Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion Boundary Layer problem: Navier-Stokes equations and Euler equations Nikolai Chemetov Joint work with F. Cipriano CMAF-UL 1 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion 1 Statement of the problem - Vanishing viscosity limit 2 First main result Estimates, independent of the viscosity Trace value of the vorticity Weak convergence 3 Second main result Estimates independent of the viscosity Strong convergence 4 Conclusion 2 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion Navier-Stokes equations. Dirichlet condition v t + div ( v ⊗ v ) − ▽ p = ν ∆ v in Ω T := Ω × (0 , T ) , div v = 0 in Ω T := Ω × (0 , T ) , v ( x , 0) = v 0 ( x ) , x ∈ Ω . Homogeneous Dirichlet boundary condition v = 0 on Γ T := Γ × (0 , T ) . 3 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion Bucur D., Feireisl E., Necasova S., Boundary Behavior of Viscous Fluids: influence of wall roughness ..., Arch. Rational Mech. Anal. , 197 , (2010). Priezjev N.V., Troian S.M., Influence of periodic wall roughness on the slip behavior at liquid/solid interfaces: ..., J. Fluid Mech. , 554 , (2006). 4 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion Navier-slip condition Flux of the fluid through the boundary Γ � v · n = a on Γ T with a d x = 0 , ∀ t ∈ [0 , T ] , Γ Navier´s slip boundary condition 2 D ( v )n · s + α v · s = β on Γ T , D ( v ) = 1 2 [ ∇ v + ( ∇ v ) T ] - the rate-of-strain tensor; n and s - the external normal and the tangent vector to Γ. α, β describe physical properties of Γ . 5 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion Vorticity formulation The Navier-Stokes equations in terms of the vorticity ω = ∂ x 1 v 2 − ∂ x 2 v 1 and the velocity v ∂ t ω + div ( v ω ) = ν ∆ ω, div v = 0 in Ω T , ω ( x , 0) = ω 0 ( x ) , x ∈ Ω with ω 0 := rot ( v 0 ) , v · n = a on Γ T . Navier slip boundary conditions in terms of the vorticity ω = b ( v ) on Γ T with b = β − 2 a ′ b ( v ) := ( α − 2 k ) v · s + b , s . 6 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion The Euler equations are equivalent to the system ∂ t ω + v ·∇ ω = 0 in Ω T , ω ( x , 0) = ω 0 ( x ) , x ∈ Ω with ω 0 := rot ( v 0 ) , v · n = a on Γ T . The trajectories for the transport equation start at t = 0 and on the inflow region Γ − , where a < 0. The Navier-slip boundary condition is given just on Γ − ω = b ( v ) . 7 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion Homogeneous case Navier-Stokes equations with v · n = 0 , ω = ( α − 2 k ) v · s on Γ Euler equations with v · n = 0 on Γ For α = 2 k : Lions P.-L., Math. Topics in Fluid Mechanics (1996) Beir˜ ao da Veiga H., Crispo F., J. Math. Fluid Mech., 13, (2011) For α � = 2 k : Clopeau T., Mikelic A., Robert R., Nonlinearity 11 , (1998) Lopes Filho M.C., Nussenzveig Lopes H.J., Planas G., SIAM Math. Anal. , 36 , 4, (2005) Kelliher J., SIAM J. Math. Anal. 38 , 1, (2006) 8 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion Non-Homogeneous case We consider the boundary conditions � v · n = a , ω = b ( v ) on Γ for the Navier-Stokes equations and � v · n = a on Γ , Γ − ω = b ( v ) on for the Euler equations. 9 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion Solvability results of the Euler equations In L p for any p ∈ (1 , ∞ ] : Chemetov N.V., Antontsev S.N., Physica D: Nonlinear Phenomena, 237 , 1, (2008). Chemetov N.V., Cipriano F.; Gavrilyuk, S., Math. Methods Appl. Sci. 33 , 6, (2010). 10 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion First main result Theorem 1 − 1 a ∈ L ∞ (0 , T ; W 1 p (Γ)) ∩ W 1 Let 2 (0 , T ; W 2 (Γ)) , 2 b ∈ W 2 , 1 ω 0 ∈ W 2 p (Γ T ) , p (Ω) , ∀ p ∈ (2 , + ∞ ] . ∃ a subsequence of { ω ν , v ν } , solutions of the N-S: � ω ν ⇀ ω weakly-* in L ∞ (0 , T ; L p (Ω)) , v ν → v strongly in C (Ω T ) , � � � ω ( ψ t + v · ∇ ψ ) d x dt + ω 0 ψ ( x , 0) d x = a b ( v ) ψ d x dt Ω T Ω Γ − T for every ψ ∈ C 1 , 1 (Ω T ) with ψ = 0 at t = 0 and on Γ + . 11 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion Sketch of the proof 1 st step : Estimates , independent of the viscosity Lemma ∃ a weak solution { ω ν , v ν } of the Navier-Stokes equations: || ω ν || L ∞ (0 , T ; L p (Ω)) � C , || v ν || L ∞ (0 , T ; W 1 p (Ω)) � C , || ∂ t v ν || L 2 (Ω T ) � C . 12 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion 2 important estimates: � t � v ν ( t ) � 2 � v 0 � 2 f ( s ) � ω ν ( s ) � 2 � � L 2 (Ω) � C L 2 (Ω) + L p (Ω) ds + 1 0 and � t � ω ν ( t ) � 2 L p (Ω) � C ( � v ν ( t ) � 2 f ( s ) � ω ν ( s ) � 2 L 2 (Ω) + L p (Ω) ds + 1) , 0 where the constants C and f ( t ) ∈ L 1 (0 , T ) are independent of ν . 13 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion 2 nd step : Trace value of the vorticity � Let B be an extension in Ω T of the initial condition B t =0 = ω 0 � � and boundary condition B Γ T = b ( v ν ) ; � d ( x )-the distance function from x ∈ Ω to Γ. Lemma � � 1 � | ω ν − B | | ( v ν ▽ d ) | − d x dt lim lim = 0 . σ σ → 0 + ν → 0 + Ω T ∩ [0 < d <σ ] 14 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion 3 d step : Weak convergence Taking ν → 0: � ω ν ⇀ ω weakly-* in L ∞ (0 , T ; L p (Ω)) , v ν → v strongly in C (Ω T ) , we will obtain the Euler equation � � � ω ( ψ t + v · ∇ ψ ) d x dt + ω 0 ψ ( x , 0) d x = a b ( v ) ψ d x dt Ω T Ω Γ − T 15 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion Second main result Theorem 2 a ∈ L ∞ (0 , T ; W 1 Let ω 0 ∈ L p (Ω) , p (Γ)) , − 1 b ∈ L 2 (0 , T ; W 1 p (Γ) ∩ W 1 p 2 (0 , T ; W (Γ)) , ∀ p ∈ (2 , + ∞ ] . p ∃ a subsequence of { ω ν , v ν } , solutions of N-S: � ω ν → ω strongly in L r (0 , T ; L p (Ω)) , ∀ r < ∞ , L r (0 , T ; W 1 v ν → v strongly in p (Ω)) , � � � � � β ( ω ) ψ t + v ·∇ ψ d x dt + β ( ω 0 ) ψ ( x , 0) d x = a β ( b ( v )) ψ d x dt Ω T Ω Γ − T holds for any β ∈ C ( R ) and any test function ψ. 16 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion Sketch of the proof Uniform estimates Lemma ∃ a weak solution { ω ν , v ν } of N-S: || ω ν || L ∞ (0 , T ; L p (Ω)) C , � || v ν || L ∞ (0 , T ; W 1 C , || ∂ t ( v ν − ∇ A ) || L 2 (Ω T ) � C , � p (Ω)) where A is the solution of the system � − ∆ A = 0 in Ω , ∂ A ∂ n = a on Γ 17 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion Strong convergence Malek J., Necas J., Rokyta M., Ruzicka M., Weak and measure-valued solutions to evolutionary PDEs. (1996). We denote h ( x ) = min { δ, d ( x ) } for any x ∈ Ω with a fixed δ > 0 . We consider the cut-off function ξ ν ( x ) := 1 − exp( − M + ν L h ( x )) , ∀ ν > 0 , ν with � v ν � L ∞ (Ω T ) � M , 18 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion Multiplying the Navier-Stokes equations by η ′ ( ω ν ) ξ ν ψ, with η ∈ C 2 ( R ) convex and ψ ∈ C ∞ 0 ( R 2+1 ) non-negative , we obtain � η ( ω ν ) [ ψ t + v ν · ∇ ψ + ν ∆ ψ ] ξ ν + 2 ν η ( ω ν ) ( ∇ ψ · ∇ ξ ν ) d x dt Ω T � � + η ( ω 0 ) ψ ( x , 0) ξ ν d x − η ( ω ( x , T )) ψ ( x , T ) ξ ν d x Ω Ω � + ( M + ν L ) η ( b ) ψ d x dt � 0 Γ T 19 / 24
Outline Statement of the problem - Vanishing viscosity limit First main result Second main result Conclusion v ν → v strongly in L ∞ (0 , T ; L 2 (Ω)) , | ω ν − c | + ⇀ z 1 , | ω ν − c | − ⇀ z 2 , ( ω ν − c ) ⇀ ω − c = z 1 − z 2 , weakly-* in L ∞ (0 , T ; L p (Ω)) , 20 / 24
Recommend
More recommend