HIGHER INTEGRABILITY OF WEAK SOLUTION OF A NONLINEAR PROBLEM ARISING IN THE ELECTRORHEOLOGICAL FLUIDS

. In this paper, we study the Dirichlet problem arising in the electrorheological ﬂuids (cid:40) where Ω is a bounded domain in R n and div a ( x,Du ) is a p ( x )-Laplace type operator with 1 < β < γ < inf x ∈ Ω p ( x ), p ( x ) ∈ (1 , 2]. By establish a reversed H¨older inequality, we show that for any suitable γ,β , the weak solution of previous equation has bounded p ( x ) energy satisﬁes | Du | p ( x ) ∈ L δ loc with some δ > 1.

1. Introduction. In this paper, we consider the following nonlinear boundary value problem −div a(x, Du) = f (x, u) x ∈ Ω, u = 0 x ∈ ∂Ω, (1.1) where Ω ⊂ R n , (n ≥ 3) is a bounded domain with smooth boundary, p(x) > 1 and p(x) ∈ C(Ω). Let us make some remarks concerning problems with p(x)−growth in general. On one hand, the p(x)−growth problem present the borderline case between standard p−growth [1] and so called (p, q) growth conditions introduced in [13], therefore involving delicate perturbation arguments to treat the variable growth situation. On the other hand, the study of variational problems involving p(x)−growth conditions is a consequence of their applications. Such as image restoration considered in [2] and electrorheological fluids, at this point, we must mention about electrorheological fluids. These are special fluids characterized by their ability to change in a dramatic way their mechanical properties when in presence of an external electromagnetic field. According to the model proposed by Rajagopal and Růžička [10], the system can be looked as the delicate interaction 1336 ZHONG TAN AND JIANFENG ZHOU between the electromagnetic fields and the moving fluids: in Ω, u t − div a(x, ε(u), E) + div (u ⊗ u) + Dφ = f in Ω, div u = 0 in Ω, u = 0 on ∂Ω, where E(x) is the electromagnetic field, u : Ω(⊂ R 3 ) −→ R 3 is the velocity of field, ε(u) is the symmetric part of the gradient, a(x, ε(u), E) is the extra stress tensor and φ is the pressure. The main point in the previous system as just mentioned, is that the monotone vector field a : R 9 −→ R 9 depends in a nonlinear way by ε(u): For the equation like (1.1) above, Rajagopal and Růžička established an existence theory in [12], which is particularly satisfying in the steady case in [11] read as Note the connection between (1.1) and (1.3), our paper can be regard as a generalize for (1.3). Notice that the existence and multiplicity of solution for problems with p(x)−growth have been established in [14]. Based on the result in [14], then we proved higher integrability of weak solutions for equation like (1.1). For more details we consider the problem of the type where 1 < β < γ < inf x∈Ω p(x), and k > k 0 for a positive constant k 0 defined in [14].

Some notes and definition.
We recall in what follows some definitions and basic properties of the generalized Lebesgue-Sobolev spaces L p(x) (Ω) and W 1,p(x) (Ω). Let P (Ω) is the set consist with Lebesgue measurable function p : where Ω ⊂ R n (n ≥ 2) nonempty. For all measurable function u, define where Ω ∞ = {x ∈ Ω : p(x) = ∞}.
Variable exponent Lebesgue space L p(x) (Ω) consist of u that satisfies the property: ∃t 0 > 0 s.t ρ p(x) (t 0 u) < ∞. ∀u ∈ L p(x) (Ω), define then L p(x) (Ω) endowed with the norm above is a Banach space. ∀p ∈ P (Ω), define its dual exponent : which is endowed with the norm (Ω) is defined as the completion of C ∞ 0 (Ω) in the norm (2.2). Using standard arguments one can derive from the properties of the space L p(x) (Ω) that W k,p(x) (Ω) and W k,p(x) 0 (Ω) are separable, reflexive Banach spaces.
Let us now discuss the embedding properties of the generalized Lebesgue space. Firstly, we know that a.e. in Ω.
We denote by We note that if q ∈ P (Ω) and q(x) < p * (x) and for all x ∈Ω, then the embedding For more details one can see zhou 5, [3,4,5,8,9].

Definition of the weak solution. We say that
(Ω). In the proof of Theorem 1.1, we will use Gerhring lemma in a version formulated by Giaquinta [6] or Giusti [7]: Let Ω ⊂ R n , 0 < m < 1, and f ∈ L 1 loc (Ω), g ∈ L σ loc (Ω) for some σ > 1 be two nonnegative functions such that for any ball B ρ with B 3ρ ⊂⊂ Ω there holds

Choice of some global constants.
In what follows, we will repeat use some constants: a, θ. To begin with, we choosing a positive number θ > 1 which satisfies the following condition γ2−1 . From now on θ will be a constant only depends on γ 1 , γ 2 , n. Next, we put a restriction on a > 1: where .
Furthermore, Let a satisfies From (2.6), one can see that the valid set of a satisfies previous inequality is a ∈ [1, B), where 1 < B < 2 dependent on γ 1 and θ. Obviously, a depends only on n, γ 1 , γ 2 . Indeed, since I 6 we should restrict θ < γ 1 , and from (ii) one can find that I 2 is valid.
Having fixed the constants θ, a we choose an open ball B ρ such that from the definition of q i and I 5 one can see that q i ≥ 1.

A equivalent form of equation (1.4).
For the weak solution u of (1.4), we can find a function v ∈ R n such that v · Du = 0, and 1 C ≤ |v| ≤ C, with C ≥ 1 is a positive constant. Indeed, without loss of generality, we set n = 2 and Du has the form , where multiply the factor h(x 1 , x 2 ) > 0 such that the exponent of x i for v j more than 1. Since Ω is bounded, we can get the previous claim. Moreover, we can choose suitable h such that 1/c ≤ div v ≤ c for a positive constant c, for n > 2 and Du has the form different with above, we have the same argument.
From previous argument, for the weak solution of (1.4), then we have a equivalence form (Ω).

Higher integrability of weak solution.
Let η ∈ C ∞ 0 (B 2ρ ) be a cut off function for B 2ρ , i.e.
where c is a positive constant independent of ρ. Then the function is a turncated function, where (u) 2ρ := − B2ρ udx denotes the mean value of u on B 2ρ . Then we have

Now let us begin to prove the main result of this paper
Proof of Theorem 1.1 . Let λ > 0, define Then, we deduce that Indeed, observing that by I 6 we infer that for y ∈ B 2ρ , from previous inequality we obtain (3.3). Note that q 1 (2 − a) ≥ 1, applying Poincaré's inequality and Hölder's inequality we obtain (3.5) From (3.5) and taking into account (3.2), it follows that for x ∈ R n \B 3ρ where c = c(n, γ 2 ). By (3.6), we infer that Recall back the Kirszbraun extension theorem, for all λ ≥ λ 1 , there exist a Lipschitz continuous function ϕ λ : R n −→ R with the following properties: (3.7) Since sptϕ ⊂ B 2ρ we have sptϕ λ ⊂ B 3ρ . Now we choose ϕ λ as a test function in (2.9), then we obtain ]ϕ λ and β < γ. From above we arrive at Note that u ≥ 0, taking into account (1.5) and (3.7), it follows that Multiplying both side by λ −a and integrating on ( Set m(x) = max{λ 1 , M (x)}, then by Fubini theorem, we arrive at From above, it follows that Since m −1 ≤ M −1 , then we obtain that Therefore, we deduce that where c = c(k). From inequality (3.10), take into account (3.8), (3.9), then we have Note that for x ∈ J, we have and for x ∈ B ρ \J we have Observing that Inserting (3.12) into (3.11), then we obtain where c = c(n, ν, γ 1 , γ 2 , L, k), c 1 = c 1 (L, ν), c 2 = c 2 (k, ν).
Estimation of H 1 . Make use of I 3 , we can see that By the aid of Young's inequality with exponents then we obtain that where c = c(n, γ 1 , γ 2 , L, k, ν).
From above, once again using Hölder's inequality and Sobolev-Poincaré's inequality we infer that We are now coming back to estimation of H 2 in (3.22). Recalling the definition of λ 1 , choose a suitable R 0 < 1, then for 2ρ < R 0 , there holds Observing that from (1.8) 2