HAL Id: hal-03217319
https://hal-cnrs.archives-ouvertes.fr/hal-03217319
Submitted on 4 May 2021
HAL is a multi-disciplinary open access archive for the deposit and dissemination of sci- entific research documents, whether they are pub- lished or not. The documents may come from teaching and research institutions in France or
L’archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires
Polygon shortening makes (most) quadrilaterals circular.
Thierry Jecko, Jean-Christophe Léger
To cite this version:
Thierry Jecko, Jean-Christophe Léger. Polygon shortening makes (most) quadrilaterals cir- cular.. Bulletin of the Korean Mathematical Society, Korean Mathematical Society, 2002,
�10.4134/BKMS.2002.39.1.097�. �hal-03217319�
irular.
by Th. Jeko and J.C. Leger
09-10-00
Abstrat: We show that an analog of the Gage-Grayson-Hamilton The-
orem for urves moving aording to their mean urvature holds for the
motionof quadrilateralsaording to theirMenger urvature.
1
1 Introdution
The Gage-Grayson-Hamilton Theorem (see [Gag84℄, [Gra87℄ and [GH86℄)
statesthatifC 0
:S 1
!C isaC 2
Jordanurveintheplanethenthereexists
afamilyC :S 1
[0;T
[!C ofsmoothJordanurvesintheplanesuhthat
C(;0)=C 0
(),C(;t) tends to aonstant funtionast!T
and
C
t
=
2
C
s 2
: (1)
Moreover, ifweresale theurvessothattheenlosedareas areequalto ,
we have that theresaledurves tend(for exampleinthe Hausdor sense)
to airle of radius1.
Notie that the right hand side of equation (1) is the urvature vetor
oftheurveC(t)(sis thear-lengthparameterontheurveC(t): itisnot
independent with thevariable t). A familyof urves C dependingon time
tsatisfyingequation (1)is saidto be movingbyitsmean urvature.
A question(among others) thistheorem raises is whether one an nd
a disrete version of the motion by mean urvature whih preserves this
theorem: this should be a test for a measuring how good a disretization
proess forthismotionis.
In this short note, we shall present a proposition for a possible good
disretization and show that at least for quadrilaterals an analog of the
Gage-Grayson-Hamilton Theoremexists.
1
AMS lassiation (2000): 34D05, 53A15, 53C44. Keywords: Menger urvature,
evolution,oyliality,disretization,Gage-Grayson-Hamiltontheorem.
Mengerurvature ofthe triplethenumber
(x;y;z)=
x y
x y
z y
z y
1
x z
: (2)
This quantity will be our analog for polygons of the usual urvature for
smooth urves. Observe that this is a natural quantity to have a look at.
The fat is that (x;y;z) is zero if and only if the points x, y and z are
olinear, j(x;y;z)j is the inverse of the radius of the irumirle of the
triangle xyz and y +
jj 2
is the enter of that irumirle. Observe as
wellthatthisquantityhasalready beenusedforndingdisreteanalogs of
well known urvature related theorems for Jordan urves (see for instane
the introdution of [Sed97℄ and the referenes therein for a variant of the
four-vertexTheorem). Anotherbeautifuluseofitisrelatedtoboundedness
properties of singularintegral operators on subsets of C (see [MMV96℄ for
instane).
For anintegern3 letusalla n-gona mapy:Z=nZ!C suhthat,
for all i 2 Z=nZ,the points y
i 1
;y
i
;y
i+1
are distint and let us dene its
Mengerurvature by
i :=(y
i 1
;y
i
;y
i+1
); 8i2Z=nZ: (3)
Remarkthat we have
X
i2Z=nZ (y
i 1 y
i+1 )
i
= 0: (4)
For aninterval I R, we willsaythat thefamilyof n-gons y:Z=nZ
I ! C is movingbyits Menger urvature ifit satisesthe followingdier-
ential system
8i2Z=nZ; y 0
i :=
dy
i
dt
=(y
i 1
;y
i
;y
i+1 )=
i
: (5)
This kind of motion shares many ommon features with the lassial
motion by mean urvature, the rst being that it preserves the shape of
polygons whih are oylial or olinear (i.e. the verties lie on a single
irleora single line).
Anotherommon propertyis theone of lengthshrinking. Foran n-gon
y, put its length L = P
i2Z=nZ jy
i+1 y
i
j. We have then, fora familyy(t)
of n-gons moving by their Menger urvature that L(t) is a noninreasing
funtion(see Lemma1). Moreover, we have
L 0
=
X
i2Z=nZ j
i j
2
+j
i+1 j
2
2
jy
i+1 y
i
j; (6)
fortheusualmotionbymean urvature.
In thisnote we willbe interested by thease n=4, thatis the motion
ofquadrilaterals(4-gons). Inthisase, aswasalreadynotedforgeneraln's,
oylial polygonskeeptheirshapes. Itturnsoutthatparallelogramskeep
theirshapesinthe motionaswell.
Here isnow arough idea ofthemain result(f. Theorem1 and Propo-
sition 2) of this note. If we are given an initial quadrilateral, the Cauhy-
Lipshitz Theorem ensures the short-time existene and uniqueness of a
family of quadrilaterals evolving by their Menger urvature and starting
withthisquadrilateral.
Itturnsoutthatfor\most"startingquadrilaterals,themaximalinterval
ofexistene[0;T
[isboundedand,ast!T
,thefamilyollapsestoapoint
whereasthelimitingshape isa oylialquadrilateral.
We would like to stress on the fat that given a quadrilateral to start
with, we have a simple test (see setion 4) to see if the limiting shape in
theevolutionisoylial: wedonotputanyrestritions(suhasonvexity
or no self-rossing) on the type of quadrilaterals we are looking at. The
evolution in thesingularases is also desribed: thelimiting shape is then
aparallelogramora olinearquadrilateral.
Inthegeneralsituation(n4),weareabletoshowthat, ifn6=0mod-
ulo4,theonlyinvariant shapesare theoylial orolinearones whereas,
ifn=0modulo4,there isanotherinvariant shape,namelyaparallelogram
desribedn=4 times(see Proposition1).
OpenProblems: Wedonotknowwhathappensintheasen>4,exept
thepreviousdesriptionofinvariant shapes, andthismaybean interesting
subjet offurtherinvestigation.
Another path of investigation is to nd out if the motion of polygons
by Menger urvature really approximates the usual motion of urves by
urvature and ifso,inwhihsense.
2 General fats and rst result
In this setion, y : Z=nZI ! C will be a family of n-gons evolving by
theirMenger urvature. Weollet some general fats aboutsuha family.
Mostproofsare elementaryomputations.
Lemma1 (i) For any triple (u;v;w) of distint points in C, we have
2Re
(u;v;w)(u v)
=j(u;v;w)j 2
ju vj 2
:
djy
i+1 y
i j
dt
= j
i j
2
+j
i+1 j
2
2
jy
i+1 y
i j:
In partiular, (6)holds true.
(iii) If nis even, thereexist someonstant b
0
>0 suh that for all t2I,
jy
1 y
2 jjy
3 y
4 jjy
n 1 y
n j = b
0 jy
n y
1 jjy
2 y
3 jjy
n 2 y
n 1 j:
Proof. For(i),wehave,asv+
jj 2
istheenteroftheirlepassingthrough
u,v and wof radius1=jj, that
ju v
jj 2
j 2
= 1
jj 2
:
Itremainsto expand theabove formulato get(i).
For(ii), itis enoughto omputethederivativeof jy
i+1 y
i j
2
and apply
(i)to the triples(y
i 1
;y
i
;y
i+1
) and (y
i
;y
i+1
;y
i+2 ).
Finally,(iii)follows from
1
2 X
i2Z=nZ j
i j
2
= d
dt lnjy
1 y
2 jjy
3 y
4 jjy
n 1 y
n j
1
2 X
i2Z=nZ j
i j
2
= d
dt lnjy
n y
1 jjy
2 y
3 jjy
n 2 y
n 1
j:
Itis straightforwardto prove the
Lemma2 (i) If at time t
0
2 I, y(t
0
) is a polygon whose verties lie on
a single line, then for all t2I, y(t)=y(t
0 ).
(ii) Ifattimet
0
2I,y(t
0
)isapolygon whosevertieslieona singleirle
withenterx
0
andradius R
0
then for all t2I, for all i2Z=nZ,
y
i (t)=x
0
+(t)(y
i (t
0 ) x
0 )=x
0 +
p
R 2
0
2(t t
0 )
R
0
(y
i (t
0 ) x
0 ):
Lemma 2 givestwo invariantshapesunder theonsidered evolution. Is
there some other invariant shape? The following proposition answer this
question.
Proposition 1 Assume that the shapeof y(t
0
) is preserved under the evo-
lution,that is, therearetwo omplex valued, dierentiable funtionsdened
on I, a and b, abeing non-vanishing, suh that,
8t2I; 8i2Z=nZ; y
i
(t)=a(t)y
i (t
0
)+b(t);
then,
0
(ii) or, n=0 modulo 4, y
i+4
=y
i
for all i2Z=nZand y
1
;y
2
;y
3
;y
4 form
a parallelogram (i.e. y
1 y
2
=y
4 y
3 ).
Proof. Sine theshapeis preserved,
0 = d
dt jy
i+1 y
i j
jy
i+2 y
i+1 j
:
UsingLemma 1, (ii), we get j
i+2
(t)j =j
i
(t)j. Now, using(2) and (5), we
obtain,foranyi2Z=nZ,
i (t)=
da
dt (t)y
i (t
0 )+
db
dt
(t)and
i (t
0
)=a(t)
i (t);
sothat, foranyi2Z=nZ,
i (t
0
) = a(t) da
dt (t)y
i (t
0
)+a(t) db
dt (t):
From now on, we onsider only the quadrilateral y(t
0
) so we drop the
refereneto the timet
0 .
If a is onstant, the
i
's are equal. Assume they are nonzero. Denote
theenters oftheirumirlesbyz
i :=y
i +
i
=j
i j
2
. Thusthe(z
i y
i )'sare
equal. Sine y
i+1
lieson the irle entered on z
i
ontaining y
i and y
i lies
on theirleentered on z
i+1
ontainingy
i+1
,thisimpliesy
i
=y
i+1
,whih
isexluded. Thereforethe
i
's areequal to0, whih meansthatthe y
i 'slie
on thesame line.
If a is not onstant, eah y
i
is the image of
i
by some i-independent
similitude
8i2Z=nZ; y
i
=
i
+ (7)
forsome omplexnumbers 6=0 and . Letusdenote byr
0
(resp. r
1 ) the
ommonmodulusof the
2i
's(resp.
2i+1
's). Ifr
0
=r
1
then r
0
>0 and (7)
showsthatthey
i
'slieonthesameirle. Assumer
0 6=r
1
. Thisimpliesthat
nis even. From (7)and (i) inLemma1, we derive that, forall i2Z=nZ,
2Re (
2i (
2i+1
2i
))=jj 2
r 2
0 j
2i+1
2i j
2
; (8)
2Re (
2i+1 (
2i
2i+1
))=jj 2
r 2
1 j
2i+1
2i j
2
: (9)
Onone hand,the sum(8)+(9) leadsto
2Re()=(r 2
0 +r
2
1 )jj
2
(10)
and ontheother hand, thedierenegives, thanksto (10),
2sin()sin(
2i )=(r
2
0 r
2
1
)jjos(
2i
); (11)
i
angle(
i
;0;
i+1
). In thesame wayforthepoints
2i and
2i 1
,we obtain
2sin()sin(
2i 1 )=(r
2
0 r
2
1
)jjos(
2i 1
): (12)
Siney
2i 1 6=y
2i+1 ,
2i 1 6=
2i+1
by(7)and
2i 6=
2i 1
beause j
2i 1 j=
j
2i+1
j. The equations (11) and (12) say that
2i
and
2i 1
satisfy an
i-independent equation, whih solutions are equal modulo . This yields
that
2i 1 and
2i+1
are symetriw.r.t. theorigine. Similarly
2i and
2i+2
are also symetri w.r.t. the origine. Therefore the
2i
's (resp.
2i 1 's) are
equal. This means that the quadrilaterals (
2i 1
;
2i
;
2i+1
;
2i+2
) form the
same parallelogramm. This is possible onlyforn=0 modulo4. Using (7)
again,we gettheresult.
3 Spei fats for quadrilaterals
We nowspeializeto the aseof quadrilateralsor4-gons (i.e. n=4).
Lemma3 (i) There exists a onstant 2C
suh that for all t2I,
y
4 y
2
=(y
3 y
1 ):
(ii) There existsa onstant b
0
>0 suh that for all t2I,
jy
1 y
2 jjy
3 y
4 j=b
0 jy
1 y
4 jjy
3 y
2 j:
Proof. For(i),we justobservethat (5)and (4)implythat
y 0
1 (y
4 y
2 )+y
0
2 (y
1 y
3 )+y
0
3 (y
2 y
4 )+y
0
4 (y
3 y
1 )=0
sothat
(y
3 y
1 )(y
4 y
2 )
0
(y
3 y
1 )
0
(y
4 y
2 )=0:
(ii)is justa rewritingof (iii)of Lemma1.
Among the invariant shapes (see Proposition 1), we have already de-
sribed the evolution of two of them in the general ase (see Lemma 2).
Beforetreating thethird one for n=4, we want to stress on the fat that
the following lemma also gives the evolution of the periodi parallelogram
(i.e. theform (ii)inProposition1) inthegeneralase.
Lemma4 Iffor t
0
2I, y(t
0
)isa nonatparallelogramwithenterx
0 then
for any t2I,y(t) isa similar parallelogram withenterx
0
givenby
8
>
>
<
>
>
: y
1
(t) = x
0
+(t t
0 )=2;
y
3
(t) = x
0
(t t
0 )=2;
y
2
(t) = x
0
+(t t
0 )=2;
y
4
(t) = x
0
(t t
0 )=2:
(13)
= y
4 y
2
y
3 y
1 (t
0
)62f 1;1g and (t)=(0) s
j(0)j 2
t
j(0)j 2
e i
ln
j(0)j
2
t
j(0)j 2
;
where
= 8
(Im) 2
(jj 2
+1)
jj 2
j+1j 2
j 1j 2
and = 4
(Im )(Re )(jj 2
1)
jj 2
j+1j 2
j 1j 2
:
Proof. Beause of the uniqueness in the Cauhy-Lipshitz Theorem, it
suÆesto hekthat (13)is a solutionof thedierentialsystem(5).
Whatoughttobesaidaboutthisformulaisthatanonatparallelogram
suhthatjj6=1(thatis,notaretangle)evolvingaordingtoitsMenger
urvature shrinks to its enter in nite time and that, while shrinking, it
rotates more and more rapidly around that point. This is a very strong
ontrast with the evolution of a oylial quadrilateral whih shrinks to
theenter oftheirle innitetimebutdoesnotrotate at all.
As we areinterested in thepropertyof beingoylial, itis naturalto
introdue theross-ratioof aquadrilateralz. It is denedby
B
z
= (z
1 z
2 )(z
3 z
4 )
(z
1 z
4 )(z
3 z
2 )
: (14)
ThequantityB
z
annotbe0or1beausewesupposedthatthepointsz
i
aredistint. Itiswellknownthatthequadrilateralzisoylialorolinear
ifand onlyifB
z
is areal number. An important featureof B
z
isthat it is
a sale and rotation invariant quantity: thisis the quantity we will use to
haraterize oylial shapes.
What islesserknown,althougheasy,is thefollowingsale androtation
invariantharaterizationof parallelogramsintermsof
z
= z4 z2
z
3 z
1
and B
z :
z isa parallelogramifand onlyif
B
z
=
z 1
z +1
2
: (15)
Thisisdueto the followingrelation
(z
2 +z
4 ) (z
1 +z
3 )
2
= (z
3 z
1 )
2
1 B
z
(
z 1)
2
B
z (
z +1)
2
: (16)
Reall that =
y(t)
is onstant during the motion and that B(t) =
B
y(t)
is suh that jB(t)j =b
0
isonstant as well(see Lemma3). The next
lemma statesthat we know quitepreiselyhowB moves on theirle with
enter0 and radiusb
0 .
B 0
= j(y
2 +y
4 ) (y
1 +y
3 )j
2
jy
2 y
1 j
2
jy
4 y
3 j
2
B(B B): (17)
(ii) Unless we are looking at a family of evolving parallelograms or at a
family of oylial or olinear quadrilaterals, ReB is a dereasing
funtionand unlesswe arelookingata familyof oylial orolinear
quadrilaterals, ImB isof onstant sign.
(iii) Put, for a xed t
0
2I and all t2I,
(t)= Z
t
t
0 j(y
2 +y
4 ) (y
1 +y
3 )j
2
jy
2 y
1 j
2
jy
4 y
3 j
2
d: (18)
Thenthereexists a onstant C suh that for all t2I,
ReB(t)=b
0
1 Ce 4b0 (t)
1+Ce 4b0 (t)
: (19)
Proof. Equation (17) is a straightforward omputation. Notie rst that
forany i2Z=4Z,
B
y
i
=( 1) i+1
y
i+1 y
i 1
(y
i+1 y
i )(y
i 1 y
i )
B;
sothat,
B 0
= X
i2Z=4Z
i B
y
i
= X
i2Z=4Z ( 1)
i+1
1
(y
i+1 y
i )(y
i 1 y
i )
1
(y
i+1 y
i )(y
i 1 y
i )
!
B
= 2iIm
1
jy
2 y
1 j
2
y
2 y
1
y
4 y
1 y
2 y
1
y
2 y
3
+
+ 1
jy
4 y
3 j
2
y
4 y
3
y
2 y
3 y
4 y
3
y
4 y
1
B
= 2iIm
1
jy
2 y
1 j
2
y
2 y
3
y
4 y
3 B
y
4 y
1
y
4 y
3 B
+
+ 1
jy
4 y
3 j
2
y
4 y
1
y
2 y
1 B
y
2 y
3
y
2 y
1 B
B
= 2iIm
1
jy
2 y
1 j
2
y
2 y
3
y
4 y
3 B+
y
4 y
1
y
4 y
3 B
+
+ 1
jy
4 y
3 j
2
y
4 y
1
y
2 y
1 B+
y
2 y
3
y
2 y
1 B
B
= j(y
2 +y
4 ) (y
1 +y
3 )j
2
jy
2 y
1 j
2
jy
4 y
3 j
2
B(B B): (20)
