Abstract
In this paper, we consider the existence of solutions for the discrete mixed boundary value problems involving -Laplacian operator. By using critical points theory, we obtain the existence of at least two positive solutions for the boundary value problem under appropriate assumptions on the nonlinearity.
1. Introduction
In recent years, with the development of mechanical engineering, control system, computer science, and economics, the existence of solutions of difference equations has attracted wide attention (see [1–6]). For example, applying Ricceri variational principle to obtain the existence of multiple solutions [7–9], taking the invariant sets of descending flow to prove the existence of sign-changing solutions [10], making the linking theorem to get the existence and multiplicity of periodic solutions [11], and using critical point theory to obtain the existence of homoclinic solutions [12–15] and heteroclinic solutions [16].
As we know, the fixed-point method and upper and lower solution techniques are important tools to solve the existence of solutions for boundary value problems (see [17, 18]). But recently, it is more common to use critical point theory to study Dirichlet boundary value problems (see [19–23]). More result on difference equations by using critical point theories can be referred to [24–27].
In [28], D’Aguì et al. established the existence of at least two positive solutions for the following discrete Dirichlet boundary value problem:where for all .
Unlike this, D’Aguì et al. in [29] proved that there are at least two nonzero weak solutions for the following mixed boundary value problem:where with is an -Carathéodory function and λ is a real positive parameter.
As a discrete analogy of the abovementioned problem, we consider the existence of positive solutions for the following discrete mixed boundary value problem:where denote the discrete interval for any integers a and b with , N be a positive integer, is continuous in u for each , is the forward difference operator, is the r-Laplacian given by with , , for all , and λ is a positive parameter.
In this paper, under suitable assumptions on the nonlinearity f, we use the theory of two nonzero critical points (see [30]) to ensure that there are at least two nonzero solutions for problem . The two nonzero critical points theorem is an appropriate combination of local minimum theorem (see [31]) and classical Ambrosetti–Rabinowitz theorem (see [32]). An important hypothesis of mountain pass theorem is Palais–Smale condition. It satisfies the application of infinite dimensional space by requiring the condition that the nonlinear term is stronger than p-superlinearity at infinity. In order to obtain the existence of two nonzero solutions, we can assume the classical Ambrosetti–Rabinowitz condition and nonlinear algebraic condition (see (40) in Theorem 2) hold, that is, more widespread than the p-sublinearity at zero. Moreover, when we require that for all , we can use strong maximum principle to obtain the existence of positive solutions, which has been proved in Lemma 2.
Let a special case of our main result is stated as follows.
Theorem 1. Let be a continuous function such thatthen, for each , the problemadmits at least two positive solutions.
The structure of the article is as follows. In Section 2, some basic definitions and properties are given. In Section 3, we give the main results. Under suitable hypothesis, Lemma 1 is used to obtain that the problem possesses at least two positive solutions. Finally, some examples are given to illustrate our main results.
2. Preliminaries
In this section, we recall some definitions, notations, and properties. Consider the N-dimensional Banach space:and define the normand is another norm in S.
Proposition 1. The following inequality holds:
Proof. Let , then there exist such that .
SincethenIf thenthat is,If thenthat is,In summary, we havePutand consider the function for all bywhereIt is clear that and their Gâteaux derivatives at the point are given byfor all So, we haveHence, a critical point u of is a solution of problem .
Now, we recall a definition and a two nonzero critical points theorem for the reader’s convenience.
Definition 1. Let X be a real Banach space; we say that a Gâteaux differentiable function satisfies the -condition, if any sequence such that(i), as (ii) as has a convergent subsequence
Lemma 1. Let X be a real Banach space and such that Assume that there are and with such thatand for eachthe functional satisfies the -condition and it is unbounded from below.
Then, for each the functional admits at least two nonzero critical points such that
In order to obtain the positive solution of problem , we establish the following strong maximum principle.
Lemma 2. Fix such that eitherfor all Then, either in or
Proof. Let such thatIf , then it is easy know that in .
If , then by (23), we havethat is,Since is increasing in u, we haveBy the definition of we know thatBy combining (27) with (28), we get If , we have Otherwise, replacing by j, we know . Continuing in this way, we have . Similarly, we have . Thus, and
Now, putwhere .
Define and Standard arguments show that and the critical points of are precisely the solutions of the following problem:
Lemma 3. If for each , any nonzero critical point of the functional is a positive solution of problem
Proof. Since a critical point of is a solution of problem , the conclusion follows by the discrete maximum principle ([33], Proposition 1).
Next, we suppose that and for all and for all Putwe have the following result.
Lemma 4. If then satisfies -condition and it is unbounded from below for all
Proof. Let . We consider a sequence such that and as Let and for all We first prove that is bounded. On one hand, we havefor all So,On the other hand, we assume thatfor each , thenTherefore,for all which leads to as So, we have as It means that there exists an such that From (10) we know that for all
Next, we suppose that the sequence is unbounded, that is, is unbounded.
As , we know that there exists an such that . From the definition of , there is such that for all Furthermore, since is a continuous function, there exists a constant such that with Thus, for all and We can obtain thatfor all , where that is,Hence, for all such that we conclude thatSince we can get and this is absurd. Hence, satisfies -condition.
Let be such that is bounded and is unbounded. From the proof above we can see that is unbounded from below.
3. Main Results
The main results of this paper are as follows.
Theorem 2. Let be a continuous function satisfying for all If there are two constants c and d with such that
Then, for each withthe problem admits at least two positive solutions.
Proof. Put as in (18). It is clear that According to Lemma 3, we know that a nonzero critical point in S of the functional is precisely a positive solution of problem Next, we just need to prove condition (21) of Lemma 1.
We observe that from (40) and is nondegenerate. Fix , Lemma 4 ensures that satisfies -condition for all and it is unbounded from below. We let , that is, PutIf , it means that . According to (8), we obtainIf , we know , thenTo sum up, we know that for all Furthermore, we havefor all with . Hence,Now, let for all and Clearly, It is easy to account that thenConsequently, from (46), (47), and assumption (40), we can obtainMoreover, because and from (40), we obtainthat is mean that
Hence, the problem admits at least two positive solutions by Lemma 1 and Lemma 3 for all
Remark 1. If is a nonnegative function and there are two positive constants with such thatthen the result of Theorem 2 is also valid for each withThere are some consequences of Theorem 2 as follows.
Corollary 1. Let be a continuous function such that where for all . Put for all and
If there exists such thatthen the problem has at least two positive solutions for each with
Proof. Consider the function is given asso thatThen, the conclusion can be obtained by Theorem 2.
Corollary 2. Assume f be a continuous function with andfor all Put Then, for each the problem admits at least two positive solutions.
Proof. We know that from (57). Fix , and then there exists such thatFrom (56), we can also obtainand then there exists such that Therefore, Theorem 2 ensures the conclusion.
Remark 2. If is a nonnegative function for all As long as condition (56) holds for at least one , then Corollary 2 ensures that the solutions are obtained for each
Remark 3. When for all , Theorem 1 can be ensured by Corollary 2. Obviously, condition (4(a)) implies . Specially, if f is nonnegative, we only need condition (4(a)) to get the corresponding solution for each
Example 1. Let
PutThen,Let , then . So, is increasing in and decreasing in . Since and , there exists an unique such that . Thus, in increasing in and decreasing in . This means that . In fact, .
Similarly, put , we can show that there exists a unique such that In fact,
SincethenTherefore, for each the problemadmits at least two positive solutions.
Example 2. Let and f be a function as follows:From Remark 1, we can choose and . Easy calculation shows thatwhich satisfy condition (50). Thus, for each , the problemadmits at least two positive solutions.
Data Availability
No data were used to support the study.
Conflicts of Interest
The authors declare that there are no conflicts of interest regarding the publication of this paper.
Authors’ Contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
Acknowledgments
This work was supported by the National Natural Science Foundation of China (Grant no. 11971126) and Program for Changjiang Scholars and Innovative Research Team in University (Grant no. IRT16R16).