#### Abstract

We first discuss the existence and uniqueness of weak solution for the obstacle problem of the nonhomogeneous -harmonic equation with variable exponent, and then we obtain the existence of the solutions of the equation in the weighted variable exponent Sobolev space

#### 1. Introduction

In [1–5], the nonhomogeneous -harmonic equation for differential forms has received much investigation. In [6], the obstacle problem of the -harmonic equation for differential forms has been discussed. However, most of these results are developed in the space or space. Meanwhile, in the past few years the subject of variable exponent space has undergone a vast development; see [7–11]. For example, [8–10] discuss the weighted and spaces and the weak solution for obstacle problem with variable growth has been studied in [10, 11].

In this paper, we are interested in the following obstacle problem: for belonging to where , , , ; means that, for any , we have , and the variable exponent satisfies

The operators and satisfy the following growth conditions on a bounded domain :(H1) and are measurable for all with respect to and continuous for with respect to ,(H2),(H3),(H4),(H5),(H6) for ,

where are nonnegative constants. nonnegative and . We will discuss the existence and uniqueness of the solution for the abovementioned obstacle problem.

Now, we introduce the existing results and related definitions.

Throughout this paper, we assume that is a bounded domain in . Let be the set of all -forms in . A differential -form is generated by ; that is, , where is differential function, , and . Let be the space of all differential -forms on . For and , then the inner product is obtained by . We write . We denote the exterior derivative by for . Its formal adjoint operator is given by , ; here is the well-known Hodge star operator. Denote the class of infinitely differential -forms on by . A differential -form is called a closed form if in .

Next we will introduce some basic properties of weighted variable exponent Lebesgue spaces and weighted variable exponent Sobolev spaces , and we define to be the set of all *n*-dimensioned Lebesgue measurable functions . Functions are called variable exponents on . We define , . If , then we call a bounded variable exponent. If , then we define by , where . The function is called the dual variable exponent of . We denote as a weight if and a.e.; also in general . From [7, 10], we know that if satisfies (3), the weighted variable exponent Lebesgue spaces with the norm and the weighted variable exponent Sobolev spaces with the norm are Banach space and reflexive and uniformly convex. On the set of all differential forms on , we define the weighted variable exponent Lebesgue spaces of differential -forms and the weighted variable exponent Sobolev spaces of differential forms .

*Definition 1. *We denote the weighted variable exponent Lebesgue spaces of differential -forms by and we endow with the following norm:

And the spaces with the norm
are the weighted variable exponent Sobolev spaces of differential -forms; . is the completion of in . We need the following Hölder inequalities; see [7, 10].

Proposition 2. *Let , be such that for -almost every . Then
**
for all and .*

Lemma 3 (see [7]). *Let be a -finite, complete measure space; if , , and -almost everywhere, then and .*

By the inequality for any , using Lemma 3, we can easily have the following lemma.

Lemma 4. *If and , then and are equivalent, and .*

#### 2. Main Results

In this section, we will obtain the existence and uniqueness of weak solution for obstacle problem of the nonhomogeneous -harmonic equation in space .

Theorem 5. *Suppose is not empty, under conditions (H1)–(H6), and there exists a unique solution to the obstacle problem (1)-(2). That is, there is a differential form in such that
**
whenever .*

We deduce Theorem 5 from a proposition of Kinderlehrer and Stampacchia.

Proposition 6 (see [12]). *Let be a nonempty closed convex subset of and let be monotone, coercive, and weakly continuous on . Then there exists an element in such that whenever .*

Now let and be the usual pairing between and , , where is in and in . We will take as . We define a mapping by for .

Lemma 7. *If satisfies (3), then spaces and are complete and convex.*

*Proof. *From [7], we know that if satisfies (3) and , then let be Banach space and uniformly convex. If and are two -forms: and , we can easily have and , so we can immediately obtain the convexity of spaces and .

Let be a Cauchy sequence in . Then for any , converges in . Suppose that in . Now let , we have
using Lemmas 3 and 4, and we know the sequence converges to in .

For the sequence , we suppose in , and then . So converges to in the normed space . From , we have

In view of [10], for any , we know that in and in , and then . Using Lemmas 3 and 4, we get the sequence that converges to in ; it follows that . So, we prove is a closed subspace of . This ends the proof of Lemma 7.

Using Lemma 7, we can immediately obtain the following lemma.

Lemma 8. * is a closed convex set.*

Lemma 9. *For each .*

*Proof. *Using Hölder inequality (7) with and (H2), we get

Similarly, for the operator , we have

Using (12) and (13) we can easily prove

So we get , whenever , , and are constants fixed to (H2) and (H4).

Lemma 10. * is monotone and coercive on .*

*Proof. *It follows from (H6) that is monotone. To show that is coercive on , fix and using the conditions (H2)–(H5), (12), (13), and (6), then
where is the norm; taking , we have

Let ; from [7, pages 24, 73], we know that if the variable exponent satisfied , then
holds. Whenever , we have or , and using (17) we have

Substituting (18) in (16) we obtain

Then it is easy to obtain
as . It follows that is coercive on . This completes the proof of Lemma 10.

Lemma 11. * is weakly continuous on .*

*Proof. *Let be a sequence that converges to an element in . Pick a subsequence such that a.e. in . Since the mapping and are continuous for a.e. in , we have a.e. in . Under the conditions (H2) and (H4), we know that and are uniformly bounded in , and we have weakly in and weakly in .

Since the weak limit is independent of the choice of the subsequence, it follows that
for all .

Then we have

Hence is weakly continuous on . This ends the proof of Lemma 11.

* Proof of Theorem 5. *We can apply Proposition 6 and the above lemmas to obtain the existence. If there are two weak solutions , to obstacle problem (1)-(2), then we have
so

In view of (H6), we can further infer that
which means that , and now we complete the proof.

Corollary 12. *Let be a bounded domain and . Under the conditions (H1)–(H6), there is a differential form with such that
**
that is, , whenever .*

*Proof. *Choose and let be the solution to the obstacle problem (1)-(2) in . For any , since and both belong to , we have

Thus
as desired.

*Remark 13. *If , then and the Luxemburg norm reduces to the norm. So (26) is the extension of the equation in [2–5].

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.