Computer Algebra Recipes

52 CHAPTER 2. PHASE-PLANE ANALYSIS
and you should feel free to experiment with di®erent positive values.
> A[1]:=2: A[2]:=1: B[1]:=3/100: B[2]:=1/100:
The nonlinear functions P =A1 r ¡ B1 rf and Q=¡A2 f + B2 rf, corresponding
to the right-hand sides of the ODEs in (2.10), are entered as \functional
operators" using the \arrow notation."
> P:=(r,f)-> A[1]*r-B[1]*r*f;
P := (r; f ) ! A1 r ¡ B1 rf
> Q:=(r,f)-> -A[2]*f+B[2]*r*f;
Q := (r; f) !¡A2f + B2 rf
The \arrow" (->) in the above inputs is formed on the keyboard by entering a
\hyphen" followed by a \greater than" sign. The operation or \procedure" on
the right-hand side of each arrow will be applied when the two3 variables r and
f on the left-hand side are supplied as arguments to P and Q. For example,
let's take r =100andf =10andcalculateP (100; 10).
> P(100,10); #example
170
The procedure on the rhs of P has been applied with the coe±cient values
automatically substituted, namely, 2 £ 100 ¡ (3=100) £ 100 £ 10 = 170.
To classify the stationary points, we need to evaluate a, b, c, andd, which
respectively involve the partial derivatives @P=@r, @P=@f, @Q=@r, and@Q=@f
evaluated at the stationary points. A functional operator F is introduced to
di®erentiate an arbitrary quantity X(r; f ) with respect to a variable v.
> F:=(X,v)->diff(X(r,f),v):
Then F is used to calculate the four relevant partial derivatives of P and Q.
> a:=F(P,r): b:=F(P,f): c:=F(Q,r): d:=F(Q,f):
The results have been labeled a, b, c, andd, but remember that the derivatives
must still be evaluated at each stationary point. So let's locate the stationary
points by solving P (r; f) =0andQ(r; f)=0forr and f.
> sol:=solve(fP(r,f)=0,Q(r,f)=0g,fr,fg);
½
sol := fr =0;f=0g; f = 200
3 ;r=100
¾
There are two ¯xed points, one at the origin (r =0, f = 0) and the second at
r =100, f =200=3=662. By choosing one of the ¯xed points and assigning the
3
solution, the coordinates will automatically be substituted into the expressions
that follow. Let's select the nonzero ¯xed point, which is the second solution
here. As a check, the coordinates r0 and f0 of the ¯xed point are displayed.
> assign(sol[2]); r0:=r; f0:=f;
r0 := 100 f0 := 200
3
Then the values of a, b, c, anddare determined for the assigned ¯xed point.
3 Any (¯nite) number of variables may be used.