Anisotropic partial regularity criteria for the Navier-Stokes equations

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1 Anisotropic partial regularity criteria for the Navier-Stokes equations Walter Rusin Department of Mathematics Mathflows 205 Porquerolles September 7, 205

2 The question of regularity of the weak solutions of the 3D incompressible Navier-Stokes equations 3 t u u + j (u j u) + p = 0 j= div u = 0 (NSE) where u(x, t) = (u (x, t), u 2 (x, t), u 3 (x, t)) and p(x, t) denote the unknown velocity and the pressure, has remained open since 930 s (works of Leray 34). 2 of 35

3 For initial data u 0 (x) L 2 (R 3 ), the energy estimate t u(, t) 2 L + 2 u(, s) 2 2 L ds u L 2 0 plays a crucial role in establishing the existence of global weak solutions. In particular, for any T > 0 we obtain u L ((0, T ), L 2 (R 3 )) L 2 ((0, T ), Ḣ (R 3 )). 3 of 35

4 The Sobolev embedding and interpolation yield that for such u(x, t) we have u L 0/3 t,x ((0, T ) R 3 ) or any other Lebesgue space L p t L q x((0, T ) R 3 ) provided that 2 p + 3 q = of 35

5 The available regularity criteria (e.g. Ladyzhenskaya-Prodi-Serrin) require however 2 p + 3 q. Note that not only we cannot draw any conclusion but there is also a substantial gap between what is at hand and what would be necessary to conclude regularity. 5 of 35

6 The theory of partial regularity for the NSE aims at estimating the Husdorff dimension of the (possible) singular set and development of interior regularity criteria. Recall that a point is regular if there exists a neighborhood in which u is bounded (and thus Hölder continuous); otherwise, the point is called singular. 6 of 35

7 The program has been started by Scheffer in the 70 s. In a classical paper, Caffarelli, Kohn, and Nirenberg proved that the one-dimensional parabolic Hausdorff measure (parabolic Hausdorff length) of the singular set equals zero. In this case, the interior regularity criterion is as follows: There exist two constants ɛ CKN (0, ] and α (0, ) such that if then Q ( u 3 + p 3/2 ) dxdt ɛ CKN u(x, t) C α (Q /2 ) < where Q r = {(x, t) : x < r, r 2 t 0}. 7 of 35

8 Alternative proofs were given by Ladyzhenskaya and Serëgin, Kukavica, Lin, Vasseur, and Wolf. The problem of partial regularity of the solutions of the Navier Stokes equations has been since then addressed in various contexts and a variety of interior regularity criteria has been proposed. 8 of 35

9 In particular Wolf proved the following: There exists ɛ W > 0 such that if Q u 3 dxdt ɛ W then the solution u(x, t) is regular at the point (0, 0). 9 of 35

10 Fairly recently, Wang and Zhang proved an anisotropic interior regularity criterion, which states: For every M > 0 there exists ɛ WZ (M) > 0 such that if Q ( u 3 + p 3/2 ) dxdt M and Q u h 3 dxdt ɛ WZ (M) where u h = (u, u 2 ), then the solution u(x, t) is regular at the point (0, 0). 0 of 35

11 One component regularity/anisotropic results: Cao, Titi Chemin, Zhang Zhou Raugel, Sell He Mucha Nestupa, Novotny, Penel, Pokorny Kukavica, Ziane, R.... of 35

12 The purpose of this talk is to present interior regularity criteria involving only one component of the velocity. We prove, with a different argument, the following statement: For every M > 0 there exists a constant ɛ(m) > 0 such that if Q ( u 3 + p 3/2 ) dxdt M and Q u 3 3 ɛ(m) then u(x, t) is regular at the point (0, 0). 2 of 35

13 Let D be an open, bounded, and connected subset of R 3 (0, ). We work in the class of suitable weak solutions, that is (u, p) satisfies (i) u L t L 2 x(d) L 2 t Hx (D) and p L 3/2 (D), (ii) the Navier-Stokes equations (NSE) are satisfied in the weak sense (iii) the local energy inequality holds in D, i.e., u 2 φ dx + 2 u 2 φ dxdt R 3 T R 3 (,T ] ( ) u 2 ( t φ + φ) + ( u 2 + 2p)u φ dxdt R 3 (,T ] for all φ C 0 (D) such that φ 0 in D and almost all T R. 3 of 35

14 Theorem Let (u, p) be a suitable weak solution of (NSE) in a neighborhood of Q r (x 0, t 0 ) D which satisfies ( u 3 r 2 + p 3/2) dxdt M. () Q r (x 0,t 0 ) Then there exists ɛ > 0 depending on M such that if r 2 u 3 3 dxdt ɛ (2) then u is regular at (x 0, t 0 ). Q r (x 0,t 0 ) 4 of 35

15 Corollary Let (u, p) be Leray s weak solution defined in a neighborhood of [0, T ]. If T is an epoch of irregularity, then for every q 3 and M > 0, we have and (u, u 2 )(, t) L q u 3 (, t) L q for t < T sufficiently close to T. M (T t) ( 3/q)/2 ɛ(m) (T t) ( 3/q)/2, 5 of 35

16 Theorem 2 For every M > 0, there exists a constant ɛ(m) > 0 with the following property: If (u, p) is a suitable weak solution of (NSE) in a neighborhood of Q r (x 0, t 0 ) D which satisfies and either or r 2 Q r (x 0,t 0 ) then u is regular at (x 0, t 0 ). 6 of 35 ( u 3 + p 3/2 ) dxdt M (3) r 2 u 3 dxdt ɛ (4) Q r (x 0,t 0 ) r 2 p 3/2 dxdt ɛ, (5) Q r (x 0,t 0 )

17 Corollary 2 Let (u, p) be Leray s weak solution defined in a neighborhood of [0, T ]. If T is an epoch of irregularity, then for every q 3 and M > 0, we have and u(, t) L q p(, t) L q/2 for t < T sufficiently close to T. M (T t) ( 3/q)/2 ɛ(m) (T t) ( 3/q)/2, 7 of 35

18 Sketch of the proof of Theorem. We argue by contradiction. Fix r > 0 and assume that there exists a sequence of suitable weak solutions (u (n), p (n) ), with (x 0, t 0 ) a singular point for each of them with r 2 Q r (x 0,t 0 ) u (n) 3 + p (n) 3/2 dxdt M and r 2 u (n) 3 3 dxdt 0. Q r (x 0,t 0 ) 8 of 35

19 Lemma Let (u (n), p (n) ) be a sequence of suitable weak solutions such that ( u (n) r p (n) 3/2) dxdt M, (6) Q r (x 0,t 0 ) and let 0 < ρ < r. Then there exists a subsequence (u (n k), p (n k) ) such that u (n k) u strongly in L q (Q ρ (x 0, t 0 )) for all q < 0/3 and p (n k) p weakly in L 3/2 (Q ρ (x 0, t 0 )). 9 of 35

20 Note that the pair (u, p) solves the system 2 t u i u i + u j j u i + i p = 0 in D i =, 2 (LS) j= 3 p = 0 in D u + 2 u 2 = 0 in D where u(x, x 2, x 3, t) and p(x, x 2, x 3, t) are unknown. 20 of 35

21 Theorem 3 Let (u, p) be a solution of (LS). Then u is regular. 2 of 35

22 Theorem 3 implies that we may reduce r and assume M 0 = u L 6 (Q r (x 0,t 0 )) <. Using Hölder s inequality we obtain u L 3 (Q ρ(x 0,t 0 )) Cρ 5/2 u L 6 (Q ρ(x 0,t 0 )) Cρ 5/2 M 0 0 < ρ r from where ρ 2/3 u L 3 (Q ρ(x 0,t 0 )) Cρ /6 u L 6 (Q ρ(x 0,t 0 )) Cρ /6 M 0 0 < ρ r. 22 of 35

23 Let r 0 r be sufficiently small so that Then r 2/3 0 Cr /6 0 M 0 2 ɛ/3 W. u L 3 (Q r0 (x 0,t 0 )) 2 ɛ/3 W. 23 of 35

24 Since u (n) u strongly in L 3 loc (Q r 0 (x 0, t 0 )), we may choose n large enough so that Thus we obtain r 2/3 0 r 2 0 u (n) u L 3 (Q r0 (x 0,t 0 )) 2 ɛ/3 W. Q r (x 0,t 0 ) u (n) 3 dxdt ɛ W. This leads to a contradiction as the above inequality implies that the suitable weak solution u (n) is regular at (x 0, t 0 ). Note that instead of Wolf s result we may use Theorem of 35

25 Sketch of the proof of Theorem 2. First, we prove the assertion under the condition on the pressure. We argue by contradiction. Fix r > 0 and assume that there exists a sequence of suitable weak solutions (u (n), p (n) ), with (x 0, t 0 ) a singular point for each of them with r 2 Q r (x 0,t 0 ) u (n) 3 dxdt M and r 2 p (n) 3/2 dxdt 0. Q r (x 0,t 0 ) 25 of 35

26 By possibly reducing r > 0 we may assume that u (n) u strongly in L 3 (Q r (x 0, t 0 )). Note that u solves the Burgers-type system with t u i u i + Therefore u is regular. 3 u i j u j = 0 in D i =, 2, 3. j= u + 2 u u 3 = 0 in D. 26 of 35

27 Thus, there exists a constant M 0 > 0 (which depends on M) such that u L 6 (Q r (x 0,t 0 )) = M 0 <. We conclude as in the proof of Theorem. Namely, Hölder s inequality yields u L 3 (Q ρ(x 0,t 0 )) Cρ 5/2 u L 6 (Q ρ(x 0,t 0 )) Cρ 5/2 M 0 0 < ρ r which implies ρ 2/3 u L 3 (Q ρ(x 0,t 0 )) Cρ /6 u L 6 (Q ρ(x 0,t 0 )) Cρ /6 M 0 0 < ρ r. 27 of 35

28 Let r 0 r be sufficiently small so that Thus r 2/3 0 Cr /6 0 M 0 6 ɛ/3 CKN. u L 3 (Q r0 (x 0,t 0 )) 6 ɛ/3 CKN. Since u (n) u strongly in L 3 loc (Q r 0 (x 0, t 0 )), we may choose n large enough so that r 2/3 0 u (n) u L 3 (Q r0 (x 0,t 0 )) 6 ɛ/3 CKN. 28 of 35

29 Therefore r 2/3 0 u (n) L 3 (Q r (x 0,t 0 )) 2 ɛ/3 CKN. Moreover, if n is sufficiently large, we also have p (n) 3/2 dxdt 2 ɛ CKN. r 2 0 Q r (x 0,t 0 ) Hence r 2 0 Q r0 (x 0,t 0 ) ( u 3 + p 3/2 ) dxdt ɛ CKN. This leads to a contradiction as the above inequality implies that the suitable weak solution u (n) is regular at (x 0, t 0 ). 29 of 35

30 Now, we sketch the proof of the assertion under the condition (4). Without loss of generality, we may assume that (x 0, t 0 ) = (0, 0). Denote Q r = Q r (x 0, t 0 ) Assume that and u L 3 (Q r ) ɛ 0 p L 3/2 (Q r ) M2 0. We shall prove that there exists ɛ 0 > 0 sufficiently small, depending on M 0, such that (0, 0) is regular. 30 of 35

31 We may rewrite the pressure equation as (ηp) = ij (ηu i u j ) u i u j ij η + j (u i u j i η) + i (u i u j j η) p η + 2 j (( j η)p). With N = /4π x, the Newtonian potential, we obtain ηp = R i R j (ηu i u j ) N (u i u j ij η) + j N (u i u j i η) + i N (u i u j j η) N (p η) + 2 j N (( j η)p) = p + p 2 + p 3 + p 4 + p 5 + p 6. 3 of 35

32 For p, we have by the Calderón-Zygmund theorem p L 3/2 (Q r0 ) C u 2 L 3 (Q r ). For the rest of the terms, we use the fact that they all contain derivatives of η. This makes all the convolutions nonsingular when x r 0 32 of 35

33 Summarizing, we obtain r 2/3 0 p /2 L 3/2 (Q r0 ) C r 2/3 0 where C 0 is a constant. u L 3 (Q r ) +C r /3 0 r M 0 C 0 r 2/3 r 2/3 0 r /3 0 ɛ 0 +C 0 r /3 M of 35

34 Let ɛ > 0 be arbitrary. Then, fix r 0 = ɛ3 r 8C 3 0 M3 0 and ɛ 0 = ɛ ( r0 ) 2/3 ɛ 3 = 2C 0 r 8C0 3M2 0 and we get u L 3 (Q r0 ) + p /2 ɛ, which implies that (0, 0) is L 3/2 (Q r0 ) regular if ɛ = ɛ /3 CKN /C with C a sufficiently large constant. 34 of 35

35 Questions? Thank you for your attention! 35 of 35