Author Topic: Quiz 5  (Read 1037 times)

Victor Ivrii

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Quiz 5
« on: November 29, 2014, 01:03:00 PM »
For system
\begin{equation*}
\left\{\begin{aligned}
&\frac{dx}{dt} = (1 - y)(2x- y), \\
&\frac{dy}{dt}  = (2 + x)(x - 2y)
\end{aligned}\right.
\end{equation*}


(a) Determine all critical points of the given system of equations and find the corresponding linear system near each critical point.

(b) Draw phase portraits of the linear systems of ODEs from the previous item and describe completely their types of the stationary points  indicating if it is stable or unstable and in the cases of the center or focus indicate orientation (clockwise or counter-clockwise).

(c) What conclusions can you then draw (and why) about the types of the critical points and phase portraits of the nonlinear system near each critical point ?

(d) Combine answers from the previous item into a (probable) phase portrait of the original non linear system.


Wei Teoh

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Re: Quiz 5
« Reply #1 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) $

Victor Ivrii

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Re: Quiz 5
« Reply #2 on: November 30, 2014, 07:02:17 PM »
Wei Teoh, it is only half of (a) since one needs to write linearization at each stationary point

Wei Teoh

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Re: Quiz 5
« Reply #3 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.

Wei Teoh

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Re: Quiz 5
« Reply #4 on: November 30, 2014, 09:10:04 PM »
Phase planes sketches are attached below.

Yuan Bian

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Re: Quiz 5
« Reply #5 on: December 02, 2014, 03:36:27 PM »
C)
nonlinear: near (0,0), saddle point, unstable;
                 near (2,1), node, AS;
                 near (-2,1), center or spiral point, indeterminate
                 near (-2, -4), spiral point, unstable