This exam is due by 5pm on Tuesday, March 2. You may use any ...

Jim LambersMAT 461/561Spring Semester 2013-14Exam 1<strong>Th<strong>is</strong></strong> <strong>exam</strong> <strong>is</strong> <strong>due</strong> <strong>by</strong> <strong>5pm</strong> on Thursday, February 27. You may use any resource, including thetextbook, course notes, or other publ<strong>is</strong>hed information, but all sources outside of course materialmust be cited. You may not collaborate with anyone.1. Using the definition of local truncation error,(a) Show that the modified Euler method <strong>is</strong> 2nd-order accurate.(b) Show that the implicit 2-step Adams-Moulton method <strong>is</strong> 3rd-order accurate.2. The ODE system given <strong>by</strong>y ′ 1 = α − y 1 − 4y 1y 21 + y 2 1y ′ 2 = βy 1(1 − y 21 + y 2 1where α and β are parameters, represents a simplified approximation to a chemical reaction.There <strong>is</strong> a parameter value β c = 3α 5 − 25 α such that for β > β c, solution trajectories decay inamplitude and spiral in (y 1 , y 2 )-space into a stable fixed point, whereas for β < β c , trajectoriesoscillate without damping and are attracted to a stable limit cycle. [<strong>Th<strong>is</strong></strong> <strong>is</strong> called a Hopfbifurcation.](a) Set α = 10 and use any of the d<strong>is</strong>cretization methods introduced in Chapter 6 with afixed step size h = 0.01 to approximate the solution starting at y 1 (0) = 0, y 2 (0) = 2, for0 ≤ t ≤ 20. Do th<strong>is</strong> for the parameter values β = 2 and β = 4. For each case plot y 1 vs.t and y 2 vs. y 1 . Describe your observations.(b) Investigate the situation closer to the critical value β c = 3.5. [You may have to increasethe length of the integration interval beyond t = 20 to get a better look.]3. (a) Use the Taylor expansion,),y(t i+1 ) = y(t i ) + hy ′ (t i ) + h22 y′′ (t i ) + h36 y′′′ (t i )+ h424 y(iv) (t i ) + h5120 y(v) (t i ) + O(h 6 ),to derive a corresponding series expansion for the local truncation error of the forwardEuler method.(b) Manipulating the forward Euler method written for the step sizes h and h/2, applyRichardson extrapolation to obtain a 2nd-order one-step method.(c) Manipulating the forward Euler method written for the step sizes h, h/2 and h/3, applyRichardson extrapolation to obtain a 3rd-order one-step method.1

(d) Explain why your obtained methods are special cases of Runge-Kutta methods.4. To draw a circle of radius r on a screen, one may proceed to evaluate pairs of values x = r cos θ,y = r sin θ for a succession of values of θ, but th<strong>is</strong> <strong>is</strong> computationally expensive. A cheapermethod may be obtained <strong>by</strong> considering the system of ODEsx ′ = −y, x(0) = r,y ′ = x, y(0) = 0,where x ′ = dxdθ, and approximating th<strong>is</strong> using a simple d<strong>is</strong>cretization method. However, caremust be taken so as to ensure that the obtained approximate solution looks right, i.e., thatthe approximate curve closes rather than spirals.Carry out th<strong>is</strong> integration using a uniform step size h = .02 for 0 ≤ θ ≤ 120, applying Euler’smethod, the modified Euler method, and the fourth-order Runge-Kutta method. Determineif the solution spirals in, spirals out, or forms an approximate circle as desired.(a) Set the value of r in the problem statement to 1.(b) For all three methods, use the same time-stepping as in eulerforward.m from the courseweb site, in which h <strong>is</strong> set to the minimum of the prescribed value of h, and T − t n ,where T <strong>is</strong> the final time and t n <strong>is</strong> the current time after n steps have been taken. ForT = 120 and h = 0.02, th<strong>is</strong> will actually result in 6001 time steps, rather than the 6000that <strong>is</strong> expected, because of roundoff error in updating the value of t n .(c) Make all three methods compute a 2×6002 matrix, in which each column cons<strong>is</strong>ts of thex- and y-coordinates of a point of the computed solution. The first column representsthe initial point (1, 0), and the remaining columns represent the points computed <strong>by</strong> the6001 time steps described above.(d) Your code must save your solution matrices in three separate ASCII files: U1.dat forEuler’s method, U2.dat for modified Euler, and U3.dat for the fourth-order Runge-Kutta method. For <strong>exam</strong>ple, if U1 <strong>is</strong> the name of the matrix computed <strong>by</strong> forwardEuler, then it can be saved as follows:save -ascii U1.dat U1These files must then be emailed to me at James.Lambers@usm.edu.5. Suppose that we apply a numerical method to an initial value problem, with two differentstep sizes h 1 and h 2 , and obtain error estimates e(h 1 ) and e(h 2 ). Furthermore, suppose thatthe error <strong>is</strong> of the form Ch p , where the constants C and p are unknown. Show thatp = ln( e(h2 )e(h 1 )6. Derive the 2-step Adams-Moulton formula) /ln(h2h 1).y i+1 = y i + h 12 (5f i+1 + 8f i − f i−1 ).2

7. The following classical <strong>exam</strong>ple from astronomy gives a strong motivation to solve initial valueproblems with adaptive step size control.Consider two bodies of masses µ = 0.012277471 and ˆµ = 1 − µ (earth and sun) in a planarmotion, and a third body of negligible mass (moon) moving in the same plane. The motion<strong>is</strong> governed <strong>by</strong> the equationsStarting with the initial conditionsu ′′1 = u 1 + 2u ′ 2 − ˆµ u 1 + µD 1− µ u 1 − ˆµD 2,u ′′2 = u 2 − 2u ′ 1 − ˆµ u 2D 1− µ u 2D 2,D 1 = ((u 1 + µ) 2 + u 2 2) 3/2 ,D 2 = ((u 1 − ˆµ) 2 + u 2 2) 3/2 .u 1 (0) = 0.994, u 2 (0) = 0, u ′ 1(0) = 0,u ′ 2(0) = −2.00158510637908252240537862224,the solution <strong>is</strong> periodic with period < 17.1. Note that D 1 = 0 at (−µ, 0) and D 2 = 0 at (ˆµ, 0),so we need to be careful when the orbit passes near these singularity points.The orbit <strong>is</strong> depicted in Figure 1. Using the fourth-order Runge-Kutta method, integrate th<strong>is</strong>Figure 1: Solution to the system of ODEs in Problem 73

problem on [0, 17.1] with a uniform step size, using 100, 1000, 10,000, and 20,000 steps. Plotthe orbit for each case. How many uniform steps are needed before the orbit appears to bequalitatively correct?8. (Bonus) Repeat problem 4 with the backward Euler methodand the implicit trapezoidal methody n+1 = y n + hf(t n+1 , y n+1 )y n+1 = y n + h 2 [f(t n, y n ) + f(t n+1 , y n+1 )].Note that because the system of ODEs <strong>is</strong> linear, you do not need to use a method such asNewton’s method to compute x n+1 and y n+1 from x n and y n .Explain the observed results. Hints: <strong>Th<strong>is</strong></strong> has to do with a certain invariant function of xand y, rather than with the order of the methods. Also, you can use the fact that the inverseof a 2 × 2 matrix <strong>is</strong> given <strong>by</strong> the formula[ ] −1 a b=c d1ad − bc[ d −b−c a].4