Derivation of the catenary equation

Question

I’ve seen two derivations of the catenary equation: one involving Lagrange multipliers and another using a balance of forces on a segment of the cable/rope.

I’m trying to derive the catenary equation by minimizing the potential energy of the cable, which is:

$$U=\mu g \int_{x_1}^{x_2}y(x) \sqrt{1+y’(x)^2} \, \text{d}x,$$

where $\mu$ is the linear mass density of the cable and $y(x)$ is the (catenary) function we are looking for.

The next step is invoking the Euler-Lagrange equation, $$\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y’}=0$$ where $$f(y,y’,x)= y(x) \sqrt{1+y’(x)^2}.$$ Doing this and simplifying a bit, we are left with the differential equation

$$y y’’=1+y’^2.$$

BUT the correct differential equation for the catenary is $y’’= a \sqrt{1+y’^2}$, with $a$ being some constant.

I am going crazy trying to figure out where my mistake is. In fact I went to a textbook that I believe uses the same differential equation I have (picture below). Unfortunately the book does not include all the steps in going from the differential equation to the catenary function $y(x)$.

Any comment is kindly appreciated.

Answer 1

The Lagrange equation is $$ \frac 1{\sqrt{1+y’^2}}-\frac{yy’’}{(1+y’^2)^{3/2}}=0 $$ which you can simplify to your $$ \frac{yy’’}{1+y’^2}=1. $$ You can simplify by multiplying the first equation by $y’$ to get $$ \frac{d}{dx}\left(\frac{y}{\sqrt{1+y’^2}}\right)=0 $$ which leads to $$ y=\kappa \sqrt{1+y’^2}. $$ This not what you say is correct, but I’m pretty sure that my version is the correct one.

Answer 2

There is no single "correct differential equation for the catenary". The catenary satisfies (1) $yy''=1+y'^2$ AND (2) $y''=(1/a)\sqrt{1+y'^2}$ AND (3) $y=a\sqrt{1+y'^2}$ for some constant $a$ (you can easily see this by plugging in the functional form $y=a \cosh(x/a)$).

In fact you can directly check that all of these equations are equivalent.

To start, note that (1) is product of (2) and (3). So if you start with one of these and deduce another then the third follows.

Starting from (3) and differentiating you will get $1=ay''/\sqrt{1+y'^2}$ i.e. (2).

Conversely from (2) in the form $1=ay''/\sqrt{1+y'^2}$ multiplying by $y'$ and integrating we get (3).

Going from (1) to (3) is as in mike stone's answer.