Radu Dascaliuc 1. Department of Mathematics, Texas A&M University, College Station, TX 77843, USA

Ann. I. H. Poincaré AN 005) 385 40 www.elsevier.com/locate/anihpc On backward-time behavior of the solutions to the -D space periodic Navier Stokes equations Sur le comportement rétrograde en temps des solutions périodiques en espace des équations de Navier Stokes en dimension Radu Dascaliuc Department of Mathematics, Texas A&M University, College Station, TX 77843, USA Received 5 October 003; received in revised form 5 July 004; accepted 7 October 004 Available online 7 April 005 Abstract In 995, Constantin, Foias, Kukavica, Majda had shown that the -D space periodic Navier Stokes equations have a rich set of the solutions that exist for all times t R grow exponentially in Sobolev H norm when t. In the present note we show that these solutions grow exponentially when t ) in any Sobolev H m norm m ) provided the driving force is bounded in H m norm. 005 Elsevier SAS. All rights reserved. Résumé En 995 Constantin, Foias, Kukavica et Majda ont demontré que les équations de Navier Stokes périodiques dans R possèdent un ensemble ample des solutions qui existent pour tout temps t R et qui ont une croissance exponentielle pour t ) dans l espace de Sobolev H. Dans cet article nous montrons que ces solutions ont aussi une croissance exponentielle pour t ) dans tout espace de Sobolev H m m ) à condition que la force soit dans l espace de Sobolev H m. 005 Elsevier SAS. All rights reserved. MSC: primary 76D05; secondary 35B40, 35Q30, 35B05 Partially supported by the NSF grant DMS-039874. 094-449/$ see front matter 005 Elsevier SAS. All rights reserved. doi:0.06/j.anihpc.004.0.00

386 R. Dascaliuc / Ann. I. H. Poincaré AN 005) 385 40. Introduction One of the remarkable properties of the -D space periodic Navier Stokes equations is the richness of the set of initial data for which the solutions exist for all times t R increase exponentially as t. These solutions were studied in [3], where, among other results, it was proved that ut) is such a solution if only if its Dirichlet quotient A / ut) / ut) n as t here is the L -norm, A is the Stokes operator, n is one of its eigenvalues see Section for more precise definitions). The invariant set M n of all the trajectories of these solutions is proved to project entirely onto the spectral space associated with the first n eigenvalues of the Stokes operator cf. [3]). This fact implied the only known partial answer to the Bardos Tartar conjecture cf. [3]). This conjecture cf. []) affirms that the set of initial data for which solutions of the -D space periodic Navier Stokes equations exist for all times is dense in the phase space equipped with the energy norm a.e. the L -norm in this case). However, in [3] the density was proved in the norm A /. The paper [3] also raised a number of questions regarding the geometric structure of M n. For example, it would be interesting to investigate the relationship between these sets the other invariant sets of the Navier Stokes equations, namely the global attractor inertial manifolds. Another open question is whether n M n is dense in the energy norm of the phase space, which, if answered affirmatively, would solve the Bardos Tartar conjecture in the energy norm. The study of higher order quotients on the sets M n is of particular interest in this respect. In fact, a good result about boundedness of quotients of the form A α u / u β would imply the desired density result for n M n via the method presented in [3]. In this paper we prove that the quotients A α u / u 4α are bounded on any M n. cf. Theorem its Corollary ). Our bounds, however, are not sufficient to prove the density of n M n in the energy norm of the phase space. But as a corollary we show that if a solution of the -D space periodic Navier Stokes equation exists for all times increases exponentially in the energy norm as t ), than it increases exponentially in any Sobolev norm, provided the driving force is regular cf. Corollary ). In particular, the L norm of any derivative of such a solution grows at most exponentially as t. It is worth mentioning that by a slight modification of the proofs given in this paper one can prove similar results for the -D space periodic Navier Stokes α-model -D space periodic Kelvin-filtered Navier Stokes equations. Note that the analogs of the sets M n defined for these systems have very similar properties compared to the Navier Stokes case. In particular for the -D space periodic Navier Stokes α-model, n M n is dense in the L norm, which is still weaker than the energy norm for that system cf. [0]). On the other h, for the -D space periodic Kelvin-filtered Navier Stokes equations the density is proved in their energy norm cf. []). However, not all dissipative systems have the same kind of behavior for negative times. For example, in the case of the -D space periodic Kuramoto Sivashinsky equation it was established that all the solutions outside the global attractor will blow up backward in finite time cf. [6,7]). The other peculiar example is Burgers original model for turbulence. Although Burgers model has a rich set M, the solutions on it display some surprising dynamical differences from those in the Navier Stokes case cf. [4]). Still, it would be interesting to see whether the results similar to the ones presented in this note can also be proved for the set M of Burgers original model for turbulence.. Preliminaries We consider the -D space periodic Navier Stokes Equations NSE) in Ω =[0,L] : d u u + u )u + p = f, dt u = 0,

u, p Ω-periodic, Ω u = 0, R. Dascaliuc / Ann. I. H. Poincaré AN 005) 385 40 387 where ut) : R R, pt): R R are unknown functions >0, f L Ω) f is Ω-periodic, Ω f = 0) are given. Let H be the closure in L Ω) of { v L Ω) : vω-periodic trigonometric polynomial, v = 0, We denote v, w) := v w Ω v :=v, v) / Ω } v = 0. the inner product the norm in H. Let A = P L be the Stokes operator defined on DA) = H H Ω) ), where P L is the orthogonal projection from L Ω) onto H. Observe that A : DA) H is an unbounded positive self-adjoint operator with a compact inverse. Its eigenvalues are π/l) k + k ), where k,k ) N \{0, 0}. We arrange them in the increasing sequence: π/l) = < <. We will need the following fact about { n } cf. [8]). lim sup n+ n ) =. n Also, it is obvious that n+ n, n, lim n =. n Next we denote Bu,v) = P L v )w)) bu,v,w) = Bu, v), w), u, w H, v DA). Observe that bu,v,w) = bu,w,v), u H, v,w DA), bu, u, Au) = 0, u DA). We will also use the following inequality for b: bu,v,w) c 0 u / A / u / A / v w / A / w /, ) where u, v, w DA / )= H H Ω) ). Finally, denote g = P L f. Then the NSE can be written as d u + Au + Bu, u) = g. dt We denote by St)u 0 the solution of the NSE which is u 0 at t = 0. )

388 R. Dascaliuc / Ann. I. H. Poincaré AN 005) 385 40 Let { A = u 0 } St)H: supst)u 0 < 3) t 0 t R be the global attractor of Eq. ). Refer to [] or [9] for the comprehensive treatment of Eq. ). We will study the St)-invariant sets { M n = A u 0 A / St)u 0 St)H: lim sup t St)u 0 n + n+ := n }. 4) t 0 We will use the following known facts about M n cf. [3]). Theorem. The set n M n is dense in H with the topology of the norm A /. Also: If ut) M n \M n then A / ut) lim t ut) = n ; 5) ut) M n if only if ut) = Oe n t ), as t ; 6) If ut) M n \M n then ut) lim inf t e > 0. nt Moreover, if { } g u 0 γ 0 := max, then A / St)u 0 St)u 0 n, for all t 0. 7) 8) 9) 3. Main result For every θ 0 g DA θ ) define G θ = Aθ g θ+, the generalized Grashoff number. Our main goal is to prove the following 0) Theorem. Let θ = k/, k N\{0}, g DA θ ). Then for every u 0 M n such that u 0 γ 0, there exists a positive constant M θ G θ ) depending only θ, c 0 where c 0 the constant from )), G θ such that A θ u 0 u 0 4θ M θ G θ ) 4θ n θ. )

R. Dascaliuc / Ann. I. H. Poincaré AN 005) 385 40 389 Moreover, if θ> than there exists a positive constant N θ G θ / ), that depends only on θ, c 0, G θ / such that 0 A θ u u 4θ dτ <N θ G θ / ) 4θ 3 n θ, ) where ut) is a solution of the NSE satisfying ut 0 ) = u 0. Also, if θ / then A θ ut) lim t ut) 4θ = 0. Observe that ) exps the estimate 9) from Theorem to the quotients involving higher powers of the operator A. In fact, these estimates hold for any power of the operator. Corollary. Let α>/ g DA θ ), where θ = [α]+)/. Then for every u 0 M n with u 0 >γ 0, there is a constant M α depending only on θ, G [α]+)/, c 0 ) such that 3) A α u 0 u 0 4α M α 4α n α. 4) Proof. Let θ = [α]+)/. Observe that θ α. Then, by interpolation, A α u 0 u 0 4α A θ u 0 ) α )/θ ) A / u 0 u 0 4θ Mθ 4θ n θ thus, 4) holds with M α = M α )/θ ) θ. ) θ α)/θ ) u 0 ) α )/θ ) θ α)/θ ) n = Mα )/θ ) θ 4α α n, Another consequence of Theorem is that on M n any Sobolev norm of a solution will grow exponentially for negative time. Corollary. Suppose ut) M n \A g DA m/ ) then A m/ ut) Oe mnt ), t. Moreover, if m, then for any α = α,α ) with α,α 0, α + α m, we have where D α u L = Oe α +α +) n t ), t, D α ux,x ) = α +α u α x α. x In particular, when g C Ω), any solution u of the NSE which exists for all times increases exponentially as t in the phase space H, will also increase exponentially as t in any Sobolev space Hper m Ω) = Wper,m Ω) m 0). Moreover, the L norm of any space) derivative of u will also increase exponentially as t.

390 R. Dascaliuc / Ann. I. H. Poincaré AN 005) 385 40 Proof. Recall that the Sobolev norm in Hper m Ω) is equivalent to the norm m := A m/ ) /. Note that by Theorem ut) M n \A implies that ut) m grows at least exponentially as t, ut) = Oe nt ) as t. On the other h, according to Theorem, A m/ ut) M m/ m k n u m = Oe mnt ). Thus, ut) increases exponentially in Hper m Ω) as t. To prove the second part of the corollary we apply the Sobolev Embedding Theorem to obtain that D α u C D α u H Ω), for any multi-index α = α,α ) N. Here we are writing D α ux,x ) = α +α u α x α. x Observe that by the first part of the corollary, D α ut) H Ω) = Oe α +α +) n t ) as t. Consequently, we also have that D α u = Oe α +α +) n t ) as t. 4. The proof of the main result For convenience we will use the following notation: Notation. := A/ u u, µ := Au u 4, ξ := A ) u u, σ := A 3 ) A / u u, n := n+ + n, µ θ,m := Aθ u u m. First, we will prove the following useful lemma. Lemma. Let u be a solution of the NSE that exists for all times satisfies ut 0 ) >γ 0 for some t 0. Then for any t t 0 any m, 3m ut) m τ) uτ) m dτ m ut) m. 5)

R. Dascaliuc / Ann. I. H. Poincaré AN 005) 385 40 39 Also, if ut 0 ) M n \A, then τ) ξτ) τ n t) ) + g ut) 6) ) µτ) dτ O ut), for t. 7) Proof. From ) we obtain dt u + u = g, u), from which we get u m+ dt u + u m = Thus t u m dτ g, g, ) u u m+ ) u u m+. 8) dτ = m ut) m. 9) Notice that for t t 0 ) u g, u m+ dτ g dτ u m+ ) g u u m dτ since ut) γ 0 g ) t) for all t t 0. Thus, returning to 9) we get u m dτ u m dτ + u m dτ m ut) m u m dτ m ut) m for all t t 0, from which the relation 5) readily follows. In order to prove 6), we observe that dt A/ u + Au = g, Au), which, together with 8), implies that ) g dt + ξ = u,ξ, from which we obtain dt + ξ g u. u m dτ u m dτ, 0)

39 R. Dascaliuc / Ann. I. H. Poincaré AN 005) 385 40 By integrating the relation above, using 5) for m = as well as the fact that t) n as t cf. results from [3] summarized in Theorem ), we get t t) ) n + ξ dτ g ut), which implies the inequality 6) from the statement of the lemma. Finally, to prove 7) consider dt u = Au + g, Au) u 4 A / u + g, u) u u, from where g µ + dt u u µ/ + u + g u 3. Consequently d g + µ + 4 dt u u u + 4 g u 3. Thus, by integrating the previous inequality using 5) we obtain ) µτ) dτ O ut) for t. Let ut) be a solution of the NSE such that ut) M n. Our first result is Proposition. If g DA) u0) γ 0, then for every t 0 we have µt) + e 3 τ)µτ)dτ e4 n t) ) + K + 3/4) e4 n ut), where K = e 4 c 0 G 0 + G ) with c 0 the constant from the inequality ). Moreover, t 4 e 3 for any t 0. A 3/ u u 4 dτ e4 n t) ) + K + 3/4) e4 n ut), Proof. Observe that since g DA), wehave dt µ = A3/ u bau, u, Au) + Ag, Au) u 4 µ A/ u + g, u) u, so, dt µ = µ bau, u, Au) Ag, Au) g, u) 3/,4 µ) u 4 + u 4 µ u.

R. Dascaliuc / Ann. I. H. Poincaré AN 005) 385 40 393 Thus, dt µ + µ = σ + 9 3 bau, u, Au) 4 u u 4 ) Ag + u, Au ) ) u u µ g, u. Note that bau, u, Au) u 4 bau u,u,au u) = u 4 c 0 ξ A / ξ A / u u = c 0 / ξ A/ ξ = c 0 / ξ σ + µ 5 3 u 4 u thus bau, u, Au) u 4 c 0 ξ + σ + µ 5 3 4 u Now, going back to ) we get dt µ + µ σ + µ 5 3 8 u + 9 4 Observe that Ag u µ/ Ag 3 Consequently d µ + µ dt u + 3 [ 3 u 4 g, µ u. ). ) /, 3 u + c 0 ξ + Ag u µ/ µ g, )] u u µ σ + c 0 ξ + ) 3 u 4 3 + Ag 3. Note that the conditions of the proposition imply that t) n, for all t 0. Let us denote Γ n := 3 4 n + Ag 3, ) β := 3 u 4 u g, u. Then, by the Gronwall inequality, µt) µt 0 ) e t 0 β + t 0 t ) u u. σ µ + c 0 ξ + Γ ) n t u e τ β dτ. ) Observe that cf. Theorem, ut) lim inf t e > 0, t so βτ) is bounded absolutely integrable on the interval,t]. Moreover, by Lemma, c 0 /) ξ + Γ n / u is also absolutely integrable on,t]. On the other h, from 7) we conclude that there exists a sequence tn 0 such that µt0 n ) 0. Thus, by taking t 0 = tn 0 letting n, the inequality ) yields: µt) + c σ + µ ) t c0 dτ c ξ + Γ ) n u dτ, 3)

394 R. Dascaliuc / Ann. I. H. Poincaré AN 005) 385 40 where c t) = inf τ t e τ β, τ c t) = sup e β. τ t Observe that the relation 5) from Lemma implies that Γ n dτ Γ n u ut). Using this, together with 6), in the inequality 3), we obtain µt) + c σ + µ ) dτ c c 0 n t) ) ) + g ut) + c Γ n ut). 4) Observe that from 4) we can infer that µt) is bounded, while σ µ are integrable on,t], thus c A 3/ u u 4 dτ c Using 4) again to estimate c 3µ dτ, we obtain c A 3/ u u 4 dτ c c 0 n t) ) c0 g + 3 Observe that from 5) we obtain that Hence, 3 dτ u ut). c > e 0 4 g / u ) e 6 g / u0) ) e 3, since u γ 0, 3µ dτ + c c 0 n t) ) c0 g ) c + 3 + Γ n ut). c e 3 / u +4 g / u ) e / ut) +6 g / ut) ) e +3 = e 4. Finally, if we define K G 0,G ) := e 4 c 0 G 0 + G ), ) 4c + Γ n <. 5) ut) use 4) 5) we will obtain the desired estimates from the proposition. The following proposition allows us to deduce the boundedness of higher order quotients based on the boundedness on the lower ones.

R. Dascaliuc / Ann. I. H. Poincaré AN 005) 385 40 395 Proposition. Let g DA θ+/ ) with θ = k/ k N. Suppose that for every ut) M n, ut 0 ) γ 0 we have 0 A θ ut) ut) m dt <C θ,mg θ / ) m n k, where C θ,m ) is a positive increasing function. Then there exist positive increasing functions C θ,m ), K θ,m ), such that A θ ut 0 ) 0 ut 0 ) m+ + A θ+/ ut) ut) m+) dt< C θ,m G θ ) m A θ+/ ut) ut) m+4) K θ,mg θ+/ ) m+ k+ n. Moreover, A θ+/ ut) A θ ut) lim t ut) m+4) = lim = 0. t ut) m+) n k Proof. Using ) we get the following equation for the Galerkin approximations u N cf. [] for the facts about the Galerkin approximations for the NSE) dt µn θ,m+ = Aθ+/ u N + g, A θ u N ) bu N,u N,A θ u N ) u N m+) + m + µ N A / u N g, u N ) θ,m+ u N. Applying Theorem 3 from the Appendix as well as the Cauchy Schwarz inequality, we get dt µn θ,m+ µn θ+/,m+ + Aθ g u N m+)/ µn / A θ+/ u N A θ u N A / u N θ,m+ + c0 c θ u N m+) + m + µ N θ,m+ + m + g u N µn θ,m+ here c θ = 6[θ]+θ [θ]) θ ) is the constant from Theorem 3). Now, using the Jensen inequality, we obtain dt µn θ,m+ µn θ+/,m+ + A θ g u N m+) + µn θ,m+ + c 0 c θ N µ N θ,m + m + N µ N θ,m+ + m + g u N µn θ,m+. Hence, d dt µn θ,m+ + µn θ+/,m+ A θ g ) u N m+) + u N + c 0 c θ N m + )N m + ) g + u N + u N 3 µ N θ,m. Since g DA θ+/ ), we can integrate from t to t 0 t<t 0 ) pass to the limit N. Taking into the account that t) n we get

396 R. Dascaliuc / Ann. I. H. Poincaré AN 005) 385 40 Since µ θ,m+ t 0 ) + 0 0 [ c + 0 cθ n + t µ θ+/,m+ dτ µ θ,m+ t) + Aθ g ut 0 ) + m + ) n + µ θ,m dτ< C θ,mg θ / ) m n k, there exists a sequence t l such that lim µ θ,mt l ) = 0 = lim µ θ,m+t l ) ). l l Thus, by letting t = t l, we get µ θ,m+ t 0 ) + Aθ g 0 0 µ θ+/,m+ dτ dτ u m+ + [ c 0 cθ n + Hence, lim µ θ,m+t) = 0. t Moreover, since according to Lemma, 0 we obtain dτ u m+ m + µ θ,m+ t 0 ) + Aθ g 0 m + ut 0 ) m+, µ θ+/,m+ dτ γ m+ 0 0 t dτ u m+ )] m + ) g t 0 ut 0 ) t µ θ,m dτ. ut 0 ) + m + ) n + [ c + 0 cθ n + γ0 + m + ) n + )] m + ) g t 0 µ θ,m dτ. ut 0 ) )] m + ) g Cθ,m G θ / ) γ 0 m k n. Observe that by the Poincaré inequality, G θ >G θ /. Thus C θ,m G θ / ) C θ,m G θ ). Using this fact, together with the definition of γ 0, we can define the positive increasing functions C θ,m G θ ) from the statement of the proposition as follows: A θ g m + [c 0 cθ + + m + ) + g γ 0 m + ) +k γ m+ γ 0 0 = m + G θ c + 0 cθ + 3 ) m + 4 C θ,m G θ / ) m + G θ + c 0 cθ + 3 m + 4 ) C θ,m G θ ) := C θ,m G θ ). ))] C θ,m G θ / )

R. Dascaliuc / Ann. I. H. Poincaré AN 005) 385 40 397 On the other h, again for the Galerkin approximations, we have dt µn θ+/,m+4 = Aθ+ u N + g, A θ+ u N ) bu N,u N,A θ+ u N ) u N m+4) + m + 4 µ N A / u N g, u N ) θ+/,m+4 u N, similarly to what was done above we get µ θ+/,m+4 t 0 ) + [ c + 0 cθ+ n + 0 t µ θ+,m+4 dτ µ θ+/,m+4 t) + Aθ+/ g ut 0 ) + m + 4) n + g ))] 0 ut 0 ) t 0 t dτ u m+4) µ θ+/,m+ dτ here again c θ+ is the constant from Theorem 3). By the same argument as in the previous case, when t we obtain µ θ+/,m+4 t 0 ) + [ c + 0 cθ+ n + 0 µ θ+,m+4 dτ Aθ+/ g ut 0 ) 0 + m + 4) n + dτ u m+4) g ut 0 ) ))] 0 Thus lim µ θ+/,m+4t) = 0, t µ θ+/,m+4 t) K θ,m k+ m+ n, where K θ,m G θ+/ ) := m + 4 G θ+/ c + 0 cθ+ + 3 ) m + 7 C θ,m G θ+/). Observe that K θ,m satisfies conditions from the proposition, since A θ+/ g m+4 0 m + 4) 4 k+3 γ m+4 + [c cθ+ + γ + m + 4) 0 0 = m + 4 G θ+/ + + g γ 0 c 0 cθ+ + 3 m + 7 ) C θ,m G θ ) K θ,m G θ+/ ). µ θ+/,m+ dτ. ))] C θ,m G θ ) Proof of the main theorem. We will prove Theorem by induction on k = θ. When k = the theorem holds cf. 9)). When k = the theorem is valid via Proposition. Observe that this proposition allows us to choose, for example, M G ) = c 0 + )G + 5 ) e 4. 4

398 R. Dascaliuc / Ann. I. H. Poincaré AN 005) 385 40 Moreover, Proposition gives us that N 3/ = 4e 3 M. Thus, applying Proposition, we conclude that the theorem holds when k = 3. Suppose now that the theorem is true for some integer k 3. Then there exists a positive increasing function N θ ), such that 0 A θ u u 4θ dτ <N θ G θ / ) 4θ 3 n θ, where θ = k/. But according to Proposition, if g DA θ+/ ), then there exist positive increasing functions M θ+/ ) N θ+/ ) such that 0 A θ+/ ut) ut) 4θ dt< N θ+/g θ ) 4θ θ n, A θ+/ u 0 u 0 4θ+ M θ+/g θ+/ ) 4θ n θ+, A θ+/ ut) lim t ut) 4θ+ = 0, which shows that the theorem is true for the integer k +, by induction, the proof is complete. Acknowledgements The author would like to thank Professor Ciprian Foias for helpful discussions, suggestions, comments. Appendix. Estimate for the nonlinear term cf. [5]) Lemma. For each n N n ) every u DA n ): n bu, u, A n u) = ba h u, u, A n h u). 6) h= Proof. Observe first that A Bu,v) + Bv,u) ) = Bu, Av) + Bv, Au) BAu, v) BAv,u). Thus, if n is odd, then n n+)/ A BA h u, A n h u) = A BA h u, A n h u) + BA n h u, A h u) ) n+)/ = n+)/ BA h u, A n h+ u) + BA n h u, A h+ u)

R. Dascaliuc / Ann. I. H. Poincaré AN 005) 385 40 399 n+)/ n+)/ BA h+ u, A n h u) BA n h+ u, A h u) = Bu,A n+ u) BA n+ u, u). Consequently, in this case, ) n n n ba h u, u, A n h u) = ba h u, A n h u, u) = A BA h u, A n h u), A u = bu,a n+,a u) + ba n+ u, u, A u) = bu,a u, A n+ u) ba n+ u, A u, u) = 0, since bav,v,w)= bw,v,av) for every v DA) w H. If n is even, we have n n/ A BA h u, A n h u) = ABA n u, u) + A BA h u, A n h u) + BA n h u, A h u) ) = ABA n u, u) + Bu,A n u) BA n u, u). Thus, ) n n n ba h u, u, A n h u) = ba h u, A n h u, u) = A BA h u, A n h u), A u = ABA n u, u), A u ) bu,a n+ u, A u) + ba n+ u, u, A u) = ba n u, u, u) + 0 = 0. Consequently the identity from the lemma holds for all n. Theorem 3. For each n N n ) every u DA n ): bu, u, A n u) c0 c n A n/ u A n+)/ u A / u,.7) where c n := 6[n/]+n [n/]) n ). Proof. Observe that going to the Fourier coefficients: bu,v,w) = a k j)b j c l ) a k j b j c l := bu,v,w), j+k+l=0 Z j+k+l=0 Z where ux) = k Z a k e πi/l)k x), vx) = j Z b j e πi/l)j x), wx) = l Z c l e πi/l)l x), with u, w H, v V. Using the previous lemma we get bu, u, A n u) n ba h u, u, A n h u) n ba h u, u, A n h u). Observe that h= h=

400 R. Dascaliuc / Ann. I. H. Poincaré AN 005) 385 40 n h= where n A := n ba h u, u, A n h u) = n B := n C := h= j+k+l=0 Z, k min{ j, l } h= j+k+l=0 Z, j min{ k, l } h= j+k+l=0 Z, l min{ j, k } k h a k j a j a l l n h) h= j+k+l=0 Z A + B + C, k h a k j a j a l l n h), k h a k j a j a l l n h), k h a k j a j a l l n h). Because of the symmetry we have that A = C. Also, n B k h a k j n h) a j a l l =C= A), thus we get h= j+k+l=0 Z, k min{ j, l } n bu, u, A n u) 3B 3 h= [n/] = 6 j+k+l=0 Z, j l k h= j+k+l=0 Z, j l k + 6 [n/] 6 n h=[n/]+ j+k+l=0 Z, j l k h= j+k+l=0 Z, j l k + 6 n h=[n/]+ j+k+l=0 Z, j l k k h a k j a j a l l n h) k h a k j a j a l l n h) k h a k j a j a l l n h) k n a k j a j a l l n Observe that in the previous sums, k = j + l j + l l. Thus, 6 n h=[n/]+ j+k+l=0 Z, j l k 6 n h=[n/]+ j+k+l=0 Z, j l k k h a k j a j a l l n h) k h a k j a j a l l n h). k n l ) h n ak j a j a l l n h)

Also, [n/] 6 6 n [n/] ) n h= j+k+l=0 Z, j l k 6[n/]) R. Dascaliuc / Ann. I. H. Poincaré AN 005) 385 40 40 j+k+l=0 Z, j l k j+k+l=0 Z, j l k k n a k j a j a l l n Consequently, bu, u, A n u) 6 [n/]+ n [n/] ) n ) k n a k j a j a l l n. k n a k j a j a l l n. j+k+l=0 Z, j l k Let c n = 6[n/]+n [n/]) n ). From the above we conclude that bu, u, A n u) c n ba n/ u, u, A n/ u) = c n φx)ψx)ζx)dx, Ω k n a k j a j a l l n. where we denote φx) = k eπi/l)k x a k k n, ψx) = j eπi/l)j x a j j, ζx) = l eπi/l)l x a l l n. Applying Schwartz inequality we get bu, u, A n u) c n φ L 4 ψ L ζ L 4. Now apply Ladyzhenskaya inequality w L 4 c 0 w H w L to estimate φ L 4 ζ L 4 obtain bu, u, A n u) c 0 c n A n/ u A n+)/ u A / u. References [] C. Bardos, L. Tartar, Sur l unicité rétrograde des équations parabolique et quelques questiones voisines, Arch. Rational Mech. 50 973) 0 5. [] P. Constantin, C. Foias, Navier Stokes Equations, Chicago Lectures in Math., University of Chicago Press, Chicago, 988. [3] P. Constantin, C. Foias, I. Kukavica, A. Majda, Dirichlet quotients -d periodic Navier Stokes equations, J. Math. Pures Appl. 76 997) 5 53. [4] R. Dascaliuc, On the backward-time behavior of Burgers original model for turbulence, Nonlinearity 6 6) 003) 945 965. [5] C. Foias, B. Nicolaenko, Some estimates on the nonlinear term of the Navier Stokes equation, Preprint, 003. [6] I. Kukavica, On the behavior of the solutions of the Kuramoto Sivashinsky equations for negative time, J. Math. Anal. Appl. 66 99) 60 606. [7] I. Kukavica, M. Malcok, Backward behavior of solutions of Kuramoto Sivashinsky equation, 003, submitted for publication. [8] I. Richards, On the gaps between numbers which are sums of two squares, Adv. in Math. 46 98). [9] R. Temam, Navier Stokes Equations Nonlinear Functional Analysis, SIAM, Philadelphia, 983. [0] J. Vukadinovic, On the backwards behavior of the solutions of the D periodic viscous Kamassa Holm equations, J. Dynam. Differential Equations 4 ) 00). [] J. Vukadinovic, On the backwards behavior of the solutions of the Kelvin-filtered D periodic Navier Stokes equations, Ph.D. Thesis, Indiana University, 00.