Regularity of the 3D Navier–Stokes equations with viewpoint of 2D flow

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ScienceDirect J. Differential Equations ••• (••••) •••–••• www.elsevier.com/locate/jde

Regularity of the 3D Navier–Stokes equations with viewpoint of 2D flow Hyeong-Ohk Bae 1 Department of Financial Engineering, Ajou University, 206, World cup-ro, Yeogtong-gu, Suwon-si, Gyeonggi-do, 443-749, Republic of Korea Received 7 April 2017; revised 7 October 2017

Abstract The regularity of 2D Navier–Stokes flow is well known. In this article we study the relationship of 3D and 2D flow, and the regularity of the 3D Naiver–Stokes equations with viewpoint of 2D equations. We consider the problem in the Cartesian and in the cylindrical coordinates. © 2017 Elsevier Inc. All rights reserved. MSC: primary 35Q30; secondary 76D03, 76D05 Keywords: Regularity; 3D flow; 2D flow; Axisymmetric flow; Swirl

1. Introduction The regularity to the 3 dimensional (3D) Navier–Stokes system is one of the most important open problem for more than 80 years. Fortunately, it is well known that the solution of the 2D flow is regular enough and unique. Physically and mathematically, a question naturally arises what the relationship between the 3D and 2D flows is. In this sense there are several partial answers. To solve the regularity for the 3D flow, there are several approaches: Serrin type criteria ([4, 12,15,30]), vorticity direction ([6,8,14]), partial regularity [10,26,27], and so on. E-mail address: [email protected]. 1 The author was supported under the framework of international cooperation program managed by the National

Research Foundation of Korea (NRF) (NRF-2016K2A9A2A06005080), and by the Basic Science Research Program funded by the NRF, Ministry of Education, Science and Technology (NRF-2015R1D1A1A01057976). https://doi.org/10.1016/j.jde.2017.12.026 0022-0396/© 2017 Elsevier Inc. All rights reserved.

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As we mentioned, the incompressible 2D flow is regular, so we can approach to the regularity with the viewpoint of 2D. In this sense, there are several approaches during last two decades: thin domain problem [17,25], two component Serrin conditions for vorticity [11], and for velocity initially in 1997 [1] (published in 2007 [2]). Also refer to [3,5,9,7]. One-component approaches are done in [21,33]. There are similar approaches in axisymmetric cases, for example [13,22–24,29]. In [19,20, 28], it is shown that the axisymmetric flow without swirl is regular. Physically, the axisymmetric flow without swirl looks like 2D flow. The regularity for the axisymmetric flow with nonzero swirl is still open problem. The regularity for the axisymmetric flow is shown initially in [22] when the radial component vr (r, x3 ) of a weak solution belongs to Serrin class Lα (0, T ; Lβ (drdx3 )), where 2 ≤ α ≤ ∞, 3 < β ≤ ∞, 2 3 + ≤ 1, α β and in [13] when the radial and the swirl parts of a weak solution vr (r, x3 ), vθ (r, x3 ) belong to Serrin class Lα (0, T ; Lβ (drdx3 )), where 2 < α ≤ ∞ and 2 < β < ∞, α1 + β1 ≤ 12 . The regularity is also shown in [32] if the swirl part satisfies a growth condition near zero, and in [18] if the swirl or the radial part satisfies a Serrin type condition. In [22–24], there are also regularity theories for axisymmetric flow in the sense of two dimension and Serrin type conditions. Based on the above two facts, we may have a question: what is the relationship between 3D and 2D flows? More precisely, we can state it in the following way: (Cart1) If solution v depends only on two components x1 , x2 , then is the third component v3 zero? (Cart2) And reversely, if the third component v3 (x) = v3 (x1 , x2 , x3 ) of solution v(x) = (v1 (x), v2 (x), v3 (x)) is identically zero, then does the solution depend only on x1, x2 ? We also have similar questions in the cylindrical coordinate: (Cyl1) If the swirl part vθ (r, θ, x3 ) is identically zero, then is the flow axisymmetric? (Cyl2) Reversely, is any axisymmetric flow swirl free? In this article we answer these questions, and as a result we show the regularity for the 3D flow. 2. Equivalence of 2D flow and the case v3 ≡ 0 In this section, we answer the first two questions: (Cart1) and (Cart2). Let  ⊂ R3 be an arbitrary domain. For given 0 < T < +∞, we consider the incompressible Navier–Stokes equations in the space time cylinder T =  × (0, T ): ∇ ·v=0 ∂t vi + (v · ∇)vi − vi + ∂i p = 0

in T , in T

(2.1) (i = 1, 2, 3),

(2.2)

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where v(x) = (v1 (x1 , x2 , x3 ), v2 (x1 , x2 , x3 ), v3 (x1 , x2 , x3 )) denotes the unknown velocity field of the fluid, while p(x) stands for the unknown pressure. To complete the equations, we provide the initial data and the no-slip boundary condition: v(x, 0) = v0 (x) for x ∈ ,

v(x, t) = 0 for x ∈ ∂, t > 0.

(2.3)

For an arbitrary domain in R2 , the existence of strong solutions for the two-dimensional Navier–Stokes equations is well known (refer to [31]). Remark 2.1. 1. By the existence of weak solutions it is well known that v ∈ Cweak (0, T ; L2σ ()) ∩ L2 (0, T ; W 1,2 ()), where L2σ () is the set of divergence free functions in L2 (). And we remark that the weak solution v is weakly continuous from [0, T ] into L2σ () space, that is, for every φ ∈ L2σ (), the map  t →

v(x, t)φ(x)dx 

is continuous. We will use this fact for the proofs. 2. We remark that in this article, we consider the case the exterior force term is identically zero, but with nonzero exterior force term, it still works. 3. We consider the no-slip boundary condition here, but for the slip one, it also works. We consider a Leray–Hopf weak solution v ∈ L∞ (0, T ; L2 ()) ∩ L2 (0, T ; H 1 ()) for v0 ∈ satisfying (2.1)–(2.3) weakly, and the energy inequality

L2σ ()



t  |v(t)| dx + 2



 |∇v(τ )| dxdτ ≤

2

|v0 |2 dx,

2

0 



∀t ∈ [0, T ),



|v(x, t) − v0 (x)|2 dx → 0

as t → 0,



 and the function t →  v(x, t) · φ(x)dx is continuous on [0, T ] for any φ ∈ C0∞ ( × (0, T )). Here, the subscript σ means the divergence free subspace of L2 . (Cart1): Suppose Leray–Hopf weak solution (v, p) of (2.1)–(2.2) depends only on x1 , x2 , that is, v(x) = (v1 (x1 , x2 ), v2 (x1 , x2 ), v3 (x1 , x2 )), p(x) = p(x1 , x2 ), and initial data v0 (x) = (v01 (x1 , x2 ), v02 (x1 , x2 ), 0). Then, we will answer that v3 = 0. For our analysis, we denote the cross section by x3 = {(x1 , x2 ) : x = (x1 , x2 , x3 ) ∈ }. Equation (2.1) and equation of v3 in (2.2) are reduced into the followings: in T ,

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∂1 v1 + ∂2 v2 = 0, ∂t v3 − (∂12 + ∂22 )v3 + v1 ∂1 v3 + v2 ∂2 v3 = 0. In fact, the above equalities are satisfied weakly. If we take inner product to the equation with v3 and integrate, we obtain   2 |v3 | dx1 dx2 ≤ |v3 (0)|2 dx1 dx2 = 0. x3

x3

Therefore, v3 is zero. (Cart2): Suppose now that the third component of Leray–Hopf weak solution satisfies v3 (x1 , x2 , x3 , t) = 0 for (x, t) ∈ T with initial data v0 (x) = (v01 (x1 , x2 ), v02 (x1 , x2 ), 0). Then we will show that v is independent of x3 . If  = R3 , then the assumption satisfies the one-component Serrin type condition, for example [21]. Hence, the weak solution is regular and unique. Therefore, if v03 = 0, then v is independent of x3 , that is, v1 (x1 , x2 , t), v2 (x1 , x2 , t). If we look at the equation for v3 in (2.2), then we obtain ∂3 p = 0, which implies p(x1 , x2 , t). Even though we ignore the one-component criterion, we still obtain the independence of solutions v, p. Let us look at this. For our analysis, we assume that  is a cylinder 2 × (−1, 1), that is,  = 2 × (−1, 1) = {(x1 , x2 , x3 ) : −1 < x3 < 1, (x1 , x2 ) ∈ 2 }, where 2 is any two-dimensional domain of class C 2 . We furthermore assume that v has the periodic boundary condition in x3 -direction. For our proof we need the stronger assumption on the initial data v0 : v0 ∈ Wσ1,2 (2 ), where σ means the divergence free. From this smooth assumption of v0 we have the existence of the short time strong solution, that is, there is the first maximum time T ∗ < T that the equations have a strong solution for 0 < t < T ∗ . Equations (2.1)–(2.2) become the following equations: ∂1 v1 + ∂2 v2 = 0, ∂t v1 − (∂12 + ∂22 + ∂32 )v1 + v1 ∂1 v1 + v2 ∂2 v1 + ∂1 p = 0, ∂t v2 − (∂12 + ∂22 + ∂32 )v2 + v1 ∂1 v2 + v2 ∂2 v2 + ∂2 p = 0, ∂3 p = 0. So, we have clearly p(x1 , x2 , t). We also consider solutions u(x1 , x2 ), q(x1 , x2 ) of the 2D Navier–Stokes equations with the same condition on the boundary of 2 and the same initial condition u(x1 , x2 , 0) = (v01 (x1 , x2 ), v02 (x1 , x2 )). It is well known that the 2D NSEs are uniquely solvable and regular enough ([31]). We extend u in  by setting u(x1 , x2 , x3 ) = u(x1 , x2 ) for (x1 , x2 , x3 ) ∈ 2 × (−1, 1). Let w = (v1 , v2 ) − u for 0 ≤ t < T ∗ , then w0 = 0 and w satisfies the following equations weakly

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∂t w1 − (∂12 + ∂22 )w1 − ∂32 v1 + w1 ∂1 w1 + w2 ∂2 w1 + u1 ∂1 w1 + u2 ∂2 w1 + w1 ∂1 u1 + w2 ∂2 u1 + ∂1 (p − q) = 0, ∂t w2 − (∂12 + ∂22 )w2 − ∂32 v2 + w1 ∂1 w2 + w2 ∂2 w2 + u1 ∂1 w2 + u2 ∂2 w2 + w1 ∂1 u2 + w2 ∂2 u2 + ∂2 (p − q) = 0. By taking inner product with (w1 , w2 ) and noticing the boundary integrations related to terms ∂32 v1 , ∂32 v2 are zero, we obtain 1 ∂t 2



 |w1 | + |w2 | dx + 2

2





≤4

¯ 1 |2 + |∇w ¯ 2 |2 + |∂3 v1 |2 + |∂3 v2 |2 dx |∇w

 2 2 ¯ |∇u|(|w 1 | + |w2 | )dx,



where ∇¯ = (∂1 , ∂2 ). Setting w¯ = (w1 , w2 ), we have 1 ∂t 2



 |w| ¯ dx + 2

|∇¯ w| ¯ 2 + |∂3 v1 |2 + |∂3 v2 |2 dx





¯ |w| ¯ 3/2 |w| ¯ 1/2 |∇u|dx

≤4 

≤4

|∇ w| ¯ 2 dx

3/4  

|w| ¯ 2 dx

1/4  

¯ 2 dx |∇u|

1/2

    2 1 ¯ 2 dx |w| ¯ 2 dx |∇u| |∇ w| ¯ 2 dx + C 2     2  1  ¯ 2 ¯ 2 dx , |w| ¯ 2 dx |∇u| |∇w| + |∂3 v1 |2 + |∂3 v2 |2 dx + C = 2 ≤

and 

 |w| ¯ dx + 2

∂t

|∇¯ w| ¯ 2 dx ≤ C



|w| ¯ dx 2

 

¯ 2 dx |∇u|

2

.

Hence,  |w| ¯ dx ≤ C 2



  2 2 |w0 |2 dx e4t sup0
That is, w1 = 0 = w2 for 0 < t < T ∗ , therefore, v1 = u1 , v2 = u2 for 0 < t < T ∗ . Since T ∗ is the first maximum time of the strong solution v, and u is strong enough until T , the identity T ∗ = T should hold by the weak continuity in time of the solution. Therefore, v1 , v2 are independent of x3 . We conclude that v3 = 0 is equivalent to the two dimensional flow.

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3. Is swirl free equivalent to axisymmetric? In this section we answer the last two questions: (Cyl1) and (Cyl2). We consider the whole domain  = R3 . As we mentioned previously, Leray–Hopf weak solution is well known in the whole domain, too (refer to [12,16]). In the cylindrical coordinate (r, θ, x3 ), velocity v(x) is expressed of the form v(x, t) = vr (r, θ, x3 , t)er + vθ (r, θ, x3 , t)eθ + v3 (r, θ, x3 , t)e3 , where 

x2 , x1 = r cos θ, x2 = r sin θ, x1 x 1 x2 x2 x1 er = ( , , 0), eθ = (− , , 0), e3 = (0, 0, 1). r r r r

r=

x12 + x22 , tan θ =

The divergence free condition (2.1) is transformed into 1 1 0 = ∇ · v = ∂r vr + vr + ∂θ vθ + ∂x3 v3 , r r

(3.1)

and the momentum equations (2.2) are 1 1 2 1 ∂t vr − ∂r2 vr − ∂r vr − 2 ∂θ2 vr − ∂32 vr + 2 ∂θ vθ + 2 vr r r r r 1 1 2 + vr ∂r vr + vθ ∂θ vr + v3 ∂3 vr − vθ + ∂r p = 0, r r 1 1 2 1 ∂t vθ − ∂r2 vθ − ∂r vθ − 2 ∂θ2 vθ − ∂32 vθ − 2 ∂θ vr + 2 vθ r r r r 1 1 1 + vr ∂r vθ + vθ vr + vθ ∂θ vθ + v3 ∂3 vθ + ∂θ p = 0, r r r 1 1 2 2 2 ∂t v3 − ∂r v3 − ∂r v3 − 2 ∂θ v3 − ∂3 v3 r r 1 + vr ∂r v3 + vθ ∂θ v3 + v3 ∂3 v3 + ∂3 p = 0. r 3.1. Swirl free implies axisymmetric? We try to answer the question positively that swirl free flow is always axisymmetric. (Cyl1): Suppose that the initial data v0 belongs to Lrσ (R3 ), r ≥ 3, and that the flow is swirl free, that is, vθ (r, θ, x3 ) = 0. Then we show it is axisymmetric, that is, vr (r, x3 , t), v3 (r, x3 , t), p(r, x3 , t). Remark 3.1. The assumption of v0 ∈ Lrσ (R3 ) implies that there exists the first maximal time T ∗ ≤ T such that there is a unique classical solution v ∈ C([0, T ∗ ); Lrσ (R3 )) ∩Lq (0, T ∗ ; Lp (R3 )) with q2 + p3 = 3r , p, q > r (refer to [16]).

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By noticing Remark 3.1, in the same way done in the previous section, we have to show T∗ =T. Divergence free equation (3.1) for the swirl free flow becomes ∂r (rvr ) + ∂3 (rv3 ) = 0,

(3.2)

and the momentum equations become 1 1 ∂t (rvr ) − ∂r (r∂r vr ) − ∂θ2 vr + vr − r∂32 vr + rvr ∂r vr + rv3 ∂3 vr + r∂r p = 0, r r 1 2 ∂θ vr − ∂θ p = 0, r 1 ∂t (rv3 ) − ∂r (r∂r v3 ) − ∂θ2 v3 − r∂32 v3 + rvr ∂r v3 + rv3 ∂3 v3 + r∂3 p = 0. r We also consider the momentum equations for the axisymmetric Navier–Stokes equations with swirl free: uθ (r, x3 , t) = 0, ur (r, x3 , t), u3 (r, x3 , t), of which regularity is well-known ([19, 20,28]): 1 ∂t (rur ) − ∂r (r∂r ur ) + ur − r∂32 ur + rur ∂r ur + ru3 ∂3 ur + r∂r q = 0, r ∂t (ru3 ) − ∂r (r∂r u3 ) − r∂32 u3 + rur ∂r u3 + ru3 ∂3 u3 + r∂3 q = 0. Setting w = v − u for 0 < t < T ∗ , we have the following equations weakly 1 1 ∂t (rwr ) − ∂r (r∂r wr ) − ∂θ2 vr + wr − ∂32 (rwr ) + rwr ∂r wr r r + rwr ∂r ur + rur ∂r wr + rw3 ∂3 wr + rw3 ∂3 ur + ru3 ∂3 wr + r∂r (p − q) = 0, 1 ∂t (rw3 ) − ∂r (r∂r w3 ) − ∂θ2 v3 − r∂32 w3 + rwr ∂r w3 r + rwr ∂r u3 + rur ∂r w3 + rw3 ∂3 w3 + rw3 ∂3 u3 + ru3 ∂3 w3 + r∂3 (p − q) = 0. Notice that vr (r, 2π, x3 , t) = vr (r, 0, x3 , t), v3 (r, 2π, x3 , t) = v3 (r, 0, x3 , t) and dx = rdrdθdx3 . Since ur , u3 are independent of θ , by the divergence free condition, taking inner product the above with wr , w3 , we obtain   1 ∂t (rwr2 + rw32 ) + r|∂r wr |2 + r|∂r w3 |2 + r|∂3 w3 |2 2  1 1 1 + r|∂3 wr |2 + |∂θ wr |2 + |∂θ w3 |2 + |wr |2 drdθdx3 r r r    ≤ |wr |2 + |w3 |2 + |w3 | |wr | + |wr | |w3 | |∇u| dx.

(3.3)

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Since the relation of cartesian derivatives and cylindrical derivatives are 1 |∂1 vr |, |∂2 vr | ≤ |∂r vr | + |∂θ vr |, r

1 |∂1 vθ |, |∂2 vθ | ≤ |∂r vθ | + |∂θ vθ |, r

we have that 

 |wr | |∇u|dx = 2

≤ ≤

 

+ε

|wr |2 dx

|wr |1/2 |wr |3/2 |∇u|dx

1/4  

r|wr |2 drdθdx3

|∇wr |2 dx  

3/4  

|∇u|2 dx

|∇u|2 dx

1/2

2

   1 r|∂r wr |2 + r|∂3 wr |2 + |∂θ wr |2 drdθdx3 . r

(3.4)

In the same way, we obtain 

 r|w3 |2 |∇u|drdθdx3 +

≤C



r|w3 | |∇u| |wr |drdθdx3

r|w3 | + r|wr | drdθdx3 2

2

 

|∇u|2 dx

2



1 r|∂r w3 |2 + r|∂3 w3 |2 + |∂θ w3 |2 drdθdx3 r    1 +ε r|∂r wr |2 + r|∂3 wr |2 + |∂θ wr |2 drdθdx3 . r +ε

(3.5)

Therefore, combining (3.3), (3.4) and (3.5) we have  ∂t

r(wr2 + w32 )drdθ dx3 ≤ C



r(w32 + wr2 )drdθ dx3

 

|∇u|2 dx

2

,

which is of the form: Y ≤ AY, where A(τ ) =



|∇u|2 dx

2

. Since we already know the swirl free axisymmetric solution u is  regular enough, if we solve this, we conclude r(wr2 + w32 )drdθ dx3 = Y (t) = 0 for 0 < t < T ∗ . From this, we say T ∗ = T , and therefore, v is axisymmetric.

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3.2. Axisymmetric flow imply swirl free motion? We now answer the final question. (Cyl2): Suppose that the initial data v0 belongs to L6σ (R3 ), and that the flow is axisymmetric and initially swirl free: vr (r, x3 , t), vθ (r, x3 , t), v3 (r, x3 , t), p(r, x3 , t) and v0θ = 0. Then, the axisymmetric flow is swirl free. Therefore, the flow is regular. For axisymmetric case with swirl, we concentrate the swirl part 1 ∂t (rvθ ) − ∂r (r∂r vθ ) − r∂32 vθ + vθ + rvr ∂r vθ + vθ vr + rv3 ∂3 vθ = 0. r By the divergence free condition (3.2) we obtain that for 0 < t < T ∗ ,  ∂t

rvθ2 drdx3 + 2

   1  r(|∂r vθ |2 + |∂3 vθ |2 ) + vθ2 drdx3 ≤ 2 vθ2 vr drdx3 . r

Since 0 ≤ θ ≤ 2π , considering constants we may treat the integration with respect to r, x3 as r, θ, x3 . Consider vr vθ2 : for 0 < a < 1, 



1 vθa ( vθ )vθ1−a vr dx r   1−a a  1 1  2 2 2 2 2 vθ rdrdθdx3 |∇vθ |2 dx ≤ vθ drdθdx3 r   1−a 3 3 vr1−a rdrdθdx3 × vθ2 vr drdθdx3 =

≤C



vθ2 rdrdθdx3

+ε

 

3

vr1−a rdrθdx3

 1−a 2 3

a

 1    |∇vθ |2 dx . vθ2 drdθdx3 + ε r

Therefore, we have that for 0 < a < 1,  ∂t

 

1  r|∂r vθ |2 + r|∂3 vθ |2 + vθ2 drdθdx3 r 2(1−a)     3 3a rvθ2 drdθdx3 vr1−a dx ≤C .

rvθ2 drdθdx3 +

◦ Swirl Free until Blowup Time: From the second part of (3.6), we have that for 0 < a ≤ 1/2, 

3

vr1−a dx

 2(1−a) 3a

=



3(1−2a)

3+6a

vr2(1−a) vr2(1−a) dx

 2(1−a) 3a

(3.6)

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≤ ≤



vr2 dx



 1−2a   2a

|v|2 dx

vr6 dx

 1−2a   2a

 1+2a 6a

|v|6 dx

 1+2a 6a

.

Therefore, ignoring θ , we have  ∂t

  1  r|∂r vθ |2 + r|∂3 vθ |2 + vθ2 drdx3 r   1−2a    1+2a   2a 6a rvθ2 drdx3 |v|2 dx |v|6 dx ≤ . rvθ2 drdx3 +

If we set  Y (t) =

rvθ2 drdx3 ,

then we have Y ≤ Y



|v|2 dx

 1−2a   2a

|v|6 dx

 1+2a 6a

.

If we solve it, we obtain Y (t) ≤ Y (0) exp





sup

 |v| dx 2

 1−2a 2a

0
t  

|v|6 dx

 1+2a 6a

dt

 .

0

Notice that for any weak solution v, t  

|v|6 dx

1

3

dt < ∞.

0

Since, however,

1+2a 6a

=

1 3

+

1 6a

> 13 , we do not know that t  

|v|6 dx

 1+2a 6a

dt < ∞.

0

Therefore, we need a higher regularity. By Remark 3.1 for r = 6, there exists the first maximal time T ∗ ≤ T such that the system has a strong solution v ∈ C([0, T ∗ ); L6σ (R3 )) ∩ Lq (0, T ∗ ; Lp (R3 )) with q2 + p3 = 12 , p, q > 6. Therefore, we have that for 0 < t < T ∗ ,  sup 0
|v|6 dx < ∞.

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Therefore, we have Y (t) = 0, which implies that if there is no swirl initially then swirl does not happen for any short time. ◦ Swirl free at the blow-up time Since the swirl part vθ is also weakly continuous from [0, T ] into L2σ (), and we have just observed vθ is zero for 0 < t < T ∗ , we conclude that it is zero up to T ∗ . That means the swirl part is zero until T ∗ , and the velocity is axisymmetric and swirl free. Therefore, it is regular up to T ∗ and T ∗ cannot be the blow-up time. Finally, we conclude that the solution is regular until T . Acknowledgment The author thanks the reviewer for helpful comments and improvement of the proofs, and Pigong Han for discussion. References [1] H.-O. Bae, H.J. Choe, L∞ -bound of weak solutions to Navier–Stokes equations, in: Proceedings of the Korea–Japan Partial Differential Equations Conference, Taejon, 1996, in: Lecture Notes Ser., vol. 39, Seoul Nat. Univ., Seoul, 1997. [2] H.-O. Bae, H.J. Choe, A regularity criterion for the Navier–Stokes equations, Comm. Partial Differential Equations 32 (7–9) (2007) 1173–1187. [3] H.-O. Bae, J. Wolf, A local regularity condition involving two velocity components of Serrin-type for the Navier– Stokes equations, C. R. Math. Acad. Sci. Paris 354 (2) (2016) 167–174. [4] H. Beirao da Veiga, A new regularity class for the Navier–Stokes equations in R n . A Chinese summary appears in Chinese Ann. Math. Ser. A 16 (6) (1995) 797; Chin. Ann. Math. Ser. B 16 (4) (1995) 407–412. [5] H. Beirao da Veiga, On the smoothness of a class of weak solutions to the Navier–Stokes equations, J. Math. Fluid Mech. 2 (2000) 315–323. [6] H. Beirao da Veiga, Direction of vorticity and regularity up to the boundary: on the Lipschitz-continuous case, J. Math. Fluid Mech. 15 (1) (2013) 55–63. [7] H. Beirao da Veiga, On the extension to slip boundary conditions of a Bae and Choe regularity criterion for the Navier–Stokes equations. The half-space case, J. Math. Anal. Appl. 453 (1) (2017) 212–220. [8] H. Beirao da Veiga, L. Berselli, Navier–Stokes equations: Green’s matrices, vorticity direction, and regularity up to the boundary, J. Differential Equations 246 (2) (2009) 597–628. [9] L. Berselli, A note on regularity of weak solutions of the Navier–Stokes equations in R n , Jpn. J. Math. (N.S.) 28 (1) (2002) 51–60. [10] L. Caffarelli, R. Kohn, L. Nirenberg, Partial regularity of suitable weak solutions of the Navier–Stokes equations, Comm. Pure Appl. Math. 35 (1982) 771–831. [11] D. Chae, H.J. Choe, Regularity of solutions to the Navier–Stokes equation, Electron. J. Differential Equations (05) (1999) 1–7. [12] D. Chae, J. Lee, Regularity criterion in terms of pressure for the Navier–Stokes equations, Nonlinear Anal. 46 (2001) 727–735. [13] D. Chae, J. Lee, On the regularity of the axisymmetric solutions of the Navier–Stokes equations, Math. Z. 239 (4) (2002) 645–671. [14] P. Constantin, C. Fefferman, Direction of vorticity and the problem of global regularity for the Navier–Stokes equations, Indiana Univ. Math. J. 42 (3) (1993) 775–789. [15] E.B. Fabes, B.F. Jones, N.M. Riviere, The initial value problem for the Navier–Stokes equations with data in Lp , Arch. Ration. Mech. Anal. 45 (1972) 222–240. [16] Y. Giga, Solutions for semilinear parabolic equations in Lp and regularity of weak solutions of the Navier–Stokes system, J. Differential Equations 62 (1986) 186–212. [17] N. Kim, M. Kwak, Global existence for 3D Navier–Stokes equations in a long periodic domain, J. Korean Math. Soc. 49 (2) (2012) 315–324.

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