A Negative Result. We consider the resolvent problem for the scalar Oseen equation

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1 O Osee Resolvet Estimates: A Negative Result Paul Deurig Werer Varhor 2 Uiversité Lille 2 Uiversität Kassel Laboratoire de Mathématiques BP 699, Calais cédex Frace paul.deurig@lmpa.uiv-littoral.fr Fachbereich Mathematik D-3432 Kassel Deutschlad varhor@mathematik.ui-kassel.de We cosider the resolvet problem for the scalar Osee equatio i the whole space R 3. We show that for small values of the resolvet parameter it is impossible to obtai a L 2 -estimate aalogous to the oe which is valid for the Stokes resolvet, eve if the resolvet parameter has positive real part. Key words. Osee equatios, resolvet estimate AMS subject classificatios. 35Q30, 65N30, 76D05

2 Itroductio Let Ω R 3 be a domai, either bouded or exterior (i. e. a domai with compact complemet) or a half-space or the whole space R 3. Set p > ad 0 < ϑ < π. The the velocity part u of the solutio (u, q) of the Stokes resolvet problem u + λu + q = f, div u = 0 i Ω, () with Dirichlet boudary coditio u = 0 o Ω (Ω R 3 ) (2) satisfies the estimate u p C λ f p, f L p (Ω) 3, 0 λ C with arg λ ϑ (3) with a costat C depedig oly o Ω, p ad ϑ. This so-called resolvet estimate is a crucial auxiliary result whe a semigroup approach is applied to the ostatioary Navier-Stokes system. The estimate (3) was proved by Giga [9] i the case that Ω is a bouded or a exterior domai ad λ is large, by McCracke [] if Ω = R 3 or if Ω is a half-space i R 3, by Borchers, Sohr [] if Ω R 3 ( 3) is a exterior domai ad λ is small, ad by Borchers, Varhor [2] if Ω R 2 is a exterior domai ad λ is small. Deurig [3], [4], [5] gave aother proof of (3) if Ω is a exterior domai, as did Soloikov [3] uder the additioal assumptio that λ is large. With iequality (3) i mid, we wat to cosider a differet situatio here. I fact, the costat motio of a rigid body i a icompressible viscous fluid is 2

3 usually modeled by the Navier-Stokes system with a Osee term. Thus, i order to apply a semigroup approach to that system, equatio () has to be replaced by the Osee resolvet problem, which reads as follows: u + τ u + λu + q = f, div u = 0 i Ω. (4) Here τ > 0 is the Reyolds umber. Agai boudary coditio (2) has to be imposed if Ω R 3. Accordig to [0, Theorem 4.4], a estimate as i (3) holds for the velocity part u of a solutio to the Osee system (4), (2) if Re λ 0, λ is large ad Ω R 3 is a exterior domai, with C depedig o Ω, p, τ ad a lower boud for λ. I [7, Theorem 4.4] this result is geeralized to the case of a exterior domai i R with 3. It is further show i [7] that for a give ϑ ( π 2, π), a sufficietly large value R 0 > 0 may be chose such that (3) holds for λ C with λ R 0 ad arg λ ϑ ([7, Lemma 4.5]). The results from [0] ad [7] are exploited i [2] ad [8], respectively, i order to solve the time-depedet Navier-Stokes system with Osee term, ad to prove decay results for solutios of this system. The precedig observatios give rise to the questio whether i the Osee case a estimate as i (3) holds for small values of λ. Of course, it must be take ito accout that the spectrum of the Osee operator touches the imagiary axis from the left, see [0, p. 7] for more details. Therefore it caot be expected that i the Osee case iequality (3) is valid for λ C with λ small, Im λ < 0 ad arg λ ϑ, for some ϑ ( π, π). However, oe might hope that (3) holds 2 for 0 λ C with Re λ 0 ad λ small. We are aware of oly oe result i this directio: Accordig to [6, Theorem 0, (3.5)], the solutio u of the scalar 3

4 Osee equatio with resolvet term u + τ u + λu = f i R 3 (5) satisfies the estimate u p C λ 2 f p, f L p (R 3 ), p 2, 0 λ C, λ ( τ 2) 2, Re λ 0. (6) I view of (3), it may be asked whether it is possible to improve iequality (6) by replacig the factor λ 2 by λ (ote that small values of λ are cosidered i (6)). It is the purpose of the preset paper to show that such a improved versio of (6) does ot hold. More precisely, we show that, give ay 0 < α < 2, there is o costat C > 0 with ( u 2 C λ 3/2+α f 2 for f L 2 (R 3 ), λ = r 2 + ir with r (0, ) 0, τ 2 8 ). Here u H 2 (R 3 ) is the solutio of (5) for give f ad λ, see Theorem 2 below, where a slightly more geeral result is preseted. This leaves ope the questio for the expoets γ [ 3 2, 2) such that the iequality u p C λ γ f p is satisfied, at least i the case p = 2, if λ is small ad Re λ 0. Nevertheless, ad this is the poit we wat to make with the preset paper, our result is sufficiet to idicate that eve for λ i the right complex half-plae, the Stokes resolvet estimate (3) does ot carry over to the Osee case if λ is small. 4

5 2 Notatios ad kow results For R > 0, let B R deote the ope ball i R 3 with radius R ad ceter i the origi. If A R 3, by χ A we deote the characteristic fuctio of A. For g L (R 3 ) let ĝ deote the Fourier trasform of g, defied by: ĝ(ξ) := (2π) 3 R 3 i ξ g(y) e y dy, ξ R 3. The iverse Fourier trasform of g is give by ǧ(ξ) := (2π) 3 g(y) e i ξ y dy, ξ R 3. R 3 The Fourier trasform ĝ of a fuctio g L 2 (R 3 ) is to be defied i the usual way ad i accordace with the precedig choice of the Fourier trasform of fuctios i L (R 3 ). If g, h are measurable fuctios with R 3 g(x y) h(y) dy < for x R 3, we deote by g h the covolutio of g ad h, defied by, (g h)(x) := R 3 g(x y) h(y) dy, x R 3. Let τ > 0. The we defie the fudametal solutio E (λ) of the scalar Osee resolvet equatio (5) by E (λ) (z) := q 4π z e λ+ τ2 4 z + τz 2, 0 z R 3, λ C. Usig this fudametal solutio, we may solve (5) i the followig sese: 5

6 Theorem. Let 0 λ C with λ ( τ 2 )2, Re λ 0. The E (λ) L (R 3 ) L 2 (R 3 ) such that, i particular, E (λ) (x y) f(y) dy < for f L 2 (R 3 ), x R 3. R 3 Moreover, for f L 2 (R 3 ), we have E (λ) f H 2 (R 3 ), ad u := E (λ) f solves (5). I additio, Ê (λ) (ξ) = (2π)3 (λ + ξ 2 + iτξ ), ξ R3. (7) Of course, equatio (5) admits a uique solutio for ay 0 λ C with Re λ 0 if f L 2 (R 3 ). But we restrict our existece result to the case λ ( τ 2 )2 sice this is sufficiet for our purposes, ad because a existece result uder this assumptio may easily be deduced from the results i [6]. This latter poit becomes clear by the Proof of Theorem : By [6, Theorem 9] there are costats C (λ), C 2 > 0 such that ( E (λ) χ(0,) ( z ) (z) C (λ) z + χ [, )( z ) e C 2 λ 2 z ), 0 z R 3. This implies E (λ) L 2 (R 3 ) L (R 3 ). Accordig to [6, (3.5)] there is some costat C(λ) > 0 with E (λ) f 2 C(λ) f 2, f L 2 (R 3 ). (8) 6

7 Moreover, usig [6, Theorem 3] we fid E (λ) f H 2 (R 3 ) ad l m (E (λ) f) 2 C(λ) f 2, f L 2 (R 3 ), l, m 3. (9) The costat C(λ) > 0 is idepedet of f. I additio, the relatios [6, (3.2)] ad [6, (3.7)] yield l (E (λ) f) 5 C(λ) f 2, (0) agai with a costat C(λ) > 0 idepedet of f. Accordig to [6, Corollary ] the fuctio u := E (λ) f belogs to C (R 3 ) ad satisfies (5) if f C 0 (R 3 ). Now it follows from (8) (0) that equatio (5) is valid also for f L 2 (R 3 ). The uiqueess result stated i the above theorem holds accordig to [6, Theorem 9]. Equatio (7) is stated i [0, p. 9]. For the coveiece of the reader, we idicate a proof here. To this ed, let 0 ξ R 3. The itegral over R 3 appearig i the defiitio of the Fourier trasform may be writte as a itegral with respect to (r, η) (0, ) B. This trasformatio gives rise to a factor r 2, which is removed by a partial itegratio. The, itegratig with respect to r we obtai Ê (λ) (ξ) = 27 π 5 B ( λ + κ2 + η (iξ κe ) ) 2 do η, () with e := (, 0, 0) ad κ := τ. Next we choose a orthoormal matrix A R3 3 2 such that ( A( κe ) ) = (Aξ) 2 ad (Aξ) = ( A( κe ) ) 2. I other words, the vectors κe ad ξ are simultaeously rotated i such a way that their projectio oto the x - x 2 - plae verifies the precedig relatios. Set a := A( κe ) ad 7

8 b := Aξ such that a = b 2 ad a 2 = b. The ( λ + κ2 + η (iξ κe ) ) 2 do η = ( λ + κ2 + iη b + η a ) 2 do η B = π/2 cos ϑ 2π B ( λ + κ2 + (si ϑ)a 3 + i(si ϑ)b 3 + (cos ϑ)e iφ (b 2 + ib ) ) 2 dφ dϑ. π/2 0 By applyig Cauchy s formula to the itegral with respect to φ, we may coclude ( λ + κ2 + η (iξ κe ) ) 2 do η (2) B = 2π π/2 π/2 cos ϑ ( λ + κ 2 + (si ϑ)a 3 + i(si ϑ)b 3 ) 2 dϑ. Fially, itegratig with respect to ϑ we obtai from () ad (2) Ê (λ) (ξ) = = ( 25 π 3 (a 3 + ib 3 ) (2π)3 (λ + κ 2 (a 3 + ib 3 ) 2 ). λ + κ2 + a 3 + ib 3 + ) λ + κ2 a 3 ib 3 Sice a 3 b 3 = ab = κξ ad a b 2 3 = a 2 + b 2 = κ 2 + ξ 2, equatio (7) follows. 8

9 3 Mai theorem I this sectio, we prove the mai result of this article, stated i the esuig theorem. Theorem 2. Let 0 α <. For N let 0 r 2 ad set λ 2 := r + i. The there is o costat C > 0 such that u 2 C λ 3/2+α f 2 for f L 2 (R 3 ) ad u H 2 (R 3 ) with u + τu + λ u = f i R 3. Proof of Theorem 2: For N we set I := ( τ, ) ( 0, 2, g (ξ) := 2τ ) 2τ 3 χ I (ξ) for ξ R 3. Obviously g L 2 (R 3 ) L (R 3 ), so we may defie f := ǧ, ad we obtai f L 2 (R 3 ), f 2 = g 2 ( N). The latter relatio ad the defiitio of g imply f 2 = 2τ 3 I = ( N). (3) Now choose N with τ 2 8 such that λ ( τ 2 )2. We defie u := E (λ) f. The we kow by Theorem that u H 2 (R 3 ) with u + τ u + λ u = f, (4) 9

10 ad that u is the oly fuctio i H 2 (R 3 ) satisfyig (4). Actually, it is the oly fuctio i the larger class metioed i Theorem. O the other had, u 2 2 = û 2 2 = (2π) 3 Ê(λ) ˆf 2 2 = (2π) 3 Ê(λ) (ξ) 2 2τ 3 dξ I = 2τ 3 i + r I + ξ 2 2 dξ + iτξ = 2τ 3 ( ξ 2 + r ) 2 + ( + τξ ) dξ 2 = 2τ 3 I /(τ) /(2τ) (0, /) 2 ( (ξ 2, ξ 3 ) 2 + ξ 2 + r ) 2 + ( τξ ) 2 d(ξ 2, ξ 3 ) dξ. But r 2 ad (ξ 2, ξ 3 ) 2 for (ξ 2, ξ 3 ) (0, )2, so we may coclude u 2 2 2τ 3 = 2τ 2τ /(τ) /(2τ) (0, /) 2 /(τ) /(2τ) /(τ) /(2τ) ( ξ ) 2 + ( τξ ) 2 d(ξ 2, ξ 3 ) dξ ( ξ 2 ) 2 + ( τξ ) 2 dξ 8 + 2ξ ( τξ ) dξ, 2 0

11 where we used the relatio (a + b) 2 2(a 2 + b 2 ) for a, b 0 i the last iequality. Next we perform the chage of variable ξ = r τ to obtai u /2 = r4 + ( dr r 4 4 )2 /2 /2 / r4 + ( r) dr ( q ( r) 2 dr ) dr, 2 + r where the last iequality holds because a 2 + b 2 (a + b) 2 for a, b 0. By itegratig with respect to r we ow obtai u τ 4 q Thus for N with max{ 8, } we get τ , hece u

12 It follows with (3) that u 2 q 3 Sice λ we obtai f 2 for N with max{ 8, 4 τ }. u 2 α λ 3/2+α f for as above. Therefore there ca be o costat C > 0 such that u 2 C λ 3/2+α f 2 for N. This implies the theorem. Refereces [] Borchers, W., Sohr,H.: O the semigroup of the Stokes-operator for exterior domais i L q -spaces. Math. Z. 96 (987), [2] Borchers, W., Varhor, W.: O the Boudedess of the Stokes Semigroup i Two-Dimesioal Exterior Domais. Math. Z. 23, (993) [3] Deurig, P.: A itegral operator related to the Stokes system i exterior domais. Math. Methods Appl. Sci. 3 (990), Addedum. Math. Methods Appl. Sci. 4 (99), 445. [4] Deurig, P.: The resolvet problem for the Stokes system i exterior domais: a elemetary approach. Math. Methods Appl. Sci. 3 (990), [5] Deurig, P.: The Stokes system i exterior domais: L p -estimates for small 2

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