# Boundary Value Problems for a Class of Sequential Integrodifferential Equations of Fractional Order

Authors: Bashir Ahmad; Juan J. Nieto
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

We investigate the existence of solutions for a sequential integrodifferential equation of fractional order with some boundary conditions. The existence results are established by means of some standard tools of fixed point theory. An illustrative example is also presented.


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## Body

## 1. Introduction

Nonlinear boundary value problems of fractional differential equations have received a considerable attention in the last few decades. One can easily find a variety of results ranging from theoretical analysis to asymptotic behavior and numerical methods for fractional equations in the literature on the topic. The interest in the subject has been mainly due to the extensive applications of fractional calculus in the mathematical modeling of several real-world phenomena occurring in physical and technical sciences; see, for example, [1–4]. An important feature of a fractional order differential operator, distinguishing it from an integer-order differential operator, is that it is nonlocal in nature. It means that the future state of a dynamical system or process based on a fractional operator depends on its current state as well as its past states. Thus, differential equations of arbitrary order are capable of describing memory and hereditary properties of some important and useful materials and processes. This feature has fascinated many researchers, and they have shifted their focus to fractional order models from the classical integer-order models. For some recent work on the topic, we refer, for instance, to [5–9]. Recently, in [10], the authors studied sequential fractional differential equations with three-point boundary conditions.In this paper, we consider a nonlinear Dirichlet boundary value problem of sequential fractional integrodifferential equations given by(1)(Dcα+kDcα-1)u(t)=pf(t,u(t))+qIβg(t,u(t)),(Dcα+kDcα-1)u(t)nnn+qIβg(t,u(t)),0<t<1,(2)u(0)=0,u(1)=0,
where cDα denotes the Caputo fractional derivative of order α,1<α≤2, Iβ(·) denotes Riemann-Liouville integral with 0<β<1,  f,g are given continuous functions, k≠0, and p,q are real constants. We also study the fractional integro-differential equation (1) subject to the following boundary conditions:
(3)u′(0)+ku(0)=a,u(1)=b,a,b∈ℝ,(4)u(0)=a,u′(0)=u′(1),a∈ℝ.


## 2. Linear Fractional Differential Equations

Forα∈(1,2], we consider the following linear fractional differential equation:
(5)(Dcα+kDcα-1)u(t)=h(t),
where Dcα denotes the Caputo fractional derivative of order α. Rewriting (1) as cDα(u(t)+kcD-1u(t))=h(t), we can write its solution as
(6)u(t)+kDc-1u(t)=1Γ(α)∫0t‍(t-s)α-1h(s)ds+c0+c1t,
where c0,c1 are arbitrary constants. Now, (6) can be expressed as
(7)u(t)=-k∫0t‍u(s)ds+1Γ(α)∫0t‍(t-s)α-1h(s)ds+c0+c1t.
Differentiating (7), we obtain
(8)u′(t)=-ku(t)+1Γ(α-1)∫0t‍(t-s)α-2h(s)ds+c1,
which can alternatively be written as
(9)(u(t)ekt)′=ekt(1Γ(α-1)∫0t‍(t-s)α-2h(s)ds+c1).Integrating from0 to t, we have
(10)u(t)=Ae-kt+∫0t‍e-k(t-s)Iα-1h(s)ds+B,
where A and B are arbitrary constants, and
(11)Iα-1h(t)=∫0t‍(t-x)α-2Γ(α-1)h(x)dx.Lemma 1.
The unique solution of the linear equation (5) subject to the Dirichlet boundary conditions (2) is given by
(12)u(t)=(1-e-kt)(e-k-1)∫01‍e-k(1-s)(∫0s‍(s-x)α-2Γ(α-1)h(x)dx)ds+∫0t‍e-k(t-s)(∫0s‍(s-x)α-2Γ(α-1)h(x)dx)ds.Proof.
Observe that the general solution of (5) is given by (10). Using the given boundary conditions in (10), we find that
(13)A=-B=1(1-e-k)∫01‍e-k(1-s)(∫0s‍(s-x)α-2Γ(α-1)h(x)dx)ds.
Substituting the values of A and B in (10) yields the solution (12). This completes the proof.In the next two lemmas, we present the unique solutions of (5) with different kinds of boundary conditions. We do not provide the proofs for these lemmas as they are similar to that of Lemma 1.Lemma 2.
The unique solution of the problem (5)–(3) is given by
(14)u(t)=ek(1-t)×[(bk-a)k-∫01‍e-k(1-s)(∫0s‍(s-x)α-2Γ(α-1)h(x)dx)ds]+∫0t‍e-k(t-s)(∫0s‍(s-x)α-2Γ(α-1)h(x)dx)ds+ak.Lemma 3.
The unique solution of (5) with the boundary conditions (4) is
(15)u(t)=-(1-e-kt)k(1-e-k)×[k∫01‍e-k(1-s)(∫0s‍(s-x)α-2Γ(α-1)h(x)dx)ds-∫01‍(1-s)α-2Γ(α-1)h(s)ds]+∫0t‍e-k(t-s)(∫0s‍(s-x)α-2Γ(α-1)h(x)dx)ds+a.


## 3. Existence Results for the Nonlinear Problems

Let𝒫=C([0,1],ℝ) denote the Banach space of all continuous functions from [0,1] into ℝ endowed with the usual norm defined by ∥x∥=sup{|x(t)|,t∈[0,1]}.In view of Lemma1, we transform problem (1)-(2) to an equivalent fixed point problem as
(16)u=𝒱u,
where 𝒱:𝒫→𝒫 is defined by
(17)(𝒱u)(t)=(1-e-kt)(e-k-1)∫01‍e-k(1-s)(p∫0s‍(s-x)α-2Γ(α-1)f(x,u(x))dx(t)(t)(t)(1-e-kt)(e-k-1)∫01‍e-k(1-s)++q∫0s‍(s-x)α+β-2Γ(α+β-1)(t)(t)(t)(1-e-kt)(e-k-1)∫01‍e-k(1-s)+××××g(x,u(x))dx∫0s‍(s-x)α+β-2Γ(α+β-1))ds+∫0t‍e-k(t-s)(p∫0s‍(s-x)α-2Γ(α-1)f(x,u(x))dx∫0t‍e-k(t-s)(t)(t)(t)(t)+q∫0s‍(s-x)α+β-2Γ(α+β-1)∫0t‍e-k(t-s)(t)(t)(t)(t)×g××g(x,u(x))dx∫0s‍(s-x)α-2Γ(α-1))ds.In a similar manner, we can define a fixed point operator𝒱1:𝒫→𝒫 for the nonlinear problem (1)–(3) as follows:
(18)(𝒱1u)(t)=ek(1-t)[(bk-a)k-∫01‍e-k(1-s)[(bk-a)k-[(bk-a)k-×(p∫0s‍(s-x)α-2Γ(α-1)f(x,u(x))dx[(bk-a)k-[(bk-a)k-+q++q∫0s‍(s-x)α+β-2Γ(α+β-1)[(bk-a)k-[(bk-a)k-+q+×g××g(x,u(x))dx∫0s‍)ds(bk-a)k]+∫0t‍e-k(t-s)(p∫0s‍(s-x)α-2Γ(α-1)f(x,u(x))dx∫0t‍e-k(t-s)nnnn+q+q∫0s‍(s-x)α+β-2Γ(α+β-1)∫0t‍e-k(t-s)nnnn+q×g××g(x,u(x))dx∫0s‍(s-x)α-2Γ(α-1))ds+ak.
A fixed point operator 𝒱2:𝒫→𝒫 for the nonlinear problem (1)–(4) is defined by
(19)(𝒱2u)(t)=-(1-e-kt)k(1-e-k)×[k∫01‍e-k(1-s)(p∫0s‍(s-x)α-2Γ(α-1)××××(p∫0s‍(s-x)α-2Γ(α-1)×f(x,u(x))dx××××(p∫0s‍(s-x)α-2Γ(α-1)+q∫0s‍(s-x)α+β-2Γ(α+β-1)××××(p∫0s‍(s-x)α-2Γ(α-1)×××g(x,u(x))dx∫0s‍(s-x)α-2Γ(α-1))ds-p∫01‍(1-s)α-2Γ(α-1)f(s,u(s))ds-q∫01‍(1-s)α+β-2Γ(α+β-1)g(s,u(s))ds]+∫0t‍e-k(t-s)(p∫0s‍(s-x)α-2Γ(α-1)f(x,u(x))dx+V+∫0t‍e-k(t-s)+q∫0s‍(s-x)α+β-2Γ(α+β-1)+VVV+∫0t‍e-k(t-s)×××g(x,u(x))dx∫0s‍)ds+a.We only present the existence results for the problem (1)-(2). Observe that problem (1)-(2) has solutions if the operator equation (16) has fixed points.For computational convenience, we introduce the following constant:(20)Q=2|1-e-k|[|p|Γ(α+β)+|q|Γ(α)]|k|Γ(α+β)Γ(α).Theorem 4.
Assume thatf,g:[0,1]×ℝ→ℝ are continuous functions satisfying the following condition:
(21)(A1)|f(t,u)-f(t,v)|≤L1|u-v|,|g(t,u)-g(t,v)|≤L2|u-v|,∀t∈[0,1],|gvg(t,v)|≤L2|u-v|L1,L2>0,u,v∈ℝ.
Then, the boundary value problem (1)-(2) has a unique solution if L<1/Q, where L=max{L1,L2} and Q is given by (20).Proof.
Let us defineM=max{M1,M2}, where M1,M2 are finite numbers given by supt∈[0,1]|f(t,0)|=M1,supt∈[0,1]|g(t,0)|=M2. Selecting r≥(QM)/(1-LQ), we show that 𝒱Br⊂Br, where Br={u∈𝒫:∥u∥≤r}. For u∈Br, we have
(22)∥(𝒱u)∥≤supt∈[0,1]{|1-e-kt1-e-k|×∫01‍e-k(1-s)(|p|∫0s‍(s-x)α-2Γ(α-1)××|f(x,u(x))|×|f(x,u(x))|dx×++|q|∫0s‍+|q|∫0s‍(s-x)α+β-2Γ(α+β-1)×|g(x,u(x))|dx××|g(x,u(x))|dx∫0s‍)ds×+∫0t‍e-k(t-s)××××(|p|∫0s‍(s-x)α-2Γ(α-1)×|f(x,u(x))|dx××××|f(x,u(x))|dx×××++×+|q|∫0s‍(s-x)α+β-2Γ(α+β-1)×××++××××|g(x,u(x))|dx∫0s‍)ds|1-e-kt1-e-k|}≤supt∈[0,1]{|1-e-kt1-e-k|∫01‍e-k(1-s)≤supt∈[0,1]<|1-e-kt1-e-k|××(|p|∫0s‍(s-x)α-2Γ(α-1)×≤supt∈[0,1]<|1-e-kt1-e-k|××××(|f(x,u(x))-f(x,0)|×≤supt∈[0,1]vvvv<|1-e-kt1-e-k|××××+|f(x,0)|f(x,u(x)))dx××≤supt∈[0,1]<|1-e-kt1-e-k|××+|q|∫0s‍(s-x)α+β-2Γ(α+β-1)××≤supt∈[0,1]×<|1-e-kt1-e-k|×(|g(x,u(x))-g(x,0)|××+|mmmmmmmmbq|×++|g(x,0)|)dx∫0s‍)ds+∫0t‍e-k(t-s)××(|p|∫0s‍(s-x)α-2Γ(α-1)××××××(|f(x,u(x))-f(x,0)|×××××××+|f(x,0)|f(x,u(x)))dx××+|q|∫0s‍(s-x)α+β-2Γ(α+β-1)××+|q|××(|g(x,0)||g(x,u(x))-g(x,0)|××++×××+|g(x,0)|g(x,u(x)))dx∫0s‍)ds|1-e-kt1-e-k|}≤|p|(L1r+M1)×supt∈[0,1]{|1-e-kt1-e-k|×∫01‍e-k(1-s)(∫0s‍(s-x)α-2Γ(α-1)dx)ds+∫0t‍e-k(t-s)(∫0s‍(s-x)α-2Γ(α-1)dx)ds}+|q|(L2r+M2)×supt∈[0,1]{|1-e-kt1-e-k|×∫01‍e-k(1-s)(∫0s‍(s-x)α+β-2Γ(α+β-1)dx)ds+∫0t‍e-k(t-s)×(∫0s‍(s-x)α+β-2Γ(α+β-1)dx)ds}≤(Lr+M)2|1-e-k|[|p|Γ(α+β)+|q|Γ(α)]|k|Γ(α+β)Γ(α)=(Lr+M)Q≤r,
which means that 𝒱Br⊂Br.
Now, foru,v∈𝒫, we obtain
(23)∥𝒱u-𝒱v∥≤supt∈[0,1]{|1-e-kt1-e-k|×∫01‍e-k(1-s)(|p|∫0s‍(s-x)α-2Γ(α-1)f(s,u(s))×××|f(s,u(s))f(s,u(s))××-f-f(s,v(s))|dx+|q|+|q|+|q|∫0s‍(s-x)α+β-2Γ(α+β-1)+|q|+|q|+|q|××|g(s,u(s))+|q|+|q|+|q|×-g-g(s,v(s))|dx∫0s‍(s-x)α-2Γ(α-1))ds+∫0t‍e-k(t-s)×+×(|p|∫0s‍(s-x)α-2Γ(α-1)××+×|f(s,u(s))-f(s,v(s))|dx××+|q|∫0s‍(s-x)α+β-2Γ(α+β-1)××××××|g(s,u(s))-g(s,v(s))|dx∫0s‍)ds∫01‍}≤L2|1-e-k|[|p|Γ(α+β)+|q|Γ(α)]|k|Γ(α+β)Γ(α)∥u-v∥=LQ∥u-v∥.
By the given assumption, L<1/Q,  𝒱 is a contraction. Thus, the conclusion of the theorem follows by the contraction mapping principle (Banach fixed point theorem).Our next existence result relies on Krasnoselskii's fixed point theorem.Lemma 5 (Krasnoselskii, see [11]).
LetM be a closed, convex, bounded, 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, and (iii) B is a contraction mapping. Then, there exists z∈M such that z=Az+Bz.Theorem 6.
Letf,g:[0,1]×ℝ→ℝ be continuous functions satisfying assumption (A1), and
(24)(A2)|f(t,u)|≤μ1(t),|g(t,u)|≤μ2(t),∀(t,u)∈[0,1]×ℝ,μ1,μ2∈C([0,1],ℝ+).
Then, the problem (1)-(2) has at least one solution on [0,1] provided that
(25)|1-e-k|[|p|∥μ1∥Γ(α+β)+|q|∥μ2∥Γ(α)]|k|Γ(α+β)Γ(α)<1,
where supt∈[0,1]|μi(t)|=∥μi∥,i=1,2.Proof.
Let us fix(26)r¯≥2|1-e-k|[|p|∥μ1∥Γ(α+β)+|q|∥μ2∥Γ(α)]|k|Γ(α+β)Γ(α)
and consider Br¯={u∈𝒫:∥u∥≤r¯}. We define the operators ψ1 and ψ2 on Br¯ as
(27)(ψ1u)(t)=∫0t‍e-k(t-s)(p∫0s‍(s-x)α-2Γ(α-1)f(x,u(x))dxmmmmmmm+q∫0s‍(s-x)α+β-2Γ(α+β-1)mmmmmmmmm×g(x,u(x))dx∫0s‍(s-x)α-2Γ(α-1))ds,mmmmmm×g(x,u(x))dx∫0s‍(s-x)α-2Γ(α-1))t∈[0,1],(ψ2u)(t)=(1-e-kt)(e-k-1)×∫01e-k(1-s)(p∫0s‍(s-x)α-2Γ(α-1)f(x,u(x))dx‍mmmmmmmmm+q∫0s‍(s-x)α+β-2Γ(α+β-1)mmmmmmmmmmm×g(x,u(x))dx∫0s‍(s-x)α-2Γ(α-1))ds,mmmmbmm×g(x,u(x))dx∫0s‍(s-x)α-2Γ(α-1))ds,t∈[0,1].
For u,v∈Br¯, we find that
(28)∥ψ1u+ψ2v∥≤2|1-e-k|[|p|∥μ1∥Γ(α+β)+|q|∥μ2∥Γ(α)]|k|Γ(α+β)Γ(α)≤r¯.
Thus,ψ1u+ψ2v∈Br¯. It follows from assumption (A1) together with (25) that ψ2 is a contraction mapping. Continuities of f and g imply that the operator ψ1 is continuous. Also, ψ1 is uniformly bounded on Br¯  as
(29)∥ψ1u∥≤|1-e-k|[|p|∥μ1∥Γ(α+β)+|q|∥μ2∥Γ(α)]|k|Γ(α+β)Γ(α).
Now, we prove the compactness of the operator ψ1. In view of (A1), we define
(30)sup(t,u)∈[0,1]×Br¯|f(t,u)|=f¯,sup(t,u)∈[0,1]×Br¯|g(t,u)|=g¯.
Consequently, we have
(31)∥(ψ1u)(t2)-(ψ1u)(t1)∥≤|e-kt2-e-kt1|∫0t1‍eks(|p|f¯∫0s‍(s-x)α-2Γ(α-1)dx≤|e-kt2-e-kt1|∫0t1‍eks+++|q|g¯∫0s‍(s-x)α+β-2Γ(α+β-1)dx)ds≤∫t1t2‍e-k(t2-s)×(|p|f¯∫0s‍(s-x)α-2Γ(α-1)dx+|q|g¯∫0s‍(s-x)α+β-2Γ(α+β-1)dx)ds
which is independent of u and tends to zero as t2→t1. Thus, ψ1 is relatively compact on Br¯. Hence, by the Arzelá-Ascoli theorem, ψ1 is compact on Br¯. Thus, all the assumptions of Lemma 5 are satisfied. So, by the conclusion of Lemma 5, problem (1)-(2) has at least one solution on [0,1].Now, we show the existence of solutions for the problem (1)-(2) via Leray-Schauder alternative.Lemma 7 (nonlinear alternative for single valued maps, see [12]).
LetE be a Banach space, C a closed, convex subset of E,U an open subset of C, and 0∈U. Suppose that F:U¯→C is a continuous, compact (that is, F(U¯) is a relatively compact subset of C) map. Then, either(i)
Fhas a fixed point in U¯, or(ii)
there is au∈∂U (the boundary of U in C) and λ∈(0,1) with u=λF(u).Theorem 8.
Letf,g:[0,1]×ℝ→ℝ be continuous functions and the following assumptions hold.(A3)
There exist functionsσ1,σ2∈C([0,1],ℝ+), and nondecreasing functions ψ1,ψ2:ℝ+→ℝ+ such that |f(t,u)|≤σ1(t)ψ1(∥u∥),|g(t,u)|≤σ2(t)ψ2(∥u∥),  for  all  (t,u)∈[0,1]×ℝ.(A4)
There exists a constantM>0 such that
(32)M×((2|1-e-k|[|p|ψ1(∥u∥)Γ(α+β)∥σ1∥|p|×+|q|ψ2+++|q|ψ2(∥u∥)∥σ2∥Γ(α)]|1-e-k|)×(|k|Γ(α+β)Γ(α))-1)-1>1.Then, the boundary value problem (1)-(2) has at least one solution on [0,1].Proof.
Consider the operator𝒱:𝒫→𝒫 with u=𝒱u, where
(33)(𝒱u)(t)=(1-e-kt)(e-k-1)×∫01‍e-k(1-s)(p∫0s‍(s-x)α-2Γ(α-1)∫01‍e-k(1-s)×××××f(x,u(x))dx∫01‍e-k(1-s)××++q∫0s‍(s-x)α+β-2Γ(α+β-1)∫01‍e-k(1-s)××+q×××g(x,u(x))dx∫0s‍)ds+∫0t‍e-k(t-s)(p∫0s‍(s-x)α-2Γ(α-1)f(x,u(x))dx++∫0t‍e-k(t-s)+q∫0s‍(s-x)α+β-2Γ(α+β-1)++∫0t‍e-k(t-s)××××g(x,u(x))dx∫0s‍(s-x)α-2Γ(α-1))ds.
We show that 𝒱 maps bounded sets into bounded sets in C([0,1],ℝ). For a positive number r, let Br={u∈C([0,1],ℝ):∥u∥≤r} be a bounded set in C([0,1],ℝ). Then,
(34)|(𝒱u)(t)|≤|1-e-kt1-e-k|∫01‍e-k(1-s)≤|1-e-kt1-e-k|×(|p|∫0s‍(s-x)α-2Γ(α-1)|f(x,u(x))|dx≤|1-e-kt1-e-k|+++|q|∫0s‍(s-x)α+β-2Γ(α+β-1)|g(x,u(x))|dx)ds+∫0t‍e-k(t-s)(|p|∫0s‍(s-x)α-2Γ(α-1)|f(x,u(x))|dx+∫0t‍e-k(t-s)++|q|∫0s‍(s-x)α+β-2Γ(α+β-1)+∫0t‍e-k(t-s)+++|q|×|g(x,u(x))|dx∫0s‍)ds≤|1-e-kt1-e-k|×∫01‍e-k(1-s)(|p|∫0s‍(s-x)α-2Γ(α-1)σ1(x)ψ1(∥u∥)dx×∫01‍e-k(1-s)+|q|+|q|∫0s‍(s-x)α+β-2Γ(α+β-1)σ2(x)ψ2(∥u∥)dx)ds+∫0t‍e-k(t-s)(|p|∫0s‍(s-x)α-2Γ(α-1)σ1(x)ψ1(∥u∥)dx∫0s‍(s-x)α+β-2Γ(α+β-1)++|q|∫0s‍(s-x)α+β-2Γ(α+β-1)∫0s‍(s-x)α+β-2Γ(α+β-1)++|q|×σ2(x)ψ2(∥u∥)dx∫0s‍)ds≤|p|ψ1(r)∥σ1∥×[|1-e-kt1-e-k|∫01‍e-k(1-s)(∫0s‍(s-x)α-2Γ(α-1)dx)ds+∫0t‍e-k(t-s)(∫0s‍(s-x)α-2Γ(α-1)dx)ds]+|q|∥σ2∥ψ2(r)[|1-e-kt1-e-k|×∫01‍e-k(1-s)(∫0s‍(s-x)α+β-2Γ(α+β-1)dx)ds+∫0t‍e-k(t-s)(∫0s‍(s-x)α+β-2Γ(α+β-1)dx)ds]≤(2|1-e-k|[|p|ψ1(∥u∥)Γ(α+β)∥σ1∥×2|1-e-k|+|q|ψ2(∥u∥)∥σ2∥Γ(α)]|1-e-k|)×(|k|Γ(α+β)Γ(α))-1.
Consequently,
(35)∥𝒱x∥≤(2|1-ek|××[|p|ψ1(∥u∥)Γ(α+β)∥σ1∥××+|q|ψ2(∥u∥)∥σ2∥Γ(α)]|1-ek|)×(|k|Γ(α+β)Γ(α))-1.
Next, we show that𝒱 maps bounded sets into equicontinuous sets of C([0,1],ℝ). Let t1,t2∈[0,1] with t1<t2 and u∈Br, where Br is a bounded set of C([0,1],ℝ). Then, we obtain(36)∥(𝒱u)(t2)-(𝒱u)(t1)∥≤|-e-kt2+e-kt11-e-k|×∫01‍e-k(1-s)(∫0s‍(s-x)α+β-2Γ(α+β-1)|p|ψ1(r)∥σ1∥××××××××∫0s(s-x)α-2Γ(α-1)dx‍×××××××+|q|ψ2(r)∥σ2∥××××××××∫0s‍(s-x)α+β-2Γ(α+β-1)dx)ds+|e-kt2-e-kt1|×∫0t1‍eks(∫0s‍(s-x)α+β-2Γ(α+β-1)|p|ψ1(r)∥σ1∥××××××∫0s(s-x)α-2Γ(α-1)‍dx×××××++|q|ψ2(r)∥σ2∥××××××∫0s‍(s-x)α+β-2Γ(α+β-1)dx)ds+∫t1t2‍e-k(t2-s)(|p|ψ1(r)∥σ1∥∫0s‍(s-x)α-2Γ(α-1)dx++|q|ψ2(r)Pσ2P∫0s‍+|q|ψ2(r)∥σ2∥++|q|ψ2(r)Pσ2P∫0s‍×∫0s‍(s-x)α+β-2Γ(α+β-1)dx)ds.
Obviously, the right hand side of the previous inequality tends to zero independently of u∈Br as t2-t1→0. As 𝒱 satisfies the previous assumptions, therefore it follows by the Arzelá-Ascoli theorem that 𝒱:C([0,1],ℝ)→C([0,1],ℝ) is completely continuous.
The proof will be complete by the application of the Leray-Schauder nonlinear alternative (Lemma7) once we establish the boundedness of the set of all solutions to equations u=λ𝒱u for λ∈(0,1).
Letu be a solution. Then, for t∈[0,1], and using the computations in proving that 𝒱 is bounded, we have
(37)|u(t)|=|λ(𝒱u)(t)|≤(2|1-e-k|×[|p|ψ1(∥u∥)Γ(α+β)∥σ1∥+++|q|ψ2(∥u∥)∥σ2∥Γ(α)]|1-e-k|)×(|k|Γ(α+β)Γ(α))-1.
Consequently, we have
(38)∥u∥×((2|1-e-k|[|p|ψ1(∥u∥)Γ(α+β)∥σ1∥×××+|q|ψ2+++|q|ψ2(∥u∥)∥σ2∥Γ(α)]|1-e-k|)×(|k|Γ(α+β)Γ(α))-1)-1≤1.
In view of (A4), there exists M such that ∥u∥≠M. Let us set
(39)U={u∈C([0,1],ℝ):∥u∥<M}.
Note that the operator 𝒱:U¯→C([0,1],ℝ) is continuous and completely continuous. From the choice of U, there is no u∈∂U such that u=λ𝒱(u) for some λ∈(0,1). Consequently, by the nonlinear alternative of Leray-Schauder type (Lemma 7), we deduce that 𝒱 has a fixed point u∈U¯ which is a solution of the problem (1)-(2). This completes the proof.Example 9.
Consider a boundary value problem of integro-differential equations of fractional order given by(40)(D3/2c+2D1/2c)u(t)=12f(t,u(t))+I1/2g(t,u(t)),0<t<1,u(0)=0,u(1)=0,
where α=3/2,  k=2,  p=1/2,  q=1,  β=1/2,  f(t,u)=(|u|(2+|u|))/(3(1+|u|))+4t,  g(t,u)=(1/4)tan-1u+cos2t+t3+5. With the given data, it is found that L1=2/3,L2=1/4 as |f(t,u)-f(t,v)|≤(2/3)|u-v|,  |g(t,u)-g(t,v)|≤(1/4)|u-v|, and
(41)Q=2|1-e-k|[|p|Γ(α+β)+|q|Γ(α)]|k|Γ(α+β)Γ(α)≃1.3525.
Clearly, L=max{L1,L2}=2/3 and L<1/Q. Thus, all the assumptions of Theorem 4 are satisfied. Hence, by the conclusion of Theorem 4, the problem (40) has a unique solution.



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*Source: 149659-2013-04-18.xml*