Acta Math Vietnam (2013) 38:487–516 DOI 10.1007/s40306-013-0030-3
WAVE-FRONT TRACKING FOR THE EQUATIONS OF NON-ISENTROPIC GAS DYNAMICS—BASIC LEMMAS Fumioki Asakura
Received: 16 March 2012 / Accepted: 17 July 2012 / Published online: 23 October 2013 © Institute of Mathematics, Vietnam Academy of Science and Technology (VAST) and Springer Science+Business Media Singapore 2013
Abstract In the random choice and its alternative wave-front tracking methods, approximate solutions are constructed by solving exactly or approximately the Riemann problem in each neighborhood of a certain finite set of jump points depending on the time. In order to obtain global in time BV solutions, one has to get a priori estimates for the total variation in x at the time t of approximate solutions, which is, roughly speaking, the summation of amplitudes of waves at t constituting the solutions to the Riemann problems. Since amplitudes may increase through the interaction of neighboring waves, the crucial point is to estimate the amplitudes of outgoing waves by those of incoming waves in a single Riemann solution, which is called the local interaction estimates. The aim of this note is to provide a detailed description of the Riemann problem to the equations of polytropic gas dynamics and a complete proof of the basic lemmas on which the local interaction estimates are based. Although all of them, except for Lemmas 4.2 and 5.1, are presented in T.-P. Liu (Indiana Univ. Math. J. 26:147–177, 1977), that paper is difficult and not well understood even at the present day in spite of its importance. For the sake of completeness, this note includes proofs of the local interaction estimates. Keywords Hyperbolic partial differential equations · Conservation laws (mathematics) · Riemann problem Mathematics Subject Classification (2000) 35L65 · 35L67
1 Introduction The equations of 1-dimensional gas dynamics in Lagrangian coordinates are given by
B
F. Asakura ( ) Department of Engineering Science, Osaka Electro-Communication University, 18-8 Hatucho, Neyagawa, Osaka 572-8530, Japan e-mail:
[email protected]
488
F. ASAKURA
⎧ ⎨ vt − ux = 0, ut + px = 0, ⎩ (E + 12 u2 )t + (pu)x = 0,
(1.1) (x, t) ∈ R × R+ ,
where u is the velocity, p the pressure, v the specific volume and E the internal energy; p and v are positive quantities. Temperature and entropy are denoted by Θ and S, respectively, and satisfy the first and second laws: d E = ΘdS − pdv. We are concerned in this note only with the simplest non-isentropic model gas: ideal i.e. pv = RΘ and polytropic i.e. E = Cv Θ + E0 . Then S is expressed as R log v + const S = Cv log p + 1 + Cv and we have pv + E0 , E= γ −1
2 −γ
p=a v
e
γ −1 R S
R >1 . γ =1+ Cv
(1.2)
2
Setting E0 = − γa−1 and γ → 1, one sees that (1.1) and (1.2) coincide with the equations of isothermal gas dynamics. We shall confine ourselves to the initial value problem in BV-space: Determine solutions from their initial data
v(x, 0), u(x, 0), S(x, 0) = v0 (x), u0 (x), S0 (x) , where v0 (x) ≥ v∗ > 0 and v0 (x), u0 (x), S0 (x) ∈ BV(R), the space of functions having bounded total variation. The following results are now classic. Theorem 1.1 (Liu [6]) Suppose that 1 < γ ≤ 53 and v0 (x) ≥ v∗ > 0. If (γ − 1)TVv0 , (γ − 1)TVu0 , (γ − 1)TVη0 are sufficiently small, then there exists a global in time BV-solution. Theorem 1.2 (Nishida [9]) Suppose that γ = 1 and v0 (x) ≥ v∗ > 0. If the total variation of the initial data TVv0 , TVu0 , TVη0 are finite, then there exists a global in time BV-solution. The global BV-solutions in the above theorems are constructed by employing the random choice method initiated by Glimm [5]. Recently, Asakura and Corli [1] have succeeded in giving an alternative proof for the above theorems by using the wave-front tracking method which was initiated by Dafermos [4] for single conservation laws, and developed by Bressan [2, 3] and Risebro [10] for hyperbolic systems of conservation laws. In both random choice and wave-front tracking methods, approximate solutions are constructed by solving exactly or approximately the Riemann problem in each neighborhood of countably many jump points P(xj , tk ), k ∈ N and j ∈ Z, or of a certain finite set depending on k. The initial data are
v(x, tk ), u(x, tk ), S(x, tk ) (v(xj − 0, tk − 0), u(xj − 0, tk − 0), S(xj − 0, tk − 0)) for x < xj , = (v(xj + 0, tk − 0), u(xj + 0, tk − 0), S(xj + 0, tk − 0)) for x > xj . A full account of the Riemann problem for the equations of ideal polytropic gas dynamics is given in Smoller [11, Chap. 18]. Solutions to the Riemann problem are composed of
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constant states separated by shock, (approximate) rarefaction and entropy waves; and the total variation in x at t of approximate solutions is, roughly speaking, the summation of the amplitudes of those waves. Since amplitudes may increase through the interaction of neighboring waves, the crucial point for proving the above existence theorems is to estimate the amplitudes of outgoing waves by those of incoming waves, which is called the local interaction estimates. The aim of this note is to provide a complete proof of the lemmas on which the local interaction estimates are based. All of them, except for Lemmas 4.2 and 5.1, are presented in [6]. These lemmas also form the basis for [1] and are useful for studying the fluid flow of real materials (Menikoff and Plohr [8]). However, in the author’s opinion, the lemmas are difficult and not well understood even in the present day in spite of their importance. For these reasons, the author has decided to publish their proof in the present form. For the sake of completeness, this note includes proofs of the local interaction estimates. The author wishes to thank Andrea Corli for careful reading and valuable comments for completing the manuscript in the present form.
2 Riemann problem The dimensionless entropy is denoted by η = RS . Because the entropy wave is a stationary wave with p = const, u = const as we shall see in (2.5), it is useful to choose the quantities p, u, η as independent variables denoted simply by U . Note that v is expressed by p and η as follows: 2
v =aγ e
γ −1 γ η
1
p− γ
and
1
1
−vp (p, η) = γ − 2 a γ e
γ −1 2γ η
p
+1 − γ2γ
.
The associated quasi-linear equations are pt −
ux = 0, vp
ut + px = 0,
ηt = 0
and we find by direct computation that the characteristic speeds1 are λ1 (U ) = −
1 , −vp (p, η)
λ2 (U ) =
1 , −vp (p, η)
λ0 (U ) = 0
and the corresponding characteristic vector fields are
R1 (U ) = t 1, − −vp (p, η), 0 ,
R2 (U ) = t 1, −vp (p, η), 0 ,
The integral curves of Rj (U ) (j = 0, 1, 2) are represented by
−vp dp = const,
η = const,
R0 (U ) : p = const, u = const, −vp dp = const, R2 (U ) : u −
η = const.
R1 (U ) : u +
1 [λ ] = [x][t]−1 = ML−2 T −1 (j = 1, 2). j
R0 (U ) = t (0, 0, 1).
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F. ASAKURA
Setting =
γ −1 , 2
we have rarefaction curves issuing from U0 in the following form:
1-rarefaction curve: u − u0 = − η − η0 = 0
√
1
γa γ
√ γ1 γa
2-rarefaction curve: u − u0 = η − η0 = 0
e γ η0 (p γ − p0γ ),
p ≤ p0 , (2.1)
e γ η0 (p γ − p0γ ),
p ≥ p0 .
It is sometimes useful to pay attention to the dimension of physical quantities: [x] = ML−2 ,
[t] = T ,
[u] = LT −1 ,
[p] = ML−1 T −2 ,
[v] = M −1 L3 ,
[R] = [S] = [E ][Θ]−1 = ML2 T −2 K −1 . The dimensions of the both sides of an equality are equal and mathematical conditions have to be represented by dimensionless quantities. A self-similar jump discontinuity having the form U (x, t) =
U− U+
for x < st, for x > st
(2.2)
is a weak solution, if and only if the constants satisfy the Rankine–Hugoniot condition:
E+ − E− + 12 (p+ + p− )(v+ − v− ) = 0,
(2.3)
(u+ − u− )2 = −(p+ − p− )(v+ − v− ). The shock speed s satisfies s2 = −
p + − p− . v+ − v−
For the polytropic gas, (2.3) is equivalent to ⎧ ⎨ e(γ −1)(η+ −η− ) = ( p+ ){ (γ −1)p+ +(γ +1)p− }γ , p− (γ +1)p+ +(γ −1)p− ⎩ (u+ − u− )2 =
(2.4)
2v− (p+ −p− )2 . (γ +1)p+ +(γ −1)p−
If p+ = p− , we have two branches of solutions 2v− u+ − u− = ± (p+ − p− ), (γ + 1)p+ + (γ − 1)p−
s=± −
p+ − p− . v+ − v−
Let us fix p− and η− . When considering η+ as a function of p+ , we have dη+ (γ + 1)(p+ − p− )2 > 0, = dp+ p+ {(γ − 1)p+ + (γ + 1)p− }{(γ + 1)p+ + (γ − 1)p− } which shows that η+ > η− if and only if p+ > p− . Since the entropy must increase as time goes on, physically relevant branches are p + > p−
for s < 0 and
p+ < p −
for s > 0.
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491
The jump discontinuity (2.2) for these branches is simply called a shock wave. The line of discontinuity x = st is often referred to as a shock front. Note that the above branches complement the rarefaction curves (2.1). If p+ = p− , we have an entropy wave u+ = u− ,
p+ = p− ,
η+ = η− ,
s=0
(2.5)
which coincides with the integral curve R0 (U− ). This type of discontinuity is also called a contact discontinuity. The Riemann problem is the initial value problem whose initial data are U (x, 0) =
U− U+
for x < 0, for x > 0,
(2.6)
where U± are constant states. In order to solve the Riemann problem, we define the forward 1F (UL ) = R F1 (UL )∪ S1F (UL ) (U = U+ , UL = U− ) and the backward 2-wave 1-wave curve W B2 (UR ) ∪ S2B (UR ) (U = U− , UR = U+ ), where R F1 (UL ) is the forward 2B (UR ) = R curve W 1-rarefaction curve for p ≤ pL and S1F (UL ) the forward 1-shock curve S1F (UL ) for p > pL ; B2 (UR ) is the backward 2-rarefaction curve for p ≤ pR and S2B (UR ) the backward 2-shock R curve for p > pR . We have
u − uL = 1F (UL ) : W
⎧ √ 1 η γa γ γ ⎪ ⎪ ⎨ − e γ L (p γ − pL ) ⎪ ⎪ ⎩−
η − ηL =
u − uR = 2B (UR ) : W
η − ηR =
(p > pL ),
0
(p ≤ pL ),
1 L γ log[( ppL ){ p+(1+)p } ] 2 (1+)p+pL
(p > pL ),
⎧√ 1 η γa γ γ ⎪ ⎪ ⎨ e γ R (p γ − pR ) ⎪ ⎪ ⎩
(p ≤ pL ),
1 η a γ e γ L (p−pL ) 1 1 2γ pL {(1+)p+pL } 2
1 η a γ e γ R (p−pR ) 1 1 2γ pR {(1+)p+pR } 2
(p ≤ pR ), (p > pR ),
(2.8)
(p ≤ pR ),
0 1 2
(2.7)
R γ log[( ppR ){ p+(1+)p } ] (1+)p+pR
(p > pR ).
Each wave curve constitutes a C 2 -curve with Lipschitz continuous second derivative and represents all realizable rarefaction waves, shock waves and entropy waves. If (p, u, η) ∈ 1F (UL ), then there is a 1-rarefaction wave or shock wave connecting (pL , uL , ηL ) and W 2B (UR ), then there is a 2-rarefaction wave or a (p, u, η). If, on the other hand, (p, u, η) ∈ W shock wave connecting (p, u, η) and (pR , uR , ηR ). 1F (UL ) and W 2B (UR ), respectively, Let W1F (UL ) and W2B (UR ) be the projections of W onto the pu-plane. The Riemann problem is solved in the following way. Let (pL , uL , ηL ) and (pR , uR , ηR ) be given Riemann data. If two curves, W1F (UL ) and W2B (UR ), have an − + 1F (UL ) and (pm , um , ηm intersection point (pm , um ), then the states (pm , um , ηm )∈W )∈ B 2 (UL ) are connected by an entropy wave (see Smoller [11, Chap. 18] for the details). W
492
F. ASAKURA
Theorem 2.1 Let c =
√
γpv be the sound speed. If uR − uL <
cL + cR
then there is a unique solution to the Riemann problem.
3 Interaction of two incoming waves We always assume that all waves considered will be in the region 0 < p∗ ≤ p ≤ p ∗ ,
0 ≤ η − η∗ ≤ H
(3.1)
and that H is sufficiently small. We introduce the Riemann invariants2 corresponding to η∗ = min η0 (x): √ w=u−
√
1
γ a γ η∗ e γ p γ − p∗γ ,
z=u+
1
γ a γ η∗ e γ p γ − p∗γ
(3.2)
and set √ γ1
γ a η∗ z−w z+w = u, τ= = e γ p γ − p∗γ . 2 2 Strengths of shock waves and rarefaction waves will be measured by w and z. The pressure and its derivative with respect to τ are computed as σ=
γ
p(τ ) = p∗ + √
γ
τ 1 γ
γa e
and
γ η∗
dp √ − γ1 − γ η∗ 1+ = γa e p γ . dτ
(3.3)
Since w +z±e γ (η0 −η∗ ) (z−w) are constant along rarefaction curves, we find that the forward B2 (U0 ) are represented as F1 (U0 ) and backward 2-rarefaction curve R 1-rarefaction curve R
F1 (U0 ) : R
z − z0 =
e γ (η0 −η∗ ) − 1
e γ (η0 −η∗ ) + 1
(w − w0 ),
w ≥ w0 , (3.4)
B2 (U0 ) R Note that 0 ≤ We set
:
(η −η ) 0 ∗ −1 (η −η ) 0 ∗ +1 γ e
eγ
w − w0 =
e γ (η0 −η∗ ) − 1
e γ (η0 −η∗ ) + 1
(z − z0 ),
z ≤ z0 .
< 1. 1
g(τ0 , τ ; η0 ) =
a γ e γ η0 (p − p0 )
, 1 1 p02γ {(1 + )p + p0 } 2 p 1 p + (1 + )p0 γ h(τ0 , τ ) = log , 2 p0 (1 + )p + p0
2 [w] = [z] = LT −1 .
EQUATIONS OF NON-ISENTROPIC GAS DYNAMICS
493
where p = p(τ ) and p0 = p(τ0 ). By using these functions, the Hugoniot curves through (p0 , u0 , η0 ) are expressed as σ − σ0 = ∓g(τ0 , τ ; η0 ),
η − η0 = h(τ0 , τ ).
The forward 1-shock curve S1F (U0 ) and the backward 2-shock curve S2B (U0 ), issuing from U0 = (p0 , u0 , η0 ), are also expressed by using the Riemann invariant coordinates. Recall that 0 < ≤ 13 is equivalent to 1 < γ ≤ 53 . Lemma 3.1 If 0 < ≤ 13 , then there are functions z = z1 (w; η0 ) and w = w2 (z; η0 ) such that 1F (U0 ) = (w, z, η); z = z1 (w; η0 ), η = η(τ ), w < w0 , S 2B (U0 ) = (w, z, η); w = w2 (z; η0 ), η = η(τ ), z > z0 . S Moreover, there is a constant B > 1 such that
e γ (η0 −η∗ ) − 1
e γ (η0 −η∗ ) + 1
=
z1 (w0 ; η0 )
≤
z1 (w; η0 )
≤
Be γ (η0 −η∗ ) − 1
e γ (η0 −η∗ ) − 1 e
γ (η0 −η∗ )
+1
Be γ (η0 −η∗ ) + 1
,
(3.5)
,
(3.6)
= w2 (z0 ; η0 ) ≤ w2 (z; η0 ) ≤
Be γ (η0 −η∗ ) − 1
Be γ (η0 −η∗ ) + 1
z1 (w; η0 ) < 0 < w2 (z; η0 ),
(3.7)
η1 (w) < η1 (w0 ) = 0 = η2 (z0 ) < η2 (z).
(3.8)
Proof Let us study S1 (U0 ) = S1F (U0 ): 1-Hugoniot curve issuing from U0 . Denote for simplicity z1 (w) = z(w). By direct computation, we have w0 − w = g(τ0 , τ ; η0 ) + (τ − τ0 ), z0 − z = g(τ0 , τ ; η0 ) − (τ − τ0 ) and dw = −(1 + gτ ) < 0, dτ Hence z is a function of w and gτ − 1 dz = dw gτ + 1
dz = 1 − gτ . dτ
d 2z 2gτ τ =− . dw2 (gτ + 1)3
and
(3.9)
Also we have 1 a γ e γ η0 {(1 + )p + (1 + 3)p0 } dp gτ (τ0 , τ ; η0 ) = 1 3 dτ 2p02γ {(1 + )p + p0 } 2
√ =
γ e γ (η0 −η∗ ) {(1 + )p + (1 + 3)p0 }p 1
3
2p02γ {(1 + )p + p0 } 2
≤ Be γ (η0 −η∗ ) ,
1+ γ
494
F. ASAKURA
where √ B = max
γ {(1 + )p + (1 + 3)p0 }p
0<≤1/3 p∗ ≤p≤p ∗
1
1+ γ
.
3
2p02γ {(1 + )p + p0 } 2
In a similar manner, we find that gτ τ (τ0 , τ ; η0 ) =
(1 + )e γ (η0 −η∗ ) p − γ (p − p0 ) dp 1 5 dτ √ 4 γ p02γ {(1 + )p + p0 } 2 × (1 + )(p − p0 ) + (1 − 3)(1 + 2)p0 .
(3.10)
Since gτ τ > 0 for p > p0 and gτ |p=p0 = e γ (η0 −η∗ ) , we have
e γ (η0 −η∗ ) ≤ gτ ≤ Be γ (η0 −η∗ ) . dz Since dw is increasing with respect to gτ , we have the desired estimates. Next we also have by direct computation
hτ (τ0 , τ ) =
(1 + )(p − p0 )2 dp . 2p{p + (1 + )p0 }{(1 + )p + p0 } dτ
(3.11)
Finally, 2gτ τ d 2z =− ≤ 0, 2 dw (gτ + 1)3
hτ dη =− ≤ 0. dw gτ + 1
In the course of the above proof, we have proved the following: Proposition 3.1 Suppose that 0 < ≤ 13 . Then gτ τ (τ0 , τ ; η0 ) > 0
for p > p0 and ητ (τ0 , τ ; η0 ) > 0.
By direct computation, we have 1
gη0 (τ0 , τ ; η0 ) =
a γ e γ η0 (p − p0 ) 1
> 0, 1
γp02γ {(1 + )p + p0 } 2
which implies Lemma 3.2 If 0 < ≤
1 3
and η1 > η0 , then
z1 (w; η1 ) ≤ z1 (w; η0 ),
w2 (z; η1 ) ≤ w2 (z; η0 ).
We use later on
gτ η0 (τ0 , τ ; η0 ) =
e γ (η0 −η∗ ) {(1 + )p + (1 + 3)p0 }p 1 3 √ 2 γ p02γ {(1 + )p + p0 } 2
1+ γ
> 0.
(3.12)
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495
We denote by α, β, δ, ξ, π , respectively, the strengths of 1-shock wave, 2-shock wave, entropy wave, 1-rarefaction wave, 2-rarefaction wave. They are defined by
β = z − z0
if (p, u, η) ∈ S1F (p0 , u0 , η0 ), if (p, u, η) ∈ S2B (p0 , u0 , η0 ),
δ = ηL − ηR
if (p0 , u0 , ηL ) and (p0 , u0 , ηR )
ξ = w − w0
constitute an entropy wave, F1 (p0 , u0 , η0 ), if (p, u, η0 ) ∈ R
π = z0 − z
B2 (p0 , u0 , η0 ). if (p, u, η0 ) ∈ R
α = w0 − w
In order to measure the increase of the entropy across shock waves, we define the quantities δα , δβ to be δα = η − η0
if (p, u, η) ∈ S1F (p0 , u0 , η0 ),
δβ = η − η0
if (p, u, η) ∈ S2B (p0 , u0 , η0 ).
From now on, we also denote by α, β, δ, ξ, π the corresponding waves themselves. Suppose that UL and UM are connected by a 2-wave θ2 (or an entropy wave), and UM and UR by a 1-wave θ1 (or an entropy wave); these two waves are assumed to be incoming and interact. Then, under the assumption of Theorem 2.1, there exists a unique solution to the Riemann problem connecting the states UL and UR . This solution is composed of a 1-wave θ connecting the states UL and U − , an entropy wave δ connecting U − and U + , and a 2-wave θ2 connecting U + and UR , which are outgoing waves. This interaction is simply denoted by θ2 + θ1 → θ1 + δ + θ2 . Note that δ = δ + δα − δα − δβ + δβ .
(3.13)
Local interaction estimates are carried out in the same way as in [6, 7]. Lemma 3.3 Assume that 0 < ≤ 13 . Assume also that p and η satisfy (3.1) and H0 is sufficiently small. Then there are constants 0 < D0 < 1 and D, D1 , D2 > 0 such that the following estimates hold:3 (1) β + α → α + δ + β : δ is generated, α ≤ α + Dαβ,
β ≤ β + Dαβ,
δα ≥ δα − D2 αβ, δβ ≥ δβ − D2 αβ,
|δ | ≤ D2 αβ; (2) π + α → α + δ + π : δ is generated, α ≤ α + Dαπ, 3 [D] = L−1 T , [D ] = 1, [D ] = [D 2 ] = L−2 T 2 . 1 2
δα ≥ δα − D2 απ,
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F. ASAKURA
π ≤ π + Dαπ, |δ | ≤ D2 απ; (3) α1 + α2 → α + δ + π : π and δ are generated, α ≤ α1 + α2 + Dα1 α2 ,
δα ≥ δα1 + δα2 − D2 α1 α2 ,
π ≤ Dα1 α2 , |δ | ≤ D2 α1 α2 ; (4) δ + α → α + δ + θ : θ = β or π , and θ is generated, α ≤ α + D1 α|δ|,
δα ≥ δα − D2 α1 α2 ,
|δ | ≤ |δ| + Dα|δ|, θ ≤ D1 α|δ|; (5) δ + ξ → ξ + δ + θ : θ = β or π , and θ is generated, ξ ≤ ξ + D1 ξ |δ|, |δ | ≤ |δ| + Dξ |δ|, θ ≤ D1 ξ |δ|; (6) ξ + α → α + δ + β : β and δ are generated, α ≤ α − ξ,
δα ≥ δα − D(α − α ) − D2 αξ,
β ≤ D0 (α − α ) + Dα ξ,
δβ ≥ 0,
|δ | ≤ D(α − α ) + D2 αξ ; (7) ξ + α → ξ + δ + β : β and δ are generated, ξ ≤ ξ, β ≤ D0 α,
δβ ≥ δα − D2 αξ
|δ | ≤ Dα; (8) α + ξ → α + δ + β : β and δ are generated, α ≤ α − ξ,
δα ≥ δα − D(α − α ) − D2 αξ,
β ≤ D0 (α − α ),
δβ ≥ 0,
|δ | ≤ D(α − α ); (9) α + ξ → ξ + δ + β : β and δ are generated, ξ ≤ ξ, β ≤ D0 α, |δ | ≤ Dα;
δβ ≥ δα − D2 αξ
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497
(10) α + ξ → ξ + δ + π : π and δ are generated, ξ ≤ ξ − α, π ≤ Dαξ, |δ | = δα ≤ Dα; (11) π + ξ → ξ + π : ξ = ξ,
π = π.
4 Proof of basic lemmas All the lemmas and the propositions in this section are proposed in Liu [6, 7]. Lemma 4.1 Suppose that (p2 , u2 , η2 ) ∈ S1F (p1 , u1 , η0 ), (p4 , u4 , η4 ) ∈ S1F (p3 , u3 , η0 ), τ3 > τ1 and u1 − u2 = u3 − u4 . Then there is a constant C2 such that [C2 ] = L−1 T and C2 (τ3 − τ1 )(τ2 − τ1 ), 0 < (τ4 − τ3 ) − (τ2 − τ1 ) ≤ (4.1) C2 (τ3 − τ1 )(τ4 − τ3 ). Proof Since uj +1 − uj = −g(τj , τj +1 ; η0 ), the last condition is equivalent to g(τ1 , τ2 ; η0 ) = g(τ3 , τ4 ; η0 ). We set τ = (τ4 − τ3 ) − (τ2 − τ1 ),
τ2 = τ2 − τ1 ,
τ3 = τ3 − τ1 .
Note that τ = τ4 − τ 2 in Fig. 1. Then it follows that τ2 = τ1 + τ2 ,
τ3 = τ1 + τ3 ,
τ4 = τ1 + τ2 + τ3 + τ
and the condition is g(τ1 , τ1 + τ2 ; η0 ) = g(τ1 + τ3 , τ1 + τ2 + τ3 + τ ; η0 ). Fig. 1 Lemma 4.1
498
F. ASAKURA
We have (τ4 − τ3 ) − (τ2 − τ1 ) =
=
τ g(τ1 + τ3 , τ1 + τ2 + τ3 + τ ; η0 ) − g(τ1 + τ3 , τ1 + τ2 + τ3 ; η0 ) × g(τ1 , τ1 + τ2 ; η0 ) − g(τ1 + τ3 , τ1 + τ2 + τ3 ; η0 ) g(τ1 , τ1 + τ2 ; η0 ) − g(τ1 + τ3 , τ1 + τ2 + τ3 ; η0 ) gτ (τ1 + τ3 , τ1 + τ2 + τ3 + θ τ ; η0 )
(0 < θ < 1).
Taking account of g(τ1 , τ1 ; η0 ) = g(τ1 + τ3 , τ1 + τ3 ; η0 ) = 0, we have g(τ1 , τ1 + τ2 ; η0 ) − g(τ1 + τ3 , τ1 + τ2 + τ3 ; η0 )
1 d = g(τ1 , τ1 + θ2 τ2 ; η0 ) − g(τ1 + τ3 , τ1 + θ2 τ2 + τ3 ; η0 ) dθ2 0 dθ2
1 = τ2 gτ (τ1 , τ1 + θ2 τ2 ; η0 ) − gτ (τ1 + τ3 , τ1 + θ2 τ2 + τ3 ; η0 ) dθ2 0
1
1
d gτ (τ1 + θ3 τ3 , τ1 + θ2 τ2 + θ3 τ3 ; η0 ) dθ2 dθ3 0 0 dθ3
1 1 = −τ2 τ3 (gτ τ0 + gτ τ )(τ1 + θ3 τ3 , τ1 + θ2 τ2 + θ3 τ3 ; η0 ) dθ2 dθ3 . = −τ2
0
0
By direct computation, we have [gτ τ0 ] = [gτ τ ] = L−1 T and − (1 + )e γ (η0 −η∗ ) p0 γ (p − p0 ) dp gτ τ0 (τ0 , τ ; η0 ) = − (1 + )p − 2(1 + 3)p0 . 1 5 dτ √ 4 γ p02γ {(1 + )p + p0 } 2
Consequently,
(1 + )e γ (η0 −η∗ ) (p − p0 ) gτ τ (τ0 , τ ; η0 ) + gτ τ0 (τ0 , τ ; η0 ) = 1 5 √ 4 γ p02γ {(1 + )p + p0 } 2
dp dτ
− × p − γ − p0 γ (1 + )(p − p0 ) + (1 − 3)(1 + 2)p0 and hence gτ τ0 + gτ τ < 0 for p > p0 . Note that p
−α
− p0−α
= −α p
−α1
log p
−α − p0 2
By setting C2 =
sup 0<≤1/3 p∗ ≤p, p0 ≤p ∗
log p0
(4.2)
1 0 < α1 , α2 ≤ . 2
gτ τ0 + gτ τ < ∞,
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499
Fig. 2 Lemma 4.2
the first inequality of (4.1) is obtained. Since τ2 − τ1 < τ4 − τ3 , the second one comes from the first one. Remark 4.1 The above lemma holds for the states on the backward 2-shock curves under the conditions: (p2 , u2 , η2 ) ∈ S2B (p1 , u1 , η0 ), (p4 , u4 , η4 ) ∈ S2B (p3 , u3 , η0 ), τ3 > τ1 and u1 − u2 = u3 − u4 . Remark 4.2 By Proposition 3.1, we have gτ τ > 0 for p > p0 . Then, by (4.2), gτ τ0 (τ0 , τ ; η0 ) < 0 for p > p0 . Lemma 4.2 Suppose that (p2 , u2 , η2 ) ∈ S1F (p1 , u1 , η1 ), (p4 , u4 , η4 ) ∈ S1F (p1 , u1 , η3 ), and u2 = u4 , η1 > η3 . Then there is a constant C3 such that [C3 ] = 1 and 0 < τ4 − τ2 ≤ C3 (τ2 − τ1 )(η1 − η3 ). Proof Note that u2 − u1 = −g(τ1 , τ2 ; η1 ), u4 − u1 = −g(τ1 , τ4 ; η3 ). Hence by setting β = τ4 − τ2 , τ2 = τ2 − τ1 , η = η1 − η3 , the condition u2 = u4 (see Fig. 2) is equivalent to g(τ1 , τ2 ; η1 ) = g(τ1 , τ4 ; η3 ) g(τ1 , τ1 + τ2 ; η3 + η) = g(τ1 , τ1 + τ2 + β; η3 ). We have, for some 0 < θ < 1, that τ4 − τ2 = = = =
{g(τ1 , τ1 + τ2 ; η3 + η) − g(τ1 , τ1 + τ2 ; η3 )}β g(τ1 , τ1 + τ2 + β; η3 ) − g(τ1 , τ1 + τ2 ; η3 ) 1 d 0 dθ2 {g(τ1 , τ1 + θ2 τ2 ; η3 + η) − g(τ1 , τ1 + θ2 τ2 ; η3 )} dθ2 1
gτ (τ1 , τ1 + τ2 + θ β; η3 )
{gτ (τ1 , τ1 + θ2 τ2 ; η3 + η) − gτ (τ1 , τ1 + θ2 τ2 ; η3 )} dθ2 gτ (τ1 , τ1 + τ2 + θ β; η3 ) 11 d τ2 0 0 dθ3 gτ (τ1 , τ1 + θ2 τ2 ; η3 + θ3 η) dθ2 dθ3
τ2
0
gτ (τ1 , τ1 + τ2 + θ β; η3 ) 11 τ2 η 0 0 gτ η0 (τ1 , τ1 + θ2 τ2 ; η3 + θ3 η) dθ2 dθ3 = . gτ (τ1 , τ1 + τ2 + θ β; η3 )
500
F. ASAKURA
Fig. 3 Lemma 4.3
By (3.12) we have [gτ η0 ] = 1 and gτ η0 > 0. Setting 11 C3 =
0
sup
0
0<≤1/3 p∗ ≤p, p0 ≤p ∗
gτ η0 (τ1 , τ1 + θ2 τ2 ; η3 + θ3 η) dθ2 dθ3 , gτ (τ1 , τ1 + τ2 + θ β; η3 )
we obtain [C3 ] = 1 and 0 < τ4 − τ2 ≤ C3 τ2 η = C3 (τ2 − τ1 )(η1 − η3 ).
Lemma 4.3 Suppose that (p1 , u1 , η1 ), (p2 , u2 , η2 ) ∈ S1F (p0 , u0 , η0 ), p2 > p1 , (p3 , u3 , η3 ) ∈ S1F (p1 , u1 , η0 ), and τ2 = τ3 . Then there are constants C4 and C5 such that [C4 ] = L−1 T , [C5 ] = L−2 T 2 and 0 < u3 − u2 ≤ C4 (τ1 − τ0 )(τ2 − τ1 ),
(4.3)
0 < (η2 − η1 ) − (η3 − η0 ) ≤ C5 (τ1 − τ0 )(τ2 − τ1 ).
(4.4)
Proof Note that uj − ui = −g(τi , τj ; η0 ). Since τ2 = τ3 (see Fig. 3), we have u3 − u2 = (u3 − u1 ) − (u2 − u0 ) + (u1 − u0 ) = −g(τ1 , τ3 ; η0 ) + g(τ0 , τ2 ; η0 ) − g(τ0 , τ1 ; η0 ) = g(τ0 , τ2 ; η0 ) − g(τ1 , τ2 ; η0 ) − g(τ0 , τ1 ; η0 ) − g(τ1 , τ1 ; η0 ) . Setting τ1 = τ1 − τ0 ,
τ2 = τ2 − τ1 ,
we have τ2 − τ0 = τ1 + τ2 and the above right-hand side is g(τ0 , τ1 + τ2 ; η0 ) − g(τ0 + τ1 , τ1 + τ2 ; η0 ) − g(τ0 , τ1 ; η0 ) − g(τ0 + τ1 , τ1 ; η0 )
1 d = g(τ0 , τ1 + θ2 τ2 ; η0 ) − g(τ0 + τ1 , τ1 + θ2 τ2 ; η0 ) dθ2 0 dθ2
1 = τ2 gτ (τ0 , τ1 + θ2 τ2 ; η0 ) − gτ (τ0 + τ1 , τ1 + θ2 τ2 ; η0 ) dθ2 0
1
1
d gτ (τ0 + θ1 τ1 , τ1 + θ2 τ2 ; η0 ) dθ2 dθ1 dθ 1 0 0
1 1 = −τ1 τ2 gτ τ0 (τ0 + θ1 τ1 , τ1 + θ2 τ2 ; η0 ) dθ2 dθ1 . = −τ2
0
0
Using Remark 4.2, we define C4 by C4 =
sup 0<≤1/3 p∗ ≤p, p0 ≤p ∗
|gτ τ0 |.
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501
Thus we obtain [C4 ] = L−1 T and 0 < u3 − u2 ≤ C4 τ1 τ2 showing (4.3). Similarly, we have ηi − η0 = h(τ0 , τi ) (i = 1, 2), η3 − η0 = h(τ1 , τ3 ). Hence, (η2 − η1 ) − (η3 − η0 ) = h(τ0 , τ2 ) − h(τ0 , τ1 ) − h(τ1 , τ3 ) = h(τ0 , τ2 ) − h(τ0 , τ1 ) − h(τ1 , τ2 ) − h(τ1 , τ1 )
1 d =− h(τ0 + θ1 τ1 , τ1 + τ2 ) − h(τ0 + θ1 τ1 , τ1 ) dθ1 0 dθ1
1 = −τ1 hτ0 (τ0 + θ1 τ1 , τ1 + τ2 ) − hτ0 (τ0 + θ1 τ1 , τ1 ) dθ1 0
1
1
d hτ0 (τ0 + θ1 τ1 , τ1 + θ2 τ2 ) dθ2 dθ1 0 0 dθ2
1 1 = −τ1 τ2 hτ0 τ (τ0 + θ1 τ1 , τ1 + θ2 τ2 ) dθ2 dθ1 . = −τ1
0
0
By (3.11) we have γ 2 (1 + )(p + p0 )(p − p0 ) dp dp0 hτ τ0 (τ0 , τ ) = − 2{p + (1 + )p0 }2 {(1 + )p + p0 }2 dτ dτ0 <0
for p > p0
and C5 is defined by C5 =
sup 0<≤1/3 p∗ ≤p, p0 ≤p ∗
|hτ τ0 |.
Then we obtain [C5 ] = L−2 T 2 and 0 < (η2 − η1 ) − (η3 − η0 ) ≤ C5 τ1 τ2 showing (4.4). Lemma 4.4 Suppose that (p2 , u2 , η2 ) ∈ S1F (p1 , u1 , η1 ), (p4 , u4 , η4 ) ∈ S1F (p3 , u3 , η3 ), τ2 − τ1 = τ4 − τ3 . Then there are constants C6 and C7 such that [C6 ] = L−2 T 2 , [C7 ] = 1 and 0 < (η3 − η1 ) − (η4 − η2 ) ≤ C6 (τ2 − τ1 )|τ1 − τ3 |,
(u1 − u2 ) − (u3 − u4 ) ≤ (τ2 − τ1 ) C2 |τ1 − τ3 | + C7 |η1 − η3 | .
(4.5) (4.6)
Moreover, if τ1 = τ3 and η1 > η3 , then 0 < (u1 − u2 ) − (u3 − u4 ) ≤ C7 (τ2 − τ1 )(η1 − η3 ).
(4.7)
Proof Let us prove the first part. Since ηi+1 − ηi = h(τi , τi+1 ), we have (η3 − η1 ) − (η4 − η2 ) = (η3 − η4 ) − (η1 − η2 ) = −h(τ3 , τ4 ) + h(τ1 , τ2 ). We set τ2 = τ2 − τ1 = τ4 − τ3 (see Fig. 4) and τ3 = τ3 − τ1 . Since h(τ3 , τ3 ) = h(τ1 , τ1 ) = 0, we obtain
502
F. ASAKURA
Fig. 4 Lemma 4.4
−h(τ3 , τ4 ) + h(τ1 , τ2 ) = −h(τ3 , τ3 + τ2 ) + h(τ1 , τ1 + τ2 )
1 d =− h(τ3 , τ3 + θ2 τ2 ) − h(τ1 , τ1 + θ2 τ2 ) dθ2 dθ 2 0
1 = −τ2 hτ (τ3 , τ3 + θ2 τ2 ) − hτ (τ1 , τ1 + θ2 τ2 ) dθ2 0
1
1
d hτ (τ1 + θ3 τ3 , τ1 + θ3 τ3 + θ2 τ2 ) dθ2 dθ3 dθ 3 0 0
1 1 = −τ2 τ3 (hτ τ0 + hτ τ )(τ1 + θ3 τ3 , τ1 + θ3 τ3 + θ2 τ2 ) dθ2 dθ3 . = −τ2
0
0
As h(τ, τ0 ) = −h(τ0 , τ ), we find that hτ0 (τ0 , τ ) = −
∂ h(τ, τ0 ). ∂τ0
Hence, by (3.11), 1 − √ (1 + ) γ a − γ e− γ η∗ (p0 γ − p − γ )(p − p0 )2 hτ (τ0 , τ ) + hτ0 (τ0 , τ ) = − 2{p + (1 + )p0 }{(1 + )p + p0 }
<0
for p > p0 .
By direct computation, we have 1 p −1− γ (p − p0 )2 √ − γ1 − γ η∗ hτ τ (τ0 , τ ) + hτ τ0 (τ0 , τ ) = − (1 + ) γ a e 2 γ {p + (1 + )p0 }{(1 + )p + p0 }
−
γ 2 (p0 γ − p − γ )p0 (p + p0 )(p − p0 ) + {p + (1 + )p0 }2 {(1 + )p + p0 }2 <0
dp dτ
for p > p0 .
Defining C6 by C6 =
sup 0<≤1/3 p∗ ≤p, p0 ≤p ∗
h τ τ 0 + hτ τ ,
we have [C6 ] = L−2 T 2 and 0 < (η3 − η1 ) − (η4 − η2 ) ≤ C6 τ2 τ3 showing (4.5).
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503
In order to prove (4.6), we note that ui − uj = g(τi , τj ; ηi ). By setting τ2 = τ2 − τ1 = τ4 − τ3 , τ3 = τ3 − τ1 and η = η1 − η3 , we have (u1 − u2 ) − (u3 − u4 ) = −g(τ3 , τ4 ; η3 ) + g(τ1 , τ2 ; η1 ) = −g(τ3 , τ3 + τ2 ; η3 ) + g(τ1 , τ1 + τ2 ; η1 )
1 d =− g(τ3 , τ3 + θ2 τ2 ; η3 ) − g(τ1 , τ1 + θ2 τ2 ; η1 ) dθ2 dθ 2 0
1 = −τ2 gτ (τ3 , τ3 + θ2 τ2 ; η3 ) − gτ (τ1 , τ1 + θ2 τ2 ; η1 ) dθ2 0
gτ (τ1 + τ3 , τ1 + τ3 + θ2 τ2 ; η3 ) − gτ (τ1 , τ1 + θ2 τ2 ; η3 ) dθ2
1
= −τ2 0
gτ (τ1 , τ1 + θ2 τ2 ; η3 + η) − gτ (τ1 , τ1 + θ2 τ2 ; η3 ) dθ2
1
+ τ2 0
= −τ2
1 0
1
0
1
d gτ (τ1 + θ3 τ3 , τ1 + θ3 τ3 + θ2 τ2 ; η3 ) dθ2 dθ3 dθ3
1
d gτ (τ1 , τ1 + θ2 τ2 ; η3 + θ η) dθ2 dθ dθ
1 1 = −τ2 τ3 (gτ0 τ + gτ τ )(τ1 + θ3 τ3 , τ1 + θ3 τ3 + θ2 τ2 ; η3 ) dθ2 dθ3 + τ2
0
0
0
+ τ2 η
1 0
0
0
1
gτ η0 (τ1 , τ1 + θ2 τ2 ; η3 + θ η) dθ2 dθ.
Recall that gτ τ0 + gτ τ < 0, p > p0 and C2 =
sup 0<≤1/3 p∗ ≤p, p0 ≤p ∗
gτ τ0 + gτ τ < ∞.
It follows from (3.12) that the quantity C7 =
sup 0<≤1/3 p∗ ≤p, p0 ≤p ∗
gτ η0
is bounded. Then we obtain [C7 ] = 1 and (u1 − u2 ) − (u3 − u4 ) ≤ C2 τ2 |τ3 | + C7 τ2 |η| showing the inequality (4.6). Since gτ η0 > 0, we have (u1 − u2 ) − (u3 − u4 ) > 0 for τ3 = 0. Then we have (4.7).
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F. ASAKURA
Fig. 5 Lemma 4.5
Lemma 4.5 Suppose that (p1 , u1 , η1 ), (p2 , u2 , η2 ) ∈ S1F (p0 , u0 , η0 ), τ2 > τ1 (p2 > p1 ), (p3 , u3 , η3 ) ∈ S1F (p1 , u1 , η1 ), and u3 = u2 . Then there is a constant C8 such that [C8 ] = L−2 T 2 and η2 − η3 + C8 (τ1 − τ0 )(τ2 − τ1 ) ≥ 0.
(4.8)
Proof Since η2 − η0 = h(τ0 , τ2 ), η3 − η1 = h(τ1 , τ3 ) and η1 − η0 = h(τ0 , τ1 ), we have η2 − η3 = h(τ0 , τ2 ) − h(τ0 , τ1 ) − h(τ1 , τ3 ). Similarly, u2 = u3 (see Fig. 5) is equivalent to u0 + g(τ0 , τ2 ; η0 ) = u1 + g(τ1 , τ3 ; η1 ). Then it follows that g(τ0 , τ2 ; η0 ) = g(τ1 , τ3 ; η1 ) + g(τ0 , τ1 ; η0 ).
(4.9)
Thus our task is to estimate η2 − η3 under the above constraint. Isothermal equations: = 0. We have τ = a log pp∗ and τ − τ0 , 2a τ − τ0 τ1 − τ0 − η − η0 = sinh a a
u − u0 = ∓ 2a sinh
(see Liu [6], Nishida [9]). Then the condition (4.9) is equivalent to sinh
τ1 − τ0 τ3 − τ1 τ2 − τ0 = sinh + sinh , 2a 2a 2a
τ0 < τ1 < τ2 .
Simple formulas yield sinh
τ2 − τ1 τ2 − τ0 τ1 − τ0 τ2 − τ1 τ3 − τ1 − sinh = sinh − sinh − sinh 2a 2a 2a 2a 2a τ1 − τ0 τ2 − τ1 τ2 − τ0 sinh sinh . = 4 sinh 4a 4a 4a
−τ1 −τ1 − sinh τ22a > 0. Hence τ3 > τ2 . Since τ0 < τ1 < τ2 , we have sinh τ32a We find by direct computation,
η2 − η3 = sinh
τ1 − τ0 τ3 − τ1 τ3 − τ2 τ2 − τ0 − sinh − sinh + . a a a a
(4.10)
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505
Note that τ1 − τ0 τ3 − τ1 τ2 − τ0 − sinh − sinh a a a τ1 − τ0 τ1 − τ0 τ2 − τ0 = 2 sinh − cosh cosh 2a 2a 2a τ2 − τ0 τ3 − τ1 τ3 − τ1 cosh − cosh . + 2 sinh 2a 2a 2a
sinh
Since τ2 − τ0 > τ1 − τ0 and τ2 − τ0 > τ3 − τ1 , we find that the above quantity is positive. Thus we obtain η2 − η3 > a1 (τ3 − τ2 ) > 0, which is (4.8) for = 0. Non-isentropic equations: > 0. We consider the non-isentropic equations as a perturbation of the isothermal equations. This lemma says that the quantity η2 − η3 is different from that in the case γ = 1 by O(1)(τ1 − τ0 )(τ2 − τ1 ). We begin with the expression 1
a γ (p − p0 )e γ (η0 +log p0 ) g(τ0 , τ ; η0 ) = √ . 1 pp0 {1 + (1 + pp0 )} 2 By this, the constraint (4.9) is expressed as p1 − p0 1 + (1 + p2 − p 0 = √ √ p2 p0 p1 p0 1 + (1 +
p0 p2
)
p0 p1
)
p3 − p1 1 + (1 + + √ p3 p1 1 + (1 +
12
p0 p2
)
p1 p3
)
12
(η −η +log p1 ) 1 0 p0
eγ
p 1 − p0 p3 − p1 = √ + √ + O(1)(p2 − p1 )(p1 − p0 ) p1 p0 p3 p1 + O(1)(η1 − η0 )(p3 − p1 ) + O(1)(p3 − p1 )(p1 − p0 ) + O(1)(p3 − p2 )(p3 − p1 ),
(4.11)
where O(1) is uniformly bounded for 0 < ≤ 1/3 and p∗ ≤ p ≤ p ∗ . We find that p − p is equivalent to τ − τ , because (3.3) yields √
1
γ a − γ e− γ η∗ (p∗ )
1+ γ
≤
dp √ − γ1 − γ η∗ ∗ 1+ ≤ γa e (p ) γ , dτ
which gives uniform bounds with respect to 0 < ≤ 13 . By setting τ1 = τ1 − τ0 , τ3 = τ3 − τ2 , η = η1 − η0 , we can rewrite the condition (4.9) as follows: g(τ0 + τ1 , τ2 + τ3 ; η0 + η) − g(τ0 , τ2 ; η0 ) = −g(τ0 , τ0 + τ1 ; η0 ) = − g(τ0 , τ0 + τ1 ; η0 ) − g(τ0 , τ0 ; η0 ) . By applying the mean value theorem for functions G1 (t) = g(τ0 + tτ1 , τ2 + tτ3 ; η0 + tη) and G2 (t) = g(τ0 , τ0 + tτ1 ; η0 ), we have gτ0 (τˆ0 , τˆ2 ; η)τ ˆ ˆ ˆ = −gτ (τ0 , τˆ0 ; η0 )τ1 , 1 + gτ (τˆ0 , τˆ2 ; η)τ 3 + gη (τˆ0 , τˆ2 ; η)η
506
F. ASAKURA
where (τˆ0 , τˆ2 ; η) ˆ = (τ0 + θ1 τ1 , τ2 + θ1 τ3 ; η0 + θ1 η) and τˆ0 = θ2 τ1 for appropriate 0 < θ1 , θ2 < 1. Since gτ > 0, we obtain τ3 − τ2 = O(1)(τ1 − τ0 ) + O(1)(η1 − η0 ). Similarly, by rewriting the condition (4.9) as g(τ0 , τ2 ; η0 ) − g(τ0 , τ1 ; η0 ) = g(τ1 , τ3 ; η1 ) = g(τ1 , τ3 ; η1 ) − g(τ1 , τ1 ; η1 ), we obtain τ3 − τ1 = O(1)(τ2 − τ1 ). We also have η1 − η0 = O(1)(p1 − p0 ) by the expression
η − η0 = log
p0 p
+
1+ γ log 2 1+
( p ) 1+ p0 . p0 ( ) 1+ p
Let us introduce a new parameter4 q = log p. Since q − q0 = log
p , p0
q − q0 is equivalent to p − p0 for p∗ ≤ p ≤ p ∗ . By using this parameter, (4.11) is equivalent to q2 − q0 q1 − q0 q3 − q 1 sinh = sinh + sinh + O(1)(p2 − p1 )(p1 − p0 ) 2 2 2 which is an analog of (4.10). Since q0 < q1 , we have sinh
q 2 − q0 q 3 − q1 ≥ sinh + O(1)(p2 − p1 )(p1 − p0 ). 2 2
Hence q2 − q0 ≥ q3 − q1 + O(1)(p2 − p1 )(p1 − p0 ). The same computation as in the isothermal case yields q2 − q 1 q 3 − q1 − sinh 2 2 q 1 − q0 q 2 − q1 q2 − q0 sinh sinh + O(1)(p2 − p1 )(p1 − p0 ) = 4 sinh 4 4 4
sinh
for q0 < q1 < q2 . A simple formula yields sinh
q 3 − q1 q2 − q 1 q3 + q2 − 2q1 q3 − q2 − sinh = 2 cosh sinh 2 2 4 4
4 The next expression shows that the difference q − q is a dimensionless quantity. 0
EQUATIONS OF NON-ISENTROPIC GAS DYNAMICS
and 1 ≤ cosh( q3 +q24−2q1 ) ≤
p∗ . p∗
507
Thus we conclude that
q3 − q2 ≥ O(1)(p2 − p1 )(p1 − p0 ). Now, let us estimate the quantity η2 − η3 . By employing the Taylor expansion of log(1 + x), we find that there is a smooth function Φ(p0 , p) such that h(τ0 , τ ) = log
p0 γ + p 2
p0 p − p0 p
+ (p − p0 )Φ(p0 , p).
Using the above expression, we have η2 − η3 = h(τ0 , τ2 ) − h(τ0 , τ1 ) − h(τ1 , τ3 ) p3 γ p 1 p0 p 3 p1 p2 p0 = log + − − − − − p2 2 p0 p2 p0 p1 p1 p3 + (p2 − p1 ) Φ(p0 , p2 ) − Φ(p1 , p2 ) + (p1 − p0 ) Φ(p0 , p2 ) − Φ(p0 , p1 ) + (p2 − p3 )Φ(p1 , p2 ) + (p3 − p1 ) Φ(p1 , p2 ) − Φ(p1 , p3 ) . The term p2 − p3 is estimated by O(1)(p2 − p1 )(p1 − p0 ), hence all O() terms are estimated by O(1)(p2 − p1 )(p1 − p0 ). Thus we obtain γ sinh(q2 − q0 ) − sinh(q1 − q0 ) − sinh(q3 − q1 ) 2 +O(1)(p2 − p1 )(p1 − p0 ),
η2 − η3 = (q3 − q2 ) +
where q = log p. As before, we find that sinh(q2 − q0 ) − sinh(q1 − q0 ) − sinh(q3 − q1 ) q1 − q0 q1 − q 0 q 2 − q0 − cosh = 2 sinh cosh 2 2 2 q3 − q1 q3 − q1 q 2 − q0 − cosh + 2 sinh cosh . 2 2 2 Since q2 − q0 > q1 − q0 and q2 − q0 ≥ q3 − q1 + O(1)(p2 − p1 )(p1 − p0 ), we find that the above quantity is larger than O(1)(p2 − p1 )(p1 − p0 ). Thus we conclude that η2 − η3 > (q3 − q2 ) + O(1)(p2 − p1 )(p1 − p0 ) > O(1)(τ2 − τ1 )(τ1 − τ0 ).
5 Proof of interaction estimates In this section we shall give a detailed proof of the local interaction estimates. The proof will be carried out for the most complicated cases (1), (3), (4) and (6). In the following propositions, the constant D represents a positive constant uniformly bounded for 0 < ≤ 1/3 and p∗ ≤ p ≤ p ∗ . We begin with a useful lemma that is obvious by the sine theorem. Let |PQ| denote the Euclidean distance between two points P and Q.
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F. ASAKURA
Fig. 6 β + α → α + δ + β
Lemma 5.1 (Triangle Lemma) If the segments PQ, QR, RP form a proper triangle, then |PQ|, |QR| ≤
|RP| , sin θQ
θQ = ∠PQR.
Proposition 5.1 Case (1): β + α → α + δ + β α ≤ α + Dαβ,
β ≤ β + Dαβ,
δα ≥ δα − D2 αβ, δβ ≥ δβ − D2 αβ,
|δ | ≤ D2 αβ. Proof Since computations will be carried out on the wz-plane, the projections of forward and backward shock curves are denoted by S1F (w0 , z0 ; η0 ) and S2B (w0 , z0 ; η0 ). In this case, the configuration of the shock curves is expressed as (wL , zL ) ∈ S2B (wM , zM ; ηM ),
(wR , zR ) ∈ S1F (wM , zM ; ηM )
and (w± , z± ) ∈ S1F (wL , zL ; ηL ) ∩ S2B (wR , zR ; ηR ). We set 1F = S1F (wL , zL ; ηM ) S
and
2B = S2B (wR , zR ; ηM ). S
F
B
We also use S 1 : the curve congruent to S1F (wM , zM ; ηM ) and issuing from (wL , zL ); S 2 : congruent to S2B (wM , zM ; ηM ) and issuing from (wR , zR ). By Remark 4.2 we have gτ τ0 < 0 F which indicates that the slope of S1F is gentler than that of S 1 ; S2B and S2B have a similar property. Constant states are often denoted by PL = P(wL , zL ), PR = P(wR , zR ), etc., as the points are in the wz-plane. Four points P, P, P1 and P2 are defined by P = S1F ∩ S2B ,
F
B
P = S1 ∩ S2 ,
F P1 = S 1 ∩ (w, z); u = u ,
B P2 = S 2 ∩ (w, z); u = u ,
where u denotes the velocity corresponding to the point P. Note that the three points P, P1 , P2 are collinear and τL > τM , τR > τM (see Fig. 6).
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509
Suppose that P2 lies between P and P1 (see the picture on the left of Fig. 6). In this case, F P1 lies between P and PL . The point P∗1 ∈ S1F corresponding to P1 ∈ S 1 is chosen so that u = uM − u∗1 , uL − which is an assumption of Lemma 4.1. This lemma yields
τ − τL ) − (τ1 − τL ) = ( τ − τL ) − τ1∗ − τM τ − τ1 = (
≤ C2 (τL − τM ) τ1∗ − τM ≤ C2 (τL − τM )(τR − τM ) and hence τ − τ1 > 0. Clearly, 0 < τR − τM < wM − wR ,
0 < τL − τM < zL − zM .
Thus, we have | PP1 | = τ − τ1 ≤ C2 (wM − wR )(zL − zM ) = C2 αβ. Hence, | PP2 |, |P2 P1 | ≤ C2 αβ. Since it is assumed that p∗ ≤ p ≤ p ∗ and H is sufficiently small (see (3.1)), we find that the slope of PP1 is smaller than 45◦ and that of PP2 bigger than 45◦ . Thus we conclude from H /γ the triangle lemma for P2 P, PP1 , P1 P2 that |P1 P| ≤ sin1 θ |P2 P1 |, θ ≥ tan−1 B 22Be , which e2H /γ −1 implies | P P| ≤ O(1)αβ.
(5.1)
In case P1 lies between P and P2 , we obtain the same estimates as above by using 2-shock curves instead of 1-shock curves (choose P∗2 ∈ S2B ). In a similar manner, we set S 1 = S1F (wL , zL ; ηL ) F
and
S 2 = S2B (wR , zR ; ηR ) B
and S1F and S2B as before (see the picture on the right of Fig. 6). By (3.12), we have gτ η0 > 0 which indicates that the slope of S F1 is steeper than that of S1F ; S B2 and S2B have a similar property. u}; P2 = We denote P± = P(w± , z± ). P1 and P2 are defined by P1 = S F1 ∩ {(w, z); u = B S 2 ∩ {(w, z); u = u}. Note that ηL > ηM and ηR > ηM . P and P1 . By Lemma 4.2, we have Suppose again that P2 lies between
τ − τ1 ≤ C3 τ1 − τL (ηL − ηM ) ≤ C1 C3 ( τ − τL )(τL − τM ), where C1 =
sup
hτ .
0<≤1/3 p∗ ≤p, p0 ≤p ∗
Hence, )(zL − zM ). | PP1 | ≤ C1 C3 (wL − w
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F. ASAKURA
Fig. 7 β + α → α + δ + β (locations of P∗ and P∗ )
Then it follows from the triangle lemma for P2 P± , P± P1 , P1 P2 that |P1 P± | ≤ θ± ≥ tan
−1 2BeH /γ B 2 eH /γ −1
1 |P P |, sin θ± 2 1
, which implies | PP± | ≤ O(1)(wL − w )β.
(5.2)
P and P2 , we use 2-shock curves instead of 1-shock curves to obtain the If P1 lies between same estimates. Combining (5.1) and (5.2), we have |α − α| + |β − β| = |w± − w| + |z± − z| ≤ 2|P± P| ≤ 2|P P| + 2| PP± |
≤ O(1)αβ + O(1) (wL − w) + (w − w ) β ≤ O(1)αβ + O(1) α + O(1)αβ β = O(1) 1 + O(1)β αβ. Thus there is a constant D such that |α − α| + |β − β| ≤ Dαβ showing the first part of the proposition. Now we carry out the estimates of the entropy. For simplicity, we will treat the case α < α, β < β (see Fig. 7). The remaining cases are mentioned in Remark 5.1. Let P∗ ∈ F S1F (wM , zM ; ηM ) such that τ∗ − τM = τ± − τL . Let P∗ ∈ S 1 (issuing from (wL , zL ) and congruent to S1F (wM , zM ; ηM )) such that τ ∗ = τ± . It follows from Lemma 4.4 that (η− − ηL ) − (η∗ − ηM ) ≤ C6 (τ∗ − τM )|τL − τM |. Since p∗ ≤ p ≤ p ∗ and H is sufficiently small, the triangle lemma for P P∗ , P∗ P± , P± P 1 yields |P P∗ | ≤ sin1θ |PP± |, θ ∗ ≥ tan−1 Be2H /γ . Hence we have ∗
|PR P∗ | = |P P∗ | ≤ C |α − α| + |β − β| ≤ CDαβ.
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511
Fig. 8 α > α, β > β and α < α, β > β
Thus,
|δα − δα | = (η− − ηL ) − (ηR − ηM ) ≤ (η− − ηL ) − (η∗ − ηM ) + |ηR − η∗ | ≤ C6 (τ∗ − τM )|τL − τM | + C|τR − τ∗ |
≤ C6 (τR − τM )|τL − τM | + C + C6 |τL − τM | |τR − τ∗ | ≤ O(1)(wM − wR )(zL − zM ) + O(1)|PR P∗ | ≤ O(1)αβ.
In a similar manner, we obtain |δβ − δβ | ≤ O(1)αβ. Finally, the above estimates imply |δ − δ| ≤ |δα − δα | + |δβ − δβ | ≤ O(1)αβ
showing the last inequality.
Remark 5.1 The configuration of Fig. 7 is the case α < α, β < β. There are other configurations as shown in Fig. 8. We note that Lemma 4.4 covers these cases, as long as P∗ is located on S1F and P∗ on S F1 . Proposition 5.2 Case (3): α1 + α2 → α + δ + π α ≤ α1 + α2 + Dα1 α2 ,
δα ≥ δα1 + δα2 − D2 α1 α2 ,
π ≤ Dα1 α2 , |δ | ≤ D2 α1 α2 . Proof Let us set S1F = S1F (wM , zM ; ηM )
S 1 = S1F (wL , zL ; ηL ). F
and
In this case, we have (see Fig. 9): (wM , zM ), (w± , z± ) ∈ S 1 , F
(wR , zR ) ∈ S1F .
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F. ASAKURA
Fig. 9 α1 + α2 → α + δ + π
We set as before S1F = S1F (wM , zM ; ηL ) and P = S1F ∩ RB2 (wR , zR ; ηR ). The points P∗ ∈ F F S1 , P∗ ∈ S1 are defined so that τ∗ = τ∗ = τ± . Since ηM > ηL , the point P∗ is located P∗ . By Lemma 4.3 we have between P± and 0 < u˜ ∗ − u± ≤ C4 (τM − τL )(τ± − τM ) ≤ C4 (τM − τL )(τR − τM ) ≤ O(1)α1 α2 . Hence, z − z± ≤ | PP± | ≤ zR − z± ≤
√
2(u˜ ∗ − u± ) ≤ O(1)α1 α2 .
Thus we conclude that π ≤ Dα1 α2 . Moreover, since wL − w± ≤ (wL − wM ) + (wM − wR ) + O(1)π , we obtain α ≤ α1 + α2 + Dα1 α2 . On the other hand, it follows from Lemma 4.3 that 0 < (η− − ηM ) − (η˜ ∗ − ηL ) ≤ C5 (τM − τL )(τ± − τM ) ≤ C5 (τM − τL )(τR − τM ) ≤ O(1)α1 α2 . Formula (2.4) says that the entropy difference across a jump depends only on τ ; hence η˜ ∗ − ηL = η∗ − ηM . Then we have 0 < (η− − ηM ) − (η˜ ∗ − ηL ) = (η− − ηL ) − (ηM − ηL ) − (η∗ − ηM ) = (η− − ηL ) − (ηM − ηL ) − (ηR − ηM ) + (ηR − η∗ ) = δα − δα1 − δα2 + (ηR − η∗ ) ≤ O(1)α1 α2 . Since τ∗ = τ± and π ≤ Dα1 α2 , we have ηR − η∗ ≤ O(1)α1 α2 .
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513
Fig. 10 α1 + α2 → α + δ + π (location of P∗± )
Thus we obtain δα − δα1 − δα2 ≤ O(1)α1 α2 .
(5.3)
In order to estimate the lower bound of δα − δα1 − δα2 , we choose P∗± ∈ S1F (wL , zL ; ηL ) so that u∗± = uR (see Fig. 10). Then, by Lemma 4.5,
∗ − ηR ≥ −C8 (τM − τL ) τ±∗ − τM η± ≥ −C8 (τM − τL )(τR − τM ) ≥ O(1)α1 α2 . ∗ , we have Noticing that η− > η±
η− − ηR = (η− − ηL ) − (ηR − ηM ) − (ηM − ηL ) = δα − δα1 − δα2 ∗ > η± − ηR
≥ O(1)α1 α2 . Combining this with (5.3), we finally obtain −D2 α1 α2 ≤ δα − δα1 − δα2 ≤ D2 α1 α2 . Since δ = δα − δα1 − δα2 in this case, we have |δ | ≤ Dα1 α2
which completes the proof. Proposition 5.3 Case (4): δ + α → α + δ + θ (θ = β or π ) α ≤ α + D1 α|δ|,
δα ≥ δα − D2 α1 α2 ,
|δ | ≤ |δ| + Dα|δ|, θ ≤ D1 α|δ|. Proof Let us set as before S1F = S1F (wM , zM ; ηM ) and S F1 = S1F (wL , zL ; ηL ), where wM = wL and zM = zL in this case (see Fig. 11).
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F. ASAKURA
Fig. 11 δ + α → α + δ + β (upper) and π (lower)
First assume that ηM > ηL ; then θ = β . Let P∗ ∈ S1F be such that u± = u∗ . Hence P∗ is located between PR and PM = PL , and it follows from Lemma 4.2 that 0 < τ± − τ∗ ≤ C3 (τ∗ − τM )(ηM − ηL ) ≤ C3 α|δ|. By the triangle lemma for P± PR , PR P∗ , P∗ P± we have |PR P± | ≤ tan
−1 2BeH /γ B 2 e2H /γ −1
1 |P∗ P± | sin θR
θR ≥
. Hence, |θ | ≤ O(1)α|δ|.
(5.4)
If ηL > ηM , then θ = π . We choose P∗ ∈ S F1 so that uR = u∗ . Then P∗ is located between P± and PM = PL , and the above argument goes in the same way. As the consequence (5.4) holds also in this case. Using (5.4), we obtain |α − α| ≤ O(1)|θ | ≤ O(1)α|δ|. Since the entropy depends only on τ , we have |δα − δα | ≤ O(1)|α − α| ≤ O(1)α|δ|, |δ − δ| ≤ |δα − δα | + O(1)|θ | ≤ O(1)α|δ|.
Thus we have proved the proposition. Proposition 5.4 Case (6): ξ + α → α + δ + β α ≤ α − ξ,
δα ≥ δα − D(α − α ) − D2 αξ,
β ≤ D0 (α − α ) + Dα ξ,
δβ ≥ 0,
|δ | ≤ D(α − α ) + D2 αξ, where 0 < D0 < 1 holds uniformly for 0 < ≤ 1/3 and p∗ ≤ p ≤ p ∗ . Proof By the configuration of the wave curves, we find that α + ξ ≤ α. Thus, α ≤ α − ξ. F
F
We set R1 : the curve congruent to RF1 (wL , zL ; ηL ) and issuing from (w± , z± ); S 1 : F congruent to S F1 (wL , zL ; ηL ) and issuing from (wM , zM ). The point P is defined to be R 1 ∩ F S 1 . We choose P∗ ∈ S1F (wM , zM ; ηL ) so that w = w∗ (see Fig. 12). Notice that ηL = ηM .
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515
Fig. 12 ξ + α → α + δ + β (1)
Fig. 13 ξ + α → α + δ + β (2)
Since the slope of S1F (wM , zM ; ηL ) is less than 1, there is a constant D0 with 0 < D0 < 1 such that
β ≤ (z − z∗ ) + (z∗ − zR ) ≤ (z − z∗ ) + D0 α − α . Here D0 is uniformly bounded for 0 < ≤ 1/3 and p∗ ≤ p, p0 ≤ p ∗ . We also choose P1 ∈ S1F (wM , zM ; ηL ) so that u1 = u. Then by Lemma 4.1 we have 0 < τ − τ1 = (τ± − τL ) − (τ1 − τM ) ≤ C3 (τL − τM )(τ± − τL ) ≤ O(1)α ξ. Hence, |PP1 | ≤ O(1)α ξ. Noticing that 0 < z − z∗ = |PP∗ | ≤
√
2|PP1 | ≤ O(1)α ξ,
we obtain β ≤ Dα ξ + D0 (α − α ). In order to estimate the entropy, we choose P2 ∈ S1F (wM , zM ; ηL ) so that τ2 = τ (see Fig. 13). Recall that ηL = ηM . Then τ± − τL = τ − τM = τ2 − τM and it follows from Lemma 4.4 that |η2 − η− | = (η2 − η− ) − (ηM − ηL ) ≤ C6 (τL − τM )(τ2 − τM ) ≤ O(1)α ξ. On the other hand, |δα − δα | = (ηR − ηM ) − (η− − ηL ) ≤ |ηR − η− | ≤ |ηR − η2 | + |η2 − η− |. Moreover, τR − τ2 = (τR − τM ) − (τ2 − τM ) ≤ O(1)(α − α ).
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F. ASAKURA
Thus we conclude that
|δα − δα | ≤ O(1)α ξ + O(1) α − α
|δ | = |δβ + δα − δα | ≤ O(1)α ξ + O(1) α − α . In particular, we have
δα ≥ δα − Dαξ − D α − α .
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