Calculus of Variations & Transversality

Calculus of variations asks for a function that minimizes a functional. Its central move is to perturb the entire candidate curve, compute the first-order change, and require that change to vanish for every admissible perturbation. The Euler–Lagrange equation is what remains in the interior; transversality conditions are what remain at free endpoints.

From a perturbed curve to the first variation #

Consider

$$J[y]=\int_a^b F\!\left(x,y(x),y'(x)\right)dx,$$

with fixed endpoints $y(a)=y_a$ and $y(b)=y_b$. Let

$$y_\varepsilon(x)=y(x)+\varepsilon\eta(x),$$

where $\eta$ is any smooth variation satisfying $\eta(a)=\eta(b)=0$. Define

$$g(\varepsilon)=J[y_\varepsilon].$$

At a differentiable local minimum, $g'(0)=0$. Differentiating under the integral gives the first variation

$$\delta J[y;\eta] =\left.\frac{d}{d\varepsilon}J[y+\varepsilon\eta]\right|_{\varepsilon=0} =\int_a^b\left(F_y\eta+F_{y'}\eta'\right)dx.$$

Integrate the second term by parts:

$$\delta J =\left[F_{y'}\eta\right]_a^b +\int_a^b\left(F_y-\frac{d}{dx}F_{y'}\right)\eta\,dx.$$

The boundary term vanishes for fixed endpoints. Because the remaining integral must be zero for every such $\eta$, the fundamental lemma yields

$$\boxed{\frac{d}{dx}F_{y'}-F_y=0.}$$

This is the Euler–Lagrange equation. For a vector curve $y\in\mathbb R^n$, it holds componentwise:

$$\frac{d}{dx}\frac{\partial F}{\partial y_i'} -\frac{\partial F}{\partial y_i}=0, \qquad i=1,\ldots,n.$$

What the equation does and does not prove #

The Euler–Lagrange equation is a necessary first-order condition. A solution is an extremal, not automatically a minimum. It may be a maximum, a saddle, or inadmissible once constraints and endpoints are enforced.

For example,

$$J[y]=\frac12\int_0^T\left[(y')^2+\omega^2y^2\right]dt$$

has

$$F_y=\omega^2y,\qquad F_{y'}=y',$$

so every extremal satisfies

$$y''-\omega^2y=0,$$

and therefore

$$y(t)=c_1e^{\omega t}+c_2e^{-\omega t}.$$

The endpoint values determine $c_1,c_2$. Because the integrand is jointly convex in $(y,y')$, that feasible extremal is also the unique global minimizer. Without convexity, the differential equation alone would not justify that conclusion.

Free endpoints create natural boundary conditions #

If $y(b)$ is free while $b$ remains fixed, $\eta(b)$ is arbitrary. The boundary term requires

$$F_{y'}(b)=0.$$

If $y(a)$ is also free, then $F_{y'}(a)=0$ as well. These are natural boundary conditions: they are not specified by the user but forced by stationarity.

When the terminal point may move in both coordinates, write its variation as $(\delta b,\delta y_b)$. The terminal boundary contribution becomes

$$F_{y'}(b)\,\delta y_b +\left[F-y'F_{y'}\right]_{x=b}\delta b.$$

For a completely free terminal point,

$$F_{y'}(b)=0, \qquad F(b)-y'(b)F_{y'}(b)=0.$$

If the endpoint must lie on a curve $g(b,y_b)=0$, then allowable endpoint variations satisfy

$$g_x\,\delta b+g_y\,\delta y_b=0.$$

Stationarity means the variational boundary covector is normal to that target curve. Equivalently, for some multiplier $\nu$,

$$F_{y'}(b)=\nu g_y, \qquad F-y'F_{y'}=\nu g_x.$$

First integrals from missing coordinates #

Euler–Lagrange contains useful conservation laws.

If $F$ does not depend explicitly on $y$, then $F_y=0$ and

$$F_{y'}=\text{constant}.$$

If $F$ does not depend explicitly on $x$, differentiate

$$F-y'F_{y'}$$

along an extremal. Euler–Lagrange cancels the remaining terms, giving the Beltrami identity

$$\boxed{F-y'F_{y'}=\text{constant}.}$$

This is the variational ancestor of Hamiltonian conservation for autonomous optimal-control problems.

Integral and pointwise constraints #

An isoperimetric constraint

$$\int_a^bG(x,y,y')dx=c$$

introduces a constant multiplier $\lambda$. Extremals satisfy Euler–Lagrange for the augmented integrand

$$\widetilde F=F+\lambda G.$$

A pointwise equality $h(x,y)=0$ requires a multiplier function $\lambda(x)$, because the restriction applies at every $x$:

$$\widetilde F=F+\lambda(x)h(x,y).$$

Inequality constraints require complementary-slackness logic and can create boundary arcs. Simply assigning a large penalty to a violated constraint changes the problem; it does not reproduce an exact hard constraint.

Second variation and minimum tests #

The second variation is

$$\delta^2J[y;\eta] =\int_a^b \left( F_{yy}\eta^2 +2F_{yy'}\eta\eta' +F_{y'y'}(\eta')^2 \right)dx.$$

A necessary local-minimum condition is the Legendre condition

$$F_{y'y'}\ge0$$

along the extremal; the strengthened form uses $F_{y'y'}>0$. This still may not be sufficient because the mixed and $F_{yy}$ terms can make the quadratic form negative for some extended variation. Conjugate-point or Jacobi analysis checks that global-in-the-interval failure mode.

For convex $F(x,y,y')$ and a convex feasible set, every stationary feasible curve is a global minimizer. Convexity is therefore much stronger than a local second-order test, but many control problems with nonlinear dynamics or nonconvex constraints do not have it.

From curves to controlled dynamics #

Introduce a state $x_1=y$, control $u=y'$, and dynamics

$$\dot x_1=u.$$

The variational integrand becomes a running cost. More general state equations constrain the permissible relation between a trajectory and its derivative, and bounded or discontinuous controls invalidate the naive assumption that every variation of $u$ is small and smooth. Needle variations solve that problem and lead to the Pontryagin maximum principle.

The central lesson survives the generalization: interior differential equations and endpoint transversality are two parts of one stationarity calculation. Dropping either part leaves the boundary-value problem incomplete.