Regularity results for the solutions of a non-local model of traffic

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Regularity results for the solutions of a non-local model

of traﬀic

Florent Berthelin, Paola Goatin

To cite this version:

Florent Berthelin, Paola Goatin. Regularity results for the solutions of a non-local model of traﬀic. Discrete and Continuous Dynamical Systems - Series A, American Institute of Mathematical Sciences, 2019, 39 (6), pp.3197-3213. �hal-01813760�

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Regularity results for the solutions of a non-local model of traffic

1

flow

2

Florent Berthelin∗ Paola Goatin† 3

June 12, 2018 4

Abstract

5

We consider a non-local traffic model involving a convolution product. Unlike other studies,

6

the considered kernel is discontinuous on R. We prove Sobolev estimates and prove the

conver-7

gence of approximate solutions solving a viscous and regularized non-local equation. It leads to

8

weak, C([0, T ], L2

(R)), and smooth, W2,2N

([0, T ] × R), solutions for the non-local traffic model.

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Key words: Scalar conservation laws; Anisotropic non-local flux; Traffic flow models; Viscous

10

approximation; Sobolev estimates.

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1 Introduction

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We consider the non-local traffic model introduced in [4, 8] to account for the reaction of drivers to downstream traffic conditions. It consists in the following scalar conservation law, where the traffic velocity depends on a weighted mean of the density:

∂tρ + ∂x(ρv(ρ ∗ ω)) = 0, (1.1) where (ρ ∗ ω)(t, x) = Z η 0 ρ(t, x + y)ω(y) dy = Z x+η x ρ(t, y)ω(y − x) dy. (1.2) We make the following assumptions for k = 1, 2, 3:

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(Ak_ω) ω ∈ Ck([0, η]) is non-negative with support in [0, η] and is non-increasing on [0, η].

14

(Ak_v) v ∈ Ck(R+) with v0, . . . , v(k) bounded.

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For traffic flow applications, it is reasonable to assume that v is non-increasing, even if

mono-16

tonicity is not required in this paper. We also recall that a similar model, considering a weighted

17

mean of downstream speeds, has been recently introduced in [7]. More generally, model (1.1)

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belongs to the class of conservation laws with non-local flux functions, which appear in several

19

applications, see for example [3, 6, 9, 10, 15]. We remark that most of the available well-posedness

20

results concern equations involving smooth convolution kernels [1, 2], and are based on the

construc-21

tion of finite-volume approximations and the use of Kruˇzkov’s doubling of variable technique [12].

22

In particular, these results rely on the concept of entropy solutions. Only recently, alternative

23

proofs based on fixed point theorems have been proposed for specific cases [11, 14], allowing to get

24

rid of the entropy requirement.

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In general, solutions to non-local equations may be discontinuous [13], despite the expected

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regularizing effect of the convolution product. Therefore, given any initial datum ρ0 ∈ L∞_{(R) ∩}

27

L1(R), the solutions to the Cauchy problem for (1.1) are usually intended in the following weak

28

form

29 ∗

Laboratoire J. A. Dieudonné, UMR 7351 CNRS, Université Côte d’Azur, LJAD, CNRS, Inria, Université de Nice Sophia-Antipolis, Parc Valrose, 06108 Nice cedex 2, France. E-mail: Florent.Berthelin@unice.fr

†

Inria Sophia Antipolis - Méditerranée, Université Côte d’Azur, Inria, CNRS, LJAD, 2004 route des Lucioles - BP 93, 06902 Sophia Antipolis Cedex, France. E-mail: paola.goatin@inria.fr

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Definition 1. A function ρ ∈ L∞∩ L1_(R+_{× R) is a solution of (1.1) with initial datum ρ}0 _if Z +∞ 0 Z +∞ −∞ ρ∂tϕ + ρv(ρ ∗ ω)∂xϕ (t, x) dxdt + Z +∞ −∞ ρ0(x)ϕ(0, x)dx = 0, (1.3) for all ϕ ∈ C∞_c (R2). 1

In this paper, we are interested in deriving regularity properties of solutions to (1.1). To this end, we will consider approximate solutions satisfying the viscous and regularized non-local equation

∂tρε+ ∂x(ρεv(ρε∗ ωε)) = ε∂xx2 ρε, (1.4)

where, for any ε ∈ ]0, 1], the smooth function ωεis an extension of ω with the following regularities:

2

(Ak_ω_ε) ωε ∈ Ck(R) is non-negative with a support in [−ε, η + ε], is non-decreasing on [−ε, xε], for

some xε∈ ] − ε, 0], is non-increasing on [xε, η + ε] and ωε= ω on [0, η].

We set Wε := ωε(xε) and we assume that limε→0Wε = ω(0). Without loss of generality we

can assume

Wε≤ 2ω(0). (1.5)

(Bk_ω_ε) ωε(j)(−ε) = ωε(j)(η + ε) = 0 for j = 1, . . . , k and |ωε0(u)| ≤ 2Wε/ε on [−ε, 0] and |ω0ε(u)| ≤

3

2ω(η)/ε on [η, η + ε].

4

Remark 1. Given ω satisfying (Ak_ω), we can construct a function ωε satisfying (Akωε) and (B

k ωε). 5

To construct such extensions, for example in the simplest case where the derivatives of ω vanish at

6

0 and η, we use the function ϕ which is zero for x ≤ −1, 1 for x ≥ 0, non-decreasing and of class

7

C∞ and we define ωε(x) = ω(0)ϕ(x/ε) for x < 0 and ω(0)ϕ((η − x)/ε) for x > η.

8

Notice that a similar approximation was used in [5] to establish a convergence property for the

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singular limit where the (smooth) convolution kernel is replaced by a Dirac delta, in the viscous

10

case. Here, we will study the properties of smooth solutions ρε of this equation corresponding to a

11

fixed initial datum ρ0, and then we will recover properties for ρ passing to the limit as ε → 0.

12

We have the following result.

13

Theorem 1. We assume (A2_ω)-(A3_v). Let ρε be a solution of (1.4) with initial datum ρ0. We

14

assume ρ0 ∈ W1,4_{(R) ∩ H}2_{(R). Then, for T > 0 sufficiently small, ρ}

ε converges in L2loc([0, T ] × R)

15

to a solution ρ ∈ C([0, T ], L2(R)) to equation (1.1) with initial datum ρ0. Furthermore, if ρ0 ∈

16

W1,2N(R), N ∈ N∗, then ρ ∈ W1,2N([0, T ] × R), and if ρ0 ∈ W1,4N_{(R) ∩ H}1_{(R) ∩ W}2,2N_{(R), then}

17

ρ ∈ W2,2N([0, T ] × R).

18

In particular, this provides an alternative proof of existence of weak solutions, locally in time.

19

To prove this result, in Section 2 we first establish estimates on the non-local term and we derive

20

Lp(R), p > 1, estimates for ρε, then we get estimates in W1,2N(R) for ρε with respect to x. This

21

allows to prove that there exists T > 0 such that the sequence ρεis uniformly bounded with respect

22

to ε in L∞(R) on [0, T ]. Then we prove uniform space estimates in W2,2N(R) for ρε, which allows

23

to derive estimates on ∂tρε. The proof of Theorem 1 is deferred to Section 3.

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2 Estimates

1

Here and in the following sections, we will denote by

kρk_∞:= kρk_L∞_{([0,T ]×R)} and by ρ(t, ·) _∞:= ρ(t, ·) L∞_(R).

Moreover, notice that we have

2 (ρ ∗ ωε)(t, x) = Z R ρ(t, x + y)ωε(y) dy = Z R ρ(t, y)ωε(y − x) dy = Z η+ε −ε ρ(t, x + y)ωε(y) dy = Z x+η+ε x−ε ρ(t, y)ωε(y − x) dy.

2.1 Estimates of the non-local term 3

We start by proving the following estimates on the non-local term.

4

Proposition 1. 1. We assume (A1

ωε) and that ρ is a continuous function. For any (t, x) ∈

R+× R, we have

|∂_x(ρ ∗ ωε)(t, x)| ≤ 2

ρ(t, ·) _∞W_ε. (2.1)

2. We assume (A2_ω_ε) − (B1_ω_ε) and that ρ is a C1 function. For any t ≥ 0, p > 1, we have

5 Z R |∂_xx2 (ρ ∗ ωε)|p(t, x) dx 1/p ≤ η1/p Z η 0 |ω00(u)|p/(p−1)du 1−1/p Z R ρp(t, y) dy 1/p + |ω0(η−)| + |ω0(0+)| Z R ρp(t, x) dx 1/p (2.2) +2 ω(η) + Wε Z R |∂xρ(t, x)|pdx 1/p .

3. We assume (A2_ω) − (B2_ω_ε) that ρ is a C2 function. For any t ≥ 0, p > 1, we have

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we obtain |∂_x(ρ ∗ ωε)(t, x)| ≤ kρ(t, ·)k∞ Z η+ε −ε |ω_ε0(u)| du ≤ kρ(t, ·)k∞ Z xε −ε ω_ε0(u) du − Z η+ε xε ω_ε0(u) du ! ≤ 2kρ(t, ·)k∞Wε. 2. From ∂_xx2 (ρ ∗ ωε)(t, x) = Z x+η+ε x−ε ρ(t, y)ω00_ε(y − x) dy − ρ(t, x + η + ε)ω0_ε(η + ε) + ρ(t, x − ε)ω_ε0(−ε) = Z x+η+ε x−ε ρ(t, y)ω00_ε(y − x) dy = Z η+ε −ε ρ(t, x + u)ω_ε00(u) du = Z 0 −ε ρ(t, x + u)ω00_ε(u) du + Z η 0 ρ(t, x + u)ω00_ε(u) du + Z η+ε η ρ(t, x + u)ω_ε00(u) du = ρ(t, x)ω_ε0(0) − ρ(t, x − ε)ω_ε0(−ε) − Z 0 −ε ∂xρ(t, x + u)ωε0(u) du + Z η 0 ρ(t, x + u)ω00_ε(u) du + ρ(t, x + η + ε)ω_ε0(η + ε) − ρ(t, x + η)ω_ε0(η) − Z η+ε η ∂xρ(t, x + u)ω0ε(u) du = ρ(t, x)ω0(0+) − ρ(t, x + η)ω0(η−) + Z η 0 ρ(t, x + u)ω_ε00(u) du − Z 0 −ε ∂xρ(t, x + u)ω0ε(u) du − Z η+ε η

∂xρ(t, x + u)ωε0(u) du,

we have Z R |∂2_xx(ρ ∗ ωε)|p(t, x) dx 1/p ≤   Z R Z 0 −ε ∂xρ(t, x + u)ωε0(u) du p dx   1/p +   Z R Z η+ε η ∂xρ(t, x + u)ω0ε(u) du p dx   1/p + Z R Z η 0 ρ(t, x + u)ω00_ε(u) du p dx !1/p + Z R ρ(t, x)p|ω0(0+)|pdx 1/p + Z R ρ(t, x + η)p|ω0(η−)|pdx 1/p . Notice that   Z R Z 0 −ε ∂xρ(t, x + u)ω0ε(u) du p dx   1/p

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then we get the announced formula.

1

3. Remark that, since ω00_ε(−ε) = ω_ε00(η + ε) = 0, we have

∂_xxx3 (ρ ∗ ωε)(t, x) = − Z η+ε −ε ρ(t, x + u)ω_ε(3)(u) du = Z η+ε −ε ∂xρ(t, x + u)ω 00 ε(u) du

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= ∂_xx2 (∂xρ ∗ ωε)(t, x),

then applying 2., we get

2.2 Lp estimates for the viscous case 3

We turn now to estimates on solutions solving the viscous and regularized non-local equation. First,

4

we deal with Lp estimates.

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Proposition 2. We assume (A1_ω)-(A1_v). Let ρε be the solution of (1.4) with initial datum ρ0 ∈

Lp(R). If ρε∈ L∞([0, T ] × R) for some T > 0, then

ρε∈ L∞([0, T ], Lp(R)) ∩ Lp([0, T ] × R).

Proof. The equation (1.4) can be rewritten as

∂tρε+ v(ρε∗ ωε)∂xρε+ ρεv0(ρε∗ ωε)∂x(ρε∗ ωε) = ε∂xx2 ρε. (2.4)

Multiplying (2.4) by ρp−1ε , then integrating with respect to x, we obtain

6 1 p d dt Z R ρp_ε(t, x) dx = − Z R ρp−1_ε (t, x)v((ρε∗ ωε)(t, x))∂xρε(t, x) dx − Z R ρp_ε(t, x)v0((ρε∗ ωε)(t, x))∂x(ρε∗ ωε)(t, x) dx +ε Z R ρp−1_ε (t, x)∂_xx2 ρε(t, x) dx. We observe that 7 Z R ρp−1_ε v(ρε∗ ωε) ∂xρεdx = Z R ∂x ρpε p ! v(ρε∗ ωε) dx = − Z R ρpε p ∂x(v(ρε∗ ωε)) dx = − Z R ρpε p v 0_(ρ ε∗ ωε)∂x(ρε∗ ωε) dx and 8 Z R ρp−1_ε ∂2_xxρεdx = −(p − 1) Z R ρp−2_ε (∂xρε)2 dx ≤ 0,

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therefore d dt Z R ρp_ε(t, x) dx ≤ (1 − p) Z R ρp_ε(t, x)v0((ρε∗ ωε)(t, x))∂x(ρε∗ ωε)(t, x) dx. (2.5)

We use (2.1) to control the right hand side of (2.5) and we get d dt Z R ρp_ε(t, x) dx ≤ C₁ε,p Z R ρp_ε(t, x) dx, (2.6) which implies Z R ρp_ε(t, x) dx ≤ eCε,p1 t Z R ρp_ε(0, x) dx, (2.7) with C₁ε,p= 2(p − 1)kρεk∞Wεkv0k∞. It gives sup t∈[0,T ] Z R ρp_ε(t, x) dx ≤ eC1ε,pT Z R ρp_ε(0, x) dx. (2.8)

By integration of (2.7) with respect to t ∈ [0, T ], we get Z T 0 Z R ρp_ε(t, x) dx dt ≤ 1 C₁ε,p eC1ε,pT − 1 Z R ρp_ε(0, x) dx. (2.9) 1

2.3 W1,p _{estimates for p = 2N in the viscous case}

2

We turn now to Sobolev estimates. Let N ∈ N∗ and set p = 2N .

3

W1,2N_{(R). If ρ}

ε∈ L∞([0, T ] × R) for some T > 0, then

ρε∈ L∞

[0, T ], W1,2N(R) and ρε, ∂xρε ∈ L∞

[0, T ], L2N(R)∩ L2N_{([0, T ] × R).}

Proof. We differentiate (2.4) with respect to x, it gives

4

∂t∂xρε+ 2v0(ρε∗ ωε) ∂x(ρε∗ ωε) ∂xρε+ v(ρε∗ ωε) ∂xx2 ρε

+ρεv00(ρε∗ ωε) (∂x(ρε∗ ωε))2+ ρεv0(ρε∗ ωε)∂xx2 (ρε∗ ωε) = ε∂xxx3 ρε. (2.10)

Multiplying this relation by (∂xρε)p−1, then integrating with respect to x, we have

5 1 p d dt Z R (∂xρε)pdx + 2 Z R v0(ρε∗ ωε) ∂x(ρε∗ ωε) (∂xρε)pdx + Z R v(ρε∗ ωε) ∂xx2 ρε(∂xρε)p−1dx + Z R ρεv00(ρε∗ ωε) (∂x(ρε∗ ωε))2(∂xρε)p−1dx + Z R ρεv0(ρε∗ ωε) ∂xx2 (ρε∗ ωε) (∂xρε)p−1dx = ε Z R (∂xρε)p−1∂xxx3 ρεdx. Notice that 6 Z R v(ρε∗ ωε) ∂xx2 ρε(∂xρε)p−1dx = 1 p Z R v(ρε∗ ωε) ∂x (∂xρε)p dx

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= −1 p

Z

R

v0(ρε∗ ωε) ∂x(ρε∗ ωε) (∂xρε)pdx

and, since p is even,

1 Z R (∂xρε)p−1∂xxx3 ρεdx = −(p − 1) Z R (∂_xx2 ρε)2(∂xρε)p−2dx ≤ 0, thus 2 d dt Z R (∂xρε)pdx ≤ (1 − 2p) Z R v0(ρε∗ ωε) ∂x(ρε∗ ωε) (∂xρε)pdx −p Z R ρεv00(ρε∗ ωε) (∂x(ρε∗ ωε))2(∂xρε)p−1dx −p Z R ρεv0(ρε∗ ωε) ∂xx2 (ρε∗ ωε) (∂xρε)p−1dx =: I₁ε+ I₂ε+ I₃ε. We estimate now each of these terms.

3 • By (2.1) we get I₁ε = (1 − 2p) Z R v0(ρε∗ ωε) ∂x(ρε∗ ωε) (∂xρε)pdx ≤ 2(2p−1)kv0k∞kρεk∞Wε Z R |∂xρε|pdx. • Again by (2.1) we get 4 I₂ε = p Z R ρεv00(ρε∗ ωε) (∂x(ρε∗ ωε))2(∂xρε)p−1dx ≤ pkv00k∞(2kρεk∞Wε)2 Z R ρε|∂xρε|p−1dx ≤ 4kv00k∞kρεk2∞Wε2 Z R ρp_εdx + (p − 1) Z R |∂xρε|pdx ,

where we have used Young’s inequality uv ≤ 1 pu p₊1 qv q _{with q = p/(p − 1).} 5 6 • Similarly, 7 I₃ε = p Z R ρεv0(ρε∗ ωε) ∂xx2 (ρε∗ ωε) (∂xρε)p−1dx ≤ pkv0k∞kρεk∞ Z R |∂_xx2 (ρε∗ ωε) |∂xρε|p−1| dx ≤ kv0k∞kρεk∞ Z R |∂_xx2 (ρε∗ ωε)|pdx + (p − 1) Z R |∂_xρε|pdx . We now observe that

(u + v + w)p ≤ 3p(up+ vp+ wp) for any u, v, w > 0 and p > 0. (2.11)

Indeed, from the binomial expansion we get

8

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= p X k=0 p k uk(v + w)p−k = p X k=0 p k uk p−k X l=0 p − k l vlwp−k−l.

Observing that ukvlwp−k−l ≤ up _{+ v}p _{+ w}p _{and that} Pp−k

l=0 p − k l = (1 + 1)p−k and 1 Pp k=0 p k

1k2p−k = 3p, we get the result.

2

Estimate (2.2) of Proposition 1 and inequality (2.11) give

3 Z R |∂_xx2 (ρε∗ ω)|pdx ≤ 3pη Z η 0 |ω00(u)|p/(p−1)du p−1 Z R ρp_ε(t, x) dx +3p |ω0(η−)| + |ω0(0+)|p Z R ρp_ε(t, x) dx +6p ω(η) + Wε p Z R |∂xρε(t, x)|pdx, (2.12) thus p Z R ρεv0(ρ ∗ ωε) ∂xx2 (ρε∗ ωε) (∂xρε)p−1dx ≤ C₂ε,p Z R ρp_εdx + C₃ε,p Z R |∂xρε|pdx, with C₂ε,p= kv0k∞kρεk∞3p " η Z η 0 |ω00(u)|p/(p−1)du p−1 + |ω0(η−)| + |ω0(0+)|p # and C₃ε,p= kv0k∞kρεk∞ h p − 1 + 6p ω(η) + Wε pi . These bounds give finally the estimate

4 d dt Z R (∂xρε(t, x))pdx ≤ C4ε,p Z R |∂xρε(t, x)|pdx + C5ε,p Z R ρp_ε(t, x) dx. with C₄ε,p= 2(2p − 1)kv0k∞kρεk∞Wε+ 4(p − 1)kv00k∞kρεk2∞Wε2+ C3ε and C₅ε,p= 4kv00k∞kρεk2∞Wε2+ C2ε. With (2.6), we get 5 d dt Z R |∂_xρε(t, x)|2Ndx + Z R ρ2N_ε (t, x) dx ≤ C₆ε,p Z R |∂xρε(t, x)|2Ndx + Z R ρ2N_ε (t, x) dx , (2.13)

with C₆ε,p= max(C₄ε,p, C₅ε,p+ C₁ε,p), which implies

6 Z R |∂xρε(t, x)|2Ndx + Z R ρ2N_ε (t, x) dx

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Integrating (2.14) with respect to t on [0, T ], we get

2 Z T 0 Z R |∂_xρε(t, x)|2Ndx dt + Z T 0 Z R ρ2N_ε (t, x) dx dt ≤ 1 C₆ε,p eC6ε,pT − 1 Z R |∂xρε(0, x)|2N dx + Z R ρ2N_ε (0, x) dx . (2.16) 3 2.4 L∞ bound on an interval [0, T ] 4

With the previous estimates, we are now able to prove an L∞ bounds for the sequence {ρε}ε on an

5

interval [0, T ].

6

H1. Then there exists a constant ¯T > 0 such that ρε ∈ L∞ [0, T ] × R for any ε > 0, T < ¯T .

Furthermore ρε∈ L∞ [0, T ], W1,2N(R) and ρε, ∂xρε∈ L∞ [0, T ], L2N(R) ∩ L2N_{([0, T ] × R)} (2.17) and this sequence is uniformly bounded in these spaces with respect to ε.

7

Proof. Let ρε be a smooth solution of (1.4) with the same initial datum ρ0 ∈ H1. The relation

8 (2.13) for N = 1 gives 9 d dt Z R |∂xρε(t, x)|2dx + Z R ρ2_ε(t, x) dx ≤ C maxn1, kρε(t, ·)k2∞ oZ R |∂xρε(t, x)|2dx + Z R ρ2_ε(t, x) dx ,

for some constant C that does not depend on ε (since Wε is uniformly bounded). If no uniform

L∞-bound on ρε is available, we can use the Sobolev injection of H1(R) in L∞(R) and get

d

dtkρε(t, ·)k

2

H1 ≤ Ckρε(t, ·)k2_H1 + Ckρε(t, ·)k4_H1,

eventually updating the constant C. We set uε(t) = kρε(t, ·)k2_H1, then u0_ε≤ C(uε+ u2ε), which leads

to u0_ε uε − u 0 ε 1 + uε ≤ C. We obtain uε(t) ≤ C0eCt 1 − C0eCt , for any 0 ≤ t < −ln C0 C ,

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with C0=

u0

1 + u0

< 1, u0 = kρ0k2_H1. Notice that the initial datum is the same for all the sequence

and then u0 and C0 do not depend on ε. Setting T < ¯T := −

ln C0

C , we have kρε(t, ·)k2_H1 ≤ Ckρ0k2_H1, for any 0 ≤ t ≤ T, ε > 0.

Therefore, by Sobolev injection, ρε∈ L∞([0, T ] × R). Using the estimates of Propositions 2 and 3,

1

we get (2.17) with bounds independents of ε.

2

2.5 W2,p _{estimate for p = 2N}

3

To pass to the limit, we need also estimates in W2,p, which will provide, in the next section, with

4

the help of the equation, the necessary regularity in time. As in Section 2.3, let N ∈ N∗ and set

5

p = 2N .

6

W1,4N(R) ∩ H1(R) ∩ W2,2N(R). Let T > 0 as in Proposition 4. Then ρε∈ L∞ [0, T ], W2,2N(R) and ρε, ∂xρε, ∂xx2 ρε∈ L∞ [0, T ], L2N(R) ∩ L2N_{([0, T ] × R)} and this sequence is bounded in these spaces with respect to ε.

7

Proof. We differentiate (2.10) with respect to x, which gives

8

∂t∂xx2 ρε+ 3v00(ρε∗ ωε) (∂x(ρε∗ ωε))2∂xρε+ 3v0(ρε∗ ωε) ∂2xx(ρε∗ ωε) ∂xρε

+3v0(ρε∗ ωε) ∂x(ρε∗ ωε) ∂xx2 ρε+ v(ρε∗ ωε) ∂xxx3 ρε+ ρεv(3)(ρε∗ ωε) (∂x(ρε∗ ωε))3 (2.18)

+3ρεv00(ρε∗ ωε) ∂x(ρε∗ ωε) ∂xx2 (ρε∗ ωε) + ρεv0(ρε∗ ωε)∂xxx3 (ρε∗ ωε) = ε∂xxxx4 ρε.

Multiplying this relation by (∂_xx2 ρε)p−1, then integrating with respect to x, we obtain

9 1 p d dt Z R (∂_xx2 ρε)pdx + 3 Z R v00(ρε∗ ωε) (∂x(ρε∗ ωε))2∂xρε(∂xx2 ρε)p−1dx +3 Z R v0(ρε∗ ωε) ∂xx2 (ρε∗ ωε) ∂xρε(∂xx2 ρε)p−1dx + 3 Z R v0(ρε∗ ωε) ∂x(ρε∗ ωε) (∂xx2 ρε)pdx + Z R v(ρε∗ ωε) ∂xxx3 ρε(∂xx2 ρε)p−1dx + Z R ρεv(3)(ρε∗ ωε) (∂x(ρε∗ ωε))3(∂xx2 ρε)p−1dx +3 Z R ρεv00(ρε∗ ωε) ∂x(ρε∗ ωε) ∂xx2 (ρε∗ ωε) (∂xx2 ρε)p−1dx + Z R ρεv0(ρε∗ ωε)∂xxx3 (ρε∗ ωε) (∂xx2 ρε)p−1dx = ε Z R ∂_xxxx4 ρε(∂xx2 ρε)p−1dx. Now 10 Z R v(ρε∗ ωε) ∂xxx3 ρε(∂xx2 ρε)p−1dx = 1 p Z R v(ρε∗ ωε) ∂x (∂_xx2 ρε)p dx = −1 p Z R v0(ρε∗ ωε) ∂x(ρε∗ ωε) (∂xx2 ρε)pdx,

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therefore 1 d dt Z R (∂_xx2 ρε)pdx = −3p Z R v00(ρε∗ ωε) (∂x(ρε∗ ωε))2∂xρε(∂xx2 ρε)p−1dx −3p Z R v0(ρε∗ ωε) ∂xx2 (ρε∗ ωε) ∂xρε(∂xx2 ρε)p−1dx +(1 − 3p) Z R v0(ρε∗ ωε) ∂x(ρε∗ ωε) (∂xx2 ρε)pdx −p Z R ρεv(3)(ρε∗ ωε) (∂x(ρε∗ ωε))3(∂xx2 ρε)p−1dx −3p Z R ρεv00(ρε∗ ωε) ∂x(ρε∗ ωε) ∂xx2 (ρε∗ ωε) (∂xx2 ρε)p−1dx −p Z R ρεv0(ρε∗ ωε)∂xxx3 (ρε∗ ωε) (∂xx2 ρε)p−1dx +εp Z R ∂_xxxx4 ρε(∂xx2 ρε)p−1dx =: J1+ J2+ J3+ J4+ J5+ J6+ J7.

We estimate now each of these terms.

2

• Using (2.1) and Young’s inequality, we get

3 |J1| = 3p Z R v00(ρε∗ ωε) (∂x(ρε∗ ωε))2∂xρε(∂xx2 ρε)p−1dx ≤ 3p kv00k∞ 2kρεk∞Wε 2 Z R |∂_xρε| |∂xx2 ρε|p−1dx ≤ 12 kv00k∞kρεk2∞Wε2 Z R |∂xρε|pdx + (p − 1) Z R |∂_xx2 ρε|pdx • |J2| = 3p Z R v0(ρε∗ ωε) ∂xx2 (ρε∗ ωε) ∂xρε(∂xx2 ρε)p−1dx ≤ 3p kv0k∞ Z R |∂_xx2 (ρε∗ ωε)| |∂xρε| |∂xx2 ρε|p−1dx ≤ 3 kv0k∞ 1 2 Z R |∂_xx2 (ρε∗ ωε)|2pdx + 1 2 Z R |∂_xρε|2pdx + (p − 1) Z R |∂_xx2 ρε|pdx

using the inequality uvw ≤ 1 p1 up1 ₊ 1 p2 vp2₊ 1 p3 wp3_, _with 1 p1 + 1 p2 + 1 p3 = 1, (2.19)

setting p1 = 2p = p2 and p3 = p/(p − 1). Estimate (2.19) can be derived applying twice the

4

classical Young’s inequality to uvw = u(vw). Using now the relation (2.12) with 2p at the

5 place of p, we get 6 |J₂| ≤ 3 2p+1 2 kv 0_k ∞η Z η 0 |ω00(u)|2p/(2p−1)du 2p−1 Z R ρ2p_ε dx

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+3 2p+1 2 kv 0_k ∞ |ω0(η−)| + |ω0(0+)| 2p Z R ρ2p_ε dx +32p+122p−1kv0k∞ ω(η) + 2ω(0) 2p Z R |∂xρε|2pdx +3 2kv 0_k ∞ Z R |∂_xρε|2pdx + 3(p − 1) kv0k∞ Z R |∂_xx2 ρε|pdx. • |J₃| = (3p − 1) Z R v0(ρε∗ ωε) ∂x(ρε∗ ωε) (∂xx2 ρε)pdx ≤ 2(3p − 1) kv0k∞kρεk∞Wε Z R |∂_xx2 ρε|pdx. • |J4| = p Z R ρεv(3)(ρε∗ ωε) (∂x(ρε∗ ωε))3(∂xx2 ρε)p−1dx ≤ p kv(3)_k ∞ 2kρεk∞Wε 3 Z R ρε|∂xx2 ρε|p−1dx ≤ 8 kv(3)k∞kρεk3∞Wε3 Z R ρp_εdx + (p − 1) Z R |∂_xx2 ρε|pdx

using Young’s inequality.

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≤ kρεk∞kv0k∞ Z R |∂_xxx3 (ρε∗ ωε)|pdx + (p − 1) Z R |∂_xx2 ρε|pdx . Estimate (2.3) of Proposition 1 and the inequality (2.11) give

The above estimates give an estimate of the form

Note that C₇p is a constant since kρεk∞Wε is bounded with respect to ε thanks to Proposition 4

5

and 1.5. This estimate, combined with (2.13) and (2.14), give then

6 d dt Z R |∂2 xxρε(t, x)|2N dx + Z R |∂_xρε(t, x)|2Ndx + Z R ρ2N_ε (t, x) dx

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Note that C₆ε,2N is bounded with respect to ε thanks to Proposition 4 and 1.5. Since an inequality of the form

u0(t) ≤ K1u(t) + K2eK3t

implies the estimate

u(t) ≤ u(0)eK1t_{+ K} 2eK1t Z t 0 e(K3−K1)s_{ds ≤ u(0)e}K1t₊K2 K3 e(K1+K3)t_{− e}K1t_,

we get the estimate

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3 Proof of Theorem 1

1

In this section, we pass to the limit as ε → 0 and we show that the limit function ρ satisfies equation

2

(1.1).

3

Using Proposition 2, the sequence {ρε}_ε is bounded in L∞([0, T ], L2(R)). Using Proposition 3,

the sequence {∂xρε}ε is bounded in L∞([0, T ], L2(R)). Using Propositions 1 and 4, the sequences

v(ρε∗ ωε) ε,v 0_(ρ ε∗ ωε) ε and ∂x(ρε∗ ωε) ε are bounded in L ∞_{([0, T ] × R). Then} ∂x(ρεv(ρε∗ ωε)) = ∂xρε · v(ρε∗ ωε) + ρεv0(ρε∗ ωε) ∂x(ρε∗ ωε)

is bounded in L∞([0, T ], L2_{(R)). Using Proposition 5, we also have a bound with respect to ε for}

∂_xx2 ρε in the space L∞([0, T ], L2(R)), then

∂tρε= ε∂xx2 ρε− ∂x(ρεv(ρε∗ ωε)) ∈ L∞([0, T ], L2(R))

uniformly with respect to ε. In particular, ρε ∈ C([0, T ], L2(R)) and the sequence is bounded in

4

this space. Since ∂tρε, ∂xρε∈ L∞([0, T ], L2(R)) with uniform bounds with respect to ε, then {ρε}ε

5

is bounded in H1_loc([0, T ] × R). Up to the extraction of a subsequence, the sequence {ρε}ε converges

6

to some ρ in L2_loc([0, T ] × R) and a.e. We have now to prove that the limit ρ ∈ C([0, T ], L2(R)) is

7 a solution of (1.1). Since 8 (ρε∗ ωε)(t, x) − (ρ ∗ ω)(t, x) = Z 0 −ε ρε(t, x + y)ωε(y) dy + Z η 0 (ρε− ρ)(t, x + y)ω(y) dy + Z η+ε η ρε(t, x + y)ωε(y) dy

tends to 0 when ε goes to zero, we have

ρεv(ρε∗ ωε) → ρv(ρ ∗ ω) a.e.

Therefore using dominated convergence Theorem, we get ρεv(ρε∗ωε) → ρv(ρ∗ω) in L1_loc([0, T ]×R),

9

implying that ρ is a solution of (1.1).

10

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