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The second variation of a time-dependent action: twice differentiability of the fixed-endpoint family and the Hessian integral formula

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 (so L is C2 and [a,b]×U×Rn⊆O), let q:[a,b]→U be a C2 curve and let η be a variation of q (a C2 map with η(a)=η(b)=0). Write ψ(ε,t):=(t,q(t)+εη(t),q′(t)+εη′(t)) and, for ∣ε∣<ε0 with ε0 as in clause 1 below,

H(ε,t):=∑j,k=1n[∂2L∂qj∂qk(ψ(ε,t))ηj(t)ηk(t)+2∂2L∂qj∂vk(ψ(ε,t))ηj(t)ηk′(t)+∂2L∂vj∂vk(ψ(ε,t))ηj′(t)ηk′(t)].

Then:

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

d2dε2S(q+εη)=∫abH(ε,t) dt;

  1. in particular the second variation of The fixed-endpoint second variation of a time-dependent action at a curve is

δ2S(q)[η,η]=∫abH(0,t) dt,

the formula of clause 2 with every second partial of L evaluated at (t,q(t),q′(t)).

Facts & Assumptions

Given: The chart data (O,L), a C2 curve q:[a,b]→U, a variation η, the map ψ(ε,t)=(t,q(t)+εη(t),q′(t)+εη′(t)), and the action S and second variation of A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations and The fixed-endpoint second variation of a time-dependent action at a curve.

[L1]

Admissibility: there is ε0>0 such that for ∣ε∣<ε0 the curve q+εη is a C2 admissible curve with values in U, hence ψ maps (−ε0,ε0)×[a,b] into O because [a,b]×U×Rn⊆O (The first variation of a time-dependent action on a finite interval: admissibility, the integral formula and continuity of the Euler-Lagrange coefficient, clause 1).

[L2]

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 is differentiable on (α,β) with derivative ∫abh(ε,t) dt.

∂∂ε[L(ψ)]=Lq(ψ)⋅η(t)+Lv(ψ)⋅η′(t),

∂∂ε[Lq(ψ)]j=∑k∂2L∂qj∂qk(ψ)ηk(t)+∑k∂2L∂qj∂vk(ψ)ηk′(t),

∂∂ε[Lv(ψ)]j=∑k∂2L∂vj∂qk(ψ)ηk(t)+∑k∂2L∂vj∂vk(ψ)ηk′(t),

and each displayed expression is continuous on every rectangle [−δ,δ]×[a,b] with 0<δ<ε0, the compositions of the continuous second partials with the continuous map ψ and the continuous functions η,η′ being continuous.

[L4]

Clairaut-Schwarz (Clairaut--Schwarz theorem for continuous second partial derivatives): the second partials of the C2 function L commute on O, so ∂2L/∂vj∂qk=∂2L/∂qk∂vj there.

[L6]

Uniform continuity and uniform integral errors (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, Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous, Integrable functions on [a,b] form a set closed under sums and scalar multiples, and ∫ab(λf+μg)=λ∫abf+μ∫abg, If f≤g on [a,b] and both are integrable then ∫abf≤∫abg; and m(b−a)≤∫abf≤M(b−a), If f,g are integrable on [a,b] then so are ∣f∣, f2, fg, max⁡(f,g) and min⁡(f,g), and ∣∫abf∣≤∫ab∣f∣): the closed bounded rectangle R=[−δ,δ]×[a,b] is compact, a continuous real function on a compact metric space is uniformly continuous; and if u,v are integrable on [a,b] with ∣u(t)−v(t)∣≤η for every t∈[a,b] and some η≥0, then u−v and ∣u−v∣ are integrable and ∣∫abu−∫abv∣=∣∫ab(u−v)∣≤∫ab∣u−v∣≤η(b−a), by linearity of the integral, the triangle inequality for the integral and the two-sided bound of monotonicity.

Proof

Proof technique: direct; apply differentiation under the integral sign twice on a compact rectangle, the first derivative being the first-variation integrand and the second being the Hessian quadratic form of L along the family.

1.1givenL1L3

Fix 0<δ<ε0 with ε0 from [L1], put R:=[−δ,δ]×[a,b], and define g(ε,t):=L(ψ(ε,t)) on R. By [L1] the values of ψ on R lie in O, so g is defined and continuous on R; by [L3] the ε-partial g1(ε,t)=Lq(ψ)⋅η(t)+Lv(ψ)⋅η′(t) exists on R and is continuous there.

2.1L2L5step 1.1

By [L2] applied to the continuous pair (g,g1) on R, the map F(ε):=S(q+εη)=∫abg(ε,t) dt is differentiable on (−δ,δ) with derivative F′(ε)=∫abg1(ε,t) dt.

2.2L2L3L5step 1.1

By [L3] the ε-partial of g1 exists on R and equals the four-term sum g2(ε,t)=[Lqq(ψ)η⋅η+Lqv(ψ)η′⋅η]+[Lvq(ψ)η⋅η′+Lvv(ψ)η′⋅η′], written with the block matrices Lqq,Lqv,Lvq,Lvv of second partials evaluated at ψ=ψ(ε,t); it is continuous on R. Applying [L2] to the pair (g1,g2) shows that ε↦∫abg1(ε,t) dt is differentiable on (−δ,δ) with derivative ∫abg2(ε,t) dt. Hence F is twice differentiable on (−δ,δ) with F′′(ε)=∫abg2(ε,t) dt, and since every 0<δ<ε0 is allowed, clause 2 holds for every ε∈(−ε0,ε0) once g2 is rewritten in the symmetric form H.

3.1L6L5step 2.2

Continuity of the second derivative: the function g2 is continuous on the compact rectangle R, so [L6] makes it uniformly continuous. Given ε~>0 choose γ>0 such that ∣g2(ε,t)−g2(ε′,t)∣<ε~/(b−a) for every t∈[a,b] whenever ε,ε′∈(−δ,δ) satisfy ∣ε−ε′∣<γ. Then for such ε,ε′, linearity of the integral and [L6] give ∣F′′(ε)−F′′(ε′)∣=∣∫ab(g2(ε,t)−g2(ε′,t)) dt∣≤ε~b−a (b−a)=ε~, so F′′ is continuous on (−δ,δ); since every 0<δ<ε0 is allowed, ε↦S(q+εη) is of class C2 on (−ε0,ε0), which is clause 1.

4.1L4step 2.2given∎

Symmetrisation: by [L4], for all j,k one has (Lvq(ψ))jk=∂2L/∂vj∂qk(ψ)=∂2L/∂qk∂vj(ψ)=(Lqv(ψ))kj, so η′⋅Lvq(ψ)η=η⋅Lqv(ψ)η′ after relabelling the summation indices; therefore g2=H and clauses 2 and 3 follow, the latter by the definition of δ2S(q)[η,η] and evaluation at ε=0.

Remarks

Depends on

Dependency tree · 2 levels

Sources