Some Results by Energy Methods on Large-Time Behavior of Viscous Gas Akitaka Matsumura (Osaka University)

Some Results by Energy Methods on Large-Time Behavior of Viscous Gas

Akitaka Matsumura (Osaka University)

”Nonlinear Evolution Equations and Related Topics”

Kyoto Universiy, March 9-11, 2015

Introduction

A model system of viscous gas

 

 

 

 

 



ρ

_t

+ ∇ · (ρu) = 0, u

_t

+ u · ∇ u + 1

ρ ∇ p = µ

ρ ∆u + µ + λ

ρ ∇ ( ∇ · u), p = p(ρ) = aρ

^γ

,

(1)

where t ≥ 0, x ∈ Rⁿ(n = 1, 2, 3), µ > 0, µ + λ > 0, a > 0, γ ≥ 1.

Consider the Cauchy problem for (1) with the initial data

(ρ, u)(0) = (ρ

₀

, u

₀

). (2)

Two papers :

• Nishida-M (1980) , J. Math. Kyoto Univ.,

asymptotic stability of constant states in R³, R²

• Nishihara-M (1985) , Japan J. Appl. Math., asymptotic stability of traveling waves in R¹

Topic 1.

(joint work with T. Maeda)

• Nishida-M (1980), x ∈ R² or R³, ρ >¯ 0

(ρ₀ − ρ, u¯ ₀) ∈ H³, small =⇒ ( ¯ρ, 0) is asymptotically stable.

Since then, Hoff ( ¯ρ > 0), Feireisl, Lions ( ¯ρ = 0), . . . , Z.Xin-J.Li- X.Huang ( ¯ρ ≥ 0), . . .

Our present aim

Consider the asymptotic stability of an unbounded state ρ = P = ρ¯

1 + t, u = U = ( x₁

1 + t, 0).

Theorem 1.

Suppose

x ∈ R

²

, p = aρ

, and

a − (2µ + λ)

¯

ρ > 0

. Then, there exists a ε₀ > 0 such that if ∥ρ₀−ρ, u¯ ₀−(x₁,0)∥H² ≤ ε₀, the Cauchy problem (1),(2) has a unique global solution in time (ρ, u), satisfying (ρ−P, u − U) ∈ C([0, +∞); H²) and

sup

x∈R²

| ρ(t, x) − ρ ¯

1 + t | ≤ C (1 + t)

⁻³²

, sup

x∈R²

| u(t, x) − ( x

₁

1 + t , 0) | ≤ C (1 + t)

⁻¹²

.

Remarks

• The proof is given by a combination of changing the variable x₁ along a characteristic curve and using a time-weighted energy method .

• For R³ , the similar results hold in H³ with the asymptotics sup

x∈R³

| ρ − P | ≤ C(1 + t)⁻², sup

x∈R³

| u − U | ≤ C(1 + t)⁻¹.

•

Open problems

– Isentropic case : p = aρ^γ, (γ > 1)

– Full system case : p = Rρθ, e = _γ^R₋₁θ ρ = P = ρ¯

1 + t, u = U = ( x₁

1 + t, 0), θ = 2µ + λ Rρ¯ .

Topic 2.

(joint work with Yang Wang)

Asymptotic stability of traveling wave solutions in R System in Lagrange coordinates :

 

 



 

 

v

_t

− u

_x

= 0,

u

_t

+ p

_x

= (µ u

_x

v )

_x

, p = p(v ) = av

^−γ

.

Traveling wave solution (viscous shock wave):

(v, u) = (V, U)(x − st), (V, U)(±∞) = (v_±, u_±) It exists under the Rankine-Hugoniot and entropy conditions.

Known results (µ : a positive constant)

• Nishihara-M (1985)

∃C(v₋, γ) > 0 with C → ∞ as γ → 1 such that if

| v

₊

− v

₋

| ≤ C (v

₋

, γ ),

then (V, U)(x − st) is asymptotically stable for small initial pertur- bations with integral zero, that is,

∫

(v₀ − V )(x) dx =

∫

(u₀ − U)(x)dx = 0.

– For γ = 1, any large viscous shock wave is OK!.

– For γ > 1, a restriction on the amplitude is imposed.

• Mascia-Zumbrun(2004), Liu-Zeng(2009)

|v₊ − v₋| : suitably small =⇒ asymptotic stability

for general initial perturbations whose integrals are not necessarily zero.

• Barker-Humpherys-Laffite-Rudd-Zumbrun (2008), Humpherys-Laffite-Zumbrun (2010)

|v₊ − v₋| : suitably large =⇒ asymptotic stability

They also carried out numerical studies which indicate the asymptotic stability for intermediate amplitude as well.

Our present aim

Consider the case µ = µ(v) > 0.

In the Chapman-Enskog expansion theory in rarefied gas dynamics (cf.

Chapman-Cowling (1970)), the viscosity coefficient is given by a func- tion of the absolute temperature θ.

Typical two examples :

{ µ = ¯µ θ¹², Hard sphere Model, µ = ¯µ θ

1

2+ ²

(s−1), Cut-off inverse power force Model, where s (≥ 5) and ¯µ (> 0) are some constants.

The above two models are unified as

µ = ¯ µ θ

^β

(β ≥ 1

2 ).

Since our model is isentropic,

p = R θ

v = a v

^−γ

, (R : gas constant)

which implies

θ = a

R v

^−(γ−1)

.

Hence,

µ = µ

₀

v

^{−(γ−1)β}

(β ≥ 1

2 , µ

₀

= ¯ µ( a

R )

^β

).

Thus, we assume

(A) µ = µ(v ) = µ

₀

v

^−α

(α ≥ 1

2 (γ − 1), µ

₀

> 0).

Cauchy problem :

 

 



 

 

v

_t

− u

_x

= 0,

u

_t

+ p

_x

= µ

₀

( u

_x

v

^α+1

)

_x

, (α ≥

¹₂

(γ − 1)) p = p(v) = av

^−γ

(3)

with the initial and far field conditions

 



(v, u)(0, x) = (v

₀

, u

₀

)(x),

x→±∞

lim (v, u)(t, x) = (v

_±

, u

_±

). (4)

Assumptions on initial data

(v₀ − V, u₀ − U) ∈ H¹ ∩ L¹,

xinf∈Rv₀(x) > 0,

∫

(v₀ − V )(x) dx =

∫

(u₀ − U)(x) dx = 0.

Setting

ϕ₀(x) =

∫ _x

−∞

(v₀ − V )(y)dy, ψ₀(x) =

∫ _x

−∞

(u₀ − U)(y) dy, we further assume

(ϕ₀, ψ₀) ∈ L². (⇒ (ϕ₀, ψ₀) ∈ H²)

Theorem 2.

Suppose α ≥ ¹₂(γ − 1). Then, there exists a ε₀ > 0 such that if ∥ϕ₀, ψ₀∥² ≤ ε₀, the Cauchy problem (3),(4) has a unique global solution in time (v, u), satisfying (v − V, u − U) ∈ C([0, ∞); H¹) and

sup

x∈R

| (v, u)(x, t) − (V, U )(x − st) | → 0 (t → ∞ ).

Remarks

• In the proof, the essetial a priori estimate is given by a technical weighted energy method, “double step weighted energy method”, developed by Mei-M (1997) and Hashimoto-M (2007).

•

Open problem

Full system case :

µ = ¯µ θ^β, λ = ¯λ θ^β, κ = ¯κ θ^β (β ≥ 1 2).

Sketch of the proof of Theorem 1

Write x = (x, y) ∈ R² and assume ¯ρ = 1, µ + λ = 0 for simplicity.

Cauchy problem :

 

 

 

 



 

 

 

 

ρ

_t

+ (ρu

₁

)

_x

+ (ρu

₂

)

_y

= 0, u

_1t

+ (u

₁

u

_1x

+ u

₂

u

_1y

) + 1

ρ p

_x

− µ

ρ (u

_1xx

+ u

_1yy

) = 0, u

_2t

+ (u

₁

u

_2x

+ u

₂

u

_2y

) + 1

ρ p

_y

− µ

ρ (u

_2xx

+ u

_2yy

) = 0, p = aρ

with the initial data

(ρ, u₁, u₂)(0, x, y) = (ρ₀, u_1.0, u_2.0)(x, y), (x, y) ∈ R².

Change the unknown variables : (ρ, u₁, u₂) → (η, ϕ, ψ) ρ = (1 + η)

1 + t , u₁ = x

1 + t + ϕ, u₂ = ψ.

System for (η, ϕ, ψ) :

























η_t + x

1 + tη_x + ((1 + η)ϕ)_x + ((1 + η)ψ)_y = 0, ϕ_t + x

1 + tϕ_x + 1

1 + tϕ + ϕϕ_x + ψϕ_y + a

1 + ηη_x − µ(1 + t)

1 + η (ϕ_xx + ϕ_yy) = 0, ψ_t + x

1 + tψ_x + ϕψ_x + ψψ_y + a

1 + ηη_y − µ(1 + t)

1 + η (ψ_xx + ψ_yy) = 0, (η, ϕ, ψ)(0, x, y) = (η₀, ϕ₀, ψ₀)(x, y).

Characteristic curve w.r.t. x







dx(t)

dt = x(t) 1 + t, x(0) = ˜x

=⇒ x = x(t) = (1 + t)˜x.

Change of variable x : x = (1 + t)˜x

∂

∂x =⇒ 1 1 + t

∂

∂x˜, ∂

∂t + x 1 + t

∂

∂x =⇒ ∂

∂t























˜

η_t + ((1 + ˜η) ˜ϕ)_x˜

1 + t + ((1 + ˜η) ˜ψ)_y = 0, ϕ˜_t +

ϕ˜

1 + t +

ϕ˜ϕ˜_x˜

1 + t + ˜ψϕ˜_y + a 1 + t

˜ η_x˜

(1 + ˜η) − µ 1 + ˜η

( ϕ˜_x˜x˜

1 + t + (1 + t)ϕ˜_yy )

= 0, ψ˜_t +

ϕ˜ψ˜_x˜

1 + t + ˜ψψ˜_y + a

1 + ˜ηη˜_y − µ 1 + ˜η

( ψ˜_x˜x˜

1 + t + (1 + t)ψ˜_yy )

= 0.

Reformulated problem :

 

 

 

 

 

 

 

 

 

 

 



η

_t

+ 1

1 + t ϕ

_x

+ ψ

_y

= N

₀

, ϕ

_t

+ 1

1 + t ϕ + a

1 + t η

_x

− µ

( 1

1 + t ϕ

_xx

+ (1 + t)ϕ

_yy

)

= N

₁

, (5) ψ

_t

+ aη

_y

− µ

( 1

1 + t ψ

_xx

+ (1 + t)ψ

_yy

)

= N

₂

, (η, ϕ, ψ )(0) = (η

₀

, ϕ

₀

, ψ

₀

) ∈ H

²

.

We look for the global solution in time of (5) such that (η, ϕ, ψ) ∈ C([0, ∞); H²), sup

(x,y)∈R²

|(η, ϕ, ψ)(t, x, y)| ≤ C(1 + t)⁻¹².

E(t) := ∥(η, ϕ, ψ)(t)∥²₂+(1 + t)²∥(η, ϕ, ψ)_y(t)∥²₁+(1 + t)⁴∥(η, ϕ, ψ)_yy(t)∥². Proposition 3 (a priori estimate). Suppose a − µ > 0. Then there exist positive constants ε₀ and C₀ such that if (η, ϕ, ψ) ∈ C([0, T]; H²) is the solution of the Cauchy problem (5) for some T > 0 and sup

t∈[0,T]

E(t) ≤ ε₀ it holds that

∥(η, ϕ, ψ)(t)∥²₂ + (1 + t)²∥(η, ϕ, ψ)_y(t)∥²₁ + (1 + t)⁴∥(η, ϕ, ψ)_yy(t)∥² +

∫ _t

0

( 1

1 + τ ∥ϕ(τ)∥² + 1

1 + τ ∥(ϕ, ψ)_x(τ)∥²₂ + (1 + τ)∥(ϕ, ψ)_y(τ)∥²₂ )

dτ +

∫ _t

0

((1 + τ)³∥(ϕ, ψ)_yy(τ)∥²₁ + (1 + τ)⁵∥(ϕ, ψ)_yyy(τ)∥²)) dτ +

∫ _t

0

( 1

1 + τ ∥η_x(τ)∥²₁ + (1 + τ)∥η_y(t)∥²₁ + (1 + τ)³∥η_yy(τ)∥² )

dτ

≤ C₀∥(η₀, ϕ₀, ψ₀)∥²₂, t ∈ [0, T].

Decay estimates sup

(x,y)∈R²

|η(t, x, y)|² ≤

∫∫

|(2ηη_x)_y| dxdy

≤ C(∥η_x∥∥η_y∥ + ∥η∥∥η_xy∥)

≤ C(1 + t)⁻¹ which implies

sup

(x,y)∈R²

| ρ(t, x, y) − 1

1 + t | ≤ C(1 + t)⁻³²,

and similarly

sup

(x,y)∈R²

| u(t, x, y) − ( x

1 + t,0) | ≤ C(1 + t)⁻¹².

Basic energy estimate

Estimate (5) · (aη, ϕ, ψ) : 1

2 d dt

∫∫

(aη

²

+ ϕ

²

+ ψ

²

)dxdy +

∫∫ ( 1

1 + t ϕ

²

+ µ

1 + t (ϕ

²_x

+ ψ

_x²

) + µ(1 + t)(ϕ

²_y

+ ψ

_y²

) )

dxdy

=

∫∫

(aηN

₀

+ ϕN

₁

+ ψN

₂

) dxdy.

(6)

Estimate (5)

₂

· η

_x

+ (5)

₃

· (1 + t)η

_y

: d

dt

∫∫ (

ϕη

_x

+ (1 + t)ψη

_y

+ 1

2a ψ

²

+ µ 2

(

η

_x²

+ (1 + t)

²

η

_y²

))

dxdy +

∫∫ ( a

1 + t η

_x²

+ (a − µ)(1 + t)η

_y²

)

dxdy

−

∫∫ ( 1

1 + t ϕ

²_x

+ 2ϕ

_x

ψ

_y

+ (1 + t)ψ

_y²

)

dxdy +

∫∫ ( µ

a(1 + t) ψ

_x²

+ µ

a (1 + t)ψ

_y²

+ 1

1 + t ϕη

_x

)

dxdy (7)

=

∫∫ (

η

_x

N

₁

+ (1 + t)η

_y

N

₂

− ϕ

_x

N

₀

− (1 + t)ψ

_y

N

₀

+ 1

a ψN

₂

+ µ (

η

_x

N

_0x

+ (1 + t)

²

η

_y

N

_0y

))

dxdy.

By (5), we can estimate

∥(η, ϕ, ψ)(t)∥² +

∫ t 0

( 1

1 + τ ∥ϕ(τ)∥² + 1

1 + τ ∥(ϕ, ψ)_x(τ)∥² + (1 + τ)∥(ϕ, ψ)_y(τ)∥² )

dτ.

By combining (5) and (6), we can estimate

∫ t 0

( 1

1 + τ ∥η_x(τ)∥² + (1 + τ)∥η_y(t)∥² )

dτ.

Proceed the time-weighted estimates by

∂

_y

(5) · (1 + t)

²

(aη, ϕ, ψ )

_y

, ∂

_y²

(5) · (1 + t)

⁴

(aη, ϕ, ψ)

_yy

,

∂

_y

(5)

₂

· (1 + t)

²

η

_xy

+ ∂

_y

(5)

₃

· (1 + t)

³

η

_yy

.

Sketch of the proof of Theorem 2

Existence of viscous shock wave

Setting ξ = x − st, and (v, u) = (V, U)(ξ), we have the ODE system





−sV_ξ − U_ξ = 0,

−sU_ξ + p(V )_ξ = µ₀( U_ξ

V ^α+1)_ξ,

with (V, U)(±∞) = (v_±, u_±). We only treat the case s > 0.

Integrating the system, we have





sV + U = sv_± + u_±,

−sU + p(V ) − µ₀ U_ξ

V ^α+1 = −su_± + p(v_±), and the “Rankine-Hugoniot condition”

{−s(v₊ − v₋) − (u₊ − u₋) = 0,

−s(u₊ − u₋) + p(v₊) − p(v₋) = 0.

Equation of V

 



 

V

_ξ

= V

^α+1

µ

₀

s h(V ), ξ ∈ R, V ( ±∞ ) = v

_±

where

h(V ) := s²(v₋ − V ) + p(v₋) − p(V ) > 0, V ∈ (v_∓, v_±), and h(v_±) = 0. Under the “entropy condition”

v₋ < v₊, (i.e. u₋ > u₊),

the solution V uniquely exists up to the shift of ξ, satisfying V_ξ(ξ) > 0, v₋ < V (ξ) < v₊, ξ ∈ R,

and U is given by U = u_± + s(v_± − V ).

Change of the variables : (x, t) → (ξ = x − st, t)

 



v

_t

− sv

_ξ

− u

_ξ

= 0,

u

_t

− su

_ξ

+ p(v)

_ξ

= µ

₀

( u

_ξ

v

^α+1

)

_ξ

.

Change of the unknown variables : Define (ϕ, ψ)(t, ξ) by

ϕ(t, ξ) =

∫ _ξ

−∞

(v − V )(t, η)dη, ψ(t, ξ) =

∫ _ξ

−∞

(u − U)(t, η)dη, that is,

v = V + ϕ

_ξ

, u = U + ψ

_ξ

.

Cauchy problem for (ϕ, ψ) :

 



 

ϕ

_t

− sϕ

_ξ

− ψ

_ξ

= 0,

ψ

_t

− sψ

_ξ

+ p(V + ϕ

_ξ

) − p(V ) = µ

₀

( (U + ψ

_ξ

)

_ξ

(V + ϕ

_ξ

)

^α+1

− U

_ξ

V

^α+1

) ,

with the initial data

(ϕ, ψ)(0) = (ϕ

₀

, ψ

₀

) ∈ H

²

.

We look for the global solution in time such that (ϕ, ψ) ∈ C([0, ∞); H²), sup

x∈R |(ϕ, ψ)(t, x)| → 0 (t → ∞).

Proposition 4 (a priori estimate).

Suppose α ≥

¹₂

(γ − 1). Then there exist positive constants ε

₀

and C

₀

such that if (ϕ, ψ) ∈ C ([0, T ]; H

²

) is the solution of the Cauchy problem for some T > 0 and

sup

t∈[0,T]

∥ (ϕ, ψ)(t) ∥

2

≤ ε

₀

, it holds that for t ∈ [0, T ]

∥ (ϕ, ψ)(t) ∥

²₂

+

∫

_t

0

( ∥ ϕ

_ξ

(τ ) ∥

²₁

+ ∥ ψ

_ξ

(τ ) ∥

²₂

) dτ ≤ C

₀

∥ ϕ

₀

, ψ

₀

∥

²₂

.

A priori estimate

Rewrite the system as

 



ϕ

_t

− sϕ

_ξ

− ψ

_ξ

= 0,

ψ

_t

− sψ

_ξ

− K (V )ϕ

_ξ

− µ

₀

V

^α+1

ψ

_ξξ

= G, (5)

where

K(V ) = −p^′(V ) + (α + 1)h(V ) V , and

G = − {p(V + ϕ_ξ) − p(V ) − p^′(V )ϕ_ξ} + µ₀{ 1

(V + ϕ_ξ)^α+1 − 1

V ^α+1}ψ_ξξ + V ^α+1h(V ){− 1

(V + ϕ_ξ)^α+1 + 1

V ^α+1 − α + 1

V ^α+2 ϕ_ξ}.

Recall

s > 0, v₋ < V < v₊, V_ξ = V ^α+1

µ₀s h(V ) > 0, p^′(V ) = −γp(V ) V , which implies

K(V ) ≥ γ p(v₊) v₊ . Basic energy estimate

There exists a positive constant C such that it holds that for t ∈ [0, T]

|| (ϕ, ψ)(t) ||² +

∫ _t

0

(|| ψ_ξ(τ) ||²+ || √

V_ξψ(τ) ||²)dτ

≤ C{|| ϕ₀, ψ₀ ||² +

∫ t 0

∫

|ψ||G| dξdτ}.

Proof

We use a technical weighted energy method, “double step weighted energy method”, developed by Mei-M (1997) and Hashimoto-M (2007).

First-Step : Introduce the functions (transform functions) χ₁ = χ₁(V ) > 0, χ₂ = χ₂(V ) > 0,

and transform the unknown variables

ϕ = χ

₁

(V ) ˜ ϕ, ψ = χ

₂

(V ) ˜ ψ.

Then we have





(χ₁ϕ)˜ _t − s(χ₁ϕ)˜ _ξ − (χ₂ψ)˜ _ξ = 0,

(χ₂ψ)˜ _t − s(χ₂ψ)˜ _ξ − K(χ₁ϕ)˜ _ξ − µ₀

V ^α+1(χ₂ψ)˜ _ξξ = G.

Second-Step :

Introduce the another set of functions (weight functions) W₁ = W₁(V ) > 0, W₂ = W₂(V ) > 0.

Multiply the first equation by W₁ϕ, the second equation by˜ W₂ψ˜ and sum them up. Then we have

d dt

∫ 1

2(W₁χ₁ϕ˜² + W₂χ₂ψ˜²) dξ +

∫ s

2{(W₁^′χ₁ − W₁χ^′₁) ˜ϕ²+(W₂^′χ₂ − W₂χ^′₂) ˜ψ²}V_ξ dξ +

∫

{(W₁^′χ₂ − KW₂χ^′₁)V_ξϕ˜ψ˜ + (W₁χ₂ − KW₂χ₁) ˜ϕ_ξψ˜} dξ +

∫

µ₀( W₂ V ^α+1

ψ)˜ _ξ(χ₂ψ)˜ _ξ dξ =

∫

W₂ψG dξ.˜

In order for the coefficients of the cross terms ˜ϕψ˜ and ˜ϕ_ξψ˜ to banish, we impose that

W₁^′χ₂ − KW₂χ^′₁ = 0, W₁χ₂ − KW₂χ₁ = 0.

Choose χ₁, χ₂, W₁ and W₂ as

χ₁(V ) = W₁(V ) = 1, χ₂(V ) = K(V )W(V ), W₂(V ) = W(V ).

Then we have d

dt

∫ 1

2( ˜ϕ² + KW²ψ˜²)dξ +

∫

(−s

2K^′W²V_ξψ˜² + µ₀{( W

V ^α+1)^′(KW)^′V_ξ²ψ˜² + (KW²

V ^α+1 )^′V_ξψ˜ψ˜_ξ + KW²

V ^α+1 ( ˜ψ_ξ)²}) dξ =

∫

WψG dξ.˜

It further deduces d

dt

∫ 1

2( ˜ϕ² + KW²ψ˜²) dξ+

∫

(A(V )V_ξψ˜² + µ₀KW²

V ^α+1 ( ˜ψ_ξ)²)dξ

=

∫

WψG dξ,˜ where

A(V ) = −s

2K^′W² + 1

s( W

V ^α+1)^′(KW)^′V ^α+1h − 1

2s((KW²

V ^α+1 )^′V ^α+1h)^′.

We show the uniform positiveness of A(V ) for V ∈ [v

₋

, v

₊

].

Recall

p^′(V ) = −γp(V )

V , h^′(V ) = −s² + γp(V ) V . Representations of K, K^′ and K^′′ in terms of h and p :

K(V ) = γp(V )

V + (α + 1)h(V ) V , K^′(V ) = −γ(γ − α)p(V )

V ² − (α + 1)s²

V − (α + 1)h(V ) V ² , K^′′(V ) = γ((γ − α)(γ + 2) − (α + 1))p(V )

V ³ + 2(α + 1) s²

V ² + 2(α + 1)h(V ) V ³ .

Reformulate A(V ) as a quadratic form of h : A(V ) = W

2s(A₀(V ) + A₁(V )h(V ) + A₂(V )h²(V )), where

A₀(V ) =γ²(γ + 1)p²

V ³ W − 2(γ²p²

V ² − γs² p

V )W^′, A₁(V ) =(γ(γ + 2)(2α + 1 − γ) p

V ³ − 2s²(α + 1) V ² )W

− 2(γ(2α + 1 − γ) p

V ² − 2s²(α + 1)

V )W^′ − 2γ p

V W^′′, A₂(V ) =2(α + 1)(−W

V ³ + W^′

V ² − W^′′

V ).

Choice of W(V ) :

W(V ) = V. ( =⇒ A₂(V ) = 0) Final form of A(V ) :

A(V ) = 1

2s{2s²γp(V ) + γ²(γ − 1)p²(V ) V +2γ²(α − 1

2(γ − 1))p(V )

V h(V ) + 2(α + 1)s²h(V )}.

By the physical assumption α ≥ ¹₂(γ − 1),

A(V ) ≥ sγp(v

₊

), V ∈ [v

₋

, v

₊

].

Thus, integrating the inequality, we have

∥( ˜ϕ,ψ)(t)˜ ∥² +

∫ _t

0

(∥ψ˜_ξ(τ)∥² + ∥√

V_ξψ(τ˜ )∥²) dτ

≤ C{∥ϕ˜₀, ψ˜₀∥² +

∫ _t

0

∫

|ψ˜||G| dξdτ}.

Recalling the transformations ϕ = ˜ϕ, ψ = V K(V ) ˜ψ, the proof is thus completed.

Thank You !

A remark

We give a remark that our arguments in the previous sections can be extended to the models with more general viscosity coefficient. In the book of Chapman-Cowling, the viscosity coefficient is given in more general form as

µ = ¯µ θ¹²F (θ).

It is noted that F(θ) = 1 for the Hard sphere Model, and F(θ) = θ

2 (s−1)

for the Cut-off inverse power force Model. As a typical model whose F is not a power function of θ, the Sutherland’s model is well known where

F(θ) = 1 1 + s₀

θ

, (s₀ > 0 : Sutherland’s constant).

Even for the above general case, if we also take

χ₁(V ) = W₁(V ) = 1, χ₂(V ) = K(V )W(V ), W₂(V ) = W(V ) = V

as in the proof of the Lemma 3, then after long calculations we can have the following :

A(V ) = 1

2s{2s²γp(V ) + γ²(γ − 1)p²(V ) V +2s²h(V )

(γ + 1

2 + (γ − 1)θF ^′

F − (γ − 1)²θ(θF ^′

F )^′)

θ=(a/R)V^1−γ

+2γ(γ − 1)h(V )p(V ) V

(

γθF ^′

F + (γ − 1)θ(θF ^′

F )^′)

θ=(a/R)V^1−γ

−(γ − 1)²h²(V ) V

(

γθ(θF^′

F )^′ + (γ − 1)θ²(θF^′

F )^′′)

θ=(a/R)V¹⁻