Solving the brachistochrone and other variational problems with soap films Available to Purchase

Johann

 

Bernoulli

challenged the mathematical world with the brachistochrone problem, published in Acta Eruditorum in June, 1696. See , edited by

D. J.

 

Struick

(, , ), pp. –.

To obtain this solution, he used Fermat’s principle and assumed that the curve of shortest descent time must be the same as the curve described by a light ray whose velocity is $v=2gy$⁠, where $y$ is the height because this velocity coincides with the velocity that a body acquires after descending a distance $y$⁠. This solution was published in Acta Eruditorum in May, , pp. –. In the same issue of this journal, there also appeared the solutions to the brachistochrone problem given by his brother Jakob Bernoulli, Leibniz, L’Hôpital, Tschirnhaus, and Newton. All except L’Hôpital found the cycloid as the solution. The method used by Jakob Bernoulli contained the roots of the calculus of variations (Ref. 1).

R.

 

Courant

and

H.

 

Robbins

, (, , ).

J. A. F.

 

Plateau

, (, , ), Vols. 1 and 2.

C. V.

 

Boys

, (, , ).

S.

 

Hildebrant

and

A.

 

Tromba

, (, , ).

C.

 

Isenberg

, (, , ).

This paper was inspired by reading the book,

D.

 

Lovett

, (, , ).

A sufficient condition for a curve $γ$ to make the integral $∫γf(y,y′,x)dx$ extremal is that it satisfies the Euler–Lagrange equation $∂f/∂y−(d/dx)(∂f/∂y′)=0$⁠. See

H.

 

Goldstein

, (, , ), p. . If $f$ does not depend on $x$⁠, the Euler–Lagrange equation reduces to $y′(∂f/∂y′)−f=constant$⁠. In our case, $f(y,y′)=1+y′2/y$⁠, and we obtain the differential equation $y′=(k−y)/y$⁠, where $k$ is a constant. It is easy to see that its solution is the cycloid.

R.

 

Osserman

, (, , ).

According to the first Plateau law (see, for example, Ref. 8, pp. –), the film must meet both surfaces $z=0$ and $z=1/y$ at an angle of 90°. Therefore, the approximation of taking a vertical surface $Σ$ becomes better as $y$ increases because the slope of the tangent to $z=1/y$ becomes smaller.

A rough estimate of the relative error when we take a vertical element of surface $ΔA$ instead of the element of surface for the soap film can be calculated by $ϵr=(ΔA′−ΔA)/ΔA=(α−tanα)Δs/tanαΔs$⁠, where $ΔA′$ is the element of surface for a cylindrical surface that is perpendicular to the two plates, $Δs$ is the line element along the curve $γ$⁠, and $tanα=dz/dy=z′$⁠. If we neglect the terms of order greater than 4 in the expansion of $α−tanα$ as function of the derivative $z′$⁠, we obtain $ϵr≃−z′2/3$⁠, which for $z=1/y$ leads to $ϵr≈−y−3/12$⁠. Therefore, for $y>2$⁠, the error is less than 1%. Similar calculations can be made for $z=ky$ and $z=1/y$ for the catenary and the Poincaré half-plane, respectively.

C.

 

Isenberg

, “,”  , – ().

See Ref. 1, pp. –.

Johann

 

Bernoulli

could easily solve this problem because of his early study on the synchronies, which are the orthogonal trajectories to the cycloids passing through O (see Fig. 6). From a mechanical point of view, these curves are formed by simultaneous positions of particles, which are released at O along the cycloids at the same time. The synchronies also correspond to the simultaneous positions of light pulses that are emitted from O at the same instant and propagate along a medium with a refractive index proportional to $1/y$⁠. The synchronies are the wavefronts. From the fact that wavefronts and light rays are perpendicular, Johann Bernoulli concluded that the synchronies are the orthogonal trajectories to the cycloids. Thus, the point of fastest descent $P$ must be the point where one of the synchronies is tangent to the line AB (see Fig. 6). For more details, see

S. B.

 

Engelsman

, (, , ), pp. –.

The parameter $a$ is determined by $a2+(L/2)2=acoshH/2a$⁠. The suspension points are symmetrical with respect to the $y$-axis and their distance to the $x$-axis is $b=a2+(L/2)2$⁠. For the proof of these properties, see

J. L.

 

Troutman

, (, , ), pp. –.

If the ratio $H/b$ of the pin separation to the distance from the pins to the $x$-axis is greater than 1.335, there is no parameter $a$ that is a solution of $b=acoshH/2a$⁠; for $H/b<1.335$⁠, there are two values of parameter $a$⁠.

The catenoid is the surface of revolution generated by the catenary, and, as Euler showed in 1744, is the only minimal surface of revolution. The reason is that the area of a surface of revolution generated by a curve $γ$ of equation $y=y(x)$ is given by $A(γ)=2π∫γyds=2π∫γy1+y′2dx$ and coincides, except for a constant factor, with Eqs. (6) and (8). The Euler–Lagrange equation reduces in this case to $y/1+y′2=1/c1$⁠, and its integration gives the catenary defined by the equation $y=c1cosh(x+c2)/c1$⁠.

The Euler–Lagrange equation reduces in this case to $1/y1+y′2=c1$⁠, and its integration gives the semicircle of $y=c1sint$⁠, $x=c2+c1cost$⁠, for $0<t<π$⁠. The Poincaré half-plane is one of the simplest models of non-Euclidean spaces. The geodesics of this space are the semicircles with centers at point C on the axis $y=0$ together with the vertical straight lines. If we think of these geodesics as the “straight lines” of this space, the associated geometry is not consistent with the fifth Euclidean postulate (the parallel postulate) because given a “straight line” and a point not on that line, there are infinitely many lines through the point that do not cut (that is, parallel to) the given straight line. Only the small step of considering semicircles as “straight lines” prevented Johann Bernoulli from discovering non-Euclidean geometry 130 years before Bolyay and Lobatchesky.

W.

 

Rindler

, (, , ), p. .

The Poincaré half-plane is a model of hyperbolic geometry used in special relativity. See, for example,

C.

Criado

and

N.

Alamo

, “,”  , – ()

Hyperbolic geometry is also used in general relativity. See, for example,

C.

 

Criado

and

N.

 

Alamo

, “,”  , – ().