Part I: Numerical Fourier analysis of quasi periodic functions

Transcription

1 Part I: Numerical Fourier analysis of quasi periodic functions I.1 The procedure I.2 Error estimation f(t) Q f (t; A, ω) DFT(f) = DFT(Q f ) Applying the above procedure, the system to be solved can be written as g(y + y) = b + b. The exact frequencies and amplitudes would be obtained if b = 0. A bound for the error is given by y Dg(y) 1 b. We will bound Dg(y) 1 and b. I.3 Applications Academic example, to show the procedure and the accuracy of the error estimation Solar System models: SS = RTBP + Perturbations (Fourier analysis) SS = RTBP + Perturbations(ω 1, ω 2,...) 3

2 Part II: The neighborhood of the collinear equilibrium points in the RTBP Final goal: To get Poincaré section representations of the flow around L 1,2,3,. 0.3 Energy Tools: Parallel algorithms for the computation of different kinds of families of periodic orbits and invariant tori. 4

3 I.1 Refined Fourier analysis procedure Given a real analytic quasi periodic function, f(t), at N equally spaced points on an interval [0, T], {f(t l )} N 1 l=0, t l = l T N, the goal is to compute a trigonometric approximation N f Q f (t) = A c 0 + (A c l cos(2πν l T t) + As l sin(2πν l T t). The Discrete Fourier Transform (DFT) of f is defined through f(t l ) = 1 2 (c f,t,n(0) + c f,t,n ( N 2 )cos(2πn 2 t l T ))+ N/2 1 (c f,t,n (j)cos( 2πjt l T ) + s f,t,n(j)sin( 2πjt l T )). where j=1 c f,t,n (j) = 2 N s f,t,n (j) = 2 N N 1 l=0 N 1 l=0 f(t l )cos(2π j N l), j = 0,..., N 2, f(t l )sin(2π j N l), j = 1,..., N

4 Algorithm 1. Set an starting thresold for collecting peaks of the modulus of the DFT of f(t). 2. Find initial approximations of the frequencies, starting from the peaks of the DFT greater than the thresold. 3. Find the amplitudes of the frequencies found in the previous step (DFT(Q f ) = DFT(f)). 4. Simultaneously refine ALL the frequencies and amplitudes of the current quasi periodic approximation of f. 5. Perform a DFT of the input signal minus the current quasi periodic approximation obtained in step 4, decrease the thresold and go back to step 2. 6-a

5 Equations of step 3 We ask DFT(Q f ) = DFT(f), being N f Q f (t) = A c 0 + (A c l cos(2πν l T t) + As l sin(2πν l T t). The system of equations to be solved is linear and (1 + 2N f ) (1 + 2N f ): Nf A c 0 cn h 1,T,N (0) + (A c l cn h ν l,n (0) + As l c n h ν l,n (0)) = cn h Nf A c 0 cn h 1,T,N (j) + (A c l cn h ν l,n (j) + As l c n h ν l,n (j)) = cn h N f (A c l sn h ν l,t (j) + Ac l s n h ν l,t (j)) = sn h where j = [ν l + 0.5], l = 1 N f, and c n h 1 (j) = cn h 1,T,N (j), c n h ν l,n (j) = cn h cos( 2πν l ),T,N(j), sn h ν l,n (j) = sn h T c n h ν l,n (j) = cn h sin( 2πν l ),T,N(j), sn h ν l,n (j) = sn h T cos( 2πν l T sin( 2πν l T f,t,n (0) f,t,n (j) f,t,n (j) ),T,N(j), ),T,N(j). 6-b

6 Equations for step 4 The system of equations is a (1 + 3N f ) (1 + 3N f ) non linear system which is solved by Newton s method: A c 0 cn h 1,T,N (0) + Nf A c 0 cn h 1,T,N (j i) + A c 0 csn h 1,T,N (j+ i ) + N f N f N f (A c l cn h ν l,n (0) + As l c n h ν l,n (0)) = cn h (A c l cn h ν l,n (j i) + A s l c n h ν l,n (j i)) = c n h (A c l sn h ν l,n (j i) + A s l s n h ν l,n (j i)) = s n h (A c l csn h ν l,n (j+ i ) + A s l cs n h ν l,n (j+ i )) = cs n h f,t,n (0) f,t,n (j i) f,t,n (j i) f,t,n (j+ i ) being j i = [ν i + 0.5], j + i j i, j + i j i = 1. 6-c

7 An example: f(t) = cos(2π0.23t) 1 sin(2π0.27t) + sin(2π0.37t) Starting thresold: 0.8 modulus of the DFT of the input data: peaks j = 61, j = Approximation of frequencies (Laskar s method): peak 61 frequency peak 189 frequency Computation of amplitudes from known frequencies: Frequency Cosine amplitude Sine amplitude modulus of the DFT of the residual Iterative refinement: Frequency Cosine amplitude Sine amplitude a

8 5. modulus of the DFT of input signal minus step 4: New threshold: Approximation of frequencies (Laskar s method): peak 138 frequency Amplitudes from known frequencies: Frequency Cosine amplitude Sine amplitude Iterative refinement: Frequency Cosine amplitude Sine amplitude modulus of the DFT of the residual: 5e-14 4e-14 3e-14 2e-14 1e b

9 I.2 Error estimation We assume f is real analytic and quasi periodic, f(t) = A c 0 + k Z m kω>0 (A c k cos(2πkωt) + As k sin(2πkωt)), and its Fourier coefficients a k satisfy the Cauchy estimates (A c k )2 + (A s k )2 Ce δ k k Z m. The frequency vector ω = (ω 1,... ω m ) satisfies a Diophantine condition with D, τ > 0. kω D k τ, We determine the frequencies {ν l } N f of order r 0 1 (ν l Tkω, 1 k r 0 1) and the corresponding amplitudes {A c l }N f l=0, {As l }N f. 8

10 Strategy Let us denote, f r0 (t) = A c 0 + (A c k cos(2πkωt) + As k sin(2πkωt)). k r 0 1 kω>0 The system of equations used for simultaneous improvement of frequencies and amplitudes can be written as A c 0 + Nf A c 0 cn h 1 (j i) + A c 0 csn h 1 (j+ i ) + N f N f N f (A c l cn h ν l,n (0) + As l c n h ν l,n (0)) = cn h f r0,t,n (0) + cn h f f r0,t,n (0) (A c l cn h ν l,n (j i) + A s l c n h ν l,n (j i)) = c n h f r0,t,n (j i) + c n h f f r0,t,n (j i) (A c l sn h ν l,n (j i) + A s l s n h ν l,n (j i)) = s n h f r0,t,n (j i) + s n h f f r0,t,n (j i) (A c l csn h ν l,n (j+ i ) + A s l cs n h ν l,n (j+ i )) }{{} g(y + y) = cs n h f r0,t,n (j+ i ) + cs n h f f r0,t,n (j+ i ). } {{ } b } {{ } b We would get the exact frequencies and amplitudes if b = 0. The error in frequencies and amplitudes is given, in the first order approximation, by y Dg(y) 1 b. We will obtain bounds for Dg(y) 1 and b. 9

11 Bound for Dg(y) 1 Preliminaries We can write Dg(y) =: M = 2 B 0,1... B 0,Nf 0 B 1,1... B 1,Nf B Nf,1... B Nf,N f. We split M = M D + M O, M = 0 B 1, B Nf,N f + 0 B 0,1... B 0,Nf B 1,Nf B Nf, The idea is to obtain bounds for M 1 D, M O and use (M D + M O ) 1 M 1 D 1 M 1 D M O. The components B i,l of M (i 0) have the following general aspect A c l cn h ν l,n (j i) + A s l cn h ν l,n (j i) c n h ν l,n (j i) c n h ν l,n (j i) A c l sn h ν l,n (j i) + A s l sn h ν l,n (j i) s n h ν l,n (j i) A c l csn h ν l,n (j+ i ) + A s l csn h ν l,n (j+ i ) cs n h ν l,n (j+ i ) cs n h s n h ν l,n (j i) ν l,n (j+ i ). 10

12 First simplification: M M The first simplification consists in the use of the Truncated Continuous Fourier Tansform (C, S), 1 T T 0 H n h T (t)f(t)e i2π j T t =: 1 2 (Cn h f,t (j) + isn h f,t (j)) instead of the DFT (c, s). We define CS n h ν (j) as in the discrete case. In this way, M M = M D + M O (substitute DFT by TCFT) = M D + M O Remark 1: The difference between cs and CS can be bounded from a discrete version of Poisson s summation formula: c n h f,t,n (j) = l= C n + ln h f,t (j ),... T Remark 2: Explicit formulae for the components of M can be obtained. 11

13 Formulae for the components of M The components of the matrix M are given by ( C n h ν (j) = K sin(2π(ν j)) n h ψ nh (ν j).. ( C n h ν (j) = K h r (ν j) n h ψ nh (ν j).. + sin(2π( ν j)) ψ nh ( ν j). h r ( ν j) ψ nh ( ν j). ), ), where K nh = ( 1)n h (n h!) 2 n h 2π ψ nh (x) = (x + l) l= n h h r (x) = 2π cos(2πx) r nh (x)sin(2πx), h i (x) = 2π sin(2πx) r nh (x)(1 cos(2πx)), r nh (x) = n h l= n h 1 x + l = ψ n h (x) ψ nh (x). 12

14 Second simplification: M M The second simplification consist in eliminating the second summands of the expressions of the components of M as given in the previous lemma. In this way, M = M D + M O (remove the second summands) M = M D + M O Remark 1: The difference between the components of M and M ( CS and cs) can be bounded. Remark 2: the expression for cs only depends on the difference ν j. 13

15 Bound for M 1 D Since M D is diagonal, to compute M 1 D to invert its diagonal blocks B i,i A c i cn h ν i,n (j i) + A s i cn h ν i,n (j i) c n h ν i,n (j i) A c i sn h ν i,n (j i) + A s i sn h ν i,n (j i) s n h ν i,n (j i) A c i csn h ν i,n (j+ i ) + A s i csn h ν i,n (j+ i ) cs n h ν i,n (j+ i ) cs n h we only need c n h ν i,n (j i) s n h ν i,n (j i) ν i,n (j+ i ) For that, we first show that, if (A c i, As i ) (0,0), B i,i is invertible either setting cs = c or cs = s.. The actual bounds of B i,i are computed numerically for each n h, by first minimizing w.r.t. cs = c, s and maximizing w.r.t. θ [0,2π], ν j 1/2 the supremum norm of c n h ν i,n (j i)cos θ + c n h ν i,n (j i)sin θ ν i,n (j i)sin θ c n h s n h ν i,n (j i) ν i,n (j i) s n h ν i,n (j i)cos θ + s n h cs n h ν i,n (j+ i )cos θ + cs n h ν i,n(j + i )sin θ cs n h ν i,n (j+ i ) cs n h We call G nh In this way, we get being c n h ν i,n (j i) s n h ν i,n (j i) ν i,n(j + i ) the obtained quantity. Some values are n h G nh (M D ) 1 max(a 1 min,1)g n h, A i = ((A c i) 2 + (A s i) 2 ) 1/2 A min = min i=1 N f {A 1,..., A Nf } 1. 14

16 Bounds for M 1 The computation of the bound for Dg(y) 1 is completed by following the following scheme: M 1 D max(a 1 min,1)g n h M 1 D M 1 D 1 M 1 D M D M D M 1 D M 1 D 1 M 1 D M D M D M O M O + M O M O + M O M O M 1 M 1 D 1 M 1 D M O 15

17 We have b 2C max j J Bound for b k =r 0 e δ k hnh N (Tkω j), where hn h N is the envelope displayed below (N = 16, n h = 0) The Diophantine condition gives a lower bound for Tkω j : Tkω j TD ( k + k j ) τ 1. For k small, hn h N (Tkω j) 1. After some order r, hn h N (Tkω j) may approach a

18 Therefore, b 2C(max j J where: r 1 k =r 0 e δ k hnh N (Tkω j) +max j J k =r e δ k ), The first term is bounded by replacing the DFT by the TCFT. This introduces an additional error term due to this approximation. All the sums are reduced to sums of the form j jα e δj, which are bounded by incomplete Gamma functions. 16-b

19 Final theorem Under the stated hypothesis, the error in frequencies and amplitudes can be bounded as where and M O (n h!) and M 1 D being ε 1 = ε 2 = π 2( N f 2( N f ( N ( 2( f 4 y M 1 b, M 1 TD M 1 D 1 M 1 D M O A l )(π + ln( 1 + n (2r 0 2) τ h ) ln( 2 n (2r 0 2) τ h )) + 2N f TD ( 1 n (2r 0 2) τ h ) 1+2n h A l )(π + ln([ν min ] + n h ) ln([ν min ] 1 n h )) + 2N f ([ν min ] n h ) 1+2n h A l )(π+ln(n Ω 0 +n h ) ln(n Ω 0 1 n h ))+2N f )(1+ 1 M 1 D 1 M 1 ε, M 1 D D 1 (N Ω 0 n h ) 1+2n h TD M 1 D 1 M 1 ε, M 1 D D 2 2n h ) G nh min(1, A min ), ( 2Amax ) 4(n h!) 2 (π+ln(n Ω 0 +n h ) ln(n Ω 0 1 n h ))+2 (1+ 1 π(n Ω 0 n h ) 1+2n h (n h!) 2 ( 2Amax (π+ln(2[ν min ]+n h ) ln(2[ν min ] n h 1)) + 2 Ω 0 = T(2r 0 2) ω +1, π(2[ν min ] n h ) 1+2n h ), 17 ) 2n h ),

20 As for b, ( b χ {r >r0 } 2m+1 C (m 1)! (n h!) 2 e δ(r 0 1) + χ {r >r0 } Error estimates: Final theorem (cont.) m 1 ( m 1 l l=0 ) ( m 2 r 0+1) m 1 l G f (2r 0 1, r 0 +r 2, l+τ(1+2n h ), δ) E π(td) 1+2n h 2(n h!) 2 e δ m 2 (1 + 1 )G 2n f (r 0 + m, r h 2 1+ m, m 1, δ) 2 π(n Ω n h ) 1+2n h + e δ m 2 G (r + m 2, m 1, δ) ), where Ω = T(r + r 0 2) ω + 1 r = max ( r 0,min E = (z 1 n h ) 1+2n h, z = z 1+2n h TD (r + r 0 2) τ, ([( TD max(( (n h!)2 ) 1 1+2n h n π h,2(1 + n h )) [ N 1 nh ]) ) r T ω ) 1 τ r0 + 2], equals 1 if condi- In the above formulas, χ {condition} tion is true and 0 otherwise. 18

21 I.3 Applications Academic example We consider the quasi periodic function f 0.9 (t) = sin(2πω 1 t + ϕ 1 ) sin(2πω 2 t + ϕ 2 ) 1 0.9cos(2πω 1 t + ϕ 1 ) 1 0.9cos(2πω 2 t + ϕ 2 ). Explicit formulae for frequencies and amplitudes can be obtained, as well as the Cauchy estimates and the Diophantine condition. We have performed Fourier analysis of this function for several T, N. µ=0.9 log10(error) log2(t) log2(t/n) µ= log10(error) log2(t/n) Solid line: actual error. Dashed line: estimated error. 19

22 Numerical test of the bounds obtained: actual error vs. predicted error In the preceeding slide, the difference between the error predicted and the actual error is big because of the Diophantine condition, which is reached for for very few orders k. The points in the plot below represent min k =const. kω for k = The curve represents the values of the Diophantine condition / k min kω If we substitute the first term of b by what is obtained just before the use of the Diophantine condition, the following figure is obtained. k log10(error) µ= log2(t/n) 13 log2(t) log10(error) µ= log2(t/n) Solid line: actual error. Dashed line: estimated error. 20