How to prove that the ellipse is a periodic orbit knowing that the orbital derivative of a function V is zero on there

Question

The question is as follows:

Show that the orbital derivative of the function $V=(1-x^2-2y^2)^2$ is zero on the ellipse $x^2+2y^2=1$, and explain why you can deduce that the ellipse is a periodic orbit.

So I've proved the first part: $$V'=2(1-x^2-2y^2)(-2xx'-4yy')$$ ($V'$ denotes the orbital derivative and $x'$ and $y'$ are the derivatives of $x$ and $y$ respectively)

Now taking the orbit: $2y^2=1-x^2$ and substituting on $V'$, we get zero.

My question is that now I don't know how to proceed to show that it is periodic...

Thank you

Answer 1

Since $$ x' = x(1-x^2-2y^2) - 2y(1+y),\quad y' = y(1-x^2-2y^2) + x(1+y), $$ you have $$ xx'+2yy'=(x^2+2y^2)(1-x^2-2y^2) $$ and so $$ V'=-4(1-x^2-2y^2)^2(x^2+2y^2). $$ Hence, for points close to the ellipse but outside the ellipse you have $V'<0$.

This implies that the ellipse is a periodic orbit since we know how the orbits behave in its neighborhood (just draw it).