Abstract

Based on truncation technique and priori estimates, we prove the existence and uniqueness of weak solution for a class of anisotropic nonlinear elliptic equations with variable exponent growth. Furthermore, we also obtain that the weak solution is locally bounded and regular; that is, the weak solution is and .

1. Introduction

Since the variable exponent spaces have reflected into a various range of applications such as non-Newtonian fluids, plasticity, image processing, and nonlinear elasticity [14], some authors began to study the various properties of variable exponent space and some nonlinear problems with variable exponent growth. Edmunds et al. [5], Fan and Zhao [6] obtained that the variable exponent space and are reflexive Banach spaces under suitable conditions on . Later, Edmunds and Rákosník [7], Fan et al. [8] proved some continuous and compact Sobolev embedding theorems for the variable exponent spaces . For the anisotropic variable exponents spaces, in 2008, Mihăilescu et al. [9] studied the eigenvalue problems for a class of anisotropic quasilinear elliptic equations with variable exponents. In 2011, Boureanu et al. [10] proved the existence of multiple solutions for a class of anisotropic elliptic equations with variable exponents. Recently, Stancu-Dumitru [11, 12] has studied the existence of nontrivial solutions for a class of nonhomogeneous anisotropic problem. In particular, Fan [13] established some embedding theorems for anisotropic variable exponent Sobolev spaces.

In this paper, we investigate the following anisotropic nonlinear elliptic equation: where () with Lipschitz conditions boundary, and () are Carathéodory functions, and satisfy suitable conditions, and and are Einstein Sum; that is, , . We usually use critical theory to obtain the existence of weak solutions. However, since the problem (1) has no variational structure, we cannot define the energy functional for the problem (1). Therefore, based on truncation technique and priori estimates, we prove the existence and uniqueness of weak solutions for the problem (1) in . Furthermore, we obtain that the weak solution for the problem (1) is locally bounded.

In particular, for the special case we obtain that the weak solution is . To our knowledge, the above two problems have not been deeply studied in the anisotropic variable exponent Sobolev spaces.

The paper is organized as follows. In Section 2, we recall some results on anisotropic variable exponent Sobolev spaces and state our main results. In Section 3, we prove the existence, uniqueness and locally bounded of weak solution for the problem (1). In Section 4, the regularity of weak solutions for the problem (2) is proved.

2. Preliminary and Main Results

This section is dedicated to a general overview on the and ; for a deeper treatment on these spaces, see [5, 7, 8, 13, 14].

Let Set for any , and we define the variable exponent Lebesgue space We define the norm of : From [6], we have the following:(1) is a Banach space;(2) is reflexive, if and only if ;(3)Hölder inequality.

For all , , where , is the conjugate space of .

Now, we recall some results on anisotropic variable exponent Sobolev space [13]; set Denote and define The anisotropic variable exponent Sobolev space is a Banach space with respect to the norm If , for , , then is reflexive. We define as the closure of with respect to the norm and is a reflexive Banach space (see [9]).

Let Hence, we have the following embedding theorem for .

Let be a bounded domain and .

Theorem 1 (see [13, Theorem 2.5]). (i) If and then . The embedding is compact.(ii) If for all , then there exists such that .
The embedding is also compact.

Theorem 2 (see [13, Theorem 2.6]). Let be a bounded domain and . Suppose that
Then one has where is a positive constant independent of .

Remark 3. From [13], we know that if , , for every , then , where is a positive constant independent of .
Assume that and are Carathéodory functions and satisfy the following:(A1) for a.e. , for all , , we have and , where ; (A2) for a.e. , , and , satisfies (A3) for a.e. , for all , , , where ;(A4) for a.e. , , for some , for all , , where ;(T1) for a.e. ,for all , , where with , are positive constants.
We enumerate the hypotheses concerning and ,(F1);(F2) for a.e. , , for some , , where ;(H1) for , ;(H2) for a.e. , , for some , , , where , , , are positive constants.

Remark 4. Now, we give a simple example. Let , , , , , , , , , . By a simple calculation, we obtain that satisfies (T1), satisfies (F2), and satisfies (H2), where .

Now, we define the weak solution of the problem (1). A function is a weak solution of the problem (1), if , where .

Theorem 5. Let () be a bounded open subset, for every , , assume , and , satisfies and , where . satisfies the hypotheses (A1), (A2), the hypotheses (F1) and (H1) hold, and satisfies (T1) and the following Lipschitz condition: where , . Then there exists a unique weak solution for the problem (1). Furthermore, .

Theorem 6. Let () be a bounded open subset. Suppose (A3), (A4) and (F2), (H2) hold, for , , and . If is a weak solution of the problem (2), then , for all .

3. The Proof of Theorem 5

We consider the following problem: where is the truncation at levels of . Due to [15], we obtain that there exists a weak solution for the problem (26).

Lemma 7. If , (A1), (A2), (T1), (F1), and (H1) hold. Assume is a weak solution of the problem (26); then one has

Proof. Let ; will be chosen later. There exist measurable subsets of and functions such that if , . For , , if , . Let , a.e. in , , in . For , if , .
Choose as test function of the problem (26) and fix . Using Young inequality, Hölder inequality, the embedding theorem of the , and the hypotheses (A1), (F1), and (H1), we have where is a constant, . By (T1), Young inequality and Hlder inequality, we obtain, for , where , , .
Combining (28) with (29), we have
Let satisfy ; then we have where is a constant. Let ; we obtain Now, we recall the following classical inequality, for are positive numbers, From (33), we have Combining (32) with (34), there exists a constant such that Furthermore, put (35) in (31) and iterate on ; we have Therefore, we obtain for some constant .

Lemma 8. If , and satisfies the hypotheses (A1), (A2), the hypotheses (T1), (F1), and (H1) hold. Then there exists at least a weak solution for the problem (1).

Proof. By (27), we obtain that there exists a sequence which is bounded in ; we have for , weakly in and strongly in . We argue as in [16] to prove
From (38), we obtain By the hypotheses (A1) and (T1), for and , we have By Vitali Theorem, we have for any . Hence, we complete the proof of Lemma 8.

Lemma 9. If , , satisfies the hypotheses (A1), (A2), the hypotheses (F1) and (H1) hold, for , , satisfies where , satisfies (T1) and (25). Then there exists a unique weak solution for the problem (1).

Proof. Suppose that there exist two distinct weak solutions and for the problem (1). Denote , and for , let . Otherwise, if , we choose and let . We choose as test function. From (A2) and (25), we have when . By Young inequality, for independent on , we obtain
Then we have where is a constant independent on . Hence, we have
Since , we have Therefore, (48) leads to a contradiction.
On the other hand, if , from (44), and Young inequality, for some constant (independent on ), we have
Argue as in (46), for some constant (independent on ), we have
Since , we obtain a contradiction.

Lemma 10. If (A1), (T1), (F1), and (H1) hold, for , and , and , and is a weak solution of the problem (1), then .

Proof. We denote . Let , , and choose as test function to the problem (1). We have Let , fix , be a ball, and let , , and . Using (51), (A1), (F1), (H1), and Hölder inequality, we obtain
For and (T1), by Young inequality, we have As and , if , there exists such that If , a similar estimate is true; just only choose a suitable .
Due to and , we have From and (52)–(55), we obtain Using Lemma 3.1 in [17], and (57), we have
By Lemma 2.4 in [18], (58), we obtain that is bounded from above on . Note that is also a weak solution of the problem (1), where and . Hence, is also bounded from above on , and . This implies that .

Proof of Theorem 5. From Lemmas 8, 9, and 10, we obtain that there exists a uniqueness weak solution for the problem (1), and what is more, . Hence we complete the proof of Theorem 5.

4. The Proof of Theorem 6

In this section, we prove the regularity of weak solutions for the problem (2).

Proof of Theorem 6. We denote . For understanding, we write , and . If is any function with compact support in , for , for small enough , we define where . Hence, if for any , has compact support in and ; from the definition of weak solution for the problem (1), we have
For the definition of and [19], we have where , , and Then, combining (60) with (61)-(62), we obtain Due to (A4), (F2), (H2), (63), Young inequality, we have that is, Choose , where , , and (65), we obtain From (66), (A3), Cauchy inequality and Young inequality, we have By (A3), we obtain Assume ; we have for , . For , let and take ; then we have Hence, . If we take , where , , . Since , , we have For , we have If we fix , , where ; let , , and take , on ; , , then we have Hence, (72) implies that is in . So we obtain . By De Giorgi-Moser regularity theorem, for any and , we have ; then , for .

Acknowledgments

The authors express their sincere thanks to the referees for their valuable suggestions. This paper was supported by Shanghai Natural Science Foundation Project (no. 11ZR1424500) and Shanghai Leading Academic Discipline Project (no. XTKX2012).