Quick classification of 2D systems

mark

(10 pts)

Open this webpage and refer to the first section on "Quick classifications" to get four different 2D linear systems. Quickly classify the origin for each as

• An unstable saddle point,
• An unstable node or spiral, or
• A stable fixed point.

You can do so using the following condensed version of theorem 4.40 from our text regarding the linear system $\vec{x}'=A\vec{x}$, where $A$ is a square matrix with $\det A \neq 0$:

• If $\det A<0$, then the origin is a saddle point (and, therefore, unstable)
• If $\det A>0$ and $\text{tr}\,A>0$, then the origin is an unstable node or spiral,
• If $\det A>0$ and $\text{tr}\,A<0$, then the origin is a stable node or spiral.

In your answers, be sure to include

1. The system,
2. The determinant and trace, and
3. The classification

Here is a template that you can use to enter a 2D system into the forum:

$$
  \begin{align}
    x' &= ax + by \\\\
    y' &= cx + dy
  \end{align}
$$

shp900

$$
\begin{align}
x' &= -3x - 3y \\
y' &= 8x + 6y
\end{align}
$$

trace = 3
determinant = 6
Unstable node or trace

$$
\begin{align}
x' &= -3x - 2y \\
y' &= 9x + 8y
\end{align}
$$

trace = 5
determinant = -6
Saddle point

$$
\begin{align}
x' &= -3x + 3y \\
y' &= -5x + 7y
\end{align}
$$

trace = 4
determinant = -6
Saddle point

$$
\begin{align}
x' &= 4x + 9y \\
y' &= 4x + 0y
\end{align}
$$

trace = 0
determinant = -36
Saddle point

Badler

$$
\begin{align}
x' &= -9x + 4y \\
y' &= -3x + -5y
\end{align}
$$
Det= 57
Tr=-14 so, the origin is a stable node/spiral.
$$
\begin{align}
x' &= 7x + 9y \\
y' &= 2x + 2y
\end{align}
$$
Det=-4 so, the origin is an unstable saddle point.
$$
\begin{align}
x' &= -5x + 7y \\
y' &= 6x +8y
\end{align}
$$
Det=-82 so, the origin is an unstable saddle point.
$$
\begin{align}
x' &= -2x - 9y \\
y' &= -4x + 6y
\end{align}
$$
Det= -48 so, origin is an unstable saddle point.

Dylan

$$
\begin{align}
x' &= 2x + 5y \\
y' &= -3x - 9y \\
tr &= -7 \\
det &=-3
\end{align}
$$
The origin is an unstable saddle point because the det<0

$$
\begin{align}
x' &= -2x - 3y \\
y' &= 9x + 5y \\
tr &= 3 \\
det &= 17
\end{align}
$$
The origin is an unstable spiral because tr>0, det>0, and 4det>tr^2

$$
\begin{align}
x' &= -10x - 10y \\
y' &= -6x + y \\
tr &= -9 \\
det &=-50
\end{align}
$$
The origin is an unstable saddle point because the det<0

$$
\begin{align}
x' &= -4x - 9y \\
y' &= -x - 4y \\
tr &= -8 \\
det &=6
\end{align}
$$
The origin is a stable node because tr<0, det<0, and 4det<tr^2

Austin

$$
\begin{align}
x' &= -1x - 7y \\
y' &= 6x +3y
\end{align}
$$
Det=39, trA=2, Unstable node or Spiral

$$
\begin{align}
x' &= 8x +4y \\
y' &= -1x -10y
\end{align}
$$
Det=-76, Saddle Point

$$
\begin{align}
x' &= -2x - 9y \\
y' &= -7x + 2y
\end{align}
$$
Det=-67, Saddle Point

$$
\begin{align}
x' &= -4x -2y \\
y' &= -4x - 5y
\end{align}
$$
Det=-3, Saddle Point

sara

My first system is
$$
\begin{align}
x' &= -6x + 1y \\
y' &= 4x + -6y
\end{align}
$$
The determinant is 40 and the trace is -12 so it is a stable node or spiral.

My second system is
$$
\begin{align}
x' &= -4x - 3y \\
y' &= 6x - 7y
\end{align}
$$
The determinant is 10 and the trace is -11 so it is a stable node or spiral.

My third system is
$$
\begin{align}
x' &= 7x - 10y \\
y' &= -3x - 7y
\end{align}
$$
The determinant is -19 so it is a saddle point.

My final system is
$$
\begin{align}
x' &= -4x + 0y \\
y' &= 3x + 3y
\end{align}
$$
The determinant is -12 so it is a saddle point.

alee

$$
\begin{align}
x' &= -3x + 0y \\
y' &= 0x - 10y
\end{align}
$$
trace = -13
determinant = 30
Stable node or spiral

$$
\begin{align}
x' &= -4x + 4y \\
y' &= -2x - 5y
\end{align}
$$
trace = -9
determinant = 28
Stable node or spiral

$$
\begin{align}
x' &= 7x - 5y \\
y' &= -7x - 3y
\end{align}
$$
trace = 4
determinant = -56
saddle

$$
\begin{align}
x' &= 3x - 8y \\
y' &= -4x - 2y
\end{align}
$$
trace = 1
determinant = -38
saddle

StephenC

$$
\begin{align}
x' &= -2x + 2y \\
y' &= -6x + 0y
\end{align}
$$
Determinant is -12
Trace is -2
Stable node or spiral
$$
\begin{align}
x' &= -6x - 10y \\
y' &= 3x + 6y
\end{align}
$$
Determinant is -6
Trace is 0
Saddle point
$$
\begin{align}
x' &= 7x + 9y \\
y' &= -1x - 7y
\end{align}
$$
Determinant is -40
Trace is 0
Saddle point
$$
\begin{align}
x' &= 5x - 8y \\
y' &= -10x + 6y
\end{align}
$$
Determinant is -50
Trace is 11
Saddle point

Alice

$$
\begin{align}
x' &= -7x + 9y \\
y' &= 8x + 1y
\end{align}
$$

Det(A)=-79
Tr(A)=-6
Saddle point

$$
\begin{align}
x' &= -4x + 7y \\
y' &= 7x -5y
\end{align}
$$

Det(A)=-4
Tr(A)=-9
Saddle point

$$
\begin{align}
x' &= -10x - 9y \\
y' &= -1x +7y
\end{align}
$$

Det(A)=-61
Tr(A)=-3
Saddle point

$$
\begin{align}
x' &= -3x - 1y \\
y' &= -5x -2y
\end{align}
$$

Det(A)=1
Tr(A)=-5
Stable Node

Angela

Determinant = 12
Trace = -7
Stable Node 



Determinant = -16
Trace = 6
Saddle Point 


Determinant = -38
Trace = 1 
Saddle Point 


Determinant = -21
Trace = 7
Saddle Point 

mcespede

$$
\begin{align}
x' &= 0x - 4y \\
y' &= 8x + 3y
\end{align}
$$

Trace = 3
Determinant = 32
Unstable Node or Trace

$$
\begin{align}
x' &= 0x - 8y \\
y' &= -6x + 7y
\end{align}
$$

Trace = 7
Determinant = -48
Saddle Point

$$
\begin{align}
x' &= 3x - 7y \\
y' &= -5x - 8y
\end{align}
$$

Trace = -5
Determinant = -59
Saddle Point

$$
\begin{align}
x' &= 8x + 8y \\
y' &= 5x + 2y
\end{align}
$$

Trace = 10
Determinant = -24
Saddle Point

EricK

1.
Determinant = -10
Trace = -7
Saddle Point

2.
Determinant = 38
Trace = -5
Stable Node

3.
Determinant = -33
Trace = 1
Saddle Point

4.
Determinant = 23
Trace = 4
Unstable Node

colepeterson

$$
\begin{align}
x' &= 5x + 7y \\
y' &= -10x
\end{align}
$$

$det(A) = -70$, so the origin is a saddle point.

$$
\begin{align}
x' &= 8x - 2y \\
y' &= -6x - 8y
\end{align}
$$

$det(A) = -76$, so the origin is a saddle point.

$$
\begin{align}
x' &= 6x + 6y \\
y' &= 6x + 1y
\end{align}
$$

$det(A) = -30$, so the origin is a saddle point.

$$
\begin{align}
x' &= -10x - 9y \\
y' &= -2x - 9y
\end{align}
$$

$det(A) = 72$ and $tr(A) = -19$, so the origin is stable node or spiral.

Holden

Determinant=3
Trace=1
unstable node

Determinant=-46
Trace=-9
saddle

Determinant=-15
Trace=8
saddle

Determinant=8
Trace=1
unstable node

Perrin

$$
\begin{align}
x' &= 8x - 2y \\
y' &= -10x - 8y
\end{align}
$$

determinant = $-44$
trace = $0$
unstable saddle point

$$
\begin{align}
x' &= -3x - 8y \\
y' &= -8x + 9y
\end{align}
$$

determinant = $-91$
trace = $6$
unstable saddle point

$$
\begin{align}
x' &= 3x + 6y \\
y' &= 4x - y
\end{align}
$$

determinant = $-27$
trace = $2$
unstable saddle point

$$
\begin{align}
x' &= 3x - y \\
y' &= -2x + 3y
\end{align}
$$

determinant = $7$
trace = $6$
unstable node or spiral

max

$$
\begin{align}
x' &= -1x - 10y \\
y' &= 0x - 5y
\end{align}
$$

det = 5
tr = -6
origin is a stable node or spiral

$$
\begin{align}
x' &= -10x - 9y \\
y' &= 7x - 5y
\end{align}
$$

det = 113
tr = -15
origin is a stable node or spiral

$$
\begin{align}
x' &= 6x - 6y \\
y' &= -4x - 7y
\end{align}
$$

det = -66
tr = -1
origin is a saddle point (and unstable)

$$
\begin{align}
x' &= 6x - 3y \\
y' &= 5x + 1y
\end{align}
$$

det = 21
tr = 7
origin is unstable node or spiral

Jules

$$ x'=-2x+y$$$$y'=-6x+4y$$
det=-2
Since the $det<0$, this is a saddle (and unstable)

$$ x'=9x-10y$$$$y'=-x-8y$$
det=-82
Since the $det<0$, this is a saddle (and unstable)

$$ x'=-7x+4y$$$$y'=-6x-y$$
det=31, 4det=124
tr=-8, $tr^2$=64
Since the $det>0$ and $tr<0$ and $4det>tr^2$, this is a stable spiral

$$ x'=-5x+5y$$$$y'=x+4y$$
det=-25
Since the $det<0$, this is a saddle (and unstable)

ChristianDonaldson

$$ x'=8x-7y$$ $$ y'=-5x-4y$$

det = -67
tr = 4
saddle; unstable

$$ x'=-x-4y$$ $$ y'=0x+4y$$

det = -4
tr = 3
saddle; unstable

$$ x'=2x+9y$$ $$ y'=6x-8y$$

det = -70
tr = -6
saddle; unstable

Wiggenout

$$
\begin{align}
x' &= 3x -10y \\
y' &= 8x + 6y
\end{align}
$$
Det>0
Tr>0
Unstable node or spiral

$$
\begin{align}
x' &= -8x + 2y \\
y' &= 6x + 6y
\end{align}
$$
Det<0
Saddle point

$$
\begin{align}
x' &= 0x -3y \\
y' &= -9x + 3y
\end{align}
$$
det<0
Saddle point

$$
\begin{align}
x' &= -8x -1y \\
y' &= -6x + 0y
\end{align}
$$
Det<0
Saddle point