1.6 Numerical Computation

§1.6 Numerical Computation

For most of initial value problem

dx  f (t, x) , x(t )  x (6.1) dt 0 0 we cannot find a closed form for the solution x(t) . Hence it is important to find the numerical solution of I.V.P. (6.1) . In the matlab there are several ODE solver for the

I.V.P.(6.1) in vector form, i.e., a system of ordinary differentiatial equations,

x1(t)  f1 (t, x1 , x2 ,, xn )

x2 (t)  f 2 (t, x1 , x2 ,, xn )

⋮ (6.2)

xn (t)  f n (t, x1 , x2 ,, xn )

xi(t0 )  xi0 , i  1,2,,n Let

 f1 (t, x)   x1 (t)       f 2 (t, x) x(t)   ⋮  , f (t, x)   ⋮       xn (t)    f n (t, x)

Then (6.2) can be written in vector form  dx    f (t, x) , dt   x(t0 )  x0 . We recommend two most popular ODE solvers in the Matlab, ode45 and ode23s. ode45 is an explicit one-step Runge-Kutter (4th to 5th order) solver. Suitable for nonstiff problems that require moderate accuracy. This is typically the first solver to try on a new problem. Ode23s is suitable for stiff problem where lower accuracy is acceptable.

Example 6.1: Write the well-known van der pol equation x  (x 2 1)x  x  0 into a system of ODE.

Let x1  x , x2  x . Then

x1  x2

2 x2  (1 x1 )x2  x1 .

In the following we show how to use the Matlab ODE solvers and sketch the graph.

Example 6.2: Use ode45 solver to solve

y  w  asin y , y(0)  2 / w , 0  t  100

clear all

format long

global w a

w  input(‘Please input w  ’);

a  w  0.001;

t0  0 ; tf  100 ;

y0  2* pi / w ;

[t y]=ode45(‘test1’, [t0 tf],y0);

plot(t, y)

x label ('t') ;

y label (' y') ; Xmin=0;

Xmax=85;

Ymin=0;

Ymax=16;

axis([Xmin,Xmax,Ymin,Ymax]);

disp(‘finish’)

function dy=test1(t,y)

global w a

dy  w  a *sin( y) ;

Example 6.3: Solve the Van der Pol equation

x  (x 2 1)x  x  0 , x(0)  2 , x(0)  0 , 0  t  20 .

function ydot =vdpol(t , y ) %VDPOL van der Pol equation.

% ydot =VDPOL(t , y )

% ydot(1)  y(2)

% ydot(2)  mu(1 y(1)^2) * y(2)  y(1)

% mu  2

mu  2 ;

ydot  [y(2);mu *(1 y(1)^2) * y(2)  y(1)];

Note that the input arguments are t and y but that the function does not use t . Note also that the output ydot must be a column vector.

Given the above ODE file, this set of ODEs is solved using the following commands.

>> t span =[0 20]; % time span to integrate over

>> y 0=[2; 0]; % initial conditions (must be a column)

>> [t,y]= ode45(@vdpol, t span, y 0);

>> size(t ) % number of time points

>>size( y ) % (i)th column is y (i) at t (i)

>> plot(t , y (:,1),t , y (:,2), ‘- -’)

>> x label(‘time’)

>> title(‘Figure 24.1: van der Pol Solution’) Figure 24.1: van der Pol Solution