MTH 264
DELTA COLLEGE
SECTION 5.1 8
Analyze the following first-order system in the first quadrant, x, y ≥ 0.
dx
dt
dy
dt
= x(10 − x − y)
= y(30 − 2x − y)
(a) Find and classify all equilibria.
(b) Sketch the phase portrait of the system near each equilibrium point.
(c) Use appropriate technology to compare the actual phase portrait to the phase portraits of the linearizations.
Solution:
(a) Equilibrium solutions are points in the xy-plane where both dx/dt and dy/dt equal 0. Since each derivative
has two factors, we can break this into four cases:
1. x = 0 and y = 0 (the intersection of two lines) which yields the equilibrium solution (x, y) = (0, 0).
2. x = 0 and 30−2x−y = 0 (the intersection of two lines) which yields the equilibrium solution (x, y) = (0, 30).
3. 10 − x − y = 0 and y = 0 (the intersection of two lines) which yields the equilibrium solution (x, y) = (10, 0).
4. 10 − x − y = 0 and 30 − 2x − y = 0 (the intersection of two lines), when we subract the latter from the
former we have −20 + x = 0, so the equilibrium solution will be (x, y) = (20, −10)
We have three equilibrium solutions in the first quadrant and we will classify them by using the Jacobian to
look at the linearization of the system at each equilibrium solution.
dx
dt
dy
dt
J(x, y) =
(x, y)
J(x, y)
(0, 0)
10 0
0 30
T
D
40
300
∂f /∂x
∂g/∂x
=
f (x, y) = 10x − x2 − xy
=
g(x, y) = 30y − 2xy − y 2
∂f /∂y
∂g/∂y
=
10 − 2x − y
−2y
−x
30 − 2x − 2y
λ2 − T λ + D = 0
eigenvalues
λ2 − 40λ + 300 = 0
10, 30
eigenvectors
1
0
classification
0
,
1
source
(λ − 10)(λ − 30) = 0
(0, 30)
−20
0
−60 −30
−50
600
λ2 + 50λ + 600 = 0
−30, −20
0
1
,
1
−6
1
0
,
1
−2
sink
(λ + 30)(λ + 20) = 0
(10, 0)
−10 −10
0
10
0
−100
λ2 − 100 = 0
(λ + 10)(λ − 10) = 0
−10, 10
saddle
(b), (c) To accomplish the sketch we can use the Jacobian alone with a computer or calculator that can sketch
the slope field of the linear system represented by that matrix. In that case we only need the Jacobian column
above. If we were to sketch the phase portraits by hand, we would need the eigenvectors to help us make the
sketch to scale.
For this problem we will present the phase portraits of the linearizations side by side with an extreme closeup
of the phase portrait of the original system. The phase portraits of the linear systems are on the left and the
magnified views of the phase portrait of the original system, centered on the equilibrium solutions are on the right.
The linear systems are essentially indistinguishable from the original system when we are near the equilibrium
solutions.