Euler-Lagrange equation — derivation: Question bank

Reminder of the symbols you'll meet (all built in the parent): $J [y] = \int_{a}^{b} L (x, y, y^{'}) d x$ is the functional (a number assigned to a whole curve), $η (x)$ is the arbitrary wiggle shape, $ε$ the small dial, and the Euler–Lagrange (E–L) equation is $\frac{\partial L}{\partial y} - \frac{d}{d x} \frac{\partial L}{\partial y ^{'}} = 0$ .

True or false — justify

E–L gives a sufficient condition for a minimum

False — it is only necessary (stationarity, "gradient zero for curves"); a solution could be a minimum, maximum, or saddle. You must test the second variation to know which.

Every solution of the E–L equation is the shortest/least curve for that

L

False — E–L finds stationary curves. For the shortest-path

L

they happen to be minima, but for other

L

an extremal can be a maximum or an inflection of the functional.

L

does not contain

y

explicitly, then

\frac{\partial L}{\partial y ^{'}}

is constant along the extremal

True — E–L reduces to

\frac{d}{d x} \frac{\partial L}{\partial y ^{'}} = 0

, so that quantity is conserved. This is the "

y

is a cyclic coordinate" first integral.

L

does not contain

x

explicitly, then

L

itself is constant along the extremal

False — it's

L - y^{'} \frac{\partial L}{\partial y ^{'}}

(the Beltrami quantity) that is constant, not

L

. See Beltrami Identity.

The endpoint condition

η (a) = η (b) = 0

is just a convenience we can drop

False — it is what kills the boundary term in Integration by Parts. Drop it and you get extra natural boundary conditions

\frac{\partial L}{\partial y ^{'}}_{a, b} = 0

instead.

Φ (ε) = \int_{a}^{b} L (x, Y, Y^{'}) d x

is an ordinary function of one real variable

True — once the wiggle shape

η

is fixed, only the dial

ε

varies, so

Φ

is a plain function whose derivative we can set to zero.

The Fundamental Lemma needs the integral to vanish for one cleverly chosen

η

False — it must vanish for every admissible

η

. A single

η

can hide a nonzero integrand via cancellation of positive and negative parts.

Adding a constant to

L

changes the extremal curve

False — a constant integrates to

C (b - a)

, independent of

y

, so both

\frac{\partial L}{\partial y}

and

\frac{\partial L}{\partial y ^{'}}

are unchanged; E–L gives the same curve.

Multiplying

L

by a positive constant leaves the extremal unchanged

True — E–L is linear/homogeneous in

L

, so scaling by

k > 0

scales the whole equation by

k

and the zero set (the extremal) is identical.

Spot the error

\frac{d}{d x}

and

\frac{\partial}{\partial y ^{'}}

commute, so E–L is

\frac{\partial L}{\partial y} - \frac{\partial L}{\partial y ^{'}} = 0

Wrong —

\frac{d}{d x} \frac{\partial L}{\partial y ^{'}}

is a total

x

-derivative that hits

x

y

, and

y^{'}

by chain rule:

\frac{\partial ^{2} L}{\partial x \partial y ^{'}} + y^{'} \frac{\partial ^{2} L}{\partial y \partial y ^{'}} + y^{''} \frac{\partial ^{2} L}{\partial y ^{'2}}

. It is not

\frac{\partial L}{\partial y ^{'}}

"In Step 4 the chain rule gives

\frac{\partial L}{\partial x} \cdot \frac{\partial x}{\partial ε}

too."

Wrong —

x

is the independent variable and does not depend on

ε

, so

\frac{\partial x}{\partial ε} = 0

. Only

Y

and

Y^{'}

carry the

ε

-dependence.

"Since

\frac{\partial L}{\partial y ^{'}} = \frac{y ^{'}}{1 + y ^{'2}} = c

in the shortest-path case, we conclude

y^{'} = c

Wrong — solving

\frac{y ^{'2}}{1 + y ^{'2}} = c^{2}

gives

y^{'}

equal to some constant

m

, not

c

itself. The point is that

y^{'}

is constant, hence a straight line;

m

is a different constant.

"For the Beltrami step we substitute

\frac{\partial L}{\partial y} = \frac{d}{d x} \frac{\partial L}{\partial y ^{'}}

which is always true."

Wrong — that substitution is E–L, valid only on the extremal. Beltrami is a statement about solutions of E–L, not an identity for arbitrary curves.

\int_{a}^{b} g η d x = 0

with

g

continuous and

g (x_{0}) > 0

somewhere is still possible."

Wrong — pick

η

a smooth positive bump concentrated near

x_{0}

(zero at endpoints); the integral becomes strictly positive, a contradiction. Hence

g \equiv 0

"Newton's law example:

\frac{\partial L}{\partial q ˙} = m \overset{q}{˙}

, and

\frac{d}{d t} (m \overset{q}{˙}) = 0

Wrong —

\frac{d}{d t} (m \overset{q}{˙}) = m \overset{q}{¨}

, which is generally nonzero. E–L gives

- V^{'} (q) - m \overset{q}{¨} = 0 \Rightarrow m \overset{q}{¨} = - V^{'} (q) = F

Why questions

Why does reducing to

Φ (ε)

solve the "infinitely many unknowns" problem

Because after fixing the wiggle shape

η

, the entire perturbation is governed by the single real dial

ε

, converting an infinite-dimensional search into ordinary 1-D calculus.

Why must the demand

Φ^{'} (0) = 0

hold for every

η

, not just one

The true curve extremizes

J

against all competitors; each admissible

η

is a different direction of competition, so stationarity must hold in every such direction.

Why is integration by parts the crucial move, not just algebra

It transfers the derivative off

η^{'}

onto

\frac{\partial L}{\partial y ^{'}}

, leaving a bare

η

everywhere so we can factor it out and invoke the Fundamental Lemma. Without it,

η

and

η^{'}

can't be separated.

Why does the boundary term vanish specifically

Because

[\frac{\partial L}{\partial y ^{'}} η]_{a}^{b}

contains

η (a)

and

η (b)

, both forced to zero by the fixed-endpoint requirement of Step 2.

Why does the Principle of Least Action "contain" Newton's laws

Extremizing

\int (\frac{1}{2} m \overset{q}{˙}^{2} - V) d t

via E–L yields

m \overset{q}{¨} = - V^{'} (q) = F

; the variational principle reproduces

F = ma

as its E–L equation. See Lagrangian Mechanics.

Why do symmetries of

L

hint at conserved quantities

L

ignores a coordinate (no explicit

x

or a cyclic

y

), E–L hands you a first integral (Beltrami or

\frac{\partial L}{\partial y ^{'}} =

const). This is the seed of Noether's Theorem.

Why is

L = 1 + y^{'2}

the right integrand for shortest path

Arc length of a curve is

\int d x^{2} + d y^{2} = \int 1 + y^{'2} d x

; minimizing total length means extremizing exactly this functional, the flat-space case of Geodesics.

Why does the Brachistochrone Problem use Beltrami rather than raw E–L

Its

L = (1 + y^{'2}) / y

has no explicit

x

, so Beltrami gives a first-order ODE directly, avoiding the messier second-order E–L equation.

Edge cases

η

is nonzero at an endpoint, is the perturbed curve still admissible

No — it would miss the required fixed endpoint

y (a)

y (b)

, so it isn't a valid competitor; that's precisely why we force

η (a) = η (b) = 0

What if the endpoints are free rather than fixed

The boundary term no longer vanishes automatically; demanding it vanish gives natural boundary conditions

\frac{\partial L}{\partial y ^{'}}_{a} = \frac{\partial L}{\partial y ^{'}}_{b} = 0

What happens when

L

is linear in

y^{'}

, say

L = P (x, y) + Q (x, y) y^{'}

Then

\frac{\partial L}{\partial y ^{'}} = Q

has no

y^{'}

, so

\frac{\partial ^{2} L}{\partial y ^{'2}} = 0

; E–L collapses to an algebraic condition

P_{y} - Q_{x} = 0

, giving no genuine ODE for

y

(a degenerate variational problem).

What if E–L is satisfied identically (E–L expression is

0

for all curves)

Then

L

is a "null Lagrangian" (a total derivative

\frac{d}{d x} F

), so

J

depends only on the endpoints and every admissible curve is stationary.

What if the brachistochrone's

y \to 0

(start point)

L = (1 + y^{'2}) / y

blows up as

y \to 0

; the cycloid handles it because the slope

y^{'} \to \infty

there in just the right way to keep

y (1 + y^{'2})

finite (constant).

Does E–L still apply if

y

is a vector

(y_{1}, \dots, y_{n})

Yes — you get one E–L equation per component,

\frac{\partial L}{\partial y _{i}} - \frac{d}{d x} \frac{\partial L}{\partial y _{i}^{'}} = 0

, since each

η_{i}

is independently arbitrary.

L

has no

y

and no explicit

x

, which shortcut is stronger

Both apply:

\frac{\partial L}{\partial y ^{'}} =

const (cyclic

y

) and Beltrami

L - y^{'} \frac{\partial L}{\partial y ^{'}} =

const. Together they often solve the curve algebraically without integrating an ODE.

Recall One-line summary of the whole trap set

E–L is necessary, not sufficient; the $\frac{d}{d x}$ is a total derivative; the boundary term dies only because of fixed endpoints; and the Fundamental Lemma needs all $η$ , not one.