Calculus of Variations: Variations and the First Variation
CalculusOfVariationsVariationsAndTheFirstVariation
2026-09-07 00:22:07
2026-09-07 00:22:07
Topic
calculus of variations
variation
first variation
Gateaux derivative
directional derivative
admissible variation
differentiation under the integral sign
Euler-Lagrange equation
% this is the default PhysicsLibrary preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics).
% If you use \includegraphics{figure} with no file extension, PhysicsLibrary
% will prefer figure.png before falling back to figure.eps.
\usepackage{graphicx}
\DeclareGraphicsExtensions{.png,.jpg,.jpeg,.pdf,.eps}
% for neatly defining theorems and propositions
%\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}
% there are many more packages, add them here as you need them
\usepackage{tikz}
\usepackage{pgfplots}
% define commands here
\renewcommand{\familydefault}{\sfdefault}
\pgfplotsset{compat=1.18}
\section*{Calculus of Variations: Variations and the First Variation}
CV01 established the setting of the classical variational problem. A
functional acts on an admissible class of functions, and local optimality must
be defined relative to a chosen neighborhood in function space. The next
question is the analogue of ordinary differentiation:
\begin{quote}
If one perturbs an admissible function slightly, how does the scalar value of
$J$ change?
\end{quote}
The central idea is to convert an infinite-dimensional problem into a
one-parameter problem. Instead of trying to vary a function in all possible
ways at once, one chooses a single admissible perturbation shape $\eta(x)$ and
studies the family
\begin{equation}
y_{\epsilon}(x)=y(x)+\epsilon\eta(x).
\end{equation}
For fixed $y$ and $\eta$, the functional becomes an ordinary scalar function of
$\epsilon$,
\begin{equation}
\Phi(\epsilon)=J[y+\epsilon\eta].
\end{equation}
The derivative of $\Phi$ at $\epsilon=0$ is the
\emph{first variation}. It plays the role that $f'(x_*)$ plays in ordinary
calculus. If $y$ is a local extremum, the first variation must vanish for
\emph{every admissible direction} $\eta$.
This entry makes that construction precise, derives the standard first
variation formula for first-order integral functionals, and works several
examples in full detail. The next two entries will then use integration by
parts and the Fundamental Lemma to pass from the vanishing of the first
variation to the Euler--Lagrange equation.
\section{Learning objectives}
After this entry, the reader should be able to
\begin{enumerate}
\item define a variation and an admissible one-parameter family
$y_{\epsilon}=y+\epsilon\eta$;
\item explain why endpoint conditions on $y$ induce endpoint conditions on
$\eta$;
\item define the scalar reduction
$\Phi(\epsilon)=J[y+\epsilon\eta]$;
\item define the first variation $\delta J[y;\eta]$ as a directional derivative;
\item derive the first variation formula for
$J[y]=\int_a^b F(x,y,y')\,dx$;
\item state clearly where differentiation under the integral sign is used;
\item prove that the first variation is linear in $\eta$;
\item compute first variations directly by expansion in $\epsilon$;
\item recognize that stationarity requires
$\delta J[y;\eta]=0$ for every admissible $\eta$;
\item distinguish the first variation from the second variation; and
\item see how the first variation formula is poised for integration by parts
and the Fundamental Lemma.
\end{enumerate}
\section{Variations as admissible perturbation directions}
Let $\mathcal A$ be an admissible class for a first-order problem. A
\emph{variation} is a comparison direction $\eta(x)$ used to perturb a
candidate function $y(x)$.
The classical perturbed family is
\begin{equation}
y_{\epsilon}(x)=y(x)+\epsilon\eta(x),
\end{equation}
where $\epsilon\in\mathbb R$ is small.
The word ``small'' has two roles. First, it keeps $y_{\epsilon}$ near $y$ in
whatever topology is relevant to the problem. Second, it allows one to use
ordinary differentiation with respect to the scalar parameter $\epsilon$.
A variation must be chosen so that, for sufficiently small $\epsilon$, the
perturbed curve remains admissible. Thus the admissibility conditions on $y$
induce admissibility conditions on $\eta$.
\subsection{Fixed endpoint problems}
Suppose the admissible class is
\begin{equation}
\mathcal A
=
\left\{
y\in C^1([a,b]):
y(a)=A,
\quad
y(b)=B
\right\}.
\end{equation}
If $y_{\epsilon}=y+\epsilon\eta$ is to satisfy the same endpoint conditions for
all sufficiently small $\epsilon$, then
\begin{equation}
y_{\epsilon}(a)=A,
\qquad
y_{\epsilon}(b)=B.
\end{equation}
Substituting the perturbed family gives
\begin{equation}
y(a)+\epsilon\eta(a)=A,
\qquad
y(b)+\epsilon\eta(b)=B.
\end{equation}
Because $y$ is already admissible, $y(a)=A$ and $y(b)=B$, so one obtains
\begin{equation}
\eta(a)=0,
\qquad
\eta(b)=0.
\end{equation}
Thus, for fixed-endpoint problems, admissible variations vanish at the
endpoints.
\begin{center}
\includegraphics{CV02_variation_family.png}
\vspace{0.45em}
\textbf{Figure.}
A typical one-parameter family $y_{\epsilon}=y+\epsilon\eta$ for a fixed-endpoint
problem. The comparison curves share the same endpoints because the variation
$\eta$ vanishes at $x=a$ and $x=b$.
\end{center}
\subsection{Other constraints}
If the admissible class includes different constraints, the variation must
respect them in the corresponding linearized sense.
\begin{itemize}
\item For periodic constraints $y(a)=y(b)$, one requires
$\eta(a)=\eta(b)$.
\item For an integral constraint such as
$\int_a^b G(x,y,y')\,dx=C$, not every $\eta$ is immediately admissible;
this issue leads to constrained variations and Lagrange multipliers in
later entries.
\item For inequality or obstacle constraints, the admissible directions may be
one-sided rather than arbitrary.
\end{itemize}
For the present entry, fixed-endpoint unconstrained problems are sufficient to
show the core mechanism.
\section{From an infinite-dimensional problem to an ordinary derivative}
The first conceptual simplification in the calculus of variations is the map
\begin{equation}
y
\longrightarrow
\bigl\{y+\epsilon\eta:\epsilon\in\mathbb R\bigr\}
\longrightarrow
\Phi(\epsilon)=J[y+\epsilon\eta].
\end{equation}
Once $y$ and $\eta$ are fixed, the functional no longer acts on an entire class
of functions. It acts on a one-parameter subfamily. Along that subfamily the
problem is ordinary calculus.
\begin{center}
\includegraphics{CV02_first_variation_reduction.png}
\vspace{0.45em}
\textbf{Figure.}
The standard reduction. A direction $\eta$ and scalar amplitude $\epsilon$
produce a comparison family $y+\epsilon\eta$. The functional then becomes the
ordinary scalar function $\Phi(\epsilon)=J[y+\epsilon\eta]$. Differentiating at
$\epsilon=0$ yields the first variation.
\end{center}
\subsection{Definition of the first variation}
Let $J$ be a functional and suppose the derivative exists. The
\emph{first variation of $J$ at $y$ in the direction $\eta$} is
\begin{equation}
\delta J[y;\eta]
=
\left.\frac{d}{d\epsilon}J[y+\epsilon\eta]\right|_{\epsilon=0}.
\end{equation}
Equivalently, if
\begin{equation}
\Phi(\epsilon)=J[y+\epsilon\eta],
\end{equation}
then
\begin{equation}
\delta J[y;\eta]=\Phi'(0).
\end{equation}
This is the classical G\^ateaux or directional derivative viewpoint. One does
not yet require the stronger uniform approximation property associated with the
Fr\'echet derivative. For deriving the Euler--Lagrange equation, the directional
viewpoint is the natural starting point.
\subsection{Stationarity as an infinite-dimensional Fermat principle}
In ordinary calculus, if $x_*$ is an interior local extremum of a differentiable
function $f$, then
\begin{equation}
f'(x_*)=0.
\end{equation}
In several variables, if $\mathbf x_*$ is an interior local extremum of a
smooth function $f(\mathbf x)$, then
\begin{equation}
\nabla f(\mathbf x_*)=\mathbf 0.
\end{equation}
The analogous variational principle is:
\begin{quote}
If $y_*$ is a local extremum of $J$, then for every admissible variation
$\eta$,
\begin{equation}
\delta J[y_*;\eta]=0.
\end{equation}
\end{quote}
The crucial phrase is ``for every admissible variation.'' A single vanishing
value of $\delta J[y;\eta]$ along one direction says very little. A local
extremum must be stationary with respect to all admissible first-order
perturbations.
\section{First variation of a first-order integral functional}
Consider the classical functional
\begin{equation}
J[y]
=
\int_a^b F\!\left(x,y(x),y'(x)\right)\,dx,
\end{equation}
where $F$ is continuously differentiable in its arguments and $y,\eta$ are
sufficiently smooth for the expressions below to make sense.
Set
\begin{equation}
y_{\epsilon}=y+\epsilon\eta,
\qquad
y_{\epsilon}'=y'+\epsilon\eta'.
\end{equation}
Then
\begin{equation}
\Phi(\epsilon)
=
J[y+\epsilon\eta]
=
\int_a^b
F\!\left(x,y+\epsilon\eta,y'+\epsilon\eta'\right)dx.
\end{equation}
\subsection{Main formula}
\textbf{Proposition.}
Assume that $F\in C^1$ and that differentiation under the integral sign is
justified for the perturbed family. Then the first variation is
\begin{equation}
\boxed{
\delta J[y;\eta]
=
\int_a^b
\left(
F_y\,\eta+F_{y'}\,\eta'
\right)dx
}
\end{equation}
where $F_y$ and $F_{y'}$ are evaluated along the unperturbed curve
$(x,y(x),y'(x))$.
\subsection{Proof}
Differentiate $\Phi(\epsilon)$ with respect to $\epsilon$:
\begin{equation}
\Phi'(\epsilon)
=
\frac{d}{d\epsilon}
\int_a^b
F\!\left(x,y+\epsilon\eta,y'+\epsilon\eta'\right)dx.
\end{equation}
Assuming differentiation may pass under the integral sign,
\begin{equation}
\Phi'(\epsilon)
=
\int_a^b
\frac{\partial}{\partial\epsilon}
F\!\left(x,y+\epsilon\eta,y'+\epsilon\eta'\right)dx.
\end{equation}
Now apply the chain rule to the integrand. Since $x$ is independent of
$\epsilon$,
\begin{equation}
\frac{\partial}{\partial\epsilon}
F\!\left(x,y+\epsilon\eta,y'+\epsilon\eta'\right)
=
F_y\!\left(x,y+\epsilon\eta,y'+\epsilon\eta'\right)\eta
+
F_{y'}\!\left(x,y+\epsilon\eta,y'+\epsilon\eta'\right)\eta'.
\end{equation}
Hence
\begin{equation}
\Phi'(\epsilon)
=
\int_a^b
\left[
F_y\!\left(x,y+\epsilon\eta,y'+\epsilon\eta'\right)\eta
+
F_{y'}\!\left(x,y+\epsilon\eta,y'+\epsilon\eta'\right)\eta'
\right]dx.
\end{equation}
Evaluating at $\epsilon=0$ gives
\begin{equation}
\delta J[y;\eta]
=
\Phi'(0)
=
\int_a^b
\left(
F_y(x,y,y')\eta
+
F_{y'}(x,y,y')\eta'
\right)dx,
\end{equation}
which is the claimed formula.
\subsection{Why the proof is important}
The proof shows exactly where the machinery comes from. The first variation is
not a mysterious formal symbol. It is an ordinary derivative with respect to
$\epsilon$ after one restricts the functional to a one-parameter family.
The proof also isolates the analytic step that needs hypotheses:
\emph{differentiation under the integral sign}. In classical coursework one
usually assumes enough smoothness for this to be valid. More advanced
functional analysis develops sharper conditions, but the basic variational idea
is already visible in the classical setting.
\section{Linearity of the first variation}
For fixed $y$, the map $\eta\mapsto\delta J[y;\eta]$ is linear.
\textbf{Proposition.}
Let $\alpha,\beta\in\mathbb R$ and let $\eta_1,\eta_2$ be admissible variations.
Then
\begin{equation}
\delta J[y;\alpha\eta_1+\beta\eta_2]
=
\alpha\,\delta J[y;\eta_1]
+
\beta\,\delta J[y;\eta_2].
\end{equation}
\textbf{Proof.}
Apply the first variation formula:
\begin{equation}
\delta J[y;\alpha\eta_1+\beta\eta_2]
=
\int_a^b
\left[
F_y(\alpha\eta_1+\beta\eta_2)
+
F_{y'}(\alpha\eta_1'+\beta\eta_2')
\right]dx.
\end{equation}
Distribute and use linearity of the integral:
\begin{equation}
\delta J[y;\alpha\eta_1+\beta\eta_2]
=
\alpha\int_a^b\left(F_y\eta_1+F_{y'}\eta_1'\right)dx
+
\beta\int_a^b\left(F_y\eta_2+F_{y'}\eta_2'\right)dx.
\end{equation}
Therefore
\begin{equation}
\delta J[y;\alpha\eta_1+\beta\eta_2]
=
\alpha\,\delta J[y;\eta_1]
+
\beta\,\delta J[y;\eta_2].
\end{equation}
This linearity is the reason one can think of $\delta J[y;\cdot]$ as an
analogue of a differential or covector acting on perturbation directions.
\section{Direct computation by $\epsilon$-expansion}
Although the general formula is fundamental, students should be able to compute
first variations directly from the definition. Expanding
$J[y+\epsilon\eta]$ in powers of $\epsilon$ is often the clearest way to see what
is happening.
\subsection{Example 1: the standard energy functional around the straight line}
Consider
\begin{equation}
J[y]
=
\int_0^1 y'(x)^2\,dx,
\end{equation}
with fixed endpoints
\begin{equation}
y(0)=0,
\qquad
y(1)=1.
\end{equation}
Take the candidate
\begin{equation}
y(x)=x,
\end{equation}
and the admissible variation
\begin{equation}
\eta(x)=\sin(\pi x),
\end{equation}
which satisfies $\eta(0)=\eta(1)=0$.
The perturbed family is
\begin{equation}
y_{\epsilon}(x)=x+\epsilon\sin(\pi x),
\end{equation}
so
\begin{equation}
y_{\epsilon}'(x)=1+\epsilon\pi\cos(\pi x).
\end{equation}
Substitute into $J$:
\begin{equation}
\Phi(\epsilon)=J[y_{\epsilon}]
=
\int_0^1\left(1+\epsilon\pi\cos(\pi x)\right)^2dx.
\end{equation}
Expand the square:
\begin{equation}
\Phi(\epsilon)
=
1
+
2\epsilon\pi\int_0^1\cos(\pi x)\,dx
+
\epsilon^2\pi^2\int_0^1\cos^2(\pi x)\,dx.
\end{equation}
The middle integral vanishes and the last integral equals $1/2$, hence
\begin{equation}
\Phi(\epsilon)=1+\frac{\pi^2}{2}\epsilon^2.
\end{equation}
Therefore
\begin{equation}
\delta J[y;\eta]=\Phi'(0)=0.
\end{equation}
This shows that the straight line is stationary in the direction
$\eta(x)=\sin(\pi x)$. In fact it is stationary for every admissible $\eta$,
which will become completely transparent after integration by parts in CV04.
\begin{center}
\includegraphics{CV02_scalar_phi_example.png}
\vspace{0.45em}
\textbf{Figure.}
For the family $y_{\epsilon}=x+\epsilon\sin(\pi x)$, the scalarized functional is
$\Phi(\epsilon)=1+(\pi^2/2)\epsilon^2$. The tangent at $\epsilon=0$ is
horizontal, so the first variation vanishes.
\end{center}
\subsection{Example 2: use the general formula}
Now compute the same first variation from the formula. Here
\begin{equation}
F(x,y,y')=y'^2.
\end{equation}
Therefore
\begin{equation}
F_y=0,
\qquad
F_{y'}=2y'.
\end{equation}
The first variation at a general admissible curve $y$ is
\begin{equation}
\delta J[y;\eta]
=
\int_0^1 2y'\eta'\,dx.
\end{equation}
For $y(x)=x$, one has $y'=1$, so
\begin{equation}
\delta J[x;\eta]
=
2\int_0^1\eta'(x)\,dx
=
2\bigl[\eta(x)\bigr]_0^1
=
0.
\end{equation}
Thus the straight line is stationary for every fixed-endpoint variation, as
expected.
\subsection{Example 3: a nonstationary admissible curve}
Still with
\begin{equation}
J[y]=\int_0^1 y'^2\,dx,
\end{equation}
consider the admissible curve
\begin{equation}
y(x)=x^2,
\end{equation}
which satisfies $y(0)=0$ and $y(1)=1$.
Choose the admissible variation
\begin{equation}
\eta(x)=x(1-x).
\end{equation}
Then
\begin{equation}
y'(x)=2x,
\qquad
\eta'(x)=1-2x.
\end{equation}
Using the formula,
\begin{equation}
\delta J[y;\eta]
=
2\int_0^1 y'\eta'\,dx
=
2\int_0^1 2x(1-2x)\,dx.
\end{equation}
Compute the integral:
\begin{equation}
\delta J[y;\eta]
=
4\int_0^1(x-2x^2)\,dx
=
4\left(\frac{1}{2}-\frac{2}{3}\right)
=
-\frac{2}{3}.
\end{equation}
Because the first variation is not zero, the curve $y=x^2$ is not stationary.
So one does not need the full Euler--Lagrange equation merely to rule out a
candidate. A single admissible direction with nonzero first variation is
already enough.
\subsection{Example 4: a functional containing both $y$ and $y'$}
Consider
\begin{equation}
J[y]=\int_0^1\left(y^2+y'^2\right)dx,
\end{equation}
with fixed endpoints $y(0)=y(1)=0$ and candidate $y(x)=0$.
Let $\eta$ be any admissible variation, so $\eta(0)=\eta(1)=0$. Then
\begin{equation}
J[\epsilon\eta]
=
\int_0^1\left(\epsilon^2\eta^2+\epsilon^2\eta'^2\right)dx
=
\epsilon^2\int_0^1\left(\eta^2+\eta'^2\right)dx.
\end{equation}
Hence
\begin{equation}
\Phi(\epsilon)
=
\epsilon^2\int_0^1\left(\eta^2+\eta'^2\right)dx,
\end{equation}
so
\begin{equation}
\delta J[0;\eta]=\Phi'(0)=0.
\end{equation}
This example is useful because it shows that the first variation alone detects
stationarity, not the full classification. Here the absence of a linear term
already suggests that $y=0$ should be minimizing, but one needs second-order
information to establish that rigorously in general.
\section{The bridge to Euler--Lagrange}
The first variation formula still contains $\eta'$:
\begin{equation}
\delta J[y;\eta]
=
\int_a^b\left(F_y\eta+F_{y'}\eta'\right)dx.
\end{equation}
The next step is to remove the derivative from the variation by integration by
parts:
\begin{equation}
\int_a^b F_{y'}\eta'\,dx
=
\bigl[F_{y'}\eta\bigr]_a^b
-
\int_a^b\frac{d}{dx}(F_{y'})\eta\,dx.
\end{equation}
Substituting gives
\begin{equation}
\delta J[y;\eta]
=
\bigl[F_{y'}\eta\bigr]_a^b
+
\int_a^b
\left(
F_y-
\frac{d}{dx}F_{y'}
\right)\eta\,dx.
\end{equation}
For fixed-endpoint variations, $\eta(a)=\eta(b)=0$, so the boundary term drops
out and one obtains
\begin{equation}
\delta J[y;\eta]
=
\int_a^b
\left(
F_y-
\frac{d}{dx}F_{y'}
\right)\eta\,dx.
\end{equation}
Now the structure is clear. If $y$ is stationary, then this integral vanishes
for every admissible variation $\eta$. The Fundamental Lemma will then imply
\begin{equation}
F_y-\frac{d}{dx}F_{y'}=0,
\end{equation}
which is the Euler--Lagrange equation.
This logical order matters:
\begin{equation}
\boxed{
\text{variation family}
\longrightarrow
\text{first variation}
\longrightarrow
\text{integration by parts}
\longrightarrow
\text{Fundamental Lemma}
\longrightarrow
\text{Euler--Lagrange}
}
\end{equation}
CV03 is devoted to the Fundamental Lemma because that final implication is a
nontrivial theorem, not a formal trick.
\section{Common misconceptions}
\subsection{Misconception 1: the variation $\eta$ is itself the new curve}
No. The perturbed curve is $y+\epsilon\eta$. The function $\eta$ provides the
shape of the perturbation, while $\epsilon$ controls its size.
\subsection{Misconception 2: if one example gives $\delta J=0$, the curve is stationary}
No. Stationarity requires
\begin{equation}
\delta J[y;\eta]=0
\qquad
\text{for every admissible }\eta.
\end{equation}
A single direction is not enough.
\subsection{Misconception 3: the first variation already proves a minimum}
Not in general. Vanishing first variation is a necessary condition for an
interior local extremum, but not a sufficient one. Classification belongs to
second-variation theory.
\subsection{Misconception 4: the first variation formula is obtained by formally ``varying'' symbols}
The formula does have a compact formal notation, but its basis is ordinary
calculus on the scalar function $\Phi(\epsilon)=J[y+\epsilon\eta]$.
\subsection{Misconception 5: differentiation under the integral sign is automatic}
One needs hypotheses to justify it. In introductory classical treatments the
required regularity is usually assumed from the start.
\section{Compact definition table}
\begin{center}
\begin{tabular}{|l|l|}
\hline
\textbf{Object} & \textbf{Meaning} \\
\hline
$\eta(x)$ & variation or perturbation direction \\
\hline
$y_{\epsilon}=y+\epsilon\eta$ & one-parameter family of nearby curves \\
\hline
$\Phi(\epsilon)=J[y+\epsilon\eta]$ & scalarized functional along one family \\
\hline
$\delta J[y;\eta]$ & first variation, equal to $\Phi'(0)$ \\
\hline
$\delta^2J$ & second variation, based on $\Phi''(0)$ \\
\hline
stationary function & one for which $\delta J[y;\eta]=0$ for all admissible $\eta$ \\
\hline
\end{tabular}
\end{center}
\section{What CV03 and CV04 add}
CV03 will prove the Fundamental Lemma of the Calculus of Variations. That
lemma tells us that if
\begin{equation}
\int_a^b g(x)\eta(x)\,dx=0
\end{equation}
for every smooth test function $\eta$ with compact support or vanishing
endpoints, then $g(x)=0$ identically.
CV04 will combine that lemma with the integration-by-parts form of the first
variation to obtain the Euler--Lagrange equation in its standard classical
form.
Thus CV02 provides the derivative computation, CV03 provides the theorem that
converts an integral identity into a pointwise statement, and CV04 assembles
those pieces into the central necessary condition of the subject.
\section{Summary}
The first variation is the directional derivative of a functional. Starting
from a candidate $y$ and an admissible variation $\eta$, one builds the family
$y+\epsilon\eta$ and defines
\begin{equation}
\delta J[y;\eta]
=
\left.\frac{d}{d\epsilon}J[y+\epsilon\eta]\right|_{\epsilon=0}.
\end{equation}
For the classical first-order functional
\begin{equation}
J[y]=\int_a^b F(x,y,y')\,dx,
\end{equation}
the first variation is
\begin{equation}
\delta J[y;\eta]
=
\int_a^b\left(F_y\eta+F_{y'}\eta'\right)dx.
\end{equation}
This quantity is linear in the variation direction and must vanish for every
admissible variation if $y$ is a stationary function.
The first variation therefore plays the same conceptual role in the calculus of
variations that the ordinary derivative plays in finite-dimensional calculus.
Its integration-by-parts form is the gateway to the Fundamental Lemma and the
Euler--Lagrange equation.
\section{References and further reading}
The following references are standard classical entry points into the subject.
They are listed here for mathematical orientation; the present article remains
self-contained.
\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 L. C. Evans, \emph{Partial Differential Equations}, for the modern weak and
functional-analytic viewpoint.
\item H. Goldstein, C. Poole, and J. Safko, \emph{Classical Mechanics}, for the
action principle connection.
\end{itemize}