Positive Solutions for a Fourth-Order Riemann–Stieltjes Integral Boundary Value Problem

Cui, Yujun; O’Regan, Donal; Xu, Jiafa

doi:https://doi.org/10.1155/2019/3748631

Mathematical Problems in Engineering

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Abstract Introduction Preliminaries Data Availability Conflicts of Interest Acknowledgments References Copyright Related Articles

Research Article | Open Access

Volume 2019 | Article ID 3748631 | https://doi.org/10.1155/2019/3748631

Positive Solutions for a Fourth-Order Riemann–Stieltjes Integral Boundary Value Problem

Yujun Cui,¹Donal O’Regan,²and Jiafa Xu³

Academic Editor: Higinio Ramos

Received04 Dec 2018

Accepted03 Dec 2019

Published18 Dec 2019

Abstract

In this paper, we use the fixed point index to study the existence of positive solutions for the fourth-order Riemann–Stieltjes integral boundary value problem , where : is a continuous function and denotes a linear function. Two existence theorems are obtained with some appropriate inequality conditions on the nonlinearity f, which involve the spectral radius of related linear operators. These conditions allow to have superlinear or sublinear growth in .

1. Introduction

In this paper, we investigate the existence of positive solutions for the following fourth-order Riemann–Stieltjes integral boundary value problem:where denotes the Riemann–Stieltjes integral with a suitable function β of bounded variation and

The deformation of an elastic beam in an equilibrium state can be described by a fourth-order ordinary equation boundary value problem [1], and there are a large number of papers in the literature in this direction; for example, see [1–30] and the references therein. In [1], the author used Krasnosel’skii’s fixed point theorem to establish one or two positive solutions for the fourth-order boundary value problem:

When and in [2], the authors studied the existence of positive solutions for the fourth-order m-point boundary value problem:where satisfies superlinear and sublinear growth conditions:where is the first eigenvalue of the relevant linear operator.

In [3], the authors studied the existence of an iterative solution for the fourth-order boundary value problem:where is continuous and satisfies some appropriate Lipschitz condition, and in [4], the authors used the method of upper and lower solution to establish existence results for the fourth-order four-point boundary value problem on time scales:where is a continuous function.

There are only a few papers in the literature which consider general nonlinearities for fourth-order boundary value problems. The difficulty lies in a priori estimates for third-order derivatives, so some authors adopted a Nagumo-type growth condition (see (H7) in Section 3) to overcome this difficulty; for example, see [15, 16, 31–34] and the references therein. In [15], the author studied the existence of positive solutions for the fourth-order boundary value problem:where satisfies some inequality conditions, where f grows both superlinearly and sublinearly about its variables . When f is superlinear, a Nagumo-type condition is used to restrict the growth of f on .

Integral boundary conditions arise in thermal conduction problems [35], semiconductor problems [36], and hydrodynamic problems [37], and there are some papers in the literature devoted to this direction (see [1, 9, 19, 34, 38–48]). In [19], the authors studied p-Laplacian fourth-order differential equations with Riemann–Stieltjes integral boundary conditions:

The authors used fixed point theory in cones to obtain the existence of positive solutions for the above problem and provided the interval ranges of the parameters for these solutions.

In [38], the authors studied the fractional differential equation with a singular decreasing nonlinearity and a p-Laplacian operator:

Using a double iterative technique, they showed that the above problem has a unique positive solution, and from an iterative technique, they established an appropriate sequence, which converges uniformly to the unique positive solution.

Motivated by the aforementioned works, the aim of this paper is to study the existence of positive solutions for the fourth-order Riemann–Stieltjes integral boundary value problem (1). The novelty is of two folds: (1) we provide some useful inequality conditions on f involving the first eigenvalue of the relevant linear operator (these conditions imply that f grows superlinearly and sublinearly) and (2) for the superlinear case, an appropriate Nagumo-type condition is used to restrict the growth of f on in (1).

2. Preliminaries

In this section, we first transform (1) into an equivalent Hammerstein-type integral equation. For this, let , for . Then, from the conditions , we have

Therefore, substituting (11) into (1) gives

Lemma 1. The problem (12) can be transformed into the Hammerstein-type integral equation:where and , for .

Proof. Using the function on to replace in (12), we consider the following problem:From the differential equation in (14), we obtainand thenThe condition implies thatUsing the condition , it enables us to obtainHence, we have As a result, substituting into (15) givesThis completes the proof.
Let , with , and . Then, is a Banach space, and P is a cone on E. From Lemma 1, we can define an operator as follows:Then, A is a completely continuous operator from the Arzelà–Ascoli theorem (this argument is standard).

Remark 1. (i) In our work, we need the nonnegativity of Green’s function , so we have the following assumption:(ii)We need some inequality conditions on the nonlinearity with respect to the variables . We consider some useful linear operators:If we know the function β, we can obtain the functions and .

Example 1. Let for . Then, , for . Letand then from (22), we findWe consider two cases:(i)Case 1: when , we have(ii)Case 2: when , we haveWe now calculate . For this, let and Then, we haveTherefore, from (22), we haveWe consider two cases:(i)Case 1: when , we have(ii)Case 2: when , we have

Example 2. Let for . Then, we haveHence, , for . Note (22) and Example 1, so we only need to calculateTherefore, we obtain

Lemma 2. (Krein–Rutman, see([49] theorem 19.3)). Let P be a reproducing cone in a real Banach space E, and let be a compact linear operator with . If , then there exists such that , where is the spectral radius of L.

Lemma 3. For not all zero numbers , we let

Then,where and .

Proof. We first give some inequalities for . Note that , for , and from the definition of , we see thatWith h and k as before, noteFor convenience, let , , and , for .
We only prove the inequalities in (35) about the spectral radius of . For convenience, let for . Then, we haveThus, we obtainFor all , we note thatHence, we can obtainGelfand’s theorem implies that Next, we introduce a conclusion in ([50], problem 2.1.4). Let , and a functional J on be asThen, we haveWe note that there exists such that . Then, in (38), for fixed t, we define a linear function:and thusThen, by the definition of the norm of linear function, we know that for all , there exists with such thatOn the contrary, note from the definition of our norm we haveConsequently, we haveFor the arbitrariness of ε, we haveAlso, for all , we obtainFrom Gelfand’s theorem, we have This completes the proof.

Lemma 4. (see [51]). Let E be a real Banach space and P be a cone on E. Suppose that is a bounded open set and that is a continuous compact operator. If there exists a such thatthen , where i denotes the fixed point index on P.

Lemma 5. (see [51]). Let E be a real Banach space and P be a cone on E. Suppose that is a bounded open set with and that is a continuous compact operator. Ifthen .

3. Main Results

In our paper, we let for . Now, and . Now, we list our assumptions on the nonlinearity f.

(H3) There exist not all zero numbers and such that , and .

(H4) There exist not all zero numbers and such that , and .

(H5) There exist not all zero numbers and such that , and .

(H6) There exist not all zero numbers and such that , and .

(H7) For any , there exists a positive continuous function on such that , and .

Remark 2. Considering Lemma 3, one can adjust the coefficients such that the spectral radii satisfy their respective conditions in (H3)–(H6).

Theorem 1. Suppose that (H0–H4) hold. Then, (1) has at least one positive solution.

Proof. Let . Now, we prove that W is a bounded set in P. If , then from (H3), we havewhere for . This implies thatNote that , , and we obtainThus,Since , we know that has a bounded inverse operator , withNote that , and we obtain . Therefore,This implies thatTherefore, we haveThat is, W is bounded. Now, we can select ( is defined in (H4)) such thatFrom Lemma 5, we haveOn the contrary, since and , it follows from Lemma 2 that there exists such that and . Now, we show thatIf this claim is false, then there exist and such that . Note that (otherwise, the theorem is proved). Then, from (H4), we havewhich implies that Let . Then, and . However, we note that , and this contradicts the definition of for . Therefore, (66) holds, as required. From Lemma 4, we haveFrom (65) and (69), we haveand hence A has at least one fixed point in , i.e., (1) has at least one positive solution. This completes the proof.

Theorem 2. Suppose that (H0–H2) and (H5–H7) hold. Then, (1) has at least one positive solution.

Proof. We show thatIf the claim is false, then there exist and such that , for . For , from (H5) we haveAlso, , for implies thatNote that and , and we have Therefore,This contradicts the fact that . Hence, (71) is true, as required. From Lemma 5, we haveOn the contrary, from Lemma 2 there exists such that , for . Let , where , for . Note that (otherwise, the theorem is proved). We shall show that U is a bounded set in P. If , then from (H6), we haveMultiplying both sides of the above inequality by and integrating from 0 to 1 yields This, together with , implies that Note that , and we haveThen, y is a concave and increasing function on . Hence, This enables us to obtainNow, note (82), and we see there is an such thatThis, together with (H7), implies thatNote that , and we obtainThis implies thatand then if we let , we haveTherefore, combining this and (H7), there exists such thatThus, U is bounded (see (82) and (88)). Taking , we haveFrom Lemma 4, we haveFrom (76) and (90), we haveand hence A has at least one fixed point in , i.e., (1) has at least one positive solution. This completes the proof.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This work was supported by the China Postdoctoral Science Foundation (grant no. 2019M652348), Technology Research Foundation of Chongqing Educational Committee (grant no. KJQN201900539), Natural Science Foundation of Chongqing Normal University (grant no. 16XYY24), and Shandong Natural Science Foundation (grant no. ZR2018MA011).

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Copyright © 2019 Yujun Cui et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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