home | index | units | counting | geometry | algebra | trigonometry & functions | calculus
analysis | sets & logic | number theory | recreational | misc | nomenclature & history | physics

Final Answers
© 2000-2017   Gérard P. Michon, Ph.D.

 Joseph-Louis Lagrange 


 border  border

Related articles:

Related Links (Outside this Site)

An Introduction to Lagrange Multipliers  by  Steuard Jensen.
Calculus of Variations and Functional Differentiation  by  Joceline Lega.
Saddlepoint Methods in Contour Integration  by  Michael Fowler  (Physics 752)
Math 302 Appendix  (Liverpool)  by  Mary Rees, FRS.
 Gaspard Monge 
 1746-1818 La brouette de Monge ou le transport optimal
(in French)  by  Yann Brenier  (2012-02-12).
Transport optimal de mesure:  Coup de neuf pour un vieux problème.
(in French)  by  Cédric Villani  (2004-10-15).
"Variétés stochastiques"  by  Anatole Khelif  &  Alain Tarica  (2015-10-15)

Wikipedia :   Plateau's problem   |   Monge-Kantorovitch transportation problem

DMOZ: Calculus of Variations



Vocabulary :  For an objective function of several variables, a "stationary point" is a set of values of those variables where all partial derivatives of the function vanish.  As illustrated below, a local extremum must be a stationary point but the converse need not hold.
Because of the inclusive meaning assigned by default to the simplest mathematical terms  (which is the exact opposite of the  exclusive  meaning often attributed to simple words in everyday language)  most mathematicians consider "stationary point" and "saddlepoint" to be synonymous:  At a saddlepoint, the relevant quantity  may  rise in some directions and fall in others but it's not required to do so...  There might very well be an extremum there!
Some authors reserve the term "saddlepoint" to  nonextremal  stationary points.  We prefer to call those  proper  saddlepoints  (thus following  normal  mathematical nomenclature).

(2015-01-30)   Optimizing before Calculus
Some historical cases have beautiful  ad hoc  solutions.

Without the general methods presented in the rest of this page, every optimization problem appears to call for a brilliant insight.  Here are a few classical such solutions, which ought to be remembered not only for the sheer beauty of the arguments leading to them but also for the simplicity in which the final results can be described.  Arguably,  ad hoc  methods are best suited for problems whose solutions are simple  (like Dido's problem).

Once guessed, such solutions may have beautiful justifications.  In general, however, correctly guessing the solution of a complicated optimization problem is all but impossible and  calculus  is required for both the discovery and the justification of the solutions.  When calculus itself becomes difficult to apply,  the latest trend is to use numerical methods instead of symbolic manipulations  (this should only be a last resort, though).

The solution of our first optimization problem depends on what can be construed as Euclid's postulated "optimization", namely:

Postulate :   In a Euclidean plane  (or Euclidean space)  the shortest path between two points is a straight line.

Single-variable optimization:  Laws of classical optics.

Heron's problem   (least length and laws of reflection):
In the plane,  we consider a straight line and two points  A  and  B  on the same side of the line.  The basic question is to find which point  M  on the line will yield the shortest two-legged path from  A  to  M  and  M  to  B.

The key observation is that any such path has the same length as a path from  A  to  M  and from  M  to the point  B',  symmetric of  B  with respect to the line.  Conversely, any path from  A  to  B'  must cross the line at a point  M.

If the path from  A  to  B'  is shortest, so is the matching path from  A  to  M  to  B.  By postulate,  the shortest path from  A  to  B'  is a straight line and the solution  M  is simply found as the intersection of two straight lines.

Heron thus showed, nearly 2000 years ago, that the main law of mirror optics  (the angle of reflection is equal to the angle of incidence)  can be deduced from the appealing postulate that light travels along the shortest available route.  A great intellectual feat for that time,  or for any time!

Fermat's principle   (least time and laws of refraction):
Postulating instead  (more generally)  that light travels along the  fastest  route, the assumption that its speed is different on either side of the line yields the basic law of refraction  (Snell's law).

Two-variable optimization:  Torricelli points.

The  Torricelli point  minimizes the sum of the distances to three given points.  This can be viewed as a two-dimensional specialization of  Plateau's laws.

Wikipedia :   Video :   Reasons for studying tensors  by  Pavel Grinfeld (2014).

(2007-09-23)   Optimizing a smooth function  f  of a single variable.
'Tis little more than finding where the function's derivative vanishes.

   Maximum within range, 
 minimum at extreme range.
For a point  x  in the  interior  of the domain of  f  and a small enough value of  e,  both points  x-e  and   x+e  are in the allowed domain and yield values of  f  which are on both sides of  f (x)  if we just assume that  f '  is nonzero.  Therefore, there can be an extremum at point  x  only when  f ' (x)  vanishes.

On the other hand, there's no such requirement for a point  x  at the border of the allowed domain, because small displacements of  only one sign  are allowed.

Away from the border,  a  saddlepoint  x  (let's use that general term to indicate that  f ' (x)  vanishes)  will be the location of a  minimum  (resp. a  maximum )  when  f '' (x)  is positive  (resp. negative).  If it's zero, further analysis is needed to determine whether  x  is the location of an extremum or not.  (It's indeed an extremum if the first nonzero derivative is of even order.)

Extreme Value Theorem   (real-valued continuous function on a compact set)

(2016-01-11)   Angle maximization problem of Regiomontanus :
An historical example of a single-variable problem of the above type.

Regiomontanus (1436-1476)  proposed to discuss how a line segment is viewed at an angle  q  from a variable viewpoint.  That angle is maximum  (and equal to 180°)  from a point inside the segment.  It is minimum  (and equal to 0°)  for a viewpoint on the same line as the segment but outside of it.  The angle is arbitrarily small from viewpoints at large distances.

Regiomontanus considers a line perpendicular to the line of the segment and crossing it at a point outside the segment.  He asks when the viewing angle is maximum from points on that line.

Wikipedia :   Regiomontanus' angle maximization problem

(E. M. of Wisconsin Rapids, WI. 2000-11-21Saddlepoints & Extrema
Determine the points where the [objective] function  z = 3x3+3y3-9xy  is maximized or minimized.   [ Check for  second-order  condition. ]

A necessary (but not sufficient) condition for a smooth function of two variables to be extremal  ( "minimized" or "maximized" )  on an  open  domain is to be at a  saddlepoint  where both partial derivatives vanish.  In this case that means:

9x2-9y = 0     and     9y2-9x = 0

So, extrema of  z  can  only  occur when  (x,y)  is either  (0,0)  or  (1,1).

To see whether a local extremum  actually  occurs, we must examine the second-order behavior of the function about each such  saddlepoint  (in the rare case where the second-order variation vanishes, further analysis would be required).

Well, if the second-order partial derivatives are  L, M and N,  then the second-order variation at point  (x+dx, y+dy)  is the quantity

½ [  L(dx)2 + 2M(dxdy) + N (dy)2 ]

We recognize the bracket as a quadratic expression whose sign remains the same  (regardless of the ratio dy/dx)  if and only if   its  (reduced)  discriminant (M2-LN)  is negative.  If it's positive, the point in question is not an extremum.

In our example, we have  L = 18x,  M = -9  and  N = 18y.  Therefore:

M2 - LN   =   81 (1 - 4xy)

At  (0,0)  this quantity is positive  (+81).  Thus, there's  no extremum  there.  On the other hand, at the point  (1,1)  this quantity is negative  (-243)  so the point  (1,1)  does correspond to  the only local extremum  of  z.

Is this a maximum or a minimum?  Well, just look at the sign of  L  (which is always the same as the sign of  N  for an extremum).  If it's positive, surrounding points yield higher values and, therefore, you've got a minimum.  If it's negative you've got a maximum.  Here, L = 18,  so a  minimum  is achieved at  (1,1).

To summarize:  z  has only  one  local extremum; it's a minimum of -3, reached at x=1 and y=1.  Does this mean we have an  absolute  minimum?  Certainly not!  (When  x  and  y  are negative, a large enough magnitude of either will make  z  fall below  any  preset threshold.)

(2007-09-22)   Unconstrained (or "absolute") saddlepoints
Saddlepoints of a function of several  independent  variables.

We're seeking saddlepoints (stationary points) of a scalar  objective function  M  of  n  variables  x1 , ... , xn  which we view as components of a vector  x.

M ( x1 , ... , xn )   =   M ( x )

If those  n  variables are  independent,  then a saddlepoint is obtained only at a point where the differential form  dM  vanishes.  This is to say:

0   =   dM   =   grad M  . dx

As this relation must hold for  any  infinitesimal vectorial displacement  dx, such  absolute  saddlepoints of  M  are thus characterized by:

grad M   =   0

That vectorial relation is shorthand for  n  separate  scalar  equations:

M     =   0

 Joseph-Louis Lagrange 
(1736-1813) (2007-09-22)   Lagrange Multipliers
Optimizing entails one  Lagrange multiplier  per constraint.

Instead of a set of independent variables  (as discussed above) we may have to deal with several variables tied to each other by several  constraints.  The method of choice was originally introduced by Lagrange  (before 1788)  as an elegant way to deal with mechanical constraints:

For example, the variables may be subject to a single constraint  (e.g.,  the cartesian equation of a surface on which a point-mass is free to move).

E ( x1 , ... , xn )   =   constant

While an  unconstrained saddlepoint  of  M  was obtained when  dM = 0.  A constrained saddlepoint is obtained when  dM  is proportional to  dE.

More generally, when  k  functions  E1 ... Ek  are given to be constant, a constrained saddlepoint of  M  is achieved when the differential form  dM  is a linear combination of the differentials of  E1 ... Ek.

dM   =   l1 dE1  +  l2 dE1  +   ...   +  lk dEk

In other words, there are  k  constants  li  (each is known as the  Lagrange multiplier  associated with the corresponding constraint)  such that the following function has an  unconstrained  (or absolute)  saddlepoint.

M  -  ( l1 E1  +  l2 E1  +   ...   +  lk Ek )

 Come back later, we're
 still working on this one...

Video:   Lecture 2 | Modern Physics: Statistical Mechanics  by  Leonard Susskind (2009-04-06)

(2007-09-23)   The  fattest  cone of given base...
What apex minimizes the lateral area of a cone of given base and volume?

The volume of a cone is one third of the product of its base area by its height.  So, by imposing the cone's volume, we're actually imposing the  height  of the cone and looking for an optimal apex somewhere in a fixed plane parallel to the base...

 Come back later, we're
 still working on this one...

(2007-09-23)   Calculus of Variations  (cf. Lagrangian mechanics)
Euler-Lagrange equations hold whenever a  path integral  is stationary.

Historically, the first problem of the type described below was the brachistochrone problem  (find the shape of the curve of fastest descent)  which had been considered by Galileo in 1638 and solved by  Johann Bernoulli  in 1696  (Bernoulli withheld his solution to turn the problem into a public challenge, which was quickly met by several prestigious mathematicians).  The subject was investigated by Euler in 1744 and by Lagrange in 1760.  It was named  calculus of variations by Euler in 1766 and made fully rigorous by Weierstrass before 1890.

Let  L  be a  smooth enough  function of  2n+1  variables:

L   =   L ( q 1 ,  q 2   ...  q n ,  v1 ,  v2   ...  vn ,  t )

We assume that the first  n  variables  (q)  are actually functions of the last one  (t)  and also that the subsequent variables  (v)  are simply their respective derivatives with respect to t:

v i (t)    =     d   q i (t)

This makes  L  a function of  t  alone and we may consider the following integral  S  (called the "action" of  L )  for prescribed  configurations  at both extremities.  In this context, a  configuration  is defined as a complete set of values for the q-variables only  (irrespective of what the v-variables may be)  so, what we are really considering now are prescribed fixed values of  all  the  2n  quantities  q i (a)  and  q i (b).

S    =    ò  b   L  dt

The fundamental problem of the  calculus of variations  is to find what local conditions make  S  stationary,  for  optimal  functions  q i .  Namely:

 Leonhard Euler 
 (1707-1783)  Joseph-Louis Lagrange 
Euler-Lagrange equations
 L     =     d  (   L  )
Vinculum Vinculum Vinculum
q i dt vi

Proof :   From a set of  optimal  functions  q i  and  arbitrary  (sufficiently smooth)  functions  h i  which vanish at both extremities  a  and  b,  let's define the following family of functions, depending on one parameter  e :

Q i (t)   =   q i (t)  +  e h i (t)   V i (t)   =   v i (t)  +  e   d   h i (t)

Those yield a value  S(e)  of the action which must be stationary at  e = 0  (since the functions  q i  are optimal).  Thus, the derivative of  S(e)  along  e  does vanish at  e = 0,   which is to say:

0    =    ò  b   å i   [  h i    L    +    d h i      L  ]  dt
Vinculum Vinculum Vinculum
a q i dt vi

Since every  h i  vanishes at both extremities  a  and  b ,  we may  integrate by parts  the second term of the square bracket to obtain:

0    =    ò  b   å i   h i   [   L    -    d  (   L  )  ]  dt
Vinculum Vinculum Vinculum
a q i dt vi

As  h i  is arbitrary, the square bracket must vanish everywhere, for every  i.  (If this wasn't so, the whole equality would be violated for a choice of  h i  vanishing wherever the square bracket doesn't have a prescribed sign).   QED

Theoretical and Practical Examples :

The above is most commonly used as the basis for variational mechanics and related aspects of theoretical physics based on a  principle of least action.  However, it's also the correct answer to more practical concerns:

On 2008-10-27, Bill Swearer wrote:       [edited summary]
As a pilot, I've always been interested in writing a  proper  piece of flight-planning software to optimize the plane's path with regard to time, fuel, or any combination thereof.  [...]
I've always felt the professionally-provided solutions were crude approximations that do not take into account the full range of possibilities, especially over long distance, as one crosses varying jet streams at different altitudes, etc.  [...] What is the correct mathematical approach?
Thanks a lot.
Bill Swearer

Well, just express carefully the  local  cost function  (L)  as a function of the position of the aicraft and of the related derivatives  (to a good approximation, the latter are only useful to compute horizontal speed).  Predicted meteorological conditions  (changing with time throughout space)  can be used for best planning.

The Euler-Lagrange equations then tell the pilot what to do at all times.

(2009-07-05)   A Proof of Noether's Theorem   (1915)
Proving Noether's theorem for continuous symmetries.

A slight modification of the above proof yields a straight derivation of one of the greatest statements of mathematical physics.  Let's just do it...

Suppose that, for specific functions  hi ,  a symmetry exists which leaves  L  unchanged,  (to first-order variations of   e  about 0)  when  q i  is replaced by

Q i   =   q i  +  e h i

Formally, this leads to a situation similar to the previous one, since we still know that the derivative of  S(e)  vanishes at  e = 0  (albeit for very different reasons).  However, the  h  functions need not vanish at the extremities  a  and  b.  So, an extra "integrated term" appears in the following expression of the derivative of  S(e which results from our integration by parts:

0   =   å i   h i    L   |  b   +   ò  b   å i   h i   [   L    -    d  (   L  )  ]  dt
Vinculum Vinculum Vinculum Vinculum
vi a a q i dt vi

Now, as the integral still vanishes  (because the previously established Euler-Lagrange equations make every square bracket vanish)  the extra term must be zero as well.  This means that the following quantity doesn't change:

Conserved  Noether Charge
å i   h i    L

Emmy Noether (1882-1935)
Video:   Lagrangian & Field Theory  by  Leonard Susskind blog

(2012-04-02)   The  geodesics  of a two-dimensional surface.
Paths of  least length.

On a 2D surface  M (u,v)  the infinitesimal distance corresponding to a displacement  du,dv  is given by the first fundamental quadratic form:

F1 (du,dv)   =     (dM)2    =   E (du)2  +  2 F du dv  +  G (dv)2

The problem of minimizing the length of the path between two points on that surface is a standard exemple of the calculus of variations.  Let's introduce notations compatible with the above:

q1 = u       q2 = v       v1 = du/dt       v2 = dv/dt
L   =   [ F1 (v1 , v2 ) ]½   =   [ E v12  +  2 F v1 v2  +  G v22 ]½

The two Euler-Lagrange equation for this Lagrangian are  (i = 1,2) :

 L     =     d  (   L  )
Vinculum Vinculum Vinculum
q i dt vi

The first of those  (i = 1, qi = u)  translates into:

E'u v12  +  2 F'u v1 v2  +  G'u v22     =     d  (  E v1  +  F v2  )
Vinculum Vinculum Vinculum
2 LdtL

Since   dX/dt  =  v1 X/u + v2 X/v   the right-hand side is equal to:

 Come back later, we're
 still working on this one...

Geodesic lines

 Bernoulli (2010-07-06)   The  brachistochrone  curve is a cycloid  (1696)
The curve of fastest descent in a uniform gravitational field.

We are now presenting the  brachistochrone  problem as a simple application of the calculus of variation.  Historically, the relation is reversed, as this problem actually helped define the need for the latter, which was formalized many years after the  brachistochrone  problem was solved.

In June 1696, Johann Bernoulli (1667-1748)  challenged the readers of Acta Eruditorum with his famous  brachistochrone  problem:

Along what curve would a point-mass fall from one prescribed point to another lower one (not directly underneath) in the least possible time?

Johann Bernoulli's own solution was published in the same journal a year later, along with 4 of the 5 solutions sent by famous contributors  (the solution of  Guillaume de l'Hôpital was only published in 1988).

We shall use a coordinate system where the origin is the starting point.  We choose to orient the y-axis downward so that  y  is positive for all target points.  Let's call  u  the slope of the trajectory  dy/dx.  In a gravitational field  g  the conservation of mechanical energy for a mass  m  dropped from the origin at zero speed tells us that:

m g y   =   ½ m [ (dx/dt)2  +  (dy/dt)2 ]
Therefore,  ( 2 g y ) ½ dt   =   ( 1 + u2 ) ½ dx

To minimize the descent time, we seek to minimize the path integral of  dt  or, equivalently, to minimize the integral of  (2g)½ dt   =   L dx

L dx   =   L (y,u,x) dx   =   ( 1 + u2 )½ / y½ dx

To minimize this integral, the relevant Euler-Lagrange equation must hold:

 L     =     d  (   L  )
Vinculum Vinculum Vinculum
y dx u

With the above expression of  L,  this translates into:

- ( 1 + u2 )1/2 / 2y3/2   =   d/dx [ u ( 1 + u2 )-1/2 / y1/2 ]

Recalling that  u = dy/dx,  we reduce this algebraically:

- ( 1 + u2 )-1/2 / 2y3/2   =   (du/dx) ( 1 + u2 )-3/2 / y1/2

- ( 1 + u2 )   =   2 y du/dx   =   2 u y du/dy

That last equality holds only when  dy/dx  doesn't vanish, which need not always be true.  We put  v = u2  and separate the variables to integrate:

d {Log y}   =   dy/y   =  - dv / (1+v)   =   - d { Log (1+v) }

Therefore,  y / 2a  =  1/(1+v)   which is to say   v  =  2a / y-1   for some  a.

( dy/dx ) 2   =   2a / y - 1

This characterizes a cycloid corresponding to a wheel of radius  a.

 Come back later, we're
 still working on this one...

Video :  The Brachistochrone  by  Grant Sanderson,  with  Steven Strogatz.
Wikipedia :  Brachistochrone curve

(2016-01-11)   Dido's Problem
Idealized legendary real-estate deal upon which Carthage was founded.

They purchased a site, which was named  'Bull's Hide'  after the bargain 
by which they should get as much land as they could enclose with a bull's hide.

  Aeneid  by  Virgil (70-19 BC)

The land so acquired by  queen Dido, founder  Carthage,  is interpreted to be whatever area could be bounded by the seashore on one side and a rope of fixed length on the other.  (The preliminary step was to cut the hide into very thin strips to make a long rope out of it.)  The problem is to maximize that area, assuming the shore line to be perfectly straight.

To prove that the solution is a half-circle, one considers the two straight lines drawn from an  arbitrary  point on the optimal boundary to either extremity of the rope (on the shoreline).  If we assume that both parts of the rope are rigidly attached to those two lines, there's a constant area between those lines and their respective ropes.  What remains is the area of the triangle formed by two sides of fixed length and part of the shore line.  This area is greatest when the two sides are perpendicular!

By the  second theorem  of Thales, a planar curve where an arbitrary point always sees the extremities at a right-angle is a  semicircle.    QED

The isometric inequality is a corollary of Dido's solution :

In 1846, the great Swiss geometer Jakob Steiner (1796-1863) showed that the above solution of Dido's problem implies the familiar  2-dimensional  isoperimetric inequality  (i.e.,  the planar curve of given perimeter which encloses the greatest area is a circle).  Three steps:

  • The optimal shape is convex  (or else its convex hull would have lesser perimeter and greater area).
  • If two points cut the perimeter of an optimal convex shape in half, the straight line joining them necessarily cuts the shape into two portions of equal area.  (Otherwise a better shape could be formed using two parts congruent to the larger portion.)
  • As solutions of  Dido's problem,  both halves must be semicircles.  Therefore, the whole shape is a circle.    QED

No such simple geometric solutions are available to justify the higher-dimensional counterparts of the isoperimetric inequality, discussed next.

(2008-11-10)   The Isoperimetric Inequality and its Generalizations
Among planar loops of unit length, the circle encloses the largest area.

The surface area  S  enclosed by a closed planar curve of given perimeter  P  is largest in the case of a circle.  This ancient statement  (proved geometrically by Steiner, in 1846)  can be expressed by the following relation, known as the  isoperimetric inequality:

S   ≤   P 2 / 4p

Generalizations to Higher Dimensions :

The three-dimensional equivalent of the isoperimetric inequality says that the closed surface of area  S  which encloses the largest volume  V  is a  sphere.

V 2   ≤   S 3 /  36 p

The above can be generalized to  n  dimensions:  No hypersurface of hyperarea  S  encloses a larger  hypervolume  V  than the hypersphere.  This makes the relations given in the  oldest Numericana article  yield:

V n-1   ≤    G ( 1 + n/2 )   [ S /  nÖp ] n

Action, Minkowski Addition & Two Proofs of the Isoperimetric Inequality (2010)  by  Jeremy Boother.
Wikipedia :  Isoperimetric inequality

(2008-11-10)   Plateau's Problem.  Minimal surfaces with given borders.
The mean curvature of a soap film is constant.

That constant vanishes when both sides of the soap film are at the same pressure, in which case the film is a surface of least area within the given border.  However, the non-vanishing case was also considered by Joseph Plateau; it arises when soap film encloses pockets of air...

Besides the familiar spherical soap bubble, the simplest case is a soap bubble merged within a large planar film.  It takes on the form of two spherical caps which form an outer angle of  120°  (inner angle of  60°)  with the planar part of the film whwere they meet.  The thickness  2h  and the equatorial diameter  d  of the lenticular bubble are proportional to the radius of curvature of the spherical parts:

2h   =    (2-Ö3) R   =   0.267949...R

The  Plateau rules  (1873)  state that the solutions are smooth surfaces of  constant mean curvature  with only two possible types of inner singularities:

  • Lines  where  3  such surfaces meet at equal angles  (120°).
  • Isolated  points  where  4  such lines meet at equal angles  q.

Because a regular tetrahedron is formed by every other vertex of a cube,  two vertices of a regular tetrahedron are seen from its center at the same angular separation as between  (1,1,1)  and  (1,-1,-1).  The cosine of that angle is  -1/3.  Chemists call it the  tetrahedral angle :

q  =  acos ( -1/3 )  =  1.910633236249...  =  109.47122063449...°

Jean Taylor  (1944-)  published a proof of Plateau's laws in 1976.

Euler's catenoid  (1741) :

By finding that the  catenoid  is the only minimal surface of revolution  (besides planes orthogonal to the axis)  Euler  solved Plateau's problem  (well before it was so named)  when the given border consists of a pair of two sufficiently close coaxial circles.

When the parallel planes of the coaxial circles are too far apart,  no catenoid can join them and the solution of Plateau's problem just consists of separate planar surfaces.

Wikipedia :   Plateau's Laws   |   Joseph Plateau (1801-1883)   |   Jean Taylor (1944-)

(2008-11-18)   Borderless embedded surfaces of minimal area
Minimal surfaces without edges or self-intersections in ordinary space.

Such surfaces are technically called  complete embedded minimal surfaces.  Before 1982, only four examples were known:  Helicoid

  • The plane.  (Arguably, just a special helicoid.) 
  • The other helicoid  (Meusnier, 1776).  Catalan remarked that the plane and the helicoids are the only  ruled  minimal surfaces. 
  • The catenoid  (Euler, 1741).  In 1776,  Meusnier  remarked that the catenoid has zero  mean curvature. 
  • A surface, found by Riemann in 1860, consisting of infinitely many horizontal planes with slanted tunnels between adjacent pairs.

   Celso da Costa
Celso José da Costa
In 1982,  Celso J. Costa  (a graduate student from Brazil)  stumbled upon a minimal surface topologically equivalent to a 3-hole torus.  Costa suspected that his surface had no self-intersections but, at first, couldn't prove it...

David Hoffman  (a 1973 Stanford Ph.D.)  undertook the task at the  University of Massachusetts at Amherst.

 Costa's Surface   David Hoffman teamed up with  William H. Meeks III . and  Jim Hoffman,  (then a graduate student)  to produce a computer visualization of the stunning symmetries of the strange surface described by Costa  (which contains two straight lines meeting at a right angle).  Dividing Costa's surface into  8  congruent pieces, they proved, indeed, that it never intersects itself!

Loosely speaking, Costa's surface features two complementary pairs of "tunnels" through the equatorial plane which connect the inside of some "catenoidal" northern half and the outside of its southern half, or vice-versa.

   Dave Hoffman
David Hoffman

 Bill Meeks
William H. Meeks, III
Subsequently, Dave Hoffman and Bill Meeks discovered that Costa's surface is just the simplest member of a whole family of  complete minimal embedded surfaces  constructed in the same way but with more "tunnels"...

 Costa-Hoffman-Meeks Surface 
 (Image courtesy of Matthias Weber)

Applied to helicoids, the idea yields yet another family of  complete minimal embedded surfaces  where tunnels provide shortcuts between slices of space which are otherwise connected only by circling around the helicoid's axis.

 Come back later, we're
 still working on this one...

Minimal Surfaces and their Classification  (pdf 4.44 MB)  by  William H. Meeks, III.
Scherk's surface (1834)   |   Heinrich Scherk (1798-1885)
Bonnet family of a minimal surface.

Denis Viala  (2008-02-28)   Connect the dots, without crossings...
Given  n  blue dots and  n  red dots in the plane  (no three aligned)  there are always  n  disjoint  segments with blue and red extremities.

Proof.  (HINT:  This puzzle appears among optimization problems.)

n=2, n=3, etc.

(2008-11-09)   Torricelli Points
Connecting three dots with lines of least total length.

The three vertices of a triangle  ABC  can be connected using just its two shortest sides.  If the angle between those two sides is  120°  or larger, this is the most economical way to do so.

Otherwise, we may define a fourth point  T  such that the three  angles  ATB, BTC and CTA  are all equal to  120°.  Then, the three llines  TA, TB and TC  form the most economical way to connect the three dots  A, B and C.

The point  T  is called the  Torricelli point  of ABC.  The results for the first and the second of the above cases are unified by defining  T  to be the vertex opposite the longest side in the first case.

Likewise, four points in a plane are best connected by introducing  0, 1 or 2  new  Torricelli points.

(2008-11-09)   The Honeycomb Theorem
In 1999, Thomas Hales finally proved an "obvious" fact.

Nobody questioned the celebrated  Honeycomb conjecture  before 1993, when Phelan and Weaire showed its 3D counterpart  (Kelvin's conjecture)  to be false!  The issue was settled in 1999  with a proper proof by  Thomas C. Hales :  Thus, the 2D  Honeycomb conjecture  is now a theorem!  Here it is:

Honeycomb Theorem   In any partition of the plane into tiles of unit area, each tile cannot have a lesser perimeter than the regular hexagon of unit area.  (Loosely speaking, the regular  honeycomb  tiling is the most economical one.)

The first proof by  Thomas C. Hales  (1999)  was surprisingly intricate.  It was crucial to get rid of the convexity restriction which earlier authors had reluctantly imposed.  (The isoperimetric inequality implies that the boundaries between optimal shapes are either circular arcs or straight lines.  Both sides of such a boundary are convex only in the latter case.)  The reason curved boundaries are ruled out is not at all obvious;  it is  false  in the  3D case.

(2015-01-10)   Highly-symmetric  monkey saddles  of minimal area. 
Surfaces whose borders have mirror-symmetry and an  odd  ternary axis.

With a vertical axis,  such a curve has a vertical plane of symmetry,  the vertical coordinate is unchanged by rotation of  2p/3  (1/3 of a turn)  and its sign changes for half of that  (p/3 rotation).

A surface containing that line and having the same symmetry will be described by an equation of the following type,  in cylindrical coordinates:

z   =   f ( r , sin 3q )       [ f  being  odd  for either argument ]

Let's find the functions  f  which make the  mean curvature  constant:

 Come back later, we're
 still working on this one...

Mean curvature (Numericana)   |   Monkey saddle (Wikipedia)   |   Monkey saddle (MathWorld)

 Kelvin's tetragon
(2008-11-09)   Foam cells of  unit volume  and minimal area. 
Kelvin's problem  &  counterexamples to Kelvin's conjecture.

 William Thomson, Lord Kelvin 
 (1824-1907) Kelvin's problem  is to partition space into unit-volume cells with least surface area.  Simon Lhuilier (1750-1840)  called this  one of the most difficult problems in geometry.

 Truncated Octahedron
The uniform truncated octahedron can  tile space  without voids.
  Lord Kelvin (1824-1907)  observed that a partition of space into  Archimedean  truncated octahedra  was quite economical.  He used the vertices as the basis for a proper foam, with  curved  faces and edges engineered to satisfy  Plateau's rules.  The tetragonal faces of Kelvin's cell are planar figures whose curved edges meet at the magic angle of  109.47122°.  The hexagonal walls are  monkey saddles  with straight diagonals.  In 1887, Kelvin conjectured that his foam was the answer to the above optimization problem.

In the straight polyhedron, the angle  q  between a square and an adjoining hexagon is slightly less than  60°.  An easy way to obtain the exact value is to consider the cube formed by the six square faces.  Seen from the center, the middles of the hexagons are in the directions of the corners of that cube.  Therefore:

q   =   acos ( 1/Ö3 )   =   54.73561...°

By  Plateau's rules,  an optimally curved hexagonal surface always meets the plane of an adjoining tetragonal face at an angle of  60°.  Therefore,  the curved surface at the middle of an edge is inclined  5.26439°  with respect to the plane of the original flat hexagon.

On the other hand, an edge between two hexagonal faces curves inward.  (It's the outward-curving edge of a tetragon in an adjacent cell.)

Although the equal-volume foams discussed below are more economical than this one,  Kelvin's foam remains best among all  monohedral  foams  (i.e., foams whose cells are all congruent to each other).

On the division of space with minimum partitional area   Sir William Thomson (Lord Kelvin).
Acta Mathematica, 11 (1887)  pp. 121-134.

When a foam  isn't  monohedral, it need not be  equal-pressure.  Actually, none of the foams below are.  In all of them, different cells may be assigned different "pressures" so a face's  mean curvature  is the difference between the pressures of the cells it separates  (not necessarily zero).

 Basic pattern of the Weaire-Phelan foam 
 (c) 2008 by Ruggero Gabbrielli

In 1993,  Denis Weaire (1942-)  and his student  Robert Phelan  disproved Kelvin's conjecture with a structure more economical than Kelvin's foam:  The  A15  Weaire-Phelan structure  is one example of a  Frank-Kasper foam.  It consists of two distinct types of cells, with pentagonal or hexagonal faces.  One cell is a squashed dodecahedron, the other is a tetradecahedron.  That basic structure occurs naturally in crystals of  b tungsten.

 Basic pattern of the p42a foam  
 (c) 2008 by Ruggero Gabbrielli

In a draft  (2008-10-25)  of a paper published on 2009-06-02Ruggero Gabbrielli  presents a new type of unit-cell foam (featuring some  tetragonal  faces)  whose cost (5.303) is intermediate between the Weaire-Phelan foam (5.288) and Kelvin's foam (5.306).  He calls this structure  "P42a".  The repeating pattern consists of  14  curved cells of  4  different types  (ten tetradecahedra and four tridecahedra).  It has an  average  of  96/7 = 13.714285...  faces per cell.

The above pictures for the  A15  and  P42a  foams were obtained by  Ruggero Gabbrielli  using the  GAVROG 3dt visualization software.  Gabbrielli has also posted several related  interactive 3D visualizations  under Java.  The 1993 optimization on A15 and Gabbrielli's recent research both used Ken Brakke's Surface Evolver (with specific code provided by John M. Sullivan, in the latter case at least).

At this writing, the  A15  foam described by Weaire and Phelan
in 1993 is still conjectured to be the answer to Kelvin's problem.
It is about 0.3% less costly than Kelvin's original foam.

The Weaire-Phelan foam was used as the basis for the design of the celebrated Water-cube  of the 2008 Olympics  (Beijing National Aquatics Center)  which is the largest structure covered with ETFE film.

The number of faces in a minimal foam  (1992)  by  Robert B. Kusner
Comparing the Weaire-Phelan foam to Kelvin's foam  (1996)  by Rob Kusner and John M. Sullivan
A new counterexample to Kelvin's conjecture on minimal surfaces  by  Ruggero Gabbrielli
Comparing the Kelvin, Gabbrielli and Weaire-Phelan structures  (Java)
The Weaire-Phelan Structure  (and the Water-Cube)  by  John C. Baez   (2008-08-21).

visits since Sept. 22, 2007
 (c) Copyright 2000-2017, Gerard P. Michon, Ph.D.