A four-point coupled boundary value problem of fractional differential
equations is studied. Based on Mawhin’s coincidence degree theory, some
existence theorems are obtained in the case of resonance.
1. Introduction
In this paper, we are concerned with the following four-point coupled boundary value problem for nonlinear fractional differential equation. Consider
(1)D0+αu(t)=f(t,u(t),D0+α-1u(t),v(t),D0+β-1v(t)),D0+βv(t)=g(t,u(t),D0+α-1u(t),v(t),D0+β-1v(t)),I0+2-αu(t)|t=0=0,u(1)=av(ξ),I0+2-βv(t)|t=0=0,v(1)=bu(η),
where 1<α,β<2, D0+α and I0+α are the standard Riemann-Liouville differentiation and integration, f,g∈C([0,1]×ℝ4,ℝ), a,b∈ℝ, ξ,η∈(0,1), and
(2)abξβ-1ηα-1=1.
The subject of fractional calculus has gained considerable popularity and importance because of its intensive development of the theory of fractional calculus itself and its varied applications in many fields of science and engineering. As a result, the subject of fractional differential equations has attracted much attention; see [1–11] for a good overview.
At the same time, we notice that coupled boundary value problems, which arise in the study of reaction-diffusion equations and Sturm-Liouville problems, have wide applications in various fields of sciences and engineering, for example, the heat equation [12–14] and mathematical biology [15, 16]. In [17], Asif and Khan used the Guo-Krasnosel’skii fixed-point theorem to show the existence of positive solutions to the nonlinear differential system with coupled four-point boundary value conditions
(3)-x′′(t)=f(t,x(t),y(t)),t∈(0,1),-y′′(t)=g(t,x(t),y(t)),t∈(0,1),x(0)=y(0)=0,x(1)=αy(ξ),y(1)=βx(η),
where ξ,η∈(0,1), 0<αβξη<1, and f,g:(0,1)×[0,+∞)×[0,+∞)→[0,+∞) are two given continuous functions.
In [18], the authors considered the existence of positive solutions of four-point coupled boundary value problem for systems of the nonlinear semipositone fractional differential equation
(4)D0+αu+λf(t,u,v)=0,t∈(0,1),λ>0,D0+αv+λg(t,u,v)=0,u(i)(0)=v(i)(0)=0,0≤i≤n-2,u(1)=av(ξ),v(1)=bu(η),
where λ is a parameter, a,b,ξ,η satisfy ξ,η∈(0,1), 0<abξη<1, α∈(n-1,n] is a real number and n≥3, and D0+αu is Riemann-Liouville’s fractional derivative.
Recently, Cui and Sun [19] showed the existence of positive solutions of singular superlinear coupled integral boundary value problems for differential systems
(5)-x′′(t)=f1(t,x(t),y(t)),t∈(0,1),-y′′(t)=f2(t,x(t),y(t)),t∈(0,1),x(0)=y(0)=0,x(1)=α[y],y(1)=β[x],
where α[x],β[x] are bounded linear functionals on C[0,1] given by
(6)α[x]=∫01x(t)dA(t),β[x]=∫01x(t)dB(t)
with A,B being functions of bounded variation with positive measures.
A key assumption in the above papers is that the case studied is not at resonance; that is, the associated fractional (or ordinary) linear differential operators are invertible. In this paper, instead, we are interested in the resonance case due to the critical condition (2). Boundary value problems at resonance have been studied by several authors including the most recent works [20–31]. In this paper, we establish new results on the existence of a solution for the couple boundary value problems at resonance. Our method is based on the coincidence degree theorem of Mawhin [32, 33].
Now, we briefly recall some notations and an abstract existence result.
Let Y,Z be real Banach spaces and let L:domL⊂Y→Z be a Fredholm operator of index zero. P:Y→Y and Q:Z→Z are continuous projectors such that
(7)ImP=KerL,KerQ=ImL,Y=KerL⊕KerP,Z=ImL⊕ImQ.
It follows that L|domL∩KerP:domL∩KerP→ImL is invertible. We denote the inverse of the mapping by KP (generalized inverse operator of L). If Ω is an open bounded subset of Y such that domL∩Ω≠∅, the mapping N:Y→Z will be called L-compact on Ω¯ if QN(Ω¯) is bounded and KP(I-Q)N:Ω¯→Y is compact.
Theorem 1 (see [32, 33]).
Let L be a Fredholm operator of index zero and let N be L-compact on Ω¯. Assume that the following conditions are satisfied.
Lx≠λNx for every (x,λ)∈[(domL∖KerL)∩∂Ω]×(0,1).
Nx∉ImL for every x∈KerL∩∂Ω.
deg(QN|KerL,KerL∩Ω,0)≠0, where Q:Z→Z is a projector as above with ImL=KerQ.
Then the equation Lx=Nx has at least one solution in domL∩Ω¯.
For convenience, let us set the following notations:
(8)Δ1=max{1+1Γ(β),(1+1Γ(α))×(1+aξβ-1Γ(α)[bΓ(α)+1Γ(β)])},Δ2=Γ(β+1)Γ(α+1)αξβ-1Γ(α+1)(ξ-1)+Γ(β+1)(η-1),Δ3=1Γ(β)max{aξβ-1(Γ(α)+1),Γ(β)+1},Δ4=max{Δ1∥a1∥1+Δ3∥a2∥1,Δ1∥b1∥1+Δ3∥a2∥1},Δ5=max{Δ1∥c1∥1+Δ3∥c2∥1,Δ1∥d1∥1+Δ3∥d2∥1}.
2. Preliminaries and Lemmas
In this section, first we provide recall of some basic definitions and lemmas of the fractional calculus, which will be used in this paper. For more details, we refer to books [1, 2, 4].
Definition 2 (see [1, 4]).
The Riemann-Liouville fractional integral of order α>0 of a function u:(0,∞)→ℝ is given by
(9)I0+αu(t)=1Γ(α)∫0t(t-s)α-1u(s)ds,
provided that the right-hand side is pointwise defined on (0,∞).
Definition 3 (see [1, 4]).
The Riemann-Liouville fractional derivative of order α>0 of a continuous function u:(0,∞)→ℝ is given by
(10)D0+αu(t)=1Γ(n-α)(ddt)n∫0tu(s)(t-s)α-n+1ds,
where n-1≤α<n, provided that the right-hand side is pointwise defined on (0,∞).
We use the classical Banach space C[0,1] with the norm ∥u∥∞=maxt∈[0,1]|u(t)| and L1[0,1] with the norm ∥u∥1=∫01|u(t)|dt. We also use the space ACn[0,1] defined by
(11)ACn[0,1]={u:[0,1]⟶ℝ∣u(n-1)areabsolutelycontinuouson[0,1]}
and the Banach space Cμ[0,1] (μ>0)
(12)Cμ[0,1]={u(t)∣u(t)=I0+μx(t)+c1tμ-1+c2tμ-2+⋯+cN-1tμ-(N-1),x∈C[0,1],t∈[0,1],ci∈ℝ,i=1,2,…,N=[μ]+1c1tμ-1}
with the norm ∥u∥Cμ=∥D0+μu∥∞+⋯+∥D0+μ-(N-1)u∥∞+∥u∥∞.
Lemma 4 (see [1]).
Let α>0, n=[α]+1. Assume that u∈L1(0,1) with a fractional integration of order n-α that belongs to ACn[0,1]. Then the equality
(13)(I0+αD0+αu)(t)=u(t)-∑i=1n((I0+n-αu)(t))(n-i)|t=0Γ(α-i+1)tα-i
holds almost everywhere on [0,1].
In the following lemma, we use the unified notation of both for fractional integrals and fractional derivatives assuming that I0+α=D0+-α for α<0.
Lemma 5 (see [1]).
Assume that α>0; then
let k∈ℕ. If Da+αu(t) and (Da+α+ku)(t) exist, then
(14)(DkDa+α)u(t)=(Da+α+ku)(t);
if β>0,α+β>1, then
(15)(Ia+αIa+β)u(t)=(Ia+α+βu)(t)
satisfies at any point on [a,b] for u∈Lp(a,b) and 1≤p≤+∞;
let u∈C[a,b]. Then (Da+αIa+α)u(t)=u(t) holds on [a,b];
note that, for λ>-1, λ≠α-1,α-2,…,α-n, one has
(16)Dαtλ=Γ(λ+1)Γ(λ-α+1)tλ-α,Dαtα-i=0,i=1,2,…,n.
Remark 6.
If 1<α<2 and u satisfies D0+αu=f(t)∈L1(0,1) and I0+2-αu|t=0=0, then u∈Cα-1[0,1]. In fact, with Lemma 4, one has
(17)u(t)=I0+αf(t)+c1tα-1+c2tα-2.
Combined with I0+2-αu|t=0=0, there is c2=0. So
(18)u(t)=I0+αf(t)+c1tα-1=I0+α-1[I0+1f(t)+c1Γ(α)].
Lemma 7 (see [34]).
F⊂Cμ[0,1] is a sequentially compact set if and only if F is uniformly bounded and equicontinuous. Here to be uniformly bounded means that there exists M>0 such that for every u∈F(19)∥u∥Cμ=∥D0+μu∥∞+⋯+∥D0+μ-(N-1)u∥∞+∥u∥∞<M
and to be equicontinuous means that for allϵ>0, ∃δ>0 and for all t1,t2∈[0,1], |t1-t2|<δ, u∈F, and i∈{1,…,[μ]}, the following holds:
(20)|u(t1)-u(t2)|<ɛ,|D0+μ-iu(t1)-D0+μ-iu(t2)|<ɛ.
We also use the following two Banach spaces Y=Cα-1[0,1]×Cβ-1[0,1] with the norm
(21)∥(x,y)∥Y=max{∥x∥Cα-1,∥y∥Cβ-1}
and Z=L1[0,1]×L1[0,1] with the norm
(22)∥(x,y)∥Z=max{∥x∥1,∥y∥1}.
Let the linear operator L:domL⊂Y→Z with
(23)domL={I0+2-β(u,v)∈Y:I0+2-αu(t)|t=0=0,u(1)=av(ξ),I0+2-βv(t)|t=0=0,v(1)=bu(η)}
be defined by
(24)L(u,v)=(L1u,L2v),
where L1:Cα-1[0,1]→L1[0,1] and L2:Cβ-1[0,1]→L1[0,1] are defined by
(25)L1u=D0+αu(t),L2v=D0+βv(t).
Let the nonlinear operator N:Y→Z be defined by
(26)(N(u,v))(t)=(N1(u,v)(t),N2(u,v)(t)),
where N1,N2:Y→L1[0,1] are defined by
(27)N1(u,v)(t)=f(t,u(t),D0+α-1u(t),v(t),D0+β-1v(t)),N2(x,y)(t)=g(t,u(t),D0+α-1u(t),v(t),D0+β-1v(t)).
Then four-point coupled boundary value problems (1) can be written as
(28)L(u,v)=N(u,v).
Lemma 8.
Let L be the linear operator defined as above. If (2) holds, then
(29)KerL={(u,v)∈domL:(u,v)=c(aξβ-1tα-1,tβ-1),c∈ℝ,t∈[0,1](aξβ-1tα-1,tβ-1)},(30)ImL={(x,y)∈Z:aI0+βy(ξ)-I0+αx(1)+aξβ-1[bI0+αx(η)-I0+βy(1)]=0}.
Proof.
Let u(t)=aξβ-1tα-1and let v(t)=tβ-1. Then by Lemma 5, we have I0+2-αu(t)|t=0=I0+2-βv(t)|t=0=0, u(1)=aξβ-1=av(ξ), v(1)=1=bu(η), and D0+αu(t)=D0+βv(t)=0. So
(31){(u,v)∈domL:(u,v)=c(aξβ-1tα-1,tβ-1),c∈ℝ}⊂KerL.
For every (u,v)∈KerL, if D0+αu(t)=D0+βv(t)=0, then
(32)u(t)=c1tα-1+c2tα-2,v(t)=c3tβ-1+c4tβ-2.
Considering that I0+2-αu(t)|t=0=I0+2-βv(t)|t=0=0, u(1)=av(ξ) and v(1)=bu(η), we can obtain that c2=c4=0 and c1:c3=aξβ-1. It yields the following:
(33)KerL⊂{(aξβ-1tα-1,tβ-1)(u,v)∈domL:(u,v)=c(aξβ-1tα-1,tβ-1),c∈ℝ}.
Let (x,y)∈ImL; then there is (u,v)∈domL such that (x,y)=L(u,v); that is, u∈Cα-1[0,1], D0+αu(t)=x(t) and v∈Cβ-1[0,1], D0+βv(t)=y(t). By Lemma 4,
(34)I0+αx(t)=u(t)-((D0+α-1u)(t))|t=0Γ(α)tα-1-((I0+2-αu)(t))|t=0Γ(α-1)tα-2,I0+βy(t)=v(t)-((D0+β-1v)(t))|t=0Γ(β)tβ-1-((I0+2-βv)(t))|t=0Γ(β-1)tβ-2
and by the couple boundary conditions, we have
(35)u(t)=I0+αx(t)+c1tα-1,v(t)=I0+βy(t)+c2tβ-1,u(1)=I0+αx(1)+c1=av(ξ)=a[I0+βy(ξ)+c2ξβ-1],v(1)=I0+βy(1)+c2=bu(η)=b[I0+αy(η)+c1ηα-1].
It yields the following:
(36)aI0+βy(ξ)-I0+αx(1)+aξβ-1[bI0+αx(η)-I0+βy(1)]=0.
On the other hand, suppose that (x,y)∈Z satisfy (36). Let u(t)=I0+αx(t)+aξβ-1tα-1 and v(t)=I0+βy(t)+[bI0+αx(η)-I0+βy(1)+1]tβ-1, and then D0+αu(t)=x(t), D0+βv(t)=y(t), and
(37)I0+2-αu(t)|t=0=I0+2-βv(t)|t=0=0,u(1)=I0+αx(1)+aξβ-1=aI0+βy(ξ)+a[bI0+αx(η)-I0+βy(1)+1]ξβ-1=av(ξ),v(1)=I0+βy(1)+[bI0+αx(η)-I0+βy(1)+1]=bI0+αx(η)+abξβ-1ηα-1=bu(η).
Therefore, (30) holds.
Lemma 9.
If (2) holds, then L is a Fredholm operator of index zero and dimKerL=
codimIm
L=1. Furthermore, the linear operator Kp:ImL→domL∩KerP can be defined by
(38)KP(u,v)(t)={I0+αu(t)-aξβ-1tα-1[bI0+αu(η)-I0+βv(1)],I0+βv(t)}.
Also
(39)∥Kp(u,v)∥Y≤Δ1∥(u,v)∥Z.
Proof.
Define operator Q:Z→Z as follows:
(40)Q(u,v)=Δ2Q1(u,v)(1,1),
where Q1:Z→ℝ is defined by
(41)Q1(u,v)=aI0+βv(ξ)-I0+αu(1)+aξβ-1[bI0+αu(η)-I0+βv(1)],Δ2=Γ(β+1)Γ(α+1)αξβ-1Γ(α+1)(ξ-1)+Γ(β+1)(η-1)≠0.
It is easy to see that Q2(u,v)=Q(u,v); that is, Q:Z→Z is a continuous linear projector. Furthermore, KerQ=ImL. For (u,v)∈Z, set (u,v)=[(u,v)-Q(u,v)]+Q(u,v). Then (u,v)-Q(u,v)∈KerQ and Q(u,v)∈ImQ. It follows from KerQ=ImL and Q2(u,v)=Q(u,v) that ImL∩ImQ=(0,0). So we have
(42)Z=ImL⊕ImQ.
Now, IndL=dimKerL-codimImL=dimKerL-dimImQ=0, and so L is a Fredholm operator of index 0.
Let P:Y→Y be continuous linear operator defined by
(43)P(u,v)=D0+β-1v(0)Γ(β)(aξβ-1tα-1,tβ-1).
Obviously, P is a linear projector and
(44)KerP={(u,v)∈Y:D0+β-1v(0)=0}.
It is easy to know that Y=KerP⊕KerL.
Define KP:ImL→domL∩KerP by
(45)KP(u,v)(t)=(I0+αu(t)-aξβ-1tα-1[bI0+αu(η)-I0+βv(1)],I0+βv(t)).
Since
(46)|bI0+αu(η)-I0+βv(1)|≤|bΓ(α)∫0η(η-s)α-1u(s)ds||bI0+αu(η)-I0+βv(1)|+|1Γ(β)∫01(1-s)α-1v(s)ds||bI0+αu(η)-I0+βv(1)|≤bΓ(α)∥u∥1+1Γ(β)∥v∥1,D0+α-1I0+αu(t)=∫0tu(s)ds,I0+αu(t)=1Γ(α)∫0t(t-s)α-1u(s)ds,D0+β-1I0+βv(t)=∫0tv(s)ds,I0+βv(t)=1Γ(β)∫0t(t-s)β-1v(s)ds,
then
(47)∥KP(u,v)∥Y≤Δ1∥(u,v)∥Z.
In fact, if (u,v)∈ImL, then
(48)LKP(u,v)(t)={D0+α[I0+αu(t)-aξβ-1tα-1[bI0+αu(η)-I0+βv(1)]],D0+βI0+βv(t)}=(u,v).
By Lemma 4, for (u,v)∈domL∩KerP,
(49)KPL(u,v)(t)=(I0+αD0+αu(t)-aξα-1tα-1[bI0+αD0+αu(η)-I0+βD0+βv(1)],I0+βD0+βv(t))=(D0+β-1v(0)Γ(β)tβ-1}u(t)-D0+α-1u(0)Γ(α)tα-1-I0+2-αu(0)Γ(α-1)tα-2-aξβ-1tα-1×[bu(η)-bD0+α-1u(0)Γ(α)ηα-1-I0+2-αu(0)Γ(α-1)ηα-2-v(1)+D0+β-1v(0)Γ(β)],v(t)-D0+β-1v(0)Γ(β)tβ-1)=(u(t)-D0+α-1u(0)Γ(α)tα-1+aξβ-1tα-1bD0+α-1u(0)Γ(α)ηα-1,v(t))=(u(t),v(t))
(D0+β-1v(0)=0 since (u,v)∈KerP and I0+2-αu(0)=0 since (u,v)∈domL). Hence,
(50)Kp=(L|domL∩KerP)-1.
The proof is complete.
3. Main Results
In this section, we will use Theorem 1 to prove the existence of solutions to BVP (1). To obtain our main theorem, we use the following assumptions.
There exist functions ai,bi,ci,di,ei∈L1[0,1](i=1,2) such that for all (x,y,z,w)∈ℝ4,t∈[0,1],
(51)|f(t,x,y,z,w)|≤e1(t)+a1(t)|x|+b1(t)|y||f(t,x,y,z,w)|+c1(t)|z|+d1(t)|w|,|g(t,x,y,z,w)|≤e2(t)+a2(t)|x|+b2(t)|y||g(t,x,y,z,w)|+c2(t)|z|+d2(t)|w|.
There exists a constant A>0 such that, for (u,v)∈domL, if |D0+β-1v(t)|>A for all t∈[0,1], then Q1(N1(u,v))≠0 or Q1(N2(u,v))≠0.
There exists a constant B>0 such that either, for each c∈ℝ:|c|>B,
(52)cN1(caξβ-1tα-1,ctβ-1)>0,cN2(caξβ-1tα-1,ctβ-1)>0
or, for each a∈ℝ:|a|>B,
(53)cN1(caξβ-1tα-1,ctβ-1)<0,cN2(caξβ-1tα-1,ctβ-1)<0.
Theorem 10.
Suppose (2) and (H1)–(H3) hold. Then (1) has at least one solution in Y, provided that
(54)max{(Δ1+Δ3)max{∥a2∥1,∥b2∥1,∥c2∥1,∥d2∥1},Δ4,Δ5}<1.
Proof.
Set
(55)Ω1={(u,v)∈domL∖KerL:L(u,v)=λN(u,v)forsomeλ∈[0,1]}.
Take (u,v)∈Ω1. Since Lu=λNu, so λ≠0 and N(u,v)∈ImL=KerQ; hence,
(56)QN(u,v)=Q1(N1(u,v),N2(u,v))(1,1)=0.
Thus, from (H2), there exist t0∈[0,1] such that
(57)|D0+β-1v(t0)|≤A.
Noticing that
(58)D0+β-1v(t)=D0+β-1v(t0)+∫t0tD0+βv(s)ds
so
(59)|D0+β-1v(0)|≤∥D0+β-1v(t)∥∞≤|D0+β-2v(t0)|+∥D0+βv(t)∥1≤A+∥L2v∥1≤A+∥N2(u,v)∥1.
Thus
(60)∥P(u,v)∥Y=∥D0+β-1v(0)Γ(β)(aξβ-1tα-1,tβ-1)∥Y=Δ3|D0+β-1v(0)|≤Δ3(A+∥N2(u,v)∥1).
For all (u,v)∈Ω1, (I-P)(u,v)∈domL∩KerP. Considering Lemma 9, we get LP(u,v)=(0,0). Together with (39), we have
(61)∥(I-P)(u,v)∥Y=∥KPL(I-P)(u,v)∥Y≤Δ1∥L(I-P)(u,v)∥Z≤Δ1∥N(u,v)∥Z.
From (60) and (61), we have
(62)∥(u,v)∥Y≤∥P(u,v)∥Y+∥(I-P)(u,v)∥Y≤Δ3(A+∥N2(u,v)∥1)+Δ1∥N(u,v)∥Z=AΔ3+max{(Δ1+Δ3)∥N2(u,v)∥1,Δ1∥N1(u,v)∥1+Δ3∥N2(u,v)∥1}.
From (62), we discuss various cases.
Case 1 (∥(u,v)∥Y≤AΔ3+(Δ1+Δ3)∥N2(u,v)∥1). From (H1), we have
(63)∥(u,v)∥Y≤AΔ3+(Δ1+Δ3)×(∥D0+β-1v∥∞∥a2∥1∥u∥∞+∥b2∥1∥D0+α-1u∥∞+∥c2∥1∥v∥∞+∥d2∥1∥D0+β-1v∥∞+∥e2∥1)≤AΔ3+(Δ1+Δ3)∥e2∥1+(Δ1+Δ3)max{∥a2∥1,∥b2∥1}∥u∥Cα-1+(Δ1+Δ3)max{∥c2∥1,∥d2∥1}∥v∥Cβ-1≤AΔ3+(Δ1+Δ3)∥e2∥1+(Δ1+Δ3)max{∥a2∥1,∥b2∥1,∥c2∥1,∥d2∥1}∥(u,v)∥Y,
which yield
(64)∥(u,v)∥Y≤AΔ2+(Δ1+Δ3)∥e2∥11-(Δ1+Δ3)max{∥a2∥1,∥b2∥1,∥c2∥1,∥d2∥1}.
Thus, Ω1 is bounded.
Case 2 (∥(u,v)∥Y≤AΔ3+Δ1∥N1(u,v)∥1+Δ3∥N2(u,v)∥1). From (H1), we have
(65)∥(u,v)∥Y≤AΔ3+Δ1(∥a1∥1∥u∥∞+∥b1∥1∥D0+α-1u∥∞+∥c1∥1∥v∥∞+∥d1∥1∥D0+β-1v∥∞+∥e1∥1)+Δ3(∥D0+β-1v∥∞∥a2∥1∥u∥∞+∥b2∥1∥D0+α-1u∥∞+∥c2∥1∥v∥∞+∥d2∥1∥D0+β-1v∥∞+∥e2∥1)≤AΔ3+Δ1∥e1∥1+Δ3∥e2∥1+Δ4∥u∥Cα-1+Δ5∥v∥Cβ-1≤AΔ3+Δ1∥e1∥1+Δ3∥e2∥1+max{Δ4,Δ5}∥(u,v)∥Y
which yield
(66)∥(u,v)∥Y≤AΔ3+Δ1∥e1∥1+Δ3∥e2∥11-max{Δ4,Δ5}.
Thus, Ω1 is bounded. Let
(67)Ω2={(u,v)∈KerL:N(u,v)∈ImL}.
For (u,v)∈Ω2 and (u,v)∈KerL, so (u,v)=c(aξβ-1tα-1,tβ-1), t∈[0,1], c∈ℝ. Noticing that ImL=KerQ, then we get QN(u,v)=0, and thus Q1N1(caξβ-1tα-1,ctβ-1)=0 and Q1N2(caξβ-1tα-1,ctβ-1)=0. From (H2), we get |c|≤A/Γ(β), and thus Ω2 is bounded.
We define the isomorphism J:KerL→ImQ by
(68)J(caξβ-1tα-1,ctβ-1)=(c,c).
If the first part of (H3) is satisfied, then let
(69)Ω3={(u,v)∈kerL:λJ(u,v)+(1-λ)QN(u,v)=0,λ∈[0,1]}.
For (u,v)=(caξβ-1tα-1,ctβ-1)∈Ω3,
(70)λ(c,c)=-(1-λ)Δ2Q1N(caξβ-1tα-1,ctβ-1)(1,1).
If λ=1, then c=0. Otherwise, if |c|>B, in view of (H3) and Δ2<0, one has
(71)(1-λ)Δ3Q1N(caξβ-1tα-1,ctβ-1)<0
which contradict λc2≥0. Thus Ω3⊂{(u,v)∈KerL∣(u,v)=(caξβ-1tα-1,ctβ-1),|c|≤B} is bounded.
If the second part of (H3) holds, then define the set
(72)Ω3={(u,v)∈KerL:-λJ(u,v)+(1-λ)QN(u,v)=0,λ∈[0,1]},
and here J is as above. Similar to the above argument, we can show that Ω3 is bounded too.
In the following, we will prove that all conditions of Theorem 1 are satisfied. Let Ω be a bounded open subset of Y such that ∪i=13Ωi¯⊂Ω. By standard arguments, we can prove that KP(I-Q)N:Ω→Y is compact, and thus N is L-compact on Ω¯. Then by the above argument we have
Lu≠λNu, for every (u,λ)∈[(domL∖KerL)∩∂Ω]×(0,1),
Nu∉ImL for u∈KerL∩∂Ω.
Finally, we will prove that (iii) of Theorem 1 is satisfied. Let H(u,λ)=±λJu+(1-λ)QNu. According to the above argument, we know that
(73)H(u,λ)≠0foru∈KerL∩∂Ω.
Thus, by the homotopy property of degree,
(74)deg(QN|KerL,KerL∩Ω,0)=deg(H(·,0),KerL∩Ω,0)=deg(H(·,1),KerL∩Ω,0)=deg(±J,KerL∩Ω,0)≠0.
Then by Theorem 1, L(u,v)=N(u,v) has at least one solution in domL∩Ω¯ so that BVP (1) has a solution in Cα-1[0,1]×Cβ-1[0,1]. The proof is complete.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
The authors are very grateful to the anonymous referees for many valuable comments and suggestions which helped to improve the presentation of the paper. The project was supported by the National Natural Science Foundation of China (11371221, 61304074), the Specialized Research Foundation for the Doctoral Program of Higher Education of China (20123705110001), the Program for Scientific Research Innovation Team in Colleges and Universities of Shandong Province, the Postdoctoral Science Foundation of Shandong Province (201303074), and Foundation of SDUST.
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