Prove solutions for $y'' = -y^3$ are periodic

Question

In physics, the equation $y''=-y^3$ represents the motion of a non-harmonic oscillator (for example, a mass between two walls with two springs that oscillates parallel to the walls). The solution is not given by elementary functions. I am trying to prove that the two solutions for this equation are periodic, without actually solving the equation. I tried to mimic the steps used to show that solutions to $y''=-y$ are periodic, but these heavily rely on the linearity of the equation (to solve by power series), as shown e.g. here.

Proving the solutions are bounded is pretty straightforward - Suppose we have initial conditions $y(0)=a, y'(0) = b$. Multiplying our ODE by $y'$ yields $$ y'y'' = -y'y^3 \Longrightarrow \frac{1}{2}\frac{d}{dx}\left({y'}^2\right) = -\frac{1}{4}\frac{d}{dx}\left(y^4\right)\Longrightarrow{y'}^2+\frac{1}{2}y^4 = C $$ where $C$ is some constant. Then by the intial conditions, $C = b^2 + \frac{1}{2}a^4$, and we can see that $y$ is bounded by $\pm\left(2C\right)^{1/4}$.

But this still doesn't mean the solution oscillates periodically between $\pm(2C)^{1/4}$.

Answer 1

Let $c$ be a positive number and let $C$ be the curve with equation $x^2+y^4/2=c$. This is a smooth, compact curve.

If $p=(x_0,y_0)$ is a point of $C$, and let $u_p$ be the maximal solution of $$u''+u^3=0,\quad u(0)=x_0, \quad u'(0)=y_0,$$ which I will call the solution that starts at $p$. It is easy to see that $u_p$ is defined on the whole of $\mathbb R$ because the function $\gamma_p:t\mapsto(u_p(t),u_p'(y))$ has bounded image: its image is contained in $C$. Moreover, that image is an open subset of $C$, because the derivative of $\gamma_p$ is everywhere nonzero.

Since the curve $C$ is connected and the images of the curves $\gamma_p$ associated to all solutions starting at points $p$ of $C$ cover $C$, this means that for all $p\in C$ the image of $\gamma_p$ is all of $C$. A continuous function $\mathbb R\to C$ that is surjective cannot be injective so it goes through some point twice, and this implies that the solutions starting at a point in $C$ are periodic.

Answer 2

For convenience, we rewrite the original equation in the form of a system: $$\tag{1} \left\{\begin{array}{lll} \dot y&=&z\\ \dot z&=&-y^3. \end{array}\right. $$ The system (1) has the first integral/conserved quantity $$ V(y,z)= \frac14 y^4+\frac12 z^2. $$ It means that it remains constant along the trajectories of (1). Indeed, let $(y(t),z(t))$ be the solution to (1). Then the derivative along the trajectories $$ \dot V= \frac{d}{dt}V(y(t),z(t))= \frac{\partial V}{\partial y}\frac{dy}{dt}+ \frac{\partial V}{\partial z}\frac{dz}{dt}= y^3z+z(-y^3)\equiv 0. $$ This implies that any solution trajectory $(y(t),z(t))$ lies on the level curve of $V$. These level curves are closed:

To prove that the solutions are periodic, one can use the uniqueness of the solution to the initial value problem. If the solution, having passed along a closed trajectory, returned to the starting point, then the next turn will be identical to the one passed due to the coincidence of the initial conditions. On the other hand, the state cannot stop halfway without going through a full turn along a closed curve, since for any fixed curve the velocity of the movement $$ \sqrt{(\dot y)^2+(\dot z)^2}=\sqrt{z^2+y^6} $$ is bounded from below, except for the equilibrium point $(0,0)$. As a result, the state travels a finite distance on the phase plane in a finite time.