Proving solutions of differential equations are periodic

Question

Given the system of differential equations:

$$ \left\{ \begin{array}{c} x'(t) = y(t) \\ y'(t) = -V'(x(t)) \end{array} \right. $$

Where $V(x) = x^2 + \epsilon x^4$, so $V'(x) = 2x + 4\epsilon x^3$.

I want to prove that every solution of this equation that doesn't go through the point $(x,y) = (0,0)$ is periodic.

Before this, I found out (as an exercise) that the fixed points of this system of equations are $(x,y) = (\sqrt{1/2\epsilon}i , 0)$ and $(x,y) = (-\sqrt{1/2\epsilon}i , 0)$.

And that $H = \frac{y^2}{2} + V(x)$ is a constant function (so for every solution $(x(t), y(t))$, H remains a constant). Perhaps this might be useful in proving that the solutions are periodic, but I couldn't figure out why, if that's the case.

I've tried proving that the solutions are periodic, but didn't get far. Also what I thought was confusing, is that the $(x,y) = (\sqrt{1/2\epsilon}i , 0)$ is a fixed point, and a solution to the system of equations. Yet it doesn't go through $(0,0)$, but it's not periodic because it's constant. What am I missing? What do I need to prove that the solutions are periodic?

Answer 1

Our OP Timon Green has expressed in the comments to Relbellos' answer that the case of his primary interest is $\epsilon > 0$; that is the case I will address here.

Let's take another look at the system

$x'(t) = y(t), \tag 1$

$y'(t) = -V'(x(t)) = -2x(t) - 4 \epsilon (x(t))^3, \tag 2$

and it's fixed points. It is evident from (1) that any fixed point $(x_0, y_0)$ must satisfy

$y_0 = 0; \tag 3$

from (2) we see that we must also have

$2x_0 + 4\epsilon x_0^3 = 0, \tag 4$

which is equivalent to

$x_0(1 + 2 \epsilon x_0^2) = 0; \tag 5$

(5) has only one real zero, at

$x_0 = 0; \tag 6$

the other two zeroes of (5) occur at those $x_0$ such that

$1 + 2\epsilon x_0^2 = 0, \tag 7$

which if $\epsilon > 0$ has the purely imaginary solutions

$x_0 = \pm \dfrac{1}{\sqrt {2 \epsilon}} i, \tag 8$

however, these will not be considered here since we are, in keeping with general convention, addressing the case of $x$, $y$, real.

Thus we see the system (1), (2) has precisely one real zero

$(x_0, y_0) = (0, 0). \tag 9$

We also see, as has been pointed out by Timon Green in his question and Rebellos in his answer, that the function

$H(x, y) = \dfrac{1}{2}y^2 + x^2 + \epsilon x^4 \tag{10}$

is constant along any trajectory $(x(t), y(t))$ obeying (1)-(2), since

$\dfrac{dH(x(t), y(t))}{dt} = \dfrac{\partial H(x(t), y(t))}{\partial x} x'(t) + \dfrac{\partial H(x(t), y(t))}{\partial y} y'(t)$ $(2x(t) + 4\epsilon (x(t))^3) y(t) + y(t)(-(2x(t) + 4 \epsilon (x(t))^3) = 0; \tag{11}$

this shows that every integral curve $(x(t), y(t))$ of (1)-(2) lying in the real $x$-$y$ plane is contained within a level surface of (10). It is easy to see, for $\epsilon \ge 0$, that these level curves are ovaloids surrounding the origin $(0, 0)$, differentiable simple closed curves. Furthermore, since the only point where $(x'(t), y'(t)) = (0, 0)$ is the origin itself, we see that the speed

$s(t) = \sqrt{(x'(t))^2 + (y'(t))^2} > 0 \tag{12}$

along each of these ovaloids. Since they are compact, this speed is bounded away from zero on each one,

$s(t) \ge s_0 > 0; \tag{13}$

since any of these level curves of $H(x, y)$ is of finite circumference $l_0$, it follows that any integral curve initialized at some point $(x_0, y_0) \ne (0, 0)$ will return to that point in time at most

$\tau_0 = \dfrac{l_0}{s_0}; \tag{14}$

having returned to $(x_0, y_0)$, a system point will again execute exactly the same path since (1)-(2) do not depend on $t$. Thus any trajectory starting at $(x_0, y_0) \ne (0, 0)$ is periodic with period at most $\tau_0$.

Answer 2

Judging from your elaboration, we have the following system of differential equations :

$$\begin{cases} x' = y \\ y' = -2x - 4εx^3 \end{cases}$$

It's obvious that the origin $(0,0)$ is a stationary point for the given system.

The matrix of the linearised system (the Jacobian), is :

$$J(x,y) \begin{bmatrix} 0 & 1 \\ -2-12εx^2 & 0\end{bmatrix}$$

and the matrix for our stationary point, is :

$$J(0,0) = \begin{bmatrix} 0 & 1 \\ -2 & 0\end{bmatrix}$$

Obviously, $\det(J(0,0))\neq 0$, which means that the origin $(0,0)$ is a non-hyperbolic critical point for the given system.

Continuing on to eigenvalues calculation :

$$\det(J(0,0) - \lambda I) = 0 \Rightarrow \lambda^2 +2 = 0 \Leftrightarrow \lambda = \pm i\sqrt2$$

Purely imaginary eigenvalues mean that the origin $(0,0)$ is a center (stationary point type) for the linearised system, which hints we're in a good track. Take account though, that the case of the center does not preserve its topological qualities via linearisation, so no clear conclusion can be yielded just yet.

Trying out the functional :

$$H(x,y) = \frac{1}{2}y^2 + x^2 + εx^4$$

by also letting $F_{\{x \space \text{or} \space y\}}(x,y)$ be the functional with the differential equation right hand sides, we will yield the following differential :

$$\nabla H(x,y)F(x,y) = H_xF_x + H_yF_y =(2x+4εx^3)y + y(-2x-4εx^3) $$

$$\Leftrightarrow$$

$$\dot{H}(x,y) = 0$$

Now, this means that we have general periodic solutions for any value of $ε>0$, since $H(x,y) > 0 \space \forall \space (x,y) \in \mathbb R^2 \setminus (0,0)$.

The need to approach such systems by dynamical analysis steams from the fact that they cannot be solved when the pertubation factor $ε$ is at place (meaning that it is non-zero). The simple case of $ε=0$ would yield the solutions :

$$x(t) = c_1\cos(\sqrt2t) + c_2 \frac{\sin(\sqrt2t)}{\sqrt2}$$

$$y(t) = c_2\cos(\sqrt2t) - \sqrt2 c_1 \sin(\sqrt2 t)$$

which can be easily found to be periodic.

Below, is a stream plot for the case of $ε=1$, which yields the expected picture of a center :

$\quad \quad \quad \space \space \space \space \quad$

Note : A more precise and exact dynamical analysis can be done via bifurcation theory, but that's more complicated. Analytically, to distinguish center from focus in a general situation one must compute the so-called Poincaré mapping that sends solutions starting at, say, polar angle $\theta=0$ and distance $r_0$ to $\theta=2\pi$. In general it has the form $$ r=f(2\pi,0,r_0)=\alpha_1 r_0+\alpha_2 r_0^2+\alpha_3r_0^3+\ldots $$

Note 2 : A polar coordinates approach is also one of the standard ways but in this case yields a very complicated expression.