Moving Frames and Conservation Laws for Euclidean Invariant Lagrangians
aa r X i v : . [ m a t h . DG ] J a n Moving frames and conservation laws for Euclideaninvariant Lagrangians
Tˆania M N Gon¸calves and Elizabeth L Mansfield
School of Mathematics, Statistics and Actuarial Science,University of Kent, Canterbury, CT2 7NZ, UKE-mail:
[email protected] , [email protected]
Abstract.
Noether’s First Theorem yields conservation laws for Lagrangianswith a variational symmetry group. The explicit formulae for the laws are wellknown and the symmetry group is known to act on the linear space generatedby the conservation laws. In recent work the authors showed the mathematicalstructure behind both the Euler-Lagrange system and the set of conservation laws,in terms of the differential invariants of the group action and a moving frame. Inthis paper we demonstrate that the knowledge of this structure considerably easesfinding the extremal curves for variational problems invariant under the specialEuclidean groups SE (2) and SE (3).PACS numbers: 02.20.Hj, 02.30.Xx, 02.40.Dr Submitted to:
J. Phys. A: Math. Gen.
1. Introduction
In 1918 Emmy Noether wrote the seminal paper “Invariante Variationsprobleme” [1],where she showed that for differential systems derived from a variational principle,conservation laws could be obtained from Lie group actions which left the functionalinvariant.Recently in [2], it was proved that for Lagrangians that are invariant under someLie symmetry group, Noether’s conservation laws can be written in terms of a movingframe and vectors of invariants. Furthermore, in [2] the authors showed that for one-dimensional Lagrangians that are invariant under a semisimple Lie symmetry group,the new format for the conservation laws could reduce considerably the calculationsneeded to solve for the extremal curves. In particular, a classification is given forvariational problems which are invariant under the three inequivalent SL (2 , C ) actionson the plane (classified by Lie [3]). In this paper we use the new structure of Noether’sconservation laws presented in [2] to simplify variational problems that are invariantunder SE (2) and SE (3); these groups are not semisimple.In Section 2, we will briefly give an overview on moving frames, on differentialinvariants of a group action and on invariant calculus of variations. ThroughoutSection 2 we will use the group action of SE (2) on the plane as our pedagogicalexample.In Section 3, we show in some detail how to compute the new version of Noether’sconservation laws presented in Theorem 2.9 for one-dimensional variational problems oving frames and conservation laws for Euclidean invariant Lagrangians SE (2), and then demonstrate how their invariantizedEuler-Lagrange equations and Noether’s conservation laws can be used to solve theintegration problem.Finally in Section 4, we present the simplified solution to the physically importantone-dimensional variational problems that are left unchanged under the SE (3) groupaction.
2. Structure of Noether’s Conservation Laws
In this section, we will give a brief overview of concepts regarding moving frames,differential invariants of a group action and the invariant calculus of variations neededto understand the statements of our result. For more information on these subjects,see Fels and Olver [4,5], Mansfield [6] and Kogan and Olver [7]. We will use the SE (2)action on the plane as our pedagogical example. Here we are using moving frames as reformulated by Fels and Olver [4, 5], adapted tothe context of differential algebra.Let X be the space of independent variables with coordinates x = ( x , ..., x p ) and U the space of dependent variables with coordinates u = ( u , ..., u q ). We will use amultiindex notation to represent the derivatives of u α , e.g. u αK = ∂ | K | u α ∂x k ∂x k · · · ∂x k p p , where the tuple K = ( k , ..., k p ), represents a multiindex of differentiation of order | K | = k + k + · · · + k p . Hence, let M = J n ( X × U ) be the n -th jet bundle withcoordinates z = ( x , ..., x p , u , ..., u q , u , ... ) . On this space, the operator ∂/∂x i extends to the total differentiation operator D i = DDx i = ∂∂x i + q X α =1 X K u αKi ∂∂u αK . In this paper we are interested in using Noether’s conservation laws to find thesolutions that extremize variational problems which are invariant under SE (2) and SE (3). Thus, consider a Lagrangian L ( z )d x that is invariant under some symmetrygroup. Let a group G act on the space M as follows G × M → M z e z = g · z , which satisfies either g · ( h · z ) = ( gh ) · z , called a left action , or g · ( h · z ) = ( hg ) · z ,called a right action . We say a Lagrangian L ( z )d x is invariant under some groupaction if L ( z )d x = L ( e z )d e x for all g ∈ G .Consider a Lie group G acting smootlhy on M such that the action is free and regular . Then for every z ∈ M there exists a neighbourhood U of z , as illustrated inFigure 1, such that oving frames and conservation laws for Euclidean invariant Lagrangians
3- the group orbits have the dimension of the group G and folliate U ;- there is a surface K ⊂ U which intersects the group orbits transversally at a singlepoint. This surface is called the cross section ;- if O ( z ) represents the group orbit through z , then the group element g ∈ G taking z ∈ U to k is unique. z k g K O ( z ) U Figure 1.
A local foliation with a transverse cross section
Under these conditions, we can define a right moving frame as the map ρ : U → G which sends z ∈ U to the unique element ρ ( z ) ∈ G such that ρ ( z ) · z = k, { k } = O ( z ) ∩ K . The element g ∈ G in Figure 1 corresponds to ρ ( z ).To obtain the right moving frame, which sends z to k , we must first define thecross section K as the locus of the set of equations ψ j ( z ) = 0, j = 1 , ..., r , where r isthe dimension of G . Normally, the cross section is chosen so as to ease the calculations.Then solving the set of equations ψ j ( e z ) = ψ j ( g · z ) = 0 , j = 1 , ..., r, (1)known as the normalization equations , for the r group parameters describing G yieldsthe right moving frame in parametric form. Hence, the frame obtained satisfies ψ j ( ρ ( z ) · z ) = 0 , j = 1 , ..., r. By the implicit function theorem, a unique solution of (1) provides an equivariantmap, i.e. for a left action ρ ( e z ) = ρ ( z ) g − and for a right action ρ ( e z ) = g − ρ ( z ) . Example 2.1.
Consider the SE (2) group acting on curves in the ( x, y ( x )) -plane asfollows, (cid:18) xy (cid:19) (cid:18) e x e y (cid:19) = (cid:18) cos θ sin θ − sin θ cos θ (cid:19) (cid:18) x − ay − b (cid:19) , (2) where θ , a and b are constants that parametrize the group action. Here we are usingthe inverse action because it simplifies the calculations.oving frames and conservation laws for Euclidean invariant Lagrangians There is an induced action on the derivatives y K , where K is the index ofdifferentiation with respect to x , called the prolonged action . The induced actionon y x is defined to be e y x = g · y x = d e y d e x = d e y/ d x d e x/ d x by the chain rule, so the action of (2) on y x is e y x = − sin θ + y x cos θ cos θ + y x sin θ . Similarly, g y xx = g · y xx = d e y d( e x ) = 1d e x/ d x dd x (cid:18) d e y d e x (cid:19) = y xx (cos θ + y x sin θ ) . (3) If we consider M to be the space with coordinates ( x, y, y x , y xx , ... ) , then the action islocally free near the identity of SE (2) . Thus, taking the normalization equations to be e x = 0 , e y = 0 and e y x = 0 , we obtain a = x, b = y, and θ = arctan y x (4) as the frame in parametric form. Remark 2.2.
In this paper we will consider all independent variables to be invariant.If these are not invariant, then we can reparametrize and set the original independentvariables as depending on the new invariant parameters.
Theorem 2.3.
Let ρ ( z ) be a right moving frame. Then the quantity I ( z ) = ρ ( z ) · z is an invariant of the group action (see [4]). Consider z = ( z , ..., z m ) ∈ M and let the normalization equations e z i = c i for i = 1 , ..., r , where r is the dimension of the group G , then ρ ( z ) · z = ( c , ..., c r , I ( z r +1 ) , ..., I ( z m )) , where I ( z l ) = g · z | g = ρ ( z ) , l = r + 1 , ..., m. Example 2.1 (cont.)
Evaluating g y xx given by (3) at the frame (4) yields g y xx | ( a = x,b = y,θ =arctan y x ) = y xx (1 + y x ) / , the Euclidean curvature . So evaluating f y K at the frame (4) yields a differentialinvariant. Definition 2.4.
For any prolonged action in the jet space J n ( X × U ) , the invariantizedjet coordinates are denoted as J i = I ( x i ) = e x i | g = ρ ( z ) , I αK = I ( u αK ) = f u αK | g = ρ ( z ) . (5) These are also known as the normalized differential invariants . Example 2.5.
Consider the group action of SE (2) as in Example 2.1. Since x isnot invariant we reparametrize ( x, y ( x )) as ( x ( s ) , y ( s )) , where s is invariant and let g ∈ SE (2) act on ( x ( s ) , y ( s )) as in Example 2.1. Solving the normalization equations e x = 0 , e y = 0 and e y s = 0 , we obtain the frame a = x, b = y, θ = arctan (cid:18) y s x s (cid:19) . (6) oving frames and conservation laws for Euclidean invariant Lagrangians We have then g · z | g = ρ ( z ) = ( e s, e x, e y, e x s , e y s , f y ss ) | g = ρ ( z ) = ( I ( s ) , I x , I y , I x , I y , I y )= s, , , p x s + y s , , x s y ss − y s x ss p x s + y s ! . (7) The second, third and fifth components of (7) correspond to the normalizationequations e x = 0 , e y = 0 and e y s = 0 respectively. The fourth and sixth components, I x and I y respectively, are the lowest order differential invariants and all higher orderinvariants can be obtained in terms of them and their derivatives. Theorem 2.6. ( Replacement Theorem [5]) If f ( z ) is an invariant, then f ( z ) = f ( I ( z )) . This theorem allows one to find the I αK in terms of historically well-knowninvariants without having to solve for the frame. Example 2.5 (cont.)
We know that SE (2) preserves | x s | , thus applying theReplacement Theorem we obtain | x s | = p x s + y s = q ( I x ) + ( I y ) = | I x | , which yields that p x s + y s = I x up to a sign. Next we know that the Euclideancurvature κ is also invariant under SE (2) , and we obtain κ = x s y ss − y s x ss ( x s + y s ) / = I x I y − I y I x (( I x ) + ( I y ) ) / = I y ( I x ) , which gives us I y in terms of κ and | x s | . Since we are considering all independent variables to be invariant, all totaldifferential operators will also be invariant. Hence, the invariantized differentialoperators D i = f D i | g = ρ ( z ) = D i . We know that ∂∂x i u αK = u αKi , although the same cannot be said about its invariantized version and in general D i I αK = I αKi ; indeed we have that D i I αK = I αKi + M αKi , (8)where M αKi is known as the correction term . We will not go into its calculation since itwould take us to far afield, but for more information on correction terms see § I αJ and I αL and letting JK = LM so that I αJK = I αLM ,then this implies that D K I αJ − M αJK = D M I αL − M αLM . (9) oving frames and conservation laws for Euclidean invariant Lagrangians syzygies or differential identities . These will play a crucialrole in the obtention of the invariantized Euler-Lagrange equations and Noether’sconservation laws. Example 2.5 (cont.)
If we set x = x ( s, t ) and y = y ( s, t ) and take the normalizationequations as before, we obtain e x t | g = ρ ( z ) = I x = x s x t + y s y t p x s + y s , e y t | g = ρ ( z ) = I y = x s y t − y s x t p x s + y s . Furthermore, since both s and t are invariant, D s and D t commute. From Figure 2,we can see that there are two ways in which we can obtain I x I x I x I x I x I x I x I x D x b b b bb bbb D t Figure 2.
Paths to I x and since both ways must be equal, we get a syzygy between I x and η = I x . The syzygyis D t η = D s I x − κηI y . (10) Similarly, we have a syzygy between I y and I y and the syzygy is D t I y = D s I y − η s η D s I y − κ η I y + 2 κη D s I x + κ s ηI x . (11) Kogan and Olver in [7] studied invariant calculus of variations from a geometric pointof view, we instead do it from a differential algebra point of view.Assume Lagrangians to be smooth functions of x , u and finitely many derivativesof the u α and denote these as L [ u ] = R L [ u ]d x . Furthermore, suppose these areinvariant under some group action and let the κ j , j = 1 , ..., N , denote the generatingdifferential invariants of that group action. Also, assume that the action leaves theindependent variables invariant so that the Lagrangians can be rewritten as R L [ κ ]d x .To obtain the invariantized Euler-Lagrange equations we proceed in a similar wayas for finding the Euler-Lagrange equations in the original variables ( x , u ).Recall that if x ( x , u ) extremizes the functional L [ u ], then for a smallperturbation of u ,0 = dd ε (cid:12)(cid:12)(cid:12)(cid:12) ε =0 L [ u + ε v ]= Z q X α =1 " E α ( L ) v α + X i DDx i (cid:18) ∂L∂u αi v α + · · · (cid:19) dx oving frames and conservation laws for Euclidean invariant Lagrangians E α = X K ( − | K | D | K | Dx k · · · Dx k p p ∂∂u αK is the Euler operator with respect to the dependent variables u α . The boundary termscorrespond to Noether’s conservation laws and the variation v to the infinitesimals.To obtain the invariantized Euler-Lagrange equations, we first introduce a dummyinvariant independent variable t and set the u α = u α ( x , t ). The introduction of thisnew independent variable results in q new invariants I αt = g · u αt | g = ρ ( z ) and a set ofsyzygies D t κ = H I ( u t ) that is D t κ ...κ N = H I t ...I qt , (12)where H is a N × q matrix of operators depending only on the D i , for i = 1 , ..., p , the κ j , for j = 1 , ..., N , and their invariant derivatives. Since all independent variables areinvariant, we have that all differential operators commute, specifically [ D i , D t ] = 0,for all i = 1 , ..., p . Remark 2.7.
Up to this moment we have represented the independent variables as x and the dependent variables as u . Since the examples in this paper only involvetwo independent variables, s and t , we will represent the coordinates of the spaceof dependent variables as x , where the dependent variables will be the usual spacecoordinates. For simplicity, consider a one-dimensional Lagrangian L ( x, y, y x , y xx , ... )d x witha finite number of arguments which is invariant under the SE (2) group action (2).Such variational problems can be rewritten in terms of the generating invariants of itsgroup action, in this case the Euclidean curvature, κ , and its derivatives with respectto s , the Euclidean arc length. We reparametrize ( x, y ( x )) as ( x ( s ) , y ( s )), and to fixparametrization as arc length, we introduce η = p x s + y s = 1 as a constraint. Thus,we consider the invariantized variational problem Z [ L ( κ, κ s , κ ss , ... ) − λ ( s )( η − s, (13)where λ ( s ) is a Lagrange multiplier. This constraint does not reduce the solution setand it will simplify the calculations. Symbolically, we know thatdd ε (cid:12)(cid:12)(cid:12)(cid:12) ε =0 L [ x + ε v ] = DDt (cid:12)(cid:12)(cid:12)(cid:12) x t = v L [ x ] . Hence, after differentiating (13) under the integral sign and integrating by parts weobtain D t Z [ L ( κ, κ s , κ ss , ... ) − λ ( s )( η − s = Z (cid:20) ∂L∂κ D t κ + ∂L∂κ s D s D t κ + ∂L∂κ ss D s D t κ + · · · − λ ( s ) D t η (cid:21) d s = Z h (cid:18) ∂L∂κ − D s (cid:18) ∂L∂κ s (cid:19) + D s (cid:18) ∂L∂κ ss (cid:19) − D s (cid:18) ∂L∂κ sss (cid:19) + · · · (cid:19) D t κ − λ ( s ) D t η oving frames and conservation laws for Euclidean invariant Lagrangians D s (cid:16) X m =1 m − X n =0 ( − n D ns (cid:18) ∂L∂κ m (cid:19) D m − − ns D t κ (cid:17)i d s = Z h E κ ( L ) D t κ − λ ( s ) D t η + D s (cid:16) X m =1 m − X n =0 ( − n D ns (cid:18) ∂L∂κ m (cid:19) D m − − ns D t κ (cid:17)i d s, where κ m = ∂ m κ∂s m . Next we substitute the circled D t η and D t κ by their respective syzygies D t (cid:18) ηκ (cid:19) = D s − κκ s D s + κ I x I y = H H I ( x t ) , (14)where we have already set η = 1; it makes no difference to the final result to do thisat this point. Hence, Z h ( E κ ( L ) H − λ ( s ) H ) I ( x t ) + D s (cid:16) X m =1 m − X n =0 ( − n D ns (cid:18) ∂L∂κ m (cid:19) D m − − ns D t κ (cid:17)i d s = Z h ( H ∗ ( E κ ( L )) − H ∗ ( λ ( s ))) I ( x t ) + D s (cid:16) − λ ( s ) I x + E κ ( L ) D s I y − D s E κ ( L ) I y + X m =1 m − X n =0 ( − n D ns (cid:18) ∂L∂κ m (cid:19) D m − − ns D t κ (cid:17)i d s, (15)after a second set of integration by parts, and where H ∗ and H ∗ are the respectiveadjoint operators of H and H . The vector I ( x t ) corresponds to the variation v andthus the coefficients of I x and I y represent respectively the Euler-Lagrange equations E x ( L ) = κ s E κ ( L ) + λ s , (16) E y ( L ) = D s E κ ( L ) + κ E κ ( L ) + λ ( s ) κ. (17)We will use E x ( L ) = 0 to eliminate λ ( s ) instead of E y ( L ) = 0, as it containsderivatives of κ of lower order. Since L does not depend on s explicitly, by the resultin page 220 of [6], κ s E κ ( L ) is a total derivative, specifically κ s E κ ( L ) = D s L − X m =1 m − X j =0 ( − j D js (cid:18) ∂L∂κ m (cid:19) κ m − j , then we obtain λ ( s ) = − L + X m =1 m − X j =0 ( − j D js (cid:18) ∂L∂κ m (cid:19) κ m − j , (18)where the constant of integration has been absorbed into λ ( s ) (see Remark 7.1.9.of [6]). Hence, we are left with one invariantized Euler-Lagrange equation in oneunknown E y ( L ) = D s E κ ( L ) + κ E κ ( L ) − κ L − X m =1 m − X j =0 ( − m D ms (cid:18) ∂L∂κ m (cid:19) κ m − j . (19)This particular equation agrees with the one appearing in Kogan and Olver [7]. oving frames and conservation laws for Euclidean invariant Lagrangians If one calculates the conservation laws for one-dimensional variational problems thatare invariant under SE (2) from Noether’s First Theorem (for the formulae of these,see Theorem 4.29 of [9]) and then rewrite these in terms of the Euclidean curvature κ and its derivatives with respect to the Euclidean arc length s , one obtains cos θ − sin θ θ cos θ a sin θ − b cos θ a cos θ + b sin θ | {z } R ( g ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) g = ρ ( z ) − − λ ( s ) − κ E κ ( L ) −D s E κ ( L ) E κ ( L ) | {z } υ ( I ) = c , (20)where ρ ( z ) − is the right moving frame (6), λ ( s ) is the Lagrange multiplier obtainedin (18) and c the constant vector.We note that R ( g ) is the Adjoint representation of SE (2) on its Lie algebra se (2).To see how calculations in Section 2.2 can yield the result in (20), we first show howthe Adjoint representation is calculated in the context we will need.Consider the SE (2) group action (cid:18) e x e y (cid:19) = (cid:18) cos θ − sin θ sin θ cos θ (cid:19) (cid:18) xy (cid:19) + (cid:18) ab (cid:19) (21)with generating infinitesimal vector fields ∂ x , ∂ y , − y∂ x + x∂ y . Let g ∈ SE (2) act on v = α∂ x + β∂ y + γ ( − y∂ x + x∂ y )as in (21), where α , β and γ are constants. Hence, g · v = α∂ e x + β∂ e y + γ ( − e y∂ e x + e x∂ e y )= (cid:0) α β γ (cid:1) cos θ − sin θ θ cos θ a sin θ − b cos θ a cos θ + b sin θ ∂ x ∂ y − y∂ x + x∂ y . Thus, R ( g ) is the Adjoint representation of SE (2), denoted as A d ( g ). Remark 2.8.
In Example 2.1 we used the right action of SE (2) on the plane tocalculate the right moving frame, as it simplified its calculation. However, to compute A d ( ρ ) − we considered the left action of SE (2) on the plane, avoiding in this way theneed to calculate the inverse of A d ( ρ ) . Next recall the boundary terms (15) obtained in the calculation of theinvariantized Euler-Lagrange equations, − λ ( s ) I x + E κ ( L ) D s I y − D s E κ ( L ) I y + X m =1 m − X n =0 ( − n D ns (cid:18) ∂L∂κ m (cid:19) D m − − ns D t κ = c, where c is a constant. Substituting D s I y and D m − − ns D t κ for all m in the aboveexpression by the differential formulae D s I y = I y − κI x , D t κ = I y − κI x , D s D t κ = I y − κ I y − κI x − κ s I x ,... oving frames and conservation laws for Euclidean invariant Lagrangians I x I x · · · ) − λ ( s ) − κ E κ ( L ) − κ ∂L∂κ s − κ s ∂L∂κ ss + · · ·− κ ∂L∂κ ss + · · · ... | {z } C x +( I y I y · · · ) −D s E κ ( L ) E κ ( L ) − κ ∂L∂κ ss + · · · ∂L∂κ s + · · · ∂L∂κ ss + · · · ... | {z } C y = k (22)yields the boundary terms in a form that is linear in the I α K .We now let t be a group parameter. If the parameters are ( a , ..., a r ) and t = a j , then from Theorem 3 of [2], it is shown that the vectors ( I α I α · · · )can be written as the product of row j of A d ( ρ ) − and the matrix of invariantizedinfinitesimals Ω α ( I ) = (cid:16)e ζ ij ( I ) (cid:17) , ζ ij = ∂ e z i ∂a j (cid:12)(cid:12)(cid:12)(cid:12) g = e , where α represents a dependent variable, a j a group parameter, and e the identityelement. The vector of invariants in (20) equals the sum of the products of the matricesof invariantized infinitesimals Ω α ( I ) with the vectors C α ,Ω x ( I ) C x + Ω y ( I ) C y = − λ ( s ) − κ E κ ( L ) −D s E κ ( L ) E κ ( L ) , whereΩ x ( I ) = x x s x ss · · · a · · · b · · · θ − κ · · · , Ω y ( I ) = y y s y ss y sss · · · a · · · b · · · θ − κ · · · . Noether’s conservation laws for one-dimensional Lagrangians invariant under SE (2) can be written as x s − y s y s x s xy s − yx s xx s + yy s − λ ( s ) − κ E κ ( L ) −D s E κ ( L ) E κ ( L ) = c , (23)where x s + y s = 1 was used to simplify the conservation laws and λ ( s ) = − L + X m =1 m − X j =0 ( − j D js (cid:18) ∂L∂κ m (cid:19) κ m − j . The following theorem generalizes what we have just seen for one-dimensionalvariational problems that are invariant under SE (2); it states that the conservationlaws from Noether’s First Theorem can be written as the divergence of the product ofa moving frame with vectors of invariants. Theorem 2.9.
Let R L ( κ , κ , ... )d x be invariant under G × M → M , where M = J n ( X × U ) , with generating invariants κ j , for j = 1 , ..., N , and let e x i = g · x i = x i ,for i = 1 , ..., p . Let ( a , ..., a r ) be coordinates of G near the identity e , and v i , for i = 1 , ..., r , the associated infinitesimal vector fields. Furthermore, let A d ( g ) be theoving frames and conservation laws for Euclidean invariant Lagrangians Adjoint representation of G with respect to these vector fields. For each dependentvariable, define the matrix of infinitesimals to be Ω α ( e z ) = (cid:16) e ζ ij (cid:17) , where ζ ij = ∂ e z i ∂a j (cid:12)(cid:12)(cid:12)(cid:12) g = e are the infinitesimals of the prolonged group action. Let Ω α ( I ) , for α = 1 , ..., q , be theinvariantized version of the above matrices.Introduce a dummy invariant variable t to effect the variation and then integrationby parts yields ∂∂t Z L ( κ , κ , ... )d x = Z h X α E α ( L ) I αt + Div ( P ) i d x , where this defines the p -tuple P , whose components are of the form P i = X α,J I αtK C αi,K , i = 1 , ..., p, and the vectors C αi = ( C αi,K ) . Let ρ ( z ) − be a right frame with canonical invariants I αtK = I ( u αtK ) , where K is the index of differentiation with respect to the independentvariables x i , for i = 1 , ..., p . Then the r conservation laws obtained via Noether’s FirstTheorem can be written in the form X i DDx i A d ( ρ ( z )) − υ i ( I ) = 0 , (24) where υ i ( I ) = X α Ω α ( I ) C αi . The proof can be found in [2].
Remark 2.10.
One can notice that none of the summands of X m =1 m − X n =0 ( − n D ns (cid:18) ∂L∂κ m (cid:19) D m − − ns D t κ in the boundary terms of (15) show up in the conservation laws (23), in other wordsall boundary terms coming from the first set of integration by parts have disappeared.This is no coincidence, it is due to the conflation of t with each group parameter; theproof of this can be found in [10]. In this paper we are interested in showing how the structure of Noether’sconservation laws can be used to solve the integration problem for variational problemsthat are invariant under SE (2) and SE (3). In the present section we have computedNoether’s conservation laws for one-dimensional Lagrangians that are invariant under SE (2). In the next section, we will see how these can be used to solve the extremizationproblems. oving frames and conservation laws for Euclidean invariant Lagrangians
3. Invariant Lagrangians under SE ( )Recall that the invariantized Euler-Lagrange equation for a one-dimensionalLagrangian invariant under SE (2) is E y ( L ) = D s E κ ( L ) + κ E κ ( L ) − κ L − X m =1 m − X j =0 ( − j D js (cid:18) ∂L∂κ m (cid:19) κ m − j , (25)and its associated conservation laws are x s − y s y s x s xy s − yx s xx s + yy s | {z } A d ( ρ ( z )) − − λ ( s ) − κ E κ ( L ) −D s E κ ( L ) E κ ( L ) | {z } υ ( I ) = c , where λ ( s ) = − L + X m =1 m − X j =0 ( − j D js (cid:18) ∂L∂κ m (cid:19) κ m − j . Multiplying both sides of A d ( ρ ( z )) − υ ( I ) = c by A d ( ρ ( z )) yields the followingsystem of equations − λ ( s ) − κ E κ ( L ) = x s c + y s c , (26) −D s E κ ( L ) = x s c − y s c , (27) E κ ( L ) = yc − xc + c . (28)Also, we obtain a first integral of the Euler-Lagrange equation as follows. Define B = , which satisfies the following equality B = A d ( ρ ) − T B A d ( ρ ) − . The first integral of the Euler-Lagrange equation is then υ T ( I ) B υ ( I ) = c T B c i.e. ( λ ( s ) + κ E κ ( L )) + ( D s E κ ( L )) = c + c . (29)Once κ is known, one can see that the integration problem has a straightforwardsolution. Hence, integrating both sides of Equation (26) with respect to s yields xc + yc = Z [ − λ ( s ) − κ E κ ( L )] d s, (30)and thus solving the two linear equations (28) and (30) with respect to x and y oneobtains x ( s ) = 1 c + c (cid:18) c Z [ − λ ( s ) − κ E κ ( L )] d s − c E κ ( L ) + c c (cid:19) , (31) y ( s ) = 1 c + c (cid:18) c Z [ − λ ( s ) − κ E κ ( L )] d s + c E κ ( L ) + c c c (cid:19) . (32) oving frames and conservation laws for Euclidean invariant Lagrangians s of (28).If we consider the Lagrangian with L = 1 and plug it in Equations (25), (29),(31) and (32), then the solution which minimizes the arc length is the equation of aline, as one would expect. Another famous Lagrangian to consider is Z κ d s. Using the equations above, we obtain, as Euler himself did [11], that the curvature ofthe minimizing curve satisfies κ ss + 12 κ = 0 , or the first integral of the Euler-Lagrange equation4 κ s + κ = c + c , which is solved by an elliptic function. Solutions are known as Euler’s elastica . For agood historical report see [12].
4. Invariant Lagrangians under SE ( )In the previous section we showed that for an invariant Lagrangian under SE (2), theinvariantized Euler-Lagrange equation and its associated conservation laws could beused to solve the extremising problem. In this section we will proceed analogously andpresent the solution to one-dimensional variational problems that are invariant underthe following SE (3) group action on the ( x ( s ) , y ( s ) , z ( s ))-space parametrized by theEuclidean arc length s x e x = R − ( x − a ) , (33)where R − represents the rotation in the three-dimensional space cos β cos γ cos β sin γ sin β − sin α sin β cos γ − cos α sin γ − sin α sin β sin γ + cos α cos γ sin α cos β − cos α sin β cos γ + sin α sin γ − cos α sin β sin γ − sin α cos γ cos α cos β and a = ( a b c ) T the translation vector with α , β , γ , a , b and c as the constantsthat parametrize the group action.To solve SE (3) invariant variational problems, we need to find the element g ∈ SE (3) that sends the tangent to the curve to the x -axis, the normal to thecurve to the y -axis and the point ( x, y, z ) to the origin, in other words which sends z = ( x, y, z, y s , z s , z ss ) to the cross section (0 , , , , , e x = 0, e y = 0, e z = 0, e y s = 0, e z s = 0 and f z ss = 0, andthus obtain the right moving frame a = x, b = y, c = z, α = tan − (cid:18) y s ( y s z ss − z s y ss ) − x s ( z s x ss − x s z ss ) √ x s + y s + z s ( x s y ss − y s x ss ) (cid:19) ,β = tan − (cid:18) z s √ x s + y s (cid:19) , γ = tan − (cid:16) y s x s (cid:17) . (34)Consider a one-dimensional variational problem L [ x ] that is invariant under thegroup action (33). To obtain the invariantized Euler-Lagrange equations, we first oving frames and conservation laws for Euclidean invariant Lagrangians L [ x ] in terms of the generating invariants of the group action, which are the Euclidean curvature κ = k x s × x ss kk x s k , and torsion τ = x sss · ( x s × x ss ) k x s × x ss k , and their derivatives with respect to s .Since s represents the Euclidean arc length, the constraint η = p x s + y s + z s = 1must be introduced into the variational problem in order to fix parametrization.Hence, the resulting invariantized functional is Z [ L ( κ, τ, κ s , τ s , κ ss , τ ss , ... ) − λ ( s )( η − s, (35)where λ ( s ) is a Lagrange multiplier. As for SE (2), this will not reduce the solutionset and will simplify the computation of the conservation laws.Next we introduce a dummy invariant independent variable t and set x = x ( s, t )to effect variation. The introduction of a new independent variable results in threenew invariants I αt , for α = x, y, z , and a set of syzygies D t ηκτ = H I x I y I z , (36)where the matrix of operators H is D s − κ κ s κ − τ + D s − τ s − τ D s τ s + 2 τ D s D s (cid:0) τ s κ (cid:1) + (cid:16) τ s κ − κ s τκ (cid:17) D s + τκ D s D s (cid:16) − τ κ (cid:17) + (cid:16) κ − τ κ (cid:17) D s − κ s κ D s + κ D s , where we have already set η = 1.As in Section 2.2, we differentiate (35) with respect to t and then integrate byparts twice to obtain the invariantized Euler-Lagrange equations E x ( L ) = κ s E κ ( L ) + τ s E τ ( L ) − D s (2 τ E τ ( L )) + λ s , E y ( L ) = D s E κ ( L ) + 2 τκ D s E τ ( L ) + (cid:18) τ s κ − τ κ s κ (cid:19) D s E τ ( L ) + ( κ − τ ) E κ ( L )+ 2 τ κ E τ ( L ) + λ ( s ) κ, E z ( L ) = − κ D s E τ ( L ) + 2 κ s κ D s E τ ( L ) + (cid:18) κ ss κ + τ κ − κ s κ − κ (cid:19) D s E τ ( L ) − κ s E τ ( L ) + 2 τ D s E κ ( L ) + τ s E κ ( L ) , and the boundary terms (2 τ E τ ( L ) − λ ) I x + (cid:18) τ s κ E τ ( L ) − τκ D s E τ ( L ) − D s E κ ( L ) (cid:19) I y + (cid:18) E κ ( L ) + 2 τκ E τ ( L ) (cid:19) D s I y + (cid:18) κ E τ ( L ) − τ E κ ( L ) − τ κ E τ ( L ) − κ s κ D s E τ ( L ) + 1 κ D s E τ ( L ) (cid:19) I z − κ D s E τ ( L ) D s I z + 1 κ E τ ( L ) D s I z = k. oving frames and conservation laws for Euclidean invariant Lagrangians t with the group parameters.Using E x ( L ) = 0 and the fact that κ s E κ ( L ) + τ s E τ ( L ) = D s (cid:18) L − X m =1 m − X i =0 ( − i D is ∂L∂κ m κ m − i − X m =1 m − X i =0 ( − i D is ∂L∂τ m τ m − i (cid:19) , (see page 220 of [6]), we can eliminate λ . Thus, λ ( s ) = 2 τ E τ ( L ) − L + X m =1 m − X i =0 ( − i D is ∂L∂κ m κ m − i + X m =1 m − X i =0 ( − i D is ∂L∂τ m τ m − i , (37)where the constant of integration has been absorbed into the Lagrange multiplier andhence we obtain two Euler-Lagrange equations in two unknowns.To calculate the conservation laws associated to the invariantized Euler-Lagrangeequations, we must compute the moving frame A d ( ρ ) − and the vector of invariants υ ( I ). To compute the former we proceed as in Section 2.3: we calculate the Adjointrepresentation A d ( g ) of SE (3) with respect to its generating infinitesimal vector fields v a = ∂ x , v b = ∂ y , v c = ∂ z , v α = y∂ z − z∂ y , v β = x∂ z − z∂ x , v γ = x∂ y − y∂ x , and evaluate it at the frame (34), which yields A d ( ρ ( z )) − = (cid:18) ρ F S DXρ
F S Dρ F S D (cid:19) , where ρ F S = (cid:0) x s x ss κ x s × x ss κ (cid:1) is the Frenet-Serret frame, D is the following diagonal matrix D = − , and X is the matrix X = − z yz − x − y x . As seen in Section 2.3, to obtain the vector of invariants we must first find theboundary terms that are linear in the I α K . To do so, consider the boundary termsobtained from the calculation of the invariantized Euler-Lagrange equations (2 τ E τ ( L ) − λ ) I x + (cid:0) τ s κ E τ ( L ) − τκ D s E τ ( L ) − D s E κ ( L ) (cid:1) I y + (cid:0) E κ ( L ) + τκ E τ ( L ) (cid:1) D s I y + (cid:16) κ E τ ( L ) − τ E κ ( L ) − τ κ E τ ( L ) − κ s κ D s E τ ( L ) + κ D s E τ ( L ) (cid:17) I z − κ D s E τ ( L ) D s I z + κ E τ ( L ) D s I z = k. (38) Substituting D s I y , D s I z and D s I z in (38) by the differential formulae D s I y = − κI x + I y + τ I z , D s I z = − τ I y + I z , D s I z = τ κI x − τ s I y − τ I y − τ I z + I z , oving frames and conservation laws for Euclidean invariant Lagrangians I α K , (cid:0) I x (cid:1) (cid:0) − κ E κ ( L ) + τ E τ ( L ) − λ ( s ) (cid:1)| {z } C x + (cid:0) I y I y (cid:1) −D s E κ ( L ) − τκ D s E τ ( L ) E κ ( L ) !| {z } C y + (cid:0) I z I z I z (cid:1) κ D s E τ ( L ) − κ s κ D s E τ ( L ) + κ E τ ( L ) − τ E κ ( L ) − κ D s E τ ( L )1 κ E τ ( L ) | {z } C z = k. Finally, adding the products of the matrices of invariantized infinitesimals Ω α ( I ) withthe vectors C α yields the vector of invariants υ ( I ) = τ E τ ( L ) − κ E κ ( L ) − λ ( s ) −D s E κ ( L ) − τκ D s E τ ( L ) κ D s E τ ( L ) − κ s κ D s E τ ( L ) + κ E τ ( L ) − τ E κ ( L ) E τ ( L ) − κ D s E τ ( L ) E κ ( L ) , where the Ω α ( I ) areΩ x ( I ) = , Ω y ( I ) = , Ω z ( I ) = κ . Thus, the conservation laws are (cid:16) ρ F S DXρ
F S Dρ F S D (cid:17) τ E τ ( L ) − κ E κ ( L ) − λ ( s ) −D s E κ ( L ) − τκ D s E τ ( L ) κ D s E τ ( L ) − κ s κ D s E τ ( L ) + κ E τ ( L ) − τ E κ ( L ) E τ ( L ) − κ D s E τ ( L ) E κ ( L ) = (cid:16) c c (cid:17) = c , (39)where c = ( c , c , c ) T , c = ( c , c , c ) T are constant vectors and λ ( s ) is equal to(37).We shall see in the remainder of this section how the conservation laws (39) canhelp reduce the integration problem.To demonstrate this, we will start by simplifying the conservation laws (39) in twosteps. These simplifications will then lead to an overdetermined system of equationsfor x , y and z , which will be solved with relative ease once κ and τ are known. Finally,we will give a reason behind the choice in the order in which we solve the equations.In the first step of the simplification, we apply an element of SE (3), say A d ( g ) − ,to both sides of A d ( ρ ( z )) − υ ( I ) = c such that it maps c and c to the z -axis. Let A d ( g ) act on c as follows A d ( g ) c = (cid:18) R DA R D R D (cid:19) (cid:18) c c (cid:19) = (cid:18) e c e c (cid:19) , oving frames and conservation laws for Euclidean invariant Lagrangians R is the three-dimensional rotation R = cos β cos γ − sin α sin β cos γ − cos α sin γ − cos α sin β cos γ + sin α sin γ cos β sin γ − sin α sin β sin γ + cos α cos γ − cos α sin β sin γ − sin α cos γ sin β sin α cos β cos α cos β ,D is the diagonal matrix, D = diag(1 , − , A the matrix, A = − c bc − a − b a . We can easily verify that the Adjoint representation of SE (3) does not act freely onthe constant vector c , since it preserves the length of c and the quantity c T D c .Indeed to prove the latter, we multiply through DA R c + D R D c = e c by c T R T D , and then obtain that e c T A e c | {z } =0 + c T D c = e c T D e c . Thus, let A d ( g ) − send c to C = (cid:16) | c | c T D c | c | (cid:17) T . Applying A d ( g ) − to the conservation laws yields A d ( g ) − A d ( ρ ( z )) − υ ( I ) = A d ( g ) − c , which reduces to A d ( ρ ( e z )) − υ ( I ) = C (40)by the equivariance of the right moving frame A d ( ρ ( z )) − .The second step of our simplification consists of applying A d ( ρ ( e z )) to (40) toobtain the following system of equations | c | e z s = υ (1) ( I ) , (41) | c | κ f z ss = υ (2) ( I ) , (42) | c | κ ( e x s f y ss − e y s f x ss ) = υ (3) ( I ) , (43) | c | ( e x e y s − e y e x s ) + c T D c | c | e z s = υ (4) ( I ) , (44) | c | κ ( f x ss e y − f y ss e x ) − c T D c κ | c | f z ss = υ (5) ( I ) , (45) | c | κ ( e x ( e z s f x ss − e x s f z ss ) − e y ( e y s f z ss − e z s f y ss )) + c T D c κ | c | ( e x s f y ss − e y s f x ss ) = υ (6) ( I ) , (46)where we have used υ ( j ) ( I ) to denote the j -th component of υ ( I ). This overdeterminedsystem of equations can now be solved more easily. oving frames and conservation laws for Euclidean invariant Lagrangians B = diag( 1 1 1 0 0 0 ), which satisfies B = A d ( ρ ) − T B A d ( ρ ) − . Thenwe obtain the first integral of the Euler-Lagrange equations, υ T ( I ) B υ ( I ) = C T B C , which is equaivalent to( τ E τ ( L ) − κ E κ ( L ) − λ ( s )) + (cid:0) −D s E κ ( L ) − τκ D s E τ ( L ) (cid:1) + (cid:0) κ D s E τ ( L ) − κ s κ D s E τ ( L ) + κ E τ ( L ) − τ E κ ( L ) (cid:1) = c + c + c . (47)For the second first integral, we define D = (cid:18) DD (cid:19) , D = diag( 1 − , which satifies D = A d ( ρ ) − T D A d ( ρ ) − . The first integral of the Euler-Lagrangeequations is then υ T ( I ) D υ ( I ) = C T D C , i.e.( τ E τ ( L ) − κ E κ ( L ) − λ ( s )) E τ ( L ) + (cid:0) −D s E κ ( L ) − τκ D s E τ ( L ) (cid:1) κ D s E τ ( L )+ (cid:0) κ D s E τ ( L ) − κ s κ D s E τ ( L ) + κ E τ ( L ) − τ E κ ( L ) (cid:1) E κ ( L ) = c c − c c + c c . (48)We can use the above first integrals to determine κ and τ . Once we have solvedfor these, we use the system of simplified conservation laws to solve for e x , e y and e z .Hence, solving Equation (41) gives g z ( s ) = 1 | c | Z υ (1) ( I )d s. (49)Next, multiplying Equation (43) by − c T D c | c | and adding it to Equation (46) yields | c | κ ( e x ( e z s f x ss − e x s f z ss ) − e y ( e y s f z ss − e z s f y ss )) = υ (6) ( I ) − c T D c | c | υ (3) ( I ) , which simplifies to | c | (cid:18) e z s (cid:18) D s ( e x · e x ) − (cid:19) − f z ss D s ( e x · e x ) (cid:19) = κυ (6) ( I ) − κ c T D c | c | υ (3) ( I ) , where D s ( e x · e x ) − e x · f x ss and D s ( e x · e x ) = e x · f x s . Setting D s ( e x · e x ) = h ( s ) andsubstituting e z s by (41) and f z ss by its derivative yields a linear equation for h D s h − D s υ (1) ( I ) υ (1) ( I ) h = 2 κ (cid:18) υ (6) ( I ) − c T D c | c | υ (3) ( I ) (cid:19)(cid:30) υ (1) ( I ) + 2 . Solving for h we obtain h ( s ) = υ (1) ( I ) Z υ (1) ( I ) (cid:18) κ (cid:18) υ (6) ( I ) − c T D c | c | υ (3) ( I ) (cid:19)(cid:30) υ (1) ( I ) + 2 (cid:19) d s. Hence, D s ( e x · e x ) = υ (1) ( I ) Z υ (1) ( I ) (cid:18) κ (cid:18) υ (6) ( I ) − c T D c | c | υ (3) ( I ) (cid:19)(cid:30) υ (1) ( I ) + 2 (cid:19) d s. (50) oving frames and conservation laws for Euclidean invariant Lagrangians g x ( s ) = r ( s ) cos θ ( s ) , g y ( s ) = r ( s ) sin θ ( s ) , g z ( s ) = g z ( s ) . Starting with Equation (50), we obtain r ( s ) = Z h ( s )d s − g z ( s ) , (51)as D s ( e x · e x ) = D s (cid:16) r ( s ) + g z ( s ) (cid:17) . After applying the change of coordinates, Equation(44) becomes r ( s ) θ s = 1 | c | (cid:18) υ (4) ( I ) − c T D c | c | υ (1) ( I ) (cid:19) , and then solving for θ yields θ ( s ) = Z r ( s ) | c | (cid:18) υ (4) ( I ) − c T D c | c | υ (1) ( I ) (cid:19) d s. (52)To recover x , y and z , we act on e x , e y and e z as follows e x x = R e x + a , where R is the three-dimensional rotation R = cos β cos γ − sin α sin β cos γ − cos α sin γ − cos α sin β cos γ + sin α sin γ cos β sin γ − sin α sin β sin γ + cos α cos γ − cos α sin β sin γ − sin α cos γ sin β sin α cos β cos α cos β , and a = ( a b c ) T is the translation vector, with α = − tan − p | c | cos β − c c ! , γ = tan − c c sin β + c p | c | cos β − c c c sin β − c p | c | cos β − c ! ,a = c c c + c | c | + c c T D c c | c | , b = c c c + c | c | − c c T D c c | c | , and where β and c are free.Although only four of the equations of the system were used to solve for x , y and z , we know that the remaining two equations have been satisfied. Indeed, if wedifferentiate A d ( ρ ( e z )) − υ ( I ) = C with respect to s and multiply by A d ( ρ ( e z )), thenwe get D s υ ( I ) = D s ( A d ( ρ ( e z )))) A d ( ρ ( e z )) − υ ( I ) , which is equivalent to D s υ ( I ) = κ − κ τ − τ − κ
00 0 − κ − τ − τ υ ( I ) . (53)The above system of equations not only forms part of an elimination ideal as it onlyinvolves invariants, but because the invariants appear only on the right-hand sides ofthe Equations (41), (42), (43), (44), (45) and (46), they also encode the relationships oving frames and conservation laws for Euclidean invariant Lagrangians D s (Equation (41)) = κ (Equation (42))and so forth. Using the Equations in (53) we can eliminate (42) and (45) from thesystem.In the following two well-known examples, we can verify that we obtain theexpected solutions. Example 4.1.
For the Lagrangian with L = 1 , Equations (49), (51) and (52) yieldthe equation of a line in parametric form as the solution that minimizes the arc length,as expected. Note that | f x s | = 1 imposes a condition on the constants of integration. Example 4.2.
Consider the Lagrangian R κ d s with torsion, τ , equal to zero. Thenwe obtain that κ satisfies the Euler-Lagrange equation κ ss + 12 κ = 0 , which is the same equation as for the SE (2) case, or the first integral κ + 4 κ s = | c | . From Equation (48) we know that c T D c = 0 , and from Equations (49), (51) and (52) we obtain g z ( s ) = − | c | Z κ d s,r ( s ) = − Z h κ Z κ d s i d s − | c | (cid:18)Z κ d s (cid:19) ,θ ( s ) = A, where A is a constant. So the solution e x lies on a plane that includes the e z -axis, asexpected.
5. Conclusion
Noether’s First Theorem is a well-known result which provides conservation lawsfor Lie group invariant variational problems. In recent work [2], the mathematicalstructure of both the Euler-Lagrange system and the set of conservation laws wasgiven in terms of the differential invariants of the group action and a moving frame.It is the knowledge of this structure that allows one to solve the invariant variationalproblems under some group action with relative ease. In this paper, we examineone-dimensional variational problems that are invariant under the actions of SE (2)and SE (3). For both cases, we obtain the invariantized Euler-Lagrange equationsand their associated conservation laws in the new format, from which we can thenobtain the solution to the variational problem by quadratures. One can verify thatthis method leads to a far simpler computational problem than the one given in theoriginal variables. oving frames and conservation laws for Euclidean invariant Lagrangians References [1] Emmy Noether. Invariante Variationsprobleme. Nachr. Ges. Wiss. G¨ottingem.
Math.-Phys.Kl. , pages 235–257, 1918. An english translation is available at arXiv:physics/0503066v1[physics.hist-ph] .[2] Tˆania M. N. Gon¸calves and Elizabeth L. Mansfield. On Moving Frames and Noether’sConservation Laws, 2011. DOI: 10.1111/j.1467-9590.2011.00522.x.[3] Sophus Lie. Klassifikation und Integration von gew¨ohnlichen Differentialgleichungen zwischenx, y, die eine Gruppe von Transformationen gestatten I, II.
Mat. Ann. , 32:213–281, 1888.[4] Mark Fels and Peter J.Olver. Moving Coframes I.
Acta Appl. Math. , 51:161–312, 1998.[5] Mark Fels and Peter J.Olver. Moving Coframes II.
Acta Appl. Math. , 55:127–208, 1999.[6] Elizabeth L. Mansfield.
A Practical Guide to the Invariant Calculus . Cambridge UniversityPress, Cambridge, 2010.[7] Irina A. Kogan and Peter J. Olver. Invariant Euler-Lagrange Equations and the InvariantVariational Bicomplex.
Acta Appl. Math. , 76:137–193, 2003.[8] Evelyn Hubert. AIDA Maple Package: Algebraic Invariants and their Differential Algebras,2007.[9] Peter J. Olver.
Applications of Lie Groups to Differential Equations, Second Edition . Springer,New York, 1993.[10] Tˆania M. N. Gon¸calves. On the Vanishing Boundary Terms of Noether’s Conservation Laws. arXiv:1112.4085v1 [math.DG] .[11] Leonhard Euler. Methodus inveniendi lineas curvas maximi minimive proprietate gaudentes,sive solutio problematis isoperimetrici lattissimo sensu accepti, 1744.[12] Raph Levien. The elastica: a mathematical history.