FINITE TIME EMERGENCE OF A SHOCK WAVE FOR SCALAR CONSERVATION LAWS VIA LAX-OLEINIK FORMULA?

Zejun WANG Qi ZHANG

Department of Mathematics,Nanjing University of Aeronautics and Astronautics,Nanjing 210016,China

E-mail:wangzejun@gmail.com;zhangqinju@gmail.com

Abstract In this paper,we use Lax-Oleinik formula to study the asymptotic behavior for the initial problem of scalar conservation law ut+F(u)x=0.First,we prove a simple but useful property of Lax-Oleinik formula(Lemma 2.7).In fact,denote the Legendre transform of F(u)as L(σ),then we can prove that the quantity F(q)? qF′(q)+L(F′(q))is a constant independent of q.As a simple application,we fi rst give the solution of Riemann problem without using of Rankine-Hugoniot condition and entropy condition.Then we study the asymptotic behavior of the problem with some special initial data and prove that the solution contains only a single shock for t>T?.Meanwhile,we can give the equation of the shock and an explicit value of T?.

Key words scalar conservation law;Lax-Oleinik formula;Riemann problem;asymptotic behavior.

1 Introduction

In this paper,we study the following scalar conservation law

where F(u):R→R is a C2function and is uniformly convex.The initial data is given by

where u0∈L∞and measurable.This problem was widely studied by many authors.The existence of global weak solution to(1.1)–(1.2)can be proved by Lax-Friendrich scheme[1,2],viscosity method[3–6],Glimm Scheme[7]and compensated compactness method[8].The uniqueness of solution to(1.1)–(1.2)is due to Oleinik[9]where the entropy condition was given(see(2.15)below).The asymptotic behavior of(1.1)–(1.2)was also studied by many authors.Earlier studies can be found in[10,11].When u0(x)is bounded and measurable,then u(x,t)decays to 0 at a rate O(t?1/2)and when the support of u0(x)is compact,u(x,t)decays to an N-wave at a rate O(t?1/2).This can be proved by characteristic method[2]or by Lax-Oleinik formula[12].Lax-Oleinik formula is an explicit formula of the solution to(1.1)–(1.2)which was initially obtained by Lax[11]and Oleinik[9],also see[12,13].

In Section 2 of this paper,we’ll recall some basic facts of Lax-Oleinik formula and give two related properties.As an example,we use these properties to study Riemann problem and give the exact solution without help of the Rankine-Hugoniot condition and entropy condition.In fact,the weak solution of this Riemann problem is well known and can be easily given(see[2,12,14]).Then in Section 3,we will go on to study the fi nite time emergence of a shock when

with u?>u+,and u0(x)is bounded measurable in[?R,R].To study this problem,an efficient method is generalized characteristics(see[15–17]).It was shown that a single shock must appear in fi nite time.For some generalization of generalized characteristics to inhomogeneous conservation laws,see[18,19].In[13],Lax-Oleinik formula was used in the special case of stationary shock solutions of Burgers equation where F(u)=u2/2.A simpli fi ed proof by using of generalized characteristics can be found in[20].In Section 3 of this paper,we use Lax-Oleinik formula to give another proof,and we also give the equation of the shock and an estimate on the time of the emergency of the shock.

2 Preliminary

Denote L as the Legendre transform of F,that is,

Here for simplicity,we denote g(y)=F′(u0(y)).This equation gives a relation of x,t and y,which is a necessary condition for y to be a minimum point of A(y;x,t).

Remark 2.1(2.9)has the same form as the characteristic of equation(1.1).However,they are not exactly the same.For fi xed y,every line satisfying(2.9)is called a characteristic.But here(2.9)is only a necessary condition that y is the minimum point of A(y;x,t)since y can also be a maximum point.

Lemma 2.2For any x,t,y satisfying(2.9),we have

ProofDi ff erentiating identical equation G(F′(u))=u with respect to u,we get G′(F′(u))F′′(u)=1.Take u=u0(y)and use(2.9),then we get(2.10). ?

Lemma 2.3If u0(y)∈C1,then for any x,t,y satisfying(2.9),we have

(i)if 1+g′(y)t>0,then y is a minimum point of A(y;x,t);

(ii)if 1+g′(y)t<0,then y is a maximum point of A(y;x,t).

ProofIf u0(y)∈C1,we can easily get

Thus,to assure that y is a minimum point of A(y;x,t),a sufficient condition is+(y)>0,which,by using of Lemma 2.2,is equivalent to 1+g′(y)t>0.Thus(i)is proved.

(ii)can be proved analogously.?

Theorem 2.4(see[12]) Assume F:R→R is smooth,uniformly convex,and u0(x)∈L∞(R).

(1)For each time t>0,there exists,for all but at most countably many values of x∈R,a unique point y(x,t)such that

(2)The mapping x→y(x,t)is nondecreasing.

(3)For each time t>0,the function u de fi ned by

is a weak solution of the initial-value(1.1)–(1.2).

(2.13)is called Lax-Oleinik formula.This theorem shows that the solution of(1.1)–(1.2)can be given as in(2.13)and y belongs to the set of minimum points of the following problem

Theorem 2.5(see[12]) Under the assumptions of Theorem 2.4,there exists a constant C such that the function u de fi ned by the Lax-Oleinik formula satis fi es the entropy inequality

for all t>0 and x,z∈R,z>0.

Theorem 2.5 shows that Lax-Oleinik formula satis fi es the entropy condition,which assure the uniqueness of the solution to(1.1)–(1.2),so we fi rmly believe that Lax-Oleinik formula is a useful tool to study the scalar conservation law.

Before going to our main results,we give two properties of Lax-Oleinik formula.Since u0(x)∈L∞,we have for some constant M,

Lemma 2.6For every y

and for every y>x?Nt,we have

ProofAssume that y

Thus we get(2.17).(2.18)can also be proved similarly. ?

This lemma shows that to study the minimum point in problem(2.14),we need only consider the values of y in between[x?Mt,x?Nt].

In this paper,for any two given constant states u?and u+,we always denote

Thus we have

The following lemma is important in our analysis below.It gives a closed relation between L and F.

Lemma 2.7For any q ∈ R,the quantity F(q)?qF′(q)+L(F′(q))is a constant.Specially,for any constant states u?and u+,we have

ProofDenote Q(q)=F(q)? qF′(q)+L(F′(q)).It is easy to verify that Q′(q)=0 with the help of(2.3)and(2.4). ?

Next,as a simple application,we use Lax-Oleinik formula to study the Riemann problem of(1.1),i.e.,the initial data is given by

Riemann problem for scalar conservation law is well known.As stated in Section 1,when u?>u+,with the help of Rankine-Hugoniot condition and the entropy condition,we can get the solution.The solution contains a single shock which divide the upper plane into two constant states(see(2.23)).When u?

The following theorem gives the solution of the Riemann problem(1.1)–(2.22)for the case u?>u+by Lax-Oleinik formula without using of Rankine-Hugoniot condition and the entropy condition.

Theorem 2.8For Riemann problem(1.1)–(2.22),if u?>u+,the solution can be written as

where

ProofLet’s fi rst calculate the value of the minimum point y in(2.14).From(2.9),and with the help of Lemma 2.6,we can get

Since u?>u+,we have σ?> σ+.See Fig.2.1 and Fig.2.2.?

Fig.2.1

Fig.2.2

Obviously,if x< σ+t(or x> σ?t),there exists a unique y?=x ? σ?t(or y+=x ? σ+t)satisfying(2.25).Since==0,from(i)of Lemma 2.3,we know that both y?and y+are the minimum points of problem(2.14).Thus

Thus when x<σ+t,from(2.13),we can get

By the same way,we can get u(x,t)=u+for x> σ?t.In summary,(2.23)holds for x> σ?t and x< σ+t.For the case of σ+t

Lemma 2.9For(x,t)satisfying σ+t

ProofWe only prove the case of“<”since the other cases can be proved similarly.When σ+t0,and we can easily know that y?and y+are both minimum points.Thus(2.27)is equivalent to

By using of(2.21)in Lemma 2.7,L(σ?)?L(σ+)=F(u+)?F(u?)+σ?u??σ+u+,then(2.28)is equivalent to

The proof of Lemma 2.9 is completed.?

From Lemma 2.9,we know that when σ+t

Remark 2.10On the line x=st,y?=x? σ?t and y+=x? σ+t are both minimum points,thus on this line,u(x,t)is double valued.As is well known,the line x=st is called shock wave,and s is its velocity,s=is the well knows called Rankine-Hugoniot condition.

Remark 2.11By similar analysis,for the case u?

In fact,since u?

For all x which are in between σ+t

3 Asymptotic Behavior

In this section,we study the asymptotic behavior of the Cauchy problem(1.1)–(1.2)where u0(x)is in L∞,measurable and satis fi es(1.3).When u?>u+,we have σ+

where α is given in(2.5).The following is the main theorem in this section which shows that a single shock must appear in fi nite time.

Theorem 3.1For Cauchy problem(1.1)–(1.2)satisfying(1.3),if u?>u+,then for t>T?,the solution can be written as

where s is given by(2.24)and

We will prove Theorem 3.1 by several lemmas.

Lemma 3.2Given(x,t)∈R×(0,+∞),then the following hold

(i)if x?σ+t>R,then for any y∈[R,+∞),we have

Remark 3.3(i)and(ii)of Lemma 3.2 show that if x?σ+t lies in[R,+∞),then it is the unique minimum point of(2.14)in[R,+∞).If x?σ+t

Lemma 3.4Given(x,t)∈R×[0,+∞)satisfying x?σ?tR,a sufficient and necessary condition of

ProofWe only prove the case of“<” since the other cases can be proved similarly.In this case,(3.8)is equivalent to

For de fi niteness,we denote three lines:l1:x=?R+σ?t,l2:x=x0+st,l3:x=R+σ+t(see Fig.3.1),then x?σ?t< ?R or x?σ+t>R means on the left side of l1or on the right side of l3respectively.

Fig.3.1

(3.19)and(3.20)imply(3.9).The proof is completed. ?

From Lemmas 3.2 to 3.5,we know that when t>T?and(x,t)is on the left side of both l1and l2,we can choose y=x?σ?t and then u=u?.When t>T?and(x,t)is on the right side of both l2and l3,we can choose y=x?σ+t and then u=u+.Thus when t>T?,the solution is given as in(3.2),and x=x0+xt is the unique the shock wave.As shown in Fig.3.1,u=u?in the region ??and u=u+in the region ?+.

Remark 3.6If R=0,we can easily know that x0=0 and T?=0.This is exactly the case in Theorem 2.8.

Remark 3.7Denote the time when l1and l2intersects as T1,and the time when l2and l3intersect as T2,then t>T?implies t>T1and t>T2since max(T1,T2)≤.