Convergence of monotone operators with respect to measures · Mohamed Mamchaoui Basque Center for...
Transcript of Convergence of monotone operators with respect to measures · Mohamed Mamchaoui Basque Center for...
Convergence of monotone operators with respect tomeasures
Mohamed Mamchaoui
Department of Mathematics, Faculty of Sciences
University Abou Bakr Belkaıd Tlemcen, Algeria
October 11, 2014
Mohamed Mamchaoui Department of Mathematics, Faculty of SciencesUniversity Abou Bakr Belkaıd Tlemcen, Algeria m [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()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 [email protected] ()Basque Center for Applied Mathematics,Bilbao,Spain. 31/31October 11, 2014 31 / 31