# Existence and Multiplicity of Solutions for Discrete Nonlinear Two-Point Boundary Value Problems

Authors: Jianmin Guo; Caixia Guo
Journal: Discrete Dynamics in Nature and Society
 (2011)

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## Abstract

By using Morse theory, the critical point theory, and the character
ofK1/2, we consider the existence and multiplicity results of solutions to the following discrete
nonlinear two-point boundary value problem −Δ2x(k−1)=f(k,x(k)),k∈ℤ(1,T) subject to x(0)=0=Δx(T), where T is a positive integer, ℤ(1,T)={1,2,…,T},Δ is the forward difference operator defined by Δx(k)=x(k+1)−x(k), and f:ℤ(1,T)×ℝ→ℝ is continuous. In argument, Morse inequalities play an important role.


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## Body

## 1. Introduction

In this paper, we discuss the existence and multiplicity results of solutions to the following discrete nonlinear two-point boundary value problem (BVP):(1.1)-Δ2x(k-1)=f(k,x(k)),k∈Z(1,T),x(0)=0=Δx(T),
where T is a positive integer, ℤ(1,T)={1,2,…,T},Δ is the forward difference operator defined by Δx(k)=x(k+1)-x(k), and f:ℤ(1,T)×ℝ→ℝ is continuous.Recent years, there have been many papers studying the existence and multiplicity of solutions for differential equations. For example, by employing the strongly monotone operator principle and the critical point theory, F. Li et al. in [1] establish some conditions on f which are able to guarantee a class of boundary value problem on differential equation has a unique solution, at least one nonzero solution and infinitely many solutions; by using the critical point theory, and Morse theory, Yang and Zhang in [2] obtained some existence results for differential equations with parameters; there are also many authors who studied the existence results of positive solutions to boundary value problem on differential equation by employing the cone expansion or compression fixed point theorem or the critical point theory, see [3–5]; Jiang and Zhou in [6] obtained variational framework of the BVP(1.1) first by virtue of Green's function and separation of linear operator and studied the existence of a unique solution or at least one nontrivial solution by employing the strongly monotone operator principle and the critical point theory respectively.In this paper, the main difference from the ordinary literatures is that we apply Morse theory and critical point theory to deal with problems on discrete systems. Then we establish some conditions onf which include sublinear, superlinear or asymptotical case to guarantee that BVP(1.1) has at least one solution, at least two nontrivial solutions, and infinitely many solutions.


## 2. Preliminary

In this section, we give some notations and lemmas.LetB={x:ℤ(1,T+1)→ℝ,x(0)=0=Δx(T)} is a T-dimensional Hilbert space with inner product (x,y)=∑k=1Tx(k)y(k), and we denote the induced norm by ∥x∥=(∑k=1T|x(k)|2)1/2. Bρ={x∈B:∥x∥⩽ρ},Sρ=∂Bρ. And Ja={x∈H:J(x)⩽a} for any J∈C(B,ℝ1) and a∈ℝ1. Let E be a real Banach space, and let C1(E,ℝ) denote the set of functionals that are Fréchet differentiable and their Fréchet derivatives are continuous on E.LetG [6] be the Green's function of the linear boundary value problem (2.1)-Δ2x(k-1)=0,k∈Z(1,T),x(0)=0=Δx(T),
where (2.2)G(k,t)=min⁡{k,t}={k,k⩽t,t,t⩽k
then the (BVP)(1.1) has a solution if and only if the following equation: (2.3)x(k)=∑t=1TG(k,t)f(t,x(t)),k∈Z(1,T)
has a solution in B. Define operators K,f:B→B by (2.4)Kx(k)=∑t=1TG(k,t)x(t),fx(k)=f(k,x(k)),k∈Z(1,T),∀x∈B.
It is easy to see that a solution of (2.3) is equivalent to a solution in B of the following operator equation: (2.5)x=Kfx.It is well know, that all eigenvalues ofK are (2.6)λt=14sin2((2t-1)π/(4T+2)),t=1,2,…,T,
which have the corresponding orthonormal eigenfunctions (2.7){et}t=1T=sin(2t-1)π2T+1(∑t=1Tsin2(2n-1)π2T+1)-1/2,t∈Z(1,T).Remark 2.1 (see [6]).
(i)
K:B→B is a linear continuous operator; furthermore, K is symmetric in B.(ii)
There arex=∑k=1T(x,ek)ek,∥x∥2=∑k=1T|x,ek|2,Kx=∑k=1Tλk(x,ek)ek,x∈B.(iii)
The square root operator ofK: K1/2:B→B is bounded linear and symmetric,(2.8)K1/2x=∑k=1Tλk1/2(x,ek)ek,x∈B.(iv)
(K1/2x,x)=∑k=1Tλk1/2|(x,ek)|2, x∈B, this implies that K1/2x≠0 for all x∈H with x≠0. Therefore, K1/2x1≠K1/2x2 for all x1,x2∈B with x1≠x2.In next section, we will use the critical point theory and Morse theory to discuss the main results. Here, we state some necessary definitions and lemmas.Definition 2.2.
LetE be a real Banach space; D is an open subset of E. Suppose the functional J:D→ℝ is C1(E,ℝ) on D. If x0∈D and the Fréchet derivative f′(x0)=0, then we call that x0 is a critical point of the functional f, and c=f(x0) is a critical value of f.Definition 2.3.
Letf∈C1(E,ℝ). If any sequence {xm}⊂E for which {f(xm)} is bounded and f′(xm)→0 as m→∞ possesses a convergent subsequence, then we say that f satisfies Palais-Smale condition (denoted by P.S. condition for short).Lemma 2.4 (see [7, Definition 1.16, page 13]).
LetE be a real reflexive Banach space. Suppose that the functional J:E→ℝ1 is C1, and is bounded below and satisfies P.S. condition, then J must have a minimum in E, that is, there exists a x*∈E such that J(x*)=infx∈EJ(x), therefore x* is a critical point of the functional J.Definition 2.5 (see [2]).
LetJ(0)=0, E=V⊕X with dimV<∞ be a real Banach space. Assume that there exists ρ>0 small, such that
(2.9)J(x)⩽0,x∈V,‖x‖⩽ρ,J(x)>0,x∈X,0<‖x‖⩽ρ
Then J has a local linking at 0.Definition 2.6.
LetE be a real Banach space, let u be an isolated critical point of J with J(u)=c and let U be a neighborhood of u, containing the unique critical point. We call Cq(J,u)=Hq(Jc∩U,Jc∩U∖{u}) the q-th critical group of J at u,q=0,1,2,…, where Hq(·,·) stands for the q-th singular relative homology group with integer coefficients. We say that u is a homological nontrivial critical point of J if at least one of its critical groups is nontrivial.Lemma 2.7 (see [8]).
Assume thatJ∈C1(E,ℝ) satisfies P.S. condition and has a local linking at 0. Then Ck(J,θ)≇0, that is, 0 is a homological nontrivial critical point of J.Lemma 2.8 (see [8]).
LetM be a Finsler manifold of C2. Assume that f∈C1(M,ℝ) satisfies the P.S. condition and a is the only critical value of f in [a,b). If connected component of Ka is only composed of isolated critical points, then fa is a deformation retract of fb∖Kb.Lemma 2.9 (see [2]).
(i) The operator equation(2.10)x=Kfx
has a solution in B if and only if the operator equation
(2.11)x=K1/2fK1/2x
has a solution in B.
(ii) The uniqueness of solution for these two above equations is also equivalent.
(iii) If (2.11) has a nonzero solution in B, then (2.10) has a nonzero solution in B. If (2.11) has infinitely many solutions in B, then (2.10) has also infinitely many solutions in B.Lemma 2.10 (see [9]).
Suppose that the functional(2.12)J(x)=12‖x‖2-∑k=1TF(k,K1/2x(k)),x∈B
has a critical point x∈B, where F(k,x)=∫0xf(k,s)ds, then (BVP)(1.1) has a solution in B.Lemma 2.11 (see [8]).
SupposeJ∈C1(B,ℝ) satisfies P.S. and (A) conditions, then, one has
(2.13)Mq-Mq-1+⋯+(-1)qM0⩾βq-βq-1+⋯+(-1)qβ0,q=0,1,2,…,
where βq=βq(a,b)=rankHq(fb,fa),q=0,1,2,…. Furthermore, if the left of the equation is convergent, then one has that
(2.14)∑q=0∞(-1)qMq=∑q=0∞(-1)qβq.
(A) Suppose that there are two regular values, thenJ has at most finite critical points and rank of critical groups of every critical point is finite.Lemma 2.12 (see [8, page 100, Theorem 3.2]).
LetJ∈C2(B,ℝ) be a functional and satisfy P.S. condition. Suppose that J′=I-A, where A is a compact map and p0 is a isolated critical point of J, then we have
(2.15)ind(J′,p0)=∑q=0∞(-1)qrank⁡Cq(J,p0).


## 3. Mail Results

Lemma 3.1.
Supposef satisfies one of the following conditions:(a)
(H1)limsup|x|→∞F(k,x)/x2<afor k∈ℤ(1,T), where a∈[0,2sin2π/(4T+2)),(b)
(H2)lim|x|→∞F(k,x)/x2=a for k∈ℤ(1,T), where a∈[0,2sin2(π/(4T+2)); and (H3)lim|x|→∞(F(k,x)-ax2)=-∞ for k∈ℤ(1,T), where a∈[0,2sin2(π/(4T+2)),(c)
(H2) and (H4)lim|x|→∞(f(k,x)x-2F(k,x))=+∞, uniformly for k∈ℤ(1,T), where a∈[0,2sin2(π/(4T+2)), then one has
(i)
J is coercive on B, that is J(x)→+∞ as ∥x∥→∞, (ii) J satisfies P.S. condition.Proof.
(i)
Let (a) hold. It follows from(H1) that there is a constant c>0 such that F(k,x)⩽ax2+c,x∈ℝ,k∈ℤ(1,T). Therefore,(3.1)J(x)=12‖x‖2-∑k=1TF(k,K1/2x(k))⩾12‖x‖2-(a|K1/2x|2+c)=12‖x‖2-a(Kx,x)-c⩾12‖x‖2-a4sin2((π/4T)+2)‖x‖2-c=12(1-a2sin2((π/4T)+2))‖x‖2-c⟶+∞,‖x‖⟶∞.Let (b) or (c) hold. By the condition (H2), write f(k,x)=2ax+g(k,x), F(k,x)=ax2+G(k,x). If (H3) holds, then lim|x|→∞G(k,x)=-∞; if (H4) holds, then lim|x|→∞G(k,x)/x2=0,
lim|x|→∞(g(k,x)x-2G(k,x))=+∞. It follows that for every M>0, there is RM>0 such that (3.2)g(k,x)x-2G(k,x)⩾M,|x|⩾RM.
Integrating the equality (3.3)ddxG(k,x)x2=g(k,x)x-2G(k,x)x3⩾Mx3
over the interval [x,y]⊂[RM,+∞), we have (3.4)G(k,y)y2-G(k,x)x2⩾-M2(1y2-1x2).
Letting y→+∞, we see that G(k,x)⩽-M/2, k∈ℤ(1,T), x⩾RM. In a similar way, we have G(k,x)⩽-M/2, k∈ℤ(1,T), x⩽-RM. Hence, lim|x|→∞G(k,x)=-∞.Let{xn}⊂B be such that ∥xn∥→∞ as n→∞ and J(xn)⩽C for some constant C∈ℝ. Taking yn=xn/∥xn∥, then yn,yn⇀y0, y0∈B and ∥y0∥≤1. So, K1/2yn→K1/2y0 in B. (3.5)C*‖yn‖2≥J(yn)‖yn‖2=12-1‖xn‖2∑k=1TF(k,K1/2xn(k))=12-1‖xn‖2∑k=1T(a(K1/2xn(k))2+G(k,K1/2xn(k)))=12-a(K1/2xn‖xn‖)2-1‖xn‖2∑k=1TG(k,K1/2xn(k))=12-2sin2π4T+2(K1/2yn)2-1‖xn‖2∑k=1TG(k,K1/2xn(k))≥12-2sin2π4T+2(K1/2yn)2-C‖xn‖2.
Then, (3.6)(Ky0,y0)=∑k=1T(K1/2y0(k))2=lim⁡n⟶∞∑k=1T(K1/2yn(k))2=lim⁡n⟶∞(K1/2yn)2≥14sin2((π/4T)+2).
Hence |K1/2yn(k)|≠0 and |K1/2xn(k)|=|K1/2(∥xn∥yn(k))|=∥xn∥|K1/2yn(k)|→∞. And G(k,K1/2xn(k))→-∞, as n→∞ for k∈ℤ(1,T). Therefore, (3.7)C*≥J(xn)=12‖xn‖2-a(K1/2xn)2-∑k=1TG(k,K1/2xn(k))≥12(‖xn‖2-a(Kxn,xn))-∑k=1TG(k,K1/2xn(k))≥-∑k=1TG(k,K1/2xn(k))⟶+∞,n⟶∞.
This is impossible, so J is coercive on B.(ii)
Let{xn}∈B,{J(xn)} is bounded and J′(xn)→θ,n→∞. By (i), {xn} is bounded on B. Clearly, {xn} possesses a convergent subsequence. Then, the P.S. condition is satisfied.Lemma 3.2.
Suppose that(H5) there exist δ,a,b∈(0,+∞) and an integer m∈ℤ(1,T), which satisfy(3.8)2sin2(2m-1)π4T+1⩽b⩽a⩽2sin2(2m+1)π4T+1,
such that by2≤F(k,y)≤ay2,|y|≤δ,k∈ℤ(1,T).
Then the functionalJ has a local linking with respect to B=V1⊕V2, where V1=span{e1,e2,…,em},V2=V1⊥.Proof.
LetV1=span{e1,e2,…,em},V2=V1⊥. If x∈V1, then (Kx,x)≥λm∥x∥2; if x∈V2, then (Kx,x)≤λm+1∥x∥2.
By Remark2.1, we can assume that for the given δ>0, there exists a ρ=δ/L>0 such that x∈V1, ∥x∥≤ρ⇒|K1/2x|≤L∥x∥≤δ, and thus, by (H5), we have(3.9)J(x)=12‖x‖2-∑k=1TF(K1/2x(k))≤12‖x‖2-∑k=1Tb(K1/2x(k))2=12‖x‖2-b(K1/2x,K1/2x)=12‖x‖2-b(Kx,x)≤(12-b4sin2((2m-1)(π/4T)+1))‖x‖2≤0,x∈V1,‖x‖≤ρ.
For x∈V2, consider the above ρ. By Remark 2.1, we still have ∥x∥≤ρ⇒|K1/2x|≤L∥x∥≤δ; thus by (H5), we have
(3.10)J(x)=12‖x‖2-∑k=1TF(K1/2x(k))≥12‖x‖2-∑k=1TA(K1/2x)2=12‖x‖2-a(K1/2x,K1/2x)=12‖x‖2-a(Kx,x)≥(12-a4sin2((2m+1)(π/4T)+1))‖x‖2≥0,x∈V2,‖x‖≤ρ.
This implies that J has a local linking at 0 with respect to B=V1⊕V2.Theorem 3.3.
If condition (H1) holds, then the BVP(1.1) has at least a solution.Proof.
We will verify that the functionalJ(x) defined in Lemma 2.10 has a critical point x∈B. By [6], we know that J:B→ℝ is C1 functional. And by Lemma 3.1, it is easy to know that J is bounded below and satisfies P.S. condition. It follows from Lemma 2.4 that J has a critical point in x∈B.Theorem 3.4.
If conditions (H2) and (H3) hold, then the BVP(1.1) has at least a solution.Theorem 3.5.
If conditions (H2) and (H4) hold, then the BVP(1.1) has at least a solution. The proof of Theorem 3.4 and Theorem 3.5 is similar to the proof of Theorem 3.3.Theorem 3.6.
If conditions (H1) and (H5) hold, then the BVP(1.1) has at least two nontrivial solutions.Proof.
By Lemma3.1, we know that J satisfies P.S. condition, and it follows from Lemmas 2.7 and 3.2 that x=θ is a homological nontrivial critical point of J, then Ck(J,0)≇0. If infx∈BJ(x)≥0, then J(x)=infx∈V1J(x)=0, x∈V1, ∥x∥<ρ, which implies that all x∈V1 with ∥x∥<ρ are solutions of the BVP(1.1). If infx∈BJ(x)<0, that is, 0 is not a minimizer of J, by Lemma 3.1, it is easy to know that J is bounded below and satisfies P.S. condition. Then by Lemma 2.4, J has a minimizer x0→B, that is, a critical point and x0≠0; Without loss of generality, we may suppose that the minimizer x0 is unique. Let c=J(x0), then Morse index of x0 is
(3.11)Cq(J,x0)=Hq(Jc∩Ux0,(Jc∖{x0})∩Ux0)={G,q=0,0,q≠0.
We suppose,J only have two critical points with θ and x0, where c1=c, c2=J(θ), c1<c2. Let b>max{c1,c2}, ε>0 is small enough such that c1+ε<c2-ε,c2+ε<b.
Ifχ(X,Y)=∑q=0∞(-1)qHq(X,Y) is finite, then χ(X,Y) satisfies additivity. So,
(3.12)χ(Jb,Jc1-ε)=χ(Jb,Jc2-ε)+χ(Jc2-ε,Jc1-ε).
It follows from the second deformation lemma, exactness, and excision that Hq(Jb,Jc2-ε)≅Hq(Jc2+ε,Jc2-ε)≅Hq(Jc2+ε,Jc2∖{θ})≅Hq(Jc2,Jc2∖{θ})≅Hq(Jc2∩Uθ,(Jc2∖{θ})∩Uθ). Thus,
(3.13)Hq(Jb,Jc2-ε)=Hq(Jc2∩Uθ,(Jc2∖{θ})∩Uθ)=Cq(J,θ)={G,q=m,0,q≠m.
Hence, χ(Jb,Jc2-ε)=(-1)m.
By the same way,(3.14)Hq(Jc2+ε,Jc1-ε)≅Hq(Jc1+ε,Jc1-ε)≅Hq(Jc1,Jc1-ε)≅Hq(Jc1,Jc1∖{x0})≅Cq(J,x0)={G,q=0,0,q≠0.
So, χ(Jc2+ε,Jc1-ε)=1. But
(3.15)χ(Jb,Jc1-ε)=∑q=0∞(-1)qHq(Jb,Jc1-ε)=∑q=0∞(-1)qHq(Jb)=1.
Then, (-1)m+1≠1 is a contradict with (3.12).
Thus,J has at least three critical points, that is, the BVP (1.1) has at least two nontrivial solutions.Theorem 3.7.
If condition (H2), (H3), and (H5) hold, then the BVP(1.1) has at least two nontrivial solutions.Theorem 3.8.
If condition (H2), (H4), and (H5) hold, then the BVP(1.1) has at least two nontrivial solutions.The proof of Theorems 3.5, and 3.6 is similar to the proof of Theorem3.6.Theorem 3.9.
Suppose that(H6) there exists m∈ℤ(1,T), such that(3.16)4sin2(2m-1)π4T+1<fx′(k,0)<4sin2(2m+1)π4T+1,k∈Z(1,T),(H7) there exist μ∈(0,1/2) and R>0 such that 0<F(k,x)=∫0xf(k,s)ds⩽μxf(k,x) for all k∈ℤ(1,T) and x∈ℝ with |x|⩾R.Then the BVP(1.1) has at least one nontrivial solution.Proof.
Obviously,dJ(θ)=θ and d2J(θ)=I-λK1/2fx′(θ)K1/2:=I-Ψ, where Ψ=K1/2fx′(θ)K1/2. It follows from (H6) that there exists a sufficiently small ε0>0 such that
(3.17)a=4sin2(2m-1)π4T+1+ε0⩽fx′(k,0)⩽4sin2(2m+1)π4T+1-ε0=b,k∈Z(1,T).
Since (Ψy,y)=(fx′(·,0)K1/2y,K1/2y) for all y∈B, a(Ky,y)⩽(Ψy,y)⩽b(Ky,y) for y∈B. So, for every m∈ℤ(1,T),
(3.18)ainf⁡Em-1sup⁡y∈Em-1⊥,‖y‖=1(Ky,y)⩽inf⁡Em-1sup⁡y∈Em-1⊥,‖y‖=1(Ψy,y)⩽binf⁡Em-1sup⁡y∈Em-1⊥,‖y‖=1(Ky,y),
where Em-1 is (m-1)-dimension subspace of B. Therefore, 4asin2((2m-1)π/(4T+1)μk⩽4bsin2((2m-1)(π/4T)+1), where m-th eigenvalues of Ψ are μm. If there exists m0 such that μm0=1, then 4asin2((2m0-1)(π/4T)+1)⩽1⩽4bsin2((2m0-1)(π/4T)+1). Thus, 1/4sin2((2m-1)π/(4T+1))+ε0⩽1/4asin2((2m0-1)π/(4T+1))⩽1/4sin2((2m+1)π/(4T+1))-ε0. It is not true. So, 1 is not eigenvalues of Ψ. Thus θ is a nondegenerate critical point of J and Morse index number is n. Then, Cq(J,θ)=δqnG.
Asumme thatJ is no other critical points, then Morse index number of (H,Ja) is(3.19)Mn=1,Mq=0,q≠n,∑q=0∞(-1)qMq=(-1)n.
It follows from (H7) that there exists C1,C2>0 such that F(k,x)⩾C1|x|1/μ-C2, (k,x)∈ℤ(1,T)×ℝ1. So,(3.20)J(τx)=12τ2‖x‖2-∑t=1TF(k,τK1/2x(k))⩽12τ2‖x‖2-C1τ1/μ‖K1/2x‖1/μ1/μ+C2,x∈B.
This implies limτ→+∞J(τx)=-∞ for all x∈S∞. So, there exists γ(x)>0 such that J(γ(x)x)=a for all x∈S∞. Then, we will prove that there exists A>0 such that a<-A, and if J(τx)⩽a, then d/dτ[J(τx)]<0.
Indeed, letA=C3, then for all x∈S∞,(3.21)ddτJ(τx)=(J′(τx),x)=τ-∑t=1TK1/2x(k)f(k,τK1/2x(k))⩽τ-1τμ∑t=1TF(k,τK1/2x(k))+C3τμ⩽1τμ(12τ2-∑t=1TF(k,τK1/2x(k)))+C3τμ=1τμ(J(τx)+C3)<0.
So, γ(x) satisfies J(γ(x)x)=a and for all x∈S∞ is unique. It follows from implicit function theorem that we have γ∈C(S∞,ℝ+1), where ℝ+1=[0,+∞).
At last,J(θ)=0 and there exists ε0>0 such that Bε0∩Ja=∅ and γ(x)⩾ε0 for all x∈S∞. We define a deformation retract η:[0,1]×(B∖Bε0)→B∖Bε0 by(3.22)η(s,x)={(1-s)x+sγ(x‖x‖)x‖x‖,x∈(B∖Bε0)∖Ja,x,x∈Ja.
This proves that B∖Bε0≃Ja. The,n S∞≃Ja. Therefore, Hq(H,Ja)≃Hq(H,S∞)≃0. Since S∞ is contractible, we have Betti number βq=0,q⩾0. Thus,
(3.23)∑q=0∞(-1)qβq=0.
Then by (3.19) and (3.23), we have ∑q=0∞(-1)qMq≠∑q=0∞(-1)qβq. Thus, according to the Morse inequality, J has at least one nontrivial critical point.Theorem 3.10.
Suppose that condition (H7) is satisfied and that (H8)f(k,x) is odd in x, that is, f(k,x)=-f(k,-x) for all (k,x)∈[0,1]×ℝ1,.Then the BVP(1.1) has infinitely many solutions.Proof.
First, it follows from (H7) that J satisfies P.S. condition in B [6].
It follows from (H8) that J is even. Asumme that J has only finitely many critical points {x1,-x1,x2,-x2,…,xm,-xm}. We choose two constant numbers a<0 and b>0, such that they satisfy a<min{J(x1),J(x2),…,J(xm),-C4} and b>min{J(x1),J(x2),…,J(xm)}.
On the one hand, it follows from Theorem3.9 that B∖Bε0≃Ja. Then, S∞≃Ja. Since S∞ is contractible, we can deduce that Betti number βq=0,q⩾0.
Therefore,(3.24)∑q=0∞(-1)qβq=0.
On the other hand, we choose enough small r>0 such that B(θ,r), B(yi,r), B(-yi,r), i=1,2,…,m is mutually disjoint. So,
(3.25)Mq=Mq(a,b)=∑i=1m[rank⁡Cq(J,yi)+rank⁡Cq(J,-yi)]+rank⁡Cq(J,θ),q⩾0.
It follows from Borsuk theorem that
(3.26)∑q=0∞(-1)qMq=∑i=1m[∑q=0∞(-1)q(ṙankCq(J,yi)+rank⁡Cq(J,-yi))]+∑q=0∞(-1)qrank⁡Cq(J,θ)=∑i=1m[deg⁡(I-A,B(yi,r),θ)+deg⁡(I-A,B(-yi,r),θ)]+deg⁡(I-A,B(θ,r),θ)=deg⁡(I-A,⋃i=1m(B(yi,r)∪B(-yi,r))∪B(θ,r),θ),
that is
(3.27)∑q=0∞(-1)qMq=anoddnumber.
By (3.24) and (3.27), we obtain that
(3.28)∑q=0∞(-1)qMq≠∑q=0∞(-1)qβq.
It is a contradiction. Thus, J has infinitely many critical points in B.Example 3.11.
Consider the (BVP)(3.29)-Δ2x(k-1)=cos⁡x1+sinx,k∈Z(1,T),x(0)=0=Δx(T),
Here, f(k,x)=cosx/(1+sinx),F(k,x)=ln(1+sinx). By simple calculation, we have lim|x|→∞(F(k,x)/x2)<2(sin2)(π/(4T+2)). Thus, the condition (H1) holds. Then by Theroem 3.3., the (BVP) has at least one solution.Example 3.12.
Consider the (BVP) of Example3.11, here
(3.30)f(k,x)={4xsin22π4T+1,-1<x<1,k∈Z(1,T);xsin2π4T+k+2+132,x⩾1,k∈Z(1,T);xsin2π4T+k+2-132,x⩽-1,k∈Z(1,T),F(k,x)={2x2sin22π4T+1,-1<x<1,k∈Z(1,T);2x2sin2π4T+k+2+132x-66,x⩾1,k∈Z(1,T);2x2sin2π4T+k+2-132x-66,x⩽-1,k∈Z(1,T).By simple calculation, we have(3.31)lim⁡|x|⟶∞⁡sup⁡F(k,x)x2=2sin2π4T+k+2<2sin2π4T+2,k∈Z(1,T),
and there exists m=1 and δ=1>0 such that (3.32)2sin22π-14T+1x2⩽F(k,x)⩽2sin22π+14T+1x2,|x|⩽1.
Let a=2sin2((2π+1)/(4T+1)), b=2sin2((2π-1)/(4T+1))  then a,b satisfy (3.33)2sin2π4T+1⩽b⩽a⩽2sin23π4T+1.
Thus, the conditions (H1) and (H5) hold. Hence, by Theorem 3.6, the (BVP) has at least two nontrivial solutions.Example 3.13.
Consider the (BVP) of Example3.11, where
(3.34)f(k,x)=ln⁡(1+x)1+x+10x5(k,x)∈Z(1,T)×R1.It is obvious thatf(k,x)∈C1(ℤ(1,T)×ℝ1,ℝ1) is odd with respect to x. On the other hand,(3.35)F(k,x)=∫0xf(k,s)ds=12ln⁡2(1+x)+53x6,k∈Z(1,T),
and we have (3.36)lim⁡|x|⟶∞F(k,x)xf(k,x)=16.
It is easy to prove that condition (H7) holds. Hence, by Theorem 3.10, the (BVP) has infinitely many solutions.



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*Source: 319829-2011-01-26.xml*