# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-7 => Topic started by: Victor Ivrii on November 30, 2018, 04:10:42 PM

Title: Q7 TUT 0801
Post by: Victor Ivrii on November 30, 2018, 04:10:42 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} = y +x(1-x^2 - y^2),\\ &\frac{dy}{dt} = -x + y(1-x^2 - y^2) \end{aligned}\right.

Bonus: Computer generated picture
Title: Re: Q7 TUT 0801
Post by: Yulin WANG on November 30, 2018, 04:40:26 PM
(a)
\begin{equation}
\left\{
\begin{array}{**lr**}
y+x-x^{3}-xy^{2}=0 &  \\
-x+y-x^{2}y-y^{3}=0\\
\end{array}
\right.
\end{equation}
\begin{equation}
\left\{
\begin{array}{**lr**}
x^{2}+y^{2}=0
\end{array}
\right.
\end{equation}
\begin{equation}
\left\{
\begin{array}{**lr**}
x=0 &  \\
y=0\\
\end{array}
\right.
\end{equation}
Therefore, the only critical point is (0,0)
(b)
The Jacobian matrix of the vector field is:
\begin{align*}
J &= \begin{bmatrix}
1-3x^{2}-y^{2} & 1-2xy \\
-1-2xy & 1-x^{2}-3y^{2}
\end{bmatrix}\\
~\\
J(0,0) &= \begin{bmatrix}
1 & 1 \\
-1 & 1
\end{bmatrix}\\
\end{align*}
(c)
\begin{align*}
For (0,0), let A&= \begin{bmatrix}
1 & 1 \\
-1 & 1
\end{bmatrix}\\
~\\
A-\lambda I &= \begin{bmatrix}
1-\lambda & 1 \\
-1 & 1-\lambda
\end{bmatrix}\\
~\\
det(A-\lambda I) &=(\lambda-1)^{2}+1=0\\
~\\
\lambda &= 1 \pm i \\
~\\
Then \ the \ system \ has \ a \ clockwise \ spiral \ outwards \ at \ (0,0) \\
\end{align*}
(d) In the attachment.
Title: Re: Q7 TUT 0801
Post by: Doris Zhuomin Jia on November 30, 2018, 07:44:34 PM
There is my solution
Title: Re: Q7 TUT 0801
Post by: Victor Ivrii on December 01, 2018, 03:35:40 AM