Research Article | Open Access
Existence of Positive Solutions of a Discrete Elastic Beam Equation
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.
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 , Yang , Del Pino and Manásevich , Galewski , Yao , and the references therein. The existence and multiplicity of positive solutions of the boundary value problem of ordinary differential equations
Recently, the existence of solutions of boundary value problems (BVPs) of difference equations has received much attention; see Agarwal and Wong , Henderson , He and Yu , Zhang et al. , 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  and Zhang et al.  dealt with that. In , 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
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.
Remark 1.3. In , 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.
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
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 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
Remark 3.3. Recently, Ma and Xu  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.
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
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).
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