Euler-Lagrange equation — derivation — Maths notes

WHAT are we optimizing?

The classic example: the shortest path between two points has $L=\sqrt{1+y'^2}$ , because arc length is $J[y]=\int_a^b\sqrt{1+y'^2}\,dx$ . We want the $y(x)$ that makes this smallest.

Recall Why can't ordinary calculus do this directly?

Because $y$ is not one number — it's infinitely many numbers (one height per $x$ ). Plain $\frac{d}{dx}$ can't search over all curves. We need a way to perturb the whole curve and demand the integral doesn't change to first order.

HOW: the variation trick (derivation from scratch)

Step 1 — Build a family of nearby curves. Suppose $y(x)$ is the true extremal. Perturb it: $Y(x,\varepsilon) \;=\; y(x) + \varepsilon\,\eta(x)$

Why this step? We need to wiggle the whole curve. $\eta(x)$ is an arbitrary smooth "wiggle shape" and $\varepsilon$ is a small dial controlling the size of the wiggle. Setting $\varepsilon=0$ recovers the true curve.

Step 2 — Respect the fixed endpoints. Since every competitor must pass through the same endpoints, $Y(a)=y_a$ and $Y(b)=y_b$ for all $\varepsilon$ . This forces $\boxed{\eta(a)=\eta(b)=0}$

Why this step? If $\eta$ were nonzero at an endpoint, the perturbed curve would miss the required endpoint — not allowed.

Step 3 — Turn the functional into an ordinary function of $\varepsilon$ . $\Phi(\varepsilon) \;=\; \int_a^b L\big(x,\,Y,\,Y'\big)\,dx$ Because the true curve is at $\varepsilon=0$ , and it extremizes $J$ , the ordinary function $\Phi(\varepsilon)$ must have a stationary point at $\varepsilon=0$ : $\left.\frac{d\Phi}{d\varepsilon}\right|_{\varepsilon=0}=0$

Why this step? We've reduced an infinite-dimensional problem to a 1-D calculus problem! Now we use ordinary $\frac{d}{d\varepsilon}$ .

Step 4 — Differentiate under the integral. $\frac{d\Phi}{d\varepsilon}=\int_a^b\left(\frac{\partial L}{\partial Y}\frac{\partial Y}{\partial\varepsilon}+\frac{\partial L}{\partial Y'}\frac{\partial Y'}{\partial\varepsilon}\right)dx$ Since $\dfrac{\partial Y}{\partial\varepsilon}=\eta$ and $\dfrac{\partial Y'}{\partial\varepsilon}=\eta'$ , and evaluating at $\varepsilon=0$ (so $Y\to y$ ): $0=\int_a^b\left(\frac{\partial L}{\partial y}\,\eta+\frac{\partial L}{\partial y'}\,\eta'\right)dx$

Why this step? Chain rule: $L$ depends on $\varepsilon$ only through $Y$ and $Y'$ .

Step 5 — Integrate the second term by parts to free $\eta$ from its derivative. $\int_a^b\frac{\partial L}{\partial y'}\,\eta'\,dx=\underbrace{\left[\frac{\partial L}{\partial y'}\,\eta\right]_a^b}_{=\,0\ \text{(endpoints!)}}-\int_a^b\frac{d}{dx}\!\left(\frac{\partial L}{\partial y'}\right)\eta\,dx$

Why this step? The boundary term vanishes because of Step 2 ( $\eta(a)=\eta(b)=0$ ). This is exactly why the endpoint condition mattered. Now every term has a bare $\eta$ .

Step 6 — Factor out $\eta$ . $0=\int_a^b\left(\frac{\partial L}{\partial y}-\frac{d}{dx}\frac{\partial L}{\partial y'}\right)\eta(x)\,dx$

Step 7 — Apply the Fundamental Lemma of the Calculus of Variations.

Why this step? $\eta$ is arbitrary. If the bracket $g(x)$ were nonzero anywhere, we could choose an $\eta$ that bumps up exactly there, making the integral nonzero — contradiction. (See worked example below for the steel-man.)

Useful shortcut: the Beltrami identity

Derive it: compute $\dfrac{dL}{dx}$ totally: $\frac{dL}{dx}=\frac{\partial L}{\partial x}+\frac{\partial L}{\partial y}y'+\frac{\partial L}{\partial y'}y''$ Substitute the E–L equation $\frac{\partial L}{\partial y}=\frac{d}{dx}\frac{\partial L}{\partial y'}$ : $\frac{dL}{dx}=\frac{\partial L}{\partial x}+y'\frac{d}{dx}\frac{\partial L}{\partial y'}+y''\frac{\partial L}{\partial y'}=\frac{\partial L}{\partial x}+\frac{d}{dx}\!\left(y'\frac{\partial L}{\partial y'}\right)$ So $\dfrac{d}{dx}\!\left(L-y'\dfrac{\partial L}{\partial y'}\right)=\dfrac{\partial L}{\partial x}$ . If $\partial L/\partial x=0$ :

Worked Examples

Worked example 1 — Shortest path is a straight line

$L=\sqrt{1+y'^2}$ . Here $\partial L/\partial y=0$ (no $y$ ).

E–L: $\dfrac{d}{dx}\dfrac{\partial L}{\partial y'}=0\Rightarrow \dfrac{\partial L}{\partial y'}=\text{const}.$ Why? If a derivative's derivative is zero, the thing itself is constant.
Compute: $\dfrac{\partial L}{\partial y'}=\dfrac{y'}{\sqrt{1+y'^2}}=c.$ Why? Power/chain rule on $\sqrt{1+y'^2}$ .
Solve: $y'^2=c^2(1+y'^2)\Rightarrow y'=\text{const}=m.$ Why? Algebra → $y'$ is constant.
Therefore $y=mx+b$ — a straight line. ✓

Worked example 2 — Brachistochrone (fastest slide), use Beltrami

Time $\propto\displaystyle\int\sqrt{\dfrac{1+y'^2}{y}}\,dx$ , so $L=\sqrt{\dfrac{1+y'^2}{y}}$ — no explicit $x$ .

Beltrami: $L-y'\dfrac{\partial L}{\partial y'}=C.$ Why? $\partial L/\partial x=0$ , shortcut applies.
Compute $\dfrac{\partial L}{\partial y'}=\dfrac{y'}{\sqrt{y}\sqrt{1+y'^2}}$ , so $\frac{\sqrt{1+y'^2}}{\sqrt y}-\frac{y'^2}{\sqrt y\sqrt{1+y'^2}}=\frac{1}{\sqrt{y}\sqrt{1+y'^2}}=C.$ Why? Combine over a common denominator: $(1+y'^2)-y'^2=1$ .
Hence $y(1+y'^2)=\text{const}$ — the differential equation of a cycloid. ✓

Common Mistakes (Steel-manned)

Common mistake "

$\dfrac{d}{dx}$ and $\dfrac{\partial}{\partial y'}$ commute, so just write $\dfrac{\partial L}{\partial y}-\dfrac{\partial L}{\partial y'}=0$ ." Why it feels right: both look like "derivatives", so you assume you can cancel/swap. The fix: $\frac{d}{dx}\frac{\partial L}{\partial y'}$ is a total derivative in $x$ — it hits $x$ , $y(x)$ , and $y'(x)$ via the chain rule: $\frac{d}{dx}\frac{\partial L}{\partial y'}=\frac{\partial^2L}{\partial x\,\partial y'}+y'\frac{\partial^2L}{\partial y\,\partial y'}+y''\frac{\partial^2L}{\partial y'^2}.$ It is a different beast from $\frac{\partial L}{\partial y'}$ . Never drop the $\frac{d}{dx}$ .

Recall Feynman: explain it to a 12-year-old

Imagine a wire bent between two nails, and you want the shape that makes a bead slide down fastest. You can't try every shape — there are infinitely many. So you take one shape and jiggle it a tiny bit. If jiggling makes the slide-time go down, that shape wasn't the best — you'd keep jiggling. The best shape is the one where tiny jiggles change nothing (like the bottom of a valley is flat). Writing "jiggle changes nothing" as a clean equation gives the Euler–Lagrange equation. It turns "try every curve" into "solve one equation".

Active-Recall Flashcards

What is a functional?

A rule

J[y]=\int_a^b L(x,y,y')\,dx

that maps each function

y(x)

to a number.

State the Euler–Lagrange equation.

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

Why must

\eta(a)=\eta(b)=0

in the derivation?

Endpoints are fixed, so every competing curve shares them; this also kills the boundary term in integration by parts.

What role does integration by parts play?

It moves the derivative off

\eta'

onto

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

\eta

appears bare and can be factored out.

State the Fundamental Lemma of calculus of variations.

\int_a^b g\,\eta\,dx=0

for all smooth

\eta

vanishing at endpoints, then

g\equiv0

[a,b]

Why is

\frac{d}{dx}\frac{\partial L}{\partial y'}

a total derivative?

Because

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

depends on

x

directly and through

y(x),y'(x)

; chain rule gives three terms.

When does the Beltrami identity apply and what is it?

When

L

has no explicit

x

: then

L-y'\frac{\partial L}{\partial y'}=\text{const}

What does E–L give for

L=\sqrt{1+y'^2}

y'=

const, i.e. a straight line (shortest path).

What does E–L give for

L=\frac12 m\dot q^2-V(q)

m\ddot q=-V'(q)

, i.e. Newton's

F=ma

What happens to the boundary term if endpoints are free?

It must vanish on its own, giving natural boundary conditions

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

Connections

Calculus of Variations — parent framework
Principle of Least Action — physics origin of $L$
Lagrangian Mechanics — E–L generalizes Newton's laws
Brachistochrone Problem — historic motivating example
Geodesics — shortest paths on surfaces
Integration by Parts — the key algebraic tool
Beltrami Identity — first-integral shortcut
Noether's Theorem — symmetries of $L$ give conservation laws

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, normal calculus mein hum poochhte hain "kaunsa number $x$ function $f(x)$ ko minimize karta hai?" Lekin yahan game alag hai — hume poora curve $y(x)$ dhoondhna hai jo ek integral (jise functional bolte hain, $J[y]=\int_a^b L\,dx$ ) ko minimize kare. Jaise do points ke beech sabse chhota path, ya wo wire-shape jispe bead sabse fast slide kare (brachistochrone). Yahan unknown ek number nahi, ek puri function hai.

Trick yeh hai: maan lo sahi curve $y(x)$ hai. Usko thoda "jiggle" karo — $Y=y+\varepsilon\,\eta(x)$ , jahan $\eta$ koi bhi wiggle-shape hai aur $\varepsilon$ chhota knob hai. Kyunki endpoints fixed hain, $\eta(a)=\eta(b)=0$ rakhna zaroori hai. Ab functional ek simple function $\Phi(\varepsilon)$ ban gaya, aur valley ke bottom ki tarah $\Phi'(0)=0$ hona chahiye. Isse infinite-dimensional problem 1-D calculus ban gaya — yahi asli jaadu hai.

Phir integration by parts lagao taaki $\eta'$ hat ke sirf $\eta$ bache. Boundary term zero ho jaata hai kyunki $\eta$ endpoints pe zero hai (isiliye fixed-endpoint condition itni important thi). Bachta hai $\int (\frac{\partial L}{\partial y}-\frac{d}{dx}\frac{\partial L}{\partial y'})\,\eta\,dx=0$ har $\eta$ ke liye. Fundamental Lemma kehta hai: agar yeh sab $\eta$ ke liye zero hai, toh bracket khud hi zero hoga. Bas — yahi Euler–Lagrange equation hai.

Yeh kyun matter karta hai? Kyunki Newton ka $F=ma$ , geodesics, optics ka Fermat's principle — sab isi ek equation se nikalte hain. Ek important galti: $\frac{d}{dx}\frac{\partial L}{\partial y'}$ ko mat cancel karo $\frac{\partial L}{\partial y'}$ se — woh total derivative hai, chain rule se teen terms deta hai. Aur agar $L$ mein $x$ explicitly nahi hai toh Beltrami identity ( $L-y'\frac{\partial L}{\partial y'}=$ const) use karke kaam aasaan ho jaata hai.

Euler-Lagrange equation — derivation

WHAT are we optimizing?

HOW: the variation trick (derivation from scratch)

Useful shortcut: the Beltrami identity

Worked Examples

Common Mistakes (Steel-manned)

Active-Recall Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Advanced Topics (Elite Level)

Connections