Period of Limit Cycle: Find B for Hopf Bifurcation

[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 $B$.

Homework Equations

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?

 

[ehild]

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 $B$.

Homework Equations

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.

 

[unscientific]

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

 

[ehild]

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

 

[unscientific]

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 $u, \dot u, \ddot u$:
$$\ddot u = A\dot v + \dot u(B-1)$$
$$\ddot u = -A(Bu+Av) + \dot u(B-1)$$

 

[ehild]

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.

 

[unscientific]

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.

 

[ehild]

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.

 

[unscientific]

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
$$-\omega^2 = i\omega(B-A-1) + A$$

 

[ehild]

But it should be periodic. No damping term. What should be B?

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 $\ddot v$ .

 

[unscientific]

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 $\ddot v = (B-A-1)\dot v - Av$.

 

[ehild]

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

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

 

[unscientific]

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 $\omega = \sqrt A = \sqrt {B-1}$?

 

[ehild]

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
$v \propto e^{i \omega t}$

 

[unscientific]

$\frac{\partial \dot x}{\partial y} = 0$ 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
$$(A-B+1)^2-4A = 0$$
$$A - B + 1 = \pm 2\sqrt A$$
$$B_H = A + 1 \pm 2\sqrt A$$

 

[ehild]

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

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?

 

[unscientific]

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 $\dot v = - Av$. $T = \frac{2\pi}{\sqrt A} =\frac{\pi}{\sqrt{B-1}}$.

 

[ehild]

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.

 

[ehild]

And A should be positive.

 

[unscientific]

And A should be positive.

Yep, it was stated in the question that both B and A are positive numbers.

 

[ehild]

You forgot to write it in the OP.

 

[unscientific]

You forgot to write it in the OP.

Sorry. One last question - how do we evaluate the stability of the fixed point?

 

[ehild]

Sorry. One last question - how do we evaluate the stability of the fixed point?

Be more specific. In this problem, or in general?

http://www.scholarpedia.org/w/image...d.gif/500px-Equilibrium_figure_summary_2d.gif

 

[unscientific]

Be more specific. In this problem, or in general?

http://www.scholarpedia.org/w/image...d.gif/500px-Equilibrium_figure_summary_2d.gif

Thanks for that link. I suppose solving for eigenvalues, then inputting $(x,y) = (1,\frac{B}{A})$ to find the stability conditions.