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Article

The Existence and Uniqueness of an Entropy Solution to Unilateral Orlicz Anisotropic Equations in an Unbounded Domain

by
Omar Benslimane
1,*,†,
Ahmed Aberqi
2,† and
Jaouad Bennouna
1,†
1
Laboratory LAMA, Department of Mathematics, Faculty of Sciences Dhar El Mahraz, B.P 1796 Atlas, Sidi Mohamed Ben Abdellah University, Fez 30050, Morocco
2
Laboratory LAMA, National School of Applied Sciences, Sidi Mohamed Ben Abdellah University, Fez 30050, Morocco
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Axioms 2020, 9(3), 109; https://doi.org/10.3390/axioms9030109
Submission received: 5 July 2020 / Revised: 5 September 2020 / Accepted: 12 September 2020 / Published: 17 September 2020
(This article belongs to the Collection Mathematical Analysis and Applications)

Abstract

:
The purpose of this work is to prove the existence and uniqueness of a class of nonlinear unilateral elliptic problem ( P ) in an arbitrary domain, managed by a low-order term and non-polynomial growth described by an N-uplet of N-function satisfying the Δ 2 -condition. The source term is merely integrable.

1. Introduction

Let Ω be an arbitrary domain of R N , ( N 2 ) . In this paper, we investigate the existence and uniqueness solution of the following problem:
( P ) A ( u ) + i = 1 N b i ( x , u , u ) = f in Ω , u ψ a . e . in Ω ,
where, A ( u ) = i = 1 N ( a i ( x , u , u ) ) x i is a Leray–Lions operator defined on W ˚ B 1 ( Ω ) (defined as the adherence space C 0 ( Ω ) ) into its dual; B ( t ) = ( B 1 ( t ) , , B N ( t ) ) are N-uplet Orlicz functions that satisfy Δ 2 -condition; the obstacle ψ is a measurable function that belongs to L ( Ω ) W ˚ B 1 ( Ω ) ; and for i = 1 , , N , b i ( x , s , ξ ) : Ω × R × R N R are Carathéodory functions (measurable with respect to x in Ω for every ( s , ξ ) in R × R N , and continuous with respect to ( s , ξ ) in R × R N for almost every x in Ω ) that do not satisfy any sign condition and the growth described by the vector N-function B ( t ) . Take f L 1 ( Ω ) too.
Statement of the problems: Suppose they have non-negative measurable functions ϕ , φ L 1 ( Ω ) ; and a ¯ , a ˜ are two constants, positive, such that for ξ = ( ξ 1 , , ξ N ) R N and ξ = ( ξ 1 , , ξ N ) R N , we have
i = 1 N a i ( x , s , ξ ) a i ( x , s , ξ ) . ( ξ i ξ i ) > 0 ,
i = 1 N a i ( x , s , ξ ) . ξ i a ¯ i = 1 N B i ( | ξ i | ) ϕ ( x ) ,
i = 1 N | a i ( x , s , ξ ) | a ˜ i = 1 N B ¯ i 1 B i ( | ξ | ) + φ ( x ) ,
and
i = 1 N | b i ( x , s , ξ ) | h ( x ) + l ( s ) . i = 1 N B i ( | ξ | ) ,
with B ¯ ( t ) being the complementary function of B ( t ) , h L 1 ( Ω ) and l : R R + being a positive continuous function such that l L 1 ( Ω ) L ( Ω ) .
We recall that in the last few decades, tremendous popularity has been achieved by the investigation of a class of nonlinear unilateral elliptic problem due to their fundamental role in describing several phenomena, such as the study of fluid filtration in porous media, constrained heating, elastoplasticity, optimal control, financial mathematics and others; for those studies, there are large numbers of mathematical articles; see [1,2,3,4] for more details.
When Ω is a bounded open set of R N , we refer to the celebrated paper by Bénilan [5], who presented the idea of entropy solutions adjusted to Boltzmann conditions. For more outcomes concerning the existence of solutions of this class in the Lebesgue Sobolev spaces (to be specific B ( t ) = | t | p ), we cite [6,7]. We cite [4,8,9] for the Sobolev space with variable exponent. In the case of Orlicz spaces, we have some difficulties due to the non-homogeneity of the N-functions B ( t ) and a rather indirect definition of the norm. It is generally difficult to move essentially L p techniques to Orlicz spaces. For more work within this framework, we quote [10,11,12,13].
On the other hand, when Ω is an unbounded domain, namely, without expecting any assumptions on the behavior when | x | + , Domanska in [14] investigated the well-posedness of nonlinear elliptic systems of equations generalizing the model equation
i = 1 N ( | u x i ( x ) | p i 2 u x i ( x ) ) x i + | u ( x ) | p 0 2 u ( x ) = f ( x ) ,
with corresponding indices of nonlinearity p i > 1 ( i = 0 , n ¯ ) . In [15] Bendahmann et al. the problem ( P ) with b ( x , u , u ) = d i v ( g ( u ) ) and g ( u ) a polynomial growth like u q in L p -spaces was solved. For more results we refer the reader to the work [16]. We mention [17,18,19], for the Sobolev space with variable exponent, and [20,21,22,23,24,25,26] for the classical anisotropic space.
The oddity of our present paper is to continue in this direction and to show the existence and uniqueness of entropy solution for equations ( P ) governed with growth and described by an N-uplet of N-functions satisfying the Δ 2 -condition, within the fulfilling of anisotropic Orlicz spaces. Besides, we address the challenges that come about due to the absence of some topological properties, such as the densities of bounded or smooth functions.
The outline of this work is as follows. In Section 2, we recall some definitions and properties of N-functions and the space of Sobolev–Orlicz anisotropic solutions. In Section 3, we prove the Theorem of the existence of the solutions in an unbounded domain with the help of some propositions; to be demonstrated later. In Section 4, we show the uniqueness of the solution to this problem, which is expected for strictly monotonic operators at least for a broad class of lower-order terms. Finally, there is Appendix A.

2. Mathematical Background and Auxiliary Results

In this section, we introduce the notation, recall some standard definitions and collect necessary propositions and facts that are used to establish our main result.
A comprehensive presentation of Sobolev–Orlicz anisotropic space can be found in the books of M.A Krasnoselskii and Ja. B. Rutickii [23] and in [20,25].
Definition 1.
We say that B : R + R + is a N-function if B is continuous, convex, with B ( z ) > 0 f o r z > 0 , B ( z ) z 0 w h e n z 0 and B ( z ) z w h e n z .
This N-function B admits the following representation: B ( z ) = 0 z b ( t ) d t , with b : R + R + which is an increasing function on the right, with b ( 0 ) = 0 in the case z > 0 and b ( z ) when z .
Its conjugate is noted by B ¯ ( z ) = 0 | z | q ( t ) d t with q also satisfying all the properties already quoted from b, with
B ¯ ( z ) = sup y 0 ( y | z | B ( y ) ) , z R .
The Young’s inequality is given as follows:
z , y R | z y | B ( y ) + B ¯ ( z ) .
Definition 2.
The N-function B ( z ) satisfies the Δ 2 -condition if c > 0 , z 0 0 such that
B ( 2 z ) c B ( z ) | z | z 0 .
This definition is equivalent to, k > 1 , c ( k ) > 0 such that
B ( k z ) c ( k ) B ( z ) f o r | z | z 0 .
Definition 3.
The N-function B ( z ) satisfies the Δ 2 -condition as long as there exist positive numbers c > 1 and z 0 0 such that for | z | z 0 we have
z b ( z ) c B ( z ) .
Additionally, each N-function B ( z ) satisfies the inequality
B ( y + z ) c B ( z ) + c B ( y ) z , y R .
Definition 4.
The N-function B ( z ) satisfies the 2 -condition if c > 2 , z 0 0 such that
B ( 2 z ) c B ( z ) | z | z 0 .
We consider the Orlicz space L B ( Ω ) provided with the norm of Luxemburg given by
| | u | | B , Ω = inf { k > 0 / Ω B u ( x ) k d x 1 } .
According with [23] we obtain the inequalities
Ω B u ( x ) | | u | | B , Ω d x 1 ,
and
| | u | | B , Ω Ω B ( u ) d x + 1 .
Moreover, the Hölder’s inequality holds and we have for all u L B ( Ω ) and v L B ¯ ( Ω )
| Ω u ( x ) v ( x ) d x | 2 | | u | | B , Ω . | | v | | B ¯ , Ω .
In [23,25], if P ( z ) and B ( z ) are two N-functions such that P ( z ) B ( z ) and m e a s Ω < , then L B ( Ω ) L P ( Ω ) ; furthermore,
| | u | | P , Ω A 0 ( m e a s Ω ) | | u | | B , Ω u L B ( Ω ) .
Additionally, for all N-functions B ( z ) , if m e a s Ω < , then L ( Ω ) L B ( Ω ) with
| | u | | B , Ω A 1 ( m e a s Ω ) | | u | | , Ω u L B ( Ω ) .
Additionally, for all N-functions B ( z ) , if m e a s Ω < , then L B ( Ω ) L 1 ( Ω ) with
| | u | | 1 , Ω A 2 | | u | | B , Ω u L B ( Ω ) .
We define for all N-functions B 1 ( z ) , , B N ( z ) the space of Sobolev–Orlicz anisotropic W ˚ B 1 ( Ω ) as the adherence space C 0 ( Ω ) under the norm
| | u | | W ˚ B 1 ( Ω ) = i = 1 N | | u x i | | B i , Ω .
Definition 5.
A sequence { u m } is said to converge modularly to u in W ˚ B 1 ( Ω ) if for some k > 0 we have
Ω B u m u k d x 0 a s m .
Remark 1.
Since B satisfies the Δ 2 -condition, the modular convergence coincides with the norm convergence.
Remark 2.
If the doubling condition is imposed on the modular function, but not on the conjugate, then the space for the solutions to exist is non-reflexive in general. For this reason we will assume in the remainder of this article that B satisfies the both conditions; the Δ 2 -condition and 2 -condition, so the Propositions 1 and 2 will remain true.
Proposition 1 ([23]).
The Sobolev–Orlicz anisotropic space W ˚ B 1 ( Ω ) is complete and reflexive.
Proposition 2 ([23]).
The Sobolev–Orlicz anisotropic W ˚ B 1 ( Ω ) is separable.
Proposition 3.
z B ( z ) = B ¯ ( B ( z ) ) + B ( z ) , z > 0 ,
with B being the right derivative of the N-function B ( z ) .
Proof. 
By (6), we take y = B ( z ) ; then we obtain
B ( z ) z B ( z ) + B ¯ ( B ( z ) ) ,
and by Ch. I [23], we get the result. □
In the following we will assume that for each N-function B i ( z ) = 0 | z | b i ( t ) d t obeys the further condition
lim α inf θ > 0 b i ( α θ ) b i ( α ) = , i = 1 , , N .
Example 1.
The function
B ( z ) = | z | b ( | l n | z | | + 1 ) ,
with b > 1 checks the Δ 2 -condition and (22).
Lemma 1.
Suppose that ( X , T , m e a s ) is a measurable set such that m e s ( X ) < . Let θ : X [ 0 , + ] be a measurable function such that m e a s { x X : θ ( x ) = 0 } = 0 . Then, for any ϵ > 0 , there exists δ > 0 such that for any bounded domain Q
Q θ ( x ) d x δ
implies that
m e a s ( Q ) ϵ .
Proof. 
See [27] (Lemma 2). □

3. The Existence of an Entropy Solution

This section is devoted to the proofs of our main results which will be split into different steps.
For m N * , we define the truncation at height m, T m ( u ) : R R by
T m ( u ) = u if | u | m , m if | u | > m .
Definition 6.
A measurable function u is said to be an entropy solution for the problem ( P ) , if u W ˚ B 1 ( Ω ) such that u ψ a.e. in Ω and
i = 1 N Ω a i ( x , u , u ) . ( u v ) d x + i = 1 N Ω b i ( x , u , u ) . ( u v ) d x + Ω m . T m ( u ψ ) . s g 1 m ( u ) . ( u v ) d x Ω f ( x ) . ( u v ) d x v K ψ L ( Ω ) ,
where, K ψ = { u W ˚ B 1 ( Ω ) / u ψ a . e . i n Ω } , and s g m ( s ) = T m ( s ) m .
We have f m f i n L 1 ( Ω ) , m , | f m ( x ) | | f ( x ) | and for i = 1 , , N , a i m ( x , u m , u m ) : ( W ˚ B 1 ( Ω ) ) N ( W ˚ B ¯ 1 ( Ω ) ) N being Carathéodory functions with
a i m ( x , u , u ) = a i ( x , T m ( u ) , u ) ,
and b i m ( x , u m , u m ) : Ω × R × R N R again being Carathéodory functions not satisfying any sign condition, with
b m ( x , u , u ) = b ( x , u , u ) 1 + 1 m | b ( x , u , u ) | ,
and
| b m ( x , u , u ) | = | b ( x , T m ( u ) , u ) | m f o r a l l m N * ,
and for all v W ˚ B 1 ( Ω ) , we consider the following approximate problem:
( P m ) : i = 1 N Ω a i m ( x , u m , u m ) . ( u m v ) d x + i = 1 N Ω b i m ( x , u m , u m ) . ( u m v ) d x + Ω m . T m ( u m ψ ) . s g 1 m ( u m ) . ( u m v ) d x = Ω f m ( x ) . ( u m v ) d x .
Theorem 1.
Assume that conditions (1)–(4) and (22) hold true, then there exists at least one solution of the approximate problem ( P m ) .
Proof. 
See Appendix A. □
Theorem 2.
Under assumptions (1)–(4), the problem ( P ) has at least one entropy solution.
Proof. 
Let R > 0 and Ω ( R ) = { x Ω : | x | R } . Note by h ( t ) = i = 1 N B i 1 ( t ) t 1 N and we assume that 0 1 h ( t ) t d t converge, so we consider the N-functions B * ( z ) defined by ( B * ) 1 ( z ) = 0 | z | h ( t ) t d t .
Lemma 2 ([20]).
Let u W ˚ B 1 ( Ω ( R ) ) . If
1 h ( t ) t d t = ,
then W ˚ B 1 ( Ω ( R ) ) L B * ( Ω ( R ) ) and | | u | | B * , Ω ( R ) N 1 N | | u | | W ˚ B 1 ( Ω ( R ) ) . If
1 h ( t ) t d t ,
then W ˚ B 1 ( Ω ( R ) ) L ( Ω ( R ) ) and | | u | | , Ω ( R ) β | | u | | W ˚ B 1 ( Ω ( R ) ) , with β = 0 h ( t ) t d t .
Step 1.
A priori estimate of { u m } :
Let v = u m η exp ( G ( u m ) ) . T k ( u m v 0 ) + where G ( s ) = 0 s l ( t ) a ¯ d t , k > 0 and η 0 ; we have v W ˚ B 1 ( Ω ) and for a small enough η we deduce that v ψ . Thus v is an admissible test function in ( P m ) and we get for all v 0 K ψ L ( Ω ) that
i = 1 N Ω a i m ( x , u m , u m ) . ( exp ( G ( u m ) ) . T k ( u m v 0 ) + ) d x + i = 1 N Ω b i m ( x , u m , u m ) . exp ( G ( u m ) ) . T k ( u m v 0 ) + d x + Ω m . T m ( u m ψ ) . s g 1 m ( u m ) . exp ( G ( u m ) ) . T k ( u m v 0 ) + d x Ω f m ( x ) . exp ( G ( u m ) ) . T k ( u m v 0 ) + d x ;
then,
i = 1 N Ω a i m ( x , u m , u m ) . u m . l ( u m ) a ¯ . exp ( G ( u m ) ) . T k ( u m v 0 ) + d x + i = 1 N Ω a i m ( x , u m , u m ) . exp ( G ( u m ) ) . T k ( u m v 0 ) + d x + Ω m . T m ( u m ψ ) . s g 1 m ( u m ) . exp ( G ( u m ) ) . T k ( u m v 0 ) + d x i = 1 N Ω | b i m ( x , u m , u m ) | . exp ( G ( u m ) ) . T k ( u m v 0 ) + d x + Ω f m ( x ) . exp ( G ( u m ) ) . T k ( u m v 0 ) + d x ,
by (2) and (4), we obtain
i = 1 N Ω a i m ( x , u m , u m ) . exp ( G ( u m ) ) . T k ( u m v 0 ) + d x + Ω m . T m ( u m ψ ) . s g 1 m ( u m ) . exp ( G ( u m ) ) . T k ( u m v 0 ) + d x Ω [ h ( x ) + f m ( x ) + ϕ ( x ) . l ( u m ) a ¯ ] . exp ( G ( u m ) ) . T k ( u m v 0 ) + d x ,
so
i = 1 N { | u m v 0 | k } a i m ( x , u m , u m ) . u m . exp ( G ( u m ) ) d x c i = 1 N { | u m v 0 | k } a i m ( x , u m , u m ) . v 0 c . exp ( G ( u m ) ) d x + Ω m . T m ( u m ψ ) . s g 1 m ( u m ) . exp ( G ( u m ) ) . T k ( u m v 0 ) + d x Ω [ h ( x ) + f m ( x ) + ϕ ( x ) . l ( u m ) a ¯ ] . exp ( G ( u m ) ) . T k ( u m v 0 ) + d x ,
where c is a constant such that 0 < c < 1 , and since h , f m , ϕ L 1 ( Ω ) we deduce that
i = 1 N { | u m v 0 | k } a i m ( x , u m , u m ) . u m . exp ( G ( u m ) d x + Ω m . T m ( u m ψ ) . s g 1 m ( u m ) . exp ( G ( u m ) ) . T k ( u m v 0 ) + d x c i = 1 N { | u m v 0 | k } a i m ( x , u m , u m ) a i m ( x , u m , v 0 c ) . ( u m v 0 c ) . exp ( G ( u m ) ) d x + c i = 1 N { | u m v 0 | k } a i m ( x , u m , u m ) . u m . exp ( G ( u m ) ) d x + c i = 1 N { | u m v 0 | k } | a i m ( x , u m , v 0 c ) | . | ( u m v 0 c ) | . exp ( G ( u m ) ) d x + c 1 ,
by (1)
( 1 c ) i = 1 N { | u m v 0 | k } a i m ( x , u m , u m ) . u m . exp ( G ( u m ) ) d x + Ω m . T m ( u m ψ ) . s g 1 m ( u m ) . exp ( G ( u m ) ) . T k ( u m v 0 ) + d x c i = 1 N { | u m v 0 | k } | a i m ( x , u m , v 0 c ) | . | u m | . exp ( G ( u m ) ) d x + c i = 1 N { | u m v 0 | k } | a i m ( x , u m , v 0 c ) | . | v 0 c | . exp ( G ( u m ) ) d x + c 1 ,
since v 0 c W ˚ B 1 ( Ω ) , and by (8), (3) and (6) and the fact that exp ( G ( ± ) ) exp | | l | | L 1 ( R ) a ¯ we have
( 1 c ) i = 1 N { | u m v 0 | k } a i m ( x , u m , u m ) . u m d x + { | u m v 0 | 0 } m . T m ( u m ψ ) . s g 1 m ( u m ) d x a ˜ ( 1 c ) 2 i = 1 N { | u m v 0 | k } B i ( u m ) d x + c 2 ( k ) . c 1 ,
where c 2 ( k ) is a positive constant which depends only on k.
Finally, by (2) we obtain
i = 1 N { | u m v 0 | k } B i ( u m ) d x c 3 . k ,
and
0 { | u m v 0 | 0 } m . T m ( u m ψ ) . s g 1 m ( u m ) d x c 1 .
Similarly, taking v = u m η . exp ( G ( u m ) ) . T k ( u m v 0 ) as a test function in ( P m ) , we obtain
i = 1 N Ω a i m ( x , u m , u m ) . ( exp ( G ( u m ) ) . T k ( u m v 0 ) ) d x + i = 1 N Ω b i m ( x , u m , u m ) . exp ( G ( u m ) ) . T k ( u m v 0 ) d x + Ω m . T m ( u m ψ ) . s g 1 m ( u m ) . exp ( G ( u m ) ) . T k ( u m v 0 ) d x Ω f m ( x ) . exp ( G ( u m ) ) . T k ( u m v 0 ) d x ,
and using same techniques, we obtain also
i = 1 N { | u m v 0 | k } B i ( u m ) d x c 4 . k ,
and
0 { | u m v 0 | 0 } m . T m ( u m ψ ) . s g 1 m ( u m ) d x c 5 .
Additionally, by (26), (27), (28) and (29) we conclude that
Ω B ( T k ( u m ) ) d x c . k ,
with c 3 , c 4 , c 5 , c 6 being positive constants.
Step 2.
Almost everywhere convergence of { u m } :
Firstly, we prove that m e a s { x Ω : | u m | k } 0 .
According to Lemma 2, we have
| | T k ( u m ) | | B * c | | T k ( u m ) | | B c . ϵ ( k ) Ω B ( T k ( u m ) ) d x c . ϵ ( k ) . k for k > 1 ,
with c being a positive constant and ϵ ( k ) 0 w h e n k . By (31) we obtain
B * k | | T k ( u m ) | | B * m e a s { x Ω : | u m | k } Ω B * T k ( u m ) | | T k ( u m ) | | B * d x Ω B * k | | T k ( u m ) | | B * d x .
Thus, we deduce that
B * k | | T k ( u m ) | | B * w h e n k .
Hence
m e a s { x Ω : | u m | k } 0 a s k f o r a l l m N .
Secondly we show that for all { u m } measurable function on Ω such that
T k ( u m ) W ˚ B 1 ( Ω ) k 1 ,
we have
m e a s { x Ω : B ( u m ) α } 0 a s α .
In the beginning
m e a s { x Ω : B ( u m ) 0 } = m e a s { { x Ω : | u m | k , B ( u m ) α } { x Ω : | u m | < k , B ( u m ) α } } ,
if we denote
g ( α , k ) = m e a s { x Ω : | u m | k , B ( u m ) α } ,
we have
m e a s { x Ω : | u m | < k , B ( u m ) α } = g ( α , 0 ) g ( α , k ) ,
then
{ x Ω : | u m | < k } B ( u m ) d x = 0 g ( α , 0 ) g ( α , k ) d α c . k ,
with α g ( α , k ) is a decreasing map; then
g ( α , 0 ) 1 α 0 α g ( s , 0 ) d s 1 α 0 α g ( s , 0 ) g ( s , k ) d s + 1 α 0 α g ( s , k ) d s 1 α 0 α g ( s , 0 ) g ( s , k ) d s + g ( 0 , k ) ,
and according to (34) and (35) we have
g ( α , 0 ) c . k α + g ( 0 , k ) ,
like [28] we obtain
lim k g ( 0 , k ) = 0 .
Hence
g ( α , 0 ) 0 a s α .
We have now to demonstrate that the almost everywhere convergence of { u m : }
u m u a l m o s t e v e r y w h e r e i n Ω .
Let g ( k ) = sup m N m e a s { x Ω : | u m | > k } 0 a s k . Since Ω is unbounded domain in R N , we define
η R ( x ) = 1 if x < R , R + 1 α if R x < R + 1 , 0 if x R + 1 .
For R , k > 0 , we have by (10)
Ω B ( η R ( | x | ) . T k ( u m ) ) d x c { x Ω : | u m | < k } B ( u m ) d x + c Ω B ( T k ( u m ) . η R ( | x | ) d x c ( k , R ) ,
which implies that the sequence { η R ( | x | ) T k ( u m ) } is bounded in W ˚ B 1 ( Ω ( R + 1 ) ) , and by embedding Theorem, for an N-function P with P B we have
W ˚ B 1 ( Ω ( R + 1 ) ) L P ( Ω ( R + 1 ) )
and since η R = 1 in Ω ( R ) , we have:
η R T k ( u m ) v k in L P ( Ω ( R + 1 ) ) a s m .
For k N * ,
T k ( u m ) v k in L P ( Ω ( R + 1 ) ) a s m
by a diagonal process, we prove that there is measurable u : Ω R such that u m u a.e. in Ω .
Lemma 3 ([29]).
Let an N-function B ¯ ( t ) satisfy the Δ 2 -condition and u m , m 1 and u be two functions of L B ( Ω ) such that
| | u m | | B c m = 1 , 2 , .
u m u a l m o s t e v e r y w h e r e i n Ω , m .
Then,
u m u w e a k l y i n L B ( Ω ) a s m .
Hence,
m e a s { x Ω : | u m | k } 0 w h e n k f o r a l l m N .
Step 3.
Weak convergence of the gradient:
Since W ˚ B 1 ( Ω ) reflexive, there exists a subsequence
T k ( u m ) v w e a k l y i n W ˚ B 1 ( Ω ) , m ,
and since
W ˚ B 1 ( Ω ) L B ( Ω ) ,
we have
T k ( u m ) u m i n L B ( Ω ) a s m ,
since
u m u a l m o s t e v e r y w h e r e i n Ω a s m ,
implies the local convergence in measure and, therefore, the local Cauchy property of u m in measure
m e a s { Ω ( R ) : | u m u n | k } 0 a s m , n f o r a n y k > 0 .
Proving that
u m u l o c a l l y i n m e a s u r e a s m .
For that, we borrow ideas from Evans [13], Demangel-Hebey [12] and Koznikova L. M. [21,22]. Let δ > 0 be given. By Egoroff’s Theorem, there exists E δ , k , α Ω such that
E δ , k , α ( R ) = { Ω ( R ) : | u m u n | < k , B ( u m ) < α , B ( u n ) < α , | u m | α , | u n | α , | ( u m u n ) | δ }
{ Ω ( R ) : | ( u m u n ) | δ } { Ω : B ( u m ) > α } { Ω : B ( u n ) > α } { Ω ( R ) : | u m u n | k } { Ω : | u m | > α } { Ω : | u n | > α } E δ , k , α ( R ) .
Then, by Lemma 3 and (33) we obtain that
m e a s { Ω ( R ) : | ( u m u n ) | δ } 4 ϵ + m e a s E δ , k , α ( R ) + m e a s { Ω ( R ) : | u m u n | k } n , m N * .
According to (1) and the fact that a continuous function on a compact set achieves the lowest value, there exists a function θ ( x ) > 0 almost everywhere in Ω , such that, for B ( ξ ) α , B ( ξ ) α , | s | α and for i = 1 , , N , | ξ i ξ i | k , we have that
i = 1 N a i m ( x , s , ξ ) a i m ( x , s , ξ ) . ( ξ i ξ i ) θ ( x ) ,
holds. Writing ( P m ) twice for { u m } and { u n } , and by subtracting the second relation from the first and according to (23), (27), (29) and (36) we obtain
i = 1 N Ω a i m ( x , u m , u m ) a i m ( x , u n , u n ) . ( u m u n v ) d x = 0 .
Consider the following test function:
v = u m u n η R ( | x | ) η α ( | u n | ) η α ( | u m | ) exp ( G ( | u m u n | ) ) T δ ( u m u n ) .
Further on, by applying (40), we get
E δ , k , α ( R ) θ ( x ) d x i = 1 N E δ , k , α ( R ) a i m ( x , u m , u m ) a i m ( x , u m , u n ) × ( η R ( | x | ) η α ( | u n | ) η α ( | u m | ) exp ( G ( | u m u n | ) ) T δ ( u m u n ) ) d x i = 1 N { Ω : | u m u n | < k } η R ( | x | ) η α ( | u n | ) η α ( | u m | ) exp ( G ( | u m u n | ) ) × a i m ( x , u m , u m ) a i m ( x , u m , u n ) . ( u m u n ) d x = A 1 m ( x ) + A 2 m ( x ) ,
with
A 1 m ( x ) = i = 1 N { | u m u n | < k } η R ( | x | ) η α ( | u n | ) η α ( | u m | ) exp ( G ( | u m u n | ) ) × a i m ( x , u m , u m ) a i m ( x , u n , u n ) . ( u m u n ) d x ,
and
A 2 m ( x ) = i = 1 N { | u m u n | < k } η R ( | x | ) η α ( | u n | ) η α ( | u m | ) exp ( G ( | u m u n | ) ) × a i m ( x , u n , u n ) a i m ( x , u m , u n ) . ( u m u n ) d x .
Since B ( u ) satisfies the Δ 2 -condition, by (14) we have
Ω B ( u ) d x c 0 | | u | | B , Ω .
According to Lemma 3, we get
| | u m | | W ˚ B 1 ( Ω ) c 1 m N * ,
and
| | B ( u m ) | | 1 c 2 m N * .
Additionally, using (14) and (3) we have
| | a ( x , u , u ) | | L B ¯ ( Ω ) = i = 1 N | | a i ( x , u , u ) | | L B ¯ i ( Ω ) i = 1 N Ω B ¯ i ( a i ( x , u , u ) ) d x + N c 3 | | B ( u ) | | 1 , Ω + | | φ | | 1 , Ω + N c 4 .
Hence,
A 1 m ( x ) = i = 1 N { Ω : | u m u n | < k , | x | < R , | u m | < α , | u n | < α } exp ( G ( | u m u n | ) ) × a i m ( x , u m , u m ) a i m ( x , u n , u n ) . ( u m u n ) d x + i = 1 N { Ω : | u m u n | < k , R | x | R + 1 , α | u m | α + 1 , α | u n | α + 1 } ( R + 1 | x | ) × ( α + 1 | u n | ) . ( α + 1 | u m | ) . exp ( G ( | u m u n | ) ) × a i m ( x , u m , u m ) a i m ( x , u n , u n ) . ( u m u n ) d x ,
since exp ( G ( ± ) ) exp | | l | | L 1 ( R a ¯ , and according to (42), (43), (44) and (15) we obtain that
A 1 m ( x ) c 1 ( R , α ) . k ;
the same for A 2 m ( x ) . We get
A 2 m ( x ) c 2 ( R , α ) . k .
Then,
E δ , k , α ( R ) θ ( x ) d x c 3 ( R , α ) . k .
For any arbitrary δ > 0 for fixed m and α , by choosing k from (45) we establish the following inequality
E δ , k , α ( R ) θ ( x ) d x < δ .
By applying Lemma 1, for any ϵ > 0 , we find
m e a s E δ , k , α ( R ) < ϵ .
In addition, according to (37), we have
m e a s { Ω ( R ) : | u m u n | k } < ϵ , m , n > 0 .
By combining (39), (46) and (47) we deduce the inequality
m e a s { Ω ( R ) : | ( u m u n ) | δ } < 6 ϵ , n , m > 0 .
Hence, the sequence { u m } is fundamental in measure on the set Ω ( R ) for any R > 0 . This implies (38) and the selective convergence,
u m u a l m o s t e v e r y w h e r e i n Ω , m .
Then, we obtain for any fixed k > 0
T k ( u m ) T k ( u ) a l m o s t e v e r y w h e r e i n Ω a s m .
Applying Lemma 3, we have the following weak convergence
T k ( u m ) T k ( u ) i n L B ( Ω ) a s m .
Proposition 4.
Suppose that Conditions (1)–(4) are satisfied and let ( u m ) m N be a sequence in W ˚ B 1 ( Ω ( R ) ) such that
(a) 
u m u i n W ˚ B 1 ( Ω ( R ) )
(b) 
a m ( x , u m , u m ) is bounded in L B ¯ ( Ω ( R ) )
(c) 
i = 1 N Ω ( R ) [ a i m ( x , u m , u m ) a i m ( x , u m , u χ ϵ ) . ( u m u χ ϵ ) d x 0 a s ϵ + ( χ ϵ the characteristic function of Ω ϵ ( R ) = { x Ω ; | u | ϵ } ). Then
B ( | u m | ) B ( | u | ) i n L 1 ( Ω ( R ) )
Proof. 
Let ϵ > 0 fixed, and η > ϵ ; then from (1) we have
0 i = 1 N Ω η ( R ) a i m ( x , u m , u m ) a i m ( x , u m , u ) . ( u m u ) d x i = 1 N Ω ϵ ( R ) a i m ( x , u m , u m ) a i m ( x , u m , u ) . ( u m u ) d x = i = 1 N Ω ϵ ( R ) a i m ( x , u m , u m ) a i m ( x , u m , u χ ϵ ) . ( u m u χ ϵ ) d x
i = 1 N Ω ( R ) a i m ( x , u m , u m ) a i m ( x , u m , u χ ϵ ) . ( u m u χ ϵ ) d x ,
using the condition ( c ) we get
lim m i = 1 N Ω ( R ) a i m ( x , u m , u m ) a i m ( x , u m , u χ ϵ ) . ( u m u χ ϵ ) d x = 0
proceeding as in [28], and we obtain u m u ; by letting ϵ we get
u m χ ϵ u ,
Thus, since
i = 1 N Ω ( R ) a i m ( x , u m , u m ) . u m d x = i = 1 N Ω ( R ) a i m ( x , u m , u m ) a i m ( x , u m , u χ ϵ ) . ( u m u χ ϵ ) d x + i = 1 N Ω ( R ) a i m ( x , u m , χ ϵ ) . ( u m u χ ϵ ) d x + i = 1 N Ω ( R ) a i m ( x , u m , u m ) . u χ ϵ d x ,
using ( b ) , we have
i = 1 N a i m ( x , u m , u m ) i = 1 N a i m ( x , u , u ) w e a k l y i n ( L B ¯ ( Ω ( R ) ) ) N .
Therefore
i = 1 N Ω ( R ) a i m ( x , u m , u m ) . u χ ϵ d x i = 1 N Ω ( R ) a i m ( x , u , u ) . u d x a s m , ϵ .
Thus,
i = 1 N Ω ( R ) a i m ( x , u m , u m ) a i m ( x , u m , u χ ϵ ) . ( u m u χ ϵ ) d x 0 a s m , ϵ .
and
i = 1 N Ω ( R ) a i m ( x , u m , u χ ϵ ) . ( u m u χ ϵ ) d x 0 a s m , ϵ .
Thus,
lim m i = 1 N Ω ( R ) a i m ( x , u m , u m ) . u m d x = i = 1 N Ω ( R ) a i m ( x , u , u ) . u d x ,
from (2), and the vitali’s Theorem, we get
a ¯ i = 1 N Ω ( R ) B i ( | u m | ) d x Ω ( R ) ϕ ( x ) d x a ¯ i = 1 N Ω ( R ) B i ( | u | ) d x Ω ( R ) ϕ ( x ) d x ,
Consequently, by Lemma 2.6 in [11] and (48), we get
B ( | u m | ) B ( | u | ) i n W ˚ B 1 ( Ω ( R ) ) ,
thanks to lemma 1 (see [20]) and (48), we have
B ( | u m | ) B ( | u | ) i n L 1 ( Ω ( R ) ) .
Step 4.
Strong convergence of the gradient:
In this step we consider again the following test function:
v = u m + η exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) ,
with, h j ( u m ) ) = 1 | T 1 ( u m T j ( u m ) ) | = 1 if { | u m | j } 0 if { | u m | j + 1 } j + 1 | u m | if { j | u m | j + 1 } and, | T k ( u m ) T k ( u ) | at the same sign when u m { | u m | > k } , where j k > 0 and η is small enough, we obtain
i = 1 N Ω a i m ( x , u m , u m ) . ( exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) ) d x + Ω m . T m ( u m ψ ) . s g 1 m ( u m ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x + i = 1 N Ω b i m ( x , u m , u m ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x Ω f m ( x ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x ,
which implies,
i = 1 N Ω a i m ( x , u m , u m ) . ( exp ( G ( | u m | ) ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x + i = 1 N Ω a i m ( x , u m , u m ) . exp ( G ( | u m | ) ) ( ( T k ( u m ) T k ( u ) ) ) η j ( | u m | ) h j ( u m ) d x + i = 1 N Ω a i m ( x , u m , u m ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x + i = 1 N Ω a i m ( x , u m , u m ) . exp ( G ( | u m | ) ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x + i = 1 N Ω b i m ( x , u m , u m ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x + Ω m . T m ( u m ψ ) . s g 1 m ( u m ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x Ω f m ( x ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x ;
then,
i = 1 N Ω a i m ( x , u m , u m ) . u m . l ( | u m | ) a ¯ . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x
+ i = 1 N Ω a i m ( x , u m , u m ) . exp ( G ( | u m | ) ) ( ( T k ( u m ) T k ( u ) ) ) η j ( | u m | ) h j ( u m ) d x + i = 1 N Ω a i m ( x , u m , u m ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x + i = 1 N Ω a i m ( x , u m , u m ) . exp ( G ( | u m | ) ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x + Ω m . T m ( u m ψ ) . s g 1 m ( u m ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x i = 1 N Ω | b i m ( x , u m , u m ) | . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x + Ω f m ( x ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x ,
by (2) and (4) we get
a ¯ i = 1 N Ω B i ( | u m | ) . l ( | u m | ) a ¯ . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x + i = 1 N Ω a i m ( x , u m , u m ) . exp ( G ( | u m | ) ) ( ( T k ( u m ) T k ( u ) ) ) η j ( | u m | ) h j ( u m ) d x + i = 1 N Ω a i m ( x , u m , u m ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x + i = 1 N Ω a i m ( x , u m , u m ) . exp ( G ( | u m | ) ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x + Ω m . T m ( u m ψ ) . s g 1 m ( u m ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x i = 1 N Ω B i ( | u m | ) l ( | u m | ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x + Ω h ( x ) + f m ( x ) + ϕ ( x ) . l ( | u m | ) a ¯ . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x ;
we then obtain
i = 1 N { | u m | j } a i m ( x , u m , u m ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) d x i = 1 N { j | u m | j + 1 } a i m ( x , u m , u m ) . u m . ( j + 1 + | u m | ) . T k ( u ) η j ( | u m | ) exp ( G ( | u m | ) ) d x + i = 1 N Ω a i m ( x , u m , u m ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x + i = 1 N { j | u m | j + 1 } a i m ( x , u m , u m ) . u m . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) d x + Ω m . T m ( u m ψ ) . s g 1 m ( u m ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x Ω h ( x ) + f m ( x ) + ϕ ( x ) . l ( | u m | ) a ¯ . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x .
By (2) we get
i = 1 N { | u m | j } a i m ( x , u m , u m ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) d x + i = 1 N Ω a i m ( x , u m , u m ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x + Ω m . T m ( u m ψ ) . s g 1 m ( u m ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x Ω h ( x ) + f m ( x ) + ϕ ( x ) . l ( | u m | ) a ¯ . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) h j ( u m ) d x i = 1 N { j | u m | j + 1 } ϕ ( x ) . ( j + 1 + | u m | ) . T k ( u ) exp ( G ( | u m | ) ) η j ( | u m | ) d x + i = 1 N { j | u m | j + 1 } ϕ ( x ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) d x + a ¯ i = 1 N { j | u m | j + 1 } B i ( | u m | ) ( j + 1 + | u m | ) . T k ( u ) exp ( G ( | u m | ) ) η j ( | u m | ) d x a ¯ i = 1 N { j | u m | j + 1 } B i ( | u m | ) exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) d x .
According to (27), (29); and T k ( u m ) T k ( u ) weakly in W ˚ B 1 ( Ω ) , h j 0 , η j ( | u m | ) 0 , and u m ( T k ( u m ) T k ( u ) ) 0 and exp ( G ( ± ) ) exp | | l | | L 1 ( R ) a ¯ , we deduce that
i = 1 N { | u m | j } a i m ( x , u m , u m ) . exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) d x C ( k , j , m ) .
Then,
i = 1 N Ω a i ( x , T k ( u m ) , T k ( u m ) ) a i ( x , T k ( u m ) , T k ( u ) ) × exp ( G ( | u m | ) ) ( T k ( u m ) T k ( u ) ) η j ( | u m | ) d x i = 1 N Ω a i ( x , T k ( u m ) , T k ( u ) ) . exp ( G ( | u m | ) ) | T k ( u m ) T k ( u ) | η j ( | u m | ) d x i = 1 N { | u m | k } a i ( x , T k ( u m ) , T k ( u ) ) . exp ( G ( | u m | ) ) T k ( u ) η j ( | u m | ) d x + C ( k , j , m ) .
By Lebesgue dominated convergence theorem, we have T k ( u m ) T k ( u ) strongly in W ˚ B , l o c 1 ( Ω ) and T k ( u m ) T k ( u ) weakly in W ˚ B 1 ( Ω ) ; then the terms on the right hand side of (50) go to zeros as k , j , m tend to infinity, which gives
i = 1 N Ω a i ( x , T k ( u m ) , T k ( u m ) ) a i ( x , T k ( u m ) , T k ( u ) ) × ( T k ( u m ) T k ( u ) ) exp ( G ( | u m | ) ) η j ( | u m | ) d x 0 .
By Proposition 4 and the diagonal process, we deduce for k that
B ( | u m | ) B ( | u | ) i n L 1 ( Ω ) .
Hence, we obtain for a subsequence
u m u a . e . i n Ω .
Step 5.
The equi-integrability of b i m ( x , u m , u m ) :
In this step we will show that
b i m ( x , u m , u m ) b i ( x , u , u ) .
Therefore, it is enough to show that b i m ( x , u m , u m ) is uniformly equi-integrable. We take the following test function
v = u m η exp ( 2 G ( | u m | ) ) η j ( | u m | ) T 1 ( u m T j ( u m ) ) .
We have
i = 1 N Ω a i m ( x , u m , u m ) . ( exp ( 2 G ( | u m | ) ) η j ( | u m | ) T 1 ( u m T j ( u m ) ) ) d x + i = 1 N Ω b i m ( x , u m , u m ) . exp ( 2 G ( | u m | ) ) η j ( | u m | ) T 1 ( u m T j ( u m ) ) d x + Ω m . T m ( u m ψ ) . s g 1 m ( u m ) . exp ( 2 G ( | u m | ) ) η j ( | u m | ) T 1 ( u m T j ( u m ) ) d x Ω f m ( x ) . exp ( 2 G ( | u m | ) ) η j ( | u m | ) T 1 ( u m T j ( u m ) ) d x ;
then,
i = 1 N Ω a i m ( x , u m , u m ) . u m . l ( | u m | ) a ¯ exp ( 2 G ( | u m | ) ) η j ( | u m | ) T 1 ( u m T j ( u m ) ) d x + i = 1 N Ω a i m ( x , u m , u m ) . exp ( 2 G ( | u m | ) ) ( η j ( | u m | ) ) T 1 ( u m T j ( u m ) ) d x + i = 1 N Ω a i m ( x , u m , u m ) . exp ( 2 G ( | u m | ) ) η j ( | u m | ) T 1 ( u m T j ( u m ) ) d x + Ω m . T m ( u m ψ ) . s g 1 m ( u m ) . exp ( 2 G ( | u m | ) ) η j ( | u m | ) T 1 ( u m T j ( u m ) ) d x i = 1 N Ω | b i m ( x , u m , u m ) | . exp ( 2 G ( | u m | ) ) η j ( | u m | ) T 1 ( u m T j ( u m ) ) d x + Ω f m ( x ) . exp ( 2 G ( | u m | ) ) η j ( | u m | ) T 1 ( u m T j ( u m ) ) d x .
By (2) and (4) we get
i = 1 N Ω a i m ( x , u m , u m ) . exp ( 2 G ( | u m | ) ) ( η j ( | u m | ) ) T 1 ( u m T j ( u m ) ) d x + i = 1 N Ω a i m ( x , u m , u m ) . exp ( 2 G ( | u m | ) ) η j ( | u m | ) T 1 ( u m T j ( u m ) ) d x + Ω m . T m ( u m ψ ) . s g 1 m ( u m ) . exp ( 2 G ( | u m | ) ) η j ( | u m | ) T 1 ( u m T j ( u m ) ) d x Ω ( h ( x ) + f m ( x ) + ϕ ( x ) . l ( | u m | ) a ¯ ) . exp ( 2 G ( | u m | ) ) η j ( | u m | ) T 1 ( u m T j ( u m ) ) d x ;
we deduce that
i = 1 N Ω a i m ( x , u m , u m ) . exp ( 2 G ( | u m | ) ) ( η j ( | u m | ) ) T 1 ( u m T j ( u m ) ) d x + i = 1 N { j | u m | j + 1 } a i m ( x , u m , u m ) . u m exp ( 2 G ( | u m | ) ) η j ( | u m | ) d x + Ω m . T m ( u m ψ ) . s g 1 m ( u m ) . exp ( 2 G ( | u m | ) ) η j ( | u m | ) T 1 ( u m T j ( u m ) ) d x Ω ( h ( x ) + f m ( x ) + ϕ ( x ) . l ( | u m | ) a ¯ ) . exp ( 2 G ( | u m | ) ) η j ( | u m | ) T 1 ( u m T j ( u m ) ) d x .
Since a i m ( x , u m , u m ) is bounded in W ˚ B 1 ( Ω ) , and η j ( | u m | ) 0 then by (27), (29) we obtain
i = 1 N { j + 1 < | u m | } B i ( | u m | ) d x exp 2 | | l | | L 1 ( R ) a ¯ . { j < | u m | } [ | C ( x ) | + ϕ ( x ) + h ( x ) + f m ( x ) + ϕ ( x ) . l ( | u m | ) a ¯ ] d x .
Thus, ϵ > 0 j ( ϵ ) > 0 such that
i = 1 N { j + 1 < | u m | } B i ( | u m | ) d x ϵ 2 j > j ( ϵ ) .
Let V ˚ ( Ω ( R ) ) be an arbitrary bounded subset for Ω ; then, for any measurable set E V ˚ ( Ω ( R ) ) we have
i = 1 N E B i ( | u m | ) d x i = 1 N E B i ( | T k ( u m ) | ) d x + i = 1 N { j + 1 < | u m | } B i ( | u m | ) d x .
We conclude that E V ˚ ( Ω ( R ) ) with m e a s ( E ) < β ( ϵ ) , and T k ( u m ) T k ( u ) in W ˚ B 1 ( Ω ) ,
i = 1 N E B i ( | T k ( u m ) | ) d x ϵ 2 .
Finally, by combining the last formulas we obtain
i = 1 N E B i ( | u m | ) d x ϵ E V ˚ ( Ω ( R ) ) s u c h t h a t m e a s ( E ) < β ( ϵ ) ,
giving the assumed results.
Step 6.
Passing to the limit:
Let φ W ˚ B 1 ( Ω ) L ( Ω ) ; we take the following test function:
v = u m ψ k T k ( u m φ ) , ψ K D ( Ω ) ,
such that
ψ k ( x ) = 1 for Ω ( R ) 0 for Ω ( R + 1 ) Ω ( R )
and | u m | | | φ | | < | u m φ | j . Then, by { | u m φ | j } { | u m | j + | | φ | | } we obtain
i = 1 N Ω ( R + 1 ) a i ( x , T m ( u m ) , u m ) ψ k T k ( u m φ ) d x + i = 1 N Ω ( R + 1 ) a i ( x , T m ( u m ) , u m ) T k ( u m φ ) ψ k d x + i = 1 N Ω ( R + 1 ) b i m ( x , u m , u m ) ψ k T k ( u m φ ) d x + Ω ( R + 1 ) m . T m ( u m ψ ) . s g 1 m ( u m ) . ψ k T k ( u m φ ) d x Ω ( R + 1 ) f m ( x ) ψ k T k ( u m φ ) d x ,
which implies that
i = 1 N Ω ( R + 1 ) a i ( x , T m ( u m ) , u m ) ψ k T k ( u m φ ) d x = i = 1 N Ω ( R + 1 ) a i ( x , T j + | | φ | | ( u m ) , T j + | | φ | | u m ) ψ k T k ( u m φ ) d x = i = 1 N Ω ( R + 1 ) a i ( x , T j + | | φ | | ( u m ) , T j + | | φ | | ( u m ) ) a i ( x , T j + | | φ | | ( u m ) , φ ) × T j + | | φ | | ( u m φ ) . χ { | u m φ | < j } d x + i = 1 N Ω ( R + 1 ) a i ( x , T j + | | φ | | ( u m ) , φ ) T j + | | φ | | ( u m φ ) . χ { | u m φ | < j } d x .
By Fatou’s Lemma we get
lim m inf i = 1 N Ω ( R + 1 ) a i ( x , T j + | | φ | | ( u m ) , φ ) T j + | | φ | | ( u m φ ) . χ { | u m φ | < j } d x = i = 1 N Ω ( R + 1 ) a i ( x , T j + | | φ | | ( u ) , φ ) T j + | | φ | | ( u φ ) . χ { | u φ | < j } d x ,
and the fact that
a i ( x , T j + | | φ | | ( u m ) , T j + | | φ | | ( u m ) ) a i ( x , T j + | | φ | | ( u ) , T j + | | φ | | ( u ) )
weakly in W ˚ B 1 ( Ω ) . Additionally, since ψ k T k ( u m φ ) ψ k T k ( u φ ) weakly in W ˚ B 1 ( Ω ) , and by (53) we obtain
i = 1 N Ω ( R + 1 ) b i m ( x , u m , u m ) ψ k T k ( u m φ ) d x i = 1 N Ω ( R + 1 ) b i ( x , u , u ) ψ k T k ( u φ ) d x ,
and
Ω ( R + 1 ) f m ( x ) ψ k T k ( u m φ ) d x Ω ( R + 1 ) f ( x ) ψ k T k ( u φ ) d x ,
and
Ω ( R + 1 ) m . T m ( u m ψ ) . s g 1 m ( u m ) . ψ k T k ( u m φ ) d x Ω ( R + 1 ) m . T m ( u ψ ) . s g 1 m ( u ) . ψ k T k ( u φ ) d x ,
so we get
i = 1 N Ω ( R + 1 ) a i ( x , u , u ) ψ k T k ( u φ ) d x + i = 1 N Ω ( R + 1 ) a i ( x , u , u ) T k ( u φ ) ψ k d x + i = 1 N Ω ( R + 1 ) b i ( x , u , u ) ψ k T k ( u φ ) d x + Ω ( R + 1 ) m . T m ( u ψ ) . s g 1 m ( u ) . ψ k T k ( u φ ) d x Ω ( R + 1 ) f ( x ) ψ k T k ( u φ ) d x ,
now passing to the limit to infinity in k , we obtain the entropy solution of the problem.

4. Uniqueness of the Entropy Solution

Theorem 3.
Suppose that conditions (1)–(3) are true, and b i ( x , u , u ) : Ω × R × R N R are strictly monotonic operators, at least for a broad class of lower order terms. Then, the problem ( P ) has a unique solution.
Proof. 
Let u and u ¯ belong to K ψ L ( Ω ) being two solutions of problem ( P ) with u u ¯ . In accordance with Definition 6, we obtain
i = 1 N Ω a i ( x , u , u ) . ( u v ) d x + i = 1 N b i ( x , u , u ) . ( u v ) d x + Ω m . T m ( u ψ ) . s g m ( u ) . ( u v ) d x Ω f ( x ) . ( u v ) d x
and
i = 1 N Ω a i ( x , u ¯ , u ¯ ) . ( u ¯ v ) d x + i = 1 N b i ( x , u ¯ , u ¯ ) . ( u ¯ v ) d x + Ω m . T m ( u ¯ ψ ) . s g m ( u ¯ ) . ( u ¯ v ) d x Ω f ( x ) . ( u ¯ v ) d x
Denote v = u μ ( x ) ( u u ¯ ) ( x ) and v = u ¯ μ ( x ) ( u u ¯ ) ( x ) with
μ ( x ) = 0 if x k , k | x | 2 k if | x | < k , 0 if x k ,
as test functions in (60) and (61) respectively. Using (1), (27), (29) and the condition of a strictly monotonic for the operator b i ( x , u , u ) , we substract the equations to obtain
i = 1 N Ω a i ( x , u , u ) a i ( x , u ¯ , u ¯ ) . ( u u ¯ ) μ ( x ) d x 0 ,
According to (6), we obtain
i = 1 N Ω B ¯ i a i ( x , u , u ) a i ( x , u ¯ , u ¯ ) d x + i = 1 N Ω B i ( u u ¯ ) . μ ( x ) d x i = 1 N Ω B ¯ i a i ( x , u , u ) a i ( x , u ¯ , u ¯ ) d x + 2 i = 1 N Ω B i ( u u ¯ ) d x 0 .
Since the N-functions B ¯ i verified the same conditions and properties of B i ; then by (10), we have
i = 1 N Ω B ¯ i a i ( x , u , u ) a i ( x , u ¯ , u ¯ ) d x c i = 1 N Ω | B ¯ i ( a i ( x , u , u ) ) | d x c i = 1 N Ω | B ¯ i ( a i ( x , u ¯ , u ¯ ) ) | d x ;
according to (3), we obtain
i = 1 N Ω B ¯ i a i ( x , u , u ) a i ( x , u ¯ , u ¯ ) d x a ˜ c i = 1 N Ω B i ( ( u u ¯ ) ) d x a ˜ c | | B ( u u ¯ ) | | 1 , Ω .
Combined with (62) and (63) we get
0 ( a ˜ c + 2 ) . | | B ( u u ¯ ) | | 1 , Ω 0 .
Finally, | | B ( u u ¯ ) | | 1 , Ω = 0 ; therefore, u = u ¯ a.e. in Ω . □

Author Contributions

All authors performed all the steps of the ideas and proofs in this research. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Conflicts of Interest

The authors declare no conflict of interest.

Appendix A

Let S be the operator defined by
S ( u ) = i = 1 N Ω a i m ( x , u , u ) d x + i = 1 N Ω b i m ( x , u , u ) d x + Ω m . T m ( u ψ ) . s g 1 m ( u ) d x Ω f m ( x ) d x ,
and for any v W ˚ B 1 ( Ω ) , u W ˚ B , l o c 1 ( Ω ( R ) ¯ ) we have
S ( u ) , v = i = 1 N Ω ( R ) a i m ( x , u , u ) . v d x + i = 1 N Ω ( R ) b i m ( x , u , u ) . v d x + Ω ( R ) m . T m ( u ψ ) . s g 1 m ( u ) . v d x Ω ( R ) f m ( x ) . v d x .
In order to show the result of the Theorem 1, it is sufficient to show that operator S is bounded, coercive and pseudo-monotonic.
Let us start by demonstrating that S is bounded. Additionally, according to (A1), (15), (27) and (29) we obtain
| S ( u ) , v | 2 | | a m ( x , u , u ) | | B ¯ , Ω ( R ) . | | v | | W ˚ B 1 ( Ω ) + 2 | | b m ( x , u , u ) | | B , Ω ( R ) . | | v | | W ˚ B 1 ( Ω ) c 0 . | | v | | W ˚ B 1 ( Ω ) .
or
| | a m ( x , u , u ) | | B ¯ , Ω ( R ) i = 1 N Ω ( R ) B ¯ i ( a i m ( x , u , u ) ) d x + N
and by diagonal process we obtain
| | a m ( x , u , u ) | | B ¯ , Ω ( R ) a ¯ | | u | | W ˚ B 1 ( Ω ) + | | φ | | 1 + N
and by (23) we deduce that
b m ( x , u , u ) b o u n d e d i n L B ( Ω )
if u with bounded support, and if support u = Ω ( R ) ¯ , then (A1) is bounded.
Next, we will move to proving that S is coercive. By (27), (29) for any u W ˚ B 1 ( Ω )
S ( u ) , u a ¯ i = 1 N Ω B i | u x i | d x Ω φ ( x ) d x + i = 1 N Ω b i m ( x , u , u ) . u d x Ω f m ( x ) . u d x ,
so
S ( u ) , u | | u | | W ˚ B 1 ( Ω ) 1 | | u | | W ˚ B 1 ( Ω ) a ¯ i = 1 N Ω B i | u x i | d x c 1 c 0 + 1 | | u | | W ˚ B 1 ( Ω ) i = 1 N Ω b i m ( x , u , u ) . u d x ,
using (23), we obtain
1 | | u | | W ˚ B 1 ( Ω ) i = 1 N Ω b i m ( x , u , u ) . u d x 2 c ( m ) .
Thus,
S ( u ) , u | | u | | W ˚ B 1 ( Ω ) 1 | | u | | W ˚ B 1 ( Ω ) a ¯ i = 1 N Ω B i | u x i | d x c 1 c 0 2 c ( m )
according to (22) we have for all k > 0 , α 0 > 0 such that
b i ( | u x i | ) > k b i | u x i | | | u x i | | B i , Ω i = 1 , , N ;
we take | | u x i | | B i , Ω > α 0 , i = 1 , , N . Additionally, since Ω is unbounded domain, then we can assume that | | u j | | W ˚ B 1 ( Ω ) as j . We suppose
| | u x 1 j | | B 1 , Ω + + | | u x N j | | B N , Ω N α 0 ,
according to (9) we get
| u j | b ( | u j | ) < c B ( u j ) ,
so,
S ( u j ) , u j | | u j | | W ˚ B 1 ( Ω ) a ¯ N α 0 i = 1 N Ω B i | u j x i | d x c 2 N α 0 2 c ( m ) a ¯ c N | | u x i j | | B i i = 1 N Ω | u x i j | . b i ( | u x i j | ) d x c 2 N α 0 2 c ( m ) a ¯ k c N i = 1 N Ω B i | u x i j | | | u x i j | | B i , Ω d x c 2 N α 0 2 c ( m ) ,
with c 2 = c 0 + c 1 . Now, by the Luxemburg norm, we have
| | u j | | B = inf { k > 0 / Ω B u j ( x ) k d x 1 }
then
i = 1 N Ω B i | u x i j | | | u x i j | | B i , Ω d x i = 1 N | | u x i j | | B i .
Hence,
S ( u j ) , u j | | u j | | W ˚ B 1 ( Ω ) a ¯ k c N | | u j | | W ˚ B 1 ( Ω ) c 2 N α 0 2 c ( m ) a s | | u j | | W ˚ B 1 ( Ω )
which gives the coercivity of the operator S.
Finally, we will end it by the demonstration of pseudo-monotonic of S. Following up this assumption, since the space W ˚ B 1 ( Ω ) is separable, then ( u j ) C 0 ( Ω ) such that
u j u i n W ˚ B 1 ( Ω ) ,
and
S ( u j ) y i n ( W ˚ B 1 ( Ω ) ) ;
according to (A2), we have for all subsequences denoted again by u j ,
| | u j | | W ˚ B 1 ( Ω ) c 2 , j N
( u j ) j N is bounded in W ˚ B 1 ( Ω ) , and since W ˚ B 1 ( Ω ) is continuously and compactly injected into L B ( Ω )
u j u i n L B ( Ω ) ,
u j u a . e . i n Ω , j N ,
and according to (53), we have
a i m ( x , u j , u j ) a i m ( x , u , u ) a . e . i n Ω , j N
and
b i m ( x , u j , u j ) b i m ( x , u , u ) a . e . i n Ω , j N
and
m . T m ( u j ψ ) . s g 1 m ( u j ) m . T m ( u ψ ) . s g 1 m ( u ) a . e . i n Ω , j N
from (A2) and (A3), a ˜ m L B ¯ ( Ω ) such that
a i m ( x , u j , u j ) a ˜ m , j N
and b ˜ m L B ( Ω ) such that
b i m ( x , u j , u j ) b ˜ m , j N ;
by (27) and (29) it is clear that for any v W ˚ B 1 ( Ω ) , we get
y , v = lim j i = 1 N Ω a i m ( x , u j , u j ) . v d x + lim j i = 1 N Ω b i m ( x , u j , u j ) . v d x = Ω a ˜ m . v d x + Ω b ˜ m . v d x
whereof
lim j sup < S ( u j ) , u j > = lim j sup { i = 1 N Ω a i m ( x , u j , u j ) u j d x + lim j i = 1 N Ω b i m ( x , u j , u j ) u j d x } Ω a ˜ m u j d x + Ω b ˜ m u j d x ,
by (A5), we have
Ω b m ( x , u j , u j ) u j d x Ω b ˜ m u d x ;
consequently,
lim j sup i = 1 N Ω a i m ( x , u j , u j ) u j d x Ω a ˜ m u j d x .
On the other hand, we have by the condition of monotony,
i = 1 N ( a i m ( x , u j , u j ) a i m ( x , u j , u ) ) . ( u j u ) 0 ,
which implies
i = 1 N ( a i ( x , T m ( u j ) , u j ) a i ( x , T j ( u j ) , u ) ) . ( u j u ) 0 ,
then;
i = 1 N a i ( x , T m ( u j ) , u j ) . u j i = 1 N a i ( x , T m ( u j ) , u ) . ( u j u ) + i = 1 N a i ( x , T m ( u j ) , u j ) . u ,
and by Step 3, we get
i = 1 N a i ( x , T m ( u j ) , u ) i = 1 N a i ( x , T m ( u ) , u ) i n L B ¯ ( Ω ) ;
according to (A4), we obtain
lim j inf i = 1 N Ω a i m ( x , u j , u j ) . u j d x Ω a ˜ m . u j d x ;
therefore, from (A9), we have
lim j i = 1 N Ω a i m ( x , u j , u j ) . u j d x = Ω a ˜ m . u j d x ;
according to (A6), (A8) and (A11) we get
< S ( u j ) , u j > < y , u > a s j .
Hence, from (A12), and (53) we obtain
lim j i = 1 N Ω ( a i m ( x , u j , u j ) a i m ( x , u j , u ) ) . ( u j u ) d x = 0 .
By (A6) we can conclude that
< y , u > = < S ( u ) , u > u W ˚ B 1 ( Ω ) .

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Benslimane, O.; Aberqi, A.; Bennouna, J. The Existence and Uniqueness of an Entropy Solution to Unilateral Orlicz Anisotropic Equations in an Unbounded Domain. Axioms 2020, 9, 109. https://doi.org/10.3390/axioms9030109

AMA Style

Benslimane O, Aberqi A, Bennouna J. The Existence and Uniqueness of an Entropy Solution to Unilateral Orlicz Anisotropic Equations in an Unbounded Domain. Axioms. 2020; 9(3):109. https://doi.org/10.3390/axioms9030109

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Benslimane, Omar, Ahmed Aberqi, and Jaouad Bennouna. 2020. "The Existence and Uniqueness of an Entropy Solution to Unilateral Orlicz Anisotropic Equations in an Unbounded Domain" Axioms 9, no. 3: 109. https://doi.org/10.3390/axioms9030109

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