Autonomous Systems
To find the critical points of autonomous system
we need to solve the algebraic system of equations
From the bottom equation, we get
that the critical points are
and
.
, and then using the top equation, we get
.
Let's have MAPLE plot the slope/direction field and the phase portrait.
First, here's the direction field.
>
5
4
3
y
2
1
0
1
2
x
>
Alternatively, we could plot the slope field.
>
3
5
4
3
y x
2
1
0
1
2
3
x
>
Notice that these two plots look very similar. These two plots will look identical
except that the arrows may possibly point in the exact opposite direction. In our
example, observe that the plots differ in the left, bottom corner. The slope field tells
us the slope of y considered as a function of x. The direction field tells us the "flow"
of x and y as a function of t. That is, the arrows describe what happens to x and y as t
increases.
Using the direction field, we can now plot the phase portrait.
>
5
4
3
y
2
1
0
1
2
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x
>
Now let's try to better understand what the critical point of the linear system
with constant coefficient matrix . Here are the possibilities for the eigenvalues
of :
(1) real and unequal with same sign,
(2) real and unequal with opposite sign,
and
(3) real and equal,
(4) complex conjugates with nonzero real part, or
(5) pure imaginary numbers.
In all of these cases (as long as
for eigenvectors
and
of
is diagonalizable), we have the general solution
and
, respectively.
CASE 1: real and unequal eigenvalues with same sign
If
and
are both positive, the critical point at (0,0) will be an (improper) nodal
source. That means the critical point will be unstable.
>
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3
y
2
1
0
1
2
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x
>
If
and
are both negative, the critical point at (0,0) will be an (improper) nodal
sink. That means the critical point will be asymptotically stable.
>
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y
2
1
0
1
2
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x
>
CASE 2: real and unequal eigenvalues with opposite sign
The critical point at (0,0) will be a saddle point. That means the critical point will be
unstable.
>
>
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y
2
1
0
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x
>
CASE 3: real and equal eigenvalues
If
and
are both positive, the critical point at (0,0) will be a (proper) nodal source.
That means the critical point will be unstable.
>
5
4
3
y
2
1
0
1
2
x
>
3
If
and
are both negative, the critical point at (0,0) will be a (proper) nodal sink.
That means the critical point will be asymptotically stable.
>
5
4
3
y
2
1
0
1
2
x
>
3
Note that if is not diagonalizable, then we will get that the critical points are
improper rather than proper.
CASE 4: compex conjugate eigenvalues
If
and
both have real part positive, the critical point at (0,0) will be an
unstable spiral point.
If
and
both have real part negative, the critical point at (0,0) will be an
asymptotically stable spiral point.
>
5
4
3
y
2
1
0
1
2
3
x
>
CASE 5: purely imaginary eigenvalues
The critical point at (0,0) will be a stable center (not asymptotically stable).
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y
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1
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x
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Alright, let's return to the almost linear system we started with--
.
To find the linearization of this system at the pritical points, we need to compute the
Jacobian:
>
(1)
>
So at the critical point (1,-2), the linearization is
.
This system has a critical point at (0,0) that corresponds to the original system's
critical point at (1,-2). Compare the the direction field with for the linearization (top
graph) with the direction field for the original system (bottom graph). What do you
notice?
>
5
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3
v
2
1
0
1
2
u
>
3
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y
2
1
0
1
2
x
>
3