NSCH phase field (mass fraction) c : J G [ 0 , 1 ] fluid moves - - PowerPoint PPT Presentation

SPECTRAL ANALYSIS FOR STEADY STATES OF

COMPRESSIBLE TWO-PHASE FLUIDS OF

NAVIER-STOKES-ALLEN-CAHN TYPE IN

BOUNDED DOMAINS

Matthias Kotschote

Department Mathematics and Statistics, University Konstanz

MathFlows 2015, Porquerolles

OUTLINE

• 1. modelling NSAC/NSCH
• 2. 2nd law of thermodynamics and classes of fluids
• 3. steady states
• 4. spectral analysis

DIFFUSE INTERFACE MODELS

Consider the flow of a binary mixture of macroscopically immiscible, viscous Newtonian fluids filling a domain G ⊂ R3. In classical models: both fluids are separated by a sharp interface Γ(t), across which certain jump conditions are prescribed. Problem: Topological transitions (e.g. due to droplet formation or coalescence) cannot be described. This motivated the development of diffuse interface models: replace the sharp interface by a narrow transition layer across which the fluids may mix.

NSAC: MODELLING

◮ fluid moves with velocity u : J × G → R3 ◮ phase field (mass fraction) c : J × G → [0, 1] corresponding to

concentration of one of two phases c x ∈ G 1 phase 1 phase 2 δ interface

◮ different apparent densities ρ1 = cρ, ρ2 = (1 − c)ρ,

ρ - total mass density, ρj satisfy mass balance equation ∂tρj + ∇ · (ρju) + Jj = 0, J1 + J2 = 0, (1) Jj - transition rates

NSAC: MODELLING

◮ fluid moves with velocity u : J × G → R3 ◮ phase field (mass fraction) c : J × G → [0, 1] corresponding to

concentration of one of two phases

◮ different apparent densities ρ1 = cρ, ρ2 = (1 − c)ρ,

ρ - total mass density, ρj satisfy mass balance equation ∂tρj + ∇ · (ρju) + Jj = 0, J1 + J2 = 0, (1) Jj - transition rates

◮ conservation of total mass ρ:

∂tρ + ∇ · (ρu) = 0

◮ equation for c:

∂t(ρc) + ∇ · (ρcu) + J1 = 0

NSAC: CONSTITUTIVE EQUATION FOR TRANSITION RATE

assume that JAC := J1 = θδΨ δc , Ψ :=

• G

ρ θF dx

F – Helmholtz energy density θ – temperature

δΨ δc - generalised chemical potential: originally proposed

F = cF1 + (1 − c)F2 + θ(W(c) + δ

2|φ|2),

φ := |∇c|2, (2) Fi – Helmholtz energy densities of phase i = 1, 2 W(c) – double-well potential

NSAC: CONSTITUTIVE EQUATION FOR TRANSITION RATE

assume that JAC := J1 = θδΨ δc , Ψ :=

• G

ρ θF dx

F – Helmholtz energy density θ – temperature

δΨ δc - generalised chemical potential: originally proposed

F = cF1 + (1 − c)F2 + θ(W(c) + δ

2|φ|2),

φ := |∇c|2, (2) Fi – Helmholtz energy densities of phase i = 1, 2 W(c) – double-well potential

HELMHOLTZ ENERGY

The restriction (1) is not necessary! Consider Helmholtz energy density F = F(ρ, θ, c, φ), φ := |∇c|2, F smooth enough.

NSAC: CONSTITUTIVE EQUATION FOR TRANSITION RATE

◮ Helmholtz energy density F = F(ρ, θ, c, φ), φ := |∇c|2 ◮ computing δΨ δc yields

δΨ δc = ∂c( ρ

θF) − ∇ · (∂∇c( ρ θF)) = ∂c( ρ θF) − ∇ · (2∂φ( ρ θF)∇c) ◮ equation for c (Allen-Cahn equation):

∂t(ρc) + ∇ · (ρcu) − θ

• ∇ · (2∂φ( ρ

θF)∇c) − ∂c( ρ θF)

• = 0

NSAC: BALANCE OF MOMENTUM

◮ So far we have

∂tρ + ∇ · (ρu) = 0, ∂t(ρc) + ∇ · (cρu) + JAC = 0, with JAC = θ

• − ∇ · (2∂φ( ρ

θF)∇c) + ∂c( ρ θF)

• .

◮ balance of momentum

∂t(ρu) + ∇ · (ρu ⊗ u) − ∇ · T = ρfext T - Cauchy stress fext - external force

NSAC: CONSTITUTIVE EQUATION FOR T

◮ assume

T = S + P S - Newtonian viscous stress, P - pressure tensor

◮ viscous stress:

S = 2ηD(u) + λ∇ · u I, D(u) = 1

2(∇u + ∇uT)

η = η(ρ, c, θ) - shear viscosity, λ = λ(ρ, c, θ) - bulk viscosity, 2η + λ > 0

◮ pressure tensor P:

P = −π I − ∇c ⊗ ∂∇c(ρF) = −π I − 2ρ∂φF ∇c ⊗ ∇c π = ρ2∂ρF thermodynamic pressure, ∇c ⊗ ∂∇c(ρF) - Ericksen’s stress represents capillarity PRESSURE TENSOR In order that the 2nd law of thermodynamics is satisfied the tensor P must have this form.

NSAC: ENERGY EQUATION

∂t(ρE) + ∇ · (ρEu) − ∇ · (β∇θ + [S + P] · u) = ρfext · u,

◮ total energy density E = E + 1 2|u|2 ◮ internal energy E = E(ρ, S, c, |∇c|2), S-Entropy

Laws of thermodynamics relate E and F through the Legendre transform, E = F + θS, S = −∂θF, Θ = ∂SE.

◮ β = β(ρ, c, θ) - heat conductivity

NSAC: MATHEMATICAL PROBLEM

Let J = [0, T] and G ⊂ R3 be a domain (with C2 boundary Γ). Consider the Navier-Stokes-Allen-Cahn system for compressible fluids ∂tρ + ∇ · (ρu) = 0, J × G, ∂t(ρu) + ∇ · (ρu ⊗ u) − ∇ · (S + P) = ρfext, J × G, ∂t(ρE) + ∇ · (ρEu) − ∇ · (β∇θ + [S + P] · u) = ρfext · u, J × G, ∂t(cρ) + ∇ · (cρu) + JAC = 0, J × G, (3) with initial data ρ(0) = ρ0, u(0) = u0, θ(0) = θ0, c(0) = c0 (4) and boundary conditions non-slip: u = 0 pure slip: (u|ν) = 0, Q(ν)S · ν = 0, Q(ν) := I − ν ⊗ ν (5) and (inhomogeneous) Dirichlet or Neumann b.c. for c and θ.

NSCH

◮ phase field (mass fraction) c : J × G → [0, 1] ◮ fluid moves with velocity u : J × G → R3 ◮ Helmholtz energy density F = F(ρ, θ, c, φ), φ := |∇c|2 ◮ ρ1 = cρ, ρ2 = (1 − c)ρ, ρ - total mass density,

ρj satisfy the conservation law ∂tρj + ∇ · (ρju) + ∇ · Jj = 0 with J1 + J2 = 0 ⇒ ∂tρ + ∇ · (ρu) = 0

◮ suppose that JCH := J1 is given by (Fick’s law)

JCH = γ∇( 1

θµ)

with mobility γ and generalised chemical potential µ,

ρ θµ := δΨ

δc , Ψ =

• Ω

ρ θF dx

NSCH: MATHEMATICAL PROBLEM

Let J = [0, T] and G ⊂ R3 be a domain (with C4 boundary Γ). Consider the compressible Navier-Stokes-Cahn-Hilliard system ∂tρ + ∇ · (ρu) = 0, J × G, ∂t(ρu) + ∇ · (ρu ⊗ u) − ∇ · (S + P) = ρfext, J × G, ∂t(ρE) + ∇ · (ρEu) − ∇ · (β∇θ + [S + P] · u) = ρfext · u, J × G, ∂t(cρ) + ∇ · (cρu) − ∇ · (γ∇ 1

θµ)

• = 0,

J × G, ∂c( ρ

θF) − ∇ ·

• 2∂φ( ρ

θF)∇c

• = ρ

θµ,

J × G, (6) with initial data ρ(0) = ρ0, u(0) = u0, c(0) = c0 (7) and boundary conditions non-slip: u = 0 pure-slip: (u|ν) = 0, Q(ν)S · ν = 0, Q := I − ν ⊗ ν Neumann: (∇µ|ν) = 0, (∇c|ν) = 0 (8)

SOME FACTS

◮ type of equations

system of hyperbolic-parabolic equations quasilinear partial differential equations, e.g. ρ∂tu, ∇ · (2∂φ(ρF)∇c) quasilinearity of highest order: ∇ · (2∂φ(ρF) ∇c ⊗ ∇c), similar to quasilinear elliptic operators: ∇ · (a(∇v)∇v) avoid vacuum, e.g. in ρ∂tu + ρ∇u · u − ∇ · S(u) + ∇ · (π I + 2ρ∂φF ∇c ⊗ ∇c) = ρfext

◮ existence and uniqueness

local and global well-posedness in 1D (strong and classical solutions) local well-posedness in any dimension existence of weak solutions (for specific Helmholtz energy and modified Allen-Cahn/Cahn-Hilliard equation)

◮ existence of travelling waves ◮ 2nd law of thermodynamics ◮ relation to Korteweg

2nd LAW OF THERMODYNAMICS

For any Helmholtz energy F(ρ, θ, c, |∇c|2) the transformation rate (JAC/JCH) is defined in such a way that both NSAC and NSCH induce an entropy balance ∂t(ρS) + ∇ · (ρSu) = ∇ · Σ + σ (9) with a net entropy production rate σ ≥ 0 and entropy exchange ∇ · Σ.

THEOREM

The thermodynamically closed systems of NSAC and NSCH are thermodynamically and mechanically consistent and σAC = σ0 +

1 ρθ|JAC|2,

ΣAC = 1

θ

• β∇θ + JAC ∂∇cF
• ,

σCH = σ0 + 1

γ |JCH|2,

ΣCH = 1

θ

• β∇θ + ∇ · JCH ∂∇cF − µJCH
• ,

σ0 := 1

θS : D + β

• ∇θ

θ

• 2 .

CLASSES OF FLUIDS

How can we construct a Helmholtz energy for mixture of fluids? Prototypical approach (for “real applications”): F(τ, θ, c, φ) = cF1 + (1 − c)F2 + Fmix(θ, c, |∇c|2), τ = 1/ρ, Fmix = W(θ, c) + δ

2|∇c|2

convex combination of energies F1 and F2 of the separate phases Fmix - mixing entropy, δ - a measure of thickness for the interface typically: W - double-well potential, e.g. W(c) = k1[c ln(c) + (1 − c) ln(1 − c)] + k2c(1 − c), k2 < 0 - interactions between different phases are more favorable k2 > 0 - separation of the phases

Question: What are the variables in the equation of state Fi?

CLASSES OF FLUIDS

Helmholtz energy: F(τ, θ, c, φ) = cF1 + (1 − c)F2 + Fmix(θ, c, |∇c|2), τ = 1/ρ, Fmix = W(θ, c) + δ

2|∇c|2

Question: What are the variables in the equation of state Fi? Answer: As F1 and F2 are the Helmholtz energies of the separate phases, the variables are the specific ones, i.e.

• 1. take a “energy-law” of fluid j = 1, 2,

Fj = Fj(˜ τj, ˜ θj) with ˜ τj = 1/˜ ρj denoting the specific volume of phase j;

CLASSES OF FLUIDS

Helmholtz energy: F(τ, θ, c, φ) = cF1 + (1 − c)F2 + Fmix(θ, c, |∇c|2), τ = 1/ρ, Fmix = W(θ, c) + δ

2|∇c|2

Question: What are the variables in the equation of state Fi? Answer: As F1 and F2 are the Helmholtz energies of the separate phases, the variables are the specific ones, i.e.

• 1. take a “energy-law” of fluid j = 1, 2,

Fj = Fj(˜ τj, ˜ θj) with ˜ τj = 1/˜ ρj denoting the specific volume of phase j;

• 2. need “mixing rules” — a constitutive law that characterizes the type of

the mixing, i.e. “laws” relating (˜ τj, ˜ θj) and “global variables” (τ, θ, c), ˜ τj = Tj(τ, θ, c), ˜ θj = Θj(τ, θ, c) with diffeomorphisms Tj, Θj.

CLASSES OF FLUIDS (TWO COMPRESSIBLE FLUIDS)

“MIXING RULES” – AN INSTANCE

• 1. ˜

θj = Θj(τ, θ, c) = θ for j = 1, 2

(Two neighbouring phases shall have the same temperature.)

• 2. Raoult’s law:

p = ˜ p1c + ˜ p2(1 − c) ˜ p1 = ˜ p2 ⇔ ˜ p1 = P1(p, θ, c) = p ˜ p2 = P2(p, θ, c) = p

(Phases shall have the same pressure at the interface.)

Better to start with Gibbs energy G = G(p, θ, c, |∇c|2), i.e. G(p, θ, c, |∇c|2) = cG1(˜ p1, ˜ θ) + (1 − c)G2(˜ p2, ˜ θ) + Gmix(c, θ, |∇c|2) (Gi – Gibbs energies of separate fluids)

CLASSES OF FLUIDS (TWO COMPRESSIBLE FLUIDS)

“MIXING RULES” – AN INSTANCE

• 1. ˜

θj = Θj(τ, θ, c) = θ for j = 1, 2

(Two neighbouring phases shall have the same temperature.)

• 2. Raoult’s law:

p = ˜ p1c + ˜ p2(1 − c) ˜ p1 = ˜ p2 ⇔ ˜ p1 = P1(p, θ, c) = p ˜ p2 = P2(p, θ, c) = p

(Phases shall have the same pressure at the interface.)

Better to start with Gibbs energy G = G(p, θ, c, |∇c|2), i.e. G(p, θ, c, |∇c|2) = cG1(p, θ) + (1 − c)G2(p, θ) + Gmix(c, θ, |∇c|2) (Gi – Gibbs energies of separate fluids)

CLASSES OF FLUIDS (TWO COMPRESSIBLE FLUIDS)

“MIXING RULES” – AN INSTANCE

• 1. ˜

θj = Θj(τ, θ, c) = θ for j = 1, 2

(Two neighbouring phases shall have the same temperature.)

• 2. Raoult’s law:

p = ˜ p1c + ˜ p2(1 − c) ˜ p1 = ˜ p2 ⇔ ˜ p1 = P1(p, θ, c) = p ˜ p2 = P2(p, θ, c) = p

(Phases shall have the same pressure at the interface.)

Better to start with Gibbs energy G = G(p, θ, c, |∇c|2), i.e. G(p, θ, c, |∇c|2) = cG1(p, θ) + (1 − c)G2(p, θ) + Gmix(c, θ, |∇c|2) (Gi – Gibbs energies of separate fluids) How does the associated Helmholtz energy F look like?

CLASSES OF FLUIDS (TWO COMPRESSIBLE FLUIDS)

So far we have the Gibbs energy G(p, θ, c, |∇c|2) = cG1(p, θ) + (1 − c)G2(p, θ) + Gmix(c, θ, |∇c|2). The Helmholtz energy F = F(τ, θ, c, |∇c|2), τ = 1/ρ, is obtained via Legendre transform. After some calculations we get F = cF1(˜ τ1, θ) + (1 − c)F2(˜ τ2, θ) + Fmix(c, θ, |∇c|2) with

• 1. Fmix ≡ Gmix
• 2. Fi is related to Gi through:

Gi = Fi + ˜ τi˜ pi, ˜ pi = −∂˜

τiFi,

˜ τj = ∂˜

pjGj

3. (∗) τ = ˜ τ1c + ˜ τ2(1 − c) ˜ p1 = ˜ p2 ⇔ τ = ˜ τ1c + ˜ τ2(1 − c) 0 = ∂˜

τ1F1(˜

τ1, θ) − ∂˜

τ1F2(˜

τ1, θ) Conclusion: The “mixing rule” called Raoult’s law is equivalent to the “mixing rule” (∗). For strictly convex functions Fi: ˜ τi = Ti(τ, c, θ), i = 1, 2.

CLASSES OF FLUIDS

MIXING OF TWO INCOMPRESSIBLE FLUIDS

In the case of two immiscible incompressible phases of different temperature-independent specific volumes, i.e. the Gibbs energies G1, G2 of both phases are affine and Raoult’s law holds which leads to G(p, θ, c, φ) = ˆ τ(c)p + Gmix(θ, c, |∇c|2), ˆ τ(c) = cτ1 + (1 − c)τ2 with constants τ1, τ2 > 0 satisfying τ1 − τ2 = 0, NSAC reduces literally to the Navier-Stokes-Korteweg system with Helmholtz energy density F and viscous stress S∗ given by F(ρ, θ, |∇ρ|2) = Gmix(θ,ˆ c(ρ), |ˆ c′(ρ)|2 |∇ρ|2), ˆ c(ρ) := 1/ρ − τ2 τ1 − τ2 , S∗ = S +

1 ρ(τ1−τ2)2 ∇ · u I.

CLASSES OF FLUIDS

MIXING RULES & EQUATIONS OF STATE

Depending on Fj and the “mixing rules” different classes of fluid mixtures can be modelled (mixture of two compressible fluids, of one compressible and one incompressible fluid, and of two incompressible fluids). These things can be found in: Freist¨ uhler, K.: Phase-Field and Korteweg-Type Models for the Time-Dependent Flow of Compressible Two-Phase Fluids, submitted.

STEADY STATES OF NSAC: ISOTHERMAL CASE

Recall the isothermal (θ = 1) NSAC system ∂tρ + ∇ · (ρu) = 0, J × G, ∂t(ρu) + ∇ · (ρu ⊗ u) − ∇ · S − ∇ · P = ρfext, J × G, ∂t(cρ) + ∇ · (cρu) + JAC = 0, J × G, plus boundary conditions and initial data. Let F = f(ρ, c) + δ

2|∇c|2,

δ = const > 0, fext = −∇ϕ(ρ), f, ϕ are smooth enough, and thus JAC = −∇ · (∂∇c(ρF)) + ∂c(ρF) = −∇ · (δρ∇c) + ∂c(ρF), P = −ρ2∂ρF I − ∇c ⊗ ∂∇c(ρF) = −ρ2∂ρ F I − ρδ∇c ⊗ ∇c, S = 2ηD(u) + λ∇ · u I.

STEADY STATES OF NSAC: ISOTHERMAL CASE

∂tρ + ∇ · (ρu) = 0, J × G ∂t(ρu) + ∇ · (ρu ⊗ u) − ∇ · S−∇ · P= −ρ∇ϕ(ρ), J × G ∂t(cρ) + ∇ · (cρu) + JAC= 0, J × G Seek stationary solutions of the form w = (0, ρ, c), i.e. −∇ · P = ∇ · (ρ2∂ρF I + ρδ∇c ⊗ ∇c) = −ρ∇ϕ(ρ), x ∈ G, JAC = −∇ · (ρδ∇c) + ∂c(ρF) = 0, x ∈ G, c = 0

• r

∂νc = 0, x ∈ ∂G,

• G

ρ dx = m0. (10)

STEADY STATES OF NSAC: ISOTHERMAL CASE

Seek stationary solutions of the form w = (0, ρ, c), i.e. −∇ · P = ∇ · (ρ2∂ρF I + ρδ∇c ⊗ ∇c) = −ρ∇ϕ(ρ), x ∈ G, JAC = −∇ · (ρδ∇c) + ∂c(ρF) = 0, x ∈ G, c = 0

• r

∂νc = 0, x ∈ ∂G,

• G

ρ dx = m0. (10) Introduce Ψ(ρ, c, ∇c) = ρF(ρ, c, |∇c|2) + Φ(ρ) = ρf(ρ, c) + ρ δ

2|∇c|2 + Φ(ρ),

Φ(ρ) = ρ ϕ(s) ds then (10) is equivalent to ρ∇Ψρ − [Ψc − ∇ · (ρδ∇c)]∇c = 0, x ∈ G, −∇ · (ρδ∇c) + Ψc = 0, x ∈ G

STEADY STATES OF NSAC: ISOTHERMAL CASE

Seek stationary solutions of the form w = (0, ρ, c), i.e. −∇ · P = ∇ · (ρ2∂ρF I + ρδ∇c ⊗ ∇c) = −ρ∇ϕ(ρ), x ∈ G, JAC = −∇ · (ρδ∇c) + ∂c(ρF) = 0, x ∈ G, c = 0

• r

∂νc = 0, x ∈ ∂G,

• G

ρ dx = m0. (10) Introduce Ψ(ρ, c, ∇c) = ρF(ρ, c, |∇c|2) + Φ(ρ) = ρf(ρ, c) + ρ δ

2|∇c|2 + Φ(ρ),

Φ(ρ) = ρ ϕ(s) ds then (10) is equivalent to Ψρ = π0, x ∈ G, −∇ · (ρδ∇c) + Ψc = 0, x ∈ G, plus b.c. for c and integral condition for ρ.

STEADY STATES OF NSAC

The complete steady state problem reads as Ψρ(ρ, c, |∇c|2) = π0, x ∈ G, −∇ · (δρ∇c) + Ψc(ρ, c, |∇c|2) = 0, x ∈ G, ∂νc = 0

• r

c = 0, x ∈ ∂G,

• G

ρ dx = m0 > 0. (11) The unknowns are ρ, c, and π0.

STEADY STATES OF NSAC

The complete steady state problem reads as Ψρ(ρ, c, |∇c|2) = π0, x ∈ G, −∇ · (δρ∇c) + Ψc(ρ, c, |∇c|2) = 0, x ∈ G, ∂νc = 0

• r

c = 0, x ∈ ∂G,

• G

ρ dx = m0 > 0. (11) The unknowns are ρ, c, and π0. Assume Ψρρ > 0. This implies ∃ P : [0, 1] × R+ × R → R+ with ρ = P(c, |∇c|2, π0) solving ∂ρΨ(P(c, |∇c|2, π0), c, ∇c) = π0, x ∈ G.

STEADY STATES OF NSAC

The complete steady state problem reads as Ψρ(ρ, c, |∇c|2) = π0, x ∈ G, −∇ · (δρ∇c) + Ψc(ρ, c, |∇c|2) = 0, x ∈ G, ∂νc = 0

• r

c = 0, x ∈ ∂G,

• G

ρ dx = m0 > 0. (11) The unknowns are ρ, c, and π0. Regard now ρ as P(c, |∇c|2, π0). This results in −M : ∇2c + δ Ψρc

Ψρρ |∇c|2 + Ψc = 0,

x ∈ G, c = 0

• r

∂νc = 0, x ∈ ∂G (12) with symmetric matrix M(ρ, c, ∇c) = δρ

• I −

δ ρΨρρ ∇c ⊗ ∇c

• .

1 −

δ ρΨρρ |∇c|2 > 0

⇒ quasilinear elliptic problem for c.

STEADY STATES OF NSAC

The problem −M : ∇2c + δ Ψρc

Ψρρ |∇c|2 + Ψc = 0,

x ∈ G, c = 0

• r

∂νc = 0, x ∈ ∂G (13) is a task of its own and more involved as one might think.

◮ ρ = P(c, |∇c|2, π0) depends on π0 which is still unknown.

⇒ the constant π0 is involved in (13)

◮ If (13) can be solved (Leray-Schauder principle) then c = cπ0 and thus

ρ = P(cπ0, |∇cπ0|2, π0). Use the condition m0 =

• G

P(cπ0, |∇cπ0|2, π0) dx =: M(π0). π0 = M−1(m0)?

◮ There are configurations that admit constant solutions. These

correspond to one phase. (uninteresting)

LINEARIZATION AROUND STEADY STATES

Let denote (0, ˆ ρ(x),ˆ c(x)) a steady state solution. The deviation w = (v, q, χ) = (u − 0, ρ − ˆ ρ, c − ˆ c) from equilibrium satisfies ˆ ρ∂tv − ∇ · ˆ S(v) + ˆ ρ∇B(q, χ) − A(q, χ)∇ˆ c = F1(w), J × G, ∂tq + ∇ · (ˆ ρv) = F2(w), J × G, ˆ ρ∂tχ + ˆ ρ∇ˆ c · v + A(q, χ) = F3(w), J × G, (v, q, χ)(0) = w0, G, (14) plus (inhomogeneous) boundary conditions as well as

• G q dx = 0, where

ˆ S(v) := 2ˆ µD(v) + ˆ λ∇ · v I, B(q, χ) := ˆ Ψρρq + ˆ Ψρcχ + δ∇ˆ c · ∇χ, A(q, χ) := −∇ · (δˆ ρ∇χ + δ∇ˆ cq) + ˆ Ψρcq + ˆ Ψccχ. Notation: ˆ a = a(ˆ ρ(x), ˆ χ(x)) with a ∈ {η, λ, Ψρρ, . . .} Assumptions: ˆ Ψρρ, ˆ Ψcc > 0, but ˆ Ψ′′ is not definite! Associate (14) with C∂tw + Lw = f(w), J × G, Bw = g(w), J × ∂G, w = w0, G. (15)

THE LINEAR PROBLEM

Consider (15) with given data, i.e. C∂tw + Lw = f, J × G, Bw = g, J × ∂G, w = w0, G. (16) One can prove maximal Lp-regularity (local well-posedness) for this linear

• problem. Investigate now the point spectrum of the linear operator

L := C−1L : D(L) ⊂ X → X, D(L) = {w ∈ H

2

p(G; Rn) × H

1

p(G) × H

2

p(G) : Bw = 0 on ∂G,

• G

q dx = 0}, X := Lp(G; Rn) × H

1

p(G) × Lp(G),

p ≥ 2.

POINT SPECTRUM OF L

Let κ ∈ C and w = (v, q, χ) ∈ D(L). Consider now κw + Lw = 0. Written

• ut

κˆ ρv − ∇ · ˆ S(v) + ˆ ρ∇B(q, χ) − A(q, χ)∇ˆ c = 0, κq + ∇ · (ˆ ρv) = 0, κˆ ρχ + ˆ ρ∇ˆ c · v + A(q, χ) = 0, (17) where ˆ S(v) := 2ˆ µD(v) + ˆ λ∇ · v I, B(q, χ) := ˆ Ψρρq + ˆ Ψρcχ + δ∇ˆ c · ∇χ, A(q, χ) := −∇ · (δˆ ρ∇χ + θδ∇ˆ cq) + ˆ Ψρcq + ˆ Ψccχ. Using energy methods leads to κ

• ˆ

ρv2

L2(G;Rn) +

• G

ˆ S(v) : D(v) dx +

• G

ˆ ρ−1|A(q, χ)|2 dx + κ

• G

(ˆ Ψ′′ · (q, χ, ∇χ)|(q, χ, ∇χ)) dx = 0. (18) The Hessian matrix ˆ Ψ′′(x) is not definite.

POINT SPECTRUM OF L

Let κ ∈ C and w = (v, q, χ) ∈ D(L). Using energy methods leads to κ

• ˆ

ρv2

L2(G;Rn) +

• G

ˆ S(v) : D(v) dx +

• G

ˆ ρ−1|A(q, χ)|2 dx + κ

• G

(ˆ Ψ′′ · (q, χ, ∇χ)|(q, χ, ∇χ)) dx = 0. (17) The Hessian matrix ˆ Ψ′′(x) is not definite. Moreover, Re κ = −

• G ˆ

S(v) : D(v) dx +

• G ˆ

ρ−1|A(q, χ)|2 dx √ˆ ρv2

L2(G;Rn) +

• G(ˆ

Ψ′′ · (q, χ, ∇χ)|(q, χ, ∇χ)) dx (18) and Im κ ·

• ˆ

ρv2

L2(G;Rn) −

• G

(ˆ Ψ′′ · (q, χ, ∇χ)|(q, χ, ∇χ)) dx

• = 0.

(19)

POINT SPECTRUM OF L

• 1. Im κ = 0: using

Re κ = −

• G ˆ

S(v) : D(v) dx +

• G ˆ

ρ−1|A(q, χ)|2 dx √ˆ ρv2

L2(G;Rn) +

• G(ˆ

Ψ′′ · (q, χ, ∇χ)|(q, χ, ∇χ)) dx Im κ ·

• ˆ

ρv2

L2(G;Rn) −

• G

(ˆ Ψ′′ · (q, χ, ∇χ)|(q, χ, ∇χ)) dx

• = 0

we get Re κ = −

• G ˆ

S(v) : D(v) dx +

• G ˆ

ρ−1|A(q, χ)|2 dx 2√ˆ ρv2

L2(G;Rn)

< 0, since Re κ = 0 implies v = 0, χ = 0, and q = 0.

POINT SPECTRUM OF L

• 2. κ = 0: the relation

κ

• ˆ

ρv2

L2(G;Rn) +

• G

ˆ S(v) : D(v) dx +

• G

ˆ ρ−1|A(q, χ)|2 dx + κ

• G

(ˆ Ψ′′ · (q, χ, ∇χ)|(q, χ, ∇χ)) dx = 0 implies v = 0 and A(q, χ) = 0. Thus −∇ · ˆ S(v) + ˆ ρ∇B(q, χ) − A(q, χ)∇ˆ c = 0, ∇ · (ˆ ρv) = 0, ˆ ρ∇ˆ c · v + A(q, χ) = 0 (20) reduces to B(q, χ) = π0, x ∈ G, A(q, χ) = 0, x ∈ G, (21) plus boundary condition for χ and

• Ω q dx = 0. The constant π0 is an

unknown!

POINT SPECTRUM OF L

• 2. κ = 0: The problem

B(q, χ) = π0, x ∈ G, A(q, χ) = 0, x ∈ G, b.c. for χ and

• G

q dx = 0 can be reformulated as an elliptic problem for χ: Sχ = 0, S : D(S) ⊂ Lp(G) → Lp(G), D(S) = {φ ∈ H

2

p(G) : φ satisfies b.c.}.

POINT SPECTRUM OF L

• 2. κ = 0: The problem

B(q, χ) = π0, x ∈ G, A(q, χ) = 0, x ∈ G, b.c. for χ and

• G

q dx = 0 can be reformulated as an elliptic problem for χ: Sχ = 0, S : D(S) ⊂ Lp(G) → Lp(G), D(S) = {φ ∈ H

2

p(G) : φ satisfies b.c.}.

Properties:

S is selfadjoint for p = 2. The unknown π0 can be expressed by χ, i.e. π0 = Π(χ). This relation will be used later. Depending on the situation (Helmholtz energy and the steady state) there is(are) non-trivial solution(s).

POINT SPECTRUM OF L

Re κ = −

• G ˆ

S(v) : D(v) dx +

• G ˆ

ρ−1|A(q, χ)|2 dx √ˆ ρv2

L2(G;Rn) +

• G(ˆ

Ψ′′ · (q, χ, ∇χ)|(q, χ, ∇χ)) dx

• 3. κ ∈ R\{0}: In this case we have to understand “the red term”.

POINT SPECTRUM OF L

Re κ = −

• G ˆ

S(v) : D(v) dx +

• G ˆ

ρ−1|A(q, χ)|2 dx √ˆ ρv2

L2(G;Rn) +

• G(ˆ

Ψ′′ · (q, χ, ∇χ)|(q, χ, ∇χ)) dx

• 3. κ ∈ R\{0}: In this case we have to understand “the red term”.

Surprisingly,

• G

(ˆ Ψ′′ · (q, χ, ∇χ)|(q, χ, ∇χ)) dx =

• G

1 ˆ Ψρρ |B(q, χ) − Π(χ)|2 dx

+ Sχ, χL2(G).

POINT SPECTRUM OF L

RELATIONS

• 1. 0 ∈ σp(−L) ⇔ 0 ∈ σp(−S)
• 2. if Imκ = 0:

Re κ = −

• G ˆ

S(v) : D(v) dx +

• G ˆ

ρ−1|A(q, χ)|2 dx 2√ˆ ρv2

L2(G;Rn)

< 0

• 3. if κ ∈ R\{0}:

κ = −

• G ˆ

S(v) : D(v) dx +

• G ˆ

ρ−1|A(q, χ)|2 dx √ˆ ρv2

L2(G;Rn) +

• G

1 ˆ Ψρρ |B(q, χ) − Π(χ)|2 dx + Sχ, χL2(G)

3.1. σp(−S) ⊂ (−∞, 0) ⇒ Re κ < 0 for all κ ∈ σp(−L) 3.2. σp(−L) ∩ (0, ∞) = ∅ ⇒ σp(−S) ∩ (0, ∞) = ∅

POINT SPECTRUM OF L

Last question: σp(−S) ∩ (0, ∞) = ∅ ⇒ σp(−L) ∩ (0, ∞) = ∅? The idea: let κ > 0

• 1. Augment

(EVP) κw + Lw = 0, w = (v, q, χ), with additional equations and “unknowns” in such a way that:

∀ε = 0: (EVP)ε κA(ε)h + B(ε)h = 0, h = (v, q, χ, ...), is equivalent to (EVP). A, B are continuous in ε. ε = 0: κA(0)h + B(0)h = 0 ⇔ κχ + Sχ = 0.

• 2. Use perturbation theorem (Chow & Hale). E.g., let κ0 > 0 be a simple

eigenvalue of −S. Then there is r > 0 such that for |ε| < r there is a unique κ(ε) ∈ C with |κ(ε) − κ0| < r, continuous in ε, κ(0) = κ0 such that κ(ε) is a simple eigenvalue of (EVP)ε.

Thank you.