A TB Model with Infection-Age-Dependent Infectivity

ccc-wxx(l) A TB Model With Infection-Age-Dependent Infectivity Carlos Castillo-Chavez and Wenxiong Xu* Biometrics Unit and Center For Applied Mathematics Cornell University 337 Warren Hall,Ithaca,N¥14853-7801, U.S.A. April 6,1996 * Visiting Scholar from Xi' an Jiaotong University ,China 1 l.The basic model and its variant equivalent forms It has been considered by Carlos Castilla-Chavez and Zhilan Feng(see[l]) that the following preliminary model for the transmission of tuberculosis(TB): d I dt S = A - {3cS N - pS d dt L I ( ) , I = {3cS N - 11 + k + TI L + {3 c TN d dt I= kL- (J-L +d) I- T2I d dt T = TIL , I + T2I- {3 c TN (1) - J-LT N=S+L+I+T Where the host population is divided into the following epidemiological classes or subgroups:Susceptibles (S) ,Latent (L, infected but not infectious) ,Infectious (I) and (effectively) Treated (T) individuals. N denotes the total population.The parameter A is the recruitment rate,{3 and {3' are respectively the average proportions of susceptible and treated individuals infected by one infectious individual per contact per unit of time, c is the per-capita contact rate, J-l is the per-capita natural death rate, k is the rate at which an individual leaves the latent class by becoming infectious, d is per-capita disease induced death rate,and T 1 and T 2 are per-capita treatment rates. It is worthwhile to pay attention that the parameters in model (1) shown above are all constant. Particularly , that k is equal to constant means that the rate at which an individual leaves the latent class by becoming infectious does not depend on the latentage. This is not reasonable. In fact, the parameter k depends on the latent-age T according to the real case (see [ 2 ]) , i.e. k = k (T) is the function of variable T • So we should improve the model (1) and further consider the following basic model: d I dt S = A - {3cS N - J-LS loo d = {3cS-I -L (p + r 1 )Lk(T)l(t, T)dT dt N 0 d dt I= k(T)l(t, T)dT- (p + d)I- T2I Joroo I + {3' cTN (2) d , I dt T =TIL+ T2I- {3 c TN - J-LT L= 1 00 l(t, T)dT N=S+L+I+T Here, the parameter A, {3, {3', c, J-L, d, Tr, T2 are all positive constants with the same epidemiological meanings as above. The function k(T) with positive value is the infectious rate, 2 l(t, r) is the density of infected but not infectious individuals. Clearly, the model (2) is equivalent to the following model in mathematical viewpiont: d dt S(t) =A- B1 (t) - p,S(t) a a <at+ ar)l(t, r) = -(p, + k(r) + rl)l(t, r) d dt I(t) = W(t) - (p, + d + r2)I(t) d dt T(t) = r1L(t) + r 2I(t)- B 2(t) - p,T(t) l(t, 0) = B1 (t) + B2(t) B1 (t) = {3 c S(t) N(t) I(t) (3) I I(t) B2(t) = {3 c T(t) N(t) W(t) = L(t) = N(t) 1 00 1 00 k(r)l(t, r)dr l(t, r)dr = S(t) + L(t) + I(t) + T(t) By integrating along characteristic lines, we can reduce (3) to the following system of integral equations: S = N° - B1 * P,_, + !1 (4a) L = B1 * P,_,+k(s)+r 1 + B2 * P,_,+k(s)+r 1 + 12 (4b) = W * P,_,+d+r + fg T = r1L * P,_, + r2I * P,_, - B2 * P,_, + !4 W = B1 * Q + B2 * Q + Is (4d) I B1 = {3 c S N (4!) I 2 I I (4e) (4g) B2 = {3 c TN N=S+L+I+T Here ,we have use the notation (B * P)(t) (4c) = 1t B(t- r)P(r)dr 3 (4h) and Pa = exp(- 1r a ds) Q = k(T)PJL+k(s)+r 1 (T) No= A J-l h h A = (S(O)- -) e-JLt J-l = (5) 1oo l(O, T- t) PJL+k(s)+rr (T) dT t PJL+k(s)+r (t- T) 1 h = I(O) e-(JL+d+r2)t !4 = T(O) e-JLt fs = 1 00 l(O,T- t)k(T) t PJL+k(s)+rr (T) dT PJL+k(S)+rr (T- t) In the following, we will respectively use the models (2),(3) and (4) for special purpose in different cases. 2.The existence of the stationary states and the basic reproductive number R 0 In this section ,we concentrate on the sdudy of the existence of the stationary states, that is, equilibria or time-indepentent solutions,that are candidates for the asymptotic behavior of the disease dynamics. Using the Laplace transform notation, we can note P(O) Q(O) 1 =1 = 1 -1r =1 -1T + 00 PJL+k(s)+rr (T)dT 00 Q(T)dT 00 = 00 (p, + k(s) + r1)ds)dT exp( (p, k(T)exp( k(s) + r 1 )ds)dT Then it can easily be checked that the equilibria of system (3) depend on the following equations: I A - (3cS- - p,S = 0 N I I ~ L- ((3cS N + (3' cT N)P(O) = 0 I ((3cS N r1L I ~ + (3' cT N )Q(O) - + r 2 I- {3 ' cT NI (p, + d + r 2 )I =0 (6) - p,T = 0 S+L+I+T=N We take a assumptation similar to [1] that the infection probabilities per contact for the treated class is the same as that of the susceptible class,i.e. ,{3 = {3' ,and pay attention to 4 the idential equation (7) then the equations (6) turn to the following equivalent equations: A-J-LN -di =0 I ~ L- f3c(N- L- I)-P(O) = 0 N I ~ f3c(N- L- I) NQ(O)- (J-L + d + r 2)I I A - {3cS- - J-LS N =0 (8) =0 I r1L + r2I- {3cT N - J-LT = 0 For solving these equations above,let q I N = f3c( N - L - I)- (9) then, clearly Q(O) I= Jl q (10) + d + r2 L = P(O)q N (11) =A_~ Q(O) Jl Jl Jl q + d + r2 A 8= + f3c fv = r1L + r2I Jl + f3c fv Jl T (12) (13) (14) substitute (10),(11),(12) into (9),the parameter q satisfy the equation By analysing (15),we get the basic reproductive number (16) and the following threshold result for existence of an endemic equilibrium. 5 THEOREM l.If R 0 < 1, there exists only the disease-free equilibrium E 0 ( S 0 , L 0 ,]0 , T 0 ), no other equilibria. If Ro > 1, there is a unique endemic equilibrium E* (S*, L *, I*, T*) except the disease-free equilibrium E 0 • Here (17) L* = P(O)q (18) Q(O) q J.L + d + rz T * _ r1L* + rzl* J• J.L + {3c N· (19) S*- (21) I*= A (20) I• J.L + {3c N• and Ro (L * + I*) Ro -1 (Ro- 1)A(J.L + d + rz) ~ ~ d(Ro- 1)Q(O) + J.LRoP(O)(J.L + d + rz) N* = q Proof. Solving (15) Case A. q = (17) from (10) Case B. q i= = (22) ~ + J.LRoQ(O) (23) for q, cause two cases. 0 as a solution of the equation (15), one finds E 0 (S 0 , L 0 , ! 0 , T 0 ) satisfied (14). 0, then the equation (15) for q is equivalent to the following equation (Ro- 1)(A - d Q(O) q)- RoP(O)q- RoQ(O) q = 0 J.L J.L J.L + d + rz J.L + d + rz (24) If Ro = 1, no nonzero q satisfy (24).Thus there are no endemic equilibria from (10)(14). If Ro < 1, no q satisfy (24) such that I, L, N expressed by (10),(11),(12) all are nonnegative and at least one of them is positive.So, there are no endemic equilibria. If Ro > 1,we can solve (24), get positive solution q expressed by {23) and find further (18)- (22) from(10)- (14).Particularly,one can check by (12) that Q(O) ~ (Ro- 1)_:\-(J.L + d + r 2 ) ~ J.L J.L J.L + d + rz d(Ro- 1)Q(O) + J.LRoP(O)(J.L + d + rz) + J.LRoQ(O) ARo[(J.L + d + rz)P(O) + Q(O)] d(Ro- 1)Q(O) + J.LRoP(O)(J.L + d + rz) + J.LRoQ(O) N* = A _ d Ro (L* +I*), Ro -1 6 (22) holds and S*, L*, I*, T*, N* all are positive. Thus the proof end. 3.Stability of the disease-free equilibrium E 0 and endemic equilibrium E* In this section,Fator's lemma and the following notation will be useful.For a bounded real-valued function f defined on [0, oo),we set foo = liminf f(t), t-+oo ! 00 =lim sup f(t), t-+oo f(oo) = lim f(t). t-+oo The following theorem connects the basic reproductive number Ro to the extinction of the disease. THEOREM 2. Let Ro < 1. Then the disease-free equilibrium E 0 is globally attractive, i.e.,E0 is globally asymptotically stable. Proof. We use the model formulation (4) and (5). First, we easily observe that fi(oo) j = 1, ... ,5. = 0, (25) Then we can apply Fatou's lemma and Lebesgue's theorem of dominated convergence to (4) to get useful estimations in inequalitys and further complete our proof. Let (26) from (4e),(4f),(4g),(4h) and the assumed condition (3' = (3, we get equation and inequality below (27) W=B*Q+fs, S+T B = (3cl N < (3cl. (28) Further,by (25),(27),(28) and (4c), we get wo~ B 00 Q(O) (29) (3cloo Bo~ / 00 ~ (30) woo roo Pp,+d+r2 ( T )dr = lo p, + ~ + r2 woo (31) Introducing (30),(31) into (29),we get W 00 ~ (32) RoW 00 , i.e., 7 (33) or, 0. wo~ Pay attention to 0 ~Wand (34) (33), we find woo= W(oo) = So, by (35),(31),(30) and 0 ~I, ! 00 B 00 0~ 0. (35) B, we find = J(oo) = 0, = B(oo) = 0. (36) (37) Similarly, we can find B1(oo) = B2(oo) = 0, (38) and L(oo) = T(oo) = 0, S(oo) (39) A = N(oo) = N° = -. J-l (40) Synthesize (36),(39),(40),we find A o (S,L,I,T)---+ (-,0,0,0) = E, t---+ oo. J-l (41) The proof end. The following theorem show that the disease may be persistent. THEOREM 3. The endemic equilibrium E* is locally asymptotically stable if c=Ro-1 is sufficiently close to 0 or k(T) =k (42) (constant). Proof. From (2),we get d - N = A - p,N - di. dt (43) So,in order to study the stability of E* ,we only need considering following equivalent 8 subsystem d a a - N dt = A - J.lN - di (at+ a7 )l(t,T) =-(J.l+k(T)+rl)l(t,T) d Set dti = W(t)- (J.l+d+r2)I l(t, 0) = {3c(N- I- L) N W(t) = 1 L(t) = I 00 1 00 (44) k(T)l(t, T)dT l(t, T)dT. N=N*+n, l = l* + u, i.e., L = L * + v, I= I*+ i, we linearize (44) around the endemic equilibrium E* as follows a a d -n dt (at+ a7 )u(t,T) = -J.ln- d"z = -(J.l+k(T) +r1 )u(t,T) -d. 'l=W- h"'l dt u(t, 0) = {3:* (n- i- v) + !~ (N*- I*- L*)i (45) - {3c .!:_(N*- I*- L*)n N* N* W = V = 1 1 00 k(T)u(t, T)dT 00 u(t, T)dT here, h = J.l + d + r2. To study the stability of this linear system (45) we look of solutions to (45) of the exponontial form n(t) = eztii, u(t, T) = eztU(T), i(t) = eztJ, 9 (46) with a complex number z and n =I= 0, u( 7) ¢. 0, i =I= 0. The endemic equilibrium will be locally asymptotically stable, provided that, for all solutions of this form (46), the real part of z is strictly negative. It will be unstable if there is at least one such solution with the real part of z being strictly positive. Substituting (46) into (45) yeilds zn = -J-£n- di + d~ u(r) = zi = w- hi u(O) = f3c(n- v-i)~* zu(r) (47a) -(J-£ + k(r) - ;* f3c(N*- L*- I*)~ L*- I*) (47d) (47e) u(r)dr w= ~i (47b) (47c) + ~*j3c(N- v = ~o= + rl)u(r) nfty0 k(r)u(r)dr (47!) Using the Laplace transform notation again, we note F(z) = looo e-zT Pp,+k(s)r 1 ( r)dr = looo exp( -loT (J-£ + k(s) + r 1 + z)ds)dr, (48) Q(z) = looo e-zTQ(r)dr = looo k(r)exp( -loT (J-£ + k(s) Solve (47b) for + r1 + z)ds)dr. (49) u and fit the result into (47e) and (47f) as follows v = u(O)F(z), w = u(O)Q(z). (50) (51) By (47c) and (51), we find 1 i = -h-u(O)Q(z), +z (52) By (47a) and (52), we find n= -d 1 ---h-u(O)Q(z). J-£+z +z 10 (53) Note that u(O) must be different from zero because otherwise both n = 0, i = 0 at least.Fitting (50),(52),(53) into (47d) and dividing by ii.(O) =/= 0 yeilds the characteristic equation -d 1 1 I* A A A 1 =f3c(J.L + z h + z Q(z)- P(z)- h + z Q(z)) N* + _.!__d___1-Q(z)f3c(N* - L*- I*)!:_ N* J.L + z h + z N* + _.!._-1-Q(z)fjc(N*- L*- I*) N* h+z (54) Substituting (18)- (23) and (42) into (54),we get -d 1 1 ch 1 = hP(O) + Q(O) ( J.L + z h + z Q(z)- h + z Q(z)- P(z)) A + A A chQ(O) d 1 Q(z) (1 + c)(hP(O) + Q(O)) J.L + z h + z Q(O) h + Q(z) h + z Q(O) (55 ) To simplify the characteristic equation further, we define the probability densities (r) = Pv+k~s)r P P(O) 1 (r) (5B) ' Q(r) q(r) = Q(O), (57) and introduce the parameters 6 = 1 6 -2 -- hP(O) hP(O) + Q(O)' (58) Q(O) hP(O) + Q(O)' (59) Then it is clear that p(z) q(z) 1 00 0 = rOO e-ZTp(r)dT = ~(z)' (60) = rOO e-ZT q(r)dT = ~(z)' (61) Jo lo p(r)dr = < 81 < 1' P(O) 0 1 00 Q(O) q(r)dr = 1, < 62 < 1' 61 + 62 = 1. We also express the complex numbers in (55) as forms ·oh -h- =ae-~ h+z ' 1 . (} --=be-t ,., J.L+z here, 11 (62) (63) (64) (65) h a = -.j;:(h=+x)2y~' b- (66) 1 (67) - J(J.L + x)2 +y2' h+x cos()h = , J(h + x)2 + y 2 . () s'tn h = y . () s'tn y = ~ , (69) ..j(h + x)2 + y2 J.L+X , J(J.L + x)2 + y2 = cos()~ (68) J(J.L + x)2 + y2 (70) , (71) and Z X+ i (72) y. Clearly, the real numbers a, b satisfy 0 < a ::; 1, 1 O<b::s;-, Jl Vz = (x, y) E R~ = {(x, y) I x 2: 0, y 2: 0}. (73) With these definitions and expressions above, (55) takes the form (74) = x +i To sdudy the position of the roots z imaginary parts as follows 1- a 100 e-xT cos(()h + yT)q(T)dT =- c 82 a y of (74), we separate (74) into real and 100 e-xT cos(()h + yT)q(T)dT - -€2- 82 b d a 100 e-xT cos()~+ 1+c 0 - c 81 a 100 e-xT sin(()h + yT)q(T)dT =c 82 a 100 100 + -€21+c + yT)q(T)dT e-XT COS(yT)p(T)dT, e-xT sin(()h (75) + yT)q(T)dT 82 b d a 100 e-xT sin()~+ 1 ()h ()h + yT)q(T)dr 0 + c 81 00 e-XT Sin(yT)p(T)dT. By the Rieman and Lebesgue Lemma, we find the facts similar to [3] below. 12 (76) Pl. If x > 0, I y I +x ~ oo, then 1 00 1 00 1 1 1 1 ~ 0, e-xT cos(yr)p(r)dr ~ e-xT sin(yr)p(r)dr 00 e-xT cos(yr)q(r)dr 00 e-xT sin(yr)q(r)dr ~ 0, ~ 0, 0, (77) e-xT cos(Bh + yr)q(r)dr ~ 0, e-xr sin(Bh + yr)q(r)dr ~ 0, 00 e-xT cos(Bp, + ()h + yr)q(r)dr ~ 0, 00 e-xT sin(Bp, + ()h + yr)q(r)dr ~ 0, 00 00 1 1 because p(T), q(T) are of bounded variation. P2. v X 2: 0, y > 0, 1 00 e-xT sin(yr)p(r)dr > 0, (78) because p(T) is nonincreasing. As we show later, the follow results can be checked step by sep. Result A.(x, y) = (0, 0) is not the root of (75), (76). Substituting x = 0, y = 0 into (75), it takes the form 1- 1o00 q(r)dr = -E: 82 100 q(r)ar- - - 82 -d 100 q(r)dr- c 8 1 p(r)dr, (79) 00 €2 o 1+c 1 J.L o o i.e., 0= €2 - (c + 1+c d 82 -)' (80) f.J, a contradiction appears in (80).So Result A is true. Result B. (x,O) with x > 0 is not the root of (75) ,(76). Substituting y = 0 into (75) , it takes the form 1- a 1 00 e-xT q(r)dr =- c 82 a 1 00 e-xT q(r)dr - -€2- 82 b d a 1+c: - c 821 13 100 e-xT q(r)dr 0 00 e-xTp(r)dr. (81) In this case, 0<a h = -h-< 1, +x 1 1 O<b= - - < -, J..L+X J..L 0< 0 1 00 1 00 < e-xr q(r)dr (82) < 1, e-xrp(r)dr < 1. So the left-hand side of (81) is strictly positive,whereas the right-hand side of (81) is negative.This contradiction in (81) show that Result B is true. Result C. There are no roots (x,y) of (75),(76) with x ~ 0 if c is sufficiently close to 0. Clearly, we only need to show roots (x,y) with x ~ 0, y > 0. IF it does not hold,we have sequences Ej, Xj, Yi satisfying (75), (76) with Cj > 0, Xj ~ Yj > 0, 0, j = 1, 2, ... , and €j ~ 0, j ~ 00. If sup{yj} = oo, then 3 subsequence Yi' of the sequence Yi with Yi' ~ oo,j' ~ oo. Substituting Xj',Yj',cj' into (75) and let j' ~ oo,it takes the form 1 = 0, a contradiction. So the sequence Yi is bounded,i.e., sup{yj} < oo. (83) Similarly,we find (84) By (83),(84), 3 subsequence (xj",Yi") of the sequence (xj,yj) with and xo > 0, y Case A. x~ forms ~ 0. + Y5 =/= 0. Substituting Xj", Yi", Ej" 14 into (75) and let j" ~ oo, it takes the i.e., (85) Paying attention to the follow inequalitys 11= ~a= e-xoT cos(th(xo, Yo)+ e-xoT q(r)dr ~a= YoT)q(r)dr I q(r)dr = 1, h a(xo, Yo) = J(h + x5)2 + Y5 < 1, we find a contradiction in (85) becauce of that the right-hand side of (85) is strictly lees than 1. Case B. x5 + Y5 = 0, i.e., xo = 0, Yo = 0. Substituting Xj", Yj", Ej" into (76) and dividing by Yj" > 0, let j" ---+ oo, we find 1 r= rq(r)dr h + Jo = 0, (86) a contradiction because of that the left-hand side of (86) is strictly positive. k(constant). So Result Cis true.Further the proof end except the case k(r) In fact,if k(r) _ k(constant),the model (2) takes the form (1),the endemic equilibrium E* is locally asymptotically stable(see [1]). References [1] Carlos Castillo-chavez and Zhilan Feng,To treat or not to treat:The case of tuberculosis, to publish; [2] (to search) [3] Horst R. Thieme and Carlos Castillo-chavez, How many infection-age-dependent infectivity affect the dynamics of HIV /AIDS? SIAM J.APPL.MATH.,Vol.53,No.5(1993). pp.1447- 1479. 15