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Positive solutions for a Riemann-Liouville-type impulsive fractional integral boundary value problem

  • In this work, we investigate a Riemann-Liouville-type impulsive fractional integral boundary value problem. Using the fixed point index, we obtain two existence theorems on positive solutions under some conditions concerning the spectral radius of the relevant linear operator. Our method improves and generalizes some results in the literature.

    Citation: Keyu Zhang, Qian Sun, Donal O'Regan, Jiafa Xu. Positive solutions for a Riemann-Liouville-type impulsive fractional integral boundary value problem[J]. AIMS Mathematics, 2024, 9(5): 10911-10925. doi: 10.3934/math.2024533

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  • In this work, we investigate a Riemann-Liouville-type impulsive fractional integral boundary value problem. Using the fixed point index, we obtain two existence theorems on positive solutions under some conditions concerning the spectral radius of the relevant linear operator. Our method improves and generalizes some results in the literature.



    In this work, we study the following Riemann-Liouville-type impulsive fractional integral boundary value problem

    {tkDβtz(t)=f(t,z(t)), ttk,ΔDβ1z(tk)=Ik(z(tk)), k=1,,m,z(0)=z(0)=0, z(1)=10g(s,z(s))dα(s), (1.1)

    where 2<β3 is a real number, tkDβt is the Riemann-Liouville fractional derivative, 0=t0<t1<<tm<tm+1=1, tkDβ1tz(t+k)=limh0+tkDβ1tz(tk+h) and tkDβ1tz(tk)=limh0tkDβ1tz(tk+h) represent the right and left limits of tkDβ1tz(t) at t=tk, respectively, tkDβ1tz(tk)=tkDβ1tz(tk), and ΔDβ1z(tk)=tkDβ1tz(t+k)tk1Dβ1tz(tk). In addition, the functions f,g,α,Ik satisfy the conditions:

    (H0) f,gC([0,1]×R+,R+), IkC(R+,R+), k=1,2,...,m, R+:=[0,+),

    (H1) α is a function of bounded variation with α(t)0, and α(t)0, t[0,1].

    In comparison to integer calculus when describing natural phenomena and objective laws, fractional calculus is more accurate and applicable in physics, chemistry, and engineering. Many scholars have applied the methods of nonlinear analysis to study fractional boundary value problems, and a large number of results have been obtained; see for example [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31] and the references therein. In [1], the authors used some fixed-point techniques to study the existence, uniqueness, and multiplicity of positive solutions for the fractional integral boundary value problem

    {0Dαtx(t)+q(t)f(t,x(t))=0, 0<t<1,x(0)=x(0)==x(n2)(0)=0, 0Dβtx(1)=10h(s,x(s))dA(s),

    where 0Dαt, 0Dβt are Riemann-Liouville fractional derivatives. In [2], the authors studied the following p-Laplacian fractional boundary value problem involving the Riemann-Stieltjes integral:

    {0Dβt(φp(0Dαtz(t)g(t,z(t),0Dγtz(t))))=f(t,z(t),0Dγtz(t)), 0<t<1,0Dαtz(0)=0Dα+1tz(0)=0Dγtz(0)=0,0Dαtz(1)=0, 0Dγtz(1)=100Dγtz(s)dA(s),

    where 0Dαt, 0Dβt, 0Dγt are Riemann-Liouville fractional derivatives. The authors used fixed point theorems on a sum operator in partial ordering Banach spaces to investigate the existence and uniqueness of positive solutions for their problem.

    In [3], the authors studied the impulsive fractional integral boundary value problem

    {tkDαtu(t)=f(t,u(t),u(t),tkDα1tu(t)), ttk,ΔDβ1u(tk)=Ik(u(tk)), k=1,,m,u(0)=u(0)=0, u(1)=η0g(s,u(s))ds,

    and they adopted the contraction mapping principle and the fixed point theorem to establish the existence and uniqueness of nontrivial solutions when the nonlinearities f,g,Ik satisfy some Lipschitz conditions. In [4], the authors studied positive solutions for the fractional integral boundary value problem

    {Dα0+χ(t)+h(t)f(t,χ(t))=0,0<t<1,χ(0)=χ(0)=χ(0)=0,χ(1)=λη0χ(s)ds,

    where fC([0,1]×R+,R+) satisfies the conditions

    (HZ1) lim infχ0+f(t,χ)χ>λ1, limsupχ+f(t,χ)χ<λ1 uniformly with respect to t[0,1],

    (HZ2) limsupχ0+f(t,χ)χ<λ1, liminfχ+f(t,χ)χ>λ1 uniformly with respect to t[0,1],

    where λ1 is the first eigenvalue of the operator (LZ1χ)(t)=10GZ(t,s)h(s)χ(s)ds and GZ is the Green's function.

    Motivated by the aforementioned works, in this paper we use the fixed point index to study positive solutions for (1.1) under some conditions concerning the spectral radius of the relevant linear operator. Note that the considered linear operator can include the Riemann-Stieltjes integral condition in (1.1) and the approach is quite different from previous works in the literature. Moreover, we also consider the effect of the impulsive term and our conditions are more general than (HZ1)–(HZ2).

    In this section, we first present the definitions of the Riemann-Liouville-type fractional integral and derivative. For the other necessary definitions and notations, we refer the reader to the books [8,13,17].

    Definition 2.1. The Riemann-Liouville fractional integral of order β>0 of a function z : (a,+)R is given by

    aIβtz(t)=1Γ(β)ta(ts)β1z(s)ds,a>0,

    provided that the right-hand side is point-wise defined on (a,+).

    Definition 2.2. The Riemann-Liouville fractional derivative of order β>0 of a continuous function z:(a,+)R is given by

    aDβtz(t)=1Γ(nβ)dndtnta(ts)nβ1z(s)ds,

    where a>0,n1<βn, provided that the right-hand side is point-wise defined on (a,+).

    Let C([0,1],R) be the Banach space of continuous functions from [0,1] to R with the norm z= sup0t1|z(t)|. Define the Banach space PC1([0,1],R) as follows

    PC1([0,1],R)={zC([0,1],R):tkDβ1tz(t+k) and tkDβ1tz(tk)exist with   tkDβ1tz(tk)=tkDβ1tz(tk),k=0,1,,m}

    with the norm zPC1=max{z, tkDβ1tz}. Let P={zC([0,1],R):z(t)0,t[0,1]} and P0={zP:z(t)tβ1z,t[0,1]}. Then P,P0 are cones on C([0,1],R).

    Lemma 2.3. (see [3, Lemma 2.4]) Let h,VC([0,1],R) and vkR, k=1,2,...,m. Then, the boundary value problem

    {tkDβtz(t)=h(t),ttk,ΔDβ1z(tk)=vk,k=1,,m,z(0)=z(0)=0,z(1)=10V(s)dα(s) (2.1)

    has a solution of the form

    z(t)=10G(t,s)h(s)ds+tβ1β110V(s)dα(s)+mk=1H(t,tk)vk, 0t1,

    where

    G(t,s)=1Γ(β){tβ1(1s)β2(ts)β1, 0st1,tβ1(1s)β2, 0ts1,

    and

    H(t,tk)=1Γ(β){tβ1, 0ttk<1,0, 0<tk<t1.

    Lemma 2.4. (see [9]) The function G has the following properties:

    (C1) G(t,s)0 for t,s[0,1];

    (C2) tβ1G(1,s)G(t,s)G(1,s) for t,s[0,1].

    From Lemma 2.3 and (H0)–(H1), we define an operator T:PP as follows:

    (Tz)(t)=10G(t,s)f(s,z(s))ds+tβ1β110g(s,z(s))dα(s)+mk=1H(t,tk)Ik(z(tk)), 0t1. (2.2)

    From Lemma 2.3 we see that if there exists zP{0} such that Tz=z, then this z is the positive solution for (1.1). Hence, in what follows we study the existence of positive fixed points of the operator T.

    Lemma 2.5. Suppose that (H0)–(H1) hold. Then, T(P)P0.

    By Lemma 2.4 and the method of [21, Lemma 2.6], we obtain the conclusion, so, we omit its proof.

    Lemma 2.6. Let

    (Lμ,νz)(t)=μ10G(t,s)z(s)ds+νtβ1β110z(s)dα(s)

    with μ,ν0 and μ2+ν20. Then Lμ,ν(P)P0 and the spectral radius of Lμ,ν, denoted by r(Lμ,ν), which satisfies the inequality

    μ10G(1,s)sβ1ds+νβ110sβ1dα(s)r(Lμ,ν)μ10G(1,s)ds+νβ110dα(s). (2.3)

    Proof. If zP, then from Lemma 2.4(C2) we have

    (Lμ,νz)(t)μ10G(1,s)z(s)ds+ν1β110z(s)dα(s),

    and

    (Lμ,νz)(t)tβ1μ10G(1,s)z(s)ds+νtβ1β110z(s)dα(s)tβ1Lμ,νz, t[0,1].

    Hence, Lμ,ν(P)P0, as required.

    Let (Lμz)(t)=μ10G(t,s)z(s)ds and (Lνz)(t)=νtβ1β110z(s)dα(s), t[0,1]. Then for all nN+ we have

    (Lnμz)(t)=μn1010nG(t,s1)G(s1,s2)G(sn1,sn)z(sn)ds1dsnμn1010ntβ1G(1,s1)sβ11G(1,s2)sβ1n1G(1,sn)z(sn)ds1dsn,

    and

    (Lnνz)(t)=(νβ1)ntβ1[10sβ1dα(s)]n110z(s)dα(s), t[0,1].

    Consequently, we have

    Lnμmaxt[0,1](Lnμ1)(t)μn[10G(1,s)sβ1ds]n110G(1,s)ds,

    and

    Lnνmaxt[0,1](Lnν1)(t)(νβ1)n[10sβ1dα(s)]n110dα(s),

    where 1(t)1, t[0,1]. Therefore, Gelfand's theorem implies that

    r(Lμ)=lim infnnLnμμ10G(1,s)sβ1ds,

    and

    r(Lν)=lim infnnLnννβ110sβ1dα(s).

    Combining the two inequalities, we get

    r(Lμ,ν)μ10G(1,s)sβ1ds+νβ110sβ1dα(s).

    On the other hand, we note that

    r(Lμ)μ10G(1,s)ds, and r(Lν)νβ110dα(s),

    and then

    r(Lμ,ν)μ10G(1,s)ds+νβ110dα(s).

    Therefore, we obtain (2.3). This completes the proof.

    From Lemma 2.6, we find r(Lμ,ν)>0. Consequently, the Krein-Rutman theorem [32] implies that there exists ζμ,νP{0} such that

    Lμ,νζμ,ν=r(Lμ,ν)ζμ,ν. (2.4)

    From [19,33], the conjugate space of C([0,1],R) is E:={γ:γ has bounded variation on [0,1]}. Moreover, the dual cone of P and the bounded linear functional on C([0,1],R) can be expressed by

    P:={γE:γ is non-decreasing on [0,1]} and γ(z)=10z(t)dγ(t),zC([0,1],R),γE.

    Note that r(Lμ,ν)>0 in Lemma 2.6, and there exists ψμ,νP{0} such that

    Lμ,νψμ,ν=r(Lμ,ν)ψμ,ν, (2.5)

    where Lμ,ν:EE is the conjugate operator of Lμ,ν, denoted by

    (Lμ,νγ)(t):=μt0ds10G(τ,s)dγ(τ)+να(t)10τβ1β1dγ(τ),γE.

    Lemma 2.7. (see [34]) Let E be a Banach space, ΩE a bounded open set, and A:¯ΩPP a completely continuous operator. If there exists z0P{0} such that zAzλz0, for all zΩP,λ0, then the fixed point index i(A,ΩP,P)=0.

    Lemma 2.8. (see [34]) Let E be a Banach space, ΩE a bounded open set with 0Ω, and A:¯ΩPP a completely continuous operator. If zλAz, for all zΩP,0λ1, then the fixed point index i(A,ΩP,P)=1.

    Consider the coefficients μi,νi0 with μ2i+ν2i0, i=1,2,3,4. From Lemma 2.6, r(Lμi,νi)>0. Then there exists ψμi,νiP{0} such that

    Lμi,νiψμi,νi=r(Lμi,νi)ψμi,νi, i=1,2,3,4. (3.1)

    Remark 3.1. Let zP. Then we have

    10z(t)dψμi,νi(t)0, 10dψμi,νi(t)>0, 10tβ1dψμi,νi(t)>0, 10H(t,tk)dψμi,νi(t)>0.

    To see this note that ψμi,νiP{0}, and from the definition of the Riemann-Stieltjes integral we have

    10z(t)dψμi,νi(t)=limρ0nj=1z(ξj)[ψμi,νi(tj)ψμi,νi(tj1)]0,

    and

    10dψμi,νi(t)=limρ0nj=1[ψμi,νi(tj)ψμi,νi(tj1)][ψμi,νi(1)ψμi,νi(0)]>0,

    for all divisions tj: 0=t0<t1<<tn1<tn<tn+1=1, ρ=max1jn(tjtj1), ξj[tj1,tj], j= 1,2,,n. The other two inequalities can be similarly proven.

    Now, we list our assumptions for the nonlinearities f,g,Ik(k=1,2,...,m):

    (H2) There exist μ1,ν10 (μ21+ν210) and lk0 (mk=1l2k0),k=1,2,...,m such that

    if r(Lμ1,ν1)<1mk=1lktβ1k10H(t,tk)dψμ1,ν1(t)>[1r(Lμ1,ν1)]10dψμ1,ν1(t),
    lim infz+f(t,z)zμ1, lim infz+g(t,z)zν1 uniformly on t[0,1], and lim infz+Ik(z)zlk,k=1,2,...,m.

    (H3) There exist μ2,ν20 (μ22+ν220) and ˜lk0 (mk=1˜l2k0),k=1,2,...,m such that

    r(Lμ2,ν2)<1[1r(Lμ2,ν2)]10tβ1dψμ2,ν2(t)>mk=1˜lk10H(t,tk)dψμ2,ν2(t),
    lim supz0+f(t,z)zμ2, lim supz0+g(t,z)zν2 uniformly on t[0,1], and lim infz0+Ik(z)z˜lk,k=1,2,...,m.

    (H4) There exist μ3,ν30 (μ23+ν230) and ¯lk0 (mk=1¯l2k0),k=1,2,...,m such that

    if r(Lμ3,ν3)<1mk=1¯lktβ1k10H(t,tk)dψμ3,ν3(t)>[1r(Lμ3,ν3)]10dψμ3,ν3(t),
    lim infz0+f(t,z)zμ3, lim infz0+g(t,z)zν3 uniformly on t[0,1], and lim infz0+Ik(z)z¯lk,k=1,2,...,m.

    (H5) There exist μ4,ν40 (μ24+ν240) and ˆlk0 (mk=1ˆl2k0),k=1,2,...,m such that

    r(Lμ4,ν4)<1(1r(Lμ4,ν4))10tβ1dψμ4,ν4(t)>mk=1ˆlk10H(t,tk)dψμ4,ν4(t),
    lim supz+f(t,z)zμ4, lim supz+g(t,z)zν4 uniformly on t[0,1], and lim infz+Ik(z)zˆlk,k=1,2,...,m.

    Theorem 3.2. Suppose that (H0)–(H3) hold. Then, (1.1) has at least one positive solution.

    Proof. Let S1={zP:zTz=λ˜z, λ0}, where ˜zP0 is a fixed element. We first prove that S1 is a bounded set in P. Note that zS1, and from Lemma 2.5 we have

    zP0, i.e., z(t)tβ1z,t[0,1] and z(tk)tβ1kz,k=1,2,...,m. (3.2)

    By (H2) there exist ˜c,˜ck>0(k=1,2,...,m) such that

    f(t,z)μ1(z˜c), g(t,z)ν1(z˜c), Ik(z)lkz˜ck, zR+,t[0,1],k=1,2,...,m.

    Consequently, if zS1, we have

    z(t)(Tz)(t)μ110G(t,s)(z(s)˜c)ds+ν1tβ1β110(z(s)˜c)dα(s)+mk=1H(t,tk)(lkz(tk)˜ck). (3.3)

    Multiplying by dψμ1,ν1(t) on both sides of (3.3) and integrating over [0,1], from (3.1) we have

    10z(t)dψμ1,ν1(t)10[μ110G(t,s)(z(s)˜c)ds+ν1tβ1β110(z(s)˜c)dα(s)]dψμ1,ν1(t)   +mk=110H(t,tk)(lkz(tk)˜ck)dψμ1,ν1(t)=10(z(s)˜c)d(μ1s0dτ10G(t,τ)dψμ1,ν1(t)+ν1α(s)10tβ1β1dψμ1,ν1(t))   +mk=110H(t,tk)dψμ1,ν1(t)(lkz(tk)˜ck)=10(z(s)˜c)d(Lμ1,ν1ψμ1,ν1)(s)+mk=110H(t,tk)dψμ1,ν1(t)(lkz(tk)˜ck)=10(z(s)˜c)d(r(Lμ1,ν1)ψμ1,ν1)(s)+mk=110H(t,tk)dψμ1,ν1(t)(lkz(tk)˜ck).

    Thus,

    10z(t)dψμ1,ν1(t)+˜cr(Lμ1,ν1)10dψμ1,ν1(t)+mk=1˜ck10H(t,tk)dψμ1,ν1(t)r(Lμ1,ν1)10z(t)dψμ1,ν1(t)+mk=1lk10H(t,tk)dψμ1,ν1(t)z(tk). (3.4)

    There are two cases to consider.

    Case 1. r(Lμ1,ν1)1. From (3.2) and (3.4) we have

     [r(Lμ1,ν1)1]z10tβ1dψμ1,ν1(t)+zmk=1lktβ1k10H(t,tk)dψμ1,ν1(t)˜cr(Lμ1,ν1)10dψμ1,ν1(t)+mk=1˜ck10H(t,tk)dψμ1,ν1(t),

    and thus

    z˜cr(Lμ1,ν1)10dψμ1,ν1(t)+mk=1˜ck10H(t,tk)dψμ1,ν1(t)[r(Lμ1,ν1)1]10tβ1dψμ1,ν1(t)+mk=1lktβ1k10H(t,tk)dψμ1,ν1(t).

    Case 2. Now r(Lμ1,ν1)<1. (H2), (3.2), and (3.4) imply that

     [r(Lμ1,ν1)1]z10dψμ1,ν1(t)+zmk=1lktβ1k10H(t,tk)dψμ1,ν1(t)˜cr(Lμ1,ν1)10dψμ1,ν1(t)+mk=1˜ck10H(t,tk)dψμ1,ν1(t),

    and then

    z˜cr(Lμ1,ν1)10dψμ1,ν1(t)+mk=1˜ck10H(t,tk)dψμ1,ν1(t)[r(Lμ1,ν1)1]10dψμ1,ν1(t)+mk=1lktβ1k10H(t,tk)dψμ1,ν1(t).

    Combining the two cases, we have proved that S1 is a bounded set, as required. Now, we choose a sufficiently large R1>supS1 such that

    zTzλ˜z, zBR1P, λ0, (3.5)

    where BR1={zP:z<R1}. Therefore, Lemma 2.7 implies that

    i(T,BR1P,P)=0. (3.6)

    By (H3) there exists r1>0 such that

    f(t,z)μ2z, g(t,z)ν2z, Ik(z)˜lkz, z[0,r1],t[0,1],k=1,2,...,m. (3.7)

    Now, we prove that

    zλTz, zBr1P, λ[0,1], (3.8)

    where Br1={zP:z<r1}. If the claim is false, then there exist a z1Br1P, λ1[0,1] such that

    z1=λ1Tz1.

    By Lemma 2.5, z1 satisfies (3.2), and from (3.7) we have

    z1(t)(Tz1)(t)μ210G(t,s)z1(s)ds+ν2tβ1β110z1(s)dα(s)+mk=1H(t,tk)˜lkz1(tk). (3.9)

    Multiplying by dψμ2,ν2(t) on both sides of (3.9) and integrating over [0,1], from (3.1) we obtain

    10z1(t)dψμ2,ν2(t)μ21010G(t,s)z1(s)dsdψμ2,ν2(t)+ν210tβ1β110z1(s)dα(s)dψμ2,ν2(t)   +mk=1˜lk10H(t,tk)dψμ2,ν2(t)z1(tk)=10z1(s)d(μ2s0dτ10G(t,τ)dψμ2,ν2(t)+ν2α(s)10tβ1β1dψμ2,ν2(t))   +mk=1˜lk10H(t,tk)dψμ2,ν2(t)z1(tk)=10z1(s)d(Lμ2,ν2ψμ2,ν2)(s)+mk=1˜lk10H(t,tk)dψμ2,ν2(t)z1(tk)=10z1(s)d(r(Lμ2,ν2)ψμ2,ν2)(s)+mk=1˜lk10H(t,tk)dψμ2,ν2(t)z1(tk).

    This, together with (3.2), implies that

    [1r(Lμ2,ν2)]z110tβ1dψμ2,ν2(t)z1mk=1˜lk10H(t,tk)dψμ2,ν2(t).

    This contradicts (H3) unless z1=0. Note that z1=0 also contradicts z1Br1P, r1>0. Therefore, we obtain that (3.8) holds, as required. From Lemma 2.8 we have

    i(T,Br1P,P)=1. (3.10)

    Note that R1 can be chosen large enough such that R1>supS1 and R1>r1. Therefore, from (3.6) and (3.10) we have

    i(T,(BR1¯Br1)P,P)=i(T,BR1P,P)i(T,Br1P,P)=1.

    Therefore, the operator T has at least one fixed point in (BR1¯Br1)P. Thus, (1.1) has at least one positive solution. This completes the proof.

    Theorem 3.3. Suppose that (H0) and (H4)–(H5) hold. Then, (1.1) has at least one positive solution.

    Proof. By (H4) there exists a sufficiently small r2>0 such that

    f(t,z)μ3z, g(t,z)ν3z, Ik(z)¯lkz, z[0,r2],t[0,1],k=1,2,...,m. (3.11)

    For this r2, we prove that

    zTzλ¯z, zBr2P, λ0, (3.12)

    where Br2={zP:z<r2}, and ¯z is a fixed element in P0. If (3.12) is false, then there exist a z2Br2P, λ20 such that

    z2Tz2=λ2¯z.

    Lemma 2.5 implies that z2 satisfies (3.2). Moreover, from (3.11) we have

    z2(t)(Tz2)(t)μ310G(t,s)z2(s)ds+ν3tβ1β110z2(s)dα(s)+mk=1H(t,tk)¯lkz2(tk). (3.13)

    Multiplying by dψμ3,ν3(t) on both sides of (3.13) and integrating over [0,1], from (3.1) we obtain

    10z2(t)dψμ3,ν3(t)μ31010G(t,s)z2(s)dsdψμ3,ν3(t)+ν310tβ1β110z2(s)dα(s)dψμ3,ν3(t)   +mk=1¯lk10H(t,tk)dψμ3,ν3(t)z2(tk)=10z2(s)d(μ3s0dτ10G(t,τ)dψμ3,ν3(t)+ν3α(s)10tβ1β1dψμ3,ν3(t))   +mk=1¯lk10H(t,tk)dψμ3,ν3(t)z2(tk)=10z2(s)d(Lμ3,ν3ψμ3,ν3)(s)+mk=1¯lk10H(t,tk)dψμ3,ν3(t)z2(tk)=10z2(s)d(r(Lμ3,ν3)ψμ3,ν3)(s)+mk=1¯lk10H(t,tk)dψμ3,ν3(t)z2(tk).

    There are two cases to consider.

    Cases 1. r(Lμ3,ν3)1. From (3.2) we obtain

    z2[(r(Lμ3,ν3)1)10tβ1dψμ3,ν3(t)+mk=1¯lktβ1k10H(t,tk)dψμ3,ν3(t)]0,

    which contradicts z2Br2P, r2>0.

    Cases 2. r(Lμ3,ν3)<1. By (3.2) we have

    z2[(r(Lμ3,ν3)1)10dψμ3,ν3(t)+mk=1¯lktβ1k10H(t,tk)dψμ3,ν3(t)]0,

    and it contradicts (H4) unless z2=0. We also have a contradiction to z2Br2P, r2>0 if z2=0.

    Therefore, we obtain that (3.12) holds, and Lemma 2.7 implies that

    i(T,Br2P,P)=0. (3.14)

    By (H5) there exist ¯c,¯ck>0(k=1,2,...,m) such that

    f(t,z)μ4(z+¯c), g(t,z)ν4(z+¯c), Ik(z)ˆlkz+¯ck, zR+,t[0,1],k=1,2,...,m. (3.15)

    Let S2={zP:z=λTz,λ[0,1]}. We now prove that S2 is bounded in P. If zS2, then by Lemma 2.5, (3.2) holds, and from (3.15) we have

    z(t)(Tz)(t)μ410G(t,s)(z(s)+¯c)ds+ν4tβ1β110(z(s)+¯c)dα(s)+mk=1H(t,tk)(ˆlkz(tk)+¯ck). (3.16)

    Multiplying by dψμ4,ν4(t) on both sides of (3.16) and integrating over [0,1], from (3.1) we obtain

    10z(t)dψμ4,ν4(t)μ41010G(t,s)(z(s)+¯c)dsdψμ4,ν4(t)+ν410tβ1β110(z(s)+¯c)dα(s)dψμ4,ν4(t)   +mk=110H(t,tk)dψμ4,ν4(t)(ˆlkz(tk)+¯ck)=10(z(s)+¯c)d(μ4s0dτ10G(t,τ)dψμ4,ν4(t)+ν4α(s)10tβ1β1dψμ4,ν4(t))   +mk=110H(t,tk)dψμ4,ν4(t)(ˆlkz(tk)+¯ck)=10(z(s)+¯c)d(Lμ4,ν4ψμ4,ν4)(s)+mk=110H(t,tk)dψμ4,ν4(t)(ˆlkz(tk)+¯ck)=10(z(s)+¯c)d(r(Lμ4,ν4)ψμ4,ν4)(s)+mk=110H(t,tk)dψμ4,ν4(t)(ˆlkz(tk)+¯ck).

    Note that (3.2) and r(Lμ4,ν4)<1, and we have

    (1r(Lμ4,ν4))z10tβ1dψμ4,ν4(t)¯cr(Lμ4,ν4)10dψμ4,ν4)(t)+mk=110H(t,tk)dψμ4,ν4(t)(ˆlkz+¯ck),

    and (H5) implies that

    z¯cr(Lμ4,ν4)10dψμ4,ν4)(t)+mk=1¯ck10H(t,tk)dψμ4,ν4(t)(1r(Lμ4,ν4))10tβ1dψμ4,ν4(t)mk=1ˆlk10H(t,tk)dψμ4,ν4(t).

    This implies that S2 is a bounded set in P, as required. Therefore, we can choose a large number R2>max{supS2,r2} such that

    zλTz, zBR2P, λ[0,1],

    where BR2={zP:z<R2}. From Lemma 2.8 we have

    i(T,BR2P,P)=1. (3.17)

    As a result, from (3.14) and (3.17) we have

    i(T,(BR2¯Br2)P,P)=i(T,BR2P,P)i(T,Br2P,P)=1.

    Therefore, the operator T has at least one fixed point in (BR2¯Br2)P. Thus, (1.1) has at least one positive solution. This completes the proof.

    In this paper, we study the existence of positive solutions for the Riemann-Liouville-type impulsive fractional integral boundary value problem (1.1). We first use the Gelfand theorem and the Krein-Rutman theorem to investigate a related positive linear operator, which can include the Riemann-Stieltjes integral condition. Then, the impulsive term is regarded as a perturbation, and we use some conditions concerning the spectral radius of the linear operator to obtain our main results. In this paper we provided a quite different method to study such problems.

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

    This research was supported by the Talent Introduction Project of Ludong University (grant No. LY2015004). The authors would like to express their heartfelt gratitude to the editors and reviewers for their constructive comments.

    The authors declare no conflicts of interest.



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