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)), t≠tk,ΔDβ−1z(tk)=−Ik(z(tk)), k=1,…,m,z(0)=z′(0)=0, z′(1)=∫10g(s,z(s))dα(s), | (1.1) |
where 2<β≤3 is a real number, tkDβt is the Riemann-Liouville fractional derivative, 0=t0<t1<⋯<tm<tm+1=1, tkDβ−1tz(t+k)=limh→0+tkDβ−1tz(tk+h) and tkDβ−1tz(t−k)=limh→0−tkDβ−1tz(tk+h) represent the right and left limits of tkDβ−1tz(t) at t=tk, respectively, tkDβ−1tz(t−k)=tkDβ−1tz(tk), and ΔDβ−1z(tk)=tkDβ−1tz(t+k)−tk−1Dβ−1tz(t−k). In addition, the functions f,g,α,Ik satisfy the conditions:
(H0) f,g∈C([0,1]×R+,R+), Ik∈C(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(n−2)(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),tkDα−1tu(t)), t≠tk,ΔDβ−1u(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)=λ∫η0χ(s)ds, |
where f∈C([0,1]×R+,R+) satisfies the conditions
(HZ1) lim infχ→0+f(t,χ)χ>λ1, limsupχ→+∞f(t,χ)χ<λ1 uniformly with respect to t∈[0,1],
(HZ2) limsupχ→0+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(t−s)β−1z(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(n−β)dndtn∫ta(t−s)n−β−1z(s)ds, |
where a>0,n−1<β≤n, 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‖= sup0≤t≤1|z(t)|. Define the Banach space PC1([0,1],R) as follows
PC1([0,1],R)={z∈C([0,1],R):tkDβ−1tz(t+k) and tkDβ−1tz(t−k)exist with tkDβ−1tz(tk)=tkDβ−1tz(t−k),k=0,1,…,m} |
with the norm ‖z‖PC1=max{‖z‖, ‖tkDβ−1tz‖}. Let P={z∈C([0,1],R):z(t)≥0,t∈[0,1]} and P0={z∈P:z(t)≥tβ−1‖z‖,t∈[0,1]}. Then P,P0 are cones on C([0,1],R).
Lemma 2.3. (see [3, Lemma 2.4]) Let h,V∈C([0,1],R) and vk∈R, k=1,2,...,m. Then, the boundary value problem
{tkDβtz(t)=−h(t),t≠tk,ΔDβ−1z(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+tβ−1β−1∫10V(s)dα(s)+m∑k=1H(t,tk)vk, 0≤t≤1, |
where
G(t,s)=1Γ(β){tβ−1(1−s)β−2−(t−s)β−1, 0≤s≤t≤1,tβ−1(1−s)β−2, 0≤t≤s≤1, |
and
H(t,tk)=1Γ(β){tβ−1, 0≤t≤tk<1,0, 0<tk<t≤1. |
Lemma 2.4. (see [9]) The function G has the following properties:
(C1) G(t,s)≥0 for t,s∈[0,1];
(C2) tβ−1G(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:P→P as follows:
(Tz)(t)=∫10G(t,s)f(s,z(s))ds+tβ−1β−1∫10g(s,z(s))dα(s)+m∑k=1H(t,tk)Ik(z(tk)), 0≤t≤1. | (2.2) |
From Lemma 2.3 we see that if there exists z∗∈P∖{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)=μ∫10G(t,s)z(s)ds+νtβ−1β−1∫10z(s)dα(s) |
with μ,ν≥0 and μ2+ν2≠0. Then Lμ,ν(P)⊂P0 and the spectral radius of Lμ,ν, denoted by r(Lμ,ν), which satisfies the inequality
μ∫10G(1,s)sβ−1ds+νβ−1∫10sβ−1dα(s)≤r(Lμ,ν)≤μ∫10G(1,s)ds+νβ−1∫10dα(s). | (2.3) |
Proof. If z∈P, then from Lemma 2.4(C2) we have
(Lμ,νz)(t)≤μ∫10G(1,s)z(s)ds+ν1β−1∫10z(s)dα(s), |
and
(Lμ,νz)(t)≥tβ−1μ∫10G(1,s)z(s)ds+νtβ−1β−1∫10z(s)dα(s)≥tβ−1‖Lμ,νz‖, t∈[0,1]. |
Hence, Lμ,ν(P)∈P0, as required.
Let (Lμz)(t)=μ∫10G(t,s)z(s)ds and (Lνz)(t)=νtβ−1β−1∫10z(s)dα(s), t∈[0,1]. Then for all n∈N+ we have
(Lnμz)(t)=μn∫10⋯∫10⏟nG(t,s1)G(s1,s2)⋯G(sn−1,sn)z(sn)ds1⋯dsn≥μn∫10⋯∫10⏟ntβ−1G(1,s1)sβ−11G(1,s2)⋯sβ−1n−1G(1,sn)z(sn)ds1⋯dsn, |
and
(Lnνz)(t)=(νβ−1)ntβ−1[∫10sβ−1dα(s)]n−1∫10z(s)dα(s), t∈[0,1]. |
Consequently, we have
‖Lnμ‖≥maxt∈[0,1](Lnμ1)(t)≥μn[∫10G(1,s)sβ−1ds]n−1∫10G(1,s)ds, |
and
‖Lnν‖≥maxt∈[0,1](Lnν1)(t)≥(νβ−1)n[∫10sβ−1dα(s)]n−1∫10dα(s), |
where 1(t)≡1, t∈[0,1]. Therefore, Gelfand's theorem implies that
r(Lμ)=lim infn→∞n√‖Lnμ‖≥μ∫10G(1,s)sβ−1ds, |
and
r(Lν)=lim infn→∞n√‖Lnν‖≥νβ−1∫10sβ−1dα(s). |
Combining the two inequalities, we get
r(Lμ,ν)≥μ∫10G(1,s)sβ−1ds+νβ−1∫10sβ−1dα(s). |
On the other hand, we note that
r(Lμ)≤μ∫10G(1,s)ds, and r(Lν)≤νβ−1∫10dα(s), |
and then
r(Lμ,ν)≤μ∫10G(1,s)ds+νβ−1∫10dα(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,ν∈P∖{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∗:={γ∈E∗:γ is non-decreasing on [0,1]} and γ(z)=∫10z(t)dγ(t),z∈C([0,1],R),γ∈E∗. |
Note that r(Lμ,ν)>0 in Lemma 2.6, and there exists ψμ,ν∈P∗∖{0} such that
L∗μ,νψμ,ν=r(Lμ,ν)ψμ,ν, | (2.5) |
where L∗μ,ν:E∗→E∗ is the conjugate operator of Lμ,ν, denoted by
(L∗μ,νγ)(t):=μ∫t0ds∫10G(τ,s)dγ(τ)+να(t)∫10τβ−1β−1dγ(τ),γ∈E∗. |
Lemma 2.7. (see [34]) Let E be a Banach space, Ω⊂E a bounded open set, and A:¯Ω∩P→P a completely continuous operator. If there exists z0∈P∖{0} such that z−Az≠λz0, for all z∈∂Ω∩P,λ≥0, then the fixed point index i(A,Ω∩P,P)=0.
Lemma 2.8. (see [34]) Let E be a Banach space, Ω⊂E a bounded open set with 0∈Ω, and A:¯Ω∩P→P a completely continuous operator. If z≠λAz, for all z∈∂Ω∩P,0≤λ≤1, then the fixed point index i(A,Ω∩P,P)=1.
Consider the coefficients μi,νi≥0 with μ2i+ν2i≠0, i=1,2,3,4. From Lemma 2.6, r(Lμi,νi)>0. Then there exists ψμi,νi∈P∗∖{0} such that
L∗μi,νiψμi,νi=r(Lμi,νi)ψμi,νi, i=1,2,3,4. | (3.1) |
Remark 3.1. Let z∈P. Then we have
∫10z(t)dψμi,νi(t)≥0, ∫10dψμi,νi(t)>0, ∫10tβ−1dψμi,νi(t)>0, ∫10H(t,tk)dψμi,νi(t)>0. |
To see this note that ψμi,νi∈P∗∖{0}, and from the definition of the Riemann-Stieltjes integral we have
∫10z(t)dψμi,νi(t)=limρ→0n∑j=1z(ξj)[ψμi,νi(tj)−ψμi,νi(tj−1)]≥0, |
and
∫10dψμi,νi(t)=limρ→0n∑j=1[ψμi,νi(tj)−ψμi,νi(tj−1)]≥[ψμi,νi(1)−ψμi,νi(0)]>0, |
for all divisions tj: 0=t0<t1<⋯<tn−1<tn<tn+1=1, ρ=max1≤j≤n(tj−tj−1), ξj∈[tj−1,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,ν1≥0 (μ21+ν21≠0) and lk≥0 (m∑k=1l2k≠0),k=1,2,...,m such that
if r(Lμ1,ν1)<1⇒m∑k=1lktβ−1k∫10H(t,tk)dψμ1,ν1(t)>[1−r(Lμ1,ν1)]∫10dψμ1,ν1(t), |
lim infz→+∞f(t,z)z≥μ1, lim infz→+∞g(t,z)z≥ν1 uniformly on t∈[0,1], and lim infz→+∞Ik(z)z≥lk,k=1,2,...,m. |
(H3) There exist μ2,ν2≥0 (μ22+ν22≠0) and ˜lk≥0 (m∑k=1˜l2k≠0),k=1,2,...,m such that
r(Lμ2,ν2)<1⇒[1−r(Lμ2,ν2)]∫10tβ−1dψμ2,ν2(t)>m∑k=1˜lk∫10H(t,tk)dψμ2,ν2(t), |
lim supz→0+f(t,z)z≤μ2, lim supz→0+g(t,z)z≤ν2 uniformly on t∈[0,1], and lim infz→0+Ik(z)z≤˜lk,k=1,2,...,m. |
(H4) There exist μ3,ν3≥0 (μ23+ν23≠0) and ¯lk≥0 (m∑k=1¯l2k≠0),k=1,2,...,m such that
if r(Lμ3,ν3)<1⇒m∑k=1¯lktβ−1k∫10H(t,tk)dψμ3,ν3(t)>[1−r(Lμ3,ν3)]∫10dψμ3,ν3(t), |
lim infz→0+f(t,z)z≥μ3, lim infz→0+g(t,z)z≥ν3 uniformly on t∈[0,1], and lim infz→0+Ik(z)z≥¯lk,k=1,2,...,m. |
(H5) There exist μ4,ν4≥0 (μ24+ν24≠0) and ˆlk≥0 (m∑k=1ˆl2k≠0),k=1,2,...,m such that
r(Lμ4,ν4)<1⇒(1−r(Lμ4,ν4))∫10tβ−1dψμ4,ν4(t)>m∑k=1ˆlk∫10H(t,tk)dψμ4,ν4(t), |
lim supz→+∞f(t,z)z≤μ4, lim supz→+∞g(t,z)z≤ν4 uniformly on t∈[0,1], and lim infz→+∞Ik(z)z≤ˆlk,k=1,2,...,m. |
Theorem 3.2. Suppose that (H0)–(H3) hold. Then, (1.1) has at least one positive solution.
Proof. Let S1={z∈P:z−Tz=λ˜z, λ≥0}, where ˜z∈P0 is a fixed element. We first prove that S1 is a bounded set in P. Note that z∈S1, and from Lemma 2.5 we have
z∈P0, i.e., z(t)≥tβ−1‖z‖,t∈[0,1] and z(tk)≥tβ−1k‖z‖,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, z∈R+,t∈[0,1],k=1,2,...,m. |
Consequently, if z∈S1, we have
z(t)≥(Tz)(t)≥μ1∫10G(t,s)(z(s)−˜c)ds+ν1tβ−1β−1∫10(z(s)−˜c)dα(s)+m∑k=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[μ1∫10G(t,s)(z(s)−˜c)ds+ν1tβ−1β−1∫10(z(s)−˜c)dα(s)]dψμ1,ν1(t) +m∑k=1∫10H(t,tk)(lkz(tk)−˜ck)dψμ1,ν1(t)=∫10(z(s)−˜c)d(μ1∫s0dτ∫10G(t,τ)dψμ1,ν1(t)+ν1α(s)∫10tβ−1β−1dψμ1,ν1(t)) +m∑k=1∫10H(t,tk)dψμ1,ν1(t)(lkz(tk)−˜ck)=∫10(z(s)−˜c)d(L∗μ1,ν1ψμ1,ν1)(s)+m∑k=1∫10H(t,tk)dψμ1,ν1(t)(lkz(tk)−˜ck)=∫10(z(s)−˜c)d(r(Lμ1,ν1)ψμ1,ν1)(s)+m∑k=1∫10H(t,tk)dψμ1,ν1(t)(lkz(tk)−˜ck). |
Thus,
∫10z(t)dψμ1,ν1(t)+˜cr(Lμ1,ν1)∫10dψμ1,ν1(t)+m∑k=1˜ck∫10H(t,tk)dψμ1,ν1(t)≥r(Lμ1,ν1)∫10z(t)dψμ1,ν1(t)+m∑k=1lk∫10H(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]‖z‖∫10tβ−1dψμ1,ν1(t)+‖z‖m∑k=1lktβ−1k∫10H(t,tk)dψμ1,ν1(t)≤˜cr(Lμ1,ν1)∫10dψμ1,ν1(t)+m∑k=1˜ck∫10H(t,tk)dψμ1,ν1(t), |
and thus
‖z‖≤˜cr(Lμ1,ν1)∫10dψμ1,ν1(t)+m∑k=1˜ck∫10H(t,tk)dψμ1,ν1(t)[r(Lμ1,ν1)−1]∫10tβ−1dψμ1,ν1(t)+m∑k=1lktβ−1k∫10H(t,tk)dψμ1,ν1(t). |
Case 2. Now r(Lμ1,ν1)<1. (H2), (3.2), and (3.4) imply that
[r(Lμ1,ν1)−1]‖z‖∫10dψμ1,ν1(t)+‖z‖m∑k=1lktβ−1k∫10H(t,tk)dψμ1,ν1(t)≤˜cr(Lμ1,ν1)∫10dψμ1,ν1(t)+m∑k=1˜ck∫10H(t,tk)dψμ1,ν1(t), |
and then
‖z‖≤˜cr(Lμ1,ν1)∫10dψμ1,ν1(t)+m∑k=1˜ck∫10H(t,tk)dψμ1,ν1(t)[r(Lμ1,ν1)−1]∫10dψμ1,ν1(t)+m∑k=1lktβ−1k∫10H(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
z−Tz≠λ˜z, z∈∂BR1∩P, λ≥0, | (3.5) |
where BR1={z∈P:‖z‖<R1}. Therefore, Lemma 2.7 implies that
i(T,BR1∩P,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
z≠λTz, z∈∂Br1∩P, λ∈[0,1], | (3.8) |
where Br1={z∈P:‖z‖<r1}. If the claim is false, then there exist a z1∈∂Br1∩P, 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)≤μ2∫10G(t,s)z1(s)ds+ν2tβ−1β−1∫10z1(s)dα(s)+m∑k=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)≤μ2∫10∫10G(t,s)z1(s)dsdψμ2,ν2(t)+ν2∫10tβ−1β−1∫10z1(s)dα(s)dψμ2,ν2(t) +m∑k=1˜lk∫10H(t,tk)dψμ2,ν2(t)z1(tk)=∫10z1(s)d(μ2∫s0dτ∫10G(t,τ)dψμ2,ν2(t)+ν2α(s)∫10tβ−1β−1dψμ2,ν2(t)) +m∑k=1˜lk∫10H(t,tk)dψμ2,ν2(t)z1(tk)=∫10z1(s)d(L∗μ2,ν2ψμ2,ν2)(s)+m∑k=1˜lk∫10H(t,tk)dψμ2,ν2(t)z1(tk)=∫10z1(s)d(r(Lμ2,ν2)ψμ2,ν2)(s)+m∑k=1˜lk∫10H(t,tk)dψμ2,ν2(t)z1(tk). |
This, together with (3.2), implies that
[1−r(Lμ2,ν2)]‖z1‖∫10tβ−1dψμ2,ν2(t)≤‖z1‖m∑k=1˜lk∫10H(t,tk)dψμ2,ν2(t). |
This contradicts (H3) unless ‖z1‖=0. Note that ‖z1‖=0 also contradicts z1∈∂Br1∩P, r1>0. Therefore, we obtain that (3.8) holds, as required. From Lemma 2.8 we have
i(T,Br1∩P,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,BR1∩P,P)−i(T,Br1∩P,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
z−Tz≠λ¯z, z∈∂Br2∩P, λ≥0, | (3.12) |
where Br2={z∈P:‖z‖<r2}, and ¯z is a fixed element in P0. If (3.12) is false, then there exist a z2∈∂Br2∩P, λ2≥0 such that
z2−Tz2=λ2¯z. |
Lemma 2.5 implies that z2 satisfies (3.2). Moreover, from (3.11) we have
z2(t)≥(Tz2)(t)≥μ3∫10G(t,s)z2(s)ds+ν3tβ−1β−1∫10z2(s)dα(s)+m∑k=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)≥μ3∫10∫10G(t,s)z2(s)dsdψμ3,ν3(t)+ν3∫10tβ−1β−1∫10z2(s)dα(s)dψμ3,ν3(t) +m∑k=1¯lk∫10H(t,tk)dψμ3,ν3(t)z2(tk)=∫10z2(s)d(μ3∫s0dτ∫10G(t,τ)dψμ3,ν3(t)+ν3α(s)∫10tβ−1β−1dψμ3,ν3(t)) +m∑k=1¯lk∫10H(t,tk)dψμ3,ν3(t)z2(tk)=∫10z2(s)d(L∗μ3,ν3ψμ3,ν3)(s)+m∑k=1¯lk∫10H(t,tk)dψμ3,ν3(t)z2(tk)=∫10z2(s)d(r(Lμ3,ν3)ψμ3,ν3)(s)+m∑k=1¯lk∫10H(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)∫10tβ−1dψμ3,ν3(t)+m∑k=1¯lktβ−1k∫10H(t,tk)dψμ3,ν3(t)]≤0, |
which contradicts z2∈∂Br2∩P, r2>0.
Cases 2. r(Lμ3,ν3)<1. By (3.2) we have
‖z2‖[(r(Lμ3,ν3)−1)∫10dψμ3,ν3(t)+m∑k=1¯lktβ−1k∫10H(t,tk)dψμ3,ν3(t)]≤0, |
and it contradicts (H4) unless ‖z2‖=0. We also have a contradiction to z2∈∂Br2∩P, r2>0 if ‖z2‖=0.
Therefore, we obtain that (3.12) holds, and Lemma 2.7 implies that
i(T,Br2∩P,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, z∈R+,t∈[0,1],k=1,2,...,m. | (3.15) |
Let S2={z∈P:z=λTz,λ∈[0,1]}. We now prove that S2 is bounded in P. If z∈S2, then by Lemma 2.5, (3.2) holds, and from (3.15) we have
z(t)≤(Tz)(t)≤μ4∫10G(t,s)(z(s)+¯c)ds+ν4tβ−1β−1∫10(z(s)+¯c)dα(s)+m∑k=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)≤μ4∫10∫10G(t,s)(z(s)+¯c)dsdψμ4,ν4(t)+ν4∫10tβ−1β−1∫10(z(s)+¯c)dα(s)dψμ4,ν4(t) +m∑k=1∫10H(t,tk)dψμ4,ν4(t)(ˆlkz(tk)+¯ck)=∫10(z(s)+¯c)d(μ4∫s0dτ∫10G(t,τ)dψμ4,ν4(t)+ν4α(s)∫10tβ−1β−1dψμ4,ν4(t)) +m∑k=1∫10H(t,tk)dψμ4,ν4(t)(ˆlkz(tk)+¯ck)=∫10(z(s)+¯c)d(L∗μ4,ν4ψμ4,ν4)(s)+m∑k=1∫10H(t,tk)dψμ4,ν4(t)(ˆlkz(tk)+¯ck)=∫10(z(s)+¯c)d(r(Lμ4,ν4)ψμ4,ν4)(s)+m∑k=1∫10H(t,tk)dψμ4,ν4(t)(ˆlkz(tk)+¯ck). |
Note that (3.2) and r(Lμ4,ν4)<1, and we have
(1−r(Lμ4,ν4))‖z‖∫10tβ−1dψμ4,ν4(t)≤¯cr(Lμ4,ν4)∫10dψμ4,ν4)(t)+m∑k=1∫10H(t,tk)dψμ4,ν4(t)(ˆlk‖z‖+¯ck), |
and (H5) implies that
‖z‖≤¯cr(Lμ4,ν4)∫10dψμ4,ν4)(t)+m∑k=1¯ck∫10H(t,tk)dψμ4,ν4(t)(1−r(Lμ4,ν4))∫10tβ−1dψμ4,ν4(t)−m∑k=1ˆlk∫10H(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
z≠λTz, z∈∂BR2∩P, λ∈[0,1], |
where BR2={z∈P:‖z‖<R2}. From Lemma 2.8 we have
i(T,BR2∩P,P)=1. | (3.17) |
As a result, from (3.14) and (3.17) we have
i(T,(BR2∖¯Br2)∩P,P)=i(T,BR2∩P,P)−i(T,Br2∩P,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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