Derivation Catenary Equation With Calculus of Variations

Sometimes it’s useful to directly apply the Calculus of Variations to physical systems. A common example is the application to the Brachistochrone problem.

I have already written an article: Derivation Catenary Equation With Calculus where I derive the catenary equation with statics and calculus.

But in a way, at least to me, it seems natural to use the Calculus of Variations on this problem.

In my previous article on the catenary, it was important to know all the forces acting on the catenary. We also made use of Newton’s Laws of Motion.

Now we don’t have to.

Derivation Catenary Equation

Stating The Problem (in the language of Variational Calculus)

The Calculus of Variations is a mathematical tool we can use to find minima and maxima of functionals (loosely meaning function of a function in this context). Whereas normal Calculus is sufficient to find extremes of functions.

First we need to answer an important question:

· What kind of variational problem is this?

We know that the length L[m] of the string or chain is fixed. Which indicates we are dealing with a variational constraint problem.

The next important question to ask is:

· What is minimized or maximized in this problem?

Well, we know that the system is in static equilibrium. There is no net force acting on the system (ΣF⃗ =0) and the system is not in motion (v⃗ = 0).

And we know that in the absence of external forces, physical systems always tend to move to a state of lowest potential energy.

From these facts we can reason that it is potential energy u[J] that is to be minimized.

So to state the problem: This is a variational problem where we want to minimize the functional for potential energy u with the constraint that the length of the cord L remains constant.

Applying The Calculus of Variations

Let’s first bring some geometry into the problem and define our axes:

Let us assume the catenary is an idealized, uniform string with a 1D linear density μ[kg/m] and let us assume that its length is greater than the horizontal distance between the poles: L > 2a

Let us start with writing down functionals for the contraint (length of the cord) and the potential energy.

Constraint:

Eq. 1

Where ds is an infinitesimally small arc section of the string. By using the Pythagorean Theorem we write the functional (eq. 1) as a function of x and y, rather than s:

Eq. 2

Now let’s write a functional for potential energy.

But first we must note that only the force of gravity contributes to the potential energy because the force of tension is always parallel to the catenary curve. On a small arc section:

Eq. 3

Doing the same as before, with the Pythagorean Theorem, we get,

Eq. 4

If you are going to solve a variational problem with a constraint (length of the cord), a good way to do so is to define a new functional:

Eq. 5

Where λ is what we call a Langrange Multiplier and this method of solving variational constraint problems is called the Method of Lagrange Multipliers.

The method states that if we want to minimize u, we need to find the stationary points of ℒ as a function of the lagrange multiplier λ.

We can expand eq.5 and write dy/dx as y’ to get:

And we can define a new functional:

To get:

You’d typically use the Euler-Lagrange Equation here but since our functional F does not explicitely depend on x, we can use the simpler Beltrami Identity to find the stationary points of ℒ,

Eq.6: Beltrami Identity

Where C is a constant.

Now let’s work out an expression for ∂F/∂y’:

Working out the eq. 6 is tedious, I know. But doing so yields:

Eq. 7

Now let’s solve eq. 7 for y’:

Eq. 8

Now all we have to do is solve this differential equation (eq. 8).

So the rest is “just maths.”

“Just Maths”
Let’s now solve eq. 8 by using seperation of variables:

The integral on the left side of above equation is quite tedious. So we will attempt to re-write it in terms of the following standard integral:

Let z = Cμgy+Cλ:

Yielding:

Now we have three unknown constants C, C₂, and λ.

But can make use of the symmetry of our defined geometry of the catenary to find one of them. Note that y = 0 at points x = -a and x = a, yielding,

As we can see, both equations are equal to eachother. So we can write:

Using the hyperbolic trig identity cosh x = cosh -x, we can conclude that C₂ = 0.

Now we can write an expression for λ:

And then finally our catenary equation:

Eq. 9: The Catenary Equation

Where C depends on the specifics of the problem.