Existence of algebraic vortex spirals and ill-posedness of inviscid - PowerPoint PPT Presentation

Existence of algebraic vortex spirals and ill-posedness of inviscid flow (Part III) Volker Elling S.I.S.S.A. Trieste, June 6–10, 2011

Birkhoff-Rott for multiple branches m4N1ell.vs Sheets p = 0 , ..., N − 1 at equal angles 2 π N , N ∈ N . z ( γ ) contour for sheet 0 � sheet p is u p z ( γ ) with u = exp 2 πi N . Redefine Γ as circulation for all branches combined ⇒ each individual branch generates Γ N � 1 N − 1 dγ ′ � ∗ � (1 − 2 µ ) γ ∂ 1 ∂γz ( γ )+ µz ( γ ) = W ∗ = � 2 πi p.v. . z ( γ ) − u p z ( γ ′ ) N R p =0 � �� � =: A p * 2

Kaden approximation (Kaden (1931), Rott (1956)): γ = 0 in center; increasing to outside Spiral turns almost circular, with almost uniform d Γ /ds . Γ Inside circle: W = 0. Outside circle: W = 2 πiZ (= point vortex). Approximation of W integral: � � ∗ 1 (1 − 2 µ ) γ ∂ γ ∂γz ( γ ) + µz ( γ ) = 2 πi · . z ( γ ) ODE for z ( γ ). Ansatz: ̺ distance from center, θ angle travelled (= 0 at infinity, → ∞ towards spiral center): z = z k = ̺ ( γ ) exp( iθ ( γ )) . 3

� 1 � ∗ (1 − 2 µ ) γ ∂ γ ∂γz + µγ = , z = ̺ ( γ ) exp( iθ ( γ )) . 2 πi z (1 − 2 µ ) γ ( ∂ ∂γ̺ + i̺ ∂ i γ ∂γθ ) e iθ + µ̺e iθ = ̺e iθ . 2 π (1 − 2 µ ) ∂ γ = ̺ 2 − 1 /µ ∂γ̺ + µ̺ = 0 ⇒ (1 − 2 µ ) ̺ ∂ γ γ = (2 πθ ) 1 − 2 µ ∂γθ = ⇒ 2 π̺ ̺ = (2 πθ ) − µ , z k = (2 πθ ) − µ e iθ . * Algebraic spirals ( µ = 1: hyperbolic). Prandtl (1920s): logarithmic spirals. Exact solutions, easier to con- struct [Saffman, Vortex Dynamics] but seem uncommon. 4

Pseudo-angle Kaden approximation: γ = (2 πθ ) 1 − 2 µ θ = (2 π ) − 1 γ 1 / (1 − 2 µ ) ⇔ More convenient: define pseudo-angle t as * t = t ( γ ) = (2 π ) − 1 γ 1 / (1 − 2 µ ) γ = (2 πt ) 1 − 2 µ ⇔ If Kaden approximation good, can expect t ≈ θ for θ ≈ ∞ . γ, t ∈ (0 , ∞ ); γ ↓ 0 means t ↑ ∞ and vice versa. Advantage: instead of ̺ ( θ ) , γ ( θ ) can use single ( C -valued) z ( t ). Left-hand side: (1 − 2 µ ) γz γ + µz = (1 − 2 µ ) γ ∂ t γz t + µz = tz t + µz New velocity integral: � ∞ � ∞ dγ ′ 1 1 z ( t ) − u p z ( s ) | dγ 1 2 πi p.v. A p z ( γ ) − u p z ( γ ′ ) = 2 πi p.v. A p dt ( s ) | ds 0 0 � ∞ 1 = − i (2 µ − 1)(2 π ) − 2 µ p.v. z ( t ) − u p z ( s ) s − 2 µ ds A p 0 5

Asymptotic series ansatz z ( t ) = z k ( t ) g ( t ) , g ( t ) = 1 + ˜ g ( t ) , z ( t ) = ˜ ˜ g ( t ) z k ( t ) . ˜ g = 0 yields z = z k (Kaden approximation). Idea: try (for example) ∞ � a j t − j g ( t ) = ˜ , a j ∈ C j =1 0 ! = F ( z ) = F 1 ( a 1 ) t − β + F 2 ( a 1 , a 2 ) t − β − 1 + F 3 ( a 1 , a 2 , a 3 ) t − β − 2 + ... Determine a 1 , then a 2 , then a 3 , ... Problem: does not seem to converge [no surprise], not even for t large. 6

Why asymptotic series rarely work: ∞ � a j t − j j =1 1. Series does not converge (coefficients a j grow too fast); 2. If it does, may converge only locally ( t ≈ ∞ ); 3. If it converges, need not converge to a solution; 4. Even if true, any of these is usually very hard to prove. Examples: du dθ + θ − 1 u = 0 u (1) = 1 , works (Frobenius) du u (1) = 1 , dθ + u = 0 fails! (reason 3.) � t ∞ s β e is ds = e it � t β i + βt β − 1 + β ( β − 1) t β − 2 � + ... fails! (reason 1.) i 2 i 3 7

Function space z ( t ) = z k ( t ) g ( t ) g ( t ) = 1 + ˜ g ( t ) z ( t ) = ˜ ˜ g ( t ) z k ( t ) . , , * Idea: use truncated asymptotic series ansatz: M � a m t − m + (1 + t ) − M r ( t ) ˜ g ( t ) = χ ( t ) j =1 M � = ˜ g m ( t ) + ˜ g M +1 ( t ) ∗ j =1 r ∈ H S a j ∈ C , ( S large) χ ( t ) smooth function, = 1 near t = ∞ , = 0 near t ∈ [ −∞ , 0]. Determine a 1 , then a 2 , then a 3 , ... as before, but r determined by iteration. For N large, will have a m , r small. 8

Approach: express problem as ( F nonlinear C 1 operator; z, F ( z ) ∈ Banach spaces) F ( z ) = 0 Solve by quasi-Newton iteration z ← K ( z ) := z − A − 1 F ( z ) A − 1 “approximate inverse” for F ′ ( z ). � � A − 1 � �� | K ( z ) − K ( w ) | = | z − w − A − 1 ( F ( z ) − F ( w )) | = � � A ( z − w ) − ( F ( z ) − F ( w ) � � � ≤ | A − 1 | � A ( z − w ) − F ′ ( z )( z − w ) + F ′ ( z )( z − w ) − ( F ( z ) − F ( w )) � � � � � ≤ | A − 1 | | A − F ′ ( z ) || z − w | + | F ( w ) − F ( z ) − F ′ ( z )( w − z ) | = | A − F ′ ( z ) | O ( z − w ) + o ( z − w ) If A ≈ F ′ ( z ) and | z − w | small, then for some 0 ≤ L< 1 ≤ L | z − w | , Iteration map K is uniform contraction, Banach fixed point theorem: converges to fixed point z = K ( z ) so that F ( z ) = 0. 9

Motivation for truncated series ansatz: M � a m t − m + t − M r ( t ) z ( t ) = z k ( t ) g ( t ) , D ∋ g = 1 + (at t ≈ ∞ ) m =1 To find “rest” r ∈ H S have to solve G ( t − M r ) = 0 for some nonlinear operator (self-similar Birkhoff-Rott) G : D → R . Need G ′ (0) isomorphism: G ′ (0) − 1 : R → D ⇒ R must enforce sufficient decay in t , say O ( t − M + O (1) ). 0 = G ( t − M r ) = G (0) + G ′ (0)[ t − M r ] + G ′′ (0) [ t − M r, t − M r ] + ... 2 For large N , G ′′ (0) integral operator with near-diagonal kernel: G ′′ (0)[ t − M r, t − M r ] = O ( t − 2 M + O (1) ) . If M large, 2 M ≫ M + O (1): nonlinear terms allow wasteful analysis. ⇒ essential difficulty is only linear ! 10

Integral expansion � ∞ � ∞ � ∞ s − 2 µ ds s − 2 µ ds s − 2 µ ds z = z k +˜ z p.v. A p z ( t ) − u p z ( s ) = p.v. A p = p.v. A p ∆ z ∆ z k + ∆˜ z 0 0 0 For N large, due to smallness | ∆˜ z | < | ∆ z k | . � ∞ ∞ z ) j (∆˜ � (∆ z k ) j +1 s − 2 µ ds ( − 1) j p.v. = A p 0 j =0 M +1 � (˜ z = ˜ z m , ˜ z k = ˜ g m z k ) m =1 � ∞ � ∞ ( � M +1 z m ) j z ) j (∆˜ m =1 ∆˜ (∆ z k ) j +1 s − 2 µ ds = p.v. s − 2 µ ds p.v. A p A p (∆ z k ) j +1 0 0 � ∞ �� M +1 z m ) α ( m ) � j m =1 (∆˜ � s − 2 µ ds = p.v. A p (∆ z k ) j +1 α 0 | α | = j 11

g m z k , | α | = � M +1 Integral expansion (2): ˜ z m = ˜ m =1 α ( m ), z m ( t ) − u p ˜ g m ( t ) z k ( t ) − u p ˜ ∆˜ z m = ˜ z m ( s ) = ˜ g m ( s ) z k ( s ) � � � � z k ( t ) − u p z ( s ) u p z ( s ) = ˜ g m ( t )∆ z k + u p z ( s )∆ g m = ˜ g m ( t ) + ˜ g m ( t ) − ˜ g m ( s ) � ∞ � M +1 z m ) α ( j ) m =1 (∆˜ s − 2 µ ds p.v. A p (∆ z k ) j +1 0 � α ( j ) � � M +1 g m ( t )∆ z k + u p z k ( s )∆˜ � ∞ ˜ g m ( s ) m =1 s − 2 µ ds = p.v. A p (∆ z k ) j +1 0 � ∞ � M +1 m =1 ( u p z k ( s )∆˜ g m ) β ( m ) (˜ g m ( t )∆ z k ) α ( m ) − β ( m ) � α � � s − 2 µ ds = p.v. A p (∆ z k ) j +1 β 0 β ≤ α � ∞ M +1 � M +1 g m ) β ( m ) m =1 (∆˜ | α | = j � g m ( t ) α ( m ) − β ( m ) p.v. A p ( u p z k ( s )) | β | s − 2 µ ds � ˜ (∆ z k ) 1+ | β | 0 m =1 12

Integral expansion (3): have expanded velocity integral into series of integrals like � ∞ � M +1 g m ) β ( m ) m =1 (∆˜ s − 2 µ ds A p ( u p z k ( s )) | β | p.v. (∆ z k ) 1+ | β | 0 | β | = 0: constant part � ∞ 1 s − 2 µ ds p.v. A p ∆ z k 0 | β | = 1: linear parts � ∞ A p u p z k ( s ) ∆˜ g m (∆ z k ) 2 s − 2 µ ds p.v. 0 | β | = 2 , 3 , ... : quadratic and higher parts like � ∞ A p ( u p z k ( s )) 2 ∆˜ g m ∆˜ g ℓ (∆ z k ) 3 s − 2 µ ds p.v. 0 13

Key difficulty: spiral rollup. With Kaden approximation z k ( t ) = (2 πt ) − µ e iθ , typical linearized operator like � ∞ ˜ z ( t ) − ˜ z ( s ) [ t − µ exp( it ) − u p s − µ exp( is )] 2 s − 2 µ ds p.v. A p 0 Key problem: near-singularity whenever (e.g.) p = 0, s ≈ t + 2 πk ( k ∈ Z ) . 1. Spiral must not self-intersect 2. delicate cancellations (left, right side of each turn) must not be upset. 14

Analogies in elliptic PDE ( a ij ) = A = A ( x ) continuous Variable coefficients = constant coefficients + small ∂ 2 ∂ 2 ∂ 2 � � � � � f = u − a ij ( x ) u = u − a ij (0) + a ij ( x ) − a ij (0) u u ∂x i ∂x j ∂x i ∂x j ∂x i ∂x j i,j i,j i,j � �� � small for x ≈ 0 � �� � L 0 u : constant coeff. ∂ 2 � ∧ � � � � � u ∧ ( ξ ) u − a ij (0) u ( ξ ) = 1 + a ij (0) ξ i ξ j ∂x i ∂x j i,j i,j � �� � σ ( ξ ) Elliptic means σ ( ξ ) invertible ⇒ Fourier transform to invert operator: : H s → H s +2 u ∧ ( ξ ) = σ ( ξ ) − 1 g ∧ ( ξ ) L − 1 L 0 u = g ⇒ , 0 ∞ ( L 0 + E ) − 1 = L − 1 � ( EL − 1 0 ) j ( L 0 + E ) u = f, 0 j =0 1 0 � so that � EL − 1 0 � ≤ � E � · � L − 1 converges if � E � < 0 � < 1. � L − 1 15