Math 2.0

Calculus of variations and optimal control; optimization Theorem

Jacobi condition

Builds on Integral functional with a Lagrangian densityWeak and strong extremumSecond variationJacobi equationEuler–Lagrange equation

Statement

Let a<ba < b and let L(x,y,p)L(x, y, p) be a C3C^3 function on [a,b]×ℝ×ℝ[a, b] \times \R \times \R. Let J(y)=∫abL(x,y,y′) dxJ(y) = \int_a^b L(x, y, y') \, dx on C1([a,b])C^1([a, b]), with fixed values y(a)y(a) and y(b)y(b). Let y0∈C2([a,b])y_0 \in C^2([a, b]) solve the Euler–Lagrange equation of JJ. Assume the strengthened Legendre condition Lpp(x,y0(x),y0′(x))>0L_{pp}(x, y_0(x), y_0'(x)) > 0 for all x∈[a,b]x \in [a, b]. A point c∈(a,b]c \in (a, b] is conjugate to aa if a solution h≠0h \ne 0 of the Jacobi equation along y0y_0 has h(a)=h(c)=0h(a) = h(c) = 0.

  1. (Necessary condition.) If y0y_0 is a weak local minimiser of JJ, then no point of the open interval (a,b)(a, b) is conjugate to aa.
  2. (Sufficient condition.) If no point of (a,b](a, b] is conjugate to aa, then y0y_0 is a strict weak local minimiser of JJ.

Remarks

Write the second variation as ∫ab(Ph′2+Qh2) dx\int_a^b (P h'^2 + Q h^2) \, dx, as in the Jacobi equation. Without conjugate points in (a,b](a, b], there is a Jacobi solution u>0u > 0 on all of [a,b][a, b]. Then w=−Pu′/uw = -P u' / u solves w′=w2/P−Qw' = w^2 / P - Q. Adding the zero term ∫ab(wh2)′ dx\int_a^b (w h^2)' \, dx turns the second variation into ∫abP(h′+wh/P)2 dx\int_a^b P (h' + w h / P)^2 \, dx, which is positive for h≠0h \ne 0. For the necessary part, a Jacobi solution on [a,c][a, c] extended by 00 gives zero second variation, and smoothing its corner makes the second variation negative.

Example: for J(y)=∫0b(y′2−y2) dxJ(y) = \int_0^b (y'^2 - y^2) \, dx and y0=0y_0 = 0, the Jacobi equation is h′′+h=0h'' + h = 0. The solution with h(0)=0h(0) = 0 is h=sin⁡xh = \sin x, so the first conjugate point is π\pi. For b<πb < \pi, the extremal y0=0y_0 = 0 is a strict weak local minimiser. For b>πb > \pi, it is not: with φ(x)=sin⁡(πx/b)\varphi(x) = \sin(\pi x / b) one gets J(tφ)=t2b2(π2/b2−1)<0J(t\varphi) = \frac{t^2 b}{2}(\pi^2 / b^2 - 1) < 0 for t≠0t \ne 0.

The strengthened Legendre condition is needed in both parts. Conjugate points here generalise conjugate points along geodesics in Riemannian geometry.

Going deeper

For a student

The Legendre condition controls the second variation for test functions that oscillate fast. The Jacobi condition controls it for all test functions, and it reduces this question to the zeros of one solution of a linear ODE. Together with the Euler–Lagrange equation and the strengthened Legendre condition, it gives the classical sufficient condition for a weak minimum. For a strong minimum one adds the Weierstrass condition on the excess function, and the proof embeds the extremal in a field of extremals.

Reading path

To understand this page from the ground up, read these 6 pages in order. The foundations come first.

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Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.
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  1. Euler–Lagrange equation concept
  2. Integral functional with a Lagrangian density concept
  3. Jacobi equation concept
  4. Second variation concept
  5. Weak and strong extremum concept
  6. Jacobi condition statement

Further reading

  1. Web https://en.wikipedia.org/wiki/Conjugate_points
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UsesIntegral functional with a Lagrangian density, Weak and strong extremum, Second variation, Jacobi equation, Euler–Lagrange equation

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