Plane curve with curvature that tends to zero

Question

Suppose you have immersed plane curve $\gamma:\mathbb{R}_+\rightarrow\mathbb{R}^2$ twice continuously differentiable and parametrized by arc-length such that its curvature $\kappa(t)=\|\gamma''(t)\|$ goes to $0$ as $t\rightarrow+\infty$. I was wondering whether it is possible for such curve to be bounded, meaning that it is contained in some ball in $\mathbb{R}^2$.

Answer 1

Yes, I think your intuition is correct.

HINT:

Recall that if $\gamma \colon I \to \mathbb{R}^2$ is parametrized by arc length ( that is $\|\gamma(t)'\| = 1$ for all $t$) then the curvature satisfies $$|\kappa(t)| = \|\gamma''(t)\|$$

Now the intuition is this: if $|\kappa|$ is small then $\gamma'$ varies little, so the direction of $\gamma$ changes little. This, coupled with the fact that the velocity is constant, implies that $\gamma$ approaches a line. Therefore, in some finite time it should get out of a given disk. Let's formalize that:

We have for any $a$, $b \in I$ $$\gamma(b) = \gamma(a) + \int_a^b \gamma'(t) dt $$

Assume that on the interval $[a,b]$ we have $|\kappa(t)|\le \epsilon$. Then we get $$\|\gamma'(t) - \gamma'(a)\| \le \epsilon(t-a) $$ Therefore $$\|(\gamma(b) - \gamma(a)) - (b-a)\gamma'(a) \|\le \epsilon \frac{(b-a)^2}{2}$$ We conclude: $$\|(\gamma(b) - \gamma(a))\| \ge (b-a)\| ( 1 - \frac{(b-a)\epsilon}{2})$$ Now assume that the interval $[a,b]$ has length $\frac{1}{\epsilon}$. Then we get $$\| \gamma(b) - \gamma(a) \| \ge \frac{1}{\epsilon} \cdot ( 1 - \frac{1}{2}) = \frac{1}{2 \epsilon}$$.

Hence $\gamma(I)$ cannot be contained in a disk of radius $< \frac{1}{4 \epsilon}$

Notice the proof works in any dimension $\ge 2$.