ESTIMATES OF LOWER ORDER DERIVATIVES OF VISCOUS FLUID FLOW PAST A ROTATING OBSTACLE
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1 REGULARITY AND OTHER ASPECTS OF THE NAVIER STOKES EQUATIONS BANACH CENTER PUBLICATIONS, VOLUME 7 INSTITUTE OF MATHEMATICS POLISH ACADEMY OF SCIENCES WARSZAWA 25 ESTIMATES OF LOWER ORDER DERIVATIVES OF VISCOUS FLUID FLOW PAST A ROTATING OBSTACLE REINHARD FARWIG Fachbereich Mathematik Darmstadt University of Technology D Darmstadt, Germany farwig@mathematik.tu-darmstadt.de Abstract. Consider the problem of time-periodic strong solutions of the Stokes system modelling viscous incompressible fluid flow past a rotating obstacle in the whole space R 3. Introducing a rotating coordinate system attached to the body yields a system of partial differential euations of second order involving an angular derivative not subordinate to the Laplacian. In a recent paper [2] the author proved L -estimates of second order derivatives uniformly in the angular and translational velocities, ω and k, of the obstacle, whereas the transport terms fails to have L -estimates independent of ω. In this paper we clarify this unexpected behavior and prove weighted L -estimates of first order terms independent of ω. 1. Introduction. Consider the Navier-Stokes euations modelling viscous flow past a rotating body K R 3 with axis of rotation ω = ωe 3 = ω(,, 1 T, ω = >, and with velocity u = ke 3, k >, at infinity. Then the velocity v and the presure p satisfy the system v t ν v + v v + = f in Ω(t, t >, div v = in Ω(t, t >, v(y, t = ω y on Ω(t, t >, v(y, t u as y, with an initial value v(y, = v (y, with constant viscosity ν > and external force f in the time-dependent exterior domain Ω(t = O ω (tω ; here O ω (t denotes the orthogonal matrix cos t sin t O ω (t = sin t cos t. 1 2 Mathematics Subject Classification: 35C15, 35Q35, 76D5, 76D99, 76U5. Key words and phrases: Oseen flow, rotating obstacles, weighted estimates. The paper is in final form and no version of it will be published elsewhere. [73]
2 74 R. FARWIG Introducing the new independent and dependent variables x = O T ω(ty, u(x, t = O T ω(t ( v(y, t u, p(x, t = (y, t, and linearizing, (u, p will satisfy the modified Stokes system u t ν u + k 3 u (ω x u + ω u + p = f, div u =, in a time-independent exterior domain Ω R 3 together with the initial-boundary condition u(x, t = ω x u for x Ω, u(x, = u, u as x. In the stationary whole space case to be analyzed in this paper we are led to the elliptic euation ν u + k 3 u (ω x u + ω u + p = f, div u = in R 3 (2 in which the term (ω x u is not subordinate to ν u. Note that a stationary solution (u, p of (2 is related to a time-periodic solution of (1. In [2] the author proved a priori estimates of strong solutions (u, p of (2 which are found in the homogeneous Sobolev spaces Ŵ 2, (R 3 3 Ŵ 1, (R 3 where Ŵ k, (Ω = {u L 1 loc(ω/π k 1 : α u L (Ω for all α N n, α = k}. Here α = α α n n for a multi-index α = (α 1,...,α n N n and Π k 1 denotes the set of all polynomials on R n of degree k 1. The space Ŵ k, (Ω consists of euivalence classes of L 1 loc -functions being uniue only up to elements from Π k 1 and is euipped with the norm α =k α u, where denotes the L -norm. However, sometimes being less careful, we will consider v Ŵ k, (Ω as a function (representative rather than an euivalence class of functions, i.e., v L 1 loc (Ω such that α v L (Ω for every multi-index α with α = k. For further results on similar problems we refer to [4], [8], [9], [1], [11], [12] and [13]. Theorem 1. (1 Let 1 < <, ν >, k >, ω = (,, ω T R 3 \{}, and let f L (R 3 3. Then the linear problem (2 has a solution (u, p Ŵ 2, (R 3 3 Ŵ 1, (R 3 satisfying the a priori estimate (1 ν 2 u + p c f (3 with a constant c independent of ν, k and ω. (2 Moreover, k 3 u + (ω x u ω u c (1 + k4 ν 2 2 f (4 with a constant c > independent of ν, k and ω. (3 Let 1 < < 4, f L (R 3 3 and let (u, p Ŵ 2, (R 3 3 Ŵ 1, (R 3 be the solution of (2. Then there exists β R such that (u βω x L r (R for all r > 1, r 1 [ 1 4, 1 ], 3 and there exists a constant C = C(ν, k, ω; r > such that (u βω x r C ( f + ν g + (ω xg kge 3. (5
3 VISCOUS FLUID FLOW 75 The proof of Theorem 1(1, see [2], is based on an explicit representation of u the estimate of which uses Fourier transforms, Hardy-Littlewood decomposition methods and maximal operators. Estimate (4 shows a surprising and crucial dependence on 1 via the term k4 ν 2. On the one hand, it is not at all clear that the terms k 2 3 u and (ω x u ω u can be estimated in L (R 3 separately from each other. On the other hand, the dependence on 1 seems to be unnatural. Note that, as, problem (2 converges formally to Oseen s euation ν u + k 3 u + p = f, div u =, the solutions of which satisfy the estimate k 3 u c f, see [1, 5]. To be more precise, a seuence of solutions (u ω converges weakly in Ŵ 2, (R 3 3 to the solution of Oseen s euation as, cf. [2] Remark 1.3(1. Concerning Theorem 1(3 note that the solutions of the homogeneous system (2 are given by βω x + αe 3, α, β R. Hence, with u also u βω x is a solution of (2. The proof of Theorem 1(3 uses an improved Sobolev embedding theorem exploiting the fact that besides 2 u also k 3 u lies in L, cf. [1]. However, the dependence on 1/ in (4 implies that the constant C in (5 also depends on 1/. Due to this dependence the above-mentioned weak convergence of (u ω in Ŵ 2, (R 3 3, i.e. of second order derivatives in L, seems not to extend to k 3 u ω in L. The aim of this paper is to clarify these unusual features. In Theorem 2 below we present an improvement of (4 and simplify the proof given in [2]. Then, for small, we prove a weighted L -estimate of k 3 u and of (ω x u ω u independent of, see Theorem 4. An example and a heuristic argument will show that the dependence in (4 k on is not a weakness of the proof. ν Theorem 2. Let 1 < <, ν >, k R, ω = (,, ω T R 3 \{} and let f L (R 3 3. Then the solution u Ŵ 2, (R 3 3 satisfies the a priori estimate 2max(1/,1 1/+ε k 3 u + (ω x u ω u c (1 + k2 f (6 ν with a constant c > independent of ν, k and ω; here ε > can be chosen arbitrarily small and ε = if = 2. Example 3. In Section 2 we will show that the term k2 ν is needed in the L 2 -case. However, it is not clear whether the exponent 2 max(1/, 1 1/ + ε > 1 is necessary if 2. We note that in [13] dealing with the nonlinear problem in exterior domains no dependence of a priori estimates on 1 occurs; the reason is that the author uses strong and weak a priori L 2 -estimates of u by assuming that even f L 6/5 (R 3 Ŵ 1,2 (R 3. Using results of a forthcoming paper [3] dealing with the weak L -theory of (2 it is obvious that ν u is bounded uniformly w.r.t. ω and k by suitable norms of f. Theorem 4. (1 Let 1 < < 2, ν >, k >, ω = (,, ω T R 3 \{}, and let f L (R 3 3. Then (2 has a solution u Ŵ 2, (R 3 3 satisfying the a priori estimates u x c ν f (7
4 76 R. FARWIG and ( (ω x u ω u 1 + x c 1 + k f (8 ν with a constant c > independent of k, ν, ω; here, for x = (x 1, x 2, x 3 R 3 the term x denotes the Euclidean length x x2 2 of x = (x 1, x 2. (2 If 1 < < 3, then and u x 1 ν f (9 ( (ω x u ω u 1 + x c 1 + k f (1 ν with a constant c > independent of k, ν, ω. (3 For all 1 < < the third component u 3 of the solution u satisfies the a priori estimate k 3 u x + (ω x u x ( c 1 + k ν f. (11 At the end of Section 2 we present a heuristic argument why L -estimates of k 3 u k and of (ω x u ω u independent of 2 ν are unlikely to hold and why weighted 1 estimates with the weight 1+ x will help. 2. Preliminaries and proofs. From [2] we recall the calculation of the explicit representation of the solution u of (2. First, we eliminate the pressure term by applying div to (2 1. Then p is seen to be the uniue weak solution of the euation p = div f satisfying the a priori estimate Hence u is the solenoidal solution of the euation p c f. (12 ν u + k 3 u (ω x u + ω u = f p (13 where f p is solenoidal. For simplicity, we will write f instead of f p and divide by ω = > to get ν u + k 3u (e 3 x u + e 3 u = 1 f. (14 Next use cylindrical coordinates (r, θ, x 3 R + [, 2π R, r = x = x x2 2, for x = (x 1, x 2, x 3 T and observe that θ u = (e 3 x u = ( x 2, x 1 u. Moreover, we introduce the Fourier transform Fu(ξ = û(ξ = (2π 3/2 R 3 e ix ξ u(x dx, ξ R 3.
5 VISCOUS FLUID FLOW 77 For the Fourier variable ξ = (ξ 1, ξ 2, ξ 3 R 3 we also use cylindrical coordinates (s, ϕ, ξ 3 R + [, 2π R, s = ξ = ξ1 2 + ξ2 2, and note that Thus û satisfies the euation θ u = ϕ û. 1 ( ν ξ 2 û + ikξ 3 ϕ û + e 3 û = 1 ˆf. (15 This inhomogeneous, linear ordinary differential euation of first order with respect to ϕ has a uniue 2π-periodic solution. An elementary calculation leads to the representation where û(ξ = 1/ D(ξ e (ν ξ 2 +ikξ 3 t/ O T (t ˆf ( O(tξ dt, (16 D(ξ = 1 e 2π(ν ξ 2 +ikξ 3 /. (17 Moreover, using the geometric series and the 2π periodicity of t OT ω(t ˆf ( O ω (tξ w.r.t. t, we get the second representation û(ξ = Note that in x-space (18 leads to the identity u(x = e (ν ξ 2 +ikξ 3 t O T ω (t ˆf ( O ω (tξ dt. (18 E t O T ω(t f ( O ω (t kte 3 (x dt (19 where E denotes the heat kernel E t (x = 1 (4πνt 3/2 e x 2 /4νt in R 3. Proof of Theorem 2. We start with the case = 2 in which we may use the Theorem of Plancherel to estimate k 3 u 2. By (16, (17, the ineuality of Cauchy-Schwarz and Fubini s Theorem, k ikξ 3 û 2 2 ξ 2 3 dξ = /2 2π D(ξ 2 e (ν ξ 2 +ikξ 3 t/ O T (t ˆf(O(tξ 2 dt dξ k 2 ξ3/ 2 2 ( = D(ξ 2 ( ( e ν ξ 2t/ ˆf(O(tξ 2 dt e ν ξ 2t/ dt dξ e ν ξ 2 t/ k2 ξ 2 3/ 2 D(ξ 2 1 e 2πν ξ 2 / ν ξ 2 / ˆf(O(tξ 2 dξ dt. In the inner integral the change of variable formula implies that the term ˆf(O(tξ 2 may be replaced by ˆf(ξ 2. Then a further application of Fubini s theorem yields ikξ 3 û 2 dξ = ˆf(ξ 2 k2 ξ3/ e 2πν ξ 2 / D(ξ 2 ν ξ 2 e ν ξ 2t/ dt / = ˆf(ξ 2 k2 ξ 2 3/ 2 D(ξ 2 (1 e 2πν ξ 2 / 2 ν 2 ξ 4 / 2 dξ.
6 78 R. FARWIG Hence it suffices to consider the multiplier function and to prove the estimate m(ξ = 1 e (2πν ξ 2 / ν ξ 2 / kξ 3 / D(ξ (2 m(ξ c ( 1 + k2, ξ R 3. (21 ν If ν ξ 2 > 1, then D(ξ is bounded below by a positive constant and m(ξ c kξ 3/ ν ξ 2 / c k ν ξ c k. ν If ν ξ 2 1, the first factor in the definition of m(ξ is uniformly bounded. To estimate the remaining term m (ξ = kξ 3/ D(ξ ν ξ we partition the ball 2 1 into infinitely many slices S n = {ξ R 3 : 1, kξ 3 n 1 4 }, n Z, and the remaining part S where dist ( kξ 3 conseuently D(ξ 1. Hence, m (ξ k ξ 3 k ν on S. For ξ S n, n Z, Taylor s expansion of 1 e z yields the lower bound D(ξ = 1 e 2π(ν ξ 2 /+i(kξ 3 / n ν ξ 2 ( n kξ3 c + i ν ξ 2, Z 1 4 and with a constant c > independent of all variables ν, ξ, k, ω, n. Hence for ξ S we get the estimate m (ξ 1 c. If ξ S n, n, then m (ξ k ξ 3 / ν ξ 2 / k ν ξ 4 3 ν, since k ξ k ξ Now (21 is proved and implies the estimate 2 ikξ 3 û 2 dξ c (1 + k2 ν ˆf 2 dξ. Then the Theorem of Plancherel completes the proof in the case = 2. For the case 2 we write (16 in the form û(ξ = 1 1 D(ξ e ν ξ 2 t/ O T (t ( Ff(O(t kte 3 / (ξdt using that e itkξ 3 S (R 3 is the Fourier transform of the shift operator f f( kte 3 on S (R 3. Hence in x-space, k 3 u(x = k 2 T t F(t, (x dt
7 VISCOUS FLUID FLOW 79 where F(t, = O T (tf(o(t kte 3 / and the operator family T t, < t < 2π, is defined by its multiplier m t (ξ = ikξ 3/ e ν ξ 2t/, (22 D(ξ i.e., T t = F 1 (m t (ξ. Note that F(t, f for all t (, 2π. Below we will prove the multiplier estimate ( max sup ξ α Dξ α k m t (ξ c k2 (1 + k2 α ξ νt ν ν with a constant c > independent of ν, ω, k and t; here α N 3 runs through the set of all multi-indices α {, 1} 3. Then Marcinkiewicz multiplier theorem [14] implies that in the operator norm on L ( k T t c k2 (1 + k2. νt ν ν Hence k 3 u c (23 T t F(t, dt c (1 + k4 ν 2 2 f (24 with a constant c > independent of ν, ω and k. Now the L 2 -result and (24 are combined by using complex multiplier theory to prove (6. Given 1 < < 2 we formally interpolate between the L 2 -result and the L 1 -result (24 using θ = 2 1 such that 1 = 1 θ 2 + θ 1. Then 2 2/ 2/ 1 3 u c (1 + (1 k2 + k4 ν ν 2 2 f yielding 3 u c(1 + k2 ν 2/ f. Since no estimate (24 holds for L 1, we have to interpolate between L 2 and L p for p > 1 arbitrarily close to 1. Therefore, the additional exponent ε > in (6 occurs. If 2 < <, we formally interpolate between L 2 and L to get (6 with the exponent 2(1 1/. Since again there is no L -result available, we choose p arbitrarily large instead of p = and have to add the exponent ε > in (6. Finally we prove (23, start with m t itself and distinguish between the cases ν ξ 2 > 1 and ν ξ 2 1. In the first case D(ξ is bounded below by a positive constant yielding m t (ξ c k ξ 3 2 e ν ξ t/ k c. νt If ν ξ 2 1, we may neglect the term e ν ξ 2t/ and conclude from the detailed estimates of m (ξ in the L 2 -case above that m t (ξ m (ξ c (1 + k2. ν
8 8 R. FARWIG Hence m t (ξ c(1 + k + k2 νt ν proving (23 for m t. Next consider where ξ 3 3 m t (ξ = m t (ξ 2νξ2 3t 3 D(ξ m t (ξ ξ 3 D(ξ m t(ξ (25 ξ 3 3 D(ξ D(ξ = 2π (2νξ2 3 + ikξ 3 / e 2π(ν ξ 2 +ikξ 3 /. (26 D(ξ Writing the exponential function e ν ξ 2t/ as e ν ξ 2 t/2 e ν ξ 2t/2, we see that the second term on the right-hand side of (25 may be estimated as m t itself. In the third term note due to properties of D(ξ proved above that 2νξ2 3 / D(ξ e 2π(ν ξ 2 +ikξ 3 / is uniformly bounded. Moreover, the estimate of kξ 3/ +ikξ 3 / D(ξ is similar to the e 2π(ν ξ 2 estimate of the multiplier function m(ξ in (2, (21 yielding k ξ 3 / e 2π( ν ξ 2 +ikξ 3 / c (1 + k2 D(ξ ν Combining the previous estimates we get (23 for ξ 3 3 m t. Concerning ξ 1 1 m t (ξ note that (26 must be replaced by ξ 1 1 D(ξ D(ξ. = 2π 2νξ2 1/ e 2π(ν ξ 2 +ikξ 3 /. D(ξ Obviously, looking at the properties of D(ξ proved above, the modulus of this term is uniformly bounded reuiring no further power of k 2 /ν. Since the same assertion holds for the derivatives ξ 2 2 and ξ 1 1 ξ 2 2 of m t, (23 is completely proved. Example 3. For fixed k >, ν > we will construct a seuence of solenoidal forces (f = (f ω L 2 (R 3 3,, such that the corresponding seuence of solutions (u = (u ω will satisfy k 3 u c k2 ν f (27 with c > independent of k, ν, ω. Given k >, ν > choose small enough such that ν k 2 < Then define f = (f, L 2 (R 3 3 such that in Fourier space { ˆf ξ (ξ = i, < ϕ < π ξ, π < ϕ < 2π, when k ξ 1 < 1 2, k ξ 3 1 < 1 2, but ˆf (ξ = elsewhere; here, as usual, ϕ is the angular part of ξ. Since ˆf(ξ = ˆf( ξ, the vector field f is real-valued and obviously solenoidal. By (16 û(ξ = eν ξ 2 +ikξ 3 ϕ/ D(ξ O(ϕ ϕ+2π ϕ e (ν ξ 2 +ikξ 3 t/ O T (t ˆf ( O(te 1 dt.
9 VISCOUS FLUID FLOW 81 Since O T (t ˆf(O(te 1 = +e 2 and = e 2 when < ϕ < π and π < ϕ < 2π, resp., a simple integration leads to the formula ikξ 3 û(ξ = ˆf(ξ +ikξ ikξ 3 / 3 / (1 (ν ξ 2 2e (π ϕ(ν ξ 2, + ikξ 3 / 1 + e π(ν ξ 2 +ikξ 3 / when < ϕ < π; for π < ϕ < 2π the exponential term e (π ϕ(ν ξ 2 +ikξ 3 / must be replaced by e (2π ϕ(ν ξ 2 +ikξ 3 /. The assumptions on k, ν, ω imply for ξ supp ˆf that kξ 3 1, ν ξ 2 e (π ϕ(ν ξ 2 +ikξ 3 / 1. Finally, the crucial term is Hence ν k 2 ; conseuently, we have ν ξ 2 + ikξ 3 / kξ 3 1 and 1 + e π(ν ξ 2 +ikξ 3 / ν ξ 2 ikξ 3 û(ξ k2 ν ˆf(ξ ν k 2 for ξ supp ˆf. for ξ supp ˆf. Since all similarity estimates can be made precise by using positive constants independent of k, ν, ω, (27 is proved. Proof of Theorem 4. (i Given a solution u Ŵ 2, (R 3 3, i.e. a function u with 2 u L (R 3, Theorems 1 and 2 yield β R such that (u β(ω x L r (R 3 6, 1 r = 1 1 4, and 3 (u β(ω x L (R 3 3. Since also u βω x solves (2, assume without loss of generality that β = implying for a.a. x 3 R that u(x, x 3 r dx <, 3 u(x, x 3 dx <. R 2 R 2 Then classical arguments show the existence of a seuence of radii (R j R + such that u(r j, θ, x 3 r dθ = o ( R 2 j, 3 u(r j, θ, x 3 dθ = o ( R 2 j as j. On the other hand, since 1 < < 2 and 2 u L (R 3, Theorem II5.1 in [6] yields for a.a. x 3 R a matrix A(x 3 R 3,3 such that ( R2 u(x, x 3 A(x 3 x dx 1/ 2 (28 ( 1/ u(x, x 3 dx. (29 R 2 Note that Theorem II5.1 in [6] is stated only for exterior domains; however, since the constant /(2 does not depend on the inner radius of the exterior domain, the estimate holds for the whole space R 2 as well. Moreover, by Lemma 5.2 in [6], u(r, θ, x 3 A(x 3 dθ = o ( R 2 (3 as R. Now (28, (3 show that A(x 3 = ; hence (29 and (3 yield (7. Then (8 is an easy conseuence of (7 and of (2. (ii If 1 < < 3, Theorem II5.1 in [6] yields the estimate ( R3 u(x A 1/ x dx ( 1/ 2 u(x dx 3 R 3
10 82 R. FARWIG with a constant matrix A R 3,3. Moreover, by Lemma 5.2 in [6], u(ry do(y = o ( R 3 y =1 as R, where y =1...do(y denotes the surface integral on the unit sphere of R3. Since u L r (R 3 6 and 3 u L (R 3 3, arguments as above imply that A vanishes. Now (9 and (1 are easy conseuences. (iii By (2 u 3 solves the problem ν u 3 + k 3 u 3 (ω x u 3 = f 3. Since (ω x u 3 = θ u 3, an integration w.r.t. θ (, 2π yields for the θ-independent function U 3 (x := 1 2π 2π u 3 ( x, θ, x 3 dθ the euation ν U 3 + k 3 U 3 = 1 2π f 3 dθ. (31 Applying Fourier transforms and using Marcinkiewicz multiplier theorem we get that U 3 satisfies the estimate k 3 U 3 c f with a constant c > independent of f, k, ν, cf. the analysis of the related Oseen problem [1], [5], [6]. By Wirtinger s ineuality there exists a constant c > such that for a.a. r = x > and x 3 R 3 u 3 (r,, x 3 L (,2π c ( θ 3 u 3 (r,, x 3 L (,2π + 3 U 3 (r,, x 3 L (,2π. Now divide by 1 + r and integrate w.r.t. r dr, r >, and dx 3, x 3 R, to get that ( 3 u 3 θ 3 u x c 1 + x + 3 U x. Since the second term on the right-hand side is bounded by 3 u 3 (c/ν f and since the third term is bounded by 3 U 3 (c/k f, we get (11. Remark. The ideas of the proof of Theorem 4 (iii do not apply to u 1 and u 2, since the term ω u does not vanish when applying the integration... dθ. Also the identity (ω x u ω u = O(θ θ (O T (θu will not help, since no a priori estimates of 3 θ (O T (θu are available except for the case when 1 < < 2. Heuristic argument. Let us motivate why estimates of (ω x u ω u and of k 3 u cannot be expected to be independent of k 2 /ν. For simplicity ignore the terms ω u and p, recall that (ω x u = θ u and let us perform a simple scaling analysis. Define the non-dimensional uantities ũ = u/a, where A R is a characteristic acceleration of the flow, and x = x /ν. Then, dividing (2 by A and omitting, (2 simplifies to the non-dimensional euation Note that u + k ν 3 u θ u = f in R 3. k is a new non-dimensional characteristic number of the flow. For fixed ν r = x let us interpret k 3 u θ u as a directional derivative defined by the unit ν
11 VISCOUS FLUID FLOW 83 vector d ω (x = ( 1 k e 3 ( x 2, x 1, T R 3 r2 + k 2 /νω ν which is tangential to the cylinder C r = {x R 3 : x = r}. Hence, defining the curve ( T k γ ω (s = r cos s, r sin s, s ν on C r with tangential vector r 2 + k 2 /νω d ω, we get that d ds u(γ ω(s = ( r 2 + k 2 /νω d ω u (γ ω (s. Obviously d ω converges to the third unit vector e 3, whereas the curve γ ω (s has no reasonable limit on the cylinder C r. In this sense, the information on the directional derivative d ω u on C r is lost in the limit ω =. This discrepancy vanishes for r =, 1 but gets larger as r. Therefore, the weight 1+ x has to occur in Theorem 1.4. References [1] R. Farwig, The stationary Navier-Stokes euations in a 3D-exterior domain. in: Recent Topics on Mathematical Theory of Viscous Incompressible Fluid, Lecture Notes in Num. Appl. Anal. 16 (1998, [2] R. Farwig, An L -analysis of viscous fluid flow past a rotating obstacle, Darmstadt University of Technology, Department of Mathematics, Preprint no (24, to appear in Tohoku Math. J. [3] R. Farwig, Weak solutions of Navier-Stokes flow past a rotating obstacle, in preparation. [4] R. Farwig, T. Hishida and D. Müller, L -theory of a singular winding integral operator arising from fluid dynamics, Pacific J. Math. 215 (24, [5] R. Farwig and H. Sohr, Weighted estimates for the Oseen euations and the Navier-Stokes euations in exterior domains, in: Theory of Navier-Stokes euations, J. Heywood et al. (eds., Series Advances Math. Appl. Sciences 47, World Scientific, 1998, [6] G. P. Galdi, An Introduction to the Mathematical Theory of the Navier-Stokes Euations, Vol. I. Linearized Steady Problems, Springer Tracts in Natural Philosophy 38, 2nd edition, [7] G. P. Galdi, On the motion of a rigid body in a viscous liuid: A mathematical analysis with applications, in: S. Friedlander and D. Serre (eds., Handbook of Mathematical Fluid Mechanics, Elsevier Science, 22, [8] G. P. Galdi, Steady flow of a Navier Stokes fluid around a rotating obstacle, J. Elasticity 71 (23, [9] T. Hishida, An existence theorem for the Navier-Stokes flow in the exterior of a rotating obstacle, Arch. Rational Mech. Anal. 15 (1999, [1] T. Hishida, The Stokes operator with rotation effect in exterior domains, Analysis 19 (1999, [11] Š. Nečasova, Some remarks on the steady fall of a body in Stokes and Oseen flow, Acad. Sciences Czech Republic, Math. Institute, Preprint 143 (21. [12] Š. Nečasova, Asymptotic properties of the steady fall of a body in viscous fluids, Acad. Sciences Czech Republic, Math. Institute, Preprint 149 (22.
12 84 R. FARWIG [13] A. L. Silvestre, On the existence of steady flows of a Navier-Stokes liuid around a moving body, Math. Meth. Appl. Sci. 27 (24, [14] E. M. Stein, Singular Integrals and Differentiability Properties of Functions, Princeton University Press, Princeton, N.J., 197.
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