Computer Algebra Recipes

70 CHAPTER 2. PHASE-PLANE ANALYSIS
> sol:=solve(fP,Qg,fx,yg);
sol := fx =0;y=0g; fy =0;x=1g; fy =0;x=1g
yields the expected root at the origin and the twofold degenerate root (x =1,
y = 0). Since we already know that the former is a vortex, let's select the
second (or third) solution and assign it for later use.
> assign(sol[2]);
Then, calculating p and q using the standard formulas,
> p:=-(a+d); q:=a*d-b*c;
p := ¡1 q := 0
we obtain p = ¡1 andq =0. Sinceqis zero, the degenerate root is a higherorder
stationary point as I predicted."
\OK, I understand your construction of a higher-order ¯xed point. But what
do the trajectories look like in the phase plane, and how does the displacement
x(t) behave in this case?"
\To answer your questions, I ¯rst have to unassign x and y from their
stationary-point values,
> unassign('x','y');
and then enter the associated ¯rst-order ODEs.
> ODEs:=fdiff(x(t),t)=y(t),diff(y(t),t)=-x(t)+2*x(t)^2-x(t)^3g;
ODEs :=
½
d
d
x(t) =y(t);
dt dt y(t) =¡x(t)+2x(t)2 ¡ x (t) 3
¾
In order to put the stationary points in the phase-plane diagram, let's create a
plot for the two points, using size-20 red circles to represent their locations.
> gr1:=plot([[0,0],[1,0]],style=point,symbol=circle,
symbolsize=20,color=red):
Employing the DEplot command, we can produce a phase-plane portrait for
di®erent initial conditions."
After some experimentation, Mike comes up with four di®erent initial conditions,
which are included in the following DEplot command line. He chooses
blue arrows with two barbs on the head (using arrows=MEDIUM) for the tangent
¯eld and colors the trajectories red with the linecolor option. The time range
is taken from t = 0 to 20 and the time step size equal to 0:05.
> gr2:=DEplot(ODEs,[x(t),y(t)],t=0..20,x(t)=-1..2.5,
y(t)=-1.5..1.5,[[x(0)=-0.4,y(0)=1],[x(0)=-0.35,y(0)=0],
[x(0)=-0.3,y(0)=0],[x(0)=-0.2,y(0)=0]],stepsize=0.05,
color=blue,linecolor=red,arrows=MEDIUM,dirgrid=[20,20]):
Putting the two graphs gr1 and gr2 together with the display command
> display(fgr1,gr2g,tickmarks=[2,3]);
and controlling the tick marks on the coordinate axes results in the phase portrait
shown in Figure 2.13. The four trajectories, corresponding to the four
initial conditions, are clearly seen.