Computer Algebra Recipes

2.1. PHASE-PLANE ANALYSIS 49
Solving for v in the ¯rst equation and substituting into the second yields the
second-order linear ODE
Äu + p _u + qu=0; with p ´¡(a + d); q ´ ad¡ bc: (2.8)
Since this ODE has constant coe±cients, a solution of the form u = e¸t is
sought. Substituting u into (2.8) yields the quadratic auxiliary equation
¸ 2 + p¸+ q =0; with two roots; ¸1;2 = ¡ p 1 p
§ p2 ¡ 4 q: (2.9)
2 2
These roots may be either real or complex. Since the general solution u is a
linear combination of e ¸1 t and e ¸2 t ,itisclearthatu ! 0ast !1if the
real parts of both ¸1 and ¸2 arenegativeandu !1if either one (or both) of
the roots has a positive real part. For the former, the stationary point will be
stable, while for the latter it will be unstable.
For simple ¯xed points, ad¡ bc ´ q 6= 0, so a zero ¸ root is not possible.
The case q = 0 corresponds to a higher-order stationary point, which will be
illustrated later. The possible roots ¸1, ¸2, which dictate the topological nature
of the ¯xed point, depend on the relative size and signs of p and q. First, let's
consider q>0andp 6= 0. Three cases have to be examined:
² For p 2 > 4q, both the roots ¸1 and ¸2 arerealandofthesamesign,
negative for p>0andpositiveforp0, the general solution
for u is of the structure u = Ae ¡j¸1j t + Be ¡j¸2j t (A, B are arbitrary
constants), while for p0
and unstable nodal points for p0isthe
critically damped SHO.
² For p2 ¡ 4 q0, and unstable focal points for p