcal{R}}_2^C}\le 1$ and $d_T-r+\frac{r{T}_3}{T_{max}}>0$. (a) Healthy epithelial cell; (b) Latent infected cells; (c) Active infected cells; (d) SARS-CoV-2 particles; (e) Antibodies; (f) CTLs.

Case 5 (Stability of$S{S}_4$): $\xi$$=0.5$, $q=1.9$ and $\sigma =0.16$. Using these data, we compute $\frac{{\mathcal{R}}_1^A}{{\mathcal{R}}_2^C}=1.1618>1,{\mathcal{R}}_2^C=1.1290>1$ and $d_T-r+\frac{r{T}_4}{T_{max}}=0.0066>0$. According to Theorem 6, $S{S}_4$ is GAS and this is clarified numerically in Fig. 5. The states of the system converge to the steady state $S{S}_4=(7.8893,0.0071,0.375,0.0147,14.1094,16.7705)$.

Fig. 5.

Solutions of system (31)–(36) with three initial conditions when ${\mathcal{R}}_1^A>{\mathcal{R}}_2^C>1$ and $d_T-r+\frac{r{T}_4}{T_{max}}>0$. (a) Healthy epithelial cell; (b) Latent infected cells; (c) Active infected cells; (d) SARS-CoV-2 particles; (e) Antibodies; (f) CTL cells.

5.2. Effect of the time delay on the SAR-CoV-2 dynamics

In this subsection, we explore the impact of time delays $\tau$ on the stability of the steady states. We observe from Eq. (37) that the parameter ${\mathcal{R}}_0$ relies on the delay parameter $\tau$ which causes a significant change in the stability of the system. To clarify this situation, we choose $\xi$ $=0.5$, $q=1.9$, $\sigma$ $=0.16$ and $\tau$ is varied. Moreover, we consider the initial condition initial-3. Fig. 6 shows the influence of time delay on the solution trajectories of the system. We notice that as time delay $\tau$ is increased the number of the healthy epithelial cells is increased, while the numbers of latent infected cells, active infected cells, SARS-CoV-2 particles, CTLs and antibodies are decreased. When all other parameters are fixed and delays are varied,${\mathcal{R}}_0$can be written as:

$${\mathcal{R}}_0\left(\tau \right)=\frac{k\xi e^{-n_4\tau }{T}_0}{d_I{d}_V}\left[\frac{\alpha \eta e^{-n_3\tau }}{\alpha +d_L}{e}^{-n_1\tau }+(1-\eta )e^{-n_2\tau }\right].$$

We can see that ${\mathcal{R}}_0$ is a decreasing function of $\tau$. When the delay length becomes sufficiently large, ${\mathcal{R}}_0$ becomes less than or equal one, which makes the healthy steady state $S{S}_0$ is GAS. From a biological viewpoint, time delays play positive roles in the SARS-CoV-2 infection process in order to eliminate the virus. Sufficiently large time delays makes the SARS-CoV-2 development slower, and the SARS-CoV-2 is controlled and disappears. This gives us some suggestions on new drugs to prolong the time of formation of latent infected epithelial cells, or the time of formation of active infected epithelial cells, or the time of activation of latent infected epithelial cells, or the time for SARS-CoV-2 particles to mature (infectious). Now we calculate the range of the time delay which makes ${\mathcal{R}}_0\le 1$. Let us compute the minimum delay length ${\tau }_{cr}$ that solves the equation ${\mathcal{R}}_0\left({\tau }_{cr}\right)=1$ and makes the system has only one steady state $S{S}_0$. Using the values of the parameters we get, ${\tau }_{cr}=0.533434$. From Fig. 6 we can see the following scenarios:

Fig. 6.

Solutions of system (31)–(36) under the influence of the time delays $\tau$. (a) Healthy epithelial cell; (b) Latent infected cells; (c) Active infected cells; (d) SARS-CoV-2 particles; (e) Antibodies; (f) CTLs.

(i) if $\tau \ge 0.533434$, then ${\mathcal{R}}_0\le 1$ and the system has a single steady state $S{S}_0$ and it is GAS;

(ii) if $0\le \tau <0.533434$, then ${\mathcal{R}}_0>1$ and one of the other steady states is GAS.

6. Conclusion

In this paper, we formulated a SARS-CoV-2 infection model with distributed and discrete delays and adaptive immune response. Four time delays are included in the model: delay in the formation of latent infected cells, delay in the formation of active infected cells, delay in the activation of latent infected cells, maturation delay of new SARS-CoV-2 particles. We consider a logistic term for the healthy epithelial cells. The non-negativity and boundedness of the solutions of the models have been shown. Further, we derived four threshold parameters, the basic infection reproductive ratio ${\mathcal{R}}_0$, the active antibody immunity reproductive ratio ${\mathcal{R}}_1^A$, the active CTL immunity reproductive ratio ${\mathcal{R}}_1^C$ and the competed CTL immunity reproductive ratio ${\mathcal{R}}_2^C$. The global stability of all steady states of the models was investigated by constructing Lyapunov functions and LaSalle’s invariance principal. We obtained that the infection-free steady state $S{S}_0$ is GAS when ${\mathcal{R}}_0\le 1$. When ${\mathcal{R}}_1^A\le 1<{\mathcal{R}}_0$, ${\mathcal{R}}_1^C\le 1$ and $d_T-r+\frac{r{T}_1}{T_{max}}>0$ the infected steady state with inactive immune response $S{S}_1$ is GAS. When ${\mathcal{R}}_1^A>1$, ${\mathcal{R}}_2^C\le 1$ and $d_T-r+\frac{r{T}_2}{T_{max}}>0$ the infected steady state with only active antibody immune response $S{S}_2$ is GAS. While, the infected steady state with only active CTL immune response $S{S}_3$ is globally asymptotically stable when ${\mathcal{R}}_1^C>1$, ${\mathcal{R}}_2^A\le 1$ and $d_T-r+\frac{r{T}_3}{T_{max}}>0$. Finally, we found that the infected steady state with both active antibody and CTL immune responses $S{S}_4$ is GAS when ${\mathcal{R}}_1^A>{\mathcal{R}}_2^C>1$ and $d_T-r+\frac{r{T}_4}{T_{max}}>0$. We performed the numerical simulations for the model and we showed that both numerical and theoretical results are consistent. Moreover, we have demonstrated that the increase of time delays period can play the same influence as the antiviral treatment.

Model (5)–(10) assumes that viruses and cells are well mixed and neglected the mobility of them. Viruses and cells can move and go from high concentration regions to low concentration regions [56], [57], [58], [59]. Taking into account the mobility of viruses and cells, model (5)–(9) can be extended as:

$$\frac{\partial T\left(x,t\right)}{\partial t}=D_T\Delta T\left(x,t\right)+\lambda -d_{T}T\left(x,t\right)+rT\left(x,t\right)\left(1-\frac{T\left(x,t\right)}{T_{max}}\right)-\xi T\left(x,t\right)V\left(x,t\right),$$

$$\frac{\partial L\left(x,t\right)}{\partial t}=D_L\Delta L\left(x,t\right)+\eta \xi {\int }_0^{{\tau }_1}f\left(\psi \right)e^{-n_1\psi }V(x,t-\psi )T(x,t-\psi )d\psi -\alpha L\left(x,t\right)-d_{L}L\left(x,t\right),$$

$$\frac{\partial I\left(x,t\right)}{\partial t}=D_I\Delta I\left(x,t\right)+(1-\eta )\xi {\int }_0^{{\tau }_2}g\left(\psi \right)e^{-n_2\psi }V(x,t-\psi )T(x,t-\psi )d\psi +\alpha e^{-n_3{\tau }_3}L(x,t-{\tau }_3)-d_{I}I\left(x,t\right)-\omega I\left(x,t\right)C\left(x,t\right),$$

$$\frac{\partial V\left(x,t\right)}{\partial t}=D_V\Delta V\left(x,t\right)+k{e}^{-n_4{\tau }_4}I(x,t-{\tau }_4)-d_{V}V\left(x,t\right)-uV\left(x,t\right)A\left(x,t\right),$$

$$\frac{\partial A\left(x,t\right)}{\partial t}=D_A\Delta A\left(x,t\right)+qV\left(x,t\right)A\left(x,t\right)-d_{A}A\left(x,t\right),$$

$$\frac{\partial C\left(x,t\right)}{\partial t}=D_C\Delta C\left(x,t\right)+\sigma I\left(x,t\right)C\left(x,t\right)-d_{C}C\left(x,t\right),$$

where $x=\left(x_1,x_2,\dots ,x_m\right)\in \Omega$, $\Delta =\frac{{\partial }^2}{\partial x^2}$ is the Laplacian operator and $D_{\chi }$ is the diffusion coefficient corresponding to compartment $\chi$ of the model.

Our model can also be extended by (i) expanding to multiscale model to get a deeper understanding of the SARS-CoV-2 dynamics [60], [61] and (ii) considering stochastic interactions [62], [63], [64], [65], [66].

CRediT authorship contribution statement

A.M. Elaiw: Conceptualization, Methodology. A.J. Alsaedi: Formal analysis, Writing – original draft. A.D. Hobiny: Investigation. S. Aly: Writing – review & editing.

Declaration of Competing Interest

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

Data availability

No data was used for the research described in the article.