Flow of heat conducting fluid in a time dependent domain Aneta - - PowerPoint PPT Presentation

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Flow of heat conducting fluid in a time dependent domain Aneta - - PowerPoint PPT Presentation

Flow of heat conducting fluid in a time dependent domain Aneta Wrblewska-Kamiska Department of Mathematics, Imperial College London Institute of Mathematics, Polish Academy of Sciences acha, and joint work with Ond rej Kreml, Vaclav


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SLIDE 1

Flow of heat conducting fluid in a time dependent domain

Aneta Wróblewska-Kamińska

Department of Mathematics, Imperial College London Institute of Mathematics, Polish Academy of Sciences

joint work with Ondˇ rej Kreml, Vaclav M´ acha, and ˇ S´ arka Neˇ casov´ a

MathFlows18, Porquerolles, September 2018 Supported by the Royal Society within the International Newton Fellowship

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SLIDE 2

Some motivations

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SLIDE 3

Some motivations

Real life motivation:

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SLIDE 4

Some motivations

Real life motivation:

◮ Motion of a piston in a cylinder filled with a heat conducting gas

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SLIDE 5

Some motivations

Real life motivation:

◮ Motion of a piston in a cylinder filled with a heat conducting gas

Mathematical development - existence of weak solutions for compressible fluids:

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SLIDE 6

Some motivations

Real life motivation:

◮ Motion of a piston in a cylinder filled with a heat conducting gas

Mathematical development - existence of weak solutions for compressible fluids:

◮ A barotropic Navier-Stokes system on a fixed domain – P.-L. Lions

(1998), E. Feireisl, H. Petzetov´ a, A. Novotn´ y (2001)

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SLIDE 7

Some motivations

Real life motivation:

◮ Motion of a piston in a cylinder filled with a heat conducting gas

Mathematical development - existence of weak solutions for compressible fluids:

◮ A barotropic Navier-Stokes system on a fixed domain – P.-L. Lions

(1998), E. Feireisl, H. Petzetov´ a, A. Novotn´ y (2001)

◮ Navier-Stokes-Fourier system with thermal energy equation – E. Feireisl

(2004)

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SLIDE 8

Some motivations

Real life motivation:

◮ Motion of a piston in a cylinder filled with a heat conducting gas

Mathematical development - existence of weak solutions for compressible fluids:

◮ A barotropic Navier-Stokes system on a fixed domain – P.-L. Lions

(1998), E. Feireisl, H. Petzetov´ a, A. Novotn´ y (2001)

◮ Navier-Stokes-Fourier system with thermal energy equation – E. Feireisl

(2004)

◮ Navier-Stokes-Fourier system with entropy inequality – E. Feireisl,

  • A. Novotn´

y (2009)

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SLIDE 9

Some motivations

Real life motivation:

◮ Motion of a piston in a cylinder filled with a heat conducting gas

Mathematical development - existence of weak solutions for compressible fluids:

◮ A barotropic Navier-Stokes system on a fixed domain – P.-L. Lions

(1998), E. Feireisl, H. Petzetov´ a, A. Novotn´ y (2001)

◮ Navier-Stokes-Fourier system with thermal energy equation – E. Feireisl

(2004)

◮ Navier-Stokes-Fourier system with entropy inequality – E. Feireisl,

  • A. Novotn´

y (2009)

◮ In a time dependent domain, barotropic case, no-slip boundary

conditions - E. Feireisl, J. Neustupa, J. Stebel - the Brinkman penelization method (2011)

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SLIDE 10

Some motivations

Real life motivation:

◮ Motion of a piston in a cylinder filled with a heat conducting gas

Mathematical development - existence of weak solutions for compressible fluids:

◮ A barotropic Navier-Stokes system on a fixed domain – P.-L. Lions

(1998), E. Feireisl, H. Petzetov´ a, A. Novotn´ y (2001)

◮ Navier-Stokes-Fourier system with thermal energy equation – E. Feireisl

(2004)

◮ Navier-Stokes-Fourier system with entropy inequality – E. Feireisl,

  • A. Novotn´

y (2009)

◮ In a time dependent domain, barotropic case, no-slip boundary

conditions - E. Feireisl, J. Neustupa, J. Stebel - the Brinkman penelization method (2011)

◮ Barotropic N-S system with a slip b.c. – E. Feireisl, O. Kreml,

ˇ

  • S. Neˇ

casov´ a, J. Neustupa, J. Stebel (2013)

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SLIDE 11

Some motivations

Real life motivation:

◮ Motion of a piston in a cylinder filled with a heat conducting gas

Mathematical development - existence of weak solutions for compressible fluids:

◮ A barotropic Navier-Stokes system on a fixed domain – P.-L. Lions

(1998), E. Feireisl, H. Petzetov´ a, A. Novotn´ y (2001)

◮ Navier-Stokes-Fourier system with thermal energy equation – E. Feireisl

(2004)

◮ Navier-Stokes-Fourier system with entropy inequality – E. Feireisl,

  • A. Novotn´

y (2009)

◮ In a time dependent domain, barotropic case, no-slip boundary

conditions - E. Feireisl, J. Neustupa, J. Stebel - the Brinkman penelization method (2011)

◮ Barotropic N-S system with a slip b.c. – E. Feireisl, O. Kreml,

ˇ

  • S. Neˇ

casov´ a, J. Neustupa, J. Stebel (2013)

◮ Navier-Stokes-Fourier system with thermal energy equation, slip b.c. –

  • O. Kreml, V. M´

acha, ˇ

  • S. Neˇ

casova, A. W-K (2017)

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SLIDE 12

The Navier-Stokes-Fourier system on time dependent domain

∂t̺ + divx(̺u) = 0, ∂t(̺u) + divx(̺u ⊗ u) + ∇xp(̺, ϑ) = divxS(∇xu), ∂t(̺s) + divx(̺su) + divx

q

ϑ

  • = σ,

d dt ̺|u|2 + ̺e dx = 0.

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The Navier-Stokes-Fourier system on time dependent domain

∂t̺ + divx(̺u) = 0, ∂t(̺u) + divx(̺u ⊗ u) + ∇xp(̺, ϑ) = divxS(∇xu), ∂t(̺s) + divx(̺su) + divx

q

ϑ

  • = σ,

d dt ̺|u|2 + ̺e dx = 0.

◮ ̺ stands for the density of the fluid, u - the velocity, ϑ - the temperature

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The Navier-Stokes-Fourier system on time dependent domain

∂t̺ + divx(̺u) = 0, ∂t(̺u) + divx(̺u ⊗ u) + ∇xp(̺, ϑ) = divxS(∇xu), ∂t(̺s) + divx(̺su) + divx

q

ϑ

  • = σ,

d dt ̺|u|2 + ̺e dx = 0.

◮ ̺ stands for the density of the fluid, u - the velocity, ϑ - the temperature ◮ p - a pressure, s - an internal entropy, e - an internal energy, S - a viscous

stress tensor, q - a heat flux

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The Navier-Stokes-Fourier system on time dependent domain

∂t̺ + divx(̺u) = 0, ∂t(̺u) + divx(̺u ⊗ u) + ∇xp(̺, ϑ) = divxS(∇xu), ∂t(̺s) + divx(̺su) + divx

q

ϑ

  • = σ,

d dt ̺|u|2 + ̺e dx = 0.

◮ ̺ stands for the density of the fluid, u - the velocity, ϑ - the temperature ◮ p - a pressure, s - an internal entropy, e - an internal energy, S - a viscous

stress tensor, q - a heat flux

◮ σ - an entropy production rate

σ 1 ϑ

  • S : ∇xu − q

ϑ · ∇xϑ

  • .
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SLIDE 16

The Navier-Stokes-Fourier system on time dependent domain

◮ We study the above system on a moving domain Ω = Ωt

The boundary of the domain Ωt occupied by the fluid is described by means of a given (regular) velocity field V(t, x), where t ∈ [0, T] and x ∈ R3,

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The Navier-Stokes-Fourier system on time dependent domain

◮ We study the above system on a moving domain Ω = Ωt

The boundary of the domain Ωt occupied by the fluid is described by means of a given (regular) velocity field V(t, x), where t ∈ [0, T] and x ∈ R3,

d dt X(t, x) = V

  • t, X(t, x)
  • , t > 0, X(0, x) = x,
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The Navier-Stokes-Fourier system on time dependent domain

◮ We study the above system on a moving domain Ω = Ωt

The boundary of the domain Ωt occupied by the fluid is described by means of a given (regular) velocity field V(t, x), where t ∈ [0, T] and x ∈ R3,

d dt X(t, x) = V

  • t, X(t, x)
  • , t > 0, X(0, x) = x,

Ωτ = X (τ, Ω0) , where Ω0 ⊂ R3 is a given domain, Γτ = ∂Ωτ, and Qτ = ∪t∈(0,τ){t} × Ωt.

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SLIDE 19

The Navier-Stokes-Fourier system on time dependent domain

◮ We study the above system on a moving domain Ω = Ωt

The boundary of the domain Ωt occupied by the fluid is described by means of a given (regular) velocity field V(t, x), where t ∈ [0, T] and x ∈ R3,

d dt X(t, x) = V

  • t, X(t, x)
  • , t > 0, X(0, x) = x,

Ωτ = X (τ, Ω0) , where Ω0 ⊂ R3 is a given domain, Γτ = ∂Ωτ, and Qτ = ∪t∈(0,τ){t} × Ωt.

◮ We assume that the volume of the domain can not degenerate in time

there exists V0 > 0 such that |Ωτ| V0 for all τ ∈ [0, T].

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The Navier-Stokes-Fourier system on time dependent domain

◮ We study the above system on a moving domain Ω = Ωt

The boundary of the domain Ωt occupied by the fluid is described by means of a given (regular) velocity field V(t, x), where t ∈ [0, T] and x ∈ R3,

d dt X(t, x) = V

  • t, X(t, x)
  • , t > 0, X(0, x) = x,

Ωτ = X (τ, Ω0) , where Ω0 ⊂ R3 is a given domain, Γτ = ∂Ωτ, and Qτ = ∪t∈(0,τ){t} × Ωt.

◮ We assume that the volume of the domain can not degenerate in time

there exists V0 > 0 such that |Ωτ| V0 for all τ ∈ [0, T].

◮ We assume that divxV = 0 on the neighborhood of Γτ,

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The Navier-Stokes-Fourier system on time dependent domain

◮ The stress tensor S is determined by the Newton rheological law

S(ϑ, ∇xu) = µ(ϑ)

  • ∇xu + ∇t

xu − 2

3divxuI

  • + ζ(ϑ)divxuI, µ > 0, ζ 0
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The Navier-Stokes-Fourier system on time dependent domain

◮ The stress tensor S is determined by the Newton rheological law

S(ϑ, ∇xu) = µ(ϑ)

  • ∇xu + ∇t

xu − 2

3divxuI

  • + ζ(ϑ)divxuI, µ > 0, ζ 0

where µ, ζ ∈ C 1[0, ∞) and satisfy 0 < µ(1 + ϑ) µ(ϑ) µ(1 + ϑ), sup

ϑ∈[0,∞)

|µ′(ϑ)| m,

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The Navier-Stokes-Fourier system on time dependent domain

◮ The stress tensor S is determined by the Newton rheological law

S(ϑ, ∇xu) = µ(ϑ)

  • ∇xu + ∇t

xu − 2

3divxuI

  • + ζ(ϑ)divxuI, µ > 0, ζ 0

where µ, ζ ∈ C 1[0, ∞) and satisfy 0 < µ(1 + ϑ) µ(ϑ) µ(1 + ϑ), sup

ϑ∈[0,∞)

|µ′(ϑ)| m, 0 ζ(ϑ) ζ(1 + ϑ).

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The Navier-Stokes-Fourier system on time dependent domain

◮ The stress tensor S is determined by the Newton rheological law

S(ϑ, ∇xu) = µ(ϑ)

  • ∇xu + ∇t

xu − 2

3divxuI

  • + ζ(ϑ)divxuI, µ > 0, ζ 0

where µ, ζ ∈ C 1[0, ∞) and satisfy 0 < µ(1 + ϑ) µ(ϑ) µ(1 + ϑ), sup

ϑ∈[0,∞)

|µ′(ϑ)| m, 0 ζ(ϑ) ζ(1 + ϑ).

◮ The heat flux q is given by the Fourier law

q = −κ(ϑ)∇xϑ, κ > 0, where 0 < κ(ϑ) ≈ 1 + ϑ + ϑ3

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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 ϑDs(̺, ϑ) = De(̺, ϑ) + p(̺, ϑ)D

  • 1

̺

  • for all ̺, ϑ > 0.
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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 ϑDs(̺, ϑ) = De(̺, ϑ) + p(̺, ϑ)D

  • 1

̺

  • for all ̺, ϑ > 0.

Further, we assume the following state equation for the pressure and the internal energy and entropy p(̺, ϑ) = pM(̺, ϑ) + pR(ϑ), pR(ϑ) = a 3ϑ4, a > 0,

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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 ϑDs(̺, ϑ) = De(̺, ϑ) + p(̺, ϑ)D

  • 1

̺

  • for all ̺, ϑ > 0.

Further, we assume the following state equation for the pressure and the internal energy and entropy p(̺, ϑ) = pM(̺, ϑ) + pR(ϑ), pR(ϑ) = a 3ϑ4, a > 0, e(̺, ϑ) = eM(̺, ϑ) + eR(̺, ϑ), ̺eR(̺, ϑ) = aϑ4, and s(̺, ϑ) = sM(̺, ϑ) + sR(̺, ϑ), ̺sR(̺, ϑ) = 4 3aϑ3.

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SLIDE 28

Hypotheses

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SLIDE 29

Hypotheses

∂pM ∂̺ > 0 for all ̺, ϑ > 0

◮ 0 < ∂eM

∂ϑ c for all ̺, ϑ > 0.

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SLIDE 30

Hypotheses

∂pM ∂̺ > 0 for all ̺, ϑ > 0

◮ 0 < ∂eM

∂ϑ c for all ̺, ϑ > 0.

◮ limϑ→0+ eM(̺, ϑ) = eM(̺) > 0 for any fixed ̺ > 0, ◮

̺ ∂eM(̺,ϑ)

∂̺

  • ceM(̺, ϑ) for all ̺, ϑ > 0.
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SLIDE 31

Hypotheses

∂pM ∂̺ > 0 for all ̺, ϑ > 0

◮ 0 < ∂eM

∂ϑ c for all ̺, ϑ > 0.

◮ limϑ→0+ eM(̺, ϑ) = eM(̺) > 0 for any fixed ̺ > 0, ◮

̺ ∂eM(̺,ϑ)

∂̺

  • ceM(̺, ϑ) for all ̺, ϑ > 0.

◮ There is a function P satisfying P ∈ C 1[0, ∞), P(0) = 0, P′(0) > 0,

such that pM(̺, ϑ) = ϑ

5 2 P

  • ̺

ϑ

3 2

  • whenever 0 < ̺ Zϑ

3 2 , or, ̺ > Zϑ 3 2

where 0 < Z < Z

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SLIDE 32

Hypotheses

∂pM ∂̺ > 0 for all ̺, ϑ > 0

◮ 0 < ∂eM

∂ϑ c for all ̺, ϑ > 0.

◮ limϑ→0+ eM(̺, ϑ) = eM(̺) > 0 for any fixed ̺ > 0, ◮

̺ ∂eM(̺,ϑ)

∂̺

  • ceM(̺, ϑ) for all ̺, ϑ > 0.

◮ There is a function P satisfying P ∈ C 1[0, ∞), P(0) = 0, P′(0) > 0,

such that pM(̺, ϑ) = ϑ

5 2 P

  • ̺

ϑ

3 2

  • whenever 0 < ̺ Zϑ

3 2 , or, ̺ > Zϑ 3 2

where 0 < Z < Z

◮ what gives

5 3 pM c

  • ϑ

5 2

if ̺ < Zϑ

3 2

̺

5 3

if ̺ > Zϑ

3 2 ,

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SLIDE 33

Hypotheses

∂pM ∂̺ > 0 for all ̺, ϑ > 0

◮ 0 < ∂eM

∂ϑ c for all ̺, ϑ > 0.

◮ limϑ→0+ eM(̺, ϑ) = eM(̺) > 0 for any fixed ̺ > 0, ◮

̺ ∂eM(̺,ϑ)

∂̺

  • ceM(̺, ϑ) for all ̺, ϑ > 0.

◮ There is a function P satisfying P ∈ C 1[0, ∞), P(0) = 0, P′(0) > 0,

such that pM(̺, ϑ) = ϑ

5 2 P

  • ̺

ϑ

3 2

  • whenever 0 < ̺ Zϑ

3 2 , or, ̺ > Zϑ 3 2

where 0 < Z < Z

◮ what gives

5 3 pM c

  • ϑ

5 2

if ̺ < Zϑ

3 2

̺

5 3

if ̺ > Zϑ

3 2 ,

◮ pM(̺, ϑ) = 2

3̺eM(̺, ϑ) for ̺ > Zϑ

3 2 .

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SLIDE 34

Boundary conditions and initial data

◮ The impermeability of the boundary of the physical domain is described

by the condition (u − V) · n|Γτ = 0 for any τ 0, where n(t, x) denotes the unit outer normal vector to the boundary Γt.

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SLIDE 35

Boundary conditions and initial data

◮ The impermeability of the boundary of the physical domain is described

by the condition (u − V) · n|Γτ = 0 for any τ 0, where n(t, x) denotes the unit outer normal vector to the boundary Γt.

◮ We assume a complete slip boundary conditions in the form

[Sn] × n = 0.

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SLIDE 36

Boundary conditions and initial data

◮ The impermeability of the boundary of the physical domain is described

by the condition (u − V) · n|Γτ = 0 for any τ 0, where n(t, x) denotes the unit outer normal vector to the boundary Γt.

◮ We assume a complete slip boundary conditions in the form

[Sn] × n = 0.

◮ For the heat flux we consider the conservative boundary conditions

q · n = 0 for all t ∈ [0, T], x ∈ Γt.

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SLIDE 37

Boundary conditions and initial data

◮ The impermeability of the boundary of the physical domain is described

by the condition (u − V) · n|Γτ = 0 for any τ 0, where n(t, x) denotes the unit outer normal vector to the boundary Γt.

◮ We assume a complete slip boundary conditions in the form

[Sn] × n = 0.

◮ For the heat flux we consider the conservative boundary conditions

q · n = 0 for all t ∈ [0, T], x ∈ Γt.

◮ Our problem is supplemented by the initial conditions

̺(0, ·) = ̺0 ∈ L

5 3 0,

(̺u)(0, ·) = (̺u)0, (̺u)0 = 0 if ̺0 = 0, ϑ(0, ·) = ϑ0 > 0, (̺s)0 = ̺0s(̺0, ϑ0) ∈ L1(Ω0), E0 =

  • Ω0
  • 1

2̺0 |(̺u)0|2 + ̺0e(̺0, ϑ0)

  • dx < ∞.
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SLIDE 38

Weak formulation

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SLIDE 39

Weak formulation

The continuity equation is satisfied in weak and renormalized sense.

T

  • Ωt

̺B(̺)(∂tϕ + u · ∇xϕ) =

T

  • Ωt

b(̺)divxuϕ −

  • Ω0

̺0B(̺0)ϕ(0) for any ϕ ∈ C 1

c ([0, T) × R3), b ∈ L∞ ∩ C[0, ∞) such that b(0) = 0 and

B(̺) = B(1) + ̺

1 b(z) z2 dz. We suppose that ̺ 0 a.e. in (0, T) × R3.

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SLIDE 40

Weak formulation

The continuity equation is satisfied in weak and renormalized sense.

T

  • Ωt

̺B(̺)(∂tϕ + u · ∇xϕ) =

T

  • Ωt

b(̺)divxuϕ −

  • Ω0

̺0B(̺0)ϕ(0) for any ϕ ∈ C 1

c ([0, T) × R3), b ∈ L∞ ∩ C[0, ∞) such that b(0) = 0 and

B(̺) = B(1) + ̺

1 b(z) z2 dz. We suppose that ̺ 0 a.e. in (0, T) × R3.

The momentum equation is satisfied in a weak sense

T

  • Ωt

(̺u · ∂tϕ + ̺[u ⊗ u] : ∇xϕ + p(̺, ϑ)divxϕ − S(ϑ, ∇xu) : ∇xϕ) = −

  • Ω0

(̺u)0 · ϕ(0, ·), for any ϕ ∈ C 1

c (QT; R3) such that ϕ(T, ·) = 0, ϕ · n|Γτ = 0 for all τ ∈ [0, T]

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SLIDE 41

Weak formulation

The continuity equation is satisfied in weak and renormalized sense.

T

  • Ωt

̺B(̺)(∂tϕ + u · ∇xϕ) =

T

  • Ωt

b(̺)divxuϕ −

  • Ω0

̺0B(̺0)ϕ(0) for any ϕ ∈ C 1

c ([0, T) × R3), b ∈ L∞ ∩ C[0, ∞) such that b(0) = 0 and

B(̺) = B(1) + ̺

1 b(z) z2 dz. We suppose that ̺ 0 a.e. in (0, T) × R3.

The momentum equation is satisfied in a weak sense

T

  • Ωt

(̺u · ∂tϕ + ̺[u ⊗ u] : ∇xϕ + p(̺, ϑ)divxϕ − S(ϑ, ∇xu) : ∇xϕ) = −

  • Ω0

(̺u)0 · ϕ(0, ·), for any ϕ ∈ C 1

c (QT; R3) such that ϕ(T, ·) = 0, ϕ · n|Γτ = 0 for all τ ∈ [0, T]

In particular, the impermeability condition is satisfied in the sense of traces, specifically, u, ∇xu ∈ L2(QT; R3) and (u − V) · n(τ, ·)|Γτ = 0 for a.a. τ ∈ [0, T].

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SLIDE 42

Weak formulation

The entropy inequality

T

  • Ωt

̺s(∂tϕ + u · ∇xϕ) −

T

  • Ωt

κ(ϑ)∇xϑ · ∇xϕ ϑ +

T

  • Ωt

ϕ ϑ

  • S : ∇xu + κ(ϑ)|∇xϑ|2

ϑ

  • Ω0

(̺s)0ϕ(0) holds for all ϕ ∈ C 1

c (QT) such that ϕ(T, ·) = 0 and ϕ 0.

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SLIDE 43

Weak formulation

The entropy inequality

T

  • Ωt

̺s(∂tϕ + u · ∇xϕ) −

T

  • Ωt

κ(ϑ)∇xϑ · ∇xϕ ϑ +

T

  • Ωt

ϕ ϑ

  • S : ∇xu + κ(ϑ)|∇xϑ|2

ϑ

  • Ω0

(̺s)0ϕ(0) holds for all ϕ ∈ C 1

c (QT) such that ϕ(T, ·) = 0 and ϕ 0.

The energy inequality for the case of a moving domain reads as

  • Ωτ

1

2̺|u|2 + ̺e

  • (τ, ·)
  • Ω0
  • 1

2 (̺u)2 ̺0 + ̺0e0−(̺u)0 · V(0)

τ

  • Ωt

(̺(u ⊗ u) : ∇xV + p divxV − S : ∇xV + ̺u · ∂tV) dxdt +

  • Ωt

̺u · V(τ, ·) for a.a. τ ∈ (0, T).

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SLIDE 44

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

  • ne.

(L.Poul, 2009)

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SLIDE 45

Main result

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SLIDE 46

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

◮ ̺ ∈ L∞(0, T; L

5 3 (R3)), ̺ 0, ̺ ∈ Lq(QT) for certain q > 5

3,

◮ u, ∇xu ∈ L2(QT), ̺u ∈ L∞(0, T; L1(R3)), ◮ ϑ > 0 a.a. on QT, ϑ ∈ L∞(0, T; L4(R3)), ϑ, ∇xϑ ∈ L2(QT), and

log ϑ, ∇x log ϑ ∈ L2(QT),

◮ ̺s, ̺su,

q ϑ ∈ L1(QT),

◮ weak formulation is satisfied.

slide-47
SLIDE 47

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

◮ ̺ ∈ L∞(0, T; L

5 3 (R3)), ̺ 0, ̺ ∈ Lq(QT) for certain q > 5

3,

◮ u, ∇xu ∈ L2(QT), ̺u ∈ L∞(0, T; L1(R3)), ◮ ϑ > 0 a.a. on QT, ϑ ∈ L∞(0, T; L4(R3)), ϑ, ∇xϑ ∈ L2(QT), and

log ϑ, ∇x log ϑ ∈ L2(QT),

◮ ̺s, ̺su,

q ϑ ∈ L1(QT),

◮ weak formulation is satisfied.

Theorem

Let Ω0 ⊂ R3 be a bounded domain of class C 2+ν with some ν > 0, and let V ∈ C 1([0, T]; C 3

c (R3; R3)) 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).

slide-48
SLIDE 48

The strategy - our penalization scheme

slide-49
SLIDE 49

The strategy - our penalization scheme

◮ Letus consider our problem on a large fixed time independent domain

B ⊃ Ωτ for all τ ∈ [0, T]

slide-50
SLIDE 50

The strategy - our penalization scheme

◮ Letus consider our problem on a large fixed time independent domain

B ⊃ Ωτ for all τ ∈ [0, T]

◮ To deal with the slip boundary condition, we introduce to the weak

formulation of the momentum equation a term 1 ε

T

  • Γt

(u − V) · n ϕ · n dSxdt (originally proposed by Stokes and Carey)

slide-51
SLIDE 51

The strategy - our penalization scheme

◮ Letus consider our problem on a large fixed time independent domain

B ⊃ Ωτ for all τ ∈ [0, T]

◮ To deal with the slip boundary condition, we introduce to the weak

formulation of the momentum equation a term 1 ε

T

  • Γt

(u − V) · n ϕ · n dSxdt (originally proposed by Stokes and Carey)

◮ In the limit ε → 0 the above term yields the boundary condition

(u − V) · n = 0

  • n Γt

and thus the large domain (0, T) × B becomes divided by an impermeable interface ∪t∈(0,T){t} × Γt to a fluid domain QT and a solid domain Qc

T.

slide-52
SLIDE 52

The strategy - our penalization scheme

◮ Letus consider our problem on a large fixed time independent domain

B ⊃ Ωτ for all τ ∈ [0, T]

◮ To deal with the slip boundary condition, we introduce to the weak

formulation of the momentum equation a term 1 ε

T

  • Γt

(u − V) · n ϕ · n dSxdt (originally proposed by Stokes and Carey)

◮ In the limit ε → 0 the above term yields the boundary condition

(u − V) · n = 0

  • n Γt

and thus the large domain (0, T) × B becomes divided by an impermeable interface ∪t∈(0,T){t} × Γt to a fluid domain QT and a solid domain Qc

T.

◮ Then we need to find a way how to get rid of terms outside of a fluid

domain.

slide-53
SLIDE 53

Penalisation in terms outside of a fluid domain

We introduce:

slide-54
SLIDE 54

Penalisation in terms outside of a fluid domain

We introduce:

◮ A variable shear viscosity coefficient µ = µω and bulk viscosity coefficient

ζ = ζω µω(ϑ, ·) ∈ C ∞

c ,

µω(ϑ, τ, ·)|Ωτ = µ(ϑ) ζω(ϑ, ·) ∈ C ∞

c ,

ζω(ϑ, τ, ·)|Ωτ = η(ϑ) and µω, ζω → 0 a.e. in ((0, T) × B) \ QT as ω → 0.

slide-55
SLIDE 55

Penalisation in terms outside of a fluid domain

We introduce:

◮ A variable shear viscosity coefficient µ = µω and bulk viscosity coefficient

ζ = ζω µω(ϑ, ·) ∈ C ∞

c ,

µω(ϑ, τ, ·)|Ωτ = µ(ϑ) ζω(ϑ, ·) ∈ C ∞

c ,

ζω(ϑ, τ, ·)|Ωτ = η(ϑ) and µω, ζω → 0 a.e. in ((0, T) × B) \ QT as ω → 0.

◮ A variable heat conductivity coefficient κν(t, x, ϑ)

κν(ϑ, t, x) = χν(t, x)κ(ϑ), where χν = 1 in QT, χν = ν in ((0, T) × B) \ QT

slide-56
SLIDE 56

Penalisation in terms outside of a fluid domain

We introduce:

◮ A variable shear viscosity coefficient µ = µω and bulk viscosity coefficient

ζ = ζω µω(ϑ, ·) ∈ C ∞

c ,

µω(ϑ, τ, ·)|Ωτ = µ(ϑ) ζω(ϑ, ·) ∈ C ∞

c ,

ζω(ϑ, τ, ·)|Ωτ = η(ϑ) and µω, ζω → 0 a.e. in ((0, T) × B) \ QT as ω → 0.

◮ A variable heat conductivity coefficient κν(t, x, ϑ)

κν(ϑ, t, x) = χν(t, x)κ(ϑ), where χν = 1 in QT, χν = ν in ((0, T) × B) \ QT

◮ A variable coefficient a := aη(t, x) which represents the radiative part of

pressure, internal energy and entropy. Namely, we assume that aη(t, x) = χη(t, x)a, a > 0, where χη = 1 in QT, χη = η in ((0, T) × B \ QT.

slide-57
SLIDE 57

Artificial pressure and extra temperature term

We introduce:

slide-58
SLIDE 58

Artificial pressure and extra temperature term

We introduce:

◮ The artificial pressure

pη,δ(̺, ϑ) = pM(̺, ϑ) + aη 3 ϑ4+δ̺β, β 4, δ > 0

slide-59
SLIDE 59

Artificial pressure and extra temperature term

We introduce:

◮ The artificial pressure

pη,δ(̺, ϑ) = pM(̺, ϑ) + aη 3 ϑ4+δ̺β, β 4, δ > 0

◮ An extra term in the energy inequality

λϑ5 and in the entropy inequality formulation λϑ4

slide-60
SLIDE 60

Existence of solutions to our new penalization on a fixed domain, uniform estimates

◮ There exists solution {̺ε, uε, ϑε}ε to our penalized problem for each fixed

ε > 0 and fixed η, ω, ν, λ, δ > 0.

slide-61
SLIDE 61

Existence of solutions to our new penalization on a fixed domain, uniform estimates

◮ There exists solution {̺ε, uε, ϑε}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)

slide-62
SLIDE 62

Existence of solutions to our new penalization on a fixed domain, uniform estimates

◮ There exists solution {̺ε, uε, ϑε}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)

◮ The term 1 ε

τ

  • Γt ((u − V) · n ϕ · n) dSxdt and

1 ε

T

  • Γt (u − V) · n u · nψdSx dt can be treated as a ”compact”

perturbation.

slide-63
SLIDE 63

Existence of solutions to our new penalization on a fixed domain, uniform estimates

◮ There exists solution {̺ε, uε, ϑε}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)

◮ The term 1 ε

τ

  • Γt ((u − V) · n ϕ · n) dSxdt and

1 ε

T

  • Γt (u − V) · n u · nψdSx dt can be treated as a ”compact”

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.

slide-64
SLIDE 64

Uniform estimates on the set (0, T) × B given by the total dissipation inequality

slide-65
SLIDE 65

Uniform estimates on the set (0, T) × B given by the total dissipation inequality T

  • Γt

|(u − V) · n|2 dSxdt εc(λ)

slide-66
SLIDE 66

Uniform estimates on the set (0, T) × B given by the total dissipation inequality T

  • Γt

|(u − V) · n|2 dSxdt εc(λ) ess sup

τ∈(0,T)

δ̺β(τ, ·)L1 + ess sup

τ∈(0,T)

√̺u(τ, ·)L2 c(λ)

slide-67
SLIDE 67

Uniform estimates on the set (0, T) × B given by the total dissipation inequality T

  • Γt

|(u − V) · n|2 dSxdt εc(λ) ess sup

τ∈(0,T)

δ̺β(τ, ·)L1 + ess sup

τ∈(0,T)

√̺u(τ, ·)L2 c(λ)

  • λϑ5
  • L1 c(λ)
slide-68
SLIDE 68

Uniform estimates on the set (0, T) × B given by the total dissipation inequality T

  • Γt

|(u − V) · n|2 dSxdt εc(λ) ess sup

τ∈(0,T)

δ̺β(τ, ·)L1 + ess sup

τ∈(0,T)

√̺u(τ, ·)L2 c(λ)

  • λϑ5
  • L1 c(λ)

uL2 + ∇xuL2 c(λ, ω)

slide-69
SLIDE 69

Uniform estimates on the set (0, T) × B given by the total dissipation inequality T

  • Γt

|(u − V) · n|2 dSxdt εc(λ) ess sup

τ∈(0,T)

δ̺β(τ, ·)L1 + ess sup

τ∈(0,T)

√̺u(τ, ·)L2 c(λ)

  • λϑ5
  • L1 c(λ)

uL2 + ∇xuL2 c(λ, ω) ess sup

τ∈(0,T)

aηϑ4(τ, ·)L1 + ess sup

τ∈(0,T)

̺(τ, ·)

L

5 3 c(λ)

slide-70
SLIDE 70

Uniform estimates on the set (0, T) × B given by the total dissipation inequality T

  • Γt

|(u − V) · n|2 dSxdt εc(λ) ess sup

τ∈(0,T)

δ̺β(τ, ·)L1 + ess sup

τ∈(0,T)

√̺u(τ, ·)L2 c(λ)

  • λϑ5
  • L1 c(λ)

uL2 + ∇xuL2 c(λ, ω) ess sup

τ∈(0,T)

aηϑ4(τ, ·)L1 + ess sup

τ∈(0,T)

̺(τ, ·)

L

5 3 c(λ)

T

  • B
  • |∇x log(ϑ)|2 + |∇xϑ

3 2 |2

dxdt c(λ)

slide-71
SLIDE 71

Uniform estimates on the set (0, T) × B given by the total dissipation inequality T

  • Γt

|(u − V) · n|2 dSxdt εc(λ) ess sup

τ∈(0,T)

δ̺β(τ, ·)L1 + ess sup

τ∈(0,T)

√̺u(τ, ·)L2 c(λ)

  • λϑ5
  • L1 c(λ)

uL2 + ∇xuL2 c(λ, ω) ess sup

τ∈(0,T)

aηϑ4(τ, ·)L1 + ess sup

τ∈(0,T)

̺(τ, ·)

L

5 3 c(λ)

T

  • B
  • |∇x log(ϑ)|2 + |∇xϑ

3 2 |2

dxdt c(λ) ̺s(̺, ϑ)Lq+̺s(̺, ϑ)uLq+

  • κν(ϑ)

ϑ ∇xϑ

  • Lq

+̺e(̺, ϑ)L1 c(λ) with some q > 1

slide-72
SLIDE 72

Passing with ε → 0

slide-73
SLIDE 73

Passing with ε → 0

◮ First of all, directly from uniform estimates we derive that

(u − V) · n(τ, ·)|Γτ = 0 for a.a. τ ∈ [0, T]. in the limit as ε → 0.

slide-74
SLIDE 74

Passing with ε → 0

◮ First of all, directly from uniform estimates we derive that

(u − V) · n(τ, ·)|Γτ = 0 for a.a. τ ∈ [0, T]. in the limit as ε → 0.

◮ More demanding (but already known strategy):

ϑε → ϑ a.a. in (0, T) × B.

̺ε → ̺ a.a. in (0, T) × B.

◮ Div-Curl Lemma, Young measure theory; effective viscous pressure,

  • scillations defect measure
slide-75
SLIDE 75

Passing with ε → 0 - delicate issues

slide-76
SLIDE 76

Passing with ε → 0 - delicate issues

◮ We have at hand only the local estimates on the pressure,

K

  • p(̺, ϑ)̺π + δ̺β+π

dxdt c(K) for certain π > 0 and for any compact K ⊂ ((0, T) × B) such that K ∩

  • ∪τ∈[0,T]
  • {τ} × Γτ
  • = ∅,
slide-77
SLIDE 77

Passing with ε → 0 - delicate issues

◮ We have at hand only the local estimates on the pressure,

K

  • p(̺, ϑ)̺π + δ̺β+π

dxdt c(K) for certain π > 0 and for any compact K ⊂ ((0, T) × B) such that K ∩

  • ∪τ∈[0,T]
  • {τ} × Γτ
  • = ∅,

◮ We have to restrict ourselves to the class of test functions for a

momentum equation ϕ ∈ C 1([0, T); W 1,∞ (B; R3)), supp[divxϕ(τ, ·)] ∩ Γτ = ∅, ϕ · n|Γτ = 0 for all τ ∈ [0, T].

slide-78
SLIDE 78

Passing with ε → 0 - delicate issues

◮ We have at hand only the local estimates on the pressure,

K

  • p(̺, ϑ)̺π + δ̺β+π

dxdt c(K) for certain π > 0 and for any compact K ⊂ ((0, T) × B) such that K ∩

  • ∪τ∈[0,T]
  • {τ} × Γτ
  • = ∅,

◮ We have to restrict ourselves to the class of test functions for a

momentum equation ϕ ∈ C 1([0, T); W 1,∞ (B; R3)), supp[divxϕ(τ, ·)] ∩ Γτ = ∅, ϕ · n|Γτ = 0 for all τ ∈ [0, T].

◮ But this can be extended to the class

ϕ ∈ C ∞

c ([0, T] × B; R3),

ϕ(τ, ·) · n|Γτ = 0 for any τ ∈ [0, T].

slide-79
SLIDE 79

Passing with ε → 0 - delicate issues

◮ Internal energy and ”pressure” terms in the energy inequality

slide-80
SLIDE 80

Passing with ε → 0 - delicate issues

◮ Internal energy and ”pressure” terms in the energy inequality ◮ To pass to the limit in

pη,δ(̺ε, ϑε)divxVψ we need to assume that divxV = 0 in the neighbourhood of Γt

slide-81
SLIDE 81

Passing with ε → 0 - delicate issues

◮ Internal energy and ”pressure” terms in the energy inequality ◮ To pass to the limit in

pη,δ(̺ε, ϑε)divxVψ we need to assume that divxV = 0 in the neighbourhood of Γt

◮ We do not have uniform integrability of the sequence

{̺εeη(̺ε, ϑε)}ε>0

slide-82
SLIDE 82

Passing with ε → 0 - delicate issues

◮ Internal energy and ”pressure” terms in the energy inequality ◮ To pass to the limit in

pη,δ(̺ε, ϑε)divxVψ we need to assume that divxV = 0 in the neighbourhood of Γt

◮ We do not have uniform integrability of the sequence

{̺εeη(̺ε, ϑε)}ε>0

◮ Since the sequence {̺εeη(̺ε, ϑε)}ε is nonnegative, the sequence is

integrable, density and temperature converges a.e., by the Fatou lemma we deduce lim sup

T

  • B

̺εeη(̺ε, ϑε)∂tψdxdt

T

  • B

̺eη(̺, ϑ)∂tψdxdt as far as ψ ∈ C 1

c ([0, T)) and ∂tψ 0.

slide-83
SLIDE 83

Vanishing density outside of fluid part

In order to get rid of the density dependent terms supported by the ”solid” part ((0, T) × B) \ QT we use result of [Feireisl, Kreml, Necasova, Neustupa, Stebel, JDE, 2013] which reads as

slide-84
SLIDE 84

Vanishing density outside of fluid part

In order to get rid of the density dependent terms supported by the ”solid” part ((0, T) × B) \ QT we use result of [Feireisl, Kreml, Necasova, Neustupa, Stebel, JDE, 2013] which reads as

Lemma

Let ̺ ∈ L∞(0, T; L2(B)), ̺ 0, u ∈ L2(0, T; W 1,2 (B; R3)) be a weak solution

  • f the equation of continuity, specifically,
  • B
  • ̺(τ, ·)ϕ(τ, ·) − ̺0ϕ(0, ·)
  • dx =

τ

  • B
  • ̺∂tϕ + ̺u · ∇xϕ
  • dxdt

for any τ ∈ [0, T] and any test function ϕ ∈ C 1

c ([0, T] × R3).

In addition, assume that (u − V)(τ, ·) · n|Γτ = 0 for a.a. τ ∈ (0, T), and that ̺0 ∈ L2(R3), ̺0 0, ̺0|B\Ω0 = 0. Then ̺(τ, ·)|B\Ωτ = 0 for any τ ∈ [0, T].

slide-85
SLIDE 85

The limit system ε → 0 and with ̺0 = 0 on B \ Ω0

slide-86
SLIDE 86

The limit system ε → 0 and with ̺0 = 0 on B \ Ω0

The continuity equation reads

T

  • Ωt

̺B(̺)(∂tϕ + u · ∇xϕ) =

T

  • Ωt

b(̺)divxuϕ −

  • Ω0

̺0,δB(̺0,δ)ϕ(0) for any ϕ ∈ C 1

c ([0, T) × R3), and any b ∈ L∞ ∩ C[0, ∞) such that b(0) = 0

and B(̺) = B(1) + ̺

1 b(z) z2 dz.

slide-87
SLIDE 87

The limit system ε → 0 and with ̺0 = 0 on B \ Ω0

The continuity equation reads

T

  • Ωt

̺B(̺)(∂tϕ + u · ∇xϕ) =

T

  • Ωt

b(̺)divxuϕ −

  • Ω0

̺0,δB(̺0,δ)ϕ(0) for any ϕ ∈ C 1

c ([0, T) × R3), and any b ∈ L∞ ∩ C[0, ∞) such that b(0) = 0

and B(̺) = B(1) + ̺

1 b(z) z2 dz.

Next the momentum equation reduces to

T

  • Ωt

(̺u · ∂tϕ + ̺[u ⊗ u] : ∇xϕ + pη,δ(̺, ϑ)divxϕ − Sω(ϑ, ∇xu) : ∇xϕ) = −

  • Ω0

(̺u)0,δ · ϕ(0, ·)+

T

  • B\Ωt

Sω(ϑ, ∇xu) : ∇ϕ −

T

  • B\Ωt

aη 3 ϑ4divxϕ for any ϕ ∈ C ∞

c ([0, T] × B; R3), ϕ(τ, ·) · n|Γτ = 0 for any τ ∈ [0, T].

slide-88
SLIDE 88

The limit system ε → 0 and with ̺0 = 0 on B \ Ω0

The balance of entropy takes the following form

T

  • Ωt

̺s(̺, ϑ)(∂tϕ + u · ∇xϕ)+

T

  • B\Ωt

4 3aηϑ3(∂tϕ + u · ∇xϕ) −

T

  • Ωt

κν(ϑ)∇xϑ · ∇xϕ ϑ −

T

  • B\Ωt

κν(ϑ)∇xϑ · ∇xϕ ϑ +

T

  • Ωt

ϕ ϑ

  • Sω : ∇xu + κν(ϑ)|∇xϑ|2

ϑ

  • +

T

  • B\Ωt

ϕ ϑ

  • Sω : ∇xu + κν(ϑ)|∇xϑ|2

ϑ

T

  • B

λϑ4ϕ −

  • Ω0

(̺s)0,δϕ(0)−

  • B\Ω0

4 3aηϑ3

0,δϕ(0)

for all ϕ ∈ C 1

c ([0, T) × B), ϕ 0.

slide-89
SLIDE 89

The limit system ε → 0 and with ̺0 = 0 on B \ Ω0

The total energy balance reads

T

  • Ωt
  • 1

2̺|u|2 + ̺e(̺, ϑ) + δ β − 1̺β

  • ∂tψ+

T

  • B\Ωt

aηϑ4∂tψ −

T

  • B

λϑ5ψ −

  • Ω0
  • 1

2 (̺u)2

0,δ

̺0,δ + ̺0,δe0,δ + δ β − 1̺β

0,δ − (̺u)0,δ · V(0, ·)

  • ψ(0)

  • B\Ω0

aηϑ4

0,δψ(0)

T

  • Ωt

(Sω : ∇xVψ − ̺u · ∂t(Vψ) − ̺(u ⊗ u) : ∇xVψ − pδ(̺, ϑ)divxVψ) −

τ

  • B\Ωt
  • Sω : ∇xV − 1

3aηϑ4divxV

  • ψ

for all ψ ∈ C 1

c ([0, T)), ∂tψ 0.

slide-90
SLIDE 90

Passing with η → 0

slide-91
SLIDE 91

Passing with η → 0

◮ Let us denote by {̺η, uη, ϑη}η>0 solutions to the system obtained as

ε → 0

slide-92
SLIDE 92

Passing with η → 0

◮ Let us denote by {̺η, uη, ϑη}η>0 solutions to the system obtained as

ε → 0

◮ Let aη = ηa on B \ Ωt

slide-93
SLIDE 93

Passing with η → 0

◮ Let us denote by {̺η, uη, ϑη}η>0 solutions to the system obtained as

ε → 0

◮ Let aη = ηa on B \ Ωt ◮ Since we have

  • λϑ5

η

  • L1 c(λ),

uηL2(0,T;W 1,2(B)) c

slide-94
SLIDE 94

Passing with η → 0

◮ Let us denote by {̺η, uη, ϑη}η>0 solutions to the system obtained as

ε → 0

◮ Let aη = ηa on B \ Ωt ◮ Since we have

  • λϑ5

η

  • L1 c(λ),

uηL2(0,T;W 1,2(B)) c

◮ for η → 0 we get

T

  • B\Ωt

1 3aηϑ4

ηdivxϕ → 0

T

  • B\Ωt

1 3aηϑ4

ηdivxV → 0

T

  • B\Ωt

4 3aηϑ3

η∂tϕ → 0

T

  • B\Ωt

4 3aηϑ3

ηuη · ∇xϕ → 0

slide-95
SLIDE 95

Passing with η → 0

slide-96
SLIDE 96

Passing with η → 0

◮ Since ϑη → ϑ weakly in L1((0, T) × B), we obtain

T

  • B

λϑ5dxdt lim inf

η→0

T

  • B

λϑ5

ηdxdt.

slide-97
SLIDE 97

Passing with η → 0

◮ Since ϑη → ϑ weakly in L1((0, T) × B), we obtain

T

  • B

λϑ5dxdt lim inf

η→0

T

  • B

λϑ5

ηdxdt.

◮ In the energy inequality we need to restrict ourself to test functions

ψ ∈ C 1

c ([0, T)),

ψ 0, ∂tψ 0

slide-98
SLIDE 98

Passing with η → 0

◮ Since ϑη → ϑ weakly in L1((0, T) × B), we obtain

T

  • B

λϑ5dxdt lim inf

η→0

T

  • B

λϑ5

ηdxdt.

◮ In the energy inequality we need to restrict ourself to test functions

ψ ∈ C 1

c ([0, T)),

ψ 0, ∂tψ 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.

slide-99
SLIDE 99

Passing with ω → 0

We would like to get rid of terms related to the viscous stress tensor outside of the fluid domain.

slide-100
SLIDE 100

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, uω, ϑω}ω>0 solutions to the system obtained as

η → 0.

slide-101
SLIDE 101

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, uω, ϑω}ω>0 solutions to the system obtained as

η → 0.

◮ Since µω and ζω ≈ ω on B \ Ωt, and we control

T

  • B\Ωt

1 ϑω Sω(ϑω, ∇xuω) : ∇xuω c,

T

  • B\Ωt

λϑ5

ω c

slide-102
SLIDE 102

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, uω, ϑω}ω>0 solutions to the system obtained as

η → 0.

◮ Since µω and ζω ≈ ω on B \ Ωt, and we control

T

  • B\Ωt

1 ϑω Sω(ϑω, ∇xuω) : ∇xuω c,

T

  • B\Ωt

λϑ5

ω c

◮ In the momentum equation

T

  • B\Ωt

Sω(ϑω, ∇xuω) : ∇xϕ → 0 as ω → 0.

slide-103
SLIDE 103

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, uω, ϑω}ω>0 solutions to the system obtained as

η → 0.

◮ Since µω and ζω ≈ ω on B \ Ωt, and we control

T

  • B\Ωt

1 ϑω Sω(ϑω, ∇xuω) : ∇xuω c,

T

  • B\Ωt

λϑ5

ω c

◮ In the momentum equation

T

  • B\Ωt

Sω(ϑω, ∇xuω) : ∇xϕ → 0 as ω → 0.

◮ In the total energy inequality

T

  • B\Ωt

Sω(ϑω, ∇xuω) : ∇xVdxdt → 0 as ω → 0.

slide-104
SLIDE 104

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, uω, ϑω}ω>0 solutions to the system obtained as

η → 0.

◮ Since µω and ζω ≈ ω on B \ Ωt, and we control

T

  • B\Ωt

1 ϑω Sω(ϑω, ∇xuω) : ∇xuω c,

T

  • B\Ωt

λϑ5

ω c

◮ In the momentum equation

T

  • B\Ωt

Sω(ϑω, ∇xuω) : ∇xϕ → 0 as ω → 0.

◮ In the total energy inequality

T

  • B\Ωt

Sω(ϑω, ∇xuω) : ∇xVdxdt → 0 as ω → 0.

◮ In the entropy inequality we skip the term T

  • B\Ωt

ϕ ϑ Sω : ∇xu

since it is positive for ϕ 0.

slide-105
SLIDE 105

Passing with ν → 0

◮ Let {̺ν, uν, ϑν}ν>0 denote solutions to the system obtained in the

previous step

slide-106
SLIDE 106

Passing with ν → 0

◮ Let {̺ν, uν, ϑν}ν>0 denote solutions to the system obtained in the

previous step

◮ For each fixed ν > 0 we still know that

  • χνκ(ϑν)|∇xϑν|2

ϑ2

ν

  • L1((0,T)×B)

c(λ) and λϑ5

νL1((0,T)×B) c(λ)

slide-107
SLIDE 107

Passing with ν → 0

◮ Let {̺ν, uν, ϑν}ν>0 denote solutions to the system obtained in the

previous step

◮ For each fixed ν > 0 we still know that

  • χνκ(ϑν)|∇xϑν|2

ϑ2

ν

  • L1((0,T)×B)

c(λ) and λϑ5

νL1((0,T)×B) c(λ)

◮ In the internal entropy inequality

T

  • B\Ωt

νκ(ϑν)∇xϑν ϑν · ∇xϕ → 0 as ν → 0

slide-108
SLIDE 108

Passing with ν → 0

◮ Let {̺ν, uν, ϑν}ν>0 denote solutions to the system obtained in the

previous step

◮ For each fixed ν > 0 we still know that

  • χνκ(ϑν)|∇xϑν|2

ϑ2

ν

  • L1((0,T)×B)

c(λ) and λϑ5

νL1((0,T)×B) c(λ)

◮ In the internal entropy inequality

T

  • B\Ωt

νκ(ϑν)∇xϑν ϑν · ∇xϕ → 0 as ν → 0

◮ Since T

  • B\Ωt

ϕ ϑν

  • κν(ϑν)|∇x ϑν|2

ϑν

  • 0 for ϕ 0, we can skip this term

as ν → 0

slide-109
SLIDE 109

Passing with λ → 0

Now we get rid the term related to coefficient λ - the only terms which are left also outside of a fluid domain.

slide-110
SLIDE 110

Passing with λ → 0

Now we get rid the term related to coefficient λ - the only terms which are left also outside of a fluid domain.

◮ Let {̺λ, uλ, ϑλ}λ∈(0,1) be solution to the limit system obtained in

previous step.

slide-111
SLIDE 111

Passing with λ → 0

Now we get rid the term related to coefficient λ - the only terms which are left also outside of a fluid domain.

◮ Let {̺λ, uλ, ϑλ}λ∈(0,1) be solution to the limit system obtained in

previous step.

◮ We need to build uniform estimates independent of λ

slide-112
SLIDE 112

Passing with λ → 0

Now we get rid the term related to coefficient λ - the only terms which are left also outside of a fluid domain.

◮ Let {̺λ, uλ, ϑλ}λ∈(0,1) be solution to the limit system obtained in

previous step.

◮ We need to build uniform estimates independent of λ ◮ We keep that

  • λϑ5
  • L1((0,T)×B) c so in the entropy inequality

T

  • B

λϑ4

λ → 0

as λ → 0.

slide-113
SLIDE 113

Passing with λ → 0

Now we get rid the term related to coefficient λ - the only terms which are left also outside of a fluid domain.

◮ Let {̺λ, uλ, ϑλ}λ∈(0,1) be solution to the limit system obtained in

previous step.

◮ We need to build uniform estimates independent of λ ◮ We keep that

  • λϑ5
  • L1((0,T)×B) c so in the entropy inequality

T

  • B

λϑ4

λ → 0

as λ → 0.

◮ Next notice that the term τ

  • B λϑ5

λψ 0 for all λ > 0 and all

ψ ∈ C 1

c ([0, T)), ψ 0 and therefore can be skipped in the total energy

inequality as λ → 0

slide-114
SLIDE 114

Passing with λ → 0

Now we get rid the term related to coefficient λ - the only terms which are left also outside of a fluid domain.

◮ Let {̺λ, uλ, ϑλ}λ∈(0,1) be solution to the limit system obtained in

previous step.

◮ We need to build uniform estimates independent of λ ◮ We keep that

  • λϑ5
  • L1((0,T)×B) c so in the entropy inequality

T

  • B

λϑ4

λ → 0

as λ → 0.

◮ Next notice that the term τ

  • B λϑ5

λψ 0 for all λ > 0 and all

ψ ∈ C 1

c ([0, T)), ψ 0 and therefore can be skipped in the total energy

inequality as λ → 0

◮ We pass to the limit with in other term in the same way as for ε → 0. The

bound ess supτ∈(0,T) aϑ4(τ, ·)L1(Ωτ ) < c provides enough information.

slide-115
SLIDE 115

Almost the final step

slide-116
SLIDE 116

Almost the final step

◮ The only term which left is artificial pressure δ̺β.

slide-117
SLIDE 117

Almost the final step

◮ The only term which left is artificial pressure δ̺β. ◮ The passage with δ → 0 is a kind of ”classic” now and we can follow

[Feireisl, Novotn´ y, 2009]

slide-118
SLIDE 118

Almost the final step step

slide-119
SLIDE 119

Almost the final step step

◮ In particular, for the energy inequality we worked with ”weak” form and

we need to choose proper test function to obtain formulation from our definition of variational solution In particular, the energy inequality has the form

T

  • Ωτ

1

2̺|u|2 + ̺e

  • ∂tψ −
  • Ω0
  • 1

2 (̺u)2 ̺0 + ̺0e0 − (̺u)0 · V(0, ·)

  • ψ(0)

T

  • Ωt

(S : ∇xVψ − ̺u · ∂t(Vψ) − ̺(u ⊗ u) : ∇xVψ − p(̺, ϑ)divxVψ) for all ψ ∈ C 1

c ([0, T)), ψ 0, ∂tψ 0.

slide-120
SLIDE 120

Almost the final step step

◮ In particular, for the energy inequality we worked with ”weak” form and

we need to choose proper test function to obtain formulation from our definition of variational solution In particular, the energy inequality has the form

T

  • Ωτ

1

2̺|u|2 + ̺e

  • ∂tψ −
  • Ω0
  • 1

2 (̺u)2 ̺0 + ̺0e0 − (̺u)0 · V(0, ·)

  • ψ(0)

T

  • Ωt

(S : ∇xVψ − ̺u · ∂t(Vψ) − ̺(u ⊗ u) : ∇xVψ − p(̺, ϑ)divxVψ) for all ψ ∈ C 1

c ([0, T)), ψ 0, ∂tψ 0.

◮ We choose ψξ such that ψξ ∈ C 1

c ([0, T)) is non-increasing function which

fulfils ψξ(t) =

  • 1 for t < τ − ξ

0 for t τ for some τ ∈ (0, T) and arbitrary ξ > 0 as a test function in the total energy inequality and we pass with ξ → 0. Then obtain the formulation required in the definition of variational solution

slide-121
SLIDE 121

What can one do next? Ongoing work

slide-122
SLIDE 122

What can one do next? Ongoing work

◮ Relative entropy inequality (like for weak-strong uniqueness)

slide-123
SLIDE 123

What can one do next? Ongoing work

◮ Relative entropy inequality (like for weak-strong uniqueness) ◮ Low Mach number limit - speed of sound dominates characteristic speed

  • f the fluid - for ill-prepared initial data the system is driven to the

incompressible Oberback-Boussinesq approximation

slide-124
SLIDE 124

What can one do next? Ongoing work

◮ Relative entropy inequality (like for weak-strong uniqueness) ◮ Low Mach number limit - speed of sound dominates characteristic speed

  • f the fluid - for ill-prepared initial data the system is driven to the

incompressible Oberback-Boussinesq approximation

◮ Barotropic case - E. Feireisl, O. Kreml, ˇ

  • S. Neˇ

casov´ a, J. Neustupa, J. Stebel (SIAM, 2014)

slide-125
SLIDE 125

What can one do next? Ongoing work

◮ Relative entropy inequality (like for weak-strong uniqueness) ◮ Low Mach number limit - speed of sound dominates characteristic speed

  • f the fluid - for ill-prepared initial data the system is driven to the

incompressible Oberback-Boussinesq approximation

◮ Barotropic case - E. Feireisl, O. Kreml, ˇ

  • S. Neˇ

casov´ a, J. Neustupa, J. Stebel (SIAM, 2014)

◮ time evolution of a Helmholtz projection operator associated with the

domain Ωτ and its gradient counterpart

slide-126
SLIDE 126

What can one do next? Ongoing work

◮ Relative entropy inequality (like for weak-strong uniqueness) ◮ Low Mach number limit - speed of sound dominates characteristic speed

  • f the fluid - for ill-prepared initial data the system is driven to the

incompressible Oberback-Boussinesq approximation

◮ Barotropic case - E. Feireisl, O. Kreml, ˇ

  • S. Neˇ

casov´ a, J. Neustupa, J. Stebel (SIAM, 2014)

◮ time evolution of a Helmholtz projection operator associated with the

domain Ωτ and its gradient counterpart

◮ propagation of the acoustic wave is governed by the Neumann Laplacian

whose spectral properties and their dependence of τ need to be examined

slide-127
SLIDE 127

Ongoing work. Low Mach number limit

◮ NSF with Ma= ε and Fr= √ε

∂t̺ + divx(̺u) = 0, ∂t(̺u) + divx(̺u ⊗ u) + 1 ε2 ∇xp = divxS + 1 ε ̺∇xF, ∂t(̺s) + divx(̺su) + divx

q

ϑ

  • = σε,

d dt ε2 2 ̺|u|2 + ̺e − ε̺F

  • dx = 0.

(NSFε)

slide-128
SLIDE 128

Ongoing work. Low Mach number limit

◮ NSF with Ma= ε and Fr= √ε

∂t̺ + divx(̺u) = 0, ∂t(̺u) + divx(̺u ⊗ u) + 1 ε2 ∇xp = divxS + 1 ε ̺∇xF, ∂t(̺s) + divx(̺su) + divx

q

ϑ

  • = σε,

d dt ε2 2 ̺|u|2 + ̺e − ε̺F

  • dx = 0.

(NSFε)

◮ OB approximation where G(t) :=

Ωt V · ∇xFdx.

divxU = 0 ̺(∂tU + divx(U ⊗ U)) + ∇xΠ − divxµ(ϑ)[∇xU + ∇T

x U] = r∇xF

̺ cp (∂tΘ + divx(ΘU)) − κ(ϑ)∆xΘ − α ρ ϑU · ∇xF = −α̺ϑG r + ̺ αΘ = 0 (OB)

◮ ̺ε → ̺, uε → U, ϑε−ϑ

ε

→ Θ, ̺ε−̺

ε

→ r in a certain sense

slide-129
SLIDE 129

Bibliography

◮ O. Kreml, V. M´

acha, ˇ

  • S. Neˇ

casova, A. Wróblewska-Kamińska. Flow of heat conducting fluid in a time dependent domain. ZAMP 2018.

◮ O. Kreml, V. M´

acha, ˇ

  • S. Neˇ

casova, A. Wróblewska-Kamińska. Weak solutions to the full Navier-Stokes-Fourier system with slip boundary conditions in time dependent domains. Journal de Math´ ematiques Pures et Appliqu´ ees, 2017.

◮ E. Feireisl and A. Novotn´

  • y. Singular limits in thermodynamics of viscous
  • fluids. Birkh¨

auser-Verlag, Basel, 2009.

◮ E. Feireisl, J. Neustupa, and J. Stebel. Convergence of a Brinkman-type

penalization for compressible fluid flows. J. Differential Equations 250(1) (2011) 596–606.

◮ E. Feireisl, O. Kreml, ˇ

  • S. Neˇ

casov´ a, J. Neustupa, J. Stebel. Weak solutions to the barotropic Navier-Stokes system with slip boundary conditions in time dependent domains. J. Differential Equations 254 (1) (2013) 125–140.

◮ O. A. Ladyzhenskaja. An initial-boundary value problem for the

Navier-Stokes equations in domains with boundary changing in time. Zap. Nauˇ

  • cn. Sem. Leningrad. Otdel. Mat. Inst. Steklov. (LOMI) 11 (1968)

97–128.