Convergence of monotone operators with respect to measures · Mohamed Mamchaoui Basque Center for Applied Mathematics,Bilbao,Spain. 10/31Department of Mathematics, Faculty of SciencesUniversity

Convergence of monotone operators with respect tomeasures

Mohamed Mamchaoui

Department of Mathematics, Faculty of Sciences

University Abou Bakr Belkaıd Tlemcen, Algeria

m [email protected]

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


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.


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)


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)


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.


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.


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.


Our Goal


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:


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 .



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.



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


∫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.



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


∫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 .



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.



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


∫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.



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.



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.


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




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



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.



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




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


∫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.



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.



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.



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


∫Ω

(a1, ϕ)dµ

=

∫Ω

(a0(x, z), ϕ)dµ

=

∫Ω

(1

2z, ϕ)dµ, ∀ϕ ∈

(C∞0 (Ω)

)2.


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.


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


= limε→0

∫Ω

(aε, ϕ)dµε.

= limε→0

∫Ω

(a(x

ε,Duε), ϕ)dµε.

=

∫Ω

(b(Du0), ϕ)dx.


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.


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


= limδ→0

∫Ω

(b(Du0,δ), ϕ)dx.

Since zε = Duε, we obtain :

limε→0

limδ→0

∫Ω

(aδ(x


= limε→0

∫Ω

(aε1 , ϕ)dµε.

= limε→0

∫Ω

(a0(x

ε,Duε), ϕ)dµε.

=

∫Ω

(b(Du0), ϕ)dx.


Conclusion

Conclusion

a(x

ε,Duε,δ)

δ→0a(

x

ε, zε)

ε↓0

ε↓0

b(Du∗,δ) δ→0

b(z∗)


Documents

Convergence of monotone operators with respect to measures · Mohamed Mamchaoui Basque Center for Applied Mathematics,Bilbao,Spain. 10/31Department of Mathematics, Faculty of SciencesUniversity