### Author Topic: Q7 TUT 0101  (Read 2997 times)

#### Victor Ivrii ##### Q7 TUT 0101
« on: November 30, 2018, 04:02:45 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} = x + x^2 + y^2, \\ &\frac{dy}{dt} = y - xy. \end{aligned}\right.

Bonus: Computer generated picture
« Last Edit: November 30, 2018, 04:05:29 PM by Victor Ivrii »

#### Chonghan Ma

• Newbie
• • Posts: 4
• Karma: 5 ##### Re: Q7 TUT 0101
« Reply #1 on: November 30, 2018, 04:29:45 PM »
(a)
Set x’ = 0 and y’=0
Then we have critical points (0,0), (-1,0)
(b)
J = \begin{bmatrix}2x+1 & 2y \\-y & 1-x \end{bmatrix}
Linear systems are shown with each critical point:
J(0,0) =  \begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}
J(-1,0) =  \begin{bmatrix}-1 & 0 \\0 & 2 \end{bmatrix}
(c)
Eigenvalues are computed by det(A - tI)= 0
So that
At (0,0): t=1
Critical point is an unstable proper node or spiral point
At (-1,0): t=-1 and 2
Critical point is an unstable saddle point

#### Jingze Wang

• Full Member
•   • Posts: 30
• Karma: 25 ##### Re: Q7 TUT 0101
« Reply #2 on: November 30, 2018, 04:35:45 PM »
This is computer generated picture #### Mengfan Zhu

• Jr. Member
•  • Posts: 10
• Karma: 5 ##### Re: Q7 TUT 0101
« Reply #3 on: December 01, 2018, 02:57:12 AM »
Hello, I solve this question step by step.
If there are any mistakes, reply to me anytime.
Thank you very much.

#### Victor Ivrii ##### Re: Q7 TUT 0101
« Reply #4 on: December 01, 2018, 02:57:45 AM »
Jingze,
You took too large range by $y$ and ...

When we have non-zero repeated eigenvalues, there is always a node (stable or unstable). Spiral may appear in this situation only when the right-hand is not smooth, which is not the case. On the picture attached you can see...
« Last Edit: December 01, 2018, 03:03:38 AM by Victor Ivrii »