Notes for AKT-140117/0:16:43

The "power-line" problem is also known as the "catenary" problem (https://en.wikipedia.org/wiki/Catenary). As stated, we want to minimize the potential energy, so we must find an expression for the total potential energy of the system, which will be done by integrating along the rope/wire/line. The infinitesimal potential energy is given by $dV = gydm$, where g is the acceleration due to gravity, y is the height, and $dm$ is the infinitesimal mass along a length of the rope. In addition, $dm = \frac{m}{s}dl$, where $dl = \sqrt{(dx)^2 + (dy)^2} = dx\sqrt{1+(\frac{dy}{dx})^2}$ and $s$ is the total arclength of the rope. If we let the rope span a length from $x = -L$ to $x = L$, the total potential gravitational energy is $V(y) = \int_{-L}^{L}\frac{mg}{s}y(x)\sqrt{1+(y'(x))^2}dx$. This is what we want to minimize. As usual, let $f(\epsilon) = V(y_c + \epsilon y_q)$, where $y_q(0) = y_q(L) = 0$, and set $\frac{d}{d\epsilon}f(\epsilon)\mid_{\epsilon = 0}$. For convenience, use dot notation for derivatives, and compute to get

$\frac{d}{d\epsilon}f(\epsilon)\mid_{\epsilon = 0} = \frac{mg}{s}\int_{-L}^{L}y_q\sqrt{1+\dot{y}_q^2}+y_c(1+\dot{y}_c^2)^{\frac{-1}{2}(\dot{y}_c\dot{y}_q)}dx = 0$

Thus, $0 = \frac{mg}{s}(I_1 + I_2)$, where $I_1 = \int_{-L}^L y_q\sqrt{1+\dot{y}_c^2}dx$ and $I_2 = \int_{-L}^L y_c\dot{y}_c\dot{y}_q(1+\dot{y}_c^2)^{\frac{-1}{2}}dx$.

Integrating $I_2$ by parts with $u = y_c\dot{y}_c(1+\dot{y}_c^2)^{\frac{-1}{2}}$ and $dv = \dot{y}_q dx$, and applying boundary conditions $y_q(-L) = y_q(L) = 0$, we obtain

$I_2 = \int_{-L}^L y_q\big(y_c\dot{y}_c^2\ddot{y}_c(1+\dot{y}_c^2)^{\frac{-3}{2}} - (\dot{y}_c^2 + y_c\ddot{y}_c)(1+\dot{y}_c^2)^{\frac{-1}{2}})\big)dx$

From $I_1 + I_2 = 0$, factoring out the $y_q$ and the fundamental lemma of the calculus of variations, we obtain an ODE (replacing $y_c$ with $y$):

$0 = (1+\dot{y}^2)^\frac{-3}{2}((1+\dot{y}^2)^2 - (\dot{y}^2 + y\ddot{y})(1+\dot{y}^2) + y\dot{y}^2\ddot{y})$

Dividing through by $(1+\dot{y}^2)^\frac{-3}{2}$, expanding, and simplifying, we obtain our final ODE:

$1 + \dot{y}^2 - y\ddot{y} = 0$

A solution to this is $y(x) = \frac1{\lambda}cosh(\lambda x + c)$, where $\lambda$ and c are determined by physical constants and the boundary values of $y$ (at $x = -L$ and $x = L$). It turns out that this ODE is the same one you get when solving the soap bubble problem (HW 2, problem 3), since the Lagrangians of the two systems are the same up to constants.