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Asymptotic profile of endemic equilibrium to a diffusive epidemic model with saturated incidence rate

  • We study the existence and asymptotic profile of endemic equilibrium (EE) of a diffusive SIS epidemic model with saturated incidence rate. By introducing the basic reproduction number R0, the existence of EE is established when R0>1. The effects of diffusion rates and the saturated coefficient on asymptotic profile of EE are investigated. Our results indicate that when the diffusion rate of susceptible individuals is small and the total population N is below a certain level, or the saturated coefficient is large, the infected population dies out, while the two populations persist if at least one of the diffusion rates of the susceptible and infected individuals is large. Finally, we illustrate the influences of the population diffusion and the saturation coefficient on this model numerically.

    Citation: Yan’e Wang , Zhiguo Wang, Chengxia Lei. Asymptotic profile of endemic equilibrium to a diffusive epidemic model with saturated incidence rate[J]. Mathematical Biosciences and Engineering, 2019, 16(5): 3885-3913. doi: 10.3934/mbe.2019192

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  • We study the existence and asymptotic profile of endemic equilibrium (EE) of a diffusive SIS epidemic model with saturated incidence rate. By introducing the basic reproduction number R0, the existence of EE is established when R0>1. The effects of diffusion rates and the saturated coefficient on asymptotic profile of EE are investigated. Our results indicate that when the diffusion rate of susceptible individuals is small and the total population N is below a certain level, or the saturated coefficient is large, the infected population dies out, while the two populations persist if at least one of the diffusion rates of the susceptible and infected individuals is large. Finally, we illustrate the influences of the population diffusion and the saturation coefficient on this model numerically.


    Let ΩR2 be a bounded convex polygonal domain covered by a thin flat plate, and Γ=Ω be the Lipschitz continuous boundary of Ω. For a given loading function fL2(Ω),f0a.e.inΩ, and an elastic obstacle ψL2(Ω), we consider the following clamped Kirchhoff plate problem with elastic unilateral obstacle (refer to [1,2]):

    {FinduV,such thatJ(u)J(v),vV, (1.1)

    where

    V=H20(Ω),J(v)=12a(v,v)+12j(v)(f,v),
    a(w,v)=Ω[ΔwΔv+(1ν)(2wxyvxywxxvyywyyvxx)]dxdy,w,vV,
    j(v)=Ωκ[(vψ)]2dxdy,(f,v)=Ωfvdxdy.

    Here J(v) is the total energy, a(v,v) represents the strain energy corresponding to displacement v of the plate, j(v) can be interpreted as the contribution from the contact with ψ and (f,v) is the potential energy. Moreover, the symbol v=min{v,0}, ν(0,1/2) is the Poisson's ratio, κL(Ω) describes the stiffness of the obstacle and satisfies κκ0>0a.e.inΩ. In the limiting case κ, the obstacle becomes rigid and the problem (1.1) reduces to a constrained minimisation:

    {FinduK,such thatu=argminvK[12a(v,v)(f,v)] (1.2)

    with K={vV:vψ in Ω}. (1.2) is a classical displacement obstacle model of clamped Kirchhoff plate (cf.[3]) and equivalent to a fourth-order variational inequality:

    {FinduK,such thata(u,vu)(f,vu),vK. (1.3)

    Different with the displacement obstacle model (1.2), the elastic obstacle problem (1.1) can be investigated based on the following weak formulation [1]:

    FinduV,such that  a(u,v)+Ωκ(uψ)vdxdy=(f,v),vV, (1.4)

    which has a unique solution [2] and is equivalent to another fourth-order variational inequality:

    {FinduV,such thatvVa(u,vu)+Ωκ2[(vψ)]2dxdyΩκ2[(uψ)]2dxdy(f,vu). (1.5)

    As we all know, analytical solutions to obstacle problems are always difficult to obtain. In this case, the study of numerical solutions has attracted a lot of attention. The finite element method (FEM) is a popular numerical method to solve obstacle problems [3–8]. In the last decades, more efforts have devoted to FEM analysis for the limit form of elastic obstacle problem (1.1), i.e., displacement obstacle problem (1.2), see [9–17] and references therein. But works focusing on FEMs of the elastic obstacle problem (1.1) are relatively rather few in the existing literature. From [1], the convergence analysis of a mixed FEM for the elastic obstacle problem (1.1) was obtained, where elements employed must satisfy C0 continuous. In [18], a stabilized FEM was constructed for (1.1) with a C1 continuous requirement. [2] provided a general framework of optimal error estimates for FEM, where the continuous requirement is relatively relaxed but continuity at each element's vertex of the subdivision is indispensable. In this situation, a natural question is whether these requirements for continuity can be completely removed. In other words, a key difficulty is how to get the optimal error estimates for a strongly discontinuous element dissatisfying above continuity requirements, which is just the motivation of this work.

    In this paper, as an attempt, we will investigate the FEM approximation for the elastic obstacle problem (1.1) by using Bergan's energy-orthogonal element. This element is constructed through an energy-orthogonal free formulation (cf. [19,20]), which is convergent for arbitrary meshes. Its degrees of freedom are all defined on the element's vertex, including the function values and the two first derivatives at the three vertices of each element, which are very simple and can be used conveniently. Moreover, the global dimension of the vector of unknowns is only 3NP (NP denotes the number of vertex in mesh subdivision), about 25 percent fewer than that of the famous triangular Morley element (about 4NP). This property is very useful for a reduction the amount of computation. But the element has strong discontinuous, that is, the shape function and its first derivatives are no longer continuous at the element's the vertex. This element does not satisfy the above mentioned continuity requirements. Despite so high discontinuity, Bergan's energy-orthogonal plate element has been applied to the FEM approximate for the displacement obstacle problem (1.2) (see [12]). The authors developed a convergence analysis method and obtained the optimal error estimate of order O(h) in [12]. We note that the convergence analysis relied on two additional introduced tools, i.e., an approximation subset and an enriching operator from Bergan's energy-orthogonal FE space to C1-conforming Bell FE space and the process is very complicated.

    This paper aims to develop a new error analysis of Bergan's energy-orthogonal element approximation for the elastic obstacle problem (1.1). Unlike the convergence analysis in [12], we make full use of some special approaches, including interpolation operator splitting and energy orthogonality, to derive the optimal error estimates of order O(h) in the broken energy norm. The theoretical analysis is very simple and clear. The numerical results demonstrate the proposed method not only enjoys one-order accuracy but also can well reflect the influence brought by the obstacle stiffness parameter κ.

    The organization of this paper is as follows: In Section 2, Bergan's energy-orthogonal plate element and its typical properties are briefly introduced. Then we propose a novel error analysis approach and obtain the optimal error estimate of order O(h) successfully in Section 3. At last, Section 4 provides some numerical results to illustrate the validity of the theoretical analysis.

    Bergan's energy-orthogonal element was first proposed by Bergan et al. in [19] using the free formulation scheme. Then Shi et al. [20] proved that this element is equivalent to a nonconforming element constructed based on a specific interpolation ΠK (see Eq (2.3) below), where ΠK is introduced to form the shape function. It is shown in [20] that the special construction of interpolation ΠK makes the two components of the shape function energy-orthogonal and the stiffness matrix consistent with that in the free formulation scheme. In the following, we will introduce Bergan's energy-orthogonal plate element briefly, and the readers can refer to [19,20] for details.

    Assume that Th is a regular triangulation of Ω with mesh size h. For a given KTh, let its diameter be hK, three vertices be pi(xi,yi) and the area coordinates be λi for i=1,2,3. Firstly, we select nine nodal parameters set as Σ(v)={v1,v1x,v1y,v2,v2x,v2y,v3,v3x,v3y}, where vi=v(pi),vix=vx(pi),viy=vy(pi),i=1,2,3. The shape function space is taken as same as that of Zienkiewicz element, i.e., ˜P(K)=span{˜N1,˜N2,,˜N9}, here ˜Nj=λj,˜N4+j=λjλj+1,˜N6+j=λ2jλj+1λj+1λ2j,j=1,2,3 and λ4=λ1. Then the associated conventional interpolation operator ˜ΠK:H3(K)˜P(K) satisfies

    ˜ΠKv=6i=1αi(v)˜Ni+9i=7αi(v)˜Ni¯rK(v)+rK(v),vH3(K), (2.1)

    where ¯rK(v) and rK(v) are the quadratic and cubic terms respectively, and the coefficients αi(v) can be written as: αj(v)=vj, α3+j(v)=c(j+2)2(vjxv(j+1)x)b(j+2)2(vjyv(j+1)y), α6+j(v)=vjv(j+1)+c(j+2)2(vjx+v(j+1)x)b(j+2)2(vjy+v(j+1)y) with bj=y(j+1)y(j+2) and cj=(x(j+1)x(j+2)) for j=1,2,3, here and later subscripts (j+i) will be replaced with (j+i)(mod3) when (j+i)>3 for i=1,2.

    Next we introduce another shape function space P(K)=span{N1,N2,,N9}, where Ni=˜Ni,i=1,2,,6,N6+j=(λjλ(j+1))3,j=1,2,3 and λ4=λ1. It is easy to verify that

    KxxNidxdy=KyyNidxdy=KxyNidxdy=0,i=7,8,9, (2.2)

    and the associated traditional interpolation operator ˆΠK:H3(K)P(K) is defined by

    ˆΠKv=6i=1βi(v)Ni+9i=7βi(v)Ni¯SK(v)+SK(v),vH3(K), (2.3)

    where βi(v),i=1,2,...,9, can be expressed as:

    βj(v)=139vj29(v(j+1)+v(j+2))+19[(c(j+2)c(j+1))vjx+c(j+2)v(j+1)xc(j+1)v(j+2)x]           +19[(b(j+1)b(j+2))vjyb(j+2)v(j+1)y+b(j+1)v(j+2)y],β3+j(v)=13(vj+v(j+1)2v(j+2))+16[(2c(j+2)cj)vjx+(c(j+1)2c(j+2))v(j+1)x           +(c(j+1)cj)v(j+2)x]+16[(bj2b(j+2))vjy+(2b(j+2)b(j+1))v(j+1)y+(bjb(j+1))v(j+2)y],β6+j(v)=29(vjv(j+1))+154[(cj6c(j+2))vjx+(c(j+1)6c(j+2))v(j+1)x+c(j+2)v(j+2)x]           +154[(6b(j+2)bj)vjy+(6b(j+2)b(j+1))v(j+1)yb(j+2)v(j+2)y],

    for j=1,2,3.

    Now employing Eq (2.1) and Eq (2.3), we introduce an interpolation operator ΠK as

    ΠK:H3(K)P(K),   ΠKv=¯rK(v)+SK(v),vH3(K). (2.4)

    Then taking ΠKv as the shape function on K and Σ(v) as its associated nodal parameters (vanishing at nodes on the boundary Γ), and defining an interpolation operator Πh for every vH3(Ω) with (Πhv)|K=ΠKv, we can obtain a piecewise cubic polynomial space on Ω denoted by Vh. It has been shown in [20] that Vh is equivalent to the Bergan's energy-orthogonal plate element first proposed by Bergan etal through the free formulation scheme. Obviously, the construction process of Vh is quite different from conventional element spaces. Herein the shape function on K is formulated through the operator ΠK involving two interpolation operators ˜ΠK and ˆΠK.

    It follows from Eqs (2.1), (2.4) and the formulas of αi(v) and βi(v)(i=1,2,,9) that

    ΠKv(pj)vj=β6+j(v)β6+(j+2)(v),
    ΠKvx(pj)vjx=12[3(bjb(j+1))β6+j(v)+3(b(j+2)bj)β6+(j+2)(v)b(j+1)α6+j(v)+b(j+2)α6+(j+2)(v)],ΠKvy(pj)vjy=12[3(cjc(j+1))β6+j(v)+3(c(j+2)cj)β6+(j+2)(v)c(j+1)α6+j(v)+c(j+2)α6+(j+2)(v)] (2.5)

    for j=1,2,3, where represents the area of K. Thus ΠKv and its two first derivatives are discontinuous at vertices pj (j=1,2,3). Moreover, the mean value of ΠKv along the element's each edge F can be calculated as

    1|F|FΠKvds=1|F|F(¯rK(v)+SK(v))ds=1|F|F¯rK(v)ds+14(β7(v)β9(v)).

    Obviously, it is not continuous neither. In spite of so high discontinuity of the interpolation ΠKv, the element space possesses the following special features (cf.[20]), which will play an important role in the follow-up convergence analysis.

    Lemma 2.1. (R1) For any vhVh, there exists a vH3(K) such that

    vh|K=ΠKv, vh=ˉvh+vh,ˉvh|K=¯rK(v),vh|K=SK(v). (2.6)

    When vh|KP2(K), there holds vh=ˉvh. Moreover, xx¯rK(v), xy¯rK(v) and yy¯rK(v) are constants, which together with Eq (2.2) imply that ¯rK(v) and SK(v) are energy-orthogonal, thus the element is called energy-orthogonal. Let 2vh be the Hessian matrix of vh, then we have K2ˉvh:2vhdxdy=0.

    (R2) For j=1,2,3, the quadratic term ¯rK(v) satisfies

    ¯rK(v)(pj)=vj, (2.7)
    ¯rK(v)(pj+p(j+1)2)=vj+v(j+1)2+(vjxv(j+1)x)c(j+2)8(vjyv(j+1)y)b(j+2)8, (2.8)

    i.e., for any vhVh, its quadratic term ˉvh is continuous at the element's vertices and midpoints of edges. In other words, ˉvhC0(Ω).

    (R3) For any vH3(K) and integer m (0m3), there holds

    |vΠK(v)|m,K+|v¯rK(v)|m,KCh3mK|v|3,K,  (2.9)
    |rK(v)|m,K+|SK(v)|m,KCh3mK|ΠKv|3,K. (2.10)

    Here and later, C denotes a positive constant independent of h and may be different at each appearance.

    We consider the Bergan's energy-orthogonal plate element discrete approximation form of the variational inequality (1.5) as:

    {FinduhVh,such thatvhVha(uh,vhuh)+KThKκ2[(vhψ)]2dxdyKThKκ2[(uhψ)]2dxdy(f,vhuh), (2.11)

    where ah(wh,vh)=KThKΔwhΔvh+(1ν)(2whxyvhxywhxxvhyywhyyvhxx)]dxdy.

    In this section, we will establish error estimates of Bergan's energy-orthogonal FEM for the elastic obstacle problem (1.1) in the energy norm.

    Firstly, using the the similar argument to [2], we have

    Theorem 3.1. The problem (2.11) is equivalent to the discrete approximation of plate problem:

    {FinduhVh,such thatvhVhah(uh,vh)+KThKκ(uhψ)vhdxdy=(f,vh), (3.1)

    which has a unique solution uh. Moreover, uhh and (uhψ)0 are uniformly bounded independently of h, where h=(KTh||22,K)12 can be shown to be a norm over Vh by using (R1) and (R2) in Lemma 2.1.

    In what follows, we will give error estimate for Eq (2.11).

    Theorem 3.2. Assume that u and uh are the solutions of Eqs (1.5) and (2.11) respectively, uH3(Ω) and f,ψL2(Ω), then we have

    uuhhCh, (3.2)

    where the constant C depends on the stiffness of the obstacle κ.

    Proof. Since uuhhuΠhuh+Πhuuhh, in view of Eq (2.9) in (R3) of Lemma 2.1, we only need to estimate the second term Πhuuhh. In fact, let wh=Πhuuh and employ Eq (3.1), we have

    wh2hCah(wh,wh)=C[ah(Πhuu,wh)+ah(u,wh)ah(uh,wh)]CΠhuuhwhh+C[ah(u,wh)+KThKκ(uhψ)whdxdy(f,wh)]. (3.3)

    Now we concentrate on the estimate of the second term in the right hand of Eq (3.3). It follows from (R1) in Lemma 2.1 that wh=ˉwh+wh and there exists a wH3(K) such that

    wh|K=ΠKw, ˉwh|K=¯rK(w),wh|K=SK(w), K2ˉwh:2whdxdy=0. (3.4)

    Then we can deduce that

    ah(u,wh)+KThKκ(uhψ)whdxdy(f,wh)=ah(u,ˉwh)(f,ˉwh)+ah(u,wh)(f,wh)+KThKκ(uhψ)whdxdy. (3.5)

    On one hand, from [2] we know that the solution u of the problem (1.5) satisfies

    ΩΔuvdxdy+Ωκ(uψ)vdxdy=(f,v)vH10(Ω). (3.6)

    On the other hand, (R2) in Lemma 2.1 implies ˉwhH10(Ω). Thus applying Green formula and Eq (3.6) yields

    ah(u,ˉwh)(f,ˉwh)=KThK(Δu(1ν)2us2)ˉwhnds+(1ν)KThK2unsˉwhsds    Ω(Δu)ˉwhdxdy(f,ˉwh)=KThK(Δu(1ν)2us2)ˉwhnds+(1ν)KThK2unsˉwhsdsΩκ(uψ)ˉwhdxdy, (3.7)

    here n and s are the unit outward normal vector and tangential vector respectively.

    Substituting Eq (3.7) into Eq (3.5) leads to

    ah(u,wh)+KThKκ(uhψ)whdxdy(f,wh)=KThK(Δu(1ν)2us2)ˉwhnds+(1ν)KThK2unsˉwhsds+ah(u,wh)(f,wh)   +KThKκ(uψ)whdxdy+KThKκ[(uhψ)(uψ)]whdxdy6j=1(Er)j. (3.8)

    In what follows, we will estimate (Er)j one by one for j=1,2,...,6.

    Firstly, applying Lemmas 3.5 and 3.6 in [21] yields

    |(Er)1|Chu3(KTh|¯rK(w)+rK(w)|22,K)12. (3.9)

    By use of Eq (3.4), we get

    |¯rK(w)|22,K+|SK(w)|22,K=|ˉwh|22,K+|wh|22,K=K2ˉwh:2ˉwhdxdy+K2wh:2whdxdy+2K2ˉwh:2whdxdy=|wh|22,K. (3.10)

    At the same time, employing Eq (2.10) in (R3) and the inverse estimate gives

    |rK(w)|2,KChK|ΠKw|3,KC|ΠKw|2,K=C|wh|2,K, (3.11)

    which in conjunction with Eq (3.10) leads to |(Er)1|Chu3whh.

    Secondly, the fact ˉwhH10(Ω) implies (Er)2=0.

    Thirdly, from Eq (2.2), we have ah(¯Πhu,wh)=0, which together with the Eq (2.10) in (R3) and Eq (3.10) gives

    |(Er)3|=|ah(u¯Πhu,wh)|u¯ΠhuhwhhCh|u|3whh. (3.12)

    Moreover, it follows from Eq (3.4), Eq (2.10) in (R3) and the inverse estimate that

    |wh|0,K=|SK(w)|0,KCh3K|ΠKw|3,KCh2K|ΠKw|2,K=Ch2K|wh|2,K, (3.13)

    which reveals

    |(Er)4|f0wh0Ch2f0whh, (3.14)
    |(Er)5|C(uψ)0wh0Ch2(uψ)0whh. (3.15)

    Finally, by use of the elementary inequality (ts)(ts)0 and Theorem 3.1, we have

    (Er)6=KThKκ[(uhψ)(uψ)](Πhuu+uuh)dxdyKThKκ[(uhψ)(uψ)](Πhuu)dxdyC((uhψ)0+(uψ)0)Πhuu0C(uψ)0Πhuu0. (3.16)

    Then combining Eq (3.8) and the above bounds of (Er)1(Er)6 results in

    ah(u,wh)+KThKκ(uhψ)whdxdy(f,wh)Ch(u3+hf0+h(uψ)0)whh+C(uψ)0Πhuu0. (3.17)

    Therefore, the desired result Eq (3.2) follows from (R3), Eqs (3.3) and (3.17) immediately.

    Remark. The analysis presented herein is also valid to the elastic obstacle problem (1.1) with j(v)=Ωκ[(vψ)+]2dxdy, v+=max{v,0} and fL2(Ω),f0a.e.inΩ.

    We consider the elastic obstacle problem (1.1) with Ω=(0,1)2, f=10, ν=0.25 and an elastic obstacle defined by the function ψ={0    if(x,y)[0.3,0.7]2,1               otherwise. The domain Ω is firstly divided into N×N rectangles. Then each rectangle is further divided along its diagonal into two equal triangles.

    Since it is not easy to derive the exact solution, we denote uN as the N-th level discrete solution and take eNh=uNuN1h as the error in the broken energy norm. The numerical results eNh for different obstacle stiffness parameter κ=10j (j=0,1,3,4) with N=16,32,64,128 are given in Table 1 and further plotted in the logarithm scales in Figure 1. We observed that the errors in the energy norm are indeed convergent at optimal order O(1N), i.e., O(h), as h=2N0. This result is consistent with the theoretical analysis in Theorem 3.2.

    Table 1.  Numerical results of eNh and Orders under different κ.
    κ N eNh Order κ N eNh Order
    κ=1 16 0.1258 κ=103 16 0.1084
    32 0.0647 0.9593 32 0.0517 1.0681
    64 0.0326 0.9889 64 0.0259 0.9972
    128 0.0163 1.0000 128 0.0128 1.0168
    κ=10 16 0.1255 κ=104 16 0.0974
    32 0.0645 0.9603 32 0.0402 1.2767
    64 0.0325 0.9889 64 0.0202 0.9928
    128 0.0162 1.0044 128 0.0100 1.0144

     | Show Table
    DownLoad: CSV
    Figure 1.  Errors with different parameter κ.

    Moreover, the discrete solution uN with N=64 is also depicted in Figures 24, where the parameter κ is chosen as 1,103and104, respectively. We can see that the bigger parameter κ, the more obvious the influence of obstacles. From Figure 2 (κ=1) and Figure 3 (κ=103), we observe that the difference of κ makes the minimum value of the discrete solution change, but the shape of the solution does not change significantly. In Figure 4 (κ=104), it is easy to see that the elastic obstacle has a very obvious effect on the solution, which is agrees with the fact that the elastic obstacle will become a rigid obstacle when κ. This phenomenon further indicates the proposed numerical method in this paper can also be used for approximated simulation of the displacement obstacle problem by taking a relatively large obstacle stiffness parameter κ.

    Figure 2.  The discrete solution uN with κ=1.
    Figure 3.  The discrete solution uN with κ=103.
    Figure 4.  The discrete solution uN with κ=104.

    In a word, the numerical results in this section confirm the theoretical analysis in Section 3 and indicate the effectiveness of the numerical method.

    This work was supported by the National Natural Science Foundation of China (No.11701523, 11801527, 11871441).

    The authors declare there is no conflict of interest.



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