# Global Stability of Delayed Viral Infection Models with Nonlinear Antibody and CTL Immune Responses and General Incidence Rate

Authors: Hui Miao; Zhidong Teng; Zhiming Li
Journal: Computational and Mathematical Methods in Medicine
 (2016)

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## Abstract

The dynamical behaviors for a five-dimensional viral infection model with three delays which describes the interactions of antibody, cytotoxic T-lymphocyte (CTL) immune responses, and nonlinear incidence rate are investigated. The threshold values for viral infection, antibody response, CTL immune response, CTL immune competition, and antibody competition, respectively, are established. Under certain assumptions, the threshold value conditions on the global stability of the infection-free, immune-free, antibody response, CTL immune response, and interior equilibria are proved by using the Lyapunov functionals method, respectively. Immune delay as a bifurcation parameter is further investigated. The numerical simulations are performed in order to illustrate the dynamical behavior of the model.


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## Body

## 1. Introduction

In recent years, many authors have formulated and studied mathematical models which describe the dynamics of virus population in vivo. These provide insights in our understanding of HIV (human immunodeficiency virus) and other viruses, such as HBV (hepatitis B virus) and HCV (hepatitis C virus) [1–34]. In particular, the global stability of steady states for these models will give us a detailed information and enhance our understanding about the viral dynamics.During viral infections, the immune system reacts against virus. The antibody and CTL play the crucial roles in preventing and modulating infections. The antibody response is implemented by the functioning of immunocompetent B lymphocytes. The CTL immune response has the ability to suppress the virus replication in vivo. Hence, in order to prevent virus infection, an effective vaccine needs both strong neutralizing antibody and CTL immune responses [1, 2, 14, 18–23, 25–32]. Based on these, it is of interest for us to investigate whether sustained oscillations are the result of delayed viral infection model. This provides us with the motivation to conduct our work. In [2], Balasubramaniam et al. developed the viral infection model by incorporating immune delays and Beddington-DeAngelis incidence rate (1)dxtdt=λ-dxt-β1-ϵrtxtvt1+mxt+nvt,dytdt=β1-ϵrtxtvt1+mxt+nvt-ayt-pytzt,dvtdt=k1-ϵpiyt-uvt-qvtwt,dwtdt=gvtwt-hwt,dztdt=cyt-τzt-τ-bzt,where x, y, v, w, and z denote the concentrations of susceptible host cells, infected cells, free virus, antibody responses, and CTL immune responses, respectively. The local and global stability of the infection-free equilibrium and infected equilibrium and the existence of Hopf bifurcation are obtained. Furthermore, by using the Nyquist criterion, the estimation of the length of the delay to preserve stability of the infected equilibrium is obtained.Motivated by the work in [1, 2, 20, 21], in the present paper we propose a general viral infection model with three time delays which describes the interactions of antibody, CTL immune responses, and nonlinear incidence rate (2)dxtdt=sx-fx,v,dytdt=e-m1τ1fxt-τ1,vt-τ1-ag1y-pg1yg4z,dvtdt=ke-m2τ2g1yt-τ2-ug2v-qg2vg3w,dztdt=cg1yt-τ3g4zt-τ3-bg4z,dwtdt=rg2vg3w-hg3w,where s(x) denotes the intrinsic growth rate of uninfected target cells accounting for both production and natural mortality. In the literature of virus dynamics, the typical forms of the growth rate are s(x)=λ-dx and s(x)=λ-dx+rx(1-x/K), where λ,d,r,K are positive real numbers [4–13, 15, 16, 18, 20–23, 26–32, 34].We assume that the incidence of new infections of target cells occurs at a ratef(x,v). This form of incident rate is general to encompass several forms such as bilinear incidence βxv [4, 13], saturated incidence βxv/(1+bv) [16], Holling type II functional response βxv/(1+ax) [15], and Crowley-Martin incidence βxv/(1+ax+bv+abxv) [12, 35], where β, a, and b are positive constants.It is also assumed that the death rates of the infected target cells, viruses, antibody, and CTLs depend on their concentrations. These rates are given byag1(y), ug2(v), hg3(w), and bg4(z), respectively. The neutralization rate of viruses and the activation rate of B cells are proportional to the product of the removal rates of the viruses and B cells. Let qg2(v)g3(w) and rg2(v)g3(w) be the neutralization rate of viruses and activation rate of B cells, respectively. The typical forms can be seen as qvw and rvw [1, 2, 20, 21, 31, 32]. Accordingly, let pg1(y)g4(z) and cg1(y)g4(z) be the killing rate of infected cells and the birth rate of the CTL cells, respectively. The typical forms are pyz and cyz that appear in several papers [1, 2, 14, 20, 22, 27, 30, 34].For model (2), based on the epidemiological background, we assume that virus production occurs after the virus entry by the time delay τ1. The probability of surviving the time period from t-τ1 to t is e-m1τ1. Let τ2 be the maturation time of the newly produced viruses. The constant e-m2τ2 denotes the surviving rate of virus during the delay period. Antigenic stimulation generating CTL cell may need a period of time τ3.In this paper, our purpose is to investigate the dynamical properties of model (2), including the local and global stability of equilibria. The reproduction numbers for viral infection, antibody response, CTL immune response, CTL immune competition, and antibody competition, respectively, are calculated. By using Lyapunov functionals and LaSalle’s invariance principle, the threshold conditions for the global asymptotic stability of infection-free equilibrium E0, immune-free equilibrium E1, infection equilibrium E2 only with antibody response, and infection equilibrium E3 only with CTL immune response and infection equilibrium E4 with both antibody and CTL immune responses when the delay τ3=0, respectively, are established. By using the linearization method, the instability of equilibria E0, E1, E2, and E3, respectively, is also established. Furthermore, by using the numerical simulation method, we will discuss the existence of the Hopf bifurcation and stability switches at equilibria E3 and E4 when τ3>0.The organization of this paper is as follows. In the next section, the basic properties of model (2) for the positivity and boundedness of solutions, the threshold values, and the existence of equilibria are discussed. In Section 3, the threshold conditions on the global stability and instability of equilibria E0, E1, and E2 are proved. When τ3=0, the threshold conditions on the global stability and instability for equilibria E3 and E4 are stated and proved. In Section 4, the numerical simulations are given to further discuss the stability of equilibria E3 and E4 when τ3>0. It is shown that the Hopf bifurcation and stability switches at these equilibria occur as τ3 increases. In the last section, we offer a brief conclusion.


## 2. Preliminaries

Letτ=max⁡{τ1,τ2,τ3} and R+5={(x1,x2,x3,x4,x5):xi≥0,i=1,2,…,5}. C([-τ,0],R+5) denotes the space of continuous functions mapping interval [-τ,0] into R+5 with norm ϕ=sup-τ≤t≤0⁡{|ϕ(t)|} for any ϕ∈C([-τ,0],R+5).The initial conditions for any solutions of model (2) are given as follows: (3)xθ,yθ,vθ,zθ,wθ=ϕ1θ,ϕ2θ,ϕ3θ,ϕ4θ,ϕ5θ,ϕiθ≥0,θ∈-τ,0,ϕi0>0,i=1,2,3,4,5,where (ϕ1(θ),ϕ2(θ),ϕ3(θ),ϕ4(θ),ϕ5(θ))∈C([-τ,0],R+5). By the fundamental theory of functional differential equation [36], model (2) admits a unique solution (x(t),y(t),v(t),z(t),w(t)) satisfying initial conditions (3).In this paper, we firstly introduce the following assumptions:(H1)
s(x) is continuously differentiable. There exists x¯>0 such that s(x¯)=0 and s′(x¯)<0.(H2)
f(x,v) is continuously differentiable; f(x,v)>0 for x∈(0,∞), v∈(0,∞); f(x,v)=0 if and only if x=0 or v=0; ∂f(x,v)/∂x≥0 and ∂f(x,v)/∂v≥0 for all x≥0 and v≥0; (d/dx)(∂f(x,0)/∂v)≥0 for all x≥0.(H3)
gi(ξ)(i=1,2,3,4) is strictly increasing on [0,∞); limξ→∞⁡gi(ξ)=+∞; and there exists ki>0 such that gi(ξ)≥kiξ for any ξ≥0; gi(0)=0 and gi′(0)=1.(H4)
f(x,v)/g2(v) is nonincreasing with respect to v for v∈(0,∞).From(H1) we easily obtain that s(x)>0 for all 0<x<x¯ and s(x)<0 for all x>x¯. Assumption (H1) shows that the number of healthy cells x has a maximum capacity x¯ in the absence of infection. When x<x¯, s(x) has a positive growth; if x>x¯ it has a negative growth. Assumption (H2) implies that there are no new infected cells (i.e., f(x,v)=0) without healthy cells (x=0) or virus (v=0). The higher the number of healthy cells x is, the higher the number of healthy cells x which are infected in the unit time will be. Similarly, the higher the amount of virus v is, the higher the number of healthy cells x which are infected in the unit time will be. Assumption (H3) assumes that the death rates of the infected target cells y, virus v, antibodies w, and CTLs z depend on their concentrations. If these numbers y,v,w,z increase, the corresponding rates ag1(y), ug2(v), hg3(w), and bg4(z) will increase, and the ratio gi(ξ)/ξ is no less than a positive constant for i=1,2,3,4. Finally, assumption (H4) indicates that both the rate of new infections of target cells and the virus clearance rate increase according to the level of virus. However, the corresponding ratio is nonincreasing.Using an argument similar to [14] we have the following result.Theorem 1.
Assume that(H1)–(H4) hold. Let (x(t),y(t),v(t),z(t),w(t)) be the solution of model (2) with initial conditions (3); then (x(t),y(t),v(t),z(t),w(t)) is positive and ultimately bounded.Next, we discuss the existence and uniqueness of equilibria of model (2). We know that any equilibrium E=(x,y,v,z,w) of model (2) satisfies (4)sx-fx,v=0,e-m1τ1fx,v-ag1y-pg1yg4z=0,ke-m2τ2g1y-ug2v-qg2vg3w=0,cg1yg4z-bg4z=0,rg2vg3w-hg3w=0.It is clear from (4) that model (2) has a unique infection-free equilibrium E0=(x¯,0,0,0,0). When y=0, from (4) we have s(x)=f(x,v), g2(v)(u+qg3(w))=0, g4(z)=0, and (rg2(v)-h)g3(w)=0. Solving these equations, we have x=x¯, v=0, z=0, and w=0. When v=0, from (4) we have s(x)=0, g1(y)(a+pg4(z))=0, g1(y)=0, g4(z)=0, and g3(w)=0. Solving these equations, we have x=x¯, v=0, z=0, and w=0. Therefore, besides equilibrium E0, model (2) only has the following four possible equilibria: E1=(x1,y1,v1,0,0), E2=(x2,y2,v2,0,w2), E3=(x3,y3,v3,z3,0), and E4=(x4,y4,v4,z4,w4).The existence of immune-free equilibriumE1=(x1,y1,v1,0,0) is equivalent to the existence of positive solution (x1,y1,v1) of the following equations: (5)sx=fx,v=aem1τ1g1y=auem1τ1+m2τ2kg2v.By (H3), the inverse function g2-1(v) exists. Solving s(x)=(auem1τ1+m2τ2/k)g2(v), we have v=φ(x)≜g2-1(ks(x)/auem1τ1+m2τ2) with φ(x¯)=0 and φ(0)=v0, where v0 is the unique positive root of equation s(0)=(auem1τ1+m2τ2/k)g2(v). Define G(x)=f(x,φ(x))-(auem1τ1+m2τ2/k)g2(φ(x)). Then G(0)=-(auem1τ1+m2τ2/k)g2(v0)<0 and G(x¯)=0.Define the basic reproduction number for viral infection(6)R0=ke-m1τ1-m2τ2au∂fx¯,0∂v.Note that (7)G′x¯=∂fx¯,0∂x+∂fx¯,0∂vφ′x¯-auem1τ1+m2τ2kg2′0φ′x¯=auem1τ1+m2τ2kφ′x¯kauem1τ1+m2τ2∂fx¯,0∂v-1=s′x¯R0-1.Thus, if R0>1, then G′(x¯)<0. This implies that there exists x1∈(0,x¯) such that G(x1)=0. The value of v1 is given by v1=φ(x1). (H3) ensures that ke-m2τ2g1(y)=ug2(v1) has a unique positive solution y1=g1-1(uem2τ2g2(v1)/k). Therefore, E1 exists if R0>1.Next we show thatE1=(x1,y1,v1,0,0) is a unique immune-free equilibrium. Otherwise, there exists another E1∗=(x1∗,y1∗,v1∗,0,0). Without of loss of generality, we assume that x1∗<x1, and then s(x1∗)>s(x1). Meanwhile, ks(x1)=auem1τ1+m2τ2g2(v1) and ks(x1∗)=auem1τ1+m2τ2g2(v1∗). By (H3) and (H4), we have v1∗>v1 and f(x1,v1∗)/g2(v1∗)≤f(x1,v1)/g2(v1). Since x1∗<x1, we obtain f(x1,v1∗)>f(x1∗,v1∗) and f(x1∗,v1∗)/g2(v1∗)<f(x1,v1)/g2(v1). For another, we have f(x1∗,v1∗)/g2(v1∗)=f(x1,v1)/g2(v1). This is a contradiction. Thus E1 is a unique equilibrium.We consider the existence of infection equilibriumE2=(x2,y2,v2,0,w2) with only antibody response. It is clear that v2=g2-1(h/r). Define F(x)=s(x)-f(x,v2). By (H1) and (H2), we obtain F′(x)<0. Since F(0)=s(0)>0 and F(x¯)=s(x¯)-f(x¯,v2)<0, there exists a unique x2∈(0,x¯) such that F(x2)=0. Then, we have y2=g1-1(e-m1τ1f(x2,v2)/a).Define the constant(8)R1=ke-m1τ1-m2τ2ufx2,v2g2v2,which is called the antibody response reproductive number of model (2). Solving w2 from (4), we obtain that (9)w2=g3-1ke-m2τ2g1y2-ug2v2qg2v2=g3-1uR1-1q>0ifR1>1.Therefore, E2 exists and is unique if R1>1.We consider the existence of infection equilibriumE3=(x3,y3,v3,z3,0) with only CTL immune response. From the third and fourth equations of (4), we obtain unique y3=g1-1(b/c) and v3=g2-1(bke-m2τ2/cu). Define F(x)=s(x)-f(x,v3). By (H1) and (H2), we obtain F′(x)<0. Since F(0)=s(0)>0 and F(x¯)=s(x¯)-f(x¯,v3)<0, there exists a unique x3∈(0,x¯) such that F(x3)=0.Define the constant(10)R2=ke-m1τ1-m2τ2aufx3,v3g2v3,which is called the CTL immune response reproductive number of model (2). Solving the second equation for z yields (11)z3=g4-1e-m1τ1fx3,v3-ag1y3pg1y3=g4-1aR2-1p>0ifR2>1.Therefore, E3 exists and is unique if R2>1.Lastly, we consider the existence of infection equilibriumE4=(x4,y4,v4,z4,w4) with both antibody and CTL immune responses. From the fourth and fifth equation of (4), we obtain unique y4=g1-1(b/c) and v4=g2-1(h/r). Define F(x)=s(x)-f(x,v4). By (H1) and (H2), we obtain F′(x)<0. Since F(0)=s(0)>0 and F(x¯)=s(x¯)-f(x¯,v4)<0, there exists a unique x4∈(0,x¯) such that F(x4)=0.Define the constants(12)R3=cfx4,v4abem1τ1,R4=kbruchem2τ2,which are called the CTL immune response competitive reproductive number and the antibody response competitive reproductive number of model (2), respectively. Solving the second equation for z yields a unique (13)z4=g4-1e-m1τ1fx4,v4-ag1y4pg1y4=g4-1aR3-1p>0ifR3>1.Solving the third equation for w, we further obtain a unique (14)w4=g3-1ke-m2τ2g1y4-ug2v4qg2v4=g3-1uR4-1q>0ifR4>1.Therefore, E4 exists and is unique if R3>1 and R4>1.Remark 2.
From(H2) and (H4), we obtain R1<R0 and R2<R0. In fact, (15)R1=ke-m1τ1-m2τ2ufx2,v2gv2≤ke-m1τ1-m2τ2ulimv→0+⁡fx2,vg2v=ke-m1τ1-m2τ2ug2′0∂fx2,0∂v<ke-m1τ1-m2τ2ug2′0∂fx¯,0∂v=R0,R2=ke-m1τ1-m2τ2aufx3,v3g2v3≤ke-m1τ1-m2τ2aulimv→0+⁡fx3,vg2v=ke-m1τ1-m2τ2aug2′0∂fx3,0∂v<ke-m1τ1-m2τ2aug2′0∂fx¯,0∂v=R0.


## 3. Stability Analysis

### 3.1. Stability of EquilibriumE0

Theorem 3.
(a) IfR0≤1, then infection-free equilibrium E0 is globally asymptotically stable.
(b) IfR0>1, then E0 is unstable.Proof.
Consider conclusion (a). Define a Lyapunov functionalV1(t) as follows: (16)V1t=xt-∫x¯xtlimv→0⁡fx¯,vfθ,vdθ+em1τ1yt+aem1τ1+m2τ2kvt+pem1τ1czt+∫-τ10fxt+s,vt+sds+aem1τ1∫-τ20g1yt+sds+pem1τ1∫-τ30g1yt+sg4zt+sds+aqem1τ1+m2τ2krwt.Calculating the time derivative of V1(t) along solutions of model (2), we obtain (17)dV1tdt=sx1-limv→0⁡fx¯,vfx,v+fx,vlimv→0⁡fx¯,vfx,v-auem1τ1+m2τ2kg2v-pbem1τ1cg4z-aqhem1τ1+m2τ2krg3w≤auem1τ1+m2τ2kg2vkauem1τ1+m2τ2fx,vg2vlimv→0⁡fx¯,vfx,v-1+sx1-limv→0⁡fx¯,vfx,v.Note that s(x)(1-limv→0⁡(fx¯,v/fx,v))≤0, and (18)fx,vg2vlimv→0⁡fx¯,vfx,v≤limv→0⁡fx,vg2v∂fx¯,0/∂v∂fx,0/∂v=∂fx¯,0∂v1g2′0.It follows that (19)dV1tdt≤auem1τ1+m2τ2kg2vR0-1.
Note thatdV1(t)/dt=0 if and only if x(t)=x¯, v(t)=0, z(t)=0, y(t)=0, and w(t)=0. So, the maximal compact invariant set in {(x,y,v,z,w)∈R+5:dV1(t)/dt=0} is singleton {E0}. By LaSalle’s invariance principle [36], E0 is globally asymptotically stable.
Next, we consider conclusion (b). By computing, the characteristic equation of the linearization system of model (2) at E0 is (20)λ+hg3′0λ+bg4′0λ-s′x¯fλ=0,where (21)fλ=λ2+a+uλ+au-k∂fx¯,0∂ve-m1+λτ1e-m2+λτ2.When R0>1, we have f(0)=au-k(∂fx¯,0/∂v)e-m1τ1e-m2τ2<0 and limλ→+∞⁡f(λ)=+∞. Hence, there is λ>0 such that f(λ)=0. Therefore, when R0>1, E0 is unstable. This completes the proof.Remark 4.
Theorem3 shows that if only equilibrium E0 exists, then it is globally asymptotically stable, and delays τ1, τ2, and τ3 do not impact the stability of E0.


### 3.2. Stability of EquilibriumE1

Firstly, we introduce two lemmas which will be used in the proof of Theorem7.Lemma 5.
Suppose that(H1)–(H4) hold and R0>1. Let x2 and v2 satisfy g2(v2)=h/r and s(x2)=f(x2,v2). Then for equilibrium E1=(x1,y1,v1,0,0), sign⁡(x2-x1)=sign⁡(v1-v2)=sign⁡(R1-1).Proof.
Sinces(x1)=f(x1,v1), we have (22)sx2-sx1=fx2,v2-fx1,v2+fx1,v2-fx1,v1.By (H1) and (H2), we get sign⁡(x2-x1)=sign⁡(v1-v2). Using (ke-m1τ1-m2τ2/au)(fx1,v1/g2v1)=1, we have (23)R1-1=kauem1τ1+m2τ21g2v2fx2,v2-fx1,v2+fx1,v2g2v2-fx1,v1g2v1.By (H2) and (H4), it follows that sign⁡(R1-1)=sign⁡(v1-v2). This completes the proof.Lemma 6.
Suppose that(H1)–(H4) hold and R0>1. Let x3, y3, and v3 satisfy g2(v3)=kbe-m2τ2/uc, g1(y3)=b/c, and s(x3)=f(x3,v3). Then, for equilibrium E1=(x1,y1,v1,0,0), sign⁡(x3-x1)=sign⁡(v1-v3)=sign⁡(y1-y3)=sign⁡(R2-1).Proof.
Sinceg1(y1)=(uem2τ2/k)g2(v1) and g1(y3)=(uem2τ2/k)g2(v3), we have sign⁡(v1-v3)=sign⁡(y1-y3). Since s(x1)=f(x1,v1), one has (24)sx3-sx1=fx3,v3-fx3,v1+fx3,v1-fx1,v1.By (H1) and (H2), we get sign⁡(x3-x1)=sign⁡(v1-v3), and (25)R2-1=kauem1τ1+m2τ2fx3,v3g2v3-fx3,v1g2v1+fx3,v1-fx1,v1g2v1.By (H2) and (H4), we further have sign⁡(R2-1)=sign⁡(x3-x1). This completes the proof.Theorem 7.
LetR0>1. (a) If R1≤1 and R2≤1, then immune-free equilibrium E1 is globally asymptotically stable. (b) If R1>1 or R2>1, then E1 is unstable.Proof.
Consider conclusion (a). DenoteH(ξ)=ξ-1-ln⁡ξ with ξ∈R+. Define a Lyapunov functional V2(t) as follows: (26)V2t=xt-∫x1xtfx1,v1fθ,v1dθ+em1τ1yt-∫y1ytg1y1g1θdθ+aem1τ1+m2τ2kvt-∫v1vtg2v1g2θdθ+pem1τ1czt+aqem1τ1+m2τ2krwt+fx1,v1∫-τ10Hfxt+s,vt+sfx1,v1ds+pem1τ1∫-τ30g1yt+sg4zt+sds+aem1τ1g1y1∫-τ20Hg1yt+sg1y1ds.Calculating the derivative of V2(t) along solutions of model (2), we obtain (27)dV2tdt=sx1-fx1,v1fx,v1+fx,vfx1,v1fx,v1-auem1τ1+m2τ2kg2v+M1+M2,where (28)M1=pem1τ1g1y1g4z-pbem1τ1cg4z+aqem1τ1+m2τ2kg2v1g3w-aqhem1τ1+m2τ2krg3w=pem1τ1g4zg1y1-g1y3+aqem1τ1+m2τ2kg3wg2v1-g2v2,M2=fx1,v12-g1y1fxt-τ1,vt-τ1g1yfx1,v1-g2v1g1yt-τ2g1y1g2v+ln⁡fxt-τ1,vt-τ1fx,v+ln⁡g1yt-τ2g1y=fx1,v1ln⁡g2vfx1,v1g2v1fx,v-fx1,v1Hg1y1fxt-τ1,vt-τ1g1yfx1,v1-fx1,v1Hg2v1g1yt-τ2g1y1g2v.Therefore, (29)dV2tdt=sx-sx11-fx1,v1fx,v1-fx1,v1Hg2v1g1yt-τ2g1y1g2v+fx1,v1g2vg2v1fx,vfx,v1-1g2v1g2v-fx,v1fx,v+M1-fx1,v1Hfx1,v1fx,v1-fx1,v1Hg2vfx,v1g2v1fx,v-fx1,v1Hg1y1fxt-τ1,vt-τ1g1yfx1,v1.Note that (s(x)-s(x1))(1-f(x1,v1)/f(x,v1))≤0, and (30)fx,vfx,v1-1g2v1g2v-fx,v1fx,v≤0fort≥0.
Lemmas5 and 6 imply that y1≤y3 and v1≤v2 if R1≤1 and R2≤1. It then follows from the monotonicity of g1 and g2 that M1≤0. We have dV2(t)/dt≤0, and dV2(t)/dt=0 if and only if x(t)=x1, y(t)=y1, v(t)=v1, z(t)=0, and w(t)=0. From LaSalle’s invariance principle [36], we finally have that equilibrium E1 of model (2) is globally asymptotically stable when R0>1, R1≤1, and R2≤1.
Next, consider conclusion (b). By computing, the characteristic equation of the linearization system of model (2) at E1 is (31)λ+h-rg2v1f1λf2λ=0,where f1(λ)=λ+b-cg1(y1)e-λτ3 and (32)f2λ=λ-s′x1+∂fx1,v1∂x0∂fx1,v1∂v-e-m1+λτ1∂fx1,v1∂xλ+ag1′y1-e-m1+λτ1∂fx1,v1∂v0-ke-m2+λτ2g1′y1λ+ug2′v1.When R1>1, we have h-rg2(v1)=r(g2(v2)-g2(v1))<0. Hence, there is a positive root λ∗=rg2(v1)-h. When R2>1, we have f1(0)=b-cg1(y1)=c(g1(y3)-g1(y1))<0 and limλ→+∞⁡f1(λ)=+∞. Hence, there is also a positive root λ∗ such that f1(λ∗)=0. Therefore, when R1>1 or R2>1, E1 is unstable. This completes the proof.Remark 8.
Theorem7 shows that if only equilibria E0 and E1 exist, then E1 is globally asymptotically stable, and delays τ1, τ2, and τ3 do not impact the stability of E1.


### 3.3. Stability of EquilibriumE2

We firstly have the following Lemma.Lemma 9.
SupposeR1>1 and R3≤1. Let E4¯=(x4¯,y4¯,v4¯,z4¯,w4¯) be the solution of equation (4) with v4¯=g2-1(h/r) and y4¯=g1-1(b/c). Then for equilibrium E2=(x2,y2,v2,0,w2),y2≤y4¯.Proof.
SinceE4¯ satisfies (4), we have y4¯=g1-1(b/c), v4¯=g2-1(h/r), and x4¯=x2. Compared with E4, we obtain x4¯=x4 and v4¯=v4. When R3≤1, we get z4¯≤0. Since (33)e-m1τ1fx2,v2=ag1y2,e-m1τ1fx4¯,v4¯=ag1y4¯+pg1y4¯g4z4¯,it follows that y2≤y4¯ if R1>1 and R3≤1. This completes the proof.Theorem 10.
LetR1>1. (a) If R3≤1, then antibody response equilibrium E2 is globally asymptotically stable.
(b) IfR3>1, then E2 is unstable.Proof.
Consider conclusion (a). Define a Lyapunov functionalV3(t) as follows: (34)V3t=xt-∫x2xtfx2,v2fθ,v2dθ+em1τ1yt-∫y2ytg1y2g1θdθ+fx2,v2g2v2u+qg3w2vt-∫v2vtg2v2g2θdθ+pem1τ1czt+qfx2,v2g2v2ru+qg3w2wt-∫w2wtg3w2g3θdθ+fx2,v2∫-τ10Hfxt+s,vt+sfx2,v2ds+fx2,v2∫-τ20Hg1yt+sg1y2ds+pem1τ1∫-τ30g1yt+sg4zt+sds.Calculating the derivative of V3(t) along solutions of model (2), we obtain (35)dV3tdt=sx1-fx2,v2fx,v2+fx,vfx2,v2fx,v2-fx2,v2g2vg2v2+M1+M2,where (36)M1=pem1τ1g1y2g4z-pbem1τ1cg4z=pem1τ1g4zg1y2-g1y4¯,M2=fx2,v2ln⁡g2vfx2,v2g2v2fx,v-fx2,v2Hg2v2g1yt-τ2g1y2g2v-fx2,v2Hg1y2fxt-τ1,vt-τ1g1yfx2,v2.Therefore, (37)dV3tdt=sx-sx21-fx2,v2fx,v2-fx2,v2Hg2v2g1yt-τ2g1y2g2v+fx2,v2g2vg2v2fx,vfx,v2-1g2v2g2v-fx,v2fx,v+pem1τ1g4zg1y2-g1y4¯-fx2,v2Hg2vfx,v2g2v2fx,v-fx2,v2Hg1y2fxt-τ1,vt-τ1g1yfx2,v2-fx2,v2Hfx2,v2fx,v2.Note that (s(x)-s(x2))(1-f(x2,v2)/f(x,v2))≤0, and (38)fx,vfx,v2-1g2v2g2v-fx,v2fx,v≤0fort≥0.
Sincey2≤y4¯, we have dV3(t)/dt≤0, and dV3(t)/dt=0 if and only if x(t)=x2, y(t)=y2, v(t)=v2, and z(t)=0. From LaSalle’s invariance principle [36], we finally have that E2 is globally asymptotically stable when R1>1 and R3≤1.
Next, consider conclusion (b). By computing, the characteristic equation of linearization system of model (2) at E2 is (39)f1λf2λ=0,where f1(λ)=λ+b-ce-λτ3g1(y2) and (40)f2λ=a110a130a21a22a2300a32a33a3400a43a44,where (41)a11=λ-s′x2+∂fx2,v2∂x,a13=∂fx2,v2∂v,a21=-e-m1+λτ1∂fx2,v2∂x,a22=λ+ag1′y2,a23=-e-m1+λτ1∂fx2,v2∂v,a32=-ke-m2+λτ2g1′y2,a33=λ+u+qg3w2g2′v2,a34=qg2v2g3′w2,a43=-rg2′v2g3w2,a44=λ+h-rg2v2g3′w2.When R3>1, we have f1(0)=b-cg1(y2)=c(g1(y4¯)-g1(y2))<0 and limλ→+∞⁡f1(λ)=+∞. Hence, there is also a positive root λ∗ such that f1(λ∗)=0. Therefore, when R3>1, E2 is unstable. This completes the proof.Remark 11.
Theorem10 shows that if only equilibria E0, E1, and E2 exist, then when R3≤1 and R1>1, E2 is globally asymptotically stable, and delays τ1, τ2, and τ3 do not impact the stability of E2.


### 3.4. Stability of EquilibriumE3

On the stability analysis of equilibriumE3, we only discuss the following case: τ1≥0, τ2≥0, and τ3=0. Other cases, τ1≥0, τ2≥0, and τ3≥0, are numerically verified for bifurcation phenomena and stability switches of E3 but the analytic analysis is left as an open problem. Before the proof of theorem, we have the following Lemma.Lemma 12.
SupposeR2>1 and R4≤1. Let E4¯=(x4¯,y4¯,v4¯,z4¯,w4¯) be the solution of (4) with v4¯=g2-1(h/r) and y4¯=g1-1(b/c). Then for equilibrium E3=(x3,y3,v3,z3,0), v3≤v4¯.Proof.
SinceE4¯ satisfies (4), we have y4¯=g1-1(b/c), v4¯=g2-1(h/r), and x4¯=x2. Compared with E4, we get y4¯=y4 and v4¯=v4. When R4≤1, we obtain w4¯<0. Since (42)ke-m2τ2g1y3=ug2v3,ke-m2τ2g1y4¯=ug2v4¯+qg2v4¯g3w4¯,it follows that v3≤v4¯ if R2>1 and R4≤1. This completes the proof.Theorem 13.
LetR2>1. (a) If R4≤1 and τ3=0, then infection equilibrium E3 with only CTL response is globally asymptotically stable.
(b) IfR4>1, then E3 is unstable.Proof.
We first consider conclusion (a). Define a Lyapunov functionalV4(t) as follows: (43)V4t=xt-∫x3xtfx3,v3fθ,v3dθ+em1τ1yt-∫y3ytg1y3g1θdθ+fx3,v3g2v3uvt-∫v3vtg2v3g2θdθ+pem1τ1czt-∫z3ztg4z3g4θdθ+qfx3,v3g2v3ruwt+fx3,v3∫-τ10Hfxt+s,vt+sfx3,v3ds+fx3,v3∫-τ20Hg1yt+sg1y3ds.Calculating the derivative of V4(t) along solutions of model (2), we obtain that (44)dV4tdt=sx1-fx3,v3fx,v3+fx,vfx3,v3fx,v3-fx3,v3g2vg2v3+qfx3,v3ug2v3g3wg2v3-g2v4¯-fx3,v3Hg2v3g1yt-τ2g1y3g2v+fx3,v3ln⁡g2vfx3,v3g2v3fx,v-fx3,v3Hg1y3fxt-τ1,vt-τ1g1yfx3,v3=sx-sx31-fx3,v3fx,v3-fx3,v3Hg2v3g1yt-τ2g1y3g2v+fx3,v3g2vg2v3fx,vfx,v3-1g2v3g2v-fx,v3fx,v-fx3,v3Hfx3,v3fx,v3-fx3,v3Hg1y3fxt-τ1,vt-τ1g1yfx3,v3-fx3,v3Hg2vfx,v3g2v3fx,v+qfx3,v3ug2v3g3wg2v3-g2v4¯.Note that (s(x)-s(x3))(1-f(x3,v3)/f(x,v3))≤0, and (45)fx,vfx,v3-1g2v3g2v-fx,v3fx,v≤0fort≥0.
Sincev3≤v4¯, we have dV4(t)/dt≤0, and dV4(t)/dt=0 if and only if x(t)=x3, y(t)=y3, v(t)=v3, and w(t)=0. From LaSalle’s invariance principle [36], we finally have that E3 is globally asymptotically stable when τ3=0, R0>1, R2>1, and R4≤1.
Next, we consider conclusion (b). By computing, the characteristic equation of the linearization system of model (2) at E3 is (46)λ+h-rg2v3fλ=0,where (47)fλ=a110a130a21a22a23a240a32a3300a420a44,where (48)a11=λ-s′x3+∂fx3,v3∂x,a13=∂fx3,v3∂v,a21=-e-m1+λτ1∂fx3,v3∂x,a22=λ+a+pg4z3g1′y3,a23=-e-m1+λτ1∂fx3,v3∂v,a24=pg1y3g4′z3,a32=-ke-m2+λτ2g1′y3,a33=λ+ug2′v3,a42=-ce-λτ3g4z3g1′y3,a44=λ+b-cg1y3e-λτ3g4′z3.When R4>1, we have h-rg2(v3)=r(g2(v4¯)-g2(v3))<0. Hence, there is a positive root λ∗=rg2(v3)-h. Therefore, when R4>1, E3 is unstable for any τ1≥0, τ2≥0, and τ3≥0. This completes the proof.Remark 14.
Theorem13 shows that if only equilibria E0, E1, E2, and E3 exist, then when R2>1, R4≤1, and τ3=0,E3 is globally asymptotically stable, and delays τ1 and τ2 do not impact the stability of E3.


### 3.5. Stability of EquilibriumE4

On the stability analysis of equilibriumE4, we here only discuss the following case: τ1≥0, τ2≥0, and τ3=0. However, for the cases τ1≥0, τ2≥0, and τ3≥0, the theoretical analysis is very complicated. We will give numerical analysis for this case in the next section.Theorem 15.
Ifτ3=0, R3>1, and R4>1, then infection equilibrium E4 with both antibody and CTL immune responses is globally asymptotically stable.Proof.
Define a Lyapunov functionalV5(t) as follows: (49)V5t=xt-∫x4xtfx4,v4fθ,v4dθ+em1τ1yt-∫y4ytg1y4g1θdθ+fx4,v4g2v4u+qg3w4vt-∫v4vtg2v4g2θdθ+pem1τ1czt-∫z4ztg4z4g4θdθ+fx4,v4∫-τ20Hg1yt+sg1y4ds+qfx4,v4g2v4u+qg3w4wt-∫w4wtg3w4g3θdθ+fx4,v4∫-τ10Hfxt+s,vt+sfx4,v4ds.Using the above similar method, we obtain (50)dV5tdt=sx-sx41-fx4,v4fx,v4-fx4,v4Hfx4,v4fx,v4+Hg2vfx,v4g2v4fx,v+Hg1y4fxt-τ1,vt-τ1g1yfx4,v4+Hg2v4g1yt-τ2g1y4g2v+fx4,v4g2vg2v4fx,vfx,v4-1g2v4g2v-fx,v4fx,v.Note that (s(x)-s(x4))(1-f(x4,v4)/f(x,v4))≤0, and (51)fx,vfx,v4-1g2v4g2v-fx,v4fx,v≤0fort≥0.Obviously, we have dV5(t)/dt≤0, and dV5(t)/dt=0 if and only if x(t)=x4, y(t)=y4, and v(t)=v4. From LaSalle’s invariance principle [36], we finally have that E4 is globally asymptotically stable when τ3=0, R3>1, and R4>1. This completes the proof.Remark 16.
Theorem15 shows that if equilibria E0, E1, E2, E3, and E4 exist, then when R3>1, R4>1, and τ3=0, E4 is globally asymptotically stable, and delays τ1 and τ2 do not impact the stability of E4.


## 3.1. Stability of EquilibriumE0

Theorem 3.
(a) IfR0≤1, then infection-free equilibrium E0 is globally asymptotically stable.
(b) IfR0>1, then E0 is unstable.Proof.
Consider conclusion (a). Define a Lyapunov functionalV1(t) as follows: (16)V1t=xt-∫x¯xtlimv→0⁡fx¯,vfθ,vdθ+em1τ1yt+aem1τ1+m2τ2kvt+pem1τ1czt+∫-τ10fxt+s,vt+sds+aem1τ1∫-τ20g1yt+sds+pem1τ1∫-τ30g1yt+sg4zt+sds+aqem1τ1+m2τ2krwt.Calculating the time derivative of V1(t) along solutions of model (2), we obtain (17)dV1tdt=sx1-limv→0⁡fx¯,vfx,v+fx,vlimv→0⁡fx¯,vfx,v-auem1τ1+m2τ2kg2v-pbem1τ1cg4z-aqhem1τ1+m2τ2krg3w≤auem1τ1+m2τ2kg2vkauem1τ1+m2τ2fx,vg2vlimv→0⁡fx¯,vfx,v-1+sx1-limv→0⁡fx¯,vfx,v.Note that s(x)(1-limv→0⁡(fx¯,v/fx,v))≤0, and (18)fx,vg2vlimv→0⁡fx¯,vfx,v≤limv→0⁡fx,vg2v∂fx¯,0/∂v∂fx,0/∂v=∂fx¯,0∂v1g2′0.It follows that (19)dV1tdt≤auem1τ1+m2τ2kg2vR0-1.
Note thatdV1(t)/dt=0 if and only if x(t)=x¯, v(t)=0, z(t)=0, y(t)=0, and w(t)=0. So, the maximal compact invariant set in {(x,y,v,z,w)∈R+5:dV1(t)/dt=0} is singleton {E0}. By LaSalle’s invariance principle [36], E0 is globally asymptotically stable.
Next, we consider conclusion (b). By computing, the characteristic equation of the linearization system of model (2) at E0 is (20)λ+hg3′0λ+bg4′0λ-s′x¯fλ=0,where (21)fλ=λ2+a+uλ+au-k∂fx¯,0∂ve-m1+λτ1e-m2+λτ2.When R0>1, we have f(0)=au-k(∂fx¯,0/∂v)e-m1τ1e-m2τ2<0 and limλ→+∞⁡f(λ)=+∞. Hence, there is λ>0 such that f(λ)=0. Therefore, when R0>1, E0 is unstable. This completes the proof.Remark 4.
Theorem3 shows that if only equilibrium E0 exists, then it is globally asymptotically stable, and delays τ1, τ2, and τ3 do not impact the stability of E0.


## 3.2. Stability of EquilibriumE1

Firstly, we introduce two lemmas which will be used in the proof of Theorem7.Lemma 5.
Suppose that(H1)–(H4) hold and R0>1. Let x2 and v2 satisfy g2(v2)=h/r and s(x2)=f(x2,v2). Then for equilibrium E1=(x1,y1,v1,0,0), sign⁡(x2-x1)=sign⁡(v1-v2)=sign⁡(R1-1).Proof.
Sinces(x1)=f(x1,v1), we have (22)sx2-sx1=fx2,v2-fx1,v2+fx1,v2-fx1,v1.By (H1) and (H2), we get sign⁡(x2-x1)=sign⁡(v1-v2). Using (ke-m1τ1-m2τ2/au)(fx1,v1/g2v1)=1, we have (23)R1-1=kauem1τ1+m2τ21g2v2fx2,v2-fx1,v2+fx1,v2g2v2-fx1,v1g2v1.By (H2) and (H4), it follows that sign⁡(R1-1)=sign⁡(v1-v2). This completes the proof.Lemma 6.
Suppose that(H1)–(H4) hold and R0>1. Let x3, y3, and v3 satisfy g2(v3)=kbe-m2τ2/uc, g1(y3)=b/c, and s(x3)=f(x3,v3). Then, for equilibrium E1=(x1,y1,v1,0,0), sign⁡(x3-x1)=sign⁡(v1-v3)=sign⁡(y1-y3)=sign⁡(R2-1).Proof.
Sinceg1(y1)=(uem2τ2/k)g2(v1) and g1(y3)=(uem2τ2/k)g2(v3), we have sign⁡(v1-v3)=sign⁡(y1-y3). Since s(x1)=f(x1,v1), one has (24)sx3-sx1=fx3,v3-fx3,v1+fx3,v1-fx1,v1.By (H1) and (H2), we get sign⁡(x3-x1)=sign⁡(v1-v3), and (25)R2-1=kauem1τ1+m2τ2fx3,v3g2v3-fx3,v1g2v1+fx3,v1-fx1,v1g2v1.By (H2) and (H4), we further have sign⁡(R2-1)=sign⁡(x3-x1). This completes the proof.Theorem 7.
LetR0>1. (a) If R1≤1 and R2≤1, then immune-free equilibrium E1 is globally asymptotically stable. (b) If R1>1 or R2>1, then E1 is unstable.Proof.
Consider conclusion (a). DenoteH(ξ)=ξ-1-ln⁡ξ with ξ∈R+. Define a Lyapunov functional V2(t) as follows: (26)V2t=xt-∫x1xtfx1,v1fθ,v1dθ+em1τ1yt-∫y1ytg1y1g1θdθ+aem1τ1+m2τ2kvt-∫v1vtg2v1g2θdθ+pem1τ1czt+aqem1τ1+m2τ2krwt+fx1,v1∫-τ10Hfxt+s,vt+sfx1,v1ds+pem1τ1∫-τ30g1yt+sg4zt+sds+aem1τ1g1y1∫-τ20Hg1yt+sg1y1ds.Calculating the derivative of V2(t) along solutions of model (2), we obtain (27)dV2tdt=sx1-fx1,v1fx,v1+fx,vfx1,v1fx,v1-auem1τ1+m2τ2kg2v+M1+M2,where (28)M1=pem1τ1g1y1g4z-pbem1τ1cg4z+aqem1τ1+m2τ2kg2v1g3w-aqhem1τ1+m2τ2krg3w=pem1τ1g4zg1y1-g1y3+aqem1τ1+m2τ2kg3wg2v1-g2v2,M2=fx1,v12-g1y1fxt-τ1,vt-τ1g1yfx1,v1-g2v1g1yt-τ2g1y1g2v+ln⁡fxt-τ1,vt-τ1fx,v+ln⁡g1yt-τ2g1y=fx1,v1ln⁡g2vfx1,v1g2v1fx,v-fx1,v1Hg1y1fxt-τ1,vt-τ1g1yfx1,v1-fx1,v1Hg2v1g1yt-τ2g1y1g2v.Therefore, (29)dV2tdt=sx-sx11-fx1,v1fx,v1-fx1,v1Hg2v1g1yt-τ2g1y1g2v+fx1,v1g2vg2v1fx,vfx,v1-1g2v1g2v-fx,v1fx,v+M1-fx1,v1Hfx1,v1fx,v1-fx1,v1Hg2vfx,v1g2v1fx,v-fx1,v1Hg1y1fxt-τ1,vt-τ1g1yfx1,v1.Note that (s(x)-s(x1))(1-f(x1,v1)/f(x,v1))≤0, and (30)fx,vfx,v1-1g2v1g2v-fx,v1fx,v≤0fort≥0.
Lemmas5 and 6 imply that y1≤y3 and v1≤v2 if R1≤1 and R2≤1. It then follows from the monotonicity of g1 and g2 that M1≤0. We have dV2(t)/dt≤0, and dV2(t)/dt=0 if and only if x(t)=x1, y(t)=y1, v(t)=v1, z(t)=0, and w(t)=0. From LaSalle’s invariance principle [36], we finally have that equilibrium E1 of model (2) is globally asymptotically stable when R0>1, R1≤1, and R2≤1.
Next, consider conclusion (b). By computing, the characteristic equation of the linearization system of model (2) at E1 is (31)λ+h-rg2v1f1λf2λ=0,where f1(λ)=λ+b-cg1(y1)e-λτ3 and (32)f2λ=λ-s′x1+∂fx1,v1∂x0∂fx1,v1∂v-e-m1+λτ1∂fx1,v1∂xλ+ag1′y1-e-m1+λτ1∂fx1,v1∂v0-ke-m2+λτ2g1′y1λ+ug2′v1.When R1>1, we have h-rg2(v1)=r(g2(v2)-g2(v1))<0. Hence, there is a positive root λ∗=rg2(v1)-h. When R2>1, we have f1(0)=b-cg1(y1)=c(g1(y3)-g1(y1))<0 and limλ→+∞⁡f1(λ)=+∞. Hence, there is also a positive root λ∗ such that f1(λ∗)=0. Therefore, when R1>1 or R2>1, E1 is unstable. This completes the proof.Remark 8.
Theorem7 shows that if only equilibria E0 and E1 exist, then E1 is globally asymptotically stable, and delays τ1, τ2, and τ3 do not impact the stability of E1.


## 3.3. Stability of EquilibriumE2

We firstly have the following Lemma.Lemma 9.
SupposeR1>1 and R3≤1. Let E4¯=(x4¯,y4¯,v4¯,z4¯,w4¯) be the solution of equation (4) with v4¯=g2-1(h/r) and y4¯=g1-1(b/c). Then for equilibrium E2=(x2,y2,v2,0,w2),y2≤y4¯.Proof.
SinceE4¯ satisfies (4), we have y4¯=g1-1(b/c), v4¯=g2-1(h/r), and x4¯=x2. Compared with E4, we obtain x4¯=x4 and v4¯=v4. When R3≤1, we get z4¯≤0. Since (33)e-m1τ1fx2,v2=ag1y2,e-m1τ1fx4¯,v4¯=ag1y4¯+pg1y4¯g4z4¯,it follows that y2≤y4¯ if R1>1 and R3≤1. This completes the proof.Theorem 10.
LetR1>1. (a) If R3≤1, then antibody response equilibrium E2 is globally asymptotically stable.
(b) IfR3>1, then E2 is unstable.Proof.
Consider conclusion (a). Define a Lyapunov functionalV3(t) as follows: (34)V3t=xt-∫x2xtfx2,v2fθ,v2dθ+em1τ1yt-∫y2ytg1y2g1θdθ+fx2,v2g2v2u+qg3w2vt-∫v2vtg2v2g2θdθ+pem1τ1czt+qfx2,v2g2v2ru+qg3w2wt-∫w2wtg3w2g3θdθ+fx2,v2∫-τ10Hfxt+s,vt+sfx2,v2ds+fx2,v2∫-τ20Hg1yt+sg1y2ds+pem1τ1∫-τ30g1yt+sg4zt+sds.Calculating the derivative of V3(t) along solutions of model (2), we obtain (35)dV3tdt=sx1-fx2,v2fx,v2+fx,vfx2,v2fx,v2-fx2,v2g2vg2v2+M1+M2,where (36)M1=pem1τ1g1y2g4z-pbem1τ1cg4z=pem1τ1g4zg1y2-g1y4¯,M2=fx2,v2ln⁡g2vfx2,v2g2v2fx,v-fx2,v2Hg2v2g1yt-τ2g1y2g2v-fx2,v2Hg1y2fxt-τ1,vt-τ1g1yfx2,v2.Therefore, (37)dV3tdt=sx-sx21-fx2,v2fx,v2-fx2,v2Hg2v2g1yt-τ2g1y2g2v+fx2,v2g2vg2v2fx,vfx,v2-1g2v2g2v-fx,v2fx,v+pem1τ1g4zg1y2-g1y4¯-fx2,v2Hg2vfx,v2g2v2fx,v-fx2,v2Hg1y2fxt-τ1,vt-τ1g1yfx2,v2-fx2,v2Hfx2,v2fx,v2.Note that (s(x)-s(x2))(1-f(x2,v2)/f(x,v2))≤0, and (38)fx,vfx,v2-1g2v2g2v-fx,v2fx,v≤0fort≥0.
Sincey2≤y4¯, we have dV3(t)/dt≤0, and dV3(t)/dt=0 if and only if x(t)=x2, y(t)=y2, v(t)=v2, and z(t)=0. From LaSalle’s invariance principle [36], we finally have that E2 is globally asymptotically stable when R1>1 and R3≤1.
Next, consider conclusion (b). By computing, the characteristic equation of linearization system of model (2) at E2 is (39)f1λf2λ=0,where f1(λ)=λ+b-ce-λτ3g1(y2) and (40)f2λ=a110a130a21a22a2300a32a33a3400a43a44,where (41)a11=λ-s′x2+∂fx2,v2∂x,a13=∂fx2,v2∂v,a21=-e-m1+λτ1∂fx2,v2∂x,a22=λ+ag1′y2,a23=-e-m1+λτ1∂fx2,v2∂v,a32=-ke-m2+λτ2g1′y2,a33=λ+u+qg3w2g2′v2,a34=qg2v2g3′w2,a43=-rg2′v2g3w2,a44=λ+h-rg2v2g3′w2.When R3>1, we have f1(0)=b-cg1(y2)=c(g1(y4¯)-g1(y2))<0 and limλ→+∞⁡f1(λ)=+∞. Hence, there is also a positive root λ∗ such that f1(λ∗)=0. Therefore, when R3>1, E2 is unstable. This completes the proof.Remark 11.
Theorem10 shows that if only equilibria E0, E1, and E2 exist, then when R3≤1 and R1>1, E2 is globally asymptotically stable, and delays τ1, τ2, and τ3 do not impact the stability of E2.


## 3.4. Stability of EquilibriumE3

On the stability analysis of equilibriumE3, we only discuss the following case: τ1≥0, τ2≥0, and τ3=0. Other cases, τ1≥0, τ2≥0, and τ3≥0, are numerically verified for bifurcation phenomena and stability switches of E3 but the analytic analysis is left as an open problem. Before the proof of theorem, we have the following Lemma.Lemma 12.
SupposeR2>1 and R4≤1. Let E4¯=(x4¯,y4¯,v4¯,z4¯,w4¯) be the solution of (4) with v4¯=g2-1(h/r) and y4¯=g1-1(b/c). Then for equilibrium E3=(x3,y3,v3,z3,0), v3≤v4¯.Proof.
SinceE4¯ satisfies (4), we have y4¯=g1-1(b/c), v4¯=g2-1(h/r), and x4¯=x2. Compared with E4, we get y4¯=y4 and v4¯=v4. When R4≤1, we obtain w4¯<0. Since (42)ke-m2τ2g1y3=ug2v3,ke-m2τ2g1y4¯=ug2v4¯+qg2v4¯g3w4¯,it follows that v3≤v4¯ if R2>1 and R4≤1. This completes the proof.Theorem 13.
LetR2>1. (a) If R4≤1 and τ3=0, then infection equilibrium E3 with only CTL response is globally asymptotically stable.
(b) IfR4>1, then E3 is unstable.Proof.
We first consider conclusion (a). Define a Lyapunov functionalV4(t) as follows: (43)V4t=xt-∫x3xtfx3,v3fθ,v3dθ+em1τ1yt-∫y3ytg1y3g1θdθ+fx3,v3g2v3uvt-∫v3vtg2v3g2θdθ+pem1τ1czt-∫z3ztg4z3g4θdθ+qfx3,v3g2v3ruwt+fx3,v3∫-τ10Hfxt+s,vt+sfx3,v3ds+fx3,v3∫-τ20Hg1yt+sg1y3ds.Calculating the derivative of V4(t) along solutions of model (2), we obtain that (44)dV4tdt=sx1-fx3,v3fx,v3+fx,vfx3,v3fx,v3-fx3,v3g2vg2v3+qfx3,v3ug2v3g3wg2v3-g2v4¯-fx3,v3Hg2v3g1yt-τ2g1y3g2v+fx3,v3ln⁡g2vfx3,v3g2v3fx,v-fx3,v3Hg1y3fxt-τ1,vt-τ1g1yfx3,v3=sx-sx31-fx3,v3fx,v3-fx3,v3Hg2v3g1yt-τ2g1y3g2v+fx3,v3g2vg2v3fx,vfx,v3-1g2v3g2v-fx,v3fx,v-fx3,v3Hfx3,v3fx,v3-fx3,v3Hg1y3fxt-τ1,vt-τ1g1yfx3,v3-fx3,v3Hg2vfx,v3g2v3fx,v+qfx3,v3ug2v3g3wg2v3-g2v4¯.Note that (s(x)-s(x3))(1-f(x3,v3)/f(x,v3))≤0, and (45)fx,vfx,v3-1g2v3g2v-fx,v3fx,v≤0fort≥0.
Sincev3≤v4¯, we have dV4(t)/dt≤0, and dV4(t)/dt=0 if and only if x(t)=x3, y(t)=y3, v(t)=v3, and w(t)=0. From LaSalle’s invariance principle [36], we finally have that E3 is globally asymptotically stable when τ3=0, R0>1, R2>1, and R4≤1.
Next, we consider conclusion (b). By computing, the characteristic equation of the linearization system of model (2) at E3 is (46)λ+h-rg2v3fλ=0,where (47)fλ=a110a130a21a22a23a240a32a3300a420a44,where (48)a11=λ-s′x3+∂fx3,v3∂x,a13=∂fx3,v3∂v,a21=-e-m1+λτ1∂fx3,v3∂x,a22=λ+a+pg4z3g1′y3,a23=-e-m1+λτ1∂fx3,v3∂v,a24=pg1y3g4′z3,a32=-ke-m2+λτ2g1′y3,a33=λ+ug2′v3,a42=-ce-λτ3g4z3g1′y3,a44=λ+b-cg1y3e-λτ3g4′z3.When R4>1, we have h-rg2(v3)=r(g2(v4¯)-g2(v3))<0. Hence, there is a positive root λ∗=rg2(v3)-h. Therefore, when R4>1, E3 is unstable for any τ1≥0, τ2≥0, and τ3≥0. This completes the proof.Remark 14.
Theorem13 shows that if only equilibria E0, E1, E2, and E3 exist, then when R2>1, R4≤1, and τ3=0,E3 is globally asymptotically stable, and delays τ1 and τ2 do not impact the stability of E3.


## 3.5. Stability of EquilibriumE4

On the stability analysis of equilibriumE4, we here only discuss the following case: τ1≥0, τ2≥0, and τ3=0. However, for the cases τ1≥0, τ2≥0, and τ3≥0, the theoretical analysis is very complicated. We will give numerical analysis for this case in the next section.Theorem 15.
Ifτ3=0, R3>1, and R4>1, then infection equilibrium E4 with both antibody and CTL immune responses is globally asymptotically stable.Proof.
Define a Lyapunov functionalV5(t) as follows: (49)V5t=xt-∫x4xtfx4,v4fθ,v4dθ+em1τ1yt-∫y4ytg1y4g1θdθ+fx4,v4g2v4u+qg3w4vt-∫v4vtg2v4g2θdθ+pem1τ1czt-∫z4ztg4z4g4θdθ+fx4,v4∫-τ20Hg1yt+sg1y4ds+qfx4,v4g2v4u+qg3w4wt-∫w4wtg3w4g3θdθ+fx4,v4∫-τ10Hfxt+s,vt+sfx4,v4ds.Using the above similar method, we obtain (50)dV5tdt=sx-sx41-fx4,v4fx,v4-fx4,v4Hfx4,v4fx,v4+Hg2vfx,v4g2v4fx,v+Hg1y4fxt-τ1,vt-τ1g1yfx4,v4+Hg2v4g1yt-τ2g1y4g2v+fx4,v4g2vg2v4fx,vfx,v4-1g2v4g2v-fx,v4fx,v.Note that (s(x)-s(x4))(1-f(x4,v4)/f(x,v4))≤0, and (51)fx,vfx,v4-1g2v4g2v-fx,v4fx,v≤0fort≥0.Obviously, we have dV5(t)/dt≤0, and dV5(t)/dt=0 if and only if x(t)=x4, y(t)=y4, and v(t)=v4. From LaSalle’s invariance principle [36], we finally have that E4 is globally asymptotically stable when τ3=0, R3>1, and R4>1. This completes the proof.Remark 16.
Theorem15 shows that if equilibria E0, E1, E2, E3, and E4 exist, then when R3>1, R4>1, and τ3=0, E4 is globally asymptotically stable, and delays τ1 and τ2 do not impact the stability of E4.


## 4. Numerical Simulations

In the above section, we obtain the global asymptotic stability of equilibriaE3 and E4 when the delay τ3=0. In this section, by using the numerical simulation, it is shown that the Hopf bifurcation and stability switches occur at equilibria E3 and E4 in the case τ3>0.Example 17.
Corresponding to model (2), we consider the following model: (52)dxtdt=λ-dxt+r1x1-xK-βxtvt-b1e-c1vt+b1,dytdt=βe-m1τ1xt-τ1vt-τ1-b1e-c1vt-τ1+b1-ayt-pytzt,dvtdt=ke-m2τ2yt-τ2-uvt-qvtwt,dztdt=cyt-τ3zt-τ3-bzt,dwtdt=rvtwt-hwt,where b1,c1>0 are constants. We have s(x)=λ-dx(t)+r1x(1-x/K), f(x,v)=βx(t)((v(t)-b1)e-c1v(t)+b1), and gi(ξ)=ξ(i=1,2,3,4). It can easily verify that (H1)–(H4) hold. Taking λ=10, d=0.01, r1=0.6, K=500, β=0.3, c1=0.01, b1=0.01, a=0.5, p=1, k=0.4, u=3, q=1, c=0.1, b=0.15, m1=m2=0.01, g=1.5, h=1, τ1=2, and τ2=5, choose τ3 as free parameter. By computing, R2=34.4139>1, R4=0.2854<1, and E3=(462.1965,1.5000,0.1902,15.3959,0). From Figures 1–4, we see that as τ3 increases the complex dynamical behaviors of equilibrium E3 occur.Figure 1
Takingτ3=0.2, we have R2=34.4139>1 and R4=0.2854<1, and the infection equilibrium E3 with only CTL response is asymptotically stable.
(a) (b) (c) (d) (e)Figure 2
Takingτ3=2, we have R2=34.4139>1 and the Hopf bifurcation at infection equilibrium E3 with only CTL response occurs.
(a) (b) (c) (d) (e)Figure 3
Takingτ3=4, we have R2=34.4139>1 and R4=0.2854<1, and the infection equilibrium E3 with only CTL response is asymptotically stable.
(a) (b) (c) (d) (e)Figure 4
Takingτ3=15, we have R2=34.4139>1 and the Hopf bifurcation at infection equilibrium E3 with only CTL response occurs.
(a) (b) (c) (d) (e)In Figures1–8, we denote by (a) the time-series of x(t), by (b) the time-series of y(t), by (c) the time-series of v(t), by (d) the time-series of z(t), and by (e) the time-series of w(t).Figure 5
Takingτ3=0.1, we have R3=1.8912>1 and R4=2.7693>1, and the infection equilibrium E4 with both CTL and antibody responses is asymptotically stable.
(a) (b) (c) (d) (e)Figure 6
Takingτ3=2.5, we have R3=1.8912>1 and R4=2.7693>1, and the Hopf bifurcation at infection equilibrium E4 with both CTL and antibody responses occurs.
(a) (b) (c) (d) (e)Figure 7
Takingτ3=6, we have R3=1.8912>1 and R4=2.7693>1, and the infection equilibrium E4 with both CTL and antibody responses is asymptotically stable.
(a) (b) (c) (d) (e)Figure 8
Takingτ3=16, we have R3=1.8912>1 and the Hopf bifurcation at infection equilibrium E4 with both CTL and antibody responses occurs.
(a) (b) (c) (d) (e)Example 18.
Corresponding to model (2), we consider the following model: (53)dxtdt=λ-dxt-βxtvt1+a1xt+b1vt+a1b1xtvt,dytdt=βe-m1τ1xt-τ1vt-τ11+a1xt-τ1+b1vt-τ1+a1b1xt-τ1vt-τ1-ayt-pytzt,dvtdt=ke-m2τ2yt-τ2-uvt-qvtwt,dztdt=cyt-τ3zt-τ3-bzt,dwtdt=rvtwt-hwt,where a1,b1>0 are constants. We have s(x)=λ-dx(t), f(x,v)=βx(t)v(t)/(1+a1xt+b1vt+a1b1xtvt), and gi(ξ)=ξ(i=1,2,3,4). It is easily verified that (H1)–(H4) hold.Takingλ=10, d=0.01, β=0.25, a1=0.01, b1=0.01, a=0.5, p=1, k=0.4, u=3, q=1, c=0.1, b=0.15, m1=m2=0.01, r=1.5, h=0.1, τ1=5, and τ2=8, choose τ3 as free parameter. By computing, R3=1.8912>1, R4=2.7693>1, and E4=(850.8857,1.5000,0.6667,0.4456,5.3039). From Figures 5–8, we see that as τ3 increases the complex dynamical behaviors of equilibrium E4 occur.


## 5. Discussion

In this paper we have considered an in-host model with intracellular delayτ1, virus replication delay τ2, and immune response delay τ3, given by (2) together with assumptions (H1)–(H4), which describes the dynamics among uninfected cells, infected cells, virus, CTL responses, and antibody responses. The model allows for general target-cell dynamics s(x), including a nonlinear incidence f(x,v), discrete delays, and state-dependent removal functions gi(i=1,2,3,4). This general model includes many existing models in the literature as special cases. Dynamical analysis of model (2) shows that τ1, τ2, and τ3 play different roles in the stability of the equilibria. Particularly, we see that τ3 may impact the stability of equilibria E3 and E4.By the analysis, we have shown that whenR0≤1, E0 is globally asymptotically stable, which means that the virus is cleared up. When R0>1, R1≤1, and R2≤1,E1 is globally asymptotically stable, which means that the infection is successful, but the establishments of both antibody and CTLs immune responses are unsuccessful. When R1>1 and R3≤1, E2 is globally asymptotically stable, which implies that the antibody response is established, but the infected cells are too weak to stimulate CTL immune response. With respect to the analysis of E3, we consider special cases τ3=0, τ1≥0, and τ2≥0; when R2>1 and R4≤1, E3 is globally asymptotically stable, which means that the CTL immune response is determined, but the viral loads are so small that it cannot activate the antibody response. About the stability of E4, we have obtained that for special case, τ3=0, τ1≥0, and τ2≥0, when R3>1 and R4>1, E4 is globally asymptotically stable, that is, susceptible cells, infected cells, free virus, CTLs, and antibodies coexist in vivo.Based on Theorems13 and 15, we obtain that the intracellular delay τ1 and virus replication delay τ2 for model (2) do not cause Hopf bifurcation. Moreover, R0 plays a crucial role in virus infection dynamics. Actually, in model (2), R0 is a decreasing function on time delay τ1. When all other parameters are fixed and delay τ1 is sufficiently large, R0 becomes less than one, only infection-free equilibrium E0 exists, and the virus is cleared in the host. By biological meanings, intracellular delay plays a positive role in virus infection process in order to eliminate virus. Sufficiently large intracellular delay makes the virus development slower and the virus has been controlled and disappeared. This gives us some suggestions on new drugs to prolong the time of infected cells producing virus. However, by the recent research of Li and Shu [37], in the case of the coexistence of mitosis rate of the target cells and an intracellular delay in the viral infection model, the intracellular delay produces Hopf bifurcation only when the mitosis rate is sufficiently large.Whenτ3>0, by numerical simulations, it is shown that the Hopf bifurcation and stability switches occur at equilibria E3 and E4 as τ3 increases. Figures 1–4 indicate that E3 remains stable as τ3>0 is small, and along with the increase of τ3, equilibrium E3 becomes unstable and periodic oscillations appear. It shows that stability switches occur as delay τ3 increases. Similarly, from Figures 5–8, we see that along with the increases of τ3>0 the dynamical behaviors of model (53) at equilibrium E4 appear as very large diversification. Particularly, when τ3 is small enough, E4 is asymptotically stable and when τ3 is increasing, the stability switches occur at equilibrium E4, and when E4 is unstable, a Hopf bifurcation occurs. Finally, when τ3 is enough large, equilibrium E4 always is unstable. Summarizing these discussions, we have the conclusion that τ3 affects markedly the stability of equilibria E3 and E4. From the numerical simulations, we observe that immune response delay τ3 can cause Hopf bifurcation. Upon primary infection, the sustained oscillations from the Hopf bifurcation imply that the pathogen may not always be cleared entirely with the CTL responses which usually occur in a few days after serum conversion. As the increase of immune delay τ3, we know that the drug prevents virus from continuing through their cell cycle, thus trapping them at some point during interphase, where the cells die from natural causes. Then susceptible cells, infected cells, free virus, CTLs, and antibodies reach a stable level in the host. When immune delay τ3 continuously increases, the activation of the immune cell is to fight against the malignant virus cells. Thus susceptible cells, infected cells, free virus, CTLs, and antibodies exhibit sustained periodic oscillations in the chronic phase of infection. This explains the fact that the immune response delay plays negative roles in controlling disease progression.Observing all obtained results in this paper, we can directly put forward the following open questions which need to be further studied in the future.For one, in addition toτ1, τ2, and τ3, antibody response delay τ4 is also considered, whether the results obtained in this paper can be extended to a virus infection model with nonlinear incidence rate and four time delays. For another, we obtain the Hopf bifurcation and stability switches at equilibria E3 and E4 for model (2) only by using the numerical simulation method for special examples (52) and (53). Up to now, the theoretical analysis and results in this aspect are few and rough. Therefore, a systemic and complete theoretical analysis and results will be a very estimable and significative subject.



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