Existence and Multiplicity of Positive Solutions of a Nonlinear Discrete Fourth-Order Boundary Value Problem

we show the existence and multiplicity of positive solutions of the nonlinear discrete fourth-order boundary value problem , , , where , is continuous, and is continuous, , . The main tool is the Dancer's global bifurcation theorem.


Introduction
It's well known that the fourth order boundary value problem can describe the stationary states of the deflection of an elastic beam with both ends hinged, it also models a rotating shaft .The existence and multiplicity of positive solutions of the boundary value problem 1.1 have been considered extensively in the literature, see 1-10 .

1.3
Zhang et al. 13 , and He and Yu 14 used the fixed point index theory in cones to study the following discrete analogue where Δ 4 u t − 2 denote the fourth forward difference operator and Δu t u t 1 − u t .It has been pointed out in 13, 14 that 1.4 , 1.5 are equivalent to the equation of the form: G 1 s, j h j f u j where

1.7
Notice that two distinct Green's functions used in 1.6 make the construction of cones and the verification of strong positivity of A 0 become more complex and difficult.Therefore, Ma and Xu 15 considered 1.4 with the boundary condition and introduced the definition of generalized positive solutions: Definition 1.1.A function y : Ì 0 → Ê is called a generalized positive solution of 1.4 , 1.8 , if y satisfies 1.4 , 1.8 , and y t ≥ 0 on Ì 1 and y t > 0 on Ì 2 .
Remark 1.2.Notice that the fact y : Ì 0 → Ê is a generalized positive solution of 1.4 , 1.8 does not means that y t ≥ 0 on Ì 0 .In fact, y satisfies 1 y t ≥ 0 for t ∈ Ì 2 ; Ma and Xu 15 also applied the fixed point theorem in cones to obtain some results on the existence of generalized positive solutions.
It is the purpose of this paper to show some new results on the existence and multiplicity of generalized positive solutions of 1.4 , 1.8 by Dancer's global bifurcation theorem.To wit, we get the following. 1.9 Assume that there exists B ∈ 0, ∞ such that f is nondecreasing on 0, B .Then

1.11
The "dual" of Theorem 1.3 is as follows.

2.3
Proof.By a simple summing computation and u 1 Δ 2 u 0 0, we can obtain

2.4
This together with u T 1 Δ 2 u T 0, it follows that
Remark 2.2.It has been pointed out in 15 that 2.1 is equivalent to the summation equation of the form

2.6
It is easy to verify that 2.2 and 2.6 are equivalent.By a similar method in 9 , it follows that K t, s satisfies where

2.9
Moreover, we have that

2.10
here Let X be a real Banach space with a cone K such that X K − K. Let us consider the equation: under the assumptions: A1 The operators L, N : X → X are compact.Furthermore, L is linear, Nx X / x X → 0 as x X → 0, and A3 L is strongly positive.
Dancer's global bifurcation theorem is the following.
x is a solution of 2.11 with x > 0 and μ > 0 .

Proof of the Main Results
Before proving Theorem 1.3, we state some preliminary results and notations.Let Then X is a Banach space under the normal:

3.3
See 22 for the detail.Let

3.4
Then K is normal and has a nonempty interior and X K − K.
Then Y is a Banach space under the norm: Define L : X → Y by setting

3.7
Then u ∈ int K.
Proof.It is enough to show that there exist two constants r 1 , r 2 ∈ 0, ∞ such that r 1 e t ≤ −Δ 2 u t − 1 ≤ r 2 e t , t ∈ Ì 1 .

3.8
In fact, we have from 3.7 that

3.9
This together with the relation t

3.10
Combining 3.9 with 3.10 and the fact that for some constants c 1 , c 2 ∈ 0, ∞ , we conclude that 3.8 is true.

3.15
Now, let J : Y → X be the linear operator: Let L, N : X → X be the operators: respectively.Then Lemma 3.1 yields that L : X → X is strongly positive.Moreover, 16, Theorem 7.c implies r L > 0. Now, it follows from Theorem 2.3 that there exists a continuum Proof .Let L i : X → X be the operator

3.23
Then Lemma 3.1 yields that L i : X → X is strongly positive.By Krein-Rutman theorem 16, Theorem 7.c the spectral radius r L i > 0 and there exist ψ i ∈ X with ψ i > 0 on Ì 2 such that

3.24
That is, the eigenvalue problems 3.22 have the principal eigenvalues λ i 1/r L i , and

3.25
Therefore, Suppose that Ì a {a 1, a 2, . . ., b − 1} is a strict subset of Ì 2 and h a denote the restriction of h on Ì a .Consider the linear eigenvalue problems:

3.26
Then we get the following result.
Proof.It is not difficult to prove that 3.26 has only one principal eigenvalue λ a > 0 by Lemma 3.1, and the corresponding eigenfunction ψ a > 0 on Ì a .So we only to verify that 0 < λ 1 < λ a .Let ψ 1 be the corresponding eigenfunction of λ 1 , we have that h t ψ a t ψ 1 t .
Proof of Theorem 1.3.We divide the proof into three steps.Let { μ n , y n } ⊂ C be such that

3.30
Step 1.We show that there exists a constant M such that μ n ∈ 0, M for all n.Suppose on the contrary that lim n → ∞ μ n ∞.

3.31
Let v n y n / y n X .Then it follows from 3.30 that there exists a constant M 0 > 0, such that f y n t y n t > M 0 > 0.

3.34
Let λ * be the principal eigenvalue of the linear eigenvalue problems:

3.37
First, we show that

3.38
Suppose on the contrary that for some M 1 > 0 independent on n .Then it follows from 3.30 and 0 and subsequently, { y n X } is bounded.This is a contradiction.So, 3.38 holds.
Next, we show that

3.41
In fact,

3.42
This together with 2.7 imply that 3.41 is valid.
Finally, we have from the facts that min{y

3.43
Consider the following linear eigenvalue problems:

3.44
By Lemma 3.3 and 3.32 , 3.44 has a positive principal eigenvalue λ a , and μ n f y n t y n t ≤ λ a , 3.45 which contradicts 3.43 .Thus lim n → ∞ μ n 0.
Step 3. Fixed λ such that Then there exists b ∈ 0, B such that We show that there is no μ, u ∈ C such that In fact, if there exists η, y ∈ C satisfying 3.48 , then y t η Proof of Theorem 1.4.We divide the proof into three steps.

3.50
Suppose on the contrary that there exists { μ n , y n } ⊂ C such that μ n → 0 , as n → ∞.

3.52
However, this contradicts with the fact that v n X 1 for all n ∈ AE.Therefore, 3.50 holds.
Step 2. We show that for any closed interval I ⊂ β, ∞ , there exists M I > 0 such that

3.53
Suppose on the contrary that there exists { μ n , y n } ⊂ C with

3.56
This together with 3.32 and f 0 ∈ 0, ∞ and f ∞ 0 that

3.57
However, this contradicts with the fact that min t∈Ì 2 v n t ≥ c 1 , n ∈ AE.
Step 3. Fixed λ such that Then there exists l ∈ 0, c 1 B such that λ > l γ * c 1 f l .

3.60
We show that there is no η, y ∈ C such that Suppose on the contrary that there exists η, y ∈ C satisfying 3.61 .Then for t ∈ Ì 2 , and subsequently, η ≤ l/c 1 γ * f l .Therefore, there is no η, y ∈ C such that 3.61 holds.Now, combining the conclusions in Steps 2 and 3, using the fact that no μ, u ∈ C satisfying 3.61 , it concludes that for every λ ∈ l/c 1 γ * f l , λ 1 /f 0 , 1.4 , 1.8 has at least two generalized positive solutions in C .For arbitrary λ ∈ inf s∈ 0,c 1 B s/c 1 γ * f s , ∞ , we may find l l λ satisfying 3.60 .So, for every λ ∈ inf s∈ 0,c 1 B s/γ * f s , λ 1 /f 0 , 1.4 , 1.8 has at least two generalized positive solutions in C .

Some Examples
In this section, we will apply our results to two examples.