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### Messages - Wei Teoh

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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 $F = (1âˆ’y)(2xâˆ’y) = 2x-y-2xy+y^2$
and $G = (2+x)(xâˆ’2y) = 2x-4y+x^2-2xy$

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: $(-2,1)$ , $(2,1)$ , $(-2,-4)$ , $(0,0)$

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