On partial regularity for the Navier-Stokes equations

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1 On partial regularity for the Navier-Stokes equations Igor Kukavica July, 2008 Department of Mathematics University of Southern California Los Angeles, CA Abstract We consider the partial regularity of suitable weak solutions of the Navier-Stokes equations in a domain D. We prove that the parabolic Hausdorff dimension of space-time singularities in D is less than or equal to provided the force f satisfies f L 2 D. Our argument simplifies the proof of a classical result of Caffarelli, Kohn, and Nirenberg, who proved the partial regularity under the assumption f L 5/2+δ where δ > 0. Keywords: Navier-Stokes equation, partial regularity, Morrey space Mathematics Subject Classification: 35Q30, 76D05, 35K55, 35K5 Introduction In this paper, we address the partial regularity of solutions of the Navier-Stokes equation t u u + j u j u + p = f u = 0.. Given an initial condition u, 0 = u 0 and boundary conditions in Ω, which is either a bounded domain or R 3, Leray and Hopf proved in [Le, H] the existence of a weak solution u of the Navier-Stokes equation satisfying a form of an energy inequality. It is not known whether such solutions may develop singularities and whether they are unique. In a series of papers [S, S2, S3], Scheffer studied the partial regularity of solutions of the Navier-Stokes equation which satisfy a local version of the energy inequality. In the classical paper [CKN], Caffarelli, supported in part by the NSF grant DMS

2 Kohn, and Nirenberg introduced a concept of a suitable weak solution and proved that for all such solutions the one-dimensional parabolic Hausdorff measure of their space-time singularities is equal to 0 if the force satisfies f L 5/2+q for some q > 0. Furthermore, they proved that given f L 2 Ω 0, T with f = 0 and a certain condition on u 0, there exists a suitable weak solution of the initial value problem with the Dirichlet boundary conditions. The purpose of the present paper is to improve the results in [CKN, LS] by relaxing the assumption on the force term f. We consider the partial regularity of solutions of the Navier-Stokes equation which are suitable in a domain D in space-time. As mentioned above, [CKN] gives the partial regularity theorem under the condition f L 5/2+q for some q > 0. This condition on the force term is needed on [CKN, p. 793] to assure that the series in 4.2 converges. In [LS], the assumption was replaced by a Morrey type condition sup Q Rx,t D R +q f 2 < Q Rx,t with q > 0, where Q R x, t is the parabolic cylinder with the top center point at x, t and radius R > 0. In this paper, we prove that the parabolic one dimensional Hausdorff measure of space-time singularities of a suitable weak solution is 0 provided f L 2 D, which we accomplish by simplifying the proof in [CKN]. The simplification also leads to a small improvement in the main statement since our parabolic cubes do not have to be centered at the point tested for regularity as in [CKN]. Our proof is based on the test function being chosen as the smoothened version of the backward Gaussian kernel, the approach used in [CKN, S, S2, S3]. By improving the energy estimates, we avoid the technical induction argument in [CKN], which in turn simplifies the proof and allows more general forces. Also, we do not rely on finding an upper bound on the quantity R 3 ess sup t R 2,t B u 2 Rx as was done in [CKN] or a Morrey-Campanato type norm in [Li, LS], but instead bound R 3/2 ess sup t R2,t B u 2 Rx. We point out that the last part of our proof relies on a Morrey-type inequality due to O Leary [O]. We also note that, following the same proof, Theorem 2. remains valid if the assumption on the force is relaxed to f L q D where q > 5/3 cf. Theorem 2.6 below or to f L 2 t H q xd, where q > /2. 2 Notation and the main theorem Fix an open connected set D R 3 0,. Let u, p be a suitable weak solution in D, which is defined as follows: i u L t L 2 xd L 2 t HxD and p L 3/2 D, ii f L 2 D is divergence-free, iii the Navier-Stokes equations. are satisfied in D in the weak sense, and iv the local energy inequality holds in D, i.e., u 2 φ T + 2 u 2 φ R 3,T ] R 3,T ] u 2 φ t + φ + u 2 + 2pu φ + 2u fφ 2. 2

3 for all φ C 0 D such that φ 0 in D and all T R. Above and in the sequel, we denote =, 2, 2. Denote by B r x 0 the standard euclidean ball with the center x 0 and the radius r, and by Q r x 0, t 0 = B r x 0 [t 0 r 2, t 0 ] the parabolic cylinder labeled by the top-center point x 0, t 0 D. For simplicity, we write Q r = Q r 0, 0 and B r = B r 0. We say that a point x 0, t 0 D is regular if u L 5 D 0 in an open neighborhood D 0 D of x 0, t 0. By [Se, So, St], this can be bootstrapped to u L t H x D L 2 t H2 x D for all D such that D D 0, which is the usual space for a definition of strong solutions [CF88]. We call a point x 0, t 0 D singular if it is not regular. For x 0, t 0 D, and all r > 0 such that Q r x 0, t 0 D, let α x0,t 0r = β x0,t 0r = r r r 2 /2 ess sup u 2 t 0 r 2,t 0 B rx 0 γ x0,t 0r = δ x0,t 0r = r 2 λ x0,t 0r = r Q rx 0,t 0 Q rx 0,t 0 Q rx 0,t 0 Q rx 0,t 0 u 2 /2 u 3 /3 p 3/2 /3 f 2 /2 If Q r x 0, t 0 D c, then the above quantities are defined by replacing Q r x 0, t 0 with Q r x 0, t 0 D. If the label x 0, t 0 is omitted, it is understood to be 0, 0, e.g. αr = α 0,0 r. The five quantities are dimensionless when following the usual convention that the dimension exponents of x, t, u, p, and f are, 2,, 2, and 3 respectively. Also, the exponents are chosen so that the the expressions are of order as far as the dependence on u is concerned; thus it is easier to track which expressions arise from linear and which from nonlinear terms. Note that the assumption f L 2 D implies. λ x0,t 0r Mr /2 2.2 for all x 0, t 0 D and all r > 0 such that Q r x 0, t 0 D, where M = f L2 D. The following is our main result. Theorem 2.. There exists a sufficiently small universal constant ɛ > 0 with the following property. If x 0, t 0 D and lim sup β x0,t 0r < ɛ 2.3 r 0+ then x 0, t 0 is a regular point. In particular, the one dimensional parabolic Hausdorff measure of the set of singular points equals 0. 3

4 The condition ii can be relaxed to f L q D divergence-free with q > 5/3 without difficulty cf. Theorem 2.6. By [CKN], the second part of the theorem follows from the first. Therefore, we only need to prove that 2.3 implies that x 0, t 0 is a regular point. Let 0 < r ρ/2, and denote κ = r/ρ. Further below, κ is going to be a fixed small enough universal constant. Denote θ x,t r = α x,t r + β x,t r + κ 4 δ x,t r 2. As before, we abbreviate θr = θ 0,0 r. Lemma 2.2. Assume 0, 0 D. Then we have θr Cκ 2/3 θρ + Cκ 5 βρθρ + Cκ /2 θρ /2 λρ /2 2.4 and θr Cκ 2/3 θρ + Cκ 5 θρ 2 + Cκ /2 θρ /2 λρ /2 2.5 for 0 < r ρ/2 such that Q ρ D. Proof of Lemma 2.2. Since βρ θρ, 2.5 follows from 2.4; therefore, it is sufficient to prove 2.4. Let Gx, t = 4πt 3/2 exp x 2 /4t be the Gaussian kernel. For r > 0 as in the statement, denote ψx, t = r 2 Gx, r 2 t, x, t R 3, 0 where the dependence of ψ on r is suppressed for the sake of simplicity. Observe that t ψ + ψ = 0 on R 3, 0]. First, we derive several bounds on ψ. In order to estimate ψ on Q r from below, observe that for a fixed t [ r 2, 0], we have ψx, t ψx, t x =r = 4πr 2 t 3/2 exp r 2 /4r 2 t. The minimum of this function for t [ r 2, 0] is at t = r 2, and we get Also, on Q ρ, we have ψ ψ x=0,t=0 = r 2 G0, r 2, i.e., ψx, t Cr, x, t Q r. 2.6 ψx, t C r, x, t Q ρ. 2.7 Also, on Q ρ we have ψx, t Cr 2 x /r 2 + t 5/2 exp x 2 /4r 2 + t which is less than or equal to Cr 2 /r 2 + t 2 since y /2 e y C for y 0. We get ψx, t C r 2, x, t Q ρ. 2.8 As shown below, we have and ψx, t Cr2 ρ 3, x, t Q ρ\q ρ/2 2.9 ψx, t Cr2 ρ 4, x, t Q ρ\q ρ/

5 The proof of 2.9 is as follows: The bound 2.9 holds on B ρ ρ 2, ρ 2 /4] since the maximum on that region is achieved at x, t = 0, ρ 2 /4, and the value of ψ at that point is less than or equal to Cr 2 /ρ 3. For x B ρ \B ρ/2 and t ρ 2 /4, 0, we have ψx, t ψx, t x =ρ/2 Cr 2 /r 2 t 3/2 expρ 2 /6r 2 t. When viewed as a function of t, the last expression is largest at t = min{r 2 ρ 2 /24, 0}. Separating the cases r 2 ρ 2 /24 and r 2 ρ 2 /24, we get that ψ at the maximum point is less than or equal to Cr 2 /ρ 3 as claimed. The proof of 2.0 is similar: On Q ρ \Q ρ/2, we have ψx, t Cr 2 ρ exp r 2 + t 5/2 x 2 4r 2 + t The bound for the right hand side is then obtained by finding the maximum of the expression on B ρ ρ 2, ρ 2 /4 which is at x = 0 and t = ρ 2 /4 and on B ρ \B ρ/2 ρ 2 /4, 0 which is at x = ρ/2 and t = max{ρ 2 /40 r 2, 0}. Now, let η : R 3 R [0, ] be a smooth cut-off function such that η on Q ρ/2 and η 0 on Q c ρ with b t α0 x η C α 0, b ρ α0 +2b, x, t R3 R, b N 0, α 0 N Substituting φx, t = ψx, tηx, t in the energy inequality 2., we get for any t 0 [ r 2, 0] u 2 ψ t0 + 2 u 2 ψ u 2 φ t + φ + u 2 u φ B r Q r Q ρ Q ρ + 2 pu φ + 2 u fφ. 2.2 Q ρ Q ρ Denote by I, I 2, I 3, and I 4 the terms on the right side of 2.2. Now, on Q ρ, φ t + φ = η t + ηψ + 2 η ψ where we used t ψ + ψ = 0 on Q r. Note that η t, η, and η all vanish on Q ρ/2. Therefore, φ t + φ vanishes on Q ρ/2, and we get sup Q ρ φ t + φ = sup Q ρ\q ρ/2 φ t + φ where we used 2.9, 2.0, and 2.. Hence, I Cr2 ρ 5. sup η t + η ψ + 2 sup η ψ Cr2 Q ρ\q ρ/2 Q ρ\q ρ/2 ρ 5 Qρ u 2 Cr2 ρ 3 ess sup u 2 Cκ 2 αρ ρ 2,0 B ρ In order to treat the second term, write A ρ g = A ρ g, t = B ρ B ρ g, t. Also denoting v L r t L q = x vx, t L q x Q ρ L r, we get t ρ 2,0 I 2 C u 2 r 2 A ρ u 2 C u Q ρ r 2 u L 3 Q ρ u 2 A ρ u 2 L 3/2 Q ρ C r 2 u L 3 Q ρ u 2 3/2 L t L x Qρ C r 2 u L 3 Q ρ u L 6 t L 2 x Qρ u L 2 t L 2 x Qρ Cρ/3 r 2 u L 3 Q ρ u L t L 2 x Qρ u L 2 t L 2 x Qρ 5

6 where we used φ η ψ + η ψ C/r 2 on Q ρ, which holds by 2.7 and 2.8. We obtain I 2 Cρ2 r 2 αρβργρ = Cκ 2 αρβργρ. 2.4 For I 3, we use the Hölder inequality and φ C/r 2 on Q ρ in order to obtain As for I 4, we have by 2.7 I 3 Cρ2 r 2 δρ2 γρ = Cκ 2 δρ 2 γρ. 2.5 I 4 C r Q ρ u f C r u L 2 t L2 x Qρ f L 2 t L 2 x Qρ Cρ r αρλρ = Cκ αρλρ. 2.6 Observe that ess sup t0 r 2,0 B r u 2 ψ t0 C αr 2 and 2 Q r u 2 ψ C βr 2 by 2.6. Using 2.2, 2.3, 2.4, 2.5, and 2.6, we thus get whence Using Cκ δργρ /2 αr 2 + βr 2 Cκ 2 αρ 2 + Cκ 2 αρβργρ + Cκ 2 δρ 2 γρ + Cκ αρλρ αr + βr Cκαρ + Cκ αρ /2 βρ /2 γρ /2 + Cκ δργρ /2 + Cκ /2 αρ /2 λρ /2. Cκ 3 δρ 2 + Cκγρ and a direct consequence of the Gagliardo-Nirenberg inequality γρ Cαρ /2 βρ /2 + Cαρ, we get αr + βr Cκαρ + Cκ αρ 3/4 βρ 3/4 + Cκ αρβρ /2 + Cκ 3 δρ 2 + Cκαρ /2 βρ /2 + Cκ /2 αρ /2 λρ / The pressure estimates follow [L] see also [CKN]. Since the argument is very short, we provide the details for the sake of completeness. Let η C0 R3 be such that η in a neighborhood of B 3ρ/5 and η 0 in a neighborhood of B4ρ/5 c with α0 ηx C α 0 ρ α0, x R 3, α 0 N 3 0. Using p = ij U ij, where U ij = u i u j A ρ u j, we get ηp = ij ηu ij + ij ηu ij j U ij i η i U ij j η p η + 2 j j ηp 2.8 which may be verified easily by expanding the right hand side. Denote by N the kernel of, and note that Nx C x for all x R 3. From 2.8, we get ηp = R i R j ηu ij + N ij ηu ij j N U ij i η i N U ij j η N p η + 2 j N j ηp = p + p 2 + p 3 + p 4 + p 5 + p 6 6

7 where R i is the i-th Riesz transform. By the Calderón-Zygmund theorem, we have for every t r 2, 0 p L 3/2 B r p L 3/2 R 3 C i,j ηu ij L 3/2 R 3 C u L2 B ρ u A ρ u L6 B ρ C u L 2 B ρ u L 2 B ρ whence p L 3/2 Q r C u L 2 B ρ u L 2 B ρ L 3/2 r 2,0 Cr /3 u L 2 B ρ u L 2 B ρ L 2 r 2,0. Therefore, /2 ρ /2 r p 4/3 L 3/2 Q ρ C αρ /2 βρ / r Next, p 2 = N ij ηu ij. Note that ij η 0 on B 3ρ/5 and on B4ρ/5 c. Using x y 4ρ/5 r 3ρ/0 if x B r and y B4ρ/5 c, we get for all t r2, 0 p 2 L B r C ρ U ij ij η L B ρ C ρ 3 U L B ρ C ρ 2 u i L 2 B ρ u j A ρ u j L 6 B ρ C ρ 2 u L 2 B ρ u L 2 B ρ i,j whence p 2 L 3/2 B r Cr 2 /ρ 2 u L2 B ρ u L2 B ρ, which gives a better bound than 2.9 for p 2 instead of p. Analogous derivations show that the upper bounds for p 3 and p 4 are the same. Now, p 5 = N p η. Since η = 0 on B 3ρ/5 and on B4ρ/5 c, we get similarly to above p 5 L B r C ρ 3 p L B ρ C ρ 2 p L 3/2 B ρ whence p 5 L 3/2 B r Cr 2 /ρ 2 p L 3/2 B ρ. This implies and thus we obtain p 5 L 3/2 Q r Cr 2 /ρ 2 p L 3/2 Q ρ /2 r p 5 4/3 L 3/2 Q ρ Cr/3 ρ δρ. /3 The same estimate holds for p 6. Collecting the above bounds leads to This inequality and 2.7 imply δr Cκ /2 αρ /2 βρ /2 + Cκ /3 δρ. θr Cκαρ + Cκ αρ 3/4 βρ 3/4 + Cκ αρβρ /2 + Cκ 3 δρ 2 + Cκαρ /2 βρ /2 + Cκ /2 αρ /2 λρ /2 + Cκ 5 αρβρ + Cκ 0/3 δρ 2 Cκθρ + Cκ βρ /2 θρ + Cκ βρ /2 θρ + Cκθρ + Cκθρ + Cκ /2 λρ /2 θρ /2 + Cκ 5 βρθρ + Cκ 2/3 θρ 7

8 and 2.4 follows. In the proof of Theorem 2., we shall also use the following continuity property of αr. Lemma 2.3. Let 0 < r < R and t < t 2 be such that B R [t, t 2 ] D. Then we have lim ess sup ux, t 2 dx ess sup ux, t 2 dx. δ 0+ t [t,t 2+δ] B r t [t,t 2] B R Proof of Lemma 2.3. It is sufficient to prove lim sup ux, t 2 dx ess sup ux, t 2 dx δ 0+ t [t 2,t 2+δ] B r t [t,t 2] B R Let L be the set of Lebesgue points of the function t B R ux, t 2 dx in t, t 2. Let ɛ > 0, and let t 2 < T < t 2 + δ be such that B R [t, T ] D. Let also T 0 L [t 2 δ, t 2 ]. Let ψ C0 R 3 be a function with a range in [0, ] such that ψ on B r and ψ 2 0 on BR c. Given h 0, t 2 t, let ψ2 ht be the continuous function which equals 0 for t T 0, which is linear between T 0 and T 0 + h, and is equal to for t T 0 + h. A sequence of approximations justifies using φx, t = ψ xψ2 h t in the local energy inequality 2.. We get whence T0+h ux, T 2 ψ x dx ux, t 2 ψ xψ2 h t dx dt B R h T 0 B R u 2 φ + u 2 + 2pu φ + 2u fφ ux, T 2 dx B r h B R T 0,T T0+h T 0 B R T 0,T ux, t 2 ψ2 h t dx dt B R u 2 ψ2 h t ψ x + u 2 + 2pψ2 h tu ψ x + 2u fψ2 h tψ x. Sending h 0+, we obtain ux, T 2 dx ux, T 0 2 dx B r B R u 2 ψ + u 2 + 2pu ψ + 2u fψ B R T 0,T t2+δ t 2 δ u 2 ψ + u 2 + 2pu ψ + 2u fψ. Now, choose δ > 0 so small that the term on far right is less than or equal to ɛ. This gives sup ux, t 2 dx sup ux, t 2 dx + ɛ t t 2,t 2+δ B r t L t 2 δ,t 2 which, since ɛ > 0 is arbitrary and since L is a subset of full measure in t, t 2, implies

9 Lemma 2.4. There exists a sufficiently small universal constant ɛ > 0 with the following property. If lim sup β x0,t 0r < ɛ r 0+ then for every ɛ 0, /2 there exist r 2, r 3 > 0 and M > 0 such that for x, t B x0,t 0r 2 and r 0, r 3. max { α x,t r, β x,t r, δ x,t r 2} Mr ɛ Above and in the sequel, we denote B x0,t 0r = {x, t : x x t t 0 2 < r 2 }. Proof of Lemma 2.4. Without loss of generality, we may assume that x 0, t 0 = 0, 0. Denote θ x,t r = θ x,t r/r ɛ and θx, t = θ 0,0 x, t. Let ɛ 0, /2. By Lemma 2.2 and 2.2, we have θr Cκ 2/3 ɛ θρ + Cκ 5 ɛ βρ θρ + CM /2 κ ɛ /2 ρ /4 ɛ/2 θρ /2 whence θr C 0 κ 2/3 ɛ θρ + C0 κ 5 ɛ βρ θρ + 6 θρ + C 0 Mκ 2ɛ ρ /2 ɛ. Similarly, the inequality 2.5 implies θ x,t r C κ 2/3 ɛ θx,t ρ + C κ 5 ɛ ρ ɛ θx,t ρ θ x,t ρ + C Mκ 2ɛ ρ /2 ɛ provided 0 < r ρ/2. We may assume without loss of generality that C = C 0. Now, fix κ = min{/2, /6C 0 /2/3 ɛ } so that we have C 0 κ 2/3 ɛ /6 and r ρ/2. Then choose ɛ = κ 5+ɛ /6C 0 so that we have C 0 ɛ κ 5+ɛ 6. By the assumptions, there exists r 4 > 0 such that Q r4 D, βr ɛ, 0 < r < r 4 and max { } C 0 Mr /2 ɛ 4 κ +2ɛ, C 0r4 ɛ κ 5+ɛ 8. For n = 0,, 2,..., denote R n = κ n r 4 and θ n = θr n. Then By induction, we obtain θ n = 2 n θ θ n+ 2 θ n +, n = 0,, 2, n, n =, 2,.... 9

10 We conclude that there exists n 0 N such that θ n0 /3, which may be rewritten as θ 0,0 κ n0 r 4 3. By the continuity of the integral and by Lemma 2.3, there exist r 2 > 0 and r 5 0, r 4 such that θ x,t κ n0 r 5 2, x, t B x 0,t 0r 2. Note that θ x,t κ n+ r 5 2 θ x,t κ n r θκ n r , n = n 0, n,... for x, t B x0,t 0r 2. By induction, we get θ x,t κ n r 5 2, n = n 0, n 0 +, for x, t B x0,t 0r 2. Monotonicity of the integral implies αρ ρ 2 /ρ /2 αρ 2 ; also, βρ ρ 2 /ρ /2 βρ 2 and δρ ρ 2 /ρ 4/3 δρ 2 provided 0 < ρ < ρ 2. Therefore, ρ2 /2+ɛ θ x,t ρ C + ρ ρ2 provided Q ρ2 x, t D. From 2.2 and 2.22, we conclude ρ 4/3+ɛ θ x,t ρ 2, 0 < ρ < ρ θ x,t r C, r 0, r 5 for all x, t B x0,t 0r 2, and the lemma is proven. In order to prove the theorem, we need the following statement due to O Leary [O]. Lemma 2.5. [O] Let V R 3 R be a bounded domain. Assume that i sup x,t V sup ρ>0 ρ λ V B ρx,t gy, s q dy ds < and ii g L m V for some m q >, and 0 λ < 5. For α > 0, define gy, s hx, t = x y + dy ds. t s 5 α V Then for all m m, such that we have h L em V. m > qα m 5 λ Proof of Theorem 2.. Without loss of generality, x 0, t 0 = 0. Using Lemma 2.4 with ɛ = /4, there exist r 2, r 3 > 0 and M > 0 such that max { α x,t r, β x,t r, δ x,t r 2} Mr /4 0

11 for x, t B x0,t 0r 2 and r 0, r 3. Without loss of generality, r 2 = r 3. Note that γ x,t r Cα x,t r /2 β x,t r /2 + Cα x,t r CMr / for x, t B x0,t 0r 2. From here on, the argument does not depend any more on the generalized energy inequality. Let η C0 Rn be a function which is identically on a neighborhood of B 0,0 3r 2 /4 and 0 on a neighborhood of B 0,0 9r 2 /0 c. Then let v k x, t = t j Gx y, t sηy, su j y, su k y, s dy ds t + + t = v x, t + v2 x, t + v3 x, t. k k k Gx y, t sηy, spy, s dy ds Gx y, t sηy, sf k y, s dy ds k Let V = B x0,t 0r 2. Clearly, u v C B 0,0 3r 2 /4. Note that Gx, t C x + t 3 and Gx, t C x + t 4 for all x, t R 3 0,. Now, u L 0/3 D. By 2.23, we have sup By Lemma 2.5, we get v L em V, where m > 2 m r 2+3ɛ u 3 <. B r V 3/2 = /4 3m where m = 0/3. Similarly, v 2 L em V with the same condition on m. By a standard estimate for the non-homogeneous heat equation, we have v 3 L 0 V. Using m = 4/3, we get v L 4/30/3 V, and thus u L 4/30/3 B 0,0 3r 2 /4. We repeat the argument with the new value of m being 4/3 0/3, and we get u L 4/32 0/3 B 0,0 5r 2 /8. After a finite number of iterations and with a proper choice of the finite decreasing sequence of radii, we obtain u L 0 B 0,0 r 2 /2. Therefore, all the points in B 0,0 r 2 /2 are regular. Theorem 2. can be easily extended to relax the assumption f L 2 D further. Theorem 2.6. Let q > 5/3. Replace the condition ii in the definition of the suitable weak solution with f L q D. There exists a sufficiently small constant ɛ = ɛ q > 0 such that if x 0, t 0 D and then x 0, t 0 is a regular point. lim sup β x0,t 0r < ɛ r 0+

12 Proof. The proof is the same as the proof of Theorem 2., but 2.6 is replaced by I 4 C r u L q/q Q ρ f Lq Q ρ Cρ0/3 5/q u L3 Q r ρ f Lq Q ρ C κ ρ3 5/q γρ f Lq Q ρ. The assumption q > 5/3 assures that 3 5/q > 0, and the rest follows as before. By estimating the terms I, I 2, I 3, and I 4 differently, we obtain the inequalities αr + βr Cκγρ + Cκ γρ 3/2 + Cκ /2 δργρ + Cκ /2 γρ /2 λρ /2 and δr Cκ 2/3 γρ + Cκ /3 δρ provided 0 < r ρ/2. From here we observe that we can make θr as small as we wish provided γρ, δρ, and λρ are sufficiently small. Then the last part of the above proof applies leading immediately to the following theorem. Theorem 2.7. For every q > 5/3, there exists a sufficiently small constant ɛ = ɛ q > 0 with the following property. If Q D and u 3 + p 3/2 + f q ɛ Q then all points in Q /2 are regular. References [CF88] [CKN] P. Constantin and C. Foias, Navier-Stokes equations, Chicago Lectures in Mathematics, University of Chicago Press, Chicago, IL, 988. L. Caffarelli, R. Kohn, and L. Nirenberg, Partial regularity of suitable weak solutions of the Navier-Stokes equations, Comm. Pure Appl. Math , no. 6, [H] E. Hopf, Über die Anfangswertaufgabe für die hydrodynamischen Grundgleichungen, Math. Nachr. 4 95, [L] P.G. Lemarié-Rieusset, Recent developments in the Navier-Stokes problem, Chapman & Hall/CRC Research Notes in Mathematics, vol. 43, Chapman & Hall/CRC, Boca Raton, FL, [Le] J. Leray, Sur le mouvement d un liquide visqueux emplissant l espace, Acta Math , no., [Li] F. Lin, A new proof of the Caffarelli-Kohn-Nirenberg theorem, Comm. Pure Appl. Math , no. 3,

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