Global Analysis of a HCV Model with CTL, Antibody Responses and Therapy

Applied Mathematical Sciences Vol 9 205 no 8 3997-4008 HIKARI Ltd wwwm-hikaricom http://dxdoiorg/02988/ams20554334 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 PO Box 46 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 PO Box 46 Mohammedia Morocco Copyright c 205 Adil Meskaf et al This article is distributed under the Creative Commons 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 antibody responses and therapy is studied e 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 e 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 responses HCV model stability therapy

3998 Adil Meskaf et al Introduction Hepatitis C is a liver disease caused by the hepatitis C virus (HCV) Approximately 30 50 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 [ 5 9] In this article we will consider the basic model presented by odarz 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 persons 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 dt = λ dx β( u )V X dy dt = β( u )V X ay py dv dt = k( u 2)Y qv d = gv h dt d = cy b dt the problem is supplemented by the initial conditions: X(0) = X 0 Y (0) = Y 0 V (0) = V 0 (0) = 0 (0) = 0 This model contains five variables that is uninfected cells (X) infected cells (Y) free virus (V) an antibody response () and a CTL response () 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 Free virus is produced by infected cells at a rate ky and decays at a rate and is neutralized by antibodies at a rate qv CTLs expand in response to viral antigen derived from infected cells at a rate cy and decay in the absence of antigenic stimulation at a rate b Antibodies develop in response to free virus at a rate gv and decay at a rate h The parameter u represents the efficiency of drug therapy in blocking new infection so that infection rate in the presence of drug is β( u ) while the parameter u 2 stands for the efficiency of drug therapy in inhibiting viral production such that the virion production rate under therapy is k( u 2 ) Clearly the system () has a basic infection reproductive number of ()

Global analysis of a HCV model with CTL antibody responses and therapy 3999 = λβk( u )( u 2 ) da 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 e 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 () describing the evolution of the cell population For biological reasons we assume that the initial data for this model satisfy : X 0 0 Y 0 0 V 0 0 0 0 0 0 Proposition All solutions with non-negative initial conditions exist for all t > 0 and remain bounded and non-negative Moreover we have i) X(t) X 0 + λ d ii) Y (t) Y 0 + max( 2 d a )X 0 + max( λ a λ d ) iii) V (t) V 0 + k( u 2) Y [ iv) (t) 0 + g max(; 2 )V q h 0 + max( k( u 2) k( u 2) ) Y h ] v) (t) 0 + c p [ ] max(; 2 d)x b 0 +Y 0 +max( λ λ)+max(0; a) Y b d b Proof For positivity we show that any solution starting in non negative orthant R 5 + = {(X Y V ) R 5 : X 0 Y 0 V 0 0 0} remains there forever In fact (X(t) Y (t) V (t) (t) (t)) R 5 + we have Ẋ X=0 = λ 0 Ẏ Y =0 = β( u )V X 0 V V =0 = k( u 2 )Y 0 Ẇ =0 = 0 0 Ż =0 = 0 0 Hence positivity of all solutions initiating in R 5 + is guaranteed [?] From the first equation of the system () we deduce that Ẋ + dx λ then d dt (Xedt ) λe dt Hence X(t) X 0 e ( dt) + λ d ( e dt )

4000 Adil Meskaf et al Since 0 e dt we deduce (i) From the first and second equation of the system () we have Thus Ẏ + ay λ dx Ẋ Y (t)e at Y 0 λ a (eat ) t 0 e(a d)s d ds (X(s)eds )ds Using the integration by parts we get Therefore t 0 e(a d)s d ds (X(s)eds )ds = [X(s)e as ] t 0 (a d) t 0 X(s)eas ds Y (t) (X 0 + Y 0 )e at + λ a ( e at ) X(t) + (a d) If a d 0 then t 0 X(s)e a(s t) ds (2) If a d 0 then According to i) we have hence Y (t) X 0 + Y 0 + λ a (3) Y (t) X 0 + Y 0 + λ a + (a d) t 0 X(s)ea(s t) ds Y (t) X 0 + Y 0 + λ a + (a d) a (X 0 + λ d )( e at ) Y (t) Y 0 + (2 d a )X 0 + λ a (4) From (3) and (4) we deduce (ii) Now we prove iii) The third equation of system and (V (t) (t)) R 2 + imply that then V (t) V 0 e t + k( u 2 ) t 0 Y (s)e (s t) ds V (t) V 0 + k( u 2) Y ( e t )

Global analysis of a HCV model with CTL antibody responses and therapy 400 Since e t we deduce (iii) Using a same technique to show (2) we get (t) = 0 e ht + g { t [ k( u2 )Y (s) + (h )V (s) ] e h(s t) ds V (t) + V 0 e ht} q 0 (5) If h 0 then (t) 0 + g ( k( u2 ) Y q h + V 0 ) (6) If h 0 using (iii) we have (t) 0 + g [ k( u2 ) Y q + (2 ] h )V 0 (7) From (6) and (7) we deduce (iv) Finally we prove v) The the fifth equation of system () implies that ] Ż + b = cy = [λ c (Ẋ + dx) (Ẏ + ay ) p Using the same techniques for the proof of (2) and (5) we get [ c (t) = p (X 0+Y 0 λ ] b )+ 0 e bt + c { λ t } p b + [(b d)x(s)+(b a)y (s)]e b(s t) ds X(t) Y (t) 0 (8) If b d 0 and b a 0 we have (t) 0 + c ( ) λ p b + X 0 + Y 0 If b d 0 and b a 0 we have (t) 0 + c p [ λ b + X 0 + Y 0 + ( a b ) Y If b d 0 and b a 0 we have (t) 0 + c [ λ p d + (2 d ] b )X 0 + Y 0 If b d 0 and b a 0 we have (t) 0 + c p From (8) and (9) we deduce (v) ] [ λ d + (2 d b )X 0 + Y 0 + ( a b ) Y ] (9)

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 () has always one disease-free equilibrium E 0 = ( λ 0 0 0 0) and four endemic equilibrium points: d E = (X Y V 0 0) where X = a kβ( u )( u 2 ) Y = λkβ( u )( u 2 ) ad V = λkβ( u )( u 2 ) ad akβ( u )( u 2 ) βa( u ) E 2 = (X 2 Y 2 V 2 0 2 ) where λc X 2 = dc + kβb( u )( u 2 ) Y 2 = b c V 2 = k( u 2)b c 2 = kβλc( u )( u 2 ) a(dc + βkb( u )( u 2 )) p[dc + βkb( u )( u 2 )] E 3 = (X 3 Y 3 V 3 0 3 ) where X 3 = λg dg + βh( u ) Y 3 = λβ( u )h a(dg + β( u )h) V 3 = h g 3 = gkλβ( u )( u 2 ) a(dg + β( u )h) aq(dg + β( u )h) E 4 = (X 4 Y 4 V 4 4 4 ) where 4 = kbg( u 2) hc cqh X 4 = λg dg + β( u )h Y 4 = b c V 4 = h g 4 = cβhλ( u ) ab(dg + β( u )h) (dg + β( u )h)pb In order to study the global stability of the points E E 2 E 3 and E 4 we define the following numbers: = λk( u 2)g ah H0 = + (0)

Global analysis of a HCV model with CTL antibody responses and therapy 4003 = H0 = λc ab + Then these equilibria can be written as follows: () E i = ( λ d QX i λ a QY i d β QV i q Q i a p Q i ) 0 i 4 where Q X 0 = Q Y 0 = Q V 0 = Q 0 = Q 0 = 0 Q X = Q Y = Q V = Q = Q = 0 Q X 2 = H 0 Q Y 2 = Q V 2 = Q 2 = 0 Q 2 = H0 Q X 3 = H 0 Q Y 3 = H 0 Q V 3 = Q 3 = H 0 Q 3 = 0 Q X 4 = H 0 Q Y 4 = Q V 4 = Q 4 = D 0 It easy to remark that Remark 2 Q 4 = D 0 H 0 If < then the point E does not exist and E = E f when = 2 If H 0 < then E 2 does not exists and E 2 = E when H 0 = 3 If H 0 < then E 3 does not exists and E 3 = E when H 0 = 4 If < or D 0 H 0 < then E 4 does not exists Moreover E 4 = E 2 when = and E 4 = E 3 when D 0 H 0 = The number represents the basic defence number by antibody response represents the basic defence number by CTL response H0 is the half harmonic mean of and and H0 is the half harmonic mean of and e put H 0 = H 0 D 0

4004 Adil Meskaf et al e remark that H0 = H 0 and H0 = H 0 The importance of these parameters is related to the following result: Proposition 3 If then the point E 0 is globally asymptotically stable 2 If > H 0 and H 0 then E is globally asymptotically stable 3 If H 0 > and > then the point E 2 is globally asymptotically stable 4 If H 0 > and H 0 then the point E 3 is globally asymptotically stable 5 If > and H 0 > then the point E 4 is globally asymptotically stable Proof For the proof we will use the same techniques given in [3 7 0] e consider the following Lyapunov function defined in R 5 + by: V (X Y V ) = x ( X X ln X X ) + Y ( Y Y ln Y Y ) + β( u )X u + q [V ( V V ln V V ) + q g ( ln )] + p c ( ln ) where E = (X Y V ) is an equilibrium of the system It easy to verifies that V (X Y V ) = λ[ + Q X i + Q Y i + Q X i + Q i λ 2 dx QX i ) β( u )λxv Q Y i λ( u 2)kY ay v Q i ) + qλ For i = 0 we have Q X i + Q i (Q V i ( QV i + Q i ) + Q i ] (dx + Q X i Q V i + ay( Q X i + Q i + Q i ) + pλz a (QY i ) + pλz a (QY i ) V = 2λ (dx + λ2 dx ) + ay ( ) λq pλ a Since dx + λ2 dx 2λ

Global analysis of a HCV model with CTL antibody responses and therapy 4005 then V < 0 and E 0 is globally asymptotically stable if For i = we have that V = λ(3 ) (dx + λ2 dx ) β( u )λxv ay + λq ( H 0 ) + pλ a ( H 0 ) ( ) λk( u 2)y ( ) uv = λ[3( ) + 2 ] (dx + λ2 R0dX ) λ2 2 dx ( ) β( u )λxv ( ) ay λk( u 2)Y ( ) + λq ( H 0 ) + pλ a ( H 0 Using the arithmetic-geometric inequality if > we have λ2 dx ( ) β( u )λxv ay ( ) λk( u 2)Y ( ) 3λ( ) Since dx + λ2 2λ R0 2 dx ) then V < 0 and E is globally asymptotically stable if > H0 H0 For i = 2 we have and V = λ(3 H 0 ) (dx + λ2 H0 ) β( u )λxv dx ay + λq = λ(3 H 0 H z 0( λk( u 2)Y ) + 2 H 0 ) [dx + λ2 dx (H 0 ) 2 ] λ2 dx H 0 + λq H 0 ( ) λk( u 2)Y H 0 H0 ( H 0 ) β( u )λxv ay Using the arithmetico-geometric inequality we have λ2 H0 dx ( H 0 ) β( u )λxv ay λk( u 2)Y H 0 3λ H 0 Since dx + λ2 dx ( H 0 ) 2 2λ H 0

4006 Adil Meskaf et al then V < 0 and E 2 is globally asymptotically stable if H0 > and > For i = 3 we have V = λ(3 H 0 λk( u 2)Y + 2 H 0 ) [dx + λ2 dx (H 0 ) 2 ] λ2 dx + λp (H a 0 ) Using the arithmetico-geometric inequality we have Since λ2 H0 dx ( H 0 ) β( u )λxv H0 ay λk( u 2)Y dx + λ2 dx ( H 0 ) 2 2λ H 0 H0 ( H 0 ) β( u )λxv ay 3λ H 0 then V < 0 and E 3 is globally asymptotically stable if H0 and H 0 < For i = 4 we have H 0 V = λ(3 H 0 ) (dx + λ2 H0 ) β( u )λxv dx ay λk( u 2)Y H 0 = λ(3 H 0 ) [dx + λ2 dx (H 0 ) 2 ] λ2 H0 ( H 0 ) β( u )λxv dx ay λk( u 2)Y H 0 Using the arithmetico-geometric inequality we have Since λ2 H0 dx ( H 0 ) β( u )λxv ay then V 0 The equality holds if and only if Let λk( u 2)Y dx + λ2 dx ( H 0 ) 2 2λ H 0 X = X and V Y = k( u 2) H 0 = V Y 3λ H 0 S = {(X Y V ) IR 5 + : V (X Y V ) = 0} The trajectory (X(t) Y (t) V (t) (t) (t)) S implies that Y = Y V = V = and = From LaSalle s invariance principle [?] we conclude that E 4 is globally asymptotically stable if > and H 0 >

Global analysis of a HCV model with CTL antibody responses and therapy 4007 Figure : The basic reproduction number as function of the drug efficacy θ for λ = 20 5 β = 20 7 k = 20 d = 0 a = 05 and = 4 4 Discussion and conclusion In this work we gave the global analysis of a HCV with CTL antibody responses and therapy The disease free equilibrium is globally asymptotically stable if the basic infection reproduction number satisfies For > 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 = u + u 2 u u 2 which represents the combined efficacy of the two drugs we will have θ = ( u )( u 2 ) which means that each drug acts independently Hence becomes = λβ( θ)k da Then from the above formula of we see that can be decreased by increasing the efficacy of therapy ie increasing u and u 2 (see Fig )

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