Critical points of a diff eq system

In summary, to calculate the critical points of the given system, we can solve the equations cx + 10x^2 = 0 and x - 2y = 0. This leads to two fixed points at (0,0) and (-c/10, -c/20). When c = 0, these points coincide and there is a bifurcation. To find the axis from the kernel of the Jacobian matrix, we can use the formula det(J - λI) = 0, where J is the Jacobian matrix, λ is the eigenvalue, and I is the identity matrix. Solving this equation will give us the axis.
  • #1
Phys pilot
30
0

Homework Statement


I have to calculate the critical points of the following system.
$$x'=cx+10x^2$$
$$y'=x-2y$$

The Attempt at a Solution


So I solve the system
$$cx+10x^2=0$$
$$x-2y=0$$
So if $$x=2y$$ I have $$2yc+10*4y^2=2yc+40y^2=y(2c+40y)=0$$ and I get $$y=0$$ and $$y=-\frac{c}{20}$$ f I substitute in $$x=2y$$ I get $$x=0$$ and $$x=-\frac{c}{10}$$
Then we have that the critical points are $$(0,0)$$ and $$(-\frac{c}{10},-\frac{c}{20})$$
Is this correct? I don't know if the procedure to get the critical points is like this.
Thanks
 
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  • #2
I'm not sure what you mean by the procedure. Critical points are where the derivatives are 0. This led you to a system of nonlinear equations, which you solved. There isn't one set way to solve a system of equations (though with nonlinear systems you'll usually end up with some form of substitution as you did), so there isn't one "procedure".

Yes, your solution appears to be correct.
 
  • #3
Phys pilot said:

Homework Statement


I have to calculate the critical points of the following system.
$$x'=cx+10x^2$$
$$y'=x-2y$$

The Attempt at a Solution


So I solve the system
$$cx+10x^2=0$$
$$x-2y=0$$
If you solve your first equation for x, you should get ##x = \frac{-c \pm \sqrt{c^2}}{10} = \frac{-c \pm |c|}{10}##.
If c ≥ 0, you get one value, but if c < 0, you get another value. What all this is saying is that there are two critical points.
Phys pilot said:
So if $$x=2y$$ I have $$2yc+10*4y^2=2yc+40y^2=y(2c+40y)=0$$ and I get $$y=0$$ and $$y=-\frac{c}{20}$$ f I substitute in $$x=2y$$ I get $$x=0$$ and $$x=-\frac{c}{10}$$
Then we have that the critical points are $$(0,0)$$ and $$(-\frac{c}{10},-\frac{c}{20})$$
Is this correct? I don't know if the procedure to get the critical points is like this.
Thanks
 
  • #4
Mark44 said:
If you solve your first equation for x, you should get ##x = \frac{-c \pm \sqrt{c^2}}{10} = \frac{-c \pm |c|}{10}##.
If c ≥ 0, you get one value, but if c < 0, you get another value. What all this is saying is that there are two critical points.
Yes, I get the critical point (0,0) and then another critical point depending on the variable c so i should study the case when c is positive, negative and zero. So i will get 3 critical points in total.
thank you
P.S. The problem is that when c=0 I obtain a null eigenvalue and I can't relate this to any type of critical point
 
Last edited:
  • #5
Phys pilot said:

Homework Statement


I have to calculate the critical points of the following system.
$$x'=cx+10x^2$$
$$y'=x-2y$$

The Attempt at a Solution


So I solve the system
$$cx+10x^2=0$$
$$x-2y=0$$
So if $$x=2y$$ I have $$2yc+10*4y^2=2yc+40y^2=y(2c+40y)=0$$ and I get $$y=0$$ and $$y=-\frac{c}{20}$$ f I substitute in $$x=2y$$ I get $$x=0$$ and $$x=-\frac{c}{10}$$
Then we have that the critical points are $$(0,0)$$ and $$(-\frac{c}{10},-\frac{c}{20})$$
Is this correct? I don't know if the procedure to get the critical points is like this.
Thanks

Substituting [itex]x = 2y[/itex] into [itex]cx + 10x^2 = 0[/itex] is not necessary. You can immediately solve [itex]cx + 10x^2 = x(c + 10x) = 0[/itex] to obtain [itex]x = 0[/itex] or [itex]x = -c/10[/itex]. Then [itex]x - 2y= 0[/itex] requires that [itex]y = \frac12 x[/itex].

Thus the fixed points are at (0,0) and (-c/10, -c/20). Now when [itex]c = 0[/itex] these fixed points coincide, and there is a bifurcation.
 
  • #6
pasmith said:
Substituting [itex]x = 2y[/itex] into [itex]cx + 10x^2 = 0[/itex] is not necessary. You can immediately solve [itex]cx + 10x^2 = x(c + 10x) = 0[/itex] to obtain [itex]x = 0[/itex] or [itex]x = -c/10[/itex]. Then [itex]x - 2y= 0[/itex] requires that [itex]y = \frac12 x[/itex].

Thus the fixed points are at (0,0) and (-c/10, -c/20). Now when [itex]c = 0[/itex] these fixed points coincide, and there is a bifurcation.
Do you know how to get the axis from the ker of the jacobian matrix?
 

Related to Critical points of a diff eq system

1. What is a critical point in a differential equation system?

A critical point in a differential equation system is a point where the derivative of the system is equal to zero. This means that at a critical point, the behavior of the system is stationary and does not change over time.

2. Why are critical points important in differential equation systems?

Critical points are important because they help us to understand the behavior and stability of a differential equation system. By analyzing the critical points, we can determine if the system will converge to a stable equilibrium or exhibit chaotic behavior.

3. How do you find the critical points of a differential equation system?

To find the critical points, you need to set the derivative of the system equal to zero and solve for the variables. This will give you the coordinates of the critical points. However, in some cases, the critical points may be complex or imaginary, making it more challenging to find them.

4. Can a differential equation system have multiple critical points?

Yes, a differential equation system can have multiple critical points. In fact, some systems may have an infinite number of critical points. It is important to analyze all of the critical points to fully understand the behavior of the system.

5. How can we determine the stability of a critical point?

The stability of a critical point can be determined by analyzing the eigenvalues of the Jacobian matrix at that point. If all the eigenvalues have negative real parts, the critical point is stable and the system will converge towards it. If any of the eigenvalues have positive real parts, the critical point is unstable and the system will move away from it.

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