# Existence of Solutions of a Discrete Fourth-Order Boundary Value Problem

Authors: Ruyun Ma; Chenghua Gao; Yongkui Chang
Journal: Discrete Dynamics in Nature and Society
 (2010)

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## Abstract

Leta,b be two integers with b-a≥5 and let 𝕋2={a+2,a+3,…,b-2}. We show the existence of solutions for nonlinear fourth-order discrete boundary value problem Δ4u(t-2)=f(t,u(t), Δ2u(t-1)), t∈𝕋2, u(a+1)=u(b-1)=Δ2u(a)=Δ2u(b-2)=0 under a nonresonance condition involving two-parameter linear eigenvalue problem. We also study the existence and multiplicity of solutions of nonlinear perturbation of a resonant linear problem.


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## Body

## 1. Introduction

The deformations of an elastic beam whose both ends are simply supported are described by a fourth-order two-point boundary value problem(1.1)y′′′′+g(x)y=e(x),0<x<1,y(0)=y(1)=y′′(0)=y′′(1)=0
See studies by Aftabizadeh [1] and Gupta in [2]. The existence of solutions of nonlinear boundary value problems of fourth-order differential equations has been studied by many authors; see [1–12] and the references therein. For example, Aftabizadeh [1] proved an existence theorem for nonlinear boundary value problems
(1.2)y′′′′=f(x,y,y′′),0<x<1,y(0)=y0,y′′(0)=y1,y(1)=y¯0,y′′(1)=y¯1,
under several conditions that f is a bounded function. Yang [3] obtained existence results of (1.2) under the following assumption.(A)
There are constantsa,b,c≥0 with a/π4+b/π2<1 such that
(1.3)|f(x,y,u)|≤a|y|+b|u|+c.Del Pino and Manásevich [4] extended Yang's result and proved the following.Theorem A.
Assume that the pair(α,β) satisfies
(1.4)α(kπ)4+β(kπ)2≠1
for all k∈ℕ and that there are positive constants a,b, and c such that
(1.5)amaxk∈ℕ{1|k4π4-α-βk2π2|}+bmaxk∈ℕ{λk|k4π4-α-βk2π2|}<1,|f(t,u,v)-(αu-βv)|≤a|u|+b|v|+c
for all x∈(0,1), u,v∈ℝ, then (1.2) possesses at least one solution.Of course, the natural question is whether or not the similar existence can be established for the corresponding discrete analog of (1.2) of the form
(1.6)Δ4u(t-2)=f(t,u(t),Δ2u(t-1)),t∈𝕋2,(1.7)u(a+1)=r1,u(b-1)=r2,Δ2u(a)=r3,Δ2u(b-2)=r4,
where 𝕋2={a+2,a+3,…,b-2}, ri∈ℝ for i∈{1,2,3,4}.The purpose of this paper is to show that the answer is yes. To this end, we state and prove a spectrum result of two-parameter linear eigenvalue problem(1.8)Δ4u(t-2)+βΔ2u(t-1)-αu(t)=0,t∈𝕋2,(1.9)u(a+1)=u(b-1)=Δ2u(a)=Δ2u(b-2)=0.
This result is a slightly generalized version of Shi and Wang [13, Theorem 2.1]. In Section 3, we use Leray-Schauder principle to study the existence of solutions of (1.6), (1.7) under some nonresonant conditions involving the spectrum of (1.8), (1.9). Section 4 is considered with some perturbations of resonant linear problems. We established some a priori bounds and used these together with bifurcation arguments to prove the existence and multiplicity of solutions.Finally, we note that the existence of solutions of second-order discrete boundary value problems has also received much attention; see studies by Agarwal and Wong in [14], Henderson in [15], and the references therein. However, relatively little is known about the existence of solutions of fourth-order discrete boundary value problems. The likely reason may be that the structure of spectrum of the corresponding linear eigenvalue problem is not very clear. To our best knowledge, only He and Yu [16] as well as Zhang et al. [17] dealt with the discrete problem of the form
(1.10)Δ4u(t-2)=f(t,u(t)),t∈𝕋2,(1.11)u(a)=u(b)=Δ2u(a)=Δ2u(b-2)=0.
As we will see in Section 2, (1.9) has more advantage than (1.11) in the study of the spectrum of two-parameter linear eigenvalue problems.


## 2. Spectrum of Two-Parameter Linear Eigenvalue Problem

Leta,b be two integers with b-a≥5. Recall (2.1)𝕋2={a+2,a+3,…,b-2}.
Let (2.2)𝕋0={a,a+1,…,b},𝕋1={a+1,a+2,…,b-1},Λ={1,2,…,b-a-3}.
Let X be the Banach space (2.3)X={u∣u:𝕋0→ℝ,u(a+1)=u(b-1)=Δ2u(a)=Δ2u(b-2)=0}
under the norm (2.4)∥u∥X:=max{|u(j)|∣j∈𝕋2}.
Let Y be the Banach space (2.5)Y={y∣y:𝕋1→ℝ,y(a+1)=y(b-1)=0}
equipped with the norm (2.6)∥y∥Y:=max{|y(j)|∣j∈𝕋2}.
Let Z be the Banach space (2.7)Z={z∣z:𝕋2→ℝ}
equipped with the norm (2.8)∥z∥Z:=max{|z(j)|∣j∈𝕋2}.Remark 2.1.
For anyz∈Z with
(2.9)z={z(a+2),z(a+3),…,z(b-2)},
it determines a unique element y∈Y by
(2.10)y={0,z(a+2),z(a+3),…,z(b-2),0}
and a unique element x∈X by
(2.11)x={-z(a+2),0,z(a+2),z(a+3),…,z(b-2),0,-z(b-2)}.
Hence, the Banach spaces X, Y, and Z are homomorphic with each other. Denote the natural homomorphism from Z to X by j.Now, we define a linear operatorL:X→Z by (2.12)Lu(t):=Δ4u(t-2)+βΔ2u(t-1)-αu(t),t∈𝕋2.
For k∈Λ, let λk be the kth-eigenvalue of the second-order linear eigenvalue problem
(2.13)Δ2u(t-1)+λu(t)=0,t∈𝕋2,u(a+1)=u(b-1)=0.
It is well known that λk is simple, and the corresponding eigenfunction (2.14)ψk(t):=sinkπ(t-a-1)b-a-2,t∈𝕋1,k∈Λ.
See the study by Kelly and Peterson in [18].The following result is considered with the spectrum of two-parameter eigenvalue problem:(2.15)Δ4u(t-2)+βΔ2u(t-1)-αu(t)=0,t∈𝕋2,(2.16)u(a+1)=u(b-1)=Δ2u(a)=Δ2u(b-2)=0.
It is a slightly generalized version of Shi and Wang [13, Theorem 2.1].Proposition 2.2.
(α,β) is an eigenvalue pair of (2.15), (2.16) if and only if
(2.17)αλk2+βλk=1
for some k∈Λ.Proof.
Letr1,r2∈ℂ such that
(2.18)r1+r2=β,r1r2=-α.
Define two second-order difference operators L1,L2:Y→Y by
(2.19)(L1y)(t):=Δ2y(t-1)+r1y(t),t∈𝕋2,(L1y)(a+1)=0,(L1y)(b-1)=0,(L2y)(t):=Δ2y(t-1)+r2y(t),t∈𝕋2,(L2y)(a+1)=0,(L2y)(b-1)=0.
Then, for y∈Y and t∈𝕋1,
(2.20)L2∘L1y(t)=L2(Δ2y(t-1)+r1y(t))=Δ2(Δ2y(t-2)+r1y(t-1))+r2(Δ2y(t-1)+r1y(t))=Δ4y(t-2)+(r1+r2)Δ2y(t-1)+r1r2y(t)=Δ4y(t-2)+βΔ2y(t-1)-αy(t)=(L|Y)y(t).
We claim that if (2.15), (2.16) possess a nontrivial solution y, then either r1=λk or r2=λk for some k∈Λ. In either case, sin(kπ(t-a-1)/(b-a-2)),(t∈𝕋1), is a nontrivial solution of (2.15), (2.16).
In fact, ifr2≠λk for all k∈Λ, then (2.20) implies that
(2.21)(L1y)(t)=0,t∈𝕋1.
This is
(2.22)Δ2y(t-1)+r1y(t)=0,t∈𝕋2,y(a+1)=y(b-1)=0.
Thus, r1=λk for some k∈Λ, and
(2.23)y(t)=sinkπ(t-a-1)b-a-2,t∈𝕋1.
Ifr2=λk for some k∈Λ, then (2.20) implies that
(2.24)L1y(t)=γψk(t),t∈𝕋1,
for some γ∈ℝ∖{0}. This is
(2.25)Δ2y(t-1)+r1y(t)=γψk(t),t∈𝕋2,(2.26)y(a+1)=y(b-1)=0.
Since γ≠0, it follows that
(2.27)r1≠λk,k∈Λ.
This implies that (2.25), (2.26) have a unique solution
(2.28)y(t):=(L1)-1γψk(t),t∈𝕋1.
We show that(2.29)r1=λk+γ,y(t)=ψk(t).
In fact, from (2.25) we have
(2.30)∑t=a+2b-2γψk2(t)=∑t=a+2b-2(Δ2y(t-1)+r1y(t))ψk(t)=∑t=a+2b-2(Δ2ψk(t-1)+r1ψk(t))y(t)=∑t=a+2b-2(r1-λk)ψk(t)y(t)=∑t=a+2b-2(r1-λk)ψk2(t),
which implies that γ=r1-λk, and, subsequently, y(t)=ψk(t).
Therefore, the claim is true.
Now, (2.17) follows by substituting this solution into (2.15), (2.16). Reciprocally, if (2.17) holds, then, clearly, sin(kπ(t-a-1)/(b-a-2)), (t∈𝕋1) is a nontrivial solution of (2.15), (2.16).Remark 2.3.
From the proof of Proposition2.2, we see that if (2.15) subjects to (1.9), then we can factor L|Y as follows:
(2.31)L|Y=L2∘L1.
However, this cannot be done if (2.15) subjects to (1.11). So, (1.9) has more advantage than (1.11) in the study of the spectrum of two-parameter linear eigenvalue problems.Next, forj∈ℕ, let us set (2.32)Lj={(α,β)∣αλj2+βλj=1}.
In view of the Proposition 2.2, we call Lj an eigenline of (2.15), (2.16). We note that an eigenvalue pair (α,β) can belong to at most two eigenlines. If (α,β) belongs to just one Lj, then the corresponding eigenspace is spanned by sin(kπ(t-a-1)/(b-a-2)). If (α,β) belongs to Lj∩Lk(j≠k), then the corresponding eigenspace is spanned by sin(jπ(t-a-1)/(b-a-2)) and sin(kπ(t-a-1)/(b-a-2)).Suppose that the pair(α,β) is not an eigenvalue pair of (2.15), (2.16), that is, (2.33)αλk2+βλk≠1
for all k∈Λ, and that h∈Z: (2.34)h(t)=∑k=1b-a-3hksinkπ(t-a-1)b-a-2,t∈𝕋2.
From the Fredholm Alternative, it follows that the boundary value problem
(2.35)Δ4u(t-2)+βΔ2u(t-1)-αu(t)=h(t),t∈𝕋2,u(a+1)=u(b-1)=Δ2u(a)=Δ2u(b-2)=0
has a unique solution for each h∈Z. Moreover, this solution admits a Fourier series expansion of the form
(2.36)u(t)=∑k=1b-a-3hksin(kπ(t-a-1)/(b-a-2))(λk2-α-βλk),t∈𝕋2,u(a+1):=0,u(b-1):=0,u(a):=-u(a+2),u(b):=-u(b-2).
Also, we have
(2.37)Δ2u(t-1)=-∑k=1b-a-3λkhksin(kπ(t-a-1)/(b-a-2))(λk2-α-βλk),t∈𝕋2,Δ2u(a):=0,Δ2u(b-2):=0.From (2.36) and (2.37), we can easily see that the operators A:Z→Z, B:Z→Z defined by (2.38)A(h)(t)=u(t),B(h)(t)=Δ2u(t-1),t∈𝕋2,
are compact linear operators. In (2.38), u is the solution of (2.35), (2.16) corresponding to h∈Z. The norms of A and B are, respectively, given by (2.39)∥A∥Z→Z=maxk∈Λ{1|λk2-α-βλk|},∥B∥Z→Z=maxk∈Λ{λk|λk2-α-βλk|}.Finally, as an immediate consequence of Proposition2.2, we have the following.Proposition 2.4.
Letγ and δ be two constants with (γ,δ)∈[0,∞)×[0,∞) and γ+δ>0. Then the generalized eigenvalues of problem
(2.40)Δ4u(t-2)+μ(δΔ2u(t-1)-γu(t))=0,t∈𝕋2,u(a+1)=u(b-1)=Δ2u(a)=Δ2u(b-2)=0
are given by
(2.41)μ1(γ,δ)<μ2(γ,δ)<⋯<μb-a-3(γ,δ),
where
(2.42)μk(γ,δ)=λk2γ+δλk,k∈Λ.
The generalized eigenfunction corresponding to μk(γ,δ) is
(2.43)ψk(t)=sinkπ(t-a-1)b-a-2,t∈𝕋1.


## 3. Existence Results for Nonresonant Problems

Theorem 3.1.
Assume that the pair(α,β) satisfies
(3.1)αλk2+βλk≠1
for all k∈Λ and that there are positive constants a*,b*, and c* such that
(3.2)a*maxk∈Λ{1|λk2-α-βλk|}+b*maxk∈Λ{λk|λk2-α-βλk|}<1,(3.3)|f(t,u,v)-(αu-βv)|≤a*|u|+b*|v|+c*
for all t∈𝕋2, u,v∈ℝ, then (1.6), (1.7) possess at least one solution.Remark 3.2.
It is not difficult to see that (3.1), (3.2) imply that
(3.4)a*|λk2-α-βλk|+b*λk|λk2-α-βλk|<1
for k∈Λ. It turns out that (3.4) is equivalent to the fact that the square [α-a*,α+a*]×[β-b*,β+b*] does not intersect any of the eigenlines Lj of (2.15), (2.16). From this point of view, (3.1), (3.2) can be thought of as a two-parameter nonresonance condition relative to the eigenlines Lj.Proof of Theorem3.1.
It is easy to check that the problem(3.5)Δ4u(t-2)=0,t∈𝕋2,u(a+1)=r1,u(b-1)=r2,Δ2u(a)=r3,Δ2u(b-2)=r4
has a unique solution l(t). Set
(3.6)y(t):=u(t)-l(t),t∈𝕋0.
Then (1.6),(1.7) can be rewritten as
(3.7)Δ4y(t-2)=f(t,y(t)+l(t),Δ2(y(t-1)+l(t-1))),t∈𝕋2,y(a+1)=y(b-1)=Δ2y(a)=Δ2y(b-2)=0.
Since
(3.8)|f(t,y(t)+l(t),Δ2(y(t-1)+l(t-1)))-[α(y(t)+l(t))-β(Δ2(y(t-1)+l(t-1)))]|≤a*|y|+b*|Δ2y(t-1)|+c**
with
(3.9)c**=a*maxt∈𝕋0|l(t)|+b*maxt∈𝕋1|Δ2l(t-1)|+c*,
it follows that (3.2) and (3.3) still hold except that c* is replaced by c**. So, we may suppose that r1=r2=r3=r4=0 in (1.7).
Let us defineT:Z×Z→Z×Z by
(3.10)T(u,v)=(A(f(·,u,v)-(αu-βv)),B(f(·,u,v)-(αu-βv))),
where A and B are the operators defined in (2.38). The growth condition (3.3) together with the compactness A and B implies that T is a completely continuous operator. By Remark 2.1, the problem
(3.11)Δ4u(t-2)=f(t,u(t),Δ2u(t-1)),t∈𝕋2,u(a+1)=u(b-1)=Δ2u(a)=Δ2u(b-2)=0
is equivalent to the fixed point problem in Z×Z:
(3.12)(u,v)=T(u,v).
We will study this fixed point problem by means of the well-known Leray-Schauder principle [18]. To do this, we show that there is a uniform bound independent of λ∈[0,1] for the solutions of the equation
(3.13)(u,v)=λT(u,v).
Thus, let (u,v) be a solution of (3.13). From the definition of T and (3.3), we obtain the result that
(3.14)∥u∥Z≤∥A∥Z→Z{a∥u∥Z+b∥v∥Z+c},(3.15)∥v∥Z≤∥B∥Z→Z{a∥u∥Z+b∥v∥Z+c}.
Combining (3.14) and (3.15) and using (3.2) and (2.38), we obtain the existence of a constant M=M(a*,b*,c*,∥A∥Z→Z,∥B∥Z→Z) such that
(3.16)∥u∥Z+∥v∥Z≤M.
By the Leray-Schauder principle [19], we conclude the existence of at least one solution of (3.12), and the theorem follows.


## 4. Existence and Multiplicity Results for Perturbations of Resonant Linear Problems

In this section, we consider the perturbations of resonant linear problems of the form(4.1)μΔ4u(t-2)+(μk+μ)(δΔ2u(t-1)-γu(t))=g(t,u(t))+h(t),t∈𝕋2,(4.1)u(a+1)=u(b-1)=Δ2u(a)=Δ2u(b-2)=0,
where (γ,δ)∈[0,∞)×[0,∞) with γ+δ>0, μk=μk(γ,δ), and g and h satisfy the following.(H1)
(Sublinear growth condition) g:𝕋2×ℝ→ℝ is continuous, and there exist α∈[0,1), C1,C2∈(0,∞) such that
(4.2)|g(t,s)|≤C1|s|α+C2,s∈ℝ,t∈𝕋2(H2)
There existsβ>0 such that
(4.3)sg(t,s)>0,fort∈𝕋2,|s|>β(H3)
h:𝕋2→ℝ satisfies
(4.4)∑t=a+2b-2h(t)ψk(t)=0.We will establish some a priori bounds and use these together with Leray-Schauder continuation and bifurcation arguments to reduce results which say that there are multiple solutions of(4.1)μ, (4.1) for μ on one side of zero and guarantee the existence of at least one solution for μ=0 and μ on the other side of zero. To wit, we have the following.Theorem 4.1.
Let (H1), (H2), and (H3) hold. Then there existμ-<0<μ+ such that (4.1)μ, (4.1) have(1)
at least one solution ifμ∈[0,μ+],(2)
at least three solutions ifμ∈[μ-,0).We have the following “dual” theorem if (H2) is replaced by the assumption(H2′)
that there existsβ>0 such that
(4.5)sg(t,s)<0,fort∈𝕋2,|s|>β.Theorem 4.2.
Let (H1), (H2'), and (H3) hold. Then there exist μ-<0<μ+ such that (4.1)μ, (4.1) have(1)
at least one solution ifμ∈[μ-,0],(2)
at least three solutions ifμ∈(0,μ+].Defineℒ:X→Z by (4.6)(ℒy)(t)=Δ4u(t-2)+μk(δΔ2u(t-1)-γu(t)),t∈𝕋2.
Define F:Z→Z by (4.7)(Fu)(t)=g(t,u(t)),t∈𝕋2.
It is easy to check that F:Z→Z is continuous. Obviously (4.1)μ, (4.1) are equivalent to (4.7)μℒu+μ(δΔ2u(t-1)-γu(t))=F(u)+h.Define an operatorP:X→X by (4.8)(Px)(t)=ψk(t)∑s=a+2b-2x(s)ψk(s),t∈𝕋0,
where (4.9)ψk(a):=-ψk(a+2),ψk(b):=-ψk(b-2).It is easy to show the following.Lemma 4.3.
P is a projection and Im(P)=Ker(ℒ).Define an operatorE:Z→Z by (4.10)(Ey)(t)=y(t)-ψk(t)∑s=a+2b-2y(s)ψk(s),t∈𝕋2.
Obviously, we have the following.Lemma 4.4.
E is a projection and Im(E)=Im(ℒ).It is clear that(4.11)X=XP⊕XI-P,Z=ZI-E⊕ZE,
where I represents the identity operator and XP,XI-P,ZI-E, and ZE are the images of P,I-P,I-E, and E, respectively.It is obvious that the restriction ofℒ to XI-P is a bijection from XI-P onto ZE, the image of ℒ. We define M:ZE→XI-P by (4.12)M:=(ℒ|XI-P)-1.
Since ker(ℒ)=span{ψk}, we see that each x∈X can be uniquely decomposed into (4.13)x=ρψk+v
for some ρ∈ℝ, and v∈XI-P. For z∈Z, we also have the decomposition (4.14)z=τψk+h¯,
with τ∈ℝ and h¯∈ZE.Lemma 4.5.
Equations(4.1)μ, (4.1) are equivalent to the system
(4.9)μℒv+μ(δΔ2v(t-1)-γv(t))=EF(ρψk+v)+h,(4.15)μ∑s=a+2b-2(δΔ2ψk(s-1)-γψk(s))ψk(s)=∑s=a+2b-2ψk(s)f(s,ρψk(s)+v(s)).Lemma 4.6.
Let (H1) and (H2) hold. Then there existsR0 such that any solution y of (4.1)μ, (4.1) satisfies
(4.16)∥y∥X<R0
as long as
(4.17)0≤μ≤δ̂:=12∥MJ∥XI-P→XI-P,
where J:X→Z is defined by
(4.18)(Jx)(t):=δΔ2x(s-1)-γx(s),t∈𝕋2.Proof.
Obviously(ℒ+μJ)|XI-P:XI-P→ZE is invertible for |μ|≤δ̂. Moreover, by (4.17),
(4.19)∥(ℒ+μJ)|XI-P-1∥ZE→XI-P=∥[(ℒ|XI-P)(I+μMJ)]-1∥ZE→XI-P=∥[(I+μMJ)]-1∥XI-P→XI-P∥M∥ZE→XI-P≤2∥M∥ZE→XI-P.
Let y=ρψk+v be any solution of (4.1)μ, (4.1). Then we have that, if ρ≠0,
(4.20)∥v∥X=∥(ℒ+μJ)|XI-P-1E(h-F(ρψk+v))∥X≤∥(ℒ+μJ)|XI-P-1∥ZE→XI-P∥E∥Z→ZE[∥h∥Z+C1(|ρ|∥ψk∥Z+∥v∥Z)α+C2]≤2∥M∥ZE→XI-P∥E∥Z→ZE[∥h∥Z+C1(|ρ|∥ψk∥Z+∥v∥Z)α+C2]≤2∥M∥ZE→XI-P∥E∥Z→ZE[∥h∥Z+C1(|ρ|∥ψk∥X+∥v∥X)α+C2]=2∥M∥ZE→XI-P∥E∥Z→ZE[∥h∥Z+C1(|ρ|∥ψk∥X)α(1+∥v∥X|ρ|∥ψk∥X)α+C2]≤2∥M∥ZE→XI-P∥E∥Z→ZE[∥h∥Z+C1(|ρ|∥ψk∥X)α(1+α∥v∥X|ρ|∥ψk∥X)+C2]=2∥M∥ZE→XI-P∥E∥Z→ZE×[∥h∥Z+C1(|ρ|∥ψk∥X)α(1+α(|ρ|∥ψk∥X)1-α∥v∥X(|ρ|∥ψk∥X)α)+C2],
and hence
(4.21)∥v∥X(|ρ|∥ψk∥X)α≤C3(|ρ|∥ψk∥X)α+C4+αC4(|ρ|∥ψk∥X)1-α∥v∥X(|ρ|∥ψk∥X)α,
where
(4.22)C3=∥M∥ZE→XI-P∥E∥Z→ZE(∥h∥Z+C2),C4=2C1∥M∥ZE→XI-P∥E∥Z→ZE.
If
(4.23)|ρ|≥(2αC4)1/(1-α)∥ψk∥X=:C̃,
then we have
(4.24)∥v∥X(|ρ|∥ψk∥X)α≤2C3(C̃∥ψk∥X)α+2C4=:C*.
If we assume that the conclusion of the lemma is false, we obtain a sequence{ηn} with 0≤ηn≤δ̂ and ηn→0, and a sequence of corresponding solutions {yn=ρnψk+vn} of (4.1)ηn, (4.1) such that ∥yn∥X→+∞. From (4.24), we conclude that it is necessary that |ρn|→+∞. We may assume that
(4.25)ρn→+∞,ρn≥C̃∀n∈ℕ
since the other case can be treated by the same way. Thus (4.24) yields that
(4.26)∥vn∥X:≤Ĉ|ρn|α
with Ĉ:=C*∥ψk∥Xα.
Now from (4.15), we get that
(4.27)ρnηn∑s=a+2b-2(δΔ2ψk(s-1)-γψk(s))ψk(s)=∑s=a+2b-2ψk(s)f(s,ρnψk(s)+vn(s)).
By (4.17) and (4.27), it follows that
(4.28)∑s=a+2b-2ψk(s)f(s,ρnψk(s)+vn(s))≤0.
Let(4.29)A+={t∣t∈{a+2,…,b-2},ψk(t)>0},A-={t∣t∈{a+2,…,b-2},ψk(t)<0}.
It is easy to see that
(4.30)A+∪A-≠∅,min{|ψk(t)|∣t∈A+∪A-}>0.
Combining (4.30) and (4.26), we conclude that there exists a positive constant Γ such that, for n∈ℕ,
(4.31)∥vn∥X≤Γ(|ρn|min{|ψk(t)|∣t∈A+∪A-})α,
which implies that
(4.32)limρn→+∞min{ρnψk(t)+vn(t)∣t∈A+}=+∞,limρn→+∞min{ρnψk(t)+vn(t)∣t∈A-}=-∞.
Applying (4.32), (4.30), and (H2), we conclude that
(4.33)∑s=a+2b-2ψk(s)f(s,ρnψk(s)+vn(s))=∑s∈A+ψk(s)f(s,ρnψk(s)+vn(s))+∑s∈A-ψk(s)f(s,ρnψk(s)+vn(s))>0,
which contracts (4.28).Using the similar arguments, we may establish the following lemma.Lemma 4.7.
Let (H1) and (H2') hold. Then there exists R0 such that any solution y of (4.1)μ, (4.1) satisfies
(4.34)∥y∥X<R0
as long as
(4.35)-δ̂≤μ≤0,
where δ̂ is given in (4.17).Lemma 4.8.
Let (H1) and (H2) hold. Then there existsR1:R1≥R0 such that, for 0≤μ≤δ̂ and R≥R1, one has
(4.36)deg(j∘(ℒ+μJ-F-h),B(R),0)=deg(j∘(ℒ+δ̂J),B(R),0)=±1,
where j:Z→X is the natural homomorphism, B(R)={u∈X∥u∥X<R}, and “deg” denotes Leray-Schauder degree when μ≠0 and coincidence degree when μ=0 (see the study by Gaines and Mawhin in [20]). Therefore (4.7)μ has a solution in B¯(R) for μ∈[0,δ̂].Proof.
By Lemma4.6 and the definition of ℒ, the degree
(4.37)deg(j∘(ℒ+μJ-F-h),B(R),0)
is well defined for μ∈[0,δ̂] and is a constant with respect to μ.
Now if(μ,y)∈[0,1]×X is a solution of
(4.38)ℒy+δ̂(δΔ2y(s-1)-γy(s))-μ(F(y)+h)=0,
then we have
(4.39)∥y∥X=μ∥(ℒ+δ̂J)-1(h+F(y))∥X≤∥(ℒ+δ̂J)-1∥ZE→XI-P(∥h∥Z+C1∥y∥Xα+C2).
Hence there exists R0′>0 such that ∥y∥X<R0′. Thus if R1=max{R0′,R0}, then we have, whenever R>R1, that
(4.40)deg(j∘(ℒ+δ̂J-F-h),B(R),0)=deg(j∘(ℒ+δ̂J),B(R),0)=±1,
which completes the proof.By a similar manner we may establish the following.Lemma 4.9.
Let (H1) and (H2′) hold. Then there exists R1:R1≥R0 such that, for 0≥μ≥-δ̂ and R≥R1, one has
(4.41)deg(j∘(ℒ+μJ-F-h),B(R),0)=deg(j∘(ℒ-δ̂J),B(R),0)=±1.
Therefore (4.7)μ has a solution in B¯(R) for μ∈[-δ̂,0].Lemma 4.10.
Let (H1) and (H2) hold. Then there existsδ1>0 such that, for -δ1<μ<0, one has
(4.42)deg(j∘(ℒ+μJ-F-h),B(R1),0)=deg(j∘(ℒ+δ1J),B(R1),0)=±1.Proof.
Let(4.43)τ0=infx∈∂B(R1)∩X∥j∘(ℒx-F(x)-h)∥X.
Then it is not difficult to check that τ0>0. Hence if we take δ1 so small that δ1R1<τ0, then for μ∈[-δ1,δ1],
(4.44)deg(j∘(ℒ+μJ-F-h),B(R1),0)=deg(j∘(ℒ-F-h),B(R1),0)=±1.Lemma 4.11.
Let (H1) and (H2') hold. Then there exists δ1>0 such that, for 0≤μ≤δ1, one has
(4.45)deg(j∘(ℒ+μJ-F-h),B(R),0)=deg(j∘(ℒ-δ1J),B(R),0)=±1.Proof of Theorem4.1.
By the study of Massabò and Pejsachowicz in [21, Theorem 1.1], (4.7)μ has a continuum 𝒞*={(μ,yμ)} of solutions with ∥yμ∥X<R1 and μ∈[-δ1,δ̂]. On the other hand, since F is ℒ-completely continuous and satisfies (H1) and since μ=0 is a simple eigenvalue, it follows from the study by Rabinowitz in [22, Theorem 1.6] that μ=0 is a bifurcation point from infinity for (4.7)μ. Moreover, there exist two continua
(4.46)𝒞∞±={(μ,yμ)}⊂ℝ×X
of solutions of (4.7)μ, bifurcating from infinity at μ=0, that is, there exists ϵ0∈(0,1/R1), such that for all ϵ:0<ϵ≤ϵ0 there exist two continua 𝒞ϵ+ and 𝒞ϵ- with
(4.47)𝒞ϵ+⊂𝒞∞+,𝒞ϵ-⊂𝒞∞-,𝒞ϵ±⊂{(μ,yμ):∥yμ∥X≥1ϵ,|μ|<ϵ}=:Uϵ(0,∞),
and 𝒞ϵ± connects (0,∞) to ∂Uϵ(0,∞). Notice that (μ,y)∈𝒞ϵ+ implies that y(a+2)>0, and (μ,y)∈𝒞ϵ- implies that y(a+2)<0. So,
(4.48)𝒞ϵ+∩𝒞ϵ-=∅.
Now, Lemma 4.6 implies that
(4.49)𝒞ϵ±⊆{(μ,yμ)∣∥yμ∥X≥1ϵ,-ϵ<μ<0}.
This completes the proof.Proof of Theorem4.2.
Using similar arguments, we may get the desired results.



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*Source: 839474-2010-06-03.xml*