# Period of Limit Cycle: Find B for Hopf Bifurcation

• unscientific
In summary: B-A-1)\ddot v = (B-A-1)\left( Av+Bu-u\right) -...- (B-A-1)\left( Av+\dot v\right) = -(Bu+Av)##. So when bifurcation occurs, the eigenvalues of the linearized jacobian matrix are imaginary, and the value of ##B = B_H## is the negative of the real eigenvalue of the jacobian matrix.In summary, when bifurcation occurs, the value of B is the negative of the real eigenvalue of the jacobian matrix.
unscientific

## Homework Statement

I'm given this system:
$$\dot x = Ax^2 y + 1 - (B+1)x$$
$$\dot y = Bx - Ax^2 y$$

(a) Find the value of B when hopf bifurcation occurs.
(b) Estimate the period of the limit cycle in terms of ##A## and ##B##.

## The Attempt at a Solution

I have found fixed point to be ##(1, \frac{B}{A})##. I have plotted the graphs of ##\dot x = 0## and ##\dot y = 0##. Does bifurcation occur at ##B_H = 0##? Because when that happens the y=intercept of the fixed point approaches ##0##.

How do I find the period of the limit cycle surrounding the trapping region?

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unscientific said:

## Homework Statement

I'm given this system:
$$\dot x = Ax^2 y + 1 - (B+1)x$$
$$\dot y = Bx - Ax^2 y$$

(a) Find the value of B when hopf bifurcation occurs.
(b) Estimate the period of the limit cycle in terms of ##A## and ##B##.

## The Attempt at a Solution

I have found fixed point to be ##(1, \frac{B}{A})##. I have plotted the graphs of ##\dot x = 0## and ##\dot y = 0##. Does bifurcation occur at ##B_H = 0##? Because when that happens the y=intercept of the fixed point approaches ##0##.

How do I find the period of the limit cycle surrounding the trapping region?

Linearise the original system by considering small deviations from the critical point, that is, replace x=1+u and y=B/A+v and solve the linearised system of equations.

ehild said:
Linearise the original system by considering small deviations from the critical point, that is, replace x=1+u and y=B/A+v and solve the linearised system of equations.

I tried that, and also tried a different approach by finding the eigenvalues of the Jacobian.

Jacobian Approach
The jacobian matrix is

Eigenvalue is ##\lambda = 0, -(B+1)##.

Linearizing approach
$$\delta \dot x = -(B+1)\delta x$$
$$\delta \dot y = B \delta x$$

You did not do the linearisation correctly. Why did you ignored y?

ehild said:
You did not do the linearisation correctly. Why did you ignored y?
Ok sorry, here's my proper working.
$$\dot u = A(1+u)^2 (\frac{B}{A}+v) - (B+1)(1+u)$$
$$\dot u = B+Av+ 2Bu - (B+1)(1+u)$$
$$\dot u = Av + u(B - 1) - 1$$

Now for ##\dot v##:
$$\dot v = B(1+u) - A(1+2u)(\frac{B}{A} + v)$$
$$\dot v = B(1+u) - A \left( \frac{B}{A} + v + 2u\frac{B}{A} \right)$$
$$\dot v = B(1+u) - B - Av - 2Bu$$
$$\dot v = -(Bu + Av)$$

Now I try to find solve the equation entirely in terms of ##u, \dot u, \ddot u##:
$$\ddot u = A\dot v + \dot u(B-1)$$
$$\ddot u = -A(Bu+Av) + \dot u(B-1)$$

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unscientific said:
Ok sorry, here's my proper working.
$$\dot u = A(1+u)^2 (\frac{B}{A}+v) - (B+1)(1+u)$$
$$\dot u = B+Av+ 2Bu - (B+1)(1+u)$$
$$\dot u = Av + u(B - 1) - 1$$
Check. That -1 should not be there.

ehild said:
Check. That -1 should not be there.
=
Yep that's right. I missed out ##+1## in the first line. The correct equations should now read:
$$\dot u =Av + Bu - u$$
$$\dot v = -(Av + Bu)$$

Solving for ##v##, I get
$$\ddot v = -B\left( Av + Bu - u \right) - A\dot v$$
$$\ddot v = +B\dot v + Bu - A\dot v$$
$$\ddot v =B\dot v - \left( Av + \dot v \right) - A\dot v$$
$$\ddot v = (B-A-1)\dot v - Av$$

Does bifurcation occur at ##B_H = 0##? Because when B=0, the trapping region disappears, i.e. no limit cycle.

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Bifurcation is at the fix point if the eigenvalues of the linearised equation are imaginary. So what should be the solutions for v?
End check the signs in the last equation.

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ehild said:
Bifurcation is at the fix point if the eigenvalues of the linearised equation are imaginary. So what should be the solutions for v?
But from the jacobian I have found the eigenvalue to be ##\lambda = 0## or ##\lambda = -(B+1)##. For positive values of B, the eigenvalue has to be real.

For ##\ddot v = (B-A-1)\dot v + Av ##, I try ##v \propto e^{i\omega t}##. We have
$$-\omega^2 = i\omega(B-A-1) + A$$

But it should be periodic. No damping term. What should be B?
unscientific said:
But from the jacobian I have found the eigenvalue to be ##\lambda = 0## or ##\lambda = -(B+1)##. For positive values of B, the eigenvalue has to be real.

For ##\ddot v = (B-A-1)\dot v + Av ##, I try ##v \propto e^{i\omega t}##. We have
$$-\omega^2 = i\omega(B-A-1) + A$$

Forget the Jacobian, Go ahead with your method and check the equation for ##\ddot v ## .

ehild said:
But it should be periodic. No damping term. What should be B?Forget the Jacobian, Go ahead with your method and check the equation for ##\ddot v ## .

The first part of this question actually wants us to find the value of ##B = B_H## when bifurcation occurs. How do I find when bifurcation occurs? The eigenvalues of the linearized jacobian matrix are never imaginary.

I checked my working again, I got ## \ddot v = (B-A-1)\dot v - Av##.

You are right, it is correct. What is the solution for v(t)?

And how did you get that Jacobian? It looks wrong.

ehild said:
You are right, it is correct. What is the solution for v(t)?

And how did you get that Jacobian? It looks wrong.

Thanks for working with me on this. I got the jacobian by finding the first row using ##\frac{\partial \dot x}{\partial x }= -(B+1)## and ##\frac{\partial \dot x}{\partial y} = 0##, and the second row using ##\frac{\partial \dot y}{\partial x } = B## and ##\frac{\partial \dot y}{\partial y } = 0##. Linearized. I should linearize the jacobian to find the eigenvalues, right? All the questions I have been doing so far are like that.

Using ANSATZ ##v \propto e^{-\omega t}## we have
$$-\omega^2 = i\omega(B-A-1) - A$$

So for it to be real, we require ##\omega = \sqrt A = \sqrt {B-1}##?

unscientific said:
Thanks for working with me on this. I got the jacobian by finding the first row using ##\frac{\partial \dot x}{\partial x }= -(B+1)## and ##\frac{\partial \dot x}{\partial y} = 0##, and the second row using ##\frac{\partial \dot y}{\partial x } = B## and ##\frac{\partial \dot y}{\partial y } = 0##. Linearized. I should linearize the jacobian to find the eigenvalues, right? All the questions I have been doing so far are like that.

Using ANSATZ ##v \propto e^{-\omega t}## we have
$$-\omega^2 = i\omega(B-A-1) - A$$

So for it to be real, we require ##\omega = \sqrt A = \sqrt {B-1}##?
##\frac{\partial \dot x}{\partial y} = 0## is wrong.

Working with the Jacobian matrix yields the same result as the method you started to follow. Go ahead with it.
By the way, you should use the Ansatz
##v \propto e^{i \omega t}##

ehild said:
##\frac{\partial \dot x}{\partial y} = 0## is wrong.

Working with the Jacobian matrix yields the same result as the method you started to follow. Go ahead with it.
By the way, you should use the Ansatz
##v \propto e^{i \omega t}##
For ##\ddot v = (B-A-1)\dot v - Av## and ##v \propto e^{i\omega t}##, we have the same result:
$$-\omega^2 = i\omega (B-A-1) - A$$

So I shouldn't linearize the jacobian from the start? So we have ##\frac{\partial \dot x}{\partial x} = 2Axy - (B+1)## and ##\frac{\partial \dot x}{\partial y} = Ax^2## and ##\frac{\partial \dot y}{\partial x} = B - 2Axy## and ##\frac{\partial \dot y}{\partial y} = -Ax^2##. This gives the jacobian matrix to be:

Expanding out to find eigenvalues, we have
$$\lambda^2 + \lambda(A-B+1) + A = 0$$
For the eigenvalues to be imaginary, we require
$$(A-B+1)^2-4A = 0$$
$$A - B + 1 = \pm 2\sqrt A$$
$$B_H = A + 1 \pm 2\sqrt A$$

unscientific said:
For ##\ddot v = (B-A-1)\dot v - Av## and ##v \propto e^{i\omega t}##, we have the same result:
$$-\omega^2 = i\omega (B-A-1) - A$$

So I shouldn't linearize the jacobian from the start? So we have ##\frac{\partial \dot x}{\partial x} = 2Axy - (B+1)## and ##\frac{\partial \dot x}{\partial y} = Ax^2## and ##\frac{\partial \dot y}{\partial x} = B - 2Axy## and ##\frac{\partial \dot y}{\partial y} = -Ax^2##. This gives the jacobian matrix to be:
Substitute x=1 and y=B/A
unscientific said:
Expanding out to find eigenvalues, we have
$$\lambda^2 + \lambda(A-B+1) + A = 0$$
For the eigenvalues to be imaginary, we require
$$(A-B+1)^2-4A = 0$$
No, it is not true. Use the quadratic formula. When do you get pure imaginary solutions?

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ehild said:
Substitute x=1 and y=B/A

No, it is not true. Use the quadratic formula. When do you get pure imaginary solutions?

For pure imaginary solutions, we require ##\frac{-b}{2a} = 0## and ##b^2-4ac <0##, so that means ##B=A+1##.

Therefore the displacement equation becomes ##\dot v = - Av##. ##T = \frac{2\pi}{\sqrt A} =\frac{\pi}{\sqrt{B-1}}##.

unscientific said:
For pure imaginary solutions, we require ##\frac{-b}{2a} = 0## and ##b^2-4ac <0##, so that means ##B=A+1##.

Therefore the displacement equation becomes ##\dot v = - Av##.
You meant ##\ddot v = - Av##
Otherwise, it is correct now.

unscientific
And A should be positive.

ehild said:
And A should be positive.
Yep, it was stated in the question that both B and A are positive numbers.

You forgot to write it in the OP.

ehild said:
You forgot to write it in the OP.
Sorry. One last question - how do we evaluate the stability of the fixed point?

## 1. What is the period of a limit cycle?

The period of a limit cycle refers to the time it takes for a system to complete one full cycle of oscillation or periodic behavior. It is often denoted as T and can be measured in seconds, minutes, or any other unit of time depending on the specific system being studied.

## 2. How is the period of a limit cycle related to Hopf bifurcation?

Hopf bifurcation is a type of bifurcation in which a stable fixed point becomes unstable and a stable limit cycle emerges. The period of this limit cycle is directly related to the Hopf bifurcation, as it is the time it takes for the system to complete one full cycle of oscillation after the bifurcation occurs.

## 3. What is B in the context of Hopf bifurcation?

In the context of Hopf bifurcation, B represents a bifurcation parameter. This parameter determines the stability of the system and can be varied to observe changes in the behavior of the system, such as the emergence of a limit cycle. For a Hopf bifurcation to occur, the value of B must fall within a certain range.

## 4. How do you find B for Hopf bifurcation?

To find B for Hopf bifurcation, one must analyze the stability of the system as the bifurcation parameter is varied. This can be done through mathematical calculations or by creating a bifurcation diagram, which shows the behavior of the system for different values of B. The value of B at which a Hopf bifurcation occurs can be determined from this analysis.

## 5. What are the implications of a Hopf bifurcation?

A Hopf bifurcation has important implications for the behavior of a system. It marks a critical point at which the system transitions from a stable fixed point to a stable limit cycle. This can lead to complex and unpredictable behavior, making it an important concept in the study of nonlinear systems and chaos theory.

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