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An explicit saddle of the oscillator action at T=2π: the two variations sin⁡(t/2) and sin⁡(3t/2)

Example

Fix ω=1 and T=2π in the chart L(q,v)=12(v2−q2) on O=R×R×R over [0,2π]×R (A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations), and consider the variations

η1(t)=sin⁡t2,η3(t)=sin⁡3t2

of the zero curve q0≡0. The proposition Second variation of the oscillator action on a finite interval: both signs for T>π/ω and the value at T=π/ω gives

δ2S(q0)[η1,η1]=−3π4<0,δ2S(q0)[η3,η3]=5π4>0,

and with the exact identity S(εη)=ε22δ2S(q0)[η,η] of that proposition,

S(εη1)=−3πε28<0<5πε28=S(εη3)(ε≠0).

Thus at the duration T=2π>π=π/ω the stationary zero curve has variations that decrease and variations that increase the action arbitrarily close to it: it is a saddle, and neither a local minimiser nor a local maximiser. Every number below is exact; no measurement or numerical integration is involved.

Facts & Assumptions

Given: The chart L(q,v)=12(v2−q2) over [0,2π]×R, the zero curve q0, the variations η1,η3, and the action S and second variation δ2S 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]

Second variation values and exact action (Second variation of the oscillator action on a finite interval: both signs for T>π/ω and the value at T=π/ω, The second variation of a time-dependent action: twice differentiability of the fixed-endpoint family and the Hessian integral formula): for this chart with parameters ω=1, T=2π one has δ2S(q0)[ηn,ηn]=T2((nπ/T)2−ω2) for the variations ηn(t)=sin⁡(nπt/T), and S(εηn)=ε22δ2S(q0)[ηn,ηn] for every real ε.

Verification

Proof technique: direct; substitute T=2π, ω=1 into the proposition and evaluate the defining integrals independently.

1.1givenL2algebra

The variations: η1 and η3 are C∞ on R with η1(0)=η1(2π)=0 and η3(0)=η3(2π)=0 by [L2], so both are variations of q0; their derivatives are η1′(t)=12cos⁡t2 and η3′(t)=32cos⁡3t2.

2.1step 1.1givenL1algebra

Second variations: substituting T=2π and ω=1 into [L1] gives δ2S(q0)[ηn,ηn]=π(n2/4−1), hence δ2S(q0)[η1,η1]=π(1/4−1)=−3π4 and δ2S(q0)[η3,η3]=π(9/4−1)=5π4.

3.1step 2.1L2algebra

Independent integral check: by [L2], ∫02πη1′(t)2dt=14∫02π1+cos⁡t2dt=14⋅π,∫02πη1(t)2dt=∫02π1−cos⁡t2dt=π, so δ2S(q0)[η1,η1]=π/4−π=−3π4; and with η3′(t)2=94cos⁡23t2 the same computation gives ∫02πη3′2=94⋅π and ∫02πη32=π, so δ2S(q0)[η3,η3]=9π4−π=5π4, in agreement with step 2.1.

4.1step 2.1step 3.1L1algebra∎

Action values and saddle: by the exact identity in [L1], S(εη1)=ε22⋅(−3π4)=−3πε28<0 and S(εη3)=5πε28>0 for every ε≠0, while S(q0)=0; since both families tend to q0 as ε→0, the stationary zero curve is neither a local minimiser nor a local maximiser, as claimed.

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