# Differential Equations

## Second Order Equations # Equation of Catenary

The catenary is a plane curve, whose shape corresponds to a hanging homogeneous flexible chain supported at its ends and sagging under the force of gravity.

The catenary is similar to parabola (Figure $$1$$).

So it was believed for a long time. In the early $$17$$th century Galileo doubted that a hanging chain is actually a parabola. However, a rigorous proof was obtained only half a century later after Isaak Newton and Gottfried Leibniz developed a framework of differential and integral calculus.

The solution of the problem about the catenary was published in $$1691$$ by Christiaan Huygens, Gottfried Leibniz, and Johann Bernoulli.

Below we derive the equation of catenary and some its variations.

Suppose that a heavy uniform chain is suspended at points $$A, B,$$ which may be at different heights (Figure $$2$$).

Consider equilibrium of a small element of the chain of length $$\Delta s.$$ The forces acting on the section of the chain are the distributed force of gravity

$\Delta P = \rho g\Delta s,$

where $$\rho$$ is the density of the chain material, $$g$$ is the acceleration of gravity, $$A$$ is the cross sectional area of the thread, and the tension forces $$T\left( x \right)$$ and $$T\left( {x + \Delta x} \right),$$ respectively, at points $$x$$ and $${x + \Delta x}.$$

The equilibrium conditions of the element of the length $$\Delta s$$ for projections on the axes $$Ox$$ and $$Oy$$ are written as

$- T\left( x \right)\cos \alpha \left( x \right) + T\left( {x + \Delta x} \right)\cos \alpha \left( {x + \Delta x} \right) = 0,$
$- T\left( x \right)\sin\alpha \left( x \right) + T\left( {x + \Delta x} \right)\sin\alpha \left( {x + \Delta x} \right) - \Delta P = 0.$

It follows from the first equation that the horizontal component of the tension force $$T\left( x \right)$$ is always a constant:

$T\left( x \right)\cos \alpha \left( x \right) = {T_0} = \text{const}.$

Using differentials in the second equation we can rewrite it as

$d\left( {T\left( x \right)\sin\alpha \left( x \right)} \right) = dP\left( x \right).$

As $$T\left( x \right) = \frac{{{T_0}}}{{\cos\alpha \left( x \right)}},$$ we have

$d\left( {{T_0}\tan\alpha \left( x \right)} \right) = dP\left( x \right),\;\; \Rightarrow {T_0}d\left( {\tan\alpha \left( x \right)} \right) = dP\left( x \right).$

Take into account that $$\tan \alpha \left( x \right) = \frac{{dy}}{{dx}} = y',$$ so the equilibrium equation is written in the differential form as

${T_0}d\left( {y'} \right) = dP\left( x \right),\;\; \Rightarrow {T_0}d\left( {y'} \right) = \rho gAds.$

The chain element of the length $$\Delta s$$ can be expressed by the formula

$ds = \sqrt {1 + {{\left( {y'} \right)}^2}} dx.$

As a result we obtain the differential equation of the catenary:

${T_0}\frac{{dy'}}{{dx}} = \rho gA\sqrt {1 + {{\left( {y'} \right)}^2}} ,\;\; \Rightarrow {T_0}y^{\prime\prime} = \rho gA\sqrt {1 + {{\left( {y'} \right)}^2}} .$

The order of this equation can be reduced. By denoting $$y^{\prime} = z,$$ we can represent it as the first order equation:

${T_0}z' = \rho gA\sqrt {1 + {z^2}} .$

The last equation can be solved by separating variables.

${T_0}dz = \rho gA\sqrt {1 + {z^2}} dx,\;\; \Rightarrow \frac{{dz}}{{\sqrt {1 + {z^2}} }} = \frac{{\rho gA}}{{{T_0}}}dx,\;\; \Rightarrow \int {\frac{{dz}}{{\sqrt {1 + {z^2}} }}} = \frac{{\rho gA}}{{{T_0}}}\int {dx} ,\;\; \Rightarrow \ln \left( {z + \sqrt {1 + {z^2}} } \right) = \frac{x}{a} + {C_1}.$

Here we denoted $$\frac{{\rho gA}}{{{T_0}}}$$ as $$\frac{1}{a}.$$

The tangent to the catenary at the lowest point is parallel to the $$x$$-axis. Hence,

$z\left( {x = 0} \right) = y'\left( {x = 0} \right) = 0.$

We can determine the constant $${C_1}$$ from here:

$\ln 1 = 0 + {C_1},\;\; \Rightarrow {C_1} = 0.$

Thus, we get the following equation:

$z + \sqrt {1 + {z^2}} = {e^{\frac{x}{a}}}.$

Multiplying both sides of the equation by the conjugate expression $$z - \sqrt {1 + {z^2}}$$ gives

$\left( {z + \sqrt {1 + {z^2}} } \right) \left( {z - \sqrt {1 + {z^2}} } \right) = \left( {z - \sqrt {1 + {z^2}} } \right){e^{\frac{x}{a}}},\;\; \Rightarrow {z^2} - \left( {1 + {z^2}} \right) = \left( {z - \sqrt {1 + {z^2}} } \right){e^{\frac{x}{a}}},\;\; \Rightarrow - 1 = \left( {z - \sqrt {1 + {z^2}} } \right){e^{\frac{x}{a}}},\;\; \Rightarrow z - \sqrt {1 + {z^2}} = - {e^{ - \frac{x}{a}}}.$

Adding to the previous equation, we find the expression for $$z = y':$$

$z + \cancel{\sqrt {1 + {z^2}}} + z - \cancel{\sqrt {1 + {z^2}}} = {e^{\frac{x}{a}}} - {e^{ - \frac{x}{a}}},\;\; \Rightarrow z = \frac{{{e^{\frac{x}{a}}} - {e^{ - \frac{x}{a}}}}}{2} = \sinh \frac{x}{a},\;\; \Rightarrow y' = \sinh \frac{x}{a}.$

Integrating once more gives the final nice expression for the shape of the catenary:

$y = a\cosh \frac{x}{a}.$

Thus, the catenary is described by the hyperbolic cosine function. Its shape is uniquely determined by the parameter $$a = \frac{{{T_0}}}{{\rho gA}}$$ as shown in Figure $$3.$$

The archs in the form of an inverted catenary (such as Saarinen's Gateway Arch in St.Louis shown in Figure $$4$$) are often used in architecture and construction.