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An Example of a Stable and Symmetric Cuve

著者 Yokoyama Misako

journal or

publication title

Reports of Faculty of Science, Shizuoka University

volume 30

page range 1‑11

year 1996‑02‑25

出版者 Shizuoka University. Faculty of Science URL http://doi.org/10.14945/00002566

(2)

Reports of the Faculty of Sciencg SHIZUOKA IINITERSITY, Vol. 30, p.

1

to p.

11 (1996)

An Example of a Stable and Symmetric Cuve

Misako Yoroyeue

Defartment of Mathematics Facalty of Science, Shizuoka Uniaersity, 836 Ohya, Sh,izuoka 422, taPan

(Receiaed Sept. 18, 1995)

Abstract.

Here we have obtained an example of

a

stable closed plane curve with

3

crossing points which is homeomorphic to a given cunre and satisfies some technical conditions. That curve is also symmetrig which means the class has a symmetric representative. To compute an example we use spline functions and nonlinear optimization techniques. The ocample has no computational error. The technical condition would be considered to be an approximation that we have done.

l.Introduction

In recent yeanr, numerical approaches have been s<tensively done in finding exact answers to analytical problems in a number of fields (e.S. f l, [9] ). In the present paper, we consider a numerical procedure for obtainingverified results on computation of the stable plane cunre, by the use of spline functions and optimrzatian techniques. The motivation of this paper is the speech about knot energy t10l . See also t3l and t8l

.

Let N be a closed plane cunre with crossing points. What is a "stable" curve F like? Is it

possibly symmetric? By "stable" we mean that F is an isolated local minimum in the class q(N) with respect to the "energy" P.

A Cz-mapping F:[0, 1] *R'z belongs to 9(N) if F satisfies some conditions (see Definition

4. 5).

The set T(N) is an infinite dimensional space. By the minimal property of spline functions we reduce the problem in an infinite dimensional space to that in a finite dimensional space.

Then we use a method in the nonlinear programming theory.

In this paper\re present asuccessful example of a closed planecunrewittr 3 crossing points.

The resulting solution of that ocample is errorfree and invariant with

a

rotation through(Z/S) n

(see Corollary 6.2).

(3)

Misako Yorov.eue

2. heparation I : Splines

We recall some basic definitions and properties on splines.

Ileftrition 2.1 (APeriodic Spline Frmction) A function f

:

[0, 1]-R is a periodi,c sflirufunction

of degree J with nodes tt, "' *o where 0 : tr

(1) f (t) is a polynomial of degree at most 3 on each subintenral [tr, tr+r] , i : 1, "', n-li (2)f e c' [0, 1]

;

(3) f(') (0) :

f(o)

0), i :0,L,2.

Prolnsition 2.2 (T\e Existence and Uniquenees [1] ) For grven te,xrcR, i:1, "' , n, with 0 :

t, < ... 1 tn :

I"

there ocists one and only one periodic spline function of degree 3 f : [0, 1]

- R with nodes tt,;" , t" which satisfies the conditions

f(t;:xr,i:1,"',tr.

The following theorem is well known. See t5l for the natural spline version.

fireorem 2.$ (Ihe Mininal Property) Letf:10, 1l -R be the periodi,c spliru function of degree 3 utith nodes tr, "' ,b(A: tr ( "' ( t" : L) which satisfies the cortditions

f (t;:xf i:1, "',tr'

Thea for any C-feriodi,c functiong: [0, L] -R whhh sati.sftes the condi'tions

g(ta)-)9,i:1,"',r,

the following holds:

J

ltrrtlF dts

J

lgttlrat

The eqaali.ty hold,s if and onU if f : g.

Ileftrition 2,4 (AParanetric Periodic Spline Mapping) A mapping F:(f, g) : [0, 1] * R2 is

a parametric

fueri,odi,c

sfline mabPtng of degree Swith nodes tr, "' ,tz where 0 : tr ("'( tz :

f. if the functions t [0,1] - B and g: [0, 1] * R are periodic spline functions of degree 3

with nodes t4 "',ta.

3. Preparartion II : Nonlinear Programming

NCIil,werecall somenonlinearprogrammingtechniques. Let o :R - Rand Fa: R" r &

i : 1, "' , m be C'tappings. Define the set O as follows:

O : {X :(Xr,... X") eR" I Er(X) 30,i:1, "',m}

.

Deftrition 3,1 (An Isolat€d Local Minimum) X* e R" is an isolated lacal minimam inthe set O with respect to o if there exists 6 > 0 such that

o (x*)<o (x),)s + v Xe,o n B(x*;0),

where B (X*; 6): {X e R" | (.> (X*,-X)T/2< 6 }

.

(4)

An Example of a Stable and Symmetic Curve

助山曖

m3。

2(翻

COnditiOnl The Kuhn‐ Tucker Condition with respect to X=

α

l,… ,XnD∈

RFand R=侭

1,…

,R″ )∈ Rtt is as follows:

R̀È∞

=0,i=1,… ,m9 R̀≧0,i=1,… ,m,

È∞

0,i=1,… ,m.

If the above condition holds9 R Is called the Kuhn‐

Tucker vector assodated with X

尭 σ ∞ +謄 lL尭=Qk=L…

,

X″

Xp(σ

+占

R̀È∞

)。

el)

Let X∈

as

I∞ = IIα

Э

= V∞

=

r/

6.カ

Ω and let R∈ Rπ bethe Kuhn‐ TuckeF VeCtOr associated with X.Dttne the sets

{i∈ {1,・

,m}IE″

=0},

{i∈ {1,"。 ,m}IR̀>0},

{Y∈

l▽

Èoo Y=Q Vi∈

I・

;

È(DY≦ Q Vi∈ {j lj∈ I(D,j¢

I100}}, where▽ È∞

=喝

Eo,¨

,光

È∞ ].

Theorem 8.8 (lbe $econd Order $ufficlent Condition) [2] Sufgose thatY (Xr, "' , X") e R"

andV (R*r, "',R*.) G Bn" satisfytheKahn-TorckerCondi,tion(3.1).LetH6",R) bethenn.matrir

whose (k, f)umiorwnt is

*,Rつ

Y>0, 0≠

VY∈

*),

thenY is an isolatd, local rninimam in the set A u,ith respect to

o .

4. Ilefinitions

Ihfinition 4.1 (A Closed Plane Cbve) A continuous mapping F:6, g): [0, 1] * R' is a closd

frtawcan)eif fe:[0,L] *BareC2-periodicfunctionswith(f(t)'+ Gi(t))'*0for anyt

[0, 1] . A point (A y) of Rz is a crossing goint of F if tlere exist tr, tz e [0, ].), tt * tz such

thatF(tr):F(tz):(:;y).ApointofF([0,1])whichisnotacrossingpointisaregalarfioint.

A crossing point (a y) is a double crossing foint if the set F*l (x,y) n [0,1) contains o<actly

two elernents.

(5)

Misako Yoroyeue

We assume that any pomt of F([0,1])iS either a renar pOint or a double crossing point v7here F([0,1])intersects transversdy(io e.fOr each double crossing pomtは

,,Of F=CD

such thatに

,=Fctl)=F(め,t■,し

[0,1),伍

≠し ,the vectors lfelゝ 」

lt.))andぱ0ご

0)

are linearly independenO。

D面

m倣

m4。 21AnArcl LetF:10,1]―

R2be a dOsed plane curve.The dosure of a connected

mponent of{F([0,1])一 {CrOSSing points of F}}is an  α π of F。

輸 価腱43 1A Specinc Pointl Let F:10,1]―

R2be a Closed plane 

α鵬亀

If the set F([t

])is an arc of F,we call the point F(

t')/21a%菱

Й例レタα

%′Of F.A point of F(10,1])iS

蒻θ夕蒻グ of F ifitis a crossing point or a middle point of F.

珈 血

441AD輌

問哺

劇れ

)A union of arcs of a closed plane ctlrve F is a蒻

磁″′ 客 απ″

if it is a circle which is the boundary of the closure of a bounded,connected component of

{R2̲F([0,11).

D面 壼億

m45(T(N))Take io∈ {2,…

,n‑1}st F(iめ

F10)and fiX it.Let N:[0,1]―

R2and F=CD:[0,1]一

R2be dOSed plane curves.

Put

{tl,・

・ ・

,tπ

}=F 1({all Specific points of F}) where O=伍 <¨

<tπ =1.F belongs to the set 

Ψω  if the following holds:

(1)t̀=(i‑1)/1n‑1),i=1,…

,n;

(atherettsts anorientationpreservinghom∞ morphismh:R2̲R2suChthathO([0,1]))=

F([0,1]);

431 the area Of each o五

ented polygon whose vertices∞ nsist of all spdic points of each

di宙

ding circle of F is equalto or more than■ 1韓

141 F o)=10,0.Put lXo,yol=Flt

).Then xO>O and yO=0.

Note thatf and g are C2̲periodicfunctions by the definitiOn Of a dosed plane curve.

D面

面仙 m461The Enewl We define the̲ρ

 lF)of a C2̲m4)ping F=C,o:

[0,1]―

R2aS fO1lows:

ρC)=∫lば'(0)2 dt+∫ lば'ctl)2&

沢物%α滋

47 LetF,Fl,F2:[0,1]‐

R2be C2̲mappings defined by

Fo=CO,2tl),

Flltl=C(0+X9go+yl,

F20=(∞

O―

Sin  θ

 gltl,Sin 

θ  fttl+∞

Ю

).

Then  ρ  C)=ρ

 Cl)=ρ

 lFD.

(6)

Am EXample of a Stable and Symmetric Curve

Ddh山

ねは4.81The Metricl Let F=cl,F2),G=(Gl,G2):[0,1]一 R2be C2̲mttpings.

We define the%ι π

d(F,G)of F and G by

d lF,G)=maxt=1,2 maXP=0,1,2 Supた Q」

IFメ

カ )ltl一

Gメ

ltl l。

Dtton4.9(Stable)Let N:[0,1]―

R2be a given dosed plane curveo A dosed plane curve F:[0,1]― R2is sれ み顔 ″

N I the following hold鋭

(1)F∈ Ψ

o);

2)F is an isolated local minimumin the set 

ΨINl宙

th reSp∝tto 

ρ

,i.e。,there exlsts ε

>O such that

ρ lF)<ρ

(G),F≠

G∈ Ψ

INl∩ B IF;ε

),

where B OL 

ε )={G:[0,1]一

R21G∈

C2[0,1],dlF,G)<ε }.

5.Aュ Exalnple

Let N be as Fな 。 1.We find a stable dosed plane curve Fx*for N as follows:

Fむ。

1 P2

P10

Sメ″ I:Putthe dmdy ordered specific points as Plメ 。,P13;See F尊1.Puttheirx‐

y∞

odinates

as follows:

P■ =P7=P13=10,0,P2=tXl,L),P3=P9=偽 ,0),

P4=偽 ,Xlo),P5=Pll=後

,X11),P6=偽

,X12),

P8=偽 ,X13),P10=lX7,X14),P12=偽,X15),

where L>0。

Put X=α l,…

,X15).

R=P7=P・

3 P3=P,

(7)

Misako Yoroyeue

S′

2:COmpute the parametnc pe■

odic spline mapping of degree 3 Fx=(fx,gxl:[0,1]一 R2v7hich satisfies

Fx((i‑1)/121=P̀,i=1,… ,13.

Put σ∞

=165/51841 

ρ

 C⇒ ,then 

σ α

)is aS A.lin Append破

sttp 3:Put

Ml∞

=lthe area of the oriented polygon<PlP12P5P4P9P3>),

M2∞ =(the area of the o五ented polygon<PlP8P9P2>),

M3∞

=(the area of the odented polygon<P9P4P5P10>), M400=(the area of the o五

ented polygon<PlP6P5P12>).

By Definition 45 131 we haVe

M′∞ ≧

1,i=1,23,4.

Put

E,(Ю

=‑2M̀(聡

+2, i=1,2,3,4,

then E′ lXl≦ 0。E′

are aS

El∞

=2+X』GO―

Xメ

13+X3Xll X4X10+X4X15 X8Xl■,

E2∞

=2‑XX9+XX■ 3,

E3∞

=2‑X』LO十

XX14 X3Xll+X4X10 X4X■ 4+X,Xll, E4∞

=2+X4X12 X4X15 X5Xll+X8Xll.

Put

Ω={X=α

l,…

,X151∈

R151Eグ

∞ ≦

0,i=1,2,3,4}。

″ イ

:The Kuhn―

Tucker Condition(3.1)宙

th respectto X=α

l,… ,X15)∈

R15,R=cl,・

・。,

R41∈ R4for σ

 lXl and E′

,i=1,2,3,4is as A2 in Appendk

5:We∞nsiderwhether there exists a sttmetric solution ofA.2.Let the three triangles

△ PlP5P9,△ P8P12P4and△Pr6P10 be all regular triangles which have the same bary centtr, and the three points P2,P8,P5be On the same line.Then we substitute the following in A.2:

Xl=X6=X4,X2=2X4,X5=2X4 X7,

草三 滋 1'=littf警

III∫

讐 F二 iJ3X3,

h this situation,we have only one solution X*and R*of A.2.Put a=21/2,b=131/2,c=

651μ,d=31ノ

then X*is as

X*1=X*4=X*6=aC/b, r2=2ac/b,r3= (aC3d/

b)+(15ac/8b),

X*5= (aC3d/40b)十

(ac/8b),XX7=(aC3d/40b)+(15ac/8b),

X*8=(aC3d/40b)+(aC/8b),X*9=(acd/4b)+(aC3/20b),

X*10=X*15= 15acd/8b)十 (ac3/4b),X*11=― (acd/b),

X*12=X*14= 15acd/8b)―

(aC3/40b),X*13=(aCd/4b)― (aC3/20b).

(8)

An D【

ample of a Stable and Symmetric Curve

Step 6: Fx* is as A3 in Appendix It contains no computational error because we did not do approximation. To verify Fx* c q(N), it is sufficient to see that Fx* is equivalent to N. Fr*

( t0, 1l ) is as Fig.2. Since Fx* is a parametric spline mapping, we can say Fr* e q(N).

Fig。

2

S″夕

7:We check whether the Second Su亜

dent Condition 13.21 holdS Or nOt̀R*is as

R*1=0,R*2=R*3=R*4=20」65。

I+α

=Iα

={2,■ 4}。

Vα *)={Y∈ R151▽ L cЮ

 Y=0,i=2,3,4}

L∞

Y=0,i=2,3,4 are aS

20Y9‑20Y13+√ 65Y2=0,

‑20Y10+20Y■4+20√ 3Y3 20√

3Y7+√ 3√

65 Yll―

65Y2+√65Y4=0,

20Y12 20Y15+20√ 3Y5 20√

3Y8+√ 3√

65 Yl■ ― √

65Y4=0.

Since

αち Rり Y≧

0,V∈

Vαり

YオHば ,P)Y=O implies Y=0,

6. Proof

Theorem 6.1 The aboue Fx*

as

a stable closd blane curae forN.

By definition

Then we have

and

We have obtained Fx*

(9)

Misako Yoroyaue

Praグ Put n=15.By Teorem 3.3 there mists  δ =O such that σ

(r)<σ

,x*≠ VX∈

Ω ∩Bσ倅

).

Put 

ε   δ

/√

n.Takeany G∈ ΨO)∩ BCx*;ε

),G≠Fx*。Letr bethevalue ofX detemined

by the specttic points of G。

If r=xt by the factthat G≠ Fx*and Theorem 24 we have

ρ

 lFxり (Gl.

rr≠

xち by Theoreln 2.4 we have

σα')=ρ (Fxり ≦ ρ

(Gl。

We daim that r∈

*;δ

).hdeed,since G∈

Bcx*;ε ),we have

lX'′

― Xち

 │ 

≦d Cx*,G)<ε ,  ∀ j=1,…

,n。

Therefore

X'一X*│12=Σ

α

,′

X*め

2

1=1

<Σ   ε

2=n 

ε2=δ

2

1=1

Since G∈

Ψω ;比holds that X∈ Ω and

ρ

(Fx*)=6184/65)σ

 αP)<1518″

651 

σ  α

')=ρ lFx*)≦

ρ

(G).

This completes the proo■

oorouary 6.2二 ι′

Nレ

助 工

T麟

潔形

Cttss ToOみ

お α s″ み ルσ Jas″

p滋

̀π

η′

Fx*ω

λ′ ε λ なれπ

ttzπ

′就

"″ 減,%厖

"箔

λ

12/31π

Rcttα tt aθ

 lf at step 5 we substitute only the fo...lula Which means that the triangle△ Pl P5P9iS aregulartriangle,we can so市 ethesystemA.2 and haveonlythesame solutton as above X*.

Ach… .

The author had done this work when she bdonged to Kyushu University.She thanks

Hiroshi Ohtsuka for his introducing her the sofhare mu MATH,Hidefumi Kawasaki and

Shunsuke Shttaishi for useful info....ation on nonlinear programmlng,Yoshihiro TakeuOhi fordiscussionsandhishelpingapartofhermonotonousanddme―nsumingwork,Mitsuhiro To Nakao for his suggestion and encouragel■ lent.

(10)

An Enanrple of a Stable and Symmetric Curye

1.Ahl挽 喀,J.H"Nilson,ENeandWalsLJoLデ

:TheTheOryofSplines andTheirApplications,″

New YoFk and London,Academic Press 1967。

2.Fiacco,A.V.and McCo..鳳ick,G.P。

,'Nonlinear Programmmg9 Sequential Unconstraind

Minimiztton Techniques,"John wiley Sons,1職

3.Fukuharap S。 ,Energy of a knot ln'切

■ le Fa℃

bf Topology,"五

じademic Press l激 、 48‑451。

4.Grebile,T◆ N.B,"Spline Functions and Applications,"Wisconsin Universiけ

,1"0。

5◆ Holladayp J◆ C。

,Sm00thest curve approxmation,in"Mathematical Tables and Aids to Computation,Ⅱ

,"1957,"鉾243.

6.Kuhn,■ 。 w.and Tuに ,Ao W"Nonlinear Programmi略

,in'Ttt of 2nd欧

dey

Symposium on Mathematical Statistics and Probabiliけ

,"Jo Nepan(ed),UniVersity of Califomia Press,1951.

7.い ,Ko Ro and Schmidt Do S。 (editOrs),"COmputer Aided PF00fS in Analysis,"SpFinger―

Verlag9 1991.

3.Sakumら

En釧

ofgeodesic links in St preprint.

9.Ullrich Co leditOrl,"COmputer Arithmetic and Sdf― Validating Numerical Methods,"

Academic Press:1990。

100■

z晦

J"Family of energy Jmctionals of knots,Topology Appl.4819%147‑161.

(11)

Misako Yoroyelae

Appendix

A.1

σα

)=‑468Xう

+1例

XIX3 104XX4+14XIX5+2XⅨ 6+14XX7+194XlX8 468XX

3 416X分

Q‑104XX5 468XX6 468XX7 lC14XX3 468Xぶ 4+194XX5+14X3X6+

2XJ酔+14XX8 468XX5 104XX6 468XX7 468XX3+194X5X6 14Xよ 7+2XX8+

194X:X7+14X6X3+194X7X3+194X9X■

0‑104Xぶ 11+14X9X12+2X9X13+14X9X14+194X

9X15 468X10X■ 1+194X10X12+14X10X13+2X10X14+14XloX■ 5 468XlIX12‑104XllX13

468X■ lX14 468XllX15+194Xl♪ Q3+14XlX14+X12X15+194XlX14+14XlX15+194X14

X15+31lX12+624X22+31lX32+α

ttx42+31lX52̲31lX62 311X72+31lX82+31lX92+

31lX102+622112+31lX122̲31lX132+31lX142+31lX151 JL2

621‑468L+194L‑1嘔 +14L+2X6+14し

+1例

=0,

一 晨渇Xl+1248X2 468X3+410L‑1

‑468X6 468X7 1創

X9R2+X10Rl―

XloR3 X13RI+X13L+X14R3=0,

194Xl一

+622X3 468L+1例 5+14X6+2X7+14L tt Xl■Rl―

XllR3=0,

IMXl+416X2‑468L+1248X4 銀

1創

‑468X7 468X8 X10Rl+X■

OR3+

X12R4 X14R3+X15RI― X■

5R4=0,

14L‑1創

+1創

‑468L+622X5+1例 +14X7+2X8 XllR4=0,

2X■ ‑468L+14L‑1創

+1例

+622X6+14X7+14L=0,

l‑468X2+2L‑468X4+14L+1例 +622X7+194X8+XllR3=0,

194Xl二

104X2+14X3 468X4+2X5+14X6+194X7+622X8 XllRl十

XllR4=0, 622X9+IMX10‑104Xll+14X12+2X13+14X14+194X15 X2R2=0,

194X9+622X10‑468X■ 1+194X12+14X13+2X14+14X15+X沢

1 X顔し ―X4Rl十

X沢

3=

0,

‑1

為 一 晨渇

X10+12蟷

Xll‑468X12 14X13 4MX■ 4 468X15+X3Rl― LR3 X5R4+

X7R3 X3RI+X8R4=0,

14聰

+194LO‑468X11+622X12+19Ⅸ 13+14脇 4+2X■

5+X4R4=0, 2X9+14X10‑1創1+1創2+622X13+12脇

4+14X15 X2Rl十

X2L=0,

14聰

+2Xl。 ‑4渇

X五 ‑14X12+19襲3+622X14+1例5+X2R3 LR3=0, 194X9+14X10‑468X11+2X12+1」

13+194X■

4+622X15+X4Rl― X4R4=0,

2Rl tt X2X10RI―

X2X13RI+X3X1lRI― LX10RI+X4X15RI― X8X1lRl=0, 2R2‑X2X8R2+X2X13R22=0,

2R3 X2XloR3+】

X14+R3 X3XllR 3 +X4X10R3 X4X14R3+X7XllR3=0, 2R4+X4X12R4 L駆 15R4 聰X1lR4+X8XllR4=0,

R′ 0,i=1,…

,4 È∞

0,i=1,…

,4.

JL3

(abcd/6)fx*│【

i‑1)/12,i/12]

=‑18t+2dc2t+36dc2t2̲144dc2t3+324t'一 LD6tt ifi=1,

(12)

An EXample of a Stable and Symmetric Curve      ll

=‑18t+2d♂

t+36d♂

‑144d♂ f+34♂ ‑1以

るご

, if i=2,

=‑23+96t一

dc2+20dFt‑72d♂

+72d♂

‑2160F+367P, if i=3,

=112‑1224t一

d♂

+20d♂t‑72d♂

+72drf+4320F‑4968F, if i=4,

=‑256+488t一 d♂

+20d♂t‑72d♂

+72d♂

ゞ‑5616ぽ +綱旺ぽ, if i=5,

=369‑2412t一 d♂

+20d♂t‑72d♂

+72d♂

+5184/‑3672f, if i=6,

=‑22+1314t+26dd‑142d♂

t+252d♂

‑144d♂

ザー

2268F+126t3, if i=7,

ニー

252+1314t+26dP‑142d♂ t+252d♂

‑14はf‑2268P+12る

, if i=8,

=1220‑5310t‑38d♂

+146d♂t‑180d♂

げ+72d♂ゞ

+7ぽ

一〔

F″

, if i=9,

=‑2425+9270t‑38d♂ +146d♂ t‑180d♂

+72d♂

ザー11777+爛駁;f, if i=lQ

=3325‑11430t‑38d♂

+146d♂t‑180d♂

+72d♂f+13068f‑4968f, if i=11,

=― 認∞+10mt― 優通♂+146d&‑180d♂

+72d♂

ザ ー

10697+3672t3, if i=12.

(abcd/61 gx*│[(1‑1)/1a i/121

=6dt‑324d♂ ‑2880df+6c2t― 野 ′

, if i=1,

=lχttt‑1116dF+2880d♂

+(冠先

‑36J′

(1(ソ

, if i=2,

=‑288dt+1368d′ ― 卿

+2に

‑144P′

+21∝

2t3+K59/助

d― ♂

,if i=3,

=252dt‑792df+792dt3+2に

̲144P′ +216ぽ

ゞ―t76/助

d―

c2, if i=4

=252dt‑792dザ +792df‑120Ft+288J′ ‑216♂

♂―176/助 d+1ば

, if i=5,

=‑1248dt+2808dP‑208d♂

‑120Ft+288P′ ‑21∝

2t3+183d+15♂ ,if i=6,

=2478dt一

i′ +2880df+4に■― 野 ′

‑4Bd‑11ば ,if i=7,

=‑802dt+5436d′

‑2880df+4た

‑3

2♂ 12116/ed‑1ガ

,if i=&

=30dt〒 4500dt2+2088dt3+ulCt‑468c■ 2+216c2t3̲r2300/31d‑76ct ifi=9,

=‑lmdt+1980dt2‑792dt3+330c%‑468c■ 2+216c%3+(1投

5/31d‑76ct ri=lQ

=‑lmdt+1980dt2‑792dt3̲570c先

+612c■2‑2Xた

3+(1&5/3D d+174ct Fi=11,

=562dt‑5940dt2+2088dt3̲570c■

+612c■

2‑216c■ 3̲1770d+174ct Fi=12.

Fig。 2 S″夕 7:We check whether the Second Su亜 dent Condition 13.21 holdS Or nOt̀R*is as R*1=0,R*2=R*3=R*4=20」 65。 I+α り =Iα り ={2,■ 4}。 Vα *)={Y∈ R151▽ L cЮ  Y=0,i=2,3,4} ▽ L∞ Y=0,i=2,3,4 are aS 20Y9‑20Y13+√ 65Y2=0,

参照

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