8Systems of differential equations
IA Differential Equations
8.2 Nonlinear dynamical systems
Consider the second-order autonomous system (i.e.
t
does not explicitly appear
in the forcing terms on the right)
˙x = f(x, y)
˙y = g(x, y)
It can be difficult to solve the equations, but we can learn a lot about phase-space
trajectories of these solutions by studying the equilibria and their stability.
Definition (Equilibrium point). An equilibrium point is a point in which
˙x = ˙y = 0 at x
0
= (x
0
, y
0
).
Clearly this occurs when
f
(
x
0
, y
0
) =
g
(
x
0
, y
0
) = 0. We solve these simultane-
ously for x
0
, y
0
.
To determine the stability, write x = x
0
+ ξ, y = y
0
+ η. Then
˙
ξ = f(x
0
+ ξ, y
0
+ η)
= f(x
0
, y
0
) + ξ
∂f
∂x
(x
0
) + η
∂f
∂y
(x
0
) + O(ξ
2
, η
2
)
So if ξ, η 1,
˙
ξ
˙η
=
f
x
f
y
g
x
g
y
ξ
η
This is a linear system, and we can determine its character from the eigensolu-
tions.
Example. (Population dynamics - predator-prey system) Suppose that there
are x prey and y predators. Then we have the following for the prey:
˙x = αx
|{z}
births - deaths
− βx
2
|{z}
natural competition
− γxy
|{z}
killed by predators
.
and the following for the predators:
˙y = εxy
|{z}
birth/survival rate
− δy
|{z}
natural death rate
For example, let
˙x = 8x − 2x
2
− 2xy
˙y = xy −y
We find the fixed points: x(8 −2x −2y) = 0 gives x = 0 or y = 4 − x.
We also want y(x − 1) = 0 which gives y = 0 or x = 1.
So the fixed points are (0, 0), (4, 0), (1, 3).
Near (0, 0), we have
˙
ξ
˙η
=
8 0
0 −1
ξ
η
We clearly have eigenvalues 8, −1 with the standard basis as the eigenvectors.
Near (4
,
0), we have
x
= 4 +
ξ
,
y
=
η
. Instead of expanding partial derivatives,
we can obtain from the equations directly:
˙
ξ = (4 + ξ)(8 − 8 − 2ξ − 2η)
= −8ξ −8η − 2ξ
2
− 2ξη
˙η = η(4 + ξ − 1)
= 3η + ξη
Ignoring the second-order terms, we have
˙
ξ
˙η
=
−8 −8
0 3
ξ
η
The eigenvalues are −8 and 3, with associated eigenvectors (1, 0), (8, −11).
Near (1, 3), we have x = 1 + ξ, y = 3 + η. So
˙
ξ = (1 + ξ)(8 − 2 − 2ξ − 6 − 2η)
≈ −2ξ − 2η
˙η = (3 + η)(1 + ξ − 1)
≈ 3η
So
˙
ξ
˙η
=
−2 −2
3 0
ξ
η
The eigenvalues are
−
1
± i
√
5
. Since it is complex with a negative real part, it
is a stable spiral.
We can determine the chirality of the spirals by considering what happens to
a small perturbation to the right
ξ
0
with
ξ >
0. We have
˙
ξ
˙η
=
−2ξ
3ξ
. So
x will head top-left, and the spiral is counter-clockwise (“positive”).
We can patch the three equilibrium points together and obtain the following
phase portrait:
0 1 2 3 4 5
0
1
2
3
4
5
We see that (1, 3) is a stable solution which almost all solutions spiral towards.