Chapter 1
Introduction
These notes are based on notes prepared by Professor David Hughes in 2018–19.
In this Chapter we shall first introduce the main idea of Calculus of Variations by looking at a
simple physical problem. We shall then discuss
iefly the topics to be covered in the course:
extensions to the basic theory, together with three areas in which we shall apply Calculus of
Variations.
Contents
1.1 The Hanging Chain Problem . . . . . . . . . . . . XXXXXXXXXX1
1.2 The Fundamental Problem in Calculus of Variations XXXXXXXXXX
1.3 A Brief Summary of the Course . . . . . . . . . . XXXXXXXXXX3
1.1 The Hanging Chain Problem
Figure 1.1: Heavy chain suspended in a vertical plane, fixed at points A and B.
Suppose that a heavy chain is hung in a vertical plane with its end points at A, with coordi-
nates (x1, y1), and B, with coordinates (x2, y2). What is the shape adopted by the chain?
1
2 1.2 The Fundamental Problem in Calculus of Variations
As we shall see in many of the problems in Calculus of Variations, we first need to consider the
physics of the problem (which, for the problems we shall look at, will be fairly straightforward)
in order to describe the problem mathematically.
Here the crucial physical consideration is that the chain, at rest, assumes a position of mini-
mum potential energy. If it did not then there would be energy available for motion (kinetic
energy) to move the chain to a position of lower potential energy.
We therefore need to express the potential energy of the chain mathematically. Consider a
small line element of the chain ds. Then the potential energy of this small element of the
chain is mgy ds, where m is the mass per unit length and g is the acceleration due to gravity.
For simplicity we shall assume that the chain is of uniform density (constant m), although
nothing fundamentally new is introduced if m varies along the chain (i.e. if m were to be a
function of s).
Hence the potential energy (PE) of the entire chain is given by
PE = mg
∫
y ds.
From Pythagoras,
ds2 = dx2 + dy2 =⇒ ds =
(
1 + y′2
)1/2
dx, where y′ =
dy
dx
.
Thus
PE = mg
∫ x2
x1
y
(
1 + y′2
)1/2
dx.
This expression gives the potential energy of the chain for any possible shape y(x). As
discussed above, the true shape will be given by the curve y(x) that minimises the potential
energy.
Thus, mathematically, the problem becomes one of finding the function y(x) that minimises
the integral
I =
∫ x2
x1
y
(
1 + y′2
)1/2
dx.
We can see immediately that this is a much more challenging problem than faced in standard
calculus, where, for example, we may wish to find the minimum value of a function y(x).
1.2 The Fundamental Problem in Calculus of Variations
The essential, most basic, problem in Calculus of Variations is to find the function y(x) that
minimises (or maximises) the integral
I =
∫ x2
x1
F (x, y, y′) dx, for some given function F. (1.1)
Whereas in standard calculus we may seek extrema (maxima or minima) of a single function
y(x), here we are looking to minimise (or maximise) I, which is a function of functions
(often refe
ed to as a functional). The functional I can also be thought of as a mapping
from the space of allowed functions y(x) to the real numbers: for every function y(x), the
integral I takes a certain value. We must therefore consider the minima of I over all possible
functions y(x).
Chapter 1 – Introduction 3
1.3 A Brief Summary of the Course
In the course we shall consider four aspects of Calculus of Variations.
(i) Basic Theory and Extensions
We shall start by considering the basic problem XXXXXXXXXXThis represents the simplest problem
in Calculus of Variations. There are several extensions to cover more complicated forms of F ,
for example:
(1) More than one dependent variable, e.g. F (x, y(x), y′(x), z(x), z′(x))
(2) More than one independent variable, e.g. F (s, t, u(s, t), ∂su, ∂tu)
(3) Higher derivatives, e.g. F (x, y, y′, y′′)
(4) Minimising integrals with constraints. Actually the hanging chain problem falls into this
category, since the minimisation of the energy must be achieved subject to the constraint that
the chain is of a given length.
We shall look at (1), (3) and (4) (but probably not (2)), together with other extensions to
the basic problem.
(ii) Classical Problems in Calculus of Variations
In order to motivate the subject, we have already considered the problem of the hanging chain
(though actually, as we shall see, this is not quite in the category of the most basic problem).
There are many other famous problems that are tackled via the Calculus of Variations. Fo
example:
• If a bead falls on a wire between two points at different heights, what is the shape of
the wire that minimises the travel time of the bead?
• What is the shortest distance between two points on a surface? On a plane surface we
know that this is a straight line (though how do we prove this?). What about two points
on a cylindrical or spherical surface, or something more complicated?
(iii) Links Between the Solution of ODEs and Minimising the Integral (1.1)
There turns out to be a strong link between the solutions of second-order ordinary differential
equations (ODEs) of the form
d
dx
(
p(x)
dy
dx
)
+ q(x)y + λr(x)y = 0
and the minimisation of integral (1.1).
This can be used, for example, to estimate the eigenvalues of certain linear operators — what
is known as the Rayleigh Ritz technique.
This is a link between two
anches of mathematics that, at least at first sight, are not
obviously connected.
(iv) A Variational Approach to Physics
The most powerful use of Calculus of Variations comes from the fact that many of the basic
laws of physics can be expressed in a variational way.
For example, Fermat’s principle states that:
“Time elapsed in the passage of light between two fixed points is an extremum with respect to
all possible paths.”
We shall see that this principle leads to all of the laws of geometrical optics.
4 1.3 A Brief Summary of the Course
Maybe even more surprisingly, the laws of mechanics can be formulated via a variational
principle.
We are used to thinking of mechanical systems as being governed by Newton’s laws of motion.
For example, the motion of a particle of mass m under the influence of gravity is governed by
the equation
m
dv
dt
= mg. (1.2)
Solving this equation will tell us everything we need to know about the particle’s velocity and
position.
However, it turns out that the laws of motion can be formulated very differently, as follows.
The path taken by the particle is that which minimises the integral
I =
∫ t2
t1
(T − V ) dt, (1.3)
where T is the kinetic energy and V the potential energy.
Clearly the philosophy of the approaches described by equations (1.2) and (1.3) is very dif-
ferent. Whereas Newton’s equation of motion (1.2) says that at each point in its trajec-
tory, the particle ‘feels’ the force and moves accordingly, the variational approach (i.e. min-
imising the integral XXXXXXXXXXsays that the particle somehow ‘checks out’ all possible paths
and then adopts the one that minimises the integral. There is a very nice discussion of
this topic in The Feynman Lectures on Physics (Vol. 2, Chap. 19); the link is https:
www.feynmanlectures.caltech.edu/II_19.html.
Chapter 2
Calculus Revision
Calculus of Variations requires a good knowledge of the ideas of Calculus that you have met
at A-level and in your first year courses. This section
iefly covers the essential topics. Please
ensure that you are on top of these.
Contents
2.1 Integration . . . . . . . . . . . . . . . . . . . . . . . XXXXXXXXXX1
2.2 Ordinary Differential Equations (ODEs) . . . . . XXXXXXXXXX3
2.3 Differentiation . . . . . . . . . . . . . . . . . . . . . XXXXXXXXXX7
2.4 Taylor’s Theorem . . . . . . . . . . . . . . . . . . . XXXXXXXXXX7
2.5 Constraints and Lagrange Multipliers . . . . . . . XXXXXXXXXX8
2.1 Integration
There are a number of standard techniques of integration. Here are some of the most useful.
2.1.1 Functions that are exact derivatives
The easiest integrals to perform are those where the integrand (the quantity to be integrated)
can be clearly identified as a derivative; in this case, since
∫
f ′(x) dx = f(x) (where f ′ denotes
df/dx), integration is straightforward. Slightly more complicatedly, the integrand may be the
derivative of a function of a function; here∫
g′(x)f ′(g(x)) dx = f(g(x)).
Examples
(i)
∫
sec2 x dx, (ii)
∫
x(1 + x2)1/3 dx, (iii)
∫
sin10 x cosx dx, (iv)
∫
x2(1 + x3)−2 dx.
When confronted with an integral, you should always first check if it falls into this category —
since it can be solved with little work