# Global Stability of a HIV-1 Model with General Nonlinear Incidence and Delays

Authors: Yaping Wang; Fuqin Sun
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

A HIV-1 model with two distributed intracellular delays and general incidence function is studied. Conditions are given under which the system exhibits the threshold behavior: the disease-free equilibriumE0 is globally asymptotically stable if R0≤1; if R0>1, then the unique endemic equilibrium E1 is globally asymptotically stable. Finally, it is shown that the given conditions are satisfied by several common forms of the incidence functions.


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## Body

## 1. Introduction

The global stability is analyzed for a general mathematical model of HIV-1 pathogenesis proposed by Nelson and Perelson [1]. The general model includes two distributed intracellular delays and a combination therapy with a reverse transcriptase inhibitor and a protease inhibitor. All incidence functions in those papers are the bilinear functions. However, there are some pieces of evidence showing that a bilinear infection rate might not be an effective assumption when the number of target cells is large enough (see [2–4] for the review of evidence and models). The aim of this paper is to establish global stability for a delay integrodifferential equation with a general incidence term f(x,v), the conditions given here are similar to those given in [4] for the ODE case.In this paper, we consider the following HIV-1 model with a side class of nonlinear incidence rates and distributed delays:(1)dx(t)dt=s-dTx(t)-(1-nrt)f(x(t),v(t)),dy(t)dt=(1-nrt)∫0∞‍G1(ξ)f(x(t-ξ),v(t-ξ))dξ-δy(t),dv(t)dt=(1-np)Nδ∫0∞‍G2(ξ)y(t-ξ)dξ-cv(t),
where x, y, and v are the concentrations of uninfected target cells (T cells), productively infected cells, and infectious virus, respectively.The form of the incidence functionf is of fundamental importance. In this paper, we want to work with a function as general as possible but still possesses the properties necessary for conclusions to be made through mathematical analysis. Because of this, we will introduce conditions on f with which it may appear technical. However, as shown in Section 5, many commonly used incidence functions satisfy these conditions. For now, we assume only the following. (A1)
f(x,v) is a nonnegative differentiable function on the nonnegative quadrant. Furthermore, f(x,v) is positive if and only if both arguments are positive. The partial derivatives of f on x and v are denoted by f1 and f2, respectively. In Sections 3 and 4, it will be shown how the extra conditions on f imply in the local and global stability of an endemic equilibrium.For the purpose of convenience, we rewrite (1) as
(2)dx(t)dt=λ-mx(t)-βf(x(t),v(t)),dy(t)dt=β1∫0∞‍g1(ξ)f(x(t-ξ),v(t-ξ))dξ-ay(t),dv(t)dt=k1∫0∞‍g2(ξ)y(t-ξ)dξ-μv(t),
where λ=s, m=dT, β=1-nrt, a=δ, μ=c, β1=(1-nrt)α1, k1=(1-np)Nδα2, gi(ξ)=Gi(ξ)/αi, αi=∫0∞‍Gi(ξ)dξ, i=1,2.The paper is organized as follows. In Section2, the basic reproduction number R0 is determined and the equilibrium is found. The local stability of the equilibrium is studied in Section 3. The global dynamics are resolved in Section 4. In Section 5, examples are given of incidence functions which satisfy the assumptions that are used throughout the paper.


## 2. Equilibria andR0

Assume the kernel functionsG1 and G2 satisfy
(3)Gi(ξ)≥0,forξ≥0,0<∫0∞‍Gi(ξ)dξ<1,i=1,2.
For any given initial condition,
(4)ϕ(θ)=(x(θ),y(θ),v(θ))∈UCg((-∞,0],R+3),
system (2) satisfies the hypotheses that are sufficient to ensure the existence, uniqueness, and continuity of solutions [5, Theorems 2.1–2.3]; for the notation of UCg, see [6, page 46]. Recall that αi=∫0∞‍Gi(ξ)dξ and gi(ξ)=Gi(ξ)/αi; thus, ∫0∞‍gi(ξ)dξ=1. Then, the basic reproduction number R0 defined in [7] can be rewritten as
(5)R0=k1β1aμf2(E0),
and for any value of the parameters, the disease-free equilibrium of (2) is given by
(6)E0=(x0,y0,v0)=(λm,0,0).
It is straightforward to show that if R0≤1, then (2) has only one nonnegative equilibrium, which is called the infection-free equilibrium E0. The presence and number of endemic equilibrium depend on the form of the nonlinearity f, as well as the values of the parameters. In searching for equilibrium, we note that the equilibria of (2) are the same as the equilibria of the corresponding ordinary differential equation system. Here we give the following result.Theorem 1.
Let(x(t),y(t),v(t))T be the unique solution to system (2) with x(0)>0, y(0)>0, and v(0)>0. Then x(t),y(t), and v(t) are positive for all t>0. Moreover, the solution is bounded and thus exists globally.Proof.
Using the variation-of-constants formula, we obtain the positivity ofy(t) and v(t).
Forx(t), we claim that if x(0)>0, then x(t)>0 for all t>0. Otherwise, there exists t′>0 such that x(t′)=0. Let t0 be the first one that satisfies x(t0)=0; then t0>0 and
(7)dx(t)dt|t=t0=λ>0,
which means that x(t) is increasing at t=t0 and it is a contradiction.
Next we will show that every solution is also bounded. It follows from the first equation of (2) that dx(t)/dt<λ-mx(t). This implies that lim-t→∞x(t)≤λ/m.
LetV(t)=(β1/β)∫0∞‍g1(ξ)x(t-ξ)dξ+y(t); then
(8)dV(t)dt|(2)=β1λβ-β1(a-m)β∫0∞‍g1(ξ)x(t-ξ)dξ-av(t).
Note that ∫0∞‍g1(ξ)x(t-ξ)dξ is ultimately bounded; then there exist positive constants C and T0 such that
(9)dV(t)dt|(2)≤β1λβ+C-aV(t),∀t≥T0.
This yields thatV(t) is eventually bounded and so is y(t). By a similar argument, we can show the boundedness of v(t). Therefore, the system (2) is point dissipative (see [1]) and hence the solution of (2) exists globally.Theorem 2.
IfR0>1, then there exists an endemic equilibrium E1=(x*,y*,v*).Proof.
We look for solution(x*,y*,v*) of the equations dx/dt=0,dy/dt=0, and dv/dt=0. We first note that (dx/dt)+(β/β1)(dy/dt)=0 implies λ-mx*-βf(x*,v*)+(β/β1)ay*=0 and k1y*-μv*=0, so x*=(λ-(aβ/β1)y*)/m, v*=k1/μ. Let H(y*)=β1f((λ-(aβ/β1)y*)/m,(k1/μ)y*)-ay*. Then, dy(t)/dt=0 whenever H(y*)=0. Thus, any zero of H in the interval (0,λβ1/aβ) corresponds to an equilibrium with x*,y*,v*>0, that is, an endemic equilibrium.
Sincef(0,v)=f(x,0)=0, it follows that H(0)=0 and H(λβ1/aβ)<0. The function H is continuous and so a sufficient condition for H to have a zero in (0,λβ1/aβ) is that H is increasing at 0. Thus, there is an endemic equilibrium if
(10)0<dHdy(0)=β1(-aβmβ1)f1(E0)+k1β1μf2(E0)-a.
Sincef(x,0)=0 for all x, it follows that f1(E0)=0 and so (10) is equivalent to R0=(k1β1/aμ)f2(E0)>1.


## 3. Local Stability of the Equilibria

Theorem 3.
IfR0<1, then E0 is locally asymptotically stable.The proof is normal, so we omitted it. We now give conditions onf that are used here and after to show the locally and globally asymptotically stability of the endemic equilibrium. As a precondition, we assume that R0>1 to guarantee the existence of endemic equilibrium E1=(x*,y*,v*). Consider the following.(A2)
sgn(f(x,v*)-f(x*,v*))=sgn(x-x*) for all x≥0.(A3)
For allx,v>0, f(x,v)/f(x,v*) is in the closed interval with endpoints at 1 and v/v*.(A4)
Eitherf1(x*,v*)>0 or f2(x*,v*)<f(x*,v*)/v*.Theorem 4.
IfR0>1, and (A2), (A3), and (A4) hold, then any endemic equilibrium which exists is locally asymptotically stable.Proof.
The characteristic equation of the linearization of (2) at endemic equilibrium E1=(x*,y*,v*) is
(11)0=(m+λ+βf1(E1))(a+λ)(μ+λ)-(m+λ)β1k1f2(E1)×∫0∞‍g1(ξ)e-λξdξ∫0∞‍g2(ξ)e-λξdξ.
We demonstrate that all zeros of it have negative real part. Since (A2) and (A3) hold, we get that 0≤βf1(E1) and 0≤f2(E1)≤aμ/β1k1. Suppose that λ has nonnegative real part, we deduce
(12)|(m+λ)β1k1f2(E1)∫0∞‍g1(ξ)e-λξdξ∫0∞‍g2(ξ)e-λξdξ|≤|m+λ|β1k1f2(E1)≤|(m+λ+βf1(E1))(a+λ)(μ+λ)|,
and so the solutions λ with nonnegative real part if and only if all of the inequalities in (12) are in fact equal. The final inequality is strict unless f1(E1)=0 (and λ=0). The second last inequality is strict unless f2(E1)=f(E1)/v*. Assumption (A4) implies at least one is strict. Thus, the endemic equilibrium E1 is locally asymptotically stable.


## 4. Global Stability of the Equilibria

Theorem 5.
IfR0≤1, then E0 is globally stable; that is,
(13)limt→∞(x(t),y(t),v(t))=E0.Proof .
LetK(t)=K1(t)+K2(t)+K3(t) with
(14)K1(t)=12(x(t)-λm)2+λβmβ1y(t)+aλβk1mβ1v(t),K2(t)=λβm∫0∞‍H1(ξ)(f(x(t-ξ),v(t-ξ))+f2(x0,0)v(t-ξ))dξ,K3(t)=aλβmβ1∫0∞‍H2(ξ)y(t-ξ)dξ,
where Hi(t)=∫t∞‍gi(ξ)dξ, i=1,2.
It is clear thatK(t)≥0 and K(t)=0 if and only if x(t)=λ/m, y(t)=v(t)=0. The derivative of K1 along the solution of (2) is
(15)dK1(t)dt|(2)=(x(t)-λm)(λ-mx(t)-βf(x(t),v(t)))-aλβmβ1y(t)-aλβμk1mβ1v(t)+λβm∫0∞‍g1(ξ)f(x(t-ξ),v(t-ξ))dξ+aλβmβ1∫0∞‍g2(ξ)y(t-ξ)dξ.
Noticing that H1(0)=1, H1(∞)=0, and dH1(t)=-g1(t)dt, and using integration by parts, we calculate the derivatives of K2 and K3(16)dK2(t)dt|(2)=λβmf(x,v)+λβmf2(x0,0)-λβm∫0∞‍g1(ξ)f(x(t-ξ),v(t-ξ))dξ-λβmf2(x0,0)∫0∞‍g1(ξ)v(t-ξ)dξ.dK3(t)dt|(2)=aλβmβ1(y(t)-∫0∞‍g2(ξ)y(t-ξ)dξ).
Thus,
(17)dK(t)dt|(2)=-m(x(t)-λm)2-βx(t)f(x(t),v(t))-aλβμmβ1k1(1-R0)v(t)-λβmf2(x0,0)∫0∞‍g1(ξ)v(t-ξ)dξ≤0.
Recalling that f is nonnegative and is positive if and only if both arguments are positive, we must have that {E0} is the largest invariant subset of {dK/dt=0}; then the global stability of E0 follows from the classical Lyapunov-LaSalle invariance principle (see [6], Theorem 2.7.1).Theorem 6.
IfR0>1, and (A2) and (A3) hold, then E1 is globally stable in D=R+3∖E0.Proof.
LetW(t)=W1(x(t),y(t),v(t))+W2(x(t),y(t),v(t))+W3(x(t),y(t),v(t)).
With(18)W1(x(t),y(t),v(t))=∫0∞‍g1(ξ)U(t,ξ)dξ,W2(x(t),y(t),v(t))=ββ1y*h(yy*)+βf(x*,v*)∫0∞‍H1(ξ)h(v(t-ξ)v*)dξ,W3(x(t),y(t),v(t))=aβk1β1F(v(t),v*)+aββ1∫0∞‍H2(ξ)F(y(t-ξ),y*)dξ,
where U(t,ξ)=x(t-ξ)-x*-∫x*x(t-ξ)‍(f(x*,v*)/f(σ,v*))dσ, h(t)=t-1-lnt, and F(u,u*)=u-u*lnu-(u*-u*lnu*); it is easy to see that h(t) has the strict global minimum h(1)=0 for t>0, F(u,u*)≥F(u*,u*)=0 for u>0 and any positive constant u*. By Theorem 1, all solutions are positive and bounded. Thus, W(t) is well defined and W(t)≥0, in which the equality holds if and only if y(t)=y*,v(t)=v* and x(t-ξ)=x*,y(t-ξ)=y*,v(t-ξ)=v* for almost all ξ∈[0,∞). For clarity, the derivatives of W1, W2, and W3 will be calculated separately and then combined to obtain dW/dt.
First,dW1/dt=∫0∞‍g1(ξ)(dU(t,ξ)/dt)dξ, where
(19)dU(t,ξ)dt=(1-f(x*,v*)f(x(t-ξ),v*))×(λ-mx(t-ξ)-βf(x(t-ξ),v(t-ξ)))=-m(x(t-ξ)-x*)(1-f(x*,v*)f(x(t-ξ),v*))+βf(x*,v*)(1-f(x(t-ξ),v(t-ξ))f(x*,v*)-f(x*,v*)f(x(t-ξ),v*)+f(x(t-ξ),v(t-ξ))f(x(t-ξ),v*)).
Since λ=mx*+βf(x*,v*). That is,
(20)dW1dt=-m∫0∞‍g1(ξ)(x(t-ξ)-x*)×(1-f(x*,v*)f(x(t-ξ),v*))dξ+βf(x*,v*)×∫0∞‍g1(ξ)(f(x*,v*)f(x(t-ξ),v*)1-f(x(t-ξ),v(t-ξ))f(x*,v*)-f(x*,v*)f(x(t-ξ),v*))dξ+βf(x*,v*)∫0∞‍g1(ξ)f(x(t-ξ),v(t-ξ))f(x(t-ξ),v*)dξ.Next, we calculatedW2/dt. Consider
(21)dW2dt=βf(x*,v*)×∫0∞‍g1(ξ)f(x(t-ξ),v(t-ξ))f(x*,v*)×(1-y*y)dξ-βaβ1y(t)+ββ1ay*+βf(x*,v*)∫0∞‍g1(ξ)×(h(vv*)-h(v(t-ξ)v*))dξ=βf(x*,v*)∫0∞‍g1(ξ)×[f(x(t-ξ),v(t-ξ))f(x*,v*)-y*yf(x(t-ξ),v(t-ξ))f(x*,v*)-y*y]dξ+βf(x*,v*)∫0∞‍g1(ξ)×[1+h(vv*)-h(v(t-ξ)v*)]dξ.Using the fact thatH2(0)=1, H2(∞)=0, and dH2=-g2(t)dt, we get
(22)dW3dt=aβk1β1(1-v*v)(k1∫0∞‍g2(ξ)y(t-ξ)dξ-μv(t))-aββ1∫0∞‍H2(ξ)dF(y(t-ξ),y*)dξdξ=-aββ1∫0∞‍g2(ξ)v*vy(t-ξ)dξ-aβμk1β1(v-v*)+aββ1y(t)+aββ1y*∫0∞‍g2(ξ)lny(t-ξ)ydξ=-βf(x*,v*)∫0∞‍g2(ξ)h(v*vy(t-ξ)y)dξ+βf(x*,v*)∫0∞‍g1(ξ)(h(yy*)-h(vv*))dξ.
The third quality holds since y*=(β1/a)f(x*,v*), v*=(k1/μ)y*. We conclude that
(23)dWdt=-m∫0∞‍g1(ξ)(x(t-ξ)-x*)×(1-f(x*,v*)f(x(t-ξ),v*))dξ-βf(x*,v*)∫0∞‍g2(ξ)h(v*vy(t-ξ)y)dξ+βf(x*,v*)∫0∞‍g1(ξ)C(ξ)dξ,
where
(24)C(ξ)=2-f(x*,v*)f(x(t-ξ),v*)+f(x(t-ξ),v(t-ξ))f(x(t-ξ),v*)-y*yf(x(t-ξ),v(t-ξ))f(x*,v*)-yy*+h(yy*)-h(v(t-ξ)v*)=-h(f(x*,v*)f(x(t-ξ),v*))-h(y*yf(x(t-ξ),v(t-ξ))f(x*,v*))+h(f(x(t-ξ),v(t-ξ))f(x(t-ξ),v*))-h(v(t-ξ)v*).
Since the function h(t) is monotone on each side of 1 and is minimized at t=1, (A3) implies that h((f(x(t-ξ),v(t-ξ)))/(f(x(t-ξ),v*)))≤h(v(t-ξ)/v*). Therefor, noting that h≥0 and (A2), we have dW/dt≤0. So the solutions must tend to M, the largest invariant subset of {dW/dt=0}.To havedW/dt equal zero, it is necessary to have (x(t-ξ)-x*)(1-(f(x*,v*)/f(x(t-ξ),v*)))=0 for almost every ξ∈[0,∞), which holds if and only if x(t)=x*; v*y(t-ξ)=v(t)y(t) and y*f(x(t-ξ),v(t-ξ))=y(t)f(x*,v*) for almost every ξ∈[0,∞).SinceM is invariant with respect to (2), we get
(25)dy(t)dt=β1∫0∞‍g1(ξ)f(x(t-ξ),v(t-ξ))dξ-ay(t)=β1∫0∞‍g1(ξ)y(t)f(x*,v*)y*dξ-ay(t)=0.
So this determines y(t) to be a constant. Using v*y(t-ξ)=v(t)y(t), we obtain v(t)=v* and this yields that y(t)=y*. Thus, each element of M satisfies x(t)=x*, y(t)=y*, and v(t)=v* for all t>0. This shows that
(26)limt→∞(x(t),y(t),v(t))=(x*,y*,v*)=E1.
This completes the proof.


## 5. Examples

We now give some examples of incidence functions for which the hypotheses are satisfied.Example 1 (mass action).
Letf(x,v)=xv. The hypotheses (A1)–(A4) are satisfied and so global dynamics are determined by the magnitude of R0. The global behavior of this model was previously studied in [1, 8] and was fully resolved in [9].Example 2 (saturating incidence).
Letf(x,v)=βx(t)v(t)/(1+αx(t)) for some constant β>0, α>0. The hypotheses (A1)–(A4) are satisfied and so global dynamics are determined by the magnitude of R0. The global properties of this model were studied in [10].Example 3 (Holling type II incidence).
Letf(x,v)=βx(t)v(t)/(1+v(t)) for some constant β>0. The hypotheses (A1)–(A4) are satisfied and so global dynamics are determined by the magnitude of R0. The global properties of this model were studied in [11].Example 4 (Beddington-DeAngelis incidence).
Letf(x,v)=x(t)v(t)/(1+ax(t)+bv(t)) for some constants a>0 and b>0. The hypotheses (A1)–(A4) are satisfied and so global dynamics are determined by the magnitude of R0. The global properties were studied in [12, 13].



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