Natural boundary conditions and transversality

First keep the interval $[a,b]$ fixed, fix the initial value, and allow the terminal value to vary without restriction. Use a smooth integrand and candidate so the classical first-variation calculation applies. Compactly supported perturbations still give the interior Euler–Lagrange equation. After imposing that equation, stationarity reduces to $F_v(b)h(b)=0$. Since the terminal perturbation can be any real number, its coefficient must vanish. This is the natural boundary condition at a free terminal value.

If an endpoint penalty $\Psi(u(b))$ is added, its derivative contributes to the same coefficient. The resulting condition is $F_v(b)+\Psi'(u(b))=0$. At the initial endpoint the integration-by-parts sign is negative, so a free initial value with an initial penalty has a different sign. These formulas assume genuinely free two-sided motions. An endpoint constrained by an inequality instead allows only a one-sided feasible motion at an active boundary and generally leads to a sign condition, not an equality.

Now let the terminal coordinate be $b_\varepsilon$ and write its total vertical displacement as $\Delta u_b$. At first order this total displacement equals the shape variation at the old terminal coordinate plus $u'(b)\Delta b$. Differentiating the upper integration limit contributes $F(b)\Delta b$. Combining this with the usual boundary momentum term gives $F_v\Delta u_b+(F-u'F_v)\Delta b$. Forgetting the distinction between total displacement and fixed-coordinate variation gives an incorrect terminal formula.

Suppose the endpoint lies on a smooth curve $u=\phi(x)$ and its horizontal coordinate is free near the candidate, with no endpoint penalty. Then $\Delta u_b=\phi'(b)\Delta b$, so the coefficient of the free horizontal displacement must vanish. This is a transversality condition. It assumes a smooth terminal curve, a smooth candidate defined sufficiently near the terminal point, and feasible endpoint motions in both directions. It does not assert that the stationary meeting point is globally optimal or even exists.

For the length integrand $\sqrt{1+(u')^2}$, terminal-curve transversality becomes $1+u'\phi'=0$. This is the dot product of the two graph tangent vectors being zero: the stationary path meets the terminal curve orthogonally. The formula is valid in a graph chart where both slopes are finite; vertical terminal lines can instead be described using a different chart or directly with tangent vectors. Coordinate formulas should not be mistaken for restrictions on the underlying geometry.

Natural conditions also make physical sense in quadratic examples. A derivative energy without terminal penalty favors zero slope at a free end. A spring-like terminal penalty balances the endpoint momentum against a restoring force. For convex energies and convex penalties, an exact expansion can certify the stationary solution as the global minimum. For general moving-endpoint problems, the feasible set itself can be nonconvex, so neither the interior equation nor an orthogonal meeting condition resolves all competing paths.