IB Variational Principles (Full)

Part IB — Variational Principles

Based on lectures by P. K. Townsend

Notes taken by Dexter Chua

Easter 2015

These notes are not endorsed by the lecturers, and I have modified them (often

significantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

Stationary points for functions on

. Necessary and sufficient conditions for minima

and maxima. Importance of convexity. Variational problems with constraints; method

of Lagrange multipliers. The Legendre Transform; need for convexity to ensure

invertibility; illustrations from thermodynamics. [4]

The idea of a functional and a functional derivative. First variation for functionals,

Euler-Lagrange equations, for both ordinary and partial differential equations. Use of

Lagrange multipliers and multiplier functions. [3]

Fermat’s principle; geodesics; least action principles, Lagrange’s and Hamilton’s equa-

tions for particles and fields. Noether theorems and first integrals, including two forms

of Noether’s theorem for ordinary differential equations (energy and momentum, for

example). Interpretation in terms of conservation laws. [3]

Second variation for functionals; associated eigenvalue problem. [2]

Contents

0 Introduction

1 Multivariate calculus

1.1 Stationary points

1.2 Convex functions

1.2.1 Convexity

1.2.2 First-order convexity condition

1.2.3 Second-order convexity condition

1.3 Legendre transform

1.4 Lagrange multipliers

2 Euler-Lagrange equation

2.1 Functional derivatives

2.2 First integrals

2.3 Constrained variation of functionals

3 Hamilton’s principle

3.1 The Lagrangian

3.2 The Hamiltonian

3.3 Symmetries and Noether’s theorem

4 Multivariate calculus of variations

5 The second variation

5.1 The second variation

5.2 Jacobi condition for local minima of F [x]

0 Introduction

Consider a light ray travelling towards a mirror and being reflected.

We see that the light ray travels towards the mirror, gets reflected at

, and hits

the (invisible) eye. What determines the path taken? The usual answer would

be that the reflected angle shall be the same as the incident angle. However,

ancient Greek mathematician Hero of Alexandria provided a different answer:

the path of the light minimizes the total distance travelled.

We can assume that light travels in a straight line except when reflected.

Then we can characterize the path by a single variable

, the point where the

light ray hits the mirror. Then we let

(

) to be the length of the path, and we

can solve for z by setting L

(z) = 0.

This principle sounds reasonable - in the absence of mirrors, light travels in

a straight line - which is the shortest path between two points. But is it always

true that the shortest path is taken? No! We only considered a plane mirror, and

this doesn’t hold if we have, say, a spherical mirror. However, it turns out that

in all cases, the path is a stationary point of the length function, i.e. L

(z) = 0.

Fermat put this principle further. Assuming that light travels at a finite

speed, the shortest path is the path that takes the minimum time. Fermat’s

principle thus states that

Light travels on the path that takes the shortest time.

This alternative formulation has the advantage that it applies to refraction as

well. Light travels at different speeds in different mediums. Hence when they

travel between mediums, they will change direction such that the total time

taken is minimized.

We usually define the refractive index

of a medium to be

= 1

, where

is the velocity of light in the medium. Then we can write the variational

principle as

minimize

path

n ds,

where d

is the path length element. This is easy to solve if we have two distinct

mediums. Since light travels in a straight line in each medium, we can simply

characterize the path by the point where the light crosses the boundary. However,

in the general case, we should be considering any possible path between two

points. In this case, we could no longer use ordinary calculus, and need new

tools - calculus of variations.

In calculus of variations, the main objective is to find a function

(

) that

minimizes an integral

(

) d

for some function

. For example, we might

want to minimize

(

) d

. This differs greatly from ordinary minimization

problems. In ordinary calculus, we minimize a function

(

) for all possible

values of

x ∈ R

. However, in calculus of variations, we will be minimizing the

integral

f(x) dt over all possible functions x(t).

1 Multivariate calculus

Before we start calculus of variations, we will first have a brief overview of

minimization problems in ordinary calculus. Unless otherwise specified,

will be a function

→ R

. For convenience, we write the argument of

= (

, ··· , x

) and

|x|

. We will also assume that

is sufficiently smooth

for our purposes.

1.1 Stationary points

The quest of minimization starts with finding stationary points.

Definition

(Stationary points)

Stationary points are points in

for which

∇f = 0, i.e.

∂f

∂x

∂f

∂x

= ··· =

∂f

∂x

= 0

All minima and maxima are stationary points, but knowing that a point is

stationary is not sufficient to determine which type it is. To know more about

the nature of a stationary point, we Taylor expand

about such a point, which

we assume is 0 for notational convenience.

f(x) = f(0) + x · ∇f +

i,j

∂

∂x

+ O(x

= f(0) +

i,j

∂

∂x

+ O(x

The second term is so important that we have a name for it:

Definition (Hessian matrix). The Hessian matrix is

(x) =

∂

∂x

Using summation notation, we can write our result as

f(x) − f(0) =

+ O(x

Since

is symmetric, it is diagonalizable. Thus after rotating our axes to a

suitable coordinate system, we have







0 ··· 0

0 λ

··· 0

0 0 ··· λ







where

are the eigenvalues of

. Since

is real symmetric, these are all real.

In our new coordinate system, we have

f(x) − f(0) =

i=1

)

This is useful information. If all eigenvalues

are positive, then

(

)

− f

(

)

must be positive (for small

). Hence our stationary point is a local minimum.

Similarly, if all eigenvalues are negative, then it is a local maximum.

If there are mixed signs, say

0 and

0, then

increases in the

direction and decreases in the

direction. In this case we say we have a saddle

point.

If some

= 0, then we have a degenerate stationary point. To identify the

nature of this point, we must look at even higher derivatives.

In the special case where

= 2, we do not need to explicitly find the

eigenvalues. We know that

det H

is the product of the two eigenvalues. Hence if

det H

is negative, the eigenvalues have different signs, and we have a saddle. If

det H is positive, then the eigenvalues are of the same sign.

To determine if it is a maximum or minimum, we can look at the trace of

, which is the sum of eigenvalues. If

tr H

is positive, then we have a local

minimum. Otherwise, it is a local maximum.

Example. Let f(x, y) = x

+ y

− 3xy. Then

∇f = 3(x

− y, y

− x).

This is zero iff

and

. This is satisfied iff

. So either

= 0,

or y = 1. So there are two stationary points: (0, 0) and (1, 1).

The Hessian matrix is

H =



6x −3

−3 6y



and we have

det H = 9(4xy − 1)

tr H = 6(x + y).

At (0

0),

det H <

0. So this is a saddle point. At (1

1),

det H >

tr H >

So this is a local minimum.

1.2 Convex functions

1.2.1 Convexity

Convex functions is an important class of functions that has a lot of nice

properties. For example, stationary points of convex functions are all minima,

and a convex function can have at most one minimum value. To define convex

functions, we need to first define a convex set.

Definition

(Convex set)

A set

S ⊆ R

is convex if for any distinct

x, y ∈

S, t ∈

1), we have (1

−t

)

ty ∈ S

. Alternatively, any line joining two points

in S lies completely within S.

non-convex convex

Definition (Convex function). A function f : R

→ R is convex if

(i) The domain D(f) is convex

(ii) The function f lies below (or on) all its chords, i.e.

f((1 − t)x + ty) ≤ (1 − t)f(x) + tf(y) (∗)

for all x, y ∈ D(f), t ∈ (0, 1).

A function is strictly convex if the inequality is strict, i.e.

f((1 − t)x + ty) < (1 − t)f(x) + tf(y).

x y

(1 − t)x + ty

(1 − t)f(x) + tf (y)

A function f is (strictly) concave iff −f is (strictly) convex.

Example.

(i) f(x) = x

is strictly convex.

(ii) f(x) = |x| is convex, but not strictly.

(iii) f(x) =

defined on x > 0 is strictly convex.

(iv) f

(

) =

defined on

∗

R \{

}

is not convex. Apart from the fact that

∗

is not a convex domain. But even if we defined, like

(0) = 0, it is

not convex by considering the line joining (

−

, −

1) and (1

1) (and in fact

f(x) =

defined on x < 0 is concave).

1.2.2 First-order convexity condition

While the definition of a convex function seems a bit difficult to work with, if

our function is differentiable, it is easy to check if it is convex.

First assume that our function is once differentiable, and we attempt to find

a first-order condition for convexity. Suppose that

is convex. For fixed

x, y

we define the function

h(t) = (1 − t)f(x) + tf(y) − f((1 − t)x + ty).

By the definition of convexity of

, we must have

(

)

≥

0. Also, trivially

h(0) = 0. So

h(t) − h(0)

≥ 0

for any t ∈ (0, 1). So

(0) ≥ 0.

On the other hand, we can also differentiate h directly and evaluate at 0:

(0) = f(y) − f(x) − (y − x) · ∇f(x).

Combining our two results, we know that

f(y) ≥ f(x) + (y − x) · ∇f(x) (†)

It is also true that this condition implies convexity, which is an easy result.

How can we interpret this result? The equation

(

) + (

y − x

)

· ∇f

(

) = 0

defines the tangent plane of

. Hence this condition is saying that a convex

differentiable function lies above all its tangent planes.

We immediately get the corollary

Corollary.

A stationary point of a convex function is a global minimum. There

can be more than one global minimum (e.g. a constant function), but there is at

most one if the function is strictly convex.

Proof. Given x

such that ∇f(x

) = 0, (†) implies that for any y,

f(y) ≥ f(x

) + (y − x

) · ∇f(x

) = f(x

We can write our first-order convexity condition in a different way. We can

rewrite (†) into the form

(y − x) · [∇f(y) − ∇f(x)] ≥ f(x) − f(y) − (x − y) · ∇f(y).

By applying (

†

) to the right hand side (with

and

swapped), we know that

the right hand side is ≥ 0. So we have another first-order condition:

(y − x) · [∇f(y) − ∇f(x)] ≥ 0,

It can be shown that this is equivalent to the other conditions.

This condition might seem a bit weird to comprehend, but all it says is that

∇f

(

) is a non-decreasing function. For example, when

= 1, the equation

states that (

y − x

)(

(

)

− f

(

))

≥

0, which is the same as saying

(

)

≥ f

(

)

whenever y > x.

1.2.3 Second-order convexity condition

We have an even nicer condition when the function is twice differentiable. We

start with the equation we just obtained:

(y − x) · [∇f(y) − ∇f(x)] ≥ 0,

Write y = x + h. Then

h · (∇f(x + h) − ∇f(x)) ≥ 0.

Expand the left in Taylor series. Using suffix notation, this becomes

∇

f + O(h

)] ≥ 0.

But ∇

∇

f = H

. So we have

+ O(h

) ≥ 0

This is true for all

if the Hessian

is positive semi-definite (or simply positive),

i.e. the eigenvalues are non-negative. If they are in fact all positive, then we say

H is positive definite.

Hence convexity implies that the Hessian matrix is positive for all

x ∈ D

(

Strict convexity implies that it is positive definite.

The converse is also true — if the Hessian is positive, then it is convex.

Example. Let f(x, y) =

for x, y > 0. Then the Hessian is

H =





The determinant is

det H =

> 0

and the trace is

tr H =





> 0.

So f is convex.

To conclude that

is convex, we only used the fact that

is positive,

instead of

and

being individually positive. Then could we relax the domain

condition to be

xy >

0 instead? The answer is no, because in this case, the

function will no longer be convex!

1.3 Legendre transform

The Legendre transform is an important tool in classical dynamics and thermo-

dynamics. In classical dynamics, it is used to transform between the Lagrangian

and the Hamiltonian. In thermodynamics, it is used to transform between

the energy, Helmholtz free energy and enthalpy. Despite its importance, the

definition is slightly awkward.

Suppose that we have a function

(

), which we’ll assume is differentiable.

For some reason, we want to transform it into a function of the conjugate variable

instead. In most applications to physics, this quantity has a particular

physical significance. For example, in classical dynamics, if

is the Lagrangian,

then

∂L

∂ ˙x

is the (conjugate) momentum.

also has a context-independent

geometric interpretation, which we will explore later. For now, we will assume

that p is more interesting than x.

Unfortunately, the obvious option

∗

(

) =

(

)) is not the transform we

want. There are various reasons for this, but the major reason is that it is ugly.

It lacks any mathematical elegance, and has almost no nice properties at all.

In particular, we want our f

∗

(p) to satisfy the property

∗

= x.

This says that if

is the conjugate of

, then

is the conjugate of

. We will

soon see how this is useful in the context of thermodynamics.

The symmetry is better revealed if we write in terms of differentials. The

differential of the function f is

df =

dx = p dx.

So we want our f

∗

to satisfy

∗

= x dp.

How can we obtain this? From the product rule, we know that

d(xp) = x dp + p dx.

So if we define

∗

xp −f

(more explicitly written as

∗

(

) =

(

)

p −f

(

))),

then we obtain the desired relation d

∗

. Alternatively, we can say

∗

= x.

The actual definition we give will not be exactly this. Instead, we define it in

a way that does not assume differentiability. We’ll also assume that the function

takes the more general form R

→ R.

Definition

(Legendre transform)

Given a function

→ R

, its Legendre

transform f

∗

(the “conjugate” function) is defined by

∗

(p) = sup

(p · x − f(x)),

The domain of

∗

is the set of

p ∈ R

such that the supremum is finite.

known as the conjugate variable.

This relation can also be written as

∗

(

) +

(

) =

, where

(

) is the

value of x that maximizes the function.

To show that this is the same as what we were just talking about, note

that the supremum of

p · x − f

(

) is obtained when its derivative is zero, i.e.

∇f

(

). In particular, in the 1D case,

∗

(

) =

px − f

(

), where

satisfies

(x) = p. So p is indeed the derivative of f with respect to x.

From the definition, we can immediately conclude that

Lemma. f

∗

is always convex.

Proof.

∗

((1 − t)p + tq) = sup



((1 − t)p · x + tq · x − f(x)



= sup



(1 − t)(p · x − f(x)) + t(q · x − f(x))



≤ (1 − t) sup

[p · x − f(x)] + t sup

[q · x − f(x)]

= (1 − t)f

∗

(p) + tf

∗

(q)

Note that we cannot immediately say that

∗

is convex, since we have to show

that the domain is convex. But by the above bounds,

∗

((1

−t

)

) is bounded

by the sum of two finite terms, which is finite. So (1

− t

)

is also in the

domain of f

∗

This transformation can be given a geometric interpretation. We will only

consider the 1D case, because drawing higher-dimensional graphs is hard. For

any fixed

, we draw the tangent line of

at the point

. Then

∗

(

) is the

intersection between the tangent line and the y axis:

slope = p

−f

∗

(p)

∗

(p) = px − f(x)

f(x)

Example.

(i)

Let

(

) =

for

a >

0. Then

at the maximum of

px −f

(

). So

∗

(p) = px − f(x) = p ·

−





So the Legendre transform maps a parabola to a parabola.

(ii) f(v) = −

√

1 − v

for |v| < 1 is a lower semi-circle. We have

p = f

(v) =

√

1 − v

v =

1 + p

and exists for all p ∈ R. So

∗

(p) = pv − f(v) =

1 + p

A circle gets mapped to a hyperbola.

(iii)

Let

for

c >

0. This is convex but not strictly convex. Then

px − f

(

) = (

p − c

)

. This has no maximum unless

. So the domain

of f

∗

is simply {c}. One point. So f

∗

(p) = 0. So a line goes to a point.

Finally, we prove that applying the Legendre transform twice gives the

original function.

Theorem.

is convex, differentiable with Legendre transform

∗

, then

∗∗

= f.

Proof.

We have

∗

(

) = (

p · x

(

)

−f

(

)) where

(

) satisfies

∇f

(

)).

Differentiating with respect to p, we have

∇

∗

(p) = x

+ p

∇

(p) − ∇

(p)∇

f(x)

= x

+ p

∇

(p) − ∇

(p)p

= x

∇f

∗

(p) = x.

This means that the conjugate variable of p is our original x. So

∗∗

(x) = (x · p − f

∗

(p))|

p=p(x)

= x · p − (p · x − f(x))

= f(x).

Note that strict convexity is not required. For example, in our last example

above with the straight line,

∗

(

) = 0 for

. So

∗∗

(

) = (

xp −f

∗

(

))

p=c

cx = f(x).

However, convexity is required. If

∗∗

is true, then

must be convex,

since it is a Legendre transform. Hence

∗∗

cannot be true for non-convex

functions.

Application to thermodynamics

Given a system of a fixed number of particles, the energy of a system is usually

given as a function of entropy and volume:

E = E(S, V ).

We can think of this as a gas inside a piston with variable volume.

There are two things that can affect the energy: we can push in the piston

and modify the volume. This corresponds to a work done of

−p

, where

the pressure. Alternatively, we can simply heat it up and create a heat change

of T dS, where T is the temperature. Then we have

dE = T dS − p dV.

Comparing with the chain rule, we have

∂E

∂S

= T, −

∂E

∂V

= p

However, the entropy is a mysterious quantity no one understands. Instead, we

like temperature, defined as

∂E

∂S

. Hence we use the (negative) Legendre

transform to obtain the conjugate function Helmholtz free energy.

F (T, V ) = inf

[E(S, V ) − T S] = E(S, V ) − S

∂E

∂S

= E − ST,

Note that the Helmholtz free energy satisfies

dF = −S dT − p dV.

Just as we could recover

and

from

via taking partial derivatives with

respect to

and

, we are able to recover

and

from

by taking partial

derivatives with respect to

and

. This would not be the case if we simply

defined F (T, V ) = E(S(T, V ), V ).

If we take the Legendre transform with respect to

, we get the enthalpy

instead, and if we take the Legendre transform with respect to both, we get the

Gibbs free energy.

1.4 Lagrange multipliers

At the beginning, we considered the problem of unconstrained maximization.

We wanted to maximize

(

x, y

) where

x, y

can be any real value. However,

sometimes we want to restrict to certain values of (

x, y

). For example, we might

want x and y to satisfy x + y = 10.

We take a simple example of a hill. We model it using the function

(

x, y

)

given by the height above the ground. The hilltop would be given by the

maximum of f, which satisfies

0 = df = ∇f · dx

for any (infinitesimal) displacement dx. So we need

∇f = 0.

This would be a case of unconstrained maximization, since we are considering all

possible values of x and y.

A problem of constrained maximization would be as follows: we have a path

p defined by p(x, y) = 0. What is the highest point along the path p?

We still need

∇f ·

= 0, but now d

is not arbitrary. We only consider

the d

parallel to the path. Alternatively,

∇f

has to be entirely perpendicular

to the path. Since we know that the normal to the path is

∇p

, our condition

becomes

∇f = λ∇p

for some lambda

. Of course, we still have the constraint

(

x, y

) = 0. So what

we have to solve is

∇f = λ∇p

p = 0

for the three variables x, y, λ.

Alternatively, we can change this into a single problem of unconstrained

extremization. We ask for the stationary points of the function

(

x, y, λ

) given

φ(x, y, λ) = f(x, y) − λp(x, y)

When we maximize against the variables

and

, we obtain the

∇f

λ∇p

condition, and maximizing against λ gives the condition p = 0.

Example.

Find the radius of the smallest circle centered on origin that intersects

y = x

− 1.

(i)

First do it the easy way: for a circle of radius

to work,

and y = x

− 1 must have a solution. So

)

− x

+ 1 − R

= 0

and

−

So R

min

√

3/2.

(ii)

We can also view this as a variational problem. We want to minimize

(

x, y

) =

subject to the constraint

(

x, y

) = 0 for

(

x, y

) =

y−x

+1.

We can solve this directly. We can solve the constraint to obtain

−

Then

(x) = f(x, y(x)) = (x

)

− x

+ 1

We look for stationary points of R

(x))

= 0 ⇒ x



−



= 0

= 0 and

= 1; or

√

and

√

. Since

√

is smaller, this is

our minimum.

(iii)

Finally, we can use Lagrange multipliers. We find stationary points of the

function

φ(x, y, λ) = f(x, y) − λp(x, y) = x

+ y

− λ(y − x

+ 1)

The partial derivatives give

∂φ

∂x

= 0 ⇒ 2x(1 + λ) = 0

∂φ

∂y

= 0 ⇒ 2y − λ = 0

∂φ

∂λ

= 0 ⇒ y − x

+ 1 = 0

The first equation gives us two choices

– x

= 0. Then the third equation gives

−

1. So

+ y

= 1.

– λ

−

1. So the second equation gives

−

and the third gives

x = ±

√

. Hence R =

√

is the minimum.

This can be generalized to problems with functions

→ R

using the same

logic.

Example. For x ∈ R

, find the minimum of the quadratic form

f(x) = x

on the surface |x|

= 1.

(i)

The constraint imposes a normalization condition on

. But if we scale up

x, f (x) scales accordingly. So if we define

Λ(x) =

f(x)

g(x)

, g(x) = |x|

the problem is equivalent to minimization of Λ(

) without constraint. Then

∇

Λ(x) =



−



So we need

Ax = Λx

So the extremal values of Λ(

) are the eigenvalues of

. So Λ

min

is the

lowest eigenvalue.

This answer is intuitively obvious if we diagonalize A.

(ii)

We can also do it with Lagrange multipliers. We want to find stationary

values of

φ(x, λ) = f(x) − λ(|x|

− 1).

0 = ∇φ ⇒ A

= λx

Differentiating with respect to λ gives

∂φ

∂λ

= 0 ⇒ |x|

= 1.

So we get the same set of equations.

Example.

Find the probability distribution

, ··· , p

}

satisfying

= 1

that maximizes the information entropy

S = −

i=1

log p

We look for stationary points of

φ(p, λ) = −

i=1

ln p

− λ

i=1

+ λ.

We have

∂φ

∂p

= −ln p

− (1 + λ) = 0.

= e

−(1+λ)

It is the same for all i. So we must have p

2 Euler-Lagrange equation

2.1 Functional derivatives

Definition

(Functional)

A functional is a function that takes in another real-

valued function as an argument. We usually write them as

[

] (square brackets),

where

(

) :

R → R

is a real function. We say that

[

] is a functional of

the function x(t).

Of course, we can also have functionals of many functions, e.g.

[

x, y

]

∈ R

for x, y : R → R. We can also have functionals of a function of many variables.

Example.

Given a medium with refractive index

(

), the time taken by a

path x(t) from x

to x

is given by the functional

T [x] =

n(x) dt.

While this is a very general definition, in reality, there is just one particular

class of functionals we care about. Given a function

(

) defined for

α ≤ t ≤ β

we study functional of the form

F [x] =

f(x, ˙x, t) dt

for some function f .

Our objective is to find a stationary point of the functional

[

]. To do so,

suppose we vary

(

) by a small amount

δx

(

). Then the corresponding change

δF [x] of F [x] is

δF [x] = F [x + δx] − F [x]



f(x + δx, ˙x + δ ˙x, t) − f(x, ˙x, t)



Taylor expand to obtain



δx

∂f

∂x

+ δ ˙x

∂f

∂ ˙x



dt + O(δx

)

Integrate the second term by parts to obtain

δF [x] =

δx



∂f

∂x

−



∂f

∂ ˙x



dt +



δx

∂f

∂ ˙x



This doesn’t seem like a very helpful equation. Life would be much easier if

the last term (known as the boundary term)

δx

∂f

∂ ˙x

vanishes. Fortunately, for

most of the cases we care about, the boundary conditions mandate that the

boundary term does indeed vanish. Most of the time, we are told that

is fixed

α, β

. So

δx

(

) =

δx

(

) = 0. But regardless of what we do, we always

choose boundary conditions such that the boundary term is 0. Then

δF [x] =



δx

δF [x]

δx(t)



where

Definition (Functional derivative).

δF [x]

δx

∂f

∂x

−



∂f

∂ ˙x



is the functional derivative of F [x].

If we want to find a stationary point of F , then we need

δF [x]

δx

= 0. So

Definition (Euler-Lagrange equation). The Euler-Lagrange equation is

∂f

∂x

−



∂f

∂ ˙x



= 0

for α ≤ t ≤ β.

There is an obvious generalization to functionals F [x] for x(t) ∈ R

∂f

∂x

−



∂f

∂ ˙x



= 0 for all i.

Example

(Geodesics of a plane)

What is the curve

of minimal length between

two points A, B in the Euclidean plane? The length is

L =

where d` =

+ dy

There are two ways we can do this:

(i)

We restrict to curves for which

(or

) is a good parameter, i.e.

can be

made a function of x. Then

d` =

1 + (y

)

dx.

Then

L[y] =

1 + (y

)

dx.

Since there is no explicit dependence on y, we know that

∂f

∂y

= 0

So the Euler-Lagrange equation says that



∂f

∂y



= 0

We can integrate once to obtain

∂f

∂y

= constant

This is known as a first integral, which will be studied more in detail later.

Plugging in our value of f, we obtain

1 + (y

)

= constant

This shows that y

must be constant. So y must be a straight line.

(ii)

We can get around the restriction to “good” curves by choosing an arbitrary

parameterization

= (

(

)

, y

(

)) for

t ∈

1] such that

(0) =

(1) =

d` =

˙x

+ ˙y

dt.

Then

L[x, y] =

˙x

+ ˙y

dt.

We have, again

∂f

∂x

∂f

∂y

= 0.

So we are left to solve



∂f

∂ ˙x





∂f

∂ ˙y



= 0.

So we obtain

˙x

+ ˙y

= c,

˙y

˙x

+ ˙y

= s

where

and

are constants. While we have two constants, they are not

independent. We must have

= 1. So we let

cos θ

sin θ

Then the two conditions are both equivalent to

( ˙x sin θ)

= ( ˙y cos θ)

Hence

˙x sin θ = ±˙y cos θ.

We can choose a θ such that we have a positive sign. So

y cos θ = x sin θ + A

for a constant A. This is a straight line with slope tan θ.

2.2 First integrals

In our example above,

did not depend on

, and hence

∂f

∂x

= 0. Then the

Euler-Lagrange equations entail



∂f

∂ ˙x



= 0.

We can integrate this to obtain

∂f

∂ ˙x

= constant.

We call this the first integral. First integrals are important in several ways. The

most immediate advantage is that it simplifies the problem a lot. We only have

to solve a first-order differential equation, instead of a second-order one. Not

needing to differentiate

∂f

∂ ˙x

also prevents a lot of mess arising from the product

and quotient rules.

This has an additional significance when applied to problems in physics.

If we have a first integral, then we get

∂f

∂ ˙x

= constant. This corresponds to

a conserved quantity of the system. When formulating physics problems as

variational problems (as we will do in Chapter 3), the conservation of energy

and momentum will arise as constants of integration from first integrals.

There is also a more complicated first integral appearing when

does not

(explicitly) depend on

. To find this out, we have to first consider the total

derivative

. By the chain rule, we have

∂f

∂t

∂f

∂x

d ˙x

∂f

∂ ˙x

∂f

∂t

+ ˙x

∂f

∂x

+ ¨x

∂f

∂ ˙x

On the other hand, the Euler-Lagrange equation says that

∂f

∂x



∂f

∂ ˙x



Substituting this into our equation for the total derivative gives

∂f

∂t

+ ˙x



∂f

∂ ˙x



+ ¨x

∂f

∂ ˙x

∂f

∂t



˙x

∂f

∂ ˙x



Then



f − ˙x

∂f

∂ ˙x



∂f

∂t

So if

∂f

∂t

= 0, then we have the first integral

f − ˙x

∂f

∂ ˙x

= constant.

Example.

Consider a light ray travelling in the vertical

plane inside a

medium with refractive index

(

) =

√

a − bz

for positive constants

a, b

. The

phase velocity of light is v =

According the Fermat’s principle, the path minimizes

T =

This is equivalent to minimizing the optical path length

cT = P =

n d`.

We specify our path by the function z(x). Then the path element is given by

d` =

+ dz

1 + z

(x)

dx,

Then

P [z] =

n(z)

1 + (z

)

dx.

Since this does not depend on x, we have the first integral

k = f − z

∂f

∂z

n(z)

1 + (z

)

for an integration constant k. Squaring and putting in the value of n gives

)

− z),

where z

= (a − k

)/b. This is integrable and we obtain

√

− z

= ±

√

dx.

√

z − z

= ±

√

(x − x

where x

is our second integration constant. Square it to obtain

z = z

−

(x − x

)

which is a parabola.

Example

(Principle of least action)

Mechanics as laid out by Newton was

expressed in terms of forces and acceleration. While this is able to describe a lot

of phenomena, it is rather unsatisfactory. For one, it is messy and difficult to

scale to large systems involving many particles. It is also ugly.

As a result, mechanics is later reformulated in terms of a variational principle.

A quantity known as the action is defined for each possible path taken by

the particle, and the actual path taken is the one that minimizes the action

(technically, it is the path that is a stationary point of the action functional).

The version we will present here is an old version proposed by Maupertuis

and Euler. While it sort-of works, it is still cumbersome to work with. The

modern version is from Hamilton who again reformulated the action principle

to something that is more powerful and general. This modern version will be

discussed more in detail in Chapter 3, and for now we will work with the old

version first.

The original definition for the action, as proposed by Maupertuis, was mass

velocity

distance. This was given a more precise mathematical definition

by Euler. For a particle with constant energy,

E =

+ U(x),

where v = |

x|. So we have

mv =

2m(E − U(x)).

Hence we can define the action to be

A =

2m(E − U(x)) d`,

where d` is the path length element. We minimize this to find the trajectory.

For a particle near the surface of the Earth, under the influence of gravity,

U = mgz. So we have

A[z] =

2mE − 2m

1 + (z

)

dx,

which is of exactly the same form as the optics problem we just solved. So the

result is again a parabola, as expected.

Example

(Brachistochrone)

The Brachistochrone problem was one of the

earliest problems in the calculus of variations. The name comes from the Greek

words br´akhistos (“shortest”) and khr´onos (“time”).

The question is as follows: suppose we have a bead sliding along a frictionless

wire, starting from rest at the origin

. What shape of wire minimizes the time

for the bead to travel to B?

The conservation of energy implies that

= mgy.

v =

2gy

We want to minimize

T =

√

+ dy

√

1 + (y

)

Since there is no explicit dependence on x, we have the first integral

f − y

∂f

∂y

y(1 + (y

)

= constant

So the solution is

y(1 + (y

)

) = c

for some positive constant c.

The solution of this ODE is, in parametric form,

x = c(θ − sin θ)

y = c(1 − cos θ).

Note that this has x = y = 0 at θ = 0. This describes a cycloid.

2.3 Constrained variation of functionals

So far, we’ve considered the problem of finding stationary values of

[

] without

any restraint on what

could be. However, sometimes there might be some

restrictions on the possible values of

. For example, we might have a surface in

defined by

(

) = 0. If we want to find the path of shortest length on the

surface (i.e. geodesics), then we want to minimize

[

] subject to the constraint

g(x(t)) = 0.

We can again use Lagrange multipliers. The problem we have to solve is

equivalent to finding stationary values (without constraints) of

[x] = F [x] −λ(P [x] − c).

with respect to the function x(t) and the variable λ.

Example (Isoperimetric problem). If we have a string of fixed length, what is

the maximum area we can enclose with it?

We first argue that the region enclosed by the curve is convex. If it is not,

we can “push out” the curve to increase the area enclosed without changing the

length. Assuming this, we can split the curve into two parts:

We have dA = [y

(x) − y

(x)] dx. So

A =

(x) − y

(x)] dx.

Alternatively,

A[y] =

y(x) dx.

and the length is

L[y] =

d` =

1 + (y

)

dx.

So we look for stationary points of

[y] =

[y(x) − λ

1 + (y

)

] dx + λL.

In this case, we can be sure that our boundary terms vanish since there is no

boundary.

Since there is no explicit dependence on x, we obtain the first integral

f − y

∂f

∂y

= constant = y

= y − λ

1 + (y

)

λ(y

)

1 + (y

)

= y −

1 + (y

)

(y − y

)

1 + (y

)

(y − y

)

− 1

(y − y

− (y − y

)

= ±1.

− (y − y

)

± x

= 0.

So we have

− (y − y

)

= (x − x

)

(x − x

)

+ (y − y

)

= λ

This is a circle of radius

. Since the perimeter of this circle will be 2

πλ

, we

must have λ = L/(2π). So the maximum area is πλ

= L

/(4π).

Example

(Sturm-Liouville problem)

The Sturm-Liouville problem is a very

general class of problems. We will develop some very general theory about these

problems without going into specific examples. It can be formulated as follows:

let

(

) and

(

) be real functions of

defined on

α ≤ x ≤ β

. We will

consider the special case where

and

are positive on

α < x < β

. Our objective

is to find stationary points of the functional

F [y] =

(ρ(x)(y

)

+ σ(x)y

) dx

subject to the condition

G[y] =

w(x)y

dx = 1.

Using the Euler-Lagrange equation, the functional derivatives of F and G are

δF [y]

δy

= 2



− (ρy

)

+ σy



δG[y]

δy

= 2(wy).

So the Euler-Lagrange equation of Φ

[y] = F [y] − λ(G[y] − 1) is

−(ρy

)

+ σy − λwy = 0.

We can write this as the eigenvalue problem

Ly = λwy.

where

L = −





+ σ

is the Sturm-Liouville operator. We call this a Sturm-Liouville eigenvalue

problem. w is called the weight function.

We can view this problem in a different way. Notice that

λwy

is linear

. Hence if

is a solution, then so is

. But if

[

] = 1, then

[

] =

Hence the condition

[

] = 1 is simply a normalization condition. We can get

around this problem by asking for the minimum of the functional

Λ[y] =

F [y]

G[y]

instead. It turns out that this Λ has some significance. To minimize Λ, we

cannot apply the Euler-Lagrange equations, since Λ is not of the form of an

integral. However, we can try to vary it directly:

δΛ =

δF −

δG =

(δF − ΛδG).

When Λ is minimized, we have

δΛ = 0 ⇔

δF

δy

= Λ

δG

δy

⇔ Ly = Λwy.

So at stationary values of Λ[y], Λ is the associated Sturm-Liouville eigenvalue.

Example

(Geodesics)

Suppose that we have a surface in

defined by

(

) = 0,

and we want to find the path of shortest distance between two points on the

surface. These paths are known as geodesics.

One possible approach is to solve

(

) = 0 directly. For example, if we have

a unit sphere, a possible solution is

cos θ cos φ

cos θ sin φ

sin θ

Then the total length of a path would be given by

D[θ, φ] =

dθ

+ sin

θdφ

We then vary θ and φ to minimize D and obtain a geodesic.

Alternatively, we can impose the condition

(

)) = 0 with a Lagrange

multiplier. However, since we want the constraint to be satisfied for all

, we

need a Lagrange multiplier function

(

). Then our problem would be to find

stationary values of

Φ[x, λ] =



x| − λ(t)g(x(t))



3 Hamilton’s principle

As mentioned before, Lagrange and Hamilton reformulated Newtonian dynamics

into a much more robust system based on an action principle.

The first important concept is the idea of a configuration space. This

configuration space is a vector space containing generalized coordinates

(

) that

specify the configuration of the system. The idea is to capture all information

about the system in one single vector.

In the simplest case of a single free particle, these generalized coordinates

would simply be the coordinates of the position of the particle. If we have

two particles given by positions

(

) = (

, x

) and

(

) = (

, y

), our

generalized coordinates might be

(

) = (

, x

, y

). In general, if we

have N different free particles, the configuration space has 3N dimensions.

The important thing is that the generalized coordinates need not be just the

usual Cartesian coordinates. If we are describing a pendulum in a plane, we do

not need to specify the

and

coordinates of the mass. Instead, the system

can be described by just the angle

between the mass and the vertical. So the

generalized coordinates is just

(

) =

(

). This is much more natural to work

with and avoids the hassle of imposing constraints on x and y.

3.1 The Lagrangian

The concept of generalized coordinates was first introduced by Lagrange in 1788.

He then showed that

(

) obeys certain complicated ODEs which are determined

by the kinetic energy and the potential energy.

In the 1830s, Hamilton made Lagrange’s mechanics much more pleasant. He

showed that the solutions of these ODEs are extremal points of a new “action”,

S[ξ] =

L dt

where

L = T − V

is the Lagrangian, with T the kinetic energy and V the potential energy.

Law

(Hamilton’s principle)

The actual path

(

) taken by a particle is the path

that makes the action S stationary.

Note that

has dimensions

−1

, which is the same as the 18th century

action (and Plank’s constant).

Example.

Suppose we have 1 particle in Euclidean 3-space. The configuration

space is simply the coordinates of the particle in space. We can choose Cartesian

coordinates x. Then

T =

, V = V (x, t)

and

S[x] =



− V (x, t)



dt.

Then the Lagrangian is

L(x,

x, t) =

− V (x, t)

We apply the Euler-Lagrange equations to obtain

0 =



∂L

∂



−

∂L

∂x

= m

x + ∇V.

x = −∇V

This is Newton’s law

with

−∇V

. This shows that Lagrangian

mechanics is “the same” as Newton’s law. However, Lagrangian mechanics has

the advantage that it does not care what coordinates you use, while Newton’s

law requires an inertial frame of reference.

Lagrangian mechanics applies even when

is time-dependent. However, if

V is independent of time, then so is L. Then we can obtain a first integral.

As before, the chain rule gives

∂L

∂t

x ·

∂L

∂x

x ·

∂

∂L

∂t

x ·



∂L

∂x

−



∂L

∂



| {z }

δS

δx

= 0

x ·



∂L

∂



x ·

∂L

∂

| {z }



x ·

∂L

∂



So we have



L −

x ·

∂L

∂



∂L

∂t

∂L

∂t

= 0, then

x ·

∂L

∂

− L = E

for some constant E. For example, for one particle,

E = m|

−

+ V = T + V = total energy.

Example.

Consider a central force field

−∇V

, where

(

) is inde-

pendent of time. We use spherical polar coordinates (r, θ, φ), where

x = r sin θ cos φ

y = r sin θ sin φ

z = r cos θ.

T =



˙r

+ r

(

+ sin

)



L =

m ˙r



+ sin



− V (r).

We’ll use the fact that motion is planar (a consequence of angular momentum

conservation). So wlog θ =

. Then

L =

m ˙r

− V (r).

Then the Euler Lagrange equations give

m¨r − mr

+ V

(r) = 0





= 0.

From the second equation, we see that

is a constant (angular momentum

per unit mass). Then

φ = h/r

. So

m¨r −

+ V

(r) = 0.

If we let

eff

= V (r) +

be the effective potential, then we have

m¨r = −V

eff

(r).

For example, in a gravitational field, V (r) = −

. Then

eff

= m



−



3.2 The Hamiltonian

In 1833, Hamilton took Lagrangian mechanics further and formulated Hamilto-

nian mechanics. The idea is to abandon

and use the conjugate momentum

∂L

∂

instead. Of course, this involves taking the Legendre transform of the

Lagrangian to obtain the Hamiltonian.

Definition

(Hamiltonian)

The Hamiltonian of a system is the Legendre trans-

form of the Lagrangian:

H(x, p) = p ·

x − L(x,

x),

where

x is a function of p that is the solution to p =

∂L

∂

is the conjugate momentum of

. The space containing the variables

x, p

is known as the phase space.

Since the Legendre transform is its self-inverse, the Lagrangian is the Legendre

transform of the Hamiltonian with respect to p. So

L = p ·

x − H(x, p)

with

x =

∂H

∂p

Hence we can write the action using the Hamiltonian as

S[x, p] =

(p ·

x − H(x, p)) dt.

This is the phase-space form of the action. The Euler-Lagrange equations for

these are

x =

∂H

∂p

p = −

∂H

∂x

Using the Hamiltonian, the Euler-Lagrange equations put

and

on a much

more equal footing, and the equations are more symmetric. There are also many

useful concepts arising from the Hamiltonian, which are explored much in-depth

in the II Classical Dynamics course.

So what does the Hamiltonian look like? Consider the case of a single particle.

The Lagrangian is given by

L =

− V (x, t).

Then the conjugate momentum is

p =

∂L

∂

= m

which happens to coincide with the usual definition of the momentum. How-

ever, the conjugate momentum is often something more interesting when we

use generalized coordinates. For example, in polar coordinates, the conjugate

momentum of the angle is the angular momentum.

Substituting this into the Hamiltonian, we obtain

H(x, p) = p ·

−





+ V (x, t)

|p|

+ V.

So H is the total energy, but expressed in terms of x, p, not x,

3.3 Symmetries and Noether’s theorem

Given

F [x] =

f(x, ˙x, t) dt,

suppose we change variables by the transformation

t 7→ t

∗

(

) and

x 7→ x

∗

(

∗

Then we have a new independent variable and a new function. This gives

F [x] 7→ F

∗

] =

∗

f(x

∗

, ˙x

∗

, t

∗

) dt

∗

with α

∗

= t

∗

(α) and β

∗

= t

∗

(β).

There are some transformations that are particularly interesting:

Definition

(Symmetry)

∗

[

∗

] =

[

] for all

and

, then the transfor-

mation ∗ is a symmetry.

This transformation could be a translation of time, space, or a rotation, or

even more fancy stuff. The exact symmetries

has depends on the form of

For example, if

only depends on the magnitudes of

˙x

and

, then rotation

of space will be a symmetry.

Example.

(i)

Consider the transformation

t 7→ t

and

x 7→ x

for some small

. Then

∗

] =

f(x + ε, ˙x, t) dx =



f(x, ˙x, t) + ε

∂f

∂x



by the chain rule. Hence this transformation is a symmetry if

∂f

∂x

= 0.

However, we also know that if

∂f

∂x

= 0, then we have the first integral



∂f

∂ ˙x



= 0.

∂f

∂ ˙x

is a conserved quantity.

(ii)

Consider the transformation

t 7→ t − ε

. For the sake of sanity, we will also

transform x 7→ x

∗

such that x

∗

) = x(t). Then

∗

] =

f(x, ˙x, t − ε) dt =



f(x, ˙x, t) − ε

∂f

∂t



dt.

Hence this is a symmetry if

∂f

∂t

= 0.

We also know that if

∂f

∂t

= 0 is true, then we obtain the first integral



f − ˙x

∂f

∂ ˙x



= 0

So we have a conserved quantity f − ˙x

∂f

∂ ˙x

We see that for each simple symmetry we have above, we can obtain a first

integral, which then gives a constant of motion. Noether’s theorem is a powerful

generalization of this.

Theorem

(Noether’s theorem)

For every continuous symmetry of

[

], the

solutions (i.e. the stationary points of

[

]) will have a corresponding conserved

quantity.

What does “continuous symmetry” mean? Intuitively, it is a symmetry we

can do “a bit of”. For example, rotation is a continuous symmetry, since we can

do a bit of rotation. However, reflection is not, since we cannot reflect by “a

bit”. We either reflect or we do not.

Note that continuity is essential. For example, if

is quadratic in

and

˙x

then

x 7→ −x

will be a symmetry. But since it is not continuous, there won’t be

a conserved quantity.

Since the theorem requires a continuous symmetry, we can just consider

infinitesimally small symmetries and ignore second-order terms. Almost every

equation will have some O(ε

) that we will not write out.

We will consider symmetries that involve only the

variable. Up to first

order, we can write the symmetry as

t 7→ t, x(t) 7→ x(t) + εh(t),

for some

(

) representing the symmetry transformation (and

a small number).

By saying that this transformation is a symmetry, it means that when we

pick

to be any (small) constant number, the functional

[

] does not change,

i.e. δF = 0.

On the other hand, since

(

) is a stationary point of

[

], we know that if

is non-constant but vanishes at the end-points, then

δF

= 0 as well. We will

combine these two information to find a conserved quantity of the system.

For the moment, we do not assume anything about

and see what happens

to F [x]. Under the transformation, the change in F [x] is given by

δF =



f(x + εh, ˙x + ε

h + ˙εh, t) − f(x, ˙x, t)





∂f

∂x

εh +

∂f

∂ ˙x

h +

∂f

∂ ˙x

˙εh





∂f

∂x

h +

∂f

∂ ˙x



dt +

˙ε



∂f

∂ ˙x



dt.

First consider the case where

is a constant. Then the second integral vanishes.

So we obtain



∂f

∂x

h +

∂f

∂ ˙x



dt = 0

This requires that

∂f

∂x

h +

∂f

∂ ˙x

h = 0

Hence we know that

δF =

˙ε



∂f

∂ ˙x



dt.

Then consider a variable

that is non-constant but vanishes at end-points. Then

we know that since x is a solution, we must have δF = 0. So we get

˙ε



∂f

∂ ˙x



dt = 0.

We can integrate by parts to obtain



∂f

∂ ˙x



dt = 0.

for any ε that vanishes at end-points. Hence we must have



∂f

∂ ˙x



= 0.

∂f

∂ ˙x

h is a conserved quantity.

Obviously, not all symmetries just involve the

variable. For example, we

might have a time translation

t 7→ t − ε

. However, we can encode this as a

transformation of the x variable only, as x(t) 7→ x(t − ε).

In general, to find the conserved quantity associated with the symmetry

(

)

7→ x

(

) +

εh

(

), we find the change

δF

assuming that

is a function of time

as opposed to a constant. Then the coefficient of ˙ε is the conserved quantity.

Example.

We can apply this to Hamiltonian mechanics. The motion of the

particle is the stationary point of

S[x, p] =

(p ·

x − H(x, p)) dt,

where

H =

|p|

+ V (x).

(i)

First consider the case where there is no potential. Since the action depends

only on

x (or p) and not x itself, it is invariant under the translation

x 7→ x + ε, p 7→ p.

For general ε that can vary with time, we have

δS =



p · (

x +

ε) − H(p)



−



p ·

x − H(p)



p ·

ε dt.

Hence p (the momentum) is a constant of motion.

(ii)

If the potential has no time-dependence, then the system is invariant under

time translation. We’ll skip the tedious computations and just state that

time translation invariance implies conservation of

itself, which is the

energy.

(iii)

The above two results can also be obtained directly from first integrals

of the Euler-Lagrange equation. However, we can do something cooler.

Suppose that we have a potential

(

|x|

) that only depends on radius. Then

this has a rotational symmetry.

Choose any favorite axis of rotational symmetry

, and make the rotation

x 7→ x + εω × x

p 7→ p + εω × p,

Then our rotation does not affect the radius

|x|

and momentum

|p|

. So

the Hamiltonian

(

x, p

) is unaffected. Noting that

ω ×p ·

= 0, we have

δS =



p ·

(x + εω × x) − p ·





p ·

(εω × x)





p ·



ω ×

(εx)



(p · [ω × ( ˙εx + ε

x)]) dt

( ˙εp ·(ω × x) + εp · (ω ×

x)) dt

Since p is parallel to

x, we are left with

( ˙εp ·(ω × x)) dt

˙εω · (x × p) dt.

ω ·

(

x ×p

) is a constant of motion. Since this is true for all

x ×p

must be a constant of motion, and this is the angular momentum.

4 Multivariate calculus of variations

So far, the function

(

) we are varying is just a function of a single variable

What if we have a more complicated function to consider?

We will consider the most general case

(

, ··· , x

)

∈ R

that maps

→ R

(we can also view this as

different functions that map

→ R

The functional will be a multiple integral of the form

F [y] =

···

f(y, ∇y, x

, ··· , x

) dx

···dx

where ∇y is the second-rank tensor defined as

∇y =



∂y

∂x

, ··· ,

∂y

∂x



In this case, instead of attempting to come up with some complicated generalized

Euler-Lagrange equation, it is often a better idea to directly consider variations

δy of y. This is best illustrated by example.

Example

(Minimal surfaces in

)

This is a natural generalization of geodesics.

A minimal surface is a surface of least area subject to some boundary conditions.

Suppose that (

x, y

) are good coordinates for a surface

, where (

x, y

) takes

values in the domain

D ⊆ R

. Then the surface is defined by

(

x, y

), where

h is the height function.

When possible, we will denote partial differentiation by suffices, i.e.

∂h

∂x

Then the area is given by

A[h] =

1 + h

+ h

dA.

Consider a variation of h(x, y): h 7→ h + δh(x, y). Then

A[h + δh] =

1 + (h

+ (δh)

)

+ (h

+ (δh)

)

= A[h] +





(δh)

+ h

(δh)

1 + h

+ h

+ O(δh

)





We integrate by parts to obtain

δA = −

δh





∂

∂x





1 + h

+ h





∂

∂y





1 + h

+ h









dA + O(δh

)

plus some boundary terms. So our minimal surface will satisfy

∂

∂x





1 + h

+ h





∂

∂y





1 + h

+ h





= 0

Simplifying, we have

(1 + h

+ (1 + h

− 2h

= 0.

This is a non-linear 2nd-order PDE, the minimal-surface equation. While it is

difficult to come up with a fully general solution to this PDE, we can consider

some special cases.

– There is an obvious solution

h(x, y) = Ax + By + C,

since the equation involves second-derivatives and this function is linear.

This represents a plane.

– If |∇h|

 1, then h

and h

are small. So we have

+ h

= 0,

∇

h = 0.

So we end up with the Laplace equation. Hence harmonic functions are

(approximately) minimal-area.

–

We might want a cylindrically-symmetric solution, i.e.

(

x, y

) =

(

), where

r =

+ y

. Then we are left with an ordinary differential equation

+ z

= 0.

The general solution is

z = A

−1

cosh(Ar) + B,

a catenoid.

Alternatively, to obtain this,this we can substitute

(

x, y

) =

(

) into

[

]

to get

A[z] = 2π

1 + (h

(r))

dr,

and we can apply the Euler-Lagrange equation.

Example

(Small amplitude oscillations of uniform string)

Suppose we have a

string with uniform constant mass density ρ with uniform tension T .

Suppose we pull the line between

= 0 and

with some tension

. Then we

set it into motion such that the amplitude is given by y(x; t). Then the kinetic

energy is

T =

ρv

dx =

˙y

dx.

The potential energy is the tension times the length. So

V = T

d` = T

1 + (y

)

dx = (T a) +

T (y

) dx.

Note that y

is the derivative wrt x while ˙y is the derivative wrt time.

The

T a

term can be seen as the ground-state energy. It is the energy initially

stored if there is no oscillation. Since this constant term doesn’t affect where

the stationary points lie, we will ignore it. Then the action is given by

S[y] =



ρ ˙y

−

T (y

)



dx dt

We apply Hamilton’s principle which says that we need

δS[y] = 0.

We have

δS[y] =



ρ ˙y

∂

∂t

δy − T y

∂

∂x

δy



dx dt.

Integrate by parts to obtain

δS[y] =

δy(ρ¨y − T y

) dx dt + boundary term.

Assuming that the boundary term vanishes, we will need

¨y − v

= 0,

where

T/ρ

. This is the wave equation in two dimensions. Note that this is a

linear PDE, which is a simplification resulting from our assuming the oscillation

is small.

The general solution to the wave equation is

y(x, t) = f

(x − vt) + f

−

(x + vt),

which is a superposition of a wave travelling rightwards and a wave travelling

leftwards.

Example

(Maxwell’s equations)

It is possible to obtain Maxwell’s equations

from an action principle, where we define a Lagrangian for the electromagnetic

field. Note that this is the Lagrangian for the field itself, and there is a separate

Lagrangian for particles moving in a field.

We have to first define several quantities. First we have the charges:

represents the electric charge density and

represents the electric current

density.

Then we have the potentials:

is the electric scalar potential and

is the

magnetic vector potential.

Finally the fields:

−∇φ −

is the electric field, and

∇ × A

is the

magnetic field.

We pick convenient units where

= 1. With these concepts in

mind, the Lagrangian is given by

S[A, φ] =



(|E|

− |B|

) + A · J − φρ



dV dt

Varying A and φ by δA and δφ respectively, we have

δS =



−E ·



∇δφ +

∂

∂t

δA



− B · ∇ × δA + δA · J − ρδφ



dV dt.

Integrate by parts to obtain

δS =



δA · (

E − ∇ × B + J) + δφ(∇ · E − ρ)



dV dt.

Since the coefficients have to be 0, we must have

∇ × B = J +

E, ∇ · E = ρ.

Also, the definitions of E and B immediately give

∇ · B = 0, ∇ × E = −

These four equations are Maxwell’s equations.

5 The second variation

5.1 The second variation

So far, we have only looked at the “first derivatives” of functionals. We can

identify stationary points, but we don’t know if it is a maximum, minimum or a

saddle. To distinguish between these, we have to look at the “second derivatives”,

or the second variation.

Suppose x(t) = x

(t) is a solution of

δF [x]

δy(x)

= 0,

i.e. F [x] is stationary at y = y

To determine what type of stationary point it is, we need to expand

[

δx

]

to second order in

δx

. For convenience, let

δx

(

) =

εξ

(

) with constant

ε 

We will also only consider functionals of the form

F [x] =

f(x, ˙x, t) dt

with fixed-end boundary conditions, i.e.

(

) =

(

) = 0. We will use both dots

( ˙x) and dashes (x

) to denote derivatives.

We consider a variation x 7→ x + δx and expand the integrand to obtain

f(x + εξ, ˙x + ε

ξ, t) − f(x, ˙x, t)

= ε



∂f

∂x

∂f

∂ ˙x





∂

∂x

+ 2ξ

∂

∂x∂ ˙x

∂

∂ ˙x



+ O(ε

)

Noting that 2ξ

ξ = (ξ

)

and integrating by parts, we obtain

= εξ



∂f

∂x

−



∂f

∂ ˙x







∂

∂x

−



∂

∂x∂ ˙x



∂f

∂ ˙x



plus some boundary terms which vanish. So

F [x + εξ] − F [x] =

εξ



∂f

∂x

−



∂f

∂ ˙x



dt +

F [x, ξ] + O(ε

where

F [x, ξ] =





∂

∂x

−



∂

∂x∂ ˙x



∂

∂ ˙x



is a functional of both x(t) and ξ(t). This is analogous to the term

δx

H(x)δx

appearing in the expansion of a regular function

(

). In the case of normal

functions, if

(

) is positive,

(

) is convex for all

, and the stationary point

is hence a global minimum. A similar result holds for functionals.

In this case, if

[

x, ξ

]

0 for all non-zero

and all allowed

, then a

solution x

(t) of

δF

δx

= 0 is an absolute minimum.

Example

(Geodesics in the plane)

We previously shown that a straight line is

a stationary point for the curve-length functional, but we didn’t show it is in

fact the shortest distance! Maybe it is a maximum, and we can get the shortest

distance by routing to the moon and back.

Recall that f =

1 + (y

)

. Then

∂f

∂y

= 0,

∂f

∂y

1 + (y

)

∂

∂y

1 + (y

)

with the other second derivatives zero. So we have

F [y, ξ] =

(1 + (y

)

3/2

dx > 0

So if we have a stationary function satisfying the boundary conditions, it is an

absolute minimum. Since the straight line is a stationary function, it is indeed

the minimum.

However, not all functions are convex

[citation needed]

. We can still ask whether

a solution

(

) of the Euler-Lagrange equation is a local minimum. For these,

we need to consider

F [x

, ξ] =

(ρ(t)

+ σ(t)ξ

) dt,

where

ρ(t) =

∂

∂ ˙x



x=x

, σ(t) =



∂

∂x

−



∂

∂x∂ ˙x



x=x

This is of the same form as the Sturm-Liouville problem. For

to minimize

F [x] locally, we need δ

F [x

, ξ] > 0. A necessary condition for this is

ρ(t) ≥ 0,

which is the Legendre condition.

The intuition behind this necessary condition is as follows: suppose that

(

) is negative in some interval

I ⊆

[

α, β

]. Then we can find a

(

) that makes

[

, ξ

] negative. We simply have to make

zero outside

, and small but

crazily oscillating inside

. Then inside

˙x

wiill be very large while

is kept

tiny. So we can make δ

F [y, ξ] arbitrarily negative.

Turning the intuition into a formal proof is not difficult but is tedious and

will be omitted.

However, this is not a sufficient condition. Even if we had a strict inequality

ρ(t) > 0 for all α < t < β, it is still not sufficient.

Of course, a sufficient (but not necessary) condition is

(

)

, σ

(

)

≥

0, but

this is not too interesting.

Example. In the Branchistochrone problem, we have

T [x] ∝

1 + ˙x

dt.

Then

ρ(t) =

∂

∂ ˙x



> 0

σ(t) =

x(1 + ˙x

)

> 0.

So the cycloid does minimize the time T .

5.2 Jacobi condition for local minima of F [x]

Legendre tried to prove that

ρ >

0 is a sufficient condition for

F >

0. This is

known as the strong Legendre condition. However, he obviously failed, since it is

indeed not a sufficient condition. Yet, it turns out that he was close.

Before we get to the actual sufficient condition, we first try to understand

why thinking ρ > 0 is sufficient isn’t as crazy as it first sounds.

ρ >

0 and

σ <

0, we would want to create a negative

[

, ξ

] by choosing

to be large but slowly varying. Then we will have a very negative

(

)

while

a small positive ρ(t)

The problem is that

has to be 0 at the end points

and

. For

to take a

large value, it must reach the value from 0, and this requires some variation of

thereby inducing some

. This is not a problem if

and

are far apart - we

simply slowly climb up to a large value of

and then slowly rappel back down,

maintaining a low

throughout the process. However, it is not unreasonable to

assume that as we make the distance

β − α

smaller and smaller, eventually all

will lead to a positive

[

, ξ

], since we cannot reach large values of

without

having large

ξ.

It turns out that the intuition is correct. As long as

and

are sufficiently

close,

[

, ξ

] will be positive. The derivation of this result is, however, rather

roundabout, involving a number of algebraic tricks.

For a solution x

to the Euler Lagrange equation, we have

F [x

, ξ] =



ρ(t)

+ σ(t)ξ



dt,

where

ρ(t) =

∂

∂ ˙x



x=x

, σ(t) =



∂

∂x

−



∂

∂x∂ ˙x



x=x

Assume

(

)

0 for

α < t < β

(the strong Legendre condition) and assume

boundary conditions ξ(α) = ξ(β) = 0. When is this sufficient for δ

F > 0?

First of all, notice that for any smooth function w(x), we have

0 =

(wξ

)

since this is a total derivative and evaluates to

wξ

(

)

− wξ

(

) = 0. So we have

0 =

(2wξ

ξ + ˙wξ

) dt.

This allows us to rewrite δ

F as

F =



+ 2wξ

ξ + (σ + ˙w)ξ



dt.

Now complete the square in ξ and

ξ. So

F =



ξ +





σ + ˙w −



This is non-negative if

= ρ(σ + ˙w). (∗)

So as long as we can find a solution to this equation, we know that

non-negative. Could it be that

= 0? Turns out not. If it were, then

ξ = −

ξ. We can solve this to obtain

ξ(x) = C exp



−

w(s)

ρ(s)



We know that ξ(α) = 0. But ξ(α) = Ce

. So C = 0. Hence equality holds only

for ξ = 0.

So all we need to do is to find a solution to (

∗

), and we are sure that

F >

Note that this is non-linear in

. We can convert this into a linear equation

by defining w in terms of a new function u by w = −ρ ˙u/u. Then (∗) becomes



˙u



= σ −



ρ ˙u



= σ −

(ρ ˙u)

+ ρ



˙u



We see that the left and right terms cancel. So we have

−(ρ ˙u)

+ σu = 0.

This is the Jacobi accessory equation, a second-order linear ODE.

There is a caveat here. Not every solution

will do. Recall that

is used

to produce

via

−ρ ˙u/u

. Hence within [

α, β

], we cannot have

= 0 since

we cannot divide by zero. If we can find a non-zero

(

) satisfying the Jacobi

accessory equation, then

F >

0 for

ξ 6

= 0, and hence

is a local minimum of

F .

A suitable solution will always exists for sufficiently small

β − α

, but may

not exist if β − α is too large, as stated at the beginning of the chapter.

Example

(Geodesics on unit sphere)

For any curve

on the sphere, we have

L =

dθ

+ sin

θ dφ

If θ is a good parameter of the curve, then

L[φ] =

1 + sin

θ(φ

)

dθ.

Alternatively, if φ is a good parameter, we have

L[θ] =

(θ

)

+ sin

θ dφ.

We will look at the second case.

We have

f(θ, θ

) =

(θ

)

+ sin

θ.

∂f

∂θ

sin θ cos θ

(θ

)

+ sin

∂f

∂θ

(θ

)

+ sin

Since

∂f

∂φ

= 0, we have the first integral

const = f − θ

∂f

∂θ

sin

(θ

)

+ sin

So a solution is

c sin

θ =

(θ

)

+ sin

θ.

Here we need c ≥ 1 for the equation to make sense.

We will consider the case where

= 1 (in fact, we can show that we can

always orient our axes such that

= 1). This occurs when

= 0, i.e.

is a

constant. Then our first integral gives

sin

sin θ

. So

sin θ

= 1 and

π/

This corresponds to a curve on the equator. (we ignore the case

sin θ

= 0 that

gives θ = 0, which is a rather silly solution)

There are two equatorial solutions to the Euler-Lagrange equations. Which,

if any, minimizes L[θ]?

We have

∂

∂(θ

)



θ=π/2

= 1

and

∂

∂θ∂θ

= −1,

∂

∂θ∂θ

= 0.

So ρ(x) = 1 and σ(x) = −1. So

F =

((ξ

)

− ξ

) dφ.

The Jacobi accessory equation is

= 0. So the general solution is

u ∝

sin φ − γ cos φ. This is equal to zero if tan φ = γ.

Looking at the graph of

tan φ

, we see that

tan

has a zero every

radians.

Hence if the domain

− φ

is greater than

(i.e. we go the long way from

the first point to the second), it will always contain some values for which

tan φ

is zero. So we cannot conclude that the longer path is a local minimum (it

is obviously not a global minimum, by definition of longer) (we also cannot

conclude that it is not a local minimum, since we tested with a sufficient and

not necessary condition). On the other hand, if

− φ

is less than

, then we

will be able to pick a γ such that u is non-zero in the domain.