Let be an integer with and let . We consider the existence of positive solutions of the nonlinear boundary value problems of fourth-order difference equations , , , where is a constant, is continuous. Our approaches are based on the Krein-Rutman theorem and the global bifurcation theorem.

1. 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

see Gupta [1, 2]. 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

have 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.

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 is an integer, , is a parameter, , 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


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, and , are used. This makes the construction of cones and the verification of strong positivity of 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

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.

Definition 1.1. Denote A function is called a positive solution of (1.3) and (1.9) if satisfies (1.3), (1.9), and on and on .

Remark 1.2. Notice that the fact is a positive solution of (1.3) and (1.9) does not mean that on . In fact, satisfies(1) for ,(2),(3).

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 and . We think it is useful in this case that is large enough.

In the rest of the paper, we will use global bifurcation technique; see Dancer [18, Theorem ] or Ma and Xu [19, Lemma ], 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.

2. Eigenvalues



Then, , and is a Banach space with the norm


Then is a Banach space with the norm


Then the operator

is a homomorphism.

In this paper, we assume that


Definition 2.1. We say that is an eigenvalue of linear problem if (2.8) 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 , with an eigenfunction satisfying(i) on ;(ii); .Moreover, there is no other eigenvalue whose eigenfunction is nonnegative on .

To prove Theorem 2.2, we need several preliminary results.

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

Proof. Let for . Then (2.9) is equivalent to the system From Kelly and Peterson [20, Theorem and Example ], it follows that Therefore, (2.10) holds.



Notice that

From the assumption , we have

Lemma 2.4. Let satisfy and where . Then

Proof. From (2.18), we get This is Combining this with the boundary conditions , it concludes that


for some . Since , from Lemma 2.4 and (2.17), we may define

For any , we have from the definition of that

It follows that

Thus, , and moreover,

Therefore, is a norm of , and is a normed linear space. Since , is actually a Banach space. Let

Then the cone is normal and has nonempty interior .

Lemma 2.5. For , where

Proof. () From the relation
it follows that
() By (2.14) and the fact that on , it follows that there exists , such that
Let Then This implies , and accordingly .

Proof of Theorem 2.2. For , define a linear operator and by Then (2.8) can be written as Since is finite dimensional, we have that is compact. Obviously, .
Next, we show that is strongly positive.
Since is positive on , there exists a constant such that on .
For , we have that
It follows that there exists such that Also, for , we have from the fact and in that for some constant . By (2.39) and (2.41), we get Thus Since Using (2.43) and (2.44), it follows that Therefore, .
Now, by the Krein-Rutman theorem [21, Theorem ; 20, Theorem ], has an algebraically simple eigenvalue with an eigenvector . Moreover, there is no other eigenvalue with an eigenfunction in .

3. The Main Result

In this section, we will make the following assumptions:

(H2) is continuous and for ;

(H3) , where

Remark 3.1. It is not difficult to see that (H2) and (H3) imply that there exists a constant such that

Theorem 3.2. Let (H1), (H2), and (H3) hold. Assume that either or 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 Our main result, Theorem 3.2, needs ; see (H3). However, in [19, Theorem ], some weaker conditions of the form are used.

Remark 3.4. The first eigenvalue of the linear problem is , and the first eigenvalue of the linear problem is It is easy to check that the function Then, for each , the condition (3.3) holds.

Remark 3.5. The condition (3.3) or (3.4) is optimal since for any , the linear problem has the unique solution . In fact, is the least eigenvalue of the linear problem

To prove Theorem 3.2, we define by


It is easy to check that is compact.

Let be such that

Obviously, (H3) implies


Then is nondecreasing and

Let us consider

as a bifurcation problem for the trivial solution . It is easy to check that (3.20) can be converted to the equivalent equation

From the proof process of Theorem 2.2, we have that for each fixed , the operator ,

is compact and strongly positive. Define by

Then we have from (3.17) and Lemma 2.5 that

locally uniformly in . Now, we have from a version of Dancer [18, Theorem ], see Ma [9, Lemma ] for details, to conclude that there exists an unbounded connected subset in the set

such that .

Proof of Theorem 3.2. It is clear that any solution of (3.20) of the form yields a solutions of (1.3) and (1.9). We will show that crosses the hyperplane in . To do this, it is enough to show that joins to . Let satisfy We note that for all , since is the only solution of (3.20) for and .Case 1 (). In this case, we show that
We divide the proof into two steps.Step 1. We show that if there exists a constant number such that then joins to .
From (3.28), we have that . We divide the equation
by and set . Since is bounded in , choosing a subsequence and relabelling if necessary, we see that for some with . Moreover, from (3.19) and the fact that is nondecreasing, we have that since Thus, where , choosing a subsequence and relabelling if necessary. Thus, By Theorem 2.2, we have Thus, joins to .
Step 2. We show that there exists a constant such that for all .
By [9, Lemma ], we only need to show that has a linear minorant and there exists a such that and .
By Remark 3.1, there exist constants such that
For , let Then is a linear minorant of . Moreover, for some constant , independent of . So, Therefore, it follows [9, Lemma ] that
Case 2 (). In this case, if is such that then and, moreover, Assume that there exists , such that for all , Applying a similar argument to that used in Step 1 of Case 1, after taking a subsequence and relabelling if necessary, if follows that Again joins to and the result follows.


The authors are very grateful to the anonymous referees for their valuable suggestions. This paper is supported by the NSFC (no. 10671158), the NSF of Gansu Province (no. 3ZS051-A25-016), NWNUKJCXGC- 03-17, the Spring-sun program (no. Z2004-1-62033), SRFDP (no. 20060736001), and the SRF for ROCS, SEM (2006[]).