# Elastica - numerical check

Following on from rmhleo's fantastic answer here, where he states that the deformation of an ideally elastic circle is a problem of the calculus of variations which may be solved with an ODE of the form

$$\frac{dy}{dx} = \frac{a^2 - c^2 + x^2}{\sqrt{(c^2 - x^2)(2a^2 - c^2 + x^2)}}$$

where

$$\int_{p_1}^{p_2} (\frac{1}{R} - k_0)^2ds$$

is the proposed integral solution.

For curve $p_1\approx\{0,-0.45\},\ p_2\approx\{0,1.957\},\ \theta_1\approx 17.227^{\circ},\ \theta_2\approx 43.627^{\circ},$ and arc length $\approx 4.251$:

with curvature plot:

approximate calculations show with $k_0\approx 1.214$, $\int_{p_1}^{p_2} (\frac{1}{R} - k_0)^2ds\approx 4.476$, whereas measurement of the arc length of the original curve is $\approx 4.251$. Does this discrepancy imply that the curve shown is not of elastica (minimal energy) type? (Data can be provided if this helps.)

• When you say "measurement of the arc length of the original curve", do you mean $\int ds$ for the original curve? In other point, the integral $\int_{P_1}^{P_2} (\frac{1}{R} - k_0)^2ds$ need not to be equal to the length of the curve, it is the functional whose minimum function solves the elastica problem, and this functional is the total elastic energy related to the bending of the curve. – rmhleo Oct 9 '14 at 13:44
• Ah! I am completely misunderstanding then! – martin Oct 9 '14 at 14:14
• re the arc length of the original curve, this was measured as $$\int_A^B \sqrt{\left(\dfrac{dx}{ds}\right)^2+\left(\dfrac{ dy}{ds}\right)^2}ds$$ of the original parametric. – martin Oct 9 '14 at 14:17
• So am I right in assuming then that the closer $\int_{p_1}^{p_2} (\frac{1}{R} - k_0)^2ds$ is to $0$, the closer the curve is to its equilibrial state? – martin Oct 9 '14 at 14:22
• @rmhleo Is it possible then that this curve could be a deformed (ideally elastic) circle with fixed-angle endpoints as discussed in the original question on PSE, or it it not possible to judge from the minimal data given? – martin Oct 9 '14 at 14:33