Solvability of the Incompressible Navier-Stokes Equations on a Moving Domain with Surface Tension. 1. Introduction

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1 Solvability of the Incompressible Navier-Stokes Equations on a Moving Domain with Surface Tension Hantaek Bae Courant Institute of Mathematical Sciences, New York University 251 Mercer Street, New York, NY, 112, USA hantaek@cims.nyu.edu Abstract: In this paper, we study the incompressible Navier-Stokes equations on the moving domain Ω t in R 3, of finite depth, bounded above by the free surface and bounded below by a solid flat bottom. We will prove that there exists a unique, global-in-time solution to the problem provided that the initial velocity field and the initial profile of the boundary are sufficiently small in Sobolev spaces. In Appendix 1, we will show the local well-posedness for the same problem without surface tension. In Appendix 2, we will study the global wellposedness of the slightly compressible Navier-Stokes equations with the free boundary under surface tension in three dimensions. As a corollary, we will prove a convergence result of the compressible Navier-Stokes equations. 1. Introduction In this paper, we study a viscous free boundary value problem with surface tension. The incompressible Navier-Stokes equation describes the evolution of the velocity field in the fluid body. With boundary conditions which are explained below, we have the following system of equations: (NSF ) v t + v v µ v + p = in Ω t v = in Ω t v = on S B η t = v 3 v 1 x η v 2 y η on S F ( η pn i = µ(v i,j + v j,i )n j + gη β ( )n ) i on S F 1 + η 2 where Ω t = {(x, y, z) : 1 < z < η(x, y, t)} with two boundaries S F = {(x, y, z) : z = η(x, y, t)} and S B = {(x, y, z) : z = 1}. Here, ˆn = (n 1, n 2, n 3 ) is the outward normal vector on S F. µ is the constant of viscosity, g is the gravitational constant, and β is the constant of surface 1

2 tension. From now, we normalize all the constants by 1. (We follow the Einstein convention where we sum upon repeated indices. Subscripts after commas denote derivatives.) The boundary condition of the velocity at the bottom is that the boundary is impenetrable: v =, which is the boundary condition of the Navier-Stokes equations on a fixed domain. Therefore, we can apply the Poincare inequality to control lower order terms by using higher order terms. On the free surface S F = {(x, y, z); z = η(x, y, t)}, we have three boundary conditions. the kinematic condition: We represent the free boundary by d(x, y, z, t) = z η(x, y, t) =. Since the free boundary moves with the fluid, ( t + v )(z η(x, y, t)) =. Therefore, η t = v 3 v 1 x η v 2 y η. the shear stress boundary condition: (ˆt v ˆn + ˆn v ˆt) =, where ˆt is any tangential 1 vector on the free boundary and ˆn = ( xη, y η, 1). 1 + η 2 η the normal force balance: pn i = (v i,j + v j,i )n j + ηn i ( )n i 1 + η 2 Since the problem is posed on a domain, we need compatibility conditions for the initial data. The initial compatibility conditions on the initial velocity field v are imposed as {(v ) i,j + (v ) j,i } tan = on S F = {(x, y, z) : z = η (x, y)} v = in Ω v = on S B = {(x, y, z) : z = b} where (tan) means the tangential component. We use the tangential part on the pressure condition on the free surface to obtain the first condition. One usually assumes that the flow is irrotational in the case of the Euler equations. The fluid motion is described by a velocity potential which is harmonic. Then, the system can be reduced into a system where all the functions are evaluated at the free surface. But, for the Navier-Stokes equations, it is impossible to assume that the flow is irroatational. The shear stress condition implies that the tangential part of the vorticity on the boundary satisfies w T = w (w ˆn)ˆn = 2ˆn v ˆn = 2(ˆn ) (ˆn v) + 2u j ((ˆn )n j ) 2

3 where (ˆn ) = (n 2 z n 3 y, n 3 x n 1 z, n 1 y n 2 x ) is a tangential derivative. This means that vorticity develops at the free surface whenever there is relative flow along a curved surface. This condition prevents a viscous flow from being irrotational. It is evident in two dimensional flow. In a local coordinate system, the vorticity w at the free surface is given by w = ˆn v ˆt ˆt v ˆn. So, from the shear stress condition, we rewrite w as w = 2ˆt v ˆn = 2 v s ˆv = 2 ˆn (v ˆn) + 2u s s = 2 s (v ˆn) + 2(v ˆt)κ where κ is the curvature of the surface. Therefore, the vorticity does not vanish at the free surface. See [11]. Here, we present some existence results. With surface tension, in [3], Beale studied the motion of a viscous incompressible fluid contained in a three dimensional ocean of infinite extent, bounded below by a solid floor and above by an atmosphere of constant pressure. He established the global existence and regularity theorem by taking into account surface tension. His approach is to transform the problem to the equilibrium domain in a way depending on the unknown η. The entire problem can be solved by iteration in K r. (For definition of parabolic-type Sobolev spaces K r, see [3].) In a bounded domain, Coutand-Shkoller [5] used energy methods to establish a priori estimate which allows to find a unique weak solution to the linearized problem in the Lagrangian coordinates. Then, they proved regularity of the weak solution, and applied the topological fixed point theorem to obtain a solution. When there is no surface tension, in [2], Beale wrote the problem in the Lagrangian formulation so that the domain of the unknowns is fixed in time. He showed the local well-posedness for arbitrary initial data with certain regularity assumptions. He also proved that for any fixed time interval, solutions exist provided the initial data are sufficiently close to equilibrium. In this case, the domain is transformed to the equilibrium domain with flat boundary. Along the same lines as [3], Lynn-Sylvester [12] showed that viscosity alone will prevent the formation of singularities so that solutions exist globally-in-time, even without surface tension. Since we are dealing with problem with the same setting in Beale s paper, we present his result here. [3] 3

4 Theorem 1.1. Suppose r is chosen with 3 < r < 7 2. There exists δ > such that for v and η satisfying η H r (R 2 ) + v H r 2 1 δ and the compatibility conditions, the problem has a (Ω ) solution v, η and p, where η K r+ 1 2 (R 2 R + ) and v and p are restrictions to the fluid domain Ω t of functions defined on R 3 R +, with v K r (R 3 R + ) and p K r 2 (R 3 R + ). Regularity of Beale s solution: Let r = 3 + δ. v K r implies that v H s 2 embedded in C t H r s x t Hx r s. This is if s > 1. Set s = 1 + ɛ. Then, r s = 2 + δ ɛ. But, the initial data is in H r 1 2 and r 1 2 > 2 + δ ɛ. Therefore, the solution does not preserve the initial regularity H r 1 2 as it evolves in time. In this paper, we want to solve the problem on the equilibrium domain. The transformation from the moving domain to the equilibrium domain will be presented in the next chapter. The transformed system of equations on the equilibrium domain is given by (LNSF ) w t w + q = f in Ω = {(x, y, z) : 1 < z < } w = in Ω w i,3 + w 3,i = g i on {z = }, η t = w 3 on {z = } q = w 3,3 + η η + g 3 on {z = } w = on {z = 1} If we solve this linearized problem, then we can solve the problem on the equilibrium domain by the contraction mapping theorem. Since the problem is posed on a fixed domain, we can use the traditional method, namely, we will prove that (LNSF) has a weak solution in L 2. Then, we will establish higher regularities of weak solutions under higher regularities of initial data and external forces. Having solved the linearized problem, we reverse our steps and obtain a solution of the original problem. Theorem 1.2. Suppose v H 2 and η H 3. If initial data are sufficiently small, then there 4

5 is a unique, global-in-time solution (v, η, p) to (NSF) such that v L t H 2 x + v L 2 t H3 x + η L t H3 x + H (η F (η)) L 2 t H1 x + p L 2 t H1 x v H 2 + η H 3 Here is the outline of the proof of Theorem 1.2. In chapter 2, we will prove the existence and uniqueness of the solution on the equilibrium domain under the assumption of the solvability of (LNSF). In chapter 3, we will solve the transformed system of equations (LNSF) on the equilibrium domain. In chapter 4, we will present the change of variables used in chapter 2. Notations: (f, g) = fgdv, ɛ > is the size of the initial data. < u, u >= 1 η (u i,j + u j,i )(u i,j + u j,i )dv, F (η) = ( 2 ) 1 + η 2 Ω H is the harmonic extension operator which extends functions defined on S F to Ḣ1 harmonic functions on Ω with zero Neumann boundary condition on S B. We also denote by H (f) = f. A B means there is a constant C such that A CB. A B + 1 2D means there is a constant C such that A C B D. 2. Existence and Uniqueness on the Equilibrium Domain The first step to prove the existence of a solution is obtaining a priori energy bounds. The basic L 2 energy estimate is easily derived by multiplying by v to the momentum equation and integrating the equation in the spatial variables. If the problem is posed on the whole space, we can obtain higher energy bounds by taking derivatives to the equation. But we cannot take usual partial derivatives on the moving domain. Traditionally, we fix the domain by the Lagrangian map. This is based on L 1 in time estimate of the velocity. But, we have L or L 2 in time estimates of the velocity when we do the energy estimates. So, we only expect local-in-time results if we fix the domain by using the Lagrangian map. In order to avoid this difficulty, we make use of Beale s method [3]: we transform the physical domain onto the equilibrium domain. Since we will project the equation on the divergence free space to obtain L 2 in-time bounds of the boundary, we need to keep the divergence free 5

6 condition to the velocity field on the equilibrium domain. Thus, the transformation is given by the change of variables in a way that the divergence free condition is preserved. We define a map θ(t) : Ω = {(x, y, z); 1 < z < } {(x, y, z ); 1 < z < η(x, y, t)} by using the harmonic extension η of η in the following way: θ(x, y, z, t) = (x, y, η(x, y, t) + z(1 η(x, y, t))) In order θ to be a diffeomorphism, η should be small for all time. This smallness condition will be achieved by higher energy estimates. We define v on θ(ω) by v i = θ i,j w j = α ij w j, = 1 η + z η(1 z), dθ = (θ i,j ) Then, v has divergence zero in θ(ω) if and only if w has the same property in Ω. We replace the system of equations of v with that of w. v i,j = ζ lj l (α ik w k ), v i,t = α ij w j,t + α ij w j + (θ 1 ) 3 3(α ij w j ) where ζ = (dθ) 1 and denotes derivatives in t. Setting q = p θ, the other three terms in the Navier-Stokes equations are of the form α jk w k ζ mj m (α il w l ) ζ kj k (ζ mj m (α il w l )) + ζ ki k q Multiplying by (α ij ) 1, we have the following equation: w t w + q = f( η, v, q) The normal boundary condition becomes ( ) ( η qn i ζ lj l (α ik w k ) + ζ mi m (α jk w k ) N j = η ( )N ) 1+ η 2 i with ˆN = ˆn θ. Let ˆT 1 = (1,, x η), ˆT2 = (, 1, y η). Taking the inner product with ˆT 1, ˆT2, and ˆN, we obtain that w i,3 + w 3,i = g i (η, w), q w 3,3 = η η + g 3 6

7 where g 3 = η F (η) + g 3, g 3 is quadratic in η and w. Finally, the evolution equation of the boundary is calculated by the definition of the new velocity field on Ω: η t = w 3 on {z = }. In sum, we have the following linearized system of equations on the equilibrium domain: (LNSF ) w t w + q = f in Ω w = in Ω w i,3 + w 3,i = g i on {z = }, η t = w 3 on {z = } q = w 3,3 + η η + g 3 on {z = } w = on {z = 1} From this system of equations on the equilibrium domain, we see that the compatibility conditions of the new initial data are the following: w = in Ω w = on {z = 1} w() i,3 + w() 3,i = g i (), q() = w() 3,3 + g 3 () on {z = } We assume that (LNSF) is solvable. (See chapter 3). Let (w, η, q) = w L t H 2 x + w L 2 t H3 x + η L t H3 x + q L 2 t H1 x Then, by Proposition 3.1, (w, η, q) ɛ + f L 2 t H 1 x + g L 2 t H 3 2 x (1) Now, we calculate nonlinear terms. Principal parts are given by f w 3 η + 2 η w + 2 w η + η q, g i η w + ( η η) 7

8 We only need to estimate the highest order terms. f L 2 t L 2 x (w 3 η) + ( 2 η w) + ( 2 w η) + ( q η) L 2 t L 2 x w 3 η L 2 t L 2 x + w 4 η L 2 t L 2 x + w 3 η L 2 t L 2 x + 2 w 2 η L 2 t L 2 x + 3 w η L 2 t L 2 x + 2 w 2 η L 2 t L 2 x + 2 q η L 2 t L 2 x + q 2 η L 2 t L 2 x w L 2 t L x 3 η L t L 2 x + w L t L x 4 η L 2 t L 2 x + w L 2 t H3 x η L t L x + w L t H 2 x 2 η L 2 t L x + 2 q L 2 t L 2 x η L t L x + q L 2 t L2 x 2 η L t L x By Lemma 3.3 in chapter 3, we can replace 4 η L 2 t L 2 x and 2 η L 2 t L x with H (η η) L 2 t H 1 x. Hence, f L 2 t H1 x (w, η, q) 2. We do the same calculation to g. 3 2 g L 2 t L 2 x (R2 ) 3 2 ( w) η + w 3 2 ( η) ( η η) L 2 t L 2 x (R2 ) 3 2 ( w) L 2 t L 2 x(r 2 ) η L t L x + w L 2 t L x (R 2 ) 5 2 η L t L 2 x + η L t L x 7 2 η L 2 t L 2 x Therefore, g L 2 t H 2 3 (w, η, q) 2. From (1), we conclude that x (R 2 ) (w, η, q) ɛ + ( (w, η, q) ) 2 (2) Since we obtain the above energy estimate on the fixed domain, we can iterate the system. The first step is to define the first iteration (w 1, η 1, q 1 ) in terms of the initial data. Then, we can define the second step (w 2, η 2, q 2 ) so on. Let ρ(t) be a nice cut-off function in time such that ρ() = 1. µ(x, y) is a C c function on R 2. Then η 1 is defined as η 1 = (η µ(x, y))ρ(t). We do the same procedure to define the first velocity field w 1. Since w is defined on the channel Ω, we restrict w interior of the domain. We choose a C function ψ supported in 3 4 < z < 1 4. Let φ(x, y, z) be a nice function such that φ = outside of the domain. Let λ(x, y, z) be a C c function on R 3. Then, we define w 1 as w 1 = ((w ψ) λ(x, y, z))φ(x, y, z)ρ(t). Finally, we define the pressure as q 1 = H (w 1 3,3 + η1 η 1 ). For the definition of H, see remark 4 in chapter 3. Then, for 8

9 n > 1, we iterate the system of equations in the following manner: (LNSF m ) wt m w m + q m = f(w m 1, η m 1, q m 1 ) in Ω w m = in Ω wi,3 m + wm 3,i = g i(w m 1, η m 1 ) on {z = } q m = w3,3 m + ηm η m + g 3 (w m 1, η m 1 ) on {z = } ηt m = w3 m on {z = }, w m = on {z = 1} Thus, from the a priori estimate in Proposition 3.1, (w m, η m, q m ) ɛ + (w m 1, η m 1, q m 1 ) 2 Therefore, we conclude that { (w m, η m, q m ) } are uniformly bounded if the initial data is small enough. By taking difference of two sequences, we show that { (w m, η m, q m ) } is a Cauchy sequence. Therefore, we obtain a unique, global-in-time solution to (LNSF) if initial data is small enough in H 2 by the contraction mapping theorem. The dependence on the initial data of the boundary η occurs when we define the first iteration. 3. Solvability of (LNSF) In this chapter, we study the linearized problem (LNSF) defined in Ω = { (x, y, z); 1 < z < }. We need to keep the divergence free condition to the velocity field from two reasons. First of all, since we only know the explicit form of the pressure at the boundary, we want to remove the interior pressure in the weak formulation. By the integration by parts, ( p, φ) = p(φ ˆn)dS (p, φ) Ω So, to remove (p, φ), we need to define test functions such that the divergence free condition holds. Secondly, we need to estimate the boundary in L 2 in-time. This can be achieved by taking the projection operator onto the divergence free space. We estimate the boundary term in L 2 in time from the L 2 in time estimate of the pressure. See Remark 4 at the end of this chapter. We will prove that (LNSF) is solvable weakly in L 2 for given initial data and for given 9

10 external forces f and g. Then, we will improve the regularity of weak solutions under higher regularities of the initial data and (f, g) in Proposition 3.1. Proposition 3.1 does not contain the L 2 in-time estimates of the boundary. We can control the boundary in L 2 in-time by projecting the equation onto the divergence free space. See Remark 4. (1) Weak Formulation First, we define a function space where weak solutions will be defined. For any fixed time interval [, T ] with T <, V (T ) = { v L 2 t H 1 x : v =, v 3 ds L t L 2 x(r 2 ), } v 3 ds L t L 2 x(r 2 ) with v = on S B. The divergence free condition is expressed in the distributional form, i.e. v is orthogonal to gradients of test functions which vanish on S F. This space is almost the same space used in the context of the Navier-Stokes equations in a fixed domain except that we include the boundary terms: v 3 ds, v 3 ds. For test functions, we define a separable space V as V = { } v Hx 1 : v =, v = on S B, v 3 L 2 (R 2 ), v 3 L 2 (R 2 ) We define a space for v t : V (T ) = { v L 2 t H 1 x : v =, v = on S B } We say that (w, w t ) V (T ) L 2 (, T ; V ) is a weak solution of (LNSF) if for all v V, w = w(, ) = w L 2 hold and ( (w t, v)+ < w, v > + ( R 2 ) ( w 3 ds)v 3 + ( R 2 ) w 3 ds) ( v 3 ) = (f, v) + (g, v) (2) Existence of Weak Solutions Here, we want to address the following existence theorem: For any w L 2, f L 2 t L 2 x(ω), g L 2 t L 2 x(r 2 ), there exists a weak solution (w, w t ) V (T ) V (T ) such that w(, ) = w. 1

11 The idea of obtaining a weak solution is clear. Since V is separable, we can use the Galerkin approximation to the equation, i.e. we solve an ODE to decide the coefficients in a fixed time interval [, T ]. Then, we do the energy estimate to these approximated equations. Since energy bounds are uniform, we can pass to the limit. By taking a cut-off function in time, we can prove that a weak solution achieves the initial data in L 2. In this section, the upper index means the third component of a vector field. Galerkin Approximation: Since V is separable, there exists an orthogonal basis { } φ k in m L 2. We approximate w by w m (t) = λ j m(t)φ j. We want to select the coefficients λ j m(t) such that λ j m() = (w, φ j ) and j=1 ( t w m, φ j )+ < w m, φ j > + wmφ 3 3 jdxdyds + wm 3 φ 3 jdxdyds = (f, φ j ) + (g, φ j ) (3) We define integrals as E mj =< φ m, φ j >, H mj = (φ 3 m)(φ 3 j)dxdy, L mj = R 2 (φ 3 m) ( φ 3 j)dxdy R 2 F j = (f, φ j ), G j = (g, φ j ) Since ( t w m, φ j ) = t λ j m, (3) is reduced to an ODE. t λ j m + E mj λ j m + H mj λ j m(s)ds + L mj λ j m(s)ds = F j + G j which is subject to the initial data λ j m() = (w, φ j ). By the standard existence theory for ODE, there exists a unique absolutely continuous function λ m (t) = { λ j m : j = 1, 2,, m }. Energy estimate: For each m = 1, 2,, w m 2 L + w m 2 t L2 x L + wmds t L2 x L + t L2 x w 2 L + 2 f 2 x L + 2 t L2 g 2 x L 2 t L2 x w 3 mds 2 L t L2 x 11

12 Proof: We multiply by λ j m(t) to (3) and sum for j = 1, 2,, m. (w m, w m )+ < w m, w m > + ( = 1 d 2 dt w m 2 L + < w 2 m, w m > + 1 d ( ( 2 dt R 2 = (f, w m ) + (g, w m ) wm)(w 3 m)dxdyds 3 + ) wmds) 3 2 dxdy ( wm) 3 (wm)dxdyds d ( ( 2 dt R 2 ) wmds) 3 2 dxdy Integrating in time, with Young s inequality, we finish the energy estimate. Passing to the limit: From the energy estimate, we know that { } w m is uniformly bounded in L t L 2 x L 2 t Hx. 1 There exists a subsequence, still denoted by { } w m, converging to w for the weak star topology in L t L 2 x and for the weak topology in L 2 t Hx. 1 Since { } w m is bounded in V (T ), for the weak star topology in L t Hx, 1 ( wmds) 3 and ( wmds) 3 converge to w 3 ds and w 3 ds, respectively. We multiply ψ D(, T ) such that ψ(t ) = to (3) and integrate in time. By the integration by parts in time, + T T T (w m, φ j ) t ψdt + R 2 ( T < w m, ψ(t)φ j > dt + ( R 2 w 3 mds) (ψ(t) φ 3 j)dxdydt = (w m (), φ j )ψ() + w 3 mds)(ψ(t)φ 3 j)dxdydt T ( ) (f, φ j ) + (g, φ j ) dt Let m. Since w m () w in L 2, + T T T (w, φ j ) t ψdt + ( T < w, ψ(t)φ j > + ( w 3 ) (ψ(t) φ 3 j)dxdydt = (w, φ j )ψ() + w 3 )(ψ(t)φ 3 j)dxdydt T ( ) (f, φ j ) + (g, φ j ) dt Since this equality holds for a finite linear combination of φ, js, it also holds for all v V + T T T (w, v) t ψdt + R 2 ( T < w, v > ψ(t)dt + ( R 2 w 3 ds) ( v 3 )ψ(t)dxdydt = (w, v)ψ() + w 3 ds)(v 3 )ψ(t)dxdydt T ( ) (f, v) + (g, v) dt (4) 12

13 Therefore, we find the following equality (w t, v)+ < w, v > + ( R 2 w 3 ds)(v 3 )dxdy + ( R 2 w 3 ds) ( v 3 )dxdy = (f, v) + (g, v) (5) in the distribution sense on (, T ). It remains to show that w() = w in L 2. We multiply by ψ(t) to (4) and integrate in time. + T T T (w, v) t ψdt + R 2 ( T < w, v > ψ(t)dt + ( R 2 w 3 ds) ( v 3 )ψ(t)dxdydt = (w(), v)ψ() + w 3 ds)(v 3 )ψ(t)dxdydt T ( ) (f, v) + (g, v) dt (6) Comparing (4) with (6), we see that (w w(), v)ψ() = for each v V. We choose ψ such that ψ(). Then w() = w. This completes the existence part. Remark: Weak solutions are not in L 2 (, T ; V ) because the trace theorem does not hold in this level of the regularity of weak solutions. So, we cannot take the difference of two weak solutions to show that a weak solution is unique. We can show uniqueness after proving the regularity result in Proposition 3.1. (3) Higher Regularity We want to improve the regularity of weak solutions under higher regularities of the initial data and external forces. Since the domain is translation-invariant in the horizontal direction, we can take tangential derivatives to the equation to obtain bounds of tangential derivatives. Other bounds are deduced from the divergence free condition and from the momentum equation. We will not try to obtain the full bounds of w in L t Hx. 2 There are two reasons. First, by compactness, w L 2 t H 3 and w t L 2 t Hx 1 imply that w L t Hx. 2 Secondly, in fact, we cannot estimate the full derivative by our method under the assumption f L 2 t Hx 1 and g L 2 t H 3 2 x. In order to estimate all derivatives of the velocity field in L t Hx, 2 we have to take one more time derivative to the equation. Then, we need to estimate f t and g t which involves much more calculations. 13

14 We will take the difference quotients of the velocity field in the horizontal direction. Since the difference quotient terms are uniformly bounded, we can say that the tangential derivatives of the velocity field and the boundary profile are bounded as well. Showing the uniform bounds of difference quotients of terms are of tedious calculations, but it is clear from the estimates below. So, in the Proposition 3.1, we skip the step of taking difference quotients and we just obtain a priori estimate. Proposition 3.1. Suppose that (w, w t ) V (T ) V (T ) is the weak solution such that the initial data satisfies the compatibility condition. Let w H 2, f L 2 t Hx, 1 g L 2 t H 3 2 x. Then, w L t Hx 2 L 2 t Hx, 3 η L t Hx 3 and q L 2 t Hx. 1 Moreover, (w, η, p) satisfies the following energy bound: w L t Hx 2 + w L 2 + η t H2 x L + q t H3 x L 2 w t H1 x H 2 + η H 3 + f L 2 t Hx 1 + g L 2 t H 2 3 x Proof: It consists of 5 steps. The basic L 2 estimate is easy. When we do higher energy estimates, we will use the Korn s inequality to control lower order terms by higher order terms. As we will see in the proof, we need to control w t to obtain estimates of the full derivatives of the velocity field. Step 1. L 2 estimate: We multiply by w to the equation and integrate over Ω. By the integration by parts. 1 d 2 dt w 2 L + < w, w > +(q w 2 3,3, w 3 ) (w i, g i ) = (f, w) From the boundary condition, (q w 3,3, w 3 ) = (η η + g 3, w 3 ) = 1 d ) ( η 2 L + η 2 2 dt 2 L + (g 2 3, w 3 ) By the trace theorem and Young s inequality, Integrating in time, (g i, w i ) g 2 L w 2 L w 2 L 2 w 2 L t + L2 w 2 + η 2 x L 2 t L2 x L + t L2 η 2 x L ɛ + t L2 f 2 + g x L 2 t L2 x L 2 t L2 x 2 w 2 L 2 t L2 x 14

15 Since w = at the bottom, we can apply Korn s inequality. w 2 L t + L2 w 2 + η 2 x L 2 t L2 x L + t L2 η 2 x L ɛ + t L2 f 2 + g 2 x L 2 t L2 x L 2 t L2 x Now, we assume that g 1 = g 2 =. We will recover these terms in Remark 2. Step 2. H 1 estimate: Next, we obtain bounds of derivatives. We multiply the equation by w and integrate in the spatial variables. Since w = at the bottom, boundary terms on z = only are involved when we do the integration by parts. By the integration by parts, From the boundary condition, d dt w 2 L 2 + < w, w > +(q w 3,3, w 3 ) f 2 L 2 d dt w 2 L 2 + < w, w > +(η η, w 3 ) + ( g 3, w 3 ) f 2 L 2 By the duality argument in ( g 3, w 3 ) and Young s inequality, d dt w 2 L + < 2 w, w > + d ) ( η 2 L + 2 dt η 2 2 L 2 g 3 2 Ḣ w 2 Ḣ f 2 L 2 g 2 H w 2 H f 2 L 2 By the trace theorem, d ) ( w 2 L + η 2 dt 2 L + 2 η 2 2 L + < 2 w, w > 1 2 w 2 L 2 + f 2 L 2 + g 2 H w 2 L 2 Integrating in time, with Korn s inequality, w 2 L t + w 2 + η 2 L2 x L 2 t L2 x L + t L2 2 η 2 x L ɛ + t L2 f 2 + g 2 x L 2 t L2 x L 2 t H 2 1 x We need to obtain bounds of w 1,33 L 2 t L 2, w 2,33 x L 2 t L 2 and w 3,33 x L 2 t L 2. Since w =, x w 3,33 L 2 2 w L 2. From the equation w i,33 = w i,jj + w i,t + i q, we can replace w L 2 by 2 w L 2, with adding w t L 2 + i q to the right-hand side. q = w t w+f implies that 3 q 2 L 2 t L2 x w 3,t 2 L 2 t L2 x + w 3 2 L 2 t L2 x + f 3 2 L 2 t L2 x w t 2 L 2 t L2 x + f 2 L 2 t L2 x + g 2 L 2 t H 1 2 x + w 2 L 2 t L2 x 15

16 We have to use a different method to bound i q because we cannot bound 33 w i L 2 from w L 2. As Coutand-Shkoller did in [5], we trade i in i q with 3 in 33 w i by the integration by parts two times. For simplicity, we assume that g 3 =. We multiply by u to the equation and integrate over Ω: (w t, u) + ( w, u) + ( q, u) = (f, u) By the integration by parts (q, u) = (w t, u)+ < w, u > +(f, u) We set u = ( 1 q, 2 q, ). Then, u 2 L = (u, f w 2 t + w) = (u, f w t ) < u, w > ( = (u, f w t ) (w k,l + w l,k) klq + 1 ) 2 (w i,3 + w 3,i ) i3 q dv ( = (u, f w t ) + (w k,kl + w l,kk) l q + 1 ) 2 (w i,i3 + w 3,ii ) 3 q dv where we do not have boundary integrals because u 3 =, g 1 = g 2 = and k, l = 1, 2. By Young s inequality, i q has the same bound as 3 q. Therefore, w 2 L t + L2 2 w 2 + η 2 x L 2 t L2 x L + t H1 q 2 x L 2 t L2 x ɛ + f 2 + g 2 + w L 2 t L2 x L 2 t H 1 t 2 2 L 2 t L2 x x Step 3. H 2 estimate: We take one more derivative. We multiply by 2 w to the equation and integrate over Ω. (w t, 2 w) + ( w, 2 w) + ( p, 2 w) = (f, 2 w) By the integration by parts, d dt w 2 L 2 + < w, w > +(q w 3,3, 2 w 3 ) f 2 L 2 Using the boundary condition, the trace theorem and the duality argument, d ) ( w 2 L + 2 η 2 dt 2 L + 3 η 2 2 L + ( 2 w) 2 L 2 t L2 f 2 x L + g 2 2 H

17 As before, we can replace ( w) 2 with 3 w 2, by adding w L 2 t L2 x L 2 t 2 to the right- t L2 x L 2 t L2 x hand side. Integrating in time, w 2 L t L2 x + 3 w 2 L 2 t L2 x + 2 η 2 L t H1 x + 2 q 2 L 2 t L2 x ɛ + f 2 L 2 t L2 x + g 2 L 2 t H 3 2 x + w t 2 L 2 t L2 x Step 4. L 2 estimate of w t : Finally, we obtain the bound w t 2 + w L 2 t 2 First, t L2 x L 2 t x. L2 we multiply by w t to the equation. (w t, w t ) + ( w, w t ) + ( q, w t ) = (f, w t ) By the integration by parts, w t 2 L + 1 d 2 2 dt < w, w > +(q w 3,3, w 3,t ) f 2 L 2 From the boundary condition, w t 2 L + 1 d 2 2 dt < w, w > +(η η + g 3, w 3,t ) f 2 L 2 We estimate boundary terms. (η η, η tt ) = d dt (η η, η t) η t 2 L 2 η t 2 L 2 By the trace theorem, (g 3, w 3,t ) g 3 2 L 2 (R 2 ) w 3,t 2 L 2 (R 2 ) g 3 2 L w 3,t 2 L w 3,t 2 L 2 Integrating in time, with the Korn s inequality, w t 2 L 2 t L2 x + w 2 L t L2 x ɛ + f 2 L 2 t L2 x + (η η, η t) + g 2 L 2 t L2 x + η t 2 L 2 t L2 x + η t 2 L 2 t L2 x + w 2 L t L2 x w 3,t 2 L 2 ɛ + f 2 L 2 t L2 x η 2 L t H2 x + η t 2 L t L2 x + g 2 L 2 t H 1 2 x + η t 2 L 2 t H1 x + w 2 L 2 t L2 x w 3,t 2 L 2 By the trace theorem and Young s inequality, η t 2 L t L2 x = w 3 2 L t L2 x(r 2 ) w 2 L t L2 x w 2 L t L2 x 17

18 η t 2 = w L t L2 x L 2 t L2 x (R2 ) w L 2 t L2 x 2 w 2 L 2 t L2 x η t 2 L 2 t L2 x = w 3 2 L 2 t L2 x(r 2 ) w L 2 t L2 x 2 2 w 2 L 2 t L2 x Therefore, we have the following energy bound: w t 2 L 2 t L2 x + w 2 L t L2 x ɛ + f 2 L 2 t L2 x η 2 L t H2 x + g 2 L 2 t H 1 2 x w 2 L 2 t H1 x + w 2 L 2 t L2 x w 3,t 2 L 2 ɛ + f 2 L 2 t L2 x η 2 L t H2 x + g 2 L 2 t H 1 2 x w 2 L 2 t H1 x + w 2 L 2 t L2 x D hw t 2 L 2 Step 5. H 1 estimate of w t : We take one more derivative. We multiply by w t to the equation. By integrating in the spatial variables, w t 2 L + 1 d 2 2 dt < w, w > +(q w 3,3, w 3,t ) = (f, w t ) By the same method used before, we obtain that w t 2 + L 2 w 2 t L2 x L ɛ + t L2 f x L 2 t L2 x 2 η 2 L + t H2 g x L 2 t H 3 2 x 2 2 w 2 L 2 t H1 x Since w t is of divergence-free, w t 2 + L 2 w 2 t L2 x L ɛ + t L2 f x L 2 t L2 x 2 η 2 L + t H2 g x L 2 t H 3 2 x 2 2 w 2 L 2 t H1 x In sum, w 2 L + w 2 t H1 x L + t L2 w 2 + η 2 x L 2 t H2 x L + t H3 q 2 x L 2 t H1 x ɛ + f 2 + g 2 L 2 t H1 x L 2 t H 3 2 x We add w t 2 L 2 t H1 x to the above estimate. We conclude that w 2 L t + H2 w 2 + w x L 2 t 2 + η 2 t H2 x L 2 t H1 x L + t H3 q 2 x L 2 t H1 x ɛ + f 2 + g 2 L 2 t H1 x L 2 t H 2 3 x We finish proof of Proposition. Remark 1: If we try to obtain the full derivative of w in L t H 2 x, we need to obtain bounds of w 1,33 L t L 2 and w x 2,33 L t L 2. From the equation, x w i,33 2 L t L2 x w i,t 2 L t L2 x + f 2 L t L2 x + iq 2 L t L2 x 18

19 We need to estimate w i,t 2 L t L2 x. In order to obtain w i,t 2 L t L2 x, we take t to the equation, multiply by w t and do the integration by parts. Then, w i,t 2 L t But, it is not clear how we estimate 1 f t and 1 2 g t. L2 1 f t x L 2 t L2 2 g t 2 x L 2 t x. L2 Remark 2: If we have g 1 and g 2, then, for k, l, m, i = 1, 2, u 2 L = (u, u) = (u, f w 2 t + w) = (u, f w t ) < u, w > + u m g m R 2 = (u, f w t ) ((w k,l + w l,k ) kl q (w i,3 + w 3,i ) i3 q) + m qg m R 2 = (u, f w t ) + ((w k,kl + w l,kk ) l q (w i,i3 + w 3,ii ) 3 q) + m qg m R 2 Therefore, i q 2 L ɛ + f 2 + g 2 + w 2 L 2 t L2 x L 2 t H 1 t L 2 t L2 x x 2 q 2 L 2 (R 2 ) We can move 1 2 q 2 L 2 (R 2 ) to the left-hand side. When we take one more derivative to the equation, we have error terms of the form: ( 2 q)( g)dxdy. By duality argument, ( 2 q)( g)dxdy q 2 L 2 (R 2 ) + g 2 H 3 2 Combining both, we have 1 2 q 2 H 1 (R 2 ) in the right-hand side. We apply the trace theorem and move it to the left-hand side. Remark 3: Under these higher regularities, solutions are unique. Remark 4: We already used the vector field method to rewrite the momentum equation to express the the pressure term as the harmonic extension of the boundary terms. Here, we repeat it. Any vector field X in Ω can be written as a sum of a divergence free vector field and a gradient : X = u + φ. This is called a Hodge decomposition. From the identity Ω u φdv + ( u)φdv = (u ˆn)φ Ω Ω 19

20 we conclude that u is of divergence free and u ˆn = on S B is L 2 orthogonal to φ with φ = on S F. We denote u by PX. Here, we list properties of the operator P. Lemma 3.2: (1) It is a bounded operator on H s. (2) If φ H 1, then P( φ) = H (π), where φ = π on S F. Proof: Let us start with the first statement. For any vector field X, X can be written as X = u + φ. Then, φ satisfies that φ = X in Ω, φ = on S F and n φ = X n on S B. By the usual elliptic theory, φ H s X H s 1 + X ˆn H s 1 X Hs. Therefore, 2 (S B ) u H s = P(X) H s X H s. We can prove the second property by using the same argument used to the first one. Now, we take the projection P to the equation: w t P( w) + P( q) = P(f) Since w and its tangential derivatives are of divergence free and is zero on the bottom boundary, for k, (I P) (D h ) k. Thus, ( w, (D h ) k w) = ( P w, (D h ) k w), ( q, (D h ) k w) = (P q, (D h ) k w) The expression H (η η) = P q H (w 3,3 ) H (g 3 ) infers that H (η η) 2 L 2 t H1 x q 2 + H (w L 2 3,3 ) 2 + H (g t H1 x L 2 3 ) 2 t H1 x L 2 t H1 x By Proposition 3.1, H (η η) 2 L 2 t H1 x ɛ + f L 2 t H 1 x + g L 2 t H 3 2 x Therefore, we can control the boundary in L 2 in-time. 2

21 Finally, we want to prove the statement used in chapter 2: we can replace 4 η L 2 t L 2 and x 2 η L 2 t L x with H (η η) L 2. Here, we present more general result. t H1 x Lemma 3.3: (η F (η)) 2 L 2 t H 1 2 x H (η F (η)) 2 L 2 t H1 x Proof: We will use the Dirichlet-Neumann operator. Suppose H (η F (η)) L 2 t H 1 x. We set f = η F (η). For i = 1, 2, G(η)f + η f R( i H (f)) = i (f) i η 1 + η 2 L 2 t H 1 2 x Therefore, by the product rule of fractional derivatives, f L 2 t H 1 2 x η L t H 3 x f L 2 t H 1 2 x + H (η F (η)) L 2 t H 1 x Here, we use the fact that G(η) is a first order pseudo-differential operator. By the smallness of η L t Hx 3, we finish proof of Lemma. 4. Change of Variables In this chapter, we present details of the change of variables which is used in chapter 2. We define a map θ(t) : Ω = {(x 1, x 2, y); 1 < y < } {(x 1, x 2, z ); 1 < z < η(x 1, x 2, t)} in terms of η, where η is the harmonic extension of η into the fluid domain. θ(x 1, x 2, y, t) = (x 1, x 2, η(x 1, x 2, t) + y(1 + η(x 1, x 2, t))) By definition, dθ = 1 1 A B, ζ = (dθ) 1 = 1 1 A B 1 A = (1 + y) η x1, B = (1 + y) η x2, = 1 + η + y η(1 + y) We define v on θ(ω) by v i = θ i,j w j = α ij w j. Then, v 1 = w 1, v 2 = w 2, v 3 = A w 1 + B w 2 + w 3 21

22 We make replacements v i,j = ζ lj l (α ik w k ). ( 1 1 w 1 + A ) w 3 A 1 3( w 1 + A ) w 3 ( 1 1 w 2 + B ) w 3 A 1 3( w 2 + B ) w 3 ( 1 2 w 1 + A ) w 3 B 1 3( w 1 + A 3) w 1 3( 1 w 1) ( 1 2 w 2 + B ) w 3 B 1 3( w 2 + B 3) w 1 3( 1 w 2) v 3,1 v 3,2 v 3,3 ( 1 v 3,1 = 1 Aw ) Bw 2 + w 3 1 ( 1 A 3 Aw ) Bw 2 + w 3 ( 1 v 3,2 = 2 Aw ) Bw 2 + w 3 1 ( 1 B 3 Aw ) Bw 2 + w 3 v 3,3 = 1 ( 1 3 Aw ) Bw 2 + w 3 First, we take time derivative. v i,t = 1 w i,t 1 ( 2,tw i + ((θ) 1 ) 1 3 w i,3 1 ),3w i for i = 1, 2 v 3,t = 1 Aw 1,t,t 2 Aw A,tw Bw 2,t,t 2 Bw B,tw 2 + w 3,t + ((θ) 1 ) 3 1 2,3Aw A,3w ) ( Aw 1,3 + ((θ) 1 ) 3 Next,,we calculate the advection terms. ( 1 2,3Bw B,3w Bw 2,3 + w 3,3 ) v v i = 1 w 1 1 ( 1 w i) + 1 w 2 2 ( 1 w i) + 1 w 3 3 ( 1 w i) for i = 1, 2 v v 3 = 1 w 1 1 ( A w 1 + B w 2 +w 3 ) + 1 w 2 2 ( A w 1 + B w 2 +w 3 ) + 1 w 3 3 ( A w 1 + B w 2 +w 3 ) Finally, we obtain the dissipation term. We calculate first. = ( 1 3) 1 ( A 3) 2 ( B 3) A 31 + A 3( 1 A 3) B 32 + B 3( 1 B 3) Therefore, v i = 11 ( 1 w i) + 22 ( 1 w i) + 1 3( 1 3( 1 w i)) 1 ( 1 A 3( 1 w i)) 2 ( 1 B 3( 1 w i)) 1 A 31( 1 w i) + 1 A 3( 1 A 3( 1 w i)) 1 B 32( 1 w i) + 1 B 3( 1 B 3( 1 w i)) for i = 1, 2 22

23 ( A v 3 = 11 w 1 + B ) ( A w 2 + w w 1 + B ) w 2 + w ( 3( A w 1 + B ) w 2 + w 3 ) ( A 1 3( A w 1 + B ) ( B w 2 + w 3 ) 2 3( A w 1 + B ) w 2 + w 3 ) + A A 3( 3( A w 1 + B w 2 + ) w 3 ) A ( A 31 w 1 + B ) w 2 + w 3 + B B 3( 3( A w 1 + B ) w 2 + w 3 ) B ( A 32 w 1 + B ) w 2 + w 3 We define the pressure as q = p θ. Then, 1 p = 1 q 1 A 3q, 2 p = 2 q 1 B 3q, 3 p = 1 3q We substitute all terms into the Navier-Stokes equations and its boundary conditions. Then, we see the quadratic nonlinear terms mentioned in chapter 2. Appendix 1. Local well-posedenss of the Incompressible Navier-Stokes equations on a moving domain without surface tension In Appendix 1, we study the incompressible Navier-Stokes equations with the free boundary in Ω t of finite depth without surface tension. We have the same equations and the free boundary conditions except that the pressure at the free surface does not contain surface tension. (NSF ) v t + v v v + p = in Ω t v = in Ω t v = on S B η t = v 3 v 1 x η v 2 y η on S F pn i = (v i,j + v j,i )n j + ηn i on S F where Ω t = {(x, y, z) : 1 < z < η(x, y, t)} with two boundaries S F = {(x, y, z) : z = η(x, y, t)} and S B = {(x, y, z) : z = 1}. Here, ˆn = (n 1, n 2, n 3 ) is the outward normal vector on S F. µ is the constant of viscosity, g is the gravitational constant, and β is the constant of surface tension. From now, we normalize all the constants by 1. (We follow the Einstein convention where we sum upon repeated indices. Subscripts after commas denote derivatives.) Here, we want to prove the following theorem. 23

24 Theorem: For arbitrary initial data v H 2 and for sufficiently small initial data η H 5 2, there is a strictly positive T > such that there is a unique local-in-time solution v, η and the pressure p on [, T ) satisfying the energy bound: v L t H 2 x + v L 2 t H3 x + η L t H 5 2 x + p L 2 t H 1 x v H 2 + η H 5 2 If we take a small initial velocity field, we recover one of the result of Beale in [2]. We take the smallness assumption to the initial profile of the boundary to show that the transformation onto the equilibrium domain defines a diffeomorphism. We will prove Theorem by the same methods used to the problem with surface tension. 1. Existence and Uniqueness on the Equilibrium Domain As we did before, let us study the problem on the equilibrium domain. First, we transform the system of equations onto those of on the equilibrium domain. We define a map θ by θ(x, y, z, t) = (x, y, η(x, y, t) + z(1 η(x, y, t))) We define the velocity field and the pressure on Ω as w = v θ, q = p θ. Let X = θ 1. Then, the acobian matrix dx is given by (X i,j ) = 1 1 A B 1 A B = 1 η z 1 η 2 η z 2 η 1 η + z η(1 z) Note that X = θ 1 exists as long as η is small enough. We cannot follow the change of variables used before due to the lack of regularities of the boundary. Instead, we simply compose v an p with θ to define the equations on the flat domain Ω. But, then, we lose the divergence free condition for the velocity field w. We replace the system of equations of v with that of w. Then, v i,t = w i,t + w i,3 X 3,t, v i,k = X j,k w i,j, v v = w k X j,k w i,j, 2 v i 2 z = X l,kx j,k w i,lj + X l,kk w i,l 24

25 On the equilibrium domain, we have the following system of equations: (LNSF ) w t w + q = f in Ω w = σ in Ω w i,3 + w 3,i = g i, q = w 3,3 + η + g 3 on {z = } w = w = v θ, η = at t = η t w 3 = w η = h on {z = }, w = on {z = 1} where nonlinear terms are given by f i = 2 A w i,13+2 B w i,23+ ( A B ) w i,33 + X 3 w i,3 w k X j,k w i,j X j,i j q+ i q w i,3 X 3,t σ = A w i,3 B ( w 2, ) w 3,3 g 1 = 1 η ( A η q + 2(w 1,1 w 1,3 )) 2 η ( B w 1,2 + w 1,3 + w A) ( 1 ) A 2,1 + w 2,3 1 w1,3 w 3,3 g 2 = 1 η ( B w 1,2 + w 1,3 + w A) 2,1 + w 2,3 2 η ( B η q 2(w 2,2 w 2,3 )) ( 1 1 ) B w2,3 w 3,3 g 3 = 1 η ( 1 w 1,3 + w A) 3,1 + w 3,3 2 η ( 1 w 2,3 + w B ) ( 1 ) 3,2 + w 3,3 2 1 w3,3 Since we lose the divergence free condition, we reformulate the system into the problem of the divergence free by introducing V such that V = σ, V = at the bottom. We will discuss the existence of such V later. Let u = w V. Then, (u, η, q) satisfies the following system of equations. (LNSF ) u t u + q = f V t + V = F in Ω u = in Ω u i,3 + u 3,i = g i V i,3 V 3,i = G i, q = u 3,3 + η + g 3 + V 3,3 on {z = } u = w V at t =, u = on {z = 1} η t u 3 = u η + V η + V 3 = h on {z = } 25

26 Let G 3 = g 3 + hds + V 3,3. Then, in terms of (u, q), u t u + q = F in Ω (LNSF ) u = in Ω u i,3 + u 3,i = G i, q = u 3,3 + u 3 ds + G 3 on {z = } u = w v at t =, u = on {z = 1} We proved the solvability of (LNSF) when we solved the problem with surface tension. The only difference is that there are no boundary terms in the above system of equations. Then, we do not need to estimate higher order terms of the boundary to prove the same result of Proposition 3.1. It is easy to show that (u, q) satisfies the following energy estimate. u L 2 t Hx 3 + u t L 2 t Hx 1 + q L 2 + u t H1 x 3 ds L t C + η H 2 x + F L 2 t H 1 x + G L 2 t H 3 2 x H 2 x C + f L 2 t Hx 1 + g L 2 t H 3 + V 2 L 2 t Hx 3 + V t L 2 t Hx 1 + ((V + u) η)ds x L 2 t H 3 2 x C + f L 2 t H 1 x + g L 2 t H 3 2 x + V L 2 t H 3 x + V t L 2 t H 1 x + T V + u L 2 t H2 x η L 2 t H 5 2 x We will estimate the boundary by solving the transport equation of the boundary. We go back to the original velocity field w = u + V. Then, (w, q) satisfies the following estimate. w L 2 t H 3 x + w t L 2 t H 1 x + q L 2 t H1 x C + f L 2 t H 1 x + g L 2 t H 3 2 x + V L 2 t H 3 x + V t L 2 t H 1 x + T w L 2 t H2 x η L 2 t H 5 2 x + T V L 2 t H 2 x η L 2 t H 5 2 x Now, we need to solve the boundary term. η satisfies a transport equation: η t + w η = w 3 Since (w 1, w 2, ) in the advection term does not satisfy the divergence free condition, we need a higher regularity to w than the Lipschitz regularity. Here, H 5 2 (R 2 ) is enough. See [6]. On 26

27 time interval [, T ], η ( η L t H H 2 + ) T w 3 L 2 t H 2 5 exp( T x w L 2 t H 5 ) 2 x We take T small enough so that, a priori, T w L 2 t H Then, we can ignore the expo- x nential term in the estimate of the boundary. w L 2 t H 3 x + w t L 2 t H 1 x + q L 2 t H1 x C + f L 2 t H 1 x + g L 2 t H 3 2 x + V L 2 t H 3 x + V t L 2 t H 1 x + T w 2 L 2 t H2 x + T V L 2 t H2 x w L 2 t H3 x We replace V in terms of the velocity field and the profile of the boundary. ( ) 2 w L 2 t Hx 3 + w t L 2 t Hx 1 + q L 2 C t H1 + T w x L 2 t Hx 3 + w t L 2 t Hx 1 + q L 2 t H1 x Therefore, by taking T small enough, we can obtain a priori estimate of the solution. The size of T is given by T 1 C. Now, we iterate the system of equations to show that there exists a unique, local-in-time solution. We define the first iteration as we did in Chapter 4. Then, the iteration is performed in the following way: (LNSF m ) wt m w m + q m = f(w m 1, η m 1, q m 1 ) in Ω w m = σ(w m 1, η m 1 ) in Ω wi,3 m + wm 3,i = g i(w m 1, η m 1 ) on {z = } q m = w3,3 m + ηm + g 3 (w m 1, η m 1 ) on {z = } w m = w, η m = η at t =, w m = on {z = 1} ηt m w3 m = wm 1 η m 1 = h(w m 1, η m 1 ) on {z = } Thus, from the a priori estimate in Proposition 3.1, (w m, η m, q m ) C + T (w m 1, η m 1, q m 1 ) 2 Therefore, we conclude that { (w m, η m, q m ) } are uniformly bounded if the time interval [, T ] is sufficiently small. By taking difference of two sequences, we show that { (w m, η m, q m ) } 27

28 is a Cauchy sequence, which implies the existence of a unique, local-in-time solution to (LNSF). The dependence on the initial data of the boundary η occurs when we define the first iteration. Having solved the linearized problem, we reverse our steps and obtain a solution of the original problem. 2. Solvability of V To complete the argument in the previous chapter, we need to prove that there exists a vector field V such that V = σ in Ω and V = at the bottom. We will look for such a V of the form V = r + τ. First, r solves the elliptic problem (E1). The existence of a solution and its regularity are well-known. Then, τ solves the elliptic problem (E2) in terms of r. (E1) r = σ in Ω n r = on S B (E2) τ = in Ω n τ = n r on S B r = on S F r = on S F By definition of V, V = σ is trivial. To show that V = at the bottom, we calculate the boundary term. ( 1 r, 2 r, 3 r) + ( 2 τ 3 3 τ 2, 3 τ 1 1 τ 3, 1 τ 2 2 τ 1 ) = ( 1 r, 2 r, 3 r) + ( 3 τ 2, 3 τ 1, ) = ( 1 r, 2 r, ) + ( 3 τ 2, 3 τ 1, ) = where we used the tangential boundary condition of τ for the first equality, used the Neumann boundary condition of r for the second equality, and used the Neumann boundary condition of τ for the last equality. In (LCNSF), we have V t V, V (), and V 3. So, we need to estimate these terms too. We calculate V t V. It satisfies that V t V = r t + τ t r = r t + τ t σ 28

29 Since r t and τ t can be estimated by taking t to (E1) and (E2), we can treat V t V. Moreover, V 3 can be estimated as V 3 = 3 r L 2 t Ḣ 5 2 x on S F by using the regularity of r and the trace theorem. Finally, we calculate the initial data. V () = ( r)() + ( τ)(). But, ( r)() and ( τ)() satisfy the same elliptic problems. Since σ() H 2, V () has the desired regularity. Appendix 2. Global well-posedenss of the Compressible Navier-Stokes equations on a moving domain with surface tension In Appendix 2, we study the free boundary value problem of the compressible Navier- Stokes equations with surface tension. We take a barotropic model so the entropy equation is not included. We consider the problem on a moving domain of finite depth, bounded above by the free surface and bounded below by a solid flat bottom. ρ(v t + v v) µ v ξ ( v) + p = in Ω t (CNSF ) ρ t + (ρv) = in Ω t v = on S B η t = v 3 v 1 x η v 2 y η on S F pn i = µ(v i,j + v j,i )n j + (gη β ( ))n 1+ η 2 i + ξ( v)n i η on S F where Ω t = {(x, y, z) : 1 < z < η(x, y, t)} with two boundaries S F = {(x, y, z) : z = η(x, y, t)} and S B = {(x, y, z) : z = 1}. Here, ˆn = (n 1, n 2, n 3 ) is the outward normal vector on S F, g is the gravitational constant, and β is the constant of surface tension. µ and ξ satisfies that µ + 2 3ξ >. From now on, we normalize all the constants by 1. Since the problem is posed on a domain, we need compatibility conditions for the initial data. The initial compatibility conditions on the initial velocity field v are imposed as {(v ) i,j + (v ) j,i } tan = on S F = {(x, y, z) : z = η (x, y)} v = on S B = {(x, y, z) : z = 1} 29

30 where (tan) means the tangential component. Since v = at the bottom, no conditions on ρ are imposed at the bottom. We do not have the density term on the moving boundary. This boundary condition is used for modeling the motion of the ocean or a star. See Lindblad [9]. There are several papers dealing with the free boundary value problem of the compressible fluid. The local-in-time existence for the general case of initial data was proved by Lindbald [9] for the free boundary problem with non-vanishing density on the boundary for the isentropic Euler equations. For a viscous compressible liquid, the free boundary problem governing the unsteady motion of an isolated liquid mass was studied in a series of papers of Solonnikov and Tani. For example, see [15]. For the incompressible Navier-Stokes equations with free boundary, see [2], [3], [5] and [12]. We easily obtain the basic energy estimate by multiplying by v to the momentum equation and integrating over the moving domain. To obtain higher energy estimates, we reformulate the problem into the problems of the Navier-Stokes equations with the external force ( v) + (1 ρ)d t v. v t + v v v + p = ( v) + (1 ρ)d t v in Ω t (CNSF ) v = on S B η t = v 3 v 1 x η v 2 y η on S F pn i = (v i,j + v j,i )n j + (η ( ))n 1+ η 2 i + ( v)n i η on S F Here, we remove the evolution equation of the density. Regularities of the density can be established by using the Lagrangian map. (ρ 1) satisfies a transport equation: (ρ 1) t + v (ρ 1) = (ρ 1) v ( v) In order to obtain global-in-time solutions, ( v) should be small in L 1 in time. Here, we give a quick answer. From the evolution equation of the density, we obtain that v = D t log(ρ). If ρ is close to 1, then D t log(ρ) is much smaller. It infers the faster decay of ( v) in time. Theorem 1. Suppose v H 2, η H 3, and ρ 1 H 2. There is a unique, global-in-time 3

31 solution v, η, ρ 1, and p such that ρ 1 L t H 2 x + v L t H2 x + v L 2 t H2 x + η L t H3 x + H (η F (η)) L 2 t H1 x + p L 2 t H1 x ɛ provided that the initial density is close to 1, and the initial data of the velocity and the boundary profile are sufficiently small, where ɛ = v H 2 + η H 3 + ρ 1 H 2. We will solve the problem on the equilibrium domain. Since we cannot project the equation onto the divergence free space, we need to solve one more problem to make the velocity field of divergence-free. In chapter 5, we study the incompressible limit for the problem. An introduction to incompressible limit problems and the main result will be addressed there. 2. Transformation onto the equilibrium domain We will solve the problem on the equilibrium domain. We transform the system of equations into that of equations on the equilibrium domain in the following way. It s exactly the same transformation used in Appendix 1, but, for the completeness, we present it again. We define a map θ(t) : Ω = {(x 1, x 2, y); 1 < y < } {(x 1, x 2, z ); 1 < z < η(x 1, x 2, t)} in a way depending on the unknown η such that θ(x 1, x 2, y, t) = (x 1, x 2, η(x 1, x 2, t) + y(1 + η(x 1, x 2, t))) We define all the quantities on the equilibrium domain by composing with θ: w = v θ, q = p θ, ϱ = ρ θ Let X = θ 1. Then, the acobian matrix dx is given by (X i,j ) = 1 1 A B 1 A B = 1 η z 1 η 2 η z 2 η 1 η + z η(1 z) Note that X = θ 1 exists as long as η is small enough. We replace the system of equations of v with that of w. 31

32 v i,t = w i,t + w i,3 X 3,t, v i,k = X j,k w i,j, v v = w k X j,k w i,j, 2 v i 2 z = X l,kx j,k w i,lj + X l,kk w i,l Then, on the equilibrium domain, we have the following system of equations. Here, we do not include the evolution equation of the density. Instead, we add ( w) to the system of equations. We will deal with the density function by using the Lagrangian map. So, we assume that the density function is given. In the iteration step, we can relate the density function with other functions. (LCNSF ) w t w + q = f + ( ) θ + ((1 ρ)d t v) θ in Ω w = σ in Ω w i,3 + w 3,i = g i, q = w 3,3 + η η + g 3 + σ on {z = } w = w = v θ, η = at t = η t w 3 = w η = h on {z = }, w = on {z = 1} where nonlinear terms are given by f i = 2 A w i,13+2 B w i,23+ ( A B ) w i,33 + X 3 w i,3 w k X j,k w i,j X j,i j q+ i q w i,3 X 3,t σ = A w i,3 B ( w 2, ) w 3,3 g 1 = 1 η ( A η q + 2(w 1,1 w 1,3 )) 2 η ( B w 1,2 + w 1,3 + w A) ( 1 ) A 2,1 + w 2,3 1 w1,3 w 3,3 g 2 = 1 η ( B w 1,2 + w 1,3 + w A) 2,1 + w 2,3 2 η ( B η q 2(w 2,2 w 2,3 )) ( 1 1 ) B w2,3 w 3,3 g 3 = 1 η ( 1 w 1,3 + w A) 3,1 + w 3,3 2 η ( 1 w 2,3 + w B ) ( 1 ) 3,2 + w 3,3 2 1 w3,3 We also need the compatibility conditions on the initial data w on the equilibrium domain. The compatibility conditions are given by w = σ() in Ω ( ) (w ) 1,3 + (w ) 3,1, (w ) 2,3 (w ) 3,2, q() 2(w ) 3,3 η T k () = g() T k () where T 1 1 = (1,, η), T 2 = 1 (, 1, η η). η 2 32

33 3. A Priori Estimate In order to apply the method used in the incompressible model, we need to replace w with a divergence-free vector field. We already proved the existence of such vector fields in Appendix 1 so that we skip it. Let u = w V. Then, (u, η, q) satisfies the following system of equations. (LCNSF u ) u t u + q = F = f V t + V in Ω u = in Ω u i,3 + u 3,i = G i = g i V i,3 V 3,i q = u 3,3 + η η + G 3 = u 3,3 + η η + g 3 + V 3,3 + σ on {z = } η t = u 3 + V 3 + h on {z = }, u = on {z = 1} u = w V, η = at t = The solvability of the above system of equations is already proved before. Moreover, (u, η, q) satisfies the following energy bound. u L t H 2 x + u L 2 t H3 x + η L t H3 x + H (η η) L 2 t H1 x + q L 2 t H1 x ɛ + F L 2 t Hx 1 + G L 2 t H V x L 2 t Ḣ 5 + h 2 x (R 2 ) L 2 t H 2 5 x (R 2 ) We have the last two terms because η t u 3. In terms of (w, η, q), w L t H 2 x + w L 2 t H3 x + η L t H3 x + q L 2 t H2 x + w t L 2 t H 1 x ɛ + f L 2 t H 1 x + g L 2 t H 3 2 x + V t V L 2 t H 1 x + V L 2 t Ḣ 5 2 x (R 2 ) + h L 2 t H 5 2 x Now, we calculate nonlinear terms. The principal term of is σ given by w = σ η w L p t H2 x for all p, 1 p 2. Moreover, σ t L 2 t L 2 x. From these estimates on σ, we can estimate V t V L 2 t H 1 x and V L 2 t Ḣ 5 2 x (R 2 ) which are quadratic. Secondly, we calculate other nonlinear terms. f w 3 η + w 2 η + 2 w η + q η + (1 ϱ)(w t + w w) + σ 33