Effect of Time Delay and Antibodies on HCV Dynamics with Cure Rate and Two Routes of Infection

In this paper we propose and analyze an HCV dynamics model taking into consideration the cure of infected hepatocytes and antibody immune response. We incorporate both virus-to-cell and cell-to-cell transmissions into the model. We incorporate a distributed-time delay to describe the time between the HCV or infected cell contacts an uninfected hepatocyte and the emission of new active HCV. We show that the solutions of the proposed model are nonnegative and ultimately bounded. We derive two threshold parameters which fully determine the existence and stability of the three steady states of the model. Using Lyapunov functionals, we established the global stability of the steady states. The theoretical results are confirmed by numerical simulations.

• HCV Infection
• Distributed Time Delay
• Global Stability
• Cell-to-Cell Transmission
• Lyapunov Function

1. Introduction

Hepatitis C virus is considered one of the dangerous human viruses that infects the liver and causes the lever cirrhosis. Mathematical modeling and analysis of within-host HCV dynamics have been studied by many authors (see e.g. [1] - [12] ). These works can help researchers for better understanding the HCV dynamical behavior and providing new suggestions for clinical treatment. Immune response plays an important role in controlling the dynamics of several viruses (see e.g. [13] [14] [15] [16] [17] ). Cytotoxic T Lymphocyte (CTL) and antibodies play a central role of immune response. CTL cells attack and kill the infected cells. The B cell produces antibodies to neutralize the viruses. Mathematical models of HCV dynamics with antibody immune response have been proposed in [18] [19] [20] . The models presented in [18] [19] [20] assume that an uninfected hepatocyte becomes infected by contacting with HCV (virus-to-cell transmission). It has been reported in [21] [22] [23] that the HCV can also spread by cell-to-cell transmission.

The “cure” of infected cells has been considered in the virus dynamics models in several works (see e.g. [24] - [39] ). In [40] , both cure and cell-to-cell transmissions have been considered in the virus dynamics model, but without taking the immune response into account. In a very recent paper, Pan and Chakrabarty [41] have proposed the following mathematical model of HCV dynamics which incorporates 1) both virus-to-cell and cell-to-cell transmissions, 2) cure of infected hepatocytes, and 3) antibody immune response:

s ˙ ( t ) = β − δ ^ s ( t ) − α 1 s ( t ) p ( t ) − α 2 s ( t ) y ( t ) + ρ y ( t ) , (1)

y ˙ ( t ) = α 1 s ( t ) p ( t ) + α 2 s ( t ) y ( t ) − ε y ( t ) − ρ y ( t ) , (2)

p ˙ ( t ) = m y ( t ) − γ p ( t ) − q z ( t ) p ( t ) , (3)

z ˙ ( t ) = r z ( t ) p ( t ) − μ z ( t ) , (4)

where, s, y, p and z represent the concentration of uninfected hepatocytes, infected hepatocytes, HCV particles and antibodies, respectively. The uninfected hepatocytes are generated at a constant rate β, die at rate δ ^ s , where δ ^ is the natural death rate constant. The infection rate due to both virus-to-cell and cell-to-cell transmissions is given by α 1 s p + α 2 s y , where α 1 and α 2 are constants. The infected hepatocytes die at rate εy and cure at rate ρy, where ε and ρ are constants. Constant m is the generation rate of the HCV from infected hepatocytes. Antibodies attack the HCV at rate q z p , proliferate at rate r z p and die at rate μz, where q, r and μ are constants.

It is assumed in model (1)-(4) that, the hepatocytes can produce HCV particles once they are contacted by HCV or infected cells. However, there is a time period from the moment of the uninfected hepatocytes that are contacted by the HCV or infected cells and the moment of producing new active HCV particles [10] [11] .

The aim of this paper is to study the qualitative behavior of an HCV dynamics model with antibody immune response. We have incorporated distributed time delay and both virus-to-cell and cell-to-cell transmissions. We derive two threshold parameters and establish the global stability of the three steady states of the model using Lyapunov method.

2. The Model

We propose the following HCV dynamics model with distributed time delay:

s ˙ ( t ) = β − δ ^ s ( t ) − α 1 s ( t ) p ( t ) − α 2 s ( t ) y ( t ) + ρ y ( t ) , (5)

y ˙ ( t ) = α 1 s ( t ) p ( t ) + α 2 s ( t ) y ( t ) − ε y ( t ) − ρ y ( t ) , (6)

p ˙ ( t ) = m ∫ 0 h ρ ( τ ) e − μ 1 τ y ( t − τ ) d τ − γ p ( t ) − q z ( t ) p ( t ) , (7)

z ˙ ( t ) = r z ( t ) p ( t ) − μ z ( t ) . (8)

We assume that, the HCV or infected cell contacts an uninfected hepatocyte at time t − τ , the cell becomes infected at time t, where τ is a distributed parameter over the time interval [ 0, h ] . The factors e − μ 1 τ represents the probability of surviving the hepatocyte during the time delay period, where μ 1 is a constant. ρ ( τ ) is a probability distribution function satisfying ρ ( τ ) > 0 and

∫ 0 h   ρ ( τ ) d τ = 1 ,     ∫ 0 h ρ ( υ ) e ς υ d υ < ∞ ,

where ς and h are positive constants. Let us denote Θ ( τ ) = ρ ( τ ) e − μ 1 τ and F = ∫ 0 h   Θ ( τ ) d τ , thus 0 < F ≤ 1 . Let the initial conditions for system (5)-(8) be given as:

s ( η ) = ζ 1 ( η ) ,     y ( η ) = ζ 2 ( η ) , p ( η ) = ζ 3 ( η ) ,     z ( η ) = ζ 4 ( η ) , ζ j ( η ) ≥ 0 ,   η ∈ [ − h , 0 ] , ζ j ∈ C ( [ − h ,0 ] , ℝ ≥ 0 4 ) ,     j = 1, ⋯ ,4, (9)

where C is the Banach space of continuous functions mapping the interval [ − h ,0 ] into ℝ ≥ 0 4 . Then, the uniqueness of the solution for t > 0 is guaranteed [42] .

2.1. Basic Properties

In this subsection, we investigate the nonnegativity and boundedness of solutions.

Proposition 1. The solutions of system (5)-(8) with the initial states (9) are nonnegative and ultimately bounded.

Proof. From Equation (5) we have s ˙ | s = 0 = β + ρ y > 0 . Hence, s ( t ) > 0 for all y ≥ 0 . Moreover, for all t ∈ [ 0, h ] we have

y ( t ) = ζ 2 ( 0 ) e − ( ε + ρ ) t + ∫ 0 t e − ( ε + ρ ) ( t − η ) [ α 1 s ( η ) p ( η ) + α 2 s ( η ) y ( η ) ] d η ≥ 0 ,

p ( t ) = ζ 3 ( 0 ) e − ∫ 0 t ( γ + q z ( ξ ) ) d ξ + m ∫ 0 t   e − ∫ η t ( γ + q z ( ξ ) ) d ξ ∫ 0 h     Θ 2 ( τ ) y ( η − τ ) d τ d η ≥ 0,

z ( t ) = ζ 4 ( 0 ) e − ∫ 0 t ( κ − r p ( ξ ) ) d ξ ≥ 0.

By recursive argument we get y ( t ) ≥ 0 , p ( t ) ≥ 0 , and z ( t ) ≥ 0 , for all t ≥ 0 .

Next, we establish the boundedness of the model’s solutions. The nonnegativity of the model’s solution implies that

s ˙ ( t ) ≤ β − δ ^ s ( t ) + ρ y ( t ) ,

We let Q 1 ( t ) = s ( t ) + y ( t ) , then

Q ˙ 1 ( t ) = β − δ ^ s ( t ) − ε y ( t ) ≤ β − σ 1 ( s ( t ) + y ( t ) ) = β − σ 1 Q 1 ( t ) ,

where σ 1 = m i n { δ ^ , ε } . Hence Q 1 ( t ) ≤ L 1 , if Q 1 ( t ) ≤ L 1 where L 1 = β σ 1 . It follows that s ( t ) ≤ L 1 and y ( t ) ≤ L 1 if s ( 0 ) + y ( 0 ) ≤ L 1 . Moreover, let Q 2 ( t ) = p ( t ) + q r z ( t ) , then

Q ˙ 2 ( t ) = m ∫ 0 h   Θ ( τ ) y ( t − τ ) d τ − γ p ( t ) − q μ r z ( t ) ≤ m L 1 F − γ p ( t ) − q μ r z ( t ) ≤ m L 1 − σ 2 ( p ( t ) + q r z ( t ) ) = m L 1 − σ 2 Q 2 ( t ) ,

where σ 2 = m i n { γ , μ } . It follows that, lim sup t → ∞ Q 2 ( t ) ≤ L 2 , where L 2 = m L 1 σ 2 . Since p ( t ) ≥ 0 and z ( t ) ≥ 0 , then lim sup t → ∞ p ( t ) ≤ L 2 and lim sup t → ∞ z ( t ) ≤ L 3 , where L 3 = r q L 2 . Therefore, s ( t ) , y ( t ) , p ( t ) and z ( t ) are ultimately bounded. ,

According to Proposition 1, we can show that the region

Δ = { ( s , y , p , z ) ∈ C 4 : ‖ s ‖ ≤ L 1 , ‖ y ‖ ≤ L 1 , ‖ p ‖ ≤ L 2 , ‖ z ‖ ≤ L 3 } ,

is positively invariant with respect to system (5)-(8).

2.2. The Steady States and Threshold Parameters

Lemma 1. For system (5)-(8) there exist two threshold parameters R 0 > 0 , and R 1 z > 0 , such that

1) if R 0 ≤ 1 , then there exists only one steady state Π 0 ,

2) if R 1 z ≤ 1 < R 0 , then there exist only two steady states Π 0 and Π 1 ,

3) if R 0 > 1 and R 1 z > 1 , then there exist three steady states Π 0 , Π 1 and Π 2 .

Proof. Let ( s , y , p , z ) be any steady state satisfying

β − δ ^ s − α 1 s p − α 2 s y + ρ y = 0 , (10)

α 1 s p + α 2 s y − ε y − ρ y = 0 , (11)

m F y − γ p − q z p = 0 , (12)

( r p − μ ) z = 0. (13)

We find that system (10)-(13) admits three steady states.

1) Infection-free steady state Π 0 = ( s 0 , 0 , 0 , 0 ) , where s 0 = β / δ ^ .

2) Chronic-infection steady state without immune response Π 1 = ( s 1 , y 1 , p 1 ,0 ) , where

s 1 = ( ε + ρ ) γ F m α 1 + γ α 2 ,

y 1 = δ ^ s 1 ε ( β ( F m α 1 + γ α 2 ) δ ^ γ ( ε + ρ ) − 1 ) ,

p 1 = F m y 1 γ .

Clearly Π 1 exists if

β ( F m α 1 + γ α 2 ) δ ^ γ ( ε + ρ ) > 1.

Let us define

R 0 = β ( F m α 1 + γ α 2 ) δ ^ γ ( ε + ρ ) ,

In terms of R 0 , we can write the steady state components for Π 1 as:

s 1 = s 0 R 0 ,           y 1 = δ ^ s 1 ε ( R 0 − 1 ) ,

p 1 = F m δ ^ s 1 γ ε ( R 0 − 1 ) .

3) Chronic-infection steady state with humoral immune Π 2 = ( s 2 , y 2 , p 2 , z 2 ) , where

s 2 = r y 2 ( ε + ρ ) μ α 1 + r y 2 α 2 ,         y 2 = − B + B 2 − 4 A C 2 A , p 2 = μ r ,       z 2 = γ q ( r m F y 2 μ γ − 1 ) . (14)

where

A = r α 2 ε , B = μ ε α 1 − r β α 2 + r δ ^ ( ε + ρ ) , C = − β μ α 1 . (15)

We note that Π 2 exists when r F 2 m y 2 μ γ > 1 . Now we define

R 1 z = r F m y 2 μ γ = F m y 2 p 2 γ . (16)

Then z 2 = γ q ( R 1 z − 1 ) . We define the basic reproduction number for the

humoral immune response R H u m which comes from the limiting (linearized) z-dynamics near z = 0 as:

R H u m z = p 1 p 2

Lemma 2 1) if R 1 z < 1 , then R H u m z < 1 ,

2) if R 1 z > 1 , then R H u m z > 1 ,

3) if R 1 z = 1 then R H u m z = 1 .

Proof. 1) Let R 1 z < 1 , then from Equation (16) we have y 2 < γ p 2 m F , and then using Equation (14) we get

− B + B 2 − 4 A C 2 A < γ p 2 F m ,

that leads to

( 2 A γ p 2 F m + B ) 2 − ( B 2 − 4 A C ) > 0.

Using Equation (15), we can get

4 α 2 ε 2 μ 2 γ ( F m α 1 + γ α 2 ) m 2 F 2 ( 1 − R H u m z ) > 0

then

R H u m z = r F m s 1 ( R 0 − 1 ) δ ^ μ ε γ < 1.

then R H u m z < 1 . Similarly, one can proof 2) and 3) ,.

3. Global Stability

The following theorems investigate the global stability of the steady states of system (5)-(8). Let us define the function H : ( 0, ∞ ) → [ 0, ∞ ) as H ( l ) = l − 1 − ln l . Denote ( s , y , p , z ) = ( s ( t ) , y ( t ) , p ( t ) , z ( t ) ) .

Theorem 1. Suppose that R 0 ≤ 1 , then the infection-free steady state Π 0 is globally asymptotically stable (GAS).

Proof. Constructing a Lyapunov functional

L 0 ( s , y , p , z ) = s 0 H ( s s 0 ) + y + α 1 s 0 γ p + q α 1 s 0 r γ z     + ρ 2 ( δ ^ + ε ) s 0 [ ( s − s 0 ) + y ] 2     + m α 1 s 0 γ ∫ 0 h   Θ ( τ ) ∫ t − τ t y ( η ) d η d τ .

We calculate d L 0 d t along the solutions of model (5)-(8) as:

d L 0 d t = ( 1 − s 0 s ) ( β − δ ^ s − α 1 s p − α 2 s y + ρ y ) + α 1 s p + α 2 s y     − ε y − ρ y + α 1 s 0 γ ( m ∫ 0 h   Θ ( τ ) y ( t − τ ) d τ − γ p − q z p )     + q α 1 s 0 r γ ( r z p − μ z ) + ρ ( δ ^ + ε ) s 0 [ ( s − s 0 ) + y ] ( β − δ ^ s − ε y )     + m α 1 s 0 γ ∫ 0 h   Θ ( τ ) [ y − y ( t − τ ) ] d τ . (17)

Collecting terms of Equation (17) and using β = δ ^ s 0 we obtain

d L 0 d t = ( 1 − s 0 s ) ( δ ^ s 0 − δ ^ s ) + α 2 s 0 y + ( 1 − s 0 s ) ρ y     − ( ε + ρ ) y + α 1 s 0 F γ m y − q α 1 s 0 r γ μ z     − ρ ( δ ^ + ε ) s 0 [ ( s − s 0 ) + y ] ( δ ^ ( s − s 0 ) + ε y ) . (18)

We note that

( 1 − s 0 s ) ρ y = − ρ y s s 0 ( s − s 0 ) 2 + ρ y s 0 ( s − s 0 ) .

Therefore

d L 0 d t = − δ ^ ( s − s 0 ) 2 s + α 2 s 0 y − ρ y s s 0 ( s − s 0 ) 2 + ρ y s 0 ( s − s 0 )     − ( ε + ρ ) y + α 1 s 0 F γ m y − q α 1 s 0 r γ μ z − ρ δ ^ ( s − s 0 ) 2 ( δ ^ + ε ) s 0     − ρ ε ( s − s 0 ) y ( δ ^ + ε ) s 0 − ρ δ ^ ( s − s 0 ) y ( δ ^ + ε ) s 0 − ρ ε y 2 ( δ ^ + ε ) s 0

= − ( δ ^ s 0 + ρ y + ρ δ ^ s δ ^ + ε ) ( s − s 0 ) 2 s s 0 − ρ ε y 2 ( δ ^ + ε ) s 0     − q α 1 s 0 r γ μ z + ( ε + ρ ) ( ( m α 1 F + γ α 2 ) s 0 γ ( ε + ρ ) − 1 ) y = − ( δ ^ s 0 + ρ y + ρ δ ^ s δ ^ + ε ) ( s − s 0 ) 2 s s 0 − ρ ε y 2 ( δ ^ + ε ) s 0     − q α 1 s 0 r γ μ z + ( ε + ρ ) ( R 0 − 1 ) y .

Since R 0 ≤ 1 , then d L 0 d t ≤ 0 for all s , y , p , z > 0 . Moreover d L 0 d t = 0 if and only if s ( t ) = s ( 0 ) ,   y ( t ) = z ( t ) = 0 . Let Γ 0 = { ( s , y , p , z ) : d L 0 d t = 0 } and Γ 0

be the largest invariant subset of Γ 0 . The solution of system (5)-(8) tend to Γ 0 . For each element of Γ 0 we have y ( t ) = 0 , then y ˙ ( t ) and Equation (6) we get

y ( t ) = 0 = α 1 s 0 p (t)

Then p ( t ) = 0 . It follows that Γ 0 contains a single point that is { Γ 0 } . Appling LaSalle’s invariance principle (LIP), we get that is GAS.

Theorem 2. Suppose that, then is GAS.

Proof. Let us define a function as:

Calculating along the trajectories of system (5)-(8), we get

Collecting terms of Equation (19), we get

Applying condition of equilibrum:

we get

thus

We note that

Then

Consider the following equalities

Simplify Equation (20) and let, in Equation(21) we get

Equation (22) can be rewrite as:

We note that

From Lemma 2 we have, then, for all and, where if and only if and. Thus, the global asymptotic stability of follows from LIP when, and. ,

Theorem 3. Suppose that and, then is GAS .

Proof. Define a function as:

Calculating as:

Collecting terms of Equation (24) and applying the equilibrium conditions for:

we get

We note that

Using equalities (21) in case, we get

Equation (25) can be simplified as:

We note that, when, where occurs at. The global asymptotic stability of follows from LIP. ,

4. Numerical Simulations

This section is devoted to performing some numerical simulations for model (5)-(8). Let us choose

where is the Dirac delta function and is constant. Let, then we obtain

Moreover,

Hence, model (5)-(8), becomes

For model (26)-(29), the threshold parameters are given by:

where y₂ is given by Equation (14). Model (26)-(29) will be solved using the values of the parameters listed in Table 1.

Now we investigate our theoretical results given in Theorem 1-3. We consider the following two cases:

Case I: Effect of α, μ and h on the asymptotic behaviors of steady states:

In this case, we have chosen three different initial conditions for model (26)-(29) as follows:

Initial-1:, (Solid lines in the figures)

Initial-2:, (Dashed lines in the figures)

Initial-3:,. (Dotted lines in the figures)

Further, we fix the value of and we use three sets of parameters and r to investigate the following five scenarios.

Scenario 1: and. For this set of parameters, we have and. From Figure 1 it can be seen that the solutions with all initial conditions converge to. This means that according to Theorem 1 is GAS. In this case the healthy state will be reached and the HCV particles will be removed.

Table 1

Some parameters and their values of model (26)-(29)

Notation	Value	Notation	Value	Notation	Value	Notation	Value
	10		0.01	q	0.1		0.1
	0.01		0.5	r	Varied
	Varied	m	10		0.1
	0.0001		3		Varied

Scenario 2: and. With such choice we get,

and exists with

. This result confirms Lemma 1. Theorem 2 states that, is GAS and this is shown in Figure 2. This case represents the

persistence of the HCV particles but with inactive antibody immune response.

Scenario 3: and. Then, we calculate, and. Lemma 1 and Theorem 3 establish that, exists and it is globally asymptotically stable. From Figure 3, we find that the numerical results agree with the theoretical one presented in Theorem 3. For all initial conditions the states reach the steady state . This case corresponds to a chronic HCV infection with active antibody immune response.

Case II: Effect of the time delays on the free HCV particles dynamics:

Let us take the initial conditions (Initial-2). We choose the values and. we assume that. Table 2 contains the values of all threshold parameters and equilibria of system (26)-(29) with different values of.

From Table 2 we can see that, the values of, and are decreased as is increased. Moreover, has a significant effect on the stability of steady states of the system. Table 2 and Figure 4 show that a high value of

Table 2

The values of the threshold parameters and the equilibria of system (26)-(29) with different values of

			The steady states
0.0	6.73	3.47
8	3.13	1.57
15	1.65	0.77
23	0.85	0.35

decreases the concentration of infected hepatocytes, free HCV particles, antibodies, and increases the population of uninfected hepatocytes. Therefore, the steady states of the system will eventually stabilized around the healthy state.

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