An introduction to the classical theory of the Navier Stokes equations. IMECC Unicamp, January 2010

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1 An introduction to the classical theory of the Navier Stokes equations IMECC Unicamp, January 2 James C. Robinson Mathematics Institute University of Warwick Coventry CV4 7AL. UK. j.c.robinson@warwick.ac.uk

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3 Introduction These notes are based on lectures give at a Summer School held in Campinas, Brazil, in January 2. I am very grateful to Helena Nussenzveig Lopez and Milton Lopez Filho for the opportunity to visit them in Campinas, and to deliver these lectures there. The intention in these notes is to give a relatively accessible overview of three major results: the existence for all time of weak solutions (due to Leray, 934, and Hopf, 95); the local regularity result of Serrin (962); and the partial regularity theory of Caffarelli, Kohn, & Nirenberg (982). The full details of the proofs are not given, but I hope that sufficient indication is given of how the proofs proceed to at least make clear the way their arguments work. The notes give a fairly complete proof of the existence of weak solutions, and sketch a proof of the existence of strong solutions and some of their properties (Chapter ). For much of this material I relied closely on the very nice lecture notes of Galdi (2), with support from the books by Temam (2) and Constantin & Foias (988). Chapter 2 covers Serrin s local regularity theory his paper relies on much external material, in particular regularity theory for parabolic equations. I give an elliptic version of the parabolic results, closely following the parabolic argument given in Lieberman (996), which does not rely on making fine estimates on the heat kernel. The proof of Campanato s Lemma that occurs in this chapter is based on online notes by Tobias Colding; and the simple proof of Young s inequality using Hölder s inequality comes from the book by Grafakos (28; his Theorem.2.2). 3

4 4 Introduction The final chapter discusses the partial regularity theory of Caffarelli, Kohn, & Nirenberg, using a very recent argument due to Kukavica (29a); I show in detail the proof of about half of one of the two theorems proved there which lead to the result that the dimension of the singular set is no larger than one; but I hope the details are sufficient to give a proper flavour of the proof as a whole. It should be possible to fill in all the gaps here with the help of Galdi s notes, Constantin & Foias (988), Serrin (962), Lieberman (996), Majda & Bertozzi (22), Caffarelli, Kohn, & Nirenberg (982), and Kukavica (29a). Of these, the key references for me have been Galdi s notes, Serrin s paper, and Kukavica s recent work. Finally, I gratefully acknowledge the support of the EPSRC, via a Leadership Fellowship, grant EP/G747/. And I cannot finish this introduction without admitting my considerable debt to José Rodrigo and Witold Sadowski, both at Warwick, with whom I have had many interesting and enlightening conversations about the topics contained in these notes. James Robinson 22nd January 2

5 Introduction 5 Some minimal functional analysis Hölder s inequality: if f L p and g L q with then fg L with p + q = fg L f L p g L q. We will often need the generalised form of Hölder s inequality with three exponents: if f L p, g L q, and h L r with then fgh L with p + q + r = fgh L f L p g L q h L r. One nice application of Hölder s inequality is to interpolation between L p spaces. A particular case that we will use frequently is that if u L 2 L 6 then u L 3 with u L 3 u L 2 u L 6. (.) L3L2L6 (To prove this, write u 3 L = u 3 = u 3/2 u 3/2, and use Hölder s inequality wityh exponents 4/3 and 3 4). This will be particularly useful, since in 3D we have the Sobolev embedding result H L 6, i.e. if u H then u L 6 c u H. (.2) L6H For functions in H (Ω), where Ω is a domain contained in a strip of width 2R, we have the Poincaré inequality u 2 dx cr 2 Du 2, dx. (.3) P Ω We also have a Poincaré-like inequality on a ball, u u BR (x) 2 cr 2 B R (x) Ω B R (x) Du 2, (.4) P2

6 6 Introduction where here u BR (x) is the average of u over B R(x), u BR (x) = ω n R n u(y) dy. B R (x) If X is a Banach space, any bounded sequence in its dual with have a weakly-* convergent subsequence. Any bounded sequence in a reflexive Banach space (in particular in a Hilbert space) will have a weakly convergent subsequence. If x n x then x lim inf n x n (and similarly if f n converges weakly-* to f). In a Hilbert space, if we have weak convergence and norm convergence, x n x, then in fact x n x: this follows from the identity (x n x, x n x) = x n 2 2(x, x n ) + x 2.

7 Existence of weak solutions We will study the incompressible Navier Stokes equations u t u + (u )u + p = div u = (.) NSE in a smooth, bounded, domain Ω R 3, with Dirichlet boundary conditions u = on Ω and a given initial condition u(x, ) = u (x). Generally speaking the equations include a viscosity coefficient in front of the Laplacian, u t ν u + (u )u + p =, (.2) NSEnu but if we are interested in questions of existence and uniqueness, we lose nothing by restricting to ν =. Indeed, if we have a solution u(x, t) of (.2) and we consider u ν (x, t) = νu(x, νt), it is easy to see that u ν (x, t) satisfies (.). So, for example, if we could prove the existence and uniqueness of solutions of (.) for all positive times, we would have the same result for (.2) for any ν >. We begin with some simple estimates, which lie at the heart of the existence of weak solutions. Take an initial condition u (x) with u 2 dx < Ω (this corresponds to finite kinetic energy) and suppose that the equations have a classical solution on [, T ]. Then we can take the dot product of the 7

8 8 Existence of weak solutions equations with u and integrate over Ω to obtain d u 2 + u 2 =. (.3) energydequality 2 dt Ω Ω We obtain the second term after integrating by parts and using the boundary condition u = on Ω. The nonlinear term vanishes this is a particular case of the more general identity [(u )v] v =, (.4) ortho Ω which holds provided that u has divergence zero and one of u and v vanishes on Ω; we will use this many times. In fact this follows from the useful antisymmetry identity [(u )v] w = [(u )w] v. (.5) antisymm Ω To obtain this identity, write the integral more explicitly in components and integrate by parts u i ( i v j )w j = ( i u i )v j w j u i v j ( i w j ) = u i ( i w j )v j. Ω Ω Ω Ω The pressure term has also vanished, since p u = p( u) =, Ω using the fact that u is divergence free (and the u = boundary condition). We can integrate (.3) in time to give an energy equality 2 u(t) 2 + t Ω Ω Du(s) 2 ds = 2 u 2, (.6) EE where now we are using the notation u for the L 2 (Ω) norm of u. This formal calculation (we are not careful, and assume that everything is as regular as it needs to be for all our manipulations to be justified), leads us to expect that if we have an initial condition in L 2, we will obtain a solution that is bounded in L 2, and whose H norm is square integrable. We write u L (, T ; L 2 ) L 2 (, T ; H ). (.7) Li2L2H

9 . Weak formulation of the equation 9 More generally, the notation u L p (, T ; X), where X is a Banach space, means that the map t u(t) X is in L p, T u(t) p X dt <. If we equip the space L p (, T ; X) with the natural L p norm it is a Banach space. A weak solution will be a function u with the regularity in (.7) that satisfies the Navier Stokes equations in a weak sense, which we now make precise. We will show that there exists at least one weak solution that also satisfies an energy inequality, i.e. (.6) with the equality replaced by.. Weak formulation of the equation Take a function ϕ(x, t) C ([, T ) Ω), i.e. a smooth function with compact support in [, T ) Ω. Multiply (.) by ϕ and integrate over the space-time domain [, T ) Ω. If we do this we obtain T (u, ϕ t ) ( u, ϕ) ((u )u, ϕ) + ( p, ϕ) = (u, ϕ); (.8) weak we use ( ) to denote the inner product in L 2 (Ω). In order to proceed, we use the (non-trivial) fact that any smooth function can be decomposed as the sum of a divergence-free function and the gradient of a scalar function (the Helmholtz decomposition, see Temam, 2), and the fact that these functions are orthogonal in L 2, (u, p) = if u = (we have already used this to in our formal derivation of the energy equality (.6). So if we write ϕ = φ + χ equation (.8) actually decomposes into two equations, one involving u alone, T (u, φ t ) ( u, φ) ((u )u, φ) = (u, φ) (.9) weak2 and one that relates the pressure p to the velocity u, T ((u )u, χ) + ( p, χ) =. (.) u2p

10 Existence of weak solutions From now on we concentrate on (.9), which must hold for all φ D([, T ) Ω), the set of all functions in C ([, T ) Ω) that are also divergence free. We define H to be the completion of D([, T ) Ω) in the L 2 (Ω) norm. This is the space of finite-energy, divergence-free, initial conditions that also satisfy the boundary condition in a weak sense (in fact u n = on Ω; the trace exists in this direction thanks to the divergence-free condition, see Temam (2), for example). Definition. Given u H, a weak solution of (.) on [, T ) is a function u L (, T ; H) L 2 (, T ; H ) such that T for every φ D([, T ) Ω). (u, φ t ) ( u, φ) ((u )u, φ) = (u, φ) In fact it more convenient to work with a formulation of the weak problem in which we test with functions that depend only on space, rather than the space-time varying functions in D([, T ) Ω). To do this, we consider a sequence of test functions in D([, T ) Ω) of the form φ h (x, t) = ψ(x)θ h (t), where ψ D(Ω = C (Ω) functions with divergence zero, and θ h (t) is a function in C ([, T )) that is identically for t t, and identically for t t + h. If we test with φ h (x, t) and let h, we obtain (u(t), ψ) (u, ψ) + t ( u, ψ) + ((u )u, ψ) = (.) weak3 for every t [, T ). (To obtain this equation for every t [, T ) we may have to redefine on a set of measure zero, so there are some subtleties see Galdi s notes, for example.) This procedure can in fact be reversed: if u L (, T ; H) L 2 (, T ; H ) satisfies (.) for every ψ D(Ω) and every t [, T ) then u satisfies (.9) for every φ D([, T ) Ω). (First show that (.9) holds for functions of the form α(t)ψ, with α(t) C ([, T )) and ψ D Ω using an integration by parts for Lebesgue functions; then use density of linear combinations of such functions in D([, T ) Ω).) So they are equivalent formulations of the problem, and we could define instead:

11 . Weak formulation of the equation Definition.2 A weak solution of (.) on [, T ) with initial condition u H is a function such that (u(t), ψ) (u, ψ) + u L (, T ; H) L 2 (, T ; H ) t for every ψ D(Ω) and for every t [, T ). ( u, ψ) + ((u )u, ψ) = (.2) weakone One immediate consequence of the definition in this form is the weak continuity of u(t): Lemma.3 Any weak solution is weakly continuous into L 2, i.e. for any v L 2, lim t t (u(t), v) = (u(t ), v). (.3) weakc Proof First we show that (.3) holds if v D(Ω). In this case we can use (.2) to write t (u(t), v) (u(t ), v) = ( u, v) + ((u )u, v) ds t Since v is smooth, v and v are bounded; since u L 2 (, T ; H ) the first of the two terms on the right-hand side is integrable; since u L (, T ; H), so is the second. It follows that the right-hand side converges to zero as t t, as required. For v L 2, we approximate v by a sequence in D(Ω) and use the fact that u L (, T ; H) to pass to the limit. Note that this gives a sense in which the initial condition is satisfied by the weak solution: u(t) u as t. We are going to prove the existence of (at least) one weak solution that satisfies two additional properties. Theorem.4 For any u H there exists at least one weak solution of the

12 2 Existence of weak solutions Navier Stokes equations. This solution is weakly continuous into L 2, and in addition satisfies the energy inequality 2 u(t) 2 + t Du(s) 2 ds 2 u 2 (.4) EI for every t [, T ). As a consequence, u(t) u as t. Note that every weak solution is weakly continuous into L 2 ; but it is not known whether every weak solution satisfies (.4). That the solution approaches the initial condition strongly is a consequence of the weak continuity and (.4), so again is not known for every weak solution, only those of the form we construct here, termed Leray Hopf weak solutions (since they were first constructed by Leray (934) and Hopf (95)). Proof We will use the Galerkin method: we construct a series of approximate solutions of the form u k (x, t) = k û k,j (t)ψ j (x), u k () = P k u = j= k (u, ψ j )ψ j, (.5) ukis where the {ψ j } j= are an orthonormal basis for L2 formed from the eigenfunctions of the Stokes operator, j= ψ j + p j = λ j ψ j ψ j Ω = ψ j =. These functions ψ j are C in Ω, are divergence-free, and satisfy the zero boundary condition. (In fact, however, the ψ j s are not elements of D(Ω), since they do not have compact support in Ω. So we are being a little careless here. Nevertheless, we could fix this by proving that we can extend the test functions used in the definition of a weak solution to include the larger class of functions in C ( Ω) that are zero on Ω.) If we use u k in place of u in (.2), and test with each of the functions {ψ,..., ψ j }, we obtain a set of k coupled ordinary differential equations (ODEs) for the coefficients û k,j : d dtûk,j + λ j û k,j + k ((ψ i )ψ l, ψ j )û k,i û k,l =. (.6) hatu i,l= Note that the third term is essentially quadratic in the û k,j s so this is a set of locally Lipschitz ODEs. The theory of existence and uniqueness for such

13 . Weak formulation of the equation 3 ODEs is well developed: a unique solution exists while the solution stays bounded, i.e. unique solutions for the û k,j (t) exist while k û k,j (t) 2 < (.7) ODEis j= (this is just the norm in R k of the vector (û k,,..., û k,k ) of coefficients). How can we show that this quantity stays bounded? The easiest way is to remember that in fact the {û k,j } k j= are the coefficients in an expansion of a function u k (x, t), and look at the equation satisfied by this function. If we multiply (.6) by ψ j, j =,..., k and sum, we obtain u k t u k + P k ((u k )u k ) =, where P k denotes the orthogonal projection (in L 2 ) onto the space spanned by the {ψ,..., ψ k } (this was defined in passing above in (.5)). Note that if P k did not appear the nonlinearity would generate terms that could not be expressed as a sum of {ψ,..., ψ k }. We can now easily estimate norms of the solution u k. Note that since all the ψ j are smooth (in space), so is u k, and therefore we can perform the manipulations that were formal at the beginning of this chapter entirely rigorously: we take the inner product with u k and integrate in time. Once again the nonlinear term vanishes, since (P k [(u k )u k ], u k ) = ((u k )u k, P k u k ) = ((u k )u k, u k ) =, and we obtain 2 u k 2 + Du k 2 =. Integrating this in time gives an energy equality for the Galerkin solutions, 2 u k(t) 2 + t Du k (s) 2 ds = 2 u k() 2. Since u k (t) 2 = k j= û k,j 2, (.7) is satisfied for all t, the approximate solution u k exists for all t. Now, since u k () = P k u, we have u k () u, and so we have a uniform bound (with respect to k) on the approximate solutions: 2 u k(t) 2 + t Du k (s) 2 ds 2 u 2.

14 4 Existence of weak solutions In other words, for any T >, u k is uniformly bounded in L (, T ; H) and in L 2 (, T ; H ). We can therefore use weak and weak-* compactness to find a subsequence (which we relabel) such that u k converges to u weakly-* in L (, T ; H) and weakly in L 2 (, T ; H ). It follows that the limit enjoys the same bounds as the approximations, 2 u(t) 2 + t Du(s) 2 ds 2 u 2 ; but note that the energy equality for the approximations has now turned into an energy inequality for the (candidate) weak solution. We may have shown that the approximations converge to a limit, but does this limit satisfy the right equation? As a first step, we need to show that for each ψ j the terms in the equation (u k (t), ψ j ) (u, ψ j ) + t ( u k, ψ j ) + ((u k )u k, ψ j ) = (.8) ukpsij converge to the same thing but with u k replaced by u. We will need some better convergence of u k to u before we can guarantee the convergence of the nonlinear term. In fact we will show that du k /dt is uniformly bounded in L 4/3 (, T ; H ), which coupled with the bounds we already have in L 2 (, T ; H ) will be enough to prove that there is a subsequence that converges that converges strongly in L 2 (, T ; L 2 ): this is the content of the Aubin Lemma (for a proof of a more general result see Temam, 2). Lemma.5 Let u k be a sequence that is bounded in L 2 (, T ; H ), and has du k /dt bounded in L p (, T ; H ) for some p >. Then u k has a subsequence that converges strongly in L 2 (, T ; L 2 ). You can think of this lemma as being a version of the familiar result that a bounded sequence in H has a subsequence that converges strongly in L 2 (i.e. that H is compact in L 2 ); things are more complicated here since for each t, u(t) lies in a space of functions instead of being a real number. We already have the bound on u k in L 2 (, T ; H ), so we need to look at the derivative of u k. We have du k dt = u k P k ((u k )u k ). Since u k L 2 (, T ; H ), we have u k L 2 (, T ; H ), so we have to estimate the nonlinear term. If we take the inner product with some v H

15 . Weak formulation of the equation 5 we obtain (P k [(u k )u k ], v) = ((u k )u k, P k v) = (u k )u k P k v u k Du k P k v u k L 3 Du k L 2 P k v L 6 c u k /2 Du k 3/2 P k v H ( c u k /2 Du k 3/2) v H, and so P k (u k )u k H c u k /2 Du k 3/2. It follows that the nonlinear term is bounded in L 4/3 (, T ; H ). So du k /dt is uniformly bounded in L 4/3 (, T ; H ), and we can use the Aubin Lemma to find a subsequence that converges strongly in L 2 (, T ; L 2 ). We are now in a position to show the convergence of all the terms in (.8). For the first term, note that if u k u in L 2 (, T ; L 2 ), then u k (t) u(t) in L 2 for almost every t (, T ). So the first term converges for almost every t (, T ). The second term doesn t depend on k, so converges trivially. The third term converges since u k u in L 2 (, T ; H ): in particular, t ( u k, v) t ( u, v) for any v L2 and all t (, T ). The nonlinear term requires the strong convergence in L 2 (, T ; L 2 ) and a little work. We have t ((u k )u k, ψ j ) ((u )u, ψ j ) ds t = (((u k u) )u k, ψ j ) + ((u )(u u k ), ψ j ) ds ( t ) /2 ( t ) /2 ψ j u k u 2 u k i= t ( i (u u k ), u i ψ j ). For the first term on the RHS we use the fact that u k is uniformly bounded in L 2 (, T ; H ) and that u k converges strongly to u in L 2 (, T ; L 2 ); for the second term we use the fact that u i ψ j L 2 and the weak convergence of u k to u in L 2 (, T ; H ).

16 6 Existence of weak solutions So for each j we have (u(t), ψ j ) (u, ψ j ) + t ( u, ψ j ) + ((u )u, ψ j ) = for all t (we get all t by adjusting on a set of measure zero). We want this equation to hold for any ψ D(Ω): this follows using the fact that finite linear combinations of the ψ j are dense in D(Ω). We have therefore obtained the existence of a weak solution, and we have already seen that this weak solution satisfies the energy inequality (.4). To see that u(t) u as t, we have from the weak continuity u lim inf t while from the energy inequality we obtain u(t), u lim sup u(t). t It follows that u(t) u as t, and since H is a Hilbert space it follows from this plus the weak continuity that u(t) u..2 Strong solutions We have shown the existence of at least one weak solution that exists for all time, but it is not known whether such weak solutions are unique. We now look (in a less rigorous way) at the existence of strong solutions. If a strong solution exists then it is unique (in the larger class of weak solutions), as we will soon see. First suppose that u H H ; take the inner product of (.) with P u and integrate over Ω. Here P is the orthogonal projection (in L 2 ) onto the space of divergence-free functions ( P is the Stokes operator ). Equivalently, take the Galerkin approximations, multiply each by λ j, and then sum; estimate each equation and take a limit as k this is the rigorous way of performing these calculations. We obtain d 2 dt Du 2 + P u 2 = ((u )u, P u) u L 6 Du L 3 P u L 2 Du 3/2 P u 3/2 2 P u 3/2 + c Du 6,

17 .2 Strong solutions 7 i.e. d dt Du 2 + P u 2 c Du 6. (.9) DI The differential inequality (.9) has a number of important consequences. Consequence (i). Local existence of strong solutions. If we drop the P u 2 term then we have dx dt cx3, where X(t) = Du(t) 2. It follows that and so i.e. dx X 3 c dt, [ 2X 2 Du(t) 4 ] X(t) X() ct, Du 4. (.2) Duuppert 2ct Du 4 It follows that if Du <, then Du(t) <, at least for t < /2c Du 4. In other words, we have small time existence of what we term a strong solution (u L (, T ; H )) if u H, with the time interval on which we can guarantee existence of such a solution tending to zero as u H. For such a strong solution, we can return to (.9) and integrate in time, keeping the Laplacian term, to deduce that Du(t) 2 + t P u(s) 2 ds Du 2 + t Du(s) 6 ds. Since we have a strong solution with u L (, T ; H ), it follows that we must also have u L 2 (, T ; H 2 ). (There is some elliptic regularity theory for the Stokes operator hidden here, of course, since what we actually show is that P u(t) is square integrable; we need an inequality that says u H 2 c P u, which is what comes from elliptic regularity.) So in fact a strong solution is a weak solution with the additional regularity u L (, T ; H ) L 2 (, T ; H 2 ).

18 8 Existence of weak solutions Strong solutions are unique in the larger class of weak solutions. This is indicated by the following argument (the argument has a flaw in, but the result is true). Lemma.6 Suppose that u is a strong solution of the Navier Stokes equations with initial condition u H H. Then u is unique in the class of weak solutions. Proof We give a formal proof (with a mistake). Take u to be a strong solution satisfying u t u + (u )u + p = u() = u and v to be a weak solution satisfying v t v + (v )v + q = v() = u. Consider w = u v: this satisfies the equation w t w + (w )u + (v )w + r = w() =. Take the inner product with w to give 2 w 2 + Dw 2 = ((w )u, w). (.2) wrong Note that one of the two nonlinear terms vanishes using the orthogonality identity ((u )v, v) = (see (.4)). In fact the equation (.2 is not valid - there is not enough regularity of w (essentially limited by v) to deduce that (w, w t ) = d 2 dt w 2. While we can pair w with w t for a.e. t [, T ), we cannot integrate in time (which we are about to do) since we only have w t L 4/3 (, T ; H ) and w L 2 (, T ; H ). But assuming that all is well... Bound the right-hand side according to ((w u), w) w Du w Dw Du /2 P u /2 w so that 2 Dw Du P u w 2, d dt w 2 + Dw 2 c Du P u w 2. Dropping the Dw 2 term and integrating gives ( t ) w(t) 2 exp Du(s) P u(s) ds w() 2.

19 .2 Strong solutions 9 Since u is a strong solution, u L (, T ; H ) L 2 (, T ; H 2 ), so the integrand on the RHS is integrable. It follows that w(t) = for all t, and hence u(t) = v(t), i.e. the solution is unique. Consequence (ii). Global existence of strong solutions for small initial data. i.e. If we return to (.9) and use Poincaré s inequality (.3), we obtain d dt Du 2 + α Du 2 c Du 6, d dt Du 2 Du 2 (c Du 4 α). It follows that if Du 4 < α/c, then d dt Du 2 <. Thus Du(t) 2 can never increase, and hence remains bounded for all t. This result is correct, but in some sense H is the wrong space in which to require that the initial data is small. This follows from an observation which will be extremely useful later when we come to consider the partial regularity results of Caffarelli, Kohn, & Nirenberg. If u(x, t) is a solution of the Navier Stokes equations (with pressure p(x, t)), then so is the rescaled solution λu(λx, λ 2 t) (and pressure λ 2 p(λx, λ 2 t)). If an initial condition u (x) is small and this implies existence for all time, we have the option of considering all the rescaled versions of u (x), λu (λx) if one of these rescaled initial conditions is small then the solution with initial condition λu (λx) exists for all time, and hence so does the solution with initial condition u (x). Ideally we choose to measure the smallness of u is a sense that is invariant under this scaling transformation. If we consider the H norm on R 3, this is not invariant under such a scaling, since u λ (x) 2 dx = λu(λx) 2 = λ 2 u(y) 2 λ 3 dy = u(y) 2 dy R 3 R 3 λ R 3 and by similar reasoning Du λ (x) 2 dx = λ Du(y) 2 dy. R 3 R 3 So the smallness condition u λ 2 + Du λ 2 = λ u 2 + λ Du 2 < ɛ

20 2 Existence of weak solutions can in some sense be optimised by choosing λ = u / Du, in which case it becomes u Du < ɛ. (.22) product This product of norms gives a measure of the size of u that is invariant under the scaling transformation. Spaces in which the norm is invariant under this transformation are termed critical for the Navier Stokes equations; examples are L 3 and H /2 (if u H, then one can bound the H /2 by interpolation between L 2 and H, u 2 H /2 u Du, c.f. (.22). Consequence (iii). Eventual regularity of weak solutions. If u is a weak solution then we know that t Du(s) 2 ds 2 u 2. It follows that if one chooses T large enough, there must be a t < T such that Du(t ) 2 < α/c, i.e. such that the existence of strong solution for all t t is guaranteed. So every weak solution is eventually strong. Consequence (iv). A lower bound on solutions that blow up. If a solution that starts with u H ceases to be a strong solution, this will be because Du(t) 2 as t T (T is the blow up time ). One can deduce that the solution must be bounded below according to 2c Du(T t) 2, (.23) lb t/2 (where c is the constant occurring in (.2)) for otherwise we would have Du(T t) 2 < 2c t /2 for some t >, and then one can use (.2) to show that in fact Du(T ) 2 < : Du(T ) 4 which is finite since 2ct Du(T t) 4 <. Du(T t) 4 2ct Du(T t) 4, Consequence (v). Limits on time singularities of weak solutions. One can use the lower bound (.23) along with the fact that Du is square integrable in time to show that for any Leray Hopf weak solution the set of

21 .2 Strong solutions 2 singular times Σ = {t R : Du(t) = } can have box-counting dimension no larger than /2. The simple argument here is due to Robinson & Sadowski (27). (An older argument due to Scheffer (976) shows that H /2 (Σ) =, where H /2 is the /2-dimensional Hausdorff measure; we will define something similar in Chapter 3 to deal with space-time singularities.) The (upper) box-counting dimension is defined as follows. Let N(X, ɛ) denote the number of balls of radius ɛ needed to cover X, and set d box (X) = lim sup ɛ log N(X, ɛ). log ɛ Essentially (but not exactly) d box (X) captures the exponent d in a relationship of the form N(X, ɛ) ɛ d as ɛ. What certainly is true is that if δ < d box (X), there exists a sequence ɛ j such that N(X, ɛ j ) > ɛ δ j. For our purposes, and equivalent definition is useful, where we replace N(X, ɛ) by N d (X, ɛ), the maximum number of disjoint balls that one can find whose centres lie in X (a relatively simple argument shows that N d (X, ɛ) is comparable to N(X, ɛ), and hence that the two definitions yields the same quantity). Suppose then that d box (X) > /2, and choose a δ with /2 < δ < d box (X). For such a δ, there exists a sequence ɛ j such that N d (X, ɛ j ) > ɛ δ j ; let the intervals in this disjoint collection be [t k ɛ j, t k + ɛ j ]. Now, consider T Du(s) 2 ds N d (X,ɛ j ) j= N d (X,ɛ j ) j= N d (X,ɛ j ) j= tk +ɛ j t k ɛ j tk Du(s) 2 ds t k ɛ j Du(s) 2 ds tk t k ɛ j 2c /2 ds ɛ d (t k s) /2 j c/2 ɛ j.

22 22 Existence of weak solutions Since d > /2, the right-hand side tends to infinity as j ; but we know that the left-hand side is bounded. It follows that d box (Σ) /2 as claimed.

23 2 Serrin s local regularity result Suppose that we return to the question of when a weak solution is unique, and rather than looking at strong solutions, assume rather than i.e. that u L s (, T ; L s (Ω)), T ( ) s /s u s <. Ω Can we find conditions on s and s that guarantee uniqueness? We ll use the same wrong argument that gave us uniqueness of strong solutions; again, we ll end up with the right answer (and essentially for the right reasons) even if the proof is less than watertight. Recall that we obtained d dt w 2 + Dw 2 = ((w )u, w). (2.) unique2 We now estimate the right-hand side differently: ((w )u, w) = ((w )w, u) w Dw u w L(2+s)/(s 2) Dw u L s, using Hölder inequality with exponents (2 + s)/(s 2), 2, and s. Now use Lebesgue space interpolation (just Hölder s inequality again) on the first 23

24 24 2 Serrin s local regularity result term to give w L (2+s)/(s 2) w (3/s) w 3/s L 6 c w (3/s) Dw 3/s after also using H L 6 and Poincaré s inequality (.3). Feeding this back in to the inequality (2.) we have d dt w 2 + Dw 2 c Dw u L s w (3/s) Dw 3/s. Integrating in time and using Hölder s inequality with exponents 2, r, and 2s/3 so that 2 + r + 3 = (2.2) rsfrom 2s we obtain w(t) 2 + t Dw(s) 2 ( t ) /2 ( t ) /r ( t ) 3/2s Dw 2 u r L s w 2 Dw 2, which after using Young s inequality (again with exponents 2, r, and 2s/3) yields w(t) 2 c t u(τ) r L s w(τ) 2 dτ. This implies that ( t ) w(t) 2 exp u(τ) r L s dτ w() 2, so that w(t) = for all t provided that u L r (, T ; L s ), where from (2.2) we must have 3 s + 2 r =. [Note that on R 3 R, the norm in L r (, T ; L s ) is scale invariant.] In this chapter we show that if u is a weak solution that satisfies the assumption u L s (T, T 2 (L s 3 (U))) s + 2 s < for some region U (T, T 2 ) of space-time, then u(t) C (U) for t (T, T 2 ), a result due to Serrin (962).

25 2. Serrin I: vorticity formulation and a representation formula Serrin I: vorticity formulation and a representation formula The key idea is to work with the vorticity form of the Navier Stokes equations. If we take the curl of the equations and define ω = u (the vorticity), we obtain ω t ω + ((u )u) = (the pressure term vanishes since curl grad= ). To simplify the nonlinear term we require two vector identities. The first is 2 u 2 = (u )u + u ω, which means that the nonlinear term becomes the second identity is [ 2 u 2 u ω] = (u ω); (a b) = a(div b) b(div a) + (b )a (a )b, from which (since both u and ω are divergence free) (u ω) = (u )ω (ω )u, and so the vorticity form of the Navier-Stokes equations is ω t ω + (u )ω (ω )u =. If we rewrite this as ω t ω = (ω )u (u ω) = div(ωu uω) (2.3) omegaheat then we have recast the equation as a heat equation for ω, although admittedly the right-hand side depends on ω and on u (which also depends on ω we will discuss recovering u from ω later). However, we know a lot about solutions of the heat equation. In particular if we set { ct 3/2 e x 2 /4t t > K(x, t) = otherwise. (K is the heat kernel ) then the convolution of K with f, K f, defined as T2 K f = K(x ξ, t τ)f(ξ, τ) dξ dτ, T U

26 26 2 Serrin s local regularity result is the solution in t of u t u = f(x, t) with u(x, t) =. We can use this to write down the following representation formula for ω in terms of g = ωu uω: ω = K (div g) + H(x, t) = K g + H(x, t), where H(x, t) is a solution of the heat equation (and hence smooth). We therefore neglect H in what follows (we should continue to include it, but it would cause no difficulty), and consider what information we can obtain from the (not quite true) representation ω = K g. (Note that there are also some subtleties here that we have ignored. For example, are u and ω (and so g) actually smooth enough to use this representation? To derive it rigorously Serrin mollified the equation and took limits.) In order to exploit this representation formula, we need some results about the smoothing properties of convolutions. 2.2 Convolutions and Young s inequality If we know something about f and something about g, we would like to be able to deduce properties of the convolution f g(x) = f(y)g(x y) dy. We start with two general such results, the first of which is a fundamental inequality due to Young, which we prove here by a carefully chosen application of Hölder s inequality. Note that r = is included, and we obtain f g L if p + q. Lemma 2. (Young s inequality) Let p, q, r satisfy r + = p + q. Then for all f L p, g L q, we have f g L r with f g r L f L p g L q.

27 2.2 Convolutions and Young s inequality 27 Proof We use p to denote the conjugate of p. Then we have q + r + p =, p r + p q =, and q r + q p =. First use Hölder s inequality with exponents q, r, and p : (f g)(x) f(y) g(x y) dy ( = f(y) p/q f(y) p/r g(x y) q/r) g(x y) q/p dy f p/q L p ( f p/q L p ( ) /r ( f(y) p g(x y) q dy ) /r f(y) p g(x y) q dy g q/p L. q ) /p g(x y) q dy Now take the L r norm (wrt x): f g L r f p/q L g q/p p L q ( ) /r f(y) p g(x y) q dy dx = f p/q L g q/p p L f p/r q L g q/r p L q = f L p g L q. We will need a space-time version of this inequality, which is provided by the following lemma. Essentially it says that we can use Young in space, and Young in time independently. Note that in the following lemma, h L x (L t ) provided that p + q and p + q. rrppqq Lemma 2.2 Let h = K g, with K L p t (Lp x) and g L q t (Lq x). h L r t (L r x) with Then where q r, q r, and h L r t (Lr x) k g, (2.4) L p t (Lp x) L q 2young t (Lq x) r = p + q, r = p +. (2.5) exponents q

28 28 2 Serrin s local regularity result Proof Use Minkowski s inequality (which works for integrals as well as simple sums) to write T2 h(, t) L r x = k(x ξ, t τ)g(ξ, τ) dξ dτ T G L r x T2 k(x ξ, t τ)g(ξ, τ) dξ dτ T G L r x k(t τ) L p x g(τ) L q x dτ, using Young s inequality in x to obtain the last line. inequality in t to obtain the result. Now use Young s If we apply this result using K as the kernel, then we obtain the following, which is fundamental to the first part of Serrin s argument. convolve Lemma 2.3 Let ω = K g, with g L q t (Lq x). Then h L r x L r t c g L q x L q t provided that q r, q r, and ( 3 q ) ( + 2 r q ) r Proof Set T = T 2 T. Clearly k(x, t) c x t (n/2) e x 2 /4t, <. (2.6) exponents2 and hence one can calculate T ( ) p /p k p c x p t p[(n/2)+] e p x 2 /4t dx dt L p xl p t U T ( ) p /p = c t p [(n/2)+] t p /2+(np /2p) y p e y 2 dy dt where α = n 2 = c T t p [(n/2)+]+(p /2)+(np /2p) dt = ct αp +, ( ) p + 2, provided that αp <, and then k L p x L p t ct αp +. U

29 2.3 Serrin II: Using convolutions to show that ω L t (L x ). 29 To apply Lemma 2.2, we must choose the exponents to satisfy (2.5). Eliminating p and p in the condition α < /p yields (2.6). 2.3 Serrin II: Using convolutions to show that ω L t (L x ). We are now in a position to give the first part of Serrin s regularity argument. Note that if u is a weak solution we know that u L 2 (, T ; H ), so in particular ω L 2 t (L 2 x). Proposition 2.4 Suppose that ω L 2 t (L 2 x) and u L s t (L s x) with Then in fact ω L t (L x ). 3 s + 2 s <. Proof Suppose that ω L ρ t (Lρ x). Then since u L s t (L s x) by assumption, and pointwise we have g 2 u ω, it follows that provided that q = s + ρ g L q xl q t and q = s + ρ (to see this, note that u q L s/q and ω q L ρ/q, so for u q ω q L we need /(s/q) + /(ρ/q) = using Hölder s inequality). We can now use Lemma 2.3 to show that ω L r t (L r x) provided that 3 ( q r Substituting for q and q we need ( 3 s + ρ r ) + 2 ) + 2 ( q r ) <. ( s + ρ ) <. r Write δ = (3/s) (2/s ): we have δ > by assumption. Note that if ρ > 5/δ then in fact ω L x L x as required. Otherwise, we can improve the regularity of ω to ω L r t (L r x) provided that ρ r < δ 5,

30 3 2 Serrin s local regularity result i.e. In particular r < 5ρ 5 δ. ω L ρ t (Lρ x) ω L αρ t (L αρ x ) with α = 5/(5 δ/2) >. Thus within a finite number of steps we obtain ω L ρ t (Lρ x) with ρ > 5/δ, from which as remarked above it follows that ω L t (L x ). We cannot go any further using the convolution results above. Instead we need to turn to the theory of regularity for parabolic equations. We will use the theory of Hölder regularity, but in order to prove this result (in fact we will prove a parallel and easier result for elliptic problems), we will need an integral characterisation of Hölder continuity which is of interest in itself. 2.4 Integral characterisation of Hölder spaces: Campanato & Morrey Lemmas. We say that f is α-hölder continuous on B R (), and write f C α (B R ()), if there exists a constant C > such that f(x) f(y) C x y α for all x, y B R (). We denote by f Br(x) the average of f over : f = Br(x) ω n r n f(y) dy, where ω n is the volume of the unit ball in R n. Lemma 2.5 Let f L (B R ()) and suppose that there exist positive constants α (, ], M >, such that f(y) f Br(x) 2 dy M 2 r n+2α (2.7) CampA for any x B R/2 () and any r (, R/2). Then f C α (B R/2 ()).

31 2.4 Integral characterisation of Hölder spaces: Campanato & Morrey Lemmas.3 Note that if we consider the integrals I c = f(y) c 2 dy = f(y) 2 2cf(y) + c 2 dy then these are minimised by the choice c = f Br(x) : simply observe that I c = f(y) 2 2cf(y) + c 2 dy, differentiate with respect to c, and set di c /dc =. Proof Choose x B R/2 () and r < R/2. We first compare f Br/2 (x) with f Br (x). We have f Br/2 (x) f 2 2 = ω n (r/2) n f(y) f dy Br(x) B r/2 (x) ( ) 2 ωn(r/2) 2 2n f(y) f dy Br(x) B r/2 (x) ( ) 2 22n ω n r 2n f(y) f dy Br(x) ( ) 22n ω n r 2n f(y) f Br(x) 2 dy ω n r n ( ) = 22n ω n r n f(y) f Br(x) 2 dy cm 2 r 2α. In particular, f Br/2 (x) f cmr α. Now consider f Br2 k (x) f. Since it follows that f Br2 k (x) f = f Br2 k (x) f k f Br2 k (x) f B r2 (k )(x), j= k cmr α 2 (j )α j= cmr α 2 jα = cmr α. (2.8) avav This shows that f Br2 k (x) forms a Cauchy sequence, and hence the averages converge for every x B r/2 (). By the Lebesgue Theorem, these averages j=

32 32 2 Serrin s local regularity result converge to f(x), and so if we let k in (2.8) we obtain an estimate for the difference between f(x) and its average, f(x) f Br(x) c Mr α. Now take another point y B R/2 (); we compare f Br(x) with f B s(y) : f f Br(x) B s(y) 2 = ω n r n ω n r n f(z) f Bs(y) dz 2 f(z) f Bs(y) 2 dz, arguing as before. Now choose r = x y and s = 2 x y, so that B r (x) B s (y), and then f f Br(x) B s(y) 2 sn r n ω n s n f(z) f Bs(y) 2 dz i.e. B s(y) 2 n M 2 2 2α x y 2α, f Br(x) f B s(y) c 2M x y α. So now (still with r = x y and s = 2 x y ) f(x) f(y) f(x) f Br(x) + f f B s(y) + f B s(y) f(y) 2c M x y α + c 2 M x y α = cm x y α. We can now prove Morrey s Lemma, which follows from Campanato s Lemma and the second form of the Poincaré inequality (.4). Corollary 2.6 Let f H (B R ()) and suppose that there exists a constant M > such that Df 2 dy Mr n 2+2α for every x B R/2 () and every r (, R/2). Then f C α (B R/2 ()).

33 2.5 Hölder regularity for parabolic problems and an elliptic proof 33 Proof Using (.4), f f Br(x) 2 dy cr 2 Df 2 dy Mr n+2α, and the result follows using the Campanato Lemma. 2.5 Hölder regularity for parabolic problems and an elliptic proof In Serrin s proof we will need the following result for the parabolic problem ω t ω = div g. We pick a point X = (x, t ) in Ω (, T ), and write Q R (X) = B R (x ) (t R 2, t ). (This parabolic cylinder is used, rather than a ball, since it obeys the right scaling for parabolic problems: if u(x, t) solves u t u = then so does u(λx, λ 2 t).) PR Theorem 2.7 Suppose that ω L (Q R (X )). Then (i) if g L (Q R (X )) then for any α (, ), ω C α (Q R/2 (X )); and (ii) if D k g C α (Q R (X )) then D k ω C α (Q R/2 (X )), for any k. We will prove an elliptic version of this result, using a version of the parabolic argument that can be found in Lieberman (996). We consider for simplicity the scalar equation w = div g. We also prove the equivalent of (ii) only for k =, but a similar argument works for higher derivatives. ER Theorem 2.8 Suppose that w L (B R (x )). Then (i) if g L (B R (x )) then for any α (, ), w C α (B R/2 (x )); and (ii) if g C α (Q R (x )) then Dw C α (B R/2 (x )).

34 34 2 Serrin s local regularity result We need an auxiliary lemma. Recall that if a function w is harmonic in a region U (i.e. w = in U) then for any r such that B r (x) U. w(x) = w Br(x) L44 Lemma 2.9 Suppose that w is harmonic in U. Then for any r < R such that B R (x) U, ( r n+2 w w Br(x) 2 = w w(x) 2 c w w R) Br(x) 2. B R (x) Proof Observe that sup w(x) w(x ) 2 r 2 x B r(x ) sup x B r(x ) Dw(x) 2. Since w is harmonic in U, so is Dw(x) = D(w(x) w(x )), from which it follows that D(w(x) w(x )) 2 2 = ω n R n D(w(y) w(x )) dx B R/2 (x) 2 = ω n R n (w(y) w(x ))n ds B R/2 c R 2 sup w(x) w(x ) 2. x B R/2(x) Now, the supremum of w(x) w(x ) 2 is attained at some x B R/2 (x), and since w(x) w(x ) is harmonic, w(x ) w(x ) = 2n ω n R n w(z) w(x ) dz, from whence w(x ) w(x ) 2 22n ωnr 2 2n 2n ω n R n 2n ω n R n B R/2 (x ) B R/2 (x ) B R/2 (x ) B R (x ) w(z) w(x ) dz w(z) w(x ) 2 dz w(z) w(x ) 2 dz. 2

35 2.5 Hölder regularity for parabolic problems and an elliptic proof 35 Putting these ingredients together we obtain w(x) w(x ) 2 dx ω n r n sup w(x) w(x ) 2 B r(x ) ω n r n+2 x B r(x ) sup x B r(x ) Dw(x) 2 ω n r n+2 c R 2 sup w(x) w(x ) 2 x B R/2(x) ω n r n+2 c 2 n R 2 ω n R n w(z) w(x ) 2 dz B R (x ) ( r n+2 = c w(z) w(x ) R) 2 dz. B R (x ) We now prove Theorem 2.8 Proof Choose x B R/2 (x ) and r (, R/2). Let v be the solution of u = in B r (x) with u = w on B r (x). Then the difference v = w u satisfies v = div g in B r (x) with v = on B r (x). If we take the inner product of this with v, then we obtain Dv 2 = g Dv, (2.9) backto since v = on B r. For case (i), we estimate this as Dv 2 ( g 2 ) /2 ( ) /2 Dv 2, so that Dv 2 g ω n r n. At this stage it is tempting to appeal to Morrey s Lemma to deduce that v C α ; but the function v is defined differently for each x and r, so we

36 36 2 Serrin s local regularity result cannot do this. Instead, we proceed as follows. First, we use the Poincaré inequality (remember that we have v = on B r ) to deduce that v 2 cr 2 Dv 2 c g ω n r n+2. (2.) vor We now use the triangle inequality to write w(y) u(x) 2 w(y) u(y) 2 dy + u(y) u(x) 2 dy. Appealing to Lemma 2.9 we have ( r n+2 u(y) u(x) 2 dy c u(y) u(x) R) 2 dy, B R (x) and hence w(y) u(x) 2 From (2.) we have ( r n+2 v(y) 2 dy + c R) ( r n+2 v(y) 2 dy + c R) ( r n+2 + R) B R (x) ( r n+2 c v(y) 2 dy + c R) ( r n+2 w(y) u(x) 2 g r n+2 + c R) = [ c g + c R n+2 B R (x) B R (x) B R (x) w(y) u(x) 2 dy B R (x) B R (x) w(y) 2 dy u(y) u(x) 2 dy w(y) u(y) 2 dy w(y) u(x) 2 dy. w(y) u(x) 2 dy ] r n+2. It is now safe to use the Campanto Lemma to deduce that w C α (B R/2 (x )). For (ii) we proceed similarly, but noting that for any x we can replace the right-hand side of our equation, div g, by div (g( ) g(x)). So we can write the right-hand side of (2.9) as [g i (y) g i (x)]d i v(y) dy

37 2.6 Serrin III: The Biot Savart law and parabolic regularity 37 and use the fact that g is Hölder continuous to estimate it as [g(y) g(x)] Dv dx g(y) g(x) 2 dy + Dv 2 2 c[g] 2 αr n+2α + Dz 2. 2 Thus we now obtain Dz 2 cr n+2α [g] 2 α. We now proceed exactly as before, but replacing v, w, and z by Dv, Dw, and Dz throughout the argument to conclude that Dw C α (B R/2 ()). 2.6 Serrin III: The Biot Savart law and parabolic regularity In the final part of the proof we will need to use the regularity we have for ω to deduce further regularity of u. Given ω = u, we can recover u using the Biot Savart law: u = 4π where A(x) is harmonic in G. G x ξ ω(ξ) dξ + A(x), x ξ 3 We will not study this here, but merely state the following for a proof see Chapter 4 of Majda & Bertozzi (22): (i) if ω L then u L ; (ii) if D k ω C α then D k u C α. We can now finish Serrin s proof. Theorem 2. Suppose that ω L 2 t (L 2 x) and u L s t (L s x) with 3 s + 2 s <. Then u(t) C (U) for all t (T, T 2 ). Proof We know from above that ω L (L ). Using the Biot Savart representation (BSL), it follows that u L (L ). Hence g L (L ).

38 38 2 Serrin s local regularity result Parabolic regularity now guarantees that ω C α. BSL implies that therefore Du C α. It follows that g C α. Parabolic regularity guarantees that Dω C α. BSL implies that therefore D 2 u C α. It follows that Dg C α. Parabolic regularity guarantees that D 2 ω C α. BSL... and so on, and therefore u C. This regularity result has been improved since Serrin s paper in a number of ways. Fabes, Jones, & Rivière (972) showed that the same result holds if 3 s + 2 s = for 3 < s <. Struwe (988) gave an alternative proof of this, also allowing the case s =. Takahashi (99) showed that one can replace the condition that the L s x norm is L s in time with the condition that it is weakly L s in time. A function f L s w if sup r> r s µ{x : f(x) > r} <. Note that /x / L, but /x L w. Takahashi shows that if 3 s + 2 s = 3 s and u L s w (L s x) is sufficiently small then u is C in space. He also proves the same result if u L t (L 3 x ) is sufficiently small. Kim & Kozono (24) extend Takahashi s result by showing that it is in fact sufficient for u to be small in L s w(l s w), i.e. in the weak Lebesgue spaces in both space and time. Finally, Escauriaza, Seregin, & Sverak (22) have recently shown that the result does still hold in the boundary case L t (L 3 x).

39 3 Partial regularity Caffarelli, Kohn, & Nirenberg (982) hereafter CKN proved that the one-dimensional parabolic measure of the set of space-time singularities of a suitable weak solution is zero. We now define all these terms. A point (x, t) is regular if u is essentially bounded in a neighbourhood of (x, t), i.e. if u L (U) where U is a neighbourhood of (x, t). A point (x, t) is singular if it is not regular. For a set X R 3 R and k, we define P k (X) = lim δ + Pδ k(x), where { Pδ k (X) = inf ri k : X } Q r i (x i, t i ) : r i < δ, i= i and Q r(x, t) = B r (x) (t r 2, t + r 2 ). We have P k (X) = if and only if for every δ > X can be covered by a collection {Q r i } such that i rk i < δ. A suitable weak solution is a weak solution that has a corresponding pressure p L 5/3 ((, T ) Ω) and satisfies a local form of the energy inequality. If assume that all terms are smooth and take the inner product of the Navier Stokes equations with uϕ, where ϕ is some C scalar cut-off function, we obtain (after an integration by parts) d u 2 ϕ u 2 ϕ t + u 2 ϕ u 2 ϕ 2 dt 2 2 u 2 (u )ϕ p(u )ϕ =. 2 39

40 4 3 Partial regularity If we integrate in time we obtain u 2 ϕ t + 2 u 2 ϕ u 2 [ϕ t + ϕ] + ( u 2 + 2p)(u )ϕ. (3.) LEI A suitable weak solution has to satisfy (3.) for every ϕ with compact support in (, T ) Ω. Note that just as we do not know whether every weak solution satisfies an energy inequality, we do not know if every weak solution is suitable. In fact, we do not know whether our Leray Hopf weak solutions are suitable. So we are required to look at a subclass of all possible weak solutions. Caffarelli, Kohn, & Nirenberg give a proof that suitable weak solutions exist in an appendix to their paper. Although P (S) = is CKN s headline result, they in fact prove two powerful theorems about local conditions for regularity of u. In what follows, we set Q r (x, t) = B r (x) [t r 2, t] (recall that Q r(x, t) = B r (x) [t r 2, t + r 2 ]). CKN Theorem 3. There exists a constant ɛ > such that if u is a suitable weak solution and for some r > r 2 u 3 + p 3/2 < ɛ (3.2) CKNcond Q r(x,t ) then u is essentially bounded on Q r/2 (x, t ). CKN s first theorem is not quite in this form; but you can find this in Lin (998) and Ladyzhenskaya & Seregin (999). CKN deduce from (3.5) that r 3 u(x, s) 2 ds C x a Recall from our discussion at the end of Chapter that any sensible smallness condition should be scale invariant, i.e. invariant under the scaling transformation u(x, t) λu(λx, λ 2 t) and p(x, t) λ 2 p(λx, λ 2 t). This is such a condition: if we consider r 2 Q r u 3 < ɛ and rescale u so that the domain of integration is Q (i.e. consider u λ (x, t) with λ = /r) we obtain r 2 Q r r u(x/r, 3 t/r2 ) dx dt = r 5 u(y, τ) 3 r 3 dy r 2 dτ = u 3 < ɛ. Q Q

41 Partial regularity 4 for every (a, s) Q r/2 (, ), and hence that u(a, s) 2 C at any Lebesgue point in Q r/2 (, ), i.e. almost everywhere in Q r/2 (, ). Lin (998) and Ladyzhenskaya & Seregin (999) show that away from the singular set u is in fact C α (Q r/2 (, )) (i.e. Hölder continuous in space and time) by showing that (3.5) implies that r 5 v v Qr 3 dx dt Cr α Q r for every r (, r/2), x Q r/2 (, ) and then using a parabolic version of the Campanato Lemma. (Lin is not explicit about the fact that this has to hold for a range of x and r; Ladyzhenskaya & Seregin are explicit about this step.) This theorem only allows one to show that P 5/3 (S) =. In fact it is also of the right form (the condition need only hold for some r > ) to allow one to show that d box (S) 5/3, using an argument similar to that used in Chapter for the set of singular times (the proof here is due to Robinson & Sadowski, 29). The argument relies in addition on the fact that the pressure associated with any weak solution satisfies p L 5/3 ((, T ) Ω), a result to due Sohr & von Wahl (986), and that any weak solution has u L /3 ((, T ) Ω). This follows using Hölder s inequality and the embedding H L 6 : indeed, we know that for any weak solution u L (, T ; L 2 ) and u L 2 (, T ; H ), so ( ) 2/3 ( ) /3 u /3 u 4/3 u 2 u 2 u 6 = u 4/3 u 2 L, 6 Ω Ω Ω Ω and so from whence T Ω Ω u /3 c u 4/3 u 2 H, ( ) ( T ) u /3 c sup u(t) 4/3 u 2 H <. (3.3) u3 <t<t Theorem 3.2 If u is a suitable weak solution of the Navier Stokes equations, then d box (S) 5/3. (Actually we should consider S K for some compact subset K of (, T ) Ω, since the parabolic cylinders must lie within (, T ) Ω for Theorem 3. to be valid, a subtlety that we have skipped in its statement.)

42 42 3 Partial regularity Proof Note that, using Hölder s inequality, we can bound the LHS of (3.5) by ( ) 9/ ( ) 9/ r 5/3 u /3 + r 5/3 p 5/3. So there is a constant ɛ such that if r 5/3 u /3 + p 5/3 < ɛ the conclusions of Theorem 3. hold. Now suppose that d box (S) > 5/3. Pick d with 5/3 < d < d box (S): then there exists a sequence ɛ j such that N(S, ɛ j ) > ɛ d j. As before, we use N(S, ɛ) to denote the maximum number of disjoint balls of radius ɛ with centres in S. It follows that on every ball u /3 + p 5/3 > ɛ ɛ 2 j, B ɛj and since these N(S, ɛ j ) ɛ d j balls are disjoint, u /3 + p 5/3 ɛ ɛ d+(5/3) j. Now if we let j, since d > 5/3 it follows that the LHS is infinite. But this is a contradiction, since we know that it is finite. It follows that d box (S) 5/3 as claimed. Kukavica (29b) has improved this bound to d box (S) 35/82. It is an open problem whether in fact d box (S). CKN obtain P (S) = by proving a second theorem: CKN2 Theorem 3.3 There exists a constant δ > such that if u is a suitable weak solution and lim sup u 2 < δ (3.4) Hlim r r Q r (x,t ) then r 2 u 3 + p 3/2 < ɛ (3.5) CKNcond Q r (x,t ) holds for some r >, and hence (x, t ) is regular.