Global well-posedness of unsteady motion of viscous incompressible capillary liquid bounded by a free surface

In this paper, we prove the global well-posedness of free boundary problems of the Navier-Stokes equations in a bounded domain with surface tension. The velocity field is obtained in the \begin{document}$L_p$\end{document} in time \begin{document}$L_q$\end{document} in space maximal regularity class, ( \begin{document}$2 , \begin{document}$N , and \begin{document}$2/p + N/q ), under the assumption that the initial domain is close to a ball and initial data are sufficiently small. The essential point of our approach is to drive the exponential decay theorem in the \begin{document}$L_p$\end{document} - \begin{document}$L_q$\end{document} framework for the linearized equations with the help of maximal \begin{document}$L_p$\end{document} - \begin{document}$L_q$\end{document} regularity theory for the Stokes equations with free boundary conditions and spectral analysis of the Stokes operator and the Laplace-Beltrami operator.

1. Introduction and results. This paper deals with the global well-posedness of free boundary problems governing the motion of a finite isolated mass of a viscous incompressible capillary liquids in the L p in time and L q in space framework. Let Ω be a domain in the N -dimensional Euclidean space R N (N ≥ 2), whose boundary consists of the compact hypersurface Γ. Our problem is to find a time dependent domain Ω t (t > 0), the velocity vector field v(x, t) = (v 1 (x, t), . . . , v N (x, t)), where M denotes the transposed M , and the pressure field p(x, t) defined for x = (x 1 , . . . , x N ) ∈ Ω t (t > 0), which satisfy the initial boundary value problem for the Navier-Stokes equations: in Ω t for t ∈ (0, T ), (µD(v) − pI)n t = σH(Γ t )n t , V N = v · n t on Γ t for t ∈ (0, T ), where Γ t is the boundary of Ω t , Γ 0 = Γ, and T ∈ (0, ∞]. As for the remaining notation in problem (1), ∂ t = ∂/∂t, ∂ j = ∂/∂x j , v 0 = (v 01 , . . . , v 0N ) is a given initial velocity, D(v) = (∇v + ∇v) the doubled strain tensor with (i, j) th component D ij (v) = ∂ i v j + ∂ j v i , I the N × N identity matrix, H(Γ t ) the N − 1 times mean curvature of Γ t given by H(Γ t )n t = ∆ Γt x (x ∈ Γ t ), where ∆ Γt is the Laplace-Beltrami operator on Γ t , V N the velocity of the evolution of Γ t in the direction of n t , and µ and σ are positive constants representing the viscous coefficient and the coefficient of surface tension, respectively. Moreover, for any matrix field K with (i, j) th component K ij , the quantity Div K is an N -vector with i th component N j=1 ∂ j K ij and for any vector of functions w = T (w 1 , . . . , w N ) we set div w = N j=1 ∂ j w j and (w · ∇)w is an N -vector with i th component N j=1 w j ∂ j w i . In the case where Ω is bounded (so called drop problem), the local well-posedness was proved by Solonnikov [26,27] and references therein, Moglilevskiȋ and Solonnikov [9], Padula and Solonnikov [13], and Schweizer [19]. On the other hand, in the case where Ω is a layer (so called ocean problem), the local well-posedness was proved by T. Beale [3], Allain [2] and Tani [30].
Solonnikov [24] proved the global well-posedness for the drop problem in the Sobolev-Slobodeckiȋ space W 2+ ,1+ /2 2 for ∈ (1/2, 1) under the assumption that the initial state Ω is close to a ball and initial data are small enough. The global well-posedness for the ocean problem was proved by Beale [4], Beale and Nishida [5,12], Tani and Tanaka [31], and Kawashima and Hataya [6,7] in the L 2 framework under the assumption that the initial state Ω is very closed to a flat space and initial data are small enough. Recently, Saito and Shibata [17,18] proved the global wellposedness for the ocean problem without bottom in the H 2,1 q,p framework, where H 2,1 q,p = H 1 p ((0, T ), L q (Ω)) ∩ L p ((0, T ), H 2 q (Ω)). We remark that without surface tension case, the local well-posedness for any initial data and the global well-posedness for small initial data were proved by Solonnikov [25], and Mucha and Zajaczkowski [10] in W 2+ ,1+ /2 2 for ∈ (1/2, 1), by Abels [1] in H 2,1 p,p (N < p < ∞), and by Shibata and Shimizu [23], and Shibata [20] in H 2,1 q,p (2 < p < ∞ and N < q < ∞). The purpose of this paper is to prove the global well-posedness of problem (1) under the assumption that the initial state Ω is very close to a ball and the initial data are sufficiently small with the help of maximal H 2,1 p,q regularity theory for the Stokes equations and spectral analysis for the Stokes operator and the Laplace-Beltrami operator, which is an extension of Solonnikov's result [24] from maximal regularity view point. As a related topic, the local and global well-posedness of the two phase viscous incomplessible flows separated by a sharp interface in a container has been extensively studied by Pruess and Simonett [14,15] (cf. also [8], [16] and references therein) with the help of maximal L p regularity theory for the Stokes equations and spectral analysis for the Laplace-Beltrami operator. Now, we formulate the problem. Let B R = {y ∈ R N | |y| < R} and S R = {y ∈ R N | |y| = R}. We assume that where |D| denotes the Lebesgue measure of a Lebesgue measurable set D in R N and ω N is the area of S 1 ; (A2) Ω x dx = 0; (A3) Γ is a normal perturbation of S R given by Γ = {x = y + ρ 0 (y)(y/|y|) | y ∈ S R } = {x = (1 + R −1 ρ 0 (y))y | y ∈ S R } with given small function ρ 0 defined on S R .

VISCOUS INCOMPRESSIBLE CAPILLARY LIQUID 119
Let Γ t be given by where ρ(y, t) is an unknown function with ρ(y, 0) = ρ 0 (y) for y ∈ S R and ξ(t) is the barycenter point of the domain Ω t defined by which is also unknown. It follows from the assumption (A2) that ξ(0) = 0. Moreover, In fact, let w(ξ, t) be the Lagrange description of the velocity field v(x, t), and then using the Lagrange transformation: x = ξ + t 0 w(ξ, s) ds (ξ ∈ Ω), we have Given a height function ρ(y, t), let H ρ (y, t) be a solution to the Dirichlet problem: We introduce the Hanzawa transformation centered at ξ(t) defined by x = h z (y, t) := (1 + R −1 H ρ (y, t))y + ξ(t) for y ∈ B R .
In our proof below, we will choose 0 ∈ (0, 1/4) suitably small in such a way that several sufficient conditions to obtain our main result hold. In the following, we set where v 0 is an initial data for problem (1). Let v and p satisfy (1) and let 120 YOSHIHIRO SHIBATA then u, q, and ρ satisfy the following equations: where and ∆ S R is the Laplace-Beltrami operator on S R . Moreover, we set < a, b >= a·b = N j=1 a j b j for any N vectors a = (a 1 , . . . , a N ) and b = (b 1 , . . . , b N ), and Π 0 a = a− < a, ω > ω. In the equations (10), f , f d , f d , k in , g and g n denote some nonlinear functions with respect to u and ρ, which will be given in Sect.2 below. We note that where ∆ S1 is the Laplace-Beltrami operator on S 1 . Before stating our main results, at this stage we explain the symbols used throughout the paper. Let N, R and C denote the sets of all natural numbers, real numbers, and complex numbers, respectively.
For any scalar function f and vector valued function g = (g 1 , . . . , g k ), we set For any domain D, let L q (D), H m q (D) (m ∈ N) and B s q,p (D) (s ∈ R \ N 0 ) be the Lebesgue space, Sobolev space and Besov space, and let · Lq(D) , · H m q (D) and · B s q,p (D) denote their respective norms. We write H 0 q (D) = L q (D) and B s qq (D) = W s q (D). Let X be a Banach space with norm · X . For an interval J = (0, T ) with T ∈ (0, ∞], L p (J, X) and H m p (J, X) (m ∈ N) denote the Xvalued Lebesgue space and X-valued Sobolev space, respectively, and · Lp(J,X) and · H m p (J,X) denote their respective norms. We set for any real number η and time interval (a, b).
. . , d)}, whose norm is written by · X instead of · X d for short. In particular, vectors and vector valued functions are denoted with bold face letters. Let where dτ is the surface element of S R andf denotes the complex conjugate of f . The letters C and c denote generic positive constants and C a,b,... denotes that the (i = 1, . . . , N ), and so, we have compatibility conditions for ρ 0 as follows: because S1 dω = ω N and S1 ω i dω = 0. Here, To obtain the decay properties of solutions, we introduce some orthogonal conditions. Let R d = {u | D(u) = 0}. We know that R d consists of constant N -vectors and any linear combinations of x i e j − x j e i (i, j = 1, . . . , N ), where e i is the unit N -vector whose i th component is one. Note that (x i e j − x j e i ) · x = 0. Let p = |B R | −1 e ( = 1, . . . , N ) and let p ( = N + 1, . . . , M ) be some linear combinations of x i e j − x j e i such that (p , p m ) B R = δ ,m , where δ ,m are the Kronecker's delta symbols, that is δ , = 1 and δ ,m = 0 for = m. Since (e , p m ) B R = 0 for = 1, . . . , N and m = N + 1, . . . , M , {p } M =1 is the orthonormal basis of R d with respect to the L 2 (B R ) innerproduct.
We know that Bx i = 0 on S R for i = 1, . . . , N . Thus, let ϕ 1 = |S R | −1 and let ϕ i (i = 2, . . . , N + 1) be some linear combinations of x 1 , . . . , x N such that which is the solution class of problem (10), and then the main result of this paper is stated as follows.
(2) By a real interpolation theorem, we see that where BU C denotes the set of bounded uniform continuous functions. Thus, the condition 2/p + N/q < 1 guarantees that v ∈ C 0 ([0, T ], C 1 (Ω t )). Therefore, we can solve the Cauchy problem: for some small interval [t 0 , t 1 ] ⊂ [0, T ], where Ω means the closure of Ω. Let x = x(ξ, t) be a unique solution of the Cauchy problem (16). And then, the kinetic condition V N = v·n t in problem (1) guarantees that the Lagrangean transformation: gives C 1 diffeomorphism from Ω onto Ω t for any t ∈ [t 0 , t 1 ]. This fact will be used to find the conservation of mass, the conservation of momentum and the conservation of angular momentum, which plays an essential role to guarantees the exponential decay properties of u and ρ.

YOSHIHIRO SHIBATA
Combining (23) and (24), we have Thus, the divergence condition in (1) is transformed to Next, we consider the kinetic condition: V N = v · n t on Γ t . For this purpose, we represent n t . By (2), the Γ t is represented by x = (1 + R −1 ρ(y, t))y + ξ(t) with y ∈ S R . Let S R be parametrized locally by y = y(u) with u = (u 1 , . . . , u N −1 ) ∈ U ⊂ R N −1 , and then Γ t is parametrized by Let n t = µ(y + N −1 k=1 a k τ k ) with τ k = ∂y(u)/∂u k (k = 1, . . . , N − 1). Since |y| 2 = R 2 , we have y · τ k = 0 for k = 1, . . . , N − 1. Since |n t | = 1, we have where g k = τ k · τ . Let G be the N × N matrix whose (k, ) element is g k , which is the first fundamental form of S R , and let G −1 be the inverse matrix of G. Let g ij be the (i, j) th component of G −1 . Since Γ t is given by (26), we have which leads to Combining (27) and (28) yields that with where η = R −1 ρ(y(u), t)), provided that 0 is sufficiently small in (6). Summing up, we have proved that holds in a local chart with ω = y/R = y/|y| ∈ S 1 and η = R −1 ρ(y(u), t). In particular, we have Since as follows from the fact that |Ω| = |B R |, (2), (3) and (20), the kinetic condition: Next, we consider the boundary condition in (1). We use the following lemma due to Solonnikov [27, p.155] Lemma 2.1. If n t · ω = 0, then for any vector d, d = 0 is equivalent to Here, Π 0 and Π t are defined by Remark 2. We can replace ω by any unit vector n with n t · n = 0. In this case, Π 0 is defined by In view of Lemma 2.1, boundary conditions in (1) are equivalent to by (19) and (31) the condition: Before turning to the second condition in (34), using the formula: H(Γ t )n t = ∆ Γt x for x ∈ Γ t we calculate < H(Γ t )n t , ω >. According to the parametrization of Γ t given above, we know that is the first fundamental form of Γ t and and . Thus, we have because < ∂y ∂u j , ω >= 0 and < y, ω >= R.
To proceed with the calculation in (37), using the parametrization in (26), we From (26) it follows that where ∇ ρ = (∂ρ/∂u 1 , . . . , ∂ρ/∂u N −1 ). Thus, by (38) and so . From the first formula in (40) we have g −1 Thus, we have ∆ Γt ρ = ∆ S R ρ +∆ Γt ρ and In fact, using the polar coordinate: x = rω ∈ R N with r = |x|, we have and so, setting r = 1 yields (44). Moreover, Thus, recalling (37), we have Finally, by (19), (31) and (45), Thus, we have 3. Decay estimate for the linearized equations. In this section and next section, we consider the following linear equations that is the linear part of problem (10): Let ι be an extension map from L 1,loc (B R ) into L 1,loc (R N ) having the following properties: b)) be the solution class defined before Theorem 1 in Sect. 1, and let T )), satisfying the compatibility condition: problem (47) admits a unique solution (u, p, ρ) ∈ S p,q ((0, T )) possessing the estimate: for any t ∈ (0, T ] with some positive constants η and C independent of t and T , where for any 0 ≤ a < b ≤ ∞ and t ∈ (a, b) we have set To prove Theorem 3.1, first we consider the following shifted equations: For the shifted equation (50), we have Theorem 3.2. Let 1 < p, q < ∞ and T > 0. Then, there exist positive constants T )) satisfy the compatibility condition (48), then problem (50) admits a unique solution (u 1 , p 1 , ρ 1 ) ∈ S p,q ((0, T )) possessing the estimate: for any t ∈ (0, T ] and η ≤ η 0 with some positive constants C independent of t and T , where are the same norms as in Theorem 3.1. Proof. Employing the argument in Shibata [20] and [21], we can prove the unique existence of u 1 , p 1 and ρ 1 possessing the estimate (51) with the help of the R bounded solution operators for the generalized resolvent problem obtained by the Laplace transform of (50) with respect to time variable t and the Weis operator valued Fourier multiplier theorem, so we may omit the detailed proof.
We consider the solutions u, p and ρ of problem (47) of the form: u = u 1 + v, p = p 1 + q and ρ = ρ 1 + h, where u 1 , p 1 and ρ 1 are solutions of the shifted equations (50), and then v, q and h should satisfy the equations: Let , the necessary and sufficient condition in order and let w = u 1 − ∇ψ. Then, w ∈ J q (B R ) and Using w and ψ, we can rewrite the first equation in (52) as follows: Thus, in what follows we may assume that To handle problem (52), in view of Duhamel's principle we consider the initial value problem: To handle equations (57) in the semigroup setting, we eliminate the pressure term q by using a unique solution K 1 = K 1 (v) of the weak Dirichlet problem: is a unique solution of the variational equation: In particular, we have On the other hand, to solve (59), let H ∈ H 3 q (B R ) be a solution to the harmonic equation: ∆H = 0 in B R subject to H| S R = −σBh, which possesses the estimate: And then, K 2 is obtained by is a solution to the variational equation: We also have the estimate: Instead of (57), we consider the equation: In fact, let t 0 ∈ (0, T ), and for any ϕ ∈ C ∞ 0 (B R ) let u be a unique solution to the reversed heat equations: Since u(·, 0) ∈ H 1 q ,0 (B R ), we have (v(·, 0), ∇u(·, 0)) B R = 0 provided that div v 0 = 0. Thus, noting that ∂ t u| S R = 0, by (58), (59) and (63), We have Theorem 3.3. Let 1 < q < ∞. Then, the operator A generates a continuous analytic semigroup {T (t)} t≥0 on H q which is exponentially stable onḢ q , that is for any t > 0 and (f , g) ∈Ḣ q with some positive constants C and η 1 .
, we see that v, q and h are solutions of the equations (57).

4.
Continuous analytic semigroup associated with (62). To treat the equations (57) and (62) in the semigroup setting, we consider the corresponding resolvent problems: As was proved in Shibata [21], we have Theorem 4.1. Let 1 < q < ∞ and 0 < < π/2. Then, there exists a positive constant λ 0 such that for any λ ∈ Σ ,λ0 , f ∈ L q (B R ) and g ∈ W Here and in the following, Σ ,λ0 denotes the subset in C defined by For any λ ∈ Σ ,λ0 the operator (λ − ∆) is a bijection from H 2 q (B R ) ∩ H 1 q ,0 (B R ) onto L q (B R ), which follows from the following Lemma 4.2, and so div v = 0 in B R . Thus, (v, h) ∈ D q . Lemma 4.2. Let 1 < q < ∞ and λ ∈ C \ (−∞, 0). Then, for any f ∈ L q (B R ), the Dirichlet problem: admits a unique solution u ∈ H 2 q (B R ). Proof. For any large λ > 0, we know the unique existence of solutions of the equations (77). From the Riesz-Schauder theorem, especially the Fredholm alternavite principle, we know that the uniqueness implies the existence. Thus, it suffices to prove the uniqueness. First we consider the case 2 ≤ q < ∞. Let u ∈ H 2 q (B R ) satisfy the homogeneous equations: Since u ∈ H 2 2 (B R ), by the divergence theorem of Gauß 0 = λ u 2 L2(B R ) + ∇u 2 L2(B R ) , and so, taking the real part and the imaginary part, we have When Im λ = 0, we have u = 0. When Re λ ≥ 0, we have ∇u = 0, which, combined with u| S R = 0, leads to u = 0. Thus, we have the unique existence theorem for 2 ≤ q < ∞. When 1 < q < 2, the uniqueness follows from the existence theorem for the problem with dual exponents 2 < q < ∞, and so we also have the unique existence theorem for 1 < q < 2, which completes the proof of Lemma 4.2.
In view of Corollary 4.1, in order to prove Theorem 3.3 it suffices to prove Then, for any λ ∈ Q λ0 and (f , g) ∈Ḣ q problem (74) admits a unique solution (v, h) ∈ D q ∩Ḣ q possessing the estimate: with some constant C independent of λ ∈ Q λ0 .

YOSHIHIRO SHIBATA
We first consider the case where 2 ≤ q < ∞. Since (f , g) ∈Ḋ q ⊂Ḋ 2 , by (87) and the divergence theorem of Gauß, For h ∈ H 2 q (S R ) and g = (g 1 , . . . , g N ), Moreover, div f = 0, because f ∈J q (B R ). Thus, noting that λg = P f · ω on S R , we have To treat (Bg, g) S R , we use the following lemma.
Before considering the case where 1 < q < 2, at this point we give a Proof of Lemma 4.5. Let {λ j } ∞ j=1 be the set of all eigen-values of the Laplace-Beltrami operator ∆ S R on S R . We may assume that λ 1 > λ 2 > λ 3 > · · · > λ j > · · · → −∞, and then λ 1 = 0 and λ 2 = −(N − 1)R −2 . Let E j be the eigen-space corresponding to λ j , and then the dimension of E j is finite (cf. Neri [11,Chapter III,Spherical Harmonics]). Let d j = dim E j , and then d 1 = 1 and d 2 = N . In particular, be the orthogonal basis of E i in L 2 (S R ), and then for any h ∈Ḣ 2 Since −λ i − (N − 1)R −1 ≥ c with some positive constant c for any i ≥ 2, we have (90), which completes the proof of Lemma 4.5.
Next, we consider the case where 1 < q < 2. Let (f , g) ∈Ḋ q satisfy the homogeneous equations (87). First, we prove that Let (u, ρ) ∈Ḋ q be a solution of the equations: Since λ ∈ Q λ0 ,λ ∈ Q λ0 , and moreover 2 < q < ∞, and so by the fact proved above we know the unique existence of (u, ρ) ∈Ḋ q . By (87), (93) and the divergence theorem of Gauß, Noting that P f · ω = λg and div f = 0 and using (88), we have On the other hand, we have Noting that P u · ω =λρ and div u = 0 and using (88), we have which, combined with (94), leads to (92). Next, we prove that (f , g) B R = 0 for any g ∈ L q (B R ) N . Given any g ∈ L q (B R ) N , let ψ ∈ H 1 q ,0 (B R ) be a solution to the variational equation: (∇ψ, ∇ϕ) B R = (g, ∇ϕ) B R for any ϕ ∈ H 1 q,0 (B R ).
To prove Theorem 5.1, we use the maximal L p -L q regularity for the linearized equations: Using a result due to Shibata [21] we have Theorem 5.2. Let 1 < p, q < ∞. Then, for any initial data w 0 ∈ B 2−2/p q,p (B R ) and ρ 0 ∈ B 3−1/p−1/q q,p (S R ) and right members: ((a, a+1)) satisfying compatibility conditions:
Proof. Since the coefficients of the partial differential operators appearing in problem (99) are constant, setting v(·, t) = u(·, t + a), p(·, t) = q(·, t + a) and h(·, t) = ρ(·, t + a), we can reduce the problem to the case where a = 0, and so Theorem 5.2 follows from a result due to Shibata [21].
To prove Theorem 5.1, we start with the estimate of the nonlinear terms given in Sect. 2. In the following, we assume that N < q < ∞, and then by the Sobolev embedding theorem, we have Moreover, we assume that H ρ and ρ satisfy the smallness condition: with some small constant 0 ∈ (0, 1/4), which is determined in the course of the proof of Theorem 5.1 below. Let f (u, ρ) be the nonlinear function given in (22). Using (100) and (101), we have ∇Φ ρ L∞(B R ) ≤ C H ρ H 2 q (B R ) ≤ C 0 , and so, choosing 0 in such a way that Moreover, using (100) and (101), we have , Let f d (u, ρ) and f d (u, ρ) be nonlinear functions defined in (25). By (100), and so, using (101) and (100), we have Here and in the following, we use the fact that H ρ is a solution of the Dirichlet problem (4) and satisfies the estimates: Thus, noting (101) and using (100) and Lemma 5.3, we have Moreover, by (100) Thus, noting (101) and using (100), we have Let k in (u, ρ) be the nonlinear function given in (33). The n t is suitably extended to B R replacing ρ and ω by H ρ and x. By (32), < ω, n t >= 1 + V 2 (ρ), and so < ω, n t > −1 = 1 + ∞ j=1 (−V 2 (ρ)) j . Thus, by (100) and (101) we have By (100),
Lemma 5.4. Let 1 < p, q < ∞ and a ≥ 0. Then, we have Here, C is a positive constant independent of a.
Proof. If we setf (·, t) = f (·, a+t) andg(·, t) = g(·, a+t), we can reduce the problem to the case a = 0. Thus,, we prove Lemma 5.4 in the case a = 0. The first assertion was proved in Shibata [20, (3.20) in p.4139] and can be proved in the same way as in the proof of the second assertion below, and so we may omit its proof. We only prove the second assertion. Given function h defined on (0, 1), h 0 denotes the zero extension of h to (−∞, 0), that is h 0 (·, t) = h(·, t) for t ∈ (0, 1) and h 0 (·, t) = 0 for t ≤ 0. Let If h| t=0 , then Let G be a function in B 3−1/q q,p (R N ) such that G| S R = g(·, 0) and , where F and F −1 denotes the Fourier transform and the invers Fourier transform on R N , and then we have Noting that g = E 1 (g − d) + d for 0 < t < 1 and using the interpolation relation: where BU C means the set of all uniformly bounded continuous functions, X 0 and X 1 are any Banach spaces such that X 1 is dense in X 0 , and 1 < p < ∞, we have Since (g − d)| t=0 = 0 on S R , by (116), which, combined with (114), leads to Choosing and 2 in such a way that (121) Let Ψ be a map defined by Ψ(u, q, ρ) = (v, p, h), and then by (121) Ψ maps H 2 into itself.
Analogously, we can prove that 1, 2). Choosing and 2 in such a way that for any (u i , q i ,q i , ρ i ) ∈ H 2 (i = 1, 2), which shows that Ψ is a contraction map on H 2 . Thus, there exists a unique element (u, q, ρ) ∈ H 2 such that Φ(u, q, ρ) = (u, q, ρ). In particular, (u, q, ρ) is a unique solution of the equations (99), which completes the proof of Theorem 5.1.