Euler–Lagrange equation and standard functionals

In one sentence

Extremising $\int F(x,y,y')\,dx$ gives the Euler–Lagrange equation, two special forms of $F$ collapse it to a first-order equation, and a solution of it is stationary rather than minimal.

Why the exam asks it

Because the machinery is mechanical — write $F$, apply one of three formulas — and because the standard functionals have named answers. And because "extremal" gets read as "minimum", which the theory does not supply.

The idea

To extremise $J[y] = \displaystyle\int_a^b F(x,y,y')\,dx$ over curves with fixed endpoints, a stationary $y$ satisfies

$$\frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) - \frac{\partial F}{\partial y} = 0 .$$

Second order in general, so two constants, fixed by the two endpoint conditions.

The two shortcuts

$F$ has no $y$. Then $\partial F/\partial y = 0$, so

$$\frac{\partial F}{\partial y'} = \text{constant}.$$

$F$ has no explicit $x$. Then the Beltrami identity holds:

$$F - y'\frac{\partial F}{\partial y'} = \text{constant}.$$

Both reduce a second-order equation to a first-order one, and almost every examinable problem uses one of them. Checking which variable is absent is the first move.

The standard three

Shortest path. $F=\sqrt{1+y'^2}$ has no $y$, so $\dfrac{y'}{\sqrt{1+y'^2}}$ is constant, hence $y'$ is constant: a straight line.

Minimal surface of revolution. $F = y\sqrt{1+y'^2}$ has no explicit $x$, so Beltrami gives $\dfrac{y}{\sqrt{1+y'^2}} = c$, whose solution is the catenary $y = c\cosh\frac{x-a}{c}$ — and the surface is the catenoid. Check with $c=1$: $\cosh x/\sqrt{1+\sinh^2x} = \cosh x/\cosh x = 1$.

Brachistochrone. Minimise descent time $\int\sqrt{\frac{1+y'^2}{2gy}}\,dx$. No explicit $x$, so Beltrami again, and the answer is a cycloid: $x = a(t-\sin t)$, $y = a(1-\cos t)$.

The figure computes both descent times numerically, and the cycloid wins despite being the longer curve — it drops steeply at the start, buying speed early and paying for it in distance. That trade is the whole content of the problem.

Stationary is not minimal

Euler–Lagrange is a necessary condition. A solution is a stationary point of the functional and may be a minimum, a maximum, or neither.

The second-order test is the Legendre condition $F_{y'y'} \ge 0$ along the extremal, necessary for a weak minimum. Sufficiency needs more — no conjugate points on the interval, the Jacobi condition.

And a minimum may not exist at all. Weierstrass's example: $\int_{-1}^{1} x^2 y'^2\,dx$ with $y(-1)=-1$, $y(1)=1$ has infimum $0$ that no admissible curve attains. Existence in the calculus of variations is a genuine question, not a formality.

Free endpoints

If an endpoint is not fixed, the extremal must satisfy the natural boundary condition $\left.\partial F/\partial y'\right| = 0$ there. That replaces the missing endpoint condition, and questions with a free end are testing whether you supply it.

Where intuition breaks

"An Euler–Lagrange solution is a minimum." Stationary only. Legendre is necessary, Jacobi is needed for sufficiency, and sometimes no minimum exists.

"The shortest path and the fastest descent are the same." Straight line against cycloid. Different functionals, different extremals.

"Beltrami applies whenever $F$ looks autonomous." It needs no explicit $x$. An $x$ hidden inside a coefficient still counts.

"Euler–Lagrange is always second order." It is, but the two shortcuts produce first-order equations directly, which is why they are worth recognising before differentiating.

"With a free endpoint you simply have one fewer condition." The natural boundary condition replaces it. There are still two conditions.

"The infimum is always attained." Weierstrass's example. The direct method needs coercivity and lower semicontinuity, which the problem may not have.

"A minimum of the integrand gives a minimum of the integral." Pointwise minimisation ignores the constraint of being a curve with fixed endpoints; the two are unrelated.

The exam's angle

1. Check which variable is absent from $F$ first. No $y$ gives a conserved momentum, no explicit $x$ gives Beltrami, and either halves the work.
2. Recognise the three standard functionals on sight — arc length, $y\sqrt{1+y'^2}$, and the brachistochrone integrand — and give the named answer.
3. If the question says "extremal", do not claim it is a minimum unless asked to verify.
4. For a free endpoint, write the natural boundary condition explicitly.
5. For several dependent variables, write one Euler–Lagrange equation per variable; for higher derivatives, alternate signs in the Euler–Poisson equation.

The night before

• $\frac{d}{dx}F_{y'} - F_y = 0$. Second order, two constants.
• No $y$: $F_{y'}$ constant. No explicit $x$: Beltrami $F - y'F_{y'}$ constant.
• Shortest path: straight line. Minimal surface of revolution: catenary, $y=c\cosh\frac{x-a}c$. Brachistochrone: cycloid.
• Euler–Lagrange is necessary only. Legendre $F_{y'y'}\ge0$ necessary; Jacobi for sufficiency.
• A minimum need not exist — Weierstrass's $\int x^2y'^2$.
• Free endpoint $\Rightarrow$ natural boundary condition $F_{y'} = 0$ there.