Jacobi fields and conjugate points

Take a scalar smooth stationary curve for a one-dimensional fixed-endpoint functional. Assume enough regularity, for example a three-times continuously differentiable integrand and a smooth curve, so the coefficients below are defined and the linearization is justified. In the second variation, integrate the mixed term by parts using zero endpoint perturbations. The quadratic form becomes the integral of $P(h')^2+Qh^2$, where $P=F_{vv}$ and $Q=F_{uu}-(F_{uv})'$, all evaluated along the stationary curve.

Applying the Euler–Lagrange procedure to this quadratic form gives the linear Jacobi equation. The same equation arises by differentiating a smoothly parameterized family of stationary curves with respect to its family parameter. Thus its solution is an infinitesimal displacement between neighboring extremals, not necessarily a finite difference that satisfies the original nonlinear equation exactly. The word field is also used in higher-dimensional geometric settings; here it is simply a scalar function along the chosen curve.

Assume $P$ is continuously differentiable and strictly positive on the interval, and $Q$ is continuous. Solve the Jacobi equation with $j(a)=0$ and $j'(a)=1$. Any later point where this nonzero solution vanishes is conjugate to the initial endpoint. Changing its nonzero initial derivative merely rescales the solution, so the zero locations do not depend on that normalization. At a conjugate terminal endpoint, the Jacobi field itself is an admissible zero direction of the quadratic form, as follows by multiplying its equation by the field and integrating.

Before the first conjugate point, a factorization explains positivity. Where the Jacobi solution does not vanish, write the quadratic form as a squared expression involving $h'-(j'/j)h$ plus an endpoint term. For smooth homogeneous perturbations the apparent singular term at the initial zero has a vanishing boundary limit. If the terminal value of the Jacobi solution is nonzero, equality in the square forces a proportionality incompatible with the terminal perturbation being zero. This is a positivity argument for the second variation; nonlinear minimality still needs the appropriate uniform estimates and remainder control.

For the quadratic action with density $[(u')^2-\omega^2u^2]/2$, the zero curve is stationary and the Jacobi equation is $j''+\omega^2j=0$. With the stated normalization its solution is $\sin(\omega x)/\omega$ when the initial endpoint is zero. The first positive conjugate point is $\pi/\omega$. The derivative coefficient remains one for every interval length, so a local Legendre condition cannot detect this global-in-the-interval change by itself.

Poincaré's sharp interval inequality makes the threshold exact: the second variation is bounded below by $[1-(\omega T/\pi)^2]\int(h')^2$ on an interval of length $T$. Below the threshold it is coercive; at the threshold a sine mode has zero energy; above it the first sine mode has negative energy. These are complete statements for this quadratic functional. In a nonlinear problem, a zero second variation alone cannot decide higher-order stability, and a geometric conjugate-point argument should not silently replace a full sufficient-condition theorem.