Existence of Positive Solutions of a Discrete Elastic Beam Equation

Let T be an integer with T ≥ 5 and let T2 {2, 3, . . . , T}. We consider the existence of positive solutions of the nonlinear boundary value problems of fourth-order difference equations Δ4u t − 2 − ra t f u t 0, t ∈ T2, u 1 u T 1 Δ2u 0 Δ2u T 0, where r is a constant, a : T2 → 0,∞ , and f : 0,∞ → 0,∞ is continuous. Our approaches are based on the KreinRutman theorem and the global bifurcation theorem.


Introduction
An elastic beam in an equilibrium state whose both ends are simply supported can be described by the fourth-order boundary value problem of the form Recently, the existence of solutions of boundary value problems BVPs of difference equations has received much attention; see Agarwal and Wong 13 , Henderson 14 , He and Yu 15 , Zhang et al. 16 , and the references therein.However, relatively little is known about the existence of positive solutions of fourth-order discrete boundary value problems.To our best knowledge, only He and Yu 15 and Zhang et al. 16 dealt with that.In 15 , He and Yu studied the existence of positive solutions of the nonlinear fourth-order discrete boundary value problem where T ≥ 3 is an integer, T 2 : {2, . . ., T}, r is a parameter, a : 0, ∞ satisfies some growth conditions which are not optimal! .The likely reason is that the spectrum structure of the linear eigenvalue problem is not clear.It has been pointed out in 15, 16 that 1.3 and 1.4 are equivalent to the integral equation of the form where and other results on the existence of positive solutions of 1.3 and 1.4 can be found in the two papers.Notice that in the integral 1.7 , two distinct Green's functions, G and G 1 , are used.This makes the construction of cones and the verification of strong positivity of A 0 more complex and difficult.Therefore, we think that the boundary condition 1.4 is not very suitable for the study of the positive solutions of fourth order difference equations.It is the purpose of this paper to assume the fourth-order difference equation 1.3 subject to a new boundary condition of the form 1.9 This will make our approaches much more simple and natural, and only one Green's function is needed.However, the classical definitions of positive solutions are useless for 1.3 and 1.9 any more.We have to adopt the following new definition of positive solutions.A function y : T 0 → R is called a positive solution of 1.3 and 1.9 if y satisfies 1.3 , 1.9 , and y t ≥ 0 on T 2 and y t / ≡ 0 on T 2 .
Remark 1.2.Notice that the fact y : T 0 → R is a positive solution of 1.3 and 1.9 does not mean that y t ≥ 0 on T 0 .In fact, y satisfies Remark 1.3.In 17 , Eleo and Henderson defined a kind of positive solutions which actually are sign-change solutions.In Definition 1.1, a positive solution may allow to take nonpositive value at t 0 and t T 2. We think it is useful in this case that T is large enough.
In the rest of the paper, we will use global bifurcation technique; see Dancer 18, Theorem 2 or Ma and Xu 19, Lemma 2.1 , to deal with 1.3 and 1.9 .To do this, we have to study the spectrum properties of 1.5 and 1.9 .This will be done in Section 2. Finally, in Section 3, we will state and prove our main result.

Eigenvalues
Recall Then, dim E T 3, and E is a Banach space with the norm Then Y is a Banach space with the norm Let Then the operator χ is a homomorphism.
Definition 2.1.We say that λ is an eigenvalue of linear problem 8 has a nontrivial solution.
In the rest of this section, we will prove the existence of the first eigenvalue of 2.8 .
Theorem 2.2.Equation 2.8 has an algebraically simple eigenvalue λ 1 , with an eigenfunction ϕ satisfying Moreover, there is no other eigenvalue whose eigenfunction is nonnegative on T 2 .
To prove Theorem 2.2, we need several preliminary results.

Lemma 2.3. For each h h 2 , . . . , h T , the linear problem
has a unique solution where

2.11
Proof.Let Δ 2 u t − 2 w t − 1 for t ∈ T 2 .Then 2.9 is equivalent to the system

2.12
From Kelly and Peterson 20, Theorem 6.8 and Example 6.12 , it follows that

2.15
Notice that From the assumption T ≥ 5, we have where γ ∈ 0, ∞ .Then Proof.From 2.18 , we get

2.21
Combining this with the boundary conditions

2.24
For any x, y ∈ X, we have from the definition of

2.25
It follows that

2.26
Thus, x y ∈ X, and moreover, x y X ≤ x X y X .

2.27
Therefore, • X is a norm of X, and X, • X is a normed linear space.Since dim X T − 1, X, • X is actually a Banach space.Let

2.28
Then the cone P is normal and has nonempty interior int P .

2.30
Proof.i From the relation

2.32
ii By 2.14 and the fact that e t > 0 on T 2 , it follows that there exists γ > 0, such that

2.33
Let H s, j a j u j , t ∈ T 1 , 2.36

2.37
Then 2.8 can be written as Since X is finite dimensional, we have that K : X → Y is compact.Obviously, J • K P ⊆ P .Next, we show that J • K : P → P is strongly positive.Since a t is positive on T 2 , there exists a constant k > 0 such that a t > k on T 2 .
For u ∈ P \ {0}, we have that H s, j u j > 0, s ∈ T 1 .

2.39
It follows that there exists k 1 > 0 such that H s, j a j u j ≥ k 1 e s .

2.40
Also, for u ∈ P \ {0}, we have from the fact J • Ku ∈ P and Ku / 0 in T 2 that for some constant k 2 > 0. By 2.39 and 2.41 , we get H s, j a j u j ≤ k 2 e s , s ∈ T 1 .

The Main Result
In this section, we will make the following assumptions: H2 f : 0, ∞ → 0, ∞ is continuous and f s > 0 for s > 0; H3 f 0 , f ∞ ∈ 0, ∞ , where It is not difficult to see that H2 and H3 imply that there exists a constant Theorem 3.2.Let H1 , H2 , and H3 hold.Assume that either Then 1.3 and 1.9 have at least one positive solution.
Remark 3.3.Recently, Ma and Xu 19 considered the nonlinear fourth-order problem under some conditions involved the generalized eigenvalues of the linear problem
Remark 3.4.The first eigenvalue μ 1 of the linear problem is μ 1 4sin 2 π/2T , and the first eigenvalue λ 1 of the linear problem It is easy to check that the function
Remark 3.5.The condition 3.3 or 3.4 is optimal since for any > 0, the linear problem has the unique solution u t ≡ 0. In fact, λ 1 is the least eigenvalue of the linear problem To prove Theorem 3.2, we define

Let us consider
Lu λra t f 0 u λra t ζ u 0, λ > 0, 3.20 as a bifurcation problem for the trivial solution u ≡ 0. It is easy to check that 3.20 can be converted to the equivalent equation

3.21
From the proof process of Theorem 2.2, we have that for each fixed λ > 0, the operator Proof of Theorem 3.2.It is clear that any solution of 3.20 of the form 1, u yields a solutions u of 1.3 and 1.9 .We will show that C crosses the hyperplane {1} × X in R × X.To do this, it is enough to show that C joins λ

3.26
We note that μ n > 0 for all n ∈ N, since 0, 0 is the only solution of 3.20 for λ 0 and C ∩ {0} × X ∅.

3.27
We divide the proof into two steps.
Step 2. We show that there exists a constant M such that μ n ∈ 0, M for all n.By 9, Lemma 2.1 , we only need to show that A has a linear minorant V and there exists a μ, y ∈ 0, ∞ × P such that y X 1 and μV y ≥ y.

3.35
For u ∈ X, let

2 . 2 Discrete
The existence of solutions of 1.3 and 1.4 has been extensively studied; see Gupta 1, 2 , Aftabizadeh 3 , Yang 4 , Del Pino and Manásevich 5 , Galewski 6 , Yao 7 , and the references therein.The existence and multiplicity of positive solutions of the boundary value problem of ordinary differential equations y x − λf x, y x 0, x ∈ 0, 1 , also been studied by many authors; see Ma and Wang 8 , Ma 9 , Bai and Wang 10 , Chai 11 , Yao and Bai 12 for some references along this line.Dynamics in Nature and Society
Now, we have from a version ofDancer 18, Theorem 2 , see Ma 9, Lemma 2.1 for details, to conclude that there exists an unbounded connected subset C in the set ∞ .From 3.28 , we have that y X → ∞.We divide the equation Ly n μ n ra t f ∞ y n μ n ra t ξ y n 0 3.29 by y n X and set y n y n / y X .Since y n is bounded in X, choosing a subsequence and relabelling if necessary, we see that y n → y for some y ∈ X with y 1.Moreover, from 3.19 and the fact that ξ is nondecreasing, we have that lim where μ : lim n → ∞ μ n , choosing a subsequence and relabelling if necessary.Thus, Case 2 λ 1 /rf 0 < 1 < λ 1 /rf ∞ .In this case, if η n , y n ∈ C is such that lim u t : Again C joins λ 1 /rf 0 , 0 to λ 1 /rf ∞ , ∞ and the result follows.