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I am learning the dynamical system. I start with Taylor-Greene-Chirikov map as follow

\begin{eqnarray*} && I_{n+1} = I_n + K\sin(\theta_n), \\ && \theta_{n+1} = \theta_n + I_{n+1} \end{eqnarray*}

According to the definition of fixed point, we should have $$ I_{n+1}=I_n, \qquad \theta_{n+1}=\theta_n $$

so we could solve for fixed point from the first equation $\sin(\theta_n)=0$ and then use the solution in the second equation to find all fixed points.

Now if we consider continuous case in Hamilton equations with general form

$$\dot{q} = \partial H/\partial p, \quad \dot{p} = -\partial H/\partial q$$

where $p$ and $q$ could be a vector. For example, in the following case, we have $p=(p_1 \ p_2)$, $q=(q_1 \ q_2)$. The problem I am dealing with give the following Hamilton equations

\begin{eqnarray*} \dot{q}_1 &=& \frac{p_1-\text{} p_2 \left(\cos \left(q_2\right)+1\right)}{3-\cos \left(2 q_2\right)}\\ \dot{q}_2 &=& \frac{2 \left(p_1 \left(-\cos \left(q_2\right)-1\right)+p_2 \left(2 \cos \left(q_2\right)+3\right)\right)}{3-\cos \left(2 q_2\right)}\\ \dot{p}_1 &=& -2 \sin \left(q_1\right)-\sin \left(q_1+q_2\right)\\ \dot{p}_2 &=& \frac{2 \sin \left(2 q_2\right) \left(-2 p_2 p_1 \left(\cos \left(q_2\right)+1\right)+p_2^2 \left(2 \cos \left(q_2\right)+3\right)+p_1^2\right)}{\left(3-\cos \left(2 q_2\right)\right){}^2}-\frac{2 \left(p_1-p_2\right) p_2 \sin \left(q_2\right)}{3-\cos \left(2 q_2\right)}-\sin \left(q_1+q_2\right) \end{eqnarray*}

I got these equations from a book, where they consider the Poincare section because the dimension is 4. There they plot the section when $q_1=0$ and $p_1>0$ so we will see two fixed points. In this sense, when I solve for the fixed point, should I impose the conditions $q_1=0$ and $p_1>0$ or I should always solve it with the following equations $$ \dot{q}_1 = 0, \qquad \dot{q}_2 = 0, \qquad \dot{p}_1 = 0, \qquad \dot{p}_2 = 0 $$

I tried the second case but seems the equations is too complicated to solve for. Any suggestion?

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