Applied Mathematical Sciences, Vol. 9, 2015, no.

81, 3997 - 4008

HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/ams.2015.54334

Global Analysis of a HCV Model with CTL,

Antibody Responses and Therapy
Adil Meskaf

Department of Mathematics, Faculty of Sciences and Techniques, Mohammedia

University Hassan II-Casablanca, P.O. Box 146, Mohammedia, Morocco

Youssef Tabit

Department of SEG, Faculty Polydisciplinary of El Jadida

University Chouaib Doukkali, El jadida, Morocco

Karam Allali

Department of Mathematics, Faculty of Sciences and Techniques, Mohammedia

University Hassan II-Casablanca, P.O. Box 146, Mohammedia, Morocco

Copyright c 2015 Adil Meskaf et al. This article is distributed under the Creative Com-
mons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.

Abstract

In this paper, the global analysis of a HCV model with CTLs, anti-
body responses and therapy is studied. We incorporate into our model
two treatments; the aim of the first one is to reduce the infected cells,
while the second is to block the virus. We prove that the solutions with
positive initial conditions are all positive and bounded. Moreover, we
establish by using some appropriate Lyapunov functions that with the
therapy the model becomes more stable than the one without treatment.

Mathematics Subject Classification: 34D23, 34K20, 92B05, 92D30

Keywords: Basic infection reproduction number, CTL and antibody re-

sponses, HCV model, stability, therapy
3998 Adil Meskaf et al.

1 Introduction
Hepatitis C is a liver disease caused by the hepatitis C virus (HCV). Approx-
imately, 130 − 150 million people globally have chronic hepatitis C infection.
Therefore, HCV infection presents a significant global public health issue [6].
For this reason, many mathematical models have been developed in order to
understand the HCV dynamics [1, 5, 9]. In this article, we will consider the
basic model presented by Wodarz in [9] and we will continue the investigation
by introducing therapy into the model. It is well known that the antiviral
therapies combined with interferon and ribavirin are successful in 90% of per-
sons with acute infection and in 50% cases of persons with chronic infection
[2]. The model of the HCV dynamics that we consider is under the following
form:

dX


 = λ − dX − β(1 − u1 )V X,


 dt
dY


= β(1 − u1 )V X − aY − pY Z,





 dt
dV

= k(1 − u2 )Y − δV − qV W, (1)

 dt
dW


= gV W − hW,





 dt
 dZ


= cY Z − bZ,

dt
the problem is supplemented by the initial conditions:

X(0) = X0 , Y (0) = Y0 , V (0) = V0 , Z(0) = Z0 , W (0) = W0 .

This model contains five variables, that is, uninfected cells (X), infected
cells (Y), free virus (V), an antibody response (W) and a CTL response (Z).
Susceptible host cells (X) are produced at a rate λ, die at a rate dX and become
infected by virus at a rate βXV . Infected cells die at a rate aY and are killed
by the CTL response at a rate pY Z. Free virus is produced by infected cells
at a rate kY and decays at a rate δV and is neutralized by antibodies at a rate
qV W . CTLs expand in response to viral antigen derived from infected cells
at a rate cY Z and decay in the absence of antigenic stimulation at a rate bZ.
Antibodies develop in response to free virus at a rate gV W and decay at a rate
hW . The parameter u1 represents the efficiency of drug therapy in blocking
new infection, so that infection rate in the presence of drug is β(1 − u1 ), while
the parameter u2 stands for the efficiency of drug therapy in inhibiting viral
production, such that the virion production rate under therapy is k(1 − u2 ).
Clearly, the system (1) has a basic infection reproductive number of
Global analysis of a HCV model with CTL, antibody responses and therapy 3999

λβk(1 − u1 )(1 − u2 )
R0 = .
daδ
The present paper is organized as follows. In section 2, we study the global
existence of solutions, followed in section 3 by the analysis of the model. We
discuss and conclude in the last section.

2 The global existence of solutions

In order to prove the global existence of solutions, we study the positivity
and boundedness of solutions of system (1) describing the evolution of the cell
population. For biological reasons, we assume that the initial data for this
model satisfy :

X0 ≥ 0, Y0 ≥ 0, V0 ≥ 0, W0 ≥ 0, Z0 ≥ 0.

Proposition 1. All solutions with non-negative initial conditions exist for all
t > 0 and remain bounded and non-negative. Moreover we have
i) X(t) ≤ X0 + λd ,

ii) Y (t) ≤ Y0 + max(1, 2 − ad )X0 + max( λa , λd ),

k(1−u2 )
iii) V (t) ≤ V0 + δ
kY k∞ ,
 
iv) W (t) ≤ W0 + g
q
max(1; 2 − δ
h
)V0 + max( k(1−u
δ
2 ) k(1−u2 )
, h ) kY k∞ ,
 
c d
v) Z(t) ≤ Z0 + p
max(1; 2 − b
)X0 + Y0 + max( λb , λd ) + max(0; 1 − ab ) kY k∞ .

Proof
For positivity, we show that any solution starting in non negative orthant
R5+ = {(X, Y, V, W, Z) ∈ R5 : X ≥ 0, Y ≥ 0, V ≥ 0, W ≥ 0, Z ≥ 0} remains
there forever. In fact, (X(t), Y (t), V (t), W (t), Z(t)) ∈ R5+ we have
Ẋ |X=0 = λ ≥ 0, Ẏ |Y =0 = β(1 − u1 )V X ≥ 0, V̇ |V =0 = k(1 − u2 )Y ≥ 0,
Ẇ |W =0 = 0 ≥ 0, Ż |Z=0 = 0 ≥ 0
Hence, positivity of all solutions initiating in R5+ is guaranteed [?].

From the first equation of the system (1), we deduce that Ẋ + dX ≤ λ,

then dtd (Xedt ) ≤ λedt .
Hence,
λ
X(t) ≤ X0 e(−dt) + (1 − e−dt ),
d
4000 Adil Meskaf et al.

Since 0 ≤ e−dt ≤ 1, we deduce (i).

From the first and second equation of the system (1), we have

Ẏ + aY ≤ λ − dX − Ẋ.

Thus,
Rt
Y (t)eat − Y0 ≤ λa (eat − 1) − 0
d
e(a−d)s ds (X(s)eds )ds.

Using the integration by parts, we get

Rt d
Rt
0
e(a−d)s ds (X(s)eds )ds = [X(s)eas ]t0 − (a − d) 0
X(s)eas ds.

Therefore,

Z t
−at λ
Y (t) ≤ (X0 + Y0 )e + (1 − e−at ) − X(t) + (a − d) X(s)ea(s−t) ds. (2)
a 0

If a − d ≤ 0, then

λ
Y (t) ≤ X0 + Y0 + . (3)
a
If a − d ≥ 0, then
λ
Rt
Y (t) ≤ X0 + Y0 + a
+ (a − d) 0
X(s)ea(s−t) ds.

According to i), we have

(a−d)
Y (t) ≤ X0 + Y0 + λ
a
+ a
(X0 + λd )(1 − e−at ),

hence,
d λ
Y (t) ≤ Y0 + (2 − )X0 + . (4)
a a
From (3) and (4), we deduce (ii).
Now, we prove iii). The third equation of system 1, and (V (t), Z(t)) ∈ R2+ ,
imply that
Z t
−δt
V (t) ≤ V0 e + k(1 − u2 ) Y (s)e(s−t) ds,
0

then
k(1−u2 )
V (t) ≤ V0 + δ
kY k∞ (1 − e−tδ ).
Global analysis of a HCV model with CTL, antibody responses and therapy 4001

Since 1 − e−tδ ≤ 1, we deduce (iii).

Using a same technique to show (2), we get

g t 
Z
−ht
k(1 − u2 )Y (s) + (h − δ)V (s) eh(s−t) ds − V (t) + V0 e−ht .

W (t) = W0 e +
q 0
(5)
If h − δ ≤ 0, then
 
g k(1 − u2 )
W (t) ≤ W0 + kY k∞ + V0 . (6)
q h

If h − δ ≥ 0, using (iii), we have

 
g k(1 − u2 ) δ
W (t) ≤ W0 + kY k∞ + (2 − )V0 . (7)
q δ h
From (6) and (7), we deduce (iv).
Finally, we prove v). The the fifth equation of system (1) implies that
 
c
Ż + bZ = cY Z = p λ − (Ẋ + dX) − (Ẏ + aY ) .

Using the same techniques for the proof of (2) and (5), we get
   Z t 
c λ −bt c λ b(s−t)
Z(t) = (X0 +Y0 − )+Z0 e + + [(b−d)X(s)+(b−a)Y (s)]e ds−X(t)−Y (t) .
p b p b 0
(8)
If b − d ≤ 0 and b − a ≤ 0, we have
 
c λ
Z(t) ≤ Z0 + + X 0 + Y0 .
p b
If b − d ≤ 0 and b − a ≥ 0, we have
 
c λ a
Z(t) ≤ Z0 + + X0 + Y0 + (1 − ) kY k∞ .
p b b
If b − d ≥ 0 and b − a ≤ 0, we have
 
c λ d
Z(t) ≤ Z0 + + (2 − )X0 + Y0 .
p d b
If b − d ≥ 0 and b − a ≥ 0, we have
 
c λ d a
Z(t) ≤ Z0 + + (2 − )X0 + Y0 + (1 − ) kY k∞ . (9)
p d b b

From (8) and (9), we deduce (v).

4002 Adil Meskaf et al.

3 Analysis of the model

In this section, we will prove that there exist a disease free equilibrium point
and four endemic equilibrium points. Next, we will study the global stability
of these equilibrium points.

By a simple calculation, the system (1) has always one disease-free equilib-
rium
E0 = ( λd , 0, 0, 0, 0) and four endemic equilibrium points:

E1 = (X1 , Y1 , V1 , 0, 0) where

aδ λkβ(1 − u1 )(1 − u2 ) − adδ λkβ(1 − u1 )(1 − u2 ) − adδ

X1 = , Y1 = , V1 = .
kβ(1 − u1 )(1 − u2 ) akβ(1 − u1 )(1 − u2 ) βδa(1 − u1 )

E2 = (X2 , Y2 , V2 , 0, Z2 ) where

λδc b k(1 − u2 )b
X2 = , Y2 = , V2 = ,
dδc + kβb(1 − u1 )(1 − u2 ) c δc
kβλc(1 − u1 )(1 − u2 ) − a(dδc + βkb(1 − u1 )(1 − u2 ))
Z2 = .
p[dδc + βkb(1 − u1 )(1 − u2 )]

E3 = (X3 , Y3 , V3 , 0, Z3 ) where

λg λβ(1 − u1 )h
X3 = , Y3 =
dg + βh(1 − u1 ) a(dg + β(1 − u1 )h)
h gkλβ(1 − u1 )(1 − u2 ) − δa(dg + β(1 − u1 )h)
V3 = , W3 = .
g aq(dg + β(1 − u1 )h)

E4 = (X4 , Y4 , V4 , W4 , Z4 ) where

λg b h
X4 = , Y4 = , V4 =
dg + β(1 − u1 )h c g
kbg(1 − u2 ) − δhc cβhλ(1 − u1 ) − ab(dg + β(1 − u1 )h)
W4 = , Z4 = .
cqh (dg + β(1 − u1 )h)pb

In order to study the global stability of the points E1 , E2 , E3 and E4 , we

define the following numbers:

λk(1 − u2 )g

W
 D0 = ,


aδh
1 (10)
W
 H0 = ,

1
+ D1W

R0 0
Global analysis of a HCV model with CTL, antibody responses and therapy 4003

λc

Z
 D0 = ,


ab
1 (11)
Z
 H0 = ,

 1 1
R0
+ D0Z

Then, these equilibria can be written as follows:

Ei = ( λd QX λ Y d V δ W a Z
i , a Qi , β Qi , q Qi , p Qi ), 0 ≤ i ≤ 4,

where

QX Y V W Z
0 = 1, Q0 = Q0 = Q0 = Q0 = 0,

1 1
QX
1 = R0
, QY1 =1− R0
, QV1 = R0 − 1, QW Z
1 = Q1 = 0,

H0Z 1 R0
QX
2 = R0
, QY2 = D0Z
, QV2 = D0Z
, QW
2 = 0, QZ2 = H0Z − 1,

H0W H0W R0
QX
3 = R0
, QY3 = D0W
, QV3 = D0W
, QW W Z
3 = H0 − 1, Q3 = 0,

H0W 1 R0 D0W D0Z

QX
4 = R0
, QY4 = D0Z
, QV4 = D0W
, QW
4 = D0Z
− 1, QZ4 = D0W
H0W − 1.

It easy to remark that

Remark 2.

1. If R0 < 1, then the point E1 does not exist and E1 = Ef when R0 = 1.

2. If H0Z < 1, then E2 does not exists and E2 = E1 when H0Z = 1.

3. If H0W < 1, then E3 does not exists and E3 = E1 when H0W = 1.

D0Z
4. If D0W < D0Z or D0W
H0W < 1, then E4 does not exists. Moreover E4 = E2
D0Z
when D0W = D0Z and E4 = E3 when D0W
H0W = 1.

The number D0W represents the basic defence number by antibody response,
Z
D0 represents the basic defence number by CTL response.
H0W is the half harmonic mean of R0 and D0W and H0Z is the half harmonic
mean of R0 and D0Z .
We put
H0W D0Z
H0W,Z = D0W
4004 Adil Meskaf et al.

We remark that H0W = H0W,W and H0Z = H0Z,Z . The importance of these
parameters is related to the following result:

Proposition 3.

1. If R0 ≤ 1, then the point E0 is globally asymptotically stable.

2. If R0 > 1, H0W ≤ 1 and H0Z ≤ 1, then E1 is globally asymptotically

stable.

3. If H0Z > 1 and D0Z > D0W , then the point E2 is globally asymptotically
stable.

4. If H0W > 1 and H0W,Z ≤ 1, then the point E3 is globally asymptotically

stable.

5. If D0W > D0Z and H0W,Z > 1, then the point E4 is globally asymptotically
stable.

Proof. For the proof, we will use the same techniques given in [3, 7, 10].
We consider the following Lyapunov function defined in R5+ by:

X X Y Y
V (X, Y, V, W, Z) = x∗ (
∗
− ln ∗ ) + Y ∗ ( ∗ − ln ∗ ) +
X X Y Y
β(1 − u1 )X ∗ ∗ V V q ∗ W W p ∗ Z Z
∗
[V ( ∗ − ln ∗ ) + W ( ∗ − ln ∗ )] + Z ( ∗ − ln ∗ ),
u + qW V V g W W c Z Z
where E ∗ = (X ∗ , Y ∗ , V ∗ , W ∗ , Z ∗ ) is an equilibrium of the system.
It easy to verifies that

R0 QXi QVi QW
i QZi
V̇ (X, Y, V, W, Z) = λ[1 + QX Y
i + Qi + ( + ) + ] − (dX +
1 + QW i R0 D0W D0Z
λ2 X β(1 − u1 )λXV Y λ(1 − u2 )kY QX i Qi
V
R0 QX i
Qi ) − Qi − + ay( −1−
dX ay δv 1 + QW i 1 + Q W
i
W qλ QX i R0 pλz Y 1 pλz Y 1
QZi ) + W
(QVi − W ) + (Qi − Z ) + (Qi − Z ).
δ 1 + Qi D0 a D0 a D0
For i = 0 we have
λ2 λqW R0 pλZ 1
V̇ = 2λ − (dX + ) + aY (R0 − 1) − W
− .
dX δ D0 a D0Z
Since
λ2
dX + ≥ 2λ,
dX
Global analysis of a HCV model with CTL, antibody responses and therapy 4005

then V̇ < 0 and E0 is globally asymptotically stable, if R0 ≤ 1.

For i = 1 we have that

1 λ2 β(1 − u1 )λXV 1 λk(1 − u2 )y 1

V̇ = λ(3 − ) − (dX + )− (1 − )− (1 − )
R0 R0 dX aY R0 uV R0
λqW 1 pλZ 1
+ (1 − W ) + (1 − Z )
δ H0 a H0
1 2 λ2 λ2 1 β(1 − u1 )λXV 1
= λ[3(1 − )+ ] − (dX + 2 )− (1 − )− (1 − )
R0 R0 R0 dX R0 dX R0 ay R0
λk(1 − u2 )Y 1 λqW 1 pλZ 1
− (1 − )+ (1 − W ) + (1 − Z ).
δV R0 δ H0 a H0

Using the arithmetic-geometric inequality, if R0 > 1, we have

2 β(1−u1 )λXV λk(1−u2 )Y
− R0λdX (1 − 1
R0
) − ay
(1 − 1
R0
) − δV
(1 − 1
R0
) ≤ −3λ(1 − 1
R0
).

Since

λ2 2λ
dX + R02 dX
≥ R0
,

then V̇ < 0 and E1 is globally asymptotically stable, if R0 > 1, H0W ≤ 1 and

H0Z ≤ 1.
For i = 2 we have

H0Z λ2 H0Z β(1 − u1 )λXV 1 λk(1 − u2 )Y H0Z

V̇ = λ(3 − ) − (dX + )− −
R0 dX R0 aY D0Z δV D0Z
λqW z 1 1
+ H0 ( Z − W )
δ D0 D0
Z Z
H H λ2 H0Z 2 λ2 H0Z HZ β(1 − u1 )λXV 1
= λ(3 0Z + 2 0 ) − [dX + ( ) ]− (1 − 0 ) −
D0 R0 dX R0 dX R0 R0 aY D0Z
λk(1 − u2 )Y H0Z λqW Z 1 1
− Z
+ H0 ( Z − W ).
δV D0 δ D0 D0

Using the arithmetico-geometric inequality, we have

λ2 H0Z H0Z β(1−u1 )λXV 1 λk(1−u2 )Y H0Z HZ

− dX R0
(1 − R0
) − aY D0Z
− δV D0Z
≤ −3λ D0Z .
0

Since
Z HZ
λ2 H0 2
dx + ( )
dX R0
≥ 2λ R00 ,
4006 Adil Meskaf et al.

then V̇ < 0 and E2 is globally asymptotically stable, if H0Z > 1 and D0Z >
D0W .
For i = 3 we have
H0W H0W λ2 H0W 2 λ2 H0W H0W β(1 − u1 )λXV H0W
V̇ = λ(3 W + 2 ) − [dX + ( ) ]− (1 − )−
D0 R0 dX R0 dx R0 R0 ay D0W
λk(1 − u2 )Y 1 λpZ
− W
+ Z
(H0W,Z − 1).
δV D0 aD0
Using the arithmetico-geometric inequality, we have
λ 2 H0W H0W β(1−u1 )λXV H0W λk(1−u2 )Y 1 HW
− dX R0
(1 − R0
)− aY D0W
− δV D0Z
≤ −3λ D0W .
0

Since
W H0W
λ2 H 0
dX + (
dX R0
)2 ≥ 2λ R0
,
then V̇ < 0 and E3 is globally asymptotically stable, if H0W ≥ 1 and
H0W,Z < 1.
For i = 4 we have

H0W λ2 H0W β(1 − u1 )λXV 1 λk(1 − u2 )Y H0W,Z

V̇ = λ(3 − ) − (dX + )− − ,
R0 dX R0 aY D0Z δV D0Z
HW λ2 H0W 2 λ2 H0W HW β(1 − u1 )λXV 1
= λ(3 − 0 ) − [dX + ( ) ]− (1 − 0 ) −
R0 dX R0 dX R0 R0 aY D0Z
λk(1 − u2 )Y H0W,Z
− .
δV D0Z
Using the arithmetico-geometric inequality, we have
W,Z
λ2 H0W H0W β(1−u1 )λXV 1 λk(1−u2 )Y H0 HW
− dX R0
(1 − R0
)− ay D0Z
− δV D0W
≤ −3λ D0W .
0

Since
W H0W
λ2 H 0
dX + (
dX R0
)2 ≥ 2λ R0
,
then V̇ ≤ 0.
The equality holds if and only if
k(1−u2 ) D0Z V∗
X = X ∗ and V
Y
= δ D0W
= Y∗
.
Let
S = {(X, Y, V, W, Z) ∈ IR5+ : V̇ (X, Y, V, W, Z) = 0}.
The trajectory (X(t), Y (t), V (t), W (t), Z(t)) ∈ S implies that
Y = Y ∗ , V = V ∗ , W = W ∗ and Z = Z ∗ .
From LaSalle’s invariance principle [?], we conclude that E4 is globally asymp-
totically stable, if D0W > D0Z and H0W,Z > 1.
Global analysis of a HCV model with CTL, antibody responses and therapy 4007

Figure 1: The basic reproduction number R0 as function of the drug efficacy

θ for λ = 2.105 , β = 2.10−7 , k = 20, d = 0.1, a = 0.5 and δ = 4.

4 Discussion and conclusion

In this work, we gave the global analysis of a HCV with CTL, antibody re-
sponses and therapy The disease free equilibrium is globally asymptotically
stable if the basic infection reproduction number satisfies R0 ≤ 1. For R0 > 1,
the global stability of the four endemic equilibrium points depends on both
the basic defence rate by antibody response and the basic defence rate by
CTL response. In addition, the goal of the therapy is to better control the
concentrations of the virus and infected cells in order to reduce the value of
the number of basic reproduction to a number below one. Indeed, if we put
θ = u1 + u2 − u1 u2 which represents the combined efficacy of the two drugs,
we will have 1 − θ = (1 − u1 )(1 − u2 ) which means that each drug acts inde-
pendently. Hence, R0 becomes

λβ(1 − θ)k
R0 = .
daδ
Then from the above formula of R0 , we see that R0 can be decreased by
increasing the efficacy of therapy, i.e. increasing u1 and u2 (see Fig. 1).
4008 Adil Meskaf et al.

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Received: April 27, 2015; Published: May 21, 2015