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

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Some Results by Energy Methods on Large-Time Behavior of Viscous Gas Akitaka Matsumura (Osaka University) Taipei, October 29 - November 2, 2012

2012 International Conference on Nonlinear Analysis Evolutionary P.D.E. and Kinetic Theory

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Introduction

A model system of viscous gas

           ρt + ∇ · (ρu) = 0, ut + u · ∇u + 1 ρ∇p = µ ρ∆u + µ + λ ρ ∇(∇ · u), p = p(ρ) = aργ, (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) = (ρ0, u0). (2)

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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, u0) ∈ H3, small = ⇒ (¯ ρ, 0) is asymptotically stable. Since then, Hoff (¯ ρ > 0), Feireisl, Lions (¯ ρ = 0), . . . , Z.Xin-J.Li-X.Huang (¯ ρ ≥ 0), . . .

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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 ∈ R2, p = aρ, and a − (2µ + λ)

¯ ρ > 0.

Then, there exists a ε0 > 0 such that if ∥ρ0−¯ ρ, u0−(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)−3

2,

sup

x∈R2

| u(t, x) − ( x1 1 + t, 0) | ≤ C(1 + t)−1

2.

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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 = aργ, (γ > 1) – Full system case : p = Rρθ, e =

R γ−1θ

ρ = P = ¯ ρ 1 + t, u = U = ( x1 1 + t, 0), θ = 2µ + λ R¯ ρ .

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Topic 2.

(joint work with Yang Wang, MAA, 2010) Asymptotic stability of traveling wave solutions in R System in Lagrange coordinates :

         vt − ux = 0, ut + px = (µux 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.

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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.

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• 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.

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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 θ

1 2,

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 θβ (β ≥ 1 2).

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Since our model is isentropic,

p = Rθ v = a v−γ, (R : gas constant)

which implies

θ = a Rv−(γ−1).

Hence,

µ = µ0 v−(γ−1)β (β ≥ 1 2, µ0 = ¯ µ( a R)β).

Thus, we assume

(A) µ = µ(v) = µ0 v−α (α ≥ 1 2(γ − 1), µ0 > 0).

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Cauchy problem :

         vt − ux = 0, ut + px = µ0( ux vα+1)x, (α ≥ 1

2(γ − 1))

p = p(v) = av−γ (3)

with the initial and far field conditions

   (v, u)(0, x) = (v0, u0)(x), lim

x→±∞(v, u)(t, x) = (v±, u±).

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Assumptions on initial data (v0 − V, u0 − U) ∈ H1 ∩ L1, inf

x∈R v0(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)

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Theorem 2.

Suppose α ≥ 1

2(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 θβ, λ = ¯ λ θβ, κ = ¯ κ θβ (β ≥ 1 2).

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Sketch of the proof of Theorem 1

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

                 ρt + (ρu1)x + (ρu2)y = 0, u1t + (u1u1x + u2u1y) + 1 ρpx − µ ρ(u1xx + u1yy) = 0, u2t + (u1u2x + u2u2y) + 1 ρpy − µ ρ(u2xx + u2yy) = 0, p = aρ

with the initial data (ρ, u1, u2)(0, x, y) = (ρ0, u1.0, u2.0)(x, y), (x, y) ∈ R2.

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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) :                        ηt + x 1 + tηx + ((1 + η)φ)x + ((1 + η)ψ)y = 0, φt + x 1 + tφx + 1 1 + tφ + φφx + ψφy + a 1 + ηηx − µ(1 + t) 1 + η (φxx + φyy) = 0, ψt + x 1 + tψx + φψx + ψψy + a 1 + ηηy − µ(1 + t) 1 + η (ψxx + ψyy) = 0, (η, φ, ψ)(0, x, y) = (η0, φ0, ψ0)(x, y).

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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                      ˜ ηt + ((1 + ˜ η) ˜ φ)˜

x

1 + t + ((1 + ˜ η) ˜ ψ)y = 0, ˜ φt + ˜ φ 1 + t + ˜ φ ˜ φ˜

x

1 + t + ˜ ψ ˜ φy + a 1 + t ˜ η˜

x

(1 + ˜ η) − µ 1 + ˜ η ( ˜ φ˜

x˜ x

1 + t + (1 + t) ˜ φyy ) = 0, ˜ ψt + ˜ φ ˜ ψ˜

x

1 + t + ˜ ψ ˜ ψy + a 1 + ˜ η ˜ ηy − µ 1 + ˜ η ( ˜ ψ˜

x˜ x

1 + t + (1 + t) ˜ ψyy ) = 0.

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Reformulated problem :

                       ηt + 1 1 + tφx + ψy = N0, φt + 1 1 + tφ + a 1 + tηx − µ ( 1 1 + tφxx + (1 + t)φyy ) = N1, (5) ψt + aηy − µ ( 1 1 + tψxx + (1 + t)ψyy ) = N2, (η, φ, ψ)(0) = (η0, φ0, ψ0) ∈ H2.

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)−1

2.

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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

• f the Cauchy problem (5) for some T > 0 and

sup

t∈[0,T ]

∥(η, φ, ψ)(t)∥2 ≤ ε0, it holds that for t ∈ [0, T ] ∥(η, φ, ψ)(t)∥2

2 + (1 + t)2∥(η, φ, ψ)y(t)∥2 1 + (1 + t)4∥(η, φ, ψ)yy(t)∥2

+ ∫ t ( 1 1 + τ ∥φ(τ)∥2 + 1 1 + τ ∥(φ, ψ)x(τ)∥2

2 + (1 + τ)∥(φ, ψ)y(τ)∥2 2

) dτ + ∫ t ( (1 + τ)3∥(φ, ψ)yy(τ)∥2

1 + (1 + τ)5∥(φ, ψ)yyy(τ)∥2)

) dτ + ∫ t ( 1 1 + τ ∥ηx(τ)∥2

1 + (1 + τ)∥ηy(t)∥2 1 + (1 + τ)3∥ηyy(τ)∥2

) dτ ≤ C0∥(η0, φ0, ψ0)∥2

2

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Decay estimates sup

(x,y)∈R2 |η(t, x, y)|2 ≤

∫∫ |(2ηηx)y| dxdy ≤ C(∥ηx∥∥ηy∥ + ∥η∥∥ηxy∥) ≤ C(1 + t)−1 which implies sup

(x,y)∈R2 | ρ(t, x, y) −

1 1 + t | ≤ C(1 + t)−3

2,

and similarly sup

(x,y)∈R2 | u(t, x, y) − (

x 1 + t, 0) | ≤ C(1 + t)−1

2.

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Basic energy estimate

Estimate (5) · (aη, φ, ψ) : 1 2 d dt ∫∫ (aη2 + φ2 + ψ2)dxdy + ∫∫ ( 1 1 + tφ2 + µ 1 + t(φ2

x + ψ2 x) + µ(1 + t)(φ2 y + ψ2 y)

) dxdy = ∫∫ (aηN0 + φN1 + ψN2) dxdy. (6)

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Estimate (5)2 · ηx + (5)3 · (1 + t)ηy : d dt ∫∫ ( φηx + (1 + t)ψηy + 1 2aψ2 + µ 2 ( η2

x + (1 + t)2η2 y

)) dxdy + ∫∫ ( a 1 + tη2

x + (a − µ)(1 + t)η2 y

) dxdy − ∫∫ ( 1 1 + tφ2

x + 2φxψy + (1 + t)ψ2 y

) dxdy + ∫∫ ( µ a(1 + t)ψ2

x + µ

a(1 + t)ψ2

y +

1 1 + tφηx ) dxdy (7) = ∫∫ ( ηxN1 + (1 + t)ηyN2 − φxN0 − (1 + t)ψyN0 +1 aψN2 + µ ( ηxN0x + (1 + t)2ηyN0y )) dxdy.

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By (5), we can estimate ∥(η, φ, ψ)(t)∥2 + ∫ t ( 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 ( 1 1 + τ ∥ηx(τ)∥2 + (1 + τ)∥ηy(t)∥2 ) dτ. Proceed the time-weighted estimates by

∂y(5) · (1 + t)2(aη, φ, ψ)y, ∂2

y(5) · (1 + t)4(aη, φ, ψ)yy,

∂y(5)2 · (1 + t)2ηxy + ∂y(5)3 · (1 + t)3ηyy.

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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    −sVξ − Uξ = 0, −sUξ + p(V )ξ = µ0( Uξ 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 Uξ 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.

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Equation of V

     Vξ = V α+1 µ0s 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 Vξ(ξ) > 0, v− < V (ξ) < v+, ξ ∈ R, and U is given by U = u± + s(v± − V ).

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

   vt − svξ − uξ = 0, ut − suξ + p(v)ξ = µ0( uξ vα+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 + φξ, u = U + ψξ.

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Cauchy problem for (φ, ψ) :

     φt − sφξ − ψξ = 0, ψt − sψξ + p(V + φξ) − p(V ) = µ0 ( (U + ψξ)ξ (V + φξ)α+1 − Uξ V α+1 ) ,

with the initial data

(φ, ψ)(0) = (φ0, ψ0) ∈ H2.

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

x∈R

|(φ, ψ)(t, x)| → 0 (t → ∞).

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Proposition 4 (a priori estimate). Suppose α ≥ 1

2(γ − 1).

Then there exist positive constants ε0 and C0 such that if (φ, ψ) ∈ C([0, T ]; H2) is the solution of the Cauchy problem for some T > 0 and sup

t∈[0,T ]

∥(φ, ψ)(t)∥2 ≤ ε0, it holds that for t ∈ [0, T ] ∥(φ, ψ)(t)∥2

2 +

∫ t (∥φξ(τ)∥2

1 + ∥ψξ(τ)∥2 2) dτ ≤ C0∥φ0, ψ0∥2 2.

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A priori estimate

Rewrite the system as

   φt − sφξ − ψξ = 0, ψt − sψξ − K(V )φξ − µ0 V α+1ψξξ = G, (5)

where K(V ) = −p′(V ) + (α + 1)h(V ) V , and G = − {p(V + φξ) − p(V ) − p′(V )φξ} + µ0{ 1 (V + φξ)α+1 − 1 V α+1}ψξξ + V α+1h(V ){− 1 (V + φξ)α+1 + 1 V α+1 − α + 1 V α+2 φξ}.

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Recall s > 0, v− < V < v+, Vξ = 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 (|| ψξ(τ) ||2+ || √ Vξψ(τ) ||2) dτ ≤ C{|| φ0, ψ0 ||2 + ∫ t ∫ |ψ||G| dξdτ}.

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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 = χ1(V ) ˜ φ, ψ = χ2(V ) ˜ ψ.

Then we have    (χ1 ˜ φ)t − s(χ1 ˜ φ)ξ − (χ2 ˜ ψ)ξ = 0, (χ2 ˜ ψ)t − s(χ2 ˜ ψ)ξ − K(χ1 ˜ φ)ξ − µ0 V α+1(χ2 ˜ ψ)ξξ = G.

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Second-Step : Introduce the another set of functions (weight functions) W1 = W1(V ) > 0, W2 = W2(V ) > 0. Multiply the first equation by W1 ˜ φ, the second equation by W2 ˜ ψ and sum them up. Then we have d dt ∫ 1 2(W1χ1 ˜ φ2 + W2χ2 ˜ ψ2) dξ + ∫ s 2{(W ′

1χ1 − W1χ′ 1) ˜

φ2+(W ′

2χ2 − W2χ′ 2) ˜

ψ2}Vξ dξ + ∫ {(W ′

1χ2 − KW2χ′ 1)Vξ ˜

φ ˜ ψ + (W1χ2 − KW2χ1) ˜ φξ ˜ ψ} dξ + ∫ µ0( W2 V α+1 ˜ ψ)ξ(χ2 ˜ ψ)ξ dξ = ∫ W2 ˜ ψG dξ.

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In order for the coefficients of the cross terms ˜ φ ˜ ψ and ˜ φξ ˜ ψ to banish, we impose that W ′

1χ2 − 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 + KW 2 ˜ ψ2) dξ + ∫ (−s 2K′W 2Vξ ˜ ψ2 + µ0{( W V α+1)′(KW )′V 2

ξ ˜

ψ2 + (KW 2 V α+1 )′Vξ ˜ ψ ˜ ψξ + KW 2 V α+1 ( ˜ ψξ)2}) dξ = ∫ W ˜ ψG dξ.

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It further deduces d dt ∫ 1 2( ˜ φ2 + KW 2 ˜ ψ2) dξ+ ∫ (A(V )Vξ ˜ ψ2 + µ0 KW 2 V α+1 ( ˜ ψξ)2) dξ = ∫ W ˜ ψG dξ, where A(V ) = −s 2K′W 2 + 1 s( W V α+1)′(KW )′V α+1h − 1 2s((KW 2 V α+1 )′V α+1h)′.

We show the uniform positiveness of A(V ) for V ∈ [v−, v+].

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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 .

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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 ).

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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 ≥ 1

2(γ − 1),

A(V ) ≥ sγp(v+), V ∈ [v−, v+].

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Thus, integrating the inequality, we have ∥( ˜ φ, ˜ ψ)(t)∥2 + ∫ t (∥ ˜ ψξ(τ)∥2 + ∥ √ Vξ ˜ ψ(τ)∥2) dτ ≤ C{∥ ˜ φ0, ˜ ψ0∥2 + ∫ t ∫ | ˜ ψ||G| dξdτ}. Recalling the transformations φ = ˜ φ, ψ = V K(V ) ˜ ψ, the proof is thus completed.

Thank You !