doi:10.7151/dmdico.1146
EXISTENCE RESULTS FOR NONLOCAL BOUNDARY VALUE PROBLEMS FOR FRACTIONAL DIFFERENTIAL
EQUATIONS AND INCLUSIONS WITH FRACTIONAL INTEGRAL BOUNDARY CONDITIONS
Sotiris K. Ntouyas Department of Mathematics
University of Ioannina 451 10 Ioannina, Greece e-mail: sntouyas@uoi.gr
Abstract
This paper studies a new class of nonlocal boundary value problems of nonlinear differential equations and inclusions of fractional order with fractional integral boundary conditions. Some new existence results are obtained by using standard fixed point theorems and Leray-Schauder degree theory. Some illustrative examples are also discussed.
Keywords: fractional differential equations, fractional differential inclu- sions, nonlocal boundary conditions, fixed point theorems, Leray-Schauder degree.
2010 Mathematics Subject Classification: 34A08, 26A33, 34A60.
1. Introduction
Fractional differential equations have aroused great interest, which is caused by both the intensive development of the theory of fractional calculus and the appli- cation of physics, mechanics, chemistry engineering. For some recent development on the topic see [1]–[18] and the references cited therein.
In this paper, we discuss the existence of solutions for a boundary value problem of nonlinear fractional differential equations and inclusions of order q ∈ (1, 2] with fractional integral boundary conditions. As a first problem, we consider the following boundary value problem of fractional differential equations
(
cD
qx(t) = f (t, x(t)), 0 < t < 1, 1 < q ≤ 2,
x(0) = 0, x(1) = αI
px(η), 0 < η < 1,
(1)
where
cD
qdenotes the Caputo fractional derivative of order q, f : [0, 1]×R → R is a given continuous function, α ∈ R is such that α 6= Γ(p + 2)/η
p+1, Γ is the Euler gamma function and I
p, 0 < p < 1 is the Riemann-Liouville fractional integral of order p.
Recently, in [25] the following boundary value problem of fractional differen- tial equations with a fractional integral condition
(
cD
q0+x(t) = f (t, x(t),
cD
0p+x(t)), 0 < t < 1, 1 < q ≤ 2, 0 < p < 1 x(0) = 0, x
0(1) = αI
0p+x(1),
(2)
was studied. Existence and uniqueness results are proved via Banach’s contrac- tion principle and Leray-Schauder Nonlinear Alternative. Also Ahmad et al. in [11] discussed existence results for nonlinear fractional differential equations with three-point integral boundary conditions
c
D
qx(t) = f (t, x(t)), 0 < t < 1, 1 < q ≤ 2, x(0) = 0, x(1) = α
Z
η 0x(s)ds, 0 < η < 1, (3)
where
cD
qdenotes the Caputo fractional derivative, and α ∈ R, α 6= 2/η
2. Motivated by the papers [25, 11] we consider in Section 3, the problem (1), for which we prove two new existence results based on Krasnoselskii’s fixed point theorem and Leray Schauder degree theory.
Afterwards we extend the results to cover the multivalued case. Thus, in Section 4, we study a boundary value problem for fractional differential inclusions with fractional integral boundary conditions given by
(
cD
qx(t) ∈ F (t, x(t)), 0 < t < 1, 1 < q ≤ 2, x(0) = 0, x(1) = αI
px(η), 0 < η < 1, (4)
where F : [0, 1] × R → P(R) is a multivalued map, P(R) is the family of all subsets of R.
We establish existence results for the problem (4), 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 selec-
tion 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.
2. Linear Problem
Let us recall some basic definitions of fractional calculus [27, 31, 32].
Definition. For at least n-times differentiable function g : [0, ∞) → R, the Caputo derivative of fractional order q is defined as
c
D
qg(t) = 1 Γ(n − q)
Z
t0
(t − s)
n−q−1g
(n)(s)ds, n − 1 < q < n, n = [q] + 1, where [q] denotes the integer part of the real number q.
Definition. The Riemann-Liouville fractional integral of order q is defined as I
qg(t) = 1
Γ(q) Z
t0
g(s)
(t − s)
1−qds, q > 0, provided the integral exists.
The following lemmas gives some properties of Riemann-Liouville fractional in- tegrals and Caputo fractional derivative [27].
Lemma 1. Let p, q ≥ 0, f ∈ L
1[a, b]. Then I
pI
qf (t) = I
p+qf (t) and
cD
qI
qf (t) = f (t), for all t ∈ [a, b].
Lemma 2. Let β > α > 0, f ∈ L
1[a, b]. Then
cD
αI
βf (t) = I
β−αf (t), for all t ∈ [a, b].
To define the solution of the boundary value problem (1) we need the following lemma, which deals with a linear variant of the problem (1).
By a solution of (1), we means a function x(t) which satisfies the equation
c
D
qx(t) = f (t, x(t)), 0 < t < 1, and the boundary conditions x(0) = 0, x(1) = αI
px(η).
Lemma 3. Let α 6=
Γ(p+2)ηp+1. Then for a given g ∈ C([0, 1], R), the solution of the fractional differential equation
c
D
qx(t) = g(t), 1 < q ≤ 2 (5)
subject to the boundary conditions
x(0) = 0, x(1) = αI
px(η) (6)
is given by
x(t) = Z
t0
(t − s)
q−1Γ(q) g(s)ds
− Γ(p + 2)t Γ(p + 2) − αη
p+1Z
10
(1 − s)
q−1Γ(q) g(s)ds
+ αΓ(p + 2)t Γ(p + 2) − αη
p+1Z
η 0(η − s)
p+q−1Γ(p + q) g(s)ds, t ∈ [0, 1].
(7)
Proof. For some constants c
0, c
1∈ R, we have [27]
x(t) = Z
t0
(t − s)
q−1Γ(q) g(s)ds − c
0− c
1t.
(8)
From x(0) = 0 we have c
0= 0. Using the Riemann-Liouville integral of order p for (8) we have
I
px(t) = Z
t0
(t − s)
p−1Γ(p)
Z
s 0(s − r)
q−1Γ(q) g(r)dr − c
1s
ds
= 1
Γ(p) 1 Γ(q)
Z
t 0Z
s 0(t − s)
p−1(s − r)
q−1g(r)drds − c
1t
p+1Γ(p + 2)
= I
pI
qg(t) − c
1t
p+1Γ(p + 2)
= I
p+qg(t) − c
1t
p+1Γ(p + 2) , where we have used Lemma 1.
Using the condition (6) in the above expression, we get
c
1= Γ(p + 2) Γ(p + 2) − αη
p+1]
"
Z
1 0(1 − s)
q−1Γ(q) g(s)ds − αI
p+qg(η)
#
= Γ(p + 2) Γ(p + 2) − αη
p+1]
"
Z
1 0(1 − s)
q−1Γ(q) g(s)ds − α Z
η0
(η − s)
p+q−1Γ(p + q) g(s)ds
# .
Substituting the values of c
0and c
1in (8), we obtain (7).
3. Existence results for single-valued case
Let C = C([0, 1], R) denotes the Banach space of all continuous functions from [0, 1] → R endowed with the norm defined by kxk = sup{|x(t)|, t ∈ [0, 1]}.
In view of Lemma 3, we define an operator F : C → C by x(t) =
Z
t 0(t − s)
q−1Γ(q) f (s, x(s))ds
− Γ(p + 2)t Γ(p + 2) − αη
p+1Z
1 0(1 − s)
q−1Γ(q) f (s, x(s))ds + αΓ(p + 2)t
Γ(p + 2) − αη
p+1Z
η0
(η − s)
p+q−1Γ(p + q) f (s, x(s))ds, t ∈ [0, 1].
(9)
Observe that the problem (1) has solutions if and only if the operator equation F x = x has fixed points.
Now, we prove the existence of solutions of (1) by applying Krasnoselskii’s fixed point theorem [28].
Theorem 4. (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 5. Let f : [0, 1] × R → R be a continuous function satisfying the following assumptions:
(A
1) |f (t, x) − f (t, y)| ≤ L|x − 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
+).
Then the boundary value problem (1) has at least one solution on [0, 1], provided that
L
1
Γ(q + 1)
Γ(p + 2)
|Γ(p + 2) − αη
p+1| + |α|η
p+qΓ(p + 2)
Γ(p + q + 1)|Γ(p + 2) − αη
p+1|
< 1.
(10)
Proof. Letting sup
t∈[0,1]|µ(t)| = kµk, we fix r ≥ kµk
1
Γ(q + 1) + 1 Γ(q + 1)
Γ(p + 2)
|Γ(p + 2) − αη
p+1| + |α|η
p+qΓ(p + 2) Γ(p + q + 1)|Γ(p + 2) − αη
p+1|
and consider B
r= {x ∈ C : kxk ≤ r}. We define the operators P and Q on B
ras
(Px)(t) = Z
t0
(t − s)
q−1Γ(q) f (s, x(s))ds, t ∈ [0, 1],
(Qx)(t) = − Γ(p + 2)t Γ(p + 2) − αη
p+1Z
1 0(1 − s)
q−1Γ(q) f (s, x(s))ds
+ αΓ(p + 2)t Γ(p + 2) − αη
p+1Z
η 0(η − s)
p+q−1Γ(p + q) f (s, x(s))ds, t ∈ [0, 1].
For x, y ∈ B
r, we find that
kPx + Qyk ≤ kµk ( Z
10
(1 − s)
q−1Γ(q) ds + Γ(p + 2)
|Γ(p + 2) − αη
p+1| Z
10
(1 − s)
q−1Γ(q) ds
+ |α|Γ(p + 2)
|Γ(p + 2) − αη
p+1| Z
η0
(η − s)
p+q−1Γ(p + q) ds
)
≤ kµk
( 1
Γ(q + 1) + 1 Γ(q + 1)
Γ(p + 2)
|Γ(p + 2) − αη
p+1|
+ 1
Γ(p + q + 1)
|α|Γ(p + 2)η
p+q|Γ(p + 2) − αη
p+1| )
≤ r.
Thus, Px + Qy ∈ B
r. It follows from the assumption (A
1) together with (10) that Q is a contraction mapping. Continuity of f implies that the operator P is continuous. Also, P is uniformly bounded on B
ras
kPxk ≤ kµk
Γ(q + 1) .
Now we prove the compactness of the operator P.
In view of (A
1), we define sup
(t,x)∈[0,1]×Br|f (t, x)| = f , and consequently we have
|(Px)(t
1) − (Px)(t
2)| =
1 Γ(q)
Z
t10
[(t
2− s)
q−1− (t
1− s)
q−1]f (s, x(s))ds
+ Z
t2t1
(t
2− s)
q−1f (s, x(s))ds
≤ f
Γ(q + 1) |2(t
2− t
1)
q+ t
q1− t
q2|,
which is independent of x. Thus, P is equicontinuous. Hence, by the Arzel´ a- Ascoli Theorem, P is compact on B
r. Thus all the assumptions of Theorem 4 are satisfied. So the conclusion of Theorem 4 implies that the boundary value problem (1) has at least one solution on [0, 1].
Our next existence result is based on Leray-Schauder degree theory.
Theorem 6. Let f : [0, 1] × R → R be a continuous function. Assume that there exist constants 0 ≤ κ <
Λ1, where
Λ = 1
Γ(q + 1) + 1 Γ(q + 1)
Γ(p + 2)
|Γ(p + 2) − αη
p+1| + |α|η
p+qΓ(p + 2)
Γ(p + q + 1)|Γ(p + 2) − αη
p+1| 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) has at least one solution.
Proof. Consider the fixed point problem x = F x, (11)
where F is defined by (9). In view of the fixed point problem (11), we just need to prove the existence of at least one solution x ∈ C[0, 1] satisfying (11). Define a suitable ball B
R⊂ C[0, 1] with radius R > 0 as
B
R= n
x ∈ C[0, 1] : max
t∈[0,1]
|x(t)| < R o ,
where R will be fixed later. Then, it is sufficient to show that F : B
R→ C[0, 1]
satisfies
x 6= λF x, ∀ x ∈ ∂B
Rand ∀ λ ∈ [0, 1].
(12)
Let us set
H(λ, x) = λF x, x ∈ C(R), λ ∈ [0, 1].
Then, by the Arzel´ a-Ascoli Theorem, h
λ(x) = x−H(λ, x) = x−λF x is completely continuous. If (12) is true, then the following Leray-Schauder degrees are well defined and by the homotopy invariance of topological degree, it follows that
deg(h
λ, B
R, 0) = deg(I − λF, B
R, 0) = deg(h
1, B
R, 0)
= deg(h
0, B
R, 0) = deg(I, B
R, 0) = 1 6= 0, 0 ∈ B
r,
where I denotes the unit operator. By the nonzero property of Leray-Schauder degree, h
1(t) = x − λF x = 0 for at least one x ∈ B
R. In order to prove (12), we assume that x = λF x, λ ∈ [0, 1]. Then for x ∈ ∂B
Rand t ∈ [0, 1] we have
|x(t)| = |λ(F x)(t)|
≤ Z
t0
(t − s)
q−1Γ(q) |f (s, x(s))|ds
+ Γ(p + 2)t
|Γ(p + 2) − αη
p+1| Z
10
(1 − s)
q−1Γ(q) |f (s, x(s))|ds
+ |α|Γ(p + 2)t
|Γ(p + 2) − αη
p+1| Z
η0
(η − s)
p+q−1Γ(p + q) |f (s, x(s))|ds
≤ (κkxk + M )
Z
1 0(1 − s)
q−1Γ(q) ds + Γ(p + 2)
|Γ(p + 2) − αη
p+1| Z
10
(1 − s)
q−1Γ(q) ds
+ |α|Γ(p + 2)
|Γ(p + 2) − αη
p+1| Z
η0
(η − s)
p+q−1Γ(p + q) ds
≤ (κkxk + M )
1
Γ(q + 1) + 1 Γ(q + 1)
Γ(p + 2)
|Γ(p + 2) − αη
p+1|
+ |α|η
p+qΓ(p + 2)
Γ(p + q + 1)|Γ(p + 2) − αη
p+1|
= (κkxk + M )Λ,
which, on taking norm (sup
t∈[0,1]|x(t)| = kxk) and solving for kxk, yields
kxk ≤ M Λ 1 − κΛ . Letting R = M Λ
1 − κΛ + 1, (12) holds. This completes the proof.
Example 7. Consider the following fractional boundary value problem
c
D
3/2x(t) = 1 (t + 2)
2|x|
1 + |x| + 1 + sin
2t, t ∈ [0, 1], x(0) = 0, x(1) = √
3I
1/2x 1 3
. (13)
Here, q = 3/2, α = √
3, p = 1/2, η = 1/3 and f (t, x) =
(t+2)1 2|x|
1+|x|
+ 1 + sin
2t.
As α = √
3 6= Γ(p + 2)/η
p+1= Γ(5/2)/(1/3)
3/2and |f (t, x) − f (t, y)| ≤
14|x − y|, therefore, (A
1) is satisfied with L =
14. Clearly, |f (t, x)| ≤ 1 = M and (A
2) is satisfied with M = 1. Since
L (
1 Γ(q + 1)
Γ(p + 2)
|Γ(p + 2) − αη
p+1| + |α|η
p+qΓ(p + 2)
Γ(p + q + 1)|Γ(p + 2) − αη
p+1| )
= 1 4
( 12
9 √
π − 4 + 3 √ 3π 2(9 √
π − 4) )
≈ 0.3474168 < 1
by the conclusion of Theorem 5, the boundary value problem (13) has a solution on [0, 1].
Example 8. Consider the following boundary value problem
c
D
3/2x(t) = 1
8π sin(2πx) + |x|
2(1 + |x|) + 1
2 , t ∈ [0, 1], 1 < q ≤ 2, x(0) = 0, x(1) = √
3I
1/2x 1 3
. (14)
Here,
f (t, x)
=
1
8π sin(2πx) + |x|
1 + |x|
≤ 1
4 |x| + 1.
Clearly M = 1 and
κ = 1 4 < 1
Λ =
"
4 3 √
π + 12 9 √
π − 4 + 3 √ 3π 2(9 √
π − 4)
#
−1= 0.4668292.
Thus, all the conditions of Theorem 6 are satisfied and consequently the problem (14) has at least one solution.
4. Existence results for multi-valued case 4.1. Preliminaries on multi-valued analysis
Let us recall some basic definitions on multi-valued maps [22], [26].
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}. A multi-valued map G : X → P(X) is convex (closed) valued if G(x) is convex (closed) for all x ∈ X.
The map G 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)}} < ∞). G 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. G is said to be completely continuous if G(B) is relatively compact for every B ∈ P
b(X). 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
∗). 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 Fix G. 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{|y − z| : z ∈ G(t)}
is measurable.
Let C([0, 1]) denote a Banach space of continuous functions from [0, 1] into R with the norm kxk = sup
t∈[0,1]|x(t)|. Let L
1([0, 1], R) be the Banach space of measurable functions x : [0, 1] → R which are Lebesgue integrable and normed by kxk
L1= R
10
|x(t)|dt.
Definition. A multivalued map F : [0, 1] × R → P(R) is said to be Carath´eodory if
(i) t 7−→ F (t, x) is measurable for each x ∈ R;
(ii) x 7−→ F (t, x) is upper semicontinuous for almost all t ∈ [0, 1];
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, 1], 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, 1].
For each y ∈ C([0, 1], R), define the set of selections of F by
S
F,y:= v ∈ L
1([0, 1], R) : v(t) ∈ F (t, y(t)) for a.e. t ∈ [0, 1] .
Let X be a nonempty closed subset of a Banach space E and 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. Let A be a subset of [0, 1] × 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, 1]
and D is Borel measurable in R. A subset A of L
1([0, 1], R) is decomposable if for all u, v ∈ A and measurable J ⊂ [0, 1] = J , the function uχ
J+ vχ
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, 1], 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, 1] × R → P(R) be a multivalued map with nonempty compact values.
Define a multivalued operator F : C([0, 1]×R) → P(L
1([0, 1], R)) associated with F as
F (x) = w ∈ L
1([0, 1], R) : w(t) ∈ F (t, x(t)) for a.e. t ∈ [0, 1] , which is called the Nemytskii operator associated with F.
Definition. Let F : [0, 1] × 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 l ower semi-continuous and has nonempty closed and decomposable values.
Let (X, d) be a metric space induced from the normed space (X; k.k). Consider H
d: P(X) × P(X) → R ∪ {∞} given by
H
d(A, B) = max n sup
a∈A
d(a, B), sup
b∈B
d(A, b) o
,
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 [29]).
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.
The following lemmas will be used in the sequel.
Lemma 9 (Nonlinear alternative for Kakutani maps) [24]. 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 a 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).
Lemma 10 ([30]). Let X be a Banach space. Let F : [0, T ] × R → P
cp,c(X) be an L
1− Carath´ eodory multivalued map and let Θ be a linear continuous mapping from L
1([0, 1], X) to C([0, 1], X). Then the operator
Θ ◦ S
F: C([0, 1], X) → P
cp,c(C([0, 1], X)), x 7→ (Θ ◦ S
F)(x) = Θ(S
F,x) is a closed graph operator in C([0, 1], X) × C([0, 1], X).
Lemma 11 ([23]). Let Y be a separable metric space and let N : Y → P(L
1([0, 1], 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, 1], R) such that g(x) ∈ N (x) for every x ∈ Y .
Lemma 12 ([21]). Let (X, d) be a complete metric space. If N : X → P
cl(X) is a contraction, then F ixN 6= ∅.
Definition. A function x ∈ C
2([0, 1], R) is a solution of the problem (4) if x(0) =
0, x(1) = αI
px(η), 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
x(t) = Z
t0
(t − s)
q−1Γ(q) f (s)ds
− Γ(p + 2)t Γ(p + 2) − αη
p+1Z
1 0(1 − s)
q−1Γ(q) f (s)ds + αΓ(p + 2)t
Γ(p + 2) − αη
p+1Z
η0
(η − s)
p+q−1Γ(p + q) f (s)ds.
(15)
4.2. The Carath´ eodory case Theorem 13. Assume that:
(H
1) F : [0, 1] × R → P(R) is Carath´eodory and has nonempty compact and convex values;
(H
2) there exists a continuous nondecreasing function ψ : [0, ∞) → (0, ∞) and a function φ ∈ L
1([0, 1], R
+) such that
kF (t, x)k
P:= sup {|y| : y ∈ F (t, x)} ≤ φ(t)ψ(kxk) for each (t, x) ∈ [0, 1] × R;
(H
3) there exists a constant M > 0 such that M ψ(M )
"
I
qφ(1) + Γ(p + 2)
|Γ(p + 2) − αη
p+1| I
qφ(1) + |α|Γ(p + 2)
|Γ(p + 2) − αη
p+1| I
p+qφ(η)
# > 1.
Then the boundary value problem (4) has at least one solution on [0, 1].
Proof. Define the operator Ω
F: C([0, 1], R) → P(C([0, 1], R)) by
Ω
F(x) =
h ∈ C([0, 1], R) :
h(t) =
Z
t0
(t − s)
q−1Γ(q) f (s)ds
− Γ(p + 2)t Γ(p + 2) − αη
p+1Z
10
(1 − s)
q−1Γ(q) f (s)ds + αΓ(p + 2)t
Γ(p + 2) − αη
p+1Z
η0
(η − s)
p+q−1Γ(p + q) f (s)ds,
for f ∈ S
F,x. We will show that Ω
Fsatisfies the assumptions of the nonlinear alternative of Leray-Schauder type. The proof consists of several steps. As a first step, we show that Ω
Fis convex for each x ∈ C([0, 1], R). This step is obvious since S
F,xis convex (F has convex values), and therefore we omit the proof.
In the second step, we show that Ω
Fmaps bounded sets (balls) into bounded sets in C([0, 1], R). For a positive number ρ, let B
ρ= {x ∈ C([0, 1], R) : kxk ≤ 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,xsuch that
x(t) = Z
t0
(t − s)
q−1Γ(q) f (s)ds − Γ(p + 2)t Γ(p + 2) − αη
p+1Z
1 0(1 − s)
q−1Γ(q) f (s)ds
+ αΓ(p + 2)t Γ(p + 2) − αη
p+1Z
η 0(η − s)
p+q−1Γ(p + q) f (s)ds.
Then for t ∈ [0, 1] we have
|h(t)| ≤ Z
t0
(t − s)
q−1Γ(q) |f (s)|ds + Γ(p + 2)
|Γ(p + 2) − αη
p+1| Z
10
(1 − s)
q−1Γ(q) |f (s)|ds + |α|Γ(p + 2)
|Γ(p + 2) − αη
p+1| Z
η0
(η − s)
p+q−1Γ(p + q) |f (s)|ds
≤ ψ(kxk)
"
Z
1 0(1 − s)
q−1Γ(q) φ(s)ds
+ Γ(p + 2)
|Γ(p + 2) − αη
p+1| Z
10
(1 − s)
q−1Γ(q) φ(s)ds
+ |α|Γ(p + 2)
|Γ(p + 2) − αη
p+1| Z
η0
(η − s)
p+q−1Γ(p + q) φ(s)ds
# . Thus,
khk ≤ ψ(ρ)
"
Z
1 0(1 − s)
q−1Γ(q) φ(s)ds + Γ(p + 2)
|Γ(p + 2) − αη
p+1| Z
10
(1 − s)
q−1Γ(q) φ(s)ds
+ |α|Γ(p + 2)
|Γ(p + 2) − αη
p+1| Z
η0
(η − s)
p+q−1Γ(p + q) φ(s)ds
#
.
Now we show that Ω
Fmaps bounded sets into equicontinuous sets of C([0, 1], R).
Let t
0, t
00∈ [0, 1] with t
0< t
00and x ∈ B
ρ. For each h ∈ Ω
F(x), we obtain
|h(t
00) − h(t
0)|
≤
ψ(ρ) Z
t00
(t
00− s)
q−1− (t
0− s)
q−1Γ(q)
φ(s)ds + ψ(ρ) Z
t00t0
(t
00− s)
q−1Γ(q) φ(s)ds
+ ψ(ρ) Γ(p + 2)|t
00− t
0|
|Γ(p + 2) − αη
p+1| Z
10
(1 − s)
q−1Γ(q) φ(s)ds
+ ψ(ρ) |α|Γ(p + 2)
|Γ(p + 2) − αη
p+1| Z
η0
(η − s)
p+q−1Γ(p + q) φ(s)ds.
Obviously the right hand side of the above inequality tends to zero independently of x ∈ B
ρas t
00− t
0→ 0. As Ω
Fsatisfies the above three assumptions, therefore it follows by the Ascoli-Arzel´ a theorem that Ω
F: C([0, 1], R) → P(C([0, 1], R)) is completely continuous.
In our next step, we show that Ω
Fhas a closed graph. Let x
n→ x
∗, h
n∈ Ω
F(x
n) and h
n→ h
∗. Then we need to show that h
∗∈ Ω
F(x
∗). Associated with h
n∈ Ω
F(x
n), there exists f
n∈ S
F,xnsuch that for each t ∈ [0, 1],
h
n(t) = Z
t0
(t − s)
q−1Γ(q) f
n(s)ds − Γ(p + 2)t Γ(p + 2) − αη
p+1Z
1 0(1 − s)
q−1Γ(q) f
n(s)ds
+ αΓ(p + 2)t Γ(p + 2) − αη
p+1Z
η 0(η − s)
p+q−1Γ(p + q) f
n(s)ds.
Thus it suffices to show that there exists f
∗∈ S
F,x∗such that for each t ∈ [0, 1],
h
∗(t) = Z
t0
(t − s)
q−1Γ(q) f
∗(s)ds − Γ(p + 2)t Γ(p + 2) − αη
p+1Z
1 0(1 − s)
q−1Γ(q) f
∗(s)ds
+ αΓ(p + 2)t Γ(p + 2) − αη
p+1Z
η 0(η − s)
p+q−1Γ(p + q) f
∗(s)ds.
Let us consider the linear operator Θ : L
1([0, 1], R) → C([0, 1], R) given by
f 7→ Θ(f )(t) = Z
t0
(t − s)
q−1Γ(q) f (s)ds − Γ(p + 2)t Γ(p + 2) − αη
p+1Z
10
(1 − s)
q−1Γ(q) f (s)ds
+ αΓ(p + 2)t Γ(p + 2) − αη
p+1Z
η0
(η − s)
p+q−1Γ(p + q) f (s)ds.
Observe that
kh
n(t) − h
∗(t)k
=
Z
t 0(t − s)
q−1Γ(q) (f
n(s) − f
∗(s))ds
− Γ(p + 2)t Γ(p + 2) − αη
p+1Z
10
(1 − s)
q−1Γ(q) (f
n(s) − f
∗(s))ds + αΓ(p + 2)t
Γ(p + 2) − αη
p+1Z
η0
(η − s)
p+q−1Γ(p + q) (f
n(r) − f
∗(r))drds
→ 0, as n → ∞.
Thus, it follows by Lemma 10 that Θ◦S
Fis a closed graph operator. Further, we have h
n(t) ∈ Θ(S
F,xn). Since x
n→ x
∗, therefore, we have
h
∗(t) = Z
t0
(t − s)
q−1Γ(q) f
∗(s)ds − Γ(p + 2)t Γ(p + 2) − αη
p+1Z
1 0(1 − s)
q−1Γ(q) f
∗(s)ds
+ αΓ(p + 2)t Γ(p + 2) − αη
p+1Z
η 0(η − s)
p+q−1Γ(p + q) f
∗(s)ds, for some f
∗∈ S
F,x∗.
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 f ∈ S
F,xsuch that, for t ∈ [0, 1], we have
x(t) = λ Z
t0
(t − s)
q−1Γ(q) f (s)ds − λ Γ(p + 2)t Γ(p + 2) − αη
p+1Z
1 0(1 − s)
q−1Γ(q) f (s)ds
+ λ αΓ(p + 2)t Γ(p + 2) − αη
p+1Z
η 0(η − s)
p+q−1Γ(p + q) f (s)ds,
and using the computations of the second step above we have
|x(t)| ≤ ψ(kxk)
"
Z
1 0(1 − s)
q−1Γ(q) φ(s)ds
+ Γ(p + 2)
|Γ(p + 2) − αη
p+1| Z
10
(1 − s)
q−1Γ(q) φ(s)ds
+ |α|Γ(p + 2)
|Γ(p + 2) − αη
p+1| Z
η0
(η − s)
p+q−1Γ(p + q) φ(s)ds
# .
Consequently, we have
kxk ψ(kxk)
"
I
qφ(1) + Γ(p + 2)
|Γ(p + 2) − αη
p+1| I
qφ(1) + |α|Γ(p + 2)
|Γ(p + 2) − αη
p+1| I
p+qφ(η)
# ≤ 1.
In view of (H
3), there exists M such that kxk 6= M . Let us set U = {x ∈ C([0, 1], R) : kxk < M }.
Note that the operator Ω
F: U → P(C([0, 1], R)) 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 Leray-Schauder type (Lemma 9), we deduce that Ω
Fhas a fixed point x ∈ U which is a solution of the problem (1). This completes the proof.
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 [19]
for lower semi-continuous maps with decomposable values.
Theorem 14. Assume that (H
2), (H
3) and the following condition holds:
(H
4) F : [0, 1] × 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, 1];
Then the boundary value problem (4) 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 11, there exists a continuous function f : C([0, 1], R) → L
1([0, 1], R) such that f (x) ∈ F (x) for all x ∈ C([0, 1], R).
Consider the problem (
cD
qx(t) = f (x(t)), t ∈ [0, 1],
x(0) = 0 x(1) = αI
px(η), 0 < η < 1.
(16)
Observe that if x ∈ C
2([0, 1], R) is a solution of (16), then x is a solution to the problem (4). In order to transform the problem (16) into a fixed point problem, we define the operator Ω
Fas
Ω
Fx(t) = Z
t0
(t − s)
q−1Γ(q) f (x(s))ds − Γ(p + 2)t Γ(p + 2) − αη
p+1Z
1 0(1 − s)
q−1Γ(q) f (x(s))ds + αΓ(p + 2)t
Γ(p + 2) − αη
p+1Z
η0
(η − s)
p+q−1Γ(p + q) f (x(s))ds.
It can easily be shown that Ω
Fis continuous and completely continuous. The remaining part of the proof is similar to that of Theorem 13. So we omit it. This completes the proof.
4.4. The Lipschitz case
Now we prove the existence of solutions for the problem (4) with a nonconvex valued right hand side by applying a fixed point theorem for multivalued map due to Covitz and Nadler [21].
Theorem 15. 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) H
d(F (t, x), F (t, ¯ x)) ≤ m(t)|x − ¯ x| for almost all t ∈ [0, 1] and x, ¯ x ∈ R 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 (4) has at least one solution on [0, 1] if I
qm(1) + Γ(p + 2)
|Γ(p + 2) − αη
p+1| I
qm(1) + |α|Γ(p + 2)
|Γ(p + 2) − αη
p+1| I
p+qm(η) < 1.
Proof. Observe that the set S
F,xis nonempty for each x ∈ C([0, 1], R) by the
assumption (H
5), so F has a measurable selection (see Theorem III.6 [20]).
Now we show that the operator Ω
F, defined in the beginning of proof of Theorem 13, satisfies the assumptions of Lemma 12. To show that Ω
F(x) ∈ P
cl((C[0, 1], R)) for each x ∈ C([0, 1], R), let {u
n}
n≥0∈ Ω
F(x) be such that u
n→ u (n → ∞) in C([0, 1], R). Then u ∈ C([0, 1], R) and there exists v
n∈ S
F,xnsuch that, for each t ∈ [0, 1],
u
n(t) = Z
t0
(t − s)
q−1Γ(q) v
n(s)ds − Γ(p + 2)t Γ(p + 2) − αη
p+1Z
1 0(1 − s)
q−1Γ(q) v
n(s)ds + αΓ(p + 2)t
Γ(p + 2) − αη
p+1Z
η0
(η − s)
p+q−1Γ(p + q) v
n(s)ds.
As F has compact values, we pass onto a subsequence to obtain that v
nconverges to v in L
1([0, 1], R). Thus, v ∈ S
F,xand for each t ∈ [0, 1],
u
n(t) → u(t) = Z
t0
(t − s)
q−1Γ(q) v(s)ds − Γ(p + 2)t Γ(p + 2) − αη
p+1Z
1 0(1 − s)
q−1Γ(q) v(s)ds + αΓ(p + 2)t
Γ(p + 2) − αη
p+1Z
η0
(η − s)
p+q−1Γ(p + q) v(s)ds.
Hence, u ∈ Ω(x).
Next we show that there exists γ < 1 such that
H
d(Ω
F(x), Ω
F(¯ x)) ≤ γkx − ¯ xk for each x, ¯ x ∈ C([0, 1], R).
Let x, ¯ x ∈ C([0, 1], R) and h
1∈ Ω(x). Then there exists v
1(t) ∈ F (t, x(t)) such that, for each t ∈ [0, 1],
h
1(t) = Z
t0
(t − s)
q−1Γ(q) v
1(s)ds − Γ(p + 2)t Γ(p + 2) − αη
p+1Z
10
(1 − s)
q−1Γ(q) v
1(s)ds + αΓ(p + 2)t
Γ(p + 2) − αη
p+1Z
η0
(η − s)
p+q−1Γ(p + q) v
1(s)ds.
By (H
6), we have
H
d(F (t, x), F (t, ¯ x)) ≤ m(t)|x(t) − ¯ x(t)|.
So, there exists w ∈ F (t, ¯ x(t)) such that
|v
1(t) − w| ≤ m(t)|x(t) − ¯ x(t)|, t ∈ [0, 1].
Define U : [0, 1] → P(R) by
U (t) = {w ∈ R : |v
1(t) − w| ≤ m(t)|x(t) − ¯ x(t)|}.
Since the multivalued operator U (t) ∩ F (t, ¯ x(t)) is measurable (Proposition III.4 [20]), 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, 1], we have |v
1(t)−v
2(t)| ≤ m(t)|x(t)− ¯ x(t)|.
For each t ∈ [0, 1], let us define h
2(t) =
Z
t 0(t − s)
q−1Γ(q) v
2(s)ds − Γ(p + 2)t Γ(p + 2) − αη
p+1Z
1 0(1 − s)
q−1Γ(q) v
2(s)ds + αΓ(p + 2)t
Γ(p + 2) − αη
p+1Z
η0
(η − s)
p+q−1Γ(p + q) v
2(s)ds.
Thus,
|h
1(t) − h
2(t)| ≤ Z
t0
(t − s)
q−1Γ(q) |v
1(s) − v
2(s)|ds + Γ(p + 2)t
Γ(p + 2) − αη
p+1Z
10
(1 − s)
q−1Γ(q) |v
1(s) − v
2(s)|ds + |α|Γ(p + 2)t
Γ(p + 2) − αη
p+1Z
η0
