Theorem: Bernoulli's Hanging Chain Problem

Consider a uniform chain $C$ whose physical properties are as follows:
:$C$ is of length $L$
:The mass per unit length of $C$ is $\lambda$
:$C$ is of zero stiffness.
Let $C$ be suspended in a vertical line from a fixed point and otherwise free to move.
Let $C$ be slightly disturbed in a vertical plane from its position of stable equilibrium.
Let $\map y t$ be the horizontal displacement at time $t$ from its position of stable equilibrium of a particle of $C$ which is a vertical distance $x$ from its point of attachment.
The $2$nd order ODE describing the motion of $y$ is:
:$\dfrac {\d^2 y} {\d t^2} = g \paren {L - x} \dfrac {\d^2 y} {\d x^2} - g \dfrac {\d y} {\d x}$
where $g$ is the acceleration due to gravity.

Proof:
{{tidy|minor stylistic stuff, no big deal}}
The following plot shows the chain at a single instant of its motion.
thumb600pxalt=The picture shows an x-y plot of the hanging [[Definition:Chain (Physics)chain. The chain is attached at the origin and hangs along the $x$-axis. The chain lies along a curving path to show its configuration at a random moment while shaking. There is a point marked with (x, y) and a pair of lines extend from it--one parallel to the $x$-axis, one parallel to the chain at that point. These two lines show the angle $\theta$ of the chain with respect to the $x$-axis.|A plot of the chain (blue) as it hangs shaking. The plot is oriented so that down is to the right along the $x$-axis. The marked point will be used for analysis.]]
The chain is fixed at the origin and free to move everywhere else.
Consider the point $\tuple {x, y}$ in the diagram above and the small bit of chain that extends to the right from that point for a length $\d x$.
The mass of this small bit of chain is $\lambda \rd x$.
The bit of chain is subject to three forces:
:$(1): \quad$ Gravity, with a force of $g \lambda \rd x$ in the positive $x$ direction
:$(2): \quad$ Tension at $x$ that pulls in the $-x$ direction: $\map T {x, t}$
:$(3): \quad$ Tension at $x + \d x$ that pulls in the $+x$ direction: $\map T {x + \d x, t}$
Because these forces are not parallel, they must be added as vectors.
The tension force act parallel to the chain at $x$ at an angle $\theta$ with the $x$-axis.
This angle is defined by
{{begin-eqn}}
{{eqn | l = \tan \theta
      | r = \frac {\d y} {\d x}
      | n = 1
}}
{{end-eqn}}
The chain is only slightly displaced, so the angle $\theta$ is small, giving the following approximations:
{{begin-eqn}}
{{eqn | l = \tan \theta
      | r = \sin \theta
      | o = \approx
      | n = 2
}}
{{end-eqn}}
and
{{begin-eqn}}
{{eqn | l = \cos \theta
      | r = 1
      | o = \approx
      | n = 3
}}
{{end-eqn}}
First, consider the $x$ direction.
By Newton's Second Law of Motion,
{{begin-eqn}}
{{eqn | l = m \mathbf a_x
      | r = \sum_i \mathbf F_{xi}
}}
{{eqn | l = \lambda \rd x \frac {\d^2 x} {\d t^2}
      | r = \lambda g \rd x + \map T {x + \d x, t} \map \cos {\theta \paren {x + \d x, t} } - \map T {x, t} \map \cos {\theta \paren {x, t} }
}}
{{eqn | r = \lambda g \rd x + \map {\frac \d {\d x} } {\map T {x, t} \cos \theta \paren {x, t} } \rd x
}}
{{eqn | r = \lambda g \rd x + \frac \d {\d x} \map T {x, t} \rd x
      | c = Substitute using $(3)$
}}
{{eqn | ll= \leadsto
      | l = \lambda \frac {\d^2 x} {\d t^2}
      | r = \lambda g + \frac \d {\d x} \map T {x, t}
      | c = dividing both sides by $\d x$
}}
{{end-eqn}}
Since the chain is only slightly disturbed and the chain is not extensible, each bit of chain cannot move in the $x$ direction.
So, according to Newton's Second Law of Motion, the acceleration in the $x$ direction, and thus the total force in the $x$ direction, must be zero.
{{begin-eqn}}
{{eqn | l = \lambda g + \frac \d {\d x} \map T {x, t}
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = \map T {x, t}
      | r = -\lambda g x + C
      | c = integrating {{WRT|Integration}} $x$
}}
{{end-eqn}}
The boundary conditions for the chain are
{{begin-eqn}}
{{eqn | l = \map T {x = L, t}
      | r = 0
      | c = The free end of the chain has no tension
}}
{{eqn | l = \map T {x = 0, t}
      | r = \lambda g L
      | c = The attachment point holds up the entire chain.
}}
{{end-eqn}}
Either one of these boundary conditions can be used solve for the integration constant.
:$\ds C = \lambda g L$
So, the tension in the chain is given by
{{begin-eqn}}
{{eqn | l = \map T x
      | r = \lambda g \paren {L - x}
      | n = 4
}}
{{end-eqn}}
Note that the tension does not have any dependence on time.
Now consider the forces in the $y$ direction.
The only forces in this direction are the two tensions pulling in nearly opposite directions.
By Newton's Second Law of Motion, we have
{{begin-eqn}}
{{eqn | l = m \mathbf a_y
      | r = \sum_i \mathbf F_{yi}
}}
{{eqn | l = \lambda \rd x \frac {\d^2 y} {\d t^2}
      | r = \map T {x + \d x} \sin \theta \paren {x + \d x, t} - \map T x \sin \theta \paren {x, t}
}}
{{eqn | r = \map {\frac \d {\d x} } {\map T x \sin \theta \paren {x, t} } \rd x
}}
{{eqn | r = \map {\frac \d {\d x} } {\lambda g \paren {L - x} \frac {\d y} {\d x} } \rd x
      | c = Substitute using $(1)$, $(2)$, and $(4)$
}}
{{eqn | ll= \leadsto
      | l = \frac {\d^2 y} {\d t^2}
      | r = \map {\frac \d {\d x} } {g \paren {L - x} \frac {\d y} {\d x} }
      | c = Divide by $\lambda \rd x$
}}
{{eqn | r = g \paren {L - x} \frac {\d^2 y} {\d x^2} - g \frac {\d y} {\d x}
      | c = Product Rule for Derivatives
}}
{{end-eqn}}
{{qed}}
{{Namedfor|Daniel Bernoulli|cat = Bernoulli, Daniel}}