It is physically intuitive that the catenary is unique? The catenary minimizes the potential energy of a cable and has equation $y - y_0 = A \cosh (\frac{x-x_0}{A})$. It is physically intuitive that the catenary is unique, but is there a mathematical (rigorous) proof that this is so?
Thanks.
 A: Assume the cable is spanned form $(p_0,q_0)$ to $(p_1,q_1)$, $p_1>p_0$, and has a certain length $L$. Then its potential energy is minimal not only globally but also locally: If this cable passes through the points $(x_0,y_0)$ and $(x_1,y_1)$ and has length $s$ in between, then in the interval $[x_0,x_1]$  it assumes the exact shape that a cable of this length suspended from $(x_0,y_0)$ and $(x_1,y_1)$ would have. It follows that minimizing the potential energy globally enforces a certain local condition which translates into a second order differential equation for the curve $x\mapsto y(x)$. The derivation of this equation happens in the first pages of any book on variational calculus, and its solutions are curves of the form $y(x)=c \cosh(a x + b)$. It turns out that the constants $a$, $b$, $c$ are uniquely determined by $(p_0,q_0)$,  $(p_1,q_1)$ and the length $L$ of the cable, as long as $L\geq\sqrt{(p_1-p_0)^2+(q_1-q_0)^2}$. 
A: The catenary is unique because it arises from the direct integration of the balance of forces on a weighted cable. Looking at a small section ${\rm d}s = \sqrt{1+y'^2}\,{\rm d}x$ with slope $y'=\tan\theta$ and weight per unit length $w$ the balance of forces is
$$ {\rm d} H = 0\\ {\rm d} V = w \, {\rm d}s  $$
where $H$ is the horizontal component of the wire tension $T$, and $V$ the vertical. Since the tension is tangential to the wire then $y' = \frac{V}{H} \Rightarrow V = H\,y'$. The derivative of the above along the cable is
$$ w = \frac{{\rm d}V}{{\rm d}s} = H \frac{{\rm d}y'}{{\rm d}s} $$
  $$ w = H \frac{1}{\sqrt{1+y'^2}} \frac{{\rm d} y'}{{\rm d} x} $$
  $$\int w \,{\rm d} x = H \int \frac{1}{\sqrt{1+y'^2}} {\rm d} y' $$
  $$ w (x-x_C) = H \sinh^{-1}(y') $$
  $$ y' = \sinh\left(\frac{x-x_C}{H/w}\right) $$
  $$ y-y_C = A \left( \cosh\left( \frac{x-x_C}{A}\right)-1\right) $$
with $A=\frac{H}{w}$ the catenary constant. The coordinates $(x_C,y_C)$ are the constants of integration and represent the lowest point on the catenary.
