Invariant Variational Problems & Integrable Curve Flows. Peter J. Olver University of Minnesota olver

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1 Invariant Variational Problems & Integrable Curve Flows Peter J. Olver University of Minnesota olver Cocoyoc, November,

2 Variational Problems x = (x 1,..., x p ) u = (u 1,..., u q ) u α J = J uα independent variables dependent variables derivatives Variational problem: I[u] = L(x, u (n) ) dx L(x, u (n) ) Lagrangian Variational derivative Euler-Lagrange equations E(L) = 0 Components: E α (L) = J ( D) J L u α J 2

3 Invariant Variational Problems G transformation group G invariant variational problem I[u] = L(x, u (n) ) dx = P (... D K I α... ) ω = Lie I 1,..., I l fundamental differential invariants D 1,..., D p invariant differential operators D K I α differentiated invariants ω = ω 1 ω p invariant volume form Invariant Euler-Lagrange equations E(L) = F (... D K I α... ) = 0 Main Problem: Construct F directly from P. (P. Griffiths, I. Anderson ) 3

4 Example. Planar Euclidean group G = SE(2) κ = u xx (1 + u 2 x )3/2 curvature ds = 1 + u 2 x dx arc length D = d ds = u 2 x d dx arc length derivative Invariant variational problem P (κ, κ s, κ ss,... ) ds Euler-Lagrange equations E(L) = F (κ, κ s, κ ss,... ) = 0 4

5 Euclidean Curve Examples Minimal curves (geodesics): I[u] = ds = 1 + u 2 x dx E(L) = κ = 0 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 5

6 General Euclidean invariant variational problem 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 Invariant Euler-Lagrange formula E(L) = (D 2 + κ 2 ) E(P ) + κ H(P ). Elastica : E(P ) = κ P = 1 2 κ2 H(P ) = P = 1 2 κ2 E(L) = κ ss κ3 = 0 6

7 z (n) = (x, u (n) ) = (... x i... u α J... ) coordinates on J n Moving Frames = Mark Fels and PJO 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 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 7

8 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). Theorem. If G acts locally effectively on subsets, then for n 0, the (prolonged) action of G is locally free on an open subset of J n. = Ovsiannikov, PJO 8

9 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) effective the only group element g G which fixes every point z M is the identity: g z = z for all z M if and only if g = e. 9

10 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. 10

11 4. 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 of order n: I k (x, u (n) ) = w k (ρ(z (n) ), z (n) )) k = r + 1,..., dim J n 11

12 Euclidean Plane Curves G = SE(2) Assume the curve is (locally) a graph: C = {u = f(x)} Prolong to J 3 via implicit differentiation y = x cos φ u sin φ + a v = x cos φ + u sin φ + b v y = sin φ + u x cos φ cos φ u x sin φ v yy = u xx (cos φ u x sin φ) 3 w = R z + c v yyy = (cos φ u x sin φ )u xxx 3u2 xx sin φ (cos φ u x sin φ ) 5. Normalization: r = dim G = 3 y = 0 v = 0 v y = 0 Right moving frame φ = tan 1 u x ρ : J 1 SE(2) a = x + uu x 1 + u 2 x b = xu x u 1 + u 2 x 12

13 Differential invariants v yy κ = v yyy dκ ds v yyyy d2 κ ds 2 + 3κ3 = u xx (1 + u 2 x )3/2 = (1 + u2 x )u xxx 3u x u2 xx (1 + u 2 x )3 Invariant one-form arc length dy = (cos φ u x sin φ) dx ds = 1 + u 2 x dx Invariant differential operator d dy = 1 cos φ u x sin φ d dx d ds = u 2 x d dx Theorem. All differential invariants are functions of the derivatives of curvature with respect to arc length: κ dκ ds d 2 κ ds 2 13

14 Euclidean Curves e 2 e 1 x Left moving frame ρ(x, u (1) ) = ρ(x, u (1) ) 1 R = ã = x b = u θ = tan 1 u x ( 1 1 ux 1 + u 2 u x x 1 ) = ( t n ) a = ( ) x u Frenet frame t = dx ds = ( ) xs y s n = t = ( ys x s ) Frenet equations = Maurer Cartan equations: dx ds = e 1 de 1 ds = κ e 2 de 2 ds = κ e 1 14

15 Invariantization The process of replacing group parameters in transformation rules by their moving frame formulae is known as invariantization: Functions Invariants ι : Forms Invariant Forms Differential Operators Invariant Differential Operators Fundamental differential invariants 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 Invariantization: ι [ F (... x i... u α J... ) ] = F (... H i... I α J... ) Replacement Theorem: If J is a differential invariant, then ι(j) = J. J(... x i... u α J... ) = J(... H i... I α J... ) 15

16 Euclidean 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,... ) Invariant one-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 ι(θ) = ϑ = θ 1 + u 2 x ι(θ x ) = ϑ 1 = (1 + u2 x ) θ x u x u xx θ (1 + u 2 x )2 16

17 Infinite jet space Local coordinates The Variational Bicomplex J = = Vinogradov, Tsujishita, I. Anderson lim n Jn z ( ) = (x, u ( ) ) = (... x i... u α 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 17

18 Bigrading of the differential forms on J Ω = M r = # of dx i Ω r,s r,s s = # of θj α Vertical and Horizontal Differentials d = d H + d V Variational Bicomplex: d H : Ω r,s Ω r+1,s d V : Ω r,s Ω r,s+1 F (x, u (n) ) differential function d H F = p i=1 d V F = α,j (D i F ) dx i total differential F u α J θ α J variation 18

19 The Simplest Example. M = R 2 x, u R 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. Differential F = F (x, u, u x, u xx,... ) df = F x dx + F u du + F u x du x + F u xx du xx + Total derivative = (D x F ) dx + F u θ + F u x θ x + F u xx θ xx + = d H F + d V F D x F = F u u x + F u x u xx + F u xx u xxx + 19

20 Lagrangian form λ = L(x, u (n) ) dx Ω 1,1 Vertical derivative variation dλ = d V λ = d V L dx = ( L u θ + L θ u x + L ) θ x u xx + dx Ω 1,1 xx Integration by parts d H (A θ) = (D x A) dx θ A θ x dx so = [ (D x A) θ + A θ x ] dx A θ x dx (D x A) θ dx mod im d H Variational derivative compute modulo im d H : dλ δλ = ( L u D x L u x + D 2 x ) L θ dx u xx = E(L) θ dx = Euler-Lagrange source form. 20

21 Variational Derivative Variation: d V : Ω p,0 Ω p,1 Integration by Parts: π : Ω p,1 F 1 = Ω p,1 / d H Ω p 1,1 = source forms Variational derivative or Euler operator: δ = π d V : Ω p,0 F 1 λ = L dx q α = 1 E α (L) θ α dx Variational Problems Source Forms 21

22 Invariant Variational Complex Fundamental differential invariants H i (x, u (n) ) = ι(x i ) IK α (x, u(l) ) = ι(u α K ) Invariant horizontal forms ϖ i = ι(dx i ) Invariant contact forms ϑ α J = ι(θα J ) Differential forms Ω = M r,s Ω r,s Differential d = d H + d V + d W d H : Ω r,s d V : Ω r,s d W : Ω r,s Ω r+1,s Ω r,s+1 Ω r 1,s+2 22

23 The Key Recurrence Formula d ι(ω) = ι(dω) + r κ = 1 µ κ ι[v κ (Ω)] v 1,..., v r basis for infinitesimal generators g µ 1,..., µ r invariantized dual Maurer Cartan forms µ k = γ k + ε k Ω 1,0 Ω 0,1 All identities, commutation formulae, etc., in the variational bicomplex can be found by applying the key formula with Ω replaced by the basic functions and differential forms! 23

24 Euclidean Curves 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 + Horizontal recurrence formula d H ι(f ) = ι(d H F ) + ι(v 1 (F )) γ 1 + ι(v 2 (F )) γ 2 + ι(v 3 (F )) γ 3 d H I = DI ϖ Use phantom invariants ι(d H F ) = ι(d x F ) ϖ = D = d/ds 0 = d H H = ι(d H x) + ι(v κ (x)) γ κ = ϖ + γ 1 0 = d H I 0 = ι(d H u) + ι(v κ (u)) γ κ = γ 2 0 = d H I 1 = ι(d H u x ) + ι(v κ (u x )) γ κ = κ ϖ + γ 3, to solve for γ 1 = ϖ γ 2 = 0 γ 3 = κ ϖ 24

25 γ 1 = ϖ γ 2 = 0 γ 3 = κ ϖ Horizontal recurrence formulae κ s ϖ = d H κ = d H (I 2 ) = ι(d H u xx ) + ι(v 3 (u xx )) γ 3 = ι(u xxx dx) ι(3u x u xx )) κ ϖ = I 3 ϖ κ ss ϖ = d H (I 3 ) = ι(d H u xxx ) + ι(v 3 (u xxx ) γ 3 = ι(u xxxx dx) ι(4u x u xxx + 3u 2 xx ) κ ϖ = I 4 3I3 2 ϖ. κ = I 2 κ s = I 3 I 2 = κ I 3 = κ s κ ss = I 4 3I 3 2 I 4 = κ ss + 3κ 3 κ sss = I 5 19I 2 2 I 3. I 5 = κ sss + 19κ 2 κ s. 25

26 Vertical recurrence formula d V ι(f ) = ι( d V F ) + ι(v 1 (F )) ε 1 + ι(v 2 (F )) ε 2 + ι(v 3 (F )) ε 3 Use phantom invariants 0 = d V H = ε 1 0 = d V I 0 = ϑ + ε 2 0 = d V I 1 = ϑ 1 + ε 3 to solve for ε 1 = 0 ε 2 = ϑ = ι(θ) ε 3 = ϑ 1 = ι(θ 1 ) d V κ = d V I 2 = ι(θ 2 ) + ι(v 3 (u xx )) ε 3 = ϑ 2 = (D 2 + κ 2 ) ϑ. Key recurrence formulae: d V κ = (D 2 + κ 2 ) ϑ d V ϖ = κ ϑ ϖ 26

27 Plane Curves Invariant Lagrangian: λ = L(x, u (n) ) dx = P (κ, κ s,...) ϖ Euler Lagrange form: d V λ E(L) ϑ ϖ Invariant Integration by Parts Formula F d V (DH) ϖ (DF ) d V H ϖ (F DH) d V ϖ d V λ = d V P ϖ + P d V ϖ = n P κ n d V κ n ϖ + P d V ϖ E(P ) d V κ ϖ + H(P ) d V ϖ Vertical differentiation formulae d V κ = A(ϑ) A Eulerian operator d V ϖ = B(ϑ) ϖ B Hamiltonian operator d V 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 27

28 Euclidean Plane Curves Eulerian operator d V κ = (D 2 + κ 2 ) ϑ A = D 2 + κ 2 A = D 2 + κ 2 Hamiltonian operator d V ϖ = κ ϑ ϖ 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 ). 28

29 Invariant Plane Curve Flows G Lie group acting on R 2 ϖ invariant horizontal form (arc length) ϑ invariant contact form C(t) parametrized family of plane curves G-invariant curve flow: Infinitesimal generator: I, J dc dt = F[C] v = I t + J n differential invariants t unit tangent n unit normal 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. 29

30 D invariant arc length derivative B invariant Hamiltonian operator d V ϖ = B(ϑ) ϖ Theorem. The curve flow generated by v = I t + J n preserves arc length if and only if B(J) + DI = 0 Proof : Cartan s formula for Lie derivatives: v(ϖ) = d(v ϖ) + v dϖ = di + v (B(ϑ) ϖ) [ DI + B(J) ] ϖ mod ϑ Corollary. [ v, D ] = 0 Every curve flow is equivalent, modulo reparametrization, to an arc length-preserving flow. 30

31 Evolution of Differential Invariants κ curvature (generating differential invariant) d V κ = A(ϑ) A invariant Eulerian operator Theorem. Under such 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... Why???? 31

32 Examples Euclidean plane curves: G = SE(2) = SO(2) R 2 d V κ = (D 2 + κ 2 ) ϑ, d V 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 32

33 Equi-affine plane curves G = SA(2) = SL(2) R 2 d V κ = A(ϑ), d V 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 33

34 Euclidean space curves in R 3 : G = SE(3) = SO(3) R 3 ( dv κ d V τ ) ) ) ϑ1 ϑ1 = A(, d ϑ V ϖ = B( ϖ 2 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 1 κ D3 s κ s 2τD s τ s κ 2D2 s + κ2 τ 2 κ D s + κ s τ 2 2κττ s κ 2 B = ( κ 0 ) Recursion operator: R = A ( κs ( ) κt τ t τ s ) = R( κs τ s D 1 B ) = vortex filament flow = nonlinear Schrödinger equation (Hasimoto) 34