Reflection: Can Two Elliptical Sheets Be Glued into a 3D Solid?

Arc Length Parametric Equation

As preparation, let us first find the arc length parametric equation of an ellipse. As shown in the figure below, when an elliptical arc starting from (a,0) has a length s, what is the corresponding y = h(s)?

Theoretically, this problem is not difficult to solve. Since the standard equation of an ellipse is: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 which means x = a\sqrt{1 - y^2/b^2}, the arc length of the ellipse can be expressed as: s = \int_0^y \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy = \int_0^y \sqrt{\frac{a^2 y^2}{b^4 - b^2 y^2} + 1} dy \label{eq:bs} This integral can be represented using the "Incomplete elliptic integral of the second kind". Within the interval [0, b], s is monotonically increasing with respect to y, so an inverse function must exist. For numerical calculation, we do not need to know the explicit form of this inverse function; we only need to calculate a set of (y, s) correspondences through numerical integration of [eq:bs]. By swapping the positions of y and s, we effectively obtain the relationship y = h(s) (defining the function point-by-point).

Equation of the Gluing Edge

Now let’s look at the original problem: assuming they can be glued, what is the equation of the edge curve? We establish a coordinate system as shown in the figure below (left):

In the figure above, we only consider the first octant. The blue line is the curve for which we want to find the parametric equation. The correspondence between the spatial curve and the original elliptical plane curve is indicated using the same colors as in the previous figure (right). Let P = (x(s), y(s), z(s)) be a point on the curve. Suppose the arc length OP is s; this arc length is actually the original arc length on the ellipse. As discussed earlier, the y-coordinate can be found from the arc length: y = h(s) Then, due to symmetry (complementarity), the z-coordinate is actually the y-coordinate corresponding to an arc length of l - s, namely: z = h(l - s) where l is 1/4 of the total circumference of the ellipse, which is the total length of the blue curve.

Now that we have y and z, we only lack x. Since s is the arc length parameter, it must satisfy dx^2 + dy^2 + dz^2 = ds^2, i.e., \left(\frac{dx(s)}{ds}\right)^2 + \left(\frac{dh(s)}{ds}\right)^2 + \left(\frac{dh(l-s)}{ds}\right)^2 = 1 Thus, x(s) can be integrated: x(s) = \int_0^s \sqrt{1 - \left(\frac{dh(s)}{ds}\right)^2 - \left(\frac{dh(l-s)}{ds}\right)^2} ds \label{eq:f} The condition for whether they can be glued is whether the following inequality holds constantly: 1 - \left(\frac{dh(s)}{ds}\right)^2 - \left(\frac{dh(l-s)}{ds}\right)^2 \geq 0 \label{eq:c}

Proof of Gluing Possibility

Below we prove theoretically that two elliptical sheets can be perfectly glued, which means verifying that [eq:c] holds constantly.

First, we need to prove a lemma:

In an ellipse x^2/a^2 + y^2/b^2 = 1 (a \geq b > 0), we have: \left.\left(\frac{dy}{dx}\right)^2\right|_{s=s_0} \leq \left.\left(\frac{dx}{dy}\right)^2\right|_{s=l-s_0} That is, the absolute value of the slope at arc length s_0 does not exceed the absolute value of the reciprocal of the slope at arc length l-s_0.

The proof process is as follows (it is not difficult; drawing it by hand makes it clear):

Using the common parametric equations of an ellipse: x = a \cos t, \quad y = b \sin t Note that the direction of increasing parameter t is the same as the direction of increasing s (both counterclockwise starting from (a,0)), so s and t have a monotonically increasing relationship.

We can obtain: \left.\left(\frac{dy}{dx}\right)^2\right|_{t=t_0} = \frac{b^2 \cos^2 t_0}{a^2 \sin^2 t_0} \leq \frac{a^2 \sin^2 (\pi/2 - t_0)}{b^2 \cos^2 (\pi/2 - t_0)} = \left.\left(\frac{dx}{dy}\right)^2\right|_{t=\pi/2-t_0} That is, the slope at t=t_0 definitely does not exceed the absolute value of the reciprocal of the slope at t=\pi/2-t_0. Obviously, the smaller t_0 is, the larger \left.\left(\frac{dx}{dy}\right)^2\right|_{t=\pi/2-t_0} becomes.

Then we consider: \frac{ds}{dt} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \sqrt{a^2 \sin^2 t + b^2 \cos^2 t} = \sqrt{a^2 - (a^2 - b^2) \cos^2 t} Note that this is a monotonically increasing function of t (considering only the interval t \in [0, \pi/2]). This means s increases as t increases, and the rate of increase also gets faster. This implies: s_{[0, t_0]} \leq s_{[\pi/2-t_0, \pi/2]} That is, the arc length of the interval [0, t_0] does not exceed the arc length of the interval [\pi/2-t_0, \pi/2]. This further implies that if s_{[0, t_0]} = s_{[\pi/2-t_1, \pi/2]}, then t_1 \leq t_0. In other words, if s=s_0 corresponds to t=t_0 and s=l-s_0 corresponds to t=t_1, then t_1 \leq t_0. According to the previous discussion: \left.\left(\frac{dy}{dx}\right)^2\right|_{s=s_0} = \left.\left(\frac{dy}{dx}\right)^2\right|_{t=t_0} \leq \left.\left(\frac{dx}{dy}\right)^2\right|_{t=\pi/2-t_1} = \left.\left(\frac{dx}{dy}\right)^2\right|_{s=l-s_0}

With this lemma, we proceed to prove: \left(\frac{dh(s)}{ds}\right)^2 + \left(\frac{dh(l-s)}{ds}\right)^2 \leq 1 Here h(s) is the original y-coordinate of the ellipse. Suppose the corresponding x-coordinate is g(s), then: \left(\frac{dh(s)}{ds}\right)^2 + \left(\frac{dg(s)}{ds}\right)^2 = 1 Then it is equivalent to proving: \left(\frac{dh(l-s)}{ds}\right)^2 \leq \left(\frac{dg(s)}{ds}\right)^2 Taking the reciprocal of both sides and using ds^2 = dg^2 + dh^2, it is essentially proving: \left.\left(\frac{dg}{dh}\right)^2\right|_{s=l-s_0} \geq \left.\left(\frac{dh}{dg}\right)^2\right|_{s=s_0} This is exactly the lemma we just proved.

Numerical Solution Results

From a mathematical perspective, [eq:f] can already be called an explicit solution (analytical solution) for the required curve. Of course, an explicit solution does not mean it is an elementary function solution, nor does it mean such a solution is easy to calculate. Although most results in differential geometry are equations with arc length as a parameter, those are mostly theoretical. In fact, except for circles and lines, very few curves have simple arc length parametric equations.

To numerically verify condition [eq:c] and solve [eq:f], I found that existing mathematical software (like Mathematica) is not very convenient to use. The easiest way (for me) is to write a piece of Python code to simulate the solution. Taking a=2, b=1 as an example, it passed the numerical verification of [eq:c] and yielded the following solution:

The code is as follows:

#! -*- coding: utf-8 -*-

import numpy as np

a = 2
b = 1
h = 1e-6 # Step size

def Int(y, x):
    """Custom numerical integration function"""
    assert len(y) == len(x)
    dx = np.diff(x)
    _x = (x[:-1] + x[1:]) / 2
    _y = (y[:-1] + y[1:]) / 2
    s = np.cumsum(_y * dx)
    s = np.interp(x, _x, s, left=0)
    return s, x

def D(y, x):
    """Custom numerical differentiation function"""
    assert len(y) == len(x)
    dy = np.diff(y) + 1e-10
    dx = np.diff(x) + 1e-10
    dy_dx = dy / dx
    _x = (x[:-1] + x[1:]) / 2
    dy_dx = np.interp(x, _x, dy_dx)
    return dy_dx, x

def s2y():
    """Mapping from arc length to ellipse y-axis (i.e., h(s))"""
    y = np.arange(0, b, h)
    ds = np.sqrt(1 + a**2 * y**2 / b**2 / (b**2 - y**2))
    return Int(ds, y)[::-1]

y, s = s2y()
l = s[-1]
z = np.interp(l - s, s, y)
dx2 = 1 - D(y, s)[0]**2 - D(z, s)[0]**2
# Check if gluing is possible: whether it's >= 0 within 10x step size error
assert (dx2 > - 10 * h).all() 
dx = np.sqrt(np.clip(dx2, 0, a))
x = Int(dx, s)[0]

# Sample 1/10 of the points for plotting
x = x[::10]
y = y[::10]
z = z[::10]

from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt

plt.clf()
fig = plt.figure(figsize=(8, 8))
ax = fig.add_subplot(111, projection='3d')
ax.plot(x, y, z, color='blue')
ax.plot(x, -y, z, color='blue')
ax.plot(x, y, -z, color='green')
ax.plot(x, -y, -z, color='green')
ax.plot(x, np.zeros_like(x), z, color='red')
ax.plot(x, np.zeros_like(x), -z, color='red')

for i in range(0, len(x), 2000):
    _ = ax.plot([x[i], x[i]], [y[i], -y[i]], [z[i], z[i]], color='blue')
    _ = ax.plot([x[i], x[i]], [y[i], -y[i]], [-z[i], -z[i]], color='green')

for i in [-1, -1000]:
    _ = ax.plot([x[i], x[i]], [y[i], -y[i]], [z[i], z[i]], color='blue')
    if i != -1:
        _ = ax.plot([x[i], x[i]], [y[i], -y[i]], [-z[i], -z[i]], color='green')

ax.set_xlim(0, a)
ax.set_ylim(-b * 1.2, b * 1.2)
ax.set_zlim(-b * 1.2, b * 1.2)

plt.show()

Summary

This article analyzed the gluing problem of two elliptical sheets. The core idea is to parameterize the curve equation using arc length. Although the article uses an ellipse as an example, it can obviously be generalized to the bending and gluing problem of any two general planar curves. One only needs to find their respective arc length parametric equations (i.e., replace h(l-s) with the arc length parameterization of the other curve) and then use the integral in [eq:f].

I have finally basically solved a quite interesting problem, proving that my mathematical interest and ability are still there!