Hamilton's principle/Applications/The brachistochrone - Revision history

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	\qquad\qquad\mathrm{(3)}</math>		\qquad\qquad\mathrm{(3)}</math>

	<math>L</math> contains no explicit <math>x</math>. ~~The Beltrami identity of~~		<math>L</math> contains no explicit <math>x</math>. [[Hamilton's principle/Mathematics/The Beltrami identity\|The Beltrami identity]] therefore applies:
	[[Hamilton's principle/Mathematics/The Beltrami identity\|~~the~~ Beltrami identity]] therefore applies:

	:<math>L - y'\,\frac{\partial L}{\partial y'} = C_1		:<math>L - y'\,\frac{\partial L}{\partial y'} = C_1

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			{{DISPLAYTITLE:The brachistochrone}}
	The '''brachistochrone''' is the curve of fastest descent: given two points at		The '''brachistochrone''' is the curve of fastest descent: given two points at
	different heights, find the wire along which a bead sliding without friction		different heights, find the wire along which a bead sliding without friction
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	<references />		<references />

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The '''brachistochrone''' is the curve of fastest descent: given two points at
different heights, find the wire along which a bead sliding without friction
under gravity travels from one to the other in the least time. Johann Bernoulli
posed it as a challenge in 1696, and the problem is conventionally taken as the
birth of the calculus of variations.

The answer is a '''cycloid''', and this page derives it.

== Setting up the functional ==

[[File:OPL-HP-F-06.png|thumb|right|400px|The brachistochrone as a cycloid.]]

Place the start at the origin and let <math>y</math> increase '''downward''', so
gravity is in the direction of increasing <math>y</math>. The bead starts from
rest. Energy conservation gives its speed after falling a height
<math>y</math>:

:<math>\tfrac{1}{2}mv^2 = mg_0 y
\qquad \Longrightarrow \qquad
v = \sqrt{2g_0 y}
\qquad\qquad\mathrm{(1)}</math>

The time to traverse an element of arc <math>ds</math> is <math>ds/v</math>, and
<math>ds = \sqrt{1 + y'^2}\, dx</math>, so the total time is

:<math>J[y] = \int_0^{x_1} \frac{\sqrt{1 + y'^2}}{\sqrt{2g_0 y}}\; dx
\qquad\qquad\mathrm{(2)}</math>

Note the sign convention doing work here. Measuring <math>y</math> downward makes
the integrand real and positive without an extra minus sign, and it is why the
brachistochrone is conventionally drawn upside down.

== Solving ==

The constant <math>\sqrt{2g_0}</math> does not affect the extremals, so take

:<math>L = \sqrt{\frac{1 + y'^2}{y}}
\qquad\qquad\mathrm{(3)}</math>

<math>L</math> contains no explicit <math>x</math>. The Beltrami identity of
[[Hamilton's principle/Mathematics/The Beltrami identity]] therefore applies:

:<math>L - y'\,\frac{\partial L}{\partial y'} = C_1
\qquad\qquad\mathrm{(4)}</math>

Compute the derivative:

:<math>\frac{\partial L}{\partial y'}
= \frac{y'}{\sqrt{y}\,\sqrt{1 + y'^2}}
\qquad\qquad\mathrm{(5)}</math>

so the left side of (4) is

:<math>\frac{\sqrt{1+y'^2}}{\sqrt{y}} - \frac{y'^2}{\sqrt{y}\sqrt{1+y'^2}}
= \frac{(1 + y'^2) - y'^2}{\sqrt{y}\,\sqrt{1 + y'^2}}
= \frac{1}{\sqrt{y}\,\sqrt{1 + y'^2}}
\qquad\qquad\mathrm{(6)}</math>

The same collapse as in the catenary on
[[Hamilton's principle/Mathematics/The Beltrami identity]]: the numerator
simplifies to a constant. Setting (6) equal to <math>C_1</math> and squaring,

:<math>y\,(1 + y'^2) = k^2, \qquad k^2 = 1/C_1^2
\qquad\qquad\mathrm{(7)}</math>

Equation (7) is the brachistochrone equation. A second-order problem has become
a first-order separable one in six lines.

== Integrating ==

Solve (7) for the slope:

:<math>y' = \sqrt{\frac{k^2 - y}{y}}
\qquad\qquad\mathrm{(8)}</math>

and separate:

:<math>dx = \sqrt{\frac{y}{k^2 - y}}\; dy
\qquad\qquad\mathrm{(9)}</math>

The substitution that works is

:<math>y = k^2 \sin^2\!\left( \frac{\theta}{2} \right)
= \frac{k^2}{2}\bigl( 1 - \cos\theta \bigr)
\qquad\qquad\mathrm{(10)}</math>

chosen because it makes the radicand a perfect square:
<math>k^2 - y = k^2\cos^2(\theta/2)</math>. Then
<math>dy = k^2 \sin(\theta/2)\cos(\theta/2)\, d\theta</math>, and (9) becomes

:<math>dx = \frac{\sin(\theta/2)}{\cos(\theta/2)}
\cdot k^2 \sin\!\left(\frac{\theta}{2}\right)\cos\!\left(\frac{\theta}{2}\right) d\theta
= k^2 \sin^2\!\left( \frac{\theta}{2} \right) d\theta
= \frac{k^2}{2}\bigl( 1 - \cos\theta \bigr)\, d\theta
\qquad\qquad\mathrm{(11)}</math>

which integrates immediately. Writing <math>a = k^2/2</math> and taking
<math>x = 0</math> at <math>\theta = 0</math>,

:<math>x = a\,(\theta - \sin\theta),
\qquad
y = a\,(1 - \cos\theta)
\qquad\qquad\mathrm{(12)}</math>

Equations (12) are the parametric equations of a '''cycloid''' — the curve traced
by a point on the rim of a circle of radius <math>a</math> rolling along the
underside of the horizontal axis.

== Fixing the constants ==

<math>a</math> and the final parameter value <math>\theta_1</math> are fixed by
requiring the curve to pass through <math>(x_1, y_1)</math>:

:<math>\frac{x_1}{y_1} = \frac{\theta_1 - \sin\theta_1}{1 - \cos\theta_1}
\qquad\qquad\mathrm{(13)}</math>

Equation (13) is transcendental in <math>\theta_1</math> and is solved
numerically; <math>a</math> then follows from either equation of (12). The
right-hand side of (13) increases monotonically from 0 at
<math>\theta_1 = 0</math>, so a solution exists and is unique for any endpoint
below and to the right of the start.

== The descent time ==

Substituting (12) into (2) gives a result worth having. With
<math>ds = a\sqrt{2(1-\cos\theta)}\,d\theta</math> and
<math>v = \sqrt{2g_0 a(1-\cos\theta)}</math>, the integrand reduces to the
constant <math>\sqrt{a/g_0}</math> and

:<math>t = \sqrt{\frac{a}{g_0}}\; \theta_1
\qquad\qquad\mathrm{(14)}</math>

The integrand being constant is the '''tautochrone''' property: the bead's
angular parameter advances uniformly in time. Huygens had found in 1659 that the
cycloid is the curve on which a bead reaches the bottom in a time independent of
where it starts, and (14) is that result. The curve of fastest descent and the
curve of equal-time descent are the same curve, discovered independently
thirty-seven years apart.

== Two features that surprise ==

'''The curve dips below the endpoint.''' When <math>x_1</math> is large relative
to <math>y_1</math>, the solution has <math>\theta_1 > \pi</math> and the cycloid
descends past <math>y_1</math> before rising to meet it. Going ''down'' first buys
speed that more than repays the extra distance. No local reasoning gives this;
it comes out of the integration.

'''The initial descent is vertical.''' As <math>\theta \to 0</math>, (12) gives
<math>x \sim a\theta^3/6</math> and <math>y \sim a\theta^2/2</math>, so
<math>y'\to\infty</math>. The bead drops straight down at the start, which is
exactly what (1) demands: at <math>y = 0</math> the speed is zero, and the only
way to keep the time integral finite is to leave that point as fast as possible.

== Is it a minimum? ==

The derivation shows only that the cycloid is '''stationary'''. Applying the
conditions of the sufficiency track:

<math>L_{y'y'} = y^{-1/2}(1+y'^2)^{-3/2} > 0</math> for <math>y > 0</math>, so
the strengthened Legendre condition of
[[Hamilton's principle/Mathematics/The Legendre condition]] holds on the open
interval. The cycloidal extremals through the origin do not refocus before
<math>\theta = 2\pi</math>, so there is no conjugate point in range and the
Jacobi condition of
[[Hamilton's principle/Mathematics/The Jacobi condition and conjugate points]]
holds. The cycloid is therefore a weak local minimum, and with the excess
function positive it is a strong one.

The endpoint <math>y = 0</math> is singular: <math>L</math> is not differentiable
there and <math>P \to \infty</math>. The problem is an improper integral at that
end, and a rigorous treatment excises a neighbourhood and takes a limit. This is
noted rather than performed.

<blockquote>
Provenance: the attributions to Johann Bernoulli 1696 and Huygens 1659 are
editorial. No primary source for either has been verified for this volume; see
[[Hamilton's principle/Sources]]. Every equation on this page is derived here and
none is transcribed from a source.
</blockquote>

== Next ==

[[Hamilton's principle/Applications/The catenary]] works the constrained
companion problem.

== References ==

<references />

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	<math>L</math> contains no explicit <math>x</math>. The Beltrami identity of		<math>L</math> contains no explicit <math>x</math>. The Beltrami identity of
	[[Hamilton's principle/Mathematics/The Beltrami identity]] therefore applies:		[[Hamilton's principle/Mathematics/The Beltrami identity\|the Beltrami identity]] therefore applies:

	:<math>L - y'\,\frac{\partial L}{\partial y'} = C_1		:<math>L - y'\,\frac{\partial L}{\partial y'} = C_1
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	The same collapse as in the catenary on		The same collapse as in the catenary on
	[[Hamilton's principle/Mathematics/The Beltrami identity]]: the numerator		[[Hamilton's principle/Mathematics/The Beltrami identity\|the Beltrami identity]]: the numerator
	simplifies to a constant. Setting (6) equal to <math>C_1</math> and squaring,		simplifies to a constant. Setting (6) equal to <math>C_1</math> and squaring,

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	<math>L_{y'y'} = y^{-1/2}(1+y'^2)^{-3/2} > 0</math> for <math>y > 0</math>, so		<math>L_{y'y'} = y^{-1/2}(1+y'^2)^{-3/2} > 0</math> for <math>y > 0</math>, so
	the strengthened Legendre condition of		the strengthened Legendre condition of
	[[Hamilton's principle/Mathematics/The Legendre condition]] holds on the open		[[Hamilton's principle/Mathematics/The Legendre condition\|the Legendre condition]] holds on the open
	interval. The cycloidal extremals through the origin do not refocus before		interval. The cycloidal extremals through the origin do not refocus before
	<math>\theta = 2\pi</math>, so there is no conjugate point in range and the		<math>\theta = 2\pi</math>, so there is no conjugate point in range and the
	Jacobi condition of		Jacobi condition of
	[[Hamilton's principle/Mathematics/The Jacobi condition and conjugate points]]		[[Hamilton's principle/Mathematics/The Jacobi condition and conjugate points\|the Jacobi condition and conjugate points]]
	holds. The cycloid is therefore a weak local minimum, and with the excess		holds. The cycloid is therefore a weak local minimum, and with the excess
	function positive it is a strong one.		function positive it is a strong one.
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	Provenance: the attributions to Johann Bernoulli 1696 and Huygens 1659 are		Provenance: the attributions to Johann Bernoulli 1696 and Huygens 1659 are
	editorial. No primary source for either has been verified for this volume; see		editorial. No primary source for either has been verified for this volume; see
	[[Hamilton's principle/Sources]]. Every equation on this page is derived here and		[[Hamilton's principle/Sources\|Sources]]. Every equation on this page is derived here and
	none is transcribed from a source.		none is transcribed from a source.
	</blockquote>		</blockquote>
Line 182:		Line 182:
	== Next ==		== Next ==

	[[Hamilton's principle/Applications/The catenary]] works the constrained		[[Hamilton's principle/Applications/The catenary\|the catenary]] works the constrained
	companion problem.		companion problem.