Barrier Solutions of Elliptic Differential Equations in Musielak-Orlicz-Sobolev Spaces

Dong, Ge; Fang, Xiaochun

doi:https://doi.org/10.1155/2021/9927898

Journal of Function Spaces

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Research Article | Open Access

Volume 2021 | Article ID 9927898 | https://doi.org/10.1155/2021/9927898

Barrier Solutions of Elliptic Differential Equations in Musielak-Orlicz-Sobolev Spaces

Ge Dong¹and Xiaochun Fang^2,3

Academic Editor: Pasquale Vetro

Received27 Mar 2021

Accepted21 Aug 2021

Published07 Oct 2021

Abstract

In this paper, we study the solution set of the following Dirichlet boundary equation: in Musielak-Orlicz-Sobolev spaces, where , , and are all Carathéodory functions. Both and depend on the solution and its gradient . By using a linear functional analysis method, we provide sufficient conditions which ensure that the solution set of the equation is nonempty, and it possesses a greatest element and a smallest element with respect to the ordering “≤,” which are called barrier solutions.

1. Introduction

Let be a bounded domain with Lipschitz boundary. Fan [1] established a sub-supersolution method for the following differential equations of divergence form: in coupled with Neumann or Dirichlet boundary condition in reflexive Musielak-Orlicz-Sobolev spaces and obtained the existence of greatest and smallest solutions within the order interval , where , , and and are subsolutions and supersolutions, respectively. Dong and Fang [2] studied the existence of weak solutions for the following equations: in coupled with Neumann or Dirichlet boundary condition in separable Musielak-Orlicz-Sobolev spaces and obtained the existence of greatest and smallest solutions within the order interval . Carl and Le [3] established the enclosure of solutions, i.e., , for multivalued elliptic variational inequalities by using the sub-supersolution method in Sobolev spaces. However, Carl [4] pointed out that there may exist some more solutions of some elliptic equations which are not within the corresponding interval . For example, the solution set of the following Dirichlet boundary value problem is given by , where is the eigenfunction belonging to , and is the first eigenvalue of the negative Laplacian (see [4]). Then, neither is norm bounded nor has the extreme solutions. Therefore, Carl [4] provided some sufficient conditions which ensure that the solution set of an elliptic variational inequality possesses a greatest element and a smallest element with respect to the natural partial ordering of functions in Sobolev spaces. Here, and are called barrier solutions (see [4]). In [5], Chipot et al. established the existence of the minimal solution to some elliptic variational inequalities in Sobolev spaces.

It is well known that different classes of differential equations correspond to different function space settings. The Musielak-Orlicz-Sobolev spaces are the extensions of Sobolev spaces, variable exponent Sobolev spaces, and Orlicz-Sobolev spaces. Gwiazda et al. [6] proved the existence of renormalized solutions to the following nonlinear elliptic equations: in coupled with Dirichlet boundary condition in Musielak-Orlicz-Sobolev spaces, where is a Carathéodory function and . The growth of is controlled by Musielak-Orlicz function : for all and , where is a complementary function to , the constant . Although they did not require nor , condition (5) is stronger than the growth condition (3.1) and the coercive condition (3.2) in [2]. In [7], Chlebicka and Karppinen studied removable sets for elliptic equations of the form in , where satisfies the nonstandard growth condition and the coercivity condition expressed by means of a Musielak-Orlicz function.

The purpose of this paper is to provide sufficient conditions which ensure that the solution set of the following Dirichlet boundary problem is nonempty and possesses barrier solutions in Musielak-Orlicz-Sobolev spaces, where , , and are Carathéodory functions. Both and depend on the solution and its gradient .

This paper is organized as follows: Section 2 contains some preliminaries and some technical lemmas which will be needed. In Section 3, we first give the sufficient conditions which ensure the existence of weak solutions for (7) in Musielak-Orlicz-Sobolev spaces. Some examples are given for the application of our results. We point that (5) is strong. Then, we provide some sufficient conditions which ensure that the solution set of (7) is bounded, compact, and direct. Finally, we provide some sufficient conditions which ensure that the solution set of (7) possesses barrier solutions in Musielak-Orlicz-Sobolev spaces.

We refer to some results in variable exponent Sobolev or Orlicz-Sobolev spaces [8–22].

2. Preliminaries

Now, we list briefly some definitions and facts about Musielak-Orlicz-Sobolev spaces; for more details, see [1, 2, 23–25].

is called a generalized -function (i.e., a Musielak-Orlicz function), denoted by , if it satisfies the following conditions: (i) is an -function of the variable for every , i.e., it is a convex, nondecreasing, and continuous function of such that , for , and there hold the conditions(ii) is a measurable function of for all

Equivalently, for all and all , admits the representation , where is the right-hand derivative of at , for a fixed and all . Then, for every , is a right-continuous and nondecreasing function of , , for , and as for every . The complementary function to is defined as follows:

Then, and is also the complementary function to . Equivalently, admits the representation , where is given by , for all .

Moreover, one has the Young inequality: , for all .

is said to satisfy the condition (, for short), if there exists a positive constant and a nonnegative function such that , for all and a.e. .

is called locally integrable, if for every .

The following assumptions will be used.

() .

() For every , there exists a such that for all , and all .

Clearly, (10) implies ().

Let . We denote by the set of all (equivalence classes of) Lebesgue measurable functions from to . The Musielak-Orlicz space (i.e., the generalized Orlicz space) is defined by with the (Luxemburg) norm

Moreover, the set will be called the Musielak-Orlicz class (i.e., the generalized Orlicz class). A function will be called a finite element of , if for every . The space of all finite elements of will be denoted by . Then, is a convex subset of , is the smallest vector subspace of containing , and is the largest vector subspace of contained in .

If is locally integrable, then is a separable space, and if and only if .

If is locally integrable and satisfy (10), then . Moreover, if is also locally integrable satisfying (11), and , then is reflexive.

The Musielak-Orlicz-Sobolev space is defined by where with nonnegative integers , , and denote the distributional derivatives.

Let for . Then, is a norm on . The pair is a Banach space if is locally integrable and satisfies ().

If is independent of , then becomes the Orlicz-Sobolev space. Let , where is continuous, . If , then becomes the variable exponent Sobolev space where . is equipped with the Luxemburg norm and is equipped with the norm

It is easy to see that

Denote and . Clearly, is equivalent to .

The space can be identified with a subspace of the product of copies of . Denote the product by . The subspace is closed. Let be the closure of the Schwartz space in .

The space can be identified with a subspace of the product of copies of . Denote the product by . Let be the (norm) closure of in .

Lemma 1 (see [23], Lemma 14.4). Let , be a complementary function to , and be the inverse functions to , respectively. Then,

By Lemma 1, we can deduce the following lemma.

Lemma 2. Let , be a complementary function to and be the inverse functions to , respectively. Then,

Lemma 3 (see [26], Lemma 2.1). If , then , and

Here, , .

3. Existence Theorems

Let be a bounded domain with Lipschitz boundary. Let , be a complementary function to with , and be locally integrable. Suppose that the embedding is compact.

For , , we use the standard notations: , , , , for a.e. . According to [1] (Remark 3.1), , for any . Clearly, , if .

Let be a Carathéodory function satisfying the following conditions:

() For a.e. , all and all , where , , , , and .

Let be a Carathéodory function satisfying the following conditions:

() For a.e. and all , where , , , , and .

Let be a Carathéodory function satisfying the following growth condition:

() For a.e. , all , and all , where , . Denote by the Nemytskii operator associated to , that is,

Consider the following Dirichlet boundary value problem:

A function is called a weak solution of (31) if , , and satisfies the equation

A function is called a subsolution (resp., supersolution) of (31) if , , and (32) holds with “” replaced by “” (resp., “”) for every nonnegative functions in (see [1]).

Theorem 4. Let , , and hold. If then there exists at least one weak solution of (31).

Proof. Denote . Define , . Then, is well defined.
Since is separable, there is a sequence such that is dense in . Let and consider .
Similar to the proof of Theorem 3.1 in [27], is continuous.

In view of (29), for every , we have for some constant .

Combining (25), (28), (33), and (35), we obtain for all , and some constant independent of .

Consequently, for all , and some constant . Using ([2], Lemma 1), we have

By [27] (Remark 2.1), there exists a Galerkin solution for every such that

Using the density of , we deduce that

Therefore, the sequence is bounded in . Consequently, there exists , and we can extract a subsequence of such that as . By (24), is bounded in , and and are bounded in . Thanks to (42), as . It follows that as .

Therefore, similar to [27], we can deduce that there exists a subsequence of still denoted by such that as . Combining (43) and (46), we obtain that as .

Using [25] (Lemma 1) and letting , passing to a subsequence if necessary, we can deduce from (40) that

By [28] (Lemma 2.5) and [25] (Lemma 7),

Thus, there exists at least one weak solution of (31).

Example 1. Let , for and , where is a measurable function such that . Then, .

Example 2. Put if and if , for a.e. , all , where is a Carathéodory function on satisfying with and for all , is the right-hand derivative of at , and is the zero vector in . Let , and , for a.e. , all and all , where is a Carathéodory function on satisfying with . Then, , , and satisfy conditions (), (), and (29), respectively.

Remark 5. Clearly, condition (5) implies that for all and . By Lemma 2, we have

It implies that

Consequently,

Hence, condition (5) is stronger than not only the growth condition (3.1) but also the coercive condition (3.2) in [2].

Let be the set of all weak solutions of (31). Under the assumptions of Theorem 4, the solution set of (31) is nonempty.

Theorem 6. Assume that the assumptions of Theorem 4 hold. Then is bounded and closed in . Moreover, is compact in .

Proof. Denote . For , define , and , then is a norm of equivalent to (see [1]). Taking in the first inequality of (36) and by [2] (Proposition 3.1), there exists , such that, for every , for all , as is large enough, where the constants and are independent of . Consequently, we can deduce that is bounded in .
Next, we prove that is closed. Indeed, assume , and in as . Since is bounded, is bounded. Therefore, by passing to a subsequence if necessary, we can assume without loss of generality that as . Similar to the proof of Theorem 4, we can obtain that is a weak solution of (31), i.e., . Consequently, is closed in .

For any subsequence in , is bounded. Repeating the proof of Theorem 4, by passing to a subsequence if necessary, we can obtain that is convergent to , and is a weak solution of (31), i.e., . Consequently, is sequentially compact. Since is closed, is compact.

Definition 7 ([4], Definition 2.1). Let (, ) be a partially ordered set. A subset of is said to be upward directed if for each pair there is a such that and . Similarly, is downward directed if for each pair there is such that and . If is both upward and downward directed, it is called directed.

If there exist such that , and , for any , then and are called barrier solutions of (31) (see [4]). Apparently, and are the least element and the greatest element with respect to the ordering “,” respectively.

To show that is directed with respect to the ordering “,” the following modulus of continuity condition on the coefficient is required:

() For a.e. , all and all , where , , and is a continuous function such that for every .

Theorem 8. Assume that the assumptions of Theorem 4 and () hold. Let . Then and are the subsolution and supersolution of (31), respectively.

Proof. Denote . Let . In view of assumption (), for any , there exists such that Define the function , which is defined by Then, for every , the function is continuous, piecewise differentiable, and the derivative is nonnegative and bounded. Consequently, is Lipschitz continuous and nondecreasing, and it satisfies as . Moreover, if , and if (see [29]).

Let . Replacing with and taking the test function in (32), respectively, and adding the resulting equations, we obtain

By applying (26), (), and the properties of , we have

The last term on the right-hand side of (60) tends to , as . Using Lebesgue’s dominated convergence theorem, we obtain that as .

Combining (59)–(63), we get that which yields that is a subsolution of (31), i.e., . Analogously, we can deduce that is a supersolution of (31).

Theorem 9. Assume that the assumptions of Theorem 4 and () hold. Then, is directed downward and upward; that is, for all , there exist such that and .

Proof. Denote . For , let . We will prove that there exists such that .
Define the cutoff function by for , . Then for any , .

Denote . For any , by Young inequality, one has

Define . By Remark 3.1 in [1], is continuous. According to the definition of , is bounded.

Thanks to (29), and by Young inequality, we have for all , and some constant independent of . From (27) and (28), we can deduce that for all , and some constant independent of .

Let us consider the following auxiliary equation of finding such that

Define ,

. Then, is well defined.

Combining (25), (28), (29), (33), (66), (67), and (68), we have

, where are two constants independent of .

Similar to the proof of Theorem 4, we can deduce that there exists such that i.e., is a weak solution of (69).

Denote . Replacing with in (69), and taking , we can obtain that

By Theorem 8, we know for any with .

Taking in (74), we have

Subtracting (73) from (75), we obtain that

Thanks to (26), the first term of the left-hand side of (76) is nonnegative. Clearly, both the second term of the left-hand side and the right-hand side of (76) equal 0. Indeed,

Therefore,

On the other hand,

It follows that .

From the definitions of , , we deduce that for a.e. . We note that (69) reduces to (32). Hence, is a weak solution of (31). Consequently, , and .

Similarly, there exists , such that . Therefore, is a directed set.

Similar to step (c) of the proof of Theorem 3.1 in [4], we have the following result.

Theorem 10. Assume that the assumptions of Theorem 4 and () hold. Let . Then (31) admits barrier solutions, i.e., there exist such that , .

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

The first author was supported by the 100 Teachers Database Project of Shanghai University of Medicine and Health Sciences (B30200203110084) and the Mandatory Plan Project of SUMHS (CZ401). The second author was supported by the National Natural Science Foundation of China (11871375).

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Copyright © 2021 Ge Dong and Xiaochun Fang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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