# Sign-Changing Solutions for Discrete Second-Order Three-Point Boundary Value Problems

Authors: Tieshan He; Wei Yang; Fengjian Yang
Journal: Discrete Dynamics in Nature and Society
 (2010)

---

## Abstract

We consider the second-order three-point discrete boundary value problem. By using the topological degree theory and the fixed point index theory, we provide sufficient conditions for the existence of sign-changing solutions, positive solutions, and negative solutions. As an application, an example is given to demonstrate our main results.


---

## Body

## 1. Introduction

In this paper, we consider the following second-order three-point discrete boundary value problem (BVP):(1.1)Δ2u(t-1)+f(t,u(t))=0,t∈[1,n],u(0)=0,u(n+1)=αu(m),
where n∈{2,3,…}, [1,n] is the discrete interval {1,2,…,n}, m∈[1,n], 0≤α≤1, Δu(t)=u(t+1)-u(t), Δ2u(t)=Δ(Δu(t)), and f:[1,n]×R→R is a continuous function.Boundary value problems for difference equations arise in different areas of applied mathematics and physics. Existence and multiplicity of positive solutions or nontrivial solutions for boundary value problems of difference equations have been extensively studied in the literature; see [1–9] and the references therein.On the other hand, in the existing literature, there are some papers studying the sign-changing solutions for boundary value problems of differential equations; for example, see [10–12]. But the problems of the existence of sign-changing solutions to discrete multipoint boundary value problems have received very little attention in the literature to the best knowledge of the authors. In this paper, motivated by [12, 13], we aim to study the existence of multiple sign-changing solutions to the second-order three-point discrete boundary value problem (1.1). Under some suitable conditions, we prove that the three-point discrete boundary value problem (1.1) has at least two sign-changing solutions, two positive solutions, and two negative solutions. The main approach is the topological degree theory and the fixed point index theory.The organization of this paper is as follows. In Section2, we present some preliminary knowledge about the topological degree theory and the fixed point index theory and use the knowledge to obtain some lemmas which are very crucial in our main results. In Section 3, by computing the topological degree and the fixed point index, we discuss the existence of multiple sign-changing solutions to BVP (1.1), and a simple example is given.


## 2. Preliminaries

As we have mentioned, we will use the theory of the Leray-Schauder degree and the fixed point index in a cone to prove our main existence results. Let us collect some results that will be used below. One can refer to [13–16] for more details.Lemma 2.1 (see [13, 14]).
LetE be a Banach space and, X⊂E be a cone in E. Assume that  Ω is a bounded open subset of E. Suppose that A:X∩Ω¯→X is a completely continuous operator. If there exists x0∈X∖{θ} such that
(2.1)x-Ax≠μx0,∀x∈X∩∂Ω,μ≥0,
then the fixed point index i(A,X∩Ω,X)=0.Lemma 2.2 (see [13, 14]).
LetE be a Banach space and let X⊂E be a cone in E. Assume that  Ω is a bounded open subset of E, θ∈Ω. Suppose that A:X∩Ω¯→X is a completely continuous operator. If
(2.2)Ax≠μx,∀x∈X∩∂Ω,μ≥1.
then the fixed point index i(A,X∩Ω,X)=1.Lemma 2.3 (see [15]).
LetE be a Banach space, let Ω be a bounded open subset of E, θ∈Ω, and A:Ω¯→E be completely continuous. Suppose that
(2.3)∥Ax∥≤∥x∥,Ax≠x,∀x∈∂Ω,
then deg(I-A,Ω,θ)=1.Lemma 2.4 (see [16]).
LetA be a completely continuous operator which is defined on a Banach space E. Let x0∈E be a fixed point of A and assume that A is defined in a neighborhood of x0 and Fréchet differentiable at x0. If 1 is not an eigenvalue of the linear operator A'(x0), then x0 is an isolated singular point of the completely continuous vector field I-A and for small enough r>0,
(2.4)deg(I-A,B(x0,r),θ)=(-1)k,
where k is the sum of the algebraic multiplicities of the real eigenvalues of A'(x0) in (1,+∞).Lemma 2.5 (see [16]).
LetA be a completely continuous operator which is defined on a Banach space E. Assume that 1 is not an eigenvalue of the asymptotic derivative. Then the completely continuous vector field I-A is nonsingular on spheres Sρ={x∈E:∥x∥=ρ} of sufficiently large radius ρ and
(2.5)deg(I-A,B(θ,ρ),θ)=(-1)k,
where k is the sum of the algebraic multiplicities of the real eigenvalues of A'(∞) in (1,+∞).From [12, Lemma 2.4], we have the following lemma.Lemma 2.6.
LetX be a solid cone of a Banach space E (X∘ is nonempty), let Ω be a relatively bounded open subset of X, and let A:X→X be a completely continuous operator. If any fixed point of A in Ω is an interior point of X, there exists an open subset O of E (O⊂Ω) such that
(2.6)deg(I-A,O,θ)=i(A,Ω,X).Now we shall consider the space(2.7)E={u:[0,n+1]→R∣u(0)=0,u(n+1)=αu(m)}equipped with the norm∥u∥=maxt∈[0,n+1]|u(t)|. Clearly E is a n-dimensional Banach space. Choose the cone P⊂E defined by(2.8)P={u∈E∣u(t)≥0,t∈[1,n]}.Obviously, the interior ofP is P∘={u∈E∣u(t)>0,t∈[1,n]}. For each u,v∈E, we write u≥v if u(t)≥v(t) for t∈[1,n]. A solution u  of BVP (1.1) is said to be a positive solution (a negative solution, resp.) if u∈P∖{θ} (u∈(-P)∖{θ}, resp.). A solution u of BVP (1.1) is said to be a sign-changing solution if u∉P∪(-P).Lemma 2.7.
Letv:[1,n]→R be fixed. Then the problem
(2.9)Δ2u(t-1)+v(t)=0,t∈[1,n],u(0)=0,u(n+1)=αu(m)
has a unique solution
(2.10)u(t)=∑k=1nG(t,k)v(k),
where G(t,k) is given by
(2.11)G(t,k)={k(n+1-αm-t+αt)n+1-αm,k∈[1,t-1]∩[1,m-1];t(n+1-αm-k+αk)n+1-αm,k∈[t,m-1];k(n+1-αm-t)+αmtn+1-αm,k∈[m,t-1];t(n+1-k)n+1-αm,k∈[t,n]∩[m,n].Proof.
We use a similar approach to that in [17, Lemmas 3.1, 3.3]. From Δ2u(t-1)+v(t)=0, we have
(2.12)Δu(1)-Δu(0)+v(1)=0,Δu(2)-Δu(1)+v(2)=0,…,Δu(t)-Δu(t-1)+v(t)=0.
Suming the above equations, one gets
(2.13)Δu(t)=Δu(0)-∑i=1tv(i),
where, and in what follows, we denote ∑k=slx(k)=0 when l<s. Again summing (2.13) from 0 to t-1, it follows that
(2.14)u(t)=u(0)+tΔu(0)-∑k=1t-1(t-k)v(k)
where t∈[0,n+1]. Since u(0)=0 and u(n+1)=αu(m), one gets
(2.15)Δu(0)=1n+1-αm∑k=1n(n+1-k)v(k)-αn+1-αm∑k=1m-1(m-k)v(k).
By (2.14) and (2.15), we have
(2.16)u(t)=tn+1-αm∑k=1n(n+1-k)v(k)-αtn+1-αm∑k=1m-1(m-k)v(k)-∑k=1t-1(t-k)v(k),t∈[0,n+1].
When t>m, it follows from (2.16) that
(2.17)u(t)=tn+1-αm(∑k=1m-1(n+1-k)v(k)+∑k=mt-1(n+1-k)v(k)+∑k=tn(n+1-k)v(k))-αtn+1-αm∑k=1m-1(m-k)v(k)-∑k=1m-1(t-k)v(k)-∑k=mt-1(t-k)v(k)=∑k=1nG(t,k)v(k).
When t≤m, it follows from (2.16) that
(2.18)u(t)=tn+1-αm(∑k=1t-1(n+1-k)v(k)+∑k=tm-1(n+1-k)v(k)+∑k=mn(n+1-k)v(k))-αtn+1-αm(∑k=1t-1(m-k)v(k)+∑k=tm-1(m-k)v(k))-∑k=1t-1(t-k)v(k)=∑k=1nG(t,k)v(k).
Then, the unique solution of (2.9) can be written as u(t)=∑k=1nG(t,k)v(k).Remark 2.8.
Green’s functionG(t,k) defined by Lemma 2.7 is positive on [1,n]×[1,n].
Define operatorsK,f,A:E→E, respectively, by
(2.19)(Ku)(t)=∑k=1nG(t,k)u(k),u∈E,t∈[1,n];(fu)(t)=f(t,u(t)),u∈E,t∈[1,n];(2.20)A=Kf.
Now from Lemma 2.7, it is easy to see that BVP (1.1) has a solution u=u(t) if and only if u is a fixed point of the operator A. It follows from the continuity of f that A:E→E is completely continuous.
We shall use the following assumptions.
(H1)
We have0≤α<1, or
(2.21)α=1,Λ=ϕ,
where Λ=A∩B, and(2.22)A={(2k-1)πn+1+m:k=1,2,…,[n+1+m2]},B={{2tπn+1-m:t=1,2,…,[n-m2]},n-m≥2,ϕ,n-m<2.[x] denotes the integer part of the real number x. (H2)
For anyt∈[1,n], f(t,0)=0; for any t∈[1,n] and x∈R, xf(t,x)≥0.(H3)
There exists an even numberk0∈[1,n] such that(2.23)1λk0<β0<1λk0+1,
where limx→0(f(t,x)/x)=β0 uniformly for t∈[1,n], and λk-1=4sin2(ξk/2),k∈[1,n], ξ1,…,ξn are given in Lemma 2.9, λn+1-1≜+∞.(H4)
There exists an even numberk1∈[1,n] such that(2.24)1λk1<β∞<1λk1+1,
where lim|x|→∞(f(t,x)/x)=β∞ uniformly for t∈[1,n], and λ1-1,…,λn-1,λn+1-1 are given in condition (H3).(H5)
There exists a constantT>0 such that for any (t,x)∈[1,n]×[-T,T],(2.25)|f(t,x)|<ρ-1T,
where ρ=[(n+1)2-αm2]2/8(n+1-αm)2.Lemma 2.9.
Suppose that (H1) holds; then there exist ξ1,ξ2,…,ξn with 0<ξ1<ξ2<⋯<ξn<π such that sin(n+1)ξi=αsinmξi, i=1,2,…,n.Proof.
First, suppose that0≤α<1. Let h(x)=sin(n+1)x-αsinmx, then we have
(2.26)h(π2(n+1))>0,h(πn+1)≤0,h(3π2(n+1))<0,h(5π2(n+1))>0,h(7π2(n+1))<0,…,(-1)nh((2n+1)π2(n+1))>0.
It follows from the intermediate value theorem that there exist ξ1∈(π/2(n+1),π/n+1] and ξi∈((2i-1)π/2(n+1),(2i+1)π/2(n+1)),i=2,…,n, such that sin(n+1)ξi=αsinmξi, i=1,2,…,n.
Now suppose thatα=1 and Λ=ϕ. Let Λ1=A∪B. It is easy to see that there exist ζi∈Λ1,i=1,2,…,n with 0<ζ1=(π/n+1+m)<ζ2<⋯<ζn<π such that sin(n+1)ζi=sinmζi. Then Lemma 2.9 is proved.Remark 2.10.
ConditionΛ=ϕ is reasonable. For example, let n=5, m=3, then Λ=ϕ. Let n=18, m=3, then ξ=π/2∈Λ≠ϕ.Lemma 2.11.
Suppose that (H1) holds; then the set of eigenvalues of the linear operator K consists of the strictly decreasing finite sequence of  λk,k=1,2,…,n, with corresponding eigenfunctions φk(t)=sin(tξk), where λk=(4sin2(ξk/2))-1, k=1,2,…,n, and ξ1,…,ξn are given in Lemma 2.9. In addition, the algebraic multiplicity of each eigenvalue λk of the linear operator K is equal to 1.Proof.
It is easy to see that(2.27)Ku(t)=λu(t),t∈[1,n],u∈E
is equivalent to the following equation:
(2.28)λΔ2u(t-1)+u(t)=0,t∈[1,n],u(0)=0,u(n+1)=αu(m).
By Lemma 2.9, we suppose that φk(t)=sin(tξk) is a nontrivial solution of (2.28). Then,
(2.29)λ(sin(t+1)ξk-2sintξk+sin(t-1)ξk)+sintξk=0.
Hence, for any k∈[1,n], λ=λk=(4sin2(ξk/2))-1is an eigenvalue of the linear operator K with the corresponding eigenfunction φk(t)=sin(tξk). Since the linear operator K is identified with a linear transformation from Rn to Rn, the set of eigenvalues of the linear operator K consists of the strictly decreasing finite sequence of λk,k=1,2,…,n. Obviously, the algebraic multiplicity of each eigenvalue λk of K is equal to 1. This completes the proof.Remark 2.12.
Whenα=0, we see that BVP (1.1) is reduced to Dirichlet boundary value problem and λk=(4sin2(kπ/2(n+1)))-1, k=1,2,…,n. When α=1 and m=n, BVP (1.1) is reduced to the focal boundary value problem and λk=(4sin2(2k-1)π/4n+2)-1, k=1,2,…,n.Lemma 2.13.
Suppose that (H2) holds, and u=(u1,u2,…,un)T∈P∖{θ} is a solution of BVP (1.1). Then u∈P∘.Proof.
Ifu(t)=0 for some t∈[1,n], then
(2.30)u(t+1)+u(t-1)=Δ2u(t-1)=-f(t,u(t))=0.
So u(t±1)=0, and it follows that if u is zero somewhere in [1,n], then it vanishes identically in [1,n].Remark 2.14.
Similarly to Lemma2.13, we know also that if (H2) holds and u∈-P∖{θ} is a solution of BVP (1.1), then u∈-P∘.Lemma 2.15.
Suppose that (H2)–(H4) hold. Then the operator A is Fréchet differentiable at θ and ∞, where operator A is defined by (2.20). Moreover, A'(θ)=β0K and A'(∞)=β∞K.Proof.
By (H3), for any ε>0, there exist δ>0 such that |f(t,x)-β0x|<ε|x| for any 0<|x|<δ, t∈[1,n]. Hence, noticing that f(t,0)=0 for any t∈[1,n], we have
(2.31)∥Au-Aθ-β0Ku∥=∥K(fu-β0u)∥≤∥K∥·maxt∈[1,n]|f(t,u(t))-β0u(t)|<ε∥K∥·∥u∥
for any u∈E with 0<∥u∥<δ, where ∥K∥=maxt∈[1,n]∑k=1n|G(t,k)|. Consequently,
(2.32)lim∥u∥→0∥Au-Aθ-β0Ku∥∥u∥=0.
This means that the nonlinear operator A is Fréchet differentiable at θ, and A'(θ)=β0K.
By (H4), for any ε>0, there exist M>0 such that |f(t,x)-β∞x|<ε|x| for any |x|>M, t∈[1,n]. Let c=max(t,x)∈[1,n]×[-M,M]|f(t,x)-β∞x|. By the continuity of f(t,x) with respect to x, we have c<+∞. Then, for any (t,x)∈[1,n]×R, |f(t,x)-β∞x|<ε|x|+c. Thus
(2.33)∥Au-β∞Ku∥≤∥K∥·maxt∈[1,n]|f(t,u(t))-β∞u(t)|<∥K∥(ε∥u∥+c)
for any u∈E. Consequently,
(2.34)lim∥u∥→∞∥Au-β∞Ku∥∥u∥=0,
which implies that operator A is Fréchet differentiable at ∞, and A'(∞)=β∞K. The proof is completed.Lemma 2.16.
LetT be given in condition (H5). Suppose that (H1)–(H4) hold. Then,  A(P)⊂P, A(-P)⊂-P. Moreover, one has the following.(i)
There exists anr0∈(0,T) such that for any 0<r≤r0,
(2.35)i(A,P∩B(θ,r),P)=0,i(A,-P∩B(θ,r),-P)=0.(ii)
There exists anR0>T such that for any R≥R0,(2.36)i(A,P∩B(θ,R),P)=0,i(A,-P∩B(θ,R),-P)=0.Proof.
By (H2) and the fact that G(t,k) is positive on [1,n]×[1,n], we get that for any t∈[1,n], f(t,P)⊂P, f(t,-P)⊂-P, and K(P)⊂P, K(-P)⊂-P. Then A(P)⊂P and A(-P)⊂-P.
We only need to prove conclusion (i). The proof of conclusion (ii) is similar and will be omitted here. Letγ0=inf∥u∥=1∥u-β0Ku∥. Condition (H3) yields γ0>0. It follows from (2.32) that there exists r0∈(0,T) such that
(2.37)∥Au-β0Ku∥<12γ0∥u∥,
where 0<∥u∥≤r0. Setting H(s,u)=sAu+(1-s)β0Ku, then H:[0,1]×E→E is completely continuous. For any s∈[0,1] and 0<∥u∥≤r0, we obtain that
(2.38)∥u-H(s,u)∥≥∥u-β0Ku∥-s∥Au-β0Ku∥≥γ0∥u∥-12γ0∥u∥>0.
According to the homotopy invariance of the fixed point index, for any 0<r≤r0, we have
(2.39)i(A,P∩B(θ,r),P)=i(β0K,P∩B(θ,r),P),(2.40)i(A,-P∩B(θ,r),-P)=i(β0K,-P∩B(θ,r),-P).
Let φ1(t)=sin(tξ1). Then Kφ1=λ1φ1 and φ1∈P (see Lemma 2.11 and the proof of Lemma 2.9). We claim
(2.41)u-β0Ku≠σφ1,∀u∈P∩∂B(θ,r),σ≥0.
Indeed, we assume that there exist u1∈P∩∂B(θ,r) and σ1≥0 such that u1-β0Ku1=σ1φ1. Obviously, u1=β0Ku1+σ1φ1≥σ1φ1. Since β0≠λk-1,k=1,2,…,n, then σ1>0. Set σmax=sup{σ:u1≥σφ1}. It is clear that σ1≤σmax<+∞ and u1≥σmaxφ1. Then
(2.42)u1=β0Ku1+σ1φ1≥β0Kσmaxφ1+σ1φ1=(β0λ1σmax+σ1)φ1.
Since β0λ1>1, then β0λ1σmax+σ1>σmax, which contradicts with the definition of σmax. This proves (2.41).
It follows from Lemma2.1 and (2.41) that
(2.43)i(β0K,P∩B(θ,r),P)=0.
Similarly to (2.43), we know also that
(2.44)i(β0K,-P∩B(θ,r),-P)=0.
By (2.39), (2.43), (2.40), and (2.44), we conclude
(2.45)i(A,P∩B(θ,r),P)=0,i(A,-P∩B(θ,r),-P)=0.


## 3. Main Results

Now with the aid of the lemmas in Section2, we are in position to state and prove our main results.Theorem 3.1.
Assume that the conditions (H1)–(H5) hold. Then BVP (1.1) has at least two sign-changing solutions. Moreover, BVP (1.1) has at least two positive solutions and two negative solutions.Proof.
From the proof of Lemma2.7, we have
(3.1)∑k=1nG(t,k)=tn+1-αm∑k=1n(n+1-k)-αtn+1-αm∑k=1m-1(m-k)-∑k=1t-1(t-k)=tn+1-αm·n(n+1)2-αtn+1-αm·m(m-1)2-t(t-1)2≤[(n+1)2-αm2]28(n+1-αm)2=ρ.
Since G(t,k) is positive on  [1,n]×[1,n], by (H5), we have for any u∈E with ∥u∥=T,
(3.2)|(Au)(t)|=|∑k=1nG(t,k)f(k,u(k))|≤∑k=1nG(t,k)|f(k,u(k))|<ρ-1T∑k=1nG(t,k)≤T,∀t∈[1,n].
This gives
(3.3)∥Au∥<T=∥u∥.
By (3.3) and Lemmas 2.3 and 2.2, we have
(3.4)deg(I-A,B(θ,T),θ)=1,(3.5)i(A,P∩B(θ,T),P)=1,(3.6)i(A,-P∩B(θ,T),-P)=1.
From (H3) and Lemma 2.11, one has that the eigenvalues of the operator A'(θ)=β0K which are larger than 1 are
(3.7)β0λ1,β0λ2,…,β0λk0,
From (H4)   and Lemma 2.11, one has that the eigenvalues of the operator A'(∞)=β∞K which are larger than 1 are
(3.8)β∞λ1,β∞λ2,…,β∞λk1.
It follows from Lemmas 2.4 and 2.5 that there exist 0<r1<r0 and R1>R0 such that
(3.9)deg(I-A,B(θ,r1),θ)=(-1)k0=1,(3.10)deg(I-A,B(θ,R1),θ)=(-1)k1=1,
where r0 and R0 are given in Lemma 2.16. Owing to Lemma 2.16, one has
(3.11)i(A,P∩B(θ,r1),P)=0,(3.12)i(A,-P∩B(θ,r1),-P)=0,(3.13)i(A,P∩B(θ,R1),P)=0,(3.14)i(A,-P∩B(θ,R1),-P)=0.
According to the additivity of the fixed point index, by (3.5), (3.11), and (3.13), we have
(3.15)i(A,P∩(B(θ,T)∖B(θ,r1)¯),P)=i(A,P∩B(θ,T),P)-i(A,P∩B(θ,r1),P)=1-0=1,(3.16)i(A,P∩(B(θ,R1)∖B(θ,T)¯),P)=i(A,P∩B(θ,R1),P)-i(A,P∩B(θ,T),P)=0-1=-1.
Hence, the nonlinear operator A has at least two fixed points u1∈P∩(B(θ,T)∖B(θ,r1)¯) and u2∈P∩(B(θ,R1)∖B(θ,T)¯), respectively. Then, u1 and u2 are positive solutions of BVP (1.1).
Using again the additivity of the fixed point index, by (3.6), (3.12), and (3.14), we get
(3.17)i(A,-P∩(B(θ,T)∖B(θ,r1)¯),-P)=1-0=1,(3.18)i(A,-P∩(B(θ,R1)∖B(θ,T)¯),-P)=0-1=-1.
Hence, the nonlinear operator A has at least two fixed points u3∈-P∩(B(θ,T)∖B(θ,r1)¯) and u4∈-P∩(B(θ,R1)∖B(θ,T)¯), respectively. Then, u3 and u4 are negative solutions of BVP (1.1).     Let
(3.19)Γ1={u∈P∩(B(θ,T)∖B(θ,r1)¯):Au=u},Γ2={u∈P∩(B(θ,R1)∖B(θ,T)¯):Au=u},Γ3={u∈-P∩(B(θ,T)∖B(θ,r1)¯):Au=u},Γ4={u∈-P∩(B(θ,R1)∖B(θ,T)¯):Au=u}.
It follows from Lemmas 2.6, 2.13, Remark 2.14, and (3.15)–(3.18) that there exist open subsets O1,O2,O3, and O4 of E such that
(3.20)Γ1⊂O1⊂P∩(B(θ,T)∖B(θ,r1)¯),Γ2⊂O2⊂P∩(B(θ,R1)∖B(θ,T)¯),Γ3⊂O3∈-P∩(B(θ,T)∖B(θ,r1)¯),Γ4⊂O4⊂-P∩(B(θ,R1)∖B(θ,T)¯),(3.21)deg(I-A,O1,θ)=1,(3.22)deg(I-A,O2,θ)=-1,(3.23)deg(I-A,O3,θ)=1,(3.24)deg(I-A,O4,θ)=-1.
By (3.4), (3.21), (3.23), (3.9), and the additivity of Leray-Schauder degree, we get
(3.25)deg(I-A,B(θ,T)∖(O1¯∪O3¯∪B(θ,r1)¯),θ)=1-1-1-1=-2,
which implies that the nonlinear operator A has at least one fixed point u5∈B(θ,T)∖(O1¯∪O3¯∪B(θ,r1)¯).
Similarly, by (3.10), (3.22), (3.24), and (3.4), we get
(3.26)deg(I-A,B(θ,R1)∖(O2¯∪O4¯∪B(θ,T)¯),θ)=1+1+1-1=2,
which implies that the nonlinear operator A has at least one fixed point u6∈B(θ,R1)∖(O2¯∪O4¯∪B(θ,T)¯). Then, u5 and u6 are two distinct sign-changing solutions of BVP (1.1). Thus, the proof of Theorem 3.1 is finished.Theorem 3.2.
Assume that the conditions (H1)–(H5) hold, and that f(t,x)=-f(t,-x) for t∈[1,n] and x∈R. Then BVP (1.1) has at least four sign-changing solutions. Moreover, BVP (1.1) has at least two positive solutions and two negative solutions.Proof.
It follows from the proof of Theorem3.1 that BVP (1.1) has at least six different nontrivial solutions ui(i=1,2,…,6) satisfying that
(3.27)u1,u2∈P∘,u3,u4∈-P∘,u5,u6∉P∪(-P),r1<∥u5∥<∥T∥<∥u6∥<R1.
By the condition that f(t,x)=-f(t,-x) for t∈[1,n] and x∈R, we know that -u5 and -u6 are also solutions of BVP (1.1). Let u7=-u5, u8=-u6, then ui(i=1,2,…,8) are different nontrivial solutions of BVP (1.1). The proof is completed.By the method used in the proof of Theorems3.1 and 3.2, we can prove the following corollaries.Corollary 3.3.
Assume that the conditions (H1)–(H3) and (H5) or (H1), (H2), (H4), and (H5) hold. Then BVP (1.1) has at least one sign-changing solution. Moreover, BVP (1.1) has at least one positive solution and one negative solution.Corollary 3.4.
Assume that the conditions (H1)–(H3) and (H5) or (H1), (H2), (H4), and (H5) hold, and that f(t,x)=-f(t,-x) for x∈R and t∈[1,n]. Then BVP (1.1) has at least two sign-changing solutions. Moreover, BVP (1.1) has at least one positive solution and one negative solution.Next, we present a simple example to which Theorem3.2 can be applied.Example 3.5.
Consider the second-order three-point discrete boundary value problem(3.28)Δ2u(t-1)+f(t,u(t))=0,t∈[1,4],u(0)=0,u(5)=u(3),
where n=4,  m=3,  α=1, and
(3.29)f(t,x)={bx1+x2,|x|≤20,(18-2b401)x+60b401-360,20≤x≤30,(18-2b401)x+360-60b401,-30≤x≤-20,x(5+1000100+x2),|x|≥30,b∈(4sin2(3π/16),4sin2(5π/16)). Obviously, β0=b,β∞=5, and f(t,x)=-f(t,-x) for all (t,x)∈[1,4]×R.
From Lemma2.11 and the proof of Lemma 2.9, we know that the set of eigenvalues of the linear operator K (see (2.19)) consists of the strictly decreasing finite sequence of λk, k=1,2,3,4, where λk=(4sin2((2k-1)π/16))-1. Then the conditions (H1)–(H4) hold. Since |f(t,x)|≤(b/2)<(3/2) for all (t,x)∈[1,4]×[-12,12], then (H5) holds with T=12 and ρ=8. Therefore, by Theorem 3.2, BVP (3.28) has at least four sign-changing solutions. Moreover, BVP (3.28) has at least two positive solutions and two negative solutions.



---

*Source: 705387-2010-03-08.xml*