# Nonlocal Integrodifferential Boundary Value Problem for Nonlinear Fractional Differential Equations on  an Unbounded Domain

Authors: Lihong Zhang; Bashir Ahmad; Guotao Wang; Ravi P. Agarwal; Maryem Al-Yami; Wafa Shammakh
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

This paper investigates the existence of nonnegative solutions for nonlinear fractional differential equations with nonlocal fractional integrodifferential boundary conditions on an unbounded domain by means of Leray-Schauder nonlinear alternative theorem. An example is discussed for the illustration of the main work.


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## Body

## 1. Introduction

Recent studies on fractional differential equations, appeared in several special issues and books, reveal an extensive development of various aspects of the subject. One of the reasons for the popularity of fractional calculus is the nonlocal behavior of fractional-order operators in contrast to the classical integer-order operators. This characteristic has motivated many experts on modelling to introduce the concept of fractional modelling by taking into account the ideas of fractional calculus. Examples include various disciplines of science and engineering such as physics, chemistry, biomathematics, dynamical processes in porous media, dynamics of earthquakes, material viscoelastic theory, and control theory of dynamical systems. Furthermore, the outcome of certain experimentations indicate that integral and derivative operators of fractional order possess some characteristics related to complex systems having long memory in time. For details and examples, we refer the reader to the works in [1–7].Boundary value problems of fractional-order differential equations have been extensively investigated during the last few years, and a variety of results on the topic have been established. A great deal of the work on fractional boundary value problems involves local/nonlocal boundary conditions on bounded and unbounded domains; for example, see [8–26].In this paper, we study a new class of problems on fractional differential equations with nonlocal boundary conditions on unbounded domains. Precisely, we consider the following problem:(1)Dαu(t)+f(t,u(t))=0,1<α≤2,t∈J=[0,+∞),I2-αu(0)=0,Dα-1u(+∞)=λIα-1u(η),0<λ,η<∞,
where Dα denotes Riemann-Liouville fractional derivative of order α, f∈C(J×ℝ,ℝ+), and ℝ+=[0,+∞).


## 2. Preliminaries

In this section, we present some useful definitions and related theorems.Definition 1 (see [4]).
The Riemann-Liouville fractional derivative of orderδ for a continuous function f is defined by
(2)Dδf(t)=1Γ(n-δ)(ddt)n∫0t‍(t-s)n-δ-1f(s)ds,n=[δ]+1,
Provided that the right hand side is pointwise defined on (0,∞) and [δ] is the integer part of δ.Definition 2 (see [4]).
The Riemann-Liouville fractional integral of orderδ for a function f is defined as
(3)Iδf(t)=1Γ(δ)∫0t‍(t-s)δ-1f(s)ds,δ>0,
provided that such integral exists.Theorem 3 (see [27) (Leray-Schauder  nonlinear alternative).
LetC be a convex subset of a Banach space, and let U be an open subset of C with 0∈U. Then every completely continuous map N:U¯→C has at least one of the following two properties:(1)
Nhas a fixed point in U¯;(2)
there is anx∈∂U and λ∈(0,1) with x=λNx.Theorem 4 (see [28]).
LetU⊂X be a bounded set. Then U is relatively compact in X if the following conditions hold:(i)
for anyu(t)∈U, u(t)/(1+tα-1) is equicontinuous on any compact interval of J;(ii)
for anyε>0, there exists a constant T=T(ε)>0 such that
(4)|u(t1)1+t1α-1-u(t2)1+t2α-1|<εfor any t1,t2≥T and u∈U.Now we list the assumptions needed in the sequel.(H1):
Γ(2α-1)>λη2α-2.(H2):
there exist nonnegative functionsa(t), b(t) defined on [0,∞) and a constant ρ>0 such that
(5)|f(t,u(t))|≤a(t)+b(t)|u(t)|ρ,∫0+∞‍a(t)dt=a*<+∞,∫0+∞‍b(t)(1+tα-1)ρdt=b*<+∞.


## 3. Some Lemmas

This section contains some preliminary works that we need to establish the main result for problem (1).Lemma 5.
Letσ(t)∈C([0,+∞)) with ∫0∞‍σ(s)ds<∞. For Γ(2α-1)≠λη2α-2, the associated linear fractional boundary value problem,
(6)Dαu(t)+σ(t)=0,1<α≤2,I2-αu(0)=0,Dα-1u(+∞)=λIα-1u(η),
has a unique solution given by
(7)u(t)=∫0+∞‍G(t,s)σ(s)ds,
where
(8)G(t,s)=1Δ{-(Γ(2α-1)-λη2α-2)(t-s)α-1+(Γ(2α-1)-λ(η-s)2α-2)tα-1,s≤t,s≤η,(Γ(2α-1)-λ(η-s)2α-2)tα-1,0≤t≤s≤η,-(Γ(2α-1)-λη2α-2)(t-s)α-1+Γ(2α-1)tα-1,0≤η≤s≤t,Γ(2α-1)tα-1,s≥t,s≥η,(9)Δ=Γ(α)(Γ(2α-1)-λη2α-2).Proof.
It is well known that the fractional equation in (6) is equivalent to the integral equation:
(10)u(t)=-Iασ(t)+c1tα-1+c0tα-2,
where c0,c1∈ℝ are arbitrary constants. From (10), we have
(11)Dα-1u(t)=-I1σ(t)+c1Γ(α)=c1Γ(α)-∫0t‍σ(s)ds,Dα-2u(t)=-I2σ(t)+c1Γ(α-1)t+c0Γ(α-1).
Using the given boundary conditions in (10), we find that c0=0 and
(12)c1=1Γ(α)-A∫0∞‍σ(s)ds-λΓ(α)-A×∫0η‍(η-s)α-2Γ(α-1)[∫0s‍(s-x)α-1Γ(α)σ(x)dx]ds=1Γ(α)-A∫0∞‍σ(s)ds-λΓ(α)-AI2α-1σ(η),
where
(13)A=λ∫0ηsα-1(η-s)α-2Γ(α-1)ds=λΓ(α)η2α-2Γ(2α-1).
Substituting the values ofc1, c2 into (10) gives
(14)u(t)=-∫0t‍(t-s)α-1Γ(α)σ(s)ds-λtα-1Γ(α)-AI2α-1σ(η)+tα-1Γ(α)-A∫0∞‍σ(s)ds=-∫0t‍(t-s)α-1Γ(α)σ(s)ds-Γ(2α-1)λtα-1Γ(α)(Γ(2α-1)-λη2α-2)×∫0η‍(η-s)2α-2Γ(2α-1)σ(s)ds+Γ(2α-1)tα-1Γ(α)(Γ(2α-1)-λη2α-2)×∫0∞‍σ(s)ds=∫0∞‍G(t,s)σ(s)ds,
where G(t,s) is defined by (8).Remark 6.
In view of the assumption(H1), Green's function G(t,s) satisfies the properties:(1)
G(t,s)≥0,(2)
(15)G(t,s)1+tα-1≤Γ(2α-1)Γ(α)[Γ(2α-1)-λη2α-2]≜L.For the forthcoming analysis, we introduce a space
(16)X={u∈C(J,ℝ):supt∈J|u(t)|1+tα-1<+∞},
equipped with the norm
(17)∥u∥X=supt∈J|u(t)|1+tα-1.
Notice that X is a Banach space.
Define an operatorT:X→X as follows:
(18)Tu(t)=∫0∞‍G(t,s)f(s,u(s))ds.
Observe that problem (1) has a solution only if the operator T has a fixed point.Lemma 7.
If(H1), (H2) hold, then the operator T:X→X is completely continuous.Proof.
We divide the proof into several steps.
(i) The operatorT:X→X is uniformly bounded. Let Ω be any bounded subset of X; then there exists a constant L1 such that ∥u∥X≤L1. By (H2), we have
(19)∥Tu∥X=supt∈J∫0∞‍G(t,s)1+tα-1|f(s,u(s))|ds≤L∫0∞‍[a(s)+b(s)(1+sα-1)ρ|u(s)|ρ(1+sα-1)ρ]ds≤L(a*+b*L1ρ)<∞.
This shows that TΩ is uniformly bounded.
(ii)T:X→X is continuous. Take un,u∈X such that ∥un∥X<∞, ∥u∥X<∞, and un→u as n→∞. Then, by (H2), we have
(20)∫0∞‍G(t,s)1+tα-1f(s,un(s))ds≤L∫0∞‍[a(s)+b(s)|un(s)|ρ]ds≤La*+Lb*∥un∥Xρ<∞,
where L is defined by (15).
By the Lebesgue dominated convergence theorem and continuity off, we obtain
(21)limn→∞∫0∞‍G(t,s)1+tα-1f(s,un(s))ds=∫0∞‍G(t,s)1+tα-1f(s,u(s))ds.
Taking the limit n→∞, we get
(22)∥Tun-Tu∥X=supt∈J∫0∞‍G(t,s)1+tα-1×|f(s,un(s))-f(s,u(s))|ds⟶0.
Therefore, T is continuous.
(iii)T:X→X is equicontinuous. We consider two cases.
(a) LetI⊂J be any compact interval, and let t1,t2∈I be such that t1<t2. Let Ω be any bounded subset of X; then for any u∈Ω, we have
(23)|Tu(t2)1+t2α-1-Tu(t1)1+t1α-1|=|∫0∞‍(G(t2,s)1+t2α-1-G(t1,s)1+t1α-1)f(s,u(s))ds|≤∫0∞‍|G(t2,s)1+t2α-1-G(t1,s)1+t1α-1|(a(s)+b(s)(1+sα-1)ρ∥u∥Xρ)ds.
Since G(t,s) is continuous on J×J, we have that G(t,s)/(1+tα-1) is a uniformly continuous function on the compact set I×I. Moreover, for s≥t, we have that this function only depends on t; in consequence it is uniformly continuous on I×(J∖I). So we have that for all s∈J and t1,t2∈I, the following property holds.
For allε>0 there is δ(ε)>0 such that if |t1-t2|<δ, then |G(t2,s)/(1+t2α-1)-G(t1,s)/(1+t1α-1)|<ε. By this, together with (23), and the fact that
(24)∫0∞‍(a(s)+b(s)(1+sα-1)ρL1)ds<∞,
we can get that TΩ is equicontinuous on I.
(b) In fact, whent→∞, we have
(25)limt→∞G(t,s)1+tα-1=1Δ{λη2α-2-λ(η-s)2α-2,s≤η,λη2α-2,0≤η≤s.
From this, it is not difficult to verify that for any givenε>0, there exists a constant T′=T′(ε)>0 such that
(26)|G(t1,s)1+t1α-1-G(t2,s)1+t2α-1|<ε
for any t1,t2≥T′ and s∈J. Hence, T is equiconvergent at ∞.
Thus the conclusion of Theorem4 applies that T is relatively compact on J. So, T:X→X is completely continuous. This completes the proof.


## 4. Main Results

Theorem 8.
Assume that(H1) and (H2) with ρ=1 hold. If there exists r>0 such that
(27)r(1-Lb*)>La*
with L given by (15), then problem (1) has a solution u(t) satisfying
(28)0≤u(t)1+tα-1≤r,fort∈J.Proof.
LetU={u∈X,∥u∥X<r}. For u∈∂U, if there exist υ∈(0,1) such that u=υTu, then we have
(29)∥u∥X=supt∈J|υ(Tu)(t)1+tα-1|≤supt∈J∫0∞‍G(t,s)1+tα-1|f(s,u(s))|ds≤L∫0∞‍|a(s)+b(s)u(s)|ds≤La*+Lb*∥u∥X.
This implies that
(30)r(1-Lb*)≤La*,
which contradicts (27). By Lemma 7 and Theorem 3, we conclude that problem (1) has a solution u(t) satisfying
(31)0≤u(t)1+tα-1≤r,t∈J.
This completes the proof.In the next, we formulate existence results for the cases0<ρ<1 and ρ>1. We do not provide the proof of these results as it is similar to that of Theorem 8. For that, we denote (H2) with 0<ρ<1 and ρ>1, respectively, by (H3) and (H4).Theorem 9.
Let the assumptions(H1) and (H3) hold. Then problem (1) has a solution u(t) satisfying
(32)0≤u(t)1+tα-1≤r,t∈J,
where r>max{2La*,(2Lb*)1/(1-ρ)} with L given by (15).Theorem 10.
Suppose that(H1) and (H4) hold and that there exists 2La*≤r≤(2Lb*)1/(1-ρ) with L given by (15). Then problem (1) has a solution u(t) such that
(33)0≤u(t)1+tα-1≤r,t∈J.


## 5. Example

Example 1.
Withα=3/2, λ=1/2, and η=1, we consider the following boundary value problem:
(34)D3/2u(t)+|u(t)|+sinu(t)8(1+t1/2)(1+t)2+4(t+4)2=0,t∈[0,+∞),I1/2u(0)=0,D1/2u(+∞)=12I1/2u(1).
Clearly the condition (H1) holds as Γ(2α-1)=Γ(2)=1, λη2α-2=1/2. Letting a(t)=4/(t+4)2,  b(t)=1/(4(1+t1/2)(1+t)2), we find that
(35)f(t,u(t))=|u(t)|+sinu(t)8(1+t1/2)(1+t)2+4(t+4)2≤a(t)+b(t)|u(t)|,∫0+∞‍a(t)dt=1<+∞,∫0+∞‍(1+tα-1)b(t)dt=14<+∞.
This shows that (H2) holds true. Finally, fixing r>1/(π-1), it can easily be verified that the condition (27) is satisfied. Thus all the conditions of Theorem 8 are satisfied. Therefore, by Theorem 8, problem (27) has a solution u(t) such that
(36)0≤u(t)1+tα-1≤r.



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