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[SOLVED] Hopf bifurcation

dwsmith

Well-known member
Feb 1, 2012
1,673
Question in red at the bottom
For $\mu\ll 1$, there are two time scales.
\begin{alignat*}{3}
x' & = & \mu x - y - xy^2\\
y' & = & x + \mu y - y^3
\end{alignat*}
Let's expand $x,y$.
\begin{alignat*}{3}
x & = & \mu^{1/2}x_0 + \mu^{3/2}x_1 + \cdots\\
y & = & \mu^{1/2}y_0 + \mu^{3/2}y_1 + \cdots
\end{alignat*}
Then we can re-write the plane autonomous system as
\begin{alignat*}{3}
\left(\frac{\partial}{\partial t} + \mu\frac{\partial}{\partial T}\right)(\mu^{1/2}x_0 + \mu^{3/2}x_1 + \cdots) & = & \mu(\mu^{1/2}x_0 + \mu^{3/2}x_1 + \cdots) - (\mu^{1/2}y_0 + \mu^{3/2}y_1 + \cdots)\\
& - & (\mu^{1/2}x_0 + \mu^{3/2}x_1 + \cdots)(\mu^{1/2}y_0 + \mu^{3/2}y_1 + \cdots)^2\\
\left(\frac{\partial}{\partial t} + \mu\frac{\partial}{\partial T}\right)(\mu^{1/2}y_0 + \mu^{3/2}y_1 + \cdots) & = & (\mu^{1/2}x_0 + \mu^{3/2}x_1 + \cdots) + \mu(\mu^{1/2}y_0 + \mu^{3/2}y_1 + \cdots)\\
& - & (\mu^{1/2}y_0 + \mu^{3/2}y_1 + \cdots)^3
\end{alignat*}
Now we can group the terms in like orders
$$
\text{order } \mu^{1/2}: \begin{cases}
x_{0t} = -y_0\\
y_{0t} = x_0
\end{cases}
$$
and
$$
\text{order } \mu^{3/2}: \begin{cases}
x_{1t} + y_1 = -x_{0T} + x_0 - x_0y^2_0\\
y_{1t} - x_1 = -y_{0T} + y_0 - y^3_0
\end{cases}
$$
Then we have that
\begin{alignat*}{3}
x_0(t,\mu) & = & r(T)\cos\theta\\
y_0(t,\mu) & = & r(T)\sin\theta
\end{alignat*}
where $\theta = t + \theta(T)$.
From order $\mu^{3/2}$, we have that $y_1 = -x_{0T} + x_0 - x_0y^2_0 - x_{1t}$.
Then
$$
y_{1t} = [-x_{0T} + x_0 - x_0y^2_0 - x_{1t}]_t.
$$
So we can re-write $y_{1t} - x_1 = -y_{0T} + y_0 - y^3_0$ as
\begin{alignat*}{3}
[-x_{0T} + x_0 - x_0y^2_0 - x_{1t}]_t - x_1 & = & -y_{0T} + y_0 - y^3_0\\
-x_{0Tt} + x_{0t} - x_{0t}y^2_0 - x_0y^2_{0t} - x_{1tt} - x_1 & = & -y_{0T} + y_0 - y^3_0\\
x_{1tt} + x_1 & = & -x_{0Tt} + x_{0t} - x_{0t}y^2_0 - x_0y^2_{0t} + y_{0T} - y_0 + y^3_0\\
& = & 2r_T\sin\theta + 2r\theta_T\cos\theta - 2r\sin\theta + 2r^3\sin^3\theta - 2r^3\sin\theta\cos^2\theta\\
& = & \sin\theta\left(2r_T - 2r + r^3\right) + 2r\theta_T\cos\theta + \text{other terms}
\end{alignat*}
In order to suppress resonance, we must have
\begin{alignat*}{3}
r_T & = & r - \frac{1}{2}r^3\\
\theta_T & = & 0
\end{alignat*}
So we have that
\begin{alignat*}{3}
x & = & \mu^{1/2}r(T)\cos\theta + \mathcal{O}(\mu^{3/2})\\
y & = & \mu^{1/2}r(T)\sin\theta + \mathcal{O}(\mu^{3/2})
\end{alignat*}
If $R = \sqrt{x^2 + y^2}$, then $R = \mu^{1/2}r$.
So $r = \frac{R}{\mu^{1/2}}$.
\begin{alignat*}{3}
\frac{R_T}{\mu^{1/2}} & = & \frac{R}{\mu^{1/2}} - \frac{1}{2}\frac{R^3}{\mu^{3/2}}\\
R_T & = & R - \frac{1}{2\mu}R^3
\end{alignat*}

But I am supposed to have
$$
R_T = \mu R -\frac{1}{2}R^3.
$$
What is wrong?
 
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