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Digital products with PNk-adjacencies and the almost fixed point property in DTCk

  • Given two digital images (Xi,ki),i{1,2}, first of all we establish a new PNk-adjacency relation in a digital product X1×X2 to obtain a relation set (X1×X2,PNk), where the term PN" means pseudo-normal". Indeed, a PN-k-adjacency is softer or broader than a normal k-adjacency. Next, the present paper initially develops both PN-k-continuity and PN-k-isomorphism. Furthermore, it proves that these new concepts, the PN-k-continuity and PN-k-isomorphism, need not be equal to the typical k-continuity and a k-isomorphism, respectively. Precisely, we prove that none of the typical k-continuity (resp. typical k-isomorphism) and the PN-k-continuity (resp. PN-k-isomorphism) implies the other. Then we prove that for each i{1,2}, the typical projection map Pi:X1×X2Xi preserves a PNk-adjacency relation in X1×X2 to the ki-adjacency relation in (Xi,ki). In particular, using a PN-k-isomorphism, we can classify digital products with PNk-adjacencies. Furthermore, in the category of digital products with PNk-adjacencies and PN-k-continuous maps between two digital products with PNk-adjacencies, denoted by DTCk, we finally study the (almost) fixed point property of (X1×X2,PNk).

    Citation: Jeong Min Kang, Sang-Eon Han, Sik Lee. Digital products with PNk-adjacencies and the almost fixed point property in DTCk[J]. AIMS Mathematics, 2021, 6(10): 11550-11567. doi: 10.3934/math.2021670

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  • Given two digital images (Xi,ki),i{1,2}, first of all we establish a new PNk-adjacency relation in a digital product X1×X2 to obtain a relation set (X1×X2,PNk), where the term PN" means pseudo-normal". Indeed, a PN-k-adjacency is softer or broader than a normal k-adjacency. Next, the present paper initially develops both PN-k-continuity and PN-k-isomorphism. Furthermore, it proves that these new concepts, the PN-k-continuity and PN-k-isomorphism, need not be equal to the typical k-continuity and a k-isomorphism, respectively. Precisely, we prove that none of the typical k-continuity (resp. typical k-isomorphism) and the PN-k-continuity (resp. PN-k-isomorphism) implies the other. Then we prove that for each i{1,2}, the typical projection map Pi:X1×X2Xi preserves a PNk-adjacency relation in X1×X2 to the ki-adjacency relation in (Xi,ki). In particular, using a PN-k-isomorphism, we can classify digital products with PNk-adjacencies. Furthermore, in the category of digital products with PNk-adjacencies and PN-k-continuous maps between two digital products with PNk-adjacencies, denoted by DTCk, we finally study the (almost) fixed point property of (X1×X2,PNk).



    In this paper, we study the following general class of functional differential equations with distributed delay and bistable nonlinearity,

    {x(t)=f(x(t))+τ0h(a)g(x(ta))da,t>0,x(t)=ϕ(t),τt0. (1.1)

    Many mathematical models issued from ecology, population dynamics and other scientific fields take the form of Eq (1.1) (see, e.g., [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15] and references therein). When a spatial diffusion is considered, many models are studied in the literature (see, e.g., [16,17,18,19,20,21,22]).

    For the case where g(x)=f(x) has not only the trivial equilibrium but also a unique positive equilibrium, Eq (1.1) is said to be a problem with monostable nonlinearity. In this framework, many authors studied problem (1.1) with various monostable nonlinearities such as Blowflies equations, where f(x)=μx and g(x)=βxeαx, and Mackey-Glass equations, where f(x)=μx and g(x)=βx/(α+xn) (see, e.g., [1,3,6,7,9,10,11,12,13,21,23,24,25,26,27]).

    When g(x) allows Eq (1.1) to have two positive equilibria x1 and x2 in addition to the trivial equilibrium, Eq (1.1) is said to be a problem with bistable nonlinearity. In this case, Huang et al. [5] investigated the following general equation:

    x(t)=f(x(t))+g(x(tτ)).

    The authors described the basins of attraction of equilibria and obtained a series of invariant intervals using the decomposition domain. Their results were applied to models with Allee effect, that is, f(x)=μx and g(x)=βx2eαx. We point out that the authors in [5] only proved the global stability for x2M, where g(M)=maxsR+g(s). In this paper, we are interested in the dynamics of the bistable nonlinearity problem (1.1). More precisely, we will present some attracting intervals which will enable us to give general conditions on f and g that ensure global asymptotic stability of equilibria x1 and x2 in the both cases x2<M and x2M.

    The paper is organized as follows: in Section 2, we give some preliminary results including existence, uniqueness and boundedness of the solution as well as a comparison result. We finish this section by proving the global asymptotic stability of the trivial equilibrium. Section 3 is devoted to establish the attractive intervals of solutions and to prove the global asymptotic stability of the positive equilibrium x2. In Section 4, we investigate an application of our results to a model with Allee effect. In Section 5, we perform numerical simulation that supports our theoretical results. Finally, Section 6 is devoted to the conclusion.

    In the whole paper, we suppose that the function h is positive and

    τ0h(a)da=1.

    We give now some standard assumptions.

    (T1) f and g are Lipschitz continuous with f(0)=g(0)=0 and there exists a number B>0 such that maxv[0,s]g(v)<f(s) for all s>B.

    (T2) f(s)>0 for all s0.

    (T3) g(s)>0 for all s>0 and there exists a unique M>0 such that g(s)>0 for 0<s<M, g(M)=0 and g(s)<0 for s>M.

    (T4) There exist two positive constants x1 and x2 such that g(x)<f(x) if x(0,x1)(x2,) and g(x)>f(x) if x(x1,x2) and g(x1)>f(x1).

    Let C:=C([τ,0],R) be the Banach space of continuous functions defined in [τ,0] with ||ϕ||=supθ[τ,0]|ϕ(θ)| and C+={ϕC;ϕ(θ)0,τθ0} is the positive cone of C. Then, (C,C+) is a strongly ordered Banach space. That is, for all ϕ,ψC, we write ϕψ if ϕψC+, ϕ>ψ if ϕψC+{0} and ϕ>>ψ if ϕψInt(C+). We define the following ordered interval:

    C[ϕ,ψ]:={ξC;ϕξψ}.

    For any χR, we write χ for the element of C satisfying χ(θ)=χ for all θ[τ,0]. The segment xtC of a solution is defined by the relation xt(θ)=x(t+θ), where θ[τ,0] and t0. In particular, x0=ϕ. The family of maps

    Φ:[0,)×C+C+,

    such that

    (t,ϕ)xt(ϕ)

    defines a continuous semiflow on C+ [28]. For each t0, the map Φ(t,.) is defined from C+ to C+ which is denoted by Φt:

    Φt(ϕ)=Φ(t,ϕ).

    The set of equilibria of the semiflow, which is generated by (1.1), is given by

    E={χC+;χRandg(χ)=f(χ)}.

    In this section, we first provide existence, uniqueness and boundedness of solution to problem (1.1). We then present a Lyapunov functional and show the global asymptotic stability of the trivial equilibrium. We begin by recalling a useful theorem related to a comparison principle (see Theorem 1.1 in page 78 in [29]).

    We consider the following problem:

    {x(t)=F(xt),t>0,x(t)=ϕ(t),τt0, (2.1)

    where F:ΩR is continuous on Ω, which is an open subset of C. We write x(t,ϕ,F) for the maximal defined solution of problem (2.1). When we need to emphasize the dependence of a solution on initial data, we write x(t,ϕ) or x(ϕ).

    Theorem 2.1. Let f1, f2 :ΩR be continuous, Lipschitz on each compact subset of Ω, and assume that either f1 or f2 is a nondecreasing function, with f1(ϕ)f2(ϕ) for all ϕΩ. Then

    x(t,ϕ,f1)x(t,ϕ,f2),

    holds for all t0, for which both are defined.

    The following lemma states existence, uniqueness and boundedness of the positive solution to problem (1.1). For the proof, see [15,30,31].

    Lemma 2.2. Suppose that (T1) holds. For any ϕC+, the problem (1.1) has a unique positive solution x(t):=x(t,ϕ) on [0,) satisfying x0=ϕ, provided ϕ(0)>0. In addition, we have the following estimate:

    0lim suptx(t)B.

    Moreover, the semiflow Φt admits a compact global attractor, which attracts every bounded set in C+.

    The following lemma can be easily proved (see also the proof of Theorem 1.1 in [29]).

    Lemma 2.3. Let ϕC be a given initial condition, x(ϕ) be the solution of problem (1.1) and xϵ(ϕ), ϵ>0 be the solution of problem (1.1) when replacing f by f±ϵ. Then

    xϵ(t,ϕ)x(t,ϕ) as ϵ0, for all t[0,).

    We focus now on the global stability of the trivial equilibrium. For this purpose, we suppose that

    f(s)>g(s) for all s(0,x1) and f(x1)=g(x1). (2.2)

    Lemma 2.4. Assume that (T1) and condition (2.2) hold. For a given ϕC[0,x1]{x1}, the solution x(ϕ) of problem (1.1) satisfies

    lim suptx(t)<x1, (2.3)

    provided that one of the following hypotheses holds:

    (i) g is a nondecreasing function over (0,x1).

    (ii) f is a nondecreasing function over (0,x1).

    Proof. Without loss of generality, we assume that ϕ(0)<x1. Suppose that (ⅰ) holds. We first claim that x(t)<x1 for all t0. Let xϵ(ϕ) be the solution of problem (1.1) by replacing f by f+ϵ. We prove that xϵ(t):=xϵ(t,ϕ)x1 for all t>0. Suppose, on the contrary, that there exists t1>0 such that xϵ(t1)=x1, xϵ(t)x1 for all tt1 and xϵ(t1)0. Then, using the equation of xϵ(t1), we have

    0xϵ(t1)=f(xϵ(t1))ϵ+τ0h(a)g(xϵ(t1a))da,<f(x1)+g(x1)=0,

    which leads to a contradiction. Now, applying Lemma 2.3, we obtain x(t)x1 for all t0. Next, set X(t)=x1x(t). We then have

    X(t)=f(x(t))f(x1)+τ0h(a)(g(x1)g(x(ta)))da.

    Since x(t)x1 for all t0 and g is a nondecreasing function, we have, for tτ,

    X(t)LX(t),

    where L is the Lipschitz constant of f. Consequently, X(t)X(0)eLt and the claim is proved. Finally, to prove inequality (2.3), we suppose, on the contrary, that there exist an increasing sequence (tn)n, tn, t0τ and a nondecreasing sequence (x(tn))n, such that x(tn)x1 as tn, x(tn)<x1 for some n and x(tn)=maxt[0,tn]x(t). Then, in view of condition (2.2), the equation of x(tn) satisfies

    0x(tn)=f(x(tn))+τ0h(a)g(x(tna))da,f(x(tn))+g(x(tn))<0,

    which is a contradiction.

    Suppose now that () holds. Let xϵ(t) be the solution of problem (1.1) by replacing f(s) by f(s)+ϵ and g(s) by g+(s)=maxσ[0,s]g(σ).

    Then, again by contradiction, suppose that there exists t1>0 such that xϵ(t1)=x1, xϵ(t)x1 for all tt1 and xϵ(t1)0. Then, arguing as above, we obtain

    0xϵ(t1)=f(xϵ(t1))ϵ+τ0h(a)g+(xϵ(ta))da,<f(x1)+g+(x1). (2.4)

    Since f is a nondecreasing function and g+ is the smallest nondecreasing function that is greater than g, we obtain f(s)>g+(s) for all s(0,x1). This contradicts with inequality (2.4). Further, by combining Theorem 2.1 and Lemma 2.3, we get x(t)x1 for all t0. Finally, following the same arguments as in the first part of this proof, the lemma is proved.

    We now prove the global asymptotic stability of the trivial equilibrium.

    Theorem 2.5. Assume that (T1) and condition (2.2) hold. Suppose also that ϕC[0,x1]{x1}. The trivial equilibrium is globally asymptotically stable if one of the following hypotheses holds:

    (i) g is a nondecreasing function over (0,x1).

    (ii) f is a nondecreasing function over (0,x1).

    Proof. Suppose that () holds. Let V be the Lyapunov functional defined by

    V(s)=s+τ0ψ(a)g(ϕ(a))da, (2.5)

    where ψ(a)=τah(σ)dσ. The derivative of V along the solution of problem (1.1) gives

    dV(xt)dt=g(x(t))f(x(t))t>0.

    From condition (2.2) and Lemma 2.4, we have dV(xt)/dt<0 and thus, the result is reached by classical Lyapunov theorem (see [6]). Next, suppose that () holds. Let the function V be defined in (2.5) by replacing g(s) by g+(s):=maxσ[0,s]g(σ). Since g+(s)<f(s) for all s(0,x1), we can employ the same argument as above to get dV(xt)/dt<0. Finally the result is obtained by applying Theorem 2.1 and Lemma 2.4. This completes the proof.

    Now, suppose that

    f(s)>g(s) for all s>0. (2.6)

    Using the same proof as in Theorem 2.5, we immediately obtain the following theorem.

    Theorem 2.6. Assume that (T1) and condition (2.6) hold. The trivial equilibrium is globally asymptotically stable for all ϕC+.

    The following theorem concerns the global stability of x2 in the case where g is a nondecreasing function.

    Theorem 3.1. Suppose that ϕC[x1,sups[τ,0]ϕ]{x1} and g is a nondecreasing function. Assume also that (T1) and (T4) hold. Then, the positive equilibrium x2 is globally asymptotically stable.

    Proof. Without loss of generality, suppose that ϕ(0)>x1. We first claim that lim inftx(t)>x1. For this, let xϵ(t):=xϵ(t;ϕ) be the solution of problem (1.1) when replacing f by fϵ. To reach the claim, we begin by proving that xϵ(t)>x1 for all t0. Otherwise, there exists t1>0 such that xϵ(t1)=x1, xϵ(t)x1 for all tt1 and xϵ(t1)0. Then, the equation of xϵ(t1) satisfies

    0xϵ(t1)=f(xϵ(t1))+ϵ+τ0h(a)g(xϵ(t1a))da>f(x1)+g(x1)=0.

    This reaches a contradiction and thus xϵ(t)>x1 for all t>0. This result, together with Lemma 2.3, gives x(t)x1 for all t0.

    Next, for X(t)=x(t)x1 and since f(x1)=g(x1), the equation of X(t) satisfies

    X(t)=f(x1)f(x(t))+τ0h(a)(g(x(ta))g(x1))da,

    this leads to

    X(t)LX(t),

    where L is the Lipschitz constant of f. Consequently x(t)>x1 for all t>0.

    Now, suppose that there exist an increasing sequence (tn)n, tn, t0τ and a nonincreasing sequence (x(tn))n, such that x(tn)x1 as tn, x(tn)>x1 for some n and x(tn)=mint[0,tn]x(t). In view of (T4), the equation of x(tn) satisfies

    0x(tn)=f(x(tn))+τ0h(a)g(x(tna))da,f(x(tn))+g(x(tn))>0,

    which is a contradiction. The claim is proved.

    To prove that x2 is globally asymptotically stable, we consider the following Lyapunov functional

    V(ϕ)=ϕ(0)x2[g(s)g(x2)]ds+12τ0ψ(a)[g(ϕ(a))g(x2)]2da,

    with ψ(a)=τah(σ)dσ. By a straightforward computation, the derivative of V along the solution of problem (1.1) gives, for all t>0,

    dV(xt)dt=[g(x(t))g(x2)]x(t)12τ0ψ(a)a[g(x(ta))g(x2)]2da=[g(x(t))g(x2)][f(x(t))+τ0h(a)g(x(ta))da]12{ψ(τ)[g(x(tτ))g(x2)]2ψ(0)[g(x(t))g(x2)]2τ0ψ(a)[g(x(ta))g(x2)]2da}=[g(x(t))g(x2)][f(x(t))+g(x(t))]+[g(x(t))g(x2)][g(x(t))+τ0h(a)g(x(ta))da]+12τ0h(a)da[g(x(t))g(x2)]212τ0h(a)[g(x(ta))g(x2)]2da=[g(x(t))g(x2)][f(x(t))+g(x(t))]+12τ0h(a){2[g(x(t))g(x2)][g(x(t))+g(x(ta))]+[g(x(t))g(x2)]2[g(x(ta))g(x2)]2}da=[g(x(t))g(x2)][f(x(t))+g(x(t))]12τ0h(a)[g(x(t))g(x(ta))]2da.

    Note that ψ(τ)=0. In view of (T4), we have dV(xt)/dt0. If g is an increasing function, then the result is reached by using a classical Lyapunov theorem (see, e.g., [30]). If g is a nondecreasing function, then the result is proved by using the same argument as in the proof of Theorem 2.6 in [31]. This completes the proof.

    We focus now on the case where g is non-monotone. Suppose that there exists ˆG(x) such that ˆG(x)=ˆg1og(x), where ˆg(.) denotes the restriction of g to the interval [M,). Then, ˆG(x)=x for x[M,) and ˆG(x)>M>x for x[0,M).

    Lemma 3.2. Assume that (T1)–(T4) hold. For a given ϕC[x1,ˆG(x1)]{x1}, let x be the solution of problem (1.1). Then, the following assertions hold:

    (i) if x2<M, then x1<x(t)<ˆG(x1) for all t>0.

    (ii) if x2M and f(ˆG(x1))>g(M), then x1<x(t)<ˆG(x1) for all t>0.

    Proof. Without loss of generality, suppose that x1<ϕ(0)ˆG(x1). First, observe that, for a given ϕC[x1,ˆG(x1)]{x1}, we have f(ˆG(x1))>f(x1)=g(x1)=g(ˆG(x1)), and thus, there exists ϵ>0 such that f(ˆG(x1))ϵ>g(ˆG(x1)). We begin by proving that x1x(t)ˆG(x1) for all t>0. To this end, in view of Lemma 2.3, we only need to prove that x1<xϵ(t):=xϵ(t;ϕ)ˆG(x1) for all t>0, where xϵ(t):=xϵ(t;ϕ) is the solution of problem (1.1) when replacing f by fϵ. Let yϵ:=yϵ(ϕ) be the solution of

    {yϵ(t)=f(yϵ(t))+ϵ+τ0h(a)g+(yϵ(ta))da,t>0,yϵ(t)=ϕ(t),τt0. (3.1)

    with g+(s)=maxσ[0,s]g(σ). Since f is an increasing function, we have f(ˆG(x1))ϵ>g+(ˆG(x1)). Accordingly, the function ˆG(x1) is a super-solution of problem (3.1). Finally, in view of Theorem 2.1, we obtain xϵ(t)yϵ(t)ˆG(x1) for all t0.

    We now prove that xϵ(t)>x1. Suppose, on the contrary, that there exists t1>0 such that xϵ(t1)=x1, xϵ(t)x1 for all tt1, and thus, xϵ(t1)0. Then, the equation of xϵ(t1) satisfies

    xϵ(t1)=f(xϵ(t1))+ϵ+τ0h(a)g(xϵ(t1a))da>f(x1)+g(x1)=0,

    since g(s)g(x1) for all s[x1,ˆG(x1)]. This reaches a contradiction. Further, from Lemma 2.3, we obtain x(t)x1. The claim is proved.

    Next, let X(t)=x(t)x1. Since f(x1)=g(x1), the equation of X(t) satisfies

    X(t)=f(x1)f(x(t))+τ0h(a)(g(x(ta))g(x1))da.

    We know that g(s)g(x1) for all s[x1,ˆG(x1)]. We then have

    X(t)LX(t),

    where L is the Lipschitz constant of f. Consequently x(t)>x1 for all t>0.

    Next, we prove that x(t)<ˆG(x1) for all t>0. Suppose, on the contrary, that there exists t1>0 such that x(t1)=ˆG(x1), x(t)ˆG(x1) for all tt1 and x(t1)0. Then

    0x(t1)=f(x(t1))+τ0h(a)g(x(t1a))da,f(ˆG(x1))+g(M). (3.2)

    Since f is an increasing function and ˆG(x1)>M, we have

    0x(t1)f(M)+g(M). (3.3)

    Now, the assertion x2<M implies that g(M)<f(M), which leads to a contradiction.

    When x2M, inequality (3.2) gives a contradiction by hypothesis. The Lemma is proved.

    Using Lemma 3.2, we next prove the following lemma.

    Lemma 3.3. Suppose that ϕC[x1,ˆG(x1)]{x1}. Assume also that (T1)–(T4) hold. Let x be the solution of problem (1.1). Then, we have

    lim inftx(t)>x1,

    provided that one of the following assertions holds:

    (i) x2<M.

    (ii) x2M and f(ˆG(x1))>g(M).

    Proof. Firstly, from Lemma 3.2, both assertions imply that x1<x(t)<ˆG(x1) for all t0. Next, observe that there exists ϵ>0 such that

    g(s)>f(x1+ϵ)for alls[x1+ϵ,θϵ],

    with θϵ(M,ˆG(x1)) and g(θϵ)=g(x1+ϵ). Indeed, in view of (T3), minσ[x1+ϵ,θϵ]g(σ)=g(θϵ)=g(x1+ϵ) and from (T4), we have g(x1+ϵ)>f(x1+ϵ).

    Now, consider the following nondecreasing function:

    g_(s)={g(s),for 0<s<ˉm,f(x1+ϵ),for ˉm<s<ˆG(x1), (3.4)

    where x1<ˉm<x1+ϵ, which is a constant satisfying g(ˉm)=f(x1+ϵ). From (T2) and (T4), we get g_(s)>f(s) for x1<s<x1+ϵ and g_(s)<f(s) for s>x1+ϵ. Let y(ϕ) be the solution of problem (1.1) when replacing g by g_. Then, according to Theorem 2.1, we have y(t;ϕ)x(t;ϕ). In addition, using Theorem 3.1, we obtain

    x1<limty(t;ϕ)=x1+ϵlim inftx(t;ϕ).

    The lemma is proved.

    Under Lemma 3.3, we prove the following theorem on the global asymptotic stability of the positive equilibrium x2<M.

    Theorem 3.4. Suppose that ϕC[x1,ˆG(x1)]{x1} and x2<M. Assume also that (T1)–(T4) hold. Then, the positive equilibrium x2 is globally asymptotically stable.

    Proof. We first claim that there exists T>0 such that x(t)M for all tT. Indeed, let xϵ:=xϵ(ϕ) be the solution of problem (1.1) when replacing f by f+ϵ. First, suppose that there exists T>0 such that xϵ(t)M for all tT. So, since x2<M, we have from (T4) that g(M)<f(M). Combining this with (T2), the equation of xϵ satisfies

    xϵ(t)f(M)ϵ+g(M)ϵ,

    which contradicts with xϵ(t)M. Hence there exists T>0 such that xϵ(T)<M. We show that xϵ(t)<M for all tT. In fact, at the contrary, if there exists t1>T such that xϵ(t1)=M and so xϵ(t1)0 then,

    xϵ(t1)=f(M)ϵ+g(M)<0,

    which is a contradiction. Further, according to Lemma 2.3, there exists T>0 such that x(t)M for all tT and the claim is proved. Finally, since g is a nondecreasing function over (0,M), the global asymptotic stability of x2 is proved by applying Theorem 3.1.

    Remark 3.5. In the case where g(s)>g(x1) for all s>x1, the above theorem holds true. In fact, it suffices to replace ˆG(x1) in C[x1,ˆG(x1)]{x1} by sups[τ,0]ϕ(s).

    We focus now on the case where x2M. To this end, suppose that there exists a unique constant A such that

    AM and g(A)=f(M). (3.5)

    Lemma 3.6. Assume that (T1)–(T4) hold. Suppose also that x2M and f(A)g(M). For a given ϕC[x1,ˆG(x1)]{x1} and the solution x:=x(ϕ) of problem (1.1), there exists T>0 such that

    Mx(t)A for all tT.

    Proof. We begin by claiming that x2A. On the contrary, suppose that x2>AM. Then, due to (T2) and (T3), we have g(x2)<g(A)=f(M)<f(x2)=g(x2), which is a contradiction. The claim is proved.

    Next, for a given ϕC[x1,ˆG(x1)]{x1}, let xϵ:=xϵ(ϕ) be the solution of problem (1.1) when replacing f by f+ϵ. Since xϵ converges to x as ϵ tends to zero, we only need to prove that there exists T>0 such that xϵ(t)<A for all tT. To this end, let yϵ:=yϵ(ϕ) be the solution of problem (1.1), when replacing f by f+ϵ and g by g+ with g+(s)=maxσ[0,s]g(σ). By Theorem 2.1, we have xϵ(t)yϵ(t) for all t0, and thus, we only need to show that yϵ(t)<A for all tT. On the contrary, we suppose that yϵ(t)A for all t>0. Then, combining the equation of yϵ and the fact that g+(s)=g(M) for all sM, we obtain

    yϵ(t)g(M)f(A)ϵ<0,

    which is a contradiction. Then, there exists T>0 such that yϵ(T)<A. We further claim that yϵ(t)<A for all tT. Otherwise, there exists t1>T such that yϵ(t1)=A and yϵ(t1)0. Substituting yϵ(t1) in Eq (1.1), we get

    0yϵ(t1)=f(yϵ(t1))ϵ+τ0h(a)g+(yϵ(t1a))da,<f(A)+g(M)0,

    which is a contradiction. Consequently, by passing to the limit in ϵ, we obtain that x(t)A for all tT.

    We focus now on the lower bound of x. First, define the function

    g_(s)={g(s),for 0<s<ˉm,f(M),for ˉm<sA, (3.6)

    where x_1^{*} < \bar{m} < M , which is the constant satisfying g(\bar{m}) = f(M) . Note that, due to (T2)–(T4) and condition (3.5), the function \underline{g} is nondecreasing over (0, A) and satisfies

    \begin{equation*} \left \{ \begin{array}{lll} \underline{g}(s)\leq g(s), \quad 0\leq s\leq A,\\ \underline{g}(s) \gt f(s),\quad x^{*}_1 \lt s \lt M,\\ \underline{g}(s) \lt f(s),\quad M \lt s\leq A. \end{array} \right. \end{equation*}

    Let y(\phi) be the solution of problem (1.1) when replacing g by \underline{g} . From Theorem 2.1, we have y(t; \phi)\leq x(t; \phi) for all t > 0 , and from Theorem 3.1, we have \lim\limits_{t\rightarrow \infty}y(t; \phi) = M . Consequently, \liminf\limits_{t\rightarrow \infty}x(t)\geq M . In addition, if, for all T > 0 , there exists (t_n)_n such that t_n > T, y(t_n) = M, \bar{m} < y(s)\leq A for all 0 < s\leq t_n and y'(t_n) < 0 , then, for t_n > T+\tau ,

    \begin{equation*} y'(t_n) = -f(M)+f(M) = 0, \end{equation*}

    which is a contradiction. The lemma is established.

    Denote \bar{f} = f|_{[M, A]} the restriction function of f over [M, A] and G(s): = \bar{f}^{-1}(g(s)) for s\in[M, A] . Now, we are ready to state our main theorem related to x^{*}_2\geq M .

    Theorem 3.7. Under the assumptions of Lemma 3.6, the positive equilibrium x_2^{*} of problem (1.1) is globally asymptotically stable, provided that one of the following conditions holds:

    (H1) fg is a nondecreasing function on [M, A] .

    (H2) f+g is a nondecreasing function over [M, A] .

    (H3) \dfrac{(GoG)(s)}{s} is a nonincreasing function over [M, x_2^*],

    (H4) \dfrac{(GoG)(s)}{s} is a nonincreasing function over [x_2^*, A],

    Proof. Denote x_{\infty}: = \liminf_{t\rightarrow \infty}x(t) and x^{\infty}: = \limsup_{t\rightarrow \infty}x(t) . First, suppose that either x^{\infty}\leq x_2^{*} or x_{\infty}\geq x_2^{*} . For x^{\infty}\leq x_2^{*} , we introduce the following function:

    \begin{equation*} \underline{g}(s) = \left\{ \begin{array}{lll} \min\limits_{\sigma\in [s,x_2^{*}]} g(\sigma)&\quad \mbox{for} &\quad x^{*}_1 \lt s \lt x_2^{*},\\ g(x_2^{*}) &\quad \mbox{for} &\quad x_2^{*}\leq s\leq A. \end{array} \right. \end{equation*}

    Let y(\phi) be the solution of problem (1.1) when replacing g by \underline{g}. Since \underline{g}(s)\leq g(s) for all x_1^{*}\leq s\leq x_2^{*} , we have y(t)\leq x(t) for all t > 0. Further, in view of Theorem 3.1, we obtain

    \lim\limits_{t\rightarrow\infty}y(t) = x_2^{*}\leq \limsup\limits_{t \to \infty} x(t) = x^{\infty}\leq x_2^{*}.

    The local stability is obtained by using the same idea as in the proof of Theorem 2.6 in [31].

    Now, for x_{\infty}\geq x_2^{*} , we introduce the function

    \begin{equation} \bar{g}(s) = \left\{ \begin{array}{lll} \min\limits_{\sigma\in [s,x_2^{*}]} g(\sigma)&\quad \mbox{for} &\quad x_1^{*} \lt s\leq x_2^{*},\\ \max\limits_{\sigma\in [x_2^{*},s]} g(\sigma)&\quad \mbox{for} &\quad x_2^{*} \lt s\leq A, \end{array} \right. \end{equation} (3.7)

    and let y(\phi) be the solution of problem (1.1) when replacing g by \bar{g}. As above, we have x(t)\leq y(t) for all t > 0 and y(t) converges to x_2^{*} as t goes to infinity. Next, we suppose that x_{\infty} < x_2^{*} < x^{\infty} and we prove that it is impossible. Indeed, according to Lemma 3.6, we know that, for every solution x of problem (1.1), we have M\leq x_{\infty}\leq x^{\infty}\leq A .

    Now, using the fluctuation method (see [28,32]), there exist two sequences t_n\rightarrow \infty and s_n\rightarrow \infty such that

    \begin{equation*} \lim\limits_{n\rightarrow \infty}x(t_n) = x^{\infty}, \;\;\ x'(t_n) = 0, \;\ \forall n\geq 1, \end{equation*}

    and

    \begin{equation*} \lim\limits_{n\rightarrow \infty}x(s_n) = x_{\infty}, \;\;\ x'(s_n) = 0, \;\ \forall n\geq 1. \end{equation*}

    Substituting x(t_n) in problem (1.1), it follows that

    \begin{eqnarray} 0 = -f(x(t_n))+\int_0^{\tau}h(a)g(x(t_n-a))da. \end{eqnarray} (3.8)

    Since g is nonincreasing over [M, A] , we obtain, by passing to the limit in Eq (3.8), that

    \begin{equation} f(x^{\infty}) \leq g(x_{\infty}). \end{equation} (3.9)

    Similarly, we obtain

    \begin{equation} f(x_{\infty}) \geq g(x^{\infty}). \end{equation} (3.10)

    Multiplying the expression (3.9) by g(x^{\infty}) and combining with inequality (3.10), we get

    \begin{equation*} f(x^{\infty})g(x^{\infty}) \leq f(x_{\infty})g(x_{\infty}). \end{equation*}

    This fact, together with the hypothesis (H1) , gives x^{\infty}\leq x_{\infty} , which is a contradiction. In a similar way, we can conclude the contradiction for (H2) . Now, suppose that (H3) holds. First, notice that G makes sense, that is, for all s\in[M, A] , the range of g is contained in [\bar{f}(M), \bar{f}(A)] since \bar{f} is strictly increasing over [M, A]. In fact, for all s\in[M, A] and since g is non-increasing over [M, A] , we have g(A)\leq g(s)\leq g(M). Now, using the fact that f(A)\geq f(M) , we show that

    \begin{equation*} \bar{f}(M)\leq g(s)\leq \bar{f}(A) \quad \mbox{for all} \;\; s\in [M,A]. \end{equation*}

    Therefore, the function G is nonincreasing and maps [M, A] to [M, A] . In view of inequalities (3.9) and (3.10) and the monotonicity of \bar{f} , we arrive at

    \begin{equation} x^{\infty} \leq G(x_{\infty}), \end{equation} (3.11)

    and

    \begin{equation} x_{\infty} \geq G(x^{\infty}), \end{equation} (3.12)

    with G(s) = \bar{f}^{-1}(g(s)). Now, applying the function G to inequalities (3.11) and (3.12), we find

    \begin{equation*} x_{\infty} \geq G(x^{\infty}) \geq (GoG)(x_{\infty}), \end{equation*}

    which gives

    \begin{equation} \dfrac{(GoG)(x_{\infty})}{x_{\infty}}\leq 1 = \dfrac{(GoG)(x_2^{*})}{x_2^{*}}. \end{equation} (3.13)

    Due to (H3) , it ensures that x_2^{*}\leq x_{\infty} , which is impossible. Using the same arguments as above, we obtain a contradiction for (H4). The theorem is proved.

    Remark 3.8. In the case where g(s) > g(A) for all s > M, the two above results hold true. In fact, it suffices to replace A in Lemma 3.6 and Theorem 3.7 by B defined in (T1).

    For the tangential case where two positive equilibria x^{*}_1 and x^{*}_2 are equal, we have the following theorem

    Theorem 3.9. Suppose that (T1)–(T3) hold. Suppose that, in addition to the trivial equilibrium, problem (1.1) has a unique positive equilibrium x^{*}_1 . Then

    (i) for \phi\in C_{+}, there exists T > 0 such that 0\leq x(t)\leq M for all t\geq T.

    (ii) \phi\in C_{[x^{*}_1, \hat{G}(x_1^{*})]}\setminus\{x_1^{*}\} implies that x^{*}_1 < x(t)\leq M for all t\geq T.

    (iii) if \phi\in C_{[x^{*}_1, \hat{G}(x_1^{*})]} , then x^{*}_1 attracts every solution of problem (1.1) and x^{*}_1 is unstable.

    Proof. The uniqueness of the positive equilibrium implies that x^{*}_1\leq M and g(x) < f(x) for all x\neq x^{*}_1 . For (), suppose that there exists t_0 > 0 such that x(t)\geq M for all t\geq t_0. Then, by substituting x in Eq (1.1), we get

    \begin{equation*} x'(t)\leq -f(M)+g(M) \lt 0. \end{equation*}

    This is impossible and then there exists T > 0 such that x(T) < M . Next, if there exists t_1 > T such that x(t_1) = M and x(t)\leq M for all t\leq t_1 , then

    \begin{equation*} 0\leq x'(t_1)\leq -f(M)+g(M) \lt 0. \end{equation*}

    Consequently () holds. We argue as in the proof of Lemma 3.2 (ⅰ), to show (). Concerning (), we consider the following Lyapunov functional

    \begin{equation*} \begin{array}{lll} V(\phi)& = & \int_{x^{*}_1}^{\phi(0)}(g(s)-g(x_1^{*}))ds+\dfrac{1}{2} \int_0^{\tau}\psi(a)(g(\phi(-a))-g(x_1^{*}))^2da, \end{array} \end{equation*}

    where \psi(a) = \int_{a}^{\tau}h(\sigma)d\sigma . As in the proof of Theorem 3.1, the derivative of V along the solution of problem (1.1) gives

    \begin{array}{lll} \dfrac{dV({x}_{t})}{dt} = -\dfrac{1}{2} \int_{0}^{\tau}h(a)\big[ g(x(t))-g(x(t-a))\big]^2da+\big[g(x(t))-g(x^{*}_1)\big] \big[ g(x(t))-f(x(t))\big], \quad \forall t \gt 0. \end{array}

    Since g(s) < f(s) for all s\neq x^{*}_1 and g(s)\geq g(x^{*}_1) for all s\in [x^{*}_1, M] , we have dV(x_t)/dt \leq 0 . By LaSalle invariance theorem, x^{*}_1 attracts every solution x(\phi) of problem (1.1) with \phi \in C_{[x^{*}_1, \hat{G}(x_1^{*})]}. From Theorems 2.5 and 3.4, we easily show that x_1 is unstable.

    In this section, we apply our results to the following distributed delay differential equation:

    \begin{equation} x'(t) = -\mu x(t)+ \int_0^{\tau}h(a)\dfrac{kx^{2}(t-a)}{1+2x^3(t-a)}da, \;\ \mbox{for} \;\ t\geq 0, \end{equation} (4.1)

    where \mu, k are positive constants. The variable x(t) stands for the maturated population at time t and \tau > 0 is the maximal maturation time of the species under consideration. h(a) is the maturity rate at age a . In this model, the death function f(x) = \mu x and the birth function g(x) = kx^{2}/(1+2x^3) reflect the so called Allee effect. Obviously, the functions f and g satisfy the assumptions (T1)–(T3) and g reaches the maximum value k/3 at the point M = 1. The equilibria of Eq (4.1) satisfies the following equation:

    \begin{equation} \mu x = \dfrac{kx^{2}}{1+2x^3}. \end{equation} (4.2)

    Analyzing Eq (4.2), we obtain the following proposition.

    Proposition 4.1. Equation (4.1) has a trivial equilibrium x = 0 . In addition

    (i) if \mu > \dfrac{2^{\frac{1}{3}}}{3}k , then Eq (4.1) has no positive equilibrium;

    (ii) if \mu = \dfrac{2^{\frac{1}{3}}}{3}k , then Eq (4.1) has exactly one positive equilibrium x_1^{*} ;

    (iii) if 0 < \mu < \dfrac{2^{\frac{1}{3}}}{3}k , then Eq (4.1) has exactly two positive equilibria x^{*}_1 < x^{*}_2 . Moreover,

    (iii)-a if \dfrac{k}{3} < \mu < \dfrac{2^{\frac{1}{3}}}{3}k, then 0 < x^{*}_1 < x_2^{*} < 1;

    (iii)-b if 0 < \mu\leq \dfrac{k}{3}, then 0 < x^{*}_1 < 1\leq x^{*}_2.

    Using Theorem 2.6, we obtain

    Theorem 4.2. If \mu > \dfrac{2^{\frac{1}{3}}}{3}k, then the trivial equilibrium of Eq (4.1) is globally asymptotically stable for all \phi \in C_{+}.

    For the case (ⅲ)-a in Proposition 4.1 we have the following result

    Theorem 4.3. Suppose that \dfrac{k}{3} < \mu < \dfrac{2^{\frac{1}{3}}}{3}k. Then

    (i) The trivial equilibrium of Eq (4.1) is globally asymptotically stable for all \phi \in C_{[0, x^{*}_1]}\setminus\{x^{*}_1\}.

    (ii) The positive equilibrium x_1^{*} is unstable and the positive equilibrium x^{*}_2 is globally asymptotically stable for all \phi \in C_{[x^{*}_1, \hat{G}(x^{*}_1)]}\setminus\{x^{*}_1\} , where \hat{G}(x^{*}_1)\in [1, \infty) satisfies g(\hat{G}(x^{*}_1)) = g(x^{*}_1).

    (iii) There exists two heteroclinic orbits X^{(1)} and X^{(2)} connecting 0 to x^{*}_1 and x^{*}_1 to x^{*}_2 , respectively.

    Proof. (ⅰ) and (ⅱ) directly follow from Proposition 4.1, (ⅲ)-a and Theorems 2.5 and 3.4.

    For (ⅲ), we follow the same proof as in Theorem 4.2 in [5]. See also [18]. For the sake of completeness, we rewrite it. Let K = \{x^{*}_1\}. Clearly, K is an isolated and unstable compact invariant set in C_{[0, x^{*}_1]} and C_{[x^{*}_1, \hat{G}(x^{*}_1)]}.

    By applying Corollary 2.9 in [33] to \Phi_{|{\bf{R}}^{+}\times C[0, x^{*}_1]} and \Phi_{|{\bf{R}}^{+}\times C[x^{*}_1, \hat{G}(x^{*}_1)]}, respectively, there exist two pre-compact full orbits X^{(1)}:{\bf{R}}\rightarrow C_{[0, x^{*}_1]}\setminus\{0\} and X^{(2)}:{\bf{R}}\rightarrow C_{[x^{*}_1, \hat{G}(x^{*}_1)]}\setminus\{0\} such that \alpha(X^{1}) = \alpha(X^{2}) = K. This together with statements (i) and (ii) gives \omega{(X^{(1)})} = \{0\} and \omega{(X^{(2)})} = \{x^{*}_2\} . In other words, there exist two heteroclinic orbits X^{(1)} and X^{(2)} , which connect 0 to x^{*}_1 and x^{*}_1 to x^{*}_2 , respectively. This completes the proof.

    For the case (ⅲ)-b in Proposition 4.1, we first show that

    \begin{equation} f(A)\geq g(M) \;\ \mbox{with} \;\ g(A) = f(M)\;\ \mbox{and} \;\ A\geq M. \end{equation} (4.3)

    Lemma 4.4. Suppose that f(s) = \mu s . Condition (4.3) holds if and only if M\leq \dfrac{1}{\mu}g(\dfrac{1}{\mu}g(M)).

    Proof. It follows that f(A)\geq g(M) if and only if A\geq f^{-1}(g(M))\geq M . Since g is a decreasing function over [M, \infty) , we have g(A)\leq g(f^{-1}(g(M))) = g(\dfrac{1}{\mu}(g(M))) . Moreover, since g(A) = f(M) , we have M\leq \dfrac{1}{\mu}g(\dfrac{1}{\mu}g(M)) . The lemma is proved.

    Lemma 4.5. For Eq (4.1), condition (4.3) holds if and only if

    \begin{equation} 0 \lt \mu\leq \dfrac{k}{3}. \end{equation} (4.4)

    Proof. By a straightforward computation, we have f^{-1}(x) = x/\mu . For G(x) = g(x)/\mu , we obtain

    \begin{equation*} (GoG)(x) = \dfrac{x(2px^{6}+px^{3})}{8x^{9}+(12+2p)x^{6}+6x^{3}+1}, \end{equation*}

    where p = k^{3}/\mu^{3}. By applying Lemma 4.4, it is readily to see that 1\leq (GoG)(1) holds if and only if

    \begin{equation*} 1\leq \dfrac{3p}{2p+27}, \end{equation*}

    and thus, 0 < \mu\leq k/3.

    Theorem 4.6. Suppose that 0 < \mu \leq \dfrac{k}{3}. Then, the positive equilibrium x^{*}_2 of Eq (4.1) is globally asymptotically stable for all \phi \in C_{[x^{*}_1, \hat{G}(x^{*}_1)]}\setminus\{x^{*}_1\} , where \hat{G}(x^{*}_1)\in [1, \infty) satisfies g(\hat{G}(x^{*}_1)) = g(x^{*}_1).

    Proof. Observe that, in view of Lemmas 4.4 and 4.5, all hypotheses of Lemma 3.6 are satisfied. Now, in order to apply Theorem 3.7, we only show that \dfrac{(GoG)(x)}{x} is nonincreasing over [M, A]. Indeed, by a simple computation, we have

    \begin{equation*} \bigg(\dfrac{(GoG)(x)}{x}\bigg)' = \dfrac{-48px^{14}-48px^{11}-6p^{2}x^{8}+12px^{5}+3px^{2}}{[8x^{9}+(12+2p)x^{6}+6x^{3}+1]^{2}}, \end{equation*}

    where p = k^{3}/\mu^{3}. Finally, observe that

    \begin{equation*} \bigg(\dfrac{(GoG)(x)}{x}\bigg)' \lt 0, \end{equation*}

    for x\geq 1. In this case, both of (H3) and (H4) in Theorem 3.7 hold. This completes the proof.

    Finally, for the tangential case, we have

    Theorem 4.7. Suppose that \mu = \dfrac{2^{\frac{1}{3}}}{3}k. Then

    (i) The trivial equilibrium of Eq (4.1) is globally asymptotically stable for all \phi \in C_{[0, x^{*}_1]}\setminus\{x^{*}_1\}.

    (ii) If \phi \in C_{[x^{*}_1, \hat{G}(x^{*}_1)]}\setminus\{x^{*}_1\} , then the unique positive equilibrium x_1^{*} is unstable and attracts every solution of Eq (4.1).

    Proof. The proof of this theorem follows immediately from Theorems 2.5 and 3.9.

    In this section, we perform numerical simulation that supports our theoretical results. As in Section 4, we assume that

    f(x) = \mu x, \quad g(x) = \frac{kx^2}{1+2x^3},

    and confirm the validity of Theorems 4.2, 4.3, 4.6 and 4.7. In what follows, we fix

    \begin{equation} \tau = 1, \quad k = 1, \quad h(a) = \frac{e^{-(a-0.5\tau)^2 \times 10^{2}}}{\int_0^\tau e^{-(\sigma-0.5\tau)^2 \times 10^{2}} d\sigma} \end{equation} (5.1)

    and change \mu and \phi . Note that \frac{2^{\frac{1}{3}}}{3}k \approx 0.42 and h is a Gaussian-like distribution as shown in Figure 1.

    Figure 1.  Function h in definition (5.1).

    First, let \mu = 0.5 . In this case, \mu > \dfrac{2^{\frac{1}{3}}}{3} k and thus, by Theorem 4.2, the trivial equilibrium is globally asymptotically stable. In fact, Figure 2 shows that x(t) converges to 0 as t increases for different initial data.

    Figure 2.  Time variation of function x for \mu = 0.5 > \dfrac{2^{\frac{1}{3}}}{3} k \approx 0.42 .

    Next, let \mu = 0.37 . In this case, \dfrac{k}{3} < \mu < \dfrac{2^{\frac{1}{3}}}{3}k and x_1^* \approx 0.43 , x_2^* \approx 0.89 and \hat{G}(x_1^*) \approx 3.09 (Figure 3). Hence, by Theorem 4.3, we see that the trivial equilibrium is globally asymptotically stable for \phi \in C_{[0, x_1^*]} \setminus \{ x_1^* \} , whereas the positive equilibrium x_2^* is so for \phi \in C_{[x_1^*, \hat{G}(x_1^*)]} \setminus \{ x_1^* \} . In fact, Figure 4 shows such a bistable situation. Moreover, heteroclinic orbits X^{(1)} and X^{(2)} , which were stated in Theorem 4.3 (ⅲ), are shown in Figure 5.

    Figure 3.  Functions f and g for \mu = 0.37 .
    Figure 4.  Time variation of function x for \mu = 0.37 \in \left(\dfrac{k}{3}, \dfrac{2^{\frac{1}{3}}}{3} k \right) \approx (0.33, 0.42) .
    Figure 5.  Heteroclinic orbits X^{(1)} and X^{(2)} in the x - x' plane for \mu = 0.37 .

    Thirdly, let \mu = 0.27 < \dfrac{k}{3} \approx 0.33 . In this case, we have x_1^* \approx 0.28 , x_2^* \approx 1.2 and \hat{G}(x_1^*) \approx 6.52 (Figure 6). Thus, by Theorem 4.6, we see that the positive equilibrium x_2^* is globally asymptotically stable for \phi \in C_{[x_1^*, \hat{G}(x_1^*)]} \setminus \{ x_1^* \} . In fact, Figure 7 shows that x(t) converges to x_2^* as t increases for different initial data.

    Figure 6.  Functions f and g for \mu = 0.27 .
    Figure 7.  Time variation of function x for \mu = 0.27 < \dfrac{k}{3} \approx 0.33 .

    Finally, let \mu = \dfrac{2^{\frac{1}{3}}}{3}k \approx 0.42 . In this case, we have x_1^* \approx 0.63 and \hat{G}(x_1^*) \approx 1.72 (Figure 8). By Theorem 4.7, we see that the trivial equilibrium is globally asymptotically stable for \phi \in C_{[0, x_1^*]} \setminus \{ x_1^* \} , whereas x_1^* is unstable and attracts all solutions for \phi \in C_{[x_1^*, \hat{G}(x_1^*)]} \setminus \{ x_1^* \} . In fact, Figure 9 shows such two situations.

    Figure 8.  Functions f and g for \mu = \dfrac{2^{\frac{1}{3}}}{3}k \approx 0.42 .
    Figure 9.  Time variation of function x for \mu = \dfrac{2^{\frac{1}{3}}}{3}k \approx 0.42 .

    In this paper, we studied the bistable nonlinearity problem for a general class of functional differential equations with distributed delay, which includes many mathematical models in biology and ecology. In contrast to the previous work [5], we considered both cases x_2^* < M and x_2^* \geq M , and obtained sufficient conditions for the global asymptotic stability of each equilibrium. The general results were applied to a model with Allee effect in Section 4, and numerical simulation was performed in Section 5. It should be pointed out that our theoretical results are robust for the variation of the form of the distribution function h . This might suggest us that the distributed delay is not essential for the dynamical system of our model. We conjecture that our results still hold for \tau = \infty , and we leave it for a future study. Extension of our results to a reaction-diffusion equation could also be an interesting future problem.

    We deeply appreciate the editor and the anonymous reviewers for their helpful comments to the earlier version of our manuscript. The first author is partially supported by the JSPS Grant-in-Aid for Early-Career Scientists (No.19K14594). The second author is partially supported by the DGRSDT, ALGERIA.

    All authors declare no conflicts of interest in this paper.



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