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Given $$\frac{dA}{d\tau}=\sigma A-\beta A|A|^2, $$ where $\sigma=\sigma_r+i\sigma_i$, $\beta$ is real and $A(\tau)=\rho(\tau)\exp(i\theta(\tau))$.

Draw the orbits in the {$\rho,\theta$}-plane and describe the evolution in each cases($\sigma_r$ and $\beta$ positive and negative), classfying subcritical and supercritical if appropriate.

Here is what I have done so far: Substitute all into the ODE we have two equations: $$\rho_\tau=\sigma_r \rho-\beta\rho^3 $$

$$\theta_\tau=\sigma_i.$$ The first one looks like a pitchfork bifurcation with parameter $\sigma_r/\beta$. I am not sure what to do next. Can anyone help?enter image description here

DDaren
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1 Answers1

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You will have to study all cases. For example, when $\beta>0$, we consider $\sigma_r\rho-\beta\rho^3=0$. So $\rho(\sigma_r-\beta\rho^2)=0$. This is a cubic curve which goes to $\infty$ on the left and $-\infty$ on the right. It always passes through $0$. The other root depends on the sign of $\sigma_r$. If $\sigma_r<0$, there's no other roots. If $\sigma_r>0$, the other two roots are $\pm\sqrt{\frac{\sigma_r}{\beta}}$.

Draw the bifurcation diagram as follows: enter image description here

The arrows indicate the direction of flow of $\rho$ on either sides of the fixed points. The solid dot means stable fixed point, and hollow dot means unstable fixed point.

You can see this is the case of supercritical bifurcation.

Now to draw the orbit in $\rho,\theta$ plane, again separate these three different cases. In the first case where $\beta>0,\sigma_r<0$, there is one stable fixed point for $\rho$, which is zero. Notice that there is no fixed point for $\theta$, and it has a constant rate. As in your edit, the orbits should be like spirals toward the origin.

You can try all the other cases in a similar way. Be careful when there are more than one fixed points. The orbits would look different.

KittyL
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  • Could you explain how you draw the bifurcation diagram? Because I have not done anything bifurcation problem with two parameters. – DDaren May 27 '15 at 14:19
  • @DDaren: For two parameters, we have to fix one parameter and let the other one move. The diagram in my answer is when $\beta>0$ is fixed. I will explain a little bit more. – KittyL May 27 '15 at 14:25
  • Thank you. Now I understand the first part. I am a bit confused about the second diagram. Shouldn't the orbit be the diagram illustrated in my updated post? Because it rotates in constant speed (say counterclockwise) and $\rho$ is decreasing to zero. – DDaren May 27 '15 at 15:14
  • @DDaren: You are right. Sorry for the confusion. – KittyL May 27 '15 at 15:46