# Existence of Solutions of Fractional Differential Equation withp-Laplacian Operator at Resonance

Authors: Zhigang Hu; Wenbin Liu; Jiaying Liu
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

By using the extension of Mawhin’s continuation theorem due to Ge, we consider boundary value problems for fractionalp-Laplacian equation. A new result on the existence of solutions for the fractional boundary value problem is obtained, which generalizes  and enriches some known results to some extent from the literature.


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## Body

## 1. Introduction and Preliminaries

Recently, fractional differential equations have played an important role in many fields such as physics, electrical circuits, and control theory (see [1–9]). Many scholars have paid more attention to boundary value problems for fractional differential equations (see [10–25]).By using a fixed point theorem on a cone, Agarwal et al. (see [10]) considered a two-point boundary value problem at nonresonance given by
(1)D0+αx(t)+f(t,x(t),D0+μx(t))=0,x(0)=x(1)=0,
where 1<α<2, μ>0 are real numbers, α-μ≥1, and D0+α is the Riemann-Liouville fractional derivative.By using the coincidence degree theory, Bai (see [20]) considered the following m-point fractional boundary value problems:
(2)D0+αu(t)=f(t,u(t),D0+α-1u(t))+e(t),0<t<1,I0+2-αu(t)|t=0=0,u(1)=∑i=1m-2‍βiu(ηi),
where 1<α≤2 is a real number, βi∈ℝ, ηi∈(0,1) are given constants such that ∑i=1m-2‍βiηim-1=1, and D0+α,I0+α are the Riemann-Liouville differentiation and integration.The turbulent flow in a porous medium is a fundamental mechanics problem. For studying this type of problems, Leibenson (see [26]) introduced the p-Laplacian equation as follows:
(3)(ϕp(x′(t)))′=f(t,x(t),x′(t)),
where ϕp(s)=|s|p-2s, p>1. Obviously, ϕp is invertible and its inverse operator is ϕq, where q>1 is a constant such that 1/p+1/q=1.In the past few decades, many important results relative to (3) with certain boundary value conditions have been obtained. We refer the reader to [27–31] and the references cited therein. However, to the best of our knowledge, there are relatively few results on boundary value problems for fractional p-Laplacian equations.Motivated by the work above, in this paper, we investigate the existence of solutions for boundary value problem (BVP for short) of fractionalp-Laplacian equation with the following form:
(4)D0+βϕp(D0+αx(t))=f(t,x(t),D0+αx(t)),t∈[0,1],D0+αx(0)=D0+αx(1)=x′(0)=0,
where 0<β≤1, 1<α≤2, D0+α is Caputo fractional derivative, and f:[0,1]×ℝ2→ℝ is continuous.BVP (4) happens to be at resonance in the sense that its associated linear homogeneous boundary value problem
(5)D0+βϕp(D0+αx(t))=0,t∈[0,1],D0+αx(0)=D0+αx(1)=x′(0)=0
has a nontrivial solution x(t)=c, where c∈ℝ.For the convenience of the reader, we present here some necessary basic knowledge and definitions about fractional calculus theory, which can be found, for instance, in [32–35].Definition 1.
The Riemann-Liouville fractional integral operator of orderα>0 of a function x:(0,+∞)→ℝ is given by
(6)I0+αx(t)=1Γ(α)∫0t‍(t-s)α-1x(s)ds,
provided that the right side integral is pointwise defined on (0,+∞).Definition 2.
The Caputo fractional derivative of orderα>0 of a continuous function x:(0,+∞)→ℝ is given by
(7)D0+αx(t)=I0+n-αdnx(t)dtn=1Γ(n-α)∫0t‍(t-s)n-α-1x(n)(s)ds,
where n is the smallest integer greater than or equal to α, provided that the right side integral is pointwise defined on (0,+∞).Lemma 3.
Assume thatD0+αx∈C[0,1], α>0. Then
(8)I0+αD0+αx(t)=x(t)+c0+c1t+c2t2+⋯+cn-1tn-1,
where ci=-x(i)(0)/i!, i=0,1,2,…,n-1, and here n is the smallest integer greater than or equal to α.
Now, one briefly recalls some notations and an abstract existence result, which can be found in [36].Definition 4.
LetX and Y be two Banach spaces with norms ∥·∥X and ∥·∥Y, respectively. A continuous operator
(9)M:X∩dom⁡M⟶Y
is said to be quasilinear if(i)
ImM∶=M(X∩dom⁡M) is a closed subset of Y,(ii)
KerM∶={X∩dom⁡M:Mu=0} is linearly homeomorphic to ℝn, n<∞.Definition 5.
LetX be a real Banach space and let X^⊂X. The operator P:X→X^ is said to be a projector provided that P2=P, P(λ1x1+λ2x2)=λ1P(x1)+λ2P(x2) for x1,x2∈X and λ1,λ2∈ℝ. The operator Q:X→X^ is said to be a semiprojector provided Q2=Q.Definition 6 (see [36]).
LetX^=Ker⁡M and let X~ be the complement space of X^ in X, and then X=X^⊕X~. On the other hand, suppose that Y^ is a subspace of Y and Y~ is the complement space of Y^ in Y so that Y=Y^⊕Y~. Let P:X→X^ be a projector, let Q:Y→Y^ be a semiprojector, and let Ω⊂X be an open and bounded set with origin θ∈Ω, where θ is the origin of a linear space.
Suppose thatNλ:Ω¯→Y, λ∈[0,1], is a continuous operator. Denote N1 by N. Let Σλ={x∈Ω¯:Mx=Nλx}. Nλ is said to be M-compact in Ω¯ if there is Y^⊂Y with dimY^ = dimX^ and an operator R:Ω¯×[0,1]→X continuous and compact such that, for λ∈[0,1],
(10)(I-Q)Nλ(Ω¯)⊂Im⁡M⊂(I-Q)Y,(11)QNλx=θ,λ∈(0,1)⟺QNx=θ,(12)R(·,0)isthezerooperatorandR(·,λ)|Σλ=(I-P)|Σλ,(13)M[P+R(·,λ)]=(I-Q)Nλ.Lemma 7 (see [36] Ge-Mawhin’s continuation theorem).
LetX and Y be two Banach spaces with norms ∥·∥X and ∥·∥Y, respectively. Ω⊂X is an open and bounded nonempty set. Suppose that
(14)M:X∩dom⁡M⟶Y
is a quasilinear operator and
(15)Nλ:Ω¯⟶Y,λ∈[0,1]
is M-compact in Ω¯. In addition, if(C1)
Mx≠Nλx, ∀(x,λ)∈(dom⁡M∩∂Ω)×(0,1),(C2)
QNx≠0, for x∈dom⁡M∩∂Ω,(C3)
deg⁡(JQN,Ker⁡M∩Ω,0)≠0,where N=N1, then the equation Mx=Nx has at least one solution in Ω-.In this paper, we takeY=C[0,1] with the norm ∥x∥∞=max⁡t∈[0,1]|x(t)| and X={x∣x,D0+αx∈Y} with the norm ∥x∥X=max⁡{∥x∥∞,∥D0+αx∥∞}. By means of the linear functional analysis theory, we can prove that X is a Banach space.Define the operatorM:dom⁡M⊂X→Y by
(16)Mx=D0+βϕp(D0+αx),
where
(17)dom⁡M={x∈X∣D0+βϕp(D0+αx)∈Y,sD0+αx(0)=D0+αx(1)=x′(0)=0}.
Define the operator N:X→Y by
(18)Nx(t)=f(t,x(t),D0+αx(t)),∀t∈[0,1].
Then BVP (4) is equivalent to the operator equation. Consider
(19)Mx=Nx,x∈dom⁡M.


## 2. Main Result

We will always assume that the nonlinearityf(t,u,v) will be retained:(H1)
there exist nonnegative functionsa,b,c∈Y such that
(20)|f(t,u,v)|≤a(t)+b(t)|u|p-1+c(t)|v|p-1,∀t∈[0,1],a(t)+b(t)|u|p-1+c(t)|v|p-1ddffs(u,v)∈ℝ2;(H2)
there exists a constantB>0 such thateither
(21)uf(t,u,v)>0,∀t∈[0,1],v∈ℝ,|u|>B,
or
(22)uf(t,u,v)<0,∀t∈[0,1],v∈ℝ,|u|>B.
Moreover, we will always assume that f:[0,1]×ℝ2→ℝ is continuous and
(23)1Γ(β+1)(2∥b∥∞(Γ(α+1))p-1+∥c∥∞)<1.Now, we begin with some lemmas below.Lemma 8.
LetM be defined by (16), and then
(24)Ker⁡M={x∈X∣x(t)=c∈ℝ,∀t∈[0,1]},(25)Im⁡M={y∈Y∣∫01‍(1-s)β-1y(s)ds=0},
and M is a quasilinear operatorProof.
By Lemma3, D0+βϕp(D0+αx(t))=0 has solution:
(26)x(t)=c0+c1t+I0+αϕq(c2)=c0+c1t+ϕq(c2)Γ(α+1)tα,c0,c1,c2∈ℝ,
which satisfies
(27)D0+αx(t)=ϕq(c2).
Combining with the boundary value condition D0+αx(0)=0 and x′(0)=0, we can get that (24) holds.
Ify∈Im⁡M, then there exists a function x∈dom⁡M such that y(t)=D0+βϕp(D0+αx(t)). Based on Lemma 3, we have
(28)D0+αx(t)=ϕq(I0+βy(t)+c)=ϕq(1Γ(β)∫0t‍(t-s)β-1y(s)ds+c),c∈ℝ.
From condition D0+αx(0)=0, one has c=0. By the condition D0+αx(1)=0, we obtain that
(29)∫01‍(1-s)β-1y(s)ds=0.
Thus, we get (25).
Then we have dim KerM=1 and M(dom⁡M∩X)⊂Y closed. Therefore, M is a quasilinear operator.Lemma 9.
LetΩ⊂X be an open and bounded set; then Nλ is M-compact in Ω¯.Proof.
Define the continuous projectorP:X→X^ and the semiprojector Q:Y→Y^:
(30)Px(t)=x(0),∀t∈[0,1],Qy(t)=β∫01‍(1-s)β-1y(s)ds,∀t∈[0,1].
where X^=Ker⁡M and Y^=Im⁡Q.
Obviously,Im⁡P=Ker⁡M and P2x(t)=Px(t). It follows from x=(x-Px)+Px that X=Ker⁡P+Ker⁡M. By a simple calculation, we can get Ker⁡M∩Ker⁡P={0}. Then we get
(31)X=Ker⁡M⊕Ker⁡P=X^⊕X~.
By the definition ofQ, we can get
(32)Q2y=Qy·β∫01‍(1-s)β-1ds=Qy.
Let y=(y-Qy)+Qy, where y-Qy∈Ker⁡Q=Im⁡M, Qy∈Im⁡Q. It follows from Ker⁡Q=Im⁡M and Q2y=Qy that Im⁡Q∩Im⁡M={0}. Then, we have
(33)Y=Im⁡Q⊕Im⁡M=Y^⊕Y~.
Thus
(34)dim⁡X^=dimKerM=dimImQ=dim⁡Y^.
LetΩ⊂X be an open and bounded set with θ∈Ω. For each x∈Ω, we can get Q[(I-Q)Nλx]=0. Thus, (I-Q)Nλx∈Im⁡M=Ker⁡Q. Take any y∈Im⁡M in the type y=(y-Qy)+Qy. Since Qy=0, we can get (I-Q)y∈Y. So (10) holds. It is easy to verify (11).
Furthermore, defineR:Ω¯×[0,1]→X~ by
(35)R(x,λ)(t)=1Γ(α)∫0t(t-s)α-1ϕqcccccccccxx×(1Γ(β)∫0s‍(s-τ)β-1cccccccccccccccccccccc×((I-Q)Nλx(τ))dτ1Γ(β)∫0s‍)ds.
By the continuity off, it is easy to get that R(x,λ) is continuous on Ω¯×[0,1]. Moreover, for all x∈Ω¯, there exists a constant T>0 such that |I0+β(I-Q)Nλx(τ))|≤T, so we can easily obtain that R(Ω¯,λ) is uniformly bounded. By the Arzelà-Ascoli theorem, we just need to prove that R:Ω¯×[0,1]→X~ is equicontinuous. Furthermore, for 0≤t1<t2≤1, (x,λ)∈Ω¯×[0,1], we have
(36)|R(x,λ)(t2)-R(x,λ)(t1)|=|I0+αϕq(I0+β(I-Q)Nλx(t2))mmmml-I0+αϕq(I0+β(I-Q)Nλx(t1))|.
By|I0+β(I-Q)Nλx|≤T, we have
(37)|I0+αϕq(I0+β(I-Q)Nλx(t2))-I0+αϕq(I0+β(I-Q)Nλx(t1))|≤1Γ(α)|∫0t2‍(t2-s)α-1ϕq(I0+β(I-Q)Nλx(s))dsvvvvvv-∫0t1‍(t1-s)α-1ϕq(I0+β(I-Q)Nλx(s))ds|≤ϕq(T)Γ(α)[∫0t1‍(t2-s)α-1-(t1-s)α-1dsvvvvvvv+∫t1t2‍(t2-s)α-1ds]=ϕq(T)Γ(α+1)(t2α-t1α).
Since tα is uniformly continuous on [0,1], so R(Ω¯,λ) is equicontinuous. Similarly, we can get that I0+β((I-Q)Nλx(τ))⊂C[0,1] is equicontinuous, and considering that ϕq(s) is uniformly continuous on [-T,T], we get that D0+αR(Ω¯,λ)=I0+β((I-Q)Nλ(Ω¯)) is also equicontinuous. So we can obtain that R(Ω¯,λ)→X~ is compact.
For eachx∈Σλ={x∈Ω¯:Mx=Nλx}, we have D0+βϕp(D0+αx(t))=Nλx(t)∈Im⁡M. Thus,
(38)R(x,λ)(t)=1Γ(α)∫0t(t-s)α-1ϕqcccccccccxx×(1Γ(β)∫0s‍(s-τ)β-1cccccccccccccccccccccc×((I-Q)Nλx(τ))dτ1Γ(β)∫0s‍)ds.=1Γ(α)∫0t(t-s)α-1ϕqxxxxxxxx×(1Γ(β)∫0s‍(s-τ)β-1xxxxxxxxxxxxxxxxxx×D0+βϕp(D0+αx(τ))dτ1Γ(β)∫0s‍)ds,
which together with D0+αx(0)=x′(0)=0 yields that
(39)R(x,λ)(t)=x(t)-x(0)=[(I-P)x](t).
It is easy to verify that R(x,0)(t) is the zero operator. So (12) holds.
On the other hand, consider(40)M[Px+R(x,λ)](t)=M[1Γ(α)222222×∫0t‍(t-s)α-1ϕq(1Γ(β)∫0s‍(s-τ)β-12222222222222222222222222×((I-Q)Nλx(τ))dτ1Γ(β)∫0s‍)ds2222222+x(0)1Γ(α)]=[((I-Q)Nλ)x](t).
So (13) holds. Then we get that Nλ is M-compact in Ω¯. The proof is complete.Lemma 10.
Suppose that(H1), (H2) hold; then the set
(41)Ω1={x∈dom⁡M∖Ker⁡M∣Mx=λNx,λ∈(0,1)}
is bounded.Proof.
Takex∈Ω1; then Mx=λNx, D0+αx(0)=x′(0)=0, and Nx∈Im⁡M. By (25), we have
(42)∫01‍(1-s)β-1f(s,x(s),D0+αx(s))ds=0.
Then, by the integral mean value theorem, there exists a constant ξ∈(0,1) such that f(ξ,x(ξ),D0+αx(ξ))=0. So, from (H2), we get |x(ξ)|≤B. By x′(0)=0, we get
(43)x(t)=x(0)+x′(0)t+I0+αD0+αx(t)=x(0)+1Γ(α)∫0t‍(t-s)α-1D0+αx(s)ds.
Take t=ξ; we have
(44)x(ξ)=x(0)+1Γ(α)∫0ξ‍(ξ-s)α-1D0+αx(s)ds.
Then we have
(45)|x(0)|≤|x(ξ)|+1Γ(α)∫0ξ‍(ξ-s)α-1|D0+αx(s)|ds≤|x(ξ)|+1Γ(α)∥D0+αx∥∞·1αξα≤B+1Γ(α+1)∥D0+αx∥∞.
So we get
(46)|x(t)|≤|x(0)|+1Γ(α)∫0t‍(t-s)α-1|D0+αx(s)|ds≤|x(0)|+1Γ(α)∥D0+αx∥∞·1αtα≤B+2Γ(α+1)∥D0+αx∥∞,∀t∈[0,1].
That is,
(47)∥x∥∞≤B+2Γ(α+1)∥D0+αx∥∞.
By Mx=λNx and D0+αx(0)=0, we get
(48)ϕp(D0+αx(t))=λI0+βNx(t)=λΓ(β)∫0t‍(t-s)β-1f(s,x(s),D0+αx(s))ds.
So, from (H1), we have
(49)|ϕp(D0+αx(t))|≤1Γ(β)∫0t‍(t-s)β-1|f(s,x(s),D0+αx(s))|ds≤1Γ(β)∫0t(t-s)β-1‍(|D0+αx(s)|p-1a(s)+b(s)|x(s)|p-1vvvvvvvvvvvvvvvvvv+c(s)|D0+αx(s)|p-1)ds≤1Γ(β)(∥D0+αx∥∞p-1∥a∥∞+∥b∥∞∥x∥∞p-1vvvvvvv+∥c∥∞∥D0+αx∥∞p-1)·1βtβ≤1Γ(β+1)(∥D0+αx∥∞p-1∥a∥∞+∥b∥∞∥x∥∞p-1vvvvvvvvvv+∥c∥∞∥D0+αx∥∞p-1),∀t∈[0,1],
which together with |ϕp(D0+αx(t))|=|D0+αx(t)|p-1 and (47) yields that
(50)∥D0+αx∥∞p-1≤1Γ(β+1)[∥D0+αx∥∞p-1∥a∥∞+∥b∥∞2222222222×(B+2Γ(α+1)∥D0+αx∥∞)p-12222222222+∥c∥∞∥D0+αx∥∞p-1].
In view of (23), we can see that there exists a constant M1>0 such that
(51)∥D0+αx∥∞≤M1.
Thus, from (47), we get
(52)∥x∥∞≤B+2M1Γ(α+1):=M2.
Combining (51) with (52), we have
(53)∥x∥X=max⁡{∥x∥∞,∥D0+αx∥∞}≤max⁡{M1,M2}∶=M.
Therefore, Ω1 is bounded. The proof is complete.Lemma 11.
Suppose that(H2) holds; then the set
(54)Ω2={x∈Ker⁡M∣Nx∈Im⁡M}
is bounded.Proof.
Forx∈Ω2, we have x(t)=c, c∈ℝ and Nx∈Im⁡M. Then we get
(55)∫01‍(1-s)β-1f(s,c,0)ds=0,
which together with (H2) implies |c|≤B. Thus, we have
(56)∥x∥X≤max⁡{B,0}=B.
Hence, Ω2 is bounded. The proof is complete.Lemma 12.
Suppose that the first part of(H2) holds; then the set
(57)Ω3={x∈Ker⁡M∣λx+(1-λ)QNx=0,λ∈[0,1]}
is bounded.Proof.
Forx∈Ω3, we have x(t)=c, c∈ℝ, and
(58)λc+(1-λ)β∫01‍(1-s)β-1f(s,c,0)ds=0.
If λ=0, then |c|≤B because of the first part of (H2). If λ∈(0,1], we can also obtain |c|≤B. Otherwise, if |c|>B, in view of the first part of (H2), one has
(59)λc2+(1-λ)β∫01‍(1-s)β-1cf(s,c,0)ds>0,
which contradicts (58). Therefore, Ω3 is bounded. The proof is complete.Remark 13.
If the second part of(H2) holds, then the set
(60)Ω3′={x∈Ker⁡M-λx+(1-λ)QNx=0,λ∈[0,1]}
is bounded.Theorem 14.
Letf:[0,1]×ℝ2→ℝ be continuous. Suppose that (H1), (H2) hold. Then BVP (4) has at least one solution.Proof.
SetΩ={x∈X|∥x∥X<max⁡{M,B}+1}. It follows from Lemmas 8 and 9 that M is a quasilinear operator and Nλ is M-compact on Ω¯. By Lemmas 10 and 11, we get that the following two conditions are satisfied:(C1)
Mx≠Nλx,∀(x,λ)∈(dom⁡M∩∂Ω)×(0,1),(C2)
QNx≠0, for x∈dom⁡M∩∂Ω.Take
(61)H(x,λ)=±λx+(1-λ)QNx.
According to Lemma 12 (or Remark 13), we know that H(x,λ)≠0 for x∈Ker⁡M∩∂Ω. Therefore
(62)deg⁡(QN|Ker⁡M,Ω∩Ker⁡M,0)=deg⁡(H(·,0),Ω∩Ker⁡M,0)=deg⁡(H(·,1),Ω∩Ker⁡M,0)=deg⁡(±I,Ω∩Ker⁡M,0)≠0.
So the condition (C3) of Lemma 7 is satisfied. By Lemma 7, we can get that Mx=Nx has at least one solution in dom⁡M∩Ω¯. Therefore BVP (4) has at least one solution. The proof is complete.


## 3. Example

In this section, we will give an example to illustrate our main result.Example 1.
Consider the following BVP:(63)D0+3/4ϕ3(D0+3/2x(t))=-253+13x2(t)+te-|D0+3/2x(t)|,t∈[0,1],D0+3/2x(0)=D0+3/2x(1)=x′(0)=0.
Corresponding to BVP (4), we get that p=3, α=3/2, β=3/4, and
(64)f(t,u,v)=-253+13u2+te-|v|.
Choose a(t)=10, b(t)=1/3, c(t)=0, B=5. By a simple calculation, we can get that ∥b∥∞=1/3,∥c∥∞=0 and
(65)1Γ(3/4+1)(2/3(Γ(3/2+1))2+0)<1.
Obviously, BVP (63) satisfies all conditions of Theorem 14. Hence, it has at least one solution.


## 4. Conclusions

In this paper, the boundary value problem forp-Laplacian equation at resonance is investigated. In view of the boundary value problem (4) is equivalent to the operator equation (19); we only need to find a fixed point of the operator equation (19). Firstly, we established the sufficient conditions of existence of boundary value problem for p-Laplacian equation. Then, by using the extension of Mawhin’s continuation theorem due to Ge, we got the fixed point of operator equation (19).



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