GLOBAL SOLUTIONS OF NAVIER-STOKES EQUATIONS WITH LARGE L 2 NORMS IN A NEW FUNCTION SPACE

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1 Advanes in Differential Equations Volume 9, Numbers 5-6, May/June 24, Pages GLOBAL SOLUTIONS OF NAVIER-STOKES EQUATIONS WITH LARGE L 2 NORMS IN A NEW FUNCTION SPACE Qi S. Zhang Department of Mathematis, University of California, Riverside, CA 9252 (Submitted by: J. Goldstein) Abstrat. First we prove ertain pointwise bounds for the fundamental solutions of the perturbed linearized Navier-Stokes equation (Theorem.). Next, utilizing a new framework with very little L p theory or Fourier analysis, we prove existene of global lassial solutions for the full Navier-Stokes equation when the initial value has a small norm in a new funtion lass of Kato type (Theorem.2). The smallness in this funtion lass does not require smallness in L 2 norm. Furthermore we prove that a Leray-Hopf solution is regular if it lies in this lass, whih allows muh more singular funtions then before (Corollary ). For instane this inludes the well-known result in [25]. A further regularity ondition (form boundedness) was given in Setion 5. We also give a different proof about the L 2 deay of Leray-Hopf solutions and prove pointwise deay of solutions for the three-dimensional Navier- Stokes equations (Corollary 2, Theorem.2). Whether suh a method exists was asked in a survey paper [2].. Introdution There are two goals for the paper. The first is to establish ertain pointwise bounds for the fundamental solutions of the perturbed linearized Navier- Stokes equation u(x, t) b(x, t) u(x, t) P (x, t) t u(x, t) =, (x, t) (, ), (.) div u =,n 3, u(x, ) = u (x). Here is the standard Laplaian, u(x, t),u (x),b(x, t), P (x, t) R and b u = n i= b i xi u. This linear system is also known as the Oseen flow. The bounds we will prove were previously known only for the ase of Stokes flow, i.e., when b in (.). See [7], e.g. Aepted for publiation: January 24. AMS Subjet Classifiations: 35K4, 76D5. 587

2 588 Qi S. Zhang The seond goal is to establish more general onditions whih imply regularity of weak solutions to Navier-Stokes equations (.2) and to prove existene of global lassial solutions, when the initial value has a small norm in a ertain funtion lass of Kato type. Sine smallness in this lass does not require smallness in L 2 norm, we have thus proven the existene of global lassial solutions for some initial values with arbitrarily large L 2 norms. Reall that the Cauhy problem for Navier-Stokes equations is u(x, t) u u(x, t) P (x, t) t u(x, t) =, (x, t) R n (, ), (.2) div u =,n 3, u(x, ) = u (x). Here and always u is a vetor-valued funtion, whih means a funtion whose range is a subset of. Let us reall some of the reent advanes in the problem of finding global (strong) solutions for (.2). Due to the large number of pertinent papers, we may miss some of them. Kiselev and Ladyzhenskaya [2] proved that (.2) for n = 3 has a global solution provided that u W 2,2 (R 3 ) is suffiiently small. See also the work of Kato and Pone [4]. Fabes, Jones, and Riviére [7] showed that (.2) has a global solution when u L n+ɛ ( ) + u L n ɛ ( ) is suffiiently small. Here ɛ is a small positive onstant. Kato [] proved that (.2) has a global strong solution when u is small in the L n ( ) sense. Later Giga and Miyakawa [9] and M. Taylor [28] proved the same result for small u in a ertain Morrey spae. A similar result was obtained by Cannone [3] and Planhon [2] for initial data in ertain Besov spaes. Related results an also be found in papers by Iftimie [] and Lions and Masmoudi [8]. Most reently Koh and Tataru [5] proved global existene when u is a small funtion in the so-alled BMO lass ([5, page 24]). A funtion is in this lass if its onvolution with the heat kernel is in a type of Morrey spae. The result in [5] reovers all the above-mentioned results on global existene of strong solutions. In the proofs, many authors have applied quite involved tools in harmoni analysis. In this paper we find that the Navier-Stokes equation has ertain global solutions in another natural funtion lass of Kato type. In Remark.2 below, we will see that this lass is different from the X lass in [5] and hene different from all the spaes preeding [5]. This lass has some interesting qualities, whih makes it rather promising. The first is that these funtions an be quite singular. For example, they do not have to be in any L p lo lass for any p>. The singular set ould be dimension n, et. (see Remark.2 below). Yet Leray-Hopf solutions in this lass must be

3 global solutions of Navier-Stokes equations 589 smooth. The seond is that this lass gives rise to pointwise bounds for the global solutions in a natural manner. The third is that within this lass, the proof of global existene is distint and self-ontained, using very little L p theory or Fourier analysis, whih have been vital omponents in all previous arguments. Diretness of the proof indiates that it may be useful in future studies. As another appliation we give a different proof about the L 2 deay of Leray-Hopf solutions of three-dimensional Navier-Stokes equations. (See Corollary 2 below.) Roughly speaking, a funtion is in a Kato, or Shehter, or Stummel lass if its onvolution with ertain kernel funtions satisfies suitable boundedness or smallness assumptions. The kernel funtions are usually hosen as the fundamental solutions of some ellipti or paraboli equations. It is well known that Kato-lass funtions are natural hoies of funtion spaes in the regularity theory of solutions of ellipti and paraboli equations. This fat has been well doumented in the paper [26]. In this paper we show that a suitable Kato lass is also a natural funtion spae in the study of Navier-Stokes equation. Now let us introdue the main funtion lass for Theorem., whih will be alled K. Then we will explain its many nie properties. For instane it is known that a Leray-Hopf solution (see Remark. for a definition) of the Navier-Stokes equations in L p,q ( n p + 2 q < ) spae is regular ([25]). We will show that the funtion lass defined below properly ontains this L p,q spae. Moreover Leray-Hopf solutions in this lass are also regular (Corollary ). Definition.. A vetor-valued funtion b = b(x, t) L lo (Rn+ ) is in lass K if it satisfies the following ondition: t lim sup [K (x, t; y, s)+k (x, s; y, l)] b(y, s) dy ds =, (.3) t l x,l> where l K (x, t; y, s) = [ x y +, t s, x y. (.4) t s] n+ For onveniene, we introdue the notation t B(b, l, t) sup x l [K (x, t; y, s)+k (x, s; y, l)] b(y, s) dy ds, (.5) t B(b, l, ) sup x,t>l l [K (x, t; y, s)+k (x, s; y, l)] b(y, s) dy ds. (.6)

4 59 Qi S. Zhang These quantities will serve as replaements of the L p and other norms used by other authors. The reader may wonder whether the appearane of two K s(k and its onjugate) is neessary in the definition. At this time we do not know the answer. However, a similar lass was defined by the author for the heat equation [3]. There K is replaed by the gradient of the heat kernel. As pointed out in [8], using one kernel will result in a different lass. Remark.. In ase b is independent of time, an easy omputation shows B(b,, ) 2 sup K (x, s; y, ) b(y) dy ds x = n sup x b(y) dy. x y n This last quantity was introdued in [22] and more expliitly in [4]. See also [3] for the time-dependent ase. As in [3], it is also easy to see that for a time-independent funtion b, b K if lim sup r x and B(b,, ) <. x y r b(y) dy = x y n The main results of the paper are the next two theorems and orollaries. Theorem.. Suppose b is in lass K. Then (.) has a fundamental solution (matrix) E = E(x, t; y, s) in the following sense. (i) Let u(x, t) = E(x, t; y, )u (y)dy, where u C (Rn ) is vetor valued and divergene free. For any vetorvalued φ C (Rn (, )) with div φ =, there holds u, t φ + φ dx dt b u, φ dx dt R n = u (x),φ(x, ) dx. Furthermore, n xi E ij (x, t; y, s) = for all j =,...,n. (ii) i= lim t E(x, t; y, )φ(y)dy = φ(x).

5 global solutions of Navier-Stokes equations 59 Here φ is a smooth, vetor-valued funtion in with div φ =. (iii) There exists δ> depending only on b and n suh that C δ E(x, t; y, s) ( x y +, when <t s δ. t s ) n Suppose in addition that t lim sup [K (x, t; y, s)+k (x, s; y, T)] b(y, s) dy ds µ, (.7) T x,t>t T where µ is a small, positive onstant depending only on n. Then there exists T > depending only on b and n suh that { C δ ( x y +, when <t s δ or T t s ) E(x, t; y, s) n s t, C δ,ɛ ( x y +, otherwise. t s ) n ɛ Here ɛ> is any suffiiently small number and C δ,ɛ depends only on b, ɛ, and n. Next we turn to the Navier-Stokes equations (.2). Definition.2. Following standard pratie, we say that u is a (weak) solution of (.2) if the following holds: For any vetor-valued φ C (Rn (, )) with div φ =, u satisfies div u =and u, t φ + φ dx dt u u, φ dx dt (.8) R n = u (x),φ(x, ) dx. We will use this definition for solutions of (.2) throughout the paper, unless stated otherwise. Note that we need only that the above integrals make sense and that there are no a priori assumptions about whih spaes u and u lie in. Remark.. Solutions thus defined are more general than Leray-Hopf solutions, whih in addition require also u(,t) L 2 ( ) < for all t> and u L 2 ( (, )) <. Theorem.2. There exists a positive number η depending only on the dimension n suh that, if sup (y) u x R dy < η and div u n x y n =, then the Navier-Stokes equations have a global solution u satisfying B(u,, ) < η. Moreover, there exists C> suh that u (y) u(x, t) C ( x y + t) dy. n

6 592 Qi S. Zhang If in addition u L 2 ( ), then u is a lassial solution when t>. Remark.2. Let b = b(x,x 2,x 3 ) be a funtion from R 3 to R 3. If b is ompatly supported and b with δ>2, it is easy to hek x ln x δ that b satisfies (.3) and hene is in lass K. The proof is given at the end of Setion 2. The lass of funtions u satisfying B(u,, ) < is different from the funtion spae X in [5], where the solutions of the Navier-Stokes equations in that paper reside. It is easy to find a funtion in X but not in that lass. On the other hand, the above funtion b is in lass K and B(b,, ) < but it is not in X. Funtions in X must be loally square integrable (see p. 24 in [5]). However b is not in L p lo for any p>. We should mention that the above funtion b is in the BMO lass, the spae of initial values in [5]. The next two orollaries are diret onsequenes of a small part of Theorem. (part (iii)). Their proofs, depending on Lemma 3. below, are independent of the rest of Theorem. and Theorem.2. Corollary. Let u K be a Leray-Hopf solution of the Navier-Stokes equations. Then u is lassial when t >. By the example in Remark.2, we see that the funtion lass K permits solutions whih apparently are muh more singular then previously known. In ase the spatial dimension is 3, a solution an have an apparent singularity of ertain type that is not L p lo for any p> and of dimension. One an also onstrut time-dependent funtions in K with quite singular behavior. Nonetheless the solution is regular. Notie also that there is no smallness assumption on the solution u as long as it is in K. By Proposition 2. below, Corollary ontains the well-known regularity result of [25]. The result here also differs from the borderline ase in [27] sine K and the spae L p,q with n/p +2/q = are different. By a diret omputation, one an also show that it ontains the Morrey-type lass in [2]. In Setion 5, we will propose a form-bounded ondition on u, ontaining this borderline ase when n = 3, whih will imply the regularity of u. As explained there, this ondition seems to be one of the widest possible to date. Corollary 2. Let u be any Leray-Hopf solution of the three-dimensional Navier-Stokes equations. Suppose in addition that u L ( ); then, for n =3and C = C(u,n), [ ] /2 u(x, t) 2 C [ ] dx t n/4 u (x) dx + u (x) 2 dx.

7 global solutions of Navier-Stokes equations 593 Remark.3. In the interesting papers [23, 24] and [3], the result in Corollary 2 (and more) was proved for some Leray-Hopf solutions or for all admissible solutions in the sense of [] (see also [7]). This extra restrition was removed in [3]. Fourier analysis is the main tool in the proof of the deay. The orollary provides an alternative proof without using the Fourier transform. However, in the papers [24], [3], and [3] a similar deay property has been established for some or all Leray-Hopf solutions under the weaker assumption that u L r ( ) when <r<2. We are not able to do it for all solutions for r 5/4. However the above method provides even pointwise deay of solutions (Theorem.2). Whether suh a method exists was asked in a survey paper [2]. Remark.4. Sine u an be quite singular, at the first glane the solution u in Theorem.2 may be too singular to satisfy (.8). However we will show that all terms in the definition are justified. The uniqueness question of the solution is also interesting. We will not address the problem in this paper. Some reent development an be found in [5]. Let us outline the proof of Theorem.2. The strategy is to use a fixedpoint argument. The novelty is a number of new inequalities involving the kernel funtion K and its relatives. We will use these inequalities to show that the Navier-Stokes equations are globally well posed in lass under the norm B(u,, ), provided the initial value u satisfies that B(u,, ) is smaller than a dimensional onstant. We will use C,,,... to denote positive onstants whih may hange from line to line. The rest of the paper is organized as follows. In Setion 2 we present some elementary properties of the funtion lass K. Theorems. and.2 will be proven in Setions 3 and 4 respetively. The proofs are independent, exept for the use of Lemma 3.. The orollaries will be proven at the end of Setion Preliminaries Proposition 2.. Suppose b L p,q ( R) with n p + 2 q <. Then b K ; i.e., lim B(b, t h, t) = h uniformly for all t. In partiular, if b L n+ɛ ( ) for ɛ>, then the above holds. Proof. For ompleteness, we reall that [ ( b L p,q = b(y, s) p dy ) ] q/p /q. ds R

8 594 Qi S. Zhang Using Hölder s inequality twie on (.5) (taking l = t h), we obtain [ t ( B(b, t h, t) 2 b(y, s) p dy ) ] q/p /q ds t h [ t ( ] sup x t h ( x y + t s) p (n+) dy) /q q /p ds. Here p = p/(p ) and q = q/(q ). Sine dy ( x y + t s) p (n+) = we see that ( t B(b, t h, t) b L p,q By the assumption that n p + 2 q = (t s)n/2 (t s) p (n+)/2 (t s) (p (n+) n)/2, t h dy ( + y ) p (n+) ds ) /q. (t s) (p (n+) n)q /(2p ) <, one has ( p )n +( )2 <. q Simple omputation then leads to µ (p (n +) n)q /(2p ) <. This implies B(b, t h, t) b L p,qh ( µ)/q. Remark. One an also define the lass K for funtions in various domains, as suggested by the referee. Proposition 2.2. Suppose, for ɛ>, b L n+ɛ, + b L n ɛ, <. Then B(b,, ) C( b L n+ɛ, + b L n ɛ, ). Proof. This is similar to that of Proposition 2., so we will be very brief. We write t K (x, t; y, s) b(y, s) dy ds R n t t = K (x, t; y, s) b(y, s) dy ds + K (x, t; y, s) b(y, s) dy ds. t As in the proof of Proposition 2., using the assumption that b L n+ɛ, < and applying Hölder s inequality, we see that the first integral on the righthand side is finite. Using the assumption that b L n ɛ, < and applying Hölder s inequality, we see that the seond integral on the right-hand side is also finite. This finishes the proof.

9 global solutions of Navier-Stokes equations 595 Here we give a proof that the funtion in Remark.2 is in lass K. Sine b is independent of time and b L lo (R3 ), by a simple loalization of the integral in (.3), it is enough to prove that lim r sup x B(x,r) b(y) dy =. (2.) x y 2 Here dy = dy dy 2 dy 3. Sine b = b(y) depends only on y, by diret omputation, one sees that b(y) x y 2 dy B(x,r) x +r x r x +r x r b(y ) x2 +r x3 +r x 2 r x 3 r ln x y y ln y δ dy. (x y ) 2 +(x 2 y 2 ) 2 +(x 3 y 3 ) 2 dy 2dy 3 dy By the assumption that δ>2, it is easy to show that (2.) holds. 3. Proof of Theorem.. Bounds for fundamental solutions The proof of Theorem. is divided into several parts. We begin with some lemmas. At the end of the setion, the orollaries will be proven. The proof of the orollaries depends only on Lemma 3.. So it an be read separately from the proof of Theorem.. Lemma 3.. The following inequalities hold for all x, y, z and t> τ>. t K bk ( x z + b(z,τ) t τ) n ( z y + dz dτ τ) n+ B(b,, t) C ( x y + t), (3.) n t K bk ( x z + b(z,τ) t τ) n+ ( z y + dz dτ τ) n+ B(b,, t) C ( x y +. (3.2) t) n+ Here and later K (x, t; y, s) ( x y +. t s) n Proof. Sine x z + t τ + y z + τ x y + t,

10 596 Qi S. Zhang we have either or x z + t τ 2 ( x y + t), (3.3) z y + τ 2 ( x y + t). (3.4) Suppose (3.3) holds; then 2 n K bk ( x y + t) n That is, t b(z,τ) ( z y + dz dτ. τ) n+ K bk 2n B(b,,t) ( x y + t). (3.5) n Suppose (3.4) holds but (3.3) fails; then z y + τ 2 ( x y + t) x z + t τ. Therefore, ( x z + t τ) n ( z y + τ) n+ ( x z + t τ) n+ ( z y + τ) n. This shows ( x z + t τ) n ( z y + τ) n+ 2 n ( x z + t τ) n+ ( x y + t). n Substituting this into (3.), we obtain 2 n t K bk ( x y + b(z,τ) t) n ( x z + dz dτ. t τ) n+ That is, K bk 2n B(b,,t) ( x y + t). n Clearly, the only remaining ase to onsider is when both (3.3) and (3.4) hold. However, this ase is already overed by (3.5). Thus (3.) is proven. Similarly (3.2) is proven.

11 global solutions of Navier-Stokes equations 597 Remark 3.. By the same argument, one an prove that, for f = f(x, t), t K fk (x, t; y, s) K (x, t; z,τ)f(z,τ)k (z,τ; y, s)dz dτ s R n t sup [K (x, t; z,τ)+k (x, τ; z,s)] f(z,τ) dz dτ K (x, t; y, s) x,t s for all x and t>. Lemma 3.2. Suppose lim h sup t B(b, t h, t) =. Define formally a funtion E(x, t; y, s) =E (x, t; y, s) t n E (x, t; z,τ) b i (z,τ) zi E(z,τ; y, s)dz dτ, (3.6) s i= where E is the fundamental solution of the Stokes flow, i.e., (.) with b. Then there exists δ> and C = C(δ) suh that E(x, t; y, s) C ( x y + t s) n, C x E(x, t; y, s) + y E(x, t; y, s) ( x y + t s) n+ when <t s δ. Furthermore, the above two inequalities hold for all t>s> provided B(b,, ) is smaller than a suitable positive onstant depending only on n. Proof. We an write (3.6) suintly as E(x, t; y, s) =E (x, t; y, s) E (b E)(x, t; y, s). Expanding this formally, we obtain E(x, t; y, s) =E (x, t; y, s)+ ( ) j E (b E ) j (x, t; y, s). (3.7) j= Here means onvolution and b E n i= b i(z,τ) zi E (z,τ; y, s). It is well known that (see [7] e.g.), E (x, t; y, s) ( x y + t s) n, x E (x, t; y, s) C ( x y + t s) n+.

12 598 Qi S. Zhang Using these bounds on E and E together with Lemma 3., we dedue B(b, s, t) E b E (x, t; y, s) ( x y + t s) n. Here (> ) depends only on n. By indution, it is easy to see that It follows from (3.7) that E (b E ) j (x, t; y, s) E(x, t; y, s) ( x y + t s) n [ B(b, s, t)] j ( x y + t s) n. [ B(b, s, t)] j. Using the ondition on b, we an hoose δ suffiiently small so that B(b, s, t) <. Then (3.7) is uniformly onvergent. Moreover, there exists C > suh that C E(x, t; y, s) ( x y + t s) n when <t s<δ. This proves the first inequality in the lemma. The inequality on x E an be derived similarly. The only hange is to use the inequality B(b, s, t) E b E (x, t; y, s) ( x y + t s) n+, whih is part of Lemma 3.. In order to prove the estimate on y E, we use the assumption that div b =. This implies, after integration by parts and a standard limiting argument, E(x, t; y, s) =E (x, t; y, s) t n + zi E (x, t; z,τ)b i (z,τ)e(z,τ; y, s)ndzdτ. s i= We remark that the integration by parts is rigorous by the just-proven bounds on x E. This shows y E(x, t; y, s) = y E (x, t; y, s) t n + zi E (x, t; z,τ)b i (z,τ) y E(z,τ; y, s)dz dτ. s i= From here the bound on y E an be obtained just like that for x E. j=

13 global solutions of Navier-Stokes equations 599 If B(b,, ) is suffiiently small, then all the above arguments hold for all t>s>. This proves the last statement in the lemma. Lemma 3.3. Suppose t lim sup B(b, T, t) = lim sup K (x, t; y, s) b(y, s) dy ds T t T x,t>t T is suffiiently small. Then there exists T > and C = C(T ) suh that E(x, t; y, s) x E(x, t; y, s) + y E(x, t; y, s) when t>s>t. Here E is defined in (3.6). C ( x y + t s) n, C ( x y + t s) n+ Proof. The proof is almost idential to that of Lemma 3.2. The only differene is that all integration takes plae in the region (T, ). Now we are ready to give a Proof of Theorem.. Let us assume that b is a smooth funtion with ompat support. This does not redue any generality sine all onstants below depend on b only in terms of the quantity in Definition.. The smoothness or the size of the support of b is irrelevant. In the sequel we an apply the limiting argument at the end of Setion 4 when b is not smooth. We mention that the limiting argument, designed for the full Navier-Stokes equations, is more than enough to over the linear ase. We are going to prove that the funtion E = E(x, t; y, s) defined in (3.6) and extended by using a reproduing formula is the right hoie for the fundamental solution of (.). The order of the proof is (iii), then (ii), then (i). Proof of (iii). By Lemmas 3.2 and 3.3, all we need to onsider is the ase t s δ and s T. Here δ and T are the ontrol onstants in the above lemmas. For larity we divide the ase into two separate parts. Part. We assume that s<t 2T and t s δ, s T. Sine δ is fixed, we an use the semigroup property of E to write E(x, t; y, s) = E(x, t; z,t δ)e(z,t δ; z 2,t 2δ) E(z k,t kδ; y, s)dz dz 2 dz k. (3.8) Here all integration takes plae in and k is an integer suh that <t kδ s δ. We are going to show that the integrals in the above reproduing

14 6 Qi S. Zhang formula are atually absolutely onvergent. Here is how. Considering the integral J ( x z + δ) n ( z y + δ) dz, n learly, J dz + dz J + J 2. x z x y /2 z y x y /2 When x z x y /2, we have z y x y + x z 3 x z. Hene x z z y /3. This shows J ( x y + dz δ) n ɛ ( z y + δ). n+ɛ Here ɛ(> ) is an arbitrary small onstant. This shows, by diret omputation, δ ɛ/2 J ( x y + δ). n ɛ Similarly, δ ɛ/2 J 2 ( x y + δ). n ɛ These imply that δ ɛ/2 J ( x y +. (3.9) δ) n ɛ Applying (3.9) to (3.8) k times, we obtain E(x, t; y, s) k δ (k )ɛ/2 ( x z k + dz k δ) n (k )ɛ ( z k y + t kδ s). n Without loss of generality we an redue δ suitably so that δ/2 t kδ s δ. So applying the same tehnique as in the proof of (3.9), we know that E(x, t; y, s) k δ kɛ/2 ( x y + δ). n kɛ Sine k is finite, we an hoose ɛ suffiiently small so that n ɛk is as lose to n as possible. Note also that t s 2T. So, in ase s<t 2T

15 and s T, one has global solutions of Navier-Stokes equations 6 E(x, t; y, s) k δ kɛ/2 ( x y +. (3.) t s) n ɛ Here we have renamed kɛ as ɛ. This finishes Part. Part 2. We assume that t 2T, s T, and t s δ. By the semigroup property again E(x, t; y, s) = E(x, t; z,.5t ) E(z,.5T ; y, t s)dz. (3.) By Lemma 3.3 and Part just above, we have E(x, t; z,.5t ) ( x z + t.5t ) n, E(z,.5T ; y, s) ( z y +.5T s) n ɛ. Substituting the above into (3.), we see that E(x, t; y, s) = As before we split (3.2) as follows E(x, t; y, s) dz+ ( x z + t.5t ) n ( z y + dz..5t s) n ɛ (3.2) x z x y /2 z y x y /2 dz E +E 2. (3.3) When x z x y /2, we have x z z y /3. Note also t.5t (.5T s)/3. Hene, E ( x y + dz t.5t ) n 2ɛ ( z y +.5T s) n+ɛ, whih implies E ( x y + t.5t ) n 2ɛ. (3.4) (T s) ɛ/2 Finally, we estimate E 2. When z y x y /2, we have z y x z /3. Sine t 2T and s T, it is lear that ( x z + t.5t )( z y +.5T s) 8 ( x z + t)( z y + T ). By elementary omputation, using the urrent assumptions on x, y, z, t, and T, ( x z + t)( z y + T ) ( x z + T )( z y + t)

16 62 Qi S. Zhang for some (, ). Combining the last two inequalities we know that E 2 (x, t; y, s) z y x y /2 ( x z + t.5t ) n ( z y + dz.5t s) n ɛ z y x y /2 ( x z + t) n ( z y + dz T ) n ɛ = z y x y /2 ( x z + dz t) ɛ [( x z + t)( z y + T )] n ɛ z y x y /2 ( x z + dz t) ɛ [( x z + T )( z y + t)] n ɛ ( x y + t) n 2ɛ z y x y /2 ( x z + dz t) ɛ ( x z + T ) n. Therefore, E 2 (x, t; y, s) ( x y + dz t) n 2ɛ ( x z + T ) n+ɛ, whih shows T ɛ/2 E 2 (x, t; y, s) ( x y +. (3.5) t s) n 2ɛ Combining (3.4) and (3.5), we have C E(x, t; y, s) ( x y + t s) n 2ɛ (3.6) when t 2T, s T, and t s δ. This proves (iii). Proof of (ii). Sine, for small t, E(x, t; y, ) = t n E (x, t; y, ) E (x, t; z,τ) b i (z,τ) zi E(z,τ; y, )dz dτ. i= From part (iii) of the theorem and Lemma 3.2, there exists > suh that t E(x, t; y, ) E (x, t; y, ) K (x, t; z,τ) b(z,τ) K (z,τ; y, )dz dτ when t is suffiiently small. By Lemma 3., E(x, t; y, ) E (x, t; y, ) B(b,,t)K (x, t; y, ).

17 global solutions of Navier-Stokes equations 63 Let φ = φ(x) be a divergene-free, smooth, vetor-valued funtion. Using the above inequality, we know E(x, t; y, )φ(y)dy E (x, t; y, )φ(y)dy R n B(b,,t) K (x, t; y, ) φ(y) dy. Sine φ is divergene-free, the above shows E(x, t; y, )φ(y)dy G(x, t; y, )φ(y)dy R n B(b,,t) K (x, t; y, ) φ(y) dy. Here G is the fundamental solution of the heat equation. Using our main assumption that B(b,,t) when t and the property of G, we obtain lim E(x, t; y, )φ(y)dy = φ(x). t This proves part (ii). Proof of (i). Let u = u (x) be an initial value suh that div u = in the weak sense. We need to prove that u(x, t) E(x, t; y, )u (y)dy (3.7) is a solution to (.). Using a partition of unity, we an assume that the support of the test funtion is suffiiently narrow in the time diretion. Hene, by the semigroup property, it is enough to prove (i) when t is suffiiently small. From formula (3.6) for E its lear that u(x, t) = E (x, t; y, )u (y)dy R n t n E (x, t; z,τ) b l (z,τ) zl u(z,τ)dz dτ. The proof, following an argument in [7], goes as follows. For simpliity we write n f = f(z,τ) =b(z,τ) z u(z,τ) b l (z,τ) zl u(z,τ). l= l=

18 64 Qi S. Zhang By Lemma 3.2 f(z,τ) b(z,τ) ( z w + τ) n+ u (w) dw. Using Young s inequality and the extra assumption that b is smooth and ompatly supported, f(,t)isinl p ( ) for some p> and almost all t. Let R =(R i R j ) n n, where R i is the Riesz transform. It is known that (see [7] e.g.) t n F (x, t) E (x, t; z,τ) b l (z,τ) zl u(z,τ)dz dτ (3.8) l= t = E (x, t; z,τ)f(z,τ)dz dτ R n t t = G(x, t; z,τ)f(z,τ)dz dτ G(x, t; z,τ)(rf)(z,τ)dz dτ, where G is the fundamental solution of the heat equation. We mention that the last term in (3.8) is valid sine, by the extra assumption that b is smooth and ompatly supported, f(,t)isinl p ( ) for some p> and almost all t. However, as shown below, this term will be integrated out. Therefore the argument will be independent of the extra assumption on b eventually. Hene, ( t )F (x, t) = f(x, t)+(rf)(x, t). (3.9) Let φ be any suitable vetor-valued test funtion; i.e., φ C (Rn R) and div φ =. By (3.9) and elementary properties of the heat equation, we have, for any T>, T F, t φ + φ (x, t)dx dy R n T T (3.2) = f,φ (x, t)dx dt + Rf, φ (x, t)dx dt. Here, means the inner produt in. Sine div φ =, quoting [7], we know that T T Rf, φ (x, t)dx dt = f,rφ (x, t)dx dt =. From (3.2), the above shows, sine f = b u, T T F, t φ + φ (x, t)dx dy = b u, φ (x, t)dx dt. (3.2)

19 global solutions of Navier-Stokes equations 65 Aording to (3.7), (3.6), and (3.8), u(x, t) = E (x, t; y, )u (y)dy F (x, t) = G(x, t; y, )u (y)dy F (x, t). Using this, we an ompute as follows: T u, t φ + φ (x, t)dx dy R n T = Gu, t φ + φ (x, t)dx dy R n T F, t φ + φ (x, t)dx dt R n T = u (x),φ(x, ) dx + b u, φ (x, t)dx dt. Here we just used (3.2)and an obvious property for Gu. Therefore, T T u, t φ + φ (x, t)dx dy b u, φ (x, t)dx dt R n = u (x),φ(x, ) dx. In addition, sine n i= x i (E ) ij (x, t; y, s) =, we know from (3.6) that n xi E ij (x, t; y, s) =. i= This shows that div u =. Hene u is a solution of (.). This proves part (i). The proof of Theorem. is omplete. Next we prove the two orollaries. The proof is in fat independent of that of Theorem.. It relies only on Lemma 3. in the setion. Proof of Corollary. It is well known that a Leray-Hopf solution is lassial at least in a finite time interval. Let T be the time suh that u(,t) is finite when t (,T ) but lim t T u(,t) =. Let E be the fundamental solution of the Stokes flow. Given t (,T ), by [7] (Theorem 2.), we have, for t>t, u(x, t) = E (x, t; y, t )u(y, t )dy R n t + b(y, s) y E (x, t; y, s)u(y, s) dy ds, (3.22) t

20 66 Qi S. Zhang where b(y, s) = u(y, s). Here we remark that the lass of solutions in Theorem 2. of [7] ontains Leray-Hopf solutions. Hene (3.22) is valid for all Leray-Hopf solutions. The seond integral in the last equality is absolutely onvergent when t (t,t ɛ) for ɛ>. This is so beause u L ( (t,t ɛ)) and b = u K. Iterating (3.22), we obtain, as in the proof of Lemma 3.2, u(x, t) C (K b ) k K (x, t; y, t ) u (y) dy. k= Here t (K b ) K (x, t; y, ) b(z,τ) K (x, t; z,τ)k (z,τ; y, t )dz dτ. t By Lemma 3., (K b ) K (x, t; y, t ) B(b, t,t)k (x, t; y, t ). Using indution, there holds u(y, t ) u(x, t) C ( x y + t t ) n dy [ B(b, t,t)] j. (3.23) j= Here depends only on n. The above series is onvergent if B(b, t,t) <. (3.24) Sine b = u K, there exists a δ > depending only on the rate of onvergene of (.3) suh that (3.24) holds when <t t <δ. In this ase we have u(y, t ) u(x, t) C (δ) ( x y + dy. (3.25) t t ) n Next, we hoose, for the above δ, t = T (δ/2). Then for t (t,t), (3.25) implies [ u(x, t) C (δ) Hene Letting t T, we see that dy ] /2[ ] /2. ( x y + t t ) 2n u(y, t ) 2 dy u(x, t) u(,t ) L C (δ) (t t ) n/4 u L 2. C (δ) (T t ) n/4 u L 2 <.

21 global solutions of Navier-Stokes equations 67 This ontradition shows that u is a bounded and hene a lassial solution when t>. Proof of Corollary 2. It suffies to prove the orollary when t is suffiiently large. Let u be a Leray-Hopf solution to (.2) with n = 3. It is well known that u beomes a bounded, lassial solution when t is suffiiently large. Moreover, the L 2 norm of u(,t) tends to zero as t ([]). Hene for large t, the norm u(,t) L n+ɛ ( ) + u(,t) L n ɛ ( ) is suffiiently small by interpolation between L 2 and L norms. By Propositions 2. and 2.2, this implies that B(u, t, ) is small when t is large. Hene, by the same arguments from (3.22) to (3.23), we have u(y, t ) u(x, t) C ( x y + t t ) n dy [ B(b, t,t)] j (3.26) for all t>t. Sine B(b, t,t) B(b, t, ) and the latter is suffiiently small, (3.26) shows u(y, t ) u(x, t) C ( x y + dy (3.27) t t ) n for all t>t. Applying Young s inequality on (3.27), we obtain [ dy ] /2 u(,t) L 2 sup x ( x y + t t ) 2n u(y, t ) dy. Hene, for t>t, u(,t) L 2 (t t ) n/4 u(y, t ) dy. (3.28) From (3.22), sine div u =, one has, from [7], t u(x, t) = G(x, t; y, )u (y)dy+ u(y, s) y E (x, t; y, s)u(y, s)dy ds, where G is the heat kernel. Therefore, u(x, t) dx u (y) dy R n t + u(y, s) 2 dy ( x y + dx dy ds. t s) (n+) This implies t u(x, t ) dx u (y) dy + u (y) 2 dy t s ds. j=

22 68 Qi S. Zhang Substituting this into (3.28), we obtain [ u(,t) L 2 (t t ) n/4 u (y) dy + t u (y) 2 dy ]. 4. Proof of Theorem.2. Existene of global solutions Throughout the setion and for a funtion u = u(x, t), we will use the global norm t u K sup [K (x, t; y, s)+k (x, s; y, )] u(y, s) dy ds. (4.) x,t> If it is finite. Here, as before, K (x, t; y, s) = ( x y + t s) n+. We will also use the related kernel funtion K (x, t; y, s) = ( x y + t s) n. If u is independent of time, then by a simple omputation, we see that u(y) u K = sup dy. (4.2) x x y n We also use the following onvention in the use of onvolutions, #,and. If f is a funtion depending both on x and t, then, for i =, 2, t K i f(x, t) K i (x, t; y, s)f(y, s) dy ds; R n t K i #f(x, t) K i (x, s; y, )f(y, s) dy ds. If f is an initial value depending only on x, then K i f(x, t) K i (x, t; y, )f(y)dy. The proof of Theorem.2 is divided into three steps. The only prerequisite in the previous setion is Lemma 3.. Step. Four basi inequalities. In this step we prove the following four inequalities. t K (x, t; y, s) K (y, s; z,) u (z) dz dy ds R n u (z) x z n dz u K. (4.3)

23 global solutions of Navier-Stokes equations 69 t K (x, t; y, s) K (y, s; z,) u (z) dz dy ds R n u (z) x z n dz u K. (4.4) t K (x, s; y, ) K (y, s; z,) u (z) dz dy ds u K. (4.3 ) R n t K (x, s; y, ) K (y, s; z,) u (z) dz dy ds u K. (4.4 ) Sometimes, for simpliity, we also write (4.3) (4.4 ) as K K u (x, t) u K, K K u (x, t) u K K #K u (x, t) u K, K #K u (x, t) u K respetively. Here is a proof of (4.3). Denote the left-hand side of (4.3) by I = I(x, t). It is lear that t I(x, t) = Jds u (z) dz, (4.5) where J ( x y + t s) n+ ( y z + s) n dy. Obviously J dy + dy J + J 2. (4.6) x y x z /2 y z x z /2 When x y x z /2, we have x y y z /3. Hene J ( x z + t s) n ( x y + t s)( y z + s) n dy ( x z + t s) n This implies { J Therefore, x y x z /2 x y x z /2 ( x z + t s) x y x z /2 n ( x z + t s) x y x z /2 n ( y z + t s)( y z + s) n dy. ( y z + s) n+ dy, ( y z + t s) n+ dy, s (,t/2) s [t/2,t]. { ( x z + J t s) n s, s (,t/2) ( x z + t s) n t s, s [t/2,t]. (4.7)

24 6 Qi S. Zhang Integrating (4.7) from to t, we obtain t t/2 ds t J ds ( x z + s) n s + ds t/2 ( x z + t s) n t s. By diret omputation t J ds. (4.8) x z n Next we estimate t J 2 ds. When y z x z /2, from (4.6), we have J 2 ( x z + s) n ( y z + dy. t s) n+ y z x z /2 Hene J 2 ( x z + s) n t s. By a simple omputation, we see that t J 2 ds t ( x z + s) n t s ds. (4.9) x z n Substituting (4.8) and (4.9) into (4.5), we dedue u (z) I(x, t) x z n dz u K. This is (4.3). The proof for (4.4), (4.3 ) and (4.4 ) is similar. Step 2. Solving an integral equation. The spae of divergene-free, vetor-valued funtions equipped with the norm K (defined in (4.)) is denoted by S. Given vetor-valued funtions b = b(x, t) and u = u(x), let us onsider the integral equation in S t u(x, t) = E (x, t; y, )u (y)dy E (x, t; y, s)b(y, s) u(y, s)dy ds. (4.) We will prove that (4.) has a solution in S provided that b K <ηand u K. Here η is a suffiiently small number depending only on dimension. It suffies to prove that the following series, obtained by iterating (4.), is norm onvergent in S. u(x, t) = E (x, t; y, )u (y)dy R n t E (x, t; y, s)b(y, s) y E (y, s; z,)u (z)dz dy ds +

25 global solutions of Navier-Stokes equations 6 By indution after exhanging the order of integration, we obtain u(x, t) = E (x, t; z,)u (z)dz [ + ( ) j E (b E ) j] (x, t; z,)u (z)dz. (4.) j= Here t [E b E ](x, t; z,) E (x, t; y, s)b(y, s) y E (y, s; z,)dy ds. By Lemma 3., we have K b K C B(b,, )K. This shows, by indution, that E (b E ) j] (x, t; z,) [C B(b,,t)] j K (x, t; z,). Substituting this into (4.), we dedue u(x, t) [C B(b,,t)] R j K (x, t; z,) u (z) dz. (4.2) n j= Note that B(b,, ) = b K. Hene, by (4.2), t (K u )(x, t) [C b K ] j K (x, t; y, s) j= K (y, s; z,) u (z) dz dy ds. (4.3) This and (4.3) imply (K u )(x, t) u K [C b K ] j. j= By (4.2) and (4.3 ), t (K # u )(x, t) [C b K ] j K (x, s; y, ) j= K (y, s; z,) u (z) dz dy ds K #K u (x, t) [C b K ] j u K [C b K ] j. j= j=

26 62 Qi S. Zhang Hene, by (4.), u K sup(k u + K # u ) u K j= [C b K ] j. Therefore, (4.) is norm onvergent in S when C b K <. In this ase u K u K. C b K Let η = 2C and S K,η {u S : u K <η}. Given b S K,η, by the above, u defined by (4.4) also belongs to S K,η provided that u K <η/(2). The mapping T defined by Tb = u maps S K,η to itself. Next we prove that T is ontration if η is suffiiently small. Given b,b 2 S K,η, let Tb = u and Tb 2 = u 2. Then it is easy to see that t t u u 2 = E (b b 2 ) u + E b 2 (u u 2 ). (4.4) We denote A = t E (b b 2 ) u and let E be the fundamental solution of (.) with b replaed by b. Then u (x, t) x E (x, t; z,) u, (z) dz, where u, (z) =u (z,). Using the gradient estimate on E (Lemma 3.2), we have, sine b k is suffiiently small, u (x, t) K (x, t; z,) u, (z) dz. Hene, t A(x, t) K (x, t; y, s) b b 2 (y, s)k (y, s; z,)dy ds u, (z) dz. By Lemma 3., A(x, t) B(b b 2,,t) K (x, t; z,) u, (z) dz. (4.5) Similarly, x A(x, t) B(b b 2,,t) K (x, t; z,) u, (z) dz. (4.6)

27 global solutions of Navier-Stokes equations 63 Substituting (4.5) and (4.6) into (4.4) and using indution, we find u u 2 (x, t) B(b b 2,,t) [ B(b 2,,t)] R j K (x, t; z,) u, (z) dz. n It follows that j= (K u u 2 )(x, t) b b 2 K [ b 2 K ] j (K K u, )(x, t), (K# u u 2 )(x, t) b b 2 K Using (4.3) and (4.3 ), we have j= j= [ b 2 K ] j (K #K u, )(x, t). u u 2 K b b 2 K u, K b b 2 k η. This implies that T is a ontration when η is suffiiently small. Therefore T has a fixed point S K,η, whih satisfies t u(x, t) = E (x, t; y, )u (y)dy E (x, t; y, s)u(y, s) y u(y, s)dy. (4.7) Step 3. Proving the solution of the integral equation is a solution of the Navier-Stokes equations. If we knew that the funtion u(,t) u(,t) is in L p ( ) for some p>, then by the argument of [7], reprodued in the proof of Theorem. (i) in Setion 3, we would know that a solution to (4.7) is a solution to the Navier-Stokes equations. The L p bound is needed sine one needs to apply the Riesz transform on u u. However, we do not have this information at the moment. In fat we do not even know whether u u is integrable for the moment. To overome this diffiulty, we arry out an approximation proess. In order to proeed, let us summarize the main properties obtained in the last step for the solution of (4.7). (a) When u S K,η and η is suffiiently small, the equation (4.7) has a solution in S K,η. (b) There exists = (η) > suh that u(x, t) K (x, t; y, ) u (y) dy, (4.8) u(x, t) K (x, t; y, ) u (y) dy. (4.9)

28 64 Qi S. Zhang The first estimate is just (4.2), and the seond an be obtained similarly by iterating the derivative of (4.). Here we emphasize that all iterations are valid under the assumption of the smallness of the K norm of all relevant funtions. The rest of the proof is divided into three parts. Part. Stability. Let u i ( S K,η ), i =, 2, be the solutions to (4.7) with the initial values of u,i. We want to prove that the u u 2 K is dominated by u, u,2 K. From the onstrution of u i by (4.7) (replaing u by u,i ), we see that u (x, t) u 2 (x, t) t (4.2) J(x, t)+ E (x, t; y, s)u 2 y (u u 2 )(y, s) dy ds, where J(x, t) E (x, t; y, )(u, (y) u,2 (y))dy R n t + E (x, t; y, s)(u u 2 ) y u (y, s)dy ds. Applying (4.9), whih obviously holds for u,wehave J(x, t) K (x, t; y, ) u, (y) u,2 (y) dy R n t + K (x, t; y, s) u u 2 K (y, s; z,)u, (y, s)dy ds. Using Lemma 3., we dedue J(x, t) K (x, t; y, ) u, (y) u,2 (y) dy + B(u u 2,, ) K (x, t; y, ) u, (y) dy. For simpliity we write the above as J(x, t) K u, u,2 (x, t)+ u u 2 K K u, (x, t). (4.2) Similarly, J(x, t) K u, u,2 (x, t)+ u u 2 K K u, (x, t). (4.22) Substituting (4.2) and (4.22) into (4.2) and iterating, we obtain u u 2 (x, t) K ( u 2 K ) j u, u,2 (x, t) j=

29 global solutions of Navier-Stokes equations 65 + u u 2 K K ( u 2 K ) j u, (x, t) j= [ B(u 2,, )] j( K u, u,2 (x, t)+ u u 2 K K u, (x, t) ). j= Here we just used Lemma 3. again. Similarly, taking the gradient on both sides of (4.2) and integrating, there holds (u u 2 ) (x, t) [ B(u 2,, )] j( K u, u,2 (x, t)+ u u 2 K K u, (x, t) ). j= Sine u and u 2 are suffiiently small in the K norm, the above imply u u 2 (x, t) ( K u, u,2 (x, t)+ u u 2 K K u, (x, t) ), (4.23) (u u 2 ) (x, t) ( K u, u,2 (x, t)+ u u 2 K K u, (x, t) ). (4.24) These show that K u u 2 (x, t) ( K K u, u,2 (x, t)+ u u 2 K K K u, (x, t) ), (4.25) K (u u 2 ) (x, t) ( K K u, u,2 (x, t)+ u u 2 K K K u, (x, t) ), (4.26) K # u u 2 (x, t) ( K #K u, u,2 (x, t)+ u u 2 K K #K u, (x, t) ), (4.25 ) K # (u u 2 ) (x, t) ( K #K u, u,2 (x, t)+ u u 2 K K #K u, (x, t) ). (4.26 ) Applying (4.3) and (4.4) to (4.25) and (4.26) respetively, we reah K u u 2 (x, t) ( u, u,2 K + u u 2 K u, K ), (4.27) K (u u 2 ) (x, t) ( u, u,2 K + u u 2 K u, K ). (4.28) Similarly, applying (4.3 ) and (4.4 ) to (4.25 ) and (4.26 ) respetively, we get K # u u 2 (x, t) ( u, u,2 K + u u 2 K u, K ), (4.27 ) K # (u u 2 ) (x, t) ( u, u,2 K + u u 2 K u, K ). (4.28 )

30 66 Qi S. Zhang Sine u, K is small, (4.27) and (4.27 ) imply u u 2 K = sup(k u u 2 + K # u u 2 )(x, t) u, u,2 K. (4.29) Substituting (4.29) into (4.28), we see that K (u u 2 ) (x, t) u, u,2 K. (4.3) Part 2. Approximation. Given u S K,η, let {u,k } S K,η C (Rn ) be a sequene suh that u,k (x) u (x) and lim u,k u K =. (4.3) k Let u be a solution to the integral equation (4.7) and u k be a solution to u k (x, t) = E (x, t; y, )u,k (y)dy R n t E (x, t; y, s)u k (y, s) y u k (y, s)dy. (4.32) Aording to (4.29) and (4.3) (replaing u by u k and u 2 by u), u k u K u,k u K (4.33) K (u k u) (x, t) u,k u K (4.34) K # (u k u) (x, t) u,k u K. (4.34 ) Sine u,k is smooth and ompatly supported, by (4.8) and (4.9), u k u k L p ( [,l)) for some p>. Here l>. Hene by the argument in the proof of Theorem. (i), whih is borrowed from [7], we know that u k is a solution to the Navier-Stokes equations with initial value u,k, i.e., for any vetor-valued φ C (Rn (, )) with div φ =, u k, t φ + φ dx dt R n u k u k,φ dx dt = u,k (x),φ(x, ) dx. (4.35) Part 3. Taking the limit. We are going to show that eah term in (4.35) onverges as k. Sine K (x, t; y, s) is bounded away from when t and x, y are finite, by (4.33), we see that u k u L (Ω) C Ω u u k K, (4.36) when k. Here Ω is any bounded domain of [, ).

31 global solutions of Navier-Stokes equations 67 Next, notie that, by (4.9), t (K u k u u k )(x, t) = K (x, t; y, s) u k u u k (y, s)dy ds R n t K (x, t; y, s) u k u (y, s) K (y, s; z,) u,k (z) dz dy ds R n t K (x, t; y, s) u k u (y, s)k (y, s; z,) dy ds u,k (z) dz u k u K K (x, t; z,) u (z) dz. (4.37) Here we just used Lemma 3. and the fat that u,k (x) u (x). By (4.4), we know that for almost every (x, t), K (x, t; z,) u (z) dz <. Let Ω be any bounded domain of [, ); we fix (x, t) Ω so that the above integral is finite. Sine K (x, t; y, s) is stritly positive when (y, s) Ω, we have, by (4.37) and (4.33), lim k Ω u k u u k (y, s) dy ds C Ω lim (K u k u u k )(x, t) =. k (4.38) Similarly, by (4.8), K u (u u k ) K ( (u u k ) K u ). Using (4.34), (4.34 ), and Remark 3. after Lemma 3., one has K u (u u k ) (x, t) sup(k (u k u) + K # (u k u) ) K (x, t; z,) u (z) dz. C u k u K K (x, t; z,) u (z) dz. By (4.4), the last integral in the above is finite for almost all (x, t). Therefore, as before lim u (u u k ) (y, s) dy ds =. (4.39) k Ω

32 68 Qi S. Zhang Note that u k u k,φ u u, φ dx dt Ω (u k u) u k,φ dx dt + Ω Ω <u (u k u),φ dx dt. Choosing an Ω that ontains the support of φ we have, by (4.38), (4.39), and (4.35), u, t φ + φ dx dt R n u u, φ dx dt = u (x),φ(x, ) dx. This shows that u is a solution to the Navier-Stokes equations. Finally, let us prove the last statement in Theorem.2. This is easy, now that we also assume u L 2 ( ). By (4.8) and Hölder s inequality u(x, t) 2 ( x y + t) dy u 2 2n L = 2 t n/2 u 2 L. 2 Therefore u(x, t) is finite for all x and t>. Hene u is lassial when t>. Note that no smallness of u 2 L is required here Improved suffiient ondition for regularity In this setion we prove that a ertain form-boundedness ondition on the veloity is suffiient to imply regularity. Throughout the years, various onditions on u that imply regularity have been proposed. One of them is the Prodi-Serrin ondition, whih requires that u L p,q with 3 p + 2 q for some 3 <p and q 2. See ([25, 27] e.g.) Reently the authors in [6] showed that the ondition p = 3 and q = also implies regularity. In another development the author of [9] improved the Prodi-Serrin ondition by a log fator, i.e., by requiring T u(,t) q p + log + dt <, u(,t) p where 3/p +2/q = and 3 <p<, 2<q<. The form-boundedness ondition, with its root in the perturbation theory of ellipti operators and mathematial physis, seems to be different from all the previous onditions. It seems to be one of the most general onditions under the available tools. This fat has been well doumented in the theory of linear ellipti equations. See [26] e.g. Here, first of all it allows singularity

33 global solutions of Navier-Stokes equations 69 of the form (t)/ x, whih does not belong to any of the previous regularity lasses. Here (t) is bounded. It also has the advantage of just requiring L 2 integrability of the veloity. Moreover, it ontains the Prodi-Serrin ondition exept when p or q are infinite. It also ontains suitable Morrey-Compamatotype spaes. However, we are not sure this ondition ontains the one in [9]. More preisely we have Theorem 5.. Let u be a Leray-Hopf solution to the three-dimensional Navier-Stokes equation (.2) in R 3 (, ). Suppose for every (x,t ) R 3 (, ), there exists a ube Q r = B(x,r) [t r 2,t ] suh that u satisfies the form-bounded ondition u 2 φ 2 dy ds (5.) Q r ( ) φ 2 dy ds + sup φ 2 (y, s)dy + B( φ 24 L 2 (Q r)). Q r s [t r 2,t ] B(x,r) Here φ is any smooth funtion vanishing on the paraboli side of Q r and B = B(t) is any given funtion whih is bounded when t is bounded. Then u is a lassial solution when t>. The next orollary shows that the form-boundedness ondition ontains the famous Prodi-Serrin ondition in the whole spae, exept for p = and q =. Corollary 3. Suppose u L p,q (Q r ) with 3/p +2/q =and neither p nor q is infinity. Then u satisfies (5.) in Q r where r (<r) is suffiiently small. Proof. Let φ be as in Theorem 5.. Then, by Hölder s inequality (( b u 2 φ 2 dy ds φ dy) /a ) /b (( b/ads ) /b. 2a ds u dy) 2a Q r Q r Here 2a = p, 2b = q, and a and b are the onjugates of a and b respetively. Hene u 2 φ 2 dy ds u 2 L p,q (Q r ) φ 2 L 2a,2b (Q r ). Q r Note that 3 2a + 2 2b = 3 2 ( a )+ b = 5 2 ( 3 2a + 2 2b )=5 2 (3 p + 2 q )=3 2. By [6, p. 52, Example 6.2], we have ( φ 2 L 2a,2b (Q r ) C φ 2 dy ds + Q r sup s [t (r ) 2,t ] Q r B(x,r ) ) φ 2 (y, s)dy.

34 62 Qi S. Zhang Choosing r suffiiently small, we see that u satisfies (5.). Proof of Theorem 5.. Let t be the first moment of singularity formation. We will reah a ontradition. It is lear that we need only to prove that u is bounded in Q r/8 = Q r/8 (x,t ) for some r>. In fat the number 8 is not essential. Any number greater than would work. We will follow the idea in [27]. Consider the the equation for vortiity w = u. It is well known that, in the interior of Q r, w is a lassial solution to the paraboli system with singular oeffiients w u w + w u w t =. (5.) Let ψ = ψ(y, s) be a standard ut-off funtion suh that ψ =inq r/2, ψ =inq r, and suh that ψ, ψ C/r, and ψ t C/r 2. We an use wψ 2 as a test funtion on (5.) to obtain (wψ) 2 dy ds + wψ 2 (y, t )dy Q r 2 B(x,r) C r Q 2 w 2 dy ds + u w wψ 2 dy ds + w u wψ 2 dy ds r Q r Q r I + I 2 + I 3. (5.2) The term I is already in good shape. Next, using integration by parts and the divergene-free ondition on u, wehave I 2 = u ψψ w 2 dy ds. 2 Q r Hene I 2 ɛ u 2 wψ 2 dy ds + C ɛ ψ 2 w 2 dy ds. (5.3) Q r Q r Here ɛ> is arbitrary. Next I 3 = Σ i w i ψ i u wψ dy ds Q r = Σ i i (w i ψ)u wψ dy ds + Σ i w i ψu i (wψ) dy ds Q r Q r (wψ) 2 dy ds + 3 u 2 wψ 2 dy ds 2 Q r 2 Q r + (wψ) 2 dy ds + u 2 wψ 2 dy ds. 4 Q r Q r

35 global solutions of Navier-Stokes equations 62 Substituting this and (5.3) into (5.2) and simplifying we obtain (wψ) 2 dy ds + wψ 2 (y, t )dy 4 Q r 2 B(x,r) C + C ɛ r 2 w 2 dy ds + 5+ɛ u 2 (wψ) 2 dy ds. Q r 2 Q r After repeating the above in Q r {(y, s) :s<t} for all t (t r 2,t ), one has (wψ) 2 dy ds + sup wψ 2 (y, s)dy 4 Q r 2 t r 2 s t B(x,r) C r Q 2 w 2 dy ds +(5+ɛ) u 2 (wψ) 2 dy ds. r Q r By the form-boundedness assumption on u, wehave (wψ) 2 dy ds + sup wψ 2 (y, s)dy Q r t r 2 s t B(x,r) C ɛ r 2 w 2 dy ds + 4(5 + ɛ) ( (wψ) 2 dy ds Q r 24 Q r ) + sup wψ 2 (y, s)dy + CB( w L 2 (Q r)). (5.4) t r 2 s t B(x,r) Hene, we an hoose ɛ so small that (wψ) 2 dy ds + sup wψ 2 (y, s)dy Q r t r 2 s t B(x,r) C ɛ r 2 w L 2 (Q r) + CB( w L 2 (Q r)). (5.5) Using standard results, we know that (5.5) implies that u is regular. Here is the proof. From (5.5), it is lear that Q r url (wψ) 2 dy ds C. Hene, sine ψ = in Q r/2, u 2 dy ds C. (5.6) Q r/2 Let η = η(y) be a ut-off funtion suh that η =inb(x,r/4) and η =in B(x,r/2). Then for eah s [t (r/4) 2,t ], we have, in the weak sense, (uη) =η u +2 u η + u η f

36 622 Qi S. Zhang in Q r/2. By standard ellipti estimates, using the fat that uη = on the boundary, This shows that D 2 u(,s) L 2 (B(x,r/4)) C f(,s) L 2 (B(x,r/2)). D 2 u L 2 (Q r/4 ) C u L 2 (Q r/2 ) + C u L 2 (Q r/2 ). By Sobolev imbedding, u L 6,2 (Q r/4 ). (5.7) Next, from (.22) on p. 36 of [29], u(,s)η W,2 C ( u(,s)η L 2 + div(uη)(,s) L 2 + url(u(,s)η) L 2). Here all norms are over the ball B(x,r/2). Therefore, uη(,s) W,2 ( ) C u(,s)η L 2 + u η(,s) L 2 + wη(,s) L 2 + u(,s) η L 2. It follows that u(,s) W,2 (B(x,r/4)) C. From Sobolev imbedding we know that u L 6, (Q r/4 ). (5.8) We treat u and u as oeffiients in equation (5.). By (5.6) and (5.7), standard paraboli theory (see [6] e.g.) shows that w is bounded and Hölder ontinuous in Q r/8. Here the bound depends only on the L 2 norm of w in Q r and r. This is so beause of the relation 3/6 +2/ < for the norm of u and 3/6+2/2 < 2 for the norm of u. Now a standard bootstrapping argument shows that u is smooth. Note that one an also use the Prodi-Serrin ondition, whih is implied by (5.8), to onlude that u is bounded and hene regular. Remark. Currently we are not able to prove a loal version of Theorem 5. due to a diffiulty in obtaining an approximation argument for weak solutions under the form-boundedness ondition. Under the Prodi-Serrin ondition, this is done in [27]. Aknowledgment. I thank Professors Maria Shonbek and Vitor Shapiro for helpful suggestions and Professor Hailiang Liu for the introdution of Navier-Stokes equations. Thanks also go to the anonymous referees for some helpful advie.

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