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Volume 2009, Article ID 730484, 10 pagesdoi:10.1155/2009/730484 Research Article for Difference Equation via Variational Method Qingrong Zou and Peixuan Weng School of Mathematics, South

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Volume 2009, Article ID 730484, 10 pages

doi:10.1155/2009/730484

Research Article

for Difference Equation via Variational Method

Qingrong Zou and Peixuan Weng

School of Mathematics, South China Normal University, Guangzhou 510631, China

Correspondence should be addressed to Peixuan Weng,wengpx@scnu.edu.cn

Received 7 July 2009; Accepted 15 October 2009

Recommended by Kanishka Perera

The variational method and critical point theory are employed to investigate the existence of

solutions for 2nth-order diﬀerence equation Δ n p k −nΔn y k −n −1n1fk, y k 0 for k ∈ 1, N with boundary value condition y1−n y2−n · · · y0 0, y N1 · · · y N n 0 by constructing

a functional, which transforms the existence of solutions of the boundary value problemBVP

to the existence of critical points for the functional Some criteria for the existence of at least one solution and two solutions are established which is the generalization for BVP of the even-order diﬀerence equations

Copyrightq 2009 Q Zou and P Weng This 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

1 Introduction

Diﬀerence equations have been applied as models in vast areas such as finance insurance, biological populations, disease control, genetic study, physical field, and computer applica-tion technology Because of their importance, many literature deals with its existence and uniqueness problems For example, see1 10

We notice that the existing results are usually obtained by various analytical techniques, for example, the conical shell fixed point theorem 1, 6 , Banach contraction map method 7 , Leray-Schauder fixed point theorem 2, 10 , and the upper and lower solution method 3 It seems that the variational technique combining with the critical point theory11 developed in the recent decades is one of the effective ways to study the boundary value problems of difference equations However because the variational method requires a “symmetrical” functional, it is hard for the odd-order difference equations to create

a functional satisfying the “symmetrical” property Therefore, the even-order diﬀerence equations have been investigated in most references

Let a, b, N > 1, n ≥ 1, k be integers, and a < b, a, b : {a, a 1, , b} be a discrete

interval inZ Inspired by 5,8 , in this paper, we try to investigate the following 2nth-order boundary value problem BVP of diﬀerence equation via variational method combining

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with some traditional analytical skills:

Δn

p k −nΔn y k −n

−1n1f

k, y k

 0 k ∈ 1, N , 1.1

y1−n y2−n · · · y0 0, y N1 · · · y N n 0, 1.2

whereΔn y k Δn−1y k1− Δn−1y k n 1 is the forward diﬀerence operator; p k ∈ R for k ∈

1 − n, N and f ∈ C1, N × R, R A variational functional for BVP 1.1-1.2 is constructed

which transforms the existence of solutions of the boundary value problem BVP to the existence of critical points of this functional In order to prove the existence criteria of critical points of the functional, some lemmas are given inSection 2 Two criteria for the existence of

at least one solution and two solutions for BVP1.1-1.2 are established inSection 3which

is the generalization for BVP of the even-order diﬀerence equations The existence results obtained in this paper are not found in the references, to the best of our knowledge

For convenience, we will use the following notations in the following sections:

F k, u 

u

0

f k, sds, p max

k ∈1−n,N p k, p min

k ∈1−n,N p k. 1.3

2 Variational Structure and Preliminaries

We need two lemmas from12 or 11

Lemma 2.1 Let H be a real reflexive Banach space with a norm · , and let φ be a weakly lower

(upper) semicontinuous functional, such that

lim

x → ∞ φ x ∞

or lim

x → ∞ φ x −∞

then there exists x0∈ H such that

φ x0 inf

x ∈H φ x0

or φ x0 sup

x ∈H φ x0

Furthermore, if φ has bounded linear Gˆateaux derivative, then φx0 0.

Lemma 2.2 mountain-pass lemma Let H be a real Banach space, and let φ : H →

R be continuously diﬀerential, satisfying the P-S condition Assume that x0, x1 ∈ H and Ω

is an open neighborhood of x0, but x1/ ∈ Ω If max{φx0, φx1} < inf x ∈∂Ω φ x, then c

infh∈Γmaxt ∈0,1 φ ht is the critical value of φ, where

Γ {h | h : 0, 1 −→ H, h is continuous, h0 x0, h 1 x1}. 2.3

This means that there exists x2∈ H, s.t φx2 0, φx2 c.

The following lemma will be used in the proof ofLemma 2.4

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Lemma 2.3 If A m ×m is a symmetric and positive-defined real matrix, B m ×n is a real matrix, B T is the transposed matrix of B Then B T AB is positive defined if and only if rank B n.

Proof Since A is positive defined, then

B T AB is positive-defined ⇐⇒ ∀x / 0, x T B T ABx > 0

⇐⇒ ∀x / 0, Bx T A Bx > 0 ⇐⇒ ∀x / 0, Bx / 0 ⇐⇒ rank B n. 2.4 Let H be a Hilbert space defined by

H y : 1 − n, N n −→ R | y1−n y2−n · · · y0 0, y N1 · · · y N n 0 2.5 with the norm

y N

k1

y2

k , y ∈ H. 2.6

Hence H is an N-dimensional Hilbert space For any q > 1, let y q N

k1y2

k1/q , then one

can show that there exist constants q1, q2 > 0, s.t q1y y q q2y; that is, · q is an equivalent norm of · see 9, page 68

Lemma 2.4 There is

λ x2 N

k 1−n

Δn x k2 4n x2, x ∈ H, where λ is a postive constant. 2.7

Proof Since x ∈ H, Δ n −j x N1 Δn −j x j −n 0, j 1, 2, , n By using the inequality a − b2 2a2 b2, a, b ∈ R, we have

N

k 1−n

Δn x k2 N

k 1−n

Δn−1x k1− Δn−1x k2 ≤ 2N

k 1−n

Δn−1x k12

Δn−1x k2

2

N1

k 2−n

Δn−1x k2

N

k 2−n

Δn−1x k2

4 N

k 2−n

Δn−1x k2

≤ 4 × 2

N

k 2−n

Δn−2x k12

N

k 2−n

Δn−2x k2

4 × 2

N1

k 3−n

Δn−2x k2

N

k 2−n

Δn−2x k2

42 N

k 3−n

Δn−2x k2.

2.8

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Repeating the above process, we obtain

N

k 1−n

Δn x k2 4nN

k1

x k2 4n x2. 2.9

On the other hand, define b k Δn x kn

j0−1j C j n x k n−j , k ∈ 1 − n, N , where C j

nis the combination number, then we can rewrite{b k}N

1−nin a vector form, that is, b Bx, where

b b1−n, b2−n, , b NT , x x1, x2, , x NT, and

B

⎛

⎜

⎝

1

c1 1

.

c n c n−1 · · · 1

c n · · · c1 1

c n c n−1 · · · 1

c n · · · c1 1

c n c n−1

c n

⎞

⎟

⎠

Nn×N

, 2.10

where c i −1i C i

n Hence rank B N Note that

N

k 1−n

Δn x k2 b T b Bx T Bx x T B T Bx, 2.11

and byLemma 2.3with A m ×m I Nn×Nn , we know that B T B is positive defined Hence

all eigenvalues of B T B are real and positive Let λ be the minimal eigenvalue of these N

eigenvalues, then λ > 0 Therefore x T B T Bx λx T x, that is,N

1−nΔn x k2 λx2.

However, how to find the λ in Lemma 2.4 is a skillful and challenging task The following lemma from13 oﬀers some help for the estimation of λ

Lemma 2.5 Brualdi 13 If A aij ∈ M N is weak irreducible, then each eigenvalue is contained

in the set

γ ∈CA

⎧

⎨

⎩z∈ C :

"

P i∈γ

|z − a ii| "

P i∈γ

Ri

⎫

⎬

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InLemma 2.5,C is the complex number set, and the denotations CA, γ, P i , Ri can

be found in 13 Since BT B is positive defined, all eigenvalues are positive real numbers.

Therefore, byLemma 2.5, let

B

γ ∈CB T B

⎧

⎨

⎩z∈ R :

"

P i∈γ

|z − b ii| "

P i∈γ

Ri

⎫

⎬

⎭, 2.13

where B T B b ij B is a subset of R and can be calculated directly from B T B Define λ max{0, min{B}} If λ > 0, we can use this λ as λ inLemma 2.4 If λ 0, then one needs to calculate the eigenvalues directly

Define the functional φ on H by

φ

y

N

k 1−n

1

2p kΔn y k2− Fk, y k

Then φ is C1with

φ

y

, x

N

k 1−n

&

p k

Δn y k

Δn x k − x k f

k, y k

'

, 2.15

where x {x k}N n

k 1−n ∈ H and ·, · is the inner product in H In fact, we have

φ

y x− φy

N

k1−n

1

2p k

(

Δn y k Δn x k

2

− Δn y k2)

−&F

k, y k x k

− Fk, y k

'

N

k1−n

p k

Δn y k

Δn x k 1

2p kΔn x k2− fk, y k θx k

x k

, θ ∈ 0, 1.

2.16

The continuity of f and the right-hand side of the inequality inLemma 2.4lead to2.15

Furthermore, for any x ∈ H, we have Δ n −j x N1 Δn −j x1−n 0, j 1, 2, , n By using

the following formulae.g., see 14, page 28 :

n2

k n

&

g k Δf k f k1Δg k

'

f k g kn2 1

n1 , 2.17

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we have

N

1−n

p k

Δn y k

Δn x k p k−1

Δn y k−1

Δn−1x kN1

1−n −N

1−n

Δp k−1Δn y k−1

Δn−1x k

−N

1−n

Δp k−1Δn y k−1

Δn−1x k

−Δp k−2Δn y k−2

Δn−2x kN1

1−n N

1−n

Δ2

p k−2Δn y k−2

Δn−2x k

N

1−n

Δ2

p k−2Δn y k−2

Δn−2x k

2.18

Repeating the above process, we obtain

N

1−n

p k

Δn y k

Δn x k −1nN

1−n

Δn

x k 2.19

Letφy, x 0, that is,

N

1−n

&

−1nΔn

− fk, y k

'

x k

−1nN

1−n

(

Δn

−1n1f

k, y k)

x k

 0.

2.20

Since x ∈ H is arbitrary, we know that the solution of BVP 1.1-1.2 corresponds to the critical point of φ.

3 Main Results

Now we present our main results of this paper

Theorem 3.1 If there exist M1> 0, a1> 0, a2∈ R, and σ > 2 s.t.

F k, u a1|u| σ a2, ∀ |u| > M1, 3.1

then BVP1.1-1.2 has at least one solution

Proof In fact, we can choose a suitable a2< 0 such that

F k, u a1|u| σ a2, ∀u ∈ R. 3.2

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Since there exists σ1> 0 with σ1y y σ , we have

N

1−n

F

k, y k

a1

N

1

y kσ a2N n a1σ σ

1 y σ a2N n. 3.3 Then byLemma 2.4, we obtain

φ

y

p

24

n y 2− a1σ σ1 y σ − a2N n. 3.4

Noticing that σ > 2, we have lim y → ∞ φ y −∞ FromLemma 2.1, the conclusion of this lemma follows

Corollary 3.2 If there exists M2 > 0 s.t uf k, u > 0 for all |u| > M2, and

inf

k ∈1−n,N lim

u→ ∞

f k, u

|u| r r1, 3.5

where r, r1 satisfy either r 1, r1 > 4 n p or r > 1, r1 > 0, then BVP1.1-1.2 has at least one

solution.

Proof Assume that r 1, r1 > 4 n p Then for 1  r1− 4n p /2 > 0, there exists M3 > M2,

such that|fk, y| r1− 1|y| as |y| > M3 We have from the continuity of fk, u that there

is a K > 0 such that −K ≤ fk, u ≤ K for all k ∈ 1, N , |u| ≤ M3 When y > 0, one has

f k, y r1− 1y > 0 for y ∈ M3, ∞, then

y

0

f k, sds 

M3

0

f k, sds 

y

M3

f k, sds −KM3 r1− 1

2 y

2−r1− 1

2 M

2

3; 3.6

when y < 0, one has f k, y r1− 1y < 0 for y ∈ −∞, −M3, then

y

0

f k, sds 

−M3

0

f k, sds 

y

−M3

f k, sds −KM3r1− 1

2 y

2−r1− 1

2 M

2

3. 3.7

Let c : −KM3− r1− 1/2M2

3, then we have*y

0f k, sds r1− 1/2y2 c for y ∈ R.

Therefore, we have

φ

y

p

24

n y 2−r1− 1

2

N

1

y k2− c 4n p − r1 1

2 y 2− c − 1

2 y 2− c, 3.8

which implies limy → ∞ φ y −∞, and byLemma 2.1, the conclusion of this lemma follows

Assume that r > 1, r1 > 0 Then for 2  r1/2 > 0, there exists M4 > M2, such

that|fk, y| r1/2 |y| as |y| > M4 We have from the continuity of fk, u that there is

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a K > 0 such that −K ≤ fk, u ≤ K for all k ∈ 1, N , |u| ≤ M4 When y > 0, one has

f k, y r1/2 y r > 0, y ∈ M4, ∞, then we have

y

0

f k, sds 

M4

0

f k, sds 

y

M4

f k, sds

−KM4 r1

2r 1y r1−

r1

2r 1M r41,

3.9

when y < 0, one has f k, y −r1/2 |y| r −r1/2 −y r < 0, y ∈ −∞, −M4, then we have

y

0

f k, sds 

−M4

0

f k, sds 

y

−M4

f k, sds

−KM4r1

2

y

−M4

−s r d −s

 −KM4 r1

2r 1yr1− r1

2r 1M4r1.

3.10

Let d : −KM4− r1/2 r 1M r1

4 , then we have*y

0f k, sds r1/2 r 1|y| r1 d for

|y| > M4 Therefore, byTheorem 3.1, the conclusion of this lemma follows

Theorem 3.3 Assume that p k > 0, k 1 − n, , N, and

i supk ∈1−n,N limu→ 0fk, u/u r2< pλ, λ > 0 is defined in Lemma 2.4 ;

ii F satisfies 3.1 in Theorem 3.1 or f satisfies the assumptions in Corollary 3.2

Then BVP1.1-1.2 has at least two solutions

Proof We first show that φ satisfies the P-S condition Let {y m}∞

m1⊂ H satisfy that {φy m}

is bounded and limm→ ∞φy m 0 If {y m} is unbounded, it possesses a divergent

subseries, say y mk → ∞ as k → ∞ However from ii, we get 3.4 or 3.8, hence

φ y mk → −∞ as k → ∞, which is contradictory to the the fact that {φy m} is bounded Next we use the mountain-pass lemma to finish the proof Byi, for 3  pλ − r2/2 >

0, there exists R1 > 0 such that f k, y/y r2 3 for|y| R1 Then*y

0f k, sds r2

3/2y2for|y| R1 Now together withLemma 2.3, we have

φ

y

p

2λ y 2−r2 3

2

N

1−n

|y k|2 y 2

pλ − r

2− 3

2

3

2 y 2> 0 fory R1,

3.11 which implies that

φ

y

3

2R

2

where θ is the zero element in H, and Ω {y ∈ H | y < R1} Since we have from 3.4

or3.8 that limy → ∞ φ y −∞, there exists y1 ∈ H with y1 > R1, that is, y1/ ∈ Ω, but

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φ y1 < φθ 0 Using Lemma 2.2, we have shown that ξ infh∈Γmaxt ∈0,1 φ ht is the critical value of φ, withΓ defined as

Γ h | h : 0, 1 −→ H, h is continuous, h0 θ, h1 y1

. 3.13

We denote y as its corresponding critical point.

On the other hand, byTheorem 3.1orCorollary 3.2, we know that there exists y∗∈ H, s.t φy∗ supy ∈H φ y If y∗/ y, the theorem is proved If on the contrary, y∗  y, that

is, supy ∈H φ y inf h∈Γmaxt ∈0,1 φ ht, that implies for any h ∈ Γ, max t ∈0,1 φ ht

supy ∈H φ y Taking h1/ h2 inΓ with maxt ∈0,1 φ h1t max t ∈0,1 φ h2t sup y ∈H φ y,

by the continuity of φht, there exist t1, t2 ∈ 0, 1 s.t φh1t1 maxt ∈0,1 φ h1t,

φ h2t2 maxt ∈0,1 φ h2t Hence h1t1, h2t2 are two diﬀerent critical points of φ, that

is, BVP1.1-1.2 has at least two diﬀerent solutions

4 An Example

Consider the 6th-order boundary value problem for diﬀerence equation

Δ6y k−3 y3

k e y2−9 0, k ∈ 1, 300 ,

y−2 y−1 y0 0, y301 y302 y303 0. 4.1

Let fk, u u3e u2 −9, we have limu→ 0fk, u/u 0, lim u→ ∞fk, u/u ∞ Hence

f k, u satisfies the conditions inTheorem 3.3, the boundary value problem4.1 has at least two solutions

Acknowledgments

This research is partially supported by the NSF of China and NSF of Guangdong Province

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is, BVP1.1-1.2 has at least two diﬀerent solutions

4 An Example

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