14 & 15 July, 2026
To do!
Poincaré Map
The Poincaré Map, that brings elements of a Poincaré section \(T:\sufsec \rightarrow \sufsec\).
From our equations of motion, that I numerically compute I can find an expression for this map. We derive by numerical integration the approximation of the trajectory \(\mathbf{x}(\mathbf{x}_0, t)\). The trajectory is a continuouse curve, which gets approximated by a set of discrete positions \(\mathbf{x}(\mathbf{x}_0, t)\approx\{\mathbf{x}(\mathbf{x}_0, 0),\dots, \mathbf{x}(\mathbf{x}_0, t'),\mathbf{x}(\mathbf{x}_0, t'+\delta t),\dots\}\). The For this we define a numerical integration, the integration over small steps \(\delta t\). Together with iteration function \(\mathbf{iter}:\R^n \times I \rightarrow \R^n\times I\) which gives the next position
\[\mathbf{iter}(\mathbf{x}(t'),\delta t)=\mathbf{x}(t'+\delta t).\]
Allowing the set approximating the trajectory to take the form of \(\mathbf{x}(\mathbf{x}_0, t)\approx\{\mathbf{iter}^k(\mathbf{x}_0) | \,k\in\N\}\).
The integration with the iteration represents the numerical integration method, in the abstract (i.e. the Runge–Kutta algorithm), every iteration results in a position of the trajectory of the initial conditions. The integration function for an small timestep \(\delta t\) reads
\[\mathbf{I}(\mathbf{x}(t'),\delta t) = \int_{t'}^{t'+\delta t} \mathbf{f}(\mathbf{x}(t))dt,\]
Thus the iteration function becomes
\[\mathbf{iter}(\mathbf{x}(t'),\delta t)= \big(\mathbb{1}+\mathbf{I}(\cdot,\delta t)\big)(\mathbf{x}(t))=\mathbf{x}(t')+\mathbf{I}(\mathbf{x}(t'),\delta t)=\mathbf{x}(t'+\delta t).\]
So how does the k-fold composition of \(\mathbf{iter}^k\) look like, i.e. \(k=2\) \(\mathbf{iter}^2=\mathbf{iter}\circ\mathbf{iter}\)
\[\begin{align} (\mathbf{iter}\circ \mathbf{iter})(\mathbf{x}(t'), \delta t)&= \mathbf{iter}\Big(\mathbf{iter}\big(\,\mathbf{x}(t'),\delta t\big), \delta t \Big)\\ &=\mathbf{iter}\Big(\,\mathbf{x}(t')+\mathbf{I}\big(\,\mathbf{x}(t'),\delta t\big), \delta t\Big)\\ &= \underbrace{\mathbf{iter}(\mathbf{x}(t'))}_{=\mathbf{x}(t'+\delta t)} + \mathbf{iter}\Big(\,\mathbf{I}\big(\mathbf{x}(t'),\delta t\big), \delta t\Big) \\ &=\underbrace{\mathbf{x}(t')+\mathbf{I}\big(\mathbf{x}(t'),\delta t\big) }_{=\mathbf{x}(t'+\delta t)} + \mathbf{I}\big(\mathbf{x}(t'), \delta t\big) + (\mathbf{I}\circ\mathbf{I})(\mathbf{x}(t'),\delta t) \\ &= \mathbf{x}(t'+\delta t) + \mathbf{I}\big(\mathbf{x}(t'), \delta t\big) + \mathbf{I}\Big(\,\mathbf{I}\big(\,x(t'),\delta t\big), \delta t\Big) \\ &= \mathbf{x}(t'+\delta t) + \mathbf{x}(t'+\delta t) - \mathbf{x}(t') + \mathbf{I}\Big(\,\mathbf{x}(t'+\delta t)-\mathbf{x}(t'),\delta t\Big) \\ &= \mathbf{x}(t'+\delta t) + \mathbf{x}(t'+\delta t) - \mathbf{x}(t') + \mathbf{I}\Big(\,\mathbf{x}(t'+\delta t) -\mathbf{x}(t')\Big) \\ &= \mathbf{x}(t'+\delta t) + \mathbf{x}(t'+\delta t) - \mathbf{x}(t') + \mathbf{I}\Big(\,\mathbf{x}(t'+\delta t)\Big) -\mathbf{I}\Big(\mathbf{x}(t') \Big) \\ &= \mathbf{x}(t'+\delta t) + \mathbf{x}(t'+\delta t) - \mathbf{x}(t') + \mathbf{x}(t'+2\delta t)-\mathbf{x}(t'+\delta t) -\Big(\mathbf{x}(t'+\delta t)-\mathbf{x}(t') \Big) \\ &= \mathbf{x}(t'+2\delta t) \end{align}\]
With this I can define the Poincaré Map
\[T(\mathbf{x})= \mathbf{iter}^k(\mathbf{x}(t_0))\]