Entropy Solution for Doubly Nonlinear Elliptic Anisotropic Problems ...

Hindawi Publishing Corporation International Journal of Differential Equations Volume 2015, Article ID 919608, 15 pages http://dx.doi.org/10.1155/2015/919608

Research Article Entropy Solution for Doubly Nonlinear Elliptic Anisotropic Problems with Robin Boundary Conditions I. Ibrango1 and S. Ouaro2 1

Laboratoire de Math´ematiques et Informatique (LAMI), UFR, Sciences et Techniques, Universit´e Polytechnique de Bobo-Dioulasso, 01 BP 1091, Bobo, Bobo-Dioulasso 01, Burkina Faso 2 Laboratoire de Math´ematiques et Informatique (LAMI), UFR, Sciences Exactes et Appliqu´ees, Universit´e de Ouagadougou, 03 BP 7021, Ouaga, Ouagadougou 03, Burkina Faso Correspondence should be addressed to S. Ouaro; [email protected] Received 4 July 2015; Accepted 21 October 2015 Academic Editor: M. El-Gebeily Copyright © 2015 I. Ibrango and S. Ouaro. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We study in this paper nonlinear anisotropic problems with Robin boundary conditions. We prove, by using the technic of monotone operators in Banach spaces, the existence of a sequence of weak solutions of approximation problems associated with the anisotropic Robin boundary value problem. For the existence and uniqueness of entropy solutions, we prove that the sequence of weak solutions converges to a measurable function which is the entropy solution of the anisotropic Robin boundary value problem.

1. Introduction The aim of this paper is to study the following nonlinear anisotropic elliptic Robin boundary value problem:

All papers on problems like (1) considered particular cases of function 𝑏. Indeed, in [1], Bonzi et al. studied the following problems: 𝑁

𝜕 𝜕𝑢 𝑎𝑖 (𝑥, ) + |𝑢|𝑝𝑀 (𝑥)−2 𝑢 = 𝑓 𝜕𝑥𝑖 𝑖=1 𝜕𝑥𝑖

−∑ 𝑁

𝜕 𝜕𝑢 𝑎𝑖 (𝑥, ) + 𝑏 (𝑢) = 𝑓 −∑ 𝜕𝑥 𝜕𝑥 𝑖 𝑖 𝑖=1

𝑁

∑𝑎𝑖 (𝑥,

in Ω,

𝑖=1

(1) 𝑁

𝜕𝑢 ) 𝜂𝑖 = −𝛾 (𝑢) ∑𝑎𝑖 (𝑥, 𝜕𝑥 𝑖 𝑖=1

on 𝜕Ω,

where Ω is an open bounded domain of R𝑁 (𝑁 ≥ 3) with smooth boundary and meas(Ω) > 0, 𝑓 ∈ 𝐿1 (Ω), 𝜂 = (𝜂1 , . . . , 𝜂𝑁) is the unit outward normal on 𝜕Ω, and 𝛾(𝑢) = |𝑢|𝑟(𝑥)−2 𝑢.

in Ω,

𝜕𝑢 ) 𝜂 = − |𝑢|𝑟(𝑥)−2 𝑢 𝜕𝑥𝑖 𝑖

(2)

on 𝜕Ω, where 𝑓 ∈ 𝐿1 (Ω). The authors use minimization technics used in [2] or [3] (see also [4, 5]) to prove the existence and uniqueness of entropy solution. The Robin type boundary conditions in the variable exponents setting are new and interesting problems and were for the first time studied by Boureanu and Radulescu in [3]. The main difficulty for the study of problem in [3] was the definition of an admissible space of solutions. The authors defined the appropriate space and obtained its properties

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International Journal of Differential Equations

which permit them to use minimization method to prove the existence of weak solutions to the following problem: 𝑁

𝜕 𝜕𝑢 𝑎𝑖 (𝑥, ) + 𝑏 (𝑥) |𝑢|𝑝𝑀 (𝑥)−2 𝑢 = 𝑓 (𝑥, 𝑢) 𝜕𝑥 𝜕𝑥 𝑖 𝑖 𝑖=1

𝑏 : R 󳨀→ R is continuous, surjective, nondecreasing

−∑

with 𝑏 (0) = 0.

in Ω, 𝑢≥0

in Ω,

The hypotheses on 𝑎𝑖 are classical in the study of nonlinear problems (see [1, 3]). The function 𝑏 is such that

(3)

Throughout this paper, for any 𝑖 = 1, . . . , 𝑁, we assume that 𝑝 (𝑁 − 1) 𝑝 (𝑁 − 1) < 𝑝𝑖− < , 𝑁−𝑝 𝑁 (𝑝 − 1)

𝑁

𝜕𝑢 ∑𝑎𝑖 (𝑥, ) 𝜂𝑖 = 𝑔 (𝑥, 𝑢) 𝜕𝑥 𝑖 𝑖=1

𝑝𝑖+ − 𝑝𝑖− − 1 𝑝−𝑁 < , 𝑝𝑖− 𝑝 (𝑁 − 1)

on 𝜕Ω. Since we consider 𝐿1 -data 𝑓 instead of the function 𝑓(𝑥, 𝑢) considered in [3], the suitable notion of solution is the entropy solution introduced by B´enilan et al. in [6] (see also [7]). With the Robin type boundary conditions, the values of the solutions at the boundary must be precise and the notion of solutions considered must include the boundary condition. In this paper, as the function 𝑏 is more general, it is not possible to use minimization technic to get the existence of solution. Therefore, we used the technic of monotone operators in Banach spaces (see [8]) to get the existence of entropy solutions of (1). For presenting our main result, we first have to describe the data involved in our problem. Let Ω be a bounded domain ⃗ = in R𝑁 (𝑁 ≥ 3) with smooth boundary domain 𝜕Ω and 𝑝(⋅) (𝑝1 (⋅), . . . , 𝑝𝑁(⋅)) such that, for any 𝑖 = 1, . . . , 𝑁, 𝑝𝑖 (⋅) : Ω → [2; 𝑁) is a continuous function with 1
1, 𝑝 𝑖=1 𝑖

∑

− where 𝑁/𝑝 = ∑𝑁 𝑖=1 (1/𝑝𝑖 ). We put for all 𝑥 ∈ Ω,

𝑝𝑀 (𝑥) max {𝑝1 (𝑥) , . . . , 𝑝𝑁 (𝑥)} , 𝑝𝑚 (𝑥) min {𝑝1 (𝑥) , . . . , 𝑝𝑁 (𝑥)}

(10)

and for all 𝑥 ∈ 𝜕Ω, (𝑁 − 1) 𝑝 (𝑥) { 𝑝𝜕 (𝑥) = { 𝑁 − 𝑝 (𝑥) {+∞

if 𝑝 (𝑥) < 𝑁, if 𝑝 (𝑥) ≥ 𝑁.

(11)

We make the following assumption: 𝑟 ∈ 𝐶 (Ω) 𝜕 with 1 < 𝑟− ≤ 𝑟+ < min {𝑝1𝜕 (𝑥) , . . . , 𝑝𝑁 (𝑥)} .

(12)

𝑥∈𝜕Ω

Note that the function 𝛾 is continuous, defined on R with 𝛾(𝑡)𝑡 ≥ 0 for all 𝑡 in R and 𝛾(0) = 0. A prototype example that is covered by our assumptions ⃗ is the following anisotropic 𝑝(⋅)-harmonic system: 𝜕 󵄨󵄨󵄨󵄨 𝜕𝑢 󵄨󵄨󵄨󵄨𝑝𝑖 (𝑥)−2 𝜕𝑢 (󵄨󵄨 ) = 𝑓, 󵄨󵄨 󵄨 󵄨 𝜕𝑥𝑖 𝑖=1 𝜕𝑥𝑖 󵄨 𝜕𝑥𝑖 󵄨 𝑁

−∑

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which, in the particular case when 𝑝𝑖 = 𝑝 for any 𝑖 = 1, . . . , 𝑁, is the 𝑝-Laplace equation. The rest of the paper is organized as follows. In Section 2, we present some preliminary results. In Section 3, we study the existence and uniqueness of entropy solution.

(6)

2. Preliminaries We recall in this section some definitions and basic properties of anisotropic Lebesgue and Sobolev spaces with variable exponent. Set

(iii) There exists a positive constant 𝐶3 such that 󵄨 󵄨𝑝 (𝑥) 𝑎𝑖 (𝑥, 𝜉) ⋅ 𝜉 ≥ 𝐶3 󵄨󵄨󵄨𝜉󵄨󵄨󵄨 𝑖 ,

(8)

(7) 𝐶+ (Ω) = {𝑝 ∈ 𝐶 (Ω) such that min 𝑝 (𝑥) > 1} . 𝑥∈Ω

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International Journal of Differential Equations

3

For any 𝑝 ∈ 𝐶+ (Ω), the variable exponent Lebesgue space is defined by

ess inf (𝑝𝑀 (𝑥) − 𝑞 (𝑥)) > 0,

𝐿𝑝(⋅) (Ω) {𝑢 : Ω

(15) 󳨀→ R, measurable such that ∫ |𝑢| Ω

𝑑𝑥 < ∞} ,

⃗

|𝑢|𝑝(⋅) 󵄨󵄨 𝑢 (𝑥) 󵄨󵄨𝑝(𝑥) 󵄨 󵄨󵄨 inf {𝜆 > 0 such that ∫ 󵄨󵄨󵄨 𝑑𝑥 ≤ 1} . 󵄨󵄨 Ω 󵄨󵄨 𝜆 󵄨󵄨

(16)

The 𝑝(⋅)-modular of the 𝐿𝑝(⋅) (Ω) space is the mapping 𝜌𝑝(⋅) : 𝐿𝑝(⋅) (Ω) → R defined by 𝜌𝑝(⋅) (𝑢) ∫ |𝑢|𝑝(𝑥) 𝑑𝑥.

(17)

Ω

(Ω), the following inequality (see [9, 10]) will 𝑝−

𝑝+

𝑝−

𝑝+

(18)

𝑞(⋅)

For any 𝑢 ∈ 𝐿 (Ω) and V ∈ 𝐿 (Ω) with (1/𝑝(𝑥)) + (1/𝑞(𝑥)) = 1 in Ω, we have the following H¨older type inequality: 󵄨 󵄨󵄨 󵄨󵄨∫ 𝑢V 𝑑𝑥󵄨󵄨󵄨 ≤ ( 1 + 1 ) |𝑢| |V| . 󵄨󵄨 󵄨󵄨 𝑝(⋅) 𝑞(⋅) 𝑝− 𝑞− 󵄨 󵄨 Ω

𝜕 1 ≤ 𝑟 (𝑥) < min {𝑝1𝜕 (𝑥) , . . . , 𝑝𝑁 (𝑥)} , ∀𝑥 ∈ 𝜕Ω.

(24)

Then, there is a compact boundary trace embedding ⃗

𝑊1,𝑝(⋅) (Ω) 󳨅→ 𝐿𝑟(⋅) (𝜕Ω) .

(25)

We introduce the numbers 𝑞=

𝑁 (𝑝 − 1) , 𝑁−1

(26)

⃗

𝑊1,𝑝(⋅) (Ω) {𝑢 ∈ 𝐿𝑝𝑀 (⋅) (Ω) such that

𝜕𝑢 𝜕𝑥𝑖

(19)

(20)

(Ω) , 𝑖 = 1, . . . , 𝑁} .

Theorem 3. Let 𝑝1 , . . . , 𝑝𝑁 ∈ [1, +∞); 𝑔 ∈ 𝑊1,(𝑝1 ,...,𝑝𝑁 ) (Ω) and ∗

𝑞 = (𝑝)

∗

𝑖𝑓 (𝑝) < 𝑁,

(27)

∗

𝑖𝑓 (𝑝) ≥ 𝑁.

Then, there exists a constant 𝐶4 > 0 depending on 𝑁, 𝑝1 , . . . , 𝑝𝑁 if 𝑝 < 𝑁 and also on 𝑞 and meas(Ω) if 𝑝 ≥ 𝑁 such that 1/𝑁 𝑁 󵄩󵄩 𝜕𝑔 󵄩󵄩 󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩 󵄩 ] . 󵄩󵄩 󵄩󵄩𝑔󵄩󵄩𝐿𝑞 (Ω) ≤ 𝐶4 ∏ [󵄩󵄩󵄩𝑔󵄩󵄩󵄩𝐿𝑝𝑀 (Ω) + 󵄩󵄩󵄩 󵄩󵄩 𝜕𝑥𝑖 󵄩󵄩𝐿𝑝𝑖 (Ω) 𝑖=1

(28)

In this paper, we will use the Marcinkiewicz space M𝑞 (Ω) (1 < 𝑞 < +∞) as the set of measurable functions 𝑔 : Ω → R for which the distribution function 󵄨 󵄨 𝜆 𝑔 (𝑘) = meas ({𝑥 ∈ Ω : 󵄨󵄨󵄨𝑔 (𝑥)󵄨󵄨󵄨 > 𝑘}) ,

𝑘≥0

(29)

satisfies an estimate of the form

⃗ (Ω) is a separable and reflexive Banach space (see [2]) 𝑊1,𝑝(⋅) under the norm 𝑁 󵄨󵄨 󵄨 󵄨󵄨 𝜕𝑢 󵄨󵄨󵄨 ‖𝑢‖𝑝(⋅) 󵄨󵄨 . ⃗ = |𝑢|𝑝𝑀 (⋅) + ∑ 󵄨󵄨 󵄨 󵄨 𝑖=1 󵄨 𝜕𝑥𝑖 󵄨𝑝𝑖 (⋅)

The following result is due to Troisi (see [12]).

𝑞 ∈ [1, + ∞)

If Ω is bounded and 𝑝, 𝑞 ∈ 𝐶+ (Ω) such that 𝑝(𝑥) ≤ 𝑞(𝑥) for any 𝑥 ∈ Ω, then the embedding 𝐿𝑞(⋅) (Ω) 󳨅→ 𝐿𝑝(⋅) (Ω) is continuous (see [11, Theorem 2.8]). Herein, we need the following anisotropic Sobolev space with variable exponent:

∈𝐿

Theorem 2 (see [3, Theorem 6]). Let Ω ⊂ R𝑁 (𝑁 ≥ 2) ⃗ ∈ be a bounded open set with smooth boundary and let 𝑝(⋅) (𝐶+ (Ω))𝑁, 𝑟 ∈ 𝐶(Ω) satisfy the condition

𝑁 (𝑝 − 1) 𝑁𝑞 𝑞 = = . 𝑁−𝑝 𝑁−𝑞

≤ max {|𝑢|𝑝(⋅) ; |𝑢|𝑝(⋅) } .

𝑝𝑖 (⋅)

(23)

∗

min {|𝑢|𝑝(⋅) ; |𝑢|𝑝(⋅) } ≤ 𝜌𝑝(⋅) (𝑢)

𝑝(⋅)

one has the compact embedding 𝑊1,𝑝(⋅) (Ω) 󳨅→ 𝐿𝑞(⋅) (Ω) .

endowed with the so-called Luxemburg norm

For any 𝑢 ∈ 𝐿 be used later:

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𝑥∈Ω

𝑝(𝑥)

𝑝(⋅)

𝑝𝑖 (𝑥) ≥ 1 a.e. in Ω. Then, for any 𝑞 ∈ 𝐿∞ (Ω) with 𝑞(𝑥) ≥ 1 a.e. in Ω such that

(21)

We need the following embedding and trace results. Theorem 1 (see [9, Corollary 2.1]). Let Ω ⊂ R𝑁 (𝑁 ≥ 3) be a bounded open set and for all 𝑖 = 1, . . . , 𝑁, 𝑝𝑖 ∈ 𝐿∞ (Ω),

𝜆 𝑔 (𝑘) ≤ 𝐶𝑘−𝑞 , for some finite constant 𝐶 > 0.

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We will use the following pseudonorm in M𝑞 (Ω): 󵄩󵄩 󵄩󵄩 −𝑞 󵄩󵄩𝑔󵄩󵄩M𝑞 (Ω) inf {𝐶 > 0 : 𝜆 𝑔 (𝑘) ≤ 𝐶𝑘 , ∀𝑘 > 0} .

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For any 𝑘 > 0, the truncation function 𝑇𝑘 is defined by 𝑇𝑘 (𝑠) = max {−𝑘; min {𝑘; 𝑠}} .

(32)

It is clear that lim𝑘 → +∞ 𝑇𝑘 (𝑠) = 𝑠 and |𝑇𝑘 (𝑠)| = min{|𝑠|; 𝑘}.

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International Journal of Differential Equations

⃗ In order to simplify the notation, for any V ∈ 𝑊1,𝑝(⋅) (Ω), we use V instead of V|𝜕Ω for the trace of V on 𝜕Ω. ⃗ Set T1,𝑝(⋅) (Ω) as the set of the measurable functions 𝑢 : ⃗ Ω → R such that, for any 𝑘 > 0, 𝑇𝑘 (𝑢) ∈ 𝑊1,𝑝(⋅) (Ω). ⃗ 1,𝑝(⋅) We define the space Ttr (Ω) as the set of functions 𝑢 ∈ ⃗ ⃗ T1,𝑝(⋅) (Ω) such that there exists a sequence (𝑢𝑛 )𝑛 ⊂ 𝑊1,𝑝(⋅) (Ω) satisfying

𝜕𝑇𝑘 (𝑢𝑛 ) 𝜕𝑇 (𝑢) 󳨀→ 𝑘 𝜕𝑥𝑖 𝜕𝑥𝑖

in 𝐿1 (Ω) , ∀𝑘 > 0,

We need the following lemma proved in [13]. ⃗ Lemma 4. Let 𝑔 be a nonnegative function in 𝑊1,𝑝(⋅) (Ω). Assume 𝑝 < 𝑁 and there exists a constant 𝐶 > 0 such that

𝑋0 = {(𝑢, V) ∈ 𝐸 : V = 𝜏 (𝑢)} ,

󵄨󵄨 𝜕𝑔 󵄨󵄨𝑝𝑖− 󵄨󵄨 󵄨󵄨 𝑑𝑥 + ∑ ∫ 󵄨󵄨 󵄨󵄨 𝑑𝑥 ≤ 𝐶 (1 + 𝑘) , 󵄨 𝜕𝑥 {|𝑔|≤𝑘} 󵄨 𝑖 󵄨󵄨 (34) 𝑖=1 𝑁

𝑁

∑ ∫ 𝑎𝑖 (𝑥, 𝑖=1 Ω

Ω

(35)

where 𝑞∗ = 𝑁(𝑝 − 1)/(𝑁 − 𝑝).

= ∫ 𝑓𝑛 V 𝑑𝑥, Ω

where 𝑓𝑛 = 𝑇𝑛 (𝑓), and we define an operator 𝐴 𝑛 by

Ω

⃗ 1,𝑝(⋅)

Definition 5. A measurable function 𝑢 ∈ Ttr (Ω) is an entropy solution of problem (1) if 𝑏(𝑢) ∈ 𝐿1 (Ω), 𝛾(𝑢) ∈ 𝐿1 (𝜕Ω), and, for every 𝑘 > 0,

𝜕Ω

𝑁

⟨𝐴 (𝑢) , V⟩ = ∑ ∫ 𝑎𝑖 (𝑥, 𝑖=1 Ω

+ 𝜀 ∫ |𝑢| Ω

(36)

(41)

𝑢V 𝑑𝑥.

Note that

+ ∫ 𝛾 (𝑢) 𝑇𝑘 (𝑢 − 𝜑) 𝑑𝜎 ≤ ∫ 𝑓𝑇𝑘 (𝑢 − 𝜑) 𝑑𝑥, 𝜕Ω

𝜕𝑢 𝜕V ) 𝑑𝑥 𝜕𝑥𝑖 𝜕𝑥𝑖

𝑝𝑀 (𝑥)−2

Ω

∀𝑢, V ∈ 𝑋0 ,

where

𝜕𝑢 𝜕 ) 𝑇 (𝑢 − 𝜑) 𝑑𝑥 𝜕𝑥𝑖 𝜕𝑥𝑖 𝑘

+ ∫ 𝑏 (𝑢) 𝑇𝑘 (𝑢 − 𝜑) 𝑑𝑥

(40)

+ ∫ 𝑇𝑛 (𝛾 (𝑢)) V 𝑑𝜎

The notion of entropy solutions to problem (1) where the data 𝑓 belongs to 𝐿1 (Ω) is the following.

𝑖=1 Ω

(39)

𝜕Ω

⟨𝐴 𝑛 (𝑢) , V⟩ = ⟨𝐴 (𝑢) , V⟩ + ∫ 𝑇𝑛 (𝑏 (𝑢)) V 𝑑𝑥

3. Entropy Solutions

∑ ∫ 𝑎𝑖 (𝑥,

𝜕𝑢𝑛 𝜕V 󵄨 󵄨𝑝 (𝑥)−2 ) 𝑑𝑥 + 𝜀 ∫ 󵄨󵄨󵄨𝑢𝑛 󵄨󵄨󵄨 𝑀 𝑢𝑛 V 𝑑𝑥 𝜕𝑥𝑖 𝜕𝑥𝑖 Ω

+ ∫ 𝑇𝑛 (𝑏 (𝑢𝑛 )) V 𝑑𝑥 + ∫ 𝑇𝑛 (𝛾 (𝑢𝑛 )) V 𝑑𝜎

Then, there exists a constant 𝐷, depending on 𝐶, such that 󵄩󵄩 󵄩󵄩 ∗ 󵄩󵄩𝑔󵄩󵄩M𝑞 (Ω) ≤ 𝐷,

(38)

where 𝜏(𝑢) is the trace of 𝑢 ∈ Ttr (Ω) in the usual sense, ⃗ since 𝑢 ∈ 𝑊1,𝑝(⋅) (Ω). In the sequel, we will identify an element ⃗ (𝑢, V) ∈ 𝑋0 with its representative 𝑢 ∈ 𝑊1,𝑝(⋅) (Ω). For any 𝑛 ∈ N and 𝜀 > 0, we consider the sequence of approximate problems:

∀𝑘 > 0.

𝑁

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1,𝑝(⋅)

a.e. on 𝜕Ω.

󵄨 󵄨 ∫ 󵄨󵄨󵄨𝑇𝑘 (𝑔)󵄨󵄨󵄨 Ω

⃗

𝐸 = 𝑊1,𝑝(⋅) (Ω) × 𝐿𝑝𝑀 (⋅) (𝜕Ω) .

(33)

there exists a measurable function V on 𝜕Ω such that 𝑢𝑛

− 𝑝𝑀

Step 1 (the approximate problem). We define the reflexive space

Let 𝑋0 be the subspace of 𝐸 defined by

𝑢𝑛 󳨀→ 𝑢 a.e. in Ω,

󳨀→ V

Proof. The proof is done in three steps.

Ω

⃗ for every 𝜑 ∈ 𝑊1,𝑝(⋅) (Ω) ∩ 𝐿∞ (Ω).

The existence result is the following theorem. Theorem 6. Assume that (4)–(12) hold. Then, problem (1) admits at least one entropy solution.

󵄩󵄩 󵄩󵄩 󵄩󵄩𝑓𝑛 󵄩󵄩∞ ≤

󵄩󵄩 󵄩󵄩 󵄩󵄩𝑓󵄩󵄩1 , meas (Ω)

󵄩 󵄩 󵄩󵄩 󵄩󵄩 󵄨 󵄨 󵄨 󵄨 󵄩󵄩𝑓𝑛 󵄩󵄩1 = ∫ 󵄨󵄨󵄨𝑓𝑛 󵄨󵄨󵄨 𝑑𝑥 ≤ ∫ 󵄨󵄨󵄨𝑓󵄨󵄨󵄨 𝑑𝑥 = 󵄩󵄩󵄩𝑓󵄩󵄩󵄩1 , Ω Ω 𝑓𝑛 󳨀󳨀󳨀󳨀󳨀󳨀→ 𝑓 in 𝐿1 (Ω) , a.e. in Ω. 𝑛 → +∞

(42)

International Journal of Differential Equations

5

Assertion 1 (the operator 𝐴 is of type M). (i) The operator 𝐴 ⃗ is monotone. Indeed, for 𝑢, V ∈ 𝑊1,𝑝(⋅) (Ω), we have ⟨𝐴 (𝑢) − 𝐴 (V) , 𝑢 − V⟩ = ⟨𝐴 (𝑢) , 𝑢 − V⟩ + ⟨𝐴 (V) , V 𝑁

− 𝑢⟩ = ∫ ∑𝑎𝑖 (𝑥, Ω 𝑖=1

𝜕𝑢 𝜕 (𝑢 − V) ) 𝑑𝑥 𝜕𝑥𝑖 𝜕𝑥𝑖

Since the operator 𝐴 is monotone and hemicontinuous, then according to Lemma 2.1 in [8], 𝐴 is of type M. Assertion 2 (the operator 𝐴 𝑛 is of type M). Indeed, let (𝑢𝑘 )𝑘∈N be a sequence in 𝑋0 such that

+ 𝜀 ∫ |𝑢|𝑝𝑀 (𝑥)−2 𝑢 (𝑢 − V) 𝑑𝑥 Ω

𝑁

+ ∫ ∑𝑎𝑖 (𝑥, Ω 𝑖=1

Using assumption (5) and [11, Theorems 4.1 and 4.2] 󸀠 we have 𝜓𝑖0 (𝑥, 𝑤) → 𝜓𝑖0 (𝑥, 𝑤0 ) in 𝐿𝑝𝑖0 (⋅) (Ω). Then, we deduce that 𝜑 is continuous; namely, the operator 𝐴 is hemicontinuous.

𝜕V 𝜕 (V − 𝑢) ) 𝑑𝑥 + 𝜀 ∫ |V|𝑝𝑀 (𝑥)−2 𝜕𝑥𝑖 𝜕𝑥𝑖 Ω

𝑢𝑘 ⇀ 𝑢 weakly in 𝑋0 , (43)

𝐴 𝑛 𝑢𝑘 ⇀ 𝜒 weakly in 𝑋0󸀠 ,

⋅ V (V − 𝑢) 𝑑𝑥

lim sup ⟨𝐴 𝑛 (𝑢𝑘 ) , 𝑢𝑘 ⟩ ≤ ⟨𝜒, 𝑢⟩ . 𝑘 → +∞

𝑁

𝜕𝑢 𝜕V = ∫ ∑ [𝑎𝑖 (𝑥, ) − 𝑎𝑖 (𝑥, )] 𝜕𝑥𝑖 𝜕𝑥𝑖 Ω 𝑖=1 ⋅(

Since 𝑇𝑛 (𝑏 (𝑢)) 𝑢 ≥ 0,

𝜕𝑢 𝜕V − ) 𝑑𝑥 + 𝜀 ∫ (|𝑢|𝑝𝑀 (𝑥)−2 𝑢 𝜕𝑥𝑖 𝜕𝑥𝑖 Ω

by Fatou’s lemma, we obtain that

Therefore, ⟨𝐴 (𝑢) − 𝐴 (V) , 𝑢 − V⟩ ≥ 0,

(44)

since for 𝑖 = 1, . . . , 𝑁, for almost every 𝑥 ∈ Ω, 𝑎𝑖 (𝑥, ⋅) and 𝑡 󳨃→ |𝑡|𝑝𝑀 (𝑥)−2 𝑡 are monotone. (ii) The operator 𝐴 is hemicontinuous. Indeed, for every ⃗ 𝑢, V in 𝑊1,𝑝(⋅) (Ω), let 𝜑 : 𝑡 ∈ R 󳨃󳨀→ 𝜑 (𝑡) = ⟨𝐴 (𝑢 + 𝑡V) , V⟩

(45)

and let 𝑡, 𝑡0 ∈ R be such that 𝑡 → 𝑡0 . We have 𝑤 = 𝑢 + 𝑡V → ⃗ 𝑤0 = 𝑢 + 𝑡0 V in 𝑊1,𝑝(⋅) (Ω). Using the H¨older type inequality, there exists 𝑖0 ∈ {1, . . . , 𝑁} such that 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨𝜑 (𝑡) − 𝜑 (𝑡0 )󵄨󵄨 = 󵄨󵄨⟨𝐴 (𝑢 + 𝑡V) , V⟩ − ⟨𝐴 (𝑢 + 𝑡0 V) , V⟩󵄨󵄨 󵄨󵄨 𝜕𝑤 󵄨󵄨󵄨 󵄨󵄨󵄨 𝜕V 󵄨󵄨󵄨󵄨 𝜕𝑤 󵄨 ≤ ∑ ∫ 󵄨󵄨󵄨𝑎𝑖 (𝑥, ) − 𝑎𝑖 (𝑥, 0 )󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨 𝑑𝑥 󵄨 𝜕𝑥𝑖 𝜕𝑥𝑖 󵄨󵄨 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨󵄨 𝑖=1 Ω 󵄨 𝑁

1 1 + ) 𝑝𝑖−0 (𝑝󸀠 )− 𝑖0

0

1 1 ) − + 󸀠 𝑝𝑀 (𝑝𝑀)−

󵄨󵄨 󵄨 󵄨𝑝 (𝑥)−2 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨|𝑤|𝑝𝑀 (𝑥)−2 𝑤 − 󵄨󵄨󵄨𝑤0 󵄨󵄨󵄨 𝑀 𝑤0 󵄨󵄨 󸀠 |V|𝑝𝑀 (⋅) . 󵄨𝑝𝑀 (⋅) 󵄨 Let us denote 𝜓𝑖0 (𝑥, 𝑤) = 𝑎𝑖0 (𝑥, 𝜕𝑤/𝜕𝑥𝑖0 ).

𝑘 → +∞

Ω

𝜕Ω

(49)

≥ ∫ 𝑇𝑛 (𝑏 (𝑢)) 𝑢 𝑑𝑥 + ∫ 𝑇𝑛 (𝛾 (𝑢)) 𝑢 𝑑𝜎 Ω

𝜕Ω

and thanks to the Lebesgue dominated convergence theorem, we have lim (∫ 𝑇𝑛 (𝑏 (𝑢𝑘 )) V 𝑑𝑥 + ∫ 𝑇𝑛 (𝛾 (𝑢𝑘 )) V 𝑑𝜎)

𝑘 → +∞

Ω

𝜕Ω

(50)

= ∫ 𝑇𝑛 (𝑏 (𝑢)) V 𝑑𝑥 + ∫ 𝑇𝑛 (𝛾 (𝑢)) V 𝑑𝜎, Ω

𝜕Ω

for all V in 𝑋0 . Consequently, 𝑇𝑛 (𝑏 (𝑢𝑘 )) + 𝑇𝑛 (𝛾 (𝑢𝑘 )) ⇀ 𝑇𝑛 (𝑏 (𝑢)) + 𝑇𝑛 (𝛾 (𝑢))

(51)

Therefore, we deduce that

𝐴𝑢𝑘 ⇀ 𝜒 − (𝑇𝑛 (𝑏 (𝑢)) + 𝑇𝑛 (𝛾 (𝑢))) (46)

󵄨󵄨 󵄨󵄨 󵄨 󵄨 𝜕𝑤 󵄨󵄨 𝜕𝑤 󵄨 󵄨󵄨 𝜕V 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 ⋅ 󵄨󵄨󵄨󵄨𝑎𝑖0 (𝑥, ) − 𝑎𝑖0 (𝑥, 0 )󵄨󵄨󵄨󵄨 𝜕𝑥𝑖0 𝜕𝑥𝑖0 󵄨󵄨𝑝󸀠 (⋅) 󵄨󵄨󵄨 𝜕𝑥𝑖0 󵄨󵄨󵄨𝑝 (⋅) 󵄨󵄨 𝑖0 𝑖

lim inf (∫ 𝑇𝑛 (𝑏 (𝑢𝑘 )) 𝑢𝑘 𝑑𝑥 + ∫ 𝑇𝑛 (𝛾 (𝑢𝑘 )) 𝑢𝑘 𝑑𝜎)

weakly in 𝑋0󸀠 .

󵄨󵄨 󵄨 󵄨𝑝 (𝑥)−2 󵄨󵄨󵄨 + 𝜀 ∫ 󵄨󵄨󵄨|𝑤|𝑝𝑀 (𝑥)−2 𝑤 − 󵄨󵄨󵄨𝑤0 󵄨󵄨󵄨 𝑀 𝑤0 󵄨󵄨 |V| 𝑑𝑥 󵄨 Ω󵄨

+ 𝜀(

(48)

𝑇𝑛 (𝛾 (𝑢)) 𝑢 ≥ 0,

− |V|𝑝𝑀 (𝑥)−2 V) (𝑢 − V) 𝑑𝑥.

≤ 𝑁(

(47)

weakly in 𝑋0󸀠 . (52)

As in Assertion 1, we prove that the operator 𝐴 is of type M, so we have 𝐴𝑢 = 𝜒 − (𝑇𝑛 (𝑏 (𝑢)) + 𝑇𝑛 (𝛾 (𝑢))) .

(53)

Thus, it follows that 𝐴 𝑛 𝑢 = 𝜒.

(54)

Hence, 𝐴 𝑛 is of type M. Assertion 3 (the operator 𝐴 𝑛 is coercive). Indeed, since 𝑇𝑛 (𝑏 (𝑢)) 𝑢 + 𝑇𝑛 (𝛾 (𝑢)) 𝑢 ≥ 0,

(55)

6

International Journal of Differential Equations

then

So, combining (56) and (61) we get ⟨𝐴 𝑛 (𝑢) , 𝑢⟩ ≥ ⟨𝐴 (𝑢) , 𝑢⟩ .

(56)

According to (7), we have

−

𝑝𝑚 󵄨󵄨 𝜕𝑢 󵄨󵄨 − 𝑝𝑚 󵄨󵄨 󵄨󵄨 [ ] ⟨𝐴 𝑛 (𝑢) , 𝑢⟩ ≥ 𝐶 (∑ 󵄨󵄨 󵄨󵄨 ) + |𝑢|𝑝𝑀 (⋅) 󵄨 󵄨 𝑖=1 󵄨 𝜕𝑥𝑖 󵄨𝑝𝑖 (⋅) [ ] 𝑁

󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑖 (𝑥) 󵄨 󵄨󵄨 𝑑𝑥 + 𝜀 ∫ |𝑢|𝑝𝑀 (𝑥) 𝑑𝑥. (57) ⟨𝐴 (𝑢) , 𝑢⟩ ≥ 𝐶3 ∫ ∑ 󵄨󵄨󵄨 󵄨󵄨 󵄨 󵄨 𝜕𝑥 Ω 𝑖=1 󵄨 Ω 𝑖󵄨 𝑁

− 𝐶3 𝑁 ≥

𝐶 − 2𝑝𝑚 −1

where 𝐶 = min {

Denote

(58)

󵄨󵄨 𝜕𝑢 󵄨󵄨 󵄨 󵄨󵄨 J = {𝑖 ∈ {1, . . . , 𝑁} : 󵄨󵄨󵄨 > 1} . 󵄨󵄨 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨𝑝𝑖 (⋅)

−

− 𝐶3 𝑁 + 𝜀 ∫ |𝑢|𝑝𝑀 (𝑥) 𝑑𝑥 ≥

󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑖 (𝑥) 󵄨 󵄨󵄨 𝑑𝑥 + ∑ ∫ 󵄨󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 𝜕𝑥 Ω 󵄨 𝑖 𝑖∈J

𝐶 ‖𝑢‖𝑝(⋅) − ⃗ − 1 − 𝐶3 𝑁, 2𝑝𝑚 −1 − 𝑝𝑚

where 𝐶 = min {

󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑖+ 󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑖− 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 + ∑ 󵄨󵄨󵄨 ≥ ∑ 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 𝜕𝑥 𝜕𝑥 𝑖 󵄨𝑝𝑖 (⋅) 𝑖 󵄨𝑝𝑖 (⋅) 𝑖∈J 󵄨 𝑖∈I 󵄨 󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑖− 󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑚− 󵄨 󵄨󵄨 󵄨 󵄨󵄨 ≥ ∑ 󵄨󵄨󵄨 ≥ ∑ 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 𝜕𝑥 𝜕𝑥 󵄨 󵄨 󵄨 𝑖 𝑝𝑖 (⋅) 𝑖 󵄨𝑝𝑖 (⋅) 𝑖∈J 𝑖∈J

𝐶3 ; 1} . − 𝑁𝑝𝑚 −1

− Consequently, since 𝑝𝑚 > 1, the operator 𝐴 𝑛 is coercive.

(59)

Besides, the operator 𝐴 𝑛 is bounded and hemicontinuous. Then, for any 𝐹𝑛 = 𝑇𝑛 (𝑓) ∈ 𝐸󸀠 ⊂ 𝑋0󸀠 , we can deduce the existence of a function 𝑢𝑛 ∈ 𝑋0 such that ⟨𝐴 𝑛 (𝑢𝑛 ) , V⟩ = ⟨𝐹𝑛 , V⟩ , ∀V ∈ 𝑋0 .

󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑚− 󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑚− 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 ≥ ∑ 󵄨󵄨 − ∑ 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨𝑝 (⋅) 𝜕𝑥 𝜕𝑥 󵄨 󵄨 󵄨 𝑖 𝑖 𝑝𝑖 (⋅) 𝑖=1 𝑖∈I 𝑖 𝑁

󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑚− 󵄨 󵄨󵄨 ≥ ∑ 󵄨󵄨󵄨 − 𝑁. 󵄨󵄨 󵄨 󵄨 𝜕𝑥 𝑖 󵄨𝑝𝑖 (⋅) 𝑖=1 󵄨 −

− Using the convexity of the application 𝑡 ∈ R+ 󳨃→ 𝑡𝑝𝑚 , 𝑝𝑚 > 1, we obtain −

𝑝𝑚 𝑁 󵄨󵄨 󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑖 (𝑥) 󵄨 1 󵄨󵄨 󵄨󵄨 󵄨󵄨 𝜕𝑢 󵄨󵄨󵄨 𝑑𝑥 ≥ 𝑝− −1 (∑ 󵄨󵄨 ∫ ∑ 󵄨󵄨 󵄨󵄨 󵄨󵄨 ) − 𝑁. (60) 󵄨 󵄨 𝑁 𝑚 Ω 𝑖=1 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨 𝑖=1 󵄨 𝜕𝑥𝑖 󵄨𝑝𝑖 (⋅)

We are now going to prove that these approximated solutions 𝑢𝑛 tend, as 𝑛 goes to infinity, to a measurable function 𝑢 which is an entropy solution of problem (1). To start with, we establish some a priori estimates. Step 2 (a priori estimates). Assume (4)–(12) and let 𝑢𝑛 be a solution of problem (39). We have the following results. Lemma 7. There exists a constant 𝐶5 > 0 such that 𝑁

󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑖− 󵄨󵄨 𝑛 󵄨󵄨 󵄨󵄨 󵄨 𝑑𝑥 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨󵄨 |≤𝑘} 𝑛

󵄨 󵄨𝑝− ∫ 󵄨󵄨󵄨𝑇𝑘 (𝑢𝑛 )󵄨󵄨󵄨 𝑀 𝑑𝑥 + ∑ ∫ Ω {|𝑢

Then,

𝑖=1

⟨𝐴 𝑛 (𝑢) , 𝑢⟩ ≥

− 𝑝𝑚

󵄨󵄨 𝜕𝑢 󵄨󵄨 𝐶3 󵄨󵄨 󵄨󵄨 − −1 (∑ 󵄨 󵄨󵄨 𝜕𝑥 󵄨󵄨󵄨 ) 𝑝 𝑁 𝑚 󵄨 𝑖 󵄨𝑝𝑖 (⋅) 𝑖=1 𝑁

𝑝𝑀 (𝑥)

+ 𝜀 ∫ |𝑢| Ω

(65)

≤ 𝐶5 (𝑘 + 1) . (61)

𝑑𝑥 − 𝐶3 𝑁.

(i) Assume |𝑢|𝑝𝑀 (⋅) > 1. Then, (18) gives ∫Ω |𝑢|𝑝𝑀 (𝑥) 𝑑𝑥 ≥ .

(64)

Namely, 𝑢𝑛 is a weak solution of problem (39).

𝑁

𝑀 (⋅)

(63)

Ω

󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑖 (𝑥) 󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑖 (𝑥) 󵄨 󵄨󵄨 󵄨 󵄨󵄨 𝑑𝑥 = ∑ ∫ 󵄨󵄨󵄨 𝑑𝑥 ∫ ∑ 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 𝜕𝑥 𝜕𝑥 Ω 𝑖=1 󵄨 𝑖󵄨 𝑖󵄨 𝑖∈I Ω 󵄨 𝑁

|𝑢|𝑝

𝐶3 ; 𝜀} . − 𝑁𝑝𝑚 −1

𝑝𝑚 󵄨󵄨 𝜕𝑢 󵄨󵄨 − 𝑝𝑚 󵄨󵄨 󵄨󵄨 [ ] ⟨𝐴 𝑛 (𝑢) , 𝑢⟩ ≥ 𝐶 (∑ 󵄨󵄨 󵄨󵄨 ) + |𝑢|𝑝𝑀 (⋅) − 1 󵄨 󵄨 𝑖=1 󵄨 𝜕𝑥𝑖 󵄨𝑝𝑖 (⋅) [ ] 𝑁

We have

𝑁

(62)

(ii) Assume |𝑢|𝑝𝑀 (⋅) ≤ 1. Then, combining (56) and (61) we get

󵄨󵄨 𝜕𝑢 󵄨󵄨 󵄨 󵄨󵄨 I = {𝑖 ∈ {1, . . . , 𝑁} : 󵄨󵄨󵄨 ≤ 1} , 󵄨 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨󵄨𝑝 (⋅) 𝑖

− 𝑝𝑚

𝑝−

𝑚 ‖𝑢‖𝑝(⋅) ⃗ − 𝐶3 𝑁,

Proof. Let us take 𝑇𝑘 (𝑢𝑛 ) as a test function in (39). Since 󵄨 󵄨𝑝 (𝑥)−2 𝑢𝑛 𝑇𝑘 (𝑢𝑛 ) ∫ 𝑇𝑛 (𝑏 (𝑢𝑛 )) 𝑇𝑘 (𝑢𝑛 ) 𝑑𝑥 + 𝜀 ∫ 󵄨󵄨󵄨𝑢𝑛 󵄨󵄨󵄨 𝑀 Ω Ω + ∫ 𝑇𝑛 (𝛾 (𝑢𝑛 )) 𝑇𝑘 (𝑢𝑛 ) 𝑑𝜎 ≥ 0, 𝜕Ω

(66)

International Journal of Differential Equations

7

using relation (7), we obtain

Proof. (i) This is a consequence of Lemmas 4 and 7. (ii)

󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑖 (𝑥) 󵄨󵄨 𝑛 󵄨󵄨 󵄩 󵄩 𝑑𝑥 ≤ 𝑘 󵄩󵄩󵄩𝑓󵄩󵄩󵄩1 . 󵄨󵄨 󵄨󵄨 {|𝑢𝑛 |≤𝑘} 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨

𝑁

𝐶3 ∑ ∫ 𝑖=1

(67)

Then, we have 𝑁

󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑖− 󵄨󵄨 𝑛 󵄨󵄨 󵄨󵄨 󵄨󵄨 𝑑𝑥 {|𝑢𝑛 |≤𝑘} 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨

∑∫ 𝑖=1

󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑖− 󵄨󵄨 𝑛 󵄨󵄨 = ∑∫ 󵄨󵄨 󵄨󵄨 𝑑𝑥 󵄨 󵄨 𝑖=1 {|𝑢𝑛 |≤𝑘;|𝜕𝑢𝑛 /𝜕𝑥𝑖 |>1} 󵄨 𝜕𝑥𝑖 󵄨 𝑁

󵄨󵄨 𝜕𝑢 󵄨󵄨 󵄨 󵄨 𝜆 𝜕𝑢𝑛 /𝜕𝑥𝑖 (𝛼) = meas ({𝑥 ∈ Ω : 󵄨󵄨󵄨 𝑛 󵄨󵄨󵄨 > 𝛼}) 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨 󵄨󵄨 𝜕𝑢 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 = meas ({󵄨󵄨󵄨 𝑛 󵄨󵄨󵄨 > 𝛼; 󵄨󵄨󵄨𝑢𝑛 󵄨󵄨󵄨 ≤ 𝑘}) 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨 󵄨󵄨 𝜕𝑢 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 + meas ({󵄨󵄨󵄨 𝑛 󵄨󵄨󵄨 > 𝛼; 󵄨󵄨󵄨𝑢𝑛 󵄨󵄨󵄨 > 𝑘}) 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨 ≤∫

󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑖− 󵄨󵄨 𝑛 󵄨󵄨 + ∑∫ 󵄨󵄨 󵄨󵄨 𝑑𝑥 {|𝑢 |≤𝑘;|𝜕𝑢 /𝜕𝑥 |≤1} 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨 𝑛 𝑛 𝑖 𝑖=1

− 1 󵄨󵄨󵄨󵄨 𝜕𝑢𝑛 󵄨󵄨󵄨󵄨 𝑝𝑖 ( 󵄨󵄨 ≤∫ 󵄨󵄨) 𝑑𝑥 + 𝜆 𝑢𝑛 (𝑘) {|𝑢𝑛 |≤𝑘} 𝛼 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨

(68)

󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑖 (𝑥) 󵄨󵄨 𝑛 󵄨󵄨 𝑑𝑥 + 𝑁 ⋅ meas (Ω) 󵄨󵄨 󵄨󵄨 {|𝑢𝑛 |≤𝑘} 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨

−

Then, there exists a positive constant 𝐶6 such that

𝑘 󵄩󵄩 󵄩󵄩 ≤ 󵄩𝑓󵄩 + 𝑁 ⋅ meas (Ω) . 𝐶3 󵄩 󵄩1

−

󵄨 󵄨 ∫ 󵄨󵄨󵄨𝑇𝑘 (𝑢𝑛 )󵄨󵄨󵄨 Ω

∗

𝜆 𝜕𝑢𝑛 /𝜕𝑥𝑖 (𝛼) ≤ 𝐶6 (𝑘𝛼−𝑝𝑖 + 𝑘−𝑞 ) .

(72)

Let us consider the function

Moreover, we have − 𝑝𝑀

∗

≤ 𝛼−𝑝𝑖 𝐶󸀠 𝑘 + 𝐶𝑘−𝑞 .

≤ ∑∫ 𝑖=1

(71)

𝑑𝑥 + 𝜆 𝑢𝑛 (𝑘)

{|𝜕𝑢𝑛 /𝜕𝑥𝑖 |>𝛼;|𝑢𝑛 |≤𝑘}

𝑁

𝑁

(a) Let 𝛼 ≥ 1. For any 𝑘 ≥ 1, we have

𝑔 : [1, +∞) 󳨀→ R,

𝑑𝑥 = ∫

{|𝑇𝑘 (𝑢𝑛

󵄨󵄨 󵄨 󵄨󵄨𝑇𝑘 (𝑢𝑛 )󵄨󵄨󵄨 )|≤1}

+∫

{|𝑇𝑘 (𝑢𝑛

− 𝑝𝑀

𝑑𝑥

𝑡 󳨃󳨀→ 𝑔 (𝑡) =

󵄨󵄨 󵄨𝑝− 󵄨󵄨𝑇𝑘 (𝑢𝑛 )󵄨󵄨󵄨 𝑀 𝑑𝑥 )|>1}

≤ meas (Ω) + ∫

{|𝑇𝑘 (𝑢𝑛 )|>1}

𝑘

− 𝑝𝑀

−

(69)

𝑑𝑥

(73)

∗

We have 𝑔󸀠 (𝑡) = 0 for 𝑡 = (𝑞∗ 𝛼𝑝𝑖 )1/(𝑞 +1) . Thus, if we ∗ − take 𝑘 = (𝑞∗ 𝛼𝑝𝑖 )1/(𝑞 +1) ≥ 1 in (72) we get 𝜆 𝜕𝑢𝑛 /𝜕𝑥𝑖 (𝛼) ≤ 𝐶6 𝑘 (

−

≤ meas (Ω) (1 + 𝑘𝑝𝑀 ) .

𝑞∗ + 1 1 󸀠 −(𝑞∗ /(𝑞∗ +1))𝑝𝑖− − ) ≤ 𝐶6 𝛼 ∗ 𝑝 𝑞 𝛼 𝑖 (74)

−

≤ 𝐶6󸀠 𝛼−𝑝𝑖 𝑞/𝑝 , ∀𝛼 ≥ 1, where 𝐶6󸀠 is a positive constant.

Therefore, we get 𝑁

(b) If 0 ≤ 𝛼 < 1, we have

󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑖− 󵄨󵄨 𝑛 󵄨󵄨 󵄨󵄨 󵄨󵄨 𝑑𝑥 {|𝑢𝑛 |≤𝑘} 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨

󵄨 󵄨𝑝− ∫ 󵄨󵄨󵄨𝑇𝑘 (𝑢𝑛 )󵄨󵄨󵄨 𝑀 𝑑𝑥 + ∑ ∫ Ω

𝑡 −𝑞∗ . − + 𝑡 𝑝 𝑖 𝛼

𝑖=1

−

≤ meas (Ω) (1 + 𝑁 + 𝑘𝑝𝑀 ) +

𝑘 󵄩󵄩 󵄩󵄩 󵄩𝑓󵄩 𝐶3 󵄩 󵄩1

(70)

≤ 𝐶5 (1 + 𝑘) ,

󵄨󵄨 𝜕𝑢 󵄨󵄨 󵄨 󵄨 𝜆 𝜕𝑢𝑛 /𝜕𝑥𝑖 (𝛼) = meas ({󵄨󵄨󵄨 𝑛 󵄨󵄨󵄨 > 𝛼}) ≤ meas (Ω) 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨 ≤ meas (Ω) 𝛼

−𝑝𝑖− 𝑞/𝑝

.

Then −

where 𝐶5 = max{meas(Ω)(1 + 𝑁 + 𝑘

(75)

− 𝑝𝑀

); (1/𝐶3 )‖𝑓‖1 }.

Lemma 8. For any 𝑘 > 0, there exist two constants 𝐶7 > 0 and 𝐶8 > 0 such that (i) ‖𝑢𝑛 ‖M𝑞∗ (Ω) ≤ 𝐶7 , (ii) ‖𝜕𝑢𝑛 /𝜕𝑥𝑖 ‖M𝑝𝑖− 𝑞/𝑝 (Ω) ≤ 𝐶8 , ∀𝑖 = 1, . . . , 𝑁.

𝜆 𝜕𝑢𝑛 /𝜕𝑥𝑖 (𝛼) ≤ (𝐶6󸀠 + meas (Ω)) 𝛼−𝑝𝑖 𝑞/𝑝 ,

∀𝛼 ≥ 0.

(76)

Therefore, we deduce that there exists a positive constant 𝐶8 such that 󵄩󵄩 𝜕𝑢 󵄩󵄩 󵄩󵄩 𝑛 󵄩󵄩 ≤ 𝐶8 , 󵄩󵄩 󵄩 󵄩󵄩 𝜕𝑥𝑖 󵄩󵄩󵄩M𝑝𝑖− 𝑞/𝑝 (Ω)

∀𝑖 = 1, . . . , 𝑁.

(77)

8

International Journal of Differential Equations

Step 3 (existence of entropy solution). Using Lemma 8, we have the following useful lemma (see [13]).

We now focus our attention on the second term of (79). We have

Lemma 9. For 𝑖 = 1, . . . , 𝑁, as 𝑛 → +∞, one has 𝑎𝑖 (𝑥,

𝜕𝑢𝑛 𝜕𝑢 ) 󳨀→ 𝑎𝑖 (𝑥, ) 𝜕𝑥𝑖 𝜕𝑥𝑖

𝑖𝑛 𝐿1 (Ω) 𝑎.𝑒. 𝑥 ∈ Ω. (78)

In order to pass to the limit in relation (39), we also need the following convergence results which can be proved as in [7] (see also [1, 13]). ⃗ (Ω) is a Proposition 10. Assume (4)–(12). If 𝑢𝑛 ∈ 𝑊1,𝑝(⋅) weak solution of (𝑃𝑛 ) then the sequence (𝑢𝑛 )𝑛∈N∗ is Cauchy in measure. In particular, there exist a measurable function u and a subsequence still denoted by 𝑢𝑛 such that 𝑢𝑛 → 𝑢 in measure.

𝑇𝑛 (𝑏 (𝑢𝑛 )) 𝑇𝑘 (𝑢𝑛 − 𝜑) 󳨀→ 𝑏 (𝑢) 𝑇𝑘 (𝑢 − 𝜑) a.e. 𝑥 ∈ Ω, (82) 󵄨 󵄨󵄨 󵄨 󵄨 󵄨󵄨𝑇𝑛 (𝑏 (𝑢𝑛 )) 𝑇𝑘 (𝑢𝑛 − 𝜑)󵄨󵄨󵄨 ≤ 𝑘 󵄨󵄨󵄨𝑏 (𝑢𝑛 )󵄨󵄨󵄨 .

Now we show that |𝑏(𝑢𝑛 )| ≤ ‖𝑓‖1 /meas(Ω). Indeed, let us denote by

⃗ (Ω) is a weak Proposition 11. Assume (4)–(12). If 𝑢𝑛 ∈ 𝑊1,𝑝(⋅) solution of (𝑃𝑛 ) then

(i) there exists 𝑠 > 1 such that 𝑢𝑛 → 𝑢 a.e. in Ω and moreover 𝑢𝑛 ⇀ 𝑢 in 𝑊1,𝑠 (Ω), (ii) for all 𝑖 = 1, . . . , 𝑁, 𝜕𝑢𝑛 /𝜕𝑥𝑖 converges strongly in 𝐿1 (Ω). Moreover 𝑎𝑖 (𝑥, 𝜕𝑢𝑛 /𝜕𝑥𝑖 ) converges to 󸀠 𝑎𝑖 (𝑥, 𝜕𝑢/𝜕𝑥𝑖 ) in 𝐿1 (Ω) strongly and in 𝐿𝑝𝑖 (⋅) (Ω) weakly for all 𝑖 = 1, . . . , 𝑁, (iii) 𝑢𝑛 converges to some measurable function V a.e. in 𝜕Ω. We can now pass to the limit in relation (39). ⃗ (Ω) ∩ 𝐿∞ (Ω) and choosing 𝑇𝑘 (𝑢𝑛 − 𝜑) as a Let 𝜑 ∈ 𝑊1,𝑝(⋅) test function in (39), we get 𝑁

𝜕𝑢 𝜕 𝑇 (𝑢 − 𝜑) 𝑑𝑥 ∑ ∫ 𝑎𝑖 (𝑥, 𝑛 ) 𝜕𝑥𝑖 𝜕𝑥𝑖 𝑘 𝑛 𝑖=1 Ω

sign+0

𝑠+ ; 1) , 𝛿

{1 if 𝑠 > 0, (𝑠) = { 0 if 𝑠 ≤ 0. {

(83)

If 𝜃 is a maximal monotone operator defined on R, we denote by 𝜃0 the main section of 𝜃; that is,

𝜃0 (𝑠) minimal absolute value of 𝜃 (𝑠) if 𝜃 (𝑠) ≠ 0, { { { (84) { = {+∞ if [𝑠, +∞) ∩ 𝐷 (𝜃) = 0, { { { if (−∞, 𝑠] ∩ 𝐷 (𝜃) = 0. {−∞

+ ∫ 𝑇𝑛 (𝑏 (𝑢𝑛 )) 𝑇𝑘 (𝑢𝑛 − 𝜑) 𝑑𝑥 Ω

󵄨 󵄨𝑝 (𝑥)−2 + 𝜀 ∫ 󵄨󵄨󵄨𝑢𝑛 󵄨󵄨󵄨 𝑀 𝑢𝑛 𝑇𝑘 (𝑢𝑛 − 𝜑) 𝑑𝑥 Ω

𝐻𝛿 (𝑠) = min (

(79) Remark that as 𝛿 goes to 0, 𝐻𝛿 (𝑠) goes to sign+0 (𝑠). We take 𝜑 = 𝐻𝛿 (𝑢𝑛 − 𝑀) as a test function in (39) for the weak solution 𝑢𝑛 and 𝑀 > 0 (a constant to be chosen later) to get

+ ∫ 𝑇𝑛 (𝛾 (𝑢𝑛 )) 𝑇𝑘 (𝑢𝑛 − 𝜑) 𝑑𝜎 𝜕Ω

= ∫ 𝑓𝑛 𝑇𝑘 (𝑢𝑛 − 𝜑) 𝑑𝑥. Ω

For the right-hand side of (79), we have

𝑁

∫ 𝑓𝑛 (𝑥) 𝑇𝑘 (𝑢𝑛 − 𝜑) 𝑑𝑥 󳨀→ ∫ 𝑓 (𝑥) 𝑇𝑘 (𝑢 − 𝜑) 𝑑𝑥, (80) Ω

Ω

since 𝑓𝑛 converges strongly to 𝑓 in 𝐿1 (Ω) and 𝑇𝑘 (𝑢𝑛 − 𝜑) converges weakly-∗ to 𝑇𝑘 (𝑢 − 𝜑) in 𝐿∞ (Ω) and a.e. in Ω. For the first term of (79), we have (see [13]) 𝑁

lim inf ∑ ∫ 𝑎𝑖 (𝑥, 𝑛 → +∞

𝑖=1 Ω

𝑁

𝜕𝑢𝑛 𝜕 ) 𝑇 (𝑢 − 𝜑) 𝑑𝑥 𝜕𝑥𝑖 𝜕𝑥𝑖 𝑘 𝑛

𝜕𝑢 𝜕 ) 𝑇𝑘 (𝑢 − 𝜑) 𝑑𝑥. ≥ ∑ ∫ 𝑎𝑖 (𝑥, 𝜕𝑥 𝜕𝑥 𝑖 𝑖 𝑖=1 Ω

∑ ∫ 𝑎𝑖 (𝑥, 𝑖=1 Ω

𝜕𝑢𝑛 𝜕 ) 𝐻 (𝑢 − 𝑀) 𝑑𝑥 𝜕𝑥𝑖 𝜕𝑥𝑖 𝛿 𝑛

󵄨 󵄨𝑝 (𝑥)−2 + 𝜀 ∫ 󵄨󵄨󵄨𝑢𝑛 󵄨󵄨󵄨 𝑀 𝑢𝑛 𝐻𝛿 (𝑢𝑛 − 𝑀) 𝑑𝑥 Ω + ∫ 𝑇𝑛 (𝑏 (𝑢𝑛 )) 𝐻𝛿 (𝑢𝑛 − 𝑀) 𝑑𝑥 Ω

(81)

+ ∫ 𝑇𝑛 (𝛾 (𝑢𝑛 )) 𝐻𝛿 (𝑢𝑛 − 𝑀) 𝑑𝜎 𝜕Ω

= ∫ 𝑓𝑛 𝐻𝛿 (𝑢𝑛 − 𝑀) 𝑑𝑥. Ω

(85)

International Journal of Differential Equations

9

We have

which is equivalent to saying

𝑁

∑ ∫ 𝑎𝑖 (𝑥, 𝑖=1 Ω

𝜕𝑢𝑛 𝜕 ) 𝐻 (𝑢 − 𝑀) 𝑑𝑥 𝜕𝑥𝑖 𝜕𝑥𝑖 𝛿 𝑛

∫ (𝑇𝑛 (𝑏 (𝑢𝑛 )) − 𝑇𝑛 (𝑏 (𝑀))) 𝐻𝛿 (𝑢𝑛 − 𝑀) 𝑑𝑥 Ω

𝑁

𝜕𝑢 𝜕 1 + 𝑎𝑖 (𝑥, 𝑛 ) (𝑢 − 𝑀) 𝑑𝑥 = ∑∫ 𝛿 𝑖=1 {(𝑢𝑛 −𝑀)+ /𝛿𝑘}

(109)

{𝑢𝑘}

Combining (80), (81), (99), (100), and (109) we obtain 𝜕𝑢 𝜕 ) 𝑇𝑘 (𝑢 − 𝜑) 𝑑𝑥 ∑ ∫ 𝑎𝑖 (𝑥, 𝜕𝑥 𝜕𝑥 𝑖 𝑖 𝑖=1 Ω

∫ |𝑏 (𝑢)| 𝑑𝑥 = ∫ Ω

{|𝑢|≤𝑘}

(110)

≤∫

Ω

{|𝑢|≤𝑘}

(116)

󵄩 󵄩 |𝑏 (𝑢)| 𝑑𝑥 ≤ 󵄩󵄩󵄩𝑓󵄩󵄩󵄩1 .

(117)

|𝑏 (𝑢)| 𝑑𝑥 + ∫

{|𝑢|>𝑘}

|𝑏 (𝑢)| 𝑑𝑥

󵄩 󵄩 |𝑏 (𝑢)| 𝑑𝑥 + 󵄩󵄩󵄩𝑓󵄩󵄩󵄩1 .

(118)

Since the function 𝑏 is nondecreasing, then

+ ∫ 𝛾 (𝑢) 𝑇𝑘 (𝑢 − 𝜑) 𝑑𝜎 ≤ ∫ 𝑓𝑇𝑘 (𝑢 − 𝜑) 𝑑𝑥. 𝜕Ω

󵄩 󵄩 −𝑏 (𝑢) 𝑑𝑥 ≤ 󵄩󵄩󵄩𝑓󵄩󵄩󵄩1 .

So, we obtain

𝑁

+ ∫ 𝑏 (𝑢) 𝑇𝑘 (𝑢 − 𝜑) 𝑑𝑥

𝑏 (𝑢) 𝑑𝑥 + ∫

Therefore,

for 𝜀 󳨀→ 0.

Ω

Then, 𝑢 is an entropy solution of problem (1). That completes the proof of Theorem 6. We now state the uniqueness result of entropy solution. Theorem 12. Assume that (4)–(12) hold true and let 𝑢 be an entropy solution of (1). Then, 𝑢 is unique.

Step 1 (a priori estimates). We consider the following. Lemma 13. Assume (4)–(12) and 𝑓 ∈ 𝐿1 (Ω). Let 𝑢 be an entropy solution of (1). Then 𝑁

󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑖 (𝑥) 𝑘 󵄩󵄩 󵄩󵄩 󵄨󵄨 󵄨󵄨 𝑑𝑥 ≤ 󵄨󵄨 󵄨󵄨 󵄩𝑓󵄩 𝐶3 󵄩 󵄩1 {|𝑢|≤𝑘} 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨

∑∫ 𝑖=1

and there exists a positive constant 𝐶9 such that 󵄩 󵄩 ‖𝑏 (𝑢)‖1 ≤ 𝐶9 ⋅ meas (Ω) + 󵄩󵄩󵄩𝑓󵄩󵄩󵄩1 .

(111)

(112)

Proof. Let us take 𝜑 = 0 in the entropy inequality (36). (i) By the fact that ∫Ω 𝑏(𝑢)𝑇𝑘 (𝑢)𝑑𝑥 + ∫𝜕Ω 𝛾(𝑢)𝑇𝑘 (𝑢)𝑑𝜎 ≥ 0, using (7), we get (111). (ii) Also, using the fact that ∑𝑁 𝑖=1 ∫Ω 𝑎𝑖 (𝑥, 𝜕𝑢/𝜕𝑥𝑖 )(𝜕/ 𝜕𝑥𝑖 )𝑇𝑘 (𝑢)𝑑𝑥+∫𝜕Ω 𝛾(𝑢)𝑇𝑘 (𝑢)𝑑𝜎 ≥ 0, relation (36) gives ∫ 𝑏 (𝑢) 𝑇𝑘 (𝑢) 𝑑𝑥 ≤ ∫ 𝑓 (𝑥) 𝑇𝑘 (𝑢) 𝑑𝑥. Ω

∫

{|𝑢|≤𝑘}

|𝑏 (𝑢)| 𝑑𝑥 ≤ max {𝑏 (𝑘) ; |𝑏 (−𝑘)|} ⋅ meas (Ω) . (119)

Consequently, there exists a constant 𝐶9 = max{𝑏(𝑘); |𝑏(−𝑘)|} such that 󵄩 󵄩 ‖𝑏 (𝑢)‖1 ≤ 𝐶9 ⋅ meas (Ω) + 󵄩󵄩󵄩𝑓󵄩󵄩󵄩1 . (120) Lemma 14. Assume (4)–(12) and let 𝑓 ∈ 𝐿1 (Ω). If 𝑢 is an entropy solution of (1), then there exists a constant 𝐷 which depends on 𝑓 and Ω such that

Proof. The proof is done in two steps.

Ω

{|𝑢|>𝑘}

(113)

meas ({|𝑢| > 𝑘}) ≤

𝐷 , ∀𝑘 > 0 min {𝑏 (𝑘) , |𝑏 (−𝑘)|}

(121)

and a constant 𝐷󸀠 > 0 which depends on 𝑓 and Ω such that 󵄨󵄨 𝜕𝑢 󵄨󵄨 𝐷󸀠 󵄨 󵄨󵄨 meas ({󵄨󵄨󵄨 󵄨󵄨 > 𝑘}) ≤ 1/(𝑝− )󸀠 , 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨 𝑘 𝑀

∀𝑘 ≥ 1.

(122)

Proof. (i) For any 𝑘 > 0, relation (112) gives ∫

{|𝑢|>𝑘}

min {𝑏 (𝑘) , |𝑏 (−𝑘)|} 𝑑𝑥 ≤ ∫

{|𝑢|>𝑘}

|𝑏 (𝑢)| 𝑑𝑥

󵄩 󵄩 ≤ 𝐶9 ⋅ meas (Ω) + 󵄩󵄩󵄩𝑓󵄩󵄩󵄩1 .

(123)

Therefore, min {𝑏 (𝑘) , |𝑏 (−𝑘)|} ⋅ meas ({|𝑢| > 𝑘}) 󵄩 󵄩 ≤ 𝐶9 ⋅ meas (Ω) + 󵄩󵄩󵄩𝑓󵄩󵄩󵄩1 = 𝐷;

(124)

12

International Journal of Differential Equations

that is,

Consequently, meas ({|𝑢| > 𝑘}) ≤

𝐷 . min {𝑏 (𝑘) , |𝑏 (−𝑘)|}

(125)

(ii) See [7] for the poof of (122). Lemma 15. Assume (4)–(12) and let 𝑓 ∈ 𝐿1 (Ω). If 𝑢 is an entropy solution of (1), then 󵄨 󵄨 lim ∫ 󵄨󵄨𝑓󵄨󵄨 𝜒{|𝑢|>ℎ−𝑡} 𝑑𝑥 = 0, ℎ → +∞ Ω 󵄨 󵄨

(126)

where ℎ > 0 and 𝑡 > 0.

󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑖 (𝑥)−1 󵄨 󵄨󵄨 𝜒𝐹 ) ≤ 𝐾, ∀𝑖 = 1, . . . , 𝑁. 𝜌𝑝𝑖󸀠 (⋅) (󵄨󵄨󵄨 󵄨 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨󵄨

Step 2 (uniqueness of entropy solution). Let ℎ > 0 and 𝑢, V be two entropy solutions of (1). We write the entropy inequality corresponding to the solution 𝑢, with 𝑇ℎ (V) as test function, and to the solution V, with 𝑇ℎ (𝑢) as test function to get

Proof. Since the function 𝑏 is surjective, according to Lemma 14-(121), we have lim meas ({|𝑢| > ℎ − 𝑡}) = 0

ℎ → +∞

𝑁

∑ ∫ 𝑎𝑖 (𝑥, 𝑖=1 Ω

(127)

Ω

and as 𝑓 ∈ 𝐿 (Ω), it follows by using the Lebesgue dominated convergence theorem that ℎ → +∞ Ω

𝜕𝑢 𝜕 ) 𝑇 (𝑢 − 𝑇ℎ (V)) 𝑑𝑥 𝜕𝑥𝑖 𝜕𝑥𝑖 𝑘

+ ∫ 𝑏 (𝑢) 𝑇𝑘 (𝑢 − 𝑇ℎ (V)) 𝑑𝑥

1

󵄨 󵄨 lim ∫ 󵄨󵄨󵄨𝑓󵄨󵄨󵄨 𝜒{|𝑢|>ℎ−𝑡} 𝑑𝑥 = 0.

+ ∫ 𝛾 (𝑢) 𝑇𝑘 (𝑢 − 𝑇ℎ (V)) 𝑑𝜎 𝜕Ω

(128)

≤ ∫ 𝑓𝑇𝑘 (𝑢 − 𝑇ℎ (V)) 𝑑𝑥, Ω

(134)

𝑁

𝜕V 𝜕 ) 𝑇𝑘 (V − 𝑇ℎ (𝑢)) 𝑑𝑥 ∑ ∫ 𝑎𝑖 (𝑥, 𝜕𝑥 𝜕𝑥 Ω 𝑖 𝑖 𝑖=1

Lemma 16. Assume (4)–(12) and let 𝑓 ∈ 𝐿1 (Ω). If 𝑢 is an entropy solution of (1), then there exists a positive constant 𝐾 such that 󵄨󵄨 𝜕𝑢 󵄨󵄨𝑝𝑖 (𝑥)−1 󵄨 󵄨󵄨 𝜒𝐹 ) ≤ 𝐾, ∀𝑖 = 1, . . . , 𝑁, 𝜌𝑝𝑖󸀠 (⋅) (󵄨󵄨󵄨 󵄨 󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨󵄨

+ ∫ 𝑏 (V) 𝑇𝑘 (V − 𝑇ℎ (𝑢)) 𝑑𝑥 Ω

(129)

+ ∫ 𝛾 (V) 𝑇𝑘 (V − 𝑇ℎ (𝑢)) 𝑑𝜎 𝜕Ω

where 𝐹 = {ℎ < |𝑢| ≤ ℎ + 𝑘}, ℎ > 0, 𝑘 > 0. Proof. Let 𝜑 = 𝑇ℎ (𝑢) as test function in the entropy inequality (36). We get 𝑁

∑ ∫ 𝑎𝑖 (𝑥, 𝑖=1 Ω

𝜕𝑢 𝜕 ) 𝑇 (𝑢 − 𝑇ℎ (𝑢)) 𝑑𝑥 𝜕𝑥𝑖 𝜕𝑥𝑖 𝑘

+ ∫ 𝑏 (𝑢) 𝑇𝑘 (𝑢 − 𝑇ℎ (𝑢)) 𝑑𝑥 Ω

≤ ∫ 𝑓𝑇𝑘 (V − 𝑇ℎ (𝑢)) 𝑑𝑥. Ω

Upon addition, we get 𝑁

∑ ∫ 𝑎𝑖 (𝑥, 𝑖=1 Ω

(130)

+ ∫ 𝛾 (𝑢) 𝑇𝑘 (𝑢 − 𝑇ℎ (𝑢)) 𝑑𝜎

𝜕𝑢 𝜕 ) 𝑇 (𝑢 − 𝑇ℎ (V)) 𝑑𝑥 𝜕𝑥𝑖 𝜕𝑥𝑖 𝑘

𝑁

+ ∑ ∫ 𝑎𝑖 (𝑥, 𝑖=1 Ω

𝜕Ω

𝜕V 𝜕 ) 𝑇 (V − 𝑇ℎ (𝑢)) 𝑑𝑥 𝜕𝑥𝑖 𝜕𝑥𝑖 𝑘

+ ∫ 𝑏 (𝑢) 𝑇𝑘 (𝑢 − 𝑇ℎ (V)) 𝑑𝑥

≤ ∫ 𝑓𝑇𝑘 (𝑢 − 𝑇ℎ (𝑢)) 𝑑𝑥.

Ω

Ω

+ ∫ 𝑏 (V) 𝑇𝑘 (V − 𝑇ℎ (𝑢)) 𝑑𝑥

Thus,

Ω

𝑁

∑∫

𝑖=1 {ℎ ℎ} .

󵄨󵄨󵄨󵄨 󵄨 󵄨󵄨 𝜕𝑢 󵄨󵄨󵄨󵄨𝑝𝑖 (𝑥)−1 󵄨󵄨󵄨 󵄨 󵄨 󵄨󵄨 ≤ 𝐶1 ∑ (󵄨󵄨󵄨𝑗𝑖 󵄨󵄨󵄨𝑝󸀠 (⋅) + 󵄨󵄨󵄨󵄨󵄨󵄨󵄨 ) 󵄨󵄨 󵄨󵄨 󸀠 𝑖 󵄨󵄨󵄨󵄨 𝜕𝑥𝑖 󵄨󵄨 󵄨𝑝𝑖 (⋅),{ℎ