On the Evolution of the Binormal Flow for a Regular Polygon

Transcription

1 On the Evolution of the Binormal Flow for a Regular Polygon Francisco de la Hoz Méndez Department of Applied Mathematics, Statistics and Operations Research University of the Basque Country UPV/EHU Joint work with Luis Vega (UPV/EHU & BCAM Second BCAM Workshop on Computational Mathematics Bilbao, October 8nd, 23

2 Vortex filaments Assume that the vorticity w of a given fluid of velocity u is supported along the filament X(s, t, i.e., w is the singular vectorial measure w = Γds, where Γ stands for the constant circulation, = X s, s is the arc-length parameter. he velocity can be recovered through the Biot-Savart integral: u(p = Γ 4π + X(s P X(s P 3 (sds. Under the Localized Induction Approximation (LIA, the dynamics of the vortex filament can be approximated by the vortex filament equation (VFE: X t = X s X ss, where is the usual cross product. his is equivalent to the binormal flow, X t = cb, where c is the curvature. Since = X s is constant, we assume S 2. Differentiating LIA, we get the Schrödinger map equation on the sphere: t = ss.

3 Corner-shaped initial data (Gutiérrez, Rivas & Vega Given A ± (c = (A ±, A± 2, A± 3 S2, we want to solve { X t = X s X ss, t >, s R, X(s, = A + sχ [,+ (s A sχ (,] (s, where χ is the characteristic function. Without lost of generality, we assume A + = A, A+ 2 = A 2, A+ 3 = A 3. Let us consider the self-similar solutions of the binormal flow: X(s, t solution = λ X(λs, λ 2 t. aking λ = t /2 and defining G(s = X(s,, they are of the form X(s, t = t /2 X(t /2 s, = tg(s/ t. Hence, the family of solutions in which we are interested is defined by X c (s, t = tg(s/ t, where c is the family parameter and G (s = (s, is the solution of

4 with initial conditions For c = we define c n = s c 2 n, b s b 2 s s G( = 2c (,,, ( = (,,, n( = (,,, b( = (,,. X (s, t = s(,,. For an arbitrary t >, we have c t n = c s t 2t n, b s b 2t which corresponds to c(s, t = c t, τ(s, t = s 2t.

5 heorem (Gutiérrez, Rivas & Vega Given c, X c is a C solution of the binormal flow, t >. Moreover, there are A ± (c = (A ±, A± 2, A± 3 S2, B ± (c, such that (i X c (s, t A + s(c χ [,+ (s A s(c χ (,] (s c t. (ii We have the following asymptotics: ( G(s = A ± (c s + 2 c2 n 4c s s + 2 O(/s3, (s = A ± b (c 2c s + O(/s2, s ± ; (n ib = B ± (c e is2 /4 e ic2 log s + O(/s, s ±. s ± ; (iii A + = A = e (c2 /2π, A + 2 = A 2, A+ 3 = A 3, A± B ± =. Conversely, since the equations for X and are time-reversible, we can solve { X t = X s X ss, t >, s R, X(s, = A + sχ [,+ (s A sχ (,] (s.

6 Figure : Left: (s, t, t >. Right: X(s, t, for different t

7 Figure : Positively Buoyant Jet: Cigarette Smoke (Perry & Lim: JFM, Vol. 88.

8 Figure : Left: X(s, t, for different t. Right: Vortices above an inclined triangular wing (ONERA photograph, Werlé 963.

9 Regular polygonal initial data (De la Hoz & Vega We want to understand the evolution of the binormal flow for polygonal initial data. We study the simplest case, a regular planar polygon of M sides, whose vertices are at X(s k, = iπeiπ(2k /M M sin(π/m. X(s,, s (s k, s k+, is in the segment that joints X(s k, and X(s k+,. Its corresponding tangent vector is: (s, = e 2πik/M, for s k < s < s k+. Symmetries of the initial data are extremely important:

10 Hasimoto transformation & Galilean invariance We consider an alternative version of the Frenet-Serret formulae α β e = α e, e 2 β e 2 s he Hasimoto-type transformation ψ α + iβ transforms t = ss into the NLS equation ( ψ t = iψ ss + i 2 ( ψ 2 + A(t ψ, his equation is Galilean-invariant: if ψ is a solution of NLS, so is ψ k (s, t e iks ik2t ψ(s 2kt, t, k, t R. If we choose ψ(s, such that ψ k (s, = ψ(s,, k R, then ψ(s, t = e iks ik2t ψ(s 2kt, t, k, t R. We are assuming uniqueness all the time!

11 An M-sided regular polygon X(s, can be regarded as a planar curve whose curvature is a sum of Dirac deltas: c(s = 2π M k= δ(s 2πk M. Since the torsion τ(s is zero, we have ψ(s, c(s. ψ(s, satisfies ψ(s, = e imks ψ(s, = ψ Mk (s,, k Z. Hence, the Galilean transformations give ψ(s, t = e imks i(mk2t ψ(s 2Mkt, t, k Z, t R. herefore, the j-th Fourier coefficient of ψ(s, t satisfies: ˆψ(j, t = e i(mk2 t im(j k(2mkt ˆψ(j k, t, j, k Z, t R. Evaluating both sides at j = k, ˆψ(k, t = e i(mk2 t ˆψ(, t.

12 Hence, ψ can be expressed as ψ(s, t = ˆψ(, t k= e i(mk2 t+imks, suggesting that ψ is periodical in time with period 2π/M 2. For a given t, ˆψ(, t has to be chosen so X(s, t and (s, t are 2π-periodic. For instance, when t =, ˆψ(, =, and we get the following well-known identity: k= e i(mks 2π M k= δ(s 2πk M. We evaluate ψ(s, t at t = t pq = (2π/M 2 (p/q, gcd(p, q =, ψ(s, t pq = ˆψ(, t pq k= q = ˆψ(, t pq l= e i(mk2 2πp/(M 2 q+imks e 2πi(p/ql2 +imls e iks. k=

13 Using the previous identity, ψ(s, t pq = 2π q ˆψ(, t pq l= = 2π q ˆψ(, t pq = 2π ˆψ(, t pq = 2π ˆψ(, t pq e 2πi(p/ql2 +imls k= δ(s 2πk e 2πi(p/ql2 +iml(2πk/ δ(s 2πk l= k= [ q q e 2πi(p/ql2 +2πi(m/ql k= m= k= m= l= q G( p, m, qδ(s 2πk M ] δ(s 2πk M 2πm, 2πm where G(a, b, c = c l= e2πi(al2 +bl/c denotes a generalized quadratic Gauß sum. An important property is that q, if q is odd, G( p, m, q = 2q, if q is even and q/2 m mod 2,, if q is even and q/2 m mod 2.

14 herefore, we conclude that, in s [, 2π, M q (α m + iβ mδ(s 2πm m= q/2 ψ(s, t pq = m= q/2 m=, if q odd, (α 2m+ + iβ 2m+ δ(s 4πm+2π, if q/2 odd, (α 2m + iβ 2m δ(s 4πm, if q/2 even. Moreover, if we write α m + iβ m = ρ me iθm, then 2π M ˆψ(, t q pq, if q mod 2, 2π α m + iβ m = ρ m ρ = q ˆψ(, t pq, if q is even and q/2 m mod 2, M 2, if q is even and q/2 m mod 2, Summarizing, at t = t pq: X(s, t pq is a skew polygon of sides (q odd or /2 sides (q even. he angle ρ between two adjacent sides is constant. he structure of the polygon is completely determined by the angles θ m appearing in the generalized quadratic Gaussian sum. We have to choose ˆψ(, t pq, in order that X(s, t pq is closed.

15 Recovering X and from ψ at t = t pq o recover X and, we have to understand the transition from one side of the polygon to the next one. Without loss of generality, we reduce ourselves to ψ(s = (a + ibδ(s, with constant a, b. Hence, we have to integrate e e 2 s = aδ(s δ(s aδ(s. bδ(s e e 2 Defining a + ib = ρe iθ, ( + e ( + e 2 ( + = exp(a ( e ( e 2 (, where exp(a is a rotation matrix: exp(a = ( cos(ρ sin(ρ cos(θ sin(ρ sin(θ sin(ρ cos(θ cos(ρ cos 2 (θ + sin 2 (θ [cos(ρ ] cos(θ sin(θ sin(ρ sin(θ [cos(ρ ] cos(θ sin(θ cos(ρ sin 2 (θ + cos 2 (θ.

16 Let be M m the rotation matrix corresponding to (α m + iβ mδ. If α m + iβ m, M m is an identity matrix. Otherwise, ( M m =. cos(ρ sin(ρ cos(θ m sin(ρ sin(θ m sin(ρ cos(θ m cos(ρ cos 2 (θ m + sin 2 (θ m [cos(ρ ] cos(θ m sin(θ m sin(ρ sin(θ m [cos(ρ ] cos(θ m sin(θ m cos(ρ sin 2 (θ m + cos 2 (θ m Bearing in mind that, e and e 2 are piecewise constant, we have ( 2π e ( 2π e 2 ( 2π ( 4π e ( 4π e 2 ( 4π ( 6π e ( 6π e 2 ( 6π = = = ( + e ( + e 2 ( + ( 2π + e ( 2π + + e 2 ( 2π ( 4π + e ( 4π + + e 2 ( 4π and so forth, i.e., there is a jump at s = 2πk. = M = M = M 2 ( e ( e 2 ( ( 2π e ( 2π e 2 ( 2π ( 4π e ( 4π e 2 ( 4π,

17 his is equivalent to writing ( 2πk + e ( 2πk e 2 ( 2πk + + = M k M k... M M ( e ( e 2 ( In order that the polygon is closed, we have to choose the angle ρ between two adjacent sides in such a way that, e and e 2 are periodic: (2π ( e (2π e 2 (2π = e ( e 2 (, which is equivalent to imposing that M M 2... M M I, where I is the identity matrix. Let us define M, an M-th root of I: M M q M q 2... M M. M induces a rotation of 2π/M degrees around a certain rotation axis. herefore, we have to choose ρ in order that r(m = + 2 cos( 2π M or λ(m = {, e2πi/m, e 2πi/M }.,

18 For small q, r(m can be easily computed using symbolic manipulation: ( + cos(ρ q, if q is odd, r(m = 2 q 2 ( + cos(ρ q/2, if q is even. 2 q/2 2 Numerically, it is immediate to check that this is valid q. Hence { 2 cos 2/q ( π, if q is odd, M cos(ρ = 2 cos 4/q ( π, if q is even. M so M ( q arccos 2 cos 2/q ( π, if q is odd, 2π M ˆψ(, t pq = q M ( 2 arccos 2 cos 4/q ( π, if q is even. 2π M

19 Getting the correct rotation We obtain, up to a rotation, the piecewise constant vectors, e and e 2, which we denote, ẽ and ẽ 2. X, is computed recursively from : { X( = (,,, X( 2πk+2π 2πk = X( + 2π 2πk + (. he symmetries allow to compute the correct rotation of X and at t = t pq. In particular, we use that For any time t, X(2πk/M, k =,..., M, have to be coplanar and lay on a plane orthogonal to the z-axis. X(2π/M X( is a positive multiple of (,,.

20 An efficient algorithm is Compute v + = X(2π/M X( X(2π/M X(, v = 2 Compute w = v v +. X( X( 2π/M X( X( 2π/M. 3 Compute the scalar product between w and (,,, w (,, = w 3. 4 If w 3 =, R is the identity matrix. If not, R is the rotation matrix that induces a rotation of arccos(w 3 degrees around the axis given by the vector w (,, w (,,. 5 Compute v + new = R v +. 6 Compute the rotation matrix R 2 that induces a rotation of arccos(v + new (,, degrees around the z-axis. 7 Compute the sought rotation matrix, R = R 2 R. 8 Update = R and X = R X. We have calculated and, up to a vertical movement, X.

21 alg.2.2 z s.2.5 y.5.5 x.5 Figure : Algebraically constructed (left and X (right, for M = 3, at t,3 = 2π 27. (, 2, 3. appears in blue, 2 in green, 3 in red.

22 Numerical method We simulate together X and : { X t = s, t = ss. We use a pseudo-spectral discretization in space, s j = 2πj/N, and a 4 th -order Runge-Kutta scheme in time: A X = (n (n s, A = (n (n ss, B X = (A (A s, B = (A (A ss, (A = (n + t 2 A, (B = (n + t 2 B, C X = (B (B s, C = (B (B ss, (C = (n + tc, D X = (C (C s, D = (C (C ss, X (n+ = X (n + t 6 (A X + 2B X + 2C X + D X, = (n + t 6 (A + 2B + 2C + D, (n+ =.

23 aking advantage of symmetries Denoting Z (s j = X (s j, t + ix 2 (s j, t N/M N Ẑ (k = Z (s j e 2πijk/N M Z (s j e 2πijk/N, if k mod M, = j= j=, if k mod M, where, Ẑ (Mk + = M N/M j= [ ] e 2πij/N Z (s j e 2πijk/(N/M. Similarly, for X 3, N/M N ˆX 3 (k = X 3 (s j, te 2πijk/N M X 3 (s j, te 2πijk/N, if k mod M, = j= j=, if k mod M, where N/M ˆX 3 (Mk = M 3 (s j, te 2πijk/(N/M. j=

24 Comparison between numerics and algebra num alg s s Figure : Comparison between num, with N/M = 496, and alg, for M = 3, at t,3 = 2π. In num, the Gibbs phenomenon is clearly visible. Otherwise, the maximum 27 discrepancy between num and alg at the 27 points indicated by the black circles is

25 Very recent result: determining the correct vertical movement of X here is strong numerical evidence that the center of gravity moves upward with constant velocity: h(t = mean(x 3 c M t, c M = h(2π/m2 2π/M 2, where lim M c M =. Moreover, choosing X num such that mean(x num,3 =, there is a good numerical agreement between X num c M t(,, and X alg. he previous equation is similar to affirming that d t mean(x 3,t dt = d t mean(x,s X 2,ss X,ss X 2,s dt = c M. dt dt We claim that Important: at t =, X is planar, so lim mean(x,sx 2,ss X,ss X 2,s = c M. t mean(x,s X 2,ss X,ss X 2,s (t = = 2π 2π c(s, ds =.

26 here is strong numerical evidence that, for infinitesimal t, the multiple corner problem can be explained as a superposition of single-corner problems. Let us choose c = [ 2 π ln(sin( π 2 π M ]/2 and integrate the corresponding single-corner X c, together with its associated, n and b = (b, b 2, b 3. Remember that lim (s = (A, A 2, A 3, s We rotate X,, n and b, in such a way that lim rot(s = (,,, lim s lim (s = (A, A 2, A 3. s rot(s = (cos(2π/m, sin(2π/m,. s his is achieved by defining X,rot cos( π sin( π M M X 2,rot = sin( π cos( π A M M 2 A 3 A 2 2 +A2 3 A X 3,rot 2 We claim that c M = M c 2π + b 3,rot (sds = M c 2π + A 3 A 2 2 +A2 3 A 2 2 +A2 3 A 2 2 +A2 3 X X 2. X 3 A 3 b 2 (s + A 2 b 3 (s ds. A A2 3

27 Figure : alg (black, for M = 5, t pq = 2π 5 2 ; against 4 rot (red, at t =.

28 Fractality phenomena It is very interesting to study the evolution of X(, t: M = 4.5 M = Figure : X(, t (left and X(, t c M t X(, (right, for M = 4.

29 he previous picture, conveniently scaled, is strikingly similar to φ(t = π 6 k= e πik2 t, t [, 2], πk 2 whose imaginary part is precisely Riemann s non-differentiable function. Moreover, there is convergence to φ(t as M. his strongly suggests that X(, t is a multifractal, M..4 φ(t.5 M = Figure : Comparison between φ(t (left and X(, t c M t X(, (right, for M = 4.

30 What happens if we consider t pq with big q?.4.2 X X X.5 Figure : X alg and alg, at t = 2π 9 (

31 .3 X X X Figure : X alg and alg, at t = 2π 9 ( = 2π

32 Figure : Stereographic projection of the right-hand side of the previous figure.

33 Open questions and on-going work Obtain an explicit value for c M. Prove analytically that X(, t is a multifractal. Give sense to our solutions from an analytical point of view. Give sense to (s, t at irrational times. Extend these ideas to arbitrary polygons.

34 hank you very much for your attention! Eskerrik asko zuen arretagatik! Muchas gracias por su atención!