# Monotone and Concave Positive Solutions to a Boundary Value Problem for Higher-Order Fractional Differential Equation

Authors: Jinhua Wang; Hongjun Xiang; Yuling Zhao
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We consider boundary value problem for nonlinear fractional differential equationD0+αu(t)+f(t,u(t))=0,0<t<1,n-1<α≤n,n>3,u(0)=u′(1)=u′′(0)=⋯=u(n-1)(0)=0, where D0+α denotes the Caputo fractional derivative. By using fixed point theorem, we obtain some new results for the existence and multiplicity of solutions to a higher-order fractional boundary value problem. The interesting point lies in the fact that the solutions here are positive, monotone, and concave.


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## Body

## 1. Introduction

In this paper, we deal with the following boundary value problem for higher-order fractional differential equation:(1.1)D0+αu(t)+f(t,u(t))=0,0<t<1,n-1<α⩽n,n>3,u(0)=u′(1)=u′′(0)=⋯=u(n-1)(0)=0,
where D0+α denotes the Caputo fractional derivative and f:[0,1]×[0,+∞)→[0,+∞) is a real function. By using fixed point theorem, some sufficient conditions for existence and multiplicity of solutions to the above boundary value problem are obtained. Moreover, we will show that the solutions obtained here are positive, monotone, and concave.Fractional differential equations are valuable tools in the modelling of many phenomena in various fields of science and engineering [1–5]. Due to their applications, fractional differential equations have gained considerable attentions and there has been a significant development in the study of existence of solutions, and positive solutions to boundary value problems for fractional differential equations (e.g., [6–9] and references therein).Some papers are devoted to study the existence of solutions for higher-order fractional boundary value problem. Salem [10] investigated the existence of pseudosolutions for the nonlinear m-point boundary value problem of fractional type (1.2)Dαx(t)+q(t)f(t,x(t))=0,0<t<1,α∈(n-1,n],n⩾2,x(0)=x′(0)=x′′(0)=⋯=x(n-2)(0)=0,x(1)=∑i=1m-2ξix(ηi).
Zhang [11] considered the existence of positive solutions to the singular boundary value problem for fractional differential equations (1.3)D0+αu(t)+q(t)f(u,u′,…,u(n-2))=0,0<t<1,α∈(n-1,n],n⩾2,u(0)=u′′(0)=⋯=u(n-2)(0)=u(n-2)(1)=0,
where D0+α is the Riemann-Liouville fractional derivative of order α. In another paper, Zhang [9] studied the existence, multiplicity, and nonexistence of positive solutions for the following higher-order fractional boundary value problem:
(1.4)Dαu+λh(t)f(u)=0,0<t<1,α∈(n-1,n],n⩾2,u(1)=u′(0)=⋯=u(n-2)(0)=u(n-1)(0)=0,
where Dα is the Caputo fractional derivative of order α.It seems that the authors of the papers only studied the existence of the solutions or positive solutions. No one consider the qualities of the solutions for boundary value problems of fractional differential equation. Motivated by all the above works, the aim of this paper is to study the monotone, concave, and positive solutions of a fractional differential equation.The rest of the paper is organized as follows. In Section2, we will introduce some lemmas and definitions which will be used later. In Section 3, the existence and multiplicity of positive solutions for the boundary value problem (1.1) will be discussed. In Section 4, examples are given to check our results.


## 2. Basic Definitions and Preliminaries

In this section, we introduce some necessary definitions and lemmas, which will be used in the proofs of our main results.Definition 2.1 (see [12]).
The integral(2.1)I0+αy(t)=1Γ(α)∫0t(t-s)α-1y(s)ds,
where α>0 is called the Riemann-Liouville fractional integral of order α.Definition 2.2 (see [12]).
The Caputo fractional derivative for a functiony:(0,∞)→R can be written as
(2.2)D0+αy(t)=1Γ(n-α)∫0t(t-s)n-α-1y(n)(s)ds,
where n=[α]+1, [α] denotes the integer part of real number α.According to the definitions of fractional calculus, we can obtain that the fractional integral and the Caputo fractional derivative satisfy the following Lemma.Lemma 2.3 (see [13]).
Assume thatu∈Cm[0,1] and ρ∈(m-1,m),m∈N and v∈C1[0,1]. Then, for t∈[0,1],(a)
D0+ρI0+ρv(t)=v(t),(b)
I0+ρD0+ρu(t)=u(t)-∑k=0m-1((u(k)(0))/k!)tk,(c)
limt→0+D0+ρu(t)=limt→0+I0+ρu(t)=0.Definition 2.4.
LetE be a real Banach space over R. A nonempty convex closed set p⊂E is said to be a cone, provided that(a)
au∈P,forallu∈P,a⩾0,(b)
u,-u∈P,impliesu=0.Definition 2.5.
LetE be a real Banach space and P⊂E a cone. A function φ:P→[0,∞) is called a nonnegative continuous concave functional if φ is continuous and
(2.3)φ(λx+(1-λ)y)⩾λφ(x)+(1-λ)φ(y),
for all x,y∈P and 0⩽λ⩽1.Lemma 2.6 (see [14]).
LetE be a Banach space, K⊆E a cone in E, and Ω1, Ω2 two bounded open subsets of E with 0∈Ω1 and Ω¯1⊂Ω2. Suppose that T:K∩(Ω¯2∖Ω1)→K is continuous and completely continuous such that either(i)
∥Tu∥⩽∥u∥foru∈K∩∂Ω1,∥Tu∥⩾∥u∥foru∈K∩∂Ω2,or(ii)
∥Tu∥⩾∥u∥foru∈K∩∂Ω1,∥Tu∥⩽∥u∥foru∈K∩∂Ω2holds. Then, T has a fixed point in K∩(Ω¯2∖Ω1).
Letb,d,r>0 be constants, Pr={u∈P:∥u∥<r}, P(φ,b,d)={u∈P:b⩽φ(u),∥u∥⩽d}.Lemma 2.7 (see [15]).
LetP be a cone in real Banach space E. Let  T:P¯c→P¯c be a completely continuous map and φ  a nonnegative continuous concave functional on P such that φ(u)⩽∥u∥, for all u∈P¯c. Suppose that there exist constants a,b,d with 0<a<b<d⩽c such that
(2.4)(i){u∈P(φ,b,d):φ(u)>b}≠∅,φ(Tu)>b∀u∈P(φ,b,d),(ii)‖Tu‖<a∀u∈P¯a,(iii)φ(Tu)>b,∀u∈P(φ,b,c)with‖Tu‖>d.
Then, T has at least three fixed points u1, u2, and u3 satisfying
(2.5)‖u1‖<a,b<φ(u2),‖u3‖>a,φ(u3)<b.Lemma 2.8.
Assume thatf(t,u)∈C([0,1]×[0,+∞),[0,+∞)), then u∈C[0,1] be a solution of fractional boundary value problem (1.1) if and only if u∈C[0,1] is a solution of integral equation
(2.6)u(t)=∫01G(t,s)f(s,u(s))ds,
where
(2.7)G(t,s)={(α-1)t(1-s)α-2-(t-s)α-1Γ(α),0⩽s⩽t⩽1,(α-1)t(1-s)α-2Γ(α),0⩽t⩽s⩽1.Proof.
Firstly, we prove the necessity. Letu∈C[0,1] is a solution of fractional boundary value problem (1.1). By Lemma 2.3, we have
(2.8)u(t)=-I0+αf(t,u(t))+u(0)+u′(0)t+⋯+u(n-1)(0)(n-1)!tn-1=-1Γ(α)∫0t(t-s)α-1f(s,u(s))ds+u′(0)t.
Therefore,
(2.9)u′(t)=-1Γ(α)∫0t(α-1)(t-s)α-2f(s,u(s))ds+u′(0).
By the boundary value condition u′(1)=0, we have
(2.10)u′(0)=1Γ(α)∫01(α-1)(1-s)α-2f(s,u(s))ds.
Hence, we obtain
(2.11)u(t)=-1Γ(α)∫0t(t-s)α-1f(s,u(s))ds+1Γ(α)∫01(α-1)t(1-s)α-2f(s,u(s))ds=∫01G(t,s)f(s,u(s))ds.
The necessity is proved.
Now, we prove the sufficiency. Letu∈C[0,1] be a solution of integral equation (2.6). Then, we have
(2.12)D0+αu(t)=-D0+α(∫0t(t-s)α-1Γ(α)f(s,u(s))ds)+(∫01(α-1)(1-s)α-2Γ(α)f(s,u(s))ds)D0+αt=-D0+αI0+αf(t,u(t))=-f(t,u(t)).
By direct computation, we obtain that
(2.13)u(0)=u′(1)=u′′(0)=⋯=u(n-1)(0)=0.
That is to say, u is a solution of fractional boundary value problem (1.1). Thus, the sufficiency is proved.Lemma 2.9.
Letf(t,u(t))∈C([0,1]×[0,∞),[0,∞)). Then, the solution u(t) of fractional boundary value problem (1.1) satisfies(1)
u(t)is concave on(0,1),(2)
u(t)⩾0is increasing fort∈[0,1].Proof.
Suppose thatu(t) is a solution of fractional boundary value problem (1.1). By (2.11), we know that
(2.14)u(t)=-1Γ(α)∫0t(t-s)α-1f(s,u(s))ds+1Γ(α)∫01(α-1)t(1-s)α-2f(s,u(s))ds.
Therefore,
(2.15)u′(t)=-1Γ(α)∫0t(α-1)(t-s)α-2f(s,u(s))ds+1Γ(α)∫01(α-1)(1-s)α-2f(s,u(s))ds,(2.16)u′′(t)=-1Γ(α)∫0t(α-1)(α-2)(t-s)α-3f(s,u(s))ds⩽0,fort∈[0,1],n-1<α⩽n,n>3,
which implies that u(t) is concave on (0,1). The statement (1) is proved.
Sinceu′′(t)⩽0, we know that u′(t) is nonincreasing. By u′(1)=0, we have u′(t)⩾0, t∈[0,1]. Thus, u(t) is increasing. Noting u(0)=0, we obtain that u(t)⩾0 for t∈[0,1]. The statement (2) is proved.Lemma 2.10.
The Green's functionG(t,s), defined by (2.7), satisfies(1)
max0⩽t⩽1G(t,s)=G(1,s),s∈[0,1],(2)
G(t,s)⩾0,t,s∈[0,1],(3)
minξ⩽t⩽ηG(t,s)⩾ξα-1G(1,s),s∈[0,1],forallξ,η∈(0,1),ξ<η.Proof.
By (2.7), we have
(2.17)Gt′(t,s)={(α-1)(1-s)α-2-(α-1)(t-s)α-2Γ(α),0⩽s⩽t⩽1,(α-1)(1-s)α-2Γ(α),0⩽t⩽s⩽1.
It is clear that Gt′(t,s)⩾0,t,s∈[0,1]. Therefore, G(t,s) is increasing respect to t for s∈[0,1]. Thus, max0⩽t⩽1G(t,s)=G(1,s). The statement (1) holds.
If0⩽t⩽s⩽1, then,
(2.18)G(t,s)=(α-1)t(1-s)α-2Γ(α)⩾0.
Since G(t,s) is increasing respect to t for s∈[0,1], it is easy to see that G(t,s)⩾0 for 0⩽s⩽t⩽1. We get the statement (2).
On the other hand, we have(2.19)min⁡ξ⩽t⩽ηG(t,s)={(α-1)ξ(1-s)α-2-(ξ-s)α-1Γ(α),s∈[0,ξ],(α-1)ξ(1-s)α-2Γ(α),s∈[ξ,η],(α-1)ξ(1-s)α-2Γ(α),s∈[η,1]={(α-1)ξ(1-s)α-2-(ξ-s)α-1Γ(α),s∈[0,ξ],(α-1)ξ(1-s)α-2Γ(α),s∈[ξ,1].
If s∈[0,ξ], then
(2.20)(α-1)ξ(1-s)α-2-(ξ-s)α-1=ξ(α-1)(1-s)α-2-ξα-1(1-(s/ξ))α-1⩾ξα-1(α-1)(1-s)α-2-ξα-1(1-s)α-1=ξα-1[(α-1)(1-s)α-2-(1-s)α-1].
If s∈[ξ,1], then
(2.21)(α-1)ξ(1-s)α-2⩾(α-1)ξα-1(1-s)α-2⩾ξα-1[(α-1)(1-s)α-2-(1-s)α-1].
Thus,
(2.22)min⁡ξ⩽t⩽ηG(t,s)⩾ξα-1[(α-1)(1-s)α-2-(1-s)α-1]Γ(α)=ξα-1G(1,s),s∈[0,1].
This yields the statement (3). The proof is finished.


## 3. Main Results

In this section, we establish the results for the existence and multiplicity of monotone and concave positive solutions for fractional boundary value problem (1.1).LetE=C[0,1] with ∥u∥=max0⩽t⩽1|u(t)|. We define the cone P⊂E by
(3.1)P={u∈E:u(t)⩾0isconcaveon⁡[0,1],min⁡ξ⩽t⩽ηu(t)⩾ξα-1‖u‖}.And denote the operatorT by
(3.2)Tu(t)=∫01G(t,s)f(s,u(s))ds.Lemma 3.1.
Assume thatf(t,u)∈C([0,1]×[0,+∞),[0,+∞)), then T:P→P is completely continuous.Proof.
In view of non-negativeness and continuity ofG(t,s) and f(t,u(t)), we know that the operator T is continuous and Tu(t)⩾0, for u∈P.
By (2.16), we have
(3.3)(Tu)′′(t)=-1Γ(α)∫0t(α-1)(α-2)(t-s)α-3f(s,u(s))ds⩽0,fort∈[0,1],n-1<α⩽n,n>3.
Moreover, it follows from Lemma2.10 that for u∈P,
(3.4)min⁡ξ⩽t⩽η⁡Tu(t)=min⁡ξ⩽t⩽η⁡∫01G(t,s)f(s,u(s))ds⩾ξα-1∫01G(1,s)f(s,u(s))ds=ξα-1‖Tu‖.
Therefore, the operator T:P→P is well defined.
Assume thatΩ∈P is bounded; that is, there exists a positive constant M>0 such that ∥u∥⩽M for all u∈Ω. Let N=max0⩽t⩽1,∥u∥⩽M|f(t,u(t))|+1. For all u∈Ω, we have
(3.5)|Tu(t)|=|∫01G(t,s)f(s,u(s))ds|⩽N∫01G(1,s)ds,
which shows that T(Ω) is uniformly bounded.
In addition, for eachu∈Ω, t1,t2∈[0,1] such that t1<t2, we have
(3.6)|Tu(t2)-Tu(t1)|=|∫01G(t2,s)f(s,u(s))ds-∫01G(t1,s)f(s,u(s))ds|⩽|∫0t1(t1-s)α-1-(t2-s)α-1Γ(α)f(s,u(s))ds-∫t1t2(t2-s)α-1Γ(α)f(s,u(s))ds|+|∫01(α-1)(t2-t1)(1-s)α-2Γ(α)f(s,u(s))ds|⩽NΓ(α+1)(t2α-t1α)+NΓ(α)(t2-t1).
Thus, by the standard arguments, we obtain that T(Ω) is equicontinuous. The Arzela-Ascoli theorem implies that T:P→P is completely continuous. The proof is completed.Let(3.7)M1=∫01G(1,s)ds,N1=ξα-1∫ξηG(1,s)ds.Theorem 3.2.
Letf(t,u)∈C([0,1]×[0,∞),[0,∞)). Assume that there exist two positive constants r2>r1>0 such that(H1)
f(t,u)≤r2/M1for(t,u)∈[0,1]×[0,r2],(H2)
f(t,u)≥r1/N1for(t,u)∈[0,1]×[0,r1].Then, fractional boundary value problem (1.1) has at least one positive, increasing, and concave solution u such that r1≤∥u∥≤r2.Proof.
Lemmas2.8 and 3.1 imply that T:P→P is completely continuous and fractional boundary problem (1.1) has a solution u=u(t) if and only if u satisfies the operator equation u=Tu.
LetΩ1:={u∈P:∥u∥<r1}. By (H2), for u∈∂Ω1 and t∈[0,1], we have
(3.8)Tu(t)=∫01G(t,s)f(s,u(s))ds⩾r1N1∫ξηmin⁡ξ≤t≤η⁡G(t,s)ds⩾r1N1ξα-1∫ξηG(1,s)ds=r1=‖u‖.
So,
(3.9)‖Tu‖⩾‖u‖,foru∈∂Ω1.
LetΩ2:={u∈P:∥u∥<r2}. For u∈∂Ω2, and t∈[0,1], it follows from (H1) that
(3.10)‖Tu(t)‖=max⁡0≤t≤1|∫01G(t,s)f(s,u(s))ds|⩽r2M1∫01G(1,s)ds=r2M1∫01G(1,s)ds=r2=‖u‖.
Lemma 2.6 implies that the fractional boundary value problem (1.1) has at least one positive solution u such that r1≤∥u∥≤r2. By Lemma 2.9, the solution is also increasing and concave.In order to use Lemma2.7, we define the nonnegative continuous concave functional φ by φ(u)=minξ⩽t⩽ηu(t),forallu∈P.Theorem 3.3.
Suppose thatf(t,u)∈C([0,1]×[0,∞),[0,∞)) and there exist constants 0<a<b<c such that the following conditions hold:(H3)
f(t,u)<aM1for(t,u)∈[0,1]×[0,a],(H4)
f(t,u)>bN1for(t,u)∈[ξ,η]×[b,c],(H5)
f(t,u)⩽cM1for(t,u)∈[0,1]×[0,c].Then, the fractional boundary problem (1.1) has at least three positive, increasing, and concave solutions u1, u2, and u3 such that
(3.11)max⁡0⩽t⩽1|u1(t)|<a,b<min⁡ξ⩽t⩽η|u2(t)|<max⁡0⩽t⩽1|u2(t)|⩽c,a<max⁡0⩽t⩽1|u3(t)|⩽c,min⁡ξ⩽t⩽η|u3(t)|<b.Proof.
By Lemmas2.8 and 3.1, T:P→P is completely continuous, and fractional boundary value problem (1.1) has a solution u=u(t) if and only if u satisfies the operator equation u=Tu.
First of all, we will prove the following assertions.
Assertion 3.4 (T(P¯c)⊆P¯c and T(P¯a)⊆P¯a).
Firstly, Lemma3.1 guarantees T(P¯c)⊆P. Secondly, for all u∈P¯c, we have ∥u∥⩽c. By (H5),
(3.12)‖Tu(t)‖=max⁡0≤t≤1|∫01G(t,s)f(s,u(s))ds|⩽cM1∫01G(1,s)ds=c,
which implies that T(P¯c)⊆P¯c. In the same way, T(P¯a)⊆P¯a.Assertion 3.5 ({u∈P(φ,b,d)∣φ(u)>b}≠∅, and φ(Tu)>b, for all u∈P(φ,b,d)).
Letd=c, u=(b+c)/2. Then, ∥u∥<d and φ(u)=φ((b+c)/2)>b. Consequently, {u∈P(φ,b,d)∣φ(u)>b}≠∅.
Ifu∈P(φ,b,d), then from (H4) and Lemma 2.10, we obtain that
(3.13)φ(Tu)=min⁡ξ⩽t⩽ηTu(t)⩾∫ξηmin⁡ξ⩽t⩽ηG(t,s)f(s,u(s))ds>bN1ξα-1∫ξηG(1,s)ds=b.
That is, φ(Tu)>b, for all u∈P(φ,b,d).Assertion 3.6 (φ(u)>b, for all u∈P(φ,b,c) with ∥u∥>d).
Ifu∈P(φ,b,c)  and ∥Tu∥>d=c, similar to the above, we also have φ(Tu)>b.
Assertions1~3 imply that all conditions of Lemma 2.7 hold. Therefore, the fractional boundary value problem (1.1) has at least three positive solutions u1,u2, and u3 satisfying
(3.14)max⁡0⩽t⩽1⁡|u1(t)|<a,b<min⁡ξ⩽t⩽η|u2(t)|<max⁡0⩽t⩽1|u2(t)|⩽c,a<max⁡0⩽t⩽1|u3(t)|⩽c,min⁡ξ⩽t⩽η|u3(t)|<b.
By Lemma 2.9, the positive solutions are also increasing and concave. The proof is completed.


## 4. Example

In this section, we will present some examples to show the effectiveness of our work.Example 4.1.
Consider the fractional boundary value problem(4.1)D0+7/2u(t)+t2+u+4=0,0<t<1,u(0)=u′′(0)=u′′′(0)=u′(1)=0.
Setting ξ=1/2,η=3/4, we obtain
(4.2)M1=∫01G(1,s)ds=(7/2)-1Γ(7/2)∫01(1-s)(7/2)-2ds-1Γ(7/2)∫01(1-s)(7/2)-1ds=1(5/2)Γ(5/2)-1(7/2)Γ(7/2)≈0.2149,N1=(12)(7/2)-1∫1/23/4G(1,s)ds=(1/2)5/2Γ(7/2)(∫1/23/452(1-s)(7/2)-2ds-∫1/23/4(1-s)(7/2)-1ds)≈0.0065.
Let r1=1/40,r2=4. We have
(4.3)f(t,u)=t2+u+4⩽r2M1≈18.613,for(t,u)∈[0,1]×[0,4],f(t,u)=t2+u+4⩾r1N1≈3.846,for(t,u)∈[0,1]×[0,140].
Theorem 3.2 implies that fractional boundary value problem (4.1) has at least one positive, increasing, and concave solution. The approximate solution is obtained by the Adams-type predictor-corrector method [16], which is displayed in Figure 1 for the step size h=0.01.Figure 1
Transient response of state variableu(t).Example 4.2.
Consider the fractional boundary value problem(4.4)D0+7/2u(t)+f(t,u)=0,0<t<1,u(0)=u′′(0)=u′′′(0)=u′(1)=0,
where
(4.5)f(t,u)={t10+155u3,0⩽u⩽1,t∈[0,1],t10+u+154,u>1,t∈[0,1].
Settingξ=1/2,η=3/4, we know that M1≈0.2149,N1≈0.0065. Choosing a=1/10,b=1,c=45, we obtain that
(4.6)f(t,u)=t10+155u3<aM1≈0.466for(t,u)∈[0,1]×[0,110],f(t,u)=t10+u+154>bN1≈153.846for(t,u)∈[12,34]×[1,45],f(t,u)=t10+u+154<cM1≈209.4for(t,u)∈[0,1]×[0,45].
Theorem 3.3 implies that fractional boundary value problem (4.4) has three positive, increasing, and concave solutions such that
(4.7)max⁡0⩽t⩽1|u1(t)|<110,1<min⁡1/2⩽t⩽3/4|u2(t)|<max⁡0⩽t⩽1|u2(t)|⩽45,110<max⁡0⩽t⩽1|u3(t)|⩽45,min⁡1/2⩽t⩽3/4|u3(t)|<1.
For numerical simulation case 1, Figure 2 depicts the phase responses state variables of u(t) with the step size h=0.01.Figure 2
Transient response of state variableu(t).



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