mechanics lagrangian and hamilitonian

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Lecture notes

An introduction to Lagrangian and Hamiltonian mechanics

Simon J.A. Malham

Simon J.A. Malham (6th March 2014)Maxwell Institute for Mathematical Sciencesand School of Mathematical and Computer SciencesHeriot-Watt University, Edinburgh EH14 4AS, UKTel.: +44-131-4513200Fax: +44-131-4513249E-mail: [email protected]


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2 Simon J.A. Malham

1 Introduction

Newtonian mechanics took the Apollo astronauts to the moon. It also took the voy-

ager spacecraft to the far reaches of the solar system. However Newtonian mechanics

is a consequence of a more general scheme. One that brought us quantum mechanics,

and thus the digital age. Indeed it has pointed us beyond that as well. The scheme is

Lagrangian and Hamiltonian mechanics. Its original prescription rested on two prin-

ciples. First that we should try to express the state of the mechanical system using

the minimum representation possible and which reflects the fact that the physics of

the problem is coordinate-invariant. Second, a mechanical system tries to optimize its

action from one split second to the next; often this corresponds to optimizing its to-

tal energy as it evolves from one state to the next. These notes are intended as an

elementary introduction into these ideas and the basic prescription of Lagrangian and

Hamiltonian mechanics. A pre-requisite is the thorough understanding of the calculus

of variations, which is where we begin.

2 Calculus of variations

Many physical problems involve the minimization (or maximization) of a quantity that

is expressed as an integral.

2.1 Example (Euclidean geodesic)

Consider the path that gives the shortest distance between two points in the plane, say

(x1, y1) and (x2, y2). Suppose that the general curve joining these two points is given

byy = y(x). Then our goal is to find the function y(x) that minimizes the arclength:

J(y) =

(x2,y2)(x1,y1)

ds

=

x2x1

1 + (yx)2 dx.

Here we have used that for a curve y = y(x), if we make a small increment in x,

say x, and the corresponding change in y is y, then by Pythagoras theorem the

corresponding change in length along the curve iss = +

(x)2 + (y)2. Hence we

see that

s= s

xx = 1 +

y

x2

x.

Note further that here, and hereafter, we use yx = yx(x) to denote the derivative ofy ,

i.e. yx(x) = y(x) for each x for which the derivative is defined.

2.2 Example (Brachistochrome problem; John and James Bernoulli 1697)

Suppose a particle/bead is allowed to slide freely along a wire under gravity (force

F = gk wherek is the unit upward vertical vector) from a point (x1, y1) to the origin


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Lagrangian and Hamiltonian mechanics 3

1

2(x , y )

x , y )1

2

y=y(x)

Fig. 1 In the Euclidean geodesic problem, the goal is to find the path with minimum totallength between points (x1, y1) and (x2, y2).

(0,0)

(x , y )1 1

y=y(x)mgk

v

Fig. 2 In the Brachistochrome problem, a bead can slide freely under gravity along the wire.The goal is to find the shape of the wire that minimizes the time of descent of the bead.

(0, 0). Find the curve y = y(x) that minimizes the time of descent:

J(y) = (0,0)

(x1,y1)

1

vds

=

0x1

1 + (yx)2

2g(y1 y)dx.

Here we have used that the total energy, which is the sum of the kinetic and potential

energies,

E= 12mv2 + mgy,

is constant. Assume the initial condition isv = 0 when y = y1, i.e. the bead starts with

zero velocity at the top end of the wire. Since its total energy is constant, its energy at

any timet later, when its height is y and its velocity is v , is equal to its initial energy.

Hence we have

12mv

2 + mgy = 0 + mgy1 v = +

2g(y1 y).

3 EulerLagrange equation

We can see that the two examples above are special cases of a more general problem

scenario.


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4 Simon J.A. Malham

3.1 Classical variational problem

Suppose the given functionF(, , ) is twice continuously differentiable with respect toall of its arguments. Among all functions/pathsy = y(x), which are twice continuously

differentiable on the interval [a, b] with y(a) and y(b) specified, find the one which

extremizes thefunctional defined by

J(y):=

ba

F(x ,y,yx) dx.

Theorem 1 (EulerLagrange equations) The function u = u(x) that extremizes the

functionalJ necessarily satisfies the EulerLagrange equation

F

u d

dxF

ux = 0,on[a, b].

ProofConsider the family of functions on [a, b] given by

y(x; ):=u(x) + (x),

where the functions = (x) are twice continuously differentiable and satisfy (a) =

(b) = 0. Here is a small real parameter, and of course, the functionu = u(x) is our

candidate extremizing function. We set (see for example Evans [7, Section 3.3])

():=J(u + ).

If the functional J has a local maximum or minimum at u, then u is a stationary

function forJ, and for all we must have

(0) = 0.

To evaluate this condition for the functional given in integral form above, we need to

first determine(). By direct calculation,

() = d

dJ(u + )

= d

d

ba

F(x, u + ,ux+ x) dx

=

ba

F(x, y(x; ), yx(x; )) dx

= b

a

Fy

y

+ Fyx

yx

dx

=

ba

F

y(x) +

F

yx(x)

dx.

The integration by parts formula tells us that

ba

F

yx(x) dx=

F

yx(x)

x=bx=a

ba

d

dx

F

yx

(x) dx.


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Recall that (a) = (b) = 0, so the boundary term (first term on the right) vanishes

in this last formula, and we see that

() =

ba

F

y d

dx

F

yx

(x) dx.

If we now set = 0, the condition for u to be a critical point ofJ is

ba

F

u d

dx

F

ux

(x) dx= 0.

Now we note that the functions = (x) are arbitrary, using Lemma 1 below, we

can deduce that pointwise, i.e. for all x [a, b], necessarily u must satisfy the EulerLagrange equation shown.

3.2 Useful lemma

Lemma 1 (Useful lemma) If(x) is continuous in[a, b] and

ba

(x) (x) dx= 0

for all continuously differentiable functions (x) which satisfy (a) = (b) = 0, then

(x) 0 in[a, b].

Proof Assume(z)> 0, say, at some pointa < z < b. Then since is continuous, we

must have that(x)> 0 in some open neighbourhood ofz, say in a < z < z < z < b.The choice

(x) =

(x z)2(z x)2, for x [z, z],0, otherwise,

which is a continuously differentiable function, implies

ba

(x) (x) dx=

zz

(x) (x z)2(z x)2 dx > 0,

a contradiction.

3.3 Remarks

Some important theoretical and practical points to keep in mind are as follows.

1. The EulerLagrange equation is a necessary condition: if such a u = u(x) exists

that extremizesJ, thenu satisfies the EulerLagrange equation. Such au is known

as a stationary functionof the functional J.

2. The EulerLagrange equation above is an ordinary differential equation foru; this

will be clearer once we consider some examples presently.


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6 Simon J.A. Malham

3. Note that the extremal solution u is independent of the coordinate system you

choose to represent it (see Arnold [3, Page 59]). For example, in the Euclideangeodesic problem, we could have used polar coordinates (r, ), instead of Cartesian

coordinates (x, y), to express the total arclength J. Formulating the EulerLagrange

equations in these coordinates and then solving them will tell us that the extrem-

izing solution is a straight line (only it will be expressed in polar coordinates).

4. Let Y denote a function space; in the context above Y was the space of twice

continuously differentiable functions on [a, b] which are fixed at x = a and x = b.

A functionalis a real-valued map and here J: Y R.5. We define thefirst variation J(u, ) of the functional J, at u in the direction ,

to be J(u, ):=(0).

6. Is u a maximum, minimum or saddle point for J? The physical context should

hint towards what to expect. Higher order variations will give you the appropriate

mathematical determination.

7. The functional J has a local minimum at u iff there is an open neighbourhoodU Y of u such that J(y) J(u) for all y U. The functional J has a localmaximum atu when this inequality is reversed.

8. We will generalize all these notions to multidimensions and systems later.

3.4 Solution (Euclidean geodesic)

Recall, this variational problem concerns finding the shortest distance between the two

points (x1, y1) and (x2, y2) in the plane. This is equivalent to minimizing the total

arclength functional

J(y) =

x2x1

1 + (yx)2 dx.

Hence in this case, theintegrand is F(x ,y,yx) =

1 + (yx)2. From the general theory

outlined above, we know that the extremizing solution satisfies the EulerLagrange

equationF

y d

dx

F

yx

= 0.

We substitute the actual form forFwe have in this case which gives

ddx

yx

1 + (yx)

2 12 = 0 d

dx

yx

1 + (yx)2

1

2

= 0

yxx1 + (yx)2

12

(yx)2

yxx1 + (yx)2

32

= 0

1 + (yx)2

yxx1 + (yx)2

32

(yx)2yxx

1 + (yx)2 32

= 0

yxx1 + (yx)2

32

= 0

yxx = 0.


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Hence y(x) = c1 + c2x for some constants c1 and c2. Using the initial and starting

point data we see that the solution is the straight line function (as we should expect)

y=

y2 y1x2 x1

(x x1) + y1.

Note that this calculation might have been a bit shorter if we had recognised that this

example corresponds to the third special case in the next section.

4 Alternative form and special cases

4.1 Alternative form

The EulerLagrange equations given by

F

y d

dx F

yx

= 0

are equivalent to the following alternative formulation of the EulerLagrange equations

F

x d

dx

F yx F

yx

= 0.

4.2 Special cases

When the functional Fdoes not explicitly depend on one or more variables, then the

EulerLagrange equations simplify considerably. We have the following three notable

cases:

1. IfF = F(y, yx) only, i.e. it does not explicitly depend on x, then the alternative

form for the EulerLagrange equation impliesd

dx

F yx F

yx

= 0 F yx F

yx=c,

for some arbitrary constant c.

2. IfF =F(x, yx) only, i.e. it does not explicitlydepend ony, then the EulerLagrange

equation impliesd

dx

F

yx

= 0 F

yx=c,

for some arbitrary constant c.

3. IfF =F(yx) only, then the EulerLagrange equation implies

0 = F

y d

dxF

yx=

F

y

2F

xyx+ yx

2F

yyx+ yxx

2F

yxyx

= yxx 2F

yxyx.

Hence yxx = 0, i.e. y = y(x) is a linear function ofx and has the form

y= c1+ c2x,

for some constants c1 and c2.


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8 Simon J.A. Malham

4.3 Solution (Brachistochrome problem)

Recall, this variational problem concerns a particle/bead which can freely slide along

a wire under the force of gravity. The wire is represented by a curve y = y(x) from

(x1, y1) to the origin (0, 0). The goal is to find the shape of the wire, i.e. y = y(x),

which minimizes the time of descent of the bead, which is given by the functional

J(y) =

0x1

1 + (yx)2

2g(y1 y)dx= 1

2g

0x1

1 + (yx)2

(y1 y) dx.

Hence in this case, the integrand is

F(x ,y,yx) = 1 + (yx)2

(y1

y) .

From the general theory, we know that the extremizing solution satisfies the Euler

Lagrange equation. Note that the multiplicative constant factor 1/

2gshould not affect

the extremizing solution path; indeed it divides out of the EulerLagrange equations.

Noting that the integrand Fdoes not explicitly depend onx, then the alternative form

for the EulerLagrange equation may be easier:

F

x d

dx

F yx F

yx

= 0.

This immediately implies that for some constantc, the EulerLagrange equation is

F

yx

F

yx=c.

Now substituting the form for Finto this gives

1 + (yx)

2 12

(y1 y) 12 yx

yx

1 + (yx)

2 12

(y1 y) 12

=c

1 + (yx)2 12

(y1 y) 12 yx

12 2 yx

(y1 y) 12

1 + (yx)2 12

=c

1 + (yx)2 12

(y1 y) 12 (yx)

2

(y1 y) 12

1 + (yx)2

1

2

=c

1 + (yx)2

(y1 y) 12

1 + (yx)2 12

(yx)2

(y1 y) 12

1 + (yx)2 12

=c

1(y1 y)

1 + (yx)2

=c2.We can now rearrange this equation so that

(yx)2 =

1

c2(y1 y) 1 (yx)2 =

1 c2y1 c2yc2y1 c2y .


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If we seta = y1 and b = 1c2

y1, then this equation becomes

yx =

b + y

a y

12

.

To find the solution to this ordinary differential equation we make the change of variable

y= 12 (a b) 12 (a + b)cos .

If we substitute this into the ordinary differential equation above and use the chain

rule we get

12 (a + b)sin

d

dx =

1 cos 1 + cos

12

.

Now we use that 1/(d/dx) = dx/d, and that sin = 1 cos2

, to get

dx

d = 12 (a + b)sin

1 + cos

1 cos 1

2

dxd

= 12 (a + b) (1 cos2 )1

2

1 + cos

1 cos 1

2

dxd

= 12 (a + b) (1 + cos )1

2 (1 cos ) 12

1 + cos

1 cos 1

2

dxd

= 12 (a + b)(1 + cos ).

We can directly integrate this last equation to findx as a function of. In other words

we can find the solution to the ordinary differential equation for y = y(x) above inparametric form, which with some minor rearrangement, can expressed as (here d is

an arbitrary constant of integration)

x + d= 12 (a + b)(+ sin ),

y+ b= 12 (a + b)(1 cos ).

This is the parametric representation of a cycloid.

5 Multivariable systems

We consider possible generalizations of functionals to be extremized. For more details

see for example Keener [9, Chapter 5].

5.1 Higher derivatives

Suppose we are asked to find the curve y = y(x) R that extremizes the functional

J(y):=

ba

F(x ,y,yx, yxx) dx,


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10 Simon J.A. Malham

subject to y (a), y (b), yx(a) and yx(b) being fixed. Here the functional quantity to be

extremized depends on the curvatureyxxof the path. Necessarily the extremizing curvey satisfies the EulerLagrange equation (which is an ordinary differential equation)

F

y d

dx

F

yx

+

d2

dx2

F

yxx

= 0.

Note this follows by analogous arguments to those used for the classical variational

problem above.

5.2 Multiple dependent variables

Suppose we are asked to find the multidimensional curve y = y(x) RN that extrem-izes the functional

J(y):=

b

aF(x,y,yx) dx,

subject to y(a) and y(b) being fixed. Note x [a, b] but here y = y(x) is a curve inN-dimensional space and is thus a vector so that (here we use the notation d/dx)

y =

y1...

yN

and yx =

y1...

yN

.

Necessarily the extremizing curvey satisfies a set of EulerLagrange equations, which

are equivalent to a system of ordinary differential equations, given fori = 1, . . . , N by:

Fyi

ddx

Fy i

= 0.

5.3 Multiple independent variables

Suppose we are asked to find the field y = y(x) that, for x Rn, extremizes thefunctional

J(y):=

F(x, y, y) dx,

subject to y being fixed at the boundary of the domain . Note herey xyis simply the gradient ofy , i.e. it is the vector of partial derivatives ofy with respect

to each of the components xi (i= 1, . . . , n) ofx:

y =

y/x1...

y/xn

.

Necessarily y satisfies an EulerLagrange equation which is a partial differential equa-

tion given byF

y yxF = 0.


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Here x is the usual divergence gradient operator with respect tox, and

yxF =

F/yx1...

F/yxn

,

where to keep the formula readable, withy the usual gradient of y, we have setyx y so thatyxi = (y)i = y/xi for i = 1, . . . , n.

Example (Laplaces equation)

The variational problem here is to find the field = (x1, x2, x3), for x= (x1, x2, x3)T

R3, that extremizes the mean-square gradient average

J():=

||2 dx.

In this case the integrand of the functionalJ is

F(x, , ) = ||2 (x1)2 + (x2)2 + (x3)2.

Note that the integrand F depends on the partial derivatives of only. Using the form

for the EulerLagrange equation above we get

x xF = 0

/x1/x2/x3

F/x1F/x2F/x3

= 0

x1

2x1

+

x2

2x2

+

x3

2x3

= 0

x1x1+ x2x2+ x3x3 = 0 2 = 0.

This is Laplaces equation for in the domain ; the solutions are called harmonicfunctions. Note that implicit in writing down the EulerLagrange partial differential

equation above, we assumed that was fixed at the boundary, i.e. Dirichlet bound-

ary conditions were specified.

Example (stretched vibrating string)

Suppose a string is tied between the two fixed points x = 0 and x = . Let y = y(x, t)

be the small displacement of the string at position x[0, ] and time t > 0 from theequilibrium positiony = 0. If is the uniform mass per unit length of the string which

is stretched to a tensionK, the kinetic and potential energy of the string are given by

T := 12

0

y2t dx, and V :=K

0

1 + (yx)

21/2 dx


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respectively, where subscripts indicate partial derivatives and the effect of gravity is

neglected. If the oscillations of the string are quite small, we can replace the expressionfor V , using the binomial expansion

1 + (yx)

21/2 1 + (yx)2, by

V = K

2

0

y2xdx.

We seek a solution y = y(x, t) that extremizes the functional (this is the action func-

tional as we see later)

J(y):=

t2t1

(T V) dt,

where t1 and t2 are two arbitrary times. In this case, with

x=

x

t

and y =

y/x

y/t

the integrand is

F(x, y, y) 12(yt)2 12K(yx)2,

i.e. F =F(y) only. The EulerLagrange equation is thus

x yxF = 0

/x

/t

F/yxF/yt

= 0

x

Kyx t yt = 0

c2 yxx ytt = 0,

where c2 = K/. This is the partial differential equation ytt = c2 yxx known as the

wave equation. It admits travelling wave solutions of speedc.

6 Lagrange multipliers

For the moment, let us temporarily put aside variational calculus and consider a prob-

lem in standard multivariable calculus.

6.1 Constrained optimization problem

Consider the following problem.

Find the stationary points of the scalar functionf(x) where x = (x1, . . . , xN)T

subject to the constraintsgk(x) = 0, wherek = 1, . . . , m, with m < N.


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Note that the graph y = f(x) of the function f represents a hyper-surface in

(N + 1)-dimensional space. The constraints are given implicitly; each one also rep-resents a hyper-surface in (N+ 1)-dimensional space. In principle we could solve the

system ofmconstraint equations, say forx1, . . . , xmin terms of the remaining variables

xm+1, . . . , xN. We could then substitute these into f, which would now be a function

of (xm+1, . . . , xN) only. (We could solve the constraints for any subset ofm variables

xi and substitute those in if we wished or this was easier, or avoided singularities, and

so forth.) We would then proceed in the usual way to find the stationary points off

by considering the partial derivative off with respect to all the remaining variables

xm+1, . . . , xN, setting those partial derivatives equal to zero, and then solving that

system of equations. However solving the constraint equations may be very difficult,

and the method ofLagrange multipliersprovides an elegant alternative (see McCallumet. al. [14, Section 14.3]).

6.2 Method of Lagrange multipliers

The idea is to convert the constrained optimization problem to an unconstrained one

as follows. Form theLagrangian function

L(x,):=f(x) +mk=1

kgk(x),

where the parameter variables = (1, . . . , m)T are known as the Lagrange multipli-

ers. Note thatL is a function of both x and , i.e. ofN+ m variables in total. Thepartial derivatives ofL with respect to all of its dependent variables are:

Lxj

= f

xj+

mk=1

kgkxj

,

Lk

=gk,

wherej = 1, . . . , N and k = 1, . . . , m. At the stationary points ofL(x,), necessarily allthese partial derivatives must be zero, and we must solve the following unconstrained

problem:

f

xj+

mk=1

kgkxj

= 0,

gk = 0,

where j = 1, . . . , N and k = 1, . . . , m. Note we have N +m equations in N + m

unknowns. Assuming that we can solve this system to find a stationary point (x,)

ofL (there could be none, one, or more) then x is also a stationary point of theoriginal constrainedproblem. Recall: what is important about the formulation of theLagrangian functionL we introduced above, is that the given constraints mean that(on the constraint manifold) we haveL =f+ 0 and therefore the stationary points off andL coincide.


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g=0

v

v

f=5

f=4

f=3

f=2

f=1

g

f

f

g

*(x , y )*

Fig. 3 At the constrained extremum f and g are parallel. (This is a rough reproductionof the figure on page 199 of McCallum et. al. [14])

6.3 Geometric intuition

Suppose we wish to extremize (find a local maximum or minimum) the objective func-

tion f(x, y) subject to the constraint g(x, y) = 0. We can think of this as follows. The

graph z = f(x, y) represents a surface in three dimensional (x ,y,z ) space, while the

constraint represents a curve in the x-y plane to which our movements are restricted.

Constrained extrema occur at points where the contours off are tangent to the

contours ofg(and can also occur at the endpoints of the constraint). This can be seen as

follows. At any point (x, y) in the plane fpoints in the direction of maximum increaseof f and thus perpendicular to the level contours of f. Suppose that the vector v is

tangent to the constraining curveg(x, y) = 0. If the directional derivativefv = f vis positive at some point, then moving in the direction ofv means thatfincreases (if

the directional derivative is negative then fdecreases in that direction). Thus at the

point (x, y) where fhas a constrained extremum we must havef v = 0 and sobothf andg are perpendicular to v and therefore parallel. Hence for some scalarparameter (the Lagrange multiplier) we have at the constrained extremum:

f = g and g = 0.Notice that here we have three equations in three unknowns x,y, .

7 Constrained variational problems

A common optimization problem is to extremize a functional Jwith respect to paths

y which are constrained in some way.

7.1 Constrained variational problem

We consider the following formulation.


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Find the extrema of the functional

J(y):=

ba

F(x,y,yx) dx,

where y = (y1, . . . , yN)T RN subject to the set of m constraints, for k =

1, . . . , m < N :

Gk(x,y) = 0.

To solve this constrained variational problem we generalize the method of Lagrange

multipliers as follows. Note that them constraint equations above imply

b

a

k(x) Gk(x,y) dx= 0,

for each k = 1, . . . , m. Note that thek s are the Lagrange multipliers, which with the

constraints expressed in this integral form, can in general be functions ofx. We now

form the equivalent of the Lagrangian function which here is the functional

J(y,):=

ba

F(x,y,yx) dx +

mk=1

ba

k(x) Gk(x,y) dx

J(y,) = ba

F(x,y,yx) +

mk=1

k(x) Gk(x,y)

dx.

The integrand of this functional is

F(x,y,yx,):=F(x,y,yx) +

mk=1

k(x) Gk(x,y),

where = (1, . . . , m)T. We know from the classical variational problem that if (y,)

extremize Jthen necessarily they must satisfy the EulerLagrange equations:

F

yi d

dx

F

y i

= 0,

F

k d

dx

F

k

= 0,

which simplify to

F

yi d

dx

F

y i

= 0,

Gk(x,y) = 0,

for i = 1, . . . , N and k = 1, . . . , m. This is a system ofdifferential-algebraic equations:

the first set of relations are ordinary differential equations, while the constraints are

algebraic relations.


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(x,y)=(x(t),y(t))c:

Fig. 4 For the isoperimetrical problem, the closed curve C has a fixed length , and the goalis to choose the shape that maximizes the area it encloses.

7.2 Integral constraints

If the constraints are (already) in integral form so that we have ba

Gk(x,y) dx= 0,

for k = 1, . . . , m, then set

J(y):=J(y) +

mk=1

k

ba

Gk(x,y) dx

=

ba

F(x,y,yx) +

mk=1

kGk(x,y)

dx.

Note we now have a variational problem with respect to the curve y = y(x) but the

constraints are classical in the sense that the Lagrangian multipliers here are simplyvariables. Proceeding as above, with this hybrid variational constraint problem, we

generate the EulerLagrange equations as above, together with the integral constraint

equations.

7.3 Example (Didos/isoperimetrical problem)

The goal of this classical constrained variational problem is to find the shape of the

closed curve, of a given fixed length , that encloses the maximum possible area.

Suppose that the curve is given in parametric coordinates

x(), y()

where the pa-

rameter [0, 2]. By Stokes theorem, the area enclosed by a closed contourC is

12C x dy y dx.

Hence our goal is to extremize the functional

J(x, y):= 12

20

(xy yx) d,

subject to the constraint 20

x2 + y2 d =.


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Note that the constraint can be expressed in standard integral form as follows. Since

is fixed we have 20

12 d=.

Hence the constraint can be expressed in the form 20

x2 + y2 12 d= 0.

To solve this variational constraint problem we use the method of Lagrange multipliers

and form the functional

J(x, y):=J(x, y) +

20

x2 + y2 12 d

= 20

F(x,y, x, y, ) d .

where the integrand F is given by

F(x,y, x, y, ) = 12 (xy yx) + (x2 + y2)1

2 12 .

Note there are two dependent variables x and y (here the parameter is the inde-

pendent variable). By the theory above we know that the extremizing solution (x, y)

necessarily satisfies an EulerLagrange system of equations, which are the pair of or-

dinary differential equations

F

x d

d

F

x

= 0,

Fy

dd

Fy

= 0,

together with the integral constraint condition. Substituting the form for Fabove, the

pair of ordinary differential equations are

12y

d

d

12y+

x

(x2 + y2)1

2

= 0,

12x d

d

12x +

y

(x2 + y2)1

2

= 0.

Integrating both these equations with respect towe get

y x

(x2 + y2)12 =c2 and x

y

(x2 + y2)12 =c1,

for arbitrary constants c1 and c2. Combining these last two equations reveals

(x c1)2 + (y c2)2 = 2y2

x2 + y2 +

2x2

x2 + y2 =2.

Hence the solution curve is given by

(x c1)2 + (y c2)2 =2,


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which is the equation for a circle with radius and centre (c1, c2). The constraint

condition implies = /2 and c1 and c2 can be determined from the initial or endpoints of the closed contour/path.

The isoperimetrical problem has quite a history. It was formulated in Virgils poem

the Aeneid, one account of the beginnings of Rome; see Wikipedia [18] or Mont-

gomery [16]. Quoting from Wikipedia (Dido was also known as Elissa):

Eventually Elissa and her followers arrived on the coast of North Africa where

Elissa asked the local inhabitants for a small bit of land for a temporary refuge

until she could continue her journeying, only as much land as could be en-

compassed by an oxhide. They agreed. Elissa cut the oxhide into fine strips

so that she had enough to encircle an entire nearby hill, which was therefore

afterwards named Byrsa hide. (This event is commemorated in modern math-

ematics: The isoperimetric problem of enclosing the maximum area within

a fixed boundary is often called the Dido Problem in modern calculus ofvariations.)

Dido found the solutionin her case a half-circleand the semi-circular city of

Carthage was founded.

Example (Helmholtzs equation)

This is a constrained variational version of the problem that generated the Laplace

equation. The goal is to find the field = (x) that extremizes the functional

J():=

||2 dx,

subject to the constraint

2 dx= 1.

This constraint corresponds to saying that the total energy is bounded and in fact

renormalized to unity. We assume zero boundary conditions, (x) = 0 for x Rn. Using the method of Lagrange multipliers (for integral constraints) we form the

functional

J():=

||2 dx +

2 dx 1

.

We can re-write this in the form

J():= |

|2 + 2

1

|| dx,where|| is the volume of the domain . Hence the integrand in this case is

F(x, , , ):= ||2 +

2 1||

.

The extremzing solution satisfies the EulerLagrange partial differential equation

F

xF = 0,


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together with the constraint equation. Note, directly computing, we have

xF = 2 and F

= 2.

Substituting these two results into the EulerLagrange partial differential equation we

find

2 (2) = 0 2 = .

This is Helmholtzs equation on. The Lagrange multiplier also represents an eigen-

value parameter.

8 Optimal linear-quadratic control

We can use calculus of variations techniques to derive the solution to an important

problem in control theory. Suppose that a system state at any time t 0 is recorded

in the vector q = q(t). Suppose further that the state evolves according to a linear

system of differential equations and we can control the system via a set of inputs orcontrolsu= u(t), i.e. the system evolution is

dq

dt =Aq + Bu.

Here A is a matrix, which for convenience we will assume is constant, and B = B(t) is

another matrix mediating the controls.

Goal: The goal is, starting in the state q(0) = q0, to bring this initial state to a final

state q(T) at time t = T >0, as expediently as possible.

There are many criteria as to what constitutes expediency. Here we will measure

the cost on the system of our actions over [0, T] by the quadratic utility

J(u):=

T0

qT(t)C(t)q(t) + uT(t)D(t)u(t) dt + qT(T)Eq(T).

Here C = C(t), D = D(t) and Eare non-negative definite symmetric matrices. The

final term represents a final state achievement cost. Note that we can in principle solve

the system of linear differential equations above for the stateq = q(t) in terms of the

controlu= u(t) so that J=J(u) is a functional ofu only (we see this presently).

Thus our goal is to find the control u = u(t) that minimizes the cost J =J(u)

whilst respecting the constraint which is the linear evolution of the system state. We

proceed as before. Suppose u = u(t) exists. Consider perturbations to u on [0, T]of the form

u= u+ u.

Changing the control/input changes the stateq= q(t) of the system so that

q= q+ q.

where we suppose here q = q(t) to be the system evolution corresponding to the

optimal control u = u(t). Note that linear system perturbations q are linear in .


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Substituting these last two perturbation expressions into the differential system for the

state evolution we find dq

dt =Aq + Bu,

where we have used that (d/dt)q = Aq+ Bu. The initial condition is q(0) = 0.

We can solve this system of differential equations forq in terms ofu by the variation

of constants formula (using an integrating factor) as follows. Since the matrix A is

constant we observe

d

dt

exp(At)q(t) = exp(At)B(t)u(t)

exp(At)q(t) = t0

exp(As)B(s)u(s) ds

q(t) =

t

0

expA(t s)B(s)u(s) ds.By the calculus of variations, if we set

():=J(u+ u),

then if the functional Jhas a minimum we have, for all u,

(0) = 0.

Note that we can substitute our expression forq(t) in terms ofuintoJ(u+u). Since

u= u+ uand q = q+ q are linear in and the functional J= J(u) is quadratic

in u and q , then J(u+ u) must be quadratic in so that

() = 0+ 1+ 22,

for some functionals0 = 0(u), 1 = 1(u) and 2 = 2(u) independent of. Since

() = 1 + 22, we see (0) = 1. This term in () = J(u + u) by direct

computation is thus

(0) = 2

T0

qT(t)C(t)q(t) +u

T(t)D(t)u(t) dt + 2qT(T)Eq(T),

where we used thatC, D and Eare symmetric. We now substitute our expression for

q in terms ofu above into this formula for(0), this gives

12

(0) =

T0

qT(t)C(t)q(t) dt +

T0

uT(s)D(s)u(s) ds + q

T(T)Eq(T)

= T0

t0

exp

A(t s)B(s)u(s)TC(t)q(t) +uT(s)D(s)u(s) ds dt+

T0

exp

A(T s)B(s)u(s)T ds Eq(T)=

T0

Ts

exp

A(t s)B(s)u(s)TC(t)q(t) +uT(s)D(s)u(s) dt ds+

T0

exp

A(T s)B(s)u(s)T ds Eq(T),


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where we have swapped the order of integration in the first term. Now we use the

transpose of the product of two matrices is the product of their transposes in reverseorder to get

12

(0) =

T0

uT(s)BT(s)

Ts

exp

AT(t s)C(t)q(t) dt +uT(s)D(s)u(s)+u(s)TBT(s)exp

AT(T s)Eq(T)

ds

=

T0

uT(s)BT(s)p(s) +uT(s)D(s)u(s) ds,

where for all s [0, T] we set

p(s):= T

s

expAT(t s)C(t)q(t) dt + expA

T(T s)Eq(T).We see the condition for a minimum, (0) = 0, is equivalent to T

0uT(s)

BT(s)p(s) + D(s)u(s)

ds= 0,

for all u. Hence a necessary condition for the minimum is that for all t [0, T] we haveu(t) = D1(t)BT(t)p(t).

Note that p depends solely on the optimal state q. To elucidate this relationship

further, note by definition p(T) = Eq(T) and differentiating p with respect to t we

finddp

dt = Cq ATp.

Using the expression for the optimal control u in terms of p(t) we derived, we seethat q and p satisfy the system of differential equations

d

dt

qp

=

A BD1(t)BT

C(t) AT

qp

.

Define S= S(t) to be the map S: q p, i.e. so that p(t) =S(t)q(t). Thenu(t) = D1(t)BT(t)S(t)q(t),

and S(t) we see characterizes the optimal current state feedback control, it tells how

to choose the current optimal controlu(t) in terms of the current state q(t). Finally

we observe that sincep(t) =S(t)q(t), we have

dp

dt =

dS

dtq+ S

dqdt

.

Thus we see that

dS

dtq =

dp

dt Sdq

dt

= (Cq ATSq) S(Aq BD1ATSq).Hence S= S(t) satisfies theRiccati equation

dS

dt = C ATS SA SBD1ATS.


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Remark 1 We can easily generalize the argument above to the case when the coefficient

matrix A = A(t) is not constant. This is achieved by carefully replacing the flow mapexp

A(t s) for the linear constant coefficient system (d/dt)q= Aq, by the flow mapfor the corresponding linear nonautonomous system withA = A(t), and carrying that

through the rest of the computation.

9 Lagrangian dynamics

9.1 Newtons equations

Consider a system ofN particles in three dimensional space, each with position vector

ri(t) for i = 1, . . . , N . Note that each ri(t) R3 is a 3-vector. We thus need 3Ncoordinates to specify the system, this is the configuration space. Newtons 2nd law

tells us that the equation of motion for the ith particle is

pi = Fexti + F

coni ,

for i = 1, . . . , N . Here pi = mivi is the linear momentum of the ith particle and

vi = ri is its velocity. We decompose the total force on theith particle into an external

force Fexti and a constraint force Fconi . By external forces we imagine forces due to

gravitational attraction or an electro-magnetic field, and so forth.

9.2 Holonomic constraints

By a constraint on a particles we imagine that the particles motion is limited in some

rigid way. For example the particle/bead may be constrained to move along a wire or

its motion is constrained to a given surface. If the system ofNparticles constitute a

rigid body, then the distances between all the particles are rigidly fixed and we have

the constraint

|ri(t) rj(t)| =cij ,

for some constants cij , for all i, j = 1, . . . , N . All of these are examples ofholonomicconstraints.

Definition 1 (Holonomic constraints) For a system of particles with positions given

byri(t) for i = 1, . . . , N , constraints that can be expressed in the form

g(r1, . . . , rN, t) = 0,

are said to be holonomic. Note they only involve the configuration coordinates.

We willonlyconsider systems for which the constraints are holonomic. Systems withconstraints that are non-holonomic are: gas molecules in a container (the constraint

is only expressible as an inequality); or a sphere rolling on a rough surface without

slipping (the constraint condition is one of matched velocities).


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9.3 Degrees of freedom

Let us suppose that for the Nparticles there are m holonomic constraints given by

gk(r1, . . . , rN, t) = 0,

for k = 1, . . . , m. The positions ri(t) of all Nparticles are determined by 3N coordi-

nates. However due to the constraints, the positions ri(t) are not all independent. In

principle, we can use themholonomic constraints to eliminatem of the 3N coordinates

and we would be left with 3N m independent coordinates, i.e. the dimension of theconfiguration space is actually 3N m.

Definition 2 (Degrees of freedom) The dimension of the configuration space is called

the number of degrees of freedom, see Arnold [3, Page 76].

Thus we can transform from the old coordinates r1, . . . , rN to new generalized

coordinates q1, . . . , q n where n = 3N m:

r1 = r1(q1, . . . , q n, t),

...

rN = rN(q1, . . . , q n, t).

9.4 DAlemberts principle

We will restrict ourselves to systems for which the net work of the constraint forces is

zero, i.e. we suppose

Ni=1

Fconi dri = 0,

for every small change dri of the configuration of the system (fortfixed). If we combine

this with Newtons 2nd law we see that we get

Ni=1

( pi Fexti ) dri = 0.

This isDAlemberts principle. In particular, no forces of constraint are present.

The assumption that the constraint force does no net work is quite general. It is

true in particular for holonomic constraints. For example, for the case of a rigid body,

the internal forces of constraint do no work as the distances |rirj | between particlesis fixed, then d(ri rj) is perpendicular to ri rj and hence perpendicular to theforce between them which is parallel to ri rj . Similarly for the case of the bead ona wire or particle constrained to move on a surfacethe normal reaction forces are

perpendicular to dri.


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9.5 Lagranges equations (from DAlemberts principle)

In his Mecanique Analytique [1788], Lagrange sought a coordinate-invariant expres-

sion for mass times acceleration, see Marsden and Ratiu [15, Page 231]. This lead to

Lagranges equations of motion. Consider the transformation to generalized coordinates

ri = ri(q1, . . . , q n, t),

for i = 1, . . . , N . If we consider a small increment in the displacements dri then the

corresponding increment in the work done by the external forces is

Ni=1

Fexti dri =

N,ni,j=1

Fexti riqj dqj =

nj=1

Qjdqj .

Here we have set for j = 1, . . . , n,

Qj =Ni=1

Fexti riqj ,

and think of these as generalized forces.

Further, the change in kinetic energy mediated through the momentum (the first

term in DAlemberts principle), due to the increment in the displacements dri, is given

byNi=1

pi dri =Ni=1

mivi dri =N,ni,j=1

mivi riqj dqj .

From the product rule we know that

d

dtvi

ri

qj vi

ri

qj+ vi

d

dtri

qj

vi riqj + vi viqj

.

Also, by differentiating the transformation to generalized coordinates (keeping t fixed),

we see that

vin

j=1

riqj

qj viqj riqj

.

Using these last two identities we see that

Ni=1

pi dri =n

j=1

Ni=1

mivi riqj

dqj

=nj=1

Ni=1

ddt

mivi ri

qj

mivi vi

qj

dqj

=

nj=1

Ni=1

d

dt

mivi vi

qj

mivi vi

qj

dqj

=

nj=1

d

dt

qj

Ni=1

12mi|vi|2

qj

Ni=1

12mi|vi|2

dqj .


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Hence if the kinetic energy is defined to be

T :=

Ni=1

12mi|vi|2,

then we see that DAlemberts principle is equivalent to

nj=1

d

dt

T

qj

T

qj Qj

dqj = 0.

Since the qj for j = 1, . . . , n, where n = 3N m, are all independent, we have

d

dt

T

qj

T

qj Qj = 0,

for j = 1, . . . , n. If we now assume that the work done depends on the initial and final

configurations only and not on the path between them, then there exists a potential

function V(q1, . . . , q n) such that

Qj = Vqj

for j = 1, . . . , n (such forces are said to be conservative). If we define the Lagrange

function or Lagrangian to be

L= T V,then we see that DAlemberts principle is equivalent to

d

dt

L

qj

L

qj= 0,

for j = 1, . . . , n. These are known as Lagranges equations. As already noted the n-

dimensional subsurface of 3N-dimensional space, on which the solutions to Lagranges

equations lie (wheren = 3N m), is called the configuration space. It is parameterizedby the n generalized coordinates q1, . . . , q n.

Remark 2 (Non-conservative forces) If the system has non-conservative forces it may

still be possible to find a generalized potential function V such that

Qj = Vqj

+ d

dt

V

qj

,

for j = 1, . . . , n. From such potentials we can still deduce Lagranges equations of

motion. Examples of such generalized potentials are velocity dependent potentials due

to electro-magnetic fields, for example the Lorentz force on a charged particle.


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10 Hamiltons principle

10.1 Action

We consider mechanical systems with holonomic constraints and with all other forces

conservative. Recall, we define theLagrange function or Lagrangian to be

L= T V,where

T =

Ni=1

12mi|vi|2

is the total kinetic energy for the system, andVis its potential energy.

Definition 3 (Action)If the LagrangianL is the difference of the kinetic and potential

energies for a system, i.e.L = T V, we define the action A from time t1 to t2 to bethe functional

A(q):=

t2t1

L(q, q, t) dt,

where q= (q1, . . . , q n)T.

10.2 Hamiltons principle of least action

Hamilton [1834] realized that Lagranges equations of motion were equivalent to a

variational principle (see Marsden and Ratiu [15, Page 231]).

Theorem 2 (Hamiltons principle of least action) The correct path of motion of a

mechanical system with holonomic constraints and conservative external forces, from

time t1 to t2, is a stationary solution of the action. Indeed, the correct path of mo-tion q = q(t), with q = (q1, . . . , q n)

T, necessarily and sufficiently satisfies Lagranges

equations of motiond

dt

L

qj

L

qj= 0,

forj = 1, . . . , n.

Quoting from Arnold [3, Page 60], it is Hamiltons form of the principle of least

action because in many cases the action ofq= q(t) is not only an extremal but also

a minimum value of the action functional.

10.3 Example (Simple harmonic motion)

Consider a particle of massm moving in a one dimensional Hookeian force fieldkx,where k is a constant. The potential function V = V(x) corresponding to this force

field satisfies

Vx

= kx

V(x) V(0) = x0

k d

V(x) = 12x2.


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The Lagrangian L = T V is thus given by

L(x, x) = 12mx2 12kx2.

From Hamiltons principle the equations of motion are given by Lagranges equations.

Here, taking the generalized coordinate to be q= x, the single Lagrange equation is

d

dt

L

x

L

x = 0.

Substituting the form for the Lagrangian above this Lagrange equation becomes

mx + kx = 0.

10.4 Example (Kepler problem)

Consider a particle of mass m moving in an inverse square law force field,m/r2,such as a small planet or asteroid in the gravitational field of a star or larger planet.

Hence the corresponding potential function satisfies

Vr

= mr2

V() V(r) = r

m

2 d

V(r) =m

r .

The Lagrangian L = T V is thus given by

L(r,r,, ) = 12m(r2 + r22) +

m

r .

From Hamiltons principle the equations of motion are given by Lagranges equations,

which here, taking the generalized coordinates to beq1 = r and q2= , are the pair of

ordinary differential equations

d

dt

L

r

L

r = 0,

ddt

L

L

= 0,

which on substituting the form for the Lagrangian above, become

mr mr2 + mr2

= 0,

d

dt

mr2

= 0.


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10.5 Non-uniqueness of the Lagrangian

Two LagrangiansL1 and L2 that differ by the total time derivative of any function of

q= (q1, . . . , q n)T and t generate the same equations of motion. In fact if

L2(q, q, t) = L1(q,q, t) + d

dt

f(q, t)

,

then for j = 1, . . . , n direct calculation reveals that

d

dt

L2qj

L2

qj=

d

dt

L1qj

L1

qj.

11 Constraints

Given a Lagrangian L(q, q, t) for a system, suppose we realize the system has some

constraints (so theqj are not all independent).

11.1 Holonomic constraints

Suppose we havem holonomic constraints of the form

Gk(q1, . . . , q n, t) = 0,

for k = 1, . . . , m < n. We can now use the method of Lagrange multipliers with

Hamiltons principle to deduce that the equations of motion are given by

ddt

Lqj

L

qj=

mk=1

k(t) Gk

qj,

Gk(q1, . . . , q n, t) = 0,

for j = 1, . . . , n and k = 1, . . . , m. We call the quantities on the right above

mk=1

k(t)Gkqj

,

the generalized forces of constraint.

11.2 Example (simple pendulum)

Consider the motion of a simple pendulum bob of mass m that swings at the end of

a light rod of length a. The other end is attached so that the rod and bob can swing

freely in a plane. Ifg is the acceleration due to gravity, then the Lagrangian L = TVis given by

L(r,r,, ) = 12m(r2 + r22) + mgr cos ,

together with the constraint

r a= 0.


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r

Fig. 5 The mechanical problem for the simple pendulum, can be thought of as a particle ofmass m moving in a vertical plane, that is constrained to always be a distance a from a fixedpoint. In polar coordinates, the position of the mass is (r, ) and the constraint is r= a.

We could just substituter = ainto the Lagrangian, obtaining a system with one degree

of freedom, and proceed from there. However, we will consider the system as one with

two degrees of freedom,q1 = r and q2 = , together with a constraintG(r) = 0, where

G(r) =r a. Hamiltons principle and the method of Lagrange multipliers imply thatthe system evolves according to the pair of ordinary differential equations together

with the algebraic constraint given by

d

dt

L

r

L

r =

G

r,

d

dt

L

L

=

G

,

G= 0.

Substituting the form for the Lagrangian above, the two ordinary differential equations

together with the algebraic constraint become

mr mr2 mg cos = ,d

dt

mr2

+ mgr sin = 0,

r a= 0.

Note that the constraint is of courser = a, which implies r = 0. Using this, the system

of differential algebraic equations thus reduces to

ma2+ mga sin = 0,

which comes from the second equation above. The first equation tells us that the

Lagrange multiplier is given by

(t) = ma2 mg cos .

The Lagrange multiplier has a physical interpretation, it is the normal reaction force,

which here is the tension in the rod.


30/56


Remark 3 (Non-holonomic constraints)Mechanical systems with certain types of non-

holonomic constraints can also be treated, in particular constraints of the formn

j=1

A(q, t)kjqj + bk(q, t) = 0,

for k = 1, . . . , m, where q = (q1, . . . , q n)T. Note the assumption is that these equa-

tions are not integrable, in particular not exact, otherwise the constraints would be

holonomic.

12 Hamiltonian dynamics

12.1 Hamiltonian

We consider mechanical systems that are holonomic and and conservative (or for which

the applied forces have a generalized potential). For such a system we can construct

a Lagrangian L(q, q, t), where q = (q1, . . . , q n)T, which is the difference of the total

kinetic T and potential Venergies. These mechanical systems evolve according to the

n Lagrange equationsd

dt

L

qj

L

qj= 0,

forj = 1, . . . , n. These are each second order ordinary differential equations and so the

system is determined for all time once 2n initial conditionsq(t0),q(t0)

are specified

(or n conditions at two different times). The state of the system is represented by a

point q= (q1, . . . , q n)T in configuration space.

Definition 4 (Generalized momenta) We define the generalized momenta for a La-grangian mechanical system to be

pj = L

qj,

for j = 1, . . . , n. Note that we have pj =pj(q,q, t) in general, where q = (q1, . . . , q n)T

and q= (q1, . . . , qn)T.

In terms of the generalized momenta, Lagranges equations become, forj = 1, . . . , n:

pj = L

qj.

Further, in principle, we can solve the relations above which define the generalized

momenta, to find functional expressions for the qj in terms ofqi,piandt. In other wordswe can solve the relations defining the generalized momenta to find qj = qj(q,p, t)

where q= (q1, . . . , q n)T and p= (p1, . . . , pn)

T.

Definition 5 (Hamiltonian)We define theHamiltonian functionas theLegendre trans-

formof the Lagrangian function, i.e. the Hamiltonian is defined by

H(q,p, t):= q p L(q,q, t),

where q= (q1, . . . , q n)T and p= (p1, . . . , pn)

T and we suppose q= q(q,p, t).


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Note that in this definition we used the notation for the dot product

q p=n

j=1

qjpj .

The Legendre transform is nicely explained in Arnold [3, Page 61] and Evans [7,

Page 121]. In practice, as far as solving example problems herein, it simply involves the

task of solving the equations for the generalized momenta above, to find qj in terms of

qi, pi and t.

12.2 Hamiltons equations of motion

From Lagranges equation of motion, we can, using the definitions for the generalized

momenta

pj = L

qj,

and Hamiltonian

H=

nj=1

qjpj L,

above, deduce Hamiltons equations of motion.

Theorem 3 (Hamiltons equations of motion) With the Hamiltonian defined as the

Legendre transform of the Lagrangian, Lagranges equations of motion imply

qi = H

pi,

pi = Hqi ,

fori = 1, . . . , n. These are Hamiltons canonical equations, consisting of2n first orderequations of motion.

ProofUsing the chain rule and the definition for the generalized momenta we have

H

pi=

nj=1

qjpi

pj+ qi n

j=1

L

qj

qjpi

qi,

and

Hqi=

nj=1

qjqipj Lqi

nj=1

Lqjqjqi

Lqi .

Now using Lagranges equations, in the form pj =L/qj , the last relation reveals

pi = Hqi

,

for i = 1, . . . , n. Collecting these relations together, we see that Lagranges equations

of motion imply Hamiltons canonical equations as shown.


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Two further observations are also useful. First, if the Lagrangian L = L(q, q) is

independent of explicitt, then when we solve the equations that define the generalizedmomenta we find q= q(q,p). Hence we see that

H= q(q,p) p Lq, q(q,p),i.e. the Hamiltonian H= H(q,p) is also independent oft explicitly. Second, using the

chain rule and Hamiltons equations we see that

dH

dt =

ni=1

H

qiqi+

ni=1

H

pipi+

H

t

dHdt

= H

t .

Hence ifHdoes notexplicitly depend on t then

H is a

constant of the motion,

conserved quantity,

integral of the motion.

Hence the absence of explicit t dependence in the Hamiltonian H could serve as a

more general definition of a conservative system, though in general H may not be

the total energy. However for simple mechanical systems for which the kinetic energyT = T(q,q) is a homogeneous quadratic function in q, and the potential V = V(q),

then the Hamiltonian H will bethe total energy. To see this, suppose

T =

n

i,j=1cij(q) qiqj ,

i.e. a homogeneous quadratic function in q; necessarilycij =cji . Then we have

T

qk=

nj=1

ckj(q) qj+

ni=1

cik(q) qi = 2

ni=1

cik(q) qi

which impliesn

k=1

qkT

qk= 2T.

Thus the HamiltonianH= 2T (T V) = T+ V, i.e. the total energy.

13 Hamiltonian formulation

13.1 Summary

To construct Hamiltons canonical equations for a mechanical system proceed as fol-

lows:

1. Choose your generalized coordinates q= (q1, . . . , q n)T and construct

L(q, q, t) = T V.


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2. Define and compute the generalized momenta

pi = L

qi,

for i = 1, . . . , n. Solve these relations to find qi = qi(q,p, t).

3. Construct and compute the Hamiltonian function

H=

nj=1

qjpj L,

4. Write down Hamiltons equations of motion

qi = H

pi,

pi = H

qi ,

for i = 1, . . . , n, and evaluate the partial derivatives of the Hamiltonian on the

right.

13.2 Example (simple harmonic oscillator)

The Lagrangian for the simple harmonic oscillator, which consists of a mass m moving

in a quadratic potential field with characteristic coefficient k , is

L(x, x) = 12mx2 12kx2.

The corresponding generalized momentum is

p= L

x =mx

which is the usual momentum. This implies x= p/m and so the Hamiltonian is given

by

H(x, p) = x p L(x, x)

= p

mp

12mx

2 12kx2

= p2

m

12m

p

m

2 12kx2

=

1

2

p2

m +

1

2kx

2

.

Note that is last expression is just the sum of the kinetic and potential energies and so

His the total energy. Hamiltons equations of motion are thus given by

x= H/p,

p= H/x, x= p/m,

p= kx.Note that combining these two equations, we get the usual equation for a harmonic

oscillator: mx= kx.


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Fig. 6 The mechanical problem for the simple harmonic oscillator consists of a particle movingin a quadratic potential field. As shown, we can think of a ball of mass m sliding freely back andforth in a vertical plane, without energy loss, inside a parabolic shaped bowl. The horizontalposition x(t) is its displacement.


Recall the Kepler problem for a mass m moving in an inverse-square central force field

with characteristic coefficient. The Lagrangian L = T V isL(r,r,, ) = 12m(r

2 + r22) +m

r .

Hence the generalized momenta are

pr = L

r =mr and p =

L

=mr2.

These imply r= pr/m and = p/mr2 and so the Hamiltonian is given by

H(r,,pr , p) = r pr+ p L(r,r,, )

= 1

m

p2r+

p2r2

12m

p2rm2

+ r2 p2m2r4

+

m

r

= 1

2m

p2r+

p2r2

m

r ,

which in this case is also the total energy. Hamiltons equations of motion are

r= H/pr ,

= H/p,

pr = H/r,

p = H/,

r= pr/m,

= p/mr2,

pr =p2/mr

3 m/r2,

p = 0.Note that p = 0, i.e. we have that p is constant for the motion. This property

corresponds to theconservation of angular momentum.

Remark 4 The LagrangianL= TVmay change its functional form if we use differentvariables (Q, Q) instead of (q,q), but its magnitude will not change. However, the

functional form and magnitude of the Hamiltonian both depend on the generalized

coordinates chosen. In particular, the HamiltonianHmay be conserved for one set of

coordinates, but not for another.


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m

0QUt

q

Fig. 7 Mass-spring system on a massless cart.

13.4 Example (harmonic oscillator on a moving platform)

Consider a mass-spring system, massmand spring stiffnessk, contained within a mass-

less cart which is translating horizontally with a fixed velocity Usee Figure 7. The

constant velocity Uof the cart is maintained by an external agency. The Lagrangian

L= T V for this system is

L(q, q, t) = 12mq2 12k(q U t)2.

The resulting equation of motion of the mass is

mq= k(q Ut).

If we setQ = q

U t, then the equation of motion is

m Q= kQ.

Let us now consider the Hamiltonian formulation using two different sets of coordinates.

First, using the generalized coordinateq, the corresponding generalized momentum

is p = mqand the Hamiltonian is

H(q ,p ,t) = p2

2m+

k

2(q U t)2.

Here the HamiltonianHisthe total energy, but it isnotconserved (there is an external

energy input maintaining U constant).

Second, using the generalized coordinateQ, the Lagrangian L= T V is

L(Q, Q, t) = 12mQ2 + m QU+ 12mU

2 12kQ2.

Here, the generalized momentum is P =m Q + mUand the Hamiltonian is

H(Q, P) = (P mU)2

2m +

k

2Q2 m

2U2.

Note that Hdoes not explicitly depend on t . Hence H isconserved, but it is not the

total energy.


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14 Symmetries and conservation laws

14.1 Cyclic coordinates

We have already seen that

dH

dt =

H

t .

Hence if the Hamiltonian does not depend explicitly on t, then it is aconstantor integral

of the motion; sometimes called Jacobis integral. Itmaybe the total energy. Further

from the definition of the generalized momenta pi = L/qi, Lagranges equations,

and Hamiltons equations for the generalized momenta, we have

pi =

H

qi

= L

qi

.

From these relations we can see that ifqi is explicitly absent from the Lagrangian L,

then it is explicitly absent from the Hamiltonian H, and

pi = 0.

Hence pi is a conserved quantity, i.e. constant of the motion. Such aqi is calledcyclic

or ignorable. Note that for such coordinates qi, the transformation

t t + t,qi qi+ qi,

leave the Lagrangian/Hamiltonian unchanged. This invariance signifies a fundamental

symmetryof the system.


Recall, the LagrangianL = T V for the Kepler problem is

L(r,r,, ) = 12m(r2 + r22) +

m

r ,

and the Hamiltonian is

H(r,,pr , p) = 1

2m

p2r+

p2r2

m

r .

Note that L, Hare independent oft and therefore H is conserved (and here it is the

total energy). Further,L,Hare independent of and thereforep = mr2is conserved

also. The original problem has two degrees of freedom (r, ). However, we have just

established two integrals of the motion, namelyH and p. This system is said to be

integrable.


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mg

a

Fig. 8 Simple axisymmetric top: this consists of a body of massm that spins around its axis of

symmetry moving under gravity. It has a fixed point on its axis of symmetryhere this is thepivot at the narrow pointed end that touches the ground. The centre of mass is a distance afrom the fixed point. The configuration of the top is given in terms of the Euler angles ( ,,)shown. Typically the top also precesses around the vertical axis = 0.

14.3 Example (axisymmetric top)

Consider a simple axisymmetric top of massm with a fixed point on its axis of sym-

metry; see Figure 8. Suppose the centre of mass is a distance a from the fixed point,

and the principle moments of inertia are A = B=C. We assume there are no torquesabout the symmetry or vertical axes. The configuration of the top is given in terms of

the Euler angles (,,) as shown in Figure 8. The LagrangianL = T V is

L= 12A(2 + 2 sin2 ) + 12C(+ cos )2 mga cos .The generalized momenta are

p = A,

p = A sin2 + Ccos (+ cos ),

p = C(+ cos ).

Using these the Hamiltonian is given by

H= p22A

+(p pcos )2

2A sin2 +

p22C

+ mga cos .

We see that L, H are both independent of t, and . Hence H, p and p areconserved. Respectively, they represent the total energy, the angular momentum about

the symmetry axis and the angular momentum about the vertical.

Remark 5 (Noethers Theorem) To accept only those symmetries which leave the

Lagrangian unchanged is needlessly restrictive. When searching for conservation laws

(integrals of the motion), we can in general consider transformations that leave the

action integralinvariant enough so that we get the same equations of motion. This is

the idea underlying (Emmy) Noethers theorem; see Arnold [3, Page 88].


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15 Poisson brackets

15.1 Definition and properties

Definition 6 (Poisson brackets) For two functions u = u(q,p) and v = v(q,p), where

q= (q1, . . . , q n)T and p= (p1, . . . , pn)

T, we define their Poisson bracket to be

[u, v]:=

ni=1

u

qi

v

pi u

pi

v

qi

.

The Poisson bracket satisfies some simple properties that can be checked directly. For

example, for any function u = u(q,p) and constant c we have [u, c] = 0. We also

observe that if v = v(q,p) and w = w(q,p) then [uv,w] = u[v, w] +v[u, w]. Three

crucial properties are summarized in the following lemma.

Lemma 2 (Lie algebra properties)The bracket satisfies the following properties for al lfunctionsu = u(q,p), v = v(q,p) andw= w(q,p) and scalars and:

1. Skew-symmetry: [v, u] = [u, v];2. Bilinearity:[u + v,w] =[u, w] + [v, w];

3. Jacobis identity:[u, [v, w]] + [v, [w, u]] + [w, [u, v]] = 0.

These three properties define a non-associative algebra known as aLie algebra.

Two further simple examples of Lie algebras are: the vector space of vectors equipped

with the wedge or vector product [u,v] = u v and the vector space of matricesequipped with the matrix commutator product [A, B] =AB BA.Corollary 1 (Properties from the definition)Using that all the coordinates(q1, . . . , q n)and(p1, . . . , pn)are independent, we immediately deduce by direct substitution into the

definition, the following results for anyu = u(q,p) and alli, j = 1, . . . , n:

u

qi= [u, pi],

u

pi= [u, qi], [qi, qj ] = 0, [pi, pj ] = 0 and [qi, pj ] =ij .

Hereij is the Kronecker delta which is zero wheni =j and unity wheni = j .Corollary 2 (Poisson bracket for canonical variables) If q = (q1, . . . , q n)

T and p =

(p1, . . . , pn)T are canonical Hamilton variables, i.e. they satisfy Hamiltons equations

for some HamiltonianH, then for any functionf=f(q,p) we have

df

dt =

f

t + [f, H].

ProofUsing Hamiltons equations of motion (in shorter vector notation) we know

dq

dt = pH and dp

dt = qH.

Thus by the chain rule and then Hamiltons equations we find

df

dt =

dq

dt qH+ dp

dt pH= pH qH+ (qH) pH

which is zero by the commutative property of the scalar (dot) product. All the properties above are essential for the next two main results of this section.


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15.2 Invariance under canonical transformations

Remarkably the the Poisson bracket is invariant under canonical transformations. By

this we mean the following. Suppose we make a transformation of the canonical co-

ordinates (q,p), satisfying Hamiltons equations with respect to a HamiltonianH, to

new canonical coordinates (Q,P), satisfying Hamiltons equations with respect to a

Hamiltonian K, where

Q= Q(q,p) and P = P(q,p).

For two functions u = u(q,p) and v = v(q,p) define U =U(Q,P) and V =V(Q,P)

by the identities

U(Q,P) =u(q,p) and V(Q,P) =v(q,p).

Then we have the following resultwhose proof we leave as an exercise in the chain

rule! (Hint: start with [u, v]q,p and immediately substitute the definitions for U andV; also see Arnold [3, Page 216].)

Lemma 3 (Invariance under canonical transformations) The Poisson bracket is in-variant under a canonical transformations, i.e. we have

[U, V]Q,P = [u, v]q,p.

15.3 Generation of constants of the motion

Using Corollary 2 on the Poisson bracket of canonical variables we see Hamiltons

equations of motion are equivalent to the system of equations

dq

dt = [q, H] and

dp

dt = [p, H].

Here, by [q, H] we mean the vector with components [qi, H], and observe we have

simply substituted the components ofq and p for f in Corollary 2. Ifu = u(q,p, t) is

a constant of the motion thendu

dt = 0,

and by Corollary 2 we see thatu

t = [H, u].

If u = u(q,p) only and does not depend explicitly on t then it must must Poisson

commute with the HamiltonianH, i.e. it must satisfy

[H, u] = 0.

Now suppose we have two constants of the motionu = u(q,p) andv = v(q,p), so that[H, u] = 0 and [H, v] = 0. Then by Jacobis identity we find

[H, [u, v]] = [u, [v, H]] [v, [H, u]] = 0.In other words it appears as though [u, v] is another constant of the motion. (This result

also extends to the case when u and v explicitly depend on t.) Indeed, in principle, we

can generate a sequence of constants of the motionu,v, [u, v], [u, [u, v]], .. . . Sometimes

we generate new constants of the motion by this procedure, i.e. we get new information,

but often we generate a constant of the motion we already know about.


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Consider the Kepler problem of a particle of mass m in a three-dimensional central

force field with Hamiltonian

H= 1

2m(p21+p

22+p

23) + V(r)

where V = V(r) is the potential with r =

q21+ q22+ q

23. Known constants of the

motion are u = q2p3 q3p2 and v = q3p1 q1p3 and their Poisson bracket [u, v] =q1p2 q2p1 is another constant of the motion (not too surprisingly in this case).

16 Geodesic flow

We have already seen the simple example of the curve that minimizes the distance

between two points in Euclidean spaceunsurprisingly a straight line. What if the

two points lie on a surface or manifold? Here we wish to determine the characterizing

properties of curves that minimize the distance between two such points. To begin to

answer this question, we need to set out the essential components we require. Indeed,

the concepts we have discussed so far and our examples above have hinted at the need

to consider the notion of a manifold and its concomitant components. We assume in

this section the reader is familiar with the notions of Hausdorff topology, manifolds,

tangent spaces and tangent bundles. A comprehensive introduction to these can be

found in Abraham and Marsden [1, Chapter 1] or Marsden and Ratiu [15]. Thorough

treatments of the overall material in this section can, for example, be found in Abraham

and Marsden [1], Marsden and Ratiu [15] and Jost [8].

16.1 Riemannian metric and manifold

We now introduce a mechanism to measure the distance between two points on a

manifold. To achieve this we need to be able to measure the length of, and angles

between, tangent vectorssee for example Jost [8, Section 1.4] or Tao [17]. The mech-

anism for this in Euclidean space is the scalar product between vectors.

Definition 7 (Riemannian metric and manifold) On a smooth manifoldMwe assign,to any point q M and pair of vectors u,v TqM, an inner product

u,vg(q).

This assignment is assumed to be smooth with respect to the base pointq M, andg= g(q) is known as theRiemannian metric. The length of a tangent vectoru TqMis then

ug(q):= u,u1/2g(q).

ARiemannian manifoldis a smooth manifold M equipped with a Riemannian metric.

Remark 6 Every smooth manifold can be equipped with a Riemannian metric; see

Jost [8, Theorem 1.4.1].


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In local coordinates with q = (q1, . . . , q n)T M, u = (u1, . . . , un)T TqM and

v = (v1, . . . , vn)T

TqM, the Riemannian metric g = g(q) is a real symmetricinvertible positive definite matrix and the inner product above is given by

u,vg(q):=n

i,j=1

gij(q)uivj .

16.2 Length and energy

We are interested in minimizing the length of a smooth curve onM, however as wewill see, computations based on the energy of a smooth curve are simpler.

Definition 8 (Length and energy of a curve) Let q : [a, b] M be a smooth curve.Then we define the lengthof this curve by

(q):=

ba

q(t)g(q(t))

dt.

We define the energyof the curve to be

E(q):= 12

ba

q(t)2g(q(t))

dt.

Remark 7 The energy E(q) is the actionassociated with the curve q = q(t) on [a, b].

Remark 8 (Distance function)Thedistancebetween two points qa, qb M, assumingq(a) = qa and q(b) = qb, can be defined as d(a, b) := infq

(q)

. This distance

function satisfies the usual axioms (positive definiteness, symmetry in its arguments

and the triangle inequality); see Jost [8, pp. 1516].

By the CauchySchwarz inequality we know that

ba

q(t)g(q(t))

dt |b a| 12 b

a

q(t)2g(q(t))

dt

12

.

Using the definitions of(q) and E(q) this implies

(q)2

2 |b a| E(q).

We have equality in this last statement, i.e. we have

(q)2

= 2 |b a| E(q), if andonly ifq is constant. The length (q) of a smooth curve q = q(t) is invariant to

reparameterization; see for example Jost [8, p. 17]. Hence we can always parameterize

a curve so as to arrange for q to be constant (this is also known as parameterization

proportional to arclength). After such a parameterization, minimizing the energy is

equivalent to minimizing the length, and this is how we proceed henceforth.


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16.3 Geodesic equations

We are now in a position to derive the equations that characterize the curves that

minimize the length/energy between two arbitrary points on a smooth manifoldM.For more background and further reading see Abraham and Marsden [1, p, 2245],

Marsden and Ratiu [15, Section 7.5], Montgomery [16, pp. 610] and Tao [17]. Our

goal is thus to minimize the total energy of a path q = q(t) where q = (q1, . . . , q n)T.

The total energy of a path q= q(t), betweent = a and t = b, as we have seen, can be

defined in terms of a Lagrangian function L = L(q, q) as follows

E(q):=

t2t1

Lq(t),q(t)

dt.

The path q = q(t) that minimizes the total energy E= E(q) necessarily satisfies the

EulerLagrange equations. Here these take the form of Lagranges equations of motion

d

dt

L

qj

L

qj= 0,

for each j = 1, . . . , n. In the following, we use gij to denote the inverse matrix ofgij(where i, j = 1, . . . , n) so that

nk=1

gikgkj =ij ,

where ij is the Kronecker delta function that is 1 wheni = j and zero otherwise.

Lemma 4 (Geodesic equations) Lagranges equations of motion for the Lagrangian

L(q,q):= 12

ni,k=1

gik(q) qiqk,

are given in local coordinates by the system of ordinary differential equations

qi+

nj,k=1

ijk qjqk = 0,

where the quantitiesijk are known as the Christoffel symbolsand fori, j, k= 1, . . . , nare given by

ijk :=

n

=1

12g

igjqk

+ gk

qj gjk

q.

ProofWe complete the proof in three steps as follows.

Step 1. For the Lagrangian L = L(q, q) defined in the statement of the lemma, using

the product and chain rules, we find that

L

qj= 12

ni,k=1

gikqj

qiqk


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and

L

qj= 12

nk=1

gjk qk+ 12

ni=1

gij qi

= 12

nk=1

gjk qk+ 12

nk=1

gkj qk

=

nk=1

gjk qk,

where in the last step we utilized the symmetry ofg . Using this last expression we see

d

dt

L

qj

=

nk=1

d

dt

gjk qk

=

nk=1

dgjkdt

qk+

nk=1

gjk qk

=

nk,=1

gjkdq

qqk+

nk=1

gjk qk

.

Substituting the expressions above into Lagranges equations of motion, using that the

summation indicesi and k can be relabelled and that g is symmetric, we find

nk=1

gjk qk+

nk,=1

gjkq

12gkqj

qkq = 0,

for each j = 1, . . . , n.

Step 2. We multiply the last equations from Step 1 by g ij and sum over the index j .

This gives

n

j,k=1 gijgjk qk+

n

j,k,=1 gij

gjkq

12gkqj qkq = 0

qi+n

j,k,=1

gijgjk

q 12

gkqj

qkq = 0

qi+n

j,k,=1

gigj

qk 12

gjkq

qjqk = 0,

where in the last step we relabelled summation indices.


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Step 3. By using the symmetry ofg and performing some further relabelling of sum-

mation indices, we seen

j,k,=1

gi

gjqk

12gjkq

qjqk =

nj,k,=1

12g

i

gjqk

+ gj

qk gjk

q

qjqk

=

nj,k,=1

12g

i

gjqk

+ gk

qj gjk

q

qjqk

=

nj,k=1

iij qjqk,

where the iij are identified with their definition in the statement of the lemma.

Remark 9 On a smooth Riemannian manifold there is unique torsion free connection or covariant derivative defined on the tangent bundle known as the LeviCivitaconnection. A curve q : [a, b] M is autoparallel or geodesicwith respect to if

q q 0,

i.e. the tangent field ofq = q(t) is parallel alongq = q(t); see Jost [8, Definition 3.1.6].

In local coordinates, the coordinates ofq q are given by

(q q)i = qi+n

j,k=1

ijkqjqk,

for i = 1, . . . , n, where the quantities ijk are equivalent to the Christoffel symbols

above.

Now suppose thatM is an embedded submanifold in a higher dimensional Euclideanspace RN. Then M is a Riemannian manifold. (Nashs Theorem says that any Rieman-nian manifold can be embedded in a higher dimensional Euclidean space RN.) For an

embedded submanifoldM, its Riemannian metric is that naturally induced from theembedding Euclidean space using the map from intrinsic to extrinsic coordinates. For

example, in our discussion of DAlemberts principle, we saw that the dynamics of the

system evolved in the Euclidean space R3N, parameterized by the extrinsic coordinates

r1, . . . , rN, though subject to m


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equations of motion for this Lagrangian, we can trace the direction of implication back

through to DAlemberts principle and thus deduce that

q TqM.

As we argued in the lead up to DAlemberts principle, this makes sense on a intuitive

and physical level. The system is free to evolve on the manifold of constraint and,

assuming no external forces, the forces of constraint must act normally to the manifold

(normal to the tangent plane at every pointq M) in order to maintain the evolutionon the manifold. In other words, as Tao [17] eminently puts it,

... the normal quantity q then corresponds to the centripetal force necessaryto keep the particle lying inM (otherwise it would fly off along a tangent linetoM, as per Newtons first law).

16.4 Example: Euler equations for incompressible flow

We formally show here how the Euler equations for incompressible flow represent a

geodesic submanifold flow. The argument is an extension of that above for finite di-

mensional geodesic submanifold flow to an infinite dimensional context and is originally

due to Arnold. Our arguments essentially combine those in Daneri and Figalli [6] and

Tao [17] and we cite Chorin and Marsden [5] and Malham [13] for further background

material on incompressible fluid mechanics.

The Euler equations for incompressible flow in a bounded Lipschitz domain D Rdand on an arbitrary finite time interval [0, T] are prescribed as follows. The velocity

field u = u(x, t) for (x, t) D [0, T] evolves according to the system of nonlinearpartial differential equations

u

t + (u )u= p,

u= 0.

Here the field p = p(x, t) is the pressure field. We need to augment this system of

equations with the zero flux boundary condition for all (x, t) D [0, T], i.e. we haveu n = 0, where n is the unit outer normal to D. We also need to prescribe initialdata u(x, 0) =u0(x) for some given function u0 = u0(x) onD.

From the Lagrangian viewpoint we can describe the flow through the swarm of

particle trajectories prescribed by integral curves of the local velocity fieldu = u(x, t).

Let q = q(x, t) Rd represent the position at time t 0 of the particle that startedat position x D at time t = 0. The flow q= q(x, t) satisfies the system of equations

q(x, t) =uq(x, t), t

,

q(x, 0) =x,

for all (x, t) D [0, T]. A straightforward though lengthy calculation (see Chorinand Marsden [5] or Malham [13]) reveals that the Jacobian of the flow det q =det x q(x, t) satisfies the system of equations

d

dt

det q = u(q, t) det q.


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Since we assume the flow is incompressible so u= 0, andq(x, 0) = x, we deduce forallt [0, T] we have det q 1. This means that for allt [0, T], q( , , t) SDiff(D),the group of measure preserving diffeomorphisms onD . If we now differentiate theevolution equations for q = q(x, t) above with respect to time t and compare this

to the Euler equations, we obtain the following equivalent formulation for the Euler

equations of incompressible flow in terms ofqt(x):= q(x, t) for all t [0, T]:qt = p(qt, t),

with the constraint

qt SDiff(D).We now demonstrate formally how these equations can be realized as a geodesic

flow on SDiff(D) equipped with the L2(D;Rd) metric induced from the inner product

v,L2

(D;Rd

)

:= D

v

dx.

We consider SDiff(D) as an infinite dimensional submanifold of Diff(D), the group ofdiffeomorphisms on D. By analogy with the finite dimensional case, the curve q : t qton SDiff(D) is geodesic if

q TqSDiff(D).Remark 10 Suppose the flow is not constrained and evolves on Diff(D). Then a geodesicwould be characterized by q = 0, i.e. Newtons Second Law, as the metric is flat. For

the intrinsic velocity field (see below) we would have u/t +(u )u= 0; see Tao [17].Our goal now is to characterize the vector space orthogonal to TqSDiff(D). Using thedefinition of the flow above, we define the intrinsic velocity field ut(x) = u(x, t) by

ut := qt q1t . Using the result for the Jacobian detq above we note that the innerproduct

,

above is invariant to transformations x

qt(x)

SDiff(

D). Further,

we can thus characterize

TqSDiff(D) =u(qt, t) : u(qt, t) = 0 and u n(qt, t) = 0 on D

.

Equivalently, since it is the Lie algebra of SDiff(D) which is the vector space of allsmooth divergence-free vector fields parallel to the b oundary, we have

TqSDiff(D) =ut : ut = 0 and ut n= 0 on D

.

The HelmholtzHodge Decomposition Theorem tells us that any vector L2(D,Rd)can be orthogonally decomposed into a divergence-free vector u parallel to D and asolenoidal vector p. In other words we can write

= u + p,

where u = 0 and p =p for some scalar field p. See for example Chorin andMarsden [5, p. 37]. In terms of tangent spaces, we have

TqDiff(D) = TqSDiff(D)

TqSDiff(D)

where TqSDiff(D)

=p L2(D,Rd) : p = p.

Hence we deduce thatqt = p(qt, t) for qt SDiff(D), which are the Euler equationsfor incompressible flow.


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xx

y=y(x)

x0 0x

ds

0

surface of

revolution

Fig. 9 Soap film stretched between two concentric rings. The radius of the surface of revolutionis given by y = y(x) for x [x0, x0].

17 Exercises

Exercise (EulerLagrange alternative form)

Show that the EulerLagrange equation is equivalent to

F

x d

dx

F yx F

mechanics lagrangian and hamilitonian

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