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

(0) 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

(1) Introduction A model system of viscous gas  ρ t + ∇ · ( ρu ) = 0 ,     u t + u · ∇ u + 1 ρ ∆ u + µ + λ ρ ∇ p = µ  (1) ∇ ( ∇ · u ) , ρ    p = p ( ρ ) = aρ γ ,   where t ≥ 0 , x ∈ R n ( 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)

(2) Two papers : • Nishida-M (1980) , J. Math. Kyoto Univ., asymptotic stability of constant states in R 3 , R 2 • Nishihara-M (1985) , Japan J. Appl. Math., asymptotic stability of traveling waves in R 1 Topic 1. (joint work with T. Maeda) x ∈ R 2 or R 3 , ¯ • Nishida-M (1980), ρ > 0 ρ, u 0 ) ∈ H 3 , small = ( ρ 0 − ¯ ⇒ (¯ ρ, 0) is asymptotically stable. Since then, Hoff (¯ ρ > 0), Feireisl, Lions (¯ ρ = 0), . . . , Z.Xin-J.Li-X.Huang (¯ ρ ≥ 0), . . .

(3) Our present aim Consider the asymptotic stability of an unbounded state ¯ ρ u = U = ( x 1 ρ = P = 1 + t, 0) . 1 + t, Suppose x ∈ R 2 , p = aρ , and a − (2 µ + λ ) Theorem 1. > 0 . ¯ ρ Then, there exists a ε 0 > 0 such that if ∥ ρ 0 − ¯ ρ, u 0 − ( x 1 , 0) ∥ H 2 ≤ ε 0 , the Cauchy problem (1),(2) has a unique global solution in time ( ρ, u ) , satisfying ( ρ − P, u − U ) ∈ C ([0 , + ∞ ); H 2 ) and ¯ ρ 1 + t | ≤ C (1 + t ) − 3 sup | ρ ( t, x ) − 2 , x ∈ R 2 | u ( t, x ) − ( x 1 1 + t, 0) | ≤ C (1 + t ) − 1 sup 2 . x ∈ R 2

(4) Remarks • The proof is given by a combination of changing the variable x 1 along a characteristic curve and using a time-weighted energy method . • For R 3 , the similar results hold in H 3 with the asymptotics x ∈ R 3 | ρ − P | ≤ C (1 + t ) − 2 , x ∈ R 3 | u − U | ≤ C (1 + t ) − 1 . sup sup • Open problems – Isentropic case : p = aρ γ , ( γ > 1) R – Full system case : p = Rρθ, e = γ − 1 θ ¯ θ = 2 µ + λ ρ u = U = ( x 1 ρ = P = 1 + t, 1 + t, 0) , . R ¯ ρ

(5) Topic 2. (joint work with Yang Wang, MAA, 2010) Asymptotic stability of traveling wave solutions in R System in Lagrange coordinates :  v t − u x = 0 ,   u t + p x = ( µu 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.

(6) 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, ∫ ∫ ( v 0 − V )( x ) dx = ( u 0 − U )( x ) dx = 0 . – For γ = 1, any large viscous shock wave is OK!. – For γ > 1, a restriction on the amplitude is imposed.

(7) • 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.

(8) 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 : 1 { µ = ¯ Hard sphere Model , µ θ 2 , 1 2 2 + µ = ¯ Cut-off inverse power force Model , µ θ ( s − 1) , where s ( ≥ 5) and ¯ µ ( > 0) are some constants. The above two models are unified as ( β ≥ 1 µ θ β µ = ¯ 2) .

(9) Since our model is isentropic, p = Rθ v = a v − γ , ( R : gas constant ) which implies θ = a Rv − ( γ − 1) . Hence, ( β ≥ 1 µ ( a µ = µ 0 v − ( γ − 1) β R ) β ) . 2 , µ 0 = ¯ Thus, we assume ( α ≥ 1 µ = µ ( v ) = µ 0 v − α ( A ) 2( γ − 1) , µ 0 > 0) .

(10) Cauchy problem :  v t − u x = 0 ,   u t + p x = µ 0 ( u x  ( α ≥ 1  v α +1 ) x , 2 ( γ − 1)) (3)   p = p ( v ) = av − γ   with the initial and far field conditions  ( v, u )(0 , x ) = ( v 0 , u 0 )( x ) ,  (4) x →±∞ ( v, u )( t, x ) = ( v ± , u ± ) . lim 

(11) Assumptions on initial data ( v 0 − V, u 0 − U ) ∈ H 1 ∩ L 1 , x ∈ R v 0 ( x ) > 0 , inf ∫ ∫ ( v 0 − V )( x ) dx = ( u 0 − U )( x ) dx = 0 . Setting ∫ x ∫ x φ 0 ( x ) = ( v 0 − V )( y ) dy, ψ 0 ( x ) = ( u 0 − U )( y ) dy, −∞ −∞ we further assume ( φ 0 , ψ 0 ) ∈ L 2 . ( ⇒ ( φ 0 , ψ 0 ) ∈ H 2 )

(12) Suppose α ≥ 1 Theorem 2. 2 ( γ − 1) . Then, there exists a ε 0 > 0 such that if ∥ φ 0 , ψ 0 ∥ 2 ≤ ε 0 , the Cauchy problem (3),(4) has a unique global solution in time ( v, u ) , satisfying ( v − V, u − U ) ∈ C ([0 , ∞ ); H 1 ) and sup | ( v, u )( x, t ) − ( V, U )( x − st ) | → 0 ( t → ∞ ) . x ∈ R 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 : ( β ≥ 1 µ θ β , λ = ¯ λ θ β , κ = ¯ κ θ β µ = ¯ 2) .

(13) Sketch of the proof of Theorem 1 Write x = ( x, y ) ∈ R 2 and assume ¯ ρ = 1, µ + λ = 0 for simplicity. Cauchy problem :  ρ t + ( ρu 1 ) x + ( ρu 2 ) y = 0 ,    u 1 t + ( u 1 u 1 x + u 2 u 1 y ) + 1 ρp x − µ   ρ ( u 1 xx + u 1 yy ) = 0 ,    u 2 t + ( u 1 u 2 x + u 2 u 2 y ) + 1 ρp y − µ  ρ ( u 2 xx + u 2 yy ) = 0 ,       p = aρ  with the initial data ( x, y ) ∈ R 2 . ( ρ, u 1 , u 2 )(0 , x, y ) = ( ρ 0 , u 1 . 0 , u 2 . 0 )( x, y ) ,

(14) Change the unknown variables : ( ρ, u 1 , u 2 ) → ( η, φ, ψ ) ρ = (1 + η ) x u 1 = 1 + t + φ, u 2 = ψ. 1 + t , System for ( η, φ, ψ ) : x  η t + 1 + tη x + ((1 + η ) φ ) x + ((1 + η ) ψ ) y = 0 ,       1 1 + ηη x − µ (1 + t ) x a    φ t + 1 + tφ x + 1 + tφ + φφ x + ψφ y + 1 + η ( φ xx + φ yy ) = 0 ,   1 + ηη y − µ (1 + t ) x a   ψ t + 1 + tψ x + φψ x + ψψ y + 1 + η ( ψ xx + ψ yy ) = 0 ,         ( η, φ, ψ )(0 , x, y ) = ( η 0 , φ 0 , ψ 0 )( x, y ) . 

(15) Characteristic curve w.r.t. x  dx ( t ) = x ( t ) 1 + t,   dt = ⇒ x = x ( t ) = (1 + t )˜ x.  x (0) = ˜ x  Change of variable x : x = (1 + t )˜ x 1 ∂ ∂ ∂ x ∂ ⇒ ∂ ∂x = ⇒ x, ∂t + ∂x = 1 + t ∂ ˜ 1 + t ∂t  η ) ˜ η t + ((1 + ˜ φ ) ˜ x  η ) ˜ ˜ + ((1 + ˜ ψ ) y = 0 ,    1 + t ( ˜    ˜ φ ˜ ˜ )  ˜ φ φ ˜ a η ˜ µ φ ˜  x x x ˜ x ˜ 1 + t + ˜ ψ ˜ 1 + t + (1 + t ) ˜  φ t + 1 + t + φ y + η ) − = 0 , φ yy 1 + t (1 + ˜ 1 + ˜ η ( ˜   φ ˜ ˜ )  ψ ˜ a µ ψ ˜  x ˜ x x  ˜ 1 + t + ˜ ψ ˜ 1 + t + (1 + t ) ˜  ψ t + ψ y + η ˜ η y − = 0 . ψ yy   1 + ˜ 1 + ˜ η  

(16) Reformulated problem :  1 η t + 1 + tφ x + ψ y = N 0 ,        1 1 a ( )   φ t + 1 + tφ + 1 + tη x − µ 1 + tφ xx + (1 + t ) φ yy = N 1 , (5)   1 ( )   ψ t + aη y − µ 1 + tψ 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 ( x,y ) ∈ R 2 | ( η, φ, ψ )( t, x, y ) | ≤ C (1 + t ) − 1 ( η, φ, ψ ) ∈ C ([0 , ∞ ); H 2 ) , sup 2 .

(17) Proposition 3 ( a priori estimate). Suppose a − µ > 0 . Then there exist positive constants ε 0 and C 0 such that if ( η, φ, ψ ) ∈ C ([0 , T ]; H 2 ) is the solution of the Cauchy problem (5) for some T > 0 and sup ∥ ( η, φ, ψ )( t ) ∥ 2 ≤ ε 0 , t ∈ [0 ,T ] 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 ( ) 1 + τ ∥ φ ( τ ) ∥ 2 + 1 + τ ∥ ( φ, ψ ) x ( τ ) ∥ 2 2 + (1 + τ ) ∥ ( φ, ψ ) y ( τ ) ∥ 2 + dτ 2 0 ∫ t (1 + τ ) 3 ∥ ( φ, ψ ) yy ( τ ) ∥ 2 1 + (1 + τ ) 5 ∥ ( φ, ψ ) yyy ( τ ) ∥ 2 ) ( ) + dτ 0 ∫ t 1 ( ) 1 + τ ∥ η x ( τ ) ∥ 2 1 + (1 + τ ) ∥ η y ( t ) ∥ 2 1 + (1 + τ ) 3 ∥ η yy ( τ ) ∥ 2 + dτ 0 ≤ C 0 ∥ ( η 0 , φ 0 , ψ 0 ) ∥ 2 2

(18) Decay estimates ∫∫ ( x,y ) ∈ R 2 | η ( t, x, y ) | 2 ≤ sup | (2 ηη x ) y | dxdy ≤ C ( ∥ η x ∥∥ η y ∥ + ∥ η ∥∥ η xy ∥ ) ≤ C (1 + t ) − 1 which implies 1 1 + t | ≤ C (1 + t ) − 3 ( x,y ) ∈ R 2 | ρ ( t, x, y ) − sup 2 , and similarly x 1 + t, 0) | ≤ C (1 + t ) − 1 ( x,y ) ∈ R 2 | u ( t, x, y ) − ( sup 2 .