# Solvability of Some Two-Point Fractional Boundary Value Problems under Barrier Strip Conditions

Authors: Limei He; Xiaoyu Dong; Zhanbing Bai; Bo Chen
Journal: Journal of Function Spaces
 (2017)

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## Abstract

Topological techniques are used to establish existence results for a class of fractional differential equationsDαx(t)=f(t,x(t),Dα-1x(t)), with one of the following boundary value conditions: x(0)=A and Dα-1x(1)=B or Dα-1x(0)=A and x(1)=B, where 1<α≤2 is a real number, Dαx(t) is the conformable fractional derivative, and f:[0,1]×R2→R is continuous. The main conditions on the nonlinear term f are sign conditions (i.e., the barrier strip conditions). The topological arguments are based on the topological transversality theorem.


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## Body

## 1. Introduction

Recently, boundary value problems of nonlinear fractional differential equations have been addressed by several researchers. Fractional differential equations arise in many engineering and scientific disciplines as the mathematical modeling of systems and processes in the fields of physics, chemistry, control theory, biology, economics, blood flow phenomena, signal and image processing, biophysics, aerodynamics, fitting of experimental data, and so forth. For example, in 2006, by using the fixed point theorem in cones, the existence and multiplicity of solutions to the following problems are obtained [1]:(1)CD0+αut=ft,ut,u0+u′0=0,u1+u′1=0,where 1<α≤2 is a real number, CD0+αu(t) is Caputo fractional derivative, and f:[0,1]×[0,∞)→[0,∞) is a continuous function. In 2010, by the use of the Lipschitz condition and the compression mapping principle, the existence of solutions of the following problem is obtained [2]:(2)D0+αut+ft,ut=0,u0=0,βuη=u1,where 1<α≤2 is a real number, 0<βηα-1<1,0<η<1, D0+αu(t) is the standard Riemann-Liouville derivative, and f:[0,1]×[0,∞)→[0,∞) is a continuous function. We refer the readers to other contributions in this line ([1–30], etc.).The technique of barrier strips has been used by Kelevedjiev and Tersian in [31, 32] to study the solvability of integer order BVPs. Recently, we use it to study the fractional differential equation with Dirichlet boundary value condition [20]. In this paper, by using the topological transversality theorem, we consider the following equation:(3)Dαxt=ft,xt,Dα-1xt,with one of the following boundary value conditions:(4)x0=A,Dα-1x1=B;or(5)Dα-1x0=A,x1=B,where 1<α⩽2 is a real number, Dαx(t) is the conformable fractional order derivative, and f:[0,1]×R2→R is a continuous function. The existence results of solutions to the problem are obtained under f which satisfies some barrier strip conditions.


## 2. Definitions and Lemmas about Fractional Calculus

Letα∈(n,n+1).Definition 1 (see [14]).
Supposeu is n-order differentiable for t≥0; the α-order fractional derivative of u is defined as(6)Dαut=limε→0⁡unt+εtn+1-α-untε,provided the limits of the right side exist.Lemma 2 (see [12]).
Lett>0. Function u(t) is α-order differentiable if and only if u is (n+1)-order differentiable; moreover, the following relation holds:(7)Dαut=tn+1-αun+1t.Definition 3 (see [14]).
Theα-order fractional integral is defined as(8)J0+αut=In+1tα-n-1ut=1n!∫0tt-snsα-n-1usds,where In+1 is the (n+1)-order integral.Lemma 4 (see [20]).
Fori=0,1,…,n, there holds(9)Dα-iJ0+αut=tn+1-αDn+1-iIn+1tα-n-1ut=tn+1-αIitα-n-1ut.Lemma 5 (see [14]).
Leta≥0,f:[a,b]→R be continuous on [a,b] and α-order differentiable on (a,b). Then, there exists d∈(a,b) such that(10)fb-fa=Dαfdbα-aαα.


## 3. Topological Preliminaries and a New Function Space

We begin with a brief review of the topological results to be used in this paper; see [33]. Let C be a convex subset of a Banach space E, X a metric space, and F:X→C a continuous map. We say that F is compact if F(X) is contained in a compact subset of C. F is completely continuous if it maps bounded subsets in X into compact subsets of C. A homotopy {Ht:X→C}0≤t≤1 is said to be compact provided that H:X×[0,1]→C given by H(x,t)=Ht(x) for (x,t) in X×[0,1] is compact.LetU⊂C be open in C. A compact map F:U¯→C is called admissible if it is fixed point free on the boundary, ∂U, of U. The set of all such maps will be denoted by ℵ∂U(U¯,C).Definition 6.
A mapF in ℵ∂U(U¯,C) is inessential if there is a fixed point free compact map G:U¯→C such that G∂U=F∂U. A map F in ℵ∂U(U¯,C) which is not inessential is called essential.Lemma 7 (see [33]).
Letp be an arbitrary value in U and F be in ℵ∂U(U¯,C) and be the constant map  F(x)≡p for x in U¯. Then F is essential.Definition 8.
Two mapsF and G in ℵ∂U(U¯,C) are called homotopic (F~G) if there is a compact homotopy Ht:U¯→C such that F=H0 and G=H1 and Ht is admissible for each t in [0,1].The following theorem called topological transversality theorem which is very important to our results.Lemma 9 (see [33]).
LetF and G be in ℵ∂U(U¯,C) and be homotopic maps, F~G. Then one of these maps is essential if and only if the other is.Now, we construct a function space. Givenα>0, let α∈(n,n+1]. Define(11)Cα0,1=0,1,…,n,xt∈C0,1u∣ut=J0+αxt+Cntn+⋯+C1t+C0,Ci∈R,i=0,1,…,n,xt∈C0,1.By the linearity of integral operatorJ0+α, Cα[0,1] is a linear space. For u∈Cα[0,1], according to Lemma 4, there are Dα-iu(t)∈C[0,1],i=0,1,…,n. Define(12)uα=Dαu0+Dα-1u0+⋯+Dα-nu0+u0,where u0=maxt∈[0,1]ut. Next, we prove ·α is a norm in the linear space Cα[0,1], and Cα[0,1] is complete with this norm.Theorem 10.
(
C
α
[
0,1
]
,
·
α
) is a Banach space.Proof.
It is easy to verify that·α satisfies the norm axioms. The following proof is about the completeness of Cα[0,1].
Let{uk}k=1∞⊂Cα[0,1] be a Cauchy sequence in Cα[0,1]:(13)ukt=J0+αykt+Ck,ntn+⋯+Ck,1t+Ck,0,where yk(t)∈C[0,1] and Ci,j∈R(i=1,2,…,k;j=0,1,…,n). Then(14)ukt-umt=J0+αykt-ymt+Ck,n-Cm,ntn+⋯+Ck,1-Cm,1t+Ck,0-Cm,0.Because {uk} is a Cauchy sequence of the Banach space Cα[0,1],(15)uk-umα=Dαuk-um0+Dα-1uk-um0+⋯+Dα-nuk-um0+uk-um0⟶0,whenk,m⟶∞.Thus every term of the above formula converges to 0. By(16)Dαuk-um0=yk-ym0⟶0,ask,m⟶∞,we know {yk}k=1∞ is a Cauchy sequence in C[0,1]. By the completeness of C[0,1], there exists y∈C[0,1] such that yk→y. The second term is(17)Dα-1uk-um0=Dα-1J0+αyk-ym+Ck,n-Cm,nDα-1tn0=tn+1-α∫0tsα-n-1yks-ymsds+Ck,n-Cm,nn!·tn+1-α0⟶0,ask,m⟶∞.Taking into account that {yk} is a Cauchy sequence in C[0,1], there is(18)Ck,n-Cm,nn!·tn+1-α0≤tn+1-α∫0tsα-n-1yks-ymsds+Ck,n-Cm,nn!·tn+1-α0+tn+1-α∫0tsα-n-1yks-ymsds0⟶0.That is to say, {Ck,n}k=1∞ is a Cauchy sequence in R. Denote its limit as Cn. Continue this process, and we can prove that sequences {Ck,i}k=1∞,i=0,1,…,n, all are Cauchy sequences. Denote their limits as Cn-1,Cn-2,…,C1,C0, respectively.
Letu(t)=J0+αy(t)+Cntn+⋯+C1t+C0, and then u(t)∈Cα[0,1] and uk-uα→0,(k→∞). The completeness of Cα[0,1] is proved.


## 4. Main Results

Denote(19)G1=max⁡L1,L4,G2=A+G1α-1,G3=max0,1×-G2,G2×-G1,G1⁡ft,x,y.Theorem 11.
Supposef:[0,1]×R2→R is continuous. If there exist four constants Li,i=1,2,3,4, L2>L1≥B and L3<L4≤B, such that(20)ft,x,y≥0,fort,x,y∈0,1×R×L1,L2,ft,x,y≤0,fort,x,y∈0,1×R×L3,L4,then Problem (3)-(4) has a solution x∈Cα[0,1] such that(21)xt≤G2,Dα-1xt≤G1,Dαxt≤G3,fort∈0,1.Proof.
Consider the family of boundary value problems(22)Dαxt=λft,xt,Dα-1xt,t∈0,1,λ∈0,1,(23)x0=A,Dα-1x1=B.Denote(24)CBα0,1=x∈Cα0,1∣x0=A,Dα-1x1=B.Define an operator Tλ:CBα[0,1]→CBα[0,1] as(25)Tλxt=λ∫01Gt,sfs,xs,Dα-1xsds+ψt,where(26)Gt,s=s·sα-2,0≤s≤t≤1;t·sα-2,0≤t≤s≤1,ψt=A+Bt.Then standard arguments yield that Tλ:CBα[0,1]→CBα[0,1] is completely continuous. Clearly, the fixed point of operator Tλ is the solution for Problem (22)-(23).
GivenM>0, let(27)U=u∈CBα0,1∣uα<M+A+2B+1.We will prove that, for some positive number M, the map Tλ:U¯→CBα[0,1] is admissible.
We claim that all the possible solutions of Problem (22)-(23) on Cα[0,1] have a priori bound without dependence on λ∈[0,1]. Assume that the sets(28)E0=t∈0,1∣L1<Dα-1xt≤L2,E1=t∈0,1∣L3≤Dα-1xt<L4are not empty.
Chooset0∈E0. Assume there exists t0′∈(t0,1] such that(29)Dα-1xt0′<Dα-1xt0.Taking into account the fact that Dα-1x(t) is continuous, we can take t0′∈(t0,1]∩E0. The assumption on f yields that Dαx(t)=λf(t,x(t) and Dα-1x(t))≥0 for t∈E0. Consequently, Dα-1x(t) is increasing on E0. Thus(30)Dα-1xt0′≥Dα-1xt0,which is a contradiction to (29). So, there holds(31)Dα-1xt≥Dα-1xt0,fort∈t0,1,and, in particular,(32)Dα-1x1≥Dα-1xt0>L1≥B,a contradiction to (23). This shows that E0=⌀. The similar arguments can show that E1=⌀.
ByDα-1x(1)=B and Dα-1x(t)∈C[0,1], there holds L4≤Dα-1x(t)≤L1 for t∈[0,1], and then |Dα-1x(t)|≤G1. On the other hand, by Lemma 5, for each t∈(0,1], there exists d∈(0,t) such that(33)xt-x0=Dα-1xd·1α-1tα-1.Thus, |x(t)|≤G2. Finally (22) together with the continuity of f and a priori estimations of Dα-1x(t) and x(t) show that |Dαx(t)|≤G3.
SetM=G1+G2+G3, and one has(34)xtα=Dαxt0+Dα-1xt0+xt0<M<∞.
The above results indicate thatTλ is completely continuous and has no fixed point on ∂U and Tλ is admissible on U¯ for each λ∈[0,1]. By Lemma 7, the constant map T0(x)≡ψ(t) for x∈U¯ is essential. Combined with Lemma 9, the topological transversality theorem, T1, is essential. So Problem (3)-(4) has a solution x∈U¯. Moreover, the above arguments show that the solution x satisfies(35)xt≤G2,Dα-1xt≤G1,Dαxt≤G3,fort∈0,1.The proof is completed.The next theorem can be proved by similar arguments.Theorem 12.
Letf:[0,1]×R2→R be continuous. Suppose there exist four constants Li, i=1,2,3,4, such that L2>L1≥A and L3<L4≤A:(36)ft,x,y≤0,fort,x,y∈0,1×R×L1,L2,ft,x,y≥0,fort,x,y∈0,1×R×L3,L4.Then Problem (3)-(4) has a solution x∈Cα[0,1] such that(37)xt≤G2,Dα-1xt≤G1,Dαxt≤G3,fort∈0,1.Example 13.
Consider the following fractional BVPs:(38)D3/2xt=sin⁡xt+D1/2xt5-52D1/2xt4+32D1/2xt3,x0=0,D1/2x1=1.Let L1=2, L2=4, L3=-2, and L4=-1, and then the simple computation show that G1=4, G2=8, G3=1.08252, and(39)ft,x,y≥0,fort,x,y∈0,1×R×2,4,ft,x,y≤0,fort,x,y∈0,1×R×-2,-1.By the use of Theorem 11, Problem (38) has a solution x∈C3/2[0,1] such that(40)xt≤8,D1/2xt≤4,D3/2xt≤1.08252,fort∈0,1.


## 5. Conclusion

By the use of the topological transversality theorem, some existence results for a class of fractional differential equations with certain boundary value conditions are obtained. The main condition is sign condition which is easy to be satisfied and checked.



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