Positive solutions to singular Dirichlettype boundary value problems of nonlinear fractional differential equations
 Jiqiang Jiang^{1}Email authorView ORCID ID profile,
 Weiwei Liu^{1} and
 Hongchuan Wang^{1}
https://doi.org/10.1186/s1366201816276
© The Author(s) 2018
Received: 3 January 2018
Accepted: 2 May 2018
Published: 8 May 2018
Abstract
In this paper, by using the properties of the Green function, \(u_{0}\)positive operator and Gelfand’s formula, some properties of the first eigenvalue corresponding to the relevant operator are obtained. Based on these properties, the fixed point index of the nonlinear operator is calculated explicitly and some sufficient conditions for the existence and uniqueness results of positive solution are established.
Keywords
MSC
1 Introduction
The boundary value problems (BVPs) for fractional differential equations arise from the studies of models of aerodynamics, fluid flows, electrodynamics of complex medium, electrical networks, rheology, polymer rheology, economics, biology chemical physics, control theory, signal and image processing. Recently, the study of such kind of problems has received considerable attention both in theory and applications, see [1–30] and the references therein, such as resonant BVP [3, 18, 25], singular BVP [7, 15, 20, 30], nonlocal BVP [8, 14, 17, 29] and semipositone BVP [11, 16, 24, 28].
Our work presented in this paper has the following features. First of all, we discuss the equation contain two term, i.e., \(D^{\alpha}_{0+}x(t)+bx(t)=f(t,x(t))\), this is different from [4, 10], in other words, Ref. [4, 10] and this article discuss different problems. The second new feature is that BVP (1.2) possesses a singularity, that is, the nonlinear term may be singular at \(t=0,1\); this is different from [4, 10]. Thirdly, we discuss BVP (1.2), by using the properties of the Green function, \(u_{0}\)positive operator and Gelfand’s formula, some sufficient conditions for the existence and uniqueness results of positive solution are established. The method used differs significantly from [4, 10, 26]. So our results are new and meaningful.
The rest of this paper is organized as follows. In Sect. 2, we present some lemmas that are used to prove our main results. In Sect. 3, under the assumption that \(f(t,y)\) is a Lipschitz continuous function, by use of \(u_{0}\)positive operator, the uniqueness of positive solutions is established. The interesting point is that the Lipschitz constant is related to the first eigenvalues corresponding to the relevant operators. In Sect. 4, the existence results of positive solutions are obtained by use of the fixed poind index and spectral radius of some related linear operator.
2 Basic definitions and preliminaries
For the convenience of the reader, we present here some notation and lemmas which will be used in the proofs of our results.
Definition 2.1
Definition 2.2
 (\(\mathrm{H}_{0}\)):

\(b\in(0,\tilde{b}]\).
 (\(\mathrm{H}_{1}\)):

\(f:[0,1]\times[0,\infty)\to[0,\infty)\) is continuous.
 (\(\mathrm{H}_{2}\)):

\(a:(0,1)\to[0,\infty)\) is continuous and not identical zero on any closed subinterval of \((0,1)\) with$$0< \int_{0}^{1} s^{\alphaj}(1s)^{\alphaj}a(s) \,ds< +\infty, \quad j=1,2. $$
The following three lemmas can be found in [26].
Lemma 2.3
Lemma 2.4
 (1)
\(G(t,s)>0\), \(\forall t,s\in(0,1)\);
 (2)
\(G(t,s)=G(1s,1t)\), \(\forall t,s\in [0,1]\);
 (3)
\(G(t,s)\leq H(1)s(1s)^{\alpha1}t^{\alpha2}\), \(\forall t,s\in[0,1]\);
 (4)\(G(t,s)\geq\Lambda s(1s)^{\alpha1}(1t)t^{\alpha1}\), \(\forall t,s\in[0,1]\), where$$\Lambda=\min \biggl\{ (\alpha1)^{2}H(1), \frac{1}{H(1)(\Gamma(\alpha))^{2}} \biggr\} . $$
Lemma 2.5
 (1)
\(K(t,s)>0\), \(\forall t,s\in(0,1)\);
 (2)
\(K(t,s)\leq H(1)(1s)^{\alpha2}t(1t)\), \(\forall t,s\in[0,1]\);
 (3)
\(K(t,s)\leq H(1)s(1s)^{\alpha1}\), \(\forall t,s\in[0,1]\);
 (4)
\(K(t,s)\geq\Lambda s(1s)^{\alpha1}t(1t)\), \(\forall t,s\in[0,1]\).
Let \(E=C[0,1]\), then E is a Banach space with the norm \(\ x\ = \max_{0\leq t\leq1} x(t)\), for any \(x\in E\). Set \(P= \{x\in E : x(t)\geq0 \mbox{ for } t\in[0,1] \}\). P is a positive cone in E.
The following concept is due to Krasnosel’skii [31].
Definition 2.6
Lemma 2.7
L is \(u_{0}\)positive operator with \(u_{0}(t)=t(1t)\).
Proof
Lemma 2.8
(Krein–Rutmann [32])
Let \(L: E\to E\) be a continuous linear operator, P be a total cone and \(L(P)\subset P\). If there exist \(\psi\in E\setminus(P)\) and a positive constant c such that \(cL(\psi)\geq\psi\), then the spectral radius \(r(L)\neq0\) and L has a positive eigenfunction corresponding to its first eigenvalue \(\lambda=(r(L))^{1}\).
It follows from Lemma 2.7 and Lemma 2.8 that the spectral radius \(r(L)\neq0\), moreover L has a positive eigenfunction \(\varphi^{*}(t)\) corresponding to its first eigenvalue \(\lambda_{1}=(r(L))^{1}\).
Remark 2.9
Lemma 2.10
([31])
Let L be a completely continuous \(u_{0}\)bounded operator, \(\lambda_{1}>0\) is the first eigenvalue of L, \(y_{0}\) is a positive eigenfunction which belongs to \(P\setminus\{\theta\}\). Then, for any \(y\in P\setminus\{\theta\}\) with \(y\neq\mu y_{0}\) (\(\mu\geq0\)), \(\lambda_{1}Ly\nleqslant y\) and \(\lambda_{1}Ly\ngeqslant y\).
3 Uniqueness of positive solution
We now wish to show that under certain conditions, BVP (1.2) has a unique positive solution.
Theorem 3.1
Proof
It is not difficult to prove that \(T: P\to P\) is completely continuous and the existence of positive solution \(t^{\alpha 2}x\) for BVP (1.2) is equivalent to that of fixed point x of T in P.
Then \(x^{*}\) is the unique fixed point of T in P and \(\bar{x}=t^{\alpha2}x^{*}\) is the unique positive solution of BVP (1.2). The proof is completed. □
Remark 3.2
Remark 3.3
In Theorem 3.1 and Remark 3.2, we not only give the condition of the existence of a unique positive solution, but also establish an iterative sequence of solution and error estimation. In particular, since \(u_{0}(t) = t(1t)\in P\), and the initial value of the iterative sequence can begin from \(x_{0}=u_{0}=t(1t)\), this is simpler and helpful for computation.
4 Existence of a positive solution
Lemma 4.1
([33])
 (i)
If there exists \(y_{0}\in Q\setminus\{\theta\}\) such that \(yTy\neq\mu y_{0}\) for any \(y\in\partial Q_{r}\) and \(\mu\geq0\), then \(i(T, Q_{r}, Q) = 0\).
 (ii)
If \(Ty\neq\mu y\) for any \(y\in\partial Q_{r}\) and \(\mu\geq1\), then \(i(T, Q_{r}, Q) = 1\).
Lemma 4.2
([33])
 (i)
if \(Ty\ngeq y\), \(\forall y\in\partial Q\cap\Omega_{1}\); \(Ty\nleq y\), \(\forall y\in\partial Q\cap\Omega_{2}\) and
 (ii)
if \(Ty\nleq y\), \(\forall y\in\partial Q\cap\Omega_{1}\); \(Ty\ngeq y\), \(\forall y\in\partial Q\cap\Omega_{2}\) is satisfied.
Lemma 4.3
Assume that (\(\mathrm{H}_{0}\)), (\(\mathrm{H}_{1}\)) and (\(\mathrm{H}_{2}\)) hold, then \(T: Q\to Q\) is completely continuous.
Proof
Theorem 4.4
Proof
Lemma 4.5
Proof
It follows from the uniform continuity of the function \(K(t,s)\) that \(\{x_{\tau_{n}}\}\) is equicontinuous and uniformly bounded. By the Arzela–Ascoli theorem, \(\{x_{\tau_{n}}\}\) has a subsequence which converges to some \(\widetilde{x}_{0}\) as \(n\to\infty\). Without loss of generality, we may assume \(x_{\tau_{n}}\to\widetilde{x}_{0}\) as well. Obviously, \(\widetilde{x}_{0}\geq\theta\) and \(\\widetilde{x}_{0}\=1\).
Theorem 4.6
Proof
Example 4.7
Therefore, all conditions of Theorem 4.4 are satisfied. Thus Theorem 4.4 ensures that BVP (4.21) has at least one positive solution.
Declarations
Acknowledgements
The authors would like to thank the referee for his/her very important comments, which improved the results and the quality of the paper. This work was supported financially by Project funded by China Postdoctoral Science Foundation (2017M612231), Doctoral Scientific Research Foundation of Qufu Normal University and Youth Foundation of Qufu Normal University (BSQD20130140, XKJ201303).
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
Competing interests
The authors declare that they have no competing interests.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
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