Computer Algebra Recipes

98 CHAPTER 2. PHASE-PLANE ANALYSIS
After loading the plots package, specifying the number of digits, and recording
the beginning time,
> restart: with(plots): Digits:=10: begin:=time():
the parameter values are entered and N = 1000 steps of size h =0:05 are
considered.
> t[0]:=0: x[0]:=3: y[0]:=3: a:=0.05: b:=0.5: N:=1000: h:=0.05:
The functional, or arrow, operator is used to produce the right-hand side of
each ODE. Whatever forms are inserted for x and y in the entries X(x,y) and
Y(x,y) will be operated on as indicated on the right-hand side of the arrow.
> X:=(x,y)-> -x+a*y+x^2*y;
X := (x; y) !¡x + ay+ x2 y
> Y:=(x,y)->b-a*y-x^2*y;
Y := (x; y) ! b ¡ ay¡ x2 y
Similarly, operators are introduced to perform the four function evaluations,
where f will be taken to be X and Y . Note that X and Y do not depend
explicitly on time here.
> k1:=f->h*f(x[n],y[n]):
> k2:=f->h*f(x[n]+K1[n]/2,y[n]+L1[n]/2):
> k3:=f->h*f(x[n]+K2[n]/2,y[n]+L2[n]/2):
> k4:=f->h*f(x[n]+K3[n],y[n]+L3[n]):
Using the above operators, the fourth-order Runge{Kutta algorithm is iterated
from n =0toN = 1000 and a plotting point produced at each step.
> for n from 0 to N do
> K1[n]:=k1(X); L1[n]:=k1(Y);
> K2[n]:=k2(X); L2[n]:=k2(Y);
> K3[n]:=k3(X); L3[n]:=k3(Y);
> K4[n]:=k4(X); L4[n]:=k4(Y);
> x[n+1]:=x[n]+(K1[n]+2*K2[n]+2*K3[n]+K4[n])/6;
> y[n+1]:=y[n]+(L1[n]+2*L2[n]+2*L3[n]+L4[n])/6;
> t[n+1]:=t[n]+h;
> pnt[n]:=[t[n],x[n],y[n]];
> end do:
Using the spacecurve command, the sequence of points is plotted as a solid
blue line with normal axes. The tickmarks are controlled, axis labels added,
and a particular orientation of the 3-dimensional viewing box chosen.
> spacecurve([seq(pnt[n],n=0..N)],color=blue,axes=normal,
tickmarks=[2,2,2],labels=["t","x","y"],
orientation=[23,71]);
The resulting picture is shown in Figure 2.24. It may be rotated on the computer
screen by dragging with the mouse.