2 phase problem for the Navier-Stokes equations in the whole space.

Transcription

1 2 phase problem for the Navier-Stokes equations in the whole space Yoshihiro Shibata Department of Mathematics & Research Institute of Sciences and Engineerings, Waseda University, Japan Department of Mechanical Engineering and Materials Science, University of Pittsburgh, USA Partially supported by Top Global University Project, and JSPS Grant-in-aid for Scientific Research (A) 17H0109 Mathflows 2018 at Porquerolles Organizers: F Charve, R Danchin, P Mucha Suponsored by ANR INFAMIE, Université Paris Universit Paris-Est Crteil and Uniwersytet Warszawski YShibata (Waseda) 2 phase problem Sept3-7, / 16

2 Roughly speaking, the free boundary problem is formulated as the Navier-Stokes equations in unknown time dependent domain Ω t with nonlinear boundary conditions on Γ t Since Ω t is unknown, we transform Ω t to a known reference domain Ω, and so the problem becomes a quasilinear equation in Ω with nonlinear boundary conditions on Ω Thus, to solve the problem at least locally in time, we need the maximal regularity for the Stokes equations with free boundary conditions YShibata (Waseda) 2 phase problem Sept3-7, / 16

3 Problem Let Ω + be a bounded domain in R N and Γ its boundary that is a smooth compact hypersurface Let Ω = R N \ Ω + and two different incompressible viscous fluids occupy Ω ±, respectively Let Ω t± and Γ t be time evolutions of Ω ± and Γ for t > 0 Let n t be the unit outer normal to Γ t Problem is to find domains Ω t±, velocity v ± = (v 1±,, v N± ) and pressure p ± satisfying (1) t v ± + (v ± )v ± m 1 ± Div (µ ±D(v ± ) p ± I) = 0, div v ± = 0 in 0<t<T [[µd(v) pi]]n t = σh(γ t )n t, [[v]] = 0 V Γt = v n t on v ± t=0 = v 0±, Ω t± t=0 = Ω ± 0<t<T Ω t± {t}, Γ t {t}, h = h ± for x Ω t±, m ±, µ ± : positive constants, mass densities and viscosity coefficients, σ: positive constant, coefficient of surface tension, H(Γ t ): N 1 times mean curvature of Γ t, V Γt : evolution speed of Γ t in the n t direction, I is the N N identity matrix, D(v) = v + v = doubled deformation tensor whose (i, j) component is i v j + j v i, [[f]](x 0 ) = x x0 lim f(x) x x0 lim f(x) jump of f at x 0 Γ t x Ω + x Ω YShibata (Waseda) 2 phase problem Sept3-7, / 16

4 I V Denisova and V A Solonnikov, Solvability of a linerized problem on the motion of a drop in a fluid flow, Mat Inst Steklov (LOMI) 171 (1989), I V Denisova, Global solvability of aproblem on two fluid motion without surface tension, Mat Inst Steklov (POMI) 248 (2007), 19 39, I V Denisova and V A Solonnikov, Global solvability of a problem governing the motion of two incompressible capillary fluids in a container, Zamp Nauk Semi POMI 397 (2011), I V Denisova, Global L 2 -solvability of a problem govering two-phase fluid motion without surface tension, Potrugal Math (NS) 71, 2014, 1 24 N Tanaka, Two-phase free boundary problem for viscous incompressible termocapillary convection, Japan J Math (NS) 21 (1) 1995, 1 42 J Pruess and G Simonett, On the two-phase Navier-Stokes equations with surface tension, Interfaces Free Bound 12(3), 2010, J Pruess and G Simonett, Analytic solutions for the two-phase Navier-Stokes equtions with surface tension and gravity, In Parabolic problems, Vol 80 of Progr Nonlinear Differential Equations Appl pages , 2011 YShibata (Waseda) 2 phase problem Sept3-7, / 16

5 Maximal Regularity for Stokes problem (2) t u ± m 1 ± Div (µ ±D(u ± ) q ± I) = f ±, div u ± = f d± = div f d± in Ω (0, T ), t ρ n u = d on Γ (0, T ), [[µd(u) qi]]n σ( Γ ρ)n = h, [[u]] = 0 in Γ (0, T ), (u ±, ρ) t=0 = (u 0±, ρ 0 ) in Ω ± Γ Weak Neumann Problem: Let 1 < q < For any f L q (Ω) N, there exists a unique ψ Ĥ1 q (Ω) that solves the variational equation: (m 1 ψ, φ) Ω = (f, φ) Ω φ Ĥ1 q (Ω), where Ω = Ω + Ω = R N \ Γ, m = m ± for x Ω ± Let Ω ± be a uniform C 3 domain Then, L p -L q maximal regularity holds with u ± L p ((0, T ), H 2 q (Ω ± )) H 1 p((0, T ), L q (Ω ± )), ρ L p ((0, T ), W 3 1/q q (Γ)) Hp((0, 1 T ), Wq 2 1/q (Γ)) where Ĥ1 q (Ω) = {ψ L q,loc (Ω) ψ L q (Ω) N } Maximal regularity + Lagrange transformation If weak Neumann problem is uniquely solvable, then Local wellposedness for two phase Navier-Stokes equations holds YShibata (Waseda) 2 phase problem Sept3-7, / 16

6 Global well-posedness with surface tension Let Ω + be a bounded domain in R N and Ω = R N \ Ω + Ω t = Ω +t Ω t = R N \ Γ t ( t v + (v )v) m 1 ± Div (µd(v) pi) = 0, div v = 0 in Ω t {t}, 0<t<T [[µd(v) pi]]n t = σh(γ t )n t, [[v]] = 0 on Γ t {t}, (3) 0<t<T V Γt = v n t on Γ t {t}, v t=0 = v 0, Ω ±t t=0 = Ω ± 0<t<T Let B R = {x R N x < R} and S R = {x R N x = r} Assumption 1 Ω + = B R = R N ω N /N, where ω N = S 1 Assumption 2 x dx = 0 Ω Assumption 3 Γ = {x = (R + ρ 0 (Rω))ω ω S 1 } with given small function ρ 0 defined on S R Γ t = {x = (R + ρ(rω, t))ω + ξ(t) ω S 1 } where ρ is a unknown function and ξ(t) is the barycenter point of the domain Ω t defined by ξ(t) = 1 x dx Ω + Let w(ξ, t) be the velocity field in the Lagrange coordinate, and then ξ (t) = d 1 x dx = d 1 t (ξ + w(ξ, s) ds) dξ = 1 dt Ω + dt Ω Ω+t + Ω+ 0 Ω + Ω+ Ω+t w dξ = 1 v dx Ω + Ω+t YShibata (Waseda) 2 phase problem Sept3-7, / 16

7 Given ρ Wq 3 1/q (S R ), let H ρ (ξ, t) Hq 3 (ḂL) be a function such that H ρ SR = R 1 ρ, H ρ H 2 (ḂL) C ρ q Wq 2 1/q (S, H R) ρ H 3 (ḂL) C ρ q Wq 3 1/q (S R) L 3R + 3 and φ C 0 (R N ) such that φ(x) = 0 for x L 1 and φ(x) = 1 for x L 2 Ω t± = {x = y + φ(y)h(y, t)y + ξ(t) y B R± }, B R+ = B R, B R = {x R N x > R}, Γ t = {x = (R + ρ(rω, t))ω + ξ(t) ω S 1 } (4) t u ± m 1 ± Div (µ ±D(u ± ) q ± I) = F ± (u ±, H ρ ) in B R± (0, T ), div u = F d (u, H) = div F d (u, H) in B R± (0, T ), [[µd(u) q]]ω σ(b R ρ)n = G(u, ρ), [[u]] = 0 in S R (0, T ), t ρ n P u = D(u, ρ) on S R (0, T ), (u ±, ρ) t=0 = (u 0±, ρ 0 ) on B R± S R R N = B R+ B R S R S1 : Laplace-Beltrami operator on S 1 B R ρ = R 2 ( S1 + N 1)ρ, P u = u 1 B R B R u(y) dy YShibata (Waseda) 2 phase problem Sept3-7, / 16

8 Theorem Let N 3 Let q 1, q 2 and p be exponents such that max(n, 2N N 2 ) < q 2 <, 1 = q 1 q 2 N, 2 p + N < 1, q 2 N q p > 1 2 Let δ be a number such that N q p > δ > 1 p For example, setting p = 2q 2(1 + σ), we have q 2 N p > 2, 2 p + N q 2 < 1, N q p > 1 2 with small σ > 0 YShibata (Waseda) 2 phase problem Sept3-7, / 16

9 Then, there exists an ϵ > 0 such that initial data u 0± Bq 2(1 1/p) 2,p (B R± ) B 2(1 1/p) (B R± ) = D p,q1,q 2 (B R± ) and q 1 /2,p ρ 0 B 3 1/p 1/q 2 q 2,p (S R ) satisfy the smallness condition: u 0± Dp,q1,q 2 (B R± ) + ρ 0 3 1/p 1/q ϵ B 2 q 2,p (S R ) ± and the compatibility condition: div u 0± = 0 in B R±, [[µd(u 0 )]]ω < [[µd(u 0 )]]ω, ω > ω = 0 on S R, then problem (7) admits unique solutions u ± and ρ ± with u ± L p ((0, ), H 2 q 2 (B R± ) H 2 q 1 /2 (B R±)) H 1 p((0, ), L q2 (B R± ) L q1 /2(B R± )), ρ L p ((0, ), W 3 1/q 2 q 2 (S R )) H 1 p((0, ), W 2 1/q 2 q 2 (S R )) YShibata (Waseda) 2 phase problem Sept3-7, / 16

10 possessing the estimate: E(u, ρ)(0, ) Cϵ Here, E(u, ρ)(0, T ) = < t > N 2q 2 +1 δ Ψρ Lp((0,T ),H 3 q 2 (B L ) + < t > N 2q 2 +1 δ t Ψ ρ Lp((0,T ),H 2 q 2 (B L )) + < t > N q 1 Ψ L (0,T ),H 2 (B L )) + < t > N q 1 t Ψ ρ L ((0,T ),H 1 (B L )) + < t > N 2q 2 +1 δ u Lp((0,T ),H 2 q 2 (ṘN )) < t > N 2q 2 +1 δ t u Lp((0,T ),L q2 (ṘN )) + < t > N 2q 1 δ u Lp((0,T ),H 2 q 1 (ṘN )) < t > N 2q 1 δ t u Lp((0,T ),L q1 (ṘN )) + < t > δ u Lp((0,T ),H 2 q 1 /2 (ṘN )) < t > δ t u Lp((0,T ),L q1 /2(ṘN )) + < t > N q 1 u L ((0,T ),H 1 (ṘN )) + < t > N 2q 2 +1 u L ((0,T ),H 1 q 2 (ṘN )) + < t > N 2q 1 u L ((0,T ),H 1 q 1 (ṘN )) + u L ((0,T ),H 1 q 1 /2 (ṘN )) Here, Ṙ N = B R+ B R = R N \ S R, B L = { x < L}, B L = { x > L}, Ψ ρ is a suitable extension of ρ to R N times a cut off function for B L, and < t >= (1 + t 2 ) 1/2 YShibata (Waseda) 2 phase problem Sept3-7, / 16

11 Decay estimate for the linearized equations Let u and ρ be solutions to the linearized equations: t u ± m 1 ± Div (µ ±D(u ± ) q ± I) = f ± in B R± (0, T ), (5) div u ± = g ± = div g ± in B R± (0, T ), [[µd(u) q]]ω σ(b R ρ)ω = h, [[u]] = 0 in S R (0, T ), t ρ ω P u = d on S R (0, T ), (u ±, ρ) t=0 = (u 0±, ρ 0 ) on B R± S R Then, we have, [ E(u, ρ)(0, T ) C { < t > b f ± Lp((0,T ),L q(b R±)) + < t > b g ± Lp(R,H q 1 (B R±)) q=q 1/2,q 1,q 2 + < t > b g ± 1/2 H p (R,L + < t >b g ± q(b R±)) H 1 p (R,L q(b R±)) + < t > b h ± Lp(R,W q 1(BR±)) + < t > b h ± 1/2 H p (R,L + < t >b d q(b R±)) Lp((0,T ),Wq 2(ḂL))} N { T } 1/p ] + (< t > b (ρ(t), ψ j ) SR ) p dt j=0 0 with b = 3N 2q 1 δ for q = q 1, b = N q 1 + N 2q δ for q = q 2, and b = N q 1 δ for q = q 1 /2 Here, ψ 0 = 1/ S R 1/2 ; ψ j (j = 1,, N) are some linear combinations of x 1,, x N such that {ψ j } N j=0 satisfy the orthogonal conditions: (ψ i, ψ j ) SR = δ ij ; δ ij = 1 for i = j and δ ij = 0 for i j; and (f, g) SR = S R f(x)g(x) dσ YShibata (Waseda) 2 phase problem Sept3-7, / 16

12 L p -L q decay estimate (6) t u ± m± 1 Div (µ ±D(u ± ) q ± I) = 0 in B R± (0, T ), div u ± = 0 in B R± (0, T ), [[µd(u) q]]ω σ(b R ρ)ω = 0, [[u]] = 0 in S R (0, T ), t ρ ω P u = 0 on S R (0, T ), (u ±, ρ) t=0 = (u 0±, ρ 0 ) on B R± S R Let H q ={(f, h) f L q (ṘN ) N, (f, φ)ṙn = 0 for any φ Ĥ1 q (RN ), h Wq 2 1/q (S R ), (h, 1) SR = 0, (h, x i ) SR = 0 (i = 1,, N)} Here, we set h = h + for x B R+ and h = h for x B R YShibata (Waseda) 2 phase problem Sept3-7, / 16

13 After eliminating the pressure term q, C 0 analytic semigroup {T (t)} t 0 associated with (6) is generated and we have the following L p -L q decay estimate: Let (u(t), ρ(t)) = T (t)(u 0, ρ 0 ) for (u 0, ρ 0 ) H q, and then (u(t), ρ(t)) H 2 p (ḂL) W 3 1/p (S R ) C p,qt N 2q (u 0, ρ 0 ) Lq (ṘN ) W 2 1/q p ( ) 1 q 1 p for t 1 provided that 1 < q p and q q (S R ), u(t) Lp(ṘN ) C p,qt N 2 (u 0, ρ 0 ) Lq(ṘN ) Wq 2 1/q (S R ), ( ) u(t) Lp(Ω) C p,q t min( N 1 2 q 1, N p 2q ) (u 0, ρ 0 ) Lq (ṘN ) Wq 2 1/q (S R ) YShibata (Waseda) 2 phase problem Sept3-7, / 16

14 Proof of Existence Theorem Let T > 0 and let (u, ρ) be a solution of t u ± m 1 ± Div (µ ±D(u ± ) q ± I) = F (u ±, H) in B R± (0, T ), div u ± = F d (u ±, H) = div F d (u ±, H) in B R± (0, T ), (7) [[µd(u) q]]ω σ(b R ρ)n = G(u, ρ), [[u]] = 0 in S R (0, T ), To prolong (u ±, ρ) beyond T > 0, it sufficient to prove that t ρ n P u = D(u, ρ) on S R (0, T ), (u ±, ρ) t=0 = (u 0±, ρ 0 ) on B R± S R E(u, ρ) C(I + E(u, ρ) 2 ) where I = u 0 Dp,q1,q 2 + ρ 0 3 1/p 1/q W 2 By the decay estimate, q 2 (S R) What we have to prove is that E(u, ρ) C(I + E(u, ρ) 2 + N { T 1/p) (< t > b (ρ, ψ j ) SR ) dx} p j=0 0 N { T 1/p (< t > b (ρ, ψ j ) SR ) dx} p C(I + E(u, ρ) 2 ) j=0 0 YShibata (Waseda) 2 phase problem Sept3-7, / 16

15 By the conservation of mass (div v = 0 in Ω t± ), Ω t+ = Ω + = B R, and so recalling that Ω t+ = {x = (R + ρ(rω, t))ω + ξ(t) ω S 1 }, we have B R = S 1 Thus, (R+ρ) Since ξ(t) = 1 Ω Ω t x dx, which leads to Thus, 0 r N 1 dr dω = 1 N (R + ρ) N dω = B R + R N 1 (ρ, 1) SR + NC j R N j (ρ j, 1) SR N S 1 j=2 (ρ, 1) SR C R 0 = x i, dx ξ i (t) Ω = Ω t = RN+1 N + 1 S 1 ω dω i =0 N j=2 (ρ j, 1) SR (x i ξ i (t)) dx = Ω t + R(ρ, x i ) SR + (ρ, x i ) SR C R N j=2 N j=2 S 1 R+ρ 0 r N ω i dr dω N+1C j N + 1 R2 j (ρ j, x i ) SR, (ρ j, x i ) SR 2 N { T 1/p (< t > b (ρ, ψ j ) SR ) dx} p C(I + E(u, ρ) 2 ) j=0 0 YShibata (Waseda) 2 phase problem Sept3-7, / 16

16 Roll of the barycenter point Note that u = u + on S R because [[u]] = 0 on S R t ρ ω P u + = d on S R (0, T ) Resolvent equations = λρ ω P u + = d on S R (d, 1) SR = 0 = (λρ, 1) SR = (ω P u, 1) SR = (div (u + 1 B R B R u + dy), 1) SR = 0 Thus, (ρ, 1) SR = 0 (d, x i ) SR = 0 = (λρ, x i ) SR = (ω P u +, x i ) SR = (div (x i (u + 1 u + dy)), 1) SR B R B R = (u i+ 1 u i+ dy, 1) BR = 0 B R B R Thus, (ρ, x i ) SR = 0 It is a role of P to keep the condition (ρ, 1) SR = 0 and (ρ, x i ) SR = 0 YShibata (Waseda) 2 phase problem Sept3-7, / 16