Basic for the proofs are the L p? L q -estimates for the Stokes operator A =?P 4 of Iwashita [6], rened by Maremonti-Solonnikov [8] and Dan-Kobayashi-

Transcription

1 Decay estimates for strong solutions of the Navier-Stokes equations in exterior domains M. WIEGNER Department of Mathematics I, RWTH Aachen, 5256 Aachen, Germany 1. Introduction: Let IR n ; n 3, denote an exterior domain. We are interested in strong solutions of the Navier-Stokes equations u t? 4u + (ur)u + rp = div u = u = with some initial value u(x; ) = a(x). Since the fundamental results on L p? L q -estimates by Iwashita 1989 [6] for the Stokes semigroup on exterior domains it is known, that there exists a global unique smooth solution provided the L n -norm of a is suciently small. It is known that Leray-Hopf solutions fulll this requirement after a certain time, provided n = 3 or 4; hence all asymptotic estimates are valid for these. On the other hand, this strong solution (with small L n -data) coincides with a Leray-Hopf solution from the beginning, if a is additionally in L 2. In this case a lot of results on the asymptotic behaviour of the energy (i.e. the L 2 -norm) is known; for the Cauchy-problem see e.g. Schonbek [9], Wiegner [1]; and Borchers-Miyakawa [2], [3] for exterior domains. We are now interested in the problem, how the various L p -norms of these solutions behave asymptotically in time. Estimates for the Cauchy-problem and p > n were given e.g. in Wiegner [11], [12]. Further let us mention in this context some estimates for p > n ; n = 3; 4 n?1 by Kozono, Ogawa and Sohr [7]; we shall improve and generalize their results. As an application, these estimates have turned out to be useful in deriving estimates for higher-order derivatives in the L 2 -norm (Wiegner [14]), thereby allowing to get pointwise estimates in space-time (Amrouche, Girault, M.E. and T.P. Schonbek [1]).

2 Basic for the proofs are the L p? L q -estimates for the Stokes operator A =?P 4 of Iwashita [6], rened by Maremonti-Solonnikov [8] and Dan-Kobayashi-Shibata [5]: ke?ta ak q ckak p t?n=2(1=p?1=q) and kre?ta ak q ckak p t?n=2p+(n=2q?1=2) for t 1 resp. ckak p t?n=2p+(n=2q?1=2) + for t 1 for 1 p < q 1 and for p = q, if 1 < p < 1. In the case of the Cauchyproblem, we may forget the "+" in the exponent, which improves the gradient estimates. The main result of this paper are the following asymptotic estimates. Main Theorem: If 1 p n; a 2 L p \ L n and kak n "(n; ), there is a unique global strong solution, which fullls for p q 1 the estimates ku(t)k q " p (t)t?n=2(1=p?1=q) t 1=2 kru(t)k q " p (t)t?n=2(1=p?1=q)+n=2(1=n?1=q) + tkau(t)k q " p (t)t?n=2(1=p?1=q) with " p (t) ckak p and (for p > 1) lim t!1 " p(t) =. 2. L q -estimates for q n. The global strong solution u(t) is given as the limit of the iteration u j+1 (t) = u (t)? e?(t?s)a P ((u j (s) r)u j (s))ds with u (t) = e?ta a, where a = P a 2 L n is suciently small. In fact in Wiegner [13], it was shown that the smallness of R := maxfsup t> t 1=2 kru (t)k n ; sup t (1?)=2 ku (t)k n= g t> is sucient, where 2 (; 1) may be choosen arbitrarily. Note, that by the L p? L q -estimates, we always have R ckak n. In the case of = IR n (and IR n +) the improved L p? L q -estimates allow for a slight modication in the proof, reducing the smallness assumption to the second quantity of R, which is nothing else than a certain Besov space norm (for an elaboration of this, see e.g. Cannone, Planchon, Schonbek [4].) In order to simplify the statements, let R 1 := kak n ; = 1. Then from [13], x 6, we know 2 the existence of a strong solution u(t), provided R 1 "(n; ) is small (independent of p). 2

3 All other constants may depend on n; p and and are denoted by c, though they might be dierent. For all t >, we have further the following estimates (compare [13]): (2:1) ku(t)k n cr 1 (2:2) t 1=4 ju(t)k 2n cr 1 (2:3) t 1=2 kru(t)k n cr 1 (2:4) t 3=8 ku(t + h)? u(t)k 2n cr 1 h 1=8 (2:5) t 5=8 kru(t + h)? ru(t)k n cr 1 h 1=8 Moreover for T, we may represent the solution for t T by (2:6) u(t) = e?(t?t )A u(t )? The rst lemma generalizes now (2.1), (2.2) and (2.4). T e?(t?s)a P ((u(s)r)u(s))ds Lemma 1: There holds (2:7) t 1=2 ku(t)k1 cr 1 and (2:8) t 5=8 ku(t + h)? u(t)k1 cr 1 h 1=8 Proof: Let T = in (2.6) and estimate ku(t)k1 ct?1=2 ku()k n (t? s)?3=4 ku(s)ru(s)k 2n=3 ds ct?1=2 R 1 (t? s)?3=4 ku(s)k 2n kru(s)k n ds ct?1=2 R 1 (t? s)?3=4 s?3=4 R 2 1ds ct?1=2 R 1 ; 3

4 ku(t)k1 cr 1 for all t. Next, with f(s) := P ((u(s)r)u(s)), we have the represen- hence t tation u(t + h)? u(t) = (u (t + h)? u (t))? For t h, we estimate h e?(t+h?s)a f(s)ds? ku(t + h)? u(t)k1 ku (t + h)? u (t)k1 h (t + h? s)?3=4 kf(s)k 2n=3 ds The second term is bounded by R cr1t 2?3=4 h 1=4 s?3=4 ds cr1t 2?1=2 h t cr 2 1t?5=8 h 1=8 and the third term with (2.2) up to (2.5) by cr 2 1 (t? s)?3=4 kf(s + h)? f(s)k 2n=3 ds (t? s)?3=4 s?7=8 h 1=8 ds = cr 2 1t?5=8 h 1=8 e?(t?s)a (f(s + h)? f(s))ds as f(s + h)? f(s) = P (u(s)(ru(s + h)? ru(s)) + P ((u(s + h)? u(s))ru(s + h)). Further ku (t + h)? u (t)k1 ct?1=4 k(e?ha? I)e?A ak 2n ct?1=4 hkae?a ak 2n ct?5=4 hke?t=4a ak 2n ct?3=2 hr 1 cr 1 t?5=8 h 1=8 On the other hand, if t h, the claimed estimate follows simply by (2.7). By interpolation, we get the following Corollary 1: (2:9) t n=2(1=n?1=q) ku(t)k q cr 1 for n q 1: We shall need the Corollary 2: For f(s) = P ((u(s) r)u(s)), we have (2:1) tkf(t)k n cr 2 1 4

5 (2:11) t kf(t + h)? f(t)k n cr 1 h Next, we want to improve (2.11) for large times. Lemma 2: If < 1, then (2:12) kf(t + h)? f(t)k n c t?1?=2 h =2 R 2 1 for t T Proof: Given < 1, let " := (1? )=2 and := + " < 1. Fix h; T >. Let As S (u) : = sup t (1?")=2 h?=2 ku(t + h)? u(t)k n= ; <tt S 1 (u) : = sup t (1+)=2 h?=2 kru(t + h)? ru(t)k n and <tt S(u) = maxfs (u); S 1 (u)g u(t + h)? u(t) = u (t + h)? u (t) + + h e?(t+h?s)a f(s)ds e?(t?s)a (f(s + h)? f(s))ds; we get with kf gk n=(1+) kfk n= kgk n for h t T ku(t + h)? u(t)k n= ku (t + h)? u (t)k n= h (t + h? s)?1=2 ku(s)k n= kru(s)k n ds (t? s)?1=2 kf(s + h)? f(s)k n=(1+) ds h ku (t + h)? u (t)k n= t?1=2 R 2 1 s?1+=2 ds (t? s)?1=2 S(u)R 1 h =2 s?1+"=2 ds ku (t + h)? u (t)k n= R1t 2?1=2+"=2 h =2 S(u)R 1 h =2 t?1=2+"=2 5

6 For t h, we simply have t (1?")=2 ku(t + h)? u(t)k n= 2 sup t cr 1 h =2 ; n ku(t)kn=t (1?)=2 o t =2 hence S (u) S (u ) R 1 R 1 S(u) Similar estimates are valid for ru, the dierence being only, that the order of the singularity is?(1 + )=2 instead of?1=2 (which does not allow to take = 1 at this place). Hence S(u) S(u ) R 1 R 1 S(u): Now, for h t, ku (t + h)? u (t)k n= hkae?ta ak n= h t ke?a ak n= cht?3=2+=2 R 1 c h t! 1?=2 h =2 t?(1?")=2 R 1 and for t h t (1?")=2 ku (t + h)? u (t)k n= c t (1?")=2 t?(1?)=2 kak n c h =2 R 1 : Similarly, kre?(t+h)a a? re?ta ak n kre?a (e?(+h)a a? e?a a)k n c t?1=2 hkae?a ak n c t?3=2 h R 1 for t h and again a trivial estimate for t h. Thus S(u ) c R 1, and the inequality c R 1 < 1 2 implies S(u) c R 1. As h and T was arbitrary, we get for all t; h ku(t + h)? u(t)k n= c R 1 h =2 t?(1?")=2 kru(t + h)? ru(t)k n c R 1 h =2 t?(1+)=2 provided c R 1 < 1. In the same way, as (2.8) followed from (2.4) and (2.5), these two 2 estimates imply now ku(t + h)? u(t)k1 c R 1 h =2 t?(1+)=2 and we conclude kf(t + h)? f(t)k n c R 2 1h =2 t?1?=2 provided c R 1 < 1 2. As ku(t)k n! (see x 3), we may have started at ~a = u(t ) with some T > in order to 6

7 fulll this smallness condition, and the claim follows. Now applying the gradient to (2.6), we may generalize (2.3) and provide additionally a Hlderestimate, complementary to (2.12). Lemma 3: For n q 1, we have (2:13) t 1=2 kru(t)k q c R 1 maxf1; t?1=2(1?n=q) g (2:14) kf(t + h)? f(t)k q c R 2 1h 1=8 t?13=8+n=2q for t T Proof: For t 2, we get for q < 1 kru(t)k q c t?1=2 ku()k n t?1 t?1 (t? s)?1=2 kf(s)k n ds (t? s)?1+n=2q kf(s)k n ds c t?1=2 R 1 by (2:1); while for t 2 kru(t)k q c t?1+n=2q R 1 (t? s)?1+n=2q kf(s)k n ds c R 1 t?1+n=2q If q = 1, we may use (2.13) for q = 2n and estimate instead by for t 2, while for t 2 (t? s)?3=4 kf(s)k 2n ds cr 1 t?1 t?1 (t? s)?3=4 s?1 ds cr 1 t?1 (t? s)?3=4 kf(s)k 2n ds cr 1 (t? s)?3=4 s?1=2 s?3=4 ds cr 1 t?1 In order to prove (2.14), we get for h t by (2.6) and (2.11) kru(t + h)? ru(t)k q c t?1+n=2q k e?(t=4+h)a? e?t=4a u()kn 7

8 ( + h? s)?1+n=2q kf( + s)k n ds (t? s)?1+n=2q kf(s + h)? f(s)k n ds c t?2+n=2q hku()k n R 2 1h t?2+n=2q R 2 1 (t? s)?1+n=2q t?9=8 h 1=8 ds c R 1 h 1=8 t?9=8+n=2q and the same estimate by (2.13) for t h. Again, if q = 1, we estimate instead for t T by c (t? s)?3=4 kf(s + h)? f(s)k 2n ds c R 2 1h 1=8 c R 2 1h 1=8 t?9=8 (t? s)?3=4 s?13=8+1=4 ds Recalling (2.8) and (2.13), the claim follows. Lemma 4: For n q 1 Proof: We have the representation kau(t)k q ct?1 ku(t=8)k n (t?1=2(1?n=q) + t 1=2 kru(t)k q ): Au(t) = Ae?3=4tA u(t=4)? Ae?(t?s)A (f(s)? f(t))ds? Ae?(t?s)A f(s)ds? (I? e?a )f(t) t=4 We estimate the L q -norms seperately. If q < 1, the rst term is bounded by ct?1 ku(t=4)k q ct?3=2+n=2q ku(t=8)k n ; the third by ct?1 ke?(t?s)=2a f(s)k q ds t=4 8

9 ct?3=2+n=2q kf(s)k n ds t=4 ct?3=2+n=2q ku(t=8)k 2 n and the last by ckf(t)k q ct?1=2 ku(t=8)k n kru(t)k q : For the second term we distinguish whether t is large or small. If t is large, we use (2.12) and bound the term by c cr 2 1 (t? s)?1 ke?(t?s)=2a (f(s)? f(t))k q ds (t? s)?1?1=2(1?n=q) t?1?=2 (t? s) =2 ds cr 2 1t?3=2+n=2q ; if we choose > 1? n q : Note, that we may again change R 1 to ku(t=8)k n. For t small we use (2.14) and estimate by c cr 2 1 (t? s)?1 kf(s)? f(t)k q ds (t? s)?7=8 t?3=2+n=2q?1=8 ds cr 2 1t?3=2+n=2q : If q = 1, we use the estimate tkae?ta vk1 t?n=2p kvk p from [8], theorem 1.2, and proceed similarly. We collect all estimates in the Theorem 1: If kak n "(n; ), there exists a unique global smooth solution u, which fullls for n q 1 u 2 C 1 ((; 1); L q ) \ C ([; 1); L n ), Au 2 C ((; 1); L q ) and ku(t)k q ct?1=2(1?n=q) R 1 t 1=2 kru(t)k q cr 1 maxf1; t?1=2(1?n=q) g tkau(t)k q ct?1=2(1?n=q) R 1 t 1=2 kru(t)k q R 1 with R 1 = ku(t=8)k n. 9

10 3. L p -estimates for 1 < p < n. If we want to derive further estimates for the solutions above, e.g. L p -estimates for p < n, we have to assume that a 2 L p, as is unbounded. The following theorem shows, that then the L p -norm of the solution even tends to zero, thereby answering an old question of Leray in a general context. Theorem 2: Let 1 < p n, and assume additionally a 2 L p. Then the solution fullls (3:1) ku(t)k p ckak p for t > and (3:2) lim t!1 ku(t)k p = Proof: I. Suppose rst, that p n, which implies that np n + p. If u n?1 j(t) denote the functions from the iteration procedure, we put M j := sup ku j (t)k p, with M ckak p. Then with (2.3), t> holding for u j as well, ku j+1 (t)k p ku (t)k p ku (t)k p M R 1 M j : (t? s)?1=2 ku j (s)ru j (s)k np=(n+p) ds (t? s)?1=2 s?1=2 R 1 ku j (s)k p ds As we may assume cr 1 < 1, we get by induction M 2 j 2M for all j. Similarly, for D j = sup ku j+1 (t)? u j (t)k p, one gets t> D j+1 cr D j M supft 1=2 kru j+1 (t)? ru j (t)k n g t> and from the proof of theorem 6.1 in [13], we know, that D j+1 cr D j M 2?j. By induction D cj2?j!, hence u j (t) converges in L p, and we conclude Similarly for t T 1 : ku(t)k p ku (t)k p ku(t)k p ckak p for all t: T 1 (t? s)?1=2 s?1=2 R 1 M ds 1

11 T1 (t? s)?1= 2 s?1= 2 R 1 ds sup tt 1 ku(t)k p Let M(T ) = sup tt ku (t)k p (T 1 =(t? T 1 )) 1=2 M R 1 R 1 sup ku(t)k p tt 1 ku(t)k p ; M (T ) = sup ku (t]k p and k >. The above estimate implies tt M((k + 1)T 1 ) M (T 1 ) k?1= M(T 1) With T 1 = (k + 1) j T and induction, we conclude As M (t)!, this implies M(t)! too. M((k + 1) j T ) 2M (T ) + 2ck?1=2 + 2?j M(T ) II. Consider now the case 1 < p < n=(n? 1). As especially p < 2, we know that u is a Leray-Hopf solution with energy-inequality Hence ku(t)k ku(t)k p ckak p ckak p kak 2 2( kru(s)k 2 2ds kak 2 2: (t? s)?n=2(1?1=p) ku(s)k 2 kru(s)k 2 ds (t? s)?n(1?1=p) ds) 1=2 ckak p kak 2 2t 1=2?n=2(1?1=p) ; as n(1? 1=p) < 1. By interpolation, kak 2 kak p kak 1? n ckak p with (1=p? 1=n) = 1=2? 1=n, and we see, that for t t := kak 2p=(n?p) p, we get kak 2 2t 1=2?n=2(1?1=p) ckak p, hence (3.1) for t t. Next let T = t in (2.6) to get for t t ku(t)k p cku(t )k p t ku(s)k 2p kru(s)k 2p ds: As n=(n? 1) 2p n for p < n=(n? 1); n 3, we may use the estimates based on part I; by (3.5), (3.4) below we have kru(s)k 2p cs?1=2 ku(s=2)k 2p and ku(s)k 2p cs?n=2((n?1)=n?1=2p) ku(s)k n=(n?1) for some < 1: 11

12 As t s?1=2?n((n?1)=n?1=2p) ds ct 3=2?n+n=2p due to 1=2 + n((n? 1)=n? 1=2p) > 1, we get ku(t)k p cku(t )k p ku(~t )k 2 n=(n?1)t 3=2?n+n=2p : Again by interpolation, the denition of t and the estimate for t t, we get (3.1). Precisely the same reasoning for arbitrary t 1 implies from which (3.2) follows. ku(t)k p ke?(t?t 1)A u(t 1 )k p t 3=2?n+n=2p 1 for t t 1 Remark: If the Stokes semigroup is also bounded in L 1 (e.g. for = IR n ), the same proof works for p = 1 too (instead of n=n? 1, one has to take now some q with 1 < q < n=(n? 1) in (3.4). The additional information on the initial value implies stronger asymptotic decay in higher norms. Generalizing (2.9), we shall prove Theorem 3: In the situation of theorem 2, the following holds: For each q; p q 1, there is some ; < < 1, such that (3:3) ku(t)k q cku(t)k p t?n=2(1=p?1=q) Remark: As a typical application, an energy decay of the type ku(t)k 2 c(1 + t)? implies ku(t)k q ct??n=2(1=2?1=q) for q 2. Proof: If p n=(n? 1), let k j = p(n=(n? 1)) j and k k. Then with T = in (2.6): ku(t)k kn=(n?1) cku()k k t?1=2k (t? s)?n=2((n+k)=nk?(n?1)=nk) ku(s)ru(s)k nk=(n+k) ds cku()k k t?1=2k (t? s)?(k+1)=2k t?1=2 ku(s)k k ds By induction c sup ku(s)k k t?1=2k st ku(t)k kj c j sup ku(s)k p t?n=2(1=p?1=k j ) tst2?j 12

13 X X as 1=2k i = 1=2p? 1=n) i= i=(1 i = n=2(1=p? 1=k j ): As k j > n for some j, interpolation and (3.1) implies (3.4) ku(t)k q cku(s)k p t?n=2(1=p?1=q) for all q with n=(n? 1) p q n. Now assume p < n=(n? 1), dene Then from (2.6) L := sup ku( + r)k n=(n?1)r n=2(1=p?(n?1)=n) : r ku( + r)k n=(n?1) cku()k p r?n=2(1=p?(n?1)=n) =2+r ( + r? s)?1=2 kru(s)k n ku(s)k n=(n?1)ds cku()k p r?n=2(1=p?(n?1)=n) t?1=2 R 1 r (r? x)?1=2 Lx?n=2(1=p?(n?1)=n) dx implying L cku()k p R 1 L with some c not depending on p. Hence we may assume c R 1 < 1=2 and get especially ku(t)k n=n?1 ct?n=2(1=p?(n?1)=n) ku()k p which may be combined with (3.4). Finally, for q > n, we use (2.9) to get the full result. Theorem 3': If a 2 L 1, then ku(t)k q ckak 1 t?n=2(1?1=q) for 1 < q 1: Remark: If e?ta is uniformly bounded in L 1, we may take ku()k 1 instead of kak 1. Proof: By theorem 3 it suces to prove the claim for some p < n=(n?1). Proceeding similar, we get for t t := kak 2=(n?1) 1. ku(t)k p t n=2(1?1=p) ckak 1 kak 2 2t 1=2 ckak 1 kak (n?2)=(n?1) 1 t 1=2 ckak 1 Next for t t, ku(t)k p t n=2(1?1=p) ckak 1 ku(s)ru(s)k 1 ds 13

14 t n=2(1?1=p) ku(s)ru(s)k p ds ckak 1 t 1=2 kak 2 2 t =2 t n=2(1?1=p) s?1=2 ku(s)k 2 ku(s=2)k 2 ds s?1=2 ku(s)k 2p ku(s=2)k 2p ds All four terms are bounded by ckak 1. To see this, choose some r with 1 < r < 2 and 1=2 + n(1=r? 1=2) > 1 and estimate the third term by c t =2 s?1=2?n(1=r?1=2) dskak 2 r ct 1=2?n(1=r?1=2) kak 2 r ckak 1 as 2(1=2? n(1=r? 1=2))=(n? 1) + 2 (1=r? 1)=(n? 1) = 1. The last term is bounded by ct n=2(1?1=p) kak 2 n=(n?1) s?1=2?n((n?1)=n?1=2p)ds ct (3?n)=2 kak 2 n=(n?1) ct (3?n)=2 kak 2 n=(n?1) ckak 1 ; as (2=(n? 1))((3? n)=2) + 2(n? 2)=(n? 1) = 1. This proves the claim. We may also derive estimates for ru(t) and Au(t) in L q ; q < n. Theorem 4: If a 2 L p ; 1 < p n, we have for p q n (3:4) t 1=2 kru(t)k q cku()k q (3:5) tkau(t)k q cku()k q Proof: Consider the representation u( + r) = e?ra u()? +r e?(+r?s)a f(s)ds: 14

15 Due to kf(s)k q kru(s)k n ku(s)k nq=(n?q) for p q < n and nq=(n? q) > q, we see that ru(t) 2 L q for t >, as Hence kru( + r)k q cs?1=2 ku()k q Q := +r sup r 1=2 kru( + r)k q r is nite. Estimating now kf(s)k q ckru(s)k q s?1=2 R 1, we get Q cku()k q R 1 Q sup cku()k q R 1 Q 8 < r : r1=2 r This implies (3.6) Q cku()k q, hence also (3.4). Next let ^Q := r 5=8 h?1=8 kru( + r + h)? ru()k q : sup r h We want to prove ^Q cku()k q. By (3.4), we may assume h r. Applying the gradient to ( + r? s)?1=2 kf(s)k q ds: (r? s)?1=2 s?1=2 (s + )?1=2 ds 9 = ; we get u( + r + h)? u( + r) = e?r=2a (e?ha? I)e?r=2A u()?? h e?(+h?s)a f( + s)ds e?(?s)a (f( + s + h)? f( + s))ds; kru( + r + h)? ru( + r)k q cr?1=2 ( h r )ku()k q h ( + h + s)?1=2 ku( + s)k1kru( + s)k q ds (? s)?1=2 ku( + s + h)? u + s)k1kru( + s + h)k q ds (? s)?1=2 ku( + s)k1kru( + s + h)? ru( + s)k q ds: In the second and third term, we use (3.6) und (2.7), resp. (2.8), and estimate by h ct?1=2 ( + s)?1=2 R 1 s?1=2 Qds ct?1 h 1=2 Q 15

16 resp. by c (? s)?1=2 ( + s)?5=8 h?1=8 s?1=2 Qds ct?5=8 h 1=8 Q: The last term is bounded by c (? s)?1=2 ( + s)?1=2 R 1 s?5=8 h 1=8 ^Qds ct?5=8 h 1=8 R 1 ^Q: As h r t, we get ^Q cku()k q R 1 ^Q; hence the claim. The consequence is (3:7) kf(s)? f(t)k q c(t? s) 1=8 t?9=8 R 1 ku()k q for s t. Proceeding now as in lemma 4, we arrive at (3.5). Putting all estimates together (using 2=n < 1=p + 1=q, when estimating Au(t) in case of p < n < q) implies the main theorem. Remark: In order to estimate all second order space-derivatives, we need (see e.g. [7]) kd 2 uk q ckauk q, if q < n 2 and kd 2 uk q c(kauk q + kruk r ); r > q, if q n 2. As an example, for q = 2 (if a 2 L 2 \ L n ), we conclude tkd 2 u(t)k 2 cku()k 2 for n > 4 tkd 2 u(t)k 2 cku()k 2 tkru(t)k n cku()k 2 t 1=2 ku(2t=3)k n for n = 3; 4 (by (2.3) cku()k 2 (1 + t 1?n=4 ) by (3.3): References [1] Amrouche, C., Girault, V., Schonbek, M.E., Schonbek, T.P., Pointwise decay of solutions and of higher derivatives to Navier-Stokes equations, Preprint [2] Borchers, W., Miyakawa, T., Algebraic L 2 decay for Navier-Stokes ows in Exterior domains. Acta Math., 165, (199) [3] Borchers, W., Miyakawa, T., Algebraic L 2 decay for Navier-Stokes ows in Exterior domains II. Hiroshima Math. J. 21, (1991) 16

17 [4] Cannone, M., Planchon, F., Schonbek, M.E., Strong solutions to the incompressible Navier-Stokes equations in the half-space. Preprint [5] Dan, W., Kobayashi, T., Shibata, Y., On the local energy decay approach to some uid ow in an exterior domain. Lecture Notes in Num. Appl. Anal., 16, 1-51 (1998) [6] Iwashita, H., L q? L r Estimates for Solutions of Nonstationary Stokes Eqauation in an Exterior Domain and the Navier-Stokes Initial Value Problems in L q Spaces. Math. Ann. 285, (1989) [7] Kozono, H., Ogawa, T., Sohr, H., Asymptotic Behaviour in L n for Turbulent Solutions of the Navier-Stokes Equations in Exterior Domains. Manuscripta Math. 74, (1992) [8] Maremonti, P., Solonnikov, V.A., On nonstationary Stokes problem in exterior domains. Preprint No. 4, Universita Potenza Dep. di Mat. (1996) [9] Schonbek, M.E., Large-time behavior of solutions of the Navier-Stokes equations. Comm. P.D.E. 11, (1986) [1] Wiegner, M., Decay results for weak solutions of the Navier-Stokes equations in R n. London Math. Soc. 35, (1987) [11] Wiegner, M., Decay and Stability in L p for strong solutions to the Cauchy problem for the Navier-Stokes equations. Proc. Oberwolfach 1988, Springer Lecture Notes 1431, (199) [12] Wiegner, M., Decay of the L1-norm of solutions to the Navier-Stokes equation in unbounded domains. Acta Appl. Math. 37, (1994) [13] Wiegner, M., The Navier-Stokes Equations - a Neverending Challenge? Jber.d.Dt. Math.-Verein. 11, 1-25 (1999) [14] Wiegner, M., Higher order estimates in further dimensions for the solutions of Navier Stokes equations. Preprint 17