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ECA-TFUnet: A U-shaped CNN-Transformer network with efficient channel attention for organ segmentation in anatomical sectional images of canines


  • Automated organ segmentation in anatomical sectional images of canines is crucial for clinical applications and the study of sectional anatomy. The manual delineation of organ boundaries by experts is a time-consuming and laborious task. However, semi-automatic segmentation methods have shown low segmentation accuracy. Deep learning-based CNN models lack the ability to establish long-range dependencies, leading to limited segmentation performance. Although Transformer-based models excel at establishing long-range dependencies, they face a limitation in capturing local detail information. To address these challenges, we propose a novel ECA-TFUnet model for organ segmentation in anatomical sectional images of canines. ECA-TFUnet model is a U-shaped CNN-Transformer network with Efficient Channel Attention, which fully combines the strengths of the Unet network and Transformer block. Specifically, The U-Net network is excellent at capturing detailed local information. The Transformer block is equipped in the first skip connection layer of the Unet network to effectively learn the global dependencies of different regions, which improves the representation ability of the model. Additionally, the Efficient Channel Attention Block is introduced to the Unet network to focus on more important channel information, further improving the robustness of the model. Furthermore, the mixed loss strategy is incorporated to alleviate the problem of class imbalance. Experimental results showed that the ECA-TFUnet model yielded 92.63% IoU, outperforming 11 state-of-the-art methods. To comprehensively evaluate the model performance, we also conducted experiments on a public dataset, which achieved 87.93% IoU, still superior to 11 state-of-the-art methods. Finally, we explored the use of a transfer learning strategy to provide good initialization parameters for the ECA-TFUnet model. We demonstrated that the ECA-TFUnet model exhibits superior segmentation performance on anatomical sectional images of canines, which has the potential for application in medical clinical diagnosis.

    Citation: Yunling Liu, Yaxiong Liu, Jingsong Li, Yaoxing Chen, Fengjuan Xu, Yifa Xu, Jing Cao, Yuntao Ma. ECA-TFUnet: A U-shaped CNN-Transformer network with efficient channel attention for organ segmentation in anatomical sectional images of canines[J]. Mathematical Biosciences and Engineering, 2023, 20(10): 18650-18669. doi: 10.3934/mbe.2023827

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  • Automated organ segmentation in anatomical sectional images of canines is crucial for clinical applications and the study of sectional anatomy. The manual delineation of organ boundaries by experts is a time-consuming and laborious task. However, semi-automatic segmentation methods have shown low segmentation accuracy. Deep learning-based CNN models lack the ability to establish long-range dependencies, leading to limited segmentation performance. Although Transformer-based models excel at establishing long-range dependencies, they face a limitation in capturing local detail information. To address these challenges, we propose a novel ECA-TFUnet model for organ segmentation in anatomical sectional images of canines. ECA-TFUnet model is a U-shaped CNN-Transformer network with Efficient Channel Attention, which fully combines the strengths of the Unet network and Transformer block. Specifically, The U-Net network is excellent at capturing detailed local information. The Transformer block is equipped in the first skip connection layer of the Unet network to effectively learn the global dependencies of different regions, which improves the representation ability of the model. Additionally, the Efficient Channel Attention Block is introduced to the Unet network to focus on more important channel information, further improving the robustness of the model. Furthermore, the mixed loss strategy is incorporated to alleviate the problem of class imbalance. Experimental results showed that the ECA-TFUnet model yielded 92.63% IoU, outperforming 11 state-of-the-art methods. To comprehensively evaluate the model performance, we also conducted experiments on a public dataset, which achieved 87.93% IoU, still superior to 11 state-of-the-art methods. Finally, we explored the use of a transfer learning strategy to provide good initialization parameters for the ECA-TFUnet model. We demonstrated that the ECA-TFUnet model exhibits superior segmentation performance on anatomical sectional images of canines, which has the potential for application in medical clinical diagnosis.



    Mathematical models for the spread of epidemic infectious diseases in populations have been studied for a long time [1]. One of the most classical epidemic models is the SIR epidemic model in which the total population is divided into three classes called susceptible, infected and removed [2]. Some types of SIR epidemic models without age structure are nonlinear systems of ordinary differential equations, and it is relatively easy to show that the long time behavior of its solution is completely determined by a threshold value R0: if R0<1, then the trivial equilibrium in which there is no infected individual (the disease-free equilibrium) is globally asymptotically stable, whereas if R0>1, then the nontrivial equilibrium in which the infected individuals persist (the endemic equilibrium) is globally asymptotically stable [3,4]. The threshold value R0 is called the basic reproduction number and it plays an important role as an indicator of the intensity of diseases since it implies the expected number of secondary cases produced by a typical infected individual during its entire period of infectiousness in a fully susceptible population [3,4].

    Some types of SIR epidemic models with age structure are nonlinear systems of partial differential equations, and the mathematical analysis for them is generally more difficult than that for the models without age structure. The most classical SIR epidemic model studied by Kermack and McKendrick [2] has the structure of infection age (time elapsed since the infection), and the complete global stability analysis for an infection age-structured SIR epidemic model was recently done by Magal et al, [5]. That is, they showed that the disease-free equilibrium in their model is globally asymptotically stable if R01, whereas the endemic equilibrium is so if R0>1. Their method of constructing a suitable Lyapunov function in a compact global attractor has been applied for the analysis of various epidemic models with age structure [6,7,8,9,10]. Recently, the global behavior of various kinds of age-structured models in biology has been studied [11,12,13].

    In basic epidemic models, the incidence rate is often assumed to take the bilinear form such as βSI, where β>0 is the disease-transmission coefficient and S and I are the populations of susceptible and infected individuals, respectively. As a generalization, Capasso and Serio [14] studied the incidence rate with the form Sg(I), where g() is a continuous bounded function that takes into account either the saturation phenomenon or the psychological effects. After their work, epidemic models with more general incidence rates such as f(S)g(I) and f(S,I) have been studied by many authors [15,16,17,18,19,20,21,22,23,24,25,26].

    In [27], Bentout and Touaoula established an infection age-structured SIR epidemic model with a general incidence rate. They proved for their model that if R01, then the disease-free equilibrium is globally asymptotically stable, whereas if R0>1, then the endemic equilibrium is globally asymptotically stable. The purpose of this study is to generalize their results for a new epidemic model. Specifically, we assume for our model that the removed individuals do not stay permanently in the removed class and can return to the infected class. This kind of assumption can be seen in models for diseases with relapse [7,10,25,26,28], drug epidemics [15,29,31,30,32,33] and (although it is a fiction) zombie epidemics [34]. Because of the complexity in the model structure, we need more advanced techniques for the mathematical analysis of our model. In this paper, we obtain the basic reproduction number R0 for our model and show that the disease-free equilibrium is globally asymptotically stable if R0<1, whereas the endemic equilibrium uniquely exists and it is globally asymptotically stable if R0>1.

    The organization of this paper is as follows. In Section 2, we establish our main model. In Section 3, we define the basic reproduction number R0 and show that the disease-free equilibrium is globally asymptotically stable if R0<1. In Section 4, we define the semiflow generated by the solution of our model and show the existence of a compact attractor. In Section 5, we prove the existence and uniqueness of the endemic equilibrium, uniform persistence of the system and global asymptotic stability of the endemic equilibrium for R0>1. In Section 6, we perform numerical simulations that illustrate our theoretical results. Finally, Section 7 is devoted to the summary.

    Let S(t), i(t,a) and R(t) denote the population of susceptible, infected with infection age a and removed individuals at time t, respectively. The function θ(a) is the age-dependent per-capita removal rate of infected individuals with age a. The parameters μ, δ and k are respectively the natural death rate, the relapse rate and the fraction at which removed individuals directly return to the infected class. The parameter A represents the entering flux into the class S. The main model of this paper is formulated as the following system with infection age and with a general class class of nonlinear incidence rate, for t>0,

    {dS(t)dt=AμS(t)f(S(t),J(t)),i(t,a)t+i(t,a)a=(μ+θ(a))i(t,a),a>0,i(t,0)=f(S(t),J(t))+k+0θ(a)i(t,a)da+δR(t),dR(t)dt=(1k)+0θ(a)i(t,a)da(μ+δ)R(t), (2.1)

    where J is the contact function given by,

    J(t)=+0β(a)i(t,a)da,t>0.

    The system (2.1) is completed by the following boundary and initial conditions,

    {i(0,)=i0()L1(R+,R+),S(0)=S0R+andR(0)=r0R+. (2.2)

    For instance, diseases with relapse such as herpes simplex virus type 2 (HSV-2) [28] can be modeled by system (2.1). Throughout this paper, we make the following hypotheses on β and θ.

    ● The function βCBU(R+,R+) (where CBU(R+,R+) is the set of all bounded and uniformly continuous function from R+ to R+) and the function θL+(R+) (where L+(R+) is the set of all essentially bounded functions from R+ to R+).

    ● The parameters A and μ are positive and k[0,1).

    The boundedness of β implies that there is a maximum of infectivity of infected individuals. The uniform continuity of β is sufficient for the practical application since any continuous functions with a closed bounded support is uniformly continuous (see Section 6). For the above system, we make also the following assumptions on the infection response function f.

    ● (H0) The function f(,J) is increasing for J>0 and f(S,) is increasing for S>0 with f(0,J)=f(S,0)=0 for all S,J0.

    ● (H1) For all S>0 the function f(S,J) is concave with respect to J.

    ● (H2) The function f(,0)/J is continuous positive on every compact set KR+.

    ● (H3) The function f is locally Lipschitz continuous in S and J, that is, for every C>0 there exists some L:=LC>0 such that

    |f(S2,J2)f(S1,J1)|L(|S2S1|+|J2J1|), (2.3)

    whenever 0S2,S1,J2,J1C.

    Now, let us define the functional space X:=R×L1(R+)×R equipped with the norm

    (S,i,R)X=|S|++0|i(a)|da+|R|,S,RR,iL1(R+).

    We put X+=R+×L1+(R+)×R+, which denotes the positive cone of X. The problem of existence, positivity and uniqueness of the solution is classical and can be treated by using the Banach-Picard fixed point theorem in an appropriate Banach space on X (see, for instance, [27], [39]). Then, we have the following theorem.

    Theorem 2.1. Let consider an initial condition belonging to X+. Then, the system (2.1) and (2.2) has a unique nonnegative solution (S,i,R)C1(R+)×C(R+,L1(R+))×C1(R+).

    We omit the proof of Theorem 2.1 as it is similar to the proof of [27,Theorem 2.2] except for a simple modification. We set, N(t)=S(t)++0i(t,a)da+R(t). Then, from the system (2.1), we have, for t>0,

    N(t)=AμN(t).

    So, for t>0, we obtain

    N(t)max{N(0),Aμ},

    with N(0)=S0++0i0(a)da+r0, and

    limt+N(t)=Aμ. (2.4)

    On the other hand, we can check that (L is the Lipschitz constant defined in (2.3)),

    lim inft+S(t)Aμ+L.

    In this section, we show the local and the global stability of the disease-free equilibrium. First, we begin by studying the local stability. We denote by

    π(a)=ea0(μ+θ(s))ds,a0. (3.1)

    We can estimate the basic reproduction number by renewal process, which is the spectral radius of the next generation matrix. For more details, we refer the reader to [3]. The basic reproduction number R0 for our model (2.1) is given by

    R0=fJ(Aμ,0)+0β(a)π(a)da+((1k)δμ+δ+k)+0θ(a)π(a)da.

    For the system (2.1), it is easy to see that the disease-free steady state always exists and it is given by E0=(A/μ,0,0)X+. The linearization of system (2.1) near the equilibrium E0=(A/μ,0,0) is given by, for t>0,

    {dS(t)dt=μS(t)S(t)fS(Aμ,0)J(t)fJ(Aμ,0),i(t,a)t+i(t,a)a=(μ+θ(a))i(t,a),a>0,i(t,0)=S(t)fS(Aμ,0)+J(t)fJ(Aμ,0)+k+0θ(a)i(t,a)da+δR(t),dR(t)dt=(1k)+0θ(a)i(t,a)da(μ+δ)R(t). (3.2)

    Next, the characteristic equation of (3.2) at E0 is given by the following equation,

    |λ+μ+fS(Aμ,0)P(λ)0fS(Aμ,0)Q(λ)δ0G(λ)λ+μ+δ|=0,

    where

    P(λ):=fJ(Aμ,0)+0β(a)π(a)eλada,
    Q(λ):=1fJ(Aμ,0)+0β(a)π(a)eλadak+0θ(a)π(a)eλada,

    and

    G(λ):=(1k)+0θ(a)π(a)eλada.

    We have the following theorem.

    Theorem 3.1. If R0<1, then the disease-free equilibrium E0=(A/μ,0,0) is locally asymptotically stable and unstable for R0>1.

    Proof. The characteristic equation of the disease free equilibrium E0 is given by

    (λ+μ)H(λ)=0, (3.3)

    where

    H(λ)=(λ+μ+δ)(1fJ(Aμ,0)+0β(a)π(a)eλadak+0θ(a)π(a)eλada)δ(1k)+0θ(a)π(a)eλada.

    Obviously, we can see that μ is a negative real root of (3.3) and all other roots are obtained from H(λ)=0. We assume first that R0<1. We claim that H(λ)=0 has no root with nonnegative real part. By contradiction, we suppose that there exists λ0C with Re(λ0)0 such that H(λ0)=0, then,

    |λ0+μ+δ|=|δ(1k)+0θ(a)π(a)eλ0ada1+0β(a)π(a)eλ0adafJ(Aμ,0)k+0θ(a)π(a)eλ0ada|.

    Since Re(λ0)0 and R0<1, so

    μ+δ|δ(1k)+0θ(a)π(a)eλ0ada||1+0β(a)π(a)eλ0adafJ(Aμ,0)k+0θ(a)π(a)eλ0ada|.

    Then,

    δ(1k)μ+δ|+0θ(a)π(a)eλ0ada|1|+0β(a)π(a)eλ0adafJ(Aμ,0)k+0θ(a)π(a)eλ0ada|,1|+0β(a)π(a)eλ0adafJ(Aμ,0)|k|+0θ(a)π(a)eλ0ada|.

    Therefore,

    R0|+0β(a)π(a)eλ0adafJ(Aμ,0)|+(k+δ(1k)μ+δ)|+0θ(a)π(a)eλ0ada|1.

    We obtain that R01 and which is a contradiction with the hypothesis. Consequently, the disease-free equilibrium E0 is locally asymptotically stable for R0<1. Next, we suppose that R0>1. Then, for λR, we have

    H(0)=(μ+δ)(1R0)<0andlimλ+H(λ)=+.

    This ensures to the existence of a positive real root of (3.3). Hence, E0 is unstable if R0>1.

    Now, we focus on the global stability of E0. Let ˆi(h):=i(t+h,a+h) and ˆθ(h):=θ(a+h). By the second equation of (2.1), we have

    dˆi(h)dh=[μ+ˆθ(h)]ˆi(h).

    Hence, ˆi(h)=ˆi(0)eh0(μ+ˆθ(s))ds and thus,

    i(t+h,a+h)=i(t,a)eh0(μ+θ(a+s))ds.

    Considering two cases a<t and at, we have

    i(t,a)={i(ta,0)ea0(μ+θ(s))ds,a<t,i(0,at)et0(μ+θ(at+s))ds,at.

    Hence, we obtain

    i(t,a)={π(a)B(ta),a<t,π(a)π(at)i0(at),at, (3.4)

    with B(t):=i(t,0). Using (3.4), we obtain

    B(t)=f(S(t),J(t))+kt0θ(a)π(a)B(ta)da+tθ(a)π(a)π(at)i0(at)da+δR(t), (3.5)
    J(t)=t0β(a)π(a)B(ta)da+tβ(a)π(a)π(at)i0(at)da, (3.6)

    and

    R(t)=(1k)t0θ(a)π(a)B(ta)da+(1k)tθ(a)π(a)π(at)i0(at)da(μ+δ)R(t). (3.7)

    We prove the following theorem.

    Theorem 3.2. The disease-free equilibrium E0 is globally asymptotically stable whenever R0<1.

    Proof. It suffices to prove the global attractivity of E0 because of the local stability of this disease-free equilibrium. We set

    lim supt+(S(t),B(t),R(t))=(S,B,R)andlim supt+J(t)=J.

    By using the fluctuation lemma (see [37], Lemma A.14), there exist tn,sn,rn tend to infinity as n tends to infinity such that S(tn)S, S(tn)0, B(sn)B and R(rn)R and R(rn)0. Next substituting rn in (3.7), and letting n tends to infinity we find,

    R(1k)+0θ(a)π(a)daμ+δB. (3.8)

    Similarly using (3.5), (3.6) and the hypothesis (H0), we obtain

    {Bf(S,J)+kB+0θ(a)π(a)da+δR,JB+0β(a)π(a)da. (3.9)

    Now, combining (3.8), (3.9) and the fact that f(S,J)f(A/μ,J), we get

    Bf(S,J)+kB+0θ(a)π(a)da+δ(1k)+0θ(a)π(a)daμ+δB.

    Since SA/μ and f is an increasing function with respect to S, then

    Bf(Aμ,J)+kB+0θ(a)π(a)da+δ(1k)+0θ(a)π(a)daμ+δB. (3.10)

    Next, from (H1), we easily obtain

    f(Aμ,J)JfJ(Aμ,0). (3.11)

    By combining (3.10) with (3.11) and the second equation of (3.9), we get

    BR0B.

    Since R0<1, then necessarily B=0. This implies that J=R=0. Hence, f(S(t),J(t))0 as t+ and the asymptotic behavior of S obyes the equation dS(t)/dt=AμS(t). This implies that S(t)A/μ as t+. The theorem is proved.

    In this section, we prove the existence of a compact attractor of all bounded subset of X+. First, we define the semiflow Φ:R+×X+X+ such that

    Φ(t,(S0,i0(),r0))=(S(t),i(t,),R(t)),(S0,i0(),r0)X+, (4.1)

    which is generated by the unique solution of system (2.1). So, it is not difficult to show that this semiflow is continuous.

    Theorem 4.1. The semiflow Φ has a compact attractor A of bounded subsets of X+.

    Proof. Following Theorem 2.33 in [37], we need to check some properties of the semiflow Φ, namely, the point-dissipativity, eventually bounded on bounded sets on X and asymptotically smoothness properties. By theorem 2.1, we can show that the first two properties are satisfied. Now, we apply Theorem 2.46 of [37] to show the asymptotic smoothness. For this, we define

    Ψ1(t,(S0,i0(),r0))=(0,u(t,),0) and Ψ2(t,(S0,i0(),r0))=(S(t),v(t,),R(t)),

    where

    u(t,a)={0,a<t,π(a)π(at)i0(at),a>t,

    and

    v(t,a)={π(a)B(ta),a<t,0,a>t.

    Let C be a bounded closed subset of initial data in X+, that is forward invariant under Φ. We set

    M1:=sup{S0+i0L1+r0,(S0,i0,r0)C} and M2:=max{M1,Aμ}. (4.2)

    Using the same arguments as in Theorem 2.1 in [18], we have Ψ10 as t+ uniformly for all initial data in C. Next, we have to prove that Ψ2 has a compact closure. We need only to check the condition (ⅲ) of Theorem B.2 in [37] or Theorem 3.1 in [27] because the other conditions are readily checked. Thus,

    +0|v(t,a+h)v(t,a)|da=th0|π(a+h)B(tah)π(a)B(ta)|da+tth|π(a)B(ta)|da. (4.3)

    Notice that, for t0, we have

    |B(t)|f(S(t),J(t))+k+0θ(a)i(t,a)da+δR(t),f(M2,βM2)+kθM2+δM2, (4.4)

    for all initial data in C. Then, the second term of (4.3) tends to 0 when h0 uniformly in C. We set

    Ih(t):=th0|π(a+h)B(tah)π(a)B(ta)|da,th0|π(a)(B(tah)B(ta))|da+th0|B(tah)(π(a+h)π(a)|da. (4.5)

    From the system (2.1) and (4.2), we have, for t0,

    |S(t)|A+μM2+f(M2,βM2), (4.6)

    and

    |R(t)|(1k)θM2+(μ+δ)M2. (4.7)

    Now, let us define Ah for h>0 as

    Ah(t,a):=|B(tah)B(ta)||f(S(tah),J(tah))f(S(ta),J(ta))|+k|+0θ(σ)i(tah,σ)dσ+0θ(σ)i(ta,σ)dσ|+δ|R(tah)R(ta)|.

    Using the fact that the function f is Lipschitz, we get

    AhL|S(tah)S(ta)|+L|J(tah)J(ta))|+k|+0θ(σ)i(tah,σ)dσ+0θ(σ)i(ta,σ)dσ|+δ|R(tah)R(ta)|.

    By (4.6) and (4.7), we can easily show that the first and the last term of Ah tend to 0 as h tends to 0+ uniformly for all initial data in C. Now we focus on the second term of Ah. Denote by,

    A1h(t,a):=L|J(tah)J(ta)|+k|+0θ(σ)i(tah,σ)dσ+0θ(σ)i(ta,σ)dσ|. (4.8)

    For t0, we set J(t)=J1(t)+J2(t), where

    J1(t)=t0β(a)π(a)f(S(ta),J(ta))daandJ2(t)=+0β(a+t)π(a+t)π(a)i0(a)da. (4.9)

    Thus, for s,h0,

    |J(s+h)J(s)||J1(s+h)J1(s)|+|J2(s+h)J2(s)|.

    After a change of variable in (4.9), we obtain

    J1(t)=t0β(tσ)π(tσ)f(S(σ),J(σ))dσ.

    Therefore, for s,h0,

    |J1(s+h)J1(s)|s+hsβ(s+hσ)π(s+hσ)f(S(σ),J(σ))dσ+s0|β(s+hσ)π(s+hσ)β(sσ)π(sσ)|f(S(σ),J(σ))dσ,βf(M2,βM2)h+f(M2,βM2)s0|β(s+hσ)π(s+hσ)β(sσ)π(sσ)|dσ.

    Consequently, we can readily checked that these last terms tend to 0 as h0+ uniformly for all s0 and for all initial data in C. Using the definition of J2 in (4.9), we get

    |J2(s+h)J2(s)||+0β(a+s+h)π(a+s+h)π(a)i0(a)da+0β(a+s)π(a+s)π(a)i0(a)da|,+0|β(a+s+h)β(a+s)|π(a+s)π(a)i0(a)da++0β(a+s+h)i0(a)|π(a+s+h)π(a)π(a+s)π(a)|da.

    Using the fact that β is uniformly continuous and using (4.2), we can see that all above terms go to 0 as h0+ uniformly for all s0 and for all initial data on C. Finally, employing the same arguments as above, we show that,

    |+0θ(σ)i(tah,σ)dσ+0θ(σ)i(ta,σ)dσ|0,

    as h0+ uniformly for all initial data in C. The theorem is established.

    Next, we describe the total trajectories of system (2.1). Let ϕ:RX+ be the total Φ-trajectories, ϕ(t)=(S(t),i(t,),R(t)). Then, ϕ(t+r)=Φ(t,ϕ(r)),t0,rR. By using the same arguments as in [37], we have, for all tR,

    {S(t)=AμSf(S(t),J(t)),i(t,a)=π(a)B(ta),B(t)=f(S(t),J(t))+k+0θ(a)i(t,a)da+δR(t),R(t)=(1k)+0θ(a)i(t,a)da(μ+δ)R(t),J(t)=+0β(a)π(a)B(ta)da. (4.10)

    The following lemma gives some estimates on the total trajectories.

    Lemma 4.2. For all (S0,i0(),r0)A, we have the following estimates, for all tR

    S(t)>Aμ+L,S(t)++0i(t,a)da+R(t)Aμandi(t,a)ξπ(a),a0.

    with ξ:=(Lβ+kθ+δ)(A/μ) and L is Lipschitz constant of f defined in (2.3).

    Proof. We set,

    I(t):=+0i(t,a)da=+0π(a)B(ta)da.

    After a change of variable, for tR,

    I(t)=tπ(ta)B(a)da.

    By differentiating I, we get

    I(t)=B(t)+0(μ+θ(a))π(a)B(ta)da.

    Combining this equation with the system (4.10), we obtain,

    S(t)+I(t)+R(t)Aμ(S(t)+I(t)+R(t)),

    Thus, for tR

    S(t)+I(t)+R(t)Aμ. (4.11)

    Moreover, from the equation of S in (4.10) and the hypothesis (H3), we have

    S(t)AμSLS(t),

    so,

    S(t)Aμ+L,tR.

    On the other hand, using the fact that f(S,0)=0 and (H3), we find

    i(t,a)=π(a)(f(S(t),J(t))+k+0θ(a)i(t,a)da+δR(t)),π(a)(LJ(t)+kθI(t)+δR(t)),

    According to (4.11), we conclude that

    i(t,a)ξπ(a),a>0,

    where ξ:=(Lβ+kθ+δ)(A/μ).

    The main purpose of this section is to study the global asymptotic stability of the endemic equilibrium. We first need to prove the strong uniform persistence of the solution of problem (4.10).

    We begin by the following lemma which concern the existence of a positive equilibria.

    Lemma 5.1. Assume that

    limJ0+f(A/μ,J)f(S,J)>1,forS[0,A/μ).

    If R0>1, then the system (2.1) has a positive steady state.

    Proof. Let E:=(S,i(.),R) be a positive fixed point of the semiflow Φ. Then,

    Φ(t,(S,i(.),R))=(S,i(.),R), for t0.

    From (4.1) and (3.4), we have

    i(a)={π(a)i(0),0<a<t,π(a)π(at)i(at),a>t, (5.1)

    and

    {A=μS+f(S,J),J=+0β(a)i(a)da. (5.2)

    Remark that if we consider i(a) from the first expression of (5.1), it satisfies also the second one of (5.1). Indeed, for t<a<2t, we have

    i(at)=π(at)i(0),=π(at)π(a)i(a),

    and thus

    i(a)=π(a)π(at)i(at),=π(a)π(at)π(at)i(0),=π(a)i(0).

    We can proceed by iteration in order to prove the result. Therefore,

    i(a)=π(a)i(0),for alla0. (5.3)

    Combining (5.2) and (5.3), we get

    i(0)=1Df(S,J),

    with

    D=1(k+δ(1k)μ+δ)+0θ(a)π(a)da, (5.4)

    thus,

    i(a)=1Df(S,J)π(a),a0. (5.5)

    Moreover, from (5.2) and (5.5), we obtain

    {A=μS+f(S,J),J=MDf(S,J), (5.6)

    where M:=+0β(a)π(a)da and D is defined in (5.4). Following the same arguments as in [22] and [23], we prove the existence of positive steady state. The following lemma is readily to prove it

    Lemma 5.2. Under (H0) and (H1), we have following assertions,

    f(.,J)Jis a nonincreasing function with respect toJ, (5.7)

    and

    {xJ<f(S,x)f(S,J)<1,for0<x<J,1<f(S,x)f(S,J)<xJ,forx>J. (5.8)

    Now we focus on the uniform persistence of the solution of problem (4.10). We define,

    ˉθ(a):=+0θ(a+t)π(a+t)π(a)dt,

    and

    ˉβ(a):=+0β(a+t)π(a+t)π(a)dt.

    We set

    X0={(S0,i0(),r0)X+|r0+0i0(a)ˉθ(a)da+0i0(a)ˉβ(a)da>0}.

    Lemma 5.3. If (S0,i0(),r0)X0, then there exists t0>0 such that B(t):=i(t,0)>0 for all t>t0. If (S0,i0(),r0)X+X0, then B(t)=0 for all t0.

    Proof. Recall that S(t)>0 for all t>0 and that

    B(t)=f(S(t),J(t))+k+0θ(a)i(t,a)da+δR(t).

    The third equation of system (2.1) implies that, for t0,

    R(t)=r0e(μ+δ)t+(1k)t0e(μ+δ)(st)+0θ(σ)i(s,σ)dσds.

    We can rewrite B as,

    B(t)=f(S(t),J(t))+k+0θ(a)i(t,a)da+δr0e(μ+δ)t+(1k)δt0e(μ+δ)(st)+0θ(σ)i(s,σ)dσds.

    So, from (3.4),

    B(t)=f(S(t),J(t))+kt0θ(a)π(a)B(ta)da+k+tθ(a)π(a)π(at)i0(at)da+δr0e(μ+δ)t+(1k)δt0e(μ+δ)(st)(s0θ(σ)π(σ)B(sσ)dσ++sθ(σ)π(σ)π(σs)i0(σs)dσ)ds. (5.9)

    If r0>0 then we have B(t)>0 for all t0. Next, we suppose that +0i0(a)ˉθ(a)da>0, thus we set,

    I0(t):=+tθ(a)π(a)π(at)i0(at)da=+0θ(a+t)π(a+t)π(a)i0(a)da. (5.10)

    By integrating I0 it yields,

    +0I0(t)dt=+0i0(a)+0θ(a+t)π(a+t)π(a)dtda=+0i0(a)ˉθ(a)da.

    In the following, we will use translations of solutions: for r0, we define Sr(t)=S(t+r) Jr(t)=J(t+r), I0r(t)=I0(t+r) and Br(t)=B(t+r). By (5.9), (5.10),

    Br(t)kt0θ(a)π(a)Br(ta)da+kI0r(t).

    Since +0I0(t)dt>0 then for some small r>0, I0ris not zero almost everywhere on [r,+). Consequently, from Corollary B.6 in [37], there exists t0>0 such that B(t)>0 for all t>t0. Using the same arguments as above in the case where +0i0(a)ˉβ(a)da>0.

    Now, if (S0,i0(),r0)X+X0, then according to (3.4), (5.9), we have

    B(t)=f(S(t),J(t))+kt0θ(a)π(a)B(ta)da+k+tθ(a)π(a)π(at)i0(at)da+(1k)δt0e(μ+δ)(st)(s0θ(σ)π(σ)B(sσ)dσ++sθ(σ)π(σ)π(σs)i0(σs)dσ)ds. (5.11)

    From the definition of the space X+X0 and the fact that the function f is locally Lipschitz and that f(S,0)=0 for all S>0, the equation (5.11) yields,

    B(t)Lt0β(a)π(a)B(ta)da+kt0θ(a)π(a)B(ta)da+(1k)δt0e(μ+δ)(st)s0θ(σ)π(σ)B(sσ)dσds.

    We can apply the Fubini's Theorem to the last term and using the assumptions on β and θ, we get

    B(t)(Lβ+(k+(1k)δμ+δ)θ)t0B(a)da.

    By Gronwall's inequality, we obtain B(t)=0 for all t0.

    We define the persistence function ρ:X+R+ by

    ρ(S0,i0(),r0)=f(S0,J0)++0θ(a)i0(a)da+r0.

    By definition, we have,

    ρ(Φ(t,(S0,i0(),r0)))=B(t),

    Lemma 5.4. Assume that (5.7) holds. If xX0, then there exists some ϵ>0 such that

    lim supt+ρ(Φ(t,x))>ϵ,

    for all solutions of (2.1) provided that R0>1.

    Proof. We suppose that the function ρ is not uniformly weakly persistent, so, there exists an arbitrarily small ϵ>0 such that

    lim supt+ρ(Φ(t,x))<ϵ.

    By Theorem 2.1, we have S(t)>0 for all t>0, so, from the above expression there exists some small ϵ0>0 such that

    lim supt+J(t)<ϵ0.

    Let S=lim inft+S(t). Then, there exists a sequence tk+ such that S(tk)S and S(tk)0 as tk+ see ([37], Lemma A.14). Combining this with the continuity of the function f, we have (for large t)

    0AμSϵ.

    Therefore,

    SAμψ(ϵ),

    with ψ(ϵ):=ϵ/μ. By using the fact that R0>1, then we can choose a small ϵ1>0 such that

    h(ϵ1)=f(Aμψ(ϵ1),ϵ1)ϵ1+0β(a)π(a)da+(k+(1k)δϵ1+μ+δ)+0θ(a)π(a)da>1. (5.12)

    From system (2.1), we have, for all t0,

    R(t)=(1k)+0θ(a)i(t,a)da(μ+δ)R(t).

    Therefore, for t0,

    R(t)(1k)t0θ(a)π(a)B(ta)da(μ+δ)R(t).

    Next, we introduce the Laplace transform to this inequality, which is given by, for λ>0,

    λˆR(λ)R(0)(1k)ˆθ(λ)ˆB(λ)(μ+δ)ˆR(λ),

    Thus, we get

    ˆR(λ)(1k)ˆθ(λ)λ+μ+δˆB(λ). (5.13)

    where ˆB and ˆθ are respectively the Laplace transforms of B and θ, defined by, for λ>0,

    ˆB(λ)=+0B(a)eλadaandˆθ(λ)=+0θ(a)π(a)eλada.

    Moreover, since there exists T>0 such that J(t)<ϵ0 for all tT, then by (5.7), we obtain

    f(S,J)Jf(Aμψ(ϵ0),J)Jf(Aμψ(ϵ0),ϵ0)ϵ0.

    Then, for xX0, we write

    B(t)f(S(t),J(t))J(t)J(t)+kt0θ(a)π(a)B(ta)da+δR(t),

    so, for tT,

    B(t)f(Aμψ(ϵ0),ϵ0)ϵ0t0β(a)π(a)B(ta)da+kt0θ(a)π(a)B(ta)da+δR(t).

    Similarly, we apply the Laplace transform to the last inequality, we get, for λ>0,

    ˆB(λ)ˆβ(λ)ˆB(λ)f(Aμψ(ϵ0),ϵ0)ϵ0+kˆθ(λ)ˆB(λ)+δˆR(λ),

    where ˆβ is the Laplace transform of β defined as

    ˆβ(λ)=+0β(a)π(a)eλada.

    By using (5.13), we obtain

    ˆB(λ)ˆβ(λ)ˆB(λ)f(Aμψ(ϵ0),ϵ0)ϵ0+(k+δ(1k)λ+μ+δ)ˆθ(λ)ˆB(λ).

    Since, ˆB(λ)>0 (See, Lemma 5.3), then

    1ˆβ(λ)f(Aμψ(ϵ0),ϵ0)ϵ0+(k+δ(1k)λ+μ+δ)ˆθ(λ).

    We can choose λ=ϵ0 and we obtain

    1h(ϵ0).

    which contradicts (5.12). This completes the proof.

    To prove the uniform strong ρpersistence from Lemma 5.4, we will use Theorem 3.5 in [37]. To this end, we need to prove that there exists no total trajectory ϕ:RX+0, ϕ is defined in (4.10), such that ρ(ϕ(0))=0 and ρ(ϕ(r))>0 and ρ(ϕ(t))>0 for some r,tR+ (see (H1) in [37], Chapter 5). Therefore, we next prove the following lemma.

    Lemma 5.5. If ρ(ϕ(t))=0 for all t0, then ρ(ϕ(t))=0 for all t>0.

    Proof. Assume that ρ(ϕ(t))=B(t)=0 for all t0 with B(t) is defined in (4.10). Then, applying the same arguments as in the proof of Lemma 5.3, we have

    B(t)L+0β(a)i(t,a)da+k+0θ(a)i(t,a)da+δt0e(μ+δ)(st)+0θ(σ)π(σ)B(sσ)dσds,

    thus since B(t)=0 for t0,

    B(t)Lβt0B(a)da+kθt0B(a)da+δt0e(μ+δ)(st)s0θ(sσ)π(sσ)B(σ)dσds.

    So,

    B(t)Lβt0B(a)da+kθt0B(a)da+δt0θ(sσ)π(sσ)B(σ)e(μ+δ)(σt)dσ.

    Finally, according to Fubini's Theorem,

    B(t)(Lβ+kθ+δθμ+δ)t0B(a)da.

    Applying the Gronwall's inequality, we get

    B(t)=0,t>0.

    Lemma 5.6. The following alternative holds: either B is 0 everywhere on R or B is positive everywhere on R.

    Proof. From Lemma 5.5, we can deduce that for each rR, if B(t)=0 for all tr then B(t)=0 for all tr. This means that either B(t)=0 for all tR or there exists a sequence tn as n with B(tn)>0. Let, Bn(t):=B(t+tn), by Lemma 4.2 and (5.9),

    Bn(t)k0θ(a)π(a)Bn(ta)da.

    After a change of variable,

    Bn(t)ktθ(ts)π(ts)Bn(s)ds,

    where Bn(0):=B(tn)>0, Now, suppose that there exists ϵ>0 such that Bn(ϵ)=0 and Bn(t)>0 for all t[0,ϵ), so, we obtain

    0=Bn(ϵ)kϵθ(ϵs)π(ϵs)Bn(s)ds.

    Then, Bn(s)=0 for all s[0,ϵ), this leads to a contradiction. Then, Bn(t)>0 for all t>0 and since tn as n+, B(t)>0 for all tR.

    Now we are ready to prove the strong uniform persistence of the disease.

    Theorem 5.7. Assume that R0>1, Then, each solution of systems (2.1)-(2.2) is strongly uniformly ρpersistent for initial data belonging to X0, that is there exists an ϵ>0 such that lim inft+ρ(Φ(t)x0)>ϵ provided x0X0.

    Proof. By Lemmas 5.4, 5.5 and 5.6, we can apply Theorem 5.2 in [37] to conclude that uniform weak ρpersistence implies the uniform strong ρpersistence. The proof is completed.

    From Theorem 5.7 in [37], we have the following result.

    Theorem 5.8. There exists a compact attractor A1 that attracts every solution with initial condition in X0. Moreover A1 is uniformly ρpositive, i.e there exists some positive constant Γ, such that

    ρ(Φ(t,(S0,i0(),r0))Γ, for all (S0,i0(.),r0)A1. (5.14)

    We will need the following estimates later.

    Lemma 5.9. For all (S0,i0(),r0)A1, the following estimates hold

    i(t,a)i(a)>Γ0,a>0,tRandR(t)>η2,tR,

    with Γ0=ΓD/f(S,J) and η2=((1k)Γ+0θ(a)π(a)da)/(μ+δ).

    Proof. Recall that the compact attractor of bounded set is the union of bounded total trajectories, see Proposition 2.34 in [37]. Thus, there exists a total trajectory Ψ:RA1, Ψ(t)=(S(t),i(t,),R(t)), S(0)=S0,i(0,a)=i0(a). In view of (5.14), we get

    i(t,a)i(a)=π(a)B(ta)π(a)f(S,J)D>ΓDf(S,J).

    In addition, from the equation of R in (4.10), and Theorem 5.8,

    R(t)(1k)Γ+0θ(a)π(a)da(μ+δ)R(t).

    Finally, by a straightforward computation, we find

    R(t)η2,tR,

    with η2:=(1k)Γ+0θ(a)π(a)da/(μ+δ).

    Now, we are ready to prove the global asymptotic stability of the unique positive endemic equilibrium.

    Theorem 5.10. Suppose that R0>1. The problem (4.10) has a unique positive endemic equilibrium (S,i(.),R) which is globally asymptotically stable in X0.

    Proof. Let Ψ:RA1 be a total Φtrajectory such that Ψ(t)=(S(t),i(t,.),R(t)), S(0)=S0, i(0,.)=i0(.), where (S(t),i(t,a),R(t)) is the solution of problem (4.10). Let, for a0,

    ϕ(a)=+a[(δR+0θ(a)i(a)da+k)θ(σ)+β(σ)Jf(S,J)]i(σ)dσ, (5.15)

    and, for y>0,

    H(y)=yln(y)1.

    Then, for Ψ(t)=(S(t),i(t,.),R(t))A1, tR, we consider the following Lyapunov functional V(Ψ(t))=V1(Ψ(t))+V2(Ψ(t))+V3(Ψ(t)) where

    V1(Ψ(t))=S(t)SS(t)Sf(S,J)f(η,J)dη,
    V2(Ψ(t))=+0H(i(t,a)i(a))ϕ(a)da,

    and

    V3(Ψ(t))=ΩH(R(t)R),withΩ:=δR2(1k)+0θ(a)i(a)da.

    First, using the equation of S in (4.10), we get

    ddtV1(Ψ(t))=(1f(S,J)f(S(t),J))(Af(S(t),J(t))μS(t)),=μ(SS(t))(1f(S,J)f(S(t),J))+(1f(S,J)f(S(t),J))(f(S,J)f(S(t),J(t))).

    By using the same arguments as in the proof of Lemma 9.18 in [37], we find

    ddtV2(Ψ(t))=H(i(t,0)i(0))ϕ(0)++0H(i(t,a)i(0))ϕ(a)da,=H(Δ1f(S(t),J(t))f(S,J)+Δ2+0θ(a)i(t,a)da+0θ(a)i(a)da+Δ3R(t)R)ϕ(0)++0H(i(t,a)i(a))ϕ(a)da,

    where

    Δ1:=f(S,J)i(0),Δ2:=k+0θ(a)i(a)dai(0)andΔ3:=δRi(0).

    In view of the third equation of (4.10), by

    i(0)=f(S,J)+k+0θ(a)i(a)da+δR,

    we have Δ1+Δ2+Δ3=1. The convexity of the function H implies

    ddtV2(Ψ(t))[Δ1H(f(S(t),J(t))f(S,J))+Δ2H(+0θ(a)i(t,a)da+0θ(a)i(a)da)+Δ3H(R(t)R)]ϕ(0)++0H(i(t,a)i(a))ϕ(a)da.

    By adding V1 and V2 and using the expression of H, we have

    (V1+V2)(Ψ(t))μ(SS(t))(1f(S,J)f(S(t),J))f(S,J)+f(S,J)f(S,J)f(S,J)+f(S,J)(1f(S,J)f(S,J))+f(S,J)(f(S,J)f(S,J)lnf(S,J)f(S,J)1)+Δ2i(0)H(+0θ(a)i(t,a)da+0θ(a)i(a)da)+Δ3i(0)H(R(t)R)++0H(i(t,a)i(a))ϕ(a)da.

    We reorder these terms, and using the fact that

    lnf(S,J)f(S,J)=lnf(S,J)f(S,J)+lnf(S,J)f(S,J),

    we obtain

    (V1+V2)(Ψ(t))μ(SS(t))(1f(S,J)f(S(t),J))+f(S,J)(lnf(S,J)f(S,J)f(S,J)f(S,J)+1)+f(S,J)H(f(S,J)f(S,J))+Δ2i(0)H(+0θ(a)i(t,a)da+0θ(a)i(a)da)+Δ3i(0)H(R(t)R)++0H(i(t,a)i(a))ϕ(a)da.

    On the other hand,

    ddtV3(Ψ(t))=ΩRH(R(t)R)((1k)+0θ(a)i(t,a)da(μ+δ)R(t)),=Ω(μ+δ)R(1RR)(RR)+Ω(1k)RH(R(t)R)(+0θ(a)i(t,a)da+0θ(a)i(a)da).

    By adding and subtracting the same term

    Ω(1k)RH(R(t)R)+0θ(a)i(a)daR(t)R,

    it yields,

    ddtV3(Ψ(t))=Ω(μ+δ)R(1RR)(RR)+Ω(1k)RH(R(t)R)+0θ(a)i(a)(i(t,a)i(a)R(t)R)da+Ω(1k)RH(R(t)R)(R(t)R1)+0θ(a)i(a)da.

    Next, by summing V=V1+V2+V3, we get

    V(Ψ(t))μ(SS(t))(1f(S,J)f(S(t),J))+f(S,J)(lnf(S,J)f(S,J)f(S,J)f(S,J)+1)+f(S,J)H(f(S,J)f(S,J))++0H(i(t,a)i(a))ϕ(a)da+Δ2i(0)H(+0θ(a)i(t,a)da+0θ(a)i(a)da)+Δ3i(0)H(R(t)R)+Ω(1k)RH(R(t)R)+0θ(a)i(a)(i(t,a)i(a)R(t)R)da+Ω(μ+δ)R(1RR)(RR)+Ω(1k)RH(R(t)R)(R(t)R1)+0θ(a)i(a)da.

    Using the fact (μ+δ)R=(1k)+0θ(a)i(a)da, the addition of the two last terms of the above expression vanishes. Moreover, by employing the Jensen inequality, we have

    H(+0θ(a)i(t,a)da+0θ(a)i(a)da)=H(+0θ(a)i(a)+0θ(a)i(a)dai(t,a)i(a)da),+0θ(a)i(a)+0θ(a)i(a)daH(i(t,a)i(a))da.

    Thus, by combining this with the fact that

    Ω(1k)R=δR+0θ(a)i(a)da,

    we obtain

    V(Ψ(t))μ(SS(t))(1f(S,J)f(S(t),J))+f(S,J)(lnf(S,J)f(S,J)f(S,J)f(S,J)+1)+f(S,J)H(f(S,J)f(S,J))++0H(i(t,a)i(a))ϕ(a)da+Δ2i(0)+0θ(a)i(a)+0θ(a)i(a)daH(i(t,a)i(a))da+Δ3i(0)H(R(t)R)+δR+0θ(a)i(a)daH(R(t)R)+0θ(a)i(a)(i(t,a)i(a)R(t)R)da.

    Now, for the values of t such that J(t)<J. From Lemma 5.2 and Jensen inequality, we have

    H(f(S,J)f(S,J))<H(J(t)J),=H(+0β(a)i(a)+0β(a)i(a)dai(t,a)i(a)da),+0β(a)i(a)+0β(a)i(a)daH(i(t,a)i(a))da,=+0β(a)i(a)JH(i(t,a)i(a))da.

    This implies that

    V(Ψ(t))μ(SS(t))(1f(S,J)f(S(t),J))+f(S,J)(lnf(S,J)f(S,J)f(S,J)f(S,J)+1)++0H(i(t,a)i(a))[ϕ(a)+f(S,J)Jβ(a)i(a)+kθ(a)i(a)]da+δRH(R(t)R)+δR+0θ(a)i(a)daH(R(t)R)+0θ(a)i(a)(i(t,a)i(a)R(t)R)da.

    Finally, in view of the expression of ϕ in (5.15), we find

    V(Ψ(t))μ(SS(t))(1f(S,J)f(S(t),J))+f(S,J)(lnf(S,J)f(S,J)f(S,J)f(S,J)+1)++0[H(R(t)R)H(i(t,a)i(a))+H(R(t)R)(i(t,a)i(a)R(t)R)]δRθ(a)i(a)+0θ(a)i(a)dada.

    Since ln(1/x)x+10 and H(b)H(a)+(ab)H(b)0 then the second and the third terms are negative. As a consequence,

    V(Ψ(t))0.

    By applying the same arguments for the values of t satisfying J(t)>J. we conclude that V(Ψ(t))0. Now, since V is bounded on Ψ, the alpha limit and omega limit set of Ψ must be contained in M the largest invariant subset of {V=0}. Notice that V(Ψ(t))=0 implies that S(t)=S, and i(t,a)/i(a)=R(t)/R. From (4.10) and the fact that i(a)=π(a)i(0), we have

    i(t,a)i(a)=B(ta)i(0)=R(t)R, a0, (5.16)

    this implies that B(ta)=B(t) for all a0, and so i(t,a)=i(t,0) for all a0. Next, from the equation of S in (4.10) and the fact that S(t)=S, we have

    AμS=f(S,J(t)).

    Moreover, from equations of equilibrium we know that

    AμS=f(S,J),

    thus we obtain f(S,J)=f(S,J(t)). From the monotonicity of f and (4.10) we obtain i(t,0)+0β(a)π(a)da=i(0)+0β(a)π(a)da, and so, B(t):=i(t,0)=i(0) for all tR. Finally, from (5.16), we get R(t)=R and i(t,.)=i(.) for all tR. Consequently, M consists of only the endemic equilibrium.

    Now, since A1 is compact, the omega limit and alpha limit are non-empty, compact, invariant and attract Ψ(t) as t±, respectively. Since V(Ψ(t)) is nonincreasing on t, V is constant on the omega and alpha limit and these both sets contain only the endemic equilibrium. Consequently Ψ(t)(S,i(.),R) as t±, and hence

    V(Ψ(t))V(S,i(.),R), as t±.

    On the other hand, we have

    limt+V(Ψ(t))V(Ψ(t))limtV(Ψ(t)),

    for all tR, then V(Ψ(t))=V(S,i(.),R) for all tR and so Ψ(t)=(S,i(.),R) for all tR. Therefore the compact attractor A1 is reduced to the endemic equilibrium. In addition, by Theorem 2.39 in [37] the endemic equilibrium is also locally asymptotically stable. The theorem is proved.

    In this section, the results of the previous sections are illustrated by numerical simulations. We use the following numerical method: we discretize our problem by the upwind method for solving hyperbolic partial differential equation. For instance, the approximation uni=u(t,a) is given by

    (ut)nun+1iuniΔt,(ua)iuniuni1Δa.

    The equations of susceptible and recover are solving by explicit Euler method for the ODE. The non-local terms are approximated by one of the composite integration formulas.

    Let's consider the Beddington-Deangelis functional response defined by

    f(S,J)=SJ1+α1S+α2J.

    We note that this function f satisfies all the assumption of the paper. The system (2.1) has two equilibria and the condition of stability is given by the basic reproduction number R0. In the case of the Beddington-Deangelis functional response, the basic reproduction number R0 is given by

    R0=Aμ+α1A+0β(a)π(a)da+((1k)δμ+δ+k)+0θ(a)π(a)da.

    We consider the following values of parameters

    A=2.103,μ=1.102andδ=1.102,

    with the initial conditions

    S0=1.103,r0=2.104andi0(a)=8.104e0.1a.

    The functions β and θ are chosen to be

    β(a)={0, if aτ1,56.103(aτ1)2e0.2(aτ1), if a>τ1,

    and

    θ(a)={0, if aτ2,3.102(aτ2)2e0.15(aτ2); if a>τ2,

    with τ1=10 and τ2=12. The parameter τ1 represents the end of the exposed period at a=τ1 (witch means that the individual is infected but he does not transmit the disease from a=0 to a=τ1), when aτ1 the infectious period starts. The time required to leave the infected class is described by the parameter τ2.

    Figure 1.  The functions β and θ with respect to age a.

    In first time, we choose k=0.1. The basic reproduction number is computed and we have R0=0.8624<1. According to the result stated, the disease-free equilibrium is globally asymptotically stable as illustrated in Figures 2 and 3.

    Figure 2.  The evolution of solutions S (left) and R (right) with respect to time t are drawn. The case of disease-free steady state with R0=0.8624<1.
    Figure 3.  The evolution of solution i with respect to time t and age a. The case of disease-free steady state with R0=0.8624<1. As seen in the figure, the density of infected individuals disappear in large time.

    In second time, we take k=0.9 and then the basic reproduction number R0=1.1849>1, that mean that, the positive endemic equilibrium is globally asymptotically stable as illustrated in Figures 4, 5 and 6.

    Figure 4.  The evolution of solutions S (left) and R (right) with respect to time t are drawn. The case of endemic steady state with R0=1.1849>1.
    Figure 5.  The evolution of solution i with respect to time t and age a. The case of endemic steady state with R0=1.1849>1. As seen in the figure, the density of infected individuals does not disappear in large time.
    Figure 6.  The evolution of solution i(tf,a) and i(a) with respect to age a at fixed tf=800.

    In this paper, we proposed and analyzed an infection age-structured SIR epidemic model with a general incidence rate. We showed the basic characters of the solution including existence, uniqueness and positivity. We obtained the basic reproduction number R0 by renewal process and showed that the disease-free equilibrium E0 is globally asymptotically stable if R0<1. The main contribution of this paper was devoted to the global asymptotic stability of the endemic equilibrium E. We prove that the unique endemic equilibrium E exists, the system is uniformly persistent and E is globally asymptotically stable if R0>1. Our results showed that the asymptotic behavior of the solution of our model can be determined by the basic reproduction number R0, allowing to separate the situations into the extinction and the persistence of diseases.

    In this paper, we have not considered the case where R0=1. By some previous results on basic epidemic models (see for instance [4,Section 5.5.2]), we can expect that the disease-free equilibrium is globally asymptotically stable for R0=1. However, it is not trivial and a future work. From the application viewpoint, our results for R01 are thought to be sufficient since R0 estimated by real data unlikely to be just equal to 1.

    The authors would like to thank the associate editor and the anonymous reviewers for their careful reading and helpful comments. TK was supported by Grant-in-Aid for Young Scientists (B) of Japan Society for the Promotion of Science (Grant No. 15K17585).

    All authors declare no conflicts of interest in this paper.



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