Citation: Jean Luc Dimi, Texance Mbaya. Dynamics analysis of stochastic tuberculosis model transmission withimmune response[J]. AIMS Mathematics, 2018, 3(3): 391-408. doi: 10.3934/Math.2018.3.391
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We consider the following model of tuberculosis transmission:
{˙S=Λ−βSIN−μS˙E=β(1−p)SIN+r2I−(μ+k(1−r1))E˙I=βpSIN+k(1−r1)E−(μ+d+δ+r2)I | (1) |
where S(t), E(t) and I(t) denote the numbers of susceptible, exposed and infected individuals at time t, respectively, with the following parameters:
Λ is the recruitment into the population; β, the probability that a susceptible individual will be infected by infectious; μ is the probability that an individual in the population died from reasons not related to the disease; d is the probability that an infectious individual dies because of the disease.An individual leaves his region to another for a new treatment with the probability δ, thus this individual goes missing of model. New infected individual may develop the disease directly with probability p. To account for treatment, we define r1E as the fraction of population receiving effective chemoprophylaxis and r2 as the rate of effective per capita therapy. We assume that chemoprophylaxis of latently infected individuals E reduces their reactivation rate r1 and that the initiation of of therapeutics immediately removes individuals from active status I and places them into state E, the time before latently infected individuals who does not received effective chemoprophylaxis become infectious is assumed to satisfy an exponential distribution, with time 1k. Thus, individuals leave the class E to I at rate k(1−r1). Also, after receiving a therapeutic treatment, individuals leave the class I to E at rate r2 I.
In [6], we have study dynamical and stochastic models of tuberculosis desease. In this paper we extend the model (1) by introducing, in two times, the effects of immune response and also of environmental fluctuations. This paper is organized as follows.In sections 2 and 3 we introduce tthe dynamical model with immune response and notations and give conditions of stability of these different equilibria. in section 4, we give we study stability of stochastic tuberculosis model.
The transmission of tuberculosis is mainly by air, but occasionally by the oral or digestive route. This is mainly the case of pulmonary tuberculosis where the individual gets the disease by inhalation of particles (nuclei) that are in the air. Thus, the fact of the presence of these particles in the body triggers a network of immune cells, antibodies and other components of the immune response. The effectors are these organs can activate or inhibit an activity.The immune system of an organism provides an extraordinary defense against foreign attacks.Once it recongnize matter as non-self, it actives multiple chemical and physiological processes to control and eliminate the pathogen.
The immune reaction is represented by the term P representing the immune effectors and is subject to the following constraints:
1-The immune system responds to the presence of parasites by producing more immune effectors,
2-Immune effectors reduce the number of parasites. Thus the model of tuberculosis, defined below, is the SEI model augmented by the part that expresses the immune response. The new model for transmitting TB from one human to another will have two components: the first components are (S, E, I), writting with taking account of cholerae bacillus:
{˙S=Λ−βSIN−β1BSK+B−μS˙E=(1−p)(βSIN+β1BSK+B)+r2I−(μ+k(1−r1))E˙I=p(βSIN+β1BSK+B)+k(1−r1)E−(μ+d+δ+r2)I | (2) |
And second components are B and P. Where B is the amount of bacilli of Koch and P the rate of effectors of immunity..
We assume that the bacillus population is suitable for logistic growth with a carrying capacity equal to K. Then, the model on immune response can be writting as follows:
{˙B=rB(1−BK)−εBP˙P=αB−γP | (3) |
where
Proposition 3.1. Let (S(t), E(t), I(t), B(t), P(t)) be the solution of system (2)–(3) with initial conditions (S0), E(0), I(0), B(0), P(0)) and the compact set:
Δθ={(S,E,I,B,P)∈R5+,0≤(S+E+I)≤Λμ+θ;θ>0;B≤K,P≤αKγ} | (4) |
Then, under the flow described by system (2)–(3) , Δ is positively set that attracts all solutions of R5+
Proof: Let be W(t)=(W1(t),W2(t)) with W1(t)=S(t)+E(t)+I(t) and W2(t)=P
Its time derivative satisfies:
dW(t)dt=(Λ−μW1(t)−(d+δ)I;αB−γP) | (5) |
One has B≤K which gives following inequalities:
{dW1(t)dt=(Λ−μW1(t)−(d+δ)I≤Λ−μW1(t)≤0forW1(t)≥ΛμdW2(t)dt≤αK−γPforW2(t)≥αKγ | (6) |
which implies that Δθ is positively invariant set.
Solving this differential equation one has:
0≤(W1(t),W2(t))≤{Λμ+W1(0)e−μt,αKγ+W2(0)e−γt} |
where W(0) is the initial condition of W(t). Then one can conclude that Δθ is an attractive set. θ≥0.
System (3) have an extinction equilibrium E0=(0,0), an immune response free equilibrium E1=(K,0) and an unique infection equilibrium E2=(rKγrγ+αK,αrKrγ+αK)
The jacobian matrix of immune response model at E0 is:
J=(r0α−γ) |
and has its trace trace(J)=r−γ<0 and its determinant det(J)=−rγ>0, this means that there is always an eigenvalue which is positive.hence equilibrium E0 is unstable.
System (3) has the following jacobian at immune response free equilibrium:
J=(−rKεα−γ) |
We see that Trace(J)=−r−γ<0 and det(J)=rγ−Kαε>0 for Kαεrγ<1.In this case, by Routh-Hurwitz all eingenvalues are negative or have negative real parts. We can deduce that:
ˆR0=Kαεrγ |
and when ˆR0≤1, this equilibrium is locally stable.
The jacobian matrix of immune response model (3) at E2 is:
J=(r−r(2γr+εαK)¢rγ+αK−rεγK¢rγ+αKα−γ) |
This jacobian has its Trace(J(E2)=r−r(2γr+εαK)¢rγ+αK−γ<0 and its determinant det(J(E2))=r2γ2−γ2rK+2εγrγ+αKr>0. From Routh Hurwitz criterion all eingenvalues are negative or have negative real parts. Hence this equilibrium is always stable, for ˆR0≥1
system (2)–(3) two equilibria points:
the desease free equilibrium (Λμ,0,0,K,0) and the endemic equilibrium
S∗=D(1−p)+pQΛμ(1−p)D+pQI∗=(1−p)(Λ−μˉS)r2−(μ+k(1−r1))AE∗=p(μ+d+δ+2r2)ˉIpμ+k(1−r1)B∗=Krr+εKP∗=αKrγ(r+εK) |
The stability of the desease free equilibrium will be investigated using the next generator operator [11] Let be X = (E, I, S), system (2)-(3) an be writting as follows: dXdt=F−V, where :
F=((1−p)(βSIN+β1BSK+B)p(βSIN+β1BSK+B)0)etV=(−r2I+(μ+k(1−r1)E−k(1−r1)E+(μ+d+δ+r2)I−Λ+βSIN+β1BSK+B+μS) |
Jacobian matrices F and V on X0 are respectively:
DF(X0)=(F000)etDV(X0)=(V0J1J2) |
o
F=(00β(1−p)βp)etV=(μ+k(1−r1)−k(1−r1)−r2μ+d+δ+r2) |
FV−1(00β(1−p)k(1−r1)+βp(μ+d+δ+r2)(μ+d+δ)(μ+k(1−r1))+μr2β[μp+k(1−r1)](μ+d+δ)(μ+k(1−r1))+μr2) |
is the next generation matrix of system (2)-(3). The radius of FV−1 is
ρ(FV−1)=β(μp+k(1−r1))(μ+d+δ)(μ+k(1−r1))+μr2 |
Hence, the basic reproductive number of system (2)-(3) is:
R0=β(μp+k(1−r1))(μ+d+δ)(μ+k(1−r1))+μr2 |
The following result is etablished (from theorem 2 of [11])
Lemma 3.2. [11] The desease free equilibrium of system (2)-(3) is locally asymptotically stable whenever R0<1 , and unstable if R0>1
It means that in this case, tuberculosis can be eliminated from community.
Theorem 3.3. If R0>1 et ˆR0>1 endemic equilibrium is globally asymptotically stable in Δθ
Preuve : Consider the Lyapounov function V:Δ→R defined as:
V(S,E,I,B,P)=W1[S−SlnS∗S∗]+W2[I−I∗lnII∗]+W3[B−B∗lnBB∗]+W4[P−P∗lnPP∗] |
where W1, W2, W3 and W4 are positive constants to be choosen latter.
Set: V1(S;E,I,B,P)=W1[S−S∗lnSS∗]+W2[I−I∗lnII∗]
V2(S;E,I,B,P)=W3[B−B∗lnBB∗]+W4[P−P∗lnPP∗]
one has:
dV1dt=W1S−S∗S(Λ−βSIN−β1BSK+B−μS)+W2I−I∗I(βpSIN+β1BSK+B+k(1−r1)E−(μ+d+δ+r2)I) |
=W1S−S∗S(βS∗I∗N+β1B∗S∗K+B∗−β1BSK+B+μS∗−βSIN−μS)+W2I−I∗I(βpSIN+pβ1BSK+B+k(1−r1)E |
−(μ+d+δ+r2)I−βpS∗I∗N−k(1−r1)E∗+(μ+d+δ+r2)I∗)−pβ1B∗S∗K+B∗ |
=W1S−S∗S[β(S∗I∗N−SIN)+μ(S∗−S)+β1(B∗S∗K+B∗−BSK+B)]+W2I−I∗I[βp(SIN−S∗I∗N) |
+k(1−r1)(E−E∗)−(μ+d+δ+r2)(I−I∗)−pβ1(B∗S∗K+B∗−BSK+B)] |
=W1S−S∗S[β(S∗I∗N−S∗IN+S∗IN−SIN)+μ(S∗−S)+β1(B∗S∗K+B∗−B∗SK+B∗ |
+B∗SK+B∗−BSK+B)]+W2I−I∗I[βp(SIN−SI∗N+SI∗N−S∗I∗N) |
+β1p(−B∗S∗K+B∗−B∗SK+B∗+B∗SK+B∗+BSK+B)]+k(1−r1)(E−E∗)−(μ+d+δ+r2)(I−I∗)] |
≤W1S−S∗S[β[S∗N(I∗−I)+IN(S∗−S)]+μ(S∗−S)+B∗K+B∗(S∗−S)] |
+W2I−I∗I[βp[SN(I−I∗)+I∗N(S−S∗)]+k(1−r1)(E−E∗)−(μ+d+δ+r2)(I−I∗) |
+β1pB∗K+B∗(S−S∗)]≤−W1βIN(S−S∗)2S−W1βS∗N(S−S∗)S(I−I∗)−W1(μ+ |
B∗S∗K+B∗)(S−S∗)2S]+W2βpSN(I−I∗)2I+W2βpI∗IN(S−S∗)(I−I∗)+W2(1Ik(1−r1)(E−E∗) |
+β1pB∗K+B∗(S−S∗))(I−I∗)−W2(μ+d+δ+r2)I(I−I∗)2] |
=−W1β(IN+μ)(S−S∗)2S−W2(μ+d+δ+r2)I(I−I∗)2+(W2βpI∗IN |
−W1βS∗SN)(S−S∗)(I−I∗)+W2Ik(1−r1)(E−E∗)(I−I∗) |
which gives the following inequality:
≤−W1β(IN+μ)(S−S∗)2S−W1(μ+B∗S∗K+B∗)(S−S∗)2S−W2(μ+d+δ+r2)I(I−I∗)2 |
+βINS(W2−W1)[S∗(S−S∗)(I−I∗)2+I∗(S−S∗)2(I−I∗)]+W2Ik(1−r1)(E−E∗)(I−I∗) |
For W1=W2 and take W2 and for the fact that W2Ik(1−r1) will be very small we deduce that:
dV1dt≤0 |
In the same way one has:
dV2=−W3(B−B∗)2B(rk(B+B∗)−r−εP))−W3B−B∗B(εB∗(P−P∗)−W4γ(P−P∗)2P+W4αP−P∗P(B−B∗)=−W3(B−B∗)2B(rk(B+B∗)−r−εp))−W4γ(P−P∗)2P−(εPB∗W3−αBW4)P−P∗BP(B−B∗)≤0 |
and: dVdt=0 for S=S∗, I=I∗, B=B∗ and P=P∗
Hence by the LaSalle's principe [8], endemic equilibrium is globally asymptotically stable in Δθ.
Some authors take stochastic perturbations into account when they investigate the epidemic system [6,12,13,14].We assume that the perturbation is of white noise type, that is β→β+σ1dW1(t)dt, β1→β1+σ2dW2dt, then we get the following stochastic system:
{˙S=Λ−βSIN−β1BSK+B−μS−σ1SIdW1dt−σ2BSK+BdW2dt˙E=(1−p)(βSIN+β1BSK+B)+r2I−(μ+k(1−r1))E+(1−p)σ1SIdW1dt+(1−p)σ2BSK+BdW2dt˙I=p(βSIN+β1BSK+B)+k(1−r1)E−(μ+d+δ+r2)I+pσ1SIdW1dt+pσ2BSK+BdW2dt˙B=rB(1−BK)−εBP˙P=αB−γP | (7) |
where w1(t) and w2(t) are standard one dimensional Brownian motion, σi>0, i = 1..3 are the intensity of the white noise.
Through this paper, unless otherwise specified, we let (Ω,F,{F}i≥0,P) be a complete space with filtration {Fi} satisfying the usual conditions (i.e. it is right continuous and increasing while F0 contains all null sets).
The following Itô's formula will be used in the sequel of this paper.
Lemma 4.1. [14] Assume that X(t)∈R+ is an Itô's process of the form
dx(t)=f(x,t)dt+ϕ(x,t)dB(t) | (8) |
where f:Rn×[0,+∞)→Rn and ϕ(x,t):Rn×[0,+∞)→Rn are measurable functions.
Given V (x, t) is a Lyapounov function, we define the operator LV by:
LV(x,t)=Vt(x,t)+Vx(x,t)f(x,t)+12trace[ϕTVxx(x,t)ϕ(x,t)]. |
Where Vt(x,t)=dV(x,t)¢dt; Vx(x,t)=dV(x,t)dx1,…,dV(x,t)dxn), Vxx(x,t)=(∂2V(x,t)∂xi∂xj)n×n
Then the general Itô's formula is given by:
dV(x,t)=LV(x,t)+Vx(x,t)G(x,t)dW(t) |
For the sequel we need the following definitions :
Definition 4.2. The solution of system (7) is stochastically ultimately bounded a.s. if for any ϵ∈(0,1) , there exists a positive constant ϱ=ϱ(ϵ) such that for any initial value (S(0), E(0), I(0)) ∈R3+ , the solution of system (7) has the property :
lim supt→∞P{|X(t)|≥ϱ}<ϵ | (9) |
Definition 4.3. The trivial solution x(t) = 0 of (7) is said to be stable in probability if for all ε>0 ,
limx0→0P(supt≥0|x(t,x0)|≥ε)=0 |
Definition 4.4. The trivial solution x(t) = 0 of (7) is said to be asymptotically stable if it is stable in probability and moreover
limx0→0P(limt→∞x(t,x0)=0)=1 |
Definition 4.5. The trivial solution x(t) = 0 of (7) is said to be globally asymptotically stable if it is stable in probability and moreover
P(limt→∞x(t,x0)=0)=1 |
Definition 4.6. The trivial solution x(t) = 0 of (7) is said to be almost surely exponentially stable if for all x0∈Rn,
limt→∞1tln|x(t,x0)|<0a.s. |
Definition 4.7. The trivial solution x(t) = 0 of (7) is said to be exponentially p stable if there is a pair of positive constants C1 and C2 such that for all x0∈Rn, E(|x(t,x0)|p≤C1|x(t,x0)|pe−C2t on t≥0
Lemma 4.8. For any given (S(0),E(0),I(0),B(0),P(0)) ∈R5+ , there is a unique solution (S(t),E(t);I(t),B(t),P(t)) ∈Δθ, on t ≥ 0 and will remain in R5+ with probability one.
Proof Since the coefficients of model (7) satisfy the local Lipchitz condition, then there exists a unique local solution on [0,τε), where τε is the explosion time. Proposition (3.1) shows us that 0≤S(t)+E(t)+I(t)≤Λμ, B≤K et P(t)<αKγ for t∈[0,τε)
We, now, want to show that this solution is global, i.e. τε=+∞ a.s. Let n0>0 be sufficiently large for for any (S(0),E(0),I(0),B(0),P(0))) remaining in the interval [1n0,n0]. For each integer n>n0, we define the stopping time:
τn=inf{t∈[0,τε);S(t)∉(1n,n),E(t)∉(1n,n),I(t)∉(1n,n),B(t)∉(1n,n)orP(t)∉(1n,n)}
By reduction to absurdity, we suppose that τε=+∞ is false, there is a pair of constant T>0 and for any ε∈(0,1) such that P{τ∞≤T}>ε. Consequently, there is an integer n1≥n0 such that
P{τn≤T}≥ε,n≥n1 | (10) |
Define C3 V:R4+→R, lake this:
V(S,I,B,R)=(S−lnS)+(E−lnE)+(I−lnI)+(B−lnB)+(P−lnP) |
for (S(t),E(t),I(t),B(t),P(t))∈Δθ. One has:
LV=(1−1S)(Λ−βSI−β1SBK+B−μS−σ1SIdW1−σ2SBK+BdW2)+(1−1E)((1−p)(βSI+βeSBK+B)+r2I−(μ+k(1−r2)E+(1−p)σ1SIdW1+(1−p)σ2SBK+BdW2)+(1−1I)(p(βSI+βeSBK+B)+k(1−r2)E−(μ+δ+d+r2)I+pσ1SIdW1+pσ2SBK+BdW2)+(1−1B)(rB(1−BK−εBP)+(1−1P)(αB−γP)+12(σ21I2+σ22B2(B+K)2+(1−p)2σ21S2I2E2+(1−p)2σ22p2S2B2E2(B+K)2)+p2σ21S2+σ22p2S2B2I2(B+K)2) |
LV=Λ+3μ+δ+γ+βeBK+B+(βh+ξ)I−μ(S+I+R)−βhS−βeBK+BSI−δB−1BξI+12(σ21I2+σ22B2(B+K)2+(1−p)2σ21S2I2E2+(1−p)2σ22p2S2B2E2(B+K)2)+p2σ21S2+σ22p2S2B2I2(B+K)2) |
from (7) we have:
LV≤Λ+3μ+k(1−r1)+r2+δ+γ+βe+(βh+ξ)λμ+σ21(Λμ)2+pσ22=C |
Therefore, we obtain:
dV≤Cdt−((1−p)(SIE−p)σ1dW1(t)−(1−(1−p)SE−pSI)σ2BK+BdW2(t) | (11) |
By integrating both sides of (18) from 0 to τk∧T yields that:
∫t∧T0dV(S(t),E(t),I(t),B(t),P(t))≤∫t∧T0Cdt−(1−p)∫t∧T0σ1(SIE−p)σ1dW1dW1(t)−∫t∧T0(1−(1−p)SE−pSI)σ2BK+BdW2(t) |
where τn∧T=min{τn,T}. Whence taking the expectative of the above inequality leads to
EV(S(τn∧T),E(τn∧T),I(τn∧T),B(τn∧T),P(τn∧T)))≤V(S(0),E(0),I(0),B(0),P(0))+CT | (12) |
Set Ωn={τn≤T} for n>n1 by inequality (18), we have P(Ωn)≥ε. Note that every ω∈Ωn, there exists at least one of S(τn,ω), I(τn,ω), B(τn,ω) and P(τn,ω) equals either à n or 1n, hence
V(S(τn,ω),E(τn,ω),I(τnω))≥(n−1−lnn)∧(1n−1−ln1n) |
as consequence from (22) one has:
V(S(0),E(0),I(0),B(0),P(0))+CT≥E[1Ωn(ω)V(Sτn,ω),E(τn,ω),I(τn,ω),B(τn,ω),P(τn,ω)]≥ε(n−1−lnn)∧(1n−1−ln1n) |
where 1Ωn is indicator function of Ωn. Let n→+∞ leads to the following contradiction:
+∞>V(S(0),E(O),I(0),B(0),P(0))+CT=+∞ | (13) |
So we must have τ∞=∞. Therefore, the solution (S(t), E(t), I(t), B(t), P(t)) of model will not explode at a finite time with probability one. This completes the proof of lemma (4.8).
Theorem 4.9. The solutions of System (7) are stochastically ultimately bounded for any initial value (S(0),I(0),B(0),R(0))∈Δθ
Proof From lemma (4.1) we know that the solution (S(t),E(t),I(t),B(t),P(t)) will remains in R5+ for all t≥0 with probability 1. defines functions:
V1=etSθ;V2=etEθetV3=etIθpour0<θ<1. |
By Ito's formula, one has :
dV1=LV1dt+θetSθ(σ1SIdW1+σ2BSK+BdW2)dV2=LV2dt+(1−p)θetEθ(σ1SIdW1+σ2BSK+BdW2)dV3=LV3dt+pθetIθ(σ1SIdW1+σ2BSK+BdW2) | (14) |
where
LV1=etSθ[1+θ(ΛS−βIN−β1BK+B−μ)+θ(θ−1)2(σ21I2+σ22(BK+B)2)e−t]LV2=etEθ[1+θ(β(1−p)(SINE+β1SB(K+B)E)+r2IE−(μ+k(1−r1))+θ(θ−1)2(σ21S2I2E2+σ22(SB(K+B)E)2)e−t]LV3=etIθ[1+θ(βpSN+β1pSB(K+B)I+k(1−r1)EI−(μ+d+δ+r2))+θ(θ−1)2(σ21S2+σ22(SB(K+B)I)2)e−t] | (15) |
Thus, there exists C1, C2 and C3 such that:
LV1<C1et,LV2<C2etetLV3<C3et |
It follows that:
etE(Sθ(t))−E(Sθ(0))≤C1etetE(Eθ(t))−E(Eθ(0))≤C2et,etetE(Iθ(t))−E(Iθ(0))≤C3et |
We get now:
lim supt→∞ESθ(t)≤C1<∞lim supt→∞E(E)θ(t)≤C2<∞lim supt→∞EIθ(t)≤C3<∞ | (16) |
for X(t)=(S(t),E(t),I(t))∈R3+, note that
|X(t)|θ=(S2(t)+E2(t)+I2(t))θ2≤3θ2max{Sθ(t),Eθ(t),Iθ(t)}≤3θ2(Sθ(t)+Eθ(t)+Iθ(t)) | (17) |
consequently:
lim supt→∞E|X(t)|≤3θ2(C1+C2+C3) |
as result, there exists a positive δ1 sutch that
lim supt→∞E|√X(t)|<δ1 | (18) |
now for ε>0, let δ=δ21ε2, by Chebychev'inequality,
P{|X(t)|}≤E|√X(t)|√δ=ε | (19) |
wich gives the desired assertion.
In this section we study the pth moment exponentially stability of the desease free equilibrium::
Theorem 5.1. Set p≥2, if R0<1, the disease free equilibrium is pth moment exponentially stable in Δθ
The proof of this theorem needs the two next results:
Theorem 5.2. (Afanas'ev et Komanowski, [2]) Suppose that there exists a function V(t, x) ∈C1,2(R+,Rn), satisfying the following inequalities:
K1|x|≤V(t,x)≤K2|x|p |
and
LV(t,x)≤−K3|x|p,t≥0 |
where p, K1, K2 and K3 are positive constants.Then the equilibrium of (7) is pth moment exponentially stable.When p = 2, it is usually said to be exponentially stable in mean square and the disease free equilibrium is globally asymptotically stable.
Lemma 5.3. If p≥2 and ε, x, y>0. then
xp−1y≤(p−1)εpxp+1pε1−pyp |
and
xp−2y2≤(p−2)εpxp+2pε(2−p)/2yp |
Proof of theorem (5.1): Set p≥2 and (S(0),I(0),B(0),P(0))∈Δθ, from lemma (4.1), the solution of the system remains in Δθ. Let be the following Lyapounov function
V=c1(Λμ−S)p+1pIp+c2Bp |
One gets by Itô's formula
LV=−c1p(Λμ−S)p−1(Λ−β1SBK+B−βSIN−μS)+12c1p(p−1)(Λμ−S)p−2[σ21S2I2+σ22S2B2(K+B)2]+Ip−1(β1pSBK+B+βpSIN+k(1−r1)E−(μ+d+δ+r2)I]+12(p−1)[σ21IpS2+σ22Ip−2S2B2(K+B)2]+c2pBp−1(rB(1−BK)−εBp]=−c1pμ(Λμ−S)p+c1p(Λμ−S)p−1(β1SBK+B+βSI)+12C1p(p−1)(Λμ−S)p−2[σ21S2I2+σ22S2B2(K+B)2]+Ip−1[β1pSBK+B+βpSIN−(μ+d+δ+r2)I]+12(p−1)[σ21IpS2+σ22Ip−2S2B2(K+B)2]+c2pBp(r−rB2K−εP) |
of (4) gives us S≤Λμ and the fact that BK+B<1, we obtain:
LV≤−c1pμ(Λμ−S)p+c1p(Λμ−S)p−1(βΛμ+β1ΛμI)+c1μp(Λμ−S)p−1S−12c1p(p−1)(Λμ−S)p−2[σ21Λμ2I2+σ2Λμ2]+Ip−1(β1pΛμ+βpΛμI−(μ+r+δ+r2)I)+12(p−1)[σ21Λμ2IP+12(p−1)σ22Λμ2]≤−c1pμ(Λμ−S)p+c1pβΛμ(Λμ−S)p−1+c1pβ1Λμ(Λμ−S)p−1I−12c1p(p−1)(Λμ−S)p−2[σ21Λμ2I2+σ2Λμ2](βΛμ−(μ+d+δ+r2))Ip+β1ΛμIp−1+12(p−1){σ1Λμ2IP+σ2Λμ2}+c2prBp−1I−c2pδBp |
and by application of lemma (4.1), one gets now :
LV≤−c1(pμ−(p−1)ε(βΛμ+12(p−1)(p−2)σ2(Λμ)2))(Λμ−S)p+pc1β1Λμ(Λμ−S)p−1−((μ+d+δ+r2)−(βΛμ(1+c1ε1−p)+c1(p−1)σ2ε(2−p)/2+c2rε1−p+12(p−1)σ2Λμ2)+c3(d+δ+r2)ε1−p)Ip−(c2pδ−c2rε)Bp≤−c1(pμ−(p−1)ε(βΛμ+12(p−1)(p−2)σ21(Λμ)2))(Λμ−S)p−((μ+d+δ+r2)−(βΛμ(1+c1ε1−p)+c1(p−1)σ22ε(2−p)/2+(c2r+c3γ)ε1−p+12(p−1)σ2Λμ2))Ip−c2(pδ−rε)Bp |
We choose ε sufficiently small such that the coefficients of (Λμ−S)p and Bp be negative.
We, also, can choose c1, c2 and c3 positive such that the coefficient of Ip be negative. according to theorem (5.1), the proof is complete.
In this subsection, we investigate stochastic stability of the desease free equilibrium, E0=(Λμ,0,0,K,0). the following result gives the suffucient condition for almost surely exponential stability.
Theorem 5.4. If R0≤1 and β2≤2σ21μ then the disease free equilibrium is almost surely exponential stable in Δθ
Proof: Define
V=ln((Λμ−S)+E+I+B+P) |
Using Itô's formula:
LV=1Λμ−S+E+I+B+P(−dS+dE+dI+dB+dP)+1(Λμ−S+E+I+B+P)2(dSdS+dEdE+dIdI+dBdB+dPdP)+1(Λμ−S+E+I+B+P)2(dSdI+dSdB+dSdP+dIdB+dIdP+dBdP) |
one has:
LV=1(Λμ−S)+E+I+B+P(−Λ+2(β1SBK+B+βSI)+μS−μ(E+I)−(d+δ)I+(r+α)B−rB2K−εBP−γP]−1Λμ−S+E+I+B+R)2(2σ21S2I2+σ22S2B2(K+B)2))≤1Λμ−S+E+I+B+P[−Λ+βSI+μS−μ(E+I)]−2σ21S2I2(Λμ−S+E+I+B+P)2−1Λμ−S+E+I+B+P(2β1SBK+B+(r+α)B) |
define U=2σ1SIΛμ−S+E+I+B+P,
2β1U−2σ21U2−μ=−2σ2(U−β12σ1)2+(β21−2σ21μ)/2σ21 |
one has:
dV≤[−2σ2(U−β12σ1)2+(β21−2σ21μ)/2σ21]dt+2σ1UdW(t)≤(β21−2σ21μ)/2σ21)dt+2σ1UdW1(t) |
By integrating from 0 to t, we cheek:
ln(Λμ−S+E+I+B+P)≤ln(Λμ−S(0)+E(0)+I(0)+B(0)+P(0))+(β21−2σ21μ)/2σ21)t+G(t) | (20) |
with G(t)a martingale defined by :G(t)=σ1∫t0ZdW1(t), and in vertue of lemma (5.3) the solution of model(7) remains in Δ, it exists a positive constatnt C sutch that
<G,G>t=σ21∫t0Z2ds≤Ct |
finally by th strong law of large numbers for local martingales, we have:
lim supt→+∞ln(Λμ−S+E+I+B+P)≤(β21−2σ21μ)/2σ21)≤0a.s. |
The following result gives conditions for extinction of tuberculosis desease, it means that the desease dies out with probability 1.:
Theorem 5.5. If [(β+β1)−(μ+d+δ)−12σ21(Λμ)2]≤0 et R0<1, then I(t) converge almost surely exponentially to 0.
Mathematicaly, we have to show that:
lim supt→∞lnI(t)t≤0 |
Proof: One has
˙E+˙I=(βSIN+β1BSK+B)−μE−(μ+d+δ)I+σ1SIdW1+σ2BSK+BdW2 |
which gives
˙I≤(βSIN+β1BSK+B)−μE−(μ+d+δ)I+σ1SIdW1+σ2BSK+BdW2 |
Let be a Lyapounov function V(I(t))=lnI(t), By Itô's calculus:
dV(I(t))=1IdI(t)−12I2(dI(t))2 |
and then
dV(I(t))≤(βSN+β1BSI(K+B))−(μ+d+δ)−12[σ21S2+σ22B2S2I2(K+B)2]+σ1SIdW1+σ2BSK+BdW2 |
≤(β+β1)−(μ+d+δ)−12σ21(Λμ)2+σ1SIdW1+σ2BSK+BdW2 |
gives the following equation:
lnI(t)=lnI0+(β+β1)−(μ+d+δ)−12σ21(Λμ)2dt+∫T0σ1SIdW1(t)+∫t0σ2BSK+BdW2 |
lnI(t)≤lnI0+[β+β1)−(μ+d+δ)−12σ21(Λμ)2]t+G(t) | (21) |
wher G(t) is a martingale defined by:
G(t)=∫T0σ1(Λμ)2dW1(t)+∫t0σ2dW2 |
THis calculus implies that:
<G,G>t≤(σ21(Λμ)2+σ22)t. |
And by the strong law of large numbers for local martingales [12,14] we have:
lim supt→∞G(t)t=0almostsurely. |
and then:
lim supt→∞I(t)t≤[β+β1)−(μ+d+δ)−12σ21(Λμ)2]≤0a.s. | (22) |
this completes the proof.
Definition 6.1. System (7) is said to be persistent in the mean, if
limt→+∞inf1t∫t0I(s)ds>0 |
Theorem 6.2. If βΛμ(μ+d+δ+r2)>1 then (7) is persistent in the mean, moreover we have:
limt→+∞inf1t∫t0I(s)ds>(μ+d+δ+r2)(βΛμ(μ+d+δ+r2)−1)(βΛμ+(μ+d+δ))limt→+∞inf1t∫t0E(s)ds≥r2μ+k(1+r1)(μ+d+δ+r2)(βΛμ(μ+d+δ+r2)−1)(βΛμ+(μ+d+δ))limt→+∞inf1t∫t0(Λμ−S(s))ds>μ(μ+d+δ+r2)(βΛμ(μ+d+δ+r2)−1)(βΛμ+(μ+d+δ)) |
The proof is based on the following lemma:
Lemma 6.3. [1] Let g∈C([0,∞)×Ω,[0,∞)) and G∈C([0,∞)×Ω,[0,∞)). If there exists positive constants λ0 and λ such that:
lng(t)≥λ0t−λ∫t0g(s)ds+G(t)a.s. |
for all t≥0, and limt→+∞G(t)t=0 a.s., then
limt→+∞inf1t∫t0g(t)dt≥λ0λa.s. |
Proof of theorem (6.2): Consider the following Lyapounov function:
V(S,E,I)=α1(S+E+I)+α2S+lnI |
where α1 and α2 are defined below. By Ito's formula we have:
dV=α1[Λ−μ(S+E+I)−(d+δ)I]+α2[Λ−βSIN−β1BSK+B−μS−σ1SIdW1−σ2BSK+BdW2]+p(βSN+β1BSI(K+B))+k(1−r1)EI−(μ+d+δ+r2)+pσ1SdW1+pσ2BSI(K+B)dW2+(2p2−2p+2)(σ21S2I2+S2B2(K+B)2)≥α1[(Λ−μ(S+E+I)−(d+δ)I]+α2[(Λ−μS)−β1SBK+B−βΛμI)dt−σ1SIdW1−σ2SBK+BdW2]+p(β1SBI(K+B)+βΛμ−(μ+d+δ+r2)+σ21S2)dt+pσ1SdW1+pσ2SBI(K+B)dW2≥p(μ+d+δ+r2)(βΛμ(μ+d+δ+r2)−1)+[(α1+α2)μ−β](Λμ−S)−(α2−μΛ)β1SBK+B+(α1−βhΛμ)I+(1−α2I)Sσ1SdW1+(1I−α2)σ2SBK+BdW2 |
with α2=μΛ and β=(α1+α2)μ one has:
dV≥(μ+d+δ+r2)(βΛμ(μ+d+δ+r2)−1)dt−(β(Λμ+(μ+d+δ))Idt+(1−α2I)Sσ1SdW1+(1I−α2)σ2SBK+BdW2 |
and integrating both sides, one obtains:
V(S,E,I)≥V(S0,E0,I0)+(μ+d+δ+r2)(βΛμ(μ+d+δ+r2)−1)t−(β(Λμ+(μ+d+δ))∫t0Idt |
+(1−α2I)σ1∫t0SdW1+(1I−α2)σ2∫t0SBK+BdW2 |
hence
lnI≥(μ+d+δ+r2)(βΛμ(μ+d+δ+r2)−1)t−(βΛμ+(μ+d+δ))∫t0Idt+D(t) |
with
D(t)=V(S0,E0,I0)−(α1+α2)S−α1E−α1I+(1−α2I)σ1∫t0SdW1+(1I−α2)σ2∫t0SBK+BdW2 |
by the the strong low of martingale, one deduces that:
limt→+∞D(t)t=0 |
By the lemma (4.8) one has:
I(t)≥(μ+d+δ+r2)(βΛμ(μ+d+δ+r2)−1)(βΛμ+(μ+d+δ)) |
From (4)
˙E≥r2I−(k(1−r1)+μ)E |
wich gives
limt→∞infE(t)≥r2μ+k(1−r1)limt→∞inf(I(t))+limt→∞infE0−E(μ+k(1−r1)t |
≥r2μ+k(1+r1)(μ+d+δ+r2)(βΛμ(μ+d+δ+r2)−1)(βΛμ+(μ+d+δ)) |
For the last equality, we take account of the following relation::
dN=d(S+E+I)=(μ(Λμ−S)−μE−(μ+d+δ)I)dt≥(μ(Λμ−S)−μE−μI)dt |
hence
limt→∞inf(Λμ−S)≥limt→∞N−N0μt+μlimt→∞infI(t)+μlimt→∞infE(t) |
≥μ(μ+d+δ+r2)(βΛμ(μ+d+δ+r2)−1)(βΛμ+(μ+d+δ)) |
The authors declare no conflict of interest.
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