Finding critical points for phase planes

In summary, the critical point of the system X'=x-y^2 and Y'=x^2 -xy -2x is (0,0) and there are two other solutions that can be found by setting both X' and Y' to 0 and solving the resulting equations simultaneously. These solutions are not (1,1) or (4,-2).
  • #1
NiallBucks
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Homework Statement


X'=x-y^2 Y'=x^2 -xy -2x

Homework Equations


Find the critical point of this system.

The Attempt at a Solution


I know one point is (0,0).
When I tried it I got sqrt(x)= 2 and sqrt(x)= 1 but the point (1,1) doesn't fit later parts of the question. I'm also aware that since a sqrt can have positive and negative values then (4, -2) can also equal zero. Am I missing something obvious?
 
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  • #2
Frankly, I don't know what you mean by "When I tried it I got sqrt(x)= 2..." When you tried what? And why write "sqrt(x)= 2" rather than "x= 4"? In any case, a critical point of the system will be where x'= x- y^2= 0 and y'= x^2- xy- 2x= 0. From the first equation, x= y^2. Replacing x with y^2 in the second equation gives y^4- y^3- 2y^2= y^2(y^2- y- 1)= y^2(y+ 1)(y- 2)= 0.
 
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  • #3
Sorry by 'try it' I meant solved for zero and I used sqrt(x) because I was trying to solve it in x instead of y. Your way makes much more sense and solved the problem! Thanks a million
 
  • #4
NiallBucks said:

Homework Statement


X'=x-y^2 Y'=x^2 -xy -2x

Homework Equations


Find the critical point of this system.

The Attempt at a Solution


I know one point is (0,0).
When I tried it I got sqrt(x)= 2 and sqrt(x)= 1 but the point (1,1) doesn't fit later parts of the question. I'm also aware that since a sqrt can have positive and negative values then (4, -2) can also equal zero.
There are two things wrong with this sentence.
1) By definition, the square root of a nonnegative number represents a single value. For example, ##\sqrt{4} = + 2##. Period. You are mistaken to think that ##\sqrt{4} = \pm 2##.
2) "then (4, -2) can also equal zero." -- No. (4, -2) is a point in the plane. It cannot equal a single number such as zero.

BTW, (4, -2) is also NOT a solution of the equations below.
NiallBucks said:
Am I missing something obvious?
Set both X' and Y' to 0 and solve the resulting equations simultaneously.
IOW, solve
##x - y^2 = 0##
##x^2 - xy - 2x = 0##
It helps to use one of the equations to substitute in the other.
There are three solutions, one of which is (0, 0), which you already found. (1, 1) is NOT a solution, nor is (4, -2)..
 

Related to Finding critical points for phase planes

1. What are critical points in phase planes?

Critical points, also known as equilibrium points, are specific points in a phase plane that represent stable or unstable behavior of a system. At these points, the rate of change of the system is zero, meaning the system is at rest.

2. How do you find critical points in a phase plane?

To find critical points, you need to set the differential equations of the system to zero and solve for the variables. This will give you the coordinates of the critical points in the phase plane.

3. What is the significance of critical points in phase planes?

Critical points are important in phase planes because they represent the stable and unstable behavior of a system. They can also give insights into the long-term behavior of a system.

4. Can a critical point be both stable and unstable?

No, a critical point can only be either stable or unstable. In some cases, a critical point can be both stable and unstable depending on the direction of the trajectories around it, but this is not common.

5. How do you determine the stability of a critical point in a phase plane?

The stability of a critical point can be determined by analyzing the behavior of the trajectories around it. If the trajectories move towards the critical point, it is stable. If the trajectories move away from the critical point, it is unstable. The direction of the trajectories can be determined by plotting a few solutions to the differential equations in the phase plane.

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