Show that all level sets of the Hamiltonian are bounded

Question

Consider the system $$ \dot{x}=y,~~~~~~~~~~\dot{y}=x^2-x^3. $$ Its Hamiltonian is $$ H(x,y)=\frac{y^2}{2}-\frac{x^3}{3}+\frac{x^4}{4}. $$

Show that all level sets of the Hamiltonian are bounded.

As far as I see, it is meant to show that the set $$ \left\{(x,y)\in\mathbb{R}^2: H(x,y)=c\right\} $$ is a bounded set.

My computation gives that $H(x,y)=c\Leftrightarrow y=\sqrt{2c+\frac{2}{3}x^3-\frac{1}{2}x^4}$, i.e. $$ \left\{(x,y)\in\mathbb{R}^2: H(x,y)=c\right\}=\left\{(x,y): y=\sqrt{2c+\frac{2}{3}x^3-\frac{1}{2}x^4}\right\} $$

How can I see that this is a bounded set?

Answer 1

We could think about different ways to show this, and after a while we might think that it would be nice to show that any level set of $H$ is contained within some ellipse. That would definitely mean that any level set is bounded.

In formulas, this would mean showing that any $x,y$-pair satisfying

$$ \frac{y^2}{2}-\frac{x^3}{3}+\frac{x^4}{4} = c $$

satisfies

$$ ay^2 + bx^2 \leq d $$

for some $a,b,d > 0$ which depend on $c$. The first equation already has a $y^2$ term, so we might be able to get away with $a=1/2$. For now let's also take the simplest choice for $b$, namely $b=1$, and hope this doesn't complicate finding $d$.

So now we want to show that if $(x,y)$ satisfies

$$ \frac{y^2}{2}-\frac{x^3}{3}+\frac{x^4}{4} = c \tag{1} $$

then

$$ \frac{y^2}{2} + x^2 \leq d \tag{2} $$

for some $d$ depending on $c$.

With our target $(2)$ in mind, let's rearrange $(1)$ like

$$ \frac{y^2}{2} + x^2 = -\frac{x^4}{4} + \frac{x^3}{3} + x^2 + c = p(x) + c. \tag{3} $$

The polynomial $p(x) = - x^4/4 + x^3/3 + x^2$ is bounded above, and to find this bound we first calculate

$$ p'(x) = -x^3 + x^2 + 2x = -x(x+1)(x-2), $$

and then check that $p(0) = 0$, $p(-1) = 5/12$, and $p(2) = 8/3$ to conclude that $p(x) \leq 8/3$, and hence, from $(3)$, we conclude that if

$$ \frac{y^2}{2}-\frac{x^3}{3}+\frac{x^4}{4} = c $$

then

$$ \frac{y^2}{2} + x^2 \leq c + \frac{8}{3}, $$

which completes the proof.

Answer 2

Let $f(x)=\sqrt{2c+\frac 23 x^3- \frac 12 x^4}$ The domain of existence of $f$ is

$$\mathcal D:= \{x\in \mathbb{R}: 2c+\frac 23 x^3- \frac 12 x^4\geq 0\}$$

You should notice that asymptotics tell us that there exists a certain $x_0$ such that $$|\frac 12 x^4| > |2c+\frac 23 x^3|~~~~~~ \textrm{for} ~x>x_0$$ You can actually prove this through explicitly solving the inequality, although I would not recommend it, calculations get messy. This implies that for $|x|>x_0$ the argument of the square root becomes negative.

Since the domain is the preimage of the set $[0, +\infty[$ which is closed and $f$ is continuous, $\mathcal{D} $ is closed. We have previously shown that it is bounded by $B(0, x_0+\varepsilon]$. You could get better bounds, but this is enough for what we're concerned. Since $\mathcal{D}$ is closed and bounded, it is compact.

Since $f$ is continuous on its compact domain, it attains a maximum and a minimum value for Weierstrass' theorem; Since both domain and codomain are bounded, the graph of this function in $\mathbb{R}^2$ is too.