The Navier-Stokes-Fourier system on time dependent domain ◮ The stress tensor S is determined by the Newton rheological law x u − 2 � � ∇ x u + ∇ t S ( ϑ, ∇ x u ) = µ ( ϑ ) + ζ ( ϑ ) div x uI , µ > 0 , ζ � 0 3 div x uI where µ, ζ ∈ C 1 [ 0 , ∞ ) and satisfy | µ ′ ( ϑ ) | � m , 0 < µ ( 1 + ϑ ) � µ ( ϑ ) � µ ( 1 + ϑ ) , sup ϑ ∈ [ 0 , ∞ ) 0 � ζ ( ϑ ) � ζ ( 1 + ϑ ) .
The Navier-Stokes-Fourier system on time dependent domain ◮ The stress tensor S is determined by the Newton rheological law x u − 2 � � ∇ x u + ∇ t S ( ϑ, ∇ x u ) = µ ( ϑ ) + ζ ( ϑ ) div x uI , µ > 0 , ζ � 0 3 div x uI where µ, ζ ∈ C 1 [ 0 , ∞ ) and satisfy | µ ′ ( ϑ ) | � m , 0 < µ ( 1 + ϑ ) � µ ( ϑ ) � µ ( 1 + ϑ ) , sup ϑ ∈ [ 0 , ∞ ) 0 � ζ ( ϑ ) � ζ ( 1 + ϑ ) . ◮ The heat flux q is given by the Fourier law where 0 < κ ( ϑ ) ≈ 1 + ϑ + ϑ 3 q = − κ ( ϑ ) ∇ x ϑ, κ > 0 ,
The Navier-Stokes-Fourier system on time dependent domain The quantities p , e , and s are continuously differentiable functions for positive values of ̺ , ϑ and satisfy Gibbs’ equation � � 1 ϑ Ds ( ̺, ϑ ) = De ( ̺, ϑ ) + p ( ̺, ϑ ) D for all ̺, ϑ > 0 . ̺
The Navier-Stokes-Fourier system on time dependent domain The quantities p , e , and s are continuously differentiable functions for positive values of ̺ , ϑ and satisfy Gibbs’ equation � � 1 ϑ Ds ( ̺, ϑ ) = De ( ̺, ϑ ) + p ( ̺, ϑ ) D for all ̺, ϑ > 0 . ̺ Further, we assume the following state equation for the pressure and the internal energy and entropy p R ( ϑ ) = a 3 ϑ 4 , a > 0 , p ( ̺, ϑ ) = p M ( ̺, ϑ ) + p R ( ϑ ) ,
The Navier-Stokes-Fourier system on time dependent domain The quantities p , e , and s are continuously differentiable functions for positive values of ̺ , ϑ and satisfy Gibbs’ equation � � 1 ϑ Ds ( ̺, ϑ ) = De ( ̺, ϑ ) + p ( ̺, ϑ ) D for all ̺, ϑ > 0 . ̺ Further, we assume the following state equation for the pressure and the internal energy and entropy p R ( ϑ ) = a 3 ϑ 4 , a > 0 , p ( ̺, ϑ ) = p M ( ̺, ϑ ) + p R ( ϑ ) , ̺ e R ( ̺, ϑ ) = a ϑ 4 , e ( ̺, ϑ ) = e M ( ̺, ϑ ) + e R ( ̺, ϑ ) , and ̺ s R ( ̺, ϑ ) = 4 3 a ϑ 3 . s ( ̺, ϑ ) = s M ( ̺, ϑ ) + s R ( ̺, ϑ ) ,
Hypotheses
Hypotheses ∂ p M ∂̺ > 0 for all ̺, ϑ > 0 ◮ ◮ 0 < ∂ e M ∂ϑ � c for all ̺, ϑ > 0 .
Hypotheses ∂ p M ∂̺ > 0 for all ̺, ϑ > 0 ◮ ◮ 0 < ∂ e M ∂ϑ � c for all ̺, ϑ > 0 . ◮ lim ϑ → 0 + e M ( ̺, ϑ ) = e M ( ̺ ) > 0 for any fixed ̺ > 0 , � ̺ ∂ e M ( ̺,ϑ ) � � ce M ( ̺, ϑ ) for all ̺, ϑ > 0 . ◮ � � ∂̺
Hypotheses ∂ p M ∂̺ > 0 for all ̺, ϑ > 0 ◮ ◮ 0 < ∂ e M ∂ϑ � c for all ̺, ϑ > 0 . ◮ lim ϑ → 0 + e M ( ̺, ϑ ) = e M ( ̺ ) > 0 for any fixed ̺ > 0 , � ̺ ∂ e M ( ̺,ϑ ) � � ce M ( ̺, ϑ ) for all ̺, ϑ > 0 . ◮ � � ∂̺ ◮ There is a function P satisfying P ∈ C 1 [ 0 , ∞ ) , P ( 0 ) = 0 , P ′ ( 0 ) > 0 , such that � � ̺ 5 3 3 2 P 2 , or, ̺ > Z ϑ whenever 0 < ̺ � Z ϑ p M ( ̺, ϑ ) = ϑ 2 3 ϑ 2 where 0 < Z < Z
Hypotheses ∂ p M ∂̺ > 0 for all ̺, ϑ > 0 ◮ ◮ 0 < ∂ e M ∂ϑ � c for all ̺, ϑ > 0 . ◮ lim ϑ → 0 + e M ( ̺, ϑ ) = e M ( ̺ ) > 0 for any fixed ̺ > 0 , � ̺ ∂ e M ( ̺,ϑ ) � � ce M ( ̺, ϑ ) for all ̺, ϑ > 0 . ◮ � � ∂̺ ◮ There is a function P satisfying P ∈ C 1 [ 0 , ∞ ) , P ( 0 ) = 0 , P ′ ( 0 ) > 0 , such that � � ̺ 5 3 3 2 P 2 , or, ̺ > Z ϑ whenever 0 < ̺ � Z ϑ p M ( ̺, ϑ ) = ϑ 2 3 ϑ 2 where 0 < Z < Z ◮ what gives 5 3 � ϑ if ̺ < Z ϑ 5 2 2 3 � p M � c c ̺ 5 3 2 , if ̺ > Z ϑ ̺ 3
Hypotheses ∂ p M ∂̺ > 0 for all ̺, ϑ > 0 ◮ ◮ 0 < ∂ e M ∂ϑ � c for all ̺, ϑ > 0 . ◮ lim ϑ → 0 + e M ( ̺, ϑ ) = e M ( ̺ ) > 0 for any fixed ̺ > 0 , � ̺ ∂ e M ( ̺,ϑ ) � � ce M ( ̺, ϑ ) for all ̺, ϑ > 0 . ◮ � � ∂̺ ◮ There is a function P satisfying P ∈ C 1 [ 0 , ∞ ) , P ( 0 ) = 0 , P ′ ( 0 ) > 0 , such that � � ̺ 5 3 3 2 P 2 , or, ̺ > Z ϑ whenever 0 < ̺ � Z ϑ p M ( ̺, ϑ ) = ϑ 2 3 ϑ 2 where 0 < Z < Z ◮ what gives 5 3 � ϑ if ̺ < Z ϑ 5 2 2 3 � p M � c c ̺ 5 3 2 , if ̺ > Z ϑ ̺ 3 3 ◮ p M ( ̺, ϑ ) = 2 2 . 3 ̺ e M ( ̺, ϑ ) for ̺ > Z ϑ
Boundary conditions and initial data ◮ The impermeability of the boundary of the physical domain is described by the condition ( u − V ) · n | Γ τ = 0 for any τ � 0 , where n ( t , x ) denotes the unit outer normal vector to the boundary Γ t .
Boundary conditions and initial data ◮ The impermeability of the boundary of the physical domain is described by the condition ( u − V ) · n | Γ τ = 0 for any τ � 0 , where n ( t , x ) denotes the unit outer normal vector to the boundary Γ t . ◮ We assume a complete slip boundary conditions in the form [ S n ] × n = 0 .
Boundary conditions and initial data ◮ The impermeability of the boundary of the physical domain is described by the condition ( u − V ) · n | Γ τ = 0 for any τ � 0 , where n ( t , x ) denotes the unit outer normal vector to the boundary Γ t . ◮ We assume a complete slip boundary conditions in the form [ S n ] × n = 0 . ◮ For the heat flux we consider the conservative boundary conditions q · n = 0 for all t ∈ [ 0 , T ] , x ∈ Γ t .
Boundary conditions and initial data ◮ The impermeability of the boundary of the physical domain is described by the condition ( u − V ) · n | Γ τ = 0 for any τ � 0 , where n ( t , x ) denotes the unit outer normal vector to the boundary Γ t . ◮ We assume a complete slip boundary conditions in the form [ S n ] × n = 0 . ◮ For the heat flux we consider the conservative boundary conditions q · n = 0 for all t ∈ [ 0 , T ] , x ∈ Γ t . ◮ Our problem is supplemented by the initial conditions 5 3 � 0 , ̺ ( 0 , · ) = ̺ 0 ∈ L ( ̺ u )( 0 , · ) = ( ̺ u ) 0 , ( ̺ u ) 0 = 0 if ̺ 0 = 0 , ϑ ( 0 , · ) = ϑ 0 > 0 , ( ̺ s ) 0 = ̺ 0 s ( ̺ 0 , ϑ 0 ) ∈ L 1 (Ω 0 ) , � � � 1 2 ̺ 0 | ( ̺ u ) 0 | 2 + ̺ 0 e ( ̺ 0 , ϑ 0 ) E 0 = dx < ∞ . Ω 0
Weak formulation
Weak formulation The continuity equation is satisfied in weak and renormalized sense. � T � T � � � ̺ B ( ̺ )( ∂ t ϕ + u · ∇ x ϕ ) = b ( ̺ ) div x u ϕ − ̺ 0 B ( ̺ 0 ) ϕ ( 0 ) Ω t Ω t Ω 0 0 0 c ([ 0 , T ) × R 3 ) , b ∈ L ∞ ∩ C [ 0 , ∞ ) such that b ( 0 ) = 0 and for any ϕ ∈ C 1 B ( ̺ ) = B ( 1 ) + � ̺ b ( z ) z 2 d z . We suppose that ̺ � 0 a.e. in ( 0 , T ) × R 3 . 1
Weak formulation The continuity equation is satisfied in weak and renormalized sense. � T � T � � � ̺ B ( ̺ )( ∂ t ϕ + u · ∇ x ϕ ) = b ( ̺ ) div x u ϕ − ̺ 0 B ( ̺ 0 ) ϕ ( 0 ) Ω t Ω t Ω 0 0 0 c ([ 0 , T ) × R 3 ) , b ∈ L ∞ ∩ C [ 0 , ∞ ) such that b ( 0 ) = 0 and for any ϕ ∈ C 1 B ( ̺ ) = B ( 1 ) + � ̺ b ( z ) z 2 d z . We suppose that ̺ � 0 a.e. in ( 0 , T ) × R 3 . 1 The momentum equation is satisfied in a weak sense � T � ( ̺ u · ∂ t ϕ + ̺ [ u ⊗ u ] : ∇ x ϕ + p ( ̺, ϑ ) div x ϕ − S ( ϑ, ∇ x u ) : ∇ x ϕ ) Ω t 0 � = − ( ̺ u ) 0 · ϕ ( 0 , · ) , Ω 0 for any ϕ ∈ C 1 c ( Q T ; R 3 ) such that ϕ ( T , · ) = 0, ϕ · n | Γ τ = 0 for all τ ∈ [ 0 , T ]
Weak formulation The continuity equation is satisfied in weak and renormalized sense. � T � T � � � ̺ B ( ̺ )( ∂ t ϕ + u · ∇ x ϕ ) = b ( ̺ ) div x u ϕ − ̺ 0 B ( ̺ 0 ) ϕ ( 0 ) Ω t Ω t Ω 0 0 0 c ([ 0 , T ) × R 3 ) , b ∈ L ∞ ∩ C [ 0 , ∞ ) such that b ( 0 ) = 0 and for any ϕ ∈ C 1 B ( ̺ ) = B ( 1 ) + � ̺ b ( z ) z 2 d z . We suppose that ̺ � 0 a.e. in ( 0 , T ) × R 3 . 1 The momentum equation is satisfied in a weak sense � T � ( ̺ u · ∂ t ϕ + ̺ [ u ⊗ u ] : ∇ x ϕ + p ( ̺, ϑ ) div x ϕ − S ( ϑ, ∇ x u ) : ∇ x ϕ ) Ω t 0 � = − ( ̺ u ) 0 · ϕ ( 0 , · ) , Ω 0 for any ϕ ∈ C 1 c ( Q T ; R 3 ) such that ϕ ( T , · ) = 0, ϕ · n | Γ τ = 0 for all τ ∈ [ 0 , T ] In particular, the impermeability condition is satisfied in the sense of traces, specifically, u , ∇ x u ∈ L 2 ( Q T ; R 3 ) and ( u − V ) · n ( τ, · ) | Γ τ = 0 for a.a. τ ∈ [ 0 , T ] .
Weak formulation The entropy inequality � T � T � � κ ( ϑ ) ∇ x ϑ · ∇ x ϕ ̺ s ( ∂ t ϕ + u · ∇ x ϕ ) − ϑ 0 Ω t 0 Ω t � T � � � S : ∇ x u + κ ( ϑ ) |∇ x ϑ | 2 � ϕ + � − ( ̺ s ) 0 ϕ ( 0 ) ϑ ϑ Ω t Ω 0 0 holds for all ϕ ∈ C 1 c ( Q T ) such that ϕ ( T , · ) = 0 and ϕ � 0.
Weak formulation The entropy inequality � T � T � � κ ( ϑ ) ∇ x ϑ · ∇ x ϕ ̺ s ( ∂ t ϕ + u · ∇ x ϕ ) − ϑ 0 Ω t 0 Ω t � T � � � S : ∇ x u + κ ( ϑ ) |∇ x ϑ | 2 � ϕ + � − ( ̺ s ) 0 ϕ ( 0 ) ϑ ϑ Ω t Ω 0 0 holds for all ϕ ∈ C 1 c ( Q T ) such that ϕ ( T , · ) = 0 and ϕ � 0. The energy inequality for the case of a moving domain reads as � � � ( ̺ u ) 2 � � 1 1 2 ̺ | u | 2 + ̺ e � 0 ( τ, · ) � + ̺ 0 e 0 − ( ̺ u ) 0 · V ( 0 ) 2 ̺ 0 Ω τ Ω 0 � τ � � − ( ̺ ( u ⊗ u ) : ∇ x V + p div x V − S : ∇ x V + ̺ u · ∂ t V ) d x d t + ̺ u · V ( τ, · ) 0 Ω t Ω t for a.a. τ ∈ ( 0 , T ) .
Remark - weak formulation, the energy inequality In contrast to [Feireisl and Novotny, 2009] we consider energy inequality rather than energy equation. Although it seems that we are losing a lot of information our definition of weak solution is still sufficient. Namely, if the above defined weak solution is smooth enough it will be a strong one. (L.Poul, 2009)
Main result
Main result Definition We say that the trio ( ̺, u , ϑ ) is a variational solution of problem NSF with slip boundary conditions and initial conditions give above if 5 ◮ ̺ ∈ L ∞ ( 0 , T ; L 3 ( R 3 )) , ̺ � 0, ̺ ∈ L q ( Q T ) for certain q > 5 3 , ◮ u , ∇ x u ∈ L 2 ( Q T ) , ̺ u ∈ L ∞ ( 0 , T ; L 1 ( R 3 )) , ◮ ϑ > 0 a.a. on Q T , ϑ ∈ L ∞ ( 0 , T ; L 4 ( R 3 )) , ϑ, ∇ x ϑ ∈ L 2 ( Q T ) , and log ϑ, ∇ x log ϑ ∈ L 2 ( Q T ) , q ◮ ̺ s , ̺ s u , ϑ ∈ L 1 ( Q T ) , ◮ weak formulation is satisfied.
Main result Definition We say that the trio ( ̺, u , ϑ ) is a variational solution of problem NSF with slip boundary conditions and initial conditions give above if 5 ◮ ̺ ∈ L ∞ ( 0 , T ; L 3 ( R 3 )) , ̺ � 0, ̺ ∈ L q ( Q T ) for certain q > 5 3 , ◮ u , ∇ x u ∈ L 2 ( Q T ) , ̺ u ∈ L ∞ ( 0 , T ; L 1 ( R 3 )) , ◮ ϑ > 0 a.a. on Q T , ϑ ∈ L ∞ ( 0 , T ; L 4 ( R 3 )) , ϑ, ∇ x ϑ ∈ L 2 ( Q T ) , and log ϑ, ∇ x log ϑ ∈ L 2 ( Q T ) , q ◮ ̺ s , ̺ s u , ϑ ∈ L 1 ( Q T ) , ◮ weak formulation is satisfied. Theorem Let Ω 0 ⊂ R 3 be a bounded domain of class C 2 + ν with some ν > 0 , and let V ∈ C 1 ([ 0 , T ]; C 3 c ( R 3 ; R 3 )) be given. Let assumptions on V , p , e , s , µ , ζ , κ be satisfied. Let proper initial data be given. Then the problem NSF with slip boundary condition admits a variational solution in the sense of Definition 1 on any finite time interval ( 0 , T ) .
The strategy - our penalization scheme
The strategy - our penalization scheme ◮ Letus consider our problem on a large fixed time independent domain B ⊃ Ω τ for all τ ∈ [ 0 , T ]
The strategy - our penalization scheme ◮ Letus consider our problem on a large fixed time independent domain B ⊃ Ω τ for all τ ∈ [ 0 , T ] ◮ To deal with the slip boundary condition, we introduce to the weak formulation of the momentum equation a term � T � 1 ( u − V ) · n ϕ · n dS x d t ε 0 Γ t (originally proposed by Stokes and Carey)
The strategy - our penalization scheme ◮ Letus consider our problem on a large fixed time independent domain B ⊃ Ω τ for all τ ∈ [ 0 , T ] ◮ To deal with the slip boundary condition, we introduce to the weak formulation of the momentum equation a term � T � 1 ( u − V ) · n ϕ · n dS x d t ε 0 Γ t (originally proposed by Stokes and Carey) ◮ In the limit ε → 0 the above term yields the boundary condition ( u − V ) · n = 0 on Γ t and thus the large domain ( 0 , T ) × B becomes divided by an impermeable interface ∪ t ∈ ( 0 , T ) { t } × Γ t to a fluid domain Q T and a solid domain Q c T .
The strategy - our penalization scheme ◮ Letus consider our problem on a large fixed time independent domain B ⊃ Ω τ for all τ ∈ [ 0 , T ] ◮ To deal with the slip boundary condition, we introduce to the weak formulation of the momentum equation a term � T � 1 ( u − V ) · n ϕ · n dS x d t ε 0 Γ t (originally proposed by Stokes and Carey) ◮ In the limit ε → 0 the above term yields the boundary condition ( u − V ) · n = 0 on Γ t and thus the large domain ( 0 , T ) × B becomes divided by an impermeable interface ∪ t ∈ ( 0 , T ) { t } × Γ t to a fluid domain Q T and a solid domain Q c T . ◮ Then we need to find a way how to get rid of terms outside of a fluid domain.
Penalisation in terms outside of a fluid domain We introduce:
Penalisation in terms outside of a fluid domain We introduce: ◮ A variable shear viscosity coefficient µ = µ ω and bulk viscosity coefficient ζ = ζ ω µ ω ( ϑ, · ) ∈ C ∞ c , µ ω ( ϑ, τ, · ) | Ω τ = µ ( ϑ ) ζ ω ( ϑ, · ) ∈ C ∞ c , ζ ω ( ϑ, τ, · ) | Ω τ = η ( ϑ ) and µ ω , ζ ω → 0 a.e. in (( 0 , T ) × B ) \ Q T as ω → 0 .
Penalisation in terms outside of a fluid domain We introduce: ◮ A variable shear viscosity coefficient µ = µ ω and bulk viscosity coefficient ζ = ζ ω µ ω ( ϑ, · ) ∈ C ∞ c , µ ω ( ϑ, τ, · ) | Ω τ = µ ( ϑ ) ζ ω ( ϑ, · ) ∈ C ∞ c , ζ ω ( ϑ, τ, · ) | Ω τ = η ( ϑ ) and µ ω , ζ ω → 0 a.e. in (( 0 , T ) × B ) \ Q T as ω → 0 . ◮ A variable heat conductivity coefficient κ ν ( t , x , ϑ ) κ ν ( ϑ, t , x ) = χ ν ( t , x ) κ ( ϑ ) , where χ ν = 1 in Q T , χ ν = ν in (( 0 , T ) × B ) \ Q T
Penalisation in terms outside of a fluid domain We introduce: ◮ A variable shear viscosity coefficient µ = µ ω and bulk viscosity coefficient ζ = ζ ω µ ω ( ϑ, · ) ∈ C ∞ c , µ ω ( ϑ, τ, · ) | Ω τ = µ ( ϑ ) ζ ω ( ϑ, · ) ∈ C ∞ c , ζ ω ( ϑ, τ, · ) | Ω τ = η ( ϑ ) and µ ω , ζ ω → 0 a.e. in (( 0 , T ) × B ) \ Q T as ω → 0 . ◮ A variable heat conductivity coefficient κ ν ( t , x , ϑ ) κ ν ( ϑ, t , x ) = χ ν ( t , x ) κ ( ϑ ) , where χ ν = 1 in Q T , χ ν = ν in (( 0 , T ) × B ) \ Q T ◮ A variable coefficient a := a η ( t , x ) which represents the radiative part of pressure, internal energy and entropy. Namely, we assume that a η ( t , x ) = χ η ( t , x ) a , a > 0 , where χ η = 1 in Q T , χ η = η in (( 0 , T ) × B \ Q T .
Artificial pressure and extra temperature term We introduce:
Artificial pressure and extra temperature term We introduce: ◮ The artificial pressure p η,δ ( ̺, ϑ ) = p M ( ̺, ϑ ) + a η 3 ϑ 4 + δ̺ β , β � 4 , δ > 0
Artificial pressure and extra temperature term We introduce: ◮ The artificial pressure p η,δ ( ̺, ϑ ) = p M ( ̺, ϑ ) + a η 3 ϑ 4 + δ̺ β , β � 4 , δ > 0 ◮ An extra term in the energy inequality λϑ 5 and in the entropy inequality formulation λϑ 4
Existence of solutions to our new penalization on a fixed domain, uniform estimates ◮ There exists solution { ̺ ε , u ε , ϑ ε } ε to our penalized problem for each fixed ε > 0 and fixed η, ω, ν, λ, δ > 0.
Existence of solutions to our new penalization on a fixed domain, uniform estimates ◮ There exists solution { ̺ ε , u ε , ϑ ε } ε to our penalized problem for each fixed ε > 0 and fixed η, ω, ν, λ, δ > 0. ◮ Is based on the proof given by Feireisl, Novotn´ y (2009)
Existence of solutions to our new penalization on a fixed domain, uniform estimates ◮ There exists solution { ̺ ε , u ε , ϑ ε } ε to our penalized problem for each fixed ε > 0 and fixed η, ω, ν, λ, δ > 0. ◮ Is based on the proof given by Feireisl, Novotn´ y (2009) � τ ◮ The term 1 � Γ t (( u − V ) · n ϕ · n ) dS x d t and ε 0 � T 1 � Γ t ( u − V ) · n u · n ψ d S x d t can be treated as a ”compact” ε 0 perturbation.
Existence of solutions to our new penalization on a fixed domain, uniform estimates ◮ There exists solution { ̺ ε , u ε , ϑ ε } ε to our penalized problem for each fixed ε > 0 and fixed η, ω, ν, λ, δ > 0. ◮ Is based on the proof given by Feireisl, Novotn´ y (2009) � τ ◮ The term 1 � Γ t (( u − V ) · n ϕ · n ) dS x d t and ε 0 � T 1 � Γ t ( u − V ) · n u · n ψ d S x d t can be treated as a ”compact” ε 0 perturbation. ◮ On the level of the Galerkin approximation and strong solutions we need to adjust a proof to the case of variable coefficients µ , ζ , κ and a .
Uniform estimates on the set ( 0 , T ) × B given by the total dissipation inequality
Uniform estimates on the set ( 0 , T ) × B given by the total dissipation inequality � T � | ( u − V ) · n | 2 dS x d t � ε c ( λ ) 0 Γ t
Uniform estimates on the set ( 0 , T ) × B given by the total dissipation inequality � T � | ( u − V ) · n | 2 dS x d t � ε c ( λ ) 0 Γ t �√ ̺ u ( τ, · ) � L 2 � c ( λ ) � δ̺ β ( τ, · ) � L 1 + ess ess sup sup τ ∈ ( 0 , T ) τ ∈ ( 0 , T )
Uniform estimates on the set ( 0 , T ) × B given by the total dissipation inequality � T � | ( u − V ) · n | 2 dS x d t � ε c ( λ ) 0 Γ t �√ ̺ u ( τ, · ) � L 2 � c ( λ ) � δ̺ β ( τ, · ) � L 1 + ess ess sup sup τ ∈ ( 0 , T ) τ ∈ ( 0 , T ) � � λϑ 5 � L 1 � c ( λ ) �
Uniform estimates on the set ( 0 , T ) × B given by the total dissipation inequality � T � | ( u − V ) · n | 2 dS x d t � ε c ( λ ) 0 Γ t �√ ̺ u ( τ, · ) � L 2 � c ( λ ) � δ̺ β ( τ, · ) � L 1 + ess ess sup sup τ ∈ ( 0 , T ) τ ∈ ( 0 , T ) � � λϑ 5 � L 1 � c ( λ ) � � u � L 2 + �∇ x u � L 2 � c ( λ, ω )
Uniform estimates on the set ( 0 , T ) × B given by the total dissipation inequality � T � | ( u − V ) · n | 2 dS x d t � ε c ( λ ) 0 Γ t �√ ̺ u ( τ, · ) � L 2 � c ( λ ) � δ̺ β ( τ, · ) � L 1 + ess ess sup sup τ ∈ ( 0 , T ) τ ∈ ( 0 , T ) � λϑ 5 � � L 1 � c ( λ ) � � u � L 2 + �∇ x u � L 2 � c ( λ, ω ) � a η ϑ 4 ( τ, · ) � L 1 + ess ess sup sup � ̺ ( τ, · ) � 3 � c ( λ ) 5 L τ ∈ ( 0 , T ) τ ∈ ( 0 , T )
Uniform estimates on the set ( 0 , T ) × B given by the total dissipation inequality � T � | ( u − V ) · n | 2 dS x d t � ε c ( λ ) 0 Γ t �√ ̺ u ( τ, · ) � L 2 � c ( λ ) � δ̺ β ( τ, · ) � L 1 + ess ess sup sup τ ∈ ( 0 , T ) τ ∈ ( 0 , T ) � � λϑ 5 � L 1 � c ( λ ) � � u � L 2 + �∇ x u � L 2 � c ( λ, ω ) � a η ϑ 4 ( τ, · ) � L 1 + ess ess sup sup � ̺ ( τ, · ) � 3 � c ( λ ) 5 L τ ∈ ( 0 , T ) τ ∈ ( 0 , T ) � T � � 2 | 2 � |∇ x log ( ϑ ) | 2 + |∇ x ϑ 3 d x d t � c ( λ ) 0 B
Uniform estimates on the set ( 0 , T ) × B given by the total dissipation inequality � T � | ( u − V ) · n | 2 dS x d t � ε c ( λ ) 0 Γ t �√ ̺ u ( τ, · ) � L 2 � c ( λ ) � δ̺ β ( τ, · ) � L 1 + ess ess sup sup τ ∈ ( 0 , T ) τ ∈ ( 0 , T ) � � λϑ 5 � L 1 � c ( λ ) � � u � L 2 + �∇ x u � L 2 � c ( λ, ω ) � a η ϑ 4 ( τ, · ) � L 1 + ess ess sup sup � ̺ ( τ, · ) � 3 � c ( λ ) 5 L τ ∈ ( 0 , T ) τ ∈ ( 0 , T ) � T � � 2 | 2 � |∇ x log ( ϑ ) | 2 + |∇ x ϑ 3 d x d t � c ( λ ) 0 B � � κ ν ( ϑ ) � � � ̺ s ( ̺, ϑ ) � L q + � ̺ s ( ̺, ϑ ) u � L q + ∇ x ϑ + � ̺ e ( ̺, ϑ ) � L 1 � c ( λ ) with some q > 1 � � ϑ � � L q
Passing with ε → 0
Passing with ε → 0 ◮ First of all, directly from uniform estimates we derive that ( u − V ) · n ( τ, · ) | Γ τ = 0 for a.a. τ ∈ [ 0 , T ] . in the limit as ε → 0.
Passing with ε → 0 ◮ First of all, directly from uniform estimates we derive that ( u − V ) · n ( τ, · ) | Γ τ = 0 for a.a. τ ∈ [ 0 , T ] . in the limit as ε → 0. ◮ More demanding (but already known strategy): ◮ ϑ ε → ϑ a.a. in ( 0 , T ) × B . ◮ ̺ ε → ̺ a.a. in ( 0 , T ) × B . ◮ Div-Curl Lemma, Young measure theory; effective viscous pressure, oscillations defect measure
Passing with ε → 0 - delicate issues
Passing with ε → 0 - delicate issues ◮ We have at hand only the local estimates on the pressure, � � � p ( ̺, ϑ ) ̺ π + δ̺ β + π � d x d t � c ( K ) K for certain π > 0 and for any compact K ⊂ (( 0 , T ) × B ) such that � � �� K ∩ ∪ τ ∈ [ 0 , T ] { τ } × Γ τ = ∅ ,
Passing with ε → 0 - delicate issues ◮ We have at hand only the local estimates on the pressure, � � � p ( ̺, ϑ ) ̺ π + δ̺ β + π � d x d t � c ( K ) K for certain π > 0 and for any compact K ⊂ (( 0 , T ) × B ) such that � � �� K ∩ ∪ τ ∈ [ 0 , T ] { τ } × Γ τ = ∅ , ◮ We have to restrict ourselves to the class of test functions for a momentum equation ϕ ∈ C 1 ([ 0 , T ); W 1 , ∞ ( B ; R 3 )) , supp [ div x ϕ ( τ, · )] ∩ Γ τ = ∅ , 0 ϕ · n | Γ τ = 0 for all τ ∈ [ 0 , T ] .
Passing with ε → 0 - delicate issues ◮ We have at hand only the local estimates on the pressure, � � � p ( ̺, ϑ ) ̺ π + δ̺ β + π � d x d t � c ( K ) K for certain π > 0 and for any compact K ⊂ (( 0 , T ) × B ) such that � � �� K ∩ ∪ τ ∈ [ 0 , T ] { τ } × Γ τ = ∅ , ◮ We have to restrict ourselves to the class of test functions for a momentum equation ϕ ∈ C 1 ([ 0 , T ); W 1 , ∞ ( B ; R 3 )) , supp [ div x ϕ ( τ, · )] ∩ Γ τ = ∅ , 0 ϕ · n | Γ τ = 0 for all τ ∈ [ 0 , T ] . ◮ But this can be extended to the class ϕ ∈ C ∞ c ([ 0 , T ] × B ; R 3 ) , ϕ ( τ, · ) · n | Γ τ = 0 for any τ ∈ [ 0 , T ] .
Passing with ε → 0 - delicate issues ◮ Internal energy and ”pressure” terms in the energy inequality
Passing with ε → 0 - delicate issues ◮ Internal energy and ”pressure” terms in the energy inequality ◮ To pass to the limit in p η,δ ( ̺ ε , ϑ ε ) div x V ψ we need to assume that div x V = 0 in the neighbourhood of Γ t
Passing with ε → 0 - delicate issues ◮ Internal energy and ”pressure” terms in the energy inequality ◮ To pass to the limit in p η,δ ( ̺ ε , ϑ ε ) div x V ψ we need to assume that div x V = 0 in the neighbourhood of Γ t ◮ We do not have uniform integrability of the sequence { ̺ ε e η ( ̺ ε , ϑ ε ) } ε> 0
Passing with ε → 0 - delicate issues ◮ Internal energy and ”pressure” terms in the energy inequality ◮ To pass to the limit in p η,δ ( ̺ ε , ϑ ε ) div x V ψ we need to assume that div x V = 0 in the neighbourhood of Γ t ◮ We do not have uniform integrability of the sequence { ̺ ε e η ( ̺ ε , ϑ ε ) } ε> 0 ◮ Since the sequence { ̺ ε e η ( ̺ ε , ϑ ε ) } ε is nonnegative, the sequence is integrable, density and temperature converges a.e., by the Fatou lemma we deduce � T � T � � lim sup ̺ ε e η ( ̺ ε , ϑ ε ) ∂ t ψ d x d t � ̺ e η ( ̺, ϑ ) ∂ t ψ d x d t 0 B 0 B as far as ψ ∈ C 1 c ([ 0 , T )) and ∂ t ψ � 0.
Vanishing density outside of fluid part In order to get rid of the density dependent terms supported by the ”solid” part (( 0 , T ) × B ) \ Q T we use result of [Feireisl, Kreml, Necasova, Neustupa, Stebel, JDE, 2013] which reads as
Vanishing density outside of fluid part In order to get rid of the density dependent terms supported by the ”solid” part (( 0 , T ) × B ) \ Q T we use result of [Feireisl, Kreml, Necasova, Neustupa, Stebel, JDE, 2013] which reads as Lemma Let ̺ ∈ L ∞ ( 0 , T ; L 2 ( B )) , ̺ � 0 , u ∈ L 2 ( 0 , T ; W 1 , 2 ( B ; R 3 )) be a weak solution 0 of the equation of continuity, specifically, � τ � � � � � � ̺ ( τ, · ) ϕ ( τ, · ) − ̺ 0 ϕ ( 0 , · ) ̺∂ t ϕ + ̺ u · ∇ x ϕ d x = d x d t B 0 B for any τ ∈ [ 0 , T ] and any test function ϕ ∈ C 1 c ([ 0 , T ] × R 3 ) . In addition, assume that ( u − V )( τ, · ) · n | Γ τ = 0 for a.a. τ ∈ ( 0 , T ) , and that ̺ 0 ∈ L 2 ( R 3 ) , ̺ 0 � 0 , ̺ 0 | B \ Ω 0 = 0 . Then ̺ ( τ, · ) | B \ Ω τ = 0 for any τ ∈ [ 0 , T ] .
The limit system ε → 0 and with ̺ 0 = 0 on B \ Ω 0
The limit system ε → 0 and with ̺ 0 = 0 on B \ Ω 0 The continuity equation reads � T � T � � � ̺ B ( ̺ )( ∂ t ϕ + u · ∇ x ϕ ) = b ( ̺ ) div x u ϕ − ̺ 0 ,δ B ( ̺ 0 ,δ ) ϕ ( 0 ) 0 Ω t 0 Ω t Ω 0 c ([ 0 , T ) × R 3 ) , and any b ∈ L ∞ ∩ C [ 0 , ∞ ) such that b ( 0 ) = 0 for any ϕ ∈ C 1 and B ( ̺ ) = B ( 1 ) + � ̺ b ( z ) z 2 d z . 1
The limit system ε → 0 and with ̺ 0 = 0 on B \ Ω 0 The continuity equation reads � T � T � � � ̺ B ( ̺ )( ∂ t ϕ + u · ∇ x ϕ ) = b ( ̺ ) div x u ϕ − ̺ 0 ,δ B ( ̺ 0 ,δ ) ϕ ( 0 ) 0 Ω t 0 Ω t Ω 0 c ([ 0 , T ) × R 3 ) , and any b ∈ L ∞ ∩ C [ 0 , ∞ ) such that b ( 0 ) = 0 for any ϕ ∈ C 1 and B ( ̺ ) = B ( 1 ) + � ̺ b ( z ) z 2 d z . 1 Next the momentum equation reduces to � T � ( ̺ u · ∂ t ϕ + ̺ [ u ⊗ u ] : ∇ x ϕ + p η,δ ( ̺, ϑ ) div x ϕ − S ω ( ϑ, ∇ x u ) : ∇ x ϕ ) 0 Ω t � T � T � � � a η 3 ϑ 4 div x ϕ = − ( ̺ u ) 0 ,δ · ϕ ( 0 , · )+ S ω ( ϑ, ∇ x u ) : ∇ ϕ − Ω 0 0 B \ Ω t 0 B \ Ω t for any ϕ ∈ C ∞ c ([ 0 , T ] × B ; R 3 ) , ϕ ( τ, · ) · n | Γ τ = 0 for any τ ∈ [ 0 , T ] .
The limit system ε → 0 and with ̺ 0 = 0 on B \ Ω 0 The balance of entropy takes the following form � T � T � � 4 3 a η ϑ 3 ( ∂ t ϕ + u · ∇ x ϕ ) ̺ s ( ̺, ϑ )( ∂ t ϕ + u · ∇ x ϕ )+ 0 Ω t 0 B \ Ω t � T � T � � κ ν ( ϑ ) ∇ x ϑ · ∇ x ϕ κ ν ( ϑ ) ∇ x ϑ · ∇ x ϕ − − ϑ ϑ Ω t B \ Ω t 0 0 � T � � � S ω : ∇ x u + κ ν ( ϑ ) |∇ x ϑ | 2 ϕ + ϑ ϑ Ω t 0 � T � � S ω : ∇ x u + κ ν ( ϑ ) |∇ x ϑ | 2 � ϕ + ϑ ϑ 0 B \ Ω t � T � � � 4 λϑ 4 ϕ � − 3 a η ϑ 3 − ( ̺ s ) 0 ,δ ϕ ( 0 ) − 0 ,δ ϕ ( 0 ) Ω 0 B \ Ω 0 0 B for all ϕ ∈ C 1 c ([ 0 , T ) × B ) , ϕ � 0.
The limit system ε → 0 and with ̺ 0 = 0 on B \ Ω 0 The total energy balance reads � T � T � T � � � � � 1 δ 2 ̺ | u | 2 + ̺ e ( ̺, ϑ ) + β − 1 ̺ β a η ϑ 4 ∂ t ψ − λϑ 5 ψ ∂ t ψ + 0 Ω t 0 B \ Ω t 0 B ( ̺ u ) 2 � � � 1 δ 0 ,δ β − 1 ̺ β � − + ̺ 0 ,δ e 0 ,δ + 0 ,δ − ( ̺ u ) 0 ,δ · V ( 0 , · ) ψ ( 0 ) 2 ̺ 0 ,δ Ω 0 � a η ϑ 4 − 0 ,δ ψ ( 0 ) B \ Ω 0 � T � − ( S ω : ∇ x V ψ − ̺ u · ∂ t ( V ψ ) − ̺ ( u ⊗ u ) : ∇ x V ψ − p δ ( ̺, ϑ ) div x V ψ ) Ω t 0 � τ � S ω : ∇ x V − 1 � � 3 a η ϑ 4 div x V − ψ B \ Ω t 0 for all ψ ∈ C 1 c ([ 0 , T )) , ∂ t ψ � 0.
Passing with η → 0
Passing with η → 0 ◮ Let us denote by { ̺ η , u η , ϑ η } η> 0 solutions to the system obtained as ε → 0
Passing with η → 0 ◮ Let us denote by { ̺ η , u η , ϑ η } η> 0 solutions to the system obtained as ε → 0 ◮ Let a η = η a on B \ Ω t
Passing with η → 0 ◮ Let us denote by { ̺ η , u η , ϑ η } η> 0 solutions to the system obtained as ε → 0 ◮ Let a η = η a on B \ Ω t ◮ Since we have � � λϑ 5 � L 1 � c ( λ ) , � u η � L 2 ( 0 , T ; W 1 , 2 ( B )) � c η �
Passing with η → 0 ◮ Let us denote by { ̺ η , u η , ϑ η } η> 0 solutions to the system obtained as ε → 0 ◮ Let a η = η a on B \ Ω t ◮ Since we have � � λϑ 5 � L 1 � c ( λ ) , � u η � L 2 ( 0 , T ; W 1 , 2 ( B )) � c η � ◮ for η → 0 we get � T � 1 3 a η ϑ 4 η div x ϕ → 0 0 B \ Ω t � T � 1 3 a η ϑ 4 η div x V → 0 0 B \ Ω t � T � 4 3 a η ϑ 3 η ∂ t ϕ → 0 B \ Ω t 0 � T � 4 3 a η ϑ 3 η u η · ∇ x ϕ → 0 B \ Ω t 0
Passing with η → 0
Passing with η → 0 ◮ Since ϑ η → ϑ weakly in L 1 (( 0 , T ) × B ) , we obtain � T � T � � λϑ 5 d x d t � lim inf λϑ 5 η d x d t . η → 0 0 B 0 B
Passing with η → 0 ◮ Since ϑ η → ϑ weakly in L 1 (( 0 , T ) × B ) , we obtain � T � T � � λϑ 5 d x d t � lim inf λϑ 5 η d x d t . η → 0 0 B 0 B ◮ In the energy inequality we need to restrict ourself to test functions ψ ∈ C 1 c ([ 0 , T )) , ψ � 0 , ∂ t ψ � 0
Passing with η → 0 ◮ Since ϑ η → ϑ weakly in L 1 (( 0 , T ) × B ) , we obtain � T � T � � λϑ 5 d x d t � lim inf λϑ 5 η d x d t . η → 0 0 B 0 B ◮ In the energy inequality we need to restrict ourself to test functions ψ ∈ C 1 c ([ 0 , T )) , ψ � 0 , ∂ t ψ � 0 ◮ To pass to the limit in remaining terms outside of the fluid part and in all terms in the fluid part we use the same arguments as for passing with ε → 0.
Passing with ω → 0 We would like to get rid of terms related to the viscous stress tensor outside of the fluid domain.
Passing with ω → 0 We would like to get rid of terms related to the viscous stress tensor outside of the fluid domain. ◮ Let us denote by { ̺ ω , u ω , ϑ ω } ω> 0 solutions to the system obtained as η → 0.
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