doi:10.7151/dmdico.1137
NONLINEAR FRACTIONAL DIFFERENTIAL INCLUSIONS WITH ANTI-PERIODIC TYPE INTEGRAL BOUNDARY
CONDITIONS
Bashir Ahmad Department of Mathematics
Faculty of Science King Abdulaziz University
P.O. Box 80203, Jeddah 21589, Saudi Arabia e-mail: bashir
−qau@yahoo.com
and
Sotiris K. Ntouyas Department of Mathematics
University of Ioannina 451 10 Ioannina, Greece e-mail: sntouyas@uoi.gr
Abstract
This article studies a boundary value problem of nonlinear fractional differential inclusions with anti-periodic type integral boundary conditions.
Some existence results are obtained via fixed point theorems. The cases of convex-valued and nonconvex-valued right hand sides are considered. Sev- eral new results appear as a special case of the results of this paper.
Keywords: fractional differential inclusions, anti-periodic, integral bound- ary conditions, existence, fixed point theorems.
2010 Mathematics Subject Classification: 34A08, 34A60, 34B15.
1. Introduction
In this article, we discuss the existence of solutions for a boundary value prob-
lem of nonlinear fractional differential inclusions with anti-periodic type integral
boundary conditions given by
c
D
qx(t) ∈ F (t, x(t)), t ∈ [0, T ], T > 0, 1 < q ≤ 2, x(0) − λ
1x(T ) = µ
1Z
T 0g(s, x(s))ds, x
0(0) − λ
2x
0(T ) = µ
2Z
T 0h(s, x(s))ds, (1)
where
cD
qdenotes the Caputo fractional derivative of order q, F : [0, T ] × R → P(R) is a multivalued map, P(R) is the family of all subsets of R, g, h : [0, T ] × R → R are given continuous functions and λ
1, λ
2, µ
1, µ
2∈ R with λ
16= 1, λ
26= 1.
The boundary conditions in problem (1) reduce to anti-periodic boundary conditions by choosing λ
1= −1 = λ
2, µ
1= 0 = µ
2. So the boundary conditions of (1) can be thought as an extension of anti-periodic boundary conditions to integral case. For a detailed description of the integral boundary conditions and some recent work on boundary value problems of fractional order with integral boundary conditions, we refer the reader to the papers [1, 2, 3, 4, 5, 6, 8, 9, 14, 18, 19] and references therein.
The present work is motivated by a recent paper [7], in which the authors discussed the existence of solutions for nonlinear fractional differential equations with non-separated type integral boundary conditions. The aim here is to estab- lish existence results for the problem (1), when the right hand side is convex as well as nonconvex valued. In the first result (Theorem 5) we consider the case when the right hand side has convex values, and prove an existence result via nonlinear alternative for Kakutani maps. In the second result (Theorem 7), 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 semicontin- uous multivalued maps with nonempty closed and decomposable values, while in the third result (Theorem 9), we shall use the fixed point theorem for contraction multivalued maps due to Covitz and Nadler.
The method of proof employed in this article is standard, however its expo- sition in the framework of problem (1) is new.
2. Preliminaries Let us recall some basic definitions [15, 20].
Definition. For q > 0, the Riemann-Liouville fractional integral of order q is defined as
I
qx(t) = 1 Γ(q)
Z
t 0x(s)
(t − s)
1−qds,
provided the integral exists.
Definition. For an at least (n − 1)-times absolutely continuous function x : [0, ∞) → R, the Caputo derivative of fractional order q is defined as
c
D
qx(t) = 1 Γ(n − q)
Z
t 0(t − s)
n−q−1x
(n)(s)ds, n − 1 < q < n, n = [q] + 1, where [q] denotes the integer part of the real number q.
In order to define the solution for (1), we recall the following Lemma [7].
Lemma 1. For a given σ ∈ C([0, T ], R) ∩ L((0, 1), R), the unique solution of the boundary value problem
c
D
qx(t) = σ(t), t ∈ [0, T ], T > 0, 1 < q ≤ 2, x(0) − λ
1x(T ) = µ
1Z
T 0g(s, x(s))ds, x
0(0) − λ
2x
0(T ) = µ
2Z
T 0h(s, x(s))ds, (2)
is given by x(t) =
Z
t 0(t − s)
q−1Γ(q) σ(s)ds − ξ
1λ
1Z
T 0(T − s)
q−1Γ(q) σ(s)ds + ξ
2λ
2[λ
1T + (1 − λ
1)t]
Z
T 0(T − s)
q−2Γ(q − 1) σ(s)ds + ξ
2µ
2[λ
1T + (1 − λ
1)t]
Z
T 0h(s, x(s))ds − µ
1ξ
1Z
T 0g(s, x(s))ds, t ∈ [0, T ], where
ξ
1= 1
(λ
1− 1) , ξ
2= 1
(λ
2− 1)(λ
1− 1) .
3. Existence results
Let us begin this section by giving a definition of the solution for the problem (1).
Definition. A function x ∈ AC
1([0, T ], R) is a solution of the problem (1) if x(0) − λ
1x(T ) = µ
1R
T0
g(s, x(s))ds, x
0(0) − λ
2x
0(T ) = µ
2R
T0
h(s, x(s))ds, and
there exists a function f ∈ L
1([0, T ], R) such that f (t) ∈ F (t, x(t)) a.e. on [0, T ]
and
x(t) = Z
t0
(t − s)
q−1Γ(q) f (s)ds − ξ
1λ
1Z
T 0(T − s)
q−1Γ(q) f (s)ds + ξ
2λ
2[λ
1T + (1 − λ
1)t]
Z
T0
(T − s)
q−2Γ(q − 1) f (s)ds + ξ
2µ
2[λ
1T + (1 − λ
1)t]
Z
T 0h(s, x(s))ds − µ
1ξ
1Z
T 0g(s, x(s))ds.
3.1. The Carath´ eodory case
In this subsection, we are concerned with the existence of solutions for the prob- lem (1) when the right hand side has convex values. We first recall some preli- minary facts.
For a normed space (X, k · k), let P
cl(X) = {Y ∈ P(X) : Y is closed}, P
b(X) = {Y ∈ P(X) : Y is bounded}, P
cp(X) = {Y ∈ P(X) : Y is compact}, and P
cp,c(X) = {Y ∈ P(X) : Y is compact and convex}.
Definition. A multi-valued map G : X → P(X) :
(i) is convex (closed) valued if G(x) is convex (closed) for all x ∈ X;
(ii) is bounded on bounded sets if G(B) = ∪
x∈BG(x) is bounded in X for all B ∈ P
b(X) (i.e. sup
x∈B{sup{|y| : y ∈ G(x)}} < ∞);
(iii) is called upper semi-continuous (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 N
0of x
0such that G(N
0) ⊆ N ;
(v) is said to be completely continuous if G(B) is relatively compact for every B ∈ P
b(X);
(v) 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 Fix G.
Remark 2. It is known that, 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., x
n→ x
∗, y
n→ y
∗, y
n∈ G(x
n) imply y
∗∈ G(x
∗).
Definition. A multivalued map G : [0; 1] → P
cl(R) is said to be measurable if for every y ∈ R, the function
t 7−→ d(y, G(t)) = inf{ky − zk : z ∈ G(t)}
is measurable.
Definition. A multivalued map F : [0, T ]×R → P(R) is said to be Carath´eodory if
(i) t 7−→ F (t, x) is measurable for each x ∈ X;
(ii) x 7−→ F (t, x) is upper semicontinuous for almost all t ∈ [0, T ];
Further a Carath´ eodory function F is called L
1-Carath´ eodory if (iii) for each α > 0, there exists ϕ
α∈ L
1([0, T ], R
+) such that
kF (t, x)k = sup{|v| : v ∈ F (t, x)} ≤ ϕ
α(t) for all kxk ≤ α and for a.e. t ∈ [0, T ].
For each y ∈ C([0, T ], R), define the set of selections of F by
S
F,y:= {v ∈ L
1([0, T ], R) : v(t) ∈ F (t, y(t)) for a.e. t ∈ [0, T ]}.
The consideration of this subsection is based on the following fixed point Theo- rem ([13]).
Theorem 3 (Nonlinear alternative for Kakutani maps) ([13]). Let E be a Banach space, C a closed convex subset of E, U an open subset of C and 0 ∈ U. Suppose that F : U → P
c,cv(C) is an upper semicontinuous compact map; here P
c,cv(C) denotes the family of nonempty, compact convex subsets of C. Then either
(i) F has a fixed point in U , or
(ii) there is a u ∈ ∂U and λ ∈ (0, 1) with u ∈ λF (u).
The following lemma will be used in the sequel.
Lemma 4 ([17]). Let X be a Banach space. Let F : [0, T ] × X → P
cp,c(X) be an L
1-Carath´ eodory multivalued map and let Θ be a linear continuous mapping from L
1([0, T ], X) to C([0, T ], X). Then the operator
Θ ◦ S
F: C([0, T ], X) → P
cp,c(C([0, T ], X)), x 7→ (Θ ◦ S
F)(x) = Θ(S
F,x) is a closed graph operator in C([0, T ], X) × C([0, T ], X).
Theorem 5. Assume that:
(H
1) F : [0, T ] × R → P(R) is Carath´eodory and has nonempty compact convex
values;
(H
2) there exists a continuous nondecreasing function ψ : [0, ∞) → (0, ∞) and a function p ∈ C([0, T ], R
+) such that
kF (t, x)k
P:= sup{|y| : y ∈ F (t, x)} ≤ p(t)ψ(kxk) for each (t, x) ∈ [0, T ]×R;
(H
3) there exists a continuous nondecreasing function ψ
1: [0, ∞) → (0, ∞) and a function p
1∈ L
1([0, T ], R
+) such that
|h(t, x)| ≤ p
1(t)ψ
1(kxk) for each (t, x) ∈ [0, T ] × R;
(H
4) there exists a continuous nondecreasing function ψ
2: [0, ∞) → (0, ∞) and a function p
2∈ L
1([0, T ], R
+) such that
|g(t, x)| ≤ p
2(t)ψ
2(kxk) for each (t, x) ∈ [0, T ] × R;
(H
5) there exists a number M > 0 such that M ψ(M )T
qΓ(q + 1) Λ
1kpk + ψ
1(M )Λ
2kp
1k
L1+ ψ
2(M )Λ
3kp
2k
L1> 1,
where
Λ
1= 1 + |ξ
1λ
1| + q|ξ
2||λ
2|(1 + |λ
1|), Λ
2= |ξ
2||µ
2|(1 + |λ
1|)T,
Λ
3= |µ
1ξ
1|.
Then the boundary value problem (1) has at least one solution on [0, T ].
Proof. In order to transform boundary value problem (1) into a fixed point problem, consider the multivalued operator Ω : C([0, T ], R) → P(C([0, T ], R)) defined by
Ω(x) =
h ∈ C([0, T ], R) :
h(t) =
Z
t0
(t − s)
q−1Γ(q) f (s)ds − ξ
1λ
1Z
T 0(T − s)
q−1Γ(q) f (s)ds +ξ
2λ
2[λ
1T + (1 − λ
1)t]
Z
T 0(T − s)
q−2Γ(q − 1) f (s)ds +ξ
2µ
2[λ
1T + (1 − λ
1)t]
Z
T 0h(s, x(s))ds
−µ
1ξ
1Z
T0
g(s, x(s))ds, 0 ≤ t ≤ 1,
for f ∈ S
F,x. Clearly, according to Lemma 1, the fixed points of Ω are solutions to boundary value problem (1). We will show that Ω satisfies the assumptions of the nonlinear alternative of Leray-Schauder type. The proof consists of several steps.
As a first step, we show that Ω is convex for each x ∈ C([0, T ], R). For that, let h
1, h
2∈ Ω(x). Then there exist f
1, f
2∈ S
F,xsuch that for each t ∈ [0, T ], we have
h
i(t) = Z
t0
(t − s)
q−1Γ(q) f
i(s)ds − ξ
1λ
1Z
T 0(T − s)
q−1Γ(q) f
i(s)ds + ξ
2λ
2[λ
1T + (1 − λ
1)t]
Z
T 0(T − s)
q−2Γ(q − 1) f
i(s)ds + ξ
2µ
2[λ
1T + (1 − λ
1)t]
Z
T 0h(s, x(s))ds − µ
1ξ
1Z
T0
g(s, x(s))ds, i = 1, 2.
Let 0 ≤ ω ≤ 1. Then, for each t ∈ [0, T ], we have [ωh
1+ (1 − ω)h
2](t)
= 1
Γ(q) Z
t0
(t − s)
q−1[ωf
1(s) + (1 − ω)f
2(s)]ds
+ ξ
2λ
2[λ
1T + (1 − λ
1)t]
Z
T 0(T − s)
q−2Γ(q − 1) [ωf
1(s) + (1 − ω)f
2(s)]ds + ξ
2µ
2[λ
1T + (1 − λ
1)t]
Z
T 0h(s, x(s))ds − µ
1ξ
1Z
T 0g(s, x(s))ds.
Since S
F,xis convex (F has convex values), then it follows that ωh
1+ (1 − ω) h
2∈ Ω(x).
Next, we show that Ω maps bounded sets into bounded sets in C([0, T ], R).
For a positive number ρ, let B
ρ= {x ∈ C([0, T ], R) : kxk ≤ ρ} be a bounded set in C([0, T ], R). Then, for each h ∈ Ω(x), x ∈ B
ρ, there exists f ∈ S
F,xsuch that
h(t) = Z
t0
(t − s)
q−1Γ(q) f (s)ds − ξ
1λ
1Z
T 0(T − s)
q−1Γ(q) f (s)ds + ξ
2λ
2[λ
1T + (1 − λ
1)t]
Z
T0
(T − s)
q−2Γ(q − 1) f (s)ds + ξ
2µ
2[λ
1T + (1 − λ
1)t]
Z
T 0h(s, x(s))ds − µ
1ξ
1Z
T0
g(s, x(s))ds, t ∈ [0, T ].
Then
|h(t)| ≤ 1 Γ(q)
Z
t 0(t − s)
q−1|f (s)|ds + |ξ
1λ
1| Γ(q)
Z
T 0(T − s)
q−1|f (s)|ds
+ |ξ
2λ
2[λ
1T + (1 − λ
1)t]|
Γ(q − 1)
Z
T 0(T − s)
q−2|f (s)|ds
+ |ξ
2µ
2[λ
1T + (1 − λ
1)t]|
Z
T 0|h(s, x(s))|ds + |µ
1ξ
1| Z
T0
|g(s, x(s))|ds
≤ ψ(kxk)T
qΓ(q + 1) [1 + |ξ
1λ
1| + q|ξ
2||λ
2|(1 + |λ
1|)] kpk + |ξ
2||µ
2|(1 + |λ
1|)T ψ
1(kxk)
Z
T 0p
1(s)ds + |µ
1ξ
1|ψ
2(kxk) Z
T0
p
2(s)ds.
Thus,
khk ≤ ψ(ρ)T
qΓ(q + 1) [1 + |ξ
1λ
1| + q|ξ
2||λ
2|(1 + |λ
1|)] kpk
+ |ξ
2||µ
2|(1 + |λ
1|)T ψ
1(ρ)kp
1k
L1+ |µ
1ξ
1|ψ
2(ρ)kp
2k
L1.
Now we show that Ω maps bounded sets into equicontinuous sets of C([0, T ], R).
Let t
0, t
00∈ [0, T ] with t
0< t
00and x ∈ B
ρ, where B
ρis a bounded set of C([0, T ], R). For each h ∈ Ω(x), we obtain
|h(t
00) − h(t
0)| ≤
ψ(ρ)kpk Γ(q)
Z
t0 0[(t
00− s)
q−1− (t
0− s)
q−1]ds +
ψ(ρ)kpk Γ(q)
Z
t00 t0(t
00− s)
q−1f (s)ds + |ξ
2λ
2(1 − λ
1)||t
00− t
0|ψ(ρ)kpk
Γ(q − 1)
Z
T 0(T − s)
q−2ds + |ξ
2µ
2(1 − λ
1)||t
00− t
0|ψ
1(ρ)kp
1k
L1.
Obviously the right hand side of the above inequality tends to zero independently of x ∈ B
ras t
00− t
0→ 0. Since Ω satisfies the above three assumptions, then it follows by the Arzel´ a-Ascoli theorem that Ω : C([0, T ], R) → P(C([0, T ], R)) is completely continuous.
In our next step, we show that Ω has a closed graph. Let x
n→ x
∗, h
n∈ Ω(x
n)
and h
n→ h
∗. We need to show that h
∗∈ Ω(x
∗). There exist f
n∈ S
F,xnassociated
with h
n∈ Ω(x
n) and such that for each t ∈ [0, T ], h
n(t) =
Z
t 0(t − s)
q−1Γ(q) f
n(s)ds − ξ
1λ
1Z
T0
(T − s)
q−1Γ(q) f
n(s)ds + ξ
2λ
2[λ
1T + (1 − λ
1)t]
Z
T 0(T − s)
q−2Γ(q − 1) f
n(s)ds + ξ
2µ
2[λ
1T + (1 − λ
1)t]
Z
T 0h(s, x
n(s))ds − µ
1ξ
1Z
T 0g(s, x
n(s))ds.
Thus we have to show that there exists f
∗∈ S
F,x∗such that for each t ∈ [0, T ], h
∗(t) =
Z
t 0(t − s)
q−1Γ(q) f
∗(s)ds − ξ
1λ
1Z
T 0(T − s)
q−1Γ(q) f
∗(s)ds + ξ
2λ
2[λ
1T + (1 − λ
1)t]
Z
T 0(T − s)
q−2Γ(q − 1) f
∗(s)ds + ξ
2µ
2[λ
1T + (1 − λ
1)t]
Z
T 0h(s, x
∗(s))ds − µ
1ξ
1Z
T0
g(s, x
∗(s))ds.
Let us consider the continuous linear operator Θ : L
1([0, T ], R) → C([0, T ], R) given by
f 7→ Θ(f ) = Z
t0
(t − s)
q−1Γ(q) f (s)ds − ξ
1λ
1Z
T 0(T − s)
q−1Γ(q) f (s)ds + ξ
2λ
2[λ
1T + (1 − λ
1)t]
Z
T 0(T − s)
q−2Γ(q − 1) f (s)ds + ξ
2µ
2[λ
1T + (1−λ
1)t]
Z
T 0h(s, x(s))ds − µ
1ξ
1Z
T0
g(s, x(s))ds, t ∈ [0, T ].
Observe that
kh
n(t) − h
∗(t)k =
1 Γ(q)
Z
t 0(t − s)
q−1(f
n(s) − f
∗(s))ds
− ξ
1λ
1Γ(q)
Z
T 0(T − s)
q−1(f
n(s) − f
∗(s))ds
+ ξ
2λ
2[λ
1T + (1 − λ
1)t]
Γ(q − 1)
Z
T 0(T − s)
q−1(f
n(s) − f
∗(s))ds
+ ξ
2µ
2[λ
1T + (1 − λ
1)t]
Z
T 0[h(s, x
n(s)) − h(s, x
∗(s))]ds
− µ
1ξ
1Z
T 0[g(s, x
n(s)) − g(s, x
∗(s))]ds
→ 0,
as n → ∞. Thus, it follows by Lemma 4 that Θ ◦ S
Fis a closed graph operator.
Further, we have h
n(t) ∈ Θ(S
F,xn). Since x
n→ x
∗, then, we have h
∗(t) =
Z
t 0(t − s)
q−1Γ(q) f
∗(s)ds − ξ
1λ
1Z
T 0(T − s)
q−1Γ(q) f
∗(s)ds + ξ
2λ
2[λ
1T + (1 − λ
1)t]
Z
T 0(T − s)
q−2Γ(q − 1) f
∗(s)ds + ξ
2µ
2[λ
1T + (1−λ
1)t]
Z
T 0h(s, x
∗(s))ds − µ
1ξ
1Z
T0
g(s, x
∗(s))ds, t ∈ [0, T ], for some f
∗∈ S
F,x∗.
Finally, we discuss a priori bounds on solutions. Let x be a solution of (1).
Then there exists f ∈ L
1([0, T ], R) with f ∈ S
F,xsuch that, for t ∈ [0, T ], we have, using the computations on the second step
|x(t)| ≤ ψ(kxk)T
qΓ(q + 1) [1 + |ξ
1λ
1| + q|ξ
2||λ
2|(1 + |λ
1|)] kpk + |ξ
2||µ
2|(1 + |λ
1|)T ψ
1(kxk)
Z
T 0p
1(s)ds + |µ
1ξ
1|ψ
2(kxk) Z
T0
p
2(s)ds
≤ ψ(kxk)T
qΓ(q + 1) Λ
1kpk + ψ
1(kxk)Λ
2kp
1k
L1+ ψ
2(kxk)Λ
3kp
2k
L1. Consequently, we have
kxk ψ(kxk)T
qΓ(q + 1) Λ
1kpk + ψ
1(kxk)Λ
2kp
1k
L1+ ψ
2(kxk)Λ
3kp
2k
L1≤ 1.
In view of (H
5), there exists M such that kxk 6= M . Let us set U = {x ∈ C([0, T ], R) : kxk < M + 1}.
Note that the operator Ω : U → P(C([0, T ], R)) is upper semicontinuous and completely continuous. From the choice of U , there is no x ∈ ∂U such that x ∈ µΩ(x) for some µ ∈ (0, 1). Consequently, by the nonlinear alternative of Leray-Schauder type (Theorem 3), we deduce that Ω has a fixed point x ∈ U which is a solution of the problem (1). This completes the proof.
3.2. The lower semi-continuous 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 [10]
for lower semi-continuous maps with decomposable values.
Definition. Let X be a nonempty closed subset of a Banach space E and let G : X → P(E) be a multivalued operator with nonempty closed values. G is lower semi-continuous (l.s.c.) if the set {y ∈ X : G(y) ∩ B 6= ∅} is open for any open set B in E.
Definition. Let A be a subset of [0, T ] × R. A is L ⊗ B measurable if A belongs to the σ-algebra generated by all sets of the form J × D, where J is Lebesgue measurable in [0, T ] and D is Borel measurable in R.
Definition. A subset A of L
1([0, T ], R) is decomposable if for all x, y ∈ A and measurable J ⊂ [0, T ] = J , the function xχ
J+ yχ
J −J∈ A, where χ
Jstands for the characteristic function of J .
Definition. Let Y be a separable metric space and let N : Y → P(L
1([0, T ], R)) 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 F : [0, T ] × R → P(R) be a multivalued map with nonempty compact values.
Define a multivalued operator F : C([0, T ] × R) → P(L
1([0, T ], R)) associated with F as
F (x) = {w ∈ L
1([0, T ], R) : w(t) ∈ F (t, x(t)) for a.e. t ∈ [0, T ]}, which is called the Nemytskii operator associated with F.
Definition. Let F : [0, T ] × R → P(R) 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 F is lower semi-continuous and has nonempty closed and decomposable values.
Lemma 6 ([10]). Let Y be a separable metric space and let N : Y → P(L
1([0, T ], R)) be a multivalued operator satisfying the property (BC). Then N has a contin- uous selection, that is, there exists a continuous function (single-valued) g : Y → L
1([0, T ], R) such that g(x) ∈ N (x) for every x ∈ Y .
Theorem 7. Assume that (H
2)–(H
5) and the following condition holds:
(H
6) F : [0, T ] × R → P(R) is a nonempty compact-valued multivalued map such that
(a) (t, x) 7−→ F (t, x) is L ⊗ B measurable,
(b) x 7−→ F (t, x) is lower semicontinuous for each t ∈ [0, T ].
Then the boundary value problem (1) has at least one solution on [0, T ].
Proof. It follows from (H
6) and (H
2) that F is of l.s.c. type. Then from Lemma 6, there exists a continuous function f : C([0, T ], R) → L
1([0, T ], R) such that f (x) ∈ F (x) for all x ∈ C([0, T ], R).
Consider the problem
c
D
qx(t) = f (x(t)), t ∈ [0, T ], T > 0, 1 < q ≤ 2, x(0) − λ
1x(T ) = µ
1Z
T 0g(s, x(s))ds, x
0(0) − λ
2x
0(T ) = µ
2Z
T 0h(s, x(s))ds, (3)
in the space C([0, T ], R). It is clear that if x ∈ C([0, T ], R) is a solution of the problem (3), then x is a solution to the problem (1). In order to transform the problem (3) into a fixed point problem, we define the operator Ω as
Ωx(t) = Z
t0
(t − s)
q−1Γ(q) f (x(s))ds − ξ
1λ
1Z
T0
(T − s)
q−1Γ(q) f (x(s))ds + ξ
2λ
2[λ
1T + (1 − λ
1)t]
Z
T 0(T − s)
q−2Γ(q − 1) f (x(s))ds + ξ
2µ
2[λ
1T + (1 − λ
1)t]
Z
T 0h(s, x(s))ds − µ
1ξ
1Z
T0
g(s, x(s))ds, t ∈ [0, T ].
It can easily be shown that Ω is continuous and completely continuous. The remaining part of the proof is similar to that of Theorem 5. So we omit it. This completes the proof.
3.3. The Lipschitz case
Now we prove the existence of solutions for the problem (1) with a nonconvex valued right hand side by applying a fixed point theorem for multivalued map due to Covitz and Nadler [12].
Let (X, d) be a metric space induced from the normed space (X; k · k). Con- sider H
d: P(X) × P(X) → R ∪ {∞} given by
H
d(A, B) = max{sup
a∈A
d(a, B), sup
b∈B
d(A, b)},
where d(A, b) = inf
a∈Ad(a; b) and d(a, B) = inf
b∈Bd(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 [16]).
Definition. A multivalued operator N : X → P
cl(X) is called:
(a) γ-Lipschitz if and only if there exists γ > 0 such that
H
d(N (x), N (y)) ≤ γd(x, y) for each x, y ∈ X;
(b) a contraction if and only if it is γ-Lipschitz with γ < 1.
Lemma 8 (Covitz-Nadler) ([12]). Let (X, d) be a complete metric space. If N : X → P
cl(X) is a contraction, then F ixN 6= ∅.
Definition. A measurable multi-valued function F : [0, T ] → P(X) is said to be integrably bounded if there exists a function h ∈ L
1([0, T ], X) such that for all v ∈ F (t), kvk ≤ h(t) for a.e. t ∈ [0, T ].
Theorem 9. Assume that the following conditions hold:
(H
7) F : [0, T ] × R → P
cp(R) is such that F (·, x) : [0, T ] → P
cp(R) is measurable for each x ∈ R;
(H
8) H
d(F (t, x), F (t, ¯ x)) ≤ m(t)|x − ¯ x| for almost all t ∈ [0, T ] and x, ¯ x ∈ R with m ∈ C([0, T ], R
+) and d(0, F (t, 0)) ≤ m(t) for almost all t ∈ [0, T ].
(H
9) There exist constants c
1, c
2> 0 such that
|h(t, x) − h(t, ¯ x)| ≤ c
1|x − ¯ x|, |g(t, x) − g(t, ¯ x)| ≤ c
2|x − ¯ x|
for all t ∈ [0, T ] and x, ¯ x ∈ R.
Then the boundary value problem (1) has at least one solution on [0, T ] if T
qΓ(q + 1) Λ
1kmk + T (c
1Λ
2+ c
2Λ
3) < 1, where Λ
1, Λ
2and Λ
3are given by (H
5).
Proof. We transform the problem (1) into a fixed point problem. Consider the set-valued map Ω : C([0, T ], R) → P(C([0, T ], R)) defined at the beginning of the proof of Theorem 5. It is clear that points fixed point of Ω are solutions of the problem (1).
Note that, by the assumption (H
7), since the set-valued map F (·, x) is mea- surable, it admits a measurable selection f : [0, T ] → R (see Theorem III.6 [11]).
Moreover, from assumption (H
8)
|f (t)| ≤ m(t) + m(t)|x(t)|,
i.e., f (·) ∈ L
1([0, T ], X). Therefore, the set S
F,xis nonempty. Also note that since S
F,x6= ∅, then Ω(x) 6= ∅ for any x ∈ C([0, T ], R).
Now we show that the operator Ω satisfies the assumptions of Lemma 8. To
show that Ω(x) ∈ P
cl((C[0, T ], R)) for each x ∈ C([0, T ], R), let {u
n}
n≥0∈ Ω(x)
be such that u
n→ u (n → ∞) in C([0, T ], R). Then u ∈ C([0, T ], R) and there exist v
n∈ S
F,xsuch that, for each t ∈ [0, T ],
u
n(t) = Z
t0
(t − s)
q−1Γ(q) v
n(s)ds − ξ
1λ
1Z
T 0(T − s)
q−1Γ(q) v
n(s)ds + ξ
2λ
2[λ
1T + (1 − λ
1)t]
Z
T 0(T − s)
q−2Γ(q − 1) v
n(s)ds + ξ
2µ
2[λ
1T + (1 − λ
1)t]
Z
T 0h(s, x(s))ds − µ
1ξ
1Z
T0
g(s, x(s))ds.
As F has compact values, we may pass to a subsequence (if necessary) to obtain that v
nconverge to v in L
1([0, T ], R). Thus, v ∈ S
F,xand for each t ∈ [0, T ],
u
n(t) → u(t) = Z
t0
(t − s)
q−1Γ(q) v(s)ds − ξ
1λ
1Z
T0
(T − s)
q−1Γ(q) v(s)ds + ξ
2λ
2[λ
1T + (1 − λ
1)t]
Z
T 0(T − s)
q−2Γ(q − 1) v(s)ds + ξ
2µ
2[λ
1T + (1 − λ
1)t]
Z
T 0h(s, x(s))ds − µ
1ξ
1Z
T 0g(s, x(s))ds.
Hence, u ∈ Ω(x) and Ω(x) is closed.
Next we show that Ω is a contraction on C([0, T ], R), i.e. there exists γ < 1 such that
H
d(Ω(x), Ω(¯ x)) ≤ γkx − ¯ xk for each x, ¯ x ∈ C([0, T ], R).
Let x, ¯ x ∈ C([0, T ], R) and h
1∈ Ω(x). Then there exists v
1(t) ∈ F (t, x(t)) such that, for each t ∈ [0, T ],
h
1(t) = Z
t0
(t − s)
q−1Γ(q) v
1(s)ds − ξ
1λ
1Z
T 0(T − s)
q−1Γ(q) v
1(s)ds + ξ
2λ
2[λ
1T + (1 − λ
1)t]
Z
T 0(T − s)
q−2Γ(q − 1) v
1(s)ds + ξ
2µ
2[λ
1T + (1 − λ
1)t]
Z
T0
h(s, x(s))ds − µ
1ξ
1Z
T0
g(s, x(s))ds.
By (H
8), we have
H
d(F (t, x), F (t, ¯ x)) ≤ m(t)|x(t) − ¯ x(t)|.
So, there exists w(t) ∈ F (t, ¯ x(t)) such that
|v
1(t) − w| ≤ m(t)|x(t) − ¯ x(t)|, t ∈ [0, T ].
Define U : [0, T ] → P(R) by
U (t) = {w ∈ R : |v
1(t) − w(t)| ≤ m(t)|x(t) − ¯ x(t)|}.
Since the multivalued operator U (t) ∩ F (t, ¯ x(t)) is measurable (Proposition III.4 [11]), there exists a function v
2(t) which is a measurable selection for U. So v
2(t) ∈ F (t, ¯ x(t)) and for each t ∈ [0, T ], we have |v
1(t)−v
2(t)| ≤ m(t)|x(t)− ¯ x(t)|.
For each t ∈ [0, T ], let us define h
2(t) =
Z
t 0(t − s)
q−1Γ(q) v
2(s)ds − ξ
1λ
1Z
T 0(T − s)
q−1Γ(q) v
2(s)ds + ξ
2λ
2[λ
1T + (1 − λ
1)t]
Z
T 0(T − s)
q−2Γ(q − 1) v
2(s)ds + ξ
2µ
2[λ
1T + (1 − λ
1)t]
Z
T0
h(s, ¯ x(s))ds − µ
1ξ
1Z
T0
g(s, ¯ x(s))ds, t ∈ [0, T ].
Thus,
|h
1(t) − h
2(t)| ≤ 1 Γ(q)
Z
t 0(t − s)
q−1|v
1(s) − v
2(s)|ds
+ |λ
1||ξ
1| Γ(q)
Z
T 0(T − s)
q−1|v
1(s) − v
2(s)|ds
+ |ξ
2λ
2[λ
1T + (1 − λ
1)t]|
Γ(q − 1)
Z
T0
(T − s)
q−2|v
1(s) − v
2(s)|ds
+ |ξ
2µ
2[λ
1T + (1 − λ
1)t]|
Z
T 0|h(s, x(s)) − h(s, ¯ x(s))|ds
+ |µ
1ξ
1| Z
T0
|g(s, x(s)) − g(s, ¯ x(s))|ds
≤ kx − ¯ xkT
qΓ(q + 1) [1 + |ξ
1λ
1| + q|ξ
2||λ
2|(1 + |λ
1|)] kmk + |ξ
2||µ
2|(1 + |λ
1|)T
2c
1kx − ¯ xk + |µ
1ξ
1|T c
2kx − ¯ xk.
Hence,
kh
1− h
2k ≤ n T
qΓ(q + 1) Λ
1kmk + T (c
1Λ
2+ c
2Λ
3) o
kx − ¯ xk.
Analogously, interchanging the roles of x and x, we obtain H
d(Ω(x), Ω(¯ x)) ≤ γkx − ¯ xk
≤ n T
qΓ(q + 1) Λ
1kmk + T (c
1Λ
2+ c
2Λ
3) o
kx − ¯ xk.
Since Ω is a contraction, it follows by Lemma 8 that Ω has a fixed point x which is a solution of (1). This completes the proof.
Example 10. Consider the following boundary value problem
c
D
3/2x(t) ∈ F (t, x(t)), t ∈ [0, 1], x(0) − 1
2 x(1) = 1 2
Z
1 0s
4(1 + |x|)ds, x
0(0) − 2x
0(T ) = 1
8 Z
10
s
2(1 + |x|)ds.
(4)
Here, q =
32, λ
1=
12, λ
2= 2, µ
1=
12, µ
2=
18, ξ
1= −2, ξ
2= −2 and F : [0, 1] × R → P(R) is a multivalued map given by
x → F (t, x) =
|x|
38(|x|
3+ 3) , |x|
9(|x| + 1) + 1 100
. For f ∈ F, we have
|f | ≤ max
|x|
38(|x|
3+ 3) , |x|
9(|x| + 1) + 1 100
≤ 1
8 , x ∈ R.
Thus,
kF (t, x)k
P:= sup{|y| : y ∈ F (t, x)} ≤ 1
8 = p(t)ψ(kxk), x ∈ R,
with p(t) = 1, ψ(kxk) =
18. Also ψ
1(x) = 1 + x, ψ
2(x) = 1 + x, kp
1k
L1=
15, kp
2k
L1=
13, Λ
1= 11, Λ
2=
38and Λ
3= 1.
Further, using the condition (H
5) we find that M > 2.438334. Clearly, all conditions of Theorem 5 are satisfied. So there exists at least one solution of the problem (4) on [0, 1].
Remark 11. The results for an anti-periodic boundary value problem of frac- tional differential inclusions of order q ∈ (1, 2] follow as a special case by taking λ
1= −1 = λ
2, µ
1= 0 = µ
2in the results of this article. In case of λ
1= 0 = λ
2, our results become the ones for the initial-integral boundary conditions
x(0) = µ
1Z
T0
g(s, x(s))ds, x
0(0) = µ
2Z
T0
