# Solvability of Some Fractional Boundary Value Problems with a Convection Term

Authors: Yongfang Wei; Zhanbing Bai
Journal: Discrete Dynamics in Nature and Society
 (2019)

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## Abstract

This paper is devoted to the research of some Caputo’s fractional derivative boundary value problems with a convection term. By the use of some fixed-point theorems and the properties of Green function, the existence results of at least one or triple positive solutions are presented. Finally, two examples are given to illustrate the main results.


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## Body

## 1. Introduction

Fractional differential equations (FDEs) present new models for many applications in physics, biomathematics, environmental issues, control theory, image processing, chemistry, mechanics, and so on [1–17]. Recently, researchers focus on studying various aspects of fractional differential equations, such as stability analysis, existence, multiplicity, and uniqueness of solutions [1–40]. Among all these topics, the existence and multiplicity results of positive solutions represent a topic of high interest in fractional calculus.Some authors studied the existence and uniqueness of solutions for fractional differential equations with Caputo or Riemann-Liouville derivatives based on the Banach contraction principle and investigate the stability results for various fractional problems [4, 5, 16, 17]. Others studied the existence and multiplicity results of positive solutions or the iterative scheme. By the use of the Krasnoesel-skii’s fixed-point theorem, Zhang [34] obtained some existence results of positive solutions of following problem without the convection term: (1)CD0+αyx=fx,yx,0<x<1,y0+y′0=0,y1+y′1=0.Wang and Liu [29] deduced the Green function and some interesting properties for the Dirichlet BVPs (2)-RLD0+αyx+byx=ft,yt,x∈0,1,y0=y1=0,where RLD0+α is the Riemann-Liouville (R-L) fractional derivative, α∈(1,2), and b>0. And they established an iterative scheme to approximate the unique positive solution under the singular conditions.Meng and Stynes [23] considered the following linear two-point fractional differential equation BVPs with general Robin type boundary condition: (3)-CD0+αyx+by′x=fx,x∈0,1,y0-β0y′0=γ0,y1+β1y′1=γ1,where CD0+α denotes the Caputo derivative, α∈(1,2], b,β0,β1 are constant, and f∈C[0,1]. Meng used two parameter Mittag-Leffler functions to establish explicitly Green’s function for the problems. They obtained the nonnegativity of Green’s function.This paper is devoted to the research of the solvability of the following nonlinear fractional BVPs:(4)-CD0+αyx+by′x=fx,yx,x∈0,1,(5)y0-β0y′0=0,y1+β1y′1=0,where 1<α≤2 and b,β0,β1∈R are constants. CD0+α is the Caputo’s fractional derivative. Compared to the existing literature, the interesting point here is that the convection term is involved in the study of the solvability of fractional differential boundary value problems. By applying some fixed-point theorems, some existence and multiplicity results of positive solutions are given. Some examples are presented in last section to illustrate the main theorems.In the sequel, the following conditions will be used:(H1) f∈C([0,1]×[0,∞),[0,∞));(H2) the constants β1≥0,β0≥-F2(1)+F1(1)Fα(1)/Fα-1(1). The function Fc(x) is defined in Section 2.


## 2. Background Material and Definitions

In order to solve the BVPs (4), (5), the following definitions and lemmas are needed.Definition 1.
Define the two-parameter Mittag-Leffler function by(6)Eτ,υx≔∑k=0∞xkΓτk+υ,forτ>0,υ≥0,x∈R.Definition 2.
Assumeg∈Cm-1[0,1] and g(m-1)(x)∈AC[0,1]. The Caputo fractional derivative of order α∈(m-1,m] is defined as(7)CD0+αgx≔1Γm-α∫0xx-tm-α-1gmtdt,for0<x≤1.In particularly, for g∈C2[0,1](8)limα→2⁡CD0+αgx=g′′x,foreachx∈0,1.From the above equation, forx≥0 there is Eτ,υ(x)>0 and(9)Eα-1,1x≔xEα-1,αx+1,forallx∈R.Remark 3.
For simplicity, let(10)Fcx≔xc-1Eα-1,cbxα-1,forc≥0,x≥0,the particular,(11)F1x=bFαx+1.Lemma 4 (see [23]).
Suppose1<α≤2 and the conditions (H1) and (H2) hold. For h∈C[0,1], the problem(12)CD0+αyx+hx=0,1<x<1,(13)y0-β0y′0=0,y1+β1y′1=0,has a unique solution(14)yx=∫01Gx,thtdt,where G(x,t) is Green’s function (15)Gx,t=σxFα1-t+β1Fα-11-t-Fαx-t,for0≤t≤x;σxFα1-t+β1Fα-11-t,forx<t≤1,σx=β0+F2xβ0+β1F11+F21,and Fc(x) is defined by Remark 3.By the use of some interesting properties of the Mittag-Leffler function, Meng and Stynes proved the following result.Lemma 5 (see [23]).
Supposeβ1≥0. Then Green’s function G(x,t)≥0 for x,t∈[0,1] if and only if(16)β0≥-F21+F11Fα1Fα-11.We pointed out here that this lemma comes from Theorem 5.1 and Remark 5.2 of Ref. [23] with some equivalent changes.The following theorems are fundamental for proving the main results.Lemma 6 (see [41]).
LetE be a Banach space, K⊆E a cone, and Ω1,Ω2 two bounded sets of E with 0∈Ω¯1⊂Ω2. Suppose that A:K∩(Ω¯2\Ω2)→K is completely continuous such that either (R1)Ax≤x,x∈K∩∂Ω1 and Ax≥x,x∈K∩∂Ω2, or
( R 2 )   A x ≥ x , x ∈ K ∩ ∂ Ω 1 and Ax≤x,x∈K∩∂Ω2 holds.
Then the operatorA has a fixed point in K∩(Ω¯2\Ω1).Lemma 7 (see [42]).
LetP be a cone in a real Banach space E, Pc={u∈P∣u≤c}, θ is a nonnegative continuous concave functional on P such that θ(u)≤u, for u∈Pc¯, and P(θ,b,d)={u∈P∣b≤θ(u),u≤d}. Suppose A:Pc¯→Pc¯ is completely continuous and there exist four constants 0<a<b<d≤c satisfying
( C 1 )   { u ∈ P ( θ , b , d ) ∣ θ ( u ) > b } ≠ ∅ and θ(Au)>b for u∈P(θ,b,d);
( C 2 )   A u < a for u≤a;
( C 3 )   θ ( A u ) > b for u∈P(θ,b,d) with Au>b.
Then the operatorA has three fixed-points u1,u2,u3 such that(17)u1<a,b<θu2,a<u3withθu3<b.Remark 8.
Specially, ifd=c, then condition (C1) of Lemma 7 implies condition (C3).


## 3. Existence and Multiplicity

In this section, by applying Lemma6 and Lemma 7, some solvability results for BVPs (4), (5) will be obtained.LetE=C[0,1] be a Banach space; u=max0≤t≤1|u(t)|. The cone P⊂E is defined as(18)P=u∈E∣ut≥0,t∈0,1. Let the concave functional θ be defined by(19)θu=min1/4≤t≤3/4ut. Define the operator T:P→E by(20)Tyx≔∫01Gx,tft,ytdt,for0≤x≤1. It is clear that y(x) is a positive solution of BVPs (4), (5) equivalent to that y∈P is a fixed-point of T.Lemma 9.
The operatorT:P→P is completely continuous.Proof.
Taking into account that the functionsG(x,t) and f(x,y) are all continuous and nonnegative, the operator T:P→E is continuous and nonnegative, i.e., T:P→P. Suppose Ω⊂P is a bounded set, and for all y∈Ω there holds y≤M1 for some M1>0. Let M¯=max0≤x≤1,0≤y≤M1fx,y+1; then, for y∈Ω, there is (21)Tyx≤∫01Gx,tft,ytdt≤M¯∫01max0≤x≤1Gx,tdt.Thus, the set T(Ω) is bounded.
By the fact that the Green functionG(x,t) is continuous on [0,1]×[0,1], one has the fact that it is uniformly continuous. Therefor, for given ϵ>0, there exists δ>0 such that for each y∈Ω,x1,x2∈[0,1],x1<x2, and x2-x1<δ; there holds Gx2,t-Gx1,t<ϵ/M¯. Thus (22)Tyx2-Tyx1≤∫01Gx2,t-Gx1,tft,ytdt≤∫01ϵM¯·M¯dt=ϵ.Hence, the set T(Ω) is equicontinuous. Thus, using the Arzela-Ascoli theorem, we claim that T:P→P is a completely continuous operator.Let(23)M=∫01max0≤x≤1Gx,tdt-1;N=∫1/43/4min1/4≤x≤3/4Gx,tdt-1.Theorem 10.
Assume conditions(H1) and (H2) hold. If there exist two different positive constants 0<r2<r1 such that
(A1) f ( x , y ) ≤ M r 1, for (x,y)∈[0,1]×[0,r1];
(A2) f ( x , y ) ≥ N r 2, for (x,y)∈[1/4,3/4]×[0,r2],
then the BVPs (4), (5) have one positive solution y such that r2≤y≤r1.Proof.
Define two open sets(24)Ω1≔y∈P∣y<r1,Ω2≔y∈P∣y<r2.
Fory∈∂Ω1, there is 0≤y(x)≤r1 for x∈[0,1]. The condition (A1) yields that (25)Ty=max0≤x≤1∫01Gx,tft,ytdt≤Mr1∫01max0≤x≤1Gx,tdt=r1=y.So(26)Ty≤y,fory∈∂Ω1.
Fory∈∂Ω2, there is 0≤y(x)≤r2 for x∈[0,1]. According to (A2), for x∈[1/4,3/4], (27)Tyx=∫01Gx,tft,ytdt≥∫01min1/4≤x≤3/4Gx,tft,ytdt≥∫1/43/4min1/4≤x≤3/4Gx,tft,ytdt≥Nr2∫1/43/4min1/4≤x≤3/4Gx,tdt=r2=y. So (28)Ty≥y,fory∈∂Ω2.
And the proof also holds when0<r2<r1. Therefore, by Lemma 6, the proof is complete.Theorem 11.
Assume conditions(H1) and (H2) hold. If there exist constants 0<a<b<c such that
(B1) f ( x , y ) < M a, for (x,y)∈[0,1]×[0,a];
(B2) f ( x , y ) ≥ N b, for (x,y)∈[1/4,3/4]×[b,c];
(B3) f ( x , y ) ≤ M c, for (x,y)∈[0,1]×[0,c],
then the BVPs (4), (5) have three positive solutions y1,y2,y3 with (29)max0≤x≤1y1x<a,b<min1/4≤x≤3/4y2x<max0≤x≤1y2x≤c,a<max0≤x≤1y3x≤c,min1/4≤x≤3/4y3x<b.Proof.
We just need to prove that all the conditions of Lemma7 hold.
Fory∈Pc¯={y∈P∣y≤c}, there is y≤c. Assumption (B3) shows f(x,y(x))≤Mc for 0≤x≤1. Taking into account the definition of M, there is (30)Ty=max0≤x≤1∫01Gx,tft,ytdt≤∫01max0≤x≤1Gx,tft,ytdt≤Mc∫01max0≤x≤1Gx,tdt≤c.Hence T:Pc¯→Pc¯. Similarly, if y∈Pa¯, then assumption (B1) shows that f(x,y(x))<Ma for 0≤x≤1. Therefore, (C2) of Lemma 7 holds.
For0≤x≤1, choose(31)yx=c+b2,0≤x≤1. It is clear that y(x)∈P(θ,b,c), θ(y)=θ((b+c)/2)>b. Thus the set y∈P(θ,b,c)∣θ(y)>b is nonempty. If y∈P(θ,b,c), then b≤y(x)≤c for 1/4≤x≤3/4. And assumption (B2) shows f(x,y(x))≥Nb for 1/4≤x≤3/4. So (32)θTy=min1/4≤x≤3/4Tyx≥∫01min1/4≤x≤3/4Gx,tft,ytdt>Nb∫1/43/4min1/4≤x≤3/4Gx,tdt=b.That is to say that θ(Ty)>b for y∈Pθ,b,c. I.e., the condition (C1) of Lemma 7 holds. By Remark 8, one has the fact that the condition (C3) holds, too.
Then by Lemma7, the BVPs (4), (5) have three positive solutions y1,y2,y3 such that (33)max0≤x≤1y1x<a,b<min1/4≤x≤3/4y2x<max0≤x≤1y2x≤c,a<max0≤x≤1y3x≤c,min1/4≤x≤3/4y3x<b. The theorem is proven.


## 4. Two Examples

Now, we present two examples to check the main results. Readers can easily find that the following problem cannot be solved with existing literature. Chooseα=3/2,b=1,β1=1, and(34)β0=1≥-F21+F11Fα1Fα-11=0.56.Example 1.
Consider the following problem:(35)-CD0+3/2yx+y′x=yx-x22+1,x∈0,1,(36)y0-y′0=0,y1+y′1=0.Clearly, conditions (H1) and (H2) are satisfied. A simple computation shows M≈1.6,N≈2.8. Let r1=1,r2=1/4, and there are(37)fx,y=y-x22+1≤1.5<Mr1=1.6,forx,y∈0,1×0,1,fx,y=y-x22+1≥1.1>Nr2=0.7,forx,y∈14,34×0,14.By Theorem 10, the BVPs (35), (36) have one positive solution y and 1/4≤y≤1.Example 2.
Consider the following fractional problem with the convection term:(38)-CD0+3/2yx+y′x=fx,yx,x∈0,1,(39)y0-y′0=0,y1+y′1=0,where (40)fx,y=y2+14x2,fory≤1;12y+4x2+1,fory≥1. Clearly, conditions (H1) and (H2) hold. A simple computation shows that M≈1.6,N≈2.8. Choose a=1/2,b=1,c=4, and there hold (41)fx,y=y2+14x2≤0.5<Ma=0.8,forx,y∈0,1×0,12,fx,y=12y+4x2+1≥3.06>Nb=2.8,forx,y∈14,34×1,4,fx,y≤1.25+4=5.25<Nc=6.4,forx,y∈0,1×0,4. By Theorem 11, the BVPs (38), (39) have three positive solutions y1,y2,y3 such that (42)max0≤x≤1y1x<12,1<min1/4≤x≤3/4y2x<max0≤x≤1y2x≤4,12<max0≤x≤1y3x≤4,min1/4≤x≤3/4y3x<1.



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