Find the surface enclosing the maximum volume per unit surface area, . The solution is a sphere, which hasThe fact that a sphere solves the isovolume problem was only proved as recently as 1882 by Schwarz (Haas 2000).
The problem of finding the curve down which a bead placed anywhere will fall to the bottom in the same amount of time. The solution is a cycloid, a fact first discovered and published by Huygens in Horologium oscillatorium (1673). This property was also alluded to in the following passage from Moby Dick: "[The try-pot] is also a place for profound mathematical meditation. It was in the left-hand try-pot of the Pequod, with the soapstone diligently circling round me, that I was first indirectly struck by the remarkable fact, that in geometry all bodies gliding along a cycloid, my soapstone, for example, will descend from any point in precisely the same time" (Melville 1851).Huygens also constructed the first pendulum clock with a device to ensure that the pendulum was isochronous by forcing the pendulum to swing in an arc of a cycloid. This is accomplished by placing two evolutes of inverted cycloid arcs on each side of the pendulum's point..
Find the figure bounded by a line which has the maximum area for a given perimeter. The solution is a semicircle. The problem is based on a passage from Virgil's Aeneid:"The Kingdom you see is Carthage, the Tyrians, the town of Agenor;But the country around is Libya, no folk to meet in war.Dido, who left the city of Tyre to escape her brother,Rules here--a long and labyrinthine tale of wrongIs hers, but I will touch on its salient points in order....Dido, in great disquiet, organised her friends for escape.They met together, all those who harshly hated the tyrantOr keenly feared him: they seized some ships which chanced to be ready...They came to this spot, where to-day you can behold the mightyBattlements and the rising citadel of New Carthage,And purchased a site, which was named 'Bull's Hide' after the bargainBy which they should get as much land as they could enclose with a bull's hide."..
The problem in calculus of variations to find the minimal surface of a boundary with specified constraints (usually having no singularities on the surface). In general, there may be one, multiple, or no minimal surfaces spanning a given closed curve in space. The existence of a solution to the general case was independently proven by Douglas (1931) and Radó (1933), although their analysis could not exclude the possibility of singularities. Osserman (1970) and Gulliver (1973) showed that a minimizing solution cannot have singularities.The problem is named for the Belgian physicist who solved some special cases experimentally using soap films and wire frames (Isenberg 1992, Wells 1991). The illustration above shows the 13-polygon surface obtained for a cubical wire frame.
Find a closed plane curve of a given perimeter which encloses the greatest area. The solution is a circle. If the class of curves to be considered is limited to smooth curves, the isoperimetric problem can be stated symbolically as follows: find an arc with parametric equations , for such that , (where no further intersections occur) constrained bysuch thatis a maximum.Zenodorus proved that the area of the circle is larger than that of any polygon having the same perimeter, but the problem was not rigorously solved until Steiner published several proofs in 1841 (Wells 1991).
A branch of mathematics that is a sort of generalization of calculus. Calculus of variations seeks to find the path, curve, surface, etc., for which a given function has a stationary value (which, in physical problems, is usually a minimum or maximum). Mathematically, this involves finding stationary values of integrals of the form(1) has an extremum only if the Euler-Lagrange differential equation is satisfied, i.e., if(2)The fundamental lemma of calculusof variations states that, if(3)for all with continuous second partial derivatives, then(4)on .A generalization of calculus of variations known as Morse theory (and sometimes called "calculus of variations in the large") uses nonlinear techniques to address variational problems.
Bubbles can meet only at angles of (for three bubbles) and (for four bubbles), where is the supplementary angle of the tetrahedral dihedral angle. This was proved by Jean Taylor using measure theory to study area minimization. The double bubble is area minimizing, but it is not known if the triple bubble is also area minimizing. It is also unknown if empty chambers trapped inside can minimize area for bubbles.
Find the shape of the curve down which a bead sliding from rest and accelerated by gravity will slip (without friction) from one point to another in the least time. The term derives from the Greek (brachistos) "the shortest" and (chronos) "time, delay."The brachistochrone problem was one of the earliest problems posed in the calculus of variations. Newton was challenged to solve the problem in 1696, and did so the very next day (Boyer and Merzbach 1991, p. 405). In fact, the solution, which is a segment of a cycloid, was found by Leibniz, L'Hospital, Newton, and the two Bernoullis. Johann Bernoulli solved the problem using the analogous one of considering the path of light refracted by transparent layers of varying density (Mach 1893, Gardner 1984, Courant and Robbins 1996). Actually, Johann Bernoulli had originally found an incorrect proof that the curve is a cycloid, and challenged his brother Jakob to find the required..
If is continuous andfor all infinitely differentiable , thenon the open interval .
The functional derivative is a generalization of the usual derivative that arises in the calculus of variations. In a functional derivative, instead of differentiating a function with respect to a variable, one differentiates a functional with respect to a function. The definition for the univariate case isFor example, the Euler-Lagrange differential equation is the result of functional differentiation of the Hamiltonian action (functional).
An identity in calculus of variations discovered in 1868 by Beltrami. The Euler-Lagrange differential equation is(1)Now, examine the derivative of with respect to (2)Solving for the term gives(3)Now, multiplying (1) by gives(4)Substituting (3) into (4) then gives(5)(6)This form is especially useful if , since in that case(7)which immediately gives(8)where is a constant of integration (Weinstock 1974, pp. 24-25; Arfken 1985, pp. 928-929; Fox 1988, pp. 8-9).The Beltrami identity greatly simplifies the solution for the minimal area surface of revolution about a given axis between two specified points. It also allows straightforward solution of the brachistochrone problem.