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Dynamics of a diffusive age-structured HBV model with saturating incidence

  • Received: 01 October 2015 Accepted: 29 June 2018 Published: 01 July 2016
  • MSC : Primary: 35C07, 35K57, 35J61; Secondary: 37B25, 92D30.

  • In this paper, we propose and investigate an age-structured hepatitis B virus (HBV) model with saturating incidence and spatial diffusion where the viral contamination process is described by the age-since-infection. We first analyze the well-posedness of the initial-boundary values problem of the model in the bounded domain ΩRn and obtain an explicit formula for the basic reproductive number R0 of the model. Then we investigate the global behavior of the model in terms of R0: if R01, then the uninfected steady state is globally asymptotically stable, whereas if R0>1, then the infected steady state is globally asymptotically stable. In addition, when R0>1, by constructing a suitable Lyapunov-like functional decreasing along the travelling waves to show their convergence towards two steady states as t tends to ±, we prove the existence of traveling wave solutions. Numerical simulations are provided to illustrate the theoretical results.

    Citation: Xichao Duan, Sanling Yuan, Kaifa Wang. Dynamics of a diffusive age-structured HBV model with saturating incidence[J]. Mathematical Biosciences and Engineering, 2016, 13(5): 935-968. doi: 10.3934/mbe.2016024

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  • In this paper, we propose and investigate an age-structured hepatitis B virus (HBV) model with saturating incidence and spatial diffusion where the viral contamination process is described by the age-since-infection. We first analyze the well-posedness of the initial-boundary values problem of the model in the bounded domain ΩRn and obtain an explicit formula for the basic reproductive number R0 of the model. Then we investigate the global behavior of the model in terms of R0: if R01, then the uninfected steady state is globally asymptotically stable, whereas if R0>1, then the infected steady state is globally asymptotically stable. In addition, when R0>1, by constructing a suitable Lyapunov-like functional decreasing along the travelling waves to show their convergence towards two steady states as t tends to ±, we prove the existence of traveling wave solutions. Numerical simulations are provided to illustrate the theoretical results.


    Let R,Z, and N be the sets of all real numbers, integers, and natural numbers, respectively. Here and below, for a,bN with a<b, we use the notation Z(a,b):={a,a+1,,b}.

    In recent years, difference equations as mathematical models to describe a variety of practical problems in the fields of economy, biology, disease prevention and control, environmental protection, etc., have attracted great attention and there have been a lot of outstanding works in theory and actual application; see, for example, [1,2].

    In the paper, we consider the following discrete Dirichlet boundary value problem involving the mean curvature operator

    {ϕ(u(k1))+q(k)ϕ(u(k))=λf(k,u(k)),kZ(1,T),u(0)=u(T+1)=0, (1.1)

    where ϕ(s)=s1+κs2,κ>0 is a constant [3], T is a given positive integer, is the forward difference operator defined by u(k)=u(k+1)u(k),2u(k)=(u(k)),q(k)0 for all kZ(1,T),λ is a real positive parameter, and f(k,)C(R,R) for each kZ(1,T).

    When q(k)=0 for all kZ(1,T), the problem (1.1) is the discrete analogy of one-dimensional prescribed mean curvature equations with Dirichlet boundary conditions

    {(ϕ(u))=λf(t,u(t)),t(0,1),u(0)=u(1)=0.

    The existence of (positive) solutions of the above problem has been studied by using the variational method, the standard techniques from bifurcation theory, fixed point index, and Kresnoselskii's fixed point theorem (see [4,5,6] and the references therein).

    Note that when κ=0, the problem (1.1) degenerates into the classical second-order difference equation boundary value problem

    {2u(k1)+q(k)u(k)=λf(k,u(k)),kZ(1,T),u(0)=u(T+1)=0. (1.2)

    The existence of (positive) solutions to the problem (1.2) has been well known with various qualitative assumptions of nonlinearity f ([7,8,9]).

    When κ=1, the problem (1.1) is the ordinary discrete mean curvature problem

    {(u(k1)1+u2(k1))+q(k)u(k)1+u2(k)=λf(k,u(k)),kZ(1,T),u(0)=u(T+1)=0, (1.3)

    which has aroused the interest of many scholars recently [10,11,12]. Their approaches are variational. The existence of (positive) solutions to the problem (1.3) depends on the behavior at zero or infinity of the potential F(k,u)=u0f(k,s)ds. In [10,11], the oscillating behavior of F at + played an important role in obtaining an infinite number of positive solutions to the problem (1.3). Chen and Zhou [12] obtained at least three positive solutions of the problem (1.3) with q(k)=0 for all kZ(1,T), where the primitive F on the nonlinear datum satisfies that lim sups+F(k,s)|s|<α (α is a positive constant). Meanwhile, the case that the potential F satisfies lim infsF(k,s)|s|>β (β is a positive constant) has been also discussed. They obtained at least two nontrivial solutions based on a two-critical-point theorem (Theorem 2.1) established in [13]. It is an important tool in obtaining at least two positive solutions of the Laplacian or the algebraic boundary value problems (see, for instance, [15,16]). However, it is relatively less used to look for positive solutions to the mean curvature boundary value problems.

    Inspired by this, we will apply the two-critical-point theorem to establish the existence of at least two positive solutions for the problem (1.1) in this paper. As a special case of our main theorem, the existence of two positive solutions for the problem (1.1) with κ=1 and all q(k)=0 is obtained in Remark 4. The result improves Theorem 2 [12], where the author establishes the existence of only two nontrivial solutions without providing sign information for them under stronger hypotheses on the potential F. In Theorem 2 [12], the bilateral limit assumption on F(k,s)|s| at ensures that the energy functional of that problem is anticoercive, which consequently guarantees the establishment of the Palais-Smale condition—a pivotal requirement for applying critical point theorems. In our paper, the bilateral limit on the potential F is weakened to a unilateral limit at +, and then the energy functional of the problem (1.1) loses its anticoercivity. However, with the help of some inequality techniques, it is proven that the energy functional still satisfies the Palais-Smale condition. Moreover, in our main result (Theorem 3.1), the existence of at least two positive solutions is established without any asymptotic condition of the potential F at 0 and with no requiring that f(k,0)>0 for any kZ(1,T). In fact, the algebraic conditions in Theorem 3.1 are more general than the conditions that the potential F is subquadratic at 0 and superlinear at + (see Corollary 3.1).

    In 2003, Guo and Yu [17] first used the variational method to study the periodic solutions of second-order difference equations. Since then, for nonlinear difference systems, many scholars have used the variational method to study the existence of various solutions, such as periodic solutions, subharmonic solutions, and homoclinic solutions [18,19,20]. For general references on difference equations and their applications, we refer the reader to monographs [21,22] and the references therein.

    This paper is organized as follows: In section 2, some definitions and results are collected. Some estimations of equivalence of the norm are provided. Moreover, Lemma 2.1 is presented to guarantee us to obtain positive solutions rather than nontrivial solutions to the problem (1.1). Section 3 is devoted to our main result. Lemma 3.1 is given to ensure the Palais-Smale condition of the functional on this basis. Some consequences of our main result are presented together with three examples.

    Consider the T dimensional Banach space

    S={u:Z(0,T+1)Rsuchthatu(0)=u(T+1)=0}

    endowed with the norm as

    u:=(Tk=0|u(k)|2)12,uS.

    Moreover, the space S can also be equipped with the following equivalent norms:

    u:=maxkZ(1,T)|u(k)|,uS,
    u1:=Tk=1|u(k)|,uS,
    u2:=(Tk=1|u(k)|2)12,uS,

    respectively. It is easy to know that for all uS,

    u1 u2. (2.1)

    By Theorem 12.6.1 in [21], we have for all uS,

    λ1u2uλTu2, (2.2)

    where λj=4sin2jπ/2(T+1) for all jZ(1,T). By virtue of Lemma 2.2 in [23], we get for all uS,

    uT+12u. (2.3)

    Note that v(0)=v(T+1)=0 for all vS, we have

    T+1k=1ϕ(u(k1))v(k1)=Tk=1ϕ(u(k1))v(k) (2.4)

    for all u,vS.

    Let

    s+=max{s,0}ands=max{s,0}

    for all sR.

    Remark 2.1. In fact, the nonnegative solution of the following problem

    {ϕ(u(k1))+q(k)ϕ(u(k))=λf(k,u+(k)),kZ(1,T),u(0)=u(T+1)=0 (2.5)

    is the nonnegative solution of the problem (1.1). To obtain positive solutions to the problem (1.1), we only need to search for positive solutions to the problem (2.5) now.

    Consider two functionals Φ and Ψ defined respectively on S by

    Φ(u)=1κT+1k=1(1+κ|u(k1)|21)+1κTk=1q(k)(1+κ|u(k)|21) (2.6)

    and

    Ψ(u)=Tk=1F+(k,u(k)), (2.7)

    F+(k,s)=s0f(k,t+)dt for all (k,s)Z(1,T)×R. Obviously,

    F+(k,s)={s0f(k,t)dt, if s>0,f(k,0)s, if s0,

    for all kZ(1,T), and Φ,ΨC1(S,R), which means that Φ and Ψ are two continuously Gâteaux-differentiable functionals defined on S. By (2.4), we have that for any u,vS,

    Φ(u)(v)=T+1k=1ϕ(u(k1))v(k1)+Tk=1q(k)ϕ(u(k))v(k)=Tk=1[ϕ(u(k1))+q(k)ϕ(u(k))]v(k)

    and

    Ψ(u)(v)=Tk=1f(k,u+(k))v(k).

    We define Iλ:SR by

    Iλ(u)=Φ(u)λΨ(u).

    Clearly, IλC1(S,R) and for any u,vS,

    Iλ(u)(v)=Tk=1[ϕ(u(k1))+q(k)ϕ(u(k))λf(k,u+(k))]v(k).

    Remark 2.2. For any u,vS, Iλ(u)(v)=0 if and only if

    ϕ(u(k1))+q(k)ϕ(u(k))λf(k,u+(k))=0

    for all uS and kZ(1,T). In other words, a critical point of Iλ on S corresponds to a solution to the problem (2.5).

    For the problem (1.1), we are interested in the existence of positive solutions rather than nontrivial ones. To the end, we present an assumption on f(k,0) as follows.

    (H1) f(k,0)0,kZ(1,T).

    Lemma 2.1. If (H1) holds, then any nonzero critical point of the functional Iλ on S is a positive solution to the problem (1.1).

    Proof. In fact, taking Remarks 2.1 and 2.2 into account, it is sufficient to verify that any nontrivial solution u of the problem (2.5) is positive, that is, u(k)>0 for all kZ(1,T).

    Notice that ϕ is a strictly monotonically increasing and odd function on R. By standard computation, we can conclude that

    ϕ(u(k1))u(k1)ϕ(u(k1))u(k1)=|u(k1)|21+κ|u(k1)|2 (2.8)

    for all kZ(1,T+1). By a direct computation, we get

    q(k)ϕ(u(k))u(k)=q(k)ϕ(u(k))u(k)=q(k)|u(k)|21+κ|u(k)|2 (2.9)

    for all kZ(1,T). For any nontrivial solution u of the problem (2.5), combining Eqs (2.4), (2.8) and (2.9) with assumption (H1), we obtain that

    0=Tk=1[ϕ(u(k1))+q(k)ϕ(u(k))λf(k,u+(k))](u(k))=T+1k=1ϕ(u(k1))u(k1)Tk=1[q(k)ϕ(u(k))λf(k,u+(k))]u(k)T+1k=1ϕ(u(k1))u(k1)+Tk=1[q(k)ϕ(u(k))+λf(k,0)]u(k)T+1k=1|u(k1)|21+κ|u(k1)|2+Tk=1q(k)|u(k)|21+κ|u(k)|20.

    Hence u(0)=u(1)==u(T)=0. Recalling that u(0)=u(T+1)=0 for any uS, we have u(k)=0 for all kZ(1,T). Thus u is nonnegative.

    Next, we further prove that u is positive. Otherwise, there exists some k0Z(1,T) such that u(k0)=0, then

    ϕ(u(k01))=λf(k0,0)0.

    Clearly, ϕ(u(k0))ϕ(u(k01)). Because ϕ is a strictly monotonically increasing homomorphism, we get u(k0)u(k01). Hence, u(k0+1)+u(k01)0. As a result, u(k0±1)=0. Then iterating the process, we have that u(k)=0 for all kZ(1,T). In short, u is zero somewhere in Z(1,T). Then it is zero identically. This contradicts the nontriviality of u. The proof is ended.

    Let (X,) be a real Banach space and φC1(X,R). φ is said to satisfy the Palais-Smale condition ((PS) condition), if any sequence {un}X for which {φ(un)} is bounded and φ(un)0 as n possesses a convergent subsequence in X.

    The following two-critical-point theorem is given by Bonanno and D'Aguì in 2016 [13].

    Theorem 2.1. [13] Let X be a real Banach space, and let Φ,Ψ:XR be two functionals of class C1 such that infXΦ=Φ(0)=Ψ(0)=0. Assume that there are rR and ˜uX, with 0<Φ(˜u)<r, such that

    supuΦ1((,r])Ψ(u)r<Ψ(˜u)Φ(˜u) (2.10)

    for each

    λΛ=(Φ(˜u)Ψ(˜u),rsupuΦ1((,r])Ψ(u)),

    the functional Iλ=ΦλΨ satisfies the (PS) condition, and it is unbounded from below. Then for each λΛ, the functional Iλ admits at least two nonzero critical points uλ,1,uλ,2 such that Iλ(uλ,1)<0<Iλ(uλ,2).

    Let

    L(k):=lim infs+F(k,s)sandL:=minkZ(1,T)L(k),

    where F(k,s)=s0f(k,t)dt for all (k,s)Z(1,T)×R. Here and below, when L=0, we think 1L=. Before we come to a conclusion, let's give a lemma on the (PS) condition.

    Lemma 3.1. If L>0 and (H1) hold, then Iλ satisfies the (PS) condition, and it is unbounded from below for all λ((T+1)λT+QκL,+), where Q:=(Tk=1|q(k)|2)12.

    Proof. Because S is finite dimensional, it is sufficient to show that any (PS) sequence of Iλ is bounded on S. Let {un}S be a sequence such that {Iλ(un)} is bounded and Iλ(un)0 as n.

    First, we claim that {un} is bounded. According to (2.8) and (2.9), we have for all nN and kZ(1,T+1),

    ϕ(un(k1))un(k1)ϕ(un(k1))un(k1)=|un(k1)|21+κ|un(k1)|2 (3.1)

    and

    q(k)ϕ(un(k))un(k)=q(k)ϕ(un(k))un(k)=q(k)|un(k)|21+κ|un(k)|2. (3.2)

    By (3.1) and (3.2), we can estimate the derivative of Φ at un in the direction of un that

    Φ(un)(un)=Tk=1[ϕ(un(k1))+q(k)ϕ(un(k))](un(k))=T+1k=1ϕ(un(k1))un(k1)Tk=1q(k)ϕ(un(k))un(k)T+1k=1|un(k1)|21+κ|un(k1)|2+Tk=1q(k)|un(k)|21+κ|un(k)|2=1κT+1k=1κ|un(k1)|21+κ|un(k1)|2+1κTk=1q(k)κ|un(k)|21+κ|un(k)|2=1κT+1k=1κ|un(k1)|2+111+κ|un(k1)|2+1κTk=1q(k)κ|un(k)|2+111+κ|un(k)|21κT+1k=1(1+κ|un(k1)|21)+1κTk=1q(k)(1+κ|un(k)|21).

    To put it simply,

    Φ(un)(un)Φ(un),nN. (3.3)

    And, from the condition (H1), it follows that

    Ψ(un)(un)=Tk=1f(k,u+n(k))un(k)=Tk=1f(k,0))un(k)0,nN. (3.4)

    Combining (3.3) and (3.4), we have

    Φ(un)Φ(un)(un)Φ(un)(un)+λΨ(un)(un)=Iλ(un)(un) (3.5)

    for all λ>0 and nN. Moreover, by the definition of the functional Φ, we obtain, for all uS,

    Φ2(u)[1κT+1k=1(1+κ|u(k1)|21)]21κ2T+1k=1(1+κ|u(k1)|21)2=1κ2[T+1k=1(1+κ|u(k1)|2)+T+1k=112T+1k=11+κ|u(k1)|2]=1κT+1k=1|u(k1)|22κ2T+1k=1(1+κ|u(k1)|21)1κu22κΦ(u),

    which implies that

    u2κΦ2(u)+2Φ(u) (3.6)

    for all uS. Thus, it is clear that for all uS,

    1κ+1κu2+1κ2Φ(u).

    So combining (3.5) and the above inequality, we have

    01κ+1κun2+1κ2Iλ(un)(un),nN.

    Also, from limnIλ(un)=0, it follows that

    limnIλ(un)(un)un=limnIλ(un)(unun)=0.

    As a result, we obtain

    limn1κ+1κun2+1κ2un=0.

    By standard computation, the above equation is transformed into

    limn[κ+1un2+1un]=+,

    which means that limnun=0. Hence our claim is proved. So there is M>0 such that 0un(k)M for all kZ(1,T) and nN.

    Next, we prove that {un} is bounded. If {un} is not bounded, we may assume, going if necessary to a subsequence, that un(n). Taking L>0 into account, we fix λ>(T+1)λT+QκL and fix l=l(λ) such that for all kZ(1,T),

    (T+1)λT+Qλκ<l<LL(k)=lim infs+F(k,s)s,

    then for all kZ(1,T), there is a δk>0 such that

    F+(k,s)=F(k,s)>ls=l|s|,s>δk.

    Meanwhile, for all kZ(1,T) and s[M,δk],

    F+(k,s)mins[M,δk]F+(k,s)l|s|lmax{δk,M}+mins[M,δk]F+(k,s)l|s|max{lmax{δk,M}mins[M,δk]F+(k,s),0}=l|s|η(k).

    Hence, for all kZ(1,T),

    F+(k,s)l|s|η(k),sM.

    On account of (2.1), we obtain that for all nN,

    Ψ(un)=Tk=1F+(k,un(k))lTk=1|un(k)|Tk=1η(k)=lun1ηlun2η,

    where η=Tk=1η(k). Using the Cauchy-Schwarz inequality, we get that for all nN,

    Φ(un)=1κT+1k=1(1+κ|un(k1)|21)+1κTk=1q(k)(1+κ|un(k)|21)1κT+1k=1(1+κ|un(k1)|21)+1κTk=1q(k)(1+κ|un(k)|21)1κT+1k=1|un(k1)|+1κTk=1q(k)|un(k)|1κT+1un+1κQun2.

    Therefore, from the previous two inequalities and (2.2), it follows that for all nN,

    Iλ(un)=Φ(un)λΨ(un)1κT+1un+1κQun2λlun2+λη1κ(T+1)λTun2+1κQun2λlun2+λη((T+1)λT+Qκλl)1λ1un+λη.

    When un+(n), by (T+1)λT+Qκλl<0, we have limnIλ(un)=. This leads to a contradiction. Hence {un} is bound and Iλ satisfies the (PS) condition.

    Finally, we prove that Iλ is unbounded from below. Let {un} be such that un=u+n for any nN and u+n(n). Arguing as before, we obtain that

    Iλ(un)((T+1)λT+Qκλl)1λ1un+λη,nN.

    So limnIλ(un)=. Our conclusion follows.

    Below, our main result is presented.

    Theorem 3.1. Suppose that (H1) holds and there exist two positive constants c and d with d<c such that

    (H2)Tk=1maxs[0,c]F(k,s)1κ+4c2κ(T+1)+1κ2<Lκ(T+1)λT+Q

    and

    (H3)Tk=1maxs[0,c]F(k,s)1κ+4c2κ(T+1)+1κ2<κTk=1F(k,d)(2+q)(1+κd21).

    Then for each

    λΛ+:=(max{(T+1)λT+QLκ,(2+q)(1+κd21)κTk=1F(k,d)},1κ+4c2κ(T+1)+1κ2Tk=1maxs[0,c]F(k,s)),

    the problem (1.1) possesses at least two positive solutions, where q:=Tk=1q(k).

    Proof. By Lemma 2.1, it is enough to prove that Iλ has at least two nonzero critical points. We apply Theorem 2.1 by putting X=S,Iλ=ΦλΨ where Φ,Ψ are the functions introduced in (2.6) and (2.7).

    Clearly, infSΦ=Φ(0)=Ψ(0)=0. Also notice that F+(k,s)=F(k,s) for all (k,s)Z(1,T)×[0,+). From the conditions (H1),(H2) and (H3), it follows that L>0 and Λ+ is non-degenerate. Then Lemma 3.1 ensures that the function Iλ satisfies the (PS) condition and it is unbounded from below for λΛ+.

    For fixed λΛ+, there exists c>0 such that λ1κ+4c2κ(T+1)+1κ2Tk=1maxs[0,c]F(k,s). Put r:=1κ+4c2κ(T+1)+1κ2. We claim that

    {uS:Φ(u)r}{uS:uc}.

    If Φ(u)r, by (3.6), we get that u2κΦ2(u)+2Φ(u)κr2+2r=4c2T+1. Thus, from the inequality (2.3), it is clear that

    uT+12uT+124c2T+1=c.

    The assertion is verified. Therefore, we obtain

    supuΦ1([0,r])Ψ(u)rTk=1maxs[0,c]F(k,s)1κ+4c2κ(T+1)+1κ2. (3.7)

    Now, we look for ˜uS in Theorem 2.1. Define ˜uRT+2 as ˜u(k)=d(0<d<c) for all kZ(1,T) and ˜u(0)=˜u(T+1)=0. Clearly, ˜uS. It is easy to see that

    Φ(˜u)=1κ(2+q)(1+κd21), (3.8)

    and hence we get

    Ψ(˜u)Φ(˜u)=κTk=1F(k,d)(2+q)(1+κd21). (3.9)

    Therefore, from (3.7), (3.9), and (H3), it follows that

    supuΦ1([0,r])Ψ(u)rΨ(˜u)Φ(˜u)

    and Λ+Λ. Moreover, since 0<d<c, and again by virtue of the condition (H3), we obtain that

    1κ(2+q)(1+κd21)<1κ+4c2κ(T+1)+1κ2=r. (3.10)

    On account of (3.8) and (3.10), we get that 0<Φ(˜u)<r. Thus, (2.10) holds. Hence Theorem 2.1 ensures that Iλ admits two nonzero critical points. This completes the proof.

    Remark 3.1. Indeed, condition (H1) implicitly covers both f(k,0)>0 and f(k,0)=0 cases. Theorem 3.1 holds true under either sub-case of condition (H1).

    Remark 3.2. If all f(k,) are nonnegative on [0,c](c>0), then maxs[0,c]F(k,s)=F(k,c) for all kZ(1,T). According to Theorem 3.1, it is enough to assume that there is a positive constant d with d<c such that

    Tk=1F(k,c)1κ+4c2κ(T+1)+1κ2<min{Lκ(T+1)λT+Q,κTk=1F(k,d)(2+q)(1+κd21)}.

    Remark 3.3. When κ=1 and q(k)=0 for all kZ(1,T) (Q=0 and q=0), the conditions in Remark 3.2 are more general than those of Theorem 2 in [12], where f(k,s) is required to be positive for all (k,s)Z(1,T)×[c,c] rather than nonnegative, and the potential F is assumed to possess asymptotic behavior not only at + but also at . Here, the two solutions we obtain are positive, while the solutions in Theorem 2 of [12] are only nontrivial. Obviously, our main result improves Theorem 2 in [12].

    Now, we present two particular cases of Theorem 3.1.

    Corollary 3.1. Assume that (H1) holds,

    lim sups0+F(k,s)s2=+ (3.11)

    and

    lims+F(k,s)s=+ (3.12)

    for all kZ(1,T). Then for each λ(0,λ), where

    λ=supc>01κ+4c2κ(T+1)+1κ2Tk=1maxs[0,c]F(k,s),

    the problem (1.1) admits at least two positive solutions.

    Proof. Let λ(0,λ) and c>0 such that

    λ<1κ+4c2κ(T+1)+1κ2Tk=1maxs[0,c]F(k,s).

    Taking (3.11) into account, we get that lim sups0+F(k,s)1+κs21=+. As a result, there is d>0 with d<c such that κ2+qTk=1F(k,d)1+κd21>1λ. Note that L=+, so Theorem 3.1 ensures the conclusion.

    Corollary 3.2. Assume that f(k,)=f for all kZ(1,T) and f is a continuous function such that

    lims0+f(s)s=+ (3.13)

    and

    lims+f(s)=+. (3.14)

    Then, for each

    λ(0,supc>01κ+4c2κ(T+1)+1κ2Tmaxs[0,c]s0f(t)dt),

    the problem (1.1) admits at least two positive solutions.

    Proof. In fact, Corollary 3.2 is a consequence of Corollary 3.1. It is easy to see that (3.13) implies f(0)0, and (3.11) and (3.12) can be derived from (3.13) and (3.14), respectively.

    Example 3.1. For each λ(0,1κ+4κ(T+1)+1κ2(3e1)T), the problem

    {ϕ(u(k1))+q(k)ϕ(u(k))=λ(e3u(k)1),kZ(1,T),u(0)=u(T+1)=0

    admits at least two positive solutions. Here, the nonlinearity f(t)=e3t1 satisfies the conditions of Corollary 3.2 and

    supc>01κ+4c2κ(T+1)+1κ2Tc0f(t)dt1κ+4κ(T+1)+1κ2(3e1)T.

    The image of f(t) is shown in Figure 1.

    Figure 1.  The image of f(t) in Example 3.1.

    Example 3.2. For each λ(0,1κ+4κ(T+1)+1κ2T(T+1)), the problem

    {ϕ(u(k1))+q(k)ϕ(u(k))=λku(k)(cosu(k)+2),kZ(1,T),u(0)=u(T+1)=0

    admits at least two positive solutions. It is easy to see that f(k,t)=kt(cost+2) satisfies the assumptions of Corollary 3.1 and

    λ1κ+4κ(T+1)+1κ2Tk=110f(k,t)dt=1κ+4κ(T+1)+1κ2T(T+1)210t(cost+2)dt1κ+4κ(T+1)+1κ2T(T+1).

    Figure 2 displays the functional plot of f(k,t) for k(0,100).

    Figure 2.  The image of f(k,t) in Example 3.2.

    Remark 3.4. We observe that Theorem 2 in [12] cannot be applied in the previous two examples with κ=1 and q(k)=0 for all kZ(1,T), since f(k,0)>0 is assumed there.

    Example 3.3. Consider the boundary value problem (1.1) with κ=1,q(k)=0 and

    f(k,t)=f(t)={12(43t)cost4,t83π,2π1,t>83π

    for all kZ(1,T). Clearly, if t(0,43), then f(t)>0; and if t(43,2π), then f(t)<0. Such a relationship is visually apparent in Figure 3. A straightforward calculation yields

    F(k,s)=F(s)={2(4sins43ssins412coss4+12),s83π,(2π1)s4(2π1)(23π+3),t>83π
    Figure 3.  The image of f(t) in Example 3.3.

    for all kZ(1,T).

    Obviously, L(k)=lim infs+F(k,s)s=2π1 for all kZ(1,T), then

    L=minkZ(1,T)L(k)=2π15.382.

    Letting T=3,c=5, and d=1, we obtain

    (T+1)λT=4sinT2(T+1)π=4sin38π3.695,
    Tk=1maxs[0,c]F(k,s)=3k=1430f(k,t)dt=72(1cos13)3.963,

    and

    Tk=1F(k,d)=3k=110f(k,t)dt=6[sin14+12(1cos14)]3.724.

    It follows that

    (T+1)λTL3.6955.3820.687,
    2(1+d21)Tk=1F(k,d)0.8283.7240.222,

    and

    1+4c2T+1+1Tk=1maxs[0,c]F(k,s)4.0993.9631.034.

    From f(k,0)=f(0)=2>0 and the preceding relations, all conditions of Theorem 3.1 hold. Thus for each λ((T+1)λTL,1+4c2T+1+1Tk=1maxs[0,c]F(k,s))(0.687,1.034), the boundary value problem (1.1) has at least two positive solutions.

    Remark 3.5. The potential function F in the above example lacks superlinearity at +, failing to satisfy the conditions of Corollaries 3.1 and 3.2. Nevertheless, Theorem 3.1 still guarantees the existence of two positive solutions for the problem (1.1) in Example 3.3.

    In this paper, we consider a generalized difference mean curvature problem, which includes the conventional one and the classical second-order difference equation boundary value problem. The main novelties of this research are as follows:

    (1) The existence of two positive solutions rather than nontrivial solutions for our problem is established based on a two-zero critical points theorem and some inequality techniques.

    (2) Under the assumption of the unilateral limit of F(k,s)|s| at + on the potential F(k,s)=s0f(k,t)dt instead of the bilateral limit at , it is proved that without anticoercivity the energy functional associated with our problem still satisfies the Palais-Smale condition that plays a key role in the critical point theorem.

    (3) Our principal result can be applied to cases with nonlinear terms where f(k,0)>0 but also to cases where the nonlinear terms satisfy f(k,0)=0.

    (4) It is also worth mentioning that the algebraic conditions in our main result are more general than the subquadraticity at 0 and the superlinearity at + for the potential F.

    In fact, due to the previous (1)–(3), our main result extends Theorem 2 in [12].

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    This work is supported by the National Nature Science Foundation of China (No.10901071, No.11501054, and No.12301186).

    The authors declare there is no conflicts of interest.

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