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Sharp upper bounds for a singular perturbation problem related to micromagnetics

ARKADYPOLIAKOVSKY

Abstract. We construct an upper bound for the following family of functionals {Eε}ε>0, which arises in the study of micromagnetics:

Eε(u)=

ε|∇u|2+1 ε

R2|Hu|2.

Hereis a bounded domain inR2,u H1(,S1)(corresponding to the mag- netization) andHu, the demagnetizing field created byu, is given by

div(u˜+Hu)=0 inR2, curlHu=0 inR2,

whereu˜is the extension ofuby 0 inR2\. Our upper bound coincides with the lower bound obtained by Rivi`ere and Serfaty.

Mathematics Subject Classification (2000): 49J45 (primary); 35B25, 35J20 (secondary).

1.

Introduction

In this paper we study the following energy-functional, related to micromagnetics:

Eε(u):=

ε|∇u|2+1 ε

R2|Hu|2. (1.1) Here is a bounded domain in

R

2 with Lipschitz boundary, u is a unit-valued vector-field (corresponding to the magnetization) in H1(,S1) and Hu, the de- magnetizing field created byu, is given by

div(u˜+Hu)=0 in

R

2

curlHu =0 in

R

2, (1.2)

Received December 5, 2006; accepted in revised form November 22, 2007.

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whereu˜is the extension ofuby 0 in

R

2\. For the physical models related toEε, we refer to [18] and all the references therein.

We can rewrite (1.1) in the following form. Denoting by1u˜ the Newtonian potential ofu, we observe that the vector-field˜ H¯u := −∇

div(1u˜)

belongs to L2(

R

2,

R

2). Moreover,

divH¯u= −divu˜ in

R

2 curlH¯u =0 in

R

2. SoHu = ¯Huand we obtain

Eε(u)=

ε|∇u|2+ 1 ε

R2

div(1u˜)2. (1.3) In [19] T. Rivi`ere and S. Serfaty proved the following theorem, giving compactness and a lower bound for the energies Eε.

Theorem 1.1. Letbe a bounded simply connected domain in

R

2. Letεn 0 and unH1(,S1)with a liftingϕnH1(,

R

)i.e., un = eiϕn a.e., and such that

Eεn(un)C (1.4)

ϕnL()N. (1.5)

Then, up to extraction of a subsequence, there exists u andϕinq<∞Lq()such that

ϕnϕinq<∞Lq() unu inq<∞Lq() . Moreover, if we consider

Ttϕ(x):=inf ϕ(x),t Ttu(x):=ei Ttϕ(x),

thendivxTtu is a bounded Radon measure on×

R

, with t divxTtu continuous from

R

to

D

(). In addition

2

R

divxTtud x dt≤ lim

n→∞

2|∇ϕn·Hun| ≤ lim

n→∞Eεn(un) <. The main contribution of this paper is to establish the upper bound for Eε in the case whereuand its liftingϕbelong toB V. First of all we observe that ifεn →0,

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Eεn(un)C, and unu in Lq, where |un| = 1 and uB V then clearly

nlim→∞

R2|Hun|2=0, which implies

nlim→∞

R2u˜n· ∇δ= − lim

n→∞

R2Hun· ∇δ=0 ∀δ∈Cc(

R

2,

R

) . Therefore, divu˜ =0 as a distribution,i.e.,





|u| =1 a.e. in , divu=0 in , u·n=0 on∂ .

(1.6)

The main result of this paper is the following theorem.

Theorem 1.2. Let be a bounded domain in

R

2 with Lipschitz boundary. Con- sider uB V(,S1), satisfyingdivu =0inand u·n=0on∂and assume there existϕB V(,

R

), such that u= eiϕ a e. in. Then there exists a family of functions{vε} ⊂C2(

R

N,

R

)satisfying

ε→lim0+vε(x)=ϕ(x) in L1(,

R

) and

ε→lim0Eε(eivε)=2

R

divxTtud x dt. Moreover, ifϕB V(,

R

)L, then we have

ε→lim0+vε(x)=ϕ(x) in Lp(,

R

) p∈ [1,∞) .

In order to construct{vε}we take the convolution ofϕ with a varying smoothing kernel, i.e., we setvε(x) := ε2

R2ηyx

ε ,x

ϕ(y)d y where ηCc2(

R

2×

R

2) satisfies

R2η(z,x)d x = 1 for every x, and we optimize the choice of the kernel η. A similar approach was used in [16] and [17], but a new ingredient is required here, since the non-local term

R2|Hu|2gives certain difficulties.

1.1. The basic idea

We shall follow essentially the strategy of [16] and [17]. The main new ingredient here is the calculation of

ε→lim0+

1 ε

R2

div(1eivε})2. (1.7) We first calculate

L(l):= lim

ε→0+

1 ε

R2

div(1ε})2,

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where ϕε(x):=ε2

R2lyx

ε ,x

ϕ(y)d y, with lCc2(

R

2 ×,

R

2)satisfying

R2l(z,x)d x =0 for every x. Since∇

div(1ε})

has the same asymp- totic behavior asε2

R2syx

ε ,x

ϕ(y)d y, wheres(z,x):=∇z

divz(z1l(z,x)) , we can calculate the limit L(l)in a similar way to what was done in [16] and [17]

(see Lemmas 3.1 and 3.2 for the details). Using the results of [17] (see Proposition 2.2 below) it is easy to calculate

D(l):= lim

ε→0

1 ε

eivεϕεu2d x.

Now, given a fixed η and a small δ > 0, we choosel = lδ in such a way that D(lδ) < δ. Then, using the estimate

R2

div(1f})2C

|f|2, we deduce that

ε→lim0

1 ε

R2

div(1eivε})2−1 ε

R2

div(1ϕε})2

≤ lim

ε→0C 1

ε

R2

div(1(eivεϕεu)})21/2

1/2. Finally, tettingδ tend to 0, we conclude that the limit in (1.7) should be equal to limδ→0L(lδ). We follow basically this strategy in the proof of Proposition 4.1.

ACKNOWLEDGEMENTS. I am grateful to Prof. Camillo De Lellis for proposing this problem to me and for some useful suggestions. Part of this research was done dur- ing a visit at the Laboratoire J. L. Lions of the University Paris VI in the framework of the RTN-Programme Fronts-Singularities. I am indebted to Prof. Haim Brezis for the invitation and for many stimulating discussions.

2.

Preliminaries

Throughout this section we assume thatis a bounded domain in

R

2with Lipschitz boundary. We begin by introducing some notation. For every νS1 (the unit sphere in

R

2) andR>0 we denote

B+R(x,ν)= {y

R

2 : |yx|< R, (yx)·ν >0}, (2.1) BR(x,ν)= {y

R

2: |yx|< R, (yx)·ν <0}, (2.2) H+(x,ν)= {y

R

2:(yx)·ν >0}, (2.3) H(x,ν)= {y

R

2:(yx)·ν <0} (2.4)

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and

Hν0= {y

R

2 : y·ν=0}. (2.5)

Definition 2.1. Consider a function fB V(,

R

m)and a pointx.

i) We say thatx is a point ofapproximate continuityof f if there exists z

R

m such that

ρ→lim0+

Bρ(x)|f(y)z|d y

L

2Bρ(x) =0.

In this casezis called anapproximate limitof f atxand we denotezby f˜(x). The set of points of approximate continuity of f is denoted byGf.

ii) We say thatx is anapproximate jump pointof f if there exista,b

R

m and νSN1such thata=band

ρ→lim0+

Bρ+(x,ν)|f(y)a|d y

L

2Bρ(x) =0, lim

ρ→0+

Bρ(x,ν)|f(y)b|d y

L

2Bρ(x) =0. (2.6) The triple (a,b,ν), uniquely determined by (2.6) up to a permutation of (a,b) and a change of sign of ν, is denoted by(f+(x), f(x),νf(x)). We shall call νf(x)the approximate jump vectorand we shall sometimes write simplyν(x)if the reference to the function f is clear. The set of approximate jump points is denoted by Jf. A choice of ν(x)for everyxJf (which is unique up to sign) determines an orientation of Jf. At a point of approximate continuity x, we shall use the convention f+(x)= f(x)= ˜f(x).

We refer to [2] for the results on BV-functions that we shall use in the sequel.

Consider a function = 1, ϕ2, . . . , ϕd)B V(,

R

d). By [2, Proposi- tion 3.21] we may extend to a function¯ ∈ B V(

R

2,

R

d), such that¯ = a.e. in , supp¯ is compact and D(∂)¯ = 0. From the proof of Proposi- tion 3.21 in [2] it follows that ifB V(,

R

d)∩Lthen its extension¯ is also in B V(

R

2,

R

d)∩L. Consider also a matrix valued functionCc2(

R

2×

R

2,

R

l×d) For everyε >0 define a functionε(x):

R

2

R

l by

ε(x):= 1 ε2

R2yx ε ,x

· ¯(y)d y

=

R2(z,x)· ¯(x+εz)d z,x

R

2.

(2.7)

Thanks to [17, Proposition 3.2], we have the following result. It generalizes Propo- sition 3.2 from [16] and provides the key tool for the calculation of the upper bound, both in [17] and in the current paper. In the proof of Lemma 3.2 we shall also follow the general strategy of its proof in [17].

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Proposition 2.2. Let WC1(

R

l×

R

q,

R

)satisfying

aW(a,b)=0 whenever W(a,b)=0. (2.8) ConsiderB V(,

R

d)Land u B V(,

R

q)Lsatisfying

W R2(z,x)d z

·(x), u(x)

=0 for a.e. x , whereCc2(

R

2×

R

2,

R

l×d), as above. Letε be as in(2.7). Then,

ε→lim0

1 εW

ε(x),u(x) d x

=

J

0

−∞W

(t,x),u+(x) dt +

+∞

0

W

(t,x),u(x) dt

d

H

1(x) , (2.9)

where

(t,x)= t

−∞P(s,x)ds

·(x)+ +∞

t

P(s,x)ds

·+(x), (2.10) with

P(t,x)=

Hν(x)0

(tν(x)+y,x)d

H

1(y) , (2.11) ν(x)is the jump vector ofand it is assumed that the orientation of Ju coincides with the orientation of J

H

1a.e. on JuJ.

Definition 2.3. Given fL(

R

2,

R

k)with compact support, we define its New- tonian potential

1f (x):=

R2

1

2π ln|xy| f(y)d y. Than it is well known that

R22 1f

(x)2d x =

R2|f(x)|2d x, (2.12) where givenv=(v1, . . . , vk):

R

2

R

k we denote by∇2v

R

k×2×2the tensor withli j-th component∂i j2vl.

Definition 2.4. Let

V

be the class of all functionsηCc2(

R

2×

R

2,

R

)such that

R2η(z,x)d z=1 ∀x . (2.13) Let

U

be the class of all functionsl(z,x)Cc2(

R

2×,

R

2)such that

R2l(z,x)d z=0 ∀x

R

2. (2.14)

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In [17, Lemma 5.1], we proved the following statement. This statement generalize Claim 3 of Lemma 3.4 from [16] and was an essential tool in the optimizing the upper bound in [17].

Lemma 2.5. Letµ be positive finite Borel measure on andν0(x) :

R

2 a Borel measurable function with0| = 1. Let

W

1 denote the set of functions

p(t,x):

R

×

R

satisfying the following conditions:

i) p is Borel measurable and bounded,

ii) there exists M >0such that p(t,x)=0for|t|> M and any x, iii)

R p(t,x)dt =1,x.

Then for every p(t,x)

W

1, there exists a sequence of functionsn} ⊂

V

(see Definition2.4), such that the sequence of functions{pn(t,x)}defined on

R

×by

pn(t,x)=

Hν0 0(x)

ηn(tν0(x)+y,x)d

H

1(y),

has the following properties:

i) there exists C0such thatpnLC0for every n,

ii) there exist M >0such that pn(t,x)=0for|t|>M and every x, for all n.

iii) limn→∞

R|pn(t,x)p(t,x)|dt dµ(x)=0.

With the same method it is not difficult to prove

Lemma 2.6. Letµ be positive finite Borel measure on andν0(x) :

R

2 a Borel measurable function with0| = 1. Let

W

0 denote the set of functions q(t,x):

R

×

R

2satisfying the following conditions:

i) q is Borel measurable and bounded,

ii) there exists M >0such that q(t,x)=0for|t|> M and every x. iii)

Rq(t,x)dt =0,x.

Then for every q(t,x)

W

0, there exists a sequence of functions{ln} ⊂

U

(see Definition2.4), such that the sequence of functions{qn(t,x)}defined on

R

×by

qn(t,x)=

Hν0 0(x)

ln(tν0(x)+y,x)d

H

1(y), has the following properties:

i) there exists C0such thatqnLC0for every n,

ii) there exist M >0such that qn(t,x)=0for|t|> M and every x, for all n,

iii) limn→∞

R|qn(t,x)q(t,x)|dt dµ(x)=0.

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3.

First estimates

Throughout this section we assume thatis a bounded domain in

R

2with Lipschitz boundary.

Letl

U

(see Definition 2.4). Considerr(z,x) := z1l(z,x). Then rC2(

R

2 ×

R

2,

R

2) with suppr

R

2 × K, where K

. Moreover, since

R2l(z,x)d z=0, for everyk=0,1,2. . .we have the estimates

|∇kxr(z,x)| ≤ Ck

|z| +1,xk

zr(z,x)Ck

|z|2+1,kx

2zr(z,x)Ck

|z|3+1,

(3.1)

whereCk >0 does not depend onzandx.

Lemma 3.1. LetϕB V(,

R

)Land l

U

(see Definition2.4). For every ε >0consider the functionϕεC1(

R

2,

R

2)by

ϕε(x):= 1 ε2

R2l yx

ε ,x

ϕ(¯ y)d y =

R2l(z,x)ϕ(¯ x+εz)d z, (3.2) where ϕ¯ is some bounded B V extension ofϕ to

R

2 with compact support. Next consider r(z,x):=z1l(z,x)and set

ξε(x):= 1 ε2

R21

div1ryx ε ,x

ϕ(¯ y)d y

=

R2z

divzr(z,x)

¯

ϕ(x+εz)d z,

(3.3)

where1(div1r)is the gradient of divergence of r(z,x)in its first variable, namely z. Then,

R2

1 ε

div(1ϕε)

(x)2d x =oε(1)+

1

εϕε(x)·ξε(x)d x. (3.4) Proof. Sincel(z,x)=0 ifx/ K, whereK is some compact subset of, we have, in particular,ϕε(x)=0 for everyx

R

2\. Then, integrating by part two times, we conclude

R2

1 ε

div(1ϕε) (x)2d x

= −

R2

1 ε

div(1ϕε) (x)·

div(1ϕε) (x)d x

= −

R2

1

εdivϕε(x)·

div(1ϕε) (x)d x

=

1

εϕε(x)· ∇

div(1ϕε) (x)d x.

(3.5)

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Next consider the functionζεC1(

R

2,

R

2)given by ζε(x):= 1

ε2

R2r yx

ε ,x

ϕ(¯ y)d y=

R2r(z,x)ϕ(¯ x+εz)d z. (3.6) We will prove now that

ε22ζε(x)

R2z2r(z,x)ϕ(¯ x+εz)d z

2/3x . (3.7) We shall denote by∇1l and∇2lthe gradient ofl(z,x)with respect to the variables zandx respectively. We have,

ε22ζε(x)

R2z2r(z,x)ϕ(¯ x+εz)d z

=

R2x2r yx

ε ,x

ϕ(¯ y)d y− 1 ε2

R212r yx

ε ,x

ϕ(¯ y)d y

= −1 ε

R2

12r yx

ε ,x

+ ∇21r yx

ε ,x

ϕ(¯ y)d y

+

R222r yx

ε ,x

ϕ(¯ y)d y.

(3.8)

Therefore, by the H¨older inequality and the estimates in (3.1), we obtain ε22ζε(x)

R2z2r(z,x)ϕ(¯ x+εz)d z

≤ε2/3 1

ε2

R2

12r yx

ε ,x

+∇21r yx

ε ,x 6/d y5

5/6

R2| ¯ϕ(y)|6d y 1/6

+ε2/3 1

ε2

R2

22r yx

ε ,x3d y 1/3

R2| ¯ϕ(y)|3/2d y 2/3

=ε2/3

R2

12r z,x

+ ∇21r

z,x6/5d z 5/6

R2| ¯ϕ(y)|6d y 1/6

+ε2/3

R2

22r

z,x3d z 1/3

R2| ¯ϕ(y)|3/2d y 2/3

2/3 which gives (3.7). In particular,

ε2ζε(x)ϕε(x) =

ε2ζε(x)

R2zr(z,x)ϕ(¯ x+εz)d z

C0ε2/3.

(3.9)

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Next by (3.5),

R2

1 ε

div(1ϕε)

(x)2d x=

1

εϕε(x)· ∇

div(1ϕε) (x)d x

=

1

ε ϕε(x)· ∇ div

ε2ζε(x) d x

1

ε ϕε(x)· ∇ div

1

ε2ζε−ϕε (x)d x.

(3.10) The last integral can be estimated by

1

εϕε(x)· ∇ div

1

ε2ζεϕε (x)d x

1

εε(x)|2 1/2

1 ε

R2

∇ div

1

ε2ζεϕε (x)2d x

1/2

1

εε(x)|2 1/2

2 ε

R2

2 1

ε2ζεϕε (x)2d x

1/2

=

1

εε(x)|2 1/2

2

εε2ζε(x)ϕε(x)2d x 1/2

. Then, since

1

εε(x)|2C

1 εε(x)|

≤ ¯C

1 ε

BR(0)l(z,x)

¯

ϕ(x+εz)ϕ(x) d z

d x

≤ ¯C

BR(0)

1

ε|l(z,x)|

ϕ(¯ x+εz)ϕ(x)d x d z

≤ ¯CDϕ(¯

R

2)

BR(0)|l(z,x)| · |z|d z= O(1) ,

(3.11)

using (3.9), from (3.10) we infer

R2

1 ε

div(1ϕε)

(x)2d x=oε(1)+

1

εϕε(x)·∇

div

ε2ζε(x)

d x. (3.12) Next we remind thatξε is defined by (3.3). By (3.7), we have,

∇ div

ε2ζε(x)

ξε(x)≤ ¯2/3x . (3.13)

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Then as before,

1 εϕε(x)·

∇ div

ε2ζε(x)

ξε(x)

d x

1

εε(x)|2 1/2

1 ε

div

ε2ζε(x)

ξε(x)2d x 1/2

1/6.

Therefore, from (3.12) we infer (3.4).

Lemma 3.2. LetϕB V(,

R

)Land l

U

(see Definition2.4). For every ε >0and every x

R

2consider the functionϕεC1(

R

2,

R

2)as in(3.2). Then,

ε→lim0

R2

1 ε

div(1ϕε) (x)2d x

=

Jϕ

+∞

−∞+(x)−ϕ(x)|2· ν(x)·

+∞

t

q(s,x)ds 2dt

d

H

1(x) , (3.14)

where

q(t,x)=

Hν(x)0

l(tν(x)+y,x)d

H

1(y) , (3.15) andν(x)is the jump vector ofϕ.

Proof. By Lemma 3.1 we have

R2

1 ε

div(1ϕε)

(x)2d x =oε(1)+

1

εϕε(x)·ξε(x)d x. (3.16) From this point the strategy of the proof is similar to that in [16] and [17] (see also Proposition 2.2). The only difference is that here ξε is defined by a convolution of ϕ¯ with a kernel whose support is not compact. However, it turns out that this difference is not crucial and we can use almost the same approach.

Step 1. We prove a useful expression,

1

εϕε(x)·ξε(x)d x

= 1

0 R2

1 t2ε2

BRtε(y)

ξε(x)·l yx

,x yx

d x

·d[Dϕ](¯ y)

dt, (3.17)

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where ξε is as in (3.3) and R > 0 is such thatl(z,x) = 0 whenever|z| ≥ R.

As before, we shall denote by ∇1l and∇2l the gradient ofl with respect to the first and second variables respectively. Denote

ϕtε,1(x), ϕtε,2(x)

:= ϕtε(x)and l1(z,x),l2(z,x)

:=l(z,x). Then for everyt(0,1], every j ∈ {1,2}and every x

R

2we have

d

ϕtε,j(x)

dt = d

dt 1

t2ε2

R2lj yx

,x

ϕ(¯ y)d y

=− 1 t3ε2

R2

1lj yx

,x · yx

+2lj yx

,x

ϕ(¯ y)d y

= − 1 t2ε

R2divy

lj yx

,x yx

¯ ϕ(y)d y

= 1 t2ε

R2lj yx

,x yx

·d[Dϕ](¯ y) .

(3.18)

Therefore, for anyρ(0,1)we have,

1 ε

ϕε(x)ϕρε(x)

·ξε(x)d x

=

1

εξε(x)· 1

ρ

d ϕtε(x)

dt

d x

=

1 ρ ξε(x)·

1 t2ε2

R2l yx

,x

yx

·d[Dϕ](¯ y) dt

d x

= 1

ρ ξε(x)· 1

t2ε2

R2l yx

,x

yx

·d[Dϕ](¯ y) d x

dt

= 1

ρ R2

1 t2ε2

BRtε(y)

ξε(x)·l yx

,x yx

d x

·d[Dϕ](¯ y)

dt. (3.19)

From our assumptions onϕ, by (3.1), it follows that there exists a constantC >0, independent ofρ, such thatξρ(x)C for everyρ >0 and everyx. There- fore, lettingρtend to zero in (3.19), using the fact that limρ0ϕρ(x)L1() = 0 (see (3.11)), we get (3.17).

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Step 2. We prove the identity

R2

1 ε

div(1ϕε) (x)2d x

=oε(1)+ 1

0 Jϕ BR(0)

l(z,x)·ξε(x−εt z) z d z

·d

(x)

dt

+ 1

0 Gϕ BR(0)

l(z,x)·ξε(xεt z) z d z

·d

(x)

dt,

(3.20)

whereGϕis the set of approximate continuity ofϕ. By (3.16) and (3.17) we deduce

R2

1 ε

div(1ϕε) (x)2d x

=oε(1)+ 1

0 R2

1 t2ε2

BRtε(y)

ξε(x)·l yx

,x yx

d x

·d[Dϕ](¯ y)

dt

=oε(1)+ 1

0 R2

1 t2ε2

KBRtε(y)

ξε(x)·l yx

,x

yx d x

·d[Dϕ](¯ y)

dt, (3.21) whereK

is a compact set (see Definition 2.4). But, for everyε <R1dist(K, ∂) we have

R2

1 t2ε2

KBRtε(y)

ξε(x)·l yx

,x

yx d x

·d[Dϕ](¯ y)

=

1 t2ε2

KBRtε(y)

ξε(x)·l yx

,x

yx d x

·d[Dϕ](¯ y)

=

1 t2ε2

R2

ξε(x)·l yx

,x

yx d x

·d[Dϕ](¯ y) .

(14)

Therefore, by (3.21), we obtain

R2

1 ε

div(1ϕε) (x)2d x

=oε(1)+ 1 0

1 t2ε2

R2

ξε(x)·l yx

,x

yx d x

·d[Dϕ](¯ y)dt

=oε(1)+ 1

0 BR(0)

l(z,y−εt z)·ξε(y−εt z) z d z

·d

(y)

dt

=oε(1)+ 1

0 BR(0)

l(z,x)·ξε(xεt z) z d z

·d

(x)

dt,

(3.22)

where in the last equality we used the estimate|l(z,xεt z)l(z,x)| ≤Cεt|z|. Therefore we obtain (3.20).

Step 3. We will prove that the second integral in the r.h.s of (3.20) vanishes as ε→0. For everyxinGϕwe have,

ρ→lim0+

1 ρ2

Bρ(x)| ¯ϕ(y)− ˜¯ϕ(x)|d y =0. Takingρ= , for everyL >0, gives

ε→lim0+

BL(0)| ¯ϕ(x+εz)− ˜¯ϕ(x)|d z=0, forxinGϕ. (3.23) Using (3.1), since

R2z

divzr(z,xεt y)

d z =0, for everyxinGϕ,yBR(0), t ∈ [0,1]andL>0 we have,

ε(xεt y)| =

R2z

divzr(z,xεt y)

¯

ϕ(x+εzεt y)d z

=

R2z

divzr(z,xεt y)

¯

ϕ(x+εzεt y)− ˜¯ϕ(x) d z

BL(0)

z

divzr(z,xεt y)·ϕ(¯ x+εzεt y)− ˜¯ϕ(x)d z +

R2\BL(0)

z

divzr(z,xεt y)·ϕ(¯ x+εzεt y)− ˜¯ϕ(x)d z

AL

BL(0)| ¯ϕ(x+εzεt y)− ˜¯ϕ(x)|d z+B

R2\BL(0)

1

|z|3+1d z

AL

B(L+R)(0)| ¯ϕ(x+εz)− ˜¯ϕ(x)|d z+B

R2\BL(0)

1

|z|3+1d z,

(3.24)

(15)

where B > 0 is a constant and AL > 0 depends only on L. Givenδ > 0 we can takeL>0 such that

B

R2\BL(0)

1

|z|3+1d z < δ ,

Then, using (3.24) and (3.23), we infer limε→0+ε(xεt y)|< δand sinceδwas arbitrary,

ε→lim0+ξε(xεt y)=0 ∀xGϕ, yBR(0), t ∈ [0,1]. (3.25) Using (3.1), we also have|ξε(xεt y)| ≤C, and therefore, plugging (3.25) into (3.20), we obtain

R2

1 ε

div(1ϕε) (x)2d x

=oε(1)+ 1

0 Jϕ BR(0)

l(z,x)·ξε(xεt z) z d z

·d

(x)

dt. (3.26)

Step 4. Considerl¯(z,x):= ∇z

divzr(z,x)

. For everyε,t(0,1),xJϕ and zBR(0), we have

ξε(xεt z)=

R2

l¯(y,xεt z)ϕ¯

x+ε(yt z) d y

=

R2

l¯(y+t z,xεt z)ϕ¯ x+εy

d y

=

H+(0,ν(x))

l¯(y+t z,xεt z)ϕ¯ x+εy

d y

+

H(0,ν(x))

l¯(y+t z,xεt z)ϕ¯ x+εy

d y

=

H+(0,ν(x))

l¯(y+t z,xεt z)ϕ+(x)d y

+

H(0,ν(x))

l¯(y+t z,xεt z)ϕ(x)d y

+

H+(0,ν(x))

l¯(y+t z,xεt z)

¯

ϕ(x+εy)ϕ+(x) d y

+

H(0,ν(x))

l¯(y+t z,xεt z)

¯

ϕ(x+εy)ϕ(x) d y.

(3.27)

Références

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