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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[9] | Chengbo Zhai, Yuanyuan Ma, Hongyu Li . Unique positive solution for a p-Laplacian fractional differential boundary value problem involving Riemann-Stieltjes integral. AIMS Mathematics, 2020, 5(5): 4754-4769. doi: 10.3934/math.2020304 |
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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), |
\liminf\limits_{z\to 0^+} \frac{f(t, z)}{z}\ge \mu_3, \ \liminf\limits_{z\to 0^+} \frac{g(t, z)}{z}\ge \nu_3\text{ uniformly on }t\in [0, 1], \text{ and } \liminf\limits_{z\to 0^+} \frac{I_k(z)}{z}\ge \overline{l}_k, k = 1, 2, ..., m. |
(H5) There exist \mu_4, \nu_4\ge 0\ (\mu_4^2+\nu_4^2 \not = 0) and \widehat{l}_k\ge 0 \ (\sum\limits_{k = 1}^m \widehat{l}_k^2 \not = 0), k = 1, 2, ..., m such that
r\left(\mathcal L_{\mu_4, \nu_4}\right) < 1\Rightarrow (1-r\left(\mathcal L_{\mu_4, \nu_4}\right))\int_0^1 t^{\beta-1} d\psi_{\mu_4, \nu_4}(t) > \sum\limits_{k = 1}^m \widehat{l}_k \int_0^1 H(t, t_k)d\psi_{\mu_4, \nu_4}(t), |
\limsup\limits_{z\to +\infty} \frac{f(t, z)}{z}\le \mu_4, \ \limsup\limits_{z\to +\infty} \frac{g(t, z)}{z}\le \nu_4\text{ uniformly on }t\in [0, 1], \text{ and } \liminf\limits_{z\to +\infty} \frac{I_k(z)}{z}\le \widehat{l}_k, k = 1, 2, ..., m. |
Theorem 3.2. Suppose that (H0)–(H3) hold. Then, (1.1) has at least one positive solution.
Proof. Let S_1 = \{z\in P: z-\mathcal Tz = \lambda \widetilde{z}, \ \lambda\ge 0\} , where \widetilde{z}\in P_0 is a fixed element. We first prove that S_1 is a bounded set in P . Note that z\in S_1 , and from Lemma 2.5 we have
\begin{equation} z\in P_0, \text{ i.e., } z(t)\ge t^{\beta-1}\|z\|, t\in [0, 1]\text{ and }z(t_k)\ge t_k^{\beta-1}\|z\|, k = 1, 2, ..., m. \end{equation} | (3.2) |
By (H2) there exist \widetilde{c}, \widetilde{c}_k > 0(k = 1, 2, ..., m) such that
f(t, z)\ge \mu_1( z-\widetilde{c}), \ g(t, z)\ge \nu_1 (z-\widetilde{c}), \ I_k(z)\ge l_k z- \widetilde{c}_k, \ z\in \mathbb R^+, t\in [0, 1], k = 1, 2, ..., m. |
Consequently, if z\in S_1 , we have
\begin{equation} \begin{aligned} z(t)&\ge (\mathcal Tz)(t) \\ & \ge \mu_1 \int_0^1 G(t, s)( z(s)-\widetilde{c})ds+\nu_1 \frac{t^{\beta-1}}{\beta-1} \int_0^1 ( z(s)-\widetilde{c}) d\alpha(s) + \sum\limits_{k = 1}^m H(t, t_k)\left(l_kz\left(t_k\right)-\widetilde{c}_k\right).\end{aligned} \end{equation} | (3.3) |
Multiplying by d\psi_{\mu_1, \nu_1}(t) on both sides of (3.3) and integrating over [0, 1] , from (3.1) we have
\begin{aligned} \int_0^1 z(t)d \psi_{\mu_1, \nu_1}(t)& \ge \int_0^1 \left[\mu_1 \int_0^1 G(t, s)( z(s)-\widetilde{c})ds+\nu_1 \frac{t^{\beta-1}}{\beta-1} \int_0^1 ( z(s)-\widetilde{c}) d\alpha(s)\right]d \psi_{\mu_1, \nu_1}(t) \\ & \ \ \ + \sum\limits_{k = 1}^m \int_0^1 H(t, t_k)\left(l_kz\left(t_k\right)-\widetilde{c}_k\right) d \psi_{\mu_1, \nu_1}(t)\\ & = \int_0^1 ( z(s)-\widetilde{c}) d\left( \mu_1 \int_0^s d\tau \int_0^1 G(t, \tau)d \psi_{\mu_1, \nu_1}(t) +\nu_1 \alpha(s) \int_0^1 \frac{t^{\beta-1}}{\beta-1} d \psi_{\mu_1, \nu_1}(t) \right) \\ & \ \ \ + \sum\limits_{k = 1}^m \int_0^1 H(t, t_k) d \psi_{\mu_1, \nu_1}(t)\left(l_kz\left(t_k\right)-\widetilde{c}_k\right) \\ & = \int_0^1 ( z(s)-\widetilde{c}) d (\mathcal L_{\mu_1, \nu_1}^* \psi_{\mu_1, \nu_1})(s) + \sum\limits_{k = 1}^m \int_0^1 H(t, t_k) d \psi_{\mu_1, \nu_1}(t)\left(l_kz\left(t_k\right)-\widetilde{c}_k\right) \\ & = \int_0^1 ( z(s)-\widetilde{c}) d(r\left(\mathcal L_{\mu_1, \nu_1}\right) \psi_{\mu_1, \nu_1})(s)+ \sum\limits_{k = 1}^m \int_0^1 H(t, t_k) d \psi_{\mu_1, \nu_1}(t)\left(l_kz\left(t_k\right)-\widetilde{c}_k\right) . \end{aligned} |
Thus,
\begin{equation} \begin{aligned} &\int_0^1 z(t)d \psi_{\mu_1, \nu_1}(t) + \widetilde{c}r\left(\mathcal L_{\mu_1, \nu_1}\right)\int_0^1 d\psi_{\mu_1, \nu_1}(t)+ \sum\limits_{k = 1}^m \widetilde{c}_k \int_0^1 H(t, t_k) d \psi_{\mu_1, \nu_1}(t) \\ \ge& r\left(\mathcal L_{\mu_1, \nu_1}\right) \int_0^1 z(t) d \psi_{\mu_1, \nu_1}(t)+ \sum\limits_{k = 1}^m l_k \int_0^1 H(t, t_k) d \psi_{\mu_1, \nu_1}(t) z\left(t_k\right). \end{aligned} \end{equation} | (3.4) |
There are two cases to consider.
Case 1. r\left(\mathcal L_{\mu_1, \nu_1}\right)\ge 1 . From (3.2) and (3.4) we have
\begin{aligned} &\ [r\left(\mathcal L_{\mu_1, \nu_1}\right)-1] \|z\| \int_0^1 t^{\beta-1} d \psi_{\mu_1, \nu_1}(t)+ \|z\| \sum\limits_{k = 1}^m l_k t_k^{\beta-1} \int_0^1 H(t, t_k) d \psi_{\mu_1, \nu_1}(t)\\ \le& \widetilde{c}r\left(\mathcal L_{\mu_1, \nu_1}\right)\int_0^1 d\psi_{\mu_1, \nu_1}(t)+ \sum\limits_{k = 1}^m \widetilde{c}_k \int_0^1 H(t, t_k) d \psi_{\mu_1, \nu_1}(t), \end{aligned} |
and thus
\|z\|\le \frac{\widetilde{c}r\left(\mathcal L_{\mu_1, \nu_1}\right)\int_0^1 d\psi_{\mu_1, \nu_1}(t)+ \sum\limits_{k = 1}^m \widetilde{c}_k \int_0^1 H(t, t_k) d \psi_{\mu_1, \nu_1}(t)}{[r\left(\mathcal L_{\mu_1, \nu_1}\right)-1] \int_0^1 t^{\beta-1} d \psi_{\mu_1, \nu_1}(t)+ \sum\limits_{k = 1}^m l_k t_k^{\beta-1} \int_0^1 H(t, t_k) d \psi_{\mu_1, \nu_1}(t)}. |
Case 2. Now r\left(\mathcal L_{\mu_1, \nu_1}\right) < 1 . (H2), (3.2), and (3.4) imply that
\begin{aligned} &\ [r\left(\mathcal L_{\mu_1, \nu_1}\right)-1] \|z\| \int_0^1 d \psi_{\mu_1, \nu_1}(t)+ \|z\| \sum\limits_{k = 1}^m l_k t_k^{\beta-1} \int_0^1 H(t, t_k) d \psi_{\mu_1, \nu_1}(t)\\ \le& \widetilde{c}r\left(\mathcal L_{\mu_1, \nu_1}\right)\int_0^1 d\psi_{\mu_1, \nu_1}(t)+ \sum\limits_{k = 1}^m \widetilde{c}_k \int_0^1 H(t, t_k) d \psi_{\mu_1, \nu_1}(t), \end{aligned} |
and then
\|z\|\le \frac{\widetilde{c}r\left(\mathcal L_{\mu_1, \nu_1}\right)\int_0^1 d\psi_{\mu_1, \nu_1}(t)+ \sum\limits_{k = 1}^m \widetilde{c}_k \int_0^1 H(t, t_k) d \psi_{\mu_1, \nu_1}(t)}{[r\left(\mathcal L_{\mu_1, \nu_1}\right)-1] \int_0^1 d \psi_{\mu_1, \nu_1}(t)+ \sum\limits_{k = 1}^m l_k t_k^{\beta-1} \int_0^1 H(t, t_k) d \psi_{\mu_1, \nu_1}(t)}. |
Combining the two cases, we have proved that S_1 is a bounded set, as required. Now, we choose a sufficiently large R_1 > \sup S_1 such that
\begin{equation} z-\mathcal Tz \not = \lambda \widetilde{z}, \ z\in \partial B_{R_1}\cap P, \ \lambda\ge 0, \end{equation} | (3.5) |
where B_{R_1} = \{z\in P: \|z\| < R_1\} . Therefore, Lemma 2.7 implies that
\begin{equation} i(\mathcal T, B_{R_1}\cap P, P) = 0. \end{equation} | (3.6) |
By (H3) there exists r_1 > 0 such that
\begin{equation} f(t, z)\le \mu_2 z, \ g(t, z)\le \nu_2 z, \ I_k(z)\le \widetilde{l}_k z, \ z\in [0, r_1], t\in [0, 1], k = 1, 2, ..., m. \end{equation} | (3.7) |
Now, we prove that
\begin{equation} z \not = \lambda \mathcal Tz, \ z\in \partial B_{r_1}\cap P, \ \lambda\in [0, 1], \end{equation} | (3.8) |
where B_{r_1} = \{z\in P: \|z\| < r_1\} . If the claim is false, then there exist a z_1\in \partial B_{r_1}\cap P, \ \lambda_1\in [0, 1] such that
z_1 = \lambda_1 \mathcal Tz_1. |
By Lemma 2.5, z_1 satisfies (3.2), and from (3.7) we have
\begin{equation} z_1(t)\le (\mathcal Tz_1)(t)\le \mu_2 \int_0^1 G(t, s) z_1(s)ds+\nu_2 \frac{t^{\beta-1}}{\beta-1} \int_0^1 z_1(s) d\alpha(s) + \sum\limits_{k = 1}^m H(t, t_k)\widetilde{l}_k z_1\left(t_k\right). \end{equation} | (3.9) |
Multiplying by d\psi_{\mu_2, \nu_2}(t) on both sides of (3.9) and integrating over [0, 1] , from (3.1) we obtain
\begin{aligned} \int_0^1 z_1(t) d\psi_{\mu_2, \nu_2}(t) & \le \mu_2 \int_0^1 \int_0^1 G(t, s) z_1(s)ds d\psi_{\mu_2, \nu_2}(t) +\nu_2 \int_0^1 \frac{t^{\beta-1}}{\beta-1} \int_0^1 z_1(s) d\alpha(s) d\psi_{\mu_2, \nu_2}(t) \\ & \ \ \ + \sum\limits_{k = 1}^m \widetilde{l}_k\int_0^1 H(t, t_k)d\psi_{\mu_2, \nu_2}(t) z_1\left(t_k\right)\\ & = \int_0^1 z_1(s) d \left(\mu_2 \int_0^s d\tau \int_0^1 G(t, \tau)d \psi_{\mu_2, \nu_2}(t) +\nu_2 \alpha(s) \int_0^1 \frac{t^{\beta-1}}{\beta-1} d \psi_{\mu_2, \nu_2}(t)\right)\\ & \ \ \ + \sum\limits_{k = 1}^m \widetilde{l}_k \int_0^1 H(t, t_k)d\psi_{\mu_2, \nu_2}(t) z_1\left(t_k\right) \\ & = \int_0^1 z_1(s) d (\mathcal L_{\mu_2, \nu_2}^* \psi_{\mu_2, \nu_2})(s) + \sum\limits_{k = 1}^m \widetilde{l}_k \int_0^1 H(t, t_k)d\psi_{\mu_2, \nu_2}(t) z_1\left(t_k\right)\\ & = \int_0^1 z_1(s)d ( r\left(\mathcal L_{\mu_2, \nu_2}\right) \psi_{\mu_2, \nu_2})(s) + \sum\limits_{k = 1}^m \widetilde{l}_k \int_0^1 H(t, t_k)d\psi_{\mu_2, \nu_2}(t) z_1\left(t_k\right) .\end{aligned} |
This, together with (3.2), implies that
[1-r\left(\mathcal L_{\mu_2, \nu_2}\right)]\|z_1\|\int_0^1 t^{\beta-1} d\psi_{\mu_2, \nu_2}(t)\le \|z_1\| \sum\limits_{k = 1}^m \widetilde{l}_k \int_0^1 H(t, t_k)d\psi_{\mu_2, \nu_2}(t) . |
This contradicts (H3) unless \|z_1\| = 0 . Note that \|z_1\| = 0 also contradicts z_1\in \partial B_{r_1}\cap P, \ r_1 > 0 . Therefore, we obtain that (3.8) holds, as required. From Lemma 2.8 we have
\begin{equation} i(\mathcal T, B_{r_1}\cap P, P) = 1. \end{equation} | (3.10) |
Note that R_1 can be chosen large enough such that R_1 > \sup S_1 and R_1 > r_1 . Therefore, from (3.6) and (3.10) we have
i(\mathcal T, (B_{R_1}\backslash \overline{B}_{r_1})\cap P, P) = i(\mathcal T, B_{R_1}\cap P, P)-i(\mathcal T, B_{r_1}\cap P, P) = -1. |
Therefore, the operator \mathcal T has at least one fixed point in (B_{R_1}\backslash \overline{B}_{r_1})\cap 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 r_2 > 0 such that
\begin{equation} f(t, z)\ge \mu_3 z, \ g(t, z)\ge \nu_3 z, \ I_k(z)\ge \overline{l}_k z, \ z\in [0, r_2], t\in [0, 1], k = 1, 2, ..., m. \end{equation} | (3.11) |
For this r_2 , we prove that
\begin{equation} z-\mathcal Tz \not = \lambda \overline{z}, \ z\in \partial B_{r_2}\cap P, \ \lambda\ge 0, \end{equation} | (3.12) |
where B_{r_2} = \{z\in P: \|z\| < r_2\} , and \overline{z} is a fixed element in P_0 . If (3.12) is false, then there exist a z_2\in \partial B_{r_2}\cap P , \lambda_2\ge 0 such that
z_2-\mathcal Tz_2 = \lambda_2 \overline{z}. |
Lemma 2.5 implies that z_2 satisfies (3.2). Moreover, from (3.11) we have
\begin{equation} z_2(t)\ge (\mathcal Tz_2)(t) \ge \mu_3 \int_0^1 G(t, s)z_2(s)ds+\nu_3 \frac{t^{\beta-1}}{\beta-1} \int_0^1 z_2(s) d\alpha(s) + \sum\limits_{k = 1}^m H(t, t_k)\overline{l}_kz_2\left(t_k\right). \end{equation} | (3.13) |
Multiplying by d\psi_{\mu_3, \nu_3}(t) on both sides of (3.13) and integrating over [0, 1] , from (3.1) we obtain
\begin{aligned} \int_0^1 z_2(t) d\psi_{\mu_3, \nu_3}(t)& \ge \mu_3 \int_0^1 \int_0^1 G(t, s)z_2(s)dsd\psi_{\mu_3, \nu_3}(t)+\nu_3 \int_0^1 \frac{t^{\beta-1}}{\beta-1} \int_0^1 z_2(s) d\alpha(s)d\psi_{\mu_3, \nu_3}(t)\\ & \ \ \ + \sum\limits_{k = 1}^m \overline{l}_k \int_0^1 H(t, t_k)d\psi_{\mu_3, \nu_3}(t)z_2\left(t_k\right)\\ & = \int_0^1 z_2(s) d\left(\mu_3 \int_0^s d\tau \int_0^1 G(t, \tau)d \psi_{\mu_3, \nu_3}(t) +\nu_3 \alpha(s) \int_0^1 \frac{t^{\beta-1}}{\beta-1} d \psi_{\mu_3, \nu_3}(t)\right)\\ & \ \ \ + \sum\limits_{k = 1}^m \overline{l}_k \int_0^1 H(t, t_k)d\psi_{\mu_3, \nu_3}(t)z_2\left(t_k\right)\\ & = \int_0^1 z_2(s) d (\mathcal L_{\mu_3, \nu_3}^* \psi_{\mu_3, \nu_3})(s)+ \sum\limits_{k = 1}^m \overline{l}_k \int_0^1 H(t, t_k)d\psi_{\mu_3, \nu_3}(t)z_2\left(t_k\right)\\ & = \int_0^1 z_2(s) d( r\left(\mathcal L_{\mu_3, \nu_3}\right) \psi_{\mu_3, \nu_3})(s)+ \sum\limits_{k = 1}^m \overline{l}_k \int_0^1 H(t, t_k)d\psi_{\mu_3, \nu_3}(t)z_2\left(t_k\right). \end{aligned} |
There are two cases to consider.
Cases 1. r\left(\mathcal L_{\mu_3, \nu_3}\right)\ge 1 . From (3.2) we obtain
\|z_2\|\left[(r\left(\mathcal L_{\mu_3, \nu_3}\right)-1) \int_0^1 t^{\beta-1} d\psi_{\mu_3, \nu_3}(t)+ \sum\limits_{k = 1}^m \overline{l}_k t_k^{\beta-1}\int_0^1 H(t, t_k)d\psi_{\mu_3, \nu_3}(t)\right]\le 0, |
which contradicts z_2\in \partial B_{r_2}\cap P, \ r_2 > 0 .
Cases 2. r\left(\mathcal L_{\mu_3, \nu_3}\right) < 1 . By (3.2) we have
\|z_2\|\left[(r\left(\mathcal L_{\mu_3, \nu_3}\right)-1) \int_0^1 d\psi_{\mu_3, \nu_3}(t)+ \sum\limits_{k = 1}^m \overline{l}_k t_k^{\beta-1}\int_0^1 H(t, t_k)d\psi_{\mu_3, \nu_3}(t)\right]\le 0, |
and it contradicts (H4) unless \|z_2\| = 0 . We also have a contradiction to z_2\in \partial B_{r_2}\cap P, \ r_2 > 0 if \|z_2\| = 0 .
Therefore, we obtain that (3.12) holds, and Lemma 2.7 implies that
\begin{equation} i(\mathcal T, B_{r_2}\cap P, P) = 0. \end{equation} | (3.14) |
By (H5) there exist \overline{c}, \overline{c}_k > 0(k = 1, 2, ..., m) such that
\begin{equation} f(t, z)\le \mu_4( z+\overline{c}), \ g(t, z)\le \nu_4 (z+\overline{c}), \ I_k(z)\le \widehat{l}_k z+ \overline{c}_k, \ z\in \mathbb R^+, t\in [0, 1], k = 1, 2, ..., m. \end{equation} | (3.15) |
Let S_2 = \{z\in P: z = \lambda \mathcal Tz, \lambda\in [0, 1] \} . We now prove that S_2 is bounded in P . If z\in S_2 , then by Lemma 2.5, (3.2) holds, and from (3.15) we have
\begin{equation} \begin{aligned} z(t)& \le (\mathcal Tz)(t)\\ & \le \mu_4 \int_0^1 G(t, s)\left(z(s)+\overline{c}\right)ds+\nu_4 \frac{t^{\beta-1}}{\beta-1} \int_0^1 \left( z(s)+\overline{c}\right) d\alpha(s) + \sum\limits_{k = 1}^m H(t, t_k)\left(\widehat{l}_k z\left(t_k\right)+\overline{c}_k\right). \end{aligned} \end{equation} | (3.16) |
Multiplying by d\psi_{\mu_4, \nu_4}(t) on both sides of (3.16) and integrating over [0, 1] , from (3.1) we obtain
\begin{aligned} \int_0^1 z(t) d\psi_{\mu_4, \nu_4}(t)& \le \mu_4 \int_0^1 \int_0^1 G(t, s)\left(z(s)+\overline{c}\right)dsd\psi_{\mu_4, \nu_4}(t) +\nu_4 \int_0^1 \frac{t^{\beta-1}}{\beta-1} \int_0^1 \left( z(s)+\overline{c}\right) d\alpha(s) d\psi_{\mu_4, \nu_4}(t)\\ & \ \ \ + \sum\limits_{k = 1}^m \int_0^1 H(t, t_k)d\psi_{\mu_4, \nu_4}(t)\left(\widehat{l}_k z\left(t_k\right)+\overline{c}_k\right)\\ & = \int_0^1 \left(z(s)+\overline{c}\right) d \left(\mu_4 \int_0^s d\tau \int_0^1 G(t, \tau)d \psi_{\mu_4, \nu_4}(t) +\nu_4\alpha(s) \int_0^1 \frac{t^{\beta-1}}{\beta-1} d \psi_{\mu_4, \nu_4}(t)\right)\\ & \ \ \ + \sum\limits_{k = 1}^m \int_0^1 H(t, t_k)d\psi_{\mu_4, \nu_4}(t)\left(\widehat{l}_k z\left(t_k\right)+\overline{c}_k\right)\\ & = \int_0^1 \left(z(s)+\overline{c}\right) d (\mathcal L_{\mu_4, \nu_4}^* \psi_{\mu_4, \nu_4})(s)+ \sum\limits_{k = 1}^m \int_0^1 H(t, t_k)d\psi_{\mu_4, \nu_4}(t)\left(\widehat{l}_k z\left(t_k\right)+\overline{c}_k\right)\\ & = \int_0^1 \left(z(s)+\overline{c}\right) d (r\left(\mathcal L_{\mu_4, \nu_4}\right) \psi_{\mu_4, \nu_4})(s)+ \sum\limits_{k = 1}^m \int_0^1 H(t, t_k)d\psi_{\mu_4, \nu_4}(t)\left(\widehat{l}_k z\left(t_k\right)+\overline{c}_k\right). \end{aligned} |
Note that (3.2) and r\left(\mathcal L_{\mu_4, \nu_4}\right) < 1 , and we have
(1-r\left(\mathcal L_{\mu_4, \nu_4}\right))\|z\|\int_0^1 t^{\beta-1} d\psi_{\mu_4, \nu_4}(t)\le \overline{c} r\left(\mathcal L_{\mu_4, \nu_4}\right)\int_0^1 d \psi_{\mu_4, \nu_4})(t) + \sum\limits_{k = 1}^m \int_0^1 H(t, t_k)d\psi_{\mu_4, \nu_4}(t)\left(\widehat{l}_k \|z\|+\overline{c}_k\right), |
and (H5) implies that
\|z\|\le \frac{\overline{c} r\left(\mathcal L_{\mu_4, \nu_4}\right)\int_0^1 d \psi_{\mu_4, \nu_4})(t)+\sum\limits_{k = 1}^m \overline{c}_k \int_0^1 H(t, t_k)d\psi_{\mu_4, \nu_4}(t)}{(1-r\left(\mathcal L_{\mu_4, \nu_4}\right))\int_0^1 t^{\beta-1} d\psi_{\mu_4, \nu_4}(t)-\sum\limits_{k = 1}^m \widehat{l}_k \int_0^1 H(t, t_k)d\psi_{\mu_4, \nu_4}(t) }. |
This implies that S_2 is a bounded set in P , as required. Therefore, we can choose a large number R_2 > \max\{\sup S_2, r_2\} such that
z \not = \lambda \mathcal Tz, \ z\in \partial B_{R_2}\cap P, \ \lambda\in [0, 1], |
where B_{R_2} = \{z\in P: \|z\| < R_2\} . From Lemma 2.8 we have
\begin{equation} i(\mathcal T, B_{R_2}\cap P, P) = 1. \end{equation} | (3.17) |
As a result, from (3.14) and (3.17) we have
i(\mathcal T, (B_{R_2}\backslash \overline{B}_{r_2})\cap P, P) = i(\mathcal T, B_{R_2}\cap P, P)-i(\mathcal T, B_{r_2}\cap P, P) = 1. |
Therefore, the operator \mathcal T has at least one fixed point in (B_{R_2}\backslash \overline{B}_{r_2})\cap 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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