Lagrange multipliers: catenary
A uniform chain suspended between two points forms a classic curve: the catenary. Contrary to common intuition, the chain does not form a parabola under its own weight. Instead, the hyperbolic cosine function describes the equilibrium shape. In this post, I explain how this result emerges from the calculus of variations by minimizing potential energy, subject to a fixed chain length.
Given a chain with linear mass density \rho, I consider its shape y(x) in the xy-plane, anchored at x = -a and x = a, with both ends at the same height. The key idea is to minimize the potential energy functional:
I[y] = \int_{-a}^{a} \rho g y \sqrt{1 + (y^\prime)^2} \, dx
while holding the total chain length:
J[y] = \int_{-a}^{a} \sqrt{1 + (y^\prime)^2} \, dx = 2L_{total}
constant. I add this constraint using a Lagrange multiplier \lambda, leading to the modified functional:
H[y] = \int_{-a}^{a} (\rho g y + \lambda) \sqrt{1 + (y^\prime)^2} \, dx
With the Lagrangian \mathcal{L}(y, y^\prime) = (\rho g y + \lambda)\sqrt{1 + (y^\prime)^2} independent of x, I apply the Beltrami identity:
\mathcal{L} - y^\prime \frac{\partial \mathcal{L}}{\partial y^\prime} = C_1
This results in a first-order equation for y(x). After algebraic steps, I arrive at the general solution:
y(x) = \frac{C_1}{\rho g} \left( \cosh\left( \frac{\rho g x}{C_1} \right) - 1 \right)
where C_1 is a constant relating to the horizontal tension at the lowest point of the chain. The value of C_1 follows from the length constraint using:
L_{total} = \frac{C_1}{\rho g} \sinh\left( \frac{\rho g a}{C_1} \right)
Numerically, I solve this transcendental equation for C_1 given a and L_{total}. The resulting catenary describes many real-world structures, such as suspension bridges and power cables, illustrating how physics and mathematics guide the equilibrium shapes of flexible chains.
For more insights into this topic, you can find the details here.