Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 3 / 14A (Dynamical Systems)

For a positively oriented simple closed curve on which the vector field is nonzero, its Poincaré index is the total rotation of the vector field along that curve divided by $2\pi$, equivalently the winding number of its normalized direction. The Poincaré index of an isolated fixed point is that of a sufficiently small enclosing curve containing no other fixed point. The index of a planar periodic orbit is $+1$.

The equilibrium points are $(0,0)$ and $(\pm 1,\pm (b-a))$, with correlated signs. The Jacobian matrix is $\left(\begin{array}{cc}a-3b{x}^2& 1\\ 3x^2-1& 0\end{array}\right)$. At the origin its eigenvalues satisfy ${\lambda }^2-a\lambda +1=0$. It is a focus for $0<|a|<2$, a repeated-eigenvalue node for $|a|=2$, and a node for $|a|>2$; stability is determined by the sign of $a$, with stability for $a<0$. Its Poincaré index is $+1$. At either other equilibrium the determinant is $-2$, so it is a saddle point with Poincaré index $-1$.

Take $H(x,y)=y^2/2+x^2/2-x^4/4$. Along a trajectory,Integrating over a periodic orbit returns $H$ to its starting value, and henceThis is the energy obstruction for a cubic planar oscillator.

The second-order equation is equivalent to the planar system after setting $y=\dot{x}-ax+b{x}^3$. Any nonconstant periodic solution has a maximum $M$ and minimum $m$. At an extremum $\ddot{x}=x^3-x$, so $M>1$ and $m<-1$ are impossible. Equality at either endpoint would give one of the equilibrium points, and uniqueness would force a constant solution. Therefore a nonconstant periodic orbit lies in $-1<x<1$.

If $b/a<1$, throughout this striphas the sign of $a$ whenever $x\ne 0$. A nonconstant orbit spends a positive time away from $x=0$, so its integral cannot vanish, contradicting part (b). Thus there are no nonconstant periodic solutions. The constant equilibrium solutions of course remain; as usual the question's periodic-orbit assertion excludes equilibria.