Toronto Math Forum

Quote from: Victor Ivrii on April 04, 2013, 12:04:28 AM

Benny, nice try, but ...

Devangi – good, Frank—even better. But phase portrait is still missing (and bonus question too)

This is an integrable system: denoting $y=x'$ we get
\begin{equation*}
H(x,y) := \frac{1}{2}y^2 \underbrace{-\frac{1}{5}x^5+\frac{5}{3}x^3 -4x}_{V(x)}=E
\end{equation*}
wehre $V(x)=-\int f(x)\,dx$, $f(x)=x^4-5x^2+4$ is the r.h.e.  $V(x)$ is a potential.

Since $V(x)$ has non-degenerate maxima at $x=-1$ and $x=2$ and minima at $x=-2$ and $x=1$ and $\frac{1}{2}y^2$ has non-degenerate minimum at $y=0$ we have 4 non-degenerate critical points of $H(x,y)$: namely, $(-1,0)$ and $(2,0)$ are saddle points and $(-2,0)$ and $(1,0)$ are minima.

For dynamics two former are saddle points and two latter re centers (recall that integrable systems cannot have spiral or nodal points and centers are detectable!)

What is missing for everyone? Plot of $V(x)$ which shows that $ V(-1)>V(-2)>V(2)>V(1)$ (one needs just calculates them) and therefore picture must be like on Jason' computer generated and not Yeong' drawing since separatrix passing through saddle $(-1,0)$ is "higher" and therefore envelops saddle $(2,0)$ and cannot pass through it as Yeong drew. And therefore separatrix passing through $(2,0)$ stops short and goes back to the right from $(-1,0)$ (not passing through  it). If in some examples we studied before separatrices passing through different saddles are the same it is not the general rule.

To analyze Jason's picture:

a) Look at the centers
b) Find separatrix passing through $(2,0)$
c) Find separatrix passing through $(-1,0)$