Abstract
In this paper, we prove the existence of solutions for a boundary value problem involving both left RiemannLiouville and right Caputotype fractional derivatives. For this, we convert the posed problem to a sum of two integral operators, then we apply Krasnoselskii’s fixed point theorem to conclude the existence of nontrivial solutions.
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1 Introduction
The study of differential equations with forward and backward fractional derivatives is interesting since they can model some physical phenomena such as the fractional oscillator equations and the fractional EulerLagrange equations. Recently, a linear boundary value problem involving both the right Caputo and the left RiemannLiouville fractional derivatives have been studied by many authors [1–9]. Blaszczyk and Ciesielski [3–6] solved numerically a class of fractional EulerLagrange equations by transforming them into integral equations. GuezaneLakoud, Khaldi and Torres [7] used an upper and lower solutions method combined with the monotonicity of the right Caputo derivative to prove the existence of solutions for a nonlinear fractional oscillator equation and the unbounded solutions were studied in [10] by GuezaneLakoud and Kılıçman.
The aim of this work is to study of existence of solutions for a nonlinear boundary value problem involving both the right Caputo and the left RiemannLiouville fractional derivatives:
where \(0<\alpha \leq 1\), \(1<\beta \leq 2\), \({}^{C}D_{1^{}}^{\alpha }\) denotes the right Caputo derivative, \(D_{0^{+}}^{\beta }\) denotes the left RiemannLiouville, u is the unknown function and f: \([ 0,1 ] \times {R}\rightarrow {R}\) satisfies some conditions that will be specified later. For this end, we transform problem (1)(2) to an integral equation that we write as a sum of a contraction and a completely continuous operator; then we use Krasnoselskii’s fixed point theorem to prove the existence of nontrivial solutions.
Note that, according to boundary conditions (2), the Caputo derivatives \({}^{C}D_{0^{+}}^{\beta }\) and \({}^{C}D_{1^{}}^{\alpha }\) coincide respectively with the RiemannLiouville derivatives \(D_{0^{+}}^{\beta }\) and \(D_{1^{}}^{\alpha }\). So, equation (1) is reduced to the one containing only Caputo derivatives or only RiemannLiouville derivatives, i.e.,
or
Note that this study can be extended to a similar boundary value problem as follows:
see Remark 2.7.
Let us mention some interesting papers where fractional boundary value problems have been studied by different methods such as the upper and lower solutions method, fixed point theorems, successive approximations method, Mawhin coincidence degree theory and many more; see, for example, [8, 11–17].
We recall some essential definitions on fractional calculus, we refer the reader to [15, 18, 19] for more details.
Let \(p>0\), then the left and right RiemannLiouville fractional integral of a function g are defined, respectively, by
The left RiemannLiouville fractional derivative and the right Caputo fractional derivative of order \(p>0\), of a function g are, respectively,
where \(n1< p< n\). For the properties of RiemannLiouville and Caputo fractional derivatives, we mention the following.
Let \(n1< p< n\) and \(f\in L_{1} [ 0,1 ]\). Then

(1)
\(I_{0^{+}}^{pC}D_{0^{+}}^{p}f ( t ) =f ( t ) \sum_{k=0}^{n1}\frac{f^{ ( k ) } ( 0 ) }{k!}t^{k}\);

(2)
\(I_{1^{}}^{pC}D_{1^{}}^{p}f ( t ) =f ( t ) \sum_{k=0}^{n1}\frac{ ( 1 ) ^{k}f^{ ( k ) } ( 1 ) }{k!} ( 1t ) ^{k}\).
Next, we state Krasnoselskii’s fixed point theorem that can be found with its proof in [20].
Theorem 1.1
Krasnoselskii
Let M be a closed bounded convex nonempty subset of a Banach space E. Suppose that A and B map M into E such that

(i)
A is completely continuous,

(ii)
B is a contraction mapping,

(iii)
\(x,y\in M\) implies \(Ax+By\in M\).
Then there exists \(z\in M\) with \(z=Az+Bz\).
2 Existence of nontrivial solutions
We begin by solving the following linear problem:
Lemma 2.1
Assume that \(y\in L_{1} [ 0,1 ] \), then u is a solution to the linear boundary value problem (3)(4) if and only if u satisfies the integral equation
where
Proof
Firstly, we apply the righthand side fractional integral \(I_{1^{}}^{\alpha }\) to equation (3), then the fractional integral \(I_{0^{+}}^{\beta }\) to the resultant equation and take into account that \(D_{0^{+}}^{\beta }u ( t ) ={}^{C}D_{0^{+}}^{\beta }u ( t ) \) and the properties of Caputo fractional derivatives. We get
Using the boundary conditions \(u ( 0 ) =u^{\prime } ( 0 ) =0\), then \(u ( 1 ) =0\), we get
Substituting \(c_{0}\) in (7) yields
Finally, by using the Fubini theorem, we get
this achieves the proof. □
In the next lemma, we give the properties of the functions G and g.
Lemma 2.2
The functions G and g satisfy the following properties:

(1)
The functions \(G ( t,r ) \) and \(g ( r ) \) are nonnegative.

(2)
\(G(t,r)\leq \frac{1}{\Gamma ( \alpha +1 ) \Gamma ( \beta ) }\) and \(g(r)\leq \frac{1}{\Gamma ( \alpha +1 ) \Gamma ( \beta ) }\) for all \(r,t\in [ 0,1 ] \).
Proof
It is obvious that \(G(t,r)\geq 0\). Set
and
Then
Consequently, \(G(t,r)\leq \frac{1}{\Gamma ( \alpha +1 ) \Gamma ( \beta ) }\) for all \(r,t\in [ 0,1 ] \). Similarly, we prove that \(g(r)\leq \frac{1}{\Gamma ( \alpha +1 ) \Gamma ( \beta ) }\) for all \(r\in [ 0,1 ]\). The proof is complete.
Define the Banach space \(E=C ( [ 0,1 ] ,\mathbb{R} ) \) and the operators A and B on E by
Obviously, problem (1)(2) has a solution if and only if \(A+B\) has a fixed point, i.e.,
To prove this end, we make the following hypothesis:
(H) The function \(f (\cdot,0 ) \) is continuous and not identically null on \([ 0,1 ] \), and there exists a nonnegative function \(k\in L_{1} ( [ 0,1 ] ,\mathbb{R}_{+} ) \) such that
Let
where R is chosen such that
with \(L=\max \{ \vert f ( t,0 ) \vert ,0\leq t\leq 1 \}\). Clearly, M is a nonempty, bounded and convex subset of the Banach space E. □
Theorem 2.3
Under the hypothesis (H), problem (1)(2) has nontrivial solutions in M.
To prove Theorem 4, we have to prove that all the assumptions of Krasnoselskii’s fixed point theorem are satisfied, for this we need the following lemmas.
Lemma 2.4
Under the hypothesis (H), the mapping A is completely continuous on M.
Proof
The proof will be done in three steps.
Step 1. The mapping A is continuous on M. Consider the sequence \(( u_{n} ) _{n}\in M\) such that \(u_{n}\rightarrow u\) in M, then from Lemma 3 and the hypothesis (H) we get
Consequently, \(\Vert Au_{n}Au \Vert \rightarrow 0\), when n tends to ∞.
Step 2. \(( Au ) \) is uniformly bounded on M. Let \(u\in M\), then by condition (H) it yields
Step 3. \(( Au ) \) is equicontinuous on M. We have, for \(u\in M\), \(0\leq t_{1}< t_{2}\leq 1\),
So, \(\vert Au ( t_{1} ) Au ( t_{2} ) \vert \) tends to zero when \(t_{1}\rightarrow t_{2}\), thus \(( Au ) \) is equicontinuous. Finally, by ArzelaAscoli’s theorem, it follows that A is a completely continuous mapping on M. The proof is complete. □
Lemma 2.5
Under the hypothesis (H), the mapping B is a contraction on M.
Proof
Let \(u,v\in M\), then
thus B is a contraction on M. The proof is complete. □
Lemma 2.6
Under the hypothesis (H), \(Au+Bv\in M\) for all \(u,v\in M\).
Proof
Proceeding as in Step 2 of the proof of Lemma 2.5, we get
thus
so, \(Au+Bv\in M\). The proof is complete. □
Proof of Theorem 4
Since the mapping A is completely continuous by Lemma 2.5, the mapping B is a contraction by Lemma 2.6 and \(Au+Bv\in M\) for all \(u,v\in M\) by Lemma 2.6. Then all the hypotheses of Theorem 1 are satisfied. Thus there exists a nontrivial solution \(u^{\ast }\in M\) for problem (1)(2) such that \(u^{\ast }=Au^{\ast }+Bu^{\ast }\). The proof is complete. □
Remark 2.7
The present study can be extended to similar problems. For example, we can prove the existence of solutions for the following boundary value problem:
In fact, let Q be the reflection operator \(( Qf ) ( t ) =f ( 1t ) \). Since \(( QQu ) ( t ) =u ( t ) \), \({}^{C}D_{1^{}}^{\beta }Q=Q^{C}D_{0^{+}}^{\beta }\) and \({}^{C}D_{0^{+}}^{\alpha }Q=Q^{C}D_{1^{}}^{\alpha }\) (see [19]), then the boundary value problem (10)(11) is equivalent to the following one:
Set \(v(t)=Qu(t)\), then applying the operator Q, the boundary value problem (10)(11) becomes
where \(F ( t,x ) =f ( ( 1t ) ,x )\). Thus both f and F satisfy condition (H). Thanks to Theorem 4, we conclude that problem (10)(11) has a nontrivial solution.
Example 2.8
If we consider problem (1)(2) with \(\alpha =\frac{1}{2}\), \(\beta =\frac{3}{2}\) and
then it has a solution \(u^{\ast }\) such that \(\Vert u^{\ast } \Vert \leq 1\). Indeed, the hypothesis (H) holds:
Moreover, we have
Since
and
then R can be chosen as \(R=1\geq 0.754 64\). Thus, by Theorem 4, this problem has a nontrivial solution
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Acknowledgements
The authors are grateful to the anonymous referees for their valuable comments and especially grateful to the editor Prof. Bashir Ahmad for his comments and suggestions that improved this paper.
The third author is very grateful to Universiti Putra Malaysia for providing conducive environment and partial support to carry out this study.
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Lakoud, A.G., Khaldi, R. & Kılıçman, A. Existence of solutions for a mixed fractional boundary value problem. Adv Differ Equ 2017, 164 (2017). https://doi.org/10.1186/s136620171226y
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DOI: https://doi.org/10.1186/s136620171226y