# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-7 => Topic started by: Victor Ivrii on November 30, 2018, 04:05:34 PM

Title: Q7 TUT 0201
Post by: Victor Ivrii on November 30, 2018, 04:05:34 PM
(a) Determine all critical points of the given system of equations.

(b) Find the corresponding linear system near each critical point.

(c) Find the eigenvalues of each linear system. What conclusions can you then draw about the nonlinear system?

(d)  Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system.
\left\{\begin{aligned} &\frac{dx}{dt} = 1 - xy, \\ &\frac{dy}{dt} = x - y^3. \end{aligned}\right.

Bonus: Computer generated picture
Title: Re: Q7 TUT 0201
Post by: Yulin WANG on November 30, 2018, 04:39:52 PM
(a)
\begin{equation}
\left\{
\begin{array}{**lr**}
1-xy=0 &  \\
x-y^{3}=0\\
\end{array}
\right.
\end{equation}
\begin{equation}
\left\{
\begin{array}{**lr**}
xy=1 &  \\
x=y^{3}\\
\end{array}
\right.
\end{equation}
\begin{equation}
\left\{
\begin{array}{**lr**}
x=1 &  \\
y=1\\
\end{array}
\right.
\end{equation} or \begin{equation}
\left\{
\begin{array}{**lr**}
x=-1 &  \\
y=-1\\
\end{array}
\right.
\end{equation}
Therefore, the critical points are (1,1) and (-1,-1)
(b)
The Jacobian matrix of the vector field is:
\begin{align*}
J &= \begin{bmatrix}
-y & -x \\
1 & -3y^{2}
\end{bmatrix}\\
~\\
J(1,1) &= \begin{bmatrix}
-1 & -1 \\
1 & -3
\end{bmatrix}\\
~\\
J(-1,-1) &= \begin{bmatrix}
1 & 1 \\
1 & -3
\end{bmatrix}
\end{align*}
(c)
\begin{align*}
For (1,1), let A&= \begin{bmatrix}
-1 & -1 \\
1 & -3
\end{bmatrix}\\
~\\
A-\lambda I &= \begin{bmatrix}
-1-\lambda & -1 \\
1 & -3-\lambda
\end{bmatrix}\\
~\\
det(A-\lambda I) &=(\lambda+3)(\lambda+1)+1=0\\
~\\
\lambda_{1} &= \lambda_{2} = -2 \\
~\\
Then \ the \ system \ has \ a \ stable \ improper \ node \ at \ (1,1) \\
~\\
For (-1,-1), let A&= \begin{bmatrix}
1 & 1 \\
1 & -3
\end{bmatrix}\\
~\\
A-\lambda I &= \begin{bmatrix}
1-\lambda & 1 \\
1 & -3-\lambda
\end{bmatrix}\\
~\\
det(A-\lambda I) &=(\lambda+3)(\lambda-1)-1=0\\
~\\
\lambda = -1 \pm \sqrt{5} \\
~\\
Then \ the \ system \ has \ a \ unstable \ saddle \ point \ at \ (1,1) \\
\end{align*}
(d) In the attachment.
Title: Re: Q7 TUT 0201
Post by: Zhuojing Yu on November 30, 2018, 06:29:30 PM
I think when (1,1), it is node or spiral point, not IN(improper node).
Title: Re: Q7 TUT 0201
Post by: Jingze Wang on November 30, 2018, 08:31:23 PM
I think the phase portrait that Yulin drew is correct, here is computer generated picture since no one posted
Title: Re: Q7 TUT 0201
Post by: Yulin WANG on November 30, 2018, 11:13:57 PM
I think the phase portrait that Yulin drew is correct, here is computer generated picture since no one posted
Thanks for submitting the computer-generated phase portrait!!!
BTW, how do u plot the phase portrait on a computer?
Title: Re: Q7 TUT 0201
Post by: Victor Ivrii on December 01, 2018, 03:57:59 AM
Sure, it is improper node. Jingze finally got a decent computer generated picture (took a correct range of variables)

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