2
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.