Free boundary problem for the Navier-Stokes equations

Transcript

1 Free boundary problem for the Navier-Stokes equations Yoshihiro Shibata Department of Mathematics & Research Institute of Sciences and Engineerings, Waseda University Fluids under Pressure Summer School 216, Aug 29 Sept 2 Nečas Center for Mathematical Modelling, Prague, Czech Republic Organizers: S Nečasová, T Bodnár, and G P Galdi Partially supported by Top Global University Project and JSPS Grant-in-aid for Scientific Research (S) # YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

2 Some physical backgound the cavitation A cloud of bubles created by a screw propeller droplet fall, motion of the surface of the sea YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

3 One phase problem for the incompressible viscous fluid flow Let Ω be a domain in R N, which is occupied by some viscous fluid Let Γ be the boundary of Ω Let Ω t and Γ t be time evolutions of Ω and Γ for t > Let n t be the unit outer normal to Γ t Problem is to find a domain Ω t, velocity v = (v 1,, v N ) and pressure p satisfying (1) div v = in <t<t Ω t (, T ), t v + (v )v Div (µd(v) pi) = in <t<t Ω t (, T ), (µd(v) pi)n t = σh(γ t )n t, V Γt = v n t on <t<t Γ t (, T ), v t= = v, Ω t t= = Ω = Ω µ : positive constant, viscosity coefficient, σ: positive constant, coefficient of surface tension H(Γ t ): doubled mean curvature of Γ t, V Γt : evolution speed of Γ t in the n t direction D(v) = v + v = doubled deformation tensor whose (i, j) component is i v j + j v i, I is the N N identity matrix YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

4 Modelling φ t : Ω Ω t : ξ x = φ t (ξ) is the diffeomorphism Navier-Stokes equations in the isothemal case ρ t + div (ρu) = ρ(u t + u u) = Div T (eq for the mass conservation), (eq for the conservation of momentum) ρ:mass density, u = (u 1,, u N ): velocity field, T = (T ij ): stress tensor, Div T i = Reynolds tranport theorem : v(x, t) = ( t φ t )(φ 1 t (x)), J: Jacobina of the trasformation x = φ t (ξ), = t J = (div v)j Conservation of Mass d ρ dx = d ρ(φ t, t)j dξ = ((ρ t + div (ρv))j dξ = div (ρ(v u)) dx = ρ(v u) n t dσ dt Ω t dt Ω Ω Ω t Γ t (v u) n t = on Γ t = d ρ dx = Γ t : boundary of Ω t, n t : unit outer normal to Γ t dt Ω t V Γt = u n t on Γ t (kinematic condition) V Γt = v n t : evolution speed of the free surface If Γ t is given by F (x, t) = locally, then F t + u F = In fact, = d dt F (ϕ t, t) = F t + F v, and F v = F n t v = F n t u = F u N j=1 T ij x j YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

5 Modelling Conservation of Momentum: d ρu dx = d ρ(φ t, t)u(φ t, t)j dξ = (ρ t + ρ v)u + ρ(u t + v u) + ρudiv v)j dξ dt Ω t dt Ω Ω using the momentum eq = (Div [ρ(v u) u] + Div T) dx = {(ρ(v u) n t )u + Tn t } dx = Tn t dx = Ω t Γ t Γ t Tn t = div Γt T Γt = Tn t dx = = the conservation of momenutum d ρu dx = Γ t dt Ω t Here, we assume that div Γt T Γt = σh Γt n t (the surface tension only acts on Γ t H Γt n t = Γt x = div Γt n t Γt : the Laplace-Beltrami op on Γ t, x: position vector of Γ t YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

6 Modelling Incompressibility : Assume that ρ is positive constant Eq of mass conservation: ρ t + div (ρu) = = div u = Constitutive Relation : Classical Newton Law T = µd(u) pi Assume that ( t φ)(φ 1 (x)) = u(x, t), and then x = φ t (ξ) is a solution of the Cauchy problem dx dt = u(x, t) (t > ), x t= = ξ v(ξ, t) = u(φ t (ξ), t) : the Lagrange representation of the velocity field Lagrange map: x = ξ + Γ t = {x = X v (ξ, t), ξ Γ}, t v(ξ, s) ds = X v (ξ, t)(= φ t (ξ)) Ω t = {x = X v (ξ, t), ξ Ω} This expresses the fact that the free surface Γ t consists for all t > of the same fluid particles, which do not leave it and are not incident on it from Ω t Point in solving the free boundary problem : Transform time dependent domain Ω t to some fixed domain 1 Lagrange transform: Ω t Ω 2 Hanzawa transform: Ω t = {x = f(y, t) y D} with some unknow function f 3 any other transformation else??? YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

7 Equation in the Lagrange coordinate t Lagrange transform: x = ξ + u(ξ, s) ds Let A be the inverse matrix of the Jacobi matrix x t j u j = δ jk + ds ξ k ξ k t x = A ξ = ξ + V ( u(ξ, s) ds) ξ Passing to the Lagrange coordinates and setting p(x u (ξ, t), t) = q(ξ, t), t u Div (µd(u) qi) = F(u), div u = G 1 (u) = div G 2 (u) in Ω (, T ), (2) (µd(u) qi + H(u))n t σh(γ t )n t = on Γ (, T ), u t= = v(ξ) in Ω n t = T A 1 n T A 1 n 1, n : the unit outer normal to Γ t t t t F(u) = V ( u ds) t u + V 2 ( u ds) 2 u + V 3 ( u ds)( 2 u ds, u), t t t G 1 (u) = V 4 ( u ds) u, G 2 (u) = V 5 ( u ds)u, H(u) = V 6 ( u ds) u, t with some matrices V j = V j (w) of polynomials with respect to w = u(ξ, s) ds such that V j () = for j =, 2, 4, 5, and 6 YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

8 Boundary condition Let Π t d = d < d, n t > n t, Π d = d < d, n > n And then, d = is equivalent to Π Π t d = and n d = Boundary condition: (S(u, q) + H(u))n t σh(γ t )n t = with S(u, q) = µd(u) qi equivalent to Π Π t (µd(u) + H(u))n t = and t n (S(u, q) + H(u))n t σn Γt (ξ + u ds) = t we have used the fact that H(Γ t )n t = Γt X u (ξ, t) = Γt (ξ + u ds) The first condition is equivalent to Π µd(u)n = Π (Π Π t )µd(u)n t + Π µd(u)(n n t ) Π Π t H(u)n t To find the equivalent condtion to the second equations, we observe that t t σn Γt (ξ + u ds) = σn {( Γt Γ )(ξ + u ds)} σn Γ ξ H(Γ) t σn Γ u ds YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

9 t t ) = σn {( Γt Γ )(ξ + u ds)} σh(γ) σ Γ (n u ds t t } + σ {2( Γ n) Γ u ds + ( Γ n) u ds t ( t ) = σn {( Γt Γ )(ξ + u ds)} σh(γ) + (m σ Γ ) n u ds t t t } mn u ds + σ {2( Γ n) Γ u ds + ( Γ n) u ds [ t = (m σ Γ ) n u ds σ(m σ Γ ) 1( t t )] n {( Γt Γ )(ξ + u ds)} ( Γ n) u ds K(u) t t σh(γ) mn u ds + 2σ( Γ n) Γ u ds Here, we choose m so large positive numeber in such a way that (m σ Γ ) 1 exists t t Let K(u) = n ( Γ Γt ) u ds + n ( Γ Γt )ξ ( Γ n) u ds, t η = n u ds (m Γ ) 1 K(u) Thus, the second equation may be written in the form : t η n u = (m σ Γ ) 1 t K(u), < µd(u)n t, n > q < n t, n > + < H(u))n t, n > +(m σ )η t t σh(γ) mn u ds + 2σ( Γ n) Γ u ds = YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

10 The second equation is changed as follows: =< µd(u)n, n > + < µd(u)(n t n), n > q < n t, n > + < H(u))n t, n > +(m σ )η t t σh(γ) mn u ds + 2σ( Γ n) Γ u ds, so that =< µd(u)n, n > q + (m σ Γ )η + (< n t, n > 1 1)(< µd(u)n, n > +(m σ Γ )η) { t t } < n t, n > 1 σh(γ) < n t, n > 1 mn u ds 2σ( Γ n) Γ u ds, Thus, our nonlinear equations in the Lagrange coordinate is: (3) t u Div S(u, q) = F(u), div u = G 1 (u) = div G 2 (u) in Ω (, T ), t η n u = (m Γ ) 1 t K(u) on Γ (, T ), S(u, q)n + (σ(m Γ )η)n = σ < n t, n > 1 H(Γ)n + I(u, η) on Γ (, T ) u t= = v in Ω, η t= = on Γ, with (4) I(u, η) = Π ((Π Π t )µd(u)n t + µd(u)(n n t ) Π t H(u)n t ) + [(1 < n, n t > 1 )(< µd(u)n, n > +(m σ Γ )η) t t + < n, n t > 1 (mn u ds 2σ( Γ n) Γ u ds)]n, t t K(u) = n ( Γ Γt ) u ds + n ( Γ Γt )ξ ( Γ n) u ds YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

11 Example of domains Main condition for the domain Ω For any f L q (Ω), there exists a unique u Ĥ1 q,(ω) of the weak Dirichlet problem: ( u, φ) Ω = (f, φ) Ω for any φ Ĥ1 q, (Ω) with some q and q (1 < q, q < ) Ĥ 1 q,(ω) = {u L q,loc (Ω) u L q (Ω) N, u Γ = } Example of domains: (1) Ω is a bounded domain : drop problem (2) Ω is a perturbed half space: ocean problem without bottom Let φ(x ) be a function in W 3 1/r R (R N 1 ) with N < r < Let H φ = {x = (x 1,, x N ) R N x N > φ(x ) (x R N 1 )}, Γ φ = {x = (x 1,, x N ) R N x N = φ(x ) (x R N 1 )} Assume that there exists an R > such that Ω B R = H φ B R and Γ B R = Γ φ B R, where B R = {x R N x > R} and B R = {x R N x < R} for any R > In addition, we assume that Γ is a hypersurface of W 3 1/r R class (3) Ω is an exterior domain the model for cavitation problem Ω = B c 1 f(x) = log x (N = 2), f(x) = x (N 1) 1 (N 3) satisfy f = in B c 1 and f x =1 = But, C (Ω) is in general not dense in Ĥ1 q,(ω), so these function are not non-trivial solutions to the homoneneous weak Dirichlet problem weak and strong are different in the exterior domains! YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

12 Local well-posedness Theorem Let R >, N < q < and 2 < p < Then, there exists a time T depending on R such that for any initial data v with v R satisfying the compatibility B q,p 2(1 1/p) (Ω) condition: (5) div v = on Ω, D(v )n < D(v )n, n > n = on Γ, problem (3) admits unique solutions u, q and η with u L p ((, T ), H 2 q (Ω) N ) H 1 p((, T ), L q (Ω) N )), η L p ((, T ), H 3 1/q q satisfying the estimate: (Γ)) Hp((, 1 T ), Wq 2 1/q (Γ))) q L p ((, T ), H 1 q (Ω) + Ĥ1 q,(ω)), u Lp((,T ),H 2 q (Ω)) + t u Lp((,T ),Lq(Ω)) + t η Lp((,T ),W 3 1/q q with some constant C independent of R Remark (Γ)) + tη Lp((,T ),W 2 1/q (Γ)) + q Lp((,T ),Lq(Ω)) CR (1) q Hq 1 (Ω) + Ĥ1 q, (Ω) means that there exist q 1 Hq 1 (Ω) and q 2 Ĥ1 q, (Ω) such that q = q 1 + q 2 Especially, q Γ = q 1 Γ (2) Since the map: x = X u (ξ, t) is invertible, the problem with Euler coordinate is also uniquely solvable for certain time interval (, T ) q YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

13 Some references H Abels, The initial-value problem for the Navier-Stokes equations with a free surface in L q Sobolev spaces, Adv Differential Equations, 1 (25), G Allain, Small-time existence for the Navier-Stokes equations with a free surface, Appl Math Optim, 16 (1987), 37 5 J T Beale, The initial value problem for the Navier-Stokes equations with a free boundary, Comm Pure Appl Math, 34 (1981), J T Beale, Large time regularity of viscous surface waves, Arch Rat Mech Anal, 84 (1984), J T Beale and T Nishida, Large time behavior of viscous surface waves, Lecture Notes in Numer Appl Anal, 128 (1985), 1 14 Y Enomoto and Y Shibata, On the R-sectoriality and its application to some mathematical study of the viscous compressible fluids, Funk Ekvaj, 56 (213), Y Hataya and S Kawashima, Decaying solution of the Navier-Stokes flow of infinite volume without surface tension, Nonlinear Anal, 71 (29), Y Hataya, A remark on Beal-Nishida s paper, Bull Inst Math Acad Sin (NS), 6 (211), I Sh Mogilevskiĭ and V A Solonnikov, On the solvability of a free boundary problem for the Navier-Stokes equations in the Hölder spaces of functions, Nonlinear Analysis A Tribute in Honour of Giovanni Prodi, Quaderni, Pisa, (1991), P B Mucha and W Zaj aczkowski, On local existence of solutions of the free boundary problem for an incompressible viscous self-gravitating fluid motion, Applicationes Mathematicae, 27 (2), T Nishida, Equations of fluid dynamics free surface problems, Comm Pure Appl Math, 39 (1986), S221 S238 M Padula and V A Solonnikov, On the local solvability of free boundary problem for the Navier-Stokes equations, J Math Sci, 17 (21), M Padula and V A Solonnikov, On the global existence of nonsteady motions of a fluid drop and their exponential decay to a uniform rigid rotation, Quad Mat, 1 (22), YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

14 H Saito and Y Shibata, On the global well posedness of free boundary problem for the Navier-Stokes equations with surface tension, in preparation B Schweizer, Free boundary fluid systems in a semigroup approach and oscillatory behavior, SIAM J Math Anal, 28 (1997), Y Shibata, On the maximal L p-l q regularity of the Stokes equations and the one phase free boundary problem for the Navier-Stokes equations, in Mathematical Analysis on the Navier-Stokes Equations and Related Topics, Past and Future - In memory of Prof T Miyakawa (ed T Adachi et al), Gakuto International Series, 35, Math Sci Appl, 211, Y Shibata, On some free boundary problem of the Navier-Stokes equations in the maximal L p-l q regularity class, J Differential Equations, 258 (215), Y Shibata, On the R-bounded solution operators in the study of free boundary problem for the Navier-Stokes equations, submitted to proceedings of the International Conference on Mathematical Fluid Dynamics, Present and Future V A Solonnikov, Unsteady motion of a finite mass of fluid, bounded by a free surface, Zap Nauchn Sem (LOMI), 152 (1986), (in Russian); English transl J Soviet Math, 4 (1988), V A Solonnikov, On the transient motion of an isolated volume of viscous incompressible fluid, Izv Akad Nauk SSSR Ser Mat, 51 (1987), (in Russian); English transl Math USSR Izv, 31 (1988), V A Solonnikov, Solvability of the problem of evolution of a viscous incompressible fluid bounded by a free surface on a finite time interval, Algebra i Analiz, 3 (1991), (in Russian); English transl St Petersburg Math J, 3 (1992), V A Solonnikov, Lectures on evolution free boundary problems: Classical solutions, Mathematical aspects of evolving interfaces (Funchal, 2), Lecture Notes in Math, 1812, Springer, Berlin, 23, D Sylvester, Large time existence of small viscous surface waves without surface tension, Commun Partial Differential Equations, 15 (199), N Tanaka, Global existence of two phase non-homogeneous viscous incompressible weak fluid flow, Commun Partial Differential Equations, 18 (1993), A Tani, Small-time existence for the three-dimensional incompressible Navier-Stokes equations with a free surface, Arch Rat Mech Anal, 133 (1996), A Tani and N Tanaka, Large time existence of surface waves in incompressible viscous fluids with or without surface tension, Arch Rat Mech Anal, 13 (1995), YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

15 The linearized problem (6) t u Div (µd(u) qi) = f, in Ω (, T ), div u = f d = div f d in Ω (, T ), t η n u = g in Γ (, T ), (µd(u) qi)n + (σ(m Γ )η)n = h in Γ (, T ), u t= = in Ω, η t= = on Γ, Maximal L p -L q regularity theorem Let T > and 1 < p, q < Assume that 2/p + 1/q < 1 Let f L p ((, T ), L q (Ω) N ), f d L p ((, T ), Hq 1 (Ω)) Hp((, 1 T ), Wq 1 (Ω)), f d Hp((, 1 T ), L q (Ω) N ), g L p ((, T ), Wq 2 1/q (Γ)), h L p ((, T ), Hq 1 (Ω) N ) Hp((, 1 T ), Wq 1 (Ω) N ), which satisfy the compatibility condition: div f d t= = in Ω and Π h t= = on Γ YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

16 Maximal Regularity Theorem Problem (6) admits unique solutions u, q and η with u Hp((, 1 T ), L q (Ω) N ) L p ((, T ), Hq 2 (Ω) N ), q L p ((, T ), Hq 1 (Ω) + Ĥ1 q,(ω)), η L p ((, T ), Hq 3 (Ω)) Hp((, 1 T ), Hq 2 (Ω)) possessing the estimate: (7) [[(u, q, η)]] t Ce γt I((, t), f, f d, g, h) with some positive constants γ and C Here, we have set [[(u, q, η)]] t = u Lp((,t),H 2 q (Ω)) + t u Lp((,t),L q(ω)) + q Lp((,t),L q(ω)) + η Lp((,t),W 3 q (Ω)) + t η Lp((,t),W 2 q (Ω)), I((, t), f, f d, g, h) = { f Lp((,t),L q(ω)) + (f d, h) Lp((,t),H 1 q (Ω)) + g Lp((,T ),Wq 2 1/q (Γ)) + tf d Lp((,t),Lq(Ω)) + t (f d, h) Lp((,t),W q 1 (Ω)) } YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

17 Abstract Setting Let me consider the equations u t Au = f in Ω (, T ), Bu Γ = g, u t= = u X, Y UMD Banach space (Hilbert transforms on X and Y are bounded), Y X, Y is dense in X, A : Y X: closed linear op Ay X C y Y Z = [X, Y ] 1/2 (complex interpolation),and then u H 1/2 p ((, T ), Z) provided that u L p ((, T ), Y ) H 1 p((, T ), X) = M p,q Assumption for B: Bu H 1/2 p ((, T ), X) L p ((, T ), Z) provided that u M p,q Example: A =, B = / ν (Neumann boundary condition); X = L q (Ω), Y = H 2 q (Ω), Z = H 1 q (Ω) Cauchy problen : u t Au = in Ω (, ), Bu Γ =, u t= = u Assumption Assume that A generates a C analytic semigroup {T (t)} t on X such that T (t)x X Ce γt x X, t T (t)x X Ct 1 e γt x X (x X), t T (t)x X Ce γt x Y (y D = {v Y Bv Γ = }) Then, T (t)u M p,q provided that u (X, Y ) 1 1/p,p and e γt T (t)u Lp((, ),Y ) + e γt t T (t)u Lp((, ),X) C u (X,Y )1 1/p,p YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

18 ( ) Let l s p = {(a n ) (2 sn a n ) p < } (1 p < ), l s = {(a n ) sup 2 ln a n < } n= (e 2γt t u(t) X ) p dt = n= 2 n+1 (2 n/p e γ2n+1 sup t u(t) X ) p 2 n <t<2 n+1 n= 2 n (e 2γt t u(t) X ) p dt Thus, setting a n = e γ2n+1 sup t u(t) X, we have 2 n <t<2 n+1 (e 2γt t u(t) X ) p dt (2 n/p a n ) p = (a n ) 1/p l n= On the other hand, using the estimates: t u(t) X Ct 1 e γt u X and t u(t) X Ce γt u Y, we have 2 n a n u X, a n u Y, so that (a n ) l 1 u X, (a n ) l u Y Since we know that l 1/p p = (l 1, l ) 1 1/p,p, we have (a n ) 1/p l C u (X,Y )1 p,p, p so that (a n ) l 1 u X, (a n ) l u Y Since we know that l 1/p p = (l 1, l ) 1 1/p,p, we have (a n ) 1/p l C u (X,Y )1 p,p p p n Z YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

19 Operator Valued Fourier Multiplier Theorem R-boundedness A family T L(X, Y ) of operators is R-bounded if there exists a constant C > such that for all m N, (T k ) k=1,,m T, and (x k ) k=1,,m X we have m L r k T k x k C m L r k x k p ([,1];Y ) p ([,1];X) k=1 Here the Rademacher functions r k, k N, are given by r k : [, 1] { 1, 1}, t sign(sin(2 k πt)) The smallest such C is called R bound of T on L(X, Y ) which is written by R L(X,Y ) T in what follows Weis operator valued Fourier mutliplier theorem: Math Ann 319 (21), Let X and Y be two UMD Banach spaces and 1 < p < Let M be a function in C 1 (R \ {}, L(X, Y )) such that k=1 R L(X,Y ) {(ρ d dρ )l M(ρ) ρ R \ {}} = κ l < (l =, 1) Let T M be the operator defined by T M ϕ = F 1 [MF[ϕ]] for any ϕ with F[ϕ] D(R, X) Then, T M is extended to a bounded linear operator from L p (R, X) into L p (R, Y ) Moreover, denoting this extension also by T M, we have T M L(Lp(R,X),L p(r,y )) C(κ + κ 1 ) for some positive constant C depending on p, X and Y YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

20 Next, we consider t u Au = f in Ω (, T ), Bu = g on Ω (, T ) Extending f and g suitably to R, we consider Apply the Laplace transform in time t = λ Σ ϵ,λ = {λ C arg λ π ϵ, λ λ } λv Av = ˆf in Ω, Bv Γ = ĝ Γ (λ = γ + iτ) u L p (R, Y ) H 1 p(r, X) = λv X, λ 1/2 v Z = [X, Y ] 1/2,, v Y X = {F = (F 1, F 2, F 3 ) F 1, F 2 X, F 3 Z}, F 1 ˆf, F 2 λ 1/2 ĝ, F 3 ĝ Assumption There exists a solution operator S(λ) Hol (Σ ϵ,λ L(X, Y )) such that for any λ Σ ϵ,λ and ( ˆf, ĝ) X Z, v = S(λ)( ˆf, λ 1/2 ĝ, ĝ) is a unique solution and R L(X,X) ({(τ τ ) l (λs(λ)) λ Σ ϵ,λ }) C (l =, 1) R L(X,Z) ({(τ τ ) l (λ 1/2 S(λ)) λ Σ ϵ,λ }) C (l =, 1), R L(X,Y ) ({(τ τ ) l S(λ) λ Σ ϵ,λ }) C (l =, 1) Then, u = L 1 [S(λ)( ˆf(λ), λ 1/2 ĝ(λ),ˆ(λ))] is a unique solution of the time dependent problem Apply the Weis theorem and note that L[f](λ) = e λt f(t) dt = F[e γt f], L 1 [g(λ)] = 1 e λt g(λ)] = e γt F 1 [g(γ + iτ)](t) R 2π R e γt t u Lp(R,X) + e γt Λ 1/2 γ u Lp(R,Z) + e γt u Lp(R,Y ) C{ e γt f Lp(R,X) + e γt Λ 1/2 γ g Lp(R,X) + e γt g Lp(R,Z)} γ λ e γt Λ 1/2 γ f = L 1 [λ 1/2 ˆf(λ)], H 1/2 p ((, T ), Z) = {v w : e γt Λ 1/2 γ w Lp(R,Z) < and w (,T ) = v} YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

21 Reduced Stokes equations (Grubb-Solonnikov method) Let Ĥ 1 q (Ω) = {v L q,loc (Ω) v L q (Ω) N, v Γ = } Assume that the Weak Dirichlet Problem ( u, φ) Ω = (f, φ) Ω has unique solution v H 1 q,(ω) for any f L q (Ω) N ( φ H 1 q, (Ω)) Example bounded domain, half-space, perturbed half-space, layer, perturbed layer, exterior domain Exterior domain: Ω = B c 1 = { x > 1} f(x) = log x (N = 2), f(x) = x (N 2) (N 3) are non-trivial sol of the strong Dirichlet prob: f = in Ω, f x =1 = But, they are not solutions of homogeneous weak Dirichlet problem ( ) C (Ω) is not dense in H 1 q (Ω) Stokes problem λu Div (µd(u) pi) = f, div u = g in Ω, (µd(u) p)n = h Here, to solve div u = g, it is necessary to assume that G : (g, φ) Ω = (G, φ) Ω ( φ H 1 q, (Ω)) div u = g means that (u, φ) Ω = (G, φ) Ω ( φ H 1 q, (Ω)) Reduced Stokes problem λu Div (µd(u) K(u)I) = f in Ω, (µd(u) K(u)I)n Γ = Here, for u H 2 q (Ω), let K(u) be a unique solution of the weak Dirichlet problem: ( K(u), φ) Ω = (Div (µd(u) div u, φ) Ω ( φ Hq 1,(Ω)), K(u) =< µd(u)n, n > div u = on Γ Equivalence Assume that (f, φ) Ω = ( φ Hq 1,(Ω)) and < h, n >= on Γ Then, Stokes and Reduced Stokes are equivalent YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

22 Stokes = Reduced Stokes Let g be a solution of the variational problem: λ(g, φ) Ω + ( g, φ) Ω = (f, φ) Ω ( φ Hq 1,(Ω)), g = on Γ Let u and p be solutions of λu Div (µd(u) pi) = f, div u = g in Ω, (µd(u) p)n = h on Γ And then, (f, φ) Ω = λ(u, φ) Ω (µdiv D(u) div u, φ) ΩK(u) ( div u, φ) Ω + ( p, φ) Ω Since (g, φ) Ω = (λ 1 (f + g), φ) Ω div u = g yields λ(u, φ) Ω = ((f + g), φ) Ω Thus, ( (p K(u)), φ) Ω = From the boundary condition ( φ H 1 q, (Ω)), = K(u) + g p = K(u) p (( )div u = g = on Γ, ) Thus, p = K(u), so that u satisfies, =< h, n >=< µd(u)n, n > p = K(u) + div u p λu Div (µd(u) K(u)) = f in Ω, (µd(u) K(u)I)n Γ = YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

23 Reduced Stokes = Stokes Given divergence deta g, let G be a function: (G, φ) Ω = (g, φ) Ω ( φ H 1 q, (Ω)) Let L be a solution of the weak Dirichlet problem: ( L, φ) Ω = (λg g, φ) Ω ( φ Hq 1,(Ω)), L = g on Γ Let u be a solution of the Reduced Stokes eq λu Div (µd(u) K(u)) = f + L in Ω, (µd(u) K(u)I)n = h + gn on Γ ( L, φ) Ω = ( L + f, φ) Ω = λ(u, φ) Ω (µdiv D(u) div u, φ) Ω ( div u, φ) Ω + ( K(u), φ) Ω Thus, λ((u G), φ) Ω ( (div u g), φ) Ω = ( φ Ĥ1 q, (Ω)) Taking φ H 1 q, (Ω)), we have λ((div u g), φ) Ω + ( (div u g), φ) Ω = Moreover, div u g =< µd(u)n, n > K(u) g = g g = on Γ Thus, the uniqueness implies that div u = g in Ω Thus, by the equation, λ((u G), φ) Ω = ( φ Ĥ1 q, (Ω)) = ((u G), φ) Ω = ( φ Ĥ1 q,(ω)) = div u = g in Ω Let p = K(u) L, and then noting that K(u) g = K(u) + L on Γ, we have λu Div (µd(u) pi) = f, div u = g in Ω, (µd(u) p)n = h YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

24 Local estimate of K(u) Let D be any compact subset of Ω, and then we have K(u) Lq(D) C D { 2 u 1/q L q(ω) u 1 1/q L q(ω) + u L q(ω)} In fact, let f be arbitrary element of C (D), Let ψ Ĥ2 q,(ω) be a solution of the strong Dirichlet problem: ψ = f in Ω, possessing the estimate: ψ H 1 q (Ω) C φ L q (Ω) By Poincarés inequality, (f, φ) Ω f Lq (D) φ Lq(D) C D f Lq (D) φ Lq(Ω) Thus, by the Hahn-Banach, F L q : (f, φ) Ω = (F, φ) Ω ( φ Ĥ1 q,(ω)) Ĥ2 q (Ω) = {ψ Ĥ1 q (Ω) ψ H1 q (Ω)N } Thus, ψ exists Then, (K(u), f) Ω = (K(u), ψ) Ω = (K(u), n ψ) Γ ( K(u), ψ) Ω = (< µd(u)n, n > div u, n ψ) Γ (µdiv D(u) div u, ψ) Ω = (< µd(u)n, n > div u, n ψ) Γ ((µd(u) div ui)n, ψ) Γ + (µd(u) div ui, 2 ψ) Ω Since u Lq(Γ) C 2 u 1/q L q(ω) u L q(ω), (K(u), f) Ω C D { 2 u 1/q L q(ω) u 1 1/q L q(ω) + u L q(ω)} ψ H 1 q (Ω) C D { 2 u 1/q L q(ω) u 1 1/q L q(ω) + u L q(ω)} f Lq (Ω) This implies the required inequality YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

25 Existence of R bounded solution operator in R N Consider the Stokes eq in R N R t u Div (µd(u) pi) = f, div u = g = div g in R N R Consider the resolvent problem in R N λv Div (µd(u) pi) = ˆf = L[f], div u = ĝ = div ĝ in R N ˆf = λv µ u µ div u + p = λv µ v µ g + p div ˆf = λdiv v µ div v µ ĝ + p = λ ˆ div g 2µ ĝ + p Thus, p = 1 div ˆf λ 1 div ĝ + 2µĝ Thus, v = (λ ) 1 (ˆf 1 div ˆf) + λ(λ ) 1 1 div ĝ 2µ(λ ) 1 ĝ Let F 1 L q (R N ) N, F 2 L q (R N ) N and F 3 H 1 q (R N ) be variables corresponding to ˆf, λĝ and ĝ Let X q (R N ) = {(F 1, F 2, F 3 ) L q (R N ) N L q (R N ) N H 1 q (R N )} Let S(λ)(F 1, F 2, F 3 ) = F 1[ F[F 1 ](ξ) ξ 2 ξξ F[F 1 ](ξ) ] λ + ξ 2 + F 1[ ξ 2 ξξ F[F 2 ] ] λ + ξ 2 2µF 1[ iξf[f 3 ](ξ) ] λ + ξ 2 Let Σ ϵ,λ = {λ C arg λ π ϵ, λ λ } with < ϵ < π/2 and λ > We have λ + ξ 2 sin ϵ 2 ( λ + ξ 2 ) for any λ Σ ϵ, and ξ R N YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

26 Theorem (Theorem 33 in Y Enomoto and Y Shibata, Funkcialaj Ekvacioj 56 (213), ) Let m(λ, ξ) be a function that is a C function with respect to ξ R N \ {} for any λ Σ ϵ,λ and satisfies the estimate: ξ αm(λ, ξ) C α ξ α for any α N N Let M(λ)f = F 1 ξ [m(λ, ξ)f[f](ξ)] Then, M is a R bounded operator satisfying the estimate: R L(Lq(R N ))({M(λ) λ Σ ϵ,λ }) C q max C α α n/2+2 v = S(λ)(ˆf, λĝ, ĝ) is a unique solution, and R L(Xq(R N ),L q(r N )){(τ τ ) l (λs(λ)) λ Σ ϵ }) C(l =, 1, λ = γ + iτ), R L(Xq(R N ),L q(r N )){(τ τ ) l (λ 1/2 S(λ)) λ Σ ϵ }) C(l =, 1), R L(Xq(R N ),L q(r N )){(τ τ ) l ( 2 S(λ)) λ Σ ϵ }) C(l =, 1, λ = γ + iτ) u(t) = L 1 [S(λ)(L[f](λ), L[λg](λ), L[g](λ))](t) is a solution to the problem: t u Div (µd(u) pi) = f, div u = g = div g in R N R e γt (u t, Λ 1/2 γ u, 2 u) Lp(R,L q(ω) C{ e γt (f, λg) Lp(R,L q) + e γt g Lp(R,H 1 q (R N )} ( γ >, 1 < p, q < ) YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

27 Existence of R bounded solution operator in R N + λv v = in R N +, ( v/ x N ) xn = = g xn = with R N + = {x = (x 1,, x N ) R N x N > } Applying the partial Fourier transform with respect to tangential variable x = (x 1,, x N 1 ) yields that (λ + ξ 2 )F [v](ξ, x N ) 2 NF [v](ξ, x N ) = in x N >, ( N F[v])(ξ, ) = F [g](ξ, ) Let ω λ (ξ ) = λ + ξ 2 ((ξ = (ξ 1,, ξ N 1 )), and then by the inverse Fourier transform F 1, the solution formula is given by S 1 (λ)g = F 1[ e ω λ(ξ )x N F [g](ξ, )](x ) = = x N F 1 [e ω λ(ξ )(x N +y N ) F [g](ξ, y N )](x ) F 1 [ω λ (ξ )e ω λ(ξ )(x N +y N ) F [g](ξ, y N )](x ) F 1 [e ω λ(ξ )(x N +y N ) F [ N g](ξ, y N )](x ) YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

28 Definition of multiplier class Let Ξ be a domain in C and let m(ξ, λ) (λ = γ + iτ Ξ) be a function defined for (ξ, λ) (R N 1 \ {}) Ξ Assume that τ m(ξ, λ) (λ = γ + iτ) exists and that m(ξ, λ) and τ m(ξ, λ) are bothis infinitely many differentiable functions with respect to ξ R N 1 \ {} for each λ Ξ (1) m(ξ, λ) is called a multiplier of order s with type 1 on Ξ if the estimates: (8) κ ξ ((τ τ ) l m(ξ, λ) C α ( λ 1/2 + ξ ) s κ (l =, 1) hold for any multi-index κ N N 1 and (ξ, λ) Ξ and (ξ, λ) Ξ with some constant C κ depending solely on κ and Ξ (2) m(ξ, λ) is called a multiplier of order s with type 2 on Ξ if the estimates: (9) κ ξ ((τ τ ) l m(ξ, λ) C κ ( λ 1/2 + ξ ) s ξ κ (l =, 1) hold for any multi-index κ N N 1 and (ξ, λ) Ξ with some constants C κ depending solely on κ and Ξ Let M s,i (Ξ) be the set of all multipliers of order s with type i on Ξ (i = 1, 2) ω(ξ, λ) M 1,1, ξ M 1,2 YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

29 Lemma (Shibata and Shimizu, J Math Soc Japan 64(2) (212), ) Let < ϑ < π/2 and 1 < q < Given l (ξ,, λ) M,1 (Σ ϑ ) and l 1 (ξ, λ) M,2 (Σ ϑ ), we define the operators L j (λ) (j = 1, 2, 3, 4) by [L 1 (λ)h](x) = F 1 [l (ξ, λ)λ 1/2 e ωλ(ξ )(x N +y ) N F[h](ξ, y N )](x ) dy N, [L 2 (λ)h](x) = F 1 [l 1 (ξ, λ) ξ e ωλ(ξ )(x N +y ) N F [h](ξ, y N )](x ) dy N, [L 3 (λ)h](x) = F 1 [l 1 (ξ, λ) ξ )(x e ωλ(ξ N +y ) N e ξ (x N +y N ) ω λ (ξ ) ξ F [h](ξ, y N )](x ) dy N, [L 4 (λ)h](x) = F 1 [l 1 (ξ, λ) ξ e ξ (x N +y ) N F [h](ξ, y N )](x ) dy N, Then, R L(Lq(R N + )({(τ τ ) l L i (λ) λ Σ ϵ,λ }) C (l =, 1, i = 1, 2, 3, 4) Theorem There exists an operator S 2 (λ) such that v = S 2 (λ)(f, λ 1/2 g, g) is a unique solution of the equations: (1) λv v = f in R N +, ( v/ x N ) xn = = g xn = for any (f, g) L q (R N + ) H 1 q (R N + ) and it satisfies the estimate: R L(Xq(R N 2 j + ),Wq (R N ) ({(τ τ ) l (λ j/2 x α S 2 (λ) λ Σ ϵ,λ }) C (l =, 1, j =, 1, 2) with X q (R N + ) = L q (R N + ) L q (R N + ) Hq 1 (R N + ) YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

30 Construction of solution operator in Ω See: Y Shibata, Differential and Integral Equations 27 (3-4) (214), λv Div (µd(u) K(u)I) = f in Ω (µd(u) K(u)I)n = g on Γ Let {φ i } i=1 be a partition of unity on Ω and let {ψ i } i=1 be some suitable sequence of C such that ψ i (x) = 1 on supp φ i Let T (λ)(f 1, F 2, F 3 ) = φ j S j (λ)[ψ j (F 1, F 2, F 3 )] j=1 with solution operators S j for the localized equations Let v = T (λ)(f, λ 1/2 h, h), and then λv Div (µd(v) K(u)I) = (f + U 1 (λ)(f, λ 1/2 h, h)) in Ω, (µd(v) K(v)I)n = h + U 2 (λ)(f, λ 1/2 h, h)) on Γ Let U(λ) = (U 1 (λ), λ 1/2 U 2 (λ), U 2 (λ)) (remainder term) and then, R L(Xq(Ω),Lq(Ω))({(τ τ ) l (λ j/2 2 j S(λ) λ Σ ϵ,λ }) C (l =, 1, j =, 1, 2) functions R L(Xq(Ω))({(τ τ ) l U(λ) λ Σ ϵ,λ }) Cλ 1/2 (l =, 1), with X q (Ω) = L q (Ω) L q (Ω) H 1 q (Ω) Choose λ > so large that (I + U(λ)) 1 (λ Σ ϵ,λ ) exists Let S(λ) = T (λ)(i + U(λ)) 1, and then v = S(λ)(ˆf, λ 1/2 ĥ, ĥ) is a unique solution of the equations: λv Div (µd(u) K(u)I) = f in Ω (µd(u) K(u)I)n = g on Γ and S(λ) satisfies the estimate: YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

31 Global well-posedness, bounded domain closed to ball (11) ( t v + (v )v) Div (µd(v) pi) =, div v = in <t<t Ω t (, t), (µd(v) pi)n t = σh(γ t )n t, V Γt = v n t on <t<t Γ t (, t), v t= = v, Ω t t= = Ω = Ω Let Ω be a reference domain and Γ its boundary 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, ω N = S 1 Assumption 2 x dx = Ω Assumption 3 Γ = {x = (R + ρ (Rω))ω ω S 1 } with given small function ρ 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 (unknown function) and assume that ξ() = Ω Ω t Let w(ξ, t) be the velocity field in the Lagrange coordinate, and then ξ (t) = d 1 xd dx = d 1 t (ξ + w(ξ, s) ds) dξ = 1 w dξ = 1 v dx dt Ω Ω t dt Ω Ω Ω Ω Ω Ω t YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

32 Let H(ξ, t) be a solution to the Dirichlet problem: (1 )H = in B R and H SR = R 1 ρ(, t) Hanzawa transform: x = e h (y, t) = y + H(y, t)y + ξ(t) for y B R (N 1)σ Let u(ξ, t) = v e h, q(ξ, t) = p e h R Ω t = {x = ξ + H(ξ, t)ξ + ξ(t) ξ B R }, Γ t = {x = (R + ρ(rω, t))ω ω S 1 } Moreover, (12) t u Div (µd(u) qi) = F (u, H) in B R (, T ), div u = F d (u, H) = div F d (u, H) in B R (, T ), Π [µd(u)n] = G (u, ρ) in S R (, T ), < µd(u)n, n > q σ Bh = g R 2 n (u, ρ) in S R (, T ), t ρ n P u = G kin (u, ρ) on S R (, T ), (u, ρ) t= = (u, ρ ) on B R S R Here, P u = u B R 1 B R u dy, n = ω S 1, B = (N 1) +, : Laplace-Beltrami operator on S 1 Π [d] = d < d, n > n YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

33 Theorem 2 < p <, N < q <, 2/p + N/q < 1, 1/p + (N + 1)/q < 1 Then, there exists a small number ϵ (, 1) such that for any initial data u B 2 2/p q,p (B R ) and ρ Bq,p 3 1/p 1/q (S R ) satisfying the following conditions: u 2 2/p B q,p (B + ρ R) 3 1/p 1/q B q,p (S ϵ, R) div u = f d (u, h ) = div f d (u, h ) in B R, Π [µd(u )ω] = g (u, h ) on S R, (v, p l ) Ω = (l = 1, M) where, {p l } M l=1 is the orthogonal base of the rigid space: R d = {u D(u) = } = {Ax + b A + A = }, Then, problem (3) with T = admits unique solutions u, q and ρ with u H 1 p((, ), L q (B R )) L p ((, ), H 2 q (B R )), q L p ((, ), H 1 q (B R )), ρ Hp((, 1 ), Wq 2 1/q (S R )) L p ((, ), Wq 3 1/q (S R )) e ηt u t Lp((, ),L q(b R)) + e ηt u Lp((, ),H 2 q (B R)) + e ηt q Lp((, ),L q(b R)) + e ηt t ρ Lp((, ),W 2 1/q q for some positive constants C and η independent of ϵ (S + eηt ρ R)) Lp((, ),W 3 1/q (S Cϵ R)) q YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

34 Derivation of equations in B R x i x i = (1 + H)y i + ξ i (t), = (1 + H)(δ ij + 1 H y i ), y j 1 + H y j det(i + a b) = 1 + a b, (I + a b) 1 = I a b 1 + a b Thus, y x = (1 + H) 1 (δ ij y t = y x x t y i ( H/ y j ) 1 + H + y ( H) = I + Φ, J = det x y = (1 + H)N + (1 + H) N 1 y ( H) = 1 + J = (I + Φ)( H t + ξ (t)) ( ) x y = (1 + H)(I + H y) ( H = H 1 + H ) ( ) ( ) (t, x) (t, y) (t, y)) (t, x)) = 1 1 = x t x y y t y x ( ) 1 I N Thus, x = y + Φ y, t = t < (I + Φ)( H t + ξ (t)), y > (I + Φ) 1 [ t v + v x v µdiv x D(v] + y q = = t u Div (µd(u) qi) = F(u, H) in B R div x v = div u + N j,k=1 Φ kj u j y k (div v, φ) Ωt = (v, φ) Ωt = N N div x v = = (J(δ ji + Φ ji )u i ) = J(div u + y j i,j=1 i,j=1 N ( (J(δ ji + Φ ji )u i ), φ e h ) BR y j i,j=1 Φ ji u i y j ) = in Ω YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

35 Equations on the boundary x = (R + ρ(rω(u), t))ω(u) + ξ(t), n t = (1 + V 1 (ρ))ω V 1 (ρ) = N 1 j=1 1 + V 1 (ρ) R + ρ N 1 (1 + τ(r + ρ) 2 i,j=1 Π Π t [D(v)n t ] = = Π (µd(u)n) = g (u, ρ) on S R S 1 ω = ω(u) (u = (u 1,, u N 1 ) U) local chart gij ρ ω, < n t, ω >= 1 + O(( ρ) 2 ) u i u j N 1 ij ρ ρ g ) 3/2 dτ u i u j i,j=1 ij ρ ρ g = O(( ρ) 2 u i u j N 1 ( ) Let n t = µ(ω + a l τ l ), τ l = ω, and use n t = 1, n t x = (j = 1,, N 1) u l u j l=1 V N =< x t, n t >=< ρ t ω + ξ (t), n t >= ρ t + 1 u(1 + J ) dy +, B R B R ( Thus, V N = u n reads ρ t u 1 ) u dy n = k in (u, ρ) on S R B R B R d = Π Π t d = & < d, n >=, Π t d = d < d, n t > n t, Π d = d < d, n > n µd(v) pi)n t σh(γ t )n t = = Π Π t [D(v)n t ] = & < µd(v)n t, n > p < n t, n > σ < H(Γ)n t, n >= YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

36 N 1 < H(Γ t )n t, n >=< Γt x, n >, Γt f = gt 1 i,j=1 ( g t g ij f ) t u i u j g tij = x u i x u j = ((R + ρ)τ i + ρ u i ) ((R + ρ)τ j + ρ u j ) = (R + ρ) 2 g ij + ρ u i ρ u j, G t = (R + ρ) 2 G(I + (R + ρ) 2 (G 1 ρ) ( ρ)), det G t = (R + ρ) 2N det G(1 + (R + ρ) 2 (G 1 ρ) ( ρ)), G 1 t = (R + ρ) 2 (G 1 ρ) ( ρ) (I (R + ρ) 2 + (G 1 ρ) ( ρ) )G 1 g t = (R + ρ) N g + O(( ρ) 2 ), G 1 t = (R + ρ) 2 G 1 + O(( ρ) 2 ) Using ω ω/ u j =, we have < H(Γ t )n t, ω >=< Γt x, ω > = N 1 i,j=1 g ij t (R + ρ) < 2 N 1 ω, ω > + Γt ρ = (R + ρ) 1 u i u j i,j=1 2 ω g ij <, ω > + 1 u i u j R 2 ρ + O(( ρ) 2 ) < ω,ω>= (N 1) 1 R + ρ = 1 R ρ R 2 + ρ 2 (R + ρ)r 2, thus, < H(Γ t)n t, ω >= N 1 R + N 1 R 2 ρ + 1 R 2 ρ < (µd(v) pi)n t H(Γ t )n, n >=< µd(u)n, n > q < n t, n > + N 1 R N 1 R 2 ρ 1 R 2 ρ + = Since < n t, n > 1 = 1 + O(( ρ) 2 ), dividing the formula by < n t, n >, we have < µd(u)n, n > (q N 1 R ) σ R 2 Bρ = g N(u, ρ) on Γ YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

37 Decay estimate for the linearized equations (13) t u Div (µd(u) pi) = f Note that (x i e j x j e i ) n SR = div u = f div = div f div t h n P u = d in B R (, ), in B R (, ), on S R (, ), (µd(u) pi)n σ Bhn = g on S r 2 R (, ), (u, h) t= = (u, h ) on B R S R B = N 1 +, : Laplace-Beltrami operator of S 1, D(u) = u + u, P u = (u 1 u dx) ω(ω = x/ x S 1 ) B R B r is an eigen-value of of order 1 with eigen-function 1 (N 1) is the first eigen-value of of order N with eigen-functions ω i = x i / x (i = 1,, N) Let φ i (i = 1,, N + 1) be the orthonormal basis of the linear hull [1, ω 1,, ω N ] with respect to the L 2 inner product on S N 1 D(u) = u + u = u = Ax + b where A : anti-symmetric matrix: b: N vector R d = {u D(u) = } = {e i, x i e j x j e i i, j = 1,, N}, e i = (,, ith 1,, ) Let p l (l = 1,, M) be the orthonormal basis of R d with respect to the L 2 inner product on B R YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

38 Decay theorem Theorem Let 1 < p, q < and T > If initial data u Bq,p 2(1 1/p) (B R ), h Bq,p 3 1/p (B R ),and right hand-sides f L p((, T ), L q(b R ) N )), f div L p((, T ), H 1 q (B R )), f div H 1 p((, T ), L q(b R ) N ), d L p((, T ), H 2 q (B R )), g H 1 p((, T ), W 1 q (B R )) L p((, T ), H 1 q (B R ) N ) satisfying the compatibility conditions: (14) div u = div f div t= in B R,, µd(u )n < D(u )n, n > n) = (g < gn, n > n) t= on S R then problem (13) admits a unique solution (u, p, h) with possessing the estimate: u L p((, T ), H 2 q (B R ) N ) H 1 p((, T ), L q(b R ) N ), p L p((, T ), H 1 q (B R )), h L p((, T ), Wq 3 1/q (B R )) Hp((, 1 T ), Wq 2 1/q (B R )) e ηs s(u, h) Lp((,t),Lq(BR) W 2 1/q q C{ u B 2(1 1/p) q,p with some positive constants C and η (SR)) + eηs (u, h) Lp((,t),H q 2 (BR) W 3 1/q (SR)) (BR) + h B 3 1/p (BR) + eηs f Lp((,t),Lq(BR)) q,p + e ηs (f div, g) Lp((,t),H q 1 (BR)) + e ηs s(f div, g) Lp((,t),W q 1 (BR)) + eηs ( sf div, f div ) Lp((,t),Lq(BR)) M ( t ) 1/p N ( + (e ηs (u(, s), p l ) BR ) p t ) 1/p ds + (e ηs (h(, s), φ l ) SR ) p ds l=1 l=1 q YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

39 Proof of Global well-posedness A bootstrap argument is used to continue the local in time solution to (, ) Let (u, h) be solutions of the nonlinear problem defined on time interval (, T ) Let E(u, h) T = e ηs s (u, h) Lp((,T ),L q(b R ) Wq 2 1/q (S R )) + eηs (u, h) Lp((,T ),Hq 2 (B R ) Wq 3 1/q (S R )) Then, by the linear estimate E(u, h) T C{ u 2(1 1/p) B q,p (B R ) + h 3 1/p B q,p (B R ) + E(u, h)2 T M ( t ) 1/p N ( + (e ηs (u(, s), p l ) BR ) p t 1/p ds + (e ηs (h(, s), φ l ) SR ) ds) p l=1 If E(u, h) T C{ u B 2(1 1/p) q,p local solutions are prolonged beyond T provided that u B 2(1 1/p) q,p small enough Thus, the point is to prove that l=1 (B R ) + h 3 1/p B q,p (B R ) + E(u, h)2 T } is obtained, then the (B R ) + h 3 1/p B (B R ) is q,p M ( t ) 1/p N ( (e ηs (u(, s), p l ) BR ) p t 1/p ds + (e ηs (h(, s), φ l ) SR ) ds) p CE(u, h) 2 T l=1 l=1 YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

40 Conservation of Mass and Momentum We move from Lagrangian to Eulerian ξ to obtain v p l dx = (l = 1,, M) In fact, N Ωt Conservation of Momentum d v dx = (v t + v v) dξ = (Div S(v) p) dx = (S(v) pi)n t dσ = Γt x dσ = dt Ωt Ω Ωt Γt Γt v dx = v dx = from the assumtion Ωt Ω Conservation of Angular Momentum d i v j x j v i ) dx = dt Ωt(x d t t [ξ i + u i dx)v j (ξ j + u j ds)v i ]J dξ dt Ω = [(u i u j u j u i ) + x i (v jt + v v j ) x j (v it + v v i )]J dξ Ω N = [x i (v jt + v v j ) x j (v it + v v i )]J dξ = [x i k (µd jk δ jk p) x j k (µd k δ ik p)] dx Ωt Ωt k=1 k=1 N = [x i ν k (µd jk δ jk p) x j ν k (µd ik δ ik p)] dσ [(µd ji δ ji p) (µd ij δ ij p)] dx Γt k 1 k 1 Ωt = σ (x i Γt x j x j Γt x i ) dσ = σ t [x i t x j t x j t x i ] dσ = Γt Γt Thus, (x i v j x j v i ) dx = (ξ i v j ξ j v i ) dξ = In particular, Ωt Ω YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

41 Orthogonal condition for the velocity field x = ξ + H(ξ, t)ξ + ξ(t), so that = v dx = uj dξ = u dξ + u(1 J) dξ Ω t B R B R B R = u dξ = u(j 1) dξ B R B R = (x i v j x j v i ) dx = [(ξ i + H(ξ, t)ξ i )u j (ξ j + H(ξ, t)ξ j )u i ]J dξ Ω t B R = (ξ i u j ξ j u i ) dξ = (ξ i u j ξ j u i )(J 1) dξ H(ξ i u j ξ j u i )J dξ B R B R B R Thus, h(, t) L (SR ) 1 = ( t (e ηs (u(, s), p l ) BR ) p ds) 1/p C e η u Lp((,T ),H 2 q (B R )) e ηs h Lp((,T ),W 3 1/q q (S R )) YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

42 Orthogonal condition for height function Ω t = {x = ξ + H(ξ, t)ξ + ξ(t) ξ B R }, Γ t = {x = Rω + h(rω, t)ω + ξ(t) ω S 1 } R+h(Rω,t) Conservation of Mass Ω t = Ω = B R = RN N ω N = dx = dω s N 1 ds Ωt ω =1 = 1 N (R + h(rω)) N dω = B R + R N 1 h(rω, t) dω + NC k R N k h(rω, t) k dω N ω =1 ω =1 k=2 ω =1 N = h(ω, t) dω = NC k R k+1 h(ω, t) k dω ω =R k=2 ω =R Conservation of Momentum d x dx = d t (ξ + u(ξ, s) ds) dξ = u(ξ, t) dξ = v dx = dt dt Ωt Ω Ω Ωt ξ(t) = 1 x dx = x dx Ω ξ(t) = (x ξ(t)) dx ( Ω = Ω t = dx) Ω Ωt Ωt Ωt Ωt h(rω,t)+r = (x i ξ i (t)) dx = dω (sω i )s N 1 ds = 1 ω i (R + h(rω, t)) N+1 dω Ωt ω =1 N + 1 ω =1 = RN+1 N+1 ω i dω + R N N+1C k R N+1 k ω i h(rω, t) dω + ω i R N+1 k h(rω, t) k dω N + 1 ω =1 ω =1 N + 1 = k=2 ω =1 N+1 N+1C k R 1 k = ξ i h(ξ, t) dω ξ = ξ i h(ξ, t) k dσ ξ ξ =R N + 1 k=2 ξ =R ( t ) 1/p Thus, (e ηs (h(, s), φ l ) SR ) p ds C e ηs h 2 Lp((,T ),Wq 3 1/q (SR)) YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

43 Idea of Proof of Decay Theorem (1) u t Au = f in Ω (, T ), Bu Γ = g, u t= = u To obtain the decay estimate of problem (1), first we consider the shifted equation: v t + λ v Av = f in Ω (, T ), Bv Γ = g, v t= = u with λ >> 1 e γt ( t v, 2 v) Lp((,T ),Lq(Ω)) C(u, f, g) ( 2 v = { x α v α 2}) Let u = v + w, and then w satisfies the equation: w t Aw = λ v in Ω (, T ), Bw Γ =, w t= = Let {T (t)} t be a C semigroup associated with problem (1) t By Duhamel principle w(t) = λ T (t s)v(s) ds Let N = {u Au =, Bu Γ = } = {b 1,, b M }, and then T (t)f Lq(Ω) Ce γt f Lq(Ω) with some positive constants γ and C provided that f N t M Let w(t) = λ T (t s)(v(s) (v(s), b j )b j ) ds, and then, e γt w Lp((,T ),Lq(Ω)) C(u, f, g) j=1 M Moreover, w satisfies w t A w = λ (v (v, b j )b j ) in Ω (, T ), B w Γ =, w t= = j=1 M t N { w = w + (v(s), b j ) dsb j, thus, e γt T 1/p w Lp((,T ),Lq(Ω)) C(u, f, g) + C e pγs (v(s), b j ) ds} p j=1 j=1 YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

44 Uniqueness Uniqueness derives the exponential stability of semi-group (2) Div (µd(w) pi) =, div w = in B R, n P w =, (µd(w) pi)n σ R 2 Bρn = on S R = (Div (µd(w) pi), w) Ω = ((µd(w) pi)n, w) Γ + µ 2 D(w) 2 L 2(Ω) = σ R 2 (Bρ, n w) Γ + µ 2 D(w) 2 L 2(Ω) n w = n P w + 1 w dx n = 1 w dx n B R B R B R B R σ N Thus, R 2 (Bρ, n w) σ 1 Γ = R 2 w j dx(bρ, ω j ) Γ = B R B R j=1 Thus, D(u) = Thus, p = in Ω p + σ Bρ = on Γ p is a constant R2 σ(n 1) Recalling that B = N 1 +, ρ = ω i and p =, or ρ = 1 and p = R 2 Thus, if w and p satisfy the null condition (w, p l ) Ω = (l = 1,, M), (ρ, φ i ) Γ = (i = 1,, N + 1), then w =, p =, ρ = (Uniqueness) YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

45 Free boundary problem in an exterior domain Ω R N (N 3): exterior domain, Ω = R N \ O (O bounded domain), Ω t : time evolution of Ω, Γ: boundary of Ω, Γ t : boundary of Ω t (15) t v + (v v) Div (µd(v) πi) =, div v = in <t<t Ω t (, t), (νd(v) πi)n t =, V t = n t v on <t<t Γ t (, t), Ω t t= = Ω, v t= = v, in Ω φ C (R N ): φ = 1 for x B R and φ(x) = for x B2R, c where R N \ Ω = O B R/2 B L = {x R N x < L} u : the velocity field in the Lagrange coordinate x = ξ + φ(ξ) x = ξ + φ(ξ) t T u(ξ, s) ds u(ξ, s) ds φ(ξ) t (Modified Lagrange transformation) T u(ξ, s) ds YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

46 New equations (16) t u J 1 Div S(u, p) = F (u, p), in Ω (, T ), div u = G(u) = G(u) in Ω (, T ), S(u, p)n = H(u) on Γ (, T ), v t= = v, in Ω J 1 Div S(u, p) and div u are linear operators derived from Div (µd(v) qi) and div v by the change of variable: x = ξ + φ(ξ) F (u, p) : nonlinear term consisting of V ( and V 2 ( T G(u) = V 3 ( t T T u(ξ, s) ds u ds) u, V 1 ( t t t t YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6 T u ds)( T 2 u ds) u u ds) p with some polyonials V, V 1 and V 2 with V () = V 2 () = T u ds) u, G(u) = V4 ( T t t u ds)u, with some polyonials V 3 and V 4 with V 3 () = V 4 () = H(u) = V 5 ( T u ds) u with some polyonials V 5 with V 5 () =

47 Thoerem Let N 3 and let q 1 and q 2 be exponents such that N < q 2 < and 1/q 1 = 1/q 2 + 1/N and q 1 > 2 Let b, p and p = p/(p 1) be numbers satisfying the conditions: (17) N > b > 1 ( N ) q 1 p, b p > 1, q 1 ( N + 1 p 2q 2 2) < 1, bp > 1, for any T > with some C > independent of ϵ (b N 2q 2 ) p > 1, b > N 2q 1, (b N 2q 2 ) p > 1, N q p < 1 Then, there exists an ϵ > such that u 2(1 1/p) B + u Lq1 ϵ, then q 2,p /2 problem (16) admits a unique solution u with possessing the estimate: T T + + u L p ((, ), W 2 q 2 (Ω) N ) W 1 p ((, ), L q2 (Ω) N ) ((1 + t) b u(, s) W 1 (Ω)) p ds T ((1 + s) (b N ) 2q 1 u(, s) W 1 q1 (Ω)) p ds + ( sup (1 + s) N 2q 1 u(, s) q1 ) p <s<t ((1 + s) (b N 2q 2 ) ( u(, s) W 2 q2 (Ω) + t u(, s) Lq2 (Ω))) p ds Cϵ p YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

48 To explain how to obtain the decay of solutions, we write the equations symbolically as follows: u t Au = f, Bu Γ = g, u t= = u time -shifted equations have the exponential stability: u = v + w, so that v t + λ v Av = f Bv Γ = g, v t= = u w t Aw = λ v Bw Γ =, w t= = YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

49 Let {T (t)} t be semi-group associated with the equations: u t Au = Bu Γ =, u t= = g Then, by the Duhamel principle, the solution u of the equations: u t Au = f Bu Γ =, u t= = is written by u(t) = L q -L p estimate: t T (t s)f(s) ds, T (t)u Lp Ct N 2 ( 1 q 1 p ) u Lq, T (t)u Lp Ct N 2 ( 1 q 1 p ) 1 2 u Lq for any t 1 with 1 < q p and 1 < q < Remark Non-slip condition case: T (t)u Lp Ct N 2 ( 1 q 1 p ) u Lq, T (t)u Lp Ct min( N 2 ( 1 q 1 p ) 1 2, N 2q ) u Lq for any t 1 with 1 < q p and 1 < q < YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

50 Assume that T + ((1 + s) b f(, s) q1 /2) p ds + T Then, we have T ((1 + s) b f(, s) q2 ) p ds ((1 + s) N 2q 1 f(, s) q3 ) p ds = M < (1/q 3 = 1/q 1 + 1/q 2 ) T ((1 + t) b u(, s) W 1 ) p ds + + ( sup + <t<t T T (1 + t) N 2q 1 u(, t) W 1 q1 ) p ((1 + s) (b N 2q 1 ) u(, s) W 1 q1 ) p ds {(1 + s) (b N 2q 2 ) ( u(, s) W 2 q2 + t u(, s) Lq2 )} p ds CM YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

51 L p -L q decay estimate = i u(t) C C + C + C C + C + C t t/2 t 1 t/2 t t 1 t/2 t 1 t/2 t t 1 i T (t s)f(s) ds (t s) N 2 ( 2 1 q 1 ) i 2 f(s) q1 /2 ds (t s) N 2 ( 2 1 q 1 ) i 2 f(s) q1 /2 ds (t s) ( N + i 2q 2 2 +ϵ) f(s) q2 ds (t s) ( N q 1 + i 2 ) f(s) q1 /2 ds (t s) ( N q 1 + i 2 ) f(s) q1 /2 ds (t s) ( N + i 2q 2 2 +ϵ) f(s) q2 ds (i =, 1) YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

52 v (t) = thus here t/2 (t s) ( N q 1 + i 2 ) f(, s) q1/2 ds ( ) (t/2) N ( + i t/2 ) 1/p ( t/2 q 1 2 (1 + s) bp ds ((1 + s) b f(, s) q1/2) p ds t ( N q 1 + i 2 T 2 ) 2( N q 1 + i 2 ) (bp 1) 1/p M T ( (t b v (t)) p dt C t N + )p i q 1 2 b dt M 2 ( N + i ) ( N ) q 1 2 b p b p > 1 q 1 ) 1/p thus T 2 (t b v (t)) p dt CM YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

53 v 1 (t) = thus t 1 t/2 (t s) ( ( t 1 ( (t s) t/2 ( t 1 ( (t s) T T t/2 N q + i 1 2 N q + i 1 2 N q + i 1 2 ) f(, s) q1/2 ds ) ) 1/p ds ) f(, s) p q 1/2 ds ) 1/p ( N (v 1 (t)t b ) p dt + i ) p/p q ( t 1 ( (t s) t/2 change the integration order N q + i 1 2 ) ) (s b f(, s) q1/2) p ds dt ( N + i ) p/p T 1 q (s b f(, s) q1/2) p( 2s ( N q 1 + i 2 1 ) pm 1 s+1 ( ) (t s) N q + i 1 2 ) dt ds YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

54 v 2 (t) = t t b v 2 (t) C t 1 t ( (t s) N + i 2q 2 2 ) f(, +ϵ s) q2 ds t 1 t 1 ( (t s) N + i 2q 2 2 )(1 +ϵ + s) b f(, s) q2 ds ( t ( ) C (t s) N + i 2q 2 2 +ϵ ds ( t t 1 ) 1/p ( (t s) N + i 2q 2 2 )((1 +ϵ + s) b f(, s) q2 ) p ds ) 1/p Thus, T 2 (t b v 2 (t)) p dt (1 N i ) p/p T 2q 2 2 ϵ ((1 + s) b f(, s) q2 ) p( s+1 ( (t s) N ) ) + i 2q 2 2 +ϵ) dt ds (1 N i ) p T 2q 2 2 ϵ ((1 + s) b f(, s) q2 ) p ds ( 1 N 2q 2 i 2 ϵ ) pm 1 1 s YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

55 estimate of nonlinear terms (1 + t) b u(, t) u(, t) q1/2 u(, t) q1 u(, t) q1 (1 + t) b N 2q 1 u(, t) W 1 q (1 + t) N 2q 1 u(, t) W 1 q1 T φ 2 u(, t) ((1 + t) b u(, t) u(, t) q1/2) p ds CM 2 T φ 2 u q1/2 t u(, s) ds q1/2 C 2 u(, t) q2 ( T t u(, s) ds t ) 1/p (1 + s) bp ds ( T ((1 + s) b u(, s) W 1 ) p ds t T ) 1/p ( supp φ is compact) T ((1 + t) b φ 2 u(, t) u(, s) ds q1/2) p ds CM 2 t YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

56 L p -L q -decay estimate (18) t u J 1 Div S(u, p) =, div u = in Ω (, ), S(u, p)n = on Γ (, ), u t= = u in Ω Let R > be a fixed number such that R N \ Ω B R/2 Let A = A(x) = (a ij ) be the inverse matrix of x/ y with x = y + with some small positive number σ and some index r (N, ) k=1 T φ(y)u(y, s) ds N u i u j N J(x) = det( x/ y), Dij (u) = (a kj + a ki ), Sij (u, p) = Ja jk ( x j x D ik (u) δ ik p), k div u = N j,k=1 Ja kj u j x k a jk = a jk (x) = δ jk + b jk (x) and J(x) = 1 + J (x) Assumption(a) : b ij (x) =, J (x) = for x 2R, (b ij, J ) L (Ω) + (b ij, J ) Lr(Ω) σ k=1 YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

57 L p -L q Decay estimate where q = q/(q 1) Theorem J q (Ω) = {u L q (Ω) N (u, φ) Ω = for any u Ĵp(Ω) and j =, 1, 2 with 1 < p < for any φ Ĥ1 q, (Ω)}, Let N < r < and 1 < q r Then, there exists a σ > such that if assumption (a) holds, then there exists a C analytic semi-group {T (t)} t associated with problem (18) such that for any u Ĵq(Ω) N, u = T (t)u is a unique solution of problem (18) with some pressure term p L p ((, T ), H 1 q (Ω) + Ĥ1 q (Ω)) possessing the estimates: (19) j T (t)u Lq(Ω) C p,q t j 2 N 2 provided that 1 < p q with p Moreover, ( 1 p 1 q ) u Lp(Ω) for any t 1, u Ĵ(Ω) and j =, 1, (2) sup t j/2 j T (t)u Lp(Ω) C p u L p (Ω) <t<2 YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

58 Resolvent problem (21) { λu J 1 Div S(u, p) =, div u = in Ω, S(u, p)n = on Γ Let and then, we have Theorem Σ ϵ = {λ C \ {} arg λ < π ϵ}, Σ ϵ,λ = {λ Σ ϵ λ > λ }, Let N < r <, 1 < q r and < ϵ < π/2 Then, there exist σ >, λ > and a family op operators S(λ) Hol (Σ ϵ,λ, H 2 q (Ω) N ) such that if assumption (a) holds, then for any λ Σ ϵ,λ, f L q (Ω) N, u = S(λ)f is a unique solution of problem (21) with some pressure term p H 1 q (Ω) + Ĥ1 q (Ω) possessing the estimate: (22) (λs(λ)f, λ 1/2 S(λ)f, 2 S(λ)f) Lq(Ω) C q,ϵ,λ f Lq(Ω) with some constant C q,ϵ,λ dependng solely on q, ϵ and λ Moreover, for any p with N < p <, S(λ) Hol (Σ ϵ, L(L p (Ω) N, H 1 (Ω) N ) with (23) (λs(λ)f, λ 1/2 S(λ)f L (Ω) C p,ϵ λ N 2p f Lq(Ω) (λ Σ ϵ ), and for any 1 < q <, S(λ) Hol (Σ ϵ, L(L q (Ω) N, H 1 q (Ω) N ) with (24) (λs(λ)f, λ 1/2 S(λ)f) Lq(Ω) C q,ϵ f Lq(Ω) (λ Σ ϵ ) YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

59 Note that J 1 Div S(u, p) = Div S(u, p) and div u = div u for x 2R Let L q, (Ω) N = {f L q (Ω) N f(x) = for x 4R} First, assuming f L q, (Ω) N, we consider λu J 1 Div S(u, p) = f, div u = in Ω, S(u, p)n Γ = Let R (λ)f = F 1 ξ [ P (ξ)ˆf(ξ) ] [ ξ ˆf(ξ) ] λ + ξ 2, Πf = F 1 ξ ξ 2 (P (ξ) = (δ ij ξ i ξ j ξ 2 )) Let v = T f and q = Qf be solutions: J 1 Div S(v, q) = f, div v = in Ω, S(v, q)n Γ =, v S5R = Let φ C (R N ) such that φ(x) = 1 for x 2R and φ(x) = for x 3R, and set Φ(λ)f = (1 φ)r (λ)f + φr 1 f + B[( φ) (R (λ)f T f)], Ψf = (1 φ)π f + φπ 1 f Then, u = Φ(λ)f and p = Ψf satisfy λu J 1 Div S(u, p) = f + S(λ)f, div u = in Ω, S(u, p)n Γ = There exists a σ > such that (I + S(λ)) 1 L(L q, (Ω)) exists YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6

60 For general f L q (Ω) N λr (λ)f J 1 Div S(R (λ)f, Πf) = f + R 1 (λ)f, div R (λ)f = g(λ)f in Ω), S(R (λ)f, Πf)n Γ = h(λ)f Let v 1 and q 1 be solutions of the equations: λv 1 J 1 Div S(v 1, q 1 ) = R 1 (λ)f, div v1 = g(λ)f in Ω 5R, S(v 1, q 1 )n Γ = h(λ)f v S5R = R 1 (λ)f, g(λ)f, h(λ)f behaves like λ 1/2 near λ = Let v 2 = R (λ)f + φv 1 B[( φ) v 1 ] and q 2 = Π b ff q 1, and then λv 2 J 1 Div S(v 2, q 2 ) = R 2 (λ)f, div v2 = in Ω, S(v2, q 2 )n Γ = And then, v 2 = Φ(λ)(I + R(λ)) 1 R 2 (λ)f, q 2 = Ψ(I + R ( λ)) 1 R 2 f Thus, we see that v 2 H 1 (Ω) C λ N 2q f Lq(Ω) for N < q <, l v 2 Lq(Ω) C λ 1+ l 2 f Lq(Ω) (l =, 1) for 1 < q < near λ = with arg λ π ϵ ( < ϵ < π/2) YShibata (Waseda) Free Boundary Prob Aug29-Sept2, / 6