The fundamental solution of compressible and incompressible fluid flow past a rotating obstacle

Transkript

1 The fundamental solution of compressible and incompressible fluid flow past a rotating obstacle JSPS-DFG Japanese-German Graduate Externship Kickoff Meeting Waseda University, Tokyo, June 17-18, 2014 Reinhard Farwig, TU Darmstadt Milan Pokorný, Charles University Prague Toshiaki Hishida, Nagoya University, et al. 1 / 25

2 The fundamental solution of compressible and incompressible fluid flow past a rotating obstacle JSPS-DFG Japanese-German Graduate Externship Kickoff Meeting Waseda University, Tokyo, June 17-18, 2014 Reinhard Farwig, TU Darmstadt Milan Pokorný, Charles University Prague Toshiaki Hishida, Nagoya University, et al. ω= ω (0,0,1) T Ω K 1 / 25 u oo

3 The Incompressible Case u t ν u + u u + p = f in Ω(t) div u = 0 in Ω(t) u = ω x on Ω(t) u u at The exterior domain Ω(t) depends on t General assumption: u = ke 3, k = u 0 and ω = e 3 2 / 25

4 The Incompressible Case u t ν u + u u + p = f in Ω(t) div u = 0 in Ω(t) u = ω x on Ω(t) u u at The exterior domain Ω(t) depends on t General assumption: u = ke 3, k = u 0 and ω = e 3 Work in a t-independent domain coordinate transform y = O(t) x, v(y, t) = O(t) ( ) u(x, t) u,... cos t sin t 0 O(t) = sin t cos t Linearize v w.r.t. v = 0 Consider the stationary problem in R 3 t-periodic solutions of the original problem in R 3 : 2 / 25

5 The Linear Problem ν v (ω y) v + u v + ω v + p = f in R 3 div v = 0 in R 3 v 0 at 3 / 25

6 The Linear Problem ν v (ω y) v + u v + ω v + p = f in R 3 Apply the Helmholtz projection P Modified Stokes/Oseen operator div v = 0 in R 3 v 0 at A ω,u (v) = νp v (ω y) v + ω v + u v T. Hishida (1999, 2000): A ω,u does not (!) generate an analytic semigroup 3 / 25

7 The Linear Problem ν v (ω y) v + u v + ω v + p = f in R 3 div v = 0 in R 3 v 0 at Apply the Helmholtz projection P Modified Stokes/Oseen operator A ω,u (v) = νp v (ω y) v + ω v + u v T. Hishida (1999, 2000): A ω,u does not (!) generate an analytic semigroup Analysis of the fundamental solution in L q -spaces: R. F., T. Hishida, D. Müller: Pacific J. Math. (2004) R. F.: Tôhoku Math. J. (2005) G.P. Galdi, M. Kyed: Proc. Amer. Math. Soc. (2013), (2013) R. F., M. Krbec, Š. Nečasová: Ann. Univ. Ferrara (2008) R. F., M. Krbec, Š. Nečasová: Math. Methods Appl. Sci. (2008) 3 / 25

8 Further Literature Spectral properties of A ω,u : R. F., J. Neustupa: Integral Equations Operator Theory (2008) R. F., Š. Nečasová, J. Neustupa: J. Math. Soc. Japan (2011) Explicit representation of the fundamental solution: R. F., R.B. Guenther, Š. Nečasová, E.A. Thomann: DCDS-A (2014) Semigroup estimates: M. Geissert, H. Heck, M. Hieber: J. reine angew. Math. (2006) T. Hishida, Y. Shibata: Arch. Ration. Mech. Anal. (2009) Numerous results on the nonlinear problem, e.g. by F.-Hishida, F.-Galdi-Kyed, Galdi-Kyed, Galdi-Silvestre, Geissert-Heck-Hieber, Goetze, Hansel, Heck-Kim-Kozono, Hieber-Shibata, Hieber-Sawada, Hishida-Shibata, et al. 4 / 25

9 L q -estimates for the Fundamental Solution Determine the fundamental solution, prove L q -estimates Use cylindrical coordinates y (r, θ, y 3 ), r = (y 1, y 2 ) 0 (ω y) = θ Use Fourier transform and cylindrical coordinates ξ (s, ϕ, ξ 3 ) θ u(ξ) = ϕ û(ξ) 5 / 25

10 L q -estimates for the Fundamental Solution Determine the fundamental solution, prove L q -estimates Use cylindrical coordinates y (r, θ, y 3 ), r = (y 1, y 2 ) 0 (ω y) = θ Use Fourier transform and cylindrical coordinates ξ (s, ϕ, ξ 3 ) θ u(ξ) = ϕ û(ξ) We get a first order ODE w.r.t. ϕ: ν ξ 2ˆv + ikξ 3ˆv ϕˆv + ω ˆv = ˆ P f, iξ ˆv = 0, where ˆv has to be 2π-periodic in ϕ 5 / 25

11 The solution of the First Order ODE (ν = 1) ˆv(ξ) = 1 1 e 2π ξ 2 2π 0 e ξ 2 t iktξ 3 O(t) P f(o(t)ξ) dt = 0 e ξ 2t O(t) F((P f)(o(t) kte 3 )) dt 6 / 25

12 The solution of the First Order ODE (ν = 1) ˆv(ξ) = 1 1 e 2π ξ 2 2π 0 e ξ 2 t iktξ 3 O(t) P f(o(t)ξ) dt = 0 e ξ 2t O(t) F((P f)(o(t) kte 3 )) dt If k = 0 and O(t) ((P f)(o(t) ) is t-independent ˆv(ξ) = 1 ξ 2 P f(ξ) 2 v q c f q for 1 < q < 6 / 25

13 The solution of the First Order ODE (ν = 1) ˆv(ξ) = 1 1 e 2π ξ 2 2π 0 e ξ 2 t iktξ 3 O(t) P f(o(t)ξ) dt = 0 e ξ 2t O(t) F((P f)(o(t) kte 3 )) dt If k = 0 and O(t) ((P f)(o(t) ) is t-independent ˆv(ξ) = 1 ξ 2 P f(ξ) 2 v q c f q for 1 < q < For q = 2: L 2 -estimates of 2 v by Plancherel s Theorem 6 / 25

14 The solution of the First Order ODE (ν = 1) ˆv(ξ) = 1 1 e 2π ξ 2 2π 0 e ξ 2 t iktξ 3 O(t) P f(o(t)ξ) dt = 0 e ξ 2t O(t) F((P f)(o(t) kte 3 )) dt If k = 0 and O(t) ((P f)(o(t) ) is t-independent ˆv(ξ) = 1 ξ 2 P f(ξ) 2 v q c f q for 1 < q < For q = 2: L 2 -estimates of 2 v by Plancherel s Theorem Evidence that the integral kernel Γ satisfies Γ(x, y) log x y x y 6 / 25

15 Partition of Unity and Littlewood-Paley Theory (k = 0) To control 2 v v let ˆψ(ξ) = ξ 2 e ξ 2, ψ t (x) = t n/2 ψ ( x t ), t > 0 Decompose ψ = j χ j ψ =: j ψj Use Littlewood-Paley theory for q : ( f q 0 ϕ s f( ) 2 ds ) 1/2 q s 7 / 25

16 Partition of Unity and Littlewood-Paley Theory (k = 0) To control 2 v v let ˆψ(ξ) = ξ 2 e ξ 2, ψ t (x) = t n/2 ψ ( x t ), t > 0 Decompose ψ = j χ j ψ =: j ψj Use Littlewood-Paley theory for q : ( f q 0 ϕ s f( ) 2 ds ) 1/2 q s Decompose operator T : f v(x) = K(x, y)p f(y) dy, R 3 into pieces T j via ψ j, i.e. T = T j 7 / 25

17 Estimate of T j (I) where T j f q c2 j M j 1/2 SL(L (q/2) ) f q M j g(x) = sup r>0 16r r/16 ( ψ j t g )(O(t) x) dt t 8 / 25

18 Estimate of T j (I) where T j f q c2 j M j 1/2 SL(L (q/2) ) f q M j g(x) = sup r>0 16r r/16 ( ψ j t g )(O(t) x) dt t Prove that M j g(x) c2 2 j M(M θ g)(x) M is the classical Hardy-Littlewood maximal operator, bounded on L q for each 1 < q < 8 / 25

19 Estimate of T j (II) (M θ g)(x) = sup r>0 16r r/16 g(o(t) x dt t behaves like a 1D maximal operator in θ for each r = (x 1, x 2 ), x 3 and is bounded on L q for each 1 < q < 9 / 25

20 Estimate of T j (II) (M θ g)(x) = sup r>0 16r r/16 g(o(t) x dt t behaves like a 1D maximal operator in θ for each r = (x 1, x 2 ), x 3 and is bounded on L q for each 1 < q < End of Proof - Incompressible Case T j f q c2 2 j f q T is bounded on L q, q 2 Complex interpolation T is bounded on L q, 1 < q < 9 / 25

21 The Compressible Case Flow of a viscous, compressible fluid around (u = 0) or past (0 u //e 3 ) a rotating obstacle with angular velocity ω = e 3 //u : ρ u t + ρu u µ u (µ + ν) div u + p(ρ) = ρf t (0, ) ρ + div (ρu) = 0 t together with the boundary conditions u(x, t) = ω x at Ω(t) {t} u(x, t) u as x ρ(x, t) ρ as x. 10 / 25

22 The new system (I) in v, σ The exterior domain Ω(t) depends on t Use a coordinate transform as before to work in a t-independent domain: y = O(t) x, v(y, t) = O(t) ( u(x, t) u ), ρ(y, t) = ρ(x, t) Linearize v w.r.t. v = 0, linearize ρ w.r.t. to ρ = 1 : ρ = 1 + σ Consider the stationary problem in R 3 t-periodic solutions of the original problem in R 3 11 / 25

23 The new system (II) in v, σ Determine the fundamental solution, prove L q -estimates µ v (µ + ν) div v + (u ω y) v + ω v + σ = F div v + div ( σ(u ω y) ) = G 12 / 25

24 The new system (II) in v, σ Determine the fundamental solution, prove L q -estimates µ v (µ + ν) div v + (u ω y) v + ω v + σ = F div v + div ( σ(u ω y) ) = G Apply div to get for g := div v and σ the system (2µ + ν) g + (u ω y) g + σ = div F g + (u ω y) σ = G. 12 / 25

25 The new system (II) in v, σ Determine the fundamental solution, prove L q -estimates µ v (µ + ν) div v + (u ω y) v + ω v + σ = F div v + div ( σ(u ω y) ) = G Apply div to get for g := div v and σ the system (2µ + ν) g + (u ω y) g + σ = div F g + (u ω y) σ = G. Simplify to get for g = div v g ( (u ω y) ) 2 g + (u ω y) (2µ + ν) g = H := G (u ω y) (div F ). 12 / 25

26 The new system (III) in g Use cylindrical coordinates y (r, θ, y 3 ), r = (y 1, y 2 ) 0 ( ω y) = θ Use Fourier transform and cylindrical coordinates ξ (s, ϕ, ξ 3 ) θ u(ξ) = ϕ û(ξ) 13 / 25

27 The new system (III) in g Use cylindrical coordinates y (r, θ, y 3 ), r = (y 1, y 2 ) 0 ( ω y) = θ Use Fourier transform and cylindrical coordinates ξ (s, ϕ, ξ 3 ) θ u(ξ) = ϕ û(ξ) With k = u, u = ke 3 we get a 2 nd order ODE w.r.t. ϕ: 2 ϕĝ + ϕ ĝ ( (2µ + ν) ξ 2 2ikξ 3 ) + ĝ ( ξ 2 + (2µ + ν)ikξ 3 ξ 2 k 2 ξ 2 3) = Ĥ 13 / 25

28 The new system (III) in g Use cylindrical coordinates y (r, θ, y 3 ), r = (y 1, y 2 ) 0 ( ω y) = θ Use Fourier transform and cylindrical coordinates ξ (s, ϕ, ξ 3 ) θ u(ξ) = ϕ û(ξ) With k = u, u = ke 3 we get a 2 nd order ODE w.r.t. ϕ: 2 ϕĝ + ϕ ĝ ( (2µ + ν) ξ 2 2ikξ 3 ) + ĝ ( ξ 2 + (2µ + ν)ikξ 3 ξ 2 k 2 ξ 2 3) = Ĥ Characteristic polynomial χ(λ) = λ 2 ( (2µ + ν) ξ 2 ) + 2ikξ 3 λ + ( ξ 2 + (2µ + ν)ikξ 3 ξ 2 k 2 ξ3 2 ) Characteristic zeros λ 1,2 (ξ) = 1 ((2µ + ν) ξ 2 + 2ikξ 3 ± ) (2µ + ν) 2 2 ξ 4 4 ξ / 25

29 The solution g With - for simplicity 2µ + ν = 2 - and the zeros λ 1,2 (ξ) = ξ 2 + ikξ 3 ± ξ 4 ξ 2 we get a unique 2π-periodic solution ĝ: 1 2π ( e λ 2 ) t ĝ(ξ) = λ 1 λ 2 1 e 2πλ e λ1t 2 1 e 2πλ Ĥ ( O(t)ξ ) dt / 25

30 The solution g With - for simplicity 2µ + ν = 2 - and the zeros λ 1,2 (ξ) = ξ 2 + ikξ 3 ± ξ 4 ξ 2 we get a unique 2π-periodic solution ĝ: 1 2π ( e λ 2 ) t ĝ(ξ) = λ 1 λ 2 1 e 2πλ e λ1t 2 1 e 2πλ Ĥ ( O(t)ξ ) dt. 1 0 If H and Ĥ are independent of θ and φ, resp., then ĝ(ξ) = 2πĤ(ξ)( ξ 2 + 2ikξ 3 ξ 2 k 2 ξ 2 3) / 25

31 The solution g With - for simplicity 2µ + ν = 2 - and the zeros λ 1,2 (ξ) = ξ 2 + ikξ 3 ± ξ 4 ξ 2 we get a unique 2π-periodic solution ĝ: 1 2π ( e λ 2 ) t ĝ(ξ) = λ 1 λ 2 1 e 2πλ e λ1t 2 1 e 2πλ Ĥ ( O(t)ξ ) dt. 1 0 If H and Ĥ are independent of θ and φ, resp., then ĝ(ξ) = 2πĤ(ξ)( ξ 2 + 2ikξ 3 ξ 2 k 2 ξ 2 3) 1. We also have, with µ 1,2 (ξ) = ξ 2 ± ξ 4 ξ 2, ĝ(ξ) = 0 e µ 2t e µ 1t µ 1 µ 2 F ( H(O(t) kte 3 ) ) (ξ) dt 14 / 25

32 Fourier Analysis Note: µ 1,2 (ξ) is a smooth multiplier function on R 3 \ {0} \ B 1 Use a partition of unity 1 = η 0 + η 1 + η 2 where η 0 C 0 (B 1/2 ), η 1 (ξ) = 1 for 1 2 ξ 2, η 2 = 1 in (B 4 ) c 15 / 25

33 Fourier Analysis Note: µ 1,2 (ξ) is a smooth multiplier function on R 3 \ {0} \ B 1 Use a partition of unity 1 = η 0 + η 1 + η 2 where η 0 C 0 (B 1/2 ), η 1 (ξ) = 1 for 1 2 ξ 2, η 2 = 1 in (B 4 ) c For ξ > 4 all multiplier functions are smooth, Hörmander-Mikhlin multiplier theory L q -estimates, 1 < q < 15 / 25

34 Fourier Analysis Note: µ 1,2 (ξ) is a smooth multiplier function on R 3 \ {0} \ B 1 Use a partition of unity 1 = η 0 + η 1 + η 2 where η 0 C 0 (B 1/2 ), η 1 (ξ) = 1 for 1 2 ξ 2, η 2 = 1 in (B 4 ) c For ξ > 4 all multiplier functions are smooth, Hörmander-Mikhlin multiplier theory L q -estimates, 1 < q < For 0 < ξ < 1 4 all multiplier functions are smooth, Hörmander-Mikhlin multiplier theory L q -estimates, 1 < q < 15 / 25

35 Fourier Analysis Note: µ 1,2 (ξ) is a smooth multiplier function on R 3 \ {0} \ B 1 Use a partition of unity 1 = η 0 + η 1 + η 2 where η 0 C 0 (B 1/2 ), η 1 (ξ) = 1 for 1 2 ξ 2, η 2 = 1 in (B 4 ) c For ξ > 4 all multiplier functions are smooth, Hörmander-Mikhlin multiplier theory L q -estimates, 1 < q < For 0 < ξ < 1 4 all multiplier functions are smooth, Hörmander-Mikhlin multiplier theory L q -estimates, 1 < q < Analysis near ξ = 1: multiplication of smooth multiplier functions (okay) with the crucial multiplier functions η 1 (ξ) 1 ξ 2 χ B1 (ξ) (1 ξ 2 ) 1/2 +, η 1 (ξ) ξ 2 1χ B c 1 (ξ) η 1 (ξ)( ξ 2 1 ) 1/ / 25

36 Bochner-Riesz Means Theorem: The Bochner-Riesz multiplier operator B 1/2 defined by the multiplier function (1 ξ 2 ) 1/2 + on R3 can be represented by convolution with the kernel cj 2 ( x ) x 2 and is L q (R 3 )-bounded if and only if 6 5 < q < / 25

37 Bochner-Riesz Means Theorem: The Bochner-Riesz multiplier operator B 1/2 defined by the multiplier function (1 ξ 2 ) 1/2 + on R3 can be represented by convolution with the kernel cj 2 ( x ) x 2 and is L q (R 3 )-bounded if and only if 6 5 < q < 6. Let the Bochner-Riesz multiplier operator B λ on R 3 be defined by the multiplier function ( ξ 2 1) λ +η 1 (ξ) and let K λ (x) denote the corresponding integral kernel: K λ (x) = 2 x 4 1 sin( x τ)(τ 2 1) Rλ e i Iλ ln(τ 2 1) τη 1 (τ) dτ 16 / 25

38 Bochner-Riesz Means Theorem: The Bochner-Riesz multiplier operator B 1/2 defined by the multiplier function (1 ξ 2 ) 1/2 + on R3 can be represented by convolution with the kernel cj 2 ( x ) x 2 and is L q (R 3 )-bounded if and only if 6 5 < q < 6. Let the Bochner-Riesz multiplier operator B λ on R 3 be defined by the multiplier function ( ξ 2 1) λ +η 1 (ξ) and let K λ (x) denote the corresponding integral kernel: K λ (x) = 2 x 4 Theorem: For λ C, Rλ > 1 K λ (x) C 1 sin( x τ)(τ 2 1) Rλ e i Iλ ln(τ 2 1) τη 1 (τ) dτ ( 1 + Iλ ) [Rλ]+2 (1 + x ) Rλ+2 ( log(2 + x ) when Rλ N0 ). Proof: Integration by parts (with sin( x τ) to get x (Rλ+2) ) 16 / 25

39 Details of Proof ˆK λ = ( ξ 2 1) λ +η 1 (ξ) defines an L 2 -multiplier operator 17 / 25

40 Details of Proof ˆK λ = ( ξ 2 1) λ +η 1 (ξ) defines an L 2 -multiplier operator Choose a partition of unity (ψ j ) j N0 on R 3 with supp ψ 0 B 1, supp ψ j B 2 j+1 \ B 2 j 1, define K j λ = ψ jk λ so that K λ = Kλ 0 + j=1 K j λ 17 / 25

41 Details of Proof ˆK λ = ( ξ 2 1) λ +η 1 (ξ) defines an L 2 -multiplier operator Choose a partition of unity (ψ j ) j N0 on R 3 with supp ψ 0 B 1, supp ψ j B 2 j+1 \ B 2 j 1, define K j λ = ψ jk λ so that K λ = Kλ 0 + j=1 K j λ K 0 λ is a bounded convolution operator on each Lq. 17 / 25

42 Details of Proof ˆK λ = ( ξ 2 1) λ +η 1 (ξ) defines an L 2 -multiplier operator Choose a partition of unity (ψ j ) j N0 on R 3 with supp ψ 0 B 1, supp ψ j B 2 j+1 \ B 2 j 1, define K j λ = ψ jk λ so that K λ = Kλ 0 + j=1 K j λ K 0 λ is a bounded convolution operator on each Lq. Claim: For p 1 satisfying p < 1 3 (2 + Rλ) ( Rλ > 1 4 ) K j λ f p C(λ)j 2 jδ f p ; here δ = 1 2 (1 + 2Rλ) 3( 1 p 1 2 ) > 0. Consequence: K λ is a bounded convolution operator on L p with p as above 17 / 25

43 Proof of Claim, j N (Part I) It suffices to consider f with supp f cube of side length 2 j supp K j λ f lies in a cube of side length 3 2j 18 / 25

44 Proof of Claim, j N (Part I) It suffices to consider f with supp f cube of side length 2 j supp K j λ f lies in a cube of side length 3 2j K j λ and ˆK j λ are radial for such f K j λ f 2 p c 2 6j( 1 p 1 2 ) K j λ f 2 2 = c 2 6j( 1 p 1 2 ) ˆK j λ ˆf 2 2, ˆK j λ ˆf 2 2 = 0 ˆK j λ (r, 0, 0) 2( ˆf(rθ) ) 2 dθ r 2 dr. S 2 18 / 25

45 Proof of Claim, j N (Part I) It suffices to consider f with supp f cube of side length 2 j supp K j λ f lies in a cube of side length 3 2j K j λ and ˆK j λ are radial for such f K j λ f 2 p c 2 6j( 1 p 1 2 ) K j λ f 2 2 = c 2 6j( 1 p 1 2 ) ˆK j λ ˆf 2 2, ˆK j λ ˆf 2 2 = 0 ˆK j λ (r, 0, 0) 2( ˆf(rθ) ) 2 dθ r 2 dr. S 2 Fourier Restriction Theorem, 1 p 4 3 ˆf(rθ) 2 dθ Cr 6(p 1)/p f 2 p S 2 18 / 25

46 Proof of Claim, j N (Part II) ˆK j λ ˆf 2 2 C 2 p f 2 p = C 2 p f 2 p Summary: it holds 0 R 3 ˆK j λ (r, 0, 0) 2 r 2 6(p 1)/p dr ˆK j λ (ζ) 2 ζ 6/p dζ (decompose integral into parts on B 1/2 and B c 1/2, use ψ j, ˆψ j S(R 3 ) or pointwise estimate of K λ (x)) Cj 2 2 (1+2Rλ)j f 2 p K j λ f p Cj 2 {3( 1 p 1 2 ) 1 2 (1+2Rλ)}j f p

47 Proof of Claim, j N (Part II) ˆK j λ ˆf 2 2 C 2 p f 2 p = C 2 p f 2 p 0 R 3 ˆK j λ (r, 0, 0) 2 r 2 6(p 1)/p dr ˆK j λ (ζ) 2 ζ 6/p dζ (decompose integral into parts on B 1/2 and B c 1/2, use ψ j, ˆψ j S(R 3 ) or pointwise estimate of K λ (x)) Cj 2 2 (1+2Rλ)j f 2 p 19 / 25

48 Proof of Claim, j N (Part II) ˆK j λ ˆf 2 2 C 2 p f 2 p = C 2 p f 2 p Summary: it holds 0 R 3 ˆK j λ (r, 0, 0) 2 r 2 6(p 1)/p dr ˆK j λ (ζ) 2 ζ 6/p dζ (decompose integral into parts on B 1/2 and B c 1/2, use ψ j, ˆψ j S(R 3 ) or pointwise estimate of K λ (x)) Cj 2 2 (1+2Rλ)j f 2 p K j λ f p Cj 2 {3( 1 p 1 2 ) 1 2 (1+2Rλ)}j f p 19 / 25

49 Theorem I K λ defines a bounded convolution operator on each L p, p 1 satisfying p < 1 3 (2 + Rλ), i.e. for 3 2+Rλ < p 4 3 and for p = 2. Complex interpolation 3 2+Rλ < p 2 Duality 1 p 1 < 1 (1 + 2Rλ) 2 6 For λ = 1 2 this condition reduces to 6 5 < p < / 25

50 Theorem II Let 6 5 < q < 6 and u 1 2. Furthermore, let G Lq (R 3 ) and let F = div Φ where Φ, (ω y) Φ, u Φ L q (R 3 ) the linear problem has a unique solution g L q (R 3 ) such that g q c ( G q + ωφ q + (ω y) Φ q + u Φ q ) 21 / 25

51 Theorem II Let 6 5 < q < 6 and u 1 2. Furthermore, let G Lq (R 3 ) and let F = div Φ where Φ, (ω y) Φ, u Φ L q (R 3 ) the linear problem has a unique solution g L q (R 3 ) such that g q c ( G q + ωφ q + (ω y) Φ q + u Φ q ) Get v, σ from the system µ v + (u ω y) v + ω v + σ = F + (µ + ν) g div v = g 21 / 25

52 Theorem II Let 6 5 < q < 6 and u 1 2. Furthermore, let G Lq (R 3 ) and let F = div Φ where Φ, (ω y) Φ, u Φ L q (R 3 ) the linear problem has a unique solution g L q (R 3 ) such that g q c ( G q + ωφ q + (ω y) Φ q + u Φ q ) Get v, σ from the system µ v + (u ω y) v + ω v + σ = F + (µ + ν) g div v = g Theorem III Let F = div Φ, g as before, let 6 5 < q < 6. Then v, σ satisfy 2 v q + σ q c ( F + (µ + ν) g q + µ g + (ω y)g u g q ) 21 / 25

53 References R. Farwig, M. Pokorný: TU Darmstadt, Preprint (2014) S. Kračmar, Š. Nečasová, A. Novotný, A.: Ann. Univ. Ferrara (2014) R. Danchin: Invent. Math. (2000) F. Charve, R. Danchin: Arch. Ration. Mech. Anal. (2010) T. Kobayashi, Y. Shibata: Comm. Math. Phys. (1999) Y. Shibata, K. Tanaka: J. Math. Soc. Japan (2003) Y. Shibata, K. Tanaka: Math. Methods Appl. Sci. (2004) Y. Enomoto, Y. Shibata: Funkcial. Ekvac. (2013) Y. Enomoto, L. von Below, Y. Shibata, Yoshihiro. Ann. Univ. Ferrara (2014) 22 / 25

54 International Research Training Group 1529 Mathematical Fluid Dynamics Autumn School and Workshop Bad Boll, Germany October 27 30, 2014 Lecture Series Peter Constantin, Princeton PDE Problems of Hydrodynamic Origin Yasunori Maekawa, Sendai Analysis of Incompressible Flows in Unbounded Domains László Székelyhidi, Leipzig The h-principle in Fluid Mechanics and Onsager's Conjecture Confirmed Speakers K. Abe (Nagoya) M. Kyed (Kassel) H. Abels (Regensburg) M. Lopes Filho (Rio de Janeiro) D. Bothe (Darmstadt) P. Maremonti (Naples) L. Brandolese (Lyon) P. Mucha (Warsaw) R. Danchin (Paris) Š. Nečasová (Prague) K. Disser (Berlin) J. Prüss (Halle) R. Farwig (Darmstadt) M. Schoenbek (Santa Cruz) E. Feireisl (Prague) F. Sueur (Paris) T. Hishida (Nagoya) R. Takada (Sendai) Organizers: J. Kelliher (Los Angeles) W. Varnhorn (Kassel) H. Koch (Bonn) E. Zatorska (Warsaw) M. Hieber H. Kozono 23 / 25 For further information please visit: or contact: Verena Schmid, , igk@mathematik.tu-darmstadt.de

55 From Incompressibility to Compressibility 24 / 25

56 From Incompressibility to Compressibility Standard estimates for incompressible fluid flow around/past a rotating obstacle: 1 < q < 24 / 25

57 From Incompressibility to Compressibility Standard estimates for incompressible fluid flow around/past a rotating obstacle: 1 < q < Estimates for compressible fluid flow around/past a rotating obstacle: 6 Only for 5 < q < 6 24 / 25

58 Bochner-Riesz Means Let B λ denote the Bochner-Riesz Operator defined by the multiplier function (1 ξ 2 ) λ +, λ 0, on R n If λ > n 1 2, then Bλ is bounded on L q, 1 q If λ > n 1 2(n+1) and 1 q 1 2 < 1 2n + λ n, then Bλ is L q -bounded If λ > 0 and 1 q n + λ n, then Bλ is L q -unbounded 25 / 25