## The Catenary Curve – On the way to the Calculus of Variations

In the beginning of the last decade of the 17th century, Jacob Bernoulli proposed the problem of the catenary; in 1696, Johann Bernoulli proposed the problem of the brachistochrone; some years later, Daniel Bernoulli would suggest to Euler the right functional to solve the elastica problem. The Bernoulli’s family had a great influence in the development of the Calculus of Variations, which was formalized by Euler in his book of 1744. Here the catenary problem and its solution are described in detail.

## Definition and first model.

In 1690, Jacob Bernoulli stated the following problem:

Assume you have a perfectly elastic wire (it can be deformed by the action of some force, but its shape is recovered immediately after the forces are ceased) made of a material with uniform density $\lambda$.  The two extremities of the wire are attached at the top of two columns of the same height. Moreover, assume that the only forces acting on the wire are tension and gravity. What is the equation described by the wire?

Consider any two points $p$ and $q$ in the wire. Denote by $T_R$ and $T_L$ the tension forces and $W_{pq}$ the weight force on the segment $\overline{pq}$ (see the figure below).

Decompose the tension forces in its vertical and horizontal components. The tension forces must cancel horizontally otherwise there will exist an acceleration on the chain. Additionally, the resultant vertical tension force (denoted by $T_y^{\star}$) must be equal to $W_{pq}$. There is one big difficulty with this model.

Let $\Theta_R$ be the angle $T_R$ makes with a horizontal line. Define similarly $\Theta_L$. In general, $\Theta_R \neq \Theta_L$. In order to derive the resultant of the forces at each point of the curve, one must keep track of both angles and this makes the computations painful. Fortunately, there is an easy model.

## Catenary derivation.

We are going to keep the assumption on the uniform density $\lambda$ of the curve. In addition, we are going to assume that $p$ is the lowest point of the curve. Moreover, assume that $p$ is at height zero and that  $q \neq p$ is a point on the right of $p$. We are going to derive the right part of the curve from $p$ (the complete curve will follow by symmetry). In the diagram of forces below, $W_{pq}$ denotes the weight force for the segment $\overline{pq}$.

At $q$ there exists a tension force $T_q$ tangent to the curve that can be decomposed in horizontal and vertical components $T_{qx}$ and $T_{qy}$. Let $\theta$ to denote the angle $T_q$ makes with the horizontal axis.

$T_{qx} = T_q\cdot \cos{\theta} \quad \quad T_{qy} = T_q \cdot \sin{\theta}.$

Since the object is in equilibrium:

$\begin{array}{ll} \displaystyle T_{qx} &=T_0 \\[0.3in] \displaystyle T_{qy} &= W_{pq} = \lambda g s, \end{array}$

where $s$ denotes the length of the segment $\overline{pq}$. Dividing the second equation by the first, we derive the differential equation that describes the right par of the curve.

$\displaystyle \frac{ \lambda g s }{T_0} = \frac{\sin{\theta}}{\cos{\theta}} = \frac{dy}{dx}\\[0.3in] \frac{dy}{dx} = \frac{\lambda g s}{T_0} = \frac{s}{a}, \quad a = \frac{T_0}{\lambda g}. \quad \quad (1)$

Recall that

$\displaystyle s(u) = \int_{0}^{u}{\sqrt{1 + (dy/dx)^2} dx}. \quad \quad\frac{ds}{dx} (u) = \sqrt{1 + (dy/dx)^2(u)}. \quad \quad (2)$

Therefore

$\displaystyle \frac{dy/dx}{ds/dx} = \frac{s}{a\sqrt{1 + s^2/a^2}} = \frac{s}{\sqrt{a^2 + s^2}} = \frac{dy}{ds}.$

Integrating both sides

$y(s) = \sqrt{a^2 + s^2}. \quad \quad (3)$

Using (1) and (2)

$\begin{array}{ll} & \displaystyle \frac{dx}{ds} = \frac{1}{\sqrt{1 + (s/a)^2}}\\[0.3in] \Rightarrow & \displaystyle x(s) = a \cdot \sinh^{-1}{(\frac{s}{a})}\\[0.3in] \Rightarrow & \displaystyle a \cdot \sinh(x/a) = s. \end{array}$

Substituting on (3)

$\begin{array}{ll} y^2(x) &= a^2 + a^2 \cdot \sinh^2(x/a) \\[0.15in] & = a^2\cosh^2(x/a) \\[0.15in] \Rightarrow y(x) &= a\cosh(x/a). \end{array}$

The same result was achieved by Leibniz, Huygens and Johann Bernoulli and the curve received the name of catenary.  As a curiosity, Jacob Bernoulli, Johann’s brother and also the person who poses the problem, spent an year trying to prove that the solution was a parabola. In fact, the parabola is the solution for a particular case of the catenary problem.

Take any segment of the curve such that its horizontal projection equals to $c$. Further, assume that every segment with horizontal projection $c$ has equal mass. In particular, for every segment with horizontal projection $dx$, let $m$ be its mass.

$\begin{array}{ll} & \displaystyle \frac{dy}{dx} = \frac{mgx}{T_0}\\[0.15in] \Rightarrow & \displaystyle y(x) = \frac{mg}{2T_0} \cdot x^2. \end{array}$

In a suspension bridge, the curve between two same height collumns is a parabola. The road is much heavier than the wire itself, and it dominates the weight force. Therefore, one can assume the weight force is equal for every segment of the curve with the same horizontal projection.

## Further References:

[1] – Raph LEVIEN. The elastica: A mathematical history. [Link]
[2] – Patricia Radelet-de GRAVE. The Problem of the Elastica Treated by Jacob Bernoulli and the Further Development of this Study by Leonhard Euler. [Link]
[3] – C. TRUESDELL. The influence of elastica on analysis: The classical heritage. [Link]