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ISSN 0002-9920

Volume 57, Number 5

Notices of th

e Am

erican M

athem

atical Society

of the American Mathematical Society

May 2010

Equationspage 598

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Fractal origami (see page 630)

Origami and PartialDifferential EquationsBernard Dacorogna, Paolo Marcellini, and Emanuele Paolini

Origami is the ancient Japanese art offolding paper. Even if origami is mainlyan artistic product, it has received agreat deal of attention from mathe-maticians, because of its interesting

algebraic and geometrical properties.We present a new mathematical model of

origami with a double purpose. On one handwe give an analytical approach which providesa new perspective to the existing algebraic andgeometrical models. On the other hand we useorigami as a tool to exhibit explicit solutions tosome systems of partial differential equations.

Mechanical properties of paper, as a material, aresimple. A sheet of paper is rigid in tangential direc-tions. Indeed, it cannot be stretched, compressed,or sheared. If a sheet of paper is constrained ona plane, it would only be possible to achieve rigidmotions, i.e., rotations and translations of thewhole sheet. On the other hand, in the normaldirection it can be easily folded. This is due tothe infinitesimal thickness of the paper and tothe elastic properties of the fibers composing thematerial.

The physical properties we have described so farcan be defined from a mathematical point of view.To this aim we let ⁄ ⇢ R2 be a two-dimensionaldomain (usually a rectangle), which is the reference

Bernard Dacorogna is professor of mathematics at theÉcole Polytechnique Fédérale de Lausanne, Switzerland.His email address is [email protected].

Paolo Marcellini is professor of mathematics at the Uni-versity of Firenze, Italy. His email address is [email protected].

Emanuele Paolini is assistant professor of mathematicsat the University of Firenze, Italy. His email address [email protected].

Figure 1. The map u(x, y) = (x, yp2 ,|y|p

2 )describes a 90-degree folding in R3 of thesquare paper ⁄ in the left-hand side. The

gradient of u is a 3⇥ 2 orthogonal matrix.

Figure 2. The mapu(x, y)=(x, sign(y)(1�cosy), sin |y|), is not

piecewise linear. However, the gradient is a3⇥ 2 orthogonal matrix.

configuration for the sheet of paper. An origami isa suitable immersion of the sheet of paper in thethree-dimensional space. Hence it can be identifiedwith a map u (see Figures 1 and 2)

u : ⁄ ⇢ R2 ! R3.The tangential rigidity can be expressed byrequiring that u is a rigid map of ⁄ into R3.In other words the gradient Du(x) of the mapu is an orthogonal 3 ⇥ 2 matrix, meaning that,

598 Notices of the AMS Volume 57, Number 5

Figure 3. On the right: the crane is the most famous origami; on the left: the correspondingsingular set.

for every x 2 ⁄,Du(x) 2 O(3,2) := {M 2 R3⇥2 : MtM = I}.

Since Dut Du = I, the rank of Du is maximal;hence the immersion is locally one-to-one. Thus,if u is sufficiently regular, it is a local isometrybetween R2 and the manifold u(⁄) ⇢ R3 with themetric induced by the ambient space R3. Thereforethe Gauss curvature of u(⁄) coincides with theGauss curvature of ⁄, which is zero, and u(⁄) is adevelopable surface (see [15]).

Since origami is a folded paper, the mapu cannotbe everywhere smooth; it is only piecewise smooth.Folding creates discontinuities in the gradient.Since we do not allow cutting the sheet of paper,u is, however, a continuous map. The singular set÷u ⇢ ⁄, which is the set of discontinuities of thegradient Du, is called crease pattern in the origamicontext. Usually this set is composed by straightsegments (as in Figure 3), but it is also possible tomake origami with curved folds (see Figure 4).

This formulation of an origami as a mapu : ⁄ ⇢ R2 ! R3 also allows us to consider modularorigami, which corresponds to a domain ⁄ whichis not connected (several sheets of paper), as forinstance in Figure 5.

A special attention will be given to the so-calledflat origami. A flat origami is defined as a mapwhose image is contained in a plane; it can berepresented, up to a change of coordinates, as amap

u : ⁄ ⇢ R2 ! R2.In this case it can be proved that the folding linesmust be straight segments which satisfy someangle condition at every vertex of the singular set÷u . More precisely, at any vertex of the singularset ÷u, exactly 2k consecutive angles ↵1, · · · ,↵2kmeet with the property that

↵1 +↵3 + · · · +↵2k�1 = ↵2 +↵4 + · · · +↵2k = ⇡ .

We will see (cf. Theorem 4.1) that such condition isenough to guarantee, on a simply connected openset ⁄, that a given union of segments ÷ is actuallythe singular set ÷u of some rigid map u.

Our general definition allows us, in a natural way,to consider in the higher-dimensional case a mapu : ⁄ ⇢ Rn ! Rm, where the gradient Du(x) is anorthogonalm⇥nmatrix. That is,Du(x) 2 O(m,n),where

O(m,n) := {M 2 Rm⇥n : MtM = I}.When m = n we simply write O(n). In this caseour mathematical global approach also allowsus to consider flat origami in higher dimensions(u : ⁄ ⇢ Rn ! Rn, Du(x) 2 O(n)), which will also becalled hyper-origami. As an example, for n = 3, wewill describe, in the section “Fractal Construction”,how to “fold” a cube.

Closely related is the more difficult problem offinding a map with orthogonal gradient satisfyingsome boundary condition. For example, whenn =m = 2, we consider the Dirichlet problem

(1)

(Du(x) 2 O(2) for a.e. x 2 ⁄u(x) = 0 for x 2 @⁄.

At first sight one could think that, by means ofrepeated folding, a sheet of paper could in principlebe crunched to a single point, say the origin. Toconvince oneself that this is not possible it isenough to look at the equation. An origami foldedto a single point y0 (= 0) is in fact representedby the constant map u ⌘ y0 , which, however, hasgradient Du ⌘ 0 62 O(2).

In order to solve (1), we have to take advantageof the fact that the boundary does not have fulldimension. Anyway, the construction cannot bethat simple; in fact, it is impossible to find a mapsolving the system and which is smooth on someneighborhood of a point of the boundary. Indeed,

May 2010 Notices of the AMS 599

Figure 4. A circular crease can be folded toobtain a cone.

Figure 5. A picture of a modular origamirepresenting an icosahedron, built with 30identical copies of a folded square, linked

together in a symmetrical way.

let u be a smooth map which satisfies Du 2 O(2)or equivalently

|Du|2 = 2 |detDu| = 2.Since detDu = ±1, by continuity the Jacobian hasa sign, say detDu = 1. Therefore the equation|Du|2 = 2 detDu, with the notation u (x, y) =⇣v(x,y)w(x,y)

⌘, can be easily transformed into the Cauchy-

Riemann system(vx �wy = 0vy +wx = 0

with furthermore |Dv| = |Dw | = 1. Denoting by⌧,⌫, respectively, the tangent and normal unitvectors on @⁄, we thus have, up to sign,

hDv ;⌧i = hDw ;⌫i , hDw ;⌧i = hDv ;⌫i .Since v = w = 0 on @⁄, we also obtain Dv = Dw =0,which contradicts the fact that |Dv| = |Dw | = 1.

Thus any solution to the differential problem(1) is Lipschitz continuous but not of class C1 nearthe boundary; therefore it assumes in a fractalway the homogenous boundary datum ' = 0 (seeFigure 6). The map u will be explicitly defined atevery (x, y) 2 ⁄, and it will be piecewise affine,with infinitely many pieces, in accord with itsfractal nature near the boundary of ⁄.

Using this property of the singular set we are ableto construct some examples of fractal origami whichsatisfy some “nonisometric” boundary conditions.In general, solutions to (1) may be very irregular, upto the point that the singular set can be everywheredense.

In conclusion, origami is seen here as a tool forfinding simple solutions in a context of systemsof partial differential equations where infinitelymany solutions exist, none of them being of classC1, most of them having very irregular gradient.General existence theories have been developed bymeans of convex integration and by Baire categorymethod to obtain Lipschitz continuous solutions.Simpler scalar problems can be treated with theviscosity theory, obtaining explicit representationformulas. This theory cannot, however, be appliedin the context considered here, due to the lack of amaximum principle. Nevertheless, origami allowsus to present particularly simple solutions witha recursive structure. It is a challenging issue toadapt the method proposed here to more generalsystems of first-order differential systems such asthose encountered in geometry, optimal design,and nonlinear elasticity (see the section: “ImplicitPartial Differential Equations: Conclusion”).

Figure 6. On the right-hand side a solution tothe Dirichlet problem (1) actually folded with

paper; it gives rise to a fractal shape. In theleft-hand side the singular set of the same map

is represented in a region near the boundary.

Some Other Topics About OrigamiMany mathematicians interested in geometry oralgebra (for example, in group theory, Galois theory,graph theory…) have studied origami constructions.An important issue is the geometrical constructionof numbers. In this aspect origami turns out to bemore powerful than the classical rule and compassconstructions (see [1], [5], [6] or [17]). In orderto determine what can be constructed throughorigami, it is important to formalize the rules.These are known as Huzita axioms and have beenproposed by Hatori, Huzita, Justin, and Lang. Hereare the seven axioms.Axiom 1: Given two points P1 and P2, there is aunique fold passing through both of them.


Figure 7. Trisecting an angle. The resulting red angle is a third of the given green angle.

Axiom 2: Given two points P1 and P2, there is aunique fold placing P1 onto P2.Axiom 3: Given two lines L1 and L2, there is a foldplacing L1 onto L2.Axiom 4: Given a point P and a line L, there is aunique fold perpendicular to L passing through P .Axiom 5: Given two points P1 and P2 and a lineL, there is a fold placing P1 onto L and passingthrough P2.Axiom 6: Given two points P1 and P2 and two linesL1 and L2, there is a fold placing P1 onto L1 and P2

onto L2.Axiom 7: Given a point P and two lines L1 and L2,there is a fold placing P onto L1 and perpendicularto L2.

For a slightly different approach to these axiomsand for the connection to Galois theory, we referto the book by Cox [6].

Origami constructions, contrary to rule-and-compass ones, allow one to trisect any angle (seeFigure 7), to double a cube, or to construct a regularheptagon.

Another topic, more closely related to what wediscuss in this paper, is the problem of finding analgorithm to decide whether a given crease patterncan actually be folded to generate a physical origami,i.e., a rigid map without interpenetration. Theproblem turns out to be NP-hard (nondeterministicpolynomial-time hard), see [4].

Other interesting research is devoted to con-structing crease patterns to fold to some giventhree-dimensional models with elaborated shape.With suitable software, Lang [21] has been able tofold realistic models of different kinds of animals.

(Im

age(

s)co

urt

esy

of

Rob

ert

J.La

ng,

ww

w.la

ngo

riga

mi.c

om

.

Figure 8. On the left-hand side the “ballpacking technique” to create the creasepattern of the model (crab) shown on the right(by R. T. Lang).

One of the main points turns out to be that offinding the best possible packing of given disksinto the paper rectangle (see the left-hand side inFigure 8); in fact, we notice that every origami is ashort map, namely

|u (x)� u (y)| |x� y| , 8x, y 2 ⁄;

this means that the distance of two points in thefolded model cannot be larger than the distance ofthe two points in the unfolded model. This implies,for example, that the preimage of the claws ofthe crab in Figure 8 must contain the whole circlecentered in the preimage of the tip of the claw withradius given by the length of the claw.

Recently in [19] some numerical algorithms havebeen found to reproduce three-dimensional curvedmodels. Some of them have the peculiar propertythat the crease pattern is mostly composed ofcurved lines, whereas classical origami only usesstraight lines. Curved creases have also beenconsidered by Huffman [15].

Another interesting application comes fromaerospace engineering. A team of scientists at theLawrence Livermore National Laboratory in theUnited States are building a large space telescope.To put it into action in space, the telescope will befolded and unfolded following the rules of origami.The picture in Figure 9, taken from [14], representsa prototype of the telescope.

(Cou

rtes

y/LL

NL.

)

Figure 9. The unfolded space telescope and, onthe left, the corresponding singular set.

Mathematical Origami as a Rigid MapIn this presentation an origami will be modeled bythe notion of rigid map. A map u : ⁄ ⇢ Rn ! Rm issaid to be a rigid map if it is Lipschitz continuousand Du(x) 2 O(m,n) for a.e. x 2 ⁄.


http://www.langorigami.com

From the definition it follows that, given anytwo vectors � ,⌧ 2 Rn, we have

hDu(x)� ,Du(x)⌧iRm = h(Du(x))tDu(x)� ,⌧iRn= h� ,⌧iRn ,

where h·, ·iRn is the scalar product in Rn. Inparticular u leaves invariant the metric on thetangent space.

Since u is Lipschitz continuous it is, by the clas-sical Rademacher theorem, differentiable almosteverywhere in ⁄. We call the singular set of u theset ÷u where the map u is not differentiable. Wenow quote two classical results. The first one saysthat isometries are rigid maps with empty singularsets.

Theorem 3.1 (Cartan-Dieudonné). Let ⁄ ⇢ Rn bean open connected set and u : ⁄! Rm be an isome-try:

|u(x)� u(y)| = |x� y|, 8x, y 2 ⁄.Then m � n and u is affine.

The second result applies to the case m = n. Inthis case a rigid map corresponds to a flat origami.The theorem states that a flat and smooth origamican only be trivial.

Theorem 3.2 (Liouville). Let ⁄ be a connected opensubset of Rn. If u : ⁄! Rn is of class C1 and satisfiesDu 2 O(n), then u is affine.

From now on we will only consider piecewise C1

rigid maps, which are maps u : ⁄ ! Rn satisfyingthe following properties: (i) ÷u is relatively closed in⁄; (ii) u is of classC1 in every connected componentof ⁄ \ ÷u; (iii) every compact set K ⇢ ⁄ intersectsonly a finite number of connected components of⁄ \ ÷u.

Reconstruction from the Crease PatternAs above, when n =m we are in the flat case, andwe consider flat origami, i.e., piecewise C1 rigidmaps u : ⁄ ⇢ Rn ! Rn. In this context we can provethat, if ⁄ is a convex polyhedral set, then everyconnected component of ⁄ \ ÷u is also a convexpolyhedral set. The singular set ÷u is the union ofthe faces bounding the convex regions, hence it isitself a (n� 1)-dimensional polyhedral complex.

Every (n � 1)-dimensional facet of ÷u is theinterface between two different regions, say A1

and A2. On each region the map u is affine,u(x) = Mix+ qi for x 2 Ai , and it is not difficult toshow that the two affine maps are related to eachother by

(2) M2 = M1S, M1 = M2S,where S 2 Rn⇥n is the matrix representing asymmetry with respect to the (n� 1)-dimensionalplane containing the facet we are considering (thusdetS = �1). In particular we have

detM2 = �detM1,

and the sign of the determinant gives a colorationto ⁄ \ ÷u .

In particular we see that if M1 is given, M2 isuniquely determined. From these considerations itfollows that, if ⁄ is also connected, once we knowthe affine map u in a single region, we can recoverit in every other region.

However, not every polyhedral complex ÷ isthe singular set of some rigid map u. Notice, inparticular, that if e is a (n�2)-dimensional edge of÷, then the number of (n� 1)-dimensional facetscontaining e must be even. This is due to the2-coloration property of the complement of ÷u. Sowe denote with f1, · · · , f2k the faces containing e. Ifwe denote with ↵i the angles between faces fi andfi+1, by means of condition (2) we can show thatthe following angle condition holds (see Figure 10):

(3)kX

i=1

↵2i�1 =kX

i=1

↵2i = ⇡ .

This condition is well known in the study offlat origami. When n = 2 this condition was firstobserved by Kawasaki [18].

Figure 10. The angle condition. The sum ofangles with the same color is equal to ⇡ .

Hence, up to now, we know that the singular set÷u of a piecewise C1 rigid map u is a polyhedralset such that ⁄ \÷u is 2-colorable; moreover, every(n� 2)-dimensional edge is contained in an evennumber of (n � 1)-dimensional facets spaced sothat the sum of angles on the regions of one coloris equal to the sum of angles for the regions of theother color.

It is interesting to know that the reverse impli-cation, stated in the following theorem, holds (see[10]).

Theorem 4.1 (Recovery theorem). Let⁄ be a simplyconnected open subset of Rn. Let ÷ ⇢ ⁄ be a locallyfinite polyhedral set satisfying the angle condition(3) on every (n� 2)-dimensional facet. Then

(i) there exists a rigid map u such that ÷ = ÷u isthe singular set of Du;


(ii) u is uniquely determined once we fix the valuey0 = u(x0) and the Jacobian J0 = Du(x0) at a pointx0 2 ⁄ \ ÷.

The theorem means that if we construct aset ÷ satisfying the conditions just mentioned,there always exists a map u such that ÷ = ÷u .Moreover, the map u is uniquely determined up tocompositions (on the left) with rigid motions ofthe whole space.

In the literature about flat folding (see, forinstance, [2], [4]), the problem of reconstructingan origami from its singular set (which is alsocalled crease pattern) has already been considered.However, a more strict definition of origami shouldbe used to treat physical origami. In a physicalorigami the interpenetration problem is addressed,i.e., the problem that physically a noninjective mapcannot be folded without cutting the paper. To bemore precise, since the thickness of the paper cananyway be neglected, we admit noninjective mapsas long as they are the limit of injective ones.

This restriction makes the angle condition notsufficient to guarantee the global reconstructionof a physical origami. An example is given by thesingular set in Figure 11 which satisfies the anglecondition, hence it is the singular set of some mapu which, however, cannot be folded without cuttingthe paper. In this more strict setting it is shown(see [4]) that the problem of deciding whether acrease pattern can be flat-folded is NP-hard.

Figure 11. A singular set of a rigid map whichcannot actually be folded without cutting thepaper.

Fractal ConstructionWe now turn our attention to the more difficultproblem of finding a solution to the followingimplicit partial differential system

(4)

(Du(x) 2 O(n) for a.e. x 2 ⁄u(x) = 0 for x 2 @⁄

in the case when ⁄ is a rectangle in R2 (orparallelepiped inRn). Now we know that a solution is

represented by a flat origami. However, the Dirichletboundary condition is, in principle, incompatiblewith the first-order differential system, as alreadymentioned in the introduction.

We first discuss the case n = 2. As we alreadysaid, only a fractal construction can ensure theboundary condition u = 0. Figure 6 gives an ideaof the construction that we are going to develop. If⁄ is a square, we can divide it in infinitely manysquares, which are smaller and smaller, while weapproach the boundary of ⁄. Then it is enough toconsider a base map u0 defined on one of thesesquares. This map will be translated, rotated, andrescaled to fit any other square. To assure that thegluing of the squares gives a continuous map, weneed the base map to have prescribed recursiveboundary conditions. More precisely we requirethat, on the right-hand side of the base square,the map is defined so that it reproduces twice thevalues of the left-hand side, rescaled by half; i.e.,

u0(1, y) = u0(0,2y) for y 2 [0,1/2],u0(1, y) = u0(0,2y � 1) for y 2 [1/2,1];

while on the upper and lower sides we only needperiodic boundary conditions:

u0(x,0) = u0(x,1) for x 2 [0,1].If the map assumes at least once the value 0 on

every square in the net, then by its Lipschitz conti-nuity (every rigid map is 1-Lipschitz continuous)it can be extended to the boundary @⁄ with the 0value.

The base module can thus be found by using theangle condition (3) and the recovery theorem 4.1. Inthis way we are able to find solutions to the aboveproblem. More generally we might consider linearboundary conditions, such as u(x, y) = (�x, µy)on @⁄. The corresponding singular set is shown inFigure 12.

Notice that we can easily pose the problem inhigher dimension by considering n-dimensional flatorigami, namely hyper-origami. Again we subdividethe domain in smaller and smaller cubes (orparallelepipeds). On the base domain we needto find an n-dimensional origami which fulfillsthe recursive boundary condition. For n = 3we can imagine this operation as the foldingof a rigid three-dimensional box. In Figure 13,we present the “instructions” for folding thebase parallelepiped to obtain the zero boundarycondition. By gluing together the base modules weobtain the 3-dimensional singular set.

Finally we note that one can solve (4) withsymmetric rigid maps, i.e., maps u which aregradients of functions v : ⁄ ⇢ R2 ! R.The problem(4) when n = 2 then becomes

(D2v(x) 2 O(2) for a.e. x 2 ⁄Dv(x) = 0 for x 2 @⁄.


Figure 12. A fractal construction for the linear boundary data.

It turns out (see [11]) that the corresponding anglecondition (3) is, in this case, much more restrictivein the sense that at every vertex of ÷u = ÷Dv exactlyfour consecutive angles ↵+,↵�,�+,�� meet with

↵+ + �+ = ⇡ and ↵� = �� = ⇡2.

We call this property “second-order angle condition”.The singular set represented in Figure 12 satisfiesthe second-order angle condition, in contrast withthe singular set of Figure 6.

Implicit Partial Differential Equations:ConclusionWhat we have done so far is to find explicit solutionsto Dirichlet problems, as in the section “FractalConstruction”. This kind of problem belongsto a wider class, called implicit partial differentialequations. Namely we look for Lipschitz continuoussolutions to

(5)

(F(x, u (x) ,Du (x)) = 0 for a.e. x 2 ⁄

u (x) =' (x) for x 2 @⁄,where u,' : ⁄ ⇢ Rn ! Rm and F : ⁄⇥Rm⇥Rm⇥n !R (also called Hamiltonian).

Note that, as in (4), the problem is apparently illposed, since we want to solve a first-order PDE andat the same time prescribe full Dirichlet conditionon the boundary. Indeed, except for very specialcases, no C1 solution is to be expected, and only

Lipschitz solution may exist. Curiously enough, ifthere is a solution of (4) or more generally of (5),then, in general, there are infinitely many solutions.Let us review some of the techniques for solvingsuch equations and see how the constructions wehave proposed for origami may shed light on someissues related to implicit PDEs.

Let us start with the scalar case m = 1 whichhas received considerable attention since the workof E. Hopf, notably by Kruzkov, Lax, and Oleinik.In this case the equation

F(x, u,Du) = 0

is sometimes called Hamilton-Jacobi equation.Crandall-Lions further developed in this contextthe important tool of viscosity solution. The beautyand the importance of the viscosity method is that,at the same time, it gives existence of solutionsand a way of selecting one among the infinitelymany solutions of (5). It is, however, of crucialimportance in this context that the HamiltonianF be convex in the variable Du. For example, theeikonal equation

( |Du (x)| = 1 for a.e. x 2 ⁄u (x) = 0 for x 2 @⁄

has, for convex domains ⁄, the viscosity solution

u (x) = dist (x, @⁄) ,


Figure 13. A “hyper-origami”, precisely a three-dimensional flat origami which solves the Dirichletproblem (4) for n = 3 with zero boundary condition. At the top an elementary brick is represented.It has a self-similar property which allows recursively gluing it together with half-sizedhomothetic copies of itself to obtain the pattern represented at the bottom.

although, as said earlier, there are infinitely manyother solutions, one of them being

u (x) = �dist (x, @⁄) .

However, as soon as the Hamiltonian F isnonconvex (with respect to the last variable) andstill in the scalar case m = 1, or in the vectorialcontext n,m � 2 (as for origami), the viscositytheory does not apply. Two other main theorieshave been developed to deal with these cases, and,although different, they give essentially the sameresults (for detailed references see [9]). However,both theories are purely existential and do not givea simple criterion for selecting one solution amongthe infinitely many as did the viscosity theory. Letus briefly discuss both of them.

(i) The theory of convex integration proposedby Gromov is one of them. It was originallyintroduced to give a new proof of the celebratedNash-Kuiper theorem on isometric embedding,which is closely related to the problem discussed inthe present paper. The theory of convex integrationhas been developed in the context of implicit partialdifferential equations by Müller-Sverak and others.

(ii) The Baire category method is the other tooland has been introduced by Cellina, Bressan-Flores,and De Blasi-Pianigiani for the scalar case andgeneralized to the vectorial context by Dacorogna-Marcellini.

We also point out that, in the scalar casem = 1, and for Hamiltonians F, not necessarilyconvex, that do not depend on lower-order terms(i.e., F(Du) = 0), explicit solutions can easily beobtained and are known as “pyramids”, followingthe work of Cellina and Friesecke. This was a sourceof inspiration to construct explicit solutions of(4) to Cellina-Perrotta and to the work presentedhere. Contrary to the finite element method used innumerical analysis to build approximated solutions,here, by means of piecewise affine maps withappropriate grid, we are able to exhibit exactsolutions.

Since the two approaches described above areonly existential, it is then an important issue tosingle out one solution among the infinitely many.It is what we achieved by means of origami for thespecial problem (4). There are many other problemsrelated, for example, to geometry, optimal design,


or nonlinear elasticity where an actual explicitconstruction would be important.

Among them a problem which is significant inapplications to microstructures in crystallographicmodels is the one of potential wells (see, forexample, [3]). When we consider the case of twowells, it consists in finding maps u whose gradientsatisfies

(Du(x) 2 SO(n)A[ SO(n)B, a.e. in ⁄

u(x) ='(x), on @⁄,

where A and B are two given n ⇥ n matrices andSO(n) is the set of special orthogonal matrices;i.e., those with determinant equal to +1. This is anatural generalization of the problem consideredhere (see (4)), since

O(n) = SO(n)I [ SO(n)I�,where I is the identity matrix and I� =diag(�1,1, . . . ,1).

Finally, implicit partial differential equations,because of the existence of infinitely many solu-tions, are particularly challenging for numericalanalysts. In the problem studied in the section“Fractal Construction”, one could imagine, as hasbeen suggested by one of the referees, findingalgorithms that mimic the origami construction.

References[1] R. C. Alperin, A mathematical theory of origami

constructions and numbers, New York J. Math. 6

(2000), 119–133.[2] E. M. Arkin, M. A. Bender, M. A. Demaine, et al.,

When can you fold a map?, Computational Geometry29 (2004), 23–46.

[3] J. M. Ball and R. D. James, Proposed experimentaltests of a theory of fine microstructure and the twowells problem, Phil. Trans. Royal Soc. London A 338

(1991), 389–450.[4] M. Bern and B. Hayes, The complexity of flat origami,

Proceedings of the 7th Annual ACM-SIAM Symposiumon Discrete Algorithms, 1996, pp. 175–183.

[5] B. Casselman, If Euclid had been Japanese, NoticesAmer. Math. Soc. 54 (2007), 626–628.

[6] D. A. Cox, Galois Theory, Wiley-Interscience,Hoboken, NJ, 2004.

[7] M. G. Crandall, H. Ishii, and P. L. Lions, User’sguide to viscosity solutions of second order par-tial differential equations, Bull. Amer. Math. Soc. 27

(1992), 1–67.[8] B. Dacorogna and P. Marcellini, General existence

theorems for Hamilton-Jacobi equations in the scalarand vectorial case, Acta Mathematica 178 (1997),1–37.

[9] B. Dacorogna and P. Marcellini, Implicit PartialDifferential Equations, Progress in Nonlinear Dif-ferential Equations and Their Applications, vol. 37,Birkhäuser, 1999.

[10] B. Dacorogna, P. Marcellini, and E. Paolini,Lipschitz-continuous local isometric immersions:Rigid maps and origami, J. Math. Pures Appl. 90

(2008), 66–81.

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[12] M. Gromov, Partial Differential Relations, Springer-Verlag, Berlin, 1986.

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About the Cover

Fractal OrigamiThe cover image is a variant of Figure 12 in this issue’s arti-cle by Bernard Dacorogna, Paolo Marcellini, and EmanuelePaolini on some mathematical aspects of origami.

The connection between the image and origami mightnot be apparent. A partition of a planar region ! into anynumber of convex polygons is said to satisfy the Kawasakiangle condition if at every vertex there is an even numberof polygons meeting at each vertex, and the alternatingsum of angles there vanishes. A partition satisfying thiscondition determines a map from ! into the plane withthe property that maps on neighboring polyhedra areobtained by reflection. The Kawasaki condition impliesthat this specification is consistent. I call such mapsfold maps. The simplest nontrivial fold map is the basicorigami move.

In this and other figures, I follow Dacorogna et al. indistinguishing those domains that change orientationfrom those that don’t.

A fold map F is contracting in the sense that |F(P)!F(Q)| " |P ! Q|. A curious question arises: given adomain ! and a boundary path map " from its boundary!! satisfying some contraction condition, does there exista partition whose associated fold map agrees with " onthe boundary? According to Dacorogna et al., this isso if the boundary map satisfies the strict contractioncondition |F(P) ! F(Q)| " "|P ! Q| for some " < 1.Any map satisfying other than a very simple boundaryconstraint must come from an infinite partition, and onemight expect that the behaviour of the partition near theboundary will be interesting. In Figure 12 of their article,and on the cover of this issue, are illustrated partitionsthat give rise to a linear boundary map in the case where! is a rectangle.

This seems to be the only large class with an explicitconstruction. In both these cases, the partition is byrectangles of di!erent sizes but of only two basic designs.It is easy to draw the unadorned picture without muchtrouble. It is made up of rectangles of various sizes,but up to similarity exactly the two basic designs shownabove, which Dacrorogna et al. call the base regions: Itis relatively easy to draw the figure in a straightforwardway, since the pattern of rectangles is easy to lay out,and the region at the top occurs precisely on diagonals.But it is somewhat more interesting to do something that

reflects more directly the recursive branching structure.The regions can be assembled in a directed acyclic graphwith the central region as root and edges going out fromthe center to smaller and smaller regions. This can bepruned to become a tree, one associated to a finite statediagram of five states, marked by distinct colors in thisfigure:

Using this finite state machine to traverse the treemakes possible what might be called “intelligent” draw-ing. What other boundary conditions on fold maps can berealized by finite state diagrams? What can one say in gen-eral about the relationship between a boundary conditionand the structure of a partition whose fold map satisfiesit?

—Bill CasselmanGraphics Editor

([email protected])


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