Today, we will analyze the famous problem of what shape a rope of uniform mass density takes if it is hung between two points. The problem is illustrated below

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Here, we will solve the problem by using a minimization of the overall potential energy of the rope. The first step is to construct the potential energy. We start by considering the potential energy \(dV\) of a small mass \(dm\) of the rope. Suppose the rope has shape \(y = y(x)\), with boundary conditions \(y(0) = h_1\) and \(y(L) = h_2\). A small mass \(\d m\) has length \(\d s = \sqrt{1 + (y’)^2} \d x\), and thus mass \(\d m = \mu \d s\), where \(\mu\) is the linear mass density of the rope, measured in kg/m. Thus, the potential energy of this small piece of mass is \(\d V = gy \d m\). Integrating this potential energy over the entire rope, we get the potential energy functional:

\[V[y] = g\mu \int_0^L y\sqrt{1+(y')^2} \d x.\]

Defining the Lagrangian \(\mathcal{L} = y\sqrt{1+(y’)^2}\) (we ignore the constants) If we plug this into the Euler-Lagrange equation

\[\frac{\d}{\d x}\left(\frac{\partial\mathcal{L}}{\partial y'}\right) = \frac{\partial\mathcal{L}}{\partial y},\]

we get the differential equation

\[\frac{d}{\d x}\left(\frac{yy'}{\sqrt{1+(y')^2}}\right) = \sqrt{1+(y')^2}\]

If we expand out all the terms, then we get

\[\frac{(y')^2}{\sqrt{1+(y')^2}} + \frac{yy"}{\sqrt{1+(y')^2}} - \frac{y(y')^2y"}{(1+(y')^2)^{3/2}}= \sqrt{1+(y')^2}\]

If we simplify the terms and cancel, then we get

\[\frac{yy"}{1+(y')^2} = 1\]

There are multiple ways to solve this differential equation. Here, I will show only one. If, in general, we have a (linear or nonlinear) differential equation where there is no appearance of the independent variable \(x\), then we can make a substitution \(y’ = v(y)\). This means that

\[y" = \frac{\d y'}{\d x} = \frac{\d v}{\d y}\frac{\d y}{\d x} = v' y' = v'v\]

so we have defined a function \(v\) whose independent variable is \(y\). Thus the differential equation becomes

\[\frac{vv'y}{1+v^2} = 1\]

This equation is separable, so it can be written as

\[\frac{v\d v}{1+v^2} = \frac{1}{y} \d y\]

If we integrate both sides,

\[\frac{1}{2}\log(1+v^2) = \log y + C\]

Exponentiating both sides, we can substitute in our definition of \(v\):

\[\sqrt{1+v^2} = c_1 y = \sqrt{1+(y')^2}.\]

Next, we can separate variables again:

\[\int \frac{\d y}{\sqrt{c_1^2 y^2 - 1}} = \int \d x\]

Making the substitution \(c_1y = \cosh \theta\), \(\d y = \frac{1}{c_1}\sinh\theta\) allows us to evaluate the integral quickly:

\[\arccosh(c_1 y)= c_1x + c_2\]

Isolating for \(y\), we get

\[y = \frac{1}{c_1}\cosh(c_1 x + c_2)\]

Now we can plug in our boundary conditions, \(y(0) = h_1\) and \(y(L) = h_2\). This gives us

\[h_1 = \frac{1}{c_1}\cosh(c_2), \qquad h_2 = \frac{1}{c_1}\cosh(c_1 L + c_2)\]

This system of equations, in general, needs to be solved numerically.