How to show an equation gives a closed curve I am analyzing the following dynamical system: $\ddot{x} + x +\epsilon x^3 = 0$, which I have rewritten to 2D system $\dot{x} = y, \dot{y} = -x-\epsilon x^3$. I have found that this is a conservative system with conserved quantity $E(x, y) = \frac{1}{2}y^2 + \frac{1}{2}x^2 +  \frac{1}{4}\epsilon x^4$. I am now trying to show that if $\epsilon>0$, all orbits are closed. I have already found that in this case the only fixed point is the origin. I suspect that if I can show that the level curves of the conserved quantity are closed, this will imply the orbits also being closed. My first question is whether this reasoning is valid.
My second question (and my main question) is how to show that an equation defines a closed curve. The level curves of the conserved quantity that result in an orbit are given by $\frac{1}{2}y^2 + \frac{1}{2}x^2 +  \frac{1}{4}\epsilon x^4 = K$, for $K>0$. This looks similar to a circle, but not quite. How can I show that it is indeed still a closed curve?
 A: In polar coordinates $(r,\theta)$ where $x = r \cos(\theta)$, $y = r \sin(\theta)$, we have $E(x,y) = r^2/2 + \epsilon r^4 \cos^4(\theta)/4$.  For any $c>0$, this equals $c$ for exactly one value $r(\theta)$, and $r(\theta)$ is a continuous function of $\theta$.  Thus it is indeed a closed curve.
More generally, for a continuous function $f(x,y)$ where $f(x,y) \to \infty$ as $|(x,y)| \to \infty$, the level set $L_c = \{(x,y): f(x,y) = c\}$ is 
a compact set.  Moreover if $f$ is continuously differentiable  we can apply the Implicit Function Theorem: if the gradient $\nabla f$ is not $0$ at a point $(x,y)$ of the set, then the level set is a curve in the neighbourhood of that point.  If the gradient is never $0$ on  the set, these local curves can be put together to form a finite collection of disjoint closed curves.
You can have various interesting situations where the gradient is zero at a point (which is thus a fixed point of the dynamical system), e.g. homoclinic and heteroclinic orbits.
A: A slightly different interpretation takes the first integral in the form 
$$
\dot x^2+x^2(1+\tfracϵ2x^2)=R^2
$$
and interprets it as a circle equation for the point 
$$(\dot x, x\sqrt{1+\tfracϵ2x^2}).$$
Here the polar coordinates have a constant radius $R$, so in the representation
$$
\dot x=R\cos\phi\text{ and }x\sqrt{1+\tfracϵ2x^2}=R\sin\phi
$$
only $\phi$ is variable.
Notice that the function $f(x)=x\sqrt{1+\tfracϵ2x^2}$ is monotonically increasing, thus invertible. This means that the map from $(\dot x,x)$ to $(R,\phi)$ is bijective. 
Name the inverse function $g$, then the derivative of $x=g(R\sin\phi)$ is
$$
\dot x=g'(R\sin\phi)\,R\cos(\phi)\,\dot\phi
$$
and because of $f'(x)=1+O(ϵx^2) \implies g'(u)=1+O(ϵu^2)$ one finds that $\phi$ is monotonically increasing, the solution of the differential equation follows the level curve of the energy function without changing orientation.
