global stability of virus dynamics model with immune ... · global stability of virus dynamics...

J. KSIAM Vol.23, No.1, 39–63, 2019 http://dx.doi.org/10.12941/jksiam.2019.23.039

GLOBAL STABILITY OF VIRUS DYNAMICS MODEL WITH IMMUNERESPONSE, CELLULAR INFECTION AND HOLLING TYPE-II

A. M. ELAIW† AND SH. A. GHALEB

DEPARTMENT OF MATHEMATICS, KING ABDULAZIZ UNIVERSITY, SAUDI ARABIA

E-mail address: a m [email protected], [email protected]

ABSTRACT. In this paper, we study the effect of Cytotoxic T Lymphocyte (CTL) and antibodyimmune responses on the virus dynamics with both virus-to-cell and cell-to-cell transmissions.The infection rate is given by Holling type-II. We first show that the model is biologically ac-ceptable by showing that the solutions of the model are nonnegative and bounded. We find theequilibria of the model and investigate their global stability analysis. We derive five thresholdparameters which fully determine the existence and stability of the five equilibria of the model.The global stability of all equilibria of the model is proven using Lyapunov method and apply-ing LaSalle’s invariance principle. To support our theoretical results we have performed somenumerical simulations for the model. The results show the CTL and antibody immune responsecan control the disease progression.

1. INTRODUCTION

Mathematical modeling and analysis of within-host pathogen dynamics have attracted theinterest of several mathematicians during the recent decades. The basic pathogen infectionmodel has been proposed in [1] which describes the interaction between susceptible host cells,infected cells and pathogens and has been used to describe the dynamics of some types ofviruses such as human immunodeficiency virus (HIV) and hepatitis B virus HBV and is givenby

s = β − δs− αsp, (1.1)y = αsp− εy, (1.2)p = my − γp, (1.3)

where s, y and p denote the concentrations of susceptible cells, infected cells and pathogens,respectively. The susceptible (uninfected) cells are generated at rate β, die at rate δs andbecome infected at rate αsp. The infected cells die at rate εy. Parameters m and γ represent,respectively, the generation and clearance rate constants of pathogens. Several modifications

Received by the editors March 1 2019; Accepted March 11 2019; Published online March 20 2019.2000 Mathematics Subject Classification. 34D20, 34D23, 37N25, 92B05.Key words and phrases. Pathogen infection, Holling-type incidence, global stability, adaptive immune response

,Lyapunov function.† Corresponding author.

39

40 A. M. ELAIW AND SH. A. GHALEB

of model (1.1-1.3) have been done to take into account either Cytotoxic T Lymphocyte (CTL)immune response (see e.g. [2]-[7]) or humoral immune response (see e.g. [8]-[21]). Wodarz[22] has presented the following mathematical model to incorporate both humoral and CTLimmunity into the pathogen dynamics:

s = β − δs− αsp, (1.4)y = αsp− kwy − εy, (1.5)p = my − qzp− γp, (1.6)z = rzp− µz, (1.7)w = gwy − hw, (1.8)

where, w and z denote the concentrations of CTL cells and B cells, respectively. Model (1.4)-(1.8) has been modified in several works (see e.g. [23]-[29]). In these works, only the pathogen-to-cell transmission has been considered. Mathematical models of pathogen dynamics withboth pathogen-to-cell and cell-to-cell transmissions have been studied in several works (seee.g. [30]-[38]). However, both CTL and humoral immune responses have not been taken intoaccount in these works.

The aim of this paper is to propose a pathogen infection model with both humoral and CTLimmune responses. We have incorporated both pathogen-to-cell and cell-to-cell transmissions.The infection rate is given by Holling type-II. We first show that the solutions of the modelare nonnegative and bounded. To investigate the global stability of the equilibria we constructLyapunov functions using the method presented [39] and followed by [40]-[52]. Numericalsimulations is performed to confirm our theoretical results.

2. MODEL WITH HOLLING-TYPE INCIDENCE

In this section, we propose the following pathogen dynamics model with Holling type-IIincidence:

s = β − δs− α1sp

1 + ηs− α2sy

1 + ηs, (2.1)

y =α1sp

1 + ηs+

α2sy

1 + ηs− εy − kwy, (2.2)

p = my − γp− qzp, (2.3)z = rzp− µz, (2.4)w = gwy − hw, (2.5)

where, η is positive constant, α1 and α2 are infection rates of pathogen-to-cell and cell-to-celltransmissions, respectively. The other variables are parameters have the same meaning as givenin the previous section.

Proposition 2.1. There exist positive numbers Li, i = 1, 2, 3, 4 such that the compact set

Ω =(s, y, p, z, w) ∈ R5

≥0 : 0 ≤ s, , y ≤ L1 , 0 ≤ p ≤ L2 , 0 ≤ z ≤ L3 , 0 ≤ w ≤ L4

is positively invariant.

GLOBAL STABILITY OF VIRUS DYNAMICS MODELS 41

Proof. From Eqs. (2.1)-(2.5) we have

s |s=0= β > 0,

y |y=0=α1sp

1 + ηs≥ 0, for all s, p ≥ 0,

p |p=0= my ≥ 0, for all y ≥ 0,

z |z=0= 0,

w |w=0= 0.

Hence, s(t) > 0, y(t) ≥ 0, p(t) ≥ 0, z(t) ≥ 0 and w(t) ≥ 0 for all t ≥ 0. Therefore model(2.1)-(2.5) is biologically acceptable in the sense that no population goes negative.

Next we show the boundedness of the solutions. Let Q(t) = s(t) + y(t) + ε2mp(t) +

εq2rmz(t) + k

gw(t), then

Q = s+ y +ε

2mp+

εq

2rmz +

k

gw

= β − δs− ε

2y − ε

2mγp− εq

2rmµz − k

ghw

≤ β − σ

(s+ y +

ε

2mp+

εq

2rmz +

k

gw

)= β − σQ,

where σ = minδ, ε2 , γ, µ, h. Then

Q(t) ≤ e−σt

(Q(0)− β

σ

)+

β

σ.

Hence, 0 ≤ Q(t) ≤ L1, where L1 =β

σ. It follows that, 0 ≤ s(t), y(t) ≤ L1, 0 ≤ p(t) ≤ L2, 0

≤ z(t) ≤ L3 and 0 ≤ w(t) ≤ L4 for all t ≥ 0 if s(0) + y(0) + ε2mp(0) + εq

2rmz(0) + kgw(0) ≤

L1, where L2 =2mL1

ε, L3 =

2mrL1

εqand L4 =

gL1

k. Therefore, s(t), y(t), p(t), z(t) and

w(t) are bounded.

Lemma 2.2. For system (2.1)-(2.5) there exist five threshold parameters R0 > 0, Rz1 > 0,

Rw1 > 0, Rw

2 > 0 and Rz2 > 0, with Rw

1 < R0 such that(i) if R0 ≤ 1, then there exists only one equilibrium Π0,(ii) if Rz

1 ≤ 1, and Rw1 ≤ 1 < R0, then there exist only two equilibria Π0 and Π1,

(iii) if Rz1 > 1 and Rw

2 ≤ 1, then there exist only three equilibria Π0, Π1 and Π2,(iv) if Rw

1 > 1 and Rz2 ≤ 1, then there exist only three equilibria Π0, Π1 and Π3, and

(v) if Rw2 > 1 and Rz

2 > 1, then there exist five equilibria Π0, Π1, Π2, Π3 and Π4.


Proof. The equilibria of system (2.6)-(2.10) satisfying

β − δs− α1sp

1 + ηs− α2sy

1 + ηs= 0, (2.6)

α1sp

1 + ηs+

α2sy

1 + ηs− εy − kwy = 0, (2.7)

my − γp− qzp = 0, (2.8)

(rp− µ)z = 0, (2.9)

(gy − h)w = 0. (2.10)

We find that system (2.6)-(2.10) admits five equilibria.(i) Infection-free equilibrium Π0 = (s0, 0, 0, 0, 0), where s0 = β/δ.(ii) Chronic-infection equilibrium without immune response Π1 = (s1, y1, p1, 0, 0), where

s1 =β(

βη + δ)(R0 − 1) + δ

,

y1 =β(βη + δ

)ε((

βη + δ)(R0 − 1) + δ

) (R0 − 1) ,

p1 =mβ

(βη + δ

)εγ

((βη + δ

)(R0 − 1) + δ

) (R0 − 1) ,

and

R0 =β (mα1 + γα2)

εγ(βη + δ

) .

Clearly Π1 exists if R0 > 1.(iii) Chronic-infection equilibrium with only humoral immune response Π2 = (s2, y2, p2, z2, 0).Now we show that s2, y2, p2 and z2 which satisfy Eqs. (2.6)-(2.9) are positive.From Eq. (2.9) we have p2 =

µ

r. From Eqs. (2.6)-(2.7) we have

β − δs = εy ⇒ y =β − δs

ε. (2.11)

Substitute Eq. (2.11) in Eq. (2.6) and define a function G(s) as

G(s) = β − δs− α1sp21 + ηs

− α2s

1 + ηs(β − δs) = 0, (2.12)

we have

G(0) = β > 0,

G(s0) = − α1s0p21 + ηs0

< 0.


Then, there exists s2 ∈ (0, s0) such as G(s2) = 0. Now

y =β − δs

ε=

δ(β

δ− s

)ε

=δ

ε(s0 − s)

⇒ y2 =δ

ε(s0 − s2) > 0.

And from Eq. (2.8) we have z2 =γ

q

(my2p2γ

− 1

). Now we define

Rz1 =

my2p2γ

=rmy2µγ

. (2.13)

Then, z2 =γ

q(Rz

1 − 1) > 0 when Rz1 > 1.

(iv) Chronic-infection equilibrium with only CTL immune response Π3 = (s3, y3, p3, 0, w3),where

s3 =gβγ

h (mα1 + γα2) + gγδ, y3 =

h

g,

p3 =hm

gγ, w3 =

ε

k(Rw

1 − 1) ,

and

Rw1 =

gβ (mα1 + γα2)

ε(h (mα1 + γα2) + gγ

(βη + δ

)) =R0

1 +εh

gβR0

.

Clearly Rw1 < R0.

Hence, Π3 exists when Rw1 > 1.

(v) Chronic-infection equilibrium with both humoral and CTL immune responsesΠ4 = (s4, y4, p4, z4, w4), where

s4 =βrg

gµα1 + hrα2 + rgδ, y4 =

h

g, p4 =

µ

r,

w4 =ε

k(Rw

2 − 1) , z4 =γ

q(Rz

2 − 1) ,

and

Rw2 =

βg (µgα1 + rhα2)

εh(gµα1 + hrα2 + rg

(βη + δ

)) and Rz2 =

r

µp3 =

mhr

γgµ.

Hence, Π4 exists when Rw2 > 1 and Rz

2 > 1.


3. GLOBAL STABILITY

We will use the arithmetic mean- geometric mean (AM-GM) inequality [38], through thepaper. If θj ≥ 0, j = 1, 2, ..., n, it follows that

1

n

n∑j=1

θj ≥ n

√√√√ n∏j=1

θj , (3.1)

with equality holding if and only if θ1 = θ2 = ... = θn. Define the function

H(ℓ) = ℓ− 1− ln ℓ. (3.2)

Clearly H(ℓ) ≥ 0 for any ℓ > 0 and H has the global minimum H(1) = 0.

Theorem 3.1. Let R0 ≤ 1, then Π0 is globally asymptotically stable (GAS).

Proof. Consider a Lyapunov functional

L0(s, y, p, z, w) = s− s0 −s∫

s0

s0 (1 + ηθ)

θ (1 + ηs0)dθ + y +

α1s0γ (1 + ηs0)

p+qα1s0

rγ (1 + ηs0)z +

k

gw.

We note L0(s, y, p, z, w) > 0 for all s, y, p, z, w > 0 and L0(s0, 0, 0, 0, 0) = 0. We calculatedL0dt along the solutions of model (2.1)-(2.5) as:

dL0

dt=

(1− s0 (1 + ηs)

s (1 + ηs0)

)(β − δs− α1sp

1 + ηs− α2sy

1 + ηs

)+

α1sp

1 + ηs+

α2sy

1 + ηs− εy − kwy +

α1s0γ (1 + ηs0)

(my − γp− qzp)

+qα1s0

rγ (1 + ηs0)(rzp− µz) +

k

g(gwy − hw) . (3.3)

Collecting terms of Eq. (3.3) we get

dL0

dt=

(1− s0 (1 + ηs)

s (1 + ηs0)

)(β − δs

)+

α2s0y

1 + ηs0− εy

+α1s0

γ (1 + ηs0)my − qα1s0

rγ (1 + ηs0)µz − kh

gw.

Using the condition β = δs0, we obtain

dL0

dt=

(1− s0 (1 + ηs)

s (1 + ηs0)

)(δs0 − δs

)+

α2s0y

1 + ηs0− εy

+α1s0

γ (1 + ηs0)my − qα1s0


gw

= −δ(s− s0)

2

s (1 + ηs0)+ ε (R0 − 1) y − qα1s0


gw. (3.4)


Since R0 ≤ 1, then dL0dt ≤ 0 for all s, y, z, w > 0. Moreover, dL0

dt = 0 when s = s0, y =0, z = 0 and w = 0. Applying LaSalle’s invariance principle (LIP) we get that Π0 is GAS.

Theorem 3.2. Let Rz1 ≤ 1 and Rw

1 ≤ 1 < R0, then Π1 is GAS.

Proof. Let us define a function L1(s, y, p, z, w) as:

L1 = s− s1 −s∫

s1

s1 (1 + ηθ)

θ (1 + ηs1)dθ + y1H

(y

y1

)(3.5)

+α1s1p1

my1 (1 + ηs1)p1H

(p

p1

)+

qα1s1p1rmy1 (1 + ηs1)

z +k

gw.

Clearly, L1(s, y, p, z, w) > 0 for all s, y, p, z, w > 0 and L1(s1, y1, p1, 0, 0) = 0. CalculatingdL1dt along the trajectories of system (2.1)-(2.5), we obtain

dL1

dt=

(1− s1 (1 + ηs)

s (1 + ηs1)


1 + ηs− α2sy

1 + ηs

)+

(1− y1

y

)(α1sp

1 + ηs+

α2sy

1 + ηs− εy − kwy

)+

α1s1p1my1 (1 + ηs1)

(1− p1

p

)(my − γp− qzp)

+qα1s1p1

rmy1 (1 + ηs1)(rzp− µz) +

k

g(gwy − hw) . (3.6)


dL1

dt=

(1− s1 (1 + ηs)

s (1 + ηs1)

)(β − δs

)+

α1s1p

1 + ηs1+

α2s1y

1 + ηs1− εy

−(

α1sp

1 + ηs+

α2sy

1 + ηs

)y1y

+ εy1 +α1s1p11 + ηs1

y

y1− α1s1p1

my1 (1 + ηs1)γp

− α1s1p1(1 + ηs1)

yp1y1p

+α1s1p1

my1 (1 + ηs1)γp1 +

qα1s1p1my1 (1 + ηs1)

p1z

− qα1s1p1my1 (1 + ηs1)

µ

rz + ky1w − kh

gw.


=

(1− s1 (1 + ηs)

s (1 + ηs1)

)(β − δs

)+

α1s1p11 + ηs1

p

p1+

α2s1y11 + ηs1

y

y1− εy1

y

y1

− α1s1p11 + ηs1

spy1 (1 + ηs1)

s1p1y (1 + ηs)− α2s1y1

1 + ηs1

s (1 + ηs1)

s1 (1 + ηs)+ εy1 +

α1s1p11 + ηs1

y

y1

− α1s1p1my1 (1 + ηs1)

γp1p

p1− α1s1p1

(1 + ηs1)

yp1y1p

+α1s1p1

my1 (1 + ηs1)γp1

+qα1s1p1

my1 (1 + ηs1)

(p1 −

µ

r

)z + k

(y1 −

h

g

)w.

Applying the equilibrium conditions for Π1 :

β = δs1 +α1s1p11 + ηs1

+α2s1y11 + ηs1

,

εy1 =α1s1p11 + ηs1

+α2s1y11 + ηs1

and my1 = γp1,

we obtain (α1s1p11 + ηs1

− α1s1p1my1 (1 + ηs1)

γp1

)p

p1= 0

and (α2s1y11 + ηs1

+α1s1p11 + ηs1

− εy1

)y

y1= 0.

ThendL1

dt= −δ

(s− s1)2

s (1 + ηs1)+

(α1s1p11 + ηs1

+α2s1y11 + ηs1

)(1− s1 (1 + ηs)

s (1 + ηs1)

)− α1s1p1

1 + ηs1

spy1 (1 + ηs1)

s1p1y (1 + ηs)− α2s1y1

1 + ηs1

s (1 + ηs1)

s1 (1 + ηs)+ 2

α1s1p11 + ηs1

+α2s1y11 + ηs1

− α1s1p11 + ηs1

yp1y1p

+qα1s1p1

my1 (1 + ηs1)

(p1 −

µ

r

)z + k

(y1 −

h

g

)w. (3.7)

ThusdL1

dt= −δ

(s− s1)2

s (1 + ηs1)+

α1s1p11 + ηs1

(3− s1 (1 + ηs)

s (1 + ηs1)− spy1 (1 + ηs1)

s1p1y (1 + ηs)− yp1

y1p

)+

α2s1y11 + ηs1

(2− s1 (1 + ηs)

s (1 + ηs1)− s (1 + ηs1)

s1 (1 + ηs)

)+

qα1s1γ (1 + ηs1)

(p1 − p2) z

+ k (y1 − y3)w. (3.8)

Thus if Rz1 ≤ 1, then Π2 does not exist since z2 = γ

q (Rz1 − 1) ≤ 0. It follows that, dz

dt =

r(p (t)− µ

r

)z (t) ≤ 0 for all z > 0. Then p (t) ≤ µ

r = p2 and p1 ≤ p2. If Rw1 ≤ 1, then

Π3 does not exist since w3 = εk (R

w1 − 1) ≤ 0. It follows that, dw

dt = g(y (t)− h

g

)w (t) ≤ 0

for all w > 0. Hence, y (t) ≤ hg = y3 and y1 ≤ y3.


Using AM-GM inequality (3.1), with j = 2 and j = 3 we get, dL1dt ≤ 0 for all s, y, p, z, w >

0. Moreover, dL1dt = 0 when s = s1, y = y1, p = p1, z = w = 0. The solutions of system

(2.1)-(2.5) tend to Γ1 the largest invariant subset of Γ1 =(s, y, p, z, w) : dL1

dt = 0. Clearly

Γ1 = Π1 . LIP implies that Π1 is GAS.

Theorem 3.3. Let Rz1 > 1 and Rw

2 ≤ 1, then Π2 is GAS.

Proof. Define L2(s, y, p, z, w) as:

L2 = s− s2 −s∫

s2

s2 (1 + ηθ)

θ (1 + ηs2)dθ + y2H

(y

y2

)

+α1s2p2

my2 (1 + ηs2)p2H

(p

p2

)+

qα1s2p2rmy2 (1 + ηs2)

z2H

(z

z2

)+

k

gw.

We note that, L2(s, y, p, z, w) > 0 for all s, y, p, z, w > 0 and L2(s2, y2, p2, z2, 0) = 0.

Calculating dL2dt along the trajectories of system (2.1)-(2.5), we get

dL2

dt=

(1− s2 (1 + ηs)

s (1 + ηs2)


1 + ηs− α2sy

1 + ηs

)+

(1− y2

y

)(α1sp

1 + ηs+

α2sy


)+

α1s2p2my2 (1 + ηs2)

(1− p2

p


+qα1s2p2

rmy2 (1 + ηs2)

(1− z2

z

)(rzp− µz) +

k

g(gwy − hw) . (3.9)


dL2

dt=

(1− s2 (1 + ηs)

s (1 + ηs2)

)(β − δs

)+

α1s2p

1 + ηs2+

α2s2y

1 + ηs2− εy

−(

α1sp

1 + ηs+

α2sy

1 + ηs

)y2y

+ εy2 +α1s2p2

(1 + ηs2)

y

y2− α1s2p2

my2 (1 + ηs2)γp

− α1s2p2(1 + ηs2)

yp2y2p

+α1s2p2

my2 (1 + ηs2)γp2 +


p2z

− qα1s2p2my2 (1 + ηs2)

µ

rz − qα1s2p2

my2 (1 + ηs2)pz2 +


µ

rz2

+ ky2w − kh

gw


=

(1− s2 (1 + ηs)

s (1 + ηs2)

)(β − δs

)+

α1s2p21 + ηs2

p

p2+

α2s2y21 + ηs2

y

y2− εy2

y

y2

− α1s2p21 + ηs2

spy2 (1 + ηs2)

s2p2y (1 + ηs)− α2s2y2

1 + ηs

s (1 + ηs2)

s2 (1 + ηs)+ εy2 +

α1s2p2(1 + ηs2)

y

y2

− α1s2p2my2 (1 + ηs2)

γp2p

p2− α1s2p2

(1 + ηs2)

yp2y2p

+α1s2p2

my2 (1 + ηs2)γp2

+qα1s2p2

my2 (1 + ηs2)p2z −


µ

rz − qα1s2p2

my2 (1 + ηs2)p2z2

p

p2

+qα1s2p2

my2 (1 + ηs2)

µ

rz2 +

kh

g

(ghy2 − 1

)w.

Applying the equilibrium conditions for Π2:

β = δs2 +α1s2p21 + ηs2

+α2s2y21 + ηs2

, εy2 =α1s2p21 + ηs2

+α2s2y21 + ηs2

,

my2 = γp2 + qp2z2), p2 =µ

r.

we get (α1s2p21 + ηs2

− α1s2p2my2 (1 + ηs2)

γp2 −α1s2p2

my2 (1 + ηs2)qz2p2

)p

p2

=α1s2p21 + ηs2

(1− γp2 + qz2p2

my2

)= 0


+α1s2p21 + ηs2

− εy2

)y

y2= 0.

ThendL2

dt= −δ

(s− s2)2

s (1 + ηs2)+

(α1s2p21 + ηs2

+α2s2y21 + ηs2

)(1− s2 (1 + ηs)

s (1 + ηs2)

)+ 2

α1s2p21 + ηs2

+α2s2y21 + ηs2

− α1s2p21 + ηs2

spy2 (1 + ηs2)

s2p2y (1 + ηs)− α2s2y2

1 + ηs

s (1 + ηs2)

s2 (1 + ηs)

− α1s2p2(1 + ηs2)

yp2y2p

+ k

(y2 −

h

g

)w.

ThusdL2

dt= −δ

(s− s2)2

s (1 + ηs2)+

α1s2p21 + ηs2

(3− s2 (1 + ηs)

s (1 + ηs2)− spy2

s2p2y

1 + ηs21 + ηs

− y

y2

p2p

)+

α2s2y21 + ηs2

(2− s2 (1 + ηs)

s (1 + ηs2)− s (1 + ηs2)

s2 (1 + ηs)

)+ k (y2 − y3)w.

Thus, if Rw2 ≤ 1, then Π4 does not exist since w4 = ε

k (Rw2 − 1) ≤ 0. It follow that dw

dt =

g(y(t)− h

g

)w(t) ≤ 0 for all w > 0. Then, y (t) ≤ h

g and y2 ≤ hg = y3. Then by using


(AM-GM) inequality (3.1), with j = 2 and j = 3 we have dL2dt ≤ 0 for all s, y, p, w > 0. We

have dL2dt = 0 when s = s2, y = y2, p = p2 and w = 0. Let Γ2 =

(s, y, p, z, w) : dL2

dt = 0

and Γ2 be the largest invariant subset of Γ2. The solutions of system (2.1)-(2.5) tend to Γ2. Foreach element of Γ2 we have p(t) = p2 and y(t) = y2. From Eq. (2.3) we get

p(t) = 0 = my2 − γp2 − qz(t)p2.

Hence

z(t) =γ

g

(my2γp2

− 1

)= z2.

Then Γ2 contains a single point that is Π2 . LIP implies that Π2 is GAS. Theorem 3.4. If Rw

1 > 1 and Rz2 ≤ 1, then Π3 is GAS in ∆.

Proof. Define a function L3(s, y, p, z, w) as:

L3 = s− s3 −s∫

s3

s3 (1 + ηθ)

θ (1 + ηs3)dθ + y3H

(y

y3

)+

α1s3p3my3 (1 + ηs3)

p3H

(p

p3

)

+qα1s3p3

rmy3 (1 + ηs3)z +

k

gw3H

(w

w3

).

Clearly, L3(s, y, p, z, w) > 0 for all s, y, p, z, w > 0 and L3(s3, y3, p3, 0, w3) = 0. CalculatingdL3dt along the trajectories of system (2.1)-(2.5), we get

dL3

dt=

(1− s3 (1 + ηs)

s (1 + ηs3)


1 + ηs− α2sy

1 + ηs

)+

(1− y3

y

)(α1sp

1 + ηs+

α2sy


)+

α1s3p3my3 (1 + ηs3)

(1− p3

p


+qα1s3p3

rmy3 (1 + ηs3)(rzp− µz) +

k

g

(1− w3

w

)(gwy − hw) . (3.10)


dL3

dt=

(1− s3 (1 + ηs)

s (1 + ηs3)

)(β − δs

)+

α1s3p

1 + ηs3+

α2s3y

1 + ηs3− εy

−(

α1sp

1 + ηs+

α2sy

1 + ηs

)y3y

+ εy3 +α2s3p3

(1 + ηs3)

y

y3− α1s3p3

my3 (1 + ηs3)γp

− α1s3p3(1 + ηs3)

yp3y3p

+α1s3p3

my3 (1 + ηs3)γp3 +


p3z

− qα1s3p3my3 (1 + ηs3)

µ

rz + kwy3 −

kh

gw − kw3y +

kh

gw3.


=

(1− s3 (1 + ηs)

s (1 + ηs3)

)(β − δs

)+

α1s3p31 + ηs3

p

p3+

α2s3y31 + ηs3

y

y3− εy3

y

y3

− α1s3p31 + ηs3

spy3 (1 + ηs3)

s3p3y (1 + ηs)− α2s3y3

1 + ηs3

s (1 + ηs3)

s3 (1 + ηs)+ εy3 +

α1s3p3(1 + ηs3)

y

y3

− α1s3p3my3 (1 + ηs3)

γp3p

p3− α1s3p3

(1 + ηs3)

yp3y3p

+α1s3p3

my3 (1 + ηs3)γp3

+qα1s3p3

my3 (1 + ηs3)

µ

r

(r

µp3 − 1

)z + kwy3 −

kh

gw − kw3y3

y

y3+

kh

gw3.

Using the equilibrium conditions for Π3:

β = δs3 +α1s3p31 + ηs3

+α2s3y31 + ηs3

, γp3 = my3,

y3 =h

g,

α1s3p31 + ηs3

+α2s3y31 + ηs3

= εy3 + ky3w3.

we obtain (α1s3p31 + ηs3

− α1s3p3my3 (1 + ηs3)

γp3

)p

p3= 0


+α1s3p31 + ηs3

− εy3 − kw3y3

)y

y3= 0.

ThusdL3

dt= −δ

(s− s3)2

s (1 + ηs3)+

(α1s3p31 + ηs3

+α2s3y31 + ηs3

)(1− s3 (1 + ηs)

s (1 + ηs3)

)+ 2

α1s3p31 + ηs3

+α2s3y31 + ηs3

− α1s3p31 + ηs3

spy3 (1 + ηs3)

s3p3y (1 + ηs)− α2s3y3

1 + ηs

s (1 + ηs3)

s3 (1 + ηs)− α1s3p3

(1 + ηs3)

yp3y3p

+qα1s3p3

my3 (1 + ηs3)

µ

r

(r

µp3 − 1

)z. (3.11)

ThendL3

dt= −δ

(s− s3)2

s (1 + ηs3)+

α1s3p31 + ηs3

(3− s3 (1 + ηs)

s (1 + ηs3)− spy3 (1 + ηs3)

s3p3y (1 + ηs)− yp3

y3p

)+

α2s3y31 + ηs3

(2− s3 (1 + ηs)

s (1 + ηs3)− s (1 + ηs3)

s3 (1 + ηs)

)+


µ

r(Rz

2 − 1) z.

Therefore, if Rz2 ≤ 1, then using (AM-GM) inequality (3.1), with j = 2 and j = 3 we have

dL3dt ≤ 0 for all s, y, p, z > 0 and dL3

dt = 0 when s = s3, y = y3, p = p3 and z = 0. Let

Γ3 =(s, y, p, z, w) : dL3

dt = 0

and Γ3 be the largest invariant subset of Γ3. The solutions of

system (2.1)-(2.5) tend to Γ3. For each element of Γ3 we have y(t) = y3. From Eq. (2.2) weget

y(t) = 0 =α2s3y31 + ηs3

+α1s3p31 + ηs3

− εy3 − kw(t)y3.


Hence

w(t) =ε

k

((mα1 + γα2) s3εγ (1 + ηs3)

− 1

)

=ε

k

gβ (mα1 + γα2)

ε(h (mα1 + γα2) + gγ

(βη + δ

)) − 1

= w3.

Then Γ3 contains a single point that is Π3 . Hence, global stability of Π3 follows from LIP.

Theorem 3.5. For system (2.1)-(2.5), suppose that Rw2 > 1, Rz

2 > 1, then Π4 is GAS in∆.

Proof. We construct a function L4(s, y, p, z, w) as:

L4 = s− s4 −s∫

s4

s4 (1 + ηθ)

θ (1 + ηs4)dθ + y4H

(y

y4

)+

α1s4p4my4 (1 + ηs4)

p4H

(p

p4

)

+qα1s4p4

rmy4 (1 + ηs4)z4H

(z

z4

)+

k

gw4H

(w

w4

).

It is seen that, L4(s, y, p, z, w) > 0 for all s, y, p, z, w > 0 and L4(s4, y4, p4, z4, w4) = 0. WeCalculating dL4

dt along the trajectories of system (2.1)-(2.5), we get

dL4

dt=

(1− s4 (1 + ηs)

s (1 + ηs4)


1 + ηs− α2sy

1 + ηs

)+

(1− y4

y

)(α1sp

1 + ηs+

α2sy

1 + ηs− εy − kwy)

+α1s4p4

my4 (1 + ηs4)

(1− p4

p


+qα1s4p4

rmy4 (1 + ηs4)

(1− z4

z

)(rzp− µz)

+k

g

(1− w4

w

)(gwy − hw) . (3.12)



dL4

dt=

(1− s4 (1 + ηs)

s (1 + ηs4)

)(β − δs

)+

α1s4p41 + ηs4

p

p4+

α2s4y41 + ηs4

y

y4

− εy4y

y4

α1s4p41 + ηs4

spy4s4p4y

1 + ηs41 + ηs

− α2s4y41 + ηs4

s (1 + ηs4)

s4 (1 + ηs)+ εy4

+α1s4p4

(1 + ηs4)

y

y4− α1s4p4

my4 (1 + ηs4)γp4

p

p4− α1s4p4

(1 + ηs4)

yp4y4p

+α1s4p4

my4 (1 + ηs4)γp4 +


p4z −qα1s4p4

my4 (1 + ηs4)

µ

rz.

Applying the equilibrium conditions for Π4:

β = δs4 +α1s4p41 + ηs4

+α2s4y41 + ηs4

, εy4 =α1s4p41 + ηs4

+α2s4y41 + ηs4

− ky4w4,

my4 = γp4 + qp4z4, p4 =µ

r, y4 =

h

g.

we obtainα1s4p

1 + ηs4− α1s4p4

my4 (1 + ηs4)γp− α1s4p4

my4 (1 + ηs4)qz4p

=

(α1s4p41 + ηs4

− α1s4p4my4 (1 + ηs4)

γp4 −α1s4p4

my4 (1 + ηs4)qz4p4

)p

p4= 0

andα2s4y

1 + ηs4− εy +

α1s4p4(1 + ηs4) y4

y − kw4y

=

(α2s4y41 + ηs4

− εy4 +α1s4p41 + ηs4

− kw4y4

)y

y4= 0.

ThendL4

dt= −δ

(s− s4)2

s (1 + ηs4)+

(α1s4p41 + ηs4

+α2s4y41 + ηs4

)(1− s4 (1 + ηs)

s (1 + ηs4)

)+ 2

α1s4p41 + ηs4

+α2s4y41 + ηs4

− α1s4p41 + ηs4

spy4 (1 + ηs4)

s4p4y (1 + ηs)

− α1s4y41 + ηs

s (1 + ηs4)

s4 (1 + ηs)− α1s4p4

(1 + ηs4)

yp4y4p

. (3.13)

Eq. (3.13) can be simplified as:

dL4

dt= −δ

(s− s4)2

s (1 + ηs4)+

α1s4p41 + ηs4

(3− s4 (1 + ηs)

s (1 + ηs4)− spy4 (1 + ηs4)

s4p4y (1 + ηs)− yp4

y4p

)+

α2s4y41 + ηs4

(2− s4 (1 + ηs)

s (1 + ηs4)− s (1 + ηs4)

s4 (1 + ηs)

).


Using (AM-GM) inequality (3.1), with j = 2 and j = 3 we have, dL4dt ≤ 0 for all s, y, p > 0

and dL4dt = 0 when s = s4, y = y4, p = p4. Let Γ4 =

(s, y, p, z, w) : dL4

dt = 0

and Γ4 be the

largest invariant subset of Γ4. The solutions of system (2.1)-(2.5) tend to Γ4. For each elementof Γ4 we have y(t) = y4. From Eq. (2.2) we get

y(t) = 0 =α2s4y41 + ηs4

+α1s4p41 + ηs4

− εy4 − kw(t)y4.

Hence

w(t) =ε

k

((µgα1 + rhα2) s4

ε (1 + ηs3)− 1

)

=ε

k

βg (µgα1 + rhα2)

εh(gµα1 + hrα2 + rg

(βη + δ

)) − 1

= w4.

From Eq. (2.3) we get

p(t) = 0 = my4 − γp4 − qz(t)p4.

Hence

z(t) =γ

g

(mhr

γµg− 1

)= z4.

Then Γ4 contains a single point that is Π4 . LIP implies that Π4 is GAS.

4. NUMERICAL SIMULATIONS

In this section we perform some numerical simulations for model (2.1)-(2.5), with param-eters values given in Table 1. In the figures we show the evolution of the five states of thesystem s, y, p, z and w. We have used MATLAB for all computations. Now we investigate our

TABLE 1. Some parameters and their values of model (2.1)-(2.5).

Notation Value Notation Value Notation Valueβ 10 m 5 k 0.1

δ 0.01 γ 3.0 g Variedε 0.4 q 0.2 h 0.1α1 Varied r Varied η Variedα2 Varied µ 0.1

theoretical results given in Theorems 3.1 - 3.5.


4.1. Stability of equilibria. In this subsection, we take η = 0 and chose three different initialconditions for model (2.1)-(2.5) as follows:

IC1 : (s(0), y(0), p(0), z(0), w(0)) = (600, 5, 0.5, 5, 0.5), (Solid lines in the figures),IC2: (s(0), y(0), p(0), z(0), w(0)) = (400, 10, 3, 8, 1), (Dashed lines in the figures),IC3: (s(0), y(0), p(0), z(0), w(0)) = (200, 15, 4, 14, 2). (Dotted lines in the figures),Scenario 1: α1 = α2 = 0.0001, r = 0.01 and g = 0.01. For this set of parameters, we have

R0 = 0.6667 < 1. Figure 1 shows that, the solutions of the system with IC1-IC3 converge toΠ0 = (1000, 0, 0, 0, 0, 0). According to Theorem 3.1, Π0 is GAS.

Scenario 2: α1 = α2 = 0.0003, r = 0.001 and g = 0.001. With such choice we get, Rz1 =

0.33 < 1 and Rw1 = 0.22 < 1 < R0 = 2 and Π1 exists with Π1 = (500, 12.5, 20.83, 0, 0).

This result supports Lemma 1. Figure 2 support Theorem 3.2 that, Π1 is GAS.Scenario 3: α1 = α2 = 0.0003, r = 0.006 and g = 0.001. Then, we calculate R0 = 2 > 1,

Rz1 = 1.14 > 1 and Rw

2 = 0.19 < 1. Figure 3 shows that the solution of the system withdifferent intial conditions reach the equilibrium Π2 = (542.57, 11.44, 16.67, 2.15, 0). Thissupport Theorem 3.3.

Scenario 4: α1 = α2 = 0.0003, r = 0.001 and g = 0.01. Then, we calculate R0 = 2 > 1,Rw

1 = 1.11 > 1 and Rz2 = 0.17 < 1. The results presented in Lemma 2.2 and The-

orem 3.4 show that the equilibrium Π3 exists and it is GAS. Figure 4 supports the theo-retical results of Theorem 3.4, where the states of the system reach the equilibrium Π3 =(555.56, 10, 16.67, 0, 0.44), for all initial conditions.

Scenario 5: α1 = α2 = 0.0004, r = 0.01 and g = 0.01. Then, we calculate R0 = 2 > 1and Rz

2 = 1.67 > 1 and Rw2 = 1.11 > 1. According to Lemma 2.2 and Theorem 3.5, Π4 exists

and it is GAS. Figure 5, confirms the results of Theorem 3.5 where the states of the systemstarting with different initials converge to the equilibrium Π4 = (555.56, 10, 10, 10, 0.44).

4.2. Effect of the Holling type-II parameter η on the pathogen dynamics. Let us take theinitial conditions (IC2). We choose the values α1 = α2 = 0.006, r = 0.01 and g = 0.01 andη is varied. Figure 6 shows the effect of the Holling type-II incidence η on the stability of theequilibria of the system. We observe that, as η is increased, both the virus-target and infected-target infection rates are decreased, and then the concentration of the uninfected (susceptiblehost) cells is increased, while the concentrations of the infected cells and free virus particles(pathogens), B cells and CTL cells are decreased. We note that R0 is a decreasing functionof η. Now we compute ηcr such thatR0 = 1 = β(mα1+γα2)

εγ(βηcr+δ). Then ηcr = mα1+γα2

εγ − 1s0.

Therefore

ηcr = max

0,

mα1 + γα2

εγ− 1

s0

.

It follows that, if η ≥ ηcr, then Π0 is GAS. For the choice of values of the parameters givenabove we found that ηcr = 0.039.


Time

0 100 200 300 400 500 600 700 800 900 1000

Su

sc

ep

tib

le h

ost

ce

lls s

(t)

0

100

200

300

400

500

600

700

800

900

1000

IC1

IC2

IC3

(A) The susceptible host cells

Time0 10 20 30 40 50 60

Infe

cte

d c

ells y

(t)

0

2

4

6

8

10

12

14

16

(B) The infected cells

Time0 10 20 30 40 50 60

Path

ogens p

(t)

0

2

4

6

8

10

12

(C) The pathogens

Time0 50 100 150 200 250 300

B c

ells z

(t)

0

2

4

6

8

10

12

14

(D) The B cells

Time0 50 100 150 200 250 300

CT

L c

ells w

(t)

0

0.5

1

1.5

2

2.5

(E) The CTLs

FIGURE 1. The simulation of trajectories of system (2.1)-(2.5) for scenario 1.


Time

0 100 200 300 400 500 600 700 800 900 1000

Su

sc

ep

tib

le h

ost

ce

lls s

(t)

0

100

200

300

400

500

600

700

800

IC1

IC2

IC3


Time0 100 200 300 400 500 600 700 800 900 1000

Infe

cte

d c

ells y

(t)

0

10

20

30

40

50

60


Time0 100 200 300 400 500 600 700 800 900 1000

Path

ogens p

(t)

0

10

20

30

40

50

60

70

80

90

100

(C) The pathogens

Time0 50 100 150 200 250 300

B c

ells z

(t)

0

2

4

6

8

10

12

14

(D) The B cells

time0 50 100 150 200 250 300

CT

L c

ells w

(t)

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

(E) The CTLs



Time

0 100 200 300 400 500 600 700 800 900 1000

Su

sc

ep

tib

le h

ost

ce

lls s

(t)

0

100

200

300

400

500

600

700

800

IC1

IC2

IC3


Time0 100 200 300 400 500 600 700 800 900 1000

Infe

cte

d c

ells y

(t)

0

10

20

30

40

50

60


Time0 100 200 300 400 500 600 700 800 900 1000

Path

ogens p

(t)

0

10

20

30

40

50

60

70

80

90

100

(C) The pathogens

Time0 100 200 300 400 500 600 700

B c

ells z

(t)

0

2

4

6

8

10

12

14

(D) The B cells

Time0 50 100 150 200 250 300

CT

L c

ells w

(t)

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

(E) The CTLs



Time

0 200 400 600 800 1000 1200

Su

sc

ep

tib

le h

ost

ce

lls s

(t)

0

100

200

300

400

500

600

700

800

IC1

IC2

IC3


Time0 200 400 600 800 1000 1200

Infe

cte

d c

ells y

(t)

0

10

20

30

40

50

60

70

(B) The actively infected cells

Time0 200 400 600 800 1000 1200

Path

ogens p

(t)

0

20

40

60

80

100

120

(C) The pathogens

Time0 50 100 150 200 250 300

B c

ells z

(t)

0

2

4

6

8

10

12

14

(D) The B cells

Time0 200 400 600 800 1000 1200

CT

L w

(t)

0

0.5

1

1.5

2

2.5

(E) The CTLs



Time

0 100 200 300 400 500 600 700 800

Su

sc

ep

tib

le h

ost

ce

lls s

(t)

0

100

200

300

400

500

600

IC1

IC2

IC3


Time0 100 200 300 400 500 600 700

Infe

cte

d c

ells y

(t)

0

10

20

30

40

50

60


Time0 100 200 300 400 500 600 700

Path

ogens p

(t)

0

10

20

30

40

50

60

70

80

90

(C) The pathogens

Time0 100 200 300 400 500 600 700

B c

ells z

(t)

0

5

10

15

20

25

30

(D) The B cells

Time0 100 200 300 400 500 600 700

CT

L c

ells w

(t)

0

0.5

1

1.5

2

2.5

(E) The CTLs



Time0 200 400 600 800 1000

Su

sc

ep

tib

le c

ell

s

0

200

400

600

800

1000η=0η=0.01η=0.02η=0.03η=0.04

(A) The susceptible cells

Time0 200 400 600 800 1000

Infe

cte

d c

ell

s

0

50

100

150

200

250η=0η=0.01η=0.02η=0.03η=0.04


Time0 200 400 600 800 1000

Pa

tho

ge

ns

0

50

100

150

200η=0η=0.01η=0.02η=0.03η=0.04

(C) The pathogens

Time0 200 400 600 800 1000

B c

ell

s

0

10

20

30

40

50

60η=0η=0.01η=0.02η=0.03η=0.04

(D) The B cells

Time0 200 400 600 800 1000

CT

L c

ell

s

0

5

10

15

20

25η=0η=0.01η=0.02η=0.03η=0.04

(E) The CTLs

FIGURE 6. The effect of holling rate constant η on the behaviour of all trajec-tories of system (2.1)-(2.5).


5. CONCLUSION

In this paper, we have studied a virus dynamics model with both virus-to-cell and cell-to-celltransmissions. The effect of both CTL and antibodies on the virus dynamics have been stud-ied. The virus-uninfected and infected-uninfected incidence rates have been given by Hollingtype-II. We have shown that, the solutions of the model are nonnegative and bounded whichensure the well-posedness of the model. We have derived five threshold numbers which fullydetermines the existence and stability of the five equilibria of the model. We have investigatedthe global stability of the equilibria of the model by using Lyapunov method and LaSalle’sinvariance principle. We have conducted numerical simulations and have shown that both thetheoretical and numerical results are consistent. The results show that the CTL and antibodiescan control the disease progression by reducing the concentration of the free virus particlesand infected cells. Our proposed model can be extended by incorporating different types oftime delay. Moreover, following the work of Gibelli et al. [54], viral infection models adaptiveimmune response and with a stochastic parameters dynamics can also be studied.

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