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Convergence of monotone operators with respect tomeasures
Mohamed Mamchaoui
Department of Mathematics, Faculty of Sciences
University Abou Bakr Belkaıd Tlemcen, Algeria
m mamchaoui@mail.univ-tlemcen.dz
October 11, 2014
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 1/31October 11, 2014 1 / 31
Table of Contents
1 References
2 Homogenization of Monotone OperatorsSetting of the Homogenization ProblemThe Homogenized Problem
3 Convergence in the variable Lp
4 Our Goal
5 Monotonicity and convergenceFirst Case :Second Case :
6 Diagram
7 Conclusion
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 2/31October 11, 2014 2 / 31
References
References
1 G. A. Chechkin, V. V. Jikov, D. Lukassen and A. L. Piatniski, On homogenization of networks and junctions,Asymptotic Analysis, 30 (2002) 61-80.
2 A. Defranceschi. An Introduction to Homogenization and G-convergence, School on Homogenization, at the ICTP,Trieste. September 6-8, 1993.
3 V. V. Jikov, S. M. Kozlov and O. A. Oleinik, Homogenization of Differential Operators and Integral Functionals,Springer. Berlin, 1994.
4 V. V. Jikov(Zhikov): 1997. Weighted Sobolev spaces, Russian Acad. Sci. Sb. Math. 189(8) 1139-1170.
5 V. V. Zhikov: 2004. On two-scale convergence, Journal of Mathematical Sciences, Vol 120, No.3, 1328-1352.
6 D. Lukkassen and P. Wall, Two-scale convergence with respect to measures and homogenization of monotone operators,(Function Spaces and Applications), Volume 3, Number 2 (2005). 125-161.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 3/31October 11, 2014 3 / 31
Homogenization of Monotone Operators Setting of the Homogenization Problem
Setting of the Homogenization Problem
Let Ω be a bounded open set of Rn and Y = ]0, 1[n the unit cube.For a fixed ε > 0, let us consider the nonlinear monotone operator:
Aε(u) := −div(a(x
ε,Du)) , ∀u ∈ H1
0 (Ω)
where a(x, .) is Y -periodic and satisfies the following properties :
1 a : Rn × Rn −→ Rn such that : ∀ξ ∈ Rn, a(., ξ) is Lebesgue mesurable and Y -periodic.
2 (a(x, ξ1)− a(x, ξ2), ξ1 − ξ2) ≥ α |ξ1 − ξ2|2 a.e x ∈ Rn and ∀ξ1, ξ2 ∈ Rn.
3 |a(x, ξ1)− a(x, ξ2)| ≤ β |ξ1 − ξ2| a.e x ∈ Rn and ∀ξ1, ξ2 ∈ Rn.
4 a(x, 0) = 0 a.e x ∈ Rn.
If a ∈ N], then for all real positif ε and fε ∈ L2(Ω) , we consider the following problem :
−div(a( x
ε,Duε)) = fε in Ω
uε ∈ H10 (Ω)
(1)
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 4/31October 11, 2014 4 / 31
Homogenization of Monotone Operators The Homogenized Problem
The Homogenized Problem
Theorem :
Let a ∈ N] and let (εh)h be a sequence of positive real numbers converging to 0.
Assume that fε →ε→0
f strongly in H−1(Ω).
Let (uε)ε be the solution to (1). Then,
uε ε→0
u∗ in H10 (Ω)
a(x
ε,Duε)
ε→0b(Du∗) in L2(Ω; Rn)
where u∗ is the unique solution to the homogenized problem :
−div(b(Du∗)) = f in Ω
u∗ ∈ H10 (Ω)
(2)
The operator b is defined by :b : Rn → Rn
ξ → b(ξ) =
∫Y
a(y, ξ + Dwξ(y))dy (3)
where wξ is the unique solution to the local problem :
∫Y (a(y, ξ + Dwξ(y)),Dv(y))dy = 0 ∀v ∈ H1
#(Y )
wξ ∈ H1#(Y )
(4)
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 5/31October 11, 2014 5 / 31
Homogenization of Monotone Operators The Homogenized Problem
The Homogenized Problem
Proposition :
The homogenized operator b has the following properties:
1 b is strictly monotone
2 There exists a constant γ =β2
α> 0 such that:
|b(ξ1)− b(ξ2)| ≤ γ |ξ1 − ξ2| ∀ξ1, ξ2 ∈ Rn
3 b(0) = 0.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 6/31October 11, 2014 6 / 31
Convergence in the variable Lp
Let Ω be a bounded Lipschitz domain in RN . It will always assumed that p and q are conjugate indexes, i.e. 1/p + 1/q = 1and that 1 < p <∞. Let µδ and µ be Randon measures in Ω and µδ
δ→0µ. We say that a sequence uδ is bounded in
Lp(Ω, dµδ) if
lim supδ→0
∫Ω
|uδ|p dµδ <∞.
Definition
A bounded sequence uδ ∈ Lp(Ω, dµδ) is weakly convergent to u ∈ Lp(Ω, dµ), uδ u if
limδ→0
∫Ω
uδϕdµδ =
∫Ω
uϕdµ. (5)
for any test function ϕ ∈ C∞0 (Ω).
We have the following results:
each bounded sequence has a weakly convergent subsequence.
if the sequence uδ weakly converges to u, then
lim infδ→0
∫Ω
|uδ|p dµδ ≥
∫Ω
|u|p dµ.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 7/31October 11, 2014 7 / 31
Convergence in the variable Lp
Definition
A bounded sequence uδ ∈ Lp(Ω, dµδ) is called strongly convergent to u ∈ Lp(Ω, dµ), uδ u if
limδ→0
∫Ω
uδvδdµδ =
∫Ω
uvdµ, if vδ v in Lq(Ω, dµδ).
We have
strong convergence implies weak convergence.
the weak convergence uδ u, together with the relation
limδ→0
∫Ω
|uδ|p dµδ =
∫Ω
|u|p dµ,
is equivalent to the strong convergence uδ → u.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 8/31October 11, 2014 8 / 31
Our Goal
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 9/31October 11, 2014 9 / 31
Monotonicity and convergence
Let Ω be a bounded Lipschitz domain in RN , where the Lebesgue measure of the boundary is zero, Y = [0, 1)N the semi-open
cube in RN . It will always assumed that p and q are conjugate indexes, i.e. 1/p + 1/q = 1 and that 1 < p <∞. Let µδ andµ be Randon measures in Ω and µδ
δ→0µ.
Consider a family of sequence (uδ) such that
∫Ω
(|uδ|
p + |Duδ|p) dµδ ≤ c.
Let a : RN × RN → RN be a function such that a(., ξ) is Y−periodic and µδ−measurable for every ξ ∈ RN . Moreover,assume that there exists constants c1, c2 > 0 and two more constants α and β, with 0 ≤ α ≤ min 1, p − 1 andmax p, 2 ≤ β <∞ such that a satisfies the following continuity and monotonicity assumptions:
a(y, 0) = 0 (6)
|a(y, ξ1)− a(y, ξ2)| ≤ c1(1 + |ξ1| + |ξ2|)p−1−α |ξ1 − ξ2|
α (7)
(a(y, ξ1)− a(y, ξ2), ξ1 − ξ2) ≥ c2(1 + |ξ1| + |ξ2|)p−β |ξ1 − ξ2|
β, (8)
for µδ a.e. y ∈ RN and any ξ1, ξ2 ∈ RN .We will treat two cases:
a(x,Duδ) and aδ(x,Duδ).
We Suppose that
a(x, ϕ) ∈(C∞0 (Ω)
)N for all ϕ ∈(C∞0 (Ω)
)N.
As δ is small enough, we obtain the following results:
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 10/31October 11, 2014 10 / 31
Monotonicity and convergence First Case :
Lemma
There exist a subseqence of uδ still denoted by uδ and a ∈ (Lq(Ω, dµ))N such that a(x,Duδ) δ→0
a.
Proof.
|a(x,Duδ)| ≤ c1(1 + |Duδ|)p−1−α |Duδ|
α ≤ c1(1 + |Duδ|)p−1 ≤ 2p−1c1(1 + |Duδ|
p−1)
which gives that a(x,Duδ) is bounded in (Lq(Ω, dµδ))N since (Duδ) is bounded in (Lp(Ω, dµδ))N . By proposition 3 there
exist a subsequence (still denoted by δ) such that: there exists a a ∈ (Lq(Ω, dµ))N such that a(x,Duδ) δ→0
a i.e :
limδ→0
∫Ω
(a(x,Duδ), ϕ)dµδ =
∫Ω
(a, ϕ)dµ
for all ϕ ∈ (C∞0 (Ω))N .
Lemma
Assume that ∫Ω
(a − a(x, ϕ), z − ϕ)dµ ≥ 0, ∀ϕ ∈(C∞0 (Ω)
)Nfor some a and z belong to (Lq(Ω, dµ))N and (Lp(Ω, dµ))N respectivelly. Then, the inequality holds for every
ϕ ∈ (Lp(Ω, dµ))N .
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 11/31October 11, 2014 11 / 31
Monotonicity and convergence First Case :
Theorem 1:
Assume that a satisfies the conditions (6)-(7)-(8). Let (Duδ) be a bounded sequence in (Lp(Ω, dµδ))N which converges weakly
to z ∈ (Lp(Ω, dµ))N and assume that a(x,Duδ) converges weakly to a ∈ (Lq(Ω, dµ))N . Then
lim infδ→0
∫Ω
(a(x,Duδ),Duδ)dµδ ≥∫Ω
(a, z)dµ.
If equality holds, i.e :
lim infδ→0
∫Ω
(a(x,Duδ),Duδ)dµδ =
∫Ω
(a, z)dµ,
thena = a(x, z) for µ a.e x ∈ Ω.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 12/31October 11, 2014 12 / 31
Monotonicity and convergence First Case :
Assume the contradiction, i.e.
lim infδ→0
∫Ω
(a(x,Duδ),Duδ)dµδ <
∫Ω
(a, z)dµ.
Then there exists a positive constant C > 0 such that
lim infδ→0
∫Ω
(a(x,Duδ),Duδ)dµδ =
∫Ω
(a, z)dµ− C . (9)
By monotonicity we have that∫Ω
(a(x,Duδ)− a(x, ϕ),Duδ − ϕ)dµδ ≥ 0, ∀ϕ ∈(C∞0 (Ω)
)N.
Using (9) on the corresponding term and passing to the limit when δ → 0 in this inequality. Its left-handside contains fourterms. We have ∫
Ω
(a − a(x, ϕ), z − ϕ)dµ ≥ C , ∀ϕ ∈(C∞0 (Ω)
)N.
By density and continuity (see lemma 4) this inequality holds also for any ϕ ∈ (Lp(Ω, dµ))N . Let tφ = z − ϕ, with t > 0.Then after dividing by t we get ∫
Ω
(a − a(x, z − tφ), φ)dµ ≥C
t.
Letting t → 0 gives ∫Ω
(a − a(x, z), φ)dµ ≥ ∞.
Repeating the procedure for t < 0 implies that ∫Ω
(a − a(x, z), φ)dµ ≤ −∞.
We have clearly reached a contradiction.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 13/31October 11, 2014 13 / 31
Monotonicity and convergence First Case :
If equality holds we can repeat the procedure above :
∫Ω
(a(x,Duδ)− a(x, ϕ),Duδ − ϕ)dµδ ≥ 0,
for every ϕ ∈(C∞0 (Ω)
)N . Let us pass to the limit in this inequality. Its left-hand side contains four terms. We have :
limδ→0
∫Ω
(a(x,Duδ), ϕ)dµδ =
∫Ω
(a, ϕ)dµ.
limδ→0
∫Ω
(a(x, ϕ), ϕ)dµδ =
∫Ω
(a(x, ϕ), ϕ)dµ.
limδ→0
∫Ω
(a(x, ϕ),Duδ)dµδ =
∫Ω
(a(x, ϕ), z)dµ.
lim infδ→0
∫Ω
(a(x,Duδ),Duδ)dµδ =
∫Ω
(a, z)dµ.
As a result, we obtain the inequality :
∫Ω
(a − a(x, ϕ), z − ϕ)dµ ≥ 0, ∀ϕ ∈(C∞0 (Ω)
)N.
By density and continuity this inequality holds for any ϕ ∈ (Lp(Ω, dµ))N . Put ϕ = z − tφ, ∀φ ∈ (Lp(Ω, dµ))N .
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 14/31October 11, 2014 14 / 31
Monotonicity and convergence First Case :
For t > 0 we get that ∫Ω
(a − a(x, z − tφ), φ)dµ ≥ 0. (10)
For t < 0 we obtain that ∫Ω
(a − a(x, z − tφ), φ)dµ ≤ 0. (11)
By taking (10) and (11) into account and letting t tend to 0 we find that
∫Ω
(a − a(x, z), φ)dµ = 0, ∀φ ∈(Lp(Ω, dµ)
)N.
This implies that a = a(x, z) for µ a.e. x ∈ Ω. This ends the proof.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 15/31October 11, 2014 15 / 31
Monotonicity and convergence First Case :
Theorem 2:
Assume that a satisfies the conditions (6)-(7)-(8). Let (Duδ) be a bounded sequence in (Lp(Ω, dµδ))N which converges weakly
to z ∈ (Lp(Ω, dµ))N , a(x,Duδ) converges weakly to a ∈ (Lq(Ω, dµ))N , and |Duδ|p−2 Duδ converges weakly to
v ∈ (Lq(Ω, dµ))N . If
limδ→0
∫Ω
(a(x,Duδ),Duδ)dµδ =
∫Ω
(a, z)dµ, (12)
then the sequence Duδ is strongly convergent to z.
For k > 0, we have :
limδ→0
∫Ω
(a(x,Duδ),Duδ)dµδ = lim supδ→0
∫Ω
k |Duδ|p dµδ − lim sup
δ→0
∫Ω
k |Duδ|p dµδ
+ limδ→0
∫Ω
(a(x,Duδ),Duδ)dµδ
= lim supδ→0
∫Ω
k |Duδ|p dµδ + lim inf
δ→0
∫Ω
−k |Duδ|p dµδ
+ limδ→0
∫Ω
(a(x,Duδ),Duδ)dµδ.
This together with (12) give
lim supδ→0
∫Ω
k |Duδ|p dµδ + lim inf
δ→0
∫Ω
(a(x,Duδ)− kDuδ |Duδ|p−2
,Duδ)dµδ
=
∫Ω
(kv, z)dµ +
∫Ω
(a − kv, z)dµ.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 16/31October 11, 2014 16 / 31
Monotonicity and convergence First Case :
For k > 0 sufficiently small The function a(x,Duδ)− kDuδ |Duδ|p−2 satisfies the conditions in Theorem 1 therefore
lim infδ→0
∫Ω
(a(x,Duδ)− kDuδ |Duδ|p−2
,Duδ)dµδ ≥∫Ω
(a − kv, z)dµ. (13)
This and (12) imply
lim supδ→0
∫Ω
|Duδ|p dµδ ≤
∫Ω
(v, z)dµ. (14)
The function Duδ satisfies the conditions in Theorem 1 which implies that
lim infδ→0
∫Ω
|Duδ|p dµδ = lim inf
δ→0
∫Ω
(Duδ |Duδ|p−2
,Duδ)dµδ ≥∫Ω
(v, z)dµ. (15)
By (14) and (15) it follows that
limδ→0
∫Ω
|Duδ|p dµδ = lim
δ→0
∫Ω
(Duδ |Duδ|p−2
,Duδ)dµδ =
∫Ω
(v, z)dµ
Hence if follows by Theorem 1 that v = |z|p−2 z. Then, Duδ converges strongly to z.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 17/31October 11, 2014 17 / 31
Monotonicity and convergence First Case :
Theorem 3:
Assume that a satisfies the conditions (6)-(7)-(8). Let (Duδ) be a bounded sequence in (Lp(Ω, dµδ))N which converges
strongly to z ∈ (Lp(Ω, dµ))N and assume that a(x,Duδ) converges weakly to a ∈ (Lq(Ω, dµ))N . Then
a = a(x, z) for µ a.e x ∈ Ω.
Proof.
In the same spirit of the proof of Theorem 1 we show that
a = a(x, z) for µ a.e x ∈ Ω.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 18/31October 11, 2014 18 / 31
Monotonicity and convergence Second Case :
Lemma
There exists a subseqence and a0 ∈ (Lq(Ω, dµ))N such that aδ(x,Du) δ→0
a0.
Proof.
|aδ(x,Du)| ≤ c1(1 + |Du|)p−1−α |Du|α ≤ c1(1 + |Du|)p−1 ≤ 2p−1c1(1 + |Du|p−1)
which gives that aδ(x,Du) is bounded in (Lq(Ω, dµδ))N since (Du) is bounded in (Lp(Ω, dµδ))N . By proposition 3 there
exist a subsequence (still denoted by δ) such that: There exist a a0 ∈ (Lq(Ω, dµ))N such that aδ(x,Du) δ→0
a0 i.e :
limδ→0
∫Ω
(aδ(x,Du), ϕ)dµδ =
∫Ω
(a0, ϕ)dµ
for all ϕ ∈ (C∞0 (Ω))N .
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 19/31October 11, 2014 19 / 31
Monotonicity and convergence Second Case :
Theorem 4:
Assume that aδ satisfies the conditions (6)-(7)-(8) and converges weakly to a0 in sens (5). Then the limite a0 also satisfies theconditions (6)-(7)-(8).
For all ϕ ∈(C∞0 (Ω)
)N :
0 = limδ→0
∫Ω
(aδ(x, 0), ϕ)dµδ =
∫Ω
(a0(x, 0), ϕ)dµ.
For ϕ ∈ C∞0 (Ω) and for all ξ1, ξ2 ∈ Rn, we have:
∣∣∣∣∣∣∣∫Ω
(aδ(x, ξ1)− aδ(x, ξ2), ϕ(ξ1 − ξ2))dµδ
∣∣∣∣∣∣∣ ≤∫Ω
|(aδ(x, ξ1)− aδ(x, ξ2)| |ϕ(ξ1 − ξ2)| dµδ
≤ c1
∫Ω
(1 + |ξ1| + |ξ2|)p−1−α |ξ1 − ξ2|
α |ϕ(ξ1 − ξ2)| dµδ
Passing here to the limit we obtain the inequality
∣∣∣∣∣∣∣∫Ω
(a0(x, ξ1)− a0(x, ξ2), ϕ(ξ1 − ξ2))dµ
∣∣∣∣∣∣∣ ≤ c1
∫Ω
(1 + |ξ1| + |ξ2|)p−1−α |ξ1 − ξ2|
α |ϕ(ξ1 − ξ2)| dµ
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 20/31October 11, 2014 20 / 31
Monotonicity and convergence Second Case :
which shows that|(a0(x, ξ1)− a0(x, ξ2)| ≤ c1(1 + |ξ1| + |ξ2|)
p−1−α |ξ1 − ξ2|α,
for µ a.e. x ∈ Ω and any ξ1, ξ2 ∈ RN .
Finally, we have for all ϕ ∈ C∞0 (Ω), ϕ ≥ 0 :
∫Ω
(aδ(x, ξ1)− aδ(x, ξ2), ϕ(ξ1 − ξ2))dµδ =
∫Ω
(aδ(x, ξ1)− aδ(x, ξ2), (ξ1 − ξ2))ϕdµδ
≥ c2
∫Ω
(1 + |ξ1| + |ξ2|)p−β |ξ1 − ξ2|
βϕdµδ
Passing here to the limit we obtain inequality
∫Ω
(a0(x, ξ1)− a0(x, ξ2), (ξ1 − ξ2))ϕdµ ≥ c2
∫Ω
(1 + |ξ1| + |ξ2|)p−β |ξ1 − ξ2|
βϕdµ
for every ϕ ∈ C∞0 (Ω), ϕ ≥ 0, which complete the proof.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 21/31October 11, 2014 21 / 31
Monotonicity and convergence Second Case :
Lemma
There exists a subsequence and a1 ∈ (Lq(Ω, dµ))N such that aδ(x,Duδ) δ→0
a1.
Proof.
|aδ(x,Duδ)| ≤ c1(1 + |Duδ|)p−1−α |Duδ|
α ≤ c1(1 + |Duδ|)p−1 ≤ 2p−1c1(1 + |Duδ|
p−1)
which gives that aδ(x,Duδ) is bounded in (Lq(Ω, dµδ))N since (Duδ) is bounded in (Lp(Ω, dµδ))N . By proposition 3 there
exists a subsequence (still denoted by δ) such that: There exist a a1 ∈ (Lq(Ω, dµ))N such that aδ(x,Duδ) δ→0
a1 i.e :
limδ→0
∫Ω
(aδ(x,Duδ), ϕ)dµδ =
∫Ω
(a1, ϕ)dµ
for all ϕ ∈ (C∞0 (Ω))N .
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 22/31October 11, 2014 22 / 31
Monotonicity and convergence Second Case :
Theorem 5:
Assume that aδ satisfies the conditions (6)-(7)-(8). Let (Duδ) be a bounded sequence in (Lp(Ω, dµδ))N which converges
strongly to z ∈ (Lp(Ω, dµ))N and assume that aδ(x,Duδ) converges weakly to a1 ∈ (Lq(Ω, dµ))N . Then,
a1 = a0(x, z)
where a0 is the limit of aδ in sens (5).
By the monotonicity of aδ we have ∫Ω
(aδ(x,Duδ)− aδ(x, ϕ),Duδ − ϕ)dµδ ≥ 0,
for every ϕ ∈(C∞0 (Ω)
)N . Let us pass to the limit in this inequality. Its left-hand side contains four terms. We have :
limδ→0
∫Ω
(aδ(x,Duδ), ϕ)dµδ =
∫Ω
(a1, ϕ)dµ,
limδ→0
∫Ω
(aδ(x, ϕ), ϕ)dµδ =
∫Ω
(a0(x, ϕ), ϕ)dµ,
limδ→0
∫Ω
(aδ(x, ϕ),Duδ)dµδ =
∫Ω
(a0(x, ϕ), z)dµ,
and
limδ→0
∫Ω
(aδ(x,Duδ),Duδ)dµδ =
∫Ω
(a1, z)dµ.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 23/31October 11, 2014 23 / 31
Monotonicity and convergence Second Case :
As a result, we obtain the inequality∫Ω
(a1 − a0(x, ϕ), z − ϕ)dµ ≥ 0, ∀ϕ ∈(C∞0 (Ω)
)N.
By density and continuity this inequality holds for any ϕ ∈ (Lp(Ω, dµ))N . Put ϕ = z − tφ, ∀φ ∈ (Lp(Ω, dµ))N . For t > 0we get that ∫
Ω
(a1 − a0(x, z − tφ), φ)dµ ≥ 0. (16)
For t < 0 we obtain that ∫Ω
(a1 − a0(x, z − tφ), φ)dµ ≤ 0 (17)
By taking (16) and (17) into account and letting t tend to 0 we find that
∫Ω
(a1 − a0(x, z), φ)dµ = 0, ∀φ ∈(Lp(Ω, dµ)
)N.
Which implies that a1 = a0(x, z) for µ a.e. x ∈ Ω. The proof is complete.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 24/31October 11, 2014 24 / 31
Monotonicity and convergence Second Case :
ExampleN = p = 2. Let
I = x = (x1, x2) / a ≤ x1 ≤ b ; x2 = 0
be a segment in R2, and suppose that a bounded domain Ω contains I . For any sufficiently small δ > 0 consider the bar
Iδ := x = (x1, x2) / a < x1 < b ; − δ < x2 < δ ⊂ Ω.
Denote by µδ the measure in Ω, concentrated and uniformly distributed on Iδ :
µδ (dx) =χIδ
(x)
2δ(b − a)dx1dx2.
The family µδ converges weakly, as δ → 0, to a measure
µ (dx) =1
(b − a)dx1 × δ(x2)
where δ(z) stands for the Dirac mass at zero.Consider the operator aδ with the following form :
aδ(x,Duδ) =
(1 + δ|x|2 + δ|x|
)Duδ.
We can prove that aδ satisfies suitable assumptions of uniform strict monotonicity and uniform Lipschitz- continuity and
aδ(x, 0) = 0 for a.e. x ∈ R2.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 25/31October 11, 2014 25 / 31
Monotonicity and convergence Second Case :
ExampleSuppose that Duδ is strongly convergent to z.
limδ→0
∫Ω
(aδ(x, ϕ), ϕ)dµδ = limδ→0
∫Ω
((1 + δ|x|2 + δ|x|
)ϕ, ϕ)dµδ
=
∫Ω
(1
2ϕ, ϕ)dµ, ∀ϕ ∈
(C∞0 (Ω)
)2.
This implies that
a0(x, ξ) =1
2ξ
and
|a0(x, ξ1)− a0(x, ξ2)| =1
2|ξ1 − ξ2|,
for a.e. x ∈ R2 and for every ξ1, ξ2 ∈ R2.Finally, the hypothesis of Theorem 5 are satisfied then
limδ→0
∫Ω
(aδ(x,Duδ), ϕ)dµδ =
∫Ω
(a1, ϕ)dµ
=
∫Ω
(a0(x, z), ϕ)dµ
=
∫Ω
(1
2z, ϕ)dµ, ∀ϕ ∈
(C∞0 (Ω)
)2.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 26/31October 11, 2014 26 / 31
Diagram
Theorem 6:
Assume that a satisfies the conditions (6)-(7)-(8). Let (Duε,δ) be a bounded sequence in(Lp(Ω, dµε,δ)
)N which converges
weakly to zε ∈ (Lp(Ω, dµε))N and assume that a(x
ε,Duε,δ) converges weakly to aε ∈ (Lq(Ω, dµε))N . If
lim infδ→0
∫Ω
(a(x
ε,Duε,δ),Duε,δ)dµε,δ =
∫Ω
(aε, zε)dµε,
then the diagram
a(x
ε,Duε,δ)
δ → 0a(
x
ε,Duε)
ε↓0
ε↓0
b(Du0,δ) δ → 0
b(Du0)
is commutative.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 27/31October 11, 2014 27 / 31
Diagram
Proof.
From Theorem 1 it follows that
aε = a(x
ε, zε) for µ a.e x ∈ Ω.
We have for all ϕ ∈(C∞0 (Ω)
)N
limδ→0
limε→0
∫Ω
(a(x
ε,Duε,δ), ϕ)dµε,δ
= limδ→0
∫Ω
(b(Du0,δ), ϕ)dx.
Since zε = Duε, we obtain :
limε→0
limδ→0
∫Ω
(a(x
ε,Duε,δ), ϕ)dµε,δ
= limε→0
∫Ω
(aε, ϕ)dµε.
= limε→0
∫Ω
(a(x
ε,Duε), ϕ)dµε.
=
∫Ω
(b(Du0), ϕ)dx.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 28/31October 11, 2014 28 / 31
Diagram
Theorem 7:
Assume that aδ satisfies the conditions (6)-(7)-(8). Let (Duε,δ) be a bounded sequence in(Lp(Ω, dµε,δ)
)N which converges
strongly to zε ∈ (Lp(Ω, dµε))N and assume that aδ(x
ε,Duε,δ) converges weakly to aε1 ∈ (Lq(Ω, dµε))N . Then the diagram
aδ(x
ε,Duε,δ)
δ → 0a0(
x
ε,Duε))
ε↓0
ε↓0
b(Du0,δ) δ → 0
b(Du0)
is commutative.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 29/31October 11, 2014 29 / 31
Diagram
Proof.
From Theorem 5 it follows that
aε1 = a0(x
ε, zε) for µ a.e x ∈ Ω.
We have for all ϕ ∈(C∞0 (Ω)
)N
limδ→0
limε→0
∫Ω
(aδ(x
ε,Duε,δ), ϕ)dµε,δ
= limδ→0
∫Ω
(b(Du0,δ), ϕ)dx.
Since zε = Duε, we obtain :
limε→0
limδ→0
∫Ω
(aδ(x
ε,Duε,δ), ϕ)dµε,δ
= limε→0
∫Ω
(aε1 , ϕ)dµε.
= limε→0
∫Ω
(a0(x
ε,Duε), ϕ)dµε.
=
∫Ω
(b(Du0), ϕ)dx.
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 30/31October 11, 2014 30 / 31
Conclusion
Conclusion
a(x
ε,Duε,δ)
δ→0a(
x
ε, zε)
ε↓0
ε↓0
b(Du∗,δ) δ→0
b(z∗)
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz ()Basque Center for Applied Mathematics,Bilbao,Spain. 31/31October 11, 2014 31 / 31
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