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Quiz 5 / Re: Quiz 5
« on: November 30, 2014, 09:10:04 PM »
Phase planes sketches are attached below.
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Quiz 5 / Re: Quiz 5
« on: November 30, 2014, 07:57:04 PM »
Continued from (a) and (b)
To linearize the systems of equations near each equilibrium point:
let $ F = (1−y)(2x−y) = 2x-y-2xy+y^2 $
and $ G = (2+x)(x−2y) = 2x-4y+x^2-2xy $
Then $ F_x = 2-2y $ , $ F_y = -1-2x+2y $
while $ G_x = 2+2x-2y $ , $ G_y = -4-2x $
The local linear system of equation for each equilibrium point $ (x_o,y_o) $ is thus:
$$
\begin{pmatrix}
(x-x_o)' \\
(y-y_o)' \\
\end{pmatrix} = \begin{pmatrix}
2-2y_o & -1-2x_o+2y_o \\
2+2x_o-2y_o & -4-2x_o \\
\end{pmatrix} \begin{pmatrix}
x-x_o \\
y-y_o \\
\end{pmatrix}
$$
At $ (0,0) $ :
$$
\begin{pmatrix}
x' \\
y' \\
\end{pmatrix} = \begin{pmatrix}
2 & -1 \\
2 & -4 \\
\end{pmatrix} \begin{pmatrix}
x \\
y \\
\end{pmatrix}
$$
$$
\begin{vmatrix}
2-r & -1 \\
2 & -4-r \\
\end{vmatrix} = r^2+2r-6
$$
setting the determinant equal to 0 yield 2 real, distinct roots: $ r_1=\sqrt{7}-1 >0 , r_2=-\sqrt{7}-1 <0 $
So the equilibrium point is a saddle point and is unstable.
At $ (-2,1) $ :
$$
\begin{pmatrix}
(x+2)' \\
(y-1)' \\
\end{pmatrix} = \begin{pmatrix}
0 & 5 \\
-4 & 0 \\
\end{pmatrix} \begin{pmatrix}
x+2 \\
y-1 \\
\end{pmatrix}
$$
$$
\begin{vmatrix}
-r & 5 \\
-4 & -r \\
\end{vmatrix} = r^2+20
$$
setting the determinant equal to 0 yield 2 complex conjugate roots $ r_1=\sqrt{20}i , r_2=-\sqrt{20}i $
So the equilibrium point is a center and is stable. Trajectory is clockwise since the first right entry is positive.
At $ (2,1) $ :
$$
\begin{pmatrix}
(x-2)' \\
(y-1)' \\
\end{pmatrix} = \begin{pmatrix}
0 & -3 \\
4 & -8 \\
\end{pmatrix} \begin{pmatrix}
x-2 \\
y-1 \\
\end{pmatrix}
$$
$$
\begin{vmatrix}
-r & -3 \\
4 & -8-r \\
\end{vmatrix} = r^2+8r+12
$$
setting the determinant equal to 0 yield 2 distinct real roots $ r_1=-2, r_2=-6 $
So the equilibrium point is a node and is asymptotically stable.
At $ (-2,-4) $ :
$$
\begin{pmatrix}
(x+2)' \\
(y+4)' \\
\end{pmatrix} = \begin{pmatrix}
10 & -5 \\
6 & 0 \\
\end{pmatrix} \begin{pmatrix}
x+2 \\
y+4 \\
\end{pmatrix}
$$
$$
\begin{vmatrix}
10-r & -5 \\
6 & -r \\
\end{vmatrix} = r^2-10r+30
$$
setting the determinant equal to 0 yield 2 complex conjugate roots $ r_1=5+\sqrt{5}i, r_2=5-\sqrt{5}i $
So the equilibrium point is a spiral point and is unstable. The first right entry of the matrix is negative so the trajectory is anticlockwise.
To linearize the systems of equations near each equilibrium point:
let $ F = (1−y)(2x−y) = 2x-y-2xy+y^2 $
and $ G = (2+x)(x−2y) = 2x-4y+x^2-2xy $
Then $ F_x = 2-2y $ , $ F_y = -1-2x+2y $
while $ G_x = 2+2x-2y $ , $ G_y = -4-2x $
The local linear system of equation for each equilibrium point $ (x_o,y_o) $ is thus:
$$
\begin{pmatrix}
(x-x_o)' \\
(y-y_o)' \\
\end{pmatrix} = \begin{pmatrix}
2-2y_o & -1-2x_o+2y_o \\
2+2x_o-2y_o & -4-2x_o \\
\end{pmatrix} \begin{pmatrix}
x-x_o \\
y-y_o \\
\end{pmatrix}
$$
At $ (0,0) $ :
$$
\begin{pmatrix}
x' \\
y' \\
\end{pmatrix} = \begin{pmatrix}
2 & -1 \\
2 & -4 \\
\end{pmatrix} \begin{pmatrix}
x \\
y \\
\end{pmatrix}
$$
$$
\begin{vmatrix}
2-r & -1 \\
2 & -4-r \\
\end{vmatrix} = r^2+2r-6
$$
setting the determinant equal to 0 yield 2 real, distinct roots: $ r_1=\sqrt{7}-1 >0 , r_2=-\sqrt{7}-1 <0 $
So the equilibrium point is a saddle point and is unstable.
At $ (-2,1) $ :
$$
\begin{pmatrix}
(x+2)' \\
(y-1)' \\
\end{pmatrix} = \begin{pmatrix}
0 & 5 \\
-4 & 0 \\
\end{pmatrix} \begin{pmatrix}
x+2 \\
y-1 \\
\end{pmatrix}
$$
$$
\begin{vmatrix}
-r & 5 \\
-4 & -r \\
\end{vmatrix} = r^2+20
$$
setting the determinant equal to 0 yield 2 complex conjugate roots $ r_1=\sqrt{20}i , r_2=-\sqrt{20}i $
So the equilibrium point is a center and is stable. Trajectory is clockwise since the first right entry is positive.
At $ (2,1) $ :
$$
\begin{pmatrix}
(x-2)' \\
(y-1)' \\
\end{pmatrix} = \begin{pmatrix}
0 & -3 \\
4 & -8 \\
\end{pmatrix} \begin{pmatrix}
x-2 \\
y-1 \\
\end{pmatrix}
$$
$$
\begin{vmatrix}
-r & -3 \\
4 & -8-r \\
\end{vmatrix} = r^2+8r+12
$$
setting the determinant equal to 0 yield 2 distinct real roots $ r_1=-2, r_2=-6 $
So the equilibrium point is a node and is asymptotically stable.
At $ (-2,-4) $ :
$$
\begin{pmatrix}
(x+2)' \\
(y+4)' \\
\end{pmatrix} = \begin{pmatrix}
10 & -5 \\
6 & 0 \\
\end{pmatrix} \begin{pmatrix}
x+2 \\
y+4 \\
\end{pmatrix}
$$
$$
\begin{vmatrix}
10-r & -5 \\
6 & -r \\
\end{vmatrix} = r^2-10r+30
$$
setting the determinant equal to 0 yield 2 complex conjugate roots $ r_1=5+\sqrt{5}i, r_2=5-\sqrt{5}i $
So the equilibrium point is a spiral point and is unstable. The first right entry of the matrix is negative so the trajectory is anticlockwise.
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Quiz 5 / Re: Quiz 5
« on: November 30, 2014, 06:09:55 PM »
(a) Determine points $ (x,y) $ such that $ x'=0 $ and $ y'=0 $
$ x'=0 $ means $ 1-y=0 $ or $ 2x-y=0 \implies y=1 $ or $ 2x=y $
and $ y'=0 $ means $ 2+x=0 $ or $ x-2y=0 \implies x=-2 $ or $ x=2y $
The equilibrium solutions are: $ (-2,1) $ , $ (2,1) $ , $ (-2,-4) $ , $ (0,0) $
$ x'=0 $ means $ 1-y=0 $ or $ 2x-y=0 \implies y=1 $ or $ 2x=y $
and $ y'=0 $ means $ 2+x=0 $ or $ x-2y=0 \implies x=-2 $ or $ x=2y $
The equilibrium solutions are: $ (-2,1) $ , $ (2,1) $ , $ (-2,-4) $ , $ (0,0) $
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