Consider the system $$ \dot{x}=y,\qquad\dot{y}=x^2-x^3. $$
The Hamiltonian of this system is $$ H(x,y)=\frac{1}{2}y^2-\frac{1}{3}x^3+\frac{1}{4}x^4. $$
Show: All level sets of this Hamiltonian are bounded.
Let $c$ be any constant and consider the set $\left\{(x,y): H(x,y)=c\right\}$.
It is to show that this set is bounded. How can I show that?
My first idea is to solve $H(x,y)=c$ for $y$: $$ y_{1,2}=\pm\sqrt{2c+\frac{2}{3}x^3-\frac{1}{2}x^4}. $$ What could be the next step?
With $p=\dot{y}$ the Hamiltonian reads
\begin{equation*} H=\frac{1}{2}p^{2}-\frac{1}{3}x^{3}+\frac{1}{4}x^{4}=\frac{1}{2}p^{2}+V(x) \end{equation*} Note that the non-negative term $+\frac{1}{4}x^{4}$ dominates the large $x$ behaviour. In this case $H$ has a compact resolvent which leads to the result that $H$ can only have discrete eigenvalues and no continuous spectrum. I think you can find these results in the volumes "Modern Mathematical Physics " by Reed and Simon.