### Author Topic: Q7 TUT 0801  (Read 2421 times)

#### Victor Ivrii

• Elder Member
• Posts: 2606
• Karma: 0
##### 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

• Full Member
• Posts: 17
• Karma: 25
• 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

• Jr. Member
• Posts: 6
• Karma: 2
##### Re: Q7 TUT 0801
« Reply #2 on: November 30, 2018, 07:44:34 PM »
There is my solution