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28 Chapter 2. Global aspects
Riccati equation corresponding to M has no equilibrium point corresponding to [0,1] since it is outside of ψ(R). In going from Example 3.2.3 to Example 3.2.4, the transformation P interchanges which equilibrium point is at infinity and hence changes the stability property of the single equilibrium point for the Riccati equation. In going from Example 2.3.3 to Example 2.3.5, P moves the asymptotically stable equilibrium point from the point [0,1] at infinity to a point [1, w2] in the image of ψ. Thus, the Riccati equation in Example 2.3.5 has an asymptotically stable equilibrium point while the one in Example 2.3.3 does not.
a −b
Example 2.3.6. Redefine M = b a ,a,b ∈ R,b ̸= 0, we then have complex
eigenvalues λ1,2 = a ± ib. Therefore we are led naturally to complex solutions of the system (2.20). Remember, that in our representation formula we already considered the general case of complex solutions. Notice that the time parameter t may also be thought of as being complex.
Since
1 1
−i , q(t)
p(t)
i are eigenvectors of M we obtain
= e = e
c1e
−i +c2e cosbt
i +i(c1 −c2)
at at
ibt 1 −ibt 1
(c1 +c2) sinbt
with c1 , c2 ∈ C, as the general (complex) solution of the system (2.20). Consideringt∈Randchoosingc1+c2 =α∈Randi(c1−c2)=β∈R
we obtain the general (real) solution
q(t) at cosbt sinbt
p(t) =e α sinbt +β −cosbt . (2.25) Again, as was done in [Shay91], we discuss the phase portrait in RP(1). We
obtain the flow as S(t) = Im
q(t) p(t)
cosbt sinbt α = Im sinbt −cosbt β cosbt sinbt α
= sinbt −cosbt Im β , α,β∈R, α
π
which is b -periodic since it is a rotation of Im β . For normalization we may
for some t0 ∈ R. With this we obtain cos(bt + t0)
sinbt −cosbt ,
(2.24)
α cost0
also write Im β = Im −sint
S(t) = Im sin(bt + t0) .
0