<p>§1.6 Numerical Computation</p><p>For most of initial value problem</p><p> 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</p><p>I.V.P.(6.1) in vector form, i.e., a system of ordinary differentiatial equations,</p><p> x1(t)  f1 (t, x1 , x2 ,, xn )</p><p> x2 (t)  f 2 (t, x1 , x2 ,, xn )</p><p>⋮ (6.2)</p><p> xn (t)  f n (t, x1 , x2 ,, xn )</p><p> xi(t0 )  xi0 , i  1,2,,n Let</p><p> f1 (t, x)   x1 (t)       f 2 (t, x) x(t)   ⋮  , f (t, x)   ⋮       xn (t)    f n (t, x)</p><p>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.</p><p>Example 6.1: Write the well-known van der pol equation x  (x 2 1)x  x  0 into a system of ODE.</p><p>Let x1  x , x2  x . Then</p><p> x1  x2</p><p>2 x2  (1 x1 )x2  x1 .</p><p>In the following we show how to use the Matlab ODE solvers and sketch the graph.</p><p>Example 6.2: Use ode45 solver to solve</p><p> y  w  asin y , y(0)  2 / w , 0  t  100</p><p> clear all</p><p> format long</p><p> global w a</p><p> w  input(‘Please input w  ’);</p><p> a  w  0.001;</p><p> t0  0 ; tf  100 ;</p><p> y0  2* pi / w ;</p><p>[t y]=ode45(‘test1’, [t0 tf],y0);</p><p> plot(t, y)</p><p> x label ('t') ;</p><p> y label (' y') ; Xmin=0;</p><p>Xmax=85;</p><p>Ymin=0;</p><p>Ymax=16;</p><p> axis([Xmin,Xmax,Ymin,Ymax]);</p><p> disp(‘finish’)</p><p> function dy=test1(t,y)</p><p> global w a</p><p> dy  w  a *sin( y) ;</p><p>Example 6.3: Solve the Van der Pol equation</p><p> x  (x 2 1)x  x  0 , x(0)  2 , x(0)  0 , 0  t  20 .</p><p> function ydot =vdpol(t , y ) %VDPOL van der Pol equation.</p><p>% ydot =VDPOL(t , y )</p><p>% ydot(1)  y(2)</p><p>% ydot(2)  mu(1 y(1)^2) * y(2)  y(1)</p><p>% mu  2</p><p> mu  2 ;</p><p> ydot  [y(2);mu *(1 y(1)^2) * y(2)  y(1)];</p><p>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.</p><p>Given the above ODE file, this set of ODEs is solved using the following commands.</p><p>>> t span =[0 20]; % time span to integrate over</p><p>>> y 0=[2; 0]; % initial conditions (must be a column)</p><p>>> [t,y]= ode45(@vdpol, t span, y 0);</p><p>>> size(t ) % number of time points</p><p> ans =</p><p>333 1</p><p>>>size( y ) % (i)th column is y (i) at t (i)</p><p> ans =</p><p>333 2</p><p>>> plot(t , y (:,1),t , y (:,2), ‘- -’)</p><p>>> x label(‘time’)</p><p>>> title(‘Figure 24.1: van der Pol Solution’) Figure 24.1: van der Pol Solution</p>