# Solvability of Nonlocal Fractional Boundary Value Problems

Authors: Zhongmin Huang; Chengmin Hou
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

This paper is devoted to introduce a new approach to investigate the existence of solutions for a three-point boundary value problem of fractional difference equations as fllows:Δνy(t)=f(t+ν-1,y(t+ν-1),Δy(t+ν-2)), y(ν-2)=0, and [Δαy(t)]t=ν+b-α+1 = γ[Δαy(t)]t=ν+ξ-α. We present an existence result at resonance case. The proof relies on coincidence degree theory.


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## Body

## 1. Introduction

In this paper, we consider a discrete fractional boundary value problem, fort∈[0,b+1]ℕ0, of the form
(1)Δνy(t)=f(t+ν-1,y(t+ν-1),Δy(t+ν-2)),
subject to the conjugate boundary conditions
(2)y(ν-2)=0,[Δαy(t)]t=ν+b-α+1=γ[Δαy(t)]t=ν+ξ-α,
where f(t+ν-1,·,·):[ν-1,ν+b]ℕν-1×ℝ×ℝ→ℝ is a continuous function, b∈ℕ*,ν∈(1,2],α∈(0,1),γ>0, and ξ∈[0,b]ℕ0, and satisfies both ν-α-1≥0 and (ν+b-α+1)ν-α-1_=γ(ν+ξ-α)ν-α-1_.The three-point boundary value problem (1), (2) happens to be at resonance in the sense that associated linear homogeneous boundary value problem
(3)Δνy(t)=0,t∈[0,b+1]ℕ0,y(ν-2)=0,[Δαy(t)]t=ν+b-α+1=γ[Δαy(t)]t=ν+ξ-α
has nontrivial solution y(t)=Ctν-1_,C∈ℝ.The research into the boundary value problems for differential equation and fractional differential equation have been always very active subjects. Rich results has been obtained due to the various powerful devices such as coincidence degree theory and cone theory. For details, see [1–7] and the references therein. Discrete fractional calculus has generated interest in recent years. There are many literatures dealing with the discrete fractional difference equation subject to various boundary value conditions or initial value conditions. We refer to [8–18] and references therein. However, we note that these results were usually obtained by analytic techniques and various fixed point theorems. For example, in [13–16], authors investigated the existence to some boundary value problems by fixed point theorems on a cone. In [17], we given the existence of multiple solutions for a fractional difference boundary value problem with parameter by establishing the corresponding variational framework and using the mountain pass theorem, linking theorem, and Clark theorem in critical point theory. As we know, the coincidence degree theory has played an important role in dealing with the existence and multiple solutions for differential equations, which include the boundary value problems. To the best of our knowledge, it has not be used in discrete fractional boundary value problems. The aim of this paper is to establish the existence conditions for boundary value problem (1), (2). The proof relies on the coincidence degree theory.Now, we will briefly recall some notations and an abstract existence result.Let𝕐,ℤ be real Banach space, let Φ:dom(Φ)⊂𝕐→ℤ be a Fredholm map of index zero, and let P:𝕐→𝕐, Q:ℤ→ℤ be continuous projectors such that Im(P)=Ker(Φ), Ker(Q)=Im(Φ),𝕐=Ker(Φ)⊕Ker(P), and ℤ=Im(Φ)⊕Im(Q). It follows that Φ∣dom(Φ)∩Ker(P):dom(Φ)∩Ker(P)→Im(Φ) is invertible. We denote the inverse of the map by Kp. If Ω is an open bounded subset of 𝕐 such that dom(Φ)∩Ω≠∅,and the map N:𝕐→ℤ will be called L-compact on Ω¯, if QN(Ω¯) is bounded and Kp(I-Q)N:Ω¯→𝕐 is compact.We need the following known result for the sequel.Theorem 1 (Mawhin continuation theorem, see [2]).
LetL be a Fredholm operator of index zero, and let N be L-compact on Ω¯. Assume that the following conditions are satisfied:(i)
Lx≠λNx for every (x,λ)∈[(dom(L)∖Ker(L))∩∂Ω]×(0,1);(ii)
Nx∉Im(L) for every x∈Ker(L)∩∂Ω;(iii)
deg(JQN∣Ker(L),Ω∩Ker(L),0)≠0, where Q:ℤ→ℤ is a projection as above with Im(L)=Ker(Q), and J:Im(Q)→Ker(L) is any isomorphism.Then the equationLx=Nx has at least one solution in dom(L)∩Ω¯.


## 2. Preliminaries

We first collect some basic lemmas for manipulating discrete fractional operators.First, for any real numberβ, we let ℕβ={β,β+1,β+2,...}. We define tα_∶=Γ(t+1)/Γ(t+1-α), for any t and α for which the right-hand side is defined. We also appeal to the convention that if t+1-α is a pole of the Gamma function and t+1 is not a pole, then tα_=0.Definition 2 (see [13]).
Theνth fractional sum of f defined on ℕa, for ν>0, is defined to be
(4)Δa-νf(t)=Δa-νf(t;a)∶=1Γ(ν)∑s=at-ν‍(t-s-1)ν-1_f(s),
where t∈ℕa+ν. We also define the νth fractional difference, where ν>0 and 0≤N-1<ν≤N with N∈ℕ, to be Δνf(t)∶=ΔNΔ-(N-ν)f(t) where t∈ℕa+N-ν.Lemma 3 (see [13]).
Lett and ν be any numbers for which tν_ and tν-1_ are defined. Then
(5)Δtν_=νtν-1_.Lemma 4 (see [13]).
Let0≤N-1<ν≤N. Then
(6)Δ-νΔνy(t)=y(t)+C1tν-1_+C2tν-2_+⋯+CNtν-N_,
for some Ci∈ℝ, with 1≤i≤N.Lemma 5 (see [11]).
Forβ>0 and all μ∈ℝ, for which the following is defined, we find that
(7)Δβtμ_=Γ(μ+1)Γ(μ-β+1)tμ-β_.Lemma 6 (see [10]).
Letp be a positive integer, and let ν>p. Then ΔpΔ-νy(t)=Δ-(ν-p)y(t).Lemma 7 (see [17]).
A real symmetric matrixA is positive definite if there exists a real nonsingular matrix M such that A=M†M, where M† is the transpose.We define the Banach space𝕐=𝒞([ν-2,ν+b+1]ℕν-2,ℝ) with the norm ∥y∥=maxt∈[ν-2,ν+b+1]ℕν-2|y(t)| and Banach space ℤ=𝒞([0,b+1]ℕ0,ℝ) with the norm ∥z∥=maxt∈[0,b+1]ℕ0|y(t)|. For y∈𝕐, since
(8)Δνy(t)=1Γ(-ν)∑s=ν-2t+ν‍(t-s-1)-ν-1_y(s),
we can easily see that Δνy∈𝒞([0,b+1]ℕ0,ℝ).DefineΦ to be the linear operator from dom(Φ)∩𝕐 to ℤ with
(9)dom(Φ)={y∈𝕐∣Δνy∈𝒞([0,b+1]ℕ0,ℝ),y(ν-2)=0,Δαy(ν+b-α+1)=γΔαy(ν+ξ-α){([0,b+1]ℕ0,ℝ)}},Φy=Δνy,y∈dom(Φ).We defineN:𝕐→ℤ as
(10)Ny(t)=f(t+ν-1,y(t+ν-1),Δy(t+ν-2)),t∈[0,b+1]ℕ0.
Then the boundary value problem (1), (2) can be written by
(11)Φy=Ny.Lemma 8.
Φ
:
dom
(
Φ
)
∩
𝕐
→
ℤ is a Fredholm operator of index zero.Proof.
By Lemma4 and the condition y(ν-2)=0, we have Ker(Φ)={Ctν-1_∣C∈ℝ}.
Leth(t+ν-1)∈ℤ,C1∈ℝ, and
(12)y(t)=1Γ(ν)∑s=0t-ν‍(t-s-1)ν-1_h(s+ν-1)+C1tν-1_,
then Φy(t)=Δνy(t)=h(t+ν-1) and
(13)Im(Φ)={h(t+ν-1)∈ℤ∣∑s=0b+1‍G(s)h(s+ν-1)=0},
where
(14)G(s):={(ν+b-α-s)ν-α-1_-(ν+b-α+1)ν-α-1_(ν+ξ-α)ν-α-1_×(ν+ξ-α-s-1)ν-α-1_,0≤s≤ξ,(ν+b-α-s)ν-α-1_,ξ+1≤s≤b+1.
In fact, ifg∈Im(Φ), then there exists y∈dom(Φ) such that g(t+ν-1)=Φy(t)=Δνy(t).
In view of Lemma4, we have
(15)y(t)=Δ-νg(t+v-1)+C1tν-1_+C2tν-2_.
Byy(ν-2)=0, we get C2=0. By Lemmas 5 and 6, we get
(16)Δαy(t)=ΔαΔ-νg(t+v-1)+C1Δαtν-1_=Δ-(ν-α)g(t+v-1)+C1Δαtν-1_=1Γ(ν-α)∑s=0t-ν+α‍(t-s-1)ν-α-1_g(s+ν-1)+C1Γ(ν)Γ(ν-α)tν-α-1_,
then
(17)Δαy(ν+b-α+1)=1Γ(ν-α)∑s=0b+1‍(ν+b-α-s)ν-α-1_g(s+ν-1)+C1Γ(ν)Γ(ν-α)(ν+b-α+1)ν-α-1_,(18)Δαy(ν+ξ-α)=1Γ(ν-α)∑s=0ξ‍(ν+ξ-α-s-1)ν-α-1_g(s+ν-1)+C1Γ(ν)Γ(ν-α)(ν+ξ-α)ν-α-1_.
By[Δαy(t)]t=ν+b-α+1=γ[Δαy(t)]t=ν+ξ-α and (ν+b-α+1)ν-α-1_=γ(ν+ξ-α)ν-α-1_, we have
(19)∑s=0b+1‍(ν+b-α-s)ν-α-1_g(s+ν-1)-(ν+b-α+1)ν-α-1_(ν+ξ-α)ν-α-1_×∑s=0ξ‍(ν+ξ-α-s-1)ν-α-1_g(s+ν-1)=0.
Therefore,(20)∑s=0b+1‍G(s)g(s+ν-1)=0,
that is,
(21)g(t+ν-1)∈{h(t+ν-1)∈ℤ∣∑s=0b+1‍G(s)h(s+ν-1)=0}.
Next, we will show thatG(s)>0, for all s∈[0,b+1]ℕ0.
ObviouslyG(s)>0 for ξ+1≤s≤b+1. Suppose that 0≤s≤ξ, then
(22)G(s)=(ν+b-α-s)ν-α-1_-(ν+b-α+1)ν-α-1_(ν+ξ-α)ν-α-1_×(ν+ξ-α-s-1)ν-α-1_=(ν+ξ-α-s-1)ν-α-1_×[(ν+b-α-s)ν-α-1_(ν+ξ-α-s-1)ν-α-1_-(ν+b-α+1)ν-α-1_(ν+ξ-α)ν-α-1_]=(ν+ξ-α-s-1)ν-α-1_×[{Γ(ν+b-α-s+1)Γ(ξ-s+1)Γ(b-s+2)Γ(ν+ξ-α-s)}Γ(ν+b-α-s+1)Γ(ξ-s+1)Γ(b-s+2)Γ(ν+ξ-α-s)-Γ(ν+b-α+2)Γ(ξ+2)Γ(b+3)Γ(ν+ξ-α+1){Γ(ν+b-α-s+1)Γ(ξ-s+1)Γ(b-s+2)Γ(ν+ξ-α-s)}]=(ν+ξ-α-s-1)ν-α-1_×[{Γ(ν+b-α-s+1)Γ(ξ-s+1)Γ(b-s+2)Γ(ν+ξ-α-s)}(ν+b-α-s)(ν+b-α-s-1)···(ν+ξ-α-s)(b-s+1)(b-s)···(ξ-s+1)-(ν+b-α+1)(ν+b-α)···(ν+ξ-α+1)(b+2)(b+1)···(ξ+2){Γ(ν+b-α-s+1)Γ(ξ-s+1)Γ(b-s+2)Γ(ν+ξ-α-s)}].
Sinceν-α-1≥0, we may imply that
(23)ν+b-α-s-ib-s+1-i-ν+b-α+1-ib+2-i=(ν+b-α-i)(b+1-i)+(ν+b-α-i)-s(b+2-i)(b-s+1-i)(b+2-i)-(ν+b-α-i)(b+1-i)+(b+1-i)-s(ν+b-α+1-i)(b-s+1-i)(b+2-i)=(ν-α-1)(s+1)(b-s+1-i)(b+2-i)>0,
for i=0,1,2,…,b-ξ, then G(s)>0.
Leth(t+ν-1)∈ℤ, consider continuous linear mapping Q:ℤ→ℤ defined by
(24)Qh=δ0∑s=0b+1‍G(s)h(s+ν-1),
where δ0=(1/∑s=0b+1‍G(s))>0.
Note that(25)Q2h=δ0∑s=0b+1‍G(s)Qh(s+ν-1)=Qh.
That is, the mapQ is idempotent. In fact, Q is a continuous linear projector. Note that h∈Im(Φ) implies Qh=0. Conversely, if Qh=0, then h∈Im(Φ). Therefore, Im(Φ)=Ker(Q).
Takeh∈ℤ in the form h=(h-Qh)+Qh, so that h-Qh∈Im(Φ) and Qh∈Im(Q). Thus, ℤ=Im(Φ)+Im(Q). Let h∈Im(Φ)∩Im(Q), and assume that h(s+ν-1)=b≠0. Then, since h∈Im(Φ), we have b=0, which is a contradiction. Hence, Im(Q)∩Im(Φ)={θ}, thus ℤ=Im(Φ)⊕Im(Q).
Now,dimKer(Φ)=1=codimIm(Φ), and so Φ is a Fredholm operator of index zero. The proof is accomplished.Define a mappingP:𝕐→𝕐 by
(26)Py(t)=1Γ(ν)Δy(ν-2)tν-1_,t∈[ν-2,ν+b+1]ℕν-2,
we see that
(27)Ker(P)={y∈𝕐∣Δy(ν-2)=0}.Next, we will show thatKer(P)⊕Ker(Φ)=𝕐.In fact, for∀u∈Ker(P)∩Ker(Φ), there exists y∈𝕐 such that Py(t)=(1/Γ(ν))Δy(ν-2)tν-1_=u(t) and Δu(ν-2)=0, and then
(28)Δu(t)∣t=ν-2=1Γ(ν)(ν-1)Δy(ν-2)(ν-2)ν-2_=0.That is,Δy(ν-2)=0, and hence Ker(P)∩Ker(Φ)={θ}. Next, if y∈𝕐, then there exists a φ(t)∈C([0,b+1]ℕ0,ℝ) such that Δνy(t)=φ(t); therefore y(t)=(1/Γ(ν))∑s=0t-2+ν‍(t-s-1)ν-1_φ(s)+c1tν-1_+c2tν-2_. Since Φ(c1tν-1_+c2tν-2_)=0 and Δ(1/Γ(ν))∑s=0t-2+ν‍(t-s-1)ν-1_φ(s)∣t=ν-2=0, we see that Ker(P)⊕Ker(Φ)=𝕐.Note that the projectorsP and Q are exact. Define Kp:Im(Φ)→dom(Φ)∩Ker(P) by
(29)Kpy(t)=Δ-νy(t).Ify∈Im(Φ), then
(30)(ΦKp)y(t)=ΔνΔ-νy(t)=y(t).Also, ify∈dom(Φ)∩Ker(P), then
(31)(KpΦ)y(t)=Δ-νΔνy(t)=y(t)+C1tν-1_+C2tν-2_.We can easily see thatKpΦ(y)∈dom(Φ)∩Ker(P); then C2=0 and Δ(y(t)+C1tν-1)∣t=ν-2=Δy(ν-2)+C1(ν-1)(ν-2)ν-2_=0; in view of y∈Ker(P), we have Δy(ν-2)=0, then C1=0, thus
(32)(KpΦ)y(t)=y(t),
hence, Kp=(Φ∣dom(Φ)∩Ker(P))-1.Lemma 9.
K
p
(
I
-
Q
)
N
:
𝕐
⟶
𝕐 is completely continuous.Proof.
SinceKp(I-Q)N is continuous and Kp(I-Q)Nf(t+ν-1,y(t+ν-1),Δy(t+ν-2)) is a finite sum for t∈[ν-2,ν+b+1]ℕν-2, Kp(I-Q)N is completely continuous.


## 3. Existence Results

Observe by Lemma6 that for t∈[0,b+1]ℕ0,
(33)Δνy(t)=Δ1Γ(1-ν)∑s=ν-2t-(1-ν)‍(t-s-1)-ν_y(s).We lety∈dom(Φ) and
(34)z(t+ν-1)=1Γ(1-ν)∑s=ν-1t-(1-ν)‍(t-s-1)-ν_y(s),
then
(35)z(ν-1)=1Γ(1-ν)∑s=ν-10-(1-ν)‍(-s-1)-ν_y(s)=y(ν-1),(36)z(ν)=1Γ(1-ν)∑s=ν-11-(1-ν)‍(1-s-1)-ν_y(s)=(1-ν)y(ν-1)+y(ν),(37)z(ν+1)=1Γ(1-ν)∑s=ν-12-(1-ν)‍(2-s-1)-ν_y(s)=(2-ν)(1-ν)2!y(ν-1)+(1-ν)y(ν)+y(ν+1),⋮(38)z(ν+b)=1Γ(1-ν)∑s=ν-1b+1-(1-ν)‍(b+1-s-1)-ν_y(s)=(b-ν+1)(b-ν)⋯(1-ν)(b+1)!y(ν-1)+(b-ν)(b-ν-1)⋯(1-ν)b!y(ν)+⋯+(1-ν)y(b+ν-1)+y(b+ν).That is,z=By, where z=(z(ν-1),z(ν),…,z(ν+b))†,y=(y(ν-1),y(ν),…,y(ν+b))†.One has(39)B=(100⋯01-ν10⋯0(2-ν)(1-ν)2!1-ν1⋯0⋮⋮⋮⋮(b-ν+1)(b-ν)⋯(1-ν)(b+1)!(b-ν+1)(b-ν)(b-ν-1)⋯(1-ν)b!⋯⋯1)(b+2)×(b+2).By Lemma7, B†B is a positive definite matrix. Let λmin and λmax denote, respectively, the minimum and the maximum eigenvalues of B†B.Sincez=By, we may easyly see that
(40)λmin(|y(ν-1)|2+|y(ν)|2+⋯+|y(ν+b)|2)≤(|z(ν-1)|2+|z(ν)|2+⋯+|z(ν+b)|2)≤λmax(|y(ν-1)|2+|y(ν)|2+⋯+|y(ν+b)|2),
furthermore,
(41)(λminb+2)1/2maxt∈[ν-1,ν+b]ℕν-1|y(t)|≤maxt∈[0,b+1]ℕ0|z(t+ν-1)|≤(λmax(b+2))1/2maxt∈[ν-1,ν+b]ℕν-1|y(t)|.Theorem 10.
Letf:[0,b+1]ℕ0×ℝ×ℝ→ℝ be a continuous function in the second and the third variables. Assume that λ∈[0,1] and(A1)
there exist nonnegative constantsa,d,l,m,andn  and constants k,r∈[0,1) which satisfy a+2d<(1/(b+1))(λmin/(b+2))1/2 such that for all (x,y)∈ℝ2,t∈[ν-2,ν+b+1]ℕν-2,
(42)f(t,x,y)≤a|x|+d|y|+l|x|k+m|y|r+n,(A2)
there existsM1>0 such that for y∈dom(Φ), if |Δν-1y(t)|>M1 for any t∈[ν-2,ν+b+1]ℕν-2, then
(43)∑s=0b+1‍G(s)f(s+ν-1,y(s+ν-1),Δy(s+ν-2))≠0,(A3)
there existsM*>0 such that for any c∈ℝ, if |c|>M*, then either
(44)c∑s=0b+1‍G(s)f(s+v-1,c(s+v-1)ν-1_,c∑s=0b+1‍G(s)fc(ν-1)(s+v-2)ν-2_)<0,or else
(45)c∑s=0b+1‍G(s)f(s+v-1,c(s+v-1)ν-1_,c∑s=0b+1‍G(s)fc(ν-1)(s+v-2)ν-2_)>0.
Then the boundary value problem (1), (2) has at least one solution in 𝕐.Proof.
SetΩ1={y∈dom(Φ)∖Ker(Φ)∣Φy=λNyforsomeλ∈(0,1]}. Then for y∈Ω1,Φy=λNy, thus Ny∈Im(Φ)=Ker(Q), and hence
(46)∑s=0b+1‍G(s)f(s+v-1,y(s+v-1),Δy(s+v-2))=0.
From(A2), there exists t0∈[ν-2,ν+b+1]ℕν-2 such that |Δν-1y(t0)|≤M1.
Next, fory∈Ω1, we have ΔΔν-1y(t)=λf(t+ν-1,y(t+ν-1),Δy(t+ν-2)), and hence
(47)∑s=t0t-1‍ΔΔν-1y(t)=∑s=t0t-1‍λf(t+ν-1,y(t+ν-1),Δy(t+ν-2)),
then we get
(48)|z(t+ν-1)|≤|z(t0+ν-1)|+∑s=t0t-1‍|f(s+ν-1,y(s+ν-1),Δy(s+ν-2))|≤M1+∑s=t0t-1‍[a|y|+d|Δy|+l|y|k+m|Δy|r+n]≤M1+∑s=0b‍[(a+2d)|y|+l|y|k+2rm|y|r+n]≤M1+(b+1)[{(maxt∈[ν-1,ν+b+1]ℕν-1|y(t)|)r}(a+2d)maxt∈[ν-1,ν+b+1]ℕν-1|y(t)|+l(maxt∈[ν-1,ν+b+1]ℕν-1|y(t)|)k+2rm(maxt∈[ν-1,ν+b+1]ℕν-1|y(t)|)r+n{(maxt∈[ν-1,ν+b+1]ℕν-1|y(t)|)r}],
then
(49)maxt∈[0,b+1]ℕ0|z(t+ν-1)|≤M1+λ(b+1)×[{(maxt∈[ν-1,ν+b+1]ℕν-1|y(t)|)r}(a+2d)maxt∈[ν-1,ν+b+1]ℕν-1|y(t)|+l(maxt∈[ν-1,ν+b+1]ℕν-1|y(t)|)k+2rm(maxt∈[ν-1,ν+b+1]ℕν-1|y(t)|)r+n{(maxt∈[ν-1,ν+b+1]ℕν-1|y(t)|)r}],
from (41), then we can get
(50)maxt∈[ν-1,ν+b]ℕν-1|y(t)|≤M1+(b+1)[l(maxt∈[ν-1,ν+b+1]ℕν-1|y(t)|)k+2rm(maxt∈[ν-1,ν+b+1]ℕν-1|y(t)|)r+n{(maxt∈[ν-1,ν+b+1]ℕν-1|y(t)|)k}]×((λmin/(b+2))1/2-(b+1)(a+2d))-1.
Furthermore, sincey(ν-2)=0,Δαy(ν+b-α+1)=γΔαy(ν+ξ-α) and for any t∈[ν-α-2,ν+b-α+1]ℕν-α-2,
(51)Δαy(t)=ΔΔα-1y(t)=Δ[1Γ(1-α)∑s=ν-2t+α-1‍(t-s-1)-α_y(s)]=1Γ(1-α)∑s=ν-1t+α‍(t-s)-α_y(s)-1Γ(1-α)×∑s=ν-1t+α-1‍(t-s-1)-α_y(s)=-αΓ(1-α)∑s=ν-2t+α‍(t-s-1)-α-1_y(s)
holds, and we may see that y(ν+b+1) can be expressed by a linear combination with y(ν-1),y(ν),…,y(ν+b). Therefore, ∥y∥=maxt∈[ν-2,ν+b+1]ℕν-2|y(t)| is bounded. Thus, Ω1 is bounded.
LetΩ2={y∈Ker(Φ)∣Ny∈Im(Φ)}. For y∈Ω2,y∈Ker(Φ), since Im(Φ)=Ker(Q), then QNy=0, thus
(52)∑s=0b+1‍G(s)f(s+v-1,c(s+v-1)ν-1_,c(ν-1)(s+v-2)ν-2_)=0.
From Lemma5 and in view of y∈Ker(Φ), we see that |Δν-1y(t)|=|c||Δν-1tν-1_|=|c||Γ(ν)|, then from (A2) we get |c||Γ(ν)|≤M1; that is, |c|≤(M1/|Γ(ν)|). Hence, Ω2 is bounded. Next, according to (A3), for any c∈ℝ, if |c|>M*, then either
(53)cδ0∑s=0b+1‍G(s)f(s+v-1,c(s+v-1)ν-1_,cδ0∑s=0b+1‍G(s)fc(ν-1)(s+v-2)ν-2_)<0,
or(54)cδ0∑s=0b+1‍G(s)f(s+v-1,c(s+v-1)ν-1_,cδ0∑s=0b+1‍G(s)fc(ν-1)(s+v-2)ν-2_)>0.
If (53) holds, set
(55)Ω3={y∈Ker(Φ)∣-λJ-1y+(1-λ)QNy=0,λ∈[0,1]},
where J:Im(Q)→Ker(Φ) is the linear isomorphism given by J(c)=ctν-1_,∀c∈ℝ,andt∈[0,b+1]ℕ0. Since y∈c0tν-1_∈Ω3,c0∈ℝ, then
(56)λc0=(1-λ)δ0∑s=0b+1‍G(s)f(s+v-1,c(s+v-1)ν-1_,(1-λ)δ0∑s=0b+1‍G(s)fc(ν-1)(s+v-2)ν-2_).
Ifλ=1, then c0=0. Otherwise, if |c0|>M*, in view of (53), one has
(57)0≤λc02=c0(1-λ)δ0∑s=0b+1‍G(s)f(s+v-1,c(s+v-1)ν-1_,c0(1-λ)δ0∑s=0b+1‍G(s)fc(ν-1)(s+v-2)ν-2_)<0,
which is a contradiction. Thus Ω3⊂{y∈Ker(Φ)∣y=ctν-1_,|c|≤M*} is bounded.
If (54) holds, then the set
(58)Ω3={y∈Ker(Φ)∣λJ-1y+(1-λ)QNy=0,λ∈[0,1]},
where J is mentioned as above. By the analogous argument, we can show that Ω3 is bounded too.
Now, we will prove that all the conditions of Theorem1 are satisfied. Set Ω to be bounded open set of 𝕐 such that ∪i=13Ω¯i⊂Ω. By Lemma 9, Kp(I-Q)N:Ω¯→𝕐 is compact; hence N is L-compact on Ω¯; then by the above argument, we have (i)
Φx≠λNx for every (x,λ)∈[(dom(Φ)∖Ker(Φ))∩∂Ω]×(0,1),(ii)
Nx∉Im(Φ) for every x∈Ker(Φ)∩∂Ω,then the conditions (i),(ii) of Theorem 1 are hold.
At last we will prove that the condition (iii) of Theorem1 is satisfied. In fact, let H(y,λ)≠0 for all y∈Ker(Φ)∩∂Ω, thus, by the homotopy property of degree as follows:
(59)deg(JQN∣Ker(Φ),Ω∩Ker(Φ),0)=deg(H(·,0),Ω∩Ker(Φ),0)=deg(±I,Ω∩Ker(Φ),0)≠0.
Then by Theorem1, Φy=Ny has at least one solution in dom(Φ)∩Ω¯; therefore the boundary value problem (1), (2) has at least one solution in 𝕐.Remark 11.
We may obtain similarly result when the third variable off is instead by fractional difference of y(t) with the order β which satisfies 0<β≤ν-1.Example 12.
Consider the boundary value problem(60)Δ3/2y(t)=f(t+12,y(t+12),Δy(t-12)),t∈[0,2]ℕ0,y(-12)=0,Δ1/4y(134)=γΔ1/4y(94),
where f(t+ν-1,y(t+ν-1),Δy(t+ν-2))=(1/8)|y|+(1/64)Δycosy+cos(yΔy).It is clearly thatf is continuous. ν=3/2,α=1/4,b=ξ=1,andγ=13/12, then ν-α-1=(1/4)>0 and (ν+b-α+1)ν-α-1_=γ(ν+ξ-α)ν-α-1_; therefore, the boundary value problem (1), (2) is at resonance case. By ν=3/2, we have
(61)B=(100-1210-18-121)3×3.By calculating,λmin=0.4344. Choosing a=1/8,d=(1/64),l=m=r=k=0,andn=1, we can get a+2d=1/8+2×(1/64)≤(1/(b+1))(λmin/(b+2))1/2=(1/2)(0.4344/3)1/2, furthermore, for all (y,Δy)∈ℝ2,t∈[0,b+1]ℕ0,
(62)|f(t+ν-1,y(t+ν-1),Δy(t+ν-2))|≤18|y|+164|Δy|+1.Therefore, the condition(A1) of Theorem 10 holds. Moreover, we can choose M1=1620,M*=10; then the condition (A2),(A3) of Theorem 10 hold too. In fact, by the expression of f, it is clear that if |z|=|Δν-1y|>1620, then |y(ν-1)|=|z(ν-1)|>1620. Therefore, we can see
(63)∑s=02‍G(s)f(s+ν-1,y(s+ν-1),Δy(s+ν-2))=G(0)[{18}18|y(ν-1)|+164Δy(ν-2)cos(y(ν-1))+cos(y(ν-1)Δy(ν-2)){18}]+G(1)[{18}18|y(ν)|+164Δy(ν-1)cos(y(ν))+cos(y(ν)Δy(ν-1)){18}]+G(2)[{18}18|y(ν+1)|+164Δy(ν)cos(y(ν+1))+cos(y(ν+1)Δy(ν)){18}]=G(0)[{18}18|y(ν-1)|+164y(ν-1)cos(y(ν-1))+cos(y(ν-1)Δy(ν-2)){18}]+G(1)[{18}18|y(ν)|+164y(ν)cos(y(ν))-164y(ν-1)cos(y(ν))+cos(y(ν)Δy(ν-1)){18}]+G(2)[{18}18|y(ν+1)|+164y(ν+1)cos(y(ν+1))-164y(ν)cos(y(ν+1))+cos(y(ν+1)Δy(ν)){18}]≥[G(0)(18-164)-164G(1)]|y(ν-1)|+[G(1)(18-164)-164G(2)]|y(ν)|+G(2)(18-164)|y(ν+1)|-(G(0)+G(1)+G(2))>1164×6×24Γ(54)|y(ν-1)|+164×6Γ(54)|y(ν)|-(G(0)+G(1)+G(2))>1164×6×24Γ(54)|y(ν-1)|-(524×4+16+1)Γ(54)>164×18Γ(54)|y(ν-1)|-54Γ(54)>0.Thus, for everyy∈ℝ,s∈[0,2]ℕ0,∑s=02‍G(s)f(s+ν-1,y(s+ν-1),andΔy(s+ν-2))≠0, so (A2) holds, and if |c|>(19/π), then for t∈[0,b+1]ℕ0,
(64)[18(t+ν-1)ν-1_-164(ν-1)(t+ν-2)ν-1_]c-1>0forc>19π
or
(65)[18(t+ν-1)ν-1_-164(ν-1)(t+ν-2)ν-1_]c+1<0forc<-19π.Furthermore,(66)cf(t+ν-1,c(t+ν-1)ν-1_,c(ν-1)(t+ν-2)ν-2_)=18c(t+ν-1)ν-1_+164c(ν-1)×(t+ν-2)ν-2_cos(c(t+ν-1)ν-1_)+cos[(c(t+ν-1)ν-1_)(c(ν-1)(t+ν-2)ν-2_)]>0,forc>19π,
or
(67)cf(t+ν-1,c(t+ν-1)ν-1_,c(ν-1)(t+ν-2)ν-2_)<0,forc<-19π,
so (53) or (54) holds; that is, (A3) of Theorem 10 holds. Then, all the assumptions of Theorem 10 hold. Thus, with Theorem 10, the boundary value problem (60) has at least one solution in 𝕐.



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*Source: 943961-2013-05-15.xml*