# Existence of Solutions to Boundary Value Problems for the Discrete Generalized Emden-Fowler Equation

Authors: Tieshan He; Fengjian Yang
Journal: Discrete Dynamics in Nature and Society
 (2009)

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## Abstract

We consider the existence of solutions to boundary value problems for the discrete generalized Emden-Fowler equation. By means of the minimax methods in the critical point theory, some new results are obtained. Two examples are also given to illustrate the main results.


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## Body

## 1. Introduction and Statements of Main Results

LetZ and R be the sets of all integers and real numbers, respectively. For a,b∈Z, define Z[a]={a,a+1,…},  Z[a,b]={a,a+1,…,b} when a≤b.In this paper, we consider the following boundary value problem (BVP for short) consisting of the discrete generalized Emden-Fowler equation:(1.1)Δ[p(t)Δu(t-1)]+q(t)u(t)=f(t,u(t)),t∈Z[1,k],
and the boundary value conditions:
(1.2)u(0)+αu(1)=A,u(k+1)+βu(k)=B,
where k is a positive integer, α, β, A, and B are constants, and Δu(t)=u(t+1)-u(t) is the forward difference operator. We assume that p(t) is nonzero and realvalued for each t∈Z[1,k+1]. We also assume that q(t) is realvalued for each t∈Z[1,k],  f(t,x) is realvalued for each (t,x)∈Z[1,k]×R, and f(t,x) is continuous in the second variable x.Equation (1.1) has been extensively studied by many authors; for example, see [1–9] concerning its disconjugacy, disfocality, oscillation, asymptotic behaviour, existence of periodic solutions, and solutions to boundary value problem.Recently, Yu and Guo in [10] employed the critical point theory to obtain the existence of solutions to the BVP (1.1)-(1.2). Motivated by this and the results in [11], the main purpose of this paper is to give some new sufficient conditions for the existence of solutions to the BVP (1.1)-(1.2) by applying the Saddle Point Theorem and the Least Action Principle.Before giving the main results, we first set(1.3)F(t,x)=∫0xf(t,s)ds,c(t)=q(t)-p(t)-p(t+1),M=(c(1)-αp(1)p(2)0⋯00p(2)c(2)p(3)⋯000p(3)c(3)⋯00⋯⋯⋯⋯⋯⋯000⋯c(k-1)p(k)000⋯p(k)c(k)-βp(k+1)),η=(η(1),η(2),…,η(k))τ=(p(1)A0⋮0p(k+1)B).
The main results are as follows.Theorem 1.1.
Suppose thatf, M satisfy the following assumptions.(F1)
There are constantsC1>0,  C2>0,  1/2≤θ<1 such that for all (t,x)∈Z[1,k]×R,
(1.4)|f(t,x)|≤C1|x|θ+C2.(F2)
One has
(i)
either∥u∥-2θ∑t=1kF(t,u(t))→-∞  as ∥u∥→∞, or(ii)
∥u∥-2θ∑t=1kF(t,u(t))→+∞  as ∥u∥→∞,where for all u=(u(1),u(2),…,u(k))τ∈Rk, ∥u∥=(∑t=1k|u(t)|2)1/2. (P1)
The matrixM is singular.Then the BVP (1.1)-(1.2) has at least one solution.Remark 1.2.
There  are functionsp(t), q(t), and f(t,x) satisfying our Theorem 1.1 and not satisfying the results in [10]. For example, let A, B be arbitrary constants, k=4,  α=β=-1,  θ=1/2 and
(1.5)p(t)≡1,q(t)≡0,f(t,x)=g(t)x1+x24+1,
where g(t)<0 for every t∈Z[1,4]. It is easy to verify that (F1), (F2)(i), and (P1) are satisfied. Then the BVP (1.1)-(1.2) has at least one solution. And it is easy to see that this solution is a nonzero solution since f(t,0)≠0.Theorem 1.3.
Suppose thatf, M satisfy the following assumptions.(F3)
For anyt∈Z[1,k],  lim|x|→∞(f(t,x)/x)=0.(P2)
The matrixM is nonsingular.Then the BVP (1.1)-(1.2) has at least one solution.Corollary 1.4.
Suppose thatf, M satisfy (F1) and (P2). Then the BVP (1.1)-(1.2) has at least one solution.Remark 1.5.
Since  for all(t,x)∈Z[1,k]×R,  |f(t,x)|≤C2 implies |f(t,x)|≤C1|x|θ+C2, our Corollary 1.4 extends  Theorem 3.2 in [10].Theorem 1.6.
Suppose thatf, M satisfy the following assumptions.(F4)
For anyt∈Z[1,k],  lim|x|→∞(F(t,x)/x2)=0.(F5)
One has
(i)
either2F(t,x)-(f(t,x)+η(t))x→-∞  as |x|→∞ for all t∈Z[1,k], or(ii)
2F(t,x)-(f(t,x)+η(t))x→+∞  as |x|→∞ for all t∈Z[1,k].(P3)
The matrixM is singular and indefinite.Then the BVP (1.1)-(1.2) has at least one solution.Remark 1.7.
There are functionsp(t),  q(t), and f(t,x) satisfying our Theorem 1.6 and not satisfying the results in [10]. For example, let A, B be arbitrary constants, k=4, α=1, β=0 and
(1.6)p(1)=p(3)=-1,p(2)=p(4)=1,p(5)=-2,q(t)≡0,f(t,x)=g(t)(x1+x2ln(1+x2)+1+x22x1+x2+1),
where g(t)<0 for every t∈Z[1,4]. It is easy to verify that (F4), (F5)(i) and (P3) are satisfied. Then the BVP (1.1)-(1.2) has at least one solution. And it is easy to see that this solution is a nonzero solution since f(t,0)≠0.Corollary 1.8.
Assume that (F4) holds. If one of the following conditions is satisfied: (H1) the matrix M is negative semi-definite, and (F5)(i) holds,  (H2) the matrix M is positive semi-definite, and (F5)(ii) holds, then, the BVP (1.1)-(1.2) has at least one solution.


## 2. Variational Structure and Two Basic Lemmas

LetRk be the real Euclidean space with dimension k. For any u,v∈Rk, ∥u∥ and (u,v) denote the usual norm and inner product in Rk, respectively.Define the functionalJ on Rk as follows:
(2.1)J(u)=12(Mu,u)+(η,u)-∑t=1kF(t,u(t)),∀u=(u(1),u(2),…,u(k))τ∈Rk.
It is well known that u=(u(1),u(2),…,u(k))τ∈Rk is a critical point of J if and only if {u(t)}t=0k+1=(u(0),u(1),u(2),…,u(k),u(k+1))τ is a solution of the BVP (1.1)-(1.2), where u(0)=A-αu(1) and u(k+1)=B-βu(k). For details, see [10]. It follows from the continuity of f that J is continuously differentiable on Rk. Moreover, one has
(2.2)(J′(u),h)=(Mu,h)+(η,h)-∑t=1kf(t,u(t))·h(t),∀u,h∈Rk.When the matrixM is singular and indefinite, we suppose that 0<λ1≤λ2≤⋯≤λl and 0>-μ1≥-μ2≥⋯≥-μs are the positive and negative eigenvalues of M, respectively, and l+s=k-1. We also suppose that ςi, 1≤i≤l and ξj, 1≤j≤s are the eigenvectors of M corresponding to eigenvalues λi, 1≤i≤l and -μj, 1≤j≤s satisfying
(2.3)(ςi1,ςi2)={0,i1≠i2,1,i1=i2,(ςi,ξj)=0,(ξj1,ξj2)={0,j1≠j2,1,j1=j2,
where i, i1, i2∈Z[1,l],  j, j1, j2∈Z[1,s]. We denote
(2.4)R+=span{ςi∣i∈Z[1,l]},R-=span{ξj∣j∈Z[1,s]},(2.5)R0=(R+⊕R-)⊥.
Then Rk has the direct sum decomposition Rk=R+⊕R0⊕R-. So, for each u∈Rk, u can be expressed by
(2.6)u=u++u0+u-,
where u*∈R*, * is +,0, or -, respectively. Furthermore, we have the following estimates:
(2.7)λ1∥u+∥2≤(Mu+,u+)≤λl∥u+∥2,u+∈R+,-μs∥u-∥2≤(Mu-,u-)≤-μ1∥u-∥2,u-∈R-.Set(2.8)∥u∥∞=maxt∈Z[1,k]|u(t)|,∥u∥p=(∑t=1k|u(t)|p)1/p,
where u=(u(1),u(2),…,u(k))τ∈Rk and ∞>p≥1. Then for any u∈Rk,
(2.9)1k∥u∥≤∥u∥∞≤∥u∥,1k∥u∥≤∥u∥p≤k1/p∥u∥.We will make use of the least action principle and saddle point theorem to obtain the critical points ofJ. Let us first recall these theorems.Lemma 2.1 (the least action principle, see [12]).
LetX be a real Banach space, and assume that J∈C1(X,R) is bounded from below in X and satisfies the Palais-Smale condition ((PS) condition for short). Then c=infu∈XJ(u) is a critical value of J.Lemma 2.2 (saddle point theorem, see [13]).
LetX be a real Hilbert space, X=X1⊕X2, where X1≠{0} and is finite dimensional. Suppose that J∈C1(X,R) satisfies the (PS) condition and(φ1)
there exist constantsσ,ρ>0 such that J∣∂Bρ∩X1≤σ, where Bρ={u∈X∣∥u∥<ρ},  ∂Bρ denotes the boundary of Bρ;(φ2)
there existe∈Bρ∩X1 and a constant ω>σ such that J∣e+X2≥ω.Then J possesses a critical value c≥ω and
(2.10)c=infh∈Φmaxu∈Bρ∩X1J(h(u)),
where Φ={h∈C(B¯ρ∩X1,X)∣h∣∂Bρ∩X1=id}.Remark 2.3.
As shown in [14], a deformation lemma can be proved with the weaker condition (C) replacing the usual (PS) condition, and it turns out that the saddle point theorem holds under condition (C).


## 3. Proofs of the Main Results

In order to prove Theorem1.1, we need to prove the following lemma.Lemma 3.1.
Assume that conditions (F1), (F2), and (P3) hold. Then the functional J (see (2.1)) satisfies the (PS) condition; that is, for any sequence {um} such that  J(um) is bounded and J′(um)→0 as m→∞, there exists a subsequence of {um} which is convergent in Rk.Proof.
First suppose that (F1), (F2)(i), and (P3) hold. Recall that Rk is a finite dimensional Hilbert space. Consequently, in order to prove that  J satisfies the (PS) condition, we only need to prove that {um} is bounded. Let {um} be a sequence in Rk such that J(um) is bounded and J′(um)→0 as m→∞. Then there exist C3>0 and m0∈Z[1] such that
(3.1)|J(um)|≤C3,|(J′(um),h)|≤∥h∥
for all m>m0,  h∈Rk.
SinceM is singular and indefinite, we write um=um++um0+um- with um*∈R*, where *=+,0,-, respectively. By (F1), (2.9), and Hölder’s inequality (p=1/θ,q=1/(1-θ)), we have
(3.2)|∑t=1kf(t,um(t))·(um-(t)-um+(t))|≤∑t=1k(C1|um0(t)+(um+(t)+um-(t))|θ+C2)·|um-(t)-um+(t)|≤2C1∥um0∥∞θ∑t=1k|um-(t)-um+(t)|+2C1∑t=1k|um+(t)+um-(t)|θ|um-(t)-um+(t)|+C2∑t=1k|um-(t)-um+(t)|≤2C12kτ1∥um0∥∞2θ+τ12∥um--um+∥2+2C1∥um++um-∥1θ∥um--um+∥q+C2k∥um--um+∥≤2C12kτ1∥um0∥2θ+τ12∥um--um+∥2+2C1k∥um++um-∥θ∥um--um+∥+C2k∥um--um+∥,
where τ1=min{λ1,μ1}. On the other hand, by the fact that um+ and um- are mutually orthogonal, one has ∥um--um+∥=∥um++um-∥. Hence we have
(3.3)|∑t=1kf(t,um(t))·(um-(t)-um+(t))|≤2C12kτ1∥um0∥2θ+τ12∥um++um-∥2+2C1k∥um++um-∥θ+1+C2k∥um++um-∥.
Similarly to (3.3), we have
(3.4)|∑t=1k[F(t,um(t))-F(t,um0(t))]|≤2C12kμs∥um0∥2θ+μs2∥um++um-∥2+2C1k∥um++um-∥θ+1+C2k∥um++um-∥.
Takeh=um--um+ in (2.2). Then
(3.5)-(Mum,um--um+)=-(J′(um),um--um+)+(η,um--um+)-∑t=1kf(t,um(t))(um-(t)-um+(t)).
We know
(3.6)-(Mum,um--um+)≥λ1∥um+∥2+μ1∥um-∥2≥τ1∥um++um-∥2.
Thus, by (3.1) and (3.3), we have
(3.7)τ1∥um++um-∥2≤(1+∥η∥)∥um++um-∥+2C12kτ1∥um0∥2θ+τ12∥um++um-∥2+2C1k∥um++um-∥θ+1+C2k∥um++um-∥,
that is,
(3.8)τ12∥um++um-∥2-2C1k∥um++um-∥θ+1-(1+∥η∥+C2k)∥um++um-∥≤2C12kτ1∥um0∥2θ.
It follows from (3.8) and 1/2≤θ<1 that
(3.9)∥um++um-∥≤C4∥um0∥θ+C5
for all m>m0 and some positive constants C4,C5.
By (3.1), (2.9), (3.4), and (3.9), we have
(3.10)C3≥J(um)=12(Mum,um)+(η,um)-∑t=1kF(t,um(t))≥12(λ1∥um+∥2-μs∥um-∥2)-∥η∥(∥um0∥+∥um++um-∥)-∑t=1k(F(t,um(t))-F(t,um0(t)))-∑t=1kF(t,um0(t))≥-μs2∥um++um-∥2-∥η∥∥um0∥-2C12kμs∥um0∥2θ-μs2∥um++um-∥2-2C1k∥um++um-∥θ+1-(∥η∥+C2k)∥um++um-∥-∑t=1kF(t,um0(t))≥-μs(C42∥um0∥2θ+2C4C5∥um0∥θ+C52)-∥η∥∥um0∥-2C12kμs∥um0∥2θ-2C1k(C4∥um0∥θ+C5)θ+1-(∥η∥+C2k)(C4∥um0∥θ+C5)-∑t=1kF(t,um0(t))=-(μsC42+2C12kμs)∥um0∥2θ-∥η∥∥um0∥-2C1k(C4∥um0∥θ+C5)θ+1-2μsC4C5∥um0∥θ-(∥η∥+C2k)(C4∥um0∥θ+C5)-μsC52-∑t=1kF(t,um0(t)).
Since 1/2≤θ<1, we deduce
(3.11)-C3≤∥um0∥2θ(∥um0∥-2θ∑t=1TF(t,um0(t))+C6)+C7
for all m>m0 and some positive constants C6,C7. The above inequality and (F2)(i) imply that {um0(t)} is bounded. Then it follows from (3.9) that {um++um-} is bounded. Thus we conclude that {um} is bounded, and the (PS) condition is verified.
Now, suppose that (F1), (F2)(ii), and (P3) hold. By a similar argument as above, we know also that J satisfies the (PS) condition. The proof is complete.Proof of Theorem1.1.
Assume that (F1), (F2)(i), and (P1) hold. The proof for the case when (F1), (F2)(ii), and (P1) hold is similar and will be omitted here. Since p(t) is nonzero for each t∈Z, the singular symmetric matrix M has at least one nonzero eigenvalue and we will give the proof in three cases. (i)
Suppose that the matrixM is singular and indefinite. Then Rk has the direct sum decomposition: Rk=R+⊕R0⊕R-. In view of Lemma 3.1, we only to check that conditions (φ1) and (φ2) in the saddle point theorem hold. To this end, let X1=R+⊕R0, X2=R-. For any u=u++u0∈X1, by (F1), (2.9), and the mean value theorem, we have
(3.12)J(u)≥λ12∥u+∥2-∥η∥(∥u0∥+∥u+∥)-∑t=1kf(t,u0(t)+θ1(t)u+(t))·u+(t)-∑t=1kF(t,u0(t))≥λ12∥u+∥2-∥η∥(∥u0∥+∥u+∥)-2C1k∥u0∥θ∥u+∥-2C1k∥u+∥θ+1-C2k∥u+∥-∑t=1kF(t,u0(t))≥λ12∥u+∥2-∥η∥(∥u0∥+∥u+∥)-4C12kλ1∥u0∥2θ-λ14∥u+∥2-2C1k∥u+∥θ+1-C2k∥u+∥-∑t=1kF(t,u0(t))=λ14∥u+∥2-2C1k∥u+∥θ+1-(∥η∥+C2k)∥u+∥-∥u0∥2θ(∥u0∥-2θ∑t=1kF(t,u0(t))+4C12kλ1+∥η∥∥u0∥-2θ+1),
where 0<θ1(t)<1, t=1,2,…,k. Since 1/2≤θ<1 and (F2)(i), we have
(3.13)J(u)→+∞as∥u∥→∞inX1.
On the other hand, for anyu=u-∈X2, by (F1), (2.9), and the mean value theorem, we have
(3.14)J(u)≤-12μ1∥u∥2+∥η∥∥u∥-∑t=1kf(t,θ2(t)u(t))·u(t)-∑t=1kF(t,0)≤-12μ1∥u∥2+∥η∥∥u∥+∑t=1kC1|u(t)|θ+1+C2k∥u∥≤-12μ1∥u∥2+C1k∥u∥θ+1+(∥η∥+C2k)∥u∥,
where 0<θ2(t)<1,  t=1,2,…,k. Since 1/2≤θ<1, we can obtain
(3.15)J(u)→-∞as∥u∥→∞inX2.
Lete=0, then it follows from (3.13) and (3.15) that (φ1) and (φ2) are satisfied. By the saddle point theorem, J has at least one critical point.
(ii)
Suppose that0<λ1≤λ2≤⋯≤λl are the positive eigenvalues of M and l=k-1. Then Rk has the direct sum decomposition: Rk=R+⊕R0, where R+ and R0 are defined as in (2.4) and (2.5), respectively. By a similar argument as in the proof of Lemma 3.1, we see that J satisfies the (PS) condition. By (3.13), J is bounded from below. Then, by the least action principle, c1=infu∈RkJ(u) is a critical value of J.
(iii)
Suppose that0>-μ1≥-μ2≥⋯≥-μs are the negative eigenvalues of M and s=k-1. Then Rk has the direct sum decomposition: Rk=R0⊕R-. Following almost the same procedure as the proof of Lemma 3.1, we know also that J satisfies the (PS) condition in this case.
For anyu=u-∈R-, (3.15) holds. For any u=u0∈R0, since Mu=0, we have
(3.16)J(u)≥-∥η∥∥u0∥-∑t=1kF(t,u0(t))=-∥u0∥2θ(∥u0∥-2θ∑t=1kF(t,u0(t))+∥η∥∥u0∥-2θ+1).
Due to (F2)(i) and 1/2≤θ<1,
(3.17)J(u)→+∞as∥u∥→∞inR0.
By the saddle point theorem, there exists at least one critical point of J.
SinceJ has at least one critical point in all three cases, the BVP (1.1)-(1.2) has at least one solution.Proof of Theorem1.3.
Since the matrixM is nonsingular, we will give the proof in three cases.(i)
Suppose that0<λ1≤λ2≤⋯≤λl and 0>-μ1≥-μ2≥⋯≥-μs are the positive and negative eigenvalues of M, respectively, and l+s=k. Then Rk has the direct sum decomposition: Rk=R+⊕R-. Denote τ1=min{λ1,μ1}. It follows from (F3) that there exists a positive constant a1 such that
(3.18)|f(t,x)|≤14τ1|x|+a1
for any (t,x)∈Z[1,k]×R.
We now prove that the functionalJ satisfies the (PS) condition. Let {um} be a sequence in Rk such that J(um) is bounded and J′(um)→0 as m→∞. Write um=um++um-, where um+∈R+,  um-∈R-. Similarly to (3.8), we have, by (3.18),
(3.19)3τ14∥um++um-∥2≤(1+∥η∥+a1k)∥um++um-∥.
Thus {um} is bounded, and the (PS) condition is verified.
For anyu=u+∈R+, by (3.18) and the mean value theorem, we have
(3.20)J(u)≥12λ1∥u∥2-∥η∥∥u∥-∑t=1kf(t,θ3(t)u(t))·u(t)-∑t=1kF(t,0)≥(λ12-τ14)∥u∥2-∥η∥∥u∥-a1k∥u∥,
where 0<θ3(t)<1,  t=1,2,…,k. For any u=u-∈R-, Similarly to (3.20), we have
(3.21)J(u)≤-(μ12-τ14)∥u∥2+∥η∥∥u∥+a1k∥u∥.
Lete=0, then it follows from (3.20) and (3.21) that (φ1) and (φ2) are satisfied. By the saddle point theorem, J has at least one critical point.
(ii)
Suppose that0<λ1≤λ2≤⋯≤λl are the positive eigenvalues of M and l=k. Then E has the direct sum decomposition: Rk=R+. It follows from (3.20) that J satisfies the (PS) condition and is bounded from below. Then, by the least action principle, c2=infu∈RkJ(u) is a critical value of J.
(iii)
Suppose that0>-μ1≥-μ2≥⋯≥-μs are the negative eigenvalues of M and s=k. Then Rk has the direct sum decomposition: Rk=R-. It follows from (3.21) that J satisfies the (PS) condition and is bounded from above. Then, by the least action principle, c3=-infu∈RkJ(u) is a critical value of -J.
SinceJ has a critical point in all three cases, the BVP (1.1)-(1.2) has at least one solution.Proof of Corollary1.4.
This is immediate from Theorem1.3.The following lemma is useful for proving Theorem1.6 and Corollary 1.8.Lemma 3.2.
Under the condition (F5), the functional J satisfies condition (C); that is, for any sequence {um} such that J(um) is bounded and ∥J′(um)∥(1+∥um∥)→0 as m→∞, there exists a subsequence of {um} which is convergent in Rk.Proof.
First suppose that (F5)(i) holds. Let {um} be a sequence in Rk such that J(um) is bounded and ∥J′(um)∥(1+∥um∥)→0 as m→∞. Then there exists a constant L1>0 such that
(3.22)|J(um)|≤L1,∥J′(um)∥(1+∥um∥)≤L1
for all m∈Z[1]. Hence, we have
(3.23)-3L1≤-∥J′(um)∥·(1+∥um∥)-2J(um)≤(J′(um),um)-2J(um)=∑t=1k[2F(t,um(t))-(f(t,um(t))+η(t))um(t)].
Then, {um} is bounded. In fact, if {um} is unbounded, there exist a subsequence of {um} (still denoted by {um}) and t0∈Z[1,k] such that  |um(t0)|→∞asm→∞. By (F5)(i), we have
(3.24)2F(t0,um(t0))-(f(t0,um(t0))+η(t0))um(t0)→-∞asm→∞.
The continuity of 2F(t,x)-(f(t,x)+η(t))x with respect to x and (F5)(i) implies that there exists a constant L2>0 such that for any (t,x)∈Z[1,k]×R, 2F(t,x)-(f(t,x)+η(t))x≤L2. Then, we get
(3.25)∑t=1k[2F(t,um(t))-(f(t,um(t))+η(t))um(t)]≤2F(t0,um(t0))-(f(t0,um(t0))+η(t0))um(t0)+(k-1)L2.
Thus,
(3.26)∑t=1k[2F(t,um(t))-(f(t,um(t))+η(t))um(t)]→-∞asm→∞,
which contradicts (3.23). Therefore, {um} is bounded in Rk and J satisfies condition (C).
Now, suppose that (F5)(ii) holds. By a similar argument as above, we know also that J satisfies condition (C). The proof is complete.Proof of Theorem1.6.
Assume that (F4), (F5)(i), and (P3) hold. The proof for the case when (F4), (F5)(ii), and (P3) hold is similar and will be omitted here. Due to (P3), Rk has the direct sum decomposition: Rk=R+⊕R0⊕R-, where R+,R- and R0 are defined as in (2.4) and (2.5), respectively. We claim that for every t∈Z[1,k],
(3.27)F(t,x)-η(t)x→-∞as|x|→∞.
Indeed, according to (F5)(i), we can obtain that for any given ε>0, there exists a positive constant G1 such that for t∈Z[1,k],|x|>G1,
(3.28)2F(t,x)-(f(t,x)+η(t))x<-1ε.
Obviously,
(3.29)(f(t,sx)+η(t))sx-2F(t,sx)>1ε,
where s>1,  |x|>G1. We have
(3.30)dds(F(t,sx)-η(t)sxs2)=(f(t,sx)+η(t))sx-2F(t,sx)s3≥dds(-12εs2).
By integrating both sides of the above inequality from 1 to s, we get
(3.31)F(t,sx)-η(t)sxs2-F(t,x)+η(t)x≥-12εs2+12ε.
Let s→+∞ in the above inequality, and it follows from (F4) that
(3.32)F(t,x)-η(t)x≤-12ε
for t∈Z[1,k],|x|>G1. From the arbitrariness of ε, we can conclude that (3.27) holds, proving our claim.
Now we prove(3.33)J(u)→+∞as∥u∥→∞inR+⊕R0.
If (3.33) does not hold, there exist a constant L3>0 and a sequence {um} in R+⊕R0 such that ∥um∥→∞ as  m→∞ and
(3.34)J(um)≤L3
for all m. Since ∥um∥→∞ as  m→∞, there exist a subsequence of {um} (still denoted by {um}) and t0∈Z[1,k] such that |um(t0)|→∞asm→∞. By (3.27), we have
(3.35)F(t0,um(t0))-η(t0)um(t0)→-∞asm→∞.
The continuity of F(t,x)-η(t)x with respect to x and (3.27) implies that there exists a constant L4>0 such that for any (t,x)∈Z[1,k]×R, F(t,x)-η(t)x≤L4. Then, we get
(3.36)J(um)≥λ12∥um+∥2-∑t=1k[F(t,um(t))-η(t)um(t)]≥-F(t0,um(t0))+η(t0)um(t0)-(k-1)L4.
Thus,
(3.37)J(um)→+∞asm→∞,
which contradicts (3.34). Hence, (3.33) follows.
On the other hand, by (F4), there exists a positive constant a2 such that for any (t,x)∈Z[1,k]×R,
(3.38)|F(t,x)|≤14μ1|x|2+a2.
Then we have
(3.39)J(u)≤-μ12∥u-∥2+∥η∥∥u-∥+μ14∥u-∥2+a2k
for all u=u-∈R-. Thus, we can conclude that
(3.40)J(u)→-∞as∥u∥→∞inR-.
It follows from (3.33) and (3.40) that J satisfies conditions (φ1) and (φ2). Hence Theorem 1.6 follows from Lemma 3.2, the saddle point theorem, and Remark 2.3.The proof of Corollary1.8 is similar to that of Theorem 1.6 and is omitted.



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*Source: 407623-2010-01-13.xml*