Abstract
This article studies a new class of nonlocal boundary value problems of nonlinear differential equations and inclusions of fractional order with strip conditions. We extend the idea of fourpoint nonlocal boundary conditions to nonlocal strip conditions of the form: , . These strip conditions may be regarded as sixpoint boundary conditions. Some new existence and uniqueness results are obtained for this class of nonlocal problems by using standard fixed point theorems and LeraySchauder degree theory. Some illustrative examples are also discussed.
MSC 2000: 26A33; 34A12; 34A40.
Keywords:
fractional differential equations; fractional differential inclusions; nonlocal boundary conditions; fixed point theorems; LeraySchauder degree1 Introduction
The subject of fractional calculus has recently evolved as an interesting and popular field of research. A variety of results on initial and boundary value problems of fractional order can easily be found in the recent literature on the topic. These results involve the theoretical development as well as the methods of solution for the fractionalorder problems. It is mainly due to the extensive application of fractional calculus in the mathematical modeling of physical, engineering, and biological phenomena. For some recent results on the topic, see [119] and the references therein.
In this article, we discuss the existence and uniqueness of solutions for a boundary value problem of nonlinear fractional differential equations and inclusions of order q ∈ (1, 2] with nonlocal strip conditions. As a first problem, we consider the following boundary value problem of fractional differential equations
where ^{c}D^{q }denotes the Caputo fractional derivative of order q, is a given continuous function and σ, η are appropriately chosen real numbers.
The boundary conditions in the problem (1.1) can be regarded as sixpoint nonlocal boundary conditions, which reduces to the typical integral boundary conditions in the limit α, γ → 0, β, δ → 1. Integral boundary conditions have various applications in applied fields such as blood flow problems, chemical engineering, thermoelasticity, underground water flow, population dynamics, etc. For a detailed description of the integral boundary conditions, we refer the reader to the articles [20,21] and references therein. Regarding the application of the strip conditions of fixed size, we know that such conditions appear in the mathematical modeling of real world problems, for example, see [22,23].
As a second problem, we study a twostrip boundary value problem of fractional differential inclusions given by
where is a multivalued map, is the family of all subsets of ℝ.
We establish existence results for the problem (1.2), when the righthand side is convex as well as nonconvex valued. The first result relies on the nonlinear alternative of LeraySchauder type. In the second result, we shall combine the nonlinear alternative of LeraySchauder type for singlevalued maps with a selection theorem due to Bressan and Colombo for lower semicontinuous multivalued maps with nonempty closed and decomposable values, while in the third result, we shall use the fixed point theorem for contraction multivalued maps due to Covitz and Nadler.
The methods used are standard, however their exposition in the framework of problems (1.1) and (1.2) is new.
2 Linear problem
Let us recall some basic definitions of fractional calculus [2426].
Definition 2.1 For at least ntimes continuously differentiable function , the Caputo derivative of fractional order q is defined as
where [q] denotes the integer part of the real number q.
Definition 2.2 The RiemannLiouville fractional integral of order q is defined as
provided the integral exists.
By a solution of (1.1), we mean a continuous function x(t) which satisfies the equation ^{c}D^{q}x(t) = f (t, x(t)), 0 < t < 1, together with the boundary conditions of (1.1).
To define a fixed point problem associated with (1.1), we need the following lemma, which deals with the linear variant of problem (1.1).
Lemma 2.3 For a given , the solution of the fractional differential equation
subject to the boundary conditions in (1.1) is given by
where
Proof. It is well known that the solution of (2.1) can be written as [24]
where are constants. Applying the boundary conditions given in (1.1), we find that
Solving these equations simultaneously, we find that
Substituting the values of c_{0 }and c_{1 }in (2.3), we obtain the solution (2.2). □
3 Existence results for singlevalued case
Let denotes the Banach space of all continuous functions from [0, 1] → ℝ endowed with the norm defined by .
In view of Lemma 2.3, we define an operator by
Observe that the problem (1.1) has solutions if and only if the operator equation Fx = x has fixed points.
For the forthcoming analysis, we need the following assumptions:
(A_{1}) f (t, x)  f (t, y) ≤ Lx  y, ∀t ∈ [0, 1], L > 0, x, y ∈ ℝ;
(A_{2}) f (t, x) ≤ μ(t), ∀(t, x) ∈ [0, 1] × ℝ, and μ ∈ C([0, 1], ℝ^{+}).
For convenience, let us set
where
Theorem 3.1 Assume that is a jointly continuous function and satisfies the assumption (A_{1}) with L < 1/Λ, where Λ is given by (3.2). Then the boundary value problem (1.1) has a unique solution.
Proof. Setting and choosing , we show that FB_{r }⊂ B_{r}, where . For x ∈ B_{r}, we have
where Λ is given by (3.2). Observe that Λ depends only on the parameters involved in the problem. As L < 1/Λ, therefore F is a contraction. Thus, the conclusion of the theorem follows by the contraction mapping principle (Banach fixed point theorem). □
Now, we prove the existence of solutions of (1.1) by applying Krasnoselskii's fixed point theorem [27].
Theorem 3.2 (Krasnoselskii's fixed point theorem). Let M be a closed, bounded, convex, and nonempty subset of a Banach space X. Let A, B be the operators such that (i) Ax + By ∈ M whenever x, y ∈ M; (ii) A is compact and continuous; (iii) B is a contraction mapping. Then there exists z ∈ M such that z = Az + Bz.
Theorem 3.3 Let be a jointly continuous function satisfying the assumptions (A_{1}) and (A_{2}) with
Then the boundary value problem (1.1) has at least one solution on [0, 1].
and consider . We define the operators and on as
Thus, It follows from the assumption (A_{1}) together with (3.3) that is a contraction mapping. Continuity of f implies that the operator is continuous. Also, is uniformly bounded on as
Now we prove the compactness of the operator .
In view of (A_{1}), we define , and consequently we have
which is independent of x. Thus, is equicontinuous. Hence, by the ArzeláAscoli Theorem, is compact on Thus all the assumptions of Theorem 3.2 are satisfied. So the conclusion of Theorem 3.2 implies that the boundary value problem (1.1) has at least one solution on [0, 1]. □
Our next existence result is based on LeraySchauder degree theory.
Theorem 3.4 Let . Assume that there exist constants , where Λ is given by (3.2) and M > 0 such that f(t, x) ≤κx+M for all t ∈ [0, 1], x ∈ C[0, 1]. Then the boundary value problem (1.1) has at least one solution.
Proof. Consider the fixed point problem
where F is defined by (3.1). In view of the fixed point problem (3.4), we just need to prove the existence of at least one solution x ∈ C[0, 1] satisfying (3.4). Define a suitable ball B_{R }⊂ C[0, 1] with radius R > 0 as
where R will be fixed later. Then, it is sufficient to show that satisfies
Let us set
Then, by the ArzeláAscoli Theorem, h_{λ }(x) = x  H (λ, x) = x  λFx is completely continuous. If (3.5) is true, then the following LeraySchauder degrees are well defined and by the homotopy invariance of topological degree, it follows that
where I denotes the unit operator. By the nonzero property of LeraySchauder degree, h_{1}(t) = x  λFx = 0 for at least one x ∈ B_{R}. In order to prove (3.5), we assume that x = λFx, λ ∈ [0, 1]. Then for x ∈ ∂B_{R }and t ∈ [0, 1] we have
which, on taking norm and solving for ‖x‖, yields
Letting , (3.5) holds. This completes the proof. □
Example 3.5 Consider the following strip fractional boundary value problem
Here, q = 3/2, σ = 1, η = 1, α = 1/3, β = 1/2, γ = 2/3, δ = 3/4 and . As , therefore, (A_{1}) is satisfied with . Further, Δ_{1 }= 65/72, Δ_{2 }= 535/288, Δ = 4945/5184, and
Clearly, LΛ = 0.282191 < 1. Thus, by the conclusion of Theorem 3.1, the boundary value problem (3.6) has a unique solution on [0, 1].
Example 3.6 Consider the following boundary value problem
Here,
Clearly M = 1 and
Thus, all the conditions of Theorem 3.4 are satisfied and consequently the problem (3.7) has at least one solution.
4 Existence results for multivalued case
4.1 Preliminaries
Let us recall some basic definitions on multivalued maps [28,29].
For a normed space (X, ‖.‖), let , , , and . A multivalued map is convex (closed) valued if G(x) is convex (closed) for all x ∈ X. The map G is bounded on bounded sets if is bounded in X for all (i.e., . G is called upper semicontinuous (u.s.c.) on X if for each x_{0 }∈ X, the set G(x_{0}) is a nonempty closed subset of X, and if for each open set N of X containing G(x_{0}), there exists an open neighborhood of x_{0 }such that . G is said to be completely continuous if is relatively compact for every . If the multivalued map G is completely continuous with nonempty compact values, then G is u.s.c. if and only if G has a closed graph, i.e., x_{n }→ x_{*}, y_{n }→ y_{*}, y_{n }∈ G(x_{n}) imply y_{* }∈ G(x_{*}). G has a fixed point if there is x ∈ X such that x ∈ G(x). The fixed point set of the multivalued operator G will be denoted by FixG. A multivalued map is said to be measurable if for every , the function
is measurable.
Let C([0, 1]) denotes a Banach space of continuous functions from [0, 1] into ℝ with the norm . Let L^{1}([0, 1], ℝ) be the Banach space of measurable functions x : [0, 1] → ℝ which are Lebesgue integrable and normed by .
Definition 4.1 A multivalued map is said to be Carathéodory if
(i) t ↦ F (t, x) is measurable for each x ∈ ℝ;
(ii) x ↦ F (t, x) is upper semicontinuous for almost all t ∈ [0, 1];
Further a Carathéodory function F is called L^{1}Carathéodory if
(iii) for each α > 0, there exists such that
for all ‖x‖_{∞ }≤ α and for a. e. t ∈ [0, 1].
For each , define the set of selections of F by
Let X be a nonempty closed subset of a Banach space E and be a multivalued operator with nonempty closed values. G is lower semicontinuous (l.s.c.) if the set {y ∈ X : G(y) ∩ B ≠ ∅} is open for any open set B in E. Let A be a subset of [0, 1] × ℝ. A is measurable if A belongs to the σalgebra generated by all sets of the form , where is Lebesgue measurable in [0, 1] and is Borel measurable in ℝ. A subset of L^{1}([0, 1], ℝ) is decomposable if for all and measurable , the function , where stands for the characteristic function of .
Definition 4.2 Let Y be a separable metric space and let be a multivalued operator. We say N has a property (BC) if N is lower semicontinuous (l.s.c.) and has nonempty closed and decomposable values.
Let be a multivalued map with nonempty compact values. Define a multivalued operator associated with F as
which is called the Nemytskii operator associated with F.
Definition 4.3 Let be a multivalued function with nonempty compact values. We say F is of lower semicontinuous type (l.s.c. type) if its associated Nemytskii operator is lower semicontinuous and has nonempty closed and decomposable values.
Let (X, d) be a metric space induced from the normed space (X; ‖.‖). Consider given by
where d(A, b) = inf_{a∈A }d(a; b) and d(a, B) = inf_{b∈B }d(a; b). Then (P_{b,cl}(X), H_{d}) is a metric space and (P_{cl}(X), H_{d}) is a generalized metric space (see [30]).
Definition 4.4 A multivalued operator N : X → P_{cl}(X) is called:
(a) γLipschitz if and only if there exists γ > 0 such that
(b) a contraction if and only if it is γLipschitz with γ < 1.
The following lemmas will be used in the sequel.
Lemma 4.5 (Nonlinear alternative for Kakutani maps) [31]. Let E be a Banach space, C is a closed convex subset of E, U is an open subset of C and 0 ∈ U. Suppose that is a upper semicontinuous compact map; here denotes the family of nonempty, compact convex subsets of C. Then either
(i) F has a fixed point in , or
(ii) there is a u ∈ ∂U and λ ∈ (0, 1) with u ∈ λF(u).
Lemma 4.6 [32]Let X be a Banach space. Let be an L^{1}Carathéodory multivalued map and let θ be a linear continuous mapping from L^{1}([0, 1], X) to C([0, 1], X). Then the operator
is a closed graph operator in C([0, 1], X) × C([0, 1], X).
Lemma 4.7 [33]Let Y be a separable metric space and let be a multivalued operator satisfying the property (BC). Then N has a continuous selection, that is, there exists a continuous function (singlevalued) such that g(x) ∈ N(x) for every x ∈ Y .
Lemma 4.8 [34]Let (X, d) be a complete metric space. If N : X → P_{cl}(X) is a contraction, then FixN ≠ ∅.
Definition 4.9 A function x ∈ C^{2}([0, 1], ℝ) is a solution of the problem (1.2) if , and there exists a function f ∈ L^{1}([0, 1], ℝ) such that
f(t) ∈ F (t, x(t)) a.e. on [0, 1] and
4.2 The Carathéodory case
Theorem 4.10 Assume that:
(H_{1}) is Carathéodory and has nonempty compact and convex values;
(H_{2}) there exists a continuous nondecreasing function ψ : [0, ∞) → (0, ∞) and a function such that
(H_{3}) there exists a constant M > 0 such that
Then the boundary value problem (1.2) has at least one solution on [0, 1].
for f ∈ S_{F,x}. We will show that Ω_{F }satisfies the assumptions of the nonlinear alternative of LeraySchauder type. The proof consists of several steps. As a first step, we show that Ω_{F }is convex for each x ∈ C([0, 1], ℝ). This step is obvious since S_{F,x }is convex (F has convex values), and therefore we omit the proof.
In the second step, we show that Ω_{F }maps bounded sets (balls) into bounded sets in C([0, 1], ℝ). For a positive number ρ, let B_{ρ }= {x ∈ C([0, 1], ℝ): ‖x‖ ≤ ρ} be a bounded ball in C([0, 1], ℝ). Then, for each h ∈ Ω_{F }(x), x ∈ B_{ρ}, there exists f ∈ S_{F,x }such that
Then for t∈[0, 1] we have
Thus,
Now we show that Ω_{F }maps bounded sets into ^{;}equicontinuous sets of C([0, 1], ℝ).
Let t', t'' ∈ [0, 1] with t' < t'' and x ∈ B_{ρ}. For each h ∈ Ω_{F}(x), we obtain
Obviously the righthand side of the above inequality tends to zero independently of x ∈ B_{ρ }as t''  t' → 0. As Ω_{F }satisfies the above three assumptions, therefore it follows by the AscoliArzelá theorem that is completely continuous.
In our next step, we show that Ω_{F }has a closed graph. Let , and . Then we need to show that . Associated with h_{n }∈ Ω_{F}(x_{n}), there exists such that for each t ∈ [0, 1],
Thus it suffices to show that there exists such that for each t ∈ [0, 1],
Let us consider the linear operator θ: L^{1}([0, 1], ℝ) → C([0, 1], ℝ) given by
Observe that
as n → ∞.
Thus, it follows by Lemma 4.6 that θ ο S_{F }is a closed graph operator. Further, we have . Since , therefore, we have
Finally, we show there exists an open set U ⊆ C([0, 1], ℝ) with x ∉ Ω_{F}(x) for any λ ∈ (0, 1) and all x ∈ ∂U. Let λ ∈ (0, 1) and x ∈ λΩ_{F}(x). Then there exists f ∈ L^{1}([0, 1], ℝ) with such that, for t ∈ [0, 1], we have
and using the computations of the second step above we have.
Consequently, we have
In view of (A_{10}), there exists M such that x ≠ M. Let us set
Note that the operator is upper semicontinuous and completely continuous. From the choice of U, there is no x ∈ ∂U such that x ∈ λΩ_{F}(x) for some λ ∈ (0, 1). Consequently, by the nonlinear alternative of LeraySchauder type (Lemma 4.5), we deduce that Ω_{F }has a fixed point which is a solution of the problem (1.1). This completes the proof. □
Example 4.11 Consider the following strip fractional boundary value problem
Here, q = 3/2, σ = 1, η = 1, α = 1/3, β = 1/2, γ = 2/3, δ = 3/4, and is a multivalued map given by
For f ∈ F, we have
Thus,
with p(t) = 1, ψ(x) = 9.
Further, we see that (H_{3}) is satisfied with M > 20.679031. Clearly, all the conditions of Theorem 4.10 are satisfied. So there exists at least one solution of the problem (4.2) on [0, 1].
4.3 The lower semicontinuous case
As a next result, we study the case when F is not necessarily convex valued. Our strategy to deal with this problem is based on the nonlinear alternative of Leray Schauder type together with the selection theorem of Bressan and Colombo [35] for lower semicontinuous maps with decomposable values.
Theorem 4.12 Assume that (A_{10}), (H_{2}) and the following condition holds:
(H_{4}) is a nonempty compactvalued multivalued map such that
(a) (t, x) ↦ F (t, x) is Λ measurable,
(b) x ↦ F (t, x) is lower semicontinuous for each t ∈ [0, 1];
Then the boundary value problem (1.2) has at least one solution on [0, 1].
Proof. It follows from (H_{2}) and (H_{4}) that F is of l.s.c. type. Then from Lemma 4.7, there exists a continuous function f : C([0, 1], ℝ) → L^{1}([0, 1], ℝ) such that for all x ∈ C([0, 1], ℝ).
Consider the problem
Observe that if x ∈ C^{2}([0, 1], ℝ) is a solution of (4.3), then x is a solution to the problem (1.2). In order to transform the problem (4.3) into a fixed point problem, we define the operator as
It can easily be shown that is continuous and completely continuous. The remaining part of the proof is similar to that of Theorem 3.1. So we omit it. This completes the proof. □
4.4 The Lipschitz case
Now we prove the existence of solutions for the problem (1.2) with a nonconvex valued righthand side by applying a fixed point theorem for multivalued map due to Covitz and Nadler [34].
Theorem 4.13 Assume that the following conditions hold:
(H_{5}) F : [0, 1] × ℝ → P_{cp}(ℝ) is such that F(·, x): [0, 1] → P_{cp}(ℝ) is measurable for each x ∈ ℝ.
(H_{6}) for almost all t ∈ [0, 1] and x, with m ∈ L^{1}([0, 1], ℝ^{+}) and d(0, F(t, 0)) ≤ m(t) for almost all t ∈ [0, 1].
Then the boundary value problem (1.2) has at least one solution on [0, 1] if
Proof. Observe that the set is nonempty for each x ∈ C([0, 1], ℝ) by the assumption (H_{5}), so F has a measurable selection (see Theorem III.6 [36]). Now we show that the operator Ω_{F}, defined in the beginning of proof of Theorem 4.10, satisfies the assumptions of Lemma 4.8. To show that Ω_{F}(x) ∈ P_{cl}((C[0, 1], ℝ)) for each x ∈ C([0, 1], ℝ), let be such that u_{n }→ u (n → ∞) in C([0, 1], ℝ). Then u ∈ C([0, 1], ℝ) and there exists such that, for each t ∈ [0, 1],
As F has compact values, we pass onto a subsequence to obtain that v_{n }converges to v in L^{1}([0, 1], ℝ). Thus, v ∈ S_{F,x }and for each t ∈ [0, 1],
Hence, u ∈ Ω(x).
Next, we show that there exists γ < 1 such that
Let and h_{1 }∈ □(x). Then there exists v_{1}(t) ∈ F(t, x(t)) such that, for each t ∈ [0, 1],
By (H_{7}), we have
Since the multivalued operator is measurable [36, Proposition III.4], there exists a function v_{2}(t) which is a measurable selection for U. So and for each t ∈ [0, 1], we have
For each t ∈ [0, 1], let us define
Thus,
Hence,
Analogously, interchanging the roles of x and , we obtain
Since Ω_{F }is a contraction, it follows by Lemma 4.8 that □ has a fixed point x which is a solution of (1.2). This completes the proof. □
Competing interests
The authors declare that they have no competing interests.
Authors' contributions
Each of the authors, BA and SKN contributed to each part of this study equally and read and approved the final version of the manuscript.
Acknowledgements
The authors were grateful to the reviewers for their useful comments. The research of B. Ahmad was supported by Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, Saudi Arabia.
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