Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 232, pp. 1–12. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu REMARKS ON REGULARITY CRITERIA FOR THE 3D NAVIER-STOKES EQUATIONS RUIYING WEI, YIN LI Abstract. In this article, we study the regularity criteria for the 3D NavierStokes equations involving derivatives of the partial components of the velocity. It is proved that if ∇h u e belongs to Triebel-Lizorkin space, ∇u3 or u3 belongs to Morrey-Campanato space, then the solution remains smooth on [0, T ]. 1. Introduction This article is devoted to the Cauchy problem for the following incompressible 3D Navier-Stokes equation: ut + (u · ∇)u + ∇p = 4u, div u = 0, x ∈ R3 , t > 0 x ∈ R3 , t > 0 (1.1) with initial data u(x, 0) = u0 , x ∈ R3 , (1.2) where u = (u1 (x, t), u2 (x, t), u3 (x, t)) and p = p(x, t) denote the unknown velocity vector and the unknown scalar pressure, respectively. In the last century, Leray [11] and Hopf [8] proved the global existence of a weak solution u(x, t) ∈ L∞ (0, ∞; L2 (R3 )) ∩ L2 (0, ∞; H 1 (R3 )) to (1.1)-(1.3) for any given initial datum u0 (x) ∈ L2 (R3 ). However, whether or not such a weak solution is regular and unique is still a challenging open problem. From that time on, different criteria for regularity of the weak solutions has been proposed. The classical Prodi-Serrin conditions (see [16, 18, 19]) say that if 2 3 + = 1, 3 ≤ s ≤ ∞, t s then the solution is smooth. Similar results is showed by Beirão da Veiga [1] involving the velocity gradient growth condition: 2 3 3 ∇u ∈ Lt (0, T ; Ls (R3 )), + = 2, < s ≤ ∞. t s 2 Actually, whether the weak solution is smooth when a part of the velocity components is involved. As for this direction, later on, criteria just for one velocity u ∈ Lt (0, T ; Ls (R3 )), 2000 Mathematics Subject Classification. 35Q35, 76D05. Key words and phrases. Regularity criteria; Triebel-Lizorkin space; Morrey-Campanato space. c 2014 Texas State University - San Marcos. Submitted June 23, 2014. Published October 29, 2014. 1 2 R. WEI, Y. LI EJDE-2014/232 component appeared. The first result in this direction is due to Neustupa et al [15] (see also Zhou [21]), where the authors showed that if 2 3 1 u3 ∈ Lt (0, T ; Ls (R3 )), + = , s ∈ (6, ∞], t s 2 then the solution is smooth. A similar result, for the gradient of one velocity component, is independently due to Zhou [22] and Pokorný [17]. In [22], Zhou proved that if 2 3 3 ∇u3 ∈ Lt (0, T ; Ls (R3 )), + = , 3 ≤ s < ∞. t s 2 then the solution is smooth on [0, T ]. This result is extended by Zhou and Pokorný [26]; that is, 2 3 23 ∇u3 ∈ Lt (0, T ; Ls (R3 )), + = , 2 ≤ s ≤ 3. t s 12 Further criteria, including several components of the velocity gradient, pressure or other quantities, can be found, here we just list some. Zhou and Pokorný [25] proved the regularity condition 3 1 10 2 3 + = + , s> . u3 ∈ Lt (0, T ; Ls (R3 )), t s 4 2s 3 And in [10], Jia and Zhou proved that if a weak solution u satisfies one of the following two conditions: 10 u3 ∈ L∞ (0, T ; L 3 (R3 )); ∇u3 ∈ L∞ (0, T ; L30/19 (R3 )), then u is regular on [0, T ]. Dong and Zhang [5] proved that if the horizontal derivatives of the two velocity components Z T k∇h ũ(s)kḂ 0 ds < ∞, 0 ∞,∞ then the solution keeps smoothness up to time T , where ũ = (u1 , u2 , 0), and ∇h ũ = (∂1 ũ, ∂2 ũ, 0). For other kinds of regularity criteria, see [2, 6, 7, 9, 23, 24, 28, 29, 30] and the references cited therein. Throughout this paper C will denote a generic R positive constantRwhich can vary from line to line. For simplicity, we shall use f (x) dx to denote R3 f (x) dx, use k · kp to denote k · kLp . The purpose of this article is to improve and extend above known regularity criterion of weak solution for the equations (1.1), (1.2) to the Triebel-Lizorkin space and Morrey-Campanato spaces. The main results of this paper read: Theorem 1.1. Assume that u0 ∈ H 1 (R3 ) with div u0 = 0. u(x,t) is the corresponding weak solution to (1.1) and (1.2) on [0, T ). If additionally Z T 2 3 3 (1.3) k∇h u e(·, t)kpḞ 0 dt < ∞, with + = 2, < q ≤ ∞, p q 2 q, 2 q 0 3 then the solution remains smooth on [0, T ]. Theorem 1.2. Assume that u0 ∈ H 1 (R3 ) with div u0 = 0. u(x, t) is the corresponding weak solution to (1.1) and (1.2) on [0, T ). If additionally Z T 8 3 k∇u3 (·, t)k 3(2−r) (1.4) dt < ∞, with 0 < r ≤ 1, 2 ≤ p ≤ , p, 3 r Ṁ r 0 EJDE-2014/232 REMARKS ON REGULARITY 3 then the solution remains smooth on [0, T ]. Theorem 1.3. Assume that u0 ∈ H 1 (R3 ) with div u0 = 0. u(x,t) is the corresponding weak solution to (1.1) and (1.2) on [0, T ). If additionally Z T 8 3 3 ku3 (·, t)k 3−4r dt < ∞, with 0 < r < , 2 ≤ p ≤ , (1.5) p, 3 r Ṁ 4 r 0 then the solution remains smooth on [0, T ]. 0 Remark 1.4. Noticing that the classical Riesz transformation is bounded in Ḃ∞,∞ , if we take q = ∞ in Theorem 1.1, then the classical Beal-Kato-Majda criterion for the Navier-Stokes equations is obtained; that is, if Z T k∇h u e(·, t)kḂ 0 dt < ∞ ∞,∞ 0 then the solution remains smooth on [0, T ]. Remark 1.5. Since (it is proved in [12, 13]) Lq (R3 ) = Ṁ q,q (R3 ) ⊂ Ṁ p,q (R3 ), 1 < p ≤ q < ∞, 3 3 3 3 3 L r (R3 ) ⊂ Ṁ p, r (R3 ) ⊂ Ẋr (R3 ) ⊂ Ṁ 2, r (R3 ), 2 < p ≤ , 0 < r < , r 2 the result of Theorem 1.2 is an improved version of [27, Theorem 2]. Also we obtain, if Z T 8 3 3−4r dt < ∞, with 0 < r < , ku3 (·, t)kẊ r 4 0 then the solution remains smooth on [0, T ]. 2. Preliminaries In this section, we shall introduce the Littlewood-Paley decomposition theory, and then give some definitions of the homogeneous Besov space, homogeneous Triebel-Lizorkin space, Morrey-Campanato space and multiplier space as well as some relate spaces used throughout this paper. Before this, let us first recall the weak solutions of (1.1)-(1.3): Let u0 ∈ L2 (R3 ) with ∇ · u0 = 0, a measurable R3 -valued vector u is said to be a weak solution of (1.1)-(1.3) if the following conditions hold: (1) u(x, t) ∈ L∞ (0, ∞; L2 (R3 )) ∩ L2 (0, ∞; H 1 (R3 )); (2) u solves (1.1)-(1.2) in the sense of distributions; (3) the energy inequality holds; i.e, Z t kuk22 + 2 k4u(·, τ )k22 dτ ≤ ku0 k22 , 0 ≤ t ≤ T. 0 Let us choose a nonnegative radial function ϕ ∈ C ∞ (R3 ) be supported in the P∞ 3 8 3 −l annulus {ξ ∈ R : 4 ≤ |ξ| ≤ 3 }, such that l=−∞ ϕ(2 ξ) = 1, ∀ξ 6= 0. For 0 3 f ∈ S (R ), the frequency projection operators 4l is defined as 4l f = F −1 (ϕ(2−j ·)) ∗ f, where F −1 (g) is the inverse Fourier transform of g. The formal decomposition ∞ X f= 4l f. (2.1) l=−∞ 4 R. WEI, Y. LI EJDE-2014/232 is called the homogeneous Littlewood-Paley decomposition. Noticing l+1 X 4l f = 4j (4l f ) j=l−1 and using the Young inequality, we have the following class Bernstein inequality: Lemma 2.1 ([3]). Let α ∈ N , then for all 1 ≤ p ≤ q ≤ ∞, sup|α|=k k∂ α 4l f kq ≤ 1 1 C2lk+3l( p − q ) k4l f kp . and C is a constant independent of f, l. s For s ∈ R and (p, q) ∈ [1, ∞] × [1, ∞], the homogeneous Besov space Ḃp,q is defined by s Ṡp,q = {f ∈ Z 0 (R3 ) : kf kṠ s < ∞}, p,q where kf kḂ s = p,q 0 P jsq k4j f (·)kqp j∈z 2 1/q sup js j∈z 2 k4j f (·)kp , , 1 ≤ q < ∞, q = ∞. 3 and Z (R ) denote the dual space of Z 0 (R3 ) = {f ∈ S(R3 ) : Dα fˆ(0) = 0. ∀a ∈ N 3 }. On the other hand, for s ∈ R, (p, q) ∈ [1, ∞) × [1, ∞], and for s ∈ R, p = q = ∞, the homogenous Triebel-Lizorkin space is defined as kf kḞ s p,q s Ḟp,q = {f ∈ Z 0 (R3 ) : kf kḞ s < ∞}, p,q ( P jsq q 1/q k( j∈z 2 |4j f (·)| ) kp , 1 ≤ q < ∞, = k supj∈z (2js |4j f (·)|)kp , q = ∞. Notice that by Minkowski inequality, we have the following two imbedding relations: s s Ḃp,q ⊂ Ḟp,q , q ≤ p; s s Ḟp,q ⊂ Ḃp,q , p ≤ q. and the following two inclusions: s s Ḣ s = Ḃ2,2 = Ḟ2,2 , 0 0 L∞ ⊂ Ḟ∞,∞ = Ḃ∞,∞ . We refer to [20] for more properties. For 1 < q ≤ p < ∞, the homogeneous Morrey-Campanato space in R3 is 3 3 Ṁ p,q = f ∈ Lqloc (R3 ); kf kṀ p,q = sup sup R p − q kf kq(B(x,R)) < ∞ , x∈R3 R>0 For 1 ≤ p0 ≤ q 0 < ∞, we define the homogeneous space n X 0 0 0 0 Ṅ p ,q = f ∈ Lq |f = gk , where gk ∈ Lqcomp (R3 ) and k∈N X 3( 10 − 10 ) dk p q kgk kq0 o < ∞, where dk = diam(supp gk ) < ∞ . k∈N For 0 < α < 3/2, we say that a function belongs to the multiplier spaces M (Ḣ α , L2 ) if it maps, by pointwise multiplication, Ḣ α to L2 : Ẋα := M (Ḣ α , L2 ) := f ∈ S 0 ; kf · gkL2 ≤ CkgkḢ α , ∀g ∈ Ḣ α . EJDE-2014/232 REMARKS ON REGULARITY 5 Here, Ḣ α is the homogeneous Sobolev space of order α, Z 1/2 Ḣ α = f ∈ L1loc ; kf kL2 ≡ |ξ|2α |û(ξ)|2 <∞ . R3 p where L (1 ≤ p ≤ ∞) is the Lebsgue space endowed with norm k · kp . Lemma 2.2 ([4, 12]). Let 1 ≤ p0 ≤ q 0 < ∞, and p, q such that p1 + p10 = 1, 1q + q10 = 1. 0 0 Then Ṁ p,q is the dual space of Ṅ p ,q . Lemma 2.3 ([4, 7, 12]). Let 1 < p0 ≤ q 0 < 2, m ≥ 2, and p1 + p10 = 1. Denote n α = − n2 + np + m ∈ (0, 1] Then there exists a constant C > 0, such that for any u ∈ Lm (Rn ), v ∈ Ḣ α (Rn ), ku · vkṄ p0 ,q0 ≤ CkukLm kvkḢ α . Lemma 2.4 ([13]). For 0 ≤ r ≤ 23 , let the space M (Ḃ2r,1 → L2 ) be the space of functions which are locally square integrable on R3 and such that pointwise multiplication with these functions maps boundedly the Besov space Ḃ2r,1 (R3 ) to L2 (R3 ). The norm in M (Ḃ2r,1 → L2 ) is given by the operator norm of pointwise multiplication: kf kM (Ḃ r,1 →L2 ) = sup{kf gk2 : kgkḂ r,1 ≤ 1}. 2 2 3 Then, f belongs to M (Ḃ2r,1 → L2 ) if and only if f belongs to Ṁ 2, r (with equivalence of norms). 3. The proof of main results Proof of Theorem 1.1. Multiplying (1.1)1 by −4u, integrating by parts, noting that ∇ · u = 0, we have Z Z Z 1 d |∇u|2 dx + |4u|2 dx = [(u · 5)u] · 4u dx =: I. (3.1) 2 dt Next we estimate the right-hand side of (3.1), with the help of integration by parts and −∂3 u3 = ∂1 u1 + ∂2 u2 , one shows that Z 3 X I=− ∂k ui ∂i uj ∂k uj dx i,j,k=1 =− 2 X 3 Z X i,j=1 k=1 − 2 Z X 2 Z X Z 3 Z X ∂k u3 ∂3 u3 ∂k u3 dx k=1 ∂3 ui ∂i u3 ∂3 u3 dx − i=1 ≤C ∂k ui ∂i u3 ∂k u3 dx i,k=1 ∂k u3 ∂3 uj ∂k uj dx − j,k=1 − 2 Z X ∂k ui ∂i uj ∂k uj dx − 2 Z X ∂3 u3 ∂3 uj ∂3 uj dx j=1 |∇h u e||∇u|2 dx. Thus, the above inequality implies Z 1 d 2 2 k∇uk2 + k4uk2 ≤ C |∇h u e||∇u|2 dx. 2 dt (3.2) 6 R. WEI, Y. LI EJDE-2014/232 Using the Littlewood-Paley decomposition (2.1), ∇h u e can be written as ∇h u e= X N X 4j (∇h u e) + j<−N 4j (∇h u e) + j=−N X 4j (∇h u e). j>N where N is a positive integer to be chosen later. Substituting this into (3.2), one obtains 1 d k∇uk22 + k4uk22 2 dt N Z X Z X ≤C |4j (∇h u e)||∇u|2 dx + C |4j (∇h u e)||∇u|2 dx j<−N +C j=−N XZ (3.3) |4j (∇h u e)||∇u|2 dx j>N =: K1 + K2 + K3 . For Ki (i = 1, 2, 3), we now give the estimates one by one. For K1 , using the Hölder inequality, the Young inequality and Lemma 2.1, it follows that X K1 ≤ C k4j (∇h u e)k∞ k∇uk22 j<−N ≤C X ≤C X 23j/2 k4j (∇h u e)k2 k∇uk22 j<−N (3.4) 2 3j 1/2 X j<−N ≤ C2 −3N/2 k4j (∇h u e)k22 1/2 k∇uk22 j<−N k∇uk32 . s Where in the last inequality, we use the fact that for all s ∈ R, Ḣ s = Ḃ2,2 . For K2 , by the Hölder inequality and the Young inequality, one has Z X N K2 = C |4j (∇h u e)||∇u|2 dx j=−N ≤ CN 2q−3 2q Z X N |4j (∇h u e)|2q/3 3/(2q) |∇u|2 dx j=−N ≤ CN ≤ CN 2q−3 2q k∇h u ekḞ 0 k∇uk22q k∇h u ekḞ 0 k∇uk2 2q q, 3 2q−3 2q q, ≤ (3.5) q−1 2q−3 q 2q 3 2q 1 k4uk22 + CN k∇h u ekḞ2q−3 k∇uk22 . 0 2q 2 q, 3 where we used the interpolation inequality 3 kuks ≤ Ckuk2s for 2 ≤ s ≤ 6. − 12 3 3/q k4uk2 −3 2 s kukḢ , 1 EJDE-2014/232 REMARKS ON REGULARITY 7 Finally, using Hölder inequality and Lemma 2.1, K3 can be estimated as XZ |4j (∇h u e)||∇u|2 dx K3 = C j>N ≤C X k4j (∇h u e)k3 k∇uk23 j>N ≤C X j (3.6) 2 2 k4j (∇h u e)k2 k∇uk23 j>N X ≤ C( j>N ≤ C2 X 2−j )1/2 ( 22j k4j (∇h u e)k22 )1/2 k∇uk2 k4uk2 j>N −N/2 k∇uk2 k4uk22 . Substituting (3.4), (3.5) and (3.6) in (3.3), we obtain d k∇uk22 + k4uk22 dt 2q 3 k∇uk22 + C2−N/2 k∇uk2 k4uk22 . ≤ C2− 2 N k∇uk32 + CN k∇h u ekḞ2q−3 0 q, (3.7) 2q 3 Now we choose N such that C2−N/2 k∇uk2 ≤ 12 ; that is N≥ ln(k∇uk22 + e) + ln C + 2. ln 2 Thus (3.7) implies d k∇uk22 ≤ C + Ck∇h u ekpḞ 0 ln(k∇uk22 + e)k∇uk22 . 2q dt q, 3 Taking the Gronwall inequality into consideration, we obtain RT Z T k∇h u ekp 0 (τ )dτ i h 0 Ḟ 2q p 2 q, 3 ln(k∇uk2 + e) ≤ C 1 + k∇h u ekḞ 0 (τ )dτ · e . 0 q, 2q 3 The proof of Theorem 1.1 is complete under the condition (1.3). Proof of Theorem 1.2. Multiplying (1.1)1 by −4h u, integrating by parts, noting that ∇ · u = 0, we have Z Z Z 1 d |∇h u|2 dx + |∇∇h u|2 dx = [(u · 5)u] · 4h u dx =: J. (3.8) 2 dt Next we estimate the right-hand side of (3.8), with the help of integration by parts and −∂3 u3 = ∂1 u1 + ∂2 u2 , one shows that 3 X 2 Z X J =− ∂k ui ∂i uj ∂k uj dx i,j=1 k=1 =− Z 2 X ∂k ui ∂i uj ∂k uj dx − i,j,k=1 − 3 X 2 Z X 2 Z X i,k=1 ∂k u3 ∂3 uj ∂k uj dx j=1 k=1 =: J1 + J2 + J3 . ∂k ui ∂i u3 ∂k u3 dx (3.9) 8 R. WEI, Y. LI EJDE-2014/232 For J2 and J3 , we obtain Z |J2 + J3 | ≤ C |∇u3 ||∇h u||∇u| dx. (3.10) J1 is a sum of eight terms, using the fact −∂3 u3 = ∂1 u1 + ∂2 u2 , we can estimate it as Z J1 = − (∂1 u1 + ∂2 u2 )[(∂1 u1 )2 − ∂1 u1 ∂2 u2 + (∂2 u2 )2 ] dx Z − (∂1 u1 + ∂2 u2 )[(∂2 u1 )2 + ∂1 u2 ∂2 u1 + (∂1 u2 )2 ] dx Z = ∂3 u3 [(∂1 u1 )2 − ∂1 u1 ∂2 u2 + (∂2 u2 )2 + (∂2 u1 )2 + ∂1 u2 ∂2 u1 + (∂1 u2 )2 ] dx Z ≤ C |∇u3 ||∇h u||∇u| dx. (3.11) Substituting the estimates (3.9)-(3.11) in (3.8), we obtain Z 1 d 2 2 k∇h uk2 + k∇∇h uk2 ≤ C |∇u3 ||∇h u||∇u| dx =: L. 2 dt (3.12) when 2 < p ≤ 3r , using Lemmas 2.2 and 2.3, and the Young inequality, we obtain L ≤ Ck∇u3 k ≤ Ck∇u3 k ≤ Ck∇u3 k ≤ 3 Ṁ p, r k|∇u| · |∇h u|k Ṅ p0 , 3 3−r 3 k∇h ukḢ r k∇uk2 3 k∇ukL2 k∇h uk1−r k∇∇h ukr2 2 Ṁ p, r Ṁ p, r (3.13) 2 1 k∇∇h uk22 + Ck∇u3 k 2−rp, 3 k∇uk22 . Ṁ r 2 where we used the inequality kf kḢ r Z ˆ = k|ξ| f k2 = ( |ξ|2r |fˆ|2r |fˆ|2−2r dξ)1/2 ≤ kf k1−r k∇f kr2 , 2 r with 0 < r ≤ 1. In the case p = 2, using Hölders inequality, Lemma 2.4, and the Young inequality, we can estimate L as L ≤ Ck|∇u3 | · |∇h u|k2 k∇uk2 ≤ Ck∇u3 k 3 k∇h ukḂ r,1 k∇uk2 3 k∇h uk1−r k∇∇h ukr2 k∇uk2 2 Ṁ 2, r ≤ Ck∇u3 k Ṁ 2, r ≤ 2 (3.14) 2 1 k∇∇h uk22 + Ck∇u3 k 2−r2, 3 k∇uk22 . Ṁ r 2 where we used the following interpolation inequality [14]: for 0 ≤ r ≤ 1, kf kḂ r,1 ≤ 2 kf k1−r k∇f kr2 . 2 Now, gathering (3.13) and (3.14) together and substituting into (3.12), we obtain 2 d k∇h uk22 + k∇∇h uk22 ≤ Ck∇u3 k 2−rp, 3 k∇uk22 . Ṁ r dt (3.15) EJDE-2014/232 REMARKS ON REGULARITY 9 Multiplying (1.1)1 by −4u, integrating by parts, noting that ∇ · u = 0, we have (see [25]) Z 1 d k∇uk22 + k4uk22 = [(u · 5)u] · 4u dx 2 dt Z ≤ C |∇h u||∇u|2 dx ≤ Ck∇h uk2 k∇uk24 1/2 3 ≤ Ck∇h uk2 k∇uk2 k∇uk62 1/2 1/2 ≤ Ck∇h uk2 k∇uk2 k∇∇h uk2 k4uk2 . Integrating, with respect to t, yields Z t 1 k∇u(t)k22 + k4u(τ )k22 dτ 2 0 Z t 1 k∇u(τ )k22 dτ )1/4 ≤ k∇u0 k22 + C sup k∇h u(τ )k2 ( 2 0≤τ ≤t 0 Z t 1/2 Z t 1/4 2 × k∇∇h u(τ )k2 dτ k4u(τ )k22 dτ . 0 (3.16) 0 Substituting (3.15) in (3.16), using Hölders inequality and the Young inequality, we obtain Z t 1 2 k∇u(t)k2 + k4u(τ )k22 dτ 2 0 Z t Z t 1/4 2 1 ≤ k∇u0 k22 + (C + C k∇u3 k 2−rp, 3 k∇u(τ )k22 dτ ) k4u(τ )k22 dτ Ṁ r 2 0 0 Z t 4/3 2 3 1/2 ≤C +C k∇u3 k 2−rp, 3 k∇u(τ )k22 k∇u(τ )k2 dτ Ṁ r 0 Z 1 t (3.17) k4u(τ )k22 dτ + 2 0 Z t Z t 1/3 8 2 2 ≤C +C k∇u3 k 3(2−r) k∇u(τ )k dτ k∇u(τ )k dτ 3 2 2 Ṁ p, r 0 0 Z t 1 k4u(τ )k22 dτ + 2 0 Z t Z 8 1 t 2 ≤C +C k∇u3 k 3(2−r) k∇u(τ )k dτ + k4u(τ )k22 dτ. 2 p, 3 r Ṁ 2 0 0 Absorbing the last term into the left hand side, applying the Gronwall inequality and combining with the standard continuation argument, we conclude that the solu8 3 tions u can be extended beyond t = T provided that ∇u3 ∈ L 3(2−r) (0, T ; Ṁ p, r (R3 )). This completes the proof of Theorem 1.2. Proof of Theorem 1.3. We start from (3.9), we can estimate J2 and J3 as Z |J2 + J3 | ≤ C |u3 ||∇u||∇∇h u| dx. (3.18) 10 R. WEI, Y. LI EJDE-2014/232 From (3.11), we find that Z J1 ≤ C |u3 ||∇u||∇∇h u| dx. Combining (3.9),(3.17), (3.18) with (3.8), we obtain Z 1 d k∇h uk22 + k∇∇h uk22 ≤ C |u3 ||∇u||∇∇h u| dx =: V. 2 dt (3.19) (3.20) When 2 < p ≤ 3r , similarly as in the proof of L in (3.13), we obtain V ≤ Cku3 k 3 ≤ Cku3 k 3 k∇ukḢ r k∇∇h uk2 ≤ Cku3 k 3 k∇∇h ukL2 k∇uk1−r k4ukr2 2 Ṁ p, r Ṁ p, r Ṁ p, r k|∇u| · |∇∇h u|k Ṅ p0 , 3 3−r (3.21) 1 2(1−r) k∇∇h uk22 + Cku3 k2 p, 3 k∇uk2 k4uk2r 2 . Ṁ r 2 When p = 2, similarly as in the proof of L in (3.14), we have ≤ V ≤ Ck|u3 | · |∇u|k2 k∇∇h uk2 ≤ Cku3 k 3 Ṁ 2, r ≤ Cku3 k 3 Ṁ 2, r k∇ukḂ r,1 k∇∇h uk2 2 k∇uk1−r k4ukr2 k∇∇h uk2 2 (3.22) 1 2(1−r) k∇∇h uk22 + Cku3 k2 2, 3 k∇uk2 k4uk2r 2 . Ṁ r 2 Substituting (3.21) and (3.22) in (3.20), we find that d 2(1−r) k∇h uk22 + k∇∇h uk22 ≤ Cku3 k2 2, 3 k∇uk2 k4uk2r (3.23) 2 . Ṁ r dt Substituting (3.23) in (3.16), we obtain Z t 1 k∇u(t)k22 + k4u(τ )k22 dτ 2 0 Z t 1 2(1−r) 2 ≤ k∇u0 k2 + C + C ku3 k2 p, 3 k∇u(τ )k2 k4u(τ )k2r dτ 2 Ṁ r 2 0 Z t 1/4 × k4u(τ )k22 dτ 0 Z 4/3 Z t 1 t 2(1−r) 2 ≤C+ k4u(τ )k2 dτ + C k4u(τ )k2r dτ ku3 k2 p, 3 k∇u(τ )k2 2 Ṁ r 4 0 0 Z t Z t 4r/3 1 ≤C+ k4u(τ )k22 dτ + C k4u(τ )k22 dτ 4 0 0 Z t 4(1−r) 2 3 × ku3 k 1−rp, 3 k∇u(τ )k22 dτ ≤ Ṁ 0 r Z t 4(1−r) 3−4r 1 2 1 3−4r k4u(τ )k22 dτ + C ku3 k 1−rp, 3 k∇u(τ )k22(1−r) k∇u(τ )k22(1−r) dτ r Ṁ 2 0 0 Z 1 3−4r Z t Z t 8 1 t 3−4r 2 2 ≤C+ k4u(τ )k2 dτ + C ku3 k p, 3 k∇u(τ )k2 dτ k∇u(τ )k22 dτ Ṁ r 2 0 0 0 Z t Z t 8 1 2 ≤C+ k4u(τ )k22 dτ + C ku3 k 3−4r 3 k∇u(τ )k2 dτ. Ṁ p, r 2 0 0 Z ≤C+ t EJDE-2014/232 REMARKS ON REGULARITY 11 By a similar argument as in the proof of Theorem 1.2, provided that u3 ∈ 8 3 L 3−4r (0, T ; Ṁ p, r (R3 )), we complete the proof of Theorem 1.3. 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