### Author Topic: Q7 TUT 0801  (Read 1160 times)

#### Victor Ivrii

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##### Q7 TUT 0801
« 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

#### Yulin WANG

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• MAT244H1 2018F
##### Re: Q7 TUT 0801
« Reply #1 on: November 30, 2018, 04:40:26 PM »
(a)

\left\{
\begin{array}{**lr**}
y+x-x^{3}-xy^{2}=0 &  \\
-x+y-x^{2}y-y^{3}=0\\
\end{array}
\right.

\left\{
\begin{array}{**lr**}
x^{2}+y^{2}=0
\end{array}
\right.

\left\{
\begin{array}{**lr**}
x=0 &  \\
y=0\\
\end{array}
\right.

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.
« Last Edit: November 30, 2018, 05:38:42 PM by Yulin WANG »

#### Doris Zhuomin Jia

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##### Re: Q7 TUT 0801
« Reply #2 on: November 30, 2018, 07:44:34 PM »
There is my solution

#### Victor Ivrii

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##### Re: Q7 TUT 0801
« Reply #3 on: December 01, 2018, 03:35:40 AM »
Yulin, your picture sucks.

Doris, please do not post several minor pdfs

On the picture attached you can see, indeed, unstable focal point, clockwise and a stable limit cycle