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

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.


Introduction and Statements of Main Results
Let Z 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: and the boundary value conditions: 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 The main results are as follows.
Theorem 1.1.Suppose that f, M satisfy the following assumptions.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 10 .For example, let A, B be arbitrary constants, k 4, α 1, β 0 and

Variational Structure and Two Basic Lemmas
Let R k be the real Euclidean space with dimension k.For any u, v ∈ R k , u and u, v denote the usual norm and inner product in R k , respectively.Define the functional J on R k as follows: When the matrix M is singular and indefinite, we suppose that 0 are the positive and negative eigenvalues of M, respectively, and l s k − 1.We also suppose that ς i , 1 ≤ i ≤ l and ξ j , 1 ≤ j ≤ s are the eigenvectors of M corresponding to eigenvalues λ i , 1 ≤ i ≤ l and −μ j , 1 ≤ j ≤ s satisfying where u * ∈ R * , * is , 0, or −, respectively.Furthermore, we have the following estimates:

2.7
Set 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 12 .Let X be a real Banach space, and assume that J ∈ C 1 X, R is bounded from below in X and satisfies the Palais-Smale condition ((PS) condition for short).Then c inf u∈X J u is a critical value of J. Lemma 2.2 saddle point theorem, see 13 .Let X be a real Hilbert space, X X 1 ⊕ X 2 , where X 1 / {0} and is finite dimensional.Suppose that J ∈ C 1 X, R satisfies the (PS) condition and denotes the boundary of B ρ ; ϕ 2 there exist e ∈ B ρ ∩ X 1 and a constant ω > σ such that J| e X 2 ≥ ω.
Then J possesses a critical value c ≥ ω and where 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 .

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 (F 1 ), (F 2 ), and (P 3 ) hold.Then the functional J (see 2.1 ) satisfies the (PS) condition; that is, for any sequence {u m } such that J u m is bounded and J u m → 0 as m → ∞, there exists a subsequence of {u m } which is convergent in R k .
Proof.First suppose that F 1 , F 2 i , and P 3 hold.Recall that R k is a finite dimensional Hilbert space.Consequently, in order to prove that J satisfies the PS condition, we only need to prove that {u m } is bounded.Let {u m } be a sequence in R k such that J u m is bounded and J u m → 0 as m → ∞.Then there exist C 3 > 0 and m 0 ∈ Z 1 such that Since M is singular and indefinite, we write u m u m u 0 m u − m with u * m ∈ R * , where * , 0, −, respectively.By F 1 , 2.9 , and Hölder's inequality p 1/θ, q 1/ 1 − θ , we have where τ 1 min{λ 1 , μ 1 }.On the other hand, by the fact that u m and u − m are mutually orthogonal, one has

3.3
Similarly to 3.3 , we have

3.5
We know Thus, by 3.1 and 3.3 , we have

3.8
It follows from 3.8 and 1/2 ≤ θ < 1 that for all m > m 0 and some positive constants C 4 , C 5 .

3.10
Since 1/2 ≤ θ < 1, we deduce for all m > m 0 and some positive constants C 6 , C 7 .The above inequality and F 2 i imply that {u 0 m t } is bounded.Then it follows from 3.9 that {u m u − m } is bounded.Thus we conclude that {u m } is bounded, and the PS condition is verified.Now, suppose that F 1 , F 2 ii , and P 3 hold.By a similar argument as above, we know also that J satisfies the PS condition.The proof is complete.
Proof of Theorem 1.1.Assume that F 1 , F 2 i , and P 1 hold.The proof for the case when F 1 , F 2 ii , and P 1 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 matrix M is singular and indefinite.Then R k has the direct sum decomposition: R k R ⊕ R 0 ⊕ R − .In view of Lemma 3.1, we only to check that conditions ϕ 1 and ϕ 2 in the saddle point theorem hold.To this end, let X 1 R ⊕ R 0 , X 2 R − .For any u u u 0 ∈ X 1 , by F 1 , 2.9 , and the mean value theorem, we have 3.12 where 0 < θ 1 t < 1, t 1, 2, . . ., k.Since 1/2 ≤ θ < 1 and F 2 i , we have

3.13
On the other hand, for any u u − ∈ X 2 , by F 1 , 2.9 , and the mean value theorem, we have

3.15
Let e 0, then it follows from 3.13 and 3.15 that ϕ 1 and ϕ 2 are satisfied.By the saddle point theorem, J has at least one critical point.

Discrete Dynamics in Nature and Society
ii Suppose that 0 < λ 1 ≤ λ 2 ≤ • • • ≤ λ l are the positive eigenvalues of M and l k − 1.
Then R k has the direct sum decomposition: R k R ⊕ R 0 , where R and R 0 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, c 1 inf u∈R k J u is a critical value of J.
iii Suppose that 0 > −μ 1 ≥ −μ 2 ≥ • • • ≥ −μ s are the negative eigenvalues of M and s k − 1.Then R k has the direct sum decomposition: R k R 0 ⊕ 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 any u u − ∈ R − , 3.15 holds.For any u u 0 ∈ R 0 , since Mu 0, we have

3.16
Due to F 2 i and 1/2 ≤ θ < 1, 3.17 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 1.3.Since the matrix M is nonsingular, we will give the proof in three cases.
and negative eigenvalues of M, respectively, and l s k.Then R k has the direct sum decomposition: R k R ⊕ R − .Denote τ 1 min{λ 1 , μ 1 }.It follows from F 3 that there exists a positive constant a 1 such that We now prove that the functional J satisfies the PS condition.Let {u m } be a sequence in R k such that J u m is bounded and J u m → 0 as m → ∞.Write u m u m u − m , where u m ∈ R , u − m ∈ R − .Similarly to 3.8 , we have, by 3.18 ,

3.19
Thus {u m } is bounded, and the PS condition is verified.
For any u u ∈ R , by 3.18 and the mean value theorem, we have

3.21
Let e 0, then it follows from 3.20 and 3.21 that ϕ 1 and ϕ 2 are satisfied.By the saddle point theorem, J has at least one critical point.
Then E has the direct sum decomposition: R k R .It follows from 3.20 that J satisfies the PS condition and is bounded from below.Then, by the least action principle, c 2 inf u∈R k J u is a critical value of J.
iii Suppose that 0 > −μ 1 ≥ −μ 2 ≥ • • • ≥ −μ s are the negative eigenvalues of M and s k.Then R k has the direct sum decomposition: R k R − .It follows from 3.21 that J satisfies the PS condition and is bounded from above.Then, by the least action principle, c 3 −inf u∈R k J 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 1.4.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 (F 5 ), the functional J satisfies condition (C); that is, for any sequence {u m } such that J u m is bounded and J u m 1 u m → 0 as m → ∞, there exists a subsequence of {u m } which is convergent in R k .
Proof.First suppose that F 5 i holds.Let {u m } be a sequence in R k such that J u m is bounded and J u m 1 u m → 0 as m → ∞.Then there exists a constant L 1 > 0 such that for all m ∈ Z 1 .Hence, we have

3.23
Then, {u m } is bounded.In fact, if {u m } is unbounded, there exist a subsequence of {u m } still denoted by {u m } and t

3.24
The continuity of 2F t, x − f t, x η t x with respect to x and F 5 i implies that there exists a constant L 2 > 0 such that for any t

3.25
Thus, .26 which contradicts 3.23 .Therefore, {u m } is bounded in R k and J satisfies condition C .Now, suppose that F 5 ii holds.By a similar argument as above, we know also that J satisfies condition C .The proof is complete.
Proof of Theorem 1.6.Assume that F 4 , F 5 i , and P 3 hold.The proof for the case when F 4 , F 5 ii , and P 3 hold is similar and will be omitted here.Due to P 3 , R k has the direct sum decomposition: R k R ⊕ R 0 ⊕ R − , where R , R − and R 0 are defined as in 2.4 and 2.5 , respectively.We claim that for every t ∈ Z 1, k , F t, x − η t x −→ −∞ as |x| −→ ∞.

3.27
Indeed, according to F 5 i , we can obtain that for any given ε > 0, there exists a positive constant G 1 such that for t ∈ Z 1, k , |x| > G

3.35
The continuity of F t, x −η t x with respect to x and 3.27 implies that there exists a constant L 4 > 0 such that for any t, x ∈ Z 1, k × R, F t, x − η t x ≤ L 4 .Then, we get F t, u m t − η t u m t ≥ −F t 0 , u m t 0 η t 0 u m t 0 − k − 1 L 4 .

3.36
Thus, J u −→ ∞ as u −→ ∞ in R ⊕ R 0 .3.33If 3.33 does not hold, there exist a constant L 3 > 0 and a sequence{u m } in R ⊕ R 0 such that u m → ∞ as m → ∞ andJ u m ≤ L 3 3.34 for all m.Since u m → ∞ as m → ∞, there exist a subsequence of {u m } still denoted by {u m } and t 0 ∈ Z 1, k such that |u m t 0 | −→ ∞ as m −→ ∞.By 3.27 , we have F t 0 , u m t 0 − η t 0 u m t 0 −→ −∞ as m −→ ∞.
11The 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 10 .For example, let A, B be arbitrary constants, k Theorem 1.3.Suppose that f, M satisfy the following assumptions.F 3 For any t ∈ Z 1, k , lim |x| → ∞ f t,x /x 0. P 2 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 (F 1 ) and (P 2 ).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 | ≤ C 2 implies |f t, x | ≤ C 1 |x| θ C 2 , our Corollary 1.4 extends Theorem 3.2 in 10 .Theorem 1.6.Suppose that f, M satisfy the following assumptions.
Z 1, k , |x| > G 1 .From the arbitrariness of ε, we can conclude that 3.27 holds, proving our claim.Now we prove .37 which contradicts 3.34 .Hence, 3.33 follows.On the other hand, by F 4 , there exists a positive constant a 2 such that for any t, x ∈ Z 1, k × R,