Calculus of variations — functionals, functional derivative

1. WHAT is a functional?

WHY this matters: Tons of physics & geometry are "find the best curve" problems:

• Shortest path between two points (geodesic): $L=\sqrt{1+y'^2}$.
• Brachistochrone (fastest slide): $L=\sqrt{\tfrac{1+y'^2}{2gy}}$.
• Classical mechanics: $L=T-V$, action $S=\int L\,dt$ is minimised (Hamilton's principle).

2. The functional derivative — derived from scratch

WHAT we want: the analogue of "$\frac{df}{dx}=0$" for functionals.

HOW — the variation trick. Suppose $y(x)$ is the minimiser, with fixed endpoints $y(a),y(b)$. Perturb it slightly: $\tilde y(x)=y(x)+\varepsilon\,\eta(x),\qquad \eta(a)=\eta(b)=0.$ Here $\eta$ is any smooth "test function" vanishing at the ends (so endpoints stay fixed). Define $\Phi(\varepsilon)=J[y+\varepsilon\eta].$ Since $y$ minimises $J$, the ordinary function $\Phi$ has a minimum at $\varepsilon=0$, so $\boxed{\Phi'(0)=0}\quad\text{(this is the WHY of everything below).}$

Compute $\Phi'(0)$. Differentiate under the integral: $\Phi'(\varepsilon)=\int_a^b\!\Big(\frac{\partial L}{\partial y}\,\eta+\frac{\partial L}{\partial y'}\,\eta'\Big)dx.$ Why this step? We used the chain rule on $L(x,\,y+\varepsilon\eta,\,y'+\varepsilon\eta')$; $\partial/\partial\varepsilon$ brings down $\eta$ for the $y$-slot and $\eta'$ for the $y'$-slot.

Set $\varepsilon=0$ and integrate the second term by parts to free $\eta$ from its derivative: $\int_a^b \frac{\partial L}{\partial y'}\eta'\,dx=\underbrace{\Big[\frac{\partial L}{\partial y'}\eta\Big]_a^b}_{=0\ (\eta(a)=\eta(b)=0)}-\int_a^b\frac{d}{dx}\!\Big(\frac{\partial L}{\partial y'}\Big)\eta\,dx.$ Why this step? Integration by parts swaps the derivative off $\eta$ onto $L_{y'}$. The boundary term dies because $\eta$ vanishes at the endpoints — that is exactly why we required fixed endpoints.

So: $\Phi'(0)=\int_a^b\underbrace{\Big[\frac{\partial L}{\partial y}-\frac{d}{dx}\frac{\partial L}{\partial y'}\Big]}_{\text{call it }\;\frac{\delta J}{\delta y}}\,\eta(x)\,dx=0\quad\forall\,\eta.$

The Fundamental Lemma. If $\int_a^b f(x)\eta(x)\,dx=0$ for every smooth $\eta$ vanishing at the ends, then $f(x)\equiv 0$. Why true: if $f>0$ somewhere, choose a bump $\eta$ concentrated there to make the integral positive — contradiction.

3. A useful shortcut: the Beltrami identity

If $L=L(y,y')$ does not depend on $x$ explicitly, then a first integral exists: $\frac{d}{dx}\Big(L-y'\frac{\partial L}{\partial y'}\Big)=L_y y'+L_{y'}y''-y''L_{y'}-y'\frac{d}{dx}L_{y'}=y'\Big(L_y-\frac{d}{dx}L_{y'}\Big)=0.$ Why this step? Expand the total $x$-derivative; the $y''$ terms cancel, and the bracket is exactly the Euler–Lagrange expression $=0$. Hence:

4. Worked examples

5. Common mistakes

6. Active recall

Connections

• Lagrangian mechanics — action $S=\int L\,dt$, E–L gives Newton's laws.
• Geodesics and differential geometry — shortest paths via variational problems.
• Functional analysis — Gâteaux/Fréchet derivatives generalise $\delta J/\delta y$.
• Constrained optimization & Lagrange multipliers — isoperimetric problems use multipliers on functionals.
• Ordinary differential equations — E–L is the ODE you actually solve.
• Brachistochrone problem — historic origin (Bernoulli, 1696).

Concept Map

Hinglish (regional understanding)