### Author Topic: Q7 TUT 5102  (Read 2546 times)

#### Victor Ivrii ##### Q7 TUT 5102
« on: November 30, 2018, 04:12:23 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} = (2 + x)( y - x),\\ &\frac{dy}{dt} = (4 - x)( y + x). \end{aligned}\right.

Bonus: Computer generated picture

#### Chonghan Ma

• Newbie
• • Posts: 4
• Karma: 5 ##### Re: Q7 TUT 5102
« Reply #1 on: November 30, 2018, 04:19:46 PM »
(a)
Set (2+x)(y-x)=0 and (4-x)(y+x)=0
Then we have critical points (0,0), (4,4), (-2,2)
(b)
J = \begin{bmatrix}-2-2x+y & 2+x \\4-2y-2x & 4-x \end{bmatrix}
Linear systems are shown with each critical point:
J(0,0) =  \begin{bmatrix}-2 & 2 \\4 & 4 \end{bmatrix}
J(-2,2) =  \begin{bmatrix}4 & 0 \\6 & 6 \end{bmatrix}
J(4,4) =  \begin{bmatrix}-6 & 6 \\-8 & 0 \end{bmatrix}
(c)
Eigenvalues are computed by det(A - tI)= 0
So that
At (0,0): t=1±√17}
Critical point is a saddle point and it is unstable
At (-2,2): t= 4 and 6
Critical point is an unstable node
At ((4,4): t=-3±√9 i
Critical point is a stable spiral point
« Last Edit: November 30, 2018, 04:23:26 PM by Chonghan Ma »

#### Xiaoyuan Wang

• Jr. Member
•  • Posts: 8
• Karma: 9 ##### Re: Q7 TUT 5102
« Reply #2 on: November 30, 2018, 04:43:46 PM »

#### Jingze Wang

• Full Member
•   • Posts: 30
• Karma: 25 ##### Re: Q7 TUT 5102
« Reply #3 on: November 30, 2018, 05:17:58 PM »
This is computer generated picture

#### Mengfan Zhu

• Jr. Member
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• Karma: 5 ##### Re: Q7 TUT 5102
« Reply #4 on: December 01, 2018, 02:59:55 AM »
For this question, I draw the graph my hand.
If there is any problem, tell me as soon as possible. 