The system of equations I am working on is the following:
$$
\begin{align}
\dot{x}& = y \\
\dot{y}& = -\mu(x^2 + ax^4 – 1)y – x
\end{align}
$$
The question asks first to find the Hamiltonian of the system at $\mu = 0$ which I evaluate to be:
$$
\begin{equation}
H(x,y) = \frac{y^2 – x^2}{2}
\end{equation}
$$
The proceeding parts are the ones I am having trouble with. This is the body of the question:
"Find an explicit expression for the trajectories solving this Hamiltonian
system. Choose suitable initial conditions having in mind
that in the next step you will integrate over a closed orbit."
Is this done by solving the system of equations when $ \mu = 0$? Doing so with initial conditions x(0) = 1 and y(0) = 1 results in:
$$
\begin{align}
x(t)& = cos(t) + sin(t)\\
y(t)& = cos(t) – sin(t)
\end{align}
$$
After plugging this into the Hamiltonian and simplifying I arrive at:
$$
\begin{equation}
H(x,y) = -sin(2t)
\end{equation}
$$
The next part which involves using the Melnikov method is as follows:
"Now consider a small positive $\mu$. To determine the radius R and
period time T to lowest (zeroth) order in $\mu$, evaluate the change
$\Delta H$ in the Hamiltonian H(x; y) as you follow a trajectory governed
by the dynamics in the system of equations. Show that H must vanish
after following a limit-cycle trajectory one lap. Use this fact to find the radius R as a function of a."
When H completes one lap I assume that it goes from 0 to $\pi$ and $H = 0$ at $t = \pi$. I am not sure how to find the radius and the period after this step. For example, I don't know how to handle $\mu$ being non-zero.
Could someone give me some pointers as to how this should be done or if what I have done so far is correct?
Best Answer
Let $\mu=0$ and then the system becomes a Hamiltonian system $$ x'=y, y'=-x $$ with $$ H(x,y)=\frac12(x^2+y^2) $$ Clearly this Hamiltonian system has an orbit ${\bf{x}}_\gamma(t)=(\gamma \cos t,y=-\gamma \sin t)$ with period $2\pi$. So the Melnikov function is \begin{eqnarray} M(\gamma,a)&=&\int_0^{2\pi}{\bf{f(x_\gamma)}\wedge\bf{g(x_\gamma)}}dt\\ &=&-\int_0^{2\pi}\gamma \sin t(\gamma^2\cos^2t+a\gamma^4\cos^4t-1)\gamma\sin tdt\\ &=&-\frac{\pi}8\gamma(-8+2\gamma^2+a \gamma^4). \end{eqnarray} Let $$ -8+2\gamma^2+a \gamma^4=0 $$ and then $$ \gamma=γ_\pm=\sqrt{\frac{-1\pm\sqrt{1+8a}}a}=\sqrt{\frac8{1\pm\sqrt{1+8a}}}. $$ This means that if $a>0$, then there is a unique limit cycle with radius $\gamma=\gamma_++O(\mu)$. If $a\in(-\frac18,0)$, then there are two limit cycles with radius $\gamma=\gamma_{\pm}+O(\mu)$. If $a<-\frac18$, there is no limit cycle. If $a=0$, there is a unique limit cycle with radius $\gamma=2+O(\mu)$.