数理解析研究所講究録
1197 巻 2001 年 1-53

1

A Study of the Relativistic Euler Equation
E)
Tetu Makino (&Faculty of Engineering,
Yamaguchi University (山口大工)This is a joint work with Cheng-Hsiung Hsu (National Central University, Chungli, Taiwan) and Song-Sun Lin (National Chiao Tung University,
Hsinchu, Taiwan).

1

Introduction

In this article we study the Cauchy problem to the one-dimensional relativistic Euler equation$\frac{\partial}{\partial t}\frac{\rho+Pu^{2}/c^{4}}{1-u^{2}/C^{2}}+\frac{\partial}{\partial x}\frac{(\rho+P/c^{2})u}{1-u^{2}/c2}=^{\mathrm{o}}$

,

..

.

$\cdot$$\rho|t=0=\rho 0(X)$,$u|_{t=0=u}.0(X)-\mathrm{z}$

(1.1)

,

$\frac{\partial}{\partial t}\frac{(\rho+P/c^{2})u}{1-u^{2}/C^{2}}+\frac{\partial}{\partial x}\frac{P+\rho u^{2}}{1-u^{2}/c2}=^{\mathrm{o}}$

.

(1.2)

is a positive constant, the speed of light, and is a given function of. The equation (1.1) governs the one dimensional motion of a perfect gas$(1.1)$ tends to the usual Eulerin the Minkowski space-time. Whenequation of gas dynamics

Here

$P$

$c$

$\rho$$carrow\infty,$. $=0$ ,$\rho_{t}+(\rho.u)_{x}$$(\rho u)_{t}+(P+\rho u^{2})_{x}=0$

.

(1.3)

Many mathematical investigations for this non-relativistic Euler equation
were done. But the first mathematical investigation for the relativistic Euler}
equation (1.1) was done recently by
and Temple [6]. They assume, where is . positive constant $<c$ . Under this assumption, they
showed that if the initial data
and $u_{0}(x)$ satisfy$\mathrm{s}_{\mathrm{m}\mathrm{o}}$$P=\sigma^{2}\rho$$1\mathrm{e}\Gamma$$\sigma$$\mathrm{a}.$$\rho 0(x)$$T.V.\log\rho 0<\infty$,$T.V. \log\frac{c+\dot{u}_{0}}{c-u_{0}}..<\infty$,

2then there exists a global weak solution to the Cauchy problem $(1.1)(1.2)$ .
The result was obtained by Glimm’s scheme and it is the relativistic versionof Nishida’s result [5] for the non-relativistic problem.
However we would like to consider a more realistic equation of states. We
keep in mind the equation of state for a neutron stars, which is given by$P=Kc^{5}f(y)$ ,$\rho=Kc^{3}g(y)$$f(y)= \int_{0}^{y}\frac{q^{4}}{\sqrt{1+q^{2}}}dq$,$g(y)=3 \int_{0}^{y}q^{2}\sqrt{1+q^{2}}dq$.

For this equation of state, we have
as
but
as
. So we assume the following properties of the function(A):$0<d^{2}P/d\rho^{2}$$P(\rho)>0$ ,$0<dP/d\rho<c^{2}$ ,$P \sim\frac{c^{2}}{3}\rho$$P \sim\frac{1}{5}K^{2/\mathrm{s}_{\rho}5/3}$$\rhoarrow\infty$$\rhoarrow 0$$P(\rho)$:for $\rho>0$ , and$P=A\rho^{\gamma}(1+[\rho\gamma-1/c^{2}]_{1})$as$\rhoarrow 0$. Here$A$and$\gamma$are positive constants and$\gamma=1+\frac{2}{2N+1}$,

being a positive integer, anddenotes a convergent power series ofthe form.
The result which we want to generalize to the relativistic problem is thoseby G.-Q. Chen et al [2]. So we assume that the initial datasatisfy$N$$[X]_{1}$$\sum_{k\geq 1}a_{k}X^{k}$$\rho 0(x),$$0\leq\rho_{0}(X)\leq M0$,$| \frac{c}{2}\log\frac{c+u_{0}(X)}{c-u_{0}(X)}|\leq M_{0}$A weak solution of (1.1) (1.2) is defined as follows.
We write
$E$

$=$

$F$

$=$

$\frac{\rho+Pu^{2}/C^{4}}{1-u^{2}/C^{2}}$,$(\rho+P/c^{2})u$$\overline{1-u^{2}/c2}$$P+\rho u^{2}$$G$$=$$\overline{1-u^{2}/c^{2}}$

$U$

$=$

$(E, F)^{T}$,’,$f(U)=(F, G)\tau$ .

Then (1.1) can be written as$U_{t}+f(U)_{x}=0$ .

.

$u_{0}(x)$3

Let us denote by $U_{0}(x)$ the initial data. Then a weak solution $U(t, x)$ is abounded measurable function which satisfies$\int\int(U\Phi_{t}+f(U)\Phi_{x})d_{Xdt}+\int U_{0}(x)\Phi(\mathrm{O}, X)dX=0$

for any test function

2$\Phi\in C_{0}^{\infty}([0, +\infty)\cross R)$.

Riemann problems

The Riemann problem is the problem to the special initial data of theform
if $x<0$$U_{0}(x)=\{$
if $x>0$$U_{L}$$U_{R}$In order to slolve this we introduce the Riemann invariants$w=x+y$ ,$z=x-y$where$x= \frac{c}{2}\log\frac{c+u}{c-u}$.

,$y= \int_{0}^{\rho}\frac{\sqrt{P’}}{\rho+P/c^{2}}d\rho$Then (1.1) is diagonarized as$w_{t}+\lambda_{2}w_{x}=0$,$z_{t}+\lambda_{1}z_{x}=^{\mathrm{o}}$

,

where
$\lambda_{1}=\frac{u-\sqrt{P’}}{1-\sqrt{P}u/C^{2}},$’$\lambda_{2}=\frac{u+\sqrt{P’}}{1+\sqrt{P}u/C^{2}},\cdot$

the possible states $U=U_{R}$ connected towaves are

$U_{L}$

on the right by rarefaction$R_{1}$:$w=w_{L},$ $Z>z_{L}$$R_{2}$:$z=z$ .
$w>w_{LL},$

andThe Rankine Hugoniot jump condition$\sigma[U]=[f(U)]$ ,where $[U]=U_{R}-U_{L},$ $[f(U)]=f(U_{R})-f(UL)$ , gives the shock curve$\frac{(u_{RL}-u)^{2}}{(1-u_{R}^{2}/c^{2})(1-u_{L}^{2}/c^{2})}=\frac{(\rho_{R}-\rho_{L})(PR-P_{L})}{(\rho_{L}+P_{L}/c^{2})(\rho R+P_{R}/c^{2})}$

Along this curve we have shocks$S_{1}$$S_{2}$::,$\rho_{L}<\rho_{R},$$u_{R}<u_{L}$$\rho_{R}<\rho_{L},$$u_{R}<u_{L}$

.

.

4

The- Riemann problem can be solved uniquely by using these rarefaction
waves and shock waves and vaccuum state. The detailed discussion can befound in J. Chen [1].
If we look at a region of the form$\Sigma_{B}=\{(w, Z)|-B\leq Z\leq w\leq B\}$ ,we have the following
belong
Proposition 1 If the initial datafor.
then the solution of the Riemann problem is confined to$U_{L},$$s^{\mathrm{J}}o$me arge$\dot{l}$$\dot{t}\mathit{0}\Sigma_{B}$$U_{R}$$B$,$\Sigma_{B}$Moreover if we consider the image of

Proposition 2 The region$\Sigma_{B}$$\Sigma_{B}$in the $(E, F)$ -space, we haveis convex in the $(E, F)$ -plane.

Proof. Let us consider the above hedge $F=F(E)$ which corresponds to$w=B,$ $-B<z<B$ . We have to show $d^{2}F/dE^{2}<0$ . Along the hedge$w=B$ , we have$u=c \tanh\frac{1}{c}(B-\int_{0}^{\rho}\frac{\sqrt{P’}}{\rho+P/c^{2}}d\rho)$,from which$\frac{du}{d\rho}=-(1-u2/c)2\frac{\sqrt{P’}}{\rho+P/c^{2}}$.

By a direct calculation we have$\frac{dF}{dE}=\frac{u-\sqrt{P’}}{1-\sqrt{P}u/C^{2}},=\lambda_{1}$.

Differentiating once more we have$\frac{d^{2}F}{dE^{2}}=-\frac{1-u^{2}/C^{2}}{(1-\sqrt{P}u/c)^{4}2},(\frac{P’’}{2\sqrt{P’}}+(1-\frac{P’}{c^{2}})\frac{\sqrt{P’}}{\rho+P/c^{2}})<0$.

This was to be seen. QED.
From Proposition 2, we have

Proposition 3 If $U(s),$ $s\in[a, b]$ , is confined to a region
average$\Sigma_{B}$, then the$\frac{1}{b-a}\int_{a}^{b}U(s)d_{S}$belongs to$\Sigma_{B}$.

Let us look at the shock wave which connects the left state
theright statewith the shock speed .
The right state
and are parametrized by. Then we have thefollowing fact, which will be used in Section 4.
$U_{L}\mathrm{t}\mathrm{o}_{}$$U_{R}$$\sigma$$U_{R}$$\sigma$$\rho=\rho_{R}$5

Proposition 4
we have $d\sigma/d\rho>0$ .

$Al_{\mathit{0}}ngS1(\rho_{L}<\rho)$, we have $d\sigma/d\rho<0$ , and along $S_{2}(\rho<$$\rho_{L})$Proof. Without loss of generality we can assume $u_{L}=0$ . Then $u=u_{R}$is given by$u—\sqrt{\frac{[\rho][P]}{(\rho_{L}+P/c)2(\rho+PL/c^{2})}}$,

where $[\rho]=\rho-\rho_{L},$ $[P]=P-P_{L}$ . We have.

$\sigma=\frac{[F]}{[E]}=\frac{(\rho+P/c^{2})u}{\rho+Pu^{2}/C-4\rho L(1-u^{2}/c^{2})}$

By a direct but tedious compu.tations,$\frac{d\sigma}{d\rho}$$X$$\mathrm{w}.\mathrm{e}$

have

$=$

$. \frac{(\rho+P/C^{2})(\rho_{L}+P_{L}/C2)[\rho]X}{2(\rho+Pu^{2}/c^{4}-\rho L(1-u/2c2))^{2}u(\rho L+P/c2)2(\rho+PL/c^{2})2}.)$$=$$(\rho+P_{L}./C)2(\rho+P/C^{2})P’[\rho]+$$+$$(\rho+P_{L}/c^{2})(-(\rho+PL/c^{2})+[P]/c^{2})[P]+$$-$$(\rho_{L}+P/c2)[P]2/C^{2}$$r$.

Since $P”>0$ we know $[P]\leq P’[\rho]$ . Thus
$X$$\geq$$(\rho+P_{L}/c^{2})(\rho+P/c^{2})[P]+$$+$$(\rho+P_{L}/c^{2})(-(\rho L+P_{L}/c^{2})+[P]/c^{2})[P]+$..

$(\rho_{L}+P/c^{2})[P]^{2}/c^{2}$$=$$[P]((\rho+P_{L}../C^{2})([\rho]+[P]/c^{2})+([\rho]-[P]/c^{2})[P]/c^{2})$.

But$1> \frac{[\rho]-[P]/c^{2}}{[\rho]}=1-P’(\rho L+\theta(\rho-\rho_{L}))/c>02$.

Using this, itis easy to see $X>0..$ bot..h.When $[.\rho.],.>0$ and when $[\rho]<0$ .
Since $u<0$ , this completes the proof. QED.

3

Entropies

A pair of functions
the$\eta$and$q$

is called an entropy- entropy flux if it satisfies$\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\dot{\mathrm{t}}\mathrm{i}_{0}\mathrm{n}$$D_{U}q=D_{U}\eta.\dot{D}uf$

.

(3.1)

Using the Riemann invariants, we can write (3.1) as$\frac{\partial q}{\partial w}=\lambda_{2}\frac{\partial\eta}{\partial w}$,$\frac{\partial q}{\partial z}=\lambda_{1}\frac{\partial\eta}{\partial z}$.

6

By eliminating
tion:

$q$

from the equation, we get the following second order equa$\frac{\partial^{2}\eta}{\partial w\partial z}+Q(J\frac{\partial\eta}{\partial w}-\frac{1}{J}\frac{\partial\eta}{\partial z})=0$

(3.2)

,

where
$Q$

$=$

$\frac{1}{4\sqrt{P’}}(1-\frac{P’}{c^{2}}-\frac{\rho+P/c^{2}}{2P},P^{J\prime})$

$J$

$=$

$\frac{1-\sqrt{P’}u/C^{2}}{1+\sqrt{P’}u/C^{2}}$Since this equation tends to the,.
-Darboux equation$\mathrm{E}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{r}-\mathrm{p}\mathrm{o}\mathrm{i}_{\mathrm{S}}\mathrm{S}\mathrm{o}\mathrm{n}$(3.3)$\frac{\partial^{2}\eta}{\partial w\partial z}+\frac{N}{w-z}(\frac{\partial\eta}{\partial w}-\frac{\partial\eta}{\partial z})=0$, we shall call (3.2) the relativisticas-Darboux equation.
Among entropies of (3.3) when $c=\infty$ the kinetic energy$carrow\infty$$\mathrm{E}\mathrm{u}\mathrm{l}\mathrm{e}\Gamma^{-}\mathrm{p}_{0}\mathrm{i}\mathrm{S}\mathrm{s}\mathrm{o}\mathrm{n}$(3.4)$\eta=\frac{1}{2}\rho u^{2}+\frac{P}{\gamma-1}$

plays an important role. Therefore we want to find an entropy of (3.2)
which tends to (3.4) as $carrow\infty$ . Let us look for an entropy-entropy flux ofthe form$\eta=H(\rho, u^{2})$ ,$q=Q(\rho, u^{2})u$ .

Inserting this to the equation it is easy to find an entropy-entropy flux$\eta^{*}$$=$$- \frac{\Psi(\rho)}{(1-u^{2}/c^{2})^{1/}2}+c^{2}(\frac{\rho+Pu^{2}/C^{4}}{1-u^{2}/C^{2}})$$q^{*}$$=$$(- \frac{\Psi(\rho)}{(1-u^{2}/c^{2})^{1/2}}+c^{2}\frac{\rho+P/c^{2}}{1-u^{2}/C^{2}})u$$\Psi$$=$$exp( \int_{1}^{\beta}\frac{d\rho}{\rho+P/c^{2}}+\mathrm{A}’\mathrm{o})$

,

,

(3.5)

(3.6)
(3.7)

,

is determined so thattends to the kinetic energy (3.4) as$c=\infty$ . We call the entropy
defined by (3.5) the relativistic standard
entropy. The important fact iswhere$K_{0}$$\eta^{*}$$\eta^{*}$

Proposition 5 The Hessian
is positive definite. For any fixed
there is a positive constant such that
$D_{U}^{2}\eta^{*}$$k$$(\xi|D_{U\eta}^{2*}(U)\epsilon)\geq k|\xi|^{2}$for any$U\in\Sigma_{B}$and$\xi=(\xi_{0},\xi_{1})$

with

,

$|\xi|^{2}=\xi_{0}^{2}+\xi_{1}^{2}$

.

$B$

7

Proof. The proof is due to direct but tedious calculations. We note$\frac{\partial\rho}{\partial E}$$\frac{\partial u}{\partial E}$$\frac{\partial\rho}{\partial F}$$\frac{\partial u}{\partial F}$,$=$$\frac{1+u^{2}/c^{2}}{1-Pu^{2}/c^{4}},$$=$$- \frac{(1+P’/C^{2})(1-u^{2}/c)2u}{(\rho+P/C2)(1-Pu^{2}/c)4},$$=$$- \frac{2u/c^{2}}{1-Pu^{2}/c^{4}},$$=$$\frac{(1-u^{2}/c^{2})(1+P^{;}u^{2}/c^{4})}{(\rho+P/C2)(1-Pu/2c4)},\cdot$,,

Using these, we have$\frac{\partial\eta^{*}}{\partial E}$$\frac{\partial\eta^{*}}{\partial F}$$\frac{\partial^{2}\eta^{*}}{\partial E^{2}}$$\frac{\partial^{2}\eta^{*}}{\partial E\partial F}$$\frac{\partial^{2}\eta^{*}}{\partial F^{2}}$,$=$$- \frac{\Psi}{(\rho+P/c2)(1-u^{2}/c^{2})^{1/2}}+c^{2}$$=$$\frac{\Psi u/c^{2}}{(\rho+P/c2)(1-u^{2}/c^{2})^{1/2}}.$$=$$\frac{\Psi/c^{2}}{(1-P^{J}u^{2}/C^{4})(1-u^{2}/C)^{1/2}2(\rho+P/c^{2})2}(P’+2P’u^{2}/C^{2}+u^{2})$$=$$\frac{-\Psi/c^{2}}{(1-P\prime u/2C^{4})(1-u^{2}/C^{2})1/2(\rho+P/c^{2})2}(2P’/c^{2}+1+P\prime 2/uc^{4})u$$=$$\frac{\Psi/c^{2}}{(1-P’u/2C4)(1-u2/c)21/2(\rho+P/c)^{2}2}(1+3P\prime u2/c^{4})$

,
,
,

.

Therefore we get$=$$\eta_{EE}^{*}\xi_{0}^{2}+2\eta^{*}Ep\xi 0\xi_{1}+\eta F*F\xi_{1}^{2}$$=$$\frac{\Psi/c^{2}}{(1-P\prime u/2C^{4})(1-u^{2}/c)21/2(\rho+P/c)^{2}2}Z$$=$$(P’+2P’u^{2}/c^{2}+u^{2})\xi_{0}^{2}-2(2P’/c^{2}+1+P’u^{2}/c^{4})u\epsilon_{0}\xi_{1}+$$+$$(1+3P’u^{2}/c^{4})\xi_{1}^{2}$$\geq$$\frac{2P’(1-u2/C2)^{2}(1-P’u/2)C^{4}}{A+C+\sqrt{(A-C)^{2}+4B^{2}}}(\xi_{0}^{2}+\xi_{1}2)$

$A$

$=$

$P’+2P’u^{2}/c^{2}+u^{2},$

$B$

$=$

$(2P’/c^{2}+1+P’u^{2}/c^{4})u$ ,

$C$

$=$

$1+3P^{J}u^{2}/c^{4}$$(\xi|D_{U}^{2}\eta^{*}\xi)$

$Z$

,

,

’

.

This completes the proof. QED.

4

Construction of approximate solutions

Let us construct approximate solutions using the Godunov scheme. The
construction is similar if we use the Lax-Friedrichs scheme.

8

Suppose that the initial data $U_{0}(x)$ is confined to an invariant region.
Put $\Lambda_{0}=\sup\{|\lambda_{j}(U)||j=1,2, U\in\Sigma_{B}\}$ . Fixing, we take meshlengths
such that. We denote.
Let us construct the approxomate solution. First we put.
$\Sigma_{B}$$\Lambda_{1}>\Lambda_{0}$$\Delta x,$$\Delta t$$\Delta=\Delta x$$\Delta x=\Lambda_{1}\Delta$$U^{\Delta}(t, x)$$U_{0}^{\Delta}(x)=U0(X)x_{[-}1/\Delta,1/\Delta]$.

We define$U^{\Delta}(+0, x)-- \frac{1}{2\Delta x}\int_{2j\Delta x}^{(2}j+2)\Delta x_{U0^{\Delta}(X\mathrm{I}^{d}x}$

for $2j\Delta x<x\leq(2j+2)\Delta x$ . Solving the Riemann problem on each interval$[2(j-1)\Delta, 2(j+1)\Delta]$ , we define
for $0<t<\Delta t$ . Since the CourantFriedrichs-Levi condition is satisfied, the wave from the center
does notintersect. If
for $0\leq t<n\Delta t$ has been defined, then we define$U^{\Delta}(t, x)$$-1$$2j\Delta$$U^{\Delta}(t, x)$$U^{\Delta}(n \Delta t, X)=\frac{1}{2\Delta}\int_{2j\Delta}^{(2j+2})\Delta nU\Delta(\Delta t-\mathrm{o}, x)dX$

for $2j\Delta<x\leq(2j+2)\Delta$ .
for $n\Delta t\leq t<(n+1)\Delta t$ .
By.
, say,.emann$\mathrm{S}\mathrm{o}\mathrm{l}\mathrm{V}\mathrm{i}\mathrm{n}.\mathrm{g}\mathrm{t}..\mathrm{h}.\mathrm{e}\mathrm{R}.\mathrm{i}$$\mathrm{p}\mathrm{r}\mathrm{o}_{\mathrm{P}\mathrm{o}\mathrm{S}}..\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{l}\mathrm{a}\mathrm{n}.\mathrm{d}$$.\cdot.3.’$

it- isgu..a$\mathrm{i}..\mathrm{n}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\sim \mathrm{i}_{\mathrm{V}}\mathrm{e}\mathrm{l}\mathrm{y}$w.. e fin..e$\mathrm{p}\mathrm{r}_{\vee}\mathrm{o}\mathrm{b}\mathrm{l}\mathrm{e}.\mathrm{m},$$U\Delta(t, X)$$\mathrm{d}.\mathrm{e}$

$’-$

:
in

$\underline{\mathrm{r}}\mathrm{a}\mathrm{n}.\mathrm{t}\mathrm{e}..\mathrm{e}\mathrm{d}.\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}..U\Delta \mathrm{r}\mathrm{e}\mathrm{m}\mathrm{a}\mathrm{i}\mathrm{n}\mathrm{s}\wedge$$\Sigma_{B}$Proposition 6 The approximate solution
therefore,.. :$0\leq\rho(\Delta t, X)\leq M$,$U^{\Delta}(t, x)$satisfies$U^{\Delta}(t, x)\in\Sigma_{B}$$| \frac{c}{2}\log\frac{c+u^{\Delta}(t,X)}{c-u^{\Delta}(t,x)}|\leq M$,.

Moreover we shall prove
Proposition 7

For. an.y teStfun$ctio_{-}n..\Phi..i.th_{\mathit{0}\iota}$$\cdot.\vee--:\wedge^{\vee}\cdot..\cdot.d.S$-\sim ..

th-. at.

$\int\int(\Phi_{t}U^{\Delta}+\Phi_{x}f(U^{\Delta}))d_{X}dt+\int\Phi(0, x)U_{0}\Delta(X)dX=O(\Delta^{1/2})$

In order to prove Proposition 7, we prepare

Proposition 8 For any shock wave fromand for any convex entropy , we have

$U_{L}$

to$U_{R}$

with the shock speed$\eta$$\sigma[\eta]-[q]\geq 0$

,
$‘*$

’
$\mathrm{A}$

where
$i$

$[\eta]=’:\eta(U_{R}^{\backslash }t, )-\eta(U^{\backslash }L‘),[l\prime\prime \mathrm{r}^{\backslash }q]\backslash =q(’ U’\backslash \mathrm{r};‘\prime R)-1q(UiL)$$..\tau\cdot.$,.

$\sigma$9

Proof. The right state of shocks can be parametrized by$Q(\rho)=\dot{\sigma}[\eta]-[q.]$$\rho=\rho_{R}$. Putting,

we shall see $dQ/d\rho\geq 0$ along $S_{1}j[\rho]>0$ and $dQ/d\rho\leq 0$ along: $[\rho]<0$ .
Using the equation (3.1) and the differentiation of the Rankine-Hugoniot
condition, we have$S_{2}$$\frac{dQ}{d\rho}$$=$$\frac{d\sigma}{d\rho}([\eta]-D_{U\eta}(U).[U])$$=$$- \frac{d\sigma}{d\rho}\int_{0}^{1}\theta(U-UL|D_{U}^{2}\eta(U_{L}+\theta(U-U_{L}).(U-U_{L}))d\theta$

We supposed. By Proposition 4, we know $d\sigma/d\rho<0$ on$d\sigma/d\rho>0$ on. QED.
Proof of Proposition 7.
We fix to consider
on $0\leq t\leq T$ . First we shall show
$D_{U}^{2}\eta\geq 0$.

$S_{1}$

and$S_{2}$$T$$U^{\Delta}$$\sum_{j_{1}n}\int_{2j\Delta}(2j+2)\Delta|U(n\Delta t-\mathrm{O}, x)-U(n\Delta t+\mathrm{O}, (2j+1)\Delta)|^{2}dx\leq C$

Let us consider the standard entropy$\eta^{*}$

.

(4.1)

. Then we have

$0$

$=$

$\int\eta^{*}(U(T_{X},))dx-\int\eta^{*}(U(0, x))dX+L+\Sigma$ ,

$L$

$=$

$\sum_{j,n}\int_{2}^{(2j}j\Delta+2)\Delta(\eta(*U(n\Delta t-\mathrm{o}, x))-\eta*(U(n\Delta t+0, (2j+1)\Delta)))dx$$=$$\int_{0sho}^{T}\sum_{k_{S}c}(\sigma[\eta]-[q^{*}])d*t$$\Sigma$,.

We write $U_{0}=U(n\Delta t+0, (2j+1)\Delta),$ $U_{1}=U(n\Delta t-\mathrm{o}, x)$ . Since$U_{0}= \frac{1}{2\Delta}\int_{2j\Delta}^{(j+}22)\Delta U_{1}d_{X},$$.$.

we see$=$$\sum_{j,n}\int_{2j\Delta}(2j+2)\Delta\int_{0}1-(1-\theta)(U1U0|D^{2}U\eta^{*}(U_{0}+\theta(U_{1}-U\mathrm{o})).(U1-U\mathrm{o}))d\theta dx$$\geq$

$0$

$L-$

.

4.

On the other hand we have
But from$\Sigma\underline{>}0$$\mathrm{w}\mathrm{e}$$\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{p}_{0}\mathrm{S}\mathrm{i}..\mathrm{t}.\mathrm{i}\mathrm{o}\mathrm{n},5,$. .

from Proposition 8. Thus $L\leq C,$.

$\mathrm{h}\mathrm{Y}$ $\cdot \mathrm{a}.\mathrm{V}\mathrm{e}.D2\eta^{*}U:..\cdot\geq k..\mathrm{T}$$\mathrm{h}\mathrm{e}\Gamma \mathrm{e}.\mathrm{f}=\cdot\cdot|\mathrm{o}\mathrm{r}\mathrm{e}’..\backslash$’$l$$C \geq L\geq\frac{k}{2}\sum_{j,n_{7}}\int_{2}^{(2}j\Delta j+2)\Delta|U_{1}-U0|^{2}d_{X}$.

.

$\Sigma\leq C$.

10Thus we get (4.1).
Now let us consider a test function$\Phi$. Put$J= \int\int(\Phi_{t}U^{\Delta}+\Phi_{x}f(U^{\Delta}))d_{X}dt+\int\Phi(0, x)U^{\Delta}0xd$

Since
have
$J$

$U^{\Delta}$.

is a weak solution on each time strip $n\Delta t<t<(n+1)\Delta t$ , we$=$$\sum_{n}\int\Phi(n\Delta t, X)(U(n\Delta t-\mathrm{o}, X)-U(n\Delta t+0, x))dX$$=$$J_{1}+J_{2}$$J_{1}$$=$$\sum_{j,n}\int_{2j\Delta}^{(2j+)\Delta}2\Phi(n\Delta t, j\Delta)(U(n\Delta t-\mathrm{o}, X)-U(n\Delta t+\mathrm{O}, x))dX$$J_{2}$$=$$\sum_{jn}\int_{2j\Delta}^{(2j}+2)\Delta(\Phi(t, x)-\Phi(n\Delta t,j\Delta))(U(n\Delta t-\mathrm{o}, x)-U(n\Delta t+\mathrm{O}, x))dX$

,
,

Since
$U(n \Delta t+\mathrm{O}, x)=\frac{1}{2\Delta}\int_{2j\Delta}^{(2j+2})\Delta U(n\Delta t-\mathrm{O}, X)dX$

for $2j\Delta<x<(2j+2)\Delta$ , we see $J_{1}=0$ . It$|J_{2}|$$\leq$$\leq$$\dot{\mathrm{f}}\mathrm{o}$llows from (4.1) that$C \Delta^{1/2}||\Phi||_{C}1(\sum_{j,n}\int_{2j}(2j+2)\Delta\Delta|U\mathrm{f}n\Delta t-0, x)-U(n\Delta t+\mathrm{O}, x)|2dX)1/2$$C’\Delta^{1/2}$.

Here we have used $T/\Delta t=O(1/\Delta)$ . QED.
Summing up, we have the following theorem.

Theorem 1 The approximate solution$0\leq\rho^{\Delta}(t, x)\leq M$,$U^{\Delta}(t, x)$satisfies$| \frac{c}{2}\log\frac{c+u^{\Delta}(tx))}{c-u^{\Delta}(t,x)}|\leq M$

and$\int\int(\Phi_{t}U^{\Delta}+\Phi_{x}f(U^{\Delta}))dxdt+\int\Phi(0, x)U_{0}^{\Delta}(x)=O(\Delta^{1/2})$

for any test function$\Phi$.

tends to a weak solution everywhere. For the nonWe expect that
and.Chen et al
relativistic gas dynamics, this was done by[2]. In their proof the Darboux formula$U^{\Delta}$$\mathrm{D}\mathrm{i}\mathrm{p}_{\mathrm{e}\mathrm{r}\mathrm{n}}\mathrm{a}[3]$$\mathrm{G}.\mathrm{Q}$$\eta=\int_{z}^{w}((w-s)(s-z))^{N}\emptyset(s)d_{S}$which gives solutions of the Euler-Poisson-Darboux equation (3.3) , being
arbitrary, plays an important role. Section 6 will be devoted to find such an-Darboux equation (3.2).
integral formula for the relativistic$\phi$$\mathrm{E}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{r}-\mathrm{p}\mathrm{o}\mathrm{i}_{\mathrm{S}\mathrm{S}\mathrm{o}\mathrm{n}}$

.

11

5

RemarkWe note that,$\lambda_{2}-\lambda_{1}$$=$$\frac{\sqrt{P’}(1-u^{2}/c)2}{1-u^{2}P/c^{4}},>0$$\frac{\partial\lambda_{1}}{\partial z}$$=$$\frac{1-u^{2}/C^{2}}{2(1-\sqrt{P}u/c^{2})},(1-\frac{P’}{c^{2}}+\frac{(\rho+P/c2)P’\prime}{2P},)>0$$=$$\frac{1-u^{2}/C^{2}}{2(1+\sqrt{P}u/c^{2})},(1-\frac{P’}{c^{2}}+\frac{(\rho+P/c^{2})P\prime\prime}{2P},)>0$$\frac{\partial.\lambda_{2}}{\partial w}$,

for $\rho>0$ and $|u|<c$ .
This says that the system is strictly hyperbolic and genuinely nonlinear
on $\rho>0$ . Therefore the Glimm’s theory can be applied if$||U0(X)-U^{*}||_{L\infty}+^{\tau.V}.U_{0}$is sufficiently small, whereis a constant state such that $\rho^{*}>0,$ $|u^{*}|<$. But the vaccum may not be covered by this application of the general
theorem.
$U^{*}$$c$

6

Generalized Darboux formula

In this section we seek an integration formula for solutions of the relativistic Euler-Poisson-Darboux equation. Let us introduce the variables$x= \frac{c}{2}\log\frac{c+u}{c-u}$

Then the relativistic$(EPD)$,$y= \int_{0}^{\rho}\frac{\sqrt{P’}}{\rho+P/c^{2}}d\rho$$\mathrm{E}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{r}-\mathrm{p}_{0}\mathrm{i}_{\mathrm{S}}\mathrm{S}\mathrm{o}\mathrm{n}-\mathrm{D}\mathrm{a}\mathrm{r}\mathrm{b}_{\mathrm{o}\mathrm{u}\mathrm{x}}$.

equation is$\eta_{xx}-\eta_{yy}+A(x, y)\eta_{y}+B(x, y)\eta_{x}=0$

,

where
$A(x, y)$$=$$\frac{1}{\sqrt{P’}}(1-\frac{P’}{c^{2}}-\frac{\rho+P/c^{2}}{2P},P\prime\prime)\frac{1+P’u^{2}/C^{4}}{1-P\prime u^{2}/C^{4}}$$B(x, y)$$=$$- \frac{2u/c^{2}}{1-Pu^{2}/c^{4}},(1-\frac{P’}{c^{2}}-\frac{\rho+P/c^{2}}{2P},P^{J\prime})$

and

$B$

The coefficients$A$.

are of the form
,

$A$

$=$

$\frac{2N}{y}+a$

$B$

$=$

$- \frac{4N}{N+1}\frac{x}{c^{2}}(1+[X^{2}/C^{2}, y^{2}/C^{2}]_{1})$$a= \frac{y}{c^{2}}(a0+[X^{2}/C, y22/c]_{1}2)$

,

,

,

12

where $[X, Y]_{1}$ denotes a convergent power series
In order to remove the singularity in , we use the trick of Weinstein [7]. We
introduce the sequence of variables $\eta_{j},j=0,1$ , , ..., $N$ by$\sum_{j+k\geq 1}cjkX^{j}Y^{k}-\cdot$$A$$\frac{\partial\eta_{j}}{\partial y}=y\eta_{j+1}$

,

or
,

$\eta_{j}(x, y)=I\eta_{j+1}(x, y)=\int_{0}^{y}Y\eta_{j}+1(x, Y)dY$

where $\eta 0=\eta$ . The sequence of formal integro-differential operators
defined by$L_{j}V$$=$$V_{xx}-V_{yy}+( \frac{2(N-j)}{y}+a)V_{y}+BV_{x}+$$+$$j \tilde{a}V+\sum_{k=1}^{j}$$L_{j}$

isFjkl $V_{x}+ \sum_{k=1}^{j}HjkIkV$,$\mathrm{w}.\cdot..\mathrm{h}..\cdot \mathrm{e}\mathrm{r}\mathrm{e}$.
$\tilde{a}=\frac{\partial a}{\partial y}+\frac{a}{y}=\frac{1}{c^{2}}[x^{2}/C2, y^{2}/c^{2}]_{0}$The coefficients$F_{jk}$and$H_{jk}$are determined inductively by$F_{j1}+ \frac{1}{y}\frac{\partial B}{\partial y}$$F_{j+1,k}"=$$\mathrm{i}\mathrm{f}_{\mathrm{g}}h=1$$\{$$F_{jk}+ \frac{1}{y}\frac{\partial}{\partial y}F_{j,k-1}$

if $k\geq 2$
if $k=1$$H_{j1}+j \frac{1}{y}\frac{\partial\tilde{a}}{\partial y}$–$H_{j+1,k}$$\{$

if $k\geq 2$$H_{jk}+ \frac{1}{y}\frac{\partial}{\partial y}H_{jk-1}$

It is easy to see thatthe form$F_{jk}$$\frac{1}{c^{2}}[X/2c2,2y/C^{2}.]_{0}$.

are of the form.
the$\frac{x^{\iota}}{c^{2}}[x^{2}/c^{2}, y^{2}/c2]_{0}$. have$\mathrm{B},.\mathrm{y}$$\mathrm{d}...\mathrm{e}..\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{i}..\mathrm{t}\mathrm{i}_{\mathrm{o}\mathrm{n}\mathrm{w}}\mathrm{e}$and]$\mathrm{f}\mathrm{o}\mathrm{r}.\mathrm{m}\mathrm{a}\prime \mathrm{l}$$H_{jk}$are of.

$.\mathrm{y}$$\frac{1}{y}\frac{\partial}{\partial y}(L_{j\eta_{j})}=L_{j+}1\eta j+1\cdot$

Now we consider the equation $L_{N}V=0$ for $V=\eta_{N}$ with the initial conditions-$\dot{V}=0$,$\check{V}_{y}=2^{N+1}N!\emptyset(X)$,at $y=0$ .

The problem is$(Q)$$V_{yy}-V_{xx}=aV_{y,\backslash }+BV_{x}+N‘\tilde{a}V+l$,$+ \sum_{k=1}^{N}FkI^{k}Vx+\sum_{k=1}^{N}H_{k}IkV$$V=0$ ,

$V_{y}=2^{N+1}N!\emptyset(X)$at $y=0$ .

13Proposition 9 If $\phi\in C^{1}(R)$ , then the problem $(Q)$ admits a unique solution
in $C^{2}(R\cross[0, \infty))$ .
$V$

Proof. Let us denote by $H(x, y, V)$ the right hand side of the equation
Then (Q) is transformed to the integral equation$L_{N}=0$ .

$V(x, y)=2^{N}N! \int_{x-y}^{x+y}\phi(\xi)d\xi+\frac{1}{2}\int_{0}^{y}\int_{x-y+}^{x+y}-YH(x, Y, V)dxdYY^{\cdot}$

We can solve this integral equation by the iteration,$V0(x, y)$$=$$2^{N}N! \int_{x-y}^{x+y}\phi(\xi)d\xi$$V^{n+1}$ $(x. , y)$$=$$2^{N}N! \int_{xy}^{x+y}-d\phi(\epsilon)\xi+\frac{1}{2}I_{0}^{y}\int_{x-y+}^{x+Y}y-\mathrm{Y}H(X, Y, Vn)dXdY$

Fixing’$L\mathrm{a}\mathrm{r}\mathrm{b}\mathrm{i}\mathrm{t}\mathrm{r}.\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{l}\mathrm{y}\iota \mathrm{l}$we.

nsider $|x|\leq L$ . Then it is easy to get the estimates$\mathrm{c}_{1}\mathrm{o}$$|V^{n+1}(x, y)-Vn(x, y)| \leq\frac{M^{n+1}y^{n+}1}{(n+1)!}$

Therefore
tends to a limit uniformly on $|x|\leq L,$
is the unique solution of (Q). QED.
Now we put$\eta_{N}=V$,$\eta N-k=I\eta N-k+1$ .
$V^{n}$$V$.

$0\leq,$$y\leq L$. The limit

Since
and its derivatives of order
all vanish on $y=0$ for $k\geq 1$ ,we seegives a solution of the relativistic
equation (EPD).
Next we give an integral formula for the solution of (Q).
$\leq 2$$\eta N-k$$\mathrm{E}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{r}-\mathrm{P}\mathrm{o}\mathrm{i}_{\mathrm{S}\mathrm{s}}\mathrm{o}\mathrm{n}-\mathrm{D}\mathrm{a}\Gamma \mathrm{b}\mathrm{o}\mathrm{u}\mathrm{x}$$\eta=\eta_{0}$$V$

Proposition 10 There is a $C^{N+2}- functionG(X\wedge\cdot’ y, \epsilon)of|x|<:\infty,$ $y\geq 0,$ $x-$$y\leq\xi\leq x+y$ such that the solutionof $(Q)$ satisfies$V$$V(x, y)= \int_{x-y}^{x+y}c(x, y,\xi)\emptyset(\xi)d\xi$

.

(6.1)

Moreover$G$$\partial_{x^{1}\xi y}^{p}\partial^{p2}\partial \mathrm{p}3G$$=$$2^{N}N!+O(y/c^{2})$ ,$=$$O(1/c^{2})$for $1\leq p_{1}+p_{2}+p_{3}\leq N+2$

Proof. We consider the approximate solution $V^{n}(x, y)$ which appeared in
the iteration of the
Proposition 9. By writing $H$ as$\mathrm{p}.\mathrm{r}\mathrm{o}$.

$\mathrm{o}\mathrm{f}\mathrm{o}\mathrm{f}\sim$

$.$

)

,

$H=(aV)_{y}+(BV)_{x}+bV+ \sum(F_{k}I^{k}V)_{x}+\sum\overline{H}_{k}I^{k}V$

14

where
$b$

$=$

$N \tilde{a}-a_{y}-B_{x}=\frac{1}{c^{2}}[X^{2}/C^{2}, y2/C^{2}]0$$\tilde{H}_{k}$$=$$H_{k}-(F_{k})_{x}= \frac{1}{c^{2}}[X2/c^{2}, y^{2}/C^{2}]0$,,

it is easy to see inductively that there is a kernel $G^{n}(x, y,\xi)$ such that$V^{n}(x, y)= \int_{x-y}^{x+y}c^{n}(x, y, \xi)\emptyset(\xi)d\xi$

In fact $G^{0}=2$ and$G^{n+1}$$G_{I}$$G_{II}$$G^{n}$.

are determined inductively by the formula,$=$$2+ \frac{1}{2}(c_{I}^{nn_{I}}+G_{I}+GII+\sum n_{I}G_{IVk}^{n}+\sum G_{Vk}n)$$=$$\int_{(-x+y+}^{y}\xi)/2(ax-y+Y, Y)G(x-y+Y, Y,\xi)dY+$$+$$\int_{x+y-}^{y}(\xi)/2Xa(+y-Y, Y)c(X+y-Y, Y, \xi)dY$,$=$$\int_{x+\xi}^{y}(y-)/2B(X+y-Y, Y)c(X+y-Y, Y,\epsilon)dY+$$\int_{-x+}^{y}(y+\epsilon)/2XB(-y+Y, Y)G(x-y+Y, Y, \xi)dY$ ,$G_{III}$$=$,$\int\int_{D(x,y,\epsilon})b(x, Y)c(x, Y,\xi)dXdY$

where$D(x, y, \xi)$$=$$\{(X, Y)|X-Y\leq\xi\leq X+Y, x-y+Y\leq X\leq x+y-Y, 0\leq Y\leq y\}$ ,$G_{IVk}$$=$$\int_{(\epsilon}^{y}x-y+)/2Fk(x+y-Y, Y)J^{k}c(x+y-Y, Y, \epsilon)dY+$$\int_{-x+}^{y}(y+\epsilon)/2F_{k}(x-y+Y, Y)J^{k}G(X-y+Y, Y, \xi)dY$ ,where$JG(x, y, \xi)=\int_{|x-\xi}^{y}|YG(X, Y, \xi)dY$,

and$G_{Vk}= \int\int_{D(x,y,\epsilon})\tilde{H}_{k}(X, Y)JkG(x, Y, \epsilon)dXdY$

It is easy to see inductively that$|G^{n+1}(X, y, \xi)-Gn(x, y, \xi)|\leq\frac{M^{n+1}y^{n+}1}{(n+1)!}$

.

.

15

converges to a limit uniformly and (6.1) holds. Moreover wetherefore
is differentiable. In fact we havecan differentiate $G^{n+1}$ supposing that$G$

$G^{n}$$G^{n}$$G_{I,x}$$=$$\frac{1}{2}aG((x-y+\xi)/2, (-X+y+\xi)/2,\xi)$$\frac{1}{2}aG((x+y+\xi)/2, (x+y-\xi)/2,$ $\xi)+$$G_{I,\xi}$$G_{I,y}$$+$$\int_{(+}^{y}-x+y\xi)/2a(G)_{x}(x-y+Y, Y, \xi)dY$$+$$\int_{x+y-}^{y}(\xi)/2x(ac)_{x}(-Y+Y, Y, \xi)dY$$=$$- \frac{1}{2}aG((x-y+\xi)/2, (-X+y+\xi)/2,\xi)+$$+$$\frac{1}{2}aG((x+y+\xi)/2, (x+y-\xi)/2,$ $\xi)+$$+$$\int_{(+}^{y}-x+y\xi)/2(ac_{\epsilon}X-y+Y, Y, \xi)dY+$$+$$\int_{(-x+y}^{y}+\epsilon)/2ac\xi(x+y-Y, Y, \xi)dY$$=$$- \frac{1}{2}aG((x-y+\xi)/2, (-X+y+\xi)/2,\xi)+$,,$\frac{1}{2}aG((x+y+\xi)/2, (x+y-\xi)/2,$ $\xi)+$$+$$2aG(x, y, \xi)+$$\int_{(+}^{y}-x+y\xi)/2y(aG)x(X-+Y, Y, \xi)dY+$$G_{II,x}$;$+$$Y,$ $\xi dY$$\int_{(-x+}^{y}y+\epsilon)/2)(ac)_{x}(X+y-Y,$$=$$- \frac{1}{2}BG((x+y+\xi)/2, (x+y-\xi)/2,$ $\xi)+$$\frac{1}{2}BG((x-y+\xi)/2, (-X+y+\xi)/2,$ $\xi)+$$+$$\int_{x+y-}^{y}(\xi)/2(BG)x(X+y-Y, Y, \epsilon)dY+$$\backslash -\backslash \cdot$$\int_{(-x+}^{y}y+\xi)/2x(BG)x(-y+Y, Y,\xi)dY$,$G_{II,\xi}$$=$$\frac{1}{2}BG((x+y+\xi)/2, (x+y-\xi)/2,\xi)+$$+$$\frac{1}{2}BG((x-y+\xi)/2, (-X+y+\xi)/2,\xi)+$$+$$\int_{(x+\xi)/2}^{y}y-BG\epsilon(x+y-Y, Y, \xi)dY+$$\int_{(-x+}^{y}y+\xi)/2BG_{\xi}(_{X}-y+Y, Y,\xi)dY$

,

.

16

$=$

$- \frac{1}{2}BG((x+y+\xi)/2, (x+y-\xi)/2,$ $\xi)+$$+$$\frac{1}{2}BG((x-y+\xi)/2, (-X+y+\xi)/2,$ $\xi)+$$+$$\int_{x+y-}^{y}(\xi)/2x(BG)x(+y-Y, Y, \xi)dY+$$+$$\int_{(+}^{y}-x+y\xi)/2x(Bc)_{x}(-y+Y, Y,\xi)dY$$G_{III,x}$$=$$\int_{(y-}^{y}x+\xi)/2bG(x+y-Y, Y, \xi)dY-\int_{(+}^{y}-x+y\xi)/2bc(x-y+Y, Y, \xi)dY$ ,$G_{III,\xi}$$=$$\int_{0}^{(\epsilon}x+y-)/2bG(\xi+Y, Y,\xi)dY+\int_{0}^{(-x}+y+\xi)/2YbG(\xi-Y, Y,\xi)d+$$+$$\int\int_{D(y,\xi)}xx\}bG(, Y, \epsilon)dxdY$$=$$\int_{x+y-}^{y}(\xi)/2bG(x+y-Y, Y, \xi)dY+\int_{(-x+}^{y}y+\xi)/2bc(x-y+Y, Y, \xi)dY$ ;$G_{II,y}$$G_{III,y}$;,and the derivatives of $G_{IVk}$ aretothe derivatives of $G_{IVk}$
are similar to. Then it is easy to see inductively that$\mathrm{s}\mathrm{i}\acute{\mathrm{m}}\mathrm{i}\mathrm{l}\mathrm{a}\mathrm{r}$$G_{II}\mathrm{a}\mathrm{n}\dot{\mathrm{d}}$$G_{III}$$\iota$$|G_{x}^{n+1}-G_{x}^{n}|+|G^{n+1}-c^{n} \xi\xi|+|c_{y}^{n+1}-G^{n}y|\leq\frac{M^{n}y^{n}}{n!}$Thus the limit$G$$|G_{xx}n+1-G_{x}n|x$.

is differentiable. In a smilar manner we see$+$$|G_{x\xi}^{n+1}-G^{n}|x\xi+|cn+1-xyGn|xy+$$+$$|G_{\xi\xi}^{n+1}-G_{\xi\xi}n|+|G^{n+1}-c^{n}|\xi y\epsilon y+|G_{yy}^{n+}1-G_{y}n|y\leq$$M^{n-1}y^{n-1}$$\leq$$\overline{(n-1)!}$.

Thus is twice continuously differentiable. In a similar manner we see thatis $N+2$ -times continuously differentiable. The rough estimates stated in
the propositions is obvious since the coefficients are all of $O(1/c^{2})$ . QED.
The solution
enjoyes an integral representation
$G$

$G$

$\eta_{N-k}$$\eta_{N-k}=\int_{x-y}^{x+y}K_{N}-k(x, y, \xi)\emptyset(\xi)d\xi$

,

where
$K_{N-k}(_{X}, y,\xi)=JI\mathrm{f}N-k+1(_{X}, y, \xi)-rightarrow J^{k}G(x, y, \xi)$So the solution of the relativistic$\eta$.

-Darboux equation is given$\mathrm{E}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{r}-\mathrm{P}\mathrm{o}\mathrm{i}_{\mathrm{S}}\mathrm{S}\mathrm{o}\mathrm{n}$by$\eta(x, y)=\int_{x-y}^{x+y}K(x, y,\xi)\phi(\xi)d\xi$

,

17

where
$K(x, y, \xi)=J^{N}c(_{X}, y)\xi)$.

By induction we see$J^{k}G(X, y, \xi)=\frac{2^{N}N!}{2^{k}k!}(y2-(_{X}-\xi)2)^{k}(1+O(y/c)2)$.

Thus we have

Proposition 11 There is a kernal $K(x, y, \xi)$ which is of $C^{N+2}$ -class in$|x|<\infty,$ $0\leq y,$ $x-y\leq\xi\leq x+y$ such that$\eta(x, y)=\int_{x-y}^{x+y}K(x, y)\xi)\emptyset(\xi)d\xi$

gives a solution of the relativistic Euler- Poisson-Darboux equation for anysmooth . Moreover$\phi$$K(x, y,\xi)=(y^{2}-(x-\xi)^{2})^{N}(1+o(y/c^{2}))$ .

But in order to apply this integration formula, the generalized Darboux
formula, to the study of the relativistic Euler equation, more detailed estimates of the remainder are necessary.

Proposition 12 We have
$G_{y}=O(y/c^{2})$.

Proof. Since $a=O(y/c^{2})$ , it is clear that $G_{I,y}=O(y/c^{2})$ . Next we see$G_{II,y}=-B((x+y+\xi)/2, (x+y-\xi)/2)+B((x-y+\xi)/2, (-x+y+\xi)/2))+O(y/c^{2})$ .

On the other hand we can write$B= \frac{1}{c^{2}}B_{0}(X)+^{o(}y/2c)2$and$\frac{x+y+\xi}{2}=x+\frac{y+Z}{2}$,$\frac{x-y+\xi}{2}=x+\frac{-y+Z}{2}$$Z=\xi-x$ .

,Therefore we see $G_{II,y}=O(y/c^{2})$ . It is clear that $G_{III,y}=O(y/c^{2})$ and$G_{IVk,y’ k,y}c_{V}=O(y^{2}/c^{2})$ . QED.
Proposition 13 We have
$G=2^{N}N!+ \frac{1}{c^{2}}C_{0}(x, C)(\xi-x)+O(y^{2}/c)2$

where $C_{0}(x, C)$ is a function of the form$[x^{2}/c^{2}]_{0}+ \frac{x}{c^{2}}[x/2C^{2}]0$

.

,

18

Proof. It is clear that $G_{I}=O(y^{2}/c^{2})$ since $a=O(y/c^{2})$ . Next we see$G_{II}=2^{N}N! \int_{x+y-}^{y}(\xi)/2-B(x+y-Y, Y)dY2^{N}N1\int_{(-x+y}^{y}+\xi)/2(B(X-y+Y, Y)dY+^{o}y2/C^{2})$ ,since $G=2^{N}N!+O(y/c^{2})$ . If we write$B= \frac{1}{c^{2}}B_{0}(X)+^{o(}y/2c)2$,$Z=\xi-x$, then we see$YdY=$$-. \int_{-x+}^{y}(y+\epsilon)/2)B(x-y+Y,$$\int_{(x+\xi)}^{y}y-/2-B(x+yY, Y)dY$

Note $|Z|\leq y$ . It is clear that$=$$\frac{1}{c^{2}}(\int_{x}^{x\dagger}2)B0(sdS\mathrm{x}\pm^{\underline{z}}-\int_{x+-}^{x}-1\mathrm{L}^{Z}(B_{0}(s)d_{S})+Oy22/C^{2})$$=$$\frac{1}{c^{2}}B_{0}(X)z_{+O}(y^{2}/c)2$$G_{III},$ $G_{IV}k,$ $G_{Vk}=O(y^{2}/c^{2})$.

. QED.

Proposition 14 We have
$G_{x}+c_{\xi}=o(y/c^{2})$.

Proof. First we see$G_{I,x}+G_{I,\xi}$

since

$=$

$\int_{-x}^{y}(+y+\xi)/2((ac)_{x}+aG_{\xi})(x-y+Y, Y,\xi)dY+$$+$$\int_{(y-}^{y}x+\epsilon)/2((ac)_{x}+aG_{\xi})(x+y-Y, Y, \epsilon)dY$$=$$O(y^{2}/c^{2})$$a,$ $a_{x}=O(y/c^{2})$$G_{II,x}+c_{I}I,\xi$$=$,. Next we see$\int_{(x+y}^{y}-\xi)/2)((Bc)x)+BG_{\xi}(x+y-Y, Y,\epsilon)dY+$$\int_{(+}^{y}-x+y\xi)/2((Bc)_{x}+BG_{\xi})(x-y+Y, Y,\xi)dY$$=$$O(y/c^{2})$.

It is clear that $GIII,x’ GIII,\xi$ $GVk,x’ c_{Vk},\xi=O(y/c^{2})$ . $G_{IVk,x}+G_{IVk,\xi}$ is)estimated in a similar manner as $G_{II,x}+G_{II,\xi}$ . QED.
Proposition 15 We have$(G_{x}+c_{\epsilon})_{y}=o(y/c^{2})$.

19

Proof. First we see$(G_{I,x}+G_{I,\xi})_{y}$$=$$2((aG)_{x}+aG_{\xi})(x, y, \xi)+$$\frac{1}{2}((aG)_{x}+aG_{\xi})((X-y+\xi)/2, (-X+y+\xi)/2,\xi)+$$\frac{1}{2}((aG)_{x}+aG_{\xi})((X+y+\xi)/2, (x+y-\xi)/2,$ $\epsilon)+$$\int_{(+}^{y}-x+y\xi)/2((aG)x+aG_{\xi})_{x}(X-y+Y, Y, \xi)dY+$

since

$+$

$\int_{((x+\xi)}^{y}/2((aG)y-x+aG_{\xi})_{x}(x+y-Y, Y, \epsilon)dY$$=$$O(y/c^{2})$$a,$ $a_{x}=O(y/c^{2})$$(G_{II,x}+G_{II,\xi})_{y}$,. Next we see$=$$- \frac{1}{2}((BG)_{x}+BG_{\xi})((X+y+\xi)/2, (x+y-\xi)/2,$ $\xi)+$$+$$\frac{1}{2}((BG)_{x}+BG_{\xi})((X-y+\xi)/2, (-X+y+\xi)/2,$ $\xi)+$$+$$\int_{x+y-}^{y}(\xi)/2((Bc)_{x}+BG_{\xi})_{x}(X+y-Y, Y, \epsilon)dY+$$+$$\int_{(-x+y+}^{y}\epsilon)/2)((BG)_{x}+Bc\xi x(X-y+Y, Y,\xi)dY$$=$$2^{N-1}N!B_{x}((x-y+\xi)/2, (-X+y+\xi)/2)$-$2^{N-1}N!B_{x}((x+y+\xi)/2, (x+y-\xi)/2)+$$+$$O(y/c^{2})$,since $G=2^{N}N!+O(y/c^{2})$ and $G_{x}+G_{\xi}=O(y/c^{2})$ . But$B_{x}= \frac{1}{c^{2}}B_{0}^{J}(x)+^{o(}y/2c)2$and$B_{x}((x-y+\xi)/2, (-X+y+\xi)/2)$$B_{x}((x+y+\xi)/2, (x+y-\xi)/2)=$$=$$\frac{1}{c^{2}}B_{0’}’(_{X)(-y)(}+Oy^{2}/c)2$$=$$O(y/c^{2})$.

It is clear that$(G_{III,x}+G_{III,\xi})_{y}$$=$$\int_{(\xi)}^{y}x+y-/2b((G)_{x}+bG_{\xi})(x+y-Y, Y,\xi)dY+$$+$$\int_{(-x+y+}^{y}\xi)/2x((bG)x+bG_{\xi})(-y+Y, Y, \epsilon)dY$$=$$O(y/c^{2})$.

20

Similarly we can estimate $(G_{IVk,x}+G_{IVk,\xi})_{y},$ $(G_{Vk,x}+G_{Vk,\xi})_{y}$ bearing in
mind that $(JG)_{x}+(JG)_{\xi}=J(G_{x}+G_{\xi})$ . ED.
$\mathrm{Q}\backslash$

Proposition 16 We have
$G_{x}+G_{\xi}= \frac{1}{c^{2}}C_{1}(x, c)(\xi-X)+O(y^{2}/c)2$,

where $C_{1}(x, C)$ is a function of the form$[x^{2}/c^{2}]0+ \frac{x}{c^{2}}[X2/c^{2}]0$.

Proof. We already observed that $G_{Ix}+G_{I\xi}=O(y^{2}/c^{2})$ . Next we look at$G_{II,x}$$+$$G_{II,\xi}= \int_{(\epsilon}^{y}x+y-)/2((BG)x+BG_{\xi})(x+y-Y, Y,\xi)dY+$-$\int_{(-x+y+}^{y}\xi$)$/2((Bc)_{x}+Bc_{\xi})(x-y+Y, Y, \xi)dY$$=$$2^{N}N! \int_{x+y-}^{y}(\xi)/2Bx(x+y-Y, Y)dY-2^{N}N!\int_{(-x+y+}^{y}\xi)/2B_{x}(X-y+Y, Y)dY+$$+$$O(y^{2}/c^{2})$,since $G=2+O(y/c^{2})$ and $G_{x}+G_{\xi}--O(y/c^{2})$ . Bearing in mind that
$B_{y}=O(y/c^{2})$ , we see$\int_{x+y-}^{y}(\xi)/2xB_{x}(+y-Y, Y)dY$–$\int_{(-x+y}^{y}+\xi$)$=- \int_{x+}^{y}(y-\xi)/2-(B_{x}+B_{y})(x+y-Y, Y)dY$$/2B_{x}(X-y+Y, Y)dY=$$\int_{(+}^{y}-x+y\xi)/2B_{x}(+B_{y})(x-y+Y, Y)dY$$+$$O(y^{2}/c^{2})$$=-2B(x, y)$$+$$B((x+y+\xi)/2, (x+y-\xi)/2)+$$+B((x-y+\xi)/2, (-x+y+\xi)/2)$$+$$O(y^{2}/c^{2})$$= \frac{1}{c^{2}}(-2B0(X)+B_{0}(x+\frac{y+Z}{2}) + B_{0}(x+\frac{-y+Z}{2}))+O(y^{2}/c)2$$\frac{1}{c^{2}}B_{0}’(_{X)z+}o(y^{2}/c)2$$=$.

Next we look at$G_{III,x}+G_{III,\xi}$$=$$\int_{x+y-}^{y}(\xi)/2bG(x+y-Y, Y,\xi)dY-\int_{(-x+}^{y}y+\xi)/2(bGx-y+Y, Y, \xi)dY+$$+$$\int_{0}^{(\epsilon}x+y-)/2bG(\xi+Y, Y,\xi)dY-\int_{0}^{(-x+y+}\epsilon)/2)bG(\xi-Y,$$Y,\xi dY+$$+$$\int\int_{D(x,y},\xi)bc(X, Y,\xi)dXdY$.

21

Putting$b(x, y)= \frac{1}{c^{2}}b0(_{X})+o(y^{2}/c)2$,

we see$G_{III,x}+G_{III,\xi}$$=$$2^{N}N!( \int_{x}x+^{\mathrm{g}}d\pm_{2}^{\underline{z}_{b0}}(s)s-\int_{x}^{x}+-_{2}-s\mathrm{K}z)b_{0}(sds+$$+$$\int_{x+Z}x+^{\mathrm{A}\pm^{\underline{z}}}2sb\mathrm{o}(_{S})d-\int_{x+-_{2}}x+z_{\mathrm{L}Z})b\mathrm{o}(sdS)+-_{\mathrm{A}}O(y2/c)2$$=$$\frac{2^{N}N!}{c^{2}}b_{0}(x)(\frac{y+Z}{2}-\frac{y-Z}{2}+\frac{y-Z}{2}-\frac{y+Z}{2})+O(y/22)c$$o(y^{2}/c)2^{\cdot}$$=$$G_{IVk,x}+G_{IVk,\xi}$Finally.

can be estimated in a similer manner as $G_{II,x}+G_{II,\xi}$ .
since $J^{k}G=O(y^{2}/c^{2})$ for $k\geq 1$ . QED.

$G_{Vk,x’ Vk,\xi}G=O(y^{3}/c^{2})$Proposition 17 We have$(G_{x}+c_{\epsilon})_{x}+(cx+c_{\xi})_{\xi}=O(y/c^{2})$.

Proof. First we see$(G_{I,x}+G_{I,\xi})x$

since

$a,$

$a_{x},$$+$$(G_{I,x}+G_{I,\xi})\epsilon=$$=$$\int_{(-x+}^{y}y+\epsilon)/2((aG)xx+2(aG\epsilon)_{x}+ac_{\xi\xi})(x-y+Y, Y,\xi)dY+$$+$$\int_{x+y-}^{y}(\xi)/2G((a)xx+2(aG_{\xi})_{x}+aG_{\xi\xi})(x+y-Y, Y, \epsilon)dY$$=$$O(y^{2}/c^{2})$$a_{xx}=O(y/c^{2})$ .

,

Next

$(G_{II,x} + G_{II_{1}\xi})_{x}+(c_{II,x}+GII,\epsilon)_{\xi}=$$=$$\int_{x+y}^{y}(-\xi)/2(((Bc)_{x}+BG_{\xi})_{x}+((BG)_{x}+BG_{\xi})_{\xi})\langle_{X}+y-Y,$ $Y,\xi)dY+$$+$$\int_{(\epsilon)}-x+y+/2Xy(((Bc)_{x}+BG_{\xi})_{x}+((BG)_{x}+BG_{\xi})\xi)(+y-Y, Y, \xi)dY$$=$$O(y/c^{2})$.

It is easy to see$(c_{III},x+cIII,\xi)_{x}+(G_{I}II,x+GIII,\epsilon)_{\xi}=o(y/c^{2})$The estimates of $G_{IVk}$ and$G_{Vk}$can be seen similarly. QED.

.

22

Proposition 18 We have$(G_{x}+G_{\xi})_{x}+(G_{x}+G_{\xi})_{\xi}= \frac{1}{c^{2}}C_{2}(x, c)(\epsilon-x)+O(y^{2}/c)2$,

where $C_{2}(x, C)$ is a function of the form$[x^{2}/c]_{0}2+ \frac{x}{c^{2}}[_{X}2/C]_{0}2$.

Proof. We already observed that$(G_{I,x}+G_{I,\epsilon})_{x}+(cI,x+GI,\xi)_{\xi}=O(y^{2}/c^{2})$.

Next, bearing in mind that $G_{x}+G_{\xi}=O(y/c^{2})$ and $(G_{x}+G_{\xi})_{x}+(G_{x}+$$G_{\xi})_{\xi}=O(y/c^{2})$, we see$(G_{II,x} + c_{II,\epsilon})_{x}+(G_{I}I,x+GII,\xi)\epsilon=$$=$$\int_{(x+y}^{y}-\xi)/2)(Bxxc+2Bx(cx+c_{\xi}+$$+$$B((G_{x}+G_{\xi})_{x}+(G_{x}+G_{\xi})_{\xi}))(x+y-Y, Y,\xi)dY+$$\int_{(+}^{y}-x+y\xi)/2)(Bxxc+2Bx(cx+c_{\xi}+$$+$$B((G_{x}+G_{\xi})_{x}+(G_{x}+G_{\xi})_{\xi})(x-y+Y, Y,\xi)dY$$=$$2^{N}N! \int_{x+y}^{y}(-\xi)/2Bxx(X+y^{-Y,Y})dY-2^{N}N!\int_{(+}^{y}-x+y\xi)/2Bxx(x-y+Y, Y)dY+$$+$$O(y^{2}/c^{2})$.

The same discussion to that of the proof of Proposition 16 can be appliedby replacing by. Let us look at $(G_{III,x}+G_{III,\xi})_{x}+(G_{III,x}+c_{III,\xi})_{\xi}$ .
Note that$B$$B_{x}$$(bG)_{x}+bG_{\xi}$

$bG$

$=$

$b_{x}G+b(G_{x}+G_{\xi})$$=$$2^{N}N!b_{x}+^{o(y}/c^{2})$$=$$2^{N}N!b+O(y/c^{2})$ .

,Applying the discussion of the proof of Proposition 16 by replacing by
we see$b$$(G_{III,x} + G_{III,\epsilon})_{x}+(cIII,x+c_{II}I,\xi)_{\xi}=$$=$$2^{N}N!( \int_{x}^{x}+zb_{0(S)s-}+\frac{y+Z}{2}d\int_{x+}^{x+^{z_{2}}}=s\mathrm{L}^{z}(b_{0}s)d_{S})+$$+$$O(y^{2}/c^{2})$$=$$-2^{N}N!b_{0}(X)z_{+}o(y^{2}/c^{2})$The estimates of $G_{IVk},$$G_{Vk}$are paralell. QED.

.

$b_{x}$,

23

Proposition 19 We have
$G_{\xi}= \frac{1}{c^{2}}C_{3}(_{X,C})+O(y/c^{2})$.

Proof. It is sufficient to note that
$G_{II,\xi}$$=$$2^{N-1}N!(B((X+y+\xi)/2, (x+y-\xi)/2)+B((x-y+\xi)/2, (-X+y+\xi)/2))+$$+$$O\{y/c^{2})$$=$$\frac{2^{N-1}N!}{c^{2}}(B_{0}(x+\frac{y+Z}{2})+B_{0}(_{X}+\frac{-y+Z}{2}))+o(y/C^{2})$$=$$\frac{2^{N}N!}{c^{2}}B_{0}(_{X})+O(y/c)2$

.

QED.

Proposition 20 We have$(G_{x}+G_{\xi}) \xi=\frac{1}{c^{2}}C_{4}(_{X,C})+o(y/c)2$.

Proof. We see$(G_{I,x}+G_{I,\epsilon})_{\xi}=^{o(y}/c^{2})$

by

$a,$

$a_{x}=O(y/c^{2})$. Next we see$(G_{II,x} + G_{II,\xi})_{\xi}=$$=$$2^{N-1}N!(Bx((X+y+\xi)/2, (x+y-\xi)/2)+$$+$$B_{x}(\{x-y+\xi)/2,$ $\{-X+y+\xi)/2))+O(y/c^{2})$$=$$\frac{2^{N}N!}{c^{2}}B_{0}’(x)+\mathit{0}_{(y/c)}2$.

And we see$(G_{III,x}1^{+} c_{III,\xi})\epsilon=$$=$$2^{N}N!b((x-y+\xi)/2, (-X+y+\xi)/2)+O(y/c^{2})$$=$$\frac{2^{N}N!}{c^{2}}b_{0}(X)+o(y/c)2$.

Other terms can be estimated similarly. QED.

7

Estimates of the derivatives of entropies
Let us consider the entropy$\eta$

generated by$\phi$of$C^{3}$$\eta(x, y)=\int_{x-y}^{x+y}K(x, y,\epsilon)\emptyset(\xi)d\xi$.

-class, that is,

24

In this section we will find estimates of the derivatives of$E,$ $F$ . As auxiliary variables we introduce$R=y^{2N+1}$ ,$M=xy^{2N+1}$$\eta$

with respect to

.

(7.1)

We are going to prove the following
Proposition 21 We have$\frac{\partial\eta}{\partial M}$$=$$2^{2N+1} \int_{0}^{1}(s-S^{2})^{N}D\emptyset(x+(2s-1)y)ds+O(y^{2}/c)2$$\frac{\partial\eta}{\partial R}$$=$$2^{2N+1} \int_{0}^{1}(s-s)2N\phi d_{S+}$

,

(7.2)

$2^{2N+1} \int_{0}^{1}(s-s)^{N}2(-X+\frac{y}{2N+1}(2S-1))D\phi ds+$$O(y^{2}/c^{2})$$\frac{\partial^{2}\eta}{\partial M^{2}}$$\frac{\partial^{2}\eta}{\partial R\partial M}$

,

(7.4)

,

$=$

$22N+1 \overline{y}12N-\int_{0}^{1}(s-s^{2})^{N}D2\emptyset ds+o(y-2N+1/C^{2})$$=$$2^{2N+1} \overline{y}2N-1\int_{0}^{1}(s-s^{2N})(-x+\frac{y}{2N+1}(2s-1))D^{2}\emptyset ds+$$O(\overline{y}2N+1/c^{2})$$\frac{\partial^{2}\eta}{\partial R^{2}}$

(7.3)

,

(7.5)

$22N+1 \overline{y}-2N1\int_{0}^{1}(s-s)^{N}2((-X+\frac{y}{2N+1}(2s-1))2+$$=$$\frac{4}{(2N+1)^{2}}s(1-S)y2)D^{2}\phi(x+(2S-1)y)d_{S+}o(y^{-}/1c^{2}\mathrm{K}7.6)$

Proof. We write$\eta=2R^{\frac{1}{2N+1}}\int_{0}^{1}K(\frac{M}{R}, R\frac{1}{2N+1}, \frac{M}{R}+(2S-1)R\frac{1}{2N+1})\emptyset(\frac{M}{R}+(2s-1)R\frac{1}{2N+1})ds$

Differentiating$\eta$

with respect to $M$ , we have$=$(1) $+(2)$ ,

(1)

$=$

$2R^{\frac{-2N}{2N+1}} \int_{0}^{1}(K_{x}+K_{\xi})(x, y, x+(2s-1)y)\phi(x+(2s-1)y)ds$

(2)

$=$

$2R^{\frac{-2N}{2N+1}} \int_{0}^{1}K(X, y, x+(2s-1)y)D\phi(X+(2s-1)y)ds$$\frac{\partial\eta}{\partial M}$

Since

,

$K(x.’ y):\epsilon).=J^{N}.G(x,.y, \xi:.\cdot),\mathrm{i}.\mathrm{e}$

.
$.$

:

.

.

$\cdot$‘$K(x, y, \xi)=\int_{1}^{y}x-\epsilon|^{Y}N\int_{|-\xi}x\mathrm{Y}_{N}|^{Y\cdots\int}N-1|xY_{2-\epsilon}|^{Yc}1(x, Y1, \epsilon)dY_{1}\cdots Y_{N}$

,

.

25

by Proposition 16 we see$(\mathrm{A}_{x}’ + K_{\xi})(x, y, x+(2s-1)y)$$=$$\int_{|2}^{y_{S-}}1|y|\int_{2s}Y_{N}1YNY-|yN-1\ldots\int_{11}^{Y_{2}}2s-|yxY1(G+G_{\xi})(_{X}, Y_{1}, X+(2s-1)y)dY_{1}\cdots YN$$=$$\frac{C_{1}(x,C)}{2^{N}N!c^{2}}y^{2N+1}(2S-1)(1-(2s-1)2)^{N}+^{o(}y/2N+2c^{2})$$=$$- \frac{2^{N}C_{1}(_{X,C})}{(N+1)!C2}y^{2N+1}\frac{d}{ds}(s-s2)^{N}+1+O(y/2N+2c^{2})$.

Therefore by integration by part we get
(1)

$=$

$=$

$R^{\frac{-2N}{2N+1}}y^{2}N+2_{\frac{2^{N+1}c_{1}(X,C)}{(N+1)!C^{2}}} \int_{0}^{1}(s-s)^{N}+1D\phi d_{S}+^{o}(y^{2}/2c^{2})$$O(y^{2}/c^{2})$.

By Proposition 13 we see$K(x, y, \xi)$,$=$$\int_{|x-\xi|}^{y}Y_{N}\int_{|x-\xi|}^{Y_{N}}Y_{N1}-\cdots\int_{|x-\xi}^{\mathrm{Y}}2|Y1c(_{X}, Y_{1},\xi)dY_{1}\cdots Y_{N}$$=$$2^{2N}(s-S^{2})^{N}y^{2}+ \frac{2^{N}C_{0}(XC)}{N!c^{2}’}N(2s-1)(s-s2)^{N}y^{2N1}++o(y/2N+2c)2$.

Therefore by integration by parts we get
(2)

$=$

$+$

$2^{2N+1\frac{-2N}{2N+1}2N}Ry \int_{0}^{1}(s(1-s))^{N}D\phi(X+(2_{S}-1)y)ds$$R^{\frac{-2N}{2N+1}}O(y^{2}/N+2c^{2})$.

Thus we have (7.2). Next we show (7.3). We have$\frac{\partial\eta}{\partial R}$$=$(3) $+(4)+(5)$ ,

(3)

$=$

$\frac{2}{2N+1}R^{\frac{-2N}{2N+1}}\int_{0}^{1}K(X, y, x+(2s-1)y)\phi(x+(2s-1)y)ds$

(4)

$=$

$2R^{\frac{-2N}{2N+1}} \int_{0}^{1}(-X(\mathrm{A}_{x}’+\mathrm{A}’\epsilon)+\frac{1}{2N+1}y(K_{y}+(2s-1)K_{\xi}’))\cross$,$\phi(x+(2_{S}-1)y)ds$ ,

(5)

$=$

$2R^{\frac{-2N}{2N+1}} \int_{0}^{1}K(x, y, x+(2s-1)y)(-x+\frac{y}{2N+1}(2S-1))D\phi(\ldots)ds$ .

By Proposition 13 we get(3) $= \frac{2^{2N+1}}{2N+1}\int_{0}^{1}(s-S)2N\emptyset(\ldots)ds+o(y2/c^{2})$ .

As for (4) we use Proposition 16 and$K_{y}$$+$$(2s-1)K_{\xi}=yJ^{N-1}G-(2s-11).(\xi-\dot{x})\grave{G}(x, |\dot{\xi}-X|,\xi)J^{N1}-\cdot.1+(2s-1)J^{N}G_{\xi}$

26

$=$

$2^{2N+1}N(_{S}-S^{2})^{N}y^{2N1}-+ \frac{2^{N-1}C_{0(x,C})}{(N-1)!c2}(2S-1)(_{S}-S^{2})Ny+2N$$+$$\frac{2^{N}C_{3}(_{X,C})}{N!c^{2}}(2s-1)(s-s2)^{N}y^{2N}+O(y2N+1/c^{2})$(See Proposition 19). Then by integration by parts we have(4) $= \frac{2^{2N+2}N}{2N+1}\int_{0}^{1}(S-s)2N\phi(\ldots)d_{S}+o(y^{2}/c^{2})$ .

As (2) we get(5) $=2^{2N+1} \int_{0}^{1}(s-s^{2})^{N}(-x+\frac{y}{2N+1}(2s-1))D\phi(\ldots)dS+O(y^{2}/c)2$ .

Thus we get (7.3).
Next we show (7.4). We have$=$(6) $+(7)+(8)$ ,

(6)

$=$

$2R^{\frac{-4N-1}{2N+1}} \int_{0}^{1}((K_{x}+\mathrm{A}’)_{x}\epsilon+(\mathrm{A}_{x}’+\mathrm{A}_{\xi}’)\epsilon)(x, y, \ldots)\mathrm{x}\phi(\ldots)ds$

(7)

$=$

$4R^{\frac{-4N-1}{2N+1}} \int_{0}^{1}(\mathrm{A}_{x}’+\mathrm{A}_{\xi}’)(_{X}, y, \ldots)D\emptyset(\ldots)ds$

(8)

$=$

$2R^{\frac{-4N-1}{2N+1}} \int_{0}^{1}K(x, y, \ldots)D^{2}\phi(\ldots)ds$$\frac{\partial^{2}\eta}{\partial M^{2}}$

,

.

,

.

By Proposition 18 we have$((\mathrm{A}_{x}’+\mathrm{A}_{\xi}’)_{x} + (R_{x}’+K_{\xi})\xi)(_{X}, y, x+(2S-1)y)=$$=$$\frac{2^{N}C_{2}(_{X,C})}{N!c^{2}}(s-s2)^{N}(2S-1)y^{2N+1}+\mathit{0}_{(y/c^{2})}2N+2$.

Thus by integration by parts we get(6) $=O(y-2N+1/c^{2})$ .
By the same discussion as (1) we see (7) $=O(y^{-2}N+1/c^{2})$ . By the same
discussion as (2) we see(8) $=2^{2N+1}y^{-}2N-1 \int_{0}^{1}(s-s)^{N}D^{2}\emptyset(\ldots)dS+O(y^{-}/2N+1c^{2})2$.

Thus we get (7.4).
Next we show (7.5). We see$\frac{\partial^{2}\eta}{\partial M\partial R}$$=$(9) $+(10)+(11)+(12)+(13)+(14)$ ,

27

(9)

$=$

$- \frac{4N}{2N+1}R^{\frac{-4N-1}{2N+1}}\int_{0}^{1}(Ic_{x}+K_{\xi})(x, y, \ldots)\phi(\ldots)ds$

(10)

$=$

$2R^{\frac{-4N-1}{2N+1}} \int_{0}^{1}(-X(Kx+K\xi)_{x}+(K_{x}+I\{\epsilon’)\xi)+$$+$$\frac{y}{2N+1}((\mathrm{A}’+x\mathrm{A}’\epsilon)y+(2s-1)(Ic_{x}+I\backslash _{\xi}^{\nearrow})\epsilon))\phi(\ldots)ds$

(11)

$=$

$2R^{\frac{4N}{2N+1}} \int_{0}^{1}(K_{x}+K\xi)(-X+\frac{y}{2N+1}(2s-1))D\emptyset ds$

(12)

$=$

$- \frac{4N}{2N+1}R^{\frac{-4N-1}{2N+1}}\int_{0}^{1}KD\phi d_{S}$

(13)

$=$

$2R^{\frac{-4N-1}{2N+1}} \int_{0}^{1}(-X(\mathrm{A}rx+R’\xi)+\frac{y}{2N+1}(K+y(2S-1)K\xi))D\emptyset ds$

(14)

$=$

$2R^{\frac{-4N-1}{2N+1}} \int_{0}^{1}K(-X+\frac{y}{2N+1}(2S-1))D^{2}\phi ds$

,

,
,

,

.

We already know that (9) $=O(\overline{y}2N+1/c^{2})$ . (Recall (1).) Next we look at(10). The first term is $O(y^{-2}N+1/c^{2})$ . (Recall (6)). By Proposition 16 and
20 we see$(K_{x}+\mathrm{A}_{\xi}^{r})_{y}$$+$$(2s-1)(Icx+K\xi)\epsilon=$$=$$\frac{2^{N-1}c_{1}(_{X,C})}{(N-1)!c2}y^{2N}(2_{S}-1)(S-S)^{N}2+$$+$$\frac{2^{N}C_{4}(_{X,C})}{N!c^{2}}y^{2N}(2s-1)(s-s^{2})^{N}+$$\frac{2^{N-1}c_{1}(x,C)}{(N-1)!c2}y(2N2s-1)^{3}(s-s^{2})N-1o(y2N+/+1c^{2})$.

By integration by parts we see (10) $=O(y^{-}2N+1/c^{2})$ . We already know(11) $=o(\overline{y}2N+1/c^{2})$ . Clearly(12) $=- \frac{2^{2N+2}N}{2N+1}\overline{y}2N-1\int_{0}^{1}(s-s2)ND\phi d_{S+}o(y-2N+1/C^{2})$ .

We see(13) $=O( \overline{y}2N+1/C2)+\frac{2}{2N+1}R^{\frac{-4N}{2N+1}}\int_{0}^{1}(K_{y}+(2S-1)R’)\epsilon D\phi ds$ .

As (4) we have(13) $= \frac{2^{2N+2}N}{2N+1}y^{-}2N-1\int_{0}^{1}(s-s^{2})ND\phi dS+O(y-2N+1/c^{2})$ .

Finally we see(14) $=2^{2}N+12N- \overline{y}1\int_{0}^{1}(_{S}-s^{2})N(-X+(2s-1)\frac{y}{2N+1})D2\phi ds+\mathit{0}_{(\overline{y}}2N+1/c^{2})$ .

28(Recall (5)). Summing up we get (7.5).
Next we show (7.6).
$\frac{\partial^{2}\eta}{\partial R^{2}}$$\frac{\partial}{\partial R}(3)+\frac{\partial}{\partial R}(4)+\frac{\partial}{\partial R}(5)$

$=$

,

$=$

(15) $+(16)+(17)$ ,

(15)

$=$

$- \frac{4N}{(2N+1)^{2}}R^{\frac{-4N-1}{2N+1}}\int_{0}^{1}K\phi d_{S}$

(16)

$=$

$\frac{2}{2N+1}R^{\frac{-4N-1}{2N+1}}\int_{0}^{1}(-x(Kx+I\mathrm{f}_{\xi})+\frac{y}{2N+1}(K_{y}+(2s-1)\mathrm{A}^{r}\xi))\phi dS$

(17)

$=$

$\frac{2}{2N+1}R^{\frac{-4N-1}{2N+1}}\int_{0}^{1}K(-x+\frac{y}{2N+1}(2s-1))D\phi dS$$=$(18) $+(19)+(20)$ ,

(18)

$=$

$- \frac{4N}{2N+1}R^{\frac{-4N-1}{2N+1}}\int_{0}^{1}(-X(K_{x}+K_{\xi})+\frac{y}{2N+1}(I\mathfrak{i}_{y}’+(2s-1)K_{\xi}))\phi dS$

(19)

$=$

$2R^{\frac{-4N-1}{2N+1}} \int_{0}^{\mathrm{i}}K’’\phi d_{S}$$=$$x(K_{x}+I\zeta_{\xi})+x^{2}((I\zeta_{x}+K_{\xi})_{x}+(K_{x}+I\mathrm{f}_{\xi})_{\xi})+$$+$$\frac{y}{2N+1}((K_{x}+I\mathrm{f}_{\xi})_{y}+(2s-1)(Kx+K_{\xi})_{\xi})+$$\frac{\partial}{\partial R}(3)$$\frac{\partial}{\partial R}(4)$

,

,

,

where
$K”$

$\frac{xy}{2N+1}((R_{x}’+\mathrm{A}_{\xi}’)_{y}+(2s-1)(\mathrm{A}_{x}^{\nearrow}+K_{\xi})_{\xi})+$

$+$

(20)

$\frac{y^{2}}{2N+1}((I\mathrm{f}_{y}+(2s-1)\mathrm{A}_{\xi}^{r})_{y}+(2s-1)(\mathrm{A}_{y}’+(2s-1)\mathrm{A}’)_{\xi}\epsilon)+$

.
,

$+$

$\frac{y}{(2N+1)^{2}}(K_{y}+(2s-1)\mathrm{A}’\xi)$$=$$2R^{\frac{-4N-1}{2N+1}} \int_{0}^{1}(-x(Kx+\mathrm{A}_{\xi}^{\nearrow})+\frac{y}{2N+1}(K_{y}+(2s-1)K_{\xi}))$$\cross$$(-x+ \frac{y}{2N+1}(2s-1))D\phi dS$ ,$=$(21) $+(22)+(23)+(24)$ ,

(21)

$=$

$- \frac{4N}{2N+1}R^{\frac{-4N-1}{2N+1}}\int_{0}^{1}K(-x+\frac{y}{2N+1}(2s-1))D\emptyset ds,$

(22)

$=$

$2R^{\frac{-4N-1}{2N+1}} \int_{0}^{1}(-x(Kx+K_{\xi})+\frac{y}{2N+1}(K_{y}+(2s-1)IC\xi))$$\frac{\partial}{\partial R}(5)$$\mathrm{x}$

(23)

$=$

$(-x+ \frac{y}{2N+1}(2_{S}-1))D\emptyset ds$,$2R^{\frac{-4N-1}{2N+1}} \int_{0}^{1}K(x+\frac{y}{(2N+1)^{2}}(2s-1))D\phi dS$

.

.. .
,

,

,

29

(24)

.

$2R^{\frac{-4N-1}{2N+1}} \int_{0}^{1}K(-x+\frac{y}{2N+1}(2s-1))2D^{2}\emptyset dS$$=$

First we see
(15)

$=$

$- \frac{-2^{2N+2}N}{(2N+1)^{2}}\overline{y}2N-1\int_{0}^{1}(S-s^{2})N\phi ds+o(y-2N+1/C^{2})$

(16)

$=$

$\frac{2^{2N+2}N}{(2N+1)^{2}}y^{-2N-1}\int_{0}^{1}(s-S)^{N}2\emptyset d_{S}+o(\overline{y}2N+1/C^{2})$

(17)

$=$

$\frac{2^{2N+1}}{2N+1}\overline{y}2N-1\int_{0}^{1}.(s-s)^{N-2}(-x+\frac{y}{2N+1}2(2S-1))D\phi dS+O(y/N+1c^{2})$

,

,

.

Thus we have$\frac{\partial}{\partial R}(3)=\frac{2^{2N+1}}{2N+1}\overline{y}2N-1\int_{0}^{1}(s-s^{2})N(-x+\frac{y}{2N+1}(2s-1))D\phi dS+O(y^{-}/2N+1c^{2})$.

Since (18) is similar to (16), we have(18) $=- \frac{2^{2N+3}N^{2}}{(2N+1)^{2}}\overline{y}2N-1\int_{0}^{1}(_{S}-S^{2})N\phi dS+o(\overline{y}/2N+1c)2$ .
Next let us look at (19). We already know
$2R^{\frac{-4N-1}{2N+1}} \int_{0}^{1}x(K_{x}+K_{\xi})\phi dS$$=$$o(\overline{y}2N+1/c^{2})$,$2R^{\frac{-4N-1}{2N+1}} \int_{0}^{1}x^{2}((K_{x}+K_{\xi})_{x}+(K_{x}+K_{\xi})_{\xi})\emptyset d_{S}$$=$$o(\overline{y}2N+1/c^{2})$.

Recalling (10), we see$\frac{2}{2N+1}R^{\frac{-4N-1}{2N+1}}y\int_{0}^{1}((K_{x}+K_{\xi})_{y}+(2s-1)(Kx+K_{\xi})_{\xi})\phi dS$$=$$o(y^{-2N\dagger 1}/c2)$,$\frac{2}{2N+1}R^{\frac{-4N-1}{2N+1}}xy\int_{0}^{1}((K_{x}+\mathrm{A}’)\epsilon y+(2s-1)(Kx+I\zeta\xi)\xi)\emptyset dS$$=$$o(y^{-2N}/+1c^{2})$.

When $N=1$ , we have$(\mathrm{A}_{y}’ + (2s-1)I\mathrm{i}_{\xi}’)_{y}+(2s-1)(K_{y}+(2s-1)\mathrm{A}_{\xi}^{\nearrow})_{\xi}$$=$$8(s-S^{2})+ \frac{C_{3}}{c^{2}}(2s-1)y-\frac{C_{0}}{c^{2}}(2S-1)^{3}y-$$\frac{2C_{3}}{c^{2}}(2s-1)3y+O(y/2c2)$.

When $N\geq 2$ , there are bounded functions $F_{j}(x, c)$ such that$(K_{y} + (2s-1)K\epsilon)_{y}+(2_{S}-1)(K_{y}+(2s-1)K\xi)\xi=$$=$$2^{2N+1}N(2N-1)(_{S}-S^{2})N2Ny-2+ \frac{F_{1}(x,c)}{c^{2}}(2S-1)(S-S)^{N1}2-y^{2}-1+N$$+$$\frac{F_{2}(x,c)}{c^{2}}(2s-1)(S-S^{2})N-2y2N-1+\frac{F_{3}(X,C)}{c^{2}}(2s-1)^{\mathrm{s}}(s-S^{2})N-2y+2N-1$$+$$\frac{F_{4}(x,c)}{c^{2}}(2s-1)^{3}(s-S^{2})N-1y2N-1+\frac{F_{5}(x,c)}{c^{2}}(2s-1)5(_{S}-s^{2})^{N2}-y^{2}-1+N$$+$$O(y^{2N}/c^{2})$.

30

Thus we see$2R^{\frac{-4N-1}{2N+1}} \frac{y^{2}}{(2N+1)^{2}}\int_{0}^{1}((K_{y}+(2S-1)\mathrm{A}_{\xi}’)y+(2s-1)(\mathrm{A}_{y}’+(2s-1)K\xi)_{y})\phi ds$$=$$\frac{2^{2N+3}N^{2}}{(2N+1)^{2}}\overline{y}2N-1\int_{0}^{1}(s-s)N\phi d_{S}-\frac{2^{2N+2}N}{(2N+1)^{2}}2\overline{y}-2N1\int_{0}^{1}(s-S)^{N}2\phi ds$$+O(\overline{y}2N+1/c^{2})$.

We have$\frac{2}{(2N+1)^{2}}R^{\frac{-4N-1}{2N+1}}y\int_{0}^{1}(I\mathrm{i}_{y}’+(2s-1)\mathrm{A}_{\xi}’)\phi ds$$= \frac{2^{2N+2}N}{(2N+1)^{2}}y^{-2N-1}\int_{0}^{1}(s-s^{2})N\emptyset ds+o(y-2N+1/C^{2})$.

Therefore(19) $= \frac{2^{2N+3}N^{2}}{(2N+1)^{2}}\overline{y}2N-1\int_{0}^{1}(s-S)^{N2}2\phi dS+o(y-2N+1/c)$ .

We see(20) $= \frac{2^{2N+2}N}{2N+1}\overline{y}2N-1\int_{0}^{1}(s-S)^{N}2(-X+\frac{y}{2N+1}(2S-1))D\phi dS+O(y-2N+1/c)2$ .

Therefore$\frac{\partial}{\partial R}(4)=\frac{2^{2N+2}N}{2N+1}y^{-2N-1}\int_{0}^{1}(s-S)^{NN}2(-X+\frac{y}{2N+1}(2S-1))D\phi dS+O(\overline{y}2+1/C)2$.

Next we see
(21)

$=$

(22)

$=$

(23)

$=$

$2^{2N+1} \overline{y}2N-1\int_{0}^{1}(_{S}-s)2N(x+\frac{y}{(2N+1)^{2}}(2S-1))D\phi dS+o(\overline{y}N2+1/C)2$

(24)

$=$

$2^{2N+1} \overline{y}2N-1\int_{0}^{1}(s-S^{2})N(-X+\frac{y}{2N+}(2_{S}-1))^{2}D2\phi dS+O(y-2N+1/C^{2})$$- \frac{2^{2N+2}N}{2N+1}y^{-}2N-1\int_{0}^{1}(s-s)2N(-X+\frac{y}{2N+1}(2S-1))D\emptyset d_{S}+^{o(/)}\overline{y}N2+1c^{2}$,$2^{2N+2}N-2N-1 \int_{0}^{1}(s-s)N(-2x+\frac{y}{2N+1}(2_{S}-1))D\phi ds+o(y-2N+1/c)2$,$\overline{2N+1}y$,.

Therefore we get$\frac{\partial}{\partial R}(5)$$=$$22N+1y-2N-1 \int_{0}^{1}(s-S)2N(_{X}+\frac{y}{(2N+1)^{2}}(2s-1))D\phi ds+$$+$$2^{2N+1} \overline{y}2N-1\int_{0}^{1}(s-s)2N(-X+\frac{y}{2N+1}(2S-1))2D2\phi ds+O(y^{-2}N+1/c^{2})$Summing up, we have$\frac{\partial^{2}\eta}{\partial R^{2}}$$=$$2^{2N1}+y^{-2}N-1 \int_{0}^{1}(s-s^{2})^{N}(-x+\frac{y}{2N+1}(2S-1))D\emptyset ds$

.

31

$+$

$22N+1 \overline{y}2N-1\int_{0}^{1}(s-S^{2})^{N}(X+\frac{y}{(2N+1)^{2}}(2s-1))D\emptyset ds$$+$$22N+1y-2N-1 \int_{0}^{1}(s-s)2N(-x+\frac{y}{2N+1}(2s-1))2D^{2}\emptyset dS$$=$$\frac{2^{2N+2}(N+1)}{(2N+1)^{2}}y^{-2N-1}\int_{0}^{1}(s-s)^{N}2(2_{S}-1)yD\phi d_{S}+$$+$$22N+1 \overline{y}2N-1\int_{0}^{1}(s-s)2N(-X+\frac{y}{2N+1}(2s-1))2D^{2}\emptyset dS$$=$$\frac{2^{2N+3}}{(2N+1)^{2}}y^{-2N-1}\int_{0}^{1}(s-\mathit{8}^{2})N+1y^{2}D2\phi d_{S}+$$+$$22N+1y-2N-1 \int_{0}^{1}(s-s)2N(-X+\frac{y}{2N+1}(2s-1))2D^{2}\emptyset dS$

Thus we get (7.6). QED.
Let us recall the standard entropy$\eta^{*}$.

. This is generated by$\phi^{*}(x)=A’C^{2}(\frac{1}{1-u^{2}/C^{2}}-\frac{1}{\sqrt{1-u^{2}/C^{2}}})$

,

where
$A’=(2N+1)^{-2N}((2N+1)/(2N+3)A)^{\frac{2N+1}{2}(}2N-1)!!/2N+1N$ !.

We note that$D^{2} \phi^{*}(X)=A’(1+\frac{u^{2}/c^{2}}{1-u^{2}/C^{2}})(2-\sqrt{1-u^{2}/C^{2}})\geq A’$

We are going to show that the Hessian

Proposition 22 For each

fixed in$\phi$

dominates any$D_{U}^{2}\eta^{*}$$C^{3}$$|(\xi|D_{U\eta}^{2}.\xi)|\leq C(\xi|D_{U\eta}^{2}*.\epsilon)$$c$,

is sufficiently large.

By the assumption we have$R$$\frac{dR}{d\rho}$$=$$y^{2N+1}=K\rho(1+[\rho^{\frac{2}{2N+1}}/c]_{1}2)$$=$$K+[\rho^{\frac{2}{2N+1}}/C]_{1}2$

$=$

,

,

$\frac{\rho^{\frac{1-2N}{2N+1}}}{c^{2}}[\rho^{\frac{2}{2N+1}/c^{2}}]_{0}$$\frac{d^{2}R}{d\rho^{2}}$,where $K=((2N+3)(2N+1)A)^{\frac{2N+1}{2}}$ . Using these, we have$\frac{\partial R}{\partial E}$$=$,$\frac{dR}{d\rho}\frac{1+u^{2}/c^{2}}{1-Pu^{2}/c^{4}}$$D_{U}^{2}\eta$.

we have on each compact subset of$\{\rho\geq 0\}$provided that.

32$\frac{\partial R}{\partial F}$$\frac{\partial M}{\partial E}$$\frac{\partial M}{\partial F}$,$=$$K(1+u^{2}/c^{2})+^{o}(y^{2}/c^{2})$$=$$- \frac{dR}{d\rho}\frac{2u/c^{2}}{1-Pu^{2}/C^{4}}$$=$$-K \frac{2u}{c^{2}}+Q(y^{2}/c)2$$=$$- \frac{R}{\rho+P/c^{2}}\frac{1+P’/c2}{1-Pu^{2}/C^{4}},u+x\frac{dR}{d\rho}\frac{1+u^{2}/C^{2}}{1-Pu^{2}/c^{4}}$$=$$K(-u+x(1+u^{2}/c^{2}))+O(y^{2}/c^{2})$ ,$=$$\frac{R}{\rho+P/c^{2}}\frac{1+P’u^{2}/C4}{1-P’u/2C4}-\frac{dR}{d\rho}2xu/c^{2}\frac{1}{1-Pu^{2}/c^{4}}$$=$$K(1-2_{X}u/c^{2})+o(y^{2}/c^{2})$

,

,

,

,

.

(7.7)

Differentiating once more, we see$\frac{\partial^{2}R}{\partial E^{2}}$$\frac{\partial^{2}M}{\partial E^{2}}$$\frac{\partial^{2}R}{\partial E\partial F}$$\frac{\partial^{2}M}{\partial E\partial F}$$\frac{\partial^{2}R}{\partial F^{2}}$$\frac{\partial^{2}M}{\partial F^{2}}$,$=$$- \frac{K^{2}}{y^{2N+1}}2u^{2}(1-u^{2}/C^{2})/C+\mathit{0}2(y^{-}/2N+1c^{2})$$=$$\frac{I\mathrm{f}^{2}}{y^{2N+1}}u(-2u^{2}/c^{2}-2ux(1-u2/C)2/c)2(+O\overline{y}2N+1/c^{2})$$=$$\frac{I\{^{\prime 2}}{y^{2N+1}}\frac{2u}{c^{2}}(1-u^{2}/C)2+^{o(y^{-}/c^{2})}2N+1$$=$$\frac{I\mathrm{f}^{2}}{y^{2N+1}}(2u^{2}/C2+2xu(1-u2/C2)/c^{2})+O(\overline{y}/2N+1C2)$$=$$- \frac{2}{c^{2}}\frac{K^{2}}{y^{2N+1}}(1-u^{2}/C^{2})+O(y-2N+1/c)2$$=$$- \frac{K^{2}}{y^{2N+1}}2(u+x(1-u2/c^{2}))/C^{2}+^{o(}y^{-}/2N+1c^{2})$

,

,
,

,

.

(7.8)

The chain rule gives$\frac{\partial^{2}\eta}{\partial E^{2}}$and so on. Insertinghave$(\xi|D_{U}^{2}\eta.\xi)$

$=$

$=$

$( \frac{\partial R}{\partial E})^{2}\frac{\partial^{2}\eta}{\partial R^{2}}+2\frac{\partial R}{\partial E}\frac{\partial M}{\partial E}\frac{\partial^{2}\eta}{\partial R\partial M}$$+$$( \frac{\partial M}{\partial E})^{2}\frac{\partial^{2}\eta}{\partial M^{2}}+\frac{\partial^{2}R}{\partial E^{2}}\frac{\partial\eta}{\partial R}+\frac{\partial^{2}M}{\partial E^{2}}\frac{\partial\eta}{\partial M}$$(7.7)\mathrm{n}\mathrm{d}(7.8)$,

into (7.9), and using Proposition 21, we$\frac{2^{2N+1}K^{2}}{y^{2N+1}}\int_{0}^{1}(_{S}-s^{2})Nz[\xi]D^{2}.\emptyset dS+$$\frac{2K^{2}}{y^{2N+1}}\frac{1}{c^{2}}(1-u^{2}/C^{2})(u\xi 0-\xi_{1})^{2}\frac{\partial\eta}{\partial R}+$$\frac{2K^{2}}{y^{2N+1}}\frac{1}{c^{2}}(u+x(1-u2/c^{2}))(u\xi 0-\xi 1)2_{\frac{\partial\eta}{\partial M}+}$$+$$O(y^{-}2N+1/c^{2})$where$Z[\xi]$

$=$

(7.9)

$z_{00\xi_{0^{+2Z}}\xi 0}201\xi 1+Z_{11}\xi_{1}^{2}$

,

,

33

$Z_{00}$$Z_{01}$$Z_{11}$$=$$(1+u^{2}/C2)2((-x+ \frac{y}{2N+1}(2s-1))^{2}+\frac{4}{(2N+1)^{2}}s(1-S)y)2$ \dagger$+$$2(1+u^{2}/c^{2})(-u+x(1+u^{2}/c^{2}))(-X+ \frac{y}{2N+1}(2s-1))$$+$$(-u+x(1+u2/c^{2}))^{2}$ ,$=$$-2(1+u^{2}/c^{2})u/c2((-x+ \frac{y}{2N+1}(2s-1))^{2}+\frac{4}{(2N+1)^{2}}s(1-s)y)2+$$+$$(1+3u^{2}/c^{2}-4X(1+u^{2}/c^{22})u/c)(-x+ \frac{y}{2N+1}(2s-1))+$$+$$(-u+x(1+u^{2}/c^{2}))(1-2xu/c^{2})$ ,$=$$\frac{4u^{2}}{c^{4}}((-x+\frac{y}{2N+1}(2s-1))^{2}+\frac{4}{(2N+1)^{2}}S(1-s)y^{2})+$$\frac{4u}{c^{2}}(1-2_{Xu}/c^{2})(-X+\frac{y}{2N+1}(2s-1))+$$+$$(1-2_{Xu}/c^{2})^{2}$.

It can be shown that
$Z[\xi]\geq\kappa s(1-s)y^{2}$,

where is a positive constant depending on the compact subset of
In fact we see$\kappa$$Z_{00}z11-Z_{01}^{2}=(1-u^{2}/C2) \frac{4}{(2N+1)^{2}}S(1-S)y^{2}$

On the other hand, we can estimate$| \frac{2K^{2}}{y^{2N+1}}\frac{1}{c^{2}}(1-u^{2}/c)2\frac{\partial\eta}{\partial R}|$$| \frac{2\mathrm{A}^{\prime 2}}{y^{2N+1}}\frac{1}{c^{2}}(u+x(1-u/2c^{2}))\frac{\partial\eta}{\partial M}|$

where$\epsilon=K’/c^{2}$$\leq$$\frac{\epsilon}{y^{2N+1}}$$\leq$$\frac{\epsilon}{y^{2N+1}}$. Let us introduce the parameters$(0=\xi 0$,$\zeta_{1}=\xi_{1}-u\xi_{0}$.

Then we have$Z[\xi]=Q_{0}\mathrm{o}\zeta^{2}0+2Q\mathrm{o}1\zeta_{0}\zeta 1+Q_{1}1\zeta^{2}1$

’

and
$=$

$Q_{00}^{(1})(X)(2_{S}-1)y+Q(2)00(_{Xs},)y^{2}$$Q_{01}$$=$$Q_{01}^{(1}())X(2s-1)y+Q(012)(X, S)y^{2}$ ,$Q_{11}$$=$$Z_{11}=1+O(1/c^{2})>0$ .

Therefore if $|D^{2}\phi|\leq C$ , we see$|(\xi|D_{U}2\eta\epsilon)|$$\leq$,$Q_{00}$$\frac{2^{2N+1}K^{2}c}{y^{2N+1}}\int_{0}^{1}(_{S}-s)NZ[2\xi]ds$

,

,

.

$\{\rho\geq 0\}$

.

34

$+$

$\frac{12\epsilon}{y^{2N+1}}\int_{0}^{1}(s-s)^{N2}2(d_{S}+o(\overline{y}/2N+1c)2$$\leq$$\frac{2^{2N+1}K^{2}C}{y^{2N+1}}\int_{0}^{1}(_{S-}S)2N(Q_{1}1(1+\epsilon)J\zeta^{2}1+2Q_{01}\zeta 0\zeta_{1}+Q_{00}\zeta 0)2dS$$+$$O(y^{-}2N+1/c^{2})$ .

But since $Q_{00}^{()}0=Q_{01}^{()}0=0,$ $\int_{0}^{1}(s-s^{2})^{N}(2_{S}-1)d_{S}=0$ , we see$\int_{0}^{1}(s-s)2N(-2\epsilon’Q\mathrm{o}1\zeta 0\zeta_{1}-\epsilon Q’\zeta_{0}\mathrm{o}02)ds=O(\overline{y}2N+1/C^{2})$.

Therefore we get$|( \xi|D^{2}U\eta\epsilon)|\leq\frac{2^{2N+1}\mathrm{A}’2C\{1+\epsilon’)}{y^{2N+1}}\int_{0}^{1}(s-S^{2})N]dS+O(\overline{y}2z[\xi N+1/c2)$

Similarly, if$D^{2}\phi^{*}\geq\mu$.

, we have$( \xi|D_{U}^{2}\eta^{*}\xi)\geq\frac{2^{2N+1}\mathrm{A}^{\prime 2}\mu(1-\epsilon^{J\prime})}{y^{2N+1}}\int_{0}^{1}(_{S-}S)z[2N]d_{S}+O(\overline{y}1\xi 2N+/c^{2})$.

Thus we get.

$|( \xi|D_{U}^{2}\eta\xi)|\leq\frac{C(1+\epsilon’)}{\mu(1-\epsilon)\prime},(\xi|D_{U}^{2}\eta\xi*)+^{o}(y-2N+1/c)2$

But we know$(\xi|D_{U}^{2}\eta^{*}\epsilon)\geq\kappa|\xi|^{2}y^{-2N+1}$.

Hence if is sufficiently large we get the required estimate. QED.
As for the first derivatives, the following conclusion is now clear.
$c$

Proposition 23 On each compact subset of$\{\rho\geq 0\}$$| \frac{\partial\eta}{\partial E}|+|\frac{\partial\eta}{\partial F}|\leq^{c}$8, we have.

Usefull entropies
Let us consider an entropy$\eta$

generated by$\phi$, that is,$\eta(x, y)=\int_{x-y}^{x+y}K(x, y, \xi)\phi(\xi)d\xi$

The corresponding entropy flux
equations(8.1)

is given by integrating the differential$q$$\frac{\partial q}{\partial w}=\lambda_{2}\frac{\partial\eta}{\partial w}$.

,$\frac{\partial q}{\partial z}=\lambda_{1}\frac{\partial\eta}{\partial z}$.

35

We can solve these equations as
$q$

$=$

$\lambda_{2}\eta-\int_{z}^{w}\frac{\partial\lambda_{2}}{\partial w}\eta dw$$=$$\lambda_{1}\eta+\int_{z}^{w}\frac{\partial\lambda_{1}}{\partial z}\eta dz$.

Thus we get the formula$q(x, y)= \int_{x-y}^{x+y}L(x, y,\xi)\emptyset(\xi)d\xi$

(8.2)

,

where
$L(x, y, \xi)$$L_{1}(x, y, \xi)$$=$$\lambda_{1}K(x, y, \xi)+L_{1}(x, y,\xi)$$=$$\lambda_{2}K(x, y, \xi)+L2(X, y,\xi)$$=$2$\int_{(y-}^{y}x+\xi$),$/2\mu 1(x+y-Y, Y, )K(x+y-Y, Y,\xi)dY$ ,$-2 \int_{(+}^{y}-x+y\xi)/2x\mu_{2}(-y+Y, Y)K(X-y+Y, Y,\epsilon)dY$,$L_{2}(x, y, \xi)$$=$$\mu_{1}(x, y)$$=$$\frac{\partial\lambda_{1}}{\partial z}$$=$$\frac{1-u^{2}/C^{2}}{2(1-\sqrt{P}u/c^{2})},(1-\frac{P’}{c^{2}}+\frac{(\rho+P/c^{2})P^{J}\prime}{2P},)$$=$$\frac{N}{2N+1}+O(1/c)2$ ,$\mu_{2}(x, y)$$=$$\frac{\partial\lambda_{2}}{\partial w}$$=$$\frac{1-u^{2}/C^{2}}{2(1+\sqrt{P}u/c^{2})},(1-\frac{P’}{c^{2}}+\frac{(\rho+P/C^{2})PJ\prime}{2P},)$$=$$\frac{N}{2N+1}+O(1/c)2$.

In this section we will construct various kinds of usefull entropies.
1) Let us put$=$$\int_{x-y}^{x+y}K(x, y,\epsilon)k^{N}+1ek\xi d\epsilon$$\eta_{k}^{2}(x, y)$$=$$\int_{x-y}^{x+y}K(X, y,\xi)k^{N}+1e-kd\xi\xi$

Proposition 24 If$\eta_{k}^{1}$$\eta_{k}^{1}$$\eta_{k}^{2}$,$\eta_{k}^{1}(x, y)$$1/c^{2}$,.

is sufficiently small, we have$\eta_{k}^{2}>0$

for $y>0$ ,

$>$

$0$

$=$

$2^{N}N!y^{N}(1+O(y/C^{2}))ek(x+y)(1+O(1/k))$ ,$=$$2^{N}N!y^{N}(1+O(y/c^{2}))e^{-}-y)(k(x1+O(1/k))$

(8.3)
(8.4)

36

uniformly on each compact subset of $\{y>0\}$ . Moreover$q_{k}^{1}$$=$$\eta_{k}^{1}(\lambda_{2}+o(1/k))$$q_{k}^{2}$$=$$\eta_{k}^{2}(\lambda_{1}+o(1/k))$

,

(8.5)

uniformly on each compact subset of $\{y\geq 0\}$ and$\eta_{k}^{2}q_{k}^{1}-\eta_{k}q_{k}12=(2^{N}N!)^{2}y-(2(N1)\frac{1}{2N+1}+o(1/c)2)e(2kyy+^{o}(1/k))^{3}$. $(8.6)$

Proof. Since $K=(1+O(y/c^{2}))(y^{2}-(x-\xi)^{2})^{N}$ , we see$\eta_{k}^{1}$$=$$(1+O(y/c^{2})) \int_{x-y}^{x+y}(y^{2}-(x-\xi)^{2})NkN+1e^{k\xi}d\xi$$=$$(1+O(y/c2))22N+1..yefNkx(ky)$where$f(r)$$=$$r^{N+1}e^{-r} \int_{0}^{1}(_{S}(1-S))Ne2rsdS$$=$$e^{r} \int_{0}^{r}(\sigma(1-\frac{\sigma}{r}))N2\sigma_{d}\sigma e^{-}$.

It is easy to see$e^{-r}f(r)=2^{-(N+1})N!+O(1/r)$

This implies (8.4). We note$\eta^{1}$$=$$(1+O(1/c^{2}))2^{N}N!ye^{k}N-1(x+y)(y+O(1/k))$$\eta^{2}$$=$$(1+O(1/c^{2}))2^{N}N!y^{N1}-e-k(x-y)(y+O(1/k))$uniformly on $\{y\geq 0\}$ . Let us consider the flux. We have$L_{2}(x, y, \xi)$$q^{1}-\lambda_{2}\eta^{1}$$=$$-2 \int_{(+}^{y}-x+y\xi)/2\mu 2(x-y+Y, Y)K(x-y+Y, Y, \xi)dY$$=$$-2( \frac{N}{2N+1}+^{o}(1/c^{2}))\int_{-}^{y}(x+y+\xi)/2)(Y^{2}-(_{X}-y+Y-\xi 2)^{N}dY$$=$$-( \frac{N}{(2N+1)(N+1)}+O(1/c^{2}))(y-x+\xi)N(y+x-\xi)N+1$ ,$=$$-( \frac{N}{(2N+1)(N+1)}+Q(1/C)2)fx-y\epsilon(y-x+\xi)^{N}(y+X-)x+yN+1kN+1e^{k}d\xi\xi$.

But
$0$

$\leq$$=$$\int_{x-y}^{x+y}.‘(y-X+\xi)^{N}(y+x-\xi)^{N}+1kN+1e^{k}d\xi\xi$$(N+1)k^{N} \int_{x-y}^{x+y}(y-(2-X\xi)2)^{N}edk\xi\xi$37$Nk^{N} \int_{x-y}^{x+y}(y-x+\xi)^{N-1}(y+x-\xi)^{N1}+edk\xi\xi$$\leq$$(N+1) \frac{1}{k}\int_{x-y}^{x+y}(y-2(x-\xi)^{2})^{N}kN+1ek\epsilon_{d\xi}$.

Thus$q^{1}-\lambda_{2\eta^{1}}=O(1/k)\eta^{1}$

.

Since
,

$\lambda_{2}-\lambda_{1}=\frac{\sqrt{P’}(1-u^{2}/c)2}{1-P’u^{2}/c^{4}}=(\frac{1}{2N+1}+O(1/C^{2}))y$

we have$\eta^{2}q^{1}-\eta^{1}q^{2}=\eta^{1}\eta^{2}((\frac{1}{2N+1}+o(1/C^{2}))y+o(1/k))$This implies (8.6). QED.
2) Let be a function in$C_{0}^{\infty}(-1,1)$$\psi$

such that th$\geq 0,$$=$$\psi_{n}(X\mathrm{I}=n\psi(n(X-a))$$\phi_{n}^{4}(x)$$=$$-D\psi_{n}(x)$$\eta_{n}^{3}(x, y)$$=$$\int_{x-y}^{x+y}K(x, y,\xi)\emptyset 3n(\xi)d\xi$,$\eta_{n}^{4}(x, y)$$=$$\int_{x-y}^{x+y}K(x, y,\xi)\phi_{n}^{4}(\epsilon)d\xi$.

$\eta^{3}(x, y)$$=$$K(x, y, a)X$ ,$\eta^{4}(x, y)$$=$$K_{\xi}(x, y, a)X$$q^{3}(x, y)$$=$$L(x, y, a)X$ ,$q^{4}(x, y)$$=$$L_{\xi}(x, y, a)X$

$X$

$=$

1

Proposition 25 As$\eta_{n}^{3}arrow\eta^{3}$$narrow\infty$,. We put,,,$(x-y<a<x+y)$$(|x-a|=y)$$\frac{1}{2}$$(|x-a|>y)$ .

$0$

$=$

$\int\psi=1$,$\phi_{n}^{3}(x)$$=$.

, we have$q_{n}^{3}arrow q^{3}$,$\eta_{n}^{4}arrow\eta^{4}$,$q_{n}^{4}arrow q^{4}$.

Moreover$|\eta_{n}^{3}|$$|\eta_{n}^{4}|$$\eta^{3}q^{4}-\eta^{4}q^{3}$$\leq$$\underline{<}$$=$$My^{2N}$,$My^{2N-1}$$|q_{n}^{3}|\leq My^{2N}(|x|+y)$,,$|q_{n}^{4}|\leq My^{2N-1}(|x|+y)$

(8.7)
,

(8.8)

$\frac{N}{(2N+1)(N+1)}(1+O(1/C2))(y^{2}-(_{X}-a)^{2})2N(8.9)$

38

Proof. We note
$K_{\xi}$$L_{1,\xi}$$=$$-( \xi-x)c(X, |\xi-x|,\xi)\frac{1}{2^{N-1}(N-1)!}(y^{2}-(X-\xi)^{2})N-1J+NG_{\xi}$$=$$(2N(X-\xi)+O(1/C^{2})(\xi-x)^{2})(y-2(x-\xi)^{2})N-1+O(1/c^{2})(y^{2}-(x-\xi)^{2})^{N}$ ,$=$2 $\int_{x+y}^{y}(-\xi)/2\mu 1(x+y-Y, Y)\mathrm{A}_{\xi}’(x+y-Y, Y,\xi)dY$.

The estimates (8.7), (8.8) can be seen easily. Let us consider$\eta^{3}q^{4}-\eta^{4}q^{3}=(KL_{\xi\epsilon}-L\mathrm{A}’)(x, y, a)$.

Suppose $x-a\geq 0$ . Then$\frac{1}{2}(KL-\xi LI\mathrm{i}’\xi)$$=$$K \int_{(x\dagger y-}^{y}a)/2\mathrm{A}\mu 1\xi(\prime x+y-Y, Y, a)dY-$.

$I \iota_{\xi}’\int_{(x+y}^{y}-a)/2(\mu 1KX+y-Y, Y, a)dY$

We note$0 \leq\frac{x+y-a}{2}\leq x-y+Y-a\leq x-a\leq y$ .

Hence we have$\int_{(x+y)/}^{y}-a2\mu_{1}K_{\xi(}x+y-Y,$ $Y,$ $a)dY$$=$$( \frac{N}{2N+1}+O(1/c^{2}))2N\int_{x+-}^{y}(ya)/2(x+y-Y-a)(Y^{2}-(x+y^{-Y-a})^{2})N-1dY+$$+$$o(1/c^{2}) \int_{x+-}^{y}(ya)/2(Y^{2}-(x+y^{-Y-a})2)NdY$$=$$( \frac{N^{2}}{2(2N+1)}+O(1/C^{2}))(x+y-a)^{N-1}(-X+y+a)^{N}\frac{1}{N(N+1)}(y+(2N+1)(x-a))$$+$$O(1/c^{2})(y^{2}-(x-a)^{2})N$ .

Thus$K \int_{(x+y)}^{y}-a/2\mu 1I\mathrm{t}_{\xi}’dY$$=$$( \frac{N}{2(2N+1)(N+1)}+O(1/c)2)(y^{2}-(x-a)^{22N})-1(-x+y+a)(y+(2N+1)(x-a))$$+$$O(1/c^{2})(y^{2}-(x-a)2)^{2}N$ .

Also we have
$K_{\xi} \int_{(-}^{y}x+ya)/2\mu 1KdY$$=$$( \frac{N^{2}}{(2N+1)(N+1)}+O(1/C^{2}))(x-a)(-x+y+a)(y^{2}-(x-a)^{2})2N-1$$+$$O(1/C^{2})(-x+y+a)(y^{2}-(x-a)^{2})^{2N}$ .

39

Hence
$\frac{1}{2}(KL_{\xi}-LK_{\xi})=(\frac{N}{2(2N+1)(N+1)}+O(1/C^{2}))(y^{2}-(x-a)^{2})^{2}N$.

Here we have used$0$$\leq$$0$$\leq$$(x-a)(y-(x-a))\leq y^{2}-(x-a)^{2}$ ,$(y-x+a)(y+(2N+1)(x-a))$$(2N+1)(y^{2}-(_{X}-a)^{2})$$\leq$

provided that $0\leq x-a\leq y$ . When $x-a\leq 0$ , we can discuss in a similarmanner by using . QED.
such that $\int\Phi=0$ and the support
3) Let be a function in$supp\Phi$ is $[-1+\alpha, 1+\alpha]$ , whereis a small positive number. We put$L_{2}$$C_{0}^{\infty}(-1,1)$$\Phi$$\alpha$$\psi_{n}(x)$$=$$n\Phi(n(X-a))$ ,$\eta_{n}^{5}(x, y)$$=$$\int_{x-y}^{x+y}K(X, y,\xi)DN+1\psi_{n}(\xi)d\xi$$q_{n}^{5}(x, y)$$=$$\int_{x-y}^{x+y}L(x, y,\xi)D^{N+}1\psi n(\xi)d\xi$$\hat{\Phi}(x)$$=$$\frac{d}{dx}(x\int_{-1}^{x}\Phi)$$\hat{\psi}_{n}(_{X)}$$=$$n\hat{\Phi}(n(X-a))$$\eta_{n}^{6}(x, y)$$=$$\int_{x-y}^{x+y}K(X, y,\epsilon)DN+1\hat{\psi}_{n}(\xi d\xi$$q_{n}^{6}(x, y)$$=$$\int_{x-y}^{x+y}L(x, y)\xi)D^{N+\hat{\psi}_{n}}1(\xi)d\xi$$B_{n}^{3}$$=$$\eta^{3}q_{n}^{\mathrm{s}}-\eta_{n}53q$,$B_{n}^{4}$$=$$\eta^{4}q_{n}^{5}-\eta_{n}^{54}q$,$B_{n}$$=$$\eta_{n}^{5665}q_{n}-\eta_{n}q_{n}$

,

;

,
,
,

;

.

Let us divide the domain $\Sigma=\{-B\leq x-y\leq x+y\leq B\}$ into the following
5 parts.
,$\{-\frac{1}{n}<x+y-a\leq\frac{1}{n}-\frac{1}{n}\leq x-y-a<\frac{1}{n}\}\cap\Sigma$$S_{0}$$=$$S_{1}$$=$$S_{L}$$=$$\{-\frac{1}{n}<x+y-a\leq\frac{1}{n}, x-y-a<-\frac{1}{n}\}\cap\Sigma$$S_{R}$$=$$\{\frac{1}{n}<x+y-a, -\frac{1}{n}\leq x-y-a<\frac{1}{n}\}\cap\Sigma$

$S$

$=$

$\Sigma-(S_{0}\cup S_{1}\cup S_{1}\cup S_{L}\cup S_{R})$$\cap\Sigma$

,

,

.

,

40

Proposition 26 We have$|B_{n}^{3}|\leq M/n$on$\Sigma$,(8.10)$|B_{n}^{4}|\leq M$, and(8.11)$|B_{n}|\leq M/n$on

$S_{0}\cup S_{1}\cup S$. Mooeover, on$S_{L}$, we have(8.12)$B_{n}=ny^{2N}A_{1}+y^{N}A_{2}+A_{3}$ ,

where

On

$S_{R}$$A_{1}$$=$$|A_{2}|$$\leq$$|A_{3}|$$\leq$$( \frac{N(2^{N}N!)^{2}}{2N+1}+^{o}(1/c^{2}))(\int^{n}-1\Phi(x+y-a))^{2}$,$M(| \int_{-1}^{n(+y)}x-an\Phi|+|\Phi((X+y-a))|)$ ,$\frac{M}{n}$.

, we have$B_{n}$$=$$ny^{2N}C_{1}+y^{N}c2+C3$ ,$C_{1}$$=$$( \frac{N(2^{N}N!)^{2}}{2N+1}+O(1/c^{2}))(\int^{n}-1\Phi)(x-y-a)2$$|C_{2}|$$\leq$$|C_{3}|$$\leq$,$M(| \int_{-1}^{n(}x-y-a))\Phi|+|\Phi(n(X-y-a)|)$ ,$\frac{M}{n}$.

Proof. For the simplicity, we write
It is easy to see inductively that, for $G_{j}=J^{j}G=IC_{N-j}$ , we have$\eta_{n}--\eta_{n}^{5},$$q_{n}=q_{n}^{5},\hat{\eta}_{n}=\eta_{n}^{6},\hat{q}_{n}=q_{n}^{6}$$\partial_{\xi}^{p}G_{j}=J\partial_{\xi}pc_{j-1}$

for $j\geq p+1$ and$\partial^{p}G_{p}=\epsilon(-1)^{p}(\xi-x)pG(x, |\xi-x|,\xi)+J\partial\xi G_{p}p-1$.

Therefore$\partial_{\xi}^{p}K=\partial_{\xi}pG_{N}(_{X}, y,\xi)=0$

for $p\leq N-1$ and $y=|x-\xi|$ . Thus by integration by parts we have$\eta_{n}$$=$$(-1)^{NN}\partial_{\xi}K(X, y, x+y)\psi_{n}(X+y)+$$(-1)^{N}\partial_{\xi}^{N}K(X, y, x-y)\psi n(X-y)+$$F_{n}^{1}(x, y)$,$+$$F_{n}^{1}(x, y)$$=$$(-1)^{N+1} \int_{x-y}^{x+y}\partial_{\xi}N+1K(x, y,\xi)\psi n(\xi)d\xi$

.

.

41

We see$\partial_{\xi}^{p}L_{2}(x, y,\xi)=-2\int_{(-x+}^{y}y+\xi)/2\mu_{2}\partial_{\xi}PK(x-y+Y, Y,\xi)dY$

for $p\leq N-1$ . Therefore$\partial_{\xi}^{p}L_{2}(x, y, x+y)=\partial_{\xi}^{p}L_{2}(x, y, x-y)=0$

for $p\leq N-1$ . Moreover we see$\partial_{\xi}^{N}L_{2}(x, y, x+y)=0$.

Therefore by integration by parts we have$=$$q_{n}(_{X}, y)-\lambda 2\eta_{n}(_{X}, y)$$=$$-(-1)^{N}\partial\xi NL_{2}(x, y, X-y)\psi n(X-y)+$$+$$F_{n}^{2}(x, y)$$F_{n}^{2}(x, y)$$=$$(-1)^{N+1} \int_{x-y}^{x+y}\partial^{N}\xi+1L2(_{X}, y,\xi)\psi_{n}(\xi)d\xi$$\overline{\sigma}_{n}(x, y)$$=$$q_{n}(x, y)-\lambda_{1\eta_{n}}(x, y)=$$=$$(-1)N\partial_{\xi}^{N}L_{1}(x, y, x+y)\psi n(x+y)+$$+$$\overline{F}_{n}^{2}(x, y)$$=$$(-1)^{N+1} \int_{x-y}^{x+y}\partial_{\xi}N+1L1(_{X}, y,\xi)\psi_{n}(\xi)d\xi$$\sigma_{n}(x, y)$,.

Similarly$\overline{F}_{n}^{2}(x, y)$,.

We note$\partial_{\xi}^{N}K(x, y, \xi)=(-1)^{N}(\xi-x)^{N}c(X, |x-\xi|,\xi)+J\partial_{\xi}^{N}G_{n-1}$.

It is easy to see inductively that$\partial_{\xi}^{p+1}c_{p}(X, y, \xi)$$=$$(-1)^{p} \frac{p(p+1)}{2}(\xi-x)p-1G(_{X,|X}-\xi|, \xi)+$$+$$(\xi-X)^{p}H_{p}(X,\xi)+J\partial^{p}\epsilon c_{p-}1$,

where $H_{p}=O(1/c^{2})$ . Therefore$\partial_{\xi}^{N+1}K(X, y, \xi)$$=$$(-1)^{N} \frac{N(N+1)}{2}(\xi-x)N-1G(x, |\xi-x|,\xi)+$$+$$(\xi-X)^{N}HN(_{X,\xi})+J\partial\xi-NG_{N1}$1) Suppose $(x, y)\in S$ . Then it is clear thatall vanish.

$B_{\mathrm{n}}^{3},$$B_{n}^{4},$$B_{n}$$\eta^{3},$$\eta^{4},$.

$q^{3},$$q^{4},$$\eta_{n},$$q_{n},\hat{\eta}_{n},\hat{q}_{n}$,422) Suppose $(x, y)\in S_{0}$ . Then we see$\eta^{3}$$\eta^{4}$$\sigma^{3}$$\sigma^{4}$$=$$K(x, y, a)$$=$$O((y^{2}-(X-a)^{2})^{N})$$=$$O(n^{-2})N$$=$$K_{\xi}(x, y, a)$$=$$O(|x-a|(y^{2}-(x-a)^{2})^{N}-1)+O((y^{2}-(x-a)^{2})^{N})$$=$$O(n^{-2N+1})$ ,$=$$L_{2}(x, y, a)$$=$$-2 \int_{(+}^{y}-x+ya)/2\mu 2K(x-y+Y, Y, a)dY$$=$$O(n^{-}2N-1)$ ,$=$$L_{2,\xi}(_{X}, y, a)$$=$$-2 \int_{(+a)/}^{y}-x+y2K\mu 2\xi(x-y+Y, Y, a)dY$$=$$O(n^{-2})N$ .

,Since $y=O(1/n)$ and $\psi_{n}=O(n)$ , we see$(-1)^{NN}\partial_{\xi}K(X, y, x+y)\psi n(x+y)+$$(-1)^{N}\partial\xi NK(_{X}, y, x-y)\psi n(x-y)=$$=$$O(n^{-N+1})$ .

Since $F_{n}^{1}=O(1)$ , we have $\eta_{n}=O(1)$ . We see$\partial_{\xi}^{N}L_{2}(x, y, x-y)=-2\int_{0}^{y}\mu 2\partial_{\xi}^{N}K(x-y+Y, Y, x-y)dY=O(n^{-N1}-)$

Therefore$-(-1)N\partial_{\xi}NL2(x, y, X-y)\psi n(x-y)=O(n^{-N})$ .

Since$\partial_{\xi}^{N+1}L_{2}(x, y, \xi)$$=$$\mu_{2}\partial_{\xi}^{N}K((x-y+\xi)/2, (-X+y+\xi)/2,\xi)+$

2$\int_{(+}^{y}-x+y\xi$)$=$$/2x\partial^{N1}+K(-\xi y+Y, Y,\xi)dY$$O((-X+y+\xi)^{N})+O(_{X}+y-\xi)$ ,we see$F_{n}^{2}(x, y)$$=$$(-1)^{N+1} \int_{x-y}^{x+y}\partial_{\xi}^{N+1}L2(x, y,\epsilon)\psi_{n}(\epsilon)d\epsilon$$=$$O(n^{-1})$

.

.

43

Hence $\sigma_{n}=O(n^{-1})$ . Therefore$B_{n}^{3}$$=$$\eta^{3}\sigma_{n}-\eta_{n}\sigma 3=o(n^{-2}-1)N$$B_{n}^{4}$$=$$\eta^{4}\sigma_{n}-\eta_{n}\sigma=O4(n-2N))$$B_{n}$$=$$\eta_{n}\hat{\sigma}_{n}-\hat{\eta}n\sigma n=O(n^{-1})$,.

3) Suppose $(x, y)\in S_{1}$ , where $x+y>a+ \frac{1}{n}$ and $x-y<a- \frac{1}{n}$ . Then$\psi_{n}(_{X+}y)=^{\psi n}(X-y)=\hat{\psi}_{n}(X+y)=\hat{\psi}_{n}(x-y)=0$ . So,,$\eta_{n}=F_{n}^{1},$ $\sigma_{n}=F_{n}^{2}$and so on. But$F_{n}^{1}(x, y)=(-1)^{N+1} \int_{x-y}^{x+y}\partial^{N}+1K(x, y,\xi)\psi_{n}(\xi)\xi d\xi$$=$$(-1)^{N+1} \int_{-1}^{1}(\partial^{N+1}K(x, y, a+\frac{s}{n})-\partial^{N}+1K(\xi\xi yx,, a))\Phi(s)d_{S}$$=$$O(1/n)$since $\int\Phi=0$ and. Thusis Lipschitz continuous. Same estimates hold for$\partial_{\xi}^{N+1}K$$F_{n’ nn}^{212}\hat{F},\hat{F}$,$B_{n}^{3}$$=$$\eta^{3}F_{nn}^{2}-F^{1}\sigma=\mathit{0}_{(1}3/n)$$B_{n}^{4}$$=$$\eta^{4}F_{nn}^{214}-F\sigma=^{o_{(1}}/n))$$B_{n}$$=$$F_{nn}^{1212}\hat{F}-\hat{F}_{nn}F=O(1/n^{2})$.

4) Suppose $(x, y)\in S_{L}$ , where $|x+y-a|\leq 1/n$ . It is easy to see
Since$\eta^{3}=O(n^{-N}),$ $\eta^{4}=O(n^{-N+1}),$ $\sigma^{3}=O(n^{-N-})1,$ $\sigma^{4}=O(n^{-N})$ .
$\sigma_{n}=F_{n}^{2}=O(1)$.
$n(x-y-a)<-1$ , we have $\psi_{n}(x-y)=0$ . Thus $\eta_{n}=O(n),$Therefore

Let us estimate$=$$\eta^{3}\sigma_{n}-\eta_{n}\sigma 3=o(n^{-N})$$B_{n}^{4}$$=$$\eta^{44}\sigma_{n}-\eta_{n}\sigma=O(n^{1N}-)$$B_{n}=\eta_{n}\hat{\sigma}_{n}-\hat{\eta}_{n}\sigma_{N}$$\partial_{\xi}^{N+1}K$,$B_{n}^{3}$.

. Since$=$$(-1)^{N_{\frac{N(N+1)}{2}(\xi-x)(}}N-1cX,$ $|x-\xi|,\xi)+$$+$$(\xi-x)^{N}H_{N}(x, \xi)+J\partial_{\zeta}NG_{N-1}$,

we have$F_{n}^{1}$

$=$

$=$

$+$

$=$

$+$

$(-1)^{N+1} \int_{x-y}^{x+y}\partial N+1K(\xi y, \xi)x,\psi n(\xi)d\xi=$$(-1)^{N+1}((-1)N \frac{N(N+1)}{2}2NN!(a-x)^{N-}1+F’(X, a))\int_{-}^{n_{1}(a)}x+y-\Phi+$$O(1/n)=$$- \frac{N(N+1)}{2}2^{N}N!y^{N1}-(1+F’’(X, y, a))\int_{-1}^{n(a)}x+y-\Phi+$$O(1/n)$ ,44where $F’=O(1/c^{2})|x-a|^{N},$ $F”=O(1/c^{2})$ . On the other hand$\partial_{\xi}^{N}K(x, y, x+y)=(-1)^{N}y^{N}c(x, y, x+y)$

.

Hence
$\eta_{n}$$=$$ny^{N}G(x, y, x+y)\Phi(n(x+y-a))+$$\frac{N(N+1)}{2}2^{N}N!y^{N1}-(1+F\prime\prime(X, y, a))\int_{-1}^{n(a)}x+y-\Phi+$$+$$O(1/n)$ .

Since$\partial_{\xi}^{N+1}L_{2}(X, y,\xi)$$=$$\mu_{2}\partial_{\xi}^{N}K((x-y+\xi)/2, (-X+y+\xi)/2,$ $\xi)+$$2 \int_{-}^{y}(x+y+\xi)/2K\mu 2\partial^{N+1}(\xi x-y+Y, Y, \xi)dY=$$( \frac{N}{2N+1}+O(1/c2))(-1)^{N}(\frac{-x+y+\xi}{2})N\cross$$=$$G((x+y+\xi)/2, (-X+y+\xi)/2,$ $\xi)+$$\mathrm{x}$$O(_{X}+y-\epsilon)$$+$,

we see$\sigma_{n}$$=$$F_{n}^{2}=$$=$$(-1)^{N+1} \int_{x-y}^{x+y}\partial_{\xi}N+1L_{2}(X, y,\xi)\psi_{n}(\xi)d\xi=$$=$$- \frac{N}{2N+1}2^{N}N!y(N1+L’(x, y, a))\int_{-1}^{n(y)}x+-a\Phi$$+$$O(1/n)$ ,where $L’=O(1/c^{2})$ . Here we have used$( \frac{-x+y+a}{2})^{N}=(y-\frac{x+y-a}{2})^{N}=y^{N}+o(1/n)$ .

Similar estimates hold for$\hat{\eta}_{n},\hat{\sigma}_{n}$. Thus$B_{n}=ny^{2NN}A_{1}+yA_{2}+A_{3}$ ,where$A_{1}$$\beta$$=$$-G \frac{N}{2N+1}2^{N}N!(1+L’)\Phi(\beta)\int_{-1}^{\beta}\hat{\Phi}+$$+$$G \frac{N}{2N+1}2^{N}N!(1+L’)\hat{\Phi}(\beta)\int_{-1}^{\beta}\Phi=$$=$$\frac{N}{2N+1}2^{N\prime}N!G(1+L)(\int-\beta 1\Phi)^{2}$$=$$n(x+y-a)$ .

,45The estimates on. QED.
If we putcan be obtained in a similar manner considering$S_{R}$$\overline{\sigma}^{3},\overline{\sigma}^{4},\overline{\sigma}_{n}$$\hat{B}_{n}^{3}$$=$$\eta^{3}\eta_{n}^{6}-\eta_{n}^{6}q3$,$\hat{B}_{n}^{4}$$=$$\eta^{4}q_{n}^{6}-\eta_{n}^{64}q$,then the same estimates hold.

9Compactness of$\eta_{t}+q_{x}$Let us consider an entropy
generated bythrough the generalized
Darboux formula and its flux . In this section we will prove$\phi$$\eta$$q$

Lemma 1 Let
be the approximate solutions constructed in Section 4.
Thenlies in a compact subset ofbeing abounded open subset of $\{t\geq 0\}$ .
$U^{\Delta}$$\eta(U^{\Delta})_{t}+q(U^{\Delta})_{x}$Proof. Let$=$$\int\int(\eta(U\Delta)\Phi_{t}+q(U^{\Delta})\Phi x)d_{Xdt}$$=$$N+L+\Sigma$ ,

$N$

$=$

$- \int\eta(U^{\Delta}(+0, x)\Phi(\mathrm{o}, X)dX$

$L$

$=$

$\sum_{n}\int[\eta(U\Delta(t, X)]_{t=}t=n\Delta t-0_{\Phi}(n\Delta t+0n\Delta tX\rangle)d_{X}$$=$$\int\sum_{shoCk}(\sigma[\eta]-[q])\Phi dt$$\Sigma$$U^{\Delta}$$\Omega$

be a test function and we consider$\Phi$$J$

Since$H_{loc}^{-1}(\Omega),$

,

,

.

is bounded, we see$|N|\leq M||\Phi||c$ .

Let us look at$L$. We see
,

$L$

$=$

$L_{1}+L_{2}$$L_{1}$$=$$\sum_{j,n}\Phi(n\Delta t, (2j+1)\Delta x)\int 2j\Delta x[\eta(U\Delta)(2j+2)\Delta x]^{t}t=n\Delta t+0X=n\Delta t-^{0_{d}}$$L_{2}$$=$$\sum_{j,n}\int^{(}2j\Delta xt2j+2)\Delta x(\Phi(n\Delta t, x)-\Phi(n\Delta, (2j+1)\Delta x)\cross$$\cross$$[\eta(U^{\Delta})]t=n\Delta t+0Xt=n\Delta t-^{0_{d}}$

.

,

46

We note$[\eta(U^{\Delta})]_{t=n}^{t}=n\Delta\Delta tt+0-0$$=$$D_{U}\eta(U^{\Delta}(n\Delta t+0, x))[U\Delta]$$+$$\int_{0}^{1}(1-\theta)([U\Delta]|D^{2}(UU^{\Delta}(n\Delta t+0)+\theta[U^{\Delta}]).[U^{\Delta}])d\theta$.

and$\int_{2j\Delta x}^{(2j}+2)\Delta x][U\Delta dx=0$

by the scheme. Therefore$|L_{1}| \leq M||\Phi||_{C}\sum_{j,n}\int\int_{0}^{1}(1-\theta)|F(\theta, \eta)|d\theta dx$

,

where
$F(\theta, \eta)=([U^{\Delta}]|D2U\eta(U\Delta(n\Delta t+\mathrm{O})+\theta[U^{\Delta}]).[U^{\Delta}])$By Proposition 22 we know
Proposition 7 we know$|F(\theta, \eta)|\leq MF(\theta, \eta^{*})$.

. But in the proof of$\sum_{j,n}\int\int^{1}0\eta^{*}(1-\theta)F(\theta,)d\theta dX\leq C$.

Thus we know$|L_{1}|\leq M||\Phi||c$.

In the proof of Proposition 7 we know.

$\sum_{j,n}\int^{(}2j\Delta x2j+2)\Delta x|[U^{\Delta}]|2d_{X}\leq c$

Therefore$|L_{2}|$$\leq$$2^{\alpha}|| \Phi||_{C^{\alpha}}\sum n\int(\Delta_{X})^{\alpha}|[\eta(U\Delta)]|d_{X}$$\leq$$2^{\alpha-1}|| \Phi||c\alpha\sum n\int((\Delta_{X})\alpha+\frac{1}{2}+(\Delta x)\alpha-\frac{1}{2}|[\eta(U\Delta)]|^{2})dX$$\leq$$M|| \Phi||_{C^{\alpha}}((\Delta x)\alpha-\frac{1}{2}+(\Delta x)^{\alpha-\frac{1}{2}}\sum\int|[U^{\Delta}]|^{2}dx$$\leq$$M’( \Delta x)^{\alpha}-\frac{1}{2}||\Phi||_{C^{\alpha}}$where we use the boundedness ofat . Along the shock we have,
$D_{U}\eta$

and $n=O(1/(\Delta x))$ . Next we look$\Sigma$$[q(U)]$$\sigma[\eta(U)]$$=$$\int_{\rho_{L}}^{\beta R}(-\frac{d\sigma}{d\rho}\int_{0}^{1}\theta(U-UL|D_{U}^{2}\eta(UL+\theta(U-U_{L}))(U-UL))d\theta)d\rho$.

47This implies$|\sigma[\eta]-[q]|\leq M(\sigma[\eta*]-[q^{*}])$.

But we know$\int\sum_{shock}(\sigma[\eta^{*}]-[q^{*}])dt\leq^{c}$in the proof of Proposition 7. Therefore.

$|\Sigma|\leq M||\Phi||c$Summing up , we know the compactness. QED.

10

Convergence of approximate solutions

constructed in Section 4. Since
We consider the approximate solutions
a family of Young measuressequenceandais bounded, there isand for any continuous function
such that$U^{\Delta}$$U^{\Delta_{n}}$$U^{\Delta}$$\nu_{t,x}$$f$$supp\nu_{t,x}\subset\Sigma=\Sigma_{B}$$f(U^{\Delta_{n}}(t, x))arrow\overline{f}=<\nu_{t,x},$$f>$

weak star topology. By Lemma 1, we can apply the compensatedincompactness theory, and we can assume$L^{\infty}$$(\eta q’-\eta q)J(U^{\Delta}n)arrow<\nu,$weak star. Herein
we have$L^{\infty}$$\eta,$$q;\eta’,$ $q$

Lemma 2 For any pairs
the identity’$q><\nu,$ $q’>-<\nu,$ $\eta’><\nu,$ $q>$

are arbitrary Darboux entropy pairs. Thus$(\eta, q),$ $(\eta’, q’)$$<\nu,$ $\eta q’-\eta’q>=<\nu,$ $\eta><\nu,$of Darboux entropies-entropy flux;$q>-’<\nu,$ $\eta’><\nu,$ $q>$

holds a. $e.-(t, X)$ , where $\nu=\nu_{t,x}$ .

Since entropies we will use are countably many, we can assune that theabove identity holds outside a null set which is common to all . We fix. Of course supp.
$(t, x)$ at which the identity holds, and we write
be the smallest triangle. Suppose that supp. $\nu\cap\{\rho>0\}\neq\phi$ . Let$\{z_{0}\leq z\leq w\leq w_{0}\}$ such that supp.
. Let us denote by. (the Dirac measure).
. It will be verified thatthe stateFirst we show$\eta$$\nu=\nu_{t,x}$$\Sigma_{0}$$\Sigma$$\nu\cap\{\rho>0\}\subset\Sigma_{0}$$P_{0}$$\nu=\delta_{P_{0}}$$(w_{0}, z\mathrm{o})$Proposition 27$P_{0}\in supp.\nu$.

$\nu\subset$

48

Proof. Suppose. Since is the smallest triangle containing$\nu\cap\{\rho>0\},$supp.
$w=w_{0}$ and $z=z_{0}$ intersect with supp. $\nu\cap\{\rho>0\}$ . Onneighborhoods of these intersection points we have$P_{0}\not\in supp.\nu$$\Sigma_{0}$$\eta^{1}$$\eta^{2}$(See Proposition 24). Since$\mathrm{S}\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{e}P0\not\in supp.\nu$$\nu,$$\frac{1}{M}e^{k(w0}-\epsilon)$$\geq$$\frac{1}{M}e^{-k(z\mathrm{o}+)}\epsilon$$\geq$$\eta^{1},$,$\eta^{2}$$<\nu,$ $\eta^{1}>$$\geq$$<\nu,$ $\eta^{2}>$$\geq$.

are nonnegative, we see$\frac{1}{M}e^{k(w_{\mathrm{O}}}-\epsilon)$,$\frac{1}{M}e^{-k(z_{0+)}}\epsilon$.

, we have$<\nu,$ $\eta^{2}q^{1}-\eta^{1}q2>\leq Me^{k(z0}w_{\mathrm{o}}--\delta)$

Taking$2\epsilon<\delta$.

, we have$| \frac{<\nu,q^{1}>}{<\nu,\eta^{1}>}-\frac{<\nu,q^{2}>}{<\nu,\eta^{2}>}|$$=$$| \frac{<\nu,\eta^{2}q^{1}-\eta^{12}q>}{<\nu,\eta^{1}><\nu,\eta^{2}>}|$$Me^{-k(\epsilon)}\delta-2$$\leq$$arrow$

as $karrow\infty$ . Let$\beta$$0$

be a sufficiently small positive number, and we put$\Sigma_{2}$$=$$\{z_{0}\leq z\leq w<w_{0}-\beta\}$$\Sigma_{3}$$=$$\{z_{0}\leq z\leq w\leq w0, w0-\beta\leq w\}$.

Then$\eta^{1}e^{-kw}=(1+O(1/c^{2}))2NN!yN-1(y+O(1/k))$is bounded on$\Sigma_{0}$and we have$<\nu|_{\Sigma_{2},\eta^{1}}>\leq Me^{k(0\beta}w-)$

Taking$\epsilon=\beta/2$.

, we know$\frac{<\nu|\Sigma_{2},\eta^{1}>}{<\nu\eta^{1})>}\leq Me^{-\beta k}/2arrow 0$

Since$\partial\lambda_{2}/\partial w>0$

, we know$\lambda_{2}(w, Z)\geq\lambda 2(w_{0}-\beta, Z0)$

.

49

on

$\Sigma_{3}$. Therefore we have$\frac{<\nu,q^{1}>}{<\nu,\eta^{1}>}$$\frac{<\nu|_{\Sigma}2\eta^{1}\lambda_{2}>}{<\nu,\eta^{1}>},+\frac{<\nu|\Sigma_{3},\eta^{1}\lambda_{2}>}{<\nu,\eta^{1}>}+$

$=$

$+$

$O(1/k)$$\geq$$o(1)+\lambda 2(w0-\beta, z\mathrm{o})$Similarly we see$<\nu,$ $q^{2}>$$\overline{<\nu,\eta^{2}>}\leq o(1)+\lambda 1(w0, Z0+\beta)$.

Therefore we have$\lambda_{2}(w0-\beta, Z\mathrm{o})-\lambda_{1}(w0, Z0+\beta)\leq 0+o(1)$.

Passing to the limit, we know$\lambda_{2}(w_{0}, z_{0})\leq\lambda_{1}(w0, z\mathrm{o})$.

But this means $P_{0}\in\{\rho=0\}$ , a contradiction. QED.
Let us fix such that $z_{0}<a<w_{0}$ . We have$a$$<\nu,$ $B_{n}^{3}>$$=$$<\nu,$$\eta_{n}^{5}><\nu,$ $q^{3}>$,$<\nu,$ $B_{n}^{4}>$$=$$<\nu,$ $\eta^{4}><\nu,$ $q_{n}5>-<\nu,$ $\eta_{n}^{5}><\nu,$ $q^{4}>$,$<\nu,$ $\eta^{3}q^{4}-\eta^{4}q^{3}>$$=$$<\nu,$ $\eta^{3}><\nu,$ $q^{4}>-<\nu,$ $\eta><4\nu,$ $q>3$,$<\nu,$ $B_{n}>$$=$$<\nu,$ $\eta_{n}^{5}><\nu,$ $q_{n}^{6}>-<\nu,$ $\eta \mathrm{n}6><\nu,$ $q_{n}^{5}>$$q_{n}\mathrm{s}_{><\nu}5>-<\nu,$$\eta,$.

From (8.8) we know$<\nu,$ $\eta^{3}q^{4}-\eta^{4}q>\mathrm{s}_{>}0$

and from(8.10) we know$<\nu,$ $B_{n}^{3}>arrow 0$

Using these we can prove the following propositions. Proofs can be foundin Chen et al [2].

Proposition 28 As$narrow\infty,$$<\nu,$ $\eta_{n}^{5}>,$ $<\nu,$ $q_{n}^{5}>,$ $<\nu,$ $q_{n}^{6}>,$ $<\nu,$ $q_{n}^{6}>are$bounded.

Proposition 29 As$narrow\infty$, we $have<\nu,$ $B_{n}>arrow 0$ .

Now, taking$e^{-\frac{1}{1-x^{2}}}$$\Phi_{0}(X)=\{$$0$

if $|x|<1$
if $|x|\geq 1$

50

we put$\Phi(x)=\frac{1}{\beta}(\Phi_{0}(\frac{x+\beta}{\beta})-\Phi_{0}(\frac{x-\beta}{\beta}))$for the generating function of$\eta_{n}^{5}$. Here $\beta=(1-\alpha)/2$ . We put$s_{+}$$=$$\{z\leq w, |w-a|\leq\frac{1-3\alpha}{n}\}$$S_{-}$$=$$\{z\leq w, |z-a|\leq\frac{1-3\alpha}{n}\}$,.

Proposition 30 As $narrow\infty$ , we have$<\nu|s+’ ny^{2N}>+<\nu|_{S}-,$ $ny2N>arrow 0$ .

Proof. Put $S_{L}’=S_{+}\cap S_{L},$ $S_{R}’=S_{-}\cap S_{R}$ . It is sufficient to prove that$<\nu|_{s_{L}}’,$$ny2N>+<\nu|s_{R}’,$ $ny^{2N}>arrow 0$ .

From (8.11) we have$<\nu|_{S_{L}},$$ny^{2N}A_{1}+y^{N}A_{2}>+<\nu|s_{R},$ $nyC2N+1yc_{2}N>arrow 0$.

Note$A_{1}=( \frac{N(2^{N}N!)^{2}}{2N+1}+O(1/c)2)(\int^{n_{1}}-\Phi(x+y-a))^{2}\geq\frac{1}{M_{0}}>0$

on$S_{L}’$. Put$E_{n}= \{0\leq y\leq(\frac{1}{n})^{\mu}\}$

where

$\mu$

$S_{L}\cap E_{n}$,is a positive parameter. Then $|y^{N}A_{2}|\leq M(1/n)^{\mu N}=o(1)$ on
and $|y^{N}A_{2}|\leq Mny^{2N}(1/n)1-\mu N$ on $S_{L}-E_{n}$ . Choosesuch$d_{n}\searrow 0$

that$\int_{-1^{-}}^{1\alpha-d_{n}}+\alpha-\Phi=\int_{1-\alpha}^{1-\alpha}-d_{n}\Phi\geq(1/n)^{\mu 0}$Then$( \int_{-1}^{H}\Phi)^{2}\geq(1/n)^{2\mu 0}$

for $|H|\leq 1-\alpha-d_{n}$ , and$| \Phi(H)|+|\int_{-1}^{H}\Phi|=o(1)$for $1-\alpha-d_{n}\leq|H|\leq 1$ . Put$s_{+}^{n}=S_{L} \cap\{|w-a|\leq\frac{1-\alpha-d_{n}}{n}\}$Then$S_{L}’\subset S_{+}^{n}\subset S_{L}$and$|y^{N}A_{2}|=o(1)$

.

.

51

on

$S_{L}-S_{+}^{n}$and$ny^{2N}A_{1}+y^{N}A_{2}$on$S_{+}^{n}-E_{n}$$<\nu|s_{L},$$\geq$$ny^{2N-\mu N}( \frac{1}{M}(1/n)^{2\mu_{0}}-M(1/n)^{1})$$\geq$$0$. Here we take $0<2\mu_{0}<1-\mu N$ . Then

$ny2NA1+yNA_{2}>$$=$$<\nu|_{S_{L}E}\mathrm{n}nyA_{1}n’+2N>$$+$$<\nu|_{S_{L}E_{n}}-,$$+$$o(1)$$\geq$$\frac{1}{M_{0}}<\nu|_{S_{L^{\cap E}}^{\prime,ny^{2N}}}n>+$十$<\nu|_{S_{L}-s^{n_{\cap}}}En’ A+ny12N>+$

十$<\nu|_{S_{L^{-}}’}En’ y^{2N}A_{1}n+y^{N}A2>+$$+$$<\nu|_{s^{n}-^{s’}}+L^{-}E_{n},$$+$$o(1)$$\geq$$\frac{1}{M_{0}}<\nu|_{s_{L}^{l}}\cap En’ ny^{2N}>+$$+$$<\nu|s_{L}^{J}-E_{n},$ $ny^{2}(N$+$$o(1)$$\geq$$\frac{1}{2M_{0}}<\nu|_{S_{L}’,y^{2N}}n>+$$+$$o(1)$$nyA_{1}+y^{N}A2N2>+$$nyA2N1+y^{N}A_{2}>+$\frac{1}{M_{0}}-M(1/n)1-\mu N>+$.

Similarly we know$<\nu|s_{R},$ $ny^{2N}c1+yC_{2}N>>\underline{1}<\nu|s_{R}’,$$-2M_{0}$$ny2N>+o(1)$. Thus we see$<\nu|s_{L}’,$$ny2N>+<\nu|_{s_{R}’},$ $ny2N>arrow 0$

.

QED.

Proposition 31 We have$\nu|\{\rho>0\}=\delta P_{\mathrm{o}}$.

Proof. Proposition 30 says that the projectionsof the measureadmits the Lebesgue lower derivatives which vanish at any .
Therefore we can claime that
$P_{w}\tilde{\nu},$$\tilde{\nu}=y^{2N}\nu$$P_{z}\tilde{\nu}$$a$supp. $\nu\cap\{\rho>0\}=\mathrm{f}P0\}$ .

52

Since

$\nu$

is a probability measure, we have$\nu|_{\{\rho>0\}}=C\delta_{P_{0}}$.

But$C(\eta^{3}q^{4}-\eta q^{3})4=C2(\eta^{3}q4-\eta^{43}q)$at. Hence $C=1$ . QED.
Summing up we get the final$P_{0}$

Theorem 2 For any
data satisfy$M_{0}$

there is a positive number$0\leq\rho_{0}(X)\leq M_{0}$,$\epsilon_{0}$such that if the initial$| \frac{c}{2}\log\frac{c+u0(X)}{c-u_{0}(x)}|\leq M_{0}$.

and if, then a subsequence of the approximate solutions
converges a. . to a limit
which is a weak solution of the relativistic Euler
equation.
$1/c^{2}\leq\epsilon_{0}$$U^{\Delta}$

$U$

$e$

11

Acknowledgment

The author expresses his sincere thanks to Dr. Tadayoshi Kano (Osaka
University) and. Hiroshi Uesaka (Nihon University) for helpfull informations on studies of the-Darboux equation. The thesis of
Dr. Shigeharu Takeno was often reffered during the preparation of themanuscript. This work was, in part, financially supported by National Center for Theoretical Sciences Mathematics Division, R. O. C.
$\mathrm{P}\mathrm{r}$$\mathrm{E}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{r}-\mathrm{p}\mathrm{o}\mathrm{i}_{\mathrm{S}}\mathrm{S}\mathrm{o}\mathrm{n}$参考文献[1] Jing Chen, Conservation laws for the relativistic
mun.P.D.E., 20(1995),1605-1646.

$\mathrm{p}$-systems, Com-[2] Ding Xiaxi, Guiqiang Chen, Peizhu Luo, Convergence of the Lax-Friedrichs scheme for isentropic gas dynamics, (I), Acta Mathematica
Scientica, 5(1985), 415-432, (II), ibid., 5(1985), 433-472, (III), ibid.,
6 (1986),75-120,[3] R. J.
, Convergence of approximate solutions to the conservation laws, Arch. Rat. Mech. anal., 82(1983), 27-70.
$\mathrm{D}\mathrm{i}\mathrm{p}_{\mathrm{e}\mathrm{r}\mathrm{n}}\mathrm{a}$[4] J. Glimm, Solutions in the large for nonlinear hyperbolic systems ofequations, Commun. Pure Appl.Math., 18(1965), 697-715.

53[5] T. Nishida, Global solutions for an initial boundary value problem of aquasi-linear hyperbolic system, Proc. Japan Acad., 44(1968), 642-646.
[6] J. Smoller and B. Temple, Global solutions of the relativistic Euler
equations, Commun. Math. Pnys., 156(1993), 67-99.

[7] A. Weinstein, On the wave equation and the equation of Euler-Poisson,

Proc. Symp. Appl. Math., 5(1954), 137-148.