Invariant Variational Problems & Invariant Curve Flows

Transcription

1 Invariant Variational Problems & Invariant Curve Flows Peter J. Olver University of Minnesota olver Oxford, December, 2008

2 Basic Notation x = (x 1,..., x p ) independent variables u = (u 1,..., u q ) dependent variables u α J = J uα partial derivatives F (x, u (n) ) = F (... x k... u α J... ) differential function G transformation group acting on the space of independent and dependent variables

3 I[u] = Variational Problems L(x, u (n) ) dx variational problem L(x, u (n) ) Lagrangian Variational derivative Euler-Lagrange equations: E(L) = 0 components: E α (L) = J ( D) J L u α J D k F = F x k + α,j u α J,k F u α J total derivative of F with respect to x k

4 Invariant Variational Problems According to Lie, any G invariant variational problem can be written in terms of the differential invariants: I[u] = L(x, u (n) ) dx = P (... D K I α... ) ω I 1,..., I l D 1,..., D p D K I α fundamental differential invariants invariant differential operators differentiated invariants ω = ω 1 ω p invariant volume form

5 If the variational problem is G-invariant, so I[u] = L(x, u (n) ) dx = P (... D K I α... ) ω then its Euler Lagrange equations admit G as a symmetry group, and hence can also be expressed in terms of the differential invariants: E(L) F (... D K I α... ) = 0 Main Problem: Construct F directly from P. (P. Griffiths, I. Anderson )

6 Planar Euclidean group G = SE(2) κ = u xx (1 + u 2 x )3/2 curvature (differential invariant) ds = 1 + u 2 x dx arc length D = d ds = u 2 x d dx arc length derivative Euclidean invariant variational problem I[u] = L(x, u (n) ) dx = P (κ, κ s, κ ss,... ) ds Euler-Lagrange equations E(L) F (κ, κ s, κ ss,... ) = 0

7 Euclidean Curve Examples Minimal curves (geodesics): I[u] = ds = 1 + u 2 x dx E(L) = κ = 0 = straight lines The Elastica (Euler): I[u] = 1 2 κ2 ds = u 2 xx dx (1 + u 2 x )5/2 E(L) = κ ss κ3 = 0 = elliptic functions

8 General Euclidean invariant variational problem I[u] = L(x, u (n) ) dx = P (κ, κ s, κ ss,... ) ds Invariantized Euler Lagrange expression E(P ) = n = 0 ( D) n P κ n D = d ds Invariantized Hamiltonian H(P ) = i>j κ i j ( D) j P κ i P

9 General Euclidean invariant variational problem I[u] = L(x, u (n) ) dx = P (κ, κ s, κ ss,... ) ds Invariantized Euler Lagrange expression E(P ) = n = 0 ( D) n P κ n D = d ds Invariantized Hamiltonian H(P ) = i>j κ i j ( D) j P κ i P

10 General Euclidean invariant variational problem I[u] = L(x, u (n) ) dx = P (κ, κ s, κ ss,... ) ds Invariantized Euler Lagrange expression E(P ) = n = 0 ( D) n P κ n D = d ds Invariantized Hamiltonian H(P ) = i>j κ i j ( D) j P κ i P

11 I[u] = L(x, u (n) ) dx = P (κ, κ s, κ ss,... ) ds Euclidean invariant Euler-Lagrange formula E(L) = (D 2 + κ 2 ) E(P ) + κ H(P ) = 0 The Elastica: I[u] = 1 2 κ2 ds P = 1 2 κ2 E(P ) = κ H(P ) = P = 1 2 κ2 E(L) = (D 2 + κ 2 ) κ + κ ( 1 2 κ2 ) = κ ss κ3 = 0

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14 Applications of Moving Frames Differential geometry Equivalence Symmetry Differential invariants Rigidity Joint invariants and semi-differential invariants Integral invariants Symmetries of differential equations Factorization of differential operators Invariant differential forms and tensors

15 Identities and syzygies Classical invariant theory Computer vision object recognition symmetry detection structure from motion Invariant variational problems Invariant numerical methods Poisson geometry & solitons Killing tensors in relativity Invariants of Lie algebras in quantum mechanics Lie pseudo-groups

16 Moving Frames G r-dimensional Lie group acting on M J n = J n (M, p) n th order jet bundle for p-dimensional submanifolds N = {u = f(x)} M z (n) = (x, u (n) ) = (... x i... u α J... ) coordinates on Jn G acts on J n by prolongation (chain rule)

17 Definition. An n th order moving frame is a G-equivariant map ρ = ρ (n) : V J n G Equivariance: ρ(g (n) z (n) ) = g ρ(z (n) ) ρ(z (n) ) g 1 left moving frame right moving frame Note: ρ left (z (n) ) = ρ right (z (n) ) 1

18 Theorem. A moving frame exists in a neighborhood of a point z (n) J n if and only if G acts freely and regularly near z (n). free the only group element g G which fixes one point z M is the identity: g z = z if and only if g = e. locally free the orbits have the same dimension as G. regular all orbits have the same dimension and intersect sufficiently small coordinate charts only once ( irrational flow on the torus)

19 Geometric Construction z O z Normalization = choice of cross-section to the group orbits

20 Geometric Construction z O z k K Normalization = choice of cross-section to the group orbits

21 Geometric Construction z O z g = ρ left (z) k K Normalization = choice of cross-section to the group orbits

22 Geometric Construction z O z g = ρ right (z) k K Normalization = choice of cross-section to the group orbits

23 The Normalization Construction 1. Write out the explicit formulas for the prolonged group action: w (n) (g, z (n) ) = g (n) z (n) = Implicit differentiation 2. From the components of w (n), choose r = dim G normalization equations: w 1 (g, z (n) ) = c 1... w r (g, z (n) ) = c r 3. Solve the normalization equations for the group parameters g = (g 1,..., g r ): g = ρ(z (n) ) = ρ(x, u (n) ) The solution is the right moving frame.

24 4. Invariantization: substitute the moving frame formulas g = ρ(z (n) ) = ρ(x, u (n) ) for the group parameters into the un-normalized components of w (n) to produce a complete system of functionally independent differential invariants: I (n) (x, u (n) ) = ι(z (n) ) = w (n) (ρ(z (n) ), z (n) ))

25 Euclidean plane curves G = SE(2) Assume the curve is (locally) a graph: C = {u = f(x)} Write out the group transformations y = x cos φ u sin φ + a v = x cos φ + u sin φ + b w = R z + c

26 Prolong to J n via implicit differentiation y = x cos φ u sin φ + a v y = sin φ + u x cos φ cos φ u x sin φ v = x cos φ + u sin φ + b v yy = u xx (cos φ u x sin φ) 3 v yyy = (cos φ u x sin φ ) u xxx 3 u2 xx sin φ (cos φ u x sin φ ) 5. Choose a cross-section, or, equivalently a set of r = dim G = 3 normalization equations: y = 0 v = 0 v y = 0 Solve the normalization equations for the group parameters: φ = tan 1 u x a = x + uu x 1 + u 2 x b = xu x u 1 + u 2 x The result is the right moving frame ρ : J 1 SE(2)

27 Substitute into the moving frame formulas for the group parameters into the remaining prolonged transformation formulae to produce the basic differential invariants: u v yy κ = xx (1 + u 2 x )3/2 v yyy v yyyy dκ ds d2 κ ds 2 + 3κ3 = = (1 + u2 x )u xxx 3u x u2 xx (1 + u 2 x )3 Theorem. All differential invariants are functions of the derivatives of curvature with respect to arc length: κ dκ ds d 2 κ ds 2

28 The invariant differential operators and invariant differential forms are also substituting the moving frame formulas for the group parameters: Invariant one-form arc length dy = (cos φ u x sin φ) dx ds = 1 + u 2 x dx Invariant differential operator arc length derivative d dy = 1 cos φ u x sin φ d dx d ds = u 2 x d dx

29 Euclidean Curves e 2 e 1 x Left moving frame ρ(x, u (1) ) = ρ(x, u (1) ) 1 R = ã = x b = u φ = tan 1 u x ( ) ( 1 1 ux x = ( t n ) ã = 1 + u 2 u x x 1 u )

30 Invariantization The process of replacing group parameters in transformation rules by their moving frame formulae is known as invariantization. The invariantization I = ι(f ) is the unique invariant function that agrees with F on the cross-section: I K = F K. Invariantization respects algebraic operations, and provides a canonical projection that maps objects to their invariantized counterparts. Functions Invariants ι : Forms Invariant Forms Differential Operators Invariant Differential Operators

31 Fundamental differential invariants = invariantized jet coordinates H i (x, u (n) ) = ι(x i ) I α K (x, u(l) ) = ι(u α K ) The constant differential invariants, coming from the moving frame normalizations, are known as the phantom invariants. The remaining non-constant differential invariants are the basic invariants and form a complete system of functionally independent differential invariants. Invariantization of differential functions: ι [ F (... x i... u α J... ) ] = F (... Hi... I α J... ) Replacement Theorem: If J is a differential invariant, then ι(j) = J. J(... x i... u α J... ) = J(... Hi... I α J... )

32 The Infinite Jet Bundle Jet bundles Inverse limit Local coordinates M = J 0 J 1 J 2 J = lim n Jn z ( ) = (x, u ( ) ) = (... x i... u α J... ) = Taylor series

33 Differential Forms Coframe basis for the cotangent space T J : Horizontal one-forms dx 1,..., dx p Contact (vertical) one-forms θ α J = duα J p i = 1 u α J,i dxi Intrinsic definition of contact form θ j N = 0 θ = A α J θα J

34 The Variational Bicomplex = Dedecker, Vinogradov, Tsujishita, I. Anderson,... Bigrading of the differential forms on J : Ω = M r,s Ω r,s r = # horizontal forms s = # contact forms Vertical and Horizontal Differentials d = d H + d H : Ω r,s Ω r+1,s : Ω r,s Ω r,s+1

35 Vertical and Horizontal Differentials d H F = F (x, u (n) ) differential function p i=1 F = α,j (D i F ) dx i total differential F u α J d H (dx i ) = (dx i ) = 0, θ α J first variation d H (θ α J ) = p i=1 dx i θ α J,i (θ α J ) = 0

36 The Simplest Example x independent variable u dependent variable Horizontal form dx Contact (vertical) forms θ = du u x dx θ x = du x u xx dx θ xx = du xx u xxx dx.

37 θ = du u x dx, θ x = du x u xx dx, θ xx = du xx u xxx dx Differential: df = F x dx + F u du + F u x du x + F u xx du xx + = (D x F ) dx + F u θ + F u x θ x + F u xx θ xx + = d H F + F Total derivative: D x F = F x + F u u x + F u x u xx + F u xx u xxx +

38 . The Variational Bicomplex.... δ Ω 0,3 d H Ω 1,3 d H Ω 0,2 d H Ω 1,2 d H Ω 0,1 d H Ω 1,1 d H R Ω 0,0 d H Ω 1,0 d H d H Ω p 1,3 d H Ω p,3 π F 3 δ d H Ω p 1,2 d H Ω p,2 π F 2 δ d H Ω p 1,1 d H Ω p,1 π F 1 d H Ω p 1,0 d H Ω p,0 E conservation laws Lagrangians PDEs (Euler Lagrange) Helmholtz conditions

39 . The Variational Bicomplex.... δ Ω 0,3 d H Ω 1,3 d H Ω 0,2 d H Ω 1,2 d H Ω 0,1 d H Ω 1,1 d H R Ω 0,0 d H Ω 1,0 d H d H Ω p 1,3 d H Ω p,3 π F 3 δ d H Ω p 1,2 d H Ω p,2 π F 2 δ d H Ω p 1,1 d H Ω p,1 π F 1 d H Ω p 1,0 d H Ω p,0 E conservation laws Lagrangians PDEs (Euler Lagrange) Helmholtz conditions

40 . The Variational Bicomplex.... δ Ω 0,3 d H Ω 1,3 d H Ω 0,2 d H Ω 1,2 d H Ω 0,1 d H Ω 1,1 d H R Ω 0,0 d H Ω 1,0 d H d H Ω p 1,3 d H Ω p,3 π F 3 δ d H Ω p 1,2 d H Ω p,2 π F 2 δ d H Ω p 1,1 d H Ω p,1 π F 1 d H Ω p 1,0 d H Ω p,0 E conservation laws Lagrangians PDEs (Euler Lagrange) Helmholtz conditions

41 . The Variational Bicomplex.... δ Ω 0,3 d H Ω 1,3 d H Ω 0,2 d H Ω 1,2 d H Ω 0,1 d H Ω 1,1 d H R Ω 0,0 d H Ω 1,0 d H d H Ω p 1,3 d H Ω p,3 π F 3 δ d H Ω p 1,2 d H Ω p,2 π F 2 δ d H Ω p 1,1 d H Ω p,1 π F 1 d H Ω p 1,0 d H Ω p,0 E conservation laws Lagrangians PDEs (Euler Lagrange) Helmholtz conditions

42 . The Variational Bicomplex.... δ Ω 0,3 d H Ω 1,3 d H Ω 0,2 d H Ω 1,2 d H Ω 0,1 d H Ω 1,1 d H R Ω 0,0 d H Ω 1,0 d H d H Ω p 1,3 d H Ω p,3 π F 3 δ d H Ω p 1,2 d H Ω p,2 π F 2 δ d H Ω p 1,1 d H Ω p,1 π F 1 d H Ω p 1,0 d H Ω p,0 E conservation laws Lagrangians PDEs (Euler Lagrange) Helmholtz conditions

43 The Variational Derivative E = π first variation π integration by parts = mod out by image of d H Ω p,0 Ω p,1 π F 1 = Ω p,1 / d H Ω p 1,1 λ = L dx α,j Variational problem L u α J First variation θ α J dx q α = 1 E α (L) θ α dx Euler Lagrange source form

44 The Simplest Example: (x, u) M = R 2 Lagrangian form: λ = L(x, u (n) ) dx Ω 1,0 First variation vertical derivative: dλ = λ = L dx ( L = u θ + L θ u x + L θ x u xx + xx Integration by parts compute modulo im d H : dλ δλ = ( L u D x = E(L) θ dx L u x + D 2 x ) dx Ω 1,1 ) L θ dx F 1 u xx = Euler-Lagrange source form.

45 The Simplest Example: (x, u) M = R 2 Lagrangian form: λ = L(x, u (n) ) dx Ω 1,0 First variation vertical derivative: dλ = λ = L dx ( L = u θ + L θ u x + L θ x u xx + xx Integration by parts compute modulo im d H : dλ δλ = ( L u D x = E(L) θ dx L u x + D 2 x ) dx Ω 1,1 ) L θ dx F 1 u xx = Euler-Lagrange source form.

46 The Simplest Example: (x, u) M = R 2 Lagrangian form: λ = L(x, u (n) ) dx Ω 1,0 First variation vertical derivative: dλ = λ = L dx ( L = u θ + L θ u x + L θ x u xx + xx Integration by parts compute modulo im d H : dλ δλ = ( L u D x = E(L) θ dx L u x + D 2 x ) dx Ω 1,1 ) L θ dx F 1 u xx = Euler-Lagrange source form.

47 To analyze invariant variational problems, invariant conservation laws, etc., we apply the moving frame invariantization process to the variational bicomplex:

48 Differential Invariants and Invariant Differential Forms ι invariantization associated with moving frame ρ. Fundamental differential invariants H i (x, u (n) ) = ι(x i ) IK α (x, u(n) ) = ι(u α K ) Invariant horizontal forms ϖ i = ι(dx i ) Invariant contact forms ϑ α J = ι(θα J )

49 The Invariant Quasi Tricomplex Differential forms Ω = M r,s Ω r,s Differential d = d H + + d W d H : Ω r,s : Ω r,s d W : Ω r,s Ω r+1,s Ω r,s+1 Ω r 1,s+2 Key fact: invariantization and differentiation do not commute: d ι(ω) ι(dω)

50 The Universal Recurrence Formula d ι(ω) = ι(dω) + r κ=1 ν κ ι[v κ (Ω)] v 1,..., v r basis for g infinitesimal generators ν 1,..., ν r invariantized dual Maurer Cartan forms = uniquely determined by the recurrence formulae for the phantom differential invariants

51 d ι(ω) = ι(dω) + r κ=1 ν κ ι[v κ (Ω)] All identities, commutation formulae, syzygies, etc., among differential invariants and, more generally, the invariant variational bicomplex follow from this universal formula by letting Ω range over the basic functions and differential forms! Moreover, determining the structure of the differential invariant algebra and invariant variational bicomplex requires only linear differential algebra, and not any explicit formulas for the moving frame, the differential invariants, the invariant differential forms, or the group transformations!

52 Euclidean plane curves Fundamental normalized differential invariants ι(x) = H = 0 ι(u) = I 0 = 0 ι(u x ) = I 1 = 0 phantom diff. invs. ι(u xx ) = I 2 = κ ι(u xxx ) = I 3 = κ s ι(u xxxx ) = I 4 = κ ss + 3κ 3 In general: ι( F (x, u, u x, u xx, u xxx, u xxxx,... )) = F (0, 0, 0, κ, κ s, κ ss + 3κ 3,... )

53 Invariant arc length form dy = (cos φ u x sin φ) dx (sin φ) θ ϖ = ι(dx) = ω + η = 1 + u 2 x dx + u x 1 + u 2 x θ = θ = du u x dx Invariant contact forms ϑ = ι(θ) = θ 1 + u 2 x ϑ 1 = ι(θ x ) = (1 + u2 x ) θ x u x u xx θ (1 + u 2 x )2

54 Prolonged infinitesimal generators v 1 = x, v 2 = u, v 3 = u x + x u + (1 + u 2 x ) u x + 3u x u xx uxx + Basic recurrence formula dι(f ) = ι(df ) + ι(v 1 (F )) ν 1 + ι(v 2 (F )) ν 2 + ι(v 3 (F )) ν 3 Use phantom invariants 0 = dh = ι(dx) + ι(v 1 (x)) ν 1 + ι(v 2 (x)) ν 2 + ι(v 3 (x)) ν 3 = ϖ + ν 1, 0 = di 0 = ι(du) + ι(v 1 (u)) ν 1 + ι(v 2 (u)) ν 2 + ι(v 3 (u)) ν 3 = ϑ + ν 2, 0 = di 1 = ι(du x ) + ι(v 1 (u x )) ν 1 + ι(v 2 (u x )) ν 2 + ι(v 3 (u x )) ν 3 = κ ϖ + ϑ 1 + ν 3, to solve for the Maurer Cartan forms: ν 1 = ϖ, ν 2 = ϑ, ν 3 = κ ϖ ϑ 1.

55 ν 1 = ϖ, ν 2 = ϑ, ν 3 = κ ϖ ϑ 1. Recurrence formulae: dκ = dι(u xx ) = ι(du xx ) + ι(v 1 (u xx )) ν 1 + ι(v 2 (u xx )) ν 2 + ι(v 3 (u xx )) ν 3 = ι(u xxx dx + θ xx ) ι(3u x u xx ) (κ ϖ + ϑ 1 ) = I 3 ϖ + ϑ 2. Therefore, Dκ = κ s = I 3, κ = ϑ 2 = (D 2 + κ 2 ) ϑ where the final formula follows from the contact form recurrence formulae dϑ = dι(θ x ) = ϖ ϑ 1, dϑ 1 = dι(θ) = ϖ (ϑ 2 κ 2 ϑ) κ ϑ 1 ϑ which imply ϑ 1 = Dϑ, ϑ 2 = Dϑ 1 + κ 2 ϑ = (D 2 + κ 2 ) ϑ

56 Similarly, dϖ = ι(d 2 x) + ν 1 ι(v 1 (dx)) + ν 2 ι(v 2 (dx)) + ν 3 ι(v 3 (dx)) = (κ ϖ + ϑ 1 ) ι(u x dx + θ) = κ ϖ ϑ + ϑ 1 ϑ. In particular, ϖ = κ ϑ ϖ Key recurrence formulae: κ = (D 2 + κ 2 ) ϑ ϖ = κ ϑ ϖ

57 Plane Curves Invariant Lagrangian: λ = L(x, u (n) ) dx = P (κ, κ s,...) ϖ Euler Lagrange form: λ E(L) ϑ ϖ Invariant Integration by Parts Formula F (DH) ϖ (DF ) H ϖ (F DH) ϖ λ = P ϖ + P ϖ = n P κ n κ n ϖ + P ϖ E(P ) κ ϖ + H(P ) ϖ

58 Vertical differentiation formulae κ = A(ϑ) A Eulerian operator ϖ = B(ϑ) ϖ B Hamiltonian operator λ E(P ) A(ϑ) ϖ + H(P ) B(ϑ) ϖ [ A E(P ) B H(P ) ] ϑ ϖ Invariant Euler-Lagrange equation A E(P ) B H(P ) = 0

59 Euclidean Plane Curves κ = (D 2 + κ 2 ) ϑ Eulerian operator A = D 2 + κ 2 A = D 2 + κ 2 ϖ = κ ϑ ϖ Hamiltonian operator B = κ B = κ Euclidean invariant Euler-Lagrange formula E(L) = A E(P ) B H(P ) = (D 2 + κ 2 ) E(P ) + κ H(P ).

60 Invariant Plane Curve Flows G Lie group acting on R 2 C(t) parametrized family of plane curves G invariant curve flow: I, J dc dt = V = I t + J n differential invariants t unit tangent n unit normal

61 t, n basis of the invariant vector fields dual to the invariant one-forms: t ; ϖ = 1, n ; ϖ = 0, t ; ϑ = 0, n ; ϑ = 1. C t = V = I t + J n The tangential component I t only affects the underlying parametrization of the curve. Thus, we can set I to be anything we like without affecting the curve evolution. There are two principal choices of tangential component:

62 Normal Curve Flows C t = J n Examples Euclidean invariant curve flows C t = n geometric optics or grassfire flow; C t = κ n curve shortening flow; C t = κ 1/3 n equi-affine invariant curve shortening flow: C t = n equi affine ; C t = κ s n modified Korteweg devries flow; C t = κ ss n thermal grooving of metals.

63 Intrinsic Curve Flows Theorem. The curve flow generated by v = I t + J n preserves arc length if and only if B(J) + D I = 0. D invariant arc length derivative ϖ = B(ϑ) ϖ B invariant Hamiltonian operator

64 Normal Evolution of Differential Invariants Theorem. Under a normal flow C t = J n, κ t = A κ (J), κ s t = A κ s (J). Invariant variations: κ = A κ (ϑ), κ s = A κs (ϑ). A κ = A invariant linearization operator of curvature; A κs = D A κ + κ κ s invariant linearization operator of κ s.

65 Euclidean invariant Curve Evolution Normal flow: C t = J n κ t = A κ (J) = (D2 + κ 2 ) J, κ s t = A κ s (J) = (D 3 + κ 2 D + 3κ κ s ) J. Warning: For non-intrinsic flows, t and s do not commute! Grassfire flow: J = 1 κ t = κ2,... κ s t = 3κ κ s, = caustics

66 Signature Curves Definition. The signature curve S R 2 of a curve C R 2 is parametrized by the two lowest order differential invariants S = { ( κ, dκ ds ) } R 2

67 Equivalence and Signature Curves Theorem. Two curves C and C are equivalent: C = g C if and only if their signature curves are identical: S = S = object recognition

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71 Euclidean Signature Evolution Evolution of the Euclidean signature curve Grassfire flow: Curve shortening flow: Φ t Modified Korteweg-deVries flow: κ s = Φ(t, κ). = 3κ Φ κ2 Φ κ. Φ t = Φ2 Φ κκ κ 3 Φ κ + 4κ 2 Φ. Φ t = Φ3 Φ κκκ + 3Φ 2 Φ κ Φ κκ + 3κΦ 2.

72 Canine Left Ventricle Signature Original Canine Heart MRI Image Boundary of Left Ventricle

73 Smootheentricle Signature

74 Intrinsic Evolution of Differential Invariants Theorem. Under an arc-length preserving flow, κ t = R(J) where R = A κ s D 1 B ( ) In surprisingly many situations, (*) is a well-known integrable evolution equation, and R is its recursion operator! = Hasimoto = Langer, Singer, Perline = Marí Beffa, Sanders, Wang = Qu, Chou, and many more...

75 Euclidean plane curves G = SE(2) = SO(2) R 2 κ = (D 2 + κ 2 ) ϑ, ϖ = κ ϑ ϖ = A = D 2 + κ 2, B = κ R = A κ s D 1 B = D 2 + κ 2 + κ s D 1 κ κ t = R(κ s ) = κ sss κ2 κ s = modified Korteweg-deVries equation

76 Equi-affine plane curves G = SA(2) = SL(2) R 2 κ = A(ϑ), ϖ = B(ϑ) ϖ A = D κ D κ s D κ ss κ2, B = 1 3 D2 2 9 κ, R = A κ s D 1 B = D κ D κ s D κ ss κ κ s D 1 κ κ t = R(κ s ) = κ 5s + 2 κ κ ss κ2 s κ2 κ s = Sawada Kotera equation

77 Euclidean space curves ( dv κ τ ) G = SE(3) = SO(3) R 3 ) ϑ1 = A( ϑ 2 ) ϑ1 ϖ = B( ϖ ϑ 2 A = D 2 s + (κ2 τ 2 ) 2τ κ D2 s + 3κτ s 2κ s τ κ 2 D s + κτ ss κ s τ s + 2κ3 τ κ 2 2τD s τ s 1 κ D3 s κ s B = ( κ 0 ) κ 2D2 s + κ2 τ 2 κ D s + κ s τ 2 2κττ s κ 2

78 Recursion operator: R = A ( κs ( ) κt τ t τ s ) = R( κs τ s D 1 B ) = vortex filament flow = nonlinear Schrödinger equation (Hasimoto)