Toronto Math Forum

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.