Hamilton's principle/Applications/The catenary

The catenary is the shape of a uniform flexible chain hanging under its own weight between two fixed points. It is the standard worked example of an isoperimetric problem: the chain's length is prescribed, and the shape minimises potential energy subject to that constraint.

the Beltrami identity derived the curve with the constraint suppressed, and Isoperimetric problems restored it. This page completes the problem: the constants, the tension, and the physical readings.

The functional and the constraint

A catenary and the directrix from which its tension is measured.

Let the chain have uniform linear density $\rho$ and hang between $(x_0,y_0)$ and $(x_1,y_1)$, with $y$ measured upward. Its potential energy is

$J[y]=\rho g_0{\int }_{x_0}^{x_1}y\sqrt{1+y{\text{'}}^2}dx\mathrm{(}\mathrm{1}\mathrm{)}$

since the mass of an arc element is $\rho ds$ at height $y$. The chain is inextensible, so its length is prescribed:

$K[y]={\int }_{x_0}^{x_1}\sqrt{1+y{\text{'}}^2}dx=\ell \mathrm{(}\mathrm{2}\mathrm{)}$

This is one condition on the whole curve, not one at each point, so its multiplier is a constant — the distinction drawn on Lagrange multipliers. Dropping the constant factor $\rho g_0$, the augmented integrand is

$L^{\ast }=(y+\mu )\sqrt{1+y{\text{'}}^2}\mathrm{(}\mathrm{3}\mathrm{)}$

The shape

$L^{\ast }$ carries no explicit $x$, so Beltrami applies. The same collapse as in (6) of the brachistochrone gives

$\frac{y+\mu }{\sqrt{1+y{\text{'}}^2}}=C\mathrm{(}\mathrm{4}\mathrm{)}$

Solve for the slope and separate:

$\frac{dy}{\sqrt{(y+\mu {)}^2/C^2-1}}=dx\mathrm{(}\mathrm{5}\mathrm{)}$

Substituting $y+\mu =C\cosh w$ makes the radicand ${\sinh }^{2}w$ and reduces the left side to $Cdw$, so $w=(x-{x_0}^{\prime })/C$ and

$y=-\mu +C\cosh \left(\frac{x-{x_0}^{\prime }}{C}\right)\mathrm{(}\mathrm{6}\mathrm{)}$

The curve is a hyperbolic cosine. It is not a parabola, a point Galileo got wrong and Huygens, Leibniz and Johann Bernoulli corrected in 1691. The two agree to second order near the vertex, which is why the error survived so long.

Fixing the three constants

$C$, ${x_0}^{\prime }$ and $\mu$ are fixed by the two endpoint conditions and by the length (2). The length of (6) integrates cleanly, because $\sqrt{1+y{\text{'}}^2}=\cosh ((x-{x_0}^{\prime })/C)$:

$\ell =C[\sinh \left(\frac{x_1-{x_0}^{\prime }}{C}\right)-\sinh \left(\frac{x_0-{x_0}^{\prime }}{C}\right)]\mathrm{(}\mathrm{7}\mathrm{)}$

For the symmetric case, endpoints at $x=\pm d$ and equal height, symmetry gives ${x_0}^{\prime }=0$ and (7) becomes

$\ell =2C\sinh \left(\frac{d}{C}\right)\mathrm{(}\mathrm{8}\mathrm{)}$

Equation (8) determines $C$ and is transcendental. Writing $u=d/C$, it reads $\sinh u/u=\ell /2d$. The left side increases monotonically from 1, so a solution exists precisely when $\ell >2d$ — the chain must be longer than the straight-line distance — and it is then unique.

The case $\ell =2d$ is the abnormal case excluded on Isoperimetric problems: the taut straight segment is the only admissible curve, it is an extremal of the length functional alone, and the multiplier rule does not apply to it. The exclusion in the theorem and the physical degenerate case are the same case.

$\mu$ then follows from the endpoint height, and the vertex sits at $y=C-\mu$.

Reading the multiplier

The multiplier is not bookkeeping. Restoring the factor $\rho g_0$, the horizontal component of the chain's tension is

$T_x=\rho g_{0}C\mathrm{(}\mathrm{9}\mathrm{)}$

constant along the chain, as it must be since no horizontal force acts on any element. The total tension at height $y$ is

$T=\rho g_0(y+\mu )\mathrm{(}\mathrm{10}\mathrm{)}$

which is (4) multiplied through. So $-\mu$ is the height of the directrix, the level from which tension is measured, and the tension at any point is the weight of a length of chain equal to the height above it.

That is a genuinely useful result and it is invisible in the unconstrained treatment. It is the practical argument for multipliers made on Lagrange multipliers, in a case where the answer is a number an engineer wants.

Tension is least at the vertex, where it equals $T_x$, and greatest at the supports. A chain hung slack has small $C$ and low tension; pulling it taut drives $C\to \mathrm{\infty }$ and the tension with it, which is why a horizontal cable cannot be pulled truly straight.

Is it a minimum?

$L_{y^{\prime }{y}^{\prime }}^{\ast }=(y+\mu )(1+y{\text{'}}^2{)}^{-3/2}$, positive wherever $y+\mu >0$, that is, above the directrix — which is everywhere on a physical chain, since $y+\mu =T/\rho g_0>0$. So the strengthened Legendre condition holds.

The Jacobi condition requires care here, because the second variation must be tested only against variations preserving the length to first order, as noted at the end of the second variation. Testing against unrestricted variations gives the wrong answer. With the restriction imposed, the catenary is a minimum.

The related surface

Rotating (6) about the horizontal axis gives the catenoid, the minimal surface spanning two coaxial rings. That problem minimises area,

$A=2\pi \int y\sqrt{1+y{\text{'}}^2}dx\mathrm{(}\mathrm{11}\mathrm{)}$

which is (1) without the constraint — so the unconstrained catenary of the Beltrami identity is the soap-film problem, and the constrained one is the chain. The same integrand, two different problems, distinguished only by whether a constraint is present.

The catenoid problem also exhibits the failure of existence. Pull the rings sufficiently far apart and no catenoid connects them: the extremals cease to exist and the physical film snaps to two discs. That is the Goldschmidt solution, and it is a discontinuous competitor no variational condition detects, since it is not near any extremal in any norm. It is the standard warning that stationarity theory presupposes existence, as noted on the Weierstrass excess function.

Provenance: the attributions to Galileo, and to the 1691 solutions of Huygens, Leibniz and Johann Bernoulli, are editorial. No primary source for any of them has been verified for this volume. Every equation here is derived on this page. See Sources.

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Geodesics treats the shortest-path problem on a curved surface.