Hamilton's principle/Applications/The brachistochrone

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

The brachistochrone as a cycloid.

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

$\frac{1}{2}m{v}^2=m{g}_{0}y⟹v=\sqrt{2g_{0}y}\mathrm{(}\mathrm{1}\mathrm{)}$

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

$J[y]={\int }_0^{x_1}\frac{\sqrt{1+y{\text{'}}^2}}{\sqrt{2g_{0}y}}dx\mathrm{(}\mathrm{2}\mathrm{)}$

Note the sign convention doing work here. Measuring $y$ 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 $\sqrt{2g_0}$ does not affect the extremals, so take

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

$L$ contains no explicit $x$. The Beltrami identity therefore applies:

$L-y^{\prime }\frac{\partial L}{\partial y^{\prime }}=C_1\mathrm{(}\mathrm{4}\mathrm{)}$

Compute the derivative:

$\frac{\partial L}{\partial y^{\prime }}=\frac{y^{\prime }}{\sqrt{y}\sqrt{1+y{\text{'}}^2}}\mathrm{(}\mathrm{5}\mathrm{)}$

so the left side of (4) is

$\frac{\sqrt{1+y{\text{'}}^2}}{\sqrt{y}}-\frac{y{\text{'}}^2}{\sqrt{y}\sqrt{1+y{\text{'}}^2}}=\frac{(1+y{\text{'}}^2)-y{\text{'}}^2}{\sqrt{y}\sqrt{1+y{\text{'}}^2}}=\frac{1}{\sqrt{y}\sqrt{1+y{\text{'}}^2}}\mathrm{(}\mathrm{6}\mathrm{)}$

The same collapse as in the catenary on the Beltrami identity: the numerator simplifies to a constant. Setting (6) equal to $C_1$ and squaring,

$y(1+y{\text{'}}^2)=k^2,k^2=1/C_1^2\mathrm{(}\mathrm{7}\mathrm{)}$

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:

$y^{\prime }=\sqrt{\frac{k^2-y}{y}}\mathrm{(}\mathrm{8}\mathrm{)}$

and separate:

$dx=\sqrt{\frac{y}{k^2-y}}dy\mathrm{(}\mathrm{9}\mathrm{)}$

The substitution that works is

$y=k^2{\sin }^2\left(\frac{\theta }{2}\right)=\frac{k^2}{2}(1-\cos \theta )\mathrm{(}\mathrm{10}\mathrm{)}$

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

$dx=\frac{\sin (\theta /2)}{\cos (\theta /2)}\cdot k^2\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)d\theta =k^2{\sin }^2\left(\frac{\theta }{2}\right)d\theta =\frac{k^2}{2}(1-\cos \theta )d\theta \mathrm{(}\mathrm{11}\mathrm{)}$

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

$x=a(\theta -\sin \theta ),y=a(1-\cos \theta )\mathrm{(}\mathrm{12}\mathrm{)}$

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

Fixing the constants

$a$ and the final parameter value ${\theta }_1$ are fixed by requiring the curve to pass through $(x_1,y_1)$:

$\frac{x_1}{y_1}=\frac{{\theta }_1-\sin {\theta }_1}{1-\cos {\theta }_1}\mathrm{(}\mathrm{13}\mathrm{)}$

Equation (13) is transcendental in ${\theta }_1$ and is solved numerically; $a$ then follows from either equation of (12). The right-hand side of (13) increases monotonically from 0 at ${\theta }_1=0$, 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 $ds=a\sqrt{2(1-\cos \theta )}d\theta$ and $v=\sqrt{2g_{0}a(1-\cos \theta )}$, the integrand reduces to the constant $\sqrt{a/g_0}$ and

$t=\sqrt{\frac{a}{g_0}}{\theta }_1\mathrm{(}\mathrm{14}\mathrm{)}$

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 $x_1$ is large relative to $y_1$, the solution has ${\theta }_1>\pi$ and the cycloid descends past $y_1$ 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 $\theta \to 0$, (12) gives $x\sim a{\theta }^3/6$ and $y\sim a{\theta }^2/2$, so $y^{\prime }\to \mathrm{\infty }$. The bead drops straight down at the start, which is exactly what (1) demands: at $y=0$ 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:

$L_{y^{\prime }{y}^{\prime }}=y^{-1/2}(1+y{\text{'}}^2{)}^{-3/2}>0$ for $y>0$, so the strengthened Legendre condition of the Legendre condition holds on the open interval. The cycloidal extremals through the origin do not refocus before $\theta =2\pi$, so there is no conjugate point in range and the Jacobi condition of 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 $y=0$ is singular: $L$ is not differentiable there and $P\to \mathrm{\infty }$. 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.

Provenance: the attributions to Johann Bernoulli 1696 and Huygens 1659 are editorial. No primary source for either has been verified for this volume; see Sources. Every equation on this page is derived here and none is transcribed from a source.

Next

the catenary works the constrained companion problem.