Box 80203, Jeddah 21589, Saudi Arabia Full list of author information is available at the end of the article Abstract This article investigates a boundary value problem of Riemann-Liouvi
Trang 1R E S E A R C H Open Access
Riemann-Liouville fractional integro-differential equations with fractional nonlocal integral
boundary conditions
Bashir Ahmad1* and Juan J Nieto1,2
* Correspondence:
bashir_qau@yahoo.com
1 Department of Mathematics,
Faculty of Science, King Abdulaziz
University P.O Box 80203, Jeddah
21589, Saudi Arabia
Full list of author information is
available at the end of the article
Abstract This article investigates a boundary value problem of Riemann-Liouville fractional integro-differential equations with fractional nonlocal integral boundary conditions Some new existence results are obtained by applying standard fixed point theorems
2010 Mathematics Subject Classification: 26A33; 34A34; 34B15
Keywords: Riemann-Liouville calculus, fractional integro-differential equations, frac-tional boundary conditions, fixed point theorems
1 Introduction
In this article, we study the existence and uniqueness of solutions for the following nonlinear fractional integro-differential equation:
D α u(t) = f (t, u(t), ( φu)(t), (ψu)(t)), t ∈ [0, T] , α ∈ (1 , 2] , (1:1) subject to the boundary conditions of fractional order given by
D α−1 u(0+) =νI α−1 u( η), 0 < η < T, ν is a constant, (1:3) where Dadenotes the Riemann-Liouville fractional derivative of order a, f: [0, T] ×ℝ
×ℝ × ℝ ® ℝ is continuous, and
(φx)(t) =
t
0
γ (t, s)x(s)ds, (ψx)(t) =
t
0
δ(t, s)x(s)ds,
with g andδ being continuous functions on [0, T] × [0, T]
Boundary value problems for nonlinear fractional differential equations have recently been investigated by several researchers As a matter of fact, fractional derivatives provide an excellent tool for the description of memory and hereditary properties of various materials and processes (see [1]) and make the fractional-order models more realistic and practical than the classical integer-order models Fractional differential equations arise in many engineering and scientific disciplines, such as physics, chemis-try, biology, economics, control theory, signal and image processing, biophysics, blood
© 2011 Ahmad and Nieto; licensee Springer This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in
Trang 2flow phenomena, aerodynamics, fitting of experimental data, etc (see [1,2]) For some
recent development on the topic, (see [3-19] and references therein)
2 Preliminaries
Let us recall some basic definitions (see [20,21])
Definition 2.1 The Riemann-Liouville fractional integral of order a > 0 for a contin-uous function u: (0,∞) ® ℝ is defined as
I α u(t) = 1
(α)
t
0
(t − s) α−1 u(s)ds,
provided the integral exists
Definition 2.2 For a continuous function u: (0, ∞) ® ℝ, the Riemann-Liouville deri-vative of fractional order a > 0, n = [a] + 1 ([a] denotes the integer part of the real
number a) is defined as
(n − α)
1
dt
nt 0
(t − s) n −α−1 u(s)ds =
d
dt
n
I n −α u(t),
provided it exists
For a < 0, we use the convention that Dau = I-au Also for b Î [0, a), it is valid that
DbIau = Ia-bu
Note that for l >-1, l ≠ a - 1, a - 2, , a - n, we have
D α t λ= (λ + 1) (λ − α + 1) t λ−α,
and
D α t α−i= 0, i = 1, 2, , n.
In particular, for the constant function u(t) = 1, we obtain
(1 − α) t −α, α /∈N.
For a ÎN, we get, of course, Da
1 = 0 because of the poles of the gamma function at the points 0, -1, -2,
For a > 0, the general solution of the homogeneous equation
D α u(t) = 0
in C(0, T)∩ L(0, T) is
u(t) = c0t α−n + c1t α−n−1+· · · + c n−2t α−2 + c n−1t α−1, where ci, i = 1, 2, , n - 1, are arbitrary real constants
We always have DaIau= u, and
I α D α u(t) = u(t) + c0t α−n + c1t α−n−1+· · · + c n−2t α−2 + c n−1t α−1.
To define the solution for the nonlinear problem (1.1) and (1.2)-(1.3), we consider the following linear equation
Trang 3D α u(t) = σ (t), α ∈ (1 , 2] , t ∈ [0, T] , T > 0, (2:1) where s Î C[0, T]
We define
A = ν
η
0
s α−1(η − s) α−2
ν (α)η2α−2
such that A ≠ Γ(a)
The general solution of (2.1) is given by
with Iathe usual Riemann-Liouville fractional integral of order a
From (2.3), we have
Using the conditions (1.2) and (1.3) in (2.4) and (2.5), we find that c0= 0 and
(α) − A
η
0
(η − s) α−2 (α − 1)
⎛
⎝
s
0
(s − x) α−1
⎞
⎠ ds,
where A is defined by (2.2)
Substituting the values of c0 and c1 in (2.3), the unique solution of (2.1) subject to the boundary conditions (1.2)-(1.3) is given by
u(t) = t
0
(t − s) α−1 (α) σ (s)ds
+ νt α−1
(α) − A
η
0
(η − s) α−2 (α − 1)
⎛
⎝
s
0
(s − x) α−1
⎞
⎠ ds
=
t
0
(t − s) α−1 (α) σ (s)ds +
νt α−1
(α) − A I
2α−1 σ (η).
(2:6)
3 Main results
Let C = C([0, T], R)denotes the Banach space of all continuous functions from [0, T]
® ℝ endowed with the norm defined by ║u║ = sup{|u(t)|, t Î [0, T]}
If u is a solution of (1.1) and (1.2)-(1.3), then
u(t) =
t
0
(t − s) α−1 (α) f (s, u (s) , (φu) (s) , (ψu) (s)) ds
+ν1t α−1
η
(η − s)2α−2 (2α − 1) f (s, u (s) , (φu) (s) , (ψu) (s)) ds,
Trang 4(α) − A.
Define an operatorP : C → C as
(Pu) (t) =
t
0
(t − s) α−1 (α) f (s, u (s) , (φu) (s) , (ψu) (s)) ds
+ν1t α−1
η
0
(η − s)2α−2 (2α − 1) f (s, u (s) , (φu) (s) , (ψu) (s)) ds, t ∈ [0, T]
Observe that the problem (1.1) and (1.2)-(1.3) has solutions if and only if the opera-tor equationPu = uhas fixed points
Lemma 3.1 The operatorPis compact
Proof
(i) LetB be a bounded set in C[0, T] Then, there exists a constant M such that |f (t,u(t), (u)(t), (ψu)(t))| ≤ M, ∀u Î B, tÎ[0, T] Thus
| (Pu) (t)| ≤ M
t
0
(t − s) α−1 (α) ds + M |ν1|t α−1
η
0
(η − s)2α−2 (2α − 1) ds
≤ MT α−1
T (α + 1) +
|ν1|η2α−1 (2α)
, which implies that
|| (Pu) || ≤ MT α−1
T (α + 1)+
|ν1|η2α−1 (2α)
< ∞.
Hence,P(B)is uniformly bounded
(ii) For any t1, t2Î [0, T], u Î B, we have
| (Pu) (t1) − (Pu) (t2) |
=
t1
0
(t1− s) α−1 (α) f (s, u (s) , (φu) (s) , (ψu) (s)) ds
−
t2
0
(t2− s) α−1 (α) f (s, u (s) , (φu) (s) , (ψu) (s)) ds
+ν1
t α−11 − t α−1
2
η
0
(η − s)2α−s (2α − 1) f (s, u (s) , (φu) (s) , (ψu) (s))ds
≤ M
⎛
⎝ (α)1
t1
0
(t1− s) α−1 − (t2− s) α−1
ds− (α)1
t2
t1
(t2− s) α−1 ds
+
ν1
t1α−1 − t α−1
2
η
(η − s)2α−s (2α − 1) ds
⎞
⎠ → 0 as t1→ t2
Trang 5Thus,P(B)is equicontinuous Consequently, the operatorP is compact This com-pletes the proof □
We need the following known fixed point theorem to prove the existence of solu-tions for the problem at hand
Theorem 3.1 ([22]) Let E be a Banach space Assume that T: E ® E be a completely continuous operator and the set V = {x Î E| x =μTx, 0 <μ < 1} be bounded
Then, T has a fixed point in E
Theorem 3.2 Assume that there exists a constant M > 0 such that
|f (t, u (t) , (φu) (t) , (ψu) (t)) | ≤ M, ∀t ∈ [0, T] , u ∈ R.
Then, the problem (1.1) and (1.2)-(1.3) has at least one solution on [0,T]
Proof We consider the set
V = {u ∈ R|u = μPu, 0 < μ < 1} ,
and show that the set V is bounded Let u Î V, thenu = μPu, 0 <μ < 1 For any t Î [0, T], we have
|u (t) | ≤μ
⎡
⎣
t
0
(t − s) α−1 (α) |f (s, u (s) , (φu) (s) , (ψu) (s)) |ds
+|ν1|t α−1
η
0
(η − s)2α−2 (2α − 1) |f (s, u (s) , (φu) (s) , (ψu) (s)) |ds
⎤
⎦
As in part (i) of Lemma 3.1, we have
(Pu) ≤ MT α−1
T (α + 1)+
|ν1|η2α−1 (2α)
< ∞.
This implies that the set V is bounded independently of μ Î (0,1) Using Lemma 3.1 and Theorem 3.1, we obtain that the operator P has at least a fixed point, which
implies that the problem (1.1) and (1.2)-(1.3) has at least one solution This completes
the proof
Theorem 3.3 Assume that (A1) there exist positive functions L1(t), L2(t), L3(t) such that
|f (t, u (t) , (φu) (t) , (ψu) (t)) − f (t, v (t) , (φv) (t) , (ψv) (t)) |
≤ L1(t) |u − v| + L2(t) |φu − φv| + L3(t) |ψu − ψv|, ∀t ∈ [0, 1] , u, v ∈ R.
(A2)Λ = (ξ1 + |ν1|Ta-1ξ2)(1 + g0 +δ0) < 1, where
γ0= sup
t∈[0,1]|
t
0
γ (t, s) ds|, δ0= sup
t∈[0,1]|
t
0
δ (t, s) ds|,
ξ1= sup
t ∈[0,T]
|I q L1(t) |, |I q L2(t) |, |I q L3(t) |,
ξ2= max
|I2α−1 L
1(η) |, |I2α−1 L
2(η) |, |I2α−1 L
3(η) |, Then the problem (1.1) and (1.2)-(1.3) has a unique solution on C[0, T]
Proof Let us set suptÎ[0, T]|f(t,0,0,0)| = M, and choose
1− .
Trang 6Then we show thatPBr ⊂ B r, whereBr={x ∈ C : u ≤ r} For x Î Br, we have
( P u) (t) = sup
t ∈[0,T]|
t
0
(t − s) α−1
(α) f (s, u (s) , (φu) (s) , (ψu) (s)) ds
+ν1t α−1 η
0
(η − s)2α−2
(2α − 1) f (s, u (s) , (φu) (s) , (ψu) (s)) ds|
≤ sup
t ∈[0,T]
⎛
⎝
t
0
(t − s) α−1
(α)
|f (s, x (s) , (φx) (s) , (ψx) (s)) − f (s, 0, 0, 0) |
+|f (s, 0, 0, 0) | ds
+|ν1|t α−1
η
0
(η − s)2α−s
(2α − 1)
|f (s, x (s) , (φx) (s) , (ψx) (s)) − f (s, 0, 0, 0)
+|f (s, 0, 0, 0) | ds
≤ sup
t ∈[0,T]
⎛
⎝
t
0
(t − s) α−1
(α) (L1 (s) |x (s) | + L2 (s) | (φx) (s) | + L3 (s) | (ψx) (s) | + M) ds
+|ν1|t α−1
η
0
(η − s)2α−2
(2α − 1) (L1 (s) |x (s) | + L2 (s) | (φx) (s) | + L3 (s) | (ψx) (s) | + M) ds
⎞
⎠
≤ sup
t ∈[0,T]
⎛
⎝
t
0
(t − s) α−1
(α) (L1 (s) |x (s) | + γ0 L2(s) |x (s) |
+δ0 L3(s) |x (s) | + M) ds
+|ν1|t α−1
η
0
(η − s)2α−2
(2α − 1) (L1 (s) |x (s) | + γ0L2 (s) |x (s) |
+δ0L3 (s) |x (s) | + M) ds)
≤ sup
t ∈[0,T]
I α L1(t) + γ0I α L2 (t) + δ0I α L3 (t) r + Mt
q
q + 1
+|ν1|t α−1
I(2α−1) L1(η) + γ0I(2α−1) L2(η) + δ0I(2α−1) L3(η)r + Mη2α−1
(2α)
≤1 +|ν1|T α−1 ξ2 (1 + γ0+δ0) r + MT α−1
T (α + 1)+
|ν1|η α−1
(2α)
=r + Mε ≤ r
In view of (A1), for every t Î [0, T], we have
| (Pu) (t) − (Pv) (t) |
≤ supt ∈[0,T]
⎛
⎝
t
0
(t − s) α−1
(α) |f
s, u (s) , (φu) (s) , (ψu) (s) − f (s, v (s) , (φv) (s) , (ψv) (s)) |ds
+|ν1|t α−1
η
0
(η − s)2α−2
(2α − 1) |f
s, u (s) , (φu) (s) , (ψu) (s) − f (s, υ (s) , (φv) (s) , (ψv) (s)) |ds
≤ supt ∈[0,T]
⎛
⎝
t
0
(t − s) α−1
(α) (L1 (s) |u − v| + L2 (s) |φv| + L3 (s) |ψu − ψv|) ds
+|ν1|t α−1
η
0
(η − s)2α−2
(2α − 1)|
L1(s) |u − v| + L2(s) |φu − φv| + L3(s) |ψu − ψv| ds
≤ supt ∈[0,T] I α L1 (t) + γ0I α L2 (t) + δ0I α L3 (t) u − v
+|ν 1|T α−1
I(2α−1) L1(η) + γ0I(2α−1) L2(η) + δ0I(2α−1) L3(η) u − v
≤ξ1+|ν1|T α−1 ξ2 (1 + γ0+δ0) u − v = u − v
Trang 7By assumption (A2), Λ < 1, therefore, the operator P is a contraction Hence, by Banach fixed point theorem, we deduce thatP has a unique fixed point which in fact
is a unique solution of problem (1.1) and (1.2)-(1.3) This completes the proof □
Theorem 3.4 (Krasnoselskii’s fixed point theorem [22]) Let Mbe a closed convex and nonempty subset of a Banach space X Let A, B be the operators such that (i)
x, y∈Mwhenever x, y∈M; (ii) A is compact and continuous; (iii) B is a contraction
mapping Then, there existsz∈Msuch that z = Az+ Bz
Theorem 3.5 Assume that f: [0, T] × ℝ × ℝ × ℝ ® ℝ is a continuous function and the following assumptions hold:
(H1)
|f (t, u (t) , (φu) (t) , (ψu) (t)) − f (t, v (t) , (φv) (t) , (ψv) (t)) |
≤ L1(t) |u − v| + L2(t) |φu − φv| + L3(t) |ψu − ψv|, ∀t ∈ [0, T] , u, v ∈R.
(H2) |f (t,u)|≤ μ(t), ∀(t,u)Î[0, T] × ℝ, and μ Î C([0, T],ℝ+
)
If
|ν1|T α−1 η2α−1
then the boundary value problem (1.1) and (1.2)-(1.3) has at least one solution on [0, T]
Proof Letting suptÎ[0, T]|μ(t)| = ||μ||, we fix
¯r ≥ μ T α−1
T (α + 1)+
|ν1|η2α−1 (2α)
,
and considerB ¯r ={u ∈ C : u ≤ ¯r} We define the operatorsP1andP2onB ¯r as
(P1u ) (t) =
t
0
(t − s) α−1
s, u(s), ( φu)(s), (ψu)(s) ds,
(P2u) (t) = ν1t α−1
η
0
(η − s)2α−s (2α − 1) f
s, u (s) , (φu) (s), (ψu(s))ds
Foru, v ∈ B ¯r, we find that
P1u + P2v ≤ μ T α−1
T (α + 1)+
|ν1|η2α−1 (2α)
≤ ¯r.
Thus,P1u + P2v ∈ B ¯r It follows from the assumption (H1) together with (3.1) thatP2
is a contraction mapping Continuity of f implies that the operatorP1is continuous
Also,P1is uniformly bounded onB ¯ras
P1u ≤ (α + 1) u T α Now we prove the compactness of the operatorP1
In view of (H1), we define sup(t,x, φx,ψx)∈[0,T]×B r ×B r ×B r |f (t, x, φx, ψx)| = ¯f, and
conse-quently we have
Trang 8| (P1u ) (t1) − (P2u ) (t2)|
= (α)1
t1
0
(t1− s) α−1 − (t2− s) α−1
s, u(s), ( φu)(s), (ψu)(s) ds
(α)
t2
t1
(t2− s) α−1 f
s, u(s), ( φu)(s), (ψu)(s) ds
(α + 1) |2(t2− t1) + t α1− t α
2|, which is independent of u and tends to zero as t2® t1 So,P1is relatively compact
on B ¯r Hence, by the Arzelá-Ascoli Theorem, P1is compact on B ¯r Thus, all the
assumptions of Theorem 3.4 are satisfied So the conclusion of Theorem 3.4 implies
that the boundary value problem (1.1) and (1.2)-(1.3) has at least one solution on [0,
T] This completes the proof □
Acknowledgements
This study was partially supported by Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, Saudi
Arabia.
Author details
1 Department of Mathematics, Faculty of Science, King Abdulaziz University P.O Box 80203, Jeddah 21589, Saudi Arabia
2
Departamento de Análisis Matemático, Facultad de Matemáticas Universidad de Santiago de Compostela, Santiago
de Compostela 15782, Spain
Authors ’ contributions
Both authors, BA and JJN, contributed to each part of this work equally and read and approved the final version of
the manuscript.
Competing interests
The authors declare that they have no competing interests.
Received: 20 July 2011 Accepted: 17 October 2011 Published: 17 October 2011
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Trang 6Then we show thatPBr ⊂ B r, whereBr={x... γ0+δ0) u − v = u − v
Trang 7By assumption (A2), Λ < 1, therefore,