Abstract

We show the existence of two sticky particles models with the same velocity function 𝑢𝑡(𝑥) which is the entropy solution of the inviscid Burgers' equation. One of them is governed by the set of discontinuity points of 𝑢0. Thus, the trajectories 𝑡𝑋𝑡 coincide; however one has different mass distributions 𝜕𝑥𝑢𝑡=d𝑢0𝑋𝑡1 and 𝜆𝑋𝑡1. Here, 𝜆 denotes the Lebesgue measure.

1. Introduction and Main Results

The one-dimensional Burgers’ equation of viscosity 𝜎0 takes the form 𝜕𝑡𝑢+𝑢𝜕𝑥𝑢=𝜎𝜕𝑥𝑥𝑢. It is widely used in the physical literature to model various phenomenon such as shock waves in hydrodynamics turbulence and gas dynamics [1, 2]. A solution of this equation was given by Hopf [3]. When 𝜎=0, one gets the inviscid form which is solved by convergence of the solution of the general form when 𝜎 tends to 0. It is well known that the entropy solution of the inviscid equation is interpreted as the velocity function of some sticky particles model [4, 5], but this link was shown only for continuous initial data, and the connection with the trajectories is still unknown.

It is easy to see the relation between the inviscid Burgers’ equation and the so-called pressure less gas system 𝜕𝑡𝜌+𝜕𝑥(𝑢𝜌)=0, 𝜕𝑡(𝑢𝜌)+𝜕𝑥(𝑢2𝜌)=0 which are simplified forms of the classical Euler equations. Here, 𝑢 and 𝜌 are smooth functions. Indeed, a correct derivation of the second equation gives [𝜕𝑡𝑢+𝑢𝜕𝑥𝑢]𝜌+𝑢[𝜕𝑡𝜌+𝜕𝑥(𝑢𝜌)]=0. Taking account of the first equation, one gets [𝜕𝑡𝑢+𝑢𝜕𝑥𝑢]𝜌=0, which leads to 𝜕𝑡𝑢+𝑢𝜕𝑥𝑢=0 (of course if 𝜌0). The above link holds again when 𝜌 is a field of nonnegative measures, 𝑢𝐶1(×+) and the derivations are made in the sense of distributions.

In the case of measures, a solution of the gas system was given for example in [46], and more recently in [7, 8]. In [5, 7, 8], the pressureless gas system was obtained from the sticky particles model. In these works, 𝑢(𝑥,𝑡) is the velocity of the particle which occupies the position 𝑥 at time 𝑡, and 𝜌(,𝑡) is the mass distribution of the whole matter at time 𝑡. In the latest work of Moutsinga [8], the sticky particle model was constructed when 𝜌(,0) is any probability measure and 𝑢(,0) has no positive jump. The author showed that the particles trajectories 𝑡𝑋𝑡(𝑦) are such that 𝜕𝑡𝑋𝑡(𝑦)=𝑢(𝑋𝑡(𝑦),𝑡).

Unfortunately, even if 𝑢(,0) is continuous, the velocity immediately presents discontinuities in space and time as soon as the first shocks occur. Thus, it is not easy to obtain rigorously Burgers’ equation from the pressure less gas system.

Very recently, Moutsinga [9] showed that if 𝜌(,0) is the Lebesgue measure 𝜆, and 𝑢(,0) has no positive jump, then the velocity field 𝑢 is the entropy solution of the inviscid Burgers’ equation. The proof was made using the exact expression of the solution given by Hopf [3].

In this this paper, we give again the same result when 𝑢(,0) is nonincreasing and the initial c.d.f. of the matter is 𝑢(,0). We use known results on scalar conservations laws by considering the conservative form of the inviscid Burgers’ equation 𝜕𝑡𝑢(𝑥,𝑡)+𝜕𝑥𝑢2(𝑥,𝑡)2=0,𝑢(𝑥,0)=𝑢0(𝑥),(𝑥,𝑡)×+,(1.1) where 𝑢0 is a smooth function.

We show that the set of discontinuity points of 𝑢0 governs its own sticky particles model whose velocity function is again 𝑢(𝑥,𝑡).

It is well known that discontinuity lines of 𝑢 start on discontinuity points 𝑦 such that 𝑢0(𝑦0)>𝑢0(𝑦+0) which are the atoms of the measure d𝑢0 (see [1] and the illustrations of Section 2.1). For this reason we consider a nonincreasing function 𝑢0 and we define d𝑢0 as the mass initial distribution of a system of particles. We consider each 𝑦 as the position of a particle which starts with the mass 𝑢0(𝑦0)𝑢0(𝑦+0) and the velocity 𝑣0(𝑦)=(𝑢0(𝑦0)+𝑢0(𝑦+0))21. This allows to interpret 𝑢(𝑥,𝑡) as the velocity of a cluster situated at position 𝑥 at time 𝑡 with the mass 𝑢(𝑥0,𝑡)𝑢(𝑥+0,𝑡). This fact is the main result of this paper. We recall that the measure d𝑢0 is a Radon measure well-defined byd𝑢0]((𝑎,𝑏)=𝑢0(𝑏+0)𝑢0(𝑎+0),𝑎<𝑏.(1.2)

Theorem 1.1. Let (𝑥,𝑡)𝑢(𝑥,𝑡) be the entropy solution of Burgers’ equation of initial data 𝑢0 like above.(1) The measure 𝜕𝑥𝑢(,𝑡) and the function 𝑢(,𝑡) represent, respectively, at time 𝑡, the mass distribution and the velocity function of the sticky particles model whose initial mass distribution and velocity function are given by d𝑢0 and 𝑣0.(2) There exists a forward flow (𝑌𝑠,𝑡,𝑡,𝑠0) defined on the measure space (,d𝑢0) such that 𝑌𝑠,𝑡(𝑥) is the position, after having browsed the time 𝑡 of the particle that occupied the position 𝑥 at time 𝑠. Moreover,d𝑢0𝑌10,𝑡=𝜕𝑥𝑌𝑢(,𝑡);𝑡,0,𝑠+𝑡(𝑥)=𝑌𝑠,𝑡𝑌0,𝑠(𝑥);(𝑥,𝑠,𝑡)×+×+,𝜕𝑌𝜕𝑡𝑠,𝑡𝑌=𝑢𝑠,𝑡.;𝑠+𝑡(1.3)(3) Let C𝑐() be the space of real continuous functions with compact support. For all 𝑠,𝑡0,𝑓𝑌0,𝑡𝑢𝑌0,𝑡,𝑡d𝑢0=𝑓𝑌0,𝑡𝑢0𝑌0,0d𝑢0𝑓C𝑐(),d𝑢0𝑌10,𝑠𝑌a.e.,𝑢𝑠,𝑡,𝑠+𝑡=E𝑢(,𝑠)𝑌𝑠,𝑡𝑢if0<.(1.4)(4) The function (𝑥,𝑡)𝑀(𝑥,𝑡)=𝑢(𝑥,𝑡) is the unique entropy solution of the scalar conservation law𝜕𝑡𝑀+𝜕𝑥(𝐴(𝑀))=0suchthat𝑀(,0)=𝑢0,withtheux𝑚𝐴(𝑚)=0𝑚𝑣0𝑢01(𝑧)d𝑧.(1.5)
Moreover, 𝜕𝑥𝐴(𝑢(𝑥,𝑡))=𝑢(𝑥,𝑡)𝜕𝑥𝑢(𝑥,𝑡)for all 𝑡.

Remark that contrary to the model of [9], Burgers equation is recovered here from the above scalar conservation law (1.5), although the flux (representing the momentum) is not a trivial function.

The second result of this paper is an interpretation of some equations from [3], in terms of image measures of the Lebesgue measure 𝜆 by applications defined from two different sticky particles models: the latest flow and the one of [9]. Let us come back to the model of [9] which is a generalization of [7, 8] to the case of infinite total mass of the particles. For fixed (𝑥,𝑡), Lagrange coordinates 𝑦(𝑥,𝑡) and 𝑦(𝑥,𝑡) were used in [3] in order to get the solution 𝑢(𝑥,𝑡) of (1.1). In the model of [9], [𝑦(𝑥,𝑡),𝑦(𝑥,𝑡)] is the set of all the initial particles (a cluster) that occupy the position 𝑥 at time 𝑡, with the mass 𝑦(𝑥,𝑡)𝑦(𝑥,𝑡). Moreover,𝑢(𝑥0,𝑡)=lim𝑥𝑥𝑢𝑥,𝑡=𝑡1𝑥𝑦,𝑢(𝑥,𝑡)(𝑥+0,𝑡)=lim𝑥𝑥+𝑢𝑥,𝑡=𝑡1𝑥𝑦,(𝑥,𝑡)𝑢(𝑥,𝑡)=𝑢(𝑥0,𝑡)+𝑢(𝑥+0,𝑡)2=𝑦2𝑥(𝑥,𝑡)+𝑦(𝑥,𝑡).2𝑡(1.6) Hence𝑦(𝑥,𝑡)+𝑦(𝑥,𝑡)2𝜕+𝑡𝑢(𝑥,𝑡)=𝑥(𝑥,𝑡),𝑥𝑦(,𝑡)+𝑦(,𝑡)2+𝑡𝜕𝑥𝑢(,𝑡)=d𝑥=𝜆𝑡.(1.7)

Proposition 1.2. Suppose that 𝑢0 is nonincreasing and lim|𝑥|𝑢0(𝑥)𝑥1=0. There exists two forward flows (𝑥,𝑠,𝑡)×+×+𝑋𝑠,𝑡(𝑥),𝑌𝑠,𝑡(𝑥) which model the trajectories of two different sticky particles models, such that for all 𝑡0,(1)𝜕𝑥(𝑦(,𝑡)+𝑦(,𝑡))21=𝜆𝑋10,𝑡;(2)𝜕𝑥𝑢(,𝑡)=d𝑢0𝑌10,𝑡;(3)if 𝑢0 is not constant in any neighborhood of 𝑥, then for all 𝑡0, 𝑋0,𝑡(𝑥)=𝑌0,𝑡(𝑥). So 𝜆𝑋10,𝑡+𝑡d𝑢0𝑋10,𝑡=𝜆, which is equivalent to 𝜆𝑋10,𝑡𝑍𝑡𝜆0,𝑡1=𝜆𝑡,(1.8) with 𝑍0,𝑡=𝑌0,𝑡((𝑢0)1).

For any nonnegative Radon measure 𝑃 and any smooth enough function 𝑢, it was defined in [9] the sticky particles model of initial mass distribution 𝑃 and initial velocity function 𝑢. The state of the particles at each time 𝑡0 is given by the function 𝑦𝜙(𝑦,𝑡,𝑃,𝑢). In fact, the flow 𝑋 was already given in [9] by𝑋𝑠,𝑡(𝑦)=𝜙𝑦,𝑡,𝜆𝑋10,𝑠,𝑢(,𝑠)(𝑦,𝑠,𝑡)×+×+.(1.9) Then, the first equation of assertion (1) is one of the properties of 𝑋 since 𝑦(𝑥,𝑡)=𝑦(𝑥,𝑡) a.e; the second equation is given in Theorem 1.1. The end of assertion (3) comes from the fact that d𝑢0 is the image measure of 𝜆 by the generalized inverse (𝑢0)1. The coincidence of the two flows is given in Section 2 (Theorem 2.2 where we also precise the link between the two models).

Let 𝒮 be the support of d𝑢0. We will see in fact that 𝑌 is given by𝑌0,𝑡(𝑦)=𝜙𝑦,𝑡,d𝑢0,𝑣0(𝑦,𝑡)𝒮×+,𝑌𝑠,𝑡(𝑦)=𝜙𝑦,𝑡,d𝑢0𝑌10,𝑠,𝑢(,𝑠)(𝑦,𝑠,𝑡)𝑌0,𝑠(𝒮)×+×+.(1.10)

Before the proof of the Theorem 1.1, we study the link between 𝑋 and 𝑌.

2. Comparison of the Two Models

2.1. When a Diffuse Dynamics Hides a Discrete One

In this part, 𝑢0 is nonincreasing and stair size.

2.1.1. Example of a Single Shock Wave

We begin with an initial velocity of the type𝑢0(𝑦)=1,if𝑦<𝑎0,if𝑦𝑎.(2.1) In this case, 𝒮={𝑎} and 𝑌0,𝑡(𝑎)=𝑎+𝑡/2 for all 𝑡0. What about 𝑋0,𝑡(𝑎)? Let us define 𝑥=𝑌0,𝑡(𝑎). The cluster {𝑦𝑋0,𝑡(𝑦)=𝑥}=[𝑦(𝑥,𝑡),𝑦(𝑥,𝑡)] is the set on which the function 𝑦𝐺1(𝑦,𝑥,𝑡)=𝑦0[𝑧+𝑡𝑢0(𝑧)]d𝑧 reaches its absolute lower bound. This is necessarily done in at least one 𝑦 which is a continuity point of 𝑢0, and 𝑦+𝑡𝑢0(𝑦)=𝑥. Suppose that 𝑦>𝑎. Then 𝑦=𝑥, and a simple computation gives𝐺1(𝑦,𝑥,𝑡)𝐺1𝑦1𝑦,𝑥,𝑡=1+𝑡𝑥22,𝑦1<𝑎(2.2) which shows that the lower bound is also reached in 𝑦1=𝑥𝑡<𝑎. If 𝑦<𝑎, then 𝑦=𝑥𝑡 and a simple computation gives𝐺1𝑦2,𝑥,𝑡𝐺1𝑦(𝑦,𝑥,𝑡)=2𝑥22,𝑦2>𝑎(2.3) which shows that the lower bound is also reached in 𝑦2=𝑥. Thus, 𝑦(𝑥,𝑡)=𝑥𝑡<𝑎<𝑥=𝑦(𝑥,𝑡). Hence,𝑋0,𝑡(𝑦)=𝑌0,𝑡𝑡(𝑎)=𝑥=𝑎+2𝑡,𝑦𝑎2𝑡,𝑎+2.(2.4)

For the dynamics modeled by 𝑋, this means that the trajectory of 𝑎 is a straight line 𝑎(𝑡)=𝑎+𝑡/2 along which all the shocks occur continuously. At time 𝑡, the particle 𝑎 is in a cluster of center 𝑎 and of diameter 𝑡, with the velocity 1/2.

2.1.2. Stairsize Velocity

By similar arguments as above, one gets the following results when 𝑢0 is a stair size function of the type𝑢0=𝑁𝑖=0𝑤𝑖1[𝑦𝑖,𝑦𝑖+1)with=𝑦0<𝑦1<<𝑦𝑁+1=+,𝑤𝑖>𝑤𝑖+1𝑖.(2.5) Let 𝑤 be the velocity field of the discrete dynamics constructed from d𝑢0,𝑣0.

Proposition 2.1. For all 𝑥𝑌0,𝑡(𝒮), there exists a unique couple (𝑛,𝑚) such that 𝑛<𝑚,𝑦𝑛<𝑥𝑡𝑤𝑛=𝑦(𝑥,𝑡)<𝑦𝑛+1𝑦𝑚<𝑥𝑡𝑤𝑚=𝑦(𝑥,𝑡)<𝑦𝑚+1 and for all (𝑦,𝑖)[𝑥𝑡𝑤𝑛,𝑥𝑡𝑤𝑚]×[𝑛+1,𝑚]:𝑥=𝑋0,𝑡(𝑦)=𝑌0,𝑡𝑦𝑖=1𝑤𝑛𝑤𝑚𝑚𝑗=𝑛+1𝑤𝑗1𝑤𝑗𝑦𝑗𝑤+𝑡𝑗+𝑤𝑗12,𝑢𝑤(𝑥,𝑡)=𝑤(𝑥,𝑡)=𝑛+𝑤𝑚2.(2.6) If 𝑥𝑌0,𝑡(𝒮),!(𝑦,𝑛)such that 𝑦(𝑦𝑛,𝑦𝑛+1) and 𝑥=𝑋0,𝑡(𝑦)=𝑦+𝑡𝑤𝑛,𝑢(𝑥,𝑡)=𝑢0(𝑦)=𝑤𝑛.(2.7)

These results can be also seen as a simple consequence of the following general case.

2.2. The General Case

The function 𝑢0 is nonincreasing and lim|𝑥|𝑢0(𝑥)𝑥1=0.From 𝑢0 and 𝑣0(𝑥)=(𝑢0(𝑥0)+𝑢0(𝑥+0))21, we have two sticky particles models:(i)each 𝑦 is the position of an initial particle which occupies the position 𝜙(𝑦,𝑡,𝜆,𝑢0)=𝑥 at time 𝑡0, with the velocity 𝑢(𝑥,𝑡) and the mass 𝑦(𝑥,𝑡)𝑦(𝑥,𝑡); (ii)each 𝑦𝒮 is the position of an initial particle which occupies the position 𝜙(𝑦,𝑡,d𝑢0,𝑣0)=𝑥 at time 𝑡0, with the velocity 𝑤(𝑥,𝑡) and the mass 𝑢0(𝛼0)𝑢0(𝛽+0), with [𝛼,𝛽]={𝑦𝜙(𝑦,𝑡,d𝑢0,𝑣0)=𝑥}.

For all 𝑡,𝜙(,𝑡,𝜆,𝑢0)=. Then, for all 𝑥𝜙(𝒮,𝑡,d𝑢0,𝑣0), there exist clusters 𝒞1(𝑥,𝑡)=[𝑦(𝑥,𝑡),𝑦(𝑥,𝑡)],𝒞2(𝑥,𝑡) such that𝜙𝑦1,𝑡,𝜆,𝑢0𝑦=𝜙2,𝑡,d𝑢0,𝑣0𝑦=𝑥1,𝑦2𝒞1(𝑥,𝑡)×𝒞2(𝑥,𝑡).(2.8) For 𝑡>0, if we suppose that 𝑢=𝑤, then both the functions 𝑠𝑦1(𝑠)=𝜙(𝑦1,𝑡,𝜆,𝑢0), 𝑦2(𝑠)=𝜙(𝑦2,𝑡,d𝑢0,𝑣0) are solutions ofd𝑦(𝑠)=𝑢(𝑦(𝑠),𝑠)d𝑠,0<𝑡𝑠,𝑦(𝑡)=𝑥.(2.9) As 𝑢 satisfies the Œlenick type entropy condition (see [9])𝑢𝑥2𝑥,𝑠𝑢1,𝑠𝑥2𝑥11𝑠,𝑥1,𝑥2,𝑠>0,(2.10) it follows from the results of Fillipov [10] that 𝑦1(𝑠)=𝑦2(𝑠), for all 𝑠𝑡>0.

Unfortunately, this method fails when 𝑡=0. In the sequel, using other arguments, we show indeed that 𝑢(,𝑡)=𝑤(,𝑡) for 𝑡>0, and that the trajectories coincide for all 𝑡0. In fact, we show that 𝒞2(𝑥,𝑡)𝒞1(𝑥,𝑡).

Remark that the link can be established only for the initial positions 𝑦𝒮,thatis,𝑢0 is not constant in any neighborhood of 𝑦. We recall that𝑢(𝑥0,𝑡)=𝑡1𝑥𝑦(𝑥,𝑡),𝑢(𝑥+0,𝑡)=𝑡1𝑥𝑦(𝑥,𝑡),(𝑥,𝑡),𝑥=𝑦(𝑥,𝑡)+𝑡𝑢0𝑦(𝑥,𝑡)=𝑦(𝑥,𝑡)+𝑡𝑢0𝑦.(𝑥,𝑡)(2.11)

Theorem 2.2. Let 𝑢 be the entropy solution of Burgers’ equation with a nonincreasing initial data 𝑢0 such that lim|𝑥|𝑢0(𝑥)𝑥1=0. For all (𝑥,𝑡), let us define 𝒞(𝑥,𝑡)={𝑦𝜙(𝑦,𝑡,d𝑢0,𝑣0)=𝑥}=[𝛼,𝛽]𝒮.(1) One has 𝑢(𝑥0,𝑡)𝑢(𝑥+0,𝑡) if and only if𝑢0(𝑢𝛼0)>𝑢(𝑥,𝑡)=𝑤(𝑥,𝑡)=0(𝛼0)+𝑢0(𝛽+0)2>𝑢0(𝑦𝛽+0),(𝑥,𝑡)=𝑥𝑡𝑢0(𝛼0)𝛼𝛽𝑦(𝑥,𝑡)=𝑥𝑡𝑢0(𝛽+0).(2.12)(2) If 𝑢(𝑥0,𝑡)=𝑢(𝑥+0,𝑡), one has two possibilities:(i)𝒞(𝑥,𝑡) satisfies 𝑦(𝑥,𝑡)=𝑦(𝑥,𝑡)=𝛼=𝛽, 𝑢(𝑥,𝑡)=𝑤(𝑥,𝑡)=𝑢0(𝛼0)=𝑢0(𝛽+0)=𝑢0(𝛼).(2.13)(ii)𝒞(𝑥,𝑡)= and there exist 𝑎<𝑏 such that𝜙𝑎,𝑡,d𝑢0,𝑣0<𝜙𝑏,𝑡,d𝑢0,𝑣0,𝑢0(𝑎+0)=𝑢0(𝑏0),𝑎<𝑦(𝑥,𝑡)=𝑦(𝑥,𝑡)=𝑥𝑡𝑢0(𝑏0)<𝑏.(2.14)

Before giving the proof, we remark that if 𝑢0 is continuous in 𝑧1,𝑧2, then 𝑧2𝑧1𝜂+𝑡𝑢0(𝜂)𝑑𝜂=𝑡𝑧2𝑧1𝜂+𝑡𝑢0(𝜂)𝑑𝑢0𝑧(𝜂)+2+𝑡𝑢0(𝑧2)2𝑧1+𝑡𝑢0(𝑧1)22.𝑡(2.15) If 𝑢0 is not continuous in 𝑧𝑖, the formula must be corrected with the right term 𝑢0(𝑧𝑖0) or 𝑢0(𝑧𝑖+0), depending on the fact that the integration is closed or not in 𝑧𝑖. This formula is due to the integration by parts𝑧2𝑧1𝜂d𝑢0(𝜂)=𝑧2𝑢0𝑧2+𝑧1𝑢0𝑧1+𝑧2𝑧1𝑢0(𝜂)d𝜂,𝑧2𝑧1𝑣0d𝑢0=𝑢0(𝑧2)𝑢0𝑧1𝑣0𝑢01(𝑧)d𝑧=𝑢0(𝑧2)𝑢0𝑧1𝑢𝑧d𝑧=0𝑧12𝑢0𝑧222.(2.16)

So we get𝑧2𝑧1𝜂+𝑡𝑣0(𝜂)d𝑢0(𝜂)=𝑧2𝑧1𝑢0(𝜂)d𝜂𝑧2𝑢0𝑧2+𝑧1𝑢0𝑧1𝑢+𝑡0𝑧12𝑢0𝑧222=𝑧2𝑧1𝑢0𝑧(𝜂)d𝜂+1+𝑡𝑢0𝑧12𝑧2+𝑡𝑢0𝑧22𝑧21+𝑧22=12𝑡𝑡𝑧2𝑧1𝜂+𝑡𝑢0𝑧(𝜂)d𝜂+1+𝑡𝑢0𝑧12𝑧2+𝑡𝑢0𝑧22.2𝑡(2.17)

Proof of Theorem 2.2. (1) When 𝑢 is discontinuous in (𝑥,𝑡), we have 𝑦(𝑥,𝑡)<𝑦(𝑥,𝑡), and 𝑢(𝑥,𝑡) is defined as the velocity of the cluster [𝑦(𝑥,𝑡),𝑦(𝑥,𝑡)]:𝑢(𝑥,𝑡)=𝑦𝑦(𝑥,𝑡)(𝑥,𝑡)𝑢0(𝜂)d𝜂𝑦(𝑥,𝑡)𝑦.(𝑥,𝑡)(2.18) For 𝑥𝜙(𝒮,𝑡,d𝑢0,𝑣0), the cluster [𝛼,𝛽]𝒮={𝑦𝜙(𝑦,𝑡,d𝑢0,𝑣0)=𝑥} satisfies 𝛽+𝑡𝑢0(𝛽+0)𝑥𝛼+𝑡𝑢0(𝛼0).(2.19) Let 𝑧1=𝑥𝑡𝑢0(𝛼0) and suppose that 𝑦(𝑥,𝑡)𝑧1. First, we remark that 𝑥=𝑧1+𝑢0(𝑧10). It is obvious if 𝑧1=𝛼. If 𝑧1𝛼, we have 𝑥<𝛼+𝑡𝑢0(𝛼0). This occurs only when 𝛼 is not an accumulation of particles from the left; then there exists another cluster [𝑎,𝑏] such that 𝑏<𝛼 and d𝑢0(𝑏,𝛼)=0=𝑢0(𝛼0)𝑢0(𝑏+0),𝑏+𝑡𝑢0(𝑏+0)𝜙𝑏,𝑡,d𝑢0,𝑣0<𝑥<𝛼+𝑡𝑢0(𝛼0).(2.20) Thus 𝑥𝑡𝑢0(𝛼0)=𝑧1(𝑏,𝛼). As 𝑢0 is constant in (𝑏,𝛼), we get 𝑧1+𝑡𝑢0(𝑧1)=𝑥. On the other hand, it is easy to see that 𝑢0 is continuous in 𝑦(𝑥,𝑡),𝑦(𝑥,𝑡). Then, we have necessarily 𝑦(𝑥,𝑡)=𝑥𝑡𝑢0(𝑦(𝑥,𝑡))[𝑏,𝛼]. If 𝑦(𝑥,𝑡)<𝑏, 𝑧2=𝑦(𝑥,𝑡) gives in (2.15): 𝑡𝑦𝛼0(𝑥,𝑡)𝜂+𝑡𝑣0(𝜂)d𝑢0(𝜂)=𝑡𝑧1𝑦0(𝑥,𝑡)𝜂+𝑡𝑣0(𝜂)d𝑢0=(𝜂)𝑧1𝑦(𝑥,𝑡)𝜂+𝑡𝑢0(𝜂)d𝜂=𝐺𝑥,𝑧1,𝑡𝐺𝑥,𝑦𝑧(𝑥,𝑡),𝑡+𝑥1𝑦.(𝑥,𝑡)(2.21) As 𝑧1𝑦𝑢(𝑥,𝑡)=𝑡0𝑦(𝑥,𝑡)𝑢0(𝛼0)=𝑡d𝑢0𝑦(𝑥,𝑡),𝛼,(2.22) we get 𝑦𝛼0(𝑥,𝑡)𝜂+𝑡𝑣0(𝜂)𝑥d𝑢0𝐺(𝜂)=𝑥,𝑧1,𝑡𝐺𝑥,𝑦(𝑥,𝑡),𝑡𝑡.(2.23) From the definition of 𝛼 and 𝑦(𝑥,𝑡), the fist term is negative and the second is nonnegative. This is absurd, as well as the case 𝑦(𝑥,𝑡)>𝛼. We can conclude that 𝑦(𝑥,𝑡)=𝑧1=𝑥𝑡𝑢0(𝛼0). In the same way, 𝑦(𝑥,𝑡)=𝑥𝑡𝑢0(𝛽+0). Thus 𝑦(𝑥,𝑡)<𝑦(𝑥,𝑡)𝑢0(𝛼0)>𝑢0(𝛽+0),(2.24) and it comes from (2.15) that 𝑢(𝑥,𝑡)=𝑡1𝑦𝑥(𝑥,𝑡)+𝑦(𝑥,𝑡)2=𝑢0(𝛼0)+𝑢0(𝛽+0)2=𝛽+0𝛼0𝑣0(𝜂)d𝑢0(𝜂)𝑢0(𝛼0)𝑢0(𝛽+0)=𝑤(𝑥,𝑡).(2.25) (2.i) 𝑦(𝑥,𝑡)=𝑦(𝑥,𝑡)=𝛼=𝛽𝑢0(𝛼0)=𝑢0(𝛽+0)=𝑢0(𝛼),(2.26) and in this case 𝑢(𝑥,𝑡)=𝑡1(𝑥𝛼)=𝑢0(𝛼)=𝑣0(𝛼)=𝑤(𝑥,𝑡).(2.27) (2.ii) For 𝑥𝜙(𝒮,𝑡,d𝑢0,𝑣0), there exist 𝑥1,𝑥2𝜙(𝒮,𝑡,d𝑢0,𝑣0) and clusters 𝛼𝑖,𝛽𝑖𝑦𝒮=𝑦𝜙,𝑡,d𝑢0,𝑣0=𝑥𝑖(2.28) such that 𝑥1<𝑥<𝑥2 and d𝑢0𝛽1,𝛼2=0=𝑢0𝛼20𝑢0𝛽1+0,𝛽1<𝛼2.(2.29) As the function 𝑥𝑦(𝑥,𝑡) does not decrease, we have 𝑦𝑥1,𝑡𝑦(𝑥,𝑡)𝑦(𝑥,𝑡)𝑦𝑥2,𝛽,𝑡1𝑥1𝑡𝑢0𝛽1+0=𝑦𝑥1,𝑡𝑦𝑥2,𝑡=𝑥2𝑡𝑢0𝛼20𝛼2,(2.30) and this leads to 𝑦(𝑥,𝑡),𝑦(𝑥,𝑡)[𝛽1,𝛼2]. From the fact that 𝑢0(𝛼20)=𝑢0(𝛽1+0) and 𝑦(𝑥,𝑡)+𝑡𝑢0(𝑦(𝑥,𝑡))=𝑥=𝑦(𝑥,𝑡)+𝑡𝑢0(𝑦(𝑥,𝑡)), we conclude that 𝛽1<𝑦(𝑥,𝑡)=𝑦(𝑥,𝑡)=𝑥𝑡𝑢0𝛽1+0<𝛼2.(2.31)

We have shown that all 𝑦𝒮 is the position of two particles which move following two different dynamics given, at time 𝑡, by their respective positions, velocities, and masses 𝑥𝑖(𝑦,𝑡),𝑣𝑖(𝑦,𝑡),𝑚𝑖(𝑦,𝑡) (with 𝑖=1,2) such that𝑥1(𝑦,𝑡)=𝜙𝑦,𝑡,𝜆,𝑢0=𝜙𝑦,𝑡,d𝑢0,𝑣0=𝑥2𝑣(𝑦,𝑡),𝑡>0,1(𝑦,𝑡)=𝑣2(𝑦,𝑡),𝑚1(𝑦,𝑡)=𝑡𝑚2(𝑦,𝑡).(2.32) However, we had at time zero,𝑚1(𝑦,0)=0,𝑚2(𝑦,0)=𝑢0(𝑦0)𝑢0𝑣(𝑦+0),1(𝑦,0)=𝑢0(𝑦),𝑣2(𝑢𝑦,0)=0(𝑦0)+𝑢0(𝑦+0)2.(2.33)

A surprising fact is that the two dynamics have exactly the same atoms after time zero. The first dynamics is diffuse, and the second one can be discrete (if 𝒮 is discrete).

Now we show the connection between the second sticky particles model and the inviscid Burgers’ equation.

Proof of Theorem 1.1. (1) and (4) The function 𝑥𝑣0(𝑥)=(𝑢0(𝑥0)+𝑢0(𝑥+0))21 is nonincreasing. As in [8] we can define the sticky particles model of initial mass distribution and velocity function d𝑢0,𝑣0. For all 𝑡, the particles position are given by a continuous nondecreasing function (𝑦,𝑡)𝜙(𝑦,𝑡,d𝑢0,𝑣0) well defined on the support of d𝑢0 and is extended to by linear interpolation. In the following, we set 𝜙(𝑦,𝑡,d𝑢0,𝑣0)=𝜙𝑡(𝑦). For any r.v. 𝑌0 having d𝑢0 as law, define 𝑌𝑡=𝜙(𝑌0,𝑡,d𝑢0,𝑣0). The velocity field of the particles (𝑦,𝑡)𝑢(𝑦,𝑡) is such that 𝑢(𝑌𝑡,𝑡)=E[𝑣0(𝑌0)𝑌𝑡]=(d/d𝑡)𝑌𝑡. Here, the derivative holds on the right-hand side, for all 𝑡.
At time 𝑡0, consider the c.d.f. 𝐹𝑡(𝑥)=d𝑢0(𝑌𝑡𝑥). One has 𝐹0=𝑢0 and its inverse is given by 𝐹01(𝑧)=𝑢01(𝑧). On ([0,1],([0,1]),𝜆), 𝐹𝑡1 is a r.v. which is the same law as 𝑌𝑡.
For fixed (𝑥,𝑡), let 𝑀(𝑥,𝑡) be the maximum among the abscissas where the function 𝑚𝐺(𝑥,𝑚,𝑡)=𝑚0𝑢01(𝑧)+𝑡𝑣0𝑢01(𝑧)𝑥d𝑧(2.34) reaches its absolute lower bound. Using the results of [9], it is clear that the function 𝑀(,𝑡) is a c.d.f. of 𝑌0,𝑡 and it is the entropy solution of𝜕𝑡𝑀+𝜕𝑥(𝐴(𝑀))=0suchthat𝑀(𝑥,0)=𝑢0(𝑥),withtheux𝑚𝐴(𝑚)=𝑚0𝑣0𝑢01(𝑧)d𝑧=0𝑚𝑣0𝑢01(𝑧)d𝑧.(2.35) Defining the at most countable set 𝐷={𝑦𝑢0(𝑦0)>𝑢0(𝑦+0)}, and remarking that 𝑣0(𝑦)=𝑢0(𝑦)for all 𝑦𝐷, we have 𝐴(𝑚)=𝑚01{𝑢01(𝑧)𝐷}𝑣0𝑢01(𝑧)d𝑧+𝑚01{𝑢01(𝑧)𝐷}𝑣0𝑢01=(𝑧)d𝑧𝑦𝑛𝐷𝑚01{𝑢01(𝑧)=𝑦𝑛}𝑢0𝑦𝑛0+𝑢0𝑦𝑛+02d𝑧+𝑚01{𝑢01(𝑧)𝐷}𝑢0𝑢01(𝑧)d𝑧.(2.36) As this c.d.f. takes its values in 𝑢0(), one should compute the values of the flux for 𝑚𝑢0(). Without loss of generality, one can suppose that 0 is a value of continuity of 𝑢0. Thus, for 𝑦𝑛𝐷there are only two possibilities: []0,𝑚𝑧𝑢01(𝑧)=𝑦𝑛=or𝑧𝑢01(𝑧)=𝑦𝑛[].0,𝑚(2.37) As {𝑧𝑢01(𝑧)=𝑦𝑛}=[𝑢(𝑦𝑛0),𝑢(𝑦𝑛+0)], we have 1{𝑢01(𝑧)=𝑦𝑛}𝑢0𝑦𝑛0+𝑢0𝑦𝑛+02𝑢d𝑧=0𝑦𝑛02𝑢0𝑦𝑛+0221={𝑢01(𝑧)=𝑦𝑛}𝑧d𝑧(2.38) and we get 𝐴(𝑚)=𝑚01{𝑢01(𝑧)𝐷}𝑧d𝑧𝑚01{𝑢01(𝑧)𝐷}𝑧d𝑧=𝑚0𝑚𝑧d𝑧=22.(2.39) Then, 𝑀(𝑥,𝑡)=𝑢(𝑥,𝑡) is the entropy solution of Burgers’ equation with initial data 𝑢0. Thus,d𝑢0(𝑌0,𝑡)1=𝜕𝑥𝑀(𝑥,𝑡)=𝜕𝑥𝑢(𝑥,𝑡).
The fact that 𝑢(𝑥,𝑡) is the velocity (for 𝑡>0) was already given in Theorem 2.2. We give another proof here. For each discontinuity point (𝑥,𝑡) of 𝑢, 𝑥 is the position, at time 𝑡, of a massive particle; so its velocity is 𝐴(𝑢(𝑥+0,𝑡))𝐴(𝑢(𝑥0,𝑡))=𝑢(𝑥+0,𝑡)+𝑢(𝑥0,𝑡)𝑢(𝑥0,𝑡)+𝑢(𝑥+0,𝑡)2=𝑢(𝑥,𝑡).(2.40) If 𝑢 is continuous in (𝑥,𝑡) and 𝑥 is the position at time 𝑡, then it is the position of a cluster {𝑎} which moves with its initial velocity 𝑣0(𝑎). We have the equalities of masses 𝑀(𝑥0,𝑡)=𝑀(𝑥+0,𝑡)=𝐹0(𝑎0)=𝐹0(𝑎+0). This means that 𝑢(𝑥,𝑡)=𝑢0(𝑎)=𝑣0(𝑎). The velocity of this particle is then 𝑤(𝑥,𝑡)=𝑣0(𝑎)=𝑢(𝑥,𝑡).
(2) and (3) These points are given by properties of 𝜙 (see [9]). The condition 𝑢0<+ means 𝑢0𝐿1(d𝑢0).