Available online at www.tjnsa.com
J. Nonlinear Sci. Appl. 5 (2012), 64–73
Research Article
Solvability of a two-point fractional boundary value
problem
Assia Guezane-Lakouda,∗, Rabah Khaldib
a
b
Laboratory of Advanced Materials. Faculty of Sciences, Badji Mokhtar-Annaba University, P. O. Box 12, 23000 Annaba, Algeria.
Laboratory LASEA. Faculty of Sciences, Badji Mokhtar-Annaba University, P. O. Box 12, 23000 Annaba, Algeria.
This paper is dedicated to Professor Ljubomir Ćirić
Communicated by Professor V. Berinde
Abstract
The aim of this paper is the study of the existence and uniqueness of solutions for a two-point fractional
boundary value problem, by means of Banach contraction principle and Leray Schauder nonlinear alternative.
c
Some examples are given.2012
NGA. All rights reserved.
Keywords: Fractional Caputo derivative, Banach Contraction principle, Leray Schauder nonlinear
alternative.
2010 MSC: Primary 34B10; Secondary 26A33, 34B15.
1. Introduction and Preliminaries
Motived by the big numbers of applications of fractional differential equations in various fields of science
and engineering such as control, porous media, electrochemistry, viscoelasticity, electromagnetic, and other
fields, see [1,5,10,11], many fractional boundary value problems with the help of technics of nonlinear analysis,
have been investigated in the last few decades.
This paper deals with the existence and uniqueness of solutions for the boundary value problem (P1):
c
D0q+ u (t) = f (t, u(t),c D0σ+ u(t)),
0<t<1
u (0) = u(1) = u00 (1) = u00 (0) = 0,
∗
Corresponding author
Email addresses: a_guezane@yahoo.fr (Assia Guezane-Lakoud ), rkhadi@yahoo.fr (Rabah Khaldi)
Received 2011-3-15
(1.1)
(1.2)
A. Guezane-Lakoud, R. Khaldi, J. Nonlinear Sci. Appl. 5 (2012), 64–73
65
where f : [0, 1] × R × R → R is a given function, 3 < q < 4, 1 < σ < 2 and c D0q+ denotes the Caputo’s
fractional derivative. It is Shown that the Caputo’s fractional derivative, allows the utilization of physically
interpretation of boundary conditions. For more details on the geometric and physical interpretation for
Caputo fractional derivatives types, see [11]. Our results allow the function f to depend on the fractional
derivative c D0σ+ u(t) which leads to some difficulties. By using Banach contraction principle and Leray
Schauder nonlinear alternative, the existence and uniqueness of solution for the fractional boundary value
problem (P1) are investigated. No contributions exist, as far as we know, concerning the existence of
solutions of the fractional differential equation (1.1) jointly with the nonlocal condition (1.2).
Most of papers dealing with similar problems are devoted to the solvability of nonlinear fractional
differential equation or to the existence and multiplicity of positive solutions [2,6,8,9,12,13-16]. Moreover,
in [9], Sihua Liang and Jihui Zhang considered the following nonlinear fractional boundary value problem
D0q+ u (t) + f (t, u(t)) = 0,
0
0<t<1
00
u (0) = u (0) = u (0) = u00 (1) = 0
where 3 < q ≤ 4 and D0q+ is the Riemann-Liouville fractional derivative. Using lower and upper solution
method and fixed-point theorems, some results on the existence of positive solutions are obtained.
With the help of Leray–Schauder nonlinear alternative, a fixed-point theorem on cones and a mixed
monotone method, Xu, Jiang and Yuan [14] have established the positivity of multiple solutions to the
following fractional differential equation,
D0q+ u (t) + f (t, u(t)) = 0,
0<t<1
0
u (0) = u (1) = u (0) = u0 (1) = 0
where 3 < q ≤ 4 and D0q+ is the Riemann-Liouville fractional derivative.
In [2], Bai established the existence of triple positive solutions for the nonlinear fractional differential
equation boundary value problem:
D0q+ u (t) + a(t)f (t, u(t), u0 (t)) = 0,
0
0<t<1
00
u (0) = u (0) = u (0) = u00 (1) = 0,
where 3 < q ≤ 4 and D0q+ is the Riemann-Liouville fractional derivative. This paper is organized as follows,
in the next section we cite some Definitions and Lemmas needed in our proofs. Section 3 treats the existence
and uniqueness of solution by using Banach contraction principle and Leray-Schauder nonlinear alternative.
Some examples are given illustrating the previous results.
2. Preliminaries and Lemmas
In this section, we introduce definitions and preliminary facts that are used throughout this paper.
Definition 2.1. If g ∈ C([a, b]) and α > 0, then the Riemann-Liouville fractional integral is defined by
R t g(s)
1
Iaα+ g (t) = Γ(α)
a (t−s)1−α ds.
Definition 2.2. Let α = 0, n = [α] + 1. If f ∈ AC n [a, b] then the Caputo fractional derivative of order α
R t g(n) (s)
1
of f defined by c Daα+ g (t) = Γ(n−α)
a (t−s)α−n+1 ds, exist almost everywhere on [a, b] ( [α] is the entire part
of α).
A. Guezane-Lakoud, R. Khaldi, J. Nonlinear Sci. Appl. 5 (2012), 64–73
Lemma 2.3. [7] Let α, β > 0 and n = [α] + 1, then the following relations hold: c D0α+ tβ−1 =
66
Γ(β) β−α−1
,
Γ(β−α) t
β > n and c D0α+ tk = 0, k = 0, 1, 2, ..., n − 1.
Lemma 2.4. [7] For α > 0, g(t) ∈ C(0, 1), the homogenous fractional differential equation c Daα+ g (t) = 0
has a solution g(t) = c1 + c2 t + c3 t2 + ... + cn tn−1 , where, ci ∈ R, i = 0, ..., n, and n = [α] + 1, (α non-integer)
1
Denote by L
R 1 ([0, 1] , R) the Banach space of Lebesgue integrable functions from [0, 1] into R with the
norm ||y||L1 = 0 |y (t)| dt.
The following Lemmas gives some properties of Riemann-Liouville fractional integrals and Caputo fractional derivative.
q p
q
c q
Lemma 2.5. [1] Let p, q ≥ 0, f ∈ L1 [a, b]. Then I0p+ I0q+ f (t) = I0p+q
+ f (t) = I0+ I0+ f (t) and D0+ I0+ f (t) =
f (t), for all t ∈ [a, b].
Lemma 2.6. [7] Let β > α > 0. Then the formula c D0α+ I0β+ f (t) = I0β−α
+ f (t), holds almost everywhere on
t ∈ [a, b], for f ∈ L1 [a, b] and it is valid at any point x ∈ [a, b] if f ∈ C[a, b].
Now we start by solving an auxiliary problem.
Lemma 2.7. Let 3 < q < 4, 1 < σ < 2 and y ∈ C[0, 1]. The unique solution of the fractional boundary
value problem
c D q u (t) = y(t), 0 < t < 1
0+
(2.1)
u (0) = u(1) = u00 (1) = u00 (0) = 0,
is given by
u(t) =
where
(
G(t, s) =
1
Γ(q)
Z
1
G(t, s)y(s)ds,
(2.2)
0
(t − s)q−1 − t (1 − s)q−1 + q(q−1)
− t2 (1 − s)q−3 , s ≤ t
6 t 1
q−3
2
−t (1 − s)q−1 + q(q−1)
, t ≤ s.
6 t 1 − t (1 − s)
(2.3)
Proof. In view of Lemmas 2.4 and 2.5 the equation (2.1) is equivalent to the integral equation
u(t) = I0q+ y(t) + c1 + c2 t + c3 t2 + c4 t3
(2.4)
for some ci ∈ R. The boundary condition u (0) = 0 gives c1 = 0 and from the boundary condition u (1) = 0
we conclude I0q+ y(1) + c2 + c3 + c4 = 0. Differentiating twice both sides of (2.4) and using Lemma 2.6 it
00
00
yields u00 (t) = I0q−2
+ y(t) + 2c3 + 6c4 t. The boundary condition u (0) = 0 implies c3 = 0 and from u (1) = 0
we deduce that c4 =
(2.4), we get
−I q−2
y(1)
0+
,
6
hence
u(t) =
I q−2
y(1)
0+
6
I0q+ y(t)
+t
− I0q+ y(1) = c2 . Substituting ci , i = 1, ...4, by their values in
!
I0q−2
t3 I0q−2
+ y(1)
+ y(1)
q
− I0+ y(1) −
6
6
that can be written as
1
Γ(q)
Z t
(t − s)q−1 − t (1 − s)q−1 +
0
!
q (q − 1) t 1 − t2 (1 − s)q−3
y(s)ds
6
Z 1
1
q (q − 1)
q−1
q−3
2
+
−t (1 − s)
+
t 1 − t (1 − s)
y(s)ds
Γ(q) t
6
R1
1
that is equivalent to u(t) = Γ(q)
0 G(t, s)y(s)ds where G is defined by (2.3). The proof is complete.
u(t) =
(2.5)
A. Guezane-Lakoud, R. Khaldi, J. Nonlinear Sci. Appl. 5 (2012), 64–73
67
3. Existence and Uniqueness results
In this section we prove the existence and uniqueness of solution in the Banach
space E defined by
all functions u ∈ C 2 ([0, 1] , R) , equipped with the norm ||u|| = max |u (t)| + max c D0σ+ u (t). We know
t∈[0,1]
t∈[0,1]
that for 1 < σ < 2 we have c D0σ+ u ∈ C [0, 1] see [7]. Throughout this section, we suppose that f ∈
C ([0, 1] × R ×
operator T : E → E by
R 1R, R) . Define thec integral
1
σ u(s))ds.
T u(t) = Γ(q)
G(t,
s)f
(s,
u(s),
D
0
0+
Lemma 3.1. The function u ∈ E is solution of the fractional boundary value problem (P1) if and only if
T u(t) = u(t), ∀t ∈ [0, 1] .
Proof. Suppose
of (P1) and let
R 1 that u is solution
1
c
σ
v(t) = Γ(q) 0 G(t, s)f (s, u(s), D0+ u(s))ds. In view of (2.5) we have
v(t) = I0q+ f (t, u(t),c D0σ+ u(t)) −
t 1 − t2 q−2
I0+ f (1, u(1),c D0σ+ u(1)).
+
6
with the help of Lemmas 2.6 and 2.3, we obtain
c q
D0+ v(t) = c D0q+ I0q+ f (t, u(t),c D0σ+ u(t)) − c D0q+ t I0q+ f (1, u(1),c D0σ+ u(1))
c D q t 1 − t2
c σ
0+
I0q−2
+
+ f (1, u(1), D0+ u(1))
6
= f (t, u(t),c D0σ+ u(t)).
tI0q+ f (1, u(1),c D0σ+ u(1))
It is clear that v satisfies conditions (1.2), then it is a solution for the problem (P1). The proof is complete.
Theorem 3.2. Assume that there exists nonnegative functions g, h ∈ L1 ([0, 1] , R+ ) such that for all
x, y, x, y ∈ R and t ∈ [0, 1] , we have
|f (t, x, x) − f (t, y, y)| ≤ g(t) |x − y| + h(t) |x − y| ,
and
1
Cg + Ch < ,
2
1
Ag + Ah < ,
2
where
Cg =
Ag
L1
+ I0q+ g(1) + I0q−2
+ g(1) ,
= I0q−σ−1
g
+
L1
Ch =
Ah
q−1 I0+ g q−1 I0+ h
L1
= I0q−σ−1
h
+
+
I0q−2
+ g (1)
,
Γ(4 − σ)
+ I0q+ h(1) + I0q−2
+ h(1) ,
L1
+
I0q−2
+ h (1)
.
Γ(4 − σ)
Then the fractional boundary value problem(P1) has a unique solution u in E.
To prove Theorem 3.2, we use the following properties of Riemann-Liouville fractional integrals.
q I + f 1 .
Lemma 3.3. Let q > 0, f ∈ L1 ([a, b], R+ ). Then, for all t ∈ [a, b] we have I0q+1
+ f (t) ≤
0
L
(3.1)
(3.2)
A. Guezane-Lakoud, R. Khaldi, J. Nonlinear Sci. Appl. 5 (2012), 64–73
68
Proof. Let f ∈ L1 ([a, b], R+ ) , then
Z tZ r
Z 1
q f (s)
1
q
I + f 1 =
dsdr
I0+ f (r)dr ≥
0
L
Γ(q) 0 0 (r − s)1−q
0
Z t Z t
1
f (s)
=
dr ds = I0q+1
+ f (t).
1−q
Γ(q) 0
s (r − s)
Now we prove Theorem 3.2.
Proof. We shall prove that T is a contraction. In fact, for any u, v ∈ E, we have
T u(t) − T v(t)
Z 1
1
G(t, s) (f (s, u (s) ,c D0σ+ u(s)) − f (s, v (s) ,c D0σ+ v(s))) ds
=
Γ(q) 0
= I0q+ (f (t, u (t) ,c D0σ+ u(t)) − f (s, v (t) ,c D0σ+ v(t)))
−t I0q+ (f (1, u (1) ,c D0σ+ u(1)) − f (1, v (1) ,c D0σ+ v(1)))
#
t 1 − t2 q−2
+
I0+ (f (1, u (1) ,c D0σ+ u(1)) − f (1, v (1) ,c D0σ+ v(1))) .
6
Inequality (3.1) implies that
!
t 1 − t2 q−2
+
+
|T u(t) − T v(t)| ≤ max |u(t) − v(t)|
I0+ g(1)
6
!
t 1 − t2 q−2
q
q
c σ
c
σ
+ max | D0+ u(t) − D0+ v(t)| I0+ h(t) + tI0+ h(1) +
I0+ h(1) .
6
I0q+ g(t)
tI0q+ g(1)
From Lemma 3.3 we deduce
√
"
g
|T u(t) − T v(t)| ≤ ku − vk I0q−1
+
L1
+
I0q+ g(1)
+
3 q−2
I + g(1)
27 0
√
+ I0q−1
+ h
L1
#
3
+ I0q+ h(1) +
≤ ku − vk (Cg + Ch ) ,
I q−2
+ h(1)
27 0
in view of (3.2) it yields
|T u − T v| <
ku − vk
.
2
(3.3)
On the other hand we have
c
D0σ+ T u −c D0σ+ T v = I0q−σ
(f (t, u (t) ,c D0σ+ u(t)) − f (s, v (t) ,c D0σ+ v(t)))
+
t3−σ
q−2
c σ
c σ
I + (f (1, u (1) , D0+ u(1)) − f (1, v (1) , D0+ v(1))) .
−
Γ(4 − σ) 0
Consequently (3.4) becomes
|c D0σ+ T u −c D0σ+ T v|
"
≤ ku − vk I0q−σ−1
g
+
L1
I0q−2
I0q−2
+ g (1)
+ h(1)
q−σ−1 +
+ I0+
h 1 +
Γ(4 − σ)
Γ(4 − σ)
L
#
(3.4)
A. Guezane-Lakoud, R. Khaldi, J. Nonlinear Sci. Appl. 5 (2012), 64–73
69
Then
|c D0σ+ T u −c D0σ+ T v| ≤ ku − vk (Ag + Ah ) .
(3.5)
With the help of hypothesis (3.2) it yields
|c D0σ+ T u −c D0σ+ T v| <
ku − vk
.
2
(3.6)
Summing (3.3) and (3.6) we deduce kT u − T vk < ku − vk , from here, the Banach contraction mapping
principle ensures the uniqueness of solution for the fractional boundary value problem (P1). This finishes
the proof.
Now we give an existence result for the fractional boundary value problem (P1).
Theorem 3.4. Assume
that f (t, 0, 0) 6= 0 and there exists nonnegative functions k, h, g ∈ L1 ([0, 1] , R+ ) ,
φ, ψ ∈ C R+ , R∗+ nondecreasing on R+ and r > 0, such that
|f (t, x, x)| ≤ k (t) ψ (|x|) + h(t)φ (|x|) + g (t) ,
C1
(ψ (r) + φ (r) + 1)
+ C2 < r,
Γ(q)
(3.7)
(3.8)
where C1 = max {Ck , Ch , Cg }, C2 = max {Ak , Ah , Ag } , Ch and Cg are defined as in Theorem 3.2 and
q−2
q
k
Ck = I0q−1
1 + I0+ k(1) + I0+ k(1),
+
L
Ak
= I0q−σ−1
k
+
L1
+
I0q−2
+ k (1)
.
Γ(4 − σ)
Then the fractional boundary value problem (P1) has at least one nontrivial solution u∗ ∈ E.
To prove this Theorem, we apply Leray-Schauder nonlinear alternative:
Lemma 3.5. [4]. Let F be a Banach space and Ω a bounded open subset of F , 0 ∈ Ω. T : Ω → F be a
completely continuous operator. Then, either there exists x ∈ ∂Ω, λ > 1 such that T (x) = λx, or there
exists a fixed point x∗ ∈ Ω.
Proof. First let us prove that T is completely continuous. It is clear that T is continuous since f and G are
continuous. Let Br = {u ∈ E, kuk ≤ r} be a bounded subset in E. We shall prove that T (Br ) is relatively
compact.
i) We will show that T maps bounded sets into bounded sets in E. In fact, it suffices to show that for
any u ∈ Br there exists a positive constant C such that kT uk ≤ C. Using (3.7) we have
1
|T u(t)| ≤
Γ(q)
Z
1
|G(t, s)| [k (s) ψ (|u (s)|) + h(s)φ (|c D0σ+ u(s|) + g (s)] ds.
0
Since ψ and φ are nondecreasing then (3.9) implies
1
|T u(t)| ≤
Γ(q)
Z
Z
1
≤
1
Γ(q)
1
|G(t, s)| [k (s) ψ (kuk) + h(s)φ (kuk) + g (s)] ds
0
|G(t, s)| [k (s) ψ (r) + h(s)φ (r) + g (s)] ds,
0
(3.9)
A. Guezane-Lakoud, R. Khaldi, J. Nonlinear Sci. Appl. 5 (2012), 64–73
70
using similar technics as to get (3.3) it yields
1 h
q
q−2
|T u(t)| ≤
+
I
k(1)
+
I
k(1)
ψ (r) I0q−1
k
1
+
0+
0+
Γ(q)
L
q
q−2
+
I
h(1)
+
I
h(1)
+φ (r) I0q−1
h
+
0+
0+
L1
i
q
q−2
+
I
g(1)
+
I
g(1)
+ I0q−1
g
+
0+
0+
1
L
=
1
(Ck ψ (r) + Ch φ (r) + Cg ) .
Γ(q)
(3.10)
C1
[ψ (r) + φ (r) + 1] .
Γ(q)
(3.11)
Hence
|T u(t)| ≤
Moreover, we have
c
|
D0σ+ T u|
q−σ−1 ≤ ψ (r) I0+
k
L1
q−σ−1 +φ (r) I0+
h
L1
I0q−2
+ h (1)
+
Γ(4 − σ)
!
I0q−2
+ k (1)
+
Γ(4 − σ)
q−σ−1 + I0+
g
L1
!
(3.12)
I0q−2
+ g (1)
+
.
Γ(4 − σ)
Using (3.5) we obtain
|c D0σ+ T u| ≤ C2 (ψ (r) + φ (r) + 1) .
(3.13)
From (3.11) and (3.13) we get
kT uk = (ψ (r) + φ (r) + 1)
C1
+ C2
Γ(q)
= C,
then T (Br ) is uniformly bounded.
ii) We will show that T maps bounded
sets into equicontinuous
sets of E. Indeed for any t1 , t2 ∈ [0, 1] ,
t1 < t2 and u ∈ Br , let C 0 = max(f t, u (t) ,c D0σ+ u(t) , 0 ≤ t ≤ 1, kuk < r), therefore
|T u(t1 ) − T u(t2 )| ≤
Z
C0
(
Γ(q)
t1
Z
|G(t1 , s) − G(t2 , s)| ds+
0
t2
Z
1
|G(t1 , s) − G(t2 , s)| ds +
t1
|G(t1 , s) − G(t2 , s)| ds),
t2
that implies
|T u(t1 ) − T u(t2 )|
Z t1
C0
≤
(t2 − s)q−1 − (t1 − s)q−1
Γ(q) 0
+ (t2 − t1 ) (1 − s)q−1 + q (q − 1) 1 + t22 + t21 + t1 t2 (1 − s)q−3 ds
Z t2
+
(t2 − s)q−1
t1
+ (t2 − t1 ) (1 − s)q−1 + q (q − 1) 1 + t22 + t21 + t1 t2 (1 − s)q−3 ds
Z 1
+
(t2 − t1 ) (1 − s)q−1 + q (q − 1) 1 + t22 + t21 + t1 t2 (1 − s)q−3 ds,
t2
if we consider the function Φ(x) = xq−1 − (q − 1)x, we see that Φ is decreasing on [0, 1], consequently
A. Guezane-Lakoud, R. Khaldi, J. Nonlinear Sci. Appl. 5 (2012), 64–73
71
(t2 − s)q−1 −(t1 − s)q−1 ≤ (q−1) (t2 − t1 ) , from this and using some elementary computations we deduce
|T u(t1 ) − T u(t2 )| ≤
C 0 (t2 − t1 ) (102 + 49 (t2 − t1 ))
Γ(q)
(3.14)
On the other hand we have
|c D0σ+ T u(t1 ) −c D0σ+ T u(t2 )|
Z t1
1
≤ +
(t1 − s)−σ+1 − (t2 − s)−σ+1 (T u(s))00 ds
Γ(2 − σ) 0
Z t2
1
(t2 − s)−σ+1 (T u(s))00 ds.
+
Γ(2 − σ) t1
Using (3.5) and (3.12) it yields
(T u(t))00 ≤ [ψ (r) + φ (r) + 1] C20
(3.15)
then
|c D0σ+ T u(t1 ) −c D0σ+ T u(t2 )| ≤
[ψ (r) + φ (r) + 1] C20 2(t2 − t1 )2−σ + t2−σ
− t2−σ
2
1
(2 − σ)Γ(2 − σ)
(3.16)
when
(3.14) and (3.16) then |T u(t1 ) − T u(t2 )| and
c σ t1 → t2 , in
D + T u(t1 ) −c Dσ+ T u(t2 ) tend to 0. Consequently T (Br ) is equicontinuous. From Arzela-Ascoli Theorem
0
0
we deduce that T is completely continuous.
Now we apply Leray Schauder nonlinear alternative to prove that T has at least one nontrivial solution
in E.
Let Ω = {u ∈ E : kuk < r}, for any u ∈ ∂Ω, such that u = λT u, 0 < λ < 1, we get with the help of
(3.11)
C1
|u(t)| = λ |T u(t)| ≤ |T u(t)| ≤
[ψ (r) + φ (r) + 1] .
(3.17)
Γ(q)
It follows from (3.13)
|c D0σ+ u(t)| ≤ C2 (ψ (r) + φ (r) + 1) .
(3.18)
From (3.17), (3.18) and (3.8) we deduce that
kuk ≤ (ψ (r) + φ (r) + 1)
C1
+ C2
Γ(q)
<r
this contradicts the fact that u ∈ ∂Ω. Lemma 3.5 allows us to conclude that the operator T has a fixed
point u∗ ∈ Ω and then the fractional boundary value problem (P1) has a nontrivial solution u∗ ∈ E. The
proof is complete.
Example 3.6. The fractional boundary value problem
(
5
7
c D 2 u = t 3 u + t−1 2 D 4 u + (t + 1),
0 < t < 1,
+
10
20
0
0+
u (0) = u(1) = u00 (1) = u00 (0) = 0,
has a unique solution in E.
Proof. We have f (t, x, y) =
t 3
x
10
+
t−1 2
y
20
|f (t, x, x) − f (t, y, y)| ≤
+ (t + 1), 3 < q =
t
10
3
|x − x| +
7
2
< 4, σ =
t−1
20
(3.19)
5
4
2
|y − y|
< 2 and
A. Guezane-Lakoud, R. Khaldi, J. Nonlinear Sci. Appl. 5 (2012), 64–73
72
2
t 3
then g(t) = 10
. Some calculus give
and h(t) = t−1
20
q−1 q
I0+ g 1 = 3.2065 × 10−6 , I0+ g(1) = 3.2064 × 10−6 ,
L
I q−2
= 0.11463 × 10−3 , Cg = 0.24208 × 10−3 , Ag = 9.7155 × 10−4
+ g(1)
0 q−1 q
I0+ h 1 = 5.4709 × 10−4 , I0+ h(1) = 5.4709 × 10−4 ,
L
q−σ−1 −6 q−σ−1 h 1 = 5. 191 8 × 10−4
g 1 = 3.245 7 × 10 , I0+
I0+
L
q−2
I0+ h(1)
Cg + Ch
Ag + Ah
L
Ch = 0.21591 × 10−2 ,
1
= 0.45799 × 10−2 <
2
1
−3
= 2. 598 2 × 10 <
2
= 0.32239 × 10
−2
,
Ah = 2.8815 × 10−2
Thus Theorem 3.2 implies that the fractional boundary value problem (3.19) has a unique in E.
Example 3.7. The fractional boundary value problem

6
(1+t2 ) −t 2
 c 10
1
c
2
3
5
D0+ u = 100
e u + ln(2 + ( D0+ u) ) + 1+t = 0,

u (0) = u(1) = u00 (1) = u00 (0) = 0
0<t<1
has at least one nontrivial solution in E. Applying Theorem 3.4, we have q = 10
3 , σ =
x 2
ln(2 + x2 )
1 + t2
|f (t, x, x)| = 1 + t2 e−t
+
+ 1 + t2
10
100
100 (1 + t)
2
2
ln(2 + x )
|x|
≤
1 + t2 + 1 + t2
+ 1 + t2
10
100
≤ k (t) ψ (|x|) + h(t)φ (|x|) + g (t) ,
2
2)
, φ (x) = ln(2+x
where k (t) = h(t) = g (t) = 1 + t2 , ψ (x) = |x|
10
100 , f (t, 0, 0) 6= 0.
Let us find r such that (3.8) holds, for this we have
q−1 q−2
q−1
I0+ g 1 = 0.117332, I0+ g(1) = 1.0559, I0+ g(1) = 0.40979
L
q−σ−1 g 1 = 0.508 63, I0q+ g(1) = 0.11733,
I0+
6
5
(3.20)
and
L
1
1.0559
Γ 4 − 65
C1 = 0.645 3, Ag = C2 = 1.138 5
r 2
We see that (3.8) is equivalent to 1.370 8 10
+
0.508 63 +
ln(2+r2 )
100
+ 1 − r which is negative for r ≥ 2.
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