Fall 2003 Math 308/ Numerical Methods 3.6, 3.7, 5.3 Runge-Kutta Methods Mon, 27/Oct c 2003, Art Belmonte

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1 Fall Math 8/ Numerical Methods.6,.7,. Runge-Kutta Methods Mon, 7/Oct c, Art Belmonte Summary Geometrical idea Runge-Kutta methods numerically approximate the solution of y = f (t, y), y(a) = y by using weighted averages of slopes near a point instead of the single slope involved by following the tangent line at a point. These methods are much more accurate than Euler s method. The second-order Runge-Kutta (RK) method Let [a, b] be the interval over which an approximation to the solution is desired. (Thus t = a and t = b are the initial and final values of the independent variable, respectively.) Partition this interval into N subintervals each of length h = (b a)/n, called the step size. Lett =aand define t k+ = t k + h, for k =,,...,N. Notice that t N = b and the other t k so-defined are the interior endpoints of the subintervals. These collectively are the discrete values of the independent variable. The initial value of the dependent variable is given by the initial condition, y(a) = y(t ) = y. The other discrete dependent variable values are computed iteratively as follows. for k = ton s = f(t k,y k ) s = f(t k +h,y k +hs ) t k+ = t k + h y k+ = y k + h s + s This method is a single-step numerical solver since it depends only on data obtained from the preceding step. It is a fixed-step solver since the lengths of the subintervals of [a, b] are all equal. The maximum error in the approximation over the interval [a, b] satisfies the following upper bound. maximum error M ( ) e L(b a) h L Here L = max f (t,y) R y where R is a rectangle containing the solution curve. This constant is the same as in Euler s method. The constant M is different than that in Euler s method, but it also depends only on the values of f and some of its derivatives over R. The fact that h occurs to the second power in the error bound is the reason that we say that this method is a second-order method. The fourth-order Runge-Kutta (RK4) method Let the independent variable values t k be as above. The initial value of the dependent variable is given by the initial condition, y(a) = y(t ) = y. The other discrete dependent variable values are computed iteratively as follows. for k = ton s = f(t k,y k ) ( s = f t k + h,y k+ h ) s ( s = f t k + h,y k+ h ) s s 4 = f (t k +h,y k +hs ) t k+ = t k + h y k+ = y k + h s + s + s + s 4 6 This method is also a single-step numerical solver since it depends only on data obtained from the preceding step. It is a fixed-step solver since the lengths of the subintervals of [a, b]are all equal. The maximum error in the approximation over the interval [a, b] satisfies the following upper bound. maximum error M ( ) e L(b a) h 4 L Here L = max f (t,y) R y where R is a rectangle containing the solution curve. This constant is the same as in Euler s method. The constant M is different than those in Euler s and RK methods, but it also depends only on the values of f and some of its derivatives over R. The fact that h occurs to the fourth power in the error bound is the reason that we say that this method is a fourth-order method. Systems The Runge-Kutta methods carry over directly to systems of first-order equations. We simply write the algorithm in terms of vectors. Let u = f(t, u), u(t ) = u where u = [u, u,...,u n ] T is an n-dimensional column vector. With the t k defined as above, simply replace the scalar entities f, s,andywith the vector entities f, s, andu, respectively in the aforementioned algorithms.

2 MATLAB Examples In MATLAB, Polking s rk routine implements the nd order Runge-Kutta method. His rk4 routine implements the 4th order Runge-Kutta method. These work for both a single first-order equation as well as for a system of such equations. We shall use these routines exclusively (instead of doing hand work). Example A [a problem revisited] Calculate the first five iterations of the second-order Runge-Kutta method for the IVP z = z, z() = Use a step size of h =.. Here f (t, z) = z. Notice that the syntax for Polking s rk routine is identical to that of his eul routine. We simply change the name of the ODE routine we are using. Function M-file f.m function zp = f(t,z) zp = - z; ============================== Diary file s6ea.txt NSS4-.6/Example A RK method tspan = [.,.]; z = ; h =.; [trk,zrk] = rk(@f, tspan, z, h); Table of t and z values tk_zk = [trk, zrk] tk_zk = M-6/-i, iii [revisited] Find the exact solution of the initial value problem z = z cos t, z() =. Plot this solution over the interval [, 6] along with the approximate solution provided by the second-order Runge-Kutta method with a step size of h = 8 =.. Make a function M-file f.m which defines the derivative z = f (t, z), then write a script M-file m6xb.m to invoke rk. In the latter, we ll use dsolve to obtain the exact solution. Here are listings of the function M-file and diary file m6xb.txt as well as the plot. Notice the increased accuracy of this method.. Function M-file f.m function zp = f(t,z) zp = z.^.* cos(.*t); ============================== Diary file m6xb.txt M-6/b: RK method only (i) Exact solution sol = dsolve( Dz = z^ * cos(*t), z()=, t ); pretty(sol) cos(t) sin(t) + t = linspace(, 6, 6); z = eval(vectorize(sol)); (iii) RK method tspan = [,6]; z = ; h =.; [trk,zrk] = rk(@f, tspan, z, h); Plot of exact solution together with Euler approximate solution plot(t,z, trk,zrk, ro ) legend( Exact, RK ) m6xb z = z cos(t), z() = Exact RK. 4 6 Example B A function defined by the equation y = ax b,for<x <, where a > iscalledapower function. Taking the logarithm of each side gives ln y = ln a + b ln x. A plot of y versus x is a graph

3 of the power function. A plot of ln y versus ln x is a straight line with slope b and intercept ln a. Consider the power function y = x 4. Produce plots of y versus x and ln y versus ln x. What is the slope of the straight line in the latter? NSS4.6/Example B y = x 4 (log scales on both axes) y Let s answer the slope question first. Taking natural logartihms, we have ln y = ln 4lnx. Therefore, the slope of the straight line is 4. We ll actually draw three plots. The first is y versus x using standard linear scaling. The second is y versus x using logarithmic scaling on both axes (using MATLAB s loglog plotting routine). Finally, we plot ln y versus ln x using linear scaling. Here is the code followed by the plots. ln y x NSS4.6/Example B: y = x ln x. y. NSS4-.6/Example B Regular plot: y vs x x = linspace(.,, ); y =./ x.^4; plot(x,y) Loglog plot: ln(y) vs ln(x) figure; loglog(x,y) Direct plot of ln(y) vs ln(x) lnx = log(x); lny = log(y); figure; plot(lnx,lny) x 9 NSS4.6/Example B y = x 4. x Example C Use the second-order Runge-Kutta method with seven different steps sizes to approximate y(), the value of the solution of the initial value problem y = t y, y()= at t =. Use step sizes h = m, m = 4,,...,. Find the exact solution, compute y() exactly, then construct a table whose first and last columns show step size and corresponding error. Make a plot of the log of the absolute error versus the log of the step size; i.e., ln E h versus ln h. Note the approximate linear relationship; i.e., ln E h a + b ln h. Estimate a and b by performing a linear regression. This is veryeasytodoviamatlab sbasic Fitting Interface. (Click on the Tools menu in the Figure Window and follow your instincts.) Then determine the step size required to ensure an error of at most. when using the second-order Runge-Kutta method. Use the second-order Runge-Kutta method with this step size and check the error at t =. Here are the function M-file, diary file, and plot.

4 Function M-file f.m function yp = f(t,y) yp = -t.^.* y; ============================== Diary file s6ec.txt NSS4-.6/Example C (i) RK approximations tspan = [,]; y = ; h =.; hcol = []; rcol = []; for k = :7 h = h/; hcol = [hcol; h]; [trk,yrk] = rk(@f, tspan, y, h); dim = size(yrk); n = dim(); rcol = [rcol; yrk(n)]; echo off (ii) Exact solution and errors sol = dsolve( Dy = -t^*y, y()=, t ); pretty(sol) exp(- / t ) syms t yxct = double(subs(sol, t, )) yxct =.69 (iii) Plot of log of error versus log of step size; table too aerr = abs(yxct - rcol); ln_hcol = log(hcol); ln_aerr = log(aerr); plot(ln_hcol, ln_aerr, * ) format long T = [hcol, rcol,... yxct*ones(size(hcol)), aerr] T = Columns through Column format short Table (iv) A linear regression to determine the linear approximation to the straight line through the plot of error vs step size was done via MATLAB s Basic Fitting Interface. (Click on the Tools menu in the Figure Window.) Estimate the step size needed to ensure that the absolute error is is <.. ln E h f = inline(. -. * h.^.87, h ); hest = fzero(f,.) hest =.86 a = ; b = ; N = (b-a) / hest N =.4 (v) Check whether prediction in (iv) pans out. h = hest; [trk,yrk] = rk(@f, tspan, y, h); dim = size(yrk); n = dim(); rkat = yrk(n) rkat =.7 aerr = abs(yxct - rkat) aerr =.6 NSS4.6/Example C: Absolute error vs step size E h =.*h.87 ln E h =.87*ln(h).49 data linear ln h Example D [a problem revisited one last time!] Calculate the first three iterations of the fourth-order Runge-Kutta method for the IVP z = z, z() = Use a step size of h =.. Here f (t, z) = z. Notice that the syntax for Polking s rk4 routine is identical to that of his eul and rk routines. We simply change the name of the ODE routine we are using. Function M-file f.m function zp = f(t,z) zp = - z; ============================== Diary file s7ed.txt NSS4-.7/Example D 4

5 RK4 method tspan = [.,.]; z = ; h =.; [trk4,zrk4] = rk4(@f, tspan, z, h); Table of t and z values tk_zk = [trk4, zrk4] tk_zk = Example E Use the fourth-order Runge-Kutta method with seven different steps sizes to approximate y(), the value of the solution of the initial value problem y = t y, y()= at t =. Use step sizes h = m, m =,,...,7. Find the exact solution, compute y() exactly, then construct a table whose first and last columns show step size and corresponding error. Make a plot of the log of the absolute error versus the log of the step size; i.e., ln E h versus ln h. Note the approximate linear relationship; i.e., ln E h a + b ln h. Estimate a and b by performing a linear regression. Again, use MATLAB s Basic Fitting Interface. Then determine the step size required to ensure an error of at most. when using the fourth-order Runge-Kutta method. Use the fourth-order Runge-Kutta method with this step size and check the error at t =. Here are the function M-file, diary file, and plot. Function M-file f.m function yp = f(t,y) yp = -t.^.* y; ============================== Diary file s7ee.txt.txt NSS4-.7/Example E (i) RK4 approximations tspan = [,]; y = ; h = ; hcol = []; rcol = []; for k = :7 h = h/; hcol = [hcol; h]; [trk4,yrk4] = rk4(@f, tspan, y, h); dim = size(yrk4); n = dim(); rcol = [rcol; yrk4(n)]; echo off (ii) Exact solution and errors sol = dsolve( Dy = -t^*y, y()=, t ); pretty(sol) exp(- / t ) syms t yxct = double(subs(sol, t, )) yxct =.69 (iii) Plot of log of error versus log of step size; table too aerr = abs(yxct - rcol); ln_hcol = log(hcol); ln_aerr = log(aerr); plot(ln_hcol, ln_aerr, * ) format long T = [hcol, rcol,... yxct*ones(size(hcol)), aerr] T = Columns through Column format short Table (iv) A linear regression to determine the linear approximation to the straight line through the plot of error vs step size was done via MATLAB s Basic Fitting Interface. (Click on the Tools menu in the Figure Window.) Estimate the step size needed to ensure that the absolute error is is <.. f = inline( * h.^4.6, h ); hest = fzero(f,.) hest =.97 a = ; b = ; N = (b-a) / hest N =.477 (v) Check whether prediction in (iv) pans out. h = hest; [trk4,yrk4] = rk4(@f, tspan, y, h);

6 dim = size(yrk4); n = dim(); rk4at = yrk4(n) rk4at =.69 aerr = abs(yxct - rk4at) aerr = 8.77e-6 mx4a curves coalesce ln E h NSS4.7/Example E: Absolute error vs step size ln E h = 4.6*ln(h).6 E h =.96*h 4.6 data linear 6 4 ln h M-/6a Use ode4 to solve the following system numerically on [, ]. x = x, x = ( x )x x, x () =, x () = 4 M-/4a Plot solutions for each of the eleven () initial value problems y + 4y = cost +sin 4t, y() =, 4,..., Like Polking s eul, rk,andrk4 routines, MATLAB s ode4 readily handles systems. We produce both time and phase plane plots. Why solve them one at a time when you can dispatch all targets at once?! As noted in M-9/ (one of your homework problems), MATLAB s ODE solvers can handle multiple initial conditions simultaneously, provided that your function M-file is array-smart. Here we use MATLAB s built-in ode4 routine. As Polking relates, This is a sophisticated, variable-step solver...which is a very clever modification of the Runge-Kutta algorithms. It is highly recommended! Function M-file f.m function yp = f(t,y) yp =.*cos(t) + sin(4.*t) - 4.*y; ============================== Diary file mx4a.txt M-/4a tspan = [,]; y = -:; [t,y] = ode4(@f, tspan, y); plot(t,y) Function M-file f.m function xp = f(t,x) xp = zeros(,); column vector! xp() = x(); xp() = (-x().^).* x() - x(); ============================== Diary file mx6a.txt M-/6a ODE4 approximate solution tspan = [,]; x = [;4]; [t,x] = ode4(@f, tspan, x); Plot of x vs t and x vs t x = x(:,); x = x(:,); plot(t,x, t,x, r-- ) legend( x vs t, x vs t ) Plot of x vs x figure; plot(x,x) 6

7 4 mx6a Time plots x vs t x vs t mx6b Phase plane plot A =. tspan = :. : ; x = [.;]; [t,x] = ode4(@f, tspan, x); X = x(:,); XP = x(:,); A = tspan = :. : ; x = [;]; [t,x] = ode4(@f, tspan, x); X = x(:,); XP = x(:,); A = tspan = :. : ; x = [;]; [t,x] = ode4(@f, tspan, x); X = x(:,); XP = x(:,); Time plots superimposed plot(t,x, t,x, --, t,x, o ) Phase plane plots superimosed figure plot(x,xp, X,XP, --, X,XP, o ). mx Time plots A =. A = A = M-/ Change this second-order IVP to a first-order system. Solve the resulting system and produce superimposed time and phase plane plots for A =.,, over [, ]. (Lord Rayleigh devised this equation to model the motion of the reed in a clarinet.) x = x 4 ( x ) x, x() = A, x () =. Convert to a first order system (as in M-/, q.v.), then proceed. Function M-file f.m function xp = f(t,x) xp = zeros(,); column vector! xp() = x(); xp() =.*x() - 4.*x().^ -.*x(); ============================== Diary file mx.txt M-/ mx Phase plane plots A =. A = A =... M-/ The IVP y +.y + 6y = sin(6.t), y() =, y () =, represents a driven, damped oscillator. Change it into a first-order planar system. Use ode4 to solve the system. Then plot the 7

8 position y, velocity v, and time t in -D on the time interval [, ]. It s quite exotic! Note the comet last line in the code. This produces an animated -D plot. You ll love it! Let x = y and x = y.thenx =y =x and x = y = sin(6.t).y 6y = sin(6.t).x 6x. Thus x = x, x = sin(6.t).x 6x, x () =, x () = Here is the function M-file and diary file. Function M-file f.m function xp = f(t,x) xp = zeros(,); column vector! xp() = x(); xp() = sin(6..*t) -..*x() - 6.*x(); ============================== Diary file mx.txt M-/ tspan = [,]; x = [;]; [t,x] = ode4(@f, tspan, x); y = x(:,); yp = x(:,); plot(y, yp, t); grid on figure; comet(y, yp, t); mx D plot: pos vs velo vs time t v y 8