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Extra resources for Advanced Topics in Control Systems Theory: Lecture Notes from FAP 2005
Observe that the generalized Legendre-Clebsch condition is only satisﬁed in the hyperbolic case. It is now crucial to notice that since the reference curve is a one-dimensional manifold, we can normalize any independent family of Lie brackets to form a frame along it. Our assumptions allow us to pick coordinates preserving the previous normalizations and deﬁning a moving frame deﬁned by: adk F0 · F1 (γ(t)) = ∂ , k = 0, . . , n − 1, t ∈ [0, T ]. ∂xn−k Moreover, since the feedback is chosen so that u is zero along γ, we can impose the linearization condition adk F0 · F1 (γ(t)) = 0 for k > n − 2 and t ∈ [0, T ].
The ﬁrst conjugate time to 0, t1c , is the smallest t such that λt,1 = 0. If t < t1c , the only minimizer of Qt on Ct is y = 0. If t > t1c , the inﬁmum of Qt is −∞. Proof . Rather than using the standard Morse theory, we make a simple proof of the loss of optimality after the ﬁrst conjugate time based on the geometric argument of the Riemannian case . Indeed, let t1c be the ﬁrst conjugate time along the reference trajectory γ. There exists a Jacobi ﬁeld vertical at 0 and t1c corresponding to a variation of γ with δx(0) = δx(t1c ) = 0.
Therefore, P = (L , L) is a symplectic matrix and we have P −1 = − tLJ . t LJ If we make the symplectic change of coordinates x = P y, we get the Hamiltonian equation y˙ = P −1 (AP − P˙ )y and using the notation x˙ = Ax = JS where S is symmetric. Decomposing y = (u, v), we obtain the equations u˙ = 0 v˙ = − tL (SL + J L˙ )u and the solution can be computed by quadrature. Similar tensor analysis can be developed to study the standard LQ problem. 1 Introduction to Nonlinear Optimal Control 25 Geometric analysis of linear quadratic problems.