Liquid crystal flows in two dimensions

Liquid crystal flows in two dimensions Fanghua Lin Junyu Lin Changyou Wang Abstract The paper is concerned with a simplified hydrodynamic equation, proposed by Ericksen and Leslie, modeling the flow of nematic liquid crystals. In dimension two, we establish both interior and boundary regularity theorem for such a flow under smallness conditions. As a consequence, we establish the existence of global weak solutions that are smooth away from at most finitely many singular times in any bounded smooth domain of R. 1 Introduction We consider the following hydrodynamic system modeling the flow of liquid crystal materials in dimension two (see [6] [7] [11] [14] and references therein): u t + u u ν u + P = λ ( d d) in (0, + ), (1.1) u = 0 in (0, + ), (1.) d t + u d = γ( d + d d) in (0, + ), (1.3) where R is a bounded smooth domain, u(x, t) : (0, + ) R represents the velocity field of the flow, d(x, t) : (0, + ) S, the unit sphere in R 3, is a unit-vector field that represents the macroscopic molecular orientation of the liquid crystal material, and P (x, t) : (0, + ) R represents the pressure Courant Institute of Mathematical Sciences, New York University, New York, NY 1001, USA. School of Mathematics, South China Normal University, Guangzhou 510640, China. Department of Mathematics, University of Kentucky, Lexington, KY 40506, USA. The first and third author are partially supported by NSF. 1

function. The constants ν, λ, and γ are positive constants that represent viscosity, the competition between kinetic energy and potential energy, and microscopic elastic relaxation time for the molecular orientation field. denotes the divergence operator, and d d denotes the matrix whose (i, j)-the entry is given by i d j d for 1 i, j. The above system is a simplified version of the Ericksen-Leslie model, which reduces to the Ossen-Frank model in the static case, for the hydrodynamics of nematic liquid crystals developed during the period of 1958 through 1968 ([7] [6] [11]). It is a macroscopic continuum description of the time evolution of the materials under the influence of both the flow field u(x, t), and the macroscopic description of the microscopic orientation configurations d(x, t) of rod-like liquid crystals. Roughly speaking, the system (1.1)-(1.3) is a coupling between the non-homogeneous Navier- Stokes equation and the transported flow of harmonic maps. In a series of papers, Lin [13] and Lin-Liu [14, 15] initiated the mathematical analysis of (1.1)-(1.3) in 1990 s. More precisely, they considered in [14] the Leslie system of variable length, i.e. the Dirichlet energy 1 d dx for d : S n 1 is replaced by the Ginzburg-Landau energy ( 1 d + (1 d ) ) dx for d : R n (ɛ > 0), and proved existence of 4ɛ global classical and weak solutions in dimensions two or three. In [15], they proved the partial regularity theorem for suitable weak solutions, similar to the classical theorem by Caffarelli-Kohn-Nirenberg [5] for the Navier-Stokes equation. However, as pointed out in [14, 15], both their estimates and arguments depend on ɛ, and it is a challenging problem to study the convergence as ɛ tends to zero. In this paper, we are interested in the existence of global weak solutions (u, d) of (1.1)-(1.3) that may enjoy possible regularity under the initial and boundary conditions: (u(x, 0), d(x, 0)) = (u 0 (x), d 0 (x)) x, (1.4) (u(x, t), d(x, t)) = (0, d 0 (x)) (x, t) (0, + ). (1.5) Throughout this paper, we introduce H = closure of C 0 (, R ) {v : v = 0} in L (, R ),

J = closure of C 0 (, R ) {v : v = 0} in H 1 0 (, R ), We make the assumptions: H 1 (, S ) = { d H 1 (, R 3 ) : d(x) S a.e. x }, u 0 H, d 0 H 1 (, S ) and d 0 C,β (, S ) for some β (0, 1). (1.6) Definition 1.1 For 0 < T +, u L ([0, T ], H) L ([0, T ], J) and d L ([0, T ], H 1 (, S )) is a weak solution of (1.1)-(1.5), if u, ψ φ + [ u u, ψφ + ν u, ψ φ ] [0,T ] = ψ(0) [0,T ] u 0, φ + λ d, ψ φ + [0,T ] = ψ(0) d 0, φ + γ [0,T ] [0,T ] d d, ψ φ, [ u d, ψφ + γ d, ψ φ ] [0,T ] d d, ψφ, for any ψ C ([0, T ]) with ψ(t ) = 0 and φ H 1 0 (, R3 ). Moreover, (u, d) satisfies (1.5) in the sense of trace. In this paper, we establish both the regularity and existence of global weak solutions in dimension two. More precisely, we prove Theorem 1. For 0 < T < +, assume u L ([0, T ], H) L ([0, T ], J) and d L ([0, T ], H 1 (, S )) is a weak solution of (1.1)-(1.5), with (1.6). If, in addition, d L ([0, T ], H ()), then (u, d) C ( (0, T ]) C,1 β ( (0, T ]). Utilizing Theorem 1., the global and local energy inequality, and the global estimate of the pressure function P in 4 below, we establish the existence of global weak solutions that enjoy the partial smoothness property. Theorem 1.3 Under the assumption (1.6), there exist a global weak solution u L ([0, ), H) L ([0, ), J) and d L ([0, ), H 1 (, S )) of (1.1)-(1.5) such that the following properties hold: (i) There exists L N depending only on (u 0, d 0 ) and 0 < T 1 < < T L, 1 i L, such that (u, d) C ( ((0, + ) \ {T i } L i=1)) C,1 β ( ((0, + ) \ {T i} L i=1)). 3

(ii) Each singular times T i, 1 i L, can be characterized by lim inf t T i max ( u + d )(y, t)dy 8π, r > 0. (1.7) x B r(x) Moreover, there exist x i m x i 0, ti m T i, r i m 0 and a non constant smooth harmonic map ω i : R S with finite energy such that as m +, (u i m, d i m) (0, ω i ) in C loc (R [, 0]), where u i m(x, t) = r i mu(x i m + r i mx, t i m + (r i m) t), d i m(x, t) = d(x i m + r i mx, t i m + (r i m) t). (iii) Set T 0 = 0. Then, for 0 i L 1, d t + d L ( [T i, T i+1 ɛ]), u t + u L 4 3 ( [Ti, T i+1 ɛ]) for any ɛ > 0, and for any 0 < T L < T < +, d t + d L ( [T L, T ]), u t + u L 4 3 ( [TL, T ]). (iv) There exist t k + and a harmonic map d C (, S ) C,β (, S ) with d = d 0 on such that u(, t k ) 0 in H 1 (), d(, t k ) d weakly in H 1 (), and there exist l N, points {x i } l i=1, and {m i} l i=1 N such that (v) If (u 0, d 0 ) satisfies d(, t k ) dx d dx + l 8πm i δ xi. (1.8) i=1 ( u 0 + d 0 ) 8π, then (u, d) C ( (0, + )) C,1 β ( (0, + )). Moreover, there exist t k + and d C (, S ) C,β (, S ) with d = d 0 on such that (u(, t k ), d(, t k )) (0, d ) in C (). Remark 1.4 We would like to point out (i) If d is a constant map, then (1.1)-(1.3) reduces to the Navier-Stokes equation. It is well-known (see Temam [4] and Ladyzhenskaya [9]) that in dimension two, any 4

weak solution u L t L x L t H 1 to the Navier-Stokes equation is smooth. (ii) If u = 0, then (1.1)-(1.3) reduces to the heat flow of harmonic maps. The classical theorems by Struwe [3] and Chang [1] assert that there exists a unique global weak solution that is smooth away from finitely many singular points. (iii) Theorem 1.3 can be viewed as a mixture of the Navier-Stokes equation and the heat flow of harmonic maps in dimension two. (iv) For smooth initial data (u 0, d 0 ), it is also a very interesting question to ask whether the short time smooth solution to (1.1)-(1.3) can develop finite time singularity (see Chang-Ding-Ye [] for the heat flow of harmonic maps). Remark 1.5 (i) We conjecture that the global weak solution (u, d) in Theorem 1.3 is unique in the class of all weak solutions (ũ, d) of (1.1)-(1.5) that enjoy the following properties: there are K N and 0 < S 1 < < S K < + such that d satisfies d L ([0, + ), H 1 ()) and d K 1 i=0 ɛ>0 L ([S i, S i+1 ɛ], H ()) L loc ([S K, + ), H ()). (ii) We also conjecture that there are at most finitely many singular points for the weak solution constructed by Theorem 1.3. The paper is written as follows. In section, we prove both interior and boundary regularity theorems for (1.1)-(1.3) under smallness conditions and Theorem 1.. In section 3, we employ the contraction map theory to establish the existence of short time smooth solutions to (1.1)-(1.3). In section 4, we show both global and local energy inequalities for smooth solutions to (1.1)-(1.3) and global estimates for the pressure function. In section 5, we prove Theorem 1.3 by estimating the first singular time in terms of energy concentration and performing blow-up analysis near each singularity. Since the exact values of ν, λ, γ don t play a role, we henceforth assume ν = λ = γ = 1. 5

Regularity of solutions and proof of Theorem 1. This section is devoted to the proof of Theorem 1.. The proof is based on two Lemmas: (i) the interior regularity under the smallness condition, and (ii) the boundary regularity under the smallness condition. We refer to [3] for interesting results on two dimensional NSE with singular forcing. We begin with some notations. Throughout the paper, we use A B to denote A C 0 B for some universal constant C 0 > 0. For x 0 R, t 0 R, z 0 = (x 0, t 0 ), and r > 0, let B r (x 0 ) = {x R : x x 0 r} and P r (z 0 ) = B r (x 0 ) [t 0 r, t 0 ] denote the ball in R and the parabolic cylinder in R 3 respectively. Let p P r (z 0 ) (B r (x 0 ) {t 0 r }) ( B r (x 0 ) [t 0 r, t 0 ]) denote the parabolic boundary of P r (z 0 ). For x 0 = 0 and t 0 = 0, we simply denote B r = B r (0), P r = P r (0, 0), p P r = p P r (0, 0). For f L 1 (P r (z 0 )), denote by 1 f z0,r = f(x, t) dxdt and f x0,r(t) = P r (z 0 ) P r(z 0 ) as the average of f over P r (z 0 ) and B r (x 0 ) respectively. 1 f(x, t) dx B r (x 0 ) B r(x 0 ) For 1 < p, q < +, denote L p,q (P r (z 0 )) = L q ([t 0 r, t 0 ], L p (B r (x 0 ))) and f L p,q (P r(z 0 )) = ( t0 t 0 r f(, t) q L p (B r(x 0 )) dt W 1,0 p,q (P r (z 0 )) L q ([t 0 r, t 0 ], W 1,p (B r (x 0 ))), with the norm ) 1 q. f W 1,0 p,q (P r(z 0 )) = f L p,q (P r(z 0 )) + f L p,q (P r(z 0 )). W,1 p,q (P r (z 0 )) = {f W 1,0 p,q (P r (z 0 )) : f, f t L p,q (P r (z 0 ))}, with the semi-norm and the norm [f] W,1 p,q (P r(z 0 )) = f + f t L p,q (P r(z 0 )), f W,1 p,q (P r(z 0 )) = f W 1,0 p,q (P r(z 0 )) + [f] W,1 p,q (P r(z 0 )). 6

For p = q, we denote L p ( ) = L p,p ( ), W 1,0 p ( ) = W 1,0 p,p ( ), and W,1 p ( ) = W,1 p,p ( ). By scaling, the first part of Theorem 1. follows from the following Lemma. Lemma.1 For any α (0, 1), there exists ɛ 0 > 0 such that for z 0 = (x 0, t 0 ) R 3 and r > 0, if (u, d) W 1,0 (P r (z 0 )), P W 1, 4 3 (P r (z 0 )) is a weak solution to (1.1)-(1.3) and P r(z 0 ) then (u, d) C α (P r (z 0), R S ). Moreover, ( u 4 + d 4 ) ɛ 4 0, (.1) [d] C α (P r (z 0 )) C( u L 4 (P r(z 0 )) + d L 4 (P r(z 0 ))) (.) { } [u] C α (P r (z 0 )) C u L 4 (P r(z 0 )) + d L 4 (P r(z 0 )) + P 4 L 3. (.3) (Pr(z0 )) Proof. By translation and dilation, we may assume that z 0 = (0, 0) and r = 1. It then follows from the interior W,1 -estimate (see [10]) that d W,1 (P 3 ) and d L (P 34 ) u L 4 (P 1 ) + d L 4 (P 1 ). (.4) 4 Pick any point z 1 = (x 1, t 1 ) P 1 Claim 1. d 4 P r(z 1 ) Let d 1 : R 3 solve and 0 < R < 1 4. We need ( r R) 4α P R d 4, 0 < r R. (.5) d 1 t d 1 = 0 in (.6) d 1 = d on p. Then d = d d 1 : R 3 solves d t d = u d + d d in (.7) d = 0 on p. Since ( u d + d d) L (), we have that d t, d L (). Hence, multiplying the equation (.7) by d and integrating over B R (x 1 ), we obtain d d + d = (u d d d) d. dt B R (x 1 ) B R (x 1 ) B R (x 1 ) 7

Integrating over [t 1 R, t 1 ] and applying Hölder inequality yields sup d + d t 1 R t t 1 [( B R (x 1 ) u d + u 4 ) 1 + ( d 4 d 4 ) 1 ]( d 4 ) 1. By the Ladyzhenskaya inequality, we have d 4 d ( d + R d ). B R (x 1 ) B R (x 1 ) B R (x 1 ) B R (x 1 ) Hence, by integrating t over [t 1 R, t 1 ], we have d 4 ( sup d ) t 1 R t t 1 ( B R (x 1 ) d + ( u 4 + For d, we have that for any θ (0, 1), d 1 4 θ 4 d 1 4 P θr (z 1 ) θ 4 { θ 4 +( Combining (.8) with (.9) yields d 4 {Cθ 4 + C( P θr (z 1 ) (Cθ 4 + Cɛ 4 0) sup t 1 R t t 1 d 4 ) B R (x 1 ) d 4 + d 4 } d 4 d ) d 4. (.8) u 4 + d 4 ) d 4. (.9) u 4 + d 4 )} d 4 d 4. (.10) For any α (0, 1), first choose θ 0 = θ 0 (α) (0, 1 ) such that Cθ4 0 θ4α 0 and then choose ɛ 0 θ 0, we have P θ0 R(z 1 ) d 4 θ0 4α d 4. (.11) 8

Iteration of this inequality yields that for any 0 < r R, Claim. P r(z 0 ) P r(z 1 ) d 4 ( r R )4α d 4. (.1) d t C( r R )α ( d 4 ) 1, 0 < r R. (.13) P R (z 0 ) Let φ C 0 (B r(x 0 )) be such that 0 φ 1, φ 1 on B r (x 0), φ r 1. Multiplying (1.3) by d t φ and integrating over B r (x 0 ) gives B r(x 0 ) = 4 B r(x 0 ) d t φ + d d φ dt B r(x 0 ) φd t d φ (u )dd t φ B r(x 0 ) d t φ + Cr B r(x 0 ) B r(x 0 ) Let s 0 (t 0 r, t 0 r 4 ) be such that d(, s 0 ) r Then we obtain P r (z 0 ) B r(x 0 ) d t C[r P r(z 0 ) d + C P r(z 0 ) P r(z 0 ) d. u d. d + u d ] P r(z 0 ) C(1 + u L 4 (P r(z 0 )) ) d L 4 (P r(z 0 )) C( r R )α d L 4 (P r(z 0 )). Claim 1 and Claim imply that r P r (z) ( d + r d t ) Cr α ( d 4 ) 1, z P 1, 0 < r 1 P 1 4. (.14) Hence the parabolic Morrey s decay Lemma (see [4]) implies that d C α (P 1 ) and (.) holds. Now we proceed to estimate u as follows. Let u 1 : R solve u 1 t u 1 = 0 in (.15) u 1 = u on p. 9

Then u = u u 1 : R solves u t u + P = [u (u u z1,r) + d d] in (.16) u = 0 on p where denotes the tensor product. Multiply both sides of (.16) by u and integrate over B R (x 1 ), we obtain d dt 8 B R (x 1 ) B R (x 1 ) B R (x 1 ) u + u B R (x 1 ) [ d + u u u z1,r ] u + P u [ d 4 + u u u z1,r ] + 1 B R (x 1 ) u + u P. B R (x 1 ) Integrating over [t 1 R, t 1 ] and applying Hölder inequality, we have sup t 1 R t t 1 B R (x 1 ) d 4 + ( u + u 4 ) 1 ( u u u z1,r 4 ) 1 + u L 4 () P L 4 (). (.17) Applying Ladyzhenskaya inequality and Hölder inequality, we have Thus u 4 [ sup C[ t [t 1 R,t 1 ] + 1 B R (x 1 ) u ] d 4 ] + C u u u z1,r 4 u 4 + C( P 4 3 ) 3. u 4 [ d 4 ] + u u z1,r 4 u 4 + ( P 4 3 ) 3. (.18) Plugging (.17) into (.18) also gives u [ d 4 ] + u u z1,r 4 u 4 +( P 4 3 ) 3. (.19) u 4 10

For u 1 solves the heat equation, we have that for any θ (0, 1 ) P θr (z 1 ) u 1 u 1 z 1,θR 4 θ 8 u 1 u 1 z 1,θR 4 θ 4 u u z1,θr 4 + u 4, and P θr (z 1 ) u 1 θ 4 u 1 θ 4 u + u. Putting these inequalities together yields solves P θr (z 1 ) P θr (z 1 ) u u z1,θr 4 (θ 8 + u θ 4 u 4 ) + [ d 4 ] + ( u + + [ d 4 ] + ( u 4 u u z1,r 4 P 4 3 ) 3. (.0) u u z1,r 4 P 4 3 ) 3. (.1) Now we estimate P L 4 3. On B R (x 1 ), write P as P = P 1 + P, where P 1 P 1 = (u u + ( d d)) in B R (x 1 ) P 1 = 0 on B R (x 1 ) so that P is a harmonic function on B R (x 1 ). For P 1, the Calderon-Zygmund theory implies that P 1 L 4 3 () and P 1 4 L 3 () u L (P 3R4 (z 1 )) u L 4 (P 3R4 (z 1 )) + d L 4 (P 3R4 (z 1 )) d L (P 3R4 (z 1 )) u L 4 () u L () + ( u L 4 () + d L 4 ()) d L 4 (). (.) 11

For P, we have that for θ (0, 1 4 ), P θr (z 1 ) P 4 3 θ θ θ P 4 3 ( P 4 3 + P 1 4 3 ) P 4 3 + u 4 3 L 4 () u 4 3 L () + ( u 4 3 L 4 () + d 4 3 L 4 () ) d 4 3 L 4 (). (.3) Putting (.) and (.3) together, we obtain P θr (z 1 ) θ P 4 3 P 4 3 + u 4 3 L 4 () u 4 3 L () + ( u 4 3 L 4 () + d 4 3 L 4 () ) d 4 3 L 4 (). (.4) Define Φ(u, z 1, R) = Ψ(u, z 1, R) = u u z1,θr 4 + u 4, D(P, z 1, R) = [ Θ(z 1, R) = Φ(u, z 1, θr) + D(P, z 1, R). u, P 4 3 ], Putting (.0) together with (.1) and applying Claim 1, we have Φ(u, z 1, θ R) (θ 8 + Ψ(u, z 1, θr))φ(u, z 1, θr) + (θr) 8α + D(P, z 1, θr), D(P, z 1, θr) { θ 6 D(P, z 1, R) + Ψ(u, z 1, R)Φ(u, z 1, R) + (R 4α + Ψ(u, z 1, R))R 4α ) }. Adding these two inequalities, we obtain Θ(z 1, θr) C[θ 6 + Ψ(u, z 1, θr)]θ(z 1, R) + CΨ(u, z 1, R)Φ(u, z 1, R) + C(R 4α + Ψ(u, z 1, R))R 4α (.5) provide P θ 1 R(z 1 ) P 1. Since Ψ(u, z 1, θr) Ψ(u, z 1, R), we have Θ(z 1, θr) C[θ 6 + Ψ(u, z 1, R)]Θ(z 1, θ 1 R) + C(R 4α + Ψ(u, z 1, R))R 4α (.6) 1

provide P θ 1 R(z 1 ) P 1. Let α 1 = 1+α 4 and choose θ = θ(α 1 ) (0, 1 ) so that Cθ 6 (θ ) +α 1. If ɛ 0 θ 3, then P 1 u 4 + d 4 θ 6. Hence, for any z 1 P 1 we have Ψ(u, z 1, R) θ 4. and 0 < R R 0 = θ 4, Hence, for ρ = R θ and τ = θ, we have Θ(z 1, τρ) τ +α 1 [Θ(z 1, ρ) + Cρ +α 1 ], z 1 P 1, 0 < ρ 1 4. (.7) Iterating (.7) finitely many times, we obtain Θ(z 1, R) R +α 1 M 0, R P 1, 0 < R 1 4 (.8) where M 0 C[ ( u 4 + d 4 ) + ( P 4 3 ) 3 ]. P 1 P 1 Now we want to show the decay estimate for Ψ(u, z 1, R) as follows. By (.8), we have u 1 ρ u ρ Cρ 1 ( u u ρ 4 ) 1 C 4 P ρ(z 0 ) ρ Θ 1 4 (z1, ρ) CM 1 4 0 ρ α 1 1 for any z 1 P 1 and 0 < ρ 1 4. Hence, for R 0 = 1 4, we have u k R 0 u R0 CM 1 4 0 k 1 i=0 ( R 0 i ) α1 1 CM 1 4 ( ) α 1 1 R0 0 k, k N so that u ρ u R0 CM 1 4 0 ρ α 1 1, ρ (0, 1 4 ). This implies, for any 0 < ρ R 0 = 1 4, Ψ(u, z 1, ρ) Φ(u, z 1, ρ) + ρ 4 u z1,ρ 4 Φ(u, z 1, ρ) + ρ 4 u z1,ρ u z1,r 0 4 + ρ 4 u z1,r 0 4 Φ(u, z 1, ρ) + M 0 ρ 4 + M 0 ρ +α 1 M 0 ρ +α 1. (.9) 13

Substituting (.9) into (.5), we arrive Θ(z 1, θr) θ 4+4α 1 Θ(z 1, R) + C(1 + M 0 )R 4+4α 1, z 1 P 1, 0 < R 1 4. (.30) Iterating (.30) finitely many times yields Φ(u, z 1, ρ) CM 0 ρ 4+4α 1, z 1 P 1, 0 < ρ 1 4. Using the characterization by Campanato spaces, we conclude that u C α 1 (P 1 ) and (.3) holds with α = α 1. By repeating the above argument, we can show that u C α (P 1 ) and (.3) holds for any α (0, 1). Now we need to establish the boundary regularity Lemma under the smallness condition. To state it, we need some notations. Denote R + = {x = (x 1, x ) R : x > 0}. For x 0 R, t 0 R and r > 0, denote z 0 = (x 0, t 0 ) let B + r (x 0 ) = B r (x 0 ) R be the upper half ball, P + r (z 0 ) = B + r (x 0 ) [t 0 r, t 0 ] denote the half parabolic cylinder. Write B + r (x 0 ) = Γ r (x 0 ) S r (x 0 ), where Γ r (x 0 ) = B + r (x 0 ) R and S r (x 0 ) = B + r (x 0 ) R +, and p P + r (z 0 ) = (B + r (x 0 ) {t 0 r }) ( B + r (x 0 ) (t 0 r, t 0 )) be its parabolic boundary. When x 0 = 0 and t 0 = 0, we simply denote B + r = B + r (0), P + r = P + r (0, 0), B + r = B + r (0), Γ r = Γ r (0), S r = S r (0), and p P + r = p P + r (0, 0). Lemma. For any α (0, 1), there exists ɛ 0 depending on α and d 0 C,β (Γ + r (x 0)) such that if for z 0 = (z 0, t 0 ) R R and r > 0, (u, d) W 1,0 (P r + (z 0)), P W 1, 4 3 (P + r (z 0)) is a weak solution of (1.1)-(1.3), with (u, d) = (0, d 0 ) on Γ + r (x 0) [t 0 r, t 0 ] and P + r (z 0 ) ( u 4 + d 4 ) ɛ 4 0, (.31) then (u, d) C α (P + r (z 0 )). Moreover, [d] C α (P + r (z 0 )) u L 4 (P + r (z 0 )) + d L 4 (P + r (z 0 )) + d 0 C,β (Γ r(x 0 )) (.3) [u] C β (P + r (z 0 )) u L 4 (P + r (z 0 )) + d L 4 (P + r (z 0 )) + d 0 C,β (Γ r(x 0 )) + P L 4 3 (P + r (z 0 )). (.33) 14

Proof. By scalings, it suffices to consider z 0 = (0, 0) and r = 1. The argument is similar to Lemma.1, except that we have to estimate P differently. Here we only outline it. For any z 1 = (x 1, t 1 ) Γ 1 [ 1 4, 0] and 0 < R 1 4, let d1 : P + R (z 1) R 3 solve d 1 t d 1 = 0 in P + R (z 1) (.34) d 1 = d 0 on Γ R (x 1 ) [t 1 R, t 1 ] d 1 = d on S R (x 1 ) [t 1 R, t 1 ] (B + R (x 1) {t 1 R }). Then d = d d 1 : P + R (z 1) R 3 solves d t d = u d + d d in (.35) d = 0 on p P + R (z 1). Since ( u d + d d) L ( + ), we have that d t, d L (P + R (z 1)). Multiplying (.35) by d and integrating over B + R (x 1), we obtain d d + d = (u d d d) d. dt B + R (x 1) B + R (x 1) B + R (x 1) Integrating over [t 1 R, t 1 ] and applying Hölder inequality yields sup t 1 R t t 1 [( P + R (z 1) P + R (z 1) d + B + R (x 1) d + u d + d 4 u 4 ) 1 + ( P + R (z 1) P + R (z 1) d 4 ) 1 ]( This, combined with the Ladyzhenskaya inequality, yields P + R (z 1) d 4 ( sup t 1 R t t 1 ( P + R (z 1) B + R (x 1) d + ( u 4 + P + R (z 1) d ) sup t 1 R t t 1 P + R (z 1) P + R (z 1) d 4 ) d 4 ) 1. B + R (x 1) P + R (z 1) d ) d 4. (.36) Since d 1 ΓR (x 1 ) [t 1 R,t 1 ] = d 0 C,β, the boundary regularity for the heat equation 15

implies that d 1 C,β (P + R (z 1 )) and for any θ (0, 1 ), d 1 4 (θr) 4 d 1 4 P + θr (z C 1) 0 (P + R (z 1 ) (θr) 4 d 0 4 C,β (Γ 1 [ 1,0]) + θ4 [ + θ 4 [R 4 d 0 4 C,β (Γ 1 [ 1,0]) + P + R (z 1) ( u 4 + d 4 ) P + R (z 1) P + R (z 1) P + R (z 1) d 4 + d 4 ] P + R (z 1) d 4 ] Combining (.36) with (.37) yields that for any θ (0, 1 ), d 4 C 0 (θr) 4 + C[θ 4 + ( u 4 + d 4 )] P + θr (z 1) where C 0 d 0 4 C,β (Γ 1 [ 1,0]). P + R (z 1) C 0 (θr) 4 + C(θ 4 + Cɛ 4 0) d 4. (.37) P + R (z 1) P + R (z 1) d 4 d 4, (.38) For any α (0, 1), choose θ 0 = θ 0 (α) (0, 1 ) such that Cθ4 0 θ4α 0 and ɛ 0 θ 0, we obtain P + θ 0 R (z 1) d 4 C 0 (θ 0 R) 4 + θ0 4α d 4. (.39) P + R (z 1) Iteration of this inequality yields that for any 0 < r R, d 4 C 0 r 4 + ( r R )4α d 4. (.40) P + r (z 1 ) P + R (z 1) The estimate of P + r (z 1 ) d t is similar to Lemma.1. Let φ C 0 (B r(x 1 )) be such that 0 φ 1, φ 1 on B r (x 1), φ r 1. Since d t = 0 on Γ r (x 1 ) [t 1 r, t 1 ], multiplying (1.3) by d t φ and integrating over B r + (x 1 ) gives d t φ + d d φ r d + dt B + r (x 1 ) B + r (x 1 ) B + r (x 1 ) Integrate t from s 0 to t 0, where s 0 (t 0 r, t 0 r 4 ) is such that d(, s 0 ) r d, we obtain P + r (z 1 ) B + r (x 1 ) d t C[r P + r (z 1 ) P + r (z 1 ) B + r (z 1 ) d + u d ] P r + (z 1 ) u d. C[r + ( r R )α d L 4 (P + R (z ]. (.41) 1)) 16

Hence we have for any z 1 Γ + 1 r P + r (z 1 ) ( d + r d t ) C[r + r α ( P + 1 d 4 ) 1 ] (.4) [ 1 4, 0] and 0 < r 1 4. Combining this with Lemma.1 and the parabolic Morrey s decay Lemma ([4]), we conclude that d C α (P + 1 holds. ) and (.3) To estimate u and P, let u 1 : R + [ 1, 0] R and P 1 : R + [ 1, 0] R solve the non homogeneous, non-stationary Stokes equation: u 1 t u 1 + P 1 = [u u + ( d d)]λ P + R (z 1), R + ( 1, 0) (.43) u 1 = 0, ( R + [ 1, 0]) (R + { 1}), where λ P + R (z 1) is the characteristic function of P + R (z 1). Set u = u u 1 : P + R (z 1) R and P = P P 1 : P + R (z 1) R. Then (u, P ) solves u t u + P = 0, P + R (z 1) (.44) u = 0, Γ + R (x 1) [t 1 R, t 1 ]. By the boundary W,1 q -estimate of non-stationary Stokes equations (see [17] Lemma 3.1), we conclude that u 1 W,1 (R + [ 1, 0]) and P 1 W 1,0 (R + [ 1, 0]), and 4 3 4 3 u 1 W,1 4 (R + [ 1,0]) + P 1 4 L 3 (R + [ 1,0]) 3 u u + u u L 4 3 (P + R 1(z 1)) u L 4 (P + R (z 1)) u L (P + R (z 1)) + d L (P + R (z 1)) d L 4 (P + R (z 1)) (.45) By Sobolev embedding theorem (see [10]), W,1 4 (R + [ 1, 0]) W 1,0 (R + [ 1, 0]) 3 L 4 (R + [ 1, 0]), (.45) yields u 1 L (R + [ 1,0]) + u1 L 4 (R + [ 1,0]) u L 4 (P + R (z 1)) u L (P + R (z 1)) (.46) + d L (P + R (z 1)) d L 4 (P + R (z 1)) 17

For (u, P ), by the boundary W,1 p,q -estimate for homogeneous, non-stationary Stokes equation (see [17] Lemma 3.), we have u W,1 for any 4 < q < + and P + q, 4 R 3 (z 1 )), P L q, 4 3 (P + R (z 1 )) R 3 q [[u ] W,1 q, 4 (P + R (z 1 )) + P L q, 4 ] 3 (P + 3 R (z 1 )) u L (P + R (z 1)) + P 4 L 3 (P + R (z 1)) ( u L (P + R (z 1)) + P L 4 ) + 3 (P + R (z 1)) ( u1 L (P + R (z 1)) + P 1 4 L 3 (P + R (z 1)) ( u L (P + R (z 1)) + P L 3 4 ) + u (P + R (z 1)) L 4 (P + R (z 1)) u L (P + R (z 1)) + d L (P + R (z 1)) d L 4 (P + R (z 1)). By Sobolev inequality and Hölder inequality, we have that for any θ (0, 1 4 ), u u z 1,θR L 4 (P + θr (z 1)) + P L 4 3 (P + θr (z 1)) θ 3 q [R 3 q ([ u ] W,1 q, 4 (P + R (z 1 )) + P L q, 4 )] 3 (P + 3 R (z 1 )) θ 3 q [ u L (P + R (z 1)) + P L 4 3 (P + R (z 1)) ] (.47) + θ 3 q [ u L 4 (P + R (z 1)) u L (P + R (z 1)) + d L (P + R (z 1)) d L 4 (P + R (z 1)) ]. To estimate u L, let φ C 1 0 (B θr(x 1 )) be such that 0 φ 1, φ 1 on B θr (x 1 ), and φ θr. Multiplying the equation of u by (u u z 1,θR )φ and integrating over B + θr (x 1), we obtain d u u dt B + θr (x z 1,θR φ + 1) (θr) u u z 1,θR + B + θr (x 1) B + θr (x 1) B + θr (x 1) u φ P u u z 1,θR. Integrating over [s 0, t 1 ], where s 0 [t 1 (θr), t 1 (θr) ] is such that u u z 1,θR (θr) u u z 1,θR, B + θr (x 1) {s 0 } P + θr (z 1) 18

we obtain P + θr (z 1 ) (θr) u P + θr (z 1) u u z 1,θR + P u u P + θr (z z 1,θR 1) u u z 1,θR L 4 (P + θr (z 1)) + P L 4 3 (P + θr (z 1)) u u z 1,θR L 4 (P + θr (z 1)) u u z 1,θR L 4 (P + θr (z 1)) + P L 4 3 (P + θr (z 1)). (.48) Putting together (.45), (.46), (.47), (.48), applying (.39), and setting q = 8, we arrive u u z1,θr L 4 (P + θr (z 1)) + u L (P + θr (z 1)) + P L 4 3 (P + θr (z 1)) [θ 5 4 + u L 4 (P + R (z 1)) ][ u u z 1,R L 4 (P + R (z 1)) + u L (P + R (z 1)) + P L 4 3 (P + R (z 1)) ] + d L (P + R (z 1 )) d L 4 (P + R (z 1)). (.49) For d L (P + R (z 1 )), we have d L (P + R (z 1 )) u L 4 (P + R (z 1)) + d L 4 (P + R (z 1)). Thus, by choosing θ = θ 0 sufficiently small and ɛ 0 θ 0, we obtain where C 0 depends on d 0 C,β (Γ 1 ), and Θ + (z 1, θr) θθ + (z 1, R) + C 0 (θ + R α )R α, (.50) Θ + (z 1, r) u u z1,r L 4 (P + r (z 1 )) + u L (P + r (z 1 )) + P L 4 3 (P + r (z 1 )), 0 < r 1 4. The same argument as in Lemma.1 can imply r 4 u u z1,r 4 C 1 (Θ + (z 1, 1 ))4 r 4α, 0 < r 1 4. P + r (z 1 ) This, combined with Lemma.1, implies that u C α (P + 1 ) and (.33) holds. Proof of Theorem 1.: Since u L ([0, T ], L ()) L ([0, T ], H 1 ()), it follows from the Ladyzhenskaya s inequality that u L 4 ( [0, T ]). L ([0, T ], H 1 ()) and d = 1, we have Since d d d + d = 0. 19

Hence d L 4 ( [0, T ]) and u u + ( d d) L 4 3 ( [0, T ]). Hence Lemma 4.4 implies that P L 4 3 ( [0, T ]). It follows from the absolute continuity of ( u 4 + d 4 ) that (i) for any z 0 = (x 0, t 0 ) (0, T ], there exists 0 < r 0 min{dist(x 0, ), t 0 } such that P r0 (z 0 ) ( u 4 + d 4 ) ɛ 4 0, where ɛ 0 > 0 is given by Lemma.1. Hence we conclude (u, d) C α (P r 0 (z 0 )) for any α (0, 1). (ii) for any z 0 = (x 0, t 0 ) (0, T ], since is smooth, it is well known that there exists r 0 > 0 depending only on such that ( B r0 (x 0 )) [t 0 r 0, t 0] is C 3 -close to P + r 0 and ( B r0 (x 0 )) [t 0 r 0,t 0] u 4 + d 4 ) ɛ 4 0, where ɛ 0 > 0 is given by Lemma.. Hence, we can perform the standard boundary flatten technique, which is a small perturbation of the one on P + r 0, so that a slight modification of Lemma. implies that (u, d) C α (( B r 0 (x 0 )) [t 0 r 0 4, t 0 ]. The reader can consult with [17] for such details. The higher order regularity can be obtained as follows. One can follow the standard hole filling argument for (1.3) of d to show that d C α ( (0, T ]) and then apply the Schauder theory to show d Cα,1 ( (0, T ]). Substitute this regularity of d into (1.1)-(1.), we can apply the standard Cα,1 -regularity theory (see [19]) to show u Cα,1 ( (0, T ]). Once we obtain Cα,1 -regularity for (u, d), the smoothness of (u, d) on (0, T ] can be obtained by the standard boot-strap argument. Similarly, the boundary C,1 β -regularity for (u, d) can be obtained. 3 Existence of short time smooth solutions In this section, we prove the existence of short time smooth solutions to (1.1)-(1.3) for smooth initial and boundary data. We would like to point out that the proof also works for R 3. More precisely, we have 0

Theorem 3.1 For any α > 0, if u 0 C,α 0 (, R ) with u 0 = 0, and d 0 C,α (, S ), then there exists T > 0 depending on u 0 C,α (), d 0 C,α () such that there is a unique smooth solution (u, d) Cα,1 ( [0, T ), R S ) to the initialboundary value problem (1.1), (1.), (1.3), (1.4), and (1.5). Proof. The proof is based on the contraction mapping principle. For T > 0 and K > 0 to be chosen later, denote T = [0, T ], X = {(v, f) C,1 α ( T, R R 3 ) v = 0, (v, f) t=0 = (u 0, d 0 ), Equip X with the norm v u 0 C,1 α ( T ) + f d 0 C,1 α ( T ) K}. (v, f) X := v C,1 α ( T ) + f Cα,1 ( T, (v, f) X. ) It is easy to see that (X, X ) is a Banach space. Define the operator L : X C,1 α (Q T, R R 3 ) as follows. For any (v, f) X, let (u, d) = L(v, f) be the unique solution to the non homogeneous, non-stationary Stokes system: u t u + P = v v ( d d), (0, T ), (3.1) u = 0, (0, T ) (3.) d t d = f f v f, (0, T ) (3.3) (u, d) = (u 0, d 0 ), {0} (3.4) (u, t) = (0, d 0 ),. (3.5) We will prove that for T > 0 sufficiently small and K > 0 sufficiently large, L : X X is a contraction map. Lemma 3.1 There exist T > 0 and K > 0 such that L : X X. Proof. For any (v, f) X, let (u, d) = L(v, f) be the unique solution to (3.1)-(3.5). Let C 0 > 0 denote constants depending only on u 0 C,α and d 0 C,α. 1

Assume K C 0. By the Schauder theory of parabolic systems, we have d d 0 C,1 α ( T ) v f C α ( T ) + f f C α ( T ). (3.6) For the first term in the right hand side, we have v f C α ( T ) v f u 0 d 0 C α ( T ) + u 0 d 0 C α () (v u 0 ) f C α ( T ) + u 0 (f d 0 ) C α ( T ) + C 0 K[ v u 0 C 0 ( T ) + v u 0 C α ( T )] + C 0 [1 + (f d 0 ) C 0 ( T ) + (f d 0 ) C α ( T )]. Since v u 0 = f d 0 = 0 at t = 0, it is easy to see v u 0 C 0 ( T ) KT, (f d 0 ) C 0 ( T ) KT. By the interpolation inequality, we have that for any 0 < δ < 1, and v u 0 C α ( T ) 1 δ v u 0 C 0 ( T ) + δ v u 0 C,1 α ( T ) (δ + T δ )K, (d f) C α ( T ) 1 δ (d f) C 0 ( T ) + δ d f C,1 α ( T ) (δ + T δ )K. Putting these inequalities together, we obtain v f C α ( T ) (C 0 K + CK )(T + δ + T δ ) + C 0 K provided K = 16C 0, δ 1 (C 0 +C K) K, and T = δ. The second term in the right hand side of (3.6) can be estimated by 4 (3.7) f f C α ( T ) f f d 0 d 0 C α ( T ) + d 0 d 0 C α ( T ) f (f d 0 ) C α ( T ) + d 0 ( f d 0 ) C α ( T ) + C 0 = I 1 + I + C 0. I 1 f d 0 C α ( T ) f C 0 ( T ) + f d 0 C 0 ( T ) f C α ( T ) K ( f d 0 C 0 ( T ) + f d 0 C α ( T )) K [(1 + 1 δ ) f d 0 C 0 ( T ) + δ f d 0 C,1 α ( T ) ] K 3 ( T δ + δ).

Similarly, I can be estimated by I f d 0 C α ( T ) + f d 0 C 0 ( T ) d 0 C α ( T ) ( f + d 0 ) (f d 0 ) C α ( T ) + C 0 ( f + d 0 ) (f d 0 ) C 0 ( T ) (1 + C 0 )K ( ) (f d 0 ) C 0 ( T ) + (f d 0 ) C α ( T ) (1 + C 0 )K (T + δ + T δ ). Hence f f d 0 d 0 C α ( T ) K3 ( T δ + δ) + (1 + C 0)K (T + δ + T δ ) provided K = 16C0, δ 1, and T = δ. Thus (1+C 0 )K 5 d d 0 C,1 α ( T ) K. (3.8) By the Schauder theory for non homogeneous, non-stationary Stokes equations [19], we have u u 0 C,1 α ( T ) v v C α ( T ) + ( d d) C α ( T ). (3.9) For the first term of the right hand side of (3.9), we have v v C α ( T ) (v u 0 ) v C α ( T ) + u 0 (v 0 u 0 ) C α ( T ) + u 0 u 0 C α () K( v u 0 C 0 ( T ) + v u 0 C α ( T ) ) + C 0 ( (v u 0 ) C 0 ( T ) + (v u 0) C α ( T ) ) + C 0 (C 0 K + K )(T + δ + T δ ) + C 0 K 4 provided K = 8C 0, δ 1 (1+C 0 )K, and T = δ. 3

For the second term in the right hand side of (3.9), it follows from (3.8) that ( d d) C α ( T ) (d d 0 ) d C α ( T ) + d 0 (d d 0 ) C α ( T ) + d 0 d 0 C α ( T ) C 0 + d d 0 C,1 α ( T ) d Cα,1 ( T ) + C 0 d d 0 C,1 K K C 0 + (C 0 + ) + C 0 K K. α ( T ) (3.10) Combining (3.8) with (3.10), we have Therefore L maps X to X. u u 0 C,1 α ( T ) + d d 0 C,1 α ( T ) K. Lemma 3. There exist sufficiently large K > 0 and sufficiently small T > 0 such that L : X X is a contraction map. Proof. For any (v i, f i ) X, let (u i, d i ) X be such that (u i, d i ) = L(v i, f i ), i = 1,. Denote u = u 1 u, d = d 1 d, P = P 1 P, v = v 1 v, f = f 1 f. Then (u, d) solves u t u + P = G, (0, T ) (3.11) u = 0, (0, T ) (3.1) d t d = H, (0, T ) (3.13) (u, d) t=0 = (0, 0) (3.14) (u, d) = 0, (0, T ) (3.15) where G = (v 1 v 1 v v ) ( d 1 d 1 d d ) = (v v 1 + v v) ( d d 1 + d d), 4

and H = ( f 1 f 1 f f ) (v 1 f 1 v f ) = f 1 f + [ (f 1 + f )] ff (v f 1 v f). By Lemma 3.1, we have that for i = 1,, u i u 0 C,1 α ( T ) + d i d 0 C,1 α ( T ) K. Applying the Schauder theory of parabolic systems, we have d C,1 α ( T ) H C α ( T ) where we have used v f 1 + v f + f 1 f + f ( f 1 + f ) f C α ( T )) K ( v C α ( T ) + f C α ( T ) + f C α ( T )) K [δ( v C,1 α ( T ) + v C,1 α ( T ) ) + 1 δ ( v C α ( T ) + f C α ( T ))] K (δ + T δ )[ v Cα,1 ( T ) + f Cα,1 ( T ], (3.16) ) v C α ( T ) + f C α ( T ) ( v C,1 α ( T ) + f Cα,1 ( T )T. (3.17) ) Applying the Schauder theory for non homogeneous, non-stationary Stokes equations ([19]) to (3.11)-(3.1), we have u C,1 α ( T ) G C α ( T ) v v1 + v v + d d 1 + d d C α ( T ) K d C,1 α ( T ) + K( v C α () + v C α ( T )) K 3 (δ + T δ )[ v C,1 α ( T ) + f C,1 α ( T ) ] + K (δ + T δ ) v C,1 α ( T ) K 3 (δ + T δ )[ v Cα,1 ( T ) + f Cα,1 ( T ]. (3.18) ) It follows from (3.16) and (3.18) that L(v 1, f 1 ) L(v, f ) X K 3 (δ + T δ ) (v 1, f 1 ) (v, f ) X 1 (v 1, f 1 ) (v, f ) X 5

provided δ and T are sufficiently small. Therefore, L : X X is a contraction map. It follows from Lemma 3.1 and Lemma 3. that if T > 0 is sufficiently small, then there exists a unique solution (u, d) C,1 α ( [0, T ), R S ) to (1.1-1.4), (1.5). Applying the maximum principle to the equation for d, one can easily see that d = 1 in (0, T ). The proof of Theorem 3.1 is now complete. 4 Energy inequalities, estimates of pressure function This section is devoted to both global and local energy inequalities, and the estimate of the pressure function. First, we have Lemma 4.1 For 0 < T < +, suppose u L, ( [0, T ]) W 1,0 ( T ), d L ([0, T ], H 1 ()) L ([0, T ], H ()), and P L 4 3 ( T ) is a weak solution to (1.1)- (1.4), (1.5). Then, for any 0 < t T, we have ( u + d ) (, t) + ( u + d + d d ) Proof. First, by Ladyzhenskaya s inequality, we have = t ( u0 + d 0 ). (4.1) u L 4 ( T ), d L 4 ( T ). Multiply (1.1) by u and integrate over. Since u H, it is well-known ([4]) that (u u) u = 0, P u = 0. Hence we have d 1 dt u + u = d d : u. (4.) Multiplying (1.3) by d + d d and integrating over, we obtain (d t + u d) d = d + d d, where we have used the fact that d = 1 to get (d t + u d) d d = 1 ( d d t + u d d ) = 0. 6

Since d t = 0 on, by integration by parts, we have d t d = d 1 dt d. Now we claim d (u d) = d d : u. (4.3) In fact, (4.3) follows from d (u d) = d ββ u α d α ] = [(d β u α d α ) β d α d β u αβ uα ( d ) α = d α d β u α β = d d : u. Hence we obtain d dt 1 d + d + d d = d d : u. (4.4) It is now easy to see that (4.1) follows by adding (4.) and (4.4) and integrating from 0 to T. In order to prove Theorem 1.3, we also need a local energy inequality of both interior and boundary types for solutions to (1.1)-(1.4), (1.5). Lemma 4. For 0 < T < +, suppose u L, ( [0, T ]) W 1,0 ( T ), d L ([0, T ], H 1 ()) L ([0, T ], H ()), and P L 4 3 ( T ) is a weak solution to (1.1)- (1.4), (1.5). Then, for any nonnegative φ C0 () and 0 < s < t T, t φ( u + d )(t) + φ( u + d + d d ) s φ( u + d )(s) t +C φ [ u 3 + P P u + u u + d u + d t d ], (4.5) s where P is the average of P over. Proof. Multiplying (1.1) by uϕ and integrating over implies d u φ + u φ dt = [ u u uφ u u φ + (P P ) u φ + d d : (uφ)]. 7

For the first term in the right hand side, we have, by integration by parts, 1 u u uφ = u u φ. For the last term in the right hand side, we have d d : (uφ) = u d d φ + d d : uφ. Putting all these two terms into the identity above yields d u φ + u φ d d : uφ (4.6) dt + ( u 3 + u u + P P u + d u ) φ. Multiplying (1.3) by ( d + d d)φ and integrating over yields (d t + u d) ( d + d d)φ = d + d d φ. (4.7) As in Lemma 4.1, since d = 1, we have (d t + u d) d dφ = 0. (4.8) On the other hand, by integration by parts, we have d t dφ = d d φ dt d t d φ, (4.9) u d dφ = (u i d i ) j d j φ u i d i d j φ j = u i ( d ) i φ u i jd i d j φ u i d i d j φ j d = u φ d d : uφ (u d)( φ d). (4.10) Combining (4.7), (4.8), (4.9), with (4.10), we obtain d d φ + d + d d φ dt = d t d φ u : ( d d)φ + d u φ (u d)( d )φ. (4.11) 8

Adding (4.7) and (4.11), we have d ( u + d )φ + ( u + d + d d )φ dt [ u 3 + u u + P P u + 6 d u + d t d ] φ. Integrating this inequality from s to t implies (4.5). This completes the proof. We also need the boundary version of the local energy inequality. More precisely, Lemma 4.3 For 0 < T < +, suppose u L, ( [0, T ]) W 1,0 ( T ), d L ([0, T ], H 1 ()) L ([0, T ], H ()), and P L 4 3 ( T ) is a weak solution to (1.1)- (1.4), (1.5). There exists r 0 = r 0 ( ) > 0 such that for any x 0 and 0 < r r 0, if φ C0 (B r(x 0 )) is nonnegative and 0 < s < t T, then t φ( u + d )(t) + φ( u + d + d d ) B r(x 0 ) +C B r(x 0 ) t s s ( B r(x 0 )) φ( u + d )(s) (4.1) ( B r(x 0 )) where P is the average of P over. φ [ u 3 + P P u + u u + d u + d t d ], Proof. Multiply (1.1) by uφ and integrate over B r (x 0 ). Since uφ = 0 on ( B r (x 0 )), all the boundary terms vanish in the process of integration by parts. Multiply (1.3) by ( d + d d)φ and integrate over B r (x 0 ). Since both d t φ = 0 and uφ = 0 on ( B r (x 0 )), all the boundary terms vanish in the process of integration by parts. The rest of the argument is exactly same as Lemma 4.. We leave it to the readers. In order to justify the assumptions on pressure functions in Lemma.1 and Lemma., we need Lemma 4.4 For 0 < T < +, suppose u L, ( [0, T ]) W 1,0 ( T ), d L ([0, T ], H 1 ()) L ([0, T ], H ()) is a weak solution to (1.1)-(1.4), (1.5). Then P L 4 3 ( T ). Moreover, P satisfies the following estimate: { } max P 4, P P L 3 (T ) L 4, 4 3 ( T ) u L 4 ( T ) u L ( T ) + d L 4 ( T ) d L ( T ). (4.13) 9

Proof. Write u = v + w, where v solves the heat equation: v t v = 0, (0, T ) v = 0, (0, T ) v = u 0, {t = 0}, and w solves the non homogeneous, non-stationary Stokes equation: w t w + P = u u ( d d), (0, T ) (4.14) w = 0, (0, T ) (4.15) w = 0, {t = 0}. (4.16) Since f u u ( d d) L 4 3 ( T ), the L p -theory [0] of non homogeneous, non-stationary Stokes equations to (4.14)-(4.16) implies that P L 4 3 ( T ) and P L 4 3 (T ) f L 4 3 (T ) u u L 4 3 (T ) + d d L 4 3 (T ) u L 4 ( T ) u L ( T ) + d L 4 ( T ) d L ( T ). Since P P L 4, 4 3 ( T )) P L 4 3 ( T ), (4.13) follows. The proof of Lemma 4.4 is complete. 5 Global weak solutions and proof of Theorem 1.3 In this section, we will establish the existence of global weak solutions to (1.1)-(1.5) that enjoy both the regularity and uniqueness properties described as in Theorem 1.3. First, we need to recall the following version of Ladyzhenskaya s inequality. Lemma 5.1 There exist C 0 > 0 and R 0 > 0 depending only on such that for any T > 0, if u L, ( T ) W 1,0 ( T ), then for R (0, R 0 ), { u 4 C 0 sup u (, t) u + 1 } T B R (x) T R u. (5.1) T (x,t) T 30

Proof. See Struwe [3] Lemma 3.1. We now derive the life span estimate for smooth solutions in term of Sobolev space norms of initial data. Lemma 5. Let ɛ 0 > 0 be given by Lemma.1 and.. There exist 0 < ɛ 1 << ɛ 0 and θ 0 = θ 0 (ɛ 1, E 0 ) (0, 1 4 ), here E 0 = ( u 0 + d 0 ), such that if (u 0, d 0 ) C,β (, R S ) satisfies sup x B R0 (x) for some 0 < R 0 1. Then there exist T 0 θ 0 R 0 ( u 0 + d 0 ) ɛ 1 (5.) and a unique solution (u, d) C ( (0, T 0 ), R S ) C,1 β ( [0, T 0), R S ) to (1.1)-(1.5) satisfying sup ( u + d )(, t) ɛ 1. (5.3) (x,t) T0 B R0 (x) Proof. By Theorem 3.1, there exists T 0 > 0 such that there exists a unique smooth solution (u, d) C ( (0, T 0 ), R S ) C,1 δ ( [0, T 0 ), R S ) to (1.1)-(1.5). Let 0 < t 0 T 0 be the maximal time such that sup sup ( u + d )(, t) ɛ 1. (5.4) 0 t t 0 x B R0 (x) Since t 0 is the maximal time for (5.4), we have ( u + d )(, t 0 ) = ɛ 1. (5.5) sup x B R0 (x) Now we estimate the lower bound of t 0 as follows. Assume t 0 R0. For, otherwise, we are done. Set E(t) = ( u + d )(, t), E 0 = ( u 0 + d 0 ). Then Lemma 4.1 implies that for any 0 < t t 0, E(t) + ( u + d + d d ) E 0. (5.6) t Lemma 5.1 implies d 4 C 0 ( sup t (x,s) t C 0 ER 0 (t)( B R0 (x) t d + te 0 R 0 d (, s))( d + 1 Q R0 d ) t 31 ) (5.7)

where and E 1 R 0 (t) = E R 0 (t) = sup (x,s) t sup (x,s) t B R0 (x) B R0 (x) E R0 (t) = E 1 R 0 (t) + E R 0 (t). u (, s), d (, s), By (5.4), we have E R0 (t) ɛ 1, 0 t t 0. Hence Since d 4 C 0 ɛ 1( d + t 0 t0 t0 R0 E 0 ). (5.8) d ( d + d d + d 4 ), (5.6) and (5.8) imply d ( d + d d + d 4 ) E 0 + d 4 t0 t0 t0 E 0 + C 0 ɛ 1( d + t 0 t0 R0 E 0 ). Therefore we get Choose 0 < ɛ 1 1 C 0, we have (1 C 0 ɛ 1) d C 0 (1 + ɛ 1 t 0 t0 R0 )E 0. t0 d C 0 (1 + ɛ 1 t 0 R 0 )E 0 C 0 (1 + t 0 R0 )E 0 C 0 E 0. (5.9) This, combined with (5.8), also gives d 4 C 0 ɛ 1(1 + t 0 t0 R0 )E 0 C 0 ɛ 1E 0. (5.10) Similarly, we can estimate t0 u 4 as follows. u 4 C 0 ER 1 0 (t 0 )( u + 1 t0 t0 R0 u ) t0 C 0 ER 1 0 (t 0 )( u + t 0E 0 t0 R0 ) C 0 ɛ 1(1 + t 0 R0 )E 0 C 0 ɛ 1E 0. (5.11) 3

Now we estimate the quantity E R0 (t) as follows. C 0 (B R 0 (x)) be a cut-off function of B R0 (x) such that For any x, let φ 0 φ 1, φ 1 on B R0 (x), φ 0 outside B R0 (x), φ 4 R 0. Then, by the local energy inequality Lemma 4. and Lemma 4.3, we have sup 0 t t 0 sup 0 t t 0 B R0 (x) B R0 (x) ( u + d ) E R0 (0) ( u + d )φ E R0 (0) [ u 3 + u u + P P u + d u + d t d ] φ t0 ) 1 4 [ ] u 3 L 4 ( t0 ) + u L ( t0 ) u L 4 ( t0 ) + P P L 4, 3 4 u ( t0 ) L 4 ( t0 ) ( t0 + R0 ( t0 R 0 ) 1 4 [ ] d L 4 ( t0 ) u L 4 ( t0 ) + d t L ( t0 ) d L 4 ( t0 ), (5.1) where we have used t 0 R0 1. Notice u L ( t0 ) (t 0 E 0 ) 1 E 1 0. (5.13) For d t, multiplying (1.3) by d t and integrating over t0, we obtain t0 d t d 0 + 4 u d t0 E 0 + 4 u L 4 ( t0 ) d L 4 ( t0 ) C 0 E 0. (5.14) Putting (4.13), (5.9), (5.10), (5.11), (5.13) and (5.14) into (5.1), we obtain ɛ 1 = sup ( u + d ) 0 t t 0 B R0 (x) ( ) 1 t0 4 1 E R0 (0) + C 0 ɛ 1 E 3 4 0 R 0 ( ) 1 ɛ t0 4 1 1 + C 0 ɛ R0 1 E 3 4 0. (5.15) This implies t 0 ɛ6 1 C0 4 R E3 0 = θ 0 R0, with θ 0 ɛ6 1 0 C0 4. E3 0 33

Set T 0 = t 0, we have T 0 θ 0 R0 and (5.3) holds. This completes the proof. Before we prove Theorem 1.3, we need the following density property of Sobolev maps, whose proof can be found in Schoen-Uhlenbeck []. Lemma 5.3 For n = and any given map f H 1 (, S ) C,δ (, S ) with 0 < δ < 1, there exist {f k } C (, S ) C,δ (, S ) such that f k = f on for all k, and Proof of Theorem 1.3: lim f k f H k 1 () = 0. Since u 0 H, there exists u k 0 C 0 (, R ), with u k 0 = 0, such that lim k uk 0 u 0 L () = 0. Since d 0 H 1 (, S ) C,β (, S ), Lemma 5.3 implies that there exist {d k 0 } C (, S ) C,β (, S ), with d k 0 = d 0 on, such that lim k dk 0 d 0 H 1 () = 0. By the absolute continuity of ( u 0 + d 0 ), we conclude that there exists R 0 > 0 such that sup ( u 0 + d 0 ) ɛ 1 x B R0 (x), (5.16) where ɛ 1 > 0 is given by Lemma 5.. By the strong convergence of (u k 0, dk 0 ) to (u 0, d 0 ) in L (), we have that ( u k 0 + d k 0 ) ɛ 1, k >> 1. (5.17) sup x B R0 (x) For simplicity, we assume (5.17) holds for all k 1. By Lemma 5., there exist θ 0 = θ 0 (ɛ 1, E 0 ) (0, 1) and T k 0 θ 0R 0 such that there exist solutions (uk, d k ) C ( T k, R S ) C,1 0 β ( T0 k, R S ) to (1.1)-(1.3) along with the initial-boundary condition: (u k, d k ) {0} = (u k 0, d k 0), (u k, d k ) (0,T k 0 ] = (uk 0, d k 0). (5.18) Moreover we have sup (x,t) T k 0 B R0 (x) ( u k + d k )(, t) ɛ 1. (5.19) 34

By Lemma 4.1, we have sup 0 t T k 0 ( u k + d k )(, t) + ( u k + d k + d k d k ) T k 0 u k 0 + d k 0 )( 1 + E 0 ) (5.0) for sufficiently large k. Combining (5.19) and (5.0) with Lemma 5.1, we conclude that and ( u k 4 + d k 4 ) Cɛ 1E 0, (5.1) T k 0 d k t + d k CE 0. (5.) T k 0 T k 0 By Lemma 4.4, (5.19), (5.1), and (5.), we have P k 4 L 3 (T k ) 0 u k L ( T k 0 ) u k L 4 ( T k 0 ) + d k L ( T k 0 ) d k L 4 ( T k 0 ) Cɛ 1 1 E 3 4 0. (5.3) Furthermore, (1.1) implies that for any φ J, u k t, φ = u k φ + (u k u k + d k d k ) φ, where, denotes the inner product between H 1 and H 1 0, we conclude that uk t L ([0, T k 0 ], H 1 ()) and u k t CE 0. (5.4) L ([0,T0 k],h 1 ()) By Theorem.1, we conclude that for any δ > 0, (u k, d k ) C,1 β ( [δ,t k 0 ]) C(δ, E 0, u k L 4 ( T k ), d k L 4 ( 0 T k ), P k 4, d 0 L 3 (T 0 k ) C,β ( )) 0 C(ɛ 1, E 0, δ, d 0 C,β ( )). (5.5) Furthermore, for any compact sub domain K and δ > 0, (u k, d k ) C l (K [δ,t k 0 ]) C(dist(K, ), δ, l, E 0), l 1. (5.6) 35

Hence, after passing to possible subsequences, there exist T 0 θ 0 R 0, u W 1,0 ( T0, R ), and d W,1 ( T0, S ) such that u k u weakly in W 1,0 ( T0, R ), d k d weakly in W,1 ( T0, R ), lim k uk u L 4 ( T0 ) = 0, ( ) lim d k d L k 4 ( T0 ) + d k d L ( T0 ) and for any l, δ > 0, γ < β, and compact K, lim k (uk, d k ) (u, d) C l (K [δ,t 0 ]) = 0, lim k (uk, d k ) (u, d) C,1 γ ( [δ,t 0 ]) = 0. = 0, It is clear that (u, d) C ( (0, T 0 ], R S ) C,1 β ( (0, T 0], R S ) solves (1.1)-(1.3) in (0, T 0 ] and satisfies the boundary condition. It follows from (5.4) and (5.) that we can assume (u, d)(, t) (u 0, d 0 ) weakly in L () as t 0. In particular, On the other hand, (5.0) implies E(0) lim inf t 0 E(t). E(0) lim sup E(t). t 0 This implies (u, d)(, t) converges to (u 0, d 0 ) strongly in L (). Hence (u, d) satisfies the initial condition (1.4). Let T 1 (T 0, + ) be the first singular time of (u, d), i.e. (u, d) C ( (0, T 1 ), R S ) C,1 β ( (0, T 1), R S ), but (u, d) / C ( (0, T 1 ], R S ) C,1 β ( (0, T 1], R S ). Then we must have lim sup t T 1 max ( u + d )(, t) ɛ 1, R > 0. (5.7) x B R (x) 36

Now we look for an eternal extension of this weak solution in time. In order to do it, we need to define the new initial data at t = T 1. Claim 3. (u, d) C 0 ([0, T 1 ], L ()). In fact, for any φ H0 (, R3 ), (1.3) yields d t, φ = ( d φ + u d φ) d d φ d L () φ L () + ( u L () d L () + d L () ) φ C 0 () [ d L () + ( u L () d L () + d L () )] φ H (), where we have used the fact H0 () C0 () and φ C 0 () φ H (). Hence d t L ([0, T 1 ], H ()). This, combined with d L ([0, T 1 ], H 1 ()), implies d C 0 ([0, T 1 ], L ()). Similarly, for any φ H0 3(, R ) with φ = 0, (1.1) yields u t, φ = ( u φ + u u φ) d d : φ [ u L () φ L () + u L () u L () φ C 0 () + u L () φ C 0 ()] [ u L () + u L () u L () + u L () ] φ H 3 ()] where we have used the fact H 3 () C 1 () and φ C 1 () φ H 3 (). Since u t is divergence free, We have u t L ([0, T 1 ], H 3 ()). This, combined with u L ([0, T 1 ], H 1 ()), implies u C 0 ([0, T 1 ], L ()). By Claim 3, we can define (u(t 1 ), d(t 1 )) = lim t T1 (u(t), d(t)) in L (). By the energy inequality, we have that d L ([0, T 1 ], L ()). Hence d(t) d(t 1 ) weakly in L (). In particular, u(t 1 ) H and d(t 1 ) H 1 (). Moreover, since d(t) = d 0, d(t 1 ) = d 0 on. Now we can use (u(t 1 ), d(t 1 )) and (0, d 0 ) as initial and boundary data in the above procedure to obtain a continuation of (u, d) beyond T 1 as a weak solution of (1.1)-(1.5). At any further singular time, we repeat this procedure. We will prove that there are at most finitely many such singular times, afterwards we will have constructed an eternal weak solution. 37

We want to show that at any singular time there is at least a loss of energy amount of ɛ 1. By (5.7), there exist t i T 1 and x 0 such that This implies lim sup ( u + d )(, t i ) ɛ 1 for all R > 0. t i T 1 B R (x 0 ) ( u + d )(, T 1 ) = lim ( u + d )(, T 1 ) R 0 \B R (x 0 ) lim lim inf ( u + d )(, t i ) R 0 t i T 1 \B R (x 0 ) lim [lim inf ( u + d )(, t i ) lim sup R 0 t i T 1 t i T 1 lim inf ( u + d )(, t i ) ɛ 1 E 0 ɛ 1. t i T 1 B R (x 0 ) ( u + d )(, t i )] From this, we see that the number of finite singular times must be bounded by L = [ E 0 ], here [ ] denotes the largest integer part. If 0 < T ɛ L < + is the last 1 singular time, then we must have E(t L ) = ( u + d )(, T L ) < ɛ 1. Hence, if we use (u(t L ), d(t L )) and (0, d 0 ) as the initial and boundary data to construct a weak solution (u, d) to (1.1)-(1.3) as before, then (u, d) will be an eternal weak solution that we look for. It is clear that (i), (iii), and (ii) (1.7) of Theorem 1.3 has been established. Now, we want to perform the blow-up analysis at each singular time. It follows from (1.7) that there exist 0 < t 0 < T 1, t m T 1, r m 0 such that ɛ 1 = sup x,t 0 t t m B rm (x) ( u + d ). (5.8) By Lemma 5., there exist θ 0, depending only on ɛ 1 and E 0 and x m such that B rm (x m) 1 max x ( u + d )(t m θ 0 r m) B rm (x) ( u + d )(t m θ 0 rm) ɛ 1 4. (5.9) 38

By Lemma 4.1, (5.8), and the Ladyzhenskaya s inequality, we have Denote m [ t 0 t m, 0] (R, S ) by rm [t 0,t m] ( u 4 + d 4 ) C(ɛ 1, E 0 ). (5.30) = r 1 m ( \ {x m }). Define the blow-up sequence (u m, d m ) : m u m (x, t) = r m u(x m + r m x, t m + r mt), d m (x, t) = d(x m + r m x, t m + r mt). Then (u m, d m ) solves (1.1)-(1.3) on m [ t 0 t m, 0]. Moreover, rm ( u m + d m )( θ 0 ) ɛ 1, m B 1 (x) m B (0) ( u m + d m )(t) ɛ 1, x m, m [ t 0 tm rm,0] t 0 t m r m u m 4 + d m C(ɛ 1, E 0 ). Assume x m x 0, we divide into two cases. t 0, Case1. x 0. We can assume r m < dist(x 0, ) and m R. Also notice that t 0 t m r m. Hence, by Theorem 1., we can assume that there exists a smooth solution (u, d ) : R (, 0] R S to (1.1)-(1.3) such that (u m, d m ) (u, d ) in C loc (R [, 0]). We want to first show u 0. In fact, for any parabolic cylinder P R R [, 0], since u L 4 ( [0, T 1 ]), we have P R u 4 = lim m u m 4 = lim P m R B Rrm (x m) tm t m R r m u 4 = 0. Next we want to show d is a nontrivial, smooth harmonic map with finite energy. In fact, since ( d + d d) L ( [0, T 1 ]), we have, for any compact K R, 0 θ 0 K d + d d lim inf m = lim m tm 0 θ 0 t m θ 0 r m d m + d m d m m d + d d = 0. 39

This implies (d ) t + u d 0 on R [ θ 0, 0]. Hence (d ) t = u 0 and d C (R, S ) is a harmonic map. Since B d = lim m B ( u m + d m )( θ 0 ) ɛ 1 4, d is nontrivial. By the lower semicontinuity, we have for any ball B R R, d lim inf B m R d m ( θ 0 ) = lim inf B m R B rmr (x m) d (t m θ 0 r m) E 0 so that d has finite energy. It is well-known ([1], [3]) that d can be lifted to be a non constant harmonic map from S to S. In particular, d has nontrivial degree and R d 8π deg(d ) 8π. It follows from the above argument that for any r > 0, lim sup t T 1 max ( u + d )(t) ( u + d ) 8π. x B r(x) R Case. x 0. Then either (a) lim m x m x 0 r m =, or (b) lim m x m x 0 r m <. If (a) holds, then m R and the same argument as Case 1 yields that (u m, d m ) (0, d ) in C loc (R ), and d C (R, S ) is a nontrivial harmonic map with finite energy. Now we want to show that (b) can t happen. For, otherwise, assume that x m x 0 r m (0, a) for some a R and m R a = {(x 1, x ) R : x a}. Since d m (x) = d(x m + r m x) for x m, we can show, similar to Case 1, that u m 0 in C loc (R a), and d m d in C loc (R a), where d : R a S is a nontrivial harmonic map with finite energy and d (, a) = d(x 0 ) is a constant map. contradicts the nonexistence theorem by Lemaire [1]. Thus (ii) is established. This To show (iv). By Lemma 4.1, there exists t k + such that for (u k, d k ) = (u(, t k ), d(, t k )), u k + d k E 0, lim ( u k + d k + d k d k ) = 0. k 40

Since u k = 0, it is easy to see that u k 0 in H 1 (). d k is a H 1 -bounded sequence of approximate harmonic maps from into S, with d k = d 0 C,β ( ) and the L norm of their tension fields τ(d k ) = d k + d k d k converging to zero. By the energy identity result by Qing [18] and Lin-Wang [16], we can conclude that there exist a harmonic map d C,β (, S ) with d = d 0 on and finitely many points {x i } l i=1, {m i} l i=1 N such that This yields (iv). To show (v). First we claim d k dx d dx + (a) There exist no finite time singularities. l 8πm i δ xi. i=1 For, otherwise, (ii) implies that we can blow up near the first singular time T 1 to obtain one nontrivial harmonic map ω C (R, S ) and 8π ω lim R t T1 This, combined with Lemma 4.1, yields T1 0 ( u + d )(t) ( u + d + d d ) = 0 ( u 0 + d 0 ) 8π. so that u = d t 0 in [0, T 1 ] and hence d(, t) = d 0 C,β (, S ), 0 t T 1, is a harmonic map. This contradicts the fact that T 1 is a singular time. (b) φ(t) max x,τ t ( u + d )(x, τ) remains bounded as t +. For, otherwise, there exist t k + and x k such that λ k = φ(t k ) = ( u + d )(x k, t k ) +. Define k = λ k ( \ {x k }) and (u k, d k ) : k [ t k λ k, 0] R S by u k (x, t) = λ 1 k u(x k + λ 1 k x, t k + λ k t), d k(x, t) = d(x k + λ 1 k Then (u k, d k ) solves (1.1)-(1.3) on k [ t k λ k, 0], and x, t k + λ k t). 1 = ( u k + d k )(0, 0) ( u k + d k )(x, t), (x, t) k [ t k λ k, 0]. 41