text:  There(From A. E. Hotchner, Papa Hemingway, Random House, NY, 1966)
  vi
  Contents1 First order dierential equations 1.1 Introduction to DEs . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Initial value problems . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Existence of solutions to ODEs . . . . . . . . . . . . . . . . . . . . . 1.3.1 First order ODEs . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.2 Second order homogeneous ODEs . . . . . . . . . . . . . . . . 1.4 First order ODEs - separable and linear cases . . . . . . . . . . . . . 1.4.1 Separable DEs . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.2 Autonomous ODEs . . . . . . . . . . . . . . . . . . . . . . . . 1.4.3 Substitution methods . . . . . . . . . . . . . . . . . . . . . . 1.4.4 Linear 1st order ODEs . . . . . . . . . . . . . . . . . . . . . . 1.5 Isoclines and direction elds . . . . . . . . . . . . . . . . . . . . . . . 1.6 Numerical solutions - Eulers method and improved Eulers method . 1.6.1 Eulers Method . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6.2 Improved Eulers method . . . . . . . . . . . . . . . . . . . . 1.6.3 Eulers method for systems and higher order DEs . . . . . . . 1.7 Numerical solutions II - Runge-Kutta and other methods . . . . . . 1.7.1 Fourth-Order Runge Kutta method . . . . . . . . . . . . . . 1.7.2 Multistep methods - Adams-Bashforth . . . . . . . . . . . . . 1.7.3 Adaptive step size . . . . . . . . . . . . . . . . . . . . . . . . 1.8 Newtonian mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . 1.9 Application to mixing problems . . . . . . . . . . . . . . . . . . . . . 1.10 Application to cooling problems . . . . . . . . . . . . . . . . . . . . . 2 Second order dierential equations 2.1 Linear dierential equations . . . . . . . . . . . . . . . . 2.1.1 Solving homogeneous constant coecient ODEs . 2.2 Linear dierential equations, revisited . . . . . . . . . . 2.3 Linear dierential equations, continued . . . . . . . . . . 2.4 Undetermined coecients method . . . . . . . . . . . . 2.4.1 Simple case . . . . . . . . . . . . . . . . . . . . . 2.4.2 Non-simple case . . . . . . . . . . . . . . . . . . 2.4.3 Annihilator method . . . . . . . . . . . . . . . . vii 3 3 11 16 16 20 23 23 26 29 30 33 39 39 42 44 48 48 50 50 52 56 60 65 65 65 68 73 78 78 81 84
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  . . . . . . . .
  viii 2.5 Variation of parameters . . . . . . . . 2.5.1 The Leibniz rule . . . . . . . . 2.5.2 The method . . . . . . . . . . . Applications of DEs: Spring problems 2.6.1 Part 1 . . . . . . . . . . . . . . 2.6.2 Part 2 . . . . . . . . . . . . . . 2.6.3 Part 3 . . . . . . . . . . . . . . Applications to simple LRC circuits . The power series method . . . . . . . . 2.8.1 Part 1 . . . . . . . . . . . . . . 2.8.2 Part 2 . . . . . . . . . . . . . . The Laplace transform method . . . . 2.9.1 Part 1 . . . . . . . . . . . . . . 2.9.2 Part 2 . . . . . . . . . . . . . . 2.9.3 Part 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
  CONTENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 86 87 90 90 94 97 102 108 108 115 118 119 126 135 139 139 139 143 143 147 148 153 155 157 158 162 165 173 179 184 191 192 198 198 200 205 205 207 208 210
  2.6
  2.7 2.8
  2.9
  3 Matrix theory and systems of DEs 3.1 Quick survey of linear algebra . . . . . . . . . . . . . . . 3.1.1 Matrix arithmetic . . . . . . . . . . . . . . . . . 3.2 Row reduction and solving systems of equations . . . . . 3.2.1 The Gauss elimination game . . . . . . . . . . . 3.2.2 Solving systems using inverses . . . . . . . . . . 3.2.3 Computing inverses using row reduction . . . . . 3.2.4 Solving higher-dimensional linear systems . . . . 3.2.5 Determinants . . . . . . . . . . . . . . . . . . . . 3.2.6 Vector spaces . . . . . . . . . . . . . . . . . . . . 3.2.7 Bases, dimension, linear independence and span . 3.3 Application: Solving systems of DEs . . . . . . . . . . . 3.3.1 Modeling battles using Lanchesters equations . . 3.3.2 Romeo and Juliet . . . . . . . . . . . . . . . . . . 3.3.3 Electrical networks using Laplace transforms . . 3.4 Eigenvalue method for systems of DEs . . . . . . . . . . 3.4.1 The eigenvalue method . . . . . . . . . . . . . . 3.4.2 Examples of the eigenvalue method . . . . . . . . 3.5 Introduction to variation of parameters for systems . . . 3.5.1 Motivation . . . . . . . . . . . . . . . . . . . . . 3.5.2 The method . . . . . . . . . . . . . . . . . . . . . 3.6 Nonlinear systems . . . . . . . . . . . . . . . . . . . . . 3.6.1 Linearizing near equilibria . . . . . . . . . . . . . 3.6.2 The nonlinear pendulum . . . . . . . . . . . . . . 3.6.3 The Lorenz equations . . . . . . . . . . . . . . . 3.6.4 Zombies attack . . . . . . . . . . . . . . . . . . .
  . . . . . . . . . . . . . . . . . . . . . . . . .
  CONTENTS 4 Introduction to partial dierential equations 4.1 Introduction to separation of variables . . . . . 4.1.1 The transport or advection equation . . 4.1.2 The heat equation . . . . . . . . . . . . 4.2 The method of superposition . . . . . . . . . . 4.3 Fourier series, sine series, cosine series . . . . . 4.3.1 Brief history . . . . . . . . . . . . . . . 4.3.2 Motivation . . . . . . . . . . . . . . . . 4.3.3 Denitions . . . . . . . . . . . . . . . . 4.4 The heat equation . . . . . . . . . . . . . . . . 4.4.1 Method for zero ends . . . . . . . . . . . 4.4.2 Method for insulated ends . . . . . . . . 4.4.3 Explanation via separation of variables . 4.5 The wave equation in one dimension . . . . . . 4.6 The Schr odinger equation . . . . . . . . . . . . 4.6.1 Method . . . . . . . . . . . . . . . . . . 5 Appendices 5.1 Appendix: 5.2 Appendix: 5.3 Appendix: 5.4 Appendix: 5.5 Appendix: Integral Tables . . . . . . . . . Fourier Series . . . . . . . . . . Bessel Functions . . . . . . . . Laplace Transform Tables . . . Partial fraction decomposition . . . . . . . . . . . . . . .
  ix 215 215 216 219 220 223 224 224 224 232 233 235 237 241 245 246 255 256 263 264 266 269
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  CONTENTS
  xi
  PrefaceThe majority of this book came from lecture notes David Joyner (WDJ) typed up over the years for a course on dierential equations with boundary value problems at the US Naval Academy (USNA). Though the USNA is a government institution and ocial work-related writing is in the public domain, since much of this was done at home during the night and weekends that he feels he has the right to claim copyright over some of this work. The dierential equations course at the USNA has used various editions of the following three books (in order of most common use to least common use) at various times: Dennis G. Zill and Michael R. Cullen, Dierential equations with Boundary Value Problems, 6th ed., Brooks/Cole, 2005. R. Nagle, E. Sa, and A. Snider, Fundamentals of Dierential Equations and Boundary Value Problems, 4th ed., Addison/Wesley, 2003. W. Boyce and R. DiPrima, Elementary Dierential Equations and Boundary Value Problems, 8th edition, John Wiley and Sons, 2005. WDJ has also included ideas of some of his colleagues who also have taught dierential equations at the USNA. For example, Professors Buchanan, Gaglione, and Withers have contributed directly or indirectly to these notes. You may see some similarities but, for the most part, WDJ has taught things a bit dierently and tried to impart a more computational perspective in these example-based course notes. After WDJ nished a draft of this book, he invited the second author, Marshall Hampton (MH), to revise and extend it. At the University of Minnesota Duluth, MH teaches a course on dierential equations and linear algebra. A new feature to this book is the fact that every section has at least one Sage exercise. Sage is FOSS (free and open source software), available on the most common computer platforms. Royalties for the sales of this book (if it ever makes its way to a publisher) will go to further development of Sage . Some exercises in the text are considerably more challenging than others. We have indicated these with an asterisk in front of the exercise number. This book is copyright David Joyner and Marshall Hampton, 2010. It is distributed using a free and open source license, the Attribution-ShareAlike Creative Commons license,, or the Gnu Free Documentation License (GFDL),, at your choice. The cover image was created with the following Sage code:Sage from math import cos, sin def RK4(f, t_start, y_start, t_end, steps): fourth-order Runge-Kutta solver with fixed time steps.
  xii
  CONTENTSf must be a function of t,y. step_size = (t_end - t_start)/steps t_current = t_start argn = len(y_start) y_current = [x for x in y_start] answer_table = [] answer_table.append([t_current,y_current]) for j in range(0,steps): k1=f(t_current,y_current) k2=f(t_current+step_size/2,[y_current[i] + k1[i]*step_size/2 for i in range(argn)]) k3=f(t_current+step_size/2,[y_current[i] + k2[i]*step_size/2 for i in range(argn)]) k4=f(t_current+step_size,[y_current[i] + k3[i]*step_size for i in range(argn)]) t_current += step_size y_current = [y_current[i] + (step_size/6)*(k1[i]+2*k2[i]+2*k3[i]+k4[i]) for i in range(len(k1))] answer_table.append([t_current,y_current]) return answer_table
  def e1(t, y): return [-y[1],sin(y[0])] npi = N(pi) sols = [] for v in srange(-1,1+.04,.05): p1 = RK4(e1,0.0,[0.0,v],-2.5*npi,200)[::-1] p1 = p1 + RK4(e1,0.0,[0.0,v],2.5*npi,200) sols.append(list_plot([[x[0],x[1][0]] for x in p1], plotjoined=True, rgbcolor = ((v+1)/2.01,0,(1.01-v)/2.01))) f = 2 show(sum(sols), axes = True, figsize = [f*5*npi/3,f*2.5], xmin = -7, xmax = 7, ymin = -1, ymax = 1)
  All other images included in this text are either in the public domain, or licensed under the dual license stated above. The authors donate the royalties of this book to the Sage Foundation to further promote free mathematical software for everyone.
  AcknowledgmentsIn a few cases we have made use of the excellent (public domain!) lecture notes by Sean Mauch, Introduction to methods of Applied Mathematics, available online at (as of Fall, 2009). In some cases, we have made use of the material on Wikipedia - this includes both discussion and in a few cases, diagrams or graphics. This material is licensed under the GFDL or the Attribution-ShareAlike Creative Commons license. In any case, the amount used here probably falls under the fair use clause. Software used: Most of the graphics in this text was created using Sage () and GIMP by the authors. The most important components of Sage for our purposes are: Maxima, SymPy and Matplotlib. The circuit diagrams were created using Dia and GIMP by the authors. A few diagrams were taken from Wikipedia (and acknowledged A in the appropriate place in the text). Of course, L TEX was used for the typesetting, and emacs () for editing. Many thanks to the developers of these programs for these free tools.
  Chapter 1
  1.1
  Introduction to DEs
  Roughly speaking, a dierential equation is an equation involving the derivatives of one or more unknown functions. Implicit in this vague denition is the assumption that the equation imposes a constraint on the unknown function (or functions). For example, we would not call the well-known product rule identity of dierential calculus a dierential equation. Example 1.1.1. Here is a simple example of a dierential equation: dx dt = x. A solution would be a function x = x(t) which is dierentiable on some interval and whose derivative is equal to itself. Can you guess any solutions? As far as notation goes, in this text we often drop the dependence on the independent variable when referring to a function. For example, x(t) is sometimes simply denoted by x. In calculus (dierential-, integral- and vector-), youve studied ways of analyzing functions. You might even have been convinced that functions you meet in applications arise naturally from physical principles. As we shall see, dierential equations arise naturally from general physical principles. In many cases, the functions you met in calculus in applications to physics were actually solutions to a naturally-arising dierential equation. Example 1.1.2. Consider a falling body of mass m on which exactly three forces act: gravitation, Fgrav , air resistance, Fres , an external force, Fext = f (t), where f (t) is some given function. 3
  Fres
  mass m
  Fgrav?
  Let x(t) denote the distance fallen from some xed initial position. (In other words, we measure x(t) so that down is positive.) The velocity is denoted by v = x and the acceleration by a = x . In this case, gravitational force is governed by Fgrav = mg, where g > 0 is the gravitational constant. We assume that air resistance Fres is proportional to velocity (a common assumption in physics), and write Fres = kv = kx , where k > 0 is a friction constant. The total force, Ftotal , is by hypothesis, Ftotal = Fgrav + Fres + Fext , and, by Newtons 2nd Law for a constant mass, Ftotal = (mx ) = mx = ma. Putting these together, we have mx = ma = mg kx + f (t), or mx + kx = f (t) + mg. This is a dierential equation in x = x(t). It may also be rewritten as a dierential equation in v = v (t) = x (t) as mv + kv = f (t) + mg. This is an example of a rst order dierential equation in v , which means that at most rst order derivatives of the unknown function v = v (t) occur. In fact, you have probably seen solutions to this in your calculus classes, at least when f (t) = 0 and k = 0. In that case, v (t) = g and so v (t) = g dt = gt + C . Here the constant of integration C represents the initial velocity. Dierential equations occur in other areas as well: weather prediction (more generally, uid-ow dynamics), electrical circuits, the temperature of a heated homogeneous wire, and many others (see the table below). They even arise in problems on Wall Street: the BlackScholes equation is a partial dierential equation which models the pricing of derivatives [BS-intro]. Learning to solve dierential equations helps you understand the behaviour of phenomenon present in these problems.
  phenomenonweather falling body motion of a mass attached to a spring motion of a plucked guitar string Battle of Trafalger cooling cup of coee in a room population growth
  description of DENavier-Stokes equation [NS-intro] a non-linear vector-valued higher-order PDE 1st order linear ODE Hookes spring equation 2nd order linear ODE [H-intro] Wave equation 2nd order linear PDE [W-intro] Lanchesters equations system of 2 1st order DEs [L-intro], [M-intro], [N-intro] Newtons Law of Cooling 1st order linear ODE logistic equation non-linear, separable, 1st order ODE
  Undened terms and notation will be dened below, except for the equations themselves. For those, see the references or wait until later sections when they will be introduced1 . Basic Concepts: Here are some of the concepts (and abbreviations) to be introduced below: dependent variable(s), independent variable(s), ODEs, PDEs, order, linearity, solution. It is really best to learn these concepts using examples. However, here are the general denitions anyway, with examples to follow. Notation/denitions: The term dierential equation is sometimes abbreviated DE, for brevity.1 For example, the important Navier-Stokes equation is too complicated to state here and its explanation would take us too far aeld. For details, see for example, or [A-ns].
  CHAPTER 1. FIRST ORDER DIFFERENTIAL EQUATIONS Dependent/independent variables: Put simply, a dierential equation is an equation involving derivatives of one of more unknown functions. The variables you are dierentiating with respect to are the independent variables of the DE. The variables (the unknown functions) you are dierentiating are the dependent variables of the DE. Other variables which might occur in the DE are sometimes called parameters. ODE/PDE: If none of the derivatives which occur in the dierential equation are partial derivatives (for example, if the dependent variable/unknown function is a function of a single variable) then the dierential equation is called an ordinary dierential equation or ODE. If some of the derivatives which occur in the dierential equation are partial derivatives then the dierential equation is a partial dierential equation or PDE. Order: The highest total number of derivatives you have to take in the dierential equation is its order. Linearity: This can be described in a few dierent ways. First of all, a dierential equation is linear if the only operations you perform on its terms are combinations of the following: dierentiation with respect to independent variable(s), multiplication by a function of the independent variable(s). Another way to dene linearity is as follows. A linear ODE having independent variable t and the dependent variable is y is an ODE of the form a0 (t)y (n) + ... + an1 (t)y + an (t)y = f (t), for some given functions a0 (t), . . . , an (t), and f (t). Here dn y (t) dtn denotes the n-th derivative of y = y (t) with respect to t. The terms a0 (t), . . . , an (t) are called the coecients of the dierential equation and we will call the term f (t) the non-homogeneous term or the forcing function. (In physical applications, this term usually represents an external force acting on the system. For instance, in the example above it represents the gravitational force, mg.) y (n) = y (n) (t) = A linear PDE is a dierential equation which is a linear ODE in each of its independent variables. Solution: An explicit solution to an ODE having independent variable t and the dependent variable x is simply a function x(t) for which the dierential equation is true for all values of t in an open interval. An implicit solution to an ODE is an implicitly dened (possibly multiple-valued) function which satises the ODE.
  IVP: A rst order initial value problem (abbreviated IVP) is a problem of the form x = f (t, x), x(a) = c, (1.1)
  where f (t, x) is a given function of two variables, and a, c are given constants. The equation x(a) = c is the initial condition. Here are some examples: Example 1.1.3. Here is a table of examples. As an exercise, determine which of the following are ODEs and which are PDEs.
  DEmx + kx = mg falling body mv + kv = mg falling body 2 u = u k t x2 heat equation mx + bx + kx = f (t) spring equation P P = k(1 K )P logistic population equation 2 2 u u k = x2 2 t wave equation T = k(T Troom ) Newtons Law of Cooling x = Ay , y = Bx, Lanchesters equations
  indep varst t t, x t t t, x t t
  dep varsx v u x P u T x, y
  order2 1 2 2 1 2 1 1
  linear?yes yes yes yes no yes yes yes
  Remark 1.1.1. Note that in many of these examples, the symbol used for the independent variable is not made explicit. For example, we are writing x when we really mean x (t) = dx(t) dt . This is very common shorthand notation and, in this situation, we shall usually use t as the independent variable whenever possible. Example 1.1.4. Recall a linear ODE having independent variable t and the dependent variable is y is an ODE of the form a0 (t)y (n) + ... + an1 (t)y + an (t)y = f (t), for some given functions a0 (t), . . . , an (t), and f (t). The order of this dierential equation is n. In particular, a linear 1st order ODE having independent variable t and the dependent variable is y is an ODE of the form
  a0 (t)y + a1 (t)y = f (t), for some a0 (t), a1 (t), and f (t). We can divide both sides of this equation by the leading coecient a0 (t) without changing the solution y to this dierential equation. Lets do that and rename the terms: y + p(t)y = q (t), where p(t) = a1 (t)/a0 (t) and q (t) = f (t)/a0 (t). Every linear 1st order ODE can be put into this form, for some p and q . For example, the falling body equation mv + kv = f (t) + mg has this form after dividing by m and renaming v as y .u P u )P or k What does a dierential equation like mx + kx = mg or P = k(1 K = x2 2t really mean? In mx + kx = mg, m and k and g are given constants. The only things that can vary are t and the unknown function x = x(t).2 2
  Example 1.1.5. To be specic, lets consider x + x = 1. This means for all t, x (t)+ x(t) = 1. In other words, a solution x(t) is a function which, when added to its derivative you always get the constant 1. How many functions are there with that property? Try guessing a few random functions: Guess x(t) = sin(t). Compute (sin(t)) + sin(t) = cos(t) + sin(t) = 2 sin(t + 4 ). x (t) + x(t) = 1 is false. Guess x(t) = exp(t) = et . Compute (et ) + et = 2et . x (t) + x(t) = 1 is false. Guess x(t) = t2 . Compute (t2 ) + t2 = 2t + t2 . x (t) + x(t) = 1 is false. Guess x(t) = exp(t) = et . Compute (et ) + et = 0. x (t) + x(t) = 1 is false. Guess x(t) = 1. Compute (1) + 1 = 0 + 1 = 1. x (t) + x(t) = 1 is true. We nally found a solution by considering the constant function x(t) = 1. Here a way of doing this kind of computation with the aid of the computer algebra system Sage :Sage sage: t = var(t) sage: de = lambda x: diff(x,t) + x - 1 sage: de(sin(t)) sin(t) + cos(t) - 1 sage: de(exp(t)) 2*et - 1 sage: de(t2) t2 + 2*t - 1 sage: de(exp(-t)) -1 sage: de(1) 0
  Note we have rewritten x + x = 1 as x + x 1 = 0 and then plugged various functions for x to see if we get 0 or not. Obviously, we want a more systematic method for solving such equations than guessing all the types of functions we know one-by-one. We will get to those methods in time. First, we need some more terminology. Under mild conditions on f , an IVP of the form (1.1) has a solution x = x(t) which is unique. This means that if f and a are xed but c is a parameter then the solution x = x(t) will depend on c. This is stated more precisely in the following result. Theorem 1.1.1. (Existence and uniqueness) Fix a point (t0 , x0 ) in the plane. Let f (t, x) (t,x) are continuous on some be a function of t and x for which both f (t, x) and fx (t, x) = fx rectangle a < t < b, c < x < d,
  in the plane. Here a, b, c, d are any numbers for which a < t0 < b and c < x0 < d. Then there is an h > 0 and a unique solution x = x(t) for which x = f (t, x), for all t (t0 h, t0 + h),
  and x(t0 ) = x0 .
  We will return to it in more detail in 1.3 below. (See 2.8 of Boyce and DiPrima [BD-intro] for more details on a proof.) In most cases we shall run across, it is easier to construct the solution than to prove this general theorem! Example 1.1.6. Let us try to solve x + x = 1, x(0) = 1.
  The solutions to the dierential equation x + x = 1 which we guessed at in the previous example, x(t) = 1, satises this IVP. Here a way of nding this solution with the aid of the computer algebra system Sage :Sage sage: sage: sage: sage: 1 t = var(t) x = function(x, t) de = lambda y: diff(y,t) + y - 1 desolve(de(x),[x,t],[0,1])
  (The command desolve is a dierential equation solver in Sage .) Just as an illustration, lets try another example. Let us try to solve x + x = 1, The Sage commands are similar: x(0) = 2.
  10
  sage: t = var(t) sage: x = function(x, t) sage: de = lambda y: diff(y,t) + y - 1 sage: x0 = 2 # this is for the IC x(0) = 2 sage: soln = desolve(de(x),[x,t],[0,x0]) sage: P = plot(soln,0,5) sage: soln; show(P, ymin = 0) (et + 1)*e(-t)
  This gives the solution x(t) = (et + 1)et = 1 + et and the plot given in Figure 1.1.
  Figure 1.1: Solution to IVP x + x = 1, x(0) = 2. Example 1.1.7. Now let us consider an example which does not satisfy all of the assumptions of the existence and uniqueness theorem: x = x2/3 . The function x = (t/3 + C )3 is a solution to this dierential equation for any choice of the constant C . If we consider only solutions with the initial value x(0) = 0, we see that we can choose C = 0 - i.e. x = t3 /27 satises the dierential equation and has x(0) = 0. But there is another solution with that initial condition - the solution x = 0. So solutions exist, but they are not necessarily unique. This does not violate the existence and uniqueness theorem because in this case f (x, t) = x2/3 is continuous near the initial condition (x0 , t0 ) = (0, 0) but its x-derivative f 2 1/3 is not continuous - indeed it is not even dened for x = 0. x = 3 x Some conditions on the ODE which guarantee a unique solution will be presented in 1.3. Exercises: 1. Verify that y = 2e3x is a solution to the dierential equation y = 3y . 2. Verify that x = t3 + 5 is a solution to the dierential equation x = 3t2 .
  11
  3. Subsitute x = ert into the dierential equation x + x 6x = 0 and determine all values of the parameter r which give a solution. 4. (a) Verify that, for any constant c, the function x(t) = 1 + cet solves x + x = 1. Find the c for which this function solves the IVP x + x = 1, x(0) = 3. (b) Solve x + x = 1, using Sage . 5. Write a dierential equation for a population P that is changing in time (t) such that the rate of change is proportional to the square root of P . x(0) = 3,
  1.2
  Recall, 1st order initial value problem, or IVP, is simply a 1st order ODE and an initial condition. For example, x (t) + p(t)x(t) = q (t), x(0) = x0 ,
  where p(t), q (t) and x0 are given. The analog of this for 2nd order linear dierential equations is this: a(t)x (t) + b(t)x (t) + c(t)x(t) = f (t), x(0) = x0 , x (0) = v0 ,
  where a(t), b(t), c(t), x0 , and v0 are given. This 2nd order linear dierential equation and initial conditions is an example of a 2nd order IVP. In general, in an IVP, the number of initial conditions must match the order of the dierential equation. Example 1.2.1. Consider the 2nd order dierential equation x + x = 0. (We shall run across this dierential equation many times later. As we will see, it represents the displacement of an undamped spring with a mass attached.) Suppose we know that the general solution to this dierential equation is x(t) = c1 cos(t) + c2 sin(t), for any constants c1 , c2 . This means every solution to the dierential equation must be of this form. (If you dont believe this, you can at least check it it is a solution by computing x (t) + x(t) and verifying that the terms cancel, as in the following Sage example. Later, we see how to derive this solution.) Note that there are two degrees of freedom (the constants c1 and c2 ), matching the order of the dierential equation.
  12
  sage: t = var(t) sage: c1 = var(c1) sage: c2 = var(c2) sage: de = lambda x: diff(x,t,t) + x sage: de(c1*cos(t) + c2*sin(t)) 0 sage: x = function(x, t) sage: soln = desolve(de(x),[x,t]); soln k1*sin(t) + k2*cos(t) sage: solnx = lambda s: RR(soln.subs(k1=1, k2=0, t=s)) sage: P = plot(solnx,0,2*pi) sage: show(P)
  This is displayed in Figure 1.2. Now, to solve the IVP x + x = 0, x(0) = 0, x (0) = 1.
  the problem is to solve for c1 and c2 for which the x(t) satises the initial conditions. There are two degrees of freedom in the general solution matching the number of initial conditions in the IVP. Plugging t = 0 into x(t) and x (t), we obtain 0 = x(0) = c1 cos(0) + c2 sin(0) = c1 , 1 = x (0) = c1 sin(0) + c2 cos(0) = c2 . Therefore, c1 = 0, c2 = 1 and x(t) = sin(t) is the unique solution to the IVP.
  Figure 1.2: Solution to IVP x + x = 0, x(0) = 0, x (0) = 1. In Figure 1.2, you see the solution oscillates, as t increases. Another example, Example 1.2.2. Consider the 2nd order dierential equation x + 4x + 4x = 0.
  13
  (We shall run across this dierential equation many times later as well. As we will see, it represents the displacement of a critially damped spring with a unit mass attached.) Suppose we know that the general solution to this dierential equation is x(t) = c1 exp(2t) + c2 t exp(2t) = c1 e2t + c2 te2t , for any constants c1 , c2 . This means every solution to the dierential equation must be of this form. (Again, you can at least check it is a solution by computing x (t), 4x (t), 4x(t), adding them up and verifying that the terms cancel, as in the following Sage example.)Sage sage: sage: sage: sage: sage: 0 sage: (k2*t sage: sage: t = var(t) c1 = var(c1) c2 = var(c2) de = lambda x: diff(x,t,t) + 4*diff(x,t) + 4*x de(c1*exp(-2*t) + c2*t*exp(-2*t)) desolve(de(x),[x,t]) + k1)*e(-2*t) P = plot(t*exp(-2*t),0,pi) show(P)
  The plot is displayed in Figure 1.3. Now, to solve the IVP x + 4x + 4x = 0, x(0) = 0, x (0) = 1.
  we solve for c1 and c2 using the initial conditions. Plugging t = 0 into x(t) and x (t), we obtain 0 = x(0) = c1 exp(0) + c2 0 exp(0) = c1 , 1 = x (0) = c1 exp(0) + c2 exp(0) 2c2 0 exp(0) = c1 + c2 . Therefore, c1 = 0, c1 + c2 = 1 and so x(t) = t exp(2t) is the unique solution to the IVP. In Figure 1.3, you see the solution tends to 0, as t increases. Suppose, for the moment, that for some reason you mistakenly thought that the general solution to this dierential equation was x(t) = c1 exp(2t) + c2 exp(2t) = e2t (c1 + c2 ), for arbitrary constants c1 , c2 . (Note: the extra t-factor on the second term on the right is missing from this expression.) Now, if you try to solve for the constant c1 and c2 using the initial conditions x(0) = 0, x (0) = 1 you will get the equations
  14
  These equations are impossible to solve! The moral of the story is that you must have a correct general solution to insure that you can always solve your IVP. One more quick example. Example 1.2.3. Consider the 2nd order dierential equation x x = 0. Suppose we know that the general solution to this dierential equation is x(t) = c1 exp(t) + c2 exp(t) = c1 et + c2 et , for any constants c1 , c2 . (Again, you can check it is a solution.) The solution to the IVP x x = 0,t t
  c1 + c2 = 0 2c1 2c2 = 1.
  x(0) = 1, x (0) = 0,
  e . (You can solve for c1 and c2 yourself, as in the examples above.) This paris x(t) = e + 2 ticular function is also called a hyperbolic cosine function, denoted cosh(t) (pronounced kosh). The hyperbolic sine function, denoted sinh(t) (pronounced sinch), satises the IVP x x = 0, x(0) = 0, x (0) = 1.
  The hyperbolic trig functions have many properties analogous to the usual trig functions and arise in many areas of applications [H-ivp]. For example, cosh(t) represents a catenary or hanging cable [C-ivp].
  15
  Figure 1.4: Solution to IVP x x = 0, x(0) = 0, x (0) = 1. You see in Figure 1.4 that the solution tends to innity, as t gets larger. Exercises: 1. Find the value of the constant C that makes x = Ce3t a solution to the IVP x = 3x, x(0) = 4. 2. Verify that x = (C + t) cos(t) satises the dierential equation x + x tan(t) cos(t) = 0. Find the value of C so that x(t) satises the condition x(2 ) = 0. 3. Determine a value of the constant C so that the given solution of the dierential equation satises the initial condition. (a) y = ln(x + C ) solves ey y = 1, (b) y = Cex + x 1 solves y y (0) = 1.
  = x y , y (0) = 3.
  16
  CHAPTER 1. FIRST ORDER DIFFERENTIAL EQUATIONS 4. Use Sage to check that the general solution to the falling body problem mv + kv = mg,kt/m . If v (0) = v , you can solve for c in terms of v to get is v (t) = mg 0 0 k + ce mg c = v0 k . Take m = k = v0 = 1, g = 9.8 and use Sage to plot v (t) for 0 < t < 1.
  5. Consider a spacecraft falling towards the moon at a speed of 500 m/s (meters per second) at a height of 50 km. Assume the moons acceleration on the spacecraft is a constant 1.63 m/s2 . When the spacecrafts rockets are engaged, the total acceleration on the rocket (gravity+rocket thrust) is 3 m/s2 . At what height should the rockets be engaged so that the spacecraft lands at zero velocity?
  1.3
  When do solutions to an ODE exist? When are they unique? This section gives some necessary conditions for determining existence and uniqueness.
  1.3.1
  We begin by considering the rst order initial value problem x (t) = f (t, x(t)), x(a) = c. (1.2)
  What conditions on f (and a and c) guarantee that a solution x = x(t) exists? If it exists, what (further) conditions guarantee that x = x(t) is unique? The following result addresses the rst question. Theorem 1.3.1. (Peanos existence theorem [P-intro]) Suppose f is bounded and continuous in x, and t. Then, for some value > 0, there exists a solution x = x(t) to the initial value problem (1.2) within the range [a , a + ]. Giuseppe Peano (1858 - 1932) was an Italian mathematician, who is mostly known for his important work on the logical foundations of mathematics. For example, the common notations for union and intersections rst appeared in his rst book dealing with mathematical logic, written while he was teaching at the University of Turin. Example 1.3.1. Take f (x, t) = x2/3 . This is continuous and bounded in x and t in 1 < x < 1, t R. The IVP x = f (x, t), x(0) = 0 has two solutions, x(t) = 0 and x(t) = t3 /27. You all know what continuity means but you may not be familiar with the slightly stronger notion of Lipschitz continuity. This is dened next. Denition 1.3.1. Let D R2 be a domain. A function f : D R is called Lipschitz continuous if there exists a real constant K > 0 such that, for all x1 , x2 D ,
  17
  |f (x1 ) f (x2 )| K |x1 x2 |. The smallest such K is called the Lipschitz constant of the function f on D . For example, the function f (x) = x2/3 dened on [1, 1] is not Lipschitz continuous; the function f (x) = x2 dened on [3, 7] is Lipschitz continuous, with Lipschitz constant K = 14; the function f dened by f (x) = x3/2 sin(1/x) (x = 0) and f (0) = 0 restricted to [0, 1], gives an example of a function that is dierentiable on a compact set while not being Lipschitz. Theorem 1.3.2. (Picards existence and uniqueness theorem [PL-intro]) Suppose f is bounded, Lipschitz continuous in x, and continuous in t. Then, for some value > 0, there exists a unique solution x = x(t) to the initial value problem (1.2) within the range [a , a + ]. Charles Emile Picard (1856-1941) was a leading French mathematician. Picard made his most important contributions in the elds of analysis, function theory, dierential equations, and analytic geometry. In 1885 Picard was appointed to the mathematics faculty at the Sorbonne in Paris. Picard was awarded the Poncelet Prize in 1886, the Grand Prix des Sciences Mathmatiques in 1888, the Grande Croix de la L egion dHonneur in 1932, the Mittag-Leer Gold Medal in 1937, and was made President of the International Congress of Mathematicians in 1920. He is the author of many books and his collected papers run to four volumes. The proofs of Peanos theorem or Picards theorem go well beyond the scope of this course. However, for the curious, a very brief indication of the main ideas will be given in the sketch below. For details, see an advanced text on dierential equations.sketch of the idea of the proof: A simple proof of existence of the solution is obtained by successive approximations. In this context, the method is known as Picard iteration. Set x0 (t) = c andt
  xi (t) = c +a
  It turns out that Lipschitz continuity implies that the mapping T dened byt
  T (y )(t) = c +a
  is a contraction mapping on a certain Banach space. It can then be shown, by using the Banach xed point theorem, that the sequence of Picard iterates xi is convergent and that the limit is a solution to the problem. The proof of uniqueness uses a result called Gr onwalls Lemma.
  18
  with the constant solution x(t) = 1. Computing the Picard iterates by hand is easy: t t x0 (t) = 1, x1 (t) = 1 + 0 1 x0 (s)) ds = 1, x2 (t) = 1 + 0 1 x1 (s)) ds = 1, and so on. Since each xi (t) = 1, we nd the solution x(t) = lim xi (t) = lim 1 = 1.i i
  We now try the Picard iteration method in Sage . Consider the IVP x = 1 x, which we considered earlier.Sage sage: var(t, s) sage: f = lambda t,x: 1-x sage: a = 0; c = 2 sage: x0 = lambda t: c; x0(t) 2 sage: x1 = lambda t: c + integral(f(s,x0(s)), s, 2 - t sage: x2 = lambda t: c + integral(f(s,x1(s)), s, t2/2 - t + 2 sage: x3 = lambda t: c + integral(f(s,x2(s)), s, -t3/6 + t2/2 - t + 2 sage: x4 = lambda t: c + integral(f(s,x3(s)), s, t4/24 - t3/6 + t2/2 - t + 2 sage: x5 = lambda t: c + integral(f(s,x4(s)), s, -t5/120 + t4/24 - t3/6 + t2/2 - t + 2 sage: x6 = lambda t: c + integral(f(s,x5(s)), s, t6/720 - t5/120 + t4/24 - t3/6 + t2/2 - t + sage: P1 = plot(x2(t), t, 0, 2, linestyle=--) sage: P2 = plot(x4(t), t, 0, 2, linestyle=-.) sage: P3 = plot(x6(t), t, 0, 2, linestyle=:) sage: P4 = plot(1+exp(-t), t, 0, 2) sage: (P1+P2+P3+P4).show()
  x(0) = 2,
  a, t); x1(t) a, t); x2(t) a, t); x3(t) a, t); x4(t) a, t); x5(t) a, t); x6(t) 2
  From the graph in Figure 1.5 you can see how well these iterates are (or at least appear to be) converging to the true solution x(t) = 1 + et . More generally, here is some Sage code for Picard iteration.Sage def picard_iteration(f, a, c, N): Computes the N-th Picard iterate for the IVP
  19
  x = f(t,x), x(a) = c. EXAMPLES: sage: var(x t s) (x, t, s) sage: a = 0; c = 2 sage: f = lambda t,x: 1-x sage: picard_iteration(f, a, c, 0) 2 sage: picard_iteration(f, a, c, 1) 2 - t sage: picard_iteration(f, a, c, 2) t2/2 - t + 2 sage: picard_iteration(f, a, c, 3) -t3/6 + t2/2 - t + 2 if N == 0: return c*t**0 if N == 1: x0 = lambda t: c + integral(f(s,c*s**0), s, a, t) return expand(x0(t)) for i in range(N): x_old = lambda s: picard_iteration(f, a, c, N-1).subs(t=s) x0 = lambda t: c + integral(f(s,x_old(s)), s, a, t) return expand(x0(t))
  20 Exercises:
  1. Compute the rst three Picard iterates for the IVP x = 1 + xt, using the initial function x0 (t) = 0. 2. Apply the Picard iteration method in Sage to the IVP x = (t + x)2 , and nd the rst three iterates. x(0) = 2, x(0) = 0
  1.3.2
  We begin by considering the second order2 initial value problem ax + bx + cx = 0, x(0) = d0 , x (0) = d1 , (1.3)
  where a, b, c, d0 , d1 are constants and a = 0. What conditions guarantee that a solution x = x(t) exists? If it exists, what (further) conditions guarantee that x = x(t) is unique? It turns out that no conditions are needed - a solution to 1.3 always exists and is unique. As we will see later, we can construct distinct explicit solutions, denoted x1 = x1 (t) and x2 = x2 (t) and sometimes called fundamental solutions, to ax + bx + cx = 0. If we let x = c1 x1 + c2 x2 , for any constants c1 and c2 , then we know that x is also a solution3 , sometimes called the general solution to ax + bx + cx = 0. But how do we know there exist c1 and c2 for which this general solution also satises the initial conditions x(0) = d0 and x (0) = d1 ? For this to hold, we need to be able to solve c1 x1 (0) + c2 x2 (0) = d1 , for c1 and c2 . By Cramers rule, d1 x2 (0) d2 x 2 (0) x1 (0) x2 (0) x 1 (0) x2 (0) x1 (0) d1 x 1 (0) d2 x1 (0) x2 (0) x 1 (0) x2 (0) c1 x 1 (0) + c2 x2 (0) = d2 ,
  c1 =
  c2 =
  For this solution to exist, the denominators in these quotients must be non-zero. This denominator is the value of the Wronskian [W-linear] at t = 0.2 It turns out that the reasoning in the second order case is very similar to the general reasoning for n-th order dierential equations. For simplicity of presentation, we restrict to the 2-nd order case. 3 This follows form the linearity assumption.
  21
  Denition 1.3.2. For n functions f1 , . . . , fn , which are n 1 times dierentiable on an interval I , the Wronskian W (f1 , ..., fn ) as a function on I is dened by f1 (x) (x) f1 . . . f1 for x I .(n1)
  W (f1 , . . . , fn )(x) =
  f2 (x) (x) f2 . . .(n1)
  .. .
  fn (x) (x) fn . . .(n1)
  (x) f2
  (x) fn
  (x)
  The matrix constructed by placing the functions in the rst row, the rst derivative of each function in the second row, and so on through the (n 1)-st derivative, is a square matrix sometimes called a fundamental matrix of the functions. The Wronskian is the determinant of the fundamental matrix. Theorem 1.3.3. (Abels identity) Consider a homogeneous linear second-order ordinary dierential equation dy d2 y + p(x) + q (x) y = 0 2 dx dx on the real line with a continuous function p. The Wronskian W of two solutions of the dierential equation satises the relationx
  P (s) ds .0
  Example 1.3.3. Consider x + 3x + 2x = 0. The Sage code below calculates the Wronskian of the two fundamental solutions et and e2t both directly and from Abels identity (Theorem 1.3.3).Sage sage: t = var("t") sage: x = function("x",t) sage: DE = diff(x,t,t)+3*diff(x,t)+2*x==0 sage: desolve(DE, [x,t]) k1*e(-t) + k2*e(-2*t) sage: Phi = matrix([[e(-t), e(-2*t)],[-e(-t), -2*e(-2*t)]]); Phi [ e(-t) e(-2*t)] [ -e(-t) -2*e(-2*t)] sage: W = det(Phi); W -e(-3*t) sage: Wt = e(-integral(3,t)); Wt e(-3*t) sage: W*W(t=0) == Wt e(-3*t) == e(-3*t) sage: bool(W*W(t=0) == Wt) True
  22
  Denition 1.3.3. We say n functions f1 , . . . , fn are linearly dependent over the interval I, if there are numbers a1 , , an (not all of them zero) such that a1 f1 (x) + + an fn (x) = 0, for x I . If the functions are not linearly dependent then they are called linearly independent. Theorem 1.3.4. If the Wronskian is non-zero at some point in an interval, then the associated functions are linearly independent on the interval. Example 1.3.4. If f1 (t) = et and f2 (t) = et then et et = 2. et et Indeed,Sage sage: var(t) t sage: f1 = exp(t); f2 = exp(-t) sage: wronskian(f1,f2) -2
  Therefore, the fundamental solutions x1 = et , x2 = et are linearly independent. Exercises: 1. Using Sage , verify Abels identity (b) in the example x + 2 x + 2x = 0. Determine what the existence and uniqueness theorem (Theorem 1.3.2 above) guarantees, if anything, about solutions to the following initial value problems. Note: that you do not have to nd the solutions. In some cases you may wish to use Theorem 1.1.1, which is a weaker result but it is easier to verify the assumptions. 2. dy/dx = xy , y (0) = 1. 3. dy/dx = y 1/3 , y (0) = 2. 4. dy/dx = y 1/3 , y (2) = 0. 5. dy/dx = x ln(y ), y (0) = 1. (a) in the example x x = 0,
  23
  1.41.4.1
  We know how to solve any ODE of the form y = f (t), at least in principle - just integrate both sides4 . For a more general type of ODE, such as y = f (t, y ), this fails. For instance, if y = t + y then integrating both sides gives y (t) = dy dt dt = t2 y dt = t + y dt = t dt + y (t) dt = 2 + y (t) dt. So, we have only succeeded in writing y (t) in terms of its integral. Not very helpful. However, there is a class of ODEs where this idea works, with some modication. If the ODE has the form y = then it is called separable5 . Strategy to solve a separable ODE: (1) write the ODE (1.4) asdy dt
  g(t) , h(y )
  (1.4)
  g (t) h(y ) ,
  (2) separate the ts and the y s: h(y ) dy = g(t) dt, (3) integrate both sides:
  h(y ) dy =
  g(t) dt + C
  (1.5)
  The +C is added to emphasize that a constant of integration must be included in your answer (but only on one side of the equation).4 , so by the fundamental theorem of calculus, y = dy dt = y dt = f (t) dt = Recall y really denotes dy dt dt F (t) + c, where F is the anti-derivative of f and c is a constant of integration. 5 It particular, any separable dierential equation must be a rst order, ordinary dierential equation.
  24
  The answer obtained in this manner is called an implicit solution of (1.4) since it expresses y implicitly as a function of t. Why does this work? It is easiest to understand by working backwards from the formula d h(y )dy = h(y ). (1.5). Recall that one form of the fundamental theorem of calculus is dy If we think of y as a being a function of t, and take the t-derivative, we can use the chain rule to get g(t) = d dt g(t)dt = d dt h(y )dy = ( d dy h(y )dy ) dy dy = h(y ) . dt dt
  So if we dierentiate both sides of equation (1.5) with respect to t, we recover the original dierential equation. Example 1.4.1. Are the following ODEs separable? If so, solve them. (a) (t2 + y 2 )y = 2ty , (b) y = x/y , y (0) = 1, (c) T = k (T Troom ), where k < 0 and Troom are constants, (d) ax + bx = c, where a = 0, b = 0, and c are constants (e) ax + bx = c, where a = 0, b, are constants and c = c(t) is not a constant. (f) y = (y 1)(y + 1), y (0) = 2. (g) y = y 2 + 1, y (0) = 1. Solutions: (a) not separable, (b) y dy = x dx, so y 2 /2 = x2 /2 + c, so x2 + y 2 = 2c. This is the general solution (note it does not give y explicitly as a function of x, you will have to solve for y algebraically to get that). The initial conditions say when x = 0, y = 1, so 2c = 02 + 12 = 1, which gives c = 1/2. Therefore, x2 + y 2 = 1, which is a circle. That is not a function so 2 2 cannot be the solution we want. The solution is either y = 1 x or y = 1 x , 2 but which one? Since y (0) = 1 (note the minus sign) it must be y = 1 x . (c) (d) = kdt, so ln |T Troom | = kt + c (some constant c), so T Troom = Cekt (some constant C ), so T = T (t) = Troom + Cekt .b dx c b b = (c bx)/a = a (x c c = dt, so ln |x b ), so x b a b | = a t + C , where C is a constant of integration. This is the implicit general solution of the dierential b t a , where B is a constant. equation. The explicit general solution is x = c b + Be dx dt dT T Troom
  25
  sage: a = var(a) sage: b = var(b) sage: c = var(c) sage: t = var(t) sage: x = function(x, t) sage: de = lambda y: a*diff(y,t) + b*y - c sage: desolve(de(x),[x,t]) (c*e(b*t/a)/b + c)*e(-b*t/a)
  Note: the constant functions y (t) = 1 and y (t) = 1 are also solutions to this differential equation. These solutions cannot be obtained (in an obvious way) from the general solution. The integral is easy to do using Sage :Sage sage: y = var(y) sage: integral(1/((y-1)*(y+1)),y) log(y - 1)/2 - (log(y + 1)/2)
  Now, lets try to get Sage to solve for y in terms of t in 1 2 (ln(y 1) ln(y + 1)) = t + C :Sage sage: C = var(C) sage: solve([log(y - 1)/2 - (log(y + 1)/2) == t+C],y) [log(y + 1) == -2*C + log(y - 1) - 2*t]
  This is not working. Lets try inputting the problem in a dierent form:Sage sage: C = var(C) sage: solve([log((y - 1)/(y + 1)) == 2*t+2*C],y) [y == (-e(2*C + 2*t) - 1)/(e(2*C + 2*t) - 1)]
  26
  CHAPTER 1. FIRST ORDER DIFFERENTIAL EQUATIONS This is what we want. Now lets assume the initial condition y (0) = 2 and solve for C and plot the function.Sage sage: solny=lambda t:(-e(2*C+2*t)-1)/(e(2*C+2*t)-1) sage: solve([solny(0) == 2],C) [C == log(-1/sqrt(3)), C == -log(3)/2] sage: C = -log(3)/2 sage: solny(t) (-e(2*t)/3 - 1)/(e(2*t)/3 - 1) sage: P = plot(solny(t), 0, 1/2) sage: show(P)
  This plot is shown in Figure 1.6. The solution has a singularity at t = ln(3)/2 = 0.5493....
  Figure 1.6: Plot of y = (y 1)(y + 1), y (0) = 2, for 0 < t < 1/2. (g) = dt so arctan(y ) = t + C , where C is a constant of integration. The initial condition y (0) = 1 says arctan(1) = C , so C = 4 . Therefore y = tan(t + 4 ) is the solution.dy y 2 +1
  1.4.2
  Autonomous ODEs
  A special subclass of separable ODEs is the class of autonomous ODEs, which have the form y = f (y ),
  27
  where f is a given function (i.e., the slope y only depends on the value of the dependent variable y ). The cases (c), (d), (f), and (g) above are examples. One of the simplest examples of an autonomous ODE is dy dt = ky , where k is a constant. We can divide by y and integrate to get 1 dy = log |y | = y After exponentiating, we get |y | = ekt+C1 = eC1 ekt . We can drop the absolute value if we allow positive and negative solutions: y = eC1 ekt . k dt = kt + C1 .
  Now note that eC1 can be any nonzero number, but in fact y = 0 is also a solution to the ODE so we can write the general solution as y = Cekt for an arbitrary constant C . If k is positive, solutions will grow in magnitude exponentially, and if k is negative solutions will decay to 0 exponentially. Perhaps the most famous use of this type of ODE is in carbon dating. Example 1.4.2. Carbon-14 has a half-life of about 5730 years, meaning that after that time one-half of a given amount will radioactively decay (into stable nitrogen-14). Prior to the nuclear tests of the 1950s, which raised the level of C-14 in the atmosphere, the ratio of C-14 to C-12 in the air, plants, and animals was 1015 . If this ratio is measured in an archeological sample of bone and found to be 1017 , how old is the sample? Solution: Since a constant fraction of C-14 decays per unit time, the amount of C-14 satises a dierential equation y = ky with solution y = Cekt . Since y (5730) = Cek5730 = y (0)/2 = C/2, we can compute k = log(2)/5730 1.21 105 . Let t0 denote the time of death of whatever the bone sample is from. We set our clock to start at 1950. We know that C-12 is stable, so if we devide the ratio of C-14 to C-12 at t = 0 to that ratio at t = t0 , the C-12 amounts will cancel. Therefore, we know that y (t0 )/y (0) = So t0 =log 102 k
  Here is a non-linear example. Example 1.4.3. Consider y = (y 1)(y + 1), Here is one way to solve this using Sage : y (0) = 1/2.
  28
  sage: t = var(t) sage: x = function(x, t) sage: de = lambda y: diff(y,t) == y2 - 1 sage: soln = desolve(de(x),[x,t]); soln 1/2*log(x(t) - 1) - 1/2*log(x(t) + 1) == c + t sage: # needs an abs. value ... sage: c,xt = var("c,xt") sage: solnxt = (1/2)*log(abs(xt - 1)) - (1/2)*log(abs(xt + 1)) == c + t sage: solve(solnxt.subs(t=0, xt=1/2),c) [c == -1/2*log(3/2) - 1/2*log(2)] sage: c0 = solve(solnxt.subs(t=0, xt=1/2),c)[0].rhs(); c0 -1/2*log(3/2) - 1/2*log(2) sage: soln0 = solnxt.subs(c=c0); soln0 1/2*log(abs(xt - 1)) - 1/2*log(abs(xt + 1)) == t - 1/2*log(3/2) -1/2*log(2) sage: implicit_plot(soln0,(t,-1/2,1/2),(xt,0,0.9))
  Sage cannot solve this (implicit) solution for x(t), though Im sure you can do it by hand if you want. The (implicit) plot is given in Figure 1.7.
  Figure 1.7: Plot of y = (y 1)(y + 1), y (0) = 1/2, for 1/2 < t < 1/2. A more complicated example is y = y (y 1)(y 2).
  29
  This has constant solutions y (t) = 0, y (t) = 1, and y (t) = 2. (Check this.) Several non-constant solutions are plotted in Figure 1.8.
  1.5
  0.5
  Figure 1.8: Plot of y = y (y 1)(y 2), y (0) = y0 , for 0 < t < 4, and various values of y0 . Exercise: 1. Find the general solution to y = y (y 1)(y 2) either by hand or using Sage . 2. Find the general solution to y = y 2 4 either by hand or using Sage . 3. x + x = 1, x(0) = 0.
  1.4.3
  Substitution methods
  Sometimes it is possible to solve dierential equations by substitution, just as some integrals can be simplied by substitution. Example 1.4.4. We will solve the ODE y = (y + x +3)2 with the substitution v = y + x +3. Dierentiating the substitution gives the relation v = y + 1, so the ODE can be rewritten as v 1 = v 2 , or v = v 2 + 1. This transformed ODE is separable, so we divide by v 2 + 1 and integrate to get arctan(v ) = x + C . Taking the tangent of both sides gives v = tan(x + C ) and now we substitute back to get y = tan(x + C ) x 3. There are a variety of very specialized substitution methods which we will not present in this text. However one class of dierential equations is common enough to warrant coverage here: homogeneous ODEs. Unfortunately, for historical reasons the word homogeneous has come to mean two dierent things. In the present context, we dene a homogeneous rstorder ODE to be one which can be written as y = f (y/x). For example, the following ODE is homogeneous in this sense: dy = x/y. dx
  30
  A rst-order homogeneous ODE can be simplied by using the substitution v = y/x, or equivalently y = vx. Dierentiating that relationship gives us v + xv = y . Example 1.4.5. We will solve the ODE y = y/x + 2 with the substitution v = y/x. As noted above, with this substituion y = v + xv so the original ODE becomes v + xv = v + 2. This simplies to v = 2/x which can be directly integrated to get v = 2 log(x) + C (in more dicult examples we would get a separable ODE after the substitution). Using the substitution once again we obtain y/x = 2 log(x) + C so the general solution is y = 2x log(x) + Cx.
  1.4.4
  The bottom line is that we want to solve any problem of the form x + p(t)x = q (t), (1.6)
  where p(t) and q (t) are given functions (which, lets assume, arent too horrible). Every rst order linear ODE can be written in this form. Examples of dierential equations which have this form: Falling Body problems, Newtons Law of Cooling problems, Mixing problems, certain simple Circuit problems, and so on. There are two approaches the formula, the method of integrating factors. Both lead to the exact same solution. The Formula: The general solution to (1.6) is x= ep(t) dt q (t) dt
  +C
  p(t) dt
  (1.7)
  where C is a constant. The factor e p(t) dt is called the integrating factor and is often denoted by . This formula was apparently rst discovered by Johann Bernoulli [F-1st]. Example 1.4.6. Solve xy + y = ex .1 y = We rewrite this as y + x gives ex x.
  Now compute = e
  1 x
  dx
  y=
  x ex dx + C = x
  ex dx + C ex + C = . x x
  31
  p(t) dt .
  x + p(t)x = q (t). The product rule implies that (x) = x + p(t)x = q (t). (In response to a question you are probably thinking now: No, this is not obvious. This is Bernoullis very clever idea.) Now just integrate both sides. By the fundamental theorem of calculus, x = (x) dt = q (t) dt.
  Dividing both side by gives (1.7). The integrating factor can be useful in organizing a computation. There are four main steps to solving a rst-order linear equation; the integral in the third step is usually the worst part: 1. Put the equation in the standard form x + p(t)x = q (t). 2. Compute = e 3. Computep(t) dt
  When solving initial value problems with x(t0 ) = x0 it can be convenient to do each of the integrations above from t0 to t, and then the constant C will be x0 . In other words, if x(t0 ) = x0 and x + p(t)x = q (t) then let (t) = e x=t t0t t0
  p(u) du
  and
  32
  Example 1.4.7. Solve the initial value problem t log(t)x = x + t(log(t))2 , x(e) = 1. First we add x to each side of this equation and divide by t log(t) to get it into standard form: x + So p(t) = 1 x = log(t). t log(t)
  1 and q (t) = log(t). t log(t) Next we compute the integrating factor. Since we are solving an initial value problem we will integrate from e to t: =e Now we needt tt 1 e u log(u)
  du
  = elog(log(t))log(log(e)) = log(t).
  (u)q (u) du =e e
  = t log(t) 2t +
  2t e + 1 . log(t)
  Exercises: Find the general solution to the following seperable dierential equations: 1. x = 2xt. 2. x x sin(t) = 0 3. (1 + t)x = 2x. 4. x xt x = 1 + t. 5. Find the solution y (x) to the following initial value problem: y = 2yex , y (0) = 2e2 . 6. Find the solution y (x) to the following initial value problem: y = x3 (y 2 +1), y (0) = 1. 7. Use the substitution v = y/x to solve the IVP: y =2xy , x 2 y 2
  y (0) = 2.
  Solve the following linear equations. Find the general solution if no initial condition is given. 8. x + 4x = 2te4t .
  33
  10. dy/dx + y tan(x) = sin(x). 11. xy = y + 2x, y (1) = 2. 12. y + 4y = 2xe4x , y (0) = 0. 13. y = cos(x) y cos(x), y (0) = 1. 14. In carbon-dating organic material it is assumed that the amount of carbon-14 (14 C ) 14 decays exponentially ( d dt C = k 14 C ) with rate constant of k 0.0001216 where t is measured in years. Suppose an archeological bone sample contains 1/7 as much carbon-14 as is in a recent sample (but which dates from before 1950). How old is the bone? 15. The function et does not have an anti-derivative in terms of elementary functions, but this anti-derivative is important in probability. Dene the error function by 2 erf(t) = Find the solution of xt 02
  eu du.
  16. (a) Use the command desolve in Sage to solve tx + 2x = et /t. (b) Use Sage to plot the solution to y = y 2 1, y (0) = 2.
  Recall from vector calculus the notion of a two-dimensional vector eld: F (x, y ) = (g(x, y ), h(x, y )). To plot F , you simply draw the vector F (x, y ) at each point (x, y ). The idea of the direction eld (or slope eld) associated to the rst order ODE y = f (x, y ), y (a) = c, (1.8)
  is similar. At each point (x, y ) you plot a small vector having slope f (x, y ). For example, the vector eld plot of F (x, y ) = (1, f (x, y )) or, better yet, F (x, y ) = (1, f (x, y ))/ 1 + f (x, y )2 (which is a unit vector) would work for the slope eld. How would you draw such a direction eld plot by hand? You could compute the value of f (x, y ) for lots and lots of points (x, y ) and then plot a tiny arrow of slope f (x, y ) at each of these points. Unfortunately, this would be impossible for a person to do on a practical level if the number of points was large.
  34
  For this reason, the notion of the isoclines of the ODE is very useful. An isocline of (1.8) is a level curve of the function z = f (x, y ): {(x, y ) | f (x, y ) = m},
  where the given constant m is called the slope of the isocline. In terms of the ODE, this curve represents the collection of all points (x, y ) at which the solution has slope m. In terms of the direction eld of the ODE, it represents the collection of points where the vectors have slope m. This means that once you have draw a single isocline, you can sketch ten or more tiny vectors describing your direction eld. Very useful indeed! This idea is recoded below more algorithmically. How to draw the direction eld of (1.8) by hand: Draw several isoclines, making sure to include one which contains the point (a, c). (You may want to draw these in pencil.) On each isocline, draw hatch marks or arrows along the line each having slope m. This is a crude direction eld plot. The plot of arrows form your direction eld. The isoclines, having served their usefulness, can now safely be ignored. Example 1.5.1. The direction eld, with three isoclines, for y = 5x + y 5, is given by the graph in Figure 1.9. y (0) = 1,
  Figure 1.9: Plot of y = 5x + y 5, y (0) = 1, for 1 < x < 1. The isoclines are the curves (coincidentally, lines) of the form 5x + y 5 = m. These are lines of slope 5, not to be confused with the fact that it represents an isocline of slope m.
  35
  The above example can be solved explicitly. (Indeed, y = 5x + ex solves y = 5x + y 5, y (0) = 1.) In the next example, such an explicit solution is not possible. Therefore, a numerical approximation plays a more important role. Example 1.5.2. The direction eld, with three isoclines, for y = x2 + y 2 , is given by the in Figure 1.10. y (0) = 3/2,
  Figure 1.10: Direction eld of y = x2 + y 2 , for 2 < x < 2. The isoclines are the concentric circles x2 + y 2 = m. The plot in Figure 1.10 was obtaining using the Sage code below.Sage sage: x,y = var("x,y") sage: f(x,y) = x2 + y2 sage: plot_slope_field(f(x,y), (x,-2,2),(y,-2,2)).show(aspect_ratio=1)
  There is also a way to manually draw these direction elds using Sage .Sage sage: pts = [(-2+i/5,-2+j/5) for i in range(20) \ for j in range(20)] # square [-2,2]x[-2,2] sage: f = lambda p:p[0]2+p[1]2 # x = p[0] and y = p[1] sage: arrows = [arrow(p, (p[0]+0.02,p[1]+(0.02)*f(p)), \ width=1/100, rgbcolor=(0,0,1)) for p in pts] sage: show(sum(arrows))
  36
  Figure 1.11: Direction eld for y = x2 + y 2 , y (0) = 3/2, for 2 < x < 2. The plot in Figure 1.12 was obtaining using the Sage code below (the Eulers method command is explained in the next section).Sage sage: sage: sage: sage: sage: sage: sage: sage: x,y = var("x,y") f(x,y) = x2 + y2 P1 = plot_slope_field(f(x,y), (x,-2,2),(y,-2,2)) from sage.calculus.desolvers import eulers_method x,y = PolynomialRing(RR,2,"xy").gens() pts = eulers_method(x2+y2,-1,0,1/10,1,algorithm="none") P2 = line(pts, xmin=-2,xmax=2,ymin=-2,ymax=2, thickness=2) show(P1+P2, aspect_ratio=1, dpi=300)
  1.5. ISOCLINES AND DIRECTION FIELDS Exercises: 1. Match the solution curves in Figure 1.13 to the ODEs below. I. y = y 2 1y t2
  37
  V. y = 2t + y VI. y = sin(3t) 2. Use Sage to plot the direction eld for y = x2 y 2 . 3. Using the direction eld in Figure 1.10 , sketch the solution to y = x2 + y 2 , y (0) = 3/2, for 2 < x < 2.
  38
  (a)
  (b)
  (c)
  (d)
  (e)
  (f)
  1.6
  Leonhard Euler was a Swiss mathematician who made signicant contributions to a wide range of mathematics and physics including calculus and celestial mechanics (see [Eu1-num] and [Eu2-num] for further details).
  1.6.1
  Eulers Method
  where f (x, y ) is some given function. We shall try to approximate the value of the solution at x = b, where b > a is given. Sometimes such a method is called numerically integrating (1.9). Note: the rst order dierential equation must be in the form (1.9) or the method described below does not work. A version of Eulers method for systems of 1-st order dierential equations and higher order dierential equations will also be described below. Geometric idea: The basic idea can be easily expressed in geometric terms. We know the solution, whatever it is, must go through the point (a, c) and we know, at that point, its slope is m = f (a, c). Using the point-slope form of a line, we conclude that the tangent line to the solution curve at (a, c) is (in (x, y )-coordinates, not to be confused with the dependent variable y and independent variable x of the dierential equation) y = c + (x a)f (a, c). (See the plot given in Figure 1.14.) In particular, if h > 0 is a given small number (called the increment) then taking x = a + h the tangent-line approximation from calculus I gives us: y (a + h) = c + h f (a, c). Now we know the solution passes through a point which is nearly equal to (a + h, c + h f (a, c). We now repeat this tangent-line approximation with (a, c) replaced by (a + h, c + h f (a, c). Keep repeating this number-crunching at x = a, x = a + h, x = a + 2h, ..., until you get to x = b.
  40
  Figure 1.14: Tangent line (solid line) of a solution (dotted curve) to a 1st order DE, (1.9). Algebraic idea: The basic idea can also be explained algebraically. Recall from the denition of the derivative in calculus 1 that y (x + h) y (x) , y (x) = h h > 0 is a given and small. This and the dierential equation together give f (x, y (x)) = y (x+h)y (x) . Now solve for y (x + h): h y (x + h) = y (x) + h f (x, y (x)). If we call h f (x, y (x)) the correction term (for lack of anything better), call y (x) the old value of y , and call y (x + h) the new value of y , then this approximation can be re-expressed ynew = yold + h f (x, yold ). Tabular idea: Let n > 0 be an integer, which we call the step size. This is related to the increment by ba . n This can be expressed simplest using a table. h= x a a+h a + 2h . . . b y c c + hf (a, c) . . . ??? hf (x, y ) hf (a, c) . . .
  xxx
  1.6. NUMERICAL SOLUTIONS - EULERS METHOD AND IMPROVED EULERS METHOD41 The goal is to ll out all the blanks of the table but the xxx entry and nd the ??? entry, which is the Eulers method approximation for y (b). Example 1.6.1. Use Eulers method with h = 1/2 to approximate y (1), where y y = 5x 5, y (0) = 1.
  Putting the dierential equation into the form (1.9), we see that here f (x, y ) = 5x + y 5, a = 0, c = 1. x 0 1/2 1 y 1 1 + (2) = 1 1 + (7/4) = 11/41 hf (x, y ) = 2 (5x + y 5) 2 7/4
  so y (1) = 11 4 = 2.75. This is the nal answer. Aside: For your information, y = ex 5x solves the dierential equation and y (1) = e 5 = 2.28.... You see from this that the numerical approxmation is not too far away from the actual value of the solution. Not bad, for only 2 steps! Here is one way to do this using Sage :Sage sage: x,y=PolynomialRing(QQ,2,"xy").gens() sage: eulers_method(5*x+y-5,1,1,1/3,2) x y h*f(x,y) 1 1 1/3 4/3 4/3 1 5/3 7/3 17/9 2 38/9 83/27 sage: eulers_method(5*x+y-5,0,1,1/2,1,method="none") [[0, 1], [1/2, -1], [1, -11/4], [3/2, -33/8]] sage: pts = eulers_method(5*x+y-5,0,1,1/2,1,method="none") sage: P = list_plot(pts) sage: show(P) sage: P = line(pts) sage: show(P) sage: P1 = list_plot(pts) sage: P2 = line(pts) sage: show(P1+P2)
  42
  1.6.2
  Geometric idea: The basic idea can be easily expressed in geometric terms. As in Eulers method, we know the solution must go through the point (a, c) and we know its slope there is m = f (a, c). If we went out one step using the tangent line approximation to the solution curve, the approximate slope to the tangent line at x = a + h, y = c + h f (a, c) would be m = f (a + h, c + h f (a, c)). The idea is that instead of using m = f (a, c) as the slope of the line to get our rst m approximation, use m+ 2 . The improved tangent-line approximation at (a, c) is: f (a, c) + f (a + h, c + h f (a, c)) m + m =c+h . y (a + h) =c+h 2 2 (This turns out to be a better approximation than the tangent-line approximation y (a + h) = c+hf (a, c) used in Eulers method.) Now we know the solution passes through a point which m is nearly equal to (a + h, c + h m+ 2 ). We now repeat this tangent-line approximation with (a, c) replaced by (a + h, c + h f (a, c). Keep repeating this number-crunching at x = a, x = a + h, x = a + 2h, ..., until you get to x = b. Tabular idea: The integer step size n > 0 is related to the increment by h= ba , n
  as before. The improved Eulers method can be expressed simplest using a table.
  x a a+h a + 2h . . . b
  The goal is to ll out all the blanks of the table but the xxx entry and nd the ??? entry, which is the improved Eulers method approximation for y (b). Example 1.6.2. Use the improved Eulers method with h = 1/2 to approximate y (1), where y y = 5x 5, y (0) = 1.
  Putting the dierential equation into the form (1.9), we see that here f (x, y ) = 5x + y 5, a = 0, c = 1. We rst compute the correction term:h,y +hf (x,y )) h f (x,y)+f (x+2
  = = = =
  + y 5 + 5(x + h) + (y + h f (x, y )) 5) + y 5 + 5(x + h) + (y + h (5x + y 5) 5) h 5 ((1 + h 2 )5x + (1 + 2 )y 2 )/2 (25x/4 + 5y/4 5)/2.y 20 m = 25x+5 h m+ 2 8 15/8 95/64
  1 4 (5x 1 4 (5x
  x 0 1/2 1
  so y (1) = 151 64 = 2.35... This is the nal answer. Aside: For your information, this is closer to the exact value y (1) = e 5 = 2.28... than the usual Eulers method approximation of 2.75 we obtained above. Here is one way to do this using Sage :Sage def improved_eulers_method(f,x0,y0,h,x1): print "%10s %30s %50s"%("x","y","(h/2)*(f(x,y)+f(x+h,y+h*f(x,y))") n=((RealField(max(6,RR.precision()))(1.0))*(x1-x0)/h).integer_part() x00=x0; y00=y0 for i in range(n+ZZ(1)): s1 = f(x00,y00) s2= f(x00+h,y00+h*f(x00,y00)) corr = h*(ZZ(1)/ZZ(2))*(s1+s2)
  44
  This implements Improved Eulers method for nding numerically the solution of the 1st order ODE y = f (x, y ), y (a) = c. The x-column of the table increments from x = 0 to x = 1 by h. In the y -column, the new y -value equals the old y -value plus the corresponding entry in the last column.Sage sage: sage: sage: x 0 1/2 1 sage: sage: x 0 1/2 1 RR = RealField(sci_not=0, prec=4, rnd=RNDU) x,y = PolynomialRing(RR,2).gens() improved_eulers_method(5*x+y-5,0,1,1/2,1) y 1 -0.875 -2.25 x,y=PolynomialRing(QQ,2).gens() improved_eulers_method(5*x+y-5,0,1,1/2,1) y 1 -7/8 -151/64
  Example 1.6.3. Use both Eulers method and improved Eulers method with h = 1/2 to approximate y (2), where y = xy, x 1 3/2 2 y 1 3/2 21/8 hf (x, y ) = xy/2 1/2 9/8 xxx y 1 29/16 261/64 y (1) = 1.
  h(f (x, y ) + f (x + h, y + hf (x, y ))/2) = 9xy/16 + y/8 + x2 y/8 13/16 145/64 xxx
  These give, for Eulers method y (2) 21/8 = 2.625, for improved Eulers method y (2) 2 261/64 = 4.078..., and the exact solution y (2) = e3/2 = 4.4816... (based on y = ex /21/2 , which the reader is asked to verify for him/herself). Clearly, improved Eulers method gives the better approximation in this case.
  1.6.3
  We only sketch the idea in some simple cases. Consider the dierential equation y + p(x)y + q (x)y = f (x), y (a) = e1 , y (a) = e2 ,
  1.6. NUMERICAL SOLUTIONS - EULERS METHOD AND IMPROVED EULERS METHOD45 and the system = f (x, y , y ), y (a) = c , y1 1 1 2 1 1 = f (x, y , y ), y (a) = c . y2 2 1 2 2 2
  We can treat both cases after rst rewriting the 2nd order dierential equation as a system: create new variables y1 = y and let y2 = y . It is easy to see that =y , y1 (a) = e1 , y1 2 y2 = f (x) q (x)y1 p(x)y2 , y2 (a) = e2 .
  Tabular idea: Let n > 0 be an integer, which we call the step size. This is related to the increment by ba . n This can be expressed simplest using a table. h= x a a+h a + 2h . . . b y1 e1 e1 + hf1 (a, e1 , e2 ) . . . ??? hf1 (x, y1 , y2 ) hf1 (a, e1 , e2 ) . . . y2 e2 e2 + hf2 (a, e1 , e2 ) hf2 (x, y1 , y2 ) hf2 (a, e1 , e2 ) . . .
  The goal is to ll out all the blanks of the table but the xxx entry and nd the ??? entry, which is the Eulers method approximation for y (b). Example 1.6.4. Using 3 steps of Eulers method, estimate x(1), where x 3x + 2x = 1, x(0) = 0, x (0) = 1 First, we rewrite x 3x + 2x = 1, x(0) = 0, x (0) = 1, as a system of 1st order dierential equations with initial conditions. Let x1 = x, x2 = x , so x1 (0) = 0, x 1 = x2 , x = 1 2 x + 3 x , x 1 2 2 (0) = 1. 2 This is the dierential equation rewritten as a system in standard form. (In general, the tabular method applies to any system but it must be in standard form.) Taking h = (1 0)/3 = 1/3, we have t 0 1/3 2/3 1 x1 0 1/3 10/9 73/27 x2 /3 1/3 7/9 43/27 xxx x2 1 7/3 43/9 xxx (1 2x1 + 3x2 )/3 4/3 22/9 xxx xxx
  46
  So x(1) = x1 (1) 73/27 = 2.7.... Here is one way to do this using Sage :Sage sage: RR = RealField(sci_not=0, prec=4, rnd=RNDU) sage: t, x, y = PolynomialRing(RR,3,"txy").gens() sage: f = y; g = 1-2*x+3*y sage: L = eulers_method_2x2(f,g,0,0,1,1/3,1,method="none") sage: L [[0, 0, 1], [1/3, 0.35, 2.5], [2/3, 1.3, 5.5], [1, 3.3, 12], [4/3, 8.0, 24]] sage: eulers_method_2x2(f,g, 0, 0, 1, 1/3, 1) t x h*f(t,x,y) y h*g(t,x,y) 0 0 0.35 1 1.4 1/3 0.35 0.88 2.5 2.8 2/3 1.3 2.0 5.5 6.5 1 3.3 4.5 12 11 sage: P1 = list_plot([[p[0],p[1]] for p in L]) sage: P2 = line([[p[0],p[1]] for p in L]) sage: show(P1+P2)
  Figure 1.16: Eulers method with h = 1/3 for x 3x + 2x = 1, x(0) = 0, x (0) = 1. Next, we give a numerical version of Example 3.3.3. Example 1.6.5. Consider x = 4y, x(0) = 400, y = x, y (0) = 100.
  1.6. NUMERICAL SOLUTIONS - EULERS METHOD AND IMPROVED EULERS METHOD47 Use 2 steps of Eulers method to approximate x(1/2) and y (1/2). t 0 1/4 1/2 x 400 300 300 hf (t, x, y ) = 2y -100 0 75 y 100 0 -75 hg (t, x, y ) = x/2 -100 -75 -75
  This computation of the approximations x(1/2) 300, y (1/2) 75, suggests that the y -men lose sometime between t = 0.25 and t = 0.5. This suggestion is further supported by the following more rened approximation:Sage sage: eulers_method_2x2(f,g, 0, 400, 100, 1/10, 1/2) t x h*f(t,x,y) 0 400 -40.00 1/10 360.0 -24.00 1/5 336.0 -9.600 3/10 326.4 3.840 2/5 330.2 16.90 1/2 347.1 30.11 -75.27 -34.71
  The Example 3.3.3 given in a later chapter will tell us the exact answer. Exercises: 1. (a) Use Eulers method to estimate x(1) if x(0) = 1 and 4 steps.dx dt
  (b) Find the exact value of x(1) by solving the ODE (hint: it is a linear ODE). (c) Use improved Eulers method to estimate x(1) if x(0) = 1 and 1,2, and 4 steps.dx dt
  = x + t2 , using
  2. Use Sage and Eulers method with h = 1/3 to nd the approximate values of x(1) and y (1), where x = x + y + t, x(0) = 0, y = x y, y (0) = 0, 3. Use 2 steps of improved Euler to nd the approximate value of x(1) where x = x2 + t2 , x(0) = 1. 4. Use Eulers method with h = 1/10 to nd the approximate value of x(1) where x + x = 1, x(0) = 1, x (0) = 1.
  48
  1.7
  The methods of 1.6 are sucient for computing the solutions of many problems, but often we are given dicult cases that require more sophisticated methods. One class of methods are called the Runge-Kutta methods, particularly the fourth-order method of that class since it achieves a popular balance of eciency and simplicity. Another class, the multistep methods, use information from some number m of previous steps. Within that class, we will briey describe the Adams-Bashforth method. Finally, we will say a little bit about adaptive step sizes - i.e. changing h adaptively depending on some local estimate of the error.
  1.7.1
  To explain why we describe the method as fourth-order it is necessary to introduce a convenient (big-O) notation for the size of the errors in our approximations. We say that f (x) = O(g(x)), as x 0 if there exist a positive constants M and x0 such that for all x [x0 , x0 ] we have |f (x)| M |g(x)|. The constant M is called the implied constant. This notation is also commonly used for the case when x but in this text we will always be considering the behavior of f (x) near x = 0. For example, sin(x) = O(x), as x 0, but sin(x) = O(1), as x . If g(x) is asymptotic to 0 as x 0, then, roughly speaking, a function f (x) is O(g(x)) if f approaches 0 at a rate equal to or faster than g(x). As another example, the function f (x) = 3x2 + 6x4 is O(x2 ) (as x 0). As we approach x = 0, the higher order terms become less and less important, and eventually the 3x2 term dominates the value of the function. There are many concrete choices of x0 and the implied constant M . One choice would be x0 = 1 and M = 9. For a numerical method we are interested in how fast the error decreases as we reduce the stepsize. By the error we mean the global truncation error: for the problem of approximating y (xf ) if y = f (x, y ) and y (xi ) = yi the global truncation error is dened as E (h) = |y (xf ) yn |. Here yn is our approximate value for y at x = xf after taking n x x steps of stepsize h = f n i . The global truncation error ignores the rounding error that is inevitable using oatingpoint arithmetic. We ignore rounding error for two reasons: it is usually not a big problem, and it is implementation-specic so it is hard to make general statements about it. Rounding error increases with the number of steps taken, and can be important if there are nearcancellations between terms of the slope function. For Eulers method, E (h) = O(h) and we say that it is a rst-order method. This means that as we decrease the stepsize h, at some point our error will become linear in h. In other words, we expect that if we halve the stepsize our error will be reduced by half. The improved Euler method is a second-order method, so E (h) = O(h2 ). This is very good news, because while the improved Euler method involves roughly twice as much work per step as the Euler method, the error will eventually fall quadratically in h.
  49
  The fourth-order Runge-Kutta method involves computing four slopes and taking a weighted average of them. We denote these slopes as k1 , k2 , k3 , and k4 , and the formula are: xn+1 = yn+1 = where k1 = f (xn , yn ), k2 = f (xn + h/2, yn + hk1 /2), k3 = f (xn + h/2), yn + hk2 /2), and k4 = f (xn + h, yn + hk3 ). xn + h, yn + h(k1 + 2k2 + 2k3 + k4 )/6,
  y (y x) Example 1.7.1. Lets consider the IVP y = x (y +x) , y (1) = 1, and suppose we wish to approximate y (2). The table below shows the Euler, improved Euler, and fourth-order Runge-Kutta (RK4) approximations for various numbers of steps from 1 to 512.
  Euler 1.0 0.933333333333 0.90307164531 0.889320511452 0.882877913323 0.879775715551 0.878255683243 0.877503588678 0.877129540678 0.876943018826
  imp. Euler 0.916666666667 0.889141488073 0.880183944727 0.877654079757 0.87698599324 0.876814710289 0.876771374145 0.876760476927 0.876757744807 0.876757060805
  RK4 0.878680484793 0.876938215214 0.876770226006 0.876757721415 0.87675688939 0.876756836198 0.876756832844 0.876756832634 0.876756832621 0.876756832620
  The nal Runge-Kutta value is correct to the number of digits shown. Note that even after 512 steps, Eulers method has not acheived the accuracy of 4 steps of Runge-Kutta or 8 steps of the improved Euler method.
  50
  1.7.2
  The idea of multistep methods is to use a sequence of solution or slope values to extrapolate to the next value. One of the most popular is the Adams-Bashforth method. To start such methods the rst few values must be found by a some other means, such as a Runge-Kutta method. The fourth-order Adams-Bashforth method is:
  (1.10) (1.11)
  *1 Write a Sage function that implements the Adams-Bashforth method. To prime the method it needs to generate four initial solution values, for which you should use the fourth-order Runge-Kutta method. Compare the accuracy of your function at x = to the fourth-order Runge-Kutta method for the IVP y = y + sin(x) by rst computing the exact value of y ( ). y (0) = 1/2
  1.7.3
  In our discussion of numerical methods we have only considered a xed step size. In some applications this is sucient but usually it is better to adaptively change the stepsize to keep the local error below some tolerance. One approach for doing this is to use two dierent methods for each step, and if the methods dier by more than the tolerance we decrease the stepsize. The Sage code below implements this for the improved Euler method and the fourth-order Runge-Kutta method.Sage def RK24(xstart, ystart, xfinish, f, nsteps = 10, tol = 10(-5.0)): Simple adaptive step-size routine. This compares the improved Euler method and the fourth-order Runge-Kutta method to estimate the error. EXAMPLE: The exact solution to this IVP is y(x) = exp(x), so y(1) should equal e = 2.718281828... Initially the stepsize is 1/10 but this is decreased during the calculation: sage: esol = RK24(0.0,1.0,1.0,lambda x,y: y)
  51
  More sophisticated implementations will also increase the stepsize when the error stays small for several steps. A very popular scheme of this type is the Runge-Kutta-Fehlberg method, which combines fourth- and fth-order Runge-Kutta methods in a particularly ecient way [A-ode]. Exercises: 1. For the initial value problem y (1) = 1 and a: (a) 2-step improved Euler method (b) 1-step 4th-order Runge-Kutta method. 2. Use Eulers method and the 4th-order Runge-Kutta method to estimate x(1) if x(0) = 2 1 and dx dt = x + t , using 2 steps. For this question you can use a calculator but you should write out the steps explicitly. 3. Use all three of the numerical methods (Euler, Improved Euler, and 4th-order RungeKutta) in Sage to nd the value of y (1) to four digits of accuracy using the smallest possible number of steps if y = xy y 3 and y (0) = 1. 4. Let sinc(x) = sin(x) . xdy dx
  y x
  y x x +y
  52
  CHAPTER 1. FIRST ORDER DIFFERENTIAL EQUATIONS Compare the results of using the fourth-order Runge-Kutta and Adams-Bashforth dy = sinc(x) y , y (0) = 1 for 5, 10, and methods to approximate y (3) for the IVP dx 100 steps. Use the Runge-Kutta values to prime the Adams-Bashforth method. 5* Modify the adaptive stepsize code so that the stepsize is increased if the estimated error is below some threshold. Pick an initial value problem and time your code compared to the original version. 6* Sometimes we wish to model quantities that are subjected to random inuences, as in Brownian motion or in nancial markets. If the random inuences change on arbitrarily small timescales, the result is usually nondierentiable. In these cases, it is better to express the evolution in terms of integrals, but they are often called stochastic dierential equations nonetheless. One example from nancial markets is a model of the value of a stock, which in integral form is:t t
  S (t) = S0 +0
  S ds +0
  S dW,
  where W is a Brownian motion. For such a model there is a stochastic analogue of Eulers method called the Euler-Maruyama method (see [H-sde] for more details and references). The main subtlety is correctly scaling the increment of the Brownian motion with the stepsize: dW = dt w0,1 , where w0,1 is a sample from a normal distribution with mean 0 and standard deviation 1. So for this example, the EulerMaruyama method gives: Si+1 = Si + Si h + hSi w0,1 . Implement the Euler-Maruyama method to simulate an IVP of this example with = 2 and = 1 from t = 0 to t = 1 with 100 steps, with S (0) = S0 = 1. To generate the w0,1 in Sage you can use the normalvariate command. Compute the average trajectory of 100 simulation runs - is it equal to the deterministic ODE S = S ? (This is a reasonable guess since the expected value of the Brownian motion W is 0.)
  1.8
  Newtonian mechanics
  We briey recall how the physics of the falling body problem leads naturally to a dierential equation (this was already mentioned in the introduction and forms a part of Newtonian mechanics [M-mech]). Consider a mass m falling due to gravity. We orient coordinates to that downward is positive. Let x(t) denote the distance the mass has fallen at time t and v (t) its velocity at time t. We assume only two forces act: the force due to gravity, Fgrav , and the force due to air resistence, Fres . In other words, we assume that the total force is given by Ftotal = Fgrav + Fres .
  53
  We know that Fgrav = mg, where g > 0 is the gravitational constant, from high school physics. We assume, as is common in physics, that air resistance is proportional to velocity: Fres = kv = kx (t), where k 0 is a constant. Newtons second law [N-mech] tells us that Ftotal = ma = mx (t). Putting these all together gives mx (t) = mg kx (t), or k v (t) = g. (1.12) m This is the dierential equation governing the motion of a falling body. Equation (1.12) can be solved by various methods: separation of variables or by integrating factors. If we assume v (0) = v0 is given and if we assume k > 0 then the solution is v (t) + v (t) = mg kt/m mg + (v0 )e . k kmg k .
  (1.13)
  Example 1.8.1. Wile E. Coyote (see [W-mech] if you havent seen him before) has mass 100 kgs (with chute). The chute is released 30 seconds after the jump from a height of 2000 m. The force due to air resistence is given by Fres = kv , where k= Find (a) the distance and velocity functions during the time when the chute is closed (i.e., 0 t 30 seconds), (b) the distance and velocity functions during the time when the chute is open (i.e., 30 t seconds), (c) the time of landing, (d) the velocity of landing. (Does Wile E. Coyote survive the impact?) soln: Taking m = 100, g = 9.8, k = 15 and v (0) = 0 in (1.13), we nd 196 196 3 t e 20 . 3 3 This is the velocity with the time t starting the moment the parachutist jumps. After t = 30 196 9/2 = 64.607.... The distance fallen is seconds, this reaches the velocity v0 = 196 3 3 e v1 ( t ) = x1 (t) = =t 0 v1 (u) du 3 t 3920 20 196 3 t+ 9 e
  3920 9 .
  9/2 = 1529.283... meters. + 3920 After 30 seconds, it has fallen x1 (30) = 13720 9 9 e Taking m = 100, g = 9.8, k = 100 and v (0) = v0 , we nd
  54
  v2 (t) =
  49 + et 5
  This is the velocity with the time t starting the moment Wile E. Coyote opens his chute (i.e., 30 seconds after jumping). The distance fallen is x2 (t) = =t 0 v2 (u) du + x1 (30) 833 t 49 t 9/2 + 196 5 t 15 e 3 e e 71099 45 3332 9
  e9/2 .
  This solves for the velocity before the coyotes chute is opened, 0 < t < 30. The last number is the velocity Wile E. Coyote is traveling at the moment he opens his chute.Sage sage: t = var(t) sage: v = function(v, t) sage: m = 100; g = 98/10; k = 100 sage: de = lambda v: m*diff(v,t) + k*v - m*g sage: desolve(de(v),[v,t],[0,RR(soln1(30))]) 1/10470*(102606*et + 573835)*e(-t) sage: soln2 = lambda t: 1/10470*(102606*et + 573835)*e(-t) sage: RR(soln2(0)) 64.607545367718 sage: RR(soln1(30)) 64.607545559502 sage: P2 = plot(soln2(t),30,50,plot_points=1000) sage: show(P1+P2, dpi=300)
  This solves for the velocity after the coyotes chute is opened, t > 30. The last command
  55
  plots the velocity functions together as a single plot6 . The terms at the end of soln2 were added to insure x2 (30) = x1 (30). Next, we nd the distance traveled at time t:Sage sage: x1 = integral(soln1(t),t) - 3920/9 sage: x1(t=0) 0 sage: RR(x1(30)) 1529.2830296034
  This solves for the distance the coyote traveled before the chute was open, 0 < t < 30. The last number says that he has gone about 1965 meters when he opens his chute.Sage sage: x2 = integral(soln2(t),t)+114767/2094 sage: x2(t=0) 0 sage: RR(x2(t=42.44)+x1(t=30)) 2000.0025749711 sage: P4 = plot(x2(t),30,50) sage: show(P3+P4)
  (Again, you see a break in the graph because of the round-o error.) The terms at the end of x2 were added to insure x2 (30) = x1 (30). You know he is close to the ground at t = 30, and going quite fast (about 65 m/s!). It makes sense that he will hit the ground soon afterwards (with a large pu of smoke, if youve seen the cartoons), even though his chute will have slowed him down somewhat. The graph of the velocity 0 < t < 50 is in Figure 1.17. Notice how it drops at t = 30 when the chute is opened. The time of impact is timpact = 42.4.... This was found numerically by solving x2 (t) = 2000 x1(30). The velocity of impact is v2 (timpact ) 9.8 m/s. Exercise: Drop an object with mass 10 kgs from a height of 2000 m. Suppose the force due to air resistence is given by Fres = 10v . Find the velocity after 10 seconds using Sage . Plot this velocity function for 0 < t < 10.6 You would see a break in the graph if you omitted the Sage s plot option plot points=1000. That is because the number of samples taken of the function by default is not sucient to capture the jump the function takes at t = 30.
  56
  1.9
  Suppose that we have two chemical substances where one is soluable in the other, such as salt and water. Suppose that we have a tank containing a mixture of these substances, and the mixture of them is poured in and the resulting well-mixed solution pours out through a value at the bottom. (The term well-mixed is used to indicate that the uid being poured in is assumed to instantly dissolve into a homogeneous mixture the moment it goes into the tank.) The rough idea is depicted in Figure 1.19.
  Assume for concreteness that the chemical substances are salt and water. Let
  57
  Figure 1.19: Solution pours into a tank, mixes with another type of solution. and then pours out. A(t) denote the amount of salt at time t, FlowRateIn = the rate at which the solution pours into the tank, FlowRateOut = the rate at which the mixture pours out of the tank, Cin = concentration in = the concentration of salt in the solution being poured into the tank, Cout = concentration out = the concentration of salt in the solution being poured out of the tank, Rin = rate at which the salt is being poured into the tank = (FlowRateIn)(Cin ), Rout = rate at which the salt is being poured out of the tank = (FlowRateOut)(Cout ). Remark 1.9.1. Some things to make note of: If FlowRateIn = FlowRateOut then the water level of the tank stays the same. We can determine Cout as a function of other quantities: Cout = A(t) , T (t)
  58
  CHAPTER 1. FIRST ORDER DIFFERENTIAL EQUATIONS The rate of change of the amount of salt in the tank, A (t), more properly could be called the net rate of change. If you think if it this way then you see A (t) = Rin Rout .
  Now the dierential equation for the amount of salt arises from the above equations: A (t) = (FlowRateIn)Cin (FlowRateOut) A(t) . T (t)
  Example 1.9.1. Consider a tank with 200 liters of salt-water solution, 30 grams of which is salt. Pouring into the tank is a brine solution at a rate of 4 liters/minute and with a concentration of 1 grams per liter. The well-mixed solution pours out at a rate of 5 liters/minute. Find the amount of salt at time t. We know A (t) = (FlowRateIn)Cin (FlowRateOut) A(t) A(t) =45 , T (t) 200 t A(0) = 30.
  where = e p(t) dt = e5 200t dt = (200 t)5 is the integrating factor. This gives A(t) = 200 t + C (200 t)5 , where the initial condition implies C = 170 2005 . Here is one way to do this using Sage :Sage sage: t = var(t) sage: A = function(A, t) sage: de = lambda A: diff(A,t) + (5/(200-t))*A - 4 sage: desolve(de(A),[A,t]) (t - 200)5*(c - 1/(t - 200)4)
  This is the form of the general solution. Let us now solve this general solution for c, using the initial conditions.Sage sage: c,t = var(c,t) sage: tank = lambda t: 200-t sage: solnA = lambda t: (c + 1/tank(t)4)*tank(t)5 sage: solnA(t) (c - (1/(t - 200)4))*(t - 200)5 sage: solnA(0) -320000000000*(c - 1/1600000000)
  59
  Figure 1.20: A(t), 0 < t < 200, A = 4 5A(t)/(200 t), A(0) = 30.
  Exercises: 1. Now use Sage to solve the same problem but with the same ow rate out as 4 liters/min (note: in this case, the water level in the tank is constant). Find and plot the solution A(t), 0 < t < 200. 2. Consider a tank containing 1000 liters (L) of brine with 100 kilograms (kg) of salt dissolved. Pure water is added to the tank at a rate of 10 L/s, and stirred mixture is drained out at a rate of 10 L/s. Find the time at which only 1 kg of salt is left in the tank. 3. Consider two water tanks that are linked in a cascade - i.e. the rst tank empties into the second. Suppose the rst tank has 100 liters of water in it, and the second has 300 liters of water. Each tank initially has 50 kilograms of salt dissolved in the water. Suppose that pure water ows into the rst tank at 5 liters per minute, well-mixed water ows from the rst tank to the second at the same rate (5 liters/minute), and well-mixed water also ows out of the second tank at 5 liters/minute.
  60
  CHAPTER 1. FIRST ORDER DIFFERENTIAL EQUATIONS (a) Find the amount of salt in the rst tank x1 (t). Note that this does not depend on what is happening in the second tank. (b) Find the amount of salt in the second tank x2 (t). (c) Find the time when there is the maximum amount of salt in the second tank. 4. A 120 gallon tank initially contains 90 lb of salt dissolved in 90 gallons of water. Brine containing 2 lb/gal of salt ows into the tank at the rate of 4 gal/min and the wellstirred mixture ows out of the tank at the rate of 3 gal/min. A dierential equation for the amount x (in pounds) of salt in the tank is (a)1 dx dt + 30 x
  =8
  dx 3 dt + 90+2t x
  =4
  dx 3 dt + 90+t x
  dx dt +90x
  = 2.. At what time does the tank contain the largest amount of salt? How much salt is in the tank at that time?. (a) Find the amount, in pounds, of salt in the tank as a function of time. (b) At what time does the tank contain the largest amount of salt? 7. Curly, Moe, and Larry are incompetant butlers for Day Duck, who orders then to prepare his bath. The bathtub holds 100 gallons when lled. Curly begins pouring in a soap water solution at a rate of 1/2 gallon/minute. The soap concentration is 1/4 lb/gallon in his soap water solution. Larry, in charge of putting the plug in the drain, screwed up and now the bath water is pouring down the drain at a rate of 1/2 gallon/minute. Originally, Moe lled the bath full of pure water. (a) What is the dierential equation modeling the mixing problem for the amount A(t) of soap at time t? (b) Find the amount (in lbs) of soap in the bathtub after 200 minutes.
  1.10
  Place an object, such as a cup of coee, in a room. How does its temperature behave? Convection is described by Newtons law of cooling, which states that the rate of change of the temperature of an object is proportional to its relative temperature (i.e., the dierence in temperatures between the object and the ambient temperature of its surroundings).
  If
  61
  Figure 1.21: How fast does a cup of coee cool? T = T (t) is the temperature of an object at time t, Troom is the ambient temperature of the objects surroundings, then T (t) = k (T Troom ), (1.14)
  where k is a constant of proportionality. If Troom is a constant then the solution has the form T (t) = Troom + Aekt , (1.15)
  where A is a constant. This can be derived by the method of separation of variables. It can also be veried directly by plugging (1.15) into (1.14). Example 1.10.1. Sherlock Holmes is awoken by a phone call from a policeman at 3:30am. A body has been discovered and foul play is suspected. Sherlock tells the police to determine the temperature of the body and, when he arrives at the scene 45 minutes later, he takes the temperature again. The two readings that cold 60 degree F morning were 80 degrees F and 70 degrees F.
  First, write down the dierential equation governing the temperature: T (t) = k (T 60), since Troom = 60. Next, we nd the time of death7 . Here is one way to do this using Sage :7
  62
  Exercises:
  63
  2. The following is a somewhat more general heating and cooling DE that includes additional heating and air conditioning of a building. dT = K [M (t) T (t)] + H (t) + U (t). dt (1.16)
  (a) Suppose that M (t) = 75 + 20 sin(t/12), K = 2, H (t) = 75 , U (t) = 0. By use of Sage , nd and plot solutions, where the initial condition is (a) T (0) = 60 , (b) T (0) = 70 , (c) T (0) = 80 over a 24 hour period. (b) What do you observe about the eect of the initial condition on the solution as time progresses? (c) Solve (1.16) explicitly and determine what factor in the solution might account for this behavior? (d) Let T (0) = 60 and let the other values stay xed except let K = 2, 4, 6. Graph the three solutions for these three K values. Can you describe the eect of the time constant K has on the solutions? 3. A cup of coee at 120o F is placed in a room of constant temperature 70o F. After 10 minutes it cools to 80o F. (a) What is the ODE modeling the temperature of the coee at time t? (b) How long did it take to cool to 100o F? (Leave your answer in symbolic form.) 4. Sherlock Holmes is awoken by a phone call from a policeman at 4am. A body has been discovered and foul play is suspected. Sherlock tells the police to determine the temperature of the body and, when he arrives at the scene 30 minutes later, he takes the temperature again. The two readings that cold 50 degree F morning were 90 degrees F and 85 degrees F. Find the time of death (use 98.6 degrees F for the normal body temperature). 5. A glass of cola at 50o F is placed in a room of constant temperature 70o F. After 10 minutes it warms to 60o F. (a) What is the dierential equation modeling the temperature of the coee at time t? (b) How long did it take to cool to 65o F?
  64
  Chapter 2
  2.12.1.1
  The problem we want to address in this section is to solve a dierential equation of the form an x(n) + + a1 x + a0 x = 0, where the ai s are given constants and x = x(t) is our unknown dependent variable. Such a dierential equation expression is homogeneous (since there is no term depending only on the independent variable t), linear (since we are not operating on the dependent variable y except in legal ways), constant coecient (since the coecients of the derivatives of x are all constants, i.e., not depending on t) dierential equation. Remark 2.1.1. Sometimes it is convenient to rewrite the above dierential equation using symbolic notation: replace by x Dx x D 2 x For example, the non-homogeneous constant coecient linear second x D3 x x(n) Dnx order dierential equation 65
  66
  x + 2x = 7x = sin(t), would be written in symbolic notation as (D 2 + 2D + 7)x = D 2 x + 2Dx + 7x = sin(t). To solve this dierential equation, please memorize these rules: real r ert complex i et cos(t), et sin(t) repeated roots repeated solutions, boosted each time by t The solution process: Find the characteristic polynomial of the equation. Find the roots of the polynomial. Construct the solution by interpreting the root according to the rules above. Example 2.1.1. Find the general solution to y 2y 3y = 0. solution: The characteristic polynomial is r 3 2r 2 3r . Factor this into r (r + 1)(r 3), so the roots are 0, 1, 3. The solution is: y = c1 e0t + c2 e1t + c3 e3 t = c1 + c2 et + c3 e3t . The following fact is called the fundamental theorem of algebra. It is not easy to prove. The rst proof was given by Carl F. Gauss in the 1800s. Theorem 2.1.1. A polynomial of degree n has n rootsif you count complex roots and repeated roots. Each of these special cases gets special handling by these rules. This also means you should always end up with n independent solutions to an nth-degree equation. Example 2.1.2. Find the general solution to y 4y + 40y = 0. The characteristic polynomial is r 2 4r + 40. The roots are (using the quadratic formula:) (4) 42 4(1)(40) 4 144 = = 2 6i. r= 2(1) 2 The solution is: y = c1 e2t cos(6t) + c2 e2t sin(6t). For a repeated root, repeat the solution as many times, but boost the solution (multiply by t) each time it is repeated.
  67
  Example 2.1.3. Find the general solution to y + 6y + 12y + 8y = 0. The characteristic polynomial is r 3 + 6r 2 + 12r + 8. This factors into (r + 2)3 , so the roots are r = 2, 2, 2 (repeated three times). The solution (with boosting) is: y = c1 e2t + c2 te2t + c3 t2 e2t . Example 2.1.4. Find the general solution to y (4) + 8y + 16y = 0. The characteristic polynomial is r 4 +8r 2 +16. This factors into (r 2 +4)2 = (r +2i)2 (r 2i)2 . This is a conjugate pair of complex roots repeated twice. The solution (with boosting) is: y = c1 e0t cos(2t) + c2 e0t sin(2t) + c3 te0t cos(2t) + c4 te0t sin(2t) = c1 cos(2t) + c2 sin(2t) + c3 t cos(2t) + c4 t sin(2t). Some more quick examples, tabulated to allow for displaying more cases. The left-hand column represents the roots of the characteristic polynomial of the homogeneous, linear, constant coecient dierential equation. The right-hand column represents the general solution to tha dierential equation, where the ci s are constants. Roots: 1, 2, 3 1, 2, 2 2, 2, 2 3, 3, 0, 0, 0, 0 1 7i, 3 2i 1 7i, 1 7i 0, 1, 1, i Exercises 1. Use Eulers formula (eia = cos(a) + i sin(a)) to calculate: (a) (e+i + ei )/2, (b) (e+i ei )/(2i). Simplify your answers as much as possible. 2. Write the general solution corresponding to the given set of roots. a. 0, 0, 2, 2 4i c. 4, 7, 7, 1 5i e. 0, 1, 1, 1, 1, 9 b. 6 i, 6 i, 10 d. 1 8i, 5i f. 2, 3 5i, 3 5i, 3 5i b. c1 cos(5t) + c2 sin(5t) + c3 e10t d. c1 e4t cos t + c2 e4t sin t + c3 te4t cos t + c4 te4t sin t f. c1 e4t + c2 + c3 cos(10t) + c4 sin(10t) General solution (simplied if possible): c1 et + c2 e2t + c3 e3t c1 et + c2 e2t + c3 te2t c1 e2t + c2 te2t + c3 t2 e2t 3 t c1 e + c2 te3t + c3 + c4 t + c5 t2 + c6 t3 c1 et cos(7t) + c2 et sin(7t) + c3 e3t cos(2t) + c4 e3t sin(2t) c1 et cos(7t) + c2 et sin(7t) + c3 tet cos(7t) + c4 tet sin(7t) c1 + c2 et + c3 et + c4 cos t + c5 sin t
  3. Write the set of roots corresponding to the given general solution. a. c1 e4t + c2 te4t + c3 e8t c. c1 + c2 t + c3 t2 + c4 et + c5 tet e. c1 e2t + c2 e5t + c3 te5t
  68
  CHAPTER 2. SECOND ORDER DIFFERENTIAL EQUATIONS 4. Find the general solution to the dierential equation. (b)y (4) + 9y = 0, (a) y 2y + y = 0,
  (c) y 9y + 20y = 0 5. Solve the initial-value problem. (b) y 2y + 5y = 0, y (0) = 1, y (0) = 2. (a) y 5y + 4y = 0, y (0) = 0, y (0) = 3,
  2.2
  To begin, we want to describe the general form a solution to a linear ordinary dierential equation (ODE) can take. We want to describe the solution as a sum of terms which can be computed explicitly in some way. Before doing this, we introduce two pieces of terminology. Suppose f1 (t), f2 (t), . . . , fn (t) are given functions. A linear combination of these functions is another function of the form c1 f1 (t) + c2 f2 (t) + + cn fn (t),
  for some constants c1 , ..., cn . For example, 3 cos(t) 2 sin(t) is a linear combination of cos(t), sin(t). An arbitrary linear combination of cos(t), sin(t) would be written as c1 cos(t) + c2 sin(t), where c1 , c2 are constants (which could be also considerd as parameters). A linear dierential equation of the form y (n) + b1 (t)y (n1) + ... + bn1 (t)y + bn (t)y = f (t), (2.1)
  is called homogeneous if f (t) = 0 (i.e., f is the 0 function) and otherwise it is called non-homogeneous. Consider the n-th order dierential equation y (n) + b1 (t)y (n1) + ... + bn1 (t)y + bn (t)y = 0. Suppose there are n functions y1 (t), . . . , yn (t) such that each y = yi (t) (1 i n) satises this homogeneous dierential equation (2.2), every solution y to (2.2) is a linear combination of these functions y1 , . . . , yn : y = c1 y1 + + cn yn , (2.2)
  69
  In this case, the yi s are called fundamental solutions. This generalizes the discussion of fundamental solutions for 2nd order dierential equations in 1.3.2 above. Remark 2.2.1. If you are worried that this denition is not very practical, then dont. There is a simple condition (the Wronskian test) which will make it much easier to see if a collection of n functions form a set of fundamental solutions. See Theorem 1.3.4 above or Theorem 2.3.1 below. The following result describes the general solution to a linear dierential equation. Theorem 2.2.1. Consider a linear dierential equation of the above form (2.1), for some given continuous functions b1 (t), . . . , bn (t), and f (t). Then the following hold. There are n functions y1 (t), . . . , yn (t) (above-mentioned fundamental solutions), each y = yi (t) (1 i n) satisfying the homogeneous dierential equation, such that every solution yh to (2.2) can be written yh = c1 y1 + + cn yn , for some constants c1 , ... , cn . Suppose you know a solution yp (t) (a particular solution) to (2.1). Then every solution y = y (t) (the general solution) to the dierential equation (2.1) has the form y (t) = yh (t) + yp (t), where yh (the homogeneous part of the general solution) is as above. Conversely, every function of the form (2.4), for any constants ci as in (2.3), for 1 i n, is a solution to (2.1). Example 2.2.1. Recall Example 1.1.5 in the introduction where we looked for functions solving x + x = 1 by guessing. We found that the function xp (t) = 1 is a particular solution to x + x = 1. The function x1 (t) = et is a fundamental solution to x + x = 0. The general solution is therefore x(t) = 1 + c1 et , for a constant c1 . Example 2.2.2. Lets look for functions solving x x = 1 by guessing. Motivated by the above example, we nd that the function xp (t) = 1 is a particular solution to x x = 1. The functions x1 (t) = et , x2 (t) = et are fundamental solutions to x x = 0. The general solution is therefore x(t) = 1 + c1 et , for a constant c1 . Example 2.2.3. The charge on the capacitor of an RLC electrical circuit1 is modeled by a 2-nd order linear dierential equation (see for example, [C-linear] for details). Series RLC Circuit notations:The abbreviation RLC stands for a simple closed electrical circuit, as in Figure 2.1 which has a resistor, an inductor, and a capacitor.1
  (2.3)
  (2.4)
  70
  CHAPTER 2. SECOND ORDER DIFFERENTIAL EQUATIONS E = E (t) - the voltage of the power source (a battery or other electromotive force, measured in volts, V) q = q (t) - the current in the circuit (measured in coulombs, C) i = i(t) - the current in the circuit (measured in amperes, A) L - the inductance of the inductor (measured in henrys, H) R - the resistance of the resistor (measured in ohms, ); C - the capacitance of the capacitor (measured in farads, F)
  The charge q on the capacitor satises the linear IPV: Lq + Rq + 1 q = E (t), C q (0) = q0 , q (0) = i0 . (2.5)
  Figure 2.1: RLC circuit. The circuit example above is analogous to the spring example below. Example 2.2.4. The displacement from equilibrium of a mass attached to a spring suspended from a ceiling as in Figure 2.2 is modeled by a 2-nd order linear dierential equation [O-ivp]. Spring-mass notations: f (t) - the external force acting on the spring (if any) x = x(t) - the displacement from equilibrium of a mass attached to a spring m - the mass b - the damping constant (if, say, the spring is immersed in a uid) k - the spring constant.
  2.2. LINEAR DIFFERENTIAL EQUATIONS, REVISITED The displacement x satises the linear IPV: mx + bx + kx = f (t), x(0) = x0 , x (0) = v0 .
  71
  (2.6)
  Figure 2.2: spring-mass model. Notice that each general solution to an n-th order dierential equation has n degrees of freedom (the arbitrary constants ci ). According to this theorem, to nd the general solution of a linear dierential equation, we need only nd a particular solution yp and n fundamental solutions y1 (t), . . . , yn (t) (n = 2). Example 2.2.5. Let us try to solve x + x = 1, x(0) = c,
  where c = 1, c = 2, and c = 3. (Three dierent IVPs, three dierent solutions, nd each one.) The rst problem, x + x = 1 and x(0) = 1, is easy. The solutions to the dierential equation x + x = 1 which we guessed at in the previous example, x(t) = 1, satises this. The second problem, x + x = 1 and x(0) = 2, is not so simple. To solve this, we would like to know what the form of the general solution is. This would help use solve the IVP, as we did in 1.2. According to Theorem 2.2.1, the general solution x has the form x = xp + xh . In this case, xp (t) = 1 and xh (t) = c1 x1 (t) = c1 et , by the earlier Example 1.1.5. Therefore, every solution to the dierential equation above is of the form x(t) = 1 + c1 et , for some constant c1 . We use the initial condition to solve for c1 : x(0) = 1: 1 = x(0) = 1 + c1 e0 = 1 + c1 so c1 = 0 and x(t) = 1. x(0) = 2: 2 = x(0) = 1 + c1 e0 = 1 + c1 so c1 = 1 and x(t) = 1 + et . x(0) = 3: 3 = x(0) = 1 + c1 e0 = 1 + c1 so c1 = 2 and x(t) = 1 + 2et .
  72
  Here is one way2 to use Sage to solve for c1 . We use Sage to solve the last IVP discussed above and then to plot the solution.Sage sage: t = var(t) sage: x = function(x,t) sage: desolve(diff(x,t)+x==1,[x,t]) (c + et)*e(-t) sage: c = var(c) sage: solnx = lambda t: 1+c*exp(-t) # the soln from desolve sage: solnx(0) c + 1 sage: solve([solnx(0) == 3],c) [c == 2] sage: c0 = solve([solnx(0) == 3], c)[0].rhs() sage: solnx1 = lambda t: 1+c0*exp(-t); solnx1(t) sage: P = plot(solnx1(t),0,5) sage: show(P)
  Exercises: 1. Let q1 (t) = et cos(t) and q2 (t) = et sin(t). Check that these are fundamental solutions to q + 2q + 2q = 0. In particular, using the Wronskian, check that they are linearly independent.Of course, you can do this yourself, but this shows you the Sage syntax for solving equations. Type solve? in Sage to get more details.2
  73
  2. Let x1 (t) = et and x2 (t) = tet . Check that these are fundamental solutions to x + 2x + x = 0. In particular, using the Wronskian, check that they are linearly independent. 3. A 32 lb weight is attached to a spring having spring constant k = 49 with no damping. The mass is initially released with a 3 ft/s upward velocity 2 ft below equilibrium. (Please choose your orientation so that down is positive.) Find the dierential equation and initial conditions for the displacement x = x(t) of this spring-mass system. A 32 lb weight is attached to a spring having spring constant k = 101 with damping constant b = 20 and an external force of f (t) = 10. The mass is initially released from rest at equilibrium. (Please choose your orientation so that down is positive.) Find the IVP for the displacement x = x(t) of this spring-mass system. 4. A simple circuit with inductor L = 1, R = 6 ohm resistor, capacitor C = 1/10, and a e(t) = 1 volt emf has initial charge and current both equal to 0. Find the dierential equation and initial conditions modeling this circuit. Draw a circuit diagram representing this situation. 5. A simple circuit with inductor L = 1, no resistor and no electro-motive force, and capacitor C = 1/100, has initial charge 1 and current equal to 1. Find the IVP modeling this circuit.
  2.3
  To better describe the form a solution to a linear ODE can take, we need to better understand the nature of fundamental solutions and particular solutions. Recall that the general solution to y (n) + b1 (t)y (n1) + ... + bn1 (t)y + bn (t)y = f (t), has the form y = yp + yh , where yh is a linear combination of fundamental solutions. Suppose we are also given n initial conditions y (x0 ) = a0 , y (x0 ) = a1 , . . . , y (n1) (x0 ) = an1 . We want to be able to solve this n-th order initial value problem and obtain a unique solution. Indeed, these n conditions will nail down the arbitrary constants given in the linear combination dening yh , so that y is unique. Example 2.3.1. The general solution to the dierential equation equation x 5x + 6x = 0 has the form x = x(t) = c1 exp(2t) + c2 exp(3t), for arbitrary constants c1 and c2 . For example, we could impose the initial conditions: x(0) = 3 and x (0) = 4. Of course, no matter what x0 and v0 are are given, we want to be able to solve for the coecients c1 , c2 in x(t) = c1 exp(2t) + c2 exp(3t) to obtain a unique solution. A few questions arise.
  74
  CHAPTER 2. SECOND ORDER DIFFERENTIAL EQUATIONS How do we know this can be done? How do we know that (for some c1 , c2 ) some linear combination x(t) = c1 exp(2t) + c2 exp(3t) isnt identically 0 (which, if true, would imply that x = x(t) couldnt possibly satisfy x(0) = 3 and x (0) = 4)?
  We shall answer these questions below. The complete answer to the questions mentioned in the above example actually involves methods from linear algebra which go beyond this course. The basic idea though is not hard to understand, and it involves what is called the Wronskian3 [W-linear]. Before we motivate the idea of the Wronskian by returning to the above example, we need to recall a basic fact from linear algebra. Lemma 2.3.1. (Cramers rule) Consider the system of two equations in two unknowns x, y : ax + by = s1 , The solution to this system is det x= det Note the determinant det s1 b s2 d a b c d det , y= det a s1 c s2 a b c d cx + dy = s2 .
  a b = ad bc is in the denominator of both expressions. c d In particular, if the determinant is 0 then the formula is invalid (and in that case, the solution either does not exist or is not unique). Example 2.3.2. Write the general solution to x 5x + 6x = 0 as x(t) = c1 x1 (t) + c2 x2 (t) (we know x1 (t) = exp(2t), x2 (t) = exp(3t), but we leave it in this more abstract notation to make a point). Assume the initial conditions x(0) = 3 and x (0) = 4 hold. We can try solve for c1 , c2 but plugging t = 0 into the general solution: 3 = x(0) = c1 e0 + c2 e0 = c1 + c2 , 4 = x (0) = c1 2e0 + c2 3e0 = 2c1 + 3c2 .
  You can solve these by hand for c1 , c2 (and you encouraged to do so). However, to motivate Wronskians we shall use the initial conditions in the more abstract form of the general solution: 3 = x(0) = c1 x1 (0) + c2 x2 (0), 4 = x (0) = c1 x 1 (0) + c2 x2 (0).
  Cramers rule gives us the solution for this system of two equations in two unknowns c1 , c2 :Josef Wronski was a Polish-born French mathematician who worked in many dierent areas of applied mathematics and mechanical engineering [Wr-linear].3
  75
  det c1 = det
  det , y= det
  What you see in the denominator of these expressions, is called the Wronskian of the fundamental solutions x1 , x2 evaluated at t = 0. From the example above we see Wronskians arise naturally in the process of solving for c1 and c2 . In general terms, what is a Wronskian? It is best to explain what this means not just for two functions, but for any nite number of functions. This more general case would be useful in case we wanted to try to solve a higher order ODE by the same method. If f1 (t), f2 (t), . . . , fn (t) are given n-times dierentiable functions then their fundamental matrix is the n n matrix = (f1 , ..., fn ) = f1 f1 (t) (t) f1 . . .(n1)
  f2 (t) (t) f2 . . .(n1)
  ... ...
  fn (t) (t) fn . . .(n1)
  (t) f2
  (t) . . . fn
  (t)
  The determinant of the fundamental matrix is called the Wronskian, denoted W (f1 , ..., fn ). The Wronskian actually helps us answer both questions above simultaneously. Example 2.3.3. Take f1 (t) = sin2 (t), f2 (t) = cos2 (t), and f3 (t) = 1. Sage allows us to easily compute the Wronskian:Sage sage: t = var(t) sage: Phi = matrix([[sin(t)2,cos(t)2,1], [diff(sin(t)2,t),diff(cos(t)2,t),0], [diff(sin(t)2,t,t),diff(cos(t)2,t,t),0]]) sage: Phi [ sin(t)2 cos(t)2 [ 2*sin(t)*cos(t) -2*sin(t)*cos(t) [ -2*sin(t)2 + 2*cos(t)2 2*sin(t)2 - 2*cos(t)2 sage: det(Phi) 0
  1] 0] 0]
  Here det(Phi) is the determinant of the fundamental matrix Phi. Since it is zero, this means W (sin(t)2 , cos(t)2 , 1) = 0. Lets try another example using Sage .
  76
  sage: t = var(t) sage: Phi = matrix([[sin(t)2,cos(t)2], sage: Phi [ sin(t)2 cos(t)2 [ 2*cos(t)*sin(t) -2*cos(t)*sin(t) ] sage: Phi.det() -2*cos(t)*sin(t)3 - 2*cos(t)3*sin(t)
  [diff(sin(t)2,t),diff(cos(t)2,t)]])
  This means W (sin(t)2 , cos(t)2 ) = 2 cos(t) sin(t)3 2 cos(t)3 sin(t), which is non-zero. If there are constants c1 , ..., cn , not all zero, for which c1 f1 (t) + c2 f2 (t) + cn fn (t) = 0, for all t, (2.7)
  then the functions fi (1 i n) are called linearly dependent. If the functions fi (1 i n) are not linearly dependent then they are called linearly independent (this denition is frequently seen for linearly independent vectors [L-linear] but holds for functions as well). This condition (2.7) can be interpreted geometrically as follows. Just as c1 x + c2 y = 0 is a line through the origin in the plane and c1 x + c2 y + c3 z = 0 is a plane containing the origin in 3-space, the equation c1 x1 + c2 x2 + cn xn = 0, is a hyperplane containing the origin in n-space with coordinates (x1 , ..., xn ). This condition (2.7) says geometrically that the graph of the space curve r(t) = (f1 (t), . . . , fn (t)) lies entirely in this hyperplane. If you pick n functions at random then they are probably linearly independent, because random space curves dont lie in a hyperplane. But certainly not all collections of functions are linearly independent. Example 2.3.4. Consider just the two functions f1 (t) = sin2 (t), f2 (t) = cos2 (t). We know from the Sage computation in the example above that these functions are linearly independent.Sage sage: P = parametric_plot((sin(t)2,cos(t)2),0,5) sage: show(P)
  The Sage plot of this space curve r(t) = (sin(t)2 , cos(t)2 ) is given in Figure 2.4. It is obviously not contained in a line through the origin, therefore making it geometrically clear that these functions are linearly independent. The following two results answer the above questions.
  77
  Figure 2.4: Parametric plot of (sin(t)2 , cos(t)2 ). Theorem 2.3.1. (Wronskian test) If f1 (t), f2 (t), . . . , fn (t) are given n-times dierentiable functions with a non-zero Wronskian then they are linearly independent. As a consequence of this theorem, and the Sage computation in the example above, f1 (t) = sin2 (t), f2 (t) = cos2 (t), are linearly independent. Theorem 2.3.2. Given any homogeneous n-th linear ODE y (n) + b1 (t)y (n1) + ... + bn1 (t)y + bn (t)y = 0, with dierentiable coecients, there always exists n solutions y1 (t), ..., yn (t) which have a non-zero Wronskian. The functions y1 (t), ..., yn (t) in the above theorem are called fundamental solutions. We shall not prove either of these theorems here. Please see [BD-intro] for further details. Exercises: 1. Use Sage to compute the Wronskian of (a) f1 (t) = sin(t), f2 (t) = cos(t), (b) f1 (t) = 1, f2 (t) = t, f3 (t) = t2 , f4 (t) = t3 , (c) f1 (t) = cos(2t), f2 (t) = sin(2t), f3 (t) = 1 cos(2t). 2. Use the Wronskian test to check that (a) y1 (t) = sin(t), y2 (t) = cos(t) are fundamental solutions for y + y = 0, (b) y1 (t) = 1, y2 (t) = t, y3 (t) = t2 , y4 (t) = t3 are fundamental solutions for y (4) = y = 0.
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  2.4
  The method of undetermined coecients [U-uc] can be used to solve the following type of problem. PROBLEM: Solve ay + by + cy = f (x), (2.8)
  where a = 0, b and c are constants, and f (x) is a special type of function. (Even the case a = 0 can be handled similarly, though some of the discussion below might need to be slightly modied.) Thet assumption that f (x) is of a special form will be explained in moe detail later. More-or-less equivalent is the method of annihilating operators [A-uc] (they solve the same class of DEs), but that method will be discussed separately.
  2.4.1
  Simple case
  For the moment, let us assume f (x) has the simple form a1 p(x) ea2 x cos(a3 x), or a1 p(x) ea2 x sin(a3 x), where a1 , a2 , a3 are constants and p(x) is a polynomial. Solution: Solve the homogeneous DE ay + by + cy = 0 as follows. Let r1 and r2 denote the roots of the characteristic polynomial aD 2 + bD + c = 0. r1 = r2 real: the solution is y = c1 er1 x + c2 er2 x . r1 = r2 real: if r = r1 = r2 then the solution is y = c1 erx + c2 xerx . r1 , r2 complex: if r1 = + i , r2 = i , where and are real, then the solution is y = c1 ex cos(x) + c2 ex sin(x). Denote this solution yh (some texts use yc ) and call this the homogeneous part of the solution. (Some texts call this the complementary part of the solution.) Compute f (x), f (x), f (x), ... . Write down the list of all the dierent terms which arise (via the product rule), ignoring constant factors, plus signs, and minus signs: t1 (x), t2 (x), ..., tk (x). If any one of these agrees with y1 or y2 then multiply them all by x. (If, after this, any of them still agrees with y1 or y2 then multiply them all again by x.)
  2.4. UNDETERMINED COEFFICIENTS METHOD Let yp be a linear combination of these functions (your guess): yp = A1 t1 (x) + ... + Ak tk (x).
  79
  This is called the general form of the particular solution (when you have not solved for the constants Ai ). now.
  Diagramatically: Factor characteristic polynomial Compute yh Compute the general form of the particular, yp Compute the undetermined coecients Answer: y = yh + yp .
  ExamplesExample 2.4.1. Solve y y = cos(2x). The characteristic polynomial is r2 1 = 0, which has 1 for roots. The homogeneous solution is therefore yh = c1 ex + c2 ex . We compute f (x) = cos(2x), f (x) = 2 sin(2x), f (x) = 4 cos(2x), ... . They are all linear combinations of f1 (x) = cos(2x), f2 (x) = sin(2x). None of these agrees with y1 = ex or y2 = ex , so we do not multiply by x.
  80
  (4A1 cos(2x) 4A2 sin(2x)) (A1 cos(2x) + A2 sin(2x)) = cos(2x). Equating the coecients of cos(2x), sin(2x) on both sides gives 2 equations in 2 unknowns: 5A1 = 1 and 5A2 = 0. Solving, we get A1 = 1/5 and A2 = 0. The general solution: y = yh + yp = c1 ex + c2 ex Example 2.4.2. Solve y y = x cos(2x). The characteristic polynomial is r2 1 = 0, which has 1 for roots. The homogeneous solution is therefore yh = c1 ex + c2 ex .). None of these agrees with y1 = ex or y2 = ex , so we do not multiply by x. Let yp be a linear combination of these functions: yp = A1 cos(2x) + A2 sin(2x) + A3 x cos(2x) + A4 x sin(2x). In principle, you can compute both sides of yp yp = x cos(2x) and solve for the Ai s. (Equate coecients of x cos(2x) on both sides, equate coecients of cos(2x) on both sides, equate coecients of x sin(2x) on both sides, and equate coecients of sin(2x) on both sides. This gives 4 equations in 4 unknowns, which can be solved.) You will not be asked to solve for the Ai s for a problem this hard. 1 5
  cos(2x).
  Example 2.4.3. Solve y + 4y = x cos(2x). The characteristic polynomial is r2 + 4 = 0, which has 2i for roots. The homogeneous solution is therefore yh = c1 cos(2x) + c2 sin(2x).). Two of these agree with y1 = cos(2x) or y2 = sin(2x), so we do multiply by x: f1 (x) = x cos(2x), f2 (x) = x sin(2x), f3 (x) = x2 cos(2x), .f4 (x) = x2 sin(2x).
  81
  In principle, you can compute both sides of yp + 4yp = x cos(2x) and solve for the Ai s. You will not be asked to solve for the Ai s for a problem this hard.
  2.4.2
  Non-simple case
  More generally, suppose that you want to solve ay + by + cy = f (x), where f (x) is a sum of functions of the simple functions in the previous subsection. In other words, f (x) = f1 (x) + f2 (x) + ... + fk (x), where each fj (x) is of the form c p(x) eax cos(bx), or c p(x) eax sin(bx), where a, b, c are constants and p(x) is a polynomial. You can proceed in either one of the following ways. 1. Split up the problem by solving each of the k problems. Suppose DE ay + by + cy = f1 (x) ay + by + cy = f2 (x) ... ay + by + cy = fk (x) has solution y = y1 (x) y = y2 (x) ... y = yk (x)
  In this case, the solution to ay + by + cy = f (x) is then y = y1 + y2 + .. + yk (the superposition principle). 2. Proceed as in the examples above but with the following slight revision: Find the homogeneous solution yh to ay + by = cy = 0, yh = c1 y1 + c2 y2 .
  Compute f (x), f (x), f (x), ... . Write down the list of all the dierent terms which arise, ignoring constant factors, plus signs, and minus signs: t1 (x), t2 (x), ..., tk (x).
  Group these terms into their families. Each family is determined from its parent(s) - which are the terms in f (x) = f1 (x) + f2 (x) + ... + fk (x) which they arose form by dierentiation. For example, if f (x) = x cos(2x) + ex sin(x) + sin(2x) then the terms you get from dierentiating and ignoring constants, plus signs and minus signs, are x cos(2x), x sin(2x), cos(2x), sin(2x), ex sin(x), ex cos(x), and sin(2x), cos(2x), (from sin(2x)). (from x cos(2x)),
  (from ex sin(x)),
  82
  CHAPTER 2. SECOND ORDER DIFFERENTIAL EQUATIONS The rst group absorbes the last group, since you can only count the dierent terms. Therefore, there are only two families in this example: {x cos(2x), x sin(2x), cos(2x), sin(2x)} is a family (with parent x cos(2x) and the other terms as its children) and {ex sin(x), ex cos(x)}
  is a family (with parent ex sin(x) and the other term as its child). If any one of these terms agrees with y1 or y2 then multiply the entire family by x. In other words, if any child or parent is bad then the entire family is bad. (If, after this, any of them still agrees with y1 or y2 then multiply them all again by x.) Let yp be a linear combination of these functions (your guess): yp = A1 t1 (x) + ... + Ak tk (x). This is called the general form of the particular solution. last - after the undetermined coecients.
  So the roots of the characteristic polynomial are 1, 1, 1, which means that the homogeneous part of the solution is
  83
  yh = c1 ex + c2 xex + c3 ex .Sage sage: sage: sage: sage: 0 de = lambda y: diff(y,x,3) - diff(y,x,2) - diff(y,x,1) + y c1 = var("c1"); c2 = var("c2"); c3 = var("c3") yh = c1*ex + c2*x*ex + c3*e(-x) de(yh)
  This just conrmed that yh solves y y y +1 = 0. Using the derivatives of F (x) = 12xex , we generate the general form of the particular:Sage sage: F = 12*x*ex sage: diff(F,x,1); diff(F,x,2); diff(F,x,3) 12*x*ex + 12*ex 12*x*ex + 24*ex 12*x*ex + 36*ex sage: A1 = var("A1"); A2 = var("A2") sage: yp = A1*x2*ex + A2*x3*ex
  Now plug this into the DE and compare coecients of like terms to solve for the undertermined coecients:Sage sage: de(yp) 12*x*ex*A2 + 6*ex*A2 + 4*ex*A1 sage: solve([12*A2 == 12, 6*A2+4*A1 == 0],A1,A2) [[A1 == -3/2, A2 == 1]]
  84
  2.4.3
  Annihilator method
  We consider again the same type of dierential equation as in the above subsection, but take a slightly dierent approach here. PROBLEM: Solve ay + by + cy = f (x). (2.9)
  We assume that f (x) is of the form c p(x) eax cos(bx), or c p(x) eax sin(bx), where a, b, c are constants and p(x) is a polynomial. solution: Write the ODE in symbolic form (aD 2 + bD + c)y = f (x). Find the homogeneous solution yh to ay + by = cy = 0, yh = c1 y1 + c2 y2 . Find the dierential operator L which annihilates f (x): Lf (x) = 0. The following annihilator table may help. function xk xk eax k x x e cos(x) xk ex sin(x) annihilator D k+1 (D a)k+1 2 (D 2D + 2 + 2 )k+1 (D 2 2D + 2 + 2 )k+1
  Find the general solution to the homogeneous ODE, L (aD 2 + bD + c)y = 0. Let yp be the function you get by taking the solution you just found and subtracting from it any terms in yh . Solve for the undetermined coecients in yp as in the method of undetermined coefcients. Example 2.4.5. Solve y y = cos(2x). The DE is (D 2 1)y = cos(2x).
  85
  The characteristic polynomial is r 2 1 = 0, which has 1 for roots. The homogeneous solution is therefore yh = c1 ex + c2 ex . We nd L = D 2 + 4 annihilates cos(2x). We solve (D 2 + 4)(D 2 1)y = 0. The roots of the characteristic polynomial (r 2 + 4)(r 2 1) are 2i, 1. The solution is y = A1 cos(2x) + A2 sin(2x) + A3 ex + A4 ex . This solution agrees with yh in the last two terms, so we guess yp = A1 cos(2x) + A2 sin(2x). y = cos(2x), Now solve for A1 and A2 as before: Compute both sides of yp p
  (4A1 cos(2x) 4A2 sin(2x)) (A1 cos(2x) + A2 sin(2x)) = cos(2x). Next, equate the coecients of cos(2x), sin(2x) on both sides to get 2 equations in 2 unknowns. Solving, we get A1 = 1/5 and A2 = 0. The general solution: y = yh + yp = c1 ex + c2 ex 1 5 cos(2x). Exercises: 1. Solve the initial value problem y 4y = 0, y (0) = 4, y (0) = 2 given that y1 = e2x and y2 = e2x are both solutions to the ODE. Solve the following problems using the method of undetermined coecients: 2. (a) Find the general solution to x x 20x = 0.
  (b) Use the general solution to solve the initial value problem x x 20x = 0, x(0) = 1, x (0) = 1.
  3. Find the general solution to y + 6y = 0. 4. Find the general solution to 4y + 4y + y = 0. 5. Find the general solution to y + 10y + 25y = 0. 6. Solve the initial value problem y 6y + 25y = 0, y (0) = 6, y (0) = 2. 7. For what second-order constant coecient linear homogeneous ODE would y = C1 + C2 x be the general solution? 8. Find the general solution to x x = t.
  86
  CHAPTER 2. SECOND ORDER DIFFERENTIAL EQUATIONS 9. Consider the dierential equation y + sgn(x)y 1 1 sgn(x) = 0 = 0, where sgn(x) is the sign function: if x > 0 if x < 0 if x = 0
  10. Find the general solution to y (4) 6y (3) + 9y = 0. 11. Find the general solution of 6y (4) +5y (3) +18y +20y 24y = 0 given that y = cos(2x) is a solution. (Hint: use polynomial division on the characteristic polynomial.) 12. Find a particular solution to the ODE y y + 2y = 4x + 12. 13. Find a particular solution to the ODE y y + y = sin2 (x). (Hint: it may be helpful to use a trig identity.) 14. Find the general solution to y (3) y = ex .
  For the following two problems, determine the form of the particular solution - note that you do not have to determine the values of the coecients. You should not include terms from the homogeneous solution.
  15. Determine the form of the particular solution to y = 9x2 + 1. 16. Determine the form of the particular solution to y (4) 16y = x2 sin(4x) + sin(4x). 17. Solve the initial value problem y + 2y + 2y = cos(3x), y (0) = 0, y (0) = 2. 18. Solve the initial value problem y (4) y = 1, y (0) = y (0) = y (0) = y (3) = 0.
  2.5
  Variation of parameters
  The method of variation of parameters is originally attributed to Joseph Louis Lagrange (1736-1813), an Italian-born mathematician and astronomer, who worked much of his professional life in Berlin and Paris [L-var]. It involves, at one step, a prepeated dierentiation which is theoretically described by the so-called Leibniz rule, described next.
  2.5.1
  2.5. VARIATION OF PARAMETERS and so on, where the coecients occurring in (f g)(n) are binomial coecients computed for example4 using Pascals triangle, 1 1 1 1 1 4 3 6 ..., and so on. The general pattern is dictated by the rule ... n1 i ... Using Sage , this can be checked as follows:Sage sage: t = var(t) sage: x = function(x, t) sage: y = function(y, t) sage: diff(x(t)*y(t),t) x(t)*diff(y(t), t, 1) + y(t)*diff(x(t), t, 1) sage: diff(x(t)*y(t),t,t) x(t)*diff(y(t), t, 2) + 2*diff(x(t), t, 1)*diff(y(t), t, 1) + y(t)*diff(x(t), t, 2) sage: diff(x(t)*y(t),t,t,t) x(t)*diff(y(t), t, 3) + 3*diff(x(t), t, 1)*diff(y(t), t, 2) + 3*diff(x(t), t, 2)*diff(y(t), t, 1) + y(t)*diff(x(t), t, 3)
  87 n i
  1 2 3 4 1 1 1
  n1 i+1 n i ...
  ...
  2.5.2
  The method
  Consider an ordinary constant coecient non-homogeneous 2nd order linear dierential equation, ay + by + cy = F (x) where F (x) is a given function and a, b, and c are constants. (The varation of parameters method works even if a, b, and c depend on the independent variable x. However, for simplicity, we assume that they are constants here.)4
  For second order ODEs, we only need the rst two Leibntiz rules.
  88
  Let y1 (x), y2 (x) be fundamental solutions of the corresponding homogeneous equation ay + by + cy = 0. Starts by assuming that there is a particular solution in the form yp (x) = u1 (x)y1 (x) + u2 (x)y2 (x), where u1 (x), u2 (x) are unknown functions [V-var]. We want to solve for u1 and u2 . By assumption, yp solves the ODE, so ayp + byp + cyp = F (x).
  (2.10)
  If we assume u 1 y 1 + u2 y 2 = 0
  Cramers rule (Lemma 2.3.1) implies that the solution to this system is det u 1 = det 0 y2 F (x) y2 y1 y2 y2 y1 det , u 2 = det y1 0 y1 F (x) y1 y2 y2 y1
  (2.11)
  (Note that the Wronskian W (y1 , y2 ) of the fundamental solutions is in the denominator.) Solve these for u1 and u2 by integration and then plug them back into (2.10) to get your particular solution. Example 2.5.1. Solve y + y = tan(x). soln: The functions y1 = cos(x) and y2 = sin(x) are fundamental solutions with Wronskian W (cos(x), sin(x)) = 1. The formulas (2.11) become: det u 1 = Therefore, 0 sin(x) tan(x) cos(x) 1 det , u 2 = cos(x) 0 sin(x) tan(x) 1
  89
  u 1 =
  u 2 = sin(x).
  Therefore, using methods from integral calculus, u1 = ln | tan(x) + sec(x)| + sin(x) and u2 = cos(x). Using Sage , this can be check as follows:Sage sage: integral(-sin(t)2/cos(t),t) -log(sin(t) + 1)/2 + log(sin(t) - 1)/2 + sin(t) sage: integral(cos(t)-sec(t),t) sin(t) - log(tan(t) + sec(t)) sage: integral(sin(t),t) -cos(t)
  As you can see, there are other forms the answer can take. The particular solution is yp = ( ln | tan(x) + sec(x)| + sin(x)) cos(x) + ( cos(x)) sin(x). The homogeneous (or complementary) part of the solution is yh = c1 cos(x) + c2 sin(x), so the general solution is y = yh + yp = c1 cos(x) + c2 sin(x) +( ln | tan(x) + sec(x)| + sin(x)) cos(x) + ( cos(x)) sin(x). Using Sage , this can be carried out as follows:Sage sage: sage: sage: sage: SR = SymbolicExpressionRing() MS = MatrixSpace(SR, 2, 2) W = MS([[cos(t),sin(t)],[diff(cos(t), t),diff(sin(t), t)]]) W
  [ cos(t) sin(t)] [-sin(t) cos(t)] sage: det(W) sin(t)2 + cos(t)2 sage: U1 = MS([[0,sin(t)],[tan(t),diff(sin(t), t)]]) sage: U2 = MS([[cos(t),0],[diff(cos(t), t),tan(t)]]) sage: integral(det(U1)/det(W),t) -log(sin(t) + 1)/2 + log(sin(t) - 1)/2 + sin(t) sage: integral(det(U2)/det(W),t) -cos(t)
  90 Exercises:
  2 cos (2x)
  2. Use the variation of parameters method to nd the general solution of y 2y + y = ex /x. 3. Use Sage to solve y + y = cot(x).
  2.6.1
  Part 1
  One of the ways DEs arise is by means of modeling physical phenomenon, such as spring equations. For these problems, consider a spring suspended from a ceiling. We shall consider three cases: (1) no mass is attached at the end of the spring, (2) a mass is attached and the system is in the rest position, (3) a mass is attached and the mass has been displaced from the rest position. One can also align the springs left-to-right instead of top-to-bottom, without changing the discussion below. Notation: Consider the rst two situations above: (a) a spring at rest, without mass attached and (b) a spring at rest, with mass attached. The distance the mass pulls the spring down is sometimes called the stretch, and denoted s. (A formula for s will be given later.) Now place the mass in motion by imparting some initial velocity (tapping it upwards with a hammer, say, and start your timer). Consider the second two situations above: (a) a spring at rest, with mass attached and (b) a spring in motion. The dierence between these two positions at time t is called the displacement and is denoted x(t). Signs here will be choosen so that down is positive. Assume exactly three forces act: 1. the restoring force of the spring, Fspring , 2. an external force (driving the ceiling up and down, but may be 0), Fext , 3. a damping force (imagining the spring immersed in oil or that it is in fact a shock absorber on a car), Fdamp .5
  91
  In other words, the total force is given by Ftotal = Fspring + Fext + Fdamp . Physics tells us that the following are approximately true: 1. (Hookes law [H-intro]): Fspring = kx, for some spring constant k > 0, 2. Fext = F (t), for some (possibly zero) function F , 3. Fdamp = bv , for some damping constant b 0 (where v denotes velocity), 4. (Newtons 2nd law [N-mech]): Ftotal = ma (where a denotes acceleration). Putting this all together, we obtain mx = ma = kx + F (t) bv = kx + F (t) bx , or mx + bx + kx = F (t). This is the spring equation. When b = F (t) = 0 this is also called the equation for simple harmonic motion. The solution in the case of simple harmonic motion has the form x(t) = c1 cos(t) + c2 sin(t), where = k/m. There is a more compact and useful form of the solution, A sin(t + ), useful for graphing. This compact form is obtained using the formulas c1 cos(t) + c2 sin(t) = A sin(t + ), (2.12)
  92 where A =
  Consider again rst two gures above: (a) a spring at rest, without mass attached and (b) a spring at rest, with mass attached. The mass in the second gure is at rest, so the gravitational force on the mass, mg, is balanced by the restoring force of the spring: mg = ks, where s is the stretch. In particular, the spring constant can be computed from the stretch: k=mg s .
  Example 2.6.1. A spring at rest is suspended from the ceiling without mass. A 2 kg weight is then attached to this spring, stretching it 9.8 cm. From a position 2/3 m above equilibrium the weight is give a downward velocity of 5 m/s. (a) Find the equation of motion. (b) What is the amplitude and period of motion? (c) At what time does the mass rst cross equilibrium? (d) At what time is the mass rst exactly 1/2 m below equilibrium? We shall solve this problem using Sage below. Note m = 2, b = F (t) = 0 (since no damping or external force is even mentioned), and the stretch is s = 9.8 cm = 0.098 m. Therefore, the spring constant is given by k = mg/s = 2 9.8/(0.098) = 200. Therefore, the DE is 2x + 200x = 0. This has general solution x(t) = c1 cos(10t) + c2 sin(10t). The constants c1 and c2 can be computed from the initial conditions x(0) = 2/3 (down is positive, up is negative), x (0) = 5. Using Sage , the displacement can be computed as follows:Sage sage: t = var(t) sage: x = function(x, t) sage: m = var(m) sage: b = var(b) sage: k = var(k) sage: F = var(F) sage: de = lambda y: m*diff(y,t,t) + b*diff(y,t) + k*y - F sage: de(x) b*D[0](x)(t) + k*x(t) + m*D[0, 0](x)(t) - F sage: m = 2; b = 0; k = 200; F = 0 sage: de(x) 200.000000000000*x(t) + 2*D[0, 0](x)(t) sage: desolve(de(x),[x,t]) k1*sin(10*t)+k2*cos(10*t) sage: print desolve_laplace(de(x(t)),["t","x"],[0,-2/3,5])
  93
  Now we write this in the more compact and useful form A sin(t + ) using formula (2.12) and Sage .Sage sage: c1 = -2/3; c2 = 1/2 sage: A = sqrt(c12 + c22) sage: A 5/6 sage: phi = 2*atan(c1/(c2 + A)) sage: phi -2*atan(1/2) sage: RR(phi) -0.927295218001612 sage: sol = lambda t: c1*cos(10*t) + c2*sin(10*t) sage: sol2 = lambda t: A*sin(10*t + phi) sage: P = plot(sol(t),0,2) sage: show(P)
  Figure 2.8: Plot of 2x + 200x = 0, x(0) = 2/3, x (0) = 5, for 0 < t < 2. (You can also, if you want, type show(plot(sol2(t),0,2)) to check that these two functions are indeed the same.) Of course, the period is 2/10 = /5 0.628. To answer (c) and (d), we solve x(t) = 0 and x(t) = 1/2:
  94
  sage: solve(A*sin(10*t + phi) == 0,t) [t == atan(1/2)/5] sage: RR(atan(1/2)/5) 0.0927295218001612 sage: solve(A*sin(10*t + phi) == 1/2,t) [t == (asin(3/5) + 2*atan(1/2))/10] sage: RR((asin(3/5) + 2*atan(1/2))/10) 0.157079632679490
  In other words, x(0.0927...) 0, x(0.157...) 1/2. Exercise: Using Sage and the problem above, answer the following questions. (a) At what time does the weight pass through the equilibrium position heading down for the 2nd time? (b) At what time is the weight exactly 5/12 m below equilibrium and heading up?
  2.6.2
  Part 2
  Recall from the previous subsection, the spring equation is mx + bx + kx = F (t) where x(t) denotes the displacement at time t. Until otherwise stated, we assume there is no external force: F (t) = 0. The roots of the characteristic polynomial mD 2 + bD = k = 0 are b b2 4mk , 2m
  by the quadratic formula. There are three cases: (a) real distinct roots: in this case the discriminant b2 4mk is positive, so b2 > 4mk. In other words, b is large. This case is referred to as overdamped. In this case, the roots are negative,
  95
  r1 =
  b2 4mk < 0, 2m
  b +
  b2 4mk < 0, 2m
  so the solution x(t) = c1 er1 t + c2 er2 t is exponentially decreasing. (b) real repeated roots: in this case the discriminant b2 4mk is zero, so b = case is referred to as critically damped. 4mk . This
  If you press down with all your weight on the hood of a new car, and then release very quickly, the resulting bounce is said to behave as in this case, In other words, a new car suspension systems in cars should be modeled by this case [D-spr]. (c) Complex roots: in this case the discriminant b2 4mk is negative, so b2 < 4mk. In other words, b is small. This case is referred to as underdamped (or simple harmonic when b = 0).
  Example 2.6.2. An 8 lb weight stretches a spring 2 ft. Assume a damping force numerically equal to 2 times the instantaneous velocity acts. Find the displacement at time t, provided that it is released from the equilibrium position with an upward velocity of 3 ft/s. Find the equation of motion and classify the behaviour. We know m = 8/32 = 1/4, b = 2, k = mg/s = 8/2 = 4, x(0) = 0, and x (0) = 3. This means we must solve 1 x + 2x + 4x = 0, x(0) = 0, x (0) = 3. 4 The roots of the characteristic polynomial are 4 and 4 (so we are in the repeated real roots case), so the general solution is x(t) = c1 e4t + c2 te4t . The initial conditions imply c1 = 0, c2 = 3, so x(t) = 3te4t . Using Sage , we can compute this as well:Sage sage: t = var(t) sage: x = function(x) sage: de = lambda y: (1/4)*diff(y,t,t) + 2*diff(y,t) + 4*y sage: de(x(t)) diff(x(t), t, 2)/4 + 2*diff(x(t), t, 1) + 4*x(t) sage: desolve(de(x(t)),[x,t]) (%k2*t+%k1)*%e-(4*t) sage: desolve_laplace(de(x(t)),[t,x],[0,0,-3]) -3*t*%e-(4*t) sage: f = lambda t : -3*t*e(-4*t) sage: P = plot(f,0,2)
  96sage: show(P)
  Example 2.6.3. An 2 kg weight is attached to a spring having spring constant 10. Assume a damping force numerically equal to 4 times the instantaneous velocity acts. Find the displacement at time t, provided that it is released from 1 m below equilibrium with an upward velocity of 1 ft/s. Find the equation of motion and classify the behaviour. Using Sage , we can compute this as well:Sage sage: t = var(t) sage: x = function(x) sage: de = lambda y: 2*diff(y,t,t) + 4*diff(y,t) + 10*y sage: desolve_laplace(de(x(t)),["t","x"],[0,1,1]) %e-t*(sin(2*t)+cos(2*t)) sage: desolve_laplace(de(x(t)),["t","x"],[0,1,-1]) %e-t*cos(2*t) sage: sol = lambda t: e(-t)*cos(2*t) sage: P = plot(sol(t),0,2) sage: show(P) sage: P = plot(sol(t),0,4) sage: show(P)
  97
  Exercises: 1. By hand, solve x + 5x + 6x = 0, x(0) = 1, x (0) = 2. Find the velocity at t = 1. 2. Refer to Example 2.6.3 above. Use Sage to nd what time the weight passes through the equilibrium position heading down for the 2nd time. 3. An 2 kg weight is attached to a spring having spring constant 10. Assume a damping force numerically equal to 4 times the instantaneous velocity acts. Use Sage to nd the displacement at time t, provided that it is released from 1 m below equilibrium (with no initial velocity).
  2.6.3
  Part 3
  If the frequency of the driving force of the spring matches the frequency of the homogeneous part xh (t), in other words if x + 2 x = F0 cos(t), satises = then we say that the spring-mass system is in (pure, mechanical) resonance. For some time, it was believed that the collapse of the Tacoma Narrows Bridge was explained by this phenomenon but this is false. Theorem 2.6.1. Consider
  98
  x + 2 x = F0 cos(t), and x + 2 x = F0 sin(t). Let xp = xp (t) denote a particular solution to either (2.13) or (2.14). Then we have case (2.13): 1. if = then xp (t) = 2. if = then xp (t) = case (2.14): 1. if = then xp (t) = 2. if = then F0 sin(t), 2 2 F0 t cos(t). 2 2 F0 cos(t), 2
  (2.13)
  (2.14)
  F0 t sin(t), 2
  is a solution to (2.14). In both of these, to derive the case = , one can take limits in these expressions. Use LH opitals rule to compute the limits. (Details are left to the interested reader.) Example 2.6.4. Solve x + 2 x = cos(t), x(0) = 0, x (0) = 0, where = = 2 (ie, mechanical resonance). We use Sage for this:
  99
  100
  Figure 2.11: A forced undamped spring, with resonance. Example 2.6.5. Solve x + 2 x = cos(t), x(0) = 0, x (0) = 0, where = 2 and = 3 (ie, no mechanical resonance). We use Sage for this:Sage sage: t = var(t) sage: x = function(x, t) sage: (m,b,k,w,g,F0) = var("m,b,k,w,g,F0") sage: de = lambda y: diff(y,t,t) + w2*y - F0*cos(g*t) sage: m = 1; b = 0; k = 4; F0 = 1; w = 2; g = 3 sage: desolve(de(x),[x,t]) k1*sin(2*t) + k2*cos(2*t) - 1/5*cos(3*t) desolve(de(x),[x,t],[0,0,0]) 1/5*cos(2*t) - 1/5*cos(3*t) sage: soln = lambda t : cos(2*t)/5-cos(3*t)/5 sage: P = plot(soln(t),0,10) sage: show(P)
  101
  Figure 2.12: A forced undamped spring, no resonance. Exercises: 1. A mass-spring system with external force is governed by the dierential equation 4x + 16x = 5 cos t. The resonant angular frequency of the system is = (a) 0 (b) 1 (c) 2 (d) 4. 2. An 8 lb weight stretches a spring 2 ft upon coming to rest at equilibrium. From equilibrium the weight is raised 1 ft and released from rest. The motion of the weight is resisted by a damping force that in numerically equal to 2 times the weights instantaneous velocity. (i) Find the position of the weight as a function of time. (ii) What type of damping does this mass-spring system possess? 3. The position of a weight in a mass-spring system subject to an external force is given by x(t) = et cos(3t) + et sin(3t) + 6 cos(2t) + 4 sin(2t). (a) What are the amplitude and period of the steady-state part of the solution? (b) Write the transient part of the solution in the form Aet sin(3t + ). (c) Find the time past which the magnitude of the transient part of the solution is less than one-percent of that of the steady-state part of the solution. 4. If the displacement x = x(t) for an object attached to a spring suspended from a ceiling is governed by the dierential equation 2x + 6x + 4x = 0 then its behavior is (a) in mechanical resonance (b) under-damped (c) critically damped (d) over-damped (e) none of these 5. The amplitude of the steady-state oscillations of a damped mass-spring system with . The resonant a sinusoidal external force F = cos t is given by C ( ) = 2 1 2 2 angular frequency of this system is =( 4) +4
  6. A spring-mass system is governed by 2x + 50x = 10 sin(t). The system has mechanical resonance when = (c) 1, (d) 5, (e) none of these. (a) 7, (b) 50, 7. (a) A 4 lb weight is suspended from a spring with spring constant k = 7/8 lb/ft and damping constant b = 1/2 lb-sec/ft. If the weight is pulled 1 ft below equilibrium and given a velocity of 5 ft/sec upward, nd its displacement in the form x(t) = exp(t) A sin(t + ). (b) What value of b > 0 would give critical damping? If b = 0 and there is an external force of 10 cos(t), what value of b would give resonance?
  (hard) Verify all the cases of Theorem 2.6.1 stated above using the method of undertermined coecients. Complete the proof of Theorem 2.6.1 as sketched in the text.
  2.7
  An LRC circuit is a closed loop containing an inductor of L henries, a resistor of R ohms, a capacitor of C farads, and an EMF (electro-motive force), or battery, of E (t) volts, all connected in series. They arise in several engineering applications. For example, AM/FM radios with analog tuners typically use an LRC circuit to tune a radio frequency. Most commonly a variable capacitor is attached to the tuning knob, which allows you to change the value of C in the circuit and tune to stations on dierent frequencies [R-cir]. We use the dictionary in Figure 2.13 to translate between the
  103
  Next, we recall the circuit laws of Gustav Kircho (sometimes spelled Kirchho), a German physicist who lived from 1824 to 1887. He was born in K onigsberg, which was part of Germany but is now part of the Kaliningrad Oblast, which is an an exclave of Russia surrounded by Lithuania, Poland, and the Baltic Sea. Kircho s First Law: The algebraic sum of the currents travelling into any node is zero. Kircho s Second Law: The algebraic sum of the voltage drops around any closed loop is zero. Generally, the charge at time t on the capacitor, q (t), satises the dierential equation Lq + Rq + When there is not EMF, so Lq + Rq + then we say that the circuit is underdamped if R2 4L/C < 0, critically damped if R2 4L/C = 0, and overdamped if R2 4L/C > 0. Example 2.7.1. In this example, we model a very simple type of radio tuner, using a variable capacitor to represent the tuning dial. Consider the simple LC circuit given by the diagram in Figure 2.14. 1 q = 0, C 1 q = E (t). C (2.15)
  Figure 2.14: A simple LC circuit. According to Kirchos 2nd Law and the above dictionary, this circuit corresponds to the dierential equation
  104
  1 q = sin(2t) + sin(11t). C The homogeneous part of the solution is q + qh (t) = c1 cos(t/ C ) + c1 sin(t/ C ). If C = 1/4 and C = 1/121 then qp (t) = 1 C 1 4 sin(2t) + C 1 1 sin(11t). 121
  When C = 1/4 and the initial charge and current are both zero, the solution is 161 1 1 sin(11t) + sin(2t) t cos(2t). 117 936 4 This can be computed using Sage as well: q (t) = Sage sage: t = var("t") sage: q = function("q",t) sage: L,R,C = var("L,R,C") sage: E = lambda t:sin(2*t)+sin(11*t) sage: de = lambda y: L*diff(y,t,t) + R*diff(y,t) + (1/C)*y-E(t) sage: L,R,C=1,0,1/4 sage: de(q) -sin(2*t) - sin(11*t) + 4*q(t) + D[0, 0](q)(t) sage: desolve(de(q(t)),[q,t],[0,0,0]) -1/4*t*cos(2*t) + 161/936*sin(2*t) - 1/117*sin(11*t) sage: soln = lambda t: -sin(11*t)/117+161*sin(2*t)/936-t*cos(2*t)/4 sage: P = plot(soln,0,10) sage: show(P)
  You can see how the frequency = 2 dominates the other terms. When 0 < R < 2 L/C the homogeneous form of the charge in (2.15) has the form qh (t) = c1 et cos(t) + c2 et sin(t), where = R/2L < 0 and = 4L/C R2 /(2L). This is sometimes called the transient part of the solution. The remaining terms in the charge (those corresponding to the particular qp , when using the method of undertermined coecients to solve the ODE) are called the steady state terms.
  105
  Figure 2.15: A LC circuit, with resonance. Example 2.7.2. An LRC circuit has a 1 henry inductor, a 2 ohm resistor, 1/5 farad capacitor, and an EMF of 50 cos(t). If the initial charge and current is 0, nd the charge at time t. The IVP describing the charge q (t) is q + 2q + 5q = 50 cos(t), q (0) = q (0) = 0. The homogeneous part of the solution is qh (t) = c1 et cos(2t) + c2 et sin(2t). The general form of the particular solution using the method of undetermined coecients is qp (t) = A1 cos(t) + A2 sin(t). Solving for A1 and A2 gives A1 = 10, A2 = 5, so the initial conditions give q (t) = qh (t) + qp (t) = 10et cos(2t) The term qp (t) = 10 cos(t) + 5 sin(t) is sometimes called the steady state charge. We can also solve this using Sage .Sage sage: t = var("t") sage: q = function("q",t)
  106
  sage: L,R,C = var("L,R,C") sage: E = lambda t: 50*cos(t) sage: de = lambda y: L*diff(y,t,t) + R*diff(y,t) + (1/C)*y-E(t) sage: L,R,C = 1,2,1/5 sage: desolve(de(q),[q,t]) (k1*sin(2*t) + k2*cos(2*t))*e(-t) + 5*sin(t) + 10*cos(t) sage: desolve(de(q),[q,t], [0,0,0]) -5/2*(3*sin(2*t) + 4*cos(2*t))*e(-t) + 5*sin(t) + 10*cos(t) sage: soln = lambda t: e(-t)*(-15*sin(2*t)/2-10*cos(2*t))+5*sin(t)+10*cos(t) sage: P = plot(soln,0,10) sage: soln_ss = lambda t: 5*sin(t)+10*cos(t) sage: P_ss = plot(soln_ss,0,10,linestyle=":") sage: soln_tr = lambda t: e(-t)*(-15*sin(2*t)/2-10*cos(2*t)) sage: P_tr = plot(soln_tr,0,10,linestyle="--") sage: show(P+P_ss+P_tr)
  This plot (the solution superimposed with the transient part of the solution) is displayed in Figure 2.7.2.
  Figure 2.16: A LRC circuit, with damping, and the transient part (dashed) of the solution.
  Exercise: 1. Use Sage to solve q + in the case C = 1/121. 1 q = sin(2t) + sin(11t), q (0) = q (0) = 0, C
  107
  2. A simple circuit has an 8 ohm resistor and a 1/17 farad capacitor. A battery of sin(t) volts is attached. Assume that the initial charge is zero. Find the charge at time t. 3. A circuit has an EMF of 20 volts, an inductor of 4 henries, and a capacitor of .01 farads. If the initial current is 5 amps and initial charge on the capacitor is 1.2 coulombs, nd the charge at time t. 4. For an LRC series circuit with L = 2 henrys, R = 14 ohms, C = 0.05 farads, and an electromotive force of E = 120 volts, the steady-state charge on the capacitor is q = a) 120 b) 6 c) 60t d) 120t.
  5. An RC circuit has an electromotive force given by E (t) = 100 sin(100t) volts, a resistor of 1000 ohms and a capacitor of .0005 farads. If the initial charge on the capacitor is zero, then do the following: (a) Draw a schematic diagram of the circuit. (b) Set up and solve an initial value problem for the charge, q (t), at any time. (c) If the voltage across the capacitor is given by vc (t) = voltage v (t) across the capcitor.1 C q (t),
  (d) Express the steady state v (t) found in (c) in the form A sin(t + ) where you must nd A and . 6. An RLC circuit has a 1 henry inductor in series with a 4 ohm resistor and a 1/5 farad capacitor. A battery of sin(t) volts is attached. (a) Assume that the initial charge and current are zero. Find the current at time t. (b) Identify the transient and steady state terms. 7. An RLC circuit has a 1/2 henry inductor in series with a 3 ohm resistor and a 1/5 farad capacitor. A battery of 39 sin(t) volts is attached. (a) Assume that the initial charge and current is zero. Find the charge at time t. (b) Identify the transient and steady state terms. 8. Solve each dierential equation and say if it is underdamped, overdamped, or critically damped. q + 5q + 4q = 0, q (0) = q (0) = 1 q + 2q + 10q = 0, q (0) = 2, q (0) = 0.
  108
  2.82.8.1
  In this part, we recall some basic facts about power series and Taylor series. We will turn to solving ODEs in part 2. Roughly speaking, power series are simply innite degree polynomials f (x) = a0 + a1 x + a2 x + ... =2 k =0
  ak x k ,
  (2.16)
  for some real or complex numbers a0 , a1 , . . . . A power series is a way of expressing a complicated function f (x) as a sum of simple functions like x, x2 , . . . . The number ak is called the coecient of xk , for k = 0, 1, .... Let us ignore for the moment the precise meaning of this innite sum. (How do you associate a value to an innite sum? Does the sum converge for some values of x? If so, for which values? ...) We will return to that issue later. First, some motivation. Why study these? This type of function is convenient for several reasons. It is easy to dierentiate a power series (term-by-term): f (x) = a1 + 2a2 x + 3a3 x2 + ... = k =0
  kak xk1 =
  k =0
  (k + 1)ak+1 xk .
  1 ak xk+1 = k+1
  k =1
  1 ak 1 x k . k
  If the summands ak xk s tend to zero very quickly, then the sum of the rst few terms of the series are often a good numerical approximation for the function itself. Power series enable one to reduce the solution of certain dierential equations down to (often the much easier problem of) solving certain recurrence relations. Power series expansions arise naturally in Taylors theorem from dierential calculus. Theorem 2.8.1. (Taylors Theorem) If f (x) is n + 1 times continuously dierentiable in the open interval (a, x) then there exists a point (a, x) such that f (x) = f (a) + (x a)f (a) + (x a)2 f (a) + 2! (x a)n+1 (n+1) (x a)n (n) f (a) + f ( ). + n! (n + 1)!
  (2.17)
  2.8. THE POWER SERIES METHOD The sum (x a)2 (x a)n (n) f (a) + + f (a), 2! n!
  109
  is called the n-th degree Taylor polynomial of f centered at a. For the case n = 0, the formula is f (x) = f (a) + (x a)f ( ), which is just a rearrangement of the terms in the mean value theorem from dierential calculus, f ( ) = f (x) f (a) . xa
  The Taylor series of f centered at a is the series f (a) + (x a)f (a) + (x a)2 f (a) + . 2!
  When this series converges to f (x) at each point x in some inverval centered about a then we say f has a Taylor series expansion (or Taylor series representation) at a. A Taylor series is basically just a power series, but using powers of x a instead of powers of just x. As the examples below indicate, many of the functions you are used to seeing from calculus have a Taylor series representation. Geometric series: 1 = 1 + x + x2 + x3 + x4 + 1x =
  xn
  (2.18)
  n=0
  To see this, assume |x| < 1 and let n in the polynomial identity 1 + x + x2 + + xn 1 = For x 1, the series does not converge. The exponential function: 1 xn+1 . 1x
  110
  ex = 1 + x +
  x2 x3 x4 + + + 2 6 24 x2 x3 x4 = 1+x+ + + + 2! 3! 4! xn = n! n=0
  (2.19)
  To see this, take f (x) = ex and a = 0 in Taylors theorem (2.17), using the fact that d x x 0 dx e = e and e = 1: ex = 1 + x + xn n+1 x2 x3 + + + + , 2! 3! n! (n + 1)!
  for some between 0 and x. Perhaps it is not clear to everyone that as n becomes larger and larger (x xed), the last (remainder) term in this sum goes to 0. However, Stirlings formula tells us how large the factorial function grows, n! 2n n en
  1 (1 + O( )), n
  so we may indeed take the limit as n to get (2.19). Wikipedias entry on Power series [P1-ps] has a nice animation showing how more and more terms in the Taylor polynomials approximate ex better and better. This animation can also be constructed using Sage (see, for example, the interact at). The cosine function: cos x = 1 x6 x2 x4 + + 2 24 720 x2 x4 x6 =1 + + 2! 4! 6! x2n (1)n = (2n)!n=0
  (2.20)
  This too follows from Taylors theorem (take f (x) = cos x and a = 0). However, there is another trick: Replace x in (2.19) by ix and use the fact (Eulers formula) that eix = cos(x) + i sin(x). Taking real parts gives (2.20). Taking imaginary parts gives (2.21), below.
  2.8. THE POWER SERIES METHOD The sine function: sin x = x x3 x5 x7 + + 6 120 5040 x3 x5 x7 + + =1 3! 5! 7! x2n+1 (1)n = (2n + 1)! n=0
  111
  (2.21)
  Indeed, you can formally check (using formal term-by-term dierentiation) that d cos(x) = sin(x). dx (Alternatively, you can use this fact to deduce (2.21) from (2.20).) The logarithm function: 1 1 1 log(1 x) = x x2 x3 x4 + 2 3 4 1 n = x nn=0
  (2.22)
  1 = log(1 x). 1tSage
  sage: taylor(sin(x), x, 0, 5) x - x3/6 + x5/120 sage: P1 = plot(sin(x),0,pi) sage: P2 = plot(x,0,pi,linestyle="--") sage: P3 = plot(x-x3/6,0,pi,linestyle="-.") sage: P4 = plot(x-x3/6+x5/120,0,pi,linestyle=":") sage: T1 = text("x",(3,2.5)) sage: T2 = text("x-x3/3!",(3.5,-1)) sage: T3 = text("x-x3/3!+x5/5!",(3.7,0.8)) sage: T4 = text("sin(x)",(3.4,0.1)) sage: show(P1+P2+P3+P4+T1+T2+T3+T4)
  This is displayed in Figure 2.17. Exercise: Use Sage to plot successive Taylor polynomial approximations for cos(x).
  112
  Figure 2.17: Taylor polynomial approximations for sin(x). Finally, we turn to the rigorous meaning of these innite sums (2.16). How do you associate a value to an innite sum? Does the sum converge for some values of x? If so, for which values? We will (for the most part) answer all of these. First, consider our innite power series f (x) in (2.16), where the ak are all given and x is xed for the moment. The partial sums of this series are f0 (x) = a0 , f1 (x) = a0 + a1 x, f2 (x) = a0 + a1 x + a2 x2 , . We say that the series in (2.16) converges at x if the limit of partial sumsn
  lim fn (x)
  exists. The following fact is not proven here, as it would take us too far aeld. See for example Gleason [Gl], Theorem 15-2.2. Theorem 2.8.2. There is an R 0 (possibly ) such that the power series (2.16) for f (x) converges absolutely, if |x| < R and diverges if |x| > R. The R in this theorem is called the radius of convergence. There are several tests for determining whether or not a series converges. One of the most commonly used tests is the Theorem 2.8.3. (Root test) Assume L = lim |ak xk |1/k = |x| lim |ak |1/k exists. If L < 1 then the innite power series f (x) in (2.16) converges at x. In general, (2.16) converges for all x satisfying lim |ak |1/k < x < lim |ak |1/k .k k k k
  113
  The number limk |ak |1/k (if it exists, though it can be ) is called the radius of convergence. Example 2.8.1. The radius of convergence of ex (and cos(x) and sin(x)) is . The radius of convergence of 1/(1 x) (and log(1 x)) is 1. Example 2.8.2. The radius of convergence of f (x) = k =0
  k7 + k + 1 k x 2k + k2
  can be determined with the help of Sage . We want to compute lim | k7 + k + 1 1/k | . 2k + k2Sage sage: k = var(k) sage: limit(((k7+k+1)/(2k+k2))(-1/k),k=infinity) 2
  In other words, the series converges for all x satisfying 2 < x < 2. Another useful test for the convergence of a power series is given in the following theorem. Theorem 2.8.4. (Ratio test) Assume L = lim |k
  if L < 1 then the series converges absolutely; if L > 1 then the series does not converge; if L = 1 (or if the limit fails to exist) then the test is inconclusive. Example 2.8.3. The radius of convergence of e = is . We have 1/(k + 1)! = |x| lim (k + 1)1 = 0, k k 1/k! x for all x. By the ratio test, e converges for all x. L = |x| limx k =0
  xk k!
  114
  Representing functions by power series is a powerful technique. You should be aware, however, that there are functions which are innitely dierentiable at a point but not analytic - that is, the Taylor series does not converge to the function in any interval containing the point of expansion. One example of such a function is f (t) = e1/t 02
  if t = 0 if t = 0
  It can be shown that this function is innitely dierentiable, but all of its derivatives at t = 0 are equal to zero. This means its power series expansion around t = 0 is the zero function. The power series converges everywhere - to zero - but since f (t) > 0 for t = 0, it is not converging to the f (t) in any open interval. Exercises: 1. Find the radius of convergence of f (x) = using the root test. 2. Find the radius of convergence of f (x) = using the ratio test. 3. Find the radius of convergence of f (x) = using the root test. 4. Find the radius of convergence of f (x) = using the ratio test. 5. Find the radius of convergence of f (x) = k =0 k =0 k =0 k =0 k =0
  1 k x 2k
  1 3k x 2k
  2k xk
  2.8. THE POWER SERIES METHOD 6. Use Sage to nd the radius of convergence of f (x) = k =0
  115
  k3 + 1 2k x 3k + 1
  2.8.2
  In this part, we solve some dierential equations using power series. We want to solve a problem of the form y (x) + p(x)y (x) + y (x) = f (x), (2.23)
  in the case where p(x), q (x) and f (x) have a power series expansion. We will call a power series solution a series expansion for y (x) where we have produced some algorithm or rule which enables us to compute as many of its coecients as we like. Solution strategy: Write y (x) = a0 + a1 x + a2 x2 + ... = complex numbers a0 , a1 , .... k k =0 ak x ,
  Plug the power series expansions for y , p, q , and f into the dierential equation (2.23). Comparing coecients of like powers of x, derive relations between the aj s. Using these recurrence relations [R-ps] and the ICs, solve for the coecients of the power series of y (x).
  Example 2.8.4. Solve y y = 5, y (0) = 4, using the power series method. This is easy to solve by undetermined coecients: yh (x) = c1 ex and yp (x) = A1 . Solving for A1 gives A1 = 5 and then solving for c1 gives 4 = y (0) = 5 + c1 e0 so c1 = 1 so y = ex 5. Solving this using power series, we compute y (x) y (x) 5 = = =k a1 + 2a2 x + 3a3 x2 + ... = k =0 (k + 1)ak +1 x 2 k a0 a1 x a2 x ... = k=0 ak x k (a0 + a1 ) + (a1 + 2a2 )x + ... = k =0 (ak + (k + 1)ak +1 )x
  Comparing coecients,
  116
  CHAPTER 2. SECOND ORDER DIFFERENTIAL EQUATIONS for k = 0: 5 = a0 + a1 , for k = 1: 0 = a1 + 2a2 , for general k: 0 = ak + (k + 1)ak+1 for k > 0.
  The IC gives us 4 = y (0) = a0 , so a0 = 4, a1 = 1, a2 = 1/2, a3 = 1/6, , ak = 1/k!. This implies y (x) = 4 + x + x/2 + + xk /k! + = 5 + ex , which is in agreement from the previous discussion. Example 2.8.5. Solve the dierential equation x2 y + xy + x2 y = 0, y (0) = 1, y (0) = 0, using the power series method. This equation is called Bessels equation [B-ps] of the 0-th order. This dierential equation is so well-known (it has important applications to physics and engineering) that the coecients in the series expansion were computed long ago. Most texts on special functions or dierential equations have further details on this, but an online reference is [B-ps]. Its Taylor series expansion around 0 is: J0 (x) = m=0
  (1)m x m!2 2
  2m
  for all x. We shall next verify that y (x) = J0 (x). Let us try solving this ourselves using the power series method. We compute x2 y (x) xy (x) x2 y (x) 0 = = = =k 0 + 0 x + 2a2 x2 + 6a3 x3 + 12a4 x4 + ... = k =0 (k + 2)(k + 1)ak +2 x 0 + a1 x + 2a2 x2 + 3a3 x3 + ... = k=0 kak xk k 0 + 0 x + a0 x2 + a1 x3 + ... = k =2 ak 2 x 2 k 0 + a1 x + (a0 + 4a2 )x2 + ..... = a1 x + k =2 (ak 2 + k ak )x .
  2.8. THE POWER SERIES METHOD In general, 1 , a2k+1 = 0, k!2 for k 1. This is in agreement with the series above for J0 . Some of this computation can be formally done in Sage using power series rings. a2k = (1)k 22kSage sage: R6.<a0,a1,a2,a3,a4,a5,a6> = PolynomialRing(QQ,7) sage: R.<x> = PowerSeriesRing(R6) sage: y = a0 + a1*x + a2*x2 + a3*x3 + a4*x4 + a5*x5 +\ a6*x6 + O(x7) sage: y1 = y.derivative() sage: y2 = y1.derivative() sage: x2*y2 + x*y1 + x2*y a1*x + (a0 + 4*a2)*x2 + (a1 + 9*a3)*x3 + (a2 + 16*a4)*x4 +\ (a3 + 25*a5)*x5 + (a4 + 36*a6)*x6 + O(x7)
  117
  This is consistent with our computations above. Sage knows quite a few special functions, such as the various types of Bessel functions.Sage sage: sage: sage: sage: sage: sage: sage: b = lambda x:bessel_J(x,0) P = plot(b,0,20,thickness=1) show(P) y = lambda x: 1 - (1/2)2*x2 + (1/8)2*x4 - (1/48)2*x6 P1 = plot(y,0,4,thickness=1) P2 = plot(b,0,4,linestyle="--") show(P1+P2)
  This is displayed in Figure 2.18-2.19. Exercises: 1. Using a power series around t = 0 of the form y (t) = relation for the ci if y satises y + ty + y = 0. n n=0 cn t ,
  nd the recurrence
  2. (a) Find the recurrence relation of the coecients of the power series solution to the n ODE x = 4x3 x around x = 0, i.e. for solutions of the form x = n=0 cn t . (b) Find an explicit formula for cn in terms of n. 3. Use Sage to nd the rst 5 terms in the power series solution to x + x = 0, x(0) = 1, x (0) = 0. Plot this Taylor polynomial approximation over < t < .
  118
  Figure 2.18: The Bessel function J0 (x), for 0 < x < 20.
  4. Find two linearly independent solutions of Airys equation x tx = 0 using power series. 5. Show that the coecients of the power series solution to the initial value problem y y y = 0, y (0) = 0, y (0) = 1 have the form cn = Fn /n! where Fn is the nth Fibonacci number. (The Fibonacci numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34, . . . , satisfying the recursion relation that each number is the sum of the previous two in the sequence.) 6. Determine the power series solution and radius of convergence of the ODE y + x2 y = 0.
  Historically, the Laplace transform method of solving dierential equations is attributed to Laplace himself. However, Oliver Heaviside6 played an important role in his (independently made) rediscovery and implementation of the method.A self-taught English electrical engineer, mathematician, and physicist, who lived from 1850-1925. Heaviside won the prestigous Faraday Medal in 1922.6
  119
  2.9.1
  Pierre Simon Laplace (1749 - 1827) was a French mathematician and astronomer who is regarded as one of the greatest scientists of all time. His work was pivotal to the development of both celestial mechanics and probability, [L-lt], [LT-lt]. The Laplace transform (sometimes abbreviated LT) of a function f (t), dened for all real numbers t 0, is the function F (s), dened by: F (s) = L [f (t)] = 0
  The Laplace transform sends nice functions of t (we will be more precise later) to functions of another variable s. It has the wonderful property that it transforms constant-coecient dierential equations in t to algebraic equations in s. The Laplace transform has two very familiar properties: Just as the integral of a sum is the sum of the integrals, the Laplace transform of a sum is the sum of Laplace transforms: L [f (t) + g(t)] = L [f (t)] + L [g(t)] Just as constant factor can be taken outside of an integral, the Laplace transform of a constant times a function is that constant times the Laplace transform of that function: L [af (t)] = aL [f (t)] In other words, the Laplace transform is linear. For which functions f is the Laplace transform actually dened on? We want the indenite integral to converge, of course. A function f (t) is of exponential order if there exist constants t0 and M such that |f (t)| < M et , for all t > t0 .
  Roughly speaking, the magnitude of f (t) should eventually be bounded above by some t exponential function. If 0 0 f (t) dt exists and f (t) is of exponential order then the Laplace transform L [f ] (s) exists for s > . Example 2.9.1. Consider the Laplace transform of f (t) = 1. The Laplace transform integral converges for s > 0. L [f ] (s) = 0
  est dt 0
  1 = est s 1 = s
  120
  Example 2.9.2. Consider the Laplace transform of f (t) = eat . The Laplace transform integral converges for s > a. L [f ] (s) = 0
  e(as)t dt 0
  1 (as)t e sa 1 = sa =
  Example 2.9.3. Consider the Laplace transform of the translated unit step (or Heaviside) function, u(t c) = 0 1 for t < c for t > c,
  where c > 0. (this is sometimes also denoted H (t c).) This function is o (i.e., equal to 0) until you get to t = c, at which time it turns on. The Laplace transform of it is 0
  L[u(t c)] = =
  est H (t c) dt est dt c
  ecs = s
  est s
  for s > 0
  Using this formula, any linear combination of translates of the unit step function can also be computed. For example, if f (t) = 2u(t) 3u(t 2) + 5u(t 6), then F (s) = 2 3e2s 5e6s + . s s s
  The inverse Laplace transform in denoted f (t) = L1 [F (s)](t), where F (s) = L [f (t)] (s). Usually we compute the inverse Laplace transform by looking up the form of the transform in a table. It is possible to compute the inverse directly with complex-valued integrals, which
  121
  is a topic in a more advanced course in complex analysis. There is also the amazing (if not impractical) formula of Emil Post [Post]: f (t) = L1 [F (s)] = lim Example 2.9.4. Consider f (t) = 1, 0, for t < 2, on t 2. (1)n n n! n tn+1
  F (n)
  n t
  (2.24)
  (Incidently, this can also be written 1 u(t 2).) We show how Sage can be used to compute the Laplace transform of this.Sage sage: sage: sage: sage: 1/s sage: sage: sage: sage: sage: t = var(t) s = var(s) f = Piecewise([[(0,2),1],[(2,infinity),0]]) f.laplace(t, s) e(-(2*s))/s f1 = lambda t: 1 f2 = lambda t: 0 f = Piecewise([[(0,2),f1],[(2,10),f2]]) P = f.plot(rgbcolor=(0.7,0.1,0.5),thickness=3) show(P)
  According to Sage , L [f ] (s) = 1/s e2s /s. Note the function f was redened for plotting purposes only (the fact that it was redened over 0 < t < 10 means that Sage will plot it over that range.) The plot of this function is displayed in Figure 2.20.
  122
  Next, some properties of the Laplace transform. Dierentiate the denition of the Laplace transform with respect to s: F (s) = Repeating this: dn F (s) = (1)n dsn 0 0
  (2.25)
  In the denition of the Laplace transform, replace f (t) by its derivative f (t): L f (t) (s) = Now integrate by parts (u = 0 0
  est ,
  dv = 0
  f (t) dt):
  Therefore, if F (s) is the Laplace transform of f (t) then sF (s) f (0) is the Laplace transform of f (t): L f (t) (s) = sL [f (t)] (s) f (0). Replace f by f in (2.26), L f (t) (s) = sL f (t) (s) f (0), and apply (2.26) again: L f (t) (s) = s2 L [f (t)] (s) sf (0) f (0), (2.28) (2.27) (2.26)
  Using (2.26) and (2.28), the Laplace transform of any constant coecient ODE ax (t) + bx (t) + cx(t) = f (t) isa(s2 L [x(t)] (s) sx(0) x (0)) + b(sL [x(t)] (s) x(0)) + cL [x(t)] (s) = F (s),
  where F (s) = L [f (t)] (s). In particular, the Laplace transform of the solution, X (s) = L [x(t)] (s), satises
  123
  X (s) = (F (s) + asx(0) + ax (0) + bx(0))/(as2 + bs + c). Note that the denominator is the characteristic polynomial of the dierential equation. Moral of the story: it is generally very easy to compute the Laplace transform X (s) of the solution to any constant coecient non-homogeneous linear ODE. Computing the actual solution x(t) is usually much more work. Example 2.9.5. We know now how to compute not only the Laplace transform of f (t) = eat (its F (s) = (s a)1 ) but also the Laplace transform of any function of the form tn eat by dierentiating it: L teat = F (s) = (s a)2 , L t2 eat = F (s) = 2 (s a)3 , ... , and in general L tn eat = F (s) = n! (s a)n1 . Let us now solve a rst order ODE using Laplace transforms. Example 2.9.6. Let us solve the dierential equation x + x = t100 et , x(0) = 0. (2.29)
  using Laplace transforms. Note this would be highly impractical to solve using undetermined coecients. (You would have 101 undetermined coecients to solve for!) First, we compute the Laplace transform of the solution to the dierential equation. The Laplace transform of the left-hand side: by (2.29), L x + x = sX (s) + X (s), where F (s) = L [f (t)] (s). For the Laplace transform of the right-hand side, let F (s) = L et = By (2.25), d100 d100 1 100 t F ( s ) = L t e = . ds100 ds100 s + 1 The rst several derivatives of1 s+1
  1 . s+1
  are as follows:
  d 1 1 1 1 d2 1 d3 1 = = 2 = 62 , , , 2 2 3 3 ds s + 1 (s + 1) ds s + 1 (s + 1) ds s + 1 (s + 1)4 and so on. Therefore, the Laplace transform of the right-hand side is: 1 d100 1 = 100! . 100 ds s + 1 (s + 1)101
  124 Consequently,
  1 . (s + 1)102
  Let us now solve a second order ODE using Laplace transforms. Example 2.9.7. Let us solve the dierential equation x + 2x + 2x = e2t , x(0) = x (0) = 0,
  using Laplace transforms. The Laplace transform of the left-hand side: by (2.29) and (2.27), L x + 2x + 2x = (s2 + 2s + 2)X (s), as in the previous example. The Laplace transform of the right-hand side is: 1 . s+2 Solving for the Laplace transform of the solution algebraically: L e2t = X (s) = 1 . (s + 2)((s + 1)2 + 1)
  The inverse Laplace transform of this can be obtained from Laplace transform tables after rewriting this using partial fractions: s s+1 1 1 1 1 1 1 1 1 = + . 2 2 2 s + 2 2 (s + 1) + 1 2 s + 2 2 (s + 1) + 1 2 (s + 1)2 + 1
  X (s) =
  The inverse Laplace transform is: 1 1 2t 1 t e e cos(t) + et sin(t). 2 2 2 We show how Sage can be used to do some of this. We break the Sage solution into steps. Step 1: First, we type in the ODE and take its Laplace transform. x(t) = L1 [X (s)] =Sage sage: sage: sage: sage: s,t,X = var(s,t,X) x = function("x",t) de = diff(x,t,t)+2*diff(x,t)+2*x==e(-2*t) laplace(de,t,s)
  125
  Step 2: Now we solve this equation for X = laplace(x(t), t, s). For this, we use Python to do some string replacements. Python is the underlying language for Sage and has very powerful string manipulation functions.Sage sage: strLTde = str(LTde).replace("laplace(x(t), t, s)", sage: strLTde0 = strLTde.replace("x(0)","0") sage: strLTde00 = strLTde0.replace("D[0](x)(0)","0") sage: LTde00 = sage_eval(strLTde00,locals={"s":s,"X":X}) sage: soln = solve(LTde00,X) sage: Xs = soln[0].rhs(); Xs 1/(s3 + 4*s2 + 6*s + 4) "X")
  Step 3: Now that we have solved for the Laplace transform of the solution, we take inverse Laplace transforms to get the solution to the original ODE. There are various ways to do this. One way which is convenient if you also want to check the answer using tables, if to compute the partial fraction decomposition then take inverse Laplace transforms.Sage sage: factor(s3 + 4*s2 + 6*s + 4) (s + 2)*(s2 + 2*s + 2) sage: f = 1/((s+2)*((s+1)2+1)) sage: f.partial_fraction() 1/(2*(s + 2)) - s/(2*(s2 + 2*s + 2)) sage: f.inverse_laplace(s,t) e(-t)*(sin(t)/2 - cos(t)/2) + e(-(2*t))/2
  Exercises: 1. Compute the Laplace transform of the function v (t) = directly from the denition L(v ) = 2. Write 1 for t [0, 1] 0 for t [0, 1],
  st v (t)dt. 0 e
  1, for t [0, 2], 3, for t (2, 5], f (t) = 2, for t (5, 7], 0, for t [0, 7],
  126
  CHAPTER 2. SECOND ORDER DIFFERENTIAL EQUATIONS as a linear combination of translates of the unit step function. Compute the Laplace transform of the function using the formula in Example 2.9.3.
  3. Compute the Laplace transform of f (t) = 1 + cos(3t). 4. Compute the Laplace transform of f (t) = sin(2t + /4). 5. Compute the Laplace transform of f (t) = (t + 1)2 . 6. Compute the Laplace transform of f (t) = t 3e3t . 7. Compute the Laplace transform of f (t) = 2tet . 8. Compute the Laplace transform of f (t) = tet sin(t). 9. Use Sage to solve the dierential equation below using the Laplace transform method. x + 2x + 5x = et , as in the example above. 10. Use the Laplace transform method to solve the initial value problem x x 2x = 0, x(0) = 0, x (0) = 1. 11. Use the Laplace transform method to solve the initial value problem x = x + 2y , y = x + et , x(0) = 0, y (0) = 0. x(0) = x (0) = 0,
  2.9.2
  In this part, we shall focus on two other aspects of Laplace transforms: solving dierential equations involving unit step (Heaviside) functions, convolutions and applications. It follows from the denition of the Laplace transform that if f (t) F (s) = L[f (t)](s), f (t)u(t c) ecs L[f (t + c)](s), f (t c)u(t c) ecs F (s).L L L
  then
  (2.30)
  (2.31)
  127
  Example 2.9.8. First, consider the Laplace transform of the piecewise-dened function f (t) = (t 1)2 u(t 1). Using (2.31), this is 1 s e . s3 Second, consider the Laplace transform of the piecewise-constant function for t < 0, 0 f (t) = 1 for 0 t 2, 1 for t > 2. L[f (t)] = es L[t2 ](s) = 2
  L[f (t)] = L[u(t)] + 2L[u(t 2)] 1 1 = + 2 e2s . s s Finally, consider the Laplace transform of f (t) = sin(t)u(t ). Using (2.30), this is L[f (t)] = es L[sin(t + )](s) = es L[ sin(t)](s) = es 1 . s2 + 1
  Figure 2.21: The piecewise continuous function u(t ) sin(t). We show how Sage can be used to compute these Laplace transforms.Sage sage: t = var(t) sage: s = var(s) sage: assume(s>0)
  128
  sage: f = Piecewise([[(0,1),0],[(1,infinity),(t-1)2]]) sage: f.laplace(t, s) 2*e(-s)/s3 sage: f = Piecewise([[(0,2),-1],[(2,infinity),2]]) sage: f.laplace(t, s) 3*e(-(2*s))/s - 1/s sage: f = Piecewise([[(0,pi),0],[(pi,infinity),sin(t)]]) sage: f.laplace(t, s) -e(-(pi*s))/(s2 + 1) sage: f1 = lambda t: 0 sage: f2 = lambda t: sin(t) sage: f = Piecewise([[(0,pi),f1],[(pi,10),f2]]) sage: P = f.plot(rgbcolor=(0.7,0.1,0.5),thickness=3) sage: show(P)
  The plot given by these last few commands is displayed in Figure 2.21. Before turning to dierential equations, let us introduce convolutions. Let f (t) and g(t) be continuous (for t 0 - for t < 0, we assume f (t) = g(t) = 0). The convolution of f (t) and g(t) is dened byt t
  (f g) =
  f (z )g(t z ) dz =
  f (t z )g(z ) dz.
  The convolution theorem states L[f g(t)](s) = F (s)G(s) = L[f ](s)L[g](s). The Laplace transform of the convolution is the product of the Laplace transforms. (Or, equivalently, the inverse Laplace transform of the product is the convolution of the inverse Laplace transforms.) To show this, do a change-of-variables in the following double integral: 0 z t 0
  L[f g(t)](s) = =
  est
  f (z )g(t z ) dz dt
  est f (z )g(t z ) dt dz z
  =0
  esz f (z )
  es(tz ) g(t z ) dt dz 0
  esu f (u) du
  esv g(v ) dv
  = L[f ](s)L[g](s).
  129
  1 Example 2.9.9. Consider the inverse Laplace transform of s3 . This can be computed s2 7 using partial fractions and Laplace transform tables . However, it can also be computed using convolutions. First we factor the denominator, as follows
  1 1 1 = 2 . s3 s2 s s1 1 = et s1
  L 1
  zetz dz ett 0
  = et zez
  ez dz
  = t 1 + et Therefore, L 1 1 1 (t) = et t 1. s2 s 1
  Example 2.9.10. Here is a neat application of the convolution theorem. Consider the convolution f (t) = 1 1 1 1 1. What is it? First, take the Laplace transform. Since the Laplace transform of the convolution is the product of the Laplace transforms: L[1 1 1 1 1](s) = (1/s)5 = We know from Laplace transform tables that L1 f (t) = L1 [F (s)] (t) =4! s5
  1 = F (s). s5
  (t) = t4 , so
  1 1 4! 1 L (t) = t4 . 5 4! s 4!
  This clever computation is called computing 1 1 1 1 1 using the convolution theorem. You can also compute 1 1 1 1 1 directly in Sage :7
  There are Laplace transforms at the back of this text as well as in Table 2.9.3 below.
  130
  sage: t,z = var("t,z") sage: f(t) = 1 sage: ff = integral(f(t-z)*f(z),z,0,t); ff t sage: fff = integral(f(t-z)*ff(z),z,0,t); fff 1/2*t2 sage: ffff = integral(f(t-z)*fff(z),z,0,t); ffff 1/6*t3 sage: fffff = integral(f(t-z)*ffff(z),z,0,t); fffff 1/24*t4 sage: s = var("s") sage: (1/s5).inverse_laplace(s,t) 1/24*t4
  Example 2.9.11. Computing sin(t) cos(t) using the convolution theorem. Let f (t) = sin(t) cos(t). By the convolution theorem F (s) = (s2 s , + 1)2
  The inverse Laplace transform can be computed using Laplace transform tables: L1[F (s)](t) = 1 t sin (t) . 2
  Example 2.9.12. Here is another application of the convolution theorem. Consider the convolution f (t) = sin(t) u(t 1). What is it? First, take the Laplace transform, using the convolution theorem: es . s3 + s By the rtanslation theorem (2.31), the inverse Laplace transform of F (s) is the translation of the inverse Laplace transform of (s3 + s)1 = s/(s2 + 1) + 1/s: F (s) = f (t) = ( cos(t 1) + 1)u(t 1). A general method Now let us turn to solving a second order constant coecient dierential equation, ay + by + cy = f (t), First, take Laplace transforms of both sides: y (0) = y0 , y (0) = y1 . (2.32)
  131
  so
  asy0 + ay1 + by0 1 F (s) + . (2.33) as2 + bs + c as2 + bs + c 1 The function as2 + is sometimes called the transfer function (this is an engineering bs+c term) and its inverse Laplace transform, Y (s) = 1 (t), as2 + bs + c the weight function for the dierential equation. Roughly speaking, this represents the response to a unit impulse at t = 0. w(t) = L1 Lemma 2.9.1. If a = 0 then w(t) = 0. (The only proof we have of this is a case-by-case proof using Laplace transform tables. Case 1 is when the roots of as2 + bs + c = 0 are real and distinct, case 2 is when the roots are real and repeated, and case 3 is when the roots are complex. In each case, w(0) = 0. The verication of this is left to the reader, if he or she is interested.) By the above lemma and the rst derivative theorem, w (t) = L1 as2 s (t). + bs + c
  Using this and the convolution theorem, the inverse Laplace transform of (2.33) is y (t) = (w f )(t) + ay0 w (t) + (ay1 + by0 ) w(t). (2.34)
  This proves the following fact, sometimes referred to as the impulse-response formula. Theorem 2.9.1. The unique solution to the dierential equation (2.32) is (2.34). Example 2.9.13. Consider the dierential equation y + y = 1, y (0) = y (0) = 0. The weight function is the inverse Laplace transform of s21 +1 , so w (t) = sin(t). By (2.34),t
  y (t) = 1 sin(t) =
  (Yes, it is just that easy to solve that diferential equation.) If the impulse f (t) is piecewise-dened, sometimes the convolution term in the formula (2.34) is awkward to compute. Example 2.9.14. Consider the dierential equation y + y = u(t 1), y (0) = y (0) = 0. , so w(t) = sin(t). By (2.34), The weight function is the inverse Laplace transform of s21 +1 y (t) = sin(t) u(t 1) = ( cos(t 1) + 1)u(t 1),
  by Example 2.9.12.
  132
  Example 2.9.15. Consider the dierential equation y y = u(t 1), y (0) = y (0) = 0. s Taking Laplace transforms gives s2 Y (s) sY (s) = 1 s e , so Y (s) = We know from a previous example that L 1 s3 s3 1 es . s2
  1 (t) = et t 1, s2
  so by the translation theorem (2.31), we have 1 es (t) = (et1 (t 1) 1) u(t 1) = (et1 t) u(t 1). s2
  y (t) = L1
  s3
  Example 2.9.16. We try to solve x x = u(t 1), x(0) = x (0) = 0 using Sage . At this stage, Sage lacks the functionality to solve this dierential equation as easily as others but we can still use Sage to help with the solution. First, we initialize some variables and take the Laplace transform of f (t) = u(t 1).Sage sage: s,t,X = var(s,t,X) sage: x = function("x",t) sage: f1 = 0 sage: f2 = 1 sage: u1 = Piecewise([[(0,1),f1],[(1,Infinity),f2]]) sage: F = u1.laplace(t,s); F e(-s)/s
  Next, we take the Laplace transform of the dierential equation x x = f (t), with an arbitrary function f (t).Sage sage: ft = function("ft",t) sage: de = diff(x,t,t) - x==ft sage: LTde = laplace(de,t,s); LTde s2*laplace(x(t), t, s) - s*x(0) - laplace(x(t), t, s) - D[0](x)(0) == laplace(ft(t), t, s)
  Next, we take this equation and solve it for X (s) using Pythons string manipulation functionality:
  133
  sage: strLTde = str(LTde).replace("laplace(x(t), t, s)", "X") sage: strLTde0 = strLTde.replace("x(0)","0") sage: strLTde00 = strLTde0.replace("D[0](x)(0)","0") sage: strLTde00F = strLTde00.replace("laplace(ft(t), t, s)", "F") sage: strLTde00F s2*X - s*0 - X - 0 == F
  Next,we plug in for f (t) the function u(t 1) (denoted u1 in the above Sage code) and solve:Sage sage: LTde00F = sage_eval(strLTde00F,locals={"s":s,"X":X,"F":F}) sage: LTde00F X*s2 - X == e(-s)/s sage: soln = solve(LTde00F,X) sage: Xs = soln[0].rhs(); Xs e(-s)/(s3 - s) sage: Xs.partial_fraction() 1/2*e(-s)/(s - 1) + 1/2*e(-s)/(s + 1) - e(-s)/s
  Unfortunately, at this time, Sage cannot take the inverse Laplace transform of this. We can look up the terms in Laplace transform tables (such as Table 2.9.3) and do this by hand with ease: 1 1 x(t) = et1 u(t 1) + e(t1) u(t 1) u(t 1). 2 2 This solves x x = u(t 1), x(0) = x (0) = 0. Dirac delta functions The Dirac delta function (t) is technically not a function. Roughly speaking, it may be thought of as being analogous to a radar ping: if a continuous function f (t) represents an objects trajectory then the delta function samples its value at t = 0. The graph of (t) can be visualized as in Figure 2.22. To be precise, (t) is a linear functional (that is, a linear function which sends functions to numbers) dened by the formula
  where f (t) is any continuous function. Of course, the interval (, ) can be replaced by any interval containing 0. In particular, if a > 0 then
  134
  A special case of the above formula gives the Laplace transform of the delta function: L((t)) = (t)est dt = 1
  Though it is not obvious at rst glance, the delta function is not unrelated to the unit step function. Indeed, in some sense the delta function may be regarded as the derivative of the (discontinuous!) unit step function. Let f (t) be any function on the real line whose derivative is integrable on (0, ) and satises limt f (t) = 0. Formally integrating by parts, we obtain (t)f (t) dt = f (0) = 0 f (t) dt = u(t)f (t) dt = u(t)f (t)| u (t)f (t) dt = u (t)f (t) dt. Comparing the left-hand side with the right-hand side, we see that in some sense (which can be made precise using the theory of functional analysis, but which goes beyond the scope of these notes), we have (t) = u (t). Example 2.9.17. Consider the dierential equation x + x = (t ), x(0) = x (0) = 0. This models a unit mass attached to an undamped spring suspended from a board with no initial displacement or initial velocity. At time t , the board is hit very hard (say with a hammer blow) in the upward direction. As we will see, this will start the mass to oscillate. Taking Laplace transforms gives s2 X (s) + X (s) = es , so X (s) = s2 1 es . +1
  135
  The inverse Laplace transform is x(t) = sin(t)u(t ) (see Example 2.9.8 above, which also has a graph of this solution).
  2.9.3
  We conclude our study of the Laplace transform with the case of a periodic non-homogeneous term, i.e. ODEs of the form ax (t) + bx (t) + cx(t) = f (t) where f (t) is a periodic function of period T . This means that f (t) = f (t + T ) for all t. For this important special case we can reduce the improper integral in the Laplace transform to the nite interval [0, T ]: Lemma 2.9.2. If f (t) is a periodic function with period T , then L[f (t)](s) = 1 1 eT sT 0
  est f (t)dt.
  Proof. First we write the Laplace transform as a sum over the period intervals: L[f (t)](s) = 0
  st
  f (t)dt =
  (n+1)T
  est f (t)dt,
  n=0 nT
  and then use the substitution u = t nT in each interval. For convenience, we then relabel u as t once again, to get L[f (t)](s) = n=0 0 T
  es(tnT ) f (t) dt =0
  est f (t) dt
  n=0
  ensT .
  The sum is over a geometric series in esT since ensT = (esT )n . For s > 0, esT < 1 and 1 nsT = ), giving the desired result. the series converges ( n=0 e 1eT s Exercises: 1. Compute the convolution t t. 2. Compute the convolution et et . 3. Compute the convolution sin(t) sin(t). 4. Show that the Laplace transform of the square-wave function g(t) graphed in Figure 1 es . 4 is equal to s(1 + es )
  136
  CHAPTER 2. SECOND ORDER DIFFERENTIAL EQUATIONS Function 1 t tn eat cos (kt) sin (kt) eat sin (kt) eat cos (kt) eat f (t) tf (t)t 0 f ( )d f (t)
  Transform1 s 1 s2 n! sn+1 1 s a s s 2 +k 2 k s 2 +k 2 k (sa)2 +k 2 s (sa)2 +k 2
  F (s a) F (s) F (s)/s
  Table 2.1: Some Laplace transforms, L(f (t)) = F (s) 5. Use Sage to solve the following problems. (a) Find the Laplace transform of u(t /4) cos(t).
  (b) Compute the convolution sin(t) cos(t). Do this directly and using the convolution theorem. 6. (a) Use Laplace transforms and partial fractions to solve y + 4y + 8y = 8, y (0) = 2, y (0) = 2. (b) Sketch f (t) = 1 + cos(t)u(t ) and nd L[f (t)]1 (s2 +1)2
  8. Use Laplace transforms to solve the initial value problem y + 4y = (t ), y (0) = 0, y (0) = 1 Sketch your solution on the interval 0 t 2 . 9. Let f (t) = 1 if 0 t 3 . 0 if t 3
  137
  (a) Write f (t) in terms of the unit step function. (c) The charge on an electrical circuit is modeled by the following q + 4q + 8q = f (t), q (0) = q (0) = 0. (b) Use (a) to nd L {f (t)}.
  Use (b) to solve this initial value problem for the charge. 10. (a) Find: (i) L{4t3 4et sin(2t)}, (ii) L{t2 U (t 1)}, (iii) L{t cos t}.
  (b) Find L{f (t)} for the periodic function given over one period by f (t) = 1, 0 < t 2, 0, 2 < t < 3.
  11. (a) Find (i) L1 {1/(s2 + 9) e3s /(s 4)2 }, (ii) L1 {(s + 4)/(s2 + 6s + 25)}. 12. The convolution t t t t t = (a)1 s10
  (c) t5
  1 (s2 +1)4
  138
  Chapter 3
  In order to handle systems of dierential equations, in which there is more than one dependent variable, it is necessary to learn some linear algebra. It is best to take a full course in that subject, but for those cases where that is not possible we aim to provide the required background in this chapter. In contrast to a linear algebra textbook per se we will omit many proofs. Linear algebra is a tremendously useful subject to understand, and we encourage the reader to take a more complete course in it if at all possible. An excellent free reference for linear algebra is the text by Robert Beezer [B-rref].
  3.13.1.1
  A matrix is a two-dimensional rectangular array of numbers. We always refer to the entry in the ith row and j th column as the (i, j )-entry - that is we always refer to the row index rst. A matrix having m rows and n columns is called an m n matrix. A matrix with m = n is called a square matrix. If A is an m n matrix then the transpose of A, denoted AT , is the n m matrix where the rows of AT are the columns of A and the columns of AT are the rows of A. For example, if A= then 139 1 2 3 4 5 6
  140
  1 4 AT = 2 5 . 3 6 Two matrices can be added together if they have the same dimensions (the same numbers of rows and columns). The sum is dened to be the componentwise sum: the matrix whose (i, j ) entry is the sum of the (i, j ) entries of the summand matrices. For example: 1 0 0 1 + 1 1 2 3 = 0 1 2 2 .
  Matix multiplication is not dened in the same way. A matrix product C = AB is dened if matrix A has the same number of columns as the number of rows in B . In other words, A must have dimensions (m, p) and matrix B must have dimensions (p, n). If A has row vectors Aj (1 j m, each of which is a vector in Rp ) and B has column vectors Bk (1 k n, each of which is a vector in Rp ) then the j, k-th entry of AB is the dot product Aj Bk . Therefore, the resulting matrix C will have dimensions (m, n), with the following formula for the (i, j ) entry of C (denoted ci,j ) in terms of the entries of A and B : cj,k =i
  aj,i bi,k .
  Lemma 3.1.1. The matrix product is associative: if A, B , and C are matrices and the products AB and BC are dened, then A(BC ) = (AB )C . Proof. We can directly verify this property from the denition of the matrix product. The entry of the ith row and j th column of the product (AB )C is: (m k
  =k
  aikm
  It is extremely important to note that matrix multiplication is not commutative. Here are some examples illustrating this: 1 2 3 4 5 6 1 2 3 4 = 5 6 22 28 49 64
  141
  In fact, we say two n n matrices A, B are similar if there is a matrix C such that AC = CB. Unlike real numbers, in general, two matrices are not similar. For example, A= are similar (why?) but A= 2 0 0 3 , B= 3 0 0 4 2 0 0 3 , B= 3 0 0 2
  1 2 3 4 5 6
  1 2 3 4 5 6
  9 12 15 = 19 26 33 29 40 51
  A matrix with all entries equal to zero is called the zero matrix. Another important dierence between matrix multiplication and multiplication with scalars is that it is possible for the product of two non-zero matrices to equal the zero matrix. For example: 1 2 3 4 5 6 1 1 2 2 = 1 1 0 0 0 0
  are not similar. The analogues of the number 1 for matrices are the identity matrices. The n n identity matrix is denoted In (or sometimes just I if the context is clear) and is dened as the matrix whose diagonal entries (at positions (i, i) for i (1, . . . , n)) are 1 and whose o-diagonal entries are zero: 1 0 ... 0 0 0 1 ... 0 0 . . .. . In = . . . . . 0 0 ... 1 0 0 0 ... 0 1
  This also means that one cannot always cancel common factors in a matrix equation - i.e. for matrices A, B , and C , if AB = AC it is not necessarily true that B = C . In order to perform such a cancellation, we need the matrix A to have a stronger property than merely being nonzero - it needs to be invertible. Denition 3.1.1. A n n matrix A is invertible if there exists a matrix B such that AB = BA = In . In this case, the inverse of A is dened to be A1 = B .
  A= is invertible and A=
  2 0 0 3
  1/2 0 0 1/3
  We shall discuss matrix inverses more below. Exercises: 1. For A = a b c d , compute the product A2 = A A.
  For the following three pairs of matrices, compute whatever products between them are dened. 2. A= 1 2 1 ,B = 1 3 0 1 2 1
  3. A= 2 0 1 0 ,B = 1 0 0 2
  4. A= 3 4 1 1 ,B = 2 2 1 1
  5. Find a two by two matrix A with all non-zero entries such that A2 = 0. 6. Find a two by two matrix A with each diagonal element equal to zero such that A2 = I . (The existence of such matrices is important, as it lets us represent complex numbers by real matrices). 7. Find as many 2 by 2 matrices A such that A2 = I as you can. Do you think you have found all of them? 8. Compute the matrix product A B , where A is the 4 1 (column) matrix of all 1s and B is the 1 4 (row) matrix (1 2, 3 4).
  143 5 3 3 1
  3.2
  Row reduction is the engine that drives a lot of the machinary of matrix theory. What we call row reduction others call computing the reduced row echelon form or GaussJordan reduction. The term Gauss elimination is usually reserved for the process of solving a system of equations by rst putting them in upper-triangular form, then using substitution to solve for the unknowns.
  3.2.1
  This is actually a discussion of solving systems of equations using the method of row reduction, but its more fun to formulate it in terms of a game. To be specic, lets focus on a 2 2 system (by 2 2, we mean 2 equations in the 2 unknowns x, y ): ax + by = r1 cx + dy = r2 (3.1)
  Here a, b, c, d, r1 , r2 are given constants. Putting these two equations down together means to solve them simultaneously, not individually. In geometric terms, you may think of each equation above as a line the the plane. To solve them simultaneously, you are to nd the point of intersection (if it exists) of these two lines. Since a, b, c, d, r1 , r2 have not been specied, it is conceivable that there are no solutions (the lines are parallel but distinct), innitely many solutions (the lines are the same), exactly one solution (the lines are distinct and not parallel, as in Figure 3.1). Usually there is exactly one solution. Of course, you can solve this by simply manipulating equations since it is such a low-dimensional system but the object of this section is to show you a method of solution which is scalable to industrial-sized problems (say 1000 1000 or larger). Strategy: Step 1: Write down the augmented matrix of (3.1):
  144
  A=
  a b r1 c d r2
  This is simply a matter of stripping o the unknowns and recording the coecients in an array. Of course, the system must be written in standard form (all the terms with x get aligned together in a single column, all the terms with y get aligned . . . ) to do this correctly. Step 2: Play the Gauss elimination game (described below) to computing the row reduced echelon form of A, call it B . Step 3: Read o the solution (assuming it exists and is unique) from the right-most column of B . The Gauss Elimination Game Legal moves: These actually apply to any m n matrix A with m < n. 1. Ri Rj : You can swap row i with row j . 2. cRi Ri (c = 0): You can replace row i with row i multiplied by any non-zero constant c. 3. cRi + Rj Ri (c = 0): You can replace row i with row i multiplied by any non-zero constant c plus row j , j = i. Note that move 1 simply corresponds to reordering the system of equations (3.1)). Likewise, move 2 simply corresponds to scaling equation i in (3.1)). In general, these legal moves correspond to algebraic operations you would perform on (3.1)) to solve it. However, there are fewer symbols to push around when the augmented matrix is used. Goal: You win the game when you can achieve the following situation. Your goal is to nd a sequence of legal moves leading to a matrix B satisfying the following criteria: 1. all rows of B have leaading non-zero term equal to 1 (the position where this leading term in B occurs is called a pivot position), 2. B contains as many 0s as possible 3. all entries above and below a pivot position must be 0, 4. the pivot position of the ith row is to the left and above the pivot position of the (i + 1)st row (therefore, all entries below the diagonal of B are 0). What we call the Gauss elimination game is usually referred to as the method of row reduction. This matrix B constructed above is unique (this is a theorem which you can nd in any text on elementary matrix theory or linear algebra1 ) and is called the row reduced echelon form of A. The matrix B is sometimes written rref(A).1
  145
  Denition 3.2.1. The rank of the matrix A is the number of non-zero rows in the matrix rref(A). Remark 3.2.1. 1. There is at most one pivot position in each row and in each column.
  2. Every non-zero row has a pivot position, located at the rst non-zero entry of that row. 3. If you are your friend both start out playing this game, it is likely your choice of legal moves will dier. That is to be expected. However, you must get the same result in the end. 4. If you try the row reduction method but are unable to nish, it could be because you are creating non-zero terms in the matrix while killing others in a non-systematic way. Try killing terms only after selecting your pivot position and try to do this in a systematic left-to-right, top-to-bottom fashion. Now its time for an example. Example 3.2.1. Solve x + 2y = 3 4x + 5y = 6 using row reduction. (3.2)
  146
  One sequence of legal moves is the following: 1 2 3 4R1 + R2 R2 , which leads to 0 3 6 (1/3)R2 R2 , which leads to 2R2 + R1 R1 , which leads to 1 2 3 0 1 2
  1 0 1 0 1 2 Now we are done (we won the game of Gauss elimination), since this matrix satises all the goals for a row reduced echelon form. The latter matrix corresponds to the system of equations x + 0y = 1 0x + y = 2 (3.3)
  Since the legal moves were simply matrix analogs of algebraic manipulations youd appy to the system (3.2), the solution to (3.2) is the same as the solution to (3.3), whihc is obviously x = 1, y = 2. You can visually check this from the graph given above. To nd the row reduced echelon form of 1 2 3 4 5 6 using Sage , just type the following:Sage sage: MS = MatrixSpace(QQ,2,3) sage: A = MS([[1,2,3],[4,5,6]]) sage: A [1 2 3] [4 5 6] sage: A.echelon_form() [ 1 0 -1] [ 0 1 2]
  What if there are more variables than equations? Suppose that there are n variables and m linear (simultaneous) equations, with m < n. Since each linear equation can cut the number of degrees of freedom down by (at most) one, the set of solutions must have at least n m > 0 degrees of freedom. In other words, we must expect that there are innitely many solutions. In general, for solutions to such a system Ax = b, where A is an m n matrix, #degrees of freedom of solution = n rank(A).
  147
  Example 3.2.2. Describe the intersection L of the plane x + y + z = 1 with x +2y +3z = 1. Note L is a line in the plane, so we can ask for its parametric representation. We start by computing the row-reduced echelon form of the augmented matrix A= The row-reduced echelon form is rref(A) = 1 0 1 3 0 1 2 2 . 1 1 1 1 1 2 3 1 .
  so the resulting system is x = z + 3, y = 2z 2. Regarding z as our free variable, this intersection L can be written L = {(x, y, z ) | x = z +3, y = 2z 2} = {(z +3, 2z 2, z ) | z R} = (3, 2, 0)+(1, 2, 1)R. The parametric representation of L is x = 3 + t, where t R. See 3.2.4 for more higher-dimensional examples. y = 2 + 2t, z = t,
  3.2.2
  There is another method of solving square systems of linear equations which we discuss next. One can rewrite the system (3.1) as a single matrix equation a b c d or more compactly as AX = r, r1 x and r = r2 y (divide by A) is the right idea: where X = (3.4) x y = r1 r2 ,
  x y
  = X = A1 r.
  Here A1 is a matrix with the property that A1 A = I , the identity matrix (which satises I X = X ).
  148
  3.2.3
  If A1 exists (and it usually does), how do we compute it? There are a few ways. One, if using a formula. In the 2 2 case, the inverse is given by a b c d1
  1 ad bc
  d b c a
  There is a similar formula for larger sized matrices but it is so unwieldy that is is usually not used to compute the inverse. Example 3.2.3. In the 2 2 case, the formula above for the inverse is easy to use and we see for example, 1 2 4 5 To nd the inverse of 1 2 4 5 using Sage , just type the following:Sage sage: sage: sage: [1 2] [4 5] sage: [-5/3 [ 4/3 MS = MatrixSpace(QQ,2,2) A = MS([[1,2],[4,5]]) A1
  1 3
  5 2 4 1
  A better way to compute A1 is the following. Compute the row reduced echelon form of the matrix (A, I ), where I is the identity matrix of the same size as A. This new matrix will be (if the inverse exists) (I, A1 ). You can read o the inverse matrix from this. In other words, the following result holds. Lemma 3.2.1. Let A be an invertible n n matrix. The row reduced echelon form of the n 2n matrix (A, I ), where I is the identity matrix of the same size as A, is the n 2n matrix (I, A1 ). If A is not invertible then row reduced echelon form of the n 2n matrix (A, I ) will not have the form (I, B ), for some n n matrix B . For example,
  149
  1 2 0 0
  is not invertible and the row reduced echelon form of (A, I ) cannot be written in the form (I, B ), for some 2 2 matrix B . Here is another illustration. Example 3.2.4. Solve x + 2y = 3 4x + 5y = 6 using (a) row reduction, (b) matrix inverses. In fact, this system was solved using row-reduction in Example 3.2.1 above. The idea was that the result of the Sage commandsSage sage: MS = MatrixSpace(QQ,2,3) sage: A = MS([[1,2,3],[4,5,6]]) sage: A [1 2 3] [4 5 6] sage: A.echelon_form() [ 1 0 -1] [ 0 1 2]
  told us that x = 1 and y = 2. In other words, to get the solution, you just read o the last column of the row-reduced echelon form of the augmented matrix. This is 3 x 1 2 , = 6 y 4 5 so x y = 1 2 4 51
  3 6
  To compute the inverse matrix, so we can compute x and y , apply the Gauss elimination game to 1 2 1 0 4 5 0 1 Using the same sequence of legal moves as in Example 3.2.1, we get
  150
  4R1 + R2 R2 , which leads to (1/3)R2 R2 , which leads to 2R2 + R1 R1 , which leads to Therefore the inverse is
  1 2 1 0 0 1 4/3 1/3
  1 2
  Of course, this again tells us that x = 1 and y = 2 is the solution to the original system. We use row-reduction to compute the matrix inverse in the next example.. Example 3.2.5. To invert 1 2 3 A= 1 2 2 1 1 1
  we rst augment it with the 3 3 identity matrix, and then row-reduce until we obtain the identity matrix: 1 1 1 1 R1 +R3 0 0 1 2 3 1 2 3 1 0 0 R1 +R2 0 1 1 2 2 0 1 0 0 1 1 1 0 1 1 0 0 1 1 0 0 4 2 3 1 0 0 Swap(R2 ,R3 ) 0 1 0 3 0 1 1 1 0 0 0 1 1 1 4 1 0 1 0 0 1 0 0 1
  5 2 4 1 , 1 0
  151
  where the last step also involved a rescaling by a sign: R3 R3 . Reading o the last 4 5 2 1 3 3 block, we therefore have, A = 3 4 1 . 1 1 0 There are many other applications of row reduction: We show how to use row reduction to compute a basis for row spaces and column spaces, matrix kernels (null spaces), and orthogonal complements. These are interesting topics, but fall outside a textbook on dierential equations. We refer to Beezers text [B-rref], for details. Exercises: 1. Using Sage , solve x + 2y + z = 1 x + 2y z = 2 y + 2z = 3
  using (a) row reduction, (b) matrix inverses (i.e., convert to a matrix equation Av = b, compute v = A1 b). 2. Solve x + y z = 1, x y + z = 2, x + y + z = 3,
  using (a) row reduction, (b) matrix inverses. 3. Determine the intersection of planes
  x + y z = 1, x y + z = 2, as a line in R3 . 4. Solve the system x y z = 1, x + y z = 2, x + 2y + 3z = 3, using row reduction of the augmented matrix.
  152
  5. Find the inverse A1 of the following matrix by augmenting with the identity and computing the reduced echelon form. 5 3 6 A= 4 2 3 3 2 5 0 1 6. Find the inverse A of the matrix A = 2 0 operations on the matrix A adjoined by the 3 3 7. (a) Using row reduction, solve x + y = 1, y + z = 1, x + z = 1. Write down the augmented matrix and its row reduced echelon form. (b) Let A be the matrix of coecients of the system in (a) (so the system can be written as a matrix equation Ax = b). Compute the inverse of A. (c) Compute the 3 3 matrix A3 3A2 + 3A. 8. Let 1 1 1 1 A = 1 1 1 2 . 2 2 0 1 0 1 0 0 by using elementary row 1 2 identity matrix.
  Find the row-reduced echelon form of A. Explain your result geometrically (i.e., describe what the intersection of the planes x + y + z = 1, x y + z = 2, and 2x + 2y = 1 looks like). 9. Solve 2x + 3y = 4, 5x + 6y = 7 using the row reduced echelon form of the augmented matrix. Label all elementary row operation steps of the row-reduction process. 10. Let
  153
  0 1 1 1 1 A = 1 0 1 2 0 . 1 1 0 1 1
  using row-reduction.
  0 1 1 A = 1 0 1 , 1 1 0
  12. Verify that if A and B are invertible n n matrices, then AB is invertible with inverse B 1 A1 .
  3.2.4
  Gauss-Jordan reduction, revisited. Example 3.2.6. Solve the linear system x + 2y + 3z = 0, 4x + 5y + 6z = 3, 7x + 8y + 9z = 6 We form the augmented row-echelon form: 1 4 7 coecient matrix and row-reduce until we obtain the reduced 1 2 3 0 2 3 0 4R1 +R2 0 3 6 3 5 6 3 7 8 9 6 8 9 6 1 2 3 0 R2 /3 0 1 6 3 0 6 12 6 1 2 3 0 R2 +R3 0 1 2 1 0 0 2 1 1 0 1 2 2R2 +R1 0 1 2 1 0 0 0 0
  1 2 7R1 +R3 0 3 0 6 1 2 R3 /6 0 1 0 1
  3 0 2 1 12 6 3 0 2 1 0 0
  154
  Reinterpreting these rows as equations we have x z = 2 and y + 2z = 1. The pivot variables are x and y , and z is the only free variable. Finally, we write the solutions in parametric vector form: 2+t x(t) y (t) = 1 2t . t z (t) Exercises: 1. Reduce the following matrix to reduced echelon form. 1 3 6 7 2 6 3 9 1
  2. Write the following system as a matrix equation, and nd all the solutions. 3x + 2y + z = 2 3x + 7y 4z = 2 3. Use elementary row operations to transform the augmented coecient matrix of the system below into echelon or reduced echelon form. Use this to write down all the solutions to the system. x + 2y + 2z = 3 x y + 2z = 2
  2x + 4y 2z = 2
  x + 3y + 2z = 1
  x + 3y 2z = 0
  5. Under what conditions on a, b, and c does the following system have a unique solution, innitely many solutions, or no solutions? 2x y + 3z = a x + 2y + z = b 6x + 2y + 8z = c
  3.2. ROW REDUCTION AND SOLVING SYSTEMS OF EQUATIONS 6. The system x + y = 1, y z = 1, x + z = 1 has
  155
  (a) a unique solution (b) no solutions (c) innitely many solutions (d) exactly two solutions (e) none of these.
  3.2.5
  Determinants
  Denition 3.2.2. The (i, j )-th minor of a matrix A is the determinant of the of the submatrix obtained by deleting row i and column j from A. 1 2 1 Example 3.2.7. The (2, 2)-minor of 0 0 0 is equal to det 2 3 4
  1 1 2 4
  = 6.
  Let A is an n n matrix, and let Aij denote the (n 1) (n 1) submatrix obtained by removing the i-th row and j -th column of A (so det(Aij ) is the (i, j )-minor of A), 1 i, j n. The determinant of A, denoted det(A), is dened inductively by: case n = 1: det(A) = A (in the case n = 1, the matrix A is simply a real number). case n = 2: det case n > 2:n
  a b c d
  = ad bc.
  det(A) = a11 det(A11 )a12 det(A12 )+ +(1)n a1n det(A1n ) = This is called the Laplace expansion across the 1st row. There is an analogous Laplace expansion across any row
  j =1
  det(A) = (1)k1 (ak1 det(Ak1 ) ak2 det(Ak2 ) + + (1)n akn det(Akn )) j +k 1 a det(A ), = n kj kj j =1 (1) and down any column: det(A) = (1)k1 (a1k det(A1k ) a2k det(A2k ) + + (1)n ank det(Ank )) j +k 1 a det(A ). = n jk jk j =1 (1)
  (3.5)
  (3.6)
  It is a theorem that all these dierent formulas, (3.5) and (3.6), for the determinant yield the same number (for details, see the textbook on linear algebra by Beezer, [B-rref]).
  156
  a b , the absolute value of the determinant | det A| c d can be viewed as the area of the parallelogram with vertices at (0, 0), (a, b), (a + c, b + d), and (c, d). Geometric interpretation: If A = Exercises: 2 1 0 0 3 0 0 9 3 0 1 1 4 5 0 2 . 0 1 0 1 .
  1 0 2 1 20 2 0 1 63 3 0 1
  3. Show that if a matrix A has the property A2 = A, then either det(A) = 0 or det(A) = 1. 4. A matrix is called orthogonal if AT = A1 . Show that if A is orthogonal then det(A) = 1.
  5. (hard) Find a three by three matrix with strictly positive entries (aij > 0 for each entry aij ) whose determinant is equal to 1. Try to nd such a matrix with the smallest 3 sum of entries, that is, such that 3 i=1 i=1 aij = a11 + a12 + . . . + a33 is as low as possible. 6. The n n Vandermonde determinant is dened as:n 1 1 x1 x2 1 x1 n 1 1 x2 x2 2 x2 . . . . .. . . . . . . . . . 2 n 1 xn xn xn 1
  V (x1 , x2 , . . . , xn ) =
  Show (a) that the 2 2 Vandermonde determinant V (a, b) = b a, (b) the 3 3 Vandermonde determinant V (1, a, b) can be factored into (a 1)(b 1)(b a).
  157
  3.2.6
  Vector spaces
  So far we have worked with vectors which are n-tuples of numbers. But the essential operations we perform on such tuples can be generalized to a much wider class of objects, which we also call vectors. A collection of vectors that are compatible in the sense that they can be added together and multiplied by some sort of scalar is called a vector space. Here is a more formal denition. Denition 3.2.3. A vector space V is a set S on which there is an operation of addition which take pairs of elements of S to elements of S , together with a eld2 of numbers K for which there is an operation of scalar multiplication. These operations must have the following properties: 1. There exists a unique 0 S such that 0 + v = v for all v S . 2. v + w = w + v for all v, w S . 3. v + (w + x) = (v + w) + x for all v, w, x S . 4. (1) v + v = 0 for all v S . 5. 0 v = 0. 6. 1 v = v for all v S . 7. (s + t) (v + w) = s v + t v + s w + t w for all s, t K and all v, w S . 8. s (t v ) = (s t) v for all s, t K and all v S . For convenience, we do not usually indicate scalar multiplication with a , but by juxtaposition or sometimes a dot . For example, we write 5 times the vector v simply as 5v or 5 v , following the usual convention. For our purposes the eld K will always be either the eld of real numbers (denoted R) or the eld of complex numbers (denoted by C). There are many other elds of numbers used in mathematics but they will not be addressed here so we will not formally dene the concept of a eld. Unless indicated otherwise, we will use K = R. Here are some vector space examples: 1. For each positive integer n, the set of lists of n real numbers, Rn , is a vector space. For n = 2 and n = 3 these are the familiar real plane vectors and real 3-space vectors. 2. The set of real-valued polynomial functions forms a vector space. The addition is the usual addition of functions. 3. For each positive integer n, the set of polynomials of degree at most n forms a vector space. We will denote this space by Pn .2
  A eld will not be dened here. We shall always that the eld to be either R or C.
  158
  Denition 3.2.4. A subspace of a vector space V is a subset of elements of V that is itself a vector space. The important thing to understand about subspaces is that they must be closed under the operations of scalar multiplication and vector addition - not every subset of a vector space is a subspace. In particular, every subspace W must contain the 0 vector since if w W , then w W , and then so is w + w = 0. Here are some vector subspace examples: 1. The set W = {(x, x) | x R} is a subspace of R2 . If we think of R2 as the x, y plane, then W is simply the line y = x. 2. The set W = {(x, 2x) | x R} is a subspace of R2 . If we think of R2 as the x, y plane, then W is simply the line y = 2x. In fact every line through the origin is a subspace. Exercises: 1. Are the following sets subspaces of R3 or not? If not, explain why. (a) {(x1 , x2 , x3 ) | x1 + x2 + x3 = 0} (c) {(x1 , x2 , 0) | x1 = 5x2 }
  (b) {(x1 , x2 , x3 ) | x1 x2 x3 = 0}
  (d) The span of the vectors (1, 2, 3), (4, 5, 6) and (7, 8, 9). 2. If W is the subset of all vectors (x, y ) in R2 such that |x| = |y |, is W a vector subspace or not? 3. Is the set of all real-valued functions of one variable with the property that f (0) = 0 a vector space? 4. Is the set of all real-valued functions of one variable with the property that f (x) = f (x) for all x a vector space? 5. Is the set of all real-valued functions of one variable with the property that f (x) = f (x) + 1 for all x a vector space?
  3.2.7
  This section briey recalls some fundamental notions of linear algebra. Denition 3.2.5. A set of non-zero vectors {v1 , . . . , vm } is linearly dependent if there are constants c1 , . . . , cm which are not all zero for which c1 v1 +. . .+cm vm = 0. If {v1 , . . . , vm } is linearly dependent then we say it is linearly independent .
  159
  9 6 3 dependent. To see this, we can write the linear dependence condition c1 v1 + c2 v2 + c3 v3 = 0 as a matrix-vector equation: 0 c1 1 4 7 2 5 8 c2 = 0 0 c3 3 6 9
  Recall, the rank of an m n matrix A is the number of non-zero rows in the row-reduced echelon form of A. It turns out, this is also the number of rows of A (considered as row vectors in Rn ) which are linearly independent. An expression of the form c1 v1 + . . . + cm vm , for constants c1 , . . . , cm , is called a span or linear combination of the vectors in {v1 , . . . , vm }. The row-span of an m n matrix A is the vector space spanned by the rows of A (considered as row vectors in Rn ). The column-span of an m n matrix A is the vector space spanned by the columns of A (considered as column vectors in Rm ). 7 4 1 Example 3.2.8. The vectors v1 = 2 , v2 = 5 , and v3 = 8 are linearly
  and solve the system with row-reduction. Since it is a homogeneous system (i.e. the lefthand side is the zero vector), we only need to reduce the coecient matrix: 1 4 7 1 4 7 1 4 7 3R1 +R3 2R1 +R2 0 3 6 0 3 6 2 5 8 0 6 12 3 6 9 3 6 9 1 0 5 1 4 7 1 4 7 2R2 +R3 4R2 +R1 R2 /3 0 1 3 0 1 3 0 3 6 0 0 0 0 0 0 0 0 0
  which tells us that c3 is a free variable, so it can be chosen to be nonzero, and the vectors are linearly dependent. As a particular example, we could choose c3 = 1, and then c1 = 5 and c2 = 3, so 5v1 3v2 + v3 = 0. The above example illustrates one way of determining if a set of vectors is linearly dependent or not. The following theorem generalizes the method used: Theorem 3.2.1. Let S = {v1 , . . . , vk } be k column vectors in Rn . Let A = (v1 , . . . , vk ) denote the n k matrix whose columns are the vi s. The set S is linearly independent if and only if the matrix rref(AT ) has no all 0s row in it. Denition 3.2.6. A basis of a vector space V is a set of linearly independent vectors that span V .
  160
  Example 3.2.9. The set of vectors {(1, 1, 0), (0, 1, 1), (1, 1, 1)} is a basis for R3 . To see that they span R3 , we can nd explicit solutions to the system x c1 1 0 1 1 1 c2 = y , 1 z c3 0 1 1
  for any element (x, y, z ) R3 , by row reducing the coecient matrix (whose columns are 1 1 1 1 2 the vectors of the basis). We nd that c1 = 2 3 x 3 y 3 z , c2 = 3 x + 3 y 3 z , and 1 1 c3 = 1 3 x + 3 y + 3 z. Since c1 = c2 = c3 = 0 is the unique solution for (x, y, z ) = (0, 0, 0), the set of vectors is linearly independent. Theorem 3.2.2. Every basis of a vector space has the same number of elements, if nite. Denition 3.2.7. The dimension of a vector space V is the number of elements in a basis of V . If the bases of V are not nite, we say V is innite-dimensional. The following theorem can be found in any textbook on linear algebra (for example, Beezer [B-rref]). Theorem 3.2.3. Let S = {v1 , . . . , vk } be k column vectors in Rn . Let V denote the span of S and let A = (v1 , . . . , vk ) denote the n k matrix whose columns are the vi s. The dimension of the vector space V is the rank of the matrix A. This is the number of non-zero rows in rref(AT ). Example 3.2.10. Compute the dimension of the subspace W of R4 spanned by the vectors v1 = (1, 1, 1, 1)T , v2 = (1, 2, 1, 2)T , and v3 = (1, 3, 1, 3)T . We can use Theorem 3.2.3, which says it is the number of non-zero rows in 1 0 1 0 1 1 1 1 rref 1 2 1 2 = 0 1 0 1 . 0 0 0 0 1 3 1 3
  To compute the dimension, we need to know if v1 , v2 , and v3 form a basis of W . Since W is dened as their span, we only need to check if they are linearly dependent or not. As before, we do this by row-reducing a matrix whose columns consist of the vi : 1 1 1 1 1 1 1 1 1 1 2 3 R1 +R2 0 1 2 R1 +R3 0 1 2 1 1 1 1 1 1 0 0 0 1 2 3 1 2 3 1 2 3 1 1 1 1 1 1 0 1 2 0 1 2 R2 +R4 R1 +R4 0 0 0 0 0 0 0 0 0 0 1 2
  161
  This shows that only two of the three vectors are linearly independent, so the dimension of W is equal to two. Exercises: 1. Find a basis for the subspace dened by the following equations for (x1 , x2 , x3 , x4 , x5 ) R5 :
  2. Consider the triple of vectors v1 = (0, 2, 3, 2), v2 = (3, 1, 4, 1), and v3 = (6, 8, 1, 8). (b) What is the dimension of their span? (a) Is the set {v1 , v2 , v3 } linearly independent or dependent? (c) If the vectors are linearly independent, nd an additional vector v4 that makes {v1 , v2 , v3 , v4 } a basis for R4 . If they are linearly dependent, write v1 as a linear combination of v2 and v3 . 3. Find a basis for the subspace of R3 given by x 2y + 7z = 0. 4. Find a basis for the subspace of R3 given by x = z . 5. Find a basis for the subspace of all vectors (x1 , x2 , x3 , x4 ) in R4 such that x1 + x2 = x3 + x4 . 6. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
  Let V be the vector space spanned by its columns. Find a basis for V . 7. Let A be as above. Same problem, but let V be the vector space spanned by its rows. 8. 1 2 3 4 A = 5 6 2 4 3 4 2 1
  A=
  162 9.
  1 2 3 4 A = 0 1 2 4 3 4 2 1
  1 2 3 3 A = 5 0 2 4 3 1 2 1
  11. Determine if the rows of the matrix A form a linearly independent set, where 1 2 3 4 A = 0 1 2 4 3 4 2 1 12. Determine if the columns of the matrix A form a linearly independent set, where 1 2 3 A= 5 6 2 3 4 2
  3.3
  Suppose we have a system of dierential equations in standard form x = ax + by + f (t), x(0) = x0 , y = cx + dy + g(t), y (0) = y0 , (3.7)
  where a, b, c, d, x0 , y0 are given constants and f (t), g(t) are given nice functions. (Here nice will be left vague but basically we dont want these functions to annoy us with any bad behaviour while trying to solve the dierential equations by the method of Laplace transforms.) One way to solve this system if to take Laplace transforms of both sides. If we let X (s) = L[x(t)](s), Y (s) = L[y (t)](s), F (s) = L[f (t)](s), G(s) = L[g(t)](s), then (3.7) becomes
  163
  (3.8)
  This is now a 2 2 system of linear equations in the unknowns X (s), Y (s) with augmented matrix A= s a b F (s) + x0 c s d G(s) + y0 .
  Compute the row-reduced echelon form of A to solve for X (s) and Y (s). Take inverse Laplace transforms of these to nd x(t) and y (t). Example 3.3.1. Solve x = y + 1, x(0) = 0, y = x + t, y (0) = 0, The augmented matrix is A= The row reduced echelon form of this is 1 0 1/s2 0 1 0 . s 1 1/s 1 s 1/s2 .
  Therefore, X (s) = 1/s2 and Y (s) = 0. Taking inverse Laplace transforms, we see that the solution to the system is x(t) = t and y (t) = 0. It is easy to check that this is indeed the solution. To make Sage compute the row reduced echelon form, just type the following:Sage sage: R = PolynomialRing(QQ,"s") sage: F = FractionField(R) sage: s = F.gen() sage: MS = MatrixSpace(F,2,3) sage: A = MS([[s,1,1/s],[1,s,1/s2]]) sage: A.echelon_form() [ 1 0 1/s2] [ 0 1 0]
  To make Sage compute the Laplace transform, just type the following:
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  sage: s,t = var(s,t) sage: f(t) = 1 sage: laplace(f(t),t,s) 1/s sage: f(t) = t sage: laplace(f(t),t,s) s(-2)
  To make Sage compute the inverse Laplace transform, just type the following:Sage sage: s,t = var(s,t) sage: F(s) = 1/s2 sage: inverse_laplace(F(s),s,t) t sage: F(s) = 1/(s2+1) sage: inverse_laplace(F(s),s,t) sin(t)
  Example 3.3.2. The displacement from equilibrium (respectively) for coupled springs attached to a wall on the left (see Figure 3.2)
  Figure 3.2: A coupled spring diagram. is modeled by the system of 2nd order ODEs m1 x 1 + (k1 + k2 )x1 k2 x2 = 0, m2 x 2 + k2 (x2 x1 ) = 0,
  where x1 denotes the displacement from equilibrium of mass 1, denoted m1 , x2 denotes the displacement from equilibrium of mass 2, denoted m2 , and k1 , k2 are the respective spring constants [CS-rref]. As another illustration of solving linear systems of equations to solving systems of linear 1st order dierential equations, we use Sage to solve the above problem with m1 = 2, m2 = 1, k1 = 4, k2 = 2, x1 (0) = 3, x 1 (0) = 0, x2 (0) = 3, x2 (0) = 0.
  165
  Soln: Take Laplace transforms of the rst dierential equation, 2x 1 (t)+6x1 (t) 2x2 (t) = 0. 2 This says 2x1 (0)+2s X1 (s) 2sx1 (0) 2X2 (s)+2X1 (s) = 0 (where the Laplace transform of a lower case function is the upper case function). Take Laplace transforms of the second 2 dierential equation, 2x 2 (t)+2x2 (t) 2x1 (t) = 0. This says s X2 (s)+2X2 (s) 2X1 (s) 3s = 0. Solve these two equations:Sage sage: s,X,Y = var(s X Y) sage: eqns = [(2*s2+6)*X-2*Y == 6*s, -2*X +(s2+2)*Y == 3*s] sage: solve(eqns, X,Y) [[X == (3*s3 + 9*s)/(s4 + 5*s2 + 4), Y == (3*s3 + 15*s)/(s4 + 5*s2 + 4)]]
  This says X1 (s) = (3s3 + 9s)/(s4 + 5s2 + 4), X2 (s) = (3s3 + 15s)/(s4 + 5s2 + 4). Take inverse Laplace transforms to get the answer:Sage sage: s,t = var(s t) sage: inverse_laplace((3*s3 + 9*s)/(s4 + 5*s2 + 4),s,t) cos(2*t) + 2*cos(t) sage: inverse_laplace((3*s3 + 15*s)/(s4 + 5*s2 + 4),s,t) 4*cos(t) - cos(2*t)
  Therefore, x1 (t) = cos(2t) + 2 cos(t), x2 (t) = 4 cos(t) cos(2t). Using Sage this can be plotted parametrically using the commands below.Sage sage: P = parametric_plot([cos(2*t) + 2*cos(t),4*cos(t) - cos(2*t)],0,3) sage: show(P)
  3.3.1
  The goal of military analysis is a means of reliably predicting the outcome of military encounters, given some basic information about the forces status. The case of two combat-
  166
  Figure 3.3: Curves x(t) = cos(2t) + 2 cos(t), y (t) = 4 cos(t) cos(2t) along the t-axis. ants in a conventional directed re battle3 was solved by Frederick William Lanchester4 , a British engineer who served in the Royal Air Force during World War I. Lanchester discovered a way to model battle-eld casualties using systems of dierential equations. He assumed that if two armies ght, with x(t) troops on one side and y (t) on the other, the rate at which soldiers in one army are put out of action is proportional to the troop strength of their enemy. This give rise to the system of dierential equations x (t) = Ay (t), x(0) = x0 , y (t) = Bx(t), y (0) = y0 , where A > 0 and B > 0 are constants (called their ghting eectiveness coecients) and x0 and y0 are the intial troop strengths. For some historical examples of actual battles modeled using Lanchesters equations, please see references in the excellent paper by McKayThis is a battle where both combatants know roughly where their enemy is and res in their direction, as in a typical tank battle. This model does not cover the scenario where one side is hidden and ring from cover. 4 Lanchester was a very remarkable man in many ways, a pioneer British builder of motor boats and cars, and owner of a successful car manufacturing company. Unfortunately, describing his many accomplishments in engineering and operations research woudl take us too far aeld in this book.3
  3.3. APPLICATION: SOLVING SYSTEMS OF DES [M-intro]. We show here how to solve these using Laplace transforms. Recall the notation for the hyperbolic cosine (pronounced kosh) cosh(x) = and the hyperbolic sine (pronounced sinch), sinh(x) = ex ex , 2 ex + ex , 2
  167
  Example 3.3.3. Solve5 x = 4y, x(0) = 400, y = x, y (0) = 100. This models a battle between x-men and y -men, where the x-men die o at a higher rate than the y -men (but there are more of them to begin with too). Take Laplace transforms of both sides of both equations to obtain sX (s) 400 = 4Y (s), sY (s) 100 = X (s). The augmented matrix of this system (in the unknowns X (s), Y (s)) is A= The row reduced echelon form of this is 1 0 0 1 Therefore, 2 s 2 s 200 2 , Y (s) = 100 2 200 2 . 4 s 4 s 4 s 4 Taking inverse Laplace transforms, we see that the solution to the system is X (s) = 400 s2 x(t) = 400 cosh(2t) 200 sinh(2t) = 100e2t + 300e2t , and y (t) = 100 cosh(2t) 200 sinh(2t) = 50e2t + 150e2t .5
  s 4 400 1 s 100
  400(s1) s 2 4 100(s4) s 2 4
  168
  The x-men win and, in fact, x(0.275) = 346.4102..., y (0.275) = 0.1201... . Question: What is x(t)2 4y (t)2 ? (Hint: Its a constant. Can you explain this?) To make Sage plot this just type the following:Sage sage: f = lambda x: 400*cosh(2*x)-200*sinh(2*x) sage: g = lambda x: 100*cosh(2*x)-200*sinh(2*x) sage: P = plot(f,0,1) sage: Q = plot(g,0,1) sage: show(P+Q) sage: g(0.275) -0.12017933629675781 sage: f(0.275) 346.41024490088557
  Here is a similar battle but with dierent initial conditions. Example 3.3.4. A battle is modeled by x = 4y, x(0) = 150, y = x, y (0) = 90. (a) Write the solutions in parameteric form. (b) Who wins? When? State the losses for each side. Solution: Take Laplace transforms of both sides: sL [x (t)] (s) x (0) = 4 L [y (t)] (s), sL [x (t)] (s) x (0) = 4 L [y (t)] (s). Solving these equations gives L [x (t)] (s) = L [y (t)] (s) = sx (0) 4 y (0) 150 s 360 = , 2 s 4 s2 4
  sy (0) + x (0) 90 s + 150 = . 2 s 4 s2 4 Inverting using Laplace transform tables gives: x(t) = 150 cosh(2t) 180 sinh(2t) = 15 e2 t + 165 e2 t y (t) = 90 cosh (2 t) 75 sinh (2 t) = 15 2t 165 2t e + e . 2 2
  169
  Figure 3.4: Plot of curves x(t) = 400 cosh(2t) 200 sinh(2t), y (t) = 100 cosh(2t) 200 sinh(2t) along the t-axis. Their graph is given in Figure 3.5. The y -army wins. Solving for x(t) = 0 gives twin = log(11)/4 = .5994738182..., so the number of survivors is y (twin ) = 49.7493718, so 49 survive. Lanchesters square law: Suppose that you are more interested in y as a function of x, instead of x and y as functions of t. One can use the chain rule from calculus to derive from the system x (t) = Ay (t), y (t) = Bx(t) the single equation dy dy/dt Bx = = . dx dx/dt Ay This dierential equation can be solved by the following separation of variables trick (due to Lanchester): Aydy = Bxdx, so Ay 2 = Bx2 + C, where C is an unknown constant. (To solve for C you must be given some initial conditions.) The quantity Bx2 is called the ghting strength of the X -men and the quantity Ay 2
  170
  is called the ghting strength of the Y -men (ghting strength is not to be confused with troop strength). This relationship between the troop strengths is sometimes called Lanchesters square law and is sometimes expressed as saying the relative ght strength is a constant: Ay 2 Bx2 = constant. Remark 3.3.1. Here are some comments on those units in support capacity, from [M-intro]. Suppose your total number of troops is some number T , where x(0) are initially placed in a ghting capacity and T x(0) are in a support role. If your troops outnumber the enemy then you want to choose the number of support units to be the smallest number such that the ghting eectiveness is not decreasing (therefore is roughly constant). The remainder should be actively engaged with the enemy in battle.
  171
  where Ai > 0, Bi > 0, and Ci > 0 are constants and x0 , y0 and z0 are the intial troop strengths. Example 3.3.5. Consider the battle modeled by x(0) = 100, x (t) = y (t) z (t), y (t) = 2x(t) 3z (t), y (0) = 100, z (t) = 2x(t) 3y (t), z (0) = 100.
  x (t) = A1 y (t) A2 z (t), x(0) = x0 , y (t) = B1 x(t) B2 z (t), y (0) = y0 , z (t) = C1 x(t) C2 y (t), z (0) = z0 ,
  The Y-men and Z-men are better ghters than the X-men, in the sense that the coecient of z in the 2nd dierential equation (describing their battle with y ) is higher than that coecient of x, and the coecient of y in the 3rd dierential equation is also higher than that coecient of x. However, as we will see, the worst ghter wins! (The X-men do have the best defensive abilities, in the sense that A1 and A2 are small.) Taking Laplace transforms, we obtain the system sX (s) + Y (s) + Z (s) = 100 2X (s) + sY (s) + 3Z (s) = 100, 2X (s) + 3Y (s) + sZ (s) = 100, s 1 1 100 2 s 3 100 2 3 s 100 This has row-reduced echelon form 100s+100 s2 +3s4 100s200 s2 +3s4 100s200 s2 +3s4
  100s200 s+100 This means X (s) = 100 s2 +3s4 and Y (s) = Z (s) = s2 +3s4 . Taking inverse LTs, we get the solution: x(t) = 40et + 60e4t and y (t) = z (t) = 20et + 120e4t . In other words, the worst ghter wins! In fact, the battle is over at t = log(6)/5 = 0.35... and at this time, x(t) = 71.54.... Therefore, the worst ghters, the X-men, not only won but have lost less than 30% of their men!
  1 0 0 0 1 0 0 0 1
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  Figure 3.6: Lanchesters model for the x vs. y vs z battle. Exercises: 1. Use Sage to solve this: A battle is modeled by x = 4y, x(0) = 150, y = x, y (0) = 40. (a) Write the solutions in parameteric form. (b) Who wins? When? State the losses for each side. 2. All 300 X-men are ghting all 101 Y-men, the battle being modeled by x = 9y, x(0) = 300, y = x, y (0) = 101. Solve this system. Who wins? Compute the losses using Lanchesters Square Law. 3. All 60 X-men are ghting all 15 Y-men, the battle being modeled by x = 4y, x(0) = 60, y = x, y (0) = 15.
  3.3. APPLICATION: SOLVING SYSTEMS OF DES Solve this system. Who wins? When? 4. Suppose that in the above battle, we dont know the number of Y-men: x = 4y, x(0) = 60, y = x, y (0) = y0 . How many Y-men are needed to tie? To win? 5. Suppose that in the above battle, we dont know the number of X-men: x = 4y, x(0) = x0 , y = x, y (0) = 15. How many X-men are needed to tie? To win? 6. Suppose that the above battle is modeled instead by x = 4xy, x(0) = 60, y = xy, y (0) = 15. Solve for x in terms of y using the separation of variables trick. 7. Suppose that the above battle is modeled instead by x = 4y, x(0) = 60, y = xy, y (0) = 15. Solve for x in terms of y using the separation of variables trick. 8. Suppose that a battle between three forces is modeled by x = y 2z, x(0) = 100, y = 2x z, y (0) = 100, z = x 2y, y (0) = 100.
  173
  In other words, the Z -men ght better than the X -men, the X -men ght better than the Y -men, the Y -men ght better than the Z -men. Solve for x, y, z .
  3.3.2
  From war and death, we turn to love and romance. In Figure 3.3.2, we see the star-crossed lovers6Title: Romeo and Juliet: Act One, Scene Five. Engraver: Facius, Georg Sigmund; Facius, Johann Gottlieb Designer: Miller, William Date: 1789.6
  174 William Shakespeares play Romeo and Juliet about two young star-crossd lovers was one of his most popular. The characters Romeo and Juliet represent young lovers whose relationship is, sadly, doomed. Let r = r (t) denote the love Romeo has for Juliet at time t and let j = j (t) denote the love Juliet has for Romeo at time t. r = Aj, r (0) = r0 , j = Br + Cj, j (0) = j0 , (3.9)
  where A > 0, B > 0, C > 0, r0 , j0 are given constants. This indicates how Romeo is madly in love with Juliet. The more she loves him, the more he loves her. Juliet is more of a complex character. She has eyes for Romeo and her love for him makes her feel good about herself, which makes her love him even more. However, if she senses Romeo seems to love her too much, she reacts negatively and her love wanes. A few examples illustrate the cyclical nature of the relationship that can arise. Example 3.3.6. Solve r = 5j, r (0) = 4, j = r + 2j, j (0) = 6.
  175
  To solve this using Laplace transforms, take Laplace transforms of both sides, to obtain sR(s) 4 = 5J (s), sJ (s) 6 = R(s) + 2J (s),
  where R(s) = L[r (t)](s), J (s) = L[j (t)](s). This gives rise to the augmented matrix s 5 4 1 s2 6 Computing the row-reduced echelon form gives R(s) = 10 s(s2 2 3s 4 + , 2s + 5) s J (s) = 2 (s2 2 3s . 2s + 5 .
  176
  Figure 3.7: Romeo and Juliet plots. Example 3.3.7. Solve r = 10j, r (0) = 4, j = r + j/10, j (0) = 6. To solve this using Laplace transforms, take Laplace transforms of both sides, to obtain 1 J (s), 10 where R(s) = L[r (t)](s), J (s) = L[j (t)](s). Solving this as above gives sR(s) 4 = 10J (s), sJ (s) 6 = R(s) + r (t) = 4 3999 299 3999 sin 3999 sin 1 3999t + 3999 cos 20 1 3999t 20 1 3999t 201 e( 20 t) ,
  j (t) =
  2 3999
  1 e( 20 t) .
  sage: t = var("t")
  177
  sage: r = function("r",t) sage: j = function("j",t) sage: de1 = diff(r,t) == 10*j sage: de2 = diff(j,t) == -r+(1/10)*j sage: soln = desolve_system([de1, de2], [r,j],ics=[0,4,6]) sage: rt = soln[0].rhs(); rt 4/3999*(299*sqrt(3999)*sin(1/20*sqrt(3999)*t) + 3999*cos(1/20*sqrt(3999)*t))*e(1/20*t) sage: jt = soln[1].rhs(); jt -2/3999*(37*sqrt(3999)*sin(1/20*sqrt(3999)*t) - 11997*cos(1/20*sqrt(3999)*t))*e(1/20*t)
  Exercises: 1. Rewrite the second-order dierential equation x + 3x + 5x = t as a system of rstorder dierential equations. (You do not have to nd the solution.) 2. Find the general solution to the system x 1 = x1 + 2x2 , x2 = 3x1 + 2x2 . 3. Find the general solution to the system x 1 = x1 5x2 , x2 = x1 x2 . Sketch some of the solutions near the origin.
  178
  For the next two problems, consider two blocks of mass m1 and m2 connected by springs to each other and to walls as shown in Figure 6. The displacement of the masses from their equilibrium positions are denoted by x1 and x2 . The stiness of the three springs are k1 , k2 , and k3 as shown. Compute the natural frequencies and describe the natural modes of oscillation in each of the two following cases:
  5. k1 = k3 = 0 and k2 = 4, and m1 = m2 = 1. 6. k1 = k3 = 1 and k2 = 4, and m1 = m2 = 1. 7. Suppose a rocket is red and after its initial burn it is at an altitude of h = 5 km with a velocity of 4 km/s straight up. Since it may reach a signicant altitude, it could be too inaccurate to use a constant gravitational force. Instead, we will neglect air resistance but use the Newtonian law of gravity: gR2 d2 h = 2 dt (h + R)2 where h is the height above sea level in kilometers, R = 6378 km, and g = 0.0098 km/s2 . Convert this to a system of rst-order equations and use a numerical method to nd the maximum height of the rocket. Your answer should have two digits of accuracy (within the assumptions of the model). Use of Sage is highly encouraged for this problem. 8. Use Sage to analyze the problem r = 2j, r (0) = 4, j = r + 3j, j (0) = 6. 9. (Much harder) Use Sage to analyze the Lotka-Volterra/Predator-Prey model x = x(1 + 2y ), x(0) = 4, y = y (3x + 4y ), y (0) = 6.
  179
  3.3.3
  Suppose we have an electrical network (i.e., a series of electrical circuits) involving emfs (electromotive forces or batteries), resistors, capacitors and inductors. We use the dictionary from 2.7 to translate between the network
  Also, recall from 2.7 Kirchos Laws. Example 3.3.8. Consider the simple RC circuit given by the diagram in Figure 3.9.
  Figure 3.9: A simple circuit. According to Kirchos 2nd Law and the above dictionary, this circuit corresponds to the dierential equation q + 5q = 2. The general solution to this is q (t) = 1 + ce2t , where c is a constant which depends on the initial charge on the capacitor.
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  Remark 3.3.2. The convention of assuming that electricity ows from positive to negative on the terminals of a battery is referred to as conventional ow. The physically-correct but opposite assumption is referred to as electron ow. We shall assume the electron ow convention. Example 3.3.9. Consider the network given by the diagram in Figure 3.10.
  Figure 3.10: A network. Assume the initial charges are 0. One dierence between this circuit and the one above is that the charges on the three paths between the two nodes (labeled node 1 and node 2 for convenience) must be labeled. The charge passing through the 5 ohm resistor we label q1 , the charge on the capacitor we denote by q2 , and the charge passing through the 1 ohm resistor we label q3 . There are three closed loops in the above diagram: the top loop, the bottom loop, and the big loop. The loops will be traversed in the clockwise direction. Note the top loop looks like the simple circuit given in Example 1 but it cannot be solved in the same way, since the current passing through the 5 ohm resistor will aect the charge on the capacitor. This current is not present in the circuit of Example 3.3.8 but it does occur in the network above. Kirchos Laws and the above dictionary give q3 + 5q2 = 2, q1 (0) = 0, 5q = 0, q (0) = 0, (3.10) 5q1 2 2 5q1 + q3 = 2, q3 (0) = 0. A few things to take note of. The minus sign in front of the term associated to the capacitor (5q2 ). This is because we are going clockwise, against the direction of the current. The equation associated to the big loop (the last equation displayed above) is the sum of the rst two equations. Also, the big loop is roughly speaking the sum of the top loop and the bottom loop.
  181
  Kirchos 1st law says the sum of the currents into a node is zero. At node 1, this says + q q = 0, or q = q + q . Since q (0) = q (0) = q (0) = 0, this implies q = q + q . q1 1 2 3 3 1 2 2 1 3 3 2 After taking Laplace transforms of the 3 dierential equations above, we get sQ3 (s) + 5Q2 (s) = 2/s, 5sQ1 (s) 5Q2 (s) = 0.
  Note you dont need to take the LT of the 2nd equation since it is the dierence of the other equations. The LT of the above q1 + q2 = q3 (Kirchos law) gives Q1 (s)+ Q2 (s) Q3 (s) = 0. We therefore, taking Laplace transforms of Kirchos law and the 1st and 3rd equations in (3.10), gives this matrix equation 2/s Q1 (s) 0 5 s 5s 0 s Q2 (s) = 2/s . 0 Q3 (s) 1 1 1 0 5 s 2/s 5s 0 s 2/s 1 1 1 0 The row-reduced echelon form is Therefore Q1 (s) = This implies q1 (t) = 1/18 + e6t /18 + t/3, q2 (t) = 1/3 e6t /3,Sage sage: sage: sage: sage: [ [ [ 1 0 0 s = var("s") MS = MatrixSpace(SymbolicExpressionRing(), 3, 4) A = MS([[0,5,s,2/s],[5*s,0,s,2/s],[1,1,-1,0]]) B = A.echelon_form(); B 0 1 0 0 0 1 2/(5*s2) - (-2/(5*s) - 2/(5*s2))/(5*(-s/5 - 6/5))] 2/(5*s) - (-2/(5*s) - 2/(5*s2))*s/(5*(-s/5 - 6/5)) ] (-2/(5*s) - 2/(5*s2))/(-s/5 - 6/5) ]
  1 0 0 2/(s3 + 6s2 ) 2/(s2 + 6s) 0 1 0 3 2 0 0 1 2(s + 1)/(s + 6s ) 2 , + 6s2 Q2 (s) = 2 , + 6s Q3 (s) = 2(s + 1) . s2 (s + 6)
  s2
  182
  sage: Q1 = B[0,3] sage: t = var("t") sage: Q1.inverse_laplace(s,t) e(-(6*t))/18 + t/3 - 1/18 sage: Q2 = B[1,3] sage: Q2.inverse_laplace(s,t) 1/3 - e(-(6*t))/3 sage: Q3 = B[2,3] sage: Q3.inverse_laplace(s,t) -5*e(-(6*t))/18 + t/3 + 5/18
  Example 3.3.10. Consider the network given by the diagram in Figure 3.11.
  Figure 3.11: Another network. Assume the initial currents are 0. Using Kirchos Laws, you get a system i1 i2 i3 = 0, 2i1 + i2 + (0.2)i 1 = 6, (0.1)i3 i2 = 0.
  Take Laplace transforms of each of these three dierential equations. You get a 3 3 system in the unknowns I1 (s) = L[i1 (t)](s), I2 (s) = L[i2 (t)](s), and I3 (s) = L[i3 (t)](s). The augmented matrix of this system is
  183
  Therefore
  1 4 3 + . s + 20 s + 5 s
  I1 (s) =
  I2 (s) =
  3 3 + , s + 20 s + 5
  I3 (s) =
  This implies i1 (t) = 3 2e5t e20t , Exercises: 1. Use Sage to solve for i1 (t), i2 (t), and i3 (t) in the above problem. 2. (a) For the circuit in Figure 1 show that the charge q on the capacitor and the current i3 in the right branch satisfy the system of dierential equations q + (1/RC )q + i3 = 0, i 3 + (1/LC )q = 0. i2 (t) = 3e5t 3e20t , i3 (t) = 3 4e5t + e20t .
  184
  CHAPTER 3. MATRIX THEORY AND SYSTEMS OF DES (b) When the switch in the circuit is closed at time t = 0, the current i3 is 0 amps and the charge on the capacitor is 5 coulombs. With R = 2, L = 3, C = 1/6 use Laplace transforms to nd the charge q (t) on the capacitor.
  3.4
  First, we shall try to motivate the study of eigenvalues and eigenvectors. This section hopefully will convince you that they are very natural. Let A be any n n matrix. If v is a vector in Rn then Av is another vector in Rn . more general, we see that the multiplication by A map, A : Rn Rn denes a function. This function is by no means arbrtrary. It is a rather special type of function, with a lot of useful properties. The only property that is important now is that there are n complex numbers, and (usually) n linearly independent complex vectors, which together determine a lot of the important properties of A. These numbers are the eigenvalues of A and these vectors are the eigenvectors of A. The simplest matrices A for which the eigenvalues and eigenvectors are easy to compute are the diagonal matrices. The arithmetic of diagonal matrices: Well focus for simplicity on the 2 2 case, but everything applies to the general case of n n diagonal matrices. Addition is easy: a1 0 0 a2 Multiplication is easy: a1 0 0 a2 Powers are easy: a1 0 0 a2n
  v Av,
  b1 0 0 b2
  a 1 + b1 0 0 a2 + b2
  a 1 b1 0 0 a2 b2
  0 an 1 0 an 2
  185
  You can even exponentiate them using the usual power series expansion for exp(x) = ex = 1 + x + x2 /2 + . . . : a1 0 0 a2 1 0 0 11 2 + 2! t
  exp(t
  ) =
  +t
  a1 0 0 a22
  a1 0 0 a2 +
  1 3 3! t
  a1 0 0 a2
  + ...
  1 0 0 1 +
  ta1 0 0 ta2 0 1 2 2 2! t a2 . +1 3 3 3! t a1
  1 2 2 2! t a1
  01 3 3 3! t a2
  0 0 eta2
  eta1 0
  So, diagonal matrices are relatively easy to compute with. The eigenvalues of a digonal matrix A are the numbers on the diagonal of A (namely, a1 and a2 in the example above). The eigenvectors of a diagonal matrix A are (provided the diagonal entries are all non-zero) the standard basis vectors. Geometric interpretation of matrix conjugation. You and your friend are piloting a rocket in space. You handle the controls, your friend handles the map. To communicate, you have to change coordinates. Your coordinates are those of the rocketship (straight ahead is one direction, to the right is another). Your friends coordinates are those of the map (north and east are map directions). Changing coordinates corresponds algebraically to conjugating by a suitable matrix. Using an example, well see how this arises in a specic case. Your basis vectors are v1 = (1, 0), v2 = (0, 1),
  which we call the v -space coordinates, and the maps basis vectors are w1 = (1, 1), w2 = (1, 1),
  7 5 . to 3 2 Suppose we ip about the 45o line (the diagonal) in each coordinate system. In the v -space:
  which we call the w-space coordinates. For example, the point (7, 3) is, in v -space coordinates of course (7, 3) but in the w-space 1 1 sends coordinates, (5, 2) since 5w1 + 2w2 = 7v1 + 3v2 . Indeed, the matrix A = 1 1
  186
  1 0 . 0 1 Conjugating by the matrix A converts the ip map in w-space to the the ip map in v -space:
  187
  1 0 0 1
  A1 =
  0 1 1 0
  Motivation for eigenvalues At rst glance, matrices can seem to be complicated objects. When the matrix is a diagonal matrix then are relatively simple. In the case of diagonal matrices, the eigenvalues are the diagonal entries. It turns out that most matrices are conjugate to a diagonal matrix. The eigenvalues of a general square matrix are relatively easy to compute, tell us something about the behavior of the matrix. This is a bit vague but more details will be given. For us, as you will see, eigenvalues are very useful for solving systems of linear rst order dierential equations. Each n n matrix A has exactly n (counted according to multiplicity) eigenvalues 1 , 2 , . . . , n . These numbers are the roots of the characteristic polynomial p() = pA () = det(A I ). (3.11)
  In other words, subtract the variable from each diagonal entry of A and take the determinant. This is a polynomial of degree n in . A corresponding eigenvector is any non-zero vector v which satises the (dening) eigenvector equation Av = v, (3.12)
  where is a xed eigenvalue. In this case, we say v corresponds to . Note that you can scale an eigenvector v (i.e., replace v by cv , for some scalar c = 0) and get another eigenvector. Example 3.4.1. Consider an n n diagonal matrix. The standard basis elements (e1 = (1, 0, ..., 0), . . . , en = (0, ...., 0, 1)) are the eigenvectors and the diagonal elements are the eigenvalues. Example 3.4.2. Find the eigenvalues and eigenvectors of 0 1 1 A = 4 0 2 0 0 3
  We compute
  188
  x x 0 1 1 2 y = 3 y . 4 z z 0 0 3 +z = 3x +2z = 3y 3z = 3z.
  4x
  You can nd some non-zero solvution to these using row-reduction/Gauss elimination, for example. The row-reduced echelon form of 3 1 1 0 4 3 2 0 0 0 0 0 is 1 0 1/5 0 A = 0 1 2/5 0 , 0 0 0 0
  so z is anything (non-zero, that is), y = 2z/5, and x = z/5. One non-zero solution is x = 1, y = 2, z = 5, so 1 v3 = 2 . 5
  However, any other scalar multiply of this will also satisfy the eigenvector equation. The other eigenvectors can be computed similarly. We obtain in this way,
  189
  Since this section is only intended to be motivation, we shall not prove this here (see any text on linear algebra, for example [B-rref] or [H-rref]). Example 3.4.3. Sage can compute with complex numbers. You can use exact arithmetic:Sage sage: MS = MatrixSpace(QQ[I],2,2) sage: A = MS([[0, 1], [-1, 0]]) sage: A.charpoly() x2 + 1 sage: A.eigenspaces_right() [ (I, Vector space of degree 2 and dimension 1 over Number Field in I with defining polynomial x2 + 1 User basis matrix: [1 I]), (-I, Vector space of degree 2 and dimension 1 over Number Field in I with defining polynomial x2 + 1 User basis matrix: [ 1 -I]) ]
  1 v1 = 2 , 0
  1 v2 = 2 . 0
  190
  Exercises: Find the eigenvalues and eigenvectors of the following matrices by hand: 1. 4 2 1 1 5 6 3 4
  2.
  2 0 0 3. 5 3 2 2 0 1 3 1 1 4. 0 1 0 0 0 1 5. 0 2 1 0 0 1 1 0
  6.
  8.
  10. Show that if A is invertible and is an eigenvalue of A, then 1/ is an eigenvalue of A1 . Are the eigenvectors the same? 11. If is an eigenvalue for a matrix A and is an eigenvalue for a matrix B , is + an eigenvalue for the matrix A + B ? If not, nd a counter-example. 12. Find a matrix P such that P 1 AP = D , where D is a diagonal matrix, for the following two matrices if such a P exists.
  1 1 0 9. 4 1 0 . 0 0 13
  3.4. EIGENVALUE METHOD FOR SYSTEMS OF DES 0 1 0 1 2 0 1 1 1 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 2 13. By computing the eigenvalues and eigenvectors of A =
  191
  3 2 nd a matrix P 1 0 such that P 1 AP = D where D is a diagonal matrix. Use this diagonalization to compute A5 .
  3.4.1
  We next discuss a method for solving a linear homogenous system of dierential equations. Solution strategy PROBLEM: Solve x = ax + by, y = cx + dy, Solution: Let A= a b c d x(0) = x0 , y (0) = y0 .
  In matrix notation, the system of dierential equations becomes X = AX, X (0) = x0 y0 , (3.13)
  x(t) . Recall the manner how we solved homogeneous constant y (t) coecient 2nd order ODEs ax + bx + cx = 0. We used Eulers guess x = Cert . Similarly, for (3.13), we try to guess an exponential: X (t) = cet . Here, is used instead of r to stick with notational convention, and c in place of C since we need a constant vector. Plugging this guess into the matrix dierential equation X = AX gives cet = Acet , or (cancelling et ) where X = X (t) = Ac = c. This means that is an eigenvalue of A with eigenvector c.
  192
  CHAPTER 3. MATRIX THEORY AND SYSTEMS OF DES Find the eigenvalues. These are the roots of the characteristic polynomial p() = det a b c d = 2 (a + d) + (ad bc).
  Call them 1 , 2 (in any order you like). You can use the quadratic formula, for example to get them: 1 = a+d + 2 (a + d)2 4(ad bc) a+d , 2 = 2 2 (a + d)2 4(ad bc) . 2
  In general, you can get them by solving the eigenvector equation Av = v . Plug these into the following formulas: (a) 1 = 2 , real: x(t) y (t) (b) 1 = 2 = , real: x(t) y (t) = c1 v1 et + c2 (v1 t + p)et , = c1 v1 e1 t + c2 v2 e2 t .
  (c) 1 = + i , complex: write v1 = u1 + iu2 , where u1 and u2 are both real vectors. x(t) y (t) = c1 [et cos(t)u1 et sin(t)u2 ] +c2 [et cos(t)u2 et sin(t)u1 ].
  (3.14)
  3.4.2
  Our rst example is based on a historical incident. Example 3.4.4. In 1805, twenty-seven British ships, led by Admiral Nelson, defeated thirty-three French and Spanish ships, under French Admiral Pierre-Charles Villeneuve. See Figure 3.13.
  193
  Figure 3.13: The Battle of Trafalgar, by William Clarkson Staneld (1836) British eet lost: zero, Franco-Spanish eet lost: twenty-two ships. The battle is modeled by the following system of dierential equations: x (t) = y (t), x(0) = 27, y (t) = 25x(t), y (0) = 33. How can you use Sage to solve this? We use the method of eigenvalues and eigenvectors: if the eigenvalues 1 = 2 of A = 0 1 are distinct then the general solution can be written 25 0 x(t) y (t) where v1 , v2 are eigenvectors of A. Solving Av = v gives us the eigenvalues and eigenvectors. Using Sage , this is easy:Sage sage: A = matrix([[0, -1], [-25, 0]])
  = c1 v1 e1 t + c2 v2 e2 t ,
  194
  sage: A.eigenspaces_right() [ (5, Vector space of degree 2 and dimension 1 over Rational Field User basis matrix: [ 1 -5]), (-5, Vector space of degree 2 and dimension 1 over Rational Field User basis matrix: [1 5]) ]
  Therefore, 1 = 5, v1 = 1 5 1 5 ,
  e5t + c2
  1 5
  e5t .
  c1 + c2 5c1 + 5c2
  This Sage computation gives c1 = 51/5 and c2 = 84/5. This gives the solution to the system of dierential equations as x(t) = The solution satises x(0.033) = 26.27... (0 losses), y (0.033) = 11.07... (22 losses), consistent with the losses in the actual battle. 51 5t 84 5t e + e , 5 5 y (t) = 51e5t + 84e5t .
  3.4. EIGENVALUE METHOD FOR SYSTEMS OF DES Example 3.4.5. Solve x = 4y, x(0) = 400, y = x, y (0) = 100, using the eigenvalue method. (See also Example 3.3.3.) The associated matrix is 0 4 A= , 1 0 2 1
  195
  v1 =
  v2 =
  2 1
  e2t + c2
  e2t .
  We must now solve for c1 and c2 . The initial conditions imply 400 100 = c1 2 1 + c2 2 1 = 2c1 + 2c2 c1 + c2 .
  The solution to the equations 2c1 + 2c2 = 400, c1 + c2 = 100, is (check this!) c1 = 50, c2 = 150. Therefore, the solution to the system of dierential equations is x = 100e2t + 300e2t , The x-men win. Example 3.4.6. The eigenvalues and eigenvectors can be computed using Sage numerically as follows.Sage sage: A = matrix([[1,2],[3,4]]) sage: A.eigenvalues() [-0.3722813232690144?, 5.372281323269015?] sage: A.eigenvectors_right() [(-0.3722813232690144?, [(1, -0.6861406616345072?)], 1), (5.372281323269015?, [(1, 2.186140661634508?)], 1)]
  y = 50e2t + 150e2t .
  196
  Example 3.4.7. Solve x (t) = x(t) y (t), y (t) = 4x(t) + y (t), x(0) = 1, y (0) = 1. Let A= and so the characteristc polynomial is p(x) = det(A xI ) = x2 2x + 5. The eigenvalues are 1 = 1 + 2i, 2 = 1 2i, so = 1 and = 2. Eigenvectors v1 , v2 are given by v1 = 1 2i , v2 = 1 2i , 1 1 4 1
  though we actually only need to know v1 . The real and imaginary parts of v1 are u1 = The solution is then x(t) y (t) = c1 et cos(2t) + c2 et sin(2t) 2c1 et sin(2t) 2c2 et cos(2t), 1 0 , u2 = 0 2 .
  so x(t) = c1 et cos(2t) + c2 et sin(2t) and y (t) = 2c1 et sin(2t) 2c2 et cos(2t). Since x(0) = 1, we solve to get c1 = 1. Since y (0) = 1, we get c2 = 1/2. The solution t t t is: x(t) = et cos(2t) 1 2 e sin(2t) and y (t) = 2e sin(2t) + e cos(2t).
  3.4. EIGENVALUE METHOD FOR SYSTEMS OF DES Example 3.4.8. Solve x (t) = 2x(t) + 3y (t), y (t) = 3x(t) + 4y (t). Let A= and so the characteristc polynomial is p(x) = det(A xI ) = x2 2x + 1. The eigenvalues are 1 = 2 = 1. An eigenvector v1 is given by v1 = 3 3 . 2 3 3 4
  197
  Since we can multiply any eigenvector by a non-zero scalar and get another eigenvector, we shall use instead 1 . v1 = 1 Let p = r s be any non-zero vector satisfying (A I )p = v1 . This means 2 1 3 3 41 r s = 1 1
  There are innitely many possibly solutions but we simply take r = 0 and s = 1/3, so p= The solution is x(t) y (t) = c1 1 1 et + c2 ( 1 1 t+ 0 1/3 )et , 0 1/3 .
  Exercises: 1. Use either the Laplace transform method or the eigenvalue/eigenvector method to nd the steady state solution to the initial value problem x = x z , y = x y , z = 2x + z , x(0) = 0, y (0) = 0, z (0) = 2.
  198
  2. Use Sage and the eigenvalue/eigenvector method to solve this: A battle is modeled by x = 4y, x(0) = 150, y = x, y (0) = 40. (a) Write the solutions in parameteric form. (b) Who wins? When? State the losses for each side. 3. All 300 X-men are ghting all 101 Y-men, the battle being modeled by x = 9y, x(0) = 300, y = x, y (0) = 101. Solve this system using the eigenvalue/eigenvector method. Who wins? When? 4. All 60 X-men are ghting all 15 Y-men, the battle being modeled by x = 4y, x(0) = 60, y = x, y (0) = 15. Solve this system using the eigenvalue/eigenvector method. Who wins? When?
  3.5
  The method called variation of parameters for systems of ordinary dierential equations has no relation to the method variation of parameters for 2nd order ordinary dierential equations discussed in an earlier lecture except for their name.
  3.5.1
  Motivation
  Recall that when we solved the 1st order ordinary dierential equation y = ay, y (0) = y0 , (3.15)
  for y = y (t) using the method of separation of variables, we got the formula y = ceat = eat c, (3.16)
  where c is a constant depending on the initial condition (in fact, c = y (0)). Consider a 2 2 system of linear 1st order ordinary dierential equations in the form x = ax + by, x(0) = x0 , y = cx + dy, y (0) = y0 . This can be rewritten in the form
  199
  (3.17)
  x(0) is a constant depending on the initial conditions and etA is a matrix expoy (0) nential dened by eB = n=0
  1 n B , n!
  for any square matrix B . You might be thinking to yourself: I cant compute the matrix exponential so what good is this formula (3.18)? Good question! The answer is that the eigenvalues and eigenvectors of the matrix A enable you to compute etA . This is the basis for the formulas for the solution of a system of ordinary dierential equations using the eigenvalue method discussed in 3.4. None-the-less, we shall instead use a matrix called the fundamental matrix (dened below7 ) in place of etA . The eigenvalue method in 3.4 shows us how to write every solution to (3.17) in the form X = c1 X1 (t) + c2 X2 (t), for some vector-valued solutions X1 (t) and X2 (t) called fundamental solutions. Frequently, we call the matrix of fundamental solutions, = X1 (t), X2 (t) , (3.19)
  the fundamental matrix. The fundamental matrix is, roughly speaking, etA . It is analogous to the Wronskian of two fundamental solutions to a second order ordinary dierential equation. See examples below for more details.In fact, if the Xi s in (3.19) are carefully selected then = etA . This is not obvious and beyond the scope of this text.7
  200
  3.5.2
  Recall that we we solved the 1st order ordinary dierential equation y + p(t)y = q (t) for y = y (t) using the method of integrating factors, we got the formula y = (ep(t) dt 1
  (3.20)
  q (t) dt + c).
  (3.21)
  Consider a 2 2 system of linear 1st order ordinary dierential equations in the form x = ax + by + f (t), x(0) = x0 , y = cx + dy + g(t), y (0) = y0 . This can be rewritten in the form X = AX + F , where F = F (t) = f (t) g(t) (3.22)
  (3.20) by replacing X by y , A by p and F by q . It turns out that (3.22) can be solved in a way analogous to (3.20) as well. Here is the variation of parameters formula for systems: X = ( where c = 1 F (t) dt + c), (3.23)
  c1 is a constant vector determined by the initial conditions and is the c2 fundamental matrix. Example 3.5.1. A battle between X-men and Y-men is modeled by x = y + 1, x(0) = 100, t y = 4x + e , y (0) = 50. The non-homogeneous terms 1 and et represent reinforcements. Find out who wins, when, and the number of survivers. Here A is the matrix A= 1 et 0 1 4 0
  and F = F (t) =
  201
  In the method of variation of parameters, you must solve the homogeneous system rst. 1 , The eigenvalues of A are 1 = 2, 2 = 2, with associated eigenvectors v1 = 2 v2 = 1 2 , resp.. The general solution to the homogeneous system x = y, y = 4x, is X = c1 where X1 (t) = e2t 2e2t , X2 (t) = e2t 2e2t . 1 2 e2t + c2 1 2 e2t = c1 X1 (t) + c2 X2 (t),
  For the solution of the non-homogeneous equation, we must compute the fundamental matrix: = e2t e2t 2e2t 2e2t , so 1 = 1 4 2e2t e2t 2e2t e2t .
  Next, we compute the product, 1 F = and its integral, 1 F dt = Finally, to nish (3.23), we compute ( 1 F (t) dt + c) = = e2t e2t 2e2t 2e2t2t + 1 et + c1 1 4e 4 1 3t 1 2t e + 4 12 e + c2 2t + 1 et 1 4e 4 1 2t 1 3t e + 4 12 e
  1 4
  1 et
  1 2t t 1 2e 4e 1 3t 1 2t 2e + 4e
  This gives the general solution to the original system 1 x(t) = c1 e2t + et + c2 e2t , 3
  202 and
  1 y (t) = 1 et 2c1 e2t + 2c2 e2t . 3 We arent done! It remains to compute c1 , c2 using the ICs. For this, solve 1 + c1 + c2 = 100, 3 2 2c1 + 2c2 = 50. 3
  Figure 3.14: Solution to system x = y + 1, x(0) = 100, y = 4x + et , y (0) = 50. As you can see from Figure 3.14, the X-men win. The solution to y (t) = 0 is about t0 = 0.1279774... and x(t0 ) = 96.9458... survive. Example 3.5.2. Solve x = y + 1, y = x + cot(t).
  203
  0 . cot(t) In the method of variation of parameters, you must solve the homogeneous system rst. 1 , The eigenvalues of A are 1 = i, 2 = i, with associated eigenvectors v1 = i 1 i , resp.. Therefore, in the notation of (3.14), we have = 0, = 1, u1 = 1 0 ,
  and u2 =
  is
  X = c1 [cos(t) where
  1 0
  sin(t)
  0 1
  ]+c2 [cos(t)
  +sin(t)
  ] = c1 X1 (t)+c2 X2 (t),
  X1 (t) =
  cos(t) sin(t)
  X2 (t) =
  sin(t) cos(t)
  For the solution of the non-homogeneous equation, we must compute the fundamental matrix: = cos(t) sin(t) sin(t) cos(t) , so 1 = cos(t) sin(t) sin(t) cos(t) = ,
  1 F =
  0 cot(t)
  204
  1 F dt = Finally, we compute
  ( 1 F (t) dt + c) = = Therefore,
  c1 c2
  1 sin (t) cos (t) 1 sin (t) + , 2 (cos (t) + 1) 2 (cos (t) + 1)
  1. Use Sage and the method of variation of parameters to solve this: A battle is modeled by x = 4y, x(0) = 150, y = x + 1, y (0) = 40. (a) Write the solutions in parameteric form. (b) Who wins? 2. All 300 X-men are ghting all 101 Y-men, the battle being modeled by x = 9y + 1, x(0) = 300, y = x 1, y (0) = 101. Solve this system using the method of variation of parameters. Who wins? 3. All 60 X-men are ghting all 15 Y-men, the battle being modeled by x = 4y 1, x(0) = 60, y = x + 1, y (0) = 15. Solve this system using the method of variation of parameters. Who wins?
  205
  3.6
  Nonlinear systems
  Most dierential equations that people are interested in are nonlinear. Unfortunately these equations are usually not exactly solvable, so we must rely on a combination of numerical approximations to the solutions and methods for analyzing some aspect of the solutions. These latter methods are often lumped together under the name qualitative methods and they are part of the modern mathematical discipline of dynamical systems. In this chapter we will introduce some of these methods for rst-order nonlinear systems. Recall that a higher-order nonlinear ODE can always be converted to a rst-order system, so there is no loss of generality.
  3.6.1
  One of the simplest and most fruitful things we can do with a nonlinear system is nd its equilibria - values where all of the rst-order derivatives are zero - and study a linear approximation of the system near these points. Our rst example of the use of this technique will be the predator-prey system of Lotka and Volterra (the Lotka-Volterra equations). We are modeling two species, one of which (population y ) preys upon the other (population x). The rate of predation will be assumed to be proportional to the product xy (similar to the mass-action law in chemical kinetic modelling). As the predators cannot live without any prey, they are subjected to a death rate proportional to their population size. The prey are assumed to grow without predation at a rate proportional to their population size: x = ax bxy y = cy + dxy where a,b,c, and d are positive parameters. We can nd the equilibria by factoring each of the right hand sides of this system: we need x(a by ) = 0 and y (c + dx) = 0. The two solutions are (x, y ) = (0, 0) and (x, y ) = (c/d, a/b). Now we linearize around each of these equilibria. To linearize the system we compute the Jacobian matrix - the matrix of all rst-order derivatives of the slope functions - and use the value of this matrix at the point of linearization as our constant matrix for a linear system. For the Lotka-Volterra system this matrix is: (axbxy ) x (cy +dxy ) x (axbxy ) y (cy +dxy ) y
  a by bx dy c + dx
  206
  a 0 0 c
  u v
  Since this is a diagonal matrix we know the eigenvalues are simply a and c, so it is unstable (more precisely, its a saddle point). If we evaluate the Jacobian matrix at the other equilbrium, (c/d, a/b), we obtain the linearization: u 0 bc/d u . . v da/b 0 v where we interpet the variables u and v as approximations to the deviation from the equilibrium, that is for small u and v we should have u x c/d and v y a/b. The eigenvalues of this system are given by det I 0 bc/d da/b 0 = 2 + ac = 0
  so = aci and they are purely imaginary. Trajectories for the particular case a = b = c = d = 1 are illustrated below. Note that close to the equilibrium at (1, 1) the trajectories are close to the elliptic solutions of the linearized equations for that point, but closer to the other equilibrium at the origin the solutions appear more and more hyperbolic.6
  Exercises
  207
  1. Find the equilibrium and compute the linearized system around it for the system 2 2 x 1 = x1 x2 , x2 = x1 + 2x1 2x2 . 2. Find all of the equilibria, linearizations around them, and the corresponding eigenval ues for the system x 1 = (1 + x1 )(x1 x2 ), x2 = (2 x1 )(x1 + x2 ).
  3.6.2
  Consider an idealized pendulum with a massless rigid shaft of length L and a small sphere of mass m swinging from a point. At rst we will assume there is no friction.
  g sin()
  When the pendulum has swung through an angle of from its bottom position, the component of the gravitational acceleration g in the tangential direction is g sin( ). The angular acceleration is g sin( )/L, so the angle of the pendulum satises the ODE: g sin( ). L First we rewrite the ODE as a system of rst-order equations by introducing the velocity v = as a variable in its own right. Then we have: = v = v sin( )
  g L
  The equilibria occur when sin( ) = 0 and v = 0, so = n for some integer n. To linearize the system around these equilibria we compute the Jacobian matrix
  208
  J=
  dv d g d L sin( ) d
  dv dv g d L sin( ) dv
  0 1 cos( ) 0
  At the equilibria, the only thing that matters when evaluating this is whether is an even or odd multiple of , since that determines the sign of cos( ). v 0 g L 1 0 v
  g When n is even, cos( ) = 1, the eigenvalues of the linearized system are satisfy 2 + L =0 g so = L i. These are pure imaginary eigenvalues, so the linearized system has solutions which orbit the equilibria on elliptic trajectories. g =0 When n is odd, cos( ) = 1, the eigenvalues of the linearized system are satisfy 2 L g . This pair of positive and negative real eigenvalues means that the linearized so = L system has an unstable equilbria, a saddle point.
  3.6.3
  The Lorenz equations are a famous example of a nonlinear system in which the trajectories are chaotic. Roughly speaking this means that although the solutions are close to periodic, two solutions starting close together will diverge at an exponential rate, so that after a little while they will have quite dierent behavior. x = (y x) y = x( z ) y z = xy z Here , , and are positive parameters. Edward Lorenz introduced these equations in 1963 as a very simplied model of convection in the atmosphere. This model is in turn derived from a 12-variable model whose variables are more closely related to physical quantities. This system has a wide variety of interesting behavior for dierent parameter values. The most commonly studied case is when = 10, = 8/3, and r = 28. A plot of the x value is given below for this case for two initial conditions very close together (x0 = 5 and x0 = 5.001).
  209
  t 5 10 15 20
  -5
  -10
  -15
  In the (x,y) plane it is easier to see how a typical trajectory slowly alternates between oscillating around two equilibria.y 20
  x -15 -10 -5 5 10 15
  -20
  Exercises: 1. In 1958, Tsuneji Rikitake formulated a simple model of the Earths magnetic core to explain the oscillations in the polarity of the magnetic eld [Rikitake]. The equations for his model are: x = x + yz z = 1 xy
  y = y + (z a)x where a and are positive constants. Find the equilibria for this system for a = = 1, and compute the eigenvalues of the linearized system at those equilibria.
  210
  3.6.4
  Zombies attack
  Mathematical research does not excite the public imagination very often, but in 2009 a paper called When zombies attack! Mathematical modelling of an outbreak of zombie infection [Zom] was published that received much attention. The mathematical model they use is based on a serious model used to investigate infectious diseases. The zombie infection pattern they assume follows the one presented by George Romero in his famous (and now public-domain) 1968 movie Night of the Living Dead. Let S represents people (the susceptibles), Z is the number of zombies, and R (the removed) represents (a) deceased zombies, (b) bitten people (who are sometimes turned into zombies), or (c) dead people.
  Figure 3.15: Zombies in George Romeros Night of the Living Dead The simplest system of ODEs developed in that paper is: S = B SZ S,
  R = S + SZ R.
  Z = SZ + R SZ,
  What do these terms mean and how do they arise? We assume that people (counted by S ) have a constant birth rate B . a proportion of people die a non-zombie related death. (This accounts for the S term in the S line and the +S term in the R line.)
  211
  a proportion of dead humans (counted by R) can resurrect and become a zombie. (This accounts for the +R term in the Z line and the R term in the R line.) There are non-linear terms as well. These correspond to when a person interacts with a zombie - the SZ terms. We assume that some of these interactions results in a person killing a zombie by destroying its brain (This accounts for the SZ term in the Z line and the +SZ term in the R line.) some of these interactions results in a zombie infecting a person and turning that person into a zombie (This accounts for the SZ term in the S line and the +SZ term in the Z line.) These non-linear terms mean that the system is too complicated to solve by simple methods, such as the method of eigenvalues or Laplace transforms. However, solutions to the Zombies Attack model can be numerically approximated in Sage . Lets take = 0.005, = 0.004, = 0.0001, = 0.0001, B = 0, and solve the above system numerically. What do we get?Sage sage: from sage.calculus.desolvers import desolve_system_rk4 sage: t,s,z,r = var(t,s,z,r) sage: a,b,zeta,d,B = 0.005,0.004)
  The plot of the new solution is given in Figure 3.16. The Zombies win! Too many zombies are infecting people (this is the constant , which is relatively large). Lets make smaller = 0.005, = 0.002, = 0.0001, = 0.0001, B = 0, and solve the above system numerically. Now, what do we get?
  212
  The plot of the solutions is given Figure 3.17. Finally, the Zombies lose!
  Figure 3.17: Yet another Suseptibles, Zombies and Removed plot The moral of the story: to survive, you must protect yourself from zombies! You may wonder about the importance of learning about this model. Although it is humorous, it is also very closely related to many models of the transmission of infectious
  213
  diseases called SIR models (the initials stand for Susceptible, Infected, and Recovered). These models can be very important for developing ecient public health strategies. In the zombie model, the terms proportional to SZ represent interactions between people and zombies, and these are the only nonlinear terms. This and the SIR models are similar to the Lotka-Volterra model of predators and prey in that they model the interaction frequency of two groups by a quadratic term. Example 3.6.1. Solutions to the Zombies Attack model can be numerically approximated in Sage .Sage sage: from sage.calculus.desolvers import desolve_system_rk4 sage: t,s,z,r = var(t,s,z,r) sage: a,b,zeta,d,B = 0.005,0.0095)
  Figure 3.18: Zombies attack model. Exercises: *1 Introduce a new term in the zombie attack equations representing ghts between zombies. What eect would this have on the equilibria and their stability for parameter values = 0.005, = 0.0095, = 0.0001, and = 0.0001?
  214
  Chapter 4
  4.1
  Recall, a partial dierential equation (PDE) is an equation satised by an unknown function (called the dependent variable) and its partial derivatives. The variables you dierentiate with respect to are called the independent variables. If there is only one independent variable then it is called an ordinary dierential equation. Examples includeu + the Laplace equation x2 the independent variables,2
  2u y 2
  the heat equation ut = uxx , and the wave equation utt = c2 uxx . All these PDEs are of second order (you have to dierentiate twice to express the equation). Here, we consider a rst order PDE which arises in applications and use it to introduce the method of solution called separation of variables. 215
  216
  4.1.1
  Advection is the transport of a some conserved scalar quantity in a vector eld. A good example is the transport of pollutants or silt in a river (the motion of the water carries these impurities downstream) or trac ow. The advection equation is the PDE governing the motion of a conserved quantity as it is advected by a given velocity eld. The advection equation expressed mathematically is: u + (ua) = 0 t where is the divergence and a is the velocity eld of the uid. Frequently, it is assumed that a = 0 (this is expressed by saying that the velocity eld is solenoidal). In this case, the above equation reduces to u + a u = 0. t Assume we have horizontal pipe in which water is owing at a constant rate c in the positive x direction. Add some salt to this water and let u(x, t) denote the concentration (say in lbs/gallon) at time t. Note that the amount of salt in an interval I of the pipe is I u(x, t) dx. This concentration satises the transport (or advection) equation: ut + cux = 0. (For a derivation of this, see for example Strauss [S-pde], 1.3.) How do we solve this? Solution 1: DAlembert noticed that the directional derivative of u(x, t) in the direction 1 1 c, 1 is Dv (u) = 1+ (cux + ut ) = 0. Therefore, u(x, t) is constant along the v = 1+ c2 c2 lines in the direction of v , and so u(x, t) = f (x ct), for some function f . We will not use this method of solution in the example below but it does help visualize the shape of the solution. For instance, imagine the plot of z = f (x ct) in (x, t, z ) space. The contour lying above the line x = ct + k (k xed) is the line of constant height z = f (k). Solution 2: The method of separation of variables indicates that we start by assuming that u(x, t) can be factored: u(x, t) = X (x)T (t), for some (unknown) functions X and T . (We shall work on removing this assumption later. This assumption works because partial dierentiation of functions like x2 t3 is so much simpler that partial dierentiation of mixed functions like sin(x2 + t3 ).) Substituting this into the PDE gives X (x)T (t) + cX (x)T (t) = 0. Now separate all the xs on one side and the ts on the other (divide by X (x)T (t)): X (x) T (t) = c . T (t) X (x)
  217
  (This same trick works when you apply the separation of variables method to other linear PDEs, such as the heat equation or wave equation, as we will see in later lessons.) It is impossible for a function of an independent variable x to be identically equal to a function of an independent variable t unless both are constant. (Indeed, try taking the partial derivative (t) of T T (t) with respect to x. You get 0 since it doesnt depend on x. Therefore, the partial(x) X (x) derivative of c X X (x) is akso 0, so X (x) is a constant!) Therefore, some (unknown) constant K . So, we have two ODEs:
  T (t) T (t)
  T (t) = K, T (t)
  Therefore, we have converted the PDE into two ODEs. Solving, we get T (t) = c1 eKt , X (x) = c2 eKx/c , so, u(x, t) = AeKtKx/c = Ae c (xct) , for some constants K and A (where A is shorthand K for c1 c2 ; in terms of DAlemberts solution, f (y ) = Ae c (y) ). The general solution is a sum of these (for various As and K s). This can also be done in Sage :Sage sage: t = var("t") sage: T = function("T",t) sage: K = var("K") sage: T0 = var("T0") sage: sage: desolve(diff(T,t) == K*T, [T,t], [0,T0]) T0*e(K*t) sage: x = var("x") sage: X = function("X",x) sage: c = var("c") sage: X0 = var("X0") sage: desolve(diff(X,x) == -c(-1)*K*X, [X,x], [0,X0]) X0*e(-K*x/c) sage: solnX =desolve(diff(X,x) == -c(-1)*K*X, [X,x], [0,X0]) sage: solnX X0*e(-K*x/c) sage: solnT = desolve(diff(T,t) == K*T, [T,t], [0,T0]) sage: solnT T0*e(K*t) sage: solnT*solnX T0*X0*e(K*t - K*x/c)K
  Example 4.1.1. Assume water is owing along a horizontal pipe at 3 gal/min in the x direction and that there is an initial concentration of salt distributed in the water with
  218
  concentration of u(x, 0) = ex . Using separation of variables, nd the concentration at time t. Plot this for various values of t. Solution: The method of separation of variables gives the separated form of the solution to the transport PDE as u(x, t) = AeKtKx/c , where c = 3. The initial condition implies ex = u(x, 0) = AeK 0Kx/c = AeKx/3 , so A = 1 and K = 3. Therefore, u(x, t) = e3tx . In other words, the salt concentration is increasing in time. This makes sense if you think about it this way: freeze the water motion at time t = 0. There is a lot of salt at the beginning of the pipe and less and less salt as you move along the pipe. Now go down the pipe in the x-direction some amount where you can barely tell there is any salt in the water. Now unfreeze the water motion. Looking along the pipe, you see the concentration is increasing since the saltier water is now moving toward you.
  Figure 4.1: Transport with velocity c = 3. An interactive version of the plot in Figure 4.1 can be produced in Sage with the following commands.Sage sage: t,x = var("t,x") sage: plot3d(exp(3*t-x),[x,0,2],[t,0,2])
  What if the initial concentration was not u(x, 0) = ex but instead u(x, 0) = ex + 3e5x ? How does the solution to ut + 3ux = 0, u(x, 0) = ex + 3e5x , (4.1)
  dier from the method of solution used above? In this case, we must use the fact that (by superposition) the general solution is of the form
  219
  (4.2)
  for some constants A1 , K1 , .... To solve this PDE (4.1), we must answer the following questions: (1) How many terms from (4.2) are needed? (2) What are the constants A1 , K1 , ...? There are two terms in u(x, 0), so we can hope that we only need to use two terms and solve ex + 3e5x = u(x, 0) = A1 eK1 (0x/3) + A2 eK2 (0x/3) for A1 , K1 , A2 , K2 . Indeed, this is possible to solve: A1 = 1, K1 = 3, A2 = 3, K1 = 15. This gives u(x, t) = e3(tx/3) + 3e15(tx/3) .
  4.1.2
  Example 4.1.2. First, assume the solution to the heat equation has the factored form u(x, t) = X (x)T (t), for some (unknown) functions X, T . If this function solves the PDE then it must satisfy kX (x)T (t) = X (x)T (t), or 1 T (t) X (x) = . X (x) k T (t) Since x, t are independent variables, these quotients must be constant. (This of it this way: (x) 1 T (t) the derivative of X X (x) with respect to t is zero. Therefore, the derivative of k T (t) with respect to t is zero. This implies it is a constant. In other words, there must be a constant C such that T (t) = kC, T (t) X (x) CX (x) = 0.
  Now we have reduced the problem of solving the one PDE to two ordinary dierential equations (which is good), but with the price that we have introduced a constant which we dont know, namely C (which maybe isnt so good). The rst ordinary dierential equation is easy to solve: T (t) = A1 ekCt , for some constant A1 . It is best to analyse three cases now:
  220
  CHAPTER 4. INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS Case C = 0: This implies X (x) = A2 + A3 x, for some constants A2 , A3 . Therefore u(x, t) = A1 (A2 + A3 x) = a0 + b0 x, 2a0 2
  Case C < 0: Write (for convenience) C = r 2 , for some r > 0. The ordinary dierential equation for X implies X (x) = A2 cos(rx) + A3 sin(rx), for some constants A2 , A3 . Therefore u(x, t) = A1 ekr t (A2 cos(rx) + A3 sin(rx)) = (a cos(rx) + b sin(rx))ekr t , where A1 A2 has been renamed a and A1 A3 has been renamed b. Case C > 0: Write (for convenience) C = r 2 , for some r . The ordinary dierential equation for X implies X (x) = A2 erx + A3 e(rx , for some constants A2 , A3 . Therefore u(x, t) = ekr t (aerx + berx ), where A1 A2 has been renamed a and A1 A3 has been renamed b. These are the solutions of the heat equation which can be written in factored form. Exercises: 1. Use Sage to solve and plot the solution to the following problem. Assume water is owing along a horizontal pipe at 3 gal/min in the x direction and that there is an initial concentration of salt distributed in the water with concentration of u(x, 0) = ex . 2. Use separation of variables to nd all solutions to ut = ux + u, in factored form (where is a non-zero constant). 3. Use separation of variables to nd all solutions to ut + xux = u, in factored form. 4. Use separation of variables to nd all solutions to ut + tux = u, in factored form. 5. Use separation of variables to nd the unique solution to ut + 2ux = 3u, u(x, 0) = 7e5x .2 2 2
  4.2
  Roughly speaking, the superposition principle says that the sum of two solutions to a linear dierential equation is another solution. We give a specic example, in the following statement.
  4.2. THE METHOD OF SUPERPOSITION Theorem 4.2.1. (Superposition) If the solution to ut = kuxx , u(x, 0) = f1 (x), is u1 (x, t), and the solution to ut = kuxx , u(x, 0) = f2 (x), is u2 (x, t), then the solution to ut = kuxx , u(x, 0) = f1 (x) + f2 (x), is u1 (x, t) + u2 (x, t), We consider a few specic instances of this, but rst we need a few more facts. For the zero ends problem:
  221
  k ( is u(x, t) = bn sin( nx L )e
  Theorem 4.2.2. If n > 0 is an integer and bn is a given real number then the solution to ut = kuxx , u(x, 0) = bn sin( nx L ), u(0, t) = u(L, t) = 0,n 2 ) t L
  k ( This is not hard to verify directly: we must check that bn sin( nx L )e nx 2 1. the PDE ut = kuxx : ut = k( n L ) bn sin( L )e )2 t k ( n L
  n 2 ) t L
  satisesn 2 ) t L
  3. the boundary conditions: u(0, t) = u(L, t) = 0 (take x = 0 and note sin(0) = 0 or take x = L and note sin(n ) = 0). Example 4.2.1. The solution to is u(x, t) = 3 sin(11x)e2112t
  1 2 ut
  222
  For the insulated ends problem, the analog of Theorem 4.2.2 is:
  is u(x, t) = 5 3 cos(11x)e211
  is u(x, t) = 5. Therefore, by superposition (Theorem 4.2.1), the solution to 1 2 ut = uxx , u(x, 0) = 5 3 cos(11x), ux (0, t) = ux (, t) = 0,2t
  = 5 3 cos(11x)e242t .
  Of course, there is no need to stop with only two terms. Using superposition, we know that, the solution to the zero ends heat problem ut = kuxx , nx (4.3) u(x, 0) = n=1 bn sin( L ), u(0, t) = u(L, t) = 0,
  223
  nx k ( L ) t 0 . is u(x, t) = a n=1 an cos( L )e 2 + This last expression motivates the study of sine series and cosine series studied in the next section.
  (4.4)
  Exercises: 1. Verify Theorem 4.2.3. 2. Use superposition and the above fact (Theorem 4.2.2) to solve 3. Solve ut = 3uxx , 2x 3x u(x, 0) = 1 2 sin( 5 ) + 2 sin( 5 ), u(0, t) = u(5, t) = 0. ut = 3uxx , 1 nx u(x, 0) = n=1 n2 sin( 5 ), u(0, t) = u(5, t) = 0.
  4. Use superposition and the above fact (Theorem 4.2.3) to solve 5. Solve ut = 3uxx , 2x 3x u(x, 0) = 1 2 cos( 5 ) + 2 cos( 5 ), ux (0, t) = ux (5, t) = 0. ut = 3uxx , nx 1 u(x, 0) = 2 + n=1 n2 cos( 5 ), ux (0, t) = ux (5, t) = 0.
  4.3
  This section introduces Fourier series, background for understanding Fouriers solution to the heat equation presented in a later section.
  224
  4.3.1
  Brief history
  Fourier series were introduced by Joseph Fourier, a Frenchman who was a physicist, among other things. In fact, Fourier was Napoleans scientic advisor during Frances invasion of Egypt in the late 1800s. When Napolean returned to France, he elected (i.e., appointed) Fourier to be a Prefect - basically an important administrative post where he oversaw some large construction projects, such as highway constructions. It was during this time when Fourier worked on the theory of heat on the side. It turns out that DLambert looked at the wave equation some years earlier, using very similar techniques, but the approach was rejected (possibly due to its lack of rigor). However, Fouriers solution, explained in his 1807 memoir On the Propagation of Heat in Solid Bodies, was awarded a mathematics prize in 1811 by the Paris Institute. The gap in time is some indication of the controversy it caused at the time. Fouriers solution to the heat equation is basically what undergraduates often learn in a Dierential Equations class. The exception being that our understanding of Fourier series now is much better than what was known in the early 1800s and some of these facts, such Dirichlets theorem (see Theorem 4.3.1 below), are covered as well.
  4.3.2
  Fourier series, sine series, and cosine series are all expansions for a given function f (x) in terms of sines and/or cosines, somewhat analogous to the way that a Taylor series a0 + a1 x + a2 x2 + ... is an expansion for a given function in powers of x. Both Fourier and Taylor series can be used to approximate the function they represent. However, there are at least three important dierences between the two types of series. 1. For a function to have a Taylor series it must be dierentiable1 , whereas for a Fourier series it does not even have to be continuous. 2. Another dierence is that the Taylor series is typically not periodic (though it can be in some cases), whereas a Fourier series is always periodic. 3. Finally, the Taylor series (when it converges) always converges to the function f (x), but the Fourier series may not (see Dirichlets theorem below for a more precise description of what happens).
  4.3.3
  Denitions
  [an cos(
  n=1
  nx nx ) + bn sin( )], L Lf (n) (0) . n!
  Remember the formula for the n-th Taylor series coecient centered at x = 0: an =
  225
  where the an and the bn are given numbers. We shall implicitly assume that, for each x, this sum converges to some value, denoted F (x). The N th partial sum of this Fourier series is denoted nx nx a0 [an cos( + ) + bn sin( )], FN (x) = 2 L L n=1 for N > 0. For the remainder of these notes, we assume the an s and bn s have the property that, for each x, limN FN (x) = F (x). As a special case, if all the an = 0, then we call this Fourier series a (Fourier) sine series. Likewise, if all the bn = 0, then we call this Fourier series a (Fourier) cosine series. Fourier series Let f (x) be a function dened on an interval of the real line. We allow f (x) to be discontinuous but the points in this interval where f (x) is discontinuous must be nite in number and must be jump discontinuities. First, we discuss Fourier series. To have a Fourier series you must be given two things: a period P = 2L, a function f (x) dened on an interval of length 2L, say L < x < L (but sometimes 0 < x < 2L is used instead). The Fourier series expansion of f (x) with period 2L is a0 f (x) + 2 N
  nx nx ) + bn sin( )], L L
  (4.5)
  f (x) cos(L L
  nx ) dx, L
  (4.6)
  f (x) sin(L
  nx ) dx. L
  (4.7)
  Convergence and Dirichlets theorem The symbol is used in (4.5) above instead of = because of the fact that the Fourier series may not converge to f (x) (see Dirichlets theorem below). Do you remember right-hand and left-hand limits from Calculus I? Recall they are denoted f (x+) = lim0,>0 f (x + ) and f (x) = lim0,>0 f (x ), resp.. We know now that the Fourier series does notThese formulas were not known to Fourier. To compute the Fourier coecients an , bn he used sometimes ingenious round-about methods using large systems of equations.2
  226
  converge to the value of f (x) at every point where x is a point of jump discontinuity3 . The convergence properties are given by the theorem below. Theorem 4.3.1. (Dirichlets theorem4 ) Let f (x) be a function dened on L < x < L which is bounded and piecewise continuous with only nitely many discontinuities. The Fourier series of f (x), f (x) a0 + 2
  (where an and bn are as in the formulas (4.6), (4.7)) converges to f (x+) + f (x) . 2 In other words, the Fourier series of f (x) converges to f (x) only if f (x) is continuous at x. If f (x) is not continuous at x then then Fourier series of f (x) converges to the midpoint of the jump. Of course, since a sine series or cosine series is a special case of a Fourier series, Dirichlets theorem applies to them as well. Example 4.3.1. Find the Fourier series of f (x) = 2 + x, 2 < x < 2. The denition of L implies L = 2. Without even computing the Fourier series, we can evaluate it using Dirichlets theorem. Exercise: Using periodicity and Dirichlets theorem, nd the value that the Fourier series of f (x) converges to at x = 1, 2, 3. (Soln: f (x) is continuous at 1, so the Fourier series atx = 1 converges to f (1) = 3 by Dirichlets theorem. f (x) is not dened at 2. Its Fourier series is f (2) = 0+4 periodic with period 4, so at x = 2 the Fourier series converges to f (2+)+ 2 2 = 2. f (x) is not dened at 3. Its Fourier series is periodic with period 4, so at x = 3 the Fourier series converges f (1+) = 1+1 to f (1)+2 2 = 1.)
  The formulas (4.6) and (4.7) enable us to compute the Fourier series coecients a0 , an and bn . (We skip the details.) These formulas give that the Fourier series of f (x) is f (x) 4 + 2 n=1
  n=1
  nx cos (n ) sin( ). n 2
  S0 = 2, S1 = 2 +
  3 Fourier believed his series converged to the function in the early 1800s but we now know this is not always true. 4 Pronounced Dear-ish-lay.
  227
  2 x
  Figure 4.2: Graph of f (x) and a Fourier series approximation of f (x). Notice that f (x) is only dened from 2 < x < 2 yet the Fourier series is not only dened everywhere but is periodic with period P = 2L = 4. Also, notice that S8 is not a bad approximation to f (x). This can also be done in Sage . First, we dene the function.Sage sage: f = lambda x:x+2 sage: f = Piecewise([[(-2,2),f]])
  This can be plotted using the command f.plot().show(). Next, we compute the Fourier series coecients:Sage sage: f.fourier_series_cosine_coefficient(0,2) # a_0 4 sage: f.fourier_series_cosine_coefficient(1,2) # a_1 0 sage: f.fourier_series_cosine_coefficient(2,2) # a_2 0 sage: f.fourier_series_cosine_coefficient(3,) # a_3 0 sage: f.fourier_series_sine_coefficient(1,2) # b_1 4/pi sage: f.fourier_series_sine_coefficient(2,) # b_2 -2/pi sage: f.fourier_series_sine_coefficient(3,2) # b_3 4/(3*pi)
  228
  Finally, the partial Fourier series and its plot verses the function can be computed using the following Sage commands.Sage sage: f.fourier_series_partial_sum(3,2) -2*sin(pi*x)/pi + 4*sin(pi*x/2)/pi + 2 sage: P1 = f.plot_fourier_series_partial_sum(15,2,-5,5,linestyle=":") sage: P2 = f.plot(rgbcolor=(1,1/4,1/2)) sage: (P1+P2).show()
  The plot (which takes 15 terms of the Fourier series) is given in Figure 4.2. Cosine series Next, we discuss cosine series. To have a cosine series you must be given two things: a period P = 2L, a function f (x) dened on the interval of length L, 0 < x < L. The cosine series expansion of f (x) with period 2L is f (x) where an is given by an = 2 LL
  a0 + 2
  an cos(
  nx ), L
  cos(0
  nx )f (x) dx. L
  (4.8)
  The cosine series of f (x) is exactly the same as the Fourier series of the even extension of f (x), dened by feven (x) = f (x), 0 < x < L, f (x), L < x < 0.
  Example 4.3.2. Lets consider an example of a cosine series. In this case, we take the piecewise constant function f (x) dened on 0 < x < 3 by f (x) = 1, 0 < x < 2, 1, 2 x < 3.
  We see therefore L = 3. The formula above for the cosine series coecients gives that sin 2 1 nx 3 n 4 f (x) + cos( ). 3 n=1 n 3
  4.3. FOURIER SERIES, SINE SERIES, COSINE SERIES The rst few partial sums are
  229
  1 3 cos 3 x , 1 2 x x 3 cos 3 3 cos 3 S3 = 1/3 + 2 , ... As before, the more terms in the cosine series we take, the better the approximation is, for 0 < x < 3. Comparing the picture below with the picture above, note that even with more terms, this approximation is not as good as the previous example. The precise reason for this is rather technical but basically boils down to the following: roughly speaking, the more dierentiable the function is, the faster the Fourier series converges (and therefore the better the partial sums of the Fourier series will approximate f (x)). Also, notice that the cosine series approximation S10 is an even function but f (x) is not (its only dened from 0 < x < 3). For instance, the graph of f (x) and of S10 are given in Figure 4.3.
  S2 = 1/3 + 2
  Sine series Finally, we dene sine series. As with a cosine series, to have a sine series you must be given two things: a period P = 2L, a function f (x) dened on the interval of length L, 0 < x < L. The sine series of f (x) with period 2L is f (x) n=1
  bn sin(
  230
  where bn is given by bn = 2 LL
  sin(0
  (4.9)
  The sine series of f (x) is exactly the same as the Fourier series of the odd extension of f (x), dened by fodd (x) = f (x), 0 < x < L, f (x), L < x < 0.
  Example 4.3.3. Lets consider an example of a sine series. In this case, we take the piecewise constant function f (x) dened on 0 < x < 3 by the same expression we used in the cosine series example above. Exercise: Using periodicity and Dirichlets theorem, nd the value that the sine series of f (x) converges to at x = 1, 2, 3. (Soln: f (x) is continuous at 1, so the Fourier seriesat x = 1 converges to f (1) = 1. f (x) is not continuous at 2, so at x = 2 the SS converges to f (2) f (2+)+f (2) = f (2+)+ = 1+1 = 0. f (x) is not dened at 3. Its SS is periodic with period 6, 2 2 2 fodd (3+) = 1+1 = 0.) so at x = 3 the SS converges to fodd (3)+ 2 2
  The formula above for the sine series coecients give that f (x) = The partial sums are n=1
  cos (n ) 2 cos n
  2 3
  n + 1
  sin(
  nx ). 3
  2 x sin 3 sin (1/3 x) +3 , 2 1 sin 3 sin 3 x x sin ( x) +3 4/3 , ... S3 = 2 These partial sums Sn , as n , converge to their limit about as fast as those in the previous example. Instead of taking only 10 terms, this time we take 40. Observe from the graph in Figure 4.4 that the value of the sine series at x = 2 does seem to be approaching 0, as Dirichlets Theorem predicts. The graph of f (x) with S40 is given in Figure 4.4.
  S2 = 2
  Exercises 1. Let f (x) = x2 , 2 < x < 2 and L = 2. Use Sage to compute the rst 10 terms of the Fourier series, and plot the corresponding partial sum. Next plot the partial sum of the rst 50 terms and compare them. 2. What mathematical results do the following Sage commands give you? In other words, if you can seen someone typing these commands into a computer, explain what problem they were trying to solve.
  231
  3. A periodic function is given over one period by f (t) = |t| , < t < . Which of the nt nt 0 following statements is correct for the Fourier series a n=1 an cos L + bn sin L 2 + of f (t) (a) an = 0 for all even integers n, but not for any odd integers n. (b) an = 0 for all odd integers n, but not for any even integers n. (c) an = 0 for n = 0, 1, 2, . . . . (d) an = 0 for n = 1, 2, 3, . . . , but a0 = 0. 4. At the point t = 0 the Fourier sine series of the function f (t) = 1, 0 t 1 converges to
  232
  d) 1
  5. Let f (x) =
  (a) Find the Fourier sine series for f (x). (b) To what value does the Fourier sine series converge at 3 ? (i) x = 1? (ii) x = 2 6. Let f (x) = x( x), < x < .1 (iii) x = 2 ?
  Find the cosine series of f (x), 0 < x < , having period 2 , and graph the function to which the series converges to, 2 < x < 2 . f (x) = period 2 . (a) Find the cosine series of f (x) and graph it, 2 < x < 2 . 1, 0 t /2, 1, /2 < t < ,
  7. Let
  (b) Write down the series in summation notation and write down the sum of the rst 3 non-zero terms.
  4.4
  u(x,t) (x,t) . This is a diusion equation whose study The heat equation is the PDE k x = ut 2 is very important for many topics (even in subjects as diverse as nancial mathematics, chemistry, and modeling criminal behavior!) The heat equation with zero ends boundary conditions models the temperature of an (insulated) wire of length L:
  233
  Here u(x, t) denotes the temperature at a point x on the wire at time t. The initial temperature f (x) is a given more-or-less arbitrary function5 that is related to our temperature function u(x, t) by the condition: u(x, 0) = f (x). In this model, it is assumed no heat escapes out the middle of the wire (which, say, is coated with some kind of insulating plastic). However, the boundary conditions, u(0, t) = u(L, t) = 0 permits heat to disipate or escape out the ends.
  4.4.1
  The recipe for solving a zero ends heat problem by Fouriers method is pretty simple. Find the sine series of f (x): f (x) n=1
  bn (f ) sin(
  where the coecients can be computed from (4.9). The solution is6 u(x, t) = Example 4.4.1. Let f (x) = Then L = and bn (f ) = Thus 2 6 2 sin(x) sin(2x) + sin(3x) + .... 3 2 0 n=1
  nx n ) exp(k( )2 t). L L
  f (x) sin(nx)dx = 2
  2 cos(n ) 3 cos( 1 2 n ) + 1 . n
  It must nice enough that it has a Fourier series, so piecewise continuous will work for us. A quick reminder: exp(x) = ex .
  234
  sage: x = var("x") sage: f1(x) = -1 sage: f2(x) = 2 sage: f = Piecewise([[(0,pi/2),f1],[(pi/2,pi),f2]]) sage: P1 = f.plot() sage: b10 = [f.sine_series_coefficient(n,pi) for n in range(1,10)] sage: b10 [2/pi, -6/pi, 2/(3*pi), 0, 2/(5*pi), -2/pi, 2/(7*pi), 0, 2/(9*pi)] sage: ss10 = sum([b10[n]*sin((n+1)*x) for n in range(len(b50))]) sage: ss10 2*sin(9*x)/(9*pi) + 2*sin(7*x)/(7*pi) - 2*sin(6*x)/pi + 2*sin(5*x)/(5*pi) + 2*sin(3*x)/(3*pi) - 6*sin(2*x)/pi + 2*sin(x)/pi sage: b50 = [f.sine_series_coefficient(n,pi) for n in range(1,50)] sage: ss50 = sum([b50[n]*sin((n+1)*x) for n in range(len(b))]) sage: P2 = ss10.plot(-5,5,linestyle="--") sage: P3 = ss50.plot(-5,5,linestyle=":") sage: (P1+P2+P3).show()
  This illustrates how the series converges to the function. The function f (x), and some of the partial sums of its sine series, looks like Figure 4.5.
  Figure 4.5: f (x) and two sine series approximations. As you can see, taking more and more terms gives partial sums which better and better approximate f (x). The solution to the heat equation, therefore, is u(x, t) = n=1
  n nx ) exp(k( )2 t). L L
  Next, we see how Sage can plot the solution to the heat equation (we use k = 1):
  235
  Taking 50 terms of this series, the graph of the solution at t = 0, t = 0.5, t = 1, looks approximately like Figure 4.6.
  Figure 4.6: f (x), u(x, 0.1), u(x, 0.5), u(x, 1.0) using 60 terms of the sine series.
  4.4.2
  Here ux (x, t) denotes the partial derivative of the temperature at a point x on the wire at time t. The initial temperature f (x) is specied by the equation u(x, 0) = f (x). Find the cosine series of f (x): nx a0 an (f ) cos( f (x) + ), 2 L n=1
  The heat equation with insulated ends boundary conditions models the temperature of an (insulated) wire of length L: 2 u(x,t) (x,t) = ut k x2 u(x, 0) = f (x), ux (0, t) = ux (L, t) = 0.
  236
  CHAPTER 4. INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS where the coecients can be computed from (4.8). The solution is u(x, t) = a0 + 2 n=1
  an (f ) cos(
  nx n )) exp(k( )2 t). L L
  Example 4.4.2. Let f (x) = Then L = and an (f ) = for n > 0 and a0 = 1. Thus a0 + a1 (f ) cos(x) + a2 (f ) cos(2x) + ... 2 This can also be done in Sage : f (x) Sage sage: x = var("x") sage: f1(x) = -1 sage: f2(x) = 2 sage: f = Piecewise([[(0,pi/2),f1],[(pi/2,pi),f2]]) sage: P1 = f.plot() sage: a10 = [f.cosine_series_coefficient(n,pi) for n in range(10)] sage: a10 [1, -6/pi, 0, 2/pi, 0, -6/(5*pi), 0, 6/(7*pi), 0, -2/(3*pi)] sage: a50 = [f.cosine_series_coefficient(n,pi) for n in range(50)] sage: cs10 = a10[0]/2 + sum([a10[n]*cos(n*x) for n in range(1,len(a10))]) sage: P2 = cs10.plot(-5,5,linestyle="--") sage: cs50 = a50[0]/2 + sum([a50[n]*cos(n*x) for n in range(1,len(a50))]) sage: P3 = cs50.plot(-5,5,linestyle=":") sage: (P1+P2+P3).show()
  f (x) cos(nx)dx = 6
  sin
  n , n
  This illustrates how the series converges to the function. The piecewise constant function f (x), and some of the partial sums of its cosine series (one using 10 terms and one using 50 terms), looks like Figure 4.7. As you can see from Figure 4.7, taking more and more terms gives functions which better and better approximate f (x). Consider the heat problem
  237
  2 u(x,t) x2
  a0 nx n u(x, t) = an (f ) cos( + ) exp(k( )2 t). 2 L L n=1 Using Sage , we can plot this function:Sage sage: sage: sage: sage: sage: sage: sage: soln50a = a50[0]/2 + sum([a50[n]*cos(n*x)*e(-n2*(1/100)) for n in range(1,len(a50))]) soln50b = a50[0]/2 + sum([a50[n]*cos(n*x)*e(-n2*(1/10)) for n in range(1,len(a50))]) soln50c = a50[0]/2 + sum([a50[n]*cos(n*x)*e(-n2*(1/2)) for n in range(1,len(a50))]) P4 = soln50a.plot(0,pi) P5 = soln50b.plot(0,pi,linestyle=":") P6 = soln50c.plot(0,pi,linestyle="--") (P1+P4+P5+P6).show()
  Taking only the rst 50 terms of this series, the graph of the solution at t = 0, t = 0.01, t = 0.1,, t = 0.5, looks approximately like the graph in Figure 4.8.
  4.4.3
  Where does this solution come from? It comes from the method of separation of variables and the superposition principle. Here is a short explanation. We shall only discuss the zero ends case (the insulated ends case is similar).
  238
  Figure 4.8: f (x) = u(x, 0), u(x, 0.01), u(x, 0.1), u(x, 0.5) using 50 terms of the cosine series. We saw in Example 4.1.2 that the partial diferential equation k has solutions in factored form u(x, t) = X (x)T (t), where T (t) = A1 ekCt , for some constant A1 and X (x) was described using a case-by-case analysis. Physical intuition tells us that u(x, t) is bounded as t . This implies C 0, which rules out the third case in Example 4.1.2. By superposition, the general solution is a sum of these: u(x, t) = =a0 2 a0 2 krn t + b0 x + n=1 (an cos(rn x) + bn sin(rn x))e 2 + b0 x + (a1 cos(r1 x) + b1 sin(r1 x))ekr1 t 2 +(a2 cos(r2 x) + b2 sin(r2 x))ekr2 t + ...,2
  u(x, t) 2 u(x, t) = 2 x t
  (4.10)
  for some ai , bi , ri . We may order the ri s to be strictly increasing if we like. We have not yet used the IC u(x, 0) = f (x) or the BCs u(0, t) = u(L, t) = 0. We do that next. What do the BCs tell us? Plugging in x = 0 into (4.10) gives 0 = u(0, t) = a0 + 2 n=1
  an ekrn t =
  239
  u(x, t) = b0 x +n=1
  bn sin(rn L)ekrn t .
  Again, exponential functions are linearly independent, so b0 = 0, bn sin(rn L) for n = 1, 2, .... In other to get a non-trivial solution to the PDE, we dont want bn = 0, so sin(rn L) = 0. This forces rn L to be a multiple of , say rn = n/L. This givesn 2 2 2 2 n 2 x)ek( L ) t = b1 sin( x))ek( L ) t + b2 sin( x))ek( L ) t + ..., L L L n=1 (4.11) for some bi s. The special case t = 0 is the so-called sine series expansion of the initial temperature function u(x, 0). This was discovered by Fourier. To solve the heat eqution, it remains to solve for the sine series coecients bi . There is one remaining condition which our solution u(x, t) must satisfy. What does the IC tell us? Plugging t = 0 into (4.11) gives
  u(x, t) =
  f (x) = u(x, 0) =
  In other words, if f (x) is given as a sum of these sine functions, or if we can somehow express f (x) as a sum of sine functions, then we can solve the heat equation. In fact there is a formula7 for these coecients bn : bn = 2 LL
  f (x) cos(0
  n x)dx. L
  It is this formula which is used in the solutions above. Exercises: 1. Solve the heat equation(x,t) u(x,t) = ut 2 x 2 ux (0, t) = ux (3, t) = 0 u(x, 0) = x,
  Fourier did not know this formula at the time; it was discovered later by Dirichlet.
  240
  2. A metal bar of length 2 and thermal diusivity 3 has its left end held at 0o and its right end insulated for all times t > 0. Mathematically this is described by the partial dierential equation and boundary conditions 2u u = 3 2, t x u(0, t) = 0, u (2, t) = 0 x
  The physically reasonable solutions of the form u(x, t) = X (x)T (t) which satisfy the 2 partial dierential equation are u(x.t) = e3 t (A cos x + B sin x). The solutions of this form which also satisfy the boundary conditions are u(x, t) =n (a) e3( 2 ) t B sin nx 2 , n = 1, 2, 3, . . . n 2 (c) e3( 4 ) t B sin nx , n = 1, 3, 5, . . . 2 n (b) e3( 2 ) t A cos nx 2 , n = 1, 2, 3, . . . n 2 (d) e3( 4 ) t A cos nx , n = 1, 3, 5, . . . 2
  3. A thin bar of length 4 and thermal diusivity k = 1, which is initially has a temperature distribution given as a function of position by 5x, has its ends insulated for times t > 0. This situation is described mathematically byu 2u t = x2 u x (0, t) = u x (4, t) =
  0, 0, t > 0 u(x, 0) = 5x, 0 x 4. Find the temperature u(x, t). Show all steps of the separation of variable process clearly. What is the nal temperature distribution on the bar as t ? 4. Let f (x) = tionu u 2 x2 = t , u(x, 0) = f (x), u(0, t) = u(1, t) = 0.2
  1 if 0 < x < 1/2 Find the function u(x, t) satisfying the heat equa0 if 1/2 < x < 1.
  Write your answer in summation notation. Estimate u(1/2, 1) using the rst 3 nonzero terms of the series solution. 5. Solve the heat equation u(x, t) 1 2 u(x, t) = , t 10 x2 with insulated end points and u(x, 0) = x( x), 0 < x < . Show all steps of the separation of variables process.
  241
  6. The temperature distribution u(x, t) on a thin bar satises the following conditions uxx = ut , where 1, 0 t /2, 1, /2 < t < , ux (0, t) = ux (, t) = 0, u(x, 0) = f (x),
  f (x) =
  (a) Interpret the boundary conditions physically. (b) Find the solution u(x, t) = X (x)T (t) of uxx = ut in separated form Show all steps of the separation of variables process clearly. (b) Find the solution u(x, t) to the above heat problem. Write your answer in summation notation. (c) Use the rst three terms of your answer to (b) to nd the temperature at the middle of the bar after one second.
  4.5century. Daniel Bernoullis solution dealt with an innite briey discussing what his basic idea was. The theorem of dAlembert on the solution to the wave equation is stated roughly as follows: The partial dierential equation:2 2w 2 w = c t2 x2
  is satised by any function of the form w = w(x, t) = g(x + ct)+ h(x ct), where g and h are arbitrary functions. (This is called the dAlembert solution.) Geometrically speaking, w the idea of the proof is to observe that w t c x is a constant times the directional derivative Dv w(x, t), where v is a unit vector in the direction c, 1 . Therefore, you integrate
  242
  2 2w 2 w c =0 t2 x2 twice, once in the v+ direction, once in the v , to get the solution. Easier said than done, but still, thats the idea. The wave equation with zero ends boundary conditions models the motion of a (perfectly elastic) guitar string of length L:
  Here w(x, t) denotes the displacement from rest of a point x on the string at time t. The initial displacement f (x) and initial velocity g(x) at specied by the equations w(x, 0) = f (x), Method: Find the sine series of f (x) and g(x): nx ), bn (f ) sin( f (x) L n=1 The solution is w(x, t) =
  wt (x, 0) = g(x).
  g(x)
  bn (g) sin(
  nx ). L
  (bn (f ) cos(c
  and let g(x) = 0. Then L = , bn (g) = 0, and bn (f ) = Thus f (x) b1 (f ) sin(x) + b2 (f ) sin(2x) + ... = 6 2 2 sin(x) sin(2x) + sin(3x) + .... 3 2 0
  The function f (x), and some of the partial sums of its sine series, looks like This was computed using the following Sage commands:
  243
  As you can see, taking more and more terms gives functions which better and better approximate f (x). The solution to the wave equation, therefore, is w(x, t) =
  Taking only the rst 50 terms of this series, the graph of the solution at t = 0, t = 0.1, t = 1/5, t = 1/4, looks approximately like Figure 4.10. Figure 4.10 was produced using the Sage commands:Sage sage: t = var("t") sage: w50t1 = sum([b50[i-1]*sin(i*x)*cos(3*i*(1/10)) for i in range(1,50)]) sage: P3 = w50t1.plot(0,pi,linestyle=":")
  244
  Of course, taking terms would give a better approximation to w(x, t). Taking the rst 100 terms of this series (but with dierent times) is given in Figure 4.11.
  4.6. THE SCHRODINGER EQUATION 1. Solve the wave equation 2 (x,t) 2 (x,t) 2 w = w 2 x t2 w(0, t) = w(3, t) = 0 w(x, 0) = x w (x, 0) = 0,t
  245
  using Sage to plot approximations as above. 2. Let f (x) = (a) Graph f (x) over 0 < x < . (b) Compute the sine series of f (x). Both 1 2
  write your answer in summation notation, and (c) Find the value of the sine series of f (x) at x = 0,
  x = 2001 .
  x = 2000 ,
  x = /2,
  (d) Solve the wave equation with = 3, u(x, 0) = f (x), ut (x, 0) = 0, and zero ends.
  4.6
  The one-dimensional Schr odinger equation for a free particle is 2 (x, t) (x, t) , = x2 t where k > 0 is a constant (involving Plancks constant and the mass of the particle) and i = 1 as usual. The solution is called the wave function describing instantaneous state of the particle. For the analog in 3 dimensions (which is the one actually used by physicists - the one-dimensional version we are dealing with is a simplied mathematical model), one can interpret the square of the absolute value of the wave function as the probability density function for the particle to be found at a point in space. In other words, | (x, t)|2 dx is the probability of nding the particle in the volume dx surrounding the position x, at time t. ik
  246
  If we restrict the particle to a box then (for our simplied one-dimensional quantummechanical model) we can impose a boundary condition of the form (0, t) = (L, t) = 0, and an initial condition of the form (x, 0) = f (x), 0 < x < L.
  |f (x)|2 dx = 1.L
  If | (x, t)|2 represents a pdf of nding a particle at x at time t then 0 |f (x)|2 dx represents the probability of nding the particle somewhere in the box initially, which is of course 1.
  4.6.1
  Method
  Find the sine series of f (x): f (x) The solution is (x, t) = Each of the terms n (x, t) = bn sin( nx n ) exp(ik( )2 t). L LL 2 0 |n (x, t)| dx n=1 n=1
  n nx ) exp(ik( )2 t). L L
  is called a standing wave (though in this case sometimes bn is chosen so that 1). Example: Let f (x) = Then L = 1 and bn (f ) = 2 11
  f (x) sin(0
  4.6. THE SCHRODINGER EQUATION Thus f (x) b1 (f ) sin(x) + b2 (f ) sin(2x) + ... 1 (1 + 2 cos( n = n n 2 ) cos(n )) sin(nx) .
  247
  Taking more and more terms gives functions which better and better approximate f (x). The solution to Schr odingers equation, therefore, is (x, t) = 1 n (1 + 2 cos( ) cos(n )) sin(nx) exp(ik(n )2 t). n 2 n=1
  Explanation: Where does this solution come from? It comes from the method of separation of variables and the superposution principle. Here is a short explanation. 2 (x,t) (x,t) has the factored form First, assume the solution to the PDE ik x = t 2 (x, t) = X (x)T (t), for some (unknown) functions X, T . If this function solves the PDE then it must satisfy kX (x)T (t) = X (x)T (t), or X (x) 1 T (t) = . X (x) ik T (t) Since x, t are independent variables, these quotients must be constant. In other words, there must be a constant C such that T (t) = ikC, T (t) X (x) CX (x) = 0.
  Now we have reduced the problem of solving the one PDE to two ODEs (which is good), but with the price that we have introduced a constant which we dont know, namely C (which maybe isnt so good). The rst ODE is easy to solve: T (t) = A1 eikCt , for some constant A1 . It remains to determine C . Case C > 0: Write (for convenience) C = r 2 , for some r > 0. The ODE for X implies X (x) = A2 exp(rx) + A3 exp(rx), for some constants A2 , A3 . Therefore (x, t) = A1 eikr t (A2 exp(rx) + A3 exp(rx)) = (a exp(rx) + b exp(rx))eikr t , where A1 A2 has been renamed a and A1 A3 has been renamed b. This will not match the boundary conditions unless a and b are both 0. Case C = 0: This implies X (x) = A2 + A3 x, for some constants A2 , A3 . Therefore2 2
  248
  (x, t) = A1 (A2 + A3 x) = a + bx, where A1 A2 has been renamed a and A1 A3 has been renamed b. This will not match the boundary conditions unless a and b are both 0. Case C < 0: Write (for convenience) C = r 2 , for some r > 0. The ODE for X implies X (x) = A2 cos(rx) + A3 sin(rx), for some constants A2 , A3 . Therefore (x, t) = A1 eikr t (A2 cos(rx) + A3 sin(rx)) = (a cos(rx) + b sin(rx))eikr t , where A1 A2 has been renamed a and A1 A3 has been renamed b. This will not match the boundary conditions unless a = 0 and r = n L These are the solutions of the heat equation which can be written in factored form. By superposition, the general solution is a sum of these:ikrn t (x, t) = n=1 (an cos(rn x) + bn sin(rn x))e 2 2 = b1 sin(r1 x)eikr1 t + b2 sin(r2 x)eikr2 t + ...,2 2 2
  (4.12)
  for some bn , where rn = n L . Note the similarity with Fouriers solution to the heat equation. There is one remaining condition which our solution (x, t) must satisfy. We have not yet used the IC (x, 0) = f (x). We do that next. Plugging t = 0 into (4.12) gives f (x) = (x, 0) = n=1
  In other words, if f (x) is given as a sum of these sine functions, or if we can somehow express f (x) as a sum of sine functions, then we can solve Schr odingers equation. In fact there is a formula for these coecients bn : bn = 2 LL
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  250
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  [DF-df] Direction Field Plotter of Prof Robert Israel: [D-spr] Wikipedia entry for damped motion: [E-num] General wikipedia introduction to Eulers method:
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  [F-1st] General wikipedia introduction to First order linear dierential equations: [F1-fs] Wikipedia Fourier series article [F2-fs] MacTutor Fourier biography: [Gl] Andrew Gleason, Fundamentals of abstract analysis, Reading, Massachusetts: Addison-Wesley, 1966; corrected reprint, Boston: Jones and Bartlett, 1991. [H-rref] Jim Heeron, Linear Algebra, released under the GNU Free Documentation License, available at [H-sde] Desmond J. Higham, An Algorithmic Introduction to Numerical Simulation of Stochastic Dierential Equations, SIAM Review Vol.43, No.3 (2001), p.525-546. [HSD] Morris W. Hirsch, Stephen Smale, and Robert L. Devaney, Dierential Equations, Dynamical Systems and An Introduction to Chaos, Second Edition, Elsevier Academic Press, 2004. [H-ivp] General wikipedia introduction to the Hyperbolic trig function [H-intro] General wikipedia introduction [H-fs] General wikipedia introduction to the heat equation: [H1-spr] Wikipedia entry for Robert Hooke: [H2-spr] MacTutor entry for Hooke:
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  251
  [K-cir] Wikipedia entry for Kirchho: [L-var] Wikipedia article on Joseph Louis Lagrange: [L-sys] Wikipedia entry for Lanchester: [LA-sys] Lanchester automobile information: [L-intro] F. W. Lanchester, Mathematics in Warfare, in The World of Mathematics, J. Newman ed., vol.4, 2138-2157, Simon and Schuster (New York) 1956; now Dover 2000. (A four-volume collection of articles.) [La-sys] Frederick William Lanchester, Aviation in Warfare: The Dawn of the Fourth Arm, Constable and Co., London, 1916. [L-lt] Wikipedia entry for Laplace: [LT-lt] Wikipedia entry for Laplace transform: [L-linear] General wikipedia introduction to Linear Independence: [Lo-intro] General wikipedia introduction to the logistic function model of population growth: [M-intro] Niall J. MacKay, Lanchester combat models, May 2005.
  [M-ps] Sean Mauch, Introduction to methods of Applied Mathematics, [M] Maxima, a general purpose Computer Algebra system. [M-mech] General wikipedia introduction to Newtonian mechanics [M-fs] Wikipedia entry for the physics of music:
  252
  [Zom] P. Munz, I. Hudea, J. Imad, R. Smith?, When zombies attack! Mathematical modelling of an outbreak of zombie infection, In: Infectious Disease Modelling Research Progress, 2009 Nova Science Publishers, Inc., pp. 133-150, 2009. [N-mech] General wikipedia introduction to Newtons three laws of motion: [N-intro] David H. Nash, Dierential equations and the Battle of Trafalgar, The College Mathematics Journal, Vol. 16, No. 2 (Mar., 1985), pp. 98-102. [N-cir] Wikipedia entry for Electrical Networks: [NS-intro] General wikipedia introduction to Navier-Stokes equations: Clay Math Institute prize page: [O-ivp] General wikipedia introduction to the Harmonic oscillator [P-intro] General wikipedia introduction to the Peano existence theorem:
  [P-fs] Howard L. Penn, Computer Graphics for the Vibrating String, The College Mathematics Journal, Vol. 17, No. 1 (Jan., 1986), pp. 79-89. [PL-intro] General wikipedia introduction to the Picard existence theorem:
  [Post] Emil L. Post, Generalized dierentiation, Trans. Amer. Math. Soc. Vol. 32 (1930), pp. 723781. [P1-ps] Wikipedia entry for Power series: [P2-ps] Wikipedia entry for the power [R-ps] Wikipedia entry for the recurrence relations: [R-cir] General wikipedia introduction to LRC circuits: series method:
  [Rikitake] T. Rikitake, Oscillations of a system of disk dynamos, Proceedings of the Royal Cambridge Philosophical Society, 54 (1958), pp. 89105. [S-intro] William Stein and the Sage Developers Group, Sage : Mathematical software, version 4.7.
  253 motion:
  [S-pde] W. Strauss, Partial dierential equations, an introduction, John Wiley, 1992. [U-uc] General wikipedia introduction to undetermined coecients: [V-var] Wikipedia introduction to variation of parameters: [W-intro] General wikipedia introduction to the Wave equation:
  [W-mech] General wikipedia introduction to Wile E. Coyote and the RoadRunner: [W-linear] General wikipedia introduction to the Wronskian [Wr-linear] St. Andrews MacTutor entry for Wronski
  254
  Chapter 5
  Appendices
  255
  256
  CHAPTER 5. APPENDICES
  5.1
  A constant of integration should be added to each formula. The letters a, b, m, and n denote constants; u and v denote functions of an independent variable such as x. Standard Integrals un+1 , n+1
  I1. I2. I3. I4. I5. I6. I7. I8. I9. I10. I11. I12. I13.
  un du =
  n = 1
  du = ln |u| u eu du = eu au du = au , ln a a>0
  cos u du = sin u sin u du = cos u sec2 u du = tan u csc2 u du = cot u sec u tan u du = sec u csc u cot u du = csc u tan u du = ln | cos u| cot u du = ln | sin u| sec u du = ln | sec u + tan u|
  257
  u dv = uv
  (au + b)n du =
  1 du = ln |au + b| au + b a u du u b = 2 ln |au + b| au + b a a b 1 u du = 2 + 2 ln |au + b| 2 (au + b) a (au + b) a du 1 u = ln u(au + b) b au + b 2(3au 2b) (au + b)3/2 u au + b du = 15a2 u du 2(au 2b) au + b = 3a2 au + b u2 au + b du = 2 8b2 12abu + 15a2 u2 (au + b)3/2 105a3
  I27.
  u2
  du ua 1 ln = 2 a 2a u+a
  258 u du = 1/2 ln u2 a2 a2
  u2 du ua a = u + ln 2 2 u a 2 u+a
  u 2 a2 1 2 u a2 33/2
  u2 a2 du =
  u2 du u = 2 2 2 u a
  du = ln u + 2 u a2 u 2 a2
  u2 a2 u 2 a2
  a2 ln u + 2
  u du 1 = ln 2 2 a u u +a a + u2 + a2 du u 1 = arcsec 2 2 a a u u a u2 a2 du = a2 u u 2 u 2 a2
  u 2
  u 2 a2
  u 2 a2 u2 u2 a2 du =
  I41. I42.
  a4 u 2 a2 u 3/2 u 2 a2 ln u + u2 a2 u a2 4 8 8 u 2 + a2 u2 + a2 a + du = u2 + a2 a ln u u u u2 a2 du = u2 a2 a arcsec u a
  259
  I43.
  u2 a2
  Integrals involving u 2
  a2 u2 u a2 arcsin 2 a
  a2 u2 du =
  a2 u2 +
  u du = a2 u 2 a2 u2 u a2 u2 du = a2 u2 1 2 a u2 3 a+3/2
  a2 u 2 du = u
  a ln
  a2 u2 u
  du a+ 1 = ln 2 2 a u a u
  a2 u 2 u u2 a2 u2 du =
  3/2
  du a2 u2 = a2 u u 2 a2 u 2
  I56. I57. I58. I59. I60. I61. I62. I63. I64. I65. I66. I67. I68. I69.
  sin(au)
  u 1 sin(4au) 8 32a
  1 tan(au) u a
  1 cot2 (au) du = cot(au) u a sec3 (au) du = 1 1 sec(au) tan(au) + ln | sec(au) + tan(au) | 2a 2a 1 1 csc(au) cot(au) + ln | csc(au) cot(au) | 2a 2a 1 (sin(au) au cos(au)) a2 1 (cos(au) + au sin(au)) a2
  csc3 (au) du =
  1 2au sin(au) ( a2 u2 2 ) cos(au) a3 1 2au cos(au) + ( a2 u2 2 ) sin(au) a3 sin(a b)u sin(a + b)u , 2(a b) 2(a + b) sin(a b)u sin(a + b)u + , 2(a b) 2(a + b) cos(a b)u cos(a + b)u , 2(a b) 2(a + b) a 2 = b2 a 2 = b2 a2 = b2
  sin(au) cos(bu) du =
  sinn2 u du
  I70. I71.
  u 1 sinh(2au) 4a 2
  261
  u 1 + sinh(2au) 2 4a
  sinh(au) cosh(au) du =
  1 n au n u e a a
  a2 u2 a2 u 2
  u a ln a2 + u2 a 2
  262
  I86. I87.
  arcsinh
  du = u arcsinh
  u 2 + a2
  arccosh
  u a
  du = u arccosh = u arccosh u a
  u a u a
  u2 a2 + u2 a2
  arccosh arccosh
  > 0; < 0.
  I88.
  arctanh
  du = u arctanh
  a u + ln a2 u2 a 2
  I89. I90.
  Wallis Formulas
  I91./2 0
  sinm x dx =
  /2 cosm x dx 0
  (m1)(m3)... (2 or 1) m(m2)...(3 or 2) k,
  I92.0
  5.2. APPENDIX: FOURIER SERIES where k = /2 if both m and n are even and k = 1 otherwise. Gamma Function
  263
  (x) =0
  tx1 et dt,
  x>0
  if n is a non-negative integer
  5.2
  Fourier Series a0 + 2 n=1
  f (x) = where
  an cos
  nx nx + bn sin L L
  a0 = an = bn = 1 L 1 L
  1 LL
  f (x) dx,L
  f (x) cosL L
  nx dx, L nx dx. L
  f (x) sinL
  f (x) = where cn = 1 2L
  n= L
  cn enix/L
  f (x)enix/L dx.
  264
  5.3
  Bessel Functions = arbitrary real number; n = integer 1. Denition. x2 y + xy + (2 x2 2 )y = 0 has solutions y = y (x) = c1 J (x) + c2 J (x), where J (t) = 2. General Properties. Jn (t) = (1)n Jn (t); J0 (0) = 1; Jn (0) = 0, (n 1). m=0
  ( = n),
  2 +2m
  (1)m t +2m . m! ( + m + 1)
  3. Identities. d [t J (t)] = t J 1 (t). (a) dt (b) (c) d J (t) = 1/2 (J 1 (t) J +1 (t)) dt = J 1 (t) t J (t) = t J (t) J +1 (t) . (d) Recurrence Relation. J +1 (t) = 2 J (t) J 1 (t) t d [t J (t)] = t J +1 (t); dt d J0 (t) = J1 (t). dt
  4. Orthogonality. Solutions y (0 x), y (1 x), . . . , y (n x), . . . of the dierential system x2 y + xy + (2 x2 2 )y = 0, x1 x x2 ak y (xk ) bk d y (x) dx = 0,x =x k
  k = 1, 2
  265
  xy (n x)y (m x) dx = 0,x1
  (m = n)
  andx2 x1 2 xy (m x) dx =
  1 22 m
  2 2 2 2 y (m x) + x2 mx
  d y (m x) dx
  2 2 x
  ,x1
  (m = n) .
  5. Integrals. (a) (b) (c) (d) (e) (f) t J 1 (t) dt = t J (t) + C t J +1 (t) dt = t J (t) + C t J0 (t) dt = tJ1 (t) + C t3 J0 (t) dt = (t3 4t)J1 (t) + 2t2 J0 (t) + C t2 J1 (t) dt = 2tJ1 (t) t2 J0 (t) + C t4 J1 (t) dt = (4t3 16t)J1 (t) (t4 8t2 )J0 (t) + C
  266
  5.4
  5.4. APPENDIX: LAPLACE TRANSFORM TABLES f (t) L1 {F (s)} (t) L1. L2. L3. L4. L5. L6. L7. L8. L9. L10. L11. L12. L13. L14. tn f (t) 1, H (t), U (t) U (t a) (n = 1, 2, 3, . . . ) ta (a > 1) eat sin(t) cos(t) f (t) f (t) tn f (t) (n = 1, 2, 3, . . . ) eat f (t) eat L1 {F (s + a)} f (t + P ) = f (t)P 0
  267
  + 2 s + 2
  sF (s) f (0) s2 F (s) sf (0) f (0) (1)n F (n) (s) F (s a) F (s) est f (t) dt 1 esP
  268
  f (t)U (t a) f (t a)U (t a)t
  L18.0
  L19. L20. L21. L22. L23. L24. L25. L26. L27. L28.
  F (z ) dz
  n! (s a)n+1
  1 s (s a)
  2 3 + 2 )2 2s + 2 )2
  2s2 (s2 + 2 )2
  269
  a sin(at) b sin(bt) a2 b2 ebt sin(t) ebt cos(t) 1 cos(t) t sin(t) (t a) (t a)f (t)
  5.5
  Partial fraction decomposition (PFD) of N (x)/D (x): Let N (x) be a polynomial of degree n (the numerator), let D (x) a polynomial D (x) of degree d (the denominator), assume n < d. (If this is not the case, you must rst use long division to reduce the degree of the numerator.) PFD Method: 1. Factor D (x) into irreducible factors having real coecients. Now D (x) is a product of distinct terms of the form (ax + b)r or (ax2 + bx + c)s , for some integers r > 0, s > 0. For each term (ax + b)r the PFD of N (x)/D (x) contains a sum of terms of the A1 Ar 2 r form (ax +b) + + (ax+b)r for some contstants Ai , and for each term (ax + bx + c)
  270
  CHAPTER 5. APPENDICES the PFD of N (x)/D (x) contains a sum of terms of the form B1 x + C1 Bs x + Cs + + . (ax2 + bx + c) (ax2 + bx + c)s for some constants Bi , Cj . functions.N (x) D (x)
  (x) 2. Now you have an expression for N D (x) which is a sum of simpler rational functions. The next step is to solve for these As, B s, C s occurring in the numerators. Cross multiply both sides by D (x) and expand out the resulting polynomial identity for N (x) in terms of the As, B s, C s. Equating coecients of powers of x on both sides gives rise to a linear system of equations for the As, B s, C s which you can.