#### Abstract

We study the following nonhomogeneous -harmonic equations: , where is a bounded and convex Lipschitz domain, and satisfy some -growth conditions, respectively. We obtain the existence of weak solutions for the above equations in subspace of .

#### 1. Introduction

Spaces of differential forms have been discussed in great details (see [1, 2] and the references therein). The theory of differential forms is an approach to multivariable calculus that is independent of coordinates and provides a better definition for integrals. Differential forms have played an important role in physical laws of thermodynamics, analytical mechanics, and physical theories, in particular Maxwell's theory, and the Yang-Mills theory, the theory of relativity, see for example [3â€“6].

In recent years, the study of *A-*harmonic equations for differential forms has developed rapidly. Many interesting results concerning *A-*harmonic equation have been established recently (see [7â€“11] and the references therein). In [12], spaces and are first introduced, and they used them to study the solutions of nonlinear Dirichlet boundary value problems with -growth conditions. In [13], spaces and are first introduced and used to study the weak solutions of obstacle problems of *A-*harmonic equations with variable growth for differential forms.

Let be a bounded and convex Lipschitz domain. It is our purpose to study the following systems: where , , and , satisfy the following conditions.(H1) and are measurable with respect to for all and continuous with respect to , respectively, for a.e. .(H2), where and are constants.(H3), where is a constant and .(H4), where is a constant and .(H5)For a.e. , the mapping satisfies for each , and , where is a constant. Here is the conjugate function of . Throughout this paper we suppose (unless declare specially)

#### 2. Preliminaries

Let be the standard orthogonal basis of . The space of all -forms in is denoted by . The dual basis to is denoted by and referred to as the standard basis for 1-form . The Grassman algebra is a graded algebra with respect to the exterior products. The standard ordered basis for consists of the forms

For and , the inner product is obtained by with summation over all â€“tuples and all integers . The Hodge star operator (see [14]) is defined by the formulas Hence, the norm of is given by the formula . Notice, the Hodge star operator is an isometric isomorphism operator on . Moreover, where I is the identity map.

Let be a bounded domain. The coordinate functions in are considered to be differential forms of degree 0. The 1-forms are constant functions from into . The value of is simply , . Therefore, every form may be written uniquely as where the coefficients are distributions from , dual to the space of smooth functions with compact support on .

We use to denote the space of all differential forms. For each form , the exterior differential is expressed by

For , the vector-valued differential form consists of differential forms , where the partial differentiation is applied to the coefficients of .

The formal adjoint operator, called the Hodge codifferential, is given by By denote the space of infinitely differentiable -forms on and by denote the subspace of with compact support on .

Let be the set of all Lebesgue measurable functions . For , we put and . Given we define the conjugate function by

*Definition 2.1 (see [15]). *A Lebesgue measurable function is called globally log-HÃ¶lder continuous in if there exist and a constant such that
hold for all . is defined by

For a differential -form on , , define the functional by

The space is a reflexive Banach space endowed with the norm

The space is a reflexive Banach space endowed with the norm

Note that and are spaces of functions on . In this paper, we denote them by and .

Iwaniec and Lutoborski proved the following results in [2].

Let be a bounded and convex domain. If is defined for some , then the value of at the vectors is denoted by . Then to each , there corresponds a linear operator defined by The homotopy operator is defined by averaging over all points where is normalized so that . Then we have a pointwise estimate where further infimum is attained at , and the decomposition holds for .

*Definition 2.2. * For , define the -form by
and the Maximal operator is defined by
where .

Lemma 2.3 (see [15]). *Let satisfies (1.3). Then the inequality
**
holds for every .*

Lemma 2.4 (see [15]). *Let be a bounded convex domain, and . Then
*

Lemma 2.5 (see [15]). *Let be a CalderÃ³n-Zygmund operator with CalderÃ³n-Zygmund kernel on . Then is bounded on . Further there exists a constant C = C(n,p) such that
**
holds for every .*

Lemma 2.6. * If , then
**
Moreover, if , then
*

* Proof. *First define . From pointwise estimate (2.16) and Lemma 2.4,
In view of Lemma 2.3, we have
that is to say, (2.24) holds.

From the definition of and (2.18), we have . Therefore,
Now in (2.24) replace with , we obtain (2.25).

Lemma 2.7. * Let satisfies (1.3).*(1)* is dense in ,*(2)* is separable. *

*Proof. *(1) For any , since is dense in and for all , we can find a sequence which converges to in for each . Now let , then the sequence converges to in , since
That is to say, is dense in .

(2) Let . Since is separable, there exists a countable dense subset of . Then for any above we can extract a sequence in which converges to in . Similar to (1), the sequence converges to in . That is to say, is separable.

Let . Note that if and only if .

Lemma 2.8. * Let satisfies (1.3). Then is a closed subspace of . In particular, it is a reflexive Banach space. *

*Proof. *Set a sequence convergent to in , then . By Lemma 2.6, the operator is continuous on . Therefore, , we have . That is to say, is a closed subspace of .

In [2], Iwaniec and Lutoborski obtained where where and Further for each and , satisfies the following properties:(i),(ii),(iii) for all .

Let . Then satisfies the conditions of CalderÃ³n-Zygmund kernel on for each .

Lemma 2.9. *Let . Then
*

*Proof. *By Lemmas 2.3 and 2.4, and (2.31),
Let
we can write as
Hence,
Taking in (2.32), we obtain
Now define . Since satisfies the conditions of CalderÃ³n-Zygmund kernel on for each , in view of Lemma 2.5,
So that
By (2.30), (2.35), and (2.41), we have

Now define another norm

*Remark 2.10. *Replacing with in (2.34), we get by the definition of
Therefore is equivalent to .

In this paper we also need the following two lemmas.

Lemma 2.11 (see[15]). * Let satisfies (1.3). Then the embedding is compact.*

Lemma 2.12 (see[15]). *Suppose that . Let be bounded in . If a.e. on , then weakly in .*

*Remark 2.13. *Let be the completion of in . Then from Remark 2.10 and Lemma 2.11, the embedding is compact.

*Remark 2.14. *Suppose satisfies (1.3), Lemma 2.12 also holds on space .

#### 3. Weak Solutions of Dirichlet Problems for the *-*Harmonic Equations with Variable Growth

Theorem 3.1. *Under conditions (H1)â€“(H5), the Dirichlet problem (1.1) has at least one weak solution in , that is to say, there exists at least one satisfying
**
for all . Here, and satisfies (1.3). *

Let and . For , define in the following way: for each Now we need only to show that there exists such that for all .

Lemma 3.2. * is strong-weakly continuous on . *

*Proof. *Let be a sequence strongly convergent to an element in . Let and . Then () for some constant , () is a sequence strongly convergent to in for each .

In view of (H2) and , we know that and are uniformly bounded in and , respectively. Hence, and are uniformly bounded in . On the other hand, by , there exists a subsequence of (still denoted by ) such that
Then there exists a subsequence of (still denoted by ) such that
In view of (H1), we obtain
Let , , and , then , in the meantime
for each and .

Now by Lemma 2.12, we can show that and as . Therefore,
that is to say, is strong-weakly continuous on .

Lemma 3.3. * is coercive on , that is,
*

*Proof. *By (H3) and (H4),
By and Lemma 2.6, we have
for all . Then , as . Taking
we have
Therefore,
Then it is immediate to obtain that
That is to say, is coercive on .

Lemma 3.4 (see[16]). *Suppose is a mapping from into itself such that
**
Then the range of is the whole of . *

Lemma 3.5. *There exists a sequence and , such that
*

*Proof . *By Lemmas 2.7 and 2.8, we can choose a Schauder basis of such that the union of subspace finitely generated from is dense in . Let be the subspace of generated by . Since is topologically isomorphic to . By Lemmas 3.3, and 3.4, there exists such that
By Lemma 3.3 again, we know that , where is independent of . Since is reflexive, by Remark 2.14 and (H1), we can extract a subsequence of (still denoted by ) such that
where is in a dense subset of . For fixed , by the continuity of , we get for all . For , we have
This completes the proof of Lemma 3.5.

Set . Then Consider once more, then as . By Remark 2.13, we get In view of (3.23) and (H2), it is immediate that that is to say,

Now if we can prove that there exists a subsequence of which is strongly convergent in , then from the strong-weakly continuity of , we get weakly in as and will be a weak solution of (1.1). We need the following lemmas.

*Definition 3.6. *Let be an open subset of provided with the Lebesgue measure. The mapping is said CarathÃ©odory function if for almost all , is continuous on , for all is measurable on .

Lemma 3.7 (see[17]). *A mapping is a CarathÃ©odory function if and only if for all compact sets and all , there exists a compact subset such that for with the restriction of to is continuous.*

Lemma 3.8 (see[15]). *Let be a sequence of bounded function in . For each there exists (where is measurable and is an infinite subset of natural numbers set ) such that for each ,
**
where and are disjoint and .*

*Definition 3.9. *For , define

Lemma 3.10 (see[18]). *If , then and for all ,
**
Furthermore, if , then
**
and if , then
**
for all .*

Lemma 3.11 (see[19]). *Let and . Set
**
Then for all , we have
*

Lemma 3.12 (see[16]). *Let be a metric space, be a subspace of , and be a positive number. Then any Lipchitz mapping from into can be extended to a Lipchitz mapping from into .*

*Proof of Theorem 3.1. *We need only to show that there exists subsequence of which is strongly convergent in .

For each measurable set , define
where . Similar to the proof of Lemma 3.2, is strongly continuous on . Since is dense in , there exists such that
So we can suppose that is bounded in .

Next define
In this way, we extend the domain of to and .

Let be a continuous increasing function satisfying and for each measurable set ,
where is the constant in (H2).

Let be a positive decreasing sequence with as . For , by Lemma 3.8, we get a subsequence of , a set satisfying , and a real number such that
for each and satisfying . By Lemma 3.10, we can choose so large that for all and ,

For each and , define
In view of Lemma 3.11, we have
Form Lemma 3.12, there exists a Lipschitz function which extends outside and Lipschitz constant of is no more than . As is an open set, we have and for all , and . We can further suppose that
By the uniformly boundedness of , there exists a subsequence of