Research Article Power Geometry and Elliptic …downloads.hindawi.com/journals/ijde/2015/340715.pdfPower Geometry and Elliptic Expansions of Solutions to the Painlevé Equations AlexanderD.Bruno

Research ArticlePower Geometry and Elliptic Expansions ofSolutions to the Painleveacute Equations

Alexander D Bruno

Keldysh Institute of Applied Mathematics Miusskaya Square 4 Moscow 125047 Russia

Correspondence should be addressed to Alexander D Bruno abrunokeldyshru

Received 30 January 2014 Accepted 24 June 2014

Academic Editor Sining Zheng

Copyright copy 2015 Alexander D BrunoThis is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We consider an ordinary differential equation (ODE)which can bewritten as a polynomial in variables and derivatives Several typesof asymptotic expansions of its solutions can be found by algorithms of 2D Power Geometry They are power power-logarithmicexotic and complicated expansions Here we develop 3D Power Geometry and apply it for calculation power-elliptic expansions ofsolutions to an ODE Among them we select regular power-elliptic expansions and give a survey of all such expansions in solutionsof the Painleve equations 119875

1 119875

6

1 Universal Nonlinear Analysis

We develop a new calculus based on Power Geometry [1ndash4]Now it allows to compute local and asymptotic expansionsof solutions to nonlinear equations of three classes (A) alge-braic (B) ordinary differential and (C) partial differential aswell as to systems of such equations

Principal ideas and algorithms are common for all classesof equations Computation of asymptotic expansions ofsolutions consists of the 3 following steps (we describe themfor one equation 119891 = 0)

(1) Isolation of truncated equations 119891(119889)

119895= 0 by means

of faces of the convex polyhedron Γ(119891) which is ageneralization of the Newton polyhedron the firstterm of the expansion of a solution to the initialequation 119891 = 0 is a solution to the correspondingtruncated equation 119891

(119889)

119895= 0

(2) Finding solutions to a truncated equation 119891(119889)

119895= 0

which is quasihomogenous using power and loga-rithmic transformations of coordinates we can reducethe equation 119891

(119889)

119895= 0 to such simple form that

can be solved Among the solutions found we mustselect appropriate ones which give the first terms ofasymptotic expansions

(3) Computation of the tail of the asymptotic expansionEach term in the expansion is a solution to a linearequation which can be written down and solved

Applications

Class A (1) Sets of stability ofmultiparameter problems [5 6]

Class B (2) Asymptotic forms and expansions of solutions tothe Painleve equations [4 7 8]

(3) Periodic motions of a satellite around its mass centermoving along an elliptic orbit [9]

(4) New properties of motion of a top [10](5) Families of periodic solutions of the restricted three-

body problem and distribution of asteroids [11 12](6) Integrability of ODE systems [13]

Class C (7) Boundary layer on a needle [14](8) Evolution of the turbulent flow [15]For a survey of these applications see [16]

2 Introduction

Let 119908(119906) be a formal elliptic asymptotic form of a solution toan ODE that is it is a solution of a corresponding truncatedequation The form 119908(119906) is suitable if it can be extended into

Hindawi Publishing CorporationInternational Journal of Differential EquationsVolume 2015 Article ID 340715 13 pageshttpdxdoiorg1011552015340715

2 International Journal of Differential Equations

power asymptotic expansion V = 119908(119906) + suminfin

119895=1119887119895119906minus119895 where

119887119895

= 119887119895(119906) are some functions The expansion is regular if all

119887119895are not branching functions of 119908(119906) and its derivatives If

all functions 119887119895(119906) =

119895(119908 ) have no branching then they

are elliptic functions with the same periods as119908(119906) Selectionof such cases is our aim For given 119908(119906) and fixed point1199080 (including infinity) we can compute power-logarithmic

expansions of functions 119895(119908 ) near 119908 = 119908

0 In theseexpansions logarithmic branching can appear only if 119908

0 isa singular point and algebraic branching (of finite order)can be for subsingular points 119908

0 To each singular point 1199080

and suitable asymptotic form 119908(119906) we assign unique regularexpansion V = V(1199080 119908(119906)) so called basic andwe are lookingfor such basic expansions near singular point 119908

0 which haveno branching

We propose algorithms for (1) finding all formal ellipticasymptotic forms (2) finding all suitable elliptic asymptoticforms and (3) calculation of power-logarithmic expansionsof functions

119895(119908 ) near a singular point119908

0 and selection ofbasic expansions without branching All algorithms are basedon 3D Power Geometry

Expansions are formal their convergence is not con-sidered Application of these algorithms to the Painleveequations 119875

1 119875

6gives following

(1) 11987511198752 and119875

4have continuumof 2-parameter families

of elliptic asymptotic forms each 1198753has three 119875

5has

two of them and 1198756does not have

(2) 1198751 1198752 and 119875

4have countable sets of families of

suitable asymptotic forms each and all 5 forms of 1198753

and 1198755are suitable

(3) Basic expansions for all suitable forms have nobranching for 119875

1 for 119875

2if the independent variable

tends to infinity for 1198753if condition C is fulfilled and

for 1198755if condition D is fulfilled and 119886 = 119887 = 0 and

119889 = 0

History of calculation of elliptic expansions of solutionsto the Painleve equations is as follows

A hundred years ago Boutroux [17] found 2 familiesof elliptic asymptotic forms of solutions to the Painleveequations 119875

1and 119875

2 During the last 5 years we found 6

additional families of elliptic asymptotic forms of solutions to1198753(three) [18 19] 119875

4(one) [20] and 119875

5(two) [21] Moreover

the Painleve equations 1198751 1198752 and 119875

4have continuum of

families of elliptic asymptotic forms each and I proposed acriterion for selection suitable asymptotic forms which canbe extended as asymptotic expansions All 8 known ellipticasymptotic forms are suitable Solutions to the equation 119875

6

have no elliptic asymptotic forms at allNear infinity of the independent variable the Painleve

equations 1198751ndash1198755have 12 families of suitable asymptotic forms

and near zero of the independent variable equations 1198751

1198752 and 119875

4have countable sets of such families each Next

I extend these suitable elliptic asymptotic forms 119908(119906) intopower-elliptic expansions V = 119908(119906) + sum

infin

119895=1119887119895119906minus119895 where

coefficients 119887119895are functions of the corresponding elliptic

asymptotic forms and their derivatives To each family ofsuitable elliptic asymptotic forms I put in correspondenceunique basic formal power-elliptic expansion near 119908

0= infin

for 1198751ndash1198755 near 119908

0= 0 for 119875

3ndash1198755 and near 119908

0= 1 for

1198755 Obstacles (logarithmic branching) in calculations of these

basic expansions appeared only for 1198752if the independent

variable tends to zero for 1198754and for 119875

5if |119886| + |119887| = 0 or

119889 = 0Thus near infinity of the independent variable there

are 10 families of regular (ie without branching) ellipticexpansions of solutions to equations 119875

1ndash1198756 4 for 119875

1 2 for 119875

2

3 for 1198753 and 1 for 119875

5 Existence of these expansions for two

Boutroux families of asymptotic forms was proven in [22]and this is all knownup-to-dateNear zero of the independentvariable there is a countable set of families of such expansionsfor 1198751 The results were obtained by means of algorithms of

3D Power Geometry [18ndash24] realized in very cumbersomecalculations

Here I introduce the third variant of 3D Power GeometryThe first was in [18 20 24] and the second was in [19 21ndash23]

In more precise form main results are as in Theorems 1214 15 and 16 and Conditions C and D

Equation 1198756cannot be studied by proposed approach

3 3D Power Geometry

Let 119909 be independent and 119910 be dependent variables 119909 119910 isin CA differential monomial 119886(119909 119910) is a product of an ordinarymonomial 119888119909

11990311199101199032 where 119888 = const isin C (119903

1 1199032) isin R2 and a

finite number of derivatives of the form 119889119897119910119889119909119897 119897 isin N The

sum of differential monomials

119891 (119909 119910) = sum 119886119894(119909 119910) (1)

is called the differential sum Let 119899 be the maximal value of 119897

in 119891(119909 119910)In [2ndash4] it was shown that as 119909 rarr 0 (120596 = minus1) or as

119909 rarr infin (120596 = 1) solutions 119910 = 120593(119909) to the ODE 119891(119909 119910) = 0where 119891(119909 119910) is a differential sum can be found by means ofalgorithms of Plane (2D) Power Geometry if

119901120596

(119889119897120593

119889119909119897) = 119901

120596(120593 (119909)) minus 119897 119897 = 1 119899 (2)

where the order

119901120596

(120593 (119909)) = 120596 lim sup119909120596rarrinfin

log 1003816100381610038161003816120593 (119909)1003816100381610038161003816

120596 log |119909|(3)

on a ray arg 119909 = const and 119899 is the maximal order ofderivatives in119891(119909 119910) Order of the power function 120593(119909) = 119909

120572

with 120572 isin C is 119901120596(119909120572) = Re120572

Here we introduce algorithms which allow to calculatesolutions 119910 = 120593(119909) with the property

119901120596

(119889119897120593

119889119909119897) = 119901

120596(120593 (119909)) minus 119897120574

120596 119897 = 1 119899 (4)

where 120574120596

isin R

International Journal of Differential Equations 3

Theorem 1 120596 minus 120596120574120596

⩾ 0

For example 1205741

= 0 for 120593 = sin119909 and 120574minus1

= 2 for120593 = sin(1119909) Note that in Plane Power Geometry we had120574120596

= 1 that is 120596 minus 120596120574120596

= 0 So new interesting possibilitiescorrespond to 120596 minus 120596120574

120596gt 0

Problem 2 Select leading terms in the sum (1) after substitu-tion 119910 = 120593(119909) with property (4)

Belowwe describe algorithms for solution of the problemTo each differential monomial 119886

119894(119909 119910) we assign its (3D)

power exponent Q(119886119894) = (119902

1 1199022 1199023) isin R3 by the following

rules

1199023= sum of orders of all derivatives

1199022= order of 119910

1199021= difference of order of 119909 and 119902

3

Then the 2D vector 119876 = (1199021 1199022) is the same as in

2D Power Geometry [2ndash4] and 1199023corresponds to the total

order of derivatives The power exponent of the product ofdifferential monomials is the sum of power exponents offactorsQ(119886

11198862) = Q(119886

1) + Q(119886

2)

The set S(119891) of power exponents Q(119886119894) of all differential

monomials 119886119894(119909 119910) presented in the differential sum 119891(119909 119910)

is called the 3D support of the sum 119891(119909 119910) Obviously S(119891) sub

R3 The convex hull Γ(119891) of the support S(119891) is called thepolyhedron of the sum 119891(119909 119910) The boundary 120597Γ(119891) of thepolyhedron Γ(119891) consists of the vertices Γ(0)

119895 the edges Γ(1)

119895

and the faces Γ(2)119895 They are called (generalized) faces Γ(119889)

119895

where the upper index indicates the dimension of the faceand the lower one is its number Each face Γ(119889)

119895corresponds

to the 3D truncated sum

(119889)

119895(119909 119910) = sum 119886

119894(119909 119910) over Q (119886

119894) isin Γ(119889)

119895cap S (119891) (5)

All these definitions are applied to differential equation

119891 (119909 119910) = 0 (6)

Thus each generalized face Γ(119889)119895

corresponds to the truncatedequation

(119889)

119895(119909 119910) = 0 (7)

Let N119895

= (1198991 1198992 1198993) be the external normal to two-

dimensional face Γ(2)119895 We will consider only normals with

1198991

= 0

Example 3 Consider the second Painleve equation 1198752

119891 (119909 119910)def= minus11991010158401015840

+ 21199103

+ 119909119910 + 119886 = 0 (8)

where 119886 is the complex parameter

q2q1

q3

1

1

3

2

1Q2Q

3Q

4Q

Figure 1 3D support S(119891) and polyhedron Γ(119891) of equation 1198752(8)

The grey face is Γ(2)1 the grey edge is Γ(1)

1 Projection on the plane

(1199021 1199022) is shown by dotted lines Dashed line is the invisible edge

If 119886 = 0 the 3D support S(119891) consists of 4 points

Q1

= (minus2 1 2)

Q2

= (0 3 0)

Q3

= (1 1 0)

Q4

= 0

(9)

They are shown in Figure 1

Their convex hull Γ(119891) is a tetrahedron It has 4 verticesQ1ndashQ4 6 edges Γ(1)

119895 and 4 faces Γ(2)

119895 Face Γ(2)

1= [Q1Q2Q3]

is distinguished in Figure 1 its external normal N1

= (2 1 3)

and its truncated equation

(2)

1(119909 119910)

def= minus11991010158401015840

+ 21199103

+ 119909119910 = 0 (10)

Edge Γ(1)1

= [Q1Q2] is also distinguished in Figure 1 its

truncated equation

(1)

1(119909 119910)

def= minus11991010158401015840

+ 21199103

= 0 (11)

Example 3 is finishedLet 119910 = 120593(119909) be a solution to (6) with property (4) and

119901 = 119901120596(120593) 120574 = 120574

120596(120593) then the order of a monomial 119886(119909 119910)

withQ(119886) = (1199021 1199022 1199023) is

1199021

+ 1199022119901 + 1199023

(1 minus 120574) = ⟨119875Q⟩ (12)

where 119875 = (1 119901 1 minus 120574) and ⟨sdot sdot⟩ is the scalar productLeading terms of the sum (1) after substitution 119910 = 120593(119909)

are monomials 119886(119909 119910) for which 120596⟨119875Q⟩ = ⟨120596119875Q⟩

reaches the maximal value on the support S(119891) Here 120596119875 =

(120596 120596119901120596 120596(1minus120574

120596)) and120596(1minus120574

120596) ⩾ 0 according toTheorem 1


On the support S(119891) = Q119894 maximum of the scalar

product ⟨120596119875Q119894⟩ is achieved on a generalized face Γ(119889)

119895of the

polyhedron Γ(119891)By R3 we denote the 3D real space where we put power

exponentsQ and byR3lowastwe denote the space dual (conjugate)

toR3 We will denote points inR3lowastasR = (119903

1 1199032 1199033)Then we

have the scalar product

⟨QR⟩ = 11990211199031

+ 11990211199032

+ 11990231199033 (13)

Each face Γ(119889)119895

corresponds to its normal cone [2]

U(119889)119895

= R ⟨Q1015840R⟩ = ⟨Q10158401015840R⟩ Q1015840Q10158401015840 isin Γ(119889)

119895

⟨Q1015840R⟩ gt ⟨Q101584010158401015840R⟩ Q101584010158401015840 isin Γ Γ(119889)

119895

(14)

Thus normal cone U(2)119895

of the face Γ(2)119895

is a ray spannedon the exterior normal N

119895of the face Γ(2)

119895 normal cone U(1)

119895

of the edge Γ(1)119895

is 2D angle spanned on rays U(2)119896

and U(2)119897

where Γ(1)119895

= Γ(2)

119896cap Γ(2)


119895of the vertex Γ(0)

119895

is a 3D angle spanned on exterior normals N119896of all 2D faces

Γ(2)

119896containing the vertex Γ(0)

119895(see [2])

Thus selection of the truncated sums (119889)

119895(119909 119910) can be

made by the following method First we compute the supportS(119891) of the initial sum 119891(119909 119910) Using support S(119891) wecompute the polyhedron Γ(119891) of sum 119891(119909 119910) that is all itsvertices Γ(0)

119895 edges Γ(1)

119895 and faces Γ(2)

119895 Next we compute their

normal cones U(119889)119895

and select only such truncated equations

(119889)

119895(119909 119910) = 0 for which the intersection U(119889)

119895cap 1199013

⩾ 0 = 0

But truncated equations (119889)

119895(119909 119910) = 0 with 119901

3= 0 can be

studied by algorithms of 2D Power Geometry So 3D PowerGeometry studies truncated equations

(119889)

119895(119909 119910) = 0 with

nonempty intersection U(119889)119895

cap 1199013

gt 0

Example 4 (continuation of Example 3) Polyhedron Γ(119891) forequation 119875

2(8) has 4 following faces with exterior normal

Γ(2)

1= [Q1Q2Q3]

N1

= (2 1 3)

Γ(2)

2= [Q1Q3Q4]

N2

= (2 minus2 3)

Γ(2)

3= [Q1Q2Q4]

N3

= (minus1 0 minus1)

Γ(2)

4= [Q2Q3Q4]

N4

= (0 0 minus1)

(15)

Only two of them N1and N

2 have 119903

3gt 0 Hence all edges

exept Γ(1)6

= [Q2Q4] and all vertices Γ(0)

119895have vectors 119877 =

(1199031 1199032 1199033) with 119903

3gt 0 in their normal cones U(1)

119895and U(0)

119895

4 Power Transformations

If the face Γ(119889)119895

has the normal N119895

= (1 0 1) then the

corresponding truncation (119889)

119895(119909 119910) = 119909

119902119892(119910) where the

differential sum 119892(119910) contains 119910 and its derivatives but doesnot contain 119909 In that case the full sum 119891(119909 119910) can be writtenas 119891(119909 119910) = 119909

119902119892(119910) + 119909

119902minus119903ℎ(119909 119910) where 119903 gt 0 and ℎ(119909 119910) is

a differential sum

Remark 5 If 119910(119909) is a solution to the equation 119892(119910) = 0 withthe property

0 lt 120576 lt1003816100381610038161003816119910 (119909)

1003816100381610038161003816 100381610038161003816100381610038161199101015840(119909)

10038161003816100381610038161003816

10038161003816100381610038161003816119910(119899)

(119909)10038161003816100381610038161003816

lt 120576minus1

(16)

when 119909 rarr 0 or 119909 rarr infin then 119910(119909) can be the asymptoticform of the solutions to the full equation (6) Here 120576 is a smallreal number We call 119910(119909) as formal asymptotic form

Let the power transformation of variables 119909 119910 rarr 119906 V

119910 = 119909120572V

119906 =1

120573119909120573

(17)

transform 119891(119909 119910) into 119891lowast(119906 V) 119891

lowast(119906 V) = 119891(119909 119910)

Theorem 6 Let the face Γ(119889)119894

of Γ(119891) have the exterior normalN119894= (1198991 1198992 1198993) with

1198991

= 0

1198993

gt 0

(18)

then the power transformation (17) with 120572 = 11989921198991 120573 =

11989931198991transforms the truncation

(119889)

119894(119909 119910) of 119891(119909 119910) into the

truncation

lowast(119889)

119894(119906 V) = 119906

119902119892 (V) (19)

of 119891lowast(119906 V) corresponding to the face Γlowast(119889)

119894of Γ(119891lowast) with

the exterior normal Nlowast119894

= (1 0 1) Here lowast(119889)

119894(119906 V) equals

(119889)

119894(119909 119910) after substitution

119906[120572+119897(120573minus1)]120573

119889119897V

119889119906119897(20)

instead of 119910(119897)

= 119889119897119910119889119909119897

So if V = 120593(119906) is a solution to the equation 119892(V) = 0 and|120593(119906)| is bounded from zero and infinity as |119910| in (16) thenthe initial equation 119891(119909 119910) = 0 can have a solution with theasymptotic form

119910 sim 119909120572120593 (

119909120573

120573) 119909

120596997888rarr infin (21)


Herewith the power transformation (17) induces thefollowing formulas for derivatives

1199101015840

= 119909120572+120573minus1V + 120572119909

120572minus1V

11991010158401015840

= 1199092120573+120572minus2V + (2120572 + 120573 minus 1) 119909

120573+120572minus2V

+ 120572 (120572 minus 1) 119909120572minus2V

(22)

where V = 119889V119889119906

Theorem 7 Let an equation of order 119899

119892 (V) +

119898

sum

119895=1

ℎ119895(V) 119906minus119895

= 0 (23)

have a solution of the form

V = 119908 +

infin

sum

119895=1

119887119895(119908) 119906minus119895

(24)

where 119908 = 119908(119906) is the solution to the truncated equation

119892 (119908) = 0 (25)

with the property

0 lt 120576 lt |119908|

10038161003816100381610038161003816100381610038161003816

119889119908

119889119906

10038161003816100381610038161003816100381610038161003816

10038161003816100381610038161003816100381610038161003816

119889119899119908

119889119906119899

10038161003816100381610038161003816100381610038161003816

lt1

120576lt infin (26)

Then 119887119895(119908) satisfies the linear equation

L (119906) 119887119895(119908) + 120579

119895(119908) = 0 (27)

where L(119906) = (120575119892120575V)|V=119908 120579119895(119908) is a polynomial on 119908

(119897)

depending on 119892(119908) and ℎ119894(119908) and 119887

(119897)

119894(119908) for 119894 lt 119895 and

119897 = 0 1 2 119899 120575119892120575V is the first variation

The first variation is the formal Frechet derivative (see[3])

Solution V = 120595(119906) to the transformed equation119891lowast(119906 V) =

0 is expanded into series (24) with integer 119895 only if thetransformed equation 119891

lowast(119906 V) = 0 divided by 119906

119902 has form(23) with integer 119895 In that case solutions V = 119908(119906) to thetruncated equation 119892(V) = 0 are suitable asymptotic formsfor continuation by power expansion (24) and correspondingnormal N

119894is also suitable

External normalN119894= (1198991 1198992 1198993) to 2D face Γ(2)

119894is unique

up to positive scalar factor Hence power transformation(17) of Theorem 6 is unique and we must only check thatthe transformed equation has form (23) with integer 119895 Theexternal normal N = (119899

1 1198992 1198993) to 1D edge Γ(1)

119894belongs

to the normal cone U(1)119894 Hence in the cone U(1)

119894we must

select suitable vectors N with mentioned property of integer119895 Things for a vertex Γ(0)

119895are the same but usually solutions

V = 119908(119906) to corresponding equation 119892(V) = 0 are so simplewhich do not give interesting expansion

Let S(119891) = Q1 Q

119872 S(

(119889)

119895) = Q

1 Q

119871 0 lt 119871 lt

119872 N = (1198991 1198992 1198993) sub U(119889)119894 and 119899

1= 0 1198993

gt 0 Denote

∘

Q119897= Q119871+119897

minus Q1 119897 = 1 119872 minus 119871 (28)

and∘

N = (11989911198993 11989921198993 1)

Theorem 8 The transformed equation (23) has the propertyof integer 119895 if and only if all numbers

minus ⟨∘

N∘

Q119897⟩ 119897 = 1 119872 minus 119871 (29)

are natural

There are 8 essentially different polyhedrons for Painleveequations 119875

1ndash1198755[19] Each of them has exactly one 2D

face in which truncated equation has elliptic solutions Itwas shown [23] that all those elliptic asymptotic forms aresuitable Among 8 polyhedrons only 3 have an edge whichtruncated equation has elliptic solutionsThese are119875

11198752 and

1198754 No truncated equations corresponding to vertices of these

8 polyhedrons have elliptic solutions

Example 9 (continuation of Examples 3 and 4) PolyhedronΓ(119891) of equation 119875

2(8) has edge Γ(1)

1= [Q

1Q2] with

truncated equation 119891(1)

1(119909 119910)

def= minus119910

10158401015840+ 21199103

= 0 Its firstintegral is

11991010158402

= 1199104

+ 1198620

def= 119875 (119910) (30)

where 1198620is arbitrary constant If 119862

0= 0 solutions to (30)

are elliptic functions The same will be true after any powertransformation (17) Let us apply Theorem 8 to the edge Γ(1)

1

The edge Γ(1)1

= Γ(2)

1cap Γ(2)

3 So normal cone U(1)

1is the conic

hull of two normals N1

= (2 1 3) and N3

= (minus1 0 minus1) thatis up to positive scalar factor vectorsN isin U(1)

1have the form

N = 120600N1

+ (1 minus 120600)N3

= (3120600 minus 1 120600 4120600 minus 1) 0 lt 120600 lt 1

(31)

Here 119872 = 4 119871 = 2∘

Q1

= (3 0 minus2)∘

Q2

= minusQ1

= (2 minus1 minus2)and

∘

N = ((3120600 minus 1)(4120600 minus 1) 120600(4120600 minus 1) 1) Conditions ofTheorem 8 are

⟨∘

N∘

Q1⟩ =

3 (3120600 minus 1)

4120600 minus 1minus 2 =

120600 minus 1

4120600 minus 1= minus119896

⟨∘

N∘

Q2⟩ =

2 (3120600 minus 1)

4120600 minus 1minus

120600

4120600 minus 1minus 2 = minus

3120600

4120600 minus 1= minus119897

(32)

where 119896 and 119897 are natural numbers Hence 120600 = (119896 + 1)(4119896 +

1) = 119897(4119897 minus 3) that is 119897 = 119896 + 1 119896 = 1 2 We can write N1015840 = (2 minus 119896 119896 + 1 3) Condition (18) of

Theorem 6 means that 119896 = 2 If 119896 = 1 then 1198991

gt 0 thatis 119909 rarr infin if 119896 gt 2 then 119899

1lt 0 that is 119909 rarr 0


So there is a countable set of suitable normalsN1015840 to edge Γ(1)1

According toTheorem 6 here

120572 =119896 + 1

2 minus 119896

120573 =3

2 minus 119896= 120572 + 1

(33)

5 Computation of Expansions

Below we consider the case when the truncated equation119892(119908) = 0 has the first integral of the form

2

= 119875 (119908)def=

120582

sum

119896=0

119901119896119908119896 119901119896

= const isin C (34)

Differentiating with respect to 119906 and dividing by 2 weobtain

=1

21198751015840(119908) (35)

Here and below the prime denotes the derivative with respectto 119908

Using (34) and (35) any power series 119877 of 119908 and itsderivatives 119889

119897119908119889119906119897 can be written as the sum 119877 = 119877

lowast(119908) +

119877lowastlowast

(119908) where 119877lowast(119908) and 119877

lowastlowast(119908) are power series only of

119908 Let 119887119895(119908) = 119865

119895(119908)+119866

119895(119908) where119865

119895and119866

119895are functions

only of 119908 Then omitting the index 119895 by (34) and (35) weobtain

= 1198651015840 + 119875119866

1015840+

1

21198751015840119866

= 11987511986510158401015840

+1

211987510158401198651015840+ (119875119866

10158401015840+

3

211987510158401198661015840+

1

211987510158401015840

119866)

(36)

Further derivatives of 119887 do not need us here because weconsider only (23) of the second order In our case

L119887 = F (119908) 119865 (119908) + G (119908) 119866 (119908) (37)

Thus (27) splits in two

F (119908) 119865119895(119908) + 120579

lowast

119895(119908) = 0

G (119908) 119866119895(119908) + 120579

lowastlowast

119895(119908) = 0

(38)

where 120579119895(119908) = 120579

lowast

119895(119908)+120579

lowastlowast

119895(119908) Note that in (38) differential

operators F(119908) and G(119908) are operators on 119908 and do notdepend on 119906 If polynomial 119875(119908) in (34) does not havemultiple roots and its degree 120582 is greater than one that is

120582 gt 1 Δ (119875) = 0 (39)

where Δ(119875) is discriminant of the polynomial 119875(119908) thensolution 119908(119906) to the truncated equation (25) is periodic (if120582 = 2) or elliptic (if 120582 = 3 or 4) or hyperelliptic (if 120582 ⩾ 5)function

Near some point 119908 = 1199080 we will compute asymptotic

expansions of fundations 119865119895(119908) and 119866

119895(119908)

119865119895

=

infin

sum

119894=minus119886119895

120593119895119894

120585119894

119866119895

=

infin

sum

119894=minus119887119895

120574119895119894

120585119894

(40)

where 120585 = 119908 minus 1199080 if 119908

0= infin and 120585 = 119908

minus1 if 1199080

= infinIf initial equation (23) is a differential sum then according to[3Theorem 31] coefficients 120593

119895119894and 120574119895119894are either constants or

polynomial of log 120585 that is expansions (40) are either poweror power-logarithmic [3] Moreover according to [3Theorem34] (see proof of Theorem 172 in [4]) power expansions(40) converge for small |120585|

If the solutions 119865119895(119908) and 119866

119895(119908) to the system (38) have

no branching then they are also periodic or (hyper)ellipticfunctions Finally if for the sequence of (38) with 119895 = 1 2 there exist solutions 119865

119895(119908) and 119866

119895(119908) without branching the

solutions to (23) have a regular asymptotic expansion (24)Let operatorsFminus1(119908) andGminus1(119908) be inverse to operators

F(119908) and G(119908) respectively Then the solutions of (27) areof forms

119865119895(119908) = minusF

minus1(119908) 120579lowast

119895(119908)

119866119895(119908) = minusG

minus1(119908) 120579lowastlowast

119895(119908)

(41)

In our case the initial ODE (23) has order two HenceoperatorsF(119908) andG(119908) are of the second order Moreoverin our case factors of 119865

10158401015840 in F and of 11986610158401015840 in G are the same

Denote it as 119877(119908) Singular points 1199080 of operators F and

G are roots of 119877(119908) Indeed 119877(119908) = 119903(119908)119875(119908) where 119903(119908)

is a simple polynomial So roots 1199080 of 119903(119908) and 119908

0= infin

will be singular points of operatorsF and G but roots 1199080 of

polynomial 119875(119908) different from singular points will be theirsubsingular points

Theorem 10 If functions 120579lowast

119895(119908) and 120579

lowastlowast

119895(119908) are regular then

the solutions to (41) can have logarithmic branching only atinfinity 119908 = infin and at singular points of the operators F(119908)

andG(119908) but they can have algebraic branching and can be insingular and subsingular points only

For the existence of a regular expansion (24) we needto prove the existence of a sequence of functions 119865

119895(119908) and

119866119895(119908) that do not have branching From the other side if it

is shown that 119865119895(119908) or 119866

119895(119908) have branching then it proves

the absence of regular expansionIn [19 23] for each polyhedron of the Painleve equations

we selected suitable 2D faces for each of them we wrote (23)operators F(119908) and G(119908) and inverse ones Fminus1(119908) andGminus1(119908) We found their singular points and the conditionson the parameters of the equation and on the solution119908(119906) under which the functions 119865

1(119908) and 119866

1(119908) do not

have logarithmic branching as well as the conditions underwhich at least one of these functions has such branching


It is a wonder that for each Painleve equation 119875119897the operators

F and G are expressed in the same way in terms ofpolynomial 119875(119908) and different cases distinguish only by thispolynomial At the same time for all cases of faces Γ(119889)

119894of five

Painleve equations 1198751ndash1198755 there are only four different pairs

of operatorsF andGSingular point of operators F and G are 119908

0= infin for

1198751ndash1198755and119908

0= 1 for119875

3ndash1198755and119908 = 1 for119875

5 To each suitable

elliptic asymptotic form and to each singular point 1199080 we

assign one basic formal asymptotic expansion (24)Our aim isto show existence or nonexistence of regular basic expansionsby means of calculation of expansions (40) near the singularpoints

6 Expansions for 1198752

Details of calculation of expansions (24) will be explained forequation 119875

2

119891 (119909 119910)def= minus11991010158401015840

+ 21199103

+ 119909119910 + 119886 = 0 (42)


(1)

1(119909 119910)

def= minus11991010158401015840

+ 21199103

= 0 (43)

First according to (33) and Theorem 6 we make powertransformation 119910 = 119909

120572V 119906 = 119909120573120573 (17) using formulas (22)

and obtain equation 1198752(42) in the form (23)

119892 (V) + ℎ1

(V) 119906minus1

+ ℎ2

(V) 119906minus2

+ ℎ119896

(V) 119906minus119896

+ ℎ119896+1

(V) 119906minus119896minus1

= 0

(44)

where

119892 (V) = minusV + 2V3

ℎ1

(V) = minus3120572

120573V

ℎ2

= minus120572 (120572 minus 1)

1205732V

ℎ119896

(V) = 120573minus119896V

ℎ119896+1

(V) = 119886120573minus119896minus1

(45)

119875 (119908) = 1199084

+ 1198620 1198620

= 0 (46)

Here V = 119889V119889119906 and 1198620is arbitrary complex constant

Operators minusFminus1 and minusGminus1 (41) are

119865119895

= 11987512

int1

11987532int 120579lowast

119895119889119908 119889119908

119866119895

= int1

11987532int 11987512

120579lowastlowast

119895119889119908 119889119908

(47)

Here 119903(119908) equiv 1 [23] and singular points of operators (47)are only infinity Let us introduce a function

119867 (119908) = int 119875minus32

119889119908 = const sdot 119908minus5

+ const sdot 119908minus6

+ sdot sdot sdot (48)

Here the integral is determined by mentioned asymptoticexpansion near 119908 = infin Solutions of system (38) or (47) have4 arbitrary constants 119862

1ndash1198624

119865 = 119862111987512

+ 119862211987512

119867 + 1198650

119866 = 1198623

+ 1198624119867 + 119866

0

(49)

where 1198650 and 119866

0 are fixed solutions Here expansions near119908 = infin are

11987512

= const sdot 1199082

+ sdot sdot sdot

11987512

119867 = const sdot 119908minus3

+ sdot sdot sdot

(50)

So we will assume that power expansion for 1198650 does not

contain terms const sdot 1199082 and const sdot 119908

minus3 but expansion for 1198660

does not contain terms const and const sdot 119908minus5 If it is necessarywe can change constants 119862

1ndash1198624 Now the functions 119865

0

119895and

1198660

119895are unique and expansion (24) is called basic if there all

119887119895

= 1198650

119895+ 1198660

119895 Below we compute this basic expansion only

Lemma 11 If 1198621

= 1198624

= 0 then solutions (49) to (47) for 1198752

are regular in subsingular points (if 120579lowast

119895and 120579lowastlowast

119895are also regular

in them)

Let 120579lowast

119895(119908) and 120579

lowastlowast

119895(119908) be power series on decreasing

power exponents of119908 and119860119895119908120590119895 and let 119861

119895119908120591119895 be their terms

with maximal power exponents 120590119895and 120591

119895correspondingly

0 = 119860119895 119861119895

isin C 120590119895 120591119895

isin R 119865119895and 119866

119895contain log119908 if

120590119895

= minus1 or 4

120591119895

= minus3 or 2

(51)

So these numbers are critical for operatorsFminus1 andGminus1We will compute 120579

119895(119908) 120579

lowast

119895 and 120579

lowastlowast

119895as functions of 119887

119894=

119865119894+ 1199081015840119866119894 ℎ119894for 119894 lt 119895 and also will compute leading terms of

119865119895and 119866

119895 that is power exponents 120590

119895and 120591119895and constants

119860119895and 119861

119895

For that we will use following expansions

V = 119908 +1198871

119906+

1198872

1199062+

1198873

1199063+

1198874

1199064+ sdot sdot sdot

V = +1

119906+

2

minus 1198871

1199062+

3

minus 21198872

1199063+

4

minus 31198873


V = +1

119906+

2

minus 21

1199062+

3

minus 42

+ 21198871

1199063

+4

minus 63

+ 61198872


V3 = 1199083

+311990821198871

119906+

31199081198872

1+ 311990821198872

1199062+

311990821198873

+ 611990811988711198872

+ 1198873

1

1199063

+311990821198874

+ 611990811988711198873

+ 31199081198872

2+ 31198872

11198872


(52)


Case 119896 gt 4 According to (45) ℎ1(V) = minus(3120572120573)V hence 120579lowast

1=

0 120579lowastlowast1

= minus3120572120573 According to (46) and (47) we obtain 1198651

= 01198661

= (1205722120573)119908minus2

+ sdot sdot sdot Next

1205792

= 21

+ 61199081198872

1minus

3120572

1205731

minus120572 (120572 minus 1)

1205732119908 (53)

Hence according to (36)

120579lowast

2= (2 minus

3120572

120573) (

1

211987510158401198661

+ 1198751198661015840

1) + 6119908119866

2

1119875 minus

120572 (120572 minus 1)

1205732119908

= minus120572 (120572 + 2)

21205732119908 + sdot sdot sdot

120579lowastlowast

2= 0

(54)

According to (47)1198652

= minus(120572(120572+2)121205732)119908minus1

+sdot sdot sdot 1198662

= 0Next

1205793

= 42

minus 21198871

+ 2 (611990811988711198872

+ 1198873

1)

minus3120572

120573(2

minus 1198871) minus

120572 (120572 minus 1)

12057321198871

(55)

Hence 120579lowast

3= 0 according to (36)

120579lowastlowast

3=

120572 + 4

1205731198651015840

2minus

2 (120572 + 1)2

minus 3120572 (120572 + 1) + 120572 (120572 minus 1)

12057321198661

+ 1211990811986611198652

+ 21198751198663

1= minus

120572 (120572 + 2)

61205732119908minus2

+ sdot sdot sdot

(56)

According to (47) 1198653

= 0 1198663

= (120572(120572 + 2)241205732)119908minus4


1205794

= 63

minus 61198872

+ 2 (31199081198872

2+ 6119908119887

11198873

+ 31198872

11198872)

minus3120572

120573(3


120572 (120572 minus 1)

12057321198872

(57)


120579lowast

4=

3 (120572 + 2)

120573(

1

211987510158401198663

+ 1198751198661015840

3)

+ 1211990811987511986611198663

minus(120572 + 2) (120572 + 3)

12057321198652

+ 61199081198652

2+ 6119875119865

21198662

1= 0119908minus1

+ sdot sdot sdotdef= 1198604119908minus1

+ sdot sdot sdot

(58)

Here power exponent minus1 of leading term in 120579lowast

4is critical

for operator Fminus1 but 1198604

= 0 Hence 1198654has no logarithmic

branchingNow we take into account terms ℎ

119896(V) and ℎ

119896+1(V) from

(45) For 119895 = 4 119896 minus 1 power exponents 120590119895and 120591119895for 119865119895

and 119866119895are small enough to neglect them So

V = 119908 +1198871

119906+

1198872

1199062+

119887119896

119906119896+

119887119896+1

119906119896+1+

119887119896+2

119906119896+2+ sdot sdot sdot (59)

We can write corresponding expansions for V V and V3 Then

120579lowast

119896= 120573minus119896

119908 + sdot sdot sdot

120579lowastlowast

119896= 0 hence

119865119896

= minus1

6120573119896119908minus1

+ sdot sdot sdot

119866119896

= 0

120579119896+1

= (119896 minus 1) 119896

+ 121199081198871119887119896

+119886

120573119896+1+

1198871

120573119896 hence

120579lowast

119896+1=

119886

120573119896+1+ sdot sdot sdot

119865119896+1

= minus119886

4120573119896+1119908minus2

+ sdot sdot sdot

120579lowastlowast

119896+1= (119896 minus 1) 119865

1015840

119896+ 12119908119866

1119865119896

+1

1205731198961198661

= minus1

3120573119896119908minus2

+ sdot sdot sdot

119866119896+1

=1

12120573119896119908minus4

+ sdot sdot sdot

120579119896+2

= 2 (119896 + 1) 119896+1

minus 119896 (119896 + 1) 119887119896

+ 121199081198871119887119896+1

+ 121199081198872119887119896

+ 61198872

1119887119896

minus (119896 + 1) (119896+1

minus 119896119887119896)

minus120572 (120572 + 1)

1205732119887119896

+1

1205731198961198872

(60)

Hence

120579lowast

119896+2= (119896 + 1) (

1

21198751015840119866119896+1

+ 1198751198661015840

119896+1) minus

120572 (120572 minus 1)

1205732119865119896

+1

1205731198961198652

+ 121199081198751198661119866119896+1

+ 121199081198652119865119896

+ 61198751198662

1119865119896

= 0 sdot 119908minus1

+ sdot sdot sdot

120579lowastlowast

119896+2= (119896 + 1) 119865

1015840

119896+1minus

120572 (120572 minus 1)

1205732119866119896

+ 121199081198661119865119896+1

+ 121199081198652119866119896

+ 61198751198662

1119866119896


+ sdot sdot sdot

(61)

It means that 119865119896+2

and 119866119896+2

have no branching at 119908 = infin and120590119895

lt minus1 and 120591119895

lt minus3 for 119896 + 2 lt 119895 lt 2119896So we neglect 119887

119895for 119895 = 119896 + 2 2119896 minus 1 and consider

V = 119908 +1198871

119906+

1198872

1199062+

119887119896

119906119896+

119887119896+1

119906119896+1+

119887119896+2

119906119896+2+

1198872119896

1199062119896+ sdot sdot sdot (62)


We have

1205792119896

= 61199081198872

119896+ sdot sdot sdot (63)

Hence according to results after (59)

120579lowast

2119896= 6119908119865

2

119896+ sdot sdot sdot =

6

361205732119896119908minus1

+ sdot sdot sdot = 1198602119896

119908minus1


where 1198602119896

= 161205732119896

= 0 and 1198652119896

has the logarithmicbranching that is the regular expansion does not exist

For 119896 = 4 we must add 120573minus4

119908 to the computed value of120579lowast

4 but it does not change result on existence of logarithmic

branching in 1198658

Case 119896 = 3 is close to the case 119896 ⩾ 4 and it hasbranching in 119865

6

Case 119896 = 1 was calculated separately It has nobranching

Case 119896 = 0 corresponds to 2D face Γ(2)1 It has no

branching

Thus for equation 1198752(42) basic formal expansions are

regular for two suitable asymptotic forms with 119896 = 0 and119896 = 1 when 119909 rarr infin

Theorem12 For1198752 the regular basic families of formal power-

elliptic expansions exist only for two suitable elliptic asymptoticforms with 119896 = 0 and 119896 = 1 that is when 119909 rarr infin

It is possible to prescribe power exponents 120590119895and 120591

119895

of leading terms in 120579lowast

119895and 120579

lowastlowast

119895 So we can compute such

numbers 119895lowast and 119895

lowastlowast where 120590119895

lt minus1 for 119895 gt 119895lowast and 120591

119895lt minus3

for 119895 gt 119895lowastlowast Here minus1 and minus3 are smaller critical values (51) of

operatorsFminus1 and Gminus1 And it is enough to calculate 119865119895and

119866119895up to 119895 = max(119895

lowast 119895lowastlowast

)

7 Nonbasic Expansions for 1198752

Basic expansions (24) were defined by formulas (47) (49)with 119862

1= 1198622

= 1198623

= 1198624

= 0 According to Lemma 11condition 119862

1= 1198624

= 0 guarantees regularity of 119865119895and 119866

119895in

subsingular points Nowwe want to study cases with nonzero1198623

Example 13 Let us show that 1198623

= 0 in 119866119895gives the

logarithmic branching in 119908 = infin for 119866119895+2

For 119895 = 1

we put 1198623

= 119860 = 0 According to formulas for case 119896 ⩾ 4we obtain

1198651

= 0

1198661

= 119860 +120572

2120573119908minus2

+ sdot sdot sdot

120579lowast

2=

2 minus 120572

2120573(

1

211987510158401198661

+ 1198751198661015840

1) + 6119908119866

2

1119875 minus

120572 (120572 minus 1)


=2 minus 120572

212057321199083119860 + 6119908

5(119860 +

120572

2120573119908minus2

)

2

minus120572 (120572 minus 1)


= 611986021199085

+5120572 + 2

1205731198601199083

+1205722

+ 2120572

21205732119908 + sdot sdot sdot

(65)

Hence

1198652

= 11986021199083

minus5120572 + 2

4120573119860119908 + sdot sdot sdot

1198662

= 1198653

= 0

(66)

Next

120579lowastlowast

3=

120572 + 4

120573(311986021199082

minus5120572 + 2

4120573) minus

2

1205732(119860 +

120572

2120573119908minus2

)

+ 1211990811986611198652

+ 21198751198663

1+ sdot sdot sdot

=120572 + 4

120573311986021199082

minus(120572 + 4) (5120572 + 2)

4120573119860 minus

2

1205732119860

+ 12119908 (119860 +120572

2120573119908minus2

) (11986021199083

minus5120572 + 2

4120573119860119908)

+ 21199084

(119860 +120572

2120573119908minus2

)

3

+ sdot sdot sdot

(67)

Power exponent 2 is critical forGminus1 (see (51)) Coefficient for1199082 in 120579lowastlowast

3is minus(3(120572minus2)120573)119860

2 It is equal to zero only for 120572 = 2but 120572 = (119896 + 1)(2 minus 119896) that is 119896 = 1 But 119896 ⩾ 4 then 119866

3has

logarithmic branching

8 Equation 1198751

Equation 1198751is

119891 (119909 119910)def= minus11991010158401015840

+ 31199102

+ 119909 = 0 (68)

Support S(119891) consists of 3 points Q1

= (minus2 1 2) Q2

=

(0 2 0) and Q3

= (1 0 0) Its polyhedron Γ(119891) is a trianglewith normal N = (4 2 5) So the equation is its owntruncation The edge Γ(1)

1= [Q

1Q2] of the triangle Γ

corresponds to the truncated equation

(1)

1(119909 119910)

def= minus11991010158401015840

+ 31199102

= 0 (69)


which has the first integral

11991010158402

= 2 (1199103

+ 1198620) (70)

with elliptic solutionsSuitable normalsN to the edge Γ(1)

1areN119896

= (4 minus 119896 2(119896 +

1) 5) 119896 = 1 2 and 1198991

= 0 if 119896 = 4 Here 120572 = 2(119896 + 1)(4 minus

119896) 120573 = 5(4 minus 119896) and 120572 = 2(120573 minus 1) 120574 = 2120573 = 120572 + 2 thetransformed equation is

minusV + 3V2 minus5120572

120574V119906minus1 minus

4120572 (120572 minus 1)

1205742V119906minus2 + 2

119896120574minus119896

119906minus119896

= 0 (71)

119875 = 2(1199083

+ 1198620) operators Fminus1 and Gminus1 are again (47) and

119903(119908) equiv 1 [23] Hence there is only one singular point1199080

= infin

and Lemma 11 is true for 1198751 Here 119867(119908) = const sdot 119908

minus72+

sdot sdot sdot and integral critical numbers are 120590119895

= minus1 and 120591119895

= 1Formulas (47)ndash(49) again define basic expansions If 119896 gt 6

then

1198651

= 0

1198661

=120572

120574119908minus1

+ sdot sdot sdot

1198652

=120572 (120572 minus 8)


1198662

= 1198653

= 0

1198663

=120572 (120572 + 4)

31205743119908minus2

+ sdot sdot sdot

1198654

= minus

120572 (120572 + 4) (1205722

+ 24120572 + 48)

601205744119908minus1

+ sdot sdot sdot

1198664

= 1198655

= 0

1198665

=

120572 (120572 + 4) (31205723

+ 561205722

+ 200120572 + 192)

1801205745119908minus3

+ sdot sdot sdot

120579lowast

6= 0 sdot 119908

minus1+ sdot sdot sdot

def= 1198606119908minus1

+ sdot sdot sdot

1198606

= 0

(72)

Hence 1198656has no logarithmic branching if 119896 gt 6

Similarly to the end of Section 6 (see (59)) we obtain

120579lowast

119896=

2119896


119865119896

= minus2119896

5120574119896119908minus1

+ sdot sdot sdot

119866119896

= 119865119896+1

= 0

119866119896+1

=(119896 + 11) 2

119896

75120574119896119908minus3

+ sdot sdot sdot

120579lowast

119896+2= 0 sdot 119908


(73)

q2

q3

1

0

2

2

q1

0

11Q

2Q

3Q

4Q

5Q

6Q

Figure 2 3D support S(119891) and polyhedron Γ(119891) of equation1198753(74)

with all 119886 119887 119888 119889 = 0 The grey face is Γ(2)1 All dotted lines are in

the plane 1199021 1199022 they show projections of Γ(119891) on the plane (119902

1 1199022)

Dashed lines are invisible edges

120590119895

lt minus1 120591119895

lt 1 for 119895 gt 119896+2 and the regular expansion existsIf 4 lt 119896 lt 7 then the regular expansion exists the same istrue for 119896 = 1 2 3 Case 119896 = 0 corresponds to 2D face and toother 119875 = 2(119908

3+ 119908 + 119862

0) but 119860

6= 0 Thus equation 119875

1has

regular basic families of elliptic expansions corresponding toall suitable asymptotic forms Thus we have the following

Theorem 14 To each suitable elliptic asymptotic form of 1198751

there corresponds the basic family of formal power-ellipticexpansions which is regular

9 Equation 1198753

Equation 1198753is

119891 (119909 119910)def= minus119909119910119910

10158401015840+ 11990911991010158402

minus 1199101199101015840+ 1198861199103

+ 119887119910 + 1198881199091199104

+ 119889119909

= 0

(74)

which has 3 different polyhedrons depending on values ofcoefficients 119886 119887 119888 119889 [19 23]

Case 119888119889 = 0 See Figure 2Here only one truncated equation

minus11990911991011991010158401015840

+ 11990911991010158402

+ 1198881199091199104

+ 119889119909 = 0 (75)

corresponding to the distinguished 2D face in Figure 2 haselliptic solutions Here the power transformation (17) isidentical


Equation (74) with 119888119889 = 0 is of the form (23) with 119898 = 1where

119892 (V) def= minusVV + V2 + 119888V4 + 119889

= 0

ℎ1

= minusVV + 119886V3 + 119887V

119875 (119908) = 1198881199084

+ 11986201199082

minus 119889

Δ (119875) =

minus119888119889 (1198622

0+ 4119888119889)

2

16= 0

(76)

Solutions to (38) are of the form

119865119895

= 11987512

int1199082

11987532int

120579lowast

119895

1199083119889119908 119889119908

119866119895

= int1199082

11987532int

11987512

120579lowastlowast

119895

1199083119889119908 119889119908

(77)

Here 119903(119908) = 1199082 [23] so there are 2 singular points 119908

0= infin

and 1199080

= 0 This is true for all cases of 1198753 Near the singular

point 1199080

= infin 119867(119908) = int(119908211987532

)119889119908 = const sdot 119908minus3

+ sdot sdot sdot So11987512

= constsdot1199082+sdot sdot sdot 11987512119867 = constsdot119908minus1+sdot sdot sdot and expansionsfor 1198650

119895and 119866

0

119895do not contain terms const sdot 119908

2 const sdot 119908minus1

and const sdot 1199080 const sdot 119908

minus3 correspondingly Critical numbersfor 120579lowast

119895and 120579

lowastlowast

119895are 2 5 and 0 3 correspondingly Moreover

120579lowast

2= 0 sdot119908

2+ sdot sdot sdot 120579lowastlowast

2= 0 sdot119908+ sdot sdot sdot and 120590

119895lt 2 120591119895

lt 0 for 119895 gt 2So expansion has no logarithmic branching at 119908 = infin

Near the singular point 1199080

= 0 we have 1198670(119908) =

int(119908211987532

)119889119908 = const sdot1199083+119874(1199084) Here we have 4 constants

1198620

1 119862

0

4and basic expansion if all 119862

0

119894= 0 Here Lemma 11

is correct for 1198753

Condition C Condition C is intinfin

0(119908211987532

)119889119908 = 0

Theorem 15 If the Condition C is satisfied then basic expan-sions for 119875

3are regular

Case 119888 = 0 119886119889 = 0 After the power transformation119910 = 11990913V

119906 = (32)11990923 (74) with 119888 = 0 takes the form (23) with 119898 = 1

where

119892 (V) = minusVV + V2 minus 119886V3 + 119889

ℎ1

=3

2119887V minus VV

119875 (119908) = 21198861199083

+ 11986201199082

minus 119889

Δ (119875) = 4119889 (1198623

0minus 27119886

2119889) = 0

(78)

Formula (77) is valid here At119908 = infin 120579lowast119895and 120579lowastlowast

119895have critical

number 2 120579lowast2

= 0 sdot 1199082

+ sdot sdot sdot and orders of 120579lowast

119895 120579lowastlowast119895

are less than2 for 119895 gt 2

The same is at 1199080

= 0 Thus here formal basic expansionis regular Lemma 11 andTheorem 15 are true

Case 119888 = 119889 = 0 119886119887 = 0 After the power transformation119910 = V119906 = 2119909

12 (74) with 119888 = 119889 = 0 takes the form (23) with119898 = 1where

119892 (V) = minusVV + V2 + 119886V3 + 119887V

ℎ1

= minusVV2

119875 (119908) = 2 (1198861199083

+ 11986201199082

minus 119887119908)

Δ (119875) = 241198872

(1198622

0+ 4119886119887) = 0

(79)

At1199080

= infin critical values for 120579lowast


119895are 2 120579lowast

2= 0sdot119908

2+sdot sdot sdot

120590119895 120591119895

lt 2 for 119895 gt 2 So here basic expansion has no branchingThe same is at119908

0= 0 Lemma 11 andTheorem 15 are true

Each of 3 polyhedrons has exactly one 2D face corre-sponding to a truncated equation with elliptic solutions [1819 23] They have different first integrals ()

2= 119875(119908) but

common operators Fminus1 and Gminus1 with singularities in twopoints 119908 = 0 and 119908 = infin

10 Equation 1198754

Equation 1198754is

119891 (119909 119910)def= minus2119910119910

10158401015840+ 11991010158402

+ 31199104

+ 81199091199103

+ 4 (1199092

minus 119886) 1199102

+ 2119887

= 0

(80)

If complex parameters 119886 119887 = 0 its support S(119891) consists of 6points polyhedron Γ(119891) is a tetrahedron and has one 2D faceΓ(2)

1and one edge Γ(1)

1with truncated equations

(2)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

+ 81199091199103

+ 411990921199102

= 0

(1)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

= 0

(81)

having elliptic solutions [19 20 23] Normal to Γ(2)1

is N0

=

(1 1 2) and suitable normals to Γ(1)1areN119896

= (1 minus 119896 119896 + 1 2)119896 = 2 3 After power transformation (17) with 120572 = (119896 +

1)(1 minus 119896) 120573 = 2(1 minus 119896) = 120572 + 1 we obtain (23) with 119898 = 6

minus 2VV + V2 + 3V4 minus4120572

120573VV119906minus1 +

120572 (2 minus 120572)

1205732V2119906minus2

+8

120573119896V3119906minus119896 minus

4119886

120573119896+1V2119906minus(119896+1)

+4

1205732119896V2119906minus2119896 +

2119887

1205732(119896+1)119906minus2(119896+1)

= 0

119875 (119908) = 1199084

+ 1198620119908

1198620

= 0 119896 = 2 3

(82)


Here solutions to (38) are

119865119895

=1

211987512

int119908

11987532int

120579lowast

119895

1199082119889119908 119889119908

119866119895

=1

2int

119908

11987532int

11987512

120579lowastlowast

119895

1199082119889119908 119889119908

(83)

119903(119908) = 119908 [23] so there are two singular points 1199080

= infin

and 1199080

= 0 Near 1199080

= infin 119867 = int(11990811987532

)119889119908 = const sdot

119908minus4

+ sdot sdot sdot Critical numbers for 120579lowast and 120579

lowastlowast are 1 5 and minus1 3

correspondingly If 119896 gt 3 1198651

= 0 1198661

= (1205722120573)119908minus2

+ sdot sdot sdot 1198652

= minus(120572(120572 + 2)121205732) 119908minus1


= 0 1198653

= 0 and120579lowastlowast

3= 0 sdot 119908


Now we compute expansion of the form (59) Then 119865119896

=

minus1120573119896


= 0 119865119896+1

= (21198863120573119896+1

) 119908minus1

+ sdot sdot sdot 119866119896+1

=

(13120573119896+1

)119908minus3

+ sdot sdot sdot 120579lowast

119896+2= (4120572(2120572 minus 1)120573

119896+2) 119908 + sdot sdot sdot and

120579lowastlowast

119896+2= 0 sdot 119908

minus1+ sdot sdot sdot Thus 119860

119896+2= 4120572(2120572 minus 1)120573

119896+2= 0 only if

2120572 minus 1 = 0 that is 119896 = minus13 which is impossible Thus 119865119896+2

has logarithmic branching and the regular basic expansion isabsent The same is true for 119896 = 3 2 and for 119896 = 0 when119875 = 119908

4+ 41199083

+ 41199082

+ 1198620119908

11 Equation 1198755

Equation 1198755is

119891 (119909 119910)def= minus 119909

2119910 (119910 minus 1) 119910

10158401015840+ 1199092 3119910 minus 1

211991010158402

minus 119909119910 (119910 minus 1) 1199101015840

+ (119910 minus 1)3

(1198861199102

+ 119887) + 1198881199091199102

(119910 minus 1)

+ 11988911990921199102

(119910 + 1) = 0

(84)

where 119886 119887 119888 119889 are complex parameters having two differentpolyhedrons depending on values of parameter 119889 [21 23]Each of the polyhedrons has only one 2D face with ellipticsolutions

Case 119889 = 0 Here transformation (17) is identical119910 = V 119909 = 119906So in (23) 119898 = 2

119892 (V) = minusV (V minus 1) V +(3V minus 1) V2

2+ 119889V2 (V + 1)

ℎ1

= minusV (V minus 1) V + 119888V2 (V minus 1)

ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119889119908 [1198620

(119908 minus 1)2

+ 119908]

Δ (119875) = (2119889)4

1198622

0(1 minus 4119862

0) = 0

(85)

Solutions to (38) are

119865119895

= 11987512

int119908 (119908 minus 1)

2

11987532int

120579lowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

119866119895

= int119908 (119908 minus 1)

2

11987532int

11987512

120579lowastlowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

(86)

Here 119903(119908) = 119908(119908 minus 1)2 [23] so singular points are 119908

0=

infin 0 1 Near the singular point 1199080

= infin

119867 = int119908 (119908 minus 1)

2

11987532119889119908 = const sdot 119908

minus12+ sdot sdot sdot (87)

critical numbers for 120579lowast

119895and 120579

lowastlowast

119895are 4 and 3 correspondingly

If 119886 = 0 then 120579lowast

2contains the term minus3119886119908

4 and 1198652has

logarithmic branching If 119886 = 0 then 120590119895

lt 4 and 120591119895

lt 3 forall 119895 gt 0 Thus the basic expansion is regular Similarly basicexpansions are regular near 119908

0= 0 if and only if 119887 = 0 and

near 1199080

= 1 without restrictions

Condition D Condition D is int1

0(119908(119908 minus 1)

211987532

)119889119908 =

intinfin

1(119908(119908 minus 1)

211987532

)119889119908 = 0

Theorem 16 If in equation 1198755with 119889 = 0 and with 119886 = 119887 =

0 Condition D is fulfilled then basic expansions are regular Ifone of these conditions is violated then all basic expansions arenonregular

Case 119889 = 0 119888 = 0 After the change 119910 = V 119906 = 211990912 equation

1198755takes the form (23) with 119898 = 2 where

119892 (V) = minusV (V minus 1) V +3V minus 1

2V2 + 119888V2 (V minus 1)

ℎ1

= minusV (V minus 1) V

ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119888119908 (119908 minus 1) [1198620

(119908 minus 1) + 1]

Δ (119875) = (1198620

minus 1)2

= 0 1198620

= 0

(88)

Formulas (86) are again valid Here basic expansions near1199080

= infin are regular if and only if 119886 = 0 near 1199080

= 0 ifand only if 119887 = 0 and near 119908 = 1 are always nonregular

12 Equation 1198756

In generic case has polyhedron Γ with ten 2D faces Γ(2)119894 but

all external normal to them N = (1198991 1198992 1198993) does not satisfy

conditions (18) 1198991

= 0 1198993

gt 0 Moreover all edges Γ(1)119894

haveno suitable normal The same is true for degenerate cases

13 Summary

Thus all basic expansions are regular for 1198751without addi-

tional restrictions (Theorem 14) for 1198752

if 119909 rarr infin

(Theorem 12) for 1198753under Condition C (Theorem 15) and

for 1198755with 119886 = 119887 = 0 and 119889 = 0 under Condition D

(Theorem 16)As next step it is necessary to study convergence of found

regular formal power-elliptic expansions

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper


References

[1] A D Bruno Local Methods in Nonlinear Differential Equa-tions Springer Berlin Germany 1989 translated from NaukaMoscow Russia 1979 (Russian)

[2] AD BrunoPowerGeometry inAlgebraic andDifferential Equa-tions Elsevier Amsterdam The Netherlands 2000 translatedfrom Fizmatlit Moscow Russia 1998 (Russian)

[3] A D Bruno ldquoAsymptotics and expansions of solutions toan ordinary differential equationrdquo Uspekhi MatematicheskikhNauk vol 59 no 3 pp 31ndash80 2004 translated inRussianMath-ematical Surveys vol 59 no 3 pp 429ndash480 2004 (Russian)

[4] A D Bruno and I V Goryuchkina ldquoAsymptotic expansions ofsolutions of the sixth Painleve equationrdquo Trudy MoskovskogoMatematicheskogo Obshchestva vol 71 pp 6ndash118 2010 trans-lated in Transactions of the Moscow Mathematical Society vol71 pp 1ndash104 2010 (Russian)

[5] A D Bruno and A B Batkhin ldquoAsymptotic solution of analgebraic equationrdquoDoklady Akademii Nauk vol 440 no 3 pp295ndash300 2011 translated inDokladyMathematics vol 84 no 2pp 634ndash639 2011 (Russian)

[6] A B Batkhin A D Bruno and V P Varin ldquoStability setsof multiparameter Hamiltonian systemsrdquo Journal of AppliedMathematics and Mechanics vol 76 no 1 pp 56ndash92 2012translated from Prikladnaya Matematika i Mekhanika vol 76no 1 pp 80ndash133 (Russian)

[7] A D Bruno and A V Parusnikova ldquoLocal expansions ofsolutions to the fifth Painleve equationrdquo Doklady AkademiiNauk vol 438 no 4 pp 439ndash443 2011 translated in DokladyMathematics vol 83 no 3 pp 348ndash352 2011 (Russian)

[8] A D Bruno and A V Parusnikova ldquoExpansions of solutions ofthe fifth Painleve equation in a neighborhood of its nonsingularpointrdquo Doklady Mathematics vol 85 no 1 pp 87ndash92 2012translated from Doklady Akademii Nauk vol 442 no 5 pp583ndash588 2012 (Russian)

[9] A D Bruno ldquoFamilies of periodic solutions to the Beletskyequationrdquo Cosmic Research vol 40 no 3 pp 274ndash295 2002

[10] A D Bruno ldquoAnalysis of the Euler-Poisson equations by themethods of power geometry and the normal formrdquo Journal ofApplied Mathematics and Mechanics vol 71 no 2 pp 168ndash1992007 translated fromPrikladnayaMatematika iMekhanika vol71 no 2 pp 192ndash227 2007 (Russian)

[11] A D Bruno and V P Varin ldquoPeriodic solutions of the restrictedthree-body problem for a small mass ratiordquo PrikladnayaMatematika i Mekhanika vol 71 no 6 pp 1034ndash1066 2007translated in Journal of AppliedMathematics andMechanics vol71 no 6 pp 933ndash960 2007 (Russian)

[12] A D Bruno and V P Varin ldquoOn asteroid distributionrdquoAstronomicheskii Vestnik vol 45 no 4 pp 334ndash340 2011translated in Solar System Research vol 45 no 4 pp 451ndash4572011 (Russian)

[13] A D Bruno and V F Edneral ldquoAlgorithmic analysis of localintegrabilityrdquo Doklady Akademii Nauk vol 424 no 3 pp 299ndash303 2009 translated in Doklady Mathematics vol 79 no 1 pp48ndash52 2009 (Russian)

[14] A D Bruno and T V Shadrina ldquoAn axisymmetric bound-ary layer on a needlerdquo Trudy Moskovskogo MatematicheskogoObshchestva vol 68 pp 224ndash287 2007 translated in Transac-tions of MoscowMathematical Society vol 68 pp 201ndash259 2007(Russian)

[15] A D Bruno ldquoPower geometry in nonlinear partial differentialequationsrdquo Ukrainean Mathematical Bulletin vol 5 no 1 pp32ndash45 2008

[16] A D Bruno ldquoAsymptotic Solving Nonlinear Equations andIdempotent Mathematicsrdquo Preprint of KIAM no 56 pp 31Moscow Russia 2013 httpwwwkeldyshrupapers2013prep2013 56 engpdf

[17] P Boutroux ldquoRecherches sur les transcendantes de M Painleveet lrsquoetude asymptotique des equations differentielles du secondordrerdquo Annales Scientifiques de lrsquoEcole Normale Superieure vol30 no 3 pp 255ndash375 1913 vol 31 pp 99ndash159 1914

[18] A D Bruno and I V Goryuchkina ldquoAsymptotic forms of thesolutions of the third Painleve equationrdquo Doklady AkademiyaNauk vol 422 no 6 pp 729ndash732 2008 translated in DokladyMathematics vol 78 no 2 pp 765ndash768 2008

[19] A D Bruno ldquoSpace power geometry for one ODE and 1198751

minus

1198754 1198756rdquo in Painleve Equations and Related Topics A D Bruno

andA B Batkhin Eds pp 41ndash51 De Gruyter Berlin Germany2012

[20] A D Bruno and I V Goryuchkina ldquoAsymptotic forms of solutions to the fourth Painleve equationrdquo Doklady AkademiiNauk vol 423 no 4 pp 443ndash448 2008 translated in DokladyMathematics vol 78 no 3 868ndash873 2008 (Russion)

[21] A D Bruno and A V Parusnikova ldquoElliptic and periodicasymptotic forms of solutions to P

5rdquo in Painleve Equations and

Related Topics A D Bruno and A B Batkhin Eds pp 53ndash65De Gruyter Berlin Germany 2012

[22] A D Bruno ldquoPower-elliptic expansions of solutions to anODErdquo Computational Mathematics and Mathematical Physicsvol 52 no 12 pp 1650ndash1661 2012 translated from ZurnalVychislitelrsquonoi Matematiki i Matematicheskoi Fiziki vol 51 no12 pp 2206ndash2218 2012 (Russian)

[23] AD Bruno ldquoRegular asymptotic expansions of solutions to oneODE and 119875

1minus 1198755rdquo in Painleve Equations and Related Topics A

D Bruno andA B Batkhin Eds pp 67ndash82 De Gruyter BerlinGermany 2012

[24] I V Goryuchkina ldquoThree-dimensional analysis of asymptoticforms of the solutions to the sixth Painleve equationrdquo Preprintof KIAM no 56 pp 24 Moscow Russia 2010 (Russian)

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


power asymptotic expansion V = 119908(119906) + suminfin

119895=1119887119895119906minus119895 where

119887119895

= 119887119895(119906) are some functions The expansion is regular if all

119887119895are not branching functions of 119908(119906) and its derivatives If

all functions 119887119895(119906) =

119895(119908 ) have no branching then they

are elliptic functions with the same periods as119908(119906) Selectionof such cases is our aim For given 119908(119906) and fixed point1199080 (including infinity) we can compute power-logarithmic

expansions of functions 119895(119908 ) near 119908 = 119908

0 In theseexpansions logarithmic branching can appear only if 119908

0 isa singular point and algebraic branching (of finite order)can be for subsingular points 119908

0 To each singular point 1199080

and suitable asymptotic form 119908(119906) we assign unique regularexpansion V = V(1199080 119908(119906)) so called basic andwe are lookingfor such basic expansions near singular point 119908

0 which haveno branching

We propose algorithms for (1) finding all formal ellipticasymptotic forms (2) finding all suitable elliptic asymptoticforms and (3) calculation of power-logarithmic expansionsof functions

119895(119908 ) near a singular point119908

0 and selection ofbasic expansions without branching All algorithms are basedon 3D Power Geometry

Expansions are formal their convergence is not con-sidered Application of these algorithms to the Painleveequations 119875

1 119875

6gives following

(1) 11987511198752 and119875

4have continuumof 2-parameter families

of elliptic asymptotic forms each 1198753has three 119875

5has

two of them and 1198756does not have

(2) 1198751 1198752 and 119875

4have countable sets of families of

suitable asymptotic forms each and all 5 forms of 1198753

and 1198755are suitable

(3) Basic expansions for all suitable forms have nobranching for 119875

1 for 119875

2if the independent variable

tends to infinity for 1198753if condition C is fulfilled and

for 1198755if condition D is fulfilled and 119886 = 119887 = 0 and

119889 = 0

History of calculation of elliptic expansions of solutionsto the Painleve equations is as follows

A hundred years ago Boutroux [17] found 2 familiesof elliptic asymptotic forms of solutions to the Painleveequations 119875

1and 119875

2 During the last 5 years we found 6

additional families of elliptic asymptotic forms of solutions to1198753(three) [18 19] 119875

4(one) [20] and 119875

5(two) [21] Moreover

the Painleve equations 1198751 1198752 and 119875

4have continuum of

families of elliptic asymptotic forms each and I proposed acriterion for selection suitable asymptotic forms which canbe extended as asymptotic expansions All 8 known ellipticasymptotic forms are suitable Solutions to the equation 119875

6

have no elliptic asymptotic forms at allNear infinity of the independent variable the Painleve

equations 1198751ndash1198755have 12 families of suitable asymptotic forms

and near zero of the independent variable equations 1198751

1198752 and 119875

4have countable sets of such families each Next

I extend these suitable elliptic asymptotic forms 119908(119906) intopower-elliptic expansions V = 119908(119906) + sum

infin

119895=1119887119895119906minus119895 where

coefficients 119887119895are functions of the corresponding elliptic

asymptotic forms and their derivatives To each family ofsuitable elliptic asymptotic forms I put in correspondenceunique basic formal power-elliptic expansion near 119908

0= infin

for 1198751ndash1198755 near 119908

0= 0 for 119875

3ndash1198755 and near 119908

0= 1 for

1198755 Obstacles (logarithmic branching) in calculations of these

basic expansions appeared only for 1198752if the independent

variable tends to zero for 1198754and for 119875

5if |119886| + |119887| = 0 or

119889 = 0Thus near infinity of the independent variable there

are 10 families of regular (ie without branching) ellipticexpansions of solutions to equations 119875

1ndash1198756 4 for 119875

1 2 for 119875

2

3 for 1198753 and 1 for 119875

5 Existence of these expansions for two

Boutroux families of asymptotic forms was proven in [22]and this is all knownup-to-dateNear zero of the independentvariable there is a countable set of families of such expansionsfor 1198751 The results were obtained by means of algorithms of

3D Power Geometry [18ndash24] realized in very cumbersomecalculations

Here I introduce the third variant of 3D Power GeometryThe first was in [18 20 24] and the second was in [19 21ndash23]

In more precise form main results are as in Theorems 1214 15 and 16 and Conditions C and D

Equation 1198756cannot be studied by proposed approach

3 3D Power Geometry

Let 119909 be independent and 119910 be dependent variables 119909 119910 isin CA differential monomial 119886(119909 119910) is a product of an ordinarymonomial 119888119909

11990311199101199032 where 119888 = const isin C (119903

1 1199032) isin R2 and a

finite number of derivatives of the form 119889119897119910119889119909119897 119897 isin N The

sum of differential monomials

119891 (119909 119910) = sum 119886119894(119909 119910) (1)

is called the differential sum Let 119899 be the maximal value of 119897

in 119891(119909 119910)In [2ndash4] it was shown that as 119909 rarr 0 (120596 = minus1) or as

119909 rarr infin (120596 = 1) solutions 119910 = 120593(119909) to the ODE 119891(119909 119910) = 0where 119891(119909 119910) is a differential sum can be found by means ofalgorithms of Plane (2D) Power Geometry if

119901120596

(119889119897120593

119889119909119897) = 119901

120596(120593 (119909)) minus 119897 119897 = 1 119899 (2)

where the order

119901120596

(120593 (119909)) = 120596 lim sup119909120596rarrinfin

log 1003816100381610038161003816120593 (119909)1003816100381610038161003816

120596 log |119909|(3)

on a ray arg 119909 = const and 119899 is the maximal order ofderivatives in119891(119909 119910) Order of the power function 120593(119909) = 119909

120572

with 120572 isin C is 119901120596(119909120572) = Re120572

Here we introduce algorithms which allow to calculatesolutions 119910 = 120593(119909) with the property

119901120596

(119889119897120593

119889119909119897) = 119901

120596(120593 (119909)) minus 119897120574

120596 119897 = 1 119899 (4)

where 120574120596

isin R


Theorem 1 120596 minus 120596120574120596

⩾ 0

For example 1205741

= 0 for 120593 = sin119909 and 120574minus1


= 1 that is 120596 minus 120596120574120596


120596gt 0



119894(119909 119910) we assign its (3D)

power exponent Q(119886119894) = (119902


rules


1199022= order of 119910


3




11198862) = Q(119886

1) + Q(119886

2)






119895


119895


119895corresponds


(119889)

119895(119909 119910) = sum 119886

119894(119909 119910) over Q (119886

119894) isin Γ(119889)

119895cap S (119891) (5)


119891 (119909 119910) = 0 (6)



(119889)

119895(119909 119910) = 0 (7)

Let N119895



1198991

= 0


119891 (119909 119910)def= minus11991010158401015840

+ 21199103

+ 119909119910 + 119886 = 0 (8)


q2q1

q3

1

1

3

2

1Q2Q

3Q

4Q






Q1

= (minus2 1 2)

Q2

= (0 3 0)

Q3

= (1 1 0)

Q4

= 0

(9)



119895 and 4 faces Γ(2)

119895 Face Γ(2)

1= [Q1Q2Q3]


= (2 1 3)


(2)

1(119909 119910)

def= minus11991010158401015840

+ 21199103

+ 119909119910 = 0 (10)

Edge Γ(1)1


truncated equation

(1)

1(119909 119910)

def= minus11991010158401015840

+ 21199103

= 0 (11)


119901 = 119901120596(120593) 120574 = 120574


withQ(119886) = (1199021 1199022 1199023) is

1199021

+ 1199022119901 + 1199023

(1 minus 120574) = ⟨119875Q⟩ (12)




(120596 120596119901120596 120596(1minus120574

120596)) and120596(1minus120574





119895of the




1 1199032 1199033)Then we


⟨QR⟩ = 11990211199031

+ 11990211199032

+ 11990231199033 (13)

Each face Γ(119889)119895


U(119889)119895

= R ⟨Q1015840R⟩ = ⟨Q10158401015840R⟩ Q1015840Q10158401015840 isin Γ(119889)

119895

⟨Q1015840R⟩ gt ⟨Q101584010158401015840R⟩ Q101584010158401015840 isin Γ Γ(119889)

119895

(14)






119895



and U(2)119897

where Γ(1)119895

= Γ(2)

119896cap Γ(2)



119895


Γ(2)


119895(see [2])


119895(119909 119910) can be


119895 edges Γ(1)





(119889)


119895cap 1199013

⩾ 0 = 0


119895(119909 119910) = 0 with 119901

3= 0 can be


(119889)

119895(119909 119910) = 0 with


cap 1199013

gt 0



Γ(2)

1= [Q1Q2Q3]

N1

= (2 1 3)

Γ(2)

2= [Q1Q3Q4]

N2

= (2 minus2 3)

Γ(2)

3= [Q1Q2Q4]

N3

= (minus1 0 minus1)

Γ(2)

4= [Q2Q3Q4]

N4

= (0 0 minus1)

(15)


2 have 119903


exept Γ(1)6



(1199031 1199032 1199033) with 119903


119895and U(0)

119895


If the face Γ(119889)119895


= (1 0 1) then the


119895(119909 119910) = 119909

119902119892(119910) where the


119902119892(119910) + 119909


a differential sum


0 lt 120576 lt1003816100381610038161003816119910 (119909)

1003816100381610038161003816 100381610038161003816100381610038161199101015840(119909)

10038161003816100381610038161003816

10038161003816100381610038161003816119910(119899)

(119909)10038161003816100381610038161003816

lt 120576minus1

(16)



119910 = 119909120572V

119906 =1

120573119909120573

(17)


lowast(119906 V) = 119891(119909 119910)



1198991

= 0

1198993

gt 0

(18)



(119889)

119894(119909 119910) of 119891(119909 119910) into the

truncation

lowast(119889)

119894(119906 V) = 119906

119902119892 (V) (19)




= (1 0 1) Here lowast(119889)

119894(119906 V) equals

(119889)


119906[120572+119897(120573minus1)]120573

119889119897V

119889119906119897(20)

instead of 119910(119897)

= 119889119897119910119889119909119897


119910 sim 119909120572120593 (

119909120573

120573) 119909

120596997888rarr infin (21)



1199101015840

= 119909120572+120573minus1V + 120572119909

120572minus1V

11991010158401015840

= 1199092120573+120572minus2V + (2120572 + 120573 minus 1) 119909

120573+120572minus2V

+ 120572 (120572 minus 1) 119909120572minus2V

(22)

where V = 119889V119889119906


119892 (V) +

119898

sum

119895=1

ℎ119895(V) 119906minus119895

= 0 (23)


V = 119908 +

infin

sum

119895=1

119887119895(119908) 119906minus119895

(24)


119892 (119908) = 0 (25)

with the property

0 lt 120576 lt |119908|

10038161003816100381610038161003816100381610038161003816

119889119908

119889119906

10038161003816100381610038161003816100381610038161003816

10038161003816100381610038161003816100381610038161003816

119889119899119908

119889119906119899

10038161003816100381610038161003816100381610038161003816

lt1

120576lt infin (26)


L (119906) 119887119895(119908) + 120579

119895(119908) = 0 (27)


(119897)


(119897)

119894(119908) for 119894 lt 119895 and









119894is unique


1 1198992 1198993) to 1D edge Γ(1)

119894belongs


119894we must




Let S(119891) = Q1 Q

119872 S(

(119889)

119895) = Q

1 Q

119871 0 lt 119871 lt

119872 N = (1198991 1198992 1198993) sub U(119889)119894 and 119899

1= 0 1198993

gt 0 Denote

∘

Q119897= Q119871+119897

minus Q1 119897 = 1 119872 minus 119871 (28)

and∘

N = (11989911198993 11989921198993 1)


minus ⟨∘

N∘

Q119897⟩ 119897 = 1 119872 minus 119871 (29)

are natural




11198752 and




2(8) has edge Γ(1)

1= [Q

1Q2] with


1(119909 119910)

def= minus119910

10158401015840+ 21199103


11991010158402

= 1199104

+ 1198620

def= 119875 (119910) (30)




1

The edge Γ(1)1

= Γ(2)

1cap Γ(2)


1is the conic


= (2 1 3) and N3


1have the form

N = 120600N1

+ (1 minus 120600)N3

= (3120600 minus 1 120600 4120600 minus 1) 0 lt 120600 lt 1

(31)

Here 119872 = 4 119871 = 2∘

Q1

= (3 0 minus2)∘

Q2

= minusQ1


∘


⟨∘

N∘

Q1⟩ =

3 (3120600 minus 1)


120600 minus 1

4120600 minus 1= minus119896

⟨∘

N∘

Q2⟩ =

2 (3120600 minus 1)


120600


3120600

4120600 minus 1= minus119897

(32)









120572 =119896 + 1

2 minus 119896

120573 =3

2 minus 119896= 120572 + 1

(33)



2

= 119875 (119908)def=

120582

sum

119896=0

119901119896119908119896 119901119896

= const isin C (34)


=1

21198751015840(119908) (35)




lowast(119908) +

119877lowastlowast

(119908) where 119877lowast(119908) and 119877


119908 Let 119887119895(119908) = 119865

119895(119908)+119866

119895(119908) where119865

119895and119866

119895are functions


= 1198651015840 + 119875119866

1015840+

1

21198751015840119866

= 11987511986510158401015840

+1

211987510158401198651015840+ (119875119866

10158401015840+

3

211987510158401198661015840+

1

211987510158401015840

119866)

(36)


L119887 = F (119908) 119865 (119908) + G (119908) 119866 (119908) (37)


F (119908) 119865119895(119908) + 120579

lowast

119895(119908) = 0

G (119908) 119866119895(119908) + 120579

lowastlowast

119895(119908) = 0

(38)

where 120579119895(119908) = 120579

lowast

119895(119908)+120579

lowastlowast



120582 gt 1 Δ (119875) = 0 (39)




119895(119908)

119865119895

=

infin

sum

119894=minus119886119895

120593119895119894

120585119894

119866119895

=

infin

sum

119894=minus119887119895

120574119895119894

120585119894

(40)

where 120585 = 119908 minus 1199080 if 119908

0= infin and 120585 = 119908

minus1 if 1199080







119895(119908) and 119866




119865119895(119908) = minusF


119895(119908)

119866119895(119908) = minusG


119895(119908)

(41)






0= infin




119895(119908) and 120579

lowastlowast





119895(119908) and


is shown that 119865119895(119908) or 119866




1(119908) and 119866

1(119908) do not





119894of five



0= infin for

1198751ndash1198755and119908

0= 1 for119875

3ndash1198755and119908 = 1 for119875

5 To each suitable





2

119891 (119909 119910)def= minus11991010158401015840

+ 21199103

+ 119909119910 + 119886 = 0 (42)


(1)

1(119909 119910)

def= minus11991010158401015840

+ 21199103

= 0 (43)


120572V 119906 = 119909120573120573 (17) using formulas (22)


119892 (V) + ℎ1

(V) 119906minus1

+ ℎ2

(V) 119906minus2

+ ℎ119896

(V) 119906minus119896

+ ℎ119896+1


= 0

(44)

where

119892 (V) = minusV + 2V3

ℎ1

(V) = minus3120572

120573V

ℎ2

= minus120572 (120572 minus 1)

1205732V

ℎ119896

(V) = 120573minus119896V

ℎ119896+1

(V) = 119886120573minus119896minus1

(45)

119875 (119908) = 1199084

+ 1198620 1198620

= 0 (46)



119865119895

= 11987512

int1

11987532int 120579lowast

119895119889119908 119889119908

119866119895

= int1

11987532int 11987512

120579lowastlowast

119895119889119908 119889119908

(47)


119867 (119908) = int 119875minus32

119889119908 = const sdot 119908minus5




1ndash1198624

119865 = 119862111987512

+ 119862211987512

119867 + 1198650

119866 = 1198623

+ 1198624119867 + 119866

0

(49)

where 1198650 and 119866


11987512


+ sdot sdot sdot

11987512


+ sdot sdot sdot

(50)






0

119895and

1198660


119887119895

= 1198650

119895+ 1198660


Lemma 11 If 1198621

= 1198624





in them)

Let 120579lowast

119895(119908) and 120579

lowastlowast



119895119908120591119895 be their terms



0 = 119860119895 119861119895

isin C 120590119895 120591119895

isin R 119865119895and 119866


120590119895

= minus1 or 4

120591119895

= minus3 or 2

(51)


119895(119908) 120579

lowast

119895 and 120579

lowastlowast


119894=


119865119895and 119866



119860119895and 119861

119895


V = 119908 +1198871

119906+

1198872

1199062+

1198873

1199063+

1198874


V = +1

119906+

2

minus 1198871

1199062+

3

minus 21198872

1199063+

4

minus 31198873


V = +1

119906+

2

minus 21

1199062+

3

minus 42

+ 21198871

1199063

+4

minus 63

+ 61198872


V3 = 1199083

+311990821198871

119906+

31199081198872

1+ 311990821198872

1199062+

311990821198873

+ 611990811988711198872

+ 1198873

1

1199063

+311990821198874

+ 611990811988711198873

+ 31199081198872

2+ 31198872

11198872


(52)



1=



= 01198661

= (1205722120573)119908minus2


1205792

= 21

+ 61199081198872

1minus

3120572

1205731

minus120572 (120572 minus 1)

1205732119908 (53)


120579lowast

2= (2 minus

3120572

120573) (

1

211987510158401198661

+ 1198751198661015840

1) + 6119908119866

2

1119875 minus

120572 (120572 minus 1)

1205732119908

= minus120572 (120572 + 2)

21205732119908 + sdot sdot sdot

120579lowastlowast

2= 0

(54)


= minus(120572(120572+2)121205732)119908minus1


= 0Next

1205793

= 42

minus 21198871

+ 2 (611990811988711198872

+ 1198873

1)

minus3120572

120573(2


120572 (120572 minus 1)

12057321198871

(55)

Hence 120579lowast


120579lowastlowast

3=

120572 + 4

1205731198651015840

2minus

2 (120572 + 1)2

minus 3120572 (120572 + 1) + 120572 (120572 minus 1)

12057321198661

+ 1211990811986611198652

+ 21198751198663

1= minus

120572 (120572 + 2)

61205732119908minus2

+ sdot sdot sdot

(56)


= 0 1198663

= (120572(120572 + 2)241205732)119908minus4


1205794

= 63

minus 61198872

+ 2 (31199081198872

2+ 6119908119887

11198873

+ 31198872

11198872)

minus3120572

120573(3


120572 (120572 minus 1)

12057321198872

(57)


120579lowast

4=

3 (120572 + 2)

120573(

1

211987510158401198663

+ 1198751198661015840

3)

+ 1211990811987511986611198663

minus(120572 + 2) (120572 + 3)

12057321198652

+ 61199081198652

2+ 6119875119865

21198662

1= 0119908minus1


+ sdot sdot sdot

(58)


4is critical




119896(V) and ℎ

119896+1(V) from



V = 119908 +1198871

119906+

1198872

1199062+

119887119896

119906119896+

119887119896+1

119906119896+1+

119887119896+2

119906119896+2+ sdot sdot sdot (59)


120579lowast

119896= 120573minus119896


120579lowastlowast

119896= 0 hence

119865119896

= minus1

6120573119896119908minus1

+ sdot sdot sdot

119866119896

= 0

120579119896+1

= (119896 minus 1) 119896

+ 121199081198871119887119896

+119886

120573119896+1+

1198871

120573119896 hence

120579lowast

119896+1=

119886


119865119896+1

= minus119886

4120573119896+1119908minus2

+ sdot sdot sdot

120579lowastlowast

119896+1= (119896 minus 1) 119865

1015840

119896+ 12119908119866

1119865119896

+1

1205731198961198661

= minus1

3120573119896119908minus2

+ sdot sdot sdot

119866119896+1

=1

12120573119896119908minus4

+ sdot sdot sdot

120579119896+2

= 2 (119896 + 1) 119896+1

minus 119896 (119896 + 1) 119887119896

+ 121199081198871119887119896+1

+ 121199081198872119887119896

+ 61198872

1119887119896

minus (119896 + 1) (119896+1

minus 119896119887119896)

minus120572 (120572 + 1)

1205732119887119896

+1

1205731198961198872

(60)

Hence

120579lowast

119896+2= (119896 + 1) (

1

21198751015840119866119896+1

+ 1198751198661015840

119896+1) minus

120572 (120572 minus 1)

1205732119865119896

+1

1205731198961198652

+ 121199081198751198661119866119896+1

+ 121199081198652119865119896

+ 61198751198662

1119865119896


+ sdot sdot sdot

120579lowastlowast

119896+2= (119896 + 1) 119865

1015840

119896+1minus

120572 (120572 minus 1)

1205732119866119896

+ 121199081198661119865119896+1

+ 121199081198652119866119896

+ 61198751198662

1119866119896


+ sdot sdot sdot

(61)

It means that 119865119896+2

and 119866119896+2


lt minus1 and 120591119895



V = 119908 +1198871

119906+

1198872

1199062+

119887119896

119906119896+

119887119896+1

119906119896+1+

119887119896+2

119906119896+2+

1198872119896

1199062119896+ sdot sdot sdot (62)


We have

1205792119896

= 61199081198872



120579lowast

2119896= 6119908119865

2


6

361205732119896119908minus1

+ sdot sdot sdot = 1198602119896

119908minus1


where 1198602119896

= 161205732119896

= 0 and 1198652119896







6



branching






119895


119895and 120579

lowastlowast





119895lt minus3



119866119895up to 119895 = max(119895


)



1= 1198622

= 1198623

= 1198624


1= 1198624


119895in



= 0 in 119866119895gives the


For 119895 = 1

we put 1198623


1198651

= 0

1198661

= 119860 +120572

2120573119908minus2

+ sdot sdot sdot

120579lowast

2=

2 minus 120572

2120573(

1

211987510158401198661

+ 1198751198661015840

1) + 6119908119866

2

1119875 minus

120572 (120572 minus 1)


=2 minus 120572

212057321199083119860 + 6119908

5(119860 +

120572

2120573119908minus2

)

2

minus120572 (120572 minus 1)


= 611986021199085

+5120572 + 2

1205731198601199083

+1205722

+ 2120572

21205732119908 + sdot sdot sdot

(65)

Hence

1198652

= 11986021199083

minus5120572 + 2

4120573119860119908 + sdot sdot sdot

1198662

= 1198653

= 0

(66)

Next

120579lowastlowast

3=

120572 + 4

120573(311986021199082

minus5120572 + 2

4120573) minus

2

1205732(119860 +

120572

2120573119908minus2

)

+ 1211990811986611198652

+ 21198751198663

1+ sdot sdot sdot

=120572 + 4

120573311986021199082

minus(120572 + 4) (5120572 + 2)

4120573119860 minus

2

1205732119860

+ 12119908 (119860 +120572

2120573119908minus2

) (11986021199083

minus5120572 + 2

4120573119860119908)

+ 21199084

(119860 +120572

2120573119908minus2

)

3

+ sdot sdot sdot

(67)


3is minus(3(120572minus2)120573)119860


3has


8 Equation 1198751

Equation 1198751is

119891 (119909 119910)def= minus11991010158401015840

+ 31199102

+ 119909 = 0 (68)


= (minus2 1 2) Q2

=

(0 2 0) and Q3


1= [Q



(1)

1(119909 119910)

def= minus11991010158401015840

+ 31199102

= 0 (69)



11991010158402

= 2 (1199103

+ 1198620) (70)


1areN119896

= (4 minus 119896 2(119896 +

1) 5) 119896 = 1 2 and 1198991

= 0 if 119896 = 4 Here 120572 = 2(119896 + 1)(4 minus




4120572 (120572 minus 1)

1205742V119906minus2 + 2

119896120574minus119896

119906minus119896

= 0 (71)

119875 = 2(1199083



= infin


minus72+


= minus1 and 120591119895


then

1198651

= 0

1198661

=120572

120574119908minus1

+ sdot sdot sdot

1198652

=120572 (120572 minus 8)


1198662

= 1198653

= 0

1198663

=120572 (120572 + 4)

31205743119908minus2

+ sdot sdot sdot

1198654

= minus

120572 (120572 + 4) (1205722

+ 24120572 + 48)

601205744119908minus1

+ sdot sdot sdot

1198664

= 1198655

= 0

1198665

=

120572 (120572 + 4) (31205723

+ 561205722

+ 200120572 + 192)

1801205745119908minus3

+ sdot sdot sdot

120579lowast

6= 0 sdot 119908


def= 1198606119908minus1

+ sdot sdot sdot

1198606

= 0

(72)



120579lowast

119896=

2119896


119865119896

= minus2119896

5120574119896119908minus1

+ sdot sdot sdot

119866119896

= 119865119896+1

= 0

119866119896+1

=(119896 + 11) 2

119896

75120574119896119908minus3

+ sdot sdot sdot

120579lowast

119896+2= 0 sdot 119908


(73)

q2

q3

1

0

2

2

q1

0

11Q

2Q

3Q

4Q

5Q

6Q




1 1199022)


120590119895

lt minus1 120591119895


3+ 119908 + 119862

0) but 119860


1has




9 Equation 1198753

Equation 1198753is

119891 (119909 119910)def= minus119909119910119910

10158401015840+ 11990911991010158402

minus 1199101199101015840+ 1198861199103

+ 119887119910 + 1198881199091199104

+ 119889119909

= 0

(74)



minus11990911991011991010158401015840

+ 11990911991010158402

+ 1198881199091199104

+ 119889119909 = 0 (75)




119892 (V) def= minusVV + V2 + 119888V4 + 119889

= 0

ℎ1

= minusVV + 119886V3 + 119887V

119875 (119908) = 1198881199084

+ 11986201199082

minus 119889

Δ (119875) =

minus119888119889 (1198622

0+ 4119888119889)

2

16= 0

(76)


119865119895

= 11987512

int1199082

11987532int

120579lowast

119895

1199083119889119908 119889119908

119866119895

= int1199082

11987532int

11987512

120579lowastlowast

119895

1199083119889119908 119889119908

(77)


0= infin

and 1199080


point 1199080

= infin 119867(119908) = int(119908211987532

)119889119908 = const sdot 119908minus3



119895and 119866

0





119895and 120579

lowastlowast


120579lowast

2= 0 sdot119908



119895lt 2 120591119895



= 0 we have 1198670(119908) =

int(119908211987532


1198620

1 119862

0


0

119894= 0 Here Lemma 11



0(119908211987532

)119889119908 = 0


3are regular



where


ℎ1

=3

2119887V minus VV

119875 (119908) = 21198861199083

+ 11986201199082

minus 119889

Δ (119875) = 4119889 (1198623

0minus 27119886

2119889) = 0

(78)


119895have critical


= 0 sdot 1199082


119895 120579lowastlowast119895






119892 (V) = minusVV + V2 + 119886V3 + 119887V

ℎ1

= minusVV2

119875 (119908) = 2 (1198861199083

+ 11986201199082

minus 119887119908)

Δ (119875) = 241198872

(1198622

0+ 4119886119887) = 0

(79)

At1199080



119895are 2 120579lowast

2= 0sdot119908

2+sdot sdot sdot

120590119895 120591119895




2= 119875(119908) but


10 Equation 1198754

Equation 1198754is

119891 (119909 119910)def= minus2119910119910

10158401015840+ 11991010158402

+ 31199104

+ 81199091199103

+ 4 (1199092

minus 119886) 1199102

+ 2119887

= 0

(80)


1and one edge Γ(1)


(2)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

+ 81199091199103

+ 411990921199102

= 0

(1)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

= 0

(81)


is N0

=





120573VV119906minus1 +

120572 (2 minus 120572)

1205732V2119906minus2

+8

120573119896V3119906minus119896 minus

4119886

120573119896+1V2119906minus(119896+1)

+4

1205732119896V2119906minus2119896 +

2119887

1205732(119896+1)119906minus2(119896+1)

= 0

119875 (119908) = 1199084

+ 1198620119908

1198620

= 0 119896 = 2 3

(82)



119865119895

=1

211987512

int119908

11987532int

120579lowast

119895

1199082119889119908 119889119908

119866119895

=1

2int

119908

11987532int

11987512

120579lowastlowast

119895

1199082119889119908 119889119908

(83)


= infin

and 1199080

= 0 Near 1199080

= infin 119867 = int(11990811987532

)119889119908 = const sdot

119908minus4




= 0 1198661

= (1205722120573)119908minus2


= minus(120572(120572 + 2)121205732) 119908minus1


= 0 1198653


3= 0 sdot 119908



=

minus1120573119896


= 0 119865119896+1

= (21198863120573119896+1

) 119908minus1

+ sdot sdot sdot 119866119896+1

=

(13120573119896+1

)119908minus3


119896+2= (4120572(2120572 minus 1)120573


120579lowastlowast

119896+2= 0 sdot 119908


119896+2= 4120572(2120572 minus 1)120573

119896+2= 0 only if



4+ 41199083

+ 41199082

+ 1198620119908

11 Equation 1198755

Equation 1198755is

119891 (119909 119910)def= minus 119909

2119910 (119910 minus 1) 119910

10158401015840+ 1199092 3119910 minus 1

211991010158402

minus 119909119910 (119910 minus 1) 1199101015840

+ (119910 minus 1)3

(1198861199102

+ 119887) + 1198881199091199102

(119910 minus 1)

+ 11988911990921199102

(119910 + 1) = 0

(84)




2+ 119889V2 (V + 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119889119908 [1198620

(119908 minus 1)2

+ 119908]

Δ (119875) = (2119889)4

1198622

0(1 minus 4119862

0) = 0

(85)


119865119895

= 11987512

int119908 (119908 minus 1)

2

11987532int

120579lowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

119866119895

= int119908 (119908 minus 1)

2

11987532int

11987512

120579lowastlowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

(86)


0=


= infin

119867 = int119908 (119908 minus 1)

2

11987532119889119908 = const sdot 119908



119895and 120579

lowastlowast


If 119886 = 0 then 120579lowast


4 and 1198652has


lt 4 and 120591119895



near 1199080



0(119908(119908 minus 1)

211987532

)119889119908 =

intinfin

1(119908(119908 minus 1)

211987532

)119889119908 = 0






2V2 + 119888V2 (V minus 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119888119908 (119908 minus 1) [1198620

(119908 minus 1) + 1]

Δ (119875) = (1198620

minus 1)2

= 0 1198620

= 0

(88)




12 Equation 1198756




= 0 1198993



13 Summary











References




















minus



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Theorem 1 120596 minus 120596120574120596

⩾ 0

For example 1205741

= 0 for 120593 = sin119909 and 120574minus1


= 1 that is 120596 minus 120596120574120596


120596gt 0



119894(119909 119910) we assign its (3D)

power exponent Q(119886119894) = (119902


rules


1199022= order of 119910


3




11198862) = Q(119886

1) + Q(119886

2)






119895


119895


119895corresponds


(119889)

119895(119909 119910) = sum 119886

119894(119909 119910) over Q (119886

119894) isin Γ(119889)

119895cap S (119891) (5)


119891 (119909 119910) = 0 (6)



(119889)

119895(119909 119910) = 0 (7)

Let N119895



1198991

= 0


119891 (119909 119910)def= minus11991010158401015840

+ 21199103

+ 119909119910 + 119886 = 0 (8)


q2q1

q3

1

1

3

2

1Q2Q

3Q

4Q






Q1

= (minus2 1 2)

Q2

= (0 3 0)

Q3

= (1 1 0)

Q4

= 0

(9)



119895 and 4 faces Γ(2)

119895 Face Γ(2)

1= [Q1Q2Q3]


= (2 1 3)


(2)

1(119909 119910)

def= minus11991010158401015840

+ 21199103

+ 119909119910 = 0 (10)

Edge Γ(1)1


truncated equation

(1)

1(119909 119910)

def= minus11991010158401015840

+ 21199103

= 0 (11)


119901 = 119901120596(120593) 120574 = 120574


withQ(119886) = (1199021 1199022 1199023) is

1199021

+ 1199022119901 + 1199023

(1 minus 120574) = ⟨119875Q⟩ (12)




(120596 120596119901120596 120596(1minus120574

120596)) and120596(1minus120574





119895of the




1 1199032 1199033)Then we


⟨QR⟩ = 11990211199031

+ 11990211199032

+ 11990231199033 (13)

Each face Γ(119889)119895


U(119889)119895

= R ⟨Q1015840R⟩ = ⟨Q10158401015840R⟩ Q1015840Q10158401015840 isin Γ(119889)

119895

⟨Q1015840R⟩ gt ⟨Q101584010158401015840R⟩ Q101584010158401015840 isin Γ Γ(119889)

119895

(14)






119895



and U(2)119897

where Γ(1)119895

= Γ(2)

119896cap Γ(2)



119895


Γ(2)


119895(see [2])


119895(119909 119910) can be


119895 edges Γ(1)





(119889)


119895cap 1199013

⩾ 0 = 0


119895(119909 119910) = 0 with 119901

3= 0 can be


(119889)

119895(119909 119910) = 0 with


cap 1199013

gt 0



Γ(2)

1= [Q1Q2Q3]

N1

= (2 1 3)

Γ(2)

2= [Q1Q3Q4]

N2

= (2 minus2 3)

Γ(2)

3= [Q1Q2Q4]

N3

= (minus1 0 minus1)

Γ(2)

4= [Q2Q3Q4]

N4

= (0 0 minus1)

(15)


2 have 119903


exept Γ(1)6



(1199031 1199032 1199033) with 119903


119895and U(0)

119895


If the face Γ(119889)119895


= (1 0 1) then the


119895(119909 119910) = 119909

119902119892(119910) where the


119902119892(119910) + 119909


a differential sum


0 lt 120576 lt1003816100381610038161003816119910 (119909)

1003816100381610038161003816 100381610038161003816100381610038161199101015840(119909)

10038161003816100381610038161003816

10038161003816100381610038161003816119910(119899)

(119909)10038161003816100381610038161003816

lt 120576minus1

(16)



119910 = 119909120572V

119906 =1

120573119909120573

(17)


lowast(119906 V) = 119891(119909 119910)



1198991

= 0

1198993

gt 0

(18)



(119889)

119894(119909 119910) of 119891(119909 119910) into the

truncation

lowast(119889)

119894(119906 V) = 119906

119902119892 (V) (19)




= (1 0 1) Here lowast(119889)

119894(119906 V) equals

(119889)


119906[120572+119897(120573minus1)]120573

119889119897V

119889119906119897(20)

instead of 119910(119897)

= 119889119897119910119889119909119897


119910 sim 119909120572120593 (

119909120573

120573) 119909

120596997888rarr infin (21)



1199101015840

= 119909120572+120573minus1V + 120572119909

120572minus1V

11991010158401015840

= 1199092120573+120572minus2V + (2120572 + 120573 minus 1) 119909

120573+120572minus2V

+ 120572 (120572 minus 1) 119909120572minus2V

(22)

where V = 119889V119889119906


119892 (V) +

119898

sum

119895=1

ℎ119895(V) 119906minus119895

= 0 (23)


V = 119908 +

infin

sum

119895=1

119887119895(119908) 119906minus119895

(24)


119892 (119908) = 0 (25)

with the property

0 lt 120576 lt |119908|

10038161003816100381610038161003816100381610038161003816

119889119908

119889119906

10038161003816100381610038161003816100381610038161003816

10038161003816100381610038161003816100381610038161003816

119889119899119908

119889119906119899

10038161003816100381610038161003816100381610038161003816

lt1

120576lt infin (26)


L (119906) 119887119895(119908) + 120579

119895(119908) = 0 (27)


(119897)


(119897)

119894(119908) for 119894 lt 119895 and









119894is unique


1 1198992 1198993) to 1D edge Γ(1)

119894belongs


119894we must




Let S(119891) = Q1 Q

119872 S(

(119889)

119895) = Q

1 Q

119871 0 lt 119871 lt

119872 N = (1198991 1198992 1198993) sub U(119889)119894 and 119899

1= 0 1198993

gt 0 Denote

∘

Q119897= Q119871+119897

minus Q1 119897 = 1 119872 minus 119871 (28)

and∘

N = (11989911198993 11989921198993 1)


minus ⟨∘

N∘

Q119897⟩ 119897 = 1 119872 minus 119871 (29)

are natural




11198752 and




2(8) has edge Γ(1)

1= [Q

1Q2] with


1(119909 119910)

def= minus119910

10158401015840+ 21199103


11991010158402

= 1199104

+ 1198620

def= 119875 (119910) (30)




1

The edge Γ(1)1

= Γ(2)

1cap Γ(2)


1is the conic


= (2 1 3) and N3


1have the form

N = 120600N1

+ (1 minus 120600)N3

= (3120600 minus 1 120600 4120600 minus 1) 0 lt 120600 lt 1

(31)

Here 119872 = 4 119871 = 2∘

Q1

= (3 0 minus2)∘

Q2

= minusQ1


∘


⟨∘

N∘

Q1⟩ =

3 (3120600 minus 1)


120600 minus 1

4120600 minus 1= minus119896

⟨∘

N∘

Q2⟩ =

2 (3120600 minus 1)


120600


3120600

4120600 minus 1= minus119897

(32)









120572 =119896 + 1

2 minus 119896

120573 =3

2 minus 119896= 120572 + 1

(33)



2

= 119875 (119908)def=

120582

sum

119896=0

119901119896119908119896 119901119896

= const isin C (34)


=1

21198751015840(119908) (35)




lowast(119908) +

119877lowastlowast

(119908) where 119877lowast(119908) and 119877


119908 Let 119887119895(119908) = 119865

119895(119908)+119866

119895(119908) where119865

119895and119866

119895are functions


= 1198651015840 + 119875119866

1015840+

1

21198751015840119866

= 11987511986510158401015840

+1

211987510158401198651015840+ (119875119866

10158401015840+

3

211987510158401198661015840+

1

211987510158401015840

119866)

(36)


L119887 = F (119908) 119865 (119908) + G (119908) 119866 (119908) (37)


F (119908) 119865119895(119908) + 120579

lowast

119895(119908) = 0

G (119908) 119866119895(119908) + 120579

lowastlowast

119895(119908) = 0

(38)

where 120579119895(119908) = 120579

lowast

119895(119908)+120579

lowastlowast



120582 gt 1 Δ (119875) = 0 (39)




119895(119908)

119865119895

=

infin

sum

119894=minus119886119895

120593119895119894

120585119894

119866119895

=

infin

sum

119894=minus119887119895

120574119895119894

120585119894

(40)

where 120585 = 119908 minus 1199080 if 119908

0= infin and 120585 = 119908

minus1 if 1199080







119895(119908) and 119866




119865119895(119908) = minusF


119895(119908)

119866119895(119908) = minusG


119895(119908)

(41)






0= infin




119895(119908) and 120579

lowastlowast





119895(119908) and


is shown that 119865119895(119908) or 119866




1(119908) and 119866

1(119908) do not





119894of five



0= infin for

1198751ndash1198755and119908

0= 1 for119875

3ndash1198755and119908 = 1 for119875

5 To each suitable





2

119891 (119909 119910)def= minus11991010158401015840

+ 21199103

+ 119909119910 + 119886 = 0 (42)


(1)

1(119909 119910)

def= minus11991010158401015840

+ 21199103

= 0 (43)


120572V 119906 = 119909120573120573 (17) using formulas (22)


119892 (V) + ℎ1

(V) 119906minus1

+ ℎ2

(V) 119906minus2

+ ℎ119896

(V) 119906minus119896

+ ℎ119896+1


= 0

(44)

where

119892 (V) = minusV + 2V3

ℎ1

(V) = minus3120572

120573V

ℎ2

= minus120572 (120572 minus 1)

1205732V

ℎ119896

(V) = 120573minus119896V

ℎ119896+1

(V) = 119886120573minus119896minus1

(45)

119875 (119908) = 1199084

+ 1198620 1198620

= 0 (46)



119865119895

= 11987512

int1

11987532int 120579lowast

119895119889119908 119889119908

119866119895

= int1

11987532int 11987512

120579lowastlowast

119895119889119908 119889119908

(47)


119867 (119908) = int 119875minus32

119889119908 = const sdot 119908minus5




1ndash1198624

119865 = 119862111987512

+ 119862211987512

119867 + 1198650

119866 = 1198623

+ 1198624119867 + 119866

0

(49)

where 1198650 and 119866


11987512


+ sdot sdot sdot

11987512


+ sdot sdot sdot

(50)






0

119895and

1198660


119887119895

= 1198650

119895+ 1198660


Lemma 11 If 1198621

= 1198624





in them)

Let 120579lowast

119895(119908) and 120579

lowastlowast



119895119908120591119895 be their terms



0 = 119860119895 119861119895

isin C 120590119895 120591119895

isin R 119865119895and 119866


120590119895

= minus1 or 4

120591119895

= minus3 or 2

(51)


119895(119908) 120579

lowast

119895 and 120579

lowastlowast


119894=


119865119895and 119866



119860119895and 119861

119895


V = 119908 +1198871

119906+

1198872

1199062+

1198873

1199063+

1198874


V = +1

119906+

2

minus 1198871

1199062+

3

minus 21198872

1199063+

4

minus 31198873


V = +1

119906+

2

minus 21

1199062+

3

minus 42

+ 21198871

1199063

+4

minus 63

+ 61198872


V3 = 1199083

+311990821198871

119906+

31199081198872

1+ 311990821198872

1199062+

311990821198873

+ 611990811988711198872

+ 1198873

1

1199063

+311990821198874

+ 611990811988711198873

+ 31199081198872

2+ 31198872

11198872


(52)



1=



= 01198661

= (1205722120573)119908minus2


1205792

= 21

+ 61199081198872

1minus

3120572

1205731

minus120572 (120572 minus 1)

1205732119908 (53)


120579lowast

2= (2 minus

3120572

120573) (

1

211987510158401198661

+ 1198751198661015840

1) + 6119908119866

2

1119875 minus

120572 (120572 minus 1)

1205732119908

= minus120572 (120572 + 2)

21205732119908 + sdot sdot sdot

120579lowastlowast

2= 0

(54)


= minus(120572(120572+2)121205732)119908minus1


= 0Next

1205793

= 42

minus 21198871

+ 2 (611990811988711198872

+ 1198873

1)

minus3120572

120573(2


120572 (120572 minus 1)

12057321198871

(55)

Hence 120579lowast


120579lowastlowast

3=

120572 + 4

1205731198651015840

2minus

2 (120572 + 1)2

minus 3120572 (120572 + 1) + 120572 (120572 minus 1)

12057321198661

+ 1211990811986611198652

+ 21198751198663

1= minus

120572 (120572 + 2)

61205732119908minus2

+ sdot sdot sdot

(56)


= 0 1198663

= (120572(120572 + 2)241205732)119908minus4


1205794

= 63

minus 61198872

+ 2 (31199081198872

2+ 6119908119887

11198873

+ 31198872

11198872)

minus3120572

120573(3


120572 (120572 minus 1)

12057321198872

(57)


120579lowast

4=

3 (120572 + 2)

120573(

1

211987510158401198663

+ 1198751198661015840

3)

+ 1211990811987511986611198663

minus(120572 + 2) (120572 + 3)

12057321198652

+ 61199081198652

2+ 6119875119865

21198662

1= 0119908minus1


+ sdot sdot sdot

(58)


4is critical




119896(V) and ℎ

119896+1(V) from



V = 119908 +1198871

119906+

1198872

1199062+

119887119896

119906119896+

119887119896+1

119906119896+1+

119887119896+2

119906119896+2+ sdot sdot sdot (59)


120579lowast

119896= 120573minus119896


120579lowastlowast

119896= 0 hence

119865119896

= minus1

6120573119896119908minus1

+ sdot sdot sdot

119866119896

= 0

120579119896+1

= (119896 minus 1) 119896

+ 121199081198871119887119896

+119886

120573119896+1+

1198871

120573119896 hence

120579lowast

119896+1=

119886


119865119896+1

= minus119886

4120573119896+1119908minus2

+ sdot sdot sdot

120579lowastlowast

119896+1= (119896 minus 1) 119865

1015840

119896+ 12119908119866

1119865119896

+1

1205731198961198661

= minus1

3120573119896119908minus2

+ sdot sdot sdot

119866119896+1

=1

12120573119896119908minus4

+ sdot sdot sdot

120579119896+2

= 2 (119896 + 1) 119896+1

minus 119896 (119896 + 1) 119887119896

+ 121199081198871119887119896+1

+ 121199081198872119887119896

+ 61198872

1119887119896

minus (119896 + 1) (119896+1

minus 119896119887119896)

minus120572 (120572 + 1)

1205732119887119896

+1

1205731198961198872

(60)

Hence

120579lowast

119896+2= (119896 + 1) (

1

21198751015840119866119896+1

+ 1198751198661015840

119896+1) minus

120572 (120572 minus 1)

1205732119865119896

+1

1205731198961198652

+ 121199081198751198661119866119896+1

+ 121199081198652119865119896

+ 61198751198662

1119865119896


+ sdot sdot sdot

120579lowastlowast

119896+2= (119896 + 1) 119865

1015840

119896+1minus

120572 (120572 minus 1)

1205732119866119896

+ 121199081198661119865119896+1

+ 121199081198652119866119896

+ 61198751198662

1119866119896


+ sdot sdot sdot

(61)

It means that 119865119896+2

and 119866119896+2


lt minus1 and 120591119895



V = 119908 +1198871

119906+

1198872

1199062+

119887119896

119906119896+

119887119896+1

119906119896+1+

119887119896+2

119906119896+2+

1198872119896

1199062119896+ sdot sdot sdot (62)


We have

1205792119896

= 61199081198872



120579lowast

2119896= 6119908119865

2


6

361205732119896119908minus1

+ sdot sdot sdot = 1198602119896

119908minus1


where 1198602119896

= 161205732119896

= 0 and 1198652119896







6



branching






119895


119895and 120579

lowastlowast





119895lt minus3



119866119895up to 119895 = max(119895


)



1= 1198622

= 1198623

= 1198624


1= 1198624


119895in



= 0 in 119866119895gives the


For 119895 = 1

we put 1198623


1198651

= 0

1198661

= 119860 +120572

2120573119908minus2

+ sdot sdot sdot

120579lowast

2=

2 minus 120572

2120573(

1

211987510158401198661

+ 1198751198661015840

1) + 6119908119866

2

1119875 minus

120572 (120572 minus 1)


=2 minus 120572

212057321199083119860 + 6119908

5(119860 +

120572

2120573119908minus2

)

2

minus120572 (120572 minus 1)


= 611986021199085

+5120572 + 2

1205731198601199083

+1205722

+ 2120572

21205732119908 + sdot sdot sdot

(65)

Hence

1198652

= 11986021199083

minus5120572 + 2

4120573119860119908 + sdot sdot sdot

1198662

= 1198653

= 0

(66)

Next

120579lowastlowast

3=

120572 + 4

120573(311986021199082

minus5120572 + 2

4120573) minus

2

1205732(119860 +

120572

2120573119908minus2

)

+ 1211990811986611198652

+ 21198751198663

1+ sdot sdot sdot

=120572 + 4

120573311986021199082

minus(120572 + 4) (5120572 + 2)

4120573119860 minus

2

1205732119860

+ 12119908 (119860 +120572

2120573119908minus2

) (11986021199083

minus5120572 + 2

4120573119860119908)

+ 21199084

(119860 +120572

2120573119908minus2

)

3

+ sdot sdot sdot

(67)


3is minus(3(120572minus2)120573)119860


3has


8 Equation 1198751

Equation 1198751is

119891 (119909 119910)def= minus11991010158401015840

+ 31199102

+ 119909 = 0 (68)


= (minus2 1 2) Q2

=

(0 2 0) and Q3


1= [Q



(1)

1(119909 119910)

def= minus11991010158401015840

+ 31199102

= 0 (69)



11991010158402

= 2 (1199103

+ 1198620) (70)


1areN119896

= (4 minus 119896 2(119896 +

1) 5) 119896 = 1 2 and 1198991

= 0 if 119896 = 4 Here 120572 = 2(119896 + 1)(4 minus




4120572 (120572 minus 1)

1205742V119906minus2 + 2

119896120574minus119896

119906minus119896

= 0 (71)

119875 = 2(1199083



= infin


minus72+


= minus1 and 120591119895


then

1198651

= 0

1198661

=120572

120574119908minus1

+ sdot sdot sdot

1198652

=120572 (120572 minus 8)


1198662

= 1198653

= 0

1198663

=120572 (120572 + 4)

31205743119908minus2

+ sdot sdot sdot

1198654

= minus

120572 (120572 + 4) (1205722

+ 24120572 + 48)

601205744119908minus1

+ sdot sdot sdot

1198664

= 1198655

= 0

1198665

=

120572 (120572 + 4) (31205723

+ 561205722

+ 200120572 + 192)

1801205745119908minus3

+ sdot sdot sdot

120579lowast

6= 0 sdot 119908


def= 1198606119908minus1

+ sdot sdot sdot

1198606

= 0

(72)



120579lowast

119896=

2119896


119865119896

= minus2119896

5120574119896119908minus1

+ sdot sdot sdot

119866119896

= 119865119896+1

= 0

119866119896+1

=(119896 + 11) 2

119896

75120574119896119908minus3

+ sdot sdot sdot

120579lowast

119896+2= 0 sdot 119908


(73)

q2

q3

1

0

2

2

q1

0

11Q

2Q

3Q

4Q

5Q

6Q




1 1199022)


120590119895

lt minus1 120591119895


3+ 119908 + 119862

0) but 119860


1has




9 Equation 1198753

Equation 1198753is

119891 (119909 119910)def= minus119909119910119910

10158401015840+ 11990911991010158402

minus 1199101199101015840+ 1198861199103

+ 119887119910 + 1198881199091199104

+ 119889119909

= 0

(74)



minus11990911991011991010158401015840

+ 11990911991010158402

+ 1198881199091199104

+ 119889119909 = 0 (75)




119892 (V) def= minusVV + V2 + 119888V4 + 119889

= 0

ℎ1

= minusVV + 119886V3 + 119887V

119875 (119908) = 1198881199084

+ 11986201199082

minus 119889

Δ (119875) =

minus119888119889 (1198622

0+ 4119888119889)

2

16= 0

(76)


119865119895

= 11987512

int1199082

11987532int

120579lowast

119895

1199083119889119908 119889119908

119866119895

= int1199082

11987532int

11987512

120579lowastlowast

119895

1199083119889119908 119889119908

(77)


0= infin

and 1199080


point 1199080

= infin 119867(119908) = int(119908211987532

)119889119908 = const sdot 119908minus3



119895and 119866

0





119895and 120579

lowastlowast


120579lowast

2= 0 sdot119908



119895lt 2 120591119895



= 0 we have 1198670(119908) =

int(119908211987532


1198620

1 119862

0


0

119894= 0 Here Lemma 11



0(119908211987532

)119889119908 = 0


3are regular



where


ℎ1

=3

2119887V minus VV

119875 (119908) = 21198861199083

+ 11986201199082

minus 119889

Δ (119875) = 4119889 (1198623

0minus 27119886

2119889) = 0

(78)


119895have critical


= 0 sdot 1199082


119895 120579lowastlowast119895






119892 (V) = minusVV + V2 + 119886V3 + 119887V

ℎ1

= minusVV2

119875 (119908) = 2 (1198861199083

+ 11986201199082

minus 119887119908)

Δ (119875) = 241198872

(1198622

0+ 4119886119887) = 0

(79)

At1199080



119895are 2 120579lowast

2= 0sdot119908

2+sdot sdot sdot

120590119895 120591119895




2= 119875(119908) but


10 Equation 1198754

Equation 1198754is

119891 (119909 119910)def= minus2119910119910

10158401015840+ 11991010158402

+ 31199104

+ 81199091199103

+ 4 (1199092

minus 119886) 1199102

+ 2119887

= 0

(80)


1and one edge Γ(1)


(2)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

+ 81199091199103

+ 411990921199102

= 0

(1)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

= 0

(81)


is N0

=





120573VV119906minus1 +

120572 (2 minus 120572)

1205732V2119906minus2

+8

120573119896V3119906minus119896 minus

4119886

120573119896+1V2119906minus(119896+1)

+4

1205732119896V2119906minus2119896 +

2119887

1205732(119896+1)119906minus2(119896+1)

= 0

119875 (119908) = 1199084

+ 1198620119908

1198620

= 0 119896 = 2 3

(82)



119865119895

=1

211987512

int119908

11987532int

120579lowast

119895

1199082119889119908 119889119908

119866119895

=1

2int

119908

11987532int

11987512

120579lowastlowast

119895

1199082119889119908 119889119908

(83)


= infin

and 1199080

= 0 Near 1199080

= infin 119867 = int(11990811987532

)119889119908 = const sdot

119908minus4




= 0 1198661

= (1205722120573)119908minus2


= minus(120572(120572 + 2)121205732) 119908minus1


= 0 1198653


3= 0 sdot 119908



=

minus1120573119896


= 0 119865119896+1

= (21198863120573119896+1

) 119908minus1

+ sdot sdot sdot 119866119896+1

=

(13120573119896+1

)119908minus3


119896+2= (4120572(2120572 minus 1)120573


120579lowastlowast

119896+2= 0 sdot 119908


119896+2= 4120572(2120572 minus 1)120573

119896+2= 0 only if



4+ 41199083

+ 41199082

+ 1198620119908

11 Equation 1198755

Equation 1198755is

119891 (119909 119910)def= minus 119909

2119910 (119910 minus 1) 119910

10158401015840+ 1199092 3119910 minus 1

211991010158402

minus 119909119910 (119910 minus 1) 1199101015840

+ (119910 minus 1)3

(1198861199102

+ 119887) + 1198881199091199102

(119910 minus 1)

+ 11988911990921199102

(119910 + 1) = 0

(84)




2+ 119889V2 (V + 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119889119908 [1198620

(119908 minus 1)2

+ 119908]

Δ (119875) = (2119889)4

1198622

0(1 minus 4119862

0) = 0

(85)


119865119895

= 11987512

int119908 (119908 minus 1)

2

11987532int

120579lowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

119866119895

= int119908 (119908 minus 1)

2

11987532int

11987512

120579lowastlowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

(86)


0=


= infin

119867 = int119908 (119908 minus 1)

2

11987532119889119908 = const sdot 119908



119895and 120579

lowastlowast


If 119886 = 0 then 120579lowast


4 and 1198652has


lt 4 and 120591119895



near 1199080



0(119908(119908 minus 1)

211987532

)119889119908 =

intinfin

1(119908(119908 minus 1)

211987532

)119889119908 = 0






2V2 + 119888V2 (V minus 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119888119908 (119908 minus 1) [1198620

(119908 minus 1) + 1]

Δ (119875) = (1198620

minus 1)2

= 0 1198620

= 0

(88)




12 Equation 1198756




= 0 1198993



13 Summary











References




















minus



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119895of the




1 1199032 1199033)Then we


⟨QR⟩ = 11990211199031

+ 11990211199032

+ 11990231199033 (13)

Each face Γ(119889)119895


U(119889)119895

= R ⟨Q1015840R⟩ = ⟨Q10158401015840R⟩ Q1015840Q10158401015840 isin Γ(119889)

119895

⟨Q1015840R⟩ gt ⟨Q101584010158401015840R⟩ Q101584010158401015840 isin Γ Γ(119889)

119895

(14)






119895



and U(2)119897

where Γ(1)119895

= Γ(2)

119896cap Γ(2)



119895


Γ(2)


119895(see [2])


119895(119909 119910) can be


119895 edges Γ(1)





(119889)


119895cap 1199013

⩾ 0 = 0


119895(119909 119910) = 0 with 119901

3= 0 can be


(119889)

119895(119909 119910) = 0 with


cap 1199013

gt 0



Γ(2)

1= [Q1Q2Q3]

N1

= (2 1 3)

Γ(2)

2= [Q1Q3Q4]

N2

= (2 minus2 3)

Γ(2)

3= [Q1Q2Q4]

N3

= (minus1 0 minus1)

Γ(2)

4= [Q2Q3Q4]

N4

= (0 0 minus1)

(15)


2 have 119903


exept Γ(1)6



(1199031 1199032 1199033) with 119903


119895and U(0)

119895


If the face Γ(119889)119895


= (1 0 1) then the


119895(119909 119910) = 119909

119902119892(119910) where the


119902119892(119910) + 119909


a differential sum


0 lt 120576 lt1003816100381610038161003816119910 (119909)

1003816100381610038161003816 100381610038161003816100381610038161199101015840(119909)

10038161003816100381610038161003816

10038161003816100381610038161003816119910(119899)

(119909)10038161003816100381610038161003816

lt 120576minus1

(16)



119910 = 119909120572V

119906 =1

120573119909120573

(17)


lowast(119906 V) = 119891(119909 119910)



1198991

= 0

1198993

gt 0

(18)



(119889)

119894(119909 119910) of 119891(119909 119910) into the

truncation

lowast(119889)

119894(119906 V) = 119906

119902119892 (V) (19)




= (1 0 1) Here lowast(119889)

119894(119906 V) equals

(119889)


119906[120572+119897(120573minus1)]120573

119889119897V

119889119906119897(20)

instead of 119910(119897)

= 119889119897119910119889119909119897


119910 sim 119909120572120593 (

119909120573

120573) 119909

120596997888rarr infin (21)



1199101015840

= 119909120572+120573minus1V + 120572119909

120572minus1V

11991010158401015840

= 1199092120573+120572minus2V + (2120572 + 120573 minus 1) 119909

120573+120572minus2V

+ 120572 (120572 minus 1) 119909120572minus2V

(22)

where V = 119889V119889119906


119892 (V) +

119898

sum

119895=1

ℎ119895(V) 119906minus119895

= 0 (23)


V = 119908 +

infin

sum

119895=1

119887119895(119908) 119906minus119895

(24)


119892 (119908) = 0 (25)

with the property

0 lt 120576 lt |119908|

10038161003816100381610038161003816100381610038161003816

119889119908

119889119906

10038161003816100381610038161003816100381610038161003816

10038161003816100381610038161003816100381610038161003816

119889119899119908

119889119906119899

10038161003816100381610038161003816100381610038161003816

lt1

120576lt infin (26)


L (119906) 119887119895(119908) + 120579

119895(119908) = 0 (27)


(119897)


(119897)

119894(119908) for 119894 lt 119895 and









119894is unique


1 1198992 1198993) to 1D edge Γ(1)

119894belongs


119894we must




Let S(119891) = Q1 Q

119872 S(

(119889)

119895) = Q

1 Q

119871 0 lt 119871 lt

119872 N = (1198991 1198992 1198993) sub U(119889)119894 and 119899

1= 0 1198993

gt 0 Denote

∘

Q119897= Q119871+119897

minus Q1 119897 = 1 119872 minus 119871 (28)

and∘

N = (11989911198993 11989921198993 1)


minus ⟨∘

N∘

Q119897⟩ 119897 = 1 119872 minus 119871 (29)

are natural




11198752 and




2(8) has edge Γ(1)

1= [Q

1Q2] with


1(119909 119910)

def= minus119910

10158401015840+ 21199103


11991010158402

= 1199104

+ 1198620

def= 119875 (119910) (30)




1

The edge Γ(1)1

= Γ(2)

1cap Γ(2)


1is the conic


= (2 1 3) and N3


1have the form

N = 120600N1

+ (1 minus 120600)N3

= (3120600 minus 1 120600 4120600 minus 1) 0 lt 120600 lt 1

(31)

Here 119872 = 4 119871 = 2∘

Q1

= (3 0 minus2)∘

Q2

= minusQ1


∘


⟨∘

N∘

Q1⟩ =

3 (3120600 minus 1)


120600 minus 1

4120600 minus 1= minus119896

⟨∘

N∘

Q2⟩ =

2 (3120600 minus 1)


120600


3120600

4120600 minus 1= minus119897

(32)









120572 =119896 + 1

2 minus 119896

120573 =3

2 minus 119896= 120572 + 1

(33)



2

= 119875 (119908)def=

120582

sum

119896=0

119901119896119908119896 119901119896

= const isin C (34)


=1

21198751015840(119908) (35)




lowast(119908) +

119877lowastlowast

(119908) where 119877lowast(119908) and 119877


119908 Let 119887119895(119908) = 119865

119895(119908)+119866

119895(119908) where119865

119895and119866

119895are functions


= 1198651015840 + 119875119866

1015840+

1

21198751015840119866

= 11987511986510158401015840

+1

211987510158401198651015840+ (119875119866

10158401015840+

3

211987510158401198661015840+

1

211987510158401015840

119866)

(36)


L119887 = F (119908) 119865 (119908) + G (119908) 119866 (119908) (37)


F (119908) 119865119895(119908) + 120579

lowast

119895(119908) = 0

G (119908) 119866119895(119908) + 120579

lowastlowast

119895(119908) = 0

(38)

where 120579119895(119908) = 120579

lowast

119895(119908)+120579

lowastlowast



120582 gt 1 Δ (119875) = 0 (39)




119895(119908)

119865119895

=

infin

sum

119894=minus119886119895

120593119895119894

120585119894

119866119895

=

infin

sum

119894=minus119887119895

120574119895119894

120585119894

(40)

where 120585 = 119908 minus 1199080 if 119908

0= infin and 120585 = 119908

minus1 if 1199080







119895(119908) and 119866




119865119895(119908) = minusF


119895(119908)

119866119895(119908) = minusG


119895(119908)

(41)






0= infin




119895(119908) and 120579

lowastlowast





119895(119908) and


is shown that 119865119895(119908) or 119866




1(119908) and 119866

1(119908) do not





119894of five



0= infin for

1198751ndash1198755and119908

0= 1 for119875

3ndash1198755and119908 = 1 for119875

5 To each suitable





2

119891 (119909 119910)def= minus11991010158401015840

+ 21199103

+ 119909119910 + 119886 = 0 (42)


(1)

1(119909 119910)

def= minus11991010158401015840

+ 21199103

= 0 (43)


120572V 119906 = 119909120573120573 (17) using formulas (22)


119892 (V) + ℎ1

(V) 119906minus1

+ ℎ2

(V) 119906minus2

+ ℎ119896

(V) 119906minus119896

+ ℎ119896+1


= 0

(44)

where

119892 (V) = minusV + 2V3

ℎ1

(V) = minus3120572

120573V

ℎ2

= minus120572 (120572 minus 1)

1205732V

ℎ119896

(V) = 120573minus119896V

ℎ119896+1

(V) = 119886120573minus119896minus1

(45)

119875 (119908) = 1199084

+ 1198620 1198620

= 0 (46)



119865119895

= 11987512

int1

11987532int 120579lowast

119895119889119908 119889119908

119866119895

= int1

11987532int 11987512

120579lowastlowast

119895119889119908 119889119908

(47)


119867 (119908) = int 119875minus32

119889119908 = const sdot 119908minus5




1ndash1198624

119865 = 119862111987512

+ 119862211987512

119867 + 1198650

119866 = 1198623

+ 1198624119867 + 119866

0

(49)

where 1198650 and 119866


11987512


+ sdot sdot sdot

11987512


+ sdot sdot sdot

(50)






0

119895and

1198660


119887119895

= 1198650

119895+ 1198660


Lemma 11 If 1198621

= 1198624





in them)

Let 120579lowast

119895(119908) and 120579

lowastlowast



119895119908120591119895 be their terms



0 = 119860119895 119861119895

isin C 120590119895 120591119895

isin R 119865119895and 119866


120590119895

= minus1 or 4

120591119895

= minus3 or 2

(51)


119895(119908) 120579

lowast

119895 and 120579

lowastlowast


119894=


119865119895and 119866



119860119895and 119861

119895


V = 119908 +1198871

119906+

1198872

1199062+

1198873

1199063+

1198874


V = +1

119906+

2

minus 1198871

1199062+

3

minus 21198872

1199063+

4

minus 31198873


V = +1

119906+

2

minus 21

1199062+

3

minus 42

+ 21198871

1199063

+4

minus 63

+ 61198872


V3 = 1199083

+311990821198871

119906+

31199081198872

1+ 311990821198872

1199062+

311990821198873

+ 611990811988711198872

+ 1198873

1

1199063

+311990821198874

+ 611990811988711198873

+ 31199081198872

2+ 31198872

11198872


(52)



1=



= 01198661

= (1205722120573)119908minus2


1205792

= 21

+ 61199081198872

1minus

3120572

1205731

minus120572 (120572 minus 1)

1205732119908 (53)


120579lowast

2= (2 minus

3120572

120573) (

1

211987510158401198661

+ 1198751198661015840

1) + 6119908119866

2

1119875 minus

120572 (120572 minus 1)

1205732119908

= minus120572 (120572 + 2)

21205732119908 + sdot sdot sdot

120579lowastlowast

2= 0

(54)


= minus(120572(120572+2)121205732)119908minus1


= 0Next

1205793

= 42

minus 21198871

+ 2 (611990811988711198872

+ 1198873

1)

minus3120572

120573(2


120572 (120572 minus 1)

12057321198871

(55)

Hence 120579lowast


120579lowastlowast

3=

120572 + 4

1205731198651015840

2minus

2 (120572 + 1)2

minus 3120572 (120572 + 1) + 120572 (120572 minus 1)

12057321198661

+ 1211990811986611198652

+ 21198751198663

1= minus

120572 (120572 + 2)

61205732119908minus2

+ sdot sdot sdot

(56)


= 0 1198663

= (120572(120572 + 2)241205732)119908minus4


1205794

= 63

minus 61198872

+ 2 (31199081198872

2+ 6119908119887

11198873

+ 31198872

11198872)

minus3120572

120573(3


120572 (120572 minus 1)

12057321198872

(57)


120579lowast

4=

3 (120572 + 2)

120573(

1

211987510158401198663

+ 1198751198661015840

3)

+ 1211990811987511986611198663

minus(120572 + 2) (120572 + 3)

12057321198652

+ 61199081198652

2+ 6119875119865

21198662

1= 0119908minus1


+ sdot sdot sdot

(58)


4is critical




119896(V) and ℎ

119896+1(V) from



V = 119908 +1198871

119906+

1198872

1199062+

119887119896

119906119896+

119887119896+1

119906119896+1+

119887119896+2

119906119896+2+ sdot sdot sdot (59)


120579lowast

119896= 120573minus119896


120579lowastlowast

119896= 0 hence

119865119896

= minus1

6120573119896119908minus1

+ sdot sdot sdot

119866119896

= 0

120579119896+1

= (119896 minus 1) 119896

+ 121199081198871119887119896

+119886

120573119896+1+

1198871

120573119896 hence

120579lowast

119896+1=

119886


119865119896+1

= minus119886

4120573119896+1119908minus2

+ sdot sdot sdot

120579lowastlowast

119896+1= (119896 minus 1) 119865

1015840

119896+ 12119908119866

1119865119896

+1

1205731198961198661

= minus1

3120573119896119908minus2

+ sdot sdot sdot

119866119896+1

=1

12120573119896119908minus4

+ sdot sdot sdot

120579119896+2

= 2 (119896 + 1) 119896+1

minus 119896 (119896 + 1) 119887119896

+ 121199081198871119887119896+1

+ 121199081198872119887119896

+ 61198872

1119887119896

minus (119896 + 1) (119896+1

minus 119896119887119896)

minus120572 (120572 + 1)

1205732119887119896

+1

1205731198961198872

(60)

Hence

120579lowast

119896+2= (119896 + 1) (

1

21198751015840119866119896+1

+ 1198751198661015840

119896+1) minus

120572 (120572 minus 1)

1205732119865119896

+1

1205731198961198652

+ 121199081198751198661119866119896+1

+ 121199081198652119865119896

+ 61198751198662

1119865119896


+ sdot sdot sdot

120579lowastlowast

119896+2= (119896 + 1) 119865

1015840

119896+1minus

120572 (120572 minus 1)

1205732119866119896

+ 121199081198661119865119896+1

+ 121199081198652119866119896

+ 61198751198662

1119866119896


+ sdot sdot sdot

(61)

It means that 119865119896+2

and 119866119896+2


lt minus1 and 120591119895



V = 119908 +1198871

119906+

1198872

1199062+

119887119896

119906119896+

119887119896+1

119906119896+1+

119887119896+2

119906119896+2+

1198872119896

1199062119896+ sdot sdot sdot (62)


We have

1205792119896

= 61199081198872



120579lowast

2119896= 6119908119865

2


6

361205732119896119908minus1

+ sdot sdot sdot = 1198602119896

119908minus1


where 1198602119896

= 161205732119896

= 0 and 1198652119896







6



branching






119895


119895and 120579

lowastlowast





119895lt minus3



119866119895up to 119895 = max(119895


)



1= 1198622

= 1198623

= 1198624


1= 1198624


119895in



= 0 in 119866119895gives the


For 119895 = 1

we put 1198623


1198651

= 0

1198661

= 119860 +120572

2120573119908minus2

+ sdot sdot sdot

120579lowast

2=

2 minus 120572

2120573(

1

211987510158401198661

+ 1198751198661015840

1) + 6119908119866

2

1119875 minus

120572 (120572 minus 1)


=2 minus 120572

212057321199083119860 + 6119908

5(119860 +

120572

2120573119908minus2

)

2

minus120572 (120572 minus 1)


= 611986021199085

+5120572 + 2

1205731198601199083

+1205722

+ 2120572

21205732119908 + sdot sdot sdot

(65)

Hence

1198652

= 11986021199083

minus5120572 + 2

4120573119860119908 + sdot sdot sdot

1198662

= 1198653

= 0

(66)

Next

120579lowastlowast

3=

120572 + 4

120573(311986021199082

minus5120572 + 2

4120573) minus

2

1205732(119860 +

120572

2120573119908minus2

)

+ 1211990811986611198652

+ 21198751198663

1+ sdot sdot sdot

=120572 + 4

120573311986021199082

minus(120572 + 4) (5120572 + 2)

4120573119860 minus

2

1205732119860

+ 12119908 (119860 +120572

2120573119908minus2

) (11986021199083

minus5120572 + 2

4120573119860119908)

+ 21199084

(119860 +120572

2120573119908minus2

)

3

+ sdot sdot sdot

(67)


3is minus(3(120572minus2)120573)119860


3has


8 Equation 1198751

Equation 1198751is

119891 (119909 119910)def= minus11991010158401015840

+ 31199102

+ 119909 = 0 (68)


= (minus2 1 2) Q2

=

(0 2 0) and Q3


1= [Q



(1)

1(119909 119910)

def= minus11991010158401015840

+ 31199102

= 0 (69)



11991010158402

= 2 (1199103

+ 1198620) (70)


1areN119896

= (4 minus 119896 2(119896 +

1) 5) 119896 = 1 2 and 1198991

= 0 if 119896 = 4 Here 120572 = 2(119896 + 1)(4 minus




4120572 (120572 minus 1)

1205742V119906minus2 + 2

119896120574minus119896

119906minus119896

= 0 (71)

119875 = 2(1199083



= infin


minus72+


= minus1 and 120591119895


then

1198651

= 0

1198661

=120572

120574119908minus1

+ sdot sdot sdot

1198652

=120572 (120572 minus 8)


1198662

= 1198653

= 0

1198663

=120572 (120572 + 4)

31205743119908minus2

+ sdot sdot sdot

1198654

= minus

120572 (120572 + 4) (1205722

+ 24120572 + 48)

601205744119908minus1

+ sdot sdot sdot

1198664

= 1198655

= 0

1198665

=

120572 (120572 + 4) (31205723

+ 561205722

+ 200120572 + 192)

1801205745119908minus3

+ sdot sdot sdot

120579lowast

6= 0 sdot 119908


def= 1198606119908minus1

+ sdot sdot sdot

1198606

= 0

(72)



120579lowast

119896=

2119896


119865119896

= minus2119896

5120574119896119908minus1

+ sdot sdot sdot

119866119896

= 119865119896+1

= 0

119866119896+1

=(119896 + 11) 2

119896

75120574119896119908minus3

+ sdot sdot sdot

120579lowast

119896+2= 0 sdot 119908


(73)

q2

q3

1

0

2

2

q1

0

11Q

2Q

3Q

4Q

5Q

6Q




1 1199022)


120590119895

lt minus1 120591119895


3+ 119908 + 119862

0) but 119860


1has




9 Equation 1198753

Equation 1198753is

119891 (119909 119910)def= minus119909119910119910

10158401015840+ 11990911991010158402

minus 1199101199101015840+ 1198861199103

+ 119887119910 + 1198881199091199104

+ 119889119909

= 0

(74)



minus11990911991011991010158401015840

+ 11990911991010158402

+ 1198881199091199104

+ 119889119909 = 0 (75)




119892 (V) def= minusVV + V2 + 119888V4 + 119889

= 0

ℎ1

= minusVV + 119886V3 + 119887V

119875 (119908) = 1198881199084

+ 11986201199082

minus 119889

Δ (119875) =

minus119888119889 (1198622

0+ 4119888119889)

2

16= 0

(76)


119865119895

= 11987512

int1199082

11987532int

120579lowast

119895

1199083119889119908 119889119908

119866119895

= int1199082

11987532int

11987512

120579lowastlowast

119895

1199083119889119908 119889119908

(77)


0= infin

and 1199080


point 1199080

= infin 119867(119908) = int(119908211987532

)119889119908 = const sdot 119908minus3



119895and 119866

0





119895and 120579

lowastlowast


120579lowast

2= 0 sdot119908



119895lt 2 120591119895



= 0 we have 1198670(119908) =

int(119908211987532


1198620

1 119862

0


0

119894= 0 Here Lemma 11



0(119908211987532

)119889119908 = 0


3are regular



where


ℎ1

=3

2119887V minus VV

119875 (119908) = 21198861199083

+ 11986201199082

minus 119889

Δ (119875) = 4119889 (1198623

0minus 27119886

2119889) = 0

(78)


119895have critical


= 0 sdot 1199082


119895 120579lowastlowast119895






119892 (V) = minusVV + V2 + 119886V3 + 119887V

ℎ1

= minusVV2

119875 (119908) = 2 (1198861199083

+ 11986201199082

minus 119887119908)

Δ (119875) = 241198872

(1198622

0+ 4119886119887) = 0

(79)

At1199080



119895are 2 120579lowast

2= 0sdot119908

2+sdot sdot sdot

120590119895 120591119895




2= 119875(119908) but


10 Equation 1198754

Equation 1198754is

119891 (119909 119910)def= minus2119910119910

10158401015840+ 11991010158402

+ 31199104

+ 81199091199103

+ 4 (1199092

minus 119886) 1199102

+ 2119887

= 0

(80)


1and one edge Γ(1)


(2)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

+ 81199091199103

+ 411990921199102

= 0

(1)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

= 0

(81)


is N0

=





120573VV119906minus1 +

120572 (2 minus 120572)

1205732V2119906minus2

+8

120573119896V3119906minus119896 minus

4119886

120573119896+1V2119906minus(119896+1)

+4

1205732119896V2119906minus2119896 +

2119887

1205732(119896+1)119906minus2(119896+1)

= 0

119875 (119908) = 1199084

+ 1198620119908

1198620

= 0 119896 = 2 3

(82)



119865119895

=1

211987512

int119908

11987532int

120579lowast

119895

1199082119889119908 119889119908

119866119895

=1

2int

119908

11987532int

11987512

120579lowastlowast

119895

1199082119889119908 119889119908

(83)


= infin

and 1199080

= 0 Near 1199080

= infin 119867 = int(11990811987532

)119889119908 = const sdot

119908minus4




= 0 1198661

= (1205722120573)119908minus2


= minus(120572(120572 + 2)121205732) 119908minus1


= 0 1198653


3= 0 sdot 119908



=

minus1120573119896


= 0 119865119896+1

= (21198863120573119896+1

) 119908minus1

+ sdot sdot sdot 119866119896+1

=

(13120573119896+1

)119908minus3


119896+2= (4120572(2120572 minus 1)120573


120579lowastlowast

119896+2= 0 sdot 119908


119896+2= 4120572(2120572 minus 1)120573

119896+2= 0 only if



4+ 41199083

+ 41199082

+ 1198620119908

11 Equation 1198755

Equation 1198755is

119891 (119909 119910)def= minus 119909

2119910 (119910 minus 1) 119910

10158401015840+ 1199092 3119910 minus 1

211991010158402

minus 119909119910 (119910 minus 1) 1199101015840

+ (119910 minus 1)3

(1198861199102

+ 119887) + 1198881199091199102

(119910 minus 1)

+ 11988911990921199102

(119910 + 1) = 0

(84)




2+ 119889V2 (V + 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119889119908 [1198620

(119908 minus 1)2

+ 119908]

Δ (119875) = (2119889)4

1198622

0(1 minus 4119862

0) = 0

(85)


119865119895

= 11987512

int119908 (119908 minus 1)

2

11987532int

120579lowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

119866119895

= int119908 (119908 minus 1)

2

11987532int

11987512

120579lowastlowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

(86)


0=


= infin

119867 = int119908 (119908 minus 1)

2

11987532119889119908 = const sdot 119908



119895and 120579

lowastlowast


If 119886 = 0 then 120579lowast


4 and 1198652has


lt 4 and 120591119895



near 1199080



0(119908(119908 minus 1)

211987532

)119889119908 =

intinfin

1(119908(119908 minus 1)

211987532

)119889119908 = 0






2V2 + 119888V2 (V minus 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119888119908 (119908 minus 1) [1198620

(119908 minus 1) + 1]

Δ (119875) = (1198620

minus 1)2

= 0 1198620

= 0

(88)




12 Equation 1198756




= 0 1198993



13 Summary











References




















minus



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra











1199101015840

= 119909120572+120573minus1V + 120572119909

120572minus1V

11991010158401015840

= 1199092120573+120572minus2V + (2120572 + 120573 minus 1) 119909

120573+120572minus2V

+ 120572 (120572 minus 1) 119909120572minus2V

(22)

where V = 119889V119889119906


119892 (V) +

119898

sum

119895=1

ℎ119895(V) 119906minus119895

= 0 (23)


V = 119908 +

infin

sum

119895=1

119887119895(119908) 119906minus119895

(24)


119892 (119908) = 0 (25)

with the property

0 lt 120576 lt |119908|

10038161003816100381610038161003816100381610038161003816

119889119908

119889119906

10038161003816100381610038161003816100381610038161003816

10038161003816100381610038161003816100381610038161003816

119889119899119908

119889119906119899

10038161003816100381610038161003816100381610038161003816

lt1

120576lt infin (26)


L (119906) 119887119895(119908) + 120579

119895(119908) = 0 (27)


(119897)


(119897)

119894(119908) for 119894 lt 119895 and









119894is unique


1 1198992 1198993) to 1D edge Γ(1)

119894belongs


119894we must




Let S(119891) = Q1 Q

119872 S(

(119889)

119895) = Q

1 Q

119871 0 lt 119871 lt

119872 N = (1198991 1198992 1198993) sub U(119889)119894 and 119899

1= 0 1198993

gt 0 Denote

∘

Q119897= Q119871+119897

minus Q1 119897 = 1 119872 minus 119871 (28)

and∘

N = (11989911198993 11989921198993 1)


minus ⟨∘

N∘

Q119897⟩ 119897 = 1 119872 minus 119871 (29)

are natural




11198752 and




2(8) has edge Γ(1)

1= [Q

1Q2] with


1(119909 119910)

def= minus119910

10158401015840+ 21199103


11991010158402

= 1199104

+ 1198620

def= 119875 (119910) (30)




1

The edge Γ(1)1

= Γ(2)

1cap Γ(2)


1is the conic


= (2 1 3) and N3


1have the form

N = 120600N1

+ (1 minus 120600)N3

= (3120600 minus 1 120600 4120600 minus 1) 0 lt 120600 lt 1

(31)

Here 119872 = 4 119871 = 2∘

Q1

= (3 0 minus2)∘

Q2

= minusQ1


∘


⟨∘

N∘

Q1⟩ =

3 (3120600 minus 1)


120600 minus 1

4120600 minus 1= minus119896

⟨∘

N∘

Q2⟩ =

2 (3120600 minus 1)


120600


3120600

4120600 minus 1= minus119897

(32)









120572 =119896 + 1

2 minus 119896

120573 =3

2 minus 119896= 120572 + 1

(33)



2

= 119875 (119908)def=

120582

sum

119896=0

119901119896119908119896 119901119896

= const isin C (34)


=1

21198751015840(119908) (35)




lowast(119908) +

119877lowastlowast

(119908) where 119877lowast(119908) and 119877


119908 Let 119887119895(119908) = 119865

119895(119908)+119866

119895(119908) where119865

119895and119866

119895are functions


= 1198651015840 + 119875119866

1015840+

1

21198751015840119866

= 11987511986510158401015840

+1

211987510158401198651015840+ (119875119866

10158401015840+

3

211987510158401198661015840+

1

211987510158401015840

119866)

(36)


L119887 = F (119908) 119865 (119908) + G (119908) 119866 (119908) (37)


F (119908) 119865119895(119908) + 120579

lowast

119895(119908) = 0

G (119908) 119866119895(119908) + 120579

lowastlowast

119895(119908) = 0

(38)

where 120579119895(119908) = 120579

lowast

119895(119908)+120579

lowastlowast



120582 gt 1 Δ (119875) = 0 (39)




119895(119908)

119865119895

=

infin

sum

119894=minus119886119895

120593119895119894

120585119894

119866119895

=

infin

sum

119894=minus119887119895

120574119895119894

120585119894

(40)

where 120585 = 119908 minus 1199080 if 119908

0= infin and 120585 = 119908

minus1 if 1199080







119895(119908) and 119866




119865119895(119908) = minusF


119895(119908)

119866119895(119908) = minusG


119895(119908)

(41)






0= infin




119895(119908) and 120579

lowastlowast





119895(119908) and


is shown that 119865119895(119908) or 119866




1(119908) and 119866

1(119908) do not





119894of five



0= infin for

1198751ndash1198755and119908

0= 1 for119875

3ndash1198755and119908 = 1 for119875

5 To each suitable





2

119891 (119909 119910)def= minus11991010158401015840

+ 21199103

+ 119909119910 + 119886 = 0 (42)


(1)

1(119909 119910)

def= minus11991010158401015840

+ 21199103

= 0 (43)


120572V 119906 = 119909120573120573 (17) using formulas (22)


119892 (V) + ℎ1

(V) 119906minus1

+ ℎ2

(V) 119906minus2

+ ℎ119896

(V) 119906minus119896

+ ℎ119896+1


= 0

(44)

where

119892 (V) = minusV + 2V3

ℎ1

(V) = minus3120572

120573V

ℎ2

= minus120572 (120572 minus 1)

1205732V

ℎ119896

(V) = 120573minus119896V

ℎ119896+1

(V) = 119886120573minus119896minus1

(45)

119875 (119908) = 1199084

+ 1198620 1198620

= 0 (46)



119865119895

= 11987512

int1

11987532int 120579lowast

119895119889119908 119889119908

119866119895

= int1

11987532int 11987512

120579lowastlowast

119895119889119908 119889119908

(47)


119867 (119908) = int 119875minus32

119889119908 = const sdot 119908minus5




1ndash1198624

119865 = 119862111987512

+ 119862211987512

119867 + 1198650

119866 = 1198623

+ 1198624119867 + 119866

0

(49)

where 1198650 and 119866


11987512


+ sdot sdot sdot

11987512


+ sdot sdot sdot

(50)






0

119895and

1198660


119887119895

= 1198650

119895+ 1198660


Lemma 11 If 1198621

= 1198624





in them)

Let 120579lowast

119895(119908) and 120579

lowastlowast



119895119908120591119895 be their terms



0 = 119860119895 119861119895

isin C 120590119895 120591119895

isin R 119865119895and 119866


120590119895

= minus1 or 4

120591119895

= minus3 or 2

(51)


119895(119908) 120579

lowast

119895 and 120579

lowastlowast


119894=


119865119895and 119866



119860119895and 119861

119895


V = 119908 +1198871

119906+

1198872

1199062+

1198873

1199063+

1198874


V = +1

119906+

2

minus 1198871

1199062+

3

minus 21198872

1199063+

4

minus 31198873


V = +1

119906+

2

minus 21

1199062+

3

minus 42

+ 21198871

1199063

+4

minus 63

+ 61198872


V3 = 1199083

+311990821198871

119906+

31199081198872

1+ 311990821198872

1199062+

311990821198873

+ 611990811988711198872

+ 1198873

1

1199063

+311990821198874

+ 611990811988711198873

+ 31199081198872

2+ 31198872

11198872


(52)



1=



= 01198661

= (1205722120573)119908minus2


1205792

= 21

+ 61199081198872

1minus

3120572

1205731

minus120572 (120572 minus 1)

1205732119908 (53)


120579lowast

2= (2 minus

3120572

120573) (

1

211987510158401198661

+ 1198751198661015840

1) + 6119908119866

2

1119875 minus

120572 (120572 minus 1)

1205732119908

= minus120572 (120572 + 2)

21205732119908 + sdot sdot sdot

120579lowastlowast

2= 0

(54)


= minus(120572(120572+2)121205732)119908minus1


= 0Next

1205793

= 42

minus 21198871

+ 2 (611990811988711198872

+ 1198873

1)

minus3120572

120573(2


120572 (120572 minus 1)

12057321198871

(55)

Hence 120579lowast


120579lowastlowast

3=

120572 + 4

1205731198651015840

2minus

2 (120572 + 1)2

minus 3120572 (120572 + 1) + 120572 (120572 minus 1)

12057321198661

+ 1211990811986611198652

+ 21198751198663

1= minus

120572 (120572 + 2)

61205732119908minus2

+ sdot sdot sdot

(56)


= 0 1198663

= (120572(120572 + 2)241205732)119908minus4


1205794

= 63

minus 61198872

+ 2 (31199081198872

2+ 6119908119887

11198873

+ 31198872

11198872)

minus3120572

120573(3


120572 (120572 minus 1)

12057321198872

(57)


120579lowast

4=

3 (120572 + 2)

120573(

1

211987510158401198663

+ 1198751198661015840

3)

+ 1211990811987511986611198663

minus(120572 + 2) (120572 + 3)

12057321198652

+ 61199081198652

2+ 6119875119865

21198662

1= 0119908minus1


+ sdot sdot sdot

(58)


4is critical




119896(V) and ℎ

119896+1(V) from



V = 119908 +1198871

119906+

1198872

1199062+

119887119896

119906119896+

119887119896+1

119906119896+1+

119887119896+2

119906119896+2+ sdot sdot sdot (59)


120579lowast

119896= 120573minus119896


120579lowastlowast

119896= 0 hence

119865119896

= minus1

6120573119896119908minus1

+ sdot sdot sdot

119866119896

= 0

120579119896+1

= (119896 minus 1) 119896

+ 121199081198871119887119896

+119886

120573119896+1+

1198871

120573119896 hence

120579lowast

119896+1=

119886


119865119896+1

= minus119886

4120573119896+1119908minus2

+ sdot sdot sdot

120579lowastlowast

119896+1= (119896 minus 1) 119865

1015840

119896+ 12119908119866

1119865119896

+1

1205731198961198661

= minus1

3120573119896119908minus2

+ sdot sdot sdot

119866119896+1

=1

12120573119896119908minus4

+ sdot sdot sdot

120579119896+2

= 2 (119896 + 1) 119896+1

minus 119896 (119896 + 1) 119887119896

+ 121199081198871119887119896+1

+ 121199081198872119887119896

+ 61198872

1119887119896

minus (119896 + 1) (119896+1

minus 119896119887119896)

minus120572 (120572 + 1)

1205732119887119896

+1

1205731198961198872

(60)

Hence

120579lowast

119896+2= (119896 + 1) (

1

21198751015840119866119896+1

+ 1198751198661015840

119896+1) minus

120572 (120572 minus 1)

1205732119865119896

+1

1205731198961198652

+ 121199081198751198661119866119896+1

+ 121199081198652119865119896

+ 61198751198662

1119865119896


+ sdot sdot sdot

120579lowastlowast

119896+2= (119896 + 1) 119865

1015840

119896+1minus

120572 (120572 minus 1)

1205732119866119896

+ 121199081198661119865119896+1

+ 121199081198652119866119896

+ 61198751198662

1119866119896


+ sdot sdot sdot

(61)

It means that 119865119896+2

and 119866119896+2


lt minus1 and 120591119895



V = 119908 +1198871

119906+

1198872

1199062+

119887119896

119906119896+

119887119896+1

119906119896+1+

119887119896+2

119906119896+2+

1198872119896

1199062119896+ sdot sdot sdot (62)


We have

1205792119896

= 61199081198872



120579lowast

2119896= 6119908119865

2


6

361205732119896119908minus1

+ sdot sdot sdot = 1198602119896

119908minus1


where 1198602119896

= 161205732119896

= 0 and 1198652119896







6



branching






119895


119895and 120579

lowastlowast





119895lt minus3



119866119895up to 119895 = max(119895


)



1= 1198622

= 1198623

= 1198624


1= 1198624


119895in



= 0 in 119866119895gives the


For 119895 = 1

we put 1198623


1198651

= 0

1198661

= 119860 +120572

2120573119908minus2

+ sdot sdot sdot

120579lowast

2=

2 minus 120572

2120573(

1

211987510158401198661

+ 1198751198661015840

1) + 6119908119866

2

1119875 minus

120572 (120572 minus 1)


=2 minus 120572

212057321199083119860 + 6119908

5(119860 +

120572

2120573119908minus2

)

2

minus120572 (120572 minus 1)


= 611986021199085

+5120572 + 2

1205731198601199083

+1205722

+ 2120572

21205732119908 + sdot sdot sdot

(65)

Hence

1198652

= 11986021199083

minus5120572 + 2

4120573119860119908 + sdot sdot sdot

1198662

= 1198653

= 0

(66)

Next

120579lowastlowast

3=

120572 + 4

120573(311986021199082

minus5120572 + 2

4120573) minus

2

1205732(119860 +

120572

2120573119908minus2

)

+ 1211990811986611198652

+ 21198751198663

1+ sdot sdot sdot

=120572 + 4

120573311986021199082

minus(120572 + 4) (5120572 + 2)

4120573119860 minus

2

1205732119860

+ 12119908 (119860 +120572

2120573119908minus2

) (11986021199083

minus5120572 + 2

4120573119860119908)

+ 21199084

(119860 +120572

2120573119908minus2

)

3

+ sdot sdot sdot

(67)


3is minus(3(120572minus2)120573)119860


3has


8 Equation 1198751

Equation 1198751is

119891 (119909 119910)def= minus11991010158401015840

+ 31199102

+ 119909 = 0 (68)


= (minus2 1 2) Q2

=

(0 2 0) and Q3


1= [Q



(1)

1(119909 119910)

def= minus11991010158401015840

+ 31199102

= 0 (69)



11991010158402

= 2 (1199103

+ 1198620) (70)


1areN119896

= (4 minus 119896 2(119896 +

1) 5) 119896 = 1 2 and 1198991

= 0 if 119896 = 4 Here 120572 = 2(119896 + 1)(4 minus




4120572 (120572 minus 1)

1205742V119906minus2 + 2

119896120574minus119896

119906minus119896

= 0 (71)

119875 = 2(1199083



= infin


minus72+


= minus1 and 120591119895


then

1198651

= 0

1198661

=120572

120574119908minus1

+ sdot sdot sdot

1198652

=120572 (120572 minus 8)


1198662

= 1198653

= 0

1198663

=120572 (120572 + 4)

31205743119908minus2

+ sdot sdot sdot

1198654

= minus

120572 (120572 + 4) (1205722

+ 24120572 + 48)

601205744119908minus1

+ sdot sdot sdot

1198664

= 1198655

= 0

1198665

=

120572 (120572 + 4) (31205723

+ 561205722

+ 200120572 + 192)

1801205745119908minus3

+ sdot sdot sdot

120579lowast

6= 0 sdot 119908


def= 1198606119908minus1

+ sdot sdot sdot

1198606

= 0

(72)



120579lowast

119896=

2119896


119865119896

= minus2119896

5120574119896119908minus1

+ sdot sdot sdot

119866119896

= 119865119896+1

= 0

119866119896+1

=(119896 + 11) 2

119896

75120574119896119908minus3

+ sdot sdot sdot

120579lowast

119896+2= 0 sdot 119908


(73)

q2

q3

1

0

2

2

q1

0

11Q

2Q

3Q

4Q

5Q

6Q




1 1199022)


120590119895

lt minus1 120591119895


3+ 119908 + 119862

0) but 119860


1has




9 Equation 1198753

Equation 1198753is

119891 (119909 119910)def= minus119909119910119910

10158401015840+ 11990911991010158402

minus 1199101199101015840+ 1198861199103

+ 119887119910 + 1198881199091199104

+ 119889119909

= 0

(74)



minus11990911991011991010158401015840

+ 11990911991010158402

+ 1198881199091199104

+ 119889119909 = 0 (75)




119892 (V) def= minusVV + V2 + 119888V4 + 119889

= 0

ℎ1

= minusVV + 119886V3 + 119887V

119875 (119908) = 1198881199084

+ 11986201199082

minus 119889

Δ (119875) =

minus119888119889 (1198622

0+ 4119888119889)

2

16= 0

(76)


119865119895

= 11987512

int1199082

11987532int

120579lowast

119895

1199083119889119908 119889119908

119866119895

= int1199082

11987532int

11987512

120579lowastlowast

119895

1199083119889119908 119889119908

(77)


0= infin

and 1199080


point 1199080

= infin 119867(119908) = int(119908211987532

)119889119908 = const sdot 119908minus3



119895and 119866

0





119895and 120579

lowastlowast


120579lowast

2= 0 sdot119908



119895lt 2 120591119895



= 0 we have 1198670(119908) =

int(119908211987532


1198620

1 119862

0


0

119894= 0 Here Lemma 11



0(119908211987532

)119889119908 = 0


3are regular



where


ℎ1

=3

2119887V minus VV

119875 (119908) = 21198861199083

+ 11986201199082

minus 119889

Δ (119875) = 4119889 (1198623

0minus 27119886

2119889) = 0

(78)


119895have critical


= 0 sdot 1199082


119895 120579lowastlowast119895






119892 (V) = minusVV + V2 + 119886V3 + 119887V

ℎ1

= minusVV2

119875 (119908) = 2 (1198861199083

+ 11986201199082

minus 119887119908)

Δ (119875) = 241198872

(1198622

0+ 4119886119887) = 0

(79)

At1199080



119895are 2 120579lowast

2= 0sdot119908

2+sdot sdot sdot

120590119895 120591119895




2= 119875(119908) but


10 Equation 1198754

Equation 1198754is

119891 (119909 119910)def= minus2119910119910

10158401015840+ 11991010158402

+ 31199104

+ 81199091199103

+ 4 (1199092

minus 119886) 1199102

+ 2119887

= 0

(80)


1and one edge Γ(1)


(2)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

+ 81199091199103

+ 411990921199102

= 0

(1)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

= 0

(81)


is N0

=





120573VV119906minus1 +

120572 (2 minus 120572)

1205732V2119906minus2

+8

120573119896V3119906minus119896 minus

4119886

120573119896+1V2119906minus(119896+1)

+4

1205732119896V2119906minus2119896 +

2119887

1205732(119896+1)119906minus2(119896+1)

= 0

119875 (119908) = 1199084

+ 1198620119908

1198620

= 0 119896 = 2 3

(82)



119865119895

=1

211987512

int119908

11987532int

120579lowast

119895

1199082119889119908 119889119908

119866119895

=1

2int

119908

11987532int

11987512

120579lowastlowast

119895

1199082119889119908 119889119908

(83)


= infin

and 1199080

= 0 Near 1199080

= infin 119867 = int(11990811987532

)119889119908 = const sdot

119908minus4




= 0 1198661

= (1205722120573)119908minus2


= minus(120572(120572 + 2)121205732) 119908minus1


= 0 1198653


3= 0 sdot 119908



=

minus1120573119896


= 0 119865119896+1

= (21198863120573119896+1

) 119908minus1

+ sdot sdot sdot 119866119896+1

=

(13120573119896+1

)119908minus3


119896+2= (4120572(2120572 minus 1)120573


120579lowastlowast

119896+2= 0 sdot 119908


119896+2= 4120572(2120572 minus 1)120573

119896+2= 0 only if



4+ 41199083

+ 41199082

+ 1198620119908

11 Equation 1198755

Equation 1198755is

119891 (119909 119910)def= minus 119909

2119910 (119910 minus 1) 119910

10158401015840+ 1199092 3119910 minus 1

211991010158402

minus 119909119910 (119910 minus 1) 1199101015840

+ (119910 minus 1)3

(1198861199102

+ 119887) + 1198881199091199102

(119910 minus 1)

+ 11988911990921199102

(119910 + 1) = 0

(84)




2+ 119889V2 (V + 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119889119908 [1198620

(119908 minus 1)2

+ 119908]

Δ (119875) = (2119889)4

1198622

0(1 minus 4119862

0) = 0

(85)


119865119895

= 11987512

int119908 (119908 minus 1)

2

11987532int

120579lowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

119866119895

= int119908 (119908 minus 1)

2

11987532int

11987512

120579lowastlowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

(86)


0=


= infin

119867 = int119908 (119908 minus 1)

2

11987532119889119908 = const sdot 119908



119895and 120579

lowastlowast


If 119886 = 0 then 120579lowast


4 and 1198652has


lt 4 and 120591119895



near 1199080



0(119908(119908 minus 1)

211987532

)119889119908 =

intinfin

1(119908(119908 minus 1)

211987532

)119889119908 = 0






2V2 + 119888V2 (V minus 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119888119908 (119908 minus 1) [1198620

(119908 minus 1) + 1]

Δ (119875) = (1198620

minus 1)2

= 0 1198620

= 0

(88)




12 Equation 1198756




= 0 1198993



13 Summary











References




















minus



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra












120572 =119896 + 1

2 minus 119896

120573 =3

2 minus 119896= 120572 + 1

(33)



2

= 119875 (119908)def=

120582

sum

119896=0

119901119896119908119896 119901119896

= const isin C (34)


=1

21198751015840(119908) (35)




lowast(119908) +

119877lowastlowast

(119908) where 119877lowast(119908) and 119877


119908 Let 119887119895(119908) = 119865

119895(119908)+119866

119895(119908) where119865

119895and119866

119895are functions


= 1198651015840 + 119875119866

1015840+

1

21198751015840119866

= 11987511986510158401015840

+1

211987510158401198651015840+ (119875119866

10158401015840+

3

211987510158401198661015840+

1

211987510158401015840

119866)

(36)


L119887 = F (119908) 119865 (119908) + G (119908) 119866 (119908) (37)


F (119908) 119865119895(119908) + 120579

lowast

119895(119908) = 0

G (119908) 119866119895(119908) + 120579

lowastlowast

119895(119908) = 0

(38)

where 120579119895(119908) = 120579

lowast

119895(119908)+120579

lowastlowast



120582 gt 1 Δ (119875) = 0 (39)




119895(119908)

119865119895

=

infin

sum

119894=minus119886119895

120593119895119894

120585119894

119866119895

=

infin

sum

119894=minus119887119895

120574119895119894

120585119894

(40)

where 120585 = 119908 minus 1199080 if 119908

0= infin and 120585 = 119908

minus1 if 1199080







119895(119908) and 119866




119865119895(119908) = minusF


119895(119908)

119866119895(119908) = minusG


119895(119908)

(41)






0= infin




119895(119908) and 120579

lowastlowast





119895(119908) and


is shown that 119865119895(119908) or 119866




1(119908) and 119866

1(119908) do not





119894of five



0= infin for

1198751ndash1198755and119908

0= 1 for119875

3ndash1198755and119908 = 1 for119875

5 To each suitable





2

119891 (119909 119910)def= minus11991010158401015840

+ 21199103

+ 119909119910 + 119886 = 0 (42)


(1)

1(119909 119910)

def= minus11991010158401015840

+ 21199103

= 0 (43)


120572V 119906 = 119909120573120573 (17) using formulas (22)


119892 (V) + ℎ1

(V) 119906minus1

+ ℎ2

(V) 119906minus2

+ ℎ119896

(V) 119906minus119896

+ ℎ119896+1


= 0

(44)

where

119892 (V) = minusV + 2V3

ℎ1

(V) = minus3120572

120573V

ℎ2

= minus120572 (120572 minus 1)

1205732V

ℎ119896

(V) = 120573minus119896V

ℎ119896+1

(V) = 119886120573minus119896minus1

(45)

119875 (119908) = 1199084

+ 1198620 1198620

= 0 (46)



119865119895

= 11987512

int1

11987532int 120579lowast

119895119889119908 119889119908

119866119895

= int1

11987532int 11987512

120579lowastlowast

119895119889119908 119889119908

(47)


119867 (119908) = int 119875minus32

119889119908 = const sdot 119908minus5




1ndash1198624

119865 = 119862111987512

+ 119862211987512

119867 + 1198650

119866 = 1198623

+ 1198624119867 + 119866

0

(49)

where 1198650 and 119866


11987512


+ sdot sdot sdot

11987512


+ sdot sdot sdot

(50)






0

119895and

1198660


119887119895

= 1198650

119895+ 1198660


Lemma 11 If 1198621

= 1198624





in them)

Let 120579lowast

119895(119908) and 120579

lowastlowast



119895119908120591119895 be their terms



0 = 119860119895 119861119895

isin C 120590119895 120591119895

isin R 119865119895and 119866


120590119895

= minus1 or 4

120591119895

= minus3 or 2

(51)


119895(119908) 120579

lowast

119895 and 120579

lowastlowast


119894=


119865119895and 119866



119860119895and 119861

119895


V = 119908 +1198871

119906+

1198872

1199062+

1198873

1199063+

1198874


V = +1

119906+

2

minus 1198871

1199062+

3

minus 21198872

1199063+

4

minus 31198873


V = +1

119906+

2

minus 21

1199062+

3

minus 42

+ 21198871

1199063

+4

minus 63

+ 61198872


V3 = 1199083

+311990821198871

119906+

31199081198872

1+ 311990821198872

1199062+

311990821198873

+ 611990811988711198872

+ 1198873

1

1199063

+311990821198874

+ 611990811988711198873

+ 31199081198872

2+ 31198872

11198872


(52)



1=



= 01198661

= (1205722120573)119908minus2


1205792

= 21

+ 61199081198872

1minus

3120572

1205731

minus120572 (120572 minus 1)

1205732119908 (53)


120579lowast

2= (2 minus

3120572

120573) (

1

211987510158401198661

+ 1198751198661015840

1) + 6119908119866

2

1119875 minus

120572 (120572 minus 1)

1205732119908

= minus120572 (120572 + 2)

21205732119908 + sdot sdot sdot

120579lowastlowast

2= 0

(54)


= minus(120572(120572+2)121205732)119908minus1


= 0Next

1205793

= 42

minus 21198871

+ 2 (611990811988711198872

+ 1198873

1)

minus3120572

120573(2


120572 (120572 minus 1)

12057321198871

(55)

Hence 120579lowast


120579lowastlowast

3=

120572 + 4

1205731198651015840

2minus

2 (120572 + 1)2

minus 3120572 (120572 + 1) + 120572 (120572 minus 1)

12057321198661

+ 1211990811986611198652

+ 21198751198663

1= minus

120572 (120572 + 2)

61205732119908minus2

+ sdot sdot sdot

(56)


= 0 1198663

= (120572(120572 + 2)241205732)119908minus4


1205794

= 63

minus 61198872

+ 2 (31199081198872

2+ 6119908119887

11198873

+ 31198872

11198872)

minus3120572

120573(3


120572 (120572 minus 1)

12057321198872

(57)


120579lowast

4=

3 (120572 + 2)

120573(

1

211987510158401198663

+ 1198751198661015840

3)

+ 1211990811987511986611198663

minus(120572 + 2) (120572 + 3)

12057321198652

+ 61199081198652

2+ 6119875119865

21198662

1= 0119908minus1


+ sdot sdot sdot

(58)


4is critical




119896(V) and ℎ

119896+1(V) from



V = 119908 +1198871

119906+

1198872

1199062+

119887119896

119906119896+

119887119896+1

119906119896+1+

119887119896+2

119906119896+2+ sdot sdot sdot (59)


120579lowast

119896= 120573minus119896


120579lowastlowast

119896= 0 hence

119865119896

= minus1

6120573119896119908minus1

+ sdot sdot sdot

119866119896

= 0

120579119896+1

= (119896 minus 1) 119896

+ 121199081198871119887119896

+119886

120573119896+1+

1198871

120573119896 hence

120579lowast

119896+1=

119886


119865119896+1

= minus119886

4120573119896+1119908minus2

+ sdot sdot sdot

120579lowastlowast

119896+1= (119896 minus 1) 119865

1015840

119896+ 12119908119866

1119865119896

+1

1205731198961198661

= minus1

3120573119896119908minus2

+ sdot sdot sdot

119866119896+1

=1

12120573119896119908minus4

+ sdot sdot sdot

120579119896+2

= 2 (119896 + 1) 119896+1

minus 119896 (119896 + 1) 119887119896

+ 121199081198871119887119896+1

+ 121199081198872119887119896

+ 61198872

1119887119896

minus (119896 + 1) (119896+1

minus 119896119887119896)

minus120572 (120572 + 1)

1205732119887119896

+1

1205731198961198872

(60)

Hence

120579lowast

119896+2= (119896 + 1) (

1

21198751015840119866119896+1

+ 1198751198661015840

119896+1) minus

120572 (120572 minus 1)

1205732119865119896

+1

1205731198961198652

+ 121199081198751198661119866119896+1

+ 121199081198652119865119896

+ 61198751198662

1119865119896


+ sdot sdot sdot

120579lowastlowast

119896+2= (119896 + 1) 119865

1015840

119896+1minus

120572 (120572 minus 1)

1205732119866119896

+ 121199081198661119865119896+1

+ 121199081198652119866119896

+ 61198751198662

1119866119896


+ sdot sdot sdot

(61)

It means that 119865119896+2

and 119866119896+2


lt minus1 and 120591119895



V = 119908 +1198871

119906+

1198872

1199062+

119887119896

119906119896+

119887119896+1

119906119896+1+

119887119896+2

119906119896+2+

1198872119896

1199062119896+ sdot sdot sdot (62)


We have

1205792119896

= 61199081198872



120579lowast

2119896= 6119908119865

2


6

361205732119896119908minus1

+ sdot sdot sdot = 1198602119896

119908minus1


where 1198602119896

= 161205732119896

= 0 and 1198652119896







6



branching






119895


119895and 120579

lowastlowast





119895lt minus3



119866119895up to 119895 = max(119895


)



1= 1198622

= 1198623

= 1198624


1= 1198624


119895in



= 0 in 119866119895gives the


For 119895 = 1

we put 1198623


1198651

= 0

1198661

= 119860 +120572

2120573119908minus2

+ sdot sdot sdot

120579lowast

2=

2 minus 120572

2120573(

1

211987510158401198661

+ 1198751198661015840

1) + 6119908119866

2

1119875 minus

120572 (120572 minus 1)


=2 minus 120572

212057321199083119860 + 6119908

5(119860 +

120572

2120573119908minus2

)

2

minus120572 (120572 minus 1)


= 611986021199085

+5120572 + 2

1205731198601199083

+1205722

+ 2120572

21205732119908 + sdot sdot sdot

(65)

Hence

1198652

= 11986021199083

minus5120572 + 2

4120573119860119908 + sdot sdot sdot

1198662

= 1198653

= 0

(66)

Next

120579lowastlowast

3=

120572 + 4

120573(311986021199082

minus5120572 + 2

4120573) minus

2

1205732(119860 +

120572

2120573119908minus2

)

+ 1211990811986611198652

+ 21198751198663

1+ sdot sdot sdot

=120572 + 4

120573311986021199082

minus(120572 + 4) (5120572 + 2)

4120573119860 minus

2

1205732119860

+ 12119908 (119860 +120572

2120573119908minus2

) (11986021199083

minus5120572 + 2

4120573119860119908)

+ 21199084

(119860 +120572

2120573119908minus2

)

3

+ sdot sdot sdot

(67)


3is minus(3(120572minus2)120573)119860


3has


8 Equation 1198751

Equation 1198751is

119891 (119909 119910)def= minus11991010158401015840

+ 31199102

+ 119909 = 0 (68)


= (minus2 1 2) Q2

=

(0 2 0) and Q3


1= [Q



(1)

1(119909 119910)

def= minus11991010158401015840

+ 31199102

= 0 (69)



11991010158402

= 2 (1199103

+ 1198620) (70)


1areN119896

= (4 minus 119896 2(119896 +

1) 5) 119896 = 1 2 and 1198991

= 0 if 119896 = 4 Here 120572 = 2(119896 + 1)(4 minus




4120572 (120572 minus 1)

1205742V119906minus2 + 2

119896120574minus119896

119906minus119896

= 0 (71)

119875 = 2(1199083



= infin


minus72+


= minus1 and 120591119895


then

1198651

= 0

1198661

=120572

120574119908minus1

+ sdot sdot sdot

1198652

=120572 (120572 minus 8)


1198662

= 1198653

= 0

1198663

=120572 (120572 + 4)

31205743119908minus2

+ sdot sdot sdot

1198654

= minus

120572 (120572 + 4) (1205722

+ 24120572 + 48)

601205744119908minus1

+ sdot sdot sdot

1198664

= 1198655

= 0

1198665

=

120572 (120572 + 4) (31205723

+ 561205722

+ 200120572 + 192)

1801205745119908minus3

+ sdot sdot sdot

120579lowast

6= 0 sdot 119908


def= 1198606119908minus1

+ sdot sdot sdot

1198606

= 0

(72)



120579lowast

119896=

2119896


119865119896

= minus2119896

5120574119896119908minus1

+ sdot sdot sdot

119866119896

= 119865119896+1

= 0

119866119896+1

=(119896 + 11) 2

119896

75120574119896119908minus3

+ sdot sdot sdot

120579lowast

119896+2= 0 sdot 119908


(73)

q2

q3

1

0

2

2

q1

0

11Q

2Q

3Q

4Q

5Q

6Q




1 1199022)


120590119895

lt minus1 120591119895


3+ 119908 + 119862

0) but 119860


1has




9 Equation 1198753

Equation 1198753is

119891 (119909 119910)def= minus119909119910119910

10158401015840+ 11990911991010158402

minus 1199101199101015840+ 1198861199103

+ 119887119910 + 1198881199091199104

+ 119889119909

= 0

(74)



minus11990911991011991010158401015840

+ 11990911991010158402

+ 1198881199091199104

+ 119889119909 = 0 (75)




119892 (V) def= minusVV + V2 + 119888V4 + 119889

= 0

ℎ1

= minusVV + 119886V3 + 119887V

119875 (119908) = 1198881199084

+ 11986201199082

minus 119889

Δ (119875) =

minus119888119889 (1198622

0+ 4119888119889)

2

16= 0

(76)


119865119895

= 11987512

int1199082

11987532int

120579lowast

119895

1199083119889119908 119889119908

119866119895

= int1199082

11987532int

11987512

120579lowastlowast

119895

1199083119889119908 119889119908

(77)


0= infin

and 1199080


point 1199080

= infin 119867(119908) = int(119908211987532

)119889119908 = const sdot 119908minus3



119895and 119866

0





119895and 120579

lowastlowast


120579lowast

2= 0 sdot119908



119895lt 2 120591119895



= 0 we have 1198670(119908) =

int(119908211987532


1198620

1 119862

0


0

119894= 0 Here Lemma 11



0(119908211987532

)119889119908 = 0


3are regular



where


ℎ1

=3

2119887V minus VV

119875 (119908) = 21198861199083

+ 11986201199082

minus 119889

Δ (119875) = 4119889 (1198623

0minus 27119886

2119889) = 0

(78)


119895have critical


= 0 sdot 1199082


119895 120579lowastlowast119895






119892 (V) = minusVV + V2 + 119886V3 + 119887V

ℎ1

= minusVV2

119875 (119908) = 2 (1198861199083

+ 11986201199082

minus 119887119908)

Δ (119875) = 241198872

(1198622

0+ 4119886119887) = 0

(79)

At1199080



119895are 2 120579lowast

2= 0sdot119908

2+sdot sdot sdot

120590119895 120591119895




2= 119875(119908) but


10 Equation 1198754

Equation 1198754is

119891 (119909 119910)def= minus2119910119910

10158401015840+ 11991010158402

+ 31199104

+ 81199091199103

+ 4 (1199092

minus 119886) 1199102

+ 2119887

= 0

(80)


1and one edge Γ(1)


(2)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

+ 81199091199103

+ 411990921199102

= 0

(1)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

= 0

(81)


is N0

=





120573VV119906minus1 +

120572 (2 minus 120572)

1205732V2119906minus2

+8

120573119896V3119906minus119896 minus

4119886

120573119896+1V2119906minus(119896+1)

+4

1205732119896V2119906minus2119896 +

2119887

1205732(119896+1)119906minus2(119896+1)

= 0

119875 (119908) = 1199084

+ 1198620119908

1198620

= 0 119896 = 2 3

(82)



119865119895

=1

211987512

int119908

11987532int

120579lowast

119895

1199082119889119908 119889119908

119866119895

=1

2int

119908

11987532int

11987512

120579lowastlowast

119895

1199082119889119908 119889119908

(83)


= infin

and 1199080

= 0 Near 1199080

= infin 119867 = int(11990811987532

)119889119908 = const sdot

119908minus4




= 0 1198661

= (1205722120573)119908minus2


= minus(120572(120572 + 2)121205732) 119908minus1


= 0 1198653


3= 0 sdot 119908



=

minus1120573119896


= 0 119865119896+1

= (21198863120573119896+1

) 119908minus1

+ sdot sdot sdot 119866119896+1

=

(13120573119896+1

)119908minus3


119896+2= (4120572(2120572 minus 1)120573


120579lowastlowast

119896+2= 0 sdot 119908


119896+2= 4120572(2120572 minus 1)120573

119896+2= 0 only if



4+ 41199083

+ 41199082

+ 1198620119908

11 Equation 1198755

Equation 1198755is

119891 (119909 119910)def= minus 119909

2119910 (119910 minus 1) 119910

10158401015840+ 1199092 3119910 minus 1

211991010158402

minus 119909119910 (119910 minus 1) 1199101015840

+ (119910 minus 1)3

(1198861199102

+ 119887) + 1198881199091199102

(119910 minus 1)

+ 11988911990921199102

(119910 + 1) = 0

(84)




2+ 119889V2 (V + 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119889119908 [1198620

(119908 minus 1)2

+ 119908]

Δ (119875) = (2119889)4

1198622

0(1 minus 4119862

0) = 0

(85)


119865119895

= 11987512

int119908 (119908 minus 1)

2

11987532int

120579lowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

119866119895

= int119908 (119908 minus 1)

2

11987532int

11987512

120579lowastlowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

(86)


0=


= infin

119867 = int119908 (119908 minus 1)

2

11987532119889119908 = const sdot 119908



119895and 120579

lowastlowast


If 119886 = 0 then 120579lowast


4 and 1198652has


lt 4 and 120591119895



near 1199080



0(119908(119908 minus 1)

211987532

)119889119908 =

intinfin

1(119908(119908 minus 1)

211987532

)119889119908 = 0






2V2 + 119888V2 (V minus 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119888119908 (119908 minus 1) [1198620

(119908 minus 1) + 1]

Δ (119875) = (1198620

minus 1)2

= 0 1198620

= 0

(88)




12 Equation 1198756




= 0 1198993



13 Summary











References




















minus



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119894of five



0= infin for

1198751ndash1198755and119908

0= 1 for119875

3ndash1198755and119908 = 1 for119875

5 To each suitable





2

119891 (119909 119910)def= minus11991010158401015840

+ 21199103

+ 119909119910 + 119886 = 0 (42)


(1)

1(119909 119910)

def= minus11991010158401015840

+ 21199103

= 0 (43)


120572V 119906 = 119909120573120573 (17) using formulas (22)


119892 (V) + ℎ1

(V) 119906minus1

+ ℎ2

(V) 119906minus2

+ ℎ119896

(V) 119906minus119896

+ ℎ119896+1


= 0

(44)

where

119892 (V) = minusV + 2V3

ℎ1

(V) = minus3120572

120573V

ℎ2

= minus120572 (120572 minus 1)

1205732V

ℎ119896

(V) = 120573minus119896V

ℎ119896+1

(V) = 119886120573minus119896minus1

(45)

119875 (119908) = 1199084

+ 1198620 1198620

= 0 (46)



119865119895

= 11987512

int1

11987532int 120579lowast

119895119889119908 119889119908

119866119895

= int1

11987532int 11987512

120579lowastlowast

119895119889119908 119889119908

(47)


119867 (119908) = int 119875minus32

119889119908 = const sdot 119908minus5




1ndash1198624

119865 = 119862111987512

+ 119862211987512

119867 + 1198650

119866 = 1198623

+ 1198624119867 + 119866

0

(49)

where 1198650 and 119866


11987512


+ sdot sdot sdot

11987512


+ sdot sdot sdot

(50)






0

119895and

1198660


119887119895

= 1198650

119895+ 1198660


Lemma 11 If 1198621

= 1198624





in them)

Let 120579lowast

119895(119908) and 120579

lowastlowast



119895119908120591119895 be their terms



0 = 119860119895 119861119895

isin C 120590119895 120591119895

isin R 119865119895and 119866


120590119895

= minus1 or 4

120591119895

= minus3 or 2

(51)


119895(119908) 120579

lowast

119895 and 120579

lowastlowast


119894=


119865119895and 119866



119860119895and 119861

119895


V = 119908 +1198871

119906+

1198872

1199062+

1198873

1199063+

1198874


V = +1

119906+

2

minus 1198871

1199062+

3

minus 21198872

1199063+

4

minus 31198873


V = +1

119906+

2

minus 21

1199062+

3

minus 42

+ 21198871

1199063

+4

minus 63

+ 61198872


V3 = 1199083

+311990821198871

119906+

31199081198872

1+ 311990821198872

1199062+

311990821198873

+ 611990811988711198872

+ 1198873

1

1199063

+311990821198874

+ 611990811988711198873

+ 31199081198872

2+ 31198872

11198872


(52)



1=



= 01198661

= (1205722120573)119908minus2


1205792

= 21

+ 61199081198872

1minus

3120572

1205731

minus120572 (120572 minus 1)

1205732119908 (53)


120579lowast

2= (2 minus

3120572

120573) (

1

211987510158401198661

+ 1198751198661015840

1) + 6119908119866

2

1119875 minus

120572 (120572 minus 1)

1205732119908

= minus120572 (120572 + 2)

21205732119908 + sdot sdot sdot

120579lowastlowast

2= 0

(54)


= minus(120572(120572+2)121205732)119908minus1


= 0Next

1205793

= 42

minus 21198871

+ 2 (611990811988711198872

+ 1198873

1)

minus3120572

120573(2


120572 (120572 minus 1)

12057321198871

(55)

Hence 120579lowast


120579lowastlowast

3=

120572 + 4

1205731198651015840

2minus

2 (120572 + 1)2

minus 3120572 (120572 + 1) + 120572 (120572 minus 1)

12057321198661

+ 1211990811986611198652

+ 21198751198663

1= minus

120572 (120572 + 2)

61205732119908minus2

+ sdot sdot sdot

(56)


= 0 1198663

= (120572(120572 + 2)241205732)119908minus4


1205794

= 63

minus 61198872

+ 2 (31199081198872

2+ 6119908119887

11198873

+ 31198872

11198872)

minus3120572

120573(3


120572 (120572 minus 1)

12057321198872

(57)


120579lowast

4=

3 (120572 + 2)

120573(

1

211987510158401198663

+ 1198751198661015840

3)

+ 1211990811987511986611198663

minus(120572 + 2) (120572 + 3)

12057321198652

+ 61199081198652

2+ 6119875119865

21198662

1= 0119908minus1


+ sdot sdot sdot

(58)


4is critical




119896(V) and ℎ

119896+1(V) from



V = 119908 +1198871

119906+

1198872

1199062+

119887119896

119906119896+

119887119896+1

119906119896+1+

119887119896+2

119906119896+2+ sdot sdot sdot (59)


120579lowast

119896= 120573minus119896


120579lowastlowast

119896= 0 hence

119865119896

= minus1

6120573119896119908minus1

+ sdot sdot sdot

119866119896

= 0

120579119896+1

= (119896 minus 1) 119896

+ 121199081198871119887119896

+119886

120573119896+1+

1198871

120573119896 hence

120579lowast

119896+1=

119886


119865119896+1

= minus119886

4120573119896+1119908minus2

+ sdot sdot sdot

120579lowastlowast

119896+1= (119896 minus 1) 119865

1015840

119896+ 12119908119866

1119865119896

+1

1205731198961198661

= minus1

3120573119896119908minus2

+ sdot sdot sdot

119866119896+1

=1

12120573119896119908minus4

+ sdot sdot sdot

120579119896+2

= 2 (119896 + 1) 119896+1

minus 119896 (119896 + 1) 119887119896

+ 121199081198871119887119896+1

+ 121199081198872119887119896

+ 61198872

1119887119896

minus (119896 + 1) (119896+1

minus 119896119887119896)

minus120572 (120572 + 1)

1205732119887119896

+1

1205731198961198872

(60)

Hence

120579lowast

119896+2= (119896 + 1) (

1

21198751015840119866119896+1

+ 1198751198661015840

119896+1) minus

120572 (120572 minus 1)

1205732119865119896

+1

1205731198961198652

+ 121199081198751198661119866119896+1

+ 121199081198652119865119896

+ 61198751198662

1119865119896


+ sdot sdot sdot

120579lowastlowast

119896+2= (119896 + 1) 119865

1015840

119896+1minus

120572 (120572 minus 1)

1205732119866119896

+ 121199081198661119865119896+1

+ 121199081198652119866119896

+ 61198751198662

1119866119896


+ sdot sdot sdot

(61)

It means that 119865119896+2

and 119866119896+2


lt minus1 and 120591119895



V = 119908 +1198871

119906+

1198872

1199062+

119887119896

119906119896+

119887119896+1

119906119896+1+

119887119896+2

119906119896+2+

1198872119896

1199062119896+ sdot sdot sdot (62)


We have

1205792119896

= 61199081198872



120579lowast

2119896= 6119908119865

2


6

361205732119896119908minus1

+ sdot sdot sdot = 1198602119896

119908minus1


where 1198602119896

= 161205732119896

= 0 and 1198652119896







6



branching






119895


119895and 120579

lowastlowast





119895lt minus3



119866119895up to 119895 = max(119895


)



1= 1198622

= 1198623

= 1198624


1= 1198624


119895in



= 0 in 119866119895gives the


For 119895 = 1

we put 1198623


1198651

= 0

1198661

= 119860 +120572

2120573119908minus2

+ sdot sdot sdot

120579lowast

2=

2 minus 120572

2120573(

1

211987510158401198661

+ 1198751198661015840

1) + 6119908119866

2

1119875 minus

120572 (120572 minus 1)


=2 minus 120572

212057321199083119860 + 6119908

5(119860 +

120572

2120573119908minus2

)

2

minus120572 (120572 minus 1)


= 611986021199085

+5120572 + 2

1205731198601199083

+1205722

+ 2120572

21205732119908 + sdot sdot sdot

(65)

Hence

1198652

= 11986021199083

minus5120572 + 2

4120573119860119908 + sdot sdot sdot

1198662

= 1198653

= 0

(66)

Next

120579lowastlowast

3=

120572 + 4

120573(311986021199082

minus5120572 + 2

4120573) minus

2

1205732(119860 +

120572

2120573119908minus2

)

+ 1211990811986611198652

+ 21198751198663

1+ sdot sdot sdot

=120572 + 4

120573311986021199082

minus(120572 + 4) (5120572 + 2)

4120573119860 minus

2

1205732119860

+ 12119908 (119860 +120572

2120573119908minus2

) (11986021199083

minus5120572 + 2

4120573119860119908)

+ 21199084

(119860 +120572

2120573119908minus2

)

3

+ sdot sdot sdot

(67)


3is minus(3(120572minus2)120573)119860


3has


8 Equation 1198751

Equation 1198751is

119891 (119909 119910)def= minus11991010158401015840

+ 31199102

+ 119909 = 0 (68)


= (minus2 1 2) Q2

=

(0 2 0) and Q3


1= [Q



(1)

1(119909 119910)

def= minus11991010158401015840

+ 31199102

= 0 (69)



11991010158402

= 2 (1199103

+ 1198620) (70)


1areN119896

= (4 minus 119896 2(119896 +

1) 5) 119896 = 1 2 and 1198991

= 0 if 119896 = 4 Here 120572 = 2(119896 + 1)(4 minus




4120572 (120572 minus 1)

1205742V119906minus2 + 2

119896120574minus119896

119906minus119896

= 0 (71)

119875 = 2(1199083



= infin


minus72+


= minus1 and 120591119895


then

1198651

= 0

1198661

=120572

120574119908minus1

+ sdot sdot sdot

1198652

=120572 (120572 minus 8)


1198662

= 1198653

= 0

1198663

=120572 (120572 + 4)

31205743119908minus2

+ sdot sdot sdot

1198654

= minus

120572 (120572 + 4) (1205722

+ 24120572 + 48)

601205744119908minus1

+ sdot sdot sdot

1198664

= 1198655

= 0

1198665

=

120572 (120572 + 4) (31205723

+ 561205722

+ 200120572 + 192)

1801205745119908minus3

+ sdot sdot sdot

120579lowast

6= 0 sdot 119908


def= 1198606119908minus1

+ sdot sdot sdot

1198606

= 0

(72)



120579lowast

119896=

2119896


119865119896

= minus2119896

5120574119896119908minus1

+ sdot sdot sdot

119866119896

= 119865119896+1

= 0

119866119896+1

=(119896 + 11) 2

119896

75120574119896119908minus3

+ sdot sdot sdot

120579lowast

119896+2= 0 sdot 119908


(73)

q2

q3

1

0

2

2

q1

0

11Q

2Q

3Q

4Q

5Q

6Q




1 1199022)


120590119895

lt minus1 120591119895


3+ 119908 + 119862

0) but 119860


1has




9 Equation 1198753

Equation 1198753is

119891 (119909 119910)def= minus119909119910119910

10158401015840+ 11990911991010158402

minus 1199101199101015840+ 1198861199103

+ 119887119910 + 1198881199091199104

+ 119889119909

= 0

(74)



minus11990911991011991010158401015840

+ 11990911991010158402

+ 1198881199091199104

+ 119889119909 = 0 (75)




119892 (V) def= minusVV + V2 + 119888V4 + 119889

= 0

ℎ1

= minusVV + 119886V3 + 119887V

119875 (119908) = 1198881199084

+ 11986201199082

minus 119889

Δ (119875) =

minus119888119889 (1198622

0+ 4119888119889)

2

16= 0

(76)


119865119895

= 11987512

int1199082

11987532int

120579lowast

119895

1199083119889119908 119889119908

119866119895

= int1199082

11987532int

11987512

120579lowastlowast

119895

1199083119889119908 119889119908

(77)


0= infin

and 1199080


point 1199080

= infin 119867(119908) = int(119908211987532

)119889119908 = const sdot 119908minus3



119895and 119866

0





119895and 120579

lowastlowast


120579lowast

2= 0 sdot119908



119895lt 2 120591119895



= 0 we have 1198670(119908) =

int(119908211987532


1198620

1 119862

0


0

119894= 0 Here Lemma 11



0(119908211987532

)119889119908 = 0


3are regular



where


ℎ1

=3

2119887V minus VV

119875 (119908) = 21198861199083

+ 11986201199082

minus 119889

Δ (119875) = 4119889 (1198623

0minus 27119886

2119889) = 0

(78)


119895have critical


= 0 sdot 1199082


119895 120579lowastlowast119895






119892 (V) = minusVV + V2 + 119886V3 + 119887V

ℎ1

= minusVV2

119875 (119908) = 2 (1198861199083

+ 11986201199082

minus 119887119908)

Δ (119875) = 241198872

(1198622

0+ 4119886119887) = 0

(79)

At1199080



119895are 2 120579lowast

2= 0sdot119908

2+sdot sdot sdot

120590119895 120591119895




2= 119875(119908) but


10 Equation 1198754

Equation 1198754is

119891 (119909 119910)def= minus2119910119910

10158401015840+ 11991010158402

+ 31199104

+ 81199091199103

+ 4 (1199092

minus 119886) 1199102

+ 2119887

= 0

(80)


1and one edge Γ(1)


(2)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

+ 81199091199103

+ 411990921199102

= 0

(1)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

= 0

(81)


is N0

=





120573VV119906minus1 +

120572 (2 minus 120572)

1205732V2119906minus2

+8

120573119896V3119906minus119896 minus

4119886

120573119896+1V2119906minus(119896+1)

+4

1205732119896V2119906minus2119896 +

2119887

1205732(119896+1)119906minus2(119896+1)

= 0

119875 (119908) = 1199084

+ 1198620119908

1198620

= 0 119896 = 2 3

(82)



119865119895

=1

211987512

int119908

11987532int

120579lowast

119895

1199082119889119908 119889119908

119866119895

=1

2int

119908

11987532int

11987512

120579lowastlowast

119895

1199082119889119908 119889119908

(83)


= infin

and 1199080

= 0 Near 1199080

= infin 119867 = int(11990811987532

)119889119908 = const sdot

119908minus4




= 0 1198661

= (1205722120573)119908minus2


= minus(120572(120572 + 2)121205732) 119908minus1


= 0 1198653


3= 0 sdot 119908



=

minus1120573119896


= 0 119865119896+1

= (21198863120573119896+1

) 119908minus1

+ sdot sdot sdot 119866119896+1

=

(13120573119896+1

)119908minus3


119896+2= (4120572(2120572 minus 1)120573


120579lowastlowast

119896+2= 0 sdot 119908


119896+2= 4120572(2120572 minus 1)120573

119896+2= 0 only if



4+ 41199083

+ 41199082

+ 1198620119908

11 Equation 1198755

Equation 1198755is

119891 (119909 119910)def= minus 119909

2119910 (119910 minus 1) 119910

10158401015840+ 1199092 3119910 minus 1

211991010158402

minus 119909119910 (119910 minus 1) 1199101015840

+ (119910 minus 1)3

(1198861199102

+ 119887) + 1198881199091199102

(119910 minus 1)

+ 11988911990921199102

(119910 + 1) = 0

(84)




2+ 119889V2 (V + 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119889119908 [1198620

(119908 minus 1)2

+ 119908]

Δ (119875) = (2119889)4

1198622

0(1 minus 4119862

0) = 0

(85)


119865119895

= 11987512

int119908 (119908 minus 1)

2

11987532int

120579lowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

119866119895

= int119908 (119908 minus 1)

2

11987532int

11987512

120579lowastlowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

(86)


0=


= infin

119867 = int119908 (119908 minus 1)

2

11987532119889119908 = const sdot 119908



119895and 120579

lowastlowast


If 119886 = 0 then 120579lowast


4 and 1198652has


lt 4 and 120591119895



near 1199080



0(119908(119908 minus 1)

211987532

)119889119908 =

intinfin

1(119908(119908 minus 1)

211987532

)119889119908 = 0






2V2 + 119888V2 (V minus 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119888119908 (119908 minus 1) [1198620

(119908 minus 1) + 1]

Δ (119875) = (1198620

minus 1)2

= 0 1198620

= 0

(88)




12 Equation 1198756




= 0 1198993



13 Summary











References




















minus



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra











1=



= 01198661

= (1205722120573)119908minus2


1205792

= 21

+ 61199081198872

1minus

3120572

1205731

minus120572 (120572 minus 1)

1205732119908 (53)


120579lowast

2= (2 minus

3120572

120573) (

1

211987510158401198661

+ 1198751198661015840

1) + 6119908119866

2

1119875 minus

120572 (120572 minus 1)

1205732119908

= minus120572 (120572 + 2)

21205732119908 + sdot sdot sdot

120579lowastlowast

2= 0

(54)


= minus(120572(120572+2)121205732)119908minus1


= 0Next

1205793

= 42

minus 21198871

+ 2 (611990811988711198872

+ 1198873

1)

minus3120572

120573(2


120572 (120572 minus 1)

12057321198871

(55)

Hence 120579lowast


120579lowastlowast

3=

120572 + 4

1205731198651015840

2minus

2 (120572 + 1)2

minus 3120572 (120572 + 1) + 120572 (120572 minus 1)

12057321198661

+ 1211990811986611198652

+ 21198751198663

1= minus

120572 (120572 + 2)

61205732119908minus2

+ sdot sdot sdot

(56)


= 0 1198663

= (120572(120572 + 2)241205732)119908minus4


1205794

= 63

minus 61198872

+ 2 (31199081198872

2+ 6119908119887

11198873

+ 31198872

11198872)

minus3120572

120573(3


120572 (120572 minus 1)

12057321198872

(57)


120579lowast

4=

3 (120572 + 2)

120573(

1

211987510158401198663

+ 1198751198661015840

3)

+ 1211990811987511986611198663

minus(120572 + 2) (120572 + 3)

12057321198652

+ 61199081198652

2+ 6119875119865

21198662

1= 0119908minus1


+ sdot sdot sdot

(58)


4is critical




119896(V) and ℎ

119896+1(V) from



V = 119908 +1198871

119906+

1198872

1199062+

119887119896

119906119896+

119887119896+1

119906119896+1+

119887119896+2

119906119896+2+ sdot sdot sdot (59)


120579lowast

119896= 120573minus119896


120579lowastlowast

119896= 0 hence

119865119896

= minus1

6120573119896119908minus1

+ sdot sdot sdot

119866119896

= 0

120579119896+1

= (119896 minus 1) 119896

+ 121199081198871119887119896

+119886

120573119896+1+

1198871

120573119896 hence

120579lowast

119896+1=

119886


119865119896+1

= minus119886

4120573119896+1119908minus2

+ sdot sdot sdot

120579lowastlowast

119896+1= (119896 minus 1) 119865

1015840

119896+ 12119908119866

1119865119896

+1

1205731198961198661

= minus1

3120573119896119908minus2

+ sdot sdot sdot

119866119896+1

=1

12120573119896119908minus4

+ sdot sdot sdot

120579119896+2

= 2 (119896 + 1) 119896+1

minus 119896 (119896 + 1) 119887119896

+ 121199081198871119887119896+1

+ 121199081198872119887119896

+ 61198872

1119887119896

minus (119896 + 1) (119896+1

minus 119896119887119896)

minus120572 (120572 + 1)

1205732119887119896

+1

1205731198961198872

(60)

Hence

120579lowast

119896+2= (119896 + 1) (

1

21198751015840119866119896+1

+ 1198751198661015840

119896+1) minus

120572 (120572 minus 1)

1205732119865119896

+1

1205731198961198652

+ 121199081198751198661119866119896+1

+ 121199081198652119865119896

+ 61198751198662

1119865119896


+ sdot sdot sdot

120579lowastlowast

119896+2= (119896 + 1) 119865

1015840

119896+1minus

120572 (120572 minus 1)

1205732119866119896

+ 121199081198661119865119896+1

+ 121199081198652119866119896

+ 61198751198662

1119866119896


+ sdot sdot sdot

(61)

It means that 119865119896+2

and 119866119896+2


lt minus1 and 120591119895



V = 119908 +1198871

119906+

1198872

1199062+

119887119896

119906119896+

119887119896+1

119906119896+1+

119887119896+2

119906119896+2+

1198872119896

1199062119896+ sdot sdot sdot (62)


We have

1205792119896

= 61199081198872



120579lowast

2119896= 6119908119865

2


6

361205732119896119908minus1

+ sdot sdot sdot = 1198602119896

119908minus1


where 1198602119896

= 161205732119896

= 0 and 1198652119896







6



branching






119895


119895and 120579

lowastlowast





119895lt minus3



119866119895up to 119895 = max(119895


)



1= 1198622

= 1198623

= 1198624


1= 1198624


119895in



= 0 in 119866119895gives the


For 119895 = 1

we put 1198623


1198651

= 0

1198661

= 119860 +120572

2120573119908minus2

+ sdot sdot sdot

120579lowast

2=

2 minus 120572

2120573(

1

211987510158401198661

+ 1198751198661015840

1) + 6119908119866

2

1119875 minus

120572 (120572 minus 1)


=2 minus 120572

212057321199083119860 + 6119908

5(119860 +

120572

2120573119908minus2

)

2

minus120572 (120572 minus 1)


= 611986021199085

+5120572 + 2

1205731198601199083

+1205722

+ 2120572

21205732119908 + sdot sdot sdot

(65)

Hence

1198652

= 11986021199083

minus5120572 + 2

4120573119860119908 + sdot sdot sdot

1198662

= 1198653

= 0

(66)

Next

120579lowastlowast

3=

120572 + 4

120573(311986021199082

minus5120572 + 2

4120573) minus

2

1205732(119860 +

120572

2120573119908minus2

)

+ 1211990811986611198652

+ 21198751198663

1+ sdot sdot sdot

=120572 + 4

120573311986021199082

minus(120572 + 4) (5120572 + 2)

4120573119860 minus

2

1205732119860

+ 12119908 (119860 +120572

2120573119908minus2

) (11986021199083

minus5120572 + 2

4120573119860119908)

+ 21199084

(119860 +120572

2120573119908minus2

)

3

+ sdot sdot sdot

(67)


3is minus(3(120572minus2)120573)119860


3has


8 Equation 1198751

Equation 1198751is

119891 (119909 119910)def= minus11991010158401015840

+ 31199102

+ 119909 = 0 (68)


= (minus2 1 2) Q2

=

(0 2 0) and Q3


1= [Q



(1)

1(119909 119910)

def= minus11991010158401015840

+ 31199102

= 0 (69)



11991010158402

= 2 (1199103

+ 1198620) (70)


1areN119896

= (4 minus 119896 2(119896 +

1) 5) 119896 = 1 2 and 1198991

= 0 if 119896 = 4 Here 120572 = 2(119896 + 1)(4 minus




4120572 (120572 minus 1)

1205742V119906minus2 + 2

119896120574minus119896

119906minus119896

= 0 (71)

119875 = 2(1199083



= infin


minus72+


= minus1 and 120591119895


then

1198651

= 0

1198661

=120572

120574119908minus1

+ sdot sdot sdot

1198652

=120572 (120572 minus 8)


1198662

= 1198653

= 0

1198663

=120572 (120572 + 4)

31205743119908minus2

+ sdot sdot sdot

1198654

= minus

120572 (120572 + 4) (1205722

+ 24120572 + 48)

601205744119908minus1

+ sdot sdot sdot

1198664

= 1198655

= 0

1198665

=

120572 (120572 + 4) (31205723

+ 561205722

+ 200120572 + 192)

1801205745119908minus3

+ sdot sdot sdot

120579lowast

6= 0 sdot 119908


def= 1198606119908minus1

+ sdot sdot sdot

1198606

= 0

(72)



120579lowast

119896=

2119896


119865119896

= minus2119896

5120574119896119908minus1

+ sdot sdot sdot

119866119896

= 119865119896+1

= 0

119866119896+1

=(119896 + 11) 2

119896

75120574119896119908minus3

+ sdot sdot sdot

120579lowast

119896+2= 0 sdot 119908


(73)

q2

q3

1

0

2

2

q1

0

11Q

2Q

3Q

4Q

5Q

6Q




1 1199022)


120590119895

lt minus1 120591119895


3+ 119908 + 119862

0) but 119860


1has




9 Equation 1198753

Equation 1198753is

119891 (119909 119910)def= minus119909119910119910

10158401015840+ 11990911991010158402

minus 1199101199101015840+ 1198861199103

+ 119887119910 + 1198881199091199104

+ 119889119909

= 0

(74)



minus11990911991011991010158401015840

+ 11990911991010158402

+ 1198881199091199104

+ 119889119909 = 0 (75)




119892 (V) def= minusVV + V2 + 119888V4 + 119889

= 0

ℎ1

= minusVV + 119886V3 + 119887V

119875 (119908) = 1198881199084

+ 11986201199082

minus 119889

Δ (119875) =

minus119888119889 (1198622

0+ 4119888119889)

2

16= 0

(76)


119865119895

= 11987512

int1199082

11987532int

120579lowast

119895

1199083119889119908 119889119908

119866119895

= int1199082

11987532int

11987512

120579lowastlowast

119895

1199083119889119908 119889119908

(77)


0= infin

and 1199080


point 1199080

= infin 119867(119908) = int(119908211987532

)119889119908 = const sdot 119908minus3



119895and 119866

0





119895and 120579

lowastlowast


120579lowast

2= 0 sdot119908



119895lt 2 120591119895



= 0 we have 1198670(119908) =

int(119908211987532


1198620

1 119862

0


0

119894= 0 Here Lemma 11



0(119908211987532

)119889119908 = 0


3are regular



where


ℎ1

=3

2119887V minus VV

119875 (119908) = 21198861199083

+ 11986201199082

minus 119889

Δ (119875) = 4119889 (1198623

0minus 27119886

2119889) = 0

(78)


119895have critical


= 0 sdot 1199082


119895 120579lowastlowast119895






119892 (V) = minusVV + V2 + 119886V3 + 119887V

ℎ1

= minusVV2

119875 (119908) = 2 (1198861199083

+ 11986201199082

minus 119887119908)

Δ (119875) = 241198872

(1198622

0+ 4119886119887) = 0

(79)

At1199080



119895are 2 120579lowast

2= 0sdot119908

2+sdot sdot sdot

120590119895 120591119895




2= 119875(119908) but


10 Equation 1198754

Equation 1198754is

119891 (119909 119910)def= minus2119910119910

10158401015840+ 11991010158402

+ 31199104

+ 81199091199103

+ 4 (1199092

minus 119886) 1199102

+ 2119887

= 0

(80)


1and one edge Γ(1)


(2)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

+ 81199091199103

+ 411990921199102

= 0

(1)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

= 0

(81)


is N0

=





120573VV119906minus1 +

120572 (2 minus 120572)

1205732V2119906minus2

+8

120573119896V3119906minus119896 minus

4119886

120573119896+1V2119906minus(119896+1)

+4

1205732119896V2119906minus2119896 +

2119887

1205732(119896+1)119906minus2(119896+1)

= 0

119875 (119908) = 1199084

+ 1198620119908

1198620

= 0 119896 = 2 3

(82)



119865119895

=1

211987512

int119908

11987532int

120579lowast

119895

1199082119889119908 119889119908

119866119895

=1

2int

119908

11987532int

11987512

120579lowastlowast

119895

1199082119889119908 119889119908

(83)


= infin

and 1199080

= 0 Near 1199080

= infin 119867 = int(11990811987532

)119889119908 = const sdot

119908minus4




= 0 1198661

= (1205722120573)119908minus2


= minus(120572(120572 + 2)121205732) 119908minus1


= 0 1198653


3= 0 sdot 119908



=

minus1120573119896


= 0 119865119896+1

= (21198863120573119896+1

) 119908minus1

+ sdot sdot sdot 119866119896+1

=

(13120573119896+1

)119908minus3


119896+2= (4120572(2120572 minus 1)120573


120579lowastlowast

119896+2= 0 sdot 119908


119896+2= 4120572(2120572 minus 1)120573

119896+2= 0 only if



4+ 41199083

+ 41199082

+ 1198620119908

11 Equation 1198755

Equation 1198755is

119891 (119909 119910)def= minus 119909

2119910 (119910 minus 1) 119910

10158401015840+ 1199092 3119910 minus 1

211991010158402

minus 119909119910 (119910 minus 1) 1199101015840

+ (119910 minus 1)3

(1198861199102

+ 119887) + 1198881199091199102

(119910 minus 1)

+ 11988911990921199102

(119910 + 1) = 0

(84)




2+ 119889V2 (V + 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119889119908 [1198620

(119908 minus 1)2

+ 119908]

Δ (119875) = (2119889)4

1198622

0(1 minus 4119862

0) = 0

(85)


119865119895

= 11987512

int119908 (119908 minus 1)

2

11987532int

120579lowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

119866119895

= int119908 (119908 minus 1)

2

11987532int

11987512

120579lowastlowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

(86)


0=


= infin

119867 = int119908 (119908 minus 1)

2

11987532119889119908 = const sdot 119908



119895and 120579

lowastlowast


If 119886 = 0 then 120579lowast


4 and 1198652has


lt 4 and 120591119895



near 1199080



0(119908(119908 minus 1)

211987532

)119889119908 =

intinfin

1(119908(119908 minus 1)

211987532

)119889119908 = 0






2V2 + 119888V2 (V minus 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119888119908 (119908 minus 1) [1198620

(119908 minus 1) + 1]

Δ (119875) = (1198620

minus 1)2

= 0 1198620

= 0

(88)




12 Equation 1198756




= 0 1198993



13 Summary











References




















minus



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










We have

1205792119896

= 61199081198872



120579lowast

2119896= 6119908119865

2


6

361205732119896119908minus1

+ sdot sdot sdot = 1198602119896

119908minus1


where 1198602119896

= 161205732119896

= 0 and 1198652119896







6



branching






119895


119895and 120579

lowastlowast





119895lt minus3



119866119895up to 119895 = max(119895


)



1= 1198622

= 1198623

= 1198624


1= 1198624


119895in



= 0 in 119866119895gives the


For 119895 = 1

we put 1198623


1198651

= 0

1198661

= 119860 +120572

2120573119908minus2

+ sdot sdot sdot

120579lowast

2=

2 minus 120572

2120573(

1

211987510158401198661

+ 1198751198661015840

1) + 6119908119866

2

1119875 minus

120572 (120572 minus 1)


=2 minus 120572

212057321199083119860 + 6119908

5(119860 +

120572

2120573119908minus2

)

2

minus120572 (120572 minus 1)


= 611986021199085

+5120572 + 2

1205731198601199083

+1205722

+ 2120572

21205732119908 + sdot sdot sdot

(65)

Hence

1198652

= 11986021199083

minus5120572 + 2

4120573119860119908 + sdot sdot sdot

1198662

= 1198653

= 0

(66)

Next

120579lowastlowast

3=

120572 + 4

120573(311986021199082

minus5120572 + 2

4120573) minus

2

1205732(119860 +

120572

2120573119908minus2

)

+ 1211990811986611198652

+ 21198751198663

1+ sdot sdot sdot

=120572 + 4

120573311986021199082

minus(120572 + 4) (5120572 + 2)

4120573119860 minus

2

1205732119860

+ 12119908 (119860 +120572

2120573119908minus2

) (11986021199083

minus5120572 + 2

4120573119860119908)

+ 21199084

(119860 +120572

2120573119908minus2

)

3

+ sdot sdot sdot

(67)


3is minus(3(120572minus2)120573)119860


3has


8 Equation 1198751

Equation 1198751is

119891 (119909 119910)def= minus11991010158401015840

+ 31199102

+ 119909 = 0 (68)


= (minus2 1 2) Q2

=

(0 2 0) and Q3


1= [Q



(1)

1(119909 119910)

def= minus11991010158401015840

+ 31199102

= 0 (69)



11991010158402

= 2 (1199103

+ 1198620) (70)


1areN119896

= (4 minus 119896 2(119896 +

1) 5) 119896 = 1 2 and 1198991

= 0 if 119896 = 4 Here 120572 = 2(119896 + 1)(4 minus




4120572 (120572 minus 1)

1205742V119906minus2 + 2

119896120574minus119896

119906minus119896

= 0 (71)

119875 = 2(1199083



= infin


minus72+


= minus1 and 120591119895


then

1198651

= 0

1198661

=120572

120574119908minus1

+ sdot sdot sdot

1198652

=120572 (120572 minus 8)


1198662

= 1198653

= 0

1198663

=120572 (120572 + 4)

31205743119908minus2

+ sdot sdot sdot

1198654

= minus

120572 (120572 + 4) (1205722

+ 24120572 + 48)

601205744119908minus1

+ sdot sdot sdot

1198664

= 1198655

= 0

1198665

=

120572 (120572 + 4) (31205723

+ 561205722

+ 200120572 + 192)

1801205745119908minus3

+ sdot sdot sdot

120579lowast

6= 0 sdot 119908


def= 1198606119908minus1

+ sdot sdot sdot

1198606

= 0

(72)



120579lowast

119896=

2119896


119865119896

= minus2119896

5120574119896119908minus1

+ sdot sdot sdot

119866119896

= 119865119896+1

= 0

119866119896+1

=(119896 + 11) 2

119896

75120574119896119908minus3

+ sdot sdot sdot

120579lowast

119896+2= 0 sdot 119908


(73)

q2

q3

1

0

2

2

q1

0

11Q

2Q

3Q

4Q

5Q

6Q




1 1199022)


120590119895

lt minus1 120591119895


3+ 119908 + 119862

0) but 119860


1has




9 Equation 1198753

Equation 1198753is

119891 (119909 119910)def= minus119909119910119910

10158401015840+ 11990911991010158402

minus 1199101199101015840+ 1198861199103

+ 119887119910 + 1198881199091199104

+ 119889119909

= 0

(74)



minus11990911991011991010158401015840

+ 11990911991010158402

+ 1198881199091199104

+ 119889119909 = 0 (75)




119892 (V) def= minusVV + V2 + 119888V4 + 119889

= 0

ℎ1

= minusVV + 119886V3 + 119887V

119875 (119908) = 1198881199084

+ 11986201199082

minus 119889

Δ (119875) =

minus119888119889 (1198622

0+ 4119888119889)

2

16= 0

(76)


119865119895

= 11987512

int1199082

11987532int

120579lowast

119895

1199083119889119908 119889119908

119866119895

= int1199082

11987532int

11987512

120579lowastlowast

119895

1199083119889119908 119889119908

(77)


0= infin

and 1199080


point 1199080

= infin 119867(119908) = int(119908211987532

)119889119908 = const sdot 119908minus3



119895and 119866

0





119895and 120579

lowastlowast


120579lowast

2= 0 sdot119908



119895lt 2 120591119895



= 0 we have 1198670(119908) =

int(119908211987532


1198620

1 119862

0


0

119894= 0 Here Lemma 11



0(119908211987532

)119889119908 = 0


3are regular



where


ℎ1

=3

2119887V minus VV

119875 (119908) = 21198861199083

+ 11986201199082

minus 119889

Δ (119875) = 4119889 (1198623

0minus 27119886

2119889) = 0

(78)


119895have critical


= 0 sdot 1199082


119895 120579lowastlowast119895






119892 (V) = minusVV + V2 + 119886V3 + 119887V

ℎ1

= minusVV2

119875 (119908) = 2 (1198861199083

+ 11986201199082

minus 119887119908)

Δ (119875) = 241198872

(1198622

0+ 4119886119887) = 0

(79)

At1199080



119895are 2 120579lowast

2= 0sdot119908

2+sdot sdot sdot

120590119895 120591119895




2= 119875(119908) but


10 Equation 1198754

Equation 1198754is

119891 (119909 119910)def= minus2119910119910

10158401015840+ 11991010158402

+ 31199104

+ 81199091199103

+ 4 (1199092

minus 119886) 1199102

+ 2119887

= 0

(80)


1and one edge Γ(1)


(2)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

+ 81199091199103

+ 411990921199102

= 0

(1)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

= 0

(81)


is N0

=





120573VV119906minus1 +

120572 (2 minus 120572)

1205732V2119906minus2

+8

120573119896V3119906minus119896 minus

4119886

120573119896+1V2119906minus(119896+1)

+4

1205732119896V2119906minus2119896 +

2119887

1205732(119896+1)119906minus2(119896+1)

= 0

119875 (119908) = 1199084

+ 1198620119908

1198620

= 0 119896 = 2 3

(82)



119865119895

=1

211987512

int119908

11987532int

120579lowast

119895

1199082119889119908 119889119908

119866119895

=1

2int

119908

11987532int

11987512

120579lowastlowast

119895

1199082119889119908 119889119908

(83)


= infin

and 1199080

= 0 Near 1199080

= infin 119867 = int(11990811987532

)119889119908 = const sdot

119908minus4




= 0 1198661

= (1205722120573)119908minus2


= minus(120572(120572 + 2)121205732) 119908minus1


= 0 1198653


3= 0 sdot 119908



=

minus1120573119896


= 0 119865119896+1

= (21198863120573119896+1

) 119908minus1

+ sdot sdot sdot 119866119896+1

=

(13120573119896+1

)119908minus3


119896+2= (4120572(2120572 minus 1)120573


120579lowastlowast

119896+2= 0 sdot 119908


119896+2= 4120572(2120572 minus 1)120573

119896+2= 0 only if



4+ 41199083

+ 41199082

+ 1198620119908

11 Equation 1198755

Equation 1198755is

119891 (119909 119910)def= minus 119909

2119910 (119910 minus 1) 119910

10158401015840+ 1199092 3119910 minus 1

211991010158402

minus 119909119910 (119910 minus 1) 1199101015840

+ (119910 minus 1)3

(1198861199102

+ 119887) + 1198881199091199102

(119910 minus 1)

+ 11988911990921199102

(119910 + 1) = 0

(84)




2+ 119889V2 (V + 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119889119908 [1198620

(119908 minus 1)2

+ 119908]

Δ (119875) = (2119889)4

1198622

0(1 minus 4119862

0) = 0

(85)


119865119895

= 11987512

int119908 (119908 minus 1)

2

11987532int

120579lowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

119866119895

= int119908 (119908 minus 1)

2

11987532int

11987512

120579lowastlowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

(86)


0=


= infin

119867 = int119908 (119908 minus 1)

2

11987532119889119908 = const sdot 119908



119895and 120579

lowastlowast


If 119886 = 0 then 120579lowast


4 and 1198652has


lt 4 and 120591119895



near 1199080



0(119908(119908 minus 1)

211987532

)119889119908 =

intinfin

1(119908(119908 minus 1)

211987532

)119889119908 = 0






2V2 + 119888V2 (V minus 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119888119908 (119908 minus 1) [1198620

(119908 minus 1) + 1]

Δ (119875) = (1198620

minus 1)2

= 0 1198620

= 0

(88)




12 Equation 1198756




= 0 1198993



13 Summary











References




















minus



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra











11991010158402

= 2 (1199103

+ 1198620) (70)


1areN119896

= (4 minus 119896 2(119896 +

1) 5) 119896 = 1 2 and 1198991

= 0 if 119896 = 4 Here 120572 = 2(119896 + 1)(4 minus




4120572 (120572 minus 1)

1205742V119906minus2 + 2

119896120574minus119896

119906minus119896

= 0 (71)

119875 = 2(1199083



= infin


minus72+


= minus1 and 120591119895


then

1198651

= 0

1198661

=120572

120574119908minus1

+ sdot sdot sdot

1198652

=120572 (120572 minus 8)


1198662

= 1198653

= 0

1198663

=120572 (120572 + 4)

31205743119908minus2

+ sdot sdot sdot

1198654

= minus

120572 (120572 + 4) (1205722

+ 24120572 + 48)

601205744119908minus1

+ sdot sdot sdot

1198664

= 1198655

= 0

1198665

=

120572 (120572 + 4) (31205723

+ 561205722

+ 200120572 + 192)

1801205745119908minus3

+ sdot sdot sdot

120579lowast

6= 0 sdot 119908


def= 1198606119908minus1

+ sdot sdot sdot

1198606

= 0

(72)



120579lowast

119896=

2119896


119865119896

= minus2119896

5120574119896119908minus1

+ sdot sdot sdot

119866119896

= 119865119896+1

= 0

119866119896+1

=(119896 + 11) 2

119896

75120574119896119908minus3

+ sdot sdot sdot

120579lowast

119896+2= 0 sdot 119908


(73)

q2

q3

1

0

2

2

q1

0

11Q

2Q

3Q

4Q

5Q

6Q




1 1199022)


120590119895

lt minus1 120591119895


3+ 119908 + 119862

0) but 119860


1has




9 Equation 1198753

Equation 1198753is

119891 (119909 119910)def= minus119909119910119910

10158401015840+ 11990911991010158402

minus 1199101199101015840+ 1198861199103

+ 119887119910 + 1198881199091199104

+ 119889119909

= 0

(74)



minus11990911991011991010158401015840

+ 11990911991010158402

+ 1198881199091199104

+ 119889119909 = 0 (75)




119892 (V) def= minusVV + V2 + 119888V4 + 119889

= 0

ℎ1

= minusVV + 119886V3 + 119887V

119875 (119908) = 1198881199084

+ 11986201199082

minus 119889

Δ (119875) =

minus119888119889 (1198622

0+ 4119888119889)

2

16= 0

(76)


119865119895

= 11987512

int1199082

11987532int

120579lowast

119895

1199083119889119908 119889119908

119866119895

= int1199082

11987532int

11987512

120579lowastlowast

119895

1199083119889119908 119889119908

(77)


0= infin

and 1199080


point 1199080

= infin 119867(119908) = int(119908211987532

)119889119908 = const sdot 119908minus3



119895and 119866

0





119895and 120579

lowastlowast


120579lowast

2= 0 sdot119908



119895lt 2 120591119895



= 0 we have 1198670(119908) =

int(119908211987532


1198620

1 119862

0


0

119894= 0 Here Lemma 11



0(119908211987532

)119889119908 = 0


3are regular



where


ℎ1

=3

2119887V minus VV

119875 (119908) = 21198861199083

+ 11986201199082

minus 119889

Δ (119875) = 4119889 (1198623

0minus 27119886

2119889) = 0

(78)


119895have critical


= 0 sdot 1199082


119895 120579lowastlowast119895






119892 (V) = minusVV + V2 + 119886V3 + 119887V

ℎ1

= minusVV2

119875 (119908) = 2 (1198861199083

+ 11986201199082

minus 119887119908)

Δ (119875) = 241198872

(1198622

0+ 4119886119887) = 0

(79)

At1199080



119895are 2 120579lowast

2= 0sdot119908

2+sdot sdot sdot

120590119895 120591119895




2= 119875(119908) but


10 Equation 1198754

Equation 1198754is

119891 (119909 119910)def= minus2119910119910

10158401015840+ 11991010158402

+ 31199104

+ 81199091199103

+ 4 (1199092

minus 119886) 1199102

+ 2119887

= 0

(80)


1and one edge Γ(1)


(2)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

+ 81199091199103

+ 411990921199102

= 0

(1)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

= 0

(81)


is N0

=





120573VV119906minus1 +

120572 (2 minus 120572)

1205732V2119906minus2

+8

120573119896V3119906minus119896 minus

4119886

120573119896+1V2119906minus(119896+1)

+4

1205732119896V2119906minus2119896 +

2119887

1205732(119896+1)119906minus2(119896+1)

= 0

119875 (119908) = 1199084

+ 1198620119908

1198620

= 0 119896 = 2 3

(82)



119865119895

=1

211987512

int119908

11987532int

120579lowast

119895

1199082119889119908 119889119908

119866119895

=1

2int

119908

11987532int

11987512

120579lowastlowast

119895

1199082119889119908 119889119908

(83)


= infin

and 1199080

= 0 Near 1199080

= infin 119867 = int(11990811987532

)119889119908 = const sdot

119908minus4




= 0 1198661

= (1205722120573)119908minus2


= minus(120572(120572 + 2)121205732) 119908minus1


= 0 1198653


3= 0 sdot 119908



=

minus1120573119896


= 0 119865119896+1

= (21198863120573119896+1

) 119908minus1

+ sdot sdot sdot 119866119896+1

=

(13120573119896+1

)119908minus3


119896+2= (4120572(2120572 minus 1)120573


120579lowastlowast

119896+2= 0 sdot 119908


119896+2= 4120572(2120572 minus 1)120573

119896+2= 0 only if



4+ 41199083

+ 41199082

+ 1198620119908

11 Equation 1198755

Equation 1198755is

119891 (119909 119910)def= minus 119909

2119910 (119910 minus 1) 119910

10158401015840+ 1199092 3119910 minus 1

211991010158402

minus 119909119910 (119910 minus 1) 1199101015840

+ (119910 minus 1)3

(1198861199102

+ 119887) + 1198881199091199102

(119910 minus 1)

+ 11988911990921199102

(119910 + 1) = 0

(84)




2+ 119889V2 (V + 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119889119908 [1198620

(119908 minus 1)2

+ 119908]

Δ (119875) = (2119889)4

1198622

0(1 minus 4119862

0) = 0

(85)


119865119895

= 11987512

int119908 (119908 minus 1)

2

11987532int

120579lowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

119866119895

= int119908 (119908 minus 1)

2

11987532int

11987512

120579lowastlowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

(86)


0=


= infin

119867 = int119908 (119908 minus 1)

2

11987532119889119908 = const sdot 119908



119895and 120579

lowastlowast


If 119886 = 0 then 120579lowast


4 and 1198652has


lt 4 and 120591119895



near 1199080



0(119908(119908 minus 1)

211987532

)119889119908 =

intinfin

1(119908(119908 minus 1)

211987532

)119889119908 = 0






2V2 + 119888V2 (V minus 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119888119908 (119908 minus 1) [1198620

(119908 minus 1) + 1]

Δ (119875) = (1198620

minus 1)2

= 0 1198620

= 0

(88)




12 Equation 1198756




= 0 1198993



13 Summary











References




















minus



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119892 (V) def= minusVV + V2 + 119888V4 + 119889

= 0

ℎ1

= minusVV + 119886V3 + 119887V

119875 (119908) = 1198881199084

+ 11986201199082

minus 119889

Δ (119875) =

minus119888119889 (1198622

0+ 4119888119889)

2

16= 0

(76)


119865119895

= 11987512

int1199082

11987532int

120579lowast

119895

1199083119889119908 119889119908

119866119895

= int1199082

11987532int

11987512

120579lowastlowast

119895

1199083119889119908 119889119908

(77)


0= infin

and 1199080


point 1199080

= infin 119867(119908) = int(119908211987532

)119889119908 = const sdot 119908minus3



119895and 119866

0





119895and 120579

lowastlowast


120579lowast

2= 0 sdot119908



119895lt 2 120591119895



= 0 we have 1198670(119908) =

int(119908211987532


1198620

1 119862

0


0

119894= 0 Here Lemma 11



0(119908211987532

)119889119908 = 0


3are regular



where


ℎ1

=3

2119887V minus VV

119875 (119908) = 21198861199083

+ 11986201199082

minus 119889

Δ (119875) = 4119889 (1198623

0minus 27119886

2119889) = 0

(78)


119895have critical


= 0 sdot 1199082


119895 120579lowastlowast119895






119892 (V) = minusVV + V2 + 119886V3 + 119887V

ℎ1

= minusVV2

119875 (119908) = 2 (1198861199083

+ 11986201199082

minus 119887119908)

Δ (119875) = 241198872

(1198622

0+ 4119886119887) = 0

(79)

At1199080



119895are 2 120579lowast

2= 0sdot119908

2+sdot sdot sdot

120590119895 120591119895




2= 119875(119908) but


10 Equation 1198754

Equation 1198754is

119891 (119909 119910)def= minus2119910119910

10158401015840+ 11991010158402

+ 31199104

+ 81199091199103

+ 4 (1199092

minus 119886) 1199102

+ 2119887

= 0

(80)


1and one edge Γ(1)


(2)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

+ 81199091199103

+ 411990921199102

= 0

(1)

1

def= minus2119910119910

10158401015840+ (1199101015840)2

+ 31199104

= 0

(81)


is N0

=





120573VV119906minus1 +

120572 (2 minus 120572)

1205732V2119906minus2

+8

120573119896V3119906minus119896 minus

4119886

120573119896+1V2119906minus(119896+1)

+4

1205732119896V2119906minus2119896 +

2119887

1205732(119896+1)119906minus2(119896+1)

= 0

119875 (119908) = 1199084

+ 1198620119908

1198620

= 0 119896 = 2 3

(82)



119865119895

=1

211987512

int119908

11987532int

120579lowast

119895

1199082119889119908 119889119908

119866119895

=1

2int

119908

11987532int

11987512

120579lowastlowast

119895

1199082119889119908 119889119908

(83)


= infin

and 1199080

= 0 Near 1199080

= infin 119867 = int(11990811987532

)119889119908 = const sdot

119908minus4




= 0 1198661

= (1205722120573)119908minus2


= minus(120572(120572 + 2)121205732) 119908minus1


= 0 1198653


3= 0 sdot 119908



=

minus1120573119896


= 0 119865119896+1

= (21198863120573119896+1

) 119908minus1

+ sdot sdot sdot 119866119896+1

=

(13120573119896+1

)119908minus3


119896+2= (4120572(2120572 minus 1)120573


120579lowastlowast

119896+2= 0 sdot 119908


119896+2= 4120572(2120572 minus 1)120573

119896+2= 0 only if



4+ 41199083

+ 41199082

+ 1198620119908

11 Equation 1198755

Equation 1198755is

119891 (119909 119910)def= minus 119909

2119910 (119910 minus 1) 119910

10158401015840+ 1199092 3119910 minus 1

211991010158402

minus 119909119910 (119910 minus 1) 1199101015840

+ (119910 minus 1)3

(1198861199102

+ 119887) + 1198881199091199102

(119910 minus 1)

+ 11988911990921199102

(119910 + 1) = 0

(84)




2+ 119889V2 (V + 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119889119908 [1198620

(119908 minus 1)2

+ 119908]

Δ (119875) = (2119889)4

1198622

0(1 minus 4119862

0) = 0

(85)


119865119895

= 11987512

int119908 (119908 minus 1)

2

11987532int

120579lowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

119866119895

= int119908 (119908 minus 1)

2

11987532int

11987512

120579lowastlowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

(86)


0=


= infin

119867 = int119908 (119908 minus 1)

2

11987532119889119908 = const sdot 119908



119895and 120579

lowastlowast


If 119886 = 0 then 120579lowast


4 and 1198652has


lt 4 and 120591119895



near 1199080



0(119908(119908 minus 1)

211987532

)119889119908 =

intinfin

1(119908(119908 minus 1)

211987532

)119889119908 = 0






2V2 + 119888V2 (V minus 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119888119908 (119908 minus 1) [1198620

(119908 minus 1) + 1]

Δ (119875) = (1198620

minus 1)2

= 0 1198620

= 0

(88)




12 Equation 1198756




= 0 1198993



13 Summary











References




















minus



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119865119895

=1

211987512

int119908

11987532int

120579lowast

119895

1199082119889119908 119889119908

119866119895

=1

2int

119908

11987532int

11987512

120579lowastlowast

119895

1199082119889119908 119889119908

(83)


= infin

and 1199080

= 0 Near 1199080

= infin 119867 = int(11990811987532

)119889119908 = const sdot

119908minus4




= 0 1198661

= (1205722120573)119908minus2


= minus(120572(120572 + 2)121205732) 119908minus1


= 0 1198653


3= 0 sdot 119908



=

minus1120573119896


= 0 119865119896+1

= (21198863120573119896+1

) 119908minus1

+ sdot sdot sdot 119866119896+1

=

(13120573119896+1

)119908minus3


119896+2= (4120572(2120572 minus 1)120573


120579lowastlowast

119896+2= 0 sdot 119908


119896+2= 4120572(2120572 minus 1)120573

119896+2= 0 only if



4+ 41199083

+ 41199082

+ 1198620119908

11 Equation 1198755

Equation 1198755is

119891 (119909 119910)def= minus 119909

2119910 (119910 minus 1) 119910

10158401015840+ 1199092 3119910 minus 1

211991010158402

minus 119909119910 (119910 minus 1) 1199101015840

+ (119910 minus 1)3

(1198861199102

+ 119887) + 1198881199091199102

(119910 minus 1)

+ 11988911990921199102

(119910 + 1) = 0

(84)




2+ 119889V2 (V + 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119889119908 [1198620

(119908 minus 1)2

+ 119908]

Δ (119875) = (2119889)4

1198622

0(1 minus 4119862

0) = 0

(85)


119865119895

= 11987512

int119908 (119908 minus 1)

2

11987532int

120579lowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

119866119895

= int119908 (119908 minus 1)

2

11987532int

11987512

120579lowastlowast

119895

1199082 (119908 minus 1)3119889119908 119889119908

(86)


0=


= infin

119867 = int119908 (119908 minus 1)

2

11987532119889119908 = const sdot 119908



119895and 120579

lowastlowast


If 119886 = 0 then 120579lowast


4 and 1198652has


lt 4 and 120591119895



near 1199080



0(119908(119908 minus 1)

211987532

)119889119908 =

intinfin

1(119908(119908 minus 1)

211987532

)119889119908 = 0






2V2 + 119888V2 (V minus 1)

ℎ1


ℎ2

= (V minus 1)3

(119886V2 + 119887)

119875 = minus2119888119908 (119908 minus 1) [1198620

(119908 minus 1) + 1]

Δ (119875) = (1198620

minus 1)2

= 0 1198620

= 0

(88)




12 Equation 1198756




= 0 1198993



13 Summary











References




















minus



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










References




















minus



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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