#### Abstract

In this paper, we study the following quasilinear equation: , where and . In the Orlicz-Sobolev space, by variational methods and a minimax theorem, we prove the equation has a nontrivial solution.

#### 1. Introduction

In this paper, we consider the following quasilinear elliptic equation: where is the dimension of space and satisfies the following conditions:

(*Ï•*_{1}) , , , for

(*Ï•*_{2}) There exist , such that , for

is a continuous function on defined by and -Laplace is defined by

Let

The quasilinear elliptic equation is an important partial differential equation; in recent years, many researchers have studied the following equation: where is an open set, . Under some growth conditions, many people proved the existence and multiplicity of solutions to (5) and several mathematicians also obtained the bifurcation and asymptotic properties. By variational method and maximal principle, Guo [1] and Guo-Webb [2] obtained the existence and uniqueness of the solution to (5) and they also considered the partial symmetric properties of the solutions. Guedda-Veron [3] used topology and spectrum analysis to study the bifurcation and multiplicity of the solutions. We point out that by constructing the pseudogradient vectors of -Laplace operator, Zhang-Li [4] firstly obtained the sign-changing solutions to (5) and see also Zhang-Chen-Li [5], for more results of -Laplace equations, one can see [6â€“8] and the references therein. By variational method and minimax theorem, Li-Guo [9] and Li-Liang [10] studied the -Laplace equation and they obtained the existence and multiple solutions.

In [11], Franchi-Lanconelli-Serrin studied the quasilinear equation (1), and in weighted Sobolev space, they considered the existence and uniqueness of the solution to (1). In fact, the function is a general elliptic equation. For example, if , then (1) is the Laplace equation; if , then (1) is the -Laplace equation; , then (1) is the curvature equation. The Orlicz-Sobolev space is a kind of general norm space and one can study the quasilinear equations in this space. Carvalho-Silva-Goulart [12] and Carvalho-Silva-Goncalves-Goulart [13] considered (1) in the Orlicz-Sobolev space, by variational method and concentration-compactness theorem. They obtained the existence of (1), and they also studied the famous problem, i.e., concave and convex terms.

In this paper, let satisfy the following conditions:

(*f*_{1}) . For , and for

(*f*_{2}) ,

(*f*_{3}) There exists a constant , such that , with for

(*f*_{4}) for all

(*f*_{5}) satisfies the Ambrosetti-Rabinowitz condition: there exists a such that

Under the preceding conditions, by a variational method, we obtain the existence of a nontrivial solution to (1). We follow the idea in [10] to obtain the existence of nontrivial solution. Our conditions - are slightly different from what is in [10]. Conditions and are standard. Our condition is the improvement of in [10], which is used to obtain the superlinear growth of at . The condition is to guarantee that is a Finsler manifold which is used to obtain a special minimizing sequence (see Lemma 14â€“Lemma 16). Condition is the compactness condition.

*Remark 1. *The condition implies that for ,

*Remark 2. *By and (7), we have that
This paper is organized as follows. In Section 2, we recall some results of the Orlicz-Sobolev space; in Section 3, we list and prove some preliminary results; and in Section 4, we prove our main theorem.

#### 2. Orlicz-Sobolev Spaces

In this section, we recall some useful knowledge for the Orlicz space and Orlicz-Sobolev space and give some inequalities on . The reader can refer to [14, 15] for more details.

By condition and the definition of , the function is a -function (see [14] for the definition of -function). The complementary of is defined by for . It is easy to see that and are complementary -functions. By (7) (Chapter 8 [14]), the function and satisfies the following inequality:

By Lemma 4, Lemma 5, and Lemma 8 below, one can see that , , and satisfy -condition (see [14] for the definition of -condition).

For any , under the assumptions and , the Orlicz space contains all measurable functions which satisfy and the Luxemburg norm on is defined by

For any integer , the corresponding Orlicz-Sobolev space is defined by and the norm on is defined by

Let denote the closure in of function u which are bounded on and have bounded support in . The space is defined by

In the following, for simplicity, we denote and by and . Then, the Orlicz-Sobolev space has the following properties.

Theorem 3 (Theorem 8.20 and Theorem 8.31 [14]). (a)*The spaces is reflexive if and only if both and are -regular*(b)*If and are -regular, then is reflexive*(c)* is dense in . Thus, **Since and are complementary -functions, the following generalize HÃ¶lder inequality (see [14]) holds:
**Next, we recall some inequalities on .*

Lemma 4 (Lemma 2.1 [15]). *Let , , . Then,
*

Lemma 5 (Lemma 2.2 [15]). *Let , , . Then,
*

Lemma 6 (Lemma 2.3 [15]). * increases essentially more slowly than near infinity, i.e.,
for every constant .*

Lemma 7 (Lemma 2.4 [15]). *(19) is equivalent to
*

Lemma 8 (Lemma 2.5 [15]). *Let , , . Then,
**We recall some imbedding results in the Orlicz-Sobolev space.*

Theorem 9 (Theorem 8.12 [14]). *The imbedding holds if and only if dominates globally, i.e., there exists a constant such that holds for all .*

Theorem 10 (Theorem 8.36 [14]). *Let be an arbitrary domain in . If the -function satisfies
then , for any integer . Moreover, if is a bounded subdomain of , then the imbedding exists and is compact for any -function increasing essentially more slowly than near infinity.*

Since and satisfy -condition, so by Theorem 3, is reflexive. Using Lemma 4 and Theorem 10, one can see that

By (7), (11), Lemma 4, and Lemma 8, we can see that for and , there exists some such that

#### 3. Preliminary Results

In this section, we prove some preliminary results for future use.

To study the existence of solution to (1), we first study its energy functional. It is clear that the functional defining on is given by

It is easy to see that under the assumptions - and ()-(), the functionals is of . For any , we have

The is the dual pairing between and its dual space .

By condition and , we have that for any , there exists a constant such that

Hence, it follows that for any , there exists a constant such that

Lemma 11. *Suppose holds. Then, there exists a positive constant such that for any , ,
*

*Proof. *By , one has that, for ,
Integrating the last inequality from to , we get the result in the lemma. This ends the proof.

Set
Then, we show that is not an empty set.

Lemma 12. *Suppose that - and - hold. Then, .*

*Proof. *By and , . Hence, there exists a constant such that . Notice that . We choose such that , , and . Let
It is easy to show that is continuous. For , by (7), , and Lemma 11, we have
Using the definition of , one has that . It follows that
For , by (7), we have
By (9), we can see
It follows that, for any , there exists some such that, for any ,
Hence,
Therefore, there exists some such that . This ends the proof.

Lemma 13. *For any , there exists some such that
for any .*

*Proof. *By Lemma 7 and (7), for It follows from Youngâ€™s inequality that, for any , there exists such that
Hence, for ,
This ends the proof.

Set

Lemma 14. *Suppose - and - hold. Then, .*

*Proof. *For any , let . Using (7) and , we obtain that
Hence,
By (7), (29), (30), and Lemma 13, for any , one has that there exist two positive constants and such that
Choose small enough such that Then, we obtain that
By (25), Lemma 4, and Lemma 5, we have that
By the definition of and , we have
Combining (49)â€“(51), there exists a positive constant , such that
which means for some . It follows that . This completes the proof.

For , set

It is easy to see that is a -functional. For any , one has that . If , then by . Noting that , one has that , which contradicts with . By and

Hence, is a closed and a complete submanifold of with the natural Finsler structure (see [16]). Using Lemma 2.14 [10] with , one has the following.

Lemma 15. *For any , , we have
with is the projection from to .*

By Lemma 12, is not empty, and by Lemma 14, is bounded from below. Hence, by Lemma 2.15 of [10] and Lemma 15, we can get the following result.

Lemma 16. *There exists a sequence , such that,
*

Lemma 17. *Let be a bounded sequence and , then there exists a subsequence, still denoted by , such that one of the following two possibilities occurs:
*(1)*(Vanishing): for all *(2)*(Nonvanishing): there exists , and , such that *

Lemma 18. *Suppose is a bounded sequence in , and for some ,
**Then, *

*Proof. *By Lemma 7, (7), and HÃ¶lder inequality, for any ,
Noting that is bounded in , there exists a constant such that
Then, it follows from imbedding theorem (25) and Lemma 4 that
for some positive constants and . Thus, by Lemma 5,
for some positive constants and . Letting , it follows from (58) and (61) that
for some . Then, coving by balls of radius , in such a way that each point of is contained in at most balls, we have
as . Noting that is bounded in , by (2.3), we get that is a bounded sequence. If , letting , then
as . By imbedding theorem (25) and Lemma 7, one sees that is a bounded sequence. If , setting , then
as . This ends the proof.

Theorem 19 (Vitaliâ€™s convergence theorem [17]). *Let be a measure space and let . Let be a sequence in and let be an -measurable function such that is finite -a.e. and -a.e. Then, and if and only if
*(i)

*For each , there exists a set such that and for all*(ii)

*uniformly in , i.e., for each , there is a such that and imply for all .*

#### 4. Main Result

Theorem 20. *Suppose conditions ()-() and ()-() hold. Then, (1) has a nonnegative weak solution.*

*Proof. *By Lemma 16, there exists a sequence , such that

*Step 1. *There exists some constant such that By (66) and (7), one has
In the last step, we use condition (). Hence, by Lemma 4, there exists some constant such that

*Step 2. *There exist , , and , such that

By Step 1, is bounded in . Lemma 17 shows that one of two situations (vanishing and nonvanishing) occur. We need only to show that the vanishing case is impossible. If vanishing occurs, by Lemma 18, It follows from (29) and (30) that

Notice , we obtain

This is a contradiction. Hence, only nonvanishing is possible, i.e., there exist , , and , such that

*Step 3. *There exists , such that a.e. in , where .

For any , set . Then, , and as . Since , so is a bounded sequence in . Hence, there exists and such that some sequence of , still denoted by ,