Lecture notes for Introduction to Dynamical Systems: CM131A
Based on notes by A. Annibale, R. K¨uhnand H.C. Rae
2018-2019
Lecturer: G. Watts
1
Contents
1 Overview of the course 4 1.1 Revision Exercises ...... 9
I Differential Equations 10
2 First order Differential Equations 11 2.1 Basic Ideas ...... 11 2.2 First order differential equations ...... 12 2.3 General solution of specific equations ...... 13 2.3.1 First order, explicit ...... 13 2.3.2 First order, variables separable ...... 14 2.3.3 First order, linear ...... 16 2.3.4 First order, homogeneous ...... 18 2.4 Initial value problems ...... 20 2.5 Existence and Uniqueness of Solutions — Picard’s theorem ...... 23 2.5.1 Picard iterates ...... 24 2.6 Exercises ...... 28
3 Second order Differential Equations 31 3.1 Second order differential equations ...... 31 3.1.1 Second order linear, with constant coefficients ...... 32 3.2 Existence and Uniqueness of Solutions — Picard’s theorem ...... 38 3.3 Exercises ...... 41
II Dynamical Systems 44
4 Introduction to Dynamical Systems 45
Version of Mar 14, 2019 2
5 First Order Autonomous Systems 50 5.1 Trajectories, orbits and phase portraits ...... 50 5.2 Termination of Motion ...... 55 5.3 Estimating times of Motion ...... 60 5.4 Stability — A More General Discussion ...... 66 5.4.1 Stability of Fixed Points ...... 66 5.4.2 Structural Stability ...... 68 5.4.3 Stability of Motion ...... 71 5.5 Asymptotic Analysis ...... 72 5.5.1 Asymptotic Analysis and Dynamical Systems ...... 74 5.6 Exercises ...... 80
6 Second Order Autonomous Systems 85 6.1 Phase Space and Phase Portraits ...... 85 6.2 Separable Systems ...... 95 6.3 The structure of orbits and Phase Space ...... 98 6.4 Limit cycles ...... 99 6.5 Fixed points of second-order autonomous systems ...... 101 6.6 Linear Stability Analysis ...... 101 6.6.1 Step 1 — Taylor Expansion of Velocity Functions ...... 102 6.6.2 Step 2 — Finding the Jordan canonical form of the Jacobian . . . . . 104 6.6.3 Step 4 — Exploring the Consequences for Dynamics ...... 105 6.7 Beyond linear stability analysis ...... 119 6.8 Exercises ...... 120
III Application to Classical Mechanics 132
7 Elements of Newtonian Mechanics 133 7.1 Motion of a particle ...... 133 7.2 Newton’s Laws of motion ...... 138 7.2.1 Newton’s First Law (N1) ...... 138 7.2.2 Newton’s Second Law (N2) ...... 138 7.2.3 Newton’s Third Law (N3) ...... 138 7.3 Newton’s Law of Gravitation ...... 140 7.4 Motion in a Straight Line; the Energy Equation ...... 143
Version of Mar 14, 2019 3
7.5 Equilibrium and Stability ...... 147 7.6 Exercises ...... 153
8 Hamiltonian Systems 165 8.1 Hamilton’s equations for motion in a potential ...... 166 8.2 Stability problems ...... 173 8.3 Summary: how to analyse motion in a potential ...... 176 8.4 Exercises ...... 181
A Functions of two variables 187 A.1 The partial derivative ...... 187 A.2 Continuity of a function of two variables ...... 189
B Taylor’s Theorem 190 B.1 Taylor Expansion for Functions of One Variable ...... 190 B.2 Taylor Expansion for Functions of two variables ...... 191 B.3 Vector functions ...... 192
C Basic Linear Algebra 195 C.1 Fundamental ideas ...... 195 C.2 Invariance of Eigenvalues Under Similarity Transformations ...... 197 C.3 Jordan Forms ...... 197 C.4 Basis Transformation ...... 199 C.5 Rotations ...... 200 C.6 Area Preserving Transformations ...... 200 C.7 Examples and Exercises ...... 201
D Jacobians and Change of variables 205 D.1 Change of variables ...... 205
Version of Mar 14, 2019 Chapter 1: Overview of the course 4
Chapter 1
Overview of the course
This course is about the study of quantities which vary in time. The quantities are meant to represent some “system” and the variation in time is called the “dynamics”, hence the name. These ideas can be very general - the quantities could take real values, such as the coordinates of bodies in a classical mechanics problems, they could take integer values such as the number of individuals in a population - and the variation in time can be continuous (as the position of a body is defined for all values of t) or discrete (maybe we measure the population only once each day). The study of variations of systems with time obviously has gone on for a long time, but the development of the branch of mathematics known as “Dynamical Systems” really started in the 1890s with Poincar´e,inspired by the problems of the motion of planets. It is partly pure mathematics and partly applied and that is reflected in the way this course is organised: there are some theorems, some study of properties of differential equations and some applications. To give a flavour of what’s to come we’ll look at three models of population size and one physical system which are all typical examples. The simplest model of a population is that of Thomas Malthus who proposed (in 1798) that if there are no constraints on the resources available then the rate of growth of a population will be proportional to its size - the number of births in any time period will be proportional to the number of people. If the size of the population is x and the birth rate is r per unit time then this can be modelled by the differential equations dx = r x . (1.1) dt The solutions of this equation are r t x(t) = x0 e . (1.2) This exponential growth of course leads to problems, as Malthus knew, and so is not a good model for long term population sizes. A better model is the logistic equation (devised by Pierre-Fran¸coisVerhulst in 1844) in which limits of resources mean there is a maximum size c for a stable population: dx x = f(x) , f(x) = r x 1 − , (1.3) dt c
Version of Mar 14, 2019 Chapter 1: Overview of the course 5 for which we can also find the general solution as the function x(t) c c x = −r t , where A = 1 − (1.4) 1 − Ae x0 The exponential solution (1.2) is one we are familiar with, the solutions (1.4) may be harder to visualise. One of the aims of the course is to see how to find out properties of the dynamics of a model – exact or approximate – using graphical methods. The simplest method is if one can find the exact solutions and then plot these, so for the exponential growth model we can plot x vs. t or t vs. x, as in figure 1.1 x 1.0 t
4
0.5
2
-2 2 4 t
-1.0 -0.5 0.5 1.0 x -0.5
-2
-1.0
Figure 1.1: Sketches of the solutions (1.2) for r = 0.1 together with the phase portrait showing the fixed point at x = 0.
Also included is a useful guide to the way the system behaves, known as the “phase portrait” which shows whether x increases or decreases with t and also that x = 0 is a “fixed point”. For the more complicated Verhulst model, the corresponding plots are in figure 1.2 x t
2.0 4
1.5
2 1.0
0.5 -1.0 -0.5 0.5 1.0 1.5 2.0 x
-4 -2 2 4 t -2 -0.5
-1.0 -4
Figure 1.2: Sketches of the solutions (1.4) for r = 0.1 and c = 1 together with the phase portrait showing the fixed points at x = 0 and x = 1.
In both these cases we can solve the equations exactly but the behaviour of the solutions is captured in the phase portrait showing the way x changes with t and the fixed points.
Version of Mar 14, 2019 Chapter 1: Overview of the course 6
These models can be generalised to a system with two species, a predator and a prey, where simple models for their interactions lead to two coupled equations dx dy = x(a − by) , = y(cx − d) . (1.5) dt dt This is the Lotka-Volterra model, also called the hare-lynx or rabbit-fox model for two situ- ations it describes. This is a case where we cannot find an exact solution in general; We can only find exact solutions if x = 0 (no predators), if y = 0 (no prey) or if the populations take the constant values x = d/c, y = a/b. However, we can study other aspects of the system - we can sketch the solutions (technically, sketch the phase curves) and see that there are going to be solutions that stay close to the fixed point at (d/c, a/b), that these will have periodic oscillations and we can find the approximate period of these oscillations. There is another fixed point at (0, 0) which is of a different nature; it is not stable, solutions close to this point do not have to stay close and we can again analyse motion near here.
2.0
1.5
1.0 Predator
0.5
0.0
0 1 2 3 4 5 6 Prey
Figure 1.3: Sketch of the time-evolution of prey (x-axis) and predator (y- axis) populations in the case a = 1, b = 2, c = 1, d = 2. The fixed point at (2, 1/2) is the isolated dot, and typical time evolution is a periodic orbit around the fixed point.
The first two examples are completely solved using the theory of differential equations; this third example takes us into the study of dynamical systems proper where we pay attention to structural properties of the solutions without necessarily solving them. Finally, as an example of a discrete dynamical system, another very well-known model is the “logistic map” which states how a variable xn changes as n increases in integer steps:
xn+1 = rxn(1 − xn) . (1.6)
This is most famous because it exhibits chaotic behaviour when r > 3.56995... Not only is
Version of Mar 14, 2019 Chapter 1: Overview of the course 7
there no exact solution, but the tiniest change in the starting value of x1 leads to unpredictable changes in the time evolution. While this discrete chaotic system is interesting, it is also requires very sophisticated mathe- matical analysis and in this course we will stick to the simpler case of time being a continuous parameter and the variables in the system having real values, so that the systems are de- scribed by one or more functions of t and the dependence on t is defined through differential equations. We can now describe the structure of the course: the course is split into three parts: differential equations, dynamical systems and classical mechanics. The first part is on differential equations and their solutions. We first consider first order differential equations, give methods to solve some standard classes and state a theorem guar- anteeing the existence and uniqueness of solutions - it is easy to write down problems which have no solution or which have an infinite number of solutions and it is good to have a test to decide what the answer is. We then turn to second order differential equations, state a similar theorem on existence and uniqueness of solutions and consider the case of linear equations with constant coefficients in some detail. The second part is on dynamical systems. We introduce the ideas of the order of a dynamical system, fixed points, stability, phase portraits, and then study these in turn for first and second order dynamical systems and how to sketch and analyse phase portraits such as figure 1.3, how to classify fixed points and discuss motion in their neighbourhoods. The third and final part is on applications of this theory in classical mechanics. We introduce Newton’s laws of motion as second order differential equations for the position of massive bodies and give his law of gravity. (1.7). We use this as an example of the idea of a potential and analyse motion in a potential using the methods of dynamical systems. A typical example of a classical mechanical system is that of a weight suspended from a spring. If a weight with mass m is held up by such a spring then the height z of the weight could satisfy a differential equation such as d2z m = −kz − lz3 − mg . (1.7) dt2 When l = 0 this is an ideal spring satisfying Hooke’s law and we will see how to find the general solution k z = z + A sin(ωt) + B cos(ωt) , where z = −mg/k and ω2 = . (1.8) 0 0 m When l 6= 0 as in a more realistic model of a spring, we cannot solve the differential equation in terms of elementary functions and cannot find z as an explicit function of t, but the system shares many of the same properties as the ideal spring equation (1.7), for example the frequency of small oscillations is approximately the same as that when l = 0. In the case of the weight on the spring, the potential is k l dV V (z) = mgz + z2 + z4 with the equation of motion mz¨ = − . (1.9) 2 4 dz The final topic in this section is on Hamilton’s reformulation of classical mechanical systems which is both very closely related to the ideas of dynamical systems but also central to modern
Version of Mar 14, 2019 Chapter 1: Overview of the course 8 algebraic approaches and leads on to the more advanced modules “Classical Mechanics” in year 2 and “Quantum Mechanics” in year 3. The equivalent reformulation by Hamilton defines a function H(p, q) where q = z and p = mz˙ and equations of motion given by
p2 dq ∂H dp ∂H H(p, q) = + V (q) , = , = − , (1.10) 2m dt ∂p dt ∂q which has the advantage of showing immediately that the function H(p, q) is constant in time. This formulation also makes it very easy to find the equilibrium points of the mechanical system and to the sketch the phase portrait of the system. The phase portrait showing how the values of q and p vary with time for an ideal spring and a non-ideal spring are shown in figure 1.4. They look very much the same - the differences are that the position of the fixed point has moved and that the orbits are circles for the ideal spring and squashed circles for the non-ideal spring.
2 2
1 1
0 0
-1 -1
-2 -2
-3 -2 -1 0 1 -3 -2 -1 0 1
Figure 1.4: Sketches of the phase portrait for the system solutions (1.10) for m=g=k=1 with l = 0 (ideal spring) with fixed point at (−1, 0) on the left and l = 16/27 (non-ideal spring) with fixed point at (−3/4, 0) on the right.
Version of Mar 14, 2019 Chapter 1: Overview of the course 9
1.1 Revision Exercises
This problem-set is about prerequisites, based on material covered in Calculus 1 and Linear Methods/Linear Algebra and Geometry I.
Exercise 1.1 Evaluate the following integrals:
Z 1 x2 Z ∞ x3 1. dx 6 2. dx 8 0 (1 + x ) 0 (1 + x ) Z 1 x2 Z 1 3. dx 4. dx 2 1/2 p 0 (1 − x ) x(1 − x) Z Z 5. dx x2 exp(2x) 6. dx xn ln x , n a positive integer.
Exercise 1.2 Give the Taylor expansion up to third order in x about x = 0 for the following functions 1 √ 1. f(x) = 2. f(x) = 1 + x 3. f(x) = ln(1 + x) 4. f(x) = sinh(x) 1 − x
Exercise 1.3 Discuss the salient properties (find zeros, poles, extrema, inflection points, asymptotes ...) and produce a qualitatively correct graph of the following functions
1. f(x) = x(x − 1)(x − 2) 2. f(x) = x2(x − 1)(x − 2) x2 x2 3. f(x) = 4. f(x) = 1 − x2 1 + x2 x 5. f(x) = x exp(−x2) 6. f(x) = tan(x) x 7. f(x) = cosh(x)
Exercise 1.4 This exercise recalls properties of differential operators and linear maps.
d Consider L = dx defined as a map acting on differentiable functions of x, df df L : f 7→ L(f) = ,L(f)(x) = (x) . dx dx Show that L is a linear map, i.e. show that for any real constants α and β and any differentiable functions f(x) and g(x), L(αf + βg) = αL(f) + βL(g) , Which of the following is a linear map on differentiable functions of x?
df d2f df 1. M : f 7→ f + , 2. M : f 7→ − x2f , 3. M : f 7→ x + 1 dx 2 dx2 3 dx
Version of Mar 14, 2019 Part I
Differential Equations
10 Chapter 2: Differential Equations 11
Chapter 2
First order Differential Equations
This chapter starts with some elementary ideas about differential equations. We then consider first order differential equations, discuss different classes and give methods to solve some of these. We then discuss initial value problems, show that there are initial value problems which can have no solutions or contrastingly an infinite number of solutions and finish with a theorem which gives conditions for the existence and uniqueness of a solution to an initial value problem.
2.1 Basic Ideas
A differential equation, or more correctly an ordinary differential equation (ODE), is a relation between a function of one variable and one or more of its derivatives (evaluated at the same point). We will often think of a function x(t) where t we can think of as time and x as a position, a population, a share price... We will also often use y(x). In general, we can use a dash to denote differentiation, so that y0(x) = dy/dx, y00(x) = d2y/dx2; if we use t as the variable then we can denote derivatives using dots,x ˙ = dx/dt,x ¨ = d2x/dt2. Here are some examples: dx = 3x2 sin(x + t) (2.1) dt x¨ = cos(2t) (2.2) d2y dy 3 = − y (2.3) dx2 dx
One of the most important properties of an ODE is its order Definition 2.1. The order of an ODE is the order of the highest derivative appearing in the ODE
In the three examples above, the orders are 1, 2 and 2 respectively. Even though (dy/dx)3 appears in (2.3), the highest derivative is the second derivative d2y/dx2 and so the order of the ODE is two. We shall start in this chapter by considering first order differential equations and then turn to second order differential equations in chapter 3.
Version of Mar 14, 2019 Chapter 2: Differential Equations 12
2.2 First order differential equations
We start by considering first order (ordinary) differential equations of the form
d x = f(t, x) , (2.4) dt in which f is a suitably defined function of two real variables.
Definition 2.2. Solution of an ODE We say that a function g is a solution of the differential equation (2.4) if equation (2.4) is satisfied identically when we substitute x(t) = g(t), i.e.
d g(t) ≡ f(t, g(t)) (2.5) dt
This allows a lot of freedom in the choice of the solution, for example for the extremely elementary ODE with f(t, x) = 0, d f = 0 , (2.6) dt then f(x) = 0 and f(x) = 1 are both solutions and indeed f(x) = c is a solution for any value of c. This leads to the idea of a general solution
Definition 2.3. General Solution We say that the family of functions {h(·,C)} is a general solution of equation (2.4), if
d h(t, C) ≡ f(t, h(t, C)) , (2.7) dt whatever C. [Rem: The notation is to be understood in the sense that x = h(·,C) is a function with parameter C, i.e. x(t) = h(t, C).]
d at E.g.: the ODE dt x = ax has general solution x(t) = Ce , with C = x(0). One way to pick out a particular solution is to choose an initial value for x(t), that is we choose an initial time t0 and require that x(t0) = x0. This is then called an initial value problem
Definition 2.4. An initial value problem for a first order ODE We say the problem of finding a function x(t) such that
d x = f(t, x) (2.8) dt and x(t0) = x0 (2.9) is an initial value problem.
Version of Mar 14, 2019 Chapter 2: Differential Equations 13
Definition 2.5. A solution of an initial value problem We say that a function g is a solution of the initial value problem (2.8,2.9) if equation (2.8) is satisfied identically when we substitute x(t) = g(t), i.e.
d g(t) ≡ f(t, g(t)) (2.10) dt and the initial condition (2.9) g(t0) = x0 . (2.11) is also satisfied.
Thus the function g must satisfy the differential equation and the initial condition.
If we plot the points {(t, g(t)); t ∈ IR} in the Cartesian plane we generate a trajectory or solution curve of (2.4) which passes through the point (t0, x0) determined by the initial condition x(t0) = x0. As we change the initial condition (i.e. the value of x0, the prescribed value of x at t = t0, and possibly t0) we generate a family of solution curves of (2.4) also called the flow corresponding to (2.4).
We will now review methods for finding the general solution of a few simple types of ODE (only for a few types do general solutions in terms of elementary functions exist!), and then turn to the question of initial value problems and briefly enquire into the question of uniqueness of solutions.
2.3 General solution of specific equations
We now summarise some well known methods for finding the general solutions of certain types of first and second order differential equations. Recall that the general solution of a first order differential equation involves one parameter. We proceed from simple to more complicated situations, trying to reduce the more complicated situations to the simpler ones dealt with before.
2.3.1 First order, explicit
Explicit equations are those which can be solved by direct explicit integration; they are of the form dx = f(t) (2.12) dt so that the r.h.s. does not depend on the unknown function x; thus explicit first order equations are not at all interesting ODEs. Integrating with respect to time t one obtains Z x = dt f(t) = F (t) + C, (2.13)
Version of Mar 14, 2019 Chapter 2: Differential Equations 14 with F (t) + C denoting the indefinite integral of the function f. Thus one has
x = x(t) = F (t) + C (2.14) as the general solution of (2.13).
2.3.2 First order, variables separable
Equations of the type dx = f(x) g(t) (2.15) dt may be integrated formally by separation of variables: dx Z dx Z = g(t) dt ⇒ = g(t) dt , (2.16) f(x) f(x)
R dx R Denoting the indefinite integrals involved by f(x) = F (x) + C, and dt g(t) = G(t) + C and combining the two integration constants into one, one obtains
F (x) = G(t) + C, (2.17) which — on (formally) solving for x — gives
x = x(t) = F −1(G(t) + C) (2.18)
with arbitrary C as the general solution of (2.15). It is not always possible to find an explicit expression for F −1 and one has sometimes just to be happy with the relation (2.17)
Example 2.1. Find the general solution of the differential equation dx = αx , (2.19) dt
where α is a constant. This differential equation will arise frequently in our later work. We have dx Z dx Z = α dt , ⇒ = α dt ⇒ ln |x| = αt + C (2.20) x x where C is a constant, so that |x| = eαt eC , (2.21) This is an expression for |x(t)|, but choosing the free parameter differently we can find x(t) explicitly as, x(t) = A eαt , (2.22) where A = ±eC is now the free parameter.
Example 2.2. Find the general solution of the differential equation dx x = , (2.23) dt t
Version of Mar 14, 2019 Chapter 2: Differential Equations 15
This is clearly separable, and so dx dt Z dx Z dt = , ⇒ = ⇒ ln |x| = ln |t| + C (2.24) x t x t where C is a constant, so that changing constants,
x = At , (2.25) for some A.
Example 2.3. Find the general solution of the differential equation dx 2t + 1 = , (2.26) dt 3x2 + exp(x)
This is clearly separable, and so Z Z (3x2+exp(x))dx = (2t+1)dt , ⇒ (3x2+exp(x))dx = (2t+1)dt ⇒ x3+exp(x) = t2+t+C, (2.27) where C is a constant. In this case there is no way to solve equation (2.27) to find x as a function of t but it is still possible to plot solutions by solving for x(t) numerically as in figure 2.1
2.0
1.5
1.0
0.5
-3 -2 -1 1 2 3
-0.5
-1.0
Figure 2.1: Sketch of three solutions to equation (2.27) with C = −1 (solid line), C = 0 (dotted line) and C = 1 (dashed line).
Version of Mar 14, 2019 Chapter 2: Differential Equations 16
2.3.3 First order, linear
First order, linear ordinary differential equations are of the form dx = f(t) x(t) + g(t) , (2.28) dt with f and g given functions of t. They are called “linear” because, when written as dx − f(t)x(t) − g(t) = 0 , (2.29) dt the left-hand side is a linear function of x and its derivative. It also means that the differential equation can be written in terms of a differential operator L which is a linear map when acting on the vector space of differentiable functions:
d dx L(x) = g , where L = − f(t) ,L(x) = − f(t)x(t) . (2.30) dt dt
We can check that L is a linear map by considering two functions x1 and x2 and arbitrary constants α, β and seeing that
L(αx1 + βx2) = αL(x1) + βL(x2) . (2.31)
This automatically has implications for the space of solutions which we will look at in more detail later in this section. For the moment we will just look at how to solve these equations. Equations of the form (2.28) can be solved using the method of integrating factors. This is a method that reduces the problem to that of solving two simpler equations, one of which can be solved by separation of variables, the other by direct integration. It works as follows. Multiply through by µ = µ(t) where µ(t) is a function of t which we choose later. We have
dx µ − µ f x = µg , (2.32) dt which can be transformed into d dµ (µx) + − µf − x = µg . (2.33) dt dt
By choosing µ such that dµ = −µf , (2.34) dt a choice that requires solving a separable first order ODE to actually determine µ, one finds that the transformed ODE simplifies to d (µx) = µg . (2.35) dt This is a first order explicit equation for the product µx, once µ is known from the solution of the separable ODE defining it. A solution of the ODE defining µ is
R µ(t) = exp− f(t) dt . (2.36)
Version of Mar 14, 2019 Chapter 2: Differential Equations 17
Note that the indefinite integral in the exponent is only defined up to an arbitrary additive constant. As we need to know µ only up to an arbitrary multiplicative constant, we can set any additive constant in the exponential to zero (also we have implicitly used this freedom to assume that µ is positive !). With this choice of µ the explicit equation for the product µx is integrated to give Z 1 Z µ(t)x(t) = µ(t)g(t) dt , x(t) = µ(t)g(t) dt . (2.37) µ Note again that the indefinite integral here comes with an arbitrary integration constant C as a free parameter of the general solution. This can not be ignored but is used to satisfy initial conditions as usual. The function µ is referred to as an integrating factor for equation (2.28). To return now to the fact that the equation can be written as
L(x) = g . (2.38)
We know from the theory of linear algebra that the solution to a linear equation such as (2.38) is only defined up to an element of the kernel of L, that is a function xcf (t) which satisfies
L(xcf ) = 0 . (2.39)
The kernel of a linear operator is a vector space so the solutions to (2.39) form a vector space - for a first order ODE this will be a one-dimensional vector space. An equation such as (2.39) is known as a homogeneous linear differential equation, a solution xcf (t) of this equation is commonly known as a “complementary function”. The original equation (2.38) is known as an inhomogeneous linear differential equation, the term g on the right-hand-side is an called the inhomogeneous term and any solution xpi(t) of (2.38) is called a particular integral. Hence the result is that the general solution of (2.38) is
x(t) = xpi(t) + xcf (t) , where L(xpi) = g and L(xcf ) = 0 . (2.40)
This also means that the difference of any two solutions of the inhomogeneous equation (2.38) will be a solution of the homogeneous equation (2.39). We will return to these ideas in chapter 3 when we treat second order linear differential equations in more detail. Now for some examples of the solution of linear differential equations.
Example 2.4. Find the general solution of the ODE
dx dx (t + 2) (2t − 3) t(t + 1) − (t + 2)x = t3(2t − 3) ⇔ − x = t2 (2.41) dt dt t(t + 1) (t + 1)
We can see that this a first order linear ODE. The functions f(t) and g(t) can be read off from the equation as t + 2 (2t − 3) f = − , g = t2 . (2.42) t(t + 1) t + 1 The integrating factor is therefore
Z (t + 2) µ(t) = exp − dt (2.43) t(t + 1)
Version of Mar 14, 2019 Chapter 2: Differential Equations 18
Using (t + 2) 2(t + 1) − t 2 1 = = − (2.44) t(t + 1) t(t + 1) t t + 1 one obtains t + 1 µ(t) = exp (−2 ln t + ln(t + 1)) = , (2.45) t2 so Z (2t − 3) Z µx = µ(t) t2 dt = (2t − 3) dt = t2 − 3t + C (2.46) (t + 1) Hence t2 x(t) = (t2 − 3t + C) (2.47) t + 1
Example 2.5. Find the general solution of the differential equation dx x = , (2.48) dt t
This is clearly linear, and if we write it as dx x − = 0 , (2.49) dt t then we read off that f = 1/t and g = 0 and so the integrating factor is
Z dt 1 µ = exp − = exp(− ln(t)) = (2.50) t t and so the differential equation becomes
d(µx) C = 0 ⇒ µx = C, so that the general solution is x(t) = = C t . (2.51) dt µ This is the same differential equation as in example 2.2, the solution method is different but the solution is the same.
2.3.4 First order, homogeneous
The ODE is called homogeneous if it is of the form dx = F (x/t) (2.52) dt Homogeneous is being used in a different way to the previous section; that is just how it is. This equation can be solved by the substitution x(t) = tv(t), or v = x/t. We then have
dx d d dv 1 = (tv) = t v + v = F (v) , = (F (v) − v) , (2.53) dt dt dt dt t which is separable with the solution dv dt Z dv = ⇒ = ln |t| + C. (2.54) F (v) − v t F (v) − v
Version of Mar 14, 2019 Chapter 2: Differential Equations 19
After carrying out the v integration we then have a relation between x(t) and t which with luck we can simplify to find x(t) - but this is often not possible. The most common ODEs of this form we will encounter are when R(x/t) is a ratio of poly- nomials of the same degree, that is
P (t, x) F (x/t) = , (2.55) Q(t, x) where P,Q are homogeneous polynomials of (the same) degree m, i.e. P and Q are sums of terms of the form cxptm−p of the same total degree m.
Example 2.6. Consider the differential equation
dx (x2 − t2) = . (2.56) dt 2xt
The function on the right hand side is the ratio of homogeneous quadratic polynomials so we can divide top and bottom by t2 to get instead
dx ((x/t)2 − 1) (v2 − 1) = = . (2.57) dt 2(x/t) 2v where x = tv(t). We then obtain dv (v2 − 1) t + v = , (2.58) dt 2v and so dv (v2 − 1) 1 + v2 t = − v = − . (2.59) dt 2v 2v Separation of variables now gives Z 2v Z dt dv = − =⇒ ln(1 + v2) = − ln |t| + C, (2.60) 1 + v2 t where C is a parameter. Exponentiation now yields
|t|(1 + v2) = eC ⇔ t(1 + v2) = A, (2.61) where A = ±eC is a free parameter. It follows that
t(1 + x2/t2) = A ⇔ x2 + t2 = At . (2.62)
This equation may be re-written as
(t − A/2)2 + x2 = (A/2)2 , (2.63) from which we see that the solution curves are a family of circles in the x − t plane with centres at (0, A/2) and radii A/2. Notice that each of the circles passes through (0, 0), whatever the value of A, so the solution is not unique in the vicinity of (0, 0).
Example 2.7. Find the general solution of the differential equation dx x = , (2.64) dt t
Version of Mar 14, 2019 Chapter 2: Differential Equations 20
This is clearly homogeneous. With the substitution xv t we get dv t + v = v , (2.65) dt or simply dv = 0 , (2.66) dt with the general solution v = C ⇒ x = C t . (2.67) This is again the same differential equation as in example 2.2, the solution method is again different but the solution is the same. As in the previous example, all the solutions go through (0, 0).
2.4 Initial value problems
So far we have been concerned with finding the general solution x(t) = h(t, C) to an ODE. Finding a solution to an initial value problem usually just amounts to find the value(s) of C for which the general solution satisfies the condition, that is finding C such that
x(t0) = h(t0,C) = x0 . (2.68) For many of the examples we have looked at so far this is straightforward. Example 2.8. Solve the initial value problem dx = αx , x(t ) = x . (2.69) dt 0 0
This is the ODE in example 2.1 with general solution x = A exp(α t). Substituting the initial conditions we find the equation αt0 −αt0 A e = x0 ⇒ A = x0 e , (2.70) so that the solution to the problem is
α(t−t0) x(t) = x0 e . (2.71)
Example 2.9. Find the solution to the initial value problem dx t(t + 1) − (t + 2)x = t3(2t − 3) , x(t ) = x . (2.72) dt 0 0
This is the differential equation in example 2.4 which has the general solution t2 x(t) = (t2 − 3t + C) (2.73) t + 1 Substituting in the initial conditions we find
x0(t0 + 1) 2 C = 2 − t0 + 3t0 . (2.74) t0
However, we have already seen differential equations for which it is easy to state initial value problems which have either no solutions or an infinite number of solutions. One of the simplest is the ivp
Version of Mar 14, 2019 Chapter 2: Differential Equations 21
Example 2.10. Consider the initial value problem (2.23),
dx x = , x(t ) = x . (2.75) dt t 0 0
This differential equation was considered many times already and the general solution is (2.51), x(t) = C t . (2.76)
If we choose an initial time t0 6= 0 then the constant C is easily seen to be C = x0/t0 and the unique solution satisfying t x(t) = x0 . (2.77) t0
If by contrast we take t0 = 0 then x(0) = 0 for all the solutions (2.76) and so Proposition 2.1. The initial value problem dx x = , x(0) = x , (2.78) dt t 0 has no solutions if x0 6= 0 and has an infinite number of solutions if x0 = 0.
A second well-known case where an initial value problem has more than one solution is
Example 2.11. Consider the initial value problem dx = 3x2/3 , x(0) = 0 . (2.79) dt
This is a separable ODE and following the method in the notes gives 1 Z 1 Z x−2/3dx = dt ⇒ x−2/3dx = dt ⇒ x1/3 = t + C. (2.80) 3 3
With the initial condition that x(0) = 0 we C = 0 and the solution
x = t3 . (2.81)
There is, however, another different and obvious solution to the problem (2.79), which is
x(t) = 0 . (2.82)
This means that we have found two solutions to the i.v.p.,
1. x(t) = t3 (2.83) 2. x(t) = 0 (2.84)
The situation is worse, however, since we can also consider the combined solutions ( 0 t ≤ 0 3. x(t) = (2.85) t3 t ≥ 0
Version of Mar 14, 2019 Chapter 2: Differential Equations 22
( t3 t ≤ 0 4. x(t) = (2.86) 0 t ≥ 0
and in fact we can find an infinite set of solutions. For each positive number a then this is a solution ( 0 t ≤ a 5. x(t) = (2.87) (t − a)3 t ≥ a
and so is this ( (t + a)3 t ≤ −a 6. x(t) = (2.88) 0 t ≥ −a
These functions are all continuous and their derivatives exist and are also continuous. They certainly satisfy the differential equation and the initial conditions. Initial value problems for the ODE in example 2.6 can also have no solutions or an infinite number of solutions since, as in the case of proposition 2.1, all the solutions satisfy x(0) = 0. So we now have found several initial value problems which either have no solution or which have an infinite number of solutions. It is very helpful to have a test which enables us to determine if there is a solution and if it is unique. This is answered by Picard’s theorem which gives the conditions under which a solutions exists and is unique and is covered in the next section.
Version of Mar 14, 2019 Chapter 2: Differential Equations 23
2.5 Existence and Uniqueness of Solutions — Picard’s theorem
There are many theorems on the existence of solutions to an initial value problem and on whether it is unique or not. Some of these require a lot of Analysis to be able to state, so we will here look at one which is fairly easy to state, even if it is not the strongest:
Theorem 2.1 (Picard) Let A denote the set {(t, x): |t − t0| ≤ δ, |x − x0| ≤ δ} (2.89)
(a square centre the initial point (t0, x0) and side of length 2δ). Suppose that f : A → IR is 1 2 ∂f continuous and that the partial derivative fx = ∂x is also continuous on A. Then the initial value problem dx = f(t, x) , x(t ) = x , (2.90) dt 0 0
has a unique solution which is defined on the interval [t0 − δ1, t0 + δ1], for some δ1 ≤ δ. See figure 2.2.
A x0+δ
x0
x0 - δ
t0-δ t0-δ1 t0 t0+δ1 t0+δ
Figure 2.2: Sketch of the region A surrounding the point (t0, x0) on which f and ∂f/∂x are continuous and the smaller interval |t − t0| ≤ δ1 on which the solution exists and is unique. The ivp example illustrated is example 2.12 with t0 = 2.5, x0 = 2, δ = 1.5, δ1 = 0.3. The solution diverges at t = 2 which is outside the dotted interval.
1The definition of continuity of a function of two variables is in appendix A 2The definition of the partial derivative is also in appendix A
Version of Mar 14, 2019 Chapter 2: Differential Equations 24
This isn’t the “best” version of Picard’s Theorem, there will be initial value problems which do not satisfy the conditions of this theorem and still have unique solutions, but this version has the advantage that it can be stated in a form which is likely to make reasonable sense to first year students. We will not give a full proof, but indicate its heuristics (“how the theorem works”) based on Picard iterates, a practical procedure for constructing the solution. Note that we cannot assume that the solution exists for all t. A very simple example is this: Example 2.12. Consider the initial value problem dx = −x2 , x(t ) = x . (2.91) dt 0 0
Here the function f and its derivative are ∂f f(t, x) = −x2 (t, x) = −2x . (2.92) ∂x These are perfectly well behaved functions for all values of t and x and so we can take the constant δ to be as large as we like. The solution to this problem is easy to find as the ODE is separable and it is x x(t) = 0 . (2.93) 1 + x0(t − t0) This has a singularity at 1 t − t0 = − , (2.94) x0
and so the maximum value of δ1 is |1/x0|; the ivp satisfies the conditions of Picard’s theorem for any value of δ and so there is a unique solution around the point (t0, x0), but the maximum interval (of the sort in the theorem) for which it is defined must satisfy 1 |t − t0| ≤ δ1 < . (2.95) |x0| An illustration of how this works is shown in figure 2.2. Choosing the initial values to be t0 = 2.5, x0 = 2, we can easily take δ = 1.5 as f and fx are continuous for all x, t. The solution diverges for t = 2 and so the theorem just says there is some value of δ1 for which the function is defined on the interval [2 − δ1, 2 + δ1]; from the explicit solution we know that δ1 < 0.5, the figure shows the interval for δ1 = 0.3 on which the solution does exist and is unique.
2.5.1 Picard iterates
Assuming that eqs.(2.4), (2.9) have a solution we may integrate both sides with respect to t to obtain Z t dx Z t x(t) = x0 + du = x0 + du f(u, x(u)) (2.96) t0 du t0 Eq. (2.96) is fully equivalent to the pair (2.4), (2.9). Note that we cannot really evaluate the r.h.s. as we do not know x(u) for u 6= t0; in other words Eq. (2.96) is an integral equation. A constructive way of solving such an equation is by iteration
Version of Mar 14, 2019 Chapter 2: Differential Equations 25
• Start with the best guess at our disposal:
x(u) = x0(u) = x0 , t0 ≤ u ≤ t . (2.97)
• Then define a sequence of functions {xn(t)} by
Z t xn+1(t) = x0 + du f(u, xn(u)) , n = 0, 1, 2,.... (2.98) t0
To actually prove Picard’s theorem we would have to show that, under the stated conditions, the sequence of functions {xn(t)} converges (uniformly) to a limit function x(t), and then prove that x(t) is the unique solution of the differential equation.
Intuitively, convergence of the iteration rests on the observation that, for |t − t0| small, the contribution of the integral to the r.h.s. of (2.96) is small under the stated condition, because f(u, x(u)) is bounded for u ∈ [t0, t]. Because of this, a small (uniform) error
δn = max |xn(s) − x(s)| , (2.99) s∈[t0,t]
will be diminished under iteration. For, we have Z s δn+1 = max du f(u, xn(u)) − f(u, x(u)) s∈[t ,t] 0 t0 Z s ≤ max du f(u, xn(u)) − f(u, x(u)) s∈[t ,t] 0 t0 Z s = max du fx(u, x(u)) (xn(u)) − x(u)) + o(xn(u)) − x(u)) s∈[t ,t] 0 t0 ≤ C|t − t0| δn ,
where C is finite owing to the continuity of fx(u, x). (Inequalities above are due to triangle-inequalities for absolute values of sums; we have exploited existence and con-
tinuity of fx(u, x)). In conclusion we have: δn+1 < δn for sufficiently small |t − t0|. By the same reasoning one would show that there cannot be two different solutions,
say x and y of the ODE with identical initial conditions x(t0) = y(t0) = x0; if they
were different, so ∆ = maxs∈[t0,t] |y(s) − x(s)| > 0, and both solutions of (2.98), then by going through the same reasoning as for the error δn we would have to conclude ∆ < ∆, a contradiction — which finally proves the point.
Let’s see how the Picard method works in a specific case.
Example 2.13. Consider the differential equation dx = f(t, x) = x , x(0) = 1 . (2.100) dt Integrating both sides of the equation with respect to t we obtain
Z t Z t x(t) = 1 + du f(u, x(u)) = 1 + du x(u) . (2.101) 0 0
Version of Mar 14, 2019 Chapter 2: Differential Equations 26
Following the method described above we now define the Picard iterates by Z t xn+1(t) = 1 + du xn(u) , x0(t) = 1 . (2.102) 0 Setting n = 0 we obtain Z t x1(t) = 1 + du x0(u) = 1 + t . (2.103) 0 Substituting n = 1 we obtain Z t Z t t2 x2(t) = 1 + du x1(u) = 1 + du (1 + u) = 1 + t + . (2.104) 0 0 2! Continuing in this way, with n = 2, 3, ··· we find that t2 t3 tn x (t) = 1 + t + + + ··· + . (2.105) n 2! 3! n! P∞ tr t We see that as n → ∞ xn(t) → r=0 r! = e , which is indeed the unique solution of the given equation.
Note 1: If the conditions of Picard’s theorem are not satisfied by a particular differential equation, the equation may not have a solution or, if it does, the solution may not be unique; in some cases there may even be an infinite number of solutions, all satisfying the same initial condition, as we have seen. Let us look again at two of these examples, 2.10 (where the solution does not exist) and 2.79 (where the solution is not unique). Example 2.14. Consider the initial value problem (2.23), dx x = , x(0) = 1 . (2.106) dt t
Here f(t, x) = x/t. If we try to construct the Picard iterates, we start with x0(t) = 1 then try to define Z t Z t 1 x1(t) = x0 + f(u, x0) du = 1 + du , (2.107) 0 0 u and the integral diverges and x1(t) does not exist. The method of constructing the Picard iterates fails. We see that f(t, x) is not defined for t = 0 and there is no region A around the initial values (t0, x0) = (0, 1) in which f and fx are defined and continuous. The ivp does not satisfy the conditions of Picard’s theorem, there is no guarantee that a solution exists and indeed there is no solution. Example 2.15. Consider the initial value problem in example 2.11 dx = 3x2/3 , x(0) = 0 . (2.108) dt In this case we have ∂f 2 f(t, x) = 3x2/3 , = . (2.109) ∂x x1/3 f is a well behaved function for all x and t but fx is not defined for x = 0, and in particular is not defined in any region Z surrounding the initial values (t0, x0) = (0, 0). If we try to construct the Picard iterates, we will actually find the solution x(t) = 0 but the proof that this is unique fails.
Version of Mar 14, 2019 Chapter 2: Differential Equations 27
Note 2: Picard’s theorem is a local existence theorem; it says that Eqs. (2.4), (2.96) have a unique solution sufficiently near t0. It might be thought that if the function f which occurs in Eq. (2.4) is well behaved and defined over the whole of IR2 that we could extend our solution arbitrarily far from t0; for example, we might begin by computing x(t0 + δ1) and use this value of x as an initial value at t = t0 + δ1; application of Picard’s theorem then allows us to extend our solution as far as t0 + δ1 + δ2, for some δ2 > 0 and by repeating the process we could extend our solution to t = t0 + δ1 + δ2 + δ3 + ··· + δn, for some δi > 0 (i = 1, 2, 3, ··· n). However, this does not prove that the solution could be extended arbitrarily far from t0, as one might find that
n X tn = t0 + δk −→ t∞ < ∞ , as n → ∞ , (2.110) k=1 exactly as happens in example 2.12.
Version of Mar 14, 2019 Chapter 2: Differential Equations 28
2.6 Exercises
Shorter Exercises Exercise 2.1
Classify the following ODEs in terms of the classification given in class [(1) first order, explicit; (2) first order, separable (3) first order, linear; (4) first order, homogeneous; (5) none of these]. An equation may belong to more than one class. You are not asked to solve these equations; occasionally, you will need some reshuffling of terms to ‘see’ which class is involved
dx dy (a) = −10x + 4 (b) = sin(x) dt dx dx x2 dx (c) = 2 − (d) t − 3x = 2 dt t2 dt dN dx x3 + t3 (e) + N = Ntet+2 (f) = dt dt 2xt2 dy dx (g) = x5y − sin x (h) tx + 3x − 4 = 0 dx dt dy (i) y00 − 4y0 + 24y = 0 (j) + y ln x = e−x ln x dx
Exercise 2.2
Find the general solutions to the following differential equations:
dx (a) y0 = x2y2 (b) + x ln t = e−t ln t dt dy (c) x = 3 (d) y0 + xy = x dx dy x y (e) cos(x) y0 − sin(x) y = 4 (f) = + dx y x
Exercise 2.3
Find the solutions to the following initial value problems
dx x + 1 dy (a) = with x(0) = 1 (b) x + y = exp(x) with y(1) = 0 dt t + 1 dx dy (c) = 3x + 5 with y(0) = 2 (d)x ˙ = t2 with x(0) = 1 dx dy (e) y0 + y = exp(x) with y(1) = 0 (f) − y = y2 with y(0) = 1 dx
Version of Mar 14, 2019 Chapter 2: Differential Equations 29
Longer Exercises
Exercise 2.4
Find solutions of the initial value problem dx = x1/3 sin(2t), x(0) = 0 dt other than the trivial solution x(t) = 0 for all t. Comment on your results in the light of Picard’s theorem.
Exercise 2.5 (taken from the 2008 exam)
Suppose that a population doubles its original size in 100 years, and triples it in 200 years. Show that this population cannot satisfy the Malthusian law of population growth which is
dp(t) = a p(t) with a = constant dt
Exercise 2.6
Find the general solution to: dx x t = + dt 2t 2x Show that this defines a family of 2-sided hyperbolae in the (t, x) plane, symmetric about the point (C, 0), where C is a parameter, one side always going through the point (0, 0). Discuss the solutions in the vicinity of the point (0, 0).
Version of Mar 14, 2019 Chapter 2: Differential Equations 30
Supplementary Exercises
Supplementary Exercise 2.1 We have looked at differential equations where y0(x) is independent of x, y0(x) = f(x) , or a linear function of x y0(x) = f(x) + g(x)y(x) . The next step would be quadratic functions of y, y0(x) = f(x) + g(x)y(x) + h(x)y(x)2 . This called the Riccati equation and cannot be solved in general. There are certain circum- stances in which it can, though, and one is when the “constant term” f(x) is zero; this is called the Bernoulli equation: y0(x) = g(x)y(x) + h(x)y(x)2 . Make the substitution u(x) = 1/y(x) and show that the resulting equation for u0(x) is linear. Use this method to solve the initial value problem y0(x) = y(x) + y2(x) , y(0) = 1 . Note: this equation can also be solved directly as the equation is separable [problem 2.3(f)]
Supplementary Exercise 2.2 Suppose that we know that there is a relation between y(x) and x, which we can write as M(x, y(x)) = 0 . (2.111) We can differentiate this equation with respect to x using the chain rule for partial differen- tiation (see appendix A) to get ∂M ∂M + y0(x) = 0 . ∂x ∂y This can also be written as ∂M ∂M dx + dy = 0 , ∂x ∂y in which case it is known as an “exact” differential equation. Show that the differential equation f(x, y) + g(x, y) y0(x) = 0 comes from a relation of the form (2.111) if ∂f ∂g = . ∂y ∂x The function M can then be found by the fundamental theorem of calculus applied to path integrals in two dimensions (see chapter 6 in the lecture notes for Calculus II), or by a suitable guess for M. Use this method to solve the initial value problem (y + 2x) + (2y + x)y0 = 0 , y(0) = 1 .
Version of Mar 14, 2019 Chapter 3: Second order differential Equations 31
Chapter 3
Second order Differential Equations
3.1 Second order differential equations
Second order ordinary differential equations have the form
d2x dx = g t, x, . (3.1) dt2 dt
The general solution of a second order differential equation is of the form x(t) = h(t, C1,C2), involving two parameters C1,C2. These can be fixed in innumerable ways, for example for some problems it would be natural to ask that x takes fixed values at two different times,
x(t0) = x0 and x(t1) = x1 , (3.2)
but such conditions are very hard to analyse. Much simpler, and also very relevant in many problems, are initial conditions, asking that x andx ˙ take particular values at some initial time t0, x(t0) = x0 andx ˙(t0) = v0 . (3.3) We will call the equation (3.1) and the conditions (3.3) an initial value problem for a second 3 order differential equation. The initial conditions define a point (t0, x0, v0) ∈ IR ; it is assumed that g is defined in some neighbourhood of this point, for example in the cube
A = {(t, x, v): |t − t0| ≤ δ, |x − x0| ≤ δ, |v − v0| ≤ δ} , for some δ > 0 (3.4)
centred at (t0, x0, v0) and with side-length 2δ.
A second order ODE can always be rewritten in terms of a system of two first order ODEs. This is achieved by setting x1(t) = x(t), x2(t) =x ˙(t) and equation (3.1) then becomes the two coupled equations dx dx 1 = = x , dt dt 2 dx d2x dx 2 = = g t, x, = g(t, x , x ) . dt d t2 dt 1 2
In terms of x1, x2 the initial conditions become x1(t0) = x0, x2(t0) = v0. We may therefore replace the second order equation (3.1) and their initial conditions by the two coupled first
Version of Mar 14, 2019 Chapter 3: Second order differential Equations 32 order differential equations and initial conditions dx 1 = x , ( dt 2 x1(t0) = x0 , dx , (3.5) 2 = g(t, x , x ) x2(t0) = v0 . dt 1 2 Equations (3.5) together define an initival value problem for the two coupled first order differential equations. The use of the word coupled reflects the fact that the equations forx ˙ 1 andx ˙ 2 do not just depends on x1 and x2 separately. Note: The above procedure to rewrite a second order ODE in terms of two ODEs of first order can be generalised to n-th order ODEs of the form dnx dkx = g t, x, x(1), x(2) . . . , x(n−1) , with x(k) ≡ , dtn dtk and with initial conditions of the form
(1) (2) (n−1) x(t0) = a1 , x (t0) = a2 , x (t0) = a3 , . . . , x (t0) = an .
(1) (n−1) Introducing x1 = x, x2 = x , . . . , xn = x , one obtains the equivalent system of first order equations d x = x dt 1 2 d x = x dt 2 3 . . . . d x = g(t, x , x , . . . , x ) dt n 1 2 n
with initial conditions xk(t0) = ak, for k = 1, . . . , n.
Suppose that by some means we have found a solution of equations (3.5), x1(t) = φ(t), x2(t) = ψ(t), satisfying the initial conditions, so that φ(t0) = x0, ψ(t0) = v0. If we plot the points 2 (φ(t), ψ(t)) in IR we obtain a solution curve which passes through the point (x0, v0) defined by the initial conditions; as we vary (x0, v0) we generate a family of solution curves, each labelled by the two parameters x0 and v0. Second order differential equations (3.1) and systems of two coupled first order differential equations (3.5) are typically much less likely to be solvable than first order differential equa- tions. We have seen that any linear first order ODE can be solved [just by using integration] by the method of integrating factors: this is not true for second order linear differential equa- tions. The study of such equations could be a module in its own right, and there are books written on special classes of equations. We will just look at one special class which can be solved: second order linear ODEs with constant coefficients
3.1.1 Second order linear, with constant coefficients
These are ODEs of the form d2x dx a + b + cx = φ(t) , (3.6) dt2 dt
Version of Mar 14, 2019 Chapter 3: Second order differential Equations 33 where a, b, c are constants and φ is a given function of time. Just as with first order linear equations, these can be written in terms of a linear differential operator L: d2 d L(x) = φ , L = a + b + c . (3.7) dt2 dt
Homogeneous equations: We will first consider the special case of the homogeneous equation, d2x dx a + b + cx = 0 L(x) = 0 . (3.8) dt2 dt Linear and homogeneous ODEs have the important properties that (i) a solution x(t) can be multiplied with an arbitrary constant, and remains a solution; (ii) for any pair x1 and x2 of independent solutions (which are not proportional to each other), any linear combination of the solutions x1 and x2 with arbitrary constant coefficients is also a solution of 3.8):
L(x1) = 0 and L(x2) = 0 ⇒ L(αx1 + βx2) = 0 for any constants α, β . (3.9) This means that the solutions of the homogeneous equation form a vector space, the kernel of the operator L. For a second order linear ODE this is a two-dimensional vector space.
Inhomogeneous equations: The general case is L(x) = φ . (3.10)
If we have two solutions x1 and x2 of the inhomogeneous equation then their difference satisfies the homogeneous equation:
L(x1) = φ and L(x2) = φ ⇒ L(x1 − x2) = L(x1) − L(x2) = φ − φ = 0 . (3.11)
Conversely, if we know one solution xpi to the inhomogeneous equation then the combina- tion xpi + xcf is also a solution to the inhomogeneous equation for any solution xcf to the homogeneous equation:
L(xpi) = φ and L(xcf ) = 0 ⇒ L(xpi + xcf ) = L(xpi) + L(xcf ) = φ + 0 = φ . (3.12)
Solving a second order linear ODE with constant coefficients: One can always solve the ODEs (3.6) and (3.8) by simple integration. This is definitely not the case of linear second order ODEs where the coefficients are not constant, it is a very special feature. There are several methods one can use to find the solutions: they all boil down to the same in the end, they just organise the steps differently. One way to explain why the fact that the coefficients are constants means the equation can be solved is that it lets one factorise the operator L into the product of two first order linear differential operators. It is simpler if we choose to normalise (3.6) by dividing by a so that we are considering the operator d2 d L = + α + β . (3.13) dt2 dt
Version of Mar 14, 2019 Chapter 3: Second order differential Equations 34
We can then write this as a product
d d d2 d L = − λ − λ = − (λ + λ ) + λ λ , (3.14) dt 1 dt 2 dt2 1 2 dt 1 2 where λ1 and λ2 satisfy λ1 + λ2 = −α , λ1λ2 = β , (3.15) which are equivalent to the equation
λ2 + αλ + β = 0 . (3.16)
This is call the auxiliary or characteristic equation. This helps us because we have now turned our second order equation into two first order linear equations which we can solve by using the method of integrating factors (twice).
As the first step, we write L2(x) = u and so the original equation becomes d L(x) = L (L (x)) = L (u) = − λ u = φ , (3.17) 1 2 1 dt 1 for which we know the solution: Z t λ1t −λ1u u = e e φ(u)du + C1 . (3.18)
Now, using the expression for u we then have
d Z t u = L (x) = − λ x = eλ1t e−λ1uφ(u)du + C . (3.19) 2 dt 2 1 This is again a first order linear ODE which we again solve by the method of integrating factors to find Z t Z u λ2t −λ2u λ1u −λ1v x = e e e e φ(v)dv + C1 du + C2 . (3.20)
As expected, this solution depends on two arbitrary constants. Pulling out these two arbitrary constants, we get the general solution as
Z t Z u 1 λ2t −λ2u λ1u −λ1v λ1t λ2t x = e e e e φ(v)dv du + C1e + C2e , (3.21) λ1 − λ2
provided λ1 6= λ2. Note that we have not said which of the two solutions of the characteristic equation is which: it will not affect the space of solutions to the problem.
If λ1 = λ2 then instead we get Z t Z u λt −λv λt λt x = e e φ(v)dv du + C1 t e + C2e . (3.22)
The functions appearing with the arbitrary constants are the solutions of the homogeneous equation. There is a simper way to find these as follows:
Version of Mar 14, 2019 Chapter 3: Second order differential Equations 35
Solving the homogeneous equation: simpler method The homogeneous equation is solved with the help of solutions of the so-called auxiliary polynomial, or auxiliary equation. This method is based on the observation that Eq. (3.8) is solved by functions of the form x(t) = eλt (3.23) provided λ is properly chosen. The condition on λ is obtained by inserting the exponential ansatz into (3.8). Using dx d2x = λ eλt and = λ2 eλt (3.24) dt dt2 one gets aλ2 + bλ + ceλt = 0 (3.25) Since eλt 6= 0, we see that eλt is a solution of Eq. (3.8) provided P (λ) = aλ2 + bλ + c = 0. (3.26) Eq. (3.26) is referred to as the auxiliary equation. It is a quadratic equation in λ and as such there are three possibilities:
λ t λ t (i) Equ. (3.26) has two distinct real solutions λ1, λ2. It follows that x1 = e 1 and x2 = e 2 are independent solutions of the given differential equation. The general solution of Eq. (3.8) is now obtained by forming an arbitrary linear combination of these two solutions: it is x(t) = αeλ1t + βeλ2t, (3.27) where α, β are parameters. One can also show that these combinations exhaust the set of possibilities (i.e. there are no solutions of a form other than (3.27)).
(ii) The auxiliary equation has two complex conjugate roots λ± = µ ± iν. This means that (µ±iν)t x±(t) = e are solutions of Eq. (3.8), as are all linear combinations thereof. Now (µ±iν)t µt x±(t) = e = e (cos νt ± i sin νt). Using special linear combinations of these, we conclude that x (t) + x (t) x (t) − x (t) x (t) = + − = eµt cos νt and x (t) = + − = eµt sin νt (3.28) 1 2 2 2i are two (independent) solutions of Eq. (3.8). Forming linear combinations of these, we see that the general solution of Eq. (3.8) is given by x(t) = eµt(α cos νt + β sin νt) (3.29) in this case; here α, β are parameters.
(iii) The third possibility is that Eq. (3.26) has two coincident real roots, λ1,2 = λ, say. In this case we get initially only a single solution
λt x1(t) = α e (3.30)
Version of Mar 14, 2019 Chapter 3: Second order differential Equations 36 where α is an arbitrary constant. An independent solution is in this case obtained by the method of varying constants. That is one tries to find an independent solution of the form λt x2(t) = F (t)e , in which the constant α above is replaced by an as yet unknown function F . The function F is determined by inserting this ansatz into (3.8), which gives h i aF¨ + (2aλ + b)F˙ + (aλ2 + bλ + c)F eλt = 0 (3.31)
As the coefficient of F in this equation is P (λ) and thus vanishes, and the coefficient of F˙ vanishes because λ = −b/2a was assumed to be the unique solution of P (λ) = 0, we are left with the condition F¨ = 0 ⇒ F (t) = α + β t . (3.32) The general solution of F¨ = 0 was obtained by two explicit integrations involving two inte- gration constants α and β. Summarising: the general solution of (3.8) in this case is again an arbitrary linear combination of the two independent solutions just found, and can be written as x(t) = (α + βt) eλt (3.33) with arbitrary parameters α and β.
Solving the Inhomogeneous equations: method of varying constants We now consider the inhomogeneous Eq. (3.6) with φ(t) 6= 0.
The first thing to note is the following: Suppose x0(t) is any particular solution of the inho- mogeneous equation. Then a solution of the form
x(t) = x0(t) + xh(t) , (3.34)
with xh(t) a general solution of the corresponding homogeneous equation is also a solution of the inhomogeneous equation, and this exhausts the possibilities, i.e. there is no solution of the inhomogeneous equation which is not of this form. We already know how to find xh(t). So it is sufficient to find a special solution of the inhomogeneous equation. A particular solution of an inhomogeneous equation can be found by the method of varying λt constants from a solution x1(t) = α e of the homogeneous equation. Thus we attempt a λt solution x0(t) of the inhomogeneous equation of the form x0(t) = F (t)e with an unknown function F to be determined by inserting the ansatz into (3.6). Following the reasoning in case (iii) above, this gives h i aF¨ + (2aλ + b)F˙ + (aλ2 + bλ + c)F eλt = φ(t) (3.35)
Once more the coefficient of F in this equation , being P (λ), vanishes. So we get aF¨ + (2aλ + b)F˙ = φ(t) e−λt . (3.36) This ODE for F does only involve first and second order derivatives of F and not the function F itself. This can be used to reduce the order of the ODE. Setting ψ(t) = F˙ (t), one obtains aψ˙ + (2aλ + b)ψ = φ(t) e−λt . (3.37) This is a first order linear equation and we know how to solve it using integrating factors (see Sec 2.4.3 above). Once ψ(t) is obtained along those lines, F (t) follows by a further integration λt w.r.t. time t, thus finally allowing to write down x0(t) = F (t)e .
Version of Mar 14, 2019 Chapter 3: Second order differential Equations 37
Example 3.1. Find a particular solution of the differential equation d2x − 9x = t, (3.38) dt2
The characteristic equation is λ2 −9 = 0 so that λ = ±3 and the solution to the homogeneous equation is x = ae3t + be−3t . (3.39) We can take as a starting point for the method of varying constants the function x = F (t)e3t , (3.40) so that x˙ = (F˙ + 3F )e3t , x¨ = (F¨ + 6F˙ + 9F )e3t , (3.41) and the equations for F and for ψ = F˙ become x¨ − 9x = (F¨ + 6F˙ )e3t = t , F¨ + 6F˙ = te−3t , ψ˙ + 6ψ = te−3t . (3.42) Using the integrating factor exp(6t), this becomes d (e6tψ) = te3t , (3.43) dt which can be integrated directly to give Z t 1 t 1 e6tψ = te3tdt = ( − )e3t + C , ψ = ( − )e−3t + Ce−6t , (3.44) 3 9 3 9 which can again be integrated to give Z t t F = ψdt = − e−3t + C0e−6t + D , x = F e3t = − + C0e−3t + De3t . (3.45) 9 9 This method will always work, but quite often one can be lucky enough to find a special solution faster by using inspired guesswork.
Solving the Inhomogeneous equations: guessing The method of guessing is perfectly valid in mathematics: if we try to guess a solution and get it right, then we have solved the problem. A guess is often called a “trial function” or an “ansatz” (plural “ans¨atze”)to make it sound better This is often the fastest way. If the inhomogeneous term involves trigonometric functions, we can try a combination of trigonometric functions; if it is a polynomial, we can try a polynomial. Example 3.2. Find a particular solution of the differential equation d2x − 9x = t, (3.46) dt2 Since the inhomogeneous term is a polynomial, we could guess x = a + bt + ct2 , (3.47) which gives x¨ − 9x = (−9c)t2 + (−9b)t + (−9a + 2c) . (3.48) The solution is clearly a = 0, b = −1/9, c = 0 or 1 x(t) = − t . (3.49) 9
Version of Mar 14, 2019 Chapter 3: Second order differential Equations 38
Example 3.3. Find a particular solution of the differential equation d2x + x = sin(t), (3.50) dt2
Since the inhomogeneous term is a trigonometric function, we could guess
x = a sin(t) + b cos(t) , (3.51)
but this just gives x¨ + x = 0 . (3.52) We can then try a polynomial times a trigonometric function,
x = (at + bt2) sin(t) + (ct + dt2) cos(t) , (3.53)
which gives x˙ = (c + (a + 2d)t + bt2) cos(t) + (a + (2b − c)t − dt2) sin(t) , (3.54) x¨ = (2b − 2c + (−a − 4d)t − bt2) sin(t) + (2a + 2d + (4b − c)t − dt2) cos(t) , (3.55) so that x¨ + x = (2b − 2c − 4dt) sin t + (2a + 2d + 4bt) cos t , (3.56) so that the solution is a = b = d = 0, c = −1/2, 1 x = − t cos(t) . (3.57) 2
3.2 Existence and Uniqueness of Solutions — Picard’s theorem
Picard’s theorem for this system of equations now takes the form:
Picard’s theorem Let A denote the cube
{(t, x1, x2): |t − t0| ≤ δ, |x1 − a1| ≤ δ, |x2 − a2| ≤ δ, for some δ > 0}
and suppose that f1, f2 : A → IR are continuous. Suppose also that the partial derivatives w.r.t. the second and third arguments, ∂fi and ∂fi for i = 1, 2 are continuous on A. Then ∂x1 ∂x2 the system of differential equations dx dx 1 = f (t, x , x ) , 2 = f (t, x , x ) , x (t ) = a , x (t ) = a dt 1 1 2 dt 2 1 2 1 0 1 2 0 2
has a unique solution x1 = φ(t), x2 = ψ(t), t ∈ [t0 − δ1, t0 + δ1], for some δ1 ≤ δ.
No full proof of this result will be given here; the underlying ideas are however, the same as in the case of first order equations, relying on convergence of an iterative solution of a system of integral equations which together are equivalent to the system of ODEs of first order with and their initial conditions. They are obtained by integration of Eqs (3.5) with respect to t, which gives Z t xi(t) = ai + du fi(u, x1(u), x2(u)) , (i = 1, 2) (3.58) t0
Version of Mar 14, 2019 Chapter 3: Second order differential Equations 39
These equations are solved by means of Picard iteration in full analogy to the first order case.
Define xi,n(t), for i = 1, 2 and n = 0, 1, 2, 3,... by
xi,0(t) = ai, (3.59) Z t xi,n+1(t) = ai + du fi(u, x1,n(t), x2,n(t)) , (n = 0, 1, 2, 3,... ) (3.60) t0
Subject to the stated conditions one can then prove that the sequences of functions {x1,n(t)}, {x2,n(t)} converge to the unique solutions x1 = φ(t), x2 = ψ(t) as n → ∞, provided that δ is sufficiently small. Let’s see by an example how this works out in practice.
Example 3.4. Simple Harmonic Motion: The equation describing simple harmonic motion given by
d2x + x = 0 dt2 with initial conditions dx x(0) = 0 , (0) = 1 . dt
dx As above put x1(t) = x(t), x2(t) = dt so that dx dx 1 = x , 2 = −x , dt 2 dt 1 with x1(0) = 0 and x2(0) = 1. The corresponding integral equations for x1 and x2 are Z t Z t x1(t) = du x2(u) , x2(t) = 1 − du x1(u) . 0 0 The Picard iterates are defined by
Z t x1,n+1(t) = du x2,n(u) , x1,0(t) = 0 , 0 Z t x2,n+1(t) = 1 − du x1,n(u) , x2,0(t) = 1 . 0 Setting n = 0, 1, 2, 3,... in turn we find that
Z t Z t x1,1(t) = du x2(0, u) = du = t , 0 0 Z t Z t x2,1(t) = 1 − du x1(0, u) = 1 − du 0 = 1 , 0 0 Z t Z t x1,2(t) = du x2,1(u) = du 1 = t , 0 0 Z t Z t t2 x2,2(t) = 1 − du x1,1(u) = 1 − du u = 1 − , 0 0 2! Z t Z t u2 t3 x1,3(t) = du x2,2(u) = du 1 − = t − , 0 0 2! 3! Z t Z t t2 x2,3(t) = 1 − du x1,2(u) = 1 − du u = 1 − , 0 0 2!
Version of Mar 14, 2019 Chapter 3: Second order differential Equations 40
Z t Z t u2 t3 x1,4(t) = du x2,3(u) = du 1 − = t − , 0 0 2! 3! Z t Z t u3 t2 t4 x2,4(t) = 1 − du x1,3(u) = 1 − du u − = 1 − + , 0 0 3! 2! 4! Z t Z t u2 u4 t3 t5 x1,5(t) = du x2,4(u) = du 1 − + = t − + , 0 0 2! 4! 3! 5! and so on. The pattern is clear, and we see the Maclaurin series for sin(t) appearing for x1,n(t) as n → ∞, so t3 t5 t7 t9 x (t) → x(t) = t − + − + − · · · = sin(t) , 1,n 3! 5! 7! 9! which is the solution of the differential equation satisfying the given initial conditions.
Version of Mar 14, 2019 Chapter 3: Second order differential Equations 41
3.3 Exercises
Shorter Exercises Exercise 3.1 Find the general solutions to the following differential equations
(a) y00 + 3y0 + 2y = 0 (b) y00 + y0 + y = 0 (c) y00 + y = exp(x) (d) y00 − y = exp(−x)
Exercise 3.2 Find the solutions to the following initial value problems
(a) y00 + y0 + y = 0 with y(0) = 0, y0(0) = 1 (b) y00 − 3y0 + 2y = x with y(0) = 0, y0(0) = 0 (c) y00 − 2y0 + y = x with y(0) = 0, y0(0) = 0 (d) y00 + 4y = sin(x) with y(0) = 0, y0(0) = 0
Longer Exercises Exercise 3.3 (taken from the 2008 exam) Three solutions of a certain 2-nd order inhomogeneous linear differential equation, with con- stant coefficients are
t t ψ1(t) = t ψ2(t) = t + e ψ3(t) = 1 + t + e . Find the general solution of the inhomogeneous differential equation. Exercise 3.4 The motion of an idealised pendulum is described by y¨ + 2y = 0 , in which y denotes the angle of deviation from the vertical orientation. Solve this differential equation, with initial conditions given by π y(0) = , y˙(0) = 0 . 12 Suppose there is now friction which dampens the motion, so that it is described by y¨ + 2y = −2y ˙ . Solve this differential equation with the same initial conditions.
Version of Mar 14, 2019 Chapter 3: Second order differential Equations 42
Exercise 3.5 The distance d(t) that a parachutist falls satisfies the equation
d¨= g − αd˙ , where g > 0 is the acceleration due to gravity and α > 0 represents the effects of air resistance due to the parachute. Solve this equation subject to the initial conditions
d(0) = 0 , d˙(0) = 0 .
You may find it easier to first solve for the speed, v(t) = d˙(t) which satisfies
v˙ = g − αv .
What is the limiting speed (terminal velocity) of the parachutist? How could you predict this without solving the equation?
Exercise 3.6 Consider the initial value problem
x¨ + 3x ˙ + 2x = 1 + t , x(0) = 0 , x˙(0) = 0 .
Solve this problem using the method of varying constants with trial functions
(i) x(t) = F (t)e−t , (ii) x(t) = F (t)e−2t .
Check that both methods give the same solution.
Version of Mar 14, 2019 Chapter 3: Second order differential Equations 43
Supplementary Exercises
Supplementary Exercise 3.1 If y1 and y2 are functions, then we can define a combination W (x) called the Wronskian as the determinant y1 y2 0 0 W = det 0 0 = y1y2 − y2y1 (3.61) y1 y2 Suppose that y1 and y2 both satisfy the differential equation y00(x) + a(x)y0(x) + b(x)y(x) = 0 , (3.62) (a) Show that dW (x) + a(x)W (x) = 0 . (3.63) dx This means that W (x) can be found very easily by the method of integrating factors as Z W (x) = exp(− a(x)dx) . (3.64)
Suppose that Y (x) is known to be a solution of (3.62) (b) show that any other solution satisfies Y 0 W y0 − y = . (3.65) Y Y Consider the ODE 1 1 y00 + −1 − y0 + y = 0 (3.66) x x (c) Show that exp(x) is a solution of equation (3.66). (d) Use the Wronskian method to find a second solution of (3.66). Supplementary Exercise 3.2 The Riccati equation from problem 2.1 cannot always be solved, but it can always be turned into a second order linear differential equation which may make its analysis easier. Recall that the general Riccati equation is y0(x) = f(x) + g(x)y(x) + h(x)y(x)2 . (a) Show that the substitution u0 y = − , hu turns the non-linear first order equation for y(x) into a second order linear differential equa- tion for u(x). Any solution u(x) of this second order linear differential equation will give a solution y(x) of the non-linear Riccati equation. (b) Apply this method to find the second order linear differential equation for u correspond- ing to the Riccati equation y0 = y + y2 , in problem 2.1 and show that it leads to the same solution. (c) How can it be that the two parameter families of solutions to the second order equation for u leads to a one parameter family of solutions for y?
Version of Mar 14, 2019 Part II
Dynamical Systems
44 Chapter 4: Introduction to Dynamical Systems 45
Chapter 4
Introduction to Dynamical Systems
First we give a definition of a dynamical system as we will consider it in this course.
Definition 4.1. Dynamical System (DS)
A DS is any system described by a set of variables {x1, . . . , xn} which depend on time t and satisfy a set of ordinary differential equations (ODEs) of the form
d x = f (t, x , . . . , x ) , i = 1, . . . , n . (4.1) dt i i 1 n
The functions fi are called velocity functions. If we write the variables as a vector, then its velocity f is given by the velocity functions, x1 x˙ 1 f1 x2 d x˙ 2 f2 x = , x = f(t, x) = = . (4.2) . . . . dt . . xn x˙ n fn
If the variables x take a value x(t1) = a at time t1 then we say that the system is in the state a. The possible allowed values for the variables x is called the phase space
Definition 4.2. Phase Space Γ The phase space Γ is defined as the set of admissible values for x. E.g., if x = (x) describes some population density, or the concentration of a chemical substance, then it cannot be negative and so Γ = IR+ = {x ∈ IR|x ≥ 0}. If x = (x, y) describe the percentages of a population of people who have either never caught a disease or who have recovered from it, then each of these must be between 0 and 100 and together cannot add up to more than 100 so that the phase space is the triangular set Γ = {(x, y)|0 ≤ 0 ≤ 100, 0 ≤ y ≤ 100, x + y ≤ 100}. We could even say that x is an angle on a circle so that the phase space is Γ = {x|x ≡ x+2π} where the angle x is identified with the angle x + 2π.
Version of Mar 14, 2019 Chapter 4: Introduction to Dynamical Systems 46
We next define the order of a DS and whether it is autonomous or not autonomous:
Definition 4.3. Order If the description involves n variables, the dynamical system is said to be of order n.
Definition 4.4. Autonomous and Non-Autonomous Systems
If the time t does not explicitly appear in the velocity functions fi, the system is said to be autonomous, otherwise it is non-autonomous. Thus for autonomous systems, we have d x = f(x) (4.3) dt instead of (4.2). (Physically, an autonomous system is a system in which the laws governing it are time-independent, or one where there are no time-dependent ‘external’ perturbations of the dynamics). .
An n-th order non-autonomous system is equivalent to an (n+1)-the order autonomous system if we just introduce an extra variable x0 which is equal to t, d x = t , x = 1 , (4.4) 0 dt 0 so that (4.1) becomes
d x = f (x , x , . . . , x ) , i = 0, . . . , n . (4.5) dt i i 0 1 n
with f0(x0, x1, . . . , xn) = 1. Since we can always turn a non-autonomous system into an autonomous system (of higher order) we will almost always deal simply with autonomous systems in this course. We treat first order autonomous DS in chapter 5 and second order autonomous DS in chapter 6. First order systems are very simple, second order systems are more complicated but their behaviour can still be characterised fairly easily; Third order systems can be much more complicated. There are two useful ways to think about the way a system evolves in time. For example, if we want to think how someone moves around in two-dimensions (e.g. walking around London), we can plot a graph of their position against time. Alternatively, we can just trace on a map where they went and in which order. The first option keeps all the information there is, including the speed they are going at; second option is easier to draw (it is a line on a map) and sometimes that is all we want to know about. This leads to the following two sets of definitions:
Version of Mar 14, 2019 Chapter 4: Introduction to Dynamical Systems 47
Definition 4.5. Trajectories, Flow
The set of points {(t, x(t)); t ∈ IR} which solve the ODEs (4.2), and for which x(t0) = x0 is called the trajectory or solution curve of the DS passing through x0. The set of all trajectories obtained by varying t0 and x0 through all physically allowed values is called the flow of the DS.
Definition 4.6. Orbits, Phase-flow, Phase portrait
The set of points {x(t); t ∈ IR} which solve the ODEs (4.2), and for which x(t0) = x0 is called the orbit of the DS passing through x0. We think of the DS moving along an orbit. The set of all orbits obtained by varying t0 and x0 through all physically allowed values is called the phase-flow of the DS. A picture which gives an illustrative selection of the orbits and the direction of motion is called a phase portrait
Note the difference between phase-flow and flow; the flow contains more dynamical informa- tion than the phase-flow, but the phase-flow is easier to visualise as it is one dimension smaller and we will often use phase portraits to understand the behaviour of dynamical systems. To give an example, we can think about the DS with the velocity functions y x x˙ = − , y˙ = , (4.6) x2 + y2 x2 + y2 which describe circular motion where the speed depends on the distance from the origin: the further out, the slower the motion. The flow is the collection of trajectories (t, x(t)) which are in IR3; the phase-flow is the collection of orbits which are in IR2; these are shown in figure 4.1.
Phase-Flow
1.0
0.5
0.0
-0.5
-1.0
-1.0 -0.5 0.0 0.5 1.0
Figure 4.1: Sketches of the flow (in IR3) and a phase portrait illustrating the phase-flow (in IR2) for the simple 2nd order autonomous system (4.6).
Version of Mar 14, 2019 Chapter 4: Introduction to Dynamical Systems 48
The phase portrait of the phase-flow shows directly that the orbits are circles, the flow shows that the motion nearer the centre is faster but makes it harder to see that the orbits are really circles. This example also shows why orbits get their name: we think of planets moving along orbits, which are curves in space. It is obvious that if the system starts at some point on a circle then it will stay in the same circle for all time. This is an example of an invariant set.
Definition 4.7. Invariant Set An invariant set S is a set such that if x(t0) ∈ S then x(t) ∈ S for all t
Particularly important examples of invariant sets are when the set is just a single point, in which case it is called a fixed point or an equilibrium point
Definition 4.8. Fixed Point (FP) or Equilibrium point t A point a = (a1, . . . an) is called a fixed point of the (autonomous) DS if d f (a) = 0 ⇔ x = 0 at x = a , i = 1, . . . , n . i dt i A system which is at a FP will stay there forever, unless perturbed.
It is also very important to know whether a fixed point is stable or not: will motion that starts near a fixed point stay near the fixed point for all time or can it move away?
Definition 4.9. Stability of a FP
• A FP a of a DS is called strongly stable, if all trajectories starting (sufficiently) close to a will approach a under the dynamics.
• A FP a of a DS is called unstable, if there are trajectories starting close to a which evolve away from a under the dynamics. (Note that in n-th order DS, mixed situations exist, i.e. some subset of the trajectories starting sufficiently close to a will approach a, whereas there exist others which will evolve away from a, no matter how close to a they start.
• A FP a of a DS is called (marginally) stable, if all trajectories starting (sufficiently) close to a will neither approach a under the dynamics, nor will they evolve away from it, but rather ‘keep circling around’.
Fig. 4.2 illustrates the definitions for 2nd order autonomous systems:
Version of Mar 14, 2019 Chapter 4: Introduction to Dynamical Systems 49
Figure 4.2: Illustration of the possible flows in the vicinity of a FP. From left to right: strongly stable, unstable, unstable (mixed), marginally stable.
We can give a formal definition of a stable fixed point and a strongly stable fixed point in terms of and δ but will rarely if ever use this: Definition 4.10. Stability of a FP0 A FP a of a DS is called stable, if for any > 0 we can find a δ > 0 such that ||x(0) − a|| < δ implies that ||x(t) − a|| < for all time. i.e. the motion never moves away from the FP. A FP a of a DS is called strongly stable, if for any δ > > 0 we can find a time T such that ||x(0) − a|| < δ implies that ||x(t) − a|| < for all times t > T , i.e. the motion must move in towards the FP
These two ideas are captured in figure 4.3
Figure 4.3: Illustration of the formal definitions of stability and strong stability. On the left, stability means that whatever larger circle we define, there is a smaller circle such that every flow starting inside the smaller circle stay insides the large circle for all time. On the right, strong stability means that whatever smaller and larger circles we define, there is a time after which every motion starting inside the larger circle remains inside the smaller circle.
We now turn to 1st order autonomous DS and consider how to characterise their dynamics, find the fixed points and determine their stability and decide if a flow carries on for all time or if it terminates after a finite time.
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 50
Chapter 5
First Order Autonomous Systems
5.1 Trajectories, orbits and phase portraits
In this chapter we look in detail at first order autonomous dynamical systems. The state at time t of a first order autonomous dynamical system is described by one variable x(t) (which we will assume is in IR unless we say otherwise) and which satisfies a differential equation of the form dx = f(x). (5.1) dt The velocity function f(x) does not depend on t because the system is autonomous. This means that the differential equation (5.1) is separable and so we can immediately write down the solution as an integral:
Z x dx Z t F (x) = = dt = t − t0 . (5.2) x0 f(x) t0 The trajectories of the dynamical system are the solutions x(t) which pass through the points (t0, x0). The fixed points of a first order autonomous system (5.1) are values of x for which f(x) = 0. The phase portrait is a sketch of the phase space of the DS (which will be the real line or some subset) together with arrows indicating the direction of the motion along the orbit. The direction is to the right if x is increasing, i.e. if f(x) > 0 and to the left if x is decreasing, i.e. f(x) < 0. It is usual to combine a plot of f(x) with the phase portrait. It is then straightforward to read off from the phase portrait whether any given fixed point is stable or unstable. Sometimes we perform the integration and solve to find x(t) explicitly, sometimes we can only find an implicit definition of x(t) and sometimes we cannot even perform the integral exactly – but often we do not need to perform the integral exactly to be able to work our properties of the dynamical system. Most of the rest of this chapter is on how to deduce properties of the dynamical system and its trajectories without even necessarily being able to do the integral. Let us first consider a few simple examples to illustrate the concepts of trajectories, orbits and phase portraits.
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 51
Example 5.1. Sketch the trajectories and phase portrait for the DS with phase space the real line and velocity function x˙ = f(x) = 1 . (5.3)
This differential equation has the general solution
x = a + t , (5.4) and so the trajectories are straight lines. There are no fixed points. The orbits are the values of x which a trajectory passes through, and these are the whole real line. To complete the phase portrait we indicate the direction of flow along the orbit by sketching arrows to the right. It is also often convenient to plot the function f(x) along with the phase portrait.
x(t) f x 3 ( )
2
1 1
t -2 -1 1 2 x -1
Figure 5.1: Sketch of the trajectories (5.4) for the DS (5.3),x ˙ = 1, together with a sketch of the velocity function f(x) = 1 and the phase portrait.
Example 5.2. Sketch the trajectories and the phase-portrait for the DS with phase space the real line and velocity function
x˙ = f(x) = r (x − 1) . (5.5) This has the general solution x = 1 + x0 exp(r t) . (5.6) There is now one fixed point, x = 1 at which f(1) = 0. The orbits are now the sets {x|x < 1}, {x = 1} and {x|x > 1} each of which is an invariant set. The nature of the trajectories and of the fixed point depends on the sign of r. If r > 0 then the trajectories move away from the point x = 1 and fixed point is unstable; if r < 0 then the trajectories move in towards from the point x = 1 and fixed point is stable. We show the flow (comprised of the trajectories) and the phase portraits (i.e. the phase-flow comprised of the orbits together with a plot of the velocity function) in figures 5.2 (for r > 0) and 5.3 (for r < 0)
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 52
t f(x) 3
2 x 1 0
-2 -1 1 2 x
-1
Figure 5.2: Sketch of the trajectories (5.6) for the DS (5.5),x ˙ = r(x − 1), for r > 0 together with a sketch of the velocity function f(x) = r (x − 1) and the phase portrait.
t f(x) 3
2
1 x 0 -2 -1 1 2 x
-1
Figure 5.3: Sketch of the trajectories (5.6) for the DS (5.5),x ˙ = r(x − 1), for r < 0 together with a sketch of the velocity function f(x) = r (x − 1) and the phase portrait.
Finally, a slightly more complicated example with several fixed points for which we can find the general solution but for which the trajectories are not easy to find given the exact solution.
Example 5.3. Sketch the trajectories and the phase-portrait for the DS with phase space the real line: x˙ = f(x) = (x − 1)(x − 2)(x − 3) . (5.7)
In this case the integral (5.2) can be found exactly but is not very informative: explicitly:
Z x dx F (x) = x0 (x − 1)(x − 2)(x − 3) Z x 1/2 1 1/2 = − + dx x0 x − 1 x − 2 x − 3 1 1 x = log |x − 1| − log |x − 2| + log |x − 3| 2 2 x0
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 53
2 1 (x − 1) (x0 − 2) (x − 3) = log 2 2 (x0 − 1) (x − 2) (x0 − 3) = t − t0 (5.8)
In this case to sketch the trajectories it is much easier first to sketch the velocity function and the phase portrait: We can easily plot the velocity function:
1.0
0.5
1 2 3 4
-0.5
-1.0
Figure 5.4: Sketch of f(x) = (x − 1)(x − 2)(x − 3).
The velocity of the system when it is at x is given by f(x). This function is negative for x < 1 and 2 < x < 3, zero at x = 1, 2, 3 and positive for 1 < x < 2 and x > 3. This means that the system is moving to the left if x < 1 or 2 < x < 3, it is stationary if x = 1, 2, 3 and is moving to the right if 1 < x < 2 or x > 3. If we sketch the real line and include arrows showing which way the system is moving and show where the fixed point are, we get the phase portrait of the system:
1 2 3
Figure 5.5: Phase portrait of the dynamical system (5.7).
It is actually more common and also more useful to combine the graph of f(x) and the phase portrait, so from now on we will always do this when we sketch the phase portrait. Here are two ways we can do this, putting the arrows on the x–axis as in figure 5.5 or displacing them above and below to make it easier to see what is going on:
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 54
f(x) f(x)
1 2 3 1 2 3
Figure 5.6: Phase portraits of the dynamical system (5.7).
Looking at the phase portrait in figure 5.6, we can see that the fixed points are at the zeroes of the velocity function f(x), and we can also see that x = 1 and x = 3 are unstable (the flows nearby move away) and that the fixed point at x = 2 is stable (the flows nearby move towards it). We can also see that if the system starts anywhere inside the interval [1, 2] then it cannot escape it. Now that we have the phase portrait, it is easy to sketch the trajectories, if we want to: the trajectories through the fixed points are straight lines; the other trajectories move in the direction shown by the arrows on the phase portrait. The only question is: do the trajectories that head towards the fixed points or ±∞ actually reach these points in a finite time? Will the flow starting at x = 3/2 reach the point x = 2? Will the flow starting at x = 7/2 reach x = ∞? These are very important questions and the subject of the next section.
3.0
2.5
2.0
1.5
1.0
-1 0 1 2 3 4 Figure 5.7: A sketch of the flow, or the set of trajectories of the dynamical system (5.7).
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 55
5.2 Termination of Motion
If a system has a stable fixed point, and initial conditions are such that the fixed point is being approached, one may ask whether the system will reach the fixed point in finite time. The same may be asked about ‘escaping’ to ±∞. These considerations lead us to the following definition. Definition 5.1. Terminating Motion The motion of a dynamical system is said to be terminating, if it reaches a fixed point or goes to ±∞ in finite time.
Here are two elementary examples Example 5.4. Consider the DS with phase space x ≥ 0 given by dx = f(x) , f(x) = −xα , α > 0,. (5.9) dt Sketch the phase portrait and find the values of α for which the motion starting at t = 0 at x0 > 0 terminates at x = 0.
The velocity function f(x) is always negative and f(0) = 0 if α > 0, leading to the simple phase portrait in figure 5.8 The differential equation is separable and so we can immediately integrate it to find Z t Z x(t) dx 1 x(t) 1 t = dt = = − x1−α = (x )1−α − x(t)1−α , (5.10) f(x) 1 − α 1 − α 0 0 x0 x0 or solving for x(t), 1−α1/(1−α) x(t) = −(1 − α)t + (x0) . (5.11) To see if the motion terminates, we could look at the solution (5.11) but it is more direct to see for which values of α the integral (5.10) is finite as x(t) → 0. The answer is that the time to reach x = 0 is finite if 1 − α > 0 i.e. α < 1. (The case α=1 is that of exponential approach which we already know does not terminate.) We illustrate this in figure 5.8 with plots of f(x) and x(t) in the cases α = 1/2, 1 and 2.
f(x) x 0.5 1.0 1.5 2.0 t 4 -0.5
-1.0 3
-1.5
2 -2.0
-2.5 1
-3.0 x (a) The velocity functions and 0.0 0.5 1.0 1.5 2.0 phase portrait (b) The trajectories
Figure 5.8: Plots of the velocity function and solutions of the DS (5.4) for α = 1/2 (solid line), α = 1 (dashed) and α = 2 (dotted) in the case x0 = 2
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 56
Example 5.5. Consider the DS with phase space x ≥ 0 given by dx = f(x) , f(x) = xα ,. (5.12) dt
Find the values of α for which the motion starting at t = 0 at x0 > 0 terminates at x = ∞.
The velocity function f(x) is now always positive. The differential equation is again separable and so we can immediately integrate it to find
Z t Z x(t) dx 1 x(t) 1 t = dt = = x1−α = x(t)1−α − (x )1−α . (5.13) f(x) 1 − α 1 − α 0 0 x0 x0 To see if the motion terminates, we see for which values of α the integral (5.13) is finite as x(t) → ∞. The answer is that the time to reach x = ∞ is finite if 1 − α < 0 i.e. α > 1. (The case α=1 is that of exponential approach which we already know does not terminate.) Note that this is exactly the opposite condition to the previous example.
Whether a motion is terminating or not can depend on the starting point of the motion. A simple example showing this is the logistic equation (1.3) with the solutions (1.4).
Example 5.6. For which values of x0 does the motion given by the logistic equation and its general solution terminate? dx x c c = r x 1 − , x = −r t ,A = 1 − (5.14) dt c 1 − Ae x0
If the motion starts at either of the two fixed points x = 0 or x = c it stays there for all time; by definition these motions terminate. If the motion starts at a point 0 < x0 < c or x0 > c then it will not terminate: as t increases then the motion will tend towards x = c but will not reach it in finite time; if however the motion starts at x0 < 0 then it will diverge to −∞ at the finite time 1 c t = log 1 − (5.15) r x0
In the example of the logistic equation we can find the solution exactly, but we can also usually answer the question without needing to solve the dynamical system. Given a dynamical system described by (5.1), dx = f(x). dt
and initial condition x(t0) = x0, the time τ01 needed for the system to reach x1 can in principle be found from (5.2): Z t1 Z x1 dx τ01 = t1 − t0 = dt = . (5.16) t0 x0 f(x) We have to be slightly careful about signs since the flow can be moving to the right (f(x) > 0, x1 > x0) or moving to the left (f(x) < 0, x1 < x0) but in both cases the integral will be a positive number.
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 57
To find out if a motion starting at x0 terminates at x1 in finite time all we have to do is find out if the integral τ01 is finite (the motion terminates) or is infinite (the motion does not terminate). Sometimes we can calculate the integral exactly, sometimes we cannot and need to find other methods
To answer the questions posed earlier: will the flow for the system in example 5.3 starting at x = 3/2 reach x = 2? Will the flow starting at x = 7/2 reach x = ∞? — we can calculate these integral exactly.
Example 5.7. Find the time it takes to go from x = 3/2 to a position ξ < 2:
Z ξ dx τ01 = 3/2 (x − 1)(x − 2)(x − 3) Z ξ 1/2 1 1/2 = − + dx 3/2 x − 1 x − 2 x − 3 1 1 ξ = log |x − 1| − log |x − 2| + log |x − 3| 2 2 3/2 1 1 1 = log |ξ − 1| − log |2 − ξ| + log |3 − ξ| − log 3 . (5.17) 2 2 2 As ξ → 2, the term (− log |2 − ξ|) → ∞; the integral diverges as ξ → 2; the time it takes for the motion to reach x = 2 is infinite; in other words, the flow does not reach x = 2 in finite time, the flow does not terminate. Let’s now find the time it takes to go from x = 7/2 to a position ξ > 7/2:
Z ξ dx τ01 = 7/2 (x − 1)(x − 2)(x − 3) Z ξ 1/2 1 1/2 = − + dx 7/2 x − 1 x − 2 x − 3 1 1 ξ = log |x − 1| − log |x − 2| + log |x − 3| 2 2 7/2
1 (ξ − 1)(ξ − 3) 1 = log + log(9/5) . (5.18) 2 (ξ − 2)2 2
As ξ → ∞, the term (ξ − 1)(ξ − 3)/(ξ − 2)2 → 1; the integral is finite as ξ → ∞; the time it takes for the motion to reach x = ∞ is finite; in other words, the flow reaches x = ∞ in finite time 1/2 log(9/5) = 0.293893.., the flow terminates. It is important to realise that it is only the motion very close to the fixed point that is relevant: the flow will always reach any point near the fixed point; the question is: will it be able to get that final extra part of the way from close to the fixed point right to the fixed point.
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 58
This is not always the case: one can think up velocity functions with the same phase diagram as figure 5.6 but for which the motion starting at x = 3/2 does terminate and the motion starting at x = 7/2 does not, for example
Example 5.8. Consider the DS
Z 2 1/3 dx f(x) = [(x − 1)(x − 2)(x − 3)] , τ01 = = 0.966897... (5.19) 3/2 f(x)
1 2 3
Figure 5.9: Phase portrait for the DS (5.19).
The difference between the two DS, (5.7) and (5.19) is how the velocity function approaches zero near x = 2 and how it approaches ∞ as x → ∞. We shall some time working out how best to decide if a motion terminates or not. There is one criterion we already have which enables us to prove that under certain circum- stances a motion cannot reach a fixed point in a finite time and that is Picard’s theorem.
Corollary to Picard’s Theorem: I
Suppose that x = a is a fixed point of the DS
x˙ = f(x) (5.20) and that f(x) and f 0(x) are both continuous at x = a, then the only trajectory that terminates at x = a in finite time is the constant trajectory x(t) = a.
Proof: suppose there was a second trajectory ξ(t) which satisfied xi(t0) = a. Then Picard’s theorem says the solution to the differential equation (5.20) satisfying x(t0) = a is unique. But x(t) = a is such a solution, hence ξ(t) = x(t) = a. Hence no trajectories that approach x = a can ever reach x = a in finite time apart from the constant trajectory.
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 59
Corollary to Picard’s Theorem: II
Consider the DS defined on the real linex ˙ = f(x) and suppose that f(x) and f 0(x) are continuous on the whole real line. Then the orbits of the system consist of fixed points and open intervals. The invariant sets are unions of fixed points and open intervals; equivalently, the phase spaces is the union of fixed points and open invariant sets, that is open intervals which are invariant sets that contain no fixed points.
Proof: consider a point x = a. Consider the trajectory with x(t0) = a. By Picard’s theorem this is unique. The orbit of this trajectory is either the point a (a is a fixed point) or an open interval (since the motion cannot terminate). Hence all orbits are either fixed points or open intervals, which are also called open invariant sets.
We can check the examples we have looked at so far. Example 5.3. In this case
f(x) = (x − 1)(x − 2)(x − 3) , f 0(x) = 3x2 − 12x + 11 , (5.21)
are both continuous at x = 2 and so any motion that approaches x = 2 cannot reach x = 2 in finite time, as we showed in example 5.7. In this case the phase space (the real line) is the union of the fixed points {1}, {2}, {3} and the open invariant sets (−∞, 1), (1, 2), (2, 3), (3, ∞).
Example 5.8. In this case f(x) is continuous for all x but
1 (x − 2)(x − 3)1/3 1 (x − 1)(x − 2)1/3 1 (x − 1)(x − 3)1/3 f 0(x) = + + , (5.22) 3 (x − 1)2 3 (x − 2)2 3 (x − 3)2 diverges at x = 2. This means that it may be possible for a flow to terminate at x = 2: it is not ruled out by Picard’s theorem. In this case the open intervals (1, 2) and (2, 3) are not invariant sets since a trajectory starting in (1, 2) will reach the point x = 2 in finite time [hence leaving the open interval] and likewise a trajectory starting in (2, 3) will also reach the point x = 2 in finite time [again leaving the open interval].
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 60
5.3 Estimating times of Motion
Estimates for τ01 can be obtained without giving the full solution of the ODE describing the dynamical system in question, i.e., without actually doing the x-integral in (5.16) exactly, using the following observation: If φ and ψ are real valued functions defined on an interval [a, b] such that φ(t) ≥ ψ(t) ∀ t ∈ [a, b], then Z b Z b φ(t) dt ≥ ψ(t) dt . a a
In terms of a dynamical system, this says that if one dynamical system moves from a to b at a faster speed then a second, it will take less time; if it moves slower then it will take more time. To be concrete, let us consider two velocity functions f(x) and g(x) defined on an interval [a, b] such that f(x) ≥ g(x) > 0 on (a, b). The the time taken for the two different systems to move from a to b are Z b dx Z b dx τf = , τg = . (5.23) a f(x) a g(x) Since f(x) ≥ g(x) then 1/f(x) < 1/g(x) and so
τf ≤ τg . (5.24) This is illustrated in figure 5.10
x
τg 1 f(x) x = g(x) g (x)
τf
1 g(x) x = f(x) f (x)
x t x a b a b a b (a) Two velocity functions (b) The trajectories for the two (c) The times taken to move f(x) > g(x) velocity functions; the faster from a to b are the areas under system reaches b first. the graphs of 1/f and 1/g.
Figure 5.10: Sketches illustrating the time of motion for different velocity functions.
Using these elementary facts to estimate times of motion and determine if a motion terminates or not is a bit of an art. Given a velocity function f(x), there is an infinite choice of functions h(x) a g(x) such that h(x) > f(x) > g(x); some will be helpful, some not. We will look at this in the next few examples.
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 61
Example 5.9. Find a lower bound on the time taken for the DS 5.3 to move from a point 3 < x0 to a point x1. Use this to find lower estimates of the time T to move from x0 = 4 to x1 = ∞.
For x0 > 3, we have f(x0) > 0 and so we simply have Z x1 dx Z x1 dx τ01 = = (5.25) x0 f(x) x0 (x − 1)(x − 2)(x − 3) To find a lower bound on the time taken, we want to find a speed h(x) which is faster than f(x) so that h(x) > f(x), 1/h(x) < 1/f(x) and we find bounds Z x1 dx Z x1 dx τh = < = τ01 (5.26) x0 h(x) x0 f(x) The idea is to find a function h(x) which gives a simpler integral. Since x > 3, we can see look at each of the factors in f(x) and find simple inequalities (a) 2 < (x − 1) (5.27) (b) 1 < (x − 2) (5.28) (c) 0 < (x − 3) (5.29) (d)(x − 3) < (x − 2) < (x − 1) (5.30)
(i) The first three are not very helpful, but we can use (d) to obtain (x − 1)(x − 2)(x − 3) < (x − 1)3 , (5.31) and so Z x1 dx 1 1 1 τ01 > 3 = 2 − 2 . (5.32) x0 (x − 1) 2 (x0 − 1) (x1 − 1) This is true and helpful. As x → ∞ this is finite, 1 τ01 > 2 . (5.33) 2(x0 − 1)
and so with x0 = 4 we get a lower estimate 1 T > = 0.0555... . (5.34) 18 (ii) As another choice, we could instead have used (x − 1) < x to get (x − 1)(x − 2)(x − 3) < x3 , (5.35) and so Z x1 dx 1 1 1 τ01 > 3 = 2 − 2 . (5.36) x0 x 2 (x0) (x1) This is again true and again helpful but the bound is not as good. In the limit x1 → ∞ we find 1 τ01 > 2 , (5.37) 2(x0) leading to 1 T > = 0.03125 , (5.38) 32 which is smaller than the previous estimate.
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 62
Example 5.10. Find an upper bound on the time taken for the DS 5.3 to move from a point 3 < x0 to a point x1. Use this to show that the motion reaches x = ∞ in finite time. Find an upper estimate for the time to reach x = ∞ from x0 = 4,
For x0 > 3, we have f(x0) > 0 and so we simply have
Z x1 dx Z x1 dx τ01 = = (5.39) x0 f(x) x0 (x − 1)(x − 2)(x − 3) To find an upper bound on the time taken, we want to find a speed g(x) which is slower than f(x) so that f(x) > g(x), 1/f(x) < 1/g(x) and we find the bound
Z x1 dx Z x1 dx τ01 = < = τg (5.40) x0 f(x) x0 g(x) The idea is to find a function g(x) which gives a simpler integral. Since x > 3, we can see look at each of the factors in f(x) and find simple inequalities
(a) 2 < (x − 1) (5.41) (b) 1 < (x − 2) (5.42) (c) 0 < (x − 3) (5.43) (d)(x − 3) < (x − 2) < (x − 1) (5.44)
(i) We can use (a) and (b) to obtain
(x − 1)(x − 2)(x − 3) > 2 · 1 · (x − 3) ≡ g1(x) , (5.45)
and so Z x1 dx 1 x0 − 3 τ01 < = ln . (5.46) x0 2(x − 3) 2 x1 − 3
As x1 → ∞, this lower bound diverges, so while this inequality is true it is not helpful. (ii) We could instead have used (d) twice to find
3 (x − 1)(x − 2)(x − 3) > (x − 3) ≡ g2(x) , (5.47)
and so Z x1 dx 1 1 1 τ01 < 3 = 2 − 2 . (5.48) x0 (x − 3) 2 (x0 − 3) (x1 − 3) This is also true but now is helpful. As x → ∞ this is finite, so we have found 1 τ01 < 2 , (5.49) 2(x0 − 3)
and with x0 = 4 we get the upper estimate 1 T < = 0.5 . (5.50) 2
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 63
(iii) Finally, as another choice, we could use (b) and (d) [once] to get
2 (x − 1)(x − 2)(x − 3) < (x − 3) · 1 · (x − 3) = (x − 3) ≡ g3(x) , (5.51)
and so Z x1 dx 1 1 τ01 < 2 = − . (5.52) x0 (x − 3) (x0 − 3) (x1 − 3) This is again true and again helpful. As x → ∞ this is also finite, 1 τ01 < , (5.53) (x0 − 3)
and so we have found another upper bound for the case x0 = 4
T < 1 . (5.54)
Again this is worse than the previous upper bound
These three choices show that there is more than one way to find an upper bound, but that one has to use some care to preserve the important properties of the original velocity function.
We illustrate these choices in figure 5.11 for the choice x0 = 4.
x 120
1.0
100 f(x)
g2(x) 0.8
80
0.6 60 τg3
0.4 40 g3(x)
τg2 0.2 20 g1(x)
τf x x 0 2 4 6 8 10 0 5 10 15 20 25 30 t 0 2 4 6 8 10 (a) The velocity functions (b) The trajectories for the ve- (c) The times taken to move locity functions; three trajec- from 4 to ∞ are the areas un- tories reach x = ∞ in finite der the graphs. time, one does not.
Figure 5.11: Sketches illustrating the time of motion for different velocity functions: f(x) (solid), g1(x) (Dashed), g2(x) (dotted) and g3(x) (dot-dashed)
It is instructive to compare our estimates with the exact result which we found before. The exact answer is 1 4 T = ln ' 0.143841.. , (5.55) 2 3
compared with the estimates 0.0555 < T < 0.50 which we obtained above
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 64
Example 5.11. Find a lower bound on the time taken for the DS 5.3 to move from a point 2 < x0 < 3 to a point x1. Use this to show that the motion cannot terminate
For 2 < x0 < 3, we have f(x0) < 0; the system will approach the stable fixed point at x = 2, so x(t) < x0 for t > t0. In this case write
Z x1 dx Z x0 dx Z x0 dx τ01 = = = (5.56) x0 f(x) x1 |f(x)| x1 (x − 1)(x − 2)(3 − x) We want to find a speed which is faster than f(x) so that if the faster moving system does not reach x = 2 then the slower moving system cannot. This means we want to find a function h(x) such that h(x) > |f(x)| and for which the time
Z x0 dx τh = (5.57) x1 h(x) is easy to calculate. Since 2 < x < 3, we can see look at each of the factors in |f(x)| and find simple inequalities
(a) 1 < (x − 1) < 2 (5.58) (b) 0 < (x − 2) < 1 (5.59) (c) 1 > (3 − x) > 0 (5.60) (d)(x − 2) < (x − 1) (5.61)
(i) We can use (a) and (c) to obtain
(x − 1)(x − 2)(3 − x) < 2 · (x − 2) · 1 , (5.62)
and so
Z x1 Z x1 dx dx 1 x0 − 2 τ01 = > = ln . (5.63) x0 (x − 1)(x − 2)(3 − x) x0 2(x − 2) 2 x1 − 2
As x1 → 2, this lower bound diverges, so the system approaches 2, but will not reach it in finite time.
(ii) We could also have used (a), (b) and (d) to find
(x − 1)(x − 2)(3 − x) < 2 · 1 · 1 , (5.64)
and so Z x1 dx Z x1 dx 1 τ01 = > = (x1 − x0) . (5.65) x0 (x − 1)(x − 2)(3 − x) x0 2 2
This is true but not helpful. The time τ01 diverges because of the singularity in 1/|f(x)| at x = 2, if we remove this singularity in our estimate we will end up with a finite time and this will not help prove that the motion does not terminate.
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 65
(iii) Finally, as another choice, we could use (c) and (d) to get
(x − 1)(x − 2)(3 − x) < (x − 2) · (x − 2) · 1 , (5.66)
and so
Z x1 dx Z x1 dx 1 1 1 τ01 = > 2 = − . (5.67) x0 (x − 1)(x − 2)(3 − x) x0 (x − 2) 2 x0 − 2 x1 − 2
This is also true and helpful. As x1 → 2 the term 1/(x1 − 2) diverges and so we can also use this estimate to show that the motion does not terminate.
These three choices show that there is more than one way to prove a motion terminates, but that one has to use some care to preserve the important properties of the original velocity function.
Example 5.12. As a final example, find upper and lower bounds on the time T for the DS in example 5.8 to move from x = 3/2 to x = 2
This is the DS f(x) = [(x − 1)(x − 2)(x − 3)]1/3 (5.68) for which the exact result is T = 0.966897... Since 1 < x < 2 we have the simple inequalities on the non-negative factors (x − 1), (2 − x) and (3 − x):
(a) 0 < (x − 1) < 1 (5.69) (b) 0 < (2 − x) < 1 (5.70) (c) 1 < (3 − x) < 2 (5.71) (d) (2 − x) < (3 − x) (5.72)
We can use (a) and (c) to find
(x − 1)(2 − x)(3 − x) < 2 , (5.73)
and so find the simple lower bound on T
Z 2 dx Z 2 dx 1 T = > = = 0.39685.. . (5.74) 1/3 1/3 4/3 3/2 ((x − 1)(2 − x)(3 − x)) 3/2 2 2
To find an upper bound, (a) and (b) are not useful, but we can instead (with a bit more thought) come up with
x0 = 3/2 ⇒ x ≥ 3/2 ⇒ (x − 1) > 1/2 , (5.75) and so combining this with (c) we get 1 (x − 1)(2 − x)(3 − x) > 1/2 · (2 − x) · 1 = (2 − x) , (5.76) 2
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 66 or combining it with (d) we get 3 (x − 1)(2 − x)(3 − x) > 1/2 · (2 − x) · (2 − x) = (2 − x)2 . (5.77) 2 Together these give the two upper bounds
Z 2 dx 3 2 T < = 21/3(2 − x)2/3 = 1.19055... , 1/3 3/2 (1/2(2 − x)) 2 3/2 " #2 Z 2 dx (2 − x)1/3 T < = 3 = 3 . (5.78) 1/3 1/3 3/2 2 (1/2) (1/2(2 − x) ) 3/2 Hence we have found
0.39685.. < T < 1.19055.. , while the exact result is T = 0.966807... (5.79)
5.4 Stability — A More General Discussion
In chapter 4 we introduced the notion of fixed points and investigated their stability graph- ically by looking at the phase portrait and investigating qualitative properties of velocity functions in the vicinity of fixed points. Below we look at stability of first order autonomous dynamical systems in a more systematic and quantitative way, discussing (i) linear (and non- linear) stability analysis of fixed points, (ii) the concept of structural stability of a dynamical system, and (iii) the notion of stability of motion in turn.
5.4.1 Stability of Fixed Points
We consider the system described by the equation dx = f(x) dt and suppose that x = a is a fixed point. Assuming that f is a suitably differentiable function in a neighbourhood of x = a we may approximate f by its Taylor expansion about x = a, f 00(a) f 00(a) f(x) = f(a) + f 0(a)(x − a) + (x − a)2 + ··· = f 0(a)(x − a) + (x − a)2 + ... (5.80) 2! 2! Near x = a, i.e. for |x − a| 1, the dominant contribution to the r.h.s of (5.80) comes from the first non-vanishing derivative of f at a.
Case (i) f 0(a) 6= 0. In this case f(x) ' f 0(a)(x − a) for |x − a| 1, and the ODE is approximated by
dx d(x − a) ' f 0(a)(x − a) ⇔ ' f 0(a)(x − a) dt dt The solution is 0 x − a ' (x0 − a) exp[f (a)t];
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 67
the distance x − a thus (i) increases exponentially in time, if f 0(a) > 0, whereas (ii) it decreases exponentially in time, if f 0(a) < 0. In the first case the fixed point is called (linearly) unstable, in the second, it is called (linearly) stable. This result is in accord with the graphical stability analysis based on a phase portrait as in the figure shown below.
Case (ii) f 0(a) = 0, but f 00(a) 6= 0. In this case f has a double zero at x = a, and 1 f(x) ' f 00(a)(x − a)2 2 near x = a. The ODE is approximated by
dx 1 d(x − a) 1 ' f 00(a)(x − a)2 ⇔ ' f 00(a)(x − a)2 dt 2 dt 2 The solution (found by separation of variables), reads x − a x − a ' 0 t 00 1 − 2 f (a)(x0 − a) Supposing that f 00(a) > 0 so that f has a minimum at x = a, we find
(a) If x0 > a, then x − a increases as a function of time t.
(b) If x0 < a, then |x − a| decreases as a function of time t. Conversely, supposing that f 00(a) < 0 so that f has a maximum at x = a, we find
(a) If x0 > a, then x − a decreases as a function of time t.
(b) If x0 < a, then |x − a| increases as a function of time t. These results are once more in accordance with the graphical stability analysis based on a phase portrait as in figure 5.12.
f(x)
x a b c
Figure 5.12: Phase portrait of a system with velocity function that has fixed pints at a with f 0(a) > 0, at b with f 0(b) < 0 and at c, with f 0(c) = 0, but f 00(c) < 0.
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 68
Definition 5.2. Linear Stability Analysis The stability analysis based on (and requiring only) Taylor series expansions to 1st order is referred to as linear stability analysis of a fixed point.
Note The approximate solution of the ODEs describing the dynamical system near fixed points based on Taylor series expansions is valid only in the vicinity of fixed points. The quality of the approximate description will therefore improve, if a fixed point is approached under the dynamics, whereas the approximate description deteriorates, if the system moves away from a fixed point.
5.4.2 Structural Stability
Definition 5.3. Structural Stability A system described by the differential equation dx = f(x) dt is called structurally unstable, if the number of fixed points is changed by some arbitrarily small (continuous) perturbation f(x) → f(x) + g(x), (i.e., for some continuous g and arbitrarily small), otherwise it is structurally stable (i.e., for all continuous g there is a sufficiently small , such that the number of fixed points remains unchanged).
The simplest systems that is structurally unstable is when f has a double zero at x = a. Close to x = a the function is approximately a parabola,
f(x) ' c(x − a)2 + .... (5.81)
If we add a small constant shift to f then the graph will either intersect the axis in two points (so the number of fixed points has gone up) or it will not intersect the axis at all (the number of fixed points has gone down). Since the number of fixed points changes, the system is structurally unstable. See figure 5.13. As another example, the function f = x3 has a single fixed point at x = 0 which is a triple zero of f. If we add a constant shift to f then the function still has a single fixed point, but if we add a small linear function then it either has one or three fixed points: the number of fixed points changes so it is also structurally unstable. More generally, if f has a zero of any multiplicity higher than 1 at x = a, the system is structurally unstable, but we might have to add a more complicated function than just a constant shift. (The proof is left as an exercise to the reader.) In structurally unstable systems the collection of invariant open sets (and thus qualitative properties of the dynamics) can change drastically in response to small perturbations. This may e.g. of economic relevance as the following example demonstrates.
Example 5.13. Bistable Ecosystems Consider a population dynamics of the form
dx ax2 = rx(1 − x/c) − , dt b2 + x2
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 69
f(x)-ϵ f(x) f(x)+ ϵ
Figure 5.13: Velocity function of a structurally unstable system with a fixed point which is a double zero [centre figure]. Vertical displacement by ± increases [left figure] or decreases [right figure] the number of fixed points by one as shown.
f(x)- ϵ(x-a) f(x) f(x)+ϵ(x-a)
Figure 5.14: Velocity function of a structurally unstable system with a fixed point which is a triple zero [centre figure]. Changing f(x) by ±(x − a) leaves the number of fixed points the same [left figure] or increases it by two [right figure]. in which all parameters (r, c, a, b) are taken to be positive. The first contribution describes logistic growth, and taken by itself has x = c as a stable fixed point. The second contribution on the r.h.s. exhibits ‘threshold behaviour’. It produces a significant contribution only for x > xc ' b. This system could describe a population of larvae, preyed upon by a natural enemy (woodpeckers) if present in sufficient quantity. For sufficiently small r the system has a stable small-x fixed point an unstable fixed point and a stable fixed point at large x. The stable fixed point at low x and the unstable fixed point coalesce and disappear as r is increased beyond a certain value rc and the system would approach a stable high-x solution. Economically, this could spell disaster if the larvae destroy wood that is meant ∗ to be sold!. Only by decreasing r to a very small value r rc can the situation be reverted to a stable small x situation; (see the diagram).
Example 5.14. (King’s College, Summer 1994) A first order dynamical system is described by the differential equation
x˙ = v(x) = x(x − 1)2(x − 2)3(x − 3)5 .
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 70
f(x) f(x)
x x
Figure 5.15: Phase portrait of the bi-stable ecosystem for r < rc (left) and r > rc (right).
Explain briefly why the initial condition x(0) = 2 ⇒ x(t) = 2 ∀t ≥ 2. Sketch the phase diagram and write down the invariant open sets. Is the system structurally stable? Give your reasons. Given that x(0) = 5/2 prove that the system will not reach x = 2 in a finite time. Suppose instead that x(0) = 4. Prove that x(t) → ∞ as t → τ, where τ < 1/10.
We note that v, v0 are both continuous in any neighbourhood of the point (t = 0, x = 2); Picard’s theorem therefore applies to the given equation. Clearly x(t) = 2 satisfies the differential equation and the initial condition x(0) = 2. It follows from Picard’s theorem that x(t) = 2, t ≥ 0. Referring to the phase diagram (see below) we see that the invariant open sets are
(−∞, 0), (0, 1), (1, 2), (2, 3), (3, ∞) .
The system is not structurally stable since the transformation v(x) → v(x) − ( > 0) increases the number of fixed points from four to five (v(x) has a double zero at x = 1).
It is clear from the phase diagram that if x(0) = 5/2 the system moves towards x = 2 and the time τ1 taken to reach x = x1 < 5/2 is
Z x1 dx Z 5/2 dx τ1 = 2 3 5 = 2 3 5 . 5/2 x(x − 1) (x − 2) (x − 3) x1 x(x − 1) (x − 2) (3 − x)
Now x(x − 1)2(3 − x)5 < (5/2)(5/2 − 1)2(3 − 2)5 = 45/8 so
8 Z 5/2 dx 8 1 1 τ1 > 3 = −2 + 2 . 45 x1 (x − 2) 45 2 (x1 − 2)
We conclude that τ1 → ∞ as x1 → 2+ and that the system does not reach x = 2 in a finite time. The time τ taken by the system to go from x(0) = 4 to infinity is given by
Z ∞ dx τ = 2 3 5 . 4 x(x − 1) (x − 2) (x − 3) Since x > x − 3, x − 1 > x − 3, x − 2 > x − 3, we obtain Z ∞ dx 1 τ < 11 = . 4 (x − 3) 10 This motion is therefore terminating.
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 71
f(x)
x
Figure 5.16: Phase portrait of the system with velocity function v(x) = x(x − 1)2(x − 2)3(x − 3)5.
5.4.3 Stability of Motion
Beyond looking at stability of fixed points, one may also look at stability of motion itself in the following sense. Let dx = f(x) (5.82) dt describe a first order autonomous dynamical system, and let x(t) be a solution for some given initial condition. Suppose now, that the solution is slightly disturbed
x(t) → y(t) = x(t) + ε(t) (5.83)
at, say, t = t1, with |ε(t)| 1. Enquiring into stability of motion means asking whether the perturbation will increase or decrease under the dynamics. We study this issue only at linear order. Definition 5.4. Stability of motion The motion described by (5.82) is stable, if a small perturbation ε(t) is diminished under the dynamics; it is called unstable, if a small perturbation is amplified. Proposition 5.1. The motion of an autonomous first order dynamical system described by a velocity function f is linearly stable at x, if f 0(x) < 0, whereas it is linearly unstable at x, if f 0(x) > 0. Proof Consider the ODE dy dx dε = f(y) ⇔ + = f(x + ε) dt dt dt Expanding the r.h.s. to first order in ε gives dx dε + ' f(x) + f 0(x)ε . dt dt As x is a solution of (5.82), we get an ODE for ε, dε ' f 0(x)ε , dt
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 72 in which f 0(x) = f 0(x(t)) is a known function of time. The equation is therefore separable and solved by Z t h 0 0 0 i ε(t) ' ε(t1) exp dt f (x(t )) (5.84) t1 0 Note that for small t − t1 the exponential may be approximated by exp f (x(t1))(t − t1) in the above expression. Thus the perturbation increases at an exponential rate, and the motion is unstable whenever f 0(x) > 0, whereas the perturbation decreases at an exponential rate, and the motion is stable whenever f 0(x) < 0, as claimed. #
Note 1: The notion of dynamic stability is extremely important if one considers solving an ODE numerically on a computer, as it tells whether rounding errors will amplify when propagating a solution or whether they will diminish. Note 2: A first order dynamical system which is linearly unstable when integrated forward in time, becomes linearly stable when integrated backwards (and vice versa).
5.5 Asymptotic Analysis
Asymptotic analysis is concerned with approximate analytic investigations of dynamical sys- tems in situations where either full solutions are not readily available, or where approximate solutions are sufficient to give a reasonably precise first level of analysis of a given system. The aim is to give a solid basis to the idea of two functions “behaving the same” or “having the same leading behaviour”. We start with a key definition. Definition 5.5. Asymptotic Equivalence A function f is said to be asymptotically equivalent to another function g as x → x∗ — in symbols f(x) ∼ g(x), as x → x∗ — if they become of the same order of magnitude as x → x∗, i.e. f(x) = g(x)(1 + ε(x)) , with ε(x) → 0 , as x → x∗ .
In this definition the case x∗ = ±∞ is allowed. There are several things to note:
(i) Given a function f(x) and a point x∗, infinitely many functions will be asymptotically equivalent to f(x) as x → x∗. This is guaranteed by the huge freedom in the choice of the function ε(x). (ii) The asymptotic behaviour of a function can be very different at different points, so if f(x) ∼ g(x) as x → a, this does not imply that f(x) ∼ g(x) for any other point. (iii) If you are given a function f(x) and asked to find a simpler function which is asymp- totically equivalent to f(x) as x → x∗, it is often usual to choose a function of the form c(x − x∗)α [if this is indeed correct], and this is often called the leading asymptotic be- haviour, but this is not a unique choice and other choices may be better for the problem in hand.
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 73
Here are some examples showing that the asymptotic behaviour of a simple function can be different in different limits Example 5.15. Consider the function f(x) = −x+x2. Find the leading asymptotic behaviour of f(x) in the limits x → 0, x → 1, x → ∞ and x → 2.
(a) f(x) ∼ −x as x → 0. Proof: f(x) = −x(1 − x) and so f(x) = −x(1 + ε(x)) with ε(x) = x and the conditions are met.
(b) f(x) ∼ (x − 1) as x → 1. Proof: f(x) = (x − 1)(1 + ε(x)) where ε(x) = x − 1 and so the conditions are met.
(c) f(x) ∼ x2 as x → ∞. 2 1 Proof: f(x) = x (1 − x ) and so again the conditions are met. (d) f(x) ∼ 2 as x → 2. Proof: f(x) = 2(1 + ε(x)) with ε(x) = (x − 2)(x + 1)/2 and so again the conditions are met.
Here are some examples showing that there is no unique choice of asymptotic equivalence Example 5.16. Given f(x) = x − x3 . We then have
(a) f(x) ∼ x as x → 0.
(b) f(x) ∼ x + x2 as x → 0.
(c) f(x) ∼ −x3 as x → ∞
(d) f(x) ∼ −x3 + 5 as x → ∞
We can check the first two of these:
(a) With f(x) = x − x3, g(x) = x then (f(x)/g(x)) − 1 = 1 − x2 = 1 + ε(x) so that indeed ε(x) = −x2 → 0 as x → 0.
(b) With f(x) = x−x3, g(x) = x+x2 then (f(x)/g(x))−1 = (1−x2)/(1+x) = 1−x = 1+ε(x) so that now ε(x) = −x and again ε(x) → 0 as x → 0.
The proof of the other two statements according to the definitions is left as an exercise.
Asymptotic analysis is useful because if two velocity functions are similar then their trajec- tories will be similar. If two velocity functions f(x) and g(x) have the same fixed point x = a and f(x) ∼ g(x) as x → a , (5.85)
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 74 then if the flow terminates, the times to reach x = a from a point b near a are also asymp- totically equivalent as b → a. This means that one can investigate whether a flow terminates as it approaches a fixed point (or ±∞) or not by looking at a simpler velocity function. This is often simpler than trying to find an upper or lower bound.
5.5.1 Asymptotic Analysis and Dynamical Systems
Suppose we have a DS dx = f(x) , (5.86) dt which has a fixed point at x = a. We want to understand what happens as x → a and to know whether the time to reach x = a from x0 is finite or not. We know this is given by an integral, so we investigate Z x1 exact dx τ01 = . (5.87) x0 f(x) This can be hard, time-consuming or impossible to do exactly and again it can be hard or time-consuming to find upper and lower bounds, so we can instead consider an alternative integral: Z x1 asymp dx τ01 = , (5.88) x0 g(x) where f(x) ∼ g(x) and f(x) = g(x)(1 + ε(x)) and ε(x) → 0 as x → 0. This means we can write this asymptotic time as
Z x1 Z x1 Z x1 asymp (1 + ε(x)) 1 ε(x) τ01 = dx = dx + dx . (5.89) x0 f(x) x0 f(x) x0 f(x) | {z } | {z } exact correction τ01 We can then show that
exact asymp (a) τ01 is finite as x1 → a if and only if τ01 is finite as x1 → a. The reason is that the only important part of the integral is the region close to x = a and in that part, ε(x) is small. This means we can test for termination of motion using simpler asymptotically equiva- lent functions.
exact asymp (b) If the time is finite and we put x1 = a then τ01 ∼ τ01 as x0 → a. The reason is that now the integration range is only over a small region between x0 and a where ε(x) is small. This means we can get asymptotically good estimates of the time for a motion to terminate using asymptotically equivalent velocity functions without having to search for upper and lower bounds.
We can also go further. If we want a solution to the initial value problem dx = f(x) , x = a + , (5.90) dt 0 where is small, then we can get a good approximation near x = a by instead solving the initial value problem dxasymp = g(xasymp) , x = a + , (5.91) dt 0
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 75 the reason being that g(x) and f(x) are very close if x is near a, and the solutions x(t) and xasymp(t) will also be close so long as x is close to a. Example 5.17. We will look at this more in some examples based on the Dynamical system with phase space the positive real axis x ≥ 0,
dx √ = f(x) , f(x) = x (x − 1)(x − 2)(x − 3) . (5.92) dt If we plot f(x) we easily see that there are four fixed points {0}, {1}, {2} and {3}, that 0 and 2 are stable fixed points and 1 and 3 are unstable fixed points.
fHxL
x 1 2 3
Figure 5.17: Sketch of f(x) and the phase space for the system (5.92).
Since f and f 0 are both continuous for all x > 0, corollary II to Picard’s theorem implies the the open intervals (1, 2) and (2, 3) are open invariant sets, but since f 0(0) diverges we cannot say whether (0, 1) is an invariant set or not. Here are several questions we can analyse using asymptotic analysis:
(i) Does the motion tending towards x = 0 terminate?
(ii) Does the motion tending towards x = 2 terminate? [we know the answer is “no” but we can still check it]
(ii) Does the motion towards x = ∞ terminate?
We will use asymptotic methods to answer these questions. We can compare them with the exact answer because we can find the time of motion exactly using the substitution x = u2 and the method of partial fraction to get √ Z dx 1 1 − x 1 1 − px/3 1 1 + px/2
= log √ + √ log p + √ log p . (5.93) f(x) 2 1 + x 2 3 1 + x/3 2 1 − x/2 Finding the exact answer is quite time-consuming and so the asymptotic methods are a lot faster, which is why the method is particularly useful in this case. Note that although we can
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 76
find the time of motion explicitly, we cannot invert this to find x(t) explicitly, but we can find the asymptotic form of the motion near a fixed point and so another questions we can answer is
(iv) What is the asymptotic solution near the fixed points x = 0, x = 2 and x = ∞?x
Example 5.17(i) Near x = 0 the factors (x − 1), (x − 2) and (x − 3) are approximately constant and so the leading behaviour of f(x) is √ f(x) ∼ −6 x , (5.94) which we can check by expanding f(x) out to see √ √ f(x) = x(x3 − 9x2 + 11x − 6) = −6 x(1 + O(x)) . (5.95)
This means we can calculate the asymptotic time of motion for x near 0,
Z x1 asymp dx 1√ 1√ τ01 = √ = x0 − x1 . (5.96) x0 −6 x 3 3
This is finite as x1 → 0 and so the motion terminates. We thus also get an asymptotic estimate of the time of motion from x0 to the fixed point 0, 1√ τ asymp = x . (5.97) 3 0 We can compare this with the exact answer (5.93), which we can expand for small x as
1√ 11 17 τ exact = x 1 + x + x2 + ... . (5.98) 3 18 36
We see, τ asymp ∼ τ exact as x → 0. Example 5.17(ii) √ Near x = 2, the factors x,(x − 1) and (x − 3) are approximately constant, so that √ f(x) ∼ − 2(x − 2) . (5.99)
This means we can calculate the asymptotic time of motion for x near 2,
Z x1 x1 asymp dx 1 τ01 = √ = −√ log |x − 2| . (5.100) x0 − 2(x − 2) 2 x0
This diverges as x1 → 2 and so the asymptotic motion does not terminate in finite time and so the true motion does not terminate. We can see that the exact answer (5.93) also diverges as x → 2 because of the term log |1 − px/2|. Example 5.17(iii) As x → ∞, the constant terms in the factors (x − 1), (x − 2) and (x − 3) become irrelevant and so f(x) ∼ x7/2 . (5.101)
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 77
Again we can check explicitly since √ 1 f(x) = x(x3 − 9x2 + 11x − 6) = x7/2 1 + O . (5.102) x This means we can calculate the asymptotic time of motion when x is large,
Z x1 dx 2 τ asymp = = (x−5/2 − x−5/2) . (5.103) 01 7/2 0 1 x0 x 5
This is finite as x1 → ∞ and so the motion terminates. We also have an asymptotic estimate of the time of motion from x0 to ∞, 2 τ asymp = x−5/2 . (5.104) 5 0 We can compare this with the exact answer (5.93), which we can expand for larges x as
2 30 1 125 1 τ exactx = x−5/2 1 + + + ... . (5.105) 5 0 7 x 9 x2 We see, τ asymp ∼ τ exact as x → ∞ . (5.106)
Example 5.17(iv) From the previous calculations we find the following asymptotic solutions:
For x0 near x = 0, we can use (5.96) to get
asymp √ 2 x (t) = (3t − x0) . (5.107)
For x0 near x = 2, we can use (5.100) to get √ asymp x (t) = 2 + (x0 − 2) exp(− 2t) . (5.108)
For x0 large, we can use (5.103) to get
asymp −5/2 −2/5 x (t) = (5t/2 − x0 ) . (5.109)
It is hard to check these exactly since we cannot easily solve for the exact solution x(t) but they are numerically good. This is left as an exercise for the reader.
Example 5.18. To illustrate further the use of asymptotic analysis in finding asymptotic solution, we now consider a case where we are able to obtain the full solution in addition to the approximate expressions, and thereby fully assess the quality of the approximate solutions. Consider a dynamical system of the form dx = f(x) = −x + x2 . dt It has fixed points at x = 0 (stable) and x = 1 (unstable). The solution of the ODE is obtained by separation of variables, using partial fractions to do the x-integral:
Z x 0 dx x − 1 x0 − 1 t = 0 0 = ln − ln x0 x (x − 1) x x0
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 78
with x(0) = x0. Solving for x gives
x0 x(t) = t x0 − (x0 − 1)e
If x0 < 1, the system will approach the fixed point at 0, if x0 > 1, the system will escape to +∞, and the motion will terminate at time t = ln(x0/(x0 − 1)). Let us compare these exact results with those of the corresponding asymptotic analyses.
(a) Considering the behaviour at small x, we have f(x) ' −x, as x → 0. Solving the corresponding approximate ODE dx ' −x , dt −t assuming x0 1, gives x(t) ' x0e . This is indeed just the result one would get to lowest order in x0 in an expansion of the exact solution. (b) Considering the behaviour at large x, we have f(x) ∼ x2, as x → ∞. Solving the corresponding approximate ODE dx ∼ x2 , dt
x0 assuming x0 1, gives x(t) ' . Asymptotic equality would be guaranteed as long as 1−x0t x − x0 remains finite. In this case, as in the exact solution, we observe that x(t) diverges at finite t. I.e., the motion terminates and the time to escape to +∞ is t = 1/x0. This is indeed just the result one would get to lowest order in 1/x0 in an expansion of the the time at which the motion terminates in the exact solution.
The following figure compares exact and approximate solution in the regions of small and large x. The quality of the approximations is remarkable even in the region of the divergence, although exact and approximate solutions terminate at slightly different times (so that there the error is strictly speaking infinitely large and the approximation bad). Note that the large x solution can even be continued with reasonable precision across the divergence!
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 79
0.11 1e+06
0.1
0.09 100000
0.08
0.07 10000 0.06 x(t) x(t) 0.05 1000 0.04
0.03
100 0.02
0.01
0 10 0 2 4 6 8 10 0 0.002 0.004 0.006 0.008 0.01 t t 10000
5000
0
-5000 x(t)
-10000
-15000
-20000 0 0.005 0.01 0.015 0.02 t
Figure 5.18: (a) Full solution (full curve) and approximate solution (dashed curve) in the region of small x, with x0 = 0.1. (b) Full and approximate solution in the region of large x, with x0 = 100, approaching the divergence. (c) Full and approximate solution in the region of large x crossing the divergence.
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 80
5.6 Exercises
Shorter exercises
Exercise 5.1 Consider the following velocity functions:
(i) v(x) = x3 (ii) v(x) = x−x2 (iii) v(x) = x−x4/3 (iv) v(x) = x1/3 −x4/3
In each case answer the following questions:
(a) Sketch the graph of v(x), find the fixed points of the dynamical system and state whether they are stable or unstable.
(b) For each fixed point x = a,
(i) Does v(x) have a Taylor expansion around x = a to all orders in (x − a)? (ii) Can you write down a linear approximation to v(x) around x = a? (iii) Can you decide if the fixed point is stable or unstable based on the linear analysis?
Exercise 5.2 Prove the following statements on asymptotic behaviour:
(i) f(x) = x2 + x3 , g(x) = x3 , f(x) ∼ g(x) , x → ∞
(ii) f(x) = x2 + x3 , g(x) = x2 , f(x) ∼ g(x) , x → 0 ex (iii) f(x) = cosh(x) , g(x) = , f(x) ∼ g(x) , x → ∞ 2 x2 (iv) f(x) = cos(x) − 1 , g(x) = − , f(x) ∼ g(x) , x → 0 2 1 π (v) f(x) = tan(x) , g(x) = π , f(x) ∼ g(x) , x → 2 − x 2
Longer exercises
Exercise 5.3 Consider the first order autonomous dynamical system dx = f(x) = x2(x − 1)(x − 2)2 dt
(a) Identify the fixed points, and the invariant open sets.
(b) Sketch the phase portrait and characterize the stability of the fixed points using your sketch.
(c) Draw qualitative solution curves, one for each invariant open set. State briefly the main features of these curves, referring to properties of the phase portrait.
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 81
Exercise 5.4 Investigate the first order autonomous dynamical system described by dx = v(x) = (x − 1)(x − 2) . dt Sketch the phase portrait including the velocity function and the phase flow. Write down the fixed points of the motion and the invariant open sets. Give a sketch of qualitative solution curves, one for each invariant open set. Describe them briefly, referring to properties of the phase portrait. Show – without using the exact solution of the differential equation – that, if x(0) = 3/2, the system will not reach the nearest stable fixed point in finite time. Show also that the exact solution of this equation, with initial condition x(0) = 3/2, is
2 + exp(t) x(t) = . 1 + exp(t)
Exercise 5.5 Discuss the system described by dx = v(x) = (x − 1)(x − 2)(x − 3)(x − 4) . dt Sketch the phase portrait. Write down the fixed points of the motion and discuss their stability. Suppose that x(0) = 5. Prove that the motion is terminating. Give upper and lower limits for the time to reach infinity. Show that, if x(0) = 7/2, the system will not reach x = 3 in a finite time.
Exercise 5.6 Consider the dynamical system of problem 5.5 dx = v(x) = (x − 1)(x − 2)(x − 3)(x − 4) . dt For each fixed point find the corresponding linearised system and state the stability of each fixed point based on the linearised system. Check your results by comparing them with the phase portrait.
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 82
Exercise 5.7 (part of 2008 exam) Consider the dynamical system described by dx = f(x) = (x + 1)(2 − x)(x − a). dt where a is a parameter.
(a) State for which values of a the system is structurally stable and for which values of a the fixed point at x = −1 is stable. Give reasons for your answer.
(b) In what follows, assume a = 1.
(i) Identify the fixed points, characterize their stability, and give the invariant open sets of the dynamics. (ii) Sketch the phase portrait. (iii) Draw qualitative solution curves for the initial conditions (i) x(0) = −3/2, (ii) x(0) = 0, and (iii) x(0) = 3/2. Justify your results. (iv) Write the linearised equation about the fixed points x = 1 and x = 2. Solve these equations and comment on how your results are related to the stability of the two fixed points respectively.
Exercise 5.8 Consider the first order dynamical system described by dx √ = v(x) = − x + x , x ≥ 0 dt
(a) Sketch the phase portrait, identify the fixed point(s) and discuss their stability.
(b) Write down a simpler asymptotic equation of motion in the vicinity of x = 0, and solve it with initial condition x(0) = a with 0 < a 1. Show that the asymptotic system will reach the nearest stable fixed point in finite time and calculate this time. Does this show that the full system will also reach the fixed point in finite time?
(c) Solve the differential equation exactly, and compute the time needed to reach the nearest fixed point from x(0) = a with 0 < a 1 using the exact solution. Verify that to lowest order in a the approximate and the exact results agree. √ (d) Show that if 0 < x < 1 then |v(x)| < x. Show that this gives a lower bound on the time for the system to reach x = 0 starting from x(0) = a with 0 < a < 1. Compare this with the exact time required to reach x = 0 and the asymptotic approximation.
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 83
Exercise 5.9 Consider the first order dynamical system described by
dx x5 = f(x) = dt 1 + x2
(a) Sketch the phase portrait, identify the fixed point(s) and discuss their stability.
(b) Write down a simpler asymptotic equation of motion as x → ∞ and solve it with initial condition x(0) = a with a 0. Show that the asymptotic system will reach x = ∞ in finite time and calculate this time. Does this show that the full system will also reach x = ∞ in finite time?
(c) Solve the differential equation exactly, and compute the time needed to reach x = ∞ from x(0) = a with a 0 using the exact solution. Verify that to lowest order in 1/a the approximate and the exact results agree.
(d) Show that if x > 1 then |f(x)| > x3/2. Show that this gives an upper bound on the time for the system to reach x = ∞ starting from x(0) = a with a > 1. Compare this with the exact time required to reach x = 0 and with the asymptotic approximation.
Exercise 5.10 Consider a first order dynamical system evolving in the phase space Γ = {x ≥ 0} according to the equation dx = xγ(x − 2), γ ≥ 0 dt
(a) Write the asymptotic equation of motion valid for x 1 and solve it for the initial condition x(0) = x0, with x0 1. Use this solution to show that the motion is terminating for γ > 0.
(b) Write the asymptotic equation of motion valid in the vicinity of x = 0 and solve it for the initial condition x(0) = x0, 0 < x0 1, assuming γ 6= 1. Use this result to show that the motion is terminating for γ < 1 whereas it is not terminating for γ > 1.
Supplementary exercises
Supplementary Exercise 5.1 Sketch the phase portraits for the three systems defined on the phase space −1 ≤ x ≤ 1: p (i)x ˙ = 1 − x2 (ii)x ˙ = 1 − x2 (iii)x ˙ = (1 − x2)3/2
For which of these is the solution with x(0) = 0 terminating? Can you relate this to the conditions for Picard’s theorem to hold? Can you find the exact solution for x(t) in each case?
Version of Mar 14, 2019 Chapter 5: First Order Autonomous Systems 84
Supplementary Exercise 5.2 Consider the dynamical system defined for x ∈ R
x˙ = exp(−1/x2) .
Sketch the phase portrait for this system. Does it have a fixed point? Is the flow starting from x(0) = −1 terminating? Hint: Consider a change of variables from x to u = −1/x and use the fact that exp(a) > an/n! for all a > 0.
Supplementary Exercise 5.3 The time evolution of the magnetization m of a ferromagnetic material in the vicinity of the critical temperature Tc is well described by the first order autonomous dynamical system dm = v(m) = am − m3 , dt where a = (Tc − T )/Tc and T is the temperature. (i) Sketch the phase portrait including the velocity function and the phase flow using the arrow representation of the flow, for the cases T > Tc and T < Tc. Write down the fixed points of the motion and the invariant open sets of the dynamics for both T > Tc and T < Tc; discuss the stability of the fixed points in both cases. (ii) Show, for T < T – without using the exact solution of the differential equation – that, if √ c √ m(0) = a/2, the system will not reach m = a in finite time. (iii) Show also that for T < T the exact solution of this equation, with initial condition √ c m(0) = a/2, is r a m(t) = 1 + 3 exp(−2at)
(iv) For T > Tc and for T = Tc give solutions of the asymptotic equation of motion in the vicinity of m = 0, assuming m(0) = m0 > 0, but small.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 85
Chapter 6
Second Order Autonomous Systems
A second order autonomous system describes the time evolution of the state of a system in some two-dimensional phase-space Γ through two differential equations. We have already seen one of these in the introductory chapter, the Lotka-Volterra (or fox-rabbit or lynx-hare) system (1.5) where the phase space is the quadrant {x > 0, y > 0} of IR2. This system already exhibits many of the features of second order autonomous systems that are not possible for first order systems, for example it has periodic orbits and two new kinds of fixed points. We have also already seen its phase portrait in figure 1.3. It turns out that the possible behaviours of second order systems are much richer than that of first order systems. In this chapter we look at qualitative methods and at approximate analytic methods to study second order systems, in particular we will look at methods to help sketch phase portraits and to analyse the nature of fixed points. In chapter 4 we said that a second order DS Will describe the time evolution of a system described by two variables (x1, x2) through differential equations of the form dx i = f (x , x ) , i = 1, 2 , (6.1) dt i 1 2
where the functions f1 and f2 are referred to as velocity functions. These functions do not depend on t as that is the definition of autonomous
In this chapter we often find it convenient to use conventional x, y notation instead of x1, x2 and sometimes to write f1, f2 as F,G, so the we could write these equations as dx dx = f1(x, y) = F (x, y) dt or dt , (6.2) dy dy = f (x, y) = G(x, y) dt 2 dt or in vector notation dr d x f (x, y) x˙ F (x, y) ≡ = f(x, y) = 1 or = . (6.3) dt dt y f2(x, y) y˙ G(x, y)
6.1 Phase Space and Phase Portraits
The terminology introduced in chapter 4 applies in an obvious way to second order au- tonomous systems.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 86
Definition 6.1. Phase Space Γ The phase space Γ is defined as the set of admissible values for the coordinates x and y. The pair (x, y) denotes a point in IR2, so generally Γ ⊆ IR2. For some systems, Γ may be a proper subset of IR2: e.g., if x and y denote concentrations of two substances in a binary chemical reaction, then x ≥ 0 and y ≥ 0.
Definition 6.2. Trajectory/Solution Curve (through x0) The set of points {(t, x(t), y(t)); t ∈ IR} with (x, y) a solution of (6.3) (and x(t0) = x0 and y(t0) = y0 for some t0) is called a trajectory or solution curve (through (x0, y0)). Definition 6.3. Fixed Points 2 A point (a, b) ∈ IR is a fixed point of the dynamical system (6.3) if f f1(a, b) = 0 and f2(a, b) = 0. This means that the constant trajectory {x(t) = a, y(t) = b} is a solution of the dynamical system (6.2).
Definition 6.4. Orbit or Phase Curve (through (x0, y0)) The set of points {(x(t), y(t)); t ∈ IR} with (x, y) a solution of (6.3) (and x(t0) = x0 and y(t0) = y0 for some t0) is called orbit or phase curve (through (x0, y0)).
If we parametrise the phase curve as y(x) then we have
dy y˙ G(x, y) = = . (6.4) dx x˙ F (x, y) It is sometimes possible to solve this differential equation to find the phase curves even if the t dependence of x(t) and y(t) cannot be found. It is also sometimes useful to solve the equations for the phase curve first and then substitute y = y(x) in the equation forx ˙ to solve for x(t) explicitly. Note: Provided the functions F and G satisfy the conditions of Picard’s theorem in a region around a point then there is precisely one trajectory that starts at that point and hence one phase curve through that point. This also means that a trajectory starting at a fixed point remains there for all time; if there are several distinct phase curves passing through the same point, or if a trajectory can reach a fixed point in finite time then the Picard conditions must fail at that point. Definition 6.5. Phase Flow The collection of all possible phase curves is called the phase flow of the dynamical system (6.3). Velocity information, i.e., f(x, y) = f1(x, y)e1 + f2(x, y)e2 drawn at each (x, y) ∈ Γ is often included in representations of the phase flow. The vector f(x, y) is called the velocity of the flow (at (x, y)) The set of vectors {f(x, y); (x, y) ∈ Γ} is called the velocity field. Definition 6.6. Phase Portrait A graphical representation of the the phase space Γ, including a graph of the phase flow and/or the velocity f(x, y) evaluated at representative points. There are several ways to present the information in a phase portrait which we show in the examples below.
There is one new concept which is very helpful especially in drawing phase portraits, and that is
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 87
Definition 6.7. Null-Clines 2 The set of points (x, y) ∈ IR for which f1(x, y) = 0 is called null-cline of f1.
The equation of the null-cline, f1(x, y) = 0, is a relation that must hold between x and y for points (x, y) to be on the null-cline. This generally defines a curve in IR2. For instance if f(x, y) = x2 + y2 − r2, then the null-cline of f is a circle of radius r centred at the origin of IR2.
Note: On a null-cline of f1, the x-component of the velocity vanishes so that the flow is vertical; similarly, on a null-cline of f2, it is the y-component of the velocity which is zero so that the flow is horizontal. Fixed points are intersections of the null-clines of f1 and f2: It is common to include the null-clines of the velocity functions in the phase portrait since the null clines divide the phase space into regions on which the velocity field points in different general directions which we can call by the points of the compass: NW, NE, SW, SE:
NE If {f1 > 0, f2 > 0} then the flow is in the general direction % or in the direction NE.
NW If {f1 < 0, f2 > 0} then the flow is in the general direction - or in the direction NW.
SW If {f1 < 0, f2 < 0} then the flow is in the general direction . or in the direction SW.
SE If {f1 > 0, f2 < 0} then the flow is in the general direction & or in the direction SE.
The velocity field points N, S (vertical) or E, W (horizontal) on the null-clines themselves. We now go through several examples in increasing complexity before turning to some of the ideas they highlight.
Example 6.1. Consider the second order autonomous systems with phase space the plane defined by ( ( ( ( ( x˙ = 2 x˙ = 2 x˙ = x x˙ = y x˙ = 2xy (a) (b) (c) (d) (e) y˙ = −1 y˙ = x y˙ = y y˙ = −x y˙ = y2 − x2
The general procedure we shall use is the following:
(i) We first plot the null clines. These are the sets on which f1(x, y) and f2(x, y) vanish. The velocity field is vertical where f1 = 0 and horizontal where f2 = 0. (ii) The null-clines divide the phase space into 8 regions on which the velocity field is in one of the directions NW, NE, SE, SW. The velocity field points N, S, E or W on the null-clines themselves. We can label each of the 8 regions by the sign of the components fi and in this way come to an understanding of the direction of the velocity field.
(iii) Looking at magnitude of f, we can see where the velocity is larger or smaller; i.w. where the motion is faster or slower.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 88
(iv) We can sketch arrows at representative points to show the direction, or the direction and magnitude, of the velocity field1
(v) We try to sketch the orbits by joining up the arrows. We can check this by solving for the equations of the phase curves exactly, if that is possible
Example 6.1(a) This is the system
x˙ 2 x˙ = F (x, y) = 2 , y˙ = G(x, y) = −1 , = f = . (6.5) y˙ −1
(i), (ii) The velocity functions F (x, y) and G(x, y) are never zero, so we could say that the null-clines are the empty sets. (iii) The velocity function is a constant so it has the same magnitude everywhere. (iv) We can sketch the velocity field by plotting the vector f at representative points on a typical part of the phase plane (see figure 6.1). (v) If we join up the arrows we get straight lines. We can check by solving the equations exactly, 1 x˙ = 2 ⇒ x = 2t + a , y˙ = −1 ⇒ y = −t + b , so y = − x + c , (6.6) 2 or solving the equation for the phase curves,
dy G(x, y) 1 = = − , y = −x/2 + y . (6.7) dx F (x, y) 2 0
2 2
1 1
0 0
-1 -1
-2 -2
-2 -1 0 1 2 -2 -1 0 1 2 (a) A set of arrows represent- (b) A set of orbits joining up ing the velocity field. the arrows of the velocity field.
Figure 6.1: Two examples of a phase portrait for the system (6.5).
1Note: If the magnitudes of the velocities vary in a large range, one may choose to plot the direction field n(x, y) = f(x, y)/|f(x, y)| (instead of the velocity field) in a phase diagram; The direction field is a field of unit vectors pointing in the direction of the velocity. As with the flow, visual impression of the actual direction field can be more faithful, if the arrows representing the direction field n(x, y) are not drawn such that they are rooted in (x, y) (and as they strictly should) but rather with their midpoint at (x, y). This is the typical result of the Mathematica command VectorPlot.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 89
Example 6.1(b) This is the system x˙ 2 x˙ = F (x, y) = 2 , y˙ = G(x, y) = x , = f = . (6.8) y˙ x (i) The velocity function F (x, y) is never zero but G(x, y) = 0 on the y–axis, x = 0. (ii) F (x, y) > 0 always but G(x, y) < 0 for x < 0 and G(x, y) > 0 for x > 0, hence the flow points SE for x < 0 and NE for x > 0.√ (iii) The velocity function has |f| = 2 + x2 and so |f| is bigger the further it is from the y–axis. (iv) We can sketch the velocity field by plotting the vector f at representative points on a typical part of the phase plane (see figure 6.2). (v) If we join up the arrows we get curves. We can check by solving the equation for the phase curves, dy G(x, y) x 1 = = , y = x2 + y , (6.9) dx F (x, y) 2 4 0 and we see that these are actually parabolae, which we can also sketch (see figure 6.2).
2 2
1 1 f1>0, f2<0: SE f1>0, f2>0: NE
0 0
-1 -1
-2 -2 -2 -1 0 1 2 -2 -1 0 1 2 (a) The null-clines showing (b) The regions in which the where the velocity field is hor- velocity field points SE and izontal and vertical. NE.
2 2
1 1
0 0
-1 -1
-2 -2
-2 -1 0 1 2 -2 -1 0 1 2 (c) A set of arrows represent- (d) A set of orbits joining up ing the velocity field. the arrows of the velocity field.
Figure 6.2: Constricting the phase portrait for the system (6.8).
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 90
Example 6.1(c) This is the system
x˙ x x˙ = F (x, y) = x , y˙ = G(x, y) = y , = f = . (6.10) y˙ y
(i) The velocity function F (x, y) = 0 on the y–axis and G(x, y) = 0 on the x–axis. These intersect at the point (0, 0) and so this is a fixed point. (ii) The velocity field points NE in the first quadrant, NW in the second, SW in the third and SE in the fourth. p (iii) The velocity function has |f| = x2 + y2 and so the field is bigger and the flow faster the further from the origin. (iv) The velocity fields points radially away from the origin (see figure 6.3). (v) If we join up the arrows it looks like we get radial straight lines. We can check by solving the equations for the flows,
x˙ = x ⇒ x = aet , y˙ = y ⇒ y = bet , so y = (b/a)x , (6.11) or solving directly for the phase curves, as the equation is separable:
dy G(x, y) y = = , log y = log x + c , y = Ax , (6.12) dx F (x, y) x and we see that these are indeed straight lines.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 91
2 2
f1<0, f2>0: NW f1>0, f2>0: NE
1 1
0 0
-1 -1
f1<0, f2<0: SW f1>0, f2<0: SE
-2 -2 -2 -1 0 1 2 -2 -1 0 1 2 (a) The null-clines showing (b) The regions in which the where the velocity field is hor- velocity field points SE and izontal and vertical. NE.
2 2
1 1
0 0
-1 -1
-2 -2
-2 -1 0 1 2 -2 -1 0 1 2 (c) A set of arrows represent- (d) A set of orbits joining up ing the velocity field. the arrows of the velocity field.
Figure 6.3: Constructing the phase portrait for the system (6.10).
Example 6.1(d) This is the system x˙ −y x˙ = F (x, y) = −y , y˙ = G(x, y) = x , = f = . (6.13) y˙ x
(i) The velocity function F (x, y) = 0 on the x–axis and G(x, y) = 0 on the y–axis. These intersect at the point (0, 0) and so this is a fixed point. (ii) The velocity field points NW in the first quadrant, SW in the second, SE in the third and NE in the fourth. p (iii) The velocity function has |f| = x2 + y2 and so the field is bigger and the flow faster the further from the origin. (iv) The velocity fields points radially away from the origin (see figure 6.4). (v) If we join up the arrows it looks like we get radial straight lines. We can check by solving the equations for the flows,
x˙ = −y , y˙ = x , ⇒ x¨ = −x , ⇒ x = A cos t + B sin t , y = A sin t − B cos t ,
so x2 + y2 = A2 + Ba , (6.14)
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 92 or solving directly for the phase curves, as the equation is separable
dy G(x, y) x 1 1 = = − , y2 = − x2 + c , x2 + y2 = A, (6.15) dx F (x, y) y 2 2 and we see that these are indeed circles.
2 2
f1<0, f2<0: SW f1<0, f2>0: NW
1 1
0 0
-1 -1
f1>0, f2<0: SE f1>0, f2>0: NE
-2 -2 -2 -1 0 1 2 -2 -1 0 1 2 (a) The null-clines showing (b) The regions in which the where the velocity field is hor- velocity field points SE and izontal and vertical. NE.
2 2
1 1
0 0
-1 -1
-2 -2
-2 -1 0 1 2 -2 -1 0 1 2 (c) A set of arrows represent- (d) A set of orbits joining up ing the velocity field. the arrows of the velocity field.
Figure 6.4: Constructing the phase portrait for the system (6.13).
Example 6.1(e) Describe the phase portrait associated with the velocity field given by 2 2 f1(x, y) 2 2 f(x, y) = 2xye1 + (y − x )e2 , f(x, y) = , f1 = 2xy , f2 = y − x . (6.16) f2(x, y)
(i) f1(x, y) = 2xy vanishes on the lines x = 0 and y = 0 so that the velocity field is vertical 2 2 on these lines; f1(x, y) = y − x vanishes on the lines x = y and y = −x so that the velocity field is horizontal on these lines. The null-clines intersect at one point (0, 0) which is therefore a fixed point of the DS. (ii) The null-clines divide the phase space into 8 regions on which the velocity field is in one of the directions NW, NE, SE, SW as in figure 6.5. (iii) Looking at the fi we see that the velocity field has larger magnitude the further it is from the origin. If we
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 93 want to be more accurate, the velocity field at (x, y) makes an angle θ with the x-axis where
f y2 − x2 tan θ = 2 = f1 2xy and has magnitude proportional to
|f(x, y)| = ((4x2y2 + (y2 − x2)2)1/2 = (x2 + y2) .
(iv) The velocity field appears to be making circles symmetrically either side of the y–axis. (v) We cannot easily solve the equations for the trajectories, (6.16), but we have already solved the equation of the phase curves
dy y2 − x2 = , (6.17) dx 2xy in example 2.6 with the general solution
(x − A/2)2 + y2 = (A/2)2 . (6.18)
This describes a family of circles, with centres at (A/2, 0) and radii A/2. Note that they all pass through (0, 0), but since this is a fixed point the orbits are actually the circles with the origin omitted.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 94
2 2 f1<0, f2>0: NW f1>0, f2>0: NE
1 1
f1<0, f2<0: SW f1>0, f2<0: SE
0 0
f1>0, f2<0: SE f1<0, f2<0: SW
-1 -1
-2 -2 f1>0, f2>0: NE f1>0, f2>0: NW
-2 -1 0 1 2 -2 -1 0 1 2 (a) The null-clines showing (b) The regions in which the where the velocity field is hor- velocity field points NE, NW, izontal and vertical. SE, SW.
2 2
1 1
0 0
-1 -1
-2 -2
-2 -1 0 1 2 -2 -1 0 1 2 (c) A set of arrows represent- (d) A set of orbits joining up ing the velocity field. the arrows of the velocity field.
Figure 6.5: Steps in a construction of a sketch of the phase-portrait of velocity field (6.16).
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 95
6.2 Separable Systems
In second order dynamical systems the dynamical evolutions of the variables x and y are quite generally coupled: both velocity functions depend on x and y. If we have the special situation x˙ = F (x) , y˙ = G(y) , (6.19) then we say that the system is separable, the differential equations for x(t) and y(t) are also separable and can be solved easily to find Z x(t) dx Z y(t) dy t − t0 = = . (6.20) x0 F (x) y0 G(y) If a system is not separable in the original coordinates (x, y) it may be possible to find new coordinates (u, v) in which it is2. We have already met a system which is separable in new coordinates, which is example 6.1(d). The orbits are circles and so it could be helpful to consider polar coordinates (r, θ) defined by ( ( x = r cos θ r = px2 + y2 , y . (6.21) y = r sin θ θ = arctan x To findr ˙ and θ˙ it can be simplest to differentiate the equations y r2 = x2 + y2 , tan θ = , (6.22) x leading to xx˙ + yy˙ xF (x, y) + yG(x, y) 2rr˙ = 2xx˙ + 2yy˙ ⇒ r˙ = = , (6.23) r r xy˙ − yx˙ xy˙ − yx˙ xG(x, y) − yF (x, y) sec2 θ θ˙ = (1 + tan2 θ)θ˙ = ⇒ θ˙ = = . (6.24) x2 x2 + y2 x2 + y2 We now turn to example 6.1(d) Example 6.2. Consider the following DS in polar coordinates
x˙ = F (x, y) = −y , y˙ = G(x, y) = x . (6.25) We find x F (x, y) + y G(x, y) x G(x, y) − y F (x, y) r˙ = = 0 , θ˙ = = 1 . (6.26) r x2 + y2 These equations are both separable and very simple with the solutions ( ( r = r x = r cos(t + θ ) 0 , 0 0 . (6.27) θ = t + θ0 y = r0 sin(t + θ0) The phase portrait is in figure 6.4.
2It is always possible to find coordinates which cover a small region of phase space around any point which is not a fixed point in which the equations are separable, see supplementary exercise 6.3, but these do not have to extend to the whole of phase space.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 96
Example 6.3. Another system separable in polar coordinates Consider the system described by the equations
x˙ = αx − ωy, y˙ = ωx + αy, (6.28)
where α, ω are constants. Changing to polar coordinates (r, θ) we find
x F (x, y) + y G(x, y) x2 + y2 x G(x, y) − y F (x, y) r˙ = = α = αr , θ˙ = = ω . (6.29) r r x2 + y2 These are again very simple and can be solved to find the trajectories
αt r = r0e , θ = ωt + θ0 . (6.30)
The orbits are spirals. If α > 0 these are spiralling away form the origin and the origin is an unstable fixed point; if α < 0 they are spiralling in towards the origin which is a stable fixed point.
2 2
1 1
0 0
-1 -1
-2 -2
-2 -1 0 1 2 -2 -1 0 1 2 (a) The case α = 1, ω = 4: an (b) The case α = −1, ω = 4: a unstable fixed point. stable fixed point.
Figure 6.6: Phase portraits of two examples of the DS (6.28).
A more complicated example separable in polar coordinates.
Example 6.4. Show that the following system is separable in polar coordinates and sketch the phase portrait
x˙ = F = x − x3 − xy2 − y , y˙ = G = y − y3 − yx2 + x . (6.31) Following the previous procedure we find x F + y G x G − x F r˙ = = r − r3 , θ˙ = = 1 . (6.32) r r2
The solution for θ = θ0 + t; the equation for r can be analysed as first order autonomous system (with phase space r ≥ 0) and we see that it has two fixed points, an unstable fixed point at r = 0 and a stable fixed point at r = 1 (corresponding to a periodic trajectory of the full system with r = 1 and a closed orbit, the unit circle). The remaining orbits of the
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 97 full system either spiral out from the fixed point at (0, 0) and tend towards the unit circle or spiral in from infinity again tending towards the unit circle. The unit circle is an example of a limit cycle. This is illustrated in figure 6.7.
2
1
fHrL
0
r -1 1
-2
-2 -1 0 1 2 (a) The phase portrait of the (b) The phase portrait of the first order DSr ˙ = r(1 − r2). full system.
Figure 6.7: Phase portraits of the DS (6.31).
A final example separable in different coordinates
Example 6.5. Show that the following DS is separable in coordinates (u, v) where x = u cosh v, y = u sinh v:
x˙ = y , y˙ = x . (6.33) We can findu ˙ andv ˙ from differentiating u2 = x2 − y2 and tanh v = y/x, or instead by direct substitution x y x˙ =u ˙ cosh v + uv˙ sinh v = u˙ + yv˙ = y , y˙ =u ˙ sinh v + uv˙ cosh v = u˙ + xv˙ = x , u u u˙ = 0 , v˙ = 1 , (6.34) which is separable with trajectories
u = u0 , v = t + v0 . (6.35)
The orbits are hyperbolae, as in figure 6.8 since
2 2 2 2 2 2 x − y = u (cosh v − sinh v) = u0 . (6.36)
Note that the coordinates (u, v) only cover the region |x| ≥ |y|. Can you suggest a change of coordinates to cover the other region?
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 98
2
1
0
-1
-2 -2 -1 0 1 2
Figure 6.8: Phase portraits of the system (6.33). The coordinates (u, v) cover the shaded region only.
6.3 The structure of orbits and Phase Space
We have seen that for a well-behaved velocity function (obey the requirements of Picard’s theorem) the orbits of a first order dynamical system are either fixed points or open intervals. Each trajectory in a first order system either stays at a fixed point or approaches a fixed point (or ±∞). There are only three kinds of fixed point, a stable fixed point and two kinds of unstable fixed point as in figure 5.12. The situation is much more complicated for second order autonomous systems. In the few examples so far we have already met several different types of stable and unstable fixed point, and a new kind of limiting behaviour, the limit cycle. It is in fact possible to show that the limiting behaviour of any trajectory is either (a) a fixed point (b) a limit cycle or (c) a union of fixed points and open intervals interpolating them but this is too complicated to even state carefully here, quite apart from contemplate proving. The example (6.137) in supplementary exercise 6.2 with α > 0 exhibits just such a complicated behaviour; the phase portrait is in figure 6.8. There are fixed points at (0, 0), (−1, −1) and (1, −1) and orbits between these last two; every orbit that starts close to (0, 0) ends up approaching the other two fixed points and the orbits between them ever more closely. This might sound complicated, but it is still simple compared to the behaviour for third or higher order systems where far more random and chaotic behaviour is possible. The reason is it is still relatively simple for second order systems is that the phase space is in the plane and an orbit is a line on the plane which divides the space up into the part on one side of the line and the part on the other. A line in three dimensions does not split three-dimensional space up at all.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 99
2
1 1 Α= 4
0
-1
-2 -2 -1 0 1 2
Figure 6.9: Phase portraits of the system (6.137) with α = 1/4..
In this course we will look more deeply into two aspects of the structure of second order systems, namely the stability of limit cycles and the stability of fixed points. The first topic is relatively straightforward as it is essentially a one-dimensional problem and we summarise the results in section 6.4; the second is more complicated. We shall look in detail into fixed points in 6.5 and undertake a linear stability analysis and classify the possible fixed points that can arise in this way in section 6.6.
6.4 Limit cycles
A limit cycle of a second order dynamical system is a periodic trajectory corresponding to a closed orbit that does not pass through any fixed points. The orbits that pass close to the limit cycle can either all stay close to, or be attracted in towards, the limit cycle, in which case it is a stable limit cycle, they can all be repelled (it is an unstable limit cycle) or some can be attracted and some repelled; formally this is also unstable but can be called “half-stable”. We have already such a system in example 6.4 for which the unit circle is a stable limit cycle - all nearby trajectories spiral ever closer to the unit circle. We shall give some examples of the other types of behaviour for systems separable in polar coordinates.
Example 6.6. Consider the systems in polar coordinates with
r˙ = f(r) , θ˙ = 1 . (6.37) The condition for there to be a limit cycle is that the constant trajectory r = a is a solution, which requires f(a) = 0. From our analysis in section 5.4.1 on the stability of fixed points, if f 0(a) > 0 then this is an unstable fixed point of the first order system for r and hence an unstable limit cycle; the first order system and the limit cycle is “half-stable”. Three simple models of this are
(a) f(r) = r − r3: the fixed point at r = 1 has f(1) = 0 and f 0(1) = −2, hence it is stable.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 100
This is precisely example 6.4 and the phase portrait of the first order system for r and the full second order system are in figure 6.7
(b) f(r) = −r + r3: the fixed point at r = 1 has f(1) = 0 and f 0(1) = +2, hence it is unstable.
(c) f(r) = r(1 − r2)2: now f(1) = f 0(1) = 0 but f 00(1) = 8, so the cycle is “half-stable”.
We illustrate cases (b) and (c) in figures 6.10 and 6.11 with plots of f(r), the phase portrait of the first order system corresponding to r and the phase portrait of the full second order system.
2
1
fHrL
0
r -1 1
-2
-2 -1 0 1 2 (a) The phase portrait of the (b) The phase portrait of the first order DSr ˙ = −r(1 − r2). full system.
Figure 6.10: Phase portraits of the DS (6.6)(b).
2
1
fHrL
0
-1
r 1
-2
-2 -1 0 1 2 (a) The phase portrait of the (b) The phase portrait of the first order DSr ˙ = r(1 − r2)2. full system.
Figure 6.11: Phase portraits of the DS (6.6)(c).
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 101
6.5 Fixed points of second-order autonomous systems
Let’s consider a second order autonomous system described by dx dy = F (x, y) , = G(x, y) , (6.38) dt dt or in vector notation d x F (x, y) = f(x, y) = . (6.39) dt y G(x, y) Let’s assume that the point (a, b) is a fixed point, i.e. F (a, b) = G(a, b) = 0 and (a, b) is an intersection point of the null-clines of F and G. The constant trajectory {x = a, y = b} is a solution of the dynamical system. This means that we can perform a Taylor expansion of the functions F and G around the fixed point and the constant terms will be zero. The Taylor expansion in two dimensions is ∂f ∂f f(x, y) = f(a + x − a, b + y − b)) = f(a, b) + (x − a) + (y − b) ∂x ∂y 1 ∂2f ∂2f ∂2f + (x − a)2 + 2 (x − a)(y − b) + (y − b)2 + ..., (6.40) 2! ∂x2 ∂x∂y ∂y2 where it is understood that all partial derivatives are evaluated at (x, y) = (a, b). Only terms up to second order in the small quantities x − a and y − b are explicitly shown. Applying this expansion idea to both velocity functions F and G, and assuming that (x, y) = (a, b) is a fixed point of the system, so that the zeroth order terms vanish, F (a, b) = G(a, b) = 0, we get ∂F ∂F F (x, y) = (x − a) + (y − b) + ... ∂x ∂y ∂G ∂G G(x, y) = (x − a) + (y − b) + ... (6.41) ∂x ∂y
We saw in section 5.4.1 that if f 0(a) 6= 0 for a first order system at a fixed point f(a) = 0 then the nature of the fixed point is completely determined by the sign of f. We can try the same analysis here for a second order system: if we neglect the second order and higher terms in 6.41 and approximate the full system by a linear system of equations we can ask: under what conditions is the nature of the fixed point determined by the linear system, and what quantities are necessary to determine the nature? This is the subject of the next section.
6.6 Linear Stability Analysis
The idea is to approximate a second order DS near a fixed point by a linear DS. Sometimes this is all that is needed to determine its stability - sometimes not. We will investigate the stability of the fixed point by performing a linear stability analysis, based on a two-variable Taylor expansion (to first order) of the velocity functions F and G in the vicinity of the fixed point. This procedure involves three steps.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 102
(1) Taylor expansion of the two velocity functions in the vicinity of the fixed point (a, b) in question. At first order, the terms in this expansion can be arranged into a matrix, the so-called Jacobian J of the system.
(2) Computing the associated Jordan canonical forms C, corresponding to J; the Jordan canonical form is determined by the eigenvalues. We will see how to (almost) determined the form from analysing simple quantities calculated from the matrix J without needing to find the eigenvalues themselves.
(3) Exploring the consequences for dynamics: The linearised equations of motion formulated in a coordinate system corresponding to the Jordan forms can all be solved. The solutions allow a complete classification of the fixed points (and the nature of the phase flow in their vicinity).
6.6.1 Step 1 — Taylor Expansion of Velocity Functions
We start from the differential equations (6.38) and substitute F (x, y) and G(x, y) by their first order approximations from (6.41) and obtain the linearised system
∂F ∂F ∂G ∂G x˙ = (x − a) + (y − b) , y˙ = (x − a) + (y − b) . (6.42) ∂x ∂y ∂x ∂y It is no much more useful to write this in vector notation as ! ! ! ! x˙ (x − a) ∂F + (y − b) ∂F ∂F ∂F x − a = ∂x ∂y = ∂x ∂y , (6.43) ∂G ∂G ∂G ∂G y˙ (x − a) ∂x + (y − b) ∂y ∂x ∂y y − b | {z } The Jacobian J where J denotes the Jacobian matrix given by
∂F ∂F ! J = ∂x ∂y , (6.44) ∂G ∂G ∂x ∂y evaluated at the fixed point (a, b). Writingx ˜ = x − a, y˜ = y − b Eq (6.43) becomes
d x˜ x˜ = J . (6.45) dt y˜ y˜
The change of variable (x, y) → (˜x, y˜) represents a change of origin; the origin of thex ˜-˜y coordinates is in fact the fixed point (x = a, y = b). Linear systems of the sort (6.45) are straight forward to solve; we have already solved several in examples 6.1(a), (b), (c) and (d). Let us consider another useful example:
Example 6.7. Find the solutions of the following linear dynamical system
x˙ 2x − y 2 −1 x = = . (6.46) y˙ y 0 1 y
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 103
If we define X = x − y and Y = y then we find X˙ =x ˙ − y˙ = 2(x − y) = 2X, Y˙ =y ˙ = y = Y, (6.47) or X˙ 2 0 X = . (6.48) Y˙ 0 1 Y The equations have decoupled and we find the solutions X = a e2t ,Y = b et . (6.49) We see that (0, 0) is a strongly unstable fixed point of the type known as an “unstable node” (will give all the various types names later on). The orbits in the new coordinates are parabolae, a X = Y 2 . (6.50) b2 We can plot the orbits in both the new coordinates (X,Y ) and the old coordinates (x, y) in figure 6.12
2 2
1 1
0 0
-1 -1
-2 -2
-2 -1 0 1 2 -2 -1 0 1 2
Figure 6.12: Sketch of the trajectories (5.4) for the DS (6.46) in original coordinates (x, y) (on the left) and redefined (X,Y ) coordinates (on the right).
If we think about what we have done, we took a system of the form x˙ x = M , (6.51) y˙ y where M is a constant matrix and found new coordinates (X,Y ), defined in terms of a new constant matrix P by x X X x = P , = P −1 . (6.52) y Y Y y so that the system X˙ X = P −1 MP , (6.53) Y˙ | {z } Y C
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 104 is simpler. It turns out that the analysis works well if we take the matrix C to the Jordan canonical form of J — this is why we have called it C, for canonical. It is time to revise some linear algebra and see how to find the canonical form C of a matrix J.
6.6.2 Step 2 — Finding the Jordan canonical form of the Jacobian
The Jordan canonical form of a matrix M is determined by the eigenvalues of the matrix M. If they are both real and distinct, as in example 6.7, then the canonical form is diagonal. One way to find the eigenvalues of a 2×2 matrix M we can assume the existence of an eigenvector and solve the resulting equations. If
a b M = , (6.54) c d
1 and we assume that the vector v = is an eigenvector with eigenvalue λ then from the α equation M v = λ v , (6.55) we can eliminate α and obtain the quadratic equation for λ
p(λ) = λ2 − tλ + ∆ = 0 , (6.56)
where t is the trace of the matrix M and ∆ is the determinant of M:
t = (a + d) = Tr(M) , ∆ = ad − bc = Det(M) . (6.57)
The equation (6.56) is called the characteristic equation for λ. It is a Theorem, the Cayley-Hamilton theorem, that every matrix M obeys its own charac- teristic equation, i.e. p(M) = M 2 − t M + ∆ I = 0 , (6.58) where I is the identity matrix. Given the characteristic equation (6.56), we can find the eigenvalues as 1 1√ λ = t ± D,D = t2 − 4∆ . (6.59) pm 2 2 D is called the discriminant of the matrix M. We now write the characteristic equation as
2 p(λ) = (λ − λ+)(λ − λ−) = λ − (λ+ + λ−)λ + (λ+λ) = 0 , (6.60) and comparing this with (6.56) we see that
t = λ+ + λ− , ∆ = λ+λ− . (6.61)
Looking at the form of the eigenvalues (6.59), we see that it is the sign of the discriminant D which determines if they are real or complex. We distinguish three cases:
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 105
(1) D > 0, λ± are both real and the Jordan canonical form C is diagonal. It is possible to find a real matrix P such that the canonical form C is diagonal, λ 0 C = P −1MP = + . (6.62) 0 λ−
(2) D < 0, λ± are complex conjugate complex numbers, λ± = α ± iβ. It is possible to find a real matrix P such that the canonical form is α β C = P −1MP = . (6.63) −β α
and the Jordan canonical form is not diagonal (It has a complex diagonal canonical form but this is the real canonical form). (3) D = 0, there is only one solution for λ but two choices for the Jordan canonical form C. (3i) If M is diagonal, then it is already in canonical form,
λ 0 C = M = . (6.64) 0 λ
(3i) If M is not diagonal, then it is possible to find a real matrix P such that
λ 1 C = P −1MP = . (6.65) 0 λ
It is sometimes necessary to find a suitable matrix P – this is covered in section C.3 in appendix C. We note that P is not unique: there is an infinite choice of suitable matrices P , there is a procedure in appendix C which finds one of which works.
6.6.3 Step 4 — Exploring the Consequences for Dynamics
It turns out that the the resulting differential equations, when formulated in the X,Y coor- dinate system can be readily solved, and the stability of the fixed point determined for each of the three different variants of Jordan canonical form. Note that the origin of the new coordinates (X,Y ) is in fact the fixed point (x = a, y = b). We assume that the stability of the fixed point (a, b) of the system described by equation 6.3 may be determined by examining the behaviour of the linearised system 6.53 near X = 0,Y = 0; this is valid provided the matrix J is non-singular. If one of the eigenvalues of J were zero, and hence J singular, one would in fact have to consider higher orders in the Taylor expansion of the velocity functions to finally decide the stability of a fixed point.
We are looking at d X X = C . (6.66) dt Y Y Referring to the three different types of Jordan canonical form introduced above we can rewrite this in terms of two coupled equations for the components X and Y . We now go through the various cases systematically.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 106
D > 0 there are two real and distinct eigenvalues. λ1 6= λ2 ∈ IR we get the two equations d d X = λ X, Y = λ Y dt 1 dt 2 which are readily solved to give
λ1t λ2t X(t) = X0e ,Y (t) = Y0e
All that is left is to work out the signs of the eigenvalues which is fixed by t and ∆
D > 0, ∆ > 0 This means λ1λ2 > 0 and so both eigenvalues are of the same sign. There are two choices: D > 0, ∆ > 0, t > 0 This means both eigenval- ues are positive, λ1, λ2 > 0. The motion is away from the fixed point: we get a so- called unstable node.
D > 0, ∆ > 0, t < 0 This means both eigenval- ues are negative, λ1, λ2 < 0. The motion is towards the fixed point: we get a so- called stable node.
D > 0, ∆ < 0 This means λ1λ2 < 0 and so the eigen- values are of opposite same sign. This means the motion is out along one axis and in along the other. This fixed point is unstable but it is given a special name, it is a hyperbolic fixed point.
D < 0 This means there are a pair of complex conjugate eigenvalues: λ1,2 = µ ± iν with µ ∈ IR and ν ∈ IR. In this case, the equations of motion for the X and Y coordinate are d d X = µX + νY , Y = −νX + µY dt dt This system of equations is separable in polar coordinates X = R cos θ, Y = R sin θ. Following the line of reasoning in Section 6.2, we find a solution of the form
µt R(t) = R0 e , θ(t) = θ0 − νt
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 107
which translates back into the X,Y coordinate system as
µt µt X(t) = R0 e cos(θ0 − νt) ,Y (t) = R0 e sin(θ0 − νt)
The nature of the fixed point now depends on whether the motion is spiralling away from the fixed point (µ > 0) or spiralling in towards the fixed point (µ < 0). Since t = 2µ, we can decide this on the sign of t.
D < 0, t > 0 This means µ > 0 and so the motion is spiralling away from the fixed point. This is known as an unstable focus.
D < 0, t < 0 This means µ < 0 and so the motion is spiralling in to towards from the fixed point. This is known as a (strongly) stable focus. The word strongly is used to differentiate it from the elliptical fixed point below.
D < 0, t = 0 This means µ = 0 and so the motion is circular in the coordinates (X,Y ). When transformed back into the coor- dinates (x, y) it would be an ellipse and so this fixed point is known as an ellip- tical fixed point. It is also mid-way between the stable and unstable foci - a small change in the system would turn it into one or the other and so it is also known as a marginally stable focus Finally we come to the case
D = 0 This means there is only one solution to the characteristic equation p(λ) = 0. C takes one of two forms: λ 0 λ 1 , (6.67) 0 λ 0 λ In the first case C = λI and so M = λI as well. This means that if M 6= λI then C must be of the second form. If the matrix C = λI then the flow is diagonalisable then it is very similar to the case of two distinct real eigenvalues except that the orbits are straight lines. It only depends on whether λ > 0 or λ < 0 to decide if the fixed point is unstable or stable. If, however, C is not diagonalisable and instead takes the form
λ 1 C = 0 λ
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 108
then we need to solve the equations d d X = λX + Y, Y = λY dt dt λt Note that the Y dynamics is independent of X; the solution is Y (t) = Y0e . Inserting this into the ODE for X, one obtains a first order linear equation which is readily solved (using integrating factors). Overall, we get
λt λt X(t) = (X0 + Y0 t)e ,Y (t) = Y0e Note that the ratio of X and Y as a function of t is Y (t) Y = 0 , X(t) X0 + Y0 t and it approaches 0 for t → ±∞. This type of fixed point is called an improper node and again it only depends on the sign of λ to determine if it is stable or unstable. Since λ = t/2, we need only look at the sign of t to decide which the case is. This gives us four cases D = 0, t > 0, M, diagonal This is an unstable node.
D = 0, t > 0, M not diagonal This is an unstable improper node
D = 0, t < 0, M, diagonal This is a stable node.
D = 0, t < 0, M not diagonal This is a stable improper node]
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 109
The properties of the phase flow in the vicinity of the fixed point are finally obtained from the properties of the phase flow in the X,Y coordinate system by transforming back into the x,˜ y˜ system via x˜ X = P ; (6.68) y˜ Y see Eq. (6.52). This transformation is linear (and non-singular). It may involve rotations, reflections, and scaling of axes. We summarise the eight types of flow pattern and the various fixed points in figure 6.13 t
Δ>0, D>0 Unstable node D=0, t>0 Unstable node or unstable improper node Δ<0, D>0 or Hyperbolic fixed point =0 D , Δ =4 D<0, t>0 2t Strongly unstable focus
D<0, t=0 Marginally stable focus Δ
D<0, t<0 Strongly stable focus
or
D=0, t<0 Stable node or stable improper node Δ>0, D>0 Stable node
Figure 6.13: Summary of the eight types of fixed point in the ∆–t plane
We have given a way to find the nature of the fixed point based on the values of t, ∆ and D = t2 − 4∆ but of course this is just a way to summarise the various possibilities for the eigenvalues. We could just have easily given the method:
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 110
(1) Find the eigenvalues of M
(2) Knowing the eigenvalues of M, find the Jordan normal form C
(3) Solve the equations X˙ = CX to find the flow pattern near the fixed point
The ultimate results will be the same.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 111
We illustrate the various fixed points in a series of examples.
Example 6.8. Find the fixed points of the second order autonomous system described by the differ- ential equations x˙ = y − x2 + 2 , y˙ = 2(x2 − y2) , (6.69) and investigate their stability. The fixed points are given by solutions of
y − x2 + 2 = 0, x2 − y2 = 0 . (6.70)
The second of these equations gives y = ±x and we soon find that there are four fixed points, (-1,-1), (2,2), (1,-1), (-2,2). We have, in the above notation,
2 2 2 f1(x, y) = y − x + 2 , f2(x, y) = 2(x − y ) , and the Jacobian matrix J, evaluated at arbitrary (x, y) is given by ! ∂f1 ∂f1 −2x 1 J = ∂x ∂y = , (6.71) ∂f2 ∂f2 4x −4y ∂x ∂y ,
We look at these fixed points in turn.
Case 1 The fixed point (−1, −1) has 2 1 J = . (6.72) (−1,−1) −4 4
The matrix J(−1,−1) has trace t = 6, determinant ∆ = 12 and discriminant D = −12. From the classification we find it is a focus since D < 0 and unstable since t > 0, so it is an unstable focus.
Let us now find the eigenvalues exactly. J(−1,−1) has eigenvalues λ given by √ (λ − 2)(λ − 4) + 4 = 0 ⇒ λ = 3 ± i 3. (6.73)
It follows that there is a matrix P such that √ 3 3 P −1J P = √ . (6.74) (−1,−1) − 3 3
Near (−1, −1) the linearised equations thus take the form √ d X 3 3 X = √ , (6.75) dt Y − 3 3 Y
where X x + 1 = P −1 . (6.76) Y y + 1 This system of equations is separable in polar coordinates. Thus we change the variables from (X,Y ) to (r, θ) where X = r cos θ, Y = r sin θ. We know from the arguments following Eq (6.28) that our equations take the form √ r˙ = 3r, θ˙ = − 3 . These equations have integrals √ 3t r(t) = r0e , θ(t) = θ0 − 3t ,
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 112
where r0 = r(0), θ0 = θ(0). The phase curves have the form of ‘outgoing spirals’ indicated in the diagram and we deduce that (x = −1, y = −1) is an unstable focus. Note In discussing the stability we didn’t need to know the form of P , merely that it existed. However, let’s calculate P explicitly so as to exhibit the change of variable we’ve used. Recall the procedure for finding Jordan forms: We consider the equation
2 1 u √ u = (3 + i 3) . (6.77) −4 4 v v √ These equations aren’t independent and we need only look at v = (1 + i 3)u. Taking u = 1 we see that an eigenvector is 1 1 0 √ = + i √ . (6.78) 1 + i 3 1 3 With respect to the basis 1 0 f = , f = √ , (6.79) 1 1 2 3
we have √ J(−1,−1)(f 1 + if 2) = (3 + i 3)(f 1 + if 2), (6.80) so that √ √ J(−1,−1)f 1 = 3f 1 − 3f 2 ,J(−1,−1)f 2 = 3f 1 + 3f 2 . (6.81) It follows that √ 3 3 P −1J P = √ , (6.82) (−1,−1) − 3 3 where 1 0 P = √ . (6.83) 1 3 We see that √ √ 1 3 0 det P = 3,P −1 = √ . (6.84) 3 −1 1 The change of variables (x, y) 7→ (X,Y ) which we used in the above argument is therefore given by X x + 1 = P −1 , (6.85) Y y + 1 from which we derive y − x X = x + 1,Y = √ . (6.86) 3 Case 2 The fixed point (2, 2): A similar calculation shows the Jacobian is
−4 1 J = , (6.87) (2,2) 8 −8
which has trace t = −12, determinant ∆ = 24 and discriminant D = 48. Hence the eigenvalues are real (D > 0), of the same sign (∆ > 0) and negative (t < 0) and the fixed point is a strong stable node. We now repeat the analysis in detail: The linearised equations take the form
d X −4 1 X = P −1 P ,X = x − 2,Y = y − 2 (6.88) dt Y 8 −8 Y
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 113
where we choose P so as to bring the matrix −4 1 (6.89) 8 −8 into Jordan form. The matrix in question has eigenvalues λ given by √ (λ + 4)(λ + 8) − 8 = 0 ⇒ λ = −6 ± 2 3 , (6.90)
so that √ √ λ = λ1 = −6 + 2 3 < 0 , λ = λ2 = −6 − 2 3 < 0 . (6.91) It follows that there exists a P such that −4 1 λ 0 P −1 P = 1 . (6.92) 8 −8 0 λ2 With this choice of P our equations become
X˙ = λ1X, Y˙ = λ2Y, (6.93) from which we derive λ1t λ2t X(t) = X0e ,Y (t) = Y0e , (6.94)
where X0 = X(0),Y0 = Y (0). We see that (x = 2, y = 2) is a strongly stable fixed point (a strongly stable node). We can, of course, eliminate t to obtain the linearised form of the equations of the phase curves. We find that X λ2/λ1 √ Y = Y0 , λ2/λ1 = 2 + 3 . (6.95) X0 Case 3 The fixed point at (1, −1): Here we find that −2 1 J = , t = 2, ∆ = −12,D = 52 . (6.96) (1,−1) 4 4 Since ∆ < 0 we know immediately that the eigenvalues are real and of opposite sign and the fixed point is hyperbolic. To go through this in detail again: the system is equivalent to d X −2 1 X = P −1 P ,X = x − 1,Y = y + 1 (6.97) dt Y 4 4 Y where we choose P so as to bring the matrix −2 1 (6.98) 4 4 into Jordan form. This matrix has eigenvalues λ given by (λ + 2)(λ − 4) − 4 = 0 ⇒ √ √ λ = λ1 = 1 + 13 > 0 , λ = λ2 = 1 − 13 < 0. The argument then proceeds as in Case 2 and we obtain the equations
X˙ = λ1X, Y˙ = λ2Y, (6.99) which gives λ1t λ2t X(t) = X0e ,Y (t) = Y0e , (6.100)
λ1t in a now familiar notation. Since λ1 > 0, e → ∞ as t → ∞. We see that (x = 1, y = −1) is an unstable node, more specifically a hyperbolic fixed point.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 114
Case 4 The fixed point at (−2, 2). We see that
4 1 J = , t = −4, ∆ = −24,D = 64 (6.101) (−2,2) −8 −8
and so since ∆ < 0 the fixed point is hyperbolic. Explicitly, the eigenvalues are √ √ λ = λ1 = −2 + 2 7 > 0, λ = λ2 = −2 − 2 7 < 0 . (6.102) The argument used in Case 3 leads to the conclusion that (x = −2, y = 2) is also an unstable (hyperbolic) fixed point.
Here we give a numerical plot of the phase portrait and we see that indeed we have found the correct nature for each fixed point.
3
2
1
0
-1
-2
-3
-3 -2 -1 0 1 2 3
Figure 6.14: Phase-portrait of DS (6.69) showing that the fixed points are hyperbolic at (1, −1) and (−1, 1), a stable node at (2, 2) and an unstable focus at (−2, −2).
Example 6.9. Find the fixed point of the system of equations
x˙ = f1(x, y) = 3x + y + 1 , y˙ = f2(x, y) = −x + y + 6 , (6.103) and investigate its stability. The fixed points are given as solutions of
3x + y + 1 = 0 , −x + y − 6 = 0 , (6.104) so that (−7/4, 17/4) is the only fixed point of the system. The Jacobian matrix J, evaluated at (x, y), is given by ! ∂f1 ∂f1 3 1 J = ∂x ∂y = , (6.105) ∂f2 ∂f2 −1 1 ∂x ∂y which has t = 4, ∆ = 4, D = 0. Since D = 0 this is either a node or an improper node, but since J is not diagonal it must be an improper node and since t > 0 this is an unstable improper node.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 115
Our equations may be written
d x 3 1 x + 7/4 = . (6.106) dt y −1 1 y − 17/4
Note that this equation is exact, because the original equations are linear ; we have not made any approximations, since the second and higher order derivatives of f1, f2 are all zero. Proceeding in the usual way we obtain d X 3 1 X = P −1 P , (6.107) dt Y −1 1 Y where X x + 7/4 = P −1 , (6.108) Y y − 17/4 and P is chosen so that the matrix 3 1 , (6.109) −1 1 is reduced to Jordan form.
We find that the eigenvalues of this matrix are given by
(λ − 1)(λ − 3) + 1 = 0 , ⇒ (λ − 2)2 = 0 , (6.110) so that λ = 2 (twice). We can therefore choose P so that (see Eq (C.3))
3 1 2 1 C = P −1 P = . (6.111) −1 1 0 2
In the coordinate system (X,Y ) determined by P the differential equations become
X˙ = 2X + Y, Y˙ = 2Y. (6.112)
The second of these equations has general solution
2t Y (t) = Y0e , (6.113) and the equation for X becomes 2t X˙ − 2X = Y0e . (6.114) This equation is of linear type with integrating factor µ(t) = exp R −2 dt = e−2t. We therefore obtain
d e−2tX = Y e2te−2t = Y (6.115) dt 0 0 from which we easily derive 2t 2t X(t) = te Y0 + e X0 . (6.116) Since e2t → ∞ as t → ∞ we conclude that (x = −7/4, y = 17/4) is an unstable fixed point, more specifically an unstable improper node. This is confirmed by eliminating t: we find that the phase curves are given by 1 Y X X = Y ln + 0 Y. (6.117) 2 Y0 Y0
For each choice of the initial values (X0,Y0) we obtain a phase curve. A plot of the phase portrait is in figure 6.15
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 116
10
5
0
-5
-10
-10 -5 0 5 10
Figure 6.15: Phase-portrait of DS (6.103) showing that the fixed point at (−7/4, 17/4) is an unstable improper node.
Example 6.10. Consider the second order system described by the differential equations
x˙ = v1(x, y) = ax − bxy + sin(x − d/c) ,
y˙ = v2(x, y) = cxy − dy + sin(y − a/b) (6.118) where is constant and a, b, c, d are positive constants.
We note that v1(d/c, a/b) = 0, v2(d/c, a/b) = 0 so that (d/c, a/b) is a fixed point.
The Jacobian matrix J, evaluated at (x, y) is given by
∂v1 ∂v1 ! ∂x ∂y a − by + cos(x − d/c) −bx J = ∂v2 ∂v2 = , (6.119) ∂x ∂y , cy cx − d + cos(y − a/b) so that bd − c J(d/c,a/b) = ac . (6.120) b Near (x = d/c, y = a/b) the linearised equations become
d X X = P −1J P , (6.121) dt Y (d/c,a/b) Y where X x − d/c = P −1 , (6.122) Y y − a/b and the matrix P is chosen so as to bring J(d/c,a/b) into Jordan form. A standard calculation shows that the eigenvalues of J(d/c,a/b) are given by
λ = ± i(ad)1/2 , (6.123)
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 117 and our equations become d X (ad)1/2 X = . (6.124) dt Y −(ad)1/2 Y This system is separable in polar coordinates and we write X = r cos θ, Y = r sin θ , (6.125) to obtain r˙ = r, θ˙ = −(ad)1/2 . (6.126)
If = 0 we get r(t) = r0 which describes a family of circles with centre (X = 0,Y = 0), the fixed point. We see that if = 0 the fixed point is stable. Note When = 0 the equations reduce to the Lotka-Volterra (“hare-lynx”) equations of chapter 1. The Lotka-Volterra equations are not structurally stable in the sense that if 6= 0 we can move from a situation of stable equilibrium to one of unstable equilibrium, as we shall now see. If < 0 we have t r(t) = r0e → 0 as t → ∞, and (x = d/c, y = a/b) is strongly stable. However, if > 0 we see that t r(t) = r0e → ∞ as t → ∞, so that (x = d/c, y = a/b) is a position of unstable equilibrium. Example 6.11. Harmonic motion. Consider the differential equation of simple harmonic motion x¨ + x = 0 . (6.127) Setting y =x ˙ we obtain d x v (x, y) y x = 1 = = J , (6.128) dt y v2(x, y) −x y where 0 1 J = . (6.129) −1 0 Notice that Eq (6.128) is exact — no approximations have been made. Moreover, the matrix J is already in the Jordan form appropriate to real 2 × 2 matrices with eigenvalues ±i. We note that Eq (6.128) has one fixed point, namely (x = 0, y = 0). If we interpret the equation x¨ + x = 0 in terms of the motion of a Newtonian particle (see following chapter), the fixed point corresponds to the particle at rest at x = 0. We now demonstrate that the fixed point is stable as follows. Changing to polar coordinates (r, θ) where x = r cos θ, y = r sin θ we find in the usual way that r˙ = 0 , θ˙ = −1. which have the integrals r(t) = r0 , θ(t) = θ0 − t, in the usual notation. These are a family of circles centred on the origin (x = 0, y = 0) with radius given by the parameter r0. The fixed point is therefore stable. We can also show this directly form Eq (6.128) by noting that x˙ = y , y˙ = −x. Eliminating t we obtain dy x = − ⇒ x dx + y dy = 0. dx y It follows that 2 2 2 x + y = r0 , where r0 is a parameter. In other words, the phase curves are indeed a family of circles, with centres at (0, 0); the point (0, 0) is therefore a stable fixed point.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 118
Example 6.12. The damped harmonic oscillator
In physical situations there are often forces such as friction which lead the oscillations to decay. Adding such a term to equation (6.127) gives the differential equation
x¨ + λx˙ + x = 0 . (6.130)
With y =x ˙ this is the second order system
x˙ 0 1 x = , (6.131) y˙ −1 −λ y
a linear system and hence equivalent to one of the linearised systems studied in section 6.6 with ∆ = 1, t = −λ and D = λ2 − 4. Hence we see that for 0 < λ < 2 this is a stable focus; for λ = 2 it is a stable improper node; and for λ > 2 it is stable node. We can sketch the graph of x(t) for typical trajectories by looking at the phase portrait and following the features along an orbit. Firstly, for λ = 1/4, this is a stable focus. Starting at the point A on the phase portrait with x = 1, x˙ = 0, the motion is for x to oscillate but with ever decreasing amplitude, as in figure 6.16. This is called under damping. The exact solution is √ x = e−t/8 cos(ωt) , ω = 3 7/8 (6.132)
ω is sometimes called the damped angular frequency or pseudofrequency of the motion. For λ = 2, this is a stable improper node. The motion will tend to zero after at most one oscillation as in figure 6.17 This is called critical damping. For λ = 4, the motion is very similar to the critically damped solution but tends to zero rather slower in the final stages as in figure 6.18; this is called over damping. One can show that the motion tends to zero faster for critical damping – see exercise 6.20.
2 xHtL
1.0 A
1 D E 0.5
C E 0 A B D F
5 10 15 t F
-1 B -0.5
C
-2 -1.0
-2 -1 0 1 2 (a) The phase portrait. (b) A typical solution.
Figure 6.16: The under-damped harmonic oscillator with λ = 1/4.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 119
x t 2 H L
A 1.0 D
1 B
0.5
C D 0 C B
1 2 3 4 5 6 t
-1 A -0.5
-2
-2 -1 0 1 2 -1.0 (a) The phase portrait. (b) Two typical solutions.
Figure 6.17: The critically damped harmonic oscillator with λ = 2.
x t 2 H L
B
1 A 1.0 A
B 0 0.5
-1
1 2 3 4 5 6 t
-2
-2 -1 0 1 2 -0.5 (a) The phase portrait. (b) A typical solution.
Figure 6.18: The over-damped harmonic oscillator with λ = 4.
6.7 Beyond linear stability analysis
The methods of section 6.6 do not always work, exactly as the linear stability analysis of a first order system dies not work in f 0(a) = 0. Looking at figure 6.13, we see that the second order analysis fails exactly when ∆ = detJ = 0. In this case one or both of the eigenvalues of J is zero and we need to go to higher orders in the Taylor expansion to understand the nature of the fixed point. We have already met a case where the fixed point is not one of the eight types in figure 6.13, and that is example 6.1(e). The fixed point at (0, 0) in figure 6.5 is not one of the eight types, and the reason is that the Jacobian vanishes at the fixed point:
∂F ∂F ! 2 2 dx ∂y 2y 2x 0 0 F = 2xy , G = y − x ,J(x, y) = ∂G ∂G = ,J(0, 0) = . dx ∂y −2x 2y 0 0 (6.133) There are more examples in supplementary exercise 6.4.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 120
6.8 Exercises
Short Exercises Exercise 6.1 Sketch the phase diagram associated with the uniform velocity field given by
f(x, y) = −2e1 + e2 , Find the equation of the phase curves. Do the same for the velocity field given by
f(x, y) = −2e1 + xe2 , Exercise 6.2 For each of the following second order dynamical systems of the form dx dy = v (x, y) , = v (x, y) dt 1 dt 2 you should
(i) Find the null-clines of v1 and v2 and hence find the fixed points. (ii) Give a qualitative sketch of the phase portrait, including the null-clines and using the usual arrow representation of the phase flow.
(iii) Find an equation for dy/dx and solve it to find the phase curves explicitly
(a) v1 = y , v2 = 1
(b) v1 = x , v2 = −y
(c) v1 = −y , v2 = 2x 2 (d) v1 = y , v2 = x + x
(e) v1 = x , v2 = y (1 − x) Exercise 6.3 For each of the following second order dynamical systems of the form dx dy = v (x, y) , = v (x, y) dt 1 dt 2 you should
(i) Find the null-clines of v1 and v2 and hence find the fixed points. (ii) Give a qualitative sketch of the phase portrait, including the null-clines and using the usual arrow representation of the phase flow.
2 2 (a) v1 = 2xy , v2 = x + y
2 2 (b) v1 = 2xy , v2 = 1 − x − y
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 121
Exercise 6.4 Find the fixed points of the system described by
x˙ = −3y + xy − 4, y˙ = y2 − x2
Exercise 6.5 Find the fixed point of the system of differential equations
x˙ = exp (x + y) − y, y˙ = xy − x and investigate its stability.
Exercise 6.6 Find the fixed point of the system of differential equations
x˙ = x + 3y, y˙ = −6x + 5y and investigate its stability.
Longer Exercises
Exercise 6.7 Consider second order linear dynamical systems of the form
x˙ x = J , y˙ y where J is one of the following constant matrices. In each case you should, in turn,
(a) Find the trace, t, the determinant, ∆, of J.
(b) Find the discriminant, D = t2 − 4∆, of the characteristic equation for J.
(c) State the nature of the fixed point (0, 0).
(d) Find the eigenvalue(s) of J
(e) Give the Jordan normal form, C, of J.
(f) Find a matrix P such that P −1JP is in Jordan normal form. Is it the same as the matrix you stated in (e)? If not, why not?
(You do not need to do part (d) to do part (c), etc.)
1 3 1 −3 2 2 1 −1 (i) (ii) (iii) (iv) 1 −1 0 −2 −1 0 1 3 −3 −1 −5 10 −8 13 11 −5 (v) (vi) (vii) (viii) −1 −3 −10 11 −5 8 18 −8
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 122
Exercise 6.8 Consider the following second order dynamical systems. In each case, linearise the system around the stated fixed point(s), find the trace and determinant of the corresponding constant matrix and state the nature of the fixed point of the linear system. In which cases does this determine the nature of the fixed point of the full system? Compare with the phase portraits given in the solution sets.
(i) Exercise 6.2(b): the fixed point at (0, 0) of the system
x˙ = x , y˙ = −y
(ii) Exercise 6.2(c): the fixed point at (0, 0) of the system
x˙ = −y , y˙ = 2x
(iii) Exercise 6.2(d): the fixed points at (0, 0) and (−1, 0) of the system
x˙ = y , y˙ = x + x2
(iv) Exercise 6.2(e): the fixed point at (0, 0) of the system
x˙ = x , y˙ = y(1 − x)
(v) Exercisr 6.3(b): the fixed points at (1, 0) and (0, −1) of the system
x˙ = 2xy , y˙ = 1 − x2 − y2
(vi) Supplementary exercise 6.2: the fixed point at (0, 0) in the three cases α = −1/4, α = 1/4 and α = 0 of the system
x˙ = y + x2 + αx(1 − y + 2x2) , y˙ = −2x(1 + y)
Exercise 6.9 Write the equation x¨ + x = f(x) in standard form by putting y =x ˙. Assuming that f is suitably differentiable and that f(0) = 0 discuss the stability of the fixed point (x = 0, y = 0), distinguishing the cases f 0(0) > 1, f 0(0) < 1.
Note We could have guessed the answer by the following argument. Near x = 0 f(x) ' f(0) + xf 0(0) + ··· so that the differential equation is approximated by
x¨ + (1 − f 0(0))x ' 0.
If f 0(0) < 1 the motion is essentially Simple Harmonic and therefore the fixed point should be stable.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 123
Exercise 6.10 Write the differential equation x¨ + 2λx˙ + ω2x = 0 (the equation of damped harmonic motion —see Exercise 5.13) in standard form by putting y =x. ˙ Find the fixed point and discuss its stability. Exercise 6.11 A second order dynamical system is described by the differential equations
x˙ = −y + x sin πr, y˙ = x + y sin πr, where r = (x2 + y2)1/2. Prove that the system is separable in polar coordinates and show in particular that r˙ = r sin πr. Sketch the phase diagram associated with this equation write down the fixed points, and indicate which are stable and which are unstable.
Describe qualitatively, without computation, how the representative point (x, y) behaves as a function of time t given the initial conditions r(0) = 3/2, θ(0) = 0. What is your conclusion if r(0) = 5/2, θ(0) = 0.? Exercise 6.12 Consider the system of differential equations
x˙ = (ay − b)x + f(x), y˙ = (c − dx)y + g(y), where a, b, c, d are positive constants, and f, g are differentiable functions such that f(c/d) = 0, g(b/a) = 0. Notice that the “fox-rabbit” equations are a particular case of these.
Show that (c/d, b/a) is a fixed point of this system. Show that linearisation of these equations with respect to this fixed point gives
d X α ac/d X X = ≡ J dt Y −bd/a β Y Y where α = f 0(c/d), β = g0(b/a),X = x − c/d, Y = y − b/a.
Show that the matrix J has eigenvalues λ given by p α + β α − β λ = µ ± ν2 − bc , µ = , ν = . 2 2 Assuming that det J 6= 0, discuss the stability of the fixed point (c/d, b/a) in the following cases: (i) µ > 0, ν2 ≥ bc, (ii) µ > 0, ν2 < bc, (iii) µ < 0, ν2 > bc, αβ > −bc, (iv) µ < 0, ν2 ≤ bc, (v) µ = 0, ν2 > bc, (vi) µ = 0, ν2 < bc. Why has the case µ = 0, ν2 = bc been omitted?
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 124
Exercise 6.13 A second order dynamical system is described by the equations d x x 5 1 = J ,J = . dt y y −1 1 Solve these equations, subject to the initial conditions x(0) = 0, y(0) = 1.
Can you think of another way of doing this without finding the Jordan form of the matrix J?
Hint Try substituting x(t) = aeλt, y(t) = beλt and determine the constraints on a, b, λ. You should find that a (λI − J) = 0 , b where I is the 2 × 2 unit matrix—this should ring a bell! Consider taking appropriate linear combinations of solutions. In this context note that if the column vectors X1,X2 are solutions of d LX = JX,L = , dt then (αX1 + βX2) is also a solution since
L(αX1 + βX2) = αLX1 + βLX2 = αJX1 + βJX2 = J(αX1 + βX2). Exercise 6.14 A second order dynamical system is described by the system of differential equations x˙ = y + xf(R), y˙ = x + yf(R),R = (x2 − y2)1/2. Show that the system is separable in the coordinates (R, φ) where
x = R cosh φ, y = R sinh φ.
Suppose now that f(R) = (R − 1)(R − 2)(R − 3).
Show that the rectangular hyperbolae Hn, n = 1, 2, 3, where
2 2 2 Hn = {(x, y): x − y = n , n = 1, 2, 3}, are invariant sets for the system. Exercise 6.15 2 A second order dynamical system with phase space (x, y) ∈ R = Γ is described by the differential equations
x˙ = y + xf(x2 − y2), y˙ = x + yf(x2 − y2) .
(i) Show that the system is separable in hyperbolic coordinates (r, θ), where
x = r cosh θ, y = r sinh θ,
and in particular thatr ˙ = rf(r2) and θ˙ = 1. Which part of the phase space does this cover? [hint, use cosh2 θ − sinh2 θ = 1]
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 125
(ii) Show that the system is also separable in different hyperbolic coordinates (ρ, φ), where
x = ρ sinh φ, y = ρ cosh φ,
and in particular thatρ ˙ = ρf(−ρ2) and φ˙ = 1. Which part of the phase space does this cover?
Exercise 6.16 Consider the first order dynamical system dx dy = −x − y , = −y + x , (6.134) dt dt where x = r cos θ , y = r sin θ .
(i) Show that the system is separable in the coordinates (r, θ).
(ii) Sketch the phase portrait for the 1st order dynamical system satisfied by r(t), find the fixed points and discuss their nature.
(iii) Solve the 1st order dynamical systems for r and for θ exactly. Does the motion termi- nate?
(iv) Find the exact solution of the system (6.134) with initial conditions x(0) = 1, y(0) = 0. Sketch the phase portrait for the dynamical system (6.134) including a sketch of the exact solution.
Exercise 6.17 Consider the first order dynamical system described by dx x dy y x = f(r) − 2y , = f(r) + , (6.135) dt r dt r 2 where x = 2r cos θ , y = r sin θ , f(r) = r(r − 1)2(r − 2)(r − 3)
(i) Show that the system is separable in the coordinates (r, θ).
(ii) Sketch the phase portrait for the 1st order dynamical system satisfied by r(t), find the fixed points and discuss their nature.
(iii) Find the limit cycles for the dynamical system (6.135). Which are stable, which are unstable and which are half-stable?
(iv) Sketch the phase portrait for the dynamical system (6.135).
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 126
Exercise 6.18 [taken from 2008 exam] Suppose that a second order autonomous dynamical system has a fixed point at (x, y) = (0, 0), and that the linearised dynamics in the vicinity of this fixed point is given by
d x x = J dt y y where 1 1 J = c 1 for some real constant c.
(a) Determine the eigenvalues of J, and give the Jordan canonical form C corresponding to J for (i) c < 0, (ii) c = 0, (iii) c > 0.
(b) The Jordan canonical form C is related to J via a similarity transformation C = P −1JP , with a suitable non-singular matrix P . You are not asked to find P . X x Now we pass to the transformed variables = P −1 and consider the equations Y y of motion d X X = C . dt Y Y For the regimes (i) c < 0, (ii) c = 0 and (iii) c > 0, solve these equations and thereby investigate the nature of the fixed point.
Exercise 6.19 [Taken from 2009 exam (the exam was slightly easier than this version)] A particle of mass m is moving on a straight line under the influence of a force F (x), but also experiences friction, so that the equation of motion is
d2x dx m + 2γ = F (x) dt2 dt with γ > 0.
(a) Rewrite the equation of motion as a system of two first order equations by settingx ˙ = y.
(b) State the conditions necessary to have a fixed point of the dynamics. Let x? be the coordinate of such a point. Calculate the Jacobian J of the system dynamics in x?, and show that the eigenvalues are given by r γ γ2 F 0(x?) λ = − ± + 1,2 m m2 m
(c) Assume F 0(x?) > 0. Give the Jordan canonical form C of J.
(d) The Jordan canonical form C is related to J via a similarity transformation C = P −1JP , with a suitable non-singular matrix P . You do not need to find P .
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 127
X x − x? Now we pass to the transformed variables = P −1 and consider the Y y equations of motion d X X = C . dt Y Y Solve these equations and thereby investigate the nature of the fixed point.
(e) Find P .
(f) Discuss briefly the nature of the fixed point when F 0(x?) < 0 for the relevant regimes of γ.
Exercise 6.20 Consider the equation of damped harmonic motion,
x¨ + λx˙ + ω2x = 0 , (6.136)
where λ and ω are real positive constants. Find the general solutions for the case of under-damping (0 < λ < 2ω), critical damping (λ = 2ω) and over-damping (λ > ω) Consider the exponentially decaying part of the solution and show that the decay is fastest for critical damping.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 128
Supplementary Exercises Supplementary Exercise 6.1 Consider the second order dynamical system
x˙ = (1 − r2)x − (1 − x/r)y
y˙ = (1 − r2)y − (1 − x/r)x where r = px2 + y2.
(a) Find the equations forr ˙ and θ˙ in polar coordinates, x = r cos θ, y = r sin θ.
(b) Show that the system has two fixed points.
(c) Sketch the phase portrait of the system.
Note - all points in a small disk around the point (1, 0) eventually flow to (1, 0), but some of them go all the way round the unit circle to get there. The point (1, 0) is clearly an unstable fixed point (compare it with problem 4s.1) but all flows in the neighbourhood of (1, 0) eventually end up there, so this is again an interesting example when coming up with definitions of stability.
Supplementary Exercise 6.2 Consider the dynamical system
x˙ = y + x2 , y˙ = −2x(1 + y)
(a) Find the fixed points of this system
(b) Show that d d (1 + y) = −2x(1 + y) , (1 − y − 2x2) = 2x(1 − y − 2x2) dt dt
(c) Show that the lines y = −1 and y = 1 − 2x2 are preserved by the dynamical system.
(d) Consider 1 H = y2 + x2(1 + y) . 2 Show that H˙ = 0 . This mean that lines of H =constant are also phase curves, or unions of phase curves.
(e) Sketch the phase portrait of the dynamical system
(f) Now consider the new system (where |α| < 1/2)
x˙ = y + x2 + αx(1 − y − 2x2) , y˙ = −2x(1 + y) (6.137)
Show that H˙ = 2αx2(1 + y)(1 − y − 2x2) .
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 129
This last equation can be used to show that the lines y = −1 and y = 1 − 2x2 remain unions of phase curves for the new system but that all orbits inside the region bounded by these curves all spiral out from the origin if α > 0, and all spiral in if α < 0.
Supplementary Exercise 6.3 A problem on separability and construction on separable variables. The idea: Give new coordinates to each point (x, y) by (i) finding the trajectories through (x, y) (ii) Consider the trajectory that passes through (0, v) at time t = 0 and through (x, y) at time t = u Let the new coordinates on phase space be (u, v). The equations satisfied by (u, v) will be v˙ = 0 , u˙ = 1 . Try this for the dynamical system 1 x˙ = 1 + x , y˙ = . x − 1
(a) Sketch the phase portrait of the system.
(b) Solve the equation for x(t) as a function of t such that x(0) = 0. [ans: x(u) = eu − 1 ]
(c) Solve the equation for the phase curves through (0, v) dy 1 = , dx x2 − 1 [ans: y = v + (1/2) log |(1 − x)/(1 + x)|]
(d) Choose as new variables (v, u). State the relation between (x, y) and (u, v) [ans: u = log(1 + x), v = y − (1/2) log |(1 − x)/(1 + x)|]
(e) Show that in these variables the dynamical system becomes
v˙ = 0 , u˙ = 1 .
Supplementary Exercise 6.4 Problems when linearisation is not possible Consider the dynamical systems
(i)x ˙ = x2 − y2 , y˙ = 2xy
(ii)x ˙ = x3 − 3xy2 , y˙ = 3x2y − y3
Sketch the phase portraits of these systems. Show that these are examples of complex dynamical systems, i.e d (x + iy) = f(x + iy) , dt for some function f. Solve these two systems exactly under the conditions x(0) = 1, y(0) = 1.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 130
Supplementary Exercise 6.5 A problem on structural stability of linearisations Consider the linear dynamical system
x˙ x = M . y˙ y
Consider changing the matrix M by a small amount N. Show (by explicit computation) that
Tr(M + N) = Tr(M) + Tr(N) ,
det(M + N) = det(M) + det(M)Tr(M −1N) + ....
Consider D(M) = [Tr(M)]2 − 4 det(M) Show that
D(M + N) = D(M) + (2Tr(M)Tr(N) − 4 det(M) Tr(M −1N)) + ...
Check these statements for the particular cases
0 1 1 0 (a) A centre or marginally stable focus: M = ,N = . −1 0 0 1 What is the nature of the system for > 0 and < 0? 1 1 0 0 (b) An improper node: M = ,N = 0 1 1 0 What is the nature of the system for > 0 and < 0?
This shows that by adding a suitable small matrix N, if D(M) = 0 then one can choose D(M + N) > 0 or D(M + N) < 0, i.e. an improper node can be charged into a proper node or a focus.
Version of Mar 14, 2019 Chapter 6: Second Order Autonomous Systems 131
Supplementary Exercise 6.6 Principal directions for focii are harder to define than for nodes, but can be found as follows. We start with a marginally stable focus in normal form
x˙ x x 0 −β r˙ = = C = C r , where r = ,C = . y˙ y y β 0
This is a rotationally symmetric flow with the orbits being circular. To make the orbits ellipses, we can stretch the flows in the x direction so that the orbits are ellipses with (major and minor) axes parallel to the x– and y– axes. (i) Show that changing coordinates from (x, y) to (X, y) with X = αx leads to the system
X˙ X 0 −αβ = C0 ,C0 = . y˙ y β/α 0
(ii) show that a rotation through an angle θ from (X, y) to (u, v) with
X u cos(θ) sin(θ) = R ,R = , y v − sin(θ) cos(θ) leads to the system u˙ u = C00 ,C00 = R−1C0R, v˙ v and calculate C00 explicitly.
(iii) Suppose that the fixed point at the origin of the dynamical system
u˙ a b u = , v˙ d −a v is a marginally stable focus. By comparing with your results in part (ii), show that the orbits are ellipses with axes at an angle θ to the (x, y) axes where 2a tan(2θ) = − . b + d
(iv) Show that this agrees with the definition that the principal axes are in the direction r such that r · v(r) = 0.
Version of Mar 14, 2019 Part III
Application to Classical Mechanics
132 Chapter 7: Elements of Newtonian Mechanics 133
Chapter 7
Elements of Newtonian Mechanics
In this present and the following chapter we are going to apply the general methods introduced to study dynamical systems to problems of classical mechanics. We begin by introducing elementary facts of Newtonian Mechanics, starting with kinematics — i.e., the description of particle motion — thereafter continuing with proper dynamics, i.e. the investigation of the dynamical laws governing particle motion in mechanics. In particular, we look at Newton’s law of gravitation, and we study motion on a straight line as a special realization of second order dynamical systems. For autonomous systems in which forces depend only on position, energy is conserved, and this aspect can be used to help analysing properties of particle motion in one dimension. Fixed points and their stability are discussed in terms of potential energy. It turns out that mechanical systems in one dimension which have a conserved energy can only have two types of fixed point — unstable hyperbolic ones and (marginally) stable elliptic ones.
7.1 Motion of a particle
What is a particle? We regard a particle as a physical object, possessing a mass, whose dimensions are negligible in relation to other lengths arising in the description of its motion. For example, we may treat the Earth as a particle for the purposes of obtaining a good approximation to its path round the sun. Someone may object: “This isn’t very satisfactory; an undefined term, mass, has crept in!” It must be admitted at this stage that mass is indeed an intuitive term, but we shall see how the concept may be made precise once we have stated Newton’s Laws of motion. We assume that space is Euclidean and that it can be described in terms of a system of three dimensional Euclidean position vectors measured relative to some conveniently chosen origin. In describing the motion of a particle the term ‘position’ has no absolute significance; it is only meaningful to talk about the particle’s position relative to an observer. In order to describe a particle’s motion an observer O may choose a reference frame consisting of a right handed system of Cartesian axes Oxyz, usually fixed relative to the observer; the position of a particle P at a particular instant is then made precise by assigning its position vector r = OP. It is assumed that the observer O possesses a clock with which to measure time t. Generally speaking, the particle’s position will vary with time and we may indicate this by writing r = r(t). In classical Newtonian theory it is assumed that time is absolute in the sense that two different observers, whatever their state of relative motion, can synchronise their clocks
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 134 in such a way that they both assign the same value of time t to a particular event. (This assumption was shown to be untenable by Einstein in 1905 when he enunciated the Special Theory of Relativity.) We continue in the above notation. O denotes an observer, Oxyz an associated reference frame, and r(t) the position vector of a particle P at time t. We give the following definitions:
Definition 7.1. The velocity v(t) of the particle (as measured by O) at time t is defined by
dr r(t + δt) − r(t) v(t) = = lim . (7.1) dt δt→0 δt
The velocity is the rate of change of the position vector.
Definition 7.2. The speed of the particle (as measured by O) at time t is the magnitude of its velocity i.e. |v| = |r˙|.
We note that whereas velocity is a vector, speed is a scalar.
As the particle moves it traces out a curve Γ, say. Let Pt denote the particle’s position at time t, and Pt+δt its position at time t + δt, so that r(t), r(t + δt) are the position vectors of Pt,Pt+δt respectively. We note that
r(t + δt) − r(t) P P = t t+δt . δt δt Proceeding to the limit as δt → 0 we see that the velocity vector v(t) (assumed non-zero) points along the direction of the forward tangent to the curve Γ at Pt.
Definition 7.3. The acceleration a(t) (as measured by O) of the particle at time t is defined by dv d2r a(t) = = . (7.2) dt dt2 The acceleration is the rate of change of the velocity vector with respect to time.
Notation It is traditional in mechanics to employ dot notation to denote differentiation of a scalar or vector function with respect to time. We write
df dr d2r f˙ ≡ , r˙ ≡ , r¨ ≡ , dt dt dt2 and so on.
If the Cartesian coordinates of the particle P at time t are x(t), y(t), z(t) we may write
r = xe1 + ye2 + ze3,
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 135
where e1, e2, e3 denote the unit vectors parallel to the axes Ox, Oy, Oz respectively. The particle’s velocity is then v =x ˙ e1 +y ˙e2 +z ˙e3.
(It is assumed that e1, e2, e3 are fixed relative to the observer O—so that their time derivatives are all zero.) Similarly the acceleration of the particle at time t is
a =x ¨e1 +y ¨e2 +z ¨e3.
We may calculate the distance travelled by the particle between times t = t0 and t = t1 as follows. Using s to denote arc length measured from a fixed point on Γ we have
ds2 dx2 dy 2 dz 2 = + + . dt dt dt dt Taking square roots we see that the required distance is given by the integral
Z t1 Z t1 |r˙| dt = |v| dt. t0 t0
In other words, the distance travelled by the particle between times t0 and t1 is obtained by integrating the speed between these limits. Note that the distance travelled is (generally) not equal to |r(t1) − r(t0)|.
Note We emphasise that the concepts of velocity, speed and acceleration have been defined with respect to an observer O; different observers will assign different values to these functions, depending on their state of relative motion.
The following example shows that it is possible for a particle to move with constant speed (so that the magnitude of its velocity is constant) yet have non-zero acceleration. Example 7.1. Circular Motion A particle P moves in the plane xOy in such a way that its position vector with respect to O is
r(t) = b cos ωte1 + b sin ωte2,
where b, ω are positive constants. Find the velocity, speed and acceleration of the particle at time t.
Clearly |r(t)| = b for all t; it follows that the particle moves on a circle, centre O and radius b. At time t the position vector OP of the particle makes an angle θ = ωt with the x–axis, and since θ˙ = ω we see that the particle moves round the circle in an anti–clockwise sense at a constant angular speed θ˙ = ω.
We note that the particle’s velocity is given by
v(t) = r˙ = −bω sin ωte1 + bω cos ωte2.
It is clear that v.r = 0 so that the vector v is indeed tangential to the circle. The speed of the particle is equal to |v| = (b2ω2cos2ωt + b2ω2sin2ωt)1/2 = bω. Thus the speed is constant, and equal to bω.
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 136
The particle’s acceleration is given by
2 2 2 a = v˙ = −bω cos ωte1 − bω sin ωte2 = −ω r.
We conclude that the acceleration vector is always directed from P towards the centre O of the circle, and that it has magnitude bω2 6= 0.
We should not be too surprised by this result since, although the magnitude of the velocity vector is constant, its direction is continuously changing as the particle moves round the circle. This is why the acceleration (rate of change of the velocity vector) is non-zero, despite the fact that the speed (magnitude of the velocity vector) is constant.
Example 7.2. Uniform Motion A particle moves with constant velocity V relative to an observer O. Find its path, given that at t = 0 its position vector with respect to O is r0.
We have r˙ = V , and integration gives r(t) = tV + C, where C is a constant vector. Since r(0) = r0 we see that C = r0 and therefore
r(t) = tV + r0.
It is clear that the path of the particle, as viewed by O is a straight line which passes through the point with position vector r0, and whose direction is given by the vector V . Example 7.3. Motion with constant acceleration A particle moves so that its coordinates (x, y, z) with respect to Cartesian axes Oxyz are given by the differential equations x¨ = 0, y¨ = 0, z¨ = −g, where g is a positive constant. The initial conditions are such that
x˙(0) = U, y˙(0) = 0, z˙(0) = V, x(0) = 0, y(0) = 0, z(0) = 0, where U, V, are positive constants.
Find the coordinates of the particle at time t.
0 An observer O is coincident with O at time t = 0, and moves with constant velocity Ue1 relative to O.O0 describes the motion of the particle in terms of coordinates x0, y0, z0 with respect to Cartesian axes O0x0y0z0, where O0x0 is parallel to Ox, O0y0 is parallel to Oy, and O0z0 is parallel to Oz.
Write down a set of equations relating x, y, z and x0, y0, z0 at time t. Describe geometrically the motion of the particle as viewed by O and O0.
Integrating the given equations, and using the stated initial conditions to evaluate the constants of integration gives: x˙(t) = A =x ˙(0) = U, x(t) = Ut + A0 = Ut (x(0) = 0),
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 137
y˙(t) = B =y ˙(0) = 0, y(t) = B0 = 0 (y(0) = 0), z˙(t) = −gt + C = −gt + V, (z ˙(0) = V ), 1 1 z(t) = V t − gt2 + C0 = V t − gt2 (z(0) = 0). 2 2 Summarising we have 1 x(t) = Ut, y(t) = 0, z(t) = V t − gt2. 2 Eliminating t now gives V x gx2 g V 2 z = − = − (x − UV/g)2 + . U 2U 2 2U 2 2g 2 g 2 Writing Z = z −V /(2g),X = x−UV/g, we see that the path is given by Z = − 2U 2 X . We conclude that the path of the particle as viewed by observer O is a parabola. (See the diagram)
Z z z z’
X x P x’ U t
x O O O’ x’
Figure 7.1: Relativity of description of motion
The observer O0 see things rather differently. Since
dOO0 = Ue , dt 1 and O,O0 coincide at time t = 0, it follows that
0 OO (t) = tUe1.
Referring to the diagram we see that the equations relating x, y, z and x0, y0, z0 are
x0 = x − Ut, y0 = y, z0 = z, so that 1 x0(t) = 0, y0(t) = 0, z0(t) = V t − gt2. 2 The path of the particle as viewed by O0 is therefore a straight line.
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 138
7.2 Newton’s Laws of motion
We now discuss the laws which govern the motion of a Newtonian particle. Newton assumed the existence of a preferred class of observers and associated reference frames, called inertial frames of reference, with respect to which Newton’s laws of motion were supposed to hold.
First, every particle P possesses an inertial mass m > 0 which is normally assumed to be constant, whatever the state of motion of the particle. If r(t) is the position vector of P relative to an observer O its momentum p is defined by the equation
p = mr˙ = mv, where v is the particle’s velocity.
7.2.1 Newton’s First Law (N1)
Newton’s First Law states that with respect to an inertial frame of reference a particle moves with constant velocity in a straight line unless constrained to depart from this state by a force acting on it. The nature of the term force is made precise through the Second and Third Laws.
7.2.2 Newton’s Second Law (N2)
Newton’s Second Law states that with respect to an inertial frame of reference the rate of change of momentum of a particle is equal to the total force acting on the particle. If F 1, F 2,..., F n are the forces acting on a particle P, whose position vector is r(t) with respect to the origin O of an inertial frame of reference, we have d (mr˙) = F , dt where the total force F is given by n X F = F i. i=1 Assuming that m is constant we may write
mr¨ = F .
In words, the mass times the acceleration of the particle is equal to the total force acting on the particle.
7.2.3 Newton’s Third Law (N3)
Newton’s Third Law is often stated in the form “action and reaction are equal (in magnitude) but opposite (in direction)”. To see more clearly the implication of this law suppose that the
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 139
force exerted on a particle i by a particle j is F ij; correspondingly the force exerted by particle i on particle j is F ji and F ij + F ji = 0, or F ij = −F ji. The forces in question could be gravitational or, in the case of two particles connected by a taut string, the tension force provided by the string.
We now show, very briefly, that Newton’s laws determine a scale of mass, once an arbitrary unit of mass has been chosen. Consider two particles 1, 2 with inertial masses m1, m2 respec- tively. Suppose that particle 2 exerts a force F 12 on particle 1 and that particle 1 exerts a force F 21 on particle 2. If a1, a2 denote the accelerations of the two particles with respect to some inertial frame of reference we have, using N2,
m1a1 = F 12, m2a2 = F 21.
Adding these equations, and using Newton’s Third Law, we obtain
m1a1 + m2a2 = F 12 + F 21 = 0. (7.3)
It follows that m |a | 1 = 2 . m2 |a1|
By measuring the accelerations a1, a2 we can determine the ratio m1/m2, and having chosen particle 2 (say) to have unit (inertial) mass, the mass m1 of particle 1 is then determined. Newton’s laws therefore determine, in principle, a mass scale.
Returning to equation 7.3 we note that if r1, r2 denote the position vectors of the two particles with respect to the origin O of an inertial frame of reference then
m1r¨1 + m2r¨2 = 0,
and therefore d2 (m r + m r ) = 0. (7.4) dt2 1 1 2 2 The centre of mass of the two particle system has position vector R defined by
m1r1 + m2r2 R = . m1 + m2 It now follows from equation 7.4 that
d2R = 0. dt2 We may interpret this result by saying that the centre of mass of the two particle system is unaccelerated with respect to any inertial frame of reference, and therefore moves with constant velocity in a straight line with respect to such a frame. All this supposes, of course, that F 12, F 21 are the only forces acting.
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 140
We note that similar arguments can be applied to show that if two particles with inertial masses m1, m2 coalesce to form a third particle with inertial mass m3, then m3 = m1 + m2, so that its inertial mass is the sum of the inertial masses of its component particles.
Throughout this discussion we have used the term “inertial frame of reference” and one may reasonably ask “Are there any such frames?” Einstein and Infeld in their book “The Evolution of Physics” show that this question poses serious difficulties and it is not possible to give an example of a frame of reference which is strictly inertial! Well, one may ask: “Would a reference frame with origin on the Earth’s surface, and fixed relative to the Earth, constitute an inertial frame of reference?” Because of the Earth’s rotation this is not strictly the case; nevertheless it is a reasonable approximation—in the sense that an application of Newton’s laws in such a frame leads to fairly satisfactory predictions. By making a better choice of reference frame it is possible to take account of effects which arise from the Earth’s rotation.
Nevertheless, this is not an entirely satisfactory state of affairs; we have a set of laws but we are unable to exhibit a reference frame in which they are strictly valid! However, if one such inertial frame F1 exists, then all frames moving with constant velocity relative to F1 are also inertial. (This point is considered in more detail in an exercise at the end of this chapter.)
In order to use Newton’s laws of motion to predict the motion of a particle, given a knowledge of certain initial conditions (its initial position and velocity), we require explicit mathematical formulae for the force(s) acting on the particle; of particular importance are gravitational forces.
7.3 Newton’s Law of Gravitation
Consider two particles of masses m1, m2 respectively and suppose that the position vector of m2 with respect to m1 is r. The the gravitational force F exerted on m2 by m1 is given by
Gm1m2 F = − r, r = |r|, (7.5) r3 where G is a universal constant known as Newton’s constant of gravitation.
2 The force F has magnitude |F | = Gm1m2/r so that the magnitude of the gravitational force varies inversely as the square of the distance between the particles; the - sign indicates that m2 is attracted towards m1.
If we now work out the acceleration of m2 due to the gravitational force we get Gm m Gm m r¨ = F = − 1 2 r , r¨ = − 1 r . (7.6) 2 2 r3 2 r3
This acceleration is independent of the mass m2 and is called the “gravitational field” due to m1.
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 141
Strictly speaking, the masses which occur in equation 7.5 are gravitational masses whereas the mass appearing in N2 is the inertial mass. The fact that the acceleration (7.6) is the same for all particles m2 is an experimental fact, first demonstrated in a famous experiment conducted from the leaning tower of Pisa by Galileo and subsequently tested to great accuracy by E¨otv¨os.As a consequence, we can deduce that the ratio of inertial mass to gravitational mass is the same for all particles, and by an appropriate choice of units we can assume that the inertial and gravitational masses of a particle are equal. In what follows, therefore, we shall not make any distinction between inertial and gravitational mass.
In some of the problems which we study the motion is due to the gravitational force arising from the whole Earth. If we assume that the Earth is a uniform sphere of radius a, and total mass M, then it can be shown that the gravitational force experienced by a particle at a point above the Earth’s surface is the same as the gravitational force which would be exerted on the particle by a particle situated at the centre of the Earth, with a mass equal to the total mass of the Earth. This means that the gravitational force F exerted on a particle of mass m, whose position vector is r relative to O, the centre of the Earth, is given by GmM F = − r. (7.7) r3 and the gravitational field of the Earth is GM g = − r. (7.8) r3 If the particle remains near the Earth’s surface throughout the motion, so that r/a is always close to 1, we may replace equations (7.7) and (7.8) by GmM F = − e = −mge , g = −ge (7.9) a2 r r r 2 where er is the unit vector parallel to r and g = GM/a . In the approximation which we have described the gravitational force on the mass m has magnitude mg and is directed along the downward vertical towards the centre of the Earth. We shall employ this notation in what follows without further explanation. Example 7.4. Motion near the Earth’s Surface A particle P of mass m is projected vertically upwards from a point A on the Earth’s surface with speed V. Discuss the subsequent motion, assuming that the particle remains near the Earth’s surface throughout the motion.
We assume that we can neglect air resistance and that the only force acting on the particle is in fact gravitational.
Let O denote the Earth’s centre. We choose axes Oxyz which are fixed relative to the Earth and which are such that the z-axis points vertically upwards, through the point of projection; as noted previously such a reference frame is approximately inertial. Suppose that the particle’s coordinates at time t are (0, 0, z(t)); the initial conditions then demand that z(0) = a, z˙(0) = V.
Application of Newton’s Second Law (N2) gives
mz¨e3 = −mge3, z¨ = −g.
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 142
Integrating with respect to time t we find that
z˙ = −gt + C, where C is a constant. Sincez ˙(0) = V , it follows that z = −gt + V. A further integration with respect to t immediately gives 1 z(t) = a + V t − gt2. 2 The maximum height of the particle above the Earth’s surface is attained whenz ˙ = 0 i.e. when t = V/g. The corresponding value of z is then given by
V 2 1 V 2 V 2 z = a + − g = a + . g 2 g2 2g We conclude that the particle rises to a height V 2/(2g) above the Earth’s surface in a time V/g before falling back towards the point of projection.
We have assumed throughout this calculation that the gravitational force exerted on m by the Earth is constant; if this is to be a reasonable assumption, we would expect that the maximum height which we have just calculated should be a good deal less than a, the Earth’s radius i.e. V 2 a, V 2 2ag. 2g We shall see below that V = (2ag)1/2 does indeed have a special significance. Example 7.5. Motion Leaving the Vicinity of Earth’s Surface Suppose that our assumption is not tenable and that the speed of projection V is such that V 2 is comparable with 2ag. In this case we have to use the exact expression for the gravitational force, without any approximations. The equation of motion is GmM GM mz¨e = − e , z¨ = − . 3 z2 3 z2 Sincez ¨ = −g when z = a it follows that GM = ga2 and we may write ga2 z¨ = − . z2 Multiplying this equation byz ˙ and noting that d 1 ( z˙2) =z ˙z¨ dt 2 we obtain d 1 ga2 dz ( z˙2) = − . dt 2 z2 dt Integration with respect to t yields 1 Z ga2 ga2 z˙2 = − dz + C = + C, 2 z2 z where C is constant. Using the initial conditions z(0) = a, z˙(0) = V we obtain 1 2 C = 2 V − ag and the equation for z becomes 2ga2 z˙2 = + V 2 − 2ag. (7.10) z
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 143
The maximum value of z, z = zmax, say, is obtained by settingz ˙ = 0 and is given by 2ga2 z = . max 2ag − V 2
1/2 We see that zmax → ∞ as V → (2ag) from below; this value of V is referred to as the escape speed of the particle. Substituting the relevant values of a, g we find that the escape speed is approximately 11.2km/sec.
1/2 For values of V which are less than (2ag) the particle rises to z = zmax before falling back to the point of projection. Re-writing equation 7.10 in terms of zmax we find that 2ga2(z − z) z˙2 = max . zzmax
If we require to compute the time τ for the particle to reach zmax we extract the square root and separate the variables to obtain
Z τ 1/2 Z zmax 1/2 zmax z τ = dt = 2 dz. 0 2ga a zmax − z
2 This integral can be evaluated by means of the substitution z = zmaxsin θ.
7.4 Motion in a Straight Line; the Energy Equation
Let x(t) denote the coordinate at time t of a particle of mass m moving in a straight line along the x-axis under the influence of a force F e1. The equation of motion is
mx¨e1 = F e1, mx¨ = F.
We now make the special assumption that F depends only on x, the position coordinate of the particle so that F = F (x) and the equation of motion becomes
mx¨ = F (x).
Multiplying byx, ˙ and recalling that d 1 x˙ 2 =x ˙x¨ , (7.11) dt 2 we obtain 1 Z dx Z mx˙ 2 = F (x) dt + C = F (x) dx + C, (7.12) 2 dt where C is a constant of integration. We write 1 Z T = mx˙ 2,V (x) = − F (x) dx (7.13) 2 and refer to T as the kinetic energy of the particle and to V as the potential energy function of the particle. In terms of these definitions the last equation may be written
T + V = C = constant. (7.14)
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 144
T + V, the sum of the kinetic and potential energies of the particle, is called the total energy of the particle and equation 7.14 then states that the total energy is constant throughout the motion—it is a constant of the motion. We emphasise that this result depends on the assumption that F is a function of x only.
Note that V is defined in terms of an indefinite integral of F (x); it follows that the potential energy V is only unique up to a constant. We note that equation 7.13 implies dV F (x) = − . (7.15) dx We may say that the force acting on the particle is minus the gradient of the potential energy function—even though this is slightly sloppy, since the actual force is not F (x) but F (x)e1. Example 7.6. Potential of Constant Gravitational Force Suppose that the axis Ox points along the upward vertical and that the particle is moving along Ox under constant gravity so that the force acting is −mge1.
The potential energy function V is given by dV − = −mg, V (x) = mgx + C. dx If we arbitrarily require that V (0) = 0 we may set C = 0 and write V (x) = mgx. In this case the energy equation 7.14 reads 1 mx˙ 2 + mgx = constant. 2 Given a set of initial conditions i.e. the values of x andx ˙ at t = 0 we could evaluate the constant explicitly.
Example 7.7. Simple Harmonic Motion 2 Suppose that the particle is moving along the x-axis under the action of a force −mω xe1, where ω is constant.
First note that, whatever the sign of x, the force is always directed towards the origin O. The potential energy V is given by dV 1 − = −mω2x, V (x) = mω2x2. dx 2 (We have arbitrarily required that V (0) = 0 in order to set the constant of integration equal to zero.)
The energy equation reads 1 1 mx˙ 2 + mω2x2 = constant, 2 2 which we can re-write in the form x˙ 2 + ω2x2 = a2, where a is a constant. We could integrate this equation to find x as a function of t but it is perhaps easier to go back to Newton’s Second Law:
mx¨ = −mω2x, x¨ + ω2x = 0.
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 145
We know from our work on differential equations that this equation has general solution
x(t) = α cos ωt + β sin ωt, where α, β are constants. It is well known that the right hand side of this equation can be expressed in the form A sin(ωt + ), where a, are constants and we may therefore write the general solution as
x(t) = A sin(ωt + ). (7.16)
We may assume, without loss of generality that A is positive.
Bearing in mind the properties of the sine function it is clear that the particle executes an oscillatory motion between the points x = ±A. A is called the amplitude of the motion. The centre of the motion is the point x = 0.
The time τ taken by the particle to complete one full oscillation is given by ωτ = 2π so that 2π τ = . ω τ is called the period of the motion. The number of complete oscillations per unit time is called the frequency of the oscillation, and is usually denoted by ν. We see that 1 ω ν = = . τ 2π The constant which appears in equation 7.16 is sometimes referred to as the phase angle. The motion which we have just described is said to be Simple Harmonic and the equationx ¨ + ω2x = 0 is called the differential equation of Simple Harmonic motion.
Example 7.8. Simple Harmonic Motion II 2 Suppose instead that the force acting is −mω (x − a)e1, where a is constant.
In this case the equation of motion becomes
mx¨ = −mω2(x − a), x¨ + ω2(x − a) = 0.
We write z = x−a and obtain the equationz ¨+ω2z = 0 (The differential equation of Simple Harmonic motion). This equation has general solution
z(t) = A sin(ωt + ), where A > 0, are constants. Expressing this result in terms of x we obtain
x(t) = a + A sin(ωt + ).
We see that this motion is also simple harmonic with amplitude A, and period τ = 2π/ω. The oscillation takes place between the points x = a ± A and the centre of the motion is x = a.
How does Simple Harmonic motion arise in practice?
Example 7.9. Elastic Strings and Springs (Hooke’s Law) One end of a light elastic string, of unstretched length b, is secured to a fixed point O, whilst a particle of mass m is attached to the other end, and the system hangs in equilibrium under (constant) gravity. Subsequently the particle is set in motion by a blow which imparts to the particle an initial speed
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 146 u in the sense of the downward vertical. Assuming that the string remains stretched, determine the subsequent motion, neglecting air resistance; you may assume Hooke’s Law.
Choose the x-axis so that Ox points along the downward vertical and let x(t) denote the x-coordinate of the mass m at time t. The forces acting on the particle are
(i) The force of gravity, mge1,
(ii) The force −T e1 due to the tension in the string— which acts on the particle in the sense of the upward vertical. It is necessary to postulate this force - it is the force which balances gravity when the particle is hanging in equilibrium, and it continues to act throughout the subsequent motion.
Application of N2 gives mx¨e1 = mge1 − T e1. In order to make progress we have to make some assumption as to how T depends on x. If we assume Hooke’s law we may write T = k(x − b), where k is a positive constant which depends on the nature of the elastic material. Hooke’s law assumes that the tension is proportional to the extension of the string beyond its natural length. The equation of motion now becomes mx¨e1 = mge1 − k(x − b)e1. Writing z = x − b our equation becomes
z¨ + ω2z = g, ω2 = k/m. (7.17)
The homogeneous equationz ¨ + ω2z = 0 has general solution of the form z = A sin(ωt + ). We note that z = g/ω2 is a particular solution of equation 7.17 and we conclude that its general solution is
z(t) = A sin(ωt + ) + g/ω2.
Initially the system is in equilibrium so thatz ¨ = 0; equation 7.17 shows that z(0) = g/ω2. The system is set in motion with initial speed u in the sense of the downward vertical soz ˙(0) = u. Imposing these initial conditions on the general solution we deduce that
A sin = 0, Aω cos = u.
These equations are satisfied by = 0,A = u/ω. Hence u g z(t) = sin ωt + , ω ω2 or, written in terms of x u g x(t) = b + sin ωt + . ω ω2 g We conclude that the particle executes Simple Harmonic motion about x = b + ω2 as centre (this is the equilibrium value of x, before the blow is struck); the period is p m u τ = 2π/ω = 2π k and the amplitude of the oscillation is ω .
Note that the potential energy function V is given by dV k − = mg − k(x − b),V (x) = −mgx + (x − b)2 + C, dx 2
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 147 where C is a constant.
u g Note that the least value of x(t) throughout the motion is equal to b − ω + ω2 and that for the string to be in a state of tension throughout the entire motion this must be greater than b. This implies that u < g/ω—a constraint on the magnitude of the initial speed of the particle.
7.5 Equilibrium and Stability
Consider a particle of mass m, moving along the x-axis under a potential V (x), so that the force F acting on m is given by dV F = − e = −V 0(x)e . dx 1 1 By N2 the equation of motion is V 0(x) mx¨e = −V 0(x)e , x¨ = − (7.18) 1 1 m
Putting x1 = x, x2 =x ˙ we find that V 0(x ) x˙ = x , x˙ = − 1 . (7.19) 1 2 2 m Let x = a be a stationary point of the potential function V, so that V 0(a) = 0. Suppose also that the initial conditions are x(0) = a, x˙(0) = 0; this means that at t = 0 the particle is instantaneously at rest at x = a. In terms of x1, x2 the initial conditions become
x1(0) = a, x2(0) = 0. (7.20)
Assuming that V has continuous second order derivatives we deduce from Picard’s Theorem that the unique solution of equations 7.19 subject to the initial conditions expressed by equations 7.20 is x1(t) = a, x2(t) = 0, ∀t ≥ 0. Reverting to x, x˙ this means that the particle remains at rest at x = a for all t ≥ 0. We say that x = a is an equilibrium position for the particle; the possible equilibrium positions are the stationary points of V.
Suppose now that the particle, at rest at the equilibrium point x = a, is set in motion with velocity ue1. In the subsequent motion energy is conserved and we have 1 1 mu2 + V (a) = mx˙ 2 + V (x), 2 2 so that 1 1 mx˙ 2 = mu2 + V (a) − V (x). (7.21) 2 2 1 2 First suppose that x = a is a minimum of V . Since 2 mx˙ ≥ 0 we deduce from equation 7.21 that 1 V (x) ≤ mu2 + V (a). (7.22) 2
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 148
V(x)
V(a)
x a a a 1 2
Figure 7.2: Stable periodic motion in the vicinity of a minimum of the potential energy.
Referring to the diagram we see that, if u is sufficiently small, the particle’s motion will be confined to the interval [a1, a2], where a1, a2 are solutions of the equation 1 V (x) = mu2 + V (a). 2 Perhaps it will be helpful to describe the motion in a little more detail. To be definite, suppose that u > 0 so that the particle moves towards a2. As it does, V (x) increases, and the particle loses kinetic energy, coming to instantaneous rest at x = a2. It cannot move to the right of 1 2 a2, otherwise 2 mx˙ would become negative! Having reached x = a2 the particle starts back towards x = a, passing through x = a with speed u, and comes to a state of instantaneous rest once more at the point x = a1, before moving back to x = a2. We see that the particle oscillates between x = a1 and x = a2.
In conclusion, we see that if |u| is small enough, the particle will remain “near” x = a, in the sense described, and we are justified in saying that x = a is a position of stable equilibrium.
Suppose, on the other hand, that x = a is a maximum of the potential function V and that the particle, initially at rest at x = a is set in motion with velocity ue1.As the particle moves 1 2 away from x = a the potential V (x) decreases and therefore 2 mx˙ increases by virtue of equation 7.21. The particle does not remain “near” x = a and we are justified in saying that x = a is a position of unstable equilibrium.
The above arguments are based on the energy equation and graphical considerations. Alter- natively, we may proceed as follows. Since x = a is an equilibrium point, V 0(a) = 0. The equation of motion is (see equation 7.18)
V 0(x) x¨ = − m Near x = a Taylor’s formula gives
V 0(a)(x − a) V 00(a)(x − a)2 V (x) = V (a) + + + ··· 1! 2!
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 149
Assuming that x is “near” a, and that V 00(a) 6= 0, we have 1 V (x) ' V (a) + V 00(a)(x − a)2 2 and the equation of motion approximates to
1 d 1 V 00(a) x¨ = − V (a) + V 00(a)(x − a)2 = − (x − a). m dx 2 m
Writing z = x − a we obtain V 00(a) z¨ + z = 0. m We have to deal with two cases, depending on the sign of V 00(a).
Case 1 Suppose that V 00(a) > 0, so that x = a is a minimum of V. Writing
V 00(a) ω2 = > 0 m we obtain z¨ + ω2z = 0. This equation has general solution
z = A sin(ωt + ),
where A, are constants. Expressing z in terms of x we now have
x(t) = a + A sin(ωt + ).
The initial conditions x(0) = a, x˙(0) = u require that A sin = 0, Aω cos = u from which it follows that = 0,A = u/ω. Therefore u x(t) = a + sin ωt. ω In the approximation considered, the particle executes Simple Harmonic motion about x = a with amplitude u/ω and period 2π/ω. This result agrees with our earlier conclusions and we see that a minimum stationary point of V is a position of stable equilibrium.
Case 2 Now suppose that V 00(a) < 0, so that x = a is a maximum of V. Writing
V 00(a) Ω2 = − > 0 m we obtain z¨ − Ω2z = 0. This equation has general solution
z(t) = αeΩt + βe−Ωt,
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 150
V(x)
x 1 2
Figure 7.3: The potential that allows stable periodic motion in the vicinity of x = 2; the minimum energy needed to escape to −∞ is shown as a dashed line.
x(t) = a + αeΩt + βe−Ωt, where α, β are constants. Imposing the initial conditions x(0) = a, x˙(0) = u we find that u x(t) = a + sinh Ωt. Ω Since sinh Ωt → ∞ as t → ∞ we see that our assumption that x remains “near” a is invalid. This conclusion is in agreement with our graphical arguments using the energy equation: a maximum of V is a position of unstable equilibrium.
Example 7.10. Motion in a Cubic Potential A particle of unit mass moves along the axis Ox under a potential V (x) = 2x3 − 9x2 + 12x. Find the values of x for which the particle can remain in equilibrium. Show that if (x − 2)2 is regarded as negligible√ in comparison with (x − 2) then the period τ of small oscillations about x = 2 is given by τ = 2π/ 6. √ Suppose that x(0) = 2, x˙(0) = −u, where u > 2. Describe (qualitatively) the subsequent motion. Hint: Sketch the graph of V (x) and appeal to the principle of energy conservation.
We have V 0(x) = 6x2 − 18x + 12 = 6(x2 − 3x + 2) = 6(x − 1)(x − 2). The possible equilibrium positions are given by V 0(x) = 0 i.e. by x = 1, x = 2. The equation of motion is x¨ = −V 0(x), x¨ + 6(x − 1)(x − 2) = 0. Writing z = x − 2 we derive z¨ + 6z(z + 1) = 0. Assuming that terms in z2 are negligible in comparison with the linear terms in z we obtain as an approximation the differential equation z¨ + 6z = 0.
This describes Simple Harmonic motion√ about the point z = 0 i.e. about x = 2 and the period τ of the oscillations is given by τ = 2π/ 6.
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 151
Referring to the graph of V we see that if the particle is projected from x = 2 with velocity −ue1 it will move to the left of x = 2. As it does so energy is conserved: 1 1 x˙ 2 + V (x) = u2 + V (2). 2 2 If the particle still has positive kinetic energy when it reaches x = 1 (a maximum of V ) then it will move off to −∞, since V decreases (rapidly) for x < 1. The condition for this to happen is 1 u2 + V (2) − V (1) > 0. 2 √ Evaluating V (1),V (2) we find that the condition for the particle to move off to −∞ is u > 2.
Example 7.11. Motion in a Non-Linear Potential A particle of mass m moves along the axis Ox under the potential V given by cx V (x) = , (a2 + x2) where a, c are positive constants. Sketch the graph of V ; show that V has a minimum at x = −a and a maximum at x = a. Given that x(0) = −a, x˙(0) = u, where u is suitably small, discuss the subsequent motion. By reference to your graph describe what will happen to the particle if, instead, the initial conditions are x(0) = a, x˙(0) = u, where u can be either positive or negative.
V(x)
x −a a
Figure 7.4: Potential that allows stable periodic motion in the vicinity of x = −a; depending on initial conditions, escape to ±∞ is possible.
We find that c(a2 − x2) 2cx(x2 − 3a2) V 0(x) = ,V 00(x) = . (a2 + x2)2 (a2 + x2)3 Now V 0(x) = 0 when x = ±a. It is clear from consideration of the graph of V that x = −a is a minimum of V (and therefore a position of stable equilibrium), and that x = a is a maximum of V, and therefore a position of unstable equilibrium.
Near x = −a the equation of motion is
dV d V 0(−a)(x + a) V 00(−a)(x + a)2 mx¨ = − = − V (−a) + + + ··· dx dx 1! 2!
= −V 00(−a)(x + a),
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 152 on the assumption that the higher powers of (x + a) are negligible. Writing z = x + a we obtain
V 00(−a) c z¨ + z = 0, z¨ + z = 0. m 2ma3 This equation has general solution
z = A sin(ωt + ), x(t) = −a + A sin(ωt + ), where c ω2 = . 2ma3 In the approximation which we have described the motion is Simple Harmonic about x = −a as centre, and the period is r 2ma3 τ = 2π . c Imposing the initial conditions x(0) = −a, x˙(0) = u, we find that u x(t) = −a + sin ωt. ω Reference to the graph of V suggests that our approximation is likely to be reasonable if the amplitude u/ω a i.e. if u ωa.
Suppose now that the initial conditions are x(0) = a, x˙(0) = u. The energy equation yields 1 1 mx˙ 2 + V (x) = mu2 + V (a). 2 2
1 2 1 2 We see that in the subsequent motion 2 mx˙ > 2 mu since V (x) < V (a). We deduce that if u > 0 the particle will move off to +∞ and that if u < 0 the particle will move off to −∞.
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 153
7.6 Exercises
Some physical data
−11 3 −1 −2 Newton’s constant GN 6.67 × 10 m kg s 30 Mass of the Sun Msun 1.99 × 10 kg 24 Mass of the Earth Mearth 5.97 × 10 kg 22 Mass of the Moon Mmoon 7.35 × 10 kg 8 Radius of the Sun rsun 6.96 × 10 m 6 Radius of the Earth rearth 6.37 × 10 m 6 Radius of the Moon rmoon 1.74 × 10 m 11 Distance of Earth from Sun dearth−sun 1.50 × 10 m 8 Distance of Moon from Earth dearth−moon 3.84 × 10 m
Short Exercises
Exercise 7.1 A particle moves along a path cos(f(t)) r(t) = sin(f(t)) . 0 Calculate v = r˙ , a = v˙ . Show (by explicit calculation) that
r · r = 1 , v · r = 0 , r · a ≤ 0 .
Find condition(s) on f(t) for r · a = 0 to be true. What does this imply for the particle’s motion? Find condition(s) on f(t) for v · a = 0 to be true.
Exercise 7.2 For each of the potentials V (x) below, find the maxima, minima and points of inflection [i.e. all points where V 0(x) = 0] and sketch the graph of V (x). x (a) V (x) = (x2 − 1)2 (b) V (x) = −sech2(x) (c) V (x) = x2 + 1 4 (d) V (x) = x4 − x3 (e) V (x) = x2e−x 3
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 154
Longer Exercises Exercise 7.3 A particle moves so that its position vector with respect to the origin O of a reference frame Oxyz is r(t) = b cos ωt e1 + b sin ωt e2 + V te3, where e1, e2, e3 are unit vectors parallel to the axes Ox, Oy, Oz; b, ω, V are positive constants. Find (i) The velocity and speed of the particle, (ii) Prove that the particle moves on a circular cylinder, whose axis is Oz. (iii) Find the particle’s acceleration, and indicate its direction in a diagram in the case b = ω = V = 1. Exercise 7.4 2 A particle has position vector r(t) = e1 + te2 + t e3 relative to the origin O of a frame Oxyz. Find the velocity, the speed, and the acceleration of the particle. Find the distance travelled by the particle between t = 0 and t = 1. What is the instantaneous direction of the particle’s motion at t = 1? (Calculate the relevant unit vector.) Exercise 7.5 Two particles 1, 2 move in such a way that their position vectors r1, r2, with respect to the origin O of an inertial frame of reference, are given by 2 r1(t) = e1 + te2 + t e3, 2 3 r2(t) = te1 + t e2 + t e3, respectively. Write down expressions for the velocity, speed and acceleration of the particles at time t and show that, at the instant when they collide, the angle α between their instantaneous directions of motion is given by 8 cos α = √ . 70 Exercise 7.6 A particle moves so that its position vector r with respect to the origin O of an inertial frame of reference is 2 r = b cos ωte1 + b sin ωte2 + V t e3, where b, ω, V are positive constants. √ Verify that the acceleration vector is orthogonal to r when t = bω/( 2V ). Prove also that the acceleration vector makes a constant angle β with the direction e3, where 2V cos β = . (b2ω4 + 4V 2)1/2 Exercise 7.7 O,O0 are the origins of two reference frames F,F 0. At time t = 0 OO0 = a. Suppose that O0 moves with a constant velocity V relative to O. At time t a particle P has position vector r(t) relative to O and r0(t) relative to O0. Use a suitable vector diagram to write down an equation relating r(t) and r0(t). Deduce that if Newton’s second Law (N2) applies in the frame F, it also applies with respect to the frame F 0.
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 155
Exercise 7.8 Suppose that two particles 1, 2 of masses m1, m2 act on each other in such a way that the force on particle 1 due to particle 2 is F 12 and that on particle 2 due to particle 1 is F 21; then F 12 + F 21 = 0, by Newton’s Third Law (N3). Write down N2 for each particle in some inertial frame of reference and deduce that the centre of mass of the 2 particle system moves with constant velocity with respect to the chosen frame. (Recall that the position vector r of the centre of mass is given by
m1r1 + m2r2 r = , (m1 + m2) where r1, r2 are the position vectors of m1, m2 with respect to the chosen origin.)
Prove that the above formula for the centre of mass defines the same point in space, irrespec- tive of the choice of origin. 0 0 Hint Choose origins O,O and suppose that the two particles have position vectors ri, ri (i = 1, 2) with respect to O,O0 respectively. Then
0 0 ri = OO + ri , (i = 1, 2).
The centre of mass has position vector R, R0 with respect to the origins O,O0 where
m1r1 + m2r2 R = , (m1 + m2) 0 0 m1r1 + m2r2 R0 = . (m1 + m2) Now verify that R = OO0 + R0. Exercise 7.9 A particle of mass m moves along the x-axis under a constant force Ce1. Let x(t) denote its coordinate at time t. Given the initial conditions
x(0) = a, x˙(0) = u,
prove that 1 x˙(t) = u + ft, x(t) = a + ut + ft2, x˙ 2 = u2 + 2f(x − a), 2 where f = C/m.
Note These formulae are only applicable if the force acting on the particle is constant; this is seldom the case, yet many students treat these formulae as though they had universal validity and could be used to find the answer to any problem in mechanics! Remember, solving a problem in Newtonian mechanics involves setting up the equation of motion N2 with the relevant set of initial conditions; if you proceed in this way you can’t go wrong!
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 156
Exercise 7.10 A particle is projected vertically upwards from the surface of the Earth with speed V = (ag)1/2, where a is the Earth’s radius. Recall that the force of gravity, when the particle is 2 2 at (0, 0, z) is F = −(mga /z )e3, m being the particle’s mass.
(We’re following the notation used earlier in this chapter, choosing the origin O at the Earth’s centre, and the axis Oz vertically upwards through the point of projection (0, 0, a).) Show that the particle rises a height a above the Earth’s surface before it starts to fall, and prove that the time to fall from this position of maximum height, back to the point of projection, is τ where Z 2a τ = (ag)−1/2 z1/2 dz/(2a − z)1/2, a and evaluate this integral by the substitution z = 2asin2θ. q a (The answer is (π/2 + 1) g .)
Exercise 7.11 A particle is projected vertically upwards from the Earth’s surface with speed (3ag/2)1/2, where a denotes the Earth’s radius, regarded as a uniform sphere. Use the energy equation to show that the particle rises to a height 3a above the Earth’s surface before returning to the point of projection.
Exercise 7.12 A particle of unit mass moves along the axis Ox of an inertial frame of reference under the 3 action of a force F = (k/x )e1, where k is a positive constant. Given that the particle starts from rest at x = a show that its speed at the point x = 2a is (3k/4a2)1/2. Compute the time taken by the particle to move from x = a to x = 2a.
Exercise 7.13 A particle of unit mass moves along the x−axis under the action of a force derived from a potential V (x) = − cos x. The particle, initially in equilibrium at x = 0, is set in motion with speed u in the sense of the positive x−axis. Write down the equation of motion of the particle and, assuming that |x(t)| remains small ( π/2) in the subsequent motion, find x(t) in terms of u. (You may assume that the replacement sin x ≈ x is justified.) What is the period τ of the oscillation? Sketch V (x) and show, by considering the energy equation, that if u exceeds a certain critical value u0 (which should be found), the particle will move off to +∞. Exercise 7.14 A particle of mass m moves along the x-axis under the potential cx V (x) = , (a2 + x2)2 where a, c are positive constants. Find the period of small oscillations about the position of stable equilibrium.
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 157
Exercise 7.15 Referring to Example 7.9 suppose that we attempt to model the effect of air resistance by supposing that the air exerts a force −2mλx˙ e1 on the particle, where λ is a positive constant. The modified equation of motion is then
mx¨e1 = mge1 − k(x − b)e1 − 2mλx˙ e1,
or, writing z = x − b, z¨ + 2λz˙ + ω2z = g, ω2 = k/m. Find the general solution of this equation on the assumption that λ2 − ω2 < 0. (This means that damping due to air resistance is small). Suppose that at time t = 0 the particle is hanging in equilibrium under gravity when it is struck a blow which causes it to move with initial velocity ue1 in the sense of the downward vertical. Find the displacement of the particle at time t, assuming that the string remains taut throughout the motion. We say that the particle is executing Damped Harmonic Motion; the equation
z¨ + 2λz˙ + ω2z = 0, λ > 0
is sometimes referred to as the equation of Damped Harmonic motion.
Exercise 7.16 (King’s College, Summer 1995 exam) Consider the motion of a particle of mass m when it moves along a straight line under the action of forces F e1 derivable from potentials V (x). (a) Use Newton’s equation of motion to prove that the total energy of the particle is a constant of the motion. (b) When F = m(11x−2 − 36x−3) compute V (x). If the particle is initially at x = 1 and has initial velocity 4e1 find the maximum value of x which it can attain. (c) If x = a is an equilibrium point of a motion show that, when |x − a| is small, Newton’s equation of motion is approximated by
d2z m + V 00(a)z = 0, dt2 where z = x − a and V 00 = d2V/dx2. Let V (x) = x3 + βx2 − γx, where β and γ are positive real numbers. Identify all the stable and unstable equilibrium points and describe the motion near any stable equilibrium points.
Exercise 7.17 [Taken from 2004 exam] A particle is moving in an inertial frame of reference in such a way that its position vector r(t) with respect to the origin O is given by
at2 r(t) = cos(ωt) e + sin(ωt) e + h − e , 1 2 2 3
where the eα, α = 1,..., 3, are orthogonal unit basis vectors of a Cartesian coordinate system.
(a) Find the velocity, speed and acceleration of the particle, and compute the cosine of the angle between its velocity and acceleration as functions of time.
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 158
(b) Assume h > 0 and a > 0. Determine, at which time t the particle will hit the e1-e2 plane. Assume that, when hitting that plane, the particle is reflected by instantaneously changing the sign of the e3-component of the velocity. Determine the cosine of the angle between the directions of motion immediately before and after the reflection.
Exercise 7.18 Consider a system of 2 particles each of which exerts a force on the other. Show that the centre of mass moves uniformly with respect to an inertial frame. Show further that
µr¨ = F12 where F12 is the force on particle 1 due to particle 2, r = r1 − r2 and the reduced mass µ is given by m m µ = 1 2 m1 + m2 Consider a system of 3 particles each of which exerts a force on the other two. Denote the force on particle i due to particle j by Fij. Use Newton’s laws to show that the centre of mass of the system moves uniformly.
Exercise 7.19 3 Consider a particle moving in R with position R cos(ωt) r = R sin(ωt) , 0 where R and ω are positive constants. Calculate the velocity v and acceleration a and their magnitudes v and a. Show that v2 r = R , v = Rω , a = −ω2r , a = ω2R , a = . r These formulae show that constant velocity does not mean constant acceleration and are also very useful when dealing with motion in a circle at constant speed.
Exercise 7.20 Suppose that the Earth is at the origin of an inertial coordinate system and the moon as at position r. Where is the centre of mass of the Earth-moon system? Is it inside or outside the Earth?
Exercise 7.21 A particle of mass m is propelled from the surface of the earth, with an initial velocity v0 = (a, b, c), with c > 0. Choose the origin O of the coordinate system Oxyz to be on the surface of the earth with z axis pointing vertically upward, and x and y axes parallel to the earth’s surface. Let O be the point from which the particle is propelled.
(a) State Newton’s equation of motion for this system, assuming a constant gravitational force in the negative z direction, F = (0, 0, −mg).
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 159
(b) Solve these equations for the given initial conditions and find the position, velocity, speed and acceleration of the particle as functions of time.
(c) Find the time at which the particle reaches maximum height, and give the position, velocity, speed and acceleration of the particle at that time.
(d) Find the time at which the particle returns to the earth’s surface, and give the position, velocity, speed and acceleration of the particle at that time.
(e) State briefly which aspects of your solution you find remarkable, and why.
Exercise 7.22 A particle moves under the influence of a constant gravitational field (in the vicinity of earth’s surface) and of the air resistance (which is velocity-dependent). Newton’s equation can then be written as mx¨ = −m g − 2mλ x˙ , in which x denotes the height above ground, and λ is a positive constant.
(a) Use the equation of motion to demonstrate that energy E = (mx˙ 2/2 + mgx) is not conserved in this system. More specifically, show that energy is a non-increasing function of time (i.e. E˙ ≤ 0)
(b) Solve the equation of motion. Give the full solution for the case that at t = 0 the particle is propelled upwards with initial speed u > 0.
Exercise 7.23 A particle of unit mass moves along the x-axis under the action of a force derived from a potential V (x) = 2x exp(−x2/2).
(a) Sketch V (x), write down the equation of motion, find the fixed points of the motion and classify them as either elliptic or hyperbolic.
(b) The particle, initially in equilibrium at the stable equilibrium point, is set in motion with speed v(0) = u > 0 in the positive x-direction. Assuming that deviations from stable equilibrium remain small in the subsequent motion, find x(t) in terms of u. What is the period τ of the oscillation?
(c) Show that there is critical value uo such that the particle will move off to x = +∞ if u > u0 but will not it u < u0.
(d) If u > u0, show that its speed v(t) tends to a limit, v(t) → v∞ as t → ∞.
Show that the limiting speed has a lower bound, u∞ such that v∞ > u∞ > 0 for all initial speeds v(0) > u0. Find the best (largest) lower bound on the limiting speed. Exercise 7.24 A particle of unit mass moves along the x-axis under the action of a force derived from a x4 x2 potential V (x) = 4 − 2 .
(a) Write down the equation of motion of the particle and find the equilibrium points.
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 160
(b) Sketch V . Assume that the particle is released from a position x0 with speedx ˙ 0 such that its total energy is E = −2, with 0 < < 1/2. Find the turning points of the dynamics.
(c) For the initial condition given in (b), find the -dependent coordinate x?() at which the system reverses its motion for the first time assuming that x0 > 0 andx ˙ 0 > 0. Answer the same question for the case x0 < 0 andx ˙ 0 > 0. Exercise 7.25 A particle of unit mass moves along the x-axis under the action of a force derived from a potential V (x) = − cos x. Write down the equation of motion of the particle and find the equilibrium points. The particle, initially in equilibrium at x = 0, is set in motion with speed π u > 0 in the positive x-direction. Assuming that |x(t)| remains small ( 2 ) in the subsequent motion, find x(t) in terms of u. What is the period τ of the oscillation?. Sketch V and show that if u exceeds a certain critical value uo (which should be found), the particle will move off to x = +∞. Exercise 7.26 Consider the motion of a particle of mass m when it moves along a straight line under the action of a force derivable from a potential.
(a) If the potential V (x) = x3 + βx2 − γx where β and γ are positive real numbers, identify all the equilibrium points and describe the motion near the stable ones. (b) In the case that F = m(11x−2 − 36x−3), compute and sketch V (x). If the particle is initially at x = 1 and has initial speed u = 4 find the maximum value of x it can attain.
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 161
Supplementary Exercises
Supplementary Exercise 7.1 Consider a particle of mass 1 moving along the real axis in the potential
V (x) = x sin(x) .
Sketch the phase portrait. Repeat this for the potential V (x) = x cos(x) .
Supplementary Exercise 7.2 (a) Consider a particle moving along a path (x(t), y(t)) in the (x, y) plane. Let its coordinates in polar coordinates be (r(t), θ(t)) where
x = r cos(θ) , y = r sin(θ) .
Show that its speed is given by v where
v2 =r ˙2 + r2θ˙2 .
(b) 3 Consider a particle moving along a path (x(t), y(t), z(t)) in R . Let its coordinates in spherical polar coordinates be (r(t), θ(t), φ(t)) where
x = r cos(φ) sin(θ) , y = r sin(φ) sin(θ) , z = r cos(θ) .
Show that its speed is given by v where
v2 =r ˙2 + r2θ˙2 + r2 sin2(θ)φ˙2 .
Supplementary Exercise 7.3 Consider a particle moving in the (x, z) plane with the motion satisfying
x¨ 0 x(0) 0 x˙(0) v = , = , = , z¨ a z(0) 0 z˙(0) 0
where a ≥ 0 and v ≥ 0. Show that the distance s travelled by the particle at time t satisfies 1 1 vt + at2 ≥ s ≥ at2 . 2 2
Supplementary Exercise 7.4 Calculate the gravitational field at the surface of the moon due to the moon. Calculate the gravitational forces at the centre of the Earth due to the moon and due to the Sun. Which is stronger?
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 162
Calculate the difference between the gravitational forces due to the moon on the side of the Earth nearest the moon compared to the force at the centre of the earth (see figure)
rearth dearth−moon − rearth
dearth−moon
Moon Earth
Repeat this calculation for difference between the field at the surface of the Earth and at the centre of the the Earth due to the Sun. Which difference is stronger? It is the difference between the forces at the surface of the Earth (where the oceans are) and on the centre of the Earth (determining the planet’s motion) which is responsible for the tides, so although the gravitational pull of the Sun on the Earth is far stronger than that of the moon, it is the Moon which is principally responsible for the tides.
Supplementary Exercise 7.5 The inwards acceleration in a circular path of radius r at constant speed v is v2/r (problem 8.1) Suppose that a planet in orbit around the sun follows a circular path of radius r with the sun at the centre, show that its speed satisfies G M v2 = N sun . r Show that the period of the orbit satisfies
4π2 T 2 = r3 . GN Msun This is an example of Kepler’s third law (T 2 ∝ r3) that you will study more generally in Intermediate Dynamics.
Supplementary Exercise 7.6 (a) Consider a particles of moving along the x-axis in the gravitational fields of a body of mass M1 at x = −1 and another of mass M2 at x = 1. Write down the gravitational potential energy of the particle of mass m due to each of the two bodies separately and the total gravitational potential energy of the particle of mass m. Hence, write down the equation of motion of the particle. Sketch the two separate contributions to the gravitational potential energy and the total gravitational potential energy of the particle. Sketch the phase portrait of the associated dynamical system.
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 163
Are there any equilibrium points? Are they elliptic (stable) or hyperbolic (unstable)? If stable, find the period of small oscillatory motion around the stable point?
(b) A particle of charge q and mass m = 1 moves in the region −1 < x < 1 in the electric field of two static charges, one of charge Q1 at position x = −1 and the other of charge Q2 at x = 1. Its motion is determined by Coulomb’s law for the total potential energy qQ 1 qQ 1 V = 1 + 2 . 4π0 |x + 1| 4π0 |x − 1|
Here 0 is a constant. Note that the potential energy depends on the relative signs of the electric charges as force can be either attractive or repulsive (gravitational masses are always positive and the force is always attractive).
Consider the case where |Q1| = |Q2| = 4π0/q. How many different cases do you have to consider? For each of the separate cases in turn: Sketch the phase portrait of the associated dynamical system. Are there any equilibrium points? Are they elliptic (stable) or hyperbolic (unstable) or can you not decide? What is the period of the motion around any stable point of equilibrium.
Supplementary Exercise 7.7 Consider a massive particle attached to a pivot by a string of length l such that the string makes an angle θ with the downward vertical.
Find the initial speed v0 at which the particle has to be moving when it is vertically beneath the pivot for the string to go slack when the string makes an angle 2π/3 with the downward vertical. Find the position of the particle when the string goes taut again. (see diagram below)
String goes slack at ? 2 Π Θ = 3
2 Π 3 Θ
v0
Version of Mar 14, 2019 Chapter 7: Elements of Newtonian Mechanics 164
Supplementary Exercise 7.8 Photons moving in the gravitational field of a black hole can be described by a function r(θ) where r is a radial coordinate and θ is an angular coordinate. With u = 1/r, the function u(θ) satisfies
d2u 3 = −u + r u2 , dθ2 2 S where rS is the Schwarzschild radius, the location of the event horizon of the black hole. [see equation (293) of the lecture notes for module 6ccm334a] Find the radius at which photons can move in a periodic orbit around the black hole. Is this a stable or unstable orbit? Find a potential function V (u) for which the u(θ) evolves as though moving in the potential V (u). Sketch the phase portrait and describe the possible paths a photon can take.
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 165
Chapter 8
Hamiltonian Systems
Hamilton’s formulation of mechanics is a way to deal with mechanics more abstractly. It was formulated by Sir William Rowan Hamilton (1805-1865) (who also invented quaternions). It lets you treat more complicated systems easily, greatly simplifies the consideration of sym- metries and is of key importance in the development of quantum mechanics. It is not the first reformulation that is still in use – that was by Joseph Louis Lagrange (1788). This is based on the Lagrangian L, a function of the system’s coordinates and velocities. For a system with one degree of freedom x, L is a function of x andx ˙ with the equations of motion being given by d ∂L ∂L = , (8.1) dt ∂x˙ ∂x which come from a “least action principle”. The standard way to construct L is as
L = T − V, (8.2) where T is the kinetic energy and V is the potential energy. For simple harmonic motion, 1 mx¨ = −mω2x , V = mω2x2 , (8.3) 2 this gives 1 1 ∂L ∂L L = mx˙ 2 − mω2x2 , = −mω2x , = mx˙ , (8.4) 2 2 ∂x ∂x˙ | {z } | {z } T V and the equations of motion are correctly given through (8.1) d (mx˙) = −mω2x . (8.5) dt There is a lot more on this in the module Intermediate Dynamics/Classical Dynamics. Hamilton reformulated this by replacing the velocityx ˙ by a new variable, the momentum p, and the Lagrangian L by a new function H(x, p), the Hamiltonian, defined by ∂L p = ,H = p x˙ − L. (8.6) ∂x˙ The equations of motion are then given by Hamilton’s equations ∂H ∂H x˙ = , p˙ = − . (8.7) ∂p ∂q
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 166
Example 8.1. The Hamiltonian for the simple harmonic oscillator
The Lagrangian is 1 1 L = mx˙ 2 − mω2x2 , (8.8) 2 2 and so the momentum p and Hamiltonian are given by
∂L p2 1 p2 1 p2 1 p = = mx˙ , H = px˙ − L = − + mω2x2 = + mω2x2 . (8.9) ∂x˙ m 2 m 2 2m 2 |{z} | {z } T V In this case H is the total energy and this is in fact its general interpretation. We can check that Hamilton’s equations give back the correct dynamics for x: ∂H p x˙ = = p˙ ∂p m ⇒ x¨ = = −ω2x . (8.10) ∂H m p˙ = − = −mω2x ∂x
More generally a Hamiltonian system of one degree of freedom may be described in terms of two functions q, p which evolve in time according to Hamilton’s Eqs (8.7) for some Hamiltonian H(q, p, t). In this wider sense the system need not be mechanical, and even in a mechanical situation q need not necessarily be a linear displacement—for example, it could be the angle which a pendulum makes with the vertical. The variable q is referred to as a generalised coordinate and p is the momentum conjugate to q. q, p are said to be conjugate variables. Of course, in the wider sense which we have described p, the momentum conjugate to q, need not be a physical momentum.
In what follows we shall usually be dealing with an autonomous Hamiltonian system; in this case H = H(q, p) and H doesn’t depend on time t explicitly.
The Lagrangian and Hamiltonian formulations of mechanics allow one to find and study the equations of motions of complicated systems such as spinning bodies or interconnected particles easily, but in such cases the definition of the momentum conjugate to a generalised coordinate can be complicated and one usually first sets up the Lagrangian formalism to find the momentum and then constructing the Hamiltonian. There is one simple situation in which we can directly write down the Hamiltonian and that is for motion in a potential as we studied in the previous chapter.
8.1 Hamilton’s equations for motion in a potential
Consider a particle of mass m moving in a straight line along the q-axis (it is traditional in Hamiltonian theory to use q rather than x) in a potential V (q). If we conjecture that the Hamiltonian is given by H = T + V and that p = mx˙, we are led to
p2 H(q, p) = + V (q) . (8.11) 2m
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 167
Hamilton’s equations (8.7) are then
∂H p x˙ = = ∂p m ⇒ mx¨ =p ˙ = −V 0(x) . (8.12) ∂H dV p˙ = − = − ∂x dx This is the simplest possible application of Hamilton’s formulation. The only essential ingre- dients are the coordinate q, the conjugate momentum p, and the Hamiltonian H(q, p, t). The mechanical system which we are describing is said to have one degree of freedom (represented by the coordinate q) but in terms of our earlier definitions is a second order dynamical system which is autonomous provided V doesn’t depend on t explicitly. We note in passing that Hamilton’s equations may be expressed in the form
d q = σ(∇H), dt p where ∇ is the vector differential operator given by ∂ ∂ ∇ = e + e ∂q 1 ∂p 2 and σ is the 2 × 2 anti-symmetric matrix given by
0 1 σ = . −1 0
Suppose now that q(t), p(t) satisfy Hamilton’s equations with H = H(q, p). Then H(q, p) is constant throughout the motion—we say that H is a constant of the motion. To prove this result we note that ∂H ∂H H˙ = q˙ + p,˙ ∂q ∂p by the chain rule. Using Hamilton’s equations we obtain
∂H ∂H ∂H ∂H H˙ = + − = 0. ∂q ∂p ∂p ∂q
We conclude that H is indeed a constant of the motion. Such Hamiltonian systems are said to be conservative; the Hamiltonian is constant throughout the motion. The phase curve which passes through the point (q0, p0) of phase space is given by
H(q, p) = H(q0, p0).
In this case we have no complicated differential equations to solve!
Returning to case of motion in a potential, suppose that V = V (q) so that
p2 H = H(q, p) = + V (q). 2m
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 168
p2 1 2 Note that 2m = 2 mq˙ , so the first term in the expression for H is the kinetic energy; V is the potential energy and H is the total energy of the system in this case.
We note that the velocity v(q, p) of the flow associated with a Hamiltonian H(q, p) (see Chapter 8, Section 1.1) is given by
∂H ! ∂p v(q, p) = ∂H . − ∂q
A flow which is governed by Hamilton’s equations is referred to as a Hamiltonian flow.
Example 8.2. A Non-Standard Hamiltonian System Prove that the system of one degree of freedom whose motion is governed by the second order differ- ential equation q¨ + G(q)q ˙2 − F (q) = 0 is Hamiltonian. We note that the given equation implies that
d 1 q˙2 + G(q)q ˙2 − F (q) = 0, dq 2 so that d q˙2 + 2G(q)q ˙2 − 2F (q) = 0. dq Using the integrating factor Z µ(q) = exp 2G(q) dq we obtain d µ(q)q ˙2 − 2µ(q)F (q) = 0. dq Writing dV µ(q)F (q) = − dq we see that d µq˙2 + 2V (q) = 0, dq and therefore 1 µq˙2 + V (q) 2 is constant throughout the motion. If we writeq ˙ = p/µ, so that
p2 + V (q) 2µ is constant, it is reasonable to guess that the given differential equation is equivalent to Hamilton’s equations with Hamiltonian H(q, p) given by
p2 H(q, p) = + V (q). 2µ
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 169
First note that p ∂ p2 ∂H q˙ = = = , µ(q) ∂p 2µ(q) ∂p so that the first of Hamilton’s equations is satisfied.
Moreover, since p = µq˙ dµ dµ p˙ = µq¨ + q˙2 = µ(−Gq˙2 + F (q)) + q˙2 dq dq dV dµ = − − µGq˙2 + q˙2. dq dq dµ Now dq = 2Gµ. It follows that
dV 1 dµ dV 1 dµ p2 p˙ = − + q˙2 = − + dq 2 dq dq 2 dq µ2
dV ∂ p2 ∂H = − − = − . dq ∂q 2µ ∂q Summarising, the given equation is equivalent to the Hamiltonian equations ∂H ∂H q˙ = , p˙ = − , ∂p ∂q where p2 H(q, p) = + V (q) 2µ and Z Z µ(q) = exp 2G(q) dq, V (q) = − µ(q)F (q) dq + C.
small
Example 8.3. Divergence-Free Velocity Fields and Hamiltonian Systems Consider the equation d x v (x, y) = v(x, y) = 1 . (8.13) dt y v2(x, y)
In order for this system to be Hamiltonian with generalised coordinate x and conjugate momentum y there must exist H(x, y) such that
∂H ∂H x˙ = , y˙ = − . (8.14) ∂y ∂x This requires that ∂H ∂H v (x, y) = , v (x, y) = − . (8.15) 1 ∂y 2 ∂x
In order for these equations to be consistent v1, v2 must satisfy
∂v ∂2H ∂v 1 = = − 2 , (8.16) ∂x ∂x∂y ∂y
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 170
∂v1 ∂v2 + = 0, i.e. ∇ · v = 0; (8.17) ∂x ∂y in other words, the vector v has to be solenoidal1. One can prove that the converse is also true i.e. if v is solenoidal, then the system is Hamiltonian and one can find the Hamiltonian H. If we know that a system is Hamiltonian then we can write
∂H ∂x −v2(x, y) ∇H = ∂H = = W , (8.18) ∂y v1(x, y) and we can use the fundamental theorem of calculus for line integrals to write
Z (x,y) H(x, y) = W · dx , (8.19) (x0,y0) that is we can find the Hamiltonian by integration of the vector field W along a path from some fixed point (x0, y0). This path can be chosen freely – the result of the integration (8.19) is independent of the path. The choice of the starting point (x0, y0) is also irrelevant as changing the starting point only adds a constant to H which does not affect Hamilton’s equations. As an example: Consider the system
q˙ p − q2 = v = . (8.20) p˙ 2qp
We have ∂ ∂ ∇ · v = p − q2 + (2qp) = −2q + 2q = 0 , (8.21) ∂q ∂p and hence the system is Hamiltonian. We can reconstruct H as the line integral
Z (Q,P ) H(Q, P ) = (p − q2)dp − 2qp dq . (8.22) (q0,p0)
One way is to choose (q0, p0) = (0, 0) and the path to consist of the straight line
Qλ r(t) = , q(λ) = Qλ , p(λ) = P λ , 0 ≤ λ ≤ 1 . (8.23) P λ
The integral then becomes
Z 1 dp dp H(Q, P ) = (p(λ) − q(λ)2) dλ − 2q(λ)p(λ) dλ λ=0 dλ dλ Z 1 = (P λ − Q2λ2)P − 2QP λ2 dλ 0 Z 1 = (P 2λ − 3Q2P λ2)dλ 0 1The term solenoidal comes from the magnetic field B of a solenoid. This field satisfies ∇ · B = 0, and hence any field satisfying the same equation is called solenoidal
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 171
1 = P 2 − Q2P. (8.24) 2 We can check this does indeed give back the correct equations of motion: 1 ∂H ∂H H(q, p) = p2 − q2p ⇒ q˙ = = p − q2 , p˙ = − = 2qp . (8.25) 2 ∂p ∂q Example 8.4. Pendulum A simple pendulum consists of an inelastic string of length a with a particle P of mass m attached to one end; the other end is attached to a fixed point O and the system is free to move in a vertical plane under gravity. Discuss the motion and show in particular that the system is Hamiltonian.
x
ψ
T
P (x,y) y
mg
Figure 8.1: Coordinate system of the pendulum discussed in the present example.
Referring to the diagram suppose that the y-axis points in the direction of the downward vertical through O. Let T denote the tension in the string, and suppose that the string makes an angle ψ(t) with the downward vertical.
The particle P has coordinates (x, y) given by
x = a sin ψ, y = a cos ψ.
The equation of motion of the particle is given by N2:
m(¨xe1 +y ¨e2) = (−T sin ψe1 + (mg − T cos ψ)e2),
mx¨ = −T sin ψ, my¨ = mg − T cos ψ. It follows that mx¨ cos ψ − my¨sin ψ = −mg sin ψ, (8.26) T = mg cos ψ − mx¨ sin ψ − my¨cos ψ. (8.27) Now x˙ = aψ˙ cos ψ, x¨ = aψ¨ cos ψ − aψ˙ 2 sin ψ, y˙ = −aψ˙ sin ψ, y¨ = −aψ¨ sin ψ − aψ˙ 2 cos ψ. We deduce from Eqs (8.26) and (8.27) that g ψ¨ + sin ψ = 0,T = mg cos ψ + maψ˙ 2. a
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 172
The first of these equations is a differential equation for ψ and the second gives the tension T in terms of ψ. We note that when ψ is small (i.e. near 0), so that the string is always close to the downward vertical, the equation for ψ approximates to g ψ¨ + ψ = 0. a In this case the motion is approximately Simple Harmonic with period τ given by
ra τ = 2π . g
Returning to the exact equation for ψ we note that
ma2ψ¨ = −mga sin ψ.
We write p = ma2ψ˙ so that p ∂ p2 ψ˙ = = . ma2 ∂p 2ma2 Moreover, ∂ p˙ = ma2ψ¨ = −mga sin ψ = − −mga cos ψ . ∂ψ We conclude that ∂H ∂H ψ˙ = , p˙ = − , ∂p ∂ψ where p2 H(ψ, p) = − mga cos ψ. 2ma2 The system is therefore Hamiltonian in the generalised coordinate ψ, p being the momentum conjugate to ψ; H(ψ, p) is of course the Hamiltonian. It is interesting to note that the kinetic energy T of the particle is given by
1 1 1 p2 p2 T = m(x ˙ 2 +y ˙2) = m(cos2ψ + sin2ψ)ψ˙ 2 = m = 2 2 2 m2a4 2ma2 so that the first term in the expression for H is just the kinetic energy of the particle. The second term is in fact the potential energy V due to gravity since dV mge = − e , 2 dy 2 and therefore V = −mgy = −mga cos ψ. It follows that the Hamiltonian is the total energy of the system; since H doesn’t depend on t explicitly it is a constant of the motion and p2 H(ψ, p) = − mga cos ψ = C, 2ma2 where C is constant; the value of C depends on the initial conditions.
We note that ψ can range over the interval [0, 2π], the ends 0, 2π being identified, so that the domain of ψ is essentially the unit circle S1. On the other hand, p can range over the whole of R. The phase space of this system is the Cartesian product R × S1 which we can picture as a circular cylinder.
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 173
8.2 Stability problems
Suppose that a particle of mass m moves along the q-axis under a potential V (q). As noted above, the Hamiltonian for the motion is p2 H(q, p) = + V (q). 2m The Hamilton equations are q˙ p/m = . p˙ −V 0(q) We see that the fixed points are those points (q, p) for which p = 0 and V 0(q) = 0. This agrees with our earlier conclusion that the equilibrium points of such a system correspond to 0 the stationary points of the potential energy function V. Suppose that V (q0) = 0. Then near q0 we can approximate V by 1 V (q) = V (q ) + (q − q )V 0(q ) + (q − q )2V 00(q ) + ··· 0 0 0 2 0 0 so that the linearised Hamiltonian equations become q˙ 0 1/m q − q0 = 00 . p˙ −V (q0) 0 p The matrix 0 1/m 00 . −V (q0) 0 2 1 00 00 has eigenvalues λ given by λ = − m V (q0). If V (q0) > 0, so that q0 is a minimum of V , we have p 00 λ = ±iβ, β = V (q0)/m. We deduce that there is a matrix P such that X˙ 0 β X = , Y˙ −β 0 Y where X q − q = P −1 0 . Y p
The fixed point (q = q0, p = 0) is the origin of the new (X,Y ) coordinates. Following standard procedures we write X = r cos θ, Y = r sin θ and obtain r˙ = 0, θ˙ = −β, r(t) = r0, θ(t) = θ0 − βt. These equations describe a family of circles and the fixed point is therefore stable; the term ‘elliptic’ is sometimes used to describe such fixed points. The time τ taken by the system to move round one of these circles is precisely the time taken by the system to execute one complete Simple Harmonic oscillation about q = q0 and is given by 2π r m τ = = 2π 00 , β V (q0)
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 174 as found previously by a different method.
00 In a similar manner we find that if V (q0) < 0, so that q0 is a maximum of the potential function V , that the linearised Hamiltonian equations may be written as r V 00(0) X˙ = λ X, Y˙ = −λ Y, λ = − > 0, 1 1 1 m with integrals λ1t −λ1t X(t) = X0e ,Y (t) = Y0e . The fixed point is clearly unstable, the phase curves being a family of rectangular hyperbolas given by XY = X0Y0; such unstable fixed points are sometimes said to be ‘hyperbolic’ fixed points.
Example 8.5. Hamiltonian with Quartic Potential The Hamiltonian H of a particle of unit mass moving in a straight line along the q-axis is given by 1 1 1 H(q, p) = p2 + q4 − q2. 2 4 2 Discuss the motion.
q
Separatrix
q
Figure 8.2: Quartic potential described in the text and construction of the phase curves at different energies.
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 175
This is the Hamiltonian for a particle moving under the potential V (q) given by 1 1 V (q) = q4 − q2. 4 2 We note that for small q2 the force acting on the particle is approximately equal to d 1 − (− q2)e = qe ; dq 2 1 1 it follows that the potential generates a short range repulsion. On the other hand, for large values of 2 1 4 q the term 4 q is dominant and this generates a force equal to d 1 − ( q4)e = −q3e . dq 4 1 1 This means that the potential V produces a long range attraction which tends to pull the particle back towards the origin. The over-all effect is seen from the graph of V. Hamilton’s equations give q˙ p = . p˙ q − q3 The fixed points are given by p = 0, q − q3 = 0 i.e. by p = 0, q = 0, q = ±1. Linearising the equations in the usual way we have to consider the Jacobian J(q, p) at the fixed points. We have
0 1 J(q, p) = . 1 − 3q2 0
We see that 0 1 J| = . 0,0 1 0 This matrix has eigenvalues λ = ±1 and the now familiar arguments lead us to conclude that (0, 0) is an unstable fixed point. Similarly we find that
0 1 J| = . (±1,0) −2 0 √ This matrix has eigenvalues λ = ±i 2, so that we are dealing with a stable (elliptic type) fixed point. The Jordan form is given by the matrix √ 0 2 √ . − 2 0
A standard calculation leads to the conclusion that the system makes√ a complete circuit (in phase space) of the fixed point point (q = ±1, 0) in time τ given by τ = 2π/ 2, precisely the time taken by the particle to execute one complete Simple Harmonic oscillation about q = ±1.
The phase curves are given by 1 1 1 H(q, p) = p2 + q4 − q2 = E, 2 4 2 where E is a parameter.
A phase curve which passes through a hyperbolic (unstable) fixed point is known as a separatrix; its importance lies in the fact that it marks the boundary between motions having different properties.
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 176
In this example the phase curve which passes through the hyperbolic point (q = 0, p = 0) is given by E = 0 so that the separatrix has equation given by q p = ±√ (2 − q2)1/2. 2 The accompanying diagram showing the phase curves and the potential function V exhibits the fol- lowing symmetries:
V (−q) = V (q),H(q, p) = H(q, −p),H(−q, p) = H(q, p).
As an example, suppose that the particle starts from P1 with coordinates (q = q1, p = 0) (so that it starts from rest at q = q1). It moves round the phase curve in the sense indicated, passing through P2 with coordinates (q = −q1, p = 0) before returning to P1. The time τ taken for one complete oscillation is just four times the time taken to go from q = 0 to q = q1. The relevant phase curve has equation given by 1 1 1 1 1 p2 + q4 − q2 = q 4 − q 2 2 4 2 4 1 2 1 and Hamilton’s equations give ∂H q˙ = = p ∂p so that Z q1 dq √ Z q1 dq τ = 4 = 2 2 1 4 4 1 2 2 1/2 0 p 0 ( 4 (q1 − q ) − 2 (q1 − q )) √ Z q1 dq = 4 2 . 2 2 1/2 2 2 1/2 0 (q1 − q ) (q1 + q − 2)
8.3 Summary: how to analyse motion in a potential
We can now summarise the analysis of motion in a potential using all the tools we have developed in the course so far. A particle of mass m = moving in a potential V (x) corresponds to the Hamiltonian
p2 H = + V (x) (8.28) 2m with Hamilton’s equations being ∂H p ∂H x˙ = = , p˙ = − = −V 0(x) , (8.29) ∂p m ∂q
⇒ mx¨ = −V 0(x) . (8.30)
1. As a Hamiltonian system with H independent of t, the phase curves are H(x, p) =constant.
2. Treating (8.29) as a second order autonomous dynamical system, we can sketch the phase portrait using the standard methods: the null clines are p = 0 on whichx ˙ = 0 and the flow is vertical; x such that V 0(x) = 0 on whichp ˙ = 0 and the flow is horizontal. These lines divide the plane into regions on which we can deduce the general direction of the flow.
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 177
3. We can find the fixed points and classify their nature. A fixed point of the system (8.29) is (x = x∗, p = 0) where V 0(x∗) = 0. This will be either elliptic or hyperbolic, depending on the sign of V 00(x∗).
If V 00(x∗) > 0 then this is a minimum of V (x), a point of stable equilibrium of (8.30) and an elliptic (marginally stable) fixed point of (8.29); If V 00(x∗) < 0 then this is a maximum of V (x), a point of unstable equilibrium of (8.30) and a hyperbolic (unstable) fixed point of (8.29);
4. If there is a hyperbolic fixed point (x∗, 0) then the separatrix will go through this fixed point and its equation with be H(x, p) = H(x∗, 0).
5. Knowing H(x, p) = p2/(2m)+V (x), we can sketch V (x) and then choosing suitable values of the total energy E which satisfies H(x, p) = E, we can deduce the ranges of x to which the motion is satisfied
6. We can find the exact period of periodic orbits using the reduction of the second order system (8.29) to a first order system for x by use of the equation of conservation of energy: Z dx Z x˙ = ±p2/m(E − V (x)) , = dt . (8.31) p2/m(E − V (x)) We can find the approximate period of small periodic orbits around a fixed point by linear stability analysis resulting in 2π T ∼ . (8.32) pV 00(x∗)/m
7. We can sketch typical trajectories by following the orbits on the phase portrait.
We will apply this now to two examples. Example 8.6. Analyse the motion in the potential of a particle of mass m = 1 1 1 V (x) = x4 − x3 − x2 . (8.33) 4 3 and sketch the graph of x vs t for a particle starting at x = 0 with initial momentum p = −1.
First we plot V (x). For large x, V (x) ∼ x4/4. The equilibrium points are solutions of V 0(x) = 0. We have V 0(x) = x3 − x2 − 2x = x(x + 1)(x − 2) , (8.34) so the points of equilibrium are x = 0, x = −1 and x = 2. To find their nature we calculate V 00(x∗). We have V 00(x) = 3x2 − 2x − 2 , (8.35) so V 00(0) = −2, an unstable point of equilibrium/a hyperbolic fixed point of the Hamiltonian system; V 00(−1) = 3 so this is a stable point of equilibrium/an elliptic fixed point of the Hamiltonian system; V 00(2) = 6 so this is also a stable point of equilibrium/an elliptic fixed point of the Hamiltonian system. At the fixed points, V (0) = 0, V (−1) = −5/12 and V (2) = −8/3. Sincep ˙ = −V 0(x), the flow is horizontal for these values of x.
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 178
We can now plot V (x) and choose four representative values for the total energy, which are labelled A, B, C, and D. The phase curves are H(x, p) = E; If the total energy is E then since E = p2/2 + V (x), the motion of the particle is confined to regions in which V (x) ≤ E. Given our choices, a particle with energy A will move in an orbit around the fixed point at x = 2; a particle with energy B will orbit around either the fixed point at x = −1 or the fixed point at x = 2 but since V (0) > B it cannot pass through the point x = 0; a particle with energy C=0 moves along the separatrix which is given by p2/2 + V (x) = 0. It will tend asymptotically to the fixed point at x = 0 but will not reach it in finite time. A particle with energy D will move in a periodic orbit enclosing both fixed points. This is summarised in figure 8.3. Note that the flow is vertical on the null-cline p = 0 and horizontal on the null-clines x = −1, x = 2 and x = 2.
VHxL D C B x
A
p
x
X
Figure 8.3: Analysis of the potential (8.33).
A particle starting at x = 0 with negative momentum (such as the starting point X) will initially move off to the left then stop, return to some (larger) positive value of x and carry on in a periodic motion, something like the plot in figure 8.4. Note that the speed slows
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 179
Figure 8.4: Sketch of the motion of a particle starting at point X. slightly as the particle passes through x = 0 as this is a local maximum of the potential energy and hence a local minimum of the kinetic energy.
Example 8.7. Analyse the motion of a particle of mass m = 1 moving in the potential
x3 − 5x V (x) = . (8.36) x2 + 3 and describe the shape of the phase curves for large negative x. Can a particle escape to x = −∞ in finite time?
First we plot V (x). For large x, V (x) ∼ x, for small x, V (x) ∼ −5x/3. The equilibrium points are solutions of V 0(x) = 0. We have
(x2 − 1)(x2 + 15) V 0(x) = , (8.37) (x2 + 3)2
so the points of equilibrium are x = −1 and x = 1. To find their nature we calculate V 00(x∗). We have V 00(−1) = −2 ,V 00(1) = 2 , (8.38) so x = −1 is a local maximum of the potential/an unstable point of equilibrium/a hyperbolic fixed point of the Hamiltonian system and x = 1 is a local minimum of the potential/a stable point of equilibrium/an elliptic fixed point of the Hamiltonian system. At the fixed points, V (−1) = 1 and V (1) = −1. Sincep ˙ = −V 0(x), the flow is horizontal for these values of x. We can now plot V (x) and choose four representative values for the total energy, which are labelled A, B, C, and D. Given our choices, a particle with energy A will move in from x = −∞ but will not have enough energy to pass over the local maximum at x = −1 and will return to x = −∞;A particle with energy B will either perform a similar motion or will move in a periodic orbit around the fixed point at x = 1, depending on its initial position and momentum; a particle with energy C=1 moves along the separatrix. It will either tend asymptotically to the fixed point at x = −1 but not reach it in finite time or it will escape to x = −∞. A particle with energy D will move in from x = −∞, pass through x = −1 and x = 0, reach a maximum value of x and then change direction and escape to x = −∞. Its speed will be a local minimum at x = −1 and a local maximum at x = 0 before reaching zero at which point it will change direction.
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 180
This is summarised in figure 8.5. Note that the flow is vertical on the null-cline p = 0 and horizontal on the null-clines x = −1 and x = 1. For large negative x, V (x) ∼ x and so H(x, p) = E = p2/2 + V (x) ∼ p2/2 + x. The curves x = E − p2/2 are parabolae. √ √ √ We find p =x ˙ = − 2 E − x ∼ − 2p|x| for large negative x and so the time to reach a point x1 is asymptotically
Z x1 dx Z x1 dx p τ = ∼ = 2|x1| x˙ −p2|x|
This diverges as x1 → −∞ and so the particle cannot escape to infinity in finite time.
VHxL
D C x B A
p
x
Figure 8.5: Analysis of the potential (8.36).
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 181
8.4 Exercises
Shorter exercises
Exercise 8.1 For each of these Hamiltonian systems: Write down and solve Hamilton’s equations, write down the equations of the phase curves and sketch the phase curves (i.e. orbits in the phase space), find the equations of any separatrices, find all fixed points and describe their nature.
p2 q2 p2 q2 (i) H = + (ii) H = − (iii) H = pq . 2 2 2 2 √ √ Show that the last two systems can be related byq ˜ = (p + q)/ 2 andp ˜ = (p − q)/ 2.
Exercise 8.2 Consider the following equations of motion. In each case decide whether they can be seen as Hamilton’s equations for some Hamiltonian H(q, p). If they can, find the Hamiltonian and check that this leads to the correct equations of motion; if they cannot, explain why
(a)q ˙ = 1 , p˙ = 1 (b)q ˙ = 1 , p˙ = −1 (c)q ˙ = q , p˙ = −p (d)q ˙ = q , p˙ = p (e)q ˙ = 2p + 2q , p˙ = −2p − 1 (f)q ˙ = ep , p˙ = −peq (g)q ˙ = 1 + qep , p˙ = −ep
Recall that a necessary condition for the equations of motionq ˙ = v(q, p) , p˙ = w(q, p) be derived from a Hamiltonian is that ∂v/∂q + ∂w/∂p = 0, in which case H(q, p) = H(q0, p0) + R (q,p) wdp − vdq . (q0,pq)
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 182
Longer exercises
Exercise 8.3 Consider the system described by the Hamiltonian 1 1 H = p2 − q4 2 4
(a) Write down Hamilton’s equations and find the fixed points of the dynamics. Can you classify them as either elliptic or hyperbolic?
(b) Derive the equations for the the phase curves, in particular that of the separatrix. Solve the equation of motion along the separatrix.
(c) Sketch the phase portrait.
Exercise 8.4 The motion of a particle of unit mass moving along the x-axis is described by the Hamiltonian 1 x H(x, p) = p2 + , 2 x2 + 1 where p is the momentum conjugate to x.
Write down Hamilton’s equations, find the fixed points of the system, and classify them as either hyperbolic or elliptic fixed points.
Find the equation of the separatrix and sketch the phase diagram. The particle is released from rest at x = −3. Explain briefly, in the context of your diagram, why the subsequent motion is oscillatory. Find the maximum speed of the particle and obtain an expression for the period τ of one complete oscillation.
Exercise 8.5 Consider the system described by the Hamiltonian
1 q2 H = p2 − 2 (1 + q2)2 .
(a) Write down Hamilton’s equations, find the fixed points of the dynamics and classify them as either elliptic or hyperbolic.
(b) Derive the equations for the the phase curves, in particular that of the separatrix. Solve the equation of motion along the separatrix.
(c) Sketch the phase portrait.
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 183
Exercise 8.6 The motion of a particle of unit mass along the x axis is governed by the Hamiltonian 1 H(x, p) = p2 + V (x) 2 with ( 3π sin x , |x| ≤ 2 V (x) = 3π −sign(x) , |x| > 2
(a) Write down Hamilton’s equations of motion for this system. Find the fixed points for 3π |x| < 2 and classify them as either elliptic or hyperbolic. (b) Derive the equation for the phase curves of the system, in particular that of the separatrix.
(c) Sketch the phase portrait.
(d) The particle is released from rest at x = −1. Describe the subsequent motion for this case. Give the coordinate(s) at which the maximum speed is attained and find the value of the maximum speed.
(e) Suppose now that the particle is released from x = 0 with initial velocity u > 0 in the positive x-direction. Give the minimum value of u for which the motion will be unbounded. Give the limiting velocity that the particle will have as x → ∞, if the motion is unbounded.
(f) Find the fixed points for |x| ≥ 3π/2. Can you classify them into elliptic or hyperbolic by considering the linearisation of Hamilton’s equations? Are the fixed points stable or unstable - recall our definition of stable was that changing their initial position and velocity by a small amount would still leave their future motion close to the fixed point.
[Note sign(x) is the sign function, which evaluates to 1 when the argument of the function is positive, i.e. for x > 0 and evaluates to −1 when x < 0. It is discontinuous at x = 0.]
Exercise 8.7 Throughout the following question you may assume that two positive electric charges, of magnitude c1, c2 respectively, distance z apart, repel each other with a force 2 c1c2/z , the direction of the force being along the straight line join of the charges.
Fixed positive electric charges, each of magnitude e1 are placed on the x-axis at x = 0, x = 2a (a > 0). A third positive electric charge of magnitude e2 and mass m is constrained to move on the x-axis between the first two charges so that its coordinate is x(t) at time t. Write down N2 for the moving charge e2, and show that it can be cast in Hamiltonian form with p2 µ µ H(x, p) = + + , 2m x 2a − x where µ = e1e2.
Find the fixed points of the system and sketch the phase diagram for 0 < x < 2a. The charge e2 starts from rest at x = a(1 + k), 0 < k < 1. Find the equation of the phase curve on which
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 184 it moves in the form f(x, p) = constant (*) and show by consideration of the phase diagram that it will subsequently oscillate with period τ given by
Z a(1+k) m dx τ = 4 , a p(x) where p(x) is given by (*). Verify that the substitution x = a+ak sin φ leads to the expression
ma3(1 − k2)1/2 Z π/2 τ = 2 (1 − k2sin2φ)1/2 dφ. µ 0
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 185
Supplementary exercises
Supplementary Exercise 8.1 This problem is about a particle (of mass m) moving smoothly (with no friction) on a wire in the x–z plane described by a function z = f(x) in a constant vertical gravitational field of strength g. There are several ways to tackle the problem. At the most basic level, we can work out the forces on the particle and set up Newton’s equations that way. We could also work out the kinetic and potential energy and set up a Lagrangian for the system, which is well outside this course, and then from that obtain a Hamiltonian and describe the system in terms of the Hamiltonian and Hamilton’s equations. The test will be whether we get the same equations from the two methods. Having set the system up, we can then solve the equations - which from the Hamiltonian point of view is a simple matter of saying H = constant. (A) To begin with: obtaining the equations of motion from Newton’s equations. If the particle has speed v, then its acceleration isv ˙ and Newton’s equations say that mv˙ is given by the component of the gravitational force parallel to the motion of the particle, denoted Fk. The component perpendicular to the motion, F⊥ keeps the particle on the wire. If the inclination of the wire to the horizontal is an angle θ as in the figure below, show that
(a) |Fk| = mg sin θ , x˙ (b) v = , cos θ (c)v ˙ = −g sin θ ,
(d) tan θ = f 0(x) , f 0f 00 f 0 (e)x ¨ = −x˙ 2 − g . 1 + (f 0)2 1 + (f 0)2
z
v z = f(x) θ Fk
mg
F⊥
x
Version of Mar 14, 2019 Chapter 8: Hamiltonian Systems 186
(B) We can then consider the construction of the Hamiltonian. Just to show how it goes, the kinetic energy T , the gravitational potential energy V and Lagrangian of the particle are 1 1 1 T = mv2 = mx˙ 2(1 + (f 0)2) ,V = mgf(x) ,L = T − V = mx˙ 2(1 + (f 0)2) − mgf . 2 2 2 This means the generalised momentum p conjugate to x and the Hamiltonian H(p, x) are
∂L 1 p2 p = = mx˙(1 + (f 0)2) ,H(p, x) = px˙ − L = + mgf . ∂x˙ 2m 1 + (f 0)2
(a) Using this expression for the Hamiltonian, show that Hamilton’s equations are
p p2 f 0f 00 x˙ = (1 + (f 0)2) , p˙ = − mgf 0 . m m (1 + (f 0)2)2
(b) Hence show that f 0f 00 f 0 x¨ +x ˙ 2 = −g . 1 + (f 0)2 1 + (f 0)2 This is exactly the sort of system considered in lectures and in notes and this is the same equation as found in part (A) from Newton’s laws. (C) (a) Show that if f(x) = h is constant, this reduces to motion in the absence of a gravitational field. (b) Show that if f(x) = αx corresponding to a slope of constant gradient then the particle’s acceleration is constant. √ (c) Consider the motion on a circular wire of radius 1, ie f(x) = 1 − 1 − x2. Let the angle φ be defined by x = sin(φ) as in the figure below.
φ
Show that this reduces to the standard equation for a pendulum,
φ¨ = −g sin φ .
[Hint: substitute x = sin(φ) into the equation forx ¨. You will need to work out thatx ˙ = cos(φ)φ˙, etc. ]
Version of Mar 14, 2019 Appendix A: Functions of two variables 187
Appendix A
Functions of two variables
A.1 The partial derivative
The partial derivatives of a function of two variables f(t, x) are defined as
∂f f(t + h, x) − f(t, x) f(t, x) = lim , (A.1) ∂t h→0 h ∂f f(t, x + h) − f(t, x) f(t, x) = lim . (A.2) ∂x h→0 h They are also often written using subscripts, so that ∂f ∂f f = , f = . (A.3) t ∂t x ∂x This definition means that you can calculate a partial derivative in exactly the same way as a normal derivative just by assuming that any other variables are just constants, with all the usual rules such as the product and chain rules. For example, ∂ ∂ (exp(xt2)) = 2tx exp(xt2) , (exp(xt2)) = t2 exp(xt2) . (A.4) ∂t ∂x
One can also take the derivative of a function more than once, and so there are in principle four second partial derivatives:
∂2f ∂ ∂f ∂2f ∂ ∂f ∂2f ∂ ∂f ∂2f ∂ ∂f = , = , = , = , (A.5) ∂x2 ∂x ∂x ∂x∂t ∂x ∂t ∂t∂x ∂t ∂x ∂t2 ∂t ∂t
For well-behaved functions, it actually does not matter in which order you take the two mixed derivatives: you get the same result,
∂ ∂f ∂ ∂f = . (A.6) ∂x ∂t ∂t ∂x
It is possible to come up with functions for which this is not true, but they have to be carefully contrived and will not make an appearance in this course.
Version of Mar 14, 2019 Appendix A: Functions of two variables 188
The reason for a different notation is that sometimes one wants to think of some variables depending on others. So, for example, if x is a function of t, then f(t, x(t)) depends on t both through the explicit t dependence in the first argument but also on the t dependence of x in the second argument. This means one can defined the “total derivative” of f(t, x(t)) with respect to t and gets the result
d f(t + h, x(t + h)) − f(t, x(t)) f(t, x(t) = lim dt h→0 h ∂f dx ∂f = (t, x(t)) + (t) (t, x(t)) . (A.7) ∂t dt ∂x This will not play any role in this course, but it goes some way to explaining why we need two different notations d/dt and ∂/∂t for the two different concepts.
Version of Mar 14, 2019 Appendix A: Functions of two variables 189
A.2 Continuity of a function of two variables
There are many equivalent definitions of the continuity of a function of one variable f(x) at x = a, for example: f(x) is continuous at x = a if the two directional limits exist and are equal to f(a):
lim f(x) = lim f(x) = f(a) . (A.8) x→a− x→a+ f(x) is continuous at x = a if, for any > 0, one can find a δ > 0 such that
|x − a| < δ ⇒ |f(x) − f(a)| < . (A.9)
These definitions are necessary and useful but it is also important to understand that the fundamental idea is that one the function f is continuous if one can draw the graph of f(x) as a line. If there is a jump at x = a or if the function is not defined there then it is not continuous at x = a. For a function of two variables f(x, y), the line is replaced by a surface. The graph of a function of two variables is the surface with height z = f(x, y). The function f is continuous at (a, b) if one can approach (a, b) from any direction, in any manner, and the limit along that path is the same as the value of the function. This can again be expressed in many ways, of which just two are f(x, y) is continuous at (a, b) if the limit exists and is equal to f(a):
lim f(x, y) = f(a, b) . (A.10) (x,y)→(a,b) f(x, y) is continuous at (a, b) if, for any > 0, one can find a δ > 0 such that
|(x − a)2 + (y − b)2| < δ ⇒ |f(x) − f(a)| < . (A.11)
A simple heuristic test is that any function for which the graph is a smooth surface will be continuous. This will apply to almost all functions of two variables we consider. The simplest way for a function of two variables to not be continuous at a point is for it not to be defined there (i.e. it is “infinite there”). There are other ways which will be highlighted if they occur. For example: — The function f(x, y) = 1/(x2 + y2) is continuous everywhere except at the point (0, 0) where it is not defined. — The function arctan(y/x) is continuous everywhere except along the line x = 0.
Version of Mar 14, 2019 Appendix B: Taylor’s Theorem 190
Appendix B
Taylor’s Theorem
B.1 Taylor Expansion for Functions of One Variable
We recall Taylor’s Theorem1 for functions of one real variable: Suppose that φ is a suitably differentiable real valued function on some interval [x − δ, x + δ]. Then
h h2 φ(x + h) = φ(x) + φ(1)(x) + φ(2)(x) + ··· (1!) (2!)
hn−1 + φ(n−1)(x) + R , h ∈ [−δ, δ] (n − 1)! n
where Rn, the remainder after n terms, has the form hn R = φ(n)(ξ), n (n!) and ξ is some point between x and x + h. In applications we are often interested in situations where h is close to 0 and it may be justifiable to neglect the term Rn, for some suitable value of n. This amounts to saying that we can approximate φ(x + h) by a polynomial in h of order (n − 1). Notice that in the computation of φ(x + h) we require to know the value of φ and its derivatives at the point x. The particular case where x = 0 goes under the name of MacLaurin’s2 Theorem; for reasons which will shortly be apparent, we use t rather than h to denote the variable: t t2 tn−1 φ(t) = φ(0) + φ(1)(0) + φ(2)(0) + ··· + φ(n−1)(0) + R , t ∈ [−δ, δ] (1!) (2!) (n − 1)! n where Rn, the remainder after n terms, has the form tn R = φ(n)(ξ), n (n!) and ξ is some point between 0 and t.
1Brook Taylor (1685-1731). In 1715 he published his most important work which contained a proof of the theorem which now bears his name. 2Colin MacLaurin (1698-1746) entered Glasgow University at the age of 11; professor of mathematics at Aberdeen at the age of 19!
Version of Mar 14, 2019 Appendix B: Taylor’s Theorem 191
B.2 Taylor Expansion for Functions of two variables
Suppose now that f is a real valued function defined in some neighbourhood of the point (x, y) (for example in a square of side 2δ, centre (x, y), whose sides are parallel to the axes) and that f has partial derivatives to any required order. Motivated by our knowledge of Taylor’s Theorem for functions of one variable we aim to obtain an expression for f(x + h, y + k) in terms of powers of h, k and the partial derivatives of f evaluated at the point (x, y). Let A denote the point (x, y) and B the point (x + h, y + k). Any point P on the straight line AB has coordinates (x + ht, y + kt), 0 ≤ t ≤ 1; A corresponds to t = 0 and B corresponds to t = 1. Let φ(t) = f(x + ht, y + kt), where x, y, h, k are regarded as fixed and t is variable, with 0 ≤ t ≤ 1. We note that φ(0) = f(x, y) and that φ(1) = f(x + h, y + k). An application of MacLaurin’s formula gives
t t2 tn−1 φ(t) = φ(0) + φ(1)(0) + φ(2)(0) + ··· + φ(n−1)(0) + R . (1!) (2!) (n − 1)! n
Let’s write φ(t) = f(u, v), u = x + ht, v = y + kt. By the Chain Rule of partial differentiation
∂f ∂u ∂f ∂v ∂f ∂f φ(1)(t) = + = h + k . ∂u ∂t ∂v ∂t ∂u ∂v Thus ∂f ∂f φ(1)(0) = h + k , ∂u ∂v where the partial derivatives are evaluated at t = 0. A little thought shows that
∂f(x, y) ∂f(x, y) φ(1)(0) = h + k , ∂x ∂y where the derivatives of f are to be evaluated at (x, y). We see that symbolically
d ∂ ∂ = h + k , dt ∂x ∂y where the t derivative is to be computed at t = 0 and the x, y partial derivatives at (x, y). It is clear that ∂ ∂ n φ(n)(0) = h + k f(x, y), ∂x ∂y where ∂ ∂ n h + k f(x, y) ∂x ∂y is to be interpreted as
∂ ∂ ∂ ∂ ∂ ∂ h + k h + k ··· h + k f(x, y), ∂x ∂y ∂x ∂y ∂x ∂y
there being n factors in all. Setting t = 1 in the MacLaurin expansion of φ we obtain
2 1 ∂ ∂ 1 ∂ ∂ f(x + h, y + k) = f(x, y) + h + k f(x, y) + h + k f(x, y) + ··· 1! ∂x ∂y 2! ∂x ∂y
Version of Mar 14, 2019 Appendix B: Taylor’s Theorem 192
1 ∂ ∂ n−1 ··· + h + k f(x, y) + R . (n − 1)! ∂x ∂y n It frequently happens in applications that (x + h, y + k) is near to (x, y) so that (h, k) is near to (0, 0) and that it is justifiable to neglect R3. In this approximation we are retaining only the linear and quadratic terms in h, k in our expansion of f(x + h, y + k) about the point (x, y).
Note: The calculation of ∂ ∂ n h + k f(x, y) ∂x ∂y is straightforward. For example, with n = 2 we have
∂ ∂ 2 ∂ ∂ ∂ ∂ h + k f(x, y) = h + k h + k f(x, y) ∂x ∂y ∂x ∂y ∂x ∂y
∂2f ∂2f ∂2f ∂2f = h2 + hk + kh + k2 ∂x2 ∂x∂y ∂y∂x ∂y2 ∂2f ∂2f ∂2f = h2 + 2hk + k2 , ∂x2 ∂x∂y ∂y2 using the commutative property of partial differentiation. It’s therefore clear that we can formally square out ∂ ∂ 2 h + k ∂x ∂y provided we make the following identifications:
∂ ∂ ∂2 ∂ ∂ ∂2 ∂ ∂ ∂2 = , = , = , ∂x ∂x ∂x2 ∂y ∂y ∂y2 ∂x ∂y ∂x∂y and so on. Using this principle we see, for example, that
∂ ∂ 3 ∂3f ∂3f ∂3f ∂3f h + k f = h3 + 3h2k + 3hk2 + k3 . ∂x ∂y ∂x3 ∂x2∂y ∂x∂y2 ∂y3
Nevertheless, calculations of this sort require care; the formal expansion procedure works here because h, k are constants as far as the x, y differentiations are concerned.
B.3 Vector functions
In applications to dynamical systems we are often concerned with vector valued functions of two real variables (x, y). With each point (x, y) we associate a 2–vector
f (x, y) f(x, y) = 1 , f2(x, y)
Version of Mar 14, 2019 Appendix B: Taylor’s Theorem 193
where f1, f2 are scalar (i.e. real valued functions) of the real variables (x, y). In applications we frequently require to expand f(x+h, y +k) about the point (x, y) retaining only the linear terms in h, k. We have ∂f ∂f f (x + h, y + k) = f (x, y) + h j (x, y) + k j (x, y) + ··· (j = 1, 2). j j ∂x ∂y The dots indicate that we have neglected higher order terms in h, k. We conclude that ! h ∂f1 + k ∂f1 h f(x + h, y + k) = f(x, y) + ∂x ∂y + ··· = f(x, y) + J + ··· ∂f2 ∂f2 k h ∂x + k ∂y where J denotes the Jacobian matrix defined by f1, f2, evaluated at (x, y), i.e.
∂f1 ∂f1 ! ∂(f1, f2) ∂x ∂y J = = ∂f2 ∂f2 . ∂(x, y) ∂x ∂y Notation Given a function f of two real variables (x, y) we frequently use subscript notation to denote its partial derivatives: ∂f ∂f f ≡ , f ≡ , x ∂x y ∂y ∂2f ∂2f ∂2f f ≡ , f ≡ , f ≡ xx ∂x2 xy ∂x∂y yy ∂y2 Example B.1. Expand f(x, y) = sin(πxy) about the point (1, 1) neglecting third and higher order terms in (x − 1), (y − 1). In this case h = x − 1, k = y − 1 and Taylor’s formula gives
sin(πxy) = f(1, 1) + (x − 1)fx(1, 1) + (y − 1)fy(1, 1) 1 + (x − 1)2f (1, 1) 2! xx 2 +2(x − 1)(y − 1)fxy(1, 1) + (y − 1) fyy(1, 1) + ··· Now, fx = πy cos(πxy) , fy = πx cos(πxy), 2 2 fxx = −(πy) sin(πxy) , fyy = −(πx) sin(πxy), ∂ ∂ f = f = −(πy)(πx) sin(πxy) + π cos(πxy) = f = f xy ∂y x ∂x y yx Since f(1, 1) = 0, fx(1, 1) = −π, fy(1, 1) = −π,
fxx(1, 1) = 0, fyy(1, 1) = 0, fxy(1, 1) = −π it follows that sin(πxy) ' −π(x − 1) − π(y − 1) − π(x − 1)(y − 1) + ··· Exercise B.1 Find the expansion of arctan(xy) f(x, y) = x + sin(xy) about the point (0, 0), as far as the linear terms in x, y.
Version of Mar 14, 2019 Appendix B: Taylor’s Theorem 194
Exercise B.2 Find the (two variable) Taylor expansion of ln(1 + x2 + y2) about the point (1, 1), up to and including the quadratic terms.
Let x/(x2 + y2) f(x, y) = ln(1 + x2 + y2) Expand f about the point (1, 1), neglecting quadratic and higher order terms.
Exercise B.3 Suppose that f is a real valued function of the real variables (x, y) with con- tinuous partial derivatives to all orders. Suppose that fx(a, b) = 0, fy(a, b) = 0, so that (a, b) is a stationary point of f. Use Taylor’s Theorem to show that near (a, b) f is given by
1 x − a f(x, y) − f(a, b) = XT HX + ··· ,X = , 2 y − b
T denotes transpose, and H is the Hessian matrix given by
f f xx xy , fxy fyy
evaluated at (a, b).
We may argue that near (a, b) the sign of f(x, y) − f(a, b) is determined by the sign of the leading term in the Taylor expansion i.e. by XT HX, if this term is not identically zero. A 2 × 2 real symmetric matrix H is said to be positive definite, negative definite depending on whether XT HX > 0 (X 6= 0),XT HX < 0 (X 6= 0), respectively. We conclude that if H is positive definite then f(x, y) − f(a, b) > 0 for all (x, y) inside some circle centre (a, b), (x, y) 6= (a, b), and therefore f has a (local) minimum at (a, b). On the other hand, if H is negative definite then f(x, y) − f(a, b) < 0 for all (x, y) inside some circle centre (a, b), (x, y) 6= (a, b), and in this case f has a (local) maximum at (a, b). We learn from books on algebra that H is positive definite provided its determinant is positive and the two entries on the leading diagonal are positive; likewise H is negative definite provided its determinant is positive and the two entries on the leading diagonal are negative. We therefore conclude that
Case 1 f has a local minimum at the stationary point (a, b), if
2 fxx(a, b)fyy(a, b) − (fxy(a, b)) > 0 , fxx(a, b) > 0 , and
fyy(a, b) > 0 .
Case 2 f has a local maximum at the stationary point (a, b), if
2 fxx(a, b)fyy(a, b) − (fxy(a, b)) > 0 , fxx(a, b) < 0 , and
fyy(a, b) < 0 .
Version of Mar 14, 2019 Appendix C: Basic Linear Algebra 195
Appendix C
Basic Linear Algebra
C.1 Fundamental ideas
We consider the space IR2 of column vectors x1 X = , x1, x2 ∈ IR. x2
With the usual rule of addition of vectors and multiplication of vectors by a real scalar IR2 becomes a linear vector space. It is assumed that the geometrical interpretation of such vectors is well known. Any two linearly independent vectors constitute a basis and any vector can be written as a linear combination of basis vectors. Geometrically, any two (non–zero) vectors, which are neither parallel nor anti–parallel, are linearly independent, and therefore constitute a basis. We use e1, e2 to denote the standard basis so that
1 0 e = , e = , 1 0 2 1
A linear transformation a of the vector space IR2 is a map
a : IR2 → IR2 such that a(X + Y ) = aX + aY , a(λX) = λaX, for all vectors X, Y ∈ IR2 and for every scalar λ ∈ IR. The action of any linear transformation 2 a of IR is determined by its action on any set of basis vectors. We note that ae1, ae2 may be written as a linear combination of e1, e2 (because e1, e2 constitute a basis). We may write
2 X aei = akiek, i = 1, 2 (C.1) k=1 for some unique coefficients aik. These coefficients determine the matrix A = (aik) of the transformation a with respect to the standard basis given by e1, e2. We note that a11 a12 1 a11 Ae1 = = = a11e1 + a21e2, a21 a22 0 a21
Version of Mar 14, 2019 Appendix C: Basic Linear Algebra 196
a11 a12 0 a12 Ae2 = = = a12e1 + a22e2 . a21 a22 1 a22
Reference to equation C.1 shows that Aei = aei, (i = 1, 2) so that we may calculate the action of the transformation a on any vector X simply by calculating AX, where A is the matrix of the transformation with respect to the standard basis as defined by equations C.1. We consider below the matrix of the linear transformation a with respect to some non-standard basis, and show how it relates to the matrix of a with respect to the standard basis. Suppose that f 1, f 2 is another basis. We can therefore express f 1, f 2 as linear combinations of e1, e2 :
2 X f i = pkiek, i = 1, 2 (C.2) k=1
The coefficients pki define a matrix P = (pki).
Because f 1, f 2 is a basis we can express e1, e2 as linear combinations of f 1, f 2. Let’s write
2 X en = qmnf m, n = 1, 2, (C.3) m=1
for some unique coefficients qmn which define a matrix Q = (qmn).
We note that 2 X X X X en = qmnf m = qmn prmer = (PQ)rner. m=1 (All the sums run from 1 to 2)
Equating coefficients we see that
(PQ)rn = δrn, δrn = 1 (r = n), δrn = 0 (r 6= n). Hence PQ = I, where I denotes the 2 × 2 unit matrix. It follows that Q = P −1, as expected.
Just as we’ve considered the matrix A of the linear transformation a with respect to the basis 0 0 e1, e2 (see equation C.1) we may consider its matrix A = (aki) (say) with respect to the basis f 1, f 2. Referring to equation C.1 this is defined by
2 X 0 af i = Af i = akif k i = 1, 2 (C.4) k=1 Let’s calculate the matrix A0.
We have
2 2 2 X X X X X X Af i = A pkiek = pkiAek = pki anken = ankpki qmnf m k=1 k,n=1 m=1
Version of Mar 14, 2019 Appendix C: Basic Linear Algebra 197
2 2 X X = qmnankpkif m = (QAP )mif m k,n,m=1 m=1 We therefore see that the matrix A0 of the linear transformation a with respect to the basis f 1, f 2 is given by A0 = QAP = P −1AP. This is a simple, but fundamental result. We note in passing that the arguments are identical in IRn.
0 −1 Given a matrix A one aims to choose a basis f 1, f 2 which makes A = P AP as simple as possible. Before obtaining the Jordan canonical forms for 2 × 2 real matrices A we prove the following simple result.
C.2 Invariance of Eigenvalues Under Similarity Transformations
Let A be any n × n matrix, and let P be any non-singular n × n matrix. Then the matrices A and P −1AP have the same eigenvalues.
Proof
We note that
det(λI − P −1AP ) = det(P −1(λI − A)P ) = det P −1 det(λI − A) det P
= det(P −1P ) det(λI − A) = det I det(λI − A) = det(λI − A). It follows that det(λI − P −1AP ) = 0 ⇐⇒ det(λI − A) = 0 and therefore the matrices P −1AP and A have the same eigenvalues.
C.3 Jordan Forms
Now let A be any real 2 × 2 matrix, and regard A as the matrix of a linear transformation a of IR2. Then A is similar to one of only three fundamentally different forms of matrix — the so-called Jordan normal form corresponding to A. Which of the three Jordan forms corresponds to a given matrix A is decided by the eigenvalues of A. Thus to obtain the Jordan forms we start by calculating the eigenvalues of A. Three different cases may occur.
(a) The eigenvalues λ1, λ2 of A are real and distinct, λ1 > λ2.
Let f 1, f 2 be the eigenvectors corresponding to λ1, λ2 respectively, so that
Af 1 = λ1f 1,Af 2 = λ2f 2.
Version of Mar 14, 2019 Appendix C: Basic Linear Algebra 198
Using the same notation as employed above we see that the matrix A0 of the transforma- tion a with respect to the basis f 1, f 2 (it is a basis because eigenvectors corresponding to distinct real eigenvalues are linearly independent – proved in CM113A) is given by
λ 0 A0 = 1 , 0 λ2 and therefore −1 λ1 0 P AP = , where P = (f 1 : f 2) . (C.5) 0 λ2 (P is expressed in partitioned form)
(b) The eigenvalues λ1, λ2 are real and coincident: λ1 = λ2 = λ, say. There are two possibilities. (i) First let’s suppose that A has two linearly independent eigenvectors f 1, f 2 so that
Af 1 = λf 1,Af 2 = λf 2,
and f 1, f 2 form a basis.
With respect to the basis f 1, f 2 a has matrix λ 0 A0 = = λI, P −1AP = λI, P = (f : f ) . 0 λ 1 2
In this case A = λI, so the matrix was of Jordan canonical form (diagonal) to begin with.
(ii) The second possibility is that A has only one independent eigenvector, f 1 (say). 2 We extend this to a basis of IR by picking another vector f 2 (f 2 isn’t an eigenvector of A.) We can write Af 2 = µf 1 + νf 2, for some real µ, ν.
We have Af 1 = λf 1,Af 2 = µf 1 + νf 2
so, with respect to the basis f 1, f 2, a has matrix λ µ A0 = P −1AP = . 0 ν
Imposing the condition that A and P −1AP have the same eigenvalues (see Section C.2 above) we immediately obtain ν = λ.
1 Writing f 1 = µ F 1 we see that λ 1 M −1AM = ,M = (F : f ) . (C.6) 0 λ 1 2
Version of Mar 14, 2019 Appendix C: Basic Linear Algebra 199
(c) The remaining possibility is that A has complex eigenvalues λ = α + iβ, α − iβ.
Suppose that the eigenvalue equation has been solved in the form
A(f 1 + if 2) = (α + iβ)(f 1 + if 2).
Equating real and imaginary parts we then have
Af 1 = αf 1 − βf 2,Af 2 = βf 1 + αf 2.
One can prove without too much difficulty that f 1, f 2 are linearly independent and therefore constitute a basis of IR2.
With respect to this basis a has matrix A0 where
α β A0 = P −1AP = , (C.7) −β α
and P = (f 1 : f 2) .
We conclude by showing how the components of a vector with respect to the bases (f 1, f 2) and (e1, e2) are related.
C.4 Basis Transformation
Suppose that (f 1, f 2) and (e1, e2) are bases related by equations C.2 and let y1, y2, x1, x2, denote the components of some vector with respect to the bases (f 1, f 2), (e1, e2), respec- tively.
Writing x y X = 1 Y = 1 , x2 y2 we have 2 2 2 X X X xiei = yif i = yipjiej i=1 i=1 i,j=1 P2 and, equating coefficients, we obtain i=1 pkiyi = xk; or, in column vector notation,
X = P Y , Y = P −1X .
We frequently use such a linear change of variable in our discussions of the stability of fixed points.
Version of Mar 14, 2019 Appendix C: Basic Linear Algebra 200
C.5 Rotations
Consider a set of Cartesian axes xOy, with origin O; the vectors e1, e1 may be regarded as unit vectors parallel to the x and y axes respectively. We aim to calculate the matrix A(θ) (with respect to the standard basis (e1, e2)) of the transformation which rotates a vector through an angle θ about the origin O. Let r denote the position vector of the point P with coordinates (x1, x2) so that r = OP = x1e1 + x2e2. Suppose that under the rotation r = OP 0 0 0 0 0 maps to r = OP , where P has coordinates (x1, x2). Regarding the Cartesian plane as a copy of the Argand diagram we may write
0 0 iθ x1 + ix2 = e (x1 + ix2) = (cos θ + i sin θ)(x1 + ix2). Equating real and imaginary parts we obtain
0 0 x1 = x1 cos θ − x2 sin θ, x2 = x1 sin θ + x2 cos θ,
or, in matrix notation, r0 = A(θ)r, where
cos θ − sin θ A(θ) = . sin θ cos θ
Note: Reference to equation C.7 shows that A(θ) has complex eigenvalues cos θ ± i sin θ, a fact which may be confirmed by direct calculation.
C.6 Area Preserving Transformations
In this section we determine the class of linear transformations of the Cartesian plane which preserve areas. We employ the notation of section C.5. First we note that the origin O is left invariant by any linear transformation of the Cartesian plane. We also note that any non- singular linear transformation of the Cartesian plane maps parallel lines into parallel lines. To see this consider the straight line which passes through the point with position vector r0 and whose direction is specified by the vector n; this line has equation
r = r0 + λn, (C.8)
where λ is a real parameter. Let a be a non-singular linear transformation of the Cartesian plane whose matrix with respect to the standard basis is A. The image of the straight line described by equation C.8 is given by
Ar = Ar0 + λ(An), which has the form
0 0 0 r = r0 + λn . (C.9) We conclude that all straight lines parallel to the vector n map to straight lines which are parallel to n0 = An; An is not the zero vector because A is non-singular. It is clear therefore that non-singular transformations of the Cartesian plane map parallelograms to parallelo- grams. Let P,R be points with coordinates (x1, y1), (x2, y2) respectively. The position vectors OP, OR of P,R are given by
OP = x1e1 + y1e2, OR = x2e1 + y2e2.
Version of Mar 14, 2019 Appendix C: Basic Linear Algebra 201
Consider the parallelogram OP QR. Under the action of a linear transformation a of the Cartesian plane this parallelogram maps to an image parallelogram OP 0Q0R0, where P 0,R0 0 0 0 0 have coordinates (x1 , y1 ), (x2 , y2 ) respectively, say. If A is the matrix of the linear trans- formation a, with respect to the standard basis, where
α β A = , (C.10) γ δ
then we have the relations 0 0 x1 α β x1 x2 α β x2 0 = , 0 = . (C.11) y1 γ δ y1 y2 γ δ y2
An elementary calculation shows that the area of parallelogram OP QR is 0 0 0 0 |OP × OR| = |x1y2 − x2y1|; similarly the area of the image parallelogram O P Q R is 0 0 0 0 |x1 y2 − x2 y1 |.
It now follows from equation C.11 that
0 0 0 0 x1 y2 − x2 y1 = v(αx1 + βy1)(γx2 + δy2) − (γx1 + δy1)(αx2 + βy2)
and after a little simplification we find that
0 0 0 0 x1 y2 − x2 y1 = (αδ − βγ)(x1y2 − x2y1) = det A(x1y2 − x2y1).
We conclude that a parallelogram and its image under the linear transformation a have the same area provided | det A| = 1. Of particular interest are those transformations for which det A = 1; of course, det I = 1, and the transformations with det A = 1 are those which can be reached continuously from the identity. We shall see later that area preserving transformations are significant in Hamiltonian mechanics.
0 0 Note: Under the change of variable (x1, y1) 7→ (x1 , y1 ) described by equation C.11 area elements transform according to the rule
0 0 0 0 ∂(x1 , y1 ) dx1 dy1 = |J|dx1 dy1, where J = . ∂(x1, y1) Referring to equation C.11 and carrying out the differentiations we soon find that J = det A; as found above, the condition for areas to be preserved is therefore | det A| = 1.
C.7 Examples and Exercises
Example C.1. For each of the matrices Ai (1 = 1, 2, 3) given by 1 2 2 1 3 −1 A = ,A = ,A = 1 1 1 2 −2 4 3 1 1
−1 find the appropriate Jordan form Ji (i = 1, 2, 3) and a matrix Pi (i = 1, 2, 3) such that Pi AiPi = Ji, (i = 1, 2, 3).
Version of Mar 14, 2019 Appendix C: Basic Linear Algebra 202
Consider the third of these matrices, A3. A vector has components x1, x2 with respect to the standard basis e1, e2 and components y1, y2 with respect to the basis f 1, f 2 in which the linear transformation corresponding to A3 has the Jordan matrix J3. Write down the linear equations which connect x1, x2 and y1, y2.
The eigenvalues of A1 are given by √ det(λI − A1) = 0, so that λ = 1 ± 2.
Solving the eigenvalue equation 1 2 x x = λ , 1 1 y y we find that x = (λ − 1)y and we may take as eigenvectors
2 √ √ , λ = 1 + 2 , 2
2 √ √ , λ = 1 − 2 . − 2 It follows that √ 1 + 2 0 2 2 P −1A P = √ , where P = √ √ . 1 1 1 0 1 − 2 1 2 − 2
Similarly, the eigenvalues of A2 are given by
det(λI − A2) = 0, λ = 3 ± i.
The complex eigenvector of A2 corresponding to the eigenvalue 3 + i is given by
2 1 x x = (3 + i) . −2 4 y y
We find that 2x + y = (3 + i)x, y = (1 + i)x. Taking x = 1 we obtain y = 1 + i and the corresponding complex eigenvector is
1 f + if = 1 2 1 + i so that 1 0 f = , f = . 1 1 2 1
Clearly, f 1, f 2 are linearly independent, and since
A2f 1 = 3f 1 − f 2 ,A2f 2 = f 1 + 3f 2 we conclude that 3 1 1 0 P −1A P = ,P = (f : f ) = . 2 2 2 −1 3 2 1 2 1 1
Finally, A3 has eigenvalues given by
det(λI − A3) = 0, λ = 2 (twice) .
Version of Mar 14, 2019 Appendix C: Basic Linear Algebra 203
To obtain an eigenvector consider the equation
3 −1 x x = 2 . 1 1 y y
It follows that 3x − y = 2x, x = y, so that we may take as an eigenvector 1 f = . 1 1 Extend this to a basis of IR2 by the vector
1 f = . 2 0
Now 3 Af = = αf + βf , 2 1 1 2 from which we easily find that α = 1, β = 2.
Hence, A3f 1 = 2f 1,A3f 2 = f 1 + 2f 2 and choosing f 1, f 2 as a basis 2 1 1 1 P −1A P = ,P = (f : f ) = . 3 3 3 0 2 3 1 2 1 0
The connection between the components x1, x2 of a vector with respect to the standard basis and its components y1, y2 with respect to the basis f 1, f 2 is given by x 1 1 y 1 = 1 . x2 1 0 y2 Exercise C.1 Find the image of the rectangle OBCD with (Cartesian) coordinates (0, 0), (1, 0), (1, 2), (0, 2) under the linear transformation of the Cartesian plane whose matrix with respect to the standard basis is 1 1 . 1 2 Sketch your result; it should look like a parallelogram!
Exercise C.2 Describe the transformation of the Cartesian plane whose matrix with respect to the standard basis is √ 1 3 −1 √ . 2 1 3 Write down the matrix of the linear transformation which describes a rotation about O through an angle of π/4.
Exercise C.3 Show that the rotation matrix cos θ − sin θ A(θ) = , sin θ cos θ
Version of Mar 14, 2019 Appendix C: Basic Linear Algebra 204 is orthogonal i.e. that A(θ)T A(θ) = I.
Show, by direct calculation, that A(θ1)A(θ2) = A(θ1+θ2). What is the geometrical significance of this result? Write down, without calculation the matrix A(θ)n, where n is a positive integer.
Verify that A(θ) has eigenvalues cos θ ± i sin θ.
Exercise C.4 Find the Jordan forms appropriate to the matrices
1 1 2 1 3 1 , , . 1 2 −4 4 −1 1
Exercise C.5 Show that the set G given by
a b G = A = , a, b, c, d ∈ IR, det A 6= 0 c d forms a group under matrix multiplication. Prove also that the set H ⊆ G which consists of those matrices which have det A = 1 is a normal sub-group of G.
Exercise C.6 Recall that a 2 × 2 real matrix is orthogonal if AT A = I. Show that any such matrix represents an area preserving transformation. As noted above, the rotation matrices A(θ) are orthogonal.
Exercise C.7 Suppose that a linear transformation a of the Cartesian plane has matrix A with respect to the standard basis e1, e2. Suppose that under the transformation a OP 7→ 0 0 0 0 OP , where P has coordinates (x1, x2),P has coordinates (x1 , x2 ), and that a leaves lengths invariant i.e. 2 2 02 02 x1 + x2 = x1 + x2 ,
T for every choice of (x1, x2). Prove that A is orthogonal i.e. that A A = I. Deduce that det A = ±1.
Version of Mar 14, 2019 Appendix D: Jacobians 205
Appendix D
Jacobians and Change of variables
D.1 Change of variables
Suppose we have a second order autonomous dynamical system for two variables x(t) and y(t) which has a fixed point at (x∗, y∗). If the DS is dx dy = f(x, y) , = g(x, y) , (D.1) dt dt then we have f(x∗, y∗) = (x∗, y∗) = 0 . (D.2) We know that the nature of the fixed point is determined by the normal form of the Jacobian matrix J evaluated at the fixed point, ∂f ∗ ∂f ∗! ∂x ∂y J = ∂g ∗ ∂g ∗ , (D.3) ∂x ∂y ∂f ∗ where ∂x is the partial derivative evaluated at the fixed point. If we change from the variables (x, y) to new variables (u, v) then the nature of the fixed point shouldn’t change. If the DS in the new variables is du dv = F (u, v) , = G(u, v) , (D.4) dt dt then we know that the nature of the fixed point is determined by the Jacobian at the fixed point in these coordinates, ∂F ∗ ∂F ∗ 0 ∂u ∂v J = ∂G ∗ ∂G ∗ . (D.5) ∂u ∂v We now see how to relate J 0 to J. The first step is to find F and G. We can find these from the chain rule for partial derivatives du ∂u dx ∂u dy ∂u ∂u = + = f + g ≡ F (u, v) , dt ∂x dt ∂y dt ∂x ∂y dv ∂v dx ∂v dy ∂v ∂v = + = f + g ≡ G(u, v) . (D.6) dt ∂x dt ∂y dt ∂x ∂y We now need to calculate the matrix elements of J 0, for example ∂F ∂ ∂u ∂u = f + g ∂u ∂u ∂x ∂y
Version of Mar 14, 2019 Appendix D: Jacobians 206
∂ ∂u ∂u ∂f ∂ ∂u ∂u ∂g = f + + g + ∂u ∂x ∂x ∂u ∂u ∂y ∂y ∂u ∂ ∂u ∂u ∂f ∂x ∂f ∂y ∂ ∂u ∂u ∂g ∂x ∂g ∂y = f + + + g + + ∂u ∂x ∂x ∂x ∂u ∂y ∂u ∂u ∂y ∂y ∂x ∂u ∂y ∂u (D.7)
At the fixed point, f(x∗, y∗) = g(x∗, y∗) = 0 and so ∂F ∗ ∂u ∂f ∗ ∂x ∂u ∂f ∗ ∂y ∂u ∂g ∗ ∂x ∂u ∂g ∗ ∂y = + + + (D.8) ∂u ∂x ∂x ∂u ∂x ∂y ∂u ∂y ∂x ∂u ∂y ∂y ∂u This is the top left entry of the matrix
∗ ∗ ∂F ∗ ∂u ∂u ! ∂f ∂f ! ∂x ! ∂u · ∂x ∂y ∂x ∂y ∂u · = ∂g ∗ ∂g ∗ ∂y (D.9) ·· ·· ∂x ∂y ∂u · Calculating the top right entry, we get ∂F ∂ ∂u ∂u = f + g ∂v ∂v ∂x ∂y ∂ ∂u ∂u ∂f ∂ ∂u ∂u ∂g = f + + g + ∂v ∂x ∂x ∂v ∂v ∂y ∂y ∂v ∂ ∂u ∂u ∂f ∂x ∂f ∂y ∂ ∂u ∂u ∂g ∂x ∂g ∂y = f + + + g + + ∂v ∂x ∂x ∂x ∂v ∂y ∂v ∂v ∂y ∂y ∂x ∂v ∂y ∂v ∂F ∗ ∂u ∂f ∗ ∂x ∂u ∂f ∗ ∂y ∂u ∂g ∗ ∂x ∂u ∂g ∗ ∂y ⇒ = + + + , (D.10) ∂v ∂x ∂x ∂v ∂x ∂y ∂v ∂y ∂x ∂v ∂y ∂y ∂v and so on, so that all together we get
∗ ∗ ∂F ∗ ∂F ∗ ∂u ∂u ! ∂f ∂f ! ∂x ∂x 0 ∂u ∂v ∂x ∂y ∂x ∂y ∂u ∂v J = ∂G ∗ ∂G ∗ = ∂v ∂v ∂g ∗ ∂g ∗ ∂y ∂y (D.11) ∂u ∂v ∂x ∂y ∂x ∂y ∂u ∂v | {z } | {z } | {z } M J N However, the matrices M and N are inverses, MN = 1, as we can check
∂u ∂u ! ∂x ∂x ∂x ∂y ∂u ∂v MN = ∂v ∂v ∂y ∂y ∂x ∂y ∂u ∂v ∂u ∂x ∂u ∂y ∂u ∂x ∂u ∂y ! ∂x ∂u + ∂y ∂u ∂x ∂v + ∂y ∂v = ∂v ∂x ∂v ∂y ∂v ∂x ∂v ∂y ∂x ∂u + ∂y ∂u ∂x ∂v + ∂y ∂v ∂u ∂u ∂u ∂v 1 0 = ∂v ∂v = (D.12) ∂u ∂v 0 1 This means that J 0 and J are similar,
J 0 = MJM −1 , (D.13) and hence have the same normal form. This means that the nature of the fixed point is unchanged when we change coordinates, as of course should be the case.
Version of Mar 14, 2019