Autonomous Systems (Part II)
1) Find the linearization of
at the critical point (2,1).
Solution: First, we need to compute the Jacobian of the given system:
.
Second, plug in our critical point
into
.
.
That means that the linearization of our system at (2,1) is
.
2) Describe the interaction of two populations modeled by the system
Find the critical points of this system, and determine the type and stability of these
points.
Solution: Remember that the critical points of this system are the solutions to the
homogeneous system of algebraic equations
We can solve algebraic systems in MAPLE by using the solve command:
>
(1)
Hence, the critical points are (0,0), (0,14), (20,0), and (12,6). We'll use the linearization
at each point to classify the type and stability of each critical point.
Eigenvalues
for the
Linearized
System
Type of
Critical point
for the
Linearization
Type of Critical
Point of the
Amost Linear
System
Asympoticall Asympotically
y stable
stable improper
improper
nodal sink
nodal sink
Asympoticall Asympotically
y stable
stable node or
(proper or
spiral point
improper)
nodal sink
Unstable
saddle point
Unstable saddle
point
Unstable
improper
nodal source
Unstable nodal
source or spiral
point
Unstable
improper
nodal source
Unstable
improper nodal
source
Asympoticall Asympotically
y stable
stable spiral point
spiral point
Unstable
spiral point
Unstable spiral
point
Stable center
Stable o r
unstable, center
or spiral point
We need to compute the Jacobian of the given system:
.
>
(2)
Case 1: Our critical point is (0,0).
>
(3)
Of course, we can tell by looking at this matrix that the eigenvalues are 60 and 42.
Since the eigenvalues are positive, real, and unequal, we know (0,0) is an unstable
improper node.
Case 2: Our critical point is (0,14).
>
(4)
In this case, the eigenvalues are 4 and -42. Since the eigenvalues are real with
opposite signs, we know (0,14) is an unstable saddle point.
Case 3: Our critical point is (20,0).
>
(5)
>
In this case, the eigenvalues are 2 and -60. Since the eigenvalues are real with
opposite signs, we know (20,0) is an unstable saddle point.
Case 4: Our critical point is (12,6).
>
(6)
>
In this case, the eigenvalues are
and
. Since the eigenvalues
are real, negative, and unequal, we know (12,6) is an stable improper node.
Sketch the phase portrait out by hand. Does our classification of the critical points
look consistent? Does it agree with what you expect from the model?
To check our work, let's also have Maple plot the phase portrait.
>
20
15
y
10
5
0
5
10
15
20
25
x
Warning,
issued:
cannot
probably
Warning,
issued:
cannot
probably
Warning,
issued:
cannot
probably
plot may be incomplete, the following errors(s) were
evaluate the solution further left of -.68672918e-1,
a singularity
plot may be incomplete, the following errors(s) were
evaluate the solution further left of -.37758295e-1,
a singularity
plot may be incomplete, the following errors(s) were
evaluate the solution further left of -.54405853e-1,
a singularity
20
15
y
10
5
0
5
10
15
x
>
20
25