# Eigenvalue of Fractional Differential Equations withp-Laplacian Operator

Authors: Wenquan Wu; Xiangbing Zhou
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

We investigate the existence of positive solutions for the fractional order eigenvalue problem withp-Laplacian operator -𝒟tβ(φp(𝒟tαx))(t)=λf(t,x(t)),t∈(0,1),x(0)=0,𝒟tαx(0)=0,𝒟tγx(1)=∑j=1m-2‍aj𝒟tγx(ξj), where 𝒟tβ,𝒟tα,𝒟tγ are the standard Riemann-Liouville derivatives and p-Laplacian operator is defined as φp(s)=|s|p-2s,p>1.f:(0,1)×(0,+∞)→[0,+∞) is continuous and f can be singular at t=0,1 and x=0. By constructing upper and lower solutions, the existence of positive solutions for the eigenvalue problem of fractional differential equation is established.


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## Body

## 1. Introduction

Differential equations of fractional order have been recently proved to be valuable tools in the modeling of many phenomena arising from science and engineering, such as viscoelasticity, electrochemistry, control, porous media, and electromagnetism. For detail, see the monographs of Kilbas et al. [1], Miller and Ross [2], and Podlubny [3] and the papers [4–23] and the references therein.In [16], the authors investigated the nonlinear nonlocal boundary value problem:
(1)𝒟tβ(φp(𝒟tαx))(t)+f(t,x(t))=0,t∈(0,1),x(0)=0,𝒟tαx(0)=0,x(1)=ax(ξ),
where 0<α≤2,0<β≤1,0≤a≤1,0<ξ<1. By using Krasnoselskii's fixed point theorem and the Leggett-Williams theorem, some sufficient conditions for the existence of positive solutions to the above BVP are obtained. In [17], by using the upper and lower solutions method, under suitable monotone conditions, the authors investigated the existence of positive solutions to the following nonlocal problem:
(2)𝒟tβ(φp(𝒟tαx))(t)+f(t,x(t))=0,t∈(0,1),x(0)=0,𝒟tαx(0)=0,x(1)=ax(ξ),𝒟tαx(1)=b𝒟tαx(η),
where 0<α,β≤2,0≤a,b≤1,0<ξ,η<1. Recently, by means of the fixed point theorem on cones, Chai [18] investigated two-point boundary value problem of fractional differential equation with p-Laplacian operator:
(3)𝒟tβ(φp(𝒟tαx))(t)+f(t,x(t))=0,t∈(0,1),x(0)=0,𝒟tαx(0)=0,x(1)+a𝒟tγx(1)=0.
Some existence and multiplicity results of positive solutions are obtained.As far as we know, no result has been obtained for the existence of positive solution for the fractional order eigenvalue problem withp-Laplacian operator:
(4)-𝒟tβ(φp(𝒟tαx))(t)=λf(t,x(t)),t∈(0,1),x(0)=0,𝒟tαx(0)=0,𝒟tγx(1)=∑j=1m-2‍aj𝒟tγx(ξj),
where 𝒟tβ, 𝒟tα, 𝒟tγ are the standard Riemann-Liouville derivatives with 1<α≤2,0<β≤1,0<γ≤1,0≤α-γ-1,0<ξ1<ξ2<⋯<ξp-2<1,aj∈[0,+∞) with c=∑j=1m-2‍ajξjα-γ-1<1, p-Laplacian operator is defined as φp(s)=|s|p-2s,p>1,f can be singular at t=0,1, and x=0. In order to obtain the existence of positive solutions of the fractional order eigenvalue problem (4), we will apply the upper and lower solutions method associated with the Schauder's fixed point theorem. It is worth emphasizing that the problem (4) not only includes the well-known Sturm-Liouville boundary value problems and the nonlocal boundary value problems as special case, but also f can be singular at t=0,1 and x=0.The organization of this paper is as follows. In Section2, we present some necessary definitions and preliminary results that will be used to prove our main results. In Section 3, we put forward and prove our main results. Finally, we will give an example to demonstrate our main results.


## 2. Preliminaries and Lemmas

In this section, we introduce some preliminary facts which are used throughout this paper.Definition 1 (see [1–3]).
The Riemann-Liouville fractional integral of orderα>0 of a functionx:(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 (see [1–3]).
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 and [α] denotes the integer part of number α, provided that the right-hand side is pointwise defined on (0,+∞).Proposition 3 (see [1–3]).
(1) Ifx∈L1(0,1), ν>σ>0, then
(7)IνIσx(t)=Iν+σx(t),𝒟tσIνx(t)=Iν-σx(t),𝒟tσIσx(t)=x(t).
(2) Ifν>0, σ>0, then
(8)𝒟tνtσ-1=Γ(σ)Γ(σ-ν)tσ-ν-1.Proposition 4 (see [1–3]).
Letα>0, and f(x) is 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 functionψ(t) is called a lower solution of the BVP (4), if it satisfies
(10)-𝒟tβ(φp(𝒟tαψ))(t)≤λf(t,ψ(t)),t∈(0,1),ψ(0)≥0,𝒟tγψ(1)≥∑j=1m-2aj𝒟tγψ(ξj),𝒟tαψ(0)≥0.Definition 6.
A continuous functionϕ(t) is called an upper solution of the BVP (4), if it satisfies
(11)-𝒟tβ(φp(𝒟tαϕ))(t)≥λf(t,ϕ(t)),t∈(0,1),ϕ(0)≤0,𝒟tγϕ(1)≤∑j=1m-2‍aj𝒟tγϕ(ξj),𝒟tαϕ(0)≤0.For forthcoming analysis, we first consider the following linear fractional differential equation:(12)𝒟tαx(t)+h(t)=0,t∈(0,1),x(0)=0,𝒟tγx(1)=∑j=1m-2‍aj𝒟tγx(ξj).Lemma 7 (see [15]).
If1<α≤2 and h∈L1[0,1], then the boundary value problem (12) has the unique solution
(13)x(t)=∫01G(t,s)h(s)ds,
where
(14)G(t,s)=g1(t,s)+tα-11-∑j=1m-2ajξjα-γ-1∑j=1m-2ajg2(ξj,s)
is the Green function of the boundary value problem (12) and
(15)g1(t,s)={tα-1(1-s)α-γ-1-(t-s)α-1Γ(α),0≤s≤t≤1,tα-1(1-s)α-γ-1Γ(α),0≤t≤s≤1,g2(t,s)={(t(1-s))α-γ-1-(t-s)α-γ-1Γ(α),0≤s≤t≤1,(t(1-s))α-γ-1Γ(α),0≤t≤s≤1.Lemma 8.
The Green functionG(t,s) in Lemma 7 has the following properties:(i)
G(t,s) is continuous on [0,1]×[0,1];(ii)
G(t,s)>0 for any s,t∈(0,1);(iii)
tα-1σ1(s)≤G(t,s)≤tα-1σ2(s), for t,s∈[0,1], where
(16)σ1(s)=∑j=1m-2ajg2(ξj,s)1-∑j=1m-2ajξjα-γ-1,σ2(s)=(1-s)α-γ-1Γ(α)+σ1(s).Letq>1 satisfy the relation 1/q+1/p=1, where p is given by (4). To study BVP (4), we first consider the associated linear BVP:
(17)𝒟tβ(φp(𝒟tαx))(t)+h(t)=0,t∈(0,1),x(0)=0,𝒟tαx(0)=0,𝒟tγx(1)=∑j=1m-2‍aj𝒟tγx(ξj),
for h∈L1[0,1] and h≥0. For convenience, let
(18)b=(Γ(β))1-q,
then we have the following lemma.Lemma 9.
The associated linear BVP (17) has the unique positive solution
(19)x(t)=b∫01‍G(t,s)(∫0s‍(s-τ)β-1h(τ)dτ)q-1ds.Proof.
In fact, letw=𝒟tαx,v=φp(w). By Proposition 4, the solution of initial value problem
(20)𝒟tβv(t)+h(t)=0,t∈(0,1),v(0)=0
is given by v(t)=C1tβ-1-Iβh(t),t∈[0,1]. From the relations v(0)=0,0<β≤1, it follows that C1=0, and so
(21)v(t)=-Iβh(t),t∈[0,1].
Noting that 𝒟tαx=w,w=φp-1(v), it follows from (21) that the solution of (17) satisfies
(22)𝒟tαx(t)=φp-1(-Iβh(t)),t∈(0,1),x(0)=0,𝒟tγx(1)=∑j=1m-2‍aj𝒟tγx(ξj).By Lemma 7, the solution of (22) can be written as
(23)x(t)=-∫01‍G(t,s)φp-1(-Iβh(s))ds,t∈[0,1].
Since h(s)≥0, s∈[0,1], we have φp-1(-Iβh(s))=-(Iβh(s))q-1, s∈[0,1], which implies that the solution of (22) is given by
(24)x(t)=b∫01‍G(t,s)(∫0s(s-τ)β-1h(τ)dτ)q-1ds,∫01‍Ga(∫0s(s-τ)β-1h(τ)dτ)q-1t∈[0,1].The following lemma is a straightforward conclusion of Lemma9.Lemma 10.
Ifx∈C([0,1],R) satisfies
(25)x(0)=0,𝒟tγx(1)=∑j=1m-2‍aj𝒟tγx(ξj),
and -𝒟tαx(t)≥0 for any t∈(0,1), then x(t)≥0, for  t∈[0,1].


## 3. Main Results

Set(26)e(t)=tα-1.We present the following two assumptions.(
H
1
)
f
:
(
(
0,1
)
×
(
0
,
∞
)
→
[
0
,
+
∞
)
) is continuous and decreasing in x.(
H
2
) For any μ>0,f(t,μ)≢0, and
(27)0<∫01‍σ2(s)(∫0s‍(s-τ)β-1f(τ,μe(τ))dτ)q-1ds<+∞.LetE=C[0,1], and
(28)P={y∈E:thereexistpositivenumbers0<lx<1,Lx>1suchthatlxe(t)≤x(t)≤Lxe(t),a0<lx<1,Lx>1suchthatlxe(t)≤xdgt∈[0,1]}.
Clearly, e(t)∈P, so P is nonempty. For any x∈P, define an operator T by
(29)(Tλx)(t)=λb∫01‍G(t,s)×(∫0s‍(s-τ)β-1f(τ,x(τ))dτ)q-1ds,×(∫0s‍(s-τ)β-1(τ,x(τ))p)t∈[0,1].Theorem 11.
Suppose conditions(H1) and (H2) hold. In addition, if the following condition (S1) holds:
(
S
1
)
(30)
lim
κ
→
+
∞
κ
1
/
(
q
-
1
)
f
(
t
,
κ
x
)
=
+
∞
,
for (t,x)∈(0,1)×(0,∞)uniformly holds. Then there exists a constant λ*>0 such that the BVP (4) has at least one positive solution w for any λ∈(λ*,+∞), and there exists one positive constant n>1 such that
(31)e(t)≤w(t)≤ne(t),t∈[0,1].Proof.
The proof is divided into four steps.
Step  1. We show that Tλ is well defined on P and Tλ(P)⊂P, and Tλ is decreasing in x.
In fact, for anyx∈P, by the definition of P, there exists 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 8 and (H1)-(H2) that
(32)(Tλx)(t)=λb∫01‍G(t,s)(∫0s‍(s-τ)β-1f(τ,x(τ))dτ)q-1ds≤λbe(t)∫01σ2(s)×(∫0s‍(s-τ)β-1f(τ,lxe(τ))dτ)q-1ds<+∞.
Now takec=maxt∈[0,1]x(t), by (H2), for any s∈(0,1), f(s,c)≢0. Thus by the continuity of f(t,x) and Lemma 8 and (32), we have
(33)(Tλx)(t)≥λbe(t)∫01σ1(s)×(∫0s‍(s-τ)β-1f(τ,x(τ))dτ)q-1ds≥λbe(t)∫01‍σ1(s)(∫0s‍(s-τ)β-1f(τ,c)dτ)q-1ds>0,t∈(0,1).
Take(34)lx′=min{1,λb∫01‍σ1(s)(∫0s‍(s-τ)β-1f(τ,c)dτ)q-1ds},Lx′=max{(∫0s‍(s-τ)β-1f(τ,lxe(τ))dτ)1,λb×∫01σ2(s)(∫0s‍(s-τ)β-1f(τ,lxe(τ))dτ)q-1ds},
then by (32) and (33),
(35)lx′e(t)≤(Tλx)(t)≤Lx′e(t),
which implies that Tλ is well defined and Tλ(P)⊂P. And the operator Tλ is decreasing in x from (H1). Moreover, by direct computations, we also have
(36)-𝒟tβ(φp(𝒟tα(Tλx)))(t)=λf(t,x(t)),t∈(0,1),(Tλx)(0)=0,𝒟tα(Tλx)(0)=0,𝒟tγ(Tλx)(1)=∑j=1m-2‍aj𝒟tγ(Tλx)(ξj).
Step  2. In this step, we will focus on lower and upper solutions of the fractional boundary value problem (4).
By Lemma8, we have
(37)b∫01‍G(t,s)(∫0s‍(s-τ)β-1f(τ,e(τ))dτ)q-1dsa≥e(t)b∫01‍σ1(s)(∫0s‍(s-τ)β-1f(τ,e(τ))dτ)q-1ds,e(t)b∫01‍(∫0s‍(s-τ)β-1f(τ,e(τ))dτ)q-1∀t∈[0,1].
Let
(38)μ=(b∫01‍σ1(s)(∫0s‍(s-τ)β-1f(τ,e(τ))dτ)q-1ds)-1;
it follows from (37) that
(39)μb∫01‍G(t,s)(∫0s‍(s-τ)β-1f(τ,e(τ))dτ)q-1ds≥e(t),(∫0s‍(s-τ)β-1(τ,e(τ))dτ)q-1∀t∈[0,1].
On the other hand, take(40)ν(t)=b∫01‍G(t,s)(∫0s‍(s-τ)β-1f(τ,e(τ))dτ)q-1ds;
then by monotonicity of f in x and (37)–(40), for any λ>μ, we have
(41)∫01‍G(t,s)(∫0s‍(s-τ)β-1f(τ,λν(τ))dτ)q-1ds≤∫01‍G(t,s)(∫0s‍(s-τ)β-1f(τ,μν(τ))dτ)q-1ds≤∫01‍σ2(s)(∫0s‍(s-τ)β-1f(τ,e(τ))dτ)q-1ds<+∞.
From (S1), we have
(42)limκ→+∞κ1/(q-1)f(t,κx)=+∞,
uniformly on (t,x)∈(0,1)×(0,∞). Thus there exists large enough λ*>μ>0, such that, for any t∈(0,1),
(43)λ*1/(q-1)f(s,λ*e(s))≥βb(∫01‍σ1(s)sβ(q-1)ds)-1/(q-1),
which yields
(44)λ*b∫01‍G(t,s)(∫0s‍(s-τ)β-1f(τ,λ*e(τ))dτ)q-1ds≥β(∫01‍σ1(s)sβ(q-1)ds)-1/(q-1)a×∫01‍G(t,s)(∫0s‍(s-τ)β-1dτ)q-1ds≥β(∫01‍σ1(s)sβ(q-1)ds)-1/(q-1)ai×∫01‍σ1(s)(∫0s‍(s-τ)β-1dτ)q-1dse(t)=e(t),×∫01‍σ1(s)(∫0s‍(s-τ)β-1dτ)q-1ghjhg∀t∈[0,1].
Letting
(45)ϕ(t)=λ*b∫01‍G(t,s)(∫0s‍(s-τ)β-1f(τ,e(τ))dτ)q-1ds=λ*ν(t)=Tλ*e,ψ(t)=λ*b∫01‍G(t,s)(∫0s‍(s-τ)β-1f(τ,λ*ν(τ))dτ)q-1ds=Tλ*ϕ,
and by Lemma 9, (39), (44), and (45), one has
(46)ϕ(t)=λ*b∫01‍G(t,s)(∫0s‍(s-τ)β-1f(τ,e(τ))dτ)q-1ds≥e(t),ϕ(0)=0,𝒟tϕϕ(0)=0,𝒟tγ(ϕx)(1)=∑j=1m-2‍aj𝒟tγϕ(ξj),(47)ψ(t)=λ*b∫01‍G(t,s)(∫0s‍(s-τ)β-1f(τ,λ*ν(τ))dτ)q-1ds≥e(t),ψ(0)=0,𝒟tϕψ(0)=0,𝒟tγ(ψx)(1)=∑j=1m-2‍aj𝒟tγψ(ξj).
By Step  1 and (46), (47), we know ϕ(t),ψ(t)∈P. And it follows from (45)–(47) that
(48)e(t)≤ψ(t)≤ϕ(t),∀t∈[0,1].
Consequently, it follows from (44)–(48) that
(49)𝒟tβ(φp(𝒟tαψ))(t)+λ*f(t,ψ(t))=𝒟tβ(φp(𝒟tα(Tλ*ϕ)))(t)+λ*f(t,ψ(t))=-λ*f(t,ϕ(t))+λ*f(t,ψ(t))≥0,𝒟tβ(φp(𝒟tαϕ))(t)+λ*f(t,ϕ(t))=𝒟tβ(φp(𝒟tα(Tλ*e)))(t)+λ*f(t,ϕ(t))=-λ*f(t,e(t))+λ*f(t,ϕ(t))≤0;
that is, ϕ(t) and ψ(t) are a couple of lower and upper solutions of fractional boundary value problem (4) by (46)–(49), respectively.
Step  3. Let
(50)F(t,x)={f(t,ψ(t)),x<ψ(t),f(t,x(t)),ψ(t)≤x≤ϕ(t),f(t,ϕ(t)),x>ϕ(t).
It follows from (H1) and (46) that F:(0,1)×[0,+∞)→[0,+∞) is continuous.
We will show that the fractional boundary value problem(51)-𝒟tβ(φp(𝒟tαx))(t)=λ*F(t,x(t)),t∈(0,1),x(0)=0,𝒟tαx(0)=0,𝒟tγx(1)=∑j=1m-2‍aj𝒟tγx(ξj)
has a positive solution.
To see this, we consider the operatorAλ*:C[0,1]→C[0,1] defined as follows:
(52)(Aλ*x)(t)=λ*b∫01‍G(t,s)×(∫0s‍(s-τ)β-1F(τ,x(τ))dτ)q-1ds,∫0s‍(s-τ)β-1τ,x(τ)sdggdsit∈[0,1].
Obviously, a fixed point of the operator Aλ* is a solution of the BVP (51). 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 8, (50), and (H2) that
(53)(Aλ*x)(t)=≤λ*b∫01‍σ2(s)(∫0s‍(s-τ)β-1F(τ,x(τ))dτ)q-1ds=≤λ*b∫01‍σ2(s)(∫0s‍(s-τ)β-1f(τ,ϕ(τ))dτ)q-1ds=≤λ*b∫01‍σ2(s)(∫0s‍(s-τ)β-1f(τ,lϕe(τ))dτ)q-1ds=<+∞,
which implies that the operator Aλ* is uniformly bounded.
From the uniform continuity ofG(t,s) and the Lebesgue dominated convergence theorem, we easily obtain that A is equicontinuous. Thus by the means of the Arzela-Ascoli theorem, we have that Aλ*:E→E is completely continuous. The Schauder fixed point theorem implies that Aλ* has at least a fixed point w such that w=Aλ*w.
Step  4. We will prove that the boundary value problem (4) has at least one positive solution.
In fact, we only need to prove that(54)ψ(t)≤w(t)≤ϕ(t),t∈[0,1].
By (46), (47) and noticing that w is fixed point of Aλ*, we know that
(55)ϕ(0)=0,𝒟tγϕ(1)=∑j=1m-2‍aj𝒟tγϕ(ξj),𝒟tαϕ(0)=0,w(0)=0,𝒟tγw(1)=∑j=1m-2‍aj𝒟tγw(ξj),𝒟tαw(0)=0.
Notice that the definition of F and the function f(t,x) is nonincreasing in x, we obtain
(56)f(t,ϕ(t))≤F(t,x(t))≤f(t,ψ(t)),∀x∈E.
So by (48) and (56),
(57)f(t,ϕ(t))≤F(t,x(t))≤f(t,e(t)),∀x∈E.
Thus one has by (57)
(58)𝒟tβ(φp(𝒟tαϕ))(t)-𝒟tβ(φp(𝒟tαw))(t)=𝒟tβ(φp(𝒟tαϕ)-φp(𝒟tαw))(t)=-λ*f(t,e(t))+λ*F(t,w(t))≤0,=-λ*f(t,e(t))+λ*dfgd∀t∈[0,1].
Let z(t)=φp(𝒟tαϕ(t))-φp(𝒟tαw(t)); then
(59)𝒟tβz(t)≤0,t∈[0,1],
and (55) implies that z(0)=0. It follows from (21) that
(60)z(t)≤0,
and then
(61)φp(𝒟tαϕ(t))-φp(𝒟tαw(t))≤0.
Notice that φp is monotone increasing; we have
(62)𝒟tαϕ(t)≤𝒟tαw(t),thatis,𝒟tα(ϕ-w)(t)≤0.
It follows from Lemma 10 and (55) that
(63)ϕ(t)-w(t)≥0.
Thus we have w(t)≤ϕ(t) on [0,1]. By the same way, we also have w(t)≥ψ(t) on [0,1]. So
(64)ψ(t)≤w(t)≤ϕ(t),t∈[0,1].
Consequently, F(t,w(t))=f(t,w(t)),t∈[0,1]. Then w(t) is a positive solution of the problem (4).
Finally, by (48) and (64) and ϕ∈P, we have
(65)e(t)≤ψ(t)≤w(t)≤ϕ(t)≤lϕe(t)=ne(t),
where
(66)n=lϕ>1.In the end of this work we also remark the above results to the problem (4) with which f(t,x) is nonsingular at x=0 and t=0,1; that is, we have the following result.Theorem 12.
Iff(t,x):[0,1]×[0,+∞)→[0,+∞) is continuous, decreasing in x and f(t,μ)≢0, for any μ>0, then the boundary value problem (4) has at least one positive solution w(t) for any λ>0, and there exists a constant n>1 such that
(67)e(t)≤w(t)≤ne(t).Proof.
The proof is similar to Theorem11; we omit it here.Example 13.
Consider the following boundary value problem:(68)-𝒟t4/3(φ2(𝒟t3/2x))(t)=1t1/2x(t),t∈(0,1),x(0)=0,𝒟t1/6x(1)=18𝒟t1/6x(14)+13𝒟t1/6x(34),𝒟t3/2x(0)=0.Letα=3/2, β=4/3, γ=1/6, p=2, and
(69)f(t,x)=1xt1/2.
Firstly,
(70)c=∑j=1p-2‍ajξjα-γ-1=18(14)1/3+13(34)1/3=0.3816<1.
And, it is easy to check that (H1) holds. For any μ>0, f(t,μ)≢0 and
(71)0<∫01‍σ2(s)(∫0s‍(s-τ)β-1f(τ,μe(τ))dτ)q-1ds=∫01‍σ2(s)∫0s‍(s-τ)1/3τ-1/2μ-1/2dτds=∫01‍s5/6σ2(s)∫01‍(1-τ)1/3τ-1/2μ-1/2dτds=μ-1/2∫01‍s5/6σ2(s)dsB(43,12)<+∞,
which implies that (H2) holds.On the other hand,(72)limκ→+∞κ1/(q-1)f(t,κx)=limκ→+∞κ1κxt1/2=+∞.
Thus (S1) also holds.By Theorem11, the boundary value problem (68) has at least one positive solution.



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*Source: 137890-2013-03-25.xml*