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. 2021 May 25;2021(1):270. doi: 10.1186/s13662-021-03416-7

Stability of an HTLV-HIV coinfection model with multiple delays and CTL-mediated immunity

N H AlShamrani 1,2,
PMCID: PMC8144699  PMID: 34054935

Abstract

In the literature, several mathematical models have been formulated and developed to describe the within-host dynamics of either human immunodeficiency virus (HIV) or human T-lymphotropic virus type I (HTLV-I) monoinfections. In this paper, we formulate and analyze a novel within-host dynamics model of HTLV-HIV coinfection taking into consideration the response of cytotoxic T lymphocytes (CTLs). The uninfected CD4+T cells can be infected via HIV by two mechanisms, free-to-cell and infected-to-cell. On the other hand, the HTLV-I has two modes for transmission, (i) horizontal, via direct infected-to-cell touch, and (ii) vertical, by mitotic division of active HTLV-infected cells. It is well known that the intracellular time delays play an important role in within-host virus dynamics. In this work, we consider six types of distributed-time delays. We investigate the fundamental properties of solutions. Then, we calculate the steady states of the model in terms of threshold parameters. Moreover, we study the global stability of the steady states by using the Lyapunov method. We conduct numerical simulations to illustrate and support our theoretical results. In addition, we discuss the effect of multiple time delays on stability of the steady states of the system.

Keywords: HTLV-HIV coinfection, CTL immune response, Intracellular delay, Mitotic transmission, Global stability, Lyapunov function

Introduction

During the past several decades many human viruses and their associated diseases, such as human immunodeficiency virus (HIV), hepatitis C virus (HCV), hepatitis B virus (HBV), dengue virus, human T-lymphotropic virus type I (HTLV-I), and recently coronavirus, have been recognized. Human body can be infected by more that one virus at the same time such as HTLV-HIV, coronavirus/influenza, HCV-HIV, HBV-HIV, HCV-HBV, and malaria-HIV. HTLV and HIV are universal public health matters. HTLV and HIV are two viruses which infect most effective immune cells, CD4+T cells. Adult T-cell leukemia (ATL) and HTLV-I-associated myelopathy/tropical spastic paraparesis (HAM/TSP) are the last stage of HTLV-I infection. Chronic HIV infection leads to acquired immunodeficiency syndrome (AIDS). Both HTLV and HIV have the same ways of transmission such as sharing contaminated needles and unprotected sexual contact with infected partners. Over the last 10 years HTLV-HIV coinfection has been widely documented (see e.g. [13], and [4]).

Mathematical models

Mathematical models of HIV and HTLV-I dynamics have become efficient tools to biological and medical scientists. These models can provide a deeper understanding of within-host virus dynamics and assist in predicting the impact of antiviral drug efficacy on viral infection progression (see e.g. [516]).

  • HIV monoinfection model: The standard HIV dynamics model under the effect of cytotoxic T lymphocytes (CTLs) has been formulated by Nowak and Bangham [17] as follows:
    {dS(t)dt=ηϱS(t)ϑ1S(t)V(t),dI(t)dt=ϑ1S(t)V(t)aI(t)μ1CI(t)I(t),dV(t)dt=bI(t)εV(t),dCI(t)dt=σ1CI(t)I(t)π1CI(t), 1
    where S(t), I(t), V(t), and CI(t) are the concentrations of uninfected CD4+T cells, active HIV-infected cells, free HIV particles, and HIV-specific CTLs, respectively, and t is the time. η refers to the generation rate of the uninfected CD4+T cells. The uninfected CD4+T cells are infected via HIV particles (free-to-cell infection) at rate ϑ1SV. The HIV-infected cells produce HIV particles at rate bI. The stimulation rate of effective HIV-specific CTLs due to the presence of HIV-infected cells is defined by σ1CII. The term μ1CII accounts for the killing rate of HIV-infected cells due to its specific CTLs. The four compartments S, I, V, and CI have normal death rates ϱS, aI, εV, and π1CI, respectively. Several extensions on model (1) have been accomplished (see e.g. [1820]).
  • HTLV-I monoinfection model: The within-host dynamics of HTLV-I has been mathematically modeled in several papers [2124]. CTL immunity has been included into the HTLV-I dynamics models in many works (see e.g. [2533]). Lim and Maini [28] have formulated a model for HTLV-I dynamics under the consideration of CTL immunity and mitotic division of active HTLV-infected cells as follows:
    {dS(t)dt=ηϱS(t)ϑ3S(t)Y(t),dE(t)dt=ϑ3S(t)Y(t)+KrY(t)(ψ+ω)E(t),dY(t)dt=ψE(t)δY(t)μ2CY(t)Y(t),dCY(t)dt=σ2Y(t)π2CY(t), 2
    where E(t), Y(t), and CY(t) are the concentrations of latent HTLV-infected cells, active HTLV-infected cells, and HTLV-specific CTLs at time t, respectively. The term ϑ3SY denotes the infected-to-cell contact rate between HTLV-infected cells and uninfected CD4+T cells (horizontal transmission). The active HTLV-infected cells transmit vertically to latent compartment at rate KrY (mitotic transmission), where K(0,1). The HTLV-specific CTLs kill the active HTLV-infected cells at rate μ2CYY and are stimulated at rate σ2Y. The term ψE denotes the activation rate of latent HTLV-infected cells. The death rates of E, Y, and CY are given by ωE, δY, and π2CY, respectively.
  • HTLV-HIV coinfection model: Elaiw and AlShamrani [34] have recently formulated an HTLV-HIV coinfection model as follows:
    {dS(t)dt=ηϱS(t)ϑ1S(t)V(t)ϑ2S(t)I(t)ϑ3S(t)Y(t),dL(t)dt=(1β)(ϑ1S(t)V(t)+ϑ2S(t)I(t))(λ+γ)L(t),dI(t)dt=β(ϑ1S(t)V(t)+ϑ2S(t)I(t))+λL(t)aI(t)μ1CI(t)I(t),dE(t)dt=φϑ3S(t)Y(t)(ψ+ω)E(t),dY(t)dt=ψE(t)δY(t)μ2CY(t)Y(t),dV(t)dt=bI(t)εV(t),dCI(t)dt=σ1CI(t)I(t)π1CI(t),dCY(t)dt=σ2CY(t)Y(t)π2CY(t), 3
    where L(t) is the concentration of latent HIV-infected cells. The term ϑ2SI describes the infection rate of uninfected CD4+T cells by HIV-infected cells. λL and γL are the activation and death rates of latent HIV-infected cells. The parameter β(0,1) represents the part of newly HIV-infected cells that becomes active, and the other part 1β enters a latent stage. The parameter φ(0,1) refers to the part of newly HTLV-infected cells that become latent.

Intracellular delay plays a crucial role in within-host virus dynamics and is defined as the time lapse between viral entry a cell and its production. In case of HIV, it has been estimated that the time between the HIV enters a target cell until producing new HIV particles is about 0.9 days [35]. Time delay has also an important effect in HTLV-I infection. Several works have been devoted to developing mathematical models with time delays to describe the dynamics of HIV (see e.g. [3643]) and HTLV (see e.g. [4453]).

Our aim is to take model (3) to further destination by incorporating multiple intracellular time delays and mitotic transmission. We study the fundamental and global properties of the system, then we present numerical simulation. The outcomes of this paper will help clinicians to estimate the suitable time to start the treatment. Our model may be helpful to study different coinfections such as influenza-coronavirus, HCV-HIV, HBV-HIV, and malaria-HIV. It is interesting to note that fractional-order differential equations (FODEs) have been widely studied in several works (see e.g. [5457]). Modeling and analysis of HIV dynamics with FODEs have been investigated in many papers (see e.g. [5860]). Clinicians can use the information (in terms of behavior predictions) of fractional-order systems to fit patients’ data with the most appropriate noninteger-order index. As a future work, our coinfection model can be formulated as a system of FODEs.

The multiple delays model

In this section, we extend system (3) by taking under consideration multiple types of distributed-time delays and mitosis of active HTLV-infected cells. We achieve this goal by considering the following system of delay differential equations (DDEs):

{dS(t)dt=ηϱS(t)ϑ1S(t)V(t)ϑ2S(t)I(t)ϑ3S(t)Y(t),dL(t)dt=(1β)0κ1Λ1()e1S(t)[ϑ1V(t)+ϑ2I(t)]d(λ+γ)L(t),dI(t)dt=β0κ2Λ2()e2S(t)[ϑ1V(t)+ϑ2I(t)]ddI(t)dt=+λ0κ3Λ3()e3L(t)daI(t)μ1CI(t)I(t),dE(t)dt=φϑ30κ4Λ4()e4S(t)Y(t)d+KrY(t)(ψ+ω)E(t),dY(t)dt=ψ0κ5Λ5()e5E(t)d+(1K)rY(t)δY(t)μ2CY(t)Y(t),dV(t)dt=b0κ6Λ6()e6I(t)dεV(t),dCI(t)dt=σ1CI(t)I(t)π1CI(t),dCY(t)dt=σ2CY(t)Y(t)π2CY(t). 4

The factor Λ1()e1 represents the probability that uninfected CD4+T cells contacted by HIV particles or active HIV-infected cells at time t survived time units and become latent infected at time t. The term Λ2()e2 is the probability that uninfected CD4+T cells contacted by HIV particles or active HIV-infected cells at time t survived time units and become actively infected at time t. The term Λ3()e3 is the probability that latent HIV-infected CD4+T cells survived time units before transmitted to be active at time t. Moreover, the factor Λ4()e4 demonstrates the probability that the initial infection of uninfected CD4+T cells and the HTLV-infected cells at time t completing all the intracellular processes that are required for it to become latent HTLV-infected CD4+T cells at time t. Further, the probability that latent HTLV-infected CD4+T cells survived time units before transmitted to active HTLV-infected cells at time t is given by the factor Λ5()e5. Furthermore, the term Λ6()e6 refers to the probability that new immature HIV particles at time t lost time units and become mature at time t. Here i, i=1,2,,6, are positive constants. The delay parameter is randomly taken from a probability distribution function Λi() over the time interval [0,κi], i=1,2,,6, where κi is the limit superior of this delay period. The function Λi(), i=1,2,,6 satisfies Λi()>0 and

0κiΛi()d=1and0κiΛi()eud<,

where u>0. Let us denote

H¯i()=Λi()eiandHi=0κiH¯i()d,

where i=1,2,,6. Thus 0<Hi1, i=1,2,6.

According to [28], we assume that r<min{ϱ,ω,δ}. This yields δ(1K)r>0. Let r=Kr and δ=δ(1K)r. Then system (4) becomes

{dS(t)dt=ηϱS(t)ϑ1S(t)V(t)ϑ2S(t)I(t)ϑ3S(t)Y(t),dL(t)dt=(1β)0κ1H¯1()S(t)[ϑ1V(t)+ϑ2I(t)]d(λ+γ)L(t),dI(t)dt=β0κ2H¯2()S(t)[ϑ1V(t)+ϑ2I(t)]ddI(t)dt=+λ0κ3H¯3()L(t)daI(t)μ1CI(t)I(t),dE(t)dt=φϑ30κ4H¯4()S(t)Y(t)d+rY(t)(ψ+ω)E(t),dY(t)dt=ψ0κ5H¯5()E(t)dδY(t)μ2CY(t)Y(t),dV(t)dt=b0κ6H¯6()I(t)dεV(t),dCI(t)dt=σ1CI(t)I(t)π1CI(t),dCY(t)dt=σ2CY(t)Y(t)π2CY(t). 5

The initial conditions of system (5) are given by:

S(x)=ϵ1(x),L(x)=ϵ2(x),I(x)=ϵ3(x),E(x)=ϵ4(x),Y(x)=ϵ5(x),V(x)=ϵ6(x),CI(x)=ϵ7(x),CY(x)=ϵ8(x),ϵj(x)0,x[κ,0],j=1,2,,8,κ=max{κ1,κ2,κ3,κ4,κ5,κ6}, 6

where ϵj(x)C([κ,0],R0), j=1,2,,8, and C=C([κ,0],R0) is the Banach space of continuous functions mapping the interval [κ,0] into R0 with norm ϵj=supκq0|ϵj(q)| for ϵjC. Therefore, system (5) with initial conditions (6) has a unique solution by using the standard theory of functional differential equations [61, 62].

Well-posedness of solutions

Proposition 1

All solutions of system (5) with initial conditions (6) are nonnegative and ultimately bounded.

Proof

From the first equation of system (5), we have dS(t)dt|S=0=η>0, then S(t)>0 for all t0. Moreover, the rest of equations of system (5) give us the following:

L(t)=ϵ2(0)e(λ+γ)t+(1β)0te(λ+γ)(tϰ)0κ1H¯1()S(ϰ)[ϑ1V(ϰ)+ϑ2I(ϰ)]ddϰ,I(t)=ϵ3(0)e0t(a+μ1CI(y))dy+0teϰt(a+μ1CI(y))dy[β0κ2H¯2()S(ϰ)[ϑ1V(ϰ)+ϑ2I(ϰ)]d+λ0κ3H¯3()L(ϰ)d]dϰ,E(t)=ϵ4(0)e(ψ+ω)t+0te(ψ+ω)(tϰ)[φϑ30κ4H¯4()S(ϰ)Y(ϰ)d+rY(ϰ)]dϰ,Y(t)=ϵ5(0)e0t(δ+μ2CY(y))dy+ψ0teϰt(δ+μ2CY(y))dy0κ5H¯5()E(ϰ)ddϰ,V(t)=ϵ6(0)eεt+b0teε(tϰ)0κ6H¯6()I(ϰ)ddϰ,CI(t)=ϵ7(0)e0t(π1σ1I(y))dy,CY(t)=ϵ8(0)e0t(π2σ2Y(y))dy.

Therefore, L(t),I(t),E(t),Y(t),V(t),CI(t),CY(t)0 for all t[0,κ]. Thus, by a recursive argument, we get S(t),L(t),I(t),E(t),Y(t),V(t),CI(t),CY(t)0 for all t0. Hence, the solutions of system (5) with initial conditions (6) satisfy (S(t),L(t),I(t),E(t),Y(t),V(t),CI(t),CY(t))R08 for all t0. Next, we establish the boundedness of the model’s solutions. The nonnegativity of the model’s solution implies that lim suptS(t)ηϱ. To show the ultimate boundedness of L(t), we let

Ψ1(t)=(1β)0κ1H¯1()S(t)d+L(t).

Then

dΨ1(t)dt=(1β)[0κ1H¯1(){ηϱS(t)}dϑ30κ1H¯1()S(t)Y(t)d](λ+γ)L(t)=(1β)[ηH1ϱ0κ1H¯1()S(t)dϑ30κ1H¯1()S(t)Y(t)d](λ+γ)L(t)ηH1(1β)ϱ(1β)0κ1H¯1()S(t)d(λ+γ)L(t)η(1β)ϕ1[(1β)0κ1H¯1()S(t)d+L(t)]=η(1β)ϕ1Ψ1(t),

where ϕ1=min{ϱ,λ+γ}. It follows that lim suptΨ1(t)Ω1, where Ω1=η(1β)ϕ1. Since 0κ1H¯1()S(t)d and L(t) are nonnegative, then lim suptL(t)Ω1. Further, we let

Ψ2(t)=β0κ2H¯2()S(t)d+I(t)+μ1σ1CI(t).

Then we obtain

dΨ2(t)dt=β[0κ2H¯2(){ηϱS(t)}dϑ30κ2H¯2()S(t)Y(t)d]+λ0κ3H¯3()L(t)daI(t)μ1π1σ1CI(t)=β[ηH2ϱ0κ2H¯2()S(t)dϑ30κ2H¯2()S(t)Y(t)d]+λ0κ3H¯3()L(t)daI(t)μ1π1σ1CI(t)ηβH2+λΩ1H3ϱβ0κ2H¯2()S(t)daI(t)μ1π1σ1CI(t)ηβ+λΩ1ϕ2[β0κ2H¯2()S(t)d+I(t)+μ1σ1CI(t)]=ηβ+λΩ1ϕ2Ψ2(t),

where ϕ2=min{ϱ,a,π1}. It follows that lim suptΨ2(t)Ω2, where Ω2=ηβ+λΩ1ϕ2. Since 0κ2H¯2()S(t)d, I(t) and CI(t) are nonnegative, then lim suptI(t)Ω2 and lim suptCI(t)Ω3, where Ω3=σ1Ω2μ1. Furthermore, we let

Ψ3(t)=0κ4H¯4()S(t)d+1φ[E(t)+Y(t)]+ψφ0κ5Λ5()tte5(tϰ)E(ϰ)dϰd+μ2σ2φCY(t).

Then

dΨ3(t)dt=0κ4H¯4()[ηϱS(t)S(t){ϑ1V(t)+ϑ2I(t)+ϑ3Y(t)}]d+ϑ30κ4H¯4()S(t)Y(t)d+rφY(t)ψ+ωφE(t)+ψφ0κ5H¯5()E(t)dδφY(t)5ψφ0κ5Λ5()tte5(tϰ)E(ϰ)dϰd+ψφE(t)0κ5Λ5()dψφ0κ5Λ5()e5E(t)dμ2π2σ2φCY(t)0κ4H¯4()[ηϱS(t)]dωφE(t)+(rφδφ)Y(t)5ψφ0κ5Λ5()tte5(tϰ)E(ϰ)dϰdμ2π2σ2φCY(t)ηϱ0κ4H¯4()S(t)dωφE(t)δrφY(t)5ψφ0κ5Λ5()tte5(tϰ)E(ϰ)dϰdμ2π2σ2φCY(t).

Since δr=δr>0, then

dΨ3(t)dtηϱ0κ4H¯4()S(t)dωφE(t)δrφY(t)5ψφ0κ5Λ5()tte5(tϰ)E(ϰ)dϰdμ2π2σ2φCY(t)ηϕ3[0κ4H¯4()S(t)d+1φ{E(t)+Y(t)}+ψφ0κ5Λ5()tte5(tϰ)E(ϰ)dϰd+μ2σ2φCY(t)]=ηϕ3Ψ3(t),

where ϕ3=min{ϱ,ω,δr,5,π2}. It follows that lim suptΨ3(t)ηϕ3. Since 0κ4H¯4()S(t)d0, E(t)0, Y(t)0, and CY(t)0, then lim suptE(t)Ω4, lim suptY(t)Ω4, and lim suptCY(t)Ω5, where Ω4=ηφϕ3 and Ω5=ησ2φμ2ϕ3. Finally, from the sixth equation of system (5), we have

dV(t)dt=b0κ6H¯6()I(t)dεV(t)bH6Ω2εV(t)bΩ2εV(t).

This implies that lim suptV(t)Ω6, where Ω6=bΩ2ε. □

According to Proposition 1, we can show that the region

Θ={(S,L,I,E,Y,V,CI,CY)C08:SΩ1,LΩ1,IΩ2,EΩ4,YΩ4,VΩ6,CIΩ3,CYΩ5}

is positively invariant with respect to system (5).

Steady states analysis

In this section, we calculate all possible steady states of the model and derive the threshold parameters which guarantee the existence of the steady states. Let us define

P=λH1H3(1β)+βH2(γ+λ), 7

which will be used throughout the paper. Let (S,L,I,E,Y,V,CI,CY) be any steady state of system (5) satisfying the following equations:

0=ηϱSϑ1SVϑ2SIϑ3SY,0=H1(1β)(ϑ1SV+ϑ2SI)(λ+γ)L,0=βH2(ϑ1SV+ϑ2SI)+λH3LaIμ1CII,0=φϑ3H4SY+rY(ψ+ω)E,0=ψH5EδYμ2CYY,0=bH6IεV,0=(σ1Iπ1)CI,0=(σ2Yπ2)CY.

We find that system (5) has eight possible steady states.

(i) Infection-free steady state, Inline graphic, where S0=η/ϱ. In this case, the body is free from HTLV and HIV.

(ii) Persistent HIV monoinfection steady state with ineffective immune response, Inline graphic, where

S1=S01,L1=aεϱH1(1β)P(bϑ1H6+εϑ2)(11),I1=εϱbϑ1H6+εϑ2(11),V1=ϱbH6bϑ1H6+εϑ2(11), 8

and 1 is the basic HIV monoinfection reproduction number for system (5) and is defined as follows:

1=PS0(bϑ1H6+εϑ2)aε(γ+λ)=11+12,

where

11=PS0bϑ1H6aε(γ+λ),12=PS0ϑ2a(γ+λ).

The parameter 1 determines whether or not a persistent HIV infection can be established. In fact, 11 measures the average number of secondary HIV infected generation caused by an existing free HIV particle due to free-to-cell transmission, while 12 measures the average numbers of secondary HIV infected generation caused by living active HIV-infected cells due to infected-to-cell transmission. The steady state Inline graphic describes the case of persistent HIV monoinfection without immune response.

(iii) Persistent HTLV monoinfection steady state with ineffective immune response, Inline graphic, where

S2=S02,E2=ϱδϑ3ψH5(21),Y2=ϱϑ3(21),

and 2 is the basic HTLV monoinfection reproduction number for system (5) and is defined as follows:

2=φϑ3ψH4H5S0(δrH5)ψ+δω.

The parameter 2 decides whether or not a persistent HTLV infection can be established. The steady state Inline graphic describes a persistent HTLV monoinfection without immune response.

We mention that 1 and 2 state the threshold dynamics of infection-free equilibrium Inline graphic and can be calculated by different methods such as (a) the next-generation matrix method of van den Driessche and Watmough [63], (b) local stability of the infection-free equilibrium Inline graphic, and (c) the existence of the chronic HIV and HTLV monoinfection equilibria with inactive immune response. In the present paper, we derive 1 and 2 by method (c).

(iv) Persistent HIV monoinfection steady state with only effective HIV-specific CTL, Inline graphic, where

S3=εσ1ηπ1(bϑ1H6+εϑ2)+ϱεσ1,L3=ηπ1H1(1β)(bϑ1H6+εϑ2)(γ+λ)[π1(bϑ1H6+εϑ2)+ϱεσ1],I3=π1σ1,V3=bH6εI3=bπ1H6εσ1,C3I=aμ1(31),

and

3=σ1ηP(bϑ1H6+εϑ2)a(γ+λ)[π1(bϑ1H6+εϑ2)+ϱεσ1],

is the HIV-specific CTL immunity reproduction number in case of HIV monoinfection. The parameter 3 determines whether or not the HIV-specific CTL immune response is effective in the absence of HTLV.

(v) Persistent HTLV monoinfection steady state with only effective HTLV-specific CTL, Inline graphic, where

S4=σ2ηπ2ϑ3+ϱσ2,Y4=π2σ2,E4=π2[r(π2ϑ3+ϱσ2)+ϑ3ηφσ2H4]σ2(ψ+ω)(π2ϑ3+ϱσ2),C4Y=(δrH5)ψ+δωμ2(ψ+ω)(41),

and 4 is the HTLV-specific CTL immunity reproduction number in case of HTLV monoinfection and is stated as follows:

4=σ2ηφϑ3ψH4H5(π2ϑ3+ϱσ2)[(δrH5)ψ+δω].

The parameter 4 determines whether or not the HTLV-specific CTL immune response is effective in the absence of HIV.

(vi) Persistent HTLV-HIV coinfection steady state with only effective HIV-specific CTL, Inline graphic, where

S5=(δrH5)ψ+δωφϑ3ψH4H5=S2,I5=π1σ1=I3,V5=bπ1H6εσ1=V3,L5=π1H1(1β)(bϑ1H6+εϑ2)[(δrH5)ψ+δω]εϑ3σ1φψH4H5(γ+λ),E5=δ[π1(bϑ1H6+εϑ2)+ϱεσ1]εϑ3σ1ψH5(51),Y5=π1(bϑ1H6+εϑ2)+ϱεσ1εϑ3σ1(51),C5I=aμ1(121),

where

5=ηφεϑ3σ1ψH4H5[(δrH5)ψ+δω][π1(bϑ1H6+εϑ2)+ϱεσ1].

Here, the parameter 5 is the HTLV infection reproduction number in the presence of HIV infection and determines whether or not HIV-infected patients could be dually infected with HTLV.

(vii) Persistent HTLV-HIV coinfection steady state with only effective HTLV-specific CTL, Inline graphic, where

S6=aε(γ+λ)P(bϑ1H6+εϑ2)=S1,Y6=π2σ2=Y4,L6=aεH1(1β)(π2ϑ3+ϱσ2)σ2P(bϑ1H6+εϑ2)(61),I6=ε(π2ϑ3+ϱσ2)σ2(bϑ1H6+εϑ2)(61),E6=π2[rP(bϑ1H6+εϑ2)+aεϑ3φH4(γ+λ)]σ2P(ψ+ω)(bϑ1H6+εϑ2),V6=bH6(π2ϑ3+ϱσ2)σ2(bϑ1H6+εϑ2)(61),C6Y=(δrH5)ψ+δωμ2(ψ+ω)(211),

and

6=ησ2P(bϑ1H6+εϑ2)aε(γ+λ)(π2ϑ3+ϱσ2),

is the HIV infection reproduction number in the presence of HTLV infection. It is clear that 6 determines whether or not HTLV-infected patients could be dually infected with HIV.

(viii) Persistent HTLV-HIV coinfection steady state with effective HIV-specific CTL and HTLV-specific CTL, Inline graphic, where

S7=εσ1σ2ηπ1σ2(bϑ1H6+εϑ2)+εσ1(π2ϑ3+ϱσ2),L7=π1σ2ηH1(1β)(bϑ1H6+εϑ2)(γ+λ)[π1σ2(bϑ1H6+εϑ2)+εσ1(π2ϑ3+ϱσ2)],E7=π2[ϑ3εσ1σ2ηφH4+r{π1σ2(bϑ1H6+εϑ2)+εσ1(π2ϑ3+ϱσ2)}]σ2(ψ+ω)[π1σ2(bϑ1H6+εϑ2)+εσ1(π2ϑ3+ϱσ2)],I7=π1σ1=I3=I5,Y7=π2σ2=Y4=Y6,V7=bπ1H6εσ1=V3=V5,C7I=aμ1(71),C7Y=(δrH5)ψ+δωμ2(ψ+ω)(81),

and

7=σ1σ2ηP(bϑ1H6+εϑ2)a(γ+λ)[π1σ2(bϑ1H6+εϑ2)+εσ1(π2ϑ3+ϱσ2)],8=εϑ3ησ1σ2φψH4H5[π1σ2(bϑ1H6+εϑ2)+εσ1(π2ϑ3+ϱσ2)][(δrH5)ψ+δω].

The parameter 7 is the competed HIV-specific CTL immunity reproduction number in case of HTLV-HIV coinfection. The parameter 8 is the competed HTLV-specific CTL immunity reproduction number in case of HTLV-HIV coinfection. Clearly, Inline graphic exists when 7>1 and 8>1.

Global stability analysis

In this section, we use the Lyapunov method to show the global asymptotic stability of the model’s steady states. For formation of Lyapunov functionals, we follow the works [64, 65]. Denote U=U(t), where U(S,L,I,E,Y,V,CI,CY).

Let a function Φj(S,L,I,E,Y,V,CI,CY) and ϒj be the largest invariant subset of

ϒj={(S,L,I,E,Y,V,CI,CY):dΦjdt=0},j=0,1,2,,7.

We define a function Ϝ(x)=x1lnx.

Theorem 1

If 11 and 21, then Inline graphic is globally asymptotically stable (GAS).

Proof

We define a Lyapunov functional as follows:

Φ0=PS0Ϝ(SS0)+λH3L+(γ+λ)I+PφH4E+P(ψ+ω)φψH4H5Y+Pϑ1S0εV+μ1(γ+λ)σ1CI+μ2P(ψ+ω)σ2φψH4H5CY+λH3(1β)0κ1H¯1()ttS(ϰ)×[ϑ1V(ϰ)+ϑ2I(ϰ)]dϰd+β(γ+λ)0κ2H¯2()ttS(ϰ)[ϑ1V(ϰ)+ϑ2I(ϰ)]dϰd+λ(γ+λ)0κ3H¯3()ttL(ϰ)dϰd+Pϑ3H40κ4H¯4()ttS(ϰ)Y(ϰ)dϰd+P(ψ+ω)φH4H50κ5H¯5()ttE(ϰ)dϰd+bPϑ1S0ε0κ6H¯6()ttI(ϰ)dϰd.

Clearly, Φ0(S,L,I,E,Y,V,CI,CY)>0 for all S,L,I,E,Y,V,CI,CY>0, and Φ0(S0,0,0,0,0,0,0,0)=0. We calculate dΦ0dt along the solutions of model (5) as follows:

dΦ0dt=P(1S0S)(ηϱSϑ1SVϑ2SIϑ3SY)+λH3[(1β)0κ1H¯1()S(t)×{ϑ1V(t)+ϑ2I(t)}d(λ+γ)L]+(γ+λ)[β0κ2H¯2()S(t)×{ϑ1V(t)+ϑ2I(t)}d+λ0κ3H¯3()L(t)daIμ1CII]+PφH4[φϑ30κ4H¯4()S(t)Y(t)d+rY(ψ+ω)E]+P(ψ+ω)φψH4H5[ψ0κ5H¯5()E(t)dδYμ2CYY]+Pϑ1S0ε[b0κ6H¯6()I(t)dεV]+μ1(γ+λ)σ1(σ1CIIπ1CI)+μ2P(ψ+ω)σ2φψH4H5(σ2CYYπ2CY)+P(ϑ1SV+ϑ2SI)λH3(1β)0κ1H¯1()×S(t)[ϑ1V(t)+ϑ2I(t)]dβ(γ+λ)0κ2H¯2()S(t)×[ϑ1V(t)+ϑ2I(t)]d+λ(γ+λ)0κ3H¯3()[LL(t)]d+Pϑ3H40κ4H¯4()[SYS(t)Y(t)]d+P(ψ+ω)φH4H50κ5H¯5()×[EE(t)]d+bPϑ1S0ε0κ6H¯6()[II(t)]d. 9

Summing the terms of Eq. (9), we obtain

dΦ0dt=P(1S0S)(ηϱS)+Pϑ2S0Ia(λ+γ)I+bPϑ1H6S0εI+Pϑ3S0YP[(δrH5)ψ+δω]φψH4H5Yμ1π1(γ+λ)σ1CIμ2π2P(ψ+ω)σ2φψH4H5CY.

Using S0=η/ϱ, we obtain

dΦ0dt=ϱP(SS0)2S+a(λ+γ)(11)I+P[(δrH5)ψ+δω]φψH4H5(21)Yμ1π1(γ+λ)σ1CIμ2π2P(ψ+ω)σ2φψH4H5CY.

Since r<δ and 0<H51, then δrH5>0. Therefore, dΦ0dt0 for all S,I,Y,CI,CY>0; moreover, dΦ0dt=0 when (S(t),I(t),Y(t),CI(t),CY(t))=(S0,0,0,0,0). The solutions of system (5) converge to ϒ0. The set ϒ0 includes elements with (S(t),I(t),Y(t),CI(t),CY(t))=(S0,0,0,0,0). Then dS(t)dt=dY(t)dt=0 and the first and fifth equations of system (5) become

0=dS(t)dt=ηϱS0ϑ1S0V(t),0=dY(t)dt=ψ0κ5H¯5()E(t)d,

which give V(t)=E(t)=0 for all t. In addition, we have dI(t)dt=0, and from the third equation of system (5) we have

0=dI(t)dt=λ0κ3H¯3()L(t)d,

which yields L(t)=0 for all t and hence Inline graphic. Applying Lyapunov–LaSalle asymptotic stability (LLAS) theorem [6668], we get that Inline graphic is GAS. □

The following equalities are needed in the next theorems:

ln(S(t)V(t)SV)=[ln(S(t)V(t)LnSnVnL)+ln(SnS)+ln(VnLVLn)],ln(S(t)V(t)SV)=[ln(S(t)V(t)InSnVnI)+ln(SnS)+ln(VnIVIn)],ln(S(t)I(t)SI)=[ln(S(t)I(t)LnSnInL)+ln(SnS)+ln(InLILn)],ln(S(t)I(t)SI)=[ln(S(t)I(t)SnI)+ln(SnS)],ln(L(t)L)=ln(L(t)InLnI)+ln(LnILIn),ln(I(t)I)=ln(I(t)VnInV)+ln(InVIVn),where n=1,3,5,6,7. 10

Further,

ln(S(t)Y(t)SY)=ln(S(t)Y(t)EmSmYmE)+ln(SmS)+ln(YmEYEm),ln(E(t)E)=ln(E(t)YmEmY)+ln(YEmYmE),where m=2,4,5,6,7. 11

Theorem 2

Let 1>1, 2/11, and 31, then Inline graphic is GAS.

Proof

Define a functional as follows:

Φ1=PS1Ϝ(SS1)+λH3L1Ϝ(LL1)+(γ+λ)I1Ϝ(II1)+PφH4E+P(ψ+ω)φψH4H5Y+Pϑ1S1εV1Ϝ(VV1)+μ1(γ+λ)σ1CI+μ2P(ψ+ω)σ2φψH4H5CY+ϑ1λH3(1β)S1V10κ1H¯1()ttϜ(S(ϰ)V(ϰ)S1V1)dϰd+ϑ2λH3(1β)S1I10κ1H¯1()ttϜ(S(ϰ)I(ϰ)S1I1)dϰd+ϑ1β(γ+λ)S1V10κ2H¯2()ttϜ(S(ϰ)V(ϰ)S1V1)dϰd+ϑ2β(γ+λ)S1I10κ2H¯2()ttϜ(S(ϰ)I(ϰ)S1I1)dϰd+λ(γ+λ)L10κ3H¯3()ttϜ(L(ϰ)L1)dϰd+Pϑ3H40κ4H¯4()ttS(ϰ)Y(ϰ)dϰd+P(ψ+ω)φH4H50κ5H¯5()ttE(ϰ)dϰd+bPϑ1S1I1ε0κ6H¯6()ttϜ(I(ϰ)I1)dϰd.

Calculate dΦ1dt as follows:

dΦ1dt=P(1S1S)(ηϱSϑ1SVϑ2SIϑ3SY)+λH3(1L1L)×[(1β)0κ1H¯1()S(t){ϑ1V(t)+ϑ2I(t)}d(λ+γ)L]+(γ+λ)(1I1I)[β0κ2H¯2()S(t){ϑ1V(t)+ϑ2I(t)}d+λ0κ3H¯3()L(t)daIμ1CII]+PφH4[φϑ30κ4H¯4()S(t)Y(t)d+rY(ψ+ω)E]+P(ψ+ω)φψH4H5[ψ0κ5H¯5()E(t)dδYμ2CYY]+Pϑ1S1ε(1V1V)[b0κ6H¯6()I(t)dεV]+μ1(γ+λ)σ1(σ1CIIπ1CI)+μ2P(ψ+ω)σ2φψH4H5(σ2CYYπ2CY)+ϑ1λH3(1β)S1V1×0κ1H¯1()[SVS1V1S(t)V(t)S1V1+ln(S(t)V(t)SV)]d+ϑ2λH3(1β)S1I1×0κ1H¯1()[SIS1I1S(t)I(t)S1I1+ln(S(t)I(t)SI)]d+ϑ1β(γ+λ)S1V1×0κ2H¯2()[SVS1V1S(t)V(t)S1V1+ln(S(t)V(t)SV)]d+ϑ2β(γ+λ)S1I1×0κ2H¯2()[SIS1I1S(t)I(t)S1I1+ln(S(t)I(t)SI)]d+λ(γ+λ)L10κ3H¯3()[LL1L(t)L1+ln(L(t)L)]d+Pϑ3H40κ4H¯4()[SYS(t)Y(t)]d+P(ψ+ω)φH4H50κ5H¯5()[EE(t)]d+bPϑ1S1I1ε0κ6H¯6()[II1I(t)I1+ln(I(t)I)]d. 12

Summing the terms of Eq. (12), we get

dΦ1dt=P(1S1S)(ηϱS)+Pϑ2S1I+Pϑ3S1Yϑ1λH3(1β)0κ1H¯1()S(t)V(t)L1Ldϑ2λH3(1β)0κ1H¯1()S(t)I(t)L1Ld+λH3(λ+γ)L1a(λ+γ)Iϑ1β(γ+λ)0κ2H¯2()S(t)V(t)I1Idϑ2β(γ+λ)0κ2H¯2()S(t)I(t)I1Idλ(λ+γ)0κ3H¯3()L(t)I1Id+a(λ+γ)I1+μ1(λ+γ)CII1+PrφH4YPδ(ψ+ω)φψH4H5YPbϑ1S1ε0κ6H¯6()I(t)V1Vd+Pϑ1S1V1μ1π1(λ+γ)σ1CIμ2π2P(ψ+ω)σ2φψH4H5CY+λH3(1β)ϑ1S1V10κ1H¯1()ln(S(t)V(t)SV)d+λH3(1β)ϑ2S1I10κ1H¯1()ln(S(t)I(t)SI)d+β(γ+λ)ϑ1S1V10κ2H¯2()ln(S(t)V(t)SV)d+β(γ+λ)ϑ2S1I10κ2H¯2()ln(S(t)I(t)SI)d+λ(γ+λ)L10κ3H¯3()ln(L(t)L)d+bPϑ1S1H6εI+Pbϑ1S1I1ε0κ6H¯6()ln(I(t)I)d.

The steady state conditions for Inline graphic are given by

η=ϱS1+ϑ1S1V1+ϑ2S1I1,H1(1β)(ϑ1S1V1+ϑ2S1I1)=(λ+γ)L1,βH2(ϑ1S1V1+ϑ2S1I1)+λH3L1=aI1V1=bH6I1ε.

Then we get

P(ϑ1S1V1+ϑ2S1I1)=a(λ+γ)I1.

Further, we obtain

dΦ1dt=P(1S1S)(ϱS1ϱS)+P(ϑ1S1V1+ϑ2S1I1)(1S1S)+Pϑ3S1YλH3(1β)ϑ1S1V10κ1H¯1()S(t)V(t)L1S1V1LdλH3(1β)ϑ2S1I10κ1H¯1()S(t)I(t)L1S1I1Ld+λH1H3(1β)(ϑ1S1V1+ϑ2S1I1)β(γ+λ)ϑ1S1V10κ2H¯2()S(t)V(t)I1S1V1Idβ(γ+λ)ϑ2S1I10κ2H¯2()S(t)I(t)S1IdλH1(1β)(ϑ1S1V1+ϑ2S1I1)0κ3H¯3()L(t)I1L1Id+P(ϑ1S1V1+ϑ2S1I1)+μ1(λ+γ)CII1P[(δrH5)ψ+δω]φψH4H5YPϑ1S1V1H60κ6H¯6()I(t)V1I1Vd+Pϑ1S1V1μ1π1(λ+γ)σ1CIμ2π2P(ψ+ω)σ2φψH4H5CY+λH3(1β)ϑ1S1V10κ1H¯1()ln(S(t)V(t)SV)d+λH3(1β)ϑ2S1I10κ1H¯1()ln(S(t)I(t)SI)d+β(γ+λ)ϑ1S1V10κ2H¯2()ln(S(t)V(t)SV)d+β(γ+λ)ϑ2S1I10κ2H¯2()ln(S(t)I(t)SI)d+λH1(1β)(ϑ1S1V1+ϑ2S1I1)0κ3H¯3()ln(L(t)L)d+Pϑ1S1V1H60κ6H¯6()ln(I(t)I)d.

Using the equalities given by (10) in case of n=1, we get

dΦ1dt=ϱP(SS1)2SP(ϑ1S1V1+ϑ2S1I1)[S1S1ln(S1S)]λH3(1β)ϑ1S1V1×0κ1H¯1()[S(t)V(t)L1S1V1L1ln(S(t)V(t)L1S1V1L)]dλH3(1β)ϑ2S1I1×0κ1H¯1()[S(t)I(t)L1S1I1L1ln(S(t)I(t)L1S1I1L)]dβ(γ+λ)ϑ1S1V1×0κ2H¯2()[S(t)V(t)I1S1V1I1ln(S(t)V(t)I1S1V1I)]dβ(γ+λ)ϑ2S1I10κ2H¯2()[S(t)I(t)S1I1ln(S(t)I(t)S1I)]dλH1(1β)(ϑ1S1V1+ϑ2S1I1)×0κ3H¯3()[L(t)I1L1I1ln(L(t)I1L1I)]dPϑ1S1V1H60κ6H¯6()[I(t)V1I1V1ln(I(t)V1I1V)]d+P[(δrH5)ψ+δω]φψH4H5(ϑ3φψH4H5S1(δrH5)ψ+δω1)Y+μ1(λ+γ)(I1π1σ1)CIμ2π2P(ψ+ω)σ2φψH4H5CY. 13

Therefore, Eq. (13) becomes

dΦ1dt=ϱP(SS1)2SP(ϑ1S1V1+ϑ2S1I1)Ϝ(S1S)λH3(1β)ϑ1S1V10κ1H¯1()Ϝ(S(t)V(t)L1S1V1L)dλH3(1β)ϑ2S1I10κ1H¯1()Ϝ(S(t)I(t)L1S1I1L)dβ(γ+λ)ϑ1S1V10κ2H¯2()Ϝ(S(t)V(t)I1S1V1I)dβ(γ+λ)ϑ2S1I10κ2H¯2()Ϝ(S(t)I(t)S1I)dλH1(1β)(ϑ1S1V1+ϑ2S1I1)0κ3H¯3()Ϝ(L(t)I1L1I)dPϑ1S1V1H60κ6H¯6()Ϝ(I(t)V1I1V)d+P[(δrH5)ψ+δω]φψH4H5(211)Y+μ1(γ+λ)[π1(bϑ1H6+εϑ2)+ϱεσ1]σ1(bϑ1H6+εϑ2)(31)CIμ2π2P(ψ+ω)σ2φψH4H5CY.

Since 2/11 and 31, then dΦ1dt0 for all S,L,I,Y,V,CI,CY>0. Moreover, dΦ1dt=0 when S=S1 and Y=CI=CY=Ϝ=0. The solutions of system (5) converge to ϒ1 which includes elements that satisfy S(t)=S1 and Ϝ=0 i.e.

S(t)V(t)L1S1V1L=S(t)I(t)L1S1I1L=S(t)V(t)I1S1V1I=S(t)I(t)S1I=L(t)I1L1I=I(t)V1I1V=1 14

for all t[0,κ]. If S(t)=S1, then from Eq. (14) we get L(t)=L1, I(t)=I1, and V(t)=V1 for all t. Further, for each element of ϒ1, we have Y(t)=0 and then dY(t)dt=0. The fifth equation of system (5) becomes

0=dY(t)dt=ψ0κ5H¯5()E(t)d,

which provides E(t)=0 for all t, and hence Inline graphic. Therefore, using LLAS theorem we get that Inline graphic is GAS. □

Theorem 3

If 2>1, 1/21, and 41, then Inline graphic is GAS.

Proof

Define

Φ2=PS2Ϝ(SS2)+λH3L+(γ+λ)I+PφH4E2Ϝ(EE2)+P(ψ+ω)φψH4H5Y2Ϝ(YY2)+Pϑ1S2εV+μ1(γ+λ)σ1CI+μ2P(ψ+ω)σ2φψH4H5CY+λH3(1β)0κ1H¯1()ttS(ϰ)×[ϑ1V(ϰ)+ϑ2I(ϰ)]dϰd+β(γ+λ)0κ2H¯2()ttS(ϰ)[ϑ1V(ϰ)+ϑ2I(ϰ)]dϰd+λ(γ+λ)0κ3H¯3()ttL(ϰ)dϰd+Pϑ3S2Y2H40κ4H¯4()ttϜ(S(ϰ)Y(ϰ)S2Y2)dϰd+P(ψ+ω)E2φH4H50κ5H¯5()ttϜ(E(ϰ)E2)dϰd+bPϑ1S2ε0κ6H¯6()ttI(ϰ)dϰd.

We calculate dΦ2dt as follows:

dΦ2dt=P(1S2S)(ηϱSϑ1SVϑ2SIϑ3SY)+λH3[(1β)0κ1H¯1()S(t){ϑ1V(t)+ϑ2I(t)}d(λ+γ)L]+(γ+λ)[β0κ2H¯2()S(t){ϑ1V(t)+ϑ2I(t)}d+λ0κ3H¯3()L(t)daIμ1CII]+PφH4(1E2E)[φϑ30κ4H¯4()S(t)Y(t)d+rY(ψ+ω)E]+P(ψ+ω)φψH4H5(1Y2Y)[ψ0κ5H¯5()E(t)dδYμ2CYY]+Pϑ1S2ε×[b0κ6H¯6()I(t)dεV]+μ1(γ+λ)σ1(σ1CIIπ1CI)+μ2P(ψ+ω)σ2φψH4H5×(σ2CYYπ2CY)+P(ϑ1SV+ϑ2SI)λH3(1β)0κ1H¯1()S(t)×[ϑ1V(t)+ϑ2I(t)]dβ(γ+λ)0κ2H¯2()S(t)[ϑ1V(t)+ϑ2I(t)]d+λ(γ+λ)0κ3H¯3()[LL(t)]d+Pϑ3S2Y2H40κ4H¯4()[SYS2Y2S(t)Y(t)S2Y2+ln(S(t)Y(t)SY)]d+P(ψ+ω)E2φH4H50κ5H¯5()[EE2E(t)E2+ln(E(t)E)]d+bPϑ1S2ε0κ6H¯6()[II(t)]d. 15

By collecting the terms of Eq. (15), we get

dΦ2dt=P[(1S2S)(ηϱS)+ϑ2S2I+ϑ3S2Ya(λ+γ)PI+rφH4Yϑ3H40κ4H¯4()S(t)Y(t)E2EdrφH4YE2E+ψ+ωφH4E2δ(ψ+ω)φψH4H5Yψ+ωφH4H50κ5H¯5()E(t)Y2Yd+δ(ψ+ω)φψH4H5Y2+μ2(ψ+ω)φψH4H5CYY2μ1π1(γ+λ)σ1PCIμ2π2(ψ+ω)σ2φψH4H5CY+ϑ3S2Y2H40κ4H¯4()ln(S(t)Y(t)SY)d+(ψ+ω)E2φH4H50κ5H¯5()ln(E(t)E)d+bϑ1S2H6εI].

Using the steady state conditions for Inline graphic

η=ϱS2+ϑ3S2Y2,ϑ3S2Y2+rY2φH4=(ψ+ω)E2φH4=δ(ψ+ω)Y2φψH4H5,

we obtain

dΦ2dt=P[(1S2S)(ϱS2ϱS)+ϑ3S2Y2(1S2S)ϑ3S2Y2H40κ4H¯4()S(t)Y(t)E2S2Y2EdrY2φH4YE2Y2E+ϑ3S2Y2+rY2φH4(ϑ3S2Y2H5+rY2φH4H5)0κ5H¯5()E(t)Y2E2Yd+ϑ3S2Y2+rY2φH4+μ2(ψ+ω)φψH4H5CYY2μ1π1(γ+λ)σ1PCIμ2π2(ψ+ω)σ2φψH4H5CY+ϑ3S2Y2H40κ4H¯4()ln(S(t)Y(t)SY)d+(ϑ3S2Y2H5+rY2φH4H5)0κ5H¯5()ln(E(t)E)d+a(λ+γ)P{PS2(εϑ2+bϑ1H6)aε(λ+γ)1}I].

Using the equalities given by (11) in case of m=2, we get

dΦ2dt=P[ϱ(SS2)2S+ϑ3S2Y2{S2S1ln(S2S)}+rY2φH4{YE2Y2E1ln(YE2Y2E)}+ϑ3S2Y2H40κ4H¯4(){S(t)Y(t)E2S2Y2E1ln(S(t)Y(t)E2S2Y2E)}d+(ϑ3S2Y2H5+rY2φH4H5)0κ5H¯5(){E(t)Y2E2Y1ln(E(t)Y2E2Y)}dμ1π1(γ+λ)σ1PCI+μ2(ψ+ω)φψH4H5(Y2π2σ2)CY]+a(λ+γ)(PS2(εϑ2+bϑ1H6)aε(λ+γ)1)I. 16

Therefore, Eq. (16) becomes

dΦ2dt=P[ϱ(SS2)2S+rY2φH4Ϝ(YE2Y2E)+ϑ3S2Y2H40κ4H¯4(){Ϝ(S(t)Y(t)E2S2Y2E)+Ϝ(S2S)}d+(ϑ3S2Y2H5+rY2φH4H5)0κ5H¯5()Ϝ(E(t)Y2E2Y)dμ1π1(γ+λ)σ1PCI+μ2(ψ+ω)(ϱσ2+ϑ3π2)ϑ3σ2φψH4H5(41)CY]+a(λ+γ)(121)I.

Thus, if 1/21 and 41, then dΦ2dt0 for all S,I,E,Y,CI,CY>0. Moreover, dΦ2dt=0 when (S,E,Y,I,CI,CY)=(S2,E2,Y2,0,0,0). The solutions of system (5) converge to ϒ2 which includes elements with (S(t),E(t),Y(t),I(t),CI(t),CY(t))=(S2,E2,Y2,0,0,0). Then we have dS(t)dt=0, and the first equation of system (5) becomes

0=dS(t)dt=ηϱS2ϑ1S2V(t)ϑ3S2Y2,

which yields V(t)=0 for all t. Moreover, we have dI(t)dt=0 and from the third equation of system (5) we get

0=dI(t)dt=λ0κ3H¯3()L(t)d,

which implies that L(t)=0 for all t. Therefore, Inline graphic. Applying LLAS theorem, we get Inline graphic is GAS. □

Theorem 4

Let 3>1 and 51, then Inline graphic is GAS.

Proof

Define a functional as follows:

Φ3=PS3Ϝ(SS3)+λH3L3Ϝ(LL3)+(γ+λ)I3Ϝ(II3)+PφH4E+P(ψ+ω)φψH4H5Y+Pϑ1S3εV3Ϝ(VV3)+μ1(γ+λ)σ1C3IϜ(CIC3I)+μ2P(ψ+ω)σ2φψH4H5CY+ϑ1λH3(1β)S3V30κ1H¯1()ttϜ(S(ϰ)V(ϰ)S3V3)dϰd+ϑ2λH3(1β)S3I3×0κ1H¯1()ttϜ(S(ϰ)I(ϰ)S3I3)dϰd+ϑ1β(γ+λ)S3V30κ2H¯2()×ttϜ(S(ϰ)V(ϰ)S3V3)dϰd+ϑ2β(γ+λ)S3I30κ2H¯2()ttϜ(S(ϰ)I(ϰ)S3I3)dϰd+λ(γ+λ)L30κ3H¯3()ttϜ(L(ϰ)L3)dϰd+Pϑ3H40κ4H¯4()ttS(ϰ)Y(ϰ)dϰd+P(ψ+ω)φH4H50κ5H¯5()ttE(ϰ)dϰd+bPϑ1S3I3ε0κ6H¯6()ttϜ(I(ϰ)I3)dϰd. 17

We calculate dΦ3dt as follows:

dΦ3dt=P(1S3S)(ηϱSϑ1SVϑ2SIϑ3SY)+λH3(1L3L)×[(1β)0κ1H¯1()S(t){ϑ1V(t)+ϑ2I(t)}d(λ+γ)L]+(γ+λ)(1I3I)[β0κ2H¯2()S(t){ϑ1V(t)+ϑ2I(t)}d+λ0κ3H¯3()L(t)daIμ1CII]+PφH4[φϑ30κ4H¯4()S(t)Y(t)d+rY(ψ+ω)E]+P(ψ+ω)φψH4H5[ψ0κ5H¯5()E(t)dδYμ2CYY]+Pϑ1S3ε(1V3V)[b0κ6H¯6()I(t)dεV]+μ1(γ+λ)σ1×(1C3ICI)(σ1CIIπ1CI)+μ2P(ψ+ω)σ2φψH4H5(σ2CYYπ2CY)+ϑ1λH3(1β)S3V3×0κ1H¯1()[SVS3V3S(t)V(t)S3V3+ln(S(t)V(t)SV)]d+ϑ2λH3(1β)S3I3×0κ1H¯1()[SIS3I3S(t)I(t)S3I3+ln(S(t)I(t)SI)]d+ϑ1β(γ+λ)S3V3×0κ2H¯2()[SVS3V3S(t)V(t)S3V3+ln(S(t)V(t)SV)]d+ϑ2β(γ+λ)S3I3×0κ2H¯2()[SIS3I3S(t)I(t)S3I3+ln(S(t)I(t)SI)]d+λ(γ+λ)L30κ3H¯3()[LL3L(t)L3+ln(L(t)L)]d+Pϑ3H40κ4H¯4()[SYS(t)Y(t)]d+P(ψ+ω)φH4H50κ5H¯5()[EE(t)]d+bPϑ1S3I3ε0κ6H¯6()[II3I(t)I3+ln(I(t)I)]d. 18

Collecting the terms of Eq. (18), we derive

dΦ3dt=P(1S3S)(ηϱS)+Pϑ2S3I+Pϑ3S3Yϑ1λH3(1β)0κ1H¯1()×S(t)V(t)L3Ldϑ2λH3(1β)0κ1H¯1()S(t)I(t)L3Ld+λH3(λ+γ)L3a(λ+γ)Iϑ1β(γ+λ)0κ2H¯2()S(t)V(t)I3Idϑ2β(γ+λ)0κ2H¯2()S(t)I(t)I3Idλ(λ+γ)0κ3H¯3()L(t)I3Id+a(λ+γ)I3+μ1(λ+γ)CII3+PrφH4YPδ(ψ+ω)φψH4H5YPbϑ1S3ε0κ6H¯6()I(t)V3Vd+Pϑ1S3V3μ1π1(λ+γ)σ1CIμ1(λ+γ)C3II+μ1π1(λ+γ)σ1C3Iμ2π2P(ψ+ω)σ2φψH4H5CY+λH3(1β)ϑ1S3V30κ1H¯1()ln(S(t)V(t)SV)d+λH3(1β)ϑ2S3I30κ1H¯1()ln(S(t)I(t)SI)d+β(γ+λ)ϑ1S3V30κ2H¯2()ln(S(t)V(t)SV)d+β(γ+λ)ϑ2S3I30κ2H¯2()ln(S(t)I(t)SI)d+λ(γ+λ)L30κ3H¯3()ln(L(t)L)d+bPϑ1S3H6εI+bPϑ1S3I3ε0κ6H¯6()ln(I(t)I)d.

Using the steady state conditions for Inline graphic

η=ϱS3+ϑ1S3V3+ϑ2S3I3,H1(1β)(ϑ1S3V3+ϑ2S3I3)=(λ+γ)L3,βH2(ϑ1S3V3+ϑ2S3I3)+λH3L3=(a+μ1C3I)I3,I3=π1σ1,V3=bH6εI3,

we get

P(ϑ1S3V3+ϑ2S3I3)=(λ+γ)(a+μ1C3I)I3.

Further, we obtain

dΦ3dt=P(1S3S)(ϱS3ϱS)+P(ϑ1S3V3+ϑ2S3I3)(1S3S)+Pϑ3S3YλH3(1β)ϑ1S3V30κ1H¯1()S(t)V(t)L3S3V3LdλH3(1β)ϑ2S3I3×0κ1H¯1()S(t)I(t)L3S3I3Ld+λH1H3(1β)(ϑ1S3V3+ϑ2S3I3)β(γ+λ)ϑ1S3V30κ2H¯2()S(t)V(t)I3S3V3Idβ(γ+λ)ϑ2S3I30κ2H¯2()×S(t)I(t)S3IdλH1(1β)(ϑ1S3V3+ϑ2S3I3)0κ3H¯3()L(t)I3L3Id+P(ϑ1S3V3+ϑ2S3I3)P[(δrH5)ψ+δω]φψH4H5YPϑ1S3V3H60κ6H¯6()×I(t)V3I3Vd+Pϑ1S3V3μ2π2P(ψ+ω)σ2φψH4H5CY+λH3(1β)ϑ1S3V3×0κ1H¯1()ln(S(t)V(t)SV)d+λH3(1β)ϑ2S3I30κ1H¯1()ln(S(t)I(t)SI)d+β(γ+λ)ϑ1S3V30κ2H¯2()ln(S(t)V(t)SV)d+β(γ+λ)ϑ2S3I30κ2H¯2()ln(S(t)I(t)SI)d+λH1(1β)(ϑ1S3V3+ϑ2S3I3)0κ3H¯3()ln(L(t)L)d+Pϑ1S3V3H60κ6H¯6()ln(I(t)I)d.

Using the equalities given by (10) in case of n=3, we get

dΦ3dt=ϱP(SS3)2SP(ϑ1S3V3+ϑ2S3I3)[S3S1ln(S3S)]λH3(1β)ϑ1S3V3×0κ1H¯1()[S(t)V(t)L3S3V3L1ln(S(t)V(t)L3S3V3L)]dλH3(1β)ϑ2S3I3×0κ1H¯1()[S(t)I(t)L3S3I3L1ln(S(t)I(t)L3S3I3L)]dβ(γ+λ)ϑ1S3V3×0κ2H¯2()[S(t)V(t)I3S3V3I1ln(S(t)V(t)I3S3V3I)]dβ(γ+λ)ϑ2S3I30κ2H¯2()[S(t)I(t)S3I1ln(S(t)I(t)S3I)]dλH1(1β)(ϑ1S3V3+ϑ2S3I3)×0κ3H¯3()[L(t)I3L3I1ln(L(t)I3L3I)]dPϑ1S3V3H60κ6H¯6()[I(t)V3I3V1ln(I(t)V3I3V)]d+Pϑ3(S3(δrH5)ψ+δωϑ3φψH4H5)Yμ2π2P(ψ+ω)σ2φψH4H5CY. 19

Therefore, Eq. (19) becomes

dΦ3dt=ϱP(SS3)2SP(ϑ1S3V3+ϑ2S3I3)Ϝ(S3S)λH3(1β)ϑ1S3V30κ1H¯1()Ϝ(S(t)V(t)L3S3V3L)dλH3(1β)ϑ2S3I30κ1H¯1()Ϝ(S(t)I(t)L3S3I3L)dβ(γ+λ)ϑ1S3V30κ2H¯2()Ϝ(S(t)V(t)I3S3V3I)dβ(γ+λ)ϑ2S3I30κ2H¯2()Ϝ(S(t)I(t)S3I)dλH1(1β)(ϑ1S3V3+ϑ2S3I3)0κ3H¯3()Ϝ(L(t)I3L3I)dPϑ1S3V3H60κ6H¯6()Ϝ(I(t)V3I3V)d+Pϑ3(S3S5)Yμ2π2P(ψ+ω)σ2φψH4H5CY.

Hence, if 51, then Inline graphic does not exist since E50 and Y50. In this case

dE(t)dt=φϑ30κ4H¯4()S(t)Y(t)d+rY(t)(ψ+ω)E(t)0,dY(t)dt=ψ0κ5H¯5()E(t)dδY(t)μ2CYY0.

Now we want to find the value such that, for all 0<S(t)S¯, we get dE(t)dt0 and dY(t)dt0. Let us consider

ddt[1H4E+ψ+ωψH4H5Y+φϑ3H40κ4H¯4()ttS(ϰ)Y(ϰ)dϰd+ψ+ωH4H50κ5H¯5()ttE(ϰ)dϰd]=φϑ3SY(δrH5)ψ+δωψH4H5Yμ2(ψ+ω)ψH4H5CYY=φϑ3(S(δrH5)ψ+δωφϑ3ψH4H5)Yμ2(ψ+ω)ψH4H5CYY0for all CY,Y>0.

This happens when S3S¯=(δrH5)ψ+δωϑ3φψH4H5=S5. Clearly, dΦ3dt0 for all S,L,I,Y,V,CY>0, where dΦ3dt=0 occurs at S=S3 and Y=CY=0. The solutions of system (5) converge to ϒ3 which includes elements satisfying S(t)=S3 and Ϝ=0 i.e.

S(t)V(t)L3S3V3L=S(t)I(t)L3S3I3L=S(t)V(t)I3S3V3I=S(t)I(t)S3I=L(t)I3L3I=I(t)V3I3V=1 20

for all t[0,κ]. If S(t)=S3, then from Eq. (20) we get L(t)=L3, I(t)=I3, and V(t)=V3 for all t. Thus, ϒ3 contains elements with I(t)=I3, V(t)=V3, Y(t)=0, and then dI(t)dt=0, dY(t)dt=0. The third and fifth equations of system (5) become

0=dI(t)dt=βH2(ϑ1S3V3+ϑ2S3I3)+λH3L3aI3μ1CI(t)I3,0=dY(t)dt=ψ0κ5H¯5E(t)d,

which yield CI(t)=C3I and E(t)=0 for all t. Therefore, Inline graphic. Applying LLAS theorem, we get Inline graphic is GAS. □

Theorem 5

If 4>1 and 61, then Inline graphic is GAS.

Proof

Let

Φ4=PS4Ϝ(SS4)+λH3L+(γ+λ)I+PφH4E4Ϝ(EE4)+P(ψ+ω)φψH4H5Y4Ϝ(YY4)+Pϑ1S4εV+μ1(γ+λ)σ1CI+μ2P(ψ+ω)σ2φψH4H5C4YϜ(CYC4Y)+λH3(1β)0κ1H¯1()ttS(ϰ)[ϑ1V(ϰ)+ϑ2I(ϰ)]dϰd+β(γ+λ)0κ2H¯2()ttS(ϰ)[ϑ1V(ϰ)+ϑ2I(ϰ)]dϰd+λ(γ+λ)0κ3H¯3()ttL(ϰ)dϰd+Pϑ3S4Y4H40κ4H¯4()ttϜ(S(ϰ)Y(ϰ)S4Y4)dϰd+P(ψ+ω)E4φH4H50κ5H¯5()ttϜ(E(ϰ)E4)dϰd+bPϑ1S4ε0κ6H¯6()ttI(ϰ)dϰd.

Calculate dΦ4dt as follows:

dΦ4dt=P(1S4S)(ηϱSϑ1SVϑ2SIϑ3SY)+λH3[(1β)0κ1H¯1()S(t){ϑ1V(t)+ϑ2I(t)}d(λ+γ)L]+(γ+λ)[β0κ2H¯2()S(t){ϑ1V(t)+ϑ2I(t)}d+λ0κ3H¯3()L(t)daIμ1CII]+PφH4(1E4E)[φϑ30κ4H¯4()S(t)Y(t)d+rY(ψ+ω)E]+P(ψ+ω)φψH4H5(1Y4Y)[ψ0κ5H¯5()E(t)dδYμ2CYY]+Pϑ1S4ε[b0κ6H¯6()I(t)dεV]+μ1(γ+λ)σ1(σ1CIIπ1CI)+μ2P(ψ+ω)σ2φψH4H5(1C4YCY)(σ2CYYπ2CY)+P(ϑ1SV+ϑ2SI)λH3(1β)0κ1H¯1()S(t)[ϑ1V(t)+ϑ2I(t)]dβ(γ+λ)0κ2H¯2()S(t)[ϑ1V(t)+ϑ2I(t)]d+λ(γ+λ)0κ3H¯3()[LL(t)]d+Pϑ3S4Y4H40κ4H¯4()[SYS4Y4S(t)Y(t)S4Y4+ln(S(t)Y(t)SY)]d+P(ψ+ω)E4φH4H50κ5H¯5()[EE4E(t)E4+ln(E(t)E)]d+bPϑ1S4ε0κ6H¯6()[II(t)]d. 21

Summing the terms of Eq. (21), we get

dΦ4dt=P[(1S4S)(ηϱS)+ϑ2S4I+ϑ3S4Ya(λ+γ)PI+rφH4Yϑ3H40κ4H¯4()S(t)Y(t)E4EdrφH4YE4E+ψ+ωφH4E4δ(ψ+ω)φψH4H5Yψ+ωφH4H50κ5H¯5()E(t)Y4Yd+δ(ψ+ω)φψH4H5Y4+μ2(ψ+ω)φψH4H5CYY4μ1π1(γ+λ)σ1PCIμ2π2(ψ+ω)σ2φψH4H5CYμ2(ψ+ω)φψH4H5C4YY+μ2π2(ψ+ω)σ2φψH4H5C4Y+ϑ3S4Y4H40κ4H¯4()ln(S(t)Y(t)SY)d+(ψ+ω)E4φH4H50κ5H¯5()ln(E(t)E)d+bϑ1S4H6εI].

Using the steady state conditions for Inline graphic

η=ϱS4+ϑ3S4Y4,Y4=π2σ2,ϑ3S4Y4+rY4φH4=(ψ+ω)E4φH4=δ(ψ+ω)φψH4H5Y4+μ2(ψ+ω)φψH4H5C4YY4,

we obtain

dΦ4dt=P[(1S4S)(ϱS4ϱS)+ϑ3S4Y4(1S4S)ϑ3S4Y4H40κ4H¯4()S(t)Y(t)E4S4Y4EdrY4φH4YE4Y4E+ϑ3S4Y4+rY4φH4(ϑ3S4Y4H5+rY4φH4H5)0κ5H¯5()E(t)Y4E4Yd+ϑ3S4Y4+rY4φH4μ1π1(γ+λ)σ1PCI+ϑ3S4Y4H40κ4H¯4()ln(S(t)Y(t)SY)d+(ϑ3S4Y4H5+rY4φH4H5)0κ5H¯5()ln(E(t)E)d+a(λ+γ)P{PS4(εϑ2+bϑ1H6)aε(λ+γ)1}I].

Using the equalities given by (11) in case of m=4, we get

dΦ4dt=P[ϱ(SS4)2S+ϑ3S4Y4{S4S1ln(S4S)}+rY4φH4{YE4Y4E1ln(YE4Y4E)}+ϑ3S4Y4H40κ4H¯4(){S(t)Y(t)E4S4Y4E1ln(S(t)Y(t)E4S4Y4E)}d+(ϑ3S4Y4H5+rY4φH4H5)0κ5H¯5(){E(t)Y4E4Y1ln(E(t)Y4E4Y)}d+a(λ+γ)P{PS4(εϑ2+bϑ1H6)aε(λ+γ)1}Iμ1π1(γ+λ)σ1PCI]. 22

Therefore, Eq. (22) becomes

dΦ4dt=P[ϱ(SS4)2S+rY4φH4Ϝ(YE4Y4E)+ϑ3S4Y4H40κ4H¯4(){Ϝ(S(t)Y(t)E4S4Y4E)+Ϝ(S4S)}d+(ϑ3S4Y4H5+rY4φH4H5)0κ5H¯5()Ϝ(E(t)Y4E4Y)d+a(λ+γ)P(61)Iμ1π1(γ+λ)σ1PCI].

Hence, if 61, then dΦ4dt0 for all S,I,E,Y,V,CI>0, where dΦ4dt=0 occurs at S=S4, E=E4, Y=Y4, and I=CI=0. The trajectories of system (5) converge to ϒ4 which includes elements with S(t)=S4, E(t)=E4, Y(t)=Y4, and then dS(t)dt=dY(t)dt=0. The first and fifth equations of system (5) become

0=dS(t)dt=ηϱS4ϑ1S4V(t)ϑ3S4Y4,0=dY(t)dt=ψH5E4δY4μ2CY(t)Y4,

which imply that V(t)=0 and CY(t)=C4Y for all t. Moreover, we have dI(t)dt=0, then the third equation of system (5) becomes

0=dI(t)dt=λ0κ3H¯3()L(t)d,

which yields L(t)=0 for all t, and then Inline graphic. Applying LLAS theorem, we get Inline graphic is GAS. □

Theorem 6

If 5>1, 81, and 1/2>1, then Inline graphic is GAS.

Proof

Define

Φ5=PS5Ϝ(SS5)+λH3L5Ϝ(LL5)+(γ+λ)I5Ϝ(II5)+PφH4E5Ϝ(EE5)+P(ψ+ω)φψH4H5Y5Ϝ(YY5)+Pϑ1S5εV5Ϝ(VV5)+μ1(γ+λ)σ1C5IϜ(CIC5I)+μ2P(ψ+ω)σ2φψH4H5CY+ϑ1λH3(1β)S5V50κ1H¯1()ttϜ(S(ϰ)V(ϰ)S5V5)dϰd+ϑ2λH3(1β)S5I50κ1H¯1()ttϜ(S(ϰ)I(ϰ)S5I5)dϰd+ϑ1β(γ+λ)S5V50κ2H¯2()ttϜ(S(ϰ)V(ϰ)S5V5)dϰd+ϑ2β(γ+λ)S5I50κ2H¯2()ttϜ(S(ϰ)I(ϰ)S5I5)dϰd+λ(γ+λ)L50κ3H¯3()ttϜ(L(ϰ)L5)dϰd+Pϑ3S5Y5H40κ4H¯4()ttϜ(S(ϰ)Y(ϰ)S5Y5)dϰd+P(ψ+ω)E5φH4H50κ5H¯5()ttϜ(E(ϰ)E5)dϰd+bPϑ1S5I5ε0κ6H¯6()ttϜ(I(ϰ)I5)dϰd.

Calculate dΦ5dt as follows:

dΦ5dt=P(1S5S)(ηϱSϑ1SVϑ2SIϑ3SY)+λH3(1L5L)×[(1β)0κ1H¯1()S(t){ϑ1V(t)+ϑ2I(t)}d(λ+γ)L]+(γ+λ)(1I5I)[β0κ2H¯2()S(t){ϑ1V(t)+ϑ2I(t)}d+λ0κ3H¯3()L(t)daIμ1CII]+PφH4(1E5E)[φϑ30κ4H¯4()S(t)Y(t)d+rY(ψ+ω)E]+P(ψ+ω)φψH4H5(1Y5Y)[ψ0κ5H¯5()E(t)dδYμ2CYY]+Pϑ1S5ε(1V5V)[b0κ6H¯6()I(t)dεV]+μ1(γ+λ)σ1(1C5ICI)(σ1CIIπ1CI)+μ2P(ψ+ω)σ2φψH4H5(σ2CYYπ2CY)+ϑ1λH3(1β)S5V5×0κ1H¯1()[SVS5V5S(t)V(t)S5V5+ln(S(t)V(t)SV)]d+ϑ2λH3(1β)S5I5×0κ1H¯1()[SIS5I5S(t)I(t)S5I5+ln(S(t)I(t)SI)]d+ϑ1β(γ+λ)S5V5×0κ2H¯2()[SVS5V5S(t)V(t)S5V5+ln(S(t)V(t)SV)]d+ϑ2β(γ+λ)S5I5×0κ2H¯2()[SIS5I5S(t)I(t)S5I5+ln(S(t)I(t)SI)]d+λ(γ+λ)L50κ3H¯3()[LL5L(t)L5+ln(L(t)L)]d+Pϑ3S5Y5H40κ4H¯4()[SYS5Y5S(t)Y(t)S5Y5+ln(S(t)Y(t)SY)]d+P(ψ+ω)E5φH4H50κ5H¯5()[EE5E(t)E5+ln(E(t)E)]d+bPϑ1S5I5ε0κ6H¯6()[II5I(t)I5+ln(I(t)I)]d. 23

Summing the terms of Eq. (23), we get

dΦ5dt=P(1S5S)(ηϱS)+Pϑ2S5I+Pϑ3S5Yϑ1λH3(1β)0κ1H¯1()×S(t)V(t)L5Ldϑ2λH3(1β)0κ1H¯1()S(t)I(t)L5Ld+λH3(λ+γ)L5a(λ+γ)Iϑ1β(γ+λ)0κ2H¯2()S(t)V(t)I5Idϑ2β(γ+λ)0κ2H¯2()S(t)I(t)I5Idλ(λ+γ)0κ3H¯3()L(t)I5Id+a(λ+γ)I5+μ1(λ+γ)CII5+PrφH4YPϑ3H40κ4H¯4()S(t)Y(t)E5EdPrφH4YE5E+P(ψ+ω)φH4E5Pδ(ψ+ω)φψH4H5YP(ψ+ω)φH4H50κ5H¯5()E(t)Y5Yd+Pδ(ψ+ω)φψH4H5Y5+μ2P(ψ+ω)φψH4H5CYY5bPϑ1S5ε0κ6H¯6()I(t)V5Vd+Pϑ1S5V5μ1π1(γ+λ)σ1CIμ1(λ+γ)C5II+μ1π1(λ+γ)σ1C5Iμ2π2P(ψ+ω)σ2φψH4H5CY+λH3(1β)ϑ1S5V50κ1H¯1()ln(S(t)V(t)SV)d+λH3(1β)ϑ2S5I50κ1H¯1()ln(S(t)I(t)SI)d+β(γ+λ)ϑ1S5V50κ2H¯2()ln(S(t)V(t)SV)d+β(γ+λ)ϑ2S5I50κ2H¯2()ln(S(t)I(t)SI)d+λ(γ+λ)L50κ3H¯3()ln(L(t)L)d+Pϑ3S5Y5H40κ4H¯4()ln(S(t)Y(t)SY)d+P(ψ+ω)E5φH4H50κ5H¯5()ln(E(t)E)d+bPϑ1S5H6εI+bPϑ1S5I5ε0κ6H¯6()ln(I(t)I)d.

Using the steady state conditions for Inline graphic

η=ϱS5+ϑ1S5V5+ϑ2S5I5+ϑ3S5Y5,H1(1β)(ϑ1S5V5+ϑ2S5I5)=(λ+γ)L5,βH2(ϑ1S5V5+ϑ2S5I5)+λH3L5=(a+μ1C5I)I5I5=π1σ1,V5=bH6εI5,ϑ3S5Y5+rY5φH4=(ψ+ω)E5φH4=δ(ψ+ω)Y5φψH4H5,

we obtain

P(ϑ1S5V5+ϑ2S5I5)=(λ+γ)(a+μ1C5I)I5.

Moreover, we get

dΦ5dt=P(1S5S)(ϱS5ϱS)+P(ϑ1S5V5+ϑ2S5I5+ϑ3S5Y5)(1S5S)λH3(1β)ϑ1S5V50κ1H¯1()S(t)V(t)L5S5V5LdλH3(1β)ϑ2S5I50κ1H¯1()S(t)I(t)L5S5I5Ld+λH1H3(1β)(ϑ1S5V5+ϑ2S5I5)β(γ+λ)ϑ1S5V50κ2H¯2()S(t)V(t)I5S5V5Idβ(γ+λ)ϑ2S5I50κ2H¯2()S(t)I(t)S5IdλH1(1β)(ϑ1S5V5+ϑ2S5I5)0κ3H¯3()L(t)I5L5Id+P(ϑ1S5V5+ϑ2S5I5)Pϑ3S5Y5H40κ4H¯4()S(t)Y(t)E5S5Y5EdPrY5φH4YE5Y5E+Pϑ3S5Y5+PrY5φH4(Pϑ3S5Y5H5+PrY5φH4H5)0κ5H¯5()E(t)Y5E5Yd+Pϑ3S5Y5+PrY5φH4Pϑ1S5V5H60κ6H¯6()I(t)V5I5Vd+Pϑ1S5V5+λH3(1β)ϑ1S5V50κ1H¯1()ln(S(t)V(t)SV)d+λH3(1β)ϑ2S5I50κ1H¯1()ln(S(t)I(t)SI)d+β(γ+λ)ϑ1S5V50κ2H¯2()ln(S(t)V(t)SV)d+β(γ+λ)ϑ2S5I50κ2H¯2()ln(S(t)I(t)SI)d+λH1(1β)(ϑ1S5V5+ϑ2S5I5)0κ3H¯3()ln(L(t)L)d+Pϑ3S5Y5H40κ4H¯4()ln(S(t)Y(t)SY)d+(Pϑ3S5Y5H5+PrY5φH4H5)0κ5H¯5()ln(E(t)E)d+Pϑ1S5V5H60κ6H¯6()ln(I(t)I)d+μ2P(ψ+ω)φψH4H5(Y5π2σ2)CY.

Using the equalities given by (10) and (11) in case of n=m=5, we get

dΦ5dt=ϱP(SS5)2SP(ϑ1S5V5+ϑ2S5I5+ϑ3S5Y5)[S5S1ln(S5S)]λH3(1β)ϑ1S5V5×0κ1H¯1()[S(t)V(t)L5S5V5L1ln(S(t)V(t)L5S5V5L)]dλH3(1β)ϑ2S5I5×0κ1H¯1()[S(t)I(t)L5S5I5L1ln(S(t)I(t)L5S5I5L)]dβ(γ+λ)ϑ1S5V5×0κ2H¯2()[S(t)V(t)I5S5V5I1ln(S(t)V(t)I5S5V5I)]dβ(γ+λ)ϑ2S5I50κ2H¯2()[S(t)I(t)S5I1ln(S(t)I(t)S5I)]dλH1(1β)(ϑ1S5V5+ϑ2S5I5)×0κ3H¯3()[L(t)I5L5I1ln(L(t)I5L5I)]dPrY5φH4[YE5Y5E1ln(YE5Y5E)]Pϑ3S5Y5H40κ4H¯4()[S(t)Y(t)E5S5Y5E1ln(S(t)Y(t)E5S5Y5E)]d(Pϑ3S5Y5H5+PrY5φH4H5)0κ5H¯5()[E(t)Y5E5Y1ln(E(t)Y5E5Y)]dPϑ1S5V5H60κ6H¯6()[I(t)V5I5V1ln(I(t)V5I5V)]d+μ2P(ψ+ω)φψH4H5(Y5π2σ2)CY. 24

Therefore, Eq. (24) becomes

dΦ5dt=ϱP(SS5)2SP(ϑ1S5V5+ϑ2S5I5+ϑ3S5Y5)Ϝ(S5S)λH3(1β)ϑ1S5V50κ1H¯1()Ϝ(S(t)V(t)L5S5V5L)dλH3(1β)ϑ2S5I50κ1H¯1()Ϝ(S(t)I(t)L5S5I5L)dβ(γ+λ)ϑ1S5V50κ2H¯2()Ϝ(S(t)V(t)I5S5V5I)dβ(γ+λ)ϑ2S5I50κ2H¯2()Ϝ(S(t)I(t)S5I)dλH1(1β)(ϑ1S5V5+ϑ2S5I5)0κ3H¯3()Ϝ(L(t)I5L5I)dPrY5φH4Ϝ(YE5Y5E)Pϑ3S5Y5H40κ4H¯4()Ϝ(S(t)Y(t)E5S5Y5E)d(Pϑ3S5Y5H5+PrY5φH4H5)0κ5H¯5()Ϝ(E(t)Y5E5Y)dPϑ1S5V5H60κ6H¯6()Ϝ(I(t)V5I5V)d+μ2P(ψ+ω)[π1σ2(bϑ1H6+εϑ2)+εσ1(π2ϑ3+ϱσ2)]φψϑ3εσ1σ2H4H5(81)CY.

Hence, if 81, then dΦ5dt0 for all S,L,I,E,Y,V,CY>0. One can show that dΦ5dt=0 when (S,L,I,E,Y,V,CY)=(S5,L5,I5,E5,Y5,V5,0). The solutions of model (5) tend to ϒ5 which includes elements with (S(t),L(t),I(t),V(t))=(S5,L5,I5,V5), and then dI(t)dt=0. The third equation of system (5) becomes

0=dI(t)dt=βH2(ϑ1S5V5+ϑ2S5I5)+λH3L5aI5μ1CI(t)I5,

which yields CI(t)=C5I for all t, and hence Inline graphic. Applying LLAS theorem, we get Inline graphic is GAS. □

Theorem 7

If 6>1, 71, and 2/1>1, then Inline graphic is GAS.

Proof

Define

Φ6=PS6Ϝ(SS6)+λH3L6Ϝ(LL6)+(γ+λ)I6Ϝ(II6)+PφH4E6Ϝ(EE6)+P(ψ+ω)φψH4H5Y6Ϝ(YY6)+Pϑ1S6εV6Ϝ(VV6)+μ1(γ+λ)σ1CI+μ2P(ψ+ω)σ2φψH4H5C6YϜ(CYC6Y)+ϑ1λH3(1β)S6V60κ1H¯1()ttϜ(S(ϰ)V(ϰ)S6V6)dϰd+ϑ2λH3(1β)S6I60κ1H¯1()ttϜ(S(ϰ)I(ϰ)S6I6)dϰd+ϑ1β(γ+λ)S6V60κ2H¯2()ttϜ(S(ϰ)V(ϰ)S6V6)dϰd+ϑ2β(γ+λ)S6I60κ2H¯2()ttϜ(S(ϰ)I(ϰ)S6I6)dϰd+λ(γ+λ)L60κ3H¯3()ttϜ(L(ϰ)L6)dϰd+Pϑ3S6Y6H40κ4H¯4()ttϜ(S(ϰ)Y(ϰ)S6Y6)dϰd+P(ψ+ω)E6φH4H50κ5H¯5()ttϜ(E(ϰ)E6)dϰd+bPϑ1S6I6ε0κ6H¯6()ttϜ(I(ϰ)I6)dϰd.

Calculate dΦ6dt as follows:

dΦ6dt=P(1S6S)(ηϱSϑ1SVϑ2SIϑ3SY)+λH3(1L6L)×[(1β)0κ1H¯1()S(t){ϑ1V(t)+ϑ2I(t)}d(λ+γ)L]+(γ+λ)(1I6I)[β0κ2H¯2()S(t){ϑ1V(t)+ϑ2I(t)}d+λ0κ3H¯3()L(t)daIμ1CII]+PφH4(1E6E)[φϑ30κ4H¯4()S(t)Y(t)d+rY(ψ+ω)E]+P(ψ+ω)φψH4H5(1Y6Y)[ψ0κ5H¯5()E(t)dδYμ2CYY]+Pϑ1S6ε(1V6V)[b0κ6H¯6()I(t)dεV]+μ1(γ+λ)σ1(σ1CIIπ1CI)+μ2P(ψ+ω)σ2φψH4H5(1C6YCY)(σ2CYYπ2CY)+ϑ1λH3(1β)S6V6×0κ1H¯1()[SVS6V6S(t)V(t)S6V6+ln(S(t)V(t)SV)]d+ϑ2λH3(1β)S6I6×0κ1H¯1()[SIS6I6S(t)I(t)S6I6+ln(S(t)I(t)SI)]d+ϑ1β(γ+λ)S6V6×0κ2H¯2()[SVS6V6S(t)V(t)S6V6+ln(S(t)V(t)SV)]d+ϑ2β(γ+λ)S6I6×0κ2H¯2()[SIS6I6S(t)I(t)S6I6+ln(S(t)I(t)SI)]d+λ(γ+λ)L60κ3H¯3()[LL6L(t)L6+ln(L(t)L)]d+Pϑ3S6Y6H40κ4H¯4()[SYS6Y6S(t)Y(t)S6Y6+ln(S(t)Y(t)SY)]d+P(ψ+ω)E6φH4H50κ5H¯5()[EE6E(t)E6+ln(E(t)E)]d+bPϑ1S6I6ε0κ6H¯6()[II6I(t)I6+ln(I(t)I)]d. 25

Collecting the terms of Eq. (25), we obtain

dΦ6dt=P(1S6S)(ηϱS)+Pϑ2S6I+Pϑ3S6Yϑ1λH3(1β)0κ1H¯1()×S(t)V(t)L6Ldϑ2λH3(1β)0κ1H¯1()S(t)I(t)L6Ld+λH3(λ+γ)L6a(λ+γ)Iϑ1β(γ+λ)0κ2H¯2()S(t)V(t)I6Idϑ2β(γ+λ)0κ2H¯2()S(t)I(t)I6Idλ(λ+γ)0κ3H¯3()L(t)I6Id+a(λ+γ)I6+μ1(λ+γ)CII6+PrφH4YPϑ3H40κ4H¯4()S(t)Y(t)E6EdPrφH4YE6E+P(ψ+ω)φH4E6Pδ(ψ+ω)φψH4H5YP(ψ+ω)φH4H50κ5H¯5()E(t)Y6Yd+Pδ(ψ+ω)φψH4H5Y6+μ2P(ψ+ω)φψH4H5CYY6bPϑ1S6ε0κ6H¯6()I(t)V6Vd+Pϑ1S6V6μ1π1(λ+γ)σ1CIμ2π2P(ψ+ω)σ2φψH4H5CYμ2P(ψ+ω)φψH4H5C6YY+μ2π2P(ψ+ω)σ2φψH4H5C6Y+λH3(1β)ϑ1S6V60κ1H¯1()ln(S(t)V(t)SV)d+λH3(1β)ϑ2S6I60κ1H¯1()ln(S(t)I(t)SI)d+β(γ+λ)ϑ1S6V60κ2H¯2()ln(S(t)V(t)SV)d+β(γ+λ)ϑ2S6I60κ2H¯2()ln(S(t)I(t)SI)d+λ(γ+λ)L60κ3H¯3()ln(L(t)L)d+Pϑ3S6Y6H40κ4H¯4()ln(S(t)Y(t)SY)d+P(ψ+ω)E6φH4H50κ5H¯5()ln(E(t)E)d+bPϑ1S6H6εI+bPϑ1S6I6ε0κ6H¯6()ln(I(t)I)d.

Using the steady state conditions for Inline graphic

η=ϱS6+ϑ1S6V6+ϑ2S6I6+ϑ3S6Y6,H1(1β)(ϑ1S6V6+ϑ2S6I6)=(λ+γ)L6,βH2(ϑ1S6V6+ϑ2S6I6)+λH3L6=aI6,Y6=π2σ2,V6=bH6I6ε,ϑ3S6Y6+rY6φH4=(ψ+ω)E6φH4=δ(ψ+ω)φψH4H5Y6+μ2(ψ+ω)φψH4H5C6YY6,

we get

P(ϑ1S6V6+ϑ2S6I6)=a(λ+γ)I6.

Moreover, we get

dΦ6dt=P(1S6S)(ϱS6ϱS)+P(ϑ1S6V6+ϑ2S6I6+ϑ3S6Y6)(1S6S)λH3(1β)ϑ1S6V60κ1H¯1()S(t)V(t)L6S6V6LdλH3(1β)ϑ2S6I6×0κ1H¯1()S(t)I(t)L6S6I6Ld+λH1H3(1β)(ϑ1S6V6+ϑ2S6I6)β(γ+λ)ϑ1S6V60κ2H¯2()S(t)V(t)I6S6V6Idβ(γ+λ)ϑ2S6I60κ2H¯2()S(t)I(t)S6IdλH1(1β)(ϑ1S6V6+ϑ2S6I6)0κ3H¯3()L(t)I6L6Id+P(ϑ1S6V6+ϑ2S6I6)Pϑ3S6Y6H40κ4H¯4()S(t)Y(t)E6S6Y6EdPrY6φH4YE6Y6E+Pϑ3S6Y6+PrY6φH4(Pϑ3S6Y6H5+PrY6φH4H5)0κ5H¯5()E(t)Y6E6Yd+Pϑ3S6Y6+PrY6φH4Pϑ1S6V6H60κ6H¯6()I(t)V6I6Vd+Pϑ1S6V6+λH3(1β)ϑ1S6V60κ1H¯1()ln(S(t)V(t)SV)d+λH3(1β)ϑ2S6I60κ1H¯1()ln(S(t)I(t)SI)d+β(γ+λ)ϑ1S6V60κ2H¯2()ln(S(t)V(t)SV)d+β(γ+λ)ϑ2S6I60κ2H¯2()ln(S(t)I(t)SI)d+λH1(1β)(ϑ1S6V6+ϑ2S6I6)0κ3H¯3()ln(L(t)L)d+Pϑ3S6Y6H40κ4H¯4()ln(S(t)Y(t)SY)d+(Pϑ3S6Y6H5+PrY6φH4H5)0κ5H¯5()ln(E(t)E)d+Pϑ1S6V6H60κ6H¯6()ln(I(t)I)d+μ1(λ+γ)(I6π1σ1)CI.

Using the equalities given by (10) and (11) in case of n=m=6, we get

dΦ6dt=ϱP(SS6)2SP(ϑ1S6V6+ϑ2S6I6+ϑ3S6Y6)[S6S1ln(S6S)]λH3(1β)ϑ1S6V6×0κ1H¯1()[S(t)V(t)L6S6V6L1ln(S(t)V(t)L6S6V6L)]dλH3(1β)ϑ2S6I6×0κ1H¯1()[S(t)I(t)L6S6I6L1ln(S(t)I(t)L6S6I6L)]dβ(γ+λ)ϑ1S6V6×0κ2H¯2()[S(t)V(t)I6S6V6I1ln(S(t)V(t)I6S6V6I)]dβ(γ+λ)ϑ2S6I60κ2H¯2()[S(t)I(t)S6I1ln(S(t)I(t)S6I)]dλH1(1β)(ϑ1S6V6+ϑ2S6I6)×0κ3H¯3())[L(t)I6L6I1ln(L(t)I6L6I)]dPϑ3S6Y6H40κ4H¯4()[S(t)Y(t)E6S6Y6E1ln(S(t)Y(t)E6S6Y6E)]dPrY6φH4[YE6Y6E1ln(YE6Y6E)](Pϑ3S6Y6H5+PrY6φH4H5)0κ5H¯5()[E(t)Y6E6Y1ln(E(t)Y6E6Y)]dPϑ1S6V6H60κ6H¯6()[I(t)V6I6V1ln(I(t)V6I6V)]d+μ1(λ+γ)(I6π1σ1)CI. 26

Therefore, Eq. (26) becomes

dΦ6dt=ϱP(SS6)2SP(ϑ1S6V6+ϑ2S6I6+ϑ3S6Y6)+Ϝ(S6S)λH3(1β)ϑ1S6V60κ1H¯1()Ϝ(S(t)V(t)L6S6V6L)dλH3(1β)ϑ2S6I60κ1H¯1()Ϝ(S(t)I(t)L6S6I6L)dβ(γ+λ)ϑ1S6V60κ2H¯2()Ϝ(S(t)V(t)I6S6V6I)dβ(γ+λ)ϑ2S6I60κ2H¯2()Ϝ(S(t)I(t)S6I)dλH1(1β)(ϑ1S6V6+ϑ2S6I6)0κ3H¯3())Ϝ(L(t)I6L6I)dPϑ3S6Y6H40κ4H¯4()Ϝ(S(t)Y(t)E6S6Y6E)dPrY6φH4Ϝ(YE6Y6E)(Pϑ3S6Y6H5+PrY6φH4H5)0κ5H¯5()Ϝ(E(t)Y6E6Y)dPϑ1S6V6H60κ6H¯6()Ϝ(I(t)V6I6V)d+μ1(γ+λ)[π1σ2(bϑ1H6+εϑ2)+εσ1(π2ϑ3+ϱσ2)]σ1σ2(bϑ1H6+εϑ2)(71)CI.

Hence, if 71, then dΦ6dt0 for all S,L,I,E,Y,V,CI>0. Similar to the previous theorems, one can show that dΦ6dt=0 at (S,L,I,E,Y,V,CI)=(S6,L6,I6,E6,Y6,V6,0). The solutions of system (5) reach ϒ6 which contains elements with E(t)=E6, Y(t)=Y6, and then dY(t)dt=0. The fifth equation of system (5) becomes

0=dY(t)dt=ψH5E6δY6μ2CY(t)Y6,

which yields CY(t)=C6Y for all t, and hence Inline graphic. Applying LLAS theorem, we get Inline graphic is GAS. □

Theorem 8

If 7>1 and 8>1, then Inline graphic is GAS.

Proof

Consider

Φ7=PS7Ϝ(SS7)+λH3L7Ϝ(LL7)+(γ+λ)I7Ϝ(II7)+PφH4E7Ϝ(EE7)+P(ψ+ω)φψH4H5Y7Ϝ(YY7)+Pϑ1S7εV7Ϝ(VV7)+μ1(γ+λ)σ1C7IϜ(CIC7I)+μ2P(ψ+ω)σ2φψH4H5C7YϜ(CYC7Y)+ϑ1λH3(1β)S7V70κ1H¯1()ttϜ(S(ϰ)V(ϰ)S7V7)dϰd+ϑ2λH3(1β)S7I70κ1H¯1()ttϜ(S(ϰ)I(ϰ)S7I7)dϰd+ϑ1β(γ+λ)S7V70κ2H¯2()ttϜ(S(ϰ)V(ϰ)S7V7)dϰd+ϑ2β(γ+λ)S7I70κ2H¯2()ttϜ(S(ϰ)I(ϰ)S7I7)dϰd+λ(γ+λ)L70κ3H¯3()ttϜ(L(ϰ)L7)dϰd+Pϑ3S7Y7H40κ4H¯4()ttϜ(S(ϰ)Y(ϰ)S7Y7)dϰd+P(ψ+ω)E7φH4H50κ5H¯5()ttϜ(E(ϰ)E7)dϰd+bPϑ1S7I7ε0κ6H¯6()ttϜ(I(ϰ)I7)dϰd.

Calculate dΦ7dt as follows:

dΦ7dt=P(1S7S)(ηϱSϑ1SVϑ2SIϑ3SY)+λH3(1L7L)×[(1β)0κ1H¯1()S(t){ϑ1V(t)+ϑ2I(t)}d(λ+γ)L]+(γ+λ)(1I7I)[β0κ2H¯2()S(t){ϑ1V(t)+ϑ2I(t)}d+λ0κ3H¯3()L(t)daIμ1CII]+PφH4(1E7E)[φϑ30κ4H¯4()S(t)Y(t)d+rY(ψ+ω)E]+P(ψ+ω)φψH4H5(1Y7Y)[ψ0κ5H¯5()E(t)dδYμ2CYY]+Pϑ1S7ε(1V7V)[b0κ6H¯6()I(t)dεV]+μ1(γ+λ)σ1(1C7ICI)(σ1CIIπ1CI)+μ2P(ψ+ω)σ2φψH4H5(1C7YCY)(σ2CYYπ2CY)+ϑ1λH3(1β)S7V7×0κ1H¯1()[SVS7V7S(t)V(t)S7V7+ln(S(t)V(t)SV)]d+ϑ2λH3(1β)S7I7×0κ1H¯1()[SIS7I7S(t)I(t)S7I7+ln(S(t)I(t)SI)]d+ϑ1β(γ+λ)S7V7×0κ2H¯2()[SVS7V7S(t)V(t)S7V7+ln(S(t)V(t)SV)]d+ϑ2β(γ+λ)S7I7×0κ2H¯2()[SIS7I7S(t)I(t)S7I7+ln(S(t)I(t)SI)]d+λ(γ+λ)L70κ3H¯3()[LL7L(t)L7+ln(L(t)L)]d+Pϑ3S7Y7H40κ4H¯4()[SYS7Y7S(t)Y(t)S7Y7+ln(S(t)Y(t)SY)]d+P(ψ+ω)E7φH4H50κ5H¯5()[EE7E(t)E7+ln(E(t)E)]d+bPϑ1S7I7ε0κ6H¯6()[II7I(t)I7+ln(I(t)I)]d. 27

Summing the terms of Eq. (27), we get

dΦ7dt=P(1S7S)(ηϱS)+Pϑ2S7I+Pϑ3S7Yϑ1λH3(1β)0κ1H¯1()×S(t)V(t)L7Ldϑ2λH3(1β)0κ1H¯1()S(t)I(t)L7Ld+λH3(λ+γ)L7a(λ+γ)Iϑ1β(γ+λ)0κ2H¯2()S(t)V(t)I7Idϑ2β(γ+λ)0κ2H¯2()S(t)I(t)I7Idλ(λ+γ)0κ3H¯3()L(t)I7Id+a(λ+γ)I7+μ1(λ+γ)CII7+PrφH4YPϑ3H40κ4H¯4()S(t)Y(t)E7EdPrφH4YE7E+P(ψ+ω)φH4E7Pδ(ψ+ω)φψH4H5YP(ψ+ω)φH4H50κ5H¯5()E(t)Y7Yd+Pδ(ψ+ω)φψH4H5Y7+μ2P(ψ+ω)φψH4H5CYY7bPϑ1S7ε0κ6H¯6()I(t)V7Vd+Pϑ1S7V7μ1π1(λ+γ)σ1CIμ1(λ+γ)C7II+μ1π1(λ+γ)σ1C7Iμ2π2P(ψ+ω)σ2φψH4H5CYμ2P(ψ+ω)φψH4H5C7YY+μ2π2P(ψ+ω)σ2φψH4H5C7Y+λH3(1β)ϑ1S7V70κ1H¯1()ln(S(t)V(t)SV)d+λH3(1β)ϑ2S7I70κ1H¯1()ln(S(t)I(t)SI)d+β(γ+λ)ϑ1S7V70κ2H¯2()ln(S(t)V(t)SV)d+β(γ+λ)ϑ2S7I70κ2H¯2()ln(S(t)I(t)SI)d+λ(γ+λ)L70κ3H¯3()ln(L(t)L)d+Pϑ3S7Y7H40κ4H¯4()ln(S(t)Y(t)SY)d+P(ψ+ω)E7φH4H50κ5H¯5()ln(E(t)E)d+bPϑ1S7H6εI+bPϑ1S7I7ε0κ6H¯6()ln(I(t)I)d.

Using the steady state conditions for Inline graphic

η=ϱS7+ϑ1S7V7+ϑ2S7I7+ϑ3S7Y7,H1(1β)(ϑ1S7V7+ϑ2S7I7)=(λ+γ)L7,βH2(ϑ1S7V7+ϑ2S7I7)+λH3L7=(a+μ1C7I)I7,I7=π1σ1,Y7=π2σ2,V7=bH6I7ε,ϑ3S7Y7+rY7φH4=(ψ+ω)E7φH4=δ(ψ+ω)φψH4H5Y7+μ2(ψ+ω)φψH4H5C7YY7,

we get

P(ϑ1S7V7+ϑ2S7I7)=(λ+γ)(a+μ1C7I)I7.

Moreover, we get

dΦ7dt=P(1S7S)(ϱS7ϱS)+P(ϑ1S7V7+ϑ2S7I7+ϑ3S7Y7)(1S7S)λH3(1β)ϑ1S7V70κ1H¯1()S(t)V(t)L7S7V7LdλH3(1β)ϑ2S7I70κ1H¯1()S(t)I(t)L7S7I7Ld+λH1H3(1β)(ϑ1S7V7+ϑ2S7I7)β(γ+λ)ϑ1S7V70κ2H¯2()S(t)V(t)I7S7V7Idβ(γ+λ)ϑ2S7I70κ2H¯2()S(t)I(t)S7IdλH1(1β)(ϑ1S7V7+ϑ2S7I7)0κ3H¯3()L(t)I7L7Id+P(ϑ1S7V7+ϑ2S7I7)Pϑ3S7Y7H40κ4H¯4()S(t)Y(t)E7S7Y7EdPrY7φH4YE7Y7E+Pϑ3S7Y7+PrY7φH4(Pϑ3S7Y7H5+PrY7φH4H5)0κ5H¯5()E(t)Y7E7Yd+Pϑ3S7Y7+PrY7φH4Y7Pϑ1S7V7H60κ6H¯6()I(t)V7I7Vd+Pϑ1S7V7+λH3(1β)ϑ1S7V70κ1H¯1()ln(S(t)V(t)SV)d+λH3(1β)ϑ2S7I70κ1H¯1()ln(S(t)I(t)SI)d+β(γ+λ)ϑ1S7V70κ2H¯2()ln(S(t)V(t)SV)d+β(γ+λ)ϑ2S7I70κ2H¯2()ln(S(t)I(t)SI)d+λH1(1β)(ϑ1S7V7+ϑ2S7I7)0κ3H¯3()ln(L(t)L)d+Pϑ3S7Y7H40κ4H¯4()ln(S(t)Y(t)SY)d+(Pϑ3S7Y7H5+PrY7φH4H5)0κ5H¯5()ln(E(t)E)d+Pϑ1S7V7H60κ6H¯6()ln(I(t)I)d.

Using the equalities given by (10) and (11) in case of n=m=7, we get

dΦ7dt=ϱP(SS7)2SP(ϑ1S7V7+ϑ2S7I7+ϑ3S7Y7)[S7S1ln(S7S)]λH3(1β)ϑ1S7V7×0κ1H¯1()[S(t)V(t)L7S7V7L1ln(S(t)V(t)L7S7V7L)]dλH3(1β)ϑ2S7I7×0κ1H¯1()[S(t)I(t)L7S7I7L1ln(S(t)I(t)L7S7I7L)]dβ(γ+λ)ϑ1S7V7×0κ2H¯2()[S(t)V(t)I7S7V7I1ln(S(t)V(t)I7S7V7I)]dβ(γ+λ)ϑ2S7I70κ2H¯2()[S(t)I(t)S7I1ln(S(t)I(t)S7I)]dλH1(1β)(ϑ1S7V7+ϑ2S7I7)×0κ3H¯3())[L(t)I7L7I1ln(L(t)I7L7I)]dPϑ3S7Y7H40κ4H¯4()[S(t)Y(t)E7S7Y7E1ln(S(t)Y(t)E7S7Y7E)]dPrY7φH4[YE7Y7E1ln(YE7Y7E)](Pϑ3S7Y7H5+PrY7φH4H5)0κ5H¯5()[E(t)Y7E7Y1ln(E(t)Y7E7Y)]dPϑ1S7V7H60κ6H¯6()[I(t)V7I7V1ln(I(t)V7I7V)]d. 28

Therefore, Eq. (28) becomes

dΦ7dt=ϱP(SS7)2SP(ϑ1S7V7+ϑ2S7I7+ϑ3S7Y7)Ϝ(S7S)λH3(1β)ϑ1S7V70κ1H¯1()Ϝ(S(t)V(t)L7S7V7L)dλH3(1β)ϑ2S7I70κ1H¯1()Ϝ(S(t)I(t)L7S7I7L)dβ(γ+λ)ϑ1S7V70κ2H¯2()Ϝ(S(t)V(t)I7S7V7I)dβ(γ+λ)ϑ2S7I70κ2H¯2()Ϝ(S(t)I(t)S7I)dλH1(1β)(ϑ1S7V7+ϑ2S7I7)0κ3H¯3())Ϝ(L(t)I7L7I)dPϑ3S7Y7H40κ4H¯4()Ϝ(S(t)Y(t)E7S7Y7E)dPrY7φH4Ϝ(YE7Y7E)(Pϑ3S7Y7H5+PrY7φH4H5)0κ5H¯5()Ϝ(E(t)Y7E7Y)dPϑ1S7V7H60κ6H¯6()Ϝ(I(t)V7I7V)d.

Hence, dΦ7dt0 for all S,L,I,E,Y,V>0. Similar to the previous theorems, one can show that dΦ7dt=0 when (S,L,I,E,Y,V)=(S7,L7,I7,E7,Y7,V7). The solutions of system (5) converge to ϒ7 which includes elements with (S,L,I,E,Y,V)(t)=(S7,L7,I7,E7,Y7,V7). Then dI(t)dt=dY(t)dt=0. The third and fifth equations of system (5) become

0=dI(t)dt=βH2(ϑ1S7V7+ϑ2S7I7)+λH3L7aI7μ1CI(t)I7,0=dY(t)dt=ψH5E7δY7μ2CY(t)Y7,

which yield CI(t)=C7I and CY(t)=C7Y for all t, and hence Inline graphic. Applying LLAS theorem, we get Inline graphic is GAS. □

Numerical simulations

In this section, we perform numerical simulations to illustrate the results of Theorems 18. Moreover, we study the influence of time delays on the dynamical behavior of the system. Let us choose a Dirac delta function D() as a special form of the kernel Λi() as follows:

Λi(x)=D(xi),i[0,κi],i=1,2,,6.

Let κi, then we get

0Λj(ς)dς=1,Hj=0D(ςj)ejςdς=ejj,j=1,2,,6.

Thus, model (4) reduces to

{dS(t)dt=ηϱS(t)ϑ1S(t)V(t)ϑ2S(t)I(t)ϑ3S(t)Y(t),dL(t)dt=(1β)e11S(t1)[ϑ1V(t1)+ϑ2I(t1)](λ+γ)L(t),dI(t)dt=βe22S(t2)[ϑ1V(t2)+ϑ2I(t2)]dI(t)dt=+λe33L(t3)aI(t)μ1CI(t)I(t),dE(t)dt=φϑ3e44S(t4)Y(t4)+KrY(t)(ψ+ω)E(t),dY(t)dt=ψe55E(t5)+(1K)rY(t)δY(t)μ2CY(t)Y(t),dV(t)dt=be66I(t6)εV(t),dCI(t)dt=σ1CI(t)I(t)π1CI(t),dCY(t)dt=σ2CY(t)Y(t)π2CY(t). 29

For model (29), the threshold parameters are given by

1=PS0(bϑ1e66+εϑ2)aε(γ+λ),2=φϑ3ψe(44+55)S0(δre55)ψ+δω,3=σ1ηP(bϑ1e66+εϑ2)a(γ+λ)[π1(bϑ1e66+εϑ2)+ϱεσ1],4=σ2ηφϑ3ψe(44+55)(π2ϑ3+ϱσ2)[(δre55)ψ+δω],5=ηφεϑ3σ1ψe(44+55)[(δre55)ψ+δω][π1(bϑ1e66+εϑ2)+ϱεσ1],6=ησ2P(bϑ1e66+εϑ2)aε(γ+λ)(π2ϑ3+ϱσ2),7=σ1σ2ηP(bϑ1e66+εϑ2)a(γ+λ)[π1σ2(bϑ1e66+εϑ2)+εσ1(π2ϑ3+ϱσ2)],8=εϑ3ησ1σ2φψe(44+55)[π1σ2(bϑ1e66+εϑ2)+εσ1(π2ϑ3+ϱσ2)][(δre55)ψ+δω], 30

where

P=λe(11+33)(1β)+βe22(γ+λ). 31

To solve system (29) numerically, we fix the values of some parameters (see Table 1) and the others will be varied.

Table 1.

The data of model (29)

Parameter Value Parameter Value Parameter Value
η 10 b 5 ω 0.03
ϱ 0.01 π1 0.1 ψ 0.003
ϑi, i = 1,2,3 Varied π2 0.1 1 0.2
β 0.7 μ1 0.2 2 0.3
a 0.5 μ2 0.2 3 0.4
φ 0.2 ε 2 4 0.5
K 0.9 γ 0.1 5 0.6
r 0.008 σi, i = 1,2 Varied 6 0.9
δ 0.05 λ 0.2 ξi, i = 1,2,…,6 Varied

Stability of the steady states

In this subsection, we select the delay parameters as 1=1, 2=0.8, 3=0.6, 4=0.4, 5=0.2, 6=0.1. Besides, we choose the following three different initial conditions for system (29):

Initial-1: (S(),L(),I(),E(),Y(),V(),CI(),CY())=(800,0.5,0.7,12,0.51.52.5,0.3),

Initial-2: (S(),L(),I(),E(),Y(),V(),CI(),CY())=(700,0.8,1,8,0.3,2.5,1.5,0.2),

Initial-3: (S(),L(),I(),E(),Y(),V(),CI(),CY())=(600,1.31.2,10,0.4,3.5,0.5,0.1), where [1,0].

Choosing different values of ϑ1, ϑ2, ϑ3, ϑ4, σ1, and σ2 under the above initial conditions leads to the following sets:

Set 1 (Stability of Inline graphic): ϑ1=0.0002, ϑ2=0.0001, ϑ3=0.001, σ1=0.3, and σ2=0.5. For this set of parameters, we have 1=0.76<1 and 2=0.27<1. Figure 1 illustrates that the trajectories starting different initials converge to the steady state Inline graphic. This supports the global stability result of Theorem 1. Here, a healthy state will be reached where both viruses are absent.

Figure 1.

Figure 1

Solutions of system (29) when 11 and 21

Set 2 (Stability of Inline graphic): ϑ1=0.0005, ϑ2=0.0003, ϑ3=0.0007, σ1=0.003, and σ2=0.2. With such a choice we get 2=0.19<1<1.96=1, 3=0.34<1, and hence 2/1=0.1<1. The steady state Inline graphic exists with Inline graphic. The stability of Inline graphic given in Theorem 2 is shown in Fig. 2. This leads to the case where HIV monoinfection is chronic but with an ineffective CTL immunity.

Figure 2.

Figure 2

Solutions of system (29) when 1>1, 2/11, and 31

Set 3 (Stability of Inline graphic): ϑ1=0.0001, ϑ2=0.0003, ϑ3=0.005, σ1=0.001, and σ2=0.05. Then we calculate 1=0.72<1<1.34=2, 4=0.67<1, and then 1/2=0.54<1 and Inline graphic. We can see from Fig. 3 that the system’s solutions tend to Inline graphic, which is compatible with Theorem 3. This case means that an HTLV monoinfection is chronic with an ineffective CTL immunity.

Figure 3.

Figure 3

Solutions of system (29) when 2>1, 1/21, and 41

Set 4 (Stability of Inline graphic): ϑ1=0.0005, ϑ2=0.0003, ϑ3=0.002, σ1=0.05, and σ2=0.005. Then we calculate 3=1.52>1 and 5=0.42<1. Figure 4 shows that the trajectories starting with different states tend to Inline graphic. Therefore, Inline graphic is GAS, and this supports Theorem 4. Hence, an HIV monoinfection is chronic with effective HIV-specific CTL immunity.

Figure 4.

Figure 4

Solutions of system (29) when 3>1 and 51

Set 5 (Stability of Inline graphic): ϑ1=0.0002, ϑ2=0.0002, ϑ3=0.02, σ1=0.07, and σ2=0.4. Then we calculate 4=3.57>1 and 6=0.60<1, and Inline graphic exists with Inline graphic. We observe from Fig. 5 that the system’s trajectories tend to Inline graphic and it is GAS. Here, an HTLV monoinfection is chronic with effective HTLV-specific CTL immunity.

Figure 5.

Figure 5

Solutions of system (29) when 4>1 and 61

Set 6 (Stability of Inline graphic): ϑ1=0.001, ϑ2=0.0001, ϑ3=0.005, σ1=0.15, and σ2=0.01. Then we calculate 5=1.15>1, 8=0.22<1, and 1/2=2.42>1. The numerical solutions of the system drawn in Fig. 6 confirm that Inline graphic exists and is GAS. This case leads to a chronic coinfection with HTLV and HIV where the HIV-specific CTL immunity is effective while the HTLV-specific CTL immunity is ineffective.

Figure 6.

Figure 6

Solutions of system (29) when 5>1, 81, and 1/2>1

Set 7 (Stability of Inline graphic): ϑ1=0.0004, ϑ2=0.0002, ϑ3=0.01, σ1=0.007, and σ2=0.7. We compute 6=1.32>1, 7=0.55<1, and 2/1=1.77>1. According to these values, we obtain that Inline graphic exists. The numerical solutions of our system plotted in Fig. 7 show that Inline graphic is GAS (Theorem 7). This case leads to a chronic coinfection with HTLV and HIV where the HTLV-specific CTL immunity is effective and the HIV-specific CTL immunity is not working.

Figure 7.

Figure 7

Solutions of system (29) when 6>1, 71, and 2/1>1

Set 8 (Stability of Inline graphic): ϑ1=0.0005, ϑ2=0.0003, ϑ3=0.01, σ1=0.1, and σ2=0.3. These data give 7=1.33>1 and 8=1.81>1. According to these values, the steady state Inline graphic exists. Figure 8 illustrates that the solutions of the system initiating with three different states tend to Inline graphic. In this case, a chronic coinfection with HTLV and HIV is reached where both immune responses are well working.

Figure 8.

Figure 8

Solutions of system (29) when 7>1 and 8>1

Effect of time delays on the HTLV-HIV dynamics

In this part we vary the delay parameters i, i=1,2,,6, and fix the parameters ϑ1=0.0005, ϑ2=0.0003, ϑ3=0.01, σ1=0.04, and σ2=0.7. Since 1 and 2 given by Eqs. (30) and (31) depend on i, i=1,2,,6, then changing the parameters i will change the stability of steady states. Let us consider the following situations:

(D.P.S1)

1=2=3=4=5=6=0,

(D.P.S2)

1=0.4, 2=0.5, 3=0.6, 4=0.7, 5=0.8, and 6=0.9,

(D.P.S3)

1=0.6, 2=0.7, 3=0.8, 4=0.9, 5=1, and 6=1.2,

(D.P.S4)

1=10, 2=11, 3=12, 4=13, 5=14, and 6=15.

With these values we solve system (29) under the following initial condition:

Initial-4: (S(),L(),I(),E(),Y(),V(),CI(),CY())=(800,1,2,4,0.14,3,1,0.1), where [maxi,0], i=1,2,,6.

From Fig. 9 we observe that the presence of time delays can increase the number of uninfected CD4+ T cells and decrease the number of other compartments. Table 2 presents the values 1 and 2 for selected values of i, i=1,2,,6. It is clear that 1 and 2 are decreased when i are increased, and thus the stability of Inline graphic can be changed. Let us calculate the critical value of the time delay that changes the stability of Inline graphic. Without loss of generality, we let the parameters =1=2=3 and fix j, j=5,6, and write 1 and 2 as functions of and 4, respectively, as follows:

1()=[λe(1+3)(1β)+βe2(γ+λ)]S0(bϑ1e66+εϑ2)aε(γ+λ),2(4)=φϑ3ψe(44+55)S0(δre55)ψ+δω.

To force the threshold parameters 1 and 2 to satisfy 1()1 and 2(4)1, we choose min, where min is the solution of

[λemin(1+3)(1β)+βe2min(γ+λ)]S0(bϑ1e66+εϑ2)aε(γ+λ)=1,

and

44min,where 4min=max{0,14lnφϑ3ψe55S0(δre55)ψ+δω}.

Therefore, if min and 44min, then Inline graphic is GAS. Let us choose the value 5=0.2 and 6=0.1 and compute min, 4min as min=2.73728, 4min=2.36794. It follows that:

Figure 9.

Figure 9

Impact of delay parameters i, i=1,2,,6, on the behavior of solution trajectories of system (29)

Table 2.

The variation of 1 and 2 with respect to the delay parameters

Delay parameters 1 2
1=2=3=4=5=6=0 2.790 3.690
1=0.3, 2=0.4, 3=0.5, 4=0.6, 5=0.7, and 6=0.8 1.408 1.787
1=0.4, 2=0.5, 3=0.6, 4=0.7, 5=0.8, and 6=0.9 1.280 1.600
1=0.6, 2=0.7, 3=0.8, 4=0.9, 5=1, and 6=1.2 1.009 1.283
1=1, 2=1.5, 3=2, 4=2.5, 5=3, and 6=3.5 0.368 0.173
1=2, 2=3, 3=4, 4=5, 5=6, and 6=7 0.188 0.008
1=3, 2=4, 3=5, 4=6, 5=7, and 6=8 0.136 0.003
1=4, 2=6, 3=8, 4=9, 5=10, and 6=11 0.072 0.1 × 10−3
1=6, 2=7, 3=9, 4=10, 5=11, and 6=12 0.052 0.3 × 10−4
1=10, 2=11, 3=12, 4=13, 5=14, and 6=15 0.016 1.2 × 10−6

(i) If 2.73728 and 42.36794, then 1()1, 2(4)1 and Inline graphic is GAS.

(ii) If <2.73728 or 4<2.36794, then 1()>1 or 2(4)>1 and Inline graphic will lose its stability.

Authors’ contributions

The author was the only one contributing to the manuscript. The author read and approved the final manuscript.

Funding

Not applicable.

Availability of data and materials

Not applicable.

Competing interests

The author declares that they have no competing interests.

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