On Tra�c Flow with Nonlocal Flux: A Relaxation

Representation

Alberto Bressan and Wen Shen

Department of Mathematics, Penn State University.

University Park, PA 16802, USA.

e-mails: axb62@psu.edu, wxs27@psu.edu

December 13, 2019

Abstract

We consider a conservation law model of tra�c flow, where the velocity of each cardepends on a weighted average of the tra�c density ⇢ ahead. The averaging kernelis of exponential type: w

"

(s) = "

�1e

�s/". By a transformation of coordinates, theproblem can be reformulated as a 2 ⇥ 2 hyperbolic system with relaxation. UniformBV bounds on the solution are thus obtained, independent of the scaling parameter". Letting " ! 0, the limit yields a weak solution to the corresponding conservationlaw ⇢

t

+ (⇢v(⇢))x

= 0. In the case where the velocity v(⇢) = a � b⇢ is a�ne, usingthe Hardy-Littlewood rearrangement inequality we prove that the limit is the uniqueentropy-admissible solution to the scalar conservation law.

1 Introduction

We consider a nonlocal PDE model for tra�c flow, where the tra�c density ⇢ = ⇢(t, x)satisfies a scalar conservation law with nonlocal flux

⇢

t

+ (⇢v(q))x

= 0. (1.1)

Here ⇢ 7! v(⇢) is a decreasing function, modeling the velocity of cars depending on thetra�c density, while the integral

q(x) =

Z

+1

0

w(s) ⇢(x+ s) ds (1.2)

computes a weighted average of the car density. On the function v and the averaging kernelw, we shall always assume

(A1) The function v : [0, ⇢jam

] 7! R+

is C2, and satisfies

v(⇢jam

) = 0, v

0(⇢) � �⇤ < 0, for all ⇢ 2 [0, ⇢jam

]. (1.3)

1

(A2) The weight function w 2 C1(R+

) satisfies

w

0(s) 0,

Z

+1

0

w(s) ds = 1. (1.4)

In (A1) one can think of ⇢jam

as the maximum possible density of cars along the road,when all cars are packed bumper-to-bumper and nobody moves. At a later stage, morespecific choices for the functions w and v will be made. In particular, we shall focus on thecase where w(s) = e

�s.

The conservation equation (1.1) will be solved with initial data

⇢(0, x) = ⇢(x) 2 [0, ⇢jam

] . (1.5)

Given a weight function w satisfying (1.4), we also consider the rescaled weights

w

"

(s).

= "

�1

w(s/") . (1.6)

As " ! 0+, the weight w

"

converges to a Dirac mass at the origin, and the nonlocalequation (1.1) formally converges to the scalar conservation law

⇢

t

+ f(⇢)x

= 0, where f(⇢) = ⇢v(⇢). (1.7)

The main purpose of this paper is to analyze the convergence of solutions of the nonlocalequation (1.1) to those of (1.7).

Conservation laws with nonlocal flux have attracted much interest in recent years, becauseof their numerous applications and the analytical challenges they pose. Applications ofnonlocal models include sedimentation [6], pedestrian flow and crowd dynamics [2, 17, 18,19], tra�c flow [7, 14], synchronization of oscillators [3], slow erosion of granular matter [4],materials with fading memory [10], some biological and industrial models [20], and manyothers. Due to the nonlocal flux, the equation (1.1) behaves very di↵erently from theclassical conservation law (1.7). Its analysis faces additional di�culties and requires noveltechniques.

For a fixed weight function w, the well posedness of the nonlocal conservation laws wasproved in [7] with a Lax-Friedrich type numerical approximation, in [26] by the methodof characteristics, and in [23] using a Godunov type scheme. Traveling waves for relatednonlocal models have been recently studied in [13, 31, 32, 33, 34]. See also the results forseveral space dimensions [1], and other related results in [21, 36].

Up to this date, however, the nonlocal to local limit for (1.1) as " ! 0+ has remaineda challenging question. Namely, is it true that the solutions of the Cauchy problem ⇢

"

of (1.1)-(1.2), with averaging kernels w

"

in (1.6), as " ! 0+ converge to the entropyadmissible solutions of (1.7)? The question was already posed in [5]. For a general weightfunction w(·), whose support covers an entire neighborhood of the origin, a negative answeris provided by the counterexamples in [14]. On the other hand, the results in [14] do notapply to the physically relevant models where the velocity v is a monotone decreasingfunction and each driver only takes into account the density of tra�c ahead (not behind)the car. Indeed, existence and uniqueness result for this more realistic model are givenin [7, 12]. Furthermore, various numerical simulations [5, 7] suggest that the behavior of

2

⇢

"

should be stable in the limit " ! 0+. See also [16] for the e↵ect of numerical viscosityin the study of this limit. In the case of monotone initial data, a convergence result wasrecently proved in [25].

The main goal of the present paper is to study the limit behavior of solutions to (1.1),for the averaging kernel w

"

(s) = "

�1 exp(�s/"), as " ! 0. In this setting, we first showthat (1.1) can be treated as a 2⇥2 system with relaxation, in a suitable coordinate system.This formulation allows us to obtain a uniform bound on the total variation, independentof ". As " ! 0, a standard compactness argument yields the convergence ⇢

"

! ⇢ in L1

loc

,for a weak solution ⇢ of (1.7). Finally, in the case of a Lighthill-Whitham speed [28, 35]of the form v(⇢) = a � b⇢, we prove that the limit solution ⇢ coincides with the uniqueentropy weak solution of (1.7).

The remainder of the paper is organized as follows. Section 2 contains a short proof of globalexistence, uniqueness, and continuous dependence on the initial data, for solutions to (1.1)-(1.2) with v, w, satisfying (A1)-(A2). For Lipschitz continuous initial data, solutions areconstructed locally in time, as the fixed point of a contractive transformation. By suitablea priori estimates, we then show that these Lipschitz solutions can be extended globally intime. In turn, the semigroup of Lipschitz solutions can be continuously extended (w.r.t. theL1 distance) to a domain containing all initial data with bounded variation.

Starting with Section 3, we restrict our attention to exponential kernels: w"

(s) = "

�1

e

�s/".In this case, the conservation law with nonlocal flux can be reformulated as a hyperbolicsystem with relaxation. In Section 4, by a suitable transformation of independent anddependent coordinates, we establish a priori BV estimates which are independent of therelaxation parameter ". We assume here that the initial density is uniformly positive. Bya standard compactness argument, in Section 5 we construct the limit of a sequence ofsolutions with averaging kernels w

"

, as " ! 0. It is then an easy matter to show that anysuch limit provides a weak solution to the conservation law (1.7). A much deeper issueis whether this limit coincides with the unique entropy-admissible solution. In Section 6we prove that this is indeed true, in the special case where the velocity function is a�ne:v(⇢) = a� b⇢. This allows a detailed analysis of the convex entropy ⌘(⇢) = ⇢

2. Using theHardy-Littlewood rearrangement inequality [24, 27], we show that the entropy productionis O(1) · ". Hence, in the limit as "! 0, this entropy is dissipated.

We leave it as an open question to understand whether the same result is valid for moregeneral velocity functions v(·). Say, for v(⇢) = a � b⇢

2. Moreover, all of our techniquesheavily rely on the fact that the averaging kernel w(·) is exponential. It would be of muchinterest to understand what happens for di↵erent kind of kernels.

2 Existence of solutions

In this section we consider the Cauchy problem for (1.1)-(1.2), for a given initial datum

⇢(0, x) = ⇢(x). (2.1)

We consider the domain

D .

=n

⇢ 2 L1(R) ; Tot.Var.{⇢} < 1, ⇢(x) 2 [0, ⇢jam

] for all x 2 Ro

. (2.2)

3

Theorem 2.1 Under the assumptions (A1) and (A2), there exists a unique semigroupS : [0,+1[⇥D 7! D, continuous in L1

loc

, such that each trajectory t 7! S

t

⇢ is a weaksolution to the Cauchy problem (1.1)-(1.2), (2.1).

Proof. We first construct a family of Lipschitz solutions, and show that they dependcontinuously on time and on the initial data, in the L1 distance. By an approximationargument, we then construct solutions for general BV data ⇢ 2 D.

1. Consider the domain of Lipschitz functions

DL

.

=n

⇢ 2 D ; infx

⇢(x) > 0, supx

⇢(x) < ⇢

jam

,

|⇢(x)� ⇢(y)| L|x� y| for all x, y 2 Ro

. (2.3)

For every initial datum ⇢ 2 DL

, we will construct a solution t 7! ⇢(t, ·) 2 D2L

as the uniquefixed point of a contractive transformation, on a suitably small time interval [0, t

0

].

Given any function t 7! ⇢(t, ·) 2 D2L

, consider the corresponding integral averages

q(t, x) =

Z 1

0

w(s)⇢(t, x+ s) ds . (2.4)

We observe that

q

x

(t, x) =

Z 1

0

w(s) ⇢x

(t, x+ s) ds.

Hencekq

x

(t, ·)kL1 k⇢x

(t, ·)kL1 2L . (2.5)

Moreover, an integration by parts yields

q

xx

(t, x) =

Z 1

0

w(s) ⇢xx

(t, x+ s) ds = �w(0)⇢x

(t, x)�Z 1

0

w

0(s) ⇢x

(t, x+ s) ds ,

therefore

kqxx

(t, ·)kL1 w(0)k⇢x

(t, ·)kL1 + kw0kL1 · k⇢x

(t, ·)kL1

= 2w(0)k⇢x

(t, ·)kL1 4Lw(0). (2.6)

Consider the transformation ⇢ 7! u = �(⇢), where u is the solution to the linear Cauchyproblem

u

t

+ (v(q)u)x

= 0, u(0, x) = ⇢(x), t 2 [0, t0

], (2.7)

with q as in (2.4). In the next two steps we shall prove:

(i) The values �(u) remain uniformly bounded in the W

1,1 norm.

(ii) The map � : D2L

7! D2L

is contractive w.r.t. the C0 norm.

4

By the contraction mapping theorem, a unique fixed point will thus exist, providing thesolution to (2.7) on the time interval [0, t

0

].

2. To fix the ideas, assume

0 < �

0

⇢(x) ⇢

jam

� �

0

, (2.8)

for some �

0

. From the equation

u

t

+ v(q)ux

= � v

0(q)qx

u , u(0, x) = ⇢, (2.9)

integrating along characteristics and using (2.5), we obtain

�

0

� t ·⇣

kv0kL1 2L⌘

⇢

jam

u(t, x) ⇢

jam

� �

0

+ t ·⇣

kv0kL1 2L⌘

⇢

jam

. (2.10)

Choosing t

0

< �

0

· (kv0kL1 2L)�1 · ⇢�1

jam

, the solution u will thus remain strictly positiveand smaller than ⇢

jam

, for all t 2 [0, t0

].

3. Di↵erentiating the conservation law in (2.7) we obtain

u

xt

+ v(q)uxx

= � 2v0(q)qx

u

x

� [v00(q)q2x

+ v

0(q)qxx

]u. (2.11)

Let Z(t) be the solution to the ODE

Z = aZ + b, Z(0) = L ,

wherea

.

= 2kv0kL1 · 2L , b

.

=h

4L2kv00kL1 + 4Lw(0)kv0kL1

i

· ⇢jam

.

Sinceku

x

(0, ·)kL1 = k⇢x

kL1 L ,

in view of (2.11) and the bounds (2.5)–(2.6), a comparison argument yields

kux

(t, ·)kL1 Z(t). (2.12)

In particular, for t 2 [0, t0

] with t

0

su�ciently small, we have

kux

(t, ·)kL1 2L . (2.13)

4. Using the identity

q

x

(t, x) = � w(0)⇢(t, x)�Z 1

0

w

0(s)⇢(t, x+ s) ds

and recalling that w0(s) 0, one obtains the bound

kqx

(t, ·)kL1 2w(0)k⇢(t, ·)kL1. (2.14)

Next, consider two functions t 7! ⇢

1

(t, ·), t 7! ⇢

2

(t, ·), both taking values inside D2L

. Then,for all t 2 [0, t

0

], the corresponding weighted averages q1

, q

2

satisfy

kq1

(t, ·)� q

2

(t, ·)kW

1,1 (1 + 2w(0)) · sup⌧2[0,t0]

k⇢1

(⌧, ·)� ⇢

2

(⌧, ·)kL1. (2.15)

5

By choosing t

0

> 0 small enough, we claim that the corresponding solutions u1

, u

2

of (2.7)satisfy

ku1

(t, ·)� u

2

(t, ·)kL1 1

2sup

⌧2[0,t]k⇢

1

(⌧, ·)� ⇢

2

(⌧, ·)kL1 for all t 2 [0, t0

] . (2.16)

Indeed, consider a point (⌧, y). Call t 7! x

i

(t), i = 1, 2, the corresponding characteristics.These solve the equations

x

i

= v(qi

(t, xi

(t))), x

i

(⌧) = y. (2.17)

Hence, moving backward in time, we have

� d

dt

|x1

(t)� x

2

(t)|

�

�

�

v(q1

(t, x1

(t)))� v(q1

(t, x2

(t)))�

�

�

+�

�

�

v(q1

(t, x2

(t)))� v(q2

(t, x2

(t)))�

�

�

kv0kL1kq1,x

kL1 · |x1

(t)� x

2

(t)|+ kv0kL1kq1

� q

2

kL1.

By (2.5), the quantity kq1,x

(t, ·)kL1 remains uniformly bounded. The distance Z(t).

=|x

1

(t)� x

2

(t)| between the two characteristics thus satisfies a di↵erential inequality of theform

� d

dt

Z(t) a⇤Z(t) + b⇤kq1(t, ·)� q

2

(t, ·)kL1, Z(⌧) = 0,

for some constants a⇤, b⇤. This implies

|x1

(t)� x

2

(t)| Z

⌧

t

e

(t�s)a⇤ · b⇤kq1(s, ·)� q

2

(s, ·)kL1ds . (2.18)

The values u

i

(⌧, y), i = 1, 2, can now be obtained by integrating along characteristics.Indeed,

d

dt

u

i

(t, xi

(t)) = v

0(qi

(t, xi

(t))) · qi,x

(t, xi

(t)) · ui

(t, xi

(t)), u

i

(0, xi

(0)) = ⇢(xi

(0)).

Thanks to the a priori bounds (2.6) on kqi,xx

(t, ·)kL1 , using (2.18) for any ✏ > 0 we canchoose t

0

> 0 such that

|u1

(⌧, y)� u

2

(⌧, y)| ✏ · supt2[0,⌧ ]

kq1

(t, ·)� q

2

(t, ·)kL1,

for all ⌧ 2 [0, t0

] and y 2 R. In view of (2.15), this implies (2.16).

5. By the contraction mapping principle, there exists a unique function t 7! ⇢(t, ·) suchthat ⇢(t, ·) = u(t, ·) for all t 2 [0, t

0

]. This fixed point of the transformation � provides theunique solution to the Cauchy problem (1.1)-(1.2) with initial data (2.1).

6. In this step we show that this solution can be extended to all times t > 0. This requires(i) a priori upper and lower bounds of the form

0 < �

0

⇢(t, x) ⇢

jam

� �

0

, (2.19)

6

independent of time, and (ii) a priori estimates on the Lipschitz constant, which shouldremain uniformly bounded on bounded intervals of time.

To establish an upper bound on the solution ⇢(t, ·), t 2 [0, t0

], we analyze its behavior alonga characteristic. Fix ✏ > 0. Consider any point (⌧, ⇠) such that

⇢(⌧, ⇠) � supx2R

⇢(⌧, x)� ✏.

At the point (⌧, ⇠) one has

⇢

t

+ v(q)⇢x

= �⇢v0(q)qx

= � ⇢(⌧, ⇠)v0(q(⌧, ⇠)) · @@⇠

Z

+1

⇠

⇢(⌧, y)w(y � ⇠) dy

�

= �⇢(⌧, ⇠)v0(q(⌧, ⇠)) ·

�⇢(⌧, ⇠)w(0)�Z

+1

⇠

⇢(⌧, y)w0(y � ⇠) dy

�

= �⇢(⌧, ⇠)v0(q(⌧, ⇠)) ·Z

+1

⇠

⇥

⇢(⌧, ⇠)� ⇢(⌧, y)⇤

w

0(y � ⇠) dy

⇢

jam

· max0q⇢jam

|v0(q)| · w(0) · ✏ .

= C

0

✏. (2.20)

The above impliesd

dt

✓

supx

⇢(t, x)

◆

C

0

✏,

as long as 0 < ⇢(t, y) < ⇢

jam

for all y 2 R.

Since ⇢ satisfies (2.8) and ✏ > 0 is arbitrary, this establishes the upper bound in (2.19).The lower bound is proved in an entirely similar way.

Next, from the analysis in step 3 it follows

k⇢x

(t, ·)kL1 Z(t), (2.21)

which immediately yields the a priori bound on the Lipschitz constant.

By induction, we can thus construct a unique solution ⇢ = ⇢(t, x) on a sequence of timeintervals [0, t

0

], [t0

, t

1

], [t1

, t

2

], . . ., where the length of each interval [tk

t

k+1

] depends onlyon (i) the constant �

0

in (2.19), and (ii) the Lipschitz constant of ⇢(tk

, ·). Thanks to (2.21),this Lipschitz constant remains Z(t

k

). This implies t

k

! +1 as k ! 1, hence thesolution can be extended to all times t > 0.

We remark that, by a further di↵erentiation of the basic equation (1.1), one can prove that,if ⇢ 2 C

k, then every derivatives up to order k remains uniformly bounded on boundedintervals of time.

7. To complete the proof, it remains to show that the semigroup of solutions can beextended by continuity to all initial data ⇢ 2 D.

Toward this goal, we first prove that the total variation of the solution ⇢(t, ·) remainsuniformly bounded on bounded time intervals. Indeed, from

⇢

xt

+ (v(q)⇢x

)x

= � (v0(q)qx

⇢)x

,

7

it follows

d

dt

k⇢x

kL1 k(v0(q)qx

⇢)x

kL1

kv0kL1kqx

kL1k⇢x

kL1 + kv0kL1kqxx

kL1k⇢kL1 + kv00kL1kqx

kL1kqx

kL1k⇢kL1

Ck⇢x

kL1 . (2.22)

Above we used the estimates

kqx

kL1 k⇢x

kL1 , kqxx

kL1 2w(0) · k⇢x

kL1 . (2.23)

Note that in (2.22) the constant C depends on the velocity function v : [0, ⇢jam

] 7! R+

and the averaging kernel w, but it does not depend on the Lipschitz constant k⇢x

kL1 ofthe solution. According to (2.22), the total variation of the solution grows at most at anexponential rate. In particular, it remains bounded on bounded intervals of time.

8. Thanks to the a priori bounds (2.22) on the total variation and (2.12) on the Lipschitzconstant, the solution can be extended to an arbitrarily large time interval [0, T ]. Thisalready defines a family of trajectories t 7! S

t

⇢ defined for every L > 0, every ⇢ 2 DL

, andt � 0.

In order to extend the semigroup S by continuity to the entire domain D, we need to provethat for every t > 0 the map ⇢ 7! S

t

⇢ is Lipschitz continuous w.r.t. the L1 distance.

Indeed, consider a family of smooth solutions, say ⇢

✓(t, ·), ✓ > 0. Define the first orderperturbations

⇣

✓(t, ·) = limh!0

⇢

✓+h(t, ·)� ⇢

✓(t, ·)h

, Q

✓(t, ·) = limh!0

q

✓+h(t, ·)� q

✓(t, ·)h

.

Notice that

Q

✓(t, x) =

Z

+1

0

w(s) ⇣✓(t, x+ s) ds.

Then ⇣✓ satisfies the linearized equation

⇣

t

+ (v(q)⇣)x

+�

v

0(q)Q⇢�

x

= 0, (2.24)

where for simplicity we dropped the upper indices. Using the estimates

kQ(t, ·)kL1 k⇣(t, ·)kL1 , kQx

(t, ·)kL1 2w(0) · k⇣(t, ·)kL1 , (2.25)

kqx

(t, ·)kL1 2w(0) · ⇢jam

, kQ(t, ·)kL1 w(0) · k⇣(t, ·)kL1 , (2.26)

we compute

d

dt

k⇣(t, ·)kL1 k(v0(q)Q⇢)x

kL1

kv00kL1kqx

kL1kQkL1k⇢kL1 + kv0kL1kQx

kL1k⇢kL1 + kv0kL1kQkL1k⇢x

kL1

C(t) · k⇣(t, ·)kL1 . (2.27)

Here C(t) depends on time because the total variation k⇢x

(t, ·)kL1 may grow at an expo-nential rate. On the other hand, it is important to observe that C(t) does not depend onthe Lipschitz constant of the solutions. From (2.27) we deduce

k⇣(t, ·)kL1 exp

⇢

Z

t

0

C(⌧) d⌧

�

k⇣(0, ·)kL1 . (2.28)

8

For any two Lipschitz solutions ⇢0, ⇢1 of (1.1)-(1.2), we now construct a 1-parameter familyof solutions ⇢✓(t, ·) with initial data

⇢

✓(0, ·) = ✓⇢

1(0, ·) + (1� ✓)⇢0(0, ·).Using (2.28) one obtains

k⇢1(t, ·)� ⇢

0(t, ·)kL1 Z

1

0

k⇣✓(t, ·)kL1 d✓ Z

1

0

exp

⇢

Z

t

0

C(⌧) d⌧

�

· k⇣✓(0, ·)kL1 d✓

exp

⇢

Z

t

0

C(⌧) d⌧

�

· k⇢1(0, ·)� ⇢

0(0, ·)kL1 . (2.29)

This establishes Lipschitz continuity of the semigroup w.r.t. the initial data. Notice thatthis Lipschitz constant may well depend on time. Since every initial datum ⇢ 2 D can beapproximated in the L1 distance by a sequence of Lipschitz continuous functions ⇢

n

2 DLn

(possibly with L

n

! +1), by continuity we obtain a unique semigroup defined on theentire domain D.

Remark 2.1 By the argument in step 6 of the above proof, if the initial condition satisfies

0 a ⇢(x) b ⇢

jam

for all x 2 R,

then the solution satisfies

a ⇢(t, x) b for all t � 0, x 2 R.

3 A hyperbolic system with relaxation

From now on, we focus on the case where w(s) = e

�s, so that the rescaled kernels are

w

"

(s) = "

�1

e

�s/"

.

This yields

@

@x

Z

+1

x

⇢(t, s)1

"

e

�(s�x)/"

ds

�

= � 1

"

⇢(t, x) +1

"

Z

+1

x

⇢(t, s)1

"

e

�(s�x)/"

ds. (3.1)

Therefore, the averaged density q satisfies the ODE

q

x

= "

�1

q � "

�1

⇢ .

The conservation law with nonlocal flux (1.1)-(1.2) can thus be written as(

⇢

t

+ (⇢v(q))x

= 0,

q

x

= "

�1(q � ⇢) .(3.2)

To make further progress, we choose a constant K > v(0) and consider new independentcoordinates (⌧, y) defined by

⌧ = t� x

K

, y = x . (3.3)

9

For future use, we derive the relations between the partial derivative operators in these twosets of coordinates:

@

⌧

= @

t

, @

y

= @

x

+K

�1

@

t

, @

x

= @

y

�K

�1

@

⌧

. (3.4)

A direct computation yields

⇢

t

= ⇢

⌧

, (⇢v(q))x

= �K

�1(⇢v(q))⌧

+ (⇢v(q))y

, q

x

= �K

�1

q

⌧

+ q

y

.

In these new coordinates, the equations (3.2) take the form

8

<

:

(K⇢� ⇢v(q))⌧

+ (K⇢v(q))y

= 0,

q

⌧

�Kq

y

=K

"

(⇢� q).(3.5)

One can easily verify that the above system of balance laws is strictly hyperbolic, with twodistinct characteristic speeds

�

1

= �K, �

2

=Kv(q)

K � v(q). (3.6)

We observe that �1

< 0 < �

2

, provided that K is su�ciently large such that K > v(0).Moreover, both characteristic families are linearly degenerate.

In the zero relaxation limit, letting " ! 0+ one formally obtains q ! ⇢. Hence (3.5)formally converges to the scalar conservation law

(K⇢� ⇢v(⇢))⌧

+ (K⇢v(⇢))y

= 0. (3.7)

Recalling the function f defined in (1.7), one obtains

(K⇢� f(⇢))⌧

+ (Kf(⇢))y

= 0. (3.8)

Note that (3.8) is equivalent to the conservation law (1.7) in the original (t, x) coordinates.

The characteristic speed for (3.8) is

�

⇤ =Kf

0(⇢)

K � f

0(⇢)=

K

2

K � f

0(⇢)�K. (3.9)

Since K > v(0) � v(⇢) > f

0(⇢), we clearly have �⇤ > �K = �

1

. Furthermore, sincef

0(⇢) < v(⇢), we conclude that �⇤ < �

2

. The sub-characteristic condition

�

1

< �

⇤< �

2

(3.10)

is thus satisfied. This is a crucial condition for stability of the relaxation system, see [29].For other related general references on zero relaxation limit, we refer to [9, 11].

From (3.5) it follows

(K � v(q))⇢⌧

+Kv(q)⇢y

= ⇢

⇥

v(q)⌧

�Kv(q)y

⇤

= ⇢v

0(q)(q⌧

�Kq

y

) = ⇢v

0(q) · K"

(⇢� q).

10

We can thus write (3.5) in diagonal form:

8

>

<

>

:

⇢

⌧

+Kv(q)

K � v(q)⇢

y

=K

"

· (⇢� q) · ⇢v

0(q)

K � v(q),

q

⌧

�Kq

y

=K

"

· (⇢� q).(3.11)

To further analyze (3.11), it is convenient to introduce the new dependent variables

u = ln ⇢, z = ln(K � v(q)), (3.12)

so that⇢ = e

u

, v(q) = K � e

z

. (3.13)

Using these new variables, (3.11) becomes

8

>

<

>

:

u

⌧

+K(Ke

�z � 1)uy

=K

"

⇤(u, z),

z

⌧

�Kz

y

= �K

"

⇤(u, z),(3.14)

where the source term ⇤ is given by

⇤(u, z) = (⇢(u)� q(z))v

0(q(z))

K � v(q(z)). (3.15)

Introducing the monotone function

g(u) = ln(K � v(eu)), where g

0(u) =�v

0(eu)eu

K � v(eu)> 0, (3.16)

one checks that⇤(u, g(u)) = 0 for all u. (3.17)

Letting "! 0, we expect that z ! g(u) hence the system (3.14) formally converges to thescalar conservation law

(u+ g(u))⌧

+K(Ke

�g(u) � 1)uy

�Kg(u)y

= 0. (3.18)

Using the identities

u+ g(u) = ln(eu(K � v(eu))), e

�g(u) =1

K � v(eu), Ke

�g(u) � 1 =v(eu)

K � v(eu),

we get(eu(K � v(eu)))

⌧

e

u(K � v(eu))+

K(euv(eu))y

e

u(K � v(eu))= 0.

Writing ⇢ = e

u, we obtain once again the conservation law (3.7).

11

4 A priori BV bounds

In order to prove a rigorous convergence result, we need an a priori BV bound on thesolution to the system (3.14), independent of the relaxation parameter ". We always assumethat the velocity v satisfies the assumptions (A1).

Di↵erentiating (3.14) w.r.t. y one obtains

8

>

<

>

:

u

y⌧

+ [K(Ke

�z � 1)uy

]y

=K

"

[⇤u

u

y

+ ⇤z

z

y

],

z

y⌧

�Kz

yy

= � K

"

[⇤u

u

y

+ ⇤z

z

y

].(4.1)

A kinetic interpretation of the above system is shown in Figure 1.

Q

τ

P y

x

Figure 1: The new system of coordinates (⌧, y) defined at (3.3), is illustrated here together withthe original coordinates (t, x). The two characteristics through a point Q have speeds �1 < 0 < �2,as in (3.6). With reference to the system (4.1), one can think of z

y

as the density of backward-moving particles, with speed �1 = �K, while u

y

is the density of forward-moving particles, withspeed �2 > 0. Backward particles are transformed into forward particles at rate K⇤

z

/", whileforward particles turn into backward ones with rate �K⇤

u

/". The total number of particles doesnot increase; actually, it decreases when positive and negative particles of the same type cancel out.

We observe that

d

d⌧

Z

|uy

(⌧, y)| dy + d

d⌧

Z

|zy

(⌧, y)| dy

=K

"

Z

n

sign(uy

)[⇤u

u

y

+ ⇤z

z

y

]� sign(zy

)[⇤u

u

y

+ ⇤z

z

y

]o

dy

K

"

Z

n

⇤u

|uy

|+ |⇤z

| · |zy

|+ |⇤u

| · |uy

|� ⇤z

|zy

|o

dy.

Therefore, if⇤u

0, ⇤z

� 0, (4.2)

then the map⌧ 7! ku

y

(⌧, ·)kL1 + kzy

(⌧, ·)kL1

will be non-increasing. By (3.15), a direct computation yields

⇤u

= e

u

v

0(q(z))

K � v(q(z))< 0.

12

It remains to verify that ⇤z

� 0. Since @q

@z

> 0, it su�ces to show that ⇤q

� 0. We compute

⇤q

= (⇢� q)v

00(q)(K � v(q)) + (v0(q))2

(K � v(q))2� v

0(q)

K � v(q)

=1

K � v(q)

(⇢� q)⇣

v

00(q) +(v0(q))2

K � v(q)

⌘

� v

0(q)

�

. (4.3)

Since v

0(q) < 0, the above inequality will hold provided that

|⇢� q| ·✓

|v00(q)|+ (v0(q))2

K � v(q)

◆

�

�

v

0(q)�

� for all q. (4.4)

Notice that, by choosing K su�ciently large, the factor (v

0)

2

K�v

can be rendered as small aswe like. Hence we can always achieve the inequality (4.4) provided that:

• Either |⇢� q| remains small. This is certainly the case if the oscillation of the initialdatum is small.

• Or else, |v00| is small compared with |v0|.

As a consequence of the above analysis, we have:

Lemma 4.1 Let (u, z) be a Lipschitz solution to the relaxation system (4.1). Assume that⇢(⌧, y) = e

u(⌧,y) 2 [⇢1

, ⇢

2

] for all (⌧, y), and moreover

minq2[⇢1,⇢2]

|v0(q)| � (⇢2

� ⇢

1

) ·✓

kv00kL1 +kv0k2L1

K � kvkL1

◆

. (4.5)

Then the total variation function

⌧ 7! kuy

(⌧, ·)kL1(R) + kz

y

(⌧, ·)kL1(R) (4.6)

is non-increasing.

We observe that, in the case where v is a�ne, say

v(⇢) = a

o

� b

o

⇢ (4.7)

for some a

o

, b

o

> 0, by (1.3) we can always choose K large enough so that

⇢

jam

· kv0k2L1

K � kvkL1 min

0q⇢jam

|v0(q)| . (4.8)

Hence (4.5) is satisfied.

Our main goal is to obtain uniform BV bounds for solutions to the nonlocal conservationlaw (1.1)-(1.2). This will be achieved by working in the (⌧, y) coordinate system.

13

Theorem 4.1 Consider the Cauchy problem for (1.1)-(1.2), with kernel w(s) = "

�1

e

�s/".Assume that the velocity function v satisfies

min⇢2[0,⇢jam]

|v0(q)| > ⇢

jam

· kv00kL1([0,⇢jam])

. (4.9)

Moreover, assume that the initial density ⇢ has bounded variation and is uniformly positive.Namely,

0 < ⇢

min

⇢(x) ⇢

max

⇢

jam

for all x 2 R. (4.10)

Then the total variation remains uniformly bounded in time:

Tot.Var.{⇢(t, ·)} ⇢

max

⇢

min

· Tot.Var.{⇢} for all t � 0. (4.11)

Proof. 1. Assume first that ⇢ is Lipschitz continuous. By (4.1) it follows

div

u

y

K(Ke

�z � 1)uy

!

+ div

z

y

�Kz

y

!

= 0. (4.12)

Thanks to (4.9), we can choose a constant K large enough so that (4.2) holds. In this casewe also have

div

|uy

|K(Ke

�z � 1)|uy

|

!

+ div

|zy

|�K|z

y

|

!

0. (4.13)

In terms of the original (t, x) coordinates, by (3.4) the inequality (4.13) takes the form

@

t

⇣

�

�

�

u

x

+u

t

K

�

�

�

+�

�

�

z

x

+z

t

K

�

�

�

⌘

+

✓

1

K

@

t

+ @

x

◆

⇣

K(Ke

�z � 1)�

�

�

u

x

+u

t

K

�

�

�

�K

�

�

�

z

x

+z

t

K

�

�

�

⌘

= @

t

⇣

Ke

�z

�

�

�

u

x

+u

t

K

�

�

�

⌘

+ @

x

⇣

K(Ke

�z � 1)�

�

�

u

x

+u

t

K

�

�

�

�K

�

�

�

z

x

+z

t

K

�

�

�

⌘

0. (4.14)

2. Integrating (4.14) over any time interval [0, T ], we obtainZ

K

K � v(q(T, x))

�

�

�

u

x

(T, x) +u

t

(T, x)

K

�

�

�

dx

Z

K

K � v(q(0, x))

�

�

�

u

x

(0, x) +u

t

(0, x)

K

�

�

�

dx . (4.15)

Since we are choosing K > v(0) � v(q(t, x)) for all t, x, the above denominators remainuniformly positive and bounded. This implies

Z

�

�

�

u

x

(T, x) +u

t

(T, x)

K

�

�

�

dx C

K

·Z

�

�

�

u

x

(0, x) +u

t

(0, x)

K

�

�

�

dx , (4.16)

with C

K

.

= K

K�v(0)

.

Repeating the same argument, with K replaced by �K where � > 1, we obtainZ

�

�

�

u

x

(T, x) +u

t

(T, x)

�K

�

�

�

dx C

�K

·Z

�

�

�

u

x

(0, x) +u

t

(0, x)

�K

�

�

�

dx , (4.17)

14

where the constant is now C

�K

= �K

�K�v(0)

.

3. Next, we observe that, for any two numbers ↵,� and any number � > 1 one has

↵ =�

� � 1

✓

↵+�

�

◆

� 1

� � 1(↵+ �),

so

|↵| �

� � 1

�

�

�

�

↵+�

�

�

�

�

�

+1

� � 1|↵+ �|.

Applying the above inequality with ↵ = u

x

, � = K

�1

u

t

, from (4.16)-(4.17) one obtains

Z

�

�

u

x

(T, x)�

�

dx �C

�K

� � 1

Z

�

�

�

u

x

(0, x) +u

t

(0, x)

�K

�

�

�

dx

+C

K

� � 1

Z

�

�

�

u

x

(0, x) +u

t

(0, x)

K

�

�

�

dx . (4.18)

4. By the assumption (4.10) and Remark 2.1 it follows that

0 < ⇢

min

⇢(t, x) ⇢

max

for all t � 0, x 2 R.

By the change of variables (3.12)-(3.13), one has

|ux

| =|⇢

x

|⇢

|⇢x

|⇢

min

, |ut

| =|⇢

t

|⇢

|⇢t

|⇢

min

, |⇢x

| ⇢

max

|ux

| . (4.19)

Combining (4.19) with (4.18) we conclude

⇢

�1

max

Z

�

�

⇢

x

(T, x)�

�

dx Z

�

�

u

x

(T, x)�

�

dx

�C

�K

� � 1

Z

�

�

�

�

u

x

(0, x) +u

t

(0, x)

�K

�

�

�

�

dx +C

K

� � 1

Z

�

�

�

�

u

x

(0, x) +u

t

(0, x)

K

�

�

�

�

dx

�C

�K

(� � 1)⇢min

Z

|⇢x

(0, x)|+ |⇢t

(0, x)|�K

�

dx

+C

K

(� � 1)⇢min

Z

|⇢x

(0, x)|+ |⇢t

(0, x)|K

�

dx. (4.20)

We observe thatZ

|⇢t

(0, x)| dx Z

|⇢x

(x)v(q(0, x))|+ ��⇢(x)v0(q(0, x))qx

(0, x)�

�

dx

kvkL1 · k⇢x

kL1 + ⇢

max

· kv0kL1 · kqx

(0, ·)kL1 C

0

k⇢x

kL1 ,

where C

0

.

= kvkL1 + ⇢

max

· kv0kL1 is a bounded constant. Recalling the values of theconstants C

K

, C

�K

, from (4.20) we obtain

Z

�

�

⇢

x

(T, x)�

�

dx

⇢

max

⇢

min

· 1

� � 1

✓

�

2

K

�K � v(0)

⇣

1 +C

0

�K

⌘

+K

K � v(0)

⇣

1 +C

0

K

⌘

◆

· k⇢x

kL1 . (4.21)

15

Since the constant K can be chosen arbitrarily large, letting K ! +1 in (4.21) we obtain

k⇢x

(T, ·)kL1 ⇢

max

⇢

min

· � + 1

� � 1· k⇢

x

kL1 .

We note that asK ! 1, (4.5) reduces to (4.9). Again, since � > 1 can be chosen arbitrarilylarge, letting � ! 1 we obtain

k⇢x

(T, ·)kL1 ⇢

max

⇢

min

· k⇢x

kL1 . (4.22)

For any Lipschitz solution, this provides an a priori bound on the total variation, whichdoes not depend on time or on the relaxation parameter ". By an approximation argumentwe conclude that (4.11) holds, for every uniformly positive initial condition ⇢ with boundedvariation.

5 Existence of a limit solution

Relying on the a priori bound on the total variation, proved in Theorem 4.1, we now showthe existence of a limit ⇢ = lim

"!0+

⇢

"

, which provides a weak solution to the conservationlaw (1.7).

Theorem 5.1 Let ⇢ : R 7! [⇢min

, ⇢

max

] be a uniformly positive initial datum, with boundedvariation. Call ⇢

"

the corresponding solutions to (1.1)-(1.2), with averaging kernel w"

(s) ="

�1

e

�s/". Then, by possibly extracting a subsequence "n

! 0, one obtains the convergence⇢

"n ! ⇢ in L1

loc

(R+

⇥R). The limit function ⇢ provides a weak solution to the conservationlaw (1.7).

Proof. By Theorem 4.1, all solutions ⇢"

(t, ·) have uniformly bounded total variation. Thesame is thus true for the weighted averages q

"

(t, ·), where

q

"

(t, x) =

Z

+1

0

"

�1

e

�s/"

⇢

"

(t, x+ s) ds . (5.1)

By (1.1), this implies that the map t 7! ⇢

"

(t, ·) is uniformly Lipschitz continuous w.r.t. theL1 distance.

By a compactness argument based on Helly’s theorem (see for example Theorem 2.4 in [8]),we can select a sequence "

n

# 0 such that

⇢

"n ! ⇢ in L1

loc

(R+

⇥ R), (5.2)

⇢

"n(t, ·) ! ⇢(t, ·) in L1

loc

(R), for a.e. t � 0. (5.3)

By (5.1), it now follows

�

�

q

"

(t, ·)� ⇢

"

(t, ·)��L1 =

ZZ

x<y

"

�1

e

(x�y)/"

�

�

⇢

"

(t, y)� ⇢

"

(t, x)�

�

dy dx

ZZZ

x<s<y

"

�1

e

(x�y)/"

�

�

⇢

",x

(t, s)�

�

ds dy dx

=

Z

+1

�1

✓

Z

+1

0

Z

+1

0

"

�1

e

��/"

e

�⇠/"

d⇠ d�

◆

�

�

⇢

",x

(t, s)�

�

ds

= " · Tot.Var.{⇢"

(t, ·)},

16

where the variables � = y � s, ⇠ = s � x were used. Therefore, as "n

! 0, we have theconvergence q

"n ! ⇢ in L1

loc

. By (1.1), this implies that the limit function ⇢ = ⇢(t, x) is aweak solution to the scalar conservation law (1.7).

6 Entropy admissibility of the limit solution

In the previous section we proved that, as "! 0, any limit in L1

loc

of solutions u"

to (1.1),(1.5) with ⇢ 2 BV and q

"

given by (5.1) is a weak solution to the conservation law (1.7). Akey question is whether this limit is the unique entropy admissible solution. The followinganalysis shows that this is indeed the case when the velocity function is a�ne, namely

v(⇢) = a� b⇢ . (6.1)

Theorem 6.1 Let the velocity function v be a�ne. Consider any uniformly positive initialdatum ⇢ 2 BV . Then as " ! 0, the corresponding solutions ⇢

"

to (1.1), (5.1), (1.5)converge to the unique entropy admissible solution of (1.7).

Proof. For simplicity, we consider the case where v(⇢) = 1 � ⇢. The general case (6.1)is entirely similar. According to [22, 30], to prove uniqueness it su�ces to prove that thelimit solution dissipates one single strictly convex entropy. We thus consider the entropyand entropy flux pair

⌘(⇢) =⇢

2

2, (⇢) =

⇢

2

2� 2⇢3

3. (6.2)

When v(⇢) = 1� ⇢, the equation (1.1) can be written as

⇢

t

+ (⇢(1� ⇢))x

= (⇢(1� ⇢)� ⇢(1� q))x

= (⇢(q � ⇢))x

.

Multiplying both sides by ⌘0(⇢) = ⇢, we obtain

⌘(⇢)t

+ (⇢)x

= ⇢(⇢(q � ⇢))x

= (⇢2(q � ⇢))x

� (q � ⇢)⇢⇢x

. (6.3)

Given a test function ' 2 C1

c

(R), ' � 0, we thus need to estimate the quantity

J = J

1

� J

2

,

where

J

1

.

=

Z

(⇢2(q � ⇢))x

' dx = �Z

⇢

2(q � ⇢)'x

dx , (6.4)

J

2

.

=

Z

�

q(x)� ⇢(x)� · ⇢(x)⇢

x

(x)'(x) dx

=

Z

✓

Z

+1

x

1

"

e

(x�y)/"

✓

Z

y

x

⇢

x

(s) ds

◆

dy

◆

⇢(x) ⇢x

(x)'(x) dx . (6.5)

Our ultimate goal is to show that

J O(1) · ".

17

Since we have

|J1

| k⇢k2L1 · k'x

kL1 ·Z

|q(x)� ⇢(x)| dx = O(1) · ",

it remains to show thatJ

2

� O(1) · ". (6.6)

A key tool to achieve this estimate is

Lemma 6.1 (Hardy-Littlewood inequality). For any two functions g1

, g

2

� 0 vanish-ing at infinity, one has

Z

g

1

(x) g2

(x) dx Z

g

⇤1

(x)g⇤2

(x) dx, (6.7)

where g

⇤1

, g

⇤2

are the symmetric decreasing rearrangements of g1

, g

2

, respectively.

For a proof, see [24] or [27].

Starting from (6.5) we compute

J

2

=

Z Z Z

x<s<y

1

"

e

(x�y)/"

⇢

x

(s)⇢(x)⇢x

(x)'(x) dy ds dx

=

Z Z

x<s

e

(x�s)/"

⇢

x

(s) ⇢(x)⇢x

(x)'(x) dx ds

=

Z

✓

Z

+1

x

e

�s/"

⇢

x

(s) ds

◆

e

x/"

⇢(x)⇢x

(x)'(x) dx

= �Z

⇢

2(x)⇢x

(x)'(x) dx+1

"

Z Z

x<s

e

�s/"

⇢(s) ex/"⇢(x)⇢x

(x)'(x) dx ds

=

Z

⇢

3(x)

3'

x

(x) dx+1

"

Z

e

�s/"

⇢(s)

✓

Z

s

�1

⇣

⇢

2(x)

2

⌘

x

e

x/"

'(x) dx

◆

ds

= A+B � C �D,

where

A =

Z

⇢

3(x)

3'

x

(x) dx ,

B =1

"

Z

⇢(s)⇢

2(s)

2'(s) ds ,

C =1

"

2

Z Z

s

�1e

(x�s)/"

⇢

2(x)

2⇢(s)'(x) dx ds ,

D =1

"

Z Z

s

�1e

(x�s)/"

⇢

2(x)

2⇢(s)'

x

(x) dx ds .

To achieve some cancellations, using a Taylor expansion of the term C we obtain

C = C

1

+ C

2

+ C

3

,

18

where

C

1

=1

"

2

Z Z

s

�1e

(x�s)/"

⇢

2(x)

2⇢(s)'(s) dx ds ,

C

2

=1

"

2

Z Z

s

�1e

(x�s)/"

⇢

2(x)

2⇢(s) (x� s)'

x

(x) dx ds ,

C

3

=1

"

2

Z Z

s

�1e

(x�s)/"

⇢

2(x)

2⇢(s)

(x� s)2

2'

xx

(⇣) dx ds . (6.8)

In the integral for C

3

, it is understood that for each x, s one must choose a suitable ⇣ =⇣(x, s) 2 [x, s].

We now compare the integrals B and C

1

. Without loss of generality one can assume ' = �

3

for some � 2 C2

c

, � � 0. For any � � 0, we now apply the Hardy-Littlewood inequalitywith

g

1

(x) = ⇢

2(x)�2(x), g

2

(x) = ⇢(x+ �)�(x+ �),

and obtainZ

⇢

2(x)

2⇢(x)'(x) dx �

Z

⇢

2(x)

2�

2(x) · ⇢(x+ �)�(x+ �) dx . (6.9)

Indeed, the level sets of the two functions ⇢2�2 and ⇢� are the same. By (6.7), the integralon the right hand side of (6.9) is maximum (and coincides with

R

g

⇤1

g

⇤2

dx) when � = 0.

Performing the change of variable s = x+ �, a further integration w.r.t. s yields

B =1

"

Z

⇢

2(x)

2⇢(x)'(x) dx � 1

"

Z

⇢

2(x)

2�

2(x) · ⇢(s)�(s) dx

� 1

"

2

Z Z

s

�1e

(x�s)/"

⇢

2(x)

2�

2(x) · ⇢(s)�(s) dx ds .

= B

1

�B

2

, (6.10)

where

B

1

=1

"

2

Z Z

s

�1e

(x�s)/"

⇢

2(x)

2⇢(s)�3(s) dx ds = C

1

, (6.11)

B

2

=1

"

2

Z Z

s

�1e

(x�s)/"

⇢

2(x)

2�(x)⇢(s) [�2(s)� �

2(x)] dx ds.

To compute the last integral for B2

we use the Taylor expansion

�

2(s)� �

2(x) = 2�(x)�x

(x) · (s� x) + [2�2x

(⇣) + 2�xx

(⇣)] · (s� x)2

2,

where ⇣ = ⇣(x, s) 2 [x, s]. This yields

B

2

=1

"

2

Z Z

s

�1e

(x�s)/"(s� x) · ⇢2(x)

2⇢(s)�2(x) 2�

x

(x)dx ds

+1

"

2

Z Z

s

�1e

(x�s)/"

(s� x)2

2· ⇢

2(x)

2⇢(s)�(x) [2�2

x

(⇣) + 2�xx

(⇣)] dx ds

= B

21

+B

22

+B

23

,

19

where

B

21

.

=1

"

2

Z Z

s

�1e

(x�s)/"(s� x) · ⇢3(x)�2(x)�x

(x)dx ds ,

B

22

.

=1

"

2

Z Z

s

�1e

(x�s)/"(s� x) · ⇢2(x)✓

Z

s

x

⇢

x

(�) d�

◆

�

2(x)�x

(x)dx ds ,

B

23

.

=1

"

2

Z Z

s

�1e

(x�s)/"

(s� x)2

2· ⇢2(x) ⇢(s)�(x) [�2

x

(⇣) + �

xx

(⇣)] dx ds .

The term B

21

is computed by

B

21

=

Z

⇢

3(x)'

x

(x)

3dx = A . (6.12)

Concerning B

22

, using �, x, and ⇠ = s� x as variables of integration, we obtain

|B22

| k⇢k2L1 · 13k'

x

kL1 · 1

"

2

Z Z Z

x<�<s

e

(x�s)/"(s� x)|⇢x

(�)| dx d� ds

= k⇢k2L1 · 13k'

x

kL1 · 1

"

2

Z Z

+1

0

e

�⇠/"

⇠

✓

Z

�

��⇠

dx

◆

d⇠ |⇢x

(�)| d�

= k⇢k2L1 · 13k'

x

kL1 ·Z

Z

+1

0

e

�⇠/"

"

2

⇠

2

d⇠

!

|⇢x

(�)| d�

= k⇢k2L1 · 13k'

x

kL1 · k⇢x

kL1 · 2" . (6.13)

The term B

23

can be estimated by

|B23

| k⇢k2L1k�kL1

⇣

k�x

k2L1 + k�xx

kL1

⌘

Z

|⇢(s)|Z

s

�1e

(x�s)/"

(x� s)2

2"2dx ds

= k⇢k2L1k�kL1

⇣

k�x

k2L1 + k�xx

kL1

⌘

· k⇢kL1

Z

+1

0

e

��/"

�

2

2"2d�

= k⇢k2L1k�kL1

⇣

k�x

k2L1 + k�xx

kL1

⌘

· k⇢kL1 · " . (6.14)

An entirely similar argument shows that the integral defining C

3

at (6.8) also approacheszero as "! 0. Indeed,

|C3

| =1

"

2

�

�

�

�

Z Z

s

�1e

(x�s)/"

⇢

2(x)

2⇢(s)

(x� s)2

2'

xx

(⇣) dx ds

�

�

�

�

k'xx

kL1 · k⇢k2L1 · 1

2"2

Z

|⇢(s)|Z

s

�1e

(x�s)/"

(x� s)2

2dx ds

k'xx

kL1 · k⇢2kL1 · k⇢kL1 · 1

2"2

Z

+1

0

e

��/"

�

2

2d�

= k'xx

kL1 · k⇢k2L1 · k⇢kL1 · "2. (6.15)

20

Finally, we estimate the sum of the remaining two terms:

D + C

2

=1

"

Z Z

s

�1e

(x�s)/"

⇢

2(x)

2⇢(s)'

x

(x) dx ds

� 1

"

2

Z Z

s

�1e

(x�s)/"

⇢

2(x)

2⇢(s) (s� x)'

x

(x) dx ds

=

Z

⇢

2(x)

2'

x

(x)

✓

Z

+1

x

e

(x�s)/"

⇣1

"

� s� x

"

2

⌘

⇢(s) ds

◆

dx .

Using the identityZ

+1

x

e

(x�s)/"

⇣1

"

� s� x

"

2

⌘

ds = 0 ,

we compute

D + C

2

=

Z

⇢

2(x)

2'

x

(x)

✓

Z

+1

x

e

(x�s)/"

⇣1

"

� s� x

"

2

⌘

[⇢(s)� ⇢(x)] ds

◆

dx

=

Z

⇢

2(x)

2'

x

(x)

Z

+1

x

e

(x�s)/"

⇣1

"

� s� x

"

2

⌘

Z

s

x

⇢

x

(�) d� ds dx

=

Z

⇢

2(x)

2'

x

(x)

Z

+1

x

⇢

x

(�)

Z

+1

�

e

(x�s)/"

⇣1

"

� s� x

"

2

⌘

ds d� dx

=

Z

⇢

2(x)

2'

x

(x)

Z

+1

x

⇢

x

(�) e(x��)/"

x� �

"

d� dx

=

Z

⇢

x

(�)

Z

�

�1

⇢

2(x)

2'

x

(x)e(x��)/"

x� �

"

dx d� .

As a consequence, we obtain the following estimate

|D + C

2

| k⇢x

kL1 ·�

�

�

⇢

2

2

�

�

�

L1· k'

x

kL1 ·Z

�

�1e

(x��)/"

� � x

"

dx

= k⇢x

kL1 ·�

�

�

⇢

2

2

�

�

�

L1· k'

x

kL1 · " . (6.16)

Summarizing all the above estimates (6.8)-(6.16), we have

J

2

= A+B � C �D

� A+B

1

� (B21

+B

22

+B

23

)� (C1

+ C

2

+ C

3

)�D (6.17)

= (A�B

21

) + (B1

� C

1

)� (D + C

2

)�B

22

�B

23

� C

3

(6.18)

= O(1) · ".

Indeed, on the line (6.18) the first two terms are zero, while the remaining four terms havesize O(1) · ". Letting "! 0 we thus obtain the desired entropy inequality.

We remark that the inequality on the line (6.17), accounting for possible entropy dissipation,is due to the relation B � B

1

� B

2

in (6.10). This follows from the Hardy-Littlewoodrearrangement inequality.

21

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