# Positive Solutions of Fractional Differential Equation withp-Laplacian Operator

Authors: Teng Ren; Xiaochun Chen
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The basic assumption of ecological economics is that resource allocation exists social optimal
solution, and the social optimal solution and the optimal solution of enterprises can be complementary.
The mathematical methods and the ecological model are one of the important means in the study of
ecological economics. In this paper, we study an ecological model arising from ecological economics by
mathematical method, that is, study the existence of positive solutions for the fractional differential equation
withp-Laplacian operator Dtβ(φp(Dtαx))(t)=f(t,x(t)), t∈(0,1), x(0)=0, x(1)=ax(ξ), Dtαx(0)=0, and Dtαx1=bDtαxη, where Dtα,Dtβ are the standard Riemann-Liouville derivatives, p-Laplacian operator is defined as φps=sp-2s,p>1, and the nonlinearity f may be singular at both t=0,1 and x=0. By finding more
suitable upper and lower solutions, we omit some key conditions of some existing works, and the existence
of positive solution is established.


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## Body

## 1. Introduction

It is well known that differential equation models can describe many nonlinear phenomena such as applied mathematics, economic mathematics, and physical and biological processes. Undoubtedly, the application of differential equation in the economics, management science, and engineering that is most successful especially plays an important role in the construction of the model for the corresponding phenomenon. In fact, many economic processes such as ecological economics model, risk model, the CIR model, and the Gaussian model in [1] can be described by differential equations. Recently, fractional-order models have proved to be more accurate than integer-order models; that is, there are more degrees of freedom in the fractional-order models. So complicated dynamic phenomenon of fractional-order calculus system has received more and more attention; see [2–15].In this paper, we study an ecological model arising from ecological economics by mathematical method, that is, study the existence of positive solutions for the followingp-Laplacian fractional boundary value problem:
(1)Dtβ(φp(Dtαx))(t)=f(t,x(t)),t∈(0,1),x(0)=0,x(1)=ax(ξ),Dtαx(0)=0,Dtαx(1)=bDtαx(η),
where Dtα,Dtβ are the standard Riemann-Liouville derivatives with 1<α, β≤2, 0≤a, b≤1, 0<ξ, and η≤1, and p-Laplacian operator is defined as φp(s)=|s|p-2s, p>1, (φp)-1=φq, and (1/p)+(1/q)=1.The upper and lower solutions method is a powerful tool to achieve the existence results for boundary value problem; see [2–6]. Recently, Zhang and Liu [2] considered the existence of positive solutions for the singular fourth-order p-Laplacian equation
(2)[φp(u′′)]′′=f(t,u(t)),0<t<1,
with the four-point boundary conditions
(3)u(0)=0,u(1)=au(ξ),u′′(0)=0,u′′(1)=bu′′(η),
where φp(t)=|t|p-2t, p>1, 0<ξ, η<1, and f∈C((0,1)×(0,+∞),[0,+∞)) may be singular at t=0 and/or 1 and u=0. By using the upper and lower solutions method and fixed-point theorems, the existence of positive solutions to the boundary value problem is obtained. In [2], a upper and lower solution condition (H3) is used.There exist a continuous functiona(t) and some fixed positive number k, such that a(t)≥kt(1-t), t∈[0,1], and
(4)∫01‍G(t,r)φp-1(∫01‍H(r,s)f(s,a(s))ds)dr=b(t)≥a(t),∫01‍G(t,r)φp-1(∫01‍H(r,s)f(s,b(s))ds)dr≥a(t),
where G(t,s), H(t,s) are the associated Green’s functions for the relevant problems. And then, the condition (H3) was also adopted by Wang et al. [3] to deal with the p-Laplacian fractional boundary value problem (1). By using similar method  as [2], the existence results of at least one positive solution for the above fractional boundary value problem are established. Recently, replaced (H3) with a simple integral condition, Jia et al. [8] studied the existence, uniqueness, and asymptotic behavior of positive solutions for the higher nonlocal fractional differential equation by using upper and lower solutions method.In this paper, we restart to establish the existence of positive solutions for the BVP (1) when the nonlinearity f may be singular at both t=0,1 and x=0. By finding more suitable upper and lower solutions of (1), we completely omit the condition (H3) in [2, 3] and integral condition in [8], thus our work improves essentially the results of [2, 3, 8].


## 2. Basic Definitions and Preliminaries

In this section, we present some necessary definitions and lemmas from fractional calculus theory, which can be found in the recent literatures [7, 16, 17].Definition 1.
The Riemann-Liouville fractional integral of orderα>0 of a function x:(0,+∞)→ℝ is given by
(5)Iαx(t)=1Γ(α)∫0t‍(t-s)α-1x(s)ds
provided that the right-hand side is pointwise defined on (0,+∞).Definition 2.
The Riemann-Liouville fractional derivative of orderα>0 of a function x:(0,+∞)→ℝ is given by
(6)Dtαx(t)=1Γ(n-α)(ddt)n∫0t‍(t-s)n-α-1x(s)ds,
where n=[α]+1, [α] denotes the integer part of number α, provided that the right-hand side is pointwise defined on (0,+∞).Lemma 3.
(
1
)
If x∈L1(0,1),α>β>0, then
(7)IαIβx(t)=Iα+βx(t),DtβIαx(t)=Iα-βx(t),DtαIαx(t)=x(t).
(
2
)
If α>0,β>0, then
(8)Dtαtβ-1=Γ(β)Γ(β-α)tβ-α-1.Lemma 4.
Letα>0, and let f(x) be integrable, then
(9)IαDtαf(x)=f(x)+c1xα-1+c2xα-2+⋯+cnxα-n,
where ci∈ℝ(i=1,2,…,n), and n is the smallest integer greater than or equal to α.Definition 5.
A continuous functionϕ(t) is called a lower solution of the BVP (1), if it satisfies
(10)Dtβ(φp(Dtαϕ))(t)≤f(t,ϕ(t)),0<t<1,ϕ(0)≤0,ϕ(1)≤aϕ(ξ),Dtαϕ(0)≥0,Dtαϕ(1)≥bDtαϕ(η).Definition 6.
A continuous functionψ(t) is called an upper solution of the BVP (1), if it satisfies
(11)Dtβ(φp(Dtαψ))(t)≥f(t,ψ(t)),0<t<1ψ(0)≥0,ψ(1)≥aψ(ξ),Dtαψ(0)≤0,Dtαψ(1)≤bDtαψ(η).
For forthcoming analysis, we first consider the following linear fractional differential equation:
(12)Dtαx(t)+y(t)=0,t∈(0,1),x(0)=0,x(1)=ax(ξ).Lemma 7.
If1<α≤2 and y∈L1[0,1], then the boundary value problem (12) has the unique solution
(13)x(t)=∫01‍G(t,s)y(s)ds,
where
(14)G(t,s)=k(t,s)+ak(ξ,s)tα-11-aξα-1,(15)k(t,s)={(t(1-s))α-1-(t-s)α-1Γ(α),0≤s≤t≤1,(t(1-s))α-1Γ(α),0≤t≤s≤1.Proof.
By applying Lemma4, we may reduce (12) to an equivalent integral equation
(16)x(t)=-Iαy(t)+c1tα-1+c2tα-2,c1,c2∈ℝ.
From x(0)=0 and (16), we have c2=0. Consequently the general solution of (12) is
(17)x(t)=-Iαy(t)+c1tα-1=-∫0t‍(t-s)α-1Γ(α)y(s)ds+c1tα-1.
By (17), one has
(18)x(1)=-∫01‍(1-s)α-1Γ(α)y(s)ds+c1,x(ξ)=-∫0ξ‍(ξ-s)α-1Γ(α)y(s)ds+c1ξα-1.
And then, we have
(19)c1=∫01‍(1-s)α-1y(s)ds-a∫0ξ‍(ξ-s)α-1y(s)dsΓ(α)(1-aξα-1).
So, the unique solution of problem (12) is
(20)x(t)=-∫0t‍(t-s)α-1y(s)Γ(α)ds+tα-11-aξα-1×{∫01‍(1-s)α-1y(s)Γ(α)ds-a∫0ξ‍(ξ-s)α-1y(s)Γ(α)ds}=-∫0t‍(t-s)α-1y(s)Γ(α)ds+∫01‍(1-s)α-1tα-1y(s)Γ(α)ds+atα-11-aξα-1∫01‍(1-s)α-1ξα-1y(s)Γ(α)ds-atα-11-aξα-1∫0ξ‍(ξ-s)α-1Γ(α)y(s)ds=∫01‍(k(t,s)+atα-1y(s)1-aξα-1k(ξ,s))ds=∫01‍G(t,s)y(s)ds.
The proof is completed.Lemma 8.
Lety∈L1[0,1], 1<α, β≤2, 0<ξ, η<1, and 0≤a, b≤1. The fractional boundary value problem
(21)Dtβ(φp(Dtαx))(t)=y(t),t∈(0,1),x(0)=0,x(1)=ax(ξ),Dtαx(0)=0,Dtαx(1)=bDtαx(η),
has unique solution
(22)x(t)=∫01‍G(t,s)φq(∫01‍H(s,τ)y(τ)dτ)ds,
where
(23)H(t,s)=g(t,s)+ag(η,s)tβ-11-b1ηβ-1,g(t,s)={(t(1-s))β-1-(t-s)β-1Γ(β),0≤s≤t≤1,(t(1-s))β-1Γ(β),0≤t≤s≤1,
and b1=bp-1, and G(t,s) is defined by (14).Proof.
At first, by Lemma4, (21) is equivalent to the integral equation
(24)φp(Dtαx(t))=Iβy(t)+c3tβ-1+c4tβ-2,c3,c4∈ℝ.
From Dtαx(0)=0,Dtαx(1)=bDtαx(η), and (24), we have c4=0. Consequently the general solution of (21) is
(25)φp(Dtαx(t))=Iβy(t)+c3tβ-1=∫0t‍(t-s)β-1Γ(β)y(s)ds+c3tβ-1.
It follows from (25) that
(26)φp(Dtαx(1))=∫01‍(1-s)β-1Γ(β)y(s)ds+c3,(27)φp(Dtαx(η))=∫0η‍(η-s)β-1Γ(β)y(s)ds+c3ηα-1.
Thus (25) and (26) imply
(28)c3=-∫01‍(1-s)β-1y(s)ds-b1∫0η‍(η-s)β-1y(s)dsΓ(β)(1-b1ηβ-1),
where b1=bp-1. Similar to Lemma 7, we have
(29)φp(Dtαx(t))=-∫01‍H(t,s)y(s)ds.
Consequently, fractional boundary value problem (21) is equivalent to the following problem:
(30)Dtαx(t)+φq(∫01‍H(t,s)y(s)ds),t∈(0,1),x(0)=0,x(1)=ax(ξ).
Lemma 7 implies that fractional boundary value problem (21) has a unique solution
(31)x(t)=∫01‍G(t,s)φq(∫01‍H(s,τ)y(τ)dτ)ds.
The proof is completed.Lemma 9.
Let1<α, β≤2, 0<ξ, η<1, and 0≤a, b≤1. The functions G(t,s) and H(t,s) defined by (14) and (23), respectively, are continuous on [0,1]×[0,1] and satisfy(i)G(t,s)≥0,H(t,s)≥0, for t,s∈[0,1].(ii) For t,s∈[0,1],
(32)σ1(s)tα-1≤G(t,s)≤σ2(s)tα-1,where
(33)σ1(s)=ak(ξ,s)1-aξα-1,σ2(s)=(1-s)α-1Γ(α)+ak(ξ,s)1-aξα-1.Proof.
The proof is obvious, so we omit the proof.Set(34)e(t)=tα-1.
We present the following assumptions:(S1)f:(0,1)×(0,∞)→[0,+∞) is is continuous and decreasing in x.(S2) For anyκ>0,
(35)0<∫01‍σ2(s)φq(∫01‍H(s,τ)f(τ,κe(τ))dτ)ds<+∞.From Lemmas7 and 9, it is easy to obtain the following conclusion.Lemma 10.
Ifx∈C([0,1],R) satisfies
(36)x(0)=0,x(1)=φp(b)x(η),
and Dtβx(t)≥0 for any t∈(0,1), then x(t)≤0, for t∈[0,1].


## 3. Main Results

LetE=C[0,1], and
(37)P={{12}x∈E:thereexistspositivenumber0<lx<1,Lx>1suchthatlxe(t)≤x(t)≤Lxe(t),t∈[0,1]}.
Clearly, e(t)∈P, so P is nonempty. For any x∈P, define an operator T by
(38)(Tx)(t)=∫01‍G(t,s)φq(∫01‍H(s,τ)f(τ,x(τ))dτ)ds,t∈[0,1].Theorem 11.
Suppose (S1)-(S2) hold. Then the BVP (1) has at least one positive solution x, and there exist two positive constants 0<μ1<1, μ2>1, such that
(39)μ1e(t)≤x(t)≤μ2e(t),t∈[0,1].Proof.
We firstly assert thatT is well defined on P and T(P)⊂P, and T is decreasing in x.
In fact, for anyx∈P, by the definition of P, there exist two positive numbers 0<lx<1,Lx>1, such that lxe(t)≤x(t)≤Lxe(t) for any t∈[0,1]. It follows from Lemma 9 and (S1)–(S3) that
(40)(Tx)(t)=∫01‍G(t,s)φq(∫01‍H(s,τ)f(τ,x(τ))dτ)ds≤e(t)∫01‍σ2(s)φq(∫01‍H(s,τ)f(τ,lxe(τ))dτ)ds<+∞.
On the other hand, by Lemma 9, we also have
(41)(Tx)(t)≥e(t)∫01‍σ1(s)φq(∫01‍H(s,τ)f(τ,x(τ))dτ)ds≥e(t)∫01‍σ1(s)φq(∫01‍H(s,τ)f(τ,Lxe(τ))dτ)ds.
Take
(42)lx′=min{1,∫01‍σ1(s)φq(∫01‍H(s,τ)f(τ,Lxe(τ))dτ)ds}Lx′=max{1,∫01‍σ2(s)φq(∫01‍H(s,τ)f(τ,lxe(τ))dτ)ds},
then by (40) and (41),
(43)lx′e(t)≤(Tx)(t)≤Lx′e(t),
which implies that T is well defined, and T(P)⊂P. It follows from (S1) that the operator T is decreasing in x. And by direct computations, we have
(44)Dtβ(φp(Dtα(Tx)))(t)=f(t,(Tx)(t)),t∈(0,1),(Tx)(0)=0,(Tx)(1)=a(Tx)(ξ),Dtα(Tx)(0)=0,Dtα(Tx)(1)=bDtα(Tx)(η).
Next we focus on lower and upper solutions of the fractional boundary value problem (1). Let
(45)m(t)=min{e(t),(Te)(t)},n(t)=max{e(t),(Te)(t)},
then, if e(t)=(Te)(t), the conclusion of Theorem 11 holds. If e(t)≠(Te)(t), clearly, m(t),n(t)∈P and
(46)m(t)≤e(t)≤n(t).
We will prove that the functionsϕ(t)=Tn(t), ψ(t)=Tm(t) are a couple of lower and upper solutions of the fractional boundary value problem (1), respectively.
From (S1),T is nonincreasing relative to x. Thus it follows from (45)-(46) that
(47)ϕ(t)=Tn(t)≤Tm(t)=ψ(t),Tn(t)≤Te(t)≤n(t),Tm(t)≥Te(t)≥m(t),
and ϕ(t),ψ(t)∈P. And it follows from (44)–(47) that
(48)Dtβ(φp(Dtαϕ))(t)-f(t,ϕ(t))=Dtβ(φp(Dtα(Tn)))(t)-f(t,(Tn)(t))≤Dtβ(φp(Dtα(Tn)))(t)-f(t,n(t))=0,t∈(0,1),ϕ(0)=0,ϕ(1)=aϕ(ξ),Dtαϕ(0)=0,Dtαϕ(1)=bDtαϕ(η),Dtβ(φp(Dtαψ))(t)-f(t,ψ(t))=Dtβ(φp(Dtα(Tm)))(t)-f(t,(Tm)(t))≥Dtβ(φp(Dtα(Tm)))(t)-f(t,m(t))=0,t∈(0,1),ψ(0)=0,ψ(1)=aψ(ξ),Dtαψ(0)=0,Dtαψ(1)=bDtαψ(η).
That is, ϕ(t) and ψ(t) are a couple of lower and upper solutions of fractional boundary value problem (1), respectively.
Now let us define a function(49)F(t,x)={f(t,ϕ(t)),x<ϕ(t),f(t,x(t)),ϕ(t)≤x≤ψ(t),f(t,ψ(t)),x>ψ(t).
It follows from (S1) and (49) that F:(0,1)×[0,+∞)→[0,+∞) is continuous.
We will show that the fractional boundary value problem(50)Dtβ(φp(Dtαx))(t)=F(t,x(t)),t∈(0,1),x(0)=0,x(1)=ax(ξ),Dtαx(0)=0,Dtαx(1)=bDtαx(η),
has a positive solution. Let us consider the operator
(51)𝔅x(t)=∫01‍G(t,s)φq(∫01‍H(s,τ)F(τ,x(τ))dτ)ds.
Thus 𝔅:C[0,1]→C[0,1], and the fixed point of the operator 𝔅 is a solution of the BVP (50). Noting that Φ∈P, then there exists a constant 0<lΦ<1, such that Φ(t)≥lΦe(t), t∈[0,1]. Thus for all x∈E, it follows from Lemma 9, (49), and (S2) that
(52)𝔅x(t)=∫01‍G(t,s)φq(∫01‍H(s,τ)F(τ,x(τ))dτ)ds≤e(t)∫01‍σ2(s)φq(∫01‍H(s,τ)F(τ,x(τ))dτ)ds≤e(t)∫01‍σ2(s)φq(∫01‍H(s,τ)f(τ,lΦe(τ))dτ)ds<+∞.
That is, the operator 𝔅 is uniformly bounded.
From the uniform continuity ofG(t,s) and Lebesgue dominated convergence theorem, we easily obtain that 𝔅 is equicontinuous. Thus by the means of the Arzela-Ascoli theorem, we have 𝔅:E→E is completely continuous. The Schauder fixed point theorem implies that 𝔅 has at least a fixed point x, such that x=𝔅x.
At the end, we claim that(53)ϕ(t)≤x(t)≤ψ(t),t∈[0,1].
In fact, since x is fixed point of 𝔅 and (44), we get
(54)x(0)=0,x(1)=ax(ξ),Dtαx(0)=0,Dtαx(1)=bDtαx(η),ψ(0)=0,ψ(1)=aψ(ξ),Dtαψ(0)=0,Dtαψ(1)=bDtαψ(η).
Otherwise, suppose thatx(t)>ψ(t). According to the definition of F, we have
(55)Dtβ(φp(Dtαx))(t)=F(t,x(t))=f(t,ψ(t)).
On the other hand, it follows from ψ is an upper solution to (1) that
(56)Dtβ(φp(Dtαψ))(t)≥f(t,ψ(t)).
Let z(t)=φp(Dtαψ(t))-φp(Dtαx(t)); it follows from (55) and (56) that
(57)Dtβz(t)=Dtβ(φp(Dtαψ))(t)-Dtβ(φp(Dtαx))(t)≥0,z(0)=0,z(1)=φp(b)z(η).
It follows from Lemma 10 that
(58)z(t)≤0,
and then
(59)φp(Dtαψ(t))-φp(Dtαx(t))≤0.
Notice that φp is monotone increasing; we have
(60)Dtαψ(t)≤Dtαx(t),thatis,Dtα(ψ-x)(t)≤0.
It follows from Lemma 10 and (54) that
(61)ψ(t)-x(t)≥0.
Thus we have x(t)≤ψ(t) on [0,1], which contradicts x(t)>ψ(t). Hence, x(t)>ψ(t) is impossible.
By the same way, we also havex(t)≥ϕ(t) on [0,1]. So
(62)ϕ(t)≤x(t)≤ψ(t),t∈[0,1].
Consequently, F(t,x(t))=f(t,x(t)),t∈[0,1]. Then x(t) is a positive solution of the problem (1).
Finally, byψ,ϕ∈P, we have
(63)μ1e(t)=lϕe(t)≤ϕ(t)≤x(t)≤ψ(t)≤Lψe(t)=μ2e(t),
where
(64)0<μ1<1,μ2>1.Remark 12.
In Theorem11, we find more suitable lower and upper solutions, then we refine the proved process, and the key condition (H3) in [2, 3] is removed, but the existence of positive solution is still obtained, thus our result is essential improvement of [2, 3].Theorem 13.
Iff(t,x):[0,1]×[0,+∞)→[0,+∞) is continuous, decreasing in x and f(t,κ)≢0, for any κ>0, then the boundary value problem (1) has at least one positive solution x(t), and there exist two positive constants 0<μ1<1, μ2>1, such that
(65)μ1e(t)≤x(t)≤μ2e(t),t∈[0,1].Proof.
The proof is similar to Theorem11, so we omit it here.Example 14.
Consider the following boundary value problem:(66)Dt5/4(φp(Dt3/2x))(t)=1t1/4x2(t)3+sint,t∈(0,1),x(0)=0,x(1)=18x(14),Dt3/2(0)=0,Dt3/2x(1)=13Dt3/2x(34).
Let α=3/2, β=5/4, and
(67)f(t,x)=1t1/4x23+sint,e(t)=t1/2.
Obviously, (S1) holds.
For anyκ>0,
(68)0<∫01‍σ2(s)φq(∫01‍H(s,τ)f(τ,κe(τ))dτ)ds=∫01‍σ2(s)φq(∫01‍H(s,τ)(κ-2/3τ-7/12+sinτ)dτ)ds<+∞,
which implies that (S2) holds.
By Theorem11, that the boundary value problem (66) has at least one positive solution.At the end of this work we also remark that the extension of the pervious results to the nonlinearities depending on the time delayed differential system for energy price adjustment or impulsive differential equation in financial field requires some further nontrivial modifications, and the reader can try to obtain results in our direction. We also anticipate that the methods and concepts here can be extended to the systems with economic processes such as risk model, the CIR model, and the Gaussian model as considered by Almeida and Vicente [1].



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