Runge - Kutta Methods for first and second order models

Transcription

1 Runge - Kutta Methods for first and second order models James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University October 3, 2013

2 Outline 1 Runge - Kutte Methods 2 RK Methods: One Variable 3 RK Methods: Two Variables

3 Abstract This lecture discusses the Runge - Kutta Method for first and second order models.

4 Runge - Kutte Methods Runge - Kutta methods are based on more sophisticated ways of approximating the solution to y = f (t, y). These methods use multiple function evaluations at different time points around a given t to approximate y(t ). In more advanced classes, we can show this technique generates a sequence {ŷ n } starting at y 0 using the following recursion equation: ŷ n+1 = ŷ n + h F o (t n, ŷ n, h, f ) ŷ 0 = y 0 where h is the step size we use and F o is a fairly complicated function of the previous approximate solution, the step size and the right hand side function f. The Runge - Kutta methods are available for various choices of the superscript o which is called the order of the method.

5 Runge - Kutte Methods We will not discuss F o here, as it is best to do that in a more advanced class. We know that for Order One: Local error is h 2 and this method is the Euler Method. The global error then goes down linearly with h. Order Two: Local error is h 3, global is h 2. If the global error for a given stepsize h is E, halving the stepsize to h/2 gives a new global error E/4. The global error goes down quadratically. Order Three: Local error is h 4, global is h 3. If the global error for a given stepsize h is E, then halving the stepsize to h/2 gives a new global error E/8. The global error goes down as a cubic power. Order Four: Local error is h 5, global is h 4. If the global error for a given stepsize h is E, then halving the stepsize to h/2 gives a new global error of E/16.

6 Runge - Kutte Methods The basic code to implement the Runge-Kutta methods is broken into two pieces. The first one, RKstep.m implements the evaluation of the next approximation solution at point (t n, ŷ n ) given the old approximation at (t n 1, ŷ n 1 ). We then loop through all the steps to get to the chosen final time using the code in FixedRK.m. The details of these algorithms are beyond the scope of this text; however, they arise from carefully combined multiple tangent line approximations at each step. In the RK code, we want the dynamic functions to depend on time also. Previously, we used dynamics like f 3*x and we expected the dynamics functions to have that form in DoEuler. However, we want more complicated dynamics now so we will define our dynamics as if they depend on time. So from now on, we write f=@(t,x) 3*x even if there is no time dependence.

7 Runge - Kutte Methods DoEulerTwo.m: the functions are now time dependent. We have to change euler also. f u n c t i o n [ xhat, x, e ] = DoEulerTwo ( f, t r u e, N, x0, h ) % N = number of approximations to f i n d % f = t h e model dynamics f u n c t i o n o f ( t, x ) % t r u e = t h e t r u e s o l u t i o n % x0 = t h e i n i t i a l c o n d i t i o n % h = t h e s t e p s i z e % e u l e r t, x, h ) x + f ( t, x ) h ; x h a t ( 1 ) = x0 ; % i n i t i a l i z e x, x h a t and e x ( 1 ) = x0 ; e ( 1 ) = 0 ; f o r i =1:N x h a t ( i +1) = e u l e r ( i h, x h a t ( i ), h ) ; x ( i +1) = t r u e ( i h ) ; e ( i +1) = abs ( x ( i +1) x h a t ( i +1) ) ; end end

8 Runge - Kutte Methods RKstep.m uses the new dynamics functions. In it you see the lines like feval(fname,t,x) which means take the function fname passed in as an argument and evaluate it as the pair (t,x). Hence, fname(t,x) is the same as f(t,x). RKstep.m does all the heavy lifting. It manages all of the multiple tangent line calculations that Runge - Kutta needs at each step. We loop through all the steps to get to the chosen final time using the code in FixedRK.m.

9 Runge - Kutte Methods f u n c t i o n [ tnew, ynew, fnew ] = RKstep ( fname, tc, yc, fc, h, k ) % fname t h e name o f t h e r i g h t hand s i d e f u n c t i o n f ( t, y ) % yc a p p r o x i m a t e s o l u t i o n to y ( t ) = f ( t, y ( t ) ) a t t=t c % f c f ( tc, yc ) % h The time step, k i s RK order 1<= k <=4 % tnew i s t c+h, ynew i s a p p r o x s o l n a t tnew, fnew i s f ( tnew, ynew ) i f k==1 k1 = h f c ; ynew = yc+k1 ; e l s e i f k==2 k1 = h f c ; k2 = h f e v a l ( fname, t c +(h /2), yc+(k1 /2) ) ; ynew = yc + k2 ; e l s e i f k==3 k1 = h f c ; k2 = h f e v a l ( fname, t c +(h /2), yc+(k1 /2) ) ; k3 = h f e v a l ( fname, t c+h, yc k1+2 k2 ) ; ynew = yc+(k1+4 k2+k3 ) / 6 ; e l s e i f k==4 k1 = h f c ; k2 = h f e v a l ( fname, t c +(h /2), yc+(k1 /2) ) ; k3 = h f e v a l ( fname, t c +(h /2), yc+(k2 /2) ) ; k4 = h f e v a l ( fname, t c+h, yc+k3 ) ; ynew = yc+(k1+2 k2+2 k3+k4 ) / 6 ; e l s e d i s p ( s p r i n t f ( The RK method %2d o r d e r i s not a l l o w e d!, k ) ) ; end tnew = tc+h ; fnew = f e v a l ( fname, tnew, ynew ) ; end

10 Runge - Kutte Methods f u n c t i o n [ t v a l s, y v a l s ] = FixedRK ( fname, t0, y0, h, k, n ) % G i v e s a p p r o x i m a t e s o l u t i o n to % y ( t ) = f ( t, y ( t ) ) % y ( t0 ) = y0 % u s i n g a kth o r d e r RK method % t0 i n i t i a l time % y0 i n i t i a l s t a t e % h s t e p s i z e % k RK order 1<= k <= 4 % n Number o f s t e p s to t a k e % t v a l s time v a l u e s o f form % t v a l s ( j ) = t0 + ( j 1) h, 1 <= j <= n % y v a l s a p p r o x i m a t e s o l u t i o n % y v a l s ( : j ) = a p p r o x i m a t e s o l u t i o n a t % t v a l s ( j ), 1 <= j <= n tc = t0 ; yc = y0 ; t v a l s = t c ; y v a l s = yc ; f c = f e v a l ( fname, tc, yc ) ; f o r j =1:n 1 [ tc, yc, f c ] = RKstep ( fname, tc, yc, fc, h, k ) ; y v a l s = [ y v a l s yc ] ; t v a l s = [ t v a l s t c ] ; end end

11 RK Methods: One Variable Example: solve a model using all four RK orders and do a plot >> f t, x ).5 x. (60 x ) ; >> t r u e t ) 6 0. / ( 1 + (60/20 1) exp (.5 60 t ) ) ; >> T =. 6 ; >> time = l i n s p a c e ( 0,T, 3 1 ) ; >> h1 =. 0 6 ; >> N1 = c e i l (T/ h1 ) ; >> [ htime1, x h a t 1 ] = FixedRK ( f, 0, 2 0,. 0 6, 1, N1) ; >> [ htime2, x h a t 2 ] = FixedRK ( f, 0, 2 0,. 0 6, 2, N1) ; >> [ htime3, x h a t 3 ] = FixedRK ( f, 0, 2 0,. 0 6, 3, N1) ; >> [ htime4, x h a t 4 ] = FixedRK ( f, 0, 2 0,. 0 6, 4, N1) ; % t h e... a t t h e end o f t h e l i n e a l l o w s us to c o n t i n u e % a l o n g l i n e to t h e s t a r t o f t h e n e x t l i n e >> p l o t ( time, t r u e ( time ), htime1, xhat1,, htime2, xhat2, o,... htime3, xhat3, +, htime4, xhat4,. ) ; >> x l a b e l ( Time ) ; >> y l a b e l ( x ) ; % We want to use t h e d e r i v a t i v e symbol x so % s i n c e Matlab t r e a t as t h e s t a r t and s t o p o f the l a b e l % we w r i t e. That way Matlab w i l l t r e a t s as a s i n g l e quote % t h a t i s our d i f f e r e n t i a t i o n symbol >> t i t l e ( RK A p p r o x i m a t i o n s to x =. 5 x(60 x ), x ( 0 ) 2 0 ) ; % N o t i c e we c o n t i n u e t h i s l i n e too >> l e g e n d ( True, RK 1, h =. 0 6, RK 2, h =. 0 6, RK 3, h =. 0 6,... RK 4, h =. 0 6, L o c a t i o n, Best ) ; This generates the figure in the next slide: note RK Order 4 does a great job even with a large h.

12 RK Methods: One Variable

13 RK Methods: One Variable Homework 45 You are now ready to solve a few on your own. For these problems, (i): Write the dynamics function code as f etc. (ii): Write the true function code as usual true etc. (iii): Solve the problem with all four Runga - Kutte Methods for the given step size h and appropriate time interval. Plot all four Runga - Kutte approximate solutions and the true solution on the same plot. (iv): Do all of this in a word doc as usual.

14 RK Methods: One Variable Homework 45 Continued 45.1 h =.05 and the model is y =.07 y (75 y), y(0) = h =.06 and the model is u (t) =.13 u(t) (30 u(t)), u(0) = h =.4 and the model is x = 1.9x, x(0) = h =.5 and the model is x = 1.95x + 30, x(0) = 16.

15 RK Methods: Two Variables We can convert a second order linear model to a system easily. Here is an example. Example Convert to a matrix - vector system u (t) u (t) 5.0 u(t) = 0; u(0) = 1.0; u (0) = +1.0 Solution Let x be given by x(t) = [ ] x1 (t) x 2 (t) = [ ] u(t) u (t)

16 RK Methods: Two Variables Solution Solution continued: x 1(t) = u (t) = x 2 (t), x 2(t) = u (t) = 5u(t) (4)u (t) = 5x 1 (t) 4x 2 (t). Thus x (t) = [ ] x 1 (t) x 2 (t) = [ ] [ ] 0 1 x1 (t) 5 4 x 2 (t) Also, note that x(0) = [ ] x1 (0) x 2 (0) = [ ] u(0) u (0) = [ ] 1 1 = x(0).

17 RK Methods: Two Variables The conversion process also tells us the dynamics of the new systems model. We can convert a second order linear model to a system easily. Example Find the systems dynamics for the model u (t) u (t) 5.0 u(t) = 0; u(0) = 1.0; u (0) = +1.0 Solution Let x be given by x(t) = [ ] x1 (t) x 2 (t) = [ ] u(t) u (t)

18 RK Methods: Two Variables Solution Solution continued: x 1(t) = u (t) = x 2 (t), x 2(t) = u (t) = 5u(t) (4)u (t) = 5x 1 (t) 4x 2 (t). Thus, the new dynamics is f given by [ ] x f(x 1, x 2 ) = 2 5x 1 4x 2

19 RK Methods: Two Variables Note we can code the dynamics in Matlab as follows: The model u (t) u (t) 5.0 u(t) = 0; u(0) = 1.0; u (0) = +1.0 in Matlab becomes >> f x ) [ x ( 2 ) ; 5 x ( 1 ) 4 x ( 2 ) ] ; Note, x can be a number or a vector as the function has been coded to accept either. We can now apply Runge - Kutta methods to solve second order models.

20 RK Methods: Two Variables Consider the problem ay + by + cy = g, y(0) = y 0, y (0) = y 1 y + (b/a)y + (c/a)y = (1/a)g, y(0) = y 0, y (0) = y 1 This converts to the system x 1 x 2 = x 2 = (c/a)x 1 (b/a)x 2 + (1/a)g And letting A = (c/a), B = (b/a) and C = (1/a) we have x 1 x 2 = x 2 = Ax 1 + Bx 2 + Cg. We ll use this in the code below.

21 RK Methods: Two Variables Consider the problem y (t) y (t) 5.0 y(t) = 0; y(0) = 1.0; y (0) = +1.0 This has characteristic equation r 2 + 4r 5 = 0 with roots r 1 = 5 and r 2 = 1. Hence, the general solution is x(t) = Ae 5t + Be t. The initial conditions give A + B = 1 5A + B = 1 It is straightforward to see that A = 1/3 and B = 2/3. Then we solve the system using MatLab with this session:

22 RK Methods: Two Variables % d e f i n e t h e dynamics f o r y + 4 y 5 y = 0 >> a = 1 ; b = 4 ; c = 5; >> B = (b/a ) ; A = (c /a ) ; C = 1/ a ; >> g t ) 0. 0 ; >> f t, y ) [ y ( 2 ) ; A y ( 1 ) + B y ( 2 ) + C g ( t ) ] ; % d e f i n e t h e t r u e s o l u t i o n >> t r u t ) [ ( 1. 0 / 3. 0 ) exp( 5 t ) (2.0/3.0) exp ( t ) ;... ( 5. 0 / 3. 0 ) exp( 5 t ) (2.0/3.0) exp ( t ) ] ; >> y0 = [ 1 ; 1 ] ; >> h =. 2 ; >> T = 3 ; >> time = l i n s p a c e ( 0,T, ) ; >> N = c e i l (T/h ) ; >> [ htime1, r k a p p r o x 1 ] = FixedRK ( f, 0, y0, h, 1,N) ; >> y h a t 1 = r k a p p r o x 1 ( 1, : ) ; >> [ htime2, r k a p p r o x 2 ] = FixedRK ( f, 0, y0, h, 2,N) ; >> y h a t 2 = r k a p p r o x 2 ( 1, : ) ; >> [ htime3, r k a p p r o x 3 ] = FixedRK ( f, 0, y0, h, 3,N) ; >> y h a t 3 = r k a p p r o x 3 ( 1, : ) ; >> [ htime4, r k a p p r o x 4 ] = FixedRK ( f, 0, y0, h, 4,N) ; >> y h a t 4 = r k a p p r o x 4 ( 1, : ) ; >> p l o t ( time, t r u e ( time ), htime1, yhat1, o, htime2, yhat2,,... htime3, yhat3, +, htime4, yhat4, ) ; >> x l a b e l ( Time ) ; >> y l a b e l ( y ) ; >> t i t l e ( S o l u t i o n to x + 4x 5 x = 0, x ( 0 ) = 1, x ( 0 ) = 1 on [ 0, 3 ] ) ; >> l e g e n d ( True, RK1, RK2, RK3, RK4, L o c a t i o n, Best ) ;

23 RK Methods: Two Variables This generates the plot

24 RK Methods: Two Variables Now add an external input. We don t know true solution anymore. % d e f i n e t h e dynamics f o r y + 4 y 5 y = 10 s i n (5 t ) eˆ{.03 t } >> a = 1 ; b = 4 ; c = 5; >> B = (b/a ) ; A = (c /a ) ; C = 1/ a ; >> g t ) 10 s i n (5 t ). exp (.03 t ) ; >> f t, y ) [ y ( 2 ) ; A y ( 1 ) + B y ( 2 ) + C g ( t ) ] ; >> y0 = [ 1 ; 1 ] ; >> h =. 2 ; >> T = 3 ; >> N = c e i l (T/h ) ; >> [ htime1, r k a p p r o x 1 ] = FixedRK ( f, 0, y0, h, 1,N) ; >> y h a t 1 = r k a p p r o x 1 ( 1, : ) ; >> [ htime2, r k a p p r o x 2 ] = FixedRK ( f, 0, y0, h, 2,N) ; >> y h a t 2 = r k a p p r o x 2 ( 1, : ) ; >> [ htime3, r k a p p r o x 3 ] = FixedRK ( f, 0, y0, h, 3,N) ; >> y h a t 3 = r k a p p r o x 3 ( 1, : ) ; >> [ htime4, r k a p p r o x 4 ] = FixedRK ( f, 0, y0, h, 4,N) ; >> y h a t 4 = r k a p p r o x 4 ( 1, : ) ; >> p l o t ( htime1, yhat1, o, htime2, yhat2,,... htime3, yhat3, +, htime4, yhat4, ) ; >> x l a b e l ( Time ) ; >> y l a b e l ( Approx y ) ; >> t i t l e ( S o l u t i o n to x + 4x 5 x = 10 s i n (5 t ) eˆ{.03 t }, x ( 0 ) = 1, x ( 0 ) = 1 on [ 0, 3 ] ) ; >> l e g e n d ( RK1, RK2, RK3, RK4, L o c a t i o n, Best ) ;

25 RK Methods: Two Variables Note the plot is not quite smooth enough because of the coarseness of h. Big question is are we getting closer to the true solution?

26 RK Methods: Two Variables Homework 46 On all of these problems, for the given h and final time T, find the true solution generate the plot of the true solution and all four RK order plots as we have done in the example. Write this up with attached plots in word For h =.3 and T = 4. u (t) + u (t) 2 u(t) = 0; u(0) = 1; u (0) = For h =.4 and T = 3. x (t) + 6 x (t) + 9 x(t) = 0; x(0) = 1; x (0) = For h =.2 and T = 5. y (t) + 4 y (t) + 13 y(t) = 0; y(0) = 1; y (0) = 2

27 RK Methods: Two Variables Homework 47 On all of these problems, for the given stepsize h and final time T, generate all four RK order plots as we have done in the example. Write this up with attached plots in word For h =.4 and T = 3. 2 u (t) + 4 u (t) 3 u(t) = exp( 2t) cos(3t + 5) 47.2 For h =.2 and T = 4. u(0) = 2 u (0) = 3 u (t) 2 u (t) + 13 u(t) = exp( 5t) sin(3t 2 + 5) u(0) = 12 u (0) = 6