Solving systems of ODEs with Matlab

Transcription

1 Solving systems of ODEs with Matlab James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University October 20, 2013

2 Outline 1 Systems of ODEs 2 Setting Up the Matrix and the Vector Functions 3 Linear Second Order Problems As Systems 4 What If We Don t Know The True Solution?

3 Abstract This lecture is going to discuss solving systems of ODEs with MatLab

4 Systems of ODEs We now want to learn how to solve systems of differential equations. A typical system is the following x (t) = f (t, x(t), y(t)) (1) y (t) = g(t, x(t), y(t)) (2) x(t 0 ) = x 0 ; y(t 0 ) = y 0 (3) where x and y are our variables of interest which might represent populations to two competing species or other quantities of biological interest.

5 Systems of ODEs We now want to learn how to solve systems of differential equations. A typical system is the following x (t) = f (t, x(t), y(t)) (1) y (t) = g(t, x(t), y(t)) (2) x(t 0 ) = x 0 ; y(t 0 ) = y 0 (3) where x and y are our variables of interest which might represent populations to two competing species or other quantities of biological interest. The system starts at time t 0 (which for us is usually 0) and we specify the values the variables x and y start at as x 0 and y 0 respectively. The functions f and g are nice meaning that as functions of three arguments (t, x, y) they do not have jumps and corners. The exact nature of nice here is a bit beyond our ability to discuss in this introductory course, so we will leave it at that.

6 Systems of ODEs For example, we could be asked to solve the system x (t) = 3x(t) + 2x(t)y(t) 3y 2 (t), (4) y (t) = 2x(t) + 2x 2 (t)y(t) + 5y(t), (5) x(0) = 2; y(0) = 3. (6)

7 Systems of ODEs For example, we could be asked to solve the system x (t) = 3x(t) + 2x(t)y(t) 3y 2 (t), (4) y (t) = 2x(t) + 2x 2 (t)y(t) + 5y(t), (5) x(0) = 2; y(0) = 3. (6) or the system x (t) = 3x(t) + 2x(t)y(t) + 10 sin(t 2 + 5), (7) y (t) = 2x(t) + 2x(t)y 2 (t) + 20t 3 e 2t2, (8) x(0) = 12; y(0) = 5. (9)

8 Systems of ODEs The functions f and g in these examples are not linear in the variables x and y; hence the system Equation 4 and Equation 5 and system Equation 7 and Equation 8 are what are called a nonlinear systems. In general, arbitrary functions of time, φ and ψ to the model giving x (t) = 3x(t) + 2x(t)y(t) 3y 2 (t) + φ(t), (10) y (t) = 2x(t) + 2x 2 (t)y(t) + 5y(t) + ψ(t), (11) x(0) = 2; y(0) = 3. (12)

9 Systems of ODEs The functions f and g in these examples are not linear in the variables x and y; hence the system Equation 4 and Equation 5 and system Equation 7 and Equation 8 are what are called a nonlinear systems. In general, arbitrary functions of time, φ and ψ to the model giving x (t) = 3x(t) + 2x(t)y(t) 3y 2 (t) + φ(t), (10) y (t) = 2x(t) + 2x 2 (t)y(t) + 5y(t) + ψ(t), (11) x(0) = 2; y(0) = 3. (12) The functions φ and ψ are what we could call data functions. For example, if φ(t) = sin(t) and ψ(t) = te t, the system Equation 10 and Equation 11 would become x (t) = 3x(t) + 2x(t)y(t) 3y 2 (t) + sin(t), y (t) = 2x(t) + 2x 2 (t)y(t) + 5y(t) + te t, x(0) = 2; y(0) = 3.

10 Systems of ODEs The functions f and g would then become f (t, x, y) = 3x + 2xy 3y 2 + sin(t) g(t, x, y) = 2x + 2x 2 y + 5y + te t

11 Systems of ODEs The functions f and g would then become f (t, x, y) = 3x + 2xy 3y 2 + sin(t) g(t, x, y) = 2x + 2x 2 y + 5y + te t How do we solve such a system of differential equations? There are some things we can do if the functions f and g are linear in x and y, but many times we will be forced to look at the solutions using numerical techniques. We explored how to solve first order equations already and we will now adapt the tools developed earlier to systems of differential equations.

12 Setting Up the Matrix and the Vector Functions Example Convert to a matrix - vector system 2.0 x (t) 3.0 x (t) x(t) = 0; x(0) = 2.0; x (0) = 4.0 Solution We let the vector x be given by [ ] x1 (t) x(t) = x 2 (t) Then, = [ ] u(t) u (t) x 1(t) = u (t) = x 2 (t), x 2(t) = u (t) = (5/2)u(t) + (3/2)u (t) = (5/2)x 1 (t) + (3/2)x 2 (t).

13 Setting Up the Matrix and the Vector Functions Example Solution We then convert the above into the matrix - vector system [ ] [ ] [ ] x x (t) = 1 (t) 0 1 x1 (t) x 2 (t) = (5/2) (3/2) x 2 (t) Also, note that x(0) = [ ] x1 (0) x 2 (0) = [ ] u(0) u (0) = [ ] 2 4 = x 0. Let the matrix above be called A. Then we have converted the original system into the matrix - vector equation x (t) = A x(t), x(0) = x 0.

14 Setting Up the Matrix and the Vector Functions Example Convert to a matrix - vector system x (t) x (t) 5.0 x(t) = 0; x(0) = 1.0; x (0) = 1.0 Solution Again, we let the vector x be given by [ ] x1 (t) x(t) = = x 2 (t) [ ] u(t) u (t) Then, similar to what we did in the previous example, we find x 1(t) = u (t) = x 2 (t), x 2(t) = u (t) = 5u(t) (4)u (t) = 5x 1 (t) 4x 2 (t).

15 Setting Up the Matrix and the Vector Functions Solution We then convert the above into the matrix - vector system [ ] [ ] [ ] x x (t) = 1 (t) 0 1 x1 (t) x 2 (t) = 5 4 x 2 (t) Also, note that x(0) = [ ] x1 (0) x 2 (0) = [ ] u(0) u (0) = [ ] 1 1 = x(0). Let the matrix above be called A. Then we have converted the original system into the matrix - vector equation x (t) = A x(t), x(0), x(0) = x 0.

16 Setting Up the Matrix and the Vector Functions Example Convert to a matrix - vector system x (t) y (t) 2.0 x(t) y(t) = x (t) y (t) x(t) 8.0 y(t) = 0 x(0) = 1.0 y(0) = 2.0 Solution Let the vector u be given by u(t) = [ ] u1 (t) u 2 (t) = [ ] x(t). y(t)

17 Setting Up the Matrix and the Vector Functions Solution It is then easy to see that if we define the matrices A and B by [ ] [ ] A =, and B =, we can convert the original system into A u (t) + B u(t) = u(0) = [ ] 0, 0 [ ] 1 2

18 Setting Up the Matrix and the Vector Functions Homework Convert to a matrix - vector system X = AX 6.0 x (t) x (t) x(t) = 0; x(0) = 1.0; x (0) = Convert to a matrix - vector system AX + BX = x (t) y (t) x(t) y(t) = x (t) y (t) x(t) 4.0 y(t) = 0 x(0) = 10.0 y(0) = Convert to a matrix - vector system AX + BX = x (t) y (t) x(t) y(t) = x (t) 5.0 y (t) x(t) 2.0 y(t) = 0 x(0) = 1.0 y(0) = 2.0

19 Linear Second Order Problems As Systems Now let s consider how to adapt our previous code to handle these systems of differential equations. We will begin with the linear second order problems because we know how to do those already. First, let s consider a general second order linear problem. a u (t) + b u (t) + c u(t) = g(t), u(0) = e, u (0) = f where we assume a is not zero, so that we really do have a second order problem!

20 Linear Second Order Problems As Systems Now let s consider how to adapt our previous code to handle these systems of differential equations. We will begin with the linear second order problems because we know how to do those already. First, let s consider a general second order linear problem. a u (t) + b u (t) + c u(t) = g(t), u(0) = e, u (0) = f where we assume a is not zero, so that we really do have a second order problem! As usual, we let the vector x be given by Then, x(t) = x 1(t) = u (t) = x 2 (t), [ x1 (t) x 2 (t) ] = [ u(t) u (t) x 2(t) = u (t) = (c/a)u(t) (b/a)u (t) + (1/a) g(t) ]

21 Linear Second Order Problems As Systems We then convert the above into the matrix - vector system [ ] [ ] [ ] [ ] x x (t) = 1 (t) 0 1 x1 (t) 0 x 2 (t) = + (c/a) (b/a) x 2 (t) (1/a) g(t)

22 Linear Second Order Problems As Systems We then convert the above into the matrix - vector system [ ] [ ] [ ] [ ] x x (t) = 1 (t) 0 1 x1 (t) 0 x 2 (t) = + (c/a) (b/a) x 2 (t) (1/a) g(t) Also, note that x(0) = [ ] x1 (0) = x 2 (0) [ ] u(0) u (0) = [ ] e = x f 0.

23 Linear Second Order Problems As Systems We then convert the above into the matrix - vector system [ ] [ ] [ ] [ ] x x (t) = 1 (t) 0 1 x1 (t) 0 x 2 (t) = + (c/a) (b/a) x 2 (t) (1/a) g(t) Also, note that x(0) = [ ] x1 (0) = x 2 (0) [ ] u(0) u (0) = [ ] e = x f 0. For our purposes of using MatLab, we need to write this in terms of vectors. We have [ ] [ ] x 1 (t) x x 2 (t) = 2 (c/a) x 1 (b/a) x 2 + (1/a) g(t)

24 Linear Second Order Problems As Systems We then convert the above into the matrix - vector system [ ] [ ] [ ] [ ] x x (t) = 1 (t) 0 1 x1 (t) 0 x 2 (t) = + (c/a) (b/a) x 2 (t) (1/a) g(t) Also, note that x(0) = [ ] x1 (0) = x 2 (0) [ ] u(0) u (0) = [ ] e = x f 0. For our purposes of using MatLab, we need to write this in terms of vectors. We have [ ] [ ] x 1 (t) x x 2 (t) = 2 (c/a) x 1 (b/a) x 2 + (1/a) g(t) Now let the dynamics vector f be defined by [ ] [ ] f1 x f = = 2 (c/a) x 1 (b/a) x 2 + (1/a) g(t) f 2

25 Linear Second Order Problems As Systems For example, given the model x + 4x 5x = te.03t x(0) = 1.0 x (0) = 1.0 we write this in a MatLab session as 1 % f o r second o r d e r model % g i v e n t h e model y + 4y 5 y = t e ˆ{.03 t } >> a = 1 ; >> b = 4 ; >> c = 5; 6 >> B = (b/a ) ; >> A = (c /a ) ; >> C = 1/ a ; >> g t ) t. exp (.03 t ) ; >> f t, y ) [ y ( 2 ) ; A y ( 1 ) + B y ( 2 ) + C g ( t ) ] ;

26 Linear Second Order Problems As Systems Example Consider the problem y (t) y (t) 5.0 y(t) = 0; y(0) = 1.0; y (0) = 1.0 Solution 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:

27 Linear Second Order Problems As Systems Solution % d e f i n e t h e dynamics f o r y + 4y 5 y = 0 >> a = 1 ; b = 4 ; c = 5; >> B = (b/a ) ; A = (c /a ) ; C = 1/ a ; >> g t ) 0. 0 ; 5 >> 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 ; 10 >> 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 ) ; 15 >> 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, : ) ; 20 >> y t r u e = t r u e ( time ) ;

28 Linear Second Order Problems As Systems Solution % do t h e p l o t >> p l o t ( time, y t r u e ( 1, : ), htime1, yhat1, o,,... htime2, yhat2, *, htime3, yhat3, +,... htime4, yhat4, - ) ; 5 % s e t t h e l a b e l s >> x l a b e l ( Time ) ; >> y l a b e l ( y ) ; % s e t t h e t i t l e. T h i s goes on one l i n e but we b r e a k % i t to f i t on t h e s l i d e. The 4 q u o t e s w i l l p r i n t a s 10 % two quotes, i e x, and t h e 2 q u o t e s w i l l p r i n t a s x >> t i t l e ( Solution to x + 4 x - 5 x = 0, x (0) = -1, x (0) = 1 on [1.3] ) ; % s e t t h e l e g e n d. T h i s goes one l i n e too, but % we b r e a k to f i t on t h e s l i d e 15 >> l e g e n d ( True, RK1, RK2, RK3, RK4, Location, Best ) ;

29 Linear Second Order Problems As Systems The function true has vector values; the first component is the true solution and the second is the true solution s derivative. We want to plot only the true solution, so we need to extract it from true. We do this with the command ytrue=true(time) which saves the vector of true values into ytrue. Then in the plot command, we plot only the solution by using ytrue(1,:).

30 What If We Don t Know The True Solution? Example Now let s add an external input, g(t) = 10 sin(5 t)e.03t. In a more advanced class, we could find the true solution for this external input, but it if we changed to g(t) = 10 sin(5 t)e.03t2 we would not be able to do that. So in general, there are many models we can not find the true solution to. Solution The code is similar, but there is no true solution now. % dynamics f o r y + 4 y 5 y = 10 s i n (5 t ) e ˆ{.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 ) ; 5 >> 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 ) ;

31 What If We Don t Know The True Solution? Solution 1 >> [ 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 ) ; 6 >> 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, - ) ; 11 >> x l a b e l ( Time ) ; >> y l a b e l ( Approx y ) ; >> t i t l e ( Solution to x + 4 x - 5 x = 10 sin (5 t ) e ^{ -.03 t }, x (0) = -1, x (0) = 1 on [0,3] ) ; >> l e g e n d ( RK1, RK2, RK3, RK4, Location, Best ) ;

32 What If We Don t Know The True Solution? Homework 61 No external inputs: on all of these problems, choose an appropriate stepsize h, time interval [0, T ] for some positive T and find the true solution and write this as MatLab code. find the Runge - Kutta order 1 through 4 solutions. Write this up with attached plots u (t) + u (t) 2 u(t) = 0; u(0) = 1; u (0) = 2 x (t) + 6 x (t) + 9 x(t) = 0; x(0) = 1; x (0) = 2 y (t) + 4 y (t) + 13 y(t) = 0; y(0) = 1; y (0) = 2

33 What If We Don t Know The True Solution? Homework 61 External Inputs: for these models, find the Runge - Kutta 1 through 4 solutions and do the write up with plot as usual u (t) + 4 u (t) 3 u(t) = exp( 2t) cos(3t + 5) 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