Stability and Bifurcation Analysis of Discrete Dynamical Systems 2020View this Special Issue
Existence of Solutions for the Discrete Dirichlet Problem Involving -Mean Curvature Operator
This work is to discuss the Dirichlet boundary value problem of the difference equation with -mean curvature operator. Under some determinate growth conditions on the nonlinear term, the existence of one solution or two nontrivial solutions is obtained via variational methods and some analysis techniques. Examples are also given to illustrate our theorems in the last section.
Let , and stand for the sets of natural numbers, integer numbers, and real numbers, respectively. For , define and when .
Consider the following Dirichlet boundary value problem:where is a given positive integer, for each , is the forward difference operator defined by , and . Here, is counted as a discretization of -mean curvature operator.
We may regard problem (1) as being a discrete analog of the following differential problem in one-dimensional space:where , named -mean curvature operator , is a generalization of mean curvature operator (, page 1212). When , it reduces to mean curvature operator [3–6]. When , it reduces to Laplacian operator.
Difference equations arise in different research fields, for instance, computer science, mechanical engineering, control systems, artificial or biological neural networks, economics, and so on, see [7–12].
In order to research discrete boundary problems, different methods have been used: fixed point theorems, upper and lower solutions techniques, see [3, 5] and the references given therein. Starting from Guo and Yu , various results have been achieved via variational methods, see [14–32].
In 2011, Candito and Bisci  studied the boundary value problem:where , is continuous with respect to for every . Through variational methods, they proved that problem (3) admits two nontrivial solutions.
In 2015, Zhou and Su  discussed the following problem:where is positive integer, , and . By critical point theory and some analytical skills, two nontrivial solutions were acquired.
Motivated by the works of [15, 30, 34–43], we deal with problem (1). Compared to problem (4), problem (1) is much more difficult to handle because of -mean curvature operator. It needs more analytical skills. For the convenience of analysis, we have no option but to split the problem into two types: and .
This paper is arranged as follows. In Section 2, we establish an appropriate Banach space and a suitable functional corresponding to problem (1). In order to gain solutions of problem (1), some preliminary knowledge is introduced. In Section 3, under certain growth conditions on the antiderivative of the nonlinearity , one solution or two nontrivial solutions for problem (1) are obtained. In Section 4, we give two concrete examples to demonstrate the applicability of our main results.
2. Mathematical Background
For problem (1), we naturally select the - dimensional Banach space:equipped with the normfor all .
Its infinity norm is
Apparently, one has
Furthermore, we need to use the following inequality. When , one hassee (, page 28).
Puttingthe function is defined byfor every , where
It is not difficult to verify that , and the critical points of on are exactly the solutions of problem (1).
One can see thatwhere is a -dimensional row vector, stands for the transpose of , and
Matrix is symmetric and positive definite. It is easy to see that has distinct positive eigenvalues which are given by
We have the following lemma, see (, page 918).
Lemma 1. For every , one has
Let E be a real Banach space, the functional is said to satisfy the Palais–Smale (P. S. for short) condition if any sequence in E such that is bounded and contains a convergent subsequence.
Let denote the open ball in E of radius about 0, and let denote its boundary.
Our main tool is Mountain pass lemma, see (, Theorem 2.2).
Lemma 2. Let E be a real Banach space and satisfying the P. S. condition. Suppose and(i)There are constants such that (ii)There exists an such that Then, admits a critical value . In addition, can be characterized aswhere
3. Main Results and Their Proofs
Theorem 1. Let , for every , be continuous. Suppose that .
Then, is coercive, and problem (1) admits at least one solution.
Proof. It is obvious that .
Owing to , we choose and such that, for every and , . Since is continuous on , there exists some constant such that , for all . Consequently, we follow thatfor every.
Using (9) and (20), Hölder inequality, and Lemma 1, we follow thatNote thatand we have . So, is coercive in S. Hence, is bounded from below. Set . According to the coercivity of the functional again, there is such that . Thus, . Taking into account that is a finite dimensional space and is continuous, possesses a critical point satisfying . The proof is completed.
Theorem 2. Let , for every , be continuous. Suppose that
Then, is anticoercive, and problem (1) has at least one solution.
Theorem 3. Let , for every , be continuous. Suppose that
Then, is coercive, and problem (1) possesses at least one solution.
Theorem 4. Let , for every , be continuous. Suppose that
Then, is anticoercive, and problem (1) admits at least one solution.
Proof. In the same way, we give only the crucial steps of the proof.
By , we chooseand such thatfor every .
Given any , applying (9) and (36), Hölder inequality, and Lemma 1, we follow thatNote thatwe have
Corollary 1. Let , for every , be continuous, and satisfy one of the following conditions:
Then, problem (1) has at least one solution.
Proof. Here, we only prove that implies . The proof that implies refers to (, Corollary 2.1).
From , there exist and such thatfor each and .
Since is continuous with respect to , we also letThus, when , one followsWhen , we havePutTherefore,Hence, our argument holds, and the proof is complete.
Remark 2. We observe that there are functions that satisfy , but does not hold, see, such as example 1 in Section 4.
Similarly, we have the following result.
Corollary 2. Let , for every , be continuous, and satisfy one of the following conditions:
Then, problem (1) admits at least one solution.
The following results guarantee the existence of two nontrivial solutions to problem (1).
Theorem 5. Let , for every , be continuous, and holds. Additionally, assume thatThen, problem (1) has at least two nontrivial solutions.
Proof. Our aim is to apply Lemma 2. To this purpose, take and as in (11). Let . Using Theorem 1, it is evident that is anticoercive. Since the dimension of is finite, we can easily deduce that satisfies the P.S. condition. Next, we prove that verifies the mountain pass geometry.
From , there are and such thatfor every .
By (8), Lemma 1, and the inequality , for , we havefor with .
Writing and , then . Hence, satisfies condition (i) of Lemma 2. It is obvious that . Because , we may pick and with , which means satisfying condition (ii) of Lemma 2.
According to Lemma 2, has a critical value given by , whereSet be a critical point of and , then is a nonzero solution of problem (1). By Theorem 1, possesses a critical point with . Noting that , we have . If , then the proof is complete. If , we divide the following proof into two cases.
Case I. . In such a case, we may see that . Using again, there exists satisfying . Since and , there exists such that . This shows that is a critical point of . Hence, problem (1) admits two nontrivial solutions.
Case II. . Through the definition of , we show that for every . Considering that , there exists which is linearly independent with . For every , definethen . Thus, we have satisfying . Obviously, when , and is a sequence of pairwise distinct critical points of in . Consequently, problem (1) has infinitely many nontrivial solutions. The proof is finished.
Theorem 6. Let , for every , be continuous, and holds. Moreover, suppose that
Then, problem (1) admits at least two nontrivial solutions.
Proof. Let and as in (11). Owing to Theorem 2, is anticoercive. Noting that is a finite dimensional space, it is obvious that satisfies the P.S. condition. Now, we show that satisfies the mountain pass geometry.
By , we pick such thatfor each .
Hence, there exists , and we follow thatwhen
Thanks to (8), for with , we getFor with exploiting Lemma 1, we getBy Lemma 1 and the inequality , we follow thatfor with .
Let , then combining (56) and (58), we havewhen with .
Picking and , then for every , . The rest of the proof runs as the proof of Theorem 5.
Theorem 7. Let , for every , be continuous. Suppose that and are satisfied. Then, problem (1) has at least two nontrivial solutions.
Proof. Take and as in (11) and denote . Applying Theorem 3, we can derive that is anticoercive. Here, we only verify that satisfies condition (i) of Lemma 2.
Since , there exists such that .
Using Taylor’s formula, we have the following inequality:From (9) and the above inequality, we follow thatOwing to , there exist and such that (49) is verified.
Because of the fact thatfor the above , there is such thatwhenever .
Picking , we have for all with .
Through (49), (61), and (63), we obtainfor with .
Taking and , we follow that if . The proof is complete.
Applying the method analogous to that used above, we can draw the following conclusion.
Theorem 8. Let , for every , be continuous. Suppose that and are fulfilled. Then, problem (1) has at least two nontrivial solutions.
4.1. Example 1
Let , and be a function which is defined as follows:
Moreover, we show that condition does not hold taking into account thatfor every .
4.2. Example 2
When , consider the following problem:
Data sharing is not applicable to this article as no data sets were generated or analyzed during the current study.
Conflicts of Interest
The authors declare that they have no conflicts ofinterest.
Both the authors contributed equally to the writing of this paper, and they read and approved the final manuscript.
This work was supported by the National Natural Science Foundation of China (Grant no. 11971126) and the Program for Changjiang Scholars and Innovative Research Team in University (Grant no. IRT_16R16).
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