Mathematics 22: Lecture 11

Transcription

1 Mathematics 22: Lecture 11 Runge-Kutta Dan Sloughter Furman University January 25, 2008 Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

2 Order of approximations One may show that the error in Euler s method is bounded by the step-size h times a constant. We call Euler s method a first-order method. Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

3 Order of approximations One may show that the error in Euler s method is bounded by the step-size h times a constant. We call Euler s method a first-order method. The modified Euler method is a second-order method: the error is bounded by a constant times h 2. Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

4 Runge-Kutta method Consider the initial-value problem du dt = f (t, u), u(t 0) = u 0. Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

5 Runge-Kutta method Consider the initial-value problem du dt = f (t, u), u(t 0) = u 0. Divide [t 0, t 0 + T ] into N equal intervals of length h = T N. Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

6 Runge-Kutta method Consider the initial-value problem du dt = f (t, u), u(t 0) = u 0. Divide [t 0, t 0 + T ] into N equal intervals of length h = T N. Let t i = t 0 + ih, i = 0, 1, 2,..., N. Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

7 Runge-Kutta method Consider the initial-value problem du dt = f (t, u), u(t 0) = u 0. Divide [t 0, t 0 + T ] into N equal intervals of length h = T N. Let t i = t 0 + ih, i = 0, 1, 2,..., N. Having computed u 0, u 1,..., u i, let k 1 = f (t i, u i ) ( k 2 = f t i + h 2, u i + h ) 2 k 1 ( k 3 = f t i + h 2, u i + h ) 2 k 2 k 4 = f (t i + h, u i + hk 3 ). Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

8 Runge-Kutta (cont d) Let u i+1 = u i + h 6 (k 1 + 2k 2 + 2k 3 + k 4 ). Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

9 Runge-Kutta (cont d) Let u i+1 = u i + h 6 (k 1 + 2k 2 + 2k 3 + k 4 ). Runge-Kutta is a fourth-order method. Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

10 Example Consider the initial-value problem on the interval [0, 6]. du dt = u cos(t), u(0) = 1. Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

11 Example Consider the initial-value problem on the interval [0, 6]. Let h = 0.1 as before. du dt = u cos(t), u(0) = 1. Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

12 Example Consider the initial-value problem on the interval [0, 6]. Let h = 0.1 as before. du dt For the first step, we have k 1 = (1.0) cos(0) = 1.0 = u cos(t), u(0) = 1. k 2 = (1.0 + (0.05)(1.0)) cos(0.05) = k 3 = (1.0 + (0.05)( )) cos(0.05) = k 4 = (1.0 + (0.1)( )) cos(0.1) = Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

13 Example (cont d) And so u 1 = (1.0 + (2)( ) + (2)( ) ) 6 = Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

14 Example (cont d) And so u 1 = (1.0 + (2)( ) + (2)( ) ) 6 = Note: the exact value is u(0.1) = e sin(0.1) = Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

15 Example (cont d) And so u 1 = (1.0 + (2)( ) + (2)( ) ) 6 = Note: the exact value is u(0.1) = e sin(0.1) = Recall: with Euler s method we had u 1 = 1.1 and with the modified Euler method we had u 1 = Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

16 Using Octave Runge-Kutta in Octave: octave:1> function w = f(t,u) > w = u*cos(t); > endfunction octave:2> t = [0:0.1:6]; octave:3> u(1) = 1.0; octave:4> for i = 1:60 > k1 = f(t(i),u(i)); > k2 = f(t(i)+0.05,u(i)+0.05*k1); > k3 = f(t(i)+0.05,u(i)+0.05*k2); > k4 = f(t(i)+0.1,u(i)+0.1*k3); > u(i+1) = u(i) + (0.1/6)*(k1 + 2*k2 + 2*k3 + k4); > endfor Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

17 Using Octave Runge-Kutta in Octave: octave:1> function w = f(t,u) > w = u*cos(t); > endfunction octave:2> t = [0:0.1:6]; octave:3> u(1) = 1.0; octave:4> for i = 1:60 > k1 = f(t(i),u(i)); > k2 = f(t(i)+0.05,u(i)+0.05*k1); > k3 = f(t(i)+0.05,u(i)+0.05*k2); > k4 = f(t(i)+0.1,u(i)+0.1*k3); > u(i+1) = u(i) + (0.1/6)*(k1 + 2*k2 + 2*k3 + k4); > endfor Note: u(6) u 60 = , which is exact to 5 decimal places. Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

18 Using Octave (cont d) Comparison of exact (green) and approximate (red) solutions: Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

19 Octave: lsode The following commands use the built-in Octave function lsode to solve our equation: octave:1> function w = f(u, t) > w = u*cos(t); > endfunction octave:2> t = [0:0.1:6]; octave:3> u = lsode("f",1.0,t); Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

20 Octave: lsode The following commands use the built-in Octave function lsode to solve our equation: octave:1> function w = f(u, t) > w = u*cos(t); > endfunction octave:2> t = [0:0.1:6]; octave:3> u = lsode("f",1.0,t); Note: u and t are reversed in the definition of f from our notation. Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

21 lsode (cont d) The method used by Octave is an adaptive step-size method. Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

22 lsode (cont d) The method used by Octave is an adaptive step-size method. That is, the actual step-size (value of h) used varies as the integration proceeds based on the behavior of the function. Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

23 lsode (cont d) The method used by Octave is an adaptive step-size method. That is, the actual step-size (value of h) used varies as the integration proceeds based on the behavior of the function. In particular, the values in the t vector do not determine the step size, but are there only for evaluation and plotting purposes. Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

24 lsode (cont d) The method used by Octave is an adaptive step-size method. That is, the actual step-size (value of h) used varies as the integration proceeds based on the behavior of the function. In particular, the values in the t vector do not determine the step size, but are there only for evaluation and plotting purposes. In particular, if one only wanted to know u(6), t could be specified by t = [0:6:6], in which case u(2) is the approximation to u(6). Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

25 lsode (cont d) The method used by Octave is an adaptive step-size method. That is, the actual step-size (value of h) used varies as the integration proceeds based on the behavior of the function. In particular, the values in the t vector do not determine the step size, but are there only for evaluation and plotting purposes. In particular, if one only wanted to know u(6), t could be specified by t = [0:6:6], in which case u(2) is the approximation to u(6). Or, we could just use u = lsode("f",1.0,[0 6]); to perform the evaluation. Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

26 Runge-Kutta in Maxima To approximate a solution in Maxima using Runge-Kutta: Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

27 Runge-Kutta in Maxima To approximate a solution in Maxima using Runge-Kutta: load("dynamics") Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

28 Runge-Kutta in Maxima To approximate a solution in Maxima using Runge-Kutta: load("dynamics") u:rk(u*cos(t),u,1.0,[t,0,6,0.1])$ Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

29 Runge-Kutta in Maxima To approximate a solution in Maxima using Runge-Kutta: load("dynamics") u:rk(u*cos(t),u,1.0,[t,0,6,0.1])$ The resulting ordered pairs are in the variable u. Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11

30 Runge-Kutta in Maxima To approximate a solution in Maxima using Runge-Kutta: load("dynamics") u:rk(u*cos(t),u,1.0,[t,0,6,0.1])$ The resulting ordered pairs are in the variable u. To plot the result: wxplot2d([discrete,u]) Note: In the lab use, plot2d([discrete,u]) Dan Sloughter (Furman University) Mathematics 22: Lecture 11 January 25, / 11