# Algebra & Analysis, Problems & Solutions by G Lefort

By G Lefort

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Note, however, that if Q(h) is only positive semidefinite, then it is possible that b is conjugate to a and this gives rise to a conjugate point (b, x∗ (b)). This happens if the Jacobi equation has a nontrivial solution that vanishes at t = b. 3 (Necessary conditions for a weak local minimum). Suppose x∗ : [a, b] → R is a weak local minimum for problem [CV]. Then 1. , satisfies the Euler–Lagrange equation d dt ∂L ∂L (t, x∗ (t), x˙∗ (t)) = (t, x∗ (t), x˙∗ (t)); ∂ x˙ ∂x 2. 4 The Legendre and Jacobi Conditions 35 3.

Since y(s, ˙ a) = 1, there exists an ε > 0 such that we actually have Z ⊂ [0, 1] × [a + ε , b]. 4 The Legendre and Jacobi Conditions Fig. 10 The zero set Z 33 t=b (s0,t0) t=a s=0 s=1 (For every s ∈ [0, 1] there exists a neighborhood Us of (s, a) in [0, 1] × [a, b] such that y(s,t) > 0 for (s,t) ∈ Us and t > 0. ) If (s0 ,t0 ) ∈ Z , then y(s ˙ 0,t0 ) cannot vanish, since otherwise y(s, ·) vanishes identically in t as a solution to a second-order linear differential equation that vanishes with its derivative at t0 .

1 (Legendre condition). If x∗ is a weak local minimum for problem [CV] (see Fig. 9), then Lx˙x˙ (t, x∗ (t), x˙∗ (t)) ≥ 0 for all t ∈ [a, b]. Proof. This condition should be clear intuitively. 8 1 Fig. 9 The variation for the proof of the Legendre condition the term multiplying h˙ 2 will become dominant and thus needs to be nonnegative. We prove this by contradiction. Suppose there exists a time τ ∈ (a, b) at which Lx˙x˙ (τ , x∗ (τ ), x˙∗ (τ )) = −2β < 0. Choose ε > 0 such that Lx˙x˙ (t, x∗ (t), x˙∗ (t)) < −β for t ∈ [τ − ε , τ + ε ] ⊂ [a, b] and pick the function h(t) = sin2 0 π ε (t − τ ) for | t − τ |≤ ε , otherwise.