Consider, for $n \in \mathbb{N}$ and $n \geq 1$, the 2 dimensional system: $$ \ddot{x}+x^n=0 \quad or \quad \left\{ \begin{array}{c} \dot{x} = y \\ \dot{y} = -x^n\end{array} \right. $$ This system defines a flow $\phi (t;x,y): \mathbb{R}\times\mathbb{R}^2 \rightarrow \mathbb{R}^2$, parametrised by time t. Solutions of the system with initial condition $(x_0,y_0)$ are denoted by $(x(t;x_0,y_0),y(t;x_0,y_0))$, orbits of the flow are denoted by $\Gamma (x_0, y_0)$. Assume $n$ is even at this stage.

I need to define a set $S_+ \subset \mathbb{R}^2$ of all initial conditions such that the limit of $t\rightarrow \infty$ for solutions $(x(t;x_0,y_0),y(t;x_0,y_0))$ of the system with $(x_0,y_0) \in S_+$ exists, and to determine $S_+$ explicitly. Finally, I must do the same for the set $S_-$, this time for the limit of $t\rightarrow -\infty$.

I have no idea where to start with this particular question. Looking through my textbook (Differential Dynamical Systems by J.D. Meiss), I find no mention of these sets of initial conditions $S_\pm$, and thus am a little stuck and confused. Any guidance is extremely welcome!

If it helps, I have determined the Hamiltonian of this system to be $$H(x,y) = \frac{y^2}{2} + \frac{x^{n+1}}{n+1}$$

Thanks in advance!


The trajectories are level curves of the Hamiltonian. If $n$ is odd, these level curves are closed curves, so the solutions are periodic: they don't have limits unless they are constant. If $n$ is even, both $x$ and $y$ go to $\pm \infty$ with a couple of exceptions...

  • $\begingroup$ Thanks for the response! So I can successfully plot the countours of the Hamiltonian, and as expected they follow the shape of the phase portrait of the initial system. However, I'm not sure I fully understand what the relation is between these contours and the "sets of all initial conditions S+ and S-", as I need to determine them explicitly, and eventually plot them on the phase portrait. $\endgroup$ – Azog4 Jul 9 at 11:58

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