Symmetry Methods for Differential Equations and Conservation Laws. Peter J. Olver University of Minnesota

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Symmetry Methods for Differential Equations and Conservation Laws Peter J. Olver University of Minnesota http://www.math.umn.edu/ olver Santiago, November, 2010

Symmetry Groups of Differential Equations System of differential equations (x, u (n) ) = 0 G Lie group acting on the space of independent and dependent variables: ( x, ũ) = g (x, u) = (Ξ(x, u), Φ(x, u))

G acts on functions u = f(x) by transforming their graphs: g Definition. G is a symmetry group of the system = 0 if f = g f is a solution whenever f is.

Vector field: Infinitesimal Generators In local coordinates: v = p i=1 v (x,u) = d dε g ε (x, u) ε=0 ξ i (x, u) x i + q α=1 generates the one-parameter group dx i dε = ξi (x, u) ϕ α (x, u) du α dε = ϕα (x, u) u α

Example. The vector field v = u x + x u generates the rotation group since x = x cos ε u sin ε d x dε = ũ ũ = x sin ε + u cos ε dũ dε = x

Jet Spaces x = (x 1,...,x p ) independent variables u = (u 1,..., u q ) dependent variables u α J = k u α x j 1... x k partial derivatives (x, u (n) ) = (... x i... u α... u α J... ) Jn jet coordinates ( p + n dimj n = p + q (n) = p + q n )

Prolongation to Jet Space Since G acts on functions, it acts on their derivatives, leading to the prolonged group action on the jet space: ( x, ũ (n) ) = pr (n) g (x, u (n) ) = formulas provided by implicit differentiation Prolonged vector field or infinitesimal generator: pr v = v + ϕ α J (x, u(n) ) α,j u α J

The coefficients of the prolonged vector field are given by the explicit prolongation formula: ϕ α J = D J Qα + p i=1 ξ i u α J,i Q = (Q 1,...,Q q ) characteristic of v Q α (x, u (1) ) = ϕ α p i=1 Invariant functions are solutions to Q(x, u (1) ) = 0. ξ i uα x i

Symmetry Criterion Theorem. (Lie) A connected group of transformations G is a symmetry group of a nondegenerate system of differential equations = 0 if and only if pr v( ) = 0 whenever u is a solution to = 0 for every infinitesimal generator v of G. (*) are the determining equations of the symmetry group to = 0. For nondegenerate systems, this is equivalent to ( ) pr v( ) = A = ν A ν ν

Nondegeneracy Conditions Maximal Rank: ( rank ν x i ν u α J ) = max Local Solvability: Any each point (x 0, u (n) 0 ) such that (x 0, u (n) 0 ) = 0 there exists a solution u = f(x) with u (n) 0 = pr (n) f(x 0 ) Nondegenerate = maximal rank + locally solvable

Normal: There exists at least one non-characteristic direction at (x 0, u (n) 0 ) or, equivalently, there is a change of variable making the system into Kovalevskaya form n u α t n = Γα (x,ũ (n) ) Theorem. (Finzi) A system of q partial differential equations = 0 in q unknowns is not normal if and only if there is a nontrivial integrability condition: D = ν D ν ν = Q order Q < order D + order

Under-determined: The integrability condition follows from lower order derivatives of the equation: D 0 Example: 1 = u xx + v xy, 2 = u xy + v yy D x 2 D y 1 0 Over-determined: The integrability condition is genuine. Example: 1 = u xx + v xy v y, 2 = u xy + v yy + u y D x 2 D y 1 = u xy + v yy

A Simple O.D.E. u xx = 0 Infinitesimal symmetry generator: Second prolongation: v = ξ(x,u) x v (2) = ξ(x,u) x + ϕ(x, u) u + ϕ(x, u) u + + ϕ 1 (x, u (1) ) u x + ϕ 2 (x, u (2) ) u xx,

ϕ 1 = ϕ x + (ϕ u ξ x )u x ξ u u 2 x, ϕ 2 = ϕ xx + (2ϕ xu ξ xx )u x + (ϕ uu 2ξ xu )u 2 x ξ uu u 3 x + (ϕ u 2ξ x )u xx 3ξ u u x u xx. Symmetry criterion: ϕ 2 = 0 whenever u xx = 0.

Symmetry criterion: ϕ xx + (2ϕ xu ξ xx )u x + (ϕ uu 2ξ xu )u 2 x ξ uu u3 x = 0. Determining equations: ϕ xx = 0 2ϕ xu = ξ xx ϕ uu = 2ξ xu ξ uu = 0 General solution: ξ(x, u) = c 1 x 2 + c 2 xu + c 3 x + c 4 u + c 5 ϕ(x, u) = c 1 xu + c 2 u 2 + c 6 x + c 7 u + c 8 = Linear!

Symmetry algebra: v 1 = x v 2 = u v 3 = x x v 4 = x u v 5 = u x v 6 = u u v 7 = x 2 x + xu u v 8 = xu x + u 2 u Symmetry Group: (x, u) ( ) ax + bu + c dx + eu + f, hx + ju + k hx + ju + k = projective group

Prolongation Infinitesimal symmetry v = ξ(x, t, u) x + τ(x, t, u) t First prolongation pr (1) v = ξ x + τ t + ϕ u + ϕx Second prolongation pr (2) v = pr (1) v + ϕ xx u xx + ϕ xt + ϕ(x, t, u) u + ϕ t u x u t u xt + ϕ tt u tt

where ϕ x = D x Q + ξu xx + τu xt ϕ t = D t Q + ξu xt + τu tt ϕ xx = D 2 x Q + ξu xxt + τu xtt Characteristic Q = ϕ ξu x τu t

ϕ x = D x Q + ξu xx + τu xt = ϕ x + (ϕ u ξ x )u x τ x u t ξ u u 2 x τ u u x u t ϕ t = D t Q + ξu xt + τu tt = ϕ t ξ t u x + (ϕ u τ t )u t ξ u u x u t τ u u 2 t ϕ xx = D 2 x Q + ξu xxt + τu xtt = ϕ xx + (2φ xu ξ xx )u x τ xx u t + (φ uu 2ξ xu )u 2 x 2τ xu u x u t ξ uu u3 x τ uu u 2 x u t + (ϕ u 2ξ x )u xx 2τ x u xt 3ξ u u x u xx τ u u t u xx 2τ u u x u xt

Heat Equation u t = u xx Infinitesimal symmetry criterion ϕ t = ϕ xx whenever u t = u xx

Determining equations Coefficient Monomial 0 = 2τ u u x u xt 0 = 2τ x u xt 0 = τ uu u 2 x u xx ξ u = 2τ xu 3ξ u ϕ u τ t = τ xx + ϕ u 2ξ x u x u xx u xx 0 = ξ uu u 3 x 0 = ϕ uu 2ξ xu u 2 x ξ t = 2ϕ xu ξ xx u x ϕ t = ϕ xx 1

General solution ξ = c 1 + c 4 x + 2c 5 t + 4c 6 xt τ = c 2 + 2c 4 t + 4c 6 t 2 ϕ = (c 3 c 5 x 2c 6 t c 6 x 2 )u + α(x, t) α t = α xx

Symmetry algebra v 1 = x v 2 = t v 3 = u u v 4 = x x + 2t t v 5 = 2t x xu u v 6 = 4xt x + 4t 2 t (x 2 + 2t)u u v α = α(x, t) u space transl. time transl. scaling scaling Galilean inversions linearity

Potential Burgers equation u t = u xx + u 2 x Infinitesimal symmetry criterion ϕ t = ϕ xx + 2u x ϕ x

Determining equations Coefficient Monomial 0 = 2τ u u x u xt 0 = 2τ x u xt τ u = τ u 2τ u = τ uu 3τ u ξ u = 2τ xu 3ξ u 2τ x ϕ u τ t = τ xx + ϕ u 2ξ x τ u = τ uu 2τ u ξ u = 2τ xu ξ uu 2τ x 2ξ u ϕ u τ t = τ xx + ϕ uu 2ξ xu + 2ϕ u 2ξ x u 2 xx u 2 x u xx u x u xx u xx u 4 x u 3 x u 2 x ξ t = 2ϕ xu ξ xx + 2ϕ x ϕ t = ϕ xx 1 u x

General solution ξ = c 1 + c 4 x + 2c 5 t + 4c 6 xt τ = c 2 + 2c 4 t + 4c 6 t 2 ϕ = c 3 c 5 x 2c 6 t c 6 x 2 + α(x, t)e u α t = α xx

Symmetry algebra v 1 = x v 2 = t v 3 = u v 4 = x x + 2t t v 5 = 2t x x u v 6 = 4xt x + 4t 2 t (x 2 + 2t) u v α = α(x, t)e u u Hopf-Cole w = e u maps to heat equation.

Symmetry Based Solution Methods Ordinary Differential Equations Lie s method Solvable groups Variational and Hamiltonian systems Potential symmetries Exponential symmetries Generalized symmetries

Partial Differential Equations Group-invariant solutions Non-classical method Weak symmetry groups Clarkson-Kruskal method Partially invariant solutions Differential constraints Nonlocal Symmetries Separation of Variables

Integration of O.D.E. s First order ordinary differential equation du = F(x, u) dx Symmetry Generator: v = ξ(x, u) + ϕ(x, u) x u Determining equation ϕ x + (ϕ u ξ x )F ξ u F 2 = ξ F x + ϕ F u Trivial symmetries ϕ ξ = F

Method 1: Rectify the vector field. Introduce new coordinates near (x 0, u 0 ) so that v (x0,u 0 ) 0 y = η(x, u) w = ζ(x, u) v = w These satisfy first order p.d.e. s ξ η x + ϕ η u = 0 ξ ζ x + ϕ ζ u = 1 Solution by method of characteristics: dx ξ(x, u) = du ϕ(x, u) = dt 1

The equation in the new coordinates will be invariant if and only if it has the form dw dy = h(y) and so can clearly be integrated by quadrature.

Method 2: Integrating Factor If v = ξ x + ϕ u is a symmetry for then P(x, u) dx + Q(x, u) du = 0 R(x, u) = is an integrating factor. 1 ξ P + ϕ Q If ϕ ξ = P Q then the integratimg factor is trivial. Also, rectification of the vector field is equivalent to solving the original o.d.e.

Higher Order Ordinary Differential Equations (x, u (n) ) = 0 If we know a one-parameter symmetry group v = ξ(x, u) + ϕ(x, u) x u then we can reduce the order of the equation by 1.

Method 1: Rectify v = w. Then the equation is invariant if and only if it does not depend on w: (y, w,..., w n ) = 0 Set v = w to reduce the order.

Method 2: Differential invariants. h[pr (n) g (x, u (n) )] = h(x, u (n) ), Infinitesimal criterion: pr v(h) = 0. g G Proposition. If η, ζ are n th order differential invariants, then dη dζ = D x η D x ζ is an (n + 1) st order differential invariant. Corollary. Let y = η(x, u), w = ζ(x, u, u ) be the independent first order differential invariants

for G. Any n th order o.d.e. admitting G as a symmetry group can be written in terms of the differential invariants y, w, dw/dy,..., d n 1 w/dy n 1. In terms of the differential invariants, the n th order o.d.e. reduces to (y, w (n 1) ) = 0 For each solution w = g(y) of the reduced equation, we must solve the auxiliary equation ζ(x, u, u ) = g[η(x, u)] to find u = f(x). This first order equation admits G as a symmetry group and so can be integrated as before.

Multiparameter groups G - r-dimensional Lie group. Assume pr (r) G acts regularly with r dimensional orbits. Independent r th order differential invariants: y = η(x, u (r) ) w = ζ(x, u (r) ) Independent n th order differential invariants: y, w, dw dy,..., dn r w dy n r.

In terms of the differential invariants, the equation reduces in order by r: (y, w (n r) ) = 0 For each solution w = g(y) of the reduced equation, we must solve the auxiliary equation ζ(x, u (r) ) = g[η(x, u (r) )] to find u = f(x). In this case there is no guarantee that we can integrate this equation by quadrature.

Example. Projective group G = SL(2) a u + b (x, u) x, c u + d Infinitesimal generators: ( ), a d b c = 1. u, u u, u 2 u Differential invariants: x, w = 2 u u 3 u 2 Reduced equation (y, w (n 3) ) = 0 u 2 = Schwarzian derivative

Let w = h(x) be a solution to reduced equation. To recover u = f(x) we must solve the auxiliary equation: 2 u u 3 u 2 = u 2 h(x), which still admits the full group SL(2). Integrate using u : u = z 2 z z z 2 = z 2 h(x) Integrate using u u = z z : v = (log z) 2 v + v 2 = h(x) No further symmetries, so we are stuck with a Riccati equation to effect the solution.

Solvable Groups Basis v 1,...,v r of the symmetry algebra g such that [ v i,v j ] = c k ij v k, i < j k<j If we reduce in the correct order, then we are guaranteed a symmetry at each stage. Reduced equation for subalgebra {v 1,...,v k }: (k) (y, w (n k) ) = 0 admits a symmetry ṽ k+1 corresponding to v k+1.

Theorem. (Bianchi) If an n th order o.d.e. has a (regular) r-parameter solvable symmetry group, then its solutions can be found by quadrature from those of the (n r) th order reduced equation.

Example. x 2 u = f(x u u) Symmetry group: v = x u, w = x x, [ v,w ] = v. Reduction with respect to v: Reduced equation: z = x u u x z = h(z) still invariant under w = x x, and hence can be solved by quadrature.

Wrong way reduction with respect to w: Reduced equation: y = u, z = z(y) = x u z(z 1) = h(z y) No remaining symmetry; not clear how to integrate directly.

Group Invariant Solutions System of partial differential equations (x, u (n) ) = 0 G symmetry group Assume G acts regularly on M with r-dimensional orbits Definition. u = f(x) is a G-invariant solution if g f = f for all g G. i.e. the graph Γ f = {(x, f(x))} is a (locally) G- invariant subset of M. Similarity solutions, travelling waves,...

Proposition. Let G have infinitesimal generators v 1,...,v r with associated characteristics Q 1,...,Q r. A function u = f(x) is G-invariant if and only if it is a solution to the system of first order partial differential equations Q ν (x, u (1) ) = 0, ν = 1,..., r. Theorem. (Lie). If G has r-dimensional orbits, and acts transversally to the vertical fibers {x = const.}, then all the G-invariant solutions to = 0 can be found by solving a reduced system of differential equations /G = 0 in r fewer independent variables.

Method 1: Invariant Coordinates. The new variables are given by a complete set of functionally independent invariants of G: η α (g (x, u)) = η α (x, u) for all g G Infinitesimal criterion: v k [η α ] = 0, k = 1,..., r. New independent and dependent variables: y 1 = η 1 (x, u),..., y p r = η p r (x, u) w 1 = ζ 1 (x, u),..., w q = ζ q (x, u)

Invariant functions: w = η(y), i.e. ζ(x, u) = h[η(x, u)] Reduced equation: /G(y, w (n) ) = 0 Every solution determines a G-invariant solution to the original p.d.e.

Example. The heat equation Scaling symmetry: u t = u xx x x + 2 t t + a u u Invariants: y = x t, w = t a u u = t a w(y), u t = t a 1 ( 1 2 y w + a w ), u xx = t a w. Reduced equation w + 12yw aw = 0 Solution: w = e y2 /8 U(2a + 1 2, y/ 2 ) = parabolic cylinder function Similarity solution: u(x, t) = t a e x2 /8t U( 2 a + 1 2, x/ 2 t )

Example. The heat equation Galilean symmetry: 2 t x x u u u t = u xx Invariants: y = t w = e x2 /4t u ( u = e x2 /4t w(y), u t = e x2 /4t w + x2 4t w 2 u xx = e x2 /4t ( x 2 4t 2 1 ) w. 2 t Reduced equation: 2 y w + w = 0 Source solution: w = k y 1/2, u = k t e x2 /4t ),

Method 2: Direct substitution: Solve the combined system (x,u (n) ) = 0 Q k (x,u (1) ) = 0, k = 1,...,r as an overdetermined system of p.d.e. For a one-parameter group, we solve for Q(x,u (1) ) = 0 u α x p = ϕα p 1 ξ n i=1 ξ i u α ξ p x i Rewrite in terms of derivatives with respect to x 1,...,x p 1. The reduced equation has x p as a parameter. Dependence on x p can be found by by substituting back into the characteristic condition.

Classification of invariant solutions Let G be the full symmetry group of the system = 0. Let H G be a subgroup. If u = f(x) is an H-invariant solution, and g G is another group element, then f = g f is an invariant solution for the conjugate subgroup H = g H g 1. Classification of subgroups of G under conjugation. Classification of subalgebras of g under the adjoint action. Exploit symmetry of the reduced equation

Non-Classical Method = Bluman and Cole Here we require not invariance of the original partial differential equation, but rather invariance of the combined system (x,u (n) ) = 0 Q k (x,u (1) ) = 0, k = 1,..., r Nonlinear determining equations. Most solutions derived using this approach come from ordinary group invariance anyway.

Weak (Partial) Symmetry Groups Here we require invariance of (x,u (n) ) = 0 Q k (x,u (1) ) = 0, k = 1,..., r and all the associated integrability conditions Every group is a weak symmetry group. Every solution can be derived in this way. Compatibility of the combined system? Overdetermined systems of partial differential equations.

The Boussinesq Equation u tt + 1 2 (u2 ) xx + u xxxx = 0 Classical symmetry group: v 1 = x v 2 = t v 3 = x x + 2t t 2u u For the scaling group Q = xu x + 2tu t + 2u = 0 Invariants: y = x w = tu u = 1 ( ) x t t t w Reduced equation: w + 1 2 (w2 ) + 1 4 y2 w + 7 4 y w + 2w = 0

u tt + 1 2 (u2 ) xx + u xxxx = 0 Group classification: v 1 = x v 2 = t v 3 = x x + 2t t 2u u Note: Ad(εv 3 )v 1 = e ε v 1 Ad(εv 3 )v 2 = e 2 ε v 2 Ad(δ v 1 + εv 2 )v 3 = v 3 δ v 1 εv 2 so the one-dimensional subalgebras are classified by: {v 3 } {v 1 } {v 2 } {v 1 + v 2 } {v 1 v 2 } and we only need to determine solutions invariant under these particular subgroups to find the most general groupinvariant solution.

Non-classical: Galilean group u tt + 1 2 (u2 ) xx + u xxxx = 0 v = t x + t 2t u Not a symmetry, but the combined system u tt + 1 2 (u2 ) xx + u xxxx = 0 Q = tu x + u t + 2t = 0 does admit v as a symmetry. Invariants: y = x 1 2 t2, w = u + t 2, u(x,t) = w(y) t 2 Reduced equation: w + ww + (w ) 2 w + 2 = 0

u tt + 1 2 (u2 ) xx + u xxxx = 0 Weak Symmetry: Scaling group: x x + t t Not a symmetry of the combined system u tt + 1 2 (u2 ) xx + u xxxx = 0 Q = xu x + t u t = 0 Invariants: y = x u Invariant solution: u(x, t) = w(y) t The Boussinesq equation reduces to t 4 w + t 2 [(w + 1 y)w + (w ) 2 y w ] = 0 so we obtain an overdetermined system w = 0 (w + 1 y)w + (w ) 2 y w = 0 Solutions: w(y) = 2 3 y2 1, or w = constant Similarity solution: u(x,t) = 2 x2 3t 2 1

Symmetries and Conservation Laws

Variational problems L[u] = L(x, Ω u(n) ) dx Euler-Lagrange equations = E(L) = 0 Euler operator (variational derivative) E α (L) = δl δu α = ( D) J L J u α J Theorem. (Null Lagrangians) E(L) 0 if and only if L = Div P

Theorem. The system = 0 is the Euler-Lagrange equations for some variational problem if and only if the Fréchet derivative D is self-adjoint: D = D. = Helmholtz conditions

Fréchet derivative Given P(x, u (n) ), its Fréchet derivative or formal linearization is the differential operator D P defined by D P [w] = d dε P[u + εw] ε = 0 Example. P = u xxx + uu x D P = D 3 x + ud x + u x

Adjoint (formal) D = J A J D J D = J ( D) J A J Integration by parts formula: where A depends on P, Q. P DQ = Q D P + Div A

Conservation Laws Definition. A conservation law of a system of partial differential equations is a divergence expression Div P = 0 which vanishes on all solutions to the system. P = (P 1 (x, u (k) ),...,P p (x, u (k) )) = The integral P ds is path (surface) independent.

If one of the coordinates is time, a conservation law takes the form D t T + Div X = 0 T conserved density X flux By the divergence theorem, b b T(x, t, Ω u(k) ) dx) = X ds dt t=a a Ω depends only on the boundary behavior of the solution. If the flux X vanishes on Ω, then Ω T dx is conserved (constant).

Trivial Conservation Laws Type I If P = 0 for all solutions to = 0, then Div P = 0 on solutions too Type II (Null divergences) If Div P = 0 for all functions u = f(x), then it trivially vanishes on solutions. Examples: D x (u y ) + D y ( u x ) 0 (u, v) D x (y, z) + D (u, v) y (z, x) + D (u, v) z (x, y) 0

Theorem. Div P(x, u (k) ) 0 for all u if and only if P = Curl Q(x, u (k) ) i.e. P i = p j=1 D j Q ij Q ij = Q ji

Two conservation laws P and P are equivalent if they differ by a sum of trivial conservation laws: where P = P + P I + P II P I = 0 on solutions Div P II 0.

Proposition. Every conservation law of a system of partial differential equations is equivalent to a conservation law in characteristic form Div P = Q = ν Q ν ν Proof : Div P = ν,j Q J ν DJ ν Integrate by parts: Div P = ν,j ( D) J Q J ν ν Q ν = J ( D) J Q J ν Q is called the characteristic of the conservation law.

Theorem. Q is the characteristic of a conservation law for = 0 if and only if Proof : D Q + D Q = 0. 0 = E(Div P) = E(Q ) = D Q + D Q

Normal Systems A characteristic is trivial if it vanishes on solutions. Two characteristics are equivalent if they differ by a trivial one. Theorem. Let = 0 be a normal system of partial differential equations. Then there is a one-toone correspondence between (equivalence classes of) nontrivial conservation laws and (equivalence classes of) nontrivial characteristics.

Variational Symmetries Definition. A (restricted) variational symmetry is a transformation ( x,ũ) = g (x, u) which leaves the variational problem invariant: Ω L( x, ũ (n) ) d x = Infinitesimal criterion: Ω L(x, u(n) ) dx pr v(l) + L Div ξ = 0 Theorem. If v is a variational symmetry, then it is a symmetry of the Euler-Lagrange equations. But not conversely!

Noether s Theorem (Weak version). If v generates a one-parameter group of variational symmetries of a variational problem, then the characteristic Q of v is the characteristic of a conservation law of the Euler- Lagrange equations: Div P = Q E(L)

Elastostatics W(x, u) dx stored energy x, u R p, p = 2, 3 Frame indifference u R u + a, R SO(p) Conservation laws = path independent integrals: Div P = 0.

1. Translation invariance 2. Rotational invariance P i = u α i P i = W u α i = Euler-Lagrange equations W u β j u β i W u α j 3. Homogeneity : W = W( u) x x + a P i = p u α W j u α δj i W i = Energy-momentum tensor α=1

4. Isotropy : W( u Q) = W( u) Q SO(p) P i = p α=1 (x j u α k xk u α j ) W u α i + (δ i j xk δ i k xj )W 5. Dilation invariance : W(λ u) = λ n W( u) P i = n p n p α,j=1 5A. Divergence identity p (u α δ i j xj u α j ) W u α i Div P = p W + x i W P i = (u α δj i xj u α j ) W j=1 u α + x i W i = Knops/Stuart, Pohozaev, Pucci/Serrin

Generalized Vector Fields Allow the coefficients of the infinitesimal generator to depend on derivatives of u: v = p i=1 Characteristic : Evolutionary vector field: ξ i (x, u (k) ) x i + q α=1 Q α (x, u (k) ) = ϕ α v Q = q α=1 ϕ α (x, u (k) ) p i=1 Q α (x, u (k) ) ξ i u α i u α u α

Prolongation formula: pr v = pr v Q + pr v Q = D J Q α α,j u α J p i=1 ξ i D i D i = u α J,i α,j u α J = total derivative

Generalized Flows The one-parameter group generated by an evolutionary vector field is found by solving the Cauchy problem for an associated system of evolution equations u α ε = Q α (x, u(n) ) u ε=0 = f(x)

Example. v = generates the one-parameter x group of translations: Evolutionary form: (x, y, u) (x + ε, y, u) v Q = u x x Corresponding group: u ε = u x Solution u = f(x, y) u = f(x ε, y)

Generalized Symmetries of Differential Equations Determining equations : pr v( ) = 0 whenever = 0 For totally nondegenerate systems, this is equivalent to pr v( ) = D = ν D ν ν v is a generalized symmetry if and only if its evolutionary form v Q is. A generalized symmetry is trivial if its characteristic vanishes on solutions to. Two symmetries are equivalent if their evolutionary forms differ by a trivial symmetry.

General Variational Symmetries Definition. A generalized vector field is a variational symmetry if it leaves the variational problem invariant up to a divergence: pr v(l) + L Div ξ = Div B v is a variational symmetry if and only if its evolutionary form v Q is. pr v Q (L) = Div B

Theorem. If v is a variational symmetry, then it is a symmetry of the Euler-Lagrange equations. Proof : First, v Q is a variational symmetry if pr v Q (L) = Div P. Secondly, integration by parts shows pr v Q (L) = D L (Q) = QD L (1)+Div A = QE(L)+Div A for some A depending on Q, L. Therefore 0 = E(pr v Q (L)) = E(QE(L)) = E(Q ) = D Q + D Q = D Q + D Q = pr v Q ( ) + D Q

Noether s Theorem. Let = 0 be a normal system of Euler-Lagrange equations. Then there is a one-toone correspondence between (equivalence classes of) nontrivial conservation laws and (equivalence classes of) nontrivial variational symmetries. The characteristic of the conservation law is the characteristic of the associated symmetry. Proof : Nother s Identity: QE(L) = pr v Q (L) Div A = Div(P A)

The Kepler Problem x + µx r 3 = 0 L = 1 2 Generalized symmetries: Conservation law where x 2 µr v = (x x) x + x(x x ) 2 x( x x ) pr v(l) = D t R R = x (x x) µx r = Runge-Lenz vector

Noether s Second Theorem. A system of Euler- Lagrange equations is under-determined if and only if it admits an infinite dimensional variational symmetry group depending on an arbitrary function. The associated conservation laws are trivial. Proof : If f(x) is any function, Set f(x)d( ) = D (f) + Div P[f, ]. Q = D (f).

Example. (u x + v y ) 2 dx dy Euler-Lagrange equations: 1 = E u (L) = u xx + v xy = 0 2 = E v (L) = u xy + v yy = 0 Symmetries D x 2 D y 2 0 (u, v) (u + ϕ y, v ϕ x )