Conservation Laws for Relativistic Fluid Dynamicsweb.unbc.ca/~chenj/FluidDynamics.pdf ·...

Conservation Laws forRelativistic Fluid Dynamics

JINGING CHENHEN

Communicated by C. DAFERMOSAFERMOS

1. Introduction

In his fundamental paper of 1948, TAUBAUB [T1] derived the equations ofrelativistic ¯uid dynamics:

@

@t

�n�1ÿ v2�ÿ1=2

�� @

@x

�nv�1ÿ v2�ÿ1=2

�� 0;

@

@t

��q� p� v

1ÿ v2

�� @

@x�q� p� v2

1ÿ v2� p

� �� 0;

@

@t�q� p� v2

1ÿ v2� q

� �� @

@x

��q� p� v

1ÿ v2

�� 0:

Here n is the rest mass density, q is the proper energy density, p is the pressureand v is the particle speed. TAUBAUB then obtained the Hugoniot curve ofrelativistic shocks, and also showed that c; the ratio of speci®c heats, must beless than 5

3. He gave a more systematic description of relativistic hydrody-namics in [T2]. In 1986, THOMPSONHOMPSON [Th] established several relations on therelativistic shock curves. He observed: ``The relativistic shock equations aremuch more complicated and do not lend themselves to expressions that areboth simple and general.'' More recently, instead of working on the fullsystem, SMOLLERMOLLER & TEMPLEEMPLE [ST] consider the system of conservation laws ofenergy and momentum in special relativity. For the equation of state p � r2q;they solved the Riemann problem and the Cauchy problem for the system.CHENHEN [C] extended their results to the general relativistic p-system where theequation of state is p � p�q�, p0�q� > 0 and p00�q�3 0:

In this paper, we rigorously develop the mathematical theory of relativ-istic ¯uid dynamics. Although the system is much more complicated and theresults are much harder to obtain, we establish a complete analogy betweenclassical and relativistic hydrodynamics. We also show that the Newtonianlimits of our results reduce to the classical results for which the reader is

Arch. Rational Mech. Anal. 139 (1997) 377±398. Ó Springer-Verlag 1997

referred to [S, Chap. 18]. Because of the complexity of the relativistic system,we develop new techniques and criteria. By using the Riemann invariants

R � 1

2ln

1� v1ÿ v

� ��Zq

0

��pqp

p � qdq; S � 1

2ln

1� v1ÿ v

� �ÿZq

0

��pqp

p � qdq

we introduce a simple criterion to exclude the formation of vacuum, that is,

RL > SR

where RL is the value of R at x 2 0 and SR is the value of R at x > 0. We noticethat in relativistic p-systems, the criterion to exclude the formation of vacuumtakes the same form [C]. We also note that in nonrelativistic cases, the cri-terion to exclude the formation of vacuum can also be written as RL > SR aswe shall see later. In fact, this criterion can be used in general hyperbolicsystems to detect the loss of strict hyperbolicity. The use of the � p; v�-plane tosolve the Riemann problem, following [CF], has been restricted by theseemingly paradoxical situation that there are two distinct admissible solu-tions [STX]. We show that the � p; v�-plane can always be used to uniquelydetermine the solution of the Riemann problem, as long as we keep in mindthe correct entropy condition.

This paper is organized as follows. In Section 2, we give the backgroundof the problem. In Section 3 we determine the characteristic speeds of thesystem, which are

k1 �vÿ ��

pqp

1ÿ v��pqp ; k2 � v; k3 �

v� ��pqp

1� v��pqp :

We can recognize that k1 and k3 are the sums of sound speed and particlespeed under the Lorentz transformation. From here we obtain a simpleformula ��

pqp

for the speed of sound in the relativistic setting. This is much simpler thanTAUBAUB's [T2] original formula, which is given by��

n1� i

didn

� �s

swhere i � �e� p�=n is the rest speci®c enthalpy. We prove below that thespeed of sound is strictly less than the speed of light. We prove that thesystem is strictly hyperbolic with the ®rst and third characteristic familiesgenuinely nonlinear and the second characteristic family linearly degenerateand we ®nd the Riemann invariants. In Section 4, we show that the Laxshock inequalities are satis®ed globally and entropy is monotone along theshock curves. We also derive the condition preventing the occurrence of avacuum and then prove the existence and uniqueness of the solution of theRiemann problem.

JINGING CHENHEN378

2. Preliminaries

This section is a brief account of Section 10, Chapter 2 of WEINBERGEINBERG [W].For more details, consult that book. In the ¯at 2-dimensional space-time, theMinkovski metric takes the form

gij � ÿ1 00 1

� �:�2:1�

The speed of light is taken to be unity, i.e., c � 1: The general conservationlaw is

riT ij � 0; j � 0; 1;�2:2�where T ij is the stress-energy tensor for a perfect ¯uid. In special relativity,(see [W, ST]) it is given by

T ij � � p � q�U iU j � pgij; i; j � 0; 1;�2:3�where p is the pressure, q is the proper energy density, and Ui is the velocityvector, i.e.,

U0 � 1

�1ÿ v2�1=2; U 1 � v

�1ÿ v2�1=2;

in which v is the velocity of the moving particle. We can write down T ij

explicitly as

T 00 T 01

T 10 T 11

� ��

q�pv2

1ÿv2 � p � q� v1ÿv2

� p � q� v1ÿv2 p � � p � q� v2

1ÿv2

!:�2:4�

The conservation laws of energy and momentum thus become

@

@t�q� p� v2

1ÿ v2� q

� �� @

@x

��q� p� v

1ÿ v2

�� 0;�2:5�

@

@t

��q� p� v

1ÿ v2

�� @

@x�q� p� v2

1ÿ v2� p

� �� 0:�2:6�

Apart from energy and momentum, the rest mass is also conserved in the¯uid:

@

@t

�n�1ÿ v2�ÿ1=2

�� @

@x

�nv�1ÿ v2�ÿ1=2

�� 0 ;�2:7�

where n is the density of the rest mass.As in [Th], we assume that the equation of state is given by

qÿ n � p=�cÿ 1� ;�2:8�

p � ksncÿ1 ;�2:9�where s is the entropy and k and c; 1 < c < 5

3 ; are positive constants.

Conservation Laws for Relativistic Fluid Dynamics 379

From the conservation of energy, momentum and rest mass, as in theclassical case, we can deduce the entropy equation

@s@t� v

@s@x� 0 :�2:10�

Throughout the paper we assume that the reader is familiar with thenotions and terminology of conservation laws as discussed in [L, S], forexample.

3. Eigenvalues and Eigenvectors of the System

Our system is

@

@t

�n�1ÿ v2�ÿ1=2

�� @

@x

�nv�1ÿ v2�ÿ1=2

�� 0;

@

@t

��q� p� v

1ÿ v2

�� @

@x�q� p� v2

1ÿ v2� p

� �� 0;

@

@t�q� p� v2

1ÿ v2� q

� �� @

@x

��q� p� v

1ÿ v2

�� 0

�3:1�

where the pressure p � �cÿ 1��qÿ n�:First we will ®nd the characteristic speeds of the system. Let

A �

n��1ÿv2p

�q� p� v1ÿv2

�q� p� v21ÿv2 � q

0BB@1CCA; B �

nv��1ÿv2p

�q� p� v21ÿv2 � p

�q� p� v1ÿv2

0BB@1CCA:

The Jacobian of A with respect to �n; v; q� is

JA �

1��1ÿv2p nv

�1ÿv2�3=2 0

ÿ�cÿ1�v1ÿv2

�q�p�1ÿv2 � 2

�q�p�v2�1ÿv2�2

cv1ÿv2 � 1

ÿ �cÿ1�v21ÿv2 2 �q�p�v

1ÿv2 � 2 �q�p�v3�1ÿv2�2

gv2

1ÿv2 � 1

0BBBB@1CCCCA�3:2�

�

1��1ÿv2p nv

�1ÿv2�3=2 0

ÿ�cÿ1�v1ÿv2

�q�p��1�v2��1ÿv2�2

cv1ÿv2 � 1

ÿ �cÿ1�v21ÿv2 2 �q�p�v

�1ÿv2�2gv2

1ÿv2 � 1

0BBBB@1CCCCA:

The Jacobian of B with respect to �n; v; q� is

JB �

v��1ÿv2p n

�1ÿv2�3=2 0

ÿ cÿ11ÿv2 2 v�q�p�

�1ÿv2�2cÿ1�v2

1ÿv2

ÿ �cÿ1�v1ÿv2

�q�p��1�v2��1ÿv2�2

cv1ÿv2

0BB@1CCA:�3:3�

JINGING CHENHEN380

Thus the Jacobian of the system is

J � JB � JAÿ1

��ÿ�cÿ1��q�p�v2�cq�v

q�pÿcpv2n��cÿ1�v2�1��1ÿv2�1=2

q�pÿcpv2 ÿ cnv�1ÿv2�1=2ÿcpv2�q�p

�cÿ1��q�p��1ÿv2�3=2ÿcpv2�q�p

��cÿ1�nv2�2c�qÿp�ÿ3�cÿ1�n�vp�qÿcpv2

�cÿ1��p�q�ÿcqv2

p�qÿcpv2

0 1 0

0BB@1CCA:�3:4�

Simple calculations yield the three eigenvalues of J :

k1 � �qÿ �cÿ 1�p�vÿ ��cp�p � q�p �1ÿ v2�

p � qÿ cpv2;

k2 � v;

k3 � �qÿ �cÿ 1�p�v� ��cp�p � q�p �1ÿ v2�

p � q� cpv2:

�3:5�

We can simplify k1 and k3: Let h ��cp

p�q

q: Then

k1 � �qÿ �cÿ 1�p�vÿ ��cp�p � q�p �1ÿ v2�

p � qÿ cpv2� vÿ h1ÿ vh

:�3:6�

From p � ksnc and n � qÿ pcÿ1 ; we have

p � ks qÿ pcÿ 1

� �c

:

Thus

@

@qp�q; s� � ksc qÿ p

cÿ 1

� �cÿ11ÿ @

@qp�q; s�=�cÿ 1�

� �� cp

qÿ p=�cÿ 1� ÿcp

qÿ p=�cÿ 1�@

@qp�q; s�=�cÿ 1�:

Hence we have

�cÿ 1�� p � q��qÿ p=�cÿ 1��cÿ 1�

@

@qp�q; s� � cp

qÿ p=�cÿ 1� :

Then

@

@qp�q; s� � cp

p � q:�3:7�

Thus, from (3.6),

k1 �vÿ ��

pqp

1ÿ v��pqp :�3:8�

Similarly,

k3 �v� ��

pqp

1� v��pqp :�3:9�


If v, the speed of the ¯uid, is zero, then

k3 � ��pqp

:

This shows that the sound speed is��pqp

. Now we prove that the sound speedis less than the speed of light. We have

pq � cpp � q

� c1� q

p

:�3:10�

Since qÿ n � p=�cÿ 1�;q > p=�cÿ 1�

and then

qp> 1=cÿ 1:

Thus, from (3.10),

pq <c

1� 1=�cÿ 1� � cÿ 1:

TAUBAUB [T1, T2] showed that the kinetic theory of gases yields c 2 53 : Then

pq 2 23 ;

that is, all sound speeds are bounded by��23

qof the speed of light.

Remarks. 1. In the classical case, the sound speed is��pnp

. Since qÿ n � e, ande is very small compared with the rest mass in non-relativistic case,

q �ÿ n :

So��pnp

is the Newtonian limit of��pqp

. We see that in the relativistic case, theeigenvalues are the sums of the sound speed and the particle speed under theLorentz transformation, and their Newtonian limits are exactly what we havein the classical situation.2. In [T2], the sound speed c is given by the complicated formula

c2 � n1� i

didn

� �s

where i � e� p=n is the rest speci®c enthalpy.It is easy to see that k1 < k2 < k3: So the system (3.1) is strictly hyperbolic.

Next we show that k1 and k3 are genuinely nonlinear while k2 is linearlydegenerate.

We ®nd the eigenvector of each eigenvalue. For simplicity, instead ofusing the conservation of rest mass, we use the equation of entropy (2.10):

@s@t� v

@s@x� 0;

to replace the equation of the conservation of mass. Then the system becomes

JINGING CHENHEN382

@

@t

��q� p� v

1ÿ v2

�� @

@x�q� p� v2

1ÿ v2� p

� �� 0;

@

@t�q� p� v2

1ÿ v2� q

� �� @

@x

��q� p� v

1ÿ v2

�� 0;

@s@t� v

@s@x� 0

�3:11�

where p � p�q; s�. This system is equivalent to (3.1) when we have smoothsolutions. In a way like that for the system (3.1), we can ®nd the Jacobian ofthe system with respect to �q; v; s� as

�J �

�ÿ1�pq�vv2pqÿ1 ÿ p�q

v2pqÿ1vps�1ÿv2�

v2pqÿ1

ÿ pq�1ÿv2�2�p�q��v2pqÿ1�

�ÿ1�pq�vv2pqÿ1 ÿ �1ÿv2�2ps

�p�q��v2pqÿ1�0 0 v

0BBB@1CCCA�3:12�

with eigenvalues

k1 � vÿ ��pq�q; s�

p1ÿ v

��pq�q; s�

p ; k2 � v; k3 � v� ��pq�q; s�

p1� v

��pq�q; s�

p :�3:13�

Let

�J2 � �J ÿ k2 � I �

�1ÿpq�v1ÿv2pq

ÿ v q�p1ÿv2pq

ÿ vps�1ÿv2�1ÿv2pq

pq�1ÿv2�2�p�q��1ÿv2pq�

�1ÿpq�v1ÿv2pq

ÿ v �1ÿv2�2ps

�p�q��1ÿv2pq�0 0 0

0BBB@1CCCA

�

�ÿvpq��1ÿv2�1ÿv2pq

q�p1ÿv2pq



�ÿvpq��1ÿv2�1ÿv2pq

�1ÿv2�2ps

�p�q��1ÿv2pq�0 0 0

0BBB@1CCCA:�3:14�

If r2 is the eigenvector corresponding to k2, then r2 satis®es

�J2 � r2 � 0

with solution

r2 � �ÿps; 0; pq�:Moreover, rk2, the gradient of k2 with respect to �q; v; s�, is

rk2 � �@qk2; @vk2; @sk2� � �@qv; @vv; @sv� � �0; 1; 0�:Thus

rk2 � r2 � �0; 1; 0� � �ÿps; 0; pq� � 0;�3:15�i.e., k2 � v2 is linearly degenerate.


Now we prove that k1 is genuinely nonlinear. Let

�J1 � �J ÿ k1 � I �

��pqp �1ÿv2�1ÿv2pq

q�p1ÿv2pq



��pqp �1ÿv2�1ÿv2pq

�1ÿv2�2ps

�p�q��1ÿv2pq�

0 0��pqp �1ÿv2�1ÿv

��pqp

0BBBB@1CCCCA:

Thus r1, the eigenvector of k1 � vÿ��p�q;s�qp

1ÿv��p�q;s�qp , satis®es

J1 � r1 � 0

with solution

r1 � 1;ÿ��pqp �1ÿ v2�

p � q; 0

!:

We ®nd that

rk1 � �@qk2; @vk2; @sk2�

� ÿ 12

@2p@q2�1ÿ v2��v ��

pqp ÿ 1�2 ��

pqp

;1ÿ pq

�1ÿ v��pqp �2 ; ÿ

1

2

@2p@s@r �1ÿ v2��1ÿ v

��pqp �2 ��

pqp

!;

so

r1 � rk1 � ÿ 12�1ÿ v2�

@2p@q2 � p � q� � 2pq�1ÿ pq��1ÿ v

��pqp �2 ��

pqp � p � q� :�3:16�

From (3.7), @p@q � cp

p�q , and so

@2p@q2� @

@qcp

p � q� c�cÿ 1�p�p � q�3 n > 0:

Since @2p@q2 > 0; r1 � rk1 is always negative. Thus k1 is genuinely nonlinear.

Similarly, we can prove that k3 is genuinely nonlinear.We now ®nd the Riemann invariants of each eigenvalue. By de®nition, a

Riemann invariant w of kj satis®es

rj � �wq; wv; ws� � 0:

For k1; r1 � 1;ÿ��pqp �1ÿv2�

p�q ; 0� �

, and the equation r1 � �wq; wv; ws� � 0 isequivalent to

wq ÿ wv

��pqp �1ÿ v2�

p � q� 0:

We can easily see that s is a Riemann invariant. The other one is constantalong the curve determined by the di�erential equation

dqdv� ÿ p � q��

pqp �1ÿ v2� ;

JINGING CHENHEN384

which we solve to get

1

2ln

1� v1ÿ v

� ��Z ��

pqp

p � qdq � constant:�3:17�

Thus 12 ln�1�v

1ÿv� �R ��

pqpp�q dq is another Riemann invariant of k1: Similarly, the

Riemann invariants of k3 can be found to be

s and1

2ln

1� v1ÿ v

� �ÿZ ��

pqp

p � qdq:�3:18�

The eigenvector of k2 is �ÿps; 0; pq�. A Riemann invariant must satisfy

ÿps � wq � pq � ws � 0

which has two easy solutions v and p. So the two Riemann invariants of k2 are

v and p:�3:19�

4. The Riemann Problem

In this section, we discuss the Riemann problem of system (3.1), namelythe initial-value problem with initial data U0�x� � �q0�x�; v0�x�; p0�x�� con-sisting of a pair of constant states UL � �qL; vL; pL� and UR � �qR; vR; pR�separated by a jump discontinuity at x � 0; that is,

U0�x� � UL if x 2 0;UR if x > 0:

�To study the Riemann problem, we must ®rst investigate the properties of

the shock curves. In [T1], TAUBAUB found the Hugoniot curve of relativisticshocks, namely,

Lemma 4.1. [T1]. The Hugoniot curve of relativistic shocks is

q� pn2�q� pL� � qL � pL

n2L�qL � p�;�4:1�

where �n; q; p� represent the rest mass density, energy density and pressure atright.

Proof. From (3.1), our system is

@

@t

�n�1ÿ v2�ÿ1=2

�� @

@x

�nv�1ÿ v2�ÿ1=2

�� 0;

@

@t

��q� p� v

1ÿ v2

�� @

@x�q� p� v2

1ÿ v2� p

� �� 0;

@

@t�q� p� v2

1ÿ v2� q

� �� @

@x

��q� p� v

1ÿ v2

�� 0;

where the pressure p � �cÿ 1��qÿ n�:


Take a coordinate system in which the shock speed is zero. The Rankine-Hugoniot conditions are

�F � � s�U � � 0;�4:2�that is,

nv��1ÿ v2p � nLvL��

1ÿ v2Lp ;�4:3�

�q� p� v2

1ÿ v2� p � �qL � pL� v2L

1ÿ v2L� pL;�4:4�

�q� p� v1ÿ v2

� �qL � pL� vL

1ÿ v2L:�4:5�

De®ne

nv��1ÿ v2p � nLvL��

1ÿ v2Lp � M :�4:6�

If v � 0; then M � 0: So vL � 0; that is, the speed of both sides are equal. Thisis a contact discontinuity. In shock waves, we can assume that M40:

From (4.4), we have

nv��1ÿ v2p � q� p

nv��

1ÿ v2p ÿ nLvL��

1ÿ v2Lp � qL � pL

nL

vL��1ÿ v2L

p � ÿ� p ÿ pL�;

that is,

Mq� p

nv��

1ÿ v2p ÿ qL � pL

nL

vL��1ÿ v2L

p" #� ÿ� p ÿ pL�:�4:7�

From (4.5), we deduce that

Mq� p

n1��


nL

1��1ÿ v2L

p" #� 0:�4:8�

From (4.7), (4.8), (4.6) we obtain

Mq� p

n1��


nL

1��1ÿ v2L

p" #� 1��

1ÿ v2p

ÿMq� p

nv��


nL

vL��1ÿ v2L

p" #� v��

1ÿ v2p

� �p ÿ pL� v��1ÿ v2p � � p ÿ pL� �Mn ;

that is,

q� pnÿ qL � pL

nL

1ÿ vvL��1ÿ v2L

p ��1ÿ v2p

!� � p ÿ pL� � 1n ;�4:9�

JINGING CHENHEN386

similarly,

q� pn

1ÿ vvL��1ÿ v2p ��

1ÿ v2Lp ÿ qL � pL

nL� � p ÿ pL� 1nL

:�4:10�

From (4.9), (4.10), it follows that

q� pnÿ qL � pL

nL

1ÿ vvL��1ÿ v2L

p ��1ÿ v2p

! !� q� p

n

� q� pn

1ÿ vvL��1ÿ v2p ��

1ÿ v2Lp ÿ qL � pL

nL

!� qL � pL

nL

� q� pn

� �2ÿ qL � pL

nL

� �2

� � p ÿ pL� q� pn2� qL � pL

n2L

� �:

We rearrange terms to get

q� pn

� �2ÿ� q� p�� p ÿ pL�

n2� qL � pL

nL

� �2

��qL � pL��p ÿ pL�n2L

;�4:11�

which can be simpli®ed to

q� pn2�q� pL� � qL � pL

n2L�qL � p�:�4:12�

This de®nes the Hugoniot curve of the relativistic shocks. (

Next we show that the Lax entropy conditions hold globally along theshock curves. For the non-relativistic case, see [We, S].

Theorem 4.1. For system (3.1), the Lax entropy conditions are satis®edeverywhere along the shock curves.

Proof. We only consider 1-shocks. The case of 3-shocks can be treated sim-ilarly. Without loss of generality, we choose a coordinate system in whichvL � 0. If we write the system (3.1) in the form Ut � F �U�x � 0, the jumpcondition is r�U � � �F �, where � f � denotes the jump of f across the shock. Weassume that the jump conditions de®ne the shock curve U � U�e� with shockspeed r � r�e�; e 2 0. From the general theory of conservation laws, weknow that for e negative and small, k1�e� < r�e� < k1�0�. We show that thisinequality holds everywhere along the shock curve.

Thus, suppose that e1 is the ®rst point where k1�e� � r�e�, e1 < 0. Since

r 0�U � � rU 0 � d FU 0;�4:13�if we consider this at e � e1, and multiply it by the left eigenvector l1�e1�, weget r 0l1 � �U � � 0: Suppose now that

l1 � �U �40;�4:14�then r 0�e1� � 0; hence at e1, k1U 0 � dFU 0, so that U 0 � r1. Then


ddp�rÿ k1�

��e�e1

� ÿdk1dp

��e�e1

� ÿrk1 � r1 � ÿ1:

Thus k1 � r for some e2, e1 < e2 < 0; this contradicts the de®nition of e1. Weconclude that if (4.14) holds, then r�e� > k1�e� for all e. Furthermore, ifr�e1� � k1�0� for some e1 < 0, then there is an e2 with e1 < e2 < 0 so thatr 0�e2� � 0. Now r�e2� > k1�e2� so (4.13) at e2 gives rU 0 � dFU 0. Thus r 0�e�does not change sign. Hence r 0 > 0 so r�e� < k1; if e < 0.

To ®nish the proof, we need to show that (4.14) holds. From (3.4),

J1 � J ÿ k1 � I �

a11n��cÿ1�v2�1�

��1ÿv2p

p�qÿcpv2ÿcnv

��1ÿv2p

p�qÿcpv2

�cÿ1��p�q��1ÿv2�3=2p�qÿcpv2 a22

�cÿ1��p�q�ÿcqv2

p�qÿcpv2

0 1 ÿ vÿ ��pqp

1ÿv��pqp

0BBBB@1CCCCA

�4:15�where

a11 � ��cÿ 1�� p � q��1ÿ v2��v� �1ÿ v2� ��cp� p � q�p

p � qÿ cpv2;

a22 � ��cÿ 1�nv2 � �2ÿ c�� p � q��v� �1ÿ v2� ��cp� p � q�p

p � qÿ cpv2:

The left eigenvector l1 of J1 satis®es

l1 � J1 � 0:�4:16�Since the third element of the ®rst column is zero, we can set

l1�1� � J1�2; 1� � �cÿ 1�� p � q��1ÿ v2�3=2p � qÿ cpv2

;�4:17�

l1�2� � ÿJ1�1; 1� � �cÿ 1�� p � q��1ÿ v2�v� �1ÿ v2� ��cp� p � q�p

p � qÿ cpv2:�4:18�

Since the third element of the second column is 1,

l1�3� � ÿ �l1�1� � J1�1; 2� � l1�2� � J1�2; 2��4:19�

� ÿ �cÿ 1�� p � q��1ÿ v2�3=2p � qÿ cpv2

n��cÿ 1�v2 � 1��1ÿ v2p

p � qÿ cpv2

� �cÿ 1�� p � q��1ÿ v2�v� �1ÿ v2� ��cp� p � q�p

p � qÿ cpv2

� ��cÿ 1�nv2 � �2ÿ c�� p � q��v� �1ÿ v2� ��cp� p � q�p

p � qÿ cpv2

�ÿ �1ÿ v2��ÿ�cÿ 1�� p � q� � v��cp� p � q�p

p � qÿ cpv2:

We also have

JINGING CHENHEN388

�U � � n��1ÿ v2p ÿ nL; �q� p� � v

1ÿ v2; �q� p� v2

1ÿ v2� qÿ qL

� �:�4:20�

From (4.12) the Hugoniot curve is

� p � q�� pL � q�n2

� �qL � pL��qL � p�n2L

:

From [C], the shock curves satisfy

v2 � � p ÿ pL��qÿ qL�� p � qL�� pL � p� :�4:21�

Since p > pL in the 1-shock, we conclude that

q > qL:�4:22�Thus

� p � q�� pL � q� > � p � qL�� pL � qL�:Then from the Hugoniot curve, we obtain

n > nL:�4:23�Now

l1 � �U � � �cÿ 1�� p � q��1ÿ v2�3=2p � qÿ cpv2

� n��1ÿ v2p ÿ nL��

ÿ �cÿ 1�� p � q��1ÿ v2�v� �1ÿ v2� ��cp� p � q�p

p � qÿ cpv2� �q� p� � v

1ÿ v2

ÿ �1ÿ v2��ÿ�cÿ 1�� p � q� � v��cp� p � q�p

p � qÿ cpv2� �q� p� v2

1ÿ v2� qÿ qL

� �� cÿ 1�� p � q��1ÿ v2�3=2

p � qÿ cpv2� n��

1ÿ v2p ÿ nL

� �ÿ

��cp� p � q�p

p � qÿ cpv2� �q� p�vÿ �v

��cp� p � q�p �

p � qÿ cpv2� �q� p�v2

ÿ �1ÿ v2��ÿ�cÿ 1�� p � q� � v��cp� p � q�p �

p � qÿ cpv2� �qÿ qL�

> 0

since each of the four terms is bigger than zero. Thus

l1 � �U �4 0:

This ®nishes the proof. (

To show that the shock curves are physically relevant, we need to provethat the entropy changes monotonically along the shock curves.

Theorem 4.2. The entropy increases monotonically along 1-shock curves, anddecreases monotonically along 3-shock curves.


Proof. We consider 1-shocks ®rst. From p � ksnc, we get

s � kÿ1pnÿc;�4:24�so

dsdp� kÿ1nÿc � kÿ1p�ÿc�nÿcÿ1 dn

dp:�4:25�

We need ds=dp > 0, whence

dndp<

ncp:�4:26�

Let us compute dn=dp along the Hugoniot curve

p � qn2� pL � q� � pL � qL

n2L� p � qL�

where q � n� p=�cÿ 1�. We can rewrite this as

ccÿ 1

p � n� ��

pL � n� pcÿ 1

�n2L � � pL � qL�� p � qL�n2:

Di�erentiating both sides with respect to p, we get, after a straightforwardcalculation that

� pL � qL�� p � qL�2nÿ � pL � q�n2L ÿ � p � q�n2L� � dn

dp

� ccÿ 1

� pL � q�n2L � � p � q�n2L1

cÿ 1ÿ � pL � qL�n2;

where we replace n� p=�cÿ 1� by q in the above equation. We want to provethat the coe�cient of dn=dp is positive; i.e., that

� pL � qL�� p � qL�2nÿ � pL � q�n2L ÿ � p � q�n2L > 0:�4:27�It su�ces to show that

� pL � qL�� p � qL�nÿ � p � q�n2L > 0�4:28�since � p � q�n2L > � pL � q�n2L. But from the Hugoniot jump condition (4.12),

� pL � qL�� p � qL�nÿ � p � q�n2L �� p � q�n2L

npl � qÿ n� � > 0:

Thus

dndp�

ccÿ1 � pL � q�n2L � � p � q�n2L 1

cÿ1ÿ � pL � qL�n2� pL � qL�� p � qL�2nÿ � pL � q�n2L ÿ � p � q�n2L

:

We need to show, from (4.26), that

dndp<

ncp;

that is,

JINGING CHENHEN390

ccÿ1 � pL � q�n2L � � p � q�n2L 1

cÿ1ÿ � pL � qL�n2� pL � qL�� p � qL�2nÿ � pL � q�n2L ÿ � p � q�n2L

<ncp:

After a straightforward calculation, using (4.12), we get

ccÿ 1

� pL � q�n2L � � p � q�n2L1

cÿ 1ÿ � pL � qL�n2

� �cp

ÿ � pL � qL�� p � qL�2nÿ � pL � q�n2L ÿ � p � q�n2L� �

n

� n2LpL � qp � qL

cp�qL ÿ q� � � p � q�� p ÿ pL��

:

Thus we need to show that

pL � qp � qL

cpp � q

>p ÿ pL

qÿ qL:�4:29�

First we prove that

cpp � q

>p ÿ pL

qÿ qL:�4:30�

We ®x the point � p; q� and let � pL; qL� move along the Hugoniot curve:p � q

n2� pL � q� � pL � qL

n2L� p � qL�:

Di�erentiating both sides of this equation with respect to qL and rearrangingterms we get

dpL

dqL� 2nL� p � q�� pL � q� ÿ � p � qL�n2 ÿ � pL � qL�n2� p � qL�n2 ÿ � p � q�n2L � 2=�cÿ 1��p � q�� pL � q�nL

:

We have

� p � qL�n2 ÿ � p � q�n2L > 0�4:31�since, from (4.22),

pL � qL < pL � q

and since the Hugoniot jump condition

� p � qL�n2� pL � qL� � � p � q�n2L� pL � q�holds. Thus the denominator is bigger than zero. By the intermediate valuetheorem,

p ÿ pL

qÿ qL� dp�n�

dq�n�for some qL < q�n� < q. For convenience, we write p0 � p�n�; q0 � q�n�. Soit is su�cient to prove that

dp0dq0

<cp

p � q;


that is,

2n0� p � q�� p0 � q� ÿ � p � q0�n2 ÿ � p0 � q0�n2� p � q0�n2 ÿ � p � q�n20 � 2=�cÿ 1�� p � q�� p0 � q�n0 ÿ

cpp � q

< 0:�4:32�

After a straightforward calculation we ®nd

2n0� p � q�� p0 � q� ÿ � p � q0�n2 ÿ � p0 � q0�n2� �� p � q�

ÿ � p � q0�n2 ÿ � p � q�n20 � 2=�cÿ 1�� p � q�� p0 � q�n0� �

cp

� 2n0� p � q�� p0 � q�nÿ � p � q�� p � q0�n2 ÿ � p � q�� p0 � q0�n2ÿ � p � q0�n2 ÿ � p � q�n20� �

cp

< 2n0� p � q�� p0 � q�nÿ � p � q�� p � q0�n2 ÿ � p � q�� p0 � q0�n2� � p � q�n 2n� p0 � q� ÿ � p � q0�nÿ � p0 � q0�n� �:

We need to prove that

2n0� p0 � q� ÿ � p � q0�nÿ � p0 � q0�n < 0:�4:33�Since

� p � q0�n� � p0 � q0�n > 2�� p � q0�� p0 � q0�

pn;

we only have to show�� p � q0�� p0 � q0�

pn > n0� p0 � q� :

But from (4.12), we have

� p � q0�� p0 � q0�n2 � � p � q�� p0 � q�n20 > n20� p0 � q�2:Thus

dp0dq0

<cp

p � q

and hence

p ÿ pL

qÿ qL<

cpp � q

:�4:34�

Since the speed of sound��cp

p�q

qis less than 1, we have

p ÿ pL

qÿ qL< 1:

Hence

q� pL

qL � p> 1:

By virtue of (4.34), we arrive at (4.29). Thus

dsdp> 0

JINGING CHENHEN392

along the Hugoniot curve. The entropy increases monotonically along 1-shock curves. Similarly, we can prove that the entropy decreases monotoni-cally along 3-shock curves. (

Now we give a condition to prevent a vacuum. Since along 2-waves, thepressure p is constant, and p � snc, a vacuum does not occur when a 1-waveinteracts with a 2-wave or when a 3-wave interacts with a 2-wave. It is easy tosee that only the interaction of a 1-rarefaction and a 3-rarefaction wave maypossibly create vacuum.

From (3.17), (3.18), it follows that

R � 1

2ln

1� v1ÿ v

� ��Zq

0

��pqp

p � qdq; S � 1

2ln

1� v1ÿ v

� �ÿZq

0

��pqp

p � qdq

�4:35�are two Riemann invariants. Let the initial condition be

U�x; 0� � �qL; vL; pL�; x < 0;�qR; vR; pR�; x 3 0:

�We can thus determine �RL; SL� and �RR; SR� from (4.35), where �RL; SL� and�RR; SR� are the Riemann invariants at x < 0 and x 3 0. Along 1-rarefactionwaves, R and the entropy s are constant; along 3-rarefaction waves, S and theentropy s are constant; along 2-waves the particle speed v and the pressure pare constant. Let us denote the state at the left of the 2-wave as state 1, andthe state at the right of the 2-wave as state 2. Then

R1 � RL; S2 � SR; v1 � v2:

Then

RL ÿ SR � R1 ÿ S2

� 1

2ln

1� v11ÿ v1

� ��Zq10

��pqp

p � qdqÿ 1

2ln

1� v21ÿ v2

� �ÿZq20

��pqp

p � qdq

0@ 1A�Zq10

��pqp

p � qdq�

Zq20

��pqp

p � qdq:

�4:36�If RL 2 SR, then Zq1

0

��pqp

p � qdq�

Zq20

��pqp

p � qdq 2 0;

i.e., q 2 0. But since n � qÿ p=�cÿ 1�2 q,

n 2 0:�4:37�Hence a vacuum occurs. Thus we have proved


Theorem 4.3. The condition for a vacuum to occur is

RL 2 SR:

Remark. This approach to derive the condition for vacuum to occur is gen-eral. When two eigenvalues coalesce and the system loses strict hyperbolicity,the corresponding Riemann invariants should also coalesce. Consider clas-sical gas dynamics for example. The condition for a vacuum to occur is (see[S, p. 355])

2

cÿ 1�cL � cR�2 vR ÿ vL

where c is the sound speed and v is the particle speed. The pressure for thepolytropic gas is

p � k2�s�qc;

where q is the density and k�s� > 0 is a function of entropy s. The soundspeed is

c � ��pqp � k�s� ��cp q�cÿ1�=2:

The two Riemann invariants are

R � v� 2

cÿ 1k�s� ��cp q�cÿ1�=2 � v� 2

cÿ 1c;

S � vÿ 2

cÿ 1k�s� ��cp q�cÿ1�=2 � vÿ 2

cÿ 1c:

So RL 2 SR in this case reduces, after rearrangement, to

2

cÿ 1�cL � cR�2 vR ÿ vL;

as before. (

We now work on the existence of solutions of the Riemann problem.

Lemma 4.2.

dvdp< 0 on 1-shock curves;

dvdp> 0 on 3-shock curves:

Proof. We only consider the 1-shock case. The 3-shock case is similar. From(4.21),

v2 � � p ÿ pL��qÿ qL�� p � q�� pL � q� :

We have

dv2

dp� 2v

dvdp� d

dp� p ÿ pL��qÿ qL�� p � qL�� pL � q� :

JINGING CHENHEN394

Since v < 0 on shock curves, dv=dp < 0 is equivalent to dv2=dp > 0. But

dv2

dp� d

dp� p ÿ pL��qÿ qL�� p � q�� pL � q�

� pL � qL

� p � qL�2� pL � q�2 �qÿ qL�� pL � q� � � p ÿ pL�� p � qL�dqdp

� �:

We need to prove that

�qÿ qL�� pL � q� � � p ÿ pL�� p � qL�dqdp

> 0:�4:38�

First we ®nd the expression for dq=dp. The Hugoniot jump condition is

� p � q�� pL � q�n2L � � pL � qL�� p � qL�n2

where n � qÿ p=�cÿ 1�:Di�erentiating both sides with respect to p; we have

1� dqdp

� �� pL � q�n2L � � p � q� dq

dpn2L

� � pL � qL�n2 � � pL � qL�� p � qL�2ndqdpÿ 1

cÿ 1

� �;

which yields

dqdp�� pL � q�n2L ÿ � pL � qL�n2 � � pL � qL�� p � qL�2n 1

cÿ1� pL � qL�� p � qL�2nÿ � pL � q�n2L ÿ � p � q�n2L

:�4:39�

Thus we have

�qÿ qL�� pL � q� � � p ÿ pL�� p � qL�dqdp

� �qÿ qL�� pL � q�

� � p ÿ pL�� p � qL�� pL � q�n2L ÿ � pL � qL�n2 � � pL � qL�� p � qL�2n 1

cÿ1� pL � qL�� p � qL�2nÿ � pL � q�n2L ÿ � p � q�n2L

� �qÿ qL�� pL � q�� pL � qL�� p � qL�2nÿ � pL � q�n2L ÿ � p � q�n2L�� pL � qL�� p � qL�2nÿ � pL � q�n2L ÿ � p � q�n2L

�� p ÿ pL�� p � qL�� pL � q�n2L ÿ � pL � qL�n2 � � pL � qL�� p � qL�2n 1

cÿ1�� pL � qL�� p � qL�2nÿ � pL � q�n2L ÿ � p � q�n2L

:

From (4.27), the denominator of this above expression is bigger than zero,and the numerator is

�qÿ qL�� pL � q� � pL � qL�� p � qL�2nÿ � pL � q�n2L ÿ � p � q�n2L� �

� � p ÿ pL�� p � qL� � pL � q�n2L ÿ � pL � qL�n2 � � pL � qL�� p � qL�2n1

cÿ 1

� �


� 2�qÿ qL�� pL � q� � p � qL�� pL � qL�nÿ � p � q�n2L� �

� � p � qL�� pL � qL�2n� p � qL�� p ÿ pL� 1

cÿ 1

> 0;

since from (4.28),

� p � qL�� pL � qL�nÿ � p � q�n2L > 0:

So (4.38) is true and hence dv=dp < 0 on the 1-shock curves. (

Next we prove a simple property of rarefaction waves.

Lemma 4.3. dv=dp < 0 on 1-rarefaction waves and dv=dp > 0 on 3-rarefactionwaves.

Proof. On 1-rarefaction waves, we have

R � 1

2ln

1� v1ÿ v

� ��Zq

0

��pqp

p � q� constant:

Di�erentiating both sides with respect to p gives

dvdp� ÿ�1ÿ v2�

��pqp

p � qdqdp:

But when s � constant, which is the case on rarefaction waves, we have, from(3.7), that

dqdp� p � q

cp> 0:

Thusdvdp< 0:

Similarly, dv=dp > 0 on 3-rarefaction waves. (

As in [S], for U 2 R3, we de®ne

Si�U� � f� p; v; s� : � p; v; s� can be connected from the left by an

i-shock wave from Ug; i � 1; 3;

Ri�U� � f� p; v; s� : � p; v; s� can be connected from the left by an

i-rarefaction wave from Ug; i � 1; 3:

Let

Spi �U� � f� p; v� : � p; v� 2 projection of Si�U� on the ( p,v)-planeg; i � 1; 3;

Rpi �U� � f� p; v� : � p; v� 2 projection of Ri�U� on the ( p,v)-planeg; i � 1; 3;

T pi �U� � Sp

i �U� [ Rpi �U�; i � 1; 3:

JINGING CHENHEN396

From Lemma 4.2 and Lemma 4.3, we know that dv=dp < 0 on T p1 and

dvdp > 0 on T p

3 . So we can de®ne T p1 by

v � f1� p; pL; vL; sL�:�4:40�The T p

1 curve divides the � p; v�-plane into two parts. If � pR; vR� is below T p1 ;

i.e., if

vR < f1� pR; pL; vL; sL�;then the 3-wave of the Riemann problem should be a shock becausedv=dp > 0 on 3-waves, and on a shock curve, the particle speed at the left isalways bigger than the particle speed at the right. On a rarefaction curve, theparticle speed at the left is always smaller than the particle speed at that right.If

vR > f1� pR; pL; vL; sL�;the 3-wave of the Riemann problem is a rarefaction wave.

Having determined the 3-wave, we can investigate the 1-wave. We de®nethe inverse 3-wave as

v � f3� p; pR; vR; sR�;�4:41�where f3� p; pR; vR; sR� is the inverse 3-rarefaction wave if the third wave is ararefaction wave or the inverse 3-shock wave if the third wave is 3-shock. If

vL < f3� pL; pR; vR; sR�;the 1-wave is a rarefaction wave. Otherwise it is a 1-shock.

Solving

v � f1� p; pL; vL; sL�; v � f3� p; pR; vR; sR��4:42�we ®nd a unique solution � pM ; vM � because dv=dp < 0 on the 1-wave curveand dv=dp > 0 on the 3-wave curve. To complete the solution of the Riemannproblem, we need to ®nd the entropy level at the left and right on the contactwaves. But since the entropy is uniquely determined by the value of p onshocks and is constant on rarefaction waves, we can obtain sM1 by the thevalue of pM on 1-curves and sM2 by the value of pM on 3-curves. Thus wededuce

Theorem 4.4. Consider the system of gas dynamics (3.1) for an ideal polytropicgas whose equation of state is q � n� p=�cÿ 1� and p � ksncÿ1. Let UL and UR

be any two states. Then there is a unique solution to the Riemann problem withthese initial states, if and only if

rL < sR:�4:43�If (4.43) is violated, then a vacuum occurs in the solution. The 3-component ofthe solution is a rarefaction if

vR > f1� pR; pL; vL; sL�;and is a shock otherwise. The 1-component is a rarefaction if


vL < f3� pL; pR; vR; sR�;and is a shock otherwise, where f1, f3 are de®ned in (4.40), (4.41). The solutionis unique in the class of constant states separated by rarefaction waves andshock waves.

Acknowledgments This work is part of my Ph.D. dissertation under the guidance of

Professor JOELOEL SMOLLERMOLLER at the University of Michigan. I thank him for his constantencouragement and his very helpful discussions during the course of this work. I amespecially grateful to Professor C. DAFERMOSAFERMOS for extensive comments and suggestionsthat signi®cantly improved the paper.

References

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Interscience, 1948.[L] P. D. LAXAX, Hyperbolic systems of conservation laws, II, Comm. Pure Appl.

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[ST] J. SMOLLERMOLLER & B. TEMPLEEMPLE, Global Solutions of The Relativistic Euler Equations,Comm. Math. Phys. 156 (1993), 67±99.

[STX] J. A. SMOLLERMOLLER, J. B. TEMPLEEMPLE & Z. P. XININ, Instability of Rarefaction Shocks in

Systems of Conservation Laws, Arch. Rational Mech Anal. 112 (1990), 63±81.[T1] A. H. TAUBAUB, Relativistic Rankine-Hugoniot Equations, Physical Review 74

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[Th] K. THOMPSONHOMPSON, The special relativistic shock tube, J. Fluid Mech. 171 (1986),

365±375.[W] S. WEINBERGEINBERG, Gravitation and Cosmology: Principles and Applications of the

General Theory of Relativity, Wiley, 1972.

[We] B. WENDROFFENDROFF, The Riemann problem for materials with nonconvex equations ofstate, II. General ¯ow, J. Math. Anal. Appl. 38 (1972), 640±658.

Department of Mathematics

City University of Hong Kong

Kowloon, Hong Kong

(Accepted June 10, 1996)

JINGING CHENHEN398

Conservation Laws for Relativistic Fluid Dynamicsweb.unbc.ca/~chenj/FluidDynamics.pdf ·...

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Transcript of Conservation Laws for Relativistic Fluid Dynamicsweb.unbc.ca/~chenj/FluidDynamics.pdf ·...