The Jacobi condition and conjugate points

Plain-language version: Where the minimum fails

The Jacobi condition is the global second-order condition that Legendre's local condition cannot supply. It says that an extremal ceases to be a minimum once it extends past the first point conjugate to its start. This page carries the volume's worked counterexample to the identification of stationary with least, and it is the reason this volume is not titled with the word least.

The accessory equation

Geodesics from a pole refocusing at the antipode.

From the second variation, the second variation in Legendre form is ${\int }_a^b(P\eta {\text{'}}^2+Q{\eta }^2)dx$, and its own Euler-Lagrange equation is the Jacobi accessory equation

$\frac{d}{dx}(P(x){\eta }^{\prime })-Q(x)\eta =0\mathrm{(}\mathrm{1}\mathrm{)}$

a linear, homogeneous, second-order ordinary differential equation whose coefficients are fixed by the extremal under test. Assume throughout that the strengthened Legendre condition $P>0$ holds, so (1) is genuinely second order.

Definition. A point $c\in (a,b]$ is conjugate to $a$ if some solution $u$ of (1), not identically zero, satisfies $u(a)=0$ and $u(c)=0$.

Since (1) is linear and second order, the solution with $u(a)=0$ is unique up to scale, so conjugacy is a property of the extremal and not of a choice.

Why zeros of the accessory equation matter

Geodesic behaviour on flat, spherical and saddle surfaces.

Suppose $c$ is conjugate to $a$ with $a<c<b$, and let $u$ be the corresponding solution. Define

${\eta }^{\ast }(x)=\{\begin{array}{cc}u(x),& a\le x\le c\\ 0,& c<x\le b\end{array}\mathrm{(}\mathrm{2}\mathrm{)}$

${\eta }^{\ast }$ is continuous, vanishes at both ends, and is admissible in the piecewise-smooth class. Evaluate the second variation on it. On $[c,b]$ it contributes nothing. On $[a,c]$, integrate by parts:

${\int }_a^c(Pu{\text{'}}^2+Q{u}^2)dx=[Pu{u}^{\prime }{]}_a^c-{\int }_a^{c}u[\frac{d}{dx}(P{u}^{\prime })-Qu]dx\mathrm{(}\mathrm{3}\mathrm{)}$

The integral on the right vanishes because $u$ solves (1). The boundary term vanishes because $u(a)=u(c)=0$. Therefore

${\delta }^{2}J[y;{\eta }^{\ast }]=0\mathrm{(}\mathrm{4}\mathrm{)}$

The second variation is degenerate: a nonzero admissible variation makes it vanish. The extremal sits on the boundary between the positive and negative cases, and it can be shown that the corner at $c$ can be rounded off to give a smooth variation with ${\delta }^{2}J<0$ strictly. So beyond the conjugate point the second variation takes negative values and the extremal is not a minimum.

Theorem (Jacobi's necessary condition). If $y$ is a weak local minimum and $P>0$, then no point of the open interval $(a,b)$ is conjugate to $a$.

Theorem (sufficiency). If $y$ satisfies the Euler-Lagrange equation, $P>0$ on $[a,b]$, and no point of $(a,b]$ is conjugate to $a$, then $y$ is a weak local minimum.

The two differ only at the closed endpoint, and that half-open gap at $b$ itself is genuinely undecided by these conditions alone.

The analytic example, revisited

Rays crossing a focus.

Take $L=\frac{1}{2}(y{\text{'}}^2-k^2{y}^2)$ and the extremal $y\equiv 0$, as on the second variation. Then $P=1$, $Q=-k^2$, and (1) becomes

${\eta }^{″}+k^2\eta =0\mathrm{(}\mathrm{5}\mathrm{)}$

with solution $u=\sin (kx)$ for $u(0)=0$. Its next zero is at $x=\pi /k$. So the first conjugate point is $c=\pi /k$, which is exactly the interval length at which the direct computation gave ${\delta }^{2}J=0$.

The two calculations agree, and they were done by different routes: one by testing a chosen function, the other by solving a differential equation. The agreement is the check that the accessory machinery describes the phenomenon rather than merely accompanying it.

The geometric counterexample: geodesics on a sphere

Take the unit sphere. Its geodesics are the great circles, and every great-circle arc satisfies the geodesic equation — every one is stationary for arc length. The question is which of them are shortest.

Let $\theta$ be arc length along a great circle from a starting point $A$. The accessory equation for arc length on a unit sphere is the Jacobi equation of a space of curvature $K=1$,

$u^{″}+Ku=0⟹u^{″}+u=0\mathrm{(}\mathrm{6}\mathrm{)}$

so $u=\sin \theta$ for $u(0)=0$, and the first conjugate point is at $\theta =\pi$ — the antipode of $A$.

Read off the three regimes:

arc length	conjugate point passed	status of the arc
$\theta <\pi$	no	stationary and the unique shortest path
$\theta =\pi$	at the endpoint	stationary; infinitely many arcs of equal length
$\theta >\pi$	yes	stationary and not shortest

The middle row is where the degeneracy is visible without any calculation. From the north pole to the south pole every meridian has length $\pi$. All of them are stationary, all have the same length, and the solution is not unique. That non-uniqueness is the geometric face of equation (4).

The bottom row is the counterexample proper. Take an arc running from $A$ the long way round to a point $B$ just short of a full revolution, of length $\theta =2\pi -\epsilon$. It satisfies the geodesic equation at every point. It is stationary. And it is obviously not the shortest route from $A$ to $B$: the short arc, of length $\epsilon$, joins the same two points along the same great circle.

Nothing local distinguishes the two arcs. They satisfy the same equation, they have the same $P=1>0$, and no pointwise test can tell them apart. Only the interval length differs, and only the Jacobi condition sees it.

This is the worked counterexample that the second variation promised and that the volume's claim rests on. Stationary means the first variation vanishes. It does not mean least, and the sphere shows the difference without any appeal to exotic integrands.

Geometric reading

The accessory equation $u^{″}+Ku=0$ is the equation of geodesic deviation: $u$ measures the separation of two neighbouring geodesics leaving $A$ in slightly different directions. Positive curvature makes them converge, and a conjugate point is where they meet again.

That gives the general picture. Conjugate points are refocusing points of the family of extremals through $A$. Before the first one, the extremal is embedded in a field of non-crossing neighbouring extremals and is minimising. At it, the field degenerates. Past it, extremals cross and minimality fails.

Three consequences follow, and each recurs later in the volume:

• Flat and negatively curved spaces have no conjugate points. $u^{″}-|K|u=0$ has no second zero, so geodesics minimise indefinitely. Straight lines in the plane are the case worked on the Euler-Lagrange equation, and the minimality claim deferred there is now discharged.
• In optics, conjugate points are foci. The optical-mechanical analogy makes this literal rather than metaphorical: a lens image point is a conjugate point of the object point.
• In mechanics, past a conjugate point the action is stationary but not minimal. This is why the classical action in the Feynman path integral carries a phase correction counting the conjugate points crossed.

Relation to the field of extremals

The sufficiency proof works by embedding the extremal in a field: a family of extremals covering a neighbourhood, exactly one through each point. Such a field exists precisely up to the first conjugate point, which is the analytic content of the geometric picture above. The field construction is also what the Weierstrass excess function needs, and what connects this material to the Hamilton-Jacobi equation, where the action as a function of endpoints becomes singular at conjugate points.

Next

the Weierstrass excess function closes the track by treating strong minima, where the comparison class admits paths with wildly different slopes.