DISCRETE AND CONTINUOUS doi:10.3934/dcds.2016045 DYNAMICAL SYSTEMS Volume 36, Number 10, October 2016 pp. 5579–5594

SERRIN’S REGULARITY RESULTS FOR THE INCOMPRESSIBLE LIQUID CRYSTALS SYSTEM

Xian-Gao School of Mathematic Sciences, Fudan University Shanghai, China Jianzhong Min School of Mathematic Sciences Fudan University/Shanghai University of Medicine and Health Sciences Shanghai, China Kui School of Mathematic Sciences, Soochow University Suzhou, China Xiaotao School of Mathematic Sciences, Fudan University Shanghai, China

(Communicated by Chongchun )

Abstract. In this paper, we study the simplified system of the original Erick- sen–Leslie equations for the flow of liquid crystals [10]. Under Serrin criteria [13], we prove a partial interior regularity result of weak solutions for the three- dimensional incompressible liquid crystal system.

1. Introduction. The three-dimensional incompressible liquid crystals system are the following coupled equations   ut − ∆u + u · ∇u + ∇P = −div(∇d ∇d), div u = 0, (1.1)  dt + u · ∇d − ∆d = f(d), 3 in Q = Ω×[0,T ], for a bounded and smooth domain Ω in R . Where u = (u1, u2, u3) is the velocity field, P is the scalar pressure and d = (d1, d2, d3) is the optical 1 2 molecule direction after penalization, and f(d) = σ2 (|d| − 1)d. Here ∇d ∇d is a symmetric tensor with its component (∇d ∇d)ij given by ∇id · ∇jd. And the initial and boundary conditions are:

u(x, 0) = u0(x), with div(u0) = 0, d(x, 0) = d0(x), for x ∈ Ω

u(x, t) = 0, d(x, t) = d0(x), for (x, t) ∈ ∂Ω × [0,T ] System (1.1) is the simplified system of the original Ericksen–Leslie equations for the flow of liquid crystals. For this system, -Liu [11] proved a regularity result of the suitable weak solution under the C-K-N condition, see [1]. Shortly after,

2010 Mathematics Subject Classification. Primary: 35Q35; Secondary: 76A15. Key words and phrases. Incompressible liquid crystals, Serrin criteria, partial regularity, Calder´on-Zygmund inequality, heat kernel theory.

5579 5580 XIAN-GAO LIU, JIANZHONG MIN, KUI WANG AND XIAOTAO ZHANG

Lin-Lin-Wang [12] also studied the regularity for the nematic liquid crystal system in dimension two, which is similar to system (1.1) with the third equation replaced by 2 dt + u · ∇d − ∆d = |∇d| d. Recently, -Lin-Liu-Wang [7] studied finite time singularity of the nematic liquid crystal flow in dimension three. For more results on the nematic liquid crystal flow, we refer to nice papers [8, 10]. In this article, we mainly focus on the regularity of weak solutions to the incom- pressible liquid crystals system (1.1), which has a lot in common with Naiver-Stokes system. For Naiver-Stokes equations, there are some excellent regularity results, see for example [1,4,5, 15]. Serrin [13] proved that weak solution u is locally regular r,s in the spatial variables if in addition u ∈ Lloc(Q) with 3/r + 2/s < 1. Here and thereafter, u ∈ Lr,s(Q) means Z T Z r s ( |u| dx) r dt < ∞. 0 Ω Shortly, Jones and Riviere [4] extended Serrin’s result [13] to equality case: 3/r + 2/s = 1, see also [5, 14, 15]. We call u satisfying the Serrin condition provided that r,s u ∈ Lloc with 3/r + 2/s = 1. Here, we aim to develop Serrin’s regularity result for liquid crystal system (1.1). To begin with, we recall the definition of weak solutions to system (1.1). Let Q be a region of space-time (R3 × R) and define the space of test functions D by  ∞ 3 D := X : X ∈ C0 (Q : R ), div X = 0 . Definition 1.1. A pair (u, d) is called a weak solution of (1.1) in Q if it satisfies system (1.1) in the following distribution sense. ∞ 1. For any ϕ ∈ C0 (Q), Z , ∇ϕi dxdt = 0; (1.2) Q 2. for any Φ ∈ D, Z Z hu, Φti + hu, ∆Φi + hu, u · ∇Φi dxdt = − h∇d ∇d, ∇Φi dxdt; (1.3) Q Q ∞ 3 3. for any Ψ ∈ C0 (Q; R ), Z Z hd, Ψti + hd, ∆Ψi + hd, u · ∇Ψi dxdt = − hF (d), Ψi dxdt. (1.4) Q Q Here hu, vi denotes the ordinary product in R3. Remark 1.2. [10, Theorem A] It has been proven that with suitable initial- boundary data, there exits a weak solution (u, d) satisfying u ∈ L2,∞(Q), |∇u| ∈ L2,2(Q), |∇d| ∈ L2,∞(Q) and |∇2d| ∈ L2,2(Q). Motivated by Serrin’s interior regularity of weak solutions of Navier-stokes equa- tions, we prove the following regularity result for the weak solutions of liquid crystals system. Theorem 1.3. Let (u, d) be a weak solution defined in Definition 1.1, satisfying u ∈ L2,∞(Q), |∇u| ∈ L2,2(Q), |∇d| ∈ L2,∞(Q) and |∇2d| ∈ L2,2(Q). SERRIN’S REGULARITY RESULTS 5581

0 s,s 0 Suppose further that u ∈ Lloc (Q) with 1 < s, s < ∞ and 3 2 + ≤ 1. (1.5) s s0 Then both u and d are of class C∞ in the space variables, and each derivative is locally bounded (and hence regular) in Q. Remark 1.4. For Navier-Stokes equations, Iscauriaza, Seregin and Sverak [9] proved Serrin’s result in the critical case s = 3, s0 = ∞. We mention here that Theorem 1.3 is also valid for s = 3, s0 = ∞ provided by an extra assumption that there is an R > 0 such that Z |u(·, t)|3 dx ≤  T BR Ω uniformly for all t and for some absolute constant  > 0. The proof is same as that of Theorem 1.3. Remark 1.5. Theorem 1.3 is true for any dimension n, with corresponding Serrin condition n 2 + = 1. s s0 The paper is organized as follows. In section 2, we derive some representation 3 2 formulas of u and d. In section 3, we prove the theorem under condition s + s0 < 1. In the last section, we complete the proof of Theorem 1.3.

2. Some representation formulas. In this section we assume that (u, d) is a weak solution characterized in the previous section. To begin with, we recall the representation formula of u given in [13]. 2.1. Representation of u. We denote by w the vorticity of u, that is w(x, t) = curl u(x, t) = ∗∇u(x, t) = ∇ × u(x, t). −4 Let α be the usual mollifier in the space-time space and αh(x, t) = h α(x/h, t/h). It is easy to check that αh is a mollifier supported in Bh(0, 0), where Bh(0, 0) is the round ball centered at the origin with radius h in R3 × R. We then modify u and w by Z uh(x, t) = u(y, s) αh(x − y, t − s) dyds, Z wh(x, t) = w(y, s) αh(x − y, t − s) dyds. Then one can check easily that

wh = curl uh. Since div u(x, t) = 0 in distribution sense, it then follows that

div uh(x, t) = 0. Thereby, we have the following equation via direct calculations

∆(∗uh) = −∇wh. (2.1) Here ∗ denotes the Hodge star operator, for more details one can refer chapter 1 in [2]. From equation (2.1), one can deduce easily the following lemma. 5582 XIAN-GAO LIU, JIANZHONG MIN, KUI WANG AND XIAOTAO ZHANG

Lemma 2.1. [13, Lemma 2] Suppose that as functions of the space variables, u and |∇u| are locally of class L1 in Ω. Then in any compact G in Ω, we have Z ∗ u(x, t) = ∇H(x − ξ) ∧ w(ξ, t) dξ + A(x, t), (2.2) G 1 where H(x) = 4π|x| is the fundamental solution of Laplace equation and A(x, t) is a harmonic 2-form in G for any t. Via direct calculation, one can find (2.2) is equivalent to 1 Z u(x, t) = ∇|x − ξ|−1 × w(ξ, t) dξ + ∗A(x, t). (2.3) 4π G

2.2. Representation of w, ∇d, ∇2d.

2 2.2.1. Deriving equations of w, ∇d, ∇ d. Let√ Gh(x, y, t) be the Green function in the ball centered at the origin with radius h2 − t2 in R3. Then we define a vector T function E = (E1,E2,E3) by Z E(x, t) = √ Gh(x, y, t) ∇αh(y, t) dy, |y|≤ h2−t2 and a scalar function ρh(x, t) by Z ρh(x, t) = √ Gh(x, y, t) αh(y, t) dy. |y|≤ h2−t2 From the properties of Green function (c.f. [3]), it follows that

∆E(x, t) = ∇αh(x, t), (2.4) ∞ and E(x, t) ∈ C0 (Bh(0, 0)). Denote g1 = u ⊗ w − w ⊗ u and g2 = ∆d · ∇d, we then have the following modified equations.

2 Lemma 2.2. For modified functions uh, wh, (∇d)h and (∇ d)h, we have

(uh)t − ∆uh + (div(u ⊗ u)h) + ∇Ph = − div(∇d ∇d)h, (2.5)

1 2 (wh)t − ∆wh = − div gh − curl gh, (2.6)

(∇dh)t − ∆∇dh = −∇(u · ∇d)h + ∇f(d)h, (2.7) and 2 2 2 2 (∇ dh)t − ∆∇ dh = −∇ (u · ∇d)h + ∇ f(d)h, (2.8) where Ph(x, t) is defined by Z Ph(x, t) = (div(u · ∇u) + div div(∇d ∇d)) (ξ, τ) ρh(ξ − x, τ − t) dξdτ. (2.9) Q Proof. To prove equation (2.5), it suffices to show it holds at the origin (0, 0). T Set Φ1 = (αh, 0, 0) , and equation (2.4) gives

Φ1 − ∇E1 ∈ D. SERRIN’S REGULARITY RESULTS 5583

Then using Φ1 − ∇E1 as a testing function in (1.3), we obtain Z hu, (Φ1 − ∇E1)ti + hu, ∆(Φ1 − ∇E1)i + hu, u · ∇(Φ1 − ∇E1)i dxdt Q Z = − h∇d ∇d, ∇(Φ1 − ∇E1)i dxdt. Q Using (1.2), we compute Z Z 1 1 hu, (Φ1 − ∇E1)ti dxdt = u (αh(x, t))t dxdt = −(uh)t(0, 0), Q Q Z 1 hu, ∆(Φ1 − ∇E1)i dxdt = ∆uh(0, 0), Q Z hu, u · ∇(Φ1 − ∇E1)i dxdt Q Z 1 = − (div(u ⊗ u)h)(0, 0) − div(u · ∇u)(x, t) E1(x, t) dxdt Q Z  1 ∂ = − (div(u ⊗ u)h)(0, 0) − div(u · ∇u)(ξ, τ) ρh(ξ − x, τ − t) dξdτ (0, 0), ∂x1 Q and Z h∇d ∇d, ∇(Φ1 − ∇E1)i dxdt Q Z 1 = −div(∇d ∇d)h(0, 0) + div(∇d ∇d)(x, t) ∇E1(x, t) dxdt Q 1 = −div(∇d ∇d)h(0, 0) ∂ Z  − divdiv(∇d ∇d)(ξ, τ) ρh(ξ − x, τ − t) dξdτ (0, 0). ∂x1 Q Then we conclude from above identities that

1 1 1 ∂ 1 (uh)t − ∆uh + (div(u ⊗ u)h) + Ph = −div(∇d ∇d)h, ∂x1 holds at (0, 0) which immediately implies (2.5). To prove (2.6), taking the vorticity of equation (2.5) yields

(wh)t − ∆wh = −curl div(u ⊗ u)h − curl div(∇d ∇d)h. Thus (2.6) is valid if one can show

1 2 curl div(u ⊗ u)h = div gh and curl div(∇d ∇d)h = curl gh. (2.10) In fact if u and d are smooth, then one can show the following identities curl div(u ⊗ u) = div(u ⊗ w − w ⊗ u) and curl div(∇d ∇d) = curl (∆d · ∇d). For general case, (2.10) holds via approximations. Similarly it follows from (1.4) that

(dh)t + (u · ∇d)h − ∆(dh) = f(d)h, 2 and taking the ∇ and ∇ , equations (2.7) and (2.8) hold respectively.  5584 XIAN-GAO LIU, JIANZHONG MIN, KUI WANG AND XIAOTAO ZHANG

2.2.2. Obtaining the representations of w, ∇d, ∇2d. Let K be the fundamental so- lution of heat equation, i.e. ( 2 − 3 − |x| K(x, t) = Ct 2 e 4t , t > 0, 0, t ≤ 0. By Young inequality, one can easily prove the following lemma. Lemma 2.3. [13, Lemma 1] Let k = ∂K/∂x and h(x, t) = R k(x−ξ, x−t) g(ξ, τ) dξ dτ. Then in any region Q = Ω × (T1,T2) we have

k h kLr,r0 (Q)≤ Const · k g kLq,q0 (Q) (2.11) provided that 1 1 1 1 1 ≤ q ≤ r, 1 ≤ q0 ≤ r0, 3( − ) + 2( − ) < 1. q r q0 r0

The constant depends only on T = T2 − T1 and upon the exponents. Now we use the equations in Lemma 2.2 to deduce representation formulas. Lemma 2.4. Assume that u ∈ L2,∞(Q), |∇u| ∈ L2,2(Q), |∇d| ∈ L2,∞(Q) and |∇2d| ∈ L2,2(Q).

Then in any compact region Q0 = G × (0,T ) of Q, we have Z Z 1 2 w = ∇K(x−ξ, t−τ)· g (ξ, τ) dξdτ + ∇K(x−ξ, t−τ)×g (ξ, τ) dξdτ +B1, Q0 Q0 (2.12) Z ∇d = ∇K(x − ξ, t − τ) ⊗ (u(ξ, τ) · ∇d(ξ, τ)) dξdτ Q0 Z − ∇K(x − ξ, t − τ) ⊗ f(ξ, τ) dξdτ + B2, (2.13) Q0 and Z ∇2d = ∇K(x − ξ, t − τ) ⊗ (∇u · ∇d)(ξ, τ) dξdτ (2.14) Q0 Z + ∇K(x − ξ, t − τ) ⊗ (u · ∇2d)(ξ, τ) dξdτ Q0 Z − ∇K(x − ξ, t − τ) ⊗ ∇f(ξ, τ) dξdτ + B3, Q0 where Bi is a i-form satisfying the heat equation, i = 1, 2, 3. To avoid misrepresen- ∂uj ∂di tation, we point out that (∇u · ∇d)ik = . ∂xk ∂xj Proof. Let Z Z 1 2 B1,h = wh − ∇K(x−ξ, t−τ)· gh(ξ, τ) dξdτ − ∇K(x−ξ, t−τ)×gh(ξ, τ) dξdτ. Q0 Q0 (2.15) Then it follows from equation (2.6) that

(B1,h)t − ∆B1,h = 0. 1 2 1 2 By the assumption of u and d, we see that gh and gh converge to g and g 1,1 R in L (Q0) respectively. Then from Young inequality we deduce ∇K(x − Q0 SERRIN’S REGULARITY RESULTS 5585

ξ, t − τ) · g1(ξ, τ) dξdτ and R ∇K(x − ξ, t − τ) × g2(ξ, τ) dξdτ also converge h Q0 h to R ∇K(x − ξ, t − τ) · g1(ξ, τ) dξdτ and R ∇K(x − ξ, t − τ) × g2(ξ, τ) dξdτ in Q0 Q0 1,1 L (Q0) respectively. Meanwhile, it is obviously that B1,h is uniformly bounded 1,1 1,1 in L (Q0). Therefore there exists B1 in L (Q0) satisfying the heat equation and 1,1 B1,h converges to B1 weakly in L (Q0). Since B1,h satisfies the heat equation, then B1 also satisfies the heat equation. Finally letting h goes to zero in (2.15), we obtain formula (2.12). By analogous strategies, formulas (2.13) and (2.14) hold true. 

3 2 3 2 3. The case: s + s0 < 1. In this section, we mainly consider the case s + s0 < 1 and use Serrin’s approach in [13] to show Theorem 3.1. Let u be a weak solution defined in Definition 1.1, satisfying u ∈ L2,∞(Q), |∇u| ∈ L2,2(Q), |∇d| ∈ L2,∞(Q) and |∇2d| ∈ L2,2(Q).

0 Suppose further that u ∈ Ls,s (Q) with 3 2 + < 1. (3.1) s s0 Then u and d are of class C∞ in the space variables, and each derivative is bounded locally in Q. Proof. We divide the proof into three steps.

Step 1. Prove ∇d ∈ L∞,∞(Q). Since d is bounded (e.g. [10]) and f(d) is smooth with respect to d, then f(d) ∈ L∞,∞(Q). We assert from Lemma 2.3 that Z ∞,∞ ∇K(x − ξ, t − τ) ⊗ f(ξ, τ) dξdτ ∈ L (Q0). Q0 Now we only need to consider the first term Z ∇K(x − ξ, t − τ) ⊗ (u(ξ, τ) · ∇d(ξ, τ)) dξdτ Q0 ρ,ρ0 in the representation (2.13) of ∇d. Suppose for the moment that ∇d ∈ L (Q0) with ρ, ρ0 ≥ 2. It is evident that q,q0 u · ∇d ∈ L (Q0), which follows from H¨olderinequality and where the exponents q, q0 are given through 1 1 1 1 1 1 = + , = + . q s ρ q0 s0 ρ0 Due to condition (3.1), we define a positive κ by 1  3 2  κ = 1 − ( + ) , 6 s s0 and set 1 1 1 1 = − κ and = − κ, r ρ r0 ρ0 if κρ ≥ 1, we set r = ∞, and similarly for r0. Clearly, we have 1 1 1 1 3( − ) + 2( − ) < 1. q r q0 r0 5586 XIAN-GAO LIU, JIANZHONG MIN, KUI WANG AND XIAOTAO ZHANG

Thus using Lemma 2.3 again, we assert Z r,r0 ∇K(x − ξ, t − τ) ⊗ (u(ξ, τ) · ∇d(ξ, τ)) dξdτ ∈ L (Q0). Q0

0 Since r and r0 are larger than ρ and ρ0, then we have that ∇d is in a higher class Lr,r than originally supposed. Repeating this step until ρ, ρ0 ≥ κ−1, we then conclude ∞,∞ ∇d ∈ L (Q0).

2 ∞,∞ ∞,∞ Step 2. Prove w, ∇u, ∇ d ∈ L (Q0). Recall from step 1 that ∇d ∈ L (Q0). Therefore it is left to consider terms Z ∇K(x − ξ, t − τ) ⊗ (∇u · ∇d)(ξ, τ) dξdτ Q0 Z and ∇K(x − ξ, t − τ) ⊗ (u · ∇2d)(ξ, τ) dξdτ Q0 in the representation (2.14) of ∇2d, and Z Z ∇K(x − ξ, t − τ) · g1(ξ, τ) dξdτ and ∇K(x − ξ, t − τ) × g2(ξ, τ) dξdτ Q0 Q0 in the representation (2.12) of w. 2 ρ,ρ0 0 Similarly as in step 1, we assume that ∇ d, w, ∇u ∈ L (Q0) with ρ, ρ ≥ 2. Using H¨olderinequality, we deduce that

1 2 2 q,q0 g , g , ∇u · ∇d, u · ∇ d ∈ L (Q0) with 1 1 1 1 1 1 = + and = + . q s ρ q0 s0 ρ0 By the same process as in step 1, we obtain Z 1 1 k ∇K(x − ξ, t − τ) · g (ξ, τ) dξdτ k r,r0 ≤ C k g k q,q0 , L (Q0) L (Q0) Q0

Z 2 2 k ∇K(x − ξ, t − τ) × g (ξ, τ) dξdτ k r,r0 ≤ C k g k q,q0 , L (Q0) L (Q0) Q0

Z k ∇K(x − ξ, t − τ) · (∇u · ∇d)(ξ, τ) dξdτ k r,r0 ≤ C k ∇u · ∇d k q,q0 , L (Q0) L (Q0) Q0 and Z 2 2 k ∇K(x − ξ, t − τ) · (u · ∇ d)(ξ, τ) dξdτ k r,r0 ≤ C k u · ∇ d k q,q0 L (Q0) L (Q0) Q0

0 r,r0 2 with same exponents r, r as in step 1. Then we have w ∈ L (Q0) and ∇ d ∈ r,r0 r,r0 L (Q0). From the representation (2.3) of u, it follows that u ∈ L (Q0) and r,r0 2 ∇u ∈ L (Q0). Thus we can repeat this process, and finally we obtain w, ∇ d ∈ ∞,∞ L (Q0). SERRIN’S REGULARITY RESULTS 5587

Step 3. Prove that d and u are of class C∞ in the space variables. We 2 ∞,∞ 2 ∞,∞ already obtain in step 2 that ∇d, ∇ d ∈ L (Q0), then g ∈ L (Q0). Recall from Lemma 2.1 that 1 Z u(x, t) = ∇|x − ξ|−1 × w(ξ, t) dξ + ∗A(x, t). (3.2) 4π G ∞,∞ −1 1 Since w ∈ L (Q0) and ∇|x − ξ| ∈ L (Q0), then Young inequality yields ∞,∞ 1 ∞,∞ ∞,∞ u ∈ L (Q0), hence g ∈ L (Q0). It follows from |∇d| + |∆d| ∈ L (Q0) 2 ∞,∞ that g = ∆d · ∇d ∈ L (Q0). Therefore from formula (2.12), we deduce w is spatially H¨oldercontinuous in Q0, with arbitrary exponent θ < 1. Now from (3.2), ∇u is also H¨odercontinuous, then ∇u · ∇d ∈ L∞,∞ and with the formula (2.14) 2 ∞,∞ 2 and u · ∇ d, ∇f(d) ∈ L , we have ∇ d is spatially H¨oldercontinuous in Q0 with arbitrary exponent θ < 1. Thus g1, g2, u · ∇d, u · ∇2d, ∇u · ∇d, f(d),∇f(d) ∞,∞ ∈ L (Q0) are H¨oldercontinuous. And again by the potential theory of the heat convolution we have that wx, (∇d)x and (∆d)x are H¨oldercontinuous. Repeating ∞ this argument, we conclude that d, u ∈ C (Q0) in the space variables. 

Similarly as Navier-Stokes system, for time regularity of a weak solution we need extra assumptions on time derivatives (see the explanations in [13]). For liquid crystal system we prove

2,p Corollary 3.2. Assume additionally that ut, dt ∈ L (Q) with p ≥ 1. Then the space derivatives of u, d are absolutely continuous functions of time. Moreover, there exists a strongly differentiable function P = P (x, t) such that

ut − ∆u + u · ∇u = − div(∇d ∇d) − ∇P (3.3) almost everywhere in Q. Proof. First of all, observe that (2.6) implies

1 2 wt = ∆w − div g − curl g in the weak sense that already been proved. Differentiation of (3.2) yields Z 1 −1 ut(x, t) = ∇|x − ξ| × wt(ξ, t) dξ + ∗At(x, t). 4π G ∞ p Thus ut is of class C in the spatial variables, and each derivative is of class L in time by the assumption. Thereby

Z t2 m m m m 1− 1 p p |D u(x, t2)−D u(x, t1)| ≤ |D ut(x, t)| dt ≤k D ut(x, t) kL (t1,t2) |t2−t1| , t1 proving that the space derivatives of u are absolutely continuous functions of time. Recall from (1.1) that

dt = ∆d − u · ∇d + f(d),

∞ thus dt is of class C in the spatial variables and then the space derivatives of d are absolutely continuous functions of time. We note that (3.3) follows by letting h tend to zero in (2.6), and then P (x, t) ∈ p ∞ L ((0,T ); C (Ω)), completing the proof.  5588 XIAN-GAO LIU, JIANZHONG MIN, KUI WANG AND XIAOTAO ZHANG

3 2 4. The equality case: s + s0 = 1. From previous sections, we know w, ∇d and ∇2d weakly satisfy the following equations respectively

wt − ∆w + div(u ⊗ w − w ⊗ u) = −curl(∆d · ∇d), (4.1)

∇dt − ∆∇d + ∇(u · ∇d) = ∇f(d), (4.2) and 2 2 2 2 ∇ dt − ∆∇ d + ∇(∇u · ∇d + u · ∇ d) = ∇ f(d), (4.3) see Lemma 2.4. Firstly, we use equation (4.2) to improve regularity of ∇d. More precisely we prove Lemma 4.1. With the same assumption as in Theorem 1.3, then for each integer 0 2l 0 l ≥ 1 and any sufficient small compact subset Q b Q, we have ∇d ∈ L (Q ) with 0  k ∇d kL2l(Q0)≤ C(dist(Q , ∂Q), ε) k f(d) kL2l(Q) + k ∇d kL2(Q) where C(dist(Q0, ∂Q), ε) denotes a constant depending on dist(Q0, ∂Q) and ε. Proof. Step 1. Assume firstly that u is uniformly bounded. ∞ 2l−2 2 Let ϕ ∈ C0 (Q) be a cut-off function. Multiplying (4.2) by |∇d| ϕ ∇d (l ≥ 1) and integrating, we then have Z 2l |∇d| 2 2 2 2l−2 2 2 2 2l−4 2 ∂t( ϕ ) + |∇ d| |∇d| ϕ + |∇|∇d| | |∇d| ϕ dxdt Q 2l Z 1 2l 2l−2 2 = |∇d| ϕ ∂tϕ − |∇d| ϕ ∇|∇d| · ∇ϕ dxdt Q l Z + (u · ∇d) · div(∇d|∇d|2l−2ϕ2) − f(d) · div(∇d|∇d|2l−2ϕ2) dxdt. Q Therefore standard estimates implies Z Z sup (|∇d|lϕ)2 dx + |∇(|∇d|lϕ)|2 dxdt t Ω Q Z Z ≤ C(ϕ) |∇d|2l dxdt + C |u|2|∇d|2lϕ2 + |u||∇d|2l|ϕ||∇ϕ| dxdt Q Q Z (4.4) + C |f(d)|2|∇d|2l−2ϕ2 + |f(d)||∇d|2l−1|ϕ||∇ϕ| dxdt Q Z Z Z ≤ C(ϕ) |∇d|2l dxdt + C |u|2|∇d|2lϕ2 dxdt + C(ϕ) |f(d)|2l dxdt, Q Q Q where C(ϕ) is a constant depending on ϕ. Therefore estimates (4.4) holds provided by w, f ∈ L2l(Q) and u is bounded in Q. The last step used Young’s inequality. It follows from H¨olderinequality and Sobolev’s embedding theorem that |∇d|sϕ ∈ π,ρ 3 2 3 L (Q) with π + ρ = 2 , and moreover Z Z l 2 l 2 l 2 k |∇d| ϕ kLπ,ρ(Q)≤ C sup |∇d| ϕ| dx + C |∇(|∇d| ϕ)| dxdt, (4.5) Ω Q (see for example inequality (3.3) in [14]). Using H¨olderinequality again, we obtain Z 2 l 2 2 l 2 |u| (|∇d| ϕ) dxdt ≤k u kLs,s0 (supp ϕ) k |∇d| ϕ kLp∗,q∗ (Q), (4.6) Q SERRIN’S REGULARITY RESULTS 5589 with 1 1 1 1 1 1 = − and = − . p∗ 2 s q∗ 2 s0 Since 3/p∗ + 2/q∗ ≥ 3/2 from assumption 3/s + 2/s0 ≤ 1, then combining (4.4), (4.5) and (4.6) we obtain that l 2 l 2 2l k |∇d| ϕ kLp∗,q∗ (Q) ≤ C(ϕ) k |∇d| kL2(Q) +C(ϕ) k f(d) kL2l(Q) 2 l 2 +C k u kLs,s0 (supp ϕ)k |∇d| ϕ kLp∗,q∗ (Q) . (4.7) By absolutely continuity of the Lebesgue integral for finite p, q, for any sufficiently 2 small  we can take supp ϕ small enough such that k u k 0 ≤ ε. Then we Ls,s (supp ϕ) assert from (4.7) that l 2 l 2 2l k |∇d| ϕ kLp∗,q∗ (Q)≤ C(ϕ) k |∇d| kL2(Q) +C(ϕ) k f(d) kL2l(Q) . (4.8) Choosing π = ρ = 10/3 in (4.5), then we obtain from (4.4)-(4.8) that 5 |∇d|lϕ ∈ L2β(Q), β = > 1. 3

Letting l0 = 1, lk+1 = βlk for k ∈ N and Q0 = Q, Qk+1 = {(x, t)|ϕk(x, t) ≥ 1} and ∞ iterating (4.8) with suitable functions ϕk+1 ∈ C0 (Qk) (that means ϕk is supported in Qk−1 and ϕk ≥ 1 in the Qk), we finally get 0  k ∇d kL2l(Q0)≤ C(dist(Q , ∂Q), ε) k f(d) kL2l(Q) + k ∇d kL2(Q) (4.9) for any compact subdomain Q0 and l < ∞. Step 2. Remove the assumption that u is uniformly bounded. We now do approximations to obtain (4.9) for general u . Let Q00 is a subdomain k of Q, but we still sign it as Q for convenient. And let u = u ∗ αhk , where αhk is the usual mollifier and hk → 0 as k → ∞. Clearly, uk → u as k → ∞ in L2(Q), and for all k k k u kLs,s0 (Q)≤ C k u kLs,s0 (Q), where C is an absolutely constant. Let Dk be the solution to the equations k Dt − ∆D + ∇(u · D) = ∇f(d), (4.10) with the initial and boundary data

Dk = ∇d on ∂Ω × [0,T ] = Σ and Ω × {0}. By the well-known Courant-Lebesgue lemma (see also Lemma 2.4 in [14]), we can assume that 2,2 2 ∇d|Σ ∈ L (Σ), ∇d|Ω×{0} ∈ L (Ω × {0}). Besides

||∇d||L2(Ω×{0}) + ||∇d||L2(Σ) ≤ C||∇d||L2(Q) 0 Here, for any small , we can take suitable domain Q b Q such that k k u kLs,s0 (Q0)≤ C k u kLs,s0 (Q0)< Cε. Let w∗ be the solution of the backward heat equation ∗ ∗ ∗ − ∂tw − ∆w = f in Q (4.11) 5590 XIAN-GAO LIU, JIANZHONG MIN, KUI WANG AND XIAOTAO ZHANG with initial and boundary data w∗ = 0 on ∂Ω × [0,T ] ∪ Ω × {T }, where f ∗ ∈ L2(Q : R3 × R3). Then it is well-known that ∗ ∗ 2 ∗ ∗ sup k ∇w (t) kL2(Ω) + k ∂tw kL2(Q) + k ∇ w kL2(Q)≤ C k f kL2(Q) . (4.12) 0≤t≤T Similarly as inequality (4.5), we deduce from H¨olderinequality and Sobolev’s em- bedding theorem that ∇w∗ ∈ Lp∗,q∗(Q) with ∗ ∗ k ∇w kLp∗,q∗(Q)≤ C k f kL2(Q), (4.13) where 1/s + 1/p∗ = 1/2 and 1/s0 + 1/q∗ = 1/2. Like ∇d, by Courant-Lebesgue theorem, we obtain ∗ ∗ ∗ ∗ ||w ||L2(Ω×{0}) + ||w ||L2(Σ) ≤ C||w ||L2(Q) ≤ C||f ||L2(Q), Thus we conclude Z Z T Z ∗ ∂ ∗ ∗ ∇d : w dx − ∇d : w dσdt ≤ C(||∇d||L2,2 ) k f kL2(Q) . (4.14) Ω×{0} 0 ∂Ω ∂n Here for two matrixes A, B ∈ R3 × R3 we define A : B = tr(AB). Integrating by parts, we estimate Z ∗ Dk : f dxdt Q Z ∗ = Dk :(−∂t − ∆)w dxdt Q Z Z T Z ∗ ∂ ∗ = (∇f(d) − ∇(uk · Dk)) : w dxdt − ∇d : w dσdt Q 0 ∂Ω ∂n Z (4.15) + ∇d : w∗ dx Ω×{0} ∗ ∗ ≤ k f(d) kL2(Q)k ∇w kL2(Q) + k Dk kL2(Q)k uk kLs,s0 (Q)k ∇w kLp∗,q∗(Q) ∗ + C(||∇d||L2,2 ) k f kL2(Q) ∗ ∗ ≤C k f(d) kL2(Q)k f kL2(Q) +C k Dk kL2(Q)k uk kLs,s0 (Q)k f kL2(Q) ∗ + C(||∇d||L2,2 ) k f kL2(Q), where we used (4.13-4.15). Since inequality (4.15) holds for all f ∗ ∈ L2(Q), then

k Dk kL2(Q)≤ Cε k Dk kL2(Q) +C k f(d) kL2(Q) +C(||∇d||L2,2 ), 2 and thereby Dk is uniformly bounded in L (Q). Exactly same proof shows Dk is 2 unique. Hence Dk converges to D weakly in L (Q). 2l 0 From step 1, the L (Q ) norm of Dk is uniformly bounded by

k Dk kL2l(Q0)≤ C k f(d) kL2l(Q) +C(||∇d||L2,2 ) 1 ≤ l < ∞, (4.16) 0 for any compact Q b Q. Thus we complete the proof.  Now we are in position to prove the main theorem.

Proof of Theorem 1.3. Assuming that both u and ∇d are uniformly bounded, we prove higher regularity of w and ∇2d via equations (4.1) and (4.3) respectively. SERRIN’S REGULARITY RESULTS 5591

Similarly we choose test functions as w|w|2l−2ϕ2 and ∇2d|∇2d|2l−2ϕ2 for the equations (4.1) and (4.3) respectively. Then we obtain Z 2 2l |∇ d| 2 2 2 2l−2 2 2 2 2 2 2l−4 2 ∂t( ϕ) + |∇∇ d| |∇ d| ϕ + |∇|∇ d| | |∇ d| ϕ dxdt Q 2l Z 1 2 2l 2 2 2 2l−2 = |∇ d| ϕ∂tϕ − ∇|∇ d| |∇ d| ϕ · ∇ϕ Q l Z + (∇u · ∇d) · (∇(∇2d|∇2d|2l−2ϕ2)) + (u · ∇2d) · (∇(∇2d|∇2d|2l−2ϕ2)) Q Z + ∇f(d) · (∇(∇2d|∇2d|2l−2ϕ2)), Q and Z  2l 2  |w| ϕ 2 2l−2 2 1 2 2 2l−4 2 ∂t + |∇w| |w| ϕ + (l − 1)|∇|w| | |w| ϕ Q 2l 2 Z 2l |w| 2 2l−2 2l−2 2 = ϕ∂tϕ − ∇|w| |w| ϕ∇ϕ + (u ⊗ w − w ⊗ u): ∇(w|w| ϕ ) Q l Z − curl(∆d · ∇d)(w|w|2l−2ϕ2). Q Then by standard estimate Z Z sup (|∇2d|lϕ)2 dx + |∇(|∇2d|lϕ)|2 dxdt t Ω Q Z 2 2l 2 2 2 2l−2 2 2 2 2 2 2l−2 2 ≤C |∇ d| ϕ|∂tϕ| + |∇u| |∇d| |∇ d| ϕ + |u| |∇ d| |∇ d| ϕ Q Z + C (|∇u||∇d| + |u||∇2d|)|∇2d|2l−1ϕ|∇ϕ| Q Z (4.17) + C(ϕ) |∇f(d)|2|∇2d|2l−2 + |∇f(d)||∇2d|2l−1 Q Z Z ≤C(ϕ) |∇2d|2l + C |∇d|2(|∇u|2l + |∇2d|2l)ϕ2 Q Q Z Z + C |u|2|∇2d|2lϕ2 dxdt + C(ϕ) |∇f(d)|2l Q Q and Z Z sup (|w|lϕ)2 + |∇(|w|lϕ)|2 t Ω Q Z 2l 2l 2 2l 2 ≤C |w| |ϕ∂tϕ| + |u||w| ϕ|∇ϕ| + |u| |w| ϕ Q Z (4.18) + C |∆d · ∇d|2|w|2l−2ϕ2 Q Z Z Z ≤C(ϕ) |w|2l + C |u|2(|w|lϕ)2 + C (|∇d|l)2(|∆d|lϕ)2. Q Q Q From inequality (4.8), we deduce Z Z 2 l 2 2 l 2 2 l 2 k |∇ d| ϕ kLπ,ρ(Q)≤ C sup |∇ d| ϕ| dx + C |∇(|∇ d| ϕ)| dxdt, (4.19) t Ω Q 5592 XIAN-GAO LIU, JIANZHONG MIN, KUI WANG AND XIAOTAO ZHANG and Z Z l 2 l 2 l 2 k |w| ϕ kLπ,ρ(Q)≤ C sup | |w| ϕ| dx + C |∇(|w| ϕ)| dxdt. (4.20) t Ω Q Recall that 1 Z u(x, t) = ∇|x − ξ|−1 × w(ξ, t) dξ + A(x, t), 4π Ω where A is a harmonic function, and the other term is classical singular operator 0 applied to w. Suppose w ∈ Ls,s (Q) for any s, s0 ≥ 2. Since u ∈ L2,∞(Q), then from the representation of u and the Lp estimate of harmonic function, we conclude that   k ∇A kLs,s0 (Q0)≤ C k w kLs,s0 (Q) + k u kL2,∞(Q) .

0 where Q b Q. Then 1 Z ∇u = ∇(∇|x − ξ|−1 × w(ξ, t)) dξ + ∇A(x, t). 4π Ω By the Calder´on-Zygmund theorem (e.g. [6])

k (∇u)ϕ kLs,s0 (Q)≤ C(n, p) k w kLs,s0 (Q) +C(ϕ)C(||u||2,∞). By H¨older’sinequality, we estimate Z 2 l 2 2 l 2 |∇d| (|∇u| ϕ) dxdt ≤k ∇d kLs,s0 (supp ϕ)k |∇u| ϕ kLp∗,q∗ (Q) Q 2 l 2 ≤ C k ∇d kLs,s0 (supp ϕ) (k |w| kLp∗,q∗ (Q) +C(ϕ)C(||u||L2,∞(Q))), (4.21)

Z 2 2 l 2 2 2 l 2 |∇d| (|∇ d| ϕ) dxdt ≤k ∇d kLs,s0 (supp ϕ)k |∇ d| ϕ kLp∗,q∗ (Q), (4.22) Q Z 2 2 l 2 2 2 l 2 |u| (|∇ d| ϕ) dxdt ≤k u kLs,s0 (supp ϕ)k |∇ d| ϕ kLp∗,q∗ (Q), (4.23) Q Z 2 l 2 2 l 2 |u| (|∇u| ϕ) dxdt ≤k u kLs,s0 (supp ϕ)k |∇u| ϕ kLp∗,q∗ (Q) Q 2 l 2 ≤ C||u||Ls,s0 (supp ϕ)(k |w| kLp∗,q∗ (Q) +C(ϕ)C(k u kL2,∞(Q))), (4.24) and Z 2l 2 l 2 l 2 2 l 2 |∇d| (|∇ d| ϕ) dxdt ≤k |∇d| kLs,s0 (supp ϕ)k |∇ d| ϕ kLp∗,q∗ (Q) . (4.25) Q Where 1 1 1 1 1 1 = − , = − , p∗ 2 s q∗ 2 s0 and 3 2 3 3 2  + = + 1 − + . p∗ q∗ 2 s s0 SERRIN’S REGULARITY RESULTS 5593

In the hypothesis 1 < s, s0 < ∞ and 3/s + 2/s0 ≤ 1, then p∗ = 4/(s − 2) + 2 < 6 and 3/p∗ + 2/q∗ ≥ 3/2. So (4.17–4.25) yield that 2 l 2 l 2 k |∇ d| ϕ kLp∗,q∗(Q) + k |w| ϕ kLp∗,q∗(Q) 2 l 2 l 2 2 l 2 ∗ ∗ ≤C(ϕ) k |∇ d| kL2(Q) +C(ϕ) k |w| ϕ kL2(Q) +C k ∇d kLs,s0 (supp ϕ)k |w| kLp ,q (Q) 2 2 l 2 2 2 l 2 ∗ ∗ ∗ ∗ + C k ∇d kLs,s0 (supp ϕ)k |∇ d| ϕ kLp ,q (Q) +C k u kLs,s0 (supp ϕ)k |∇ d| ϕ kLp ,q (Q) 2 l 2 l 2 2 l 2 ∗ ∗ ∗ ∗ + C k u kLs,s0 (supp ϕ)k |w| kLp ,q (Q) +C k |∇d| kLs,s0 (supp ϕ)k |∇ d| ϕ kLp ,q (Q) 2l 2 2 + C(ϕ) k ∇f(d) kL2l +C(ϕ)C(k u kL2,∞(Q))(k u kLs,s0 (supp ϕ) + k ∇d kLs,s0 (supp ϕ)).

By absolutely continuity of the Lebesgue integral for finite s, s0, we can take 2 2 supp(ϕ) small enough such that k u k 0 ≤ ε, k ∇d k 0 ≤ ε and Ls,s (supp ϕ) Ls,s (supp ϕ) l 2 k |∇d| k 0 ≤ ε, where ε is sufficiently small. Then we obtain Ls,s (supp ϕ) 2 l 2 l 2 k |∇ d| ϕ kLp∗,q∗(Q) + k |w| ϕ kLp∗,q∗(Q)  2 l 2 l 2 2l  ≤ C(ϕ) k |∇ d| kL2(Q) + k |w| kL2(Q) + k ∇f(d) kL2l(Q) +C(k u kL2,∞(Q))

 2 l 2 l 2 2l  ≤ C(ϕ) k |∇ d| kL2(Q) + k |w| kL2(Q) + k ∇d kL2l(Q) +C(k u kL2,∞(Q)) . Therefore taking π = ρ = 10/3 in (4.19) and (4.20), we obtain 5 |∇2d|l, |∇u|l ∈ L2β(Q0), with β = . 3

3+ε0,∞ Then we do iterate similarly to ∇d to obtain that w ∈ Lloc (Q) for some ε0 > 0. For the general case, we do approximations similarly as in the step 2 of Lemma 4.1. 

Acknowledgments. The first author’s research is partially supported by NSFC No.11131005.

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