Calculus of Variations: The Euler--Lagrange equation
CalculusOfVariationsTheEulerLagrangeEquation
2026-09-02 04:06:43
2026-09-14 04:10:29
Topic
calculus of variations
Euler-Lagrange equation
first variation
Fundamental Lemma
integration by parts
stationary function
extremal
functional derivative
weak form
strong form
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\section*{Calculus of Variations: The Euler--Lagrange Equation}
The first four entries of this series have been building toward one result.
CV00 introduced functionals and admissible curves. CV01 made local optimality
precise in function space. CV02 defined the first variation,
\begin{equation}
\delta J[y;\eta]
=
\left.\frac{d}{d\epsilon}J[y+\epsilon\eta]\right|_{\epsilon=0},
\end{equation}
and showed that for
\begin{equation}
J[y]
=
\int_a^b F(x,y,y')\,dx,
\end{equation}
one has
\begin{equation}
\delta J[y;\eta]
=
\int_a^b
\left(
F_y\eta+F_{y'}\eta'
\right)dx.
\end{equation}
CV03 proved the Fundamental Lemma: if a continuous function has zero integral
against every sufficiently localized test function, then that function must
vanish pointwise.
The Euler--Lagrange equation is what results when those pieces are assembled.
For a stationary fixed-endpoint curve,
\begin{equation}
F_y-
\frac{d}{dx}F_{y'}=0.
\end{equation}
This equation is the central necessary condition of classical calculus of
variations. Its importance in physics is difficult to overstate. With the
independent variable interpreted as time and $F$ as a Lagrangian, it becomes
Lagrange's equation of motion. With $F$ representing arc length, optical path,
energy, or field action, the same structure generates geodesics, ray equations,
minimum-energy configurations, and field equations.
The purpose of this entry is not merely to memorize the formula. The goal is
to understand exactly why it follows, which assumptions are used, what it does
and does not prove, and how to apply it reliably.
\section{Learning objectives}
After this entry, the reader should be able to
\begin{enumerate}
\item state the classical fixed-endpoint Euler--Lagrange theorem with explicit
regularity hypotheses;
\item derive the theorem from local extremality, the first variation,
integration by parts, and the Fundamental Lemma;
\item identify where the fixed-endpoint condition is used;
\item distinguish the ordinary derivative $d/dx$ from the partial derivatives
$F_y$ and $F_{y'}$;
\item expand the total derivative $dF_{y'}/dx$ by the chain rule;
\item determine when the Euler--Lagrange equation can be solved explicitly for
$y''$;
\item use the equation to test and construct candidate extremals;
\item distinguish a stationary function or extremal from a proven minimum;
\item interpret the first-variation identity as a weak form and the
Euler--Lagrange ODE as a strong form under sufficient regularity;
\item use the functional-derivative notation $\delta J/\delta y$ correctly;
and
\item recognize how the surviving boundary term leads to natural boundary
conditions in CV05.
\end{enumerate}
\section{The entire logical chain}
The Euler--Lagrange equation is not obtained by differentiating a functional in
one step. The derivation consists of a sequence of logically distinct facts.
\begin{center}
\includegraphics{CV04_derivation_chain.png}
\vspace{0.45em}
\textbf{Figure.}
Logical chain for the classical fixed-endpoint Euler--Lagrange theorem. Each
arrow represents a separate argument: finite-dimensional Fermat stationarity,
the first-variation calculation, integration by parts, endpoint admissibility,
and the Fundamental Lemma.
\end{center}
The chain can be summarized as
\begin{equation}
\text{local extremum}
\Longrightarrow
\delta J[y;\eta]=0\ \text{for every admissible }\eta
\Longrightarrow
\text{integral identity}
\Longrightarrow
\text{Euler--Lagrange equation}.
\end{equation}
Only the forward implication is guaranteed. Solving the Euler--Lagrange
equation finds stationary candidates. It does not, by itself, prove that a
candidate is a minimum.
\section{Classical fixed-endpoint theorem}
We now state a clean classical version. More general versions require less
smoothness, but the following hypotheses make every step of the proof
transparent.
\textbf{Theorem: Euler--Lagrange necessary condition.}
Let
\begin{equation}
J[y]
=
\int_a^b F(x,y,y')\,dx,
\end{equation}
where $F\in C^2$ on an open set containing the values
$(x,y(x),y'(x))$ under consideration. Let
\begin{equation}
\mathcal A
=
\left\{
y\in C^2([a,b]):
y(a)=A,
\quad
y(b)=B
\right\}.
\end{equation}
Suppose $y_\ast\in\mathcal A$ is a weak local minimum or weak local maximum of
$J$. Then $y_\ast$ satisfies
\begin{equation}
F_y\!\left(x,y_\ast,y_\ast'\right)
-
\frac{d}{dx}
F_{y'}\!\left(x,y_\ast,y_\ast'\right)
=0
\end{equation}
for every $x\in(a,b)$.
A strong local extremum also satisfies the theorem because every strong local
extremum is, in particular, a weak local extremum under the classical
$C^0$/$C^1$ hierarchy developed in CV01.
\subsection{Why these hypotheses are convenient}
The proof uses several operations:
\begin{itemize}
\item differentiating $F(x,y+\epsilon\eta,y'+\epsilon\eta')$ with respect to
$\epsilon$;
\item differentiating $F_{y'}(x,y_\ast,y_\ast')$ with respect to $x$;
\item integrating by parts; and
\item applying the classical Fundamental Lemma to a continuous coefficient.
\end{itemize}
The assumptions $F\in C^2$ and $y_\ast\in C^2$ are stronger than strictly
necessary, but they ensure all four steps are classical and pointwise. Later
sections explain the weaker-form viewpoint.
\section{Proof of the Euler--Lagrange theorem}
The proof is short once the earlier machinery has been established, but every
step has a specific purpose.
\subsection{Step 1: choose an arbitrary admissible variation}
Let $\eta\in C^1([a,b])$ satisfy
\begin{equation}
\eta(a)=0,
\qquad
\eta(b)=0.
\end{equation}
Construct the family
\begin{equation}
y_\epsilon(x)
=
y_\ast(x)+\epsilon\eta(x).
\end{equation}
Because $\eta$ vanishes at the endpoints,
\begin{equation}
y_\epsilon(a)=A,
\qquad
y_\epsilon(b)=B,
\end{equation}
so the perturbed family remains in the fixed-endpoint admissible class for
sufficiently small $\epsilon$.
\subsection{Step 2: reduce to an ordinary scalar extremum}
Define
\begin{equation}
\Phi(\epsilon)
=
J[y_\ast+\epsilon\eta].
\end{equation}
Because $y_\ast$ is a local extremum of $J$, the scalar function $\Phi$ has a
local extremum at $\epsilon=0$. Ordinary Fermat stationarity therefore gives
\begin{equation}
\Phi'(0)=0.
\end{equation}
By definition of the first variation,
\begin{equation}
\delta J[y_\ast;\eta]=0.
\end{equation}
This conclusion holds for every admissible variation $\eta$.
\subsection{Step 3: insert the first-variation formula}
From CV02,
\begin{equation}
\delta J[y_\ast;\eta]
=
\int_a^b
\left(
F_y\eta
+
F_{y'}\eta'
\right)dx,
\end{equation}
where the partial derivatives of $F$ are evaluated along
$(x,y_\ast(x),y_\ast'(x))$.
Stationarity therefore implies
\begin{equation}
\int_a^b
\left(
F_y\eta
+
F_{y'}\eta'
\right)dx
=0
\end{equation}
for every admissible $\eta$.
\subsection{Step 4: integrate by parts}
The term involving $\eta'$ cannot yet be handled directly by the Fundamental
Lemma because the test function appears differentiated. Integrate that term
by parts:
\begin{equation}
\int_a^b F_{y'}\eta'\,dx
=
\left[F_{y'}\eta\right]_a^b
-
\int_a^b
\frac{d}{dx}F_{y'}\,\eta\,dx.
\end{equation}
Thus
\begin{equation}
\delta J[y_\ast;\eta]
=
\left[F_{y'}\eta\right]_a^b
+
\int_a^b
\left(
F_y-
\frac{d}{dx}F_{y'}
\right)
\eta\,dx.
\end{equation}
\begin{center}
\includegraphics{CV04_integration_by_parts_transfer.png}
\vspace{0.45em}
\textbf{Figure.}
Integration by parts transfers the derivative from the arbitrary variation
$\eta$ onto the coefficient $F_{y'}$. This creates both the interior
Euler--Lagrange expression and a boundary term. Fixed endpoints remove the
boundary term; free endpoints will not, which is the subject of CV05.
\end{center}
\subsection{Step 5: use the fixed endpoints}
Since
\begin{equation}
\eta(a)=\eta(b)=0,
\end{equation}
one has
\begin{equation}
\left[F_{y'}\eta\right]_a^b=0.
\end{equation}
Therefore stationarity reduces to
\begin{equation}
\int_a^b
\left(
F_y-
\frac{d}{dx}F_{y'}
\right)
\eta(x)\,dx
=0
\end{equation}
for every admissible variation $\eta$.
This is the exact point at which the fixed-endpoint assumption enters the
classical proof.
\subsection{Step 6: apply the Fundamental Lemma}
Define
\begin{equation}
g(x)
=
F_y\!\left(x,y_\ast,y_\ast'\right)
-
\frac{d}{dx}
F_{y'}\!\left(x,y_\ast,y_\ast'\right).
\end{equation}
Under the stated smoothness assumptions, $g$ is continuous. The stationarity
condition says
\begin{equation}
\int_a^b g(x)\eta(x)\,dx=0
\end{equation}
for every admissible test function. In particular, it holds for every smooth
compactly supported test function in $(a,b)$. The Fundamental Lemma therefore
gives
\begin{equation}
g(x)=0
\end{equation}
for all $x\in(a,b)$.
Consequently,
\begin{equation}
F_y-
\frac{d}{dx}F_{y'}=0.
\end{equation}
This completes the proof.
\section{What integration by parts is really doing}
The first-variation formula contains two independent pieces of perturbation
data: $\eta$ and $\eta'$. The Fundamental Lemma is designed for an integral of
the form
\begin{equation}
\int_a^b g(x)\eta(x)\,dx=0,
\end{equation}
not for an expression containing both $\eta$ and $\eta'$. Integration by parts
reorganizes the first variation so that the arbitrary interior perturbation
appears only as $\eta$.
This is why the operation is structural rather than cosmetic. It separates
the first variation into
\begin{equation}
\text{boundary contribution}
+
\text{interior contribution}.
\end{equation}
The interior contribution produces the Euler--Lagrange differential equation.
The boundary contribution produces endpoint conditions when the endpoints are
not fixed.
This pattern reappears throughout mathematical physics. In field theory,
multidimensional integration by parts separates bulk field equations from
boundary terms. In finite-element and weak-form methods, the same operation
reduces derivative requirements on the trial or test functions.
\section{Anatomy of the Euler--Lagrange equation}
For
\begin{equation}
F=F(x,y,y'),
\end{equation}
the equation is
\begin{equation}
F_y-
\frac{d}{dx}F_{y'}=0.
\end{equation}
There are two different kinds of derivative in this expression.
\subsection{Partial derivative with respect to $y$}
The quantity
\begin{equation}
F_y
=
\frac{\partial F}{\partial y}
\end{equation}
means differentiate the function $F(x,y,p)$ with respect to its second
argument while holding the other arguments fixed.
\subsection{Partial derivative with respect to the slope argument}
Similarly,
\begin{equation}
F_{y'}
=
\frac{\partial F}{\partial y'}
\end{equation}
means differentiate $F$ with respect to its third argument.
\subsection{Total derivative along the candidate curve}
After $F_{y'}$ is formed, it becomes a function of $x$ through
\begin{equation}
F_{y'}
=
F_{y'}\!\left(x,y(x),y'(x)\right).
\end{equation}
Its derivative in the Euler--Lagrange equation is therefore a total derivative:
\begin{equation}
\frac{d}{dx}F_{y'}
=
F_{xy'}
+
F_{yy'}y'
+
F_{y'y'}y''.
\end{equation}
Substitution gives the expanded form
\begin{equation}
F_y
-
F_{xy'}
-
F_{yy'}y'
-
F_{y'y'}y''
=0.
\end{equation}
This expansion is often useful when converting the variational condition into
an explicit differential equation.
\section{Regular and degenerate cases}
If
\begin{equation}
F_{y'y'}\neq 0,
\end{equation}
then locally the expanded Euler--Lagrange equation can be solved for $y''$:
\begin{equation}
y''
=
\frac{
F_y-F_{xy'}-F_{yy'}y'
}{F_{y'y'}}.
\end{equation}
Such a problem is often called \emph{regular} with respect to the slope
variable. The Euler--Lagrange condition is then an ordinary second-order
differential equation.
If
\begin{equation}
F_{y'y'}=0,
\end{equation}
the equation can be degenerate. It may reduce to a first-order relation or an
algebraic constraint rather than determining $y''$. This is an early glimpse
of the distinction between regular and singular variational problems.
\section{A reliable calculation workflow}
For practical calculations it is useful to separate the operations rather than
trying to write the final differential equation from memory.
\begin{center}
\includegraphics{CV04_euler_lagrange_workflow.png}
\vspace{0.45em}
\textbf{Figure.}
A robust Euler--Lagrange workflow. Treat $F$ as a function of independent
arguments, compute the two required partial derivatives, take the total
$x$-derivative only after substituting the path dependence, and then assemble
the equation.
\end{center}
The algorithm is
\begin{enumerate}
\item Identify the integrand $F(x,y,y')$.
\item Compute $F_y$.
\item Compute $F_{y'}$.
\item Compute the total derivative $d(F_{y'})/dx$.
\item Form
\begin{equation}
F_y-\frac{d}{dx}F_{y'}=0.
\end{equation}
\item Solve the resulting differential equation subject to the original
endpoint or boundary conditions.
\item After finding a stationary curve, separately ask whether it is actually
a minimum, maximum, or saddle-type stationary function.
\end{enumerate}
\section{Example 1: quadratic slope functional}
Consider
\begin{equation}
J[y]
=
\frac{1}{2}
\int_a^b y'(x)^2\,dx
\end{equation}
with fixed endpoints
\begin{equation}
y(a)=A,
\qquad
y(b)=B.
\end{equation}
The integrand is
\begin{equation}
F(x,y,y')
=
\frac{1}{2}y'^2.
\end{equation}
Therefore
\begin{equation}
F_y=0,
\qquad
F_{y'}=y'.
\end{equation}
The Euler--Lagrange equation gives
\begin{equation}
0-
\frac{d}{dx}(y')=0,
\end{equation}
so
\begin{equation}
y''=0.
\end{equation}
Integrating twice,
\begin{equation}
y(x)=C_1x+C_2.
\end{equation}
The endpoint conditions determine the straight line
\begin{equation}
y(x)
=
A+
\frac{B-A}{b-a}(x-a).
\end{equation}
CV01 already proved directly that this curve is the global minimizer of the
quadratic slope functional. Here Euler--Lagrange recovers it as the stationary
candidate.
This comparison is useful: Euler--Lagrange gives a necessary differential
equation, while the direct square-completion or convexity argument establishes
actual minimality.
\section{Example 2: shortest planar curve}
The length of a graph $y(x)$ between fixed $x$-coordinates is
\begin{equation}
J[y]
=
\int_a^b
\sqrt{1+y'^2}\,dx.
\end{equation}
Here
\begin{equation}
F(x,y,y')
=
\sqrt{1+y'^2}.
\end{equation}
Since $F$ does not depend explicitly on $y$,
\begin{equation}
F_y=0.
\end{equation}
Also,
\begin{equation}
F_{y'}
=
\frac{y'}{\sqrt{1+y'^2}}.
\end{equation}
Euler--Lagrange yields
\begin{equation}
\frac{d}{dx}
\left(
\frac{y'}{\sqrt{1+y'^2}}
\right)
=0.
\end{equation}
Differentiate explicitly:
\begin{equation}
\frac{d}{dx}
\left(
\frac{y'}{\sqrt{1+y'^2}}
\right)
=
\frac{y''}{(1+y'^2)^{3/2}}.
\end{equation}
Hence
\begin{equation}
y''=0,
\end{equation}
and the stationary curve is again a straight line.
CV04E2 will develop this problem in more detail, including the geometric
interpretation and direct length comparison.
\section{Example 3: a nonlinear integrand}
Consider
\begin{equation}
J[y]
=
\int_a^b
\left(
\frac{1}{4}y'^4+
\frac{1}{2}y^2
\right)dx.
\end{equation}
The integrand is
\begin{equation}
F
=
\frac{1}{4}y'^4+
\frac{1}{2}y^2.
\end{equation}
Therefore
\begin{equation}
F_y=y,
\qquad
F_{y'}=y'^3.
\end{equation}
Taking the total derivative,
\begin{equation}
\frac{d}{dx}F_{y'}
=
3y'^2y''.
\end{equation}
Thus the Euler--Lagrange equation is
\begin{equation}
y-3y'^2y''=0.
\end{equation}
This example illustrates why one should not assume that the Euler--Lagrange
equation is linear. A simple-looking integral functional can produce a highly
nonlinear differential equation.
\section{Example 4: preview of classical mechanics}
Let the independent variable be time $t$ and consider the harmonic-oscillator
action
\begin{equation}
S[q]
=
\int_{t_1}^{t_2}
L(q,\dot q,t)\,dt,
\end{equation}
with
\begin{equation}
L(q,\dot q)
=
\frac{1}{2}m\dot q^2
-
\frac{1}{2}kq^2.
\end{equation}
The Euler--Lagrange equation is now written
\begin{equation}
\frac{\partial L}{\partial q}
-
\frac{d}{dt}
\frac{\partial L}{\partial \dot q}
=0.
\end{equation}
Compute
\begin{equation}
\frac{\partial L}{\partial q}
=-kq,
\qquad
\frac{\partial L}{\partial \dot q}
=m\dot q.
\end{equation}
Therefore
\begin{equation}
-kq-
\frac{d}{dt}(m\dot q)=0,
\end{equation}
or
\begin{equation}
m\ddot q+kq=0.
\end{equation}
Thus the familiar differential equation for a harmonic oscillator is itself an
Euler--Lagrange equation. CV14 will derive the full connection between
Hamilton's principle and Lagrange's equations for many-degree-of-freedom
mechanical systems.
\section{Necessary does not mean sufficient}
The theorem says
\begin{equation}
\text{local extremum}
\Longrightarrow
\text{Euler--Lagrange equation}.
\end{equation}
The converse is false in general.
A solution of the Euler--Lagrange equation is commonly called an
\emph{extremal} in classical terminology, even though it may not actually be a
minimum or maximum. To avoid ambiguity, it is often helpful to say
\emph{stationary extremal} or \emph{Euler--Lagrange extremal} when classification
has not yet been established.
\subsection{Stationary maximum}
Consider
\begin{equation}
J[y]
=
-
\int_0^1 y'^2\,dx
\end{equation}
with $y(0)=y(1)=0$. The Euler--Lagrange equation is again
\begin{equation}
y''=0,
\end{equation}
so the only endpoint-compatible stationary curve is $y=0$. But now
\begin{equation}
J[y]\leq 0=J[0]
\end{equation}
for every admissible $y$. Thus $y=0$ is a global maximum, not a minimum.
\subsection{Stationary saddle-type example}
Consider
\begin{equation}
J[y]
=
\int_0^1
\left(
y'^2-2\pi^2y^2
\right)dx
\end{equation}
with $y(0)=y(1)=0$. The zero function satisfies the Euler--Lagrange equation
and is stationary.
Now compare the two admissible directions
\begin{equation}
\eta_1(x)=\sin(\pi x),
\qquad
\eta_2(x)=\sin(2\pi x).
\end{equation}
For $y=\epsilon\eta_1$,
\begin{equation}
J[\epsilon\eta_1]
=
-
\frac{\pi^2}{2}\epsilon^2
<0
\end{equation}
when $\epsilon\neq 0$, while for $y=\epsilon\eta_2$,
\begin{equation}
J[\epsilon\eta_2]
=
\pi^2\epsilon^2
>0.
\end{equation}
Therefore arbitrarily close admissible curves exist with both larger and
smaller values of $J$. The stationary curve $y=0$ is saddle-type.
\begin{center}
\begin{tikzpicture}
\begin{axis}[
width=13.4cm,height=8.1cm,
xmin=-1.05,xmax=1.05,ymin=-1.25,ymax=1.45,
xlabel={$\epsilon$},ylabel={$\Phi_\eta(\epsilon)-\Phi_\eta(0)$},
title={Stationarity does not classify the extremal},
grid=major,
legend style={at={(0.98,0.98)},anchor=north east,draw=gray!40,fill=white},
tick label style={font=\small},label style={font=\small},title style={font=\normalsize},
every axis plot/.append style={domain=-1:1,samples=250,line width=1.1pt,no marks}]
\addplot[blue] {x^2};
\addlegendentry{direction $\eta_1$: curves upward}
\addplot[orange] {-0.8*x^2};
\addlegendentry{direction $\eta_2$: curves downward}
\addplot[green!60!black,dashed] {0.35*x^4};
\addlegendentry{direction $\eta_3$: flat to second order}
\addplot[only marks,mark=*,mark size=2.7pt,black] coordinates {(0,0)};
\draw[->,black] (axis cs:0.32,0.25) -- (axis cs:0.02,0.01);
\node[anchor=west] at (axis cs:-0.14,0.4) {all have $\Phi_\eta'(0)=0$};
\end{axis}
\end{tikzpicture}
%\includegraphics{CV04_stationarity_not_classification.png}
\vspace{0.45em}
\textbf{Figure.}
Vanishing first variation means that every one-dimensional slice
$\Phi_\eta(\epsilon)=J[y_\ast+\epsilon\eta]$ has zero slope at
$\epsilon=0$. It does not determine the curvature of those slices. Different
directions can bend upward, downward, or remain flat to second order, so
stationarity alone does not classify the extremal.
\end{center}
Second-variation theory begins in CV11 precisely because this classification
question requires additional information.
\section{Weak form and strong form}
Before applying the Fundamental Lemma, stationarity can be written as
\begin{equation}
\int_a^b
\left(
F_y\eta+F_{y'}\eta'
\right)dx=0
\end{equation}
for every admissible test function $\eta$. This is a variational or
\emph{weak-form} statement.
After integration by parts and sufficient regularity, the Fundamental Lemma
produces the pointwise differential equation
\begin{equation}
F_y-
\frac{d}{dx}F_{y'}=0.
\end{equation}
This is the corresponding classical \emph{strong form}.
In modern analysis, a function may satisfy the weak variational identity even
when it does not possess enough classical derivatives for the strong equation
to make pointwise sense. Under additional regularity, weak solutions can often
be shown to satisfy the strong equation. This weak-to-strong relationship is a
major bridge from the calculus of variations to partial differential equations
and finite-element methods.
\section{Functional derivative notation}
For a first-order functional, one often writes
\begin{equation}
\frac{\delta J}{\delta y}
=
F_y-
\frac{d}{dx}F_{y'}.
\end{equation}
Then the first variation may be represented schematically as
\begin{equation}
\delta J[y;\eta]
=
\int_a^b
\frac{\delta J}{\delta y}\,\eta\,dx
\end{equation}
when the appropriate boundary term vanishes.
The notation is useful, but it hides the integration-by-parts step. In full,
\begin{equation}
\delta J[y;\eta]
=
\left[F_{y'}\eta\right]_a^b
+
\int_a^b
\frac{\delta J}{\delta y}\,\eta\,dx.
\end{equation}
Thus the functional derivative captures the interior coefficient of the
variation. Boundary conditions must still be handled separately.
This distinction becomes essential in CV05, field theory, and Hamiltonian
mechanics.
\section{Why endpoint conditions are not bookkeeping}
The fixed-endpoint proof used
\begin{equation}
\eta(a)=\eta(b)=0
\end{equation}
only after integration by parts. If an endpoint is free, the term
\begin{equation}
\left[F_{y'}\eta\right]_a^b
\end{equation}
does not automatically vanish.
Stationarity must then control both the interior integral and the boundary
contribution. The interior still yields the Euler--Lagrange equation, while the
boundary term yields a \emph{natural boundary condition}. If an endpoint is
allowed to move along a prescribed curve, still more general transversality
conditions appear.
This is why the admissible class introduced in CV01 is part of the theorem
itself. Changing the endpoint freedom changes the necessary conditions.
\section{Dimensional consistency}
In physical applications the two terms of the Euler--Lagrange equation must
have the same dimensions.
If $x$ has dimensions $[x]$, $y$ has dimensions $[y]$, and $F$ has dimensions
$[F]$, then
\begin{equation}
[F_y]
=
\frac{[F]}{[y]}.
\end{equation}
Since
\begin{equation}
[y']
=
\frac{[y]}{[x]},
\end{equation}
one has
\begin{equation}
[F_{y'}]
=
\frac{[F][x]}{[y]}.
\end{equation}
Taking $d/dx$ gives
\begin{equation}
\left[
\frac{d}{dx}F_{y'}
\right]
=
\frac{[F]}{[y]},
\end{equation}
matching $[F_y]$. This provides a quick error check in mechanics, optics, and
engineering applications.
\section{Common mistakes}
\subsection{Mistake 1: using a partial derivative where a total derivative is required}
The equation contains
\begin{equation}
\frac{d}{dx}F_{y'},
\end{equation}
not merely $\partial F_{y'}/\partial x$. The quantity $F_{y'}$ also changes
because $y(x)$ and $y'(x)$ change.
\subsection{Mistake 2: dropping the boundary term before integrating by parts}
The boundary term must first be produced. It vanishes only because of the
specific endpoint conditions of the problem.
\subsection{Mistake 3: treating $y'$ as dependent on $y$ when computing partial derivatives}
When computing $F_y$ and $F_{y'}$, treat $x$, $y$, and $y'$ as independent
arguments of $F$. Their path dependence is reintroduced when taking the total
$x$-derivative.
\subsection{Mistake 4: believing every Euler--Lagrange solution is a minimum}
Euler--Lagrange is a stationarity condition. A stationary curve can be a
minimum, maximum, or saddle-type extremal.
\subsection{Mistake 5: forgetting the original endpoint conditions}
The differential equation produces a family of candidate solutions. The
original boundary data select the admissible members of that family.
\subsection{Mistake 6: cancelling the variation from the integral}
From
\begin{equation}
\int_a^b g(x)\eta(x)\,dx=0
\end{equation}
one cannot algebraically cancel $\eta$. The conclusion $g=0$ follows from the
Fundamental Lemma and the richness of the test-function class.
\subsection{Mistake 7: confusing existence with stationarity}
Even if the Euler--Lagrange boundary-value problem has a formal solution, a
minimizer of the original variational problem need not exist under arbitrary
hypotheses. Existence is a separate question involving compactness,
coercivity, and lower semicontinuity in more advanced theory.
\section{A compact theorem checklist}
When using Euler--Lagrange, ask the following questions.
\begin{enumerate}
\item What is the admissible class?
\item Are the endpoints fixed or free?
\item Is the candidate regular enough for the classical theorem?
\item What is the integrand $F(x,y,y')$?
\item What are $F_y$ and $F_{y'}$?
\item Did I take a total derivative of $F_{y'}$?
\item What boundary conditions must the resulting ODE satisfy?
\item Have I shown only stationarity, or have I actually proved minimality?
\end{enumerate}
Keeping these questions separate prevents most common variational mistakes.
\section{What CV05 and CV06 add}
CV04 has used fixed endpoints to eliminate the boundary term
\begin{equation}
\left[F_{y'}\eta\right]_a^b.
\end{equation}
CV05 removes that simplification. Free endpoints, movable endpoints, and
endpoints constrained to curves produce natural boundary and transversality
conditions.
CV06 then studies special forms of the Euler--Lagrange equation. If the
integrand lacks explicit dependence on certain variables, the differential
equation admits first integrals such as the Beltrami identity. These
conservation-like reductions are especially important in the catenary,
brachistochrone, mechanics, and optics.
\section{Summary}
For the classical fixed-endpoint functional
\begin{equation}
J[y]
=
\int_a^b F(x,y,y')\,dx,
\end{equation}
a sufficiently smooth local extremum must satisfy the Euler--Lagrange equation
\begin{equation}
F_y-
\frac{d}{dx}F_{y'}=0.
\end{equation}
The derivation is
\begin{equation}
\text{local extremum}
\Longrightarrow
\delta J=0
\Longrightarrow
\text{integration by parts}
\Longrightarrow
\text{Fundamental Lemma}
\Longrightarrow
\text{Euler--Lagrange}.
\end{equation}
Integration by parts separates boundary and interior behavior. Fixed
endpoints remove the boundary term, while the Fundamental Lemma converts the
remaining integral identity into a pointwise differential equation.
The resulting equation is a necessary condition for stationarity, not a proof
of minimality. Classification, existence, boundary freedom, conservation
laws, and physical specialization all require additional theory developed in
the following entries.
\section{References and further reading}
\begin{itemize}
\item I. M. Gelfand and S. V. Fomin, \emph{Calculus of Variations}.
\item B. van Brunt, \emph{The Calculus of Variations}.
\item C. Fox, \emph{An Introduction to the Calculus of Variations}.
\item C. Lanczos, \emph{The Variational Principles of Mechanics}.
\item H. Goldstein, C. Poole, and J. Safko, \emph{Classical Mechanics}.
\item L. C. Evans, \emph{Partial Differential Equations}, for weak formulations
and the modern variational viewpoint.
\end{itemize}