Hamilton's principle/Mathematics/The Legendre condition

Plain-language version: Is it really a minimum

The Legendre condition is the local necessary condition for a minimum: the coefficient of $\eta {\text{'}}^2$ in the second variation must not be negative anywhere on the extremal. It is cheap to check and it is the first thing to check. It is also decisively insufficient, and this page is as much about the gap it leaves as about the condition itself.

Statement

A needle variation.

Theorem (Legendre's necessary condition). If $y$ is a weak local minimum of $J$, then

$P(x)=L_{y^{\prime }{y}^{\prime }}(x,y(x),y^{\prime }(x))\ge 0\mathrm{(}\mathrm{1}\mathrm{)}$

for every $x\in [a,b]$. For a maximum the inequality reverses.

The strengthened Legendre condition is the strict form $P(x)>0$ on $[a,b]$. It is the hypothesis every sufficiency theorem in this track assumes.

Derivation

The four conditions of the sufficiency theory.

Suppose $P(x_0)<0$ at some interior $x_0$. Build a variation that makes the second variation negative.

By continuity there is $\delta >0$ with $P(x)<\frac{1}{2}P(x_0)<0$ on $I=(x_0-\delta ,x_0+\delta )\subset (a,b)$. For $0<h\le \delta$ define the needle variation

${\eta }_h(x)=\{\begin{array}{cc}[h^2-(x-x_0{)}^2{]}^2/h^3,& |x-x_0|\le h\\ 0,& \text{otherwise}\end{array}\mathrm{(}\mathrm{2}\mathrm{)}$

This is admissible: it is $C^2$, vanishes outside a subinterval of $(a,b)$, and therefore vanishes at both endpoints.

The scaling is the point of the construction. Direct integration gives

$\int {\eta }_h^{2}dx=O(h^{-1}),\int {\eta }_h{\text{'}}^{2}dx=O(h^{-3})\mathrm{(}\mathrm{3}\mathrm{)}$

so as $h\to 0$ the derivative term grows two powers of $h$ faster than the amplitude term. In the Legendre form of the second variation,

${\delta }^{2}J[y;{\eta }_h]={\int }_I[P{\eta }_h{\text{'}}^2+Q{\eta }_h^2]dx\mathrm{(}\mathrm{4}\mathrm{)}$

the first integral is bounded above by $\frac{1}{2}P(x_0)\int {\eta }_h{\text{'}}^{2}dx$, which is negative and of order $h^{-3}$. The second is bounded in magnitude by $max|Q|\int {\eta }_h^{2}dx$, of order $h^{-1}$. For $h$ small enough the first dominates and

${\delta }^{2}J[y;{\eta }_h]<0\mathrm{(}\mathrm{5}\mathrm{)}$

contradicting the necessary condition (7) of the second variation. Hence $P\ge 0$ on the interior, and at the endpoints by continuity. $◼$

The mechanism in one sentence: a narrow, tall-sloped wiggle changes the path very little but its derivative enormously, so wherever $P$ is negative there is a variation that lowers $J$.

Reading the condition

$L_{y^{\prime }{y}^{\prime }}\ge 0$ says $L$ is convex in its $y^{\prime }$ slot along the extremal. Convexity in the velocity is exactly what makes going fast expensive, and a functional that rewards high speed cannot have a minimum: the wiggle exploits it.

Three connected facts, all consequences of the same coefficient:

$P=L_{y^{\prime }{y}^{\prime }}$	consequence	where treated
$P<0$ somewhere	no minimum	this page
$P=0$ somewhere	Euler-Lagrange degenerates; corners possible	the Euler-Lagrange equation
$P>0$ throughout	equation is genuinely second order; corners excluded	Boundary terms and natural boundary conditions

The middle row connects to the Weierstrass-Erdmann corner conditions: a corner requires $P=0$, so the strengthened condition rules corners out. In several variables $P$ becomes the Hessian matrix $W_{ik}={\partial }^{2}L/\partial {y_i}^{\prime }\partial {y_k}^{\prime }$ of Several dependent variables, and (1) becomes positive semidefiniteness of $W$.

What it does not buy

The strengthened condition $P>0$ is not sufficient, and the counterexample is already on the record. On the second variation the integrand $L=\frac{1}{2}(y{\text{'}}^2-k^2{y}^2)$ has $P=1>0$ everywhere, and yet

${\delta }^{2}J=\frac{b}{2}(\frac{{\pi }^2}{b^2}-k^2)\mathrm{(}\mathrm{6}\mathrm{)}$

is negative for $b>\pi /k$. Legendre's condition holds, uniformly and strictly, on an interval where the extremal is not a minimum.

The reason is visible in the derivation. The needle argument is local: it probes one point at a time and therefore detects only pointwise failures. When $Q<0$, the reward accumulates over the interval and can defeat the $P$ penalty using a variation spread over the whole domain — a variation the needle construction never builds.

So the condition is blind in exactly one direction, and it is a direction that matters. Detecting the global failure requires the accessory equation, which is the Jacobi condition and conjugate points.

A historical caution

Legendre proposed in 1786 that the strengthened condition was sufficient. Lagrange identified the error. The gap stood until Jacobi supplied the missing condition in 1837, and Weierstrass completed the picture for strong minima later in the century.

The mistake is instructive rather than embarrassing: it is precisely the finite-dimensional analogy, in which a positive definite Hessian does settle the question. The failure of that analogy on a function space is the reason the sufficiency theory has three named conditions instead of one.

Provenance: the attributions in this section are editorial summary. No primary source for Legendre 1786 or Jacobi 1837 has been verified for this volume; the Jacobi row of the survey on Sources is UNCOMPUTED. No equation on this page is attributed to either.

Where the three conditions stand

condition	type	status
Euler-Lagrange	local, first order	necessary
Legendre, $P\ge 0$	local, second order	necessary
Jacobi, no conjugate point in $(a,b)$	global, second order	necessary
all three, strengthened		sufficient for a weak minimum
Weierstrass excess function	strong	needed for a strong minimum

Only the fourth line is a sufficiency statement, and it is confined to weak minima. The last line is the Weierstrass excess function.

Next

the Jacobi condition and conjugate points supplies the global condition and the geometric counterexample.