文档介绍：2 Stability of solutions to ODEs
How can we address the question of stability in general?
We proceed from the example of the pendulum equation. We reduce this
second order ODE,
g
β¨ + sin β= 0,
l
to two first order ODE’s.
Write x1 = β, x2 = β˙. Then
x˙1 = x2
g
x˙= sin x
2 − l 1
The equilibrium points, or fixed points, are where the trajectories in phase
space stop, . where
x˙
ψx˙= 1 = ψ0
x˙
2
For the pendulum, this requires
x2 = 0
x = nα, n = 0, 1, 2, . . .
1 ±
Since sin x1 is periodic, the only distinct fixed points are
β 0 βα
= and =
β˙ 0 β˙ 0
Intuitively, the first is stable and the second is not.
How may we be more precise?
Linear systems
Consider the problem in general. First, assume that we have the linear system
u˙1 = a11u1 + a12u2
u˙2 = a21u1 + a22u2
14
or
ψu˙= Aψu
with
u (t) a a
ψu(t) = 1 and A = 11 12
u (t) a a
2 21 22
Assume A has an inverse and that its eigenvalues are distinct.
Then the only fixed point (where ψu˙= 0) is ψu = 0.
The solution, in general, is
1t 2t
ψu(t) = 1e cψ1 + 2e cψ2
where
, are eigenvalues of A.
• 1 2
cψ, cψ are eigenvectors of A.
• 1 2
and are constants (deriving from initial conditions).
• 1 2
What are the possibilities for stability?
1. 1 and 2 are both real.
(a) If < 0 and < 0, then u(t) 0 as