Harnack inequality for solutions of the p(x)-Laplace equation under the precise non-logarithmic Zhikov’s conditions

We prove continuity and Harnack’s inequality for bounded solutions to the equation div(∣∇u∣p(x)-2∇u)=0,p(x)=p¯+Lloglog1∣x-x0∣log1∣x-x0∣,p¯>1,L>0,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \textrm{div}\big (\mid \nabla u\mid ^{p(x)-2}\,\nabla u \big )&=0, \quad p(x)= {\bar{p}} + L\frac{\log \log \frac{1}{\mid x-x_{0}\mid }}{\log \frac{1}{\mid x-x_{0}\mid }}, \\ {\bar{p}}&>1, \quad L>0, \end{aligned}$$\end{document}under the precise non-logarithmic condition on the function p(x).


Introduction and main results
Let Ω be a bounded domain in R n , n 2. In this paper we are concerned with elliptic equations of the type divA(x, ∇u) = 0, x ∈ Ω. (1.1) We suppose that the functions A : Ω × R n → R n are such that A(•, ξ) are Lebesgue measurable for all ξ ∈ R n , and A(x, •) are continuous for almost all x ∈ Ω.We assume also that the following structure conditions are satisfied where K 1 , K 2 are positive constants, p(x) = p + p(|x and L > 0. The aim of this paper is to establish basic qualitative properties such as continuity of bounded solutions and Harnack's inequality for non-negative bounded solutions to equation (1.1).
Before formulating the main results, we say few words concerning the history of the problem.
Equations of this type and systems of such equations arise in various problems of mathematical physics (see e.g. the monographs [8,25,40,51] and references therein).
The case when condition (1.differs substantially from the logarithmic case.It turns out that such non-logarithmic condition is a precise condition for the smoothness of finite functions in the corresponding Sobolev space W 1,p(x) (Ω).Thus this case is extremely interesting to study.But to our knowledge there are only few results in this direction.Zhikov [59] obtained a generalization of the logarithmic condition which guaranteed the density of smooth functions in Sobolev space W 1,p(x) (Ω).Particularly, this result holds if 1 < p p(x) and osc Later Zhikov and Pastukhova [61] proved higher integrability of the gradient of solutions to the p(x)-Laplace equation under the same condition.Interior continuity, continuity up to the boundary and Harnack's inequality to the p(x)-Laplace equation were proved in [4], [7] and [49] under condition (1.4) and 0 exp − γ exp(µ c (r)) dr r = +∞, (1.6) with some numbers γ, c > 1.For example, the function µ(r) = L log log log 1 r stisfies conditions (1.4), (1.6), provided that L is a sufficiently small positive number.
These results were generalized in [41,47] for a wide class of elliptic and parabolic equations with non-logarithmic Orlicz growth.Particularly, it was proved in [47] that under conditions (1.4), (1.6) functions from the correspondent De Giorgi's B 1 (Ω) classes are continuous and moreover, it was shown that the solutions of the correspondent elliptic and parabolic equations with non-standard growth belong to these classes.
The exponential condition of the type (1.6) was substantially refined in [24].Particularly, the continuity of solutions to double-phase and degenerate double-phase elliptic equations div |∇u| p−2 ∇u + a(x)|∇u| q−2 ∇u = 0, q > p, and was proved under the conditions osc Br(x 0 ) a(x) A µ(r) q−p r q−p , osc Note that the function µ(r) = log 1 r satisfies the above conditions.In the present paper the continuity and the Harnack's type inequality have been proved under the conditions similar to (1.5).
Before formulating the main results, let us recall the definition of a bounded weak solution to equation (1.1).We introduce W (Ω) as a class of functions u ∈ W 1,1 (Ω), such that Ώ |∇u| p(x) dx < +∞, and holds for all non-negative test functions ϕ ∈ W 0 (Ω).
The following Theorem is the first main result of this paper.
Theorem 1.1.Let u be a bounded weak solution of equation (1.1) and let conditions (1.2), (1.3) be fulfilled, then u is Hölder continuous at point x 0 .
The next result is a weak Harnack type inequality for non-negative super-solutions.
In the present paper, we substantially refine the results of [4,7,41,46,47,49].We would like to mention the approach taken in this paper.To prove the interior continuity we use De Giorgi's approach.Let us consider the standard De Giorgi's class DG p(•) (Ω) of functions u which corresponds to equation (1.1) : B 16r (x 0 ) ⊂ Ω and ζ(x) is the correspondent cut-off function for the ball B 16r (x 0 ).Using the Young inequality, by conditions (1.4) we have This estimate leads us to condition (1.6) (see, e.g.[46,47]).It is easy to see that condition (1.6) fails for the function µ(r) = L log log 1 r .To avoid this, using the Young inequality and our choice of p(x) we rewrite inequality (1.15) as (1.16) It appears that the weight satisfies the Muckenhoupt type properties.In Section 2 we define the correspondent weighted De Giorgi's classes by inequalities (1.16) and prove the Hölder continuity at point x 0 for the functions which belong to these classes.
The main difficulty arising in the proof of the Harnack type inequalities is related to the so-called theorem on the expansion of positivity.Roughly speaking, having information on the measure of the "positivity set" of u over the ball B r (x) ⊂ B ρ (x 0 ): with some r > 0, m > 0 and γ > 1, and using the standard De Giorgi's or Moser's arguments, we inevitably arrive at the estimate with some γ, c > 1.This estimate leads us to condition (1.6) (see, e.g.[41,47]).Note that we can not use the classical approach of Krylov and Safonov [31], DiBenedetto and Trudinger [19], as it was done in [9] under the logarithmic conditions.We also can not use the local clustering lemma of DiBenedetto, Gianazza and Vespri [17] (see also [18,50] ).Difficulties arise not only due to the constant α(r) which depends on r, but also when an additional term, that couldn't be estimated, occurs during the process of iteration from B r (x) to B ρ (x 0 ).To overcome it, we use a workaround that goes back to Mazya [38] and Landis [33,34] papers.
We will demonstrate our approach on the p-Laplacian.Fix x 0 ∈ Ω and let 0 < r < ρ, v − mψ ∈ W 1,p 0 (D), (1.18) where m > 0 is some fixed number, and ψ ∈ W 1,p 0 (B 16ρ (x 0 )), ψ = 1 on E. By the well-known estimate (see e.g.[23]) we have min where C p (E) is a capacity of the set E. By the Poincare inequality from the previous we obtain min Let u be a non-negative bounded super-solution to the p-Laplace equation in Ω and construct the set Consider also a solution v of the problem (1.17), (1.18) with E replaced by E(ρ, m).Then since u v on ∂D, by the maximal principle and by (1.19) we obtain m(2ρ) := min which by standard arguments yields for any θ ∈ (0, p − 1) from which the weak Harnack type inequality follows.
In Sections 3, 4 we adapt this simple idea to the case of p(x)-Laplacian with non-logarithmic growth.The weight which naturally arises in the proof of Theorem 1.2 also satisfies the Muckenhoupt type conditions.
Remark 1.1.It was unexpected for authors that the modulus of continuity and the constants in the Harnack type inequalities do not depend on the additional term log log 1 r (usually, there is a dependency, see e.g.[7,24,47,49]).
The rest of the paper contains the proof of the above theorems.
2 Elliptic DG classes, proof of Theorem 1.1 In this Section we define the following De Giorgi's classes.
Definition 2.1.We say that a measurable function u |u| M and there exists numbers σr) −1 , the following inequalities hold: We refer to the parameters c 1 , n, p, q and M as our structural data, and we write γ if it can be quantitatively determined a priory in terms of the above quantities.The generic constant γ may change from line to line.
Our main result of this Section reads as follows: We note that the solutions of equation (1.1) belong to the corresponding DG(B R (x 0 )) classes, provided that B 2R (x 0 ) ⊂ Ω.We test identity (1.7) by ϕ = (u−k) ± ζ q (x), by the Young inequality we obtain From this, using again the Young inequality dx, from which the required (2.1) follows.

Auxiliary Propositions
For k ∈ R and 0 < r < R set w ± (x, u, k, r) , further we need the following lemmas Lemma 2.1.There exists C > 0 depending only on the data, such that for any u ∈ DG(B R (x 0 )) and for any t > 0 the following inequalities hold provided that

4)
Proof.To prove inequalities (2.2), (2.3) we just need to check , t > 0, (2.5) The left inequality in (2.5) and the right inequality in (2.6) are obvious due to the fact that ) and r is sufficiently small.Let us check the right inequality in (2.5).Integrating by parts, using the fact that log log 1 s log 2 1 s is increasing on the interval (0, r), we obtain 1, from which the required inequality follows. Similarly, r M ± (u, k, r) 1, which completes the proof of the lemma.
Choosing C 1 = t C, we arrive at the required (2.7), which completes the proof of the lemma.

De Giorgi Type Lemma
Let B 8r (x 0 ) ⊂ B R (x 0 ) and let µ + r ess sup ).Then there exists ν ∈ (0, 1) depending only on n, p, q, c 1 and M , such that if then either provided that where C 1 is the constant defined in Lemma 2.2.
Proof.We provide the proof of (2.11), while the proof of (2.14) is completely similar.For ) and set A j := B r j (x 0 ) ∩ {u k j }.Further we will assume that sup which implies , from which by standard arguments (see e.g.[32]) the required (2.11) follows, provided that ν is chosen to satisfy ν γ −1 .This completes the proof of the lemma.

Expansion of the Positivity
To prove our next result we need the following lemma.
provided that (2.17) here C > 1 is the constant, defined in Lemma 2.1.
Using the Poincare inequality and inequality (2.3) from which the required inequality (2.15) follows.The proof of (2.16) is completely similar.) and assume that with some α ∈ (0, 1) there holds (2.18) Then there exists number s * depending only on n, p, q, c 1 , M , α and ξ such that either or provided that then there exists number s * depending only on n, p, q, c 1 , M , α and ξ, such that either (2.19) provided that (2.21) holds.
Proof.We provide the proof of (2.20), while the proof of (2.23) is completely similar.We set where s * is large enough to be chosen later.We will assume that sup is violated, then Lemma 2.5 with l = k s+1 and k = k s yields From this, by inequality (2.1) we obtain By our choice and (2.21) we have ) and w + u,k s+1 ,r (B r (x 0 )) + w + u,ks,r (B 2r (x 0 )) γw + u,ks * ,r (B r (x 0 )), s = log 1 ξ , . . ., s * − 1, so from the previous relation we have Summing up these inequalities over s = log 1 ξ , . . ., s * − 1, we conclude that Using inequality (2.2) from the last inequality we arrive at Choosing s * by the condition γ(α) = ν and using Lemma 2.3 we obtain (2.20), which proves Lemma 2.5.

Proof of Theorems 1.1, 2.1
To complete the proof of Theorems 1.1 and 2.1 we fix R by the condition where s * is the number defined in Lemma 2.5, and assume that the following two alternative cases are possible: or for any 0 < r < ρ < R. Assume, for example, the first one holds.Then by Lemma 2.5 we obtain Iterating this inequality, we have This completes the proof of Theorems 1.1 and 2.1.

Upper and lower estimates of auxiliary solutions
In this Section we prove upper and lower bounds for auxiliary solutions v := v(x, m) to the where 0 < m M is some fixed number, and ψ ∈ W 0 (B 16ρ (x 0 )), ψ = 1 on E. The existence of the solutions v follows from the general theory of monotone operators.We will assume that the following integral identity holds: w(x, ρ 1 ) dx.
To formulate our results, we need the notion of the capacity.Let E ⊂ B r (x 0 ) ⊂ B ρ (x 0 ) and for any m > 0 set where the infimum is taken over the set M(E) of all functions ϕ ∈ W 0 (B 16ρ (x 0 )) with ϕ 1 on E. If m = 1, this definition leads to the standard definition of Cp(•)(E, B 16ρ (x 0 )) capacity (see, e.g.[3]).

Upper bound for the function v
We note that in the standard case (i.e. if L = 0) the upper bound for the function v was proved in [43] (see also [44,Chap. 8,Sec. 3], [45]).
Lemma 3.1.There exists positive number γ 1 depending only on the data such that if provided that (3.7) Proof.First we prove inequality on the right-hand side of (3.6).Fix σ ∈ (0, 1/4) and for any Further we will assume that since otherwise, by (3.5) inequality (3.6) is evident, moreover this inequality yields Testing (3.1) by ϕ = (v − k j+1 ) + ζ q j and using the Young inequality we obtain Using again the Young inequality, assuming that k > ε 0 M 0 , where ε 0 ∈ (0, 1) is small enough, from the previous we have 0 ) dx+ Choose γ 1 > 0 large enough, by our assumption Lemma 2.2 is applicable, therefore from this we obtain 0 , 16ρ)] 0 ) dx Hence, setting M σ := M ∞ , by standard arguments (see, e.g.[32]) and by our choice, we arrive at Let us estimate the second term on the right-hand side of (3.8).For this we set v M 0 := min{v, M 0 }, by Lemma 2.2 we have for any ε ∈ (0, 1) [w(x, ρ Collecting the last two inequalities we obtain Let us estimate the second term on the right-hand side of (3.9).Let ψ ∈ M(E) be such that 0 , 16ρ).
using the Young inequality and the previous inequality, we have 0 , 16ρ) .
This inequality, the Young inequality and (3.9) imply that Iterating the last inequality, choosing ε 0 and then ε = ε(ε 0 ) small enough, by (3.5) we arrive at which completes the proof of the lemma.Now we prove inequality on the left-hand side of (3.6) . Testing (3.1) by ϕ = v − m ζ q 1 , using the Young inequality , we obtain for any ε 1 > 0 and using the Young inequality, we estimate the first term on the right-hand side of the previous inequality as follows: Combining the last two inequalities and using the definition of capacity, we obtain Choose ε 1 from the condition ε 1 = M ( 3 2 ρ) ρ , then inequality (3.10) yields from which the required inequality follows, this completes the proof of the lemma.

Lower bound for the function v
Further we need the following lemma.
Similarly, by Lemma 3.1 Collecting estimates (3.10), (3.12), (3.13) we obtain w(x, which completes the proof of the lemma. The following lemma is the main result of this Section Lemma 3.3.There exists ε ∈ (0, 1) depending only on the data such that either or provided that inequality (3.7) holds.
Proof.Lemma 3.3 is a consequence of Lemma 3.2, for this we first estimate the capacity of the set E from below.Let ϕ ∈ W 0 (B 16ρ (x 0 )), ϕ = 1 on E , then by Lemmas 2.1, 2.2, 3.1 and using the evident inequalities γ Since ϕ is arbitrary, estimate (3.16) yields If inequality (3.14) is violated then so, if ε is sufficiently small, from the previous we arrive at And hence provided that (3.14) is violated and ε is sufficiently small.Now we use Lemma 3.2 for this we set We have by Lemmas 2.1 and 3.2 , which completes the proof of the lemma.
by Lemma 3.3 there holds

4 3 4. 1 2 For 0 4 . 1 . 2 ( 1 )
Harnack's inequality, proof of Theorems 1.2 and 1.Weak Harnack inequality, proof of Theorem 1.< m < M set E(ρ, m) := B ρ (x 0 ) : u m .As it was mentioned in Section 1 Theorem 1.2 is a simple consequence of the following lemma Lemma Let u be a non-negative bounded super-solution to equation (1.1) in Ω and let condition (1.8) be fulfilled, then there exist positive numbers C 2 , C 3 depending only on the data such that |E(ρ, m)| C 2 |B ρ (x 0 )| m 1−p ρ + min B ρ provided that B 16ρ (x 0 ) ⊂ Ω and We construct the solution v of the problem (3.1) in D = B 16ρ (x 0 ) \ E(ρ, m), since u v on ∂D, by (1.8) u v in D. First we use Lemma 3.3, if inequality (3.14) is violated, i.e. if ε m
10) is violated, then condition (2.8) holds due to (2.12) and the choice of ξ, r j sup j (u − k j ) + 1.So,by Lemma 2.2 and inequality (2.1) we have