# Gently falling functions

I wonder if it is possible to characterize the class of gently falling functions, which I would like to define as follows. Let $g(x)$ be a $C^2$ function defined on an interval $R \subseteq \mathbb{R}$ of the real line. It will be no loss of generality for my purposes to assume that $g(x) \ge 0$ on $R$, and that $\max_{x \in R} \; g(x) = 1$. Imagine placing a particle on the curve at any point and letting it frictionlessly slide down the curve under the influence of gravity (with nominal gravitation constant 1). Say that $g(x)$ is gently falling if, for every start point, the particle never "ski-jumps off" the curve in its downward descent.

Example 1. $g(x) = \sqrt{1-x^2}$ for $R=[0,1]$ is not gently falling. If I've calculated correctly, a particle starting near the max $(0,1)$ will separate from the curve after falling height $1/3$.

Example 2. $g(x) = \cos^2(x)$ for $R=[0,\pi/2]$. Under the assumptions above, I believe this is gently falling: a particle released at any point remains on the curve throughout its descent.

Example 3. (Added later.) $g(x)$ is a scaled and translated piece of $\cos^2(x)$ for $x>2$, preceded by a linear ramp tangent to $\cos^2(x)$:

Now a particle released at the max gathers enough momentum to lift off at the $x=2$ transition. So $g(x)$ is not gently falling.

I've worked out a condition that (I think!) needs to be satisfied for $g(x)$ to be gently falling, under my assumptions. Let $g=g(x)$, $\dot{g} = dg/dx$, and $\ddot{g} = d^2g /dx^2$. Then, for $g$ to be gently falling, $$2 (1-g) \ddot{g} + (1 + \dot{g}^2) \ge 0$$ should be satisfied for all $x$. For the quarter circle, the left side of this inequality becomes $$\frac{x^2}{1-x^2} - 2 \left(\frac{x^2}{\left(1-x^2\right)^{3/2}}+\frac{1}{\sqrt{1-x^2}} \right) \left(1-\sqrt{1-x^2}\right)+1$$ which simplifies to $$\frac{3}{1-x^2} -\frac{2}{\left(1-x^2\right)^{3/2}}$$ which is positive for $x$ less than the root $\sqrt{5}/3 \approx 0.745$, in accord with the picture above. When I work out the calculation for the second example, there are no roots within $R$.

I have two questions. First, has anyone seen a curve definition like this before? If so, even for tangentially related ideas, I'd appreciate a pointer. Second, if my gently-falling inequality on $g$, or some analogous inequality, is correct, are there general methods to characterize all the functions $g$ that satisfy it? Right now I can determine if any given function is gently falling, but I don't have a sense for the contours of the set of all gently falling functions. Thanks for any ideas!

(The motivation for these questions is a bit far from their form above, and it might distract from the mathematics to explain.)

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Worth getting, at least, a numerical solution with $g(0) = 1$ and $\dot{g}(0) = 0$ to the differential equation... – Will Jagy Jun 18 '11 at 2:22
Sorry if that's a silly question but in example 1), a particle starting at (0,1) won't go anywhere unless you give it some initial horizontal velocity. Are you suggesting that the separation point tends to the indicated point as that velocity tends to 0? – Alon Amit Jun 18 '11 at 6:59
@Alon: It seems that you can start a stationary particle anywhere, so the limit of separation points as your starting point approaches the top equilibrium point is what is indicated. – S. Carnahan Jun 18 '11 at 10:43
@Will: The equation is highly singular with your initial conditions. The particle is happy to drop like a rock when it has no initial speed. – S. Carnahan Jun 18 '11 at 10:49
A Joycean title! – Mariano Suárez-Alvarez Dec 10 '11 at 2:58

Let $p(t) = (x(t), y(t))$ be a parametrization of a smooth curve $(a,b) \to [0,\infty) \times (-\infty,0)$, and assume it satisfies the "zero total energy parametrization" $(x')^2 + (y')^2 = -2y$. We want the apparent force (given by combining acceleration and gravity) at any time to be in a counterclockwise direction from the velocity vector. Using cross products, we find that this is equivalent to

$$x'(t)(y''(t) + 1) \geq x''(t)y'(t).$$

If you choose initial conditions $x'(0) > 0$ and $y'(0) = 0$, and set the difference between the sides to a non-negative constant $c$, numerical integration will yield one of three types of trajectory, depending on the sign of $x'(0)-c$. If it is negative, you get a periodic sequence of narrow loops, like a frictionless roller coaster. If it is zero, you get a constant speed horizontal trajectory. If it is positive, the trajectory will fall gently but increasingly steeply.

Now, we consider the special case where the path satisfies the vertical line test, so $y(t) = f(x(t))$ for some function $f$. By the chain rule, we see that $y'(t) = f'(x(t)) x'(t)$ and $y''(t) = f''(x(t))x'(t)^2 + f'(x(t))x''(t)$. Feeding this into the inequality and simplifying, we obtain the condition

$$f''(x(t))x'(t)^2 + 1 \geq 0.$$

In order to express this purely in terms of $f$, we substitute our variables into the zero energy condition to get $x'(t)^2 + f'(x(t))^2 x'(t)^2 = -2f(x(t))$, which simplifies to $x'(t)^2 = -2\frac{f(x(t))}{1 + f'(x(t))^2}$. The inequality then becomes:

$${f'}^2 + 1 \geq 2f'' f$$

If we apply the substitution $g = f+1$, we get the equation in the question.

You can find similar equations by searching the web with terms like "roller coaster differential equation". For example the second version of equation 4.17 in this document looks like the second displayed equation, but with "real-world" terms like mass included.

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@Scott: Nice derivation! And that document to which you linked is spot-on: "Will a car fly off the rollercoaster?" Near the end: "Let us note that Eq. (4.10b) for a general form of the track (i.e., for a general $y(x)$) cannot be solved analytically. Nevertheless, it can be easily solved numerically, e.g., in Matlab." – Joseph O'Rourke Jun 20 '11 at 10:56

I will post this as an answer rather than as a series of comments.

In order to emphasize the gently part, you should focus on concave down functions ( or whatever is the nontrivial direction; I think of $e^{-x}$ as an example of the trivial direction that should be ruled out by beefing up the definition ).

I also recall that (under mild assumptions often found in high school physics courses) things often fell in parabolic curves after leaving a guided track. Roller coaster designers take this into account. So if your curve "moves no faster than a parabola" throughout the trajectory, then the object should stay on the track.

It is my belief that such curves have been studied, especially in the design of roller coasters. I have no references for you at this time, however. Perhaps a well phrased net search can give you some of what you seek.

Gerhard "Moves Slower Than Many Conics" Paseman, 2011.06.17

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@Gerhard: I like the roller-coaster suggestion! Will pursue. The inequality derives from considering the free-fall parabola. – Joseph O'Rourke Jun 18 '11 at 13:13
Apparently roller-coaster designers employ clothoids! ffden-2.phys.uaf.edu/211_fall2002.web.dir/shawna_sastamoinen/… – Joseph O'Rourke Jun 18 '11 at 13:56
If I had scrutinized your answer Joseph, I could probably have figured out the source of the inequality. Since I did not see parabola mentioned (at my first reading) I thought I would mention it. Perhaps "free fall" or "celestial mechanics" might also help, but it looks like you have good thing going with "roller coaster". Gerhard "Fears and Enjoys Roller Coasters" Paseman, 2011.06.18 – Gerhard Paseman Jun 18 '11 at 22:19