Can there be an interval where $F(x)=4 x^2-\frac{1}{2}$ is chaotic?

Question

The function $$F(x)=4 x^2-\frac{1}{2}$$ has two repelling fixed points. Now, I wonder, can there be an interval $I$ where it is chaotic? I think not, because of the repulsiveness of the fixed points. What do you think?

Answer 1

It is a chaotic function. In fact, the repulsiveness of the fixed points is key to this: unless there are at least two repulsive fixed points, this function cannot be chaotic. If both points were attractive, then every orbit would end up at one of this. If one point was attractive and the other repulsive, then every orbit would end up at the first point or infinity. Only because there are different repulsive fixed points can the orbit "dance" around.

Now, it's very true that, sufficiently far from these fixed points (such as $x>1$), the orbit will tend away and away from them and go to infinity. Examine the orbit of points within $[-1/2,1/2]$. The minimum value of $F$ on this interval is $-1/2$ (when $x=0$) and the maximum is $1/2$ (when $|x|=1/2$). Thus the interval maps to itself. Since the interval maps to itself but contains no attractive fixed points, orbits must circulate inside the interval forever, never settling down to one point.

This doesn't rule out that an orbit could settle into a periodic orbit. Proving that the function is chaotic requires showing that $F^2, F^3, F^4, ...$ have no stable fixed points either -- if they did, this would correspond to a stable periodic orbit. This is a harder proof but shows the function to be truly chaotic.

Answer 2

If you are asking whether it can be potentially chaotic -- of course, yes. Standard logisitc map is chaotic having two repelling fixed points. You would have obstructions to chaoticity if you had attracting fixed points.