Citation: Shanjing Ren. Global stability in a tuberculosis model of imperfect treatment with age-dependent latency and relapse[J]. Mathematical Biosciences and Engineering, 2017, 14(5&6): 1337-1360. doi: 10.3934/mbe.2017069
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Mathematical modeling is a very important tool in analyzing the propagation and controlling of infectious diseases. Age structure is an important characteristic in the modeling of some infectious diseases. The first formulation of a partial differential equation(PDE) for the age distribution of a population was due to McKendrick [21]. Since the seminar papers by Kermack and McKendrick [13]-[15], age structure models have been used extensively to study the transmission dynamics of infectious diseases, we refer to the monographs by Hoppensteadt [11], Iannalli [12] and Webb [30] on this topic.
As an ancient disease, TB peaked and declined by 1940's before it became curable, while the downtrend stopped in the middle 1980's and 1990's. As one of the top 3 deadly infectious diseases, TB would cause a higher death rate if not treated, while the disease would be latent in an individual body for months, years or even decades before it outbreaks. McCluskey [20] pointed out that the risk of activation can be modeled as a function of duration age, and this form can be used to describe more general latent period via introducing the duration age in the latent class as a variable.
On the other hand, for the infectious tuberculosis, the removed individuals often have a higher relapse rate. Actually, the recurrence as an important feature of some animal and human diseases has been studied extensively, see [4], [23]. For instance, van den Driessche and coauthors in [4], [5] established two models with a constant relapse period and a general relapse distribution respectively, which showed the threshold property of the basic reproduction number. It is interest to investigate the model with age-dependent relapse rate and to determine whether the threshold property can be preserved or not.
Recently, Wang et al. [24]-[27] considered the global stability of nonlinear age-structured models, Liu et al. [17] introduced age-dependent latency and relapse into an
However, most of the models assumed that TB would show neither its clinical symptoms nor its infectivity during its latent period, while in fact, TB has many early clinical symptoms such as fever, fatigability, night sweat, chest pain, hemoptysis and so on. Here we formulate and analyze an
This paper will be organized as follows: In Section 2, we formulate our general
The total population is decomposed into four disjoint subclasses, susceptible class
{dS(t)dt=Λ−bS(t)−βS(t)I(t),(∂∂t+∂∂a)e(t,a)=−(b+δe+μ(a)+σ(a))e(t,a),dI(t)dt=−(r1+r2+b+δi)I(t)+∫∞0σ(a)e(t,a)da+∫∞0k(c)r(t,c)dc,(∂∂t+∂∂c)r(t,c)=−(k(c)+b)r(t,c), | (1) |
with boundary conditions
{e(t,0)=βS(t)I(t)+r2I(t),r(t,0)=r1I(t)+∫∞0μ(a)e(t,a)da, | (2) |
and initial conditions
S(0)=S0,e(0,a)=e0(a),I(0)=I0,r(0,c)=r0(c), | (3) |
where
In order to simplify the later derivation, we make the following hypotheses about the parameters of the system (1)
(H1)
(H2)
(H3) There exists
For
ε(a)=σ(a)+μ(a)+b+δe,η(c)=k(c)+b,ρ1(a)=e−∫a0ε(s)ds,ρ2(c)=e−∫c0η(s)ds,θ1=∫+∞0σ(a)ρ1(a)da,θ2=∫+∞0k(c)ρ2(c)dc,θ3=∫+∞0μ(a)ρ1(a)da. |
According to Webb [30], by solving the PDE parts of (1) along the characteristic lines
e(t,a)={e(t−a,0)e−∫a0ε(s)ds,t>a≥0,e0(a−t)e−∫aa−tε(s)ds,a≥t≥0, | (4) |
r(t,c)={r(t−c,0)e−∫c0η(s)ds,t>c≥0,r0(c−t)e−∫cc−tη(s)ds,c≥t≥0. | (5) |
Define the space of functions
X:=R+×L1+(0,+∞)×R+×L1+(0,+∞) |
equipped with the norm
∥(x1,x2,x3,x4)∥X=|x1|+∫+∞0|x2(a)|da+|x3|+∫+∞0|x4(c)|dc. |
The norm has the biological interpretation of giving the total population size. The initial conditions (3) for system (1) can be rewritten as
S(t1)=e−bt1−∫t10βI(τ)dτ{S(0)+∫t10ebs+∫s0βI(τ)dτΛds}>0. |
Similarly, assume that
I(t2)=e−(r1+r2+b+δi)t2∫t20(∫∞0σ(a)e(t,a)da+∫∞0k(c)r(t,c)dc)e(r1+r2+b+δi)s)ds+e−(r1+r2+b+δi)t2I(0)>0. |
Thus
Let us consider a function
ddtN(t)=ddtS(t)+ddt∫+∞0e(t,a)da+ddtI(t)+ddt∫+∞0r(t,c)dc. |
due to
ddt∫+∞0e(t,a)da=ddt(∫t0e(t−a,0)ρ1(a)da+∫+∞te0(a−t)ρ1(a)ρ1(a−t)da)=ddt∫t0(βS(t−a)I(t−a)+r2I(t−a))ρ1(a)da+ddt∫+∞te0(a−t)ρ1(a)ρ1(a−t)da=ddt∫t0(βS(τ)I(τ)+r2I(τ))ρ1(t−τ)dτ+ddt∫+∞0e0(τ)ρ1(t+τ)ρ1(τ)dτ=βS(t)I(t)+r2I(t)−∫+∞0ε(a)e(t,a)da. |
Similarly, by using
ddt∫+∞0r(t,c)dc=ddt∫t0(r1I(t−c)+∫+∞0μ(a)e(t−c,a)da)ρ2(c)dc+ddt∫+∞tr0(c−t)ρ2(c)ρ2(c−t)dc=r1I(t)+∫+∞0μ(a)e(t,a)da−∫+∞0η(c)r(t,c)dc. |
Hence, we have
ddt(S(t)+∫+∞0e(t,a)da+I(t)+∫+∞0r(t,c)dc)=Λ−bS(t)−βS(t)I(t)+βS(t)I(t)+r2I(t)−∫+∞0ε(a)e(t,a)da−(r1+r2+b+δi)I(t)+∫∞0σ(a)e(t,a)da+∫∞0k(c)r(t,c)dc+r1I(t)+∫+∞0μ(a)e(t,a)da−∫+∞0η(c)r(t,c)dc=Λ−bS(t)−∫+∞0br(t,c)dc−bI(t)−∫+∞0(b+δe)e(t,a)da−δiI(t)≤Λ−bN(t). |
It follows from the variation of constants formula that
Ω={(S(t),e(t,⋅),I(t),r(t,⋅))∈R+×L1+(0,+∞)×R+×L1+(0,+∞):N(t)≤Λb} |
is positively invariant absorbing set for system (1).
Proposition 1. If
(ⅰ)0≤S(t),∫+∞0e(t,a)da,I(t),∫+∞0r(t,c)dc≤M,(ⅱ)e(t,0)≤βM2+r2M,r(t,0)≤r1M+ˉμM. |
For convenience, we rewrite (1) as follows
{(∂∂t+∂∂a)e(t,a)=−(b+δe+μ(a)+σ(a))e(t,a),(∂∂t+∂∂c)r(t,c)=−(k(c)+b)r(t,c),dV(t)dt=G(e(t,a),r(t,c),V(t))−CV(t),e(t,0)=βS(t)I(t)+r2I(t),e(0,a)=e0(a),r(t,0)=r1I(t)+∫∞0μ(a)e(t,a)da,r(0,c)=r0(c),V(0)=V0, | (6) |
where
V(t)=(S(t)I(t)), |
C=(b00r1+r2+b+δi), |
G(e(t,a),r(t,c),V(t))=(Λ−βS(t)I(t)∫∞0σ(a)e(t,a)da+∫∞0k(c)r(t,c)dc,). |
Set
Z+=Y+×R2+,Z0=Y0×R2,Z0+=Z0⋂Z+, |
where
Y+=R+×L′+(R+,R),Y0={0}×L′(R+,R). |
We define
A1(0ϕ1)=(−ϕ1(0)−ϕ′1−(b+δe+μ(a)+σ(a))ϕ1) |
with
(λI−A1)−1(θ1ψ1)=(0ϕ1), |
then we can get
ϕ1(a)=e−(λ+b+δe)aθ1+∫a0e−∫as(μ(l)+σ(l))dle−(λ+b+δe)(a−s)ψ1(s)ds. |
Similarly, we define
A2(0ϕ2)=(−ϕ2(0)−ϕ′2−(b+k(c))ϕ2) |
with
ϕ1(a)=e−(λ+b)cθ2+∫c0e−∫cs(k(l))dle−(λ+b)(c−s)ψ2(s)ds. |
Thus (6) can be rewritten as
{ddt(0e(t,⋅))=A1(0e(t,⋅))+F1((0e(t,⋅)),(0r(t,⋅)),V(t)),ddt(0r(t,⋅))=A2(0r(t,⋅))+F2((0e(t,⋅)),(0r(t,⋅)),V(t)),dV(t)dt=−CV(t)+F3((0e(t,⋅)),(0r(t,⋅)),V(t)),e(0,a)=e0(a),r(0,c)=r0(c),V(0)=V0, | (7) |
where
F1((0e(t,⋅)),(0r(t,⋅)),V(t))=(βS(t)I(t)+r2I(t)0),F2((0e(t,⋅)),(0r(t,⋅)),V(t))=(r1I(t)+∫∞0μ(a)e(t,a)da0),F3((0e(t,⋅)),(0r(t,⋅)),V(t))=(Λ−βS(t)I(t)∫∞0σ(a)e(t,a)da+∫∞0k(c)r(t,c)dc). |
Let
L(u(t))=(A1(0e(t,⋅)),A2(0r(t,⋅)),−CV(t)), |
where
F(u(t))=(F1(u(t))F2(u(t))F3(u(t))). |
Therefore (7) can be rewritten as an abstract Cauchy problem
{du(t)dt=L(u(t))+F(u(t)),u(0)=((0e0(⋅)),(0r0(⋅)),V0). | (8) |
By using the results in Magal [19] and Chen et al. [3], there exists a uniquely determined semiflow
Lemma 2.1. ([1]) Let
Lemma 2.2. ([1]) If the following two conditions hold, then the semi-flow
(ⅰ) There exists a continuous function
(ⅱ) For
In order to prove Lemma 2.2, we first decompose
y2(t,a)={0,t>a≥0,e0(a−t)ρ1(a)ρ1(a−t),a≥t≥0,y4(t,c)={0,t>c≥0,r0(c−t)ρ2(c)ρ2(c−t),c≥t≥0. | (9) |
˜y2(t,a)={e(t−a,0)ρ1(a),t>a≥0,0,a≥t≥0,˜y4(t,c)={r(t−c,0)ρ2(c),t>c≥0,0,c≥t≥0. | (10) |
In order to verify condition (ⅰ) of Lemma 2.2, we need to prove the following proposition.
Proposition 2. For
Proof. It is obvious that
‖ϕ(t,x0)‖X=|0|+∫+∞0|y2(t,a)|da+|0|+∫+∞0|y4(t,c)|dc=∫+∞t|e0(a−t)ρ1(a)ρ1(a−t)|da+∫+∞t|r0(c−t)ρ2(c)ρ2(c−t)|dc=∫+∞0|e0(τ)e−∫τ+tτε(s)ds|dτ+∫+∞0|r0(τ)e−∫τ+tτη(s)ds)|dτ≤e−(b+2b0+δe)t(|0|+∫+∞0|e0(τ)|dτ+|0|+∫+∞0|r0(τ)|dτ)=e−(b+2b0+δe)t‖x0‖X, |
by the known condition
Lemma 2.3. ([1]) Let
(ⅰ)
(ⅱ)
Proposition 3. For
Proof. According to Proposition 1(ⅰ),
˜y2(t,a)≤(βM2+r2M)e−(b+2b0+δe)a. | (11) |
∫+∞0|˜y2(t,a+h)−˜y2(t,a)|da=∫t−h0|e(t−a−h,0)ρ1(a+h)−e(t−a,0)ρ1(a)|da+∫tt−h|0−e(t−a,0)ρ1(a)|da≤∫t−h0e(t−a−h,0)|ρ1(a+h)−ρ1(a)|da+∫t−h0ρ1(a)|e(t−a−h,0)−e(t−a,0)|da+∫tt−h|e(t−a,0)ρ1(a)|da. | (12) |
Recall that
∫t−h0|ρ1(a+h)−ρ1(a)|da=∫t−h0ρ1(a)da−∫thρ1(a)da=∫t−h0ρ1(a)da−∫t−hhρ1(a)da−∫tt−hρ1(a)da=∫h0ρ1(a)da−∫tt−hρ1(a)da≤h. | (13) |
From Proposition 1 and
∫t−h0ρ1(a)|e(t−a−h,0)−e(t−a,0)|da |
≤∫t−h0ρ1(a)(|βS(t−a−h)I(t−a−h)−βS(t−a)I(t−a)|+|r2I(t−a−h)−r2I(t−a)|)da≤∫t−h0(βMSI+r2MI)(−h)e−(b+2b0+δe)ada=(βMSI+r2MI)hb+2b0+δe(1−e−(b+2b0+δe)(t−h))≤(βMSI+r2MI)hb+2b0+δe. | (14) |
From (12)-(14), we have
∫+∞0|˜y2(t,a+h)−˜y2(t,a)|da≤(βMSI+r2MIb+2b0+δe+2(βM2+r2M))h |
which converges uniformly to 0 as
From (11) we have
limh→+∞∫+∞h|˜y2(t,a)|da≤limh→+∞∫+∞h(βM2+r2M))e−(b+2b0+δe)ada=limh→+∞βM2+r2Mb+2b0+δee−(b+2b0+δe)h=0 |
which meet the condition (ii) in Lemma 2.3. Similarly,
Theorem 2.4. The semi-flow
Now we consider the existence of equilibria of system (1). The steady state
{0=Λ−bS∗−βS∗I∗,ddae∗(a)=−ε(a)e∗(a),0=−(r1+r2+b+δi)I∗+∫∞0σ(a)e∗(a)da+∫∞0k(c)r∗(c)dc,ddcr∗(c)=−η(c)r∗(c), | (15) |
with boundary conditions
{e∗(0)=βS∗I∗+r2I∗,r∗(0)=r1I∗+∫∞0μ(a)e∗(a)da. | (16) |
From the second equation of (15) and the first equation of (16), we obtain
e∗(a)=(βS∗I∗+r2I∗)e−∫a0ε(s)ds. | (17) |
Similarly, by using the fourth equation of (15) and the second equation of (16), we get
r∗(c)=(r1I∗+∫+∞0μ(a)e∗(a)da)e−∫c0η(s)ds. | (18) |
If
E0=(S0,0,0,0),whereS0=Λb. |
In order to find any endemic equilibrium, we introduce the basic reproduction number
R0=βS0(θ1+θ2θ3)r1+r2+b+δi−r1θ2−r2(θ1+θ2θ3). |
Now, if
0=−(r1+r2+b+δi)I∗+∫+∞0σ(a)(βS∗I∗+r2I∗)e−∫a0ε(s)dsda+∫+∞0k(c)(r1I∗+∫+∞0μ(a)(βS∗I∗+r2I∗)e−∫a0ε(s)dsda)e−∫c0η(s)ds=−(r1+r2+b+δi)I∗+(βS∗I∗+r2I∗)(θ1+θ2θ3)+r1I∗θ2. | (19) |
Thus, combining the first equation of (15) and the equation (19), we get
S∗=ΛbR0,I∗=bβ(R0−1). | (20) |
If
E∗=(S∗,e∗(⋅),I∗,r∗(⋅)), |
where
S∗=ΛbR0,e∗(a)=(βS∗I∗+r2I∗)ρ1(a),I∗=bβ(R0−1),r∗(c)=((βS∗I∗+r2I∗)θ3+r1I∗)ρ2(c). |
In this section, sufficient conditions for the local asymptotic stability of the equilibria will be derived.
Theorem 4.1. The disease-free equilibrium
Proof. First, we introduce the change of variables as follows
x1(t)=S(t)−S0,x2(t,a)=e(t,a),x3(t)=I(t),x4(t,c)=r(t,c). |
Linearizing the system (1) about disease-free equilibrium
{dx1(t)dt=−bx1(t)−βΛbx3(t),(∂∂t+∂∂a)x2(t,a)=−(b+δe+μ(a)+σ(a))x2(t,a),dx3(t)dt=−(r1+r2+b+δi)x3(t)+∫+∞0σ(a)x2(t,a)da+∫+∞0k(c)x4(t,c)dc,(∂∂t+∂∂c)x4(t,c)=−(k(c)+b)x4(t,c),x2(t,0)=(βΛb+r2)x3(t),x4(t,0)=r1x3(t)+∫+∞0μ(a)x2(t,a)da. | (21) |
Set
x1(t)=x01eλt,x2(t,a)=x02(a)eλt,x3(t)=x03eλt,x4(t,c)=x04(c)eλt, | (22) |
where
λx01=−bx01−βΛbx03, | (23) |
{λx02(a)+dx02(a)da=−(b+δe+μ(a)+σ(a))x02(a),x02(0)=(βΛb+r2)x03, | (24) |
λx03=−(r1+r2+b+δi)x03+∫+∞0σ(a)x02(a)da+∫+∞0k(c)x04(c)dc, | (25) |
{λx04(c)+dx04(c)dc=−(k(c)+b)x04(c),x04(0)=r1x03+∫+∞0μ(a)x02(a)da. | (26) |
Integrating the first equation of (24) from 0 to
x02(a)=(βΛb+r2)x03e−(λ+b+δe)a−∫a0(σ(s)+μ(s))ds. | (27) |
Similarly, we have from (26) that
x04(c)=(r1x03+∫+∞0μ(a)x02(a)da)e−(λ+b)c−∫c0k(s)ds. | (28) |
Substituting (27) and (28) into (25) and solving (25) gives
λ=−(r1+r2+b+δi)+∫+∞0k(c)r1e−(λ+b)c−∫c0k(s)dsdc+∫+∞0σ(a)(βΛb+r2)e−(λ+b+δe)a−∫a0(σ(s)+μ(s)dsda+∫+∞0k(c)∫+∞0μ(a)(βΛb+r2)e−(λ+b+δe)a−∫a0(σ(s)+μ(s))dsda⋅e−(λ+b)c−∫c0k(s)dsdc | (29) |
which is the characteristic equation. Let
F(λ)=∫+∞0k(c)∫+∞0μ(a)(βΛb+r2)e−(λ+b+δe)a−∫a0(σ(s)+μ(s))dsda⋅e−(λ+b)c−∫c0k(s)dsdc−λ−(r1+r2+b+δi)+∫+∞0σ(a)(βΛb+r2)e−(λ+b+δe)a−∫a0(σ(s)+μ(s)dsda+∫+∞0k(c)r1e−(λ+b)c−∫c0k(s)dsdc. |
Obviously,
F′(λ)=−(βΛb+r2)∫+∞0aσ(a)e−(λ+b+δe)a−∫a0(σ(s)+μ(s))dsda−a∫+∞0k(c)e−(λ+b)c−∫c0k(s)dsdc⋅∫+∞0μ(a)(βΛb+r2)e−(λ+b+δe)a−∫a0(σ(s)+μ(s))dsda−c∫+∞0k(c)e−(λ+b)c−∫c0k(s)dsdc⋅∫+∞0μ(a)(βΛb+r2)e−(λ+b+δe)a−∫a0(σ(s)+μ(s))dsda−r1∫+∞0ck(c)e−(λ+b)c−∫c0k(s)dsdc−1<0, |
and
limλ→+∞F(λ)=−∞,limλ→−∞F(λ)=+∞. |
Thus, we know (29) has a unique real root
F(0)=[(r1+r2+b+δi)−r1θ2−r2(θ1+θ2θ3)](R0−1), |
we have
0=F(λ)=F(x+yi)≤F(x) |
which means that
Theorem 4.2. The unique endemic equilibrium
Proof. First, we introduce the perturbation variables as follows
y1(t)=S(t)−S∗,y2(t,a)=e(t,a)−e∗(a),y3(t)=I(t)−I∗,y4(t,c)=r(t,c)−r∗(c). |
Linearizing system (1) at the endemic equilibrium
{dy1(t)dt=−by1(t)−βI∗y1(t)−βS∗y3(t),(∂∂t+∂∂a)y2(t,a)=−(b+δe+μ(a)+σ(a))y2(t,a),dy3(t)dt=−(r1+r2+b+δi)y3(t)+∫+∞0σ(a)y2(t,a)da+∫+∞0k(c)y4(t,c)dc,(∂∂t+∂∂c)y4(t,c)=−(k(c)+b)y4(t,c),y2(t,0)=βy1(t)I∗+βS∗y3(t)+r2y3(t),y4(t,0)=r1y3(t)+∫+∞0μ(a)y2(t,a)da. | (30) |
Set
y1(t)=y01eλt,y2(t,a)=y02(a)eλt,y3(t)=y03eλt,y4(t,c)=y04(c)eλt, | (31) |
Substituting (31) into (30) gives
λy01=−by01−βI∗y01−βS∗y03, | (32) |
{dy02(a)da=−(λ+b+δe+μ(a)+σ(a))y02(a),y02(0)=βI∗y01+βS∗y03+r2y03, | (33) |
λy03=−(r1+r2+b+δi)y03+∫+∞0σ(a)y02(a)da+∫+∞0k(c)y04(c)dc, | (34) |
{dy04(c)dc=−(λ+k(c)+b)y04(c),y04(0)=r1y03+∫+∞0μ(a)y02(a)da. | (35) |
Integrating the first equation of (33) and (35) from 0 to
y02(a)=(βI∗y01+βS∗y03+r2y03)e−(λ+b+δe)a−∫a0(σ(s)+μ(a))ds,y04(c)=(r1y03+∫+∞0μ(a)y02(a)da)e−(λ+b)c−∫c0k(s)ds. | (36) |
substituting the above two equations into (34) and solving (34) we get
λy03=(βI∗y01+βS∗y03+r2y03)(K1(λ)+K2(λ)K3(λ))+r1y03K2(λ)−(r1+r2+b+δi)y03, | (37) |
where
K1(λ)=∫+∞0σ(a)e−(λ+b+δe)a−∫a0(σ(s)+μ(s))dsda,K2(λ)=∫+∞0k(c)e−(λ+b)c−∫c0k(s)dsdc,K3(λ)=∫+∞0μ(a)e−(λ+b+δe)a−∫a0(σ(s)+μ(a))dsda, |
By combining (37) and (32) we obtain the characteristic equation
det(λ+b+βI∗βS∗βI∗(K1(λ)+K2(λ)K3(λ))M)=0. |
where
M=β2S∗I∗λ+b+βI∗(K1(λ)+K2(λ)K3(λ)). | (38) |
It follows from (20) that (38) can also be rewritten as
(βS0R0+r2)(K1(λ)+K2(λ)K3(λ))+r1K2(λ)=βbS0(R0−1)(λ+bR0)R0(K1(λ)+K2(λ)K3(λ))+λ+r1+r2+b+δi. | (39) |
Note that
(βS0R0+r2)∣(K1(λ)+K2(λ)K3(λ))∣+r1∣K2(λ)∣≤(βS0R0+r2)(θ1+θ2θ3)+r1θ2=r1+r2+b+δi |
which, together with (39), leads to
∣βbS0(R0−1)(λ+bR0)R0(K1(λ)+K2(λ)K3(λ))+λ+r1+r2+b+δi∣≤r1+r2+b+δi. | (40) |
Since
βbS0(R0−1)(λ+bR0)R0(K1(λ)+K2(λ)K3(λ))+λ≤0. | (41) |
that is
In this section, we study the uniform persistence of system (1). Define
Theorem 5.1.
Proof. Let
T′(t)≥−max{(r1+r2+b+δi),(b+δe+μmax)}T(t), |
where
T(t)≥e−max{(r1+r2+b+δi),(b+δe+μmax)}tT(0). |
This completes the fact that
∫∞0e(t,a)da=∫t0e(t−a,0)e−∫a0ε(s)dsda+∫∞te0(a−t)e−∫aa−tε(s)dsda=∫t0[βS(t−a)I(t−a)+r2I(t−a)]e−∫a0ε(s)dsda+∫∞te0(a−t)e−∫aa−tε(s)dsda≤e−εmint‖e0‖L1→0,t→∞. |
Similarly,
∫∞0r(t,c)dc=∫t0r(t−c,0)e−∫c0η(s)dsdc+∫∞tr0(c−t)e−∫cc−tη(s)dsdc≤e−ηmint‖r0‖L1→0,t→∞. |
Thus
{dIdt=−(r1+r2+b+δi)I(t)+∫∞0σ(a)e(t,a)da+∫∞0k(c)r(t,c)dc,(∂∂t+∂∂a)e(t,a)=−(b+δe+μ(a)+σ(a))e(t,a),e(t,0)=βSI+r2I,(∂∂t+∂∂c)r(t,c)=−(k(c)+b)r(t,c),r(t,0)=r1I(t)+∫∞0μe(t,a)da,I(0)=0,e(0,a)=e0(a),r(0,c)=r0(c). |
Since
{d˜Idt=−(r1+r2+b+δi)˜I+∫∞0σ(a)˜e(t,a)da+∫∞0k(c)˜r(t,c)dc,(∂∂t+∂∂a)˜e(t,a)=−(b+δe+μ(a)+σ(a))˜e(t,a),˜e(t,0)=βS0˜I+r2˜I,(∂∂t+∂∂c)˜r(t,c)=−(k(c)+b)˜r(t,c),˜r(t,0)=r1˜I+∫∞0μ(a)˜e(t,a)da,˜I(0)=0,˜e(0,a)=e0(a),˜r(0,c)=r0(c). | (42) |
By the formulations (4), (5), we have
˜e(t,a)={˜e(t−a,0)e−∫a0ε(s)ds,t>a≥0,e0(a−t)e−∫aa−tε(s)ds,a≥t≥0. | (43) |
˜r(t,c)={˜r(t−c,0)e−∫c0η(s)ds,t>c≥0,r0(c−t)e−∫cc−tη(s)ds,c≥t≥0. | (44) |
Substituting (43) and (44) into the first equation of (42), with the help of the third and the fifth equations of (42), we obtain
{d˜I(t)dt=(H1+H2+H3+H4)˜I(t)+Fe(t)+Fr(t)+Fer(t),˜I(0)=0, | (45) |
where
H1=−(r1+r2+b+δi),H2=∫t0σ(a)(βS0+r2)e−∫a0ε(s)dsda,H3=∫t0k(c)r1e−∫c0η(s)dsdc,H4=∫t0k(c)∫t0μ(a)(βS0+r2)e−∫a0ε(s)dsdae−∫c0η(s)dsdc,Fe(t)=∫∞tσ(a)e0(a−t)e−∫aa−tε(s)dsda,Fr(t)=∫∞tk(c)r0(c−t)e−∫cc−tη(s)dsdc,Fer(t)=∫t0k(c)∫∞tμ(a)e0(a−t)e−∫aa−tε(s)dsdae−∫c0η(s)dsdc. |
It's simple to know that, for each
Fe(t)≤e−εmint∫∞tσ(a)e0(a−t)da=0,Fr(t)≤e−ηmint∫∞tk(c)r0(c−t)dc=0,Fer(t)≤∫t0k(c)e−εmint∫∞tμ(a)e0(a−t)dae−∫c0η(s)dsdc=0. |
Then, we know that equation (45) has a unique solution
Theorem 5.2. Assume
\displaystyle \lim\limits_{t\rightarrow +\infty}inf d(U(t)y, \partial M _{0})\geq \varepsilon. |
Furthermore, there exists a compact subset
Proof. Since the infection-free equilibrium
\begin{split} \displaystyle A(a)=&\int_{a}^{\infty}(\sigma(\theta)+B(0)\mu(\theta))e^{-\int_{a}^{\theta}\varepsilon(s)ds}d\theta, \\ \displaystyle B(c)=&\int_{c}^{\infty}k(\theta)e^{-\int_{c}^{\theta}\eta(s)ds}d\theta. \end{split} | (46) |
For,
\begin{split} \displaystyle A'(a)&=-\sigma(a)+\varepsilon(a)A(a)-\theta_{2}\mu(a), \\ \displaystyle B'(c)&=-k(c)+\eta(c)B(c) . \end{split} | (47) |
Consider the function
\displaystyle\Phi(t)=I(t)+\int_{0}^{\infty}A(a)e(t, a)da+\int_{0}^{\infty}B(c)r(t, c)dc, |
which satisfies
\displaystyle\frac{d\Phi(t)}{dt}=\beta I(\theta_{1}+\theta_{2}\theta_{3})(S-\frac{ S^{0}}{R_{0}}). |
Since
Let
\displaystyle g(x)=x-1- \ln x, |
denote
Theorem 6.1. The unique endemic equilibrium
Proof. Constructing the Lyapunov functional as follows
\displaystyle V_{*}=W_{s}+W_{e}+W_{i}+W_{r}, |
where
\begin{split} \displaystyle W_{s}=&(\theta_{1}+\theta_{2}\theta_{3})S^{*}g(\frac{S}{S^{*}}), W_{i}=I^{*}g(\frac{I}{I^{*}}), \\ \displaystyle W_{e}=&\int_{0}^{+\infty}A(a)e^{*}(a)g(\frac{e(t, a)}{e^{*}(a)})da, W_{r}=\int_{0}^{+\infty}B(c)r^{*}(c)g(\frac{r(t, c)}{r^{*}(c)})dc. \end{split} |
Since
\begin{split} \frac{dW_{s}}{dt}=&(\theta_{1}+\theta_{2}\theta_{3})bS^{*}(2-\frac{S}{S^{*}}-\frac{S^{*}}{S})\\ &+(\theta_{1}+\theta_{2}\theta_{3})\beta S^{*}I^{*}(1-\frac{SI}{S^{*}I^{*}}-\frac{S^{*}}{S}+\frac{I}{I^{*}}). \end{split} |
Calculating the derivative of
\begin{split} \displaystyle \frac{dW_{e}}{dt}=&\int_{0}^{+\infty}A(a)e^{*}(a)\frac{\partial}{\partial t}g(\frac{e(t, a)}{e^{*}(a)})da\\ \displaystyle =&\int_{0}^{+\infty}A(a)e^{*}(a)\frac{\partial}{\partial t}(\frac{e(t, a)}{e^{*}(a)}-1-\ln\frac{e(t, a)}{e^{*}(a)})da\\ \displaystyle =&\int_{0}^{+\infty}A(a)e^{*}(a)(\frac{1}{e^{*}(a)}-\frac{1}{e(t, a)})\frac{\partial}{\partial t}e(t, a)da\\ \displaystyle =&\int_{0}^{+\infty}A(a)e^{*}(a)(\frac{1}{e^{*}(a)}-\frac{1}{e(t, a)})(-\frac{\partial}{\partial a}e(t, a)-\varepsilon(a)e(t, a))da\\ \displaystyle =&-\int_{0}^{+\infty}A(a)e^{*}(a)(\frac{e(t, a)}{e^{*}(a)}-1)(\frac{e_{a}(t, a)}{e(t, a)}+\varepsilon(a))da. \end{split} |
Note that
\begin{split} \displaystyle \frac{\partial}{\partial a}g(\frac{e(t, a)}{e^{*}(a)})=&\frac{e_{a}(t, a)+e(t, a)\varepsilon(a)}{e^{*}(a)}-\frac{e_{a}(t, a)}{e(t, a)}+\frac{e^{*}(a)(-\varepsilon(a))}{e^{*}(a)}\\ \displaystyle =&(\frac{e(t, a)}{e^{*}(a)}-1)(\frac{e_{a}(t, a)}{e(t, a)}+\varepsilon(a)). \end{split} |
And
\begin{split} \displaystyle \frac{dA(a)}{da}=&A(a)\varepsilon(a)-\sigma(a)-\mu(a)B(0), \\ \displaystyle \frac{de^{*}(a)}{da}=&-\varepsilon(a)e^{*}(a). \end{split} |
Hence, using integration by parts, we have
\begin{split} \displaystyle \frac{dW_{e}}{dt}=&-\int_{0}^{+\infty}A(a)e^{*}(a)\frac{\partial}{\partial a}g(\frac{e(t, a)}{e^{*}(a)})da\\ \displaystyle =&-A(a)e^{*}(a)g(\frac{e(t, a)}{e^{*}(a)})\mid_{0}^{+\infty}+\int_{0}^{+\infty}(\frac{d}{d a}A(a))e^{*}(a)g(\frac{e(t, a)}{e^{*}(a)})da \end{split} |
\begin{split}\displaystyle &+\int_{0}^{+\infty}A(a)(\frac{d}{da}e^{*}(a))g(\frac{e(t, a)}{e^{*}(a)})da\\ \displaystyle=&-A(a)e^{*}(a)g(\frac{e(t, a)}{e^{*}(a)})\mid_{+\infty}+A(0)e^{*}(0)g(\frac{e(t, 0)}{e^{*}(0)})\\ \displaystyle &-\int_{0}^{+\infty}e^{*}(a)(\sigma(a)+\mu(a) B(0))g(\frac{e(t, a)}{e^{*}(a)})da. \end{split} |
Note
\begin{split} \displaystyle\frac{dW_{e}}{dt}=&-A(a)e^{*}(a)g(\frac{e(t, a)}{e^{*}(a)})\mid_{+\infty}+(\theta_{1}+\theta_{2}\theta_{3})(\beta S^{*}I^{*}+r_{2}I^{*})g(\frac{e(t, 0)}{e^{*}(0)})\\ \displaystyle&-\int_{0}^{+\infty}e^{*}(a)(\sigma(a)+\mu(a) \theta_{2})g(\frac{e(t, a)}{e^{*}(a)})da. \end{split} |
Further, it follows from
\begin{split} \displaystyle\frac{dW_{i}}{dt}=&I^{*}(\frac{I_{t}}{I^{*}}-\frac{I_{t}}{I})\\ \displaystyle=&I^{*}(\frac{1}{I^{*}}-\frac{1}{I})[-(r_{1}+r_{2}+b+\delta_{i})I\\ \displaystyle&+\int_{0}^{+\infty}\sigma(a)e(t, a)da+\int_{0}^{+\infty}k(c)r(t, c)dc]\\ \displaystyle=&\int_{0}^{+\infty}\sigma(a)e^{*}(a)(1-\frac{I}{I^{*}}-\frac{I^{*}e(t, a)}{Ie^{*}(a)}+\frac{e(t, a)}{e^{*}(a)})da\\ \displaystyle&+\int_{0}^{+\infty}k(c)r^{*}(c)(1-\frac{I}{I^{*}}-\frac{I^{*}r(t, c)}{Ir^{*}(c)}+\frac{r(t, c)}{r^{*}(c)})dc. \end{split} |
Similar to
\begin{split} \displaystyle\frac{dW_{r}}{dt}&=\int_{0}^{+\infty}B(c)r^{*}(c)\frac{\partial}{\partial t}g(\frac{r(t, c)}{r^{*}(c)})dc\\ \displaystyle&=\int_{0}^{+\infty}B(c)r^{*}(c)\frac{\partial}{\partial t}[\frac{r(t, c)}{r^{*}(c)}-1-\ln\frac{r(t, c)}{r^{*}(c)}]dc\\ \displaystyle&=\int_{0}^{+\infty}B(c)r^{*}(c)[(\frac{1}{r^{*}(c)}-\frac{1}{r(t, c)})\frac{\partial}{\partial t}r(t, c)]dc\\ \displaystyle&=\int_{0}^{+\infty}B(c)r^{*}(c)[(\frac{1}{r^{*}(c)}-\frac{1}{r(t, c)})(-\frac{\partial}{\partial c}r(t, c)-\eta(c)r(t.c))]dc\\ \displaystyle&=-\int_{0}^{+\infty}B(c)r^{*}(c)(\frac{r(t, c)}{r^{*}(c)}-1)(\frac{r_{c}(t, c)}{r(t, c)}+\eta(c))dc. \end{split} |
Note
\begin{split} \displaystyle \frac{\partial}{\partial c}g(\frac{r(t, c)}{r^{*}(c)}) =(\frac{r(t, c)}{r^{*}(c)}-1)(\frac{r_{c}(t, c)}{r(t, c)}+\eta(c)), \end{split} |
and
\displaystyle \frac{dB(c)}{dc}=B(c)\eta(c)-k(c), \frac{dr^{*}(c)}{dc}=-\eta(c)r^{*}(c). |
Hence, using integration by parts, we have
\begin{split} \displaystyle\frac{dW_{r}}{dt}=&-\int_{0}^{+\infty}B(c)r^{*}(c)\frac{\partial}{\partial c}g(\frac{r(t, c)}{r^{*}(c)})dc\\ \displaystyle =&-B(c)r^{*}(c)g(\frac{r(t, c)}{r^{*}(c)})\mid_{+\infty}+B(0)r^{*}(0)g(\frac{r(t, 0)}{r^{*}(0)})\\ \displaystyle &-\int_{0}^{+\infty}r^{*}(c)k(c)g(\frac{r(t, c)}{r^{*}(c)})dc\\ \displaystyle =&-B(c)r^{*}(c)g(\frac{r(t, c)}{r^{*}(c)})\mid_{+\infty}-\int_{0}^{+\infty}r^{*}(c)k(c)g(\frac{r(t, c)}{r^{*}(c)})dc\\ \displaystyle &+\theta_{2}(r_{1}I^{*}+\int_{0}^{+\infty}\mu(a) e^{*}(a)da)g(\frac{r(t, 0)}{r^{*}(0)}). \end{split} |
Note
\begin{split} \displaystyle&\int_{0}^{+\infty}(\sigma(a)+\mu(a)\theta_{2})e^{*}(a)da=(\theta_{1}+\theta_{2}\theta_{3})(\beta S^{*}I^{*}+r_{2}I^{*}), \\ \displaystyle&\int_{0}^{+\infty}k(c)r^{*}(c)dc=\theta_{2}r_{1}I^{*}+\theta_{2}\theta_{3}(\beta S^{*}I^{*}+r_{2}I^{*}). \end{split} |
We derive
\begin{split} \displaystyle\frac{dV_{*}}{dt}=&(\theta_{1}+\theta_{2}\theta_{3})bS^{*}(2-\frac{S}{S^{*}}-\frac{S^{*}}{S})-A(a)e^{*}(a)g(\frac{e(t, a)}{e^{*}(a)})\mid_{+\infty}\\ \displaystyle&-B(c)r^{*}(c)g(\frac{r(t, c)}{r^{*}(c)})\mid_{+\infty}+\int_{0}^{+\infty}\sigma(a)e^{*}(a)dag(\frac{e(t, 0)}{e^{*}(0)})\\ &-\int_{0}^{+\infty}\mu(a)\theta_{2}e^{*}(a)(g(\frac{e(t, a)}{e^{*}(a)})-g(\frac{e(t, 0)}{e^{*}(0)}))da\\ \displaystyle& +\int_{0}^{+\infty}k(c)r^{*}(c)dcg(\frac{r(t, 0)}{r^{*}(0)})+H_{1}+H_{2}+H_{3} \end{split} |
where
\begin{split} \displaystyle H_{1}=&(\theta_{1}+\theta_{2}\theta_{3})\beta S^{*}I^{*}[-g(\frac{SI}{S^{*}I^{*}})-g(\frac{S^{*}}{S})+g(\frac{I}{I^{*}})]\\ \displaystyle\leq&(\theta_{1}+\theta_{2}\theta_{3})(\beta S^{*}I^{*}+r_{2}I^{*})[-g(\frac{SI}{S^{*}I^{*}})-g(\frac{S^{*}}{S})+g(\frac{I}{I^{*}})]\\ \displaystyle=&\int_{0}^{+\infty}(\sigma(a)+\mu(a)\theta_{2})e^{*}(a)[-g(\frac{SI}{S^{*}I^{*}})-g(\frac{S^{*}}{S})+g(\frac{I}{I^{*}})]da, \\ \displaystyle H_{2}=&\int_{0}^{+\infty}\sigma(a)e^{*}(a)(1-\frac{I}{I^{*}}-\frac{I^{*}e(t, a)}{Ie^{*}(a)}+\frac{e(t, a)}{e^{*}(a)}-\frac{e(t, a)}{e^{*}(a)}+1+\ln\frac{e(t, a)}{e^{*}(a)})da\\ \displaystyle=&-\int_{0}^{+\infty}\sigma(a)e^{*}(a)(g(\frac{I}{I^{*}})+g(\frac{I^{*}e(t, a)}{Ie^{*}(a)})da\\ \displaystyle\leq& -\int_{0}^{+\infty}\sigma(a)e^{*}(a)(g(\frac{I}{I^{*}})+g(\frac{e(t, a)}{e^{*}(a)})da, \end{split} |
\begin{split} \displaystyle H_{3}=&\int_{0}^{+\infty}k(c)r^{*}(c)(1-\frac{I}{I^{*}}-\frac{I^{*}r(t, c)}{Ir^{*}(c)}+\frac{r(t, c)}{r^{*}(c)}-\frac{r(t, c)}{r^{*}(c)}+1+\ln\frac{r(t, c)}{r^{*}(c)})dc\\ \displaystyle=&-\int_{0}^{+\infty}k(c)r^{*}(c)(g(\frac{I}{I^{*}})+g(\frac{I^{*}r(t, c)}{Ir^{*}(c)}))dc \end{split} |
\begin{split}\displaystyle\leq&-\int_{0}^{+\infty}k(c)r^{*}(c)(g(\frac{I}{I^{*}})+g(\frac{r(t, c)}{r^{*}(c)}))dc. \end{split} |
Hence,
In the following, we provide some numerical simulations to illustrate the global stability of the disease-free equilibrium and the endemic equilibrium for system (1). We choose parameters
\sigma(a)=\left\{ \begin{array}{ccc} 0.3&a\geq \tau\\ 0&\tau\geq a\geq 0 \end{array}\right., k(c)=\left\{ \begin{array}{ccc} 0.1 &c\geq \tau\\ 0&\tau\geq c\geq 0 \end{array}\right., \mu(a)=\left\{ \begin{array}{ccc} 0.25 &a\geq \tau\\ 0&\tau\geq a\geq 0 \end{array}\right.. |
Under the initial values
\displaystyle S(0)=30, e(0, a)=6e^{-0.3a}, I(0)=10, r(0, c)=6e^{-0.3c}. |
In Figure 2, we choose
In this section, we briefly summarize our results. First, a PDE tuberculosis model (1) is proposed here to incorporate the latent-stage progression age of latent individuals and the relapse age of removed individuals. In addition, we assumed that infectious individuals might come into the latent class
The author is very grateful to Professors Shigui Ruan and Xingan Zhang for their supervision and assistance. The author also thanks the editor and the anonymous reviewers for their constructive comments that help to improve an early version of this paper.
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