Dynamical Systems

R.S. Thorne

Preface

The course will consist of three relatively distinct parts. The first of theseis classical dynamical systems and will comprise slightly over 1/3 of the total.The topics under study will be:

An introduction to the origins and usage of Lagrangian and Hamiltonianmechanics, which are a more general way of describing systems which areequivalent to Newtonian dynamics, but with more flexibility and with amore transparent means of noting the effect of symmetries on the system.

Particular specific examples using Lagrangian and Hamiltonian mechan-ics. These will include normal modes, i.e. oscillations about stable equi-librium; particles interacting with electromagnetic fields; the Lagrangianand Hamiltonian framework for relativistic systems; an introduction toclassical field theory: and an example of a system where nonlinear effectsbecome important. This last example will lead to the next set of topics.

The second part of the course will develop the mathematical techniquesrequired to study nonlinear systems. It will also comprise slightly over 1/3 ofthe total. The topics will be:

A study of differential equations, with particular emphasis on nonlineardifferential equations written as a system of 1st order equations. We willlook at the methods of solution for such systems, mainly in one and twodimensions. In particular we will concentrate on tractable results suchas fixed points and the solutions of the equations linearised about thesepoints. Using these results we will learn how to draw phase portraitsand thus illustrate the general dynamics for a given set of equations. Wewill look at examples of physical systems, and examine the stability ofsolutions, discussing in detail what can be meant by stability.

An examination of the particular case of conservative systems, where thereis a constant quantity associated with each trajectory. This automaticallybrings us back to the subject of Hamiltonian mechanics where there arevery frequently conserved quantities, in particular the energy of the sys-tem. Again we will look at some physical examples.

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The final topic for this part of the course will be the particular featureof solutions known as bifurcations, where the type of solution can changenature qualitatively for a given value of some parameter in the equations.We will look at 4 different types of bifurcation, and again examine physicalexamples.

The final part of the course consists of the topic of chaos and discrete mapsand will comprise a little under 1/3 of the total. Chaos occurs when we havea bounded but not periodic trajectory (solution) with extreme (exponential)dependence on the initial conditions. We will illustrate this in two ways:

Using differential equations, in particular the Lorenz equations which area very simplified model for evolution of meteorological systems, and inwhich the phenomenon of chaos was first observed. In order to look atthe solutions to these equations we will make use of the techniques in thefirst part of the course.

Looking at discrete maps, i.e. we think of mapping a position from onetime xn to the position at the next time xn+1 using some rule, i.e. xn+1 =f(xn) for some function f . There are advantages to using this approachin order to exhibit chaotic behaviour.

Finally we will look at some features of the solutions to chaotic systemsby introducing the topic of fractal geometry.

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1 Lagrangian and Hamiltonian Mechanics

1.1 Reasons for Introducing Lagrangian Mechanics

We know that classical mechanics is well-described by Newtons equations, i.e.consider the case where we have particles i = 1 N , with masses mi andposition vectors ri = (xi, yi, zi). This leads to Newtons equations

miri = Fi(r1, , rN , t) (1)

where Fi is the force, which may be between the particles or some externalapplied force, or a combination of both. Hence, we have a system of 3N 2ndorder differential equations describing the system. This can quickly becomevery complicated in practice for seemingly quite simple systems. Consider thecase of two linked pendula, both of length l and with masses m1 and m2, shownopposite. The equations of motion are given by

m1x1 = T2 sin 2 T1 sin 1 (2)m1y1 = T1 cos 1 T2 cos 2 m1g (3)m2x2 = T2 sin 2 (4)m2y2 = T2 cos 2 m2g, (5)

and we have the constraints

l2 = (l y1)2 + x21 (6)= (l (y2 y1))2 + (x2 x1)2, (7)

wheresin 1 = x1/l, sin 2 = (x2 x1)/l. (8)

It is not easy to start approaching the problem if we are not able to makethe small angle approximations 1 = x1/l, 2 = (x2 x1)/l. There are twoissues which cause complications:

The natural coordinates to describe the system are the angles 1 and 2and the lengths of the pendula, not x1, y1, x2, y2.

There are constraints in the system, i.e. because the two lengths areconstrained to be l there are in fact only two degrees of freedom, not the4 Cartesian coordinates. Implicitly we have two forces of constraint.

Neither of these two issues is made obvious by the standard form of New-tons equations in eq. (1). We will now describe how to address more generalproblems. In order to do so we will introduce a new way of describing themotion of a system of particles, i.e. Lagranges equations. We will begin byderiving Lagranges equations directly from Newtons equations and then alsoin a simpler manner from a new rule of physics, the Principle of Least Action.

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T1

T2

l

l

x1

y1

x2

y2

m1

m2

m1g

m2g

1

2

O

O1

O2

1.2 Derivation of Lagranges Equation

For a general system with N particles, and hence 3N degrees of freedom in 3dimensions, we could define an alternative set of generalised coordinates for theparticles, i.e. define

qi = qi(x1, y1, z1, , yN , zN , t), i = 1 3N, (9)

e.g. the qi could be the coordinates in spherical polars. They could be com-binations of the original coordinates. If there is a force which depends onthe separation of particles 1 and 2 then a suitable choice for one of the qi isr = |r1 r2|.

As seen with the pendula we may also have constraints, e.g. 0 = l2 (l y1)

2x21. Such a constraint where we can write a constraint equation f(ri, t) = 0is known as a holonomic constraint. I will consider this form of constraint withinthe lecture course. (An example of a nonholonomic constraint is a ball rollingon the surface of a sphere of radius a which can fall off. While on the surface wehave a2 = x2+y2+z2, but after it has fallen off this becomes x2+y2+z2 > a2.)Hence, if there are k constraints there are k equations of constraint

ai = hi(x1, y1, z1, , yN , zN , t), i = 1 k. (10)

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This means that there are 3N k independent degrees of freedom. We wantequations in terms of these, which we label by qj,

qj = qj(x1, y1, z1, , yN , zN , t), j = 1 3N k. (11)In order to gain some confidence in the veracity of Lagranges equations let

us show they are in fact a consequence of Newtons equations. In eq. (11) weexpressed the generalised coordinates qi in terms of the Cartesian coordinatesof a set of particles. There also exists some inverse transformation

ri = ri(a1, , ak, q1, , q3Nk, t). (12)The relationship in eq. (12) allows us to use the chain rule. For example we seethat the fact that the ai are fixed leads to

ri =3Nkj=1

riqj

qj +rit

. (13)

Differentiating again with respect to qm and using the fact that the qj and qjare independent variables, i.e. differentiation with respect to qj is for fixed qkfor all j and k and vice versa, we get

riqm

=riqm

. (14)

We are now in a position to rewrite Newtons equations. The equations ofmotion are

Fi mri = 0, i = 1, N. (15)This automatically means that

Ni

(mri Fi) ri = 0, (16)

for small displacements ri. But from eq.(12) we have

ri =3Nkj=1

riqj

qj, (17)

so we can write the second term in eq.(16) as

Ni

Fi ri =Ni

3Nkj=1

Fi riqj

qj

=3Nkj=1

Qjqj, (18)

where we define the generalised force

Qj Ni

Fi riqj

. (19)

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This is the projection of the force into the direction of the coordinate qj (thoughthe dimension of the generalised force may be different to that of a force).

We now consider the first term term in eq.(16), i.e.

Ni=1

miri ri =Ni=1

3Nkj=1

miri riqj

qj

=Ni=1

3Nkj=1

[d

dt

(miri ri

qj

)miri d

dt

(riqj

)]qj

=Ni=1

3Nkj=1

[d

dt

(miri ri

qj

)miri

(riqj

)]qj, (20)

where we have used eq.(14) for the 1st term and and the commutation of partialderivatives in the final step. Using also the relationship (d/dx)y2 = 2y(dy/dx)this becomes

Ni=1

miri ri =3Nkj=1

[d

dt

(

qj

( Ni=1

1

2mir

2

i

))

qj

( Ni=1

1

2mir

2

i

)]qj. (21)

ButN

i=11

2mir

2i = T , where T is the total kinetic energy of the system. Putting

together eqs. (18) and (21) the equation becomes

3Nkj=1

[[d

dt

(T

qj

) T

qj

]Qj

]qj = 0. (22)

But the original equality in terms of Newtons equations were independent ofthe ri so this must be independent of the qj, and the equation of motionbecomes [

d

dt

(T

qj

) T

qj

]Qj = 0. (23)

These are formally Lagranges equations. We can make a further simplification.If we write Fi = iV (ri) then

Qj =Ni=1

iV riqj

= Vqj

. (24)

Using this we can write Lagranges equation as[d

dt

(T

qj

) (T V )

qj

]= 0. (25)

Tqj

can be defined to be a generalised momentum to go along with the gener-

alised force Qj = Vqj . As a final step we make use of the fact that V doesnot depend on the velocity of the particles, i.e. V = V (qj, t) to simplify eq.(25) further. We define the Lagrangian function L = T V , and the equationbecomes

d

dt

(L

qj

)=

L

qj. (26)

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Our system is therefore defined in terms of 3N k 2nd order differential equa-tions in terms of the generalised coordinates qj.

Let us check that these new equations look plausible. Consider the sim-plest general case of a single particle of mass m moving in one dimension withcoordinate x in potential V (x). In this case the Lagrangian takes the simpleform

L = 12mx2 V (x). (27)

In this case Lx

= mx and Lx

= Vx

so Lagranges equation becomes

d

dtmx mx = V

x. (28)

So for this simple case we recover Newtons equation that mass accelerationis equal to the applied force.

A slightly more complicated case is the single pendulum of length l with bobof mass m. In this case T = 1

2ml22 and V = mgl(1cos ). So the Lagrangian

is1

2ml22 mgl(1 cos ). (29)

Again, the simple application of Lagranges equation leads to

d

dtml2 ml2 = mgl sin , (30)

which we recognise as the correct equation. The derivation was rather simple,not relying on having to resolve forces along any axis, just on writing down thecomponents of the energy correctly.

If T were independent of the coordinates qj then Lagranges equations wouldthen simply be that the rate of change of generalised momentum is equal to thegeneralised force. However, this is very often not the case. Consider a particlemoving in two dimensions, but described in terms of circular polars rather thanCartesian coordinates x and y. The kinetic energy of the particles is

T = 12mr2 + 1

2mr22, (31)

and T/r 6= 0. this gives rise to so-called fictional forces. In fact thequantity L

qjis sometimes called the generalised force. I will try to be clear

which I mean.Note that the generalised momentum may genuinely be the momentum, as in

the case of the particle in one dimension. However, it may not be of the correctdimensions. For the single pendulum it was ml22, which has the interpretationas the angular momentum of the pendulum, linear momentum being a lessuseful quantity in this case. Similarly, for the pendulum L/ = mgl sin isthe torque rather than the force.

It is simplest, and also easiest to make generalisations (e.g. inclusion ofmagnetic fields) to derive Lagranges equations from the Principle of Least Ac-tion, which is a fundamental theorem of Classical Mechanics (and also can be

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interpreted in terms of Quantum Mechanics). In this we begin by defining aquantity called the Lagrangian by

L = T V, (32)

where T is the total kinetic energy of a system and V the total potential energy.We then consider the motion of the particle between the fixed positions qi(t1)and qi(t2) for times t1 and t2. The Action is defined by

S = t2t1

L(qi, qi, t) dt. (33)

The Principle of Least Action states that S is at an extremum, either a max-imum or a minimum, for the actual paths between the fixed positions at fixedtimes. This then provides equations for the paths of the particles. Supposeq0i (t) represents the true path, and we make infinitesimal variations about this.

qi(t) = q0

i (t) + i(t). (34)

By definition i(t1) = i(t2) = 0. Expanding to 1st order about the q0i (t) we

obtain

S = t2t1

L(qi, qi, t) dt S0 + S = t2t1

L(q0i + i, q0

i + i, t) dt

= S0(q0i , q0

i ) + t2t1

i

[(L

qi

)i +

(L

qi

)i

]dt

= S0(q0i , q0

i ) +i

[i(t)

(L

qi

)]t2t1

+ t2t1

i

[ ddt

(L

qi

)+(L

qi

)]i dt,(35)

where to obtain the second term we have integrated by parts with respect to t.This second term is identically zero from the boundary conditions, so we have

S = t2t1

i

[ ddt

(L

qi

)+(L

qi

)]i dt. (36)

This must be true for arbitrary small variations i(t), so we obtain the Euler-Lagrange equations

d

dt

(L

qj

)=

L

qj. (37)

Although we have already shown that Lagranges equations are equivalent toNewtons equations in this derivation we have made no reference at any pointto what any set of coordinates is, or the constraints Lagranges equations aretrue for any set qi which describe the independent degrees of freedom of thesystem.

We now see that in order to find the equations of motion for any physicalsystem we have to choose the most suitable set of coordinates (not alwaysobvious), write down the kinetic and potential energy, and then just use eq.

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(37). For example, considering our case of the double pendulum (taking theequilibrium position of each bob to be zero potential)

T = 12m1l

221 +1

2m2l

2(21 + 2

1 + 2 cos(1 2)12), (38)and

V = m1gl(1 cos 1) +m2gl(2 cos 1 cos 2). (39)The equations for 1 and 2 are then obtained just by differentiating. Theseequations, and some extensions of this system are left as an exercise in theproblem sheets.

As well as the ease of obtaining the equations of motion the Lagrangianformulation is also very useful for considering the effect of symmetries. If theLagrangian L is independent of one of the coordinates qk then the equation ofmotion is

d

dt

(L

qk

)=

L

qk= 0, (40)

and the generalised momentum Lqk

= pk is a conserved quantity. A simpleexample of this is illustrated by the motion of a particle in a central potentialin 2-d, where the most appropriate coordinate system is circular polars. In thiscase

T = 12mr2 + 1

2mr22, V = V (r). (41)

leading toL = 1

2mr2 + 1

2mr22 V (r). (42)

This gives L

= 0 dpdt

= 0 where p = mr2. In this case the conserved

generalised momentum p has a direct physical interpretation. It is the angularmomentum.

1.3 Hamiltonian Mechanics

Defining the generalised momentum pi =Lqi

we can choose to use the pi asthe complementary degrees of freedom to the qi rather than the generalisedvelocities qi. Using these we can define a new function, the Hamiltonian

H(pi, qi, t) =i

piqi(pi, qi) L(qi(pi, qi), qi, t). (43)

This is known as a Legendre transformation (also used in thermodynamics) andthe independent degrees of freedom are now defined to be the pi and the qi. Ifwe consider the case of the particle moving in a central potential then

pr = mr, p = mr2, (44)

and

H(p, q) = prr + p 12mr2 12mr22 + V (r)

=p2r2m

+p2

2mr2+ V (r)

[= 1

2mr2 + 1

2mr22 + V (r)

]= T + V. (45)

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Thus in this case the Hamiltonian has the physical interpretation of the energy.However, have we lost a simple set of equations of motion by making this changein definition? We can see by deriving Hamiltons equations.

H

qi=j

(qjqi

pj Lqi

Lqj

qjqi

). (46)

But Lqj

pj and the first and third terms cancel, leavingH

qi= L

qi= d

dt

(L

qi

)= pi. (47)

Also we can differentiate with respect to pi,

H

pi=j

(qjij + pj

qjpi

Lqj

qjpi

). (48)

But Lqj

pj and so the second and third terms cancel. Hence, we obtain thefull set of Hamiltons equations

H

qi= pi, H

pi= qi. (49)

These are just as simple as for the Lagrangian case, i.e. if we can write down thekinetic and potential energy it is a straightforward progression to the equationsof motion. In our example of the particle in a central potential the qi equationsgive

r =prm, =

pmr2

, (50)

and the pi equations give

pr = Vr

+p2mr3

, p = 0. (51)

Again the conservation laws are immediately apparent. If H(pi, qi) is indepen-dent of one of the qi the conjugate momentum pi is conserved. Also, we seethat the equation for pr has two terms on the right-hand side. The first, Vr ,is the conventional radial force, which may be attractive towards the origin or

repulsive away from it. The second,p2

mr3is guaranteed to be positive and is the

fictitious centripetal force. It simply reflects that a particle with conservedangular momentum cannot get too close to r = 0 else its orbital kinetic energyget very large.

The Hamiltonian and Lagrangian formulations are completely equivalent.However, certain physical aspects are more clear in the Hamiltonian framework.Consider the time dependence of the Hamiltonian

dH

dt=

i

(H

qiqi +

H

pipi

)+

H

t

i

(piqi + qipi) + Ht

=H

t L

t. (52)

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Hence, if H (or L) have no explicit time dependence then H is a conservedquantity. But what is this quantity? Suppose the Lagrangian is of the form

L =i

1

2q2i (qi, pi)fi({q}) V (qi), (53)

for any set of functions fi of the position coordinates, e.g. f = mr2 in circular

polars. This results in

pi Lqi

= qifi({q}), (54)this means that

H =i

q2i f({q})i

1

2q2i f({q}) + V (qi)

=i

1

2q2i f({q}) + V (qi)

= T + V = total energy. (55)

Hence, if all the qi are independent degrees of freedom H is the energy, and ifthere is no explicit time dependence it is a conserved quantity. In general it canbe either, both or neither.

Since the Hamiltonian is directly related to the time-dependence of the sys-tem we can illustrate one more property. Consider any function of the coor-dinates and conjugate momenta f(qi, pi, t). The time-dependence of f is givenby

df

dt=i

(f

qiqi +

f

pipi

)+

f

t. (56)

Using the Hamiltonian equations of motion this can be re-expressed as

df

dt=i

(f

qi

H

pi f

pi

H

qi

)+

f

t. (57)

Thus, up to the explicit time dependence the evolution of f is driven by theHamiltonian. In fact the quantity

{f, g} =i

(f

qi

g

pi f

pi

g

qi

)(58)

is a form which appears frequently in Hamiltonian physics and is called the Pois-son bracket of f and g. Hence, we can write eq. (57) as Using the Hamiltonianequations of motion this can be re-expressed as

df

dt= {f,H}+ f

t. (59)

This has clear parallels to quantum mechanics where the time evolution of theexpectation value of an operator f is

df

dt

=

1

ih[f,H]+

f

t

. (60)

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Indeed the Hamiltonian framework is central to the quantization of a classicalsystem, and Poisson brackets and commutators relate the two regimes.

As an example of the time evolution being governed by the Hamiltonian weconsider the example of the angular momentum component Lx when a particleis moving in a spherically symmetric potential V (r) where r =

x2 + y2 + z2.

Using the definition L = rp we have Lx = ypzzpy. We consider the Poissonbracket {Lx, H}. This is given by

{Lx, H} = Lxx

H

px Lx

px

H

x+Lxy

H

py Lx

py

H

y+Lxz

H

pz Lx

pz

H

z. (61)

Since Lx has no x or px dependence and H = p2/2m+ V (r) we obtain

{Lx, H} = pz pym

+ zV

y py pz

m yV

z. (62)

The first and third terms obviously cancel. Also

V

y=

r

y

V

r=

y

r

V

rand

V

z=

z

r

V

r, (63)

so the second and fourth terms cancel and

{Lx, H} = 0. (64)

Since Lx has no explicit time dependence this also means that

dLxdt

= 0 (65)

for the system, and the vanishing Poisson bracket of Lx and H results in thewell-known conclusion that angular momentum components are conserved ifthere is a spherically symmetric potential. This is a general rule. Any quantity(if not explicitly time dependent) will be conserved if its Poisson bracket withthe Hamiltonian is zero.

Finally for this section, one other difference between the Lagrangian andHamiltonian formulations is that in the Lagrangian description we have 3N k2nd order differential equations

d

dt

(L

qi

)=

L

qi, (66)

while in the Hamiltonian approach we have 2(3N k) 1st order differentialequations

H

qi= pi, H

pi= qi. (67)

These contain the same information and require the same number of bound-ary conditions, but the latter will be more immediately useful for analysis ofnonlinear differential equations using the techniques developed in the course.

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1.4 Liouvilles Theorem

We note that the set of coordinates and momenta {qi, pi} define the 6N dimen-sional phase space for a system of N particles (or less than this if there aresome constraints on the system). We can consider the element of volume of thisphase space as it evolves with time. Consider the change as we go from time tto time t+t. Denoting the {qi, pi} generically by xi, then under time evolutioneach satisfies

xi(t+ t) = xi(t) + tdxidt

+O(t2). (68)Under this small change in time an element of volume of the phase space un-dergoes a change

V V = V det(xi(t+ t)

xj(t)

)

= V det(ij + t

xixj

+O(t2))

(69)

Ignoring all terms of O(t2), off diagonal terms make no contribution to thedeterminant and we obtain

V = V (1 + t x). (70)

However, the 6N -dimensional divergence is given by

x = i

[

qiqi +

pipi

]

=i

[

qi

(H

pi

)+

pi

(Hqi

)]= 0. (71)

So V = V , and a Hamiltonian system preserves the volume in phase space,i.e. the space behaves like an incompressible fluid. (Note that there can be adecrease in volume in coordinate space if countered by an increase in momentumspace, or vice versa.) This is known as Liouvilles theorem. Such systems areclassified as measure preserving or non-dissipative, whereas those where thevolume is not preserved are called dissipative. We will encounter examples ofthis later.

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