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 [36].

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 [711] 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 –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-Hölder 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 Calderón-Zygmund operator with Calderón-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 Calderón-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 Calderón-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)–(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 Carathéodory function if for almost all , is continuous on , for all is measurable on .

Lemma 3.7 (see[17]). A mapping is a Carathéodory 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 su