Question: The system
$$ \dot{x} = 3y + 3x^3 + xy $$
$$\dot{y} = -3x + \mu y + 2xy^2 – y^3$$
undergoes a Hopf bifurcation at $(x, y) = (0, 0)$ as $\mu$ passes through 0. Calculate the approximated period of the periodic orbit created. Derive an approximate equation for the periodic orbit for $\mu$ small.
Context: For a related question, I placed the system in the form
$$\dot{x} = -\omega y + f(x, y)$$
$$\dot{y} = \omega x + g(x, y)$$
where $\omega = -3$, $f(x, y) = 3x^3 + xy$ and $g(x,y) = \mu y + 2xy^2 – y^3$ to calculate partial derivatives of $f$ and $g$ and hence the analytical criterion to show that the Hopf bifurcation is subcritical. Note I'm not sure if this is relevant to the question above.
The eigenvalues of the system are
$$\lambda_\pm = \frac{\mu \pm \sqrt{\mu^2 – 6^2}}{2} $$
which, for small $\mu$ becomes,
$$\lambda_\pm = \frac{\mu \pm i\sqrt{|6^2 – \mu^2|}}{2}.$$
Attempt at a solution:
Approximate the system as $\dot{x} = – \omega y$ and $\dot{y} = \omega x$. Then
$$ \frac{dy}{dx} = \frac{\dot{y}}{\dot{x}} = -\frac{x}{y} \;\text{ so }\; \frac{1}{2} \frac{d y^2}{dy} \frac{dy}{dx} = -x.$$
Integrating both side with respect to $x$ and setting $C \in \mathbb{R}$,
$x^2 + y^2 = C$.
Near the Hopf bifurcation the period of oscillation is
$$\frac{2 \pi}{\sqrt{|6^2 – \mu^2| / 2}} = \frac{\pi}{\sqrt{|6^2 – \mu^2|}}. $$
I'm not sure how to find $C$ or $\mu$, or if instead they are meant to remain variables.
What is the relevance of $\mu$ being small? In general should this be $\mu \approx \mu_c$ where $\mu_c$ is the bifurcation point? I suspect $\mu$ must be close to $\mu_c$ so that the linear terms dominate.
Best Answer
Using the analytical criterion mentioned above, we compute $a = 12$. Since $a > 0$ we have a subcritical (unstable) Hopf bifurcation at the fixed point $(x, y) = (0, 0)$. Also note that $\omega = -3$.
For small $\mu$, close to the fixed point, we can approximate the system as a pitchfork bifurcation with respect to the radius:
$$\dot{r} = \mu r + a r^3 = \mu r + 12r^3.$$ $$\dot{\theta} = \omega = -3$$
Hence there is a limit cycle at $r = \sqrt{\mu} / (2 \cdot \sqrt{3})$ with period $2 \pi / (|-3|) = (2/3) \cdot \pi$. In the coordinate system of the normal form, this becomes
$$ \dot{x} = \frac{\sqrt{\mu}}{2\sqrt{3}}\cos\left(\frac{-2}{3}\pi(t - t_0)\right)$$ $$\dot{y} = \frac{\sqrt{\mu}}{2\sqrt{3}}\sin\left(\frac{-2}{3}\pi(t - t_0)\right)$$ so that in the coordinate system of the question (undoing the transform $y \mapsto -y$) we have
$$ \dot{x} = \frac{\sqrt{\mu}}{2\sqrt{3}}\cos\left(\frac{2}{3}\pi(t - t_0)\right)$$ $$\dot{y} = \frac{\sqrt{\mu}}{2\sqrt{3}}\sin\left(\frac{2}{3}\pi(t - t_0)\right)$$