Homework Set #2-3, Math 475B

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1 Homework Set #2-3, Math 475B Part I: Matlab In the last semester you learned a number of essential features of MATLAB. 1. In this instance, you will learn to make 3D plots and contour plots of z = f(x, y). (a) At the Matlab prompt >> type help mesh for a list of commands related to mesh plots. Also check out contour. (b) Type clear, clf xa = -2:.2:2; ya = -2:.2:2; [x,y] = meshgrid(xa,ya); z = x.* exp(-x.^2 - y.^2); mesh(x,y,z) title( This is a 3-D plot of z = x * exp(-x^2 - y^2) ) xlabel( x ); ylabel( y ); zlabel( z ); You should get a plot. You can change the colormap. For example, type colormap(summer), colormap(winter), and colormap(hsv). Type help colormap for more details. (c) Type contour(z) to obtain a simple contour map. There are many features to the command. Use the help facilities for more information. Now, using the editor, create a file called showc.m with the following: clear, clf, clc, axis( square ) xm=-2:.2:2; ym=-2:.2:2; [x, y ] = meshgrid(xm,ym); z = x.* exp(-x.^2 - y.^2); zmax=max(max(z)); zmin=min(min(z)); dz = (zmax-zmin)/10; level = zmin + 0.5*dz: dz: zmax; k=input( If contour label placed manually, type 1; if not 0.\n ); 1

2 if k==1 disp The following messages are generated by clabel command disp (use mouse in the figure window): h=contour(x,y,z,level); clabel(h, manual ) title( Contour plot by contour(x,y,z,level) ) xlabel( x ); ylabel( y ) end if k==0 h=contour(x,y,z,level); clabel(h) title( Contour plot by contour(x,y,z,level) ) xlabel( x ); ylabel( y ) end (d) Now, in matlab, type showc.m and you will get prompted: If contour label placed manually, type 1; if not 0 first, try 0. Then run the code again by typing showc.m once more. Now, try entering 1 and then follow the instructions. You should be able to use the mouse to place contour levels. To exit, hit return while putting the cursor in the matlab figure window. To end, type meshc(z) and note what happens. 2. Next you will learn to use the profiler. (a) At the Matlab prompt >> type help profile for a list of commands related to the command line profiler (In Windows and on Macs there are some nice interfaces, but not so for UNIX/LINUX). Performance is an important aspect of computing. Performance is measured in terms of computing speed and storage usage. Matlab is an interpreter language and hence, not flexible enough to allow for optimizing scheme performance. Nevertheless, we can still learn some simple yet important aspects of computing performance using matlab. Here we learn how to find out where a program spends its time. This information is useful in identifying computing bottlenecks, i.e. unnecessary computations, inefficient aspects of an 2

3 algorithm, inadvertent recomputations, etc. Profilers are available in most high-level languages. (b) Specify a file to profile: profile hilb type help hilb to find out what this M-file does. (c) Execute the hilb M-file: H=hilb(400); (d) Generate a report on the functions execution: profile report, and then end profiling: profile done. (e) Hand in a concise explanation of the report. (f) Visualizing Profiler Results. Type the following: >> profile erfcore >> z=erf(0:0.01:100); >> profile report >> t=profile >> pareto(t.count) Part II: Theoretical Part. 1. Verify that the function x(t) = t 2 /4 solves the initial-value problem x = x x(0) = 0. Apply the Taylor-series method of order 1 and explain why the numerical solution differs from the solution t 2 /4. 2. Derive the Modified Euler s Method: x(t + h) = x(t) + hf (t + 12 h, x(t) + 12 ) hf(t, x(t)) by performing Richardson s extrapolation on Euler s method using step sizes h and h/2. Hint: Find the local truncation error for Euler s method. 3

4 3. Prove that when the fourth-order Runge Kutta method is applied to the problem x = λx, the formula for advancing the solution will be x(t + h) = x(t) [1 + hλ + 12 h2 λ h3 λ ] h4 λ 4 Part III: Experimental Part. 1. Using various values of λ such as 5, 5, or 10, solve the following initialvalue problem numerically using a fourth-order Runge Kutta method: x = λx + cos t λ sin t x(0) = 0. Compare the numerical solution to the analytic solution on the interval [0, 5]. Use step size h = What effect does λ have on numerical accuracy and on the convergence rate? Perform a convergence test. Show quantitatively how the accuracy and convergence depends on λ, if at all. Perform a stability analysis. 2. In this experiment we investigate the solution of (1) dy dx = 2xy, y(0) = 1, for 0 x 3. The exact solution of (1) is y(x) = exp(x 2 ). This is a stiff equation. We will learn a few things: that one could get around the stiffness (to a large degree) of this equation by scaling the time steps. We will also learn how adaptivity is a good thing, it is by no means a way to get around stiffness. Finally, we will see how an implicit A stable scheme will handle the stiffness of the equation. (a) We can use scaling to circumvent the stiffness issue: let z = x 2 /2. Show that this substitution in (1) leads to dy dz = y, y(z = 0) = 1. This equation is much less stiff and thus easier to solve numerically than the original equation. Once you solve for y(z), you then remap the solution onto y(x). 4

5 (b) Let s focus on (1). Set up a code to approximate solutions to it using (1) Adams-Bashford of order 2, and (2) Trapezoidal. (c) Set up a convergence result for both methods. The idea is to look at the global error at x = 3, between the exact and approximate results given by both methods, as h is varied. Show the plot and describe what you see. (d) Now we ll see how adaptivity plays out. Solve the equation using the adaptive Runge-Kutta available as matlab functions. Type help ode45 for information. Use a tolerance of , with h min = and h max = Since the results use an adaptive step size h, a comparison to the Adams Bashford and the Trapezoidal is hard to do, for a fixed h. So the thing to do is to fix the tolerance and compare how small h has to be to reach similar tolerances using ode45, your Adams Bashford, and your Trapezoidal. 5