Hamilton's principle/Mathematics/Boundary terms and natural boundary conditions

Plain-language version: Free ends

The derivation on the first variation discarded a boundary term because the endpoints were held fixed. When an endpoint is free, that term survives and imposes its own condition on the extremal. These are the natural boundary conditions: conditions the variational problem generates by itself, rather than conditions imposed on it.

The term that was discarded

A free endpoint and its natural boundary condition.

Integration by parts gave

$\delta J[y;\eta ]=[L_{y^{\prime }}\eta {]}_a^b+{\int }_a^b[L_y-\frac{d}{dx}{L}_{y^{\prime }}]\eta dx\mathrm{(}\mathrm{1}\mathrm{)}$

With $\eta (a)=\eta (b)=0$ the first term vanishes. Now enlarge the admissible set. Fix only the left endpoint:

${𝒜}^{\ast }=\{y\in C^2[a,b]:y(a)=y_a\},{𝒱}^{\ast }=\{\eta \in C^2[a,b]:\eta (a)=0\}\mathrm{(}\mathrm{2}\mathrm{)}$

$\eta (b)$ is now unconstrained. Equation (1) becomes

$\delta J[y;\eta ]=L_{y^{\prime }}(b,y(b),y^{\prime }(b))\eta (b)+{\int }_a^b[L_y-\frac{d}{dx}{L}_{y^{\prime }}]\eta dx\mathrm{(}\mathrm{3}\mathrm{)}$

Extracting two conditions from one requirement

The boundary term left by integration by parts.

Stationarity requires (3) to vanish for every $\eta \in {𝒱}^{\ast }$. The argument splits in two stages, and the order matters.

Stage 1. ${𝒱}^{\ast }$ contains $𝒱$ as a subset. Restrict attention to those $\eta$ that do vanish at $b$. For these the boundary term is absent and (3) reduces to the fixed-endpoint case, so the fundamental lemma applies and

$L_y-\frac{d}{dx}{L}_{y^{\prime }}=0\mathrm{(}\mathrm{4}\mathrm{)}$

on $[a,b]$. The Euler-Lagrange equation is unchanged. A free endpoint does not alter the differential equation.

Stage 2. With (4) established, the integral in (3) vanishes for every $\eta$, so stationarity now demands

$L_{y^{\prime }}(b,y(b),y^{\prime }(b))\eta (b)=0\mathrm{(}\mathrm{5}\mathrm{)}$

for every $\eta \in {𝒱}^{\ast }$. Choosing any $\eta$ with $\eta (b)\ne 0$ forces

$L_{y^{\prime }}(b,y(b),y^{\prime }(b))=0\mathrm{(}\mathrm{6}\mathrm{)}$

Equation (6) is the natural boundary condition at the free end.

The two-stage structure is worth stating plainly: a larger class of variations yields the same equation plus an extra condition. Nothing is lost by freeing an endpoint; something is gained.

The condition count still balances

An endpoint free to lie on a curve.

The Euler-Lagrange equation is second order and needs two conditions. With both endpoints fixed, the admissible set supplies both. With the right endpoint free, it supplies only $y(a)=y_a$, and (6) supplies the second. With both endpoints free, (6) holds at each end and again the count is two.

This is why the term "natural" is apt. The conditions are not an extra assumption; they are what the problem produces when the modeller declines to impose one.

Interpretation

A corner in an extremal.

$L_{y^{\prime }}$ is the conjugate momentum of the Euler-Lagrange equation. Condition (6) therefore reads: the conjugate momentum vanishes at a free end.

In mechanics, a free end carries no momentum transfer. In elasticity, where $L$ is a strain energy and $L_{y^{\prime }}$ a stress, (6) is the statement that an unloaded edge is stress free. The same algebra covers both, and the interpretation follows from what $L$ was taken to be.

Transversality: endpoints on a curve

A more general case fixes the right endpoint not to a point but to a curve. Let the right end be required to lie on $y=g(x)$, with the crossing abscissa $b$ itself free. Varying both $y$ and $b$, the boundary contribution becomes

$[L+(g^{\prime }-y^{\prime })L_{y^{\prime }}{]}_{x=b}=0\mathrm{(}\mathrm{7}\mathrm{)}$

Equation (7) is the transversality condition. It reduces to (6) when $g$ is a vertical line, and to $L=0$ when $g^{\prime }=y^{\prime }$.

For the arc-length integrand $L=\sqrt{1+y{\text{'}}^2}$ of the Euler-Lagrange equation, substituting into (7) gives $1+y^{\prime }{g}^{\prime }=0$, so $y^{\prime }{g}^{\prime }=-1$ and the extremal meets the curve at a right angle. Transversality is orthogonality for this integrand, and only for integrands of this kind; the general condition (7) is not a statement about angles.

Corner conditions

If the admissible class is widened to piecewise smooth paths, a path may have a corner at an interior point $c$. The variation then produces a boundary term at $c$ from each side, and stationarity requires both

$[L_{y^{\prime }}{]}_{c^{-}}^{c^{+}}=0\text{and}[L-y^{\prime }{L}_{y^{\prime }}{]}_{c^{-}}^{c^{+}}=0\mathrm{(}\mathrm{8}\mathrm{)}$

These are the Weierstrass-Erdmann corner conditions. Both bracketed quantities are continuous across a corner even though $y^{\prime }$ is not. The second is the Beltrami quantity of the Beltrami identity, and the pair reappears on the Weierstrass excess function.

Note that when $L_{y^{\prime }{y}^{\prime }}\ne 0$, the first condition in (8) forces $y^{\prime }$ to be continuous, so no corner can occur. Corners require degeneracy, which is the connection back to the Legendre condition.

Why this matters for Hamilton's principle

The endpoints in Hamilton's principle are configurations at two fixed times, and they are held fixed. So natural boundary conditions play no role in the basic mechanical statement. They matter in two places this volume reaches later: in continuum problems on Continuum systems: the vibrating string, where a free string end generates its own condition, and in field theory on Classical field theory, where the surface term must be argued away rather than assumed away. The Einstein-Hilbert action on the Einstein-Hilbert action is the case where it cannot be argued away and a compensating surface term must be added by hand.

Next

the Beltrami identity treats the first integral available when $L$ does not contain the independent variable.