INSTITUTE OF MATHEMATICS THE CZECH ACADEMY OF SCIENCES

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1 INSTITUTE OF MATHEMATICS THE CZECH ACADEMY OF SCIENCES A contribution to the theory of regularity of a weak solution to the Navier-Stokes equations via one component of velocity and other related quantities Jiří Neustupa Preprint No PRAHA 207

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3 A Contribution to the Theory of Regularity of a Weak Solution to the Navier Stokes Equations via One Component of Velocity and Other Related Quantities Jiří Neustupa Abstract We deal with a suitable weak solution v, p) to the Navier Stokes equations in Ω 0, T ), where Ω is a domain in R, T > 0 and v = v, v 2, v ). We show that the regularity of v, p) at a point x 0, t 0 ) Ω 0, T ) is essentially determined by the Serrin type integrability of the positive part of a certain linear combination of v 2, v 2 2, v 2 and p in a backward neighborhood of x 0, t 0 ). An appropriate choice of coefficients in the linear combination leads to the Serrin type condition on one component of v or, alternatively, on the positive part of the Bernoulli pressure 2 v 2 + p or the negative part of p, etc. AMS math. classification 200): 5Q0, 76D0, 76D05. Keywords: Navier Stokes equations, weak solution, regularity. Introduction.. The Navier Stokes system. Let Ω be either the whole space R or a half-space or a bounded or exterior domain with the boundary of the class C 2+ς ς > 0) and let T > 0. We deal with the Navier Stokes problem t v + v v = p + ν v in Ω 0, T ),.) div v = 0 in Ω 0, T ),.2) v = v 0 in Ω {0}.) for the unknown velocity v = v, v 2, v ) and pressure p. Symbol ν denotes the coefficient of viscosity, which is supposed to be a positive constant. If Ω then we consider the problem.),.2),.) with the homogeneous Dirichlet boundary condition v = 0 on Ω 0, T )..4).2. Weak and suitable weak solution, regular and singular points. The definition of a weak solution to the system.),.2) and its basic properties are explained e.g. in the books by Ladyzhenskaya 8], Temam 2], Sohr 9] and in the survey paper 6] by Galdi. Here, we only recall that the weak solution satisfies.),.2) in the sense of distributions in Ω 0, T ) and belongs to L 0, T ; L 2 Ω)) L 2 0, T ; W,2 0 Ω)). The existence of a weak solution to.),.2),.5) is known on an arbitrarily long time interval 0, T ) provided that the initial velocity v 0 is an appropriate space, see 8], 9], 2]

4 or 6]), but its regularity and uniqueness are generally open problems. Since, roughly speaking, regular solutions are unique, the question of uniqueness also leads to the question of regularity. The definition of the so called suitable weak solution to the system.),.2), with many related results, can be found e.g. in papers ], 9], 0] and 22]. Recall that a weak solution v of system.),.2) is called a suitable weak solution if an associated pressure p belongs to L /2 Ω 0, T )) and the pair v, p) satisfies the so called generalized energy inequality T T 2ν v 2 φ dx dt v 2 t φ + ν φ ) + v 2 + 2p ) v φ ] dx dt.5) 0 Ω 0 Ω for every non negative function φ from C0 Ω 0, T )). Some authors use different conditions on the pressure in their definitions. Our class L /2 Ω 0, T )) is the same as in 9], 0] and 22].) By the definition from ], the point x 0, t 0 ) Ω 0, T ) is said to be a regular point of weak solution v if there exists a neighborhood U of x 0, t 0 ) such that v L U). Points in Ω 0, T ) that are not regular are called singular. It is shown in ] that the set of singular points of a suitable weak solution has the dimensional parabolic measure which dominates the dimensional Hausdorff measure) equal to zero... On some local regularity criteria. There exist many so called local regularity criteria, saying that if a suitable weak solution a posteriori satisfies certain conditions in a backward neighborhood of point x 0, t 0 ) then x 0, t 0 ) is a regular point. See e.g. papers ], 4], 9], 0], 5], 22], etc. In this paper, we use a criterion from 22] by Wolf). The criterion is formulated more generally, but it particularly says that there exists ε > 0 such that if δ 2 t0 t 0 δ 2 B δ x 0 ) v dx dt ε.6) holds for at least one δ > 0 then x 0, t 0 ) is a regular point of the solution v. Here, B δ x 0 ) naturally denotes the ball of radius δ and center x 0.) Let us also note that Takahashi 20] proved that if the norm of a weak solution v in L r wt 0 ρ 2, t 0 ; L s B ρ x 0 )) where L r w denotes the weak L r space and 2/r + /s, < s ) is less than or equal to ε then x 0, t 0 ) is a regular point of v. Takahashi s criterion has been refined in 6] and 7]. In 7], v is supposed to be integrable with powers r, ) in time) and s, ) in space) not necessarily in some backward neighbourhood of x 0, t 0 ), but only in the intersection of such a neighbourhood with the exterior of the space time paraboloid P a : at 0 t) = x x ) Exponents r and s are required to satisfy the condition 2/r +/s and number a is supposed to satisfy the inequalities 0 < a < 4νλ S B ), where λ S B ) is the least eigenvalue of the Dirichlet Stokes operator in the unit ball B in R..4. More on one component regularity criteria. The studies of regularity of a suitable weak solution v in dependence on one component of v were started by paper 2] Neustupa, Penel), where the authors proved the regularity of v in D t, t 2 ) where D was a sub domain of Ω and 0 t < t 2 T ) under the assumption that the component v was essentially bounded in D. The condition on v has been successively improved in a series of further papers: ) 4] by Neustupa, Penel and Novotný; here, v is only assumed to be in L r t, t 2 ; L s D)) where 2/r + /s 2 ), 2) 7] by Kukavica and Ziane; the case D = R, v is assumed to be in 2

5 L r 0, T ; L s R )) where 2/r + /s = 5 8 for r 6 5, ) and s 24 5, ]), ) 2] by Cao and Titi); here the authors consider the spatially periodic problem in R and use the condition 2/r + /s < 2 + 2/s), s > 7 2 ), 4) 2] by Zhou and Pokorný; the exponents r, s are supposed to satisfy the conditions 2/r + /s 4 + /2s), s > 0 ). One can observe that none of these papers reaches the natural Serrin level 2/r + /s. This level was in a certain sense reached by Chemin, Zhang and Zhang ], where the regularity of solution v has been proven under the assumption that v L r 0, T ; Ḣ /2+2/r R )), where r 4, ). The homogeneous Sobolev space Ḣ/2+2/r R ) is continuously imbedded to L r/r 2) R ). Hence the condition v L r 0, T ; Ḣ /2+2/r R )) implies that v L r 0, T ; L r/r 2) R )), and the exponents r and s := /r 2) now satisfy Serrin s condition 2/r + /s. Nevertheless, the requirement v L r 0, T ; Ḣ /2+2/r R )) includes the condition on the fractional derivative of v and it is stronger than just the condition v L r 0, T ; L r/r 2) R )). Thus, we may conclude that, to our best knowledge, the question whether the condition v L r t, t 2 ; L s D)) for r and s, basically satisfying the condition 2/r + /s, is sufficient for regularity of solution v in D t, t 2 ), is still open..5. On the results of this paper. We provide a partial answer to the question formulated at the end of the previous subsection. Our answer concerns the regularity of a suitable weak solution v at a chosen point x 0, t 0 ) Ω 0, T ). For ρ 0, t 0 ) and a, we denote Q ρ := {x, t); x x 0 < ρ, t 0 ρ 2 < t < t 0 }, U ρ,a := { x, t); θt) < x x 0 < ρ, t 0 ρ 2 /a < t < t 0 }, where θt) := at 0 t)..8) Q ρ is a ρ backward parabolic neighborhood of point x 0, t 0 ). Set U ρ,a is separated from the interior of Q ρ U ρ,a by the space time paraboloid P a, see.7). It should be noted that parameter a can be chosen arbitrarily large. Consequently, paraboloid P a may be arbitrarily wide and set U ρ,a can be proportionally an arbitrarily small part of Q ρ. t x 0, t 0 ) U ρ,a t = t 0 Fig. : The sets Q ρ, U ρ,a and paraboloid P a. P a Q ρ t = t 0 ρ 2 /a t = t 0 ρ 2 x 0 x x 0 = ρ x We suppose that v satisfies Serrin s integrability condition in U ρ,a and the component v of v satisfies Serrin s condition in Q ρ U ρ,a, which is the major part of Q ρ. We show that these assumptions imply that x 0, t 0 ) is a regular point of solution v see Theorem ). Theorem 2 generalizes Theorem so that the assumption on v is replaced by an assumption on the positive part of a certain linear combination of v 2, v2 2, v2 and p. Our method is especially based on the

6 transformation of the system.),.2) to new coordinates x, t subsection 2.), application of the generalized energy inequality in the x, t ) space subsection 2.7), estimates of appropriate quantities and on the precise evaluation of critical integrals, where the directions of the velocity at various points also play an important role subsections.6 and.7). Although we still need the assumption on the Serrin type integrability of all components of v in set U ρ,a, we believe that the presented results shed in addition to the papers 2], ], 7], 2], 4], 2]) another light on the mechanism how the behavior of just one component of v or more generally, a linear combination of v 2, v2 2, v2 and p) influences the regularity of solution v. For r >, s >, we abbreviate L r,s Q ρ ) := L r t 0 ρ 2, t 0 ; L s B ρ x 0 )) and we denote by. r,s; Qρ the corresponding norm. More generally, if M is a measurable set in Ω 0, T ), IM) is the orthogonal projection of M into the t axis and M t := {x Ω; x, t) D} then we denote by L r,s D) the space of functions f with the finite norm f r,s; M := IM) )r ] fx, t) s s r dx dt. M t We also denote by L r,s D) the corresponding space of vector functions. The next theorem shows that the local regularity of a suitable weak solution v at a space time point x 0, t 0 ) is essentially determined just by one component of v: Theorem. Let v = v, v 2, v ) be a suitable weak solution of the system.),.2) in Ω 0, T ), x 0, t 0 ) Ω 0, T ), a and ρ 0, t 0 ). Suppose that a) there exist r, ) and s, ) satisfying 2/r + /s =, such that v L r,s U ρ,a ) and b) there exist r 2, ) and s, ] satisfying 2/r + /s =, such that v L r,s Q ρ U ρ,a ). Then x 0, t 0 ) is a regular point of solution v. Denote Fv, p, γ, γ 2, γ ] := + γ )v γ 2 )v γ )v 2 + γ + γ 2 + γ )p ] + for γ, γ 2, γ R. The subscript + denotes the positive part.) The next theorem is a generalization of Theorem : Theorem 2. Let v = v, v 2, v ) be a suitable weak solution of the system.),.2) in Ω 0, T ), p be an associated pressure, x 0, t 0 ) Ω 0, T ), a and ρ 0, t 0 ). Assume that v satisfies condition a) of Theorem and also the condition c) there exist r, ), s 2, ] satisfying 2/r + /s = 2 and γ, γ 2, γ R, such that 5π 28 γ + γ 2 + γ ) + 5π 28 γ k > 0 for k =, 2, ).9) and Fv, p, γ, γ 2, γ ] L r,s Q ρ U ρ,a ). Then x 0, t 0 ) is a regular point of solution v. 4

7 Observe that if γ = 2 and γ 2 = γ = then condition c) reduces to condition b). On the other hand, if γ = γ 2 = γ = then condition c) requires p] + L r,s Q ρ U ρ,a ). It is equivalent to the condition p L r,s Q ρ U ρ,a ), which has already been used in paper ]. Here, p denotes the negative part of p.) Thus, our Theorem 2 generalizes Theorem from ]. Finally, if γ = γ 2 = γ = 2 then condition c) just requires that the positive part of the so called Bernoulli pressure 2 v 2 + p is in L r,s Q ρ U ρ,a ). As Theorem is a special case of Theorem 2, we will further prove Theorem 2. 2 Proof of Theorem 2 part I 2.. The used regularity criterion. We will show that lim δ 0+ δ 2 v dx dt = 0 2.) U δ,a and there exists a sequence {δ n }, such that δ n 0 for n and lim v dx dt = 0, 2.2) n V δn,a δ 2 n where V δ,a := { x, t); x x 0 < θt), t 0 δ 2 /a < t < t 0 }. We will show in subsection 2.8 that 2.) and 2.2) imply.6) The proof of 2.). Applying Hölder s inequality, we get δ 2 v,; U δ,a t0 δ 2 t 0 δ 2 /a 4π ) t0 s a r θt)< x x 0 <δ t 0 δ 2 /a v s dx ) s 4πδ θt)< x x 0 <δ ) s dt )r ] v s s dx dt Since v belongs to L r,s U ρ,a ), the right hand side tends to zero as δ 0+. Hence 2.) holds. 2.. Transformation to the new coordinates x, t. In order to prove 2.2), we transform the system.),.2) to the new coordinates x and t, which are related to x and t through the formulas Then x = x x t 0 θt), t = t 0 ρ 2 /a dτ θ 2 τ) = a ln ρ 2 at 0 t). 2.) t = t 0 ρ2 a e at and θt) = ρ e 2 at. 2.4) The time interval t 0 ρ 2 /a, t 0 ) on the t axis now corresponds to the interval 0, ) on the t axis. Equations 2.) represent a one to one transformation of the parabolic region V ρ,a in the x, t space onto the infinite stripe V a := { x, t ) R 4 ; t > 0 and x < } 5 r.

8 in the x, t space. Similarly, 2.) is a one to one mapping of the set U ρ,a in the x, t space onto U a := { x, t ) R 4 ; t > 0 and < x < e 2 at } in the x, t space. We denote t δ := 2 a ln ρ δ. 2.5) t U a Fig. 2: The sets U a and V a. V a x : 0 x = e 2 at x t = t δ 2 a ln ρ δ corresponds to t = t 0 δ 2 /a) Then t = t δ corresponds to t = t 0 δ 2 /a. Numbers δ and t δ are also related through the formula δ = ρ e 2 at δ and δ 0+ corresponds to t δ. The transformation 2.) has also been used in 7]. However, while a was supposed to satisfy certain condition of smallness in 7], here it can be arbitrarily large.) If we put vx, t) = px, t) = θt) v x x 0 θt), a ln ρ 2 ), at 0 t) θ 2 t) p x x 0 θt), a ln ρ 2 ) at 0 t) then the functions v, p represent a suitable weak solution of the system of equations t v + v v = p + ν v 2 av 2 ax v, 2.6) div v = 0 2.7) in any bounded sub domain of Q a := { x, t ) R 4 ; t > 0 and x < e 2 at }. The symbols and denote the nabla operator and the Laplace operator with respect to the spatial variable x.) One can simply calculate that condition a) implies that v L r,s U a) and condition c) implies that Fv, p, γ, γ 2, γ ] L r,s V a) Notation. Let 0 < d < d 2. We denote by A d,d 2 and A d,d 2 the annuli {x R ; d < x x 0 < d 2 } and {x R ; d < x < d 2 }, respectively. We also denote by B d the ball {x R ; x < d }. The mapping x x = x x 0 )/θt) is a one to one transformation of A θt)d,θt)d 2 onto A d,d 2 and B θt)d x 0 ) onto B d at each time instant t t 0 ρ 2 /a, t 0 ) The cut off functions ψ and ϕ. Let d >. Number d will be finally specified to be sufficiently large in subsection.8.) Let ψ be an infinitely differentiable function in the interval, ), such that 6

9 ψξ) { = for ξ d, = 0 for 2d < ξ, ψ is non-increasing in d, 2d] and there exists c > 0 independent of d) such that for d ξ 2d. Put ψξ) c d and ψξ) c d 2 2.8) ϕ := ψ. We will further use ψξ) and ϕξ) with ξ = x, and we shall mostly write only ψ or ϕ instead of ψ x ) or ϕ x ), respectively The first estimate of δ 2 v,; V δ,a. Recall that t = t δ = 2a lnρ/δ) corresponds to t = t 0 δ 2 /a see formulas 2.) 2.5)). Transforming δ 2 v,; V δ,a to the variables x, t, we get δ 2 Vδ,a v dx dt = ρ2 ρ2 δ 2 t δ c 2 ρ 2 δ 2 δ 2 t δ B v dx e at dt ρ2 ϕv 2 6; B 2d ϕv 2 2; B 2d e at dt c 2 ρ 2 δ 2 ) ϕv ) 2 2; B 2d e 2 at dt 4 ϕv 6 2; B 2d t δ ) = c 2 ϕv ) 2 2; B 2d e 2 at t δ ) dt 4 ϕv 6 2; B 2d t δ t δ t δ t δ δ 2 t δ v 2 6; B v 2 2; B e at dt ϕv ) 2 2; B 2d ϕv 2 2; B 2d e at dt e 2at dt ) 4 e 2at t δ ) dt ) ) Here, c 2 is an absolute constant, coming from Sobolev s inequality. See e.g. 5, p. 54].) In order to estimate the integrals on the right hand side of 2.9), we use the generalized energy inequality in the x, t space The generalized energy inequality in the x, t space. Since v, p is a suitable weak solution to the system 2.6), 2.7), it satisfies by analogy with.5)) the generalized energy inequality 2ν v 2 φ dx dt v 2 t φ + ν φ ) + v 2 + 2p ) v φ Q a Q a + 2 a v 2 φ + 2 a x φ) v 2] dx dt 2.0) for every non negative function φ from C 0 Q a). Consider function φ in the form φx, t ) = R /m ψ] x ) e κt t δ ) R /m χ]t ), where κ R, χ is the characteristic function of the interval t δ, t ) and R /m is a one dimensional mollifier with the kernel supported in /m, /m). Then the term 2 a x φ) v 2 on the right hand side of 2.0) can be omitted, because x φ 0. The limit for m yields ϕv., t ) 2 2; B 2d e κt t δ ) + 2ν t t δ ϕv., τ)) 2 2; B 2d e κτ t δ ) dτ 7

10 a ) ϕv., t t δ ) 2 2; B 2d κ ϕv 2 t 2; B e κτ t δ ) dτ 2d δ t + 2ν ϕ 2 v 2 + v 2 + 2p ) v ψ) ] e κτ t δ ) dx dτ. 2.) t δ A d,2d A similar limit procedure has been used in 7].) Inequality 2.) holds for a.a. t t δ, where t δ is for technical reasons supposed to be greater than t := 2a ln 2d. Choosing κ = 2 a, we get ϕv., t ) 2 2; B 2d e 2 at t δ ) + a 6 + 2ν t t δ t ϕv., τ)) 2 2; B 2d e 2 aτ t δ ) dτ t δ ϕv., τ) 2 2; B 2d e 2 aτ t δ ) dτ ϕv., t δ ) 2 2; B 2d + K I δ) + K II δ), 2.2) where K I δ) := K II δ) := t δ t δ A d,2d A d,2d 2ν ϕ 2 v 2 + v 2 v ψ ) dx e 2 aτ t δ ) dτ, 2p v ψ) dx e 2 aτ t δ ) dτ. The next lemma is proven in 7]: Lemma. Assume that v L r,s U a), 0 < α r, 0 < β s, R >, t δ > 2a ln R, and at least one of the two conditions ) α = r, ω 0, 2) α < r, ω > 0 holds. Then t δ A,R )α v β dx β e ωat t δ ) dt 0 as t δ. 2.) Applying Lemma, we can show that K I δ) 0 for δ 0+. Note that in the case of the integral containing v 2 v ϕ 2, we apply Lemma with α = β = and ω = 2. Here, we use the assumption r.) As to the term K II δ), we refer to 7], where K II δ) is estimated as follows: K II δ) c δ) ϕv., t δ ) 2 2; B 2d + c 4 δ), 2.4) where c δ) 0 and c 4 δ) 0 for δ 0+. In 7], the author considers an infinitely differentiable function ϕ with values in 0, ] such that ϕ = in B and ϕ = 0 outside B 4 instead of our ϕ, but this difference plays no role.) The proof of 2.4) is relatively laborious especially because it requires to estimate the transformed pressure p. Note that both c δ) and c 4 δ) also depend on parameter a. Thus, inequalities 2.2) and 2.4) yield ϕv., t ) 2 2; B 2d e 2 at t δ ) + a 6 + 2ν t t δ t ϕv., τ)) 2 2; B 2d e 2 aτ t δ ) dτ 8 t δ ϕv., τ) 2 2; B 2d e 2 aτ t δ ) dτ

11 where c 5 δ) 0 for δ c δ)] ϕv., t δ ) 2 2; B 2d + c 5 δ), 2.5) 2.8. A conditional completion of the proof of Theorem. Suppose that i) there exists a sequence {δ n } such that δ n 0 and lim n ϕv., t δ n ) 2; B 2d = 0. Then the proof of Theorem can be completed as follows: the identity 2.) is proven in subsection 2.2. The inequalities 2.9), 2.5) and condition i) imply that lim n δn 2 v dx dt V δn,a ) ) lim c 2 ϕv ) 2 n 2; B 2d e 2 at t δn ) dt 4 ϕv 6 2; B 2d e 2at t δn ) dt 4 t δn ) c 2 lim ϕv ) 2 n t 2; B 2d e 2 at t δn ) dt 4 δn ) ) ess sup ϕv., t ) 2; B t 2d e at t δn ) ϕv ) 2 δn <t < t 2; B 2d e 2 at t δn ) dt 4 δn ) 4 6 ) ) 4 c 2 lim + c δ n )] ϕv., t n δ 2ν a n ) 2 2 2; B 2d + c 5 δ n ) = 0. This proves 2.2). For all m N and n N such that δ n aδ m, we have Q δm U δn,a V δn,a). Denote by n m the maximum of all n N such that δ nm aδ m. Then δ m implies δ nm for m ). Hence, using also 2.) and 2.2), we have lim m δ 2 m Q δm v dx dt lim m δ 2 n m t δn U δnm,a V δnm,a v dx dt = 0. This implies.6), which means that x 0, t 0 ) is a regular point of the solution v, p.. Proof of Theorem 2 part II The purpose of this section is to show that condition i) holds, provided that assumption c) of Theorem 2 is satisfied. Recall that t δ n = 2a lnρ/δ n ). We observe that δ n 0 is equivalent to t δ n. In order to simplify the notation, we further write only t n instead of t δ n. The existence of a sequence {t n} such that t n and ϕv., t n) 2; B 2d 0 for n ) will be established in this section... The integrals of v k 2 + p ) ψ k =, 2, ). Here, we show that the integrals of v k 2 + p ) ψ in B 2d are equal to certain integrals over A d,2d. Assume, for example, that k =. Let us multiply equation 2.7) by 2 x 2 ψ) x, 0, 0) ψ + 2 x 2 ψ and integrate in B 2d. Since v is a suitable weak solution to the system 2.7), 2.0) and the set of its singular points has D Hausdorff measure equal to zero, the integral has a sense for a.a. t > t. We obtain 0 = v v + p ] x, 0, 0) ψ + ] 2 x 2 ψ dx, B 2d 9

12 0 = 0 = B 2d B 2d { v j jv )x + p )x ] ψ + v j jv i) 2 x 2 i ψ + ip 2 x 2 i ψ } dx, v 2 ψ + v j v x jψ + p ψ + p x ψ + v v ix iψ + v j v i 2 x 2 i jψ + p x ψ + p 2 x 2 ψ ] dx. Similar identities also hold for k = 2 and k =. Thus, we have 0 = v k2 + p ] ψ dx + 2v k x k v ψ) + 2p x k k ψ + 2 x2 k v 2 ψ v 0 = B 2d B A d,2d + p 2 x2 k ψ ] dx, v k2 + p ] ψ dx + v k2 + p ] ψ dx + A,2d + x 2 k v 2 ψ 2 x v x ) 2 ψ x + v x ) 2 ) ψ x 2 A d,2d 2v k v x ) x k x + p x 2 k 2 2 x ψ + 2p x 2 k x )] ψ + ψ dx.) for k =, 2,. One does not sum over k in.). Moreover, the second integral is considered only in A d,2d, because the derivatives of ψ are supported in the closure of A d,2d.).2. The integral of v 2 + p in B. In this subsection, we express the integral of v 2 + p ) in B by means of some other integrals over A,2d. Define function φ in the interval, ) by the formulas = for ξ, φξ) = + 4 ) ξ ψξ) for ξ >. φ is continuous and piecewise continuously differentiable. Moreover, φξ) = 0 for ξ 2d and ξ φξ) + φξ) = for < ξ < d..2) By analogy with functions ψ and ϕ, φ will further mostly mean φ x ). We multiply equation 2.7) by x φ x ) and integrate in R. Since x φ x ) = Φ x ), where Φξ) is an antiderivative to ξ φξ), we get 0 = v v x φ + p x φ ] dx, A d,2d 0 = B 2d B 2d v 2 φ + v x ) 2 x φ + p φ + x φ )] dx. Using the concrete form of function φ and applying.2), we further obtain 0 = v 2 + p ) dx + v 2 B A + 4 ) x 4v x ) 2 ] x 5 dx,d + v 2 ψ + 4ψ ) x + v x ) 2 ψ x + 4 ψ ) x 4 v x ) 2 ] x 5 ψ A d,2d p ψ + x ψ A,d ψ p dx dx 4 ψ ) x 2 dx..) 0

13 .. Condition i) the beginning. In order to fulfill condition i), we need an information on the behavior of ϕv., t δ ) 2 2; B for t 2d δ. Therefore we multiply formula.) by α k, sum over k =, 2, and add the sum to the equation ϕv 2 2; B = ϕv 2 2d 2; B + ϕv 2 2; A,2d. Furthermore, we multiply formula.) by β := α + α 2 + α and also add the product to ϕv 2 2; B The real numbers α 2d., α 2, α will be specified later, see.5).) Due to the choice of β, the factor α + α 2 + α β in front of A p ψ dx is equal to zero. Thus, we obtain,2d = ϕv 2 2; A + + α k )v k2 + αk p ] dx + β v 2 + p ) dx,2d B B ϕv 2 2; B 2d + + k= A,d α k k= α k v k2 dx + β A d,2d k= A,d v k2 ψ + 2v k x v x k x + x 2 k 2 v 2 x + β v 2 ψ A + 4ψ d,2d β A d,2d p x Subtracting ϕv 2 2; A,2d where Ψ := Ψ 2 := Ψ := B ψ v ) x 4v x ) 2 ] x 5 dx ψ + 2p x 2 k x ψ v x ) 2 ψ x + v x ) 2 ) ψ x 2 x ) 4 ψ ) x 2 dx. + v x ) 2 x ψ + p x 2 2 k 2 x ψ + 4 ψ x ) 4 v x ) 2 x 5 ψ ψ + ψ )] dx ] dx from both sides and taking into account that ϕ = in B, we get v 2 2; B = Ψ + Ψ 2 + Ψ + Ψ 4 + Ψ 5,.4) + 2α + α 2 + α ) v α + 2α 2 + α ) v α + α 2 + 2α ) v 2 + 4α + α 2 + α ) p ] ψ dx, α k v k2 dx + β v ) x 4v x ) 2 ] x 5 dx, k= A,d α k k= + β A d,2d α k Ψ 4 := 2 k= Ψ 5 := β A d,2d A d,2d A,d v k2 ψ + 2v k x v x k x ψ + x 2 k v 2 2 x v 2 ψ + 4ψ ) x + v x ) 2 x A d,2d p x x 2 k p 6 ψ + x ψ ) ψ dx, 4 ψ ) x 2 dx. ψ v x ) 2 ψ x + v x ) 2 )] ψ x 2 dx ψ + 4 ψ ) x 4 v x ) 2 ] x 5 ψ dx,

14 Let us now choose α, α 2, α so that 2α + α 2 + α = γ, α + 2α 2 + α = γ 2 and α + α 2 + 2α = γ. Then α k = γ k 4 γ + γ 2 + γ ) for k =, 2, )..5) Thus, function Ψ satisfies Ψ = + γ ) v γ2 ) v γ ) v 2 + γ + γ 2 + γ ) p ] ψ dx B Ψ := B Fv, p, γ, γ 2, γ ] ψ dx, where t = Ψ r dt 4π ) s s r t0 4π t ) s s r x x 0 <θt) t B )r Fv, p, γ, γ 2, γ ] s dx s dt Fv, p, γ, γ 2, γ ] s dx )r s dt <.6) due to assumption c), provided that γ, γ 2, γ satisfy the restrictions formulated in this condition. Here, we denote t := t 0 ρ 2 /a) e at compare with 2.4).) Hence Ψ L r t, ). Since t Ψ 2 r 2 dt C = Cd) t A,d t0 t v 2 dx )r 2 dt Cd) v s dx θt)< x x 0 <dθt) )r t A,d v s dx )r s dt s dt <.7) due to condition a)), we observe that Ψ 2 L r/2 t, ). Function Ψ can be treated in the same way, with a small difference, i.e. that we integrate in A d,2d instead of A,d in the x space and in the region dθt) < x x 0 < 2dθt) instead of θt) < x x 0 < dθt) in the x space. However, we also deduce that Ψ L r/2 t, )..4. The estimates of Ψ 4. The functions Ψ 4 and Ψ 5 need a special treatment, because they contain the pressure p and we not have an explicit additional information on the integrability of p in A d,2d in contrast to v, which is due to assumption a) of Theorem in L r,s U a)). Nevertheless, if η is an appropriate cut off function in R then p satisfies the obvious identity ηx ) p x, t ) = 4π R x y ηp ) ] y, t ) dy.8) for x R. Concretely, we assume that 0 < κ < and choose η so that it is infinitely differentiable and satisfies = for x κ e 2 at, ηx ) 0, ] for κ e 2 at < x e 2 at, = 0 for e 2 at < x, 2

15 η 2 κ e 2 at and 2 η 8 κ) 2 e at. Integrating by parts in.8) and using the formula p = i j v i v j ), we derive that ηx ) p x, t ) = p x, t ) + p 2x, t ) + p x, t ),.9) where p x, t ) = 4π p 2x, t ) = 4π p x, t ) = 2π + 4π + 2π + 4π B 2 y i y j A, e at /2 ) x y v iv j]y, t ) dy, 2 ) x y ηv iv j]y, t ) dy, y i y j A κe at /2, e at /2 A κe at /2, e at /2 A κe at /2, e at /2 A κe at /2, e at /2 x i y i η ) x y y j v iv j y, t ) dy 2 η ) x y y i v iv y j y, t ) dy j x i y i η x y y i p ) y, t ) dy x y η p ]y, t ) dy. We can now split Ψ 4 to the sum Ψ 4 + Ψ 42 + Ψ 4, where Ψ 4l := k= α k 2 A d,2d x 2 k p l ψ x ) + 6 x ) ψ x ) dx for l =, 2,..5. Estimates of Ψ 42 and Ψ 4. Let us at first deal with the easy terms Ψ 42 and Ψ 4.. Applying the Calderon Zygmund theorem, we obtain p 2y, t ) s 2 dy C v y, t ) s dy..0) A, e at /2 A, e at /2 Using also 2.8), we can show that Ψ 42 L r/2 t, ): )r Ψ 42 r 2 dt C p 2 dx 2 dt C t C = C t t0 t t A,e at /2 A d,2d )r p 2 s s 2 dx dt C t t A,e at /2 A d,2d )r p 2 s s 2 dx d s 2 r s 2 dt v s dx )r s dt )r v s s dx dt <..) θt)< x x 0 <ρ In order to derive an analogous information on Ψ 4, we estimate p as follows: if x A d,2d then p x, t ) C e 2 at v 2 + p ] dx..2) A κe at /2, e at /2

16 Note that the generic constant C in.) and.2) depends on α, α 2, α and d.) Since the set of possible singular points of solution v has the dimensional Hausdorff measure equal to zero, we can assume without loss of generality that ρ respectively κ 0, )) are chosen so small respectively close to ) that v has no singular points in the region κρ σ < x x 0 < ρ + σ, t 0 ρ 2 σ 2 < t < t 0 + σ 2 for some σ > 0. Then v, with all its spatial derivatives, is essentially bounded in {x, t) R 4 ; κρ < x x 0 < ρ, t 0 ρ 2 < t < t 0 }. The known results on the interior regularity of pressure Lemma 2 from 7] or Theorem 4 from 8]) imply that if Ω = R then p together with all its spatial derivatives) is also essentially bounded in the same region. If Ω satisfies the assumptions from Section, but it differs from R, then p together with all its spatial derivatives) is only in L λ t 0 ρ 2, t 0 ; L A κρ,ρ ) for each λ, 2). See Lemma 2 in ] or Theorem 2 in 8]). Denote, for a while, by p t) respectively v t)) the L norm of p., t) respectively v., t)) in A κρ,ρ. Put t := 2/a) ln2d/κ) and t := t 0 ρ 2 /a) e at. Then 2d < κ e at /2 for t > t.) Since r, we have r/r ) 2. Choose µ and λ so that < µ < λ < 2. Then ) µ Ψ 4 µ dt C p dx dt t C t t A d,2d e 2 at t0 = C ρ µ t C ρ µ t0 C ρ µ t0 A κe at /2,e at /2 κρ< x x 0 <ρ v 2 + p ) dx ) µ dt v 2 + p ) ) α dx θ 2µ 2 t) dt t v 2µ t) + p µ t) ) θ 2α 2 t) dt t v 2λ t) + p λ t) ) dt ] µ λ t 0 t θ 2µ 2) ] λ µ λ λ λ µ t) dt, where C = Cα, α 2, α, d). The first integral on the last line is finite because < λ < 2 λ and the second integral is finite because the exponent 2µ 2) λ µ is greater than 2. Hence Ψ 4 L µ t, ) for each µ, 2)..6. The function Ψ 4. Here, we deal with the most difficult part of Ψ 4, which is the term Ψ 4. Function p satisfies: p x, t ) = x i y i )x j y j ) δ ) ij 4π B x y 5 x y v iv j]y, t ) dy = vy, t ) x y )] 2 4π x y 5 v y, t ) 2 ) x y v iv j]y, t ) dy. One can calculate that B vy, t ) x y )] 2 x y 5 v y, t ) 2 x y = vy, t ) x ] 2 x 5 v y, t ) 2 x + O d 4 vy, t ) 2) for x B and y A d,2d. Hence p x, t ) = 4π B vy, t ) x ] 2 x 5 v y, t ) 2 x ) dy 4

17 + O d 4 vy, t ) 2) dy..) 4π B The contribution of the second term on the right hand side to Ψ 4 let us denote it by Ψ 42 ) satisfies Ψ 42 C d 4 k= α k 2 v., t ) 2 2; B A d,2d x 2 k ψ x ) + 6 x ψ x ) dx C d v., t ) 2 2; B..4) The contribution of the first integral on the right hand side of.) to Ψ 4 we denote it by Ψ 4 ) can be split to the sum: α k Ψ 4 := 2 Ψ 4k,.5) where Ψ 4k = A d,2d x 2 k ψ x ) + 6 x k= ) ψ x ) v y, t ) x ] 2 4π B x 5 v y, t ) 2 ) x dy dx. The term Ψ 4 will be finally after we substitute for Ψ,..., Ψ 5 to.4)) compared with the left hand side of.4), i.e. with v 2 2; B. Hence we cannot just estimate Ψ 4k by a constant times v 2 2; B but we must evaluate it precisely. Assume at first that k =. The integral over A, d,2d with respect to x ) can be split by Fubini s theorem) to the iterated integral 2d d S ξ... ds ξ dξ, where S ξ is the sphere in R with the center at the point 0 and radius ξ. Furthermore, the surface integral over S ξ is equal to the iterated integral ξ ξ C ξ x )... dl dx, where C ξx ) is a circle on S ξ, corresponding to fixed x. Hence the radius h of C ξx ) is h = ξ2 x 2 ) /2.) Finally, the line integral over C ξ x ) can be expressed as the integral from 0 to 2π with respect to σ, using the parametric equations x 2 = h cos σ, x = h sin σ. Then the Cartesian coordinates of point x on C ξ x ) are x, h cos σ, h sin σ) and dl transforms to h dσ.) Thus, Ψ 4 = = 2d ξ 2π d B ξ 0 2d ξ d ξ x 2 6 ψξ) + ξ ψξ) ] v y, t ) x, h cos σ, h sin σ)]2 4π B ξ 5 v y, t ) 2 ) ξ dy h dσ dx dξ x 2 6 ψξ) + ξ ψξ) ] h 2π v y, t ) x, h cos σ, h sin σ)]2 4π ξ 5 0 v y, t ) 2 ξ ) dσ dx dξ dy. Since v v y, t ) is independent of σ, the inside integral can be explicitly calculated: 2π 0 v x, h cos σ, h sin σ)]2 ξ 5 v 2 ) ξ dσ = π 6x 2 ) ξ ξ 2 2 v 2 h 2 ) + ξ 2 2 v 2 2 h 2 ) ] + ξ 2 2 v 2 5

18 = π x 2 ) 2v 2 ξ ξ 2 v 2 2 v 2 ). We have used the formula h 2 = ξ 2 x 2.) Hence, calculating also the integral from ξ to ξ with respect to x : ξ we get where ξ x 2 h 4π Ψ 4 = Iψ) := π 64 π x 2 ) ξ ξ 2 B = Iψ) 2d d 2d d B We obtain similarly the formulas dx = ξ 4ξ x 2 ξ 2 x 2 x 2 ) ξ ξ 2 dx 6 ψξ) + ξ ψξ) ] ) πξ 2v 64 dξ 2 v 2 2 v 2 ) dy 2v 2 v 2 2 v 2 ) dy = Iψ) v 2 v 2) dy, 6 ψξ) + ξ ψξ) ] ξ dξ = π 64 Ψ 4k = 5π 64 B 2d d B d ξ ψξ) ) ψξ)] + 5 dξ v 2 v k 2 ) dy for k = 2,. = πξ 64, dξ = 5π 64. Substituting now for Ψ 4k k =, 2, ) to.5), we get Ψ 4 = 5π α + α 2 + α ) v 2 α v 2 + α2 v α v 2 ] ) dy..6) 28 B.7. The function Ψ 5. By analogy with Ψ 4, we write Ψ 5 = Ψ 5 + Ψ 52 + Ψ 5, where Ψ 5l x := β p l A d,2d ψ 4 ψ ) x 2 dx for l =, 2, ) and similarly as in the cases of Ψ 42 and Ψ 4, we can also prove that Ψ 52 L r/2 t, ) and Ψ 5 L µ t, ) for all µ, 2). The most difficult part of Ψ 5 is again Ψ 5, which contains p. If we express p by formula.), we get Ψ 5 = Ψ 5 + Ψ 52, where Ψ 5 Ψ 52 := β 4π := β 4π A d,2d A d,2d x x ψ ψ 4 ψ ) x 2 4 ψ ) x 2 B B vy, t ) x ] 2 x 5 v y, t ) 2 x ) dy dx, O d 4 vy, t ) 2) dy dx. Hence Ψ 5 = β 4π B A d,2d x ψ 4 ψ ) x 2 v x ] 2 x 5 v 2 ) x dx dy,.7) 6

19 Ψ 52 C d v 2 2; B,.8) where v vy, t ). Transforming the inside integral in.7) to the spherical coordinates R, ζ, ϑ, we calculate: where I = I 2 = = = π 2d A d,2d d 2π π/2 x ψ R ψr) 4 ψr) ) R 2 R dr, 0 π/2 2π π/2 0 π/2 π/2 π/2 4 ψ ) x 2 v x ] 2 x 5 v 2 ) x dx = I I 2, v cos ζ cos ϑ, sin ζ cos ϑ, sin ϑ)] 2 v 2] cos ϑ dϑ dζ v 2 cos 2 ζ cos ϑ + v 2 2 sin 2 ζ cos ϑ + v 2 sin 2 ϑ cos ϑ + 6v v 2 cos ζ sin ζ cos ϑ + 6v v cos ζ cos 2 ϑ sin ϑ + 6v 2v sin ζ cos 2 ϑ sin ϑ v 2 + v v 2 ) cos ϑ ] dϑ dζ v 2 cos ϑ + v 2 2 cos ϑ + 6v 2 sin 2 ϑ cos ϑ 2 v 2 + v v 2 ) cos ϑ ] dϑ. The last integral equals zero. Hence Ψ 5 = 0. This is in fact not surprising, because for each x on the sphere S R 0), the difference v x ] 2 / x 2 v 2 is equal to the second power of the component of v in the direction of x multiplied by 2) minus the second power of the component of v in the direction perpendicular to x, and when one integrates with respect to x over the sphere S R 0), it yields zero.).8. Condition i) the completion. If we now use formulas.4),.6) and the identity Ψ 5 = 0, we obtain v 2 2; B Ψ + Ψ 2 + Ψ + Ψ 4 + Ψ 42 + Ψ 42 + Ψ 4 + Ψ 52 + Ψ 52 + Ψ 5 = Ψ 42 + Ψ 52 + Ψ + 5π ] α + α 2 + α ) v 2 α k v k2 dy, 28 where Ψ = Ψ + Ψ 2 + Ψ + Ψ 42 + Ψ 4 + Ψ 52 + Ψ 5. Hence B B 5π ) 28 α + α 2 + α ) v 2 + 5π 28 k= ] α k v k2 dx Ψ 42 + Ψ 52 + Ψ. Substituting for α, α 2 and α from.5), we obtain exactly the same inequality, only with γ, γ 2, γ instead of α, α 2, α : B 5π ) 28 γ + γ 2 + γ ) v 2 + 5π 28 7 k= ] γ k v k2 dx Ψ 42 + Ψ 52 + Ψ. k=

20 Due to inequalities.9), there exists ɛ > 0 such that the left hand side is greater than or equal to ɛ v 2 2; B. Due to.4) and.8), one can choose d so large that Ψ 42 + Ψ 52 2 ɛ v 2 2; B. Then we have 2 ɛ v 2 2; B Ψ, which implies that ɛ 2 ϕv 2 2; B 2d Ψ + ɛ 2 ϕv 2 2; A,2d. Ψ is a sum of terms which are either in L r t, ) like Ψ ), or in L r/2 t, ) this concerns Ψ 2, Ψ and Ψ 42 ), or in L µ t, ) like Ψ 4 and Ψ 5 ). Moreover, by analogy with Ψ 2 or Ψ, one can show that ϕv 2 2; A L r/2 t, ), too. Consequently, there exists a sequence {t n} such,2d that t n and Ψt n) + 2 ɛ ϕv., t n) 2 2; A 0 for n ). This verifies condition i),2d from Section 2. Acknowledgments. The research was supported by the Grant Agency of the Czech Republic grant No S) and by the Czech Academy of Sciences RVO ). References ] L. Caffarelli, R. Kohn, L. Nirenberg: Partial regularity of suitable weak solutions of the Navier Stokes equations. Comm. on Pure and Appl. Math. 5, 982, ] C. Cao, E. Titi: Regularity criteria for the three dimensional NavierStokes equations. Indiana Univ. Math. J. 57, No. 6, 2008, ] J. Y. Chemin, P. Zhang, Z. Zhang: On the critical one component regularity for -D Navier Stokes system: general case. Arch. Rat. Mech. Anal. 224, 207, No., ] R. Farwig, H. Kozono and H. Sohr: Criteria of local in time regularity of the Navier Stokes equations beyond Serrin s condition. Banach Center Publ. 8, Parabolic and Navier Stokes equations, Part, Warsaw 2008, ] G. P. Galdi: An Introduction to the Mathematical Theory of the Navier Stokes Equations. 2nd edition, Springer Verlag, New York Berlin Heidelberg 20. 6] G. P. Galdi: An Introduction to the Navier Stokes initial boundary value problem. In Fundamental Directions in Mathematical Fluid Mechanics, ed. G. P. Galdi, J. Heywood, R. Rannacher, series Advances in Mathematical Fluid Mechanics. Birkhauser, Basel 2000, 98. 7] I. Kukavica, M. Ziane: One component regularity for the Navier Stokes equations. Nonlinearity 9, No. 2, 2006, ] O. A. Ladyzhenskaya: Mathematical Problems in the Dynamics of Viscous Incompressible Fluid. Gordon and Breach, New York 96. 9] O. A. Ladyzhenskaya, G. A. Seregin: On partial regularity of suitable weak solutions to the three dimensional Navier Stokes equations. J. Math. Fluid Mech., 999, ] F. Lin: A new proof of the Caffarelli Kohn Nirenberg theorem. Comm. on Pure and Appl. Math. 5, 998, ] J. Nečas, J. Neustupa: New conditions for local regularity of a suitable weak solution to the Navier Stokes equations. J. Math. Fluid Mech. 4, 2002, ] J. Neustupa, P. Penel: Regularity of a suitable weak solution of the Navier Stokes equations as a consequence of regularity of one velocity component. In Applied Nonlinear Analysis, 8

21 Powered by TCPDF ed. A. Sequeira, H. Beirao da Veiga, J. Videman, Kluwer Academic/Plenum Publishers, New York 999, ] J. Neustupa, P. Penel: Anisotropic and geometric criteria for interior regularity of weak solutions to the D Navier Stokes equations. In Mathematical Fluid Mechanics, Recent Results and Open Questions, ed. J. Neustupa and P. Penel, Birkhauser, Basel 200, ] J. Neustupa, A. Novotný, P. Penel: An interior regularity of a weak solution to the Navier Stokes equations in dependence on one component of velocity. Quaderni di Matematica Vol. 0: Topics in Mathematical Fluid Mechanics, ed. G. P. Galdi and R. Rannacher, Napoli 200, ] J. Neustupa: A note on local interior regularity of a suitable weak solution to the Navier- Stokes problem. Discr. Cont. Dyn. Systems, Ser. S. 6, No. 5, 20, ] J. Neustupa: A removable singularity in a suitable weak solution to the Navier Stokes equations. Nonlinearity 25, 202, ] J. Neustupa: A refinement of the local Serrin type regularity criterion for a suitable weak solution to the Navier Stokes equations. Arch. Rat. Mech. Anal. 24, 204, ] Z. Skalák, P. Kučera: Regularity of pressure in the neighbourhood of regular points of weak solution of the Navier Stokes equations. Appl. of Math. 48, No. 6, 200, ] H. Sohr: The Navier Stokes Equations. An Elementary Functional Analytic Approach. Birkhäuser Advanced Texts, Basel Boston Berlin ] S. Takahashi: On interior regularity criteria for weak solutions of the Navier Stokes equations. Manuscripta Math. 69, No., 990, ] R. Temam: Navier Stokes Equations. North Holland, Amsterdam New York Oxford ] J. Wolf: A new criterion for partial regularity of suitable weak solutions to the Navier Stokes equations. Advances in Mathematical Fluid Mechanics, ed. R. Rannacher, A. Sequeira, Springer, Berlin, 200, ] Y. Zhou, M. Pokorný: On the regularity of the solutions of the Navier-Stokes equations via one velocity component. Nonlinearity 2, No. 5, 200, Author s address: Czech Academy of Sciences, Institute of Mathematics Žitná 25, 5 67 Praha Czech Republic neustupa@math.cas.cz 9