7in 0in Felder c0_online.te V3 - Januar, 05 0:5 A.M. Page CHAPTER 0 Methods of Solving Ordinar Differential Equations (Online) 0.3 Phase Portraits Just as a slope field (Section.4) gives us a wa to visualize the solutions to a first-order ODE, a phase portrait is a wa of visualizing the solutions to two (or more) coupled first-order ODEs, or to a single second-order ODE. 0.3. Eplanation: Phase Portraits In Section.7 (see felderbooks.com) we introduced coupled differential equations. Such equations occur when two variables depend on each other. For instance, d dt ma depend on both and, andd dt ma also depend on and. A solution would be a pair of equations (t) and (t) that solve both equations simultaneousl. Our first eample was the math problem of Romeo and Juliet. Romeo s love grows the more Juliet loves him (dr dt = J ), while Juliet s love diminishes the more Romeo loves her (dj dt = R). The state of the sstem at an moment is the value of the two functions R and J. If ou know those two numbers at an moment in time ou can figure out how the will evolve for all future times. In Chapter we solved these equations and found that R and J oscillate sinusoidall, 90 out of phase, as indicated in Figure 0.. In this section we will arrive at the same conclusion b a graphical method. Like slope fields, the method of phase portraits allows us to visualize the possible behaviors of a sstem, in this case of two coupled ODEs. Although we re introducing them in the contet of this simple problem that we can solve eactl, phase portraits can be used to understand the behavior of sstems whose equations can t be solved analticall. The phase portrait for the Romeo and Juliet sstem is a plot with R on one ais and J on another. For instance, Figure 0. shows that Juliet loves an indifferent Romeo. Ever point in this space represents one possible state of the Romeo-and-Juliet sstem. Figure 0. is not like most graphs ou have worked with. You are accustomed to the variable on the -ais depending on.0 0.5 0.5.0 R J 4 6 8 0 t FIGURE 0. Romeo and Juliet s feelings for each other oscillate out of phase. R FIGURE 0. J Juliet loves Romeo. adapted with permission from Nonlinear Dnamics and Chaos b Steven Strogatz.
7in 0in Felder c0_online.te V3 - Januar, 05 0:5 A.M. Page Chapter 0 Methods of Solving Ordinar Differential Equations (Online) the variable on the -ais. In this case both the - and-aes represent dependent variables. The independent variable, time, does not appear in the diagram at all. So the state R = 0, J = shown on the plot could be an initial condition (t = 0) or it could occur at an other time. We can t sa. J 4 4 4 R 4 FIGURE 0.3 A phase portrait showing possible solutions to the Romeo and Juliet equations. What we can sa, based on the differential equations, is how this state will evolve. dr dt = J tells us that R will increase; dj dr = R tells us that J will hold stead. So if the sstem is ever momentaril in the state shown in Figure 0., its net shift will be to the right. From there the positive R will start causing J to decrease. If ou follow this logical progression ou will end up describing a circle back around to the point (0, ) where we started, and so on forever. This circle describes the same progression, and for the same reasons, that Figure 0. described. But that circle is onl the particular solution that starts at the point (0, ). If the sstem starts farther from the origin it will trace out a similar circle with a larger radius. We can therefore represent the sstem b Figure 0.3, a phase portrait. Just like Figure 0., this new representation shows Romeo and Juliet s loves oscillating 90 out of phase with each other in a perpetual ccle of love and hate. Each of these representations has an advantage relative to the other. The advantage of Figure 0. is that it includes the time, which Figure 0.3 does not. Looking at the phase portrait we can see that R and J will oscillate, but there s no wa to know how long each oscillation takes, or even whether it s going faster at some points of the ccle than at others. The advantage of the phase portrait is that it shows not just a single trajector, but a whole famil of possible trajectories corresponding to different initial conditions. When we look at a phase portrait that includes enough trajectories we can see in one plot all the possible behaviors of the sstem. Definition: Phase Portrait A phase portrait is a plot showing the possible solutions to a set of coupled first-order differential equations. Each dependent variable is plotted on one ais. The curves on a phase portrait, usuall called trajectories, show possible behaviors of the sstem. (We ll discuss below how to make phase portraits for second-order differential equations b writing them as sets of coupled first-order equations.) Each point on a phase portrait represents a possible state of the sstem. Since the sstem could presumabl start in an possible state, each point can also be said to represent a possible initial condition. For eample, we can see from Figure 0.3 that if Romeo and Juliet start with an combination of feelings for each other that has a combined magnitude R + J = 9 (in whatever units we might use to measure feelings) then the will oscillate with that same magnitude forever. If two trajectories crossed each other, one set of initial conditions could lead to two possible outcomes. That is technicall possible for non-linear equations, but it takes work to contrive such an eample. We will assume that phase portrait trajectories never cross.
7in 0in Felder c0_online.te V3 - Januar, 05 0:5 A.M. Page 3 0.3 Phase Portraits 3 EXAMPLE A Phase Portrait Problem: The figure below is a computer-drawn phase portrait for the equations =, = 3. Based on this drawing, describe the possible long-term behaviors of the sstem. 3 3 3 3 Solution: The overall trend is a counterclockwise rotation. Whenever is positive, is increasing; in the upper left of the plot is decreasing and in the lower right is increasing. Take a glance at the differential equations and convince ourself that this is just what we would epect. But we also see something else: no matter what initial conditions the sstem starts in, the amplitude of the oscillation increases over time. The sstem spirals out toward infinit. Critical Points and Separatrices We are going to analze the behavior of the following sstem b drawing a phase portrait. d dt =, d dt = (0.3.) The first thing we alwas look for are the points where (t) = (t) =0. If the sstem ever reaches such a critical point it will sta there forever. Phsicall these represent equilibrium states of the sstem. Definition: Critical Point A set of first-order differential equations for the functions (t), (t), has a critical point at a set of values (,, ) if all of the derivatives,, equal zero there. A critical point is stable (or attractive ) if all the trajectories near that point converge toward it. A critical point is unstable (or repulsive ) if all the trajectories near that point move awa from it. It is also possible for a critical point to be neither attractive nor repulsive either because some nearb trajectories approach it and others move awa, or because nearb trajectories orbit around the critical point as in Figure 0.3. Critical points are the onl places where a trajector can begin or end. A trajector can come in from infinit and end at a critical point, start from a critical point and go off to infinit, start and end at infinit, start and end at critical points, or make a closed loop.
7in 0in Felder c0_online.te V3 - Januar, 05 0:5 A.M. Page 4 4 Chapter 0 Methods of Solving Ordinar Differential Equations (Online) It isn t hard to determine that Equations 0.3. have a critical point at (, 0) and nowhere else. So we will start building our phase portrait out from there. Along the -ais we have = 0and =. So to the right of our critical point the trajectories point straight up, and to the left the point straight down. Along the line = wehave = and = 0. So above our critical point the trajectories point directl to the right, and below the point left. Let s draw what we have so far. 3 Based on just those arrows and a little bit of thought we can sa a surprising amount about this sstem. For instance, suppose the initial conditions are =, = 0 (directl on one of our arrows, just to make the first step eas). The initial movement will be straight up on the graph: that is, will increase ( = ) while holds stead ( = 0). So a short time later finds us at the point (, Δ). Now is still, but is now a small positive number. Our graph starts to veer slightl to the right. As we move higher the increasing causes to increase, and the increasing causes to increase. So our graph will head toward (, ). In Problem 0.37 ou will sketch in a few other curves using a similar process. You ma be able to predict much of the behavior just b looking at the drawings we ve alread done. Trajectories in the upperright-hand corner will generall head up and right, as our eample above did. Trajectories in the lower-left-hand corner will generall head down and left. 3 3 The phase portrait for =, = Where is the dividing line between these two destinations? In Problem 0.38 ou will show that the ultimate destin of an trajector in this sstem depends on what side of = the initial conditions fall on. An path above this line will eventuall head toward (, ),while an path below it will head toward (, ). We sa that = is a separatri for this phase portrait, because it separates two regions that ehibit qualitativel different behavior. If ou look closel at the phase portrait ou can tell that = is actuall made up of two separatrices, each one a trajector that comes in from infinitel far awa and asmptoticall approaches the critical point. Just like a critical point, a separatri can be attractive, repulsive, or neither. The drawing shows that the separatrices along = are repulsors : trajectories near them move awa from them over time. The drawing also shows another pair of separatrices along =. The are attractors : trajectories move toward them over time. All of the separatrices begin or end on the critical point (, 0), which is itself neither attractive nor repulsive. Drawing phase portraits b hand is not something we do a lot of. It s a worthwhile eercise to go through a few times because it gives ou a greater appreciation for the reall valuable skill, which is interpreting the behavior of a sstem from a given phase portrait.
7in 0in Felder c0_online.te V3 - Januar, 05 0:5 A.M. Page 5 0.3 Phase Portraits 5 EXAMPLE More Critical Points Problem: The drawing shows the phase portrait for d dt =, d dt = 3 +. Use this drawing to discuss the possible behaviors of the sstem. Solution: It s alwas helpful to start with the critical points. Setting the first equation equal to zero gives = 0. The second one is zero when 3 =, with solutions = 0, ±. We can see the three critical points on the drawing. Around those points the phase portrait shows two qualitativel different behaviors. The sstem can either spiral in toward the critical point at = ortheoneat =. Dividing those two possibilities are a pair of separatrices, shown in bold. One approaches the origin from the upper left and the other from the lower right. (There s another pair of separatrices connecting the origin to the other critical points, but we re going to focus on the ones shown in bold.) It s far from obvious that the initial conditions (, ) lead toward the point (, 0) and the initial conditions (3, ) lead toward the point (, 0), but it s eas to see those things looking at the phase portrait. On a technical note, it can often take a lot of trial and error with the computer to plot separatrices since ou have to find just the right initial conditions. Later in this section we ll show ou a trick that can help with that. A lot of the point of the previous two eamples is that critical points and separatrices reveal a tremendous amount about the possible behaviors of the sstem. Second-Order Equations We said above that phase portraits are for sstems of coupled first-order differential equations. The can also be used for second-order equations, however, which include most of the equations used in phsics. The trick is to write one second-order equation as two first-order equations. Suppose we have an equation of the form (t) =F ((t), t). We can alwas define a new function v(t) = (t), and now we have the two coupled equations (t) =v(t) and v (t) =F ((t), t).
7in 0in Felder c0_online.te V3 - Januar, 05 0:5 A.M. Page 6 6 Chapter 0 Methods of Solving Ordinar Differential Equations (Online) EXAMPLE Damped Simple Harmonic Oscillator Problem: A mass on a damped spring obes the differential equation ẍ = 4, where and t are measured in SI units. Find all the critical points for this sstem, draw a phase portrait for it, and describe the possible behaviors. Solution: We begin b writing this as two first-order equations. = v, v = 4 v The onl critical point occurs at the origin: = v = 0. We had a computer graph trajectories for a set of initial conditions laid out along the unit circle in the first quadrant. We can see that all of the trajectories spiral in toward a stable equilibrium at the origin. This point represents = v = 0:themassis at rest at = 0, as we would epect from a damped oscillator. v 4 4 4 4 Remember that ever point on a phase portrait represents one possible state of the sstem. For a pair of first-order equations for (t) and (t), a state is a value for each of those functions. For a second-order equation ẍ =, a state is a value for and a value for its first derivative for instance, the position and velocit of the mass at an given moment so those are the aes of the phase portrait. Some Tips for Generating Phase Portraits on a Computer In general, we can get a computer to make a phase portrait b giving it a list of initial conditions and having it numericall solve the differential equations for each one. Each solution will be a pair of functions (t), (t), and we can ask the computer to plot the parametricall defined curve (, ) for each one. Finall we ask it to show all those curves together on one plot. Coming up with a good set of initial conditions can be trick, and in the end there s no substitute for trial and error, but some guidelines can help. We begin b finding all of the critical points and make sure that the initial conditions cover the region around those critical points. Sometimes a vertical or horizontal line of points can be useful, sometimes a circle of points around a critical point can help, and sometimes using a mi of the two can help. Even with all that, the direction of the trajectories can be a problem. If all trajectories spiral in toward the origin and we start at a circle of points around the origin, our phase
7in 0in Felder c0_online.te V3 - Januar, 05 0:5 A.M. Page 7 0.3 Phase Portraits 7 portrait won t show anthing outside that initial circle. One wa to deal with that is to solve the equations backwards. If our sstem is = +, = then we can generate a valid trajector b starting at an initial point and solving =, =. That will trace the trajector backwards from that point. If we plot the trajector forwards and backwards from each point we will cover the phase portrait more effectivel. That backwards trick is especiall useful for plotting separatrices that asmptoticall approach critical points. If ou look at the figure in Eample: More Critical Points above, the separatrices are the onl two trajectories that approach the critical point at the origin. Just choosing random initial conditions it could take forever to find one that just happens to be on one of the separatrices. So instead we chose two points near the critical point and evolved the sstem backwards from those points to plot the separatrices. Stepping Back: Phase Space An phsical sstem has some number of dependent variables. Most commonl these are functions of time. The state of the sstem at an given time consists of the value of each of those variables and, if the obe second-order differential equations, of their derivatives. Given a set of differential equations for those variables, we can predict the future behavior of the sstem from the initial conditions. A phase portrait is a tool for visualizing those behaviors. Each ais is a dependent variable, and taken together those aes define all the possible states of the sstem. That space of all possible states is called phase space. In principle, an object moving in 3 dimensions has a 6 dimensional phase space, because to specif its state we need to give all three components of its position and of its velocit. For simpler situations such as an object that can onl move in dimension, however, the phase space is dimensional and a phase portrait can provide a useful visualization. 0.3. Problems: Phase Portraits 0.34 Walk-Through: Phase Portraits. Consider the sstem described b the equations d dt = +, d dt =. You re going to draw a phase portrait for this sstem. B the time the problem is done ou re going to have a lot on that drawing, so ou might want to start b drawing a big set of aes going from 6 to 6 in both directions. When we ask where a trajector begins or ends, remember that one possible answer is at infinit. (a) Find all critical points of this sstem. (That is, find all points where both (t) and (t) are zero.) (b) Along the -ais, (t) =0. i. What does that impl about all trajectories that start along that ais? Draw arrows to represent the time evolution of the sstem along that ais. You will need to think about which wa those arrows point. ii. The positive -ais is a separatri for this sstem. Is it an attractor or a repulsor? Eplain how our answer comes from the equations. Where does this trajector begin and end? (One of the answers will be at infinit. ) iii. The negative -ais is another separatri. Is it an attractor or repulsor? Where does it begin and end? (c) Analze the behavior in the first quadrant. i. At the point (, ) the derivative in the -direction is and in the -direction it s. If the sstem starts at that point, in what direction will it initiall move in the -plane? Draw an arrow at (, ) pointing in that direction. ii. What can ou sa about the direction of the trajectories everwhere in the first quadrant? Based on that, describe the behavior of an trajector that starts in the first quadrant. iii. Sketch a trajector beginning at the point (, ). You don t need to be eact, but tr to have the trajector at each point going in roughl the correct direction, and be sure to show the right behavior in the limit t.
7in 0in Felder c0_online.te V3 - Januar, 05 0:5 A.M. Page 8 8 Chapter 0 Methods of Solving Ordinar Differential Equations (Online) (d) The second quadrant is more complicated. i. If a trajector starts slightl to the left of the positive -ais, what direction will it move in initiall? Where will that trajector go at late times? Sketch one such trajector, starting at (., 4). ii. If a trajector starts at the point (, ), what direction will it move in initiall? As soon as it moves awa from that point, what will happen to its direction? Where will that trajector go at late times? Sketch the trajector starting at (, ). iii. Eplain how our last two answers impl that there must be a separatri in the second quadrant, passing somewhere in between the points (, ) and (., 4). Is this separatri attractive or repulsive? Where does it begin and end? iv. You don t have enough information et to know the eact shape of this separatri; ou ll work that out in Problem 0.35. For now just sketch in a curve that matches the answers ou ve given about it. Include arrows showing which direction the separatri trajector goes. v. You drew two trajectories that started in the second quadrant. Now etend them backwards, showing where the would have come from in order to reach the points (, ) and (., 4).As with all of these sketches our goal is to show the correct qualitative behavior, not to plot eact curves. (e) Describe the behaviors of trajectories in the third and fourth quadrants. For each one ou ll have to figure out if ou can do it with a simple argument like the first quadrant or if it needs more careful work like the second one. When ou re done ou should have drawn in one more separatri and a couple of other sample trajectories showing the possible behaviors of the sstem. (f) If there are an regions of our phase portrait where it is not et clear how the trajectories behave, sketch in enough trajectories to make it clear. (g) Is the one critical point of the sstem attractive, repulsive, or neither? (h) The separatrices divide the graph into four regions. For each region, indicate how trajectories starting in that region will evolve over time. What will the sstem approach as t in each case? 0.35 [This problem depends on Problem 0.34.] In Problem 0.34 ou found roughl where the separatrices were b looking at the behavior of the sstem. In some cases that s the best ou can do, but in this case ou can find the separatrices analticall b guessing (correctl as it turns out) that the are lines. The ke is that, for this particular sstem, the two separatrices ou sketched are the onl two trajectories that end on the critical point at the origin. (a) If the sstem d dt = +, d dt = startsoutatapoint(, ), what is the initial slope d d of its trajector? Your answer should be a function of and. (b) Suppose the sstem starts at a point on the line = m. In order for the trajector to sta on that line its initial slope would have to equal m. Using our answer to Part (a), write an equation epressing the statement Starting at the point (, m), the trajector s initial slope equals m. (c) Solve that equation for m. (d) What is the equation for the separatrices other than the positive and negative -aes? (The separatrices are two halves of the same line, so the have the same equation.) 0.36 [This problem depends on Problem 0.34.] Have a computer make a phase portrait for the sstem d dt = +, d dt =. Clearl indicate critical points and separatrices. Make sure our phase portrait has enough trajectories to see the behavior in each region, and that it includes arrows showing the directions of the trajectories. 0.37 Consider the sstem described b Equations 0.3. with initial conditions =, =. (a) Calculate d dt and d dt at that point. Based on our answers, in what direction along the phase portrait will this trajector initiall move? (b) After it moves in that direction for a short time, how will d dt and d dt change? (c) B following the curve along in this wa, trace the trajector from that point. (d) The equations =, = + represent the same sstem evolving backward in time. Starting again at (, )
7in 0in Felder c0_online.te V3 - Januar, 05 0:5 A.M. Page 9 0.3 Phase Portraits 9 trace in that direction to complete the trajector. (e) Draw a similar trajector starting at the origin. 0.38 The Eplanation (Section 0.3.) claimed that Equations 0.3. have a repulsive separatri along the line =. In this problem ou will show how the equations lead to the behavior we saw in the computergenerated phase portrait. (a) Show mathematicall that at an point along this line =. What does that impl about the time evolution of the sstem if the initial condition lies on that line? Now consider three points. The point B =( 0, 0 ) lies directl on the line =. The point A =( 0, 0 +Δ) lies directl above that point, and C =( 0, 0 Δ) directl below. (b) How does d dt at point A compare to d dt at point B? (c) How does d dt at point A compare to d dt at point B? (d) We have seen what will happen to the sstemifitstartsatpointb. Based on that and our answers, what will happen to the sstem if it starts at point A? Will it move closer or farther from the line =? (e) Repeat this analsis for point C. 0.39 [This problem depends on Problem 0.38.] Show that Equations 0.3. have an attractive separatri along the line =. In Problems 0.40 0.4 ou will be given a phase portrait. Classif all of the critical points and separatrices as attractive, repulsive, or neither. Describe how the sstem will evolve in time. (Your answer will almost alwas be of the form if it starts in this region, then ) 0.40 0.4 0.4 In Problems 0.43 0.50 sketch a phase portrait for the given set of equations. Your phase portrait should show all the critical points and enough trajectories to indicate the possible behaviors of the sstem. 0.43 d dt = 4, d dt = 0.44 d dt =, d dt = 0.45 d dt =, d dt = + 0.46 d dt =, d dt = 0.47 d dt = +, d dt = 0.48 d dt =, d dt = + 0.49 d dt =, d dt = 0.50 d dt = ρ, d dt = φ. 0.5 Given a pair of equations (t) =f (, ) and (t) =g(, ), write the equations that draw the same trajectories but in the opposite direction. 0.5 Given a pair of equations (t) =f (, ) and (t) =g(, ), write an equation for the slope d d of the trajector at the point (, ).
7in 0in Felder c0_online.te V3 - Januar, 05 0:5 A.M. Page 0 0 Chapter 0 Methods of Solving Ordinar Differential Equations (Online) In Problems 0.53 0.58 ou will be given a second-order differential equation. Rewrite it as two coupled first-order equations. Have a computer draw the phase portrait for those two equations, and use that phase portrait to predict the possible time evolution of the sstem. A good answer would look like If it starts with > 5 moving to the right it will move in a positive direction forever. If it starts at > 5 at rest or moving left slowl enough it will start moving right and continue that wa forever. If it starts at < 5 then 0.53 (t)+9(t) =0 0.54 (t)+5 (t)+6(t) =0 0.55 (t)+5 (t)+6(t) = 0.56 (t) 5 (t)+6(t) = 0.57 (t)+ 3 = 0 0.58 (t)+tan(4π) =0 0.59 If ou have a second-order differential equation for (t) ou can draw a phase portrait for it where the aes are and. Without knowing anthing else about the sstem, what can ou conclude about what the trajectories look like? In other words, what must be true about all phase portraits for second-order equations that is not necessaril true about phase portraits for coupled first-order equations? 0.60 Eplain wh a trajector cannot begin or end at an point other than a critical point. 0.6 Inflationar Cosmolog According to the theor of inflation, the earl universe went through a period during which virtuall all of the energ was in the form of a scalar field. You don t need to know what a scalar field is to solve this problem. All ou need to know is that in the simplest model of inflation the field φ obes the differential equation φ + m φ + πg( φ + m φ )=0. (a) Define v = φ and rewrite this secondorder equation as two coupled firstorder equations for φ and v. (b) What is the one critical point for this sstem? (c) Have a computer draw a phase portrait for the sstem. You can set the constants G and m equal to. There are two separatrices. Where do the begin and end? (In each case one of the answers is at infinit. ) Are the attractive or repulsive? (d) For a tpical trajector, describe the evolution of the sstem. What happens at earl times, middle times, and late times? 0.6 Eploration: The pendulum. A rigid undamped pendulum of length l obes the differential equation d θ dt + g sin θ = 0 L where θ is the pendulum s angle off the vertical and g is the gravitational constant. The usual approach is to sa that sin θ θ for small θ, which reduces the equation to the simple harmonic oscillator equation. This approimation onl works for small θ, but the equation is difficult to solve more generall. (Go ahead and tr. We dare ou.) L θ (a) Rewrite this equation as two coupled firstorder equations. What variables should go on the aes of the phase portrait for this sstem? Draw a set of aes with those labels. You ll fill in the phase portrait as ou go through the problem. (b) What are the critical points for the sstem? What phsical states do these points represent? Hint: Mathematicall there are infinitel man critical points, but there are onl two states the can represent. Draw a few of these points on our plot. (c) If the pendulum starts at the bottom (θ = 0) and ou give it a little push it will swing back and forth. Draw a set of trajectories on our plot representing this motion. Include arrows showing the direction of the trajectories. (d) If instead ou give the pendulum a large push it will swing in circles. Draw a set of trajectories on our plot showing motion
7in 0in Felder c0_online.te V3 - Januar, 05 0:5 A.M. Page 0.3 Phase Portraits in clockwise circles, and another set showing counterclockwise circles. (We said the pendulum is rigid, a rod instead of a string, so the bob alwas stas a distance L from the center, even when θ>π.) Again, include arrows on the trajectories. (e) You ve drawn trajectories showing two tpes of motion, oscillations and circles. Draw separatrices on our plot in between those tpes of motion. Where do those separatrices begin and end? What do the represent phsicall?