J. Appl. Math. Comput. https://doi.org/10.1007/s12190-018-1194-8 ORIGINAL RESEARCH

Global stability of an age-structured model for pathogen–immune interaction Tsuyoshi Kajiwara1 · Toru Sasaki1 · Yoji Otani1

Received: 1 August 2017 © Korean Society for Computational and Applied Mathematics 2018

Abstract In this paper, we present an age-structured mathematical model for infectious disease in vivo with infection age of cells. The model contains an immune variable and the effect of absorption of pathogens into uninfected cells. We construct Lyapunov functionals for the model and prove that the time derivative of the functionals are nonpositive. Using this, we prove the global stability results for the model. Especially, we present the full mathematical detail of the proof of the global stability. Keywords Lyapunov functionals · Age-structured equations · Immunity · Persistence Mathematics Subject Classification 35B15 · 45D05 · 92B05

1 Introduction Infectious diseases in vivo have been investigated extensively using ordinary differential equations, and differential equations with delay. Anderson et al. [2] and Nowak and Bangham [17] proposed models containing immune variables explicitly. Local stability analyses of the equilibria of such models are often difficult because the sizes of Jacobi matrices are large and a lot of symbolic calculations are needed [16] or transcendental equations appear as characteristic equations. On the other hand, global stability analysis using Lyapunov functions were extremely extended by Korobeinikov for ordinary differential equation models and McCluskey for ordinary differential equation models with delay [13,15]. Pang et al. [20] constructed Lyapunov functions for models of infectious disease in vivo with an

B 1

Tsuyoshi Kajiwara [email protected] Graduate School of Environmental and Life Science, Okayama University, Okayama, Japan

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immune variable, that is, a pathogen–immune interaction model, and Kajiwara et al. [12] proposed a systematic method of constructing Lyapunov functions for models of infectious disease in vivo with an immune variable. Recently, models for infectious diseases in vivo with infection age of cells has been developed. Huang et al. et al. [8] proposed a model incorporating infection age for infected cells to the model in Nowak and Bangham [17]. They constructed Lyapunov functionals for their model to show the global stability of the equilibria. Demasse and Ducrot [6], Browne and Pilyugin [3] and Browne [4] presented mathematical analysis of the global stability. Moreover they extended their models to multistrain situation. Demasse and Ducrot [6] also incorporated to their model the absorption effect of pathogens into uninfected cells. For an epidemic model with infection age, Magal et al. [14] presented a construction of a Lyapunov functional and mathematical arguments of global stability. We treat an age-structured model with a variable of humoral immunity. We use a linear immune response function in Nowak and May [18]. Anderson et al. [2] presented a model with humoral immunity for malaria infection. The local stability of ordinary differential equation models with humoral immunity and absorption effect has been treated by Murase et al. [16]. In the case of ordinary differential equations which ignore age-structures, for the analysis of the global stability of the equilibrium, the construction of a Lyapunov functional and the computation for nonpositivity of the derivative of the Lyapunov functional is of course important. But for age-structured equations, mathematical arguments for the global stability are not simple. We present mathematical details of the proof of the global stability of the models. Browne [5] considers an age-structured model with cell-mediated immunity. The dynamics of the model in Browne [5] is completely different from the dynamics of our model. The contents of the paper is as follows. In Sect. 2, we present the model and fundamental analysis, that is, existence, positivity, boundedness and arguments for compactness. In Sect. 3, we present the definitions and formal calculations of Lyapunov functionals for our model. In Sect. 4, we present the proof of the global stability of the equilibria of our model. We use the Lyapunov functionals in Sect. 3 and the arguments of ρ-persistence in Smith and Thieme [21].

2 A model with humoral immunity and fundamental analysis In this section, we present a model with humoral immunity and the absorption effect of pathogens into uninfected cells, and prove some fundamental results.

2.1 A model with an immune variable Huang et al. [8] proposed the following age-structured model for an infectious disease in vivo: dx = λ − d x − βvx, dt

123

∂y ∂y + = −(d + μ(a))y, ∂t ∂a

Global stability of an age-structured model for…

 ∞ dv = g(a)y(t, a) da − bv, dt 0 y(t, 0) = βx(t)v(t), x(0) = x0 , y(0, a) = y0 (a), v(0) = v0 .

(1)

We note that the notations are not the same as [8]. They constructed a Lyapunov functional, and showed that the time derivative of it is non-positive. In this paper, we propose the following model incorporating the absorption effect of pathogens into uninfected cells, and the variables of humoral immunity explicitly: ∂y ∂y dx = λ − d x − βvx, + = −(d + μ(a))y, dt ∂t ∂a  ∞ dv = g(a)y(t, a) da − θβxv − bv − pvz, dt 0 dz = qv − mz, dt y(t, 0) = βx(t)v(t), x(0) = x0 , y(0, a) = y0 (a), v(0) = v0 , z(0) = z 0 .

(2)

The variable x(t) denotes the amount of uninfected cells, v(t) pathogens and y(t, a) the age density of infected cells with infection age a, z(t) the amount of humoral immunity. The function μ(a) is continuous and non negative, g(a) is bounded, non negative,  continuous and positive on some non empty interval in [0, ∞). Put  a a = sup{a ∈  R+ | g(a) > 0 }. We note that a can be infinity. Put σ (a) = exp(− 0 (d + μ(b)) db), ∞ and r = 0 g(a)σ (a) da. Then σ (a) is differentiable, and r is called the burst size. The nonnegative θ expresses the amount of absorption effect and is assumed that θ < r . Demasse and Ducrot [6] considered an age structured model without immune variable which incorporates absorption effect. We use the linear immune response function qv in the Section 6.4 in Nowak and May [18], where q is a positive constant. The same immune response function is also used in Inoue et al. [10]. In this paper, we construct Lyapunov functionals for the equation (2), and show the global stability results: If the basic reproductive ratio R0 ≤ 1, the disease free equilibrium (DFE) is globally asymptotically stable, and if R0 > 1, then DFE is globally asymptotically stable for the subspace of the phase space where the disease does not exists initially, and the interior equilibrium is globally asymptotically stable for the subspace of the phase space where the disease exists initially. 2.2 Positivity and boundedness Since we can treat only classical solutions by the equation (2), we rewrite the equation (2) into the integral equation following the method in Browne and Pilyugin [3], using the Volterra formulation. Put X˜ = R × L1 ([0, ∞)) × R × R with the ordinary metric. We use the following integral equation on X˜ : 

t

x(t) = x0 +

(λ − d x(s) − βx(s)v(s)) ds,

0

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σ (a) y0 (a − t)1{a>t} , y(t, a) = βσ (a)v(t − a)x(t − a)1{t>a} + σ (a − t)  t ∞  t g(a)y(s, a) da ds − (θβx(s)v(s) + bv(s) + pv(s)z(s)) ds, v(t) = v0 + 0 0 0  t (qv(s) − mz(s)) ds, (3) z(t) = z 0 + 0

with (x0 , y0 (·), v0 , z 0 ) ∈ X˜ . We note that if a solution of (3) is differentiable with respect to t and a, it is also a classical solution of (2). Since y(t, a) in (3) is not continuous and we can not use the usual methods of integral equation, we use an integral equation of continuous functions instead of (3). For w ∈ R3 , define η(w) = |w1 | + |w2 | + |w3 |. Proposition 1 Let u 0 = (x0 , y0 (·), v0 , z 0 ) ∈ X˜ . Then there exists a T > 0 and a unique continuous solution u(t) of (3) with u(0) = u 0 for 0 ≤ t < T . Proof Substituting the the second equation of (3) into the third equation, we can get the following integral equation with three variables: 

t

x(t) = x0 +

(λ − d x(s) − βx(s)v(s)) ds,  t s v(t) = v0 + βg(a)σ (a)v(s − a)x(s − a) da ds, 0 0  t − (θβx(s)v(s) + bv(s) + pv(s)z(s)) ds 0  t ∞ g(a)σ (a) y0 (a − s) dads, + σ (a − s) 0 s  t (qv(s) − mz(s)) ds. z(t) = z 0 + 0

(4)

0

We consider (4) as an integral equation with respect to x(t), v(t) and z(t), and consider y0 (a) as a known function. We fix M > 0. Let τ be a positive constant which will be determined later. Put H = { (x, v, z) ∈ C([0, τ ])3 | |x(t) − x0 | ≤ M, |v(t) − v0 | ≤ M, |z(t) − z 0 | ≤ M for 0 ≤ t ≤ τ }. We consider the norm w for w ∈ H by w = supt∈[0,τ ] η(w(t)). Define a map G(w(t)) = (G 1 (w(t)), G 2 (w(t)), G 3 (w(t))) on H by  G 1 (w(t)) = x0 +

t

(λ − d x(s) − βx(s)v(s)) ds, 0  t s G 2 (w(t)) = v0 + βg(a)σ (a)v(s − a)x(s − a) da ds 0

123

0

Global stability of an age-structured model for…

 −

t

(θβx(s)v(s) + bv(s) + pv(s)z(s)) ds  t ∞ g(a)σ (a) + y0 (a − s) dads, σ (a − s) 0 s  t (qv(s) − mz(s)) ds. G 3 (w(t)) = z 0 + 0

0

If τ > 0 is sufficiently small, it holds that G(w) ∈ H for each w ∈ H and that there exists a constant 0 < K < 1 such that G(w1 ) − G(w2 ) ≤ K w1 − w2  for w1 , w2 ∈ H . The proof of the first statement is easy. We show the second statement. We put L = max{ |x0 − M|, |x0 + M|, |v0 − M|, |v0 + M|, |z 0 − M|, |z 0 + M| }. Then  ∞ it holds |x(t)| ≤ L, |v(t)| ≤ L and |z(t)| ≤ L for 0 ≤ t ≤ τ . Then, using 0 g(a)σ (a) da = r , it holds that for 0 ≤ t ≤ τ |G(w1 (t)) − G(w2 (t))|  t     =  {−d(x1 (s) − x2 (s))} − β {x1 (s)v1 (s) − x2 (s)v2 (s)} ds  0 t  s     + βg(a)σ (a) {v1 (s − a)x1 (s − a) − v2 (s − a)x2 (s − a)} da ds  0 t 0  +  {θβ(x1 (s)v1 (s) − x2 (s)v2 (s)) + b(v1 (s) − v2 (s)) + p(v1 (s)z 1 (s) 0

− v2 (s)z 2 (s))} ds|  t     +  {q(v1 (s) − v2 (s)) − m(z 1 (s) − z 2 (s))} ds  0

≤ dτ sup |x1 (t) − x2 (t)| + β Lτ { sup |x1 (t) − x2 (t)| + sup |v1 (t) − v2 (t)|} 0≤t≤τ

0≤t≤τ





0≤t≤τ

sup |x1 (t) − x2 (t)| + sup |v1 (t) − v2 (t)|

+ βr Lτ

0≤t≤τ

0≤t≤τ

 + θβ Lτ



sup |x1 (t) − x2 (t)| + sup |v1 (t) − v2 (t)| 0≤t≤τ

0≤t≤τ



+ bτ sup |v1 (t) − v2 (t)| + pLτ 0≤t≤τ



sup |v1 (t) − v2 (t)| + sup |z 1 (t) − z 2 (t)| 0≤t≤τ

0≤t≤τ

+ qτ sup |v1 (t) − v2 (t)| + mτ sup |z 1 (t) − z 2 (t)|. 0≤t≤τ

0≤t≤τ

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Then there exists a P > 0 which does not depend on τ , and it holds that for 0 ≤ t ≤ τ |G(w1 (t)) − G(w2 (t))| 



sup |x1 (t) − x2 (t)| + sup |v1 (t) − v2 (t)| + sup |z 1 (t) − z 2 (t)|

≤ Pτ

0≤t≤τ

0≤t≤τ

0≤t≤τ

≤ 3Pτ sup {|x1 (t) − x2 (t)| + |v1 (t) − v2 (t)| + |z 1 (t) − z 2 (t)|} 0≤t≤τ

= 3Pτ w1 − w2 . If we take τ < 1/(3P), then the second statement follows. By the Banach fixed-point theorem, there exists a unique solution w(t) of (4) for 0 ≤ t ≤ τ . We define y(t, a) by the second equation of (3). Then u(t) = (x(t), y(t, ·), v(t), z(t)) is a solution of (3) with u(0) = (x0 , y0 (·), v0 , z 0 ) for 0 ≤ t ≤ τ.   We denote by [0, T∞ (u)) the existence interval of the solution of (3) for the initial value u ∈ X˜ . We note that T∞ (u) may be infinity. Lemma 2 Let u(t) = (x(t), y(t, ·), v(t), z(t)) be a solution of (3). For 0 < t < T∞ (u), x(t), v(t) and z(t) are differentiable. Proof The solutions of the integral equation are continuous. Then the conclusion follows from the fundamental theorem of calculus. The following proposition is proved in Lemma 2.2 of Browne and Pilyugin [3] for the model without immune variables. We state the shortened proof. Proposition 3 If the elements of an initial value, x(0), v(0), z(0) and y0 (·) are nonnegative, it holds that x(t) > 0, v(t) ≥ 0, z(t) ≥ 0 and y(t, ·) ≥ 0 for 0 < t < T∞ (u). Proof Since x(t) is differentiable, the first equation of (2) is also transformed into the following equation:    t x(t) = exp − (d + βv(u)) du x(0) 0   t  t  s  + λ exp − (d + βv(u)) du exp (d + βv(u)) du ds. 0

0

0

Then it holds x(t) > 0 for 0 < t < T∞ (u). Put t1 = sup{ t0 ≥ 0 | v(t) ≥ 0 for 0 ≤ t t ≤ t0 }. We assume t1 < T∞ (u). We note that v(t1 ) = 0. Put f (t) = exp{ 0 (θβx(s)+ b + pz(s)) ds}. Then by the third equation of (2) and the second equation of (3), the following holds:  f (t)v(t) = 0

123

t

G(t, a)( f (a)v(a)) da +

 t t1

s

∞

f (s)g(a)

σ (a) y0 (a − s) dads, σ (a − s) (5)

Global stability of an age-structured model for…

where G(t, a) is the integral kernel given by ⎧ t β f (s) ⎪ ⎪ ⎪ σ (s − a)x(a) ds 0 ≤ a ≤ t1 , ⎨ f (a) t G(t, a) =  1t ⎪ β f (s) ⎪ ⎪ σ (s − a)x(a) ds t1 ≤ a ≤ t. ⎩ a f (a) We consider x(t) and z(t) as known functions and consider the above the integral equation (5) of Volterra type for the unknown function f (t)v(t). We note that x(t) > 0 for 0 < t < T∞ (u). Let M > 0 be a positive constant and 0 < ε be a positive constant with t1 + ε < T∞ (u) which will be determined later. Put H = { X (t) ∈ C([t1 , t1 + ε] | 0 ≤ X (t) ≤ M }. We do the argument using the Banach fixed-point theorem as in Proposition 1. Then we can choose an ε such that there exists a unique nonnegative solution f (t)v(t) in [0, t1 + ε] of (5). It contradicts to t1 < T∞ (u), and it holds t1 = T∞ (u). By the second equation of (3), y(t, ·) ≥ 0 for 0 ≤ t < T∞ (u). It holds dz = qv − mz ≥ −mz. (6) dt Then it holds z(t) ≥ 0 for 0 ≤ t < T∞ (u).   By Proposition 3, we can define the phase space X of the original integral equation by ◦

X = R+ × L1 (R+ )+ × R+ × R+ , ◦

where R+ = { x ∈ R | x ≥ 0 }, R+ = { x ∈ R | x > 0 } and L1 (R+ )+ = { f ∈ L1 (R+ ) | f (x) ≥ 0 for each = x∈ R+ }. ◦

We introduce a metric ξ on the first component R+ such that the distance of two points near 0 tends to ∞. For example, we put  k(x) =

x 2 − 1/x

(1 ≤ x),

ξ(x, y) = |k(x) − k(y)|.

(0 < x < 1),

◦

Although R+ is not closed subset of R, it is a complete metric space which generates ◦

the same topology as the ordinary topology of R+ . New metric is used in Proposition 5, and completeness of phase spaces is often used in the arguments for compactness, for example, in Hale and Waltman [7]. Proposition 4 There exists a positive constant M such that for every solution u(t) = (x(t), y(t, ·), v(t), z(t)) with u(0) ∈ X , there exits a T such that for each t ≥ T it

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holds that 

∞

x(t) ≤ M, v(t) ≤ M, z(t) ≤ M,

y(t, a) da ≤ M.

0

Proof Since solutions are non negative, it holds that dx = λ − d x − βxv ≤ λ − d x. dt Then there exists a T1 > 0 such that for each t ≥ T1 it holds x(t) ≤

λ + 1. d

We define V (t) as  V (t) = x(t) +

∞

y(t, a) da. 0

Then it holds 

t

V (t) = x(t) + 



0

t

t

= x(t) +

∞

βx(t − a)v(t − a)σ (a) da +  βx(t − a)v(t − a)σ (a) da +

0

0

∞

σ (a) y0 (a − t) da σ (a − t) σ (b + t) y0 (b) db. σ (b)

By differentiating V (t) it holds dV = λ − d x(t) − βx(t)v(t) + βx(0)v(0)σ (t) + dt  ∞

σ (b + t) y0 (b) db + σ (b) 0



t 0

β

∂ (x(t − a)v(t − a))σ (a) da ∂t

 t ∂ = λ − d x(t) − βx(t)v(t) + βx(0)v(0)σ (t) − β (x(t − a)v(t − a))σ (a) da ∂a 0  ∞

σ (b + t) y0 (b) db + σ (b) 0 = λ − d x(t) − βx(t)v(t) + βx(0)v(0)σ (t) − [βx(t − a)v(t − a)σ (a)]t0  t  ∞ σ (a) y0 (a − t) da + βx(t − a)v(t − a)σ (a) da + σ (a − t) 0 t = λ − d x(t) − βx(t)v(t) + βx(0)v(0)σ (t) − βx(0)v(0)σ (t) + βx(t)v(t)  ∞ − (d + μ(a))y(t, a) da 0  ∞ = λ − d x(t) − (d + μ(a))y(t, a) da ≤ λ − d V (t). 0

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Then there exists a T2 > T1 > 0 such that for each t ≥ T2 it holds V (t) ≤

λ + 1. d

(7)

Then it holds for each t ≥ T2 

∞

y(t, a) da ≤ M1 ,

0

where M1 = λ/d + 1. Using the equation of v(t), for sufficiently large t it holds dv = dt



∞ 0

g(a)y(t, a) da − θβxv − bv − pvz ≤ g∞ M1 − bv.

Then there exists a T3 > T2 such that for each t ≥ T3 it holds v(t) ≤ M2 , where M2 = (g∞ M1 /b) + 1. Hence for each t ≥ T3 , it holds dz = qv − mz ≤ q M2 − mz. dt Then there exists T4 > T3 such that for each t ≥ T4 it holds z(t) ≤ (q M2 /m) + 1. Put T = T4 , and M = max{ M1 , M2 , (q M2 /m) + 1 }. We note that M does not depend on the solutions.   By Proposition 4, the solution of the original integral equation (3) exists for all t > 0. For each u ∈ X , we define a solution operator St by St (u) = u(t) for each t ≥ 0. Then {St }t≥0 is a semigroup on X , and it is point dissipative. Proposition 5 For each bounded subset B of X , the subset ∪t≥0 St (B) is also bounded. Proof Let u(t) = (x(t), y(t, ·), v(t), z(t)) be a solution with u(0) = (x(0), y(0, ·), v(0), z(0)). Using the differential inequalities and the positive constant M in Proposition 4, it holds   ∞ λ , x(0) + y(t, a) da ≤ max y(0, a) da , v(t) ≤ max {M, v(0)} , d 0 0 x(t) ≤ max {M, x(0)} , z(t) ≤ max {M, z(0)} , (8) for each t ≥ 0. We show boundedness for x(t) from below. Put M˜ = max {M, v(0)}. Then it holds 

∞

dx ˜ ≥ λ − (d + β M)x dt

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for t ≥ 0. Then it holds λ

˜

x(t) ≥ e−(d+β M)t x(0) +

d + β M˜

˜

(1 − e−(d+β M)t )

for t ≥ 0. It follows that

x(t) ≥ min

λ d + β M˜

 , x(0) > 0

for t ≥ 0.

 

For f ∈ L1 (R+ ), we write as f (·) ≥ 0 if f (x) ≥ 0 for almost all x, f (·) = 0 if f (x) = 0 for almost all x with respect to Lebesgue measure. Let u(t) = (x(t), y(t, a), v(t), z(t)) be a solution with u(0) = (x(0), y(0, ·), v(0), z(0)) ∈ X . Lemma 6 (1) If v(0) = 0 and y(0, ·) = 0, then it holds v(t) = 0 and y(t, ·) = 0 for all t > 0. ∞ ∞ (2) If v(0) > 0 then v(t) > 0, if 0 y(0, a) da > 0 then 0 y(t, a) da > 0, and if z(0) > 0 then z(t) > 0 for all t > 0 respectively. Proof (1) It is proved by the uniqueness theorem.   t (2) As in Proposition 3, we put f (t) = exp 0 (θβx(u) + b + pz(u)) du . Then f (t) is a positive continuous function. The second equation of (3) for v(t) is transformed into the following equation: v(t) = f (t)−1 v(0) +

 t 0

∞

f (t)−1 f (s)g(a)y(s, a) da ds.

(9)

0

∞ Then if v(0) > 0, then v(t) > 0 for t > 0. If 0 y0 (a) da > 0, by the second ∞ equation of (3), it holds that t y(t, a) da > 0. The statement for z(t) holds by (6).   The following is a form of Gronwall’s inequality. Lemma 7 (Agarwal and O’Regan [1], Corollary 7.5) Let u(x), p(x) and q(x) be nonnegative continuous functions on [x0 , a], p(x) be nondecreasing, and  u(x) ≤ p(x) +

x

q(t)u(t) dt. x0

Then  u(x) ≤ p(x) exp

q(t) dt x0

123



x

for x ∈ [x0 , a].

Global stability of an age-structured model for…

The following lemma is used in the proof of the persistence result.  a  a Lemma 8 (1) If v(0) + 0 y(0, a) da = 0, then v(t) + 0 y(t, a) da = 0 for t > 0.  a  a (2) If v(0) + 0 y(0, a) da > 0, then v(t) + 0 y(t, a) da > 0 for all t > 0.  a (3) If v(0) + 0 y(0, a) da > 0, then there exists a t0 > 0 such that v(t0 ) > 0. Proof (1) If a  = ∞, (1) follows from Lemma 6. We assume a  < ∞. Let L > 0 be an arbitrary constant. Then for 0 ≤ t ≤ L, it holds using (9) and the second equation of (3),  t

−1

∞

β f (t)−1 f (s)g(a))x(s − a)v(s − a) da ds v(t) = f (t) v(0) + 0 0  t ∞ f (s)g(a)σ (a) + y(0, a − s) da ds f (t)σ (a − s) 0 s  t ∞ β f (t)−1 f (s)g(a))x(s − a)v(s − a) da ds = 0 0   t  t = β f (t)−1 f (s)g(s − b) ds x(b)v(b) db 0

b

 t 

L



−1

× β f (t) f (s)g(s − b) ds x(b)v(b) db 0 b  t ≤ β L f (L)g∞ x(b)v(b) db. 0

 a We used that 0 y(0, a − s) da = 0 if 0 ≤ s ≤ a  , and g(a)y(0, a − s) = 0 for a > s if s > a  . Then Gronwall’s inequality, it holds v(t) = 0 for 0 ≤ t ≤ L. Since L > 0 is arbitrary, v(t) = 0 for each t ≥ 0. By the second equation of (3), y(t, a) = 0 for almost all 0 ≤ a ≤ a  . (2) We assume a  < ∞. If v(0) > 0, then v(t) > 0 for each t > 0 by Lemma 6.  a Then we assume that v(0) = 0 and 0 y(0, a) da > 0. From (9), it holds  v(t) = 0

t

f (t)−1 f (s)



∞

g(a)y(0, a − s) dads.

s

 a  −ε  a y(0, a) da > 0. Since 0 y(0, a) da > 0, there exists an ε > 0 such that 0 By the definition of a  , there exist c1 and c2 with 0 < c1 < c2 , c2 > a  − ε and [c1 , c2 ] ⊂ supp g. Put δ = min{c2 − c1 , c2 − a  + ε} > 0. Moreover there exist d d1 and d2 with 0 ≤ d1 < d2 ≤ a  − ε, d2 − d1 < δ/4 and d12 y(0, a) da > 0. Put ∞ t0 = c2 −d2 −δ/4 > 0. Since [d1 +t0 , d2 +t0 ] ⊂ supp g, t0 g(a)y(0, a −t0 ) da > 0. Then it holds v(t0 ) > 0. On the other hand, it holds d2 + t0 = c2 − δ/4 < a  ,  a  −t y(0, a) da > 0 for 0 ≤ t ≤ t0 . Then it holds that then a  − t0 > d2 . Then 0  a v(t) + 0 y(t, a) da > 0 for t > 0. For the case a  = ∞, we use that there exists

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M a M > 0 such that 0 y(0, a) da > 0, and there exists a K > M such that K is contained in the interior point of supp g.  

(3) It follows from the proof of (2).

Let u i = (x0i , y0i (·), v0i , z 01 ) ∈ X (i = 1, 2). Let wi (t) = (xi (t), vi (t), z i (t)) (i = 1, 2) be the solutions of the integral equations:  x(t) = x0i +

t

(λ − d x(s) − βx(s)v(s)) ds, 0  t ∞  t v(t) = v0i + g(a)y(s, a) da ds − (θβx(s)v(s) + bv(s) + pv(s)z(s)) ds, 0 0 0  t ∞ g(a)σ (a) y0i (a − s) dads, + σ (a − s) 0 s  t (qv(s) − mz(s)) ds, (i = 1, 2). (10) z(t) = z 0i + 0

Lemma 9 We assume that (x0i , y0i (·), v0i , z 0i ) (i = 1, 2) are contained in a bounded subset B of X . Then there exists a positive constant  such that it holds   η(w1 (t) − w2 (t)) ≤ η(w1 (0) − w2 (0)) + g∞ t

∞

 |y01 (b) − y02 (b)| db · et

0

for each t > 0. The positive constant  is determined for each bounded subset B. Proof We note that it holds  t    

∞

g(a)σ (a) y01 (a − s) dads − σ (a − s) 0 s  ∞ ≤ g∞ t |y01 (b) − y02 (b)| db.

 t 0

s

∞

  g(a)σ (a) y02 (a − s) dads σ (a − s)

0

For each t ≥ 0, it holds η(w1 (t) − w2 (t)) = |x01 (t) − x02 (t)| + |v01 (t) − v02 (t)| + |(z 01 (t) − z 02 (t)|  ∞ |y01 (b) − y02 (b)| db ≤ η(w1 (0) − w2 (0)) + g∞ t 0  t + η(w1 (s) − w2 (s)) ds, 0

where the constant  can be chosen using Proposition 5. By Gronwall’s inequality, the conclusion holds.   Proposition 10 Each solution of (3) depends continuously on its initial values uniformly on compact time intervals.

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Proof By Lemma 9, the elements x(t), v(t), z(t) of the solution depend continuously on the initial value uniformly on compact time intervals. The element y(t, a) also depends on the initial value by the second equation of (3) uniformly on compact time intervals. Proposition 11 The semigroup {St }t≥0 is continuous semigroup. Proof The semigroup {St }t≥0 is time continuous and is state continuous uniformly on compact time intervals by Lemma 10. By Lemma 1.34 of [21], the semigroup {St }t≥0 is a continuous semigroup on X .   We define the basic reproductive ratio. Definition 12 We define the positive number R0 by βx R0 = b + θβx



∞

g(a)σ (a) da =

0

βr x , b + θβx

where x = λ/d. Proposition 13 The number R0 is the basic reproductive ratio for the age-structured equation (2). Proof The proof for the corresponding age-structured equation without immune variables is given in Appendix A of Demasse and Ducrot [6]. The proof for equation (2) is similar. For the calculation of the basic reproductive ratio, the linearlized equations at DFE (x, 0, 0, 0): σ (a) y0 (a − t)1{a>t} , y(t, a) = βσ (a)xv(t − a)1{t>a} + σ (a − t)  ∞ dv = g(a)y(t, a) da − (θβx + b)v dt 0

(11) (12)

are used. The quadratic term pvz in the third equation in (2) containing an immune ∞ variable z does not appear in the linearization at DFE. Put b(t) = 0 g(a)y(t, a) da. Then by using (11) and (12) we get the following renewal equation for b(t): b(t) = βx

 t 

u

−(θβx+b)(u−a)



g(a)σ (a) da  t  + βxv(0) g(a)e−(θβx+b)(t−a) da + e

0

0

0

t

b(t − u) du ∞

g(a)σ (a) y0 (a − t) da. σ (a − t)

As in Demasse and Ducrot [6] and Inaba [9], the basic reproductive ratio R0 is calculated as   ∞  u e−(θβx+b)(u−a) g(a)σ (a) da du βx 0

0

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 = βx 0

 e−(θβx+b)(u−a) du g(a)σ (a) da =

a

βx b + θβx



∞

g(a)σ (a) da.

0

  Remark 2.1 The biological interpretation of R0 is as follows. By (12), the average life span of pathogens is 1/(θβx+b) at DFE, a pathogen generates βx infected cells per unit time at DFE and the positive number r is the burst size. Then R0 = 1/(θβx +b)·βx ·r expresses the average number of pathogens which a pathogen generates in the next generation at DFE. We summarize the existence of the equilibria. Proposition 14 When R0 ≤ 1, there exists only disease free equilibrium (DFE) (x, 0, 0, 0). When R0 > 1, there exist DFE and the disease equilibrium (x ∗ , y ∗ (·), v ∗ , z ∗ ), where x ∗ > 0, v ∗ > 0, z ∗ > 0, and y ∗ (a) satisfy y ∗ (a) = y ∗ (0)σ (a) = βx ∗ v ∗ σ (a),

for a ≥ 0, z ∗ =

q ∗ v m

and y ∗ (0) > 0. Proof We use the following equations for the elements x ∗ , v ∗ and z ∗ of an equilibrium: λ − d x ∗ − βv ∗ x ∗ = 0, qv ∗ − mz ∗ = 0, βr x ∗ v ∗ − θβx ∗ v ∗ − bv ∗ − pv ∗ z ∗ = 0. Other element y ∗ (a) is calculated by y ∗ (a) = βx ∗ v ∗ σ (a). If v ∗ = 0, then z ∗ = 0, y ∗ (a) = 0 for each a ≥ 0 and x ∗ = λ/d. There always exists DFE (x, 0, 0, 0). We note that R0 > 1 if and only if x > b/(β(r − θ )). We assume that v ∗ > 0. Then it holds that   m ∗ β(r − θ ) b d(x − x ∗ ) ∗ ∗ ∗ ∗ z x . = βv , v = , z = − x∗ q p β(r − θ ) Then we consider the equation of x ∗ :   d(x − x ∗ ) β 2 m(r − θ ) b ∗ x . = − x∗ pq β(r − θ )

(13)

We assume R0 > 1, then x > b/(β(r − θ )). Put g(x ∗ ) =

  d(x − x ∗ ) β 2 m(r − θ ) b ∗ x . − − x∗ pq β(r − θ )

The function g(x ∗ ) is strictly monotone decreasing for x ∗ ∈ [b/(β(r − θ )), x], g(b/(β(r − θ ))) > 0 and g(x) < 0. Then there exists a unique x ∗ satisfying (13) in [b/(β(r − θ )), x], and v ∗ and z ∗ are positive. If R0 ≤ 1, then x ≤ b/(β(r − θ )), and   such x ∗ does not exists.

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2.3 Arguments over compactness Let B ⊂ X be an arbitrary bounded set. The semigroup {St }t≥0 is called asymptotically compact on B if for each infinite sequences (u p ) p=1,2,... ⊂ B and (t p ) p=1,2,... ⊂ R, t p → ∞, the sequence (S(t p )u p ) p=1,2,... has a convergent subsequence. The semigroup {St }t≥0 is called asymptotically smooth if {St }t≥0 is asymptotically compact on each forward invariant bounded subset B in X . The following proposition is proved in Theorem 3.1 of Demasse and Ducrot [6] for an age-structured equation without immune variable. We refer the proof to [6]. We give only a short outline of the proof. Proposition 15 The semigroup {St }t≥0 associated with the equation (2) is asymptotically smooth. Proof Let B be a forward invariant bounded subset in X , and u p = (x p , v p , z p , y p ) ( p = 1, 2, . . . , ) be an infinite sequence of B. Let t p be an increasing sequence of positive number such that t p → ∞. For t ≥ 0, we put u p (t) = St (u p ). It is a solution such that u p (0) = u p . We can show that {u p (t p )} p=1,2,... is relatively compact in the topology of X , using Ascoli-Arzela theorem on a noncompact space R and the last equation of (3), as in Theorem 3.1 (iii) in Demasse and Ducrot [6].   Definition 16 (Smith and Thieme [21]) A compact invariant subset A is called a compact attractor if A attracts each bounded subset of X . Proposition 17 The semigroup {St }t≥0 on X associated with (3) has a compact attractor A in X . Proof The semigroup {St }t≥0 is asymptotically smooth and each positive orbit of a bounded subset is bounded under {St }t≥0 . Then by Theorem 2.33 in Smith and Thieme   [21], {St }t≥0 has a compact attractor in X . The following proposition is used to show that if a compact attractor contains only one point, it attracts all points and locally stable. We present a proof for the convenience of readers. Proposition 18 (Simplified version of Lemma 23.7 in Sell and You [22]) Assume that a compact attractor consists of one equilibrium x ∗ . Then the equilibrium x ∗ is locally stable and is globally asymptotically stable. Proof Since {x ∗ } is a compact attractor, it attracts each point. Then {x ∗ } is globally attractive. It is sufficient to prove that the equilibrium x ∗ is locally stable. We show that for each neighborhood U of x ∗ there exists a neighborhood V of ∗ x such that for t ≥ 0 St (V ) ⊂ U . It is sufficient to prove it for each bounded neighborhood U . Contrary we assume that there exists a bounded neighborhood U such that for each neighborhood Vn ⊂ U of {x ∗ } contracting to {x ∗ } there exists a / U . Then it should hold that sequence {u n } and {tn } with u n ∈ Vn and Stn (u n ) ∈ tn → ∞. Put U = {u n | n ≥ 0} ⊂ U . Since U is bounded, it is attracted by {x ∗ }. Then there exists a T > 0 such that it holds St (u n ) ∈ U for each t ≥ T , n ≥ 1. It is a contradiction.  

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3 Construction of Lyapunov functionals In Sect. 3, we present formal discussion on the definition of Lyapunov functional and the calculation of the derivative of such a functional along the age-structured equation (2). In Sects. 3.1 and 3.2 we assume that the following Assumption 1 is satisfied for a solution of the age-structured equation (2). Assumption 1 The following (a) and (b) hold. (a) A solution or an entire solution u(t) = (x(t), y(t, ·), v(t), z(t)) is differentiable for each t and for each a ≥ 0. (b) For R0 > 1, y(t, a) is positive for each t and a ≥ 0. Moreover for each t, log(y(t, a)) is bounded over 0 ≤ a < ∞. We show that each entire solution in a compact attractor or a persistence attractor satisfies the Assumption 1 in Proposition 39 and Proposition 41. Moreorev, in Sect. 4 we substitute only such entire solutions to the Lyapunov functional presented in Sect. 3. We use the methods in Kajiwara et al. [11] and Otani et al. [19] for calculation of Lyapunov functionals. We use the following ordinary differential equation: dx = λ − d x − βvx, dt

dv = rβvx − θβxv − bv − pvz, dt

dz = qv − mz, (14) dt

where  r=

∞

g(a)σ (a) da.

0

The elements x ∗ , v ∗ , and z ∗ of equilibria of the equation (14) are the same as the elements of equilibria of the equation (2). Let f(u) be the vector field defined by the right hand sides of (14). We use the following lemma which appears in Smith and Thieme [21], in this section. Lemma 19 ([21]) Let h(τ ) be an integrable function satisfying h(τ ) ≥ 0 for τ ≥ 0, and ξ(t) a continuous bounded function on R. Put  α(a) =

∞

h(τ ) dτ.

a

We assume that α(a) is differentiable, and α(a), α (a) are integrable over R+ . Put  F(t) = 0

123

∞

α(a)ξ(t − a) da.

Global stability of an age-structured model for…

Then F(t) is locally absolutely continuous and differentiable for almost all t > 0 and the following holds 



F (t) = α(0)ξ(t) +

∞





∞

α (a)ξ(t − a) da =

0

h(a) (ξ(t) − ξ(t − a)) da.

0

3.1 The case R0 > 1 Let (x ∗ , v ∗ , z ∗ ) be the interior equilibrium of (14). Define V1 (u) by V1 (u) = (r − θ )(x − x ∗ log x) + v − v ∗ log v +

p (z − z ∗ )2 . 2q

We rewrite (14) as follows: dx = − d(x − x ∗ ) + β(x ∗ v ∗ − xv), dt dz = q(v − v ∗ ) − m(z − z ∗ ). dt

dv = (r − θ )β(x − x ∗ )v − p(z − z ∗ )v, dt (15)

Then the derivation of V1 (u) along (14) is as follows:   dV1 (u(t)) x∗ = (r − θ ) 1 − (−d(x − x ∗ ) + β(x ∗ v ∗ − xv)) dt x   v∗ + 1− ((r − θ )β(x − x ∗ )v − p(z − z∗)v) v p + (z − z ∗ )(q(v − v ∗ ) − m(z − z ∗ )) q     x∗ x x∗ ∗ − ∗ + (r − θ )β 1 − (x ∗ v ∗ − xv) = d x (r − θ ) 2 − x x x + (r − θ )β(x − x ∗ )(v − v ∗ ) − p(v − v ∗ )(z − z ∗ ) pm (z − z ∗ )2 + p(v − v ∗ )(z − z ∗ ) − q     x x x∗ x∗ − ∗ + (r − θ )βx ∗ v ∗ 2 − − ∗ = d x ∗ (r − θ ) 2 − x x x x pm ∗ 2 (z − z ) − q      x∗ x βv ∗ 2− − ∗ = dx∗ r − θ 1 + d x x   ∗ x pm x + rβx ∗ v ∗ 2 − (16) − ∗ − (z − z ∗ )2 . x x q

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The last expression fits to the calculation of the age-structured equation. Next, we differentiate V1 (u) along the original age-structured equation: dx = −d x−βvx, dt

dv = dt



∞

g(a)y(t, a) da−θβxv−bv− pvz, 0

dz = qv−mz. dt (17)

We rewrite the second equation as: dv = (r − θ )βxv − bv − pvz + dt



∞

g(a)y(t, a) da − rβxv.

(18)

0

Then the derivative of V1 (u) along (17) is as follows:   ∞   dV1 (u(t)) v∗ = ∇V1 (u) · f(u) + 1 − g(a)y(t, a) da − rβxv . dt v 0

(19)

We calculate the second term:   ∞   v∗ 1− g(a)y(t, a) da − βr xv v 0  ∞  v∗ ∞ v∗ g(a)y(t, a) da − βr xv − g(a)y(t, a) da + βr xv = v 0 v 0

 ∞    xv y(t, a) da − ∗ ∗ = βr x ∗ v ∗ g(a)σ (a) ∗ v ∗ σ (a) βr x x v

0     ∞ x y(t, a) da + βr x ∗ v ∗ − g(a)σ (a) x∗ βr x ∗ vσ (a) 0    ∞ xv y(t, a) da − = βx ∗ v ∗ g(a)σ (a) βx ∗ v ∗ σ (a) x ∗ v ∗ 0    ∞ x y(t, a) da. + βx ∗ v ∗ g(a)σ (a) − x∗ βx ∗ vσ (a) 0

(20)

We add the second term in (20) to the second term in (16). We note that we assume Assumption 1.      ∞ x x∗ x y(t, a) βr x ∗ v ∗ 2 − ∗ − + βx ∗ v ∗ da g(a)σ (a) − x x x∗ βx ∗ vσ (a) 0    ∞ y(t, a) x∗ − da = βx ∗ v ∗ g(a)σ (a) 2 − x βx ∗ vσ (a) 0    ∞ y(t, a) y(t, a) x∗ − + log da = βx ∗ v ∗ g(a)σ (a) 2 − x βx ∗ vσ (a) βxvσ (a) 0    ∞ y(t, a) da. − βx ∗ v ∗ g(a)σ (a) log βxvσ (a) 0

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Global stability of an age-structured model for…

We add the last term to the first term in (20). Then it holds      ∞ y(t, a) y(t, a) xv ∗ ∗ da − βx g(a)σ (a) v g(a)σ (a) log − da βx ∗ v ∗ σ (a) x ∗ v∗ βxvσ (a) 0 0

   ∞ y(t, a) y(t, a) xv g(a)σ (a) − ∗ ∗ − log da = βx ∗ v ∗ βx ∗ v ∗ σ (a) x v βxvσ (a) 0

   ∞ y(t, a) y(t, 0) y(t, a)y ∗ (0) g(a)σ (a) − ∗ + ∗ + log da. = −βx ∗ v ∗ y (a) y (0) y(t, 0)y ∗ (a) 0

βx ∗ v ∗



∞

For differentiation of the functionals of integral form, we use Lemma 19. Put α(a) as 

∞

α(a) =

g(ε)σ (ε) dε.

a

Then α(a) is integrable. We define V2 (u) by ∗ ∗



V2 (u) = βx v

0

∞

 α(a)G

y(t, a) y ∗ (a)

 da.

By Assumption 1, the right hand side of V2 (u) is integrable. Using Lemma 19, it holds dV2 (u(t)) = βx ∗ v ∗ dt



∞ 0

 

y(t, 0) y(t, a)y ∗ (0) y(t, a) + ∗ + log da. g(a)σ (a) − ∗ y (a) y (0) y(t, 0)y ∗ (a)

Put as follows: V (u) = V1 (u) + V2 (u).

(21)

Then the derivative of V (u) along (17) is as follows:      x βv ∗ dV (u(t)) x∗ = dx∗ r − θ 1 + 2− ∗ − dt d x x    ∞ ∗ y(t, a) y(t, a) x − + log da + βx ∗ v ∗ g(a)σ (a) 2 − x βx ∗ vσ (a) βxvσ (a) 0 pm (z − z ∗ )2 . − q (22) The second term is nonpositive using some extension of inequality of arithmetic and geometric means [11]. Proposition 20 We assume R0 > 1. Let u(t) = (x(t), y(t, ·), v(t), z(t)) be a solution or an entire solution of (2) satisfying Assumption 1. Then the time derivative of V (u(t)) is nonpositive if it holds r > θ (1 + βv ∗ /d).

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3.2 The case R0 ≤ 1 If R0 ≤ 1, (x, 0, 0, 0) (x = λ/δ) is the unique equilibrium of the ordinary differential equation. Put U1 (u) by U1 (u) = (r − θ )(x − x log x) + v +

p 2 z . 2q

The time derivative of U1 (u) along (14) is as follows:   dU1 (u(t)) x x pm 2 = (r − θ )d x 2 − − + ((r − θ )βx − b) v − z . dt x x q The time derivative of U1 (u) along (17) is as follows: dU1 (u(t)) = ∇U1 (u) · f(u) + dt



∞

g(a)y(t, a) da − rβxv.

0

Put 

∞

α(a) =

g(ε)σ (ε) dε.

a

Then α (a) = −g(a)σ (a). Lemma 21 α(a)σ (a)−1 is bounded for a ≥ 0. Proof It holds that  ∞ ε g(ε)σ (ε) da = g(ε)e− a (d+μ(b)) db dε σ (a) a a  ∞ g∞ −d(ε−a) . ≤ g∞ e dε = d a

α(a) = σ (a)



∞

  For u = (x, y(·), v, z), put 

∞

U2 (u) =

α(a)y(a)σ (a)−1 da.

0

Then by Lemma 21, the right hand side is integrable. Using Lemma 19, it holds dU2 (u(t)) = dt



∞

g(a)σ (a)(y(t, 0) − y(t, a)σ (a)−1 ) da 0  ∞ g(a)y(t, a) da. = rβxv − 0

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Global stability of an age-structured model for…

Put U (u) = U1 (u) + U2 (u).

(23)

Then the time derivative of U (u) along the original age-structured equation (17) is as follows:   x dU (u(t)) x pm 2 (24) = (r − θ )d x 2 − − + ((r − θ )βx − b)v − z . dt x x q By the definition, R0 ≤ 1 is equivalent to (r − θ )βx − b ≤ 0. Proposition 22 We assume R0 ≤ 1. Let u(t) = (x(t), y(t, ·), v(t), z(t)) be a solution or an entire solution of (2) satisfying Assumption 1. Then the time derivative of U (u(t)) is nonpositive.

4 Proof of the global stability We describe the global dynamics of the age-structured equation in the section. It is necessary to state the statements and proofs for the case R0 > 1 and the case R0 ≤ 1 separately. 4.1 The case R0 > 1 We assume R0 > 1. For u = (x, y(·), v, z) ∈ X , and define a function ρ from X to R+ by  ρ(u) =

a

y(a) da + v.

0

Then ρ is uniformly continuous. Put X 0 = { u ∈ X | ρ(St (u)) = 0 ∀t ≥ 0 }. The subset X 0 of X is a forward invariant closed subset of X . Put X 0 = X \X 0 , which is called the interior of X . Lemma 23 It holds X 0 = { u ∈ X | ρ(u) = 0 }.  a Proof In Lemma 8, it is shown that if 0 y0 (a) da = 0 and v(0) = 0 then  a 0 y(t, a) da = 0 and v(t) = 0 for each t ≥ 0. Since it is shown in Sect. 2.3 that there exists a compact attractor which attracts each bounded subset of X , the following assumption in Chapter 8 of Smith and Thieme [21] holds. Assumption 2 There exists a subset B of X and a constant c > 0 satisfying the following: 1. For each u ∈ X , it holds St (u) → B (t → ∞). 2. The subset B ∩ { ρ ≤ c } of X has the compact closure in X .

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Definition 24 A subset M of X 0 is called uniformly weakly ρ-repelling if there exists an ε > 0 such that for each x ∈ X with ρ(x) > 0 it holds lim sup d(t (x), M) ≥ ε. t→∞

We define a subset  of X 0 by =



ω(x).

x∈X 0

The following lemma holds. Lemma 25 DFE is globally asymptotically stable in the boundary X 0 . Then the set  is { DFE }. Proof Since v(0) +

 a 0

y(0, a) da = 0, v(t) = 0 for t ≥ 0. Since it holds dx = λ − d x, dt

dz = −mz, dt

x(t) → λ/d and z(t) → 0 as t → ∞. If a  = ∞, y(t, ·) = 0 for t ≥ 0. For a  < ∞, it hold σ (a) y(0, a − t)1{a>t} , σ (a − t)

y(t, a) = then y(t, a) = 0 for t > a  .

 

The following lemma is proved in Lemma 3.5 in Browne and Pilyugin [3] for the case of integral equation. For the case of integral inequality, the proof is similarly done using Laplace transformation. Lemma 26 Let g(t) be a nonnegative continuous bounded function on [0, ∞), c be a positive constant with 

∞

g(t) dt > c.

0

If a continuous nonnegative function Y (t) with Y (0) > 0 satisfies the following integral inequality dY ≥ dt



then Y (t) must be unbounded.

123

0

t

g(a)Y (t − a) da − cY (t),

Global stability of an age-structured model for…

Proof Contrary we assume that Y (t) is bounded. Then we can define the Laplace transformation L[Y ](s) for s > 0. Then it holds  dY (s) ≥ L[g](s)L[Y ](s) − cL[Y ](s), L dt −Y (0) ≥ (L[g](s) − s − c)L[Y ](s). 

For small s > 0, the left hand side is negative and the right hand side is positive. A contradiction follows. Lemma 27 DFE is uniformly weakly ρ-repelling in X . Proof Since R0 > 1, it holds 

∞

βx

g(a)σ (a) da > b + θβx.

0

Taking sufficiently small δ > 0, the following can hold: 

∞

β(x − δ)

g(a)σ (a) da > b + θβ(x + δ) + pδ.

(25)

0

We fix such a δ > 0. We take (x0 , y0 (·), v0 , z 0 ) ∈ X 0 in the δ-neighborhood of ˜ = (x(t), ˜ y˜ (t, ·), v(t), ˜ z˜ (t)) with DFE (x, 0, 0, 0). We assume that the solution u(t) the initial value (x0 , y0 (·), v0 , z 0 ) stays in the δ-neighborhood of DFE (x, 0, 0, 0) for ˜ 0 ) > 0. We consider each t ≥ 0. By (3) in Lemma 8, there exists t0 > 0 such that v(t the solution u(t) = (x(t), y(t, ·), v(t), z(t)) with u(0) = u(t ˜ 0 ). The following holds: dv = dt

 

∞

g(a)y(t, a) da − θβxv − bv − pvz

0 t

≥

g(a)y(t, a) da − θβxv − bv − pvz

0



t

= 

βσ (a)g(a)x(t − a)v(t − a) da − θβxv − bv − pvz

0

≥

t

β(x − δ)σ (a)g(a)v(t − a) da − (θβ(x + δ) + b + pδ)v(t).

0

Then by Lemma 26, v(t) is unbounded for t ≥ 0, and contradiction holds. Then there exists a time t1 such that the solution u(t) ˜ escapes from the δ-neighborhood of DFE, and DFE is uniformly weakly ρ-repelling.   A neighborhood V of an equilibrium E is called isolated if each compact invariant subset of V is {E}. Lemma 28 DFE has an isolated neighborhood in X .

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Proof DFE is uniformly weakly ρ-repelling by Lemma 27 and DFE is asymptotically stable in X 0 by Lemma 25. Then it has an isolated neighborhood in X by Lemma 8.22 in Smith and Thieme [21].   Proposition 29 The semigroup {St }t≥0 is uniformly weakly ρ-persistent. Proof DFE has an isolated neighborhood and there exists no homoclinic trajectory through DFE. Then  has an acyclic covering. Then  is uniformly weakly ρpersistent by Theorem 8.17 of Smith and Thieme [21]. The following lemma is used later. We note that the uniqueness of the solution does not holds in general to backward direction. We argue as in pages 243–244 in Smith and Thieme [21]. Lemma 30 Let u(t) = (x(t), y(t, ·), v(t), z(t)) be an entire solution. Then it holds for each t ∈ R, y(t, a) = βσ (a)x(t − a)v(t − a) for a ≥ 0. Proof We fix an arbitrary r ∈ R. We note that u(r + t) = St (u(r )) for each t ≥ 0. Define xr (t) = x(t + r ),

yr (t, ·) = y(t + r, ·), vr (t) = v(t + r ), zr (t) = z(t + r ).

Then u r (t) = (xr (t), yr (t, ·), vr (t, ·), zr (t)) = St (u(r )). It holds that yr (0, ·) = y(r, ·). and u r (t) satisfies  xr (t) = xr (0) +

t

(λ − d x(s) − βxr (s)vr (s)) ds,

0

σ (a) yr (t, a) = βσ (a)vr (t − a)xr (t − a)1{t>a} + yr (0, a − t)1{a>t} , σ (a − t)  t ∞  t g(a)yr (s, a) da ds − (θβxr (s)vr (s) vr (t) = vr (0) + 0

0

(26)

0

+ bvr (s) + pvr (s)zr (s)) ds,  t zr (t) = zr (0) + (qvr (s) − mzr (s)) ds. 0

Then it holds y(t + r, a) = βσ (a)v(t + r − a)x(t + r − a)1{t>a} +

σ (a) y(r, a − t))1{a>t} . σ (a − t)

For s ≥ r , with t = s − r , y(s, a) = βσ (a)v(s − a)x(s − a)1{s−r >a} σ (a) y(r, a − (s − r ))1{a>s−r } . + σ (a − (s − r ))

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Then, by letting r → −∞, it holds that for each s ∈ R, y(s, a) = βσ (a)v(s − a)x(s − a), for a ≥ 0.   In the previous section, it is shown that there exists a compact attractor A which attracts each bounded subset of X . Moreover, the following lemma holds. Lemma 31 Let u(t) be an entire solution. If v(t0 ) = 0 for some t0 ∈ R, it holds that v(t) = 0 and y(t, ·) = 0 for each t ∈ R. Proof Let t2 < t1 . By (9), if v(t2 ) > 0 then v(t1 ) > 0. Then v(t) = 0 for t ≤ t0 . Then y(t, a) = βσ (a)x(t − a)v(t − a) is zero for t ≤ t0 and a ≥ 0. Then by Lemma 6,   v(t) = 0 and y(t, a) = 0 for each t ≥ t0 and a ≥ 0. Lemma 32 There does not exist an entire solution u(t) contained in A such that ρ(u(−r )) > 0, ρ(u(0)) = 0 and ρ(u(t)) > 0 for some r > 0 and t > 0. Proof It follows from Lemma 31. Theorem 33 The semigroup {St }t≥0 is uniformly ρ-persistent. Proof It holds X 0 = ∅ and ρ ◦ St is continuous. Assumption (H0) and (H1) in Chapter 5 in Smith and Thieme [21] are satisfied. Then the conclusion follows by Theorem 5.2 in Smith and Thieme [21].   Since the assumption (H1) is satisfied, by Theorem 5.7 in Smith and Thieme [21], there exists a persistence attractor in the interior X 0 of X , and the decomposition of the compact attractor in Proposition 17 holds. The following is a part of Theorem 5.7 in [21]. Proposition 34 ([21] Theorem 5.7 (b)) We assume X 0 = ∅ and {St }t≥0 is uniformly weakly ρ-persistent. Then the compact attractor A is decomposed into A = A0 ∪ C ∪ A1 . The set A1 is a compact invariant subset and uniformly ρ-positive. The set A1 attracts each neighborhood of compact subset of X 0 , and A1 attracts each subset of X 0 which is attracted by A and which is eventually uniformly ρ-positive. Proposition 35 Let A1 be the persistence attractor of a semigroup. If A1 reduces to a singleton set { x ∗ } where x ∗ is an equilibrium in X 0 , then the equilibrium x ∗ is locally stable, and globally asymptotically stable in X 0 . Proof We assume that A1 reduces to one point. The persistence attractor A1 attracts a neighborhood of a point contained in X 0 , and the proof of Proposition 18 does work for the case.   Proposition 36 There exists positive values ε > 0 and M > 0 such that for each entire solution u(t) = (x(t), y(t, ·), v(t), z(t)) in the persistence attractor A1 it holds ε ≤ x(t) ≤ M, ε ≤ v(t) ≤ M, ε ≤

y(t, a) ≤ M, z(t) ≤ M. y ∗ (a)

for each t ∈ R.

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Proof Let A1 be the persistence attractor. Then A1 is a compact subset of X 0 which is uniformly ρ-positive. Since A1 is compact, there exist points in A1 at which x, v and z have minimum values and maximum values. There exist M1 > 0, M2 > 0 and M3 > 0 such that x(t) ≤ M1 , v(t) ≤ M2 , z(t) ≤ M3 ∀t ∈ R. The x-component has a positive minimum value ε1 because the phase space excludes points whose x-components are zero. We assume that there exists u = (x, y(·), v, z) ∈ A1 with v = 0. Then there exists an entire solution u(t) ⊂ A1 with u(t0 ) = u. By Lemma 6, it holds v(t) = 0 and y(t, ·) = 0 for each t ∈ R. Then u ∈ X 0 and the contradiction occurs. It follows that the value of v-component of each entire solution contained in A1 must be greater than some positive constant ε2 . We consider y(t, ·). For an entire solution, by Lemma 30 the following Volterra formulation y(t, a) = βx(t − a)v(t − a)σ (a) holds for each a ≥ 0. Then it holds y(t, a) x(t − a)v(t − a) . = y ∗ (a) x ∗v∗ Then there exist ε3 > 0 and M4 with ε3 ≤

y(t, a) ≤ M4 y ∗ (a)

which do not depend on each entire solution. We put ε = min{ε1 /2, ε2 , ε3 }, M =   max{M1 , M2 , M3 , M4 }. Proposition 37 For an entire solution u(t) = (x(t), y(t, ·), v(t), z(t)) in the persistence attractor A1 , there exists ε > 0 such that for each t ∈ R it holds that z(t) ≥ ε . Proof By using the variation of the constant formula, we deform the equation of z(t). Let t ∈ R and r > 0. Then it holds z(t) = e−m(t+r ) z(−r ) + qe−mt



t −r

ems v(s) ds.

For each t, we take r with −r < t. By Proposition 36, v(s) ≥ ε for each s ∈ R. It holds  t −m(t+r ) −mt z(t) ≥ e z(−r ) + qe ems ε ds −r  qε  1 − e−m(t+r ) = e−m(t+r ) z(−r ) + m

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≥

 qε  1 − e−m(t+r ) . m

Since r > 0 can be arbitrarily large, it holds z(t) ≥

qε = ε . m  

Let u(t) = (x(t), v(t), z(t), y(t, ·)) be an entire solution in the persistence attractor A1 . Then, by Lemma 30, y(t, ·) is expressed as: y(t, a) = βx(t − a)v(t − a)σ (a)

for all a ≥ 0,

(27)

and the y(t, a) is differentiable with respect to t and a. By Proposition 36, u(t) satisfies the Assumption 1. We can substitute the entire solution u(t) into the Lyapunov functional (21). We use the following simple lemma. Lemma 38 Let f (t) be a continuous function of t ∈ R, and I = [c, d] be a nonempty closed interval in R+ . If f (t) = f (t − a) holds for each t ∈ R and a ∈ I , then f (t) is a constant function. Let u ∈ A and u(t) be an entire solution contained in A with u(0) = u. For a functional V on X , we put d V˙ (u) = V (u(t))|t=0 . dt Proposition 39 We assume R0 > 1. Let u(t) be an entire solution of the original age-structured equation (2) contained in the persistence attractor A1 . Then the entire solution satisfies Assumption 1. And if r > θ (1 + βv ∗ /d), the time derivative of V (u(t)) is nonpositive. The largest invariant subset of the set {u ∈ A1 | V˙ (u) = 0} is the singleton set { (x ∗ , y ∗ , v ∗ , z ∗ ) }. Proof The first part is already proved. Let u = (x, y(·), v, z) ∈ A1 and u(t) = (x(t), y(t, ·), v(t), z(t)) be an entire solution in A1 with u(0) = u. We calculate the time derivative using (20) as follows:      βv ∗ x(t) x∗ d V (u(t)) = d x ∗ r − θ 1 + 2− ∗ − dt d x x(t)    ∞ ∗ x(t − a)v(t − a) x(t − a)v(t − a) x − + log da g(a)σ (a) 2 − + βx ∗ v ∗ x(t) x ∗ v(t) x(t)v(t) 0 pm − (z(t) − z ∗ )2 ≤ 0, (28) q

where we used the notation x(t), v(t) instead of x, v in contrast to Sect. 3. We assume that u is contained in the largest invariant subset of the set {u ∈ A1 | V˙ (u) = 0}. Then

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the equation (28) holds identically for all t ∈ R. Then, it is shown that x(t) = x ∗ and z(t) = z ∗ for each t ∈ R, and v(t) is constant for each t ∈ R using Lemma 38. By the fourth equation of (2), v(t) = v ∗ for each t ∈ R. Then y(t, a) = βx ∗ v ∗ σ (a) = y ∗ (a) holds for each t ∈ R and a ≥ 0. Then the largest invariant subset of the set   {u ∈ A1 | V˙ (u) = 0} is {(x ∗ , y ∗ , v ∗ , z ∗ )}. Using it we can show the global stability of {(x ∗ , y ∗ , v ∗ , z ∗ )} for the case R0 > 1. Theorem 40 We assume R0 > 1. If r > θ (1 + βv ∗ /d), the disease equilibrium {(x ∗ , y ∗ , v ∗ , z ∗ )} is globally asymptotically stable in X 0 . DFE is globally asymptotically stable in X 0 . Proof Let u(t) = (x(t), y(t, ·), v(t), z(t)) be an entire solution contained in the persistence attractor A1 . Since the persistence attractor A1 is compact, the alpha limit set of the entire solution u(t) contained in A1 is not empty. Since z(t) is bounded from below for t → −∞, V (u(t)) is bounded as t → −∞, and the alpha limit set of the entire solution u(t) is contained in the largest invariant subset of the set {u ∈ A1 | V˙ (u) = 0}, which is {(x ∗ , y ∗ , v ∗ , z ∗ )}. Since the Lyapunov functional is non increasing and takes minimum value at {(x ∗ , y ∗ , v ∗ , z ∗ )}, the persistence attractor reduces to the singleton set {(x ∗ , y ∗ , v ∗ , z ∗ )}. Then by Proposition 35, (x ∗ , y ∗ , v ∗ , z ∗ ) is globally asymptotically stable in X 0 . The last statement follows from Lemma 25.   4.2 The case R0 ≤ 1 There exist a compact attractor A in X , which attracts all bounded subsets in X . We fix an entire solution in A. Proposition 41 We assume R0 ≤ 1. Let u(t) be an entire solution of the original age-structured equation (2). Then the entire solution u(t) satisfies Assumption 1, and the time derivative of U (u(t)) is non-positive. Moreover, the largest invariant subset of the set {u ∈ A | U˙ (u) = 0} is singleton set { (x, 0, 0, 0) }. Proof Let u = (x, y(·), v, z) ∈ A, and u(t) = (x(t), y(t, ·), v(t), z(t)) be an entire solution in A with u(0) = u. By Lemma 30, u(t) satisfies Assumption 1 in Sect. 3. It holds that   x dU (u(t)) x(t) pm = (r − θ )d x 2 − − + ((r − θ )βx − b)v(t) − z(t)2 , dt x x(t) q where we used the same notation as in (28). If u is contained in the largest invariant subset of the set {u ∈ A | U˙ (u) = 0 }, then x(t) = x, z(t) = 0 for each t ∈ R by −(r − θ ) < 0, (r − θ )βx − b ≤ 0 and − pm/q < 0. Using the first equation of (2), v(t) = 0 for t ∈ R. Then it follows y(t, ·) = 0 for t ∈ R. The conclusion holds.   Remark 4.1 We note that the proof of Proposition 41 does work for R0 ≤ 1, especially for R0 = 1. We do not use characteristic equations for equilibria.

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Theorem 42 If R0 ≤ 1, DFE is globally asymptotically stable in X . Proof Let u(t) = (x(t), y(t, ·), v(t), z(t)) be an entire solution contained in the compact attractor A. As in the proof of Proposition 40, the alpha limit set of the entire solution u(t) is not empty and contained in the largest invariant subset of the set {u ∈ A1 | V˙ (u) = 0}, which is {(x, 0, 0, 0)}. Then as in the proof of Proposition 40, the compact attractor A reduces to a singleton set { (x, 0, 0, 0))}. Then by Proposi  tion 18, { (x, 0, 0, 0))} is globally asymptotically stable. Acknowledgements The authors would like to thank the anonymous referees for very helpful suggestions and comments which led to improvement of our original manuscripts. This work is partly supported by Grand-in-Aid Scientific Research (C) No. 17K05365 from Japan Society for the Promotion of Science.

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