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Volume 2010, Article ID 971540, 18 pagesdoi:10.1155/2010/971540 Research Article Existence and Uniqueness of Positive Solutions for Discrete Fourth-Order Lidstone Problem with a Paramete

Volume 2010, Article ID 971540, 18 pages

doi:10.1155/2010/971540

Research Article

Existence and Uniqueness of Positive Solutions for Discrete Fourth-Order Lidstone Problem with

a Parameter

Yanbin Sang,1, 2 Zhongli Wei,2, 3 and Wei Dong4

1 Department of Mathematics, North University of China, Taiyuan, Shanxi 030051, China

2 School of Mathematics, Shandong University, Jinan, Shandong 250100, China

3 Department of Mathematics, Shandong Jianzhu University, Jinan, Shandong 250101, China

4 Department of Mathematics, Hebei University of Engineering, Handan, Hebei 056021, China

Correspondence should be addressed to Yanbin Sang,sangyanbin@126.com

Received 9 January 2010; Revised 23 March 2010; Accepted 26 March 2010

Academic Editor: A Pankov

Copyrightq 2010 Yanbin Sang et al 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

This work presents sufficient conditions for the existence and uniqueness of positive solutions for

a discrete fourth-order beam equation under Lidstone boundary conditions with a parameter; the iterative sequences yielding approximate solutions are also given The main tool used is monotone iterative technique

1 Introduction

In this paper, we are interested in the existence, uniqueness, and iteration of positive solutions for the following nonlinear discrete fourth-order beam equation under Lidstone boundary

conditions with explicit parameter β given by

Δ4y t − 2 − βΔ2y t − 1  htf1



y t f2y t, t Z, 1.1

y a  0  Δ2y a − 1, y b  0  Δ2y b − 1, 1.2

whereΔ is the usual forward difference operator given by Δyt  yt  1 − yt, Δ n y t 

Δn−1

Z: {c, c  1, , d − 1, d}, and β > 0 is a real parameter

In recent years, the theory of nonlinear difference equations has been widely applied

to many fields such as economics, neural network, ecology, and cybernetics, for details, see

1 7

and multiplicity of positive solutions of fourth-order problem, for example, 8 10

particular the discrete problem with Lidstone boundary conditions 11–17

little work has been done on the uniqueness and iteration of positive solutions of discrete fourth-order equation under Lidstone boundary conditions We would like to mention some results of Anderson and Minh ´os 11 12

BVP1.1 and 1.2

In 11

equation with explicit parameters β and λ given by

Δ4y t − 2 − βΔ2y t − 1  λft, y t, t Z, 1.3

with Lidstone boundary conditions1.2, where β > 0 and λ > 0 are real parameters The

authors obtained the following result

Theorem 1.1 see 11

A1 Z → 0, ∞ withb−1

z a1 g z > 0, w :

0, ∞ → 0, ∞ is continuous and nondecreasing, and there exists θ ∈ 0, 1 such that

w κy ≥ κ θ w y for κ ∈ 0, 1 and y ∈ 0, ∞,

then, for any λ ∈ 0, ∞, the BVP 1.3 and 1.2 has a unique positive solution y λ Furthermore, such a solution y λ satisfies the following properties:

i limλ→ 0 y λ   0 and lim λ→ ∞y λ   ∞;

ii y λ is nondecreasing in λ;

iii y λ is continuous in λ, that is, if λ → λ0 , then y λ − y λ0 → 0.

Very recently, in 12

nonexistence of nontrivial solutions to the following discrete nonlinear fourth-order boundary value problem

Δ4u t − 2  ηΔ2u

u a  0  Δ2u a − 1, u b  2  0  Δ2u b  1, 1.4

whereΔ denotes the forward difference operator defined by Δut  ut  1 − ut, Δ n u t 

ΔΔn−1u

b a < b integers, η, ξ, λ are real parameters and satisfy

η < 8 sin2 π

2b − a  2, η24ξ ≥ 0, ξ4η sin2

π

2b − a  2 < 16 sin4

π

2b − a  2, λ > 0.

1.5

For the function f, the authors imposed the following assumption:

t a1 g t > 0, h :

R → 0, ∞ is continuous and nondecreasing, and there exists θ ∈ 0, 1 such that

h μx ≥ μ θ h x for μ ∈ 0, 1 and x ∈ 0, ∞.

Their main result is the following theorem.

unique positive solution u λ Furthermore, such a solution u λ satisfies the properties (i)–(iii) stated in Theorem 1.1

The aim of this work is to relax the assumptionsA1 and B1 on the nonlinear term,

without demanding the existence of upper and lower solutions, we present conditions for the BVP1.1 and 1.2 to have a unique solution and then study the convergence of the iterative sequence The ideas come from Zhai et al 18,19 20

supremum norm

y  sup

y t. 1.6

Throughout this paper, we need the following hypotheses:

H1 f i: 0, ∞ → 0, ∞ are continuous and f i y > 0 for y > 0 i  1, 2;

H2 Z → 0, ∞ withb−1

z a1 h z > 0;

H3 f1 : 0, ∞ → 0, ∞ is nondecreasing, f2 : 0, ∞ → 0, ∞ is nonincreasing, and there exist ϕ Zwith ϕ : Z → 0, 1, for all e0∈ 0, 1, there exists τ0 Zsuch that ϕτ0  e0 , and ψ τ > ϕτ, for all τ Zwhich satisfy

f1



ϕ τy≥ ψτf1y

, f2

1

ϕ τ y ≥ ψτf2y

, Z, y ≥ 0. 1.7

2 Two Lemmas

To prove the main results in this paper, we will employ two lemmas These lemmas are based

on the linear discrete fourth-order equation

Δ4y t − 2 − βΔ2y Z, 2.1 with Lidstone boundary conditions1.2

fourth-order Lidstone boundary value problem2.1, 1.2 has solution

y t b

s a

b−1

z a1

where G2t, s given by

G2t, s  1

1, 0b, a

⎧

⎨

⎩

t, ab, s: t ≤ s,

s, ab, t: s ≤ t, Z Z 2.3

with t, s  μ t −s − μ s −t for μ  β  2 β β  4/2, is the Green’s function for the second-order

discrete boundary value problem

−Δ2y t − 1 − βyt Z,

y a  0  yb, 2.4

and G1s, z given by

G1s, z  1

b − a

⎧

⎨

⎩

s − ab − z: s ≤ z,

z − ab − s: z ≤ s, Z Z 2.5

is the Green’s function for the second-order discrete boundary value problem

x a  0  xb. 2.6

Lemma 2.2 see 11

m :  1, 0b, a  1

b − a2b, a , M :

b − a2b/2, a/2

41, 0b, a . 2.7

m ≤ G2t, sG1s, z ≤ M. 2.8

3 Main Results

Theorem 3.1 Assume that H1–H3 hold Then, the BVP 1.1 and 1.2 has a unique solution

y∗t in D, where

. 3.1

Moreover, for any x0, y0 ∈ D, constructing successively the sequences

x n1t b

s a

b−1

z a1

G2t, sG1s, zhzf1xn z  f2y n z,

t Z, n  0, 1, 2, ,

y n1t b

s a

b−1

z a1

G2t, sG1s, zhzf1

y n z f2x n z,

t Z, n  0, 1, 2, ,

3.2

One has x n t, y n t converge uniformly to y∗

Z Proof First, we show that the BVP1.1 and 1.2 has a solution

It is easy to see that the BVP1.1 and 1.2 has a solution y  yt if and only if y is a

fixed point of the operator equation

A

y1, y2

t b

s a

b−1

z a1

G2t, sG1s, zhzf1

y1z f2y2z, t Z.

3.3

In view ofH3 and 3.3, Ay1 , y2 is nondecreasing in y1and nonincreasing in y2 Moreover, for any τ Z, we have

A

ϕ τy1 , 1

ϕ τ y2 t  b−1

s a1

b−1

z a1

G2t, sG1s, zhz



f1



ϕ τy1z f2

1

ϕ τ y2z

≥ ψτb−1

s a1

b−1

z a1

G2t, sG1s, zhzf1



y1z f2y2z

 ψτAy1, y2

t

3.4

for t Zand y1 , y2∈ D.

Let

L  b − a − 1 b−1

z a1

conditionH2 implies L > 0 Since f i y > 0 for y > 0 i  1, 2, byLemma 2.2, we have

A L, L  b−1

s a1

b−1

z a1

G2t, sG1s, zhzf1L  f2L

≥ mf1L  f2Lb−1

s a1

b−1

z a1

h z

 mf1L  f2LL

3.6

for m in2.1 and L in 3.5

Moreover, we obtain

A L, L ≤ Mf1L  f2LL 3.7

for M in2.1

Thus

m

f1L  f2LL ≤ AL, L ≤ Mf1L  f2LL. 3.8 Therefore, we can choose a sufficiently small number e1∈ 0, 1 such that

e1L ≤ AL, L ≤ e L

1

which together withH3 implies that there exists τ1 Zsuch that ϕτ1  e1, so

ϕ τ1L ≤ AL, L ≤ L

ϕ τ1 . 3.10 Since ψτ1/ϕτ1 > 1, we can take a sufficiently large positive integer k such that

ψ τ1

ϕ τ1

k

≥ 1

ϕ τ1 . 3.11

It is clear that

ϕ τ1

ψ τ1

k

We define

u0t 

⎧

⎪

⎨

⎪

⎩

−ϕ τ1k

L: t  a − 1, b  1,

0: t  a, b,



ϕ τ1k

v0t 

⎧

⎪

⎪

⎪

⎪

⎪

⎪

− L

ϕ τ1k : t  a − 1, b  1,

0: t  a, b,

L



ϕ τ1k: t Z.

3.13

Evidently, for t Z, u0 ≤ v0 Take any λ ∈ 0, ϕτ1 2k

0≥ λv0.

By the mixed monotonicity of A, we have Au0 , v0 ≤ Av0, u0 In addition, combining

H3 with 3.10 and 3.11, we get

A u0 , v0  A





ϕ τ1k

L, 1

ϕ τ1k L



 A



ϕ τ1ϕ τ1k−1L, 1

ϕ τ1ϕ τ1k−1L



≥ ψτ1A





ϕ τ1k−1

L, 1

ϕ τ1k−1L



≥ · · ·

≥ψ τ1k

A L, L ≥ψ τ1k

ϕ τ1L

≥ϕ τ1k

L  u0

3.14

FromH3, we have

A

y1, y2

 A

ϕ s y1

ϕ s ,

1

ϕ s ϕ sy2

≥ ψsA

y 1

ϕ s , ϕ sy2 , Z, y1, y2≥ 0,

3.15

and hence

A

y

1

ϕ s , ϕ sy2 ≤

1

ψ s A



y1, y2

, Z, y1, y2 ≥ 0. 3.16

Thus, we have

A v0 , u0  A



L



ϕ τ1k ,

ϕ τ1k L



 A



L

ϕ τ1ϕ τ1k−1, ϕ τ1ϕ τ1k−1L



≤ 1

ψ τ1 A



L



ϕ τ1k−1,

ϕ τ1k−1L



≤ · · ·

≤  1

ψ τ1k A L, L ≤  1

ψ τ1k

L

ϕ τ1 .

3.17

In accordance with3.12, we can see that

A v0 , u0 ≤  L

ϕ τ1k  v0 3.18 Construct successively the sequences

u n  Au n−1, v n−1, vn  Av n−1, u n−1, n  1, 2, 3.19

By the mixed monotonicity of A, we have u1  Au0 , v0 ≤ Av0, u0  v1 By induction, we

obtain u n ≤ v n , n  1, 2, It follows from 3.14, 3.18, and the mixed monotonicity of A

that

u0≤ u1 ≤ · · · ≤ u n ≤ · · · ≤ v n ≤ · · · ≤ v1 ≤ v0 3.20

Note that u0≥ λv0 , so we can get u n t ≥ u0t ≥ λv0t ≥ λv n Z, n  1, 2, Let

λ n  sup{λ > 0 | u n t ≥ λv n Z}, n  1, 2, 3.21 Thus, we have

u n t ≥ λ n v n Z, n  1, 2, , 3.22 and then

u n1t ≥ un t ≥ λ n v n t ≥ λ n v n1 Z, n  1, 2, 3.23

Therefore, λ n1≥ λ n , that is, {λ n } is increasing with {λ n n→ ∞λ n We can show that λ  1 In fact, if 0 < λ < 1, by H3, there exists τ2 Zsuch that ϕτ2  λ.

Consider the following two cases

i There exists an integer N such that λ N  λ In this case, we have λ n  λ for all

n ≥ N holds Hence, for n ≥ N, it follows from 3.4 and the mixed monotonicity of A that

u n1 Aun , v n  ≥ A

λv n ,1

λ u n  A

ϕ τ2v n , 1

ϕ τ2 u n ≥ ψτ2Av n , u n   ψτ2v n1.

3.24

By the definition of λ n, we have

λ n1 λ ≥ ψτ2 > ϕτ2  λ. 3.25

This is a contradiction

ii For all integer n, λ n <  λ In this case, we have 0 < λ n / λ < 1 In accordance with

H3, there exists θ n Zsuch that ϕθ n   λ n / λ Hence, combining3.4 with the

mixed monotonicity of A, we have

u n1 Au n , v n  ≥ A

λ n v n , 1

λ n u n

 A

⎛

⎜λ n

λ λv n , u n

λ n / λ

λ

⎞

⎟

⎠  A

ϕ θ n ϕτ2v n , u n

ϕ θ n ϕτ2

≥ ψθ n A

ϕ τ2v n , u n

ϕ τ2 ≥ ψθ n ψτ2Av n , u n

 ψθ n ψτ2v n1.

3.26

By the definition of λ n, we have

λ n1≥ ψθ n ψτ2 > ϕθ n ψτ2  λ n

λ ψ τ2. 3.27

Let n → ∞, we have λ ≥ λ/λψτ2 > λ/λϕτ2  ϕτ2  λ, and this is also a contradiction.

Hence, limn→ ∞λ n 1

Thus, combining3.20 with 3.22, we have

0≤ u n l t − u n t ≤ v n t − u n t ≤ v n t − λ n v n t  1 − λ n v n t ≤ 1 − λ n v0t 3.28

for t Z, where l is a nonnegative integer Thus,

u n l − u n  ≤ v n − u n  ≤ 1 − λ n v0 3.29

Therefore, there exists a function y∗∈ D such that

lim

n→ ∞u n t  lim

n→ ∞v n t  y∗

Z. 3.30

By the mixed monotonicity of A and3.20, we have

u n1t  Aun t, v n t ≤ Ay∗t, y∗t≤ Av n t, u n t  v n1t. 3.31

Let n → ∞ and we get Ay∗t, y∗t  y∗

Z That is, y∗is a nontrivial solution of the BVP1.1 and 1.2

Next, we show the uniqueness of solutions of the BVP1.1 and 1.2 Assume, to the

contrary, that there exist two nontrivial solutions y1 and y2of the BVP1.1 and 1.2 such

that Ay1t, y1t  y1t and Ay2t, y2t  y2 Z According to3.9,

we can know that there exists 0 < η ≤ 1 such that ηy2t ≤ y1t ≤ 1/ηy2 Z Let

η0 sup 0 < η ≤ 1 | ηy2 ≤ y1≤ 1

η y2

!

Then 0 < η0≤ 1 and η0 y2t ≤ y1t ≤ 1/η0y2 Z

We now show that η0  1 In fact, if 0 < η0 < 1, then, in view of H3, there exists

τ Zsuch that ϕτ  η0 Furthermore, we have

y1 Ay1, y1



≥ A

η0y2, 1

η0y2  A

ϕ τy2 , 1

ϕ τ y2 ≥ ψτAy2, y2



 ψτy2 , 3.33

y1 Ay1, y1



≤ A

y2

η0, η0y2  A

y2

ϕ τ , ϕ τy2 ≤

1

ψ τ A



y2, y2



 1

ψ τ y2. 3.34

In3.34, we used the relation formula 3.16 Since ψτ > ϕτ  η0, this contradicts the

definition of η0 Hence η0 1 Therefore, the BVP 1.1 and 1.2 has a unique solution

Finally, we show that “moreover” part of the theorem For any initial x0, y0 ∈ D, in

accordance with3.9, we can choose a sufficiently small number e2 ∈ 0, 1 such that

e2L ≤ x0≤ 1

e2L, e2L ≤ y0≤ 1

e2L. 3.35

It follows fromH3 that there exists τ3 Zsuch that ϕτ3  e2, and hence

ϕ τ3L ≤ x0≤ L

ϕ τ3 , ϕ τ3L ≤ y0≤

L

ϕ τ3 . 3.36

Thus, we can choose a sufficiently large positive integer k such that

ψ τ3

ϕ τ3

k

≥ 1

ϕ τ3 . 3.37

Define

"u0ϕ τ3k

L, "v0  L

ϕ τ3k 3.38

Obviously,"u0 < x0, y0< "v0 Let

"u n  A"u n−1, "v n−1, "vn  A"v n−1, "u n−1, n  1, 2, ,

x n t  Ax n−1, y n−1

t b

s a

b−1

z a1

G2t, sG1s, zhzf1xn−1z  f2y n−1z,

y n t  Ay n−1, x n−1

t b

s a

b−1

z a1

G2t, sG1s, zhzf1



y n−1z f2x n−1z

3.39

for t Z, n  1, 2, By induction, we get "u n ≤ x n ≤ "v n,"u n ≤ y n ≤ "v n , n  1, 2,

Similarly to the above proof, it follows that there exists"y ∈ D such that

lim

n→ ∞"u n lim

n→ ∞"v n  "y, A"y, "y "y. 3.40

By the uniqueness of fixed points A in D, we get "y  y∗ Therefore, we have

lim

n→ ∞x n lim

n→ ∞y n  y∗. 3.41 This completes the proof of the theorem

Remark 3.2 From the proof ofTheorem 3.1, we easily know that assume y  Ay, x, x 

A x, y, thus, let y0  y, x0  x, we have

y n  y, x n  x, n  1, 2, 3.42

Therefore y  x  y∗

Theorem 3.3 Assume that H2 holds, and the following conditions are satisfied:

C1 f : 0, ∞ → 0, ∞ is continuous and fy > 0 for y > 0;

C2 f : 0, ∞ → 0, ∞ is nondecreasing;

b

s a

b−1

z a1

G2t, sG1s, zhzfϕ τyz≥ ψτ, yb

s a

b−1

z a1

G2t, sG1s, zhzfy z,

3.43

there exists τ0 Zsuch that ϕ τ0  e0 , and ψ : Z× 0, ∞ →

Z, y ∈ 0, ∞;

C3 Z, one has

i ψτ, y is nonincreasing with respect to y, and there exists τ4 Zsuch that

mf L ≥ ϕτ4, ψ



τ4, L/ϕ τ4

ϕ τ4 ≥ MfL 3.44

or

ii ψτ, y is nondecreasing with respect to y, and there exists τ5 Zsuch that

mf L ≥ ϕ τ5

ψ τ5 , L,

1

ϕ τ5 ≥ MfL, 3.45

where m, M are defined in2.1, L is defined in 3.5 Then, the BVP

Δ4y t − 2 − βΔ2y t − 1  htfy t, t Z,

y a  0  Δ2y a − 1, y b  0  Δ2y b − 1 3.46

has a unique solution y∗.

Proof For convenience, we still define the operator equation A by

Ay t b

s a

b−1

z a1

G2t, sG1s, zhzfy z, t Z. 3.47

In the following, we consider the following two cases

Z, ψτ, y is nonincreasing with respect to y.

According to conditionC3 andLemma 2.2, we can know that there exists τ4 ∈ a 

1, b Zsuch that

ϕ τ4L ≤ AL ≤ ψ



τ4, L/ϕ τ4

ϕ τ4 L. 3.48 Since ψτ4 , L /ϕτ4 > 1, we can find a sufficiently large positive integer k such that

ψ τ4 , L

ϕ τ4

k

≥ 1

ϕ τ4 . 3.49

Z. 3.30

By the mixed monotonicity of A and< /i>3.20,... 3.1