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WeconsiderinthispaperthefollowingnonlinearanisotropicellipticFourierboundaryvalueproblem: 1.Introduction andStanislasOuaro IdrissaIbrango ENTROPYSOLUTIONFORDOUBLYNONLINEARELLIPTICANISOTROPICPROBLEMSWITHFOURIERBOUNDARYCONDITIONS ( 2015 ) 123–150doi:10.7151

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doi:10.7151/dmdico.1175

ENTROPY SOLUTION FOR DOUBLY NONLINEAR ELLIPTIC ANISOTROPIC PROBLEMS WITH FOURIER

BOUNDARY CONDITIONS

Idrissa Ibrango

Laboratoire de Math´ ematiques et Informatique (LAMI)

UFR. Sciences et Techniques, Universit´ e Polytechnique de Bobo-Dioulasso 01 BP 1091 Bobo 01, Bobo-Dioulasso, Burkina Faso

e-mail: ibrango2006@yahoo.fr

and

Stanislas Ouaro

Laboratoire de Math´ ematiques et Informatique (LAMI) UFR. Sciences Exactes et Appliqu´ ees, Universit´ e de Ouagadougou

03 BP 7021 Ouaga 03, Ouagadougou, Burkina Faso e-mail: souaro@univ-ouaga.bf, ouaro@yahoo.fr

Abstract

The goal of this paper is to study nonlinear anisotropic problems with Fourier boundary conditions. We first prove, by using the technic of mono- tone operators in Banach spaces, the existence of weak solutions, and by approximation methods, we prove a result of existence and uniqueness of entropy solution.

Keywords: anisotropic Sobolev spaces, variable exponent, monotone oper- ator, Fourier boundary conditions, entropy solutions.

2010 Mathematics Subject Classification: 35J20, 35J25, 35D30, 35J60.

1. Introduction

We consider in this paper the following nonlinear anisotropic elliptic Fourier

boundary value problem:

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(1.1)

 

 

 

 

 

 

N

X

i=1

∂x

i

a

i

 x, ∂u

∂x

i



+ b(u) = f in Ω

N

X

i=1

a

i

 x, ∂u

∂x

i



η

i

+ λu = g on ∂Ω,

where Ω is an open bounded domain of R

N

(N ≥ 3) with smooth boundary and meas(Ω) > 0, f ∈ L

1

(Ω), g ∈ L

1

(∂Ω), η = (η

1

, . . . , η

N

) is the unit outward normal on ∂Ω and λ > 0 is a constant.

The problem (1.1) is the anisotropic case of the nonlinear isotropic problem

(1.2)

( b(u) − div a(x, ∇u) = f in Ω a(x, ∇u)η + λu = g on ∂Ω,

studied in [14] by Nyanquini and Ouaro. The authors use the minimization tech- nic used in [12] (see also [6, 10, 15]) to prove the existence of weak solution when f and g are bounded, namely f ∈ L

(Ω), g ∈ L

(∂Ω) and by approximation methods they obtain the entropy solution when f ∈ L

1

(Ω), g ∈ L

1

(∂Ω).

The study of problems involving variable exponents has received considerable attention (see [10]–[15]) due to the fact that they can model various phenomena (see [1, 7, 13]).

All papers concerned by problems like (1.1) considered particular cases of function b. Indeed, in [3], Bonzi et al. adopted the technic used in [14] to study the following anisotropic problem:

(1.3)

 

 

 

 

 

 

N

X

i=1

∂x

i

a

i

 x, ∂u

∂x

i



+ |u|

pM(x)−2

u = f in Ω

N

X

i=1

a

i

 x, ∂u

∂x

i



η

i

+ λu = g on ∂Ω,

where f ∈ L

1

(Ω) and g ∈ L

1

(∂Ω). In the present paper, as the function b is more general, it is not possible to use minimization technic to get the existence of entropy solution. Therefore, we used the technic of monotone operators in Banach spaces (see [16]) to obtain the existence of entropy solutions of problem (1.1).

Indeed, we define an approximation problem, and we prove that this problem has a solution u

n

which converges to u, an entropy solution of (1.1).

For presenting our main result, we first have to describe the data involved

in our problem. Let Ω be a bounded domain in R

N

(N ≥ 3) with smooth

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boundary domain ∂Ω and ~ p(·) = p

1

(·), . . . , p

N

(·) such that for any i = 1, . . . , N, p

i

(·) : Ω −→ [2; N ) is a continuous function with

(1.4) 1 < p

i

:= ess inf

x∈Ω

p

i

(x) ≤ ess sup

x∈Ω

p

i

(x) := p

+i

< +∞.

For any i = 1, . . . , N , let a

i

: Ω × R −→ R be a Carath´eodory function satisfying:

• there exists a positive constant C

1

such that (1.5) |a

i

(x, ξ)| ≤ C

1



j

i

(x) + |ξ|

pi(x)−1

 ,

for almost every x ∈ Ω and for every ξ ∈ R, where j

i

is a nonnegative function lying in L

p0i(·)

(Ω), with

p1

i(x)

+

p01 i(x)

= 1;

• for every ξ, η ∈ R with ξ 6= η and for almost every x ∈ Ω, there exists a positive constant C

2

such that

(1.6) a

i

(x, ξ) − a

i

(x, η)(ξ − η) ≥

C

2

|ξ − η|

pi(x)

if |ξ − η| ≥ 1 C

2

|ξ − η|

pi

if |ξ − η| < 1;

• there exists a positive constant C

3

such that

(1.7) a

i

(x, ξ) · ξ ≥ C

3

|ξ|

pi(x)

, for ξ ∈ R, for almost every x ∈ Ω.

The function b is such that

(1.8) b : R −→ R is continuous, surjective, nondecreasing with b(0) = 0.

Throughout the paper, for any i = 1, . . . , N , we assume that (1.9) p(N − 1) ¯

N (¯ p − 1) < p

i

< p(N − 1) ¯

N − ¯ p , p

+i

− p

i

− 1

p

i

< p − N ¯

¯

p(N − 1) and

(1.10)

N

X

i=1

1 p

i

> 1, where

Np¯

= P

N

i=1 1 pi

. We put for all x ∈ Ω,

p

M

(x) := max p

1

(x), . . . , p

N

(x)

and p

m

(x) := min p

1

(x), . . . , p

N

(x)

(4)

and for all x ∈ ∂Ω,

p

(x) =

(N − 1)p(x)

N − p(x) if p(x) < N

+∞ if p(x) ≥ N.

The hypotheses on a

i

are classical in the study of nonlinear problems (see [4, 6]). A prototype example that is covered by our assumptions is the following anisotropic ~ p(·)-harmonic system

(1.11) −

N

X

i=1

∂x

i



∂u

∂x

i

pi(x)−2

∂u

∂x

i



= f,

which, in the particular case when p

i

= p for any i = 1, . . . , N , is a generalization of the classical p-Laplace equation

−div |∇u|

p−2

∇u = f.

The rest of the paper is organized as follows. We first present some basic prelim- inary results including the variable exponent in Section 2. In Section 3, we study the existence and uniqueness of entropy solution.

2. Preliminaries

We recall in this section some definitions and basic properties of anisotropic Lebesgue and Sobolev spaces with variable exponent, which will be used in the next Section. Set

C

+

(Ω) =



p ∈ C(Ω) such that min

x∈Ω

p(x) > 1

 . For any p ∈ C

+

(Ω), the variable exponent Lebesgue space is defined by

L

p(·)

(Ω) :=



u : Ω −→ R, measurable such that Z

|u|

p(x)

dx < ∞

 , endowed with the so-called Luxemburg norm

|u|

p(·)

:= inf



β > 0 such that Z

u β

p(x)

dx ≤ 1

 .

The p(·)-modular of the L

p(·)

(Ω) space is the mapping ρ

p(·)

: L

p(·)

(Ω) −→ R defined by

ρ

p(·)

(u) :=

Z

|u|

p(x)

dx.

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For any u ∈ L

p(·)

(Ω), the following inequality (see [8, 9]) will be used later.

(2.1) min |u|

pp(·)

; |u|

pp(·)+

≤ ρ

p(·)

(u) ≤ max |u|

pp(·)

; |u|

pp(·)+

.

For any u ∈ L

p(·)

(Ω) and v ∈ L

p0(·)

(Ω) with

p(x)1

+

p01(x)

= 1 in Ω, we have the H¨ older type inequality:

(2.2)

Z

uvdx

≤  1

p

+ 1 (p

0

)



|u|

p(·)

|v|

p0(·)

.

If Ω is bounded and p, q ∈ C

+

(Ω) such that p(x) ≤ q(x) for any x ∈ Ω, then the embedding L

q(·)

(Ω) ,→ L

p(·)

(Ω) is continuous (see [11], Theorem 2.8).

Herein, we need the following anisotropic Sobolev space with variable exponent.

W

1,~p(·)

(Ω) :=



u ∈ L

pM(·)

(Ω) such that ∂u

∂x

i

∈ L

pi(·)

(Ω), i = 1, . . . , N

 . W

1,~p(·)

(Ω) is a separable and reflexive Banach space (see [12]) under the norm

||u||

~p(·)

= |u|

pM(·)

+

N

X

i=1

∂u

∂x

i

pi(·)

.

We need the following embedding and trace results.

Theorem 2.1 ([8], Corollary 2.1). Let Ω ⊂ R

N

(N ≥ 3) be a bounded open set and for all i = 1, . . . , N, p

i

∈ L

(Ω), p

i

(x) ≥ 1 a.e. in Ω. Then, for any q ∈ L

(Ω) with q(x) ≥ 1 a.e. in Ω such that

ess inf

x∈Ω

p

M

(x) − q(x) > 0, we have the compact embedding

(2.3) W

1,~p(·)

(Ω) ,→ L

q(·)

(Ω).

Theorem 2.2 ([6], Theorem 6). Let Ω ⊂ R

N

(N ≥ 2) be a bounded open set with smooth boundary and let ~ p(·) ∈ C

+

(Ω) 

N

, r ∈ C(Ω) satisfy the condition (2.4) 1 ≤ r(x) < min p

1

(x), . . . , p

N

(x) , ∀ x ∈ ∂Ω.

Then, there is a compact boundary trace embedding

W

1,~p(·)

(Ω) ,→ L

r(·)

(∂Ω).

(6)

In particular case

W

1,~p(·)

(Ω) ,→ L

1

(∂Ω).

We introduce the numbers q = N (¯ p − 1)

N − 1 and q

= N (¯ p − 1)

N − ¯ p = N q N − q . The following result is due to Troisi (see [17]).

Theorem 2.3. Let p

1

, . . . , p

N

∈ [1, +∞); g ∈ W

1,(p1,...,pN)

(Ω) and ( q = (¯ p)

if (¯ p)

< N

q ∈ [1, +∞) if (¯ p)

≥ N.

Then, there exists a constant C

4

> 0 depending on N, p

1

, . . . , p

N

if ¯ p < N and also on q and meas(Ω) if ¯ p ≥ N such that

(2.5) ||g||

Lq(Ω)

≤ C

4

N

Y

i=1



||g||

LpM(Ω)

+

∂g

∂x

i

Lpi(Ω)



1/N

.

In this paper, we will use the weak Lebesgue (Marcinkiewicz) space M

q

(Ω) (1 <

q < +∞) as the set of measurable functions h : Ω −→ R for which the distribution function

(2.6) λ

h

(k) = meas {x ∈ Ω : |h(x)| > k}, k ≥ 0 satisfies an estimate of the form

(2.7) λ

h

(k) ≤ Ck

−q

, for some finite constant C > 0.

We will use the following pseudo norm in M

q

(Ω):

(2.8) ||h||

Mq(Ω)

:= inf{C > 0 : λ

h

(k) ≤ Ck

−q

, ∀ k > 0}.

For any k > 0, the truncation function T

k

is defined on R by (2.9) T

k

(s) = max{−k; min{k; s}}.

It is clear that lim

k→+∞

T

k

(s) = s and |T

k

(s)| = min{|s|; k}. In order to simplify the notation, for any v ∈ W

1,~p(·)

(Ω), we use v instead of v

|∂Ω

for the trace of v on ∂Ω.

Set T

1,~p(·)

(Ω) as the set of the measurable functions u : Ω −→ R such that for any

k > 0, T

k

(u) ∈ W

1,~p(·)

(Ω). We define the space T

tr1,~p(·)

(Ω) as the set of functions

(7)

u ∈ T

1,~p(·)

(Ω) such that there exists a sequence (u

n

)

n

⊂ W

1,~p(·)

(Ω) satisfying

(2.10) u

n

−→ u a.e. in Ω,

(2.11) ∂

∂x

i

T

k

(u

n

) −→ ∂

∂x

i

T

k

(u) in L

1

(Ω), ∀ k > 0 and there exists a measurable function v on ∂Ω such that

(2.12) u

n

−→ v a.e. on ∂Ω.

We need the following lemma proved in [5].

Lemma 2.1. Let h be a nonnegative function in W

1,~p(·)

(Ω). Assume ¯ p < N and there exists a constant C > 0 such that

(2.13)

Z

|T

k

(h)|

pM

dx +

N

X

i=1

Z

{|h|≤k}

∂h

∂x

i

pi

dx ≤ C(1 + k), ∀ k > 0.

Then, there exists a constant D, depending on C, such that

(2.14) ||h||

Mq∗(Ω)

≤ D,

where q

= N (¯ p − 1)/(N − ¯ p).

3. Entropy solution

The notion of entropy solution to problem (1.1) is the following.

Definition 3.1. A measurable function u ∈ T

tr1,~p(·)

(Ω) is an entropy solution of problem (1.1) if b(u) ∈ L

1

(Ω), u ∈ L

1

(∂Ω) and

(3.1)

 

 

 

 

N

X

i=1

Z

a

i

 x, ∂u

∂x

i

 ∂

∂x

i

T

k

(u − ϕ)dx + Z

b(u)T

k

(u − ϕ)dx

+ λ Z

∂Ω

uT

k

(u − ϕ)dσ ≤ Z

f T

k

(u − ϕ)dx + Z

∂Ω

gT

k

(u − ϕ)dσ, for every ϕ ∈ W

1,~p(·)

(Ω) ∩ L

(Ω) and for every k > 0.

Remark that as we have in the definition ϕ ∈ W

1,~p(·)

(Ω) ∩ L

(Ω) then (u − ϕ) ∈

T

tr1,~p(·)

(Ω), hence T

k

(u − ϕ) ∈ W

1,~p(·)

(Ω) ∩ L

(Ω). Consequently the first, the

second, the third and the fifth integrals in (3.1) are well defined.

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The existence result is the following theorem.

Theorem 3.1. Assume (1.4)–(1.10). There exists at least one entropy solution of the problem (1.1).

Proof. The proof is done in three steps.

Step 1. The approximate problem.

We define the reflexive space

E = W

1,~p(·)

(Ω) × L

pM(·)

(∂Ω).

Let X

0

be the subspace of E defined by

X

0

= {(u, v) ∈ E : v = τ (u)},

where τ (u) is the trace of u ∈ T

tr1,~p(·)

(Ω) in the usual sense, since u ∈ W

1,~p(·)

(Ω).

In the sequel, we will identify an element (u, v) ∈ X

0

with its representative u ∈ W

1,~p(·)

(Ω).

For any n ∈ N and ε > 0, we consider the sequence of approximate problems

(3.2)

 

 

 

 

N

X

i=1

Z

a

i

 x, ∂u

n

∂x

i

 ∂v

∂x

i

dx + ε Z

|u

n

|

pM(x)−2

u

n

vdx

+ Z

T

n

b(u

n

)vdx + λ Z

∂Ω

T

n

(u

n

)vdσ = Z

f

n

vdx + Z

∂Ω

g

n

vdσ, where f

n

= T

n

(f ) and g

n

= T

n

(g).

Remark that (f

n

)

n

and (g

n

)

n

are sequences of bounded functions which converge strongly to f ∈ L

1

(Ω) and to g ∈ L

1

(∂Ω), respectively. Moreover,

||f

n

||

L1(Ω)

≤ ||f ||

L1(Ω)

, ||g

n

||

L1(∂Ω)

≤ ||g||

L1(∂Ω)

for all n ∈ N and

kf

n

k

≤ kf k

L1(Ω)

meas(Ω) , kg

n

k

≤ kgk

L1(∂Ω)

meas(Ω) for all n ∈ N.

We define operators A

n

by hA

n

(u), vi = hA(u), vi +

Z

T

n

b(u)vdx + λ Z

∂Ω

T

n

(u)vdσ ∀ u, v ∈ X

0

, where

hA(u), vi =

N

X

i=1

Z

a

i

 x, ∂u

∂x

i

 ∂v

∂x

i

dx + ε Z

|u|

pM(x)−2

uvdx.

(9)

Assertion 1. The operator A is of type M.

• The operator A is monotone. Indeed, for u, v ∈ W

1,~p(·)

(Ω), we have hA(u) − A(v), u − vi = hA(u), u − vi + hA(v), v − ui

= Z

Ω N

X

i=1

a

i

 x, ∂u

∂x

i

 ∂(u − v)

∂x

i

dx + ε Z

|u|

pM(x)−2

u(u − v)dx

+ Z

Ω N

X

i=1

a

i

 x, ∂v

∂x

i

 ∂(v − u)

∂x

i

dx + ε Z

|v|

pM(x)−2

v(v − u)dx

= Z

Ω N

X

i=1

 a

i

 x, ∂u

∂x

i



− a

i

 x, ∂v

∂x

i

 ∂u

∂x

i

− ∂v

∂x

i

 dx

+ ε Z



|u|

pM(x)−2

u − |v|

pM(x)−2

v



(u − v)dx.

Therefore,

(3.3) hA(u) − A(v), u − vi ≥ 0,

since for i = 1, . . . , N, for almost every x ∈ Ω, a

i

(x, ·) and t 7−→ |t|

pM(x)−2

t are monotone.

• The operator A is hemicontinuous. Indeed, for every u, v in W

1,~p(·)

(Ω), let ϕ : t ∈ R 7−→ ϕ(t) = hA(u + tv), vi

and let t, t

0

∈ R such that t −→ t

0

. We have w = u + tv −→ w

0

= u + t

0

v in W

1,~p(·)

(Ω). Using the H¨ older type inequality, there exists i

0

∈ {1, . . . , N } such that

|ϕ(t) − ϕ(t

0

)| = |hA(u + tv), vi − hA(u + t

0

v), vi|

N

X

i=1

Z

a

i

 x, ∂w

∂x

i



− a

i

 x, ∂w

0

∂x

i



∂v

∂x

i

dx

+ ε Z

|w|

pM(x)−2

w − |w

0

|

pM(x)−2

w

0

|v|dx

≤ N

 1 p

i0

+ 1

(p

0i0

)



a

i0

 x, ∂w

∂x

i0



− a

i0

 x, ∂w

0

∂x

i0



p0i0(·)

∂v

∂x

i0

pi0(·)

+ ε

 1

p

M

+ 1 (p

0M

)



|w|

pM(x)−2

w − |w

0

|

pM(x)−2

w

0

p0M(·)

|v|

pM(·)

.

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Let’s denote ψ

i0

(x, w) = a

i0

x,

∂x∂w

i0

. Using assumption (1.5) and ([11], Theorems 4.1 and 4.2) we have ψ

i0

(x, w) −→ ψ

i0

(x, w

0

) in L

p0i0(·)

(Ω). Then, we deduce that ϕ is continuous, namely the operator A is hemicontinuous.

Since the operator A is monotone and hemicontinuous, then according to the Lemma 2.1 in [16], A is of type M.

Assertion 2. Operators A

n

are of type M. Indeed, let (u

k

)

k∈N

be a sequence in X

0

such that

(3.4)

 

 

u

k

* u weakly in X

0

, A

n

u

k

* χ weakly in X

00

, lim sup

k→+∞

hA

n

(u

k

), u

k

i ≤ hχ, ui.

Since

T

n

(b(u))u ≥ 0 and T

n

(u)u ≥ 0, by the Fatou’s lemma, we obtain

(3.5)

lim inf

k→+∞

 Z

T

n

(b(u

k

))u

k

dx + Z

∂Ω

λT

n

(u

k

)u

k



≥ Z

T

n

(b(u))udx + Z

∂Ω

λT

n

(u)udσ

and thanks to the Lebesgue dominated convergence theorem, we get

(3.6)

k→+∞

lim

 Z

T

n

(b(u

k

))vdx + Z

∂Ω

λT

n

(u

k

)vdσ



= Z

T

n

(b(u))vdx + Z

∂Ω

λT

n

(u)vdσ, for all v in X

0

. Consequently,

T

n

(b(u

k

)) + λT

n

(u

k

) * T

n

(b(u)) + λT

n

(u) weakly in X

00

. Therefore, we deduce that

Au

k

* χ − T

n

(b(u)) + λT

n

(u) 

weakly in X

00

.

As in Assertion 1 we prove that the operator A is of type M, so we have

Au = χ − T

n

(b(u)) + λT

n

(u).

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Thus, it follows that

A

n

u = χ.

Hence A

n

is of type M.

Assertion 3. Operators A

n

are coercive. Indeed, since T

n

(b(u))u + λT

n

(u)u ≥ 0, then

(3.7) hA

n

(u), ui ≥ hA(u), ui.

According to (1.7), we have

(3.8) hA(u), ui ≥ C

3

Z

Ω N

X

i=1

∂u

∂x

i

pi(x)

dx + ε Z

|u|

pM(x)

dx.

Denote I =



i ∈ {1, . . . , N } :

∂u

∂x

i

pi(·)

≤ 1



and J =



i ∈ {1, . . . , N } :

∂u

∂x

i

pi(·)

> 1

 . We have

Z

Ω N

X

i=1

∂u

∂x

i

pi(x)

dx = X

i∈I

Z

∂u

∂x

i

pi(x)

dx + X

i∈J

Z

∂u

∂x

i

pi(x)

dx ≥ X

i∈I

∂u

∂x

i

p+i pi(·)

+ X

i∈J

∂u

∂x

i

pi pi(·)

≥ X

i∈J

∂u

∂x

i

pi pi(·)

≥ X

i∈J

∂u

∂x

i

pm

pi(·)

N

X

i=1

∂u

∂x

i

pm

pi(·)

− X

i∈I

∂u

∂x

i

pm

pi(·)

N

X

i=1

∂u

∂x

i

pm

pi(·)

− N.

Using the convexity of the application t ∈ R

+

7−→ t

pm

, p

m

> 1, we obtain Z

Ω N

X

i=1

∂u

∂x

i

pi(x)

dx ≥ 1 N

pm−1



N

X

i=1

∂u

∂x

i

pi(·)



pm

− N.

Then

(3.9) hA

n

(u), ui ≥ C

3

N

pm−1



N

X

i=1

∂u

∂x

i

pi(·)



pm

+ ε Z

|u|

pM(x)

dx − C

3

N.

(12)

• Assume |u|

p

M(·)

> 1. Then, (2.1) gives R

|u|

pM(x)

dx ≥ |u|

p

m

pM(·)

. So, combining (3.7) and (3.9) we get

hA

n

(u), ui ≥ C



N

X

i=1

∂u

∂x

i

pi(·)



pm

+ |u|

p

m

pM(·)



− C

3

N

≥ C

2

pm−1

||u||

p~p(·)m

− C

3

N, where C = min

 C

3

N

pm−1

; ε

 .

• Assume |u|

pM(·)

≤ 1. Then, combining (3.7) and (3.9) we get

hA

n

(u), ui ≥ C



N

X

i=1

∂u

∂x

i

pi(·)



pm

+ |u|

p

m

pM(·)



− 1 − C

3

N + ε Z

|u|

pM(x)

dx

≥ C

2

pm−1

||u||

p~p(·)m

− 1 − C

3

N, where C = min

 C

3

N

pm−1

; 1

 . Consequently, since p

m

> 1, the operator A

n

is coercive.

Besides, the operators A

n

are bounded and hemicontinuous.

Then for any F

n

= T

n

(f ), T

n

(g) ∈ E

0

⊂ X

00

, we can deduce the existence of functions u

n

∈ X

0

such that

hA

n

(u

n

), vi = hF

n

, vi for all v ∈ X

0

.

Namely, every u

n

is a weak solution of the approximate problem (3.2).

Now we are going to prove that these approximated solutions u

n

tend, as n goes to infinity, to a measurable function u which is an entropy solution of the problem (1.1). To start with, we establish some a priori estimates.

Step 2. A priori estimates.

Assume (1.4)–(1.10) and let u

n

be a solution of problem (3.2). We have the following results.

Lemma 3.1. There exists a constant C

5

> 0 such that

(3.10)

Z

|T

k

(u

n

)|

pM

dx +

N

X

i=1

Z

{|un|≤k}

∂u

n

∂x

i

pi

dx ≤ C

5

(k + 1).

Proof. Let us take T

k

(u

n

) as test function in (3.2). Since Z

T

n

(b(u

n

))T

k

(u

n

)dx + ε Z

|u

n

|

pM(x)−2

u

n

T

k

(u

n

) + Z

∂Ω

λT

n

(u

n

)T

k

(u

n

)dσ ≥ 0,

(13)

using relation (1.7), we obtain

(3.11) C

3 N

X

i=1

Z

{|un|≤k}

∂u

n

∂x

i

pi(x)

dx ≤ k ||f ||

L1(Ω)

+ ||g||

L1(∂Ω)

.

Then, we have

N

X

i=1

Z

{|un|≤k}

∂u

n

∂x

i

pi

dx =

N

X

i=1

Z

{|un|≤k; |∂un

∂xi|>1}

∂u

n

∂x

i

pi

dx

+

N

X

i=1

Z

{|un|≤k; |∂un

∂xi|≤1}

∂u

n

∂x

i

pi

dx

N

X

i=1

Z

{|un|≤k}

∂u

n

∂x

i

pi(x)

dx + N · meas(Ω)

≤ k C

3

||f ||

L1(Ω)

+ ||g||

L1(∂Ω)

 + N · meas(Ω).

Moreover, we have Z

|T

k

(u

n

)|

pM

dx = Z

{|Tk(un)|≤1}

|T

k

(u

n

)|

pM

dx + Z

{|Tk(un)|>1}

|T

k

(u

n

)|

pM

dx

≤ meas(Ω) + Z

{|Tk(un)|>1}

k

pM

dx ≤ meas(Ω)(1 + k

pM

).

Therefore, we get Z

|T

k

(u

n

)|

pM

dx +

N

X

i=1

Z

{|un|≤k}

∂u

n

∂x

i

pi

dx

≤ meas(Ω)(1 + N + k

pM

) + k 1

C

3

||f ||

L1(Ω)

+ ||g||

L1(∂Ω)

 ≤ C

5

(1 + k), where C

5

= max meas(Ω)(1 + N + k

pM

);

C1

3

||f ||

L1(Ω)

+ ||g||

L1(∂Ω)

 .

Lemma 3.2. For any k > 0, there exists two constants C

7

> 0 and C

8

> 0 such that

(i) ||u

n

||

Mq∗(Ω)

≤ C

7

; (ii)

∂u

n

∂x

i

Mpiq/ ¯p(Ω)

≤ C

8

, ∀ i = 1, . . . , N.

(14)

Proof. (i) is a consequence of Lemmas 2.1 and 3.1.

(ii) • Let α ≥ 1. For any k ≥ 1, we have λ

∂un

∂xi

(α) = meas



x ∈ Ω :

∂u

n

∂x

i

> α



= meas



∂u

n

∂x

i

> α; |u

n

| ≤ k



+ meas



∂u

n

∂x

i

> α; |u

n

| > k



≤ Z



|∂un

∂xi|>α; |un|≤k

dx + λ

un

(k)

≤ Z

{|un|≤k}

 1 α

∂u

n

∂x

i



pi

dx + λ

un

(k) ≤ α

−pi

C

0

k + Ck

−q

. Then, there exists a positive constant C

6

such that

(3.12) λ

∂un

∂xi

(α) ≤ C

6

−pi

+ k

−q

.

Let us consider the function

g : [1, +∞[−→ R, t 7−→ g(t) = t

α

pi

+ t

−q

. We have g

0

(t) = 0 for t = q

α

pi



q∗+11

. Thus, if we take k = q

α

pi



q∗+11

≥ 1 in (3.12) we get

λ

∂un

∂xi

(α) ≤ C

6

k  q

+ 1 q

1 α

pi



≤ C

60

α

q∗

q∗+1pi

≤ C

60

α

−piq/ ¯p

,

∀ α ≥ 1, where C

60

is a positive constant.

• If 0 ≤ α < 1, we have λ

∂un

∂xi

(α) = meas



∂u

n

∂x

i

> α



≤ meas(Ω) ≤ meas(Ω)α

−piq/ ¯p

. Then

λ

∂un

∂xi

(α) ≤ (C

60

+ meas(Ω))α

−pi q/ ¯p

, ∀ α ≥ 0.

Therefore, we deduce that there exists a positive constant C

8

such that

∂u

n

∂x

i

Mpiq/ ¯p(Ω)

≤ C

8

, ∀ i = 1, . . . , N.

Step 3. Existence of entropy solution.

Using Lemma 3.2, we have the following useful lemma (see [5]).

(15)

Lemma 3.3. For i = 1, . . . , N , as n −→ +∞, we have (3.13) a

i

 x, ∂u

n

∂x

i



−→ a

i

 x, ∂u

∂x

i



in L

1

(Ω) a.e. x ∈ Ω.

In order to pass to the limit in relation (3.2), we also need the following conver- gence results which can be proved as in [2] (see also [4, 5]).

Proposition 3.1. Assume (1.4)–(1.10). If u

n

∈ W

1,~p(·)

(Ω) is a weak solution of (3.2) then the sequence (u

n

)

n∈N

is Cauchy in measure. In particular, there exists a measurable function u and a sub-sequence still denoted by u

n

such that u

n

−→ u in measure.

Proposition 3.2. Assume (1.4)–(1.10). If u

n

∈ W

1,~p(·)

(Ω) is a weak solution of (3.2) then

(i) there exists s > 1 such that u

n

→ u a.e. in Ω and u

n

* u in W

1,s

(Ω), (ii) for all i = 1, . . . , N,

∂u∂xn

i

converges strongly in L

1

(Ω). Moreover, a

i

x,

∂u∂xn

i

 converges to a

i

x,

∂x∂u

i

 in L

1

(Ω) strongly and in L

p0i(·)

(Ω) weakly for all i = 1, . . . , N ,

(iii) u

n

converges to some measurable function v a.e. on ∂Ω.

We can now pass to the limit in relation (3.2).

Let ϕ ∈ W

1,~p(·)

(Ω) ∩ L

(Ω) and choosing T

k

(u

n

− ϕ) as test function in (3.2), we get

(3.14)

 

 

 

 

 

 

 

 

 

 

N

X

i=1

Z

a

i

 x, ∂u

n

∂x

i

 ∂

∂x

i

T

k

(u

n

− ϕ)dx + Z

T

n

(b(u

n

))T

k

(u

n

− ϕ)dx

+ ε Z

|u

n

|

pM(x)−2

u

n

T

k

(u

n

− ϕ)dx + λ Z

∂Ω

T

n

(u

n

)T

k

(u

n

− ϕ)dσ

= Z

f

n

T

k

(u

n

− ϕ)dx + Z

∂Ω

g

n

T

k

(u

n

− ϕ)dσ.

For the right-hand side of (3.14) we have

(3.15)

Z

f

n

T

k

(u

n

− ϕ)dx + Z

∂Ω

g

n

T

k

(u

n

− ϕ)dσ

−→

Z

f T

k

(u − ϕ)dx + Z

∂Ω

gT

k

(u − ϕ)dσ,

because f

n

and g

n

converges strongly respectly to f in L

1

(Ω) and g in L

1

(∂Ω)

and T

k

(u

n

− ϕ) converges weakly-* to T

k

(u − ϕ) in L

(Ω) and a.e. in Ω.

(16)

For the first term of (3.14) we have (see [5]):

(3.16)

lim inf

n→+∞

N

X

i=1

Z

a

i

 x, ∂u

n

∂x

i

 ∂

∂x

i

T

k

(u

n

− ϕ)dx

N

X

i=1

Z

a

i

 x, ∂u

∂x

i

 ∂

∂x

i

T

k

(u − ϕ)dx.

We now focus our attention on the second and the fourth terms of (3.14). We have

(3.17) T

n

(b(u

n

))T

k

(u

n

− ϕ) −→ b(u)T

k

(u − ϕ) a.e. x ∈ Ω and

(3.18) |T

n

(b(u

n

))T

k

(u

n

− ϕ)| ≤ k|b(u

n

)|.

We will show that

|b(u

n

)| ≤ ||f

n

||

a.e. on Ω and |u

n

| ≤ 1

λ ||g

n

||

a.e. on ∂Ω.

Indeed, recall that for any δ > 0, H

δ

(s) = min  s

+

δ ; 1



and sign

+0

(s) =

( 1 if s > 0 0 if s ≤ 0.

If γ is a maximal monotone operator defined on R, we denote by γ

0

the main section of γ i.e.,

γ

0

(s) =

 

 

minimal absolute value of γ(s) if γ(s) 6= ∅

+∞ if [s, +∞) ∩ D(γ) = ∅

−∞ if (−∞, s] ∩ D(γ) = ∅.

Remark that as δ goes to 0, H

δ

(s) goes to sign

+0

(s).

We take ϕ = H

δ

(u

n

− M ) as a test function in (3.2) for the weak solution u

n

and M > 0 (a constant to be chosen later), to get

(3.19)

 

 

 

 

 

 

 

 

 

 

N

X

i=1

Z

a

i

 x, ∂u

n

∂x

i

 ∂

∂x

i

H

δ

(u

n

− M )dx + ε Z

|u

n

|

pM(x)−2

u

n

H

δ

(u

n

− M )dx

+ Z

T

n

b(u

n

)H

δ

(u

n

− M )dx + λ Z

∂Ω

T

n

(u

n

)H

δ

(u

n

− M )dσ

= Z

f

n

H

δ

(u

n

− M )dx + Z

∂Ω

g

n

H

δ

(u

n

− M )dσ.

(17)

We have

N

X

i=1

Z

a

i

 x, ∂u

n

∂x

i

 ∂

∂x

i

H

δ

(u

n

− M )dx

= 1 δ

N

X

i=1

Z



(un−M )+

δ <1

a

i

 x, ∂u

n

∂x

i

 ∂

∂x

i

(u

n

− M )

+

dx

= 1 δ

N

X

i=1

Z



0<un−M <δ

a

i

 x, ∂u

n

∂x

i

 ∂

∂x

i

u

n

dx ≥ 0 according to (1.7)

and Z

|u

n

|

pM(x)−2

u

n

H

δ

(u

n

− M )dx

= Z



(un−M )+

δ <1

|u

n

|

pM(x)−2

u

n

(u

n

− M )

+

δ dx +

Z



(un−M )+

δ ≥1

|u

n

|

pM(x)−2

u

n

dx

≥ 1 δ

Z



M <un<M +δ

|u

n

|

pM(x)−2

u

n

(u

n

− M )dx ≥ 0.

Then, (3.19) gives Z

T

n

b(u

n

)H

δ

(u

n

− M )dx + λ Z

∂Ω

T

n

(u

n

)H

δ

(u

n

− M )dσ

≤ Z

f

n

H

δ

(u

n

− M )dx + Z

∂Ω

g

n

H

δ

(u

n

− M )dσ, which is equivalent to

(3.20)

 

 

 

 

 

 

 

 

 

 

 

 

 Z



T

n

b(u

n

) − T

n

b(M ) 



H

δ

(u

n

− M )dx + λ

Z

∂Ω

T

n

(u

n

) − T

n

(M )H

δ

(u

n

− M )dσ

≤ Z



f

n

− T

n

b(M ) 



H

δ

(u

n

− M )dx +

Z

∂Ω

g

n

− λT

n

(M )H

δ

(u

n

− M )dσ.

We obtain the same inequality as in [14] (see (3.4) p. 210). Thus, we get

|b(u

n

)| ≤ ||f

n

||

a.e. in Ω and |u

n

| ≤ 1

λ ||g

n

||

a.e. in ∂Ω.

(18)

We use the Lebesgue dominated convergence theorem to get

(3.21) lim

n→+∞

Z

T

n

(b(u

n

))T

k

(u

n

− ϕ)dx = Z

b(u)T

k

(u − ϕ)dx

and

(3.22) lim

n→+∞

Z

∂Ω

T

n

(u

n

)T

k

(u

n

− ϕ)dσ = Z

∂Ω

uT

k

(u − ϕ)dσ.

For the third term of (3.14), we prove that lim inf

n→+∞

ε Z

|u

n

|

pM(x)−2

u

n

T

k

(u

n

− ϕ)dx ≥ 0 for ε → 0.

We have Z

|u

n

|

pM(x)−2

u

n

T

k

(u

n

− ϕ)dx = Z



|u

n

|

pM(x)−2

u

n

− |ϕ|

pM(x)−2

ϕ



T

k

(u

n

− ϕ)dx

+ Z

|ϕ|

pM(x)−2

ϕT

k

(u

n

− ϕ)dx.

Since the quantity |u

n

|

pM(x)−2

u

n

− |ϕ|

pM(x)−2

ϕT

k

(u

n

− ϕ) is nonnegative and for all x in Ω, the application ξ 7−→ |ξ|

pM(x)−2

ξ is continuous, then we get



|u

n

|

pM(x)−2

u

n

−|ϕ|

pM(x)−2

ϕ



T

k

(u

n

−ϕ) →



|u|

pM(x)−2

u−|ϕ|

pM(x)−2

ϕ



T

k

(u−ϕ) a.e. in Ω. It follows by Fatou’s lemma that

(3.23)

 

 

 

  lim inf

n→+∞

Z



|u

n

|

pM(x)−2

u

n

− |ϕ|

pM(x)−2

ϕ



T

k

(u

n

− ϕ)dx

≥ Z



|u|

pM(x)−2

u − |ϕ|

pM(x)−2

ϕ



T

k

(u − ϕ)dx.

We have Z

|ϕ|

pM(x)−2

ϕ dx =

Z

|ϕ|

pM(x)−1

dx ≤ Z



||ϕ||



pM(x)−1

dx

≤ Z

{||ϕ||≤1}



||ϕ||



pM(x)−1

dx + Z

{||ϕ||>1}



||ϕ||



pM(x)−1

dx

≤ meas(Ω) + ||ϕ||



p+M−1

meas(Ω) < +∞.

(19)

Hence, |ϕ|

pM(x)−2

ϕ ∈ L

1

(Ω). Since T

k

(u

n

− ϕ) converges weakly-* to T

k

(u − ϕ) in L

(Ω) and |ϕ|

pM(x)−2

ϕ ∈ L

1

(Ω), then we obtain

(3.24) lim

n→+∞

Z

|ϕ|

pM(x)−2

ϕT

k

(u

n

− ϕ)dx = Z

|ϕ|

pM(x)−2

ϕT

k

(u − ϕ)dx.

By adding (3.23) and (3.24), we get (3.25) lim inf

n→+∞

Z

|u

n

|

pM(x)−2

u

n

T

k

(u

n

− ϕ)dx ≥ Z

|u|

pM(x)−2

uT

k

(u − ϕ)dx.

Since

Z

|u|

pM(x)−2

uT

k

(u − ϕ)dx ≤ k Z

|u|

pM(x)−1

dx < +∞, thus we get

(3.26) lim inf

n→+∞

ε Z

|u

n

|

pM(x)−2

u

n

T

k

(u

n

− ϕ)dx ≥ 0 for ε → 0.

Thus, combining (3.15), (3.16), (3.21), (3.22) and (3.26) we have

(3.27)

 

 

 

 

N

X

i=1

Z

a

i

 x, ∂u

∂x

i

 ∂

∂x

i

T

k

(u − ϕ)dx + Z

b(u)T

k

(u − ϕ)dx

+ Z

∂Ω

λuT

k

(u − ϕ)dσ ≤ Z

f T

k

(u − ϕ)dx + Z

∂Ω

gT

k

(u − ϕ)dσ.

It means that, u is an entropy solution of problem (1.1).

Now we state the uniqueness result of entropy solution.

Theorem 3.2. Assume that (1.4)–(1.10) hold and let u be an entropy solution of (1.1). Then, u is unique.

Proof. The proof is done in two steps.

Step 1. A priori estimates.

Assume (1.4)–(1.10), f ∈ L

1

(Ω) and g ∈ L

1

(∂Ω).

Lemma 3.4. Let u be an entropy solution of (1.1). Then

(3.28)

N

X

i=1

Z

{|u|≤k}

∂u

∂x

i

pi(x)

dx ≤ k C

3

||f ||

L1(Ω)

+ ||g||

L1(∂Ω)



(20)

and there exists a positive constant C

9

such that

(3.29) ||b(u)||

1

≤ C

9

· meas(Ω) + ||f ||

L1(Ω)

+ ||g||

L1(∂Ω)

. Proof. Let us take ϕ = 0 in the entropy inequality (3.1).

• Since R

b(u)T

k

(u)dx + λ R

∂Ω

uT

k

(u)dσ ≥ 0, by (1.7), then we get (3.28).

• Also, using the fact that

N

X

i=1

Z

a

i

x, ∂u

∂x

i

 ∂

∂x

i

T

k

(u)dx + λ Z

∂Ω

uT

k

(u)dσ ≥ 0,

the relation (3.1) gives (3.30)

Z

b(u)T

k

(u)dx ≤ Z

f T

k

(u)dx + Z

∂Ω

gT

k

(u)dσ.

Then Z

{|u|≤k}

b(u)T

k

(u)dx + Z

{|u|>k}

b(u)T

k

(u)dx ≤ k||f ||

L1(Ω)

+ k||g||

L1(∂Ω)

, which imply that

Z

{|u|>k}

b(u)T

k

(u)dx ≤ k||f ||

L1(Ω)

+ k||g||

L1(∂Ω)

or Z

{u>k}

b(u)dx + Z

{u<−k}

−b(u)dx ≤ ||f ||

L1(Ω)

+ ||g||

L1(∂Ω)

. Therefore,

Z

{|u|>k}

|b(u)|dx ≤ ||f ||

L1(Ω)

+ ||g||

L1(∂Ω)

. So, we obtain

Z

|b(u)|dx = Z

{|u|≤k}

|b(u)|dx + Z

{|u|>k}

|b(u)|dx

≤ Z

{|u|≤k}

|b(u)|dx + ||f ||

L1(Ω)

+ ||g||

L1(∂Ω)

. Since the function b is nondecreasing, then

Z

{|u|≤k}

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

(21)

Consequently, there exists a constant C

9

= max{b(k); |b(−k)|} such that

||b(u)||

1

≤ C

9

· meas(Ω) + ||f ||

L1(Ω)

+ ||g||

L1(∂Ω)

.

Lemma 3.5. If u is an entropy solution of (1.1), then there exists a constant D which depends on f, g and Ω and such that

(3.31) meas{|u| > k} ≤ D

min(b(k), |b(−k)|) , ∀ k > 0 and a constant D

0

> 0 which depends on f, g and Ω and such that

(3.32) meas



∂u

∂x

i

> k



≤ D

0

k

1 (p M)0

, ∀ k ≥ 1.

Proof. • For any k > 0, the relation (3.29) gives Z

{|u|>k}

min(b(k), |b(−k)|)dx ≤ Z

{|u|>k}

|b(u)|dx ≤ C

9

·meas(Ω)+||f ||

L1(Ω)

+||g||

L1(∂Ω)

. Therefore,

min(b(k), |b(−k)|) · meas{|u| > k} ≤ C

9

· meas(Ω) + ||f ||

L1(Ω)

+ ||g||

L1(∂Ω)

= D;

that is

meas{|u| > k} ≤ D

min(b(k), |b(−k)|) .

• See [2] for the proof of (3.32).

Lemma 3.6. If u is an entropy solution of (1.1), then

(3.33) lim

h→+∞

Z

|f |χ

{|u|>h−t}

dx + lim

h→+∞

Z

∂Ω

|g|χ

{|u|>h−t}

dσ = 0, where h > 0 and t > 0.

Proof. Since the function b is surjective, according to (3.31), we have

h→+∞

lim meas{|u| > h − t} = 0.

As f ∈ L

1

(Ω) and g ∈ L

1

(∂Ω), it follows by the Lebesgue dominated convergence theorem that

h→+∞

lim Z

|f |χ

{|u|>h−t}

dx + lim

h→+∞

Z

∂Ω

|g|χ

{|u|>h−t}

dσ = 0.

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