# ODE system. Limit cycle; Hopf bifurcation.

#### Jason

##### New member
Problem:

The following two-dimensional system of ODEs possesses a limit-cycle solution for certain values of the parameter$a$. What is the nature of the Hopf bifurcation that occurs at the critical value of $a$ and state what the critical value is.

$\dot{x}=-y+x(a+x^2+(3/2)y^2)$

$\dot{y}=x+y(a+x^2+(3/2)y^2)$

By setting each equation to zero, i found the only equilibrium point to be $(0,0)$.

For the Jacobian matrix at $(0,0)$, I have:

$J(0,0)=\left( \begin{array}{cc} a & -1\\ 1 & a\end{array} \right)$

So:

$\tau=2a$

$\delta=a^2+1>0$

$\bigtriangleup=-4<0$

which gives:

$a<0$: $(0,0)$ is an attractor spiral.

$a>0$: $(0,0)$ is a repellor spiral.

$a=0$: $(0,0)$ is a center.

Does this mean that paths spiral into $(0,0)$ for negative $a$, and then spiral out towards a stable limit cycle? And, for positive $a$ beyond the limit cycle, paths spiral in towards it?

How do I find the critical value?

#### Jester

##### Well-known member
MHB Math Helper
If you go to polar coordinates you can really see what's going on

$\dot{r} = r\left(a + x^2 + \dfrac{3}{2}y^2\right),\;\;\; \dot{\theta} = 1$.

• Jason

#### Alexmahone

##### Active member
Does this mean that paths spiral into $(0,0)$ for negative $a$, and then spiral out towards a stable limit cycle?
Paths outside the limit cycle spiral outwards, whereas paths inside the limit cycle spiral into $(0,\ 0)$.

And, for positive $a$ beyond the limit cycle, paths spiral in towards it?
When $a$ is positive, there is no limit cycle.

How do I find the critical value?
Clearly, the critical value of $a$ is 0.

Last edited:

#### Jason

##### New member
Yeah I see I got that backwards.

So when $a$ is positive, paths spiral out from $(0,0)$. When $a$ is negative, paths spiral in toward $(0,0)$, and the limit cycle is unstable?

#### Alexmahone

##### Active member
So when $a$ is positive, paths spiral out from $(0,0)$. When $a$ is negative, paths spiral in toward $(0,0)$, and the limit cycle is unstable?
Yes, the limit cycle is unstable. (See my previous post.)

• Jason