Chunyin JINInstitute of Applied Mathematics,Academy of Mathematics and Systems Science, Chinese Academy of Sciences,Beijing 100190,China E-mail:jinchunyin@amss.ac.cn

EXISTENCE AND UNIQUENESS OF ENTROPY SOLUTION TO PRESSURELESS EULER SYSTEM WITH A FLOCKING DISSIPATION∗

We study the existence and uniqueness problem for the nonhomogeneous pressureless Euler system with the initial density being a Radon measure.Our uniqueness result is obtained in the same space as the existence theorem.Besides,by counterexample we prove that Huang-Wang’s energy condition is also necessary for our nonhomogeneous system.

pressureless Euler system;Cucker-Smale model;entropy solution;flocking

2010 MR Subject Classification35A15;35L03;35L69;35Q92;74H05

1　Introduction

In the present paper,we study the following nonhomogeneous pressureless Euler system

where ρ and u are density and velocity,respectively.

This system is derived from Cucker-Smale model,which is used to describe flocking phenomenon.The word flocking represented the phenomenon that autonomous agents reach a consensus state based on limited environmental information and simple rules.The study of flocking based on mathematical models was first started from the work of Vicsek et al.[17], and was further motivated by the hydrodynamic approach[16].Our system(1.1)can be viewed as a close to equilibrium model for the hydrodynamic Cucker-Smale model[5,9].For the detailed derivation,we refer the reader to[8,11].

Recently,Ha,Huang and Wang[8]studied this problem with the initial density ρo∈Lloc(ℝ) and ρo>o a.e.in ℝ.However,this is not natural since the solution to this system is measure in general.So we should consider this problem in measure space rather than in function spacefor the general case.In this paper,we will settle this problem and our uniqueness theorem is obtained in the same space,i.e.,Radon measure space as the existence theorem.

Now let us review some related work for the homogeneous counterpart of our system(1.1), which reads as

This model is used in plasma physics and has attracted a great deal of attention in mathematic since its solution is measure in general,which poses new challenge to the analysis of the posedness.For the existence of the global weak solution,the result was first obtained independently by Brenier and Grenier[4]and E et al.[7].Wang et al.[18]extended their results.Boudin[3] showed that the weak solution can also be obtained as limit of solutions to a viscous pressureless model.For the uniqueness of the global weak solution,the authors of[1,4]and[7]found that the Lax entropy condition was insufficient to guarantee it.E et al.[7]pointed out that the Oleinik entropy condition might be necessary.Along this line,Wang and Ding[18]proved the uniqueness of the weak solution for the case that the initial density ρo∈Lloc(ℝ)and ρo>o a.e..Similar results were also obtained by Bouchut and Jame[2].As for the general case that ρois a Radon measure,it is more subtle and difficult.Huang,Wang[14]found that besides the Oleinik entropy condition,it is also important to require the energy to be weakly continuous initially.They called it energy condition and further proved that the energy condition is necessary and sufficient for the uniqueness.Our idea in this paper mainly comes from their famous work.

The strategy is as follows:First,we manage to construct the entropy solution by the potential functional

with mo(x)=ρo([o,x)).Then we show that any entropy solution coincides with the solution we construct.Thus the uniqueness is achieved.We find that F(y;x,t)only depends on the initial data,which will play a crucial role in our proof.We shall investigate the properties of its minimizer.It is the basis of our proof.When ρois a Radon measure,we can first prove the uniqueness in the region t≥t1>o instead of t≥o.This is because the characteristic curves issued from some points of x-axis may not be unique.Then we study the convergence of the sequence(ρt1,ut1),as t1→o+o.In order to prove the limit coincides with the entropy solution we construct,we need the energy condition since ρomay have mass on some points of x-axis.Besides,we can show that the energy condition is necessary by counterexample.

The main feature of the pressureless system is the formation of δ-shock no matter how smooth the initial values are.Motivated by the previous work,we define the following potential function m(x,t)by

Due to the conserved equation(1.1)1,m(x,t)is independent of the integral path and satisfiesmx=ρ.Thus,system(1.1)is transformed into

Now we will focus our attention on this system and give the definition to the entropy solution.

Definition 1.1Let m(x,t)be of bounded variation locally in x and u(x,t)be bounded and measurable to mx.Assume that the measure mxand umxare weakly continuous in t. (ρ,u)=(mx,u)is called a weak solution of(1.1),if

Here

ℝ×ℝ+···dmdt denotes Lebesgue-Stieltjes integral.The initial value is understood in the following sense:as t→o+o,m(x,t)converges to mo(x)a.e.and the measure ρu weakly converges to ρouo,where uois bounded and measurable to ρo.

Definition 1.2Let(ρ,u)be a weak solution of(1.1).Then(ρ,u)is called an entropy solution of(1.1)if

holds for any x1o and the measure ρu2weakly converges toas t→o+o.

Our main results read as:

Theorem 1.3(Existence theorem)Let o≤ρo∈Mloc(ℝ)and uobe bounded and measurable to ρo.Then system(1.1)admits at least one entropy solution.

Theorem 1.4(Uniqueness theorem)Let o≤ρo∈Mloc(ℝ)and uobe bounded and measurable to ρo.Suppose(m1,u1)and(m2,u2)are two entropy solutions of(1.1)with the same initial data ρo,uoin the above sense.Then u1=u2a.e.with respect to the measure m1x=m2x.

The rest of the paper is arranged as follows.In Section 2,we construct the entropy solution and study its structure in detail.In Section 3,we study the general uniqueness for the system.

2　Existence of Entropy Solution

In this section,we use the generalized potential functional

to construct an entropy solution,where mo(x)=ρo([o,x))is an increasing function.Since ρo=o is trivial,we assume that there always exists ao,i.e.,the support of ρois not empty set.Before constructing the entropy solution,we give some lemmas first.

Lemma 2.1For any point(x,t),as a function of y,F(y;x,t)has a finite low bound.

ProofFor any y1y2≥x+M(1−e−t),where M:=‖uo‖L∞, we have

This implies the lemma.

Since F(y;x,t)is left continuous with respect to y,for any yo∈S(x,t),the following holds:

Thus we have

Lemma 2.2Assume that yo∈S(x,t)and[mo(yo)]=mo(yo+o)−mo(yo−o)>o,then

ProofSince uois measurable to ρo,uomust be well defined at the point yo,where ρohas a positive mass.We calculate This implies the lemma.

Lemma 2.3Let(xn,tn)and yn∈S(xn,tn)converge to(x,t)and yorespectively.Then yo∈S(x,t).

ProofBy the definition of S(x,t),we only need to prove

Without loss of generality,we suppose v(x,t)=F(¯y;x,t),then

where we assume F(y;xn,tn)achieves its minimum at yn.Since F(y;x,t)is continuous in(x,t) and ynis bounded,we have

Then the definition of S(x,t)leads to yo∈S(x,t). It is easy to see y∗(x,t),y∗(x,t)∈S(x,t)∩sptρo.For each point(xo,to),we introduce the left and right backward generalized characteristics L1,L2as follows:

We claim that y∗(x,t)=y∗(x,t)along the backward characteristics.

Lemma 2.4For any yo∈S(xo,to),y∗(x,t)=y∗(x,t)holds along the curve Furthermore,y∗(x,t)=y∗(x,t)≤y∗(xo,to)along the curve L1,and y∗(x,t)=y∗(x,t)= y∗(xo,to)along L2.

ProofWe assume v(xo,to)=F(yo;xo,to).For any point(x,t)on the curve

which implies yo∈S(x,t).By the definition of y∗(x,t)and y∗(x,t),we have y∗(x,t)=y∗(x,t) along L.Furthermore,y∗(x,t)=y∗(x,t)≤y∗(xo,to)along L1and y∗(x,t)=y∗(x,t)= y∗(xo,to)along L2.

In terms of Lemma 2.4,for each point(x,t),the backward generalized characteristics L1,L2and x-axis form a characteristic area.We denote it△(x,t).Fix the time t.As x changes,all these characteristic area never interact with each other except x-axis.Thus,we have the following lemma.

Lemma 2.5y∗(x,t),y∗(x,t)are increasingly monotonic in x.In addition,y∗(x1,t)≤y∗(x2,t)holds for any x1

ProofIf there exist two points(x1,to)and(x2,to)with to>o such that x1y∗(x2,to),then the curve

must interact with the curve at a point(xs,ts),oy∗(x2,to).

Lemma 2.6Each point(xo,to)at to>o uniquely determines a Lipschitz continuous curve L:x=x(t),xo=x(to).In addition,for any t∈{τ:τ≥to},

which contradicts the fact that all characteristic area never interact with each other above the x-axis.

Thus a(t1)=b(t1),i.e.,there is a unique point x=x(t1)on the line t=t1,which is determined by(xo,to).Since t1>tois arbitrary,we obtain a curve x=x(t)with xo=x(to) in t>to.We call it the forward generalized characteristic generated by(xo,to).Furthermore, x(t)is continuous due to(2.2),(2.3).

Next we calculate the upper derivative x′(t).Our argument is divided into two situation.

(1)y∗(x,t)

For any t′>t,we denote x′=x(t′),y′=y∗(x(t′),t′),y′=y∗(x(t′),t′).Then we claim

We prove by means of contradiction.Without loss of generality,we assume y′≤y∗(x,t). Now we choose a sequence(xn,t′)with xn→x(t′)+o,as n→∞.Then

which contradicts the definition of y∗.Therefore the claim is true.

For any t′′>t′>t,in the same way,we denote

Without loss of generality,we assume

Since v(x′′,t′′)and v(x′,t′)converge to v(x,t)as t′′,t′→t+o and

(2)y∗(x,t)=y∗(x,t).

There are four subcases in this situation.

(a)(x,t)∈V2.

The proof is the same as(1).

(b)(x,t)∈V3.

By the definition of x=x(t),we have

It is easy to calculate that

Letting t′→t+o yields

(c)(x,t)∈V4.

In this case,it is easy to see(x(t′),t′)∈Lta,t≤t′≤ta,where

Since(x,t)/∈Lta,without loss of generality,we assume

where xn→x(ta)+o,as n→∞.

By the definition of y∗,we have y∗(y(τ),τ)=y∗(y(τ),τ)=y∗(x,t)along the curve

By the definition of L,we have x≤y(t).Next we prove x=y(t)by contradiction.

Suppose x

Obviously,the curve x=x(t)in Lemma 2.6 is Lipschitz continuous.We define

Obviously,u(x,t)is well defined inSome properties of u(x,t)follow from Lemma 2.6.

(2)For a.e.t>o,u(x,t)satisfies

(3)For any x1o,u(x,t)satisfies

Proof(1),(2)are easy due to the definition of u(x,t)and Lemmas 2.5,2.6.We only prove(3).For any x1

Therefore,(3)is proved.

Now we introduce the forward generalized characteristics originated from the x-axis.For each point η∈sptρo,let

where N1={x:y∗(x,t)<η},N2={x:y∗(x,t)>η}.By the definition of˜a(η,t)and b(η,t), it is easy to see tha t(η,t)≤b(η,t)and˜a(η,o)≥η.Let

Each point(η,o)may generate at least one Lipschitz continuous curve x(t).If there is only one forward generalized characteristic x(t)from(η,o),we denote X(η,t)=x(t).If there are at least two curves from(η,o),we define

at[mo(η)]>o.Thus we have defined X(η,t)for η∈sptρo.For each point η∈ℝsptρo,let c(η,t)=η+M(1−e−t).If c(η,t)does not interact with other curves X(˜η,t),˜η∈sptρo,we denote X(η,t)=η+M(1−e−t).Otherwise,we denote

where tηois the first time that c(η,t)interact with X(ηo,t),ηo∈sptρo.

It is easy to see X(η,t)is increasing in η and Lipschitz continuous in t>o.Thus X(η,t) is well defined with respect to the measure ρo.By Lemma 2.6,we have

if|X(η,t)−y∗(X(η,t),t)|≤M(1−e−t)and η∈sptρo.Next we define the following quantities:

Obviously,m(x,t),q(x,t)and E(x,t)are of bounded variation locally in x.Besides, u(X(η,t),t)is measurable to the measure ρo.For theses functions,we have the following relations.

Lemma 2.8In the sense of Radon-Nikodym derivatives,for a.e.t>o,m(x,t),q(x,t) and E(x,t)satisfies

(1)qx=umx,

ProofFirst we consider the simplest case that there is a neighborhood U(x,r)of x such that y∗(ξ,t)keeps constant if ξ∈U(x,r).Namely,for any point ξ∈U(x,r),y∗(ξ,t)=y∗(x,t). There are two subcases in this situation.

Case 1m(ξ,t),q(ξ,t)and E(ξ,t)keep constant for ξ∈U(x,r).We note that the case |ξ−y∗(ξ,t)|>M(1−e−t)belongs to Case 1.This case is trivial.

Case 2m(ξ,t),q(ξ,t)and E(ξ,t)take two values for ξ∈U(x,r).Without loss of generality,we suppose m(x−o,t)/=m(x+o,t).It is easy to see ρohas a positive mass at (yo,o),where yo=y∗(x,t).Thus Lemmas 2.2 and 2.6 give

which imply the lemma.

Next we turn to the case that y∗is not a constant for any neighborhood of(x,t).Let x1

For y∗(x,t)=y∗(x,t),there are four subcases:(a)(x,t)∈V2,(b)(x,t)∈V3,(c)(x,t)∈V4, (d)(x,t)∈V5.

For case(a),it is similar to the case y∗(x,t)

For case(b),without loss of generality,we assume that

By the definition of v(x,t),we have

For case(c),we can prove qx=umxin the same way as in case(b).

For case(d),by the definition of x(t),we have

It is observed that ρo=o on(y∗,yta).When

Letting x1→x−o,x2→x+o,we have

where we have used the fact that y2→ytaas x2→x+o.Thus qx=umx.The first part of Lemma 2.8 is proved.

On the other hand,it is observed that

If y∗(x−o,t)o,we have

Lemma 2.9

ProofLet(z1,z2)be any subdivision of the interval(x1,x2).Without loss of generality,

Then we get(2.11)by integrating m(x,t)over(x1,x2).Similarly,let(τ1,τ2)be a subdivision of the interval(t1,t2).We assume

In the same way,we can achieve

Proof of Theorem 1.3From Lemma 2.9,we have vx=−m(x,t)and vt=q(x,t)in the sense of distributions.In fact,for any φ∈we have

Therefore,the first relation of(1.6)is proved.

In order to prove the second equation of(1.6),we need to introduce another generalized potential to get some information on q,E like Lemma 2.9.Similar to F(y;x,t),we define

where k is any constant satisfying k>M.Next we prove G(y;x,t)and F(y;x,t)achieve their minimum at the same point for a.e.[mx]x∈ℝ.By the definition of X(η,t),we know X(y∗(x,t),t)≤x for a.e.[mx]x∈ℝ.Without loss of generality,we suppose v(x,t)= F(y∗(x,t);x,t).We only need to prove

Since qx=umxfor a.e.τ∈[o,t],and y∗(x,t)∈S(X(y∗(x,t),τ),τ),o<τ≤t,we have

Then we obtain

by integrating the above equation from o to t.Therefore

It is easy to see F(y;x,t)and G(y;x,t)achieve their minimum at the same point y∗(x,t)for x∈sptmx,since k+uo(η)>o.We denote

Similar to Lemma 2.9,we have

in the sense of distributions.Since k>M is arbitrary,we have

in the sense of distributions.For any

Therefore,the second relation of(1.6)is proved.

To prove(m,u)is an entropy solution of(1.1),it is sufficient to prove m(x,t)converges to mo(x)a.e.,umxand u2mxare weakly continuous at t=o.We only study the point x, where mo(x)is continuous.If there exists tosuch that|x−y∗(x,to)|>M(1−e−to),then there must exist yto∈S(x,to)such that|x−yto|

Now we study the initial condition for the energyWe only need to prove u(X(η,t),t)→ uo(η)a.e.with respect to the measure ρo,as t→o+o.We denote

Then ρo(ℝP)=o,since uo(η)and u(X(η,t),t)are bounded measurable with respect to ρo. From qx=umx,we can derive

Therefore we get Then we have E(x,t)→Eo(x)a.e.as t→o+o.

In addition,it is easy to verify for a.e.x∈ℝ,

with any yo∈sptρo∩S(xo,to)and

3　Uniqueness of Entropy Solution

This section is devoted to the uniqueness of entropy solution.We assume(ρ,u)is any entropy solution in the sense of Definitions 1.1 and 1.2.We have

Let M=‖uo‖L∞,we have|u(x,t)|≤Me−ta.e.[mx]since u satisfies

Take|u(x,t)|≤Me−t.Next we consider the following linear equation:

Definition 3.1Suppose that m(x,t)is of bounded variation locally in x and the measure mxis weakly continuous in t.Then m(x,t)is a weak solution of(3.2)if

Some properties of these characteristics was studied in[13,18].We state them in the following lemma.

Lemma 3.2(1)There exists a subsequencefor all η and t.Furthermore,Xt1(η,t)is Lipschitz continuous with respect to t and is increasing with respect to η.

(2)If Xt1(ξ1,to)=Xt1(ξ2,to)holds for some ξ1<ξ2and to>t1>o,then Xt1(ξ1,t)= Xt1(ξ2,t)for all t≥to.Furthermore,Xt1(ξ,t),ξ∈ℝ cover the domain?(x,t):t≥t1?.

(5)Let ξt1(x,t)=sup{ξ:Xt1(ξ,t)x}.Then the set Γt1={(x,t):ξt1(x,t)/=ηt1(x,t)}consists of Lipschitz continuous curves.We call these curves shock waves of u.Furthermore,

and ξt1(x1,t)→ξt1(x,t)as x1→x−o,ηt1(x2,t)→ηt1(x,t)as x2→x+o.

(6)For any point(xo,to),to>t1,there exists at least one curve L′through(xo,to)such that ξt1(x,t)keeps constant along the curve.

In the sense of Definition 3.1,we have the following existence and uniqueness results[14].

Theorem 3.3Suppose that m(x,t1)∈BVloc(ℝ),t1>o and u(x,t)satisfies(3.1).Then (3.2)admits only one weak solution in t≥t1.Furthermore,

Lemma 3.4qx=umx,t≥t1holds in the sense of Radon-Nikodym derivatives.

ProofFor any(x1,t),(x2,t),we denote

It is easy to see ξt1(x1(τ),τ)and ξt1(x2(τ),τ)keep constant along the curves x1(τ)and x2(τ) respectively.Denote ξ1=ξt1(x1(τ),τ),ξ2=ξt1(x2(τ),τ).Let φ1∊(x),φ2∊(x)∈C∞(ℝ)satisfying

Then by(1.6)2,we haveZ

For any t1≤s

Then substituting(3.7)into(3.6)and letting δ→o yield

Direct calculation on(3.9)gives that for any t1≤s≤t,

Due to the arbitrariness of ξ1and ξ2,we have qx=umx,t≥t1.

On the other hand,we choose another integral path that is from(o,t1)to(yo,t1)along the line t=t1and from(yo,t1)to(x,t)along the curve

Since yois arbitrary,we have

On the other hand,by the existence Theorem 1.3,we can construct a standard entropy solution(mt1,s,ut1,s)satisfying mt1,s=(Φt1)x,qt1,s=−(Φt1)t,where(qt1,s)x=ut1,s(mt1,s)x. We note that the constructive procedure above starts from the line t=t1>o.Since m=(Φt1)x, q=−(Φt1)t,we have m=mt1,s,q=qt1,sa.e.in t≥t1>o.Therefore the uniqueness of the entropy solution(1.1)in the region t≥t1>o is proved.Next,we prove m(x,t),q(x,t)only depends on the initial value.For

Using(3.1o),we obtain

It is easy to see ξt1(x,t)∈St1(x,t)since Xt1(ξt1(x,t),t)=x.Now we choose a sequence of ti>o,i=1,2,···,such that ti→o as i→∞.By Lemma 2.4 and(2.14)–(2.16),m and q keep constant along the curve

It is easy to see the map from(x,t)to ξ(x,t)is not one to one.If there exists x1

For any(x,t)∈ℝ×ℝ+V,it is to see ξ(x,t)∈ℝU.Thus

Next we claim that m(x1,t1)=m(x2,t2)for any(x1,t1),(x2,t2)∈Γxo(xo∈ℝU).Suppose not.From Theorem 3.3,we can infer that there exists o

which implies xo∈U.This contradicts the assumption that xo∈ℝU.Thus we prove the claim.

Then m(x,t)is a function of ξ(x,t)for(x,t)∈ℝ×ℝ+V and there exists an increasing function w(·)such that

However,according to the result of[15],the domain of w(·)may be a zero measure set,but we can infer that it is dense in ℝU from Lemma 3.2(5).

By Theorem 3.3,we know m(x,t)is left continuous with respect to x.Thus,w(·)is also left continuous.Using this property and the density of w’s domain in ℝU,we can define w(x) for all x∈ℝU.At the continuous point of w(·),we haveξ(x,t)=x.On the other hand,from Definition 1.1,we know m(x,t)takes the initial data in the following sense:

Combining with the fact that both w(x)and mo(x)are left continuous with respect to x,then it leads to Accordingly,we define

Next we turn to the case that a whole interval maps into one point ξ(x,t),without loss of generality,we assume ξ(x,t)=o.We denote x1(τ)=X(o−o,τ),x2(τ)=X(o+o,τ), o≤τ≤t.For any x∈[x1(τ),x2(τ)],

since mx,umxand u2mxweakly converge to ρo,ρouoandin measure as t→o+o, respectively.Thus we have

If[mo(o)]=o,m(x,t)is constant in(x1(t),x2(t))and

If[mo(o)]>o,m(ξ)must be discontinuous at ξ=uo(o).From(3.14),we have ξ1≤uo(o)≤ξ2. This means m(ξ)takes two values determined by ξ=uo(o),namely,

Accordingly,we define

It is observed that the previous argument for t1>o applies to the case t1=o if m can be expressed by its initial data.Therefore,we have

where Φ(x,t)=−infyF(y;x,t).

Thus m(x,t)and q(x,t)coincides with the standard entropy solution we construct a.e.inℝ×ℝ+.Therefore the uniqueness is proved.

Finally,we show that the energy solution is necessary for the uniqueness of the entropy solution.Without the energy condition,we are able to construct at least two weak solutions satisfying the Oleinik type entropy condition with the same initial data.For example,we consider the initial dada

We can construct two weak solutions

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∗November 24,2015;revised April 11,2016.