Research Article | Open Access
Jianye Xia, Yuji Liu, "Solvability of a Class of Generalized Neumann Boundary Value Problems for Second-Order Nonlinear Difference Equations", Discrete Dynamics in Nature and Society, vol. 2011, Article ID 308362, 11 pages, 2011. https://doi.org/10.1155/2011/308362
Solvability of a Class of Generalized Neumann Boundary Value Problems for Second-Order Nonlinear Difference Equations
This paper is motivated by Rachnkovab and Tisdell (2006) and Anderson et al. (2007). New sufficient conditions for the existence of at least one solution of the generalized Neumann boundary value problems for second order nonlinear difference equations , , , , are established.
Recently, there have been many papers discussed the solvability of two-point or multipoint boundary value problems for second-order or higher-order difference equations, we refer the readers to the text books [1, 2] and papers [3–8] and the references therein.
In a recent paper , Anderson et al. studied the following problem: where The following result was proved.
Suppose that is continuous and there exist constants , such that Then BVP(1.1) has at least one solution.
The methods in  involved new inequalities on the right-hand side of the difference equation and Schaefer's Theorem in the finite-dimensional space setting.
In , the following discrete boundary value problem (BVP) involving second order difference equations and two-point boundary conditions was studied, where an integer, is continuous, scalar-valued function, the step size is with a positive constant, the grid points are for . The differences are given by
The following two results were proved in .
Let be continuous on and , and, be nonnegative constants. If there exist such that then the discrete BVP(1.3) has at least one solution.
Let be continuous on and , and nonnegative constants. If then the discrete BVP(1.3) has at least one solution.
In paper , Cabada and Otero-Espinar studied the existence of solutions of a class of nonlinear second-order difference problem with Neumann boundary conditions by using upper and lower solution methods. Assuming the existence of a pair of ordered lower and upper solutions and , they obtained optimal existence results for the case and even for .
In this paper, we study the following boundary value problem for second-order nonlinear difference equation where , an integer, and is continuous, scalar-valued function. We note that when , BVP(1.8) becomes the following BVP: which is called Neumann boundary value problem of difference equation and is a special case of BVP(1.1). When , BVP(1.8) is changed to which is the so-called Dirichlet problem for discrete difference equations and is a special case of BVP(1.3).
The purpose of this paper is to improve the assumptions (*), (1.5), and (1.6) in the results in paper [3, 5, 7–9], by using Mawhins continuation theorem of coincidence degree, to establish sufficient conditions for the existence of at least one solution of BVP(1.8). It is interesting that we allow to be sublinear, at most linear or superlinear.
This paper is organized as follows. In Section 2, we make the main results, and in Section 3, we give some examples, which cannot be solved by theorems in [5, 7, 9], to illustrate the main results presented in Section 3.
2. Main Results
To get the existence results for solutions of BVP(1.8), we need the following fixed point theorems.
Let and be Banach spaces, a Fredholm operator of index zero, and , projectors such that . It follows that is invertible; we denote the inverse of that map by .
If is an open bounded subset of , , the map will be called compact on if is bounded and is compact.
Lemma 2.1 . (see ). Let be a Fredholm operator of index zero, and let be compact on . Assume that the following conditions are satisfied:
(i) for every ;
(ii) for every ;
(iii), where is the isomorphism.
Then the equation has at least one solution in .
Lemma 2.2 . (see ). Let and be Banach spaces. Suppose is a Fredholm operator of index zero with , is compact on any open bounded subset of . If is an open bounded subset and , then there is at least one so that .
Let be endowed with the norms for and , respectively. It is easy to see that and are Banach spaces. Choose . Let , and by .
Consider the following problem: It is easy to see that problem (2.2) has a unique solution if and only if If (2.3) holds, we call BVP(1.8) at nonresonance case. If , then problem (2.2) has infinite nontrivial solutions. At this case, we call BVP(1.8) at resonance case. In this paper, we establish sufficient conditions for the existence of solutions of BVP(1.8) at nonresonance case, that is, , and at resonance case, . It is similar to get existence results for the existence of solutions at resonance case when and .
Lemma 2.3. Suppose . Then the following results are valid.
(iii) is a Fredholm operator of index zero.
(iv) There are projectors and such that ,. Furthermore, let be an open bounded subset with ; then is compact on .
(v) is a solution of which implies that is a solution of BVP(1.8).
The projectors and , the isomorphism , and the generalized inverse are as follows:
Lemma 2.4. Suppose . Then the following results are valid.
(i) is a solution of which implies that is a solution of BVP(1.8).
(iii) is a Fredholm operator of index zero, is compact on each open bounded subset of .
Suppose(A)there exist numbers , , nonnegative sequences , functions , such that and for all ;(B)there exists a constant so that for all or for all .
Suppose , and that and hold. Then BVP(1.8) has at least one solution if
Proof. To apply Lemma 2.1, we consider for .
Step 1. Let . For , we have So It is easy to see that Since , we get So, we get Then It follows that For , Holders inequality implies It follows that It follows from (2.8) that there exists a constant such that Hence for all . Hence . So is bounded.
Step 2. Prove that the set is bounded.
For , we have for . Thus we have . implies that It follows from condition that . Thus is bounded.
Step 3. Prove the set is bounded.
If the first inequality of holds, let
We will prove that is bounded. For for such that , and , we have
If , then . If , then a contradiction.
If the second inequality of holds, let Similarly, we can get a contradiction. So is bounded.
Step 4. Obtain open bounded set such (i), (ii), and (iii) of Lemma 2.1.
In the following, we will show that all conditions of Lemma 2.1 are satisfied. Set an open bounded subset of such that . We know that is a Fredholm operator of index zero and is -compact on . By the definition of , we have and , thus for and ; for .
In fact, let . According the definition of , we know , thus for , thus by homotopy property of degree, Thus by Lemma 2.1, has at least one solution in , which is a solution of BVP(1.8). The proof is completed.
Proof. To apply Lemma 2.2, we consider for . Let . For , we get (2.9) and (2.10). using the methods in the proof of Theorem LX1, we get that is bounded. Let be a nonempty open bounded subset of such that centered at zero. It is easy to see that is a Fredholm operator of index zero and is L-compact on . One can see that . Thus, from Lemma 2.2, has at least one solution , so is a solution of BVP(1.8). The proof is complete.
3. An Example
In this section, we present an example to illustrate the main results in Section 2.
Example 3.1. Consider the following problem: where are integers and , , , are sequences. Corresponding to the assumptions of Theorem L1, we set and . It is easy to see that holds, and implies that there is such that for all and .
This paper is supported by Natural Science Foundation of Hunan province, China (no. 06JJ5008) and Natural Science Foundation of Guangdong province (no. 7004569).
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