Euler-Lagrange equation — derivation: Worked examples

This page is a catalogue of cases. The parent derivation gave you the machine: $\frac{\partial L}{\partial y} - \frac{d}{d x} (\frac{\partial L}{\partial y ^{'}}) = 0.$ Here we feed it every kind of Lagrangian you can meet, so you never hit an input we did not show. First a map, then the worked cases.

The scenario matrix

Every problem in the Calculus of Variations is decided by which variables $L$ actually depends on. That single fact tells you which shortcut applies. The matrix below lists every case class.

Cell	$L$ depends on…	What simplifies	Tool of choice	Example
A	$y^{'}$ only ( $L = L (y^{'})$ )	$\partial L / \partial y = 0$	$\partial L / \partial y^{'} = const$	Ex 1
B	$y, y^{'}$ (no $x$ )	$\partial L / \partial x = 0$	Beltrami Identity	Ex 2
C	$x, y^{'}$ (no $y$ )	$\partial L / \partial y = 0$	$\partial L / \partial y^{'} = const$	Ex 3
D	full $x, y, y^{'}$	nothing skips	full E–L ODE	Ex 4
E	degenerate $L$ linear in $y^{'}$ , OR no $y^{'}$ at all	E–L collapses	check consistency	Ex 5
F	free endpoint (one end loose in $y$ or in $x$ )	boundary term survives	natural BC / transversality	Ex 6
G	word problem (real world)	model then apply B	Brachistochrone Problem-style	Ex 7
H	exam twist (higher $y^{''}$ / symmetry)	generalized E–L / Noether's Theorem	modified rule	Ex 8

Ex 1 — Cell A: $L$ depends on $y^{'}$ only

Worked example Shortest path, re-derived cleanly

Extremize $J [y] = \int_{0}^{1} 1 + y^{'2} d x$ with $y (0) = 0$ , $y (1) = 2$ .

Forecast: guess the answer's shape before reading on. (A straight line? A curve?)

Steps.

Note $\partial L / \partial y = 0$ because $y$ never appears. Why this step? Kills half the E–L equation instantly — Cell A's gift.
E–L becomes $\frac{d}{d x} (\frac{\partial L}{\partial y ^{'}}) = 0$ , so $\frac{\partial L}{\partial y ^{'}} = c_{1}$ (constant). Why? A quantity whose $x$ -derivative is $0$ cannot change with $x$ .
Compute $\frac{\partial L}{\partial y ^{'}} = \frac{y ^{'}}{1 + y ^{'2}} = c_{1}$ . Why? Chain rule on $1 + y^{'2}$ .
Solve: square both sides, $y^{'2} = c_{1}^{2} (1 + y^{'2})$ , so $y^{'2} (1 - c_{1}^{2}) = c_{1}^{2}$ , giving $y^{'} = \pm \frac{c _{1}}{1 - c _{1}^{2}} =: m$ a constant. Why the $\pm$ ? Squaring lost the sign; both branches are candidate slopes. The endpoints decide: we need $y$ to rise from $0$ to $2$ , so we take the positive branch. (Had the target been $y (1) = - 2$ , we'd take the negative one.)
Integrate: $y = m x + b$ . Endpoints $y (0) = 0 \Rightarrow b = 0$ ; $y (1) = 2 \Rightarrow m = 2 > 0$ (consistent with the positive branch). So $y = 2 x$ .

Verify: slope $2$ everywhere, $\frac{\partial L}{\partial y ^{'}} = \frac{2}{5} \approx 0.894$ — genuinely constant along the path. Straight line ✓ (see Geodesics for the curved-space version).

Figure — Euler-Lagrange equation — derivation

Ex 2 — Cell B: no explicit $x$ , use Beltrami

Worked example Minimal surface of revolution (soap film)

Spin a curve $y (x)$ about the $x$ -axis; its surface area $\propto \int y 1 + y^{'2} d x$ . So $L = y 1 + y^{'2}$ — contains $y$ and $y^{'}$ , no bare $x$ ⇒ Cell B.

Forecast: the answer is a famous curve. Guess which "hanging" shape.

Steps.

Because $\partial L / \partial x = 0$ , use Beltrami Identity: $L - y^{'} \frac{\partial L}{\partial y ^{'}} = C_{B}$ . Why? Skips one whole $\frac{d}{d x}$ — the parent note proved this shortcut.
$\frac{\partial L}{\partial y ^{'}} = \frac{y y ^{'}}{1 + y ^{'2}}$ . Why? Differentiate $y 1 + y^{'2}$ treating $y$ as a constant w.r.t. $y^{'}$ .
Substitute: $y 1 + y^{'2} - \frac{y y ^{'2}}{1 + y ^{'2}} = \frac{y}{1 + y ^{'2}} = C_{B}$ . Why? Common denominator: $(1 + y^{'2}) - y^{'2} = 1$ .
Rearrange to isolate the slope: $y = C_{B} 1 + y^{'2} \Rightarrow y^{'2} = \frac{y ^{2}}{C _{B}^{2}} - 1 \Rightarrow y^{'} = \pm \frac{1}{C _{B}} y^{2} - C_{B}^{2}$ . Why the $\pm$ ? Square-root of $y^{'2}$ ; the sign is fixed later by which way the curve climbs. Take the $+$ branch on the rising half.
Separate variables (this is the WHY of the ODE solve): the equation is now $\frac{d y}{y ^{2} - C _{B}^{2}} = \frac{d x}{C _{B}}$ . Why separate? $y^{'} = d y / d x$ depends only on $y$ , so we can collect all $y$ 's on one side and integrate each side independently.
Integrate both sides: $\int \frac{d y}{y ^{2} - C _{B}^{2}} = arccosh (\frac{y}{C _{B}})$ and $\int \frac{d x}{C _{B}} = \frac{x - x _{0}}{C _{B}}$ (with $x_{0}$ the integration constant). Why arccosh? Because $\frac{d}{d u} arccosh u = \frac{1}{u ^{2} - 1}$ — exactly our integrand after $u = y / C_{B}$ .
Equate and invert: $arccosh (\frac{y}{C _{B}}) = \frac{x - x _{0}}{C _{B}} \Rightarrow y = C_{B} cosh (\frac{x - x _{0}}{C _{B}})$ — a catenary. Why $cosh$ ? It is the inverse of $arccosh$ ; applying $cosh$ to both sides undoes the arc-function.

Verify (numeric): take $C_{B} = 1, x_{0} = 0$ . Then $y = cosh x$ , $y^{'} = sinh x$ , and $y^{2} = cosh^{2} x = 1 + sinh^{2} x = 1 + y^{'2}$ . Hence $\frac{y ^{2}}{1 + y ^{'2}} = 1$ , i.e. $\frac{y}{1 + y ^{'2}} = 1 = C_{B}$ ✓ for every $x$ .

Ex 3 — Cell C: no explicit $y$ , but $x$ present

Worked example Weighted straight-ish path

Extremize $J = \int_{1}^{2} x 1 + y^{'2} d x$ . Here $L = x 1 + y^{'2}$ : contains $x$ and $y^{'}$ , no $y$ ⇒ Cell C.

Forecast: will the answer still be a straight line, or does the $x$ -weight bend it?

Steps.

$\partial L / \partial y = 0$ , so E–L gives $\frac{\partial L}{\partial y ^{'}} = c_{3}$ (constant). Why? Cell C's gift — same conserved-quantity logic as Cell A, even though $x$ appears.
$\frac{\partial L}{\partial y ^{'}} = \frac{x y ^{'}}{1 + y ^{'2}} = c_{3}$ . Why? $x$ rides along as a constant during the $y^{'}$ -derivative.
Solve for $y^{'}$ : $x^{2} y^{'2} = c_{3}^{2} (1 + y^{'2}) \Rightarrow y^{'2} (x^{2} - c_{3}^{2}) = c_{3}^{2} \Rightarrow y^{'} = \pm \frac{c _{3}}{x ^{2} - c _{3}^{2}}$ . Why the $\pm$ , and note the domain: the square-root demands $x^{2} > c_{3}^{2}$ , i.e. $x > ∣ c_{3} ∣$ (for $x > 0$ ) — the solution is only valid on $x \geq c_{3}$ . We take the $+$ branch for an increasing curve.
Integrate: $y = c_{3} arccosh (\frac{x}{c _{3}}) + b$ on the valid range $x \geq c_{3}$ . Why? $\int \frac{d x}{x ^{2} - c _{3}^{2}} = arccosh (x / c_{3})$ , whose domain $x / c_{3} \geq 1$ is exactly the restriction found in step 3.

Verify (numeric): pick $c_{3} = 1, b = 0$ , so $y = arccosh (x)$ (valid for $x \geq 1$ ; our interval $[1, 2]$ sits inside it), $y^{'} = \frac{1}{x ^{2} - 1}$ . At $x = 2$ : $\frac{x y ^{'}}{1 + y ^{'2}} = \frac{2/ 3}{1 + 1/3} = \frac{2/ 3}{2/ 3} = 1 = c_{3}$ ✓.

Ex 4 — Cell D: full dependence, run the whole ODE

Worked example Nothing skips

Extremize $J = \int_{0}^{1} (y^{'2} + y^{2} + 2 x y) d x$ . $L = y^{'2} + y^{2} + 2 x y$ contains $x, y, y^{'}$ ⇒ Cell D, full E–L.

Forecast: the extremal solves a 2nd-order linear ODE — guess if it oscillates or grows.

Steps.

$\frac{\partial L}{\partial y} = 2 y + 2 x$ . Why? Differentiate $y^{2} + 2 x y$ in $y$ , with $x$ frozen.
$\frac{\partial L}{\partial y ^{'}} = 2 y^{'}$ , so $\frac{d}{d x} \frac{\partial L}{\partial y ^{'}} = 2 y^{''}$ . Why? Then take a total $x$ -derivative — here $2 y^{'} \to 2 y^{''}$ .
E–L: $2 y + 2 x - 2 y^{''} = 0 \Rightarrow y^{''} - y = x$ . Why? Plug both pieces into $\partial_{y} L - \frac{d}{d x} \partial_{y^{'}} L = 0$ .
Solve: homogeneous $y_{h} = A e^{x} + B e^{- x}$ ; particular $y_{p} = - x$ (since $y_{p}^{''} - y_{p} = 0 - (- x) = x$ ). So $y = A e^{x} + B e^{- x} - x$ . Why? Linear ODE ⇒ complementary + particular. It grows (exponentials), doesn't oscillate.

Verify: substitute $y = - x$ (particular part): $y^{''} - y = 0 - (- x) = x$ ✓. Add any $A e^{x} + B e^{- x}$ : each satisfies $y^{''} - y = 0$ , so the sum still gives $x$ ✓.

Ex 5 — Cell E: degenerate — two flavours

Worked example 5b:

$L$ has NO $y^{'}$ at all — $L = L (x, y)$ Extremize $J = \int_{0}^{1} (y^{2} - 2 x y) d x$ . Here $L = y^{2} - 2 x y$ contains no $y^{'}$ .

Forecast: with no $y^{'}$ there is no ODE to integrate — so what pins $y$ ?

Steps.

$\frac{\partial L}{\partial y ^{'}} = 0$ (there is no $y^{'}$ ), so $\frac{d}{d x} \frac{\partial L}{\partial y ^{'}} = 0$ . Why? You cannot differentiate what isn't there.
E–L collapses to a purely algebraic condition $\frac{\partial L}{\partial y} = 0$ : $2 y - 2 x = 0 \Rightarrow y = x$ . Why? No derivative term survives — the extremal is found point-by-point, not by integrating.
Degeneracy warning: $y = x$ is forced pointwise and cannot in general meet imposed endpoint values (here $y (0) = 0, y (1) = 1$ happen to fit, but any other endpoints make the problem inconsistent — no smooth extremal exists). Why? An algebraic (order-0) condition has no free constants to satisfy boundary data.

Verify (numeric): at $y = x$ , $\partial L / \partial y = 2 x - 2 x = 0$ ✓ for all $x$ ; and it satisfies $y (0) = 0, y (1) = 1$ ✓ (a lucky-fit case).

Ex 6 — Cell F: free endpoints (in $y$ and in $x$ )

Worked example 6a: one end let loose in

$y$ — natural boundary condition Extremize $J = \int_{0}^{1} (\frac{1}{2} y^{'2} - y) d x$ with $y (0) = 0$ but $y (1)$ free (the right end may slide vertically).

Forecast: with a fixed end you'd need a second endpoint value — what replaces it?

Steps.

Interior E–L: $\frac{\partial L}{\partial y} = - 1$ , $\frac{\partial L}{\partial y ^{'}} = y^{'}$ , so $- 1 - y^{''} = 0 \Rightarrow y^{''} = - 1$ . Why? The free end changes only the boundary, not the interior ODE.
General solution $y = - \frac{1}{2} x^{2} + A x + B$ . Apply $y (0) = 0 \Rightarrow B = 0$ .
Recall the variation $η$ (defined at the top): the Integration by Parts step in the parent left a boundary term $[\frac{\partial L}{\partial y ^{'}} η]_{0}^{1}$ , where $η$ is the wiggle of $y$ . At $x = 0$ the end is fixed, so $η (0) = 0$ kills that piece. At $x = 1$ the end is free, so $η (1)$ is arbitrary and non-zero — the term cannot be dropped. Why this matters: the only way $\frac{\partial L}{\partial y ^{'}}_{1} η (1) = 0$ for all such $η (1)$ is if the coefficient vanishes: the natural boundary condition $\frac{\partial L}{\partial y ^{'}}_{x = 1} = 0$ , i.e. $y^{'} (1) = 0$ .
Apply $y^{'} (1) = 0$ : $y^{'} = - x + A$ , so $- 1 + A = 0 \Rightarrow A = 1$ . Final: $y = - \frac{1}{2} x^{2} + x$ .

Verify: $y^{''} = - 1$ ✓, $y (0) = 0$ ✓, $y^{'} (1) = - 1 + 1 = 0$ ✓ — the curve arrives flat at the loose end, the geometric meaning of a natural BC.

Ex 7 — Cell G: real-world word problem

Worked example Fastest ski run (a Brachistochrone with numbers)

A frictionless skier starts at rest at $(0, 0)$ and reaches $(π, 2)$ , with $y$ measured downward so gravity assists.

Setting up the physics (defining every constant). By energy conservation, sliding from rest through a vertical drop $y$ gives speed $v = 2 g y$ , where $g$ is the acceleration due to gravity. An arc-length element is $d s = 1 + y^{'2} d x$ , so the time to traverse it is $d t = d s / v = \frac{1 + y ^{'2}}{2 g y} d x$ . Summing, $T = \int_{0}^{π} \frac{1 + y ^{'2}}{2 g y} d x = \frac{1}{2 g} \int_{0}^{π} \frac{1 + y ^{'2}}{y} d x .$ The prefactor $1/ 2 g$ is a constant and does not affect which curve minimises $T$ , so we extremize $L = \frac{1 + y ^{'2}}{y}$ — no bare $x$ ⇒ Cell B.

Forecast: straight ramp, or a curve that dives steeply first?

Steps.

Beltrami (since $\partial L / \partial x = 0$ ): $L - y^{'} \frac{\partial L}{\partial y ^{'}} = C_{B}$ . Why? The energy-like first integral of any $x$ -free problem.
Compute $\frac{\partial L}{\partial y ^{'}} = \frac{y ^{'}}{y 1 + y ^{'2}}$ . Substitute: $\frac{1 + y ^{'2}}{y} - \frac{y ^{'2}}{y 1 + y ^{'2}} = \frac{( 1 + y ^{'2} ) - y ^{'2}}{y 1 + y ^{'2}} = \frac{1}{y 1 + y ^{'2}} = C_{B} .$ Why? Common denominator; the numerator collapses to $1$ .
Square and rearrange to the first integral: $\frac{1}{y ( 1 + y ^{'2} )} = C_{B}^{2} \Rightarrow y (1 + y^{'2}) = \frac{1}{C _{B}^{2}} =: k$ . Why this is the "first integral": it is one order lower than E–L would give and holds along the whole path — the physical WHY is that no bare $x$ means a conserved quantity, here $k$ .
Solve $y (1 + y^{'2}) = k$ by the standard substitution $y^{'} = cot (θ /2)$ ; it yields the parametric cycloid $x = r (θ - sin θ)$ , $y = r (1 - cos θ)$ with $k = 2 r$ . Why a cycloid? It is the unique curve satisfying $y (1 + y^{'2}) = const$ ; near the start it plunges nearly vertically, banking speed early to beat the straight ramp.

Verify (numeric): for $r = 1$ so $k = 2$ , use $y^{'} = \frac{d y / d θ}{d x / d θ} = \frac{sin θ}{1 - cos θ}$ . At $θ = π /2$ : $y = 1$ , $y^{'} = 1$ , so $y (1 + y^{'2}) = 1 \cdot 2 = 2 = k$ ✓. At $θ = π$ : $y = 2$ , $y^{'} = 0$ , $y (1 + y^{'2}) = 2 \cdot 1 = 2 = k$ ✓ — constant, cycloid confirmed.

Ex 8 — Cell H: exam twist, higher derivative

Worked example When

$L$ contains $y^{''}$ Bending energy of a beam: extremize $J = \int_{0}^{1} (\frac{1}{2} (y^{''})^{2}) d x$ . The Lagrangian holds a second derivative $y^{''}$ — the standard E–L is not enough.

Forecast: the answer is the shape a thin ruler takes — guess its polynomial degree.

Steps.

Use the generalized Euler–Poisson equation: $\frac{\partial L}{\partial y} - \frac{d}{d x} \frac{\partial L}{\partial y ^{'}} + \frac{d ^{2}}{d x ^{2}} \frac{\partial L}{\partial y ^{''}} = 0$ . Why this step? Each extra derivative in $L$ contributes an extra $\pm \frac{d ^{k}}{d x ^{k}}$ term (repeat Integration by Parts $k$ times, exactly the parent's Step 5 done twice).
Here $\partial L / \partial y = 0$ , $\partial L / \partial y^{'} = 0$ , $\partial L / \partial y^{''} = y^{''}$ . Why? $L$ depends only on $y^{''}$ .
Equation: $\frac{d ^{2}}{d x ^{2}} (y^{''}) = y^{''''} = 0$ . Why? Only the third term survives.
Solve: $y^{''''} = 0 \Rightarrow y = c_{3} x^{3} + c_{2} x^{2} + c_{1} x + c_{0}$ — a cubic, the classic beam-deflection shape. Why? Fourth derivative zero ⇒ degree-3 polynomial.

Verify: $y = x^{3}$ gives $y^{''''} = 0$ ✓, and $y^{''} = 6 x$ so bending energy $\int_{0}^{1} \frac{1}{2} (6 x)^{2} d x = \int_{0}^{1} 18 x^{2} d x = 6$ (finite, well-defined) ✓.

Recall Which cell am I in? (self-quiz)

$L = 1 + y^{'2}$ — which cell and which shortcut? ::: Cell A (only $y^{'}$ ) → $\partial L / \partial y^{'} =$ const. $L = y 1 + y^{'2}$ — which cell? ::: Cell B (no bare $x$ ) → Beltrami identity. $L = x 1 + y^{'2}$ — which cell? ::: Cell C (no bare $y$ ) → $\partial L / \partial y^{'} =$ const, but path bends (valid only on $x \geq c_{3}$ ). $L = y^{'} g (y)$ gives E–L $0 = 0$ . What does that mean? ::: $L$ is a total derivative — every path is extremal (Cell E-5a, degenerate). $L = L (x, y)$ has no $y^{'}$ — what does E–L become? ::: An algebraic condition $\partial L / \partial y = 0$ ; no ODE, may clash with endpoints (Cell E-5b). Right end is free in $y$ . What replaces the missing value? ::: The natural BC $\partial L / \partial y^{'} = 0$ there (Cell F-6a). The endpoint's $x$ is also free — what condition? ::: Transversality $L - y^{'} \partial L / \partial y^{'} = 0$ at that end (Cell F-6b). $L$ contains $y^{''}$ . What equation replaces E–L? ::: Euler–Poisson: add $+ \frac{d ^{2}}{d x ^{2}} \frac{\partial L}{\partial y ^{''}}$ (Cell H).