Motion of vortex-filaments for superconductivity

Nonlinear Analysis 69 (2008) 4412–4442 www.elsevier.com/locate/na

Motion of vortex-filaments for superconductivity Zuhan Liu ∗ Department of Mathematics, Xuzhou Normal University, Xuzhou 221116, China Department of Mathematics, Yangzhou University, Yangzhou 225002, China Received 1 May 2007; accepted 29 October 2007

Abstract In this paper, we study the asymptotic behavior of solutions to the simplified Ginzburg–Landau model for superconductivity. We prove that, asymptotically, vortex-filaments evolves according to the mean curvature flow in the sense of weak formulation. This can be seen as a first attempt to understand the nature of the motion of vortex filaments in three dimensions with magnetic field. On the other hand, this paper revisits the pioneering work of Bethuel–Orlandi–Smets [F. Bethuel, G. Orlandi, D. Smets, Convergence of the parabolic Ginzburg–Landau equation to motion by mean curvature, Ann. of Math. 163 (2006) 37–163] in a slightly relaxed setting. c 2007 Elsevier Ltd. All rights reserved.

MSC: 35B40; 35K55; 35Q40 Keywords: Vortices; Mean curvature flow; Superconductivity; Geometric measure theory

1. Introduction The quantized vortex phenomena are well-known signatures of superfluidity that appear in superfluid Helium, superconductors and Bose–Einstein condensates. The nucleation, motion and pinning of vortices in superconductors play an important role in the theoretical studies and the practical applications of superconductivity. The study of Abrikosov on the vortex lattices based on the Ginzburg–Landau model was a Nobel prize winning work that has become the highlight of the impact of the Ginzburg–Landau model on the study of superconductivity and quantum superfluidity. In a suitably non-dimensionalized form, the primary variables used in the Ginzburg–Landau model [22] are the complex scalar-valued order parameter u, the real vector-valued magnetic potential A, and the real scalarvalued electric potential Ψ . After a rescaling, the reduced model equation that retains the basic features of the vortex state satisfied by u can be formulated as [5]  uε  ∂u ε + iΨε u ε = (∇ − iAε )2 u ε + 2 (1 − |u ε |2 ) on R3 × (0, T 0 ), (1.1) ∂t ε u (x, 0) = u 0 (x) for x ∈ R3 , ε

ε

where T 0 is a positive constant, Ψε (x) is a prescribed reduced scalar electric potential such that ∇Ψε represents an applied electric current. Aε (x) is a prescribed reduced vector magnetic potential such that curl Aε = h(ε) represents an applied magnetic field and div Aε = 0 in R3 , A(x) → 0 as |x| → ∞. κ = 1ε is a dimensionless constant ∗ Corresponding address: Department of Mathematics, Xuzhou Normal University, Xuzhou 221116, China.

E-mail address: [email protected]. c 2007 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2007.10.062

Z. Liu / Nonlinear Analysis 69 (2008) 4412–4442

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(the Ginzburg–Landau parameter) depending only on characteristic lengths of the material and of temperature. The complex order parameter u indicates the local state of the material. |u|2 represents the density of superconducting electron pairs, so that, |u| ' 1 corresponds to the superconducting state and |u| ' 0 corresponds to the normal state. We know that a superconductor placed in an applied magnetic field may change its phases when the field varies. When the lower critical value Hc1 is reached, there is a phase-transition from the superconducting state to the “mixedstate”, where vortices appear (zero-set of u). This motion of vortices generates an electric field and leads to currentdissipation. On the other hand, energy concentrates and blows up like the order |ln ε| on the vortex set and hence singularities happen. Define the energy of the vortices Z Z |∇u ε |2 + Vε (u ε ), (1.2) Eε (u ε ) = eε (u ε ) = 2 R3 R3 where Vε denotes the potential Vε (u ε ) = 4ε12 (1−ε 2 |Aε |2 −|u ε |2 )2 . Roughly speaking, Eε (u ε )/|ln ε| is the total length 1 with finite energy E (v 0 ). of vortex curves [9]. It is well known that (1.1) is well posed for initial data in Hloc ε ε By the standard theory of parabolic equations, we have the following energy estimate Z tZ ∂u ε 2 0 0 (1.3) ∂t + Eε (u ε (·, t)) ≤ C1 [Eε (u ε ) + C2 ] for any t ∈ [0, T ), 3 0 R where C1 and C2 are constants depending on T 0 , kΨε k L 2 and kAε k L ∞ . We assume that the initial condition u 0ε satisfies the bound (with finite length initial vortex filaments), Eε (u 0ε ) ≤ M0 |ln ε|,

(1.4)

where M0 is a fixed positive constant. The main theorem of this paper examines the asymptotic limits of the Radon measures µε defined by µε (x, t) =

eε (u ε (x, t)) dxdt, |ln ε|

(1.5)

and of their time slices µtε defined on R3 × {t} by µtε (x) =

eε (u ε (x, t)) dx, |ln ε|

(1.6)

so that µε = µtε dt. In view of (1.3) and (1.4), we may assume, up to a subsequence εn → 0, that there exists a Radon measure µ∗ defined on R3 × [0, T 0 ) such that µε * µ∗

in the sense of measures.

Passing possibly to a further subsequence, we may also assume (see Lemma 4.1) that µtε * µt∗

in the sense of measures on R3 × {t}, for all t ∈ [0, T 0 ).

We have the following theorem. Theorem 1.1. Assume that kΨε k L 2 + kΨε k L ∞ ≤ C3 , kh(ε)k H 1 (R3 ) ≤ C3 and (1.4) holds. Then, there exist a subset Σµ in R3 × (0, T 0 ), and a smooth real-valued function Φ∗ defined on R3 × (0, T 0 ) such that the following properties hold. (1) Σµ is closed in R3 × (0, T 0 ) and for any compact subset K ⊂ R3 × (0, T 0 ) \ Σµ |u ε (x, t)| → 1 (2) Set

Σµt

= Σµ

uniformly on K as ε → 0. ∩ (R3

× {t}), for any t ∈

(0, T 0 ),

(1.7) then

H1 (Σµt ) ≤ C, where C is a positive constant depending on

(1.8) T 0,

M0 , Ψε and Aε .

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(3) The function Φ∗ satisfies ∂ Φ∗ − 4Φ∗ = 0 ∂t

on R3 × (0, T 0 ).

(1.9)

(4) For any t ∈ (0, T 0 ), µt∗ can be decomposed as 1 |∇Φ∗ |2 H3 + Θ∗ (x, t)H1 bΣµt , (1.10) 2 where Θ∗ (·, t) is a bounded function. 0 t (5) There exists a positive function η defined on R+ ∗ such that, for almost every t ∈ (0, T ), the set Σµ is 1-rectifiable and µt∗ =

µt∗ (B(x, t)) ≥ η(t), r →0 ω1r

Θ∗ (x, t) = Θ1 (µt∗ , x) = lim for H1 a.e. x ∈ Σµt .

(1.11)



Define ν∗t = Θ∗ (x, t)H1 bΣµt .

(1.12)

For the evolution of the concentrated part ν∗t (i.e. limiting vortex-filaments), we have Theorem 1.2. The {ν∗t }t∈(0,T 0 ) is a mean curvature flow in the sense of Brakke [3]. Remark 1.1. Theorems 1.1 and 1.2 is identical to [2]. In this small h(ε) = O(1) scaling, the vector potential has no effect on the motion law for the filament. From [3], there exists a T ∗ > 0 such that Σµt = ∅, ∀t > T ∗ , which implies the complete annihilation of vorticity in dimension 3. This result does not hold in dimension 2. This fact is related to the so-called “slow motion of vortices” as established in [12,20]: vortices essentially move with a speed of order |ln ε|−1 . Therefore, a time of order |ln ε| is necessary to annihilate vorticity. Eq. (1.1) without Aε , Ψε has been considered in recent years. In particular, the dynamics of vortices has been described in the two-dimensional case by Lin [12] and Jerrard–Soner [8]. Concerning higher dimensions N ≥ 3, under the assumption that the initial measure is concentrated on a smooth manifold, the motion of vortex-line has been obtained by Lin [14] and Jerrard–Soner [10], in the time interval where the classical solution exists, that is only before the appearance of singularities. First convergence result past the singularities has been obtained by Ambrosio–Soner [1], under the additional density assumption for the measure µt∗ . Later, without this additional assumption, some important asymptotic properties for solutions of (1.1) in dimensions N ≤ 4 were considered by Lin–Riviere [15] and by Wang [23]. Very recently, in higher dimensions case, an important convergence result past the singularities has been obtained by Bethuel, Orlandi and Smets [2] without the additional condition in [1]. The work of [2] unifies several hard, deep results of Jerrard–Soner [8], Lin–Riviere [15], and Ambrosio–Soner [1], along with some new results that in total constitutes a tour-de-force of applied analysis. For the case with Aε , Ψε , the dynamics of vortices has been described in the two-dimensional case by Spirn [20] and by Sandier and Serfaty [19]. In this paper, we study the convergence result past the singularities in a threedimensional case. Here we use the approach of a recent important work of Bethuel, Orlandi and Smets [2]. The work in [2] presents new ideas and interesting points of departure for new research. We also use the arguments of geometric measure theory developed in [1]. The first important step in our proof is the monotonicity formula and the Clear-out Lemma, which bounds |u ε | away from zero as soon as the local energy is bounded by η|ln ε| with η small. The second important step is the uniform estimates away from vortices. When using the approach in [2], we need to analyze the effect of Ψε , Aε . In this paper, we cannot deal with the large applied magnetic h(ε) = O(|ln ε|) case. This can be seen as a first attempt to understand the nature of the motion of vortex filaments in three dimensions with magnetic field. Finally, we mention the recent results on dynamics of vortices for other Ginzburg–Landau type equations without magnetic field in [4,7,11,13,16–18].

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We organize the paper as follows: In Section 2, we derive the monotonicity formula of energy and Clear-out Lemma. In Section 3, we derive some improved estimates. In Section 4, we study the defect measures and prove the main theorems. 2. Monotonicity formula and energy estimates 2.1. Pointwise estimates In this subsection, first we give the following pointwise estimates by the maximum principle. Lemma 2.1. Let u ε be solution of (1.1) with Eε (u 0ε ) < +∞. Then there exists a constant K > 0 depending only on Aε , Ψε such that, for t ≥ ε 2 and x ∈ R3 , we have ∂u ε K ≤ K (x, t) |u ε (x, t)| ≤ 3, |∇u ε (x, t)| ≤ , (2.1) ε2 ε ∂t and 

t |u ε (x, t)| ≤ 1 + K exp − 2 2ε 2



for t ≥ ε 2 .

(2.2)

Moreover, if for some C4 ≥ 1 such that |u 0ε (x)| ≤ C4 ,

|∇u 0ε (x)| ≤

C4 , ε

|D 2 u 0ε (x)| ≤

C4 , ε2

∀x ∈ R3 .

(2.3)

Then, for any x ∈ R3 and t > 0, |u ε (x, t)| ≤ 3,

K |∇u ε (x, t)| ≤ , ε

∂u ε K ∂t (x, t) ≤ ε 2 ,

(2.4)

where K depends only on C3 , C4 . Next result provides a local lower bound on |u ε |, when we know it is away from zero on some region. Let Λα (x0 , T, R, 4T ) = B(x0 , α R) × [T + (1 − α 2 )4T, T + 4T ] ⊂ R3 × (0, T 0 ). Sometimes, 4T = R and write Λα (x0 , T, R). If this is not misleading we will simply write Λα and Λ if α = 1. Lemma 2.2. Let u ε be solution of (1.1) with Eε (u 0ε ) < +∞. Let x0 ∈ R3 , R > 0, T > 0 and 4T > 0 be given. Assume that |u ε | ≥

1 2

on Λ(x0 , T, R, 4T ),

then |1 − |u ε || ≤ C(α, Λ)ε2 (k∇ϕε k2L ∞ + kAε k2L ∞ + |ln ε|)

on Λα ,

(2.5)

where ϕε is defined on Λ, up to a multiple of 2π , by u ε = |u ε | exp(iϕε ). Proof. Without loss of generality, we assume that T ≥ ε, otherwise we consider a smaller cylinder. Let ρ = |u ε |, ζ = 1 − ρ. Then, ζt − 4ζ +

1 1 ζ = (1 − ζ )|∇ϕε |2 − 2 ζ (1 − ζ )2 + (ζ − 1)Aε · ∇ϕε + |Aε |2 (1 − ζ ). ε2 ε

By Lemma 2.1, we have ζ ≥ −K exp(− 1ε ), then   1 1 2 2 −2 ζt − 4ζ + 2 ζ ≤ 2|∇ϕε | + 6|Aε | + K ε exp − . ε ε

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Let χ be a cut-off function defined on R3 such that 0 ≤ χ ≤ 1 and χ ≡ 1 on B(x0 , α R), χ ≡ 0 on R3 \ B(x0 , 1+α 2 R). Define   1 1 t−T 2 τ (t) = − exp ln ε , ∀t ∈ [T, T + 4T ], 2 2 (1 − α 2 )4T 1 σε (x, t) = − τ (t)χ (x). 2 Then σε ≥ 0 and 1 |∂t σε | = |τ 0 (t)|χ (x) ≤ |ln ε|, (1 − α 2 )4T |4σε | ≤ τ (t)|4χ (x)| ≤ C(Λ), |∇σε | ≤ τ (t)|∇χ (x)| ≤ C(Λ). Hence, ∂σε 1 − 4σε + 2 σε ≥ −C(Λ)|ln ε|. ∂t ε Let σ = σε + 2ε 2 (k∇ϕε k2L ∞ (Λ) + 3kAε k2L ∞ (Λ) + C(Λ)|ln ε|), then ∂ζ ∂σ 1 1 − 4σ + 2 σ ≥ 2[k∇ϕε k2L ∞ (Λ) + 3kAε k2L ∞ (Λ) + C(Λ)|ln ε|] ≥ − 4ζ + 2 ζ. ∂t ∂t ε ε Note that 1 σ ≥ ≥ ζ on B(x0 , R) × {T } ∪ ∂ B(x0 , R) × [T, T + 4T ], 2 by the maximum principle, we obtain ζ ≤ σ on Λ. Since χ ≡ 1 on B(x0 , α R), we have σ (x, t) ≤ 2ε 2 (k∇ϕε k2L ∞ (Λ) + 3kAε k2L ∞ (Λ) + C(Λ)|ln ε|) which imply the conclusion.



By the standard elliptic equation theory, we have the following lemma. Lemma 2.3. Assume that kh(ε)k H 1 (R3 ) ≤ C, then we have kAε kC 1 ≤ C.

(2.6)

2.2. The monotonicity formula Let u ≡ u ε be a solution to (1.1) satisfying (1.4). For simplicity, we√will drop the subscript when this is not misleading. For (x∗ , t∗ ) ∈ R3 × (0, T 0 ) we set z ∗ = (x∗ , t∗ ). For 0 < R ≤ t∗ we define   Z |x − x∗ |2 E ω (z ∗ , R) ≡ E ω,ε (u; z ∗ , R) = dx, (2.7) eε (u(x, t∗ − R 2 )) exp − 4R 2 R3 and the corresponding scaled energy   Z 1 1 |x − x∗ |2 2 e E ω (z ∗ , R) = E ω (z ∗ , R) = eε (u(x, t∗ − R )) exp − dx. R R R3 4R 2 Proposition 2.1 (Monotonicity Formula). We have d 2 e [exp(K R)( E ω (z ∗ , R) + R + Rε)] ≥ 0, dR R=r where K depends only on kAε kC 1 , kΨε k L ∞ .

(2.8)

(2.9)

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Proof. To simplify the presentation, we take x∗ = 0, t∗ = 0, assuming that (1.1) is solved for some negative t. So we eω (R) ≡ E eω (z ∗ , R). We consider the following change of variables, for z = (x, t): will have t < 0 below. Set E z = (x, t) = (Ry, R 2 τ ) = Φ R (y, τ ) = Φ R (z 0 ). Set u R (z 0 ) = u(Ry, R 2 τ ) = u(x, t),

A R (y) = Aε (x),

Ψ R (y) = Ψε (x),

b R (y) = bε (x).

Then ∂u R 0 ∂u (z ) = R 2 (z), 4u R (z 0 ) = R 2 4u(z), ∂τ ∂t ∂u R R2 + iΨ R u R − 4u R + 2iR(A R · ∇)u R = 2 (b R − |u R |2 )u R , ∂τ ε     Z 2 |y|2 R 1 2 2 2 e E ω (R) = |∇u R (y, −1)| + 2 (b R − |u R (y, −1)| ) exp − dy. 4 4ε R3 ×{−1} 2

∇u R (z 0 ) = R∇u(z),

We compute    Z eω (R) du R dE ∇u r · ∇ = dR dR r R3 ×{−1} r      r2 du R dA R |y|2 2 − 2 (br − |u r |2 ) 2u r + ε A + 2r V (u ) exp − dy r ε r dR r dR r 4 ε "  #    Z 1 ∂u r 2 |y|2 = y · ∇u r − 2 + 2r Vε (u r ) r −3 exp − dy ∂τ 4 R3 ×{−1} 2r      Z 1 ∂u r |y|2 (ir 2 Ψr u r + 2ir (Ar · ∇)u r ) + y · ∇u r − 2 exp − dy r ∂τ 4 R3 ×{−1}   Z |y|2 + [r 2 (br − |u r |2 )Ar (y · ∇ Aε )] exp − dy. 4 R3 ×{−1} Note that, kuk L ∞ + kΨε k L ∞ ≤ C and (2.6), we have Z      1 ∂u r |y|2 2 (ir Ψ u + 2ir (A · ∇)u ) y · ∇u − 2 exp − dy r r r r r 3 r ∂τ 4 R ×{−1} # "   2  Z |y|2 1 1 ∂u r + 2r Vε (u r ) r −3 exp − y · ∇u r − 2 dy ≤ 8 R3 ×{−1} 2r ∂τ 4   Z h i |y|2 dy + Cr |Ar |2 |∇u r |2 + r 3 |Ψr |2 |u r |2 exp − 4 R3 ×{−1} "  #    Z 1 1 ∂u r 2 |y|2 −3 eω (r ) + Cr 3 , ≤ y · ∇u r − 2 + 2r Vε (u r ) r exp − dy + Cr E 8 R3 ×{−1} 2r ∂τ 4 Z     Z h i 1 |y|2 |y|2 2 2 r (br − |u r | )Ar (y · ∇ Aε ) exp − dy ≤ 2r Vε (u r ) exp − dy 3 4 8 3 4

(2.10)

(2.11)

R ×{−1}

R ×{−1}

+ Cr ε. 2

(2.12)

Combining (2.10) and (2.11) with (2.12), we have "  #    Z eω (R) dE 1 2 ∂u 2 |x|2 x · ∇u + 2t + 2r Vε (u) r −3 exp − 2 dx ≥ dR 4 R3 ×{−r 2 } r ∂t 4r r

eω (r ) − C(r 3 + r 2 ε). − Cr E

(2.13)

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Hence, d eω (R) + R 2 + Rε)) |r ≥ 0, (exp(K R)( E dR which yields (2.8).  2.3. The energy estimates Lemma 2.4. For 0 < T1 ≤ T2 < T , we have     Z T2 Z |x − x T |2 |x − x T |2 exp − dxdt eε (u) 1 + 4(T − t) 4(T − t) T1 R3 p p eω (z T , T − T1 ) − (T − T2 )1/2 E eω (z T , T − T2 ) ≤ (T − T1 )1/2 E   Z T2 Z 1 |x − x T |2 2 + [(x − x T ) · ∇u − 2(T − t)∂t u] exp − dxdt 4(T − t) T1 R3 2(T − t)   Z T2 Z |x − x T |2 − dxdt. 2i(T − t)[Ψε u + (Aε · ∇)u]∂t u exp − 4(T − t) T1 R3

(2.14)

2

3 T| Proof. Multiplying (1.1) by 2(T − t)∂t u exp(− |x−x 4(T −t) ) and integrate on R × [T1 , T2 ]. By integrating by parts in the space variable, we have   Z T2 Z |x − x T |2 2 dxdt 2(T − t)|∂t u| exp − 4(T − t) T1 R3     Z T2 Z |x − x T |2 ∂ 1 = − (T − t) |∇u|2 + 2 (bε − |u|2 )2 exp − dxdt ∂t 4(T − t) 2ε T1 R3   Z T2 Z |x − x T |2 + ((x − x T ) · ∇u)∂t u exp − dxdt 4(T − t) T1 R3   Z T2 Z |x − x T |2 − dxdt. 2i(T − t)[Ψε u + (Aε · ∇)u]∂t u exp − 4(T − t) T1 R3

Integration by parts in time variable yields   Z T2 Z |x − x T |2 dxdt 2(T − t)|∂t u|2 exp − 4(T − t) T1 R3    Z T2 Z  1 |x − x T |2 2 2 2 = − |∇u| + 2 (bε − |u| ) exp − dxdt 4(T − t) 2ε T1 R3 Z T2 Z |x − x T |2 1 |x − x T |2 − [|∇u|2 + 2 (bε − |u|2 )2 ] exp(− )dxdt 4(T − t) 2ε T1 R3 4(T − t)     Z 1 |x − x T |2 2 2 2 + (T − T1 ) |∇u| + 2 (bε − |u| ) exp − dx 4(T − t) 2ε R3 ×{T1 }     Z 1 |x − x T |2 − (T − T2 ) |∇u|2 + 2 (bε − |u|2 )2 exp − dx 4(T − t) 2ε R3 ×{T2 }   Z T2 Z |x − x T |2 + [((x − x T ) · ∇u)∂t u − 2i(T − t)(Ψε u + (Aε · ∇)u)∂t u] exp − dxdt. 4(T − t) T1 R3 Adding the integral   Z T2 Z 1 |x − x T |2 2 |(x − x T ) · ∇u| exp − dxdt 4(T − t) T1 R3 2(T − t)

(2.15)

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to (2.15) we obtain Z T2 Z

  1 |x − x T |2 2 dxdt |(x − x T ) · ∇u − 2(T − t)∂t u| exp − 4(T − t) R3 2(T − t) T1     Z T2  |x − x T |2 1 |x − x T |2 |∇u|2 + 2 (bε − |u|2 )2 exp − dxdt + 1+ 4(T − t) 4(T − t) 2ε T1     Z |x − x T |2 1 2 2 2 dx = (T − T1 ) |∇u| + 2 (bε − |u| ) exp − 4(T − t) 2ε R3 ×{T1 }     Z |x − x T |2 1 dx − (T − T2 ) |∇u|2 + 2 (bε − |u|2 )2 exp − 4(T − t) 2ε R3 ×{T2 }   Z T2 Z |x − x T |2 1 2 |(x − x T ) · ∇u| exp − dxdt + 4(T − t) T1 R3 2(T − t)   Z T2 Z |x − x T |2 − dxdt ((x − x T ) · ∇u)∂t u exp − 4(T − t) T1 R3   Z T2 Z |x − x T |2 − 2i(T − t)[Ψε u + (Aε · ∇)u]∂t u exp − dxdt. 4(T − t) T1 R3

Using ab ≤ a=

a2 4

(2.16)

+ b2 , with

  (x − x T ) · ∇u |x − x T |2 exp − , √ 8(T − t) 2(T − t)

b=

  [(x − x T ) · ∇u − 2(T − t)∂t u] |x − x T |2 exp − , √ 8(T − t) 2(T − t)

we bound the third and the fourth term on the right-hand side of (2.15), and the conclusion follows.



Lemma 2.5. For any t ∈ (0, T ), we have     Z |x − x T |2 |x − x T |2 eε (u) 1 + exp − dx 4(T − t) 4(T − t) R3 ×{t}   Z 1 |x − x T |2 2 ≤ [(x − x T ) · ∇u − 2(T − t)∂t u] exp − dx 4(T − t) R3 ×{t} 2(T − t)   Z |x − x T |2 − 2i(T − t)[Ψε u + (Aε · ∇)u]∂t u exp − dx + g 0 (T − t), 4(T − t) R3 ×{t} eω (z T , √s), where g(s) := s 3/2 E g 0 (T − t) =

(2.17)

√ √ 1 3 eω (z T , T − t) + 1 (T − t) d E eω (z T , T − t). (T − t) 2 E 2 2 ds

Proof. Let T1 = t, T2 = t + 4t, for 4t > 0. We divide by 4t and let 4t → 0 in (2.14). This yields the conclusion.  Proposition 2.2. Let 0 < t < T < T 0 . The following inequality holds, for any x T ∈ R3 ,     Z Z |x − x T |2 |x − x T |2 eε (u) exp − dx ≤ 2 eε (u) exp − dx 4(T − t) 4(T − t) R3 ×{t} A×{t}     Z 1 1 |x − x T |2 2 2 2 [(x − x ) · ∇u − 2(T − t)∂ u] exp − dx +K (b − |u| ) + T t 2 ε 2(T − t) 4(T − t) R3 ×{t} 2ε + K [(T − t)3/2 + (T − t)ε], √ where A = B(x T , r T ), r T = K (1 + T − t)(T − t), K depends only on kAε k L ∞ , kΨε k L 2 .

(2.18)

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Proof. By Lemma 2.5, we have     Z |x − x T |2 |x − x T |2 exp − dx eε (u) 1 + 4(T − t) 4(T − t) R3 ×{t}   Z 1 |x − x T |2 ≤ [(x − x T ) · ∇u − 2(T − t)∂t u]2 exp − dx 4(T − t) R3 ×{t} 2(T − t)   Z |x − x T |2 dx + 2(T − t)[iΨε u + (iAε · ∇)u]∂t u exp − 4(T − t) R3 ×{t} √ √ 3 eω (z T , T − t) + 1 (T − t) d E eω (z T , T − t). + (T − t)1/2 E 2 2 dR Note that Z

(2.19)

 |x − x T |2 (T − t)[iψε u + 2i(Aε · ∇)u]∂t u exp − dx 3 4(T − t) R ×{t} Z  1 = − (iψε u + 2i(Aε · ∇)u)((x − x T ) · ∇u − 2(T − t)∂t u) 2 R3 ×{t}    |x − x T |2 1 + (iψε u + 2i(Aε · ∇)u)(x − x T ) · ∇u exp − dx 2 4(T − t)   Z 2 |x − x T | (T − t)(1 + |∇u|2 ) exp − ≤ K dx 3 4(T − t) R ×{t}     Z 1 |x − x T |2 1 + [(x − x T ) · ∇u − 2(T − t)∂t u]2 + 2 (bε − |u|2 )2 exp − dx 4(T − t) 2ε R3 ×{t} 2(T − t)   Z 1 |x − x T |2 |x − x T |2 + |∇u|2 exp − dx, (2.20) 4 R3 ×{t} 4(T − t) 4(T − t) 

where K depends only on kAε k L ∞ , kΨε k L 2 . On the other hand, by (2.10)–(2.12), we have   Z √ d e 1 |x − x T |2 E ω (z T , T − t) ≤ K [(x − x T ) · ∇u − 2(T − t)∂t u]2 exp − dx dR 4(T − t) R3 ×{t} 2(T − t)   Z |x − x T |2 eε (u) exp − dx + K [(T − t)3/2 + (T − t)ε]. + K (T − t + 1) 4(T − t) R3 ×{t}

(2.21)

Inserting (2.19) and (2.20) into (2.18), we obtain   Z |x − x T |2 |x − x T |2 eε (u) exp − dx 4(T − t) 4(T − t) R3 ×{t}   Z |x − x T |2 ≤ K 1 (T − t + 1) eε (u) exp − dx + K [(T − t)3/2 + (T − t)ε] 4(T − t) R3 ×{t}     Z 1 1 |x − x T |2 +K [(x − x T ) · ∇u − 2(T − t)∂t u]2 + 2 (bε − |u|2 )2 exp − dx, 4(T − t) 2ε R3 ×{t} 2(T − t) (2.22) where K , K 1 depend only on Aε , Ψε . Consider the region   2 K1 3 |x − x T | A := x ∈ R : ≤ (1 + T − t) . 8(T − t) 2

Z. Liu / Nonlinear Analysis 69 (2008) 4412–4442

4421

From (2.21), we deduce that     Z Z |x − x T |2 |x − x T |2 |x − x T |2 eε (u) dx ≤ K 1 (1 + T − t) eε (u) exp − dx exp − 4(T − t) 4(T − t) 4(T − t) R3 ×{t} A×{t}   Z |x − x T |2 |x − x T |2 + eε (u) exp − dx + K [(T − t)3/2 + (T − t)ε] 8(T − t) 4(T − t) (R3 \A)×{t}     Z 1 |x − x T |2 1 2 2 2 +K dx [(x − x T ) · ∇u − 2(T − t)∂t u] + 2 (bε − |u| ) exp − 4(T − t) 2ε R3 ×{t} 2(T − t) (2.23) which imply   |x − x T |2 dx eε (u) exp − 4(T − t) (R3 \A)×{t}   Z |x − x T |2 eε (u) exp − dx + K [(T − t)3/2 + (T − t)ε] ≤ K 1 (1 + T − t) 4(T − t) A×{t}     Z 1 1 |x − x T |2 +K [(x − x T ) · ∇u − 2(T − t)∂t u]2 + 2 (bε − |u|2 )2 exp − dx. 4(T − t) 2ε R3 ×{t} 2(T − t) (2.24)

K 1 (1 + T − t)

Z

Combining (2.22) with (2.23), the conclusion follows.  √ Proposition 2.3. Let T > 0, x T ∈ R3 and R > 2ε. Assume that u ε is a solution to (1.1) satisfying (1.4). Then, for any λ > 0,   Z Z |x − x T |2 eε (u)dx eε (u) exp − dx ≤ 4R 2 B(x T ,λR)×{T } R3 ×{T } √   p √ λ 2R +√ exp(K T ) exp − (2.25) [M0 | ln ε| + T + 2R 2 + T + 2R 2 ε], 8 T + 2R 2 where K depends only on Aε , Ψε . Proof. Note that    2   |x − x T |2 λ |x − x T |2 exp − ≤ exp − exp − , 8 4R 2 8R 2

∀x ∈ R3 \ B(x T , λR)

for λ > 0. On the other hand, using the monotonicity formula at the point (x T , T + 2R 2 ), we have √   Z exp(K 2R) |x − x T |2 eε (u) exp − dx √ 8R 2 2R R3 ×{T } √ Z    p exp(K T + 2R 2 ) |x − x T |2 2 2ε ≤ eε (u) exp − dx + T + 2R + T + 2R √ 4(T + 2R 2 ) T + 2R 2 R3 ×{0} √ p exp(K T + 2R 2 ) ≤ [M0 |ln ε| + T + 2R 2 + T + 2R 2 ε]. √ T + 2R 2 Hence,   2Z    Z λ |x − x T |2 |x − x T |2 eε (u) exp − dx ≤ exp − eε (u) exp − dx 8 4R 2 8R 2 {|x−x T |≥λR}×{T } R3 ×{T } √  2 p √ 2R λ ≤ √ exp(K T ) exp − [M0 |ln ε| + T + 2R 2 + T + 2R 2 ε], 8 T + 2R 2 the conclusion follows. 

(2.26)

(2.27)

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Using Lemma 2.1 and Proposition 2.1, the same argument of Theorem 1.1 in [2] gives the following Clear-out Lemma, which bounds |u ε | away from zero as soon as the local energy is bounded by η|ln ε| with η small. Theorem 2.1 (Clearing-out Lemma). Let 0 < ε < 1, u ε be a solution of (1.1) with Eε (u 0ε ) < +∞, and σ > 0 be given. Then there exist ε0 = ε0 (σ, Aε , Ψε ) > 0 and η = η(σ, Aε , Ψε ) > 0 such that if ε ≤ ε0   Z |x|2 0 eε (u ε ) exp − dx ≤ η|ln ε| (2.28) 4 R3 then |u ε (0, 1)| ≥ 1 − σ.

(2.29)

3. Uniformly estimates away from the vortex set In this section we will give the uniformly estimate away from the vortex set. First of all, we give the following result, which is immediate from Theorem 2.1. √ Proposition 3.1. Let T > 0, x T ∈ R3 , and set z T = (x T , T ). Assume u ε is a solution to (1.1) and let R > 2ε and σ > 0 be given. There exists η1 depending only on σ, Aε , Ψε such that if eω,ε (z T , R) ≤ η1 |ln ε|, E

(3.1)

|u ε (x T , T + R 2 )| ≥ 1 − σ.

(3.2)

then

Proposition 3.2. Let u ε be a solution of (1.1) satisfying (1.4) and σ > 0 be given. Let x T ∈ R3 , T 0 > T > 0 and √ R > 2ε. There exist a positive continuous function λ defined on R+ ∗ and a small positive ε0 such that, if Z 1 η1 η(x ˇ T , T, R) ≡ eε (u ε (·, T )) ≤ R|ln ε| B(x T ,λ(T )R) 2 then |u ε (x, t)| ≥ 1 − σ

  R for t ∈ [T + T0 , T + T1 ], x ∈ B x T , , 0 < ε < ε0 , 2

where T0 and T1 are defined by  2 ! 2ηˇ T0 = max 2ε, R2 , η1 (σ, R)

T1 = R 2 .

Proof. Let x0 be any given points in B(x T , R2 ). We claim that we can find λ(T ) > 0 such that p p eω,ε ((x0 , T ), r ) ≤ η1 |ln ε|, for T0 < r < T1 = R E provided ηˇ ≤

(3.3)

η1 2.

Indeed, by Proposition 2.3, we have √ Z p √ 1 2K e E ω,ε ((x0 , T ), r ) ≤ eε (u ε ) + √ exp(K ( T + 2r 2 − 2r )) 2 r B(x0 ,λr ) T + 2r  2 p λ × exp − [|ln ε| + T + r 2 + T + r 2 ε]. 8 √ We choose ε0 such that T + r 2 + T + r 2 ε ≤ 2(T 0 + T 0 2 ) ≤ |ln ε0 |, then, choose λε (T ) such that  1    2  1  2 √ λ 2 2 λ η1 2 2 KT K exp √ M0 exp − ≤K exp(K T )M0 exp − ≤ . √ 2 2 T 8 T 8 2 T + 2r + 2r

(3.4)

(3.5)

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Set λ(T ) = max(2, 2λε (T )). Since x0 ∈ B(x T , R2 ) and r < R, it follows that B(x0 , λε (T )r ) ⊂ B(x T , λ(T )R). Therefore, Z Z R 1 R 1 ˇ ε|. eε (u ε ) ≤ eε (u ε ) ≤ √ η|ln r B(x0 ,λε (T )r )×{T } r R B(x T ,λ(T )R)×{T } T0 Choosing T0 of the form T0 = C ηˇ 2 R 2 , we obtain Z 1 1 eε (u ε ) ≤ C − 2 |ln ε|. r B(x0 ,λε (T )r )×{T }

(3.6)

Now we fix the constant C as  η −2 1 C= . 2 Combining (3.4) and (3.5) with (3.6) we have (3.3). The conclusion then follows from Proposition 3.1.



Theorem 3.1. Let B(x0 , R) be a ball in R3 and T > 0, 4T > 0 be given. Consider the cylinder Λ = B(x0 , R) × [T, T + 4T ]. There exist 0 < σ ≤ 21 , s > 0 and C(Λ) depending only on Λ, Aε , Ψε , such that |u ε | ≥ 1 − σ

on Λ,

(3.7)

then eε (u ε )(x, t) ≤ C(Λ)

Z Λ

for any (x, t) ∈ Λ 1 = B(x0 , 2

(eε (u ε ) + |Ψε | ) + M0 ε

R0 2 ) × [T

2

+

4T 4

s

 (3.8)

, T + 4T ]. Moreover,

1 |∇Φε |2 + Kε in Λ 1 , 2 2 where the functions Φε and Kε are defined on Λ 1 and satisfy eε (u ε )(x, t) =

(3.9)

2

∂Φε − 4Φε + (Aε · ∇)|u ε |2 + Ψε = 0 ∂t

in Λ 1 ,

k|Kε | + |∇|u ε ||k L ∞ (Λ 1 ) ≤ C(Λ)εs ,

k∇Φε k2L ∞ (Λ 1 ) ≤ C(Λ)M0 | ln ε|.

2

(3.10)

2

(3.11)

2

Proof of Theorem 3.1. By the assumption (3.7), there exists a function ϕε defined on Λ such that u ε = ρε exp(iϕε )

in Λ,

(3.12)

where ρε = |u ε |. Changing u ε possibly by a constant phase, we may impose the condition Z 1 ϕε = 0. |Λ| Λ

(3.13)

From (1.1), we have ρε2

∂ϕε − div(ρε2 ∇ϕε ) + ρε2 Ψε + (Aε · ∇)ρε2 = 0 ∂t

in Λ.

(3.14)

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In what follows, we write ϕ = ϕε and ρ = ρε when this is not misleading. Set e ϕ (x, t) = ϕ(x, t)χ (x, t),

(3.15)

where χ is a smooth cut-off function such that     4 5 3 R R . χ ≡ 1 on B and χ ≡ 0 on R \ B 5 6 The function e ϕ then satisfies the equation ρ2

∂e ϕ − div(ρ 2 ∇e ϕ ) = div(ρ 2 ϕ∇χ ) + ρ 2 ∇χ ∇ϕ − χ (Aε · ∇)ρ 2 − ρ 2 Ψε χ ∂t

in Λ.

(3.16)

By construction, we have   4 supp(e ϕ) ⊂ B R × [T, T + 4T ], 5 and in particular e ϕ = 0 on the vertical part of the boundary of Λ. By a mean value argument, we may choose some to [T, T + 14 4T ] such that Z Z 4 eε (u ε ) eε (u ε ) ≤ 4T Λ B(R)×{t0 } and we set Λε = B(R) × [t0 , T + 4T ] ⊃ Λ 3 = B



4

   4T 3 R × , T + 4T . 4 4

We rewrite (3.16) as ∂e ϕ ∂e ϕ − 4e ϕ = div((ρ 2 − 1)e ϕ ) + (1 − ρ 2 ) + div(ρ 2 ϕ∇χ ) ∂t ∂t + ρ 2 ∇χ∇ϕ − χ (Aε · ∇)ρ 2 − ρ 2 Ψε χ

in Λ.

We introduce the function ϕ0 defined on Λε as the solution of  ∂ϕ0   − 4ϕ0 = div(ρ 2 ϕ∇χ ) + ρ 2 ∇χ ∇ϕ − χ (Aε · ∇)ρ 2 − Ψε χ ∂t ϕ (x, t0 ) = e ϕ (x, t0 )   0 ϕ0 (x, t) = 0

(3.17)

in Λε , on B(R) × {t0 }, ∀x ∈ ∂ B(R), ∀t ≥ t0 .

(3.18)

In particular, since χ ≡ 1 on B( 45 R), we have ∂ϕ0 − 4ϕ0 = −(Aε · ∇)ρ 2 − Ψε ∂t

 in B

 4 R × [t0 , T + 4T ]. 5

(3.19)

We set ϕ1 = e ϕ − ϕ0 , that is, e ϕ = ϕ0 + ϕ1 . Now, we will prove that ϕ1 is a perturbation term. We divide the estimates into several steps. Step 1. Estimates for ϕ0 . Z k∇ϕ0 k L 2 L 2∗ (Λε ) ≤ C(Λ)

Λ

Z k∇ϕ0 k L ∞ (Λ 3 ) ≤ C(Λ) 4

Λ

 eε (u ε ) + kΨε k L 2 (Λ) ,

(3.20)



eε (u ε ) + kΨε k L 2 (Λ) .

(3.21)

Z. Liu / Nonlinear Analysis 69 (2008) 4412–4442

4425

Proof. We split ϕ0 = ϕ00 + ϕ01 , where ϕ00 satisfies  0 ∂ϕ    0 − 4ϕ00 = 0 in Λε , ∂t 0 ϕ (x, t0 ) on B(R) × {t0 }, ϕ (x, t0 ) = e    00 ϕ0 (x, t) = 0 ∀x ∈ ∂ B(R), ∀t ≥ t0 . By standard estimates for the parabolic equation, we have ϕ k L 2 (B(R)×{t0 }) ≤ C(Λ)keε (u ε )k L 1 (Λ) , k∇ 2 ϕ00 k L 2 (Λε ) ≤ C(Λ)k∇e and then by Sobolev embedding k∇ϕ00 k L 2 L 2∗ (λε ) ≤ C(Λ)keε (u ε )k L 1 (Λ) .

(3.22)

Let T be the linear mapping which, to any function f defined on λε , associates the unique solution v = T f of the following equation  ∂v   − 4v = f on Λε , ∂t v=0 on B(R) × {t0 },   v=0 ∀x ∈ ∂ B(R), ∀t ≥ t0 , the operators f 7→ ∇ 2 (T f ), f 7→ ∂∂ (T f ) and g 7→ ∇(T (div g)) are linear continuous on L p L q (Λε ). We may write ϕ01 = T f + T (div g), where f = ρ 2 ∇χ · ∇ϕ − χ (Aε · ∇)ρ 2 − Ψε χ , g = ρ 2 ϕ∇χ. It is easy to check that k f k L 2 (Λε ) ≤ C(Λ)[k∇ϕk L 2 (Λε ) + kΨε k L 2 (Λ) ]

≤ C(Λ)[keε (u ε )k L 1 (Λ) + k∇ρk L 2 (Λ) + kΨε k L 2 (Λ) ],

and since (3.13) kgk L 2 L 2∗ (Λε ) ≤ Ckϕk L 2 L 2∗ (λε ) ≤ C(Λ)k∇ϕk L 2 L 2 (λε ) ≤ C(Λ)keε (u ε )k L 1 (Λ) . Therefore, k∇ϕ01 k L 2 L 2∗ (λε ) ≤ k∇(T f )k L 2 L 2∗ (λε ) + k∇(T (div g))k L 2 L 2∗ (λε ) ≤ C(Λ)[k(∇ 2 + I )T f k L 2 L 2 (λε ) + k∇(T (div g))k L 2 L 2∗ (λε ) ] ≤ C(Λ)[k f k L 2 L 2 (λε ) + kgk L 2 L 2∗ (λε ) ] ≤ C(Λ)[keε (u ε )k L 1 (Λ) + kΨε k L 2 (Λ) ].

(3.23)

By (3.22) and (3.23) we prove (3.20). (3.21) follows from (3.19) and (3.20) and the standard estimates for the parabolic equation.  Step 2. The equation of ϕ1 .  ∂ϕ1 ∂e ϕ   − 4ϕ1 = div((ρ 2 − 1)∇e ϕ ) + (1 − ρ 2 ) − (ρ 2 − 1)Ψε χ ∂t ∂t ϕ (x, t0 ) = 0   1 ϕ1 (x, t) = 0

on Λ, on B(R) × {t0 }, ∀x ∈ ∂ B(R), ∀t ≥ t0 .

We have from [2] k∇ϕ1 k L 2 L q (λε ) ≤ C(Λ)(ε

2− p p

+ε

2∗ −q q

)(C1 (M0 + C2 ) + 1)|ln ε|.

(3.24)

Combining (3.20) with (3.24) we obtain k∇e ϕ k L 2 L q (λε ) ≤ C(Λ)(C1 (M0 + C2 ) + 1)|ln ε|.

(3.25)

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Step 3. Estimates for the modulus and potential terms. The ρ satisfies ∂ρ 1 − 4ρ + ρ|∇ϕ|2 = 2 ρ(1 − ρ 2 ) + 2(Aε · ∇)ϕ − |Aε |2 ρ. ∂t ε

(3.26)

Since χ ≡ 1 on B( 45 R), we have ϕ = e ϕ on B( 54 R). Let ξ be a nonnegative cut-off function such that ξ ≡ 1 on B( 34 R) and ξ ≡ 0 outside B( 45 R). Multiplying (3.26) by (1 − ρ 2 )ξ and integrating by parts we obtain Z Z Z Z Z (1 − ρ 2 )2 ∂ρ 2 2 2ρ|∇ρ|2 ξ + ρ = (1 − ρ )ξ + ∇ρ · ∇ξ(1 − ρ ) + ρ(1 − ρ 2 )|∇e ϕ |2 ξ ε2 λε λε λε ∂t λε λε Z Z + (ρ(Aε · ∇)ϕ)(1 − ρ 2 )ξ − |Aε |2 ρ(1 − ρ 2 )ξ. λε

Hence, since ρ ≥ Z

1 2

on Λ we obtain

|∇ρ| + Vε (u ε ) ≤ K ε 2

Λ3

λε

Z

2

1 1 Z 2 ! 12 Z 2 2 ∂ρ Vε (u ε ) + K ε · V (u ) ε ε ∂t λε λε

 1 Z 2

|∇ρ|

λε

λε

4

Z

T +4T

+K

q

B( 45 R)×{t}

t0

+ Kε

!2

Z

Z λε

|∇e ϕ|

2

 1 Z 2

λε

|∇e ϕ |q

Z B( 45 R)×{t}

(1 − ρ 2 )

q q−2

! q−2 q

1 Z 1 2 2 Vε (u ε ) + K ε Vε (u ε ) λε

so that using (3.25) we obtain Z 2 (q−2) [|∇ρ 2 |2 + Vε (u ε )] ≤ C(Λ)(C1 (M0 + C2 ) + 1)(ε q + ε)|ln ε|2 .

(3.27)

Λ3 4

We have proved that 2eε (u ε ) ≤ |∇ϕ0 |2 + rε for some rε ≥ 0 which satisfies Z rε ≤ C(Λ)C1 (M0 + C2 )εs ,

(3.28)

(3.29)

Λ3 4

for some small s > 0 depending only on Aε , Ψε . Therefore, we set Φε = ϕ0 . Step 4. Proof of (3.8) for the energy. We need the following proposition [21,18]. Proposition 3.3. Let 0 <  < 1 and v be a solution of (1.1) on the cylinder Λ0R = B(R) × [0, R 2 ] for some R > 0. √ Then there exists γε > 0 depending only on Aε , Ψε such that if R >  and Z 1 e (v ) ≤ γε (3.30) R 3 Λ0R then e (v )(x, t) ≤

Z K e (v ) R 5 Λ0R

for any (x, t) ∈ B( R2 ) × [ 43 R, R].

(3.31)

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Let Λ01

√

ε < r0 < 18 R, to be determined later, set  =

ε r0

and let (x0 , t0 ) ∈ Λ 5 be fixed. Consider the map v defined on 8

= B(1) × [0, 1] by v (x, t) = u ε

x − x0 (t − t0 ) + r02 , r0 r02

!

so that u ε (x0 , t0 ) = v (0, 1). Then, we have Z Z 1 e (v ) = 3 eε (u ε ), r0 Λr0 (x0 ,t0 ) Λ01

(3.32)

where Λr0 (x0 , t0 ) = B(x0 , r0 ) × [t0 − r02 , t0 ]. Note in particular, since r0 < 18 R, that Λr0 (x0 , t0 ) ⊂ Λ 3 . 4

By (3.28), we have Z Z 2 eε (u ε ) ≤ meas(Λr0 (x0 , t0 ))k∇Φε k L ∞ + Λr0 (x0 ,t0 )

Λ3

rε

4

≤ C(Λ)r05 [keε (u ε )k L 1 (Λ) + kΨε k L 2 (Λ) ] + C(Λ)C1 (M0 + C2 )εs . Inserting (3.33) into (3.32), we have Z e (v ) ≤ C(Λ)r02 [keε (u ε )k L 1 (Λ) + kΨε k L 2 (Λ) ] + C(Λ)C1 (M0 + C2 )r0−3 ε s . Λ01

(3.33)

(3.34)

Choosing   1) C(Λ)(keε (u ε )k L 1 (Λ) + kΨε k L 2 (Λ) ) − 2 1 r0 = inf R, , 8 γε (

we have, for ε sufficiently small, γε . 2 On the other hand, by construction, γε C(Λ)r02 keε (u ε )k L 1 (Λ) ≤ . 2 Applying Proposition 3.3 to v , together with R = 1, we obtain Z 2 r0 eε (u ε )(x0 , t0 ) = e (v )(0, 1) ≤ K e (v ) C(Λ)C1 (M0 + C2 )r0−3 ε s ≤

Λ01

≤ K C(Λ)r02 [keε (u ε )k L 1 (Λ) + kΨε k L 2 (Λ) ] + C(Λ)C1 (M0 + C2 )r0−3 ε s , which yields eε (u ε )(x0 , t0 ) ≤ C(Λ)

Z Λ

(eε (u ε ) + |Ψε |2 ) + C(Λ)C1 (M0 + C2 )εs1 ,

for some constant 0 < s1 < s. This proves (3.8), for (x0 , t0 ) ∈ Λ 5 . 8

Step 5. Set ζ = 1 − ρ. By Lemma 2.2, we have |ζ | ≤ C(Λ)ε2 (k∇ϕk2L ∞ (Λ 5 ) + |ln ε| + kAε k2L ∞ ) ≤ C(Λ)ε2 |ln ε| 8

on Λ 9 , 16

(3.35)

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where we have used (3.8) for the last inequality. Using (3.26) and (3.8), we derive that |∂t ζ − 4ζ | ≤ C(Λ)|ln ε|

on Λ 9 .

(3.36)

16

Hence, kζ kW 1,q1 (I,L q2 (B)) ≤ C(Λ)|ln ε|,

kζ k L q1 (I,W 2,q2 (B)) ≤ C(Λ)|ln ε|,

where I = [T + 12 4T, T + 4T ] and B = B(x0 , 12 R), 0 < q1 , q2 < +∞. By interpolation inequality, we have kζ kW 1/3,q1 (I,W 4/3,q2 (B)) ≤ C(Λ)|ln ε|. By the Sobolev embedding, choosing q1 , q2 sufficiently large, we have that for every 0 < γ < 1 kζ kC 0,1/4 (I,C 1,γ (B)) ≤ C(γ , Λ)|ln ε|. From (3.35), we have kζ k L ∞ (I,L ∞ (B)) ≤ C(Λ)ε2 |ln ε|, and therefore by interpolation again, kζ kC 0,1/5 (I,C 1,β (B)) ≤ C(Λ)εs

(3.37)

for some small s > 0, γ > β > 0. Hence, k∇ρk L ∞ (Λ 1 ) = k∇ζ k L ∞ (Λ 1 ) ≤ C(Λ)εs . 2

(3.38)

2

From (3.35), we have Vε (u ε ) ≤ K

ζ 2 + ε 4 |Aε |4 ≤ C(Λ)ε2 |ln ε| ε2

on Λ 1

2

then |∇ρ| + Vε (u ε ) ≤ C(Λ)εs

on Λ 1 .

(3.39)

2

Step 6. Going back to the equation for ϕ1 , we rewrite it as ∂t ϕ1 − 4ϕ1 = div((ρ 2 − 1)∇ϕ1 ) + f 0 + div g0 ,

(3.40)

where f 0 = (1 − ρ 2 )∂t e ϕ , g0 = (ρ 2 − 1)∇ϕ. Note that ϕ = e ϕ on Λ 1 , then f 0 = (1 − ρ 2 )∂t ϕ. Since ϕ satisfies 2

∂ϕε ρε2 − div(ρε2 ∇ϕε ) + ρε2 Ψε + (Aε · ∇)ρε2 = 0 ∂t

in Λ,

from the H¨older regularity bound for ρ, we obtain the H¨older regularity bounds for ∂t ϕ, of the order of |ln ε|. On the other hand, (3.38) holds, we have k f 0 kC 0,α (Λ 1 ) ≤ C(Λ)εs |ln ε|2 . 2

From (3.37) kg0 kC 0,1/5 (I,C 1,β (B)) ≤ C(Λ)εs |ln ε|. By the Schauder theory, we have k∇ϕ1 kC 0,α (Λ 1 ) ≤ C(Λ)εs 2

for some s > 0.

(3.41)

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Step 7. 2eε (u ε ) = |∇u ε |2 + Vε (u ε ) = |∇ρ|2 + ρ 2 |∇ϕ|2 + Vε (u ε ) = |∇ρ|2 + (ρ 2 − 1)|∇ϕ|2 + |∇Φε |2 + 2∇Φε ∇ϕ1 + |∇ϕ1 |2 + Vε (u ε ) ≡ |∇Φε |2 + Kε and the conclusion follows from the previous estimates. The proof of Theorem 3.1 is completed.



Proposition 3.4. There √ exists an absolute constant η2 > 0 and a positive function λ defined on R+ ∗ such that if, for 3 x ∈ R , t > 0 and r > 2ε, we have Z eε (u ε ) ≤ η2r |ln ε|, (3.42) B(x,λ(t)r )

then eε (u ε ) =

1 |∇Φε |2 + Kε 2

(3.43)

in Λ1/2 (x, t, r ) = B(x, r/4) × [t + µε :=

15 2 16 r , t

+ r 2 ], where Φε and Kε are as in Theorem 3.1. In particular,

eε (u ε ) ≤ C(t, r ) on Λ1/2 (x, t, r ). |ln ε|

Proof. This is an immediate consequence from Theorem 2.1 and Theorem 3.1.

(3.44) 

Theorem 3.2. Let u ε be a solution to (1.1) with (1.4). Let K ⊂ R3 × (0, T 0 ) be any compact set. There exist a real-valued function φε and a complex-valued function wε , both defined on a neighborhood of K, such that 1. u ε = wε exp(iφε )

on K,

2. ∂t φε − 4φε + Ψε = 0 on K, p 3. |∇φε (x, t)| ≤ C(K) |ln ε| for all (x, t) ∈ K, 4 4. k∇wε k L p (K) ≤ C( p, K) for any 1 ≤ p < . 3 Here, C(K) and C(K, p) are constants depending only on K, and p, C1 (M0 + C2 ), Aε , Ψε , respectively. In order to prove Theorem 3.2, we first choose a Λ which contains K K ⊂ Λ ≡ B × (T0 , T1 ) ⊂ R3 × (0, T 0 ). Here B is some open ball in R3 and 0 < T0 < T1 . Next, let Ω be a smooth bounded domain with simply connected boundary, such that K ⊂ Λ ⊂ Ω ⊂ R3 × (0, T 0 ). Without loss of generality, we may assume that for ε sufficiently small Z Z eε (u ε ) ≤ M1 |ln ε|, eε (u ε ) ≤ M1 |ln ε|, |u ε | ≤ 3, Ω

∂Ω

where M1 = C(K, Aε , Ψε )C1 (M0 + C2 ). Using Proposition 5.2 in [2] to u ε , then there exist a smooth function Φ, a smooth 1-form ζ , and a smooth 2-form Ψ defined on Ω , such that u ε × δu ε = δΦ + δ ∗ Ψ + ζ,

δΨ = 0 in Ω , ΨT = 0 on ∂Ω

(3.45)

and k∇Φk L 2 (Ω ) + k∇Ψ k L 2 (Ω ) ≤ C(Ω )M1 |ln ε|,

(3.46)

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where δ is the exterior differential operator on R4 and d is the exterior differential operator on R3 . Moreover, for any k 1 ≤ p ≤ k−1 we have k∇Ψ k L p (Ω ) ≤ C( p, Ω )M1 ,

kζ k L p (Ω ) ≤ C( p, Ω )M1 ε 1/2 ,

(3.47)

where C( p, Ω ) is a constant depending only on p and Ω . From (3.47), we have already obtained estimates for Ψ and ζ . In order to handle Φ, we first prove that it solves an evolution equation. Lemma 3.1. The function Φ satisfies the equation ∂t Φ − 4Φ + Ψε = d∗ (δ ∗ Ψ + ζ − Pt (δ ∗ Ψ + ζ )dt) − Pt (δ ∗ Ψ + ζ ) − Aε · ∇|u ε |2 .

(3.48)

Here, for a 1-form ω on Λ, we denote by Pt (ω) its dt component. Proof. First, by (1.1), we have u ε × ∂t u ε − div(u ε × ∇u ε ) + Ψε + Aε · ∇|u ε |2 = 0

in Λ.

On the other hand, by (3.45),  u ε × du ε = dΦ + (δ ∗ Ψ + ζ ) − Pt (δ ∗ Ψ + ζ )dt, u ε × ∂t u ε = ∂t Φ + Pt (δ ∗ Ψ + ζ )dt. Combining (3.49) with (3.50) leads to the conclusion.

(3.49)

(3.50) 

Proof of Theorem 3.2. Without loss of generality, we may assume that Z |∇x,t Φ|2 ≤ C(K)C1 (M0 + C2 )|ln ε|, ∂t Λ

(3.51)

where ∇x,t denotes the gradient with respect to both space and time coordinates. Indeed, by (3.46), k∇x,t Φk L 2 (Ω ) ≤ C(Ω )M1 |ln ε|, if (3.51) were not satisfied for our original Λ, we could shrink it to a smaller cylinder, still containing K and satisfying (3.51). In order to estimate the L p norm of the gradient of the modulus, we need the following lemma. Lemma 3.2. Set ρ = |u ε |. Then, for any compact subset K ⊂ R3 × (0, T 0 ), and any 0 ≤ p < 32 , we have Z |∇|u ε || p ≤ C(K, M0 , C1 , C2 )ε1−( p/2) |ln ε|, K

where C(K, M0 , C1 , C2 ) depends only on K, Ψε , Aε , M0 , C1 , C2 . Proof. ρ satisfies ∂t ρ 2 − 4ρ 2 + 2|∇u ε |2 =

2 2 ρ (1 − ρ 2 ) − 2Aε (u ε × ∇u ε ) + |Aε |2 ρ 2 . ε2

(3.52)

Define A = {(x, t) ∈ Ω : ρ(x, t) > 1 − ε 1/2 } and ρ¯ = max{ρ, 1 − ε 1/2 }, thus ρ¯ = ρ on A and 1 − ρ¯ ≤ ε 1/2 in Ω . Let χK be a cut-off function in D(Ω ) such that 0 ≤ χK ≤ 1 on Ω , χK ≡ 1 on K, and |∇χK | ≤ C(K). Multiplying (3.52) by χK (ρ¯ 2 − 1), and integrating over Ω we obtain Z Z Z Z 2ρ(1 − ρ 2 )(1 − ρ¯ 2 ) 2 2 ∇ρ 2 ∇ ρ¯ 2 χK + χ = χ (1 − ρ ¯ )|∇u | + ∇ρ 2 ∇χK (1 − ρ¯ 2 ) ε K K ε2 Ω Ω Ω Ω Z Z Z ∂ρ 2 2 − (ρ¯ − 1)χK + (Aε · (u ε × ∇u ε ))(ρ¯ 2 − 1)χK + |Aε |2 ρ 2 (ρ¯ 2 − 1)χK . Ω ∂t Ω Ω

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It follows that on the set AK = A ∩ K we have Z Z 2 2 |∇ρ | = ∇ρ 2 ∇ ρ¯ 2 AK AK Z Z ≤ 2ε 1/2 |∇u ε |2 + C(K) |∇ρ||1 − ρ 2 | + C(K)C1 (M0 + C2 )ε1/2 |ln ε| Ω Z Ω Z 2 +C |u ε × ∇u ε ||1 − ρ | + C(K) |1 − ρ 2 | Ω Ω Z  Z Z (1 − ρ 2 )2 |∇u ε |2 + C(K)ε |∇ρ|2 + + C(K)C1 (M0 + C2 )ε1/4 ≤ 2ε 1/2 4ε 2 Ω Ω Ω  1 Z 1− 1 Z Z 1 p p p (1 − ρ 2 )2 2 p 2 p−1 +C |u ε × ∇u ε | |1 − ρ | + C(K)ε . 4ε 2 Ω Ω Ω Since ρ ≥ 1 − ε 1/2 on AK , we have, for ε ≤ 1/4 and 1 ≤ p ≤ 32 , Z Z 1 2− 2 |∇ρ 2 |2 ≤ C(K)(ε1/4 + ε p |ln ε| + ε| ln ε| 2 ), |∇ρ|2 ≤ 4 AK

(3.53)

AK

R where we have used the inequalities (3.46) and (3.47). On the other hand, on BK = K \ A we have BK (1 − ρ 2 )2 ≤ C(K, M0 , C1 , C2 )ε2 |ln ε| and hence, since (1 − ρ) ≥ ε 1/2 on BK , it follows that |BK | ≤ C(K, M0 , C1 , C2 )ε|ln ε|. Thus Z  p/2 Z |∇ρ| p ≤ |∇ρ|2 |BK |1−( p/2) ≤ C(K, M0 , C1 , C2 )ε1−( p/2) |ln ε|. (3.54) BK

BK

Combining (3.53) with (3.54) we finish the proof of Lemma 3.2.



Now we will define ϕε . We denote ∂Λ = ϑε ∪ ϑ1 , where ϑε = (B × {T0 }) ∪ (∂ B × [T0 , T1 ]) and

ϑ1 = B × {T0 }.

Let Φ1 be the unique solution of the parabolic problem  ∂t Φ1 − 4Φ1 = d∗ (δ ∗ Ψ + ζ − Pt (δ ∗ Ψ + ζ )dt) − Pt (δ ∗ Ψ + ζ ) − Aε · ∇|u ε |2 Φ1 = 0

on Λ on ϑε .

(3.55)

By (3.46) and Lemma 3.2, we have kδ ∗ Ψ + ζ − Pt (δ ∗ Ψ + ζ )dtk L p (Λ) + k∇ρk L p (Λ) + kPt (δ ∗ Ψ + ζ )k L p (Λ) ≤ C( p, K, M0 , C1 , C2 ), thus k∇Φ1 k L p (Λ) ≤ C( p, K)C1 (M0 + C2 ). Now, we define ϕε = Φ − Φ1 and wε = u ε exp(−iϕε ). By construction, ϕε satisfies  ∂t ϕε − 4ϕε − Ψε = 0 in Λ ϕε = Φ on ϑ1 .

(3.56)

(3.57)

By standard regularity theory for the parabolic equation, we have k∇ϕε k2L ∞ (K) ≤ C(K)[k∇x,t Φk2L 2 (ϑ ) + kΨε kC 1 ] ≤ C(K)|ln ε|. ε

The third statement of Theorem 3.2 is proven. We next turn to the fourth and last one. Now we will estimate W 1, p norm for wε . Note that |wε |2 |∇wε |2 = |wε |2 |∇|wε ||2 + |wε × ∇wε |2 ,

(3.58)

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and |wε | = |u ε |, hence  Z Z Z |∇wε | p ≤ C( p) |wε × dwε | p + |∇|wε || p . K∩{|u ε |≥ 21 }

K

(3.59)

K

On the other hand, since |∇wε | ≤ |∇u ε | + K |∇ϕε | ≤ C(K)C1 (M0 + C2 )ε−1 , we have Z Z (bε − |u ε |2 )2 p 2− p |∇wε | ≤ C(K)C1 (M0 + C2 )ε ≤ C( p, K)C1 (M0 + C2 ). 4ε 2 K∩{|u ε |≤ 21 } K

(3.60)

By construction, we have wε × δwε = u ε × δu ε − |u ε |2 δϕε = δ ∗ Ψ + δΦ1 + ζ + (1 − |u ε |2 )δϕε .

(3.61)

Note that k(1 − |u ε |2 )δϕε k L p (K) ≤ Ck1 − |u ε |2 k

2p

L 2− p (K)

kδϕε k L p (K) ≤ Cε 2− p C1 (M0 + C2 )| ln ε|.

(3.62)

Combining (3.62), (3.47) and (3.56) with this inequality, we have Z |wε × δwε | p ≤ C(C1 (M0 + C2 ) + 1).

(3.63)

Combining (3.63) with Lemma 3.2 we obtain W 1, p estimate for wε . The proof of Theorem 3.2 is complete.



K

4. Analysis of the measures µ∗t 4.1. Preliminary results In this part we study the asymptotic limits, as ε → 0, of the Radon measures µε defined on R3 × [0, T 0 ) by µε (x, t) =

eε (u ε ) dxdt, | ln ε|

where for 0 < ε < 1, eε (u ε ) = 12 [|∇u ε |2 + asymptotic of the time-slices µtε defined by µtε (x) =

1 (b 2ε 2 ε

− |u ε |2 )2 ], u ε are solutions of (1.1) and (1.4). We examines the

eε (u ε ) dx. |ln ε|

In view of assumption (1.4), we have Z eε (u ε ) ≤ C|ln ε|. R3 ×{t}

We may assume that for a subsequence εn → 0, there exists a Radon measure µ∗ defined on R3 × [0, T 0 ) such that µεn * µ∗

as measures.

We may also assume weak convergence of

(4.1) µtεn

for all t ∈

(0, T 0 ),

in the sense of measures.

Lemma 4.1 ([6]). There exist a subsequence of εn (still denoted εn ) such that µtεn * µt∗

for all t ∈ (0, T 0 ),

where µt∗ is a finite Radon measure on R3 for all t ∈ (0, T 0 ). Moreover, µ∗ = µt∗ dt. From the monotonicity formula, we have

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Lemma 4.2. For each t ∈ (0, T 0 ), x ∈ R3 , the function ×µ ((x, t), ·) defined on R+ ∗ by  Z    1 |x − y|2 2 r 7→ ×µ ((x, t), r ) ≡ exp(K r ) dµ∗t−r (y) + r 2 exp − r R3 4r 2 √ is non-decreasing for 0 < r < t. Theorem 4.1. There exist an absolute constant η2 > 0, and a positive continuous function λ defined on R+ ∗ such that if, for x ∈ R3 , t ∈ (0, T 0 ) and r > 0 we have µt∗ (B(x, λ(t)r )) < η2r,

(4.2)

2 2 s then for every s ∈ [t + 15 16 r , t + r ], µ∗ is absolutely continuous with respect to the Lebesgue measure on the ball 1 B(x, 4 r ). More precisely,   1 1 µs∗ = |∇Φ∗ |2 dx on B x, r , 2 4

where Φ∗ satisfies the equation     15 1 in Λ 1 = B x0 , r × t + r 2 , t + r 2 . 4 4 16

∂t Φ∗ − 4Φ∗ = 0

Proof. In view of assumption (4.2), for ε = εn small enough, Z eε (u ε ) ≤ η2r |ln ε|, B(x,λ(t)r )

so that we may invoke Proposition 3.4. This yields 1 |∇Φε |2 + Kε in Λ 1 , 2 2 which satisfies (3.10) in Λ 1 and eε (u ε ) =

2

|∇Φε | ≤ C(Λ)|ln ε|, 2

|Kε | ≤ C(Λ)εs

in Λ 1 . 2

Extracting possibly a further subsequence we may assume that Φε → Φ∗ √ |ln ε|

uniformly on Λ 5 . 16

Note that, by (3.11), k∇|u ε |2 k L 2 (Λ 1 ) ≤ Cε s , it follows that Φ∗ satisfies the heat equation on Λ 1 . On the other hand, Kε → 0 uniformly on Λ 1 , hence

4

2

4

eε (u ε ) 1 → |∇Φ∗ |2 √ 2 |ln ε|

uniformly on Λ 5 . 16

This completes the proof of Theorem 4.1.



4.2. Densities and concentration set Definition 4.1. For m ∈ N, the m-dimensional lower density of ν at the point x is defined by Θ∗,m (ν, x) = lim inf r →0

ν(B(x, r )) , ωm r m

where ωm denotes the volume of the unit ball B m . Similarly, the m-dimensional upper density Θm∗ (ν, x) is given by Θm∗ (ν, x) = lim sup r →0

ν(B(x, r )) ωm r m

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when both quantities coincide, ν admits a m-dimensional density Θm (ν, x) at the point x, defined as the common value. Definition 4.2. Let ν be a Radon measure on R3 × [0, +∞) such that ν = ν t dt. For t > 0 and m ∈ N, the parabolic m-dimensional lower density of ν at the point (x, t) is defined by   Z 1 |x − y|2 2 p Θ∗,m (ν, (x, t)) = lim inf m exp − dν t−r (y). 2 r →0 r 4r R3 p,∗

p

The parabolic upper density and parabolic density are defined accordingly, and denoted respectively by Θm and Θm . Lemma 4.3. For all x ∈ R3 and for all t ∈ (0, T 0 ), Θ∗,1 (µt∗ , x) ≤ Θ1∗ (µt∗ , x) ≤ C(T 0 ) < +∞. √ Proof. Let x ∈ R3 and t > 0. We have for every 0 < r < 21 t,     Z 1 1 µt∗ (B(x, r )) |x − y|2 ≤ exp exp − dµt∗ (y) ω1r 4 ω1r R3 4r 2 #  "     Z √ 1 1 |x − y|2 r2 0 ≤ exp K t + exp − exp − dµ∗ (y) + 2t 1 4 4r 2 t + r2 ω1 (t + r 2 ) 2 R3 ≤ C(T 0 ), where we have used the monotonicity formula Lemma 4.2 for the second inequality.



By the monotonicity formula, we have Lemma 4.4. The function (x, t) 7→ Θ1P (µ∗ , (x, t)) is upper semi-continuous on the set R3 × (0, T 0 ). Lemma 4.5. p

Θ1 (µ∗ , (x, t)) ≥ K Θ∗,1 (µt∗ , x).

(4.3)

√ Proof. Let (x, t) ∈ R3 × (0, T 0 ) be given. Let 0 < r < t be fixed for the moment. We have, for 0 < s < t,   Z   1 1 1 t |x − y|2 µ (B(x, r )) ≤ exp dµt∗ (y) exp − r ∗ 4 r R3 4r 2 # "    Z 2  p 1 1 |x − y| 2 t−s 2 2 ≤ exp exp(K ( r 2 + s 2 − r )) exp − 2 dµ∗ (y) + r + s , 1 4 4(r + s 2 ) (r 2 + s 2 ) 2 R3 √ where we have used the Lemma 4.2 (at the point (x, t + r 2 )) for the last inequality. Next, we choose s = r . This yields " #   Z 1 t 1 |x − y|2 2 µ (B(x, r )) ≤ K exp − 2 dµt−r (4.4) ∗ (y) + r + r . 1 r ∗ 4(r + r ) (r 2 + r ) 2 R3 Set R3 = B(x, 1) ∪ (R3 \ B(x, 1)). On B(x, 1),     |x − y|2 |x − y|2 exp − 2 , ≤ K exp − 4r 4(r + r ) for some absolute constant K . On the other hand, on R3 \ B(x, 1), we have     Z |x − y|2 1 t−r exp − 2 dµ∗ (y) ≤ exp − 2 C. 4(r + r ) 4(r + r ) R3 \B(x,1)

(4.5)

(4.6)

Z. Liu / Nonlinear Analysis 69 (2008) 4412–4442

4435

Inserting (4.5) and (4.6) into (4.4), we are led to       Z 1 t |x − y|2 1 1 K 2 exp − dµt−r + C exp − + r + r . µ∗ (B(x, r )) ≤ 1 ∗ 1 r 4r 4(r 2 + r ) r 2 R3 r2 Letting r → 0, the conclusion follows.  Now we define p

Σµ = {(x, t) ∈ R3 × (0, +∞) : Θ1 (µ∗ , (x, t)) > 0}, and for t > 0, Σµt = Σµ ∩ (R3 × {t}). An obvious result of (4.3) is that Θ∗,1 (µt∗ , x) ≡ 0

on R3 \ Σµt .

Lemma 4.6. Let (x, t) ∈ Σµ and 0 < r <

(4.7) √ t. Then, we have

2

r −1 µt−r (B(x, λ(t − r 2 )r )) > η2 , ∗ where η2 is the constant in Theorem 4.1. Proof. Assume by contradiction that 2

r −1 µt−r (B(x, λ(t − r 2 )r )) ≤ η2 . ∗ 1 2 Then, by Theorem 4.1, for every τ ∈ [t − 16 r , t]   1 1 µτ∗ = |∇Φ∗ |2 dx on B x, r , 2 4

where Φ∗ is smooth. We claim that   Z |x − y|2 2 −1 s dµt−s → 0, exp − ∗ 2 4s R3

as s → 0.

(4.8)

Indeed, we write     Z Z |x − y|2 |x − y|2 t−s 2 −1 2 s −1 exp − dµ ≤ s k∇Φ k exp − dx ∗ L ∞ (B(x, 1 r )) ∗ 4s 2 4s 2 8 B(x, 18 r ) R3 ≤ K k∇Φ∗ k2 ∞  L

s2 B(x, 18 r )

→ 0,

as s → 0. On the other hand, as s → 0,     Z |x − y|2 r2 t−s 2 −1 s −1 dµ ≤ s exp − C1 M0 → 0.   exp − ∗ 4s 2 256s 2 R3 \B x, 18 r

(4.9)

(4.10)

p

Combining (4.9) and (4.10), (4.8) follows and Θ1 (µ∗ , (x, t)) = 0, that is, (x, t) 6∈ Σµ , which yields a contradiction.  4.3. Clearing-out lemma The main tool in the study of geometric properties of Σµ is the following, Clearing-out which is a consequence of Theorem 4.1.

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Theorem 4.2 (Clearing-Out). √ There exists a positive continuous function η3 defined R+ ∗ , such that for any (x, t) ∈ 3 0 R × (0, T ) and any 0 < r < t, if  Z    1 |x − y|2 t−r 2 2 ×µ ((x, t), r ) ≡ exp(K r ) exp − dµ (y) + r ≤ η3 (t − r 2 ) ∗ r R3 4r 2 then (x, t) 6∈ Σµ . √ Proof. Let (x, t) ∈ R3 × (0, T 0 ) and 0 < r < t. We have   2 λ (t − r 2 ) 2 2 r −1 µt−r (B(x, λ(t − r )r )) ≤ exp ×µ ((x, t), r ). ∗ 4 √ 2 Set η3 (s) = exp(− λ 4(s) )η2 , and assume next that, for some 0 < r < t,

(4.11)

×µ ((x, t), r ) ≤ η3 (t − r 2 ). Then, by (4.11), 2

µt−r (B(x, λ(t − r 2 )r )) ≤ η2 ∗ and the conclusion follows by Lemma 4.6.



An immediate result is Corollary 4.1. For any (x, t) ∈ Σµ , we have p

Θ1 (µ∗ , (x, t)) ≥ η3 (t). We are in position to derive the following result. Proposition 4.1. We have 1. Σµ is closed set in R3 × (0, T 0 ). 2. H1 (Σµ ) ≤ C(C1 , M0 ), ∀t ∈ (0, T 0 ). 3. For any t > 0, the measure µt∗ can be decomposed as µt∗ = g(x, t)H3 + Θ∗ (x, t)H1 bΣµt , where g is some smooth function defined on R3 × (0, T 0 ) \ Σµ and Θ∗ satisfies Θ∗ (x, t) ≤ C(T 0 ). Proof. The proof is identical to that in [2].



4.4. Regularity of Σµt In order to study the regularity we need the reverse inequality to (4.3). Proposition 4.2. For almost every t > 0, the following inequality holds p

Θ∗,1 (µt∗ , x) ≥ K Θ1 (µ∗ , (x, t))

(4.12)

for H1 almost every x ∈ R3 . Corollary 4.2. For almost every t ≥ 0, Θ∗,1 (µt∗ , x) ≥ K η3 (t),

for H1 a.e. x ∈ Σµt .

Proof. The conclusion follows from Corollary 4.1 and Proposition 4.2.

(4.13) 

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Now we will give the following Proposition 4.3. For almost every t > 0, Θ∗,1 (µt∗ , x) = Θ1∗ (µt∗ , x) ≥ K η3 (t), for H1 almost every x ∈ Σµt . Therefore, for almost every t > 0, the set Σµt is 1-rectifiable. Proof. The proof of Proposition 8 in [2] carries over word for word, using (4.3), (4.12) and (4.13).



4.5. Proof of Theorem 1.1 For m ∈ N, set Km = B(m) × [ m1 , T 0 − we may write u ε = exp(iφεm )wεm

1 m ],

then ∪m Km = R3 × (0, T 0 ). Using Theorem 3.2 to u ε and K = Km ,

on Km ,

(4.14)

where φεm solves ∂ m φ − 4φεm = −Ψε ∂t ε

(4.15)

k∇φεm k L ∞ (Km )

(4.16)

on Km , p ≤ C(m) |ln ε|

and 4 . 3 Let m ∈ N be fixed now. Extracting possibly a further subsequence of {εn }n∈N , we may assume that kwεm k L p (Km ) ≤ C(m, p)

φm → φ∗m √ ε |ln ε|

for any 1 ≤ p <

in C 2 (Km−1 ),

(4.17)

(4.18)

and we obtain that φ∗m solves the heat equations on Km−1 . Let x0 ∈ Ωµ := R3 × (0, T 0 ) \ Σµ . Since Ωµ is open, we can find a small cylindrical neighborhood Λx0 of x0 in Ωµ . There exist m 0 ∈ N such that for m ≥ m 0 , Λx0 ⊂ Km . For ε small enough, we have 1 on Λx0 2 (where σ is the constant in Theorem 3.1), so that |u ε | ≥ 1 − σ ≥

(4.19)

u ε = ρε exp(iϕε )

(4.20)

for some real-valued function ϕε (defined up to an integer multiple of 2π ). In view of (4.19) and Theorem 3.1, we have that there exists a solution Φε of the Eq. (3.10) on Λx0 such that k∇Φε − ∇ϕε k L ∞ ((Λx0 ) 1 ) ≤ Cε s .

(4.21)

2

On the other hand, we may write for m ≥ m 0 wεm = ρε exp(iψεm ) where

ψεm

on Λx0 ,

(4.22)

is a real-valued function. Combining (4.20), (4.14) and (4.22) we have

∇ϕε = ∇φεm + ∇ψεm , and using (4.21), for m fixed, we obtain ∇φεm − ∇Φε ∇ψεm √ ≤ √ + Cε s |ln ε| |ln ε|

on (Λx0 ) 1 . 2

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Using (4.17) we have

∇φεm ∇Φε

√

|ln ε| − √| ln ε| p L ((Λx Since

√∇ Φε |ln ε|

∇φ∗m

→0 ) ) 0 1 2

as ε → 0.

→ ∇Φ∗ on (Λx0 ) 1 , by (4.18) we obtain 2

= ∇Φ∗

on (Λx0 ) 1 . 2

Since Φ∗ is independent of m, changing possibly φ∗m by a constant we may assume that all the φ∗m coincide on (Λx0 ) 1 . 2 Hence, for each m ≥ m 0 , the functions {φ∗m }m≥n coincide on Kn . Letting n → +∞, we define their common value φ∗ on R3 × (0, T 0 ) and we set Φ∗ = φ∗ .

(4.23)

Combining (4.23), Theorem 4.1, Proposition 4.1, Corollary 4.2, and Proposition 4.3, we prove Theorem 1.1. 4.6. Mean curvature flows First we recall weak notions of mean curvature flow. 4.6.1. Brakke flows Let {νt }t≥0 be a family of Radon measures on R3 . For χ ∈ C02 (R3 , R+ ), we define ν t (χ ) − ν t0 (χ ) . D¯ t ν t (χ ) = lim sup t − t0 t→t0 If ν t b{χ > 0} is a k-rectifiable measure which has a first variation satisfying χ| HE |2 ∈ L 1 (ν t ), then we set Z Z B(ν t , χ ) = − χ| HE |2 dν t + ∇χ · π( HE )dν t , where π denotes Hk a.e. the orthogonal projection onto the tangent space to ν t . Otherwise, we set B(ν t , χ ) = −∞. Definition 4.3 ([3]). Let {νt }t≥0 be a family of Radon measures. We say that {νt }t≥0 is a k-dimensional Brakke flow if and only if D¯ t ν t (χ ) ≤ B(ν t , χ ), for any χ ∈

C0∞ (R3 , R+ )

(4.24)

and for all t ≥ 0.

4.6.2. Relating (1.1) to mean curvature flow By computation directly, we have, for the solution u ε to (1.1), that Z Z Z d |∂t u ε |2 −∂t u ε ∇u ε χ (x)dµtε = − χ (x) (x)dx + ∇χ (x) (x)dx dt R3 |ln ε| |ln ε| R3 ×{t} R3 ×{t} Z (iΨε u + 2i(Aε · ∇)u ε )∂t u ε − χ (x) (x)dx. 3 |ln ε| R ×{t} Set σεt ≡

−∂t u ε ∇u ε (x)dx, |ln ε|

ωεt ≡

|∂t u ε |2 (x)dx. |ln ε|

(4.25)

Z. Liu / Nonlinear Analysis 69 (2008) 4412–4442

4439

4.6.3. Convergence of σεt Consider σε = σεt dt defined on R3 × [0, T 0 ). Then σε is uniformly bounded on R3 × [0, T ] for every 0 < T < T 0 , so that passing possibly to a further subsequence, we may assume that σε * σ∗ as measures. The Radon–Nikodym derivative of |σε | with respect to µε satisfying √ d|σε | 2|∂t u ε | . ≤ √ dµε eε (u ε ) On the other hand,

Z

|∂t u ε | |∂t u ε |2

√ ≤ ≤ C(T ),

e (u ) 2 3 R3 ×[0,T ] |ln ε| ε ε L (R ×[0,T ],dµε ) 2 3 ε| so that d|σ dµε is uniformly bounded in L (R × [0, T ], dµε ). It follows that σ∗ is absolutely continuous with respect to µ∗ (arguing as in [1] Remark 2.2). Hence, we have

σ∗ = ηEµt∗ dt, where ηE ∈ L 2 (R3 × [0, T ], µt∗ dt). By Theorem 1.1, we obtain the following lemma. Lemma 4.7. We have σ∗ = σ∗t dt, where for a.e. t ≥ 0, σ∗t = −∂t Φ∗ · ∇Φ∗ dx + ηEν∗t . Next we are in a position to prove that the restriction of ηE on Σµt is equal to the mean curvature vector. Let XE ∈ C0∞ (R3 , R3 ). We have, for every t ≥ 0,   Z Z 1 ∂u ε ∂u ε ∂ XE ∂t u ε · ∇u ε dx eε (u ε )δi j − dx = XE 3 |ln ε| R3 ×{t} ∂ xi ∂ x j ∂ x j |ln ε| R ×{t} (   )   Z E · (u ε × ∇u ε ) Aε Ψ X ε 2 ∗ ∧ ∗J u ε , XE + dx − |ln ε| |ln ε| R3 ×{t} Z 1 + ( XE · ∇|Aε |2 )(b − |u ε |2 )dx, (4.26) |ln ε| R3 ×{t} where J u = 12 d(u × du) (the Jacobian of u) denotes the 2-form. Set   ∇u ε ⊗ ∇u ε αεt = Id − dµtε . eε (u ε ) Since (αεt ) ≤ K µtε , we may assume that αεt * α∗t := B · µt∗ ,

(4.27)

where B is a 3 × 3 symmetric matrix, and we have B ≤ Id.

(4.28)

On the other hand, Tr(eε (u ε )Id − ∇u ε ⊗ ∇u ε ) = eε (u ε ) + 2Vε (u ε ). Therefore, Tr(B) = 1 + 2

dV∗ , dµ∗

(4.29)

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Z. Liu / Nonlinear Analysis 69 (2008) 4412–4442

ε (u ε ) where the nonnegative measure V∗ is the limit (up to possibly a further subsequence) of V|ln ε| . Passing to the limit in (4.26), using the decomposition in Theorem 1.1, we have  Z Z  Z Z i ∂Φ∗ ∂Φ∗ ∂ X i |∇Φ∗ |2 ij ∂X t t E B dν + δi j − dx = − X · ηEdν∗ − XE · ∇Φ∗ ∂t Φ∗ dx, (4.30) ∂x j ∗ 2 ∂ xi ∂ x j ∂ x j R3 R3 R3 R3

where we have used the fact that ∇u ε ⊗ ∇u ε t dµε * (I − B)µt∗ eε (u ε )

as measures

and Z Ψε XE · (u ε × ∇u ε ) Aε → 0, dx → 0, |ln ε| |ln ε| R3 ×{t} Z 1 ( XE · ∇|Aε |2 )(b − |u ε |2 )dx → 0 as ε → 0. |ln ε| R3 ×{t}

(4.31)

Since ∂t Φ∗ − 4Φ∗ = 0. Hence, Z

 R3

∂Φ∗ ∂Φ∗ |∇Φ∗ |2 δi j − 2 ∂ xi ∂ x j



Z ∂ Xi dx = − XE · ∇Φ∗ ∂t Φ∗ dx. ∂x j R3

Combining (4.30) with (4.32), we have Z Z i t ij ∂X XE · ηEdν∗t . dν = − B ∂x j ∗ R3 R3 Lemma 4.8. For a.e. t ≥ 0, "Z # 1 B(x) ∇χ (y)dH (y) = 0 Tx Σµt

(4.32)

(4.33)

for H1 a.e. x ∈ Σµt ,

(4.34)

and for all χ ∈ C0∞ (R3 , R). Proof. Let x ∈ Σµt be such that Tx Σµt exists and such that x is a Lebesgue point for Θ∗ (with respect to H1 ) and A (with respect to ν∗t ). For r > 0, consider the vector field XE r,l (y) = χ ( x−y r )el . Inserting it to (4.34) and letting r → 0, we obtain by difference of homogeneity, that the right-hand side is neglectible with respect to the left-hand side, and the conclusion follows.  A straightforward result is Corollary 4.3. For t and x as in Lemma 4.8, (Tx Σµt )⊥ ⊆ Ker(B(x)). Corollary 4.4. For t and x as in Lemma 4.8, B = P is the orthogonal projection onto the tangential space Tx Σµt . Combining the previous results we obtain Proposition 4.4. For a.e. t ≥ 0, ν∗t has a first variation and δν∗t = ηEν∗t

(4.35)

Z. Liu / Nonlinear Analysis 69 (2008) 4412–4442

4441

i.e. the mean curvature of ν∗t HE = ηE.

(4.36)

In order to prove Theorem 1.2, we need the following result, which follows directly from the Lemma 7, Proposition 10 of Part II in [2]. Proposition 4.5. Z lim inf

Z Z |∂t u ε |2 2 t χ dx ≥ χ |E η| dν∗ + χ|∂t Φ∗ |2 dx. |ln ε| R3 ×{t} R3 ×{t} R3 ×{t}

ε→0

(4.37)

Proof of Theorem 1.2. By Theorem 4.4 in [1], it suffices to prove the integral version of (4.24). Let 0 < T0 < T1 < T 0 . We integrate (4.25) on [T0 , T1 ] and let ε → 0+ . Combining the results of Lemma 4.7, Proposition 4.4, 4.5 and Theorem 1.1, we obtain Z Z 1 1 ν∗T1 (χ ) − ν∗T0 (χ ) + χ|∇Φ∗ |2 dx − χ |∇Φ∗ |2 dx R3 ×{T0 } 2 R3 ×{T1 } 2 Z Z χ| HE |2 dν∗ − χ|∂t Φ∗ |2 dxdt ≤ − R3 ×[T0 ,T1 ]

Z + R3 ×[T0 ,T1 ]

R3 ×[T0 ,T1 ]

∇χ · π(E η)dν∗ −

Z

∇χ ∇Φ∗ ∂t Φ∗ dxdt.

R3 ×[T0 ,T1 ]

(4.38)

Since ∂t Φ∗ − 4Φ∗ = 0, we have Z R3 ×{T0 }

Z =

Z 1 1 2 χ|∇Φ∗ | dx − χ |∇Φ∗ |2 dx 2 R3 ×{T1 } 2 Z χ|∂t Φ∗ |2 dxdt + ∇χ ∇Φ∗ ∂t Φ∗ dxdt.

R3 ×[T0 ,T1 ]

R3 ×[T0 ,T1 ]

Inserting (4.39) into (4.38), we obtain Z Z χ|E η|2 dν∗ + ν∗T1 (χ ) − ν∗T0 (χ ) ≤ −

R3 ×[T0 ,T1 ]

R3 ×[T0 ,T1 ]

This completes the proof of Theorem 1.2.

(4.39)

∇χ · π(E η)dν∗ .



Acknowledgments The author acknowledges the hospitality of the Institute for Mathematics and its Applications at the University of Minnesota where parts of this work was carried out. This work is supported by the National Natural Science Foundation of China (No. 10471119, 10771181), EYTP. References [1] L. Ambrosio, H.M. Soner, A measure-theoretic approach to higher codimension mean curvature flows, Ann. Scuola Norm. Sup. Pisa Cl. Sci. 25 (1997) 27–49. [2] F. Bethuel, G. Orlandi, D. Smets, Convergence of the parabolic Ginzburg–Landau equation to motion by mean curvature, Ann. of Math. 163 (2006) 37–163. [3] K. Brakke, The Motion of a Surface by its Mean Curvature, Princeton University Press, 1978. [4] J. Colliander, R. Jerrard, Vortex dynamics for the Ginzburg–Landau–Schrodinger equation, Inter. Math. Res. Notices 7 (1998) 333–358. [5] Q. Du, P. Gray, High-kappa limit of the time dependent Ginzburg–Landau model for superconductivity, SIAM J. Appl. Math. 56 (1996) 1060–1093.

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