Consider a homogeneous, flexible string of a prescribed length hanging in a vertical plane where its ends are fixed at two points P and Q. Determine the equilibrium configuration of the string by minimising the potential energy of the string under extra conditions.
I can right away say that this is an exercise I've been given to solve so I'm not looking for full solutions, just some advice on some things that are a bit confusing to me.
So, first of all we have a fixed length $\textit{l}$ of the string, so if we let $y(x)$ denote the position of the string at the $x$-coordinate $x$ on the interval $[a,b]$, we get $$J[y] = \int_a^b \! \sqrt{1 + y'(x)^2} \, \mathrm{d}x = l.$$ So we have a constraint that we have to consider when finding $y$. This could be done by for example using the Calculus of variations variant of Lagrange multipliers.
Now, my main problem is how to find the functional we want to minimize $I[y]$, if $y$ denotes the height of the string then the natural way to formulate the integral would be
$$I[y] = \int_a^b \! mgy \, \mathrm{d}x.$$
But is it really that simple, how do we consider mass of a string, do we take the entire mass and just factor it out of the integral expression or should one consider small mass segments?
The expression $I[y] = \int_a^b \! mgy \, \mathrm{d}x$ is not quite correct. For one thing, $m$ should be replaced by linear density $\rho$, measured in mass per unit of length. Indeed, to get the mass of a short piece of the string, we multiply its length by $\rho$. The length is represented by $\sqrt{1+(y'(x))^2} \, \mathrm{d}x$. Thus, the correct formula is $$I[y] = \int_a^b \! \rho gy \sqrt{1+(y'(x))^2} \, \mathrm{d}x$$
Using the Lagrange multiplier $\lambda$ to handle the constraint, you'll be working with the functional $$ \int_a^b \! \rho gy\sqrt{1+(y'(x))^2} \, \mathrm{d}x + \lambda \int_a^b \! \sqrt{1 + y'(x)^2} \, \mathrm{d}x,\qquad y(a)=A, \ y(b)=B $$ If you get stuck, see page 199 here.