4

The question is for which $a \in \mathbb{R}$ the system:
$x'' + x - x^3 + a = 0$,
has a homoclinic orbit.
I let $y = x'$ so the system becomes:
$x' = y$
$y' = x^3 - x - a$
and determined the Hamiltonian as $H(x,y) = \frac{1}{2}(x^2 + y^2) - \frac{1}{4} x^4 + ax$.
The problem is that I don't know much about the fixed points of this system.
Does anyone have an idea to get me started? Thanks.

MrReese
  • 719
  • 1
    Where are the points where $x' = 0$ and $y' = 0$ simultaneously? – Amzoti Dec 16 '13 at 21:16
  • The problem is that I don't know when $y' = x^3 - x - a = 0$ – MrReese Dec 16 '13 at 21:26
  • http://en.wikipedia.org/wiki/Cubic_function#General_formula_for_roots. They are found in terms of $a$. However, that is a bit nasty looking. You could a qualitative analysis for $a = -2,-1, 0, 1, 2$ and explore the system that way by drawing phase portraits and you clearly see what is going on. – Amzoti Dec 16 '13 at 21:29

0 Answers0