Solving an autonomous system finding critical points

In summary, the conversation is discussing the process of finding critical points and deriving the linearized system around each critical point for a given system of equations. The process involves setting the equations equal to zero and solving for the variables, which yields multiple solutions and therefore multiple critical points. The nature of the critical points can be determined by studying the linearized system around each point. The values of p, q, and Δ can potentially be used to determine the stability of the critical points, but their relevance to the linear system is not explained.
  • #1
Rubik
97
0

Homework Statement



Find all critical points and derive the linearised system about each critical point

[itex]\dot{y}[/itex] = [3y1 + y1y2; y1 + y2 - y22]

Homework Equations





The Attempt at a Solution



Setting

0 = 3y1 + y1y2 so y2 = -3
0 = y1 + y2 - y22 so y1 = 12

= [0, 1; 12, -3]
so p = -3 (<0)
q = -12 (<0)
[itex]\Delta[/itex] = 57 (>0)

Is this right that my only critical point is (12,-3) and is that how I determine the nature of the critical point and whether my critical point is stable or unstable with my values for p, q and [itex]\Delta[/itex]?
 
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  • #2
You have quadratic equations, so there are more solutions, and so more critical points.

The nature of the critical points can be determined by studying the linearized systems asked for in the problem. Around a critical point [itex](y_1^{(c)},y_2^{(c)})[/itex], expand

[itex] y_1 = y_1^{(c)} + \epsilon_1,~~y_2=y_2^{(c)}+\epsilon_2,[/itex]

and keep the terms linear in [itex]\epsilon_{1,2}[/itex] (and derivatives). Solving the linear system tells you how the complete solution must behave in the neighborhood of the critical points.
Perhaps your [itex]p,q,\Delta[/itex] are somehow relevant to the linear system but you haven't explained how they're defined.
 
  • #3
Rubik said:

Homework Statement



Find all critical points and derive the linearised system about each critical point

[itex]\dot{y}[/itex] = [3y1 + y1y2; y1 + y2 - y22]

Homework Equations





The Attempt at a Solution



Setting

0 = 3y1 + y1y2 so y2 = -3
Either y2= -3 or y1= 0.

0 = y1 + y2 - y22 so y1 = 12
If y2= -3, then y1= 12, but if y1= 0, then y2- y22= 0.

= [0, 1; 12, -3]
so p = -3 (<0)
q = -12 (<0)
[itex]\Delta[/itex] = 57 (>0)

Is this right that my only critical point is (12,-3) and is that how I determine the nature of the critical point and whether my critical point is stable or unstable with my values for p, q and [itex]\Delta[/itex]?
 

1. What is an autonomous system?

An autonomous system is a mathematical model or system that is able to function independently, without any external input or control. It follows a set of predetermined rules or equations to make decisions and perform tasks.

2. What are critical points in an autonomous system?

Critical points are the points in an autonomous system where the behavior of the system changes. They are typically where the system reaches equilibrium or where the system's behavior shifts from one state to another.

3. How do you solve for critical points in an autonomous system?

To solve for critical points, you need to set the derivative of the system's equations to zero and solve for the variables. This will give you the values of the variables at which the system's behavior changes.

4. Why is it important to find critical points in an autonomous system?

Finding critical points allows us to understand the behavior of the system and predict its future behavior. It also helps us identify stable and unstable points, which can be useful in controlling the system or making informed decisions.

5. Are there different methods for solving an autonomous system for critical points?

Yes, there are various methods for solving autonomous systems for critical points, such as the Jacobian method, the eigenvalue method, and the gradient descent method. Each method has its advantages and is suitable for different types of systems.

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