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Research article

Global structure of positive solutions for third-order semipositone integral boundary value problems

  • Received: 20 December 2023 Revised: 31 January 2024 Accepted: 01 February 2024 Published: 20 February 2024
  • MSC : 34B10, 34B18

  • In this paper, we were concerned with the global behavior of positive solutions for third-order semipositone problems with an integral boundary condition

    y+βy+αy+λf(t,y)=0,t(0,1),y(0)=y(0)=0,y(1)=χ10y(s)ds,

    where α(0,) and β(,) are two constants, λ,χ are two positive parameters, and fC([0,1]×[0,),R) with f(t,0)<0. Our analysis mainly relied on the bifurcation theory.

    Citation: Zhonghua Bi, Sanyang Liu. Global structure of positive solutions for third-order semipositone integral boundary value problems[J]. AIMS Mathematics, 2024, 9(3): 7273-7292. doi: 10.3934/math.2024353

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  • In this paper, we were concerned with the global behavior of positive solutions for third-order semipositone problems with an integral boundary condition

    y+βy+αy+λf(t,y)=0,t(0,1),y(0)=y(0)=0,y(1)=χ10y(s)ds,

    where α(0,) and β(,) are two constants, λ,χ are two positive parameters, and fC([0,1]×[0,),R) with f(t,0)<0. Our analysis mainly relied on the bifurcation theory.



    Third-order boundary value problems emerge in applied mathematics and physics, which have been investigated via various methods; see [1,2,3,4,5,6,7]. In [5], by using the disconjugacy theory [8], Ma and Lu obtained the optimal intervals to guarantee that the Green's functions corresponding to the third order linear problem are of one sign. Such results have been extended in [1].

    Lemma 1.1. ([1,Theorem 2.1]) Suppose that there exist two constants α(0,) and β(,), which satisfy

    0<4αβ2<π2andeβ24αβ2+βsin(4αβ22)4αβ2cos(4αβ22). (1.1)

    The property of disconjugacy of y+βy+αy=0 is valid.

    To our knowledge, the behavior of positive solutions for second-order boundary value problems under the semipositone condition have been discussed extensively; see [9,10,11,12,13,14,15]. It is worth noting that based upon the bifurcation theory, excluding Ambrosetti et al. [9] who explored semipositone elliptic problems under the linear boundary condition, recently, Ma and Wang [14,15] proved the solvability for second-order problems with a nonlinear boundary condition. In comparison to second-order cases, the challenge in studying third-order cases lies in that the third-order differential operator is non-self-adjoint.

    Inspired by the aforementioned literature, we investigate the behavior of positive solutions for the following problems

    y+βy+αy+λf(t,y)=0,t(0,1),y(0)=y(0)=0,y(1)=χ10y(s)ds, (1.2)

    where α(0,) and β(,) are two constants, λ,χ are two positive parameters, and

    f:[0,1]×[0,)R

    is continuous and satisfies that the following assumption holds.

    (F1) (semipositone) f(t,0)<0,t[0,1].

    The integral boundary conditions can be decomposed into various different situations, such as multipoint and nonlocal boundary conditions. As far as we are concerned, few literatures on the solvability of third-order semipositone problems with an interval boundary condition since the maximum principle may fail. In order to overcome this difficulty, as a first step, Henderson [3] translated the third-order semipositone problems into the positone problems under the three-point boundary condition. Concurrently, we found that they only showed that the discussed problems have at least one positive solution. No detailed information about the global behavior of solutions was investigated since the spectral structure of third-order linear eigenvalue problems has not yet been found. Meanwhile, in order to use the fixed point index theory and bifurcation theory to show our results, we need to obtain a suitable cone utilizing the property of Green's function in a Banach space. Recently, Cabada used a new technique to construct a smaller positive cone (see [16,17]).

    To sum up, we first obtain the relationship between the Green's function of (1.2) and the Green's function corresponding to (1,2), (2,1)-conjugate boundary conditions. As we will see, the expression of G(t,s) of (1.2) is so complex with two parameters that it is hard to obtain the suitable cone. Fortunately, we look for a condition showed in [18] to avoid the complex computations by the form of G(t,s) of (1.2). We will be concerned with the limits of G(t,s)/G(1,s) as s=0 and s=1.

    (PG) There is a continuous function ϕ: [0,1](0,) and k1,k2C[0,1] such that 0<k1(s)<k2(s) satisfies

    ϕ(t)k1(s)G(t,s)ϕ(t)k2(s),(t,s)[0,1]×[0,1].

    Finally, relying on the linear behavior of f(t,y) as y, we are employed to consider three cases. All conclusions are derived that a global branch of solutions for (1.2) exists and bifurcates from infinity. As λ nears to the bifurcation point, we demonstrate that the solutions of large norms are still positive, enabling the application of bifurcation theory or topological methods.

    From Lemma 1.1, by means of the method introduced in [8,Page 105–106], we enunciate the following expression of the Green's function.

    Theorem 2.1. Assume that Lemma 1.1 is satisfied, then,

    y(t)+βy(t)+αy(t)=h1(t),t(0,1),y(0)=y(0)=y(1)=0, (2.1)

    has a unique solution given by

    y(t)=10G1(t,s)h1(s)ds,

    where h1X and G1(t,s) is the Green's function of (2.1), that is,

    G1(t,s)={eβ(st)g(t)g(1s)α4αβ2g(1),0ts1,eβ(st)(g(t)g(1s)g(1)g(ts))α4αβ2g(1),0st1, (2.2)

    where

    g(ξ)=4αβ24αβ2eβξ2cos(4αβ22ξ)βeβξ2sin(4αβ22ξ),ξ[0,1].

    Corollary 2.1. The following properties of the function G1(t,s) are satisfied.

    (1) G1(t,s) is negative on (t,s)(0,1)×(0,1).

    (2) G1(0,s)=G1t(0,s)=G1(1,s)=0, s(0,1).

    (3) G1(t,1)=G1s(t,1)=G1(t,0)=0, t(0,1).

    (4) 2G1t2(0,s)<0<G1t(1,s), s(0,1).

    (5) G1s(t,0)<0 and 2G1s2(t,1)<0, t(0,1).

    First of all, we point out that the following problem

    y(t)+βy(t)+αy(t)=0,t(0,1),y(0)=y(0)=0,y(1)=1, (2.3)

    has no solution if, and only if, Lemma 1.1 is fulfilled.

    In another case, (2.3) has a unique solution ω, that is,

    ω(t)=g(t)g(1). (2.4)

    Obviously, ω(t) is positive on t(0,1]. Moreover, by denoting

    D:=10ω(τ)dτ>0,

    we have that it is given by the following expression

    D=(2αβ)4αβ2+2β4αβ2eβ2cos(4αβ22)2(2αβ2)eβ2sin(4αβ22)2αg(1). (2.5)

    From the disconjugacy theory, if we investigate the following problem (h2X),

    v+βv+αv=h2(t),t(0,1),v(0)=v(1)=v(1)=0, (2.6)

    one can know that (2.6) is the adjoint of (2.1) and the Green's function of (2.6) satisfies

    G2(t,s)=G1(s,t).

    Furthermore, we have

    z(s)=10G2(s,η)dη=10G1(η,s)dη (2.7)

    as the unique solution satisfying the following problem

    z(s)+βz(s)+αz(s)=1,t(0,1),z(0)=z(1)=z(1)=0. (2.8)

    Moreover, from Lemma 1.1 and [8,Theorem 11], z(s)>0 on s(0,1), z(s)>0, and z(1)>0.

    Lemma 2.1. Suppose that Lemma 1.1 is satisfied. The problem

    y(t)+βy(t)+αy(t)+(t)=0,t(0,1),y(0)=y(0)=0,y(1)=χ10y(s)ds, (2.9)

    is equivalent to the following integral equation if, and only if, χD1, where X, i.e.,

    y(t)=10G(t,s)(s)ds,

    where G(t,s) represents the Green's function of (2.9), i.e.,

    G(t,s)=G1(t,s)χω(t)1λD10G1(η,s)dη, (2.10)

    and ω(t) and D are given in (2.4) and (2.5), respectively.

    Proof. Since (2.1) and (2.4) have the solution v and ω, respectively, then the unique solution of (2.9) is

    y=v+χω10y(s)ds.

    Furthermore,

    y=10G1(t,s)(s)ds+χω(t)10y(s)ds. (2.11)

    Put

    A:=10y(s)ds,

    then integrating (2.11) on the interval (0,1), we get

    A=1010G1(η,s)(s)dsdη+χ10ω(η)10y(s)dsdη=1010G1(η,s)(s)dsdη+χA10ω(η)dη,

    and, furthermore,

    A=10(s)10G1(η,s)dηds1χ10ω(η)dη.

    Replacing A in (2.11), we arrive at

    y=10G1(t,s)(s)dsχω(t)10(s)10G1(η,s)dηds1χ10ω(η)dη.

    Next, we give a careful analysis of G(t,s), and the following theorem is fulfilled.

    Theorem 2.2. From Lemma 1.1 and χ(0,1D), we derive that G(t,s)>0 on (t,s)[0,1]×[0,1].

    Proof. From Corollary 2.1, G1(t,s) is negative. Next, by the property of (PG), we show that there is a positive constant R and a function lX satisfying l(t)>0 on t(0,1] and l(0)=0, for which the following result is satisfied, that is,

    l(t)χ1χDz(s)G(t,s)Rχ1χDz(s),(t,s)[0,1]×[0,1]. (2.12)

    Now, set

    ψ(t,s):=G(t,s)G(1,s)=1χDχG1(t,s)10G1(η,s)dη+ω(t). (2.13)

    Obviously, ψ(t,s) is continuous and

    G(1,s)=χ1χD10G1(η,s)dη.

    By Corollary 2.1, z(s) is the solution of (2.8). From the L'Hôptial rule, we derive

    lims0+G1(t,s)10G1(η,s)dη=lims0+G1(t,s)z(s)=lims0+G1s(t,s)z(s)=G1s(t,0)z(0)>0.

    Thus,

    lims0+ψ(t,s)=1χDχ(G1s(t,0)z(0))+ω(t):=ψ1(t)>0.

    Analogously,

    lims1G1(t,s)10G1(η,s)dη=lims1G1(t,s)z(s)=lims12G1s2(t,s)z(s)=2G1s2(t,1)z(1)>0

    and

    lims1ψ(t,s)=1χDχ(2G1s2(t,1)z(1))+ω(t):=ψ2(t)>0.

    Since the limits functions ψ1(t) and ψ2(t) exist and are finite, then at s=0,1, ψ(t,s) has removable discontinuities, so it can be extended to a function ˉψC((0,1)×(0,1)). Therefore,

    l(t)=mins[0,1]ˉψ(t,s)

    is continuous such that

    l(0)=0,0<l(t)ˉψ(t,s)R:=max(t,s)[0,1]×[0,1]ˉψ(t,s).

    Corollary 2.2. From Lemma 1.1 and χ(0,1D), for all constant δ(0,1), there is a constant γ(0,1) depending on δ, such that the following result is satisfied, that is,

    γχ1χDz(s)G(t,s),(t,s)[δ,1]×[0,1]. (2.14)

    Proof. The result is satisfied from the definition of the function l.

    Define a cone

    P:={yX|y(t)0,t[0,1]andmint[δ,1]y(t)γy}, (2.15)

    where δ satisfies Corollary 2.2.

    Remark 2.1. In this paper, in view of the boundary conditions, the integral boundary condition is more widely used than the obtained third-order conjugate boundary condition in [1,5].

    The work space is X=C[0,1] with the norm

    y:=maxt[0,1]|y(t)|.

    We also set

    Br:={yX:y<r}

    with r>0.

    Next, by means of the Krein-Rutman theorem [19,Theorem 19.3 (a)], we discuss the existence of the principal eigenvalue for the linear eigenvalue problem as follows

    yβyαy=λb(t)y,t(0,1),y(0)=y(0)=0,y(1)=χ10y(s)ds. (3.1)

    Denote A: PX as the map,

    Ay(t):=λ10G(t,s)b(s)y(s)ds,t[0,1].

    Lemma 3.1. Assume that χ(0,1D) and Lemma 1.1 are satisfied, (3.1) has a principal eigenvalue λ1, and the corresponding eigenfunction ϕ1(t) is positive.

    Proof. It follows from (2.15), then P is normal and has nonempty interior, so X=¯PP. From Theorem 2.2, A is a strong positive operator and AintP. By the Krein-Rutman theorem, the spectral radius r(A) is positive, and ϕ1X exists, satisfying ϕ1>0 and Aϕ1=r(A)ϕ1. Thus,

    λ1=(r(A))1>0.

    Let A be the conjugate operator of A, then Aϕ2=r(A)ϕ2, where ϕ2X such that ϕ2>0 on (0,1), corresponding to λ1. Since

    10(Aϕ1)ϕ2dt=λ110ϕ1ϕ2dt=10ϕ1(Aϕ2)dt,

    then the algebraic multiplicity of λ1 is 1. Thus, λ1 is the principal eigenvalue of (3.1).

    Denote a nonlinear operator K: XX by

    y:=K.

    From the above notation, it follows that (1.2) is equivalent to

    yλKf(,y)=0,yX. (3.2)

    Throughout the paper, we will use the same symbol to represent both the function and the corresponding Nemytskii operator.

    We denote if there is a sequence (μn,yn) with μnλ and ynX, such that ynμnKf(yn)=0 and yn, then λ is a bifurcation from infinity for (3.2).

    In some cases, such as what we will later discuss in detail, by application of an appropriate rescaling to look for bifurcation from infinity based on the Leray-Schauder topological degree, which represents deg(,,), it is worth noting that K is continuous and compact. Thus it is reasonable to investigate the topological degree of IλKf, where I is the identity map.

    Theorem 3.1. Assume that χ(0,1D) and Lemma 1.1 are satisfied. We suppose that

    fC([0,1]×[0,),R)

    satisfies (F1) and (F2).

    (F2) There exists a positive function cX such that

    limyf(,y)y=c.

    There is a positive constant ε such that (1.2) exist positive solutions if either

    (Ⅰ) υ1>0 (possibly ) in [0,1] and λ[λε,λ),

    or

    (Ⅱ) υ2<0 (possibly ) in [0,1] and λ(λ,λ+ε],

    where

    λ:=λ1c

    and

    υ1(t):=lim infy(f(t,y)cy),υ2(t):=lim supy(f(t,y)cy).

    In order to show Theorem 3.1 is valid, we first extend f(t,) to R and set

    F(t,y):=f(t,|y|) (3.3)

    and

    Φ(λ,y):=yλKF(t,y),yX. (3.4)

    Obviously, the solution y>0 of Φ(λ,y)=0 is equivalent to a positive solution of (1.2).

    Lemma 3.2. Assume that χ(0,1D) and Lemma 1.1 are satisfied. For every compact interval [0,+){λ}, there is a positive constant r such that, for all λ, yr, Φ(λ,y)0. Moreover,

    (1) if υ1>0, then =[λ,λ], forλ>λ;

    (2) if υ2<0, then =[0,λ].

    Proof. Let μnμ0, that is, μλ and yn be such that

    yn=μnKF(t,yn).

    Setting

    wn:=ynyn,

    it follows that

    wn=μnyn1KF(t,yn).

    From (F2) and (3.3), up to a subsequence in X, wnw is fulfilled for which w satisfies the problem

    wβwαw=μc|w|,t(0,1),w(0)=w(0)=0,w(1)=χ10w(s)ds,

    and w=1. By Theorem 2.2, w0, then μc=λ1, i.e., μ=λ, which contradicts with the assumption of μλ.

    Next, we give a short illustration of Lemma 3.2 (1), and (2) follows similarly. Now, assume that there is a sequence (μn,yn)(0,)×X, where μnλ, yn, and μn>λ, such that

    Φ(μn,yn)=0. (3.5)

    Note that ynX has a unique decomposition

    yn=vn+snϕ1, (3.6)

    where snR, since yn>0, ϕ2>0, and

    10vn(t)ϕ2(t)dt=0,

    and by (3.6), we obtain

    sn=(10un(t)ϕ2(t)dt)(10ϕ1(t)ϕ2(t)dt)1>0,nN. (3.7)

    From (3.5), it follows that

    10ϕ2(t)yn(t)dt=μn10ϕ2(t)KF(t,yn(t))dt.

    Since

    10ϕ2(t)KF(t,yn(t))dt=λ110ϕ2(t)yn(t)dt,

    we obtain

    λ110ϕ2(t)yn(t)dt=10(yn+βyn+αyn)ϕ2(t)dt=10μnf(t,yn(t))ϕ2(t)dt=10μn(f(t,yn(t))cyn(t))ϕ2(t)dt10μnyn(t)cϕ2(t)dt,

    then

    (μncλ1)10ϕ2(t)yn(t)dt=10μn(f(t,yn(t))cyn(t))ϕ2(t)dt.

    For n large enough, μn>λ and

    10yn(t)ϕ2(t)dt>0

    hold, then we infer that

    10(f(t,yn(t))cyn(t))ϕ2(t)dt<0

    and from the Fatou lemma, we yield

    0lim infn10(f(t,yn)cyn)ϕ2dt10υ1ϕ2dt,

    which contradicts with υ1>0.

    Lemma 3.3. Assume that χ(0,1D) and Lemma 1.1 are satisfied. For λ(λ,), there is a positive constant r such that

    Φ(λ,y)τϕ1,forallτ0,yr.

    Proof. If there is a sequence {yn}X with yn and numbers τn0 satisfy Φ(λ,yn)=τnϕ1, then

    ynβynαyn=λF(t,yn)+τnλ1ϕ1.

    Since F(t,y)c|y| and τnλ1ϕ10, to the maximum principle, yn>0 for all t(0,1).

    Choose ϵ>0 such that

    λ<λ(1ϵ).

    From condition (F2), there is a constant R0>0 such that

    f(t,y)(1ϵ)cy,y>R0,t(0,1).

    By yn, we know there is a positive constant N such that

    yn>R0,nN

    and

    f(t,yn)(1ϵ)cyn. (3.8)

    By (3.7) and (3.8), we derive

    snλ110ϕ1(t)ϕ2(t)dt=10(yn+βyn+αyn)ϕ2(t)dt=λ10F(t,yn)ϕ2(t)dt+τλ110ϕ1(t)ϕ2(t)dtλ10F(t,yn)ϕ2(t)dtλ10(1ϵ)cyn(t)ϕ2(t)dt=λ(1ϵ)csn10ϕ1(t)ϕ2(t)dt.

    Thus,

    λλ(1ϵ),

    which is a contradiction.

    For y0, we set

    Z:=yy2.

    Set

    Ψ(λ,Z):=Φ(λ,y)y2=yλKF(t,y)y2=ZλZ2KF(t,ZZ2).

    λ is a bifurcation from infinity for (3.4) if, and only if, λ is a bifurcation from Z=0 for Ψ=0. From Lemma 3.2, for λ(0,λ), by homotopy, we have

    deg(Ψ(λ,),B1r,0)=deg(Ψ(0,),B1r,0)=deg(I,B1r,0)=1. (3.9)

    Similarly, from Lemma 3.2, for τ[0,1] and λ(λ,),

    deg(Ψ(λ,),B1r,0)=deg(Ψ(0,)τϕ2,B1r,0)=deg(Ψ(0,)ϕ1,B1r,0)=0. (3.10)

    Set

    Σ:={(λ,y)[0,)×X:y0,Φ(λ,y)=0}.

    It follows from (3.9) and (3.10) that the following lemma is fulfilled.

    Lemma 3.4. λ is a bifurcation from infinity for (3.4). To be precise, there is an unbounded, closed, connected set ΣΣ which bifurcates from infinity. Furthermore, if υ1>0, Σ bifurcates to the left (respectively, if υ2<0, Σ bifurcates to the right).

    Proof of Theorem 3.1. From Lemmas 3.2–3.4, it is sufficient to demonstrate that if μnλ and yn, for all t(0,1) and n large enough, yn is positive. Set

    wn=ynyn.

    By utilizing the preceding results up to subsequence in X, wnw and w=ϑϕ1 with ϑ>0 are satisfied, then, yn>0 for n large enough.

    Theorem 3.2. Assume that χ(0,1D) and Lemma 1.1 are satisfied. We suppose that

    fC([0,1]×[0,),R)

    satisfies (F1) and (F3).

    (F3) There is a positive function cX such that

    limyf(,y)yp=c,p(1,).

    There is a constant λ>0, such that for λ(0,λ], (1.2) exist positive solutions. More specifically, there exist a connected set of positive solutions for (1.2), which bifurcates from infinity at λ=0.

    Set

    F(t,y):=F(t,y)c|y|p,

    where F(t,y) is denoted as (3.3).

    Next, using the rescaling w=dy and λ=dp1 with d>0, shows that λ=0 is a bifurcation from infinity for

    yλKF(t,y)=0, (3.11)

    which is equivalent to (λ,y), which is a solution of (3.11) if, and only if,

    wKˆF(d,w)=0, (3.12)

    where

    ˆF(d,w):=c|w|p+dpF(d1w).

    As d=0, we set

    ˆF(0,w):=c|w|p.

    By (F3), we know that ˆF(d,w) is continuous for (d,w)[0,)×R. Let

    ˜S(d,w):=wKˆF(d,w),d(0,).

    Thus, ˜S(d,) is compact. For d=0, solution of ˜S(0,w)=0 are nothing but solutions of

    wβwαw=c|w|p,t(0,1),w(0)=w(0)=0,w(1)=χ10w(s)ds. (3.13)

    Now, we show that the following results are fulfilled:

    ˜S(0,w)0,for allwR2, (3.14)
    ˜S(0,w)0,for all0<wR1, (3.15)
    deg(˜S(0,w),PR¯Pr,0)=1,forr(0,R1],R[R2,), (3.16)

    where R1,R2 are two constants with 0<R1<R2.

    First, we claim that there exists R>0 such that for wR, ˜S(0,w)0.

    Assume that (3.13) has a sequence {wn} satisfying

    limnwn=,

    i.e.,

    wnβwnαwn=(c|wn|p1)wn,t(0,1),wn(0)=wn(0)=0,wn(1)=χ10wn(s)ds.

    Note that

    limnc|wn|p1=,t(0,1).

    By the remarks in the final paragraph on [20,Page 56], wn must change its sign in (0,1), which is a contradiction.

    Second, we prove that for 0<wR1, ˜S(0,w)0, where R1>0 is a constant.

    On the contrary, if (3.15) does not hold, then (3.13) exists a sequence of solutions wn, which satisfies

    wn0,n. (3.17)

    Let

    vn=wnwn.

    From (3.13), we have

    vnβvnαvn=c|wn|pwn,t(0,1),vn(0)=vn(0)=0,vn(1)=χ10vn(s)ds.

    From (3.17), we have

    limnvn=0

    uniformly in t(0,1).

    According to the standard argument, after taking a subsequence and relabeling if necessary, vX exists and v=1, satisfying

    vnv,n

    and

    vβvαv=0,t(0,1),v(0)=v(0)=0,v(1)=χ10v(s)ds,

    which shows that v=0 holds. However, this is a contradiction. Hence, (3.15) holds.

    In the end, we show (3.16) is valid. Denote

    Pr:={yP:y<r}.

    Now, by (3.14) and (3.15), we derive that for all wPR and wPr, ~S(0,w)0 is valid. So for all

    w(PR¯Pr), ˜S(0,w)0

    is still fulfilled. Hence, deg(˜S(0,w),PR¯Pr,0) is well defined.

    Note that ˜f(w)=|w|p, so we claim that

    deg(˜S(0,w),PR¯Pr,0)=1.

    It is easy to verify the following conditions:

    (H1) f0:=limw0+˜f(w)w=0;

    (H2) f:=limw+˜f(w)w=.

    Choose M1>0, which satisfies

    M1γχ1χD10z(s)c(s)ds>1.

    From (H2), there is a constant R2>0, such that wR2, f(w)>M1w is satisfied. Choosing R>max{R1,R2}, we claim that

    KˆF(0,w)>w

    for wPR. In fact, for wPR,

    (KˆF(0,w))(t)=10G(t,s)c(s)|w|pdsM1γχ1χDw10z(s)c(s)ds>w.

    Hence, it follows from the fixed point index theorem of [19] that

    i(KˆF(0,),PR,P)=0. (3.18)

    From (H1), there exists a constant ι>0 such that w[0,ι], and

    ˜f(w)M2w,

    for which M2>0 satisfies

    M2Rχ1χD10z(s)c(s)ds1.

    Choose 0<r<min{ι,R2}, for wPr,

    KˆF(0,w)=maxt[δ,1]10G(t,s)c(s)|w|pdsM2Rχ1χDw1δz(s)c(s)dsw.

    Obviously, KˆF(0,w)w for wPr. By the fixed point index theorem of [19],

    i(KˆF(0,),Pr,P)=1. (3.19)

    From the additivity of the fixed point index, (3.18), and (3.19), we get

    i(KˆF(0,),PR¯Pr,P)=1. (3.20)

    From (3.20) and

    ˜S(0,w):XPR¯Pr,

    we obtain

    deg(˜S(0,w),PR¯Pr,0)=1.

    Lemma 3.5. Assume that χ(0,1D) and Lemma 1.1 are fulfilled, then there is a constant d0>0 such that

    (ⅰ) deg(˜S(d,),PR¯Pr,0)=1, d[0,d0];

    (ⅱ) if ˜S(d,w)=0,d[0,d0],w[r,R], then w>0.

    Proof. On the contrary, there is a sequence (dn,wn), where dn0, wn{r,R}, and

    wn=KˆF(dn,wn).

    Since the operator K is compact, up to a subsequence, wnw and

    ˜S(0,w)=0,wn{r,R},

    which contradicts with (3.14) and (3.15). Therefore, (ⅰ) holds.

    In order to show (ⅱ) is valid, we once argue by means of contradiction. From the preceding results, we can present a sequence {wn}X with {t[0,1]:wn0}, which satisfies wnw, w{r,R} and ˜S(0,w)=0; that is, w is a solution of (3.13). From Theorem 2.2, w>0 holds. Therefore, we get a contradiction that for n large enough, wn is positive.

    Proof of Theorem 3.2. From Lemma 2.1, d[0,d0], and (3.12) exists a positive solution wd. Recalling for d>0 and λ=dp1, y=wd is a solution (λ,yλ) of (3.11) for

    0<λ<λ:=dp10.

    For wd>0, (λ,yλ) is the positive solution of (1.2). So, d[0,d0], wdd implies that

    yd=wd/d

    as d0.

    Theorem 3.3. Assume that χ(0,1D) and Lemma 1.1 are satisfied. We suppose that

    fC([0,1]×[0,),R)

    satisfies (F1) and (F4).

    (F4) There is a positive function cX, such that

    limyf(,y)yq=c,q[0,1).

    There is a constant λ>0, such that for λ[λ,), positive solutions of (1.2) exist. To be precise, there is a connected set of positive solutions of (1.2), which bifurcates from infinity for λ=.

    In this case, we will show that (1.2) exist positive solutions, which branch off from as λ=. As the same treatment as the superlinear problems, we again use w=dy, λ=dq1 with q replacing p. In the case of superlinear problems, (λ,y) is the solution of (3.11) if (d,w) satisfies (3.12). Now, by q[0,1),

    λd0. (3.21)

    Lemma 3.6. Assume that χ(0,1D) and Lemma 1.1 are satisfied. For q(0,1), the problem

    w(t)βw(t)αw(t)=c(t)wq(t),t(0,1),w(0)=w(0)=0,w(1)=χ10w(s)ds (3.22)

    has a unique positive solution w0.

    Proof. Assume that w1, w2 are two positive solutions of (3.22), i.e.,

    w1(t)βw1(t)αw1(t)=c(t)wq1,w1(0)=w1(0)=0,w1(1)=χ10w1(s)ds,w2(t)βw2(t)αw2(t)=c(t)wq2,w2(0)=w2(0)=0,w2(1)=χ10w2(s)ds.

    We will show that w1w2 and w2w1. If w1w2, we discuss that the element ˉw satisfies the following form, that is,

    ˉw(t)=w1(t)ϵw2(t),t[0,1],

    where ϵ(0,1). Let there exist a point ζ0(0,1) such that

    ˉw(ζ0)=w1(ζ0)ϵ0w2(ζ0)=0. (3.23)

    On the other hand,

    ˉw(t)βˉw(t)αˉw(t)=(w1(t)ϵ0w2(t))β(w1(t)ϵ0w2(t))α(w1(t)ϵ0w2(t))=c(t)[wq1(t)ϵ0wq2(t)]c(t)[ϵq0wq2(t)ϵ0wq2(t)]>0,ˉw(0)=ˉw(0)=0,ˉw(1)=χ10ˉw(s)ds.

    So, ˉw(t)>0, which contradicts with (3.23). Therefore, w1w2. By the same method, we may prove that w1w2.

    As the same treatment as the superlinear case, we also can obtain that the following results are fulfilled:

    ˜S(0,w)0for allwR3, (3.24)
    ˜S(0,w)0for all0<wR4, (3.25)

    and

    deg(˜S(0,w),PR¯Pr,0)=1forr(0,R3],R[R4,), (3.26)

    where R3 and R4 are two constants satisfying 0<R3<R4. Therefore, the following results can be obtained.

    Lemma 3.7. Assume that χ(0,1D) and Lemma 1.1 are satisfied. There exists d0>0 such that

    (ⅰ) deg(˜S(d,),PR¯Pr,0)=1, d[0,d0];

    (ⅱ) If ˜S(d,w)=0,d[0,d0],w[r,R], then w>0.

    Proof of Theorem 3.3. By the continuation, a connected subset Γ of solutions of ˜S(d,w)=0 satisfies (0,w0)Γ. From Lemma 3.7, we derive that there is a positive constant d0, such that for d<d0, these solutions are positive. Since λ=dq1, u=wd, then a connected subset of solutions of (1.2) exists, which it can be transformed by Γ, so for

    λ>λ:=dq10,

    these solutions are positive. Furthermore, from (3.21), Σ bifurcates from infinity for λ=+.

    Remark 3.1. Comparing with [3], we no longer consider the nonlinear term as a translation and we obtain the optimal interval of the parameter λ under the three cases.

    In order to illustrate that Theorems 3.1–3.3 are valid, we take α=1 and β=0, then Lemma 1.1 holds and

    χ(0,1cos(1)1sin(1)).

    If

    f(t,y)=2y˜p+tln(1+y)t1,˜p[0,),

    then f(t,0)=t1<0 is satisfied for t[0,1]. Let λ1 be the first positive eigenvalue corresponding to the linear problem (3.1) and ϕ1 be the positive eigenfunction corresponding to λ1. Next, we will check that all conditions in Theorems 3.1–3.3 are fulfilled. In fact,

    c=limyf(t,y)y˜p=2.

    In view of Theorem 3.1, λ=λ12, where ˜p=1,

    υ1(t)=lim infy(f(t,y)cy)=lim infy(tln(1+y)t1)>0,t[0,1].

    Thus, Theorem 3.1 is valid. In the case of the superlinear problem,

    λ:=λ12y1˜p,

    where ˜p>1, so we derive that

    limyλ=λ12y1˜p=0.

    Therefore, Theorem 3.2 is valid. Similarly, we also show that Theorem 3.3 is fulfilled.

    By virtue of bifurcation theory or topological methods, we show the global behavior of positive solutions of (1.2) in the cases of asymptotically linear, superlinear, and sublinear as y, and for λ near the bifurcation value, where the solutions norms are indeed positive. The domain of (1.2) offers potential for further studies. For example, in this kind of semipositone case, f(t,0)<0 is bounded. A natural question is whether or not there exist positive solutions for (1.2) if f(t,0). Furthermore, the methods from this paper can be used for studying neutral-type and impulse differential equations as forms in [21,22,23,24,25]. We can also further discuss the stability analysis of the obtained positive solutions.

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

    This work was supported by the National Natural Science Foundation of China (Grant No. 12271419) and Natural Science Basic Research Program of Shaanxi Province of China (No. 2023-JC-QN-0081).

    The authors declare that they have no competing interests in this paper.



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