Autonomous Systems (Part III)
1) Describe the undamped hard spring oscillations modeled by the system
or, equivalently,
Find the critical points of this system, and determine the type and stability of these
points.
Solution: Rewrite the given second-order DE as the first-order system
Solve for the critical points by solving the homogeneous system of algebraic
equations
Since
is never 0 for any real value of , the bottom equation is only zero when
, so the only critical point occurs at (0,0). We'll use the linearization at (0,0) to
classify the type and stability of this critical point.
Eigenvalues
for
the Linearized System
Type of Critical Point
of the Amost Linear
System
Stable improper node
Stable node or spiral
point
Unstable saddle point
Unstable node or
spiral point
Unstable improper
node
Stable spiral point
Unstable spiral point
Stable or unstable,
center or spiral point
We need to compute the Jacobian of the given system:
.
At (0,0), the Jacobian is just
.
It is easy to see that
eigenvalues are
always
has the characteristic polynomial
. Notice that
. That means the
and check that the eigenvalues are
in the hard spring oscillation case.
Maple tells us what the phase portrait of our autonomous DE looks like.
>
3
2
y
1
0
1
2
3
x
>
That means our critical point (0,0) is a stable center. What does this tell us about the
motion of our mass?
2) Describe the undamped soft spring oscillations modeled by the system
or, equivalently,
Find the critical points of this system, and determine the type and stability of these
points.
Solution: Rewrite the given second-order DE as the first-order system
Solve for the critical points by solving the homogeneous system of algebraic
equations
Hence, we have critical points at (0,0), (2,0), and (-2,0). We'll use the linearization at
each critical point to classify the type and stability of each critical point.
We need to compute the Jacobian of the given system:
.
Case 1: Our critical point is (0,0).
At (0,0), the Jacobian is just
.
It is easy to see that
eigenvalues are
has the characteristic polynomial
. That means the
.
Case 2:
At (
,0), the Jacobian is just
.
It is easy to see that
has the characteristic polynomial
. That means
the eigenvalues are
. Since the eigenvalues are real with opposite signs, we
know
are both unstable saddle points.
To check our work, let's also have Maple plot the phase portrait.
>
4
y
2
0
1
2
x
3
4
>
What does this tell us about the motion of our mass?
3) Describe the damped hard spring oscillations modeled by the system
or, equivalently,
Find the critical points of this system, and determine the type and stability of these
points.
Solution: Rewrite the given second-order DE as the first-order system
Solve for the critical points by solving the homogeneous system of algebraic
equations
Just as in damped hard oscillation case, the only critical point occurs at (0,0).
We need to compute the Jacobian of the given system:
.
At (0,0), the Jacobian is just
.
It is easy to see that
has the characteristic polynomial
. That means the eigenvalues are
.
Since the eigenvalues are complex conjugates with negative real part, we know (0,0) is
an stable spiral point.
>
3
2
y
1
0
1
2
3
x
>
3) Describe the damped soft spring oscillations modeled by the system
or, equivalently,
Find the critical points of this system, and determine the type and stability of these
points.
Solution: Rewrite the given second-order DE as the first-order system
Solve for the critical points by solving the homogeneous system of algebraic
equations
Just as in damped soft spring oscillation case, the critical points occurs at (0,0), (2,0),
and (-2,0).
We need to compute the Jacobian of the given system:
.
Case 1: Our critical point is (0,0).
At (0,0), the Jacobian is just
.
It is easy to see that
has the characteristic polynomial
. That means the eigenvalues are
. Since the
eigenvalues are complex conjugates with negative real part, we know (0,0) is an stable
spiral point.
Case 2:
At (
,0), the Jacobian is just
.
It is easy to see that
has the characteristic polynomial
.
That means the eigenvalues are
.
Since the eigenvalues are real with opposite signs, we know
saddle points.
To check our work, let's also have Maple plot the phase portrait.
>
are both unstable
4
y
2
0
1
2
x
3
4
>
Notice that the phase portrait is split into 5 regions. What does our phase portrait tell
us about our system?