A Journey through the World of Incompressible Viscous Flows : an - PowerPoint PPT Presentation

A Journey through the World of Incompressible Viscous Flows : an Evolution Equation Perspective Matthias Hieber TU Darmstadt, Germany Mathematics for Nonlinear Phenomena, Conference in Honor of Prof. Yoshikazu Giga on his 60th Birthday Sapporo August 17, 2015

Balance Laws for Incompressible Fluids Incompressible fluids are subject to balance laws of momentum, mass and energy ̺ ( ∂ t + u · ∇ ) u + ∇ π = div S in Ω , div u = 0 in Ω , ̺ ( ∂ t + u · ∇ ) ǫ + div q = S : ∇ u in Ω , u = q · ν = 0 on ∂ Ω , u , ̺ , π velocity, density, pressure of fluid S stress tensor ǫ internal energy q heat flux Ω ⊂ R n bounded domain with smooth boundary conservation of energy since for total energy e := | u | 2 / 2 + ǫ ρ ( ∂ t + u · ∇ ) e + div ( q + π u − Su ) = 0 in Ω � integrating yields ∂ t E( t ) = 0 where E( t ) = Ω ρ edx provided q · ν = u = 0 on ∂ Ω Hence, above boundary conditions imply : total energy is preserved, independent of choice of S and q

Entropy Considerations define free energy ψ as ψ = ψ ( ρ, θ ), θ temperature define entropy η by η = − ∂ θ ψ Clausius-Duhem equation yields : entropy production given by θ r := S : ∇ u − q · ∇ θ/θ + ( ρ 2 ∂ ρ ψ − π )( div u ) � total entropy N = Ω ρη is non-decreasing provided r ≥ 0 div u has no sign : require π = ρ 2 ∂ ρ ψ , Maxwell’s relation Further, S : ∇ u ≥ 0 and q · ∇ θ ≤ 0, classical conditions Summary : For all choices of S and q , there is conservation of energy and total entropy is non-decreasing provided classical, Maxwell’s conditions and boundary conditions hold we say : model is thermodynamically consistent Example : classical laws due to Newton and Fourier : S := S Newton := 2 µ s D + µ b div u I , 2 D = ( ∇ u + [ ∇ u ] T ), q = − α 0 ∇ θ thermodynamically consistent if µ s ≥ 0, 2 µ s + n µ b ≥ 0 and α 0 ≥ 0

Examples Isothermal and Incompressible Situation In the special case where θ = const , ̺ = const , we obtain S = 0 : Euler’s equations, S = S Newton = 2 µ D ( u ) : equations of Navier-Stokes S = S Newton + S elastic , fluids of Oldroyd-B type S = S non − Newton , non-Newtonian fluids hydrostatic approximation of pressure : primitive equations in geophysical sciences Non-Isothermal Situation S = S Ericksen + S Leslie , Ericksen-Leslie model for liquid crystals

Stage 1 : The equations of Navier-Stokes Setting ̺ = µ = 1, we have u t − ∆ u + ( u · ∇ ) u + ∇ p = f , in [0 , T ] × Ω div u = 0 , in [0 , T ] × Ω = 0 , in [0 , T ] × ∂ Ω u u (0) = u 0 , in Ω Strategy for strong well-posedness : write equations of Navier-Stokes as Evolution Equation u ′ ( t ) − Au ( t ) = − P [ u ( t ) · ∇ ) u ( t ) in Banach space L p σ (Ω), where ◮ A = P ∆, Stokes operator ◮ P , Helmholtz projection rewrite evolution equation as integral equation � t e ( t − s ) A P [( u ( s ) · ∇ ) u ( s )] ds u ( t ) = e tA u 0 − 0 solve integral equation via fixed point methods central importance : properties of Stokes operator and Stokes semigroup

The Stokes Operator, 1 < p < ∞ Define Stokes operator A p u = P p ∆ u for 1 < p < ∞ with domain D ( A p ) = W 2 , p (Ω) ∩ W 1 , p (Ω) ∩ L p σ (Ω) provided ∂ Ω smooth 0 Then A 2 selfadjoint and generator of bounded analytic semigroup on L 2 σ (Ω) Ω bounded : A 2 invertible and A − 1 has compact resolvent. 2 L p -setting, much more difficult : key result by Y. Giga ’81 : A p generates analytic semigroup on L p σ (Ω) Hence : above strategy can be made rigorous and solving Navier-Stokes is equivalent to � t e ( t − s ) A P [( u ( s ) · ∇ ) u ( s )] ds u ( t ) = e tA u 0 − 0 Find function space F in which iteration scheme ◮ u 1 ( t ) = e tA u 0 � t ◮ u n +1 ( t ) = e tA u 0 − 0 e ( t − s ) A P [( u n ( s ) ∇ ) u n ( s )] ds converges.

Unique, Strong solutions for Equations of Navier-Stokes Assume Ω ⊂ R 3 bounded, ∂ Ω smooth Fujita-Kato : if either u 0 ∈ D ( A ) 1 / 4 or interval of existence for T is sufficiently small, then there exists a unique, strong solution on [0 , T ). in particular : L 2 -situation : u 0 ∈ ˙ H 1 / 2 Extension of iteration schema on scaling invariant function spaces key results by Y. Giga ’86, T. Kato : u 0 ∈ L p σ (Ω) for p ≥ 3 Cannone-Meyer : Well-posedness for u 0 ∈ B − 1+3 / p ( R 3 ) p , ∞ Koch-Tataru : Well-posedness for u 0 ∈ BMO − 1 ( R 3 ) Bourgain-Pavlovic : Ill-posedness for u 0 ∈ B − 1 ∞ , ∞ ( R 3 ), i.e. solution operator u 0 �→ u ( t ) is not continuous with respect to � · � B − 1 ∞ , ∞ global strong solution provided n = 2

The role of the Stokes semigroup Following Y. Giga’s approach : 2 main ingredients needed in proof for u 0 ∈ L 3 ( R 3 ) : global L p − L q -smoothing : � e tA f � q ≤ Ct − n / 2(1 / p − 1 / q ) � f � p , t > 0 , 1 < p ≤ q < ∞ global gradient estimates : �∇ e tA f � p ≤ Ct − 1 / 2 � f � p , t > 0 Then : ( t n / 2(1 / p − 1 / q ) u j ) is Cauchy sequence in BC ([0 , T ); L q σ ) and ( t 1 / 2 ∇ u j ) is Cauchy sequnece in BC ([0 , T ); L q ) Hence : global solution provided u 0 small further : analyticity of e tA implies classical solution

Navier-Stokes in Geophysical Setting Geophysical situation : rotating coordinate system yields additional term : Coriolis force in [0 , T ] × R 3 u t − ∆ u + ( u · ∇ ) u + ω e 3 × u + ∇ p = f , div u = 0 , in [0 , T ] × Ω u (0) = u 0 , in Ω Babenko, Mahalov, Nikolenco : pioneering result on global well-posedness for large data provided ω is large enough global well-posedness result Chemin, Desjardins, Gallagher, Grenier : let u 0 ∈ L 2 ( R 2 ) 2 + H 1 / 2 ( R 3 ) with div u 0 = 0. Then exists ω 0 > 0 such that for all ω ≥ ω 0 the (NSC)-equation admits a unique, global mild solution surprising : no smallness condition for u 0 proof relies on dispersive estimates for linear part.

Stage 2 : The Stokes equation on L ∞ σ (Ω) Aim : well-posedness for Navier-Stokes equations for non-decaying data P = Id − ∇ ( − ∆) − 1 div is not bounded in L ∞ ( R n ) even for Ω = R n positive result on half space : Ω = R n + : Desch, H., Pr¨ uss, 2001 : A generates analytic semigroup on L ∞ σ (Ω) � u · ∇ ϕ = 0 , ϕ ∈ � W 1 , 1 (Ω) } Here : L ∞ σ (Ω) = { u ∈ L ∞ (Ω) : negative result for layers : Ω = R 2 × (0 , 1) negative results for L 1 (Ω) for Ω= half space or layer How to treat situation of domains ? Difficulties : ◮ localization procedure does not work ◮ pressure estimate for ∇ q of the form �∇ q � p ≤ C � ∆ u � p does not hold for p = ∞ .

Approaches Masuda, 1972, Stewart, 1974 : A priori estimates for various elliptic operators on L ∞ (Ω) breakthrough result by Abe-Giga, 2013 : L ∞ -estimates for Stokes system via blow-up argument for admissable domains ◮ N ( u , p )( t , x ) = | u ( t , x ) | + t 1 / 2 |∇ u ( t , x ) | + + t | ∂ t u ( t , x ) | + t |∇ 2 u ( t , x ) | + t |∇ p ( t , x ) | ◮ Then sup 0 < t < T 0 � N ( u , p ) � ∞ ( t ) ≤ C � u 0 � ∞ ◮ Idea : suppose false. ◮ rescaled blow-up sequence ( u m , p m ) on rescaled domain converning to solution ( v , q ) with v 0 ≡ 0. ◮ if convergence strong enough : N ( v , q )(0 , 0) ≥ 1 / 2 ◮ if limit unique, then v ≡ 0 ≡ ∇ q yielding contraction approach jointly with K. Abe and Y. Giga : extend Masuda-Stewart approach to Stokes

Resolvent Estimates for Stokes equation for p = ∞ Consider λ v − ∆ v + ∇ q = f in Ω div v = g in Ω = h on ∂ Ω v Aim : For p > n set | λ || v ( x ) | + | λ | 1 / 2 |∇ v ( x ) | + | λ | n / 2 p �∇ 2 v � L p (Ω x , | λ |− 1 / 2 ) M p ( v , q )( x , λ ) = + | λ | n / 2 p �∇ q � L p (Ω x , | λ |− 1 / 2 ) and show sup � M p ( v , q ) � ∞ ( λ ) ≤ C � f � ∞ λ ∈ Σ Θ Here Ω x 0 , r = B x 0 ( r ) ∩ Ω Σ Θ = sector in complex plane of angle Θ ∈ ( π/ 2 , π )

Idea of Approach Step 1 : Localize Localize equation in Ω ′ = B x 0 (( η + 1) r ) ∩ Ω by setting u = θ v , p = ( q − q c ) θ with cutoff function θ . Then ( u , q ) solves resolvent equation in Ω ′ with error terms on right hand side h and g 2 scaling parameters : η > 0 , r > 0 to be determined later Step 2 : Apply L p -estimates in Ω ′ | λ |� u � p + | λ | 1 / 2 �∇ u � p + �∇ 2 u � p + �∇ p � p ≤ C p ( � h � p + �∇ g � p + | λ |� g � W − 1 , p ) 0 Note : ∂ Ω ′ is not smooth. Step 3 : Estimates for error terms Recall : h = f θ − 2 ∇ v ∇ θ − v ∆ θ + ( q − q c ) ∇ θ estimate for the first three terms as in elliptic situation

Key estimate for pressure Step 4 : Handle pressure term by Poincar´ e-Sobolev type inequality d (Ω) for ϕ ∈ � W 1 , ∞ | ϕ − ( ϕ ) | L p (Ω x 0 , s ) ≤ Cs n / p |∇ ϕ | L ∞ (Ω) d ( ϕ ) mean value of ϕ | f | L ∞ d (Ω) = sup x ∈ Ω d Ω ( x ) | f ( x ) | and d distance function from boundary Step 5 : We call a domain Ω strictly admissible if pressure term can be estimated by the velocity, i.e. more precisely if d Ω ( x ) |∇ q ( x ) | ≤ C Ω | W ( v ) | L ∞ ( ∂ Ω) with W ( v ) = ( ∇ v − ( ∇ v ) T ) n Ω sup x ∈ Ω bounded or exterior domains with smooth boundary are admissible Combining this estimate with Poincar´ e-Sobolev type estimate in Step 4 yields | h | L p (Ω ′ ) ≤ Cr n / p � � ( η +1) n / p | f | ∞ +( η +1) − (1 − n / p ) ( r − 2 | v | ∞ + r − 1 |∇ v | ∞ )