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Cm Math Variations

1 · Prerequisites

2 · Summary

The variational layer of the non-relativistic classical mechanics collection, all of it mathematics: the time-dependent Lagrangian chart with its action and fixed-endpoint first variation, the equivalence of stationarity with the Euler–Lagrange equation, the Legendre map with its regularity and hyperregularity hypotheses, the Hamiltonian identities on an inverse branch, the point-symmetry Noether identity with its conserved charges, the covariance of the Euler–Lagrange expression under chart changes, the reduced d'Alembert equation for an immersed system, the distinction between non-holonomic d'Alembert and vakonomic evolutions, regular holonomic constraints with ideal reactions, and the phase-action and endpoint-action forms of stationarity. Every item carries domain: mathematics and depends only on mathematical suppliers; nothing here asserts that any model describes nature, and the physical reading of forces, masses and empirical apparatus remains with the consuming classical-mechanics pages.

The core of the page is the variational chart. A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations fixes the data (O,L), the admissible curves and variations, the action S and the Euler–Lagrange expression EL; The first variation of a time-dependent action on a finite interval: admissibility, the integral formula and continuity of the Euler-Lagrange coefficient proves that the first variation exists and equals ∫(Lq⋅η+Lv⋅η′)dt, and Stationarity of the time-dependent action is equivalent to the vanishing of the Euler-Lagrange expression proves the equivalence of stationarity with EL(q)=0, using the explicit C2 bumps of Explicit twice differentiable bumps detect nonzero values of continuous functions. Scope of stationarity: stationary is not minimum, and no second-variation theory is developed here separates stationarity from minimality and records the exclusion of Sobolev and broken-extremal classes.

The Legendre layer is The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity with Nonsingular velocity Hessian is equivalent to invertibility of the Legendre derivative, hence gives a local C^1 inverse (a C1 local inverse branch by the inverse function theorem), the mechanical positive example A mechanical chart Lagrangian with positive definite kinetic matrix is hyperregular, with the explicit kinetic-plus-potential Hamiltonian, the branch identities and equivalence of The Hamiltonian on an inverse Legendre branch: differential, equivalence with the Euler-Lagrange equation, and the energy identity, and the scope remarks Scope of the Legendre map: regular versus hyperregular, the meaning of the covector datum, and the limits of the branch description. The symmetry layer is Infinitesimal point symmetry of a time-dependent Lagrangian and its conserved charge, The charge of an infinitesimal point symmetry is constant along Euler-Lagrange solutions, Adding a total time derivative and rescaling a Lagrangian: the Euler-Lagrange expression and the stationary curves, the general reparametrisation identity Noether identity for point transformations with time reparametrisation: the conserved charge p⋅ξ−Eτ−B, the charges Conserved charges for translation, rotation and boost symmetries of a mechanical Lagrangian and the scope note Scope of point symmetries: symmetries belong to the Lagrangian, not to the configuration space, and time translation is the energy case.

The constraint layer separates the mathematical laws cleanly: Coordinate covariance of the Euler-Lagrange expression under time-dependent point transformations and Reduced d'Alembert equation for an immersion: generalized forces and the quadratic correction give the chart and reduced forms; The velocity constraint z′=xy′: virtual displacements, the d'Alembert law and the vakonomic law, The distribution span{∂x,∂y+x∂z} is not tangent to any regular level sets and The d'Alembert and vakonomic motions of z′=xy′ differ from the same initial data exhibit a non-integrable velocity constraint for which the d'Alembert and vakonomic evolutions differ; A regular holonomic constraint, its virtual displacements and the ideal-reaction law, Reactions annihilating the virtual displacements are transpose multiples, with a positive definite multiplier matrix and The ideal holonomic system in a possibly time-dependent chart: explicit acceleration, constraint preservation, local existence and reaction power develop the regular holonomic theory with multipliers, existence, constraint preservation and the rheonomic reaction power. The page closes with Stationarity of the phase-space action is equivalent to Hamilton's equations, and hyperregular elimination recovers the Lagrangian action, The endpoint action function of a smooth family of extremals has differential dW=p dq−H dt and A smooth local family of extremals with prescribed endpoint exists when the endpoint momentum map has invertible derivative, the phase-action and endpoint-action forms of stationarity on a hyperregular chart.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passOpen item page →

The d'Alembert and vakonomic motions of z′=xy′ differ from the same initial data

Statement refuted

Every d'Alembert motion of the constraint z′=xy′ with unit mass and zero applied force that satisfies the initial data

(x,y,z)(0)=(0,0,0),(x′,y′,z′)(0)=(1,1,0)

is a vakonomic motion of the same constraint with the same initial data.

Facts & Assumptions

Given: The definitions and notation of The velocity constraint z′=xy′: virtual displacements, the d'Alembert law and the vakonomic law: the distribution D(q)={δ:δz=xδy}, the d'Alembert condition q′′(t)⋅δ=0 for all δ∈D(q(t)) together with z′=xy′, the augmented functional Φ(q,η)=∫ab[12∣q′∣2+η(z′−xy′)]dt, and the vakonomic stationary pairs and motions defined by the vanishing of the directional derivative of Φ for every admissible pair variation.

[F2]

The first-order system with right-hand side F(x,y,z,X,Y,Z)=(X,Y,Z, 0, −xXY/(1+x2), XY/(1+x2)) is of class C1 on R6, so Picard–Lindelöf gives a unique local solution through every initial point, and the maximal-solution theorem gives the unique maximal solution (Picard-Lindelöf local existence and uniqueness for first-order systems, Every Picard–Lindelöf initial value problem has one maximal solution on an open interval, First-order systems, initial value problems, and solutions on intervals, Ck Euclidean maps and diffeomorphisms).

[F3]

Calculus: the chain rule and the power rule give ddt(1+t2)−1/2=−t(1+t2)−3/2 and ddt[t(1+t2)−1/2]=(1+t2)−3/2 for all real t (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a), Continuity and derivatives of positive-base real powers, Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0, The derivative f′(c)=lim⁡x→cf(x)−f(c)x−c of f:A→R at a point c∈A that is a limit point of A, and differentiability on a set, Absolute value in an ordered field); a differentiable function on an interval with vanishing derivative is constant (A function continuous on an interval I whose derivative vanishes at every interior point of I is constant on I; consequently two such functions with the same derivative differ by a constant), and the Euclidean inner product is that of The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, Intervals of R: the nine order-convex forms, nondegeneracy, and length).

Counterexample

technique · direct; solve the d'Alembert system explicitly for the displayed data and show that no multiplier can make the same curve stationary for the augmented functional
1.1givenF1algebra

(the d'Alembert system.) Let q be a d'Alembert motion on an open interval containing 0. Since q′′ is orthogonal to δ=(1,0,0) and to δ=(0,1,x), we get x′′=0 and y′′+xz′′=0. Put λ:=z′′, a continuous function; then y′′=−λx and z′′=λ. Differentiating the constraint z′=xy′ gives z′′=x′y′+xy′′, so λ=x′y′+x(−λx)=x′y′−λx2, hence λ=x′y′1+x2,x′′=0,y′′=−xx′y′1+x2,z′′=x′y′1+x2.

1.2F1givenalgebra

(the vakonomic equations.) Suppose q is a vakonomic motion with the same data: there are a<b in the motion's domain and a C2 function η on an open interval containing [a,b] such that (q,η) is a vakonomic stationary pair, i.e. the directional derivative of Φ vanishes for every admissible pair variation. Computing that derivative for a variation (ξ,ζ) with ξ(a)=ξ(b)=0 and integrating by parts in the terms containing ξ′ and ζ gives 0=∫ab[(−x′′−ηy′)ξx+(−y′′+(ηx)′)ξy+(−z′′−η′)ξz]dt+∫abζ (z′−xy′) dt. Taking first ζ=0 and ξ running through the bumps of [F1] in each component yields x′′=−ηy′, y′′=(ηx)′, z′′=−η′ on [a,b]; then taking ξ=0 and ζ a bump gives z′=xy′.

2.1F2step 1.1

(existence and uniqueness.) The system of step 1.1 is the first-order system of [F2]. Hence it has a unique maximal solution through the data (x,y,z,X,Y,Z)(0)=(0,0,0,1,1,0); every d'Alembert motion with these data solves it, so all such motions are restrictions of that unique maximal solution.

3.1F2F3step 2.1algebra

(the explicit solution.) Define x(t):=t, y(t):=∫0t(1+s2)−1/2 ds and z(t):=∫0ts(1+s2)−1/2 ds, using oriented integrals (The integral with oriented limits: ∫aaf:=0 and ∫baf:=−∫abf). The integrands are C1 on R, so these integrals exist on every bounded interval; additivity (For a<c<b: f is integrable on [a,b] if and only if it is integrable on [a,c] and on [c,b], and then ∫abf=∫acf+∫cbf; with the oriented form for arbitrary a,b,c) and The first fundamental theorem: if f is integrable on [a,b] and continuous at c, then F′(c)=f(c); in particular a continuous f has F as a primitive give y′(t)=(1+t2)−1/2 and z′(t)=t(1+t2)−1/2 for all t∈R, including t=0. Thus y,z are C2 and y(0)=z(0)=0. By [F3], y′′(t)=−t(1+t2)−3/2=−x(t)x′(t)y′(t)1+x(t)2,z′′(t)=(1+t2)−3/2=x′(t)y′(t)1+x(t)2, and x′′=0, while x(t)y′(t)=t(1+t2)−1/2=z′(t); the data at 0 are (0,0,0,1,1,0). Therefore this curve is a solution of the system defined on all of R, so by step 2.1 it is the unique maximal d'Alembert motion with the displayed data. Note y′(t)>0 and y′(t)<1 for t>0.

4.1F3step 3.1step 1.2

(contradiction.) For the explicit motion of step 3.1 we have x′′=0 and y′>0 on [a,b], so the first equation of step 1.2 gives η=0 on [a,b]. Then y′′=(ηx)′=η′x+ηx′=0 on [a,b], on its interior. This is impossible: every nondegenerate interval (a,b) contains a nonzero point t, where the explicit formula gives y′′(t)=−t(1+t2)−3/2≠0. Hence no such pair (q,η) exists, and the d'Alembert motion of step 3.1 is not a vakonomic motion.

5.1step 3.1step 4.1∎

The displayed data therefore give a d'Alembert motion which is not a vakonomic motion with the same initial data, refuting the statement.

Remarks

  • The counterexample separates the two prescriptions as mathematical definitions; it does not by itself say which of them, if either, models a given physical system. The empirical question is the subject of the consuming physical items and requires its own apparatus assumptions and data.

  • The specific numbers were chosen for an elementary closed-form solution; the general statement that the two dynamics need not coincide is the content of the cited treatments of vakonomic and non-holonomic mechanics, and the computation here is self-contained.

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Conserved charges for translation, rotation and boost symmetries of a mechanical Lagrangian

Statement

Let n≥1, let m>0, let a0<b0 be reals, let ε>0, let U⊆Rn be open and nonempty, let U0:(a0−ε,b0+ε)×U→R be of class C2, and let

L(t,q,v):=12m∣v∣2−U0(t,q)

be the corresponding Lagrangian chart (A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations, The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn). Let q:I→U be a C2 Euler–Lagrange solution on an open interval [a0,b0]⊆I⊆(a0−ε,b0+ε). Then:

  1. (translation) if a∈Rn and a⋅∇qU0(t,q)=0 for every (t,q), then Pa(t):=m a⋅q′(t) is constant on I;
  2. (rotation) if n=3, a∈R3 and (a×q)⋅∇U0(t,q)=0 for every (t,q), then Ja(t):=m (q(t)×q′(t))⋅a is constant on I;
  3. (boost) if a∈Rn and a⋅∇qU0=0 on the whole chart, then Ka(t):=a⋅(t m q′(t)−m q(t)) is constant on I.

Facts & Assumptions

Given: The real m>0, the C2 potential U0, the Lagrangian L(t,q,v)=12m∣v∣2−U0(t,q), a vector a (in Rn in clauses 1 and 3, in R3 in clause 2), and a C2 Euler–Lagrange solution q:I→U.

[F1]

The Noether theorem (Noether identity for point transformations with time reparametrisation: the conserved charge p⋅ξ−Eτ−B): if C2 data τ,ξ,B satisfy the Noether identity of that theorem, then C=Lv⋅ξ−Eτ−B is constant along q. Here Lv=m v and E=12m∣v∣2+U0. [given, algebraic computation of Lv and E from the definition]

Proof

technique · direct; verify the Noether identity for each of the three families and read off the charge
1.1F1F3given

Translation. Take τ=0, ξ(t,q):=a (constant) and B=0. Then Dtξ=0 and the Noether identity reduces to Lq⋅a=−a⋅∇U0=0, which holds by hypothesis of clause 1. The charge is C=Lv⋅a−E⋅0−0=m a⋅q′(t)=Pa(t), constant on I by [F1].

1.2F1F2given

Rotation. Take n=3, τ=0, ξ(t,q):=a×q and B=0. Then Dtξ=a×q′, so the identity reads Lq⋅(a×q)+Lv⋅(a×q′)=−(a×q)⋅∇U0+m q′⋅(a×q′). The first term vanishes by the hypothesis of clause 2 and the second by [F2] (the scalar triple product with two equal factors q′ vanishes). Hence the identity holds and the charge is C=m q′⋅(a×q)=m (q×q′)⋅a=Ja(t), constant on I by [F1].

1.3F1F3givenalgebra

Boost. Take τ=0, ξ(t,q):=t a and B(t,q):=m a⋅q. Then Dtξ=a and DtB=m a⋅q′, so the identity reads Lq⋅ta+Lv⋅a=−t a⋅∇U0+m a⋅q′, which equals m a⋅q′=DtB by the hypothesis a⋅∇U0=0. Hence the identity holds with charge C=m q′(t)⋅(ta)−m a⋅q(t)=a⋅(t m q′(t)−m q(t))=Ka(t), constant on I by [F1].

2.1F3given∎

Failure for a quadratic potential. For U0(q)=k2∣q∣2 with k>0 and a≠0, the hypotheses of clauses 1 and 3 fail, since a⋅∇U0=k a⋅q is not identically zero; so the translation and boost identities are not available, and the corollary gives no conserved Pa or Ka for this potential.

Remarks

  • Clause 3 is the Galilean boost charge: its defining Noether identity holds when a⋅∇qU0=0 on the chart, giving conservation along every Euler–Lagrange solution. Constancy on a particular solution alone does not imply this whole-chart symmetry. A confining potential such as k2∣q∣2 breaks the boost symmetry even though the Galilean charts of the framework still exist; this is the mathematical content of the warning that a kinematic symmetry need not be a symmetry of the chosen Lagrangian.

  • All three statements are conditional deductions inside the chart model: they identify conserved quantities of the Euler–Lagrange equation under explicitly stated symmetry hypotheses of the potential. They are mathematical statements about the model and are not claims that any particular physical system possesses the required symmetry.

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Reduced d'Alembert equation for an immersion: generalized forces and the quadratic correction

Statement

Let n,m≥1, let Ω⊆Rn be open and nonempty, and let X:Ω→Rm be of class C3 with DX(q) injective for every q∈Ω; fix an immersion in the sense of Submersions and immersions between Euclidean open sets (this forces n≤m). Let M∈Mm(R) be symmetric positive definite, meaning MT=M and uTMu>0 for every 0≠u∈Rm, let D⊆R×Rm×Rm be open and let F:D→Rm be continuous, let I⊆R be an open interval (Intervals of R: the nine order-convex forms, nondegeneracy, and length), and let q:I→Ω be a C2 curve with x(t):=X(q(t)) and (t,x(t),x′(t))∈D for all t∈I. Suppose there is a continuous R:I→Rm with

Mx′′(t)=F(t,x(t),x′(t))+R(t),R(t)⋅DX(q(t))δ=0for all δ∈Rn, t∈I.

Then, writing DX(q(t))T for the transpose and (Xqq[q′,q′])k:=∑i,j∂i∂jXk(q)qi′qj′, one has

DX(q)TMDX(q) q′′=DX(q)TF(t,x,x′)−DX(q)TM Xqq[q′,q′]

on I. The vector DX(q)TF(t,x,x′) is the generalized force of the ambient force F in the coordinates q, and DXTMXqq[q′,q′] is the acceleration correction of the parametrisation; when X is affine, the correction vanishes.

Facts & Assumptions

Given: The open set Ω, the immersion X, the constant matrix M, the field F, the curve q with x=X∘q, and the reaction R satisfying the two displayed relations.

[F1]

Differentiating x=X∘q twice by the chain rule (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a), Coordinate covariance of the Euler-Lagrange expression under time-dependent point transformations, Ck Euclidean maps and diffeomorphisms, Ck maps and multi-index derivative notation in Euclidean space): x′=DX(q)q′ and x′′=ddt(DX(q))q′+DX(q)q′′=(Xqq[q′,q′])+DX(q)q′′.

[F2]

R(t) annihilates the tangent space of the immersion: R(t)⋅DX(q(t))δ=0 for all δ; equivalently DX(q(t))TR(t)=0, because (DXTR)⋅δ=R⋅DXδ for every δ (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes, Linear map between vector spaces over the same field).

Proof

technique · direct; pull the ambient equation back by the transpose of the derivative
1.1F2F3given

Multiplying the ambient equation Mx′′=F+R on the left by DX(q(t))T gives DXTMx′′=DXTF+DXTR=DXTF, because DXTR=0 by [F2].

2.1F1step 1.1algebra∎

Substituting the second-derivative chain rule of [F1] into step 1.1 gives DXTM(Xqq[q′,q′]+DX q′′)=DXTF. By associativity of matrix products this is DXTMDX q′′=DXTF−DXTMXqq[q′,q′], which is the displayed equation.

Remarks

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A regular holonomic constraint, its virtual displacements and the ideal-reaction law

Definition

Let n≥1 and k≥1, let O⊆R×Rn be open and nonempty, let G:O→Rk be of class C2 (Ck Euclidean maps and diffeomorphisms), and write Gq(t,q) for the Jacobian matrix of q↦G(t,q) (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case), a k×n matrix. The constraint

G(t,q)=0

is regular when

Gq(t,q) has rank k for every (t,q)∈O,

equivalently when Gq is surjective at every point (Submersions and immersions between Euclidean open sets). Under this hypothesis each fibre G(t,⋅)−1(0) is a regular level set and every point of it is a regular point in the sense of Regular and critical points, regular and critical values, and level sets.

Constraint set and admissible data. Fix a symmetric positive definite matrix M∈Mn(R), meaning MT=M and uTMu>0 for every 0≠u∈Rn, and a continuous map f:O×Rn→Rn, the given force. A map q:I→Rn on an open interval I⊆R (Intervals of R: the nine order-convex forms, nondegeneracy, and length) is a constrained curve when

(t,q(t))∈O,G(t,q(t))=0(t∈I).

Virtual displacements and permitted velocities. For (t,q)∈O the virtual displacements at (t,q) are the elements of the kernel

V(t,q):=ker⁡Gq(t,q)={δ∈Rn:Gq(t,q)δ=0}⊆Rn,

the tangent space to the level set of G(t,⋅) at q in the sense of The tangent space to a regular level set; a permitted velocity at (t,q) is a vector v∈Rn with

Gt(t,q)+Gq(t,q)v=0.

For a C1 constrained curve the differentiated constraint forces Gt(t,q(t))+Gq(t,q(t))q′(t)=0, so its velocity is automatically permitted; the permitted condition is stated separately because it is the correct initial condition for the first-order system below.

Ideal-constraint motions. A C2 constrained curve q:I→Rn is an ideal-constraint motion of the data (G,M,f) when

  1. q′(t) is a permitted velocity at (t,q(t)) for every t∈I, and
  2. there is a continuous map λ:I→Rk with

Mq′′(t)=f(t,q(t),q′(t))+Gq(t,q(t))Tλ(t)(t∈I).

The vector field R(t):=Mq′′(t)−f(t,q(t),q′(t)) is the reaction and R=GqTλ is its multiplier form. Since Gq(t,q)Tλ⋅δ=λ⋅Gq(t,q)δ=0 for every δ∈V(t,q), every ideal-constraint motion satisfies the d'Alembert condition

(Mq′′(t)−f(t,q(t),q′(t)))⋅δ=0for every δ∈V(t,q(t)),

i.e. the reaction performs no virtual work; the converse implication, that every curve satisfying the d'Alembert condition is an ideal-constraint motion with a unique multiplier for its given reaction, is the content of Reactions annihilating the virtual displacements are transpose multiples, with a positive definite multiplier matrix and is not assumed here. Determination of the reaction and acceleration additionally uses the twice-differentiated constraint, as in The ideal holonomic system in a possibly time-dependent chart: explicit acceleration, constraint preservation, local existence and reaction power under its stated regularity hypotheses.

Scope. The definition is chartwise and time-dependent: the constraint may depend on t (rheonomic case), its velocity condition is the fixed-time linearisation recorded above, and no relation between the variational (vakonomic) treatment of the constraint and the present d'Alembert law is asserted. The construction and long-time behaviour of solutions is the subject of The ideal holonomic system in a possibly time-dependent chart: explicit acceleration, constraint preservation, local existence and reaction power. For a holonomic constraint written in a chart, the multiplier λ is not a measured quantity: only the reaction GqTλ enters the equations of motion.

Remarks

  • The definition covers the classical examples: a planar pendulum is n=2, k=1, G(q)=∣q∣2−l2 with l>0 on the chart {(x,y):y≠0} or on R2∖{0}, where Gq=2qT≠0; a rolling disk in the plane with configuration (x,θ) is G(q)=x−aθ, so that in that planar special case the constraint is holonomic, while general three-dimensional rolling is not reducible to a single level set. These examples are worked out on the companion page.

  • The word ideal refers to the d'Alembert condition and not to any assertion that a real contact or joint enforces it; frictionless joints, no-slip rolling with sufficient static friction, and similar mechanisms are physical idealisations modelled by this law, and the empirical qualifications of such models belong to the consuming physics items, not here.

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Infinitesimal point symmetry of a time-dependent Lagrangian and its conserved charge

Definition

Let (O,L) be a time-dependent Lagrangian chart over [a,b]×U in the sense of A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations. Let V⊆Rn and A⊆R×V be open sets containing respectively the configuration projection of O and its time-configuration projection. Let

ξ:V⟶Rn,B:A⟶R

be of class C1 (Ck Euclidean maps and diffeomorphisms); write Dξ(q)∈Mn(R) for the Jacobian matrix of ξ at q (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, Directional derivatives and partial derivatives of a map U⊆Rm→Rn), and for a C1 function φ(t,q) write

Dtφ(t,q,v):=φt(t,q)+φq(t,q)⋅v.

The pair (ξ,B) is an infinitesimal point symmetry of L when the identity

Lq(t,q,v)⋅ξ(q)+Lv(t,q,v)⋅Dξ(q)v=Bt(t,q)+Bq(t,q)⋅v

holds for every (t,q,v)∈O. The charge of the pair is the function

C(t,q,v):=Lv(t,q,v)⋅ξ(q)−B(t,q).

Reading of the identity. The left-hand side is the derivative of the Lagrangian under the infinitesimal family of point transformations

q⟼q+s ξ(q),t⟼t

with the velocity transformed accordingly by v↦v+s Dξ(q)v, and the right-hand side is the total derivative ddtB(t,q(t)) along a C1 trajectory with q′(t)=v; thus the identity says that the infinitesimal variation of L is a total time derivative of the chart function B. In particular:

  • Taking B=0 reduces the identity to Lq(t,q,v)⋅ξ(q)+Lv(t,q,v)⋅Dξ(q)v=0, the invariance of L under the infinitesimal family.
  • For a constant field ξ≡a one has Dξ=0, and the identity reads Lq⋅a=Bt+Bq⋅v; when B is independent of q this is the statement that L is unchanged by the translation q↦q+sa up to a total time derivative.

Scope. The definition is an identity imposed on the chart data; it does not assert that a given Lagrangian possesses any symmetry, nor that a symmetry of the configuration space, a frame change or any kinematic structure automatically induces one. Whether a symmetry produces a conserved quantity along the Euler–Lagrange solutions is the theorem The charge of an infinitesimal point symmetry is constant along Euler-Lagrange solutions; the general reparametrisation version is Noether identity for point transformations with time reparametrisation: the conserved charge p⋅ξ−Eτ−B, and the scope of the statement for physical models is recorded in Scope of point symmetries: symmetries belong to the Lagrangian, not to the configuration space, and time translation is the energy case.

Remarks

  • The notation Dtφ is the derivative of φ along the curve t↦(t,q(t)) when v=q′(t): for such a curve, Dtφ(t,q(t),q′(t))=ddtφ(t,q(t)). This is the reason the velocity slot of the symmetry identity is filled by v rather than by a fixed vector: the infinitesimal transformation moves q with velocity ξ(q), and along a motion the velocity changes by Dtξ−v Dtτ in the general case treated later.

  • The charge is a function on the chart O; a conserved quantity is the composite t↦C(t,q(t),q′(t)) along a solution. No claim is made that the charge is the only conserved quantity, or that it is invariant under changes of chart.

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The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity

Definition

Chart data. Let n≥1, let J⊆R and U⊆Rn be open and nonempty, and let

L:J×U×Rn⟶R

be of class C2 (Ck Euclidean maps and diffeomorphisms). Such a map is a fibre-full time-dependent Lagrangian chart; it is the special case of the chart of A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations in which the velocity block is the whole space Rn at every (t,q)∈J×U. Write Lv(t,q,v)∈Rn for the Euclidean gradient of L in its third slot and Lvv(t,q,v)∈Mn(R) for the matrix of second partial derivatives in the third slot (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, Directional derivatives and partial derivatives of a map U⊆Rm→Rn).

Legendre map. The Legendre map of L is the map

FL:J×U×Rn⟶J×U×Rn,FL(t,q,v):=(t,q,Lv(t,q,v)).

The fibre coordinate of the target is called the momentum and is written p:=Lv(t,q,v). Thus FL is the identity in the first two slots and the gradient in the velocity slot. Since L is C2, FL is C1 (Ck Euclidean maps and diffeomorphisms).

Regularity. Let (t,q,v)∈J×U×Rn. The Legendre map is regular at (t,q,v) when the velocity Hessian is nonsingular,

det⁡Lvv(t,q,v)≠0;

it is regular (on the chart) when it is regular at every point. At fixed (t,q) this is exactly the condition under which the map v↦(q,Lv(t,q,v)) is locally invertible near v, a statement made precise and proved in Nonsingular velocity Hessian is equivalent to invertibility of the Legendre derivative, hence gives a local C^1 inverse.

Hyperregularity. The Lagrangian L is hyperregular (on the chart) when for every (t,q)∈J×U the map

v⟼Lv(t,q,v)

is a bijection of Rn onto Rn and the inverse assignment

(t,q,p)⟼v(t,q,p),Lv(t,q,v(t,q,p))=p,

is defined on all of J×U×Rn and is of class C1. In that case v is called the inverse Legendre branch of L, and the Hamiltonian of the branch is

H:J×U×Rn⟶R,H(t,q,p):=p⋅v(t,q,p)−L(t,q,v(t,q,p)).

The Hamiltonian is C1 on its domain, being a sum of products of C1 maps.

The definition asserts nothing about solving the equations of motion: the equivalence of the Euler–Lagrange equation with Hamilton's equations on an inverse branch, and the differential of H, are proved in The Hamiltonian on an inverse Legendre branch: differential, equivalence with the Euler-Lagrange equation, and the energy identity.

Remarks

Scope and warnings. Hyperregularity is a global-in-the-fibre hypothesis: it requires the velocity Legendre map to be bijective onto the whole momentum fibre Rn, with globally defined inverse. Regularity is a local condition. Hyperregularity implies regularity: at fixed (t,q), differentiating v(t,q,Lv(t,q,w))=w gives Dpv Lvv=In, so the square velocity Hessian is invertible. The converse fails: a regular Lagrangian whose velocity gradient is not surjective is not hyperregular, and a fibre-full domain whose Legendre map fails to be injective has no inverse branch. Concrete witnesses — including the mechanical Lagrangian of A mechanical chart Lagrangian with positive definite kinetic matrix is hyperregular, with the explicit kinetic-plus-potential Hamiltonian, the bounded gradient L(v)=1+v2 and the identically zero Lagrangian — are collected in Regular versus hyperregular: the bounded-gradient Lagrangian 1+v2 and the degenerate zero Lagrangian ↗ and Scope of the Legendre map: regular versus hyperregular, the meaning of the covector datum, and the limits of the branch description.

  • The published bundle version of this notion — a smooth Lagrangian on a tangent bundle, its fibre derivative, and the definitions of regular and hyperregular Lagrangians by local diffeomorphism, respectively global fibre-preserving diffeomorphism, of the Legendre map — is the autonomous manifold formulation of the same concept. The present definition is the finite-dimensional chart version with explicit time dependence, which is the form consumed by the classical-mechanics items on this page; it is not a restatement of the bundle definition, because time-dependent Lagrangians and the explicit C1 inverse branch on J×U×Rn have no counterpart in the autonomous formulation.

  • When L is hyperregular, the fibre condition and the open domain J×U make H a globally defined C1 function. If only regularity holds, an inverse branch exists only locally, and the local Hamiltonian is correspondingly local; the items below state which hypothesis each conclusion uses.

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The velocity constraint z′=xy′: virtual displacements, the d'Alembert law and the vakonomic law

Definition

Work in R3 with coordinates (x,y,z) on configuration space and velocities (x′,y′,z′); inner products and norms are those of The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn. All intervals used as motion domains below are open intervals of R (Intervals of R: the nine order-convex forms, nondegeneracy, and length), and a map on such an interval is said to be Cr when all its components are (Ck Euclidean maps and diffeomorphisms).

The constraint and its virtual displacements. The constraint law is

z′=xy′.

A configuration q=(x,y,z) is written without a subscript, and its velocity is v=(x′,y′,z′). The allowed velocities at q and the virtual displacements at q are both the same linear subspace

D(q):={w=(wx,wy,wz)∈R3:wz=xwy}=span⁡{(1,0,0),(0,1,x)}⊆R3.

Thus a virtual displacement δ=(δx,δy,δz) obeys δz=x δy; this is the Chetaev prescription for the linear velocity constraint z′−xy′=0, in which the virtual displacements are computed from the constraint function frozen at the current configuration.

Sample Lagrangian and zero applied force. The adopted Lagrangian is

L0(q,v):=12∣v∣2=12(x′2+y′2+z′2),

of class C∞ on R3×R3; no applied force is introduced. Since L0 is a special case of a Lagrangian chart as in A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations (with n=3, U=R3 and O=R×R3×R3), the words stationary, action and Euler--Lagrange expression below have the meanings fixed there whenever they are used.

d'Alembert motions. A C2 curve q:I→R3 is a d'Alembert motion of the constraint z′=xy′ when

  1. z′(t)=x(t)y′(t) for every t∈I, and
  2. q′′(t)⋅δ=0 for every δ∈D(q(t)) and every t∈I.

Clause 2 is the ideal-reaction (d'Alembert) condition for unit mass and zero applied force: the acceleration has no component along any admissible virtual displacement, so the constraint force performs no virtual work.

The augmented functional and the vakonomic law. Let a<b be reals with [a,b]⊆I0 for some open interval I0. For a C2 curve q:I0→R3 and a C2 function η:I0→R put

Φ(q,η):=∫ab[12∣q′(t)∣2+η(t)(z′(t)−x(t)y′(t))]dt.

The integrand is continuous because q,q′ and η are continuous, so Φ(q,η) is a well-defined real number (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, The lower and upper Darboux integrals of a bounded f on [a,b] as sup⁡PL(f,P) and inf⁡PU(f,P), Darboux integrability as their equality, and the notation ∫abf, A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion). An admissible pair variation of a pair (q,η) is a pair of C2 maps (ξ,ζ) on I0 with ξ(a)=ξ(b)=0, and the pair (q,η) is a vakonomic stationary pair when the ordinary two-sided directional derivative

dds∣s=0Φ(q+sξ,η+sζ)

exists and equals 0 for every admissible pair variation (ξ,ζ) with ξ(a)=ξ(b)=0. A C2 curve q:I0→R3 is a vakonomic motion of the constraint when there is a C2 function η on I0 such that (q,η) is a vakonomic stationary pair on every [a,b]⊆I0 with a<b, using the same multiplier η throughout I0. The functional and its endpoint conditions are taken on each such subinterval.

Scope. Both laws are prescriptions for the same constraint, and the definition deliberately does not identify them: the d'Alembert law is a differential condition on the acceleration along the constraint distribution, while the vakonomic law arises from stationarity of the augmented functional over curves and multipliers. For this constraint the two prescriptions give different motions from the same initial data; that is the content of The d'Alembert and vakonomic motions of z′=xy′ differ from the same initial data. The constraint law z′=xy′ is not equivalent to any condition G(q)=const for a C2 function G with nonvanishing differential, as proved in The distribution span{∂x,∂y+x∂z} is not tangent to any regular level sets; this is the sense in which the constraint is non-holonomic.

Remarks

  • The two spanning fields of D are A:=∂x and B:=∂y+x∂z, in the sense that at each q the vectors A and B evaluated at q span D(q). The non-integrability computation of The distribution span{∂x,∂y+x∂z} is not tangent to any regular level sets uses exactly these two fields.

  • The stationary-pair definition quantifies over all C2 multipliers η and all C2 variations, with no condition on η at the endpoints; this is the standard augmented-Lagrangian formulation of the vakonomic equations, recorded here as a definition and not as a statement about which of the two laws describes any physical system.

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A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations

Definition

Chart data. Let n≥1, let a<b be reals and let U⊆Rn be open and nonempty. Let O⊆R×Rn×Rn be an open set with

[a,b]×U×Rn⊆O,

and let L:O→R be of class C2 (Ck Euclidean maps and diffeomorphisms). The pair (O,L) is a time-dependent Lagrangian chart over [a,b]×U. The three slots of L are written (t,q,v), so that t is the time, q the configuration and v the velocity; the open set O has the three coordinate blocks of sizes 1,n,n.

Admissible curves and variations. A map q:[a,b]→U is of class C2 when there are ε>0 and a C2 map q~:(a−ε,b+ε)→Rn with q~∣[a,b]=q and q~(t)∈U for every t∈[a,b]; the derivative q′ and the second derivative q′′ below are those of any such extension, and they are independent of the extension on [a,b]. A variation of a C2 curve q is a map η:[a,b]→Rn of class C2 in the same sense with

η(a)=η(b)=0.

Thus the admissible curves and variations are the restrictions to [a,b] of C2 maps on a common open interval. This closed-interval extension convention is stipulated here; on the open interval each component has continuous derivatives through order two in the sense of Ck maps and multi-index derivative notation in Euclidean space.

The action. For a C2 curve q:[a,b]→U the map t↦L(t,q(t),q′(t)) is continuous on [a,b], because q,q′ are continuous and L is continuous (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral); it is therefore Darboux integrable (The lower and upper Darboux integrals of a bounded f on [a,b] as sup⁡PL(f,P) and inf⁡PU(f,P), Darboux integrability as their equality, and the notation ∫abf). The action of q is the real number

S(q):=∫abL(t,q(t),q′(t)) dt.

First variation. For a C2 curve q and a variation η the curve qε:=q+εη takes values in U for all sufficiently small ∣ε∣; this is proved in The first variation of a time-dependent action on a finite interval: admissibility, the integral formula and continuity of the Euler-Lagrange coefficient ↗ and is not part of the present definition. The first variation of S at q in the direction η is the ordinary two-sided derivative at ε=0 of the function ε↦S(q+εη), written

δS(q)[η]:=ddε∣ε=0S(q+εη),

whenever this derivative exists; The first variation of a time-dependent action on a finite interval: admissibility, the integral formula and continuity of the Euler-Lagrange coefficient ↗ proves that it does exist under the present hypotheses and computes it, so the notation denotes a single real number for every pair (q,η) arising here.

Euler–Lagrange expression. Let Lq(t,q,v)∈Rn and Lv(t,q,v)∈Rn be the gradients of L in the second and third slots (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, Directional derivatives and partial derivatives of a map U⊆Rm→Rn). Since L is C2, the maps t↦Lv(t,q(t),q′(t)) are C1 and t↦Lq(t,q(t),q′(t)) is continuous on [a,b]. The Euler–Lagrange expression of L along q is the continuous map

EL(q)(t):=Lq(t,q(t),q′(t))−ddt(Lv(t,q(t),q′(t)))∈Rn.

Stationarity. A C2 curve q is stationary for S when

δS(q)[η]=0for every variation η.

Scope. The definition introduces only chart data, the action and its first variation. It asserts nothing about minimising S, nothing about the existence of stationary curves and nothing about any physical interpretation of L; the equivalence of stationarity with the vanishing of EL(q) is the theorem Stationarity of the time-dependent action is equivalent to the vanishing of the Euler-Lagrange expression, and the distinction between stationary and minimising curves is recorded in Scope of stationarity: stationary is not minimum, and no second-variation theory is developed here.

Remarks

  • The Lagrangian is allowed to depend on t, which is what makes the chart suitable for systems with prescribed time-dependent constraints, moving frames or external drifts; the chart formulation with the extension convention above is the one used by the consuming classical mechanics pages.

  • If U=Rn and L is the restriction of a C2 function on R×Rn×Rn, then O may be taken to be all of R1+2n and no extension discussion is needed. The general formulation keeps O open so that the Legendre map of The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity can be studied on the same domain.

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Explicit twice differentiable bumps detect nonzero values of continuous functions

Statement

For real c and r>0 define h:R→R by h(s):=10s3−15s4+6s5 and βc,r:R→R by

βc,r(t):=h(1−∣t−c∣r) for ∣t−c∣<r,βc,r(t):=0 for ∣t−c∣≥r.

Then:

  1. h(0)=0, h(1)=1, h′(0)=h′(1)=0, h′′(0)=h′′(1)=0, and 0≤h≤1 on [0,1] with h(s)>0 for 0<s<1;
  2. βc,r is of class C2 on R (Ck Euclidean maps and diffeomorphisms), 0≤βc,r≤1, βc,r(c)=1, βc,r(t)>0 for ∣t−c∣<r, and βc,r(t)=0 for ∣t−c∣≥r;
  3. (detection) let J⊆R be an open interval, let E:J→R be continuous (Continuity of f:A→R at a point of A and on A: the ε-δ condition, its agreement with lim⁡x→cf(x)=f(c) at a limit point, and continuity at an isolated point) and let t0∈J with E(t0)≠0. Then there is ρ>0 with [t0−ρ,t0+ρ]⊆J such that, with β:=βt0,ρ,

∫t0−ρt0+ρE(t)β(t) dt≠0.

Conventions. The displayed integral is the Darboux integral over the closed bounded interval [t0−ρ,t0+ρ] of the continuous function t↦E(t)β(t) (The lower and upper Darboux integrals of a bounded f on [a,b] as sup⁡PL(f,P) and inf⁡PU(f,P), Darboux integrability as their equality, and the notation ∫abf, A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion); note that β vanishes at both endpoints and outside the interval, so no extension convention is needed.

Facts & Assumptions

Given: The reals c,r, the polynomials and bumps displayed in the Statement, an open interval J, a continuous E:J→R and t0∈J with E(t0)≠0.

[F1]

h is a polynomial, hence differentiable on R; the product and sum rules (Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0) give h′(s)=30s2(1−s)2,h′′(s)=60s(1−s)(1−2s), so h(0)=0,h(1)=1,h′(0)=h′(1)=h′′(0)=h′′(1)=0, and h′≥0 on R; in particular h is nondecreasing with h(s)∈[0,1] for s∈[0,1] and h(s)>0 for 0<s<1.

[F3]

If E is continuous at t0 with E(t0)≠0, then for ε=∣E(t0)∣/2>0 there is δ>0 such that ∣E(t)−E(t0)∣<ε for ∣t−t0∣<δ; then E(t)>E(t0)/2>0 when E(t0)>0 and E(t)<E(t0)/2<0 when E(t0)<0 (Continuity of f:A→R at a point of A and on A: the ε-δ condition, its agreement with lim⁡x→cf(x)=f(c) at a limit point, and continuity at an isolated point, Absolute value in an ordered field). Since J is open there is ρ>0 with [t0−ρ,t0+ρ]⊆J; take ρ≤δ.

Proof

technique · direct; compute the polynomial and glue the pieces of the bump, then localise the sign of $E$ and integrate
1.1F1algebra

The derivative formulas of [F1] follow from the sum and product rules, and substituting s=0 and s=1 gives the vanishing values. Since h′≥0 on R, [F1] and the monotonicity criterion make h nondecreasing on R; hence 0=h(0)≤h(s)≤h(1)=1 for s∈[0,1], and h(s)>0 for 0<s<1 because h(s)=s3(10−15s+6s2) and 10−15s+6s2=6(s−5/4)2+5/8>0. This proves clause 1.

1.2F1F2algebra

Clause 2: on each of the three regions ∣t−c∣<r split at t=c, and ∣t−c∣≥r, the function βc,r is a polynomial in t composed with the affine map t↦1−(t−c)/r or t↦1+(t−c)/r, or is the constant 0; by [F2] and the chain rule its one-sided derivatives of order ≤2 at the gluing points c−r,c,c+r are obtained from h(0)=0, h(1)=1, h′(0)=h′(1)=0, h′′(0)=h′′(1)=0 of clause 1, and they agree from the two sides (at c the factor ∓1/r from the two affine branches multiplies h′(1)=0, and at c±r it multiplies h′(0)=0, so the first derivative is 0 from both sides; the second derivatives are h′′(0)/r2=0 and h′′(1)/r2=0). At each joining point u the polynomial values agree, so the glued function is continuous. Its two one-sided difference quotients tend to the common first derivative 0, hence its two-sided derivative exists at u. On either side this derivative is the polynomial derivative and has limit 0, so it is continuous there. Applying the same difference-quotient argument to this first derivative, whose one-sided derivatives both tend to 0, shows that the second derivative exists at u and equals 0. Its polynomial pieces tend to that same value, making it continuous. Hence βc,r is of class C2 on R; the bounds 0≤βc,r≤1, the value βc,r(c)=h(1)=1, the strict positivity on (c−r,c+r) and the support statement are read off from clause 1 and the definition.

2.1F3step 1.2

Clause 3: by [F3] choose ρ>0 with [t0−ρ,t0+ρ]⊆J and with E of the strict sign of E(t0) on (t0−ρ,t0+ρ). Put β:=βt0,ρ. On the support interval, Eβ is continuous and has the strict sign of E(t0) wherever β>0, and E(t0)β(t0)=E(t0)≠0; by clause 2, β>0 exactly on (t0−ρ,t0+ρ) and β≥0 everywhere.

3.1F4step 2.1algebra∎

If E(t0)>0 then Eβ≥0 on [t0−ρ,t0+ρ] and (Eβ)(t0)>0; were ∫t0−ρt0+ρEβ=0, [F4] would force Eβ=0 identically on the interval, contradicting (Eβ)(t0)>0. If E(t0)<0 apply the same argument to −Eβ, whose integral is −∫Eβ. In both cases the displayed integral is nonzero.

Remarks

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Reactions annihilating the virtual displacements are transpose multiples, with a positive definite multiplier matrix

Statement

Let G:O→Rk be a C2 regular holonomic constraint with Gq(t,q) of rank k on O⊆R×Rn (A regular holonomic constraint, its virtual displacements and the ideal-reaction law), and let M∈Mn(R) be symmetric positive definite. Fix (t,q)∈O and write V=ker⁡Gq(t,q) for the virtual displacements. Then:

  1. for every R∈Rn, the condition R⋅δ=0 for all δ∈V holds if and only if there is a unique λ∈Rk with

R=Gq(t,q)Tλ;

  1. the k×k matrix

C(t,q):=Gq(t,q)M−1Gq(t,q)T

is symmetric and positive definite, hence invertible; in particular for every R orthogonal to V the multiplier of clause 1 is λ=(GqM−1GqT)−1GqM−1R, and the multiplier is a C1 function of (t,q,R).

Facts & Assumptions

Given: The regular constraint G with k×n Jacobian Gq(t,q) of rank k at the fixed point, the symmetric positive definite matrix M, and a vector R∈Rn.

[F1]

The multiplier lemma (A linear functional annihilating the kernel of a surjection is a unique transpose multiple): for a surjective linear map A:Rn→Rk and a linear functional ℓ:Rn→R vanishing on ker⁡A, there is a unique λ∈Rk with ℓ(v)=λ⋅Av for all v. Here A=Gq(t,q) is surjective because its rank is k (Ck Euclidean maps and diffeomorphisms, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, Directional derivatives and partial derivatives of a map U⊆Rm→Rn).

[F3]

A linear map of a finite-dimensional space with trivial kernel is injective (The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial), and an injective linear endomorphism of Rk is invertible: Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T gives image dimension k, hence surjectivity, and the inverse of a linear bijection is linear and two-sided (Invertible matrices and the general linear group GL⁡n(F)).

Proof

technique · direct; translate the orthogonality condition into the multiplier lemma and verify positive definiteness of the multiplier matrix by a quadratic-form computation
1.1F1F2algebra

Clause 1, forward: assume R⋅δ=0 for all δ∈V=ker⁡Gq(t,q). Apply [F1] to the surjective A=Gq(t,q) and the linear functional ℓ(v):=R⋅v, which vanishes on ker⁡A=V by hypothesis: there is a unique λ∈Rk with R⋅v=λ⋅Gq(t,q)v=(Gq(t,q)Tλ)⋅v for every v∈Rn. Hence R=Gq(t,q)Tλ.

1.2F1F2algebra

Clause 1, converse: if R=GqTλ and δ∈V, then R⋅δ=λ⋅Gqδ=0. Uniqueness of λ is the uniqueness statement of [F1].

1.3F2algebra

Clause 2, symmetry: CT=(GqM−1GqT)T=Gq(M−1)TGqT=GqM−1GqT=C, because M−1 is symmetric by [F2].

1.4F2F3algebra

Clause 2, positive definiteness: let 0≠z∈Rk and put w:=GqTz. If w=0, then z⋅Gqv=(GqTz)⋅v=0 for every v∈Rn; since Gq is surjective, z⋅u=0 for every u∈Rk, hence z=0, a contradiction. Thus w≠0, and by positive definiteness of M−1, zTCz=zTGqM−1GqTz=w⋅M−1w>0. Hence C is positive definite. By the kernel argument (Cz=0 implies zTCz=0 implies z=0) and [F3], C is invertible.

2.1F2step 1.4algebra∎

The formula for λ: if R is orthogonal to V, clause 1 gives R=GqTλ; multiplying by GqM−1 on the left gives GqM−1R=Cλ, hence λ=C−1GqM−1R. The right-hand side is a composition of C1 maps of (t,q) — the entries of Gq and of C−1 are C1 by If det⁡(A) is a unit, then A−1=det⁡(A)−1adj⁡(A), because G is C2 and det⁡C≠0 — and is linear in R, so λ depends on (t,q,R) in a C1 way.

Remarks

  • The uniqueness in clause 1 is uniqueness of the multiplier for a given reaction R, not yet uniqueness of the reaction or of the acceleration: which R occurs is determined only after the twice-differentiated constraint is imposed, and that is done in The ideal holonomic system in a possibly time-dependent chart: explicit acceleration, constraint preservation, local existence and reaction power.

  • The matrix C is the inverse of the reduced mass matrix on the conormal directions; its positive definiteness is the reason the constrained acceleration is determined by the data wherever the rank hypothesis holds. At a point where the rank of Gq drops, C degenerates and the construction is not available; the proposition below makes no claim there.

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The first variation of a time-dependent action on a finite interval: admissibility, the integral formula and continuity of the Euler-Lagrange coefficient

Statement

Let (O,L) be a time-dependent Lagrangian chart over [a,b]×U in the sense of A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations, let q:[a,b]→U be a C2 admissible curve and let η be a variation of q. Then:

  1. there is ε0>0 such that for every ε∈(−ε0,ε0) the curve q+εη is again a C2 admissible curve with values in U;
  2. for every such ε the action S(q+εη) is defined, the function ε↦S(q+εη) is differentiable on (−ε0,ε0) and

δS(q)[η]=∫ab[Lq(t,q(t),q′(t))⋅η(t)+Lv(t,q(t),q′(t))⋅η′(t)]dt;

  1. the Euler–Lagrange expression EL(q):[a,b]→Rn of A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations is continuous.

Facts & Assumptions

Given: The chart data (O,L) with O open satisfying [a,b]×U×Rn⊆O, a C2 curve q:[a,b]→U, and a C2 variation η with η(a)=η(b)=0. Write ψ(ε,t):=(t,q(t)+εη(t),q′(t)+εη′(t)).

[F1]

The set [a,b]⊆R is compact by Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line (it is closed and bounded), the curve q is continuous on [a,b], and a continuous image of a compact set is compact (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset); hence q[a,b]⊆U is a nonempty compact subset. If U≠Rn, then the complement Rn∖U is a nonempty closed set disjoint from q[a,b], and A compact set and a disjoint closed set have a positive norm-distance gap supplies d>0 with ∥q(t)−c∥2≥d for all t∈[a,b] and all c∉U.

[F2]

The continuous function η:[a,b]→Rn attains a maximum of ∥η(t)∥2 (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value); write mη:=max⁡t∈[a,b]∥η(t)∥2≥0.

[F3]

The map ψ is continuous on R×[a,b] and is continuously partially differentiable in ε, with ∂ψ∂ε(ε,t)=(0,η(t),η′(t)), because η is C1; hence ψ is totally differentiable in (ε,t) with the displayed derivatives and the derivative depends continuously on (ε,t) (The total (Fréchet) derivative Df(a) as the linear first-order approximation with o(∥h∥2) remainder, If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative, Ck Euclidean maps and diffeomorphisms, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[F4]

The chain rule computes the derivative of a composite through the total derivatives (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)). In particular, if g(ε,t):=L(ψ(ε,t)) on the domain where ψ takes values in O, then ∂g∂ε(ε,t)=Lq(ψ(ε,t))⋅η(t)+Lv(ψ(ε,t))⋅η′(t), which is continuous in (ε,t) because Lq,Lv,η,η′ are continuous and ψ is continuous.

[F5]

Differentiation under the integral sign on a compact rectangle (Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral): if g,h:[α,β]×[a,b]→R are continuous and for each fixed t the map ε↦g(ε,t) is differentiable on (α,β) with derivative h(ε,t), then ε↦∫abg(ε,t) dt has derivative ∫abh(ε,t) dt at every ε∈[α,β].

Proof

Proof technique: direct; trap the curve in a positive distance shell of U, then differentiate under the integral on a compact rectangle.

1.1givenF3

Clause 1, the case U=Rn: take ε0:=1; then for every ε∈(−1,1) the map q+εη is C2 by the algebra of C2 maps on an open interval and takes values in Rn=U.

1.2F1F2algebra

Clause 1, the case U≠Rn: by [F1] choose d>0 with ∥q(t)−c∥2≥d for all t∈[a,b] and all c∉U, and by [F2] put mη:=max⁡∥η∥2. If mη=0 then η=0 identically and ε0:=1 works. Otherwise take ε0:=d/(2mη). For ∣ε∣<ε0 and t∈[a,b], ∥q(t)+εη(t)−q(t)∥2=∣ε∣ ∥η(t)∥2<d/2<d, so if q(t)+εη(t) were outside U, the distance from q(t) to the complement would be <d, contradicting the choice of d. Hence q+εη takes values in U; it is C2 by the algebra of C2 maps. This proves clause 1.

2.1F3F4F5F6step 1.2algebra

Clause 2: fix 0<δ<ε0 and consider R:=[−δ,δ]×[a,b]; by clause 1, ψ(R)⊆{(t,x,w):(t,x,w)∈O}, so g(ε,t):=L(ψ(ε,t)) is defined and continuous on R. By [F3], [F4] the partial derivative h:=∂g/∂ε exists and is continuous on R, and for each fixed t it is the derivative of ε↦g(ε,t). Hence [F5] applies and gives ddε∫abg(ε,t) dt=∫abh(ε,t) dt(∣ε∣≤δ). Evaluating at ε=0 and inserting the description of h from [F4] gives the displayed formula for δS(q)[η].

3.1F3F4givenalgebra∎

Clause 3: the map t↦ψ(0,t)=(t,q(t),q′(t)) is C1 on an open interval around [a,b] and takes values in O; since Lv is C1, the composite t↦Lv(t,q(t),q′(t)) is C1 by the chain rule, with continuous derivative because the derivative is a composite of continuous maps; and t↦Lq(t,q(t),q′(t)) is continuous. The difference EL(q) of these two continuous maps is continuous.

Remarks

  • The formula is stated for the chartwise finite-interval setting. The autonomous smooth version on a manifold, where the same computation is performed in every coordinate chart and assembled by a partition argument, is the published theorem on the Hamiltonian-mechanics page of the differential-geometry track; the present statement is the one needed when the Lagrangian depends explicitly on time and C2 data on a fixed interval are used.

  • Only the differentiability of ε↦S(q+εη) at ε=0 is used later. The proof establishes differentiability on the whole small interval, which is the natural output of the compact-rectangle theorem.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

Nonsingular velocity Hessian is equivalent to invertibility of the Legendre derivative, hence gives a local C^1 inverse

Statement

Let L:J×U×Rn→R be a fibre-full time-dependent Lagrangian chart of class C2 with Legendre map FL (The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity). Then:

  1. FL is C1 and at every (t,q,v) its total derivative acts on (τ,δq,δv)∈R×Rn×Rn by

DFL(t,q,v)(τ,δq,δv)=(τ, δq, Lvt(t,q,v)τ+Lvq(t,q,v)δq+Lvv(t,q,v)δv);

  1. DFL(t,q,v) is invertible (Invertible Euclidean linear maps) if and only if det⁡Lvv(t,q,v)≠0; consequently FL is regular at (t,q,v) in the sense of The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity exactly when its derivative there is invertible;
  2. if Lvv(t0,q0,v0) is nonsingular, then FL is a local C1 diffeomorphism at (t0,q0,v0): there are open neighbourhoods W∋(t0,q0,v0) and W′∋FL(t0,q0,v0) such that FL∣W:W→W′ is a C1 diffeomorphism (Continuously differentiable maps, local inverses, and local diffeomorphisms).

Facts & Assumptions

Given: The open sets J⊆R, U⊆Rn, the C2 Lagrangian L, a point (t0,q0,v0)∈J×U×Rn, and the notation Lvt,Lvq,Lvv for the matrices of partial derivatives of Lv with respect to the three blocks (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, Directional derivatives and partial derivatives of a map U⊆Rm→Rn, Ck maps and multi-index derivative notation in Euclidean space).

[F1]

Lv is of class C1 because L is of class C2 (Ck Euclidean maps and diffeomorphisms); hence Lv is totally differentiable everywhere with derivative the linear map given by its Jacobian matrix (Lvt Lvq Lvv), and that derivative depends continuously on the point (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative, The total (Fréchet) derivative Df(a) as the linear first-order approximation with o(∥h∥2) remainder).

[F2]

The chain rule for total derivatives (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)) computes derivatives of composites; the derivative of the identity-like map (t,q,v)↦(t,q) is the projection (τ,δq,δv)↦(τ,δq), and scalar multiplication of a differentiable map by 1 is the map itself.

[F3]

The Euclidean inverse function theorem (The Euclidean inverse function theorem): a C1 map between open subsets of Rm whose derivative at a point is invertible has a local C1 inverse there, in the sense of Continuously differentiable maps, local inverses, and local diffeomorphisms, with derivative Dg(y)=Df(g(y))−1.

[F4]

Linear algebra of block maps (Linear map between vector spaces over the same field, Invertible Euclidean linear maps, Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes): a linear map L presented in blocks is injective iff its kernel is trivial, and on finite-dimensional spaces injectivity is equivalent to invertibility for endomorphisms: Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T gives full image dimension from a zero kernel, hence a bijection whose inverse is linear and two-sided.

Proof

technique · direct; compute the block derivative, reduce invertibility to invertibility of the velocity Hessian by an explicit kernel computation, and apply the inverse function theorem
1.1F1F2algebra

By [F1] the map Lv is totally differentiable, and by [F2] the Legendre map FL(t,q,v)=(t,q,Lv(t,q,v)) is totally differentiable with the displayed action on (τ,δq,δv): the first two components reproduce τ and δq, and the third is the sum of the three block contributions. This proves clause 1.

1.2F2F4algebra

Clause 2, kernel computation: for a linear map L of the displayed block form, L(τ,δq,δv)=(0,0,0) holds iff τ=0, δq=0 and Lvvδv=0. Hence L has trivial kernel iff Lvvδv=0 implies δv=0, i.e. iff Lvv has trivial kernel. By [F4] trivial kernel is equivalent to invertibility of L, and of the square matrix Lvv; and a square matrix over R is invertible if and only if its determinant is a unit of R, i.e. nonzero (A positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit). This proves clause 2.

2.1F3step 1.1step 1.2

Clause 3: assume Lvv(t0,q0,v0) nonsingular. By clause 2 the derivative DFL(t0,q0,v0) is invertible, and by clause 1 FL is C1; [F3] therefore supplies open neighbourhoods W of (t0,q0,v0) and W′ of its image such that FL∣W:W→W′ is a C1 diffeomorphism.

3.1step 1.1step 1.2step 2.1∎

The three clauses are established.

Remarks

  • The determinant criterion in clause 2 is the reason the single scalar condition det⁡Lvv≠0 carries all the local invertibility information: the first two blocks of the derivative are already the identity. No determinant of a larger block matrix is needed, because the kernel computation of step 1.2 is performed directly.

  • Invertibility of the derivative is local. It neither asserts that FL is globally injective nor that it is surjective onto the momentum fibre; those global properties are the hyperregularity hypothesis of The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity.

PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

Coordinate covariance of the Euler-Lagrange expression under time-dependent point transformations

Statement

Let n≥1, let Ω⊆R×Rn be open and nonempty, and let X:Ω→Rn be of class C3 (Ck Euclidean maps and diffeomorphisms) with Xq(t,q) invertible for every (t,q)∈Ω (Invertible Euclidean linear maps, Linear map between vector spaces over the same field). Let Ox⊆R×Rn×Rn be open and let L:Ox→R be of class C2. Put

O:={(t,q,v)∈Ω×Rn:(t,X(t,q),Xt(t,q)+Xq(t,q)v)∈Ox},L~(t,q,v):=L(t,X(t,q),Xt(t,q)+Xq(t,q)v).

Then:

  1. O is open, O⊆R×Rn×Rn, and L~ is of class C2 on O;
  2. with w:=Xt(t,q)+Xq(t,q)v one has L~v(t,q,v)=Xq(t,q)TLv(t,X(t,q),w);
  3. L~q(t,q,v)=XqTLx+(Xtq+Xqq[v])TLv, where Xtq is the t-derivative of Xq and Xqq[v] is the matrix with entries ∑j∂j∂iXk(t,q)vj, all evaluated at (t,q) and Lx,Lv at (t,X(t,q),w);
  4. (transport) let a<b be reals and let q:[a,b]→Rn be a C2 curve with (t,q(t),q′(t))∈O for all t∈[a,b], put x(t):=X(t,q(t)), and define P(t):=L~v(t,q(t),q′(t)) and p(t):=Lv(t,x(t),x′(t)), which are C1. Then

ddtP(t)−L~q(t,q(t),q′(t))=Xq(t,q(t))T(ddtp(t)−Lx(t,x(t),x′(t)))(t∈[a,b]);

consequently, since Xq(t,q(t)) is invertible, the Euler–Lagrange expression of L~ vanishes along q if and only if the Euler–Lagrange expression of L vanishes along x.

Facts & Assumptions

Given: The open sets Ω,Ox, the C3 map X with everywhere invertible Xq, the C2 Lagrangian L, a C2 curve q:[a,b]→Rn with (t,q(t),q′(t))∈O, and the shorthand x(t):=X(t,q(t)).

[F2]

The chain rule for total derivatives (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)) gives the derivatives of all composites below; in particular x′(t)=Xt(t,q(t))+Xq(t,q(t))q′(t), so that x′(t)=w(t,q(t),q′(t)), and ddtXq(t,q(t))=Xqt(t,q(t))+Xqq(t,q(t))[q′(t)] as matrices.

[F3]

Mixed second partials of a C2 function commute (Clairaut--Schwarz theorem for continuous second partial derivatives): since X is C3, its mixed second partials are C1 and satisfy Xqt=Xtq and ∂j∂iXk=∂i∂jXk.

Proof

technique · direct; compute the transformed gradients and compare the two Euler–Lagrange expressions along a curve, using the equality of mixed partials of $X$
1.1F1F2algebra

O is the preimage under the continuous map (t,q,v)↦(t,X(t,q),Xt(t,q)+Xq(t,q)v) of the open set Ox, hence open; and L~ is the composite of the C2 map L with that map, which is C2 because X is C3 and v enters linearly. This proves clause 1.

1.2F2algebra

Clause 2: by the chain rule and the definition of w, ∂L~∂vj=∑kLvk(t,X,w)∂wk∂vj=∑kLvk(t,X,w)Xkj(t,q), which is the j-th entry of XqTLv.

1.3F2F3algebra

Clause 3: differentiating L~ in qi gives ∂L~∂qi=∑kLxk∂Xk∂qi+∑kLvk(∂iXt,k+∑j∂iXk,jvj)=(XqTLx)i+((Xtq+Xqq[v])TLv)i, as displayed.

2.1F2F3step 1.2step 1.3algebra

Clause 4: by clause 2, P(t)=Xq(t,q(t))Tp(t). Differentiating by [F2] and using Xqt=Xtq from [F3], ddtP=(Xqt(t,q)+Xqq(t,q)[q′])Tp+XqTddtp=(Xtq(t,q)+Xqq(t,q)[q′])Tp+XqTddtp. By clause 3 evaluated at (t,q(t),q′(t)), L~q(t,q,q′)=XqTLx+(Xtq+Xqq[q′])TLv. Subtracting and using p=Lv(t,x,x′) cancels the bracket terms and gives the displayed transport identity.

3.1F4step 2.1given∎

Consequence: if Xq(t,q(t)) is invertible, then by [F4] the identity XqTu=0 implies u=0; applying this to u=ddtp−Lx(t,x,x′), the vanishing of ddtP−L~q is equivalent to the vanishing of ddtp−Lx. This proves clause 4.

Remarks

  • The transformation is restricted to the configuration variables; no transformation of the time variable is performed here. The general point transformation that also changes time, together with the resulting Noether charge, is treated in Noether identity for point transformations with time reparametrisation: the conserved charge p⋅ξ−Eτ−B.

  • The C3 hypothesis makes (t,q,v)↦(t,X,Xt+Xqv) a C2 map and hence ensures the transformed Lagrangian is C2. The mixed second derivatives used in transport already commute for C2 X; that weaker assumption alone does not ensure the claimed C2 transformed chart.

PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

The Hamiltonian on an inverse Legendre branch: differential, equivalence with the Euler-Lagrange equation, and the energy identity

Statement

Let L:J×U×Rn→R be a fibre-full C2 time-dependent Lagrangian chart that is hyperregular, with inverse Legendre branch v(t,q,p) and Hamiltonian H(t,q,p)=p⋅v(t,q,p)−L(t,q,v(t,q,p)) (The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity). Then:

  1. (differential) for every (t,q,p)∈J×U×Rn,

Hp(t,q,p)=v(t,q,p),Hq(t,q,p)=−Lq(t,q,v(t,q,p)),Ht(t,q,p)=−Lt(t,q,v(t,q,p));

  1. (equivalence) for a C1 curve (q,p):I→U×Rn on an open interval I⊆J (Intervals of R: the nine order-convex forms, nondegeneracy, and length, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral) the following are equivalent: (a) q is C2 and EL(q)(t)=0 for all t∈I, with p(t)=Lv(t,q(t),q′(t)); (b) (q,p) satisfies Hamilton's equations q′(t)=Hp(t,q(t),p(t)) and p′(t)=−Hq(t,q(t),p(t)) for all t∈I;
  2. (energy) along any curve satisfying (a) or (b), the function e(t):=Lv(t,q,q′)⋅q′(t)−L(t,q(t),q′(t)) equals H(t,q(t),p(t)) and satisfies e′(t)=−Lt(t,q(t),q′(t)) on I; in particular, if Lt vanishes on the set {(s,q(s),q′(s)):s∈I}, then e is constant on I.

Facts & Assumptions

Given: The hyperregular chart data L, its inverse branch v, the Hamiltonian H, and a C1 curve (q,p):I→U×Rn.

[F2]

The Euler–Lagrange expression and the meaning of its vanishing are those of A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations: EL(q)=Lq(t,q,q′)−ddt(Lv(t,q,q′)). Here q:I→U is C2 on its open interval in the sense of Ck Euclidean maps and diffeomorphisms. On each compact subinterval [a,b]⊆I, its restriction has the extension required by the action definition, using q itself on a slightly larger interval inside I; thus the same Euler–Lagrange expression applies locally throughout I. Partial derivatives in the first and second slots are the entries of the gradient in the sense of The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case and Directional derivatives and partial derivatives of a map U⊆Rm→Rn.

Proof

technique · direct; differentiate $H$ along the inverse branch using the inverse relation, then translate between the two systems of equations
1.1F1algebra

For fixed (t,q,p) write v:=v(t,q,p). Differentiating the identity H(t,q,p)=∑jpjvj(t,q,p)−L(t,q,v(t,q,p)) with respect to pk by the product and chain rules [F1] gives Hpk=vk+∑j(pj−Lvj(t,q,v))∂vj∂pk=vk, because Lvj(t,q,v)=pj for every j by the inverse relation. Hence Hp(t,q,p)=v(t,q,p).

1.2F1algebra

Differentiating the same identity with respect to qk gives Hqk=∑j(pj−Lvj(t,q,v))∂vj∂qk−Lqk(t,q,v)=−Lqk(t,q,v), and differentiating with respect to t gives Ht=∑j(pj−Lvj)∂tvj−Lt(t,q,v)=−Lt(t,q,v). This proves clause 1.

2.1F1F2step 1.1step 1.2given

Clause 2, (a)⇒(b): assume (a). Then p(t)=Lv(t,q(t),q′(t)) is C1 as a composite of C1 maps, and applying the inverse relation of [F1] with (t,q(t),Lv(t,q(t),q′(t))) gives v(t,q(t),p(t))=q′(t). Hence by clause 1 q′(t)=v(t,q(t),p(t))=Hp(t,q(t),p(t)), and, differentiating the relation p=Lv(t,q,q′) and using (a), p′(t)=ddtLv(t,q,q′)=Lq(t,q,q′)=−Hq(t,q(t),p(t)), where the last equality is clause 1 with v(t,q,p)=q′. Thus (b) holds.

2.2F1step 1.1step 1.2given

Clause 2, (b)⇒(a): assume (b). By clause 1, q′=Hp(t,q,p)=v(t,q,p) is a composite of C1 maps, hence C1 on I; a curve with C1 derivative is C2, so q is C2. Applying Lv(t,q,⋅) to the identity q′=v(t,q,p) and using the inverse relation gives Lv(t,q,q′)=p, so p(t)=Lv(t,q(t),q′(t)). Now p′=−Hq(t,q,p)=Lq(t,q,q′) by clause 1, while also p′=ddtLv(t,q,q′); comparing gives EL(q)(t)=0 on I. Thus (a) holds, and clause 2 is proved.

3.1F1step 1.1step 1.2step 2.1

Clause 3: for a curve satisfying (a) (equivalently (b)) put v(t):=q′(t) and p(t):=Lv(t,q,q′). Then e(t)=p(t)⋅v(t)−L(t,q(t),v(t))=H(t,q(t),p(t)) by the definition of H and v(t)=v(t,q(t),p(t)). By clause 1 and the chain rule, e′(t)=Ht+Hq⋅q′+Hp⋅p′=−Lt−Lq⋅q′+v⋅p′. Substituting v=q′ and p′=Lq(t,q,q′) from (a) gives e′=−Lt−Lq⋅q′+q′⋅Lq=−Lt.

4.1F3step 3.1∎

If Lt vanishes along the trajectory, then e′=0 identically on I; by [F3] e is constant on I.

Remarks

  • The identity dH=−Lt dt−Lq⋅dq+v⋅dp is the coordinate expression of the displayed partial derivatives; it is the form in which the differential is usually quoted, and it makes manifest that the inverse-branch Hamiltonian is determined by L and the branch alone.

  • Only the fibre bijectivity and C1 inverse of hyperregularity are used. When L is merely regular, Nonsingular velocity Hessian is equivalent to invertibility of the Legendre derivative, hence gives a local C^1 inverse supplies local branches and the same computations hold on each branch; the global statements above are then local, and the page keeps the two hypotheses separate.

  • The energy e need not be bounded below or above and is not asserted to be a minimum of anything; it is conserved exactly when L has no explicit time dependence along the trajectory, in which case H is a first integral of the branch system.

PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

The ideal holonomic system in a possibly time-dependent chart: explicit acceleration, constraint preservation, local existence and reaction power

Statement

Let O⊆R×Rn be open and nonempty, let G:O→Rk be of class C3 with Gq(t,q) of rank k for every (t,q)∈O (a regular holonomic constraint, A regular holonomic constraint, its virtual displacements and the ideal-reaction law), let M∈Mn(R) be symmetric positive definite, and let f:O×Rn→Rn be of class C1. Define, for (t,q,v)∈O×Rn,

C(t,q):=Gq(t,q)M−1Gq(t,q)T,Λ(t,q,v):=−C(t,q)−1[GqM−1f+Gtt+2Gtq[v]+Gqq[v,v]],a(t,q,v):=M−1(f+GqTΛ),

where all terms in the bracket are evaluated at (t,q,v), and 2Gtq[v]:=∑i2∂t∂qiG vi and Gqq[v,v]:=∑i,j∂i∂jG vivj. Then:

  1. C is symmetric positive definite and C−1, Λ and a are of class C1 on O×Rn; in particular a is locally Lipschitz in (q,v);
  2. through every (t0,q0,v0)∈O×Rn there is a unique maximal solution (q,v):It0→R2n, with (t,q(t),v(t))∈O×Rn, of the first-order system

q′=v,v′=a(t,q,v);

  1. if the data satisfy the consistency conditions

G(t0,q0)=0,Gt(t0,q0)+Gq(t0,q0)v0=0,

then the maximal solution satisfies G(t,q(t))=0 and Gt(t,q(t))+Gq(t,q(t))q′(t)=0 for all t∈It0; consequently q is an ideal-constraint motion of (G,M,f) with multiplier λ(t)=Λ(t,q(t),v(t)), and its reaction R=GqTΛ satisfies the power identity

R(t)⋅q′(t)=−Λ(t,q(t),v(t))⋅Gt(t,q(t));

  1. conversely, every C2 ideal-constraint motion q:I→Rn of (G,M,f) whose velocity is permitted solves this first-order system with v=q′; hence the ideal-constraint motions with permitted velocity are exactly the solutions of the system with consistent data.

Facts & Assumptions

Given: The C3 regular constraint G with rank k everywhere, the symmetric positive definite M, the C1 force field f, the functions C,Λ,a, and data (t0,q0,v0)∈O×Rn.

[F1]

The multiplier lemma (Reactions annihilating the virtual displacements are transpose multiples, with a positive definite multiplier matrix): C is symmetric positive definite and invertible, R⊥ker⁡Gq iff R=GqTλ for a unique λ, and λ=C−1GqM−1R for such R.

[F2]

Regularity: G is C3, so Gq is C2, C is C2, and C−1 is C2 (Matrix inversion preserves Ck regularity where the determinant is nonzero); the bracket in Λ is C1 in (t,q,v) (its terms are sums and products of C1 and C2 maps in (t,q) with polynomials in v), and f is C1; hence Λ and a are C1 (Ck Euclidean maps are closed under componentwise algebra and composition, Ck Euclidean maps and diffeomorphisms, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, Ck maps and multi-index derivative notation in Euclidean space).

[F3]

Chain rule, twice: for a C2 curve q with (t,q(t))∈O, the function h(t):=G(t,q(t)) is C2 and h′′=Gtt+2Gtq[q′]+Gqq[q′,q′]+Gq q′′, the terms being evaluated at (t,q(t)); this follows by differentiating h′=Gt+Gq⋅q′ and commuting the mixed partials of G, which are continuous because G is C3 (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a), Clairaut--Schwarz theorem for continuous second partial derivatives, If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral).

[F4]

Proof

technique · direct; verify smoothness, take the twice-differentiated constraint, and use Picard–Lindelöf for the flow statement
1.1F1F2

Clause 1: by [F1], C is symmetric positive definite and invertible; by [F2], C−1, Λ and a are C1. A C1 right-hand side F(t,q,v):=(v,a(t,q,v)) is locally Lipschitz in the state variable: around every (t0,q0,v0) choose h,r>0 with [t0−h,t0+h]×B‾((q0,v0),r)⊆O×Rn. Its continuous state derivative has a finite bound L on this compact product. At fixed time, integration along the straight segment between two states in the ball gives ∥F(t,z)−F(t,z~)∥2≤L∥z−z~∥2. The same argument works on any such compact product (A C1 map is locally Lipschitz on compact coordinate subsets, Local Lipschitz continuity in the state variable, locally uniform in time and parameters, If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative). This proves clause 1.

2.1F2F4step 1.1

Clause 2: the system q′=v, v′=a(t,q,v) has a C1 right-hand side on the open set O×Rn, so by [F4] every initial value problem has a unique maximal solution; this is clause 2.

3.1F1F3F5step 2.1algebra

Clause 3, preservation: let (q,v) be the maximal solution through consistent data and put h(t):=G(t,q(t)) on the maximal interval. By [F3], h′′=Gtt+2Gtq[q′]+Gqq[q′,q′]+Gq a=Gtt+2Gtq[q′]+Gqq[q′,q′]+GqM−1f+GqM−1GqTΛ=Gtt+2Gtq[q′]+Gqq[q′,q′]+GqM−1f+CΛ=0, because GqM−1GqT=C and Λ=−C−1[GqM−1f+Gtt+2Gtq[q′]+Gqq[q′,q′]]. Thus h′′=0 on the interval, so h′ is constant and h is affine by [F5]: h(t)=h(t0)+h′(t0)(t−t0). The consistency conditions say h(t0)=0 and h′(t0)=Gt(t0,q0)+Gq(t0,q0)v0=0, so h=0 and h′=Gt(t,q(t))+Gq(t,q(t))q′(t)=0 on the interval. In particular q is a constrained curve with permitted velocity, and by construction Mq′′=f+GqTΛ, so q is an ideal-constraint motion with multiplier Λ along it.

4.1step 3.1algebra

Clause 3, power: since Gt+Gq q′=0 on the interval, R⋅q′=(GqTΛ)⋅q′=Λ⋅Gq q′=−Λ⋅Gt, where R=Mq′′−f=GqTΛ.

5.1F1F3step 3.1algebra∎

Clause 4: let q:I→Rn be an ideal-constraint motion with permitted velocity. Then G(t,q(t))=0 and Mq′′=f+GqTλ for some continuous λ; differentiating the constraint twice along the curve and using [F3] gives 0=Gtt+2Gtq[q′]+Gqq[q′,q′]+Gq q′′, and substituting q′′=M−1f+M−1GqTλ gives Cλ=−[GqM−1f+Gtt+2Gtq[q′]+Gqq[q′,q′]]=CΛ(t,q,q′), so λ=Λ(t,q,q′) by invertibility of C from [F1]. Hence q′′=a(t,q,q′) and the system holds; conversely every solution with consistent data is an ideal-constraint motion by step 3.1. This proves clause 4.

Remarks

  • The construction is rheonomic: G may depend on t and is never frozen. The time dependence enters the acceleration through Gt, Gtt and Gtq, and the reaction power R⋅q′=−Λ⋅Gt need not vanish, so an ideal moving constraint can do work on the system. This is the mechanism behind the rotating-hoop example of the companion page.

  • The hypothesis that Gq has rank k on all of O is used twice: to invert C and to make the vector field a everywhere defined and C1. At a point where the rank drops, C becomes singular and the construction stops; the proposition asserts nothing about continuation beyond such a point, and no continuation theorem is implied.

  • The consistency conditions are exactly the requirements that the initial position lies on the constraint and the initial velocity is permitted. Arbitrary data (t0,q0,v0) still determine a unique solution of the system, but not an ideal-constraint motion: the constraint is preserved only for consistent data.

PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

Adding a total time derivative and rescaling a Lagrangian: the Euler-Lagrange expression and the stationary curves

Statement

Let (O,L) be a time-dependent Lagrangian chart over [a,b]×U (A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations), let A⊆R×Rn be open and contain the time-configuration projection of O, and let F:A→R be of class C3, and define

L~(t,q,v):=L(t,q,v)+Ft(t,q)+Fq(t,q)⋅v.

Then:

  1. L~ is C2 on O, and for every C2 curve q:[a,b]→U,

EL~(q)(t)=EL(q)(t)(t∈[a,b]);

  1. for every such curve, S~(q)=S(q)+F(b,q(b))−F(a,q(a)), where S and S~ are the actions of L and L~;
  2. consequently q is stationary for S~ if and only if q is stationary for S;
  3. for every real c≠0, EcL(q)=c EL(q) and hence the stationary curves of cL and of L coincide; for c=0 one has E0=0, so every C2 curve is stationary for the zero Lagrangian, and setting c=0 does not preserve the stationary set unless every C2 curve was already stationary for L.

Facts & Assumptions

Given: The chart data (O,L), a C3 function F on A covering the chart projection, the modified Lagrangian L~, a real c and a C2 curve q:[a,b]→U.

[F1]

The chain rule and the equality of mixed partial derivatives (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a), Clairaut--Schwarz theorem for continuous second partial derivatives, Ck Euclidean maps and diffeomorphisms, Ck maps and multi-index derivative notation in Euclidean space): the functions Ft,Fq are C2; their mixed second derivatives commute, so ∂qi(Ft)=∂t(Fqi) and ∂qi(Fqj)=∂qj(Fqi) on the domain.

[F2]

Along a C2 curve q, the function t↦F(t,q(t)) is C2 with derivative Ft(t,q(t))+Fq(t,q(t))⋅q′(t) and second derivative Ftt+2∑iFtqiqi′+∑i,jFqiqjqi′qj′+Fq⋅q′′; this is the chain rule applied twice (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)).

Proof

Proof technique: direct; compute the Euler–Lagrange expression of the added total derivative and integrate it exactly.

1.1F1F2algebra

Put G(t,q,v):=Ft(t,q)+Fq(t,q)⋅v, so L~=L+G. Since A covers the chart projection and F is C3, G and L~ are defined and C2 throughout O. By [F1], Gq(t,q,v)=Ftq+Fqq[v],Gv(t,q,v)=Fq(t,q), where the first display abbreviates the vector with entries ∂qiFt+∑j∂qiFqjvj. Hence for any C2 curve q, ddtGv(t,q,q′)=ddtFq(t,q)=Fqt+Fqq[q′]=Ftq+Fqq[q′]=Gq(t,q,q′), using [F1] in the penultimate equality. Therefore EG(q)=Gq−ddtGv=0 along every C2 curve.

1.2F2F3algebra

Clause 2: by [F2] and [F3], S~(q)−S(q)=∫abddtF(t,q(t))dt=F(b,q(b))−F(a,q(a)), which is the displayed identity.

2.1step 1.1algebra

The Euler–Lagrange expression is additive and homogeneous in the Lagrangian: from the definition, EL+cG(q)=EL(q)+c EG(q) for every real c, because L+cG has gradients Lq+cGq and Lv+cGv and the derivative is linear. By step 1.1, EL~(q)=EL(q); this proves clause 1 and the identity in clause 4. For c≠0, EcL=cEL vanishes exactly when EL does, so the stationarity theorem gives the same stationary curves.

3.1step 2.1step 1.2given

Clause 3: stationary curves are characterised by the vanishing of the Euler–Lagrange expression (Stationarity of the time-dependent action is equivalent to the vanishing of the Euler-Lagrange expression). By clause 1 the two expressions agree, so the stationary sets of S and S~ coincide; equivalently, the boundary term of clause 2 is unchanged by the variations, whose endpoint values are zero by definition of admissible variation.

4.1step 2.1given∎

Clause 4, degenerate case: for c=0 the expression E0 vanishes identically, so by Stationarity of the time-dependent action is equivalent to the vanishing of the Euler-Lagrange expression every C2 curve is stationary for the zero Lagrangian; if L has a non-stationary curve, the stationary sets of L and 0⋅L differ.

Remarks

  • Clause 2 is the exact statement of the common phrase that two Lagrangians differing by a total time derivative give the same equations of motion: the actions differ by endpoint values, which are held fixed in the variational problem, and the Euler–Lagrange expressions agree pointwise.

  • The rescaling clause is stated for the stationary set only. Multiplying the action by c≠0 also rescales the first variation and the Euler–Lagrange expression, but it does not change which curves are stationary; the case c=0 is genuinely exceptional and is recorded because the zero Lagrangian is the degenerate witness used in the companion item on regular and hyperregular Lagrangians and in Scope of the Legendre map: regular versus hyperregular, the meaning of the covector datum, and the limits of the branch description.

PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

A smooth local family of extremals with prescribed endpoint exists when the endpoint momentum map has invertible derivative

Statement

Let n≥1, let J⊆R and U⊆Rn be open and nonempty, and let L:J×U×Rn→R be a smooth (C∞) fibre-full time-dependent Lagrangian chart that is hyperregular, with inverse Legendre branch v. Let t0,t1∈J, q0∈U, p0∗∈Rn, and suppose the maximal solution of Hamilton's equations

q′=Hp(t,q,p),p′=−Hq(t,q,p),H(t,q,p):=p⋅v(t,q,p)−L(t,q,v(t,q,p)),

with initial data (q(t0),p(t0))=(q0,p0∗) is defined at t1. Write E(t,p0) for the q-component at time t of the solution with data (q0,p0) at t0, and suppose that the derivative ∂p0E(t1,p0∗) is invertible (Invertible Euclidean linear maps, The total (Fréchet) derivative Df(a) as the linear first-order approximation with o(∥h∥2) remainder). Then there are open neighbourhoods J1⊆J of t1, Q1⊆U of q1:=E(t1,p0∗) and P1⊆Rn of p0∗, with t0∉J1, and a smooth map

p0:J1×Q1⟶P1,p0(t1,q1)=p0∗,

with E(t,p0(t,q))=q for all (t,q)∈J1×Q1; and, after possibly shrinking J1 and Q1 once more, the formula X(t,q,s):= the q-component at time s of the solution with data (q0,p0(t,q)) at t0 defines, for some open interval K⊆J containing t0 and J1, a C∞ map X:J1×Q1×K→U with X(t,q,t)=q, X(t,q,t0)=q0, whose curves s↦X(t,q,s) solve the Euler–Lagrange equation of L. Consequently the family X satisfies the hypotheses of The endpoint action function of a smooth family of extremals has differential dW=p dq−H dt on J1×Q1, and the resulting action function W(t,q)=∫t0tL(s,X,∂sX)ds has Wq=p and Wt=−H(t,q,p).

Facts & Assumptions

Given: The smooth hyperregular chart L with inverse branch v, the Hamiltonian H, the data t0,t1,q0,p0∗, and the endpoint map E.

[F1]

Since L is C∞ and hyperregular, the inverse branch v is C∞: locally, the Legendre map is a C∞ map with invertible derivative and hence has C∞ local inverses (Nonsingular velocity Hessian is equivalent to invertibility of the Legendre derivative, hence gives a local C^1 inverse, A local inverse of a Ck regular map is Ck), and the global inverse agrees with each local inverse near the corresponding point. Hence H is C∞, and Hamilton's equations are a smooth (C∞) nonautonomous system on J×U×Rn (The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity, Ck Euclidean maps are closed under componentwise algebra and composition, Ck Euclidean maps and diffeomorphisms).

[F2]

Smooth dependence on initial data (The fundamental theorem for nonautonomous smooth ODEs): for every base point of the smooth nonautonomous field XH(t,q,p)=(Hp,−Hq) there are a common time interval and a neighbourhood of initial data on which the local solution map is smooth in the initial time and initial state.

[F3]

The Euclidean inverse function theorem and its higher regularity (The Euclidean inverse function theorem, A local inverse of a Ck regular map is Ck): a C∞ map between open subsets of R1+n with invertible derivative at a point has a C∞ local inverse; the derivative of the map F(t,p0)=(t,E(t,p0)) is invertible at (t1,p0∗) if and only if ∂p0E(t1,p0∗) is invertible, by the explicit kernel computation of Nonsingular velocity Hessian is equivalent to invertibility of the Legendre derivative, hence gives a local C^1 inverse applied to the block form (10∂tE∂p0E) (The total (Fréchet) derivative Df(a) as the linear first-order approximation with o(∥h∥2) remainder, Invertible Euclidean linear maps, Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes, The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)).

[F4]

Translation between the Hamiltonian and Lagrangian descriptions on the inverse branch (The Hamiltonian on an inverse Legendre branch: differential, equivalence with the Euler-Lagrange equation, and the energy identity): a C1 curve (q,p) solves Hamilton's equations iff q solves the Euler–Lagrange equation of L with p=Lv(t,q,q′). Hence the smooth family constructed below consists of Euler–Lagrange extremals. Neighbourhoods are Euclidean balls (Open ball, closed ball and sphere in a metric space, Intervals of R: the nine order-convex forms, nondegeneracy, and length, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

Proof

Proof technique: construct the momentum selection by the inverse function theorem and read the solution family off the smooth flow.

1.1F1F2F3

By [F1] and [F2], the solution map of Hamilton's equations is smooth in the initial state near (q0,p0∗) on a common open solution interval K containing t0 and t1. The local conclusion is propagated along the reference solution as follows. Choose a compact time interval strictly inside its lifetime containing both times in its interior. Local smooth solution neighbourhoods from [F2] cover the compact reference trajectory. Choose around each reference time u a radius ru>0 so small that initial pairs along the reference curve for ∣s−u∣<2ru lie in its local smooth-solution domain, and that this local map evaluates solutions throughout that larger interval. The smaller intervals ∣s−u∣<ru cover the compact time segment, so finitely many suffice. A subdivision of mesh less than the minimum of these finitely many radii, with t0 inserted, has each short piece in one larger interval: its first endpoint lies in a smaller interval and every point of that piece differs from it by less than the corresponding radius. Compose the local solution maps, shrinking the initial-state neighbourhood at each of the finitely many joins so its image stays in the next domain. Uniqueness identifies the composition with the initial-value solution; the same construction backwards covers earlier times. This supplies smooth dependence on a common open interval containing the compact segment. Thus in particular the endpoint map E is defined and smooth on a neighbourhood of (t1,p0∗), and F(t,p0)=(t,E(t,p0)) is a smooth map whose derivative at (t1,p0∗) is invertible by the hypothesis and [F3].

2.1F3step 1.1

Since E(t0,p0)≡q0, its momentum derivative at t0 is zero; invertibility implies t1≠t0, so choose J1 to exclude t0. By the inverse function theorem with higher regularity [F3], there are open neighbourhoods J1∋t1, Q1∋q1 and P1∋p0∗ and a C∞ map p0:J1×Q1→P1 with p0(t1,q1)=p0∗ and F(t,p0(t,q))=(t,q), i.e. E(t,p0(t,q))=q, for all (t,q)∈J1×Q1.

3.1F2step 2.1

Define X(t,q,s) as the q-component at time s of the Hamiltonian solution with data (q0,p0(t,q)) at t0. By [F2] the local solution map is smooth in the initial state and in the evaluation time, so, after shrinking J1 and Q1 so that p0(t,q) remains in the region of smooth dependence and so that all these solutions exist on a common open interval K⊆J containing t0 and J1, the map X is C∞ on J1×Q1×K.

4.1step 2.1step 3.1F4

Properties of X: X(t,q,t0)=q0, because every solution starts at q0 at time t0; and X(t,q,t)=E(t,p0(t,q))=q by step 2.1. For each (t,q), the curve s↦(X(s),p(s)) solves Hamilton's equations with p the momentum component of the same solution, so by [F4] s↦X(t,q,s) solves the Euler–Lagrange equation of L with velocity ∂sX.

5.1step 3.1step 4.1F4∎

Steps 3.1 and 4.1 show that X satisfies the hypotheses of The endpoint action function of a smooth family of extremals has differential dW=p dq−H dt on J1×Q1; applying that theorem gives that W(t,q):=∫t0tL(s,X,∂sX)ds is C1 with Wq(t,q)=p(t,q) and Wt(t,q)=−H(t,q,p(t,q)), where p(t,q)=Lv(t,q,∂sX(t,q,t)).

Remarks

  • The proposition is a local existence statement: it produces the family on some neighbourhood of the chosen endpoint (t1,q1) from the invertibility of the endpoint momentum map. It proves neither that the family extends beyond that neighbourhood, nor that the endpoint problem is solvable for arbitrary data; failure of the invertibility hypothesis is exactly the degenerate case excluded here, and no boundary-value solvability is claimed.

  • The construction uses the Hamiltonian flow and the equivalence with the Euler–Lagrange equations on the inverse branch, so the hyperregularity of L is used both to define H globally and to translate trajectories. In the merely regular case the same argument works locally on each inverse branch.

PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

A mechanical chart Lagrangian with positive definite kinetic matrix is hyperregular, with the explicit kinetic-plus-potential Hamiltonian

Statement

Let n≥1, let J⊆R and U⊆Rn be open and nonempty, and let G:J×U→Mn(R), A:J×U→Rn and V:J×U→R be of class C2 (Ck Euclidean maps and diffeomorphisms). Assume that G(t,q) is symmetric and positive definite for every (t,q):

G(t,q)T=G(t,q),wTG(t,q)w>0for every 0≠w∈Rn.

Define the mechanical chart Lagrangian

L(t,q,v):=12vTG(t,q)v+A(t,q)⋅v−V(t,q),

with the Euclidean inner product of The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn. Then:

  1. L is of class C2 on J×U×Rn, with Lv(t,q,v)=G(t,q)v+A(t,q) and Lvv(t,q,v)=G(t,q); in particular L is regular on the chart;
  2. the Legendre map FL(t,q,v)=(t,q,G(t,q)v+A(t,q)) is a C2 bijection of J×U×Rn onto itself with C2 inverse

v(t,q,p)=G(t,q)−1(p−A(t,q));

hence L is hyperregular in the sense of The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity; 3. the Hamiltonian of the inverse branch is

H(t,q,p)=12(p−A(t,q))TG(t,q)−1(p−A(t,q))+V(t,q),

a C2 function on J×U×Rn.

Facts & Assumptions

Given: The C2 fields G,A,V with G(t,q) symmetric positive definite, the Lagrangian L displayed above, and the Legendre map FL(t,q,v)=(t,q,Lv(t,q,v)).

[F1]

Partial derivatives of a quadratic form: if G is symmetric, then the gradient of v↦12vTGv is v↦Gv, and the second derivative in v is the constant matrix G; this is the sum and product rule applied to 12∑i,jGijvivj together with Gij=Gji (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, Directional derivatives and partial derivatives of a map U⊆Rm→Rn, Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes).

[F2]

Finite sums and products of Cr Euclidean maps are Cr, and composites of composable Cr maps are Cr (Ck Euclidean maps are closed under componentwise algebra and composition, Ck Euclidean maps and diffeomorphisms). In particular L is C2 because G,A,V are.

[F3]

A symmetric positive definite matrix is invertible: if Gw=0 then wTGw=0, so w=0, and a linear map of Rn with trivial kernel is bijective by Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T (zero kernel gives image dimension n); its linear inverse is two-sided (Invertible matrices and the general linear group GL⁡n(F), Linear map between vector spaces over the same field).

[F4]

Inversion preserves regularity: if the entries of B:J×U→Mn(R) are Cr and det⁡B never vanishes, then the entries of B−1 are Cr (Matrix inversion preserves Ck regularity where the determinant is nonzero). For B=G, invertibility holds by [F3], and det⁡G≠0 because det⁡(G)det⁡(G−1)=det⁡(GG−1)=det⁡(In)=1 for an invertible G with inverse G−1 (For same-sized finite square matrices over a commutative ring, det⁡(AB)=det⁡(A)det⁡(B), A positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit).

Proof

technique · direct; differentiate the quadratic Lagrangian, invert the velocity--momentum relation by positive definiteness, and complete the square
1.1F1F2F3algebra

By [F1] and [F2] the map L is C2, its gradient in the velocity slot is Lv(t,q,v)=G(t,q)v+A(t,q) and the velocity Hessian is Lvv(t,q,v)=G(t,q). Since G(t,q) is positive definite, it is invertible by [F3], so det⁡Lvv(t,q,v)=det⁡G(t,q)≠0; this proves clause 1.

1.2F3algebra

Clause 2, bijectivity: fix (t,q,p)∈J×U×Rn. By [F3] the matrix G(t,q) is invertible, so the equation G(t,q)v+A(t,q)=p has the unique solution v=G(t,q)−1(p−A(t,q)); hence FL is bijective with the displayed inverse.

2.1F4step 1.2algebra

Clause 2, regularity of the inverse: the entries of G are C2 and det⁡G never vanishes, so by [F4] the entries of (t,q)↦G(t,q)−1 are C2; the formula for v is then a finite combination of C2 maps and is C2 (Ck Euclidean maps are closed under componentwise algebra and composition). By The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity this makes L hyperregular.

3.1F1F2step 2.1algebra∎

Clause 3: with v=v(t,q,p)=G−1(p−A) one has p−A=Gv, hence p⋅v−L(t,q,v)=(Gv+A)⋅v−12vTGv−A⋅v+V=12vTGv+V=12(p−A)TG−1(p−A)+V, where the last step uses (Gv)⋅v=vTGv=(p−A)TG−1(p−A) and the symmetry of G−1 (the inverse of a symmetric matrix is symmetric). Thus clause 3 holds, and H is C2 by the same closure properties.

Remarks

  • The case of a constant diagonal matrix G=diag⁡(m1,…,mn) with mi>0 and A=0 gives L=12∑imiq˙i2−V(q) and H=12∑ipi2/mi+V(q), the classical kinetic-plus-potential form. The proposition is stated for a configuration- and time-dependent positive definite G, which is what the consuming classical-mechanics items need for moving constraints and rotating frames.

  • The covector field A is a mathematical datum of the chart: it is allowed to depend on (t,q) and enters the momentum as p=Gv+A. It is not an electromagnetic potential and no physical interpretation of A is asserted here; in particular p is not mass×velocity unless G is constant and A=0. The scope of these identifications is recorded in Scope of the Legendre map: regular versus hyperregular, the meaning of the covector datum, and the limits of the branch description.

PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

The distribution span{∂x,∂y+x∂z} is not tangent to any regular level sets

Statement

Let O⊆R3 be open and nonempty, and consider the derivations A:=∂x and B:=∂y+x ∂z acting on C2 functions on O. If G:O→R is of class C2 with

AG=0andBG=0on O,

then DG=0 on O. Consequently, if G:O→R is C2 and DG(q)≠0 for every q∈O (so that every level set of G is a regular level set with tangent space ker⁡DG(q), Regular and critical points, regular and critical values, and level sets, The tangent space to a regular level set), then the distribution D(q)=span⁡{A(q),B(q)} of The velocity constraint z′=xy′: virtual displacements, the d'Alembert law and the vakonomic law cannot equal ker⁡DG(q) for every q∈O; that is, no such G has level sets whose tangent planes are everywhere the virtual displacements of the constraint z′=xy′.

Facts & Assumptions

Given: The open set O, a C2 function G:O→R, and the two derivations A=∂x, B=∂y+x∂z.

[F1]

Partial derivatives, the gradient and the Jacobian (Directional derivatives and partial derivatives of a map U⊆Rm→Rn, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, Ck maps and multi-index derivative notation in Euclidean space, Ck Euclidean maps and diffeomorphisms): AG=∂xG and BG=∂yG+x∂zG are the displayed C1 functions, and DG=(∂xG,∂yG,∂zG).

[F2]

Equality of mixed partial derivatives for a C2 function (Clairaut--Schwarz theorem for continuous second partial derivatives): ∂x∂yG=∂y∂xG and ∂x∂zG=∂z∂xG on O.

Proof

technique · direct; bracket the two defining vector fields and use the equality of mixed partials
1.1F1F3givenalgebra

The hypotheses AG=0 and BG=0 on O give, by the product rule and linearity of the derivations, A(BG)−B(AG)=0−0=0 on O; that is, [A,B]G=0 where [A,B]G:=ABG−BAG.

1.2F2F3algebra

Compute [A,B]G from [F3]: ABG−BAG=(∂x∂yG+∂zG+x∂x∂zG)−(∂y∂xG+x∂z∂xG)=∂zG, because [F2] cancels the first pair and the pair multiplied by x.

2.1F1step 1.1step 1.2algebra

Steps 1.1 and 1.2 give ∂zG=0 on O. Then BG=∂yG+x∂zG=∂yG=0, and AG=∂xG=0; hence all three partial derivatives vanish and DG=0 on O.

3.1step 2.1F1∎

Consequently, if DG(q)≠0 at every point of O, no C2 function satisfies AG=BG=0; in particular no such G has D(q)=ker⁡DG(q) for every q, since equality with the kernel would force A and B to annihilate G.

Remarks

  • The computation shows that the constraint distribution of z′=xy′ is not tangent to a two-dimensional regular level foliation: steps 1.1-1.2 identify the bracket [A,B] on C2 functions as the derivation ∂z, and the hypothesis that A and B annihilate a defining function forces ∂zG=0 and hence a vanishing gradient. The conclusion needed by the classical-mechanics development is drawn directly, without invoking a converse of the Frobenius theorem.

  • The proposition is the precise sense in which z′=xy′ is a non-holonomic constraint: it is not equivalent to G(q)=constant for any C2 function with nonvanishing differential. Nothing is asserted about constraints that merely look similar or about kinematic mechanisms; those belong to the consuming physical models.

RemarkRemark: AI-adaptedProof: Not applicableOpen item page →

Scope of point symmetries: symmetries belong to the Lagrangian, not to the configuration space, and time translation is the energy case

Remarks

RemarkRemark: AI-adaptedProof: Not applicableOpen item page →

Scope of stationarity: stationary is not minimum, and no second-variation theory is developed here

Remarks

  • Stationarity, as defined in A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations and characterised by Stationarity of the time-dependent action is equivalent to the vanishing of the Euler-Lagrange expression, is the vanishing of the first variation for all fixed-endpoint variations. It is not the statement that q minimizes the action, nor that it is a local minimum. The companion page exhibits an explicit quadratic model in which a stationary curve is not a local minimum: for the Lagrangian L(q,v)=12(v2−4q2) on [0,π] and the stationary curve q∗(t)=sin⁡2t, the exact identity S(q∗+ϵsin⁡)−S(q∗)=−3π4ϵ2 shows that every nearby element of the one-parameter family has strictly smaller action.

  • No second variation, and no convexity or coercivity hypothesis, is introduced on this page. The stationarity theorem is an equivalence between zero first variation and the Euler–Lagrange equation. For a local extremum on fixed-endpoint C2 curves, use the norm ∥q∥C2:=max⁡0≤j≤2sup⁡t∈[a,b]∥q(j)(t)∥2. Each admissible family q+sη tends to q in this norm; a local extremum therefore gives an extremum at s=0 of its differentiable action, whose derivative is zero. Thus the Euler–Lagrange equation is necessary for such extrema, while stationarity alone need not give an extremum.

  • The finite-interval chart statements on this page concern C2 data and a fixed interval [a,b]. They make no claim about minimizers in a Sobolev or absolutely continuous class, about broken extremals, or about boundary conditions other than fixed endpoints; those belong to the calculus-of-variations development of the mathematical library.

RemarkRemark: AI-adaptedProof: Not applicableOpen item page →

Scope of the Legendre map: regular versus hyperregular, the meaning of the covector datum, and the limits of the branch description

Remarks

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

The endpoint action function of a smooth family of extremals has differential dW=p dq−H dt

Statement

Let n≥1, let J⊆R and U⊆Rn be open and nonempty, and let L:J×U×Rn→R be a fibre-full C2 time-dependent Lagrangian chart that is hyperregular, with inverse Legendre branch v and Hamiltonian H(t,q,p)=p⋅v(t,q,p)−L(t,q,v(t,q,p)) (The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity, A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations). Fix t0∈J and q0∈U, let T⊆J∖{t0} be open and nonempty, and let Q⊆U be open and nonempty, let D⊆T×Q×J be open, and let X:D→U be a C2 map such that for every (t,q)∈T×Q the points (t,q,t) and (t,q,t0) lie in D, the closed interval with endpoints t0 and t is contained in the s-section Dt,q:={s∈J:(t,q,s)∈D}, which is required to be an open interval, and:

  1. X(t,q,t)=q and X(t,q,t0)=q0;
  2. s↦X(t,q,s) is a solution of the Euler–Lagrange equation of L on the open interval Dt,q, i.e. (s,X(t,q,s),∂sX(t,q,s))∈J×U×Rn and EL(X(t,q,⋅))=0 on Dt,q.

Define

p(t,q):=Lv(t,q,∂sX(t,q,t)),W(t,q):=∫t0tL(s,X(t,q,s),∂sX(t,q,s))ds((t,q)∈T×Q).

Then W is of class C1 on T×Q and

Wq(t,q)=p(t,q),Wt(t,q)=−H(t,q,p(t,q))((t,q)∈T×Q).

Facts & Assumptions

Given: The chart data L with its inverse branch v and Hamiltonian H, the fixed data t0,q0, the open set D and the C2 family X, and the functions p and W displayed in the Statement.

[F1]

The integrand G(t,q,s):=L(s,X(t,q,s),∂sX(t,q,s)) is of class C1 on D, because L is C2 and X is C2 (Ck Euclidean maps are closed under componentwise algebra and composition, The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a), Ck Euclidean maps and diffeomorphisms). Mixed second partial derivatives of X commute by Clairaut--Schwarz theorem for continuous second partial derivatives.

[F2]

Integrals with reversed limits mean ∫t0uG=−∫ut0G (The integral with oriented limits: ∫aaf:=0 and ∫baf:=−∫abf). Differentiation of an integral with a variable upper limit and parameters (The first fundamental theorem: if f is integrable on [a,b] and continuous at c, then F′(c)=f(c); in particular a continuous f has F as a primitive, For a<c<b: f is integrable on [a,b] if and only if it is integrable on [a,c] and on [c,b], and then ∫abf=∫acf+∫cbf; with the oriented form for arbitrary a,b,c, Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral, The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a), If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative): for Φ(u,v,q):=∫t0uG(v,q,s) ds one has Φu=G(v,q,u),Φv=∫t0u∂vG(v,q,s) ds,Φq=∫t0u∂qG(v,q,s) ds, locally on a parameter rectangle containing the integration segment. For the upper-limit derivative, additivity writes Φ(u,v,q) as a primitive with a fixed lower endpoint minus its value at t0; for reversed limits the parameter derivatives follow by antisymmetry. The displayed partial derivatives are continuous because their integrands are continuous on a compact rectangle. Since W(t,q)=Φ(t,t,q), the chain rule gives that W is C1 and Wt(t,q)=G(t,q,t)+∫t0t∂tG(t,q,s) ds,Wq(t,q)=∫t0t∂qG(t,q,s) ds.

[F4]

The Euler–Lagrange equation along s↦X(t,q,s) says ∂s(Lv(s,X,∂sX))=Lq(s,X,∂sX) for every (t,q,s)∈D (A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations); the notation Lq,Lv is that of The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case and The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn.

[F5]

Endpoint differentiation (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a), The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, Intervals of R: the nine order-convex forms, nondegeneracy, and length): the identities X(t,q,t)=q and X(t,q,t0)=q0 hold for every (t,q)∈T×Q, and X is C2, so they may be differentiated in q and in t; this gives ∂qX(t,q,t)=In,∂qX(t,q,t0)=0,∂tX(t,q,t0)=0,∂tX(t,q,t)+∂sX(t,q,t)=0.

Proof

technique · direct; differentiate the endpoint integral in $q$ and in $t$ and use the Euler–Lagrange equation to remove the interior terms
1.1F1F2F3F4algebra

At each (t,q)∈T×Q, the compact integration segment lies in open D. A finite cover by product neighbourhoods yields a common parameter neighbourhood and an open s-interval containing that segment where G is C1; shrinking the parameter neighbourhood also covers nearby upper limits. Thus [F2] applies locally, with the integral oriented if t<t0, and W is C1. Put Jq:=∂qX and P(s):=Lv(s,X,∂sX), both evaluated at fixed (t,q). Using column gradients, Wq=∫t0t[JqTLq+(∂sJq)TP]ds=[JqTP]t0t+∫t0tJqT(Lq−∂sP)ds=[JqTP]t0t. The middle identity is componentwise integration by parts [F3], and the last uses [F4].

1.2F1F2F3F4algebra

For Wt, [F2] gives Wt(t,q)=G(t,q,t)+∫t0t[Lq⋅∂tX+Lv⋅∂s∂tX]ds=L(t,q,∂sX(t,q,t))+[Lv⋅∂tX]s=t0s=t, where the integral was integrated by parts and cancelled by [F4].

2.1F5step 1.1givenalgebra

The boundary terms of step 1.1: differentiating X(t,q,t)=q with respect to q gives ∂qX(t,q,t)=In, so the first term is p(t,q); and differentiating X(t,q,t0)=q0 with respect to q gives ∂qX(t,q,t0)=0, so the second term vanishes. Thus Wq(t,q)=p(t,q).

3.1F5step 1.2givenalgebra∎

The boundary terms of step 1.2: differentiating X(t,q,t0)=q0 with respect to t gives ∂tX(t,q,t0)=0, so the lower term vanishes; and differentiating the identity X(t,q,t)=q with respect to t gives ∂tX(t,q,t)+∂sX(t,q,t)=0. Hence Wt(t,q)=L(t,q,v(t,q))−p(t,q)⋅v(t,q)=−H(t,q,p(t,q)), with v(t,q):=∂sX(t,q,t) and p(t,q)=Lv(t,q,v(t,q)), using H(t,q,p)=p⋅v(t,q,p)−L(t,q,v(t,q,p)) and the inverse relation v(t,q,Lv(t,q,v(t,q)))=v(t,q).

Remarks

  • The two identity clauses determine the first variation of W completely: dW=p dq−H dt at every point of T×Q, where p is the momentum of the selected extremal. This is a conditional statement: it presumes the C2 family X of extremals with the endpoint property, and its proof never asserts that such a family exists.

  • The existence of families with the required endpoint property is the subject of A smooth local family of extremals with prescribed endpoint exists when the endpoint momentum map has invertible derivative, where the hypothesis used is invertibility of the derivative of the endpoint map with respect to the initial momentum. Several extremals may share the same endpoint data while a selected branch still forms a single-valued C2 family. A singular derivative of the endpoint map prevents applying that inverse-function argument, but does not by itself prove that no such family exists. The present theorem applies to every given family satisfying its hypotheses.

  • The identities Wq=p and Wt=−H(t,q,Wq) are the Hamilton–Jacobi relations for the endpoint action on the chart. The second follows from the first by substitution, since p=Wq; the theorem proves the pair directly.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

Stationarity of the phase-space action is equivalent to Hamilton's equations, and hyperregular elimination recovers the Lagrangian action

Statement

Let n≥1, let OH⊆R×Rn×Rn be open and nonempty, let H:OH→R be of class C2, let a<b, and suppose that for a C1 path γ(t)=(q(t),p(t)) (Ck Euclidean maps and diffeomorphisms, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral) with (t,q(t),p(t))∈OH for all t∈[a,b] the phase action

A(γ):=∫ab[p(t)⋅q′(t)−H(t,q(t),p(t))]dt

is defined, the integrand being continuous (The lower and upper Darboux integrals of a bounded f on [a,b] as sup⁡PL(f,P) and inf⁡PU(f,P), Darboux integrability as their equality, and the notation ∫abf, A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion, The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn). Then:

  1. for every C1 variation pair (δq,δp) with δq(a)=δq(b)=0 and with γ+s(δq,δp) in the domain of A for small ∣s∣,

dds∣s=0A(γ+s(δq,δp))=∫ab[(q′−Hp)⋅δp+(−p′−Hq)⋅δq]dt;

  1. γ is stationary (the displayed derivative vanishes for every such variation) if and only if Hamilton's equations hold on [a,b]:

q′(t)=Hp(t,q(t),p(t)),p′(t)=−Hq(t,q(t),p(t))(t∈[a,b]);

  1. (elimination) let in addition L:J×U×Rn→R be a hyperregular C2 chart with inverse branch v(t,q,p) and Hamiltonian H(t,q,p)=p⋅v(t,q,p)−L(t,q,v(t,q,p)) (The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity), such that the phase domain of H contains the points below. For every C2 curve q:[a,b]→U with (t,q(t),q′(t)) in the domain of L, put p(t):=Lv(t,q(t),q′(t)). Then A(q,p)=S(q), where S is the action of L; consequently q is stationary for S if and only if (q,p) is stationary for the phase action.

Facts & Assumptions

Given: The C2 phase function H, a C1 path γ=(q,p), the phase action A, and admissible variation pairs (δq,δp) with δq(a)=δq(b)=0.

[F1]

Differentiation under the integral sign on a compact rectangle (Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral): the one-parameter integrand has continuous partial derivative in the parameter, so the derivative passes through the integral. Integration by parts (If u,v are differentiable on [a,b] with u′,v′ integrable, then ∫abuv′=u(b)v(b)−u(a)v(a)−∫abu′v) applies to the term ∫abp⋅δq′ dt componentwise with u=pj, w=δqj, and yields −∫abp′⋅δq dt, because the boundary term [p⋅δq]ab vanishes by δq(a)=δq(b)=0; the integral is linear (Integrable functions on [a,b] form a set closed under sums and scalar multiples, and ∫ab(λf+μg)=λ∫abf+μ∫abg).

[F2]

The bumps of Explicit twice differentiable bumps detect nonzero values of continuous functions: a continuous scalar coefficient E on (a,b) that is nonzero at some interior point admits a C2 bump β supported in (a,b) with ∫Eβ≠0. Such bumps, multiplied by the standard basis vectors of the q- and p-components, give the admissible variations used below.

[F3]

Hyperregular elimination (The Hamiltonian on an inverse Legendre branch: differential, equivalence with the Euler-Lagrange equation, and the energy identity): on the inverse branch Hp(t,q,p)=v(t,q,p), Hq=−Lq(t,q,v), and (q,p) solves Hamilton's equations iff q solves the Euler–Lagrange equation with p=Lv(t,q,q′); in the replacement p=Lv(t,q,q′) the inverse relation gives v(t,q,p)=q′, so p⋅q′−H(t,q,p)=Lv⋅q′−(Lv⋅q′−L)=L(t,q,q′).

Proof

technique · direct; differentiate the phase action, integrate by parts, and use bumps in the two components separately; then substitute the inverse Legendre branch
1.1F1givenalgebra

Clause 1: the parameter derivative of the integrand is δp⋅q′+p⋅δq′−Hq⋅δq−Hp⋅δp, continuous in (s,t); by [F1] the derivative of A is the integral of this expression. Integrating the term ∫p⋅δq′ by parts and collecting the coefficients of δq and δp gives the displayed formula.

2.1step 1.1given

Clause 2, forward direction: if Hamilton's equations hold, then q′−Hp=0 and −p′−Hq=0 identically, so the integral of step 1.1 vanishes for every admissible variation; thus γ is stationary.

2.2F2step 1.1algebra

Clause 2, converse: assume γ stationary. First take δq=0 and let δp run through the bumps of [F2] in a single component. Step 1.1 gives ∫ab(q′−Hp)jβ dt=0 for every such bump, hence the continuous function (q′−Hp)j vanishes on (a,b) and, by continuity, on [a,b]; as j was arbitrary, q′=Hp on [a,b]. Now take δp=0 and let δq run through bumps in a single component (admissible, since these vanish near the endpoints): step 1.1 gives ∫ab(−p′−Hq)jβ dt=0, hence p′=−Hq on [a,b]. This proves clause 2.

3.1F3F4step 2.1step 2.2∎

Clause 3: for a C2 curve q and p(t)=Lv(t,q,q′), the path (q,p) is C1 and [F3] gives A(q,p)=∫abL(t,q,q′)dt=S(q). Therefore the phase action along this path equals the Lagrangian action, and the two stationarity notions correspond: by clause 2, (q,p) is stationary for A iff it solves Hamilton's equations; by [F3] this holds iff q solves the Euler–Lagrange equation with p=Lv(t,q,q′); and by [F4] this holds iff q is stationary for S.

Remarks

  • The identification in clause 3 is made on the actual inverse branch: the path in phase space is the Legendre image of the Lagrangian path, so no dynamical assumption such as p=q′ is built into the statement; the elimination uses the inverse relation of the hyperregular chart, which is exactly the hypothesis under which the branch exists globally.

  • The phase action has free p endpoints and fixed q endpoints. This is why the p-variations require no boundary condition and the q-variations do: the boundary term produced by integrating ∫p⋅δq′ vanishes precisely because δq vanishes at the endpoints, while no corresponding term arises from δp.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

Noether identity for point transformations with time reparametrisation: the conserved charge p⋅ξ−Eτ−B

Statement

Let (O,L) be a time-dependent Lagrangian chart over [a,b]×U (A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations). Let A⊆R×Rn be open and contain the time-configuration projection of O. Let τ,ξ:A→R,Rn and B:A→R be of class C2, and for a C1 function φ(t,q) with scalar values write Dtφ(t,q,v):=φt(t,q)+φq(t,q)⋅v; for vector-valued ξ set Dtξ:=ξt+Dqξ v, equivalently (Dtξ)j=Dt(ξj). Assume the Noether identity

Lt τ+Lq⋅ξ+Lv⋅(Dtξ−v Dtτ)+L Dtτ=DtB

on O, where all arguments of L and its derivatives are (t,q,v). Write p:=Lv for the momentum and put

E(t,q,v):=Lv(t,q,v)⋅v−L(t,q,v),C(t,q,v):=Lv(t,q,v)⋅ξ(t,q)−E(t,q,v)τ(t,q)−B(t,q).

Then for every C2 curve q:I→U on an open interval I⊇[a,b] with (t,q(t),q′(t))∈O that solves the Euler–Lagrange equation EL(q)(t)=0, the function t↦C(t,q(t),q′(t)) is constant on I. Moreover:

  1. at τ=0 and with ξ independent of t the Noether identity is exactly the infinitesimal point symmetry identity of Infinitesimal point symmetry of a time-dependent Lagrangian and its conserved charge for the pair (ξ,B), and the charge reduces to C=Lv⋅ξ−B, so the present theorem specialises to The charge of an infinitesimal point symmetry is constant along Euler-Lagrange solutions;
  2. at ξ=0, τ=1, B=0 the Noether identity holds if and only if Lt=0 on O, and the charge is C=−E; thus the theorem contains the conservation of the energy in the time-independent case, in agreement with clause 3 of The Hamiltonian on an inverse Legendre branch: differential, equivalence with the Euler-Lagrange equation, and the energy identity.

Facts & Assumptions

Given: The chart data (O,L), the C2 data τ,ξ,B, the Noether identity, the functions E and C, and a C2 Euler–Lagrange solution q:I→U.

[F1]

Euler–Lagrange equation along q: ddtLv(t,q,q′)=Lq(t,q,q′); this is the content of EL(q)=0, which is equivalent to stationarity (A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations, Stationarity of the time-dependent action is equivalent to the vanishing of the Euler-Lagrange expression).

[F2]

Chain rule along the curve (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a), The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, Intervals of R: the nine order-convex forms, nondegeneracy, and length): for the C1 composite t↦φ(t,q(t)) the derivative is Dtφ(t,q(t),q′(t)); in particular ddtξ(t,q(t))=Dtξ,ddtτ(t,q(t))=Dtτ,ddtB(t,q(t))=DtB along the curve, with the right-hand sides evaluated at (t,q(t),q′(t)).

Proof

technique · direct; differentiate the charge along the solution and substitute the Euler–Lagrange equation and the Noether identity
1.1F1F2F3algebra

Along the solution, the energy satisfies E′=−Lt: by [F2] and the product rule [F3], E′=(ddtLv)⋅q′+Lv⋅q′′−Lt−Lq⋅q′−Lv⋅q′′=Lq⋅q′−Lt−Lq⋅q′=−Lt, where [F1] was used in the second equality.

1.2givenalgebra

Clause 1: at τ=0 and ξ(t,q)=ξ0(q) one has Dtτ=0, so the Noether identity becomes Lq⋅ξ+Lv⋅Dtξ=DtB, with Dtξ=Dqξ v, which is the identity of Infinitesimal point symmetry of a time-dependent Lagrangian and its conserved charge; the charge is C=Lv⋅ξ−B, and the conclusion is the one of The charge of an infinitesimal point symmetry is constant along Euler-Lagrange solutions.

2.1F1F2F3step 1.1algebra

The charge is differentiable along the curve, and by [F2], [F3] and step 1.1, C′=(ddtLv)⋅ξ+Lv⋅Dtξ−E′τ−E Dtτ−DtB=Lq⋅ξ+Lv⋅Dtξ+Lt τ−E Dtτ−DtB.

3.1givenstep 2.1algebra

The Noether identity, evaluated at (t,q(t),q′(t)), says DtB=Lt τ+Lq⋅ξ+Lv⋅Dtξ−Lv⋅q′ Dtτ+L Dtτ=Lt τ+Lq⋅ξ+Lv⋅Dtξ−E Dtτ, because E=Lv⋅q′−L and v=q′. Substituting this expression for DtB into step 2.1 gives C′=0 at every t∈I.

4.1F3step 3.1

By [F3] the function C is constant on I. This proves the conservation statement.

5.1givenstep 4.1algebra∎

Clause 2: at ξ=0, τ=1, B=0 one has Dtξ=0 and Dtτ=0, so the Noether identity reads Lt⋅1=Lt=DtB=0, i.e. it holds exactly when Lt=0 on O; the charge is C=Lv⋅0−E⋅1−0=−E, so E is conserved. This is the energy statement of The Hamiltonian on an inverse Legendre branch: differential, equivalence with the Euler-Lagrange equation, and the energy identity clause 3 in the autonomous case.

Remarks

  • Dimensions of the generators. If a one-parameter family of transformations is parametrised by a parameter s of physical dimension [s], then τ has dimension time/[s], ξ has dimension [q]/[s], B has dimension action/[s], and the charge C=p⋅ξ−Eτ−B has dimension action/[s]. Rescaling the generator by a constant rescales the charge by the same constant. These dimensional remarks are bookkeeping for applications; the theorem itself is the unconditional statement proved above, and no physical interpretation of the dimensions is asserted.

  • Orientation. The local transformations are meant to act on the same oriented path space, which for an infinitesimal family with time change requires dtnew/dt>0; the present theorem only uses the infinitesimal identity and the charge, and the global interpretation of the family is not needed for the conservation statement.

  • Scope. The identity is a hypothesis on the chart data. It does not follow from the existence of a kinematic transformation of the configuration space; a chosen potential may break it. The concrete charges and the failure of the symmetry for a quadratic potential are recorded in Conserved charges for translation, rotation and boost symmetries of a mechanical Lagrangian and Scope of point symmetries: symmetries belong to the Lagrangian, not to the configuration space, and time translation is the energy case.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

The charge of an infinitesimal point symmetry is constant along Euler-Lagrange solutions

Statement

Let (O,L) be a time-dependent Lagrangian chart over [a,b]×U, let (ξ,B) be an infinitesimal point symmetry of L with charge C(t,q,v)=Lv(t,q,v)⋅ξ(q)−B(t,q) (Infinitesimal point symmetry of a time-dependent Lagrangian and its conserved charge), and let q:I→U be a C2 curve on an open interval I⊇[a,b] with (t,q(t),q′(t))∈O for all t∈I that solves the Euler–Lagrange equation

EL(q)(t)=0(t∈I),

i.e. ddtLv(t,q,q′)=Lq(t,q,q′) (A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations, Stationarity of the time-dependent action is equivalent to the vanishing of the Euler-Lagrange expression). Then the function

t⟼C(t,q(t),q′(t))=Lv(t,q(t),q′(t))⋅ξ(q(t))−B(t,q(t))

is constant on I.

Facts & Assumptions

Given: The chart data (O,L), an infinitesimal point symmetry (ξ,B), and a C2 Euler–Lagrange solution q:I→U.

[F1]

The symmetry identity of Infinitesimal point symmetry of a time-dependent Lagrangian and its conserved charge: Lq(t,q,v)⋅ξ(q)+Lv(t,q,v)⋅Dξ(q)v=Bt(t,q)+Bq(t,q)⋅v for all (t,q,v)∈O, and the Euler–Lagrange equation ddtLv(t,q,q′)=Lq(t,q,q′) along q.

[F2]

Chain rule and algebra of differentiable curves (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a), The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, Intervals of R: the nine order-convex forms, nondegeneracy, and length): the curve t↦ξ(q(t)) is C1 with derivative Dξ(q(t))q′(t), and the curve t↦B(t,q(t)) is C1 with derivative Bt(t,q(t))+Bq(t,q(t))⋅q′(t).

Proof

Proof technique: direct; differentiate the charge along the solution and substitute the symmetry identity.

1.1F1F2given

The charge C(t)=C(t,q(t),q′(t)) is differentiable on I, because Lv, ξ, B and q are C1. By [F2], C′(t)=(ddtLv(t,q,q′))⋅ξ(q(t))+Lv(t,q,q′)⋅Dξ(q(t))q′(t)−Bt(t,q(t))−Bq(t,q(t))⋅q′(t).

2.1F1step 1.1algebra

Substituting the Euler–Lagrange equation of [F1] in the first term and the symmetry identity with (t,q,v)=(t,q(t),q′(t)) in the remaining three terms gives C′(t)=0 for every t∈I.

3.1F3step 2.1∎

Since C is differentiable with C′=0, [F3] makes C constant on I.

Remarks

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

Stationarity of the time-dependent action is equivalent to the vanishing of the Euler-Lagrange expression

Statement

Let (O,L) be a time-dependent Lagrangian chart over [a,b]×U in the sense of A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations and let q:[a,b]→U be a C2 admissible curve. Then q is stationary for the action S if and only if

EL(q)(t)=0for every t∈[a,b].

Facts & Assumptions

Given: The chart data (O,L), a C2 curve q:[a,b]→U, and the notation of A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations: the action S, the first variation δS(q)[η] over variations η, the gradients Lq,Lv and the Euler–Lagrange expression EL(q).

[F1]

First variation formula and continuity (The first variation of a time-dependent action on a finite interval: admissibility, the integral formula and continuity of the Euler-Lagrange coefficient): for every variation η the value δS(q)[η] exists and equals ∫ab[Lq(t,q,q′)⋅η+Lv(t,q,q′)⋅η′]dt, and EL(q) is continuous on [a,b].

[F2]

The bumps of Explicit twice differentiable bumps detect nonzero values of continuous functions: for every open interval J, continuous E:J→R and t1∈J with E(t1)≠0 there are ρ>0 with [t1−ρ,t1+ρ]⊆J and a C2 function β:R→R vanishing outside [t1−ρ,t1+ρ], 0≤β≤1 and ∫t1−ρt1+ρEβ≠0.

[F3]

Integration by parts (If u,v are differentiable on [a,b] with u′,v′ integrable, then ∫abuv′=u(b)v(b)−u(a)v(a)−∫abu′v): if u,v:[a,b]→R are differentiable at every point with integrable derivatives, then ∫abuv′=u(b)v(b)−u(a)v(a)−∫abu′v. Here u:=(Lv(t,q(t),q′(t)))i is C1 and v:=ηi is C2 for a variation η, so both derivatives are continuous, hence integrable (A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion, The lower and upper Darboux integrals of a bounded f on [a,b] as sup⁡PL(f,P) and inf⁡PU(f,P), Darboux integrability as their equality, and the notation ∫abf).

[F4]

Linearity of the integral (Integrable functions on [a,b] form a set closed under sums and scalar multiples, and ∫ab(λf+μg)=λ∫abf+μ∫abg) permits finite sums and fixed real coefficients to pass through the integral. A map on an open interval is C2 precisely when its scalar components are C2 (Ck Euclidean maps and diffeomorphisms). In step 1.3 each component of the extended variation is either the given C2 bump or the zero function, whose derivatives are zero by The derivative f′(c)=lim⁡x→cf(x)−f(c)x−c of f:A→R at a point c∈A that is a limit point of A, and differentiability on a set; hence that variation has a C2 extension as required by A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations.

[F5]

Continuity of EL(q) at a point; if EL(q)(t0)≠0 then for small δ>0 the expression is nonzero throughout (t0−δ,t0+δ)∩[a,b] (Continuity of f:A→R at a point of A and on A: the ε-δ condition, its agreement with lim⁡x→cf(x)=f(c) at a limit point, and continuity at an isolated point, Absolute value in an ordered field).

Proof

technique · direct; one direction is integration by parts, the other localises a nonzero component of the coefficient with a bump and produces a nonzero first variation
1.1F1F3F4given

Assume EL(q)=0 on [a,b] and let η be any variation. By [F1], δS(q)[η]=∑i=0n−1∫ab[(Lq)iηi+(Lv)iηi′]dt. For each i, [F3] applied to u=(Lv)i and v=ηi gives ∫ab(Lv)iηi′=−∫abddt(Lv)i ηi, because ηi(a)=ηi(b)=0. Hence each summand equals ∫ab((Lq)i−ddt(Lv)i)ηi dt, and by [F4] and the hypothesis all summands vanish. Thus δS(q)[η]=0; as η was arbitrary, q is stationary.

1.2F5given

Conversely, suppose EL(q)(t0)≠0 for some t0∈[a,b]. By [F5] the continuous expression EL(q) is nonzero at a point t1 interior to [a,b]: if t0∈(a,b) take t1:=t0; if t0=a, choose t1∈(a,min⁡(a+δ,b)) provided by [F5] applied at a (the interval is nonempty because a<b), and symmetrically for t0=b. Choose a component i with (EL(q)(t1))i≠0.

1.3F2F4

Apply [F2] to the open interval J:=(a,b), the continuous function E:=(EL(q))i and the point t1: this gives ρ>0 with [t1−ρ,t1+ρ]⊆(a,b) and a C2 bump β vanishing outside [t1−ρ,t1+ρ] with ∫Eβ≠0. Define η:[a,b]→Rn by ηi:=β and ηj:=0 for j≠i on [a,b], extended to an open interval around [a,b] by the same formula. Then η is a variation and, because the support of β lies inside (a,b), it vanishes near both endpoints.

2.1F1F2step 1.1step 1.3algebra

Evaluate the first variation of η: by [F1] and the reduction of step 1.1 applied to this single component, δS(q)[η]=∫ab[(Lq)iβ+(Lv)iβ′]dt=∫abE(t)β(t) dt≠0. Hence q is not stationary. Contraposition gives: if q is stationary, then EL(q)(t0)=0 for every t0∈[a,b].

3.1step 1.1step 2.1∎

Steps 1.1 and 2.1 prove the two implications of the statement.

Remarks

5 · Examples, experiments and counterexamples

None yet.

Sources