MATH2071: LAB 1(b): Using Matlab ODE solvers. This version of the first lab is intended only for students who have already taken Math 2070.

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1 MATH2071: LAB 1(b): Using Matlab ODE solvers 1 Introduction Introduction Exercise 1 Matlab ODE solvers Exercise 2 ODE systems Exercise 3 Goodwin s economic model Exercise 4 Higher Order ODE s Exercise 5 Phase Plane Plots Exercise 6 Stiff systems Exercise 7 Roundoff errors Exercise 8 This version of the first lab is intended only for students who have already taken Math There are two versions of the first lab. This version introduces the Matlab ODE solvers and is intended for students who took Math If you have not already taken Math 2070, please see Lab 1(a). That version of the first lab introduces the Matlab environment and programming language, and presents the general format of the work you need to hand in. This lab is concerned with solution of ordinary differential equations (ODEs) using a Matlab function for the solution. You will see it applied first to a simple scalar equation, then to a system of equations, and then to a higher order equation converted into a system. In later labs, you will be writing your own ODE solver routines so you can understand the underlying theory. A very simple ordinary differential equation (ODE) is the explicit scalar first-order initial value problem: y = f(x, y) y(x 0 ) = y 0. Here y is shorthand for the derivative dy/dx. The equation is explicit because y can be written explicitly as a function of x and y. It is scalar because we assume that y(x) is a scalar quantity. It is first-order because the highest derivative that appears is the first derivative y. It is an initial value problem (IVP) because we are given the value of the solution at some time or location x 0 and are asked to produce a formula for the solution at later times. An analytic solution of an ODE is a formula y(x), that we can evaluate, differentiate, or analyze in any way we want. However, analytic solutions can only be determined for a limited class of ODE s. A numerical solution of an ODE is simply a table of abscissæ and approximate values (x k, y k ) that approximate the value of an analytic solution. The values x k are often called steps, especially for initial value problems. In general, a numerical solution is always wrong; the important question is, how wrong is it? One way to pose this question is to determine how close the computed values (x k, y k ) are to the analytic solution, which we might write as (x k, y(x k )). This lab will take three sessions. If you print this lab, you may prefer to use the pdf version. 2 Matlab ODE solvers Matlab has a number of built-in ODE solvers. These include: 1

2 ode23 ode113 ode15s ode23s ode23t ode23tb ode45 Matlab ODE solvers non-stiff, low order non-stiff, variable order stiff, variable order, includes DAE stiff, low order trapezoid rule stiff, low order non-stiff, medium order (Runge-Kutta) Do not worry about the meaning of stiff in this context, you will see it later this term. All of these functions use the very best methods, are highly reliable, use adaptive step size control, and allow close control of errors and other parameters. As a bonus, they can be told to precisely locate interesting events such as zero crossings and will even allow user-written functions to be called when certain types of events occur. There is very little reason to write your own ODE solvers unless you are actively researching new methods. In the following exercises you will see how to use these solvers. In the following exercise, you will solve the following initial value problem (IVP). Exercise 1: (a) Copy the following code to a function m-file named ex1 ode.m. function yprime=ex1_ode(x,y) % yprime=ex1_ode(x,y) % computes the right side of the ODE y =sin(x)-y % x,y are scalars, yprime is y % your name and the date yprime=sin(x)-y; dy =sin x y (1) dx y(0) =1 (b) You can watch a solution evolve if you call the solver without any output variables. Use ode45 to solve the IVP in Equation (1) on the interval [0,15] with the following Matlab command. ode45( ex1_ode,[0,15],1) You should see that the solution becomes very much like sinx after a short period of adjustment to the initial condition that is different from sin 0. The circles on the plot indicate points at which Matlab has chosen to evaluate solution, based on its default error criteria. Please include this plot with your summary. (c) If you wish to examine the solution in more detail, or manipulate it, you need the solution values, not a picture. You can get the solution values with commands similar to the following: [x,y]=ode45( ex1_ode,[0,15],1); If you wish, you can plot the solution, compute its error, etc. Plot this solution (plot y vs x) to see that it looks the same as before. Please send me this plot with your summary also. (d) For this solution, what is the value of y at x=15? (Please use at least 6 decimal places for this answer. You can use the command format long to get more than the default amount of printed accuracy.) How many steps did it take? (The length of x is one more than the number of steps because it includes the initial value.) 2

3 (e) Suppose you want a very accurate solution. The default tolerance is.001 relative accuracy in Matlab, but suppose you want a relative accuracy of 1.e-8? There is an extra variable to provide options to the solver. It works in the following manner: myoptions=odeset( RelTol,1.e-8) [x,y]=ode45( ex1_ode,[0,15],1,myoptions); (The semicolon is left off the odeset call so you can see some of the other options and their defaults. Use help odeset for more detail.) How many steps did it take this time? What is the value of y at x=15 (to 6 decimal places)? 3 ODE systems The vast majority of ODEs that are solved are actually systems of differential equations, that is, a set of equations for which the unknown function y(x) is a vector, say of dimension n. In this case, we might write an individual component of the vector as y l or y l (x). Assuming that the right hand side function f returns a vector, then the system of equations has the same form as the scalar equation, and we can apply the same ODE solvers to this system. Here s an implementation issue: Matlab uses the variables x and y as column vectors. The value y(k) is the approximate solution at x(k). Since this as a column vector, it s possible to handle the case of a system by assuming the individual solution values are row vectors. In this setting, a numerical solution will be a table of values, an initial condition specifies the first row of that table, and each row specifies the behavior of one of the components of the solution. 4 Goodwin s economic model A model for cyclic economic behavior due to R. M. Goodwin ( A Growth Cycle, 1967, in Feinstein, editor, Socialism, Capitalism and Economic Growth) will be described below. This model results in a system of ordinary differential equations that we will be able to solve using Euler s method. The primary role of this model in the study of economics is that it predicts cyclical behavior without a cyclical driving force it just arises naturally from the model itself. It is not necessary for you to understand the discussion of the foundation of the model or its economic interpretation. We will be using it as a source of a system of differential equations to be solved. The model involves two primary dependent variables, y 1 and y 2. The variable y 1 is the share of the total economy represented by employee salaries y 1 = (wages)*(numberemployed), where wages refers to the average amount paid to each employee, numberemployed refers to the number of employed people, and refers to the gross national product, the total value of all production. Of course, the total of all wages paid to all workers is given by (wages)*(numberemployed). The employment rate, y 2 is the second primary dependent variable: y 2 = numberemployed population where population is the total number of people. The independent variable is time, t. The first assumption is that there is a constant, ρ, representing the average return on capital for all investments. Thus, the gross national product is a fixed percentage of total capital invested. totalcapital = ρ. (2) 3

4 The second assumption is that the money left over after wages have been paid is assumed to be invested, increasing the total capital available. Thus the total capital increases in the following way. dtotalcapital dt = (wages) (numberemployed) = (1 y 1 ) (3) The following expression, in terms of relative rates of change, is a consequence of (2) and (3). dtotalcapital/dt totalcapital = d/dt = (1 y 1) totalcapital = 1 y 1 ρ The third assumption is that the relative rates of change of productivity and population are both constants. Productivity is the average value of goods produced by employees productivity = and population was used above in the definition of y 2. Thus numberemployed (4) dproductivity/dt productivity dpopulation/dt population = δ (5) = ν (6) A simple consequence of (5) is that d/dt = dnumberemployed/dt numberemployed + δ. (7) The fourth assumption is that the relative change in wages is a function of employment. dwages/dt wages = α + βy 2. (8) A consequence of the derivative of definition of employment rate, e, (7), (4) and (6) is one of the two differential equations for Goodwin s model dy 2 /dt = 1 y 1 δ ν. (9) y 2 ρ A consequence of the derivative of the definition of wage share, y 1, is the other equation for Goodwin s model dy 1 /dt y 1 = dwages/dt wages In summary, Goodwin s model can be written where the Greek letters signify positive constants. + dnumberemployed/dt numberemployed d/dt = α + βy 2 δ (10) dy 1 dt = (δ + α)y 1 + βy 1 y 2 (11) dy 2 dt = (1 ρ δ ν)y 2 y 1y 2 ρ (12) 4

5 Exercise 2: (a) Copy the following code to a Matlab function m-file called goodwin ode.m, that returns the derivative function as a column vector: function yprime = goodwin_ode ( x, y ) % comments % your name and the date ALPHA =0.4; BETA =0.9; DELTA =0.05; NU =0.01; RHO =5.0; yprime=zeros(2,1); yprime(1)=-(delta+alpha)*???+beta*y(1)*y(2); yprime(2)=(1/rho-delta-nu)*???-y(1)*y(2)/rho; (b) Add comments just below the signature to indicate the calling and return sequences and to provide a summary of what the function does. (c) Replace the symbols??? with the proper quantities so goodwin ode implements Goodwin s model equations. (d) What is the role of the line yprime=zeros(2,1); What would happen if it were left out? (e) Note that the variable x, corresponding to time, is unused. Nonetheless, it must be present in the signature. (f) Now use ode45 to integrate from x = 0 to x = 100 starting from y(1)=0.5, y(2)=0.5; yinit = [ ]; ode45 ( goodwin_ode, [ 0.0, ], yinit ); (g) Why do you get two curves? Please include a copy of this plot with your summary. (h) Open a blank plot window with the command figure(2). Solve the same system, but this time use ode15s. Your solution should look the same, but you should see a slightly different distribution of circles indicating a different set of step sizes. 5 Higher Order ODE s The previous examples are for first order equations and systems. What about higher order differential equations, ones that include derivatives such as y? There is a standard trick for transforming higher order equations into first order systems. You may have seen this trick in an ODE course. It also shows up in control theory in the conversion to state-space form of a system model and, I am sure, many other places. Atkinson describes the trick in Section 6.1, page 340. The trick will be illustrated by example here. Consider a fourth-order differential equation 5z + 4z + 3z + 2z + z = sin(x) with initial conditions z ( 0) = 0, z (0) = 10, z (0) = 20, z (0) = 30. 5

6 The first step is to define new variables y 1 = z y 2 = z y 3 = z y 4 = z (Note that the original equation is fourth order and there are four new variables defined. With these new variables, the original scalar equation can be written as a system of equations y 1 = y 2 y 2 = y 3 y 3 = y 4 y 4 = (sin x y 1 2y 2 3y 3 4y 4 )/5 Exercise 3: A pendulum swings through a circular arc. At any time, the angle θ(x) the pendulum makes with the downward vertical is described by Newton s law (F = ma): with initial conditions θ + 1.5θ = 0 θ(x 0 ) = 1 θ (x 0 ) = 0 (a) Rewrite this as a first order system. (Hint: set y 1 = θ and y 2 = θ.) Write a function m-file named pendulum ode.m to represent this system. Be sure to put comments after the signature line and wherever else they are needed. Be sure that you return a column vector. (b) Use ode45 to solve for θ up to x = 25. Please include a plot of your solution with your summary file. Exercise 4: Your team has a cannon that shoots water balloons with a velocity of 100 ft/sec. You can adjust its angle of inclination, α, to change the range. The other team is in front of a wall 10 ft high, 300 ft away from you. If you hit the wall with a water balloon, you will soak the other team. Your plan is to use Matlab to solve the equations of motion and trial and error to choose the correct value of α. The trick is to pick a value of α that results in the balloon being between 0 and 10 ft above ground at 300 ft. The equations of motion of a cannonball of unit mass under the influence of gravity are: ξ = 0 η = 32 where ξ denotes distance along the ground, and η denotes height. The independent variable, x, is time. The initial conditions are ξ(0) = 0 ξ (0) = 100 cosα η(0) = 0 η (0) = 100 sinα 6

7 where α is a constant you will discover by trial and error. This is a combination of two second-order systems. Define the variables: y 1 = ξ y 2 = ξ y 3 = η y 4 = η (a) Write a function m-file called cannon ode.m to implement the system. Be sure to include comments and to have the function return a column vector. (b) Use ode45 to integrate the equations of motion and determine, by trial and error, a value of α so that the cannon ball hits the wall. You can determine success by checking that 0 η 10 when ξ = 300. Watch out: the condition is on ξ, not x. You can find the first step beyond ξ = 300 using the Matlabfind statement. The first index, k for which ξ k 300 is k=min(find(y(:,1)>=300)). Alternatively, you can plot the balloon s trajectory and see where it hits the ground or the wall. You can plot the trajectory (ξ, η) with plot ( y(:,1), y(:,3) ) You might find the axis([xmin,xmax,ymin,ymax]) command in Matlab useful to zoom in on the target, or you can use the magnifying icon on the plot window. (Click on the magnifying glass and then use the mouse to outline a square by clicking on the upper left and dragging. Warning: this might cause Matlab to crash on the computers in GSCC.) Include the file cannon ode.m and the value of alpha you used when you turn in this exercise. 6 Phase Plane Plots In many problems, a system is defined by a second order ODE whose right hand side does not depend on time. Such a system is called autonomous. The behavior of the solution doesn t depend on when you start the problem, only on the initial condition. Most of the above examples are autonomous. Especially for these kind of problems, it can be useful to look at the solution on a special phase plane plot. This simply means that instead of plotting x versus the solution y, we plot y versus y. This is an especially good way to detect periodic solutions, decaying transients, and so on. Phase plane plots are most interesting when the differential equation is not linear. Phase plane plots are valuable because they highlight periodic or almost periodic behavior. Think about it: the solution to the pendulum case is θ = cosx, and that makes θ = sin x. If you plot this in the phase plane (θ on the vertical axis, θ on the horizontal axis) you get a circle. Of course, once you go around the circle once, the autonomous nature of the system guarantees that you will continue going around forever, so the periodic behavior is highlighted. If you add a little damping to the pendulum equation, the phase plane plot becomes a slow spiral into the center. Exercise 5: An equation for motion of a pendulum including a frictional term proportional to velocity is θ +.2θ + 1.5θ = 0. (a) Copy your file pendulum ode.m to decay ode.m and modify it to include the frictional term.2θ. (b) Use ode45 to solve the resulting ODE for x between 0 and 20, starting from the same initial condition as in Exercise 3 above, [1;0]. (c) Plot it in phase space (θ along the horizontal axis and θ on the vertical axis) using the command 7

8 plot(y(:,1),y(:,2)); Please send me the plot. Exercise 6: In the previous exercise, all three coefficients on the left in the ODE are positive. If the coefficient of velocity is negative instead of positive, the physical result is that the solution grows instead of decaying. (a) Copy your filedecay ode.m togrow ode.m and modify it by changing the coefficient of the velocity term from.2 to -.2. (b) Use ode45 to solve the resulting ODE for x between 0 and 20, starting from [1;0]. (c) Plot y vs. x to see one view of what the solution looks like. You do not have to send me this plot. (d) Plot it in phase space (θ along the horizontal axis and θ on the vertical axis) using the command plot(y(:,1),y(:,2)); You should observe an outward spiral. Please send me this phase-plane plot. One simple nonlinear equation with quite complicated behavior is van der Pol s equation. This equation is written z + a(z 2 1)z + z = 0 (13) For positive a, it can be seen by analogy with decay ode.m and grow ode.m that solutions z to (13) will grow when z < 1 and shrink when z > 1. The result is a non-linear oscillator. Physical systems that behave somewhat as a van der Pol oscillator include electrical circuits with semiconductor devices such as tunnel diodes in them (see this web page from Duke, and some biological systems, such as the beating of a heart, or the firing of neurons. Exercise 7: This exercise illustrates phase plane plots using the van der Pol ODE. (a) Rewrite the van der Pol equation as a first-order system, and place the code in a function m-file named vanderpol ode.m. Use the value a = 1. Be sure you place comments where they belong. (b) Use ode45 to solve the system of ODEs for x between x=0 to x=20, starting from an initial condition of yinitial = [0; 0.25] (notice the semicolon, making it a column vector). Make a phase plane plot from your solution. (c) Now, on the same plot (use the Matlab command hold on), plot the solution starting from the initial condition [0;3]. Send me the plot and your guesses about: What would the plot look like if you started at other points close to the ones you already did? What would the plot look like if you started very near the origin? What would the plot look like if you started exactly on the origin? 7 Stiff systems ODEs can be either stiff or non-stiff. We will look at stiff ODEs again later in the term, but it is instructive to look at one practical method for distinguishing stiff from non-stiff ODEs. Basically, you try both non-stiff and stiff solvers and note which is faster. Take my word that the IVP can be regarded as a stiff system. y =1000(sinx y) (14) y(0) =1 8

9 Exercise 8: (a) Implement the right side of the above IVP, Equation (14) in a function m-file namedstiff ode.m. You may want to start from ex1 ode.m. (b) Integrate it from x=0 to x=15, starting from y=1 using ode45. Measure the time it takes with the commands tic;[x,y]=ode45( stiff_ode,[0,15],1);toc How long does it take? How many steps did it take (length(x)-1)? (c) Use the following command to plot your result, using a blue line. plot(x,y, b ) (d) Repeat the above steps using ode15s. Again, how much time does it take (should be much smaller)? How many steps does it take (length(x)-1)? Plot the results in red on the same frame as above. You should see the lines almost on top of one another, so the solutions are essentially the same. Please include this plot with your summary. 8 Roundoff errors Last term you saw some effects of roundoff errors. Later in this term you will look at roundoff errors again. Right now, though, is a good time to look at how some roundoff errors come about. In the exercise below we will have occasion to use a special matrix called the Frank matrix. Row k of the n n Frank matrix has the formula: The Frank matrix for n = 5 looks like: A k,j = 0 for j < k 2 n + 1 k for j = k 1 n + 1 j for j k The determinant of the Frank matrix is 1, but is difficult to compute numerically. This matrix has a special form called Hessenberg form wherein all elements below the first subdiagonal are zero. Matlab provides the Frank matrix in its gallery of matrices, gallery( frank,n), but we will use an m-file frank.m. The inverse of the Frank matrix also consists of integer entries and an m-file for it can be downloaded as frank inv.m. You can find more information about the Frank matrix from the Matrix Market, and the references therein. Exercise 9: Let s look carefully at the Frank matrix and its inverse. For convenience, define A to be the Frank matrix of order 6, and Ainv its inverse, computed using frank and frank inv. Similarly, let B and Binv be the Frank matrix of order 24 and its inverse. Do not use the Matlab inv function for this exercise! You know that both A*Ainv and B*Binv should equal the identity matrices of order 6 and 24 respectively. Let s just look at the top left entry. Compute A(1,:)*Ainv(:,1)= B(1,:)*Binv(:,1)= Both of these answers should equal 1. The first does and the second does not. To see what goes right, compute the terms: 9

10 A(1,6)*Ainv(6,1)= A(1,5)*Ainv(5,1)= A(1,4)*Ainv(4,1)= A(1,3)*Ainv(3,1)= A(1,2)*Ainv(2,1)= A(1,1)*Ainv(1,1)= sum = Note that the signs alternate, so that when you add them up, each term tends to cancel part of the preceeding term. Now, to see what goes wrong, compute the terms: B(1,24)*Binv(24,1)= B(1,23)*Binv(23,1)= B(1,22)*Binv(22,1)= B(1,21)*Binv(21,1)= B(1,20)*Binv(20,1)= B(1,16)*Binv(16,1)= B(1,11)*Binv(11,1)= B(1,6) *Binv(6,1) = B(1,1) *Binv(1,1) = You can see what happens to the sum. The first few terms are huge compared with the final sum of 1. Matlab uses 64-bit floating point numbers, so you can only rely on the first thirteen or fourteen significant digits in numbers like B(1,24)*Binv(24,1). Further, they are of opposing signs so that there is extensive cancellation. There simply are not enough bits in the calculation to get anything like the correct answer. (Remark: It would not have been productive to compute each of the products B(1,k)*Binv(k,1) for each k, so I had you do the five largest and then sampled the rest. I chose to sample the terms with an odd-sized interval between adjacent terms. Had I chosen an even interval say every other term the alternating sign pattern would have been obscured. When you are sampling errors or residuals for any reason, never take every other term!) 10