Variational Calculus Part 2: The Euler-Lagrange Equation

Deriving the Euler-Lagrange equation, the fundamental differential equation that extremizing functions must satisfy in variational problems, using the first variation and the fundamental lemma.

Posted May 17, 2025 Updated Mar 15, 2026

By Ji-Ha Kim

11 min read

In Part 1, we introduced functionals

\[ J[y] \]

– functions of functions – and established that a necessary condition for a function

\[ y(x) \]

to extremize

\[ J[y] \]

is that its first variation must vanish for all admissible variations

\eta(x)

\delta J[y; \eta] = \left. \frac{d}{d\epsilon} J[y + \epsilon \eta] \right\vert_{\epsilon=0} = 0

This is the variational calculus equivalent of setting

\[ f'(x)=0 \]

in ordinary calculus. Now, our goal is to transform this abstract condition into a concrete, usable tool. We will focus on a very common type of functional and derive the celebrated Euler-Lagrange equation, a differential equation that the extremizing function

\[ y(x) \]

must satisfy.

1. The Standard Functional Form

Many problems in physics, engineering, and even machine learning involve functionals of the following form:

Definition. Standard Functional

A common type of functional depends on a function
$\[ y(x) \]$
, its first derivative
$\[ y'(x) = dy/dx \]$
, and the independent variable
$\[ x \]$
, integrated over an interval
$\[ [a, b] \]$
:
$J[y] = \int_a^b F(x, y(x), y'(x)) \, dx$
Here,
$\[ F(x, y, y') \]$
is a given function of three variables, often called the Lagrangian or the integrand function. We assume
$\[ F \]$
has continuous partial derivatives with respect to its arguments. We also assume
$\[ y(x) \]$
is twice continuously differentiable.

Examples from Part 1, like the arc length functional (

F = \sqrt{1+(y')^2}

) and the Fermat’s principle functional (

F = \sqrt{1+(y')^2}/v(x)

), fit this form.

2. Calculating the First Variation Explicitly

Let’s compute

\delta J[y; \eta]

for the standard functional. Recall that

\tilde{y}(x; \epsilon) = y(x) + \epsilon \eta(x)

. Then, its derivative is

\tilde{y}'(x; \epsilon) = y'(x) + \epsilon \eta'(x)

. Substituting into the functional:

J[y + \epsilon \eta] = \int_a^b F(x, y(x) + \epsilon \eta(x), y'(x) + \epsilon \eta'(x)) \, dx

To find the first variation, we differentiate this expression with respect to

\epsilon

and then set

\epsilon = 0

. Assuming we can differentiate under the integral sign (Leibniz integral rule, valid here due to our smoothness assumptions):

\delta J[y; \eta] = \left. \frac{d}{d\epsilon} \int_a^b F(x, y + \epsilon \eta, y' + \epsilon \eta') \, dx \right\vert_{\epsilon=0}

\delta J[y; \eta] = \int_a^b \left. \frac{d}{d\epsilon} F(x, y + \epsilon \eta, y' + \epsilon \eta') \right\vert_{\epsilon=0} \, dx

Now, we apply the chain rule to the integrand

\[ F \]

. Let

Y = y + \epsilon \eta

and

Y' = y' + \epsilon \eta'

. Then

\[ F = F(x, Y, Y') \]

. So,

\frac{dF}{d\epsilon} = \frac{\partial F}{\partial Y} \frac{\partial Y}{\partial \epsilon} + \frac{\partial F}{\partial Y'} \frac{\partial Y'}{\partial \epsilon}

. We have:

\frac{\partial Y}{\partial \epsilon} = \frac{\partial}{\partial \epsilon}(y + \epsilon \eta) = \eta

\frac{\partial Y'}{\partial \epsilon} = \frac{\partial}{\partial \epsilon}(y' + \epsilon \eta') = \eta'

Therefore,

\frac{d}{d\epsilon} F(x, y + \epsilon \eta, y' + \epsilon \eta') = \frac{\partial F}{\partial (y + \epsilon \eta)} \eta + \frac{\partial F}{\partial (y' + \epsilon \eta')} \eta'

Setting

\epsilon = 0

, we get:

\left. \frac{d}{d\epsilon} F(x, y + \epsilon \eta, y' + \epsilon \eta') \right\vert_{\epsilon=0} = \frac{\partial F}{\partial y}(x, y, y') \eta(x) + \frac{\partial F}{\partial y'}(x, y, y') \eta'(x)

For brevity, we’ll write

\frac{\partial F}{\partial y}

and

\frac{\partial F}{\partial y'}

}, understanding they are evaluated at

\[ (x, y(x), y'(x)) \]

Plugging this back into the integral for

\delta J[y; \eta]

\delta J[y; \eta] = \int_a^b \left( \frac{\partial F}{\partial y} \eta(x) + \frac{\partial F}{\partial y'} \eta'(x) \right) \, dx

The necessary condition for an extremum is

\delta J[y; \eta] = 0

\int_a^b \left( \frac{\partial F}{\partial y} \eta(x) + \frac{\partial F}{\partial y'} \eta'(x) \right) \, dx = 0

This equation must hold for all admissible variation functions

\eta(x)

3. The Key Maneuver: Integration by Parts

The expression above involves both

\eta(x)

and its derivative

\eta'(x)

. To make progress, we want to factor out

\eta(x)

from the entire integrand. We can achieve this by applying integration by parts to the second term:

\int u \, dv = uv - \int v \, du

Let

u = \frac{\partial F}{\partial y'}

and

dv = \eta'(x) \, dx

. Then

du = \frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) \, dx

and

v = \eta(x)

So, the second term becomes:

\int_a^b \frac{\partial F}{\partial y'} \eta'(x) \, dx = \left[ \frac{\partial F}{\partial y'} \eta(x) \right]_a^b - \int_a^b \eta(x) \frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) \, dx

The boundary term

\left[ \frac{\partial F}{\partial y'} \eta(x) \right]_a^b = \frac{\partial F}{\partial y'}(b) \eta(b) - \frac{\partial F}{\partial y'}(a) \eta(a)

. Recall from Part 1 that for problems with fixed endpoints

\[ y(a)=y_a \]

and

\[ y(b)=y_b \]

, the admissible variations

\eta(x)

must satisfy

\eta(a) = 0

and

\eta(b) = 0

. Therefore, for such problems, the boundary term vanishes:

\left[ \frac{\partial F}{\partial y'} \eta(x) \right]_a^b = 0

Note on Boundary Conditions: If the endpoints are not fixed (so-called “natural boundary conditions”), then
$\eta(a)$
and
$\eta(b)$
are not necessarily zero, and the boundary terms must be handled differently. This leads to additional conditions on
$\frac{\partial F}{\partial y'}$
at the endpoints. We will focus on fixed endpoints for now.

Substituting the result of the integration by parts (with the vanishing boundary term) back into the equation for

\delta J = 0

\int_a^b \frac{\partial F}{\partial y} \eta(x) \, dx - \int_a^b \eta(x) \frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) \, dx = 0

Combining the integrals:

\int_a^b \left( \frac{\partial F}{\partial y} - \frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) \right) \eta(x) \, dx = 0

This equation is crucial. It states that the integral of the product of the term in the parenthesis and

\eta(x)

is zero for any admissible variation function

\eta(x)

. This leads us to a powerful lemma.

4. The Fundamental Lemma of Variational Calculus

The equation we’ve reached is of the form

\int_a^b g(x) \eta(x) \, dx = 0

, where

g(x) = \frac{\partial F}{\partial y} - \frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right)

Lemma. Fundamental Lemma of Variational Calculus (du Bois-Reymond)

If a function
$\[ g(x) \]$
is continuous on the interval
$\[ [a, b] \]$
, and if
$\int_a^b g(x) \eta(x) \, dx = 0$
for every continuously differentiable function
$\eta(x)$
such that
$\eta(a) = 0$
and
$\eta(b) = 0$
, then
$\[ g(x) = 0 \]$
for all
$x \in [a, b]$
.

Intuition behind the Lemma: Suppose, for the sake of contradiction, that

g(x_0) \neq 0

for some

x_0 \in (a, b)

. Let’s say

\[ g(x_0) > 0 \]

. Since

\[ g(x) \]

is continuous, there must be a small subinterval around

\[ x_0 \]

, say

[c, d] \subset (a, b)

, where

\[ g(x) > 0 \]

throughout this subinterval.

Now, we can construct a specific variation function

\eta(x)

that is positive within

\[ [c, d] \]

and zero outside this subinterval (and still satisfies

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

because

\[ [c,d] \]

is strictly inside

\[ (a,b) \]

). Such functions, often called “bump functions,” can be made smooth. For example, one could choose

\eta(x) = (x-c)^2(x-d)^2

for

x \in [c,d]

and

\eta(x)=0

otherwise (or a smoother version using exponentials as shown in the reference materials from the prompt).

For such an

\eta(x)

g(x) \eta(x) > 0

for

x \in (c, d)

g(x) \eta(x) = 0

for

x \notin [c, d]

Then the integral

\int_a^b g(x) \eta(x) \, dx = \int_c^d g(x) \eta(x) \, dx

would be strictly positive. This contradicts our premise that the integral is zero for all admissible

\eta(x)

. Therefore, our assumption that

g(x_0) \neq 0

must be false. Thus,

\[ g(x) = 0 \]

for all

x \in [a, b]

More on Bump Functions

A common example of a smooth bump function that is non-zero only on a finite interval, say

\[ (-1, 1) \]

, is:

B(t) = \begin{cases} \exp\left(-\frac{1}{1-t^2}\right) & \text{if } \vert t \vert < 1 \\ 0 & \text{if } \vert t \vert \ge 1 \end{cases}

This function is infinitely differentiable everywhere, including at

t=\pm 1

where all derivatives are zero. By scaling and translating

\[ t \]

, we can create such a bump function

\eta(x)

over any desired subinterval

\[ [c, d] \]

within

\[ [a, b] \]

. This rigorous construction underpins the Fundamental Lemma.

5. The Euler-Lagrange Equation

Applying the Fundamental Lemma of Variational Calculus to our equation:

\int_a^b \left( \frac{\partial F}{\partial y} - \frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) \right) \eta(x) \, dx = 0

The term in the parenthesis plays the role of

\[ g(x) \]

. If this integral is zero for all admissible

\eta(x)

, then the term itself must be identically zero:

Theorem. The Euler-Lagrange Equation

A function
$\[ y(x) \]$
that extremizes the functional
$J[y] = \int_a^b F(x, y(x), y'(x)) \, dx$
with fixed boundary conditions
$\[ y(a)=y_a \]$
and
$\[ y(b)=y_b \]$
, must satisfy the following second-order ordinary differential equation:
$\frac{\partial F}{\partial y} - \frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) = 0$
This is known as the Euler-Lagrange equation.

Understanding the terms:

\frac{\partial F}{\partial y}

: The partial derivative of

\[ F(x, y, y') \]

with respect to its second argument

\[ y \]

, treating

\[ x \]

and

\[ y' \]

as constants.

\frac{\partial F}{\partial y'}

: The partial derivative of

\[ F(x, y, y') \]

with respect to its third argument

\[ y' \]

, treating

\[ x \]

and

\[ y \]

as constants.

\frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right)

: The total derivative with respect to

\[ x \]

of the expression

\frac{\partial F}{\partial y'}

. Since

\[ y \]

and

\[ y' \]

are functions of

\[ x \]

, this derivative will generally involve

\[ y'(x) \]

and

\[ y''(x) \]

via the chain rule:

<div class="math-block" markdown="0"> \[ \frac{d}{dx} \left( \frac{\partial F}{\partial y'}(x, y(x), y'(x)) \right) = \frac{\partial^2 F}{\partial x \partial y'} + \frac{\partial^2 F}{\partial y \partial y'} y' + \frac{\partial^2 F}{\partial y'^2} y'' \]
</div>

The Euler-Lagrange equation is a differential equation for the unknown function

\[ y(x) \]

. Solving it (subject to the boundary conditions) provides the candidate functions that could extremize the functional.

6. Significance and What’s Next

The derivation of the Euler-Lagrange equation is a monumental step in variational calculus. It converts the problem of optimizing over an infinite-dimensional space of functions into the more familiar problem of solving a differential equation.

Global Criterion to Local Rule: The original problem was to minimize a global quantity (the integral

\[ J[y] \]

). The Euler-Lagrange equation provides a local condition (a differential equation) that must hold at every point

\[ x \]

Universality: This single method, encapsulated by the Euler-Lagrange equation, can tackle a vast array of problems that seek to find an optimal function, simply by identifying the correct integrand

\[ F(x, y, y') \]

In the next part of this crash course, we will:

Apply the Euler-Lagrange equation to solve some classic variational problems, such as finding the shortest path between two points (revisiting our initial example) and the brachistochrone problem.
Discuss some special cases and first integrals of the Euler-Lagrange equation (Beltrami identity).

This will demonstrate the power and utility of the machinery we’ve developed.

Crash Course, Calculus

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