DDNSDiscrete Dynamics in Nature and Society1607-887X1026-0226Hindawi Publishing Corporation40762310.1155/2009/407623407623Research ArticleExistence of Solutions to Boundary Value Problems for the Discrete Generalized Emden-Fowler EquationHeTieshan1YangFengjian1BerezanskyLeonidDepartment of Computation ScienceZhongkai University of Agriculture and EngineeringGuangzhou, Guangdong 510225Chinazhku.edu.cn2009131201020091605200930102009291120092009Copyright © 2009This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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.

1. Introduction and Statements of Main Results

Let Z and R be the sets of all integers and real numbers, respectively. For a,bZ, define Z[a]={a,a+1,},  Z[a,b]={a,a+1,,b} when ab.

In this paper, we consider the following boundary value problem (BVP for short) consisting of the discrete generalized Emden-Fowler equation:   Δ[p(t)Δu(t-1)]+q(t)u(t)=f(t,u(t)),tZ[1,k], and the boundary value conditions: 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 tZ[1,k+1]. We also assume that q(t) is realvalued for each tZ[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  concerning its disconjugacy, disfocality, oscillation, asymptotic behaviour, existence of periodic solutions, and solutions to boundary value problem.

Recently, Yu and Guo in  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 , 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 F(t,x)=0xf(t,s)ds,c(t)=q(t)-p(t)-p(t+1),M=(c(1)-αp(1)p(2)000p(2)c(2)p(3)000p(3)c(3)00000c(k-1)p(k)000p(k)c(k)-βp(k+1)),η=(η(1),η(2),,η(k))τ=(p(1)A00p(k+1)B). The main results are as follows.

Theorem 1.1.

Suppose that f, M satisfy the following assumptions.

There are constants C1>0,  C2>0,  1/2θ<1 such that for all (t,x)Z[1,k]×R, |f(t,x)|C1|x|θ+C2.

One has

either u-2θt=1kF(t,u(t))-  as u, or

u-2θt=1kF(t,u(t))+  as u,

where for all u=(u(1),u(2),,u(k))τRk, u=(t=1k|u(t)|2)1/2.

The matrix M is singular.

Then the BVP (1.1)-(1.2) has at least one solution.

Remark 1.2.

There  are functions p(t), q(t), and f(t,x) satisfying our Theorem 1.1 and not satisfying the results in . For example, let A, B be arbitrary constants, k=4,  α=β=-1,  θ=1/2 and p(t)1,q(t)0,f(t,x)=g(t)x1+x24+1, where g(t)<0 for every tZ[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 that f, M satisfy the following assumptions.

For any tZ[1,k],  lim|x|(f(t,x)/x)=0.

The matrix M is nonsingular.

Then the BVP (1.1)-(1.2) has at least one solution.

Corollary 1.4.

Suppose that f, 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 .

Theorem 1.6.

Suppose that f, M satisfy the following assumptions.

For any tZ[1,k],  lim|x|(F(t,x)/x2)  =0.

One has

either 2F(t,x)-(f(t,x)+η(t))x-  as |x| for all tZ[1,k], or

2F(t,x)-(f(t,x)+η(t))x+  as |x| for all tZ[1,k].

The matrix M is singular and indefinite.

Then the BVP (1.1)-(1.2) has at least one solution.

Remark 1.7.

There are functions p(t),  q(t), and f(t,x) satisfying our Theorem 1.6 and not satisfying the results in . For example, let A, B be arbitrary constants, k=4, α=1, β=0 and 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 tZ[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

Let Rk be the real Euclidean space with dimension k. For any u,vRk, u and (u,v) denote the usual norm and inner product in Rk, respectively.

Define the functional J on Rk as follows:   J(u)=12(Mu,u)+(η,u)-t=1kF(t,u(t)),u=(u(1),u(2),,u(k))τRk. It is well known that u=(u(1),u(2),,u(k))τRk 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 . It follows from the continuity of f that J is continuously differentiable on Rk. Moreover, one has (J(u),h)=(Mu,h)+(η,h)-t=1kf(t,u(t))·h(t),u,hRk.

When the matrix M is singular and indefinite, we suppose that 0<λ1λ2λl and 0>-μ1-μ2-μs are the positive and negative eigenvalues of M, respectively, and l+s=k-1. We also suppose that ςi, 1il and ξj, 1js are the eigenvectors of M corresponding to eigenvalues λi, 1il and -μj, 1js satisfying (ςi1,ςi2)={0,i1i2,1,i1=i2,(ςi,ξj)=0,(ξj1,ξj2)={0,j1j2,1,j1=j2, where i, i1, i2Z[1,l],  j, j1, j2Z[1,s]. We denote R+=span{ςiiZ[1,l]},R-=span{ξjjZ[1,s]},  R0=(R+R-). Then Rk has the direct sum decomposition Rk=R+R0R-. So, for each uRk, u can be expressed by u=u++u0+u-, where u*R*, * is +,0, or -, respectively. Furthermore, we have the following estimates: λ1u+2(Mu+,u+)λlu+2,u+R+,-μsu-2(Mu-,u-)-μ1u-2,u-R-.

Set u=maxtZ[1,k]|u(t)|,up=(t=1k|u(t)|p)1/p, where u=(u(1),u(2),,u(k))τRk and >p1. Then for any uRk, 1kuuu,1kuupk1/pu.

We will make use of the least action principle and saddle point theorem to obtain the critical points of J. Let us first recall these theorems.

Lemma 2.1 (the least action principle, see [<xref ref-type="bibr" rid="B12">12</xref>]).

Let X be a real Banach space, and assume that JC1(X,R) is bounded from below in X and satisfies the Palais-Smale condition ((PS) condition for short). Then c=infuXJ(u) is a critical value of J.

Lemma 2.2 (saddle point theorem, see [<xref ref-type="bibr" rid="B13">13</xref>]).

Let X be a real Hilbert space, X=X1X2, where X1{0} and is finite dimensional. Suppose that JC1(X,R) satisfies the (PS) condition and

there exist constants σ,ρ>0 such that JBρX1σ, where Bρ={uXu<ρ},  Bρ denotes the boundary of Bρ;

there exist eBρX1 and a constant ω>σ such that Je+X2ω.

Then J possesses a critical value cω and c=infhΦmaxuBρX1J(h(u)), where Φ={hC(B¯ρX1,X)hBρX1=id}.

Remark 2.3.

As shown in , 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 Theorem 1.1, 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 m0Z such that |J(um)|C3,|(J(um),h)|h for all m>m0,  hRk.

Since M 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 |t=1kf(t,um(t))·(um-(t)-um+(t))|t=1k(C1|um0(t)+(um+(t)+um-(t))|θ+C2)·|um-(t)-um+(t)|2C1um0θt=1k|um-(t)-um+(t)|+2C1t=1k|um+(t)+um-(t)|θ|um-(t)-um+(t)|+C2t=1k|um-(t)-um+(t)|2C12kτ1um02θ+τ12um--um+2+2C1um++um-1θum--um+q  +C2kum--um+2C12kτ1um02θ+τ12um--um+2+2C1kum++um-θum--um++C2kum--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 |t=1kf(t,um(t))·(um-(t)-um+(t))|2C12kτ1um02θ+τ12um++um-2+2C1kum++um-θ+1+C2kum++um-. Similarly to (3.3), we have |t=1k[F(t,um(t))-F(t,um0(t))]|2C12kμsum02θ+μs2um++um-2+2C1kum++um-θ+1+C2kum++um-.

Take h=um--um+ in (2.2). Then -(Mum,um--um+)=-(J(um),um--um+)+(η,um--um+)-t=1kf(t,um(t))(um-(t)-um+(t)). We know -(Mum,um--um+)λ1um+2+μ1um-2τ1um++um-2. Thus, by (3.1) and (3.3), we have τ1um++um-2(1+η)um++um-+2C12kτ1um02θ+τ12um++um-2+2C1kum++um-θ+1+C2kum++um-, that is, τ12um++um-2-2C1kum++um-θ+1-(1+η+C2k)um++um-2C12kτ1um02θ. It follows from (3.8) and 1/2θ<1 that um++um-C4um0θ+C5 for all m>m0 and some positive constants C4,C5.

By (3.1), (2.9), (3.4), and (3.9), we have C3J(um)=12(Mum,um)+(η,um)-t=1kF(t,um(t))12(λ1um+2-μsum-2)-η(um0+um++um-)-t=1k(F(t,um(t))-F(t,um0(t)))-t=1kF(t,um0(t))-μs2um++um-2-ηum0-2C12kμsum02θ-μs2um++um-2-2C1kum++um-θ+1-(η+C2k)um++um--t=1kF(t,um0(t))-μs(C42um02θ+2C4C5um0θ+C52)-ηum0-2C12kμsum02θ  -2C1k(C4um0θ+C5)θ+1-(η+C2k)(C4um0θ+C5)-t=1kF(t,um0(t))=-(μsC42+2C12kμs)um02θ-ηum0-2C1k(C4um0θ+C5)θ+1-2μsC4C5um0θ-(η+C2k)(C4um0θ+C5)-μsC52-t=1kF(t,um0(t)). Since 1/2θ<1, we deduce -C3um02θ(um0-2θt=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 Theorem <xref ref-type="statement" rid="thm1.1">1.1</xref>.

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 tZ, the singular symmetric matrix M has at least one nonzero eigenvalue and we will give the proof in three cases.

Suppose that the matrix M is singular and indefinite. Then Rk has the direct sum decomposition: Rk=R+R0R-. 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++u0X1, by (F1), (2.9), and the mean value theorem, we have J(u)λ12u+2-η(u0+u+)-t=1kf(t,u0(t)+θ1(t)u+(t))·u+(t)-t=1kF(t,u0(t))λ12u+2-η(u0+u+)-2C1ku0θu+-2C1ku+θ+1-C2ku+-t=1kF(t,u0(t))λ12u+2-η(u0+u+)-4C12kλ1u02θ-λ14u+2-2C1ku+θ+1-C2ku+-t=1kF(t,u0(t))=λ14u+2-2C1ku+θ+1-(η+C2k)u+-u02θ(u0-2θt=1kF(t,u0(t))+4C12kλ1+ηu0-2θ+1), where 0<θ1(t)<1, t=1,2,,k. Since 1/2θ<1 and (F2)(i), we have J(u)+as  u  in  X1.

On the other hand, for any u=u-X2, by (F1), (2.9), and the mean value theorem, we have J(u)-12μ1u2+ηu-t=1kf(t,θ2(t)u(t))·u(t)-t=1kF(t,0)-12μ1u2+ηu+t=1kC1|u(t)|θ+1+C2ku-12μ1u2+C1kuθ+1+(η+C2k)u, where 0<θ2(t)<1,  t=1,2,,k. Since 1/2θ<1, we can obtain J(u)-as  u  in  X2.

Let e=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.

Suppose that 0<λ1λ2λl 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=infuRkJ(u) is a critical value of J.

Suppose that 0>-μ1-μ2-μs are the negative eigenvalues of M and s=k-1. Then Rk has the direct sum decomposition: Rk=R0R-. 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 any u=u-R-, (3.15) holds. For any u=u0R0, since Mu=0, we have J(u)-ηu0-t=1kF(t,u0(t))=-u02θ(u0-2θt=1kF(t,u0(t))+ηu0-2θ+1). Due to (F2)(i) and 1/2θ<1, J(u)+as  u  in  R0. By the saddle point theorem, there exists at least one critical point of J.

Since J has at least one critical point in all three cases, the BVP (1.1)-(1.2) has at least one solution.

Proof of Theorem <xref ref-type="statement" rid="thm1.2">1.3</xref>.

Since the matrix M is nonsingular, we will give the proof in three cases.

Suppose that 0<λ1λ2λl 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 |f(t,x)|14τ1|x|+a1 for any (t,x)Z[1,k]×R.

We now prove that the functional J 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), 3τ14um++um-2(1+η+a1k)um++um-. Thus {um} is bounded, and the (PS) condition is verified.

For any u=u+R+, by (3.18) and the mean value theorem, we have J(u)12λ1u2-ηu-t=1kf(t,θ3(t)u(t))·u(t)-t=1kF(t,0)(λ12-τ14)u2-ηu-a1ku, where 0<θ3(t)<1,  t=1,2,,k. For any u=u-R-, Similarly to (3.20), we have J(u)-(μ12-τ14)u2+ηu+a1ku.

Let e=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.

Suppose that 0<λ1λ2λl 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=infuRkJ(u) is a critical value of J.

Suppose that 0>-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=-infuRkJ(u) is a critical value of -J.

Since J has a critical point in all three cases, the BVP (1.1)-(1.2) has at least one solution.

Proof of Corollary <xref ref-type="statement" rid="coro1.1">1.4</xref>.

This is immediate from Theorem 1.3.

The following lemma is useful for proving Theorem 1.6 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 |J(um)|L1,J(um)(1+um)L1 for all mZ. Hence, we have -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 t0Z[1,k] such that |um(t0)|  as  m. By (F5)(i), we have 2F(t0,um(t0))-(f(t0,um(t0))+η(t0))um(t0)-as  m. 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))xL2. Then, we get 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, t=1k[2F(t,um(t))-(f(t,um(t))+η(t))um(t)]-as  m, 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 Theorem <xref ref-type="statement" rid="thm1.3">1.6</xref>.

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+R0R-, where R+,R- and R0 are defined as in (2.4) and (2.5), respectively. We claim that for every tZ[1,k], 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 tZ[1,k],|x|>G1, 2F(t,x)-(f(t,x)+η(t))x<-1ε. Obviously, (f(t,sx)+η(t))sx-2F(t,sx)>1ε, where s>1,  |x|>G1. We have dds(F(t,sx)-η(t)sxs2)=(f(t,sx)+η(t))sx-2F(t,sx)s3dds(-12εs2). By integrating both sides of the above inequality from 1 to s, we get 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 F(t,x)-η(t)x-12ε for tZ[1,k],|x|>G1. From the arbitrariness of ε, we can conclude that (3.27) holds, proving our claim.

Now we prove J(u)+as  u  in  R+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 J(um)L3 for all m. Since um as m, there exist a subsequence of {um} (still denoted by {um}) and t0Z[1,k] such that |um(t0)|  as  m. By (3.27), we have F(t0,um(t0))-η(t0)um(t0)-as  m. 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)xL4. Then, we get J(um)λ12um+2-t=1k[F(t,um(t))-η(t)um(t)]-F(t0,um(t0))+η(t0)um(t0)-(k-1)L4. Thus, J(um)+as  m, 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, |F(t,x)|14μ1|x|2+a2. Then we have J(u)-μ12u-2+ηu-+μ14u-2+a2k for all u=u-R-. Thus, we can conclude that J(u)-as  u  in  R-.

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 Corollary 1.8 is similar to that of Theorem 1.6 and is omitted.

Acknowledgments

The authors thank the referees for their valuable comment and careful corrections. This project is supported by Science and Technology Plan Foundation of Guangzhou (No. 2006J1-C0341).

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