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.
It is physically intuitive that the catenary is unique?
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3"is there a mathematical (rigorous) proof that this is so?" - yes, you solve the differential equations that govern the physics of the cable. – 2011-08-11
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0Why is it physically intuitive that the catenary is unique? – 2011-08-11
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0@Quinn: Because every time one hangs a particular cable on 2 pegs/supports by it's ends, it assumes the same form. – 2011-08-11
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0This is a path integral that minimizes the energy. I can think of settings where there are more than one minimum, for example consider a springy steel ribbon also suspended between two posts and also feeling gravity. Symmetry breaking? This problem only has one minimum. Is the idea to prove that there is only one minimum? The path integral itself is solved by requiring that it is a minimum. – 2011-08-11
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0@J.M.: Thanks. However I seem a bit stuck in solving the equations. Suppose the 2 support points have coordinates (p, q) and (-p, q). Then how am I to determine A in the expression above? – 2011-08-11
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0I have always seen the DE of the catenary derived from simple physical principles: balancing the forces affecting an infinitesimal segment of the chain + using the physically "obvious" fact that at an equilibrium the force due to stress at any point of the chain must be tangential. If you are not afraid of text in Finnish, then such a derivation (and a solution of the DE) is on pages 170-171 [of my lecture notes for 1st year calculus](http://users.utu.fi/lahtonen/Analyysi2011Kevat/AnalyysiII2011.pdf) (spring term). May be this is approach is too simplistic? – 2011-08-11
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0@Lea: Giving two points is not enough. You also need to know the length of the chain/wire/cable. – 2011-08-11
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0@Lea: *Because every time one hangs a particular cable on 2 pegs/supports by it's ends, it assumes the same form*... First, for a given cable, changing the location of the pegs changes the form. Second, this proves nothing, one would have to compare the forms assumed by cables of different lengthes and weights, and once again this changes the form. – 2011-08-11
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0"...compare the forms assumed by cables of different lengthes and weights..." - and different materials. There's no reason to expect that both yarn and cord would both hang as catenaries... – 2011-08-11
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0Have you read the Planeth Math entry http://planetmath.org/encyclopedia/Catenary.html ? – 2011-08-11
2 Answers
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 $L\geq\sqrt{(p_1-p_0)^2+(q_1-q_0)^2}$.
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0If looking for a (formal) derivation of this equation that doesn't involve variational calculus, could you recommend any specific textbook? – 2018-04-02
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.