Proof of Theorem 7.1

Step 1 : Decomposition of db, db_n.

Fix η. Through all proof K will denote constants depending on η or b and C will be kept for constants depending only on d or Ω.

Decompose b as in the definition

db=m+θH^d−1|J.

J is countably rectifiable and of finite measure. Therefore decompose J = H∪J⁰ with H a finite union of rectifiable setsH_i such that

H^d−1(J⁰)< η/C. (7.2) By the definition of the Hausdorff measure (and asJ⁰ is countably rectifiable), there exists a coveringJ⁰ ⊂S

iB(x_i, r_i) s.t. a point ofR^dbelongs to at most C balls and with

X

i

r_i^d−1 <2η/C.

As b_n converges toward b, and taking n ≥N large enough, one may decom-pose accordingly dbn as

|db_n|=m_n+σ_n+r_n, m_n, σ_n, r_n≥0, with for some ˜φ with ˜φ(ξ)/ξ →+∞ as ξ→+∞

sup

n

Z

Ω

φ(m˜ _n)dx≤C, (7.3)

and

suppσ_n⊂ {x, d(x, H)≤δ}, (7.4) with finally

suppr_n ⊂[

i

B(x_i,2r_i). (7.5)

Denote ˜b_n the function equal to b_n on Ω\Ω_r with Ω_r =S

iB(x_i,4r_i) and in B(x_i,3r_i)

˜bn =bn? Lri, with Lr(x) = r^−dL(x/r)

withLaC^∞function with total mass 1 and compactly supported in B(0,1).

InB(x_i,4r_i)\B(x_i,3r_i), choose a linear interpolation between the two values

By De La Vall´ee Poussin, there exists Φ, with ψ(ξ) = Φ(ξ)/ξ increasing and converging to +∞ s.t.

provided φ ≤φ˜and recalling that X

Hence one has

|ω_r| ≤η/8. (7.6)

Because of (7.4) and Lemma 3.2 as long asx and y are on the same side of H, not inω_r and both at distance larger than δ then

|bn(Xn(t, x))−bn(Xn(t, y))|=|˜bn(Xn(t, x))−˜bn(Xn(t, y))|

≤C Z

BK(x,y)

˜ m_n

1

|x−z|^d−1 + 1

|y−z|^d−1

dz, (7.7)

with Z

Ω

Φ( ˜m_n)dx <∞.

Step 2 : Decomposition of the trajectories.

Fixw∈S^d−1. For anyxwe decompose the time interval [0, T] into segments ]s_i, t_i[ such that on such an interval X_n(t, x) and X_n,δ =X_n(t, x+δ w) are both on the same side of H and both at a distance larger than δ of H.

On the contrary on each interval [t_i, s_i+1], eitherX_n(t, x) andX_n,δ =X_n(t, x+

δ w) are on different sides of H or one of them is at a distance less than δ of H.

Notice that of courseti and si depend on x, δ, b and η (as H depends on η) and

[0, T] = [

i≤n

]si, ti[∪[ti, si+1].

The transition between the two intervals (att_i as well as at s_i) is when one ofX_n(t, x) orX_n(t, x+δw) is exactly at distanceδofH, because for the two of them to be either on both side or on the same side of H, one of then has to crossH and thus to come first at a distance δ.

The idea of the rest of the proof is to bound|X−X_δ|on ]s_i, t_i[ with arguments similar to theW^1,1 case. On [ti, si+1], the bound will be obtained simply by controlling s_i+1−t_i. This will be enough provided that the number of such intervals n (depending on x again) is not too large.

Step 3 : Bound on the number of intervals.

At s_i and t_i either X_n(t, x) or X_n,δ = X(t, x+δ w) is at distance δ of H.

Therefore as kbk∞ ≤ 2, on the intervals [s_i−δ, s_i+δ] and [t_i−δ, t_i+δ], again either X_n or X_n,δ is at distance less than 3δ of H.

So if there are n such intervals for x then either X_n stays at distance less than 3δ of H for a total time of at least nδ orXn,δ does.

Now denote by ω¹ the set of all xsuch that Xn(t, x) stays distance less than 3δ of H for a total time of at least nδ. Obviously

Z T

0

Z

ω¹

Id(Xn(t,x),H)≤3δdx dt≥nδ|ω¹|.

But on the other hand, changing variable Z T

0

Z

ω¹

Id(Xn(t,x),H)≤3δdx dt≤ Z T

0

Z

Ω

Id(Xn(t,x),H)≤3δdx dt

≤ Z T

0

Z

I^d(x,H)≤3δdx dt≤K T δH^d−1(H)< K T δ, as the mesure ofH is finite but depends on η. Thus one finds

|ω¹| ≤K/n.

Choosen= 1/(8Kη) so that except a set of measure less thanη/4 (and equal toω¹∪(ω¹ −δw)), no trajectory is decomposed on more than n intervals.

Step 4 : Control on [ti, si+1], definitions.

First of all, take anyz ∈H. Denote by b−(z) the trace of b on one side ofH and b₊(z) on the other side (again orientation is chosen locally and does not have to hold globally forH). For simplicity, we assume thatb−(z)·ν(z)≥0 (with ν the normal to H chosen according to the orientation).

Now fix some α to be chosen later.

As b is approximately continuous (see [5]) at z on either side ofH one has r^−d|{x∈B±(z, r), |b(x)−b±(z)| ≥α}| −→0,

where B±(z, r) is the subset ofB(z, r) which is on the −or + side of H.

It is even possible to be more precise. Take a function ˜η(r) → 0 as r → 0 with ˜η≤η/(6H^d−1(H)). Denote forr≤1/2

µ(z, r)± ={x∈B±(z, r), |b(x)−b±(z)| ≥α}|,

and

˜

µ(z, r)± ={x∈µ(z, r)±, d(x, H)≥rη/K},˜

with K large enough with respect to the Lipschitz constant of H. Assume that |µ(z, r)−| ≥ r^dη, then necessarily˜ |˜µ(z, r)−| ≥ r^dη/2; For any˜ x ∈

Integrate this inequality over ˜µ to find Z 1 inequality on H_r on the right hand side and on H on the left hand side to obtain

Let us bound the integral on the left hand side. As the case θ ≥ 1/2 is easy, assume θ < 1/2 and change variables locally around H such that H has equation x₁ = 0. The integral is then dominated by

K

Then changing variable from θ to θx₁, one may finally bound by 2^d−1K

Note that this last integral converges to 0 as r→0 and hence conclude that by choosing ˜η large enough in terms of r, one can ensure that

|H_r⁰| ≤χ(r⁰), χ(r⁰)→0 as r⁰ →0, χ(r⁰)≤η/C ∀r⁰ ≤r.

In addition from the strong convergence bn → b, take n0 s.t. ∀n ≥ n0,

∀z ∈H\H_r

r^0−d|{x∈B±(z, r⁰), |b_n(x)−b±(z)| ≥α}| ≤3˜η/2≤η/(4H^d−1(H))

∀|δ|< r⁰ < r. (7.8) This is the second point where n has ot be large enough in terms ofδ.

Remark eventually that one may always assume that r ≤ α (unfortunately not the opposite way) and (α r)^d< η/C.

Step 5 : Control on [t_i, s_i+1], non-crossing trajectories.

We start by taking apart the trajectories which do not cross H. So define H₀ ={z_i 6∈H_r, b−(z_i)·ν(z_i)≤2α}, and

ω² = (

x, ∃t∈[0, T] s.t. X_n(t, x)∈ [

z∈H₀

B−(z, α r)∪B₊(z, α r) )

, ω³ ={x, ∀t∈[0, T] s.t. d(X_n(t, x), H)< α r/2,

one has B(X_n, α r)∩(H\H_r) = ∅}.

Note that by (1.4) (see 2.3 in the introduction), if b−(z)·ν(z) ≤ 2α on a neighborhood of the border then b+ ·ν(z) ≤ Kα (with K function of the constant in (1.3) and the Lipschitz bound on H). Therefore in the previous definition of H₀, one only needs to putb− (in any case the proof could work the same by essentially ignoring what occurs on the side where b ·ν ≥ 0, except in a band of width δ).

Bound ω³ first. Simply notice that if

Ω³ ={x with d(x, H)< α r/2, s.t. B(x, α r)∩(H\H_r) =∅},

then forx∈Ω³,H^d−1(B(x, αr)∩Hr)≥C α^d−1r^d−1 and asH^d−1(Hr)≤η/C then

|Ω³| ≤α rη/C, |Ω³_r|=|{x, d(x,Ω³)≤αr}| ≤α r η/C.

Asω³ ={x, ∃t∈[0, T]X_n(t, x)∈Ω³}, and ifX_n(t, x)∈Ω³ thenX_n(s, x)∈ by change of variable and the bound on |Ω³_r|. This implies

|ω³| ≤T η/C. (7.9)

Let us now boundω², decompose

ω² =ω^2,1∪ω^2,2,

with ω^2,1 the subset of all x∈ω² s.t. the trajectory X_n(t, x) stays at least a time intervalr/2 at a distance less than Kα r of H. Notice that

Z On the other hand by (1.4)

Z There remains ω^2,2 which is made of those x ∈ ω² staying a time less than r/2 at a distance less than Kαr of H. Denote t₀ s.t. X_n(t₀, x)∈B−(z, α r) for some z (and of course the same analysis holds for the + side).

If this is so then the average over the interval [t₀, t₀+r/2]

TakingKlarge enough with respect to the Lipschitz bound onH, this implies 2

r

Z t0+r/2

t0

ν(z)·b_n(X_n(t, x))dt ≥K α,

and consequently 2 r

Z t0+r/2

t0

|b_n(X_n(t, x))−b₋(z)|dt ≥K α.

As 1 ≤ b₁ ≤ 2, notice that on the time interval [t₀, t₀ +r/2] then X_n(t, x) stays inside the ball B(z, r) and moreover the length of the path is of order r. Consequently, denoting ω_z^2,2 the subset of ω^2,2 corresponding to the ball B(z, r), one has by change of variable (and again the use of (1.4))

1

r^d{x∈B−(x), |b_n(x)−b−(z)| ≥α} ≥K|ω_z^2,2|, which finally givesω_z^2,2 ≤ηr^d/K and summing over all z =z_i

|ω²| ≤ |ω^2,1|+|ω^2,2| ≤η/8 +η X

i

r^d/K ≤η/4, (7.11) by choosing K large enough.

Step 6 : Control on [t_i, s_i+1], the crossing trajectories.

We now only have to take into account the trajectories passing through.

Consider for example

H₁ ={z 6∈H_r, b−(z)·ν(z)≥2α }, and ω⁰ =

(

x, ∃t∈[0, T] s.t. X_n(t, x)∈ [

z∈H₁

B−(z, α r) )

.

The same holds if one usesb₊(z_i)·ν(z)<−2αin the definition and note again that by (1.3) one has necessarily one or the other as the case |b_±·ν(z)| ≤2α was taken care of in the previous step.

As |H_r⁰| → 0, there exists an extracted sequence r_k → 0 (with r₀ = r) s.t.

for any 0< γ <1 X

k

rk <∞, X

k

|Hr_k|<∞, and so

∞

X

k=0

|Hr_k| ≤η/K,

X

k

˜

η(r_k)<∞,

+∞

X

k=0

˜

η(r_k)≤η/K.

(7.12)

For anyδ, denote k_δ s.t.

X

k≤k_δ

δ

r_k ≤η/K. (7.13)

Of courser_kδ > δ but note that k_δ →+∞ as δ→0.

Denote Ω+ the set of y ∈ S

z∈HB+(z, r) s.t. d(y, H) > δ and Ω− = S

z∈HB₊(z, r)\Ω₊.

For any scaler_k > Cδ with k≥1, define ω_k as the set of x s.t.

(i)∃t0 ∈[0, T] withd(Xn(t0, x), H)≤rkα/K,K ≥6, (ii) one never has X_n(t, x)∈Ω₊ for any t∈[t₀, t₀+r_k/3].

By (i) and kb_nk∞ ≤ 2, one has that X_n(t, x) remains in the same ball B(Xn(t0, x), rk/2) for all t∈[t0, t0+rk/K] (forK ≥4). The intersection of this ball withH has diameter larger than r_k/3 as d(X_n(t₀, x), H)≤r_kα/6.

Accordingly decompose ω_k=ω¹_k∪ω_k² with ω_k¹ the set ofx∈ω_k s.t.

B(X_n(t₀, x), r_k/2)∩(H\H_rk) = ∅, and ω²_k =ω_k\ω¹_k.

Start by bounding ω¹_k. Define

Ω¹_k={x, B(x, r_k/2)∩(H\H_rk) = ∅}.

Just as in the previous step

|Ω¹_k| ≤CH^d−1(H_rk)r_k, |Ω˜¹_k|={x, d(x,Ω¹_k)≤r_k}| ≤CH^d−1(H_rk)r_k. On the other handX_n(t, x) stays inside ˜Ω¹_k for all the interval [t₀, t₀+r_k/K]

so again

|ω_k¹|r_k/K ≤ Z T

0

Z

Ω

IXn(t,x)∈Ω˜¹_kdx dt≤ Z T

0

Z

R^d

Ix∈Ω˜¹_kdx dt≤CH^d−1(H_rk)r_k, which gives the desired bound

|ω¹_k| ≤KH^d−1(H_rk). (7.14) Now for ω_k², notice that for any x∈ ω²_k there exists z ∈ B(X_n(t₀, x), r_k/2)∩ (H\H_rk). Hence we may take N points z_i ∈H\H_rk with

N ≤KH^d−1(H)/r^d−1_k ,

and such that for any x∈ω²_k, there is z_i ∈B(X_n(t₀, x), r_k/K).

On the interval [t₀, t₀+r_k/K],X_nis in Ω−. DenoteH^δ={x, d(x, H)≤δ}, decompose again ω_k² into ω_k^2,δ∪ω˜_k² with ω_k^2,δ the set of x such that Xn stays more than a total time ∆ =αr_k/K in H^δ. Bound

|ω^2,δ_k |αr_k/K ≤ Z T

0

Z

Ω

Id(Xn(t,x),H)<δdx dt≤K δH^d−1(H).

Thus

|ω_k^2,δ| ≤K δH^d−1(H)/r_k. (7.15) For x∈ ω˜²_k, denote byI the subset of [t₀, r_k/K] s.t. X_n(t, x) 6∈H^δ. As X_n never reaches Ω₊ and stays in H^δ at most αr_k/K one has

1

|I|

Z

I

bn(Xn(t, x))·ν(Xn)dt < K

r_k (δ+αrk/K+ 2αrk/K)< α/3.

Moreover for any t ∈[t₀, t₀ +r_k/K], one has |X_n(t, x)−z_i| ≤ r_k/K and as ν is lipschitz (at least aroundH) then

1

|I| Z

I

b_n(X_n(t, x))·ν(z_i)dt < α/2,

and 1

|I|

Z

I

I^|bn(Xn(t,x))−b±(zi)|≥αdt >1.

Put ˜ω_kⁱ = {x ∈ ω˜_k², X(t₀, x) ∈ B(z_i, r_k/K). By (7.8) and integrating along the trajectories (as we did many times before), we deduce that

|˜ωⁱ_k| ≤Kη(r˜ _k)r_k^d−1. Summing over i,

|ω˜_k²| ≤Kη(r˜ _k). (7.16) Summary and conclusion of the estimate.

Define

˜

ω =ω_r∪ω¹∪ω²∪ω³∪_kω_k, ω¯ = ˜ω∪(˜ω−δw).

By the previous computations (end of steps 1, 3, step 5 (7.9) and (7.11) and step 6 (7.14), (7.15), (7.16))

|¯ω| ≤η/4 +η/4 +T η/C+η/4 +K X

k

(|H_rk|+δ/r_k+ ˜η(r_k))≤7η/8, (7.17)

by (7.12) and (7.13).

As in the proof of 4.1, we define ψ(|δ|) = inf

Now by induction let us prove that up to timeti

|Xn−Xn,δ| ≤δi, except for x∈ω¯∪ω_i⁴ with

|ω_i⁴| ≤i η/8n, and δ_i defined through

δ_i+1 = (δ_i+r_ki/K) exp(8n ψ(δ_i+r_ki/K)/η), r_ki = inf{r_k, α r_k/K > δ+δ_i}.

(7.18) This is true for i = 1. So assume it is still true up to ti and study what happens on [t_i, t_i+1]. First of all on [t_i, s_i+1] : at t_i assume for instance that d(X_n,δ, H) = δ and take k_i s.t. r_ki = inf{r_k, αr_k/K > δ +δ_i}. Then

necessarily d(X_n(t_i), H) ≤ αr_ki/K. As x 6∈ ω, by step 6, one has that¯ s_i+1 ≤t_i+r_ki/K. Therefore

|X_n(s_i)−X_n,δ(s_i)| ≤δ_i+r_ki/K.

Now on [si+1, ti+1], simply write Z

¯ ω^c∩ω_i⁴

log

δi+rki/K+|Xn(ti+1, x)−Xn,δ(ti+1, x)|

δ_i+r_ki/K+|X_n(s_i+1, x)−X_n,δ(s_i+1, x)|

≤ Z

ω^c∩ω⁴_i

Z ti+1

si+1

|b(X_n)−b(X_n,δ)|

δ_i+r_ki/K+|X_n−X_n,δ|dt dx

≤C Z

Ω

Z T

0

Z

BK(Xn, Xn,δ)

˜ m_n

1

|X_n−z|^d−1 + 1

|X_n,δ−z|^d−1

dz

≤ψ(δ_i+r_ki/K).

Therefore for t ≤ti+1, one has that

|X_n−X_n,δ| ≤δ_i+1 = (δ_i+r_ki/K) exp(8n ψ(δ_i+r_ki/K)/η), except for x∈ω¯∪ω_i+1⁴ , withω_i⁴ ⊂ω⁴_i+1 and

|ω_i+1⁴ \ω_i⁴| ≤η/8n, s.t.|ω_i+1⁴ | ≤(i+ 1)η/8n.

Finally for any t∈[0, T], one has that

|X_n−X_n,δ| ≤δ_n, for any x6∈ω with ω = ¯ω∪ωn and thus

|ω| ≤η.

This concludes the proof once one notices that δ_n → 0 as δ → 0 since ψ(δ)/|logδ| →0.

Compactness.

Let us just briefly indicate how to obtain the compactness from the estimate onX_n−X_n,δ.

Notice first that in the previous proof, n had to be taken large enough in terms of δ. So if for a given n, one considers δ⁰ << δ instead of δ, it is not true that |X_n−X_n,δ⁰| may be bounded by ε(δ⁰). Instead the best that can

be done is nothing as long as |X_n−X_n,δ⁰| ≤ δ and bound as before once

|X_n−X_n,δ⁰| ≥δ. So in the end |X_n−X_n,δ⁰| is only bounded by ε(δ).

For the compactness of X_n, recalling that X_n is uniformly lipschitz in time, we apply the usual criterion saying that X_n is compact inL¹([0, T]×Ω) iff

∀γ, ∀w, ∃δ s.t.∀δ⁰ < δ, sup

n

Z T

0

Z

Ω

|X_n(t, x)−X_n(t, x+δ⁰w)|dx dt < γ.

So fix γ and w. First choose η s.t.

η T max

Ω |x|< γ/4.

Now apply the previous quantitative estimate to obtain a functionε(δ) (recall that this depends on η). Choose δ₁ with ε(δ₁) < γ/(2|Ω|T). This gives n₁ s.t. for any n ≥n₁ and any δ < δ₁

|X_n−X_n,δ| ≤ε(δ₁)< γ/(2|Ω|T), ∀x∈Ω\ω_δ with |ω_δ| ≤η.

Now for n < n1 as Xn is regular (not uniformly in n but there are only n1

indicesn now), choose δ₂ s.t. for any n < n₁ and any δ < δ₂

|X_n−X_n,δ|< γ/(|Ω|T).

Consequently takeδ = min(δ₁, δ₂). For any n and any δ⁰ < δ, if n < n₁ then simply

Z T

0

Z

Ω

|X_n(t, x)−X_n(t, x+δ⁰w)|dx dt≤T|Ω| sup

x

|X_n−X_n,δ|< γ.

And ifn ≥n1, then decompose Z T

0

Z

Ω

|X_n(t, x)−X_n(t, x+δ⁰w)|dx dt= Z T

0

Z

ω_δ0

. . .+ Z T

0

Z

Ω\ω_δ0

. . .

≤2T |ω_δ⁰| max

Ω |x|+T|Ω| sup

Ω\ω_δ0

|X_n−X_n,δ⁰|< γ.

Hence the compactness criterion is indeed satisfied.

Dans le document Differential Equations with singular fields (Page 22-35)