Hamilton's principle/Mathematics/The second variation

Plain-language version: Is it really a minimum

Every result so far has been necessary and none has been sufficient. The Euler-Lagrange equation says the first variation vanishes; it says nothing about whether the extremal is a minimum, a maximum, or neither. The second variation is the tool that begins to answer that, and this page is where the volume's central distinction — stationary against least — acquires machinery rather than assertion.

Definition

The second variation changing sign as the interval lengthens.

With $\Phi (\epsilon )=J[y+\epsilon \eta ]$ as on Hamilton's principle/Mathematics/The first variation, the second variation is

${\delta }^{2}J[y;\eta ]={\Phi }^{″}(0)\mathrm{(}\mathrm{1}\mathrm{)}$

Taylor's theorem with the first variation vanishing gives

$J[y+\epsilon \eta ]-J[y]=\frac{1}{2}{\epsilon }^2{\delta }^{2}J[y;\eta ]+o({\epsilon }^2)\mathrm{(}\mathrm{2}\mathrm{)}$

So the sign of ${\delta }^{2}J$ controls the comparison, subject to the error term — and that qualification is where the difficulty lives.

Computing it

Local penalty against accumulated reward.

Differentiate under the integral twice and set $\epsilon =0$:

${\delta }^{2}J[y;\eta ]={\int }_a^b[L_{yy}{\eta }^2+2L_{y{y}^{\prime }}\eta {\eta }^{\prime }+L_{y^{\prime }{y}^{\prime }}\eta {\text{'}}^2]dx\mathrm{(}\mathrm{3}\mathrm{)}$

with all second partials evaluated along the extremal. This is a quadratic functional of $\eta$ — the analogue of the Hessian in finite dimensions.

The cross term can be removed. Since $2\eta {\eta }^{\prime }=({\eta }^2{)}^{\prime }$, integrate the middle term by parts; the boundary term vanishes because $\eta$ does, and

${\int }_a^{b}2L_{y{y}^{\prime }}\eta {\eta }^{\prime }dx=-{\int }_a^b\frac{d}{dx}\left(L_{y{y}^{\prime }}\right){\eta }^{2}dx\mathrm{(}\mathrm{4}\mathrm{)}$

Substituting into (3) gives the Legendre form

${\delta }^{2}J[y;\eta ]={\int }_a^b[P\eta {\text{'}}^2+Q{\eta }^2]dx\mathrm{(}\mathrm{5}\mathrm{)}$

with

$P=L_{y^{\prime }{y}^{\prime }},Q=L_{yy}-\frac{d}{dx}{L}_{y{y}^{\prime }}\mathrm{(}\mathrm{6}\mathrm{)}$

Two terms, two coefficients, both determined by the extremal.

Necessary conditions, and the trap

If $y$ is a weak local minimum then $\Phi$ has a minimum at $\epsilon =0$ for every $\eta$, so

${\delta }^{2}J[y;\eta ]\ge 0\text{for all }\eta \in 𝒱\mathrm{(}\mathrm{7}\mathrm{)}$

This is necessary. The temptation is to reverse it and conclude that ${\delta }^{2}J>0$ for all $\eta \ne 0$ makes $y$ a minimum, by analogy with the finite-dimensional second derivative test.

That inference is false in infinite dimensions. The reason is the error term in (2). In finite dimensions a positive definite Hessian is bounded below by $c\Vert v{\Vert }^2$ with $c>0$, which dominates the remainder. On a function space, positivity of a quadratic functional does not give such a bound: the infimum of ${\delta }^{2}J[y;\eta ]$ over $\Vert \eta \Vert =1$ can be zero without being attained. The remainder is then not dominated and no conclusion follows.

Repairing this is the whole content of the sufficiency theory. It needs a strengthened positivity, and that is what the Legendre and Jacobi conditions supply.

What the two coefficients contribute

$P$ and $Q$ enter (5) very differently, and seeing why motivates the next two pages.

$P$ is local and dominant. ${\eta }^{\prime }$ can be made enormous while $\eta$ stays small — a rapid, low-amplitude wiggle. Any such $\eta$ makes the $P\eta {\text{'}}^2$ term overwhelm the $Q{\eta }^2$ term. So if $P<0$ anywhere, a wiggle localised there drives ${\delta }^{2}J$ negative and (7) fails. This yields $P\ge 0$ as a necessary condition, derived properly on Hamilton's principle/Mathematics/The Legendre condition.

$Q$ is global and accumulative. With $P>0$ the first term is a penalty on bending and the second may be a reward, if $Q<0$. Over a short interval the penalty wins. Over a long enough interval the accumulated reward can win, and ${\delta }^{2}J$ turns negative even though $P>0$ throughout. The length at which this happens is set by conjugate points, on Hamilton's principle/Mathematics/The Jacobi condition and conjugate points.

Interval length is therefore part of the answer. The same extremal can be a minimum on $[a,b]$ and fail to be one on $[a,b^{\prime }]$ with $b^{\prime }>b$, with no change to $L$ and no change to the equation it satisfies. Nothing in the Euler-Lagrange equation sees this.

The accessory problem

Equation (5) is itself a variational problem, in the unknown $\eta$. Its Euler-Lagrange equation is

$\frac{d}{dx}\left(P{\eta }^{\prime }\right)-Q\eta =0\mathrm{(}\mathrm{8}\mathrm{)}$

called the Jacobi accessory equation. Analysing the second variation thus means solving a linear second-order equation derived from the original extremal — a striking reduction, and the reason the sufficiency theory is tractable at all.

Solutions of (8) are exactly the directions in which ${\delta }^{2}J$ is stationary, and its zeros are the conjugate points.

A first sight of the failure

Take $L=\frac{1}{2}(y{\text{'}}^2-k^2{y}^2)$ on $[0,b]$ with $y\equiv 0$ as the extremal, endpoints zero. Then $P=1$ and $Q=-k^2$, so

${\delta }^{2}J[0;\eta ]={\int }_0^b(\eta {\text{'}}^2-k^2{\eta }^2)dx\mathrm{(}\mathrm{9}\mathrm{)}$

Test with $\eta =\sin (\pi x/b)$, which vanishes at both ends. Using ${\int }_0^b{\sin }^2(\pi x/b)dx={\int }_0^b{\cos }^2(\pi x/b)dx=b/2$,

${\delta }^{2}J=\frac{b}{2}(\frac{{\pi }^2}{b^2}-k^2)\mathrm{(}\mathrm{10}\mathrm{)}$

This is positive for $b<\pi /k$ and negative for $b>\pi /k$. The extremal, the integrand and the Legendre coefficient $P=1>0$ are identical in both cases. Only the interval length changed.

At $b=\pi /k$ exactly, (10) vanishes for this $\eta$: the second variation is degenerate, and $x=\pi /k$ is the point conjugate to the origin. Beyond it the zero solution is stationary and is not a minimum.

Equation (10) is the volume's first exhibited demonstration that stationary does not imply least. It is a computation, not an assertion, and the geometric version of the same phenomenon — a great-circle arc past its antipode — is worked on Hamilton's principle/Mathematics/The Jacobi condition and conjugate points.

Second variation with constraints

For constrained problems the second variation is that of the augmented integrand of Hamilton's principle/Mathematics/Lagrange multipliers, evaluated with the multipliers at their solution values, and it is tested only against variations satisfying the linearised constraint. Both restrictions matter; dropping either gives the wrong sign test.

Next

Hamilton's principle/Mathematics/The Legendre condition derives the local condition $P\ge 0$ and shows exactly how much it does and does not buy.