# Solvability for a Coupled System of Fractional Integrodifferential Equations withm-Point Boundary Conditions on the Half-Line

Authors: Payam Nasertayoob; S. Mansour Vaezpour; Dumitru Baleanu
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

The aim of this paper is to study the solvability for a coupled system of fractional integrodifferential equations with multipoint fractional boundary value problems on the half-line. An example is given to demonstrate the validity of our assumptions.


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## Body

## 1. Introduction

The theory of derivatives and integrals of fractional order has undergone rapid development over the years and played a very important role in modern applied mathematical models of real processes arising in phenomena studied in physics, mechanics, engineering, and so on [1–3]. Recently, the existence of solutions for coupled systems involving fractional differential equations is one of the theoretical fields investigated by many authors [4–13].Very recently, Wang et al. [10] studied the existence of solutions for the following coupled system of nonlinear fractional differential equations by using Schauder’s fixed point theorem:
(1)Dpu(t)+f(t,v(t))=0,2<p<3,t∈J∶=[0,∞),Dqv(t)+g(t,u(t))=0,2<q<3,t∈J∶=[0,∞),u(0)=u′(0)=0,Dp-1u(∞)=∑i=1m-2‍βiu(ξi),v(0)=v′(0)=0,Dq-1v(∞)=∑i=1m-2‍γiv(ξi),
where f,g∈C(J×R,R), 0<ξ1<ξ2<⋯<ξm-2<∞, and Dp and Dq denote Riemann-Liouville fractional derivatives of order p and order q, respectively; also βi>0, γi>0 are such that 0<∑i=1m-2‍βiξip-1<Γ(p) and 0<∑i=1m-2‍γiξiq-1<Γ(q).Motivated by [10], in this paper, we consider a coupled system of nonlinear fractional integrodifferential equations on an unbounded domain and more general boundary conditions:
(2)Dαu(t)+f(t,v(t),Iαv(t))=0,t∈J=[0,∞),Dβv(t)+g(t,u(t),Iβu(t))=0,t∈J=[0,∞),u(0)=u′(0)=0,v(0)=v′(0)=0,Dα-1u(∞)=∑i=1m-2‍aiu(ξi)+∑i=1m-2‍biDα-1u(ξi),Dβ-1v(∞)=∑i=1m-2‍civ(ξi)+∑i=1m-2‍diDβ-1v(ξi),
where 2<α, β≤3, 0<ξ1<ξ2<⋯<ξm-2<∞, ai,bi,ci,di≥0 are real numbers, f,g∈C(J×R×R,R), and Ddenotes Riemann-Liouville fractional derivative. It is clear that boundary value problem (2) includes problem (1) as special case.Integrodifferential equations have become important in recent years as mathematical models of phenomena in both the physical and social sciences. In particular, some physical phenomena involving certain type of memory effects are represented by integrodifferential equations [14–18].However, to the best of our knowledge, no work has been reported on the existence results for coupled system of nonlinear fractional integrodifferential equations on an unbounded domain.The paper is organized as follows. In Section2, we recall some basic definitions, notations, and preliminary facts. Section 3 is devoted to the existence results for system of nonlinear fractional integrodifferential equations on an unbounded domain. In Section 4, an example is given to demonstrate the applicability of our results.


## 2. Preliminaries

In this section, we will first recall some basic definitions and lemmas which are used in what follows and can be found in [2, 19].Definition 1.
The Riemann-Liouville fractional integral of orderδ>0 of a function f:(0,∞)→R is given by
(3)Iδf(t)=1Γ(δ)∫0t‍(t-s)δ-1f(s)ds,
provided that the right-hand side is pointwise defined.Definition 2.
The Riemann-Liouville fractional derivative of orderδ>0 of a continuous function f:(0,∞)→R is given by
(4)Dδf(t)=1Γ(n-δ)(ddt)n∫0t‍(t-s)n-δ-1f(s)ds,
where n=[α]+1, provided that the right-hand side is pointwise defined.Remark 3.
The following properties are well known:(5)DδIδf(t)=f(t),δ>0,f(t)∈L1(0,∞),DαIδf(t)=Iδ-αf(t),δ>α>0,f(t)∈L1(0,∞),Dδtλ=Γ(λ+1)Γ(λ-δ+1)tλ-δ,λ>-1,t>0.Lemma 4.
Forδ>0, the equation Dδu(t)=0 is valid if and only if
(6)u(t)=c1tδ-1+c2tδ-2+⋯+cntδ-n,iiiiiiiiiiiiiiiiiici∈R,j=1,2,…,n,
where n is the smallest integer greater than or equal to δ.Lemma 5.
Assume thatDδu(t)∈L1(0,∞); then,
(7)IδDδu(t)=u(t)+c1tδ-1+c2tδ-2+⋯+cntδ-n,iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiici∈R,j=1,2,…,n,
where n is the smallest integer greater than or equal to δ.For anyδ>1 we can define the space
(8)Xδ={u∈C[0,∞):sup⁡t∈J|u(t)|1+tδ-1<∞},
equipped with the norm
(9)∥u∥δ=sup⁡t∈J|u(t)|1+tδ-1.
Clearly, (Xδ,∥u∥δ) is a Banach space [19]. For (u,v)∈Xα×Xβ we define
(10)∥(u,v)∥α,β=max⁡{∥u∥α,∥v∥β};
then, (Xα×Xβ,∥·∥α,β) is a Banach space.


## 3. Main Results

In this section, we prove the existence results for the boundary value problem (2). For convenience we use the following notation:
(11)Δα=Γ(α)-∑i=1m-2‍aiξiα-1-Γ(α)∑i=1m-2‍bi.
By replacing α,ai,bi with β,ci,di, respectively, we can define Δβ.Lemma 6.
Leth∈C[0,∞) and Δα>0; then, the unique solution of
(12)Dαu(t)+h(t)=0,2<α<3,t∈J=[0,∞),(13)u(0)=u′(0)=0,(14)Dα-1u(∞)=∑i=1m-2‍aiu(ξi)+∑i=1m-2‍biDα-1u(ξi)
is given by
(15)u(t)=∫0∞‍Gα(t,s)h(s)ds,
where Gα(t,s) is Green’s function given by
(16)Gα(t,s)=Kα(t,s)+tα-1Δα∑i=1m-2‍aiKα(ξi,s)+tα-1Δα∑i=1m-2‍biH(ξi,s),
with
(17)Kα(t,s)=1Γ(α){tα-1-(t-s)α-1,0≤s≤t<∞,tα-1,0≤t≤s<∞,(18)H(t,s)={0,0≤s≤t<∞,1,0≤t≤s<∞.Proof.
By Lemma5, the solution of (12) can be writen as
(19)u(t)=c1tα-1+c2tα-2+c3tα-3-Iαh(t).
Using the boundary conditions (13), we find that c2=c3=0 and
(20)Dα-1u(t)=c1Γ(α)-I1h(t).
Now considering the second boundary condition, we have
(21)c1=1Δα(∫0∞‍h(s)ds∑i=1m-2‍iiiiiiiii-∑i=1m-2‍aiΓ(α)∫0ξi‍(ξi-s)α-1h(s)dsiiiiiiiii-∑i=1m-2‍bi∫0ξi‍h(s)ds).
Therefore, the unique solution of the boundary value problem (12)–(14) is
(22)u(t)=tα-1Δα(∫0∞‍h(s)ds∑i=1m-2‍iiiiiiiiiii-∑i=1m-2‍aiΓ(α)∫0ξi‍(ξi-s)α-1h(s)dsiiiiiiiiiii-∑i=1m-2‍bi∫0ξi‍h(s)ds)-1Γ(α)∫0t‍(t-s)α-1h(s)ds=∫0∞‍Kα(t,s)h(s)ds+∑i=1m-2‍aitα-1Δα∫0∞‍ξiα-1Γ(α)h(s)ds+Γ(α)∑i=1m-2‍bitα-1Δα∫0∞‍h(s)Γ(α)ds-∑i=1m-2‍aitα-1Δα∫0ξi‍(ξi-s)α-1Γ(α)h(s)ds-Γ(α)∑i=1m-2‍bitα-1Δα∫0ξi‍h(s)Γ(α)ds=∫0∞‍Kα(t,s)h(s)ds+∑i=1m-2‍aitα-1Δα∫0∞‍Kα(ξi,s)h(s)ds+∑i=1m-2‍bitα-1Δα∫0∞‍H(ξi,s)h(s)ds=∫0∞‍Gα(t,s)h(s)ds,
where Gα(t,s), Kα(t,s), and H(t,s) are defined by (16), (17), and (18), respectively. The proof is complete.Now, we introduce the following function:(23)Gβ(t,s)=Kβ(t,s)+tβ-1Δβ∑i=1m-2‍ciKβ(ξi,s)+tβ-1Δβ∑i=1m-2‍diH(ξi,s),
where
(24)Kβ(t,s)=1Γ(β){tβ-1-(t-s)β-1,0≤s≤t<∞,tβ-1,0≤t≤s<∞.Remark 7.
From the definition ofGα(t,s) and Gβ(t,s), for any (s,t)∈[0,∞)×[0,∞), we have
(25)Gα(t,s)1+tα-1≤Q,Gβ(t,s)1+tβ-1≤Q,
where
(26)Q=max⁡{1Γ(α)+∑i=1m-2‍aiξm-2α-1Γ(α)Δα+∑i=1m-2‍biΔα,iiiiiiiiii1Γ(β)+∑i=1m-2‍ciξm-2β-1Γ(β)Δβ+∑i=1m-2‍diΔβ}.Let an operatorT:Xα×Xβ→Xα×Xβ be defined by
(27)T(u,v)=(T1(v),T2(u))=(∫0∞‍Gα(t,s)f(s,v(s),Iαv(s))ds,iiiiiiii∫0∞‍Gβ(t,s)g(s,u(s),Iβu(s))ds).
From the definition of operator T, the problem (2) has a solution if and only if the operator T has a fixed point.Theorem 8.
Assume the following.(H1)
There exist nonnegative functionsa(t),b(t),ϕ(t)∈C[0,∞) such that
(28)|f(t,x,y)|≤a(t)|x|+b(t)|y|+ϕ(t),∫0∞‍ϕ(t)dt<∞,Δα>0,∫0∞‍(1+tα-1)a(t)dt+1Γ(α)∫0∞‍[tαα+B(α,α)t2α-1]b(t)dt<1Q,
where B(α,α) is the beta-function.(H2)
There exist nonnegative functionsc(t),d(t),φ(t)∈C[0,∞) such that
(29)|g(t,x,y)|≤c(t)|x|+d(t)|y|+φ(t),∫0∞‍φ(t)dt<∞,Δβ>0,∫0∞‍(1+tβ-1)c(t)dt+1Γ(β)∫0∞‍[tββ+B(β,β)t2β-1]d(t)dt<1Q,
where B(β,β) is the beta-function.
Then, the system (2) has a solution.Proof.
Take(30)R>max⁡{(Q∫0∞‍ϕ(t)dt)∫0∞‍[tββ])-1iiiiiiiiii×(1-Q∫0∞‍(1+tα-1)a(t)dt-QΓ(α)iiiiiiiiiiiiiiiiii×∫0∞‍[tαα+B(α,α)t2α-1]b(t)dt)-1,iiiiiiiiii(Q∫0∞‍φ(t)dt)iiiiiiiiii×(1-Q∫0∞‍(1+tβ-1)c(t)dt-QΓ(β)iiiiiiiiiiiiiiiiii×∫0∞‍[tββ+B(β,β)t2β-1]d(t)dt)-1},
and define a ball
(31)BR={(u,v)∈Xα×Xβ:∥(u,v)∥α,β≤R}.
At the first step, we prove that the operator T transforms the ball BR into itself. For any (u,v)∈BR we have
(32)∥T1(v)∥α=sup⁡t∈J11+tα-1|∫0∞‍Gα(t,s)f(s,v(s),Iαv(s))ds|≤sup⁡t∈J11+tα-1×∫0∞‍Gα(t,s)[a(s)|v(s)|+b(s)|Iαv(s)|+ϕ(s)]ds≤Q∫0∞‍a(s)|v(s)|ds+QΓ(α)×∫0∞‍b(s)[∫0s‍(s-τ)α-1v(τ)dτ]ds+Q∫0∞‍ϕ(s)ds≤Q∥v∥α∫0∞‍(1+sα-1)a(s)ds+Q∥v∥αΓ(α)∫0∞‍b(s)[∫0s‍(1+τα-1)(s-τ)α-1dτ]ds+Q∫0∞‍ϕ(s)ds≤Q∥v∥α∫0∞‍(1+sα-1)a(s)ds+Q∥v∥αΓ(α)∫0∞‍b(s)[sα∫01‍(1-θ)α-1+s2α-1iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii×∫01‍θα-1(1-θ)α-1dθ]ds+Q∫0∞‍ϕ(s)ds≤Q∥v∥α∫0∞‍(1+sα-1)a(s)ds+Q∥v∥αΓ(α)∫0∞‍b(s)[sαα+B(α,α)s2α-1]ds+Q∫0∞‍ϕ(s)ds<R.
In a similar way, we can get
(33)∥T2(u)∥β≤Q∥u∥β∫0∞‍(1+sβ-1)c(s)ds+Q∥u∥βΓ(β)∫0∞‍d(s)[sββ+B(β,β)s2β-1]ds+Q∫0∞‍φ(t)ds<R.
Hence, ∥T(u,v)∥α,β≤R and this shows that TBR⊂BR.
Next, we show thatT:BR→BR is completely continuous. First, Let (un,vn)→(u,v) as n→∞ in BR. From (32) we have
(34)∫0∞‍|f(s,v(s),Iαv(s))|ds<RQ.
Then, by the Lebsegue dominated convergence theorem and continuity of f, we obtain
(35)∫0∞‍f(s,vn(s),Iαvn(s))ds⟶∫0∞‍f(s,v(s),Iαv(s))ds,
as n→∞. Therefore, by Remark 7, we have
(36)∥T1(vn)-T1(v)∥α≤Q|∫0∞‍f(s,vn(s),Iαvn(s))dsiiiiiiiiiiiii-∫0∞‍f(s,v(s),Iαv(s))ds|⟶0,
as n→∞. Similar process can be repeated for T2; thus, operator T is continuous.
Now, we show thatT:BR→BR is equicontinuous operator. Let L>0 and t1,t2∈[L,∞); without loss of generality, we may assume that t1<t2. Since Kα(ξi,s)≤ξm-2α-1/Γ(α) and H(ξi,s)≤1, for any s>0 and i=1,2,…,m-2, we have
(37)|T1(u)(t2)1+t2α-1-T1(u)(t1)1+t1α-1|≤∫0∞‍|(Gα(t2,s)1+t2α-1-Gα(t1,s)1+t1α-1)iiiiiiiiiiiiiiii(Gα(t2,s)1+t2α-1)×f(s,v(s),Iαv(s))|ds≤∫0∞‍|(Kα(t2,s)1+t2α-1-Kα(t1,s)1+t1α-1)iiiiiiiiiiiiiiii(Gα(t2,s)1+t2α-1)×f(s,v(s),Iαv(s))|ds+[∑i=1m-2‍aiξm-2α-1Γ(α)Δα+∑i=1m-2‍biΔα]×|t2α-11+t2α-1-t1α-11+t1α-1|×∫0∞‍|f(s,v(s),Iαv(s))|ds.
In view of (37), by the similar process used in [20], we can easily prove that operator T1 is equicontinuous. Similar process can be repeated for T2; thus, T is equicontinuous. On the other hand, TBR is uniformly bounded as TBR⊂BR. Therefore, T is completely continuous operator. Hence, by Schauder fixed point theorem the boundary value problem (2) has at least one solution in BR.


## 4. An Example

Consider the following boundary value problem on unbounded domain:(38)D2.25u(t)=cos⁡tsin(|v(t)|)(1+t1.25)(10+t)2+|∫0∞‍(t-s)1.25e-5tv(s)ds|Γ(2.25)(t2.25+0.248t3.5)+t2e-t1+t2,D2.5v(t)=sintarctan(|u(t)|)(1+t1.5)(8+t)2+|∫0∞‍(t-s)1.5e-4tu(s)ds|Γ(2.5)(t2.5+0.183t4)+t4e-t1+t4,u(0)=u′(0)=0,v(0)=v′(0)=0,D1.25u(∞)=12u(14)+14u(1)+110D1.25u(14)+310D1.25u(1),D1.5v(∞)=15v(14)+110v(1)+311D1.5v(14)+17D1.5v(1).
Here t∈[0,∞), α=2.25, β=2.5, ξ1=1/4, ξ2=1, a1=1/2, a2=1/4, b1=1/10, b2=3/10, c1=1/5, c2=1/10, d1=3/11, and d2=1/7. We have
(39)f(t,x,y)=cos⁡tsin(|x|)(1+t1.25)(10+t)2+e-5t|y|(t2.25+0.248t3.5)+t2e-t1+t2,g(t,x,y)=sintarctan(|x|)(1+t1.5)(8+t)2+e-4t|y|(t2.5+0.183t4)+t4e-t1+t4.
For
(40)a(t)=1(1+t1.25)(10+t)2,b(t)=e-5t(t2.25+0.248t3.5),ϕ(t)=φ(t)=e-t,c(t)=1(1+t1.5)(8+t)2,d(t)=e-4t(t2.5+0.183t4),
by direct calculation we obtain Δα=0.342, Δβ=0.645, Q=3.98, and
(41)∫0∞‍(1+t1.25)a(t)dt+1Γ(2.25)×∫0∞‍[t2.252.25+B(2.25,2.25)t3.5]b(t)dt=0.17480<1Q=0.25125,∫0∞‍(1+t1.5)c(t)dt+1Γ(2.5)×∫0∞‍[t2.52.5+B(2.5,2.5)t4]d(t)dt=0.20000<1Q=0.25125.
Thus all the conditions of Theorem 8 are satisfied and the problem (38) has at least one solution.


## 5. Conclusion

In the current paper, we have studied the existence results for a coupled system of nonlinear fractional integrodifferential equations withm-point fractional boundary conditions on an unbounded domain. The result obtained in this paper is based on Schauder’s fixed point theorem. In order to show the validity of the assumptions made in our result, we also include an illustrative example.



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