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 four-point nonlocal boundary conditions
to nonlocal strip conditions of the form:
,
. These strip conditions may be regarded as six-point boundary conditions. Some new
existence and uniqueness results are obtained for this class of nonlocal problems
by using standard fixed point theorems and Leray-Schauder 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; Leray-Schauder 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 fractional-order 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 [1-19] 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 cDq 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 six-point 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 two-strip 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 right-hand side is convex as well as non-convex valued. The first result relies on the nonlinear alternative of Leray-Schauder type. In the second result, we shall combine the nonlinear alternative of Leray-Schauder type for single-valued 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 [24-26].
Definition 2.1 For at least n-times 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 Riemann-Liouville 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 cDqx(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 c0 and c1 in (2.3), we obtain the solution (2.2). □
3 Existence results for single-valued 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:
(A1) |f (t, x) - f (t, y)| ≤ L|x - y|, ∀t ∈ [0, 1], L > 0, x, y ∈ ℝ;
(A2) |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 (A1) 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 FBr ⊂ Br, where
. For x ∈ Br, 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 (A1) and (A2) 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 (A1) 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 (A1), we define
, and consequently we have
which is independent of x. Thus,
is equicontinuous. Hence, by the Arzelá-Ascoli 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 Leray-Schauder 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)| ≤κ|x|+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 BR ⊂ 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 Arzelá-Ascoli Theorem, hλ (x) = x - H (λ, x) = x - λFx is completely continuous. If (3.5) is true, then the following Leray-Schauder 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 Leray-Schauder degree, h1(t) = x - λFx = 0 for at least one x ∈ BR. In order to prove (3.5), we assume that x = λFx, λ ∈ [0, 1]. Then for x ∈ ∂BR 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, (A1) is satisfied with
. Further, Δ1 = 65/72, Δ2 = 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 multi-valued case
4.1 Preliminaries
Let us recall some basic definitions on multi-valued maps [28,29].
For a normed space (X, ‖.‖), let
,
,
, and
. A multi-valued 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 semi-continuous (u.s.c.) on X if for each x0 ∈ X, the set G(x0) is a nonempty closed subset of X, and if for each open set N of X containing G(x0), there exists an open neighborhood
of x0 such that
. G is said to be completely continuous if
is relatively compact for every
. If the multi-valued 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., xn → x*, yn → y*, yn ∈ G(xn) 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 L1([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 L1-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 semi-continuous (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 σ-algebra generated by all sets of the form
, where
is Lebesgue measurable in [0, 1] and
is Borel measurable in ℝ. A subset
of L1([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 semi-continuous
(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 semi-continuous
type (l.s.c. type) if its associated Nemytskii operator
is lower semi-continuous 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) = infa∈A d(a; b) and d(a, B) = infb∈B d(a; b). Then (Pb,cl(X), Hd) is a metric space and (Pcl(X), Hd) is a generalized metric space (see [30]).
Definition 4.4 A multivalued operator N : X → Pcl(X) is called:
(a) γ-Lipschitz if and only if there exists γ > 0 such that
(b) a contraction if and only if it is γ-Lipschitz 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 L1-Carathéodory multivalued map and let θ be a linear continuous mapping from L1([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 (single-valued)
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 → Pcl(X) is a contraction, then FixN ≠ ∅.
Definition 4.9 A function x ∈ C2([0, 1], ℝ) is a solution of the problem (1.2) if
, and there exists a function f ∈ L1([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:
(H1)
is Carathéodory and has nonempty compact and convex values;
(H2) there exists a continuous nondecreasing function ψ : [0, ∞) → (0, ∞) and a function
such that
(H3) 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 ∈ SF,x. We will show that ΩF satisfies the assumptions of the nonlinear alternative of Leray-Schauder type. The proof consists of several steps. As a first step, we show that ΩF is convex for each x ∈ C([0, 1], ℝ). This step is obvious since SF,x is convex (F has convex values), and therefore we omit the proof.
In the second step, we show that ΩF maps bounded sets (balls) into bounded sets in C([0, 1], ℝ). For a positive number ρ, let Bρ = {x ∈ C([0, 1], ℝ): ‖x‖ ≤ ρ} be a bounded ball in C([0, 1], ℝ). Then, for each h ∈ ΩF (x), x ∈ Bρ, there exists f ∈ SF,x such that
Then for t∈[0, 1] we have
Thus,
Now we show that ΩF maps bounded sets into ;equicontinuous sets of C([0, 1], ℝ).
Let t', t'' ∈ [0, 1] with t' < t'' and x ∈ Bρ. For each h ∈ ΩF(x), we obtain
Obviously the right-hand side of the above inequality tends to zero independently
of x ∈ Bρ as t'' - t' → 0. As ΩF satisfies the above three assumptions, therefore it follows by the Ascoli-Arzelá theorem
that
is completely continuous.
In our next step, we show that ΩF has a closed graph. Let
, and
. Then we need to show that
. Associated with hn ∈ ΩF(xn), 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: L1([0, 1], ℝ) → C([0, 1], ℝ) given by
Observe that
as n → ∞.
Thus, it follows by Lemma 4.6 that θ ο SF 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 ∉ ΩF(x) for any λ ∈ (0, 1) and all x ∈ ∂U. Let λ ∈ (0, 1) and x ∈ λΩF(x). Then there exists f ∈ L1([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 (A10), 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 ∈ λΩF(x) for some λ ∈ (0, 1). Consequently, by the nonlinear alternative of Leray-Schauder
type (Lemma 4.5), we deduce that ΩF 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 (H3) 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 semi-continuous maps with decomposable values.
Theorem 4.12 Assume that (A10), (H2) and the following condition holds:
(H4)
is a nonempty compact-valued 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 (H2) and (H4) that F is of l.s.c. type. Then from Lemma 4.7, there exists a continuous function f : C([0, 1], ℝ) → L1([0, 1], ℝ) such that
for all x ∈ C([0, 1], ℝ).
Consider the problem
Observe that if x ∈ C2([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 right-hand 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:
(H5) F : [0, 1] × ℝ → Pcp(ℝ) is such that F(·, x): [0, 1] → Pcp(ℝ) is measurable for each x ∈ ℝ.
(H6)
for almost all t ∈ [0, 1] and x,
with m ∈ L1([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 (H5), so F has a measurable selection (see Theorem III.6 [36]). Now we show that the operator ΩF, defined in the beginning of proof of Theorem 4.10, satisfies the assumptions of
Lemma 4.8. To show that ΩF(x) ∈ Pcl((C[0, 1], ℝ)) for each x ∈ C([0, 1], ℝ), let
be such that un → 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 vn converges to v in L1([0, 1], ℝ). Thus, v ∈ SF,x and for each t ∈ [0, 1],
Hence, u ∈ Ω(x).
Next, we show that there exists γ < 1 such that
Let
and h1 ∈ □(x). Then there exists v1(t) ∈ F(t, x(t)) such that, for each t ∈ [0, 1],
By (H7), we have
Since the multivalued operator
is measurable [36, Proposition III.4], there exists a function v2(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 ΩF 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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