research article geometric singularities of the stokes problem

Research ArticleGeometric Singularities of the Stokes Problem

Nejmeddine Chorfi

Department of Mathematics College of Science King Saud University PO Box 2455 Riyadh 11451 Saudi Arabia

Correspondence should be addressed to Nejmeddine Chorfi nejmeddinechorfiyahoocom

Received 11 November 2013 Accepted 12 December 2013 Published 6 January 2014

Academic Editor Bessem Samet

Copyright copy 2014 Nejmeddine ChorfiThis 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

When the domain is a polygon ofR2 the solution of a partial differential equation is written as a sum of a regular part and a linearcombination of singular functions The purpose of this paper is to present explicitly the singular functions of Stokes problem Weprove the Kondratiev method in the case of the crack We finish by giving some regularity results

1 Introduction

The regularity of the solution of a partial differential equationdepends on the geometry of the domain even when the datais smooth Indeed for each corner of the polygonal domaina countable family of singular functions can be definedwhich depends only on the geometry of the domain Thenthe solution of the equation can be written as the sum of afinite number of singular functions multiplied by appropriatecoefficients and of a much more regular part We refer toKondratiev [1] and Grisvard [2] for their description

The purpose of our work is to study the singularities ofthe Stokes equation and the behavior of the solution in theneighborhood of a corner of a polygonal domain of R2 Weare interested to nonconvex domains we assume that thereexists an angle equal either to 31205872 or to 2120587 (case of thecrack) Handling the singular function is local process so thatthere is no restriction to suppose that the nonconvex corneris unique see Dauge [3] We deduce the singular functionof the velocity from those of the bilaplacian problem with ahomogenous boundary conditions by applying the curl oper-atorWe prove theKondratievmethod in the case of the crackThe singularities of the pressure are done by integration fromthe singular functions of the velocity near the corner

To approach these problems by a numerical method weneed to take into account the singular functions Severalnumeric methods have been proposed in this context see [4ndash9] Since the singular functions are developed for the Stokesproblem in this paper we intend in future work to implementStrang and Fix algorithm see [10] by the mortar spectral

element method It will be an extension of a work done onan elliptic operator [11 12]

An outline of this paper is as follows In Section 2 wepresent the geometry of the domain and the continuous pro-blem In Section 3 we give the singular functions and someregularity results The Kondratiev method is described inSection 4 Section 5 is devoted to the conclusion

2 The Continuous Problem

We suppose that Ω is a polygonal domain of R2 simplyconnected and has a connected boundary Γ Γ is the unionof vertex Γ

119895for 119895 isin 1 119869 119869 is positive integer Let 119886

119895be

the corner of Ω between Γ119895and Γ119895+1

120596119895is the measure of the

angle on 119886119895 We consider the velocity-pressure formulation of

the Stokes problem on the domainΩFind the velocity u and the pressure 119901 such that

minus]Δu + nabla119901 = f in Ω

div u = 0 in Ω

u = 0 on Γ

(1)

where ] is the viscosity of the fluid that we suppose a positifconstant and f is the data which represent a density of bodyforces Then for f in [119867

minus1(Ω)]2 the functional spaces are

[1198671

0(Ω)]2 for the velocity and 119871

2

0(Ω) for the pressure where

1198712

0(Ω) = 119902 isin 119871

2(Ω) int

Ω

119902 (119909) 119889119909 = 0 (2)

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 491326 8 pageshttpdxdoiorg1011552014491326

2 Abstract and Applied Analysis

The problem (1) is equivalent to the following variationalformulation

For f in [119867minus1(Ω)]2 find u in [119867

1

0(Ω)]2 and 119901 in 119871

2(Ω)

such that for all k in [1198671

0(Ω)]2 and for all 119902 in 119871

2(Ω)

119886 (u k) + 119887 (k 119901) = ⟨119891 k⟩

119887 (u 119902) = 0

(3)

where

119886 (u k) = ]intΩ

nablaunablak 119889119909

119887 (u 119902) = minusint

Ω

(div u) 119902 119889119909

(4)

where the ⟨sdot sdot⟩ denotes the duality pairing between 119867minus1(Ω)

and 1198671

0(Ω) The bilinear form 119886(sdot sdot) is continuous on the

space [11986710(Ω)]2times [1198671

0(Ω)]2 and elliptic on [119867

1

0(Ω)]2 also the

bilinear form 119887(sdot sdot) is continuous and verifies the followinginf-sup condition (see [13 14]) there exists a nonnull positiveconstant 120573 such that

forall119902 isin 1198712

0(Ω) sup

kisin[11986710(Ω)]2

119887 (k 119902)k[1198671(Ω)]2

ge 1205731003817100381710038171003817119902

10038171003817100381710038171198712(Ω) (5)

Then we conclude [15] that for all f in the space [119867minus1(Ω)]2

the problem (3) has a unique solution (u 119901) in [1198671

0(Ω)]2times

1198712

0(Ω)This solution verifies the following stability condition

u[1198671(Ω)]2 + 120573

100381710038171003817100381711990110038171003817100381710038171198712(Ω)

le 119862f[119867minus1(Ω)]2 (6)

where 119862 is a positive constantWe refer to the work of Pironneau [16] for themathemati-

calmodeling of problems resulting fromfluidsmechanics andGirault and Raviart [17] for the mathematical analysis of theNavier-Stokes equations

3 Singular Functions and Regularity Results

We recall that in an open simply connected ofR2 the condi-tion of incompressibility div(u) = 0 induces the existence ofa stream function 120593 in the space119867

2

0(Ω) such that

u = curl (120593) (7)

It thus brings to study the regularity of the function 120593solution of Dirichlet problem for bilaplacian

minus]Δ2120593 = curl (f) in Ω

120593 = 0 on Γ

120597120593

120597119899= 0 on Γ

(8)

The regularity of the solution of the problem is related to thegeometry of the domain and its behavior is local Let 119892 =

curl(f)We know see Cattabriga [18] and Ladyzhenskaya [14]the following theorem

Theorem 1 For 119892 in the space 119867119904(Ω) where 119904 ge minus2 then

120593 isin 119867119904+4

(Ω V) (9)

where V is the union of neighborhoods 119881119895of 119886119895for 119895 isin

1 119869

To study the function 120593 in the neighborhood of a fixedvertex 119886

119895 119895 isin 1 119869 it is convenient to introduce the polar

coordinates (119903 120579) centered at 119886119895 We start by enunciating the

characteristic equation of bilaplacian

sin2 120596119895119911 = 1199112sin2 120596

119895 (10)

Then we stated the following theorem We refer to ([17]Chapter 7Theorem 72112) andKondratiev [1] for the proof

Theorem 2 Let 120596119895in ]0 2120587[ For all 119892 in 119867

119904(Ω) 119904 ge minus2 the

solution 120593 of the problem (8) is written as

120593 = 120593119903+ 120593119904 (11)

where 120593119903is in the space (119867119904+4(Ω) cap 119867

2

0(Ω))

We set

120591119896(119903 120579) = 119903

1+119911119896120595119896(120579)

120583119896(119903 120579) = 119903

1+119896 (120590119896(120579) + log (119903) 120578

119896(120579))

(12)

120593119904is given by

120593119904= sum

0ltRe(119911119896)lt119904+2

120582119896120591119896(119903 120579) + sum

0ltRe(119896)lt119904+2

119896120583119896(119903 120579) (13)

where 120582119896and

119896are real constants120595

119896 120590119896 and 120578

119896belong to the

vectorial space Cinfin([0 120596119895]) cap 119867

2

0([0 120596119895]) of finite dimension

119911119896and 119896are respectively the simple and double roots of (10)

in the band 0 lt Re(119911119896) lt 119904 + 2 excepting 1 if 120596

119895= 119905119892 (120596

119895)

without exception if 120596119895= 119905119892(120596

119895)

We called 120596119890the unique solution of equation 120596 = 119905119892 (120596)

in the ]0 2120587[ (120596119890≃ 1430297120587) We prove see ([19] chapter

3 sect 33) that 119911 is a double root of (10) if and only if 119911 = 0 or119911 = plusmnradic(1 sin120596

2

119895) minus (1120596

2

119895) Hence the sufficient condition

on the angle 120596 for a double root is

sin(radic

1205962

119895

(sin120596119895)2minus 1) = plusmnradic(1 minus

(sin120596119895)2

1205962

119895

)

120596119895isin ]0 2120587[

(14)

Then if 120596119895is not a solution of (14) (10) takes the following

simplified form

120593119904(119903 120579) = sum

0ltRe(119911119896)lt119904+2

120582119896120591119896(119903 120579) (15)

In the following we suppose that Ω has a unique vertex 119886

where 120596 is equal to 31205872 or 2120587 and the other angles are 1205872

Abstract and Applied Analysis 3

We introduce 119881 as a neighborhood of 119886 All these angles aredifferent of 120596

119890

In the case where 120596 = 31205872 and 119904 = minus1 (10) has tworeal simple roots in the band 0 lt Re(119911) lt 1 We apply theNewton method to approximate those roots 119911

1≃ 0544484

and 1199112

≃ 0908529 The functions 1205911and 120591

2are written as

follows

1205911(119903 120579) = 119903

1+11991111205951(120579)

1205912(119903 120579) = 119903

1+11991121205952(120579)

(16)

with

120595119894(120579) = ((119911

119894minus 1)minus1 sin(

3 (119911119894minus 1) 120587

2) minus (119911

119894+ 1)minus1

times sin(3 (119911119894+ 1) 120587

2))

times (cos ((119911119894minus 1) 120579) minus cos ((119911

119894+ 1) 120579))

minus((119911119894minus 1)minus1 sin ((119911

119894minus1) 120579)minus(119911

119894+1)minus1 sin ((119911

119894+1) 120579))

times(cos(3 (119911119894minus 1) 120587

2) minus cos(

3 (119911119894+ 1) 120587

2))

(17)

Proposition 3 For all 120598 ge 0 the functions 120591119894 119894 in 1 2 are

belonging to the space 119867(2+119911119894)minus120598

(119881) and are solutions of theproblem

Δ2120591119894= 0 119894119899 119881

120591119894=

120597120591119894

120597119899= 0 119900119899 Γ cap 120597119881

(18)

where Γ is the boundary of Ω

Proof Since 120595119894is inC0(]0 31205872[) we find 119904 such that

1199031+119911119894120595119894(120579) isin 119867

119904(119881) (19)

Then we find (119898 119902) in N timesR119898 gt 119904 and 1 lt 119902 le 2 such that

1199031+119911119894120595119894(120579) isin 119882

119898

119902(119881) sub 119867

119904(119881) (20)

From the Sobolev injection theorem we have 119898 minus 2 = 119904 minus 1If 1199031+119911119894120595

119894(120579) isin 119882

119898

119902(119881) then 1 + 119911

119894gt 119898 minus 2119902 Since 1 lt 119902 le

2 we can take only the value 119864(3 + 119911119894) = 3 (0 lt 119911

119894lt 1)

Consequently 119904 lt (2 + 119911119894) for 119894 isin 1 2 And by construction

we obtain that 120591119894 for 119894 isin 1 2 is the solution of problem

(18)

Corollary 4 For all 119891 in [1198712(Ω)]2 the velocity u is written as

u = u119903+ u119904 (21)

where u119903is in [119867

2(Ω)cap119867

1

0(Ω)]2 and there exists a constant 120582

119894

such that

u119904= sum

1le119894le2

120582119894119904119894 (22)

where 119904119894(119903 120579) = curl(120591

119894(119903 120579)) The functions 119904

119894belong to the

space [119867(1+119911119894)minus120598(119881)]2 119894 in 1 2 for all 120598 gt 0

4 Kondratievrsquos Method Case of the Crack

In the case of the crack there are no known results we recallthe method of kondratiev We extend this method to the caseof crack We consider the polar coordinates and a truncationfunction 120594with compact support that does not intersect withthe boundary Γ We define

119866 = 119903119890119894120579 0 lt 120579 lt 2120587 119903 gt 0 (23)

Let 120595 = 120594120593 then 120595 is in the space 1198672

0(119866) with a compact

support included in 119866 then 120595 is the solution of the problem

]Δ2120595 = 119892 in 119867119904(119866)

120595 isin 1198672

0(119866) with compact support

(24)

Definition 5 For 119904 ge 0 one defines the weight Sobolev space

119885119904

2(119866) = 120595 isin 119871

2(119866) 119903

minus119904+|120572|119863120572120595 isin 119871

2(119866) |120572| le 119904

119885minus1

2(119866) = 120595 isin D

1015840(119866) 120595 =

1198920

119903+ 1205971199091198921+ 1205971199101198922

such that 1198920 1198921and 119892

2isin 1198712(Ω)

(25)

Remark 6 We remark that 1198850

2(119866) = 119885

2(119866) and 119867

119904

0(119866) sub

119885119904

2(119866) sub 119867

119904(119866)

In the following we suppose that 119892 is in the space 119885119904

2(119866)

for 119904 in minus1 0 The solution 120595 of the problem

]Δ2120595 = 119892 in 119867119904(119866)

119906 isin 1198852

2(119866)

(26)

is with a compact support then there exists a real number 119877such that

120595 (119903 120579) = 0 for 119903 ge 119877 (27)

The Kondratievrsquos method consists to change the variable119903 = 119890

119905 then we replace the domain 119866 by the domain 119861 =

Rtimes]0 2120587[ and the weight Sobolev space by the ordinaryone We apply the Fourier transformation relative to the firstvariable of the problem (26)

Proposition 7 If 120595 is in the space 1198752

2(119866) the function V =

119890minus119905120595(119890119905 cos 120579 119890119905 sin 120579) belongs to 119867

2(119861)

For 119904 isin minus1 0 if 119892 belongs to 119875119904

2(119866) and let ℎ =

1198903119905120595(119890119905 cos 120579 119890119905 sin 120579) then the function 119890

minus(119904+2)119905ℎ is in the space

119867119904(119861)

Proof Let 119904 = minus1


Since 119892 isin 119875minus1

2(119866) then 119892 = 119892

0119903+1205971199091198921+1205971199101198922with 119892

0 1198921

and 1198922in 1198712(119866)

We denote

119865 (119905 120579) = 119892 (119890119905 cos 120579 119890119905 sin 120579)

119866119895(119905 120579) = 119892

119895(119890119905 cos 120579 119890119905 sin 120579) for 119895 isin 0 1 2

119865 (119905 120579) = 119890minus1199051198660+ (119890minus119905 cos 120579120597119866

1

120597119905minus 119890minus119905 sin 120579

1205971198661

120597120579)

+ (119890minus119905 sin 120579

1205971198662

120597119905minus 119890minus119905 cos 120579120597119866

2

120597120579)

(28)

We have

119890minus119905ℎ = 1198902119905119865 (119905 120579) = 119890

1199051198660+ 120597119905(119890119905 cos 120579119866

1+ sin 120579119890

1199051198662)

+ 120597120579(minus119890119905 sin 120579119866

1+ cos 120579119890119905119866

2)

(29)

and if

1198700= 1198660

1198701= cos 120579119866

1+ sin 120579119866

2

1198702= minus sin 120579119866

1+ cos 120579119866

2

(30)

then we conclude

119890minus119905ℎ = 1198901199051198700+ 1205971199051198901199051198701 + 1205971205791198901199051198702 (31)

since 119892119895

isin 1198712(119866) et 119890119905119866

119895isin 1198712(119861) then 119890119905119870

119895isin 1198712(119861) for

119895 isin 0 1 2 So

119890minus119905ℎ isin 119867

minus1(119861) (32)

We end the proof

As 120595 is the solution of the problem (26) then for ℎ = 1198903119905119892

the function V = 119890minus119905120595(119890119905 cos 120579 119890119905 sin 120579) is a solution of the

problem

(1198634

119905minus 1198632

119905+ 1) V + 2 (119863

2

119905+ 1)119863

2

120579V + 119863

4

120579V = ℎ

V (119905 0) = V (119905 2120587) = 120597119905V (119905 0) = 120597

120579V (119905 2120587) = 0

V isin 1198672(119861)

(33)

We have V(119905 120579) = 0 for 119905 ge log(119877) = 1199050

The Fourierrsquos transformation on the variable 119905 of thefunction V is

V (119911 119905) =1

radic2120587

int

R

119890minus119894119905119911V (119905 119911) 119889119905 (34)

119911 is a complex variable we denote

119871 (119911119863120579) = (119911

4+ 21199112+ 1) + 2 (1 minus 119911

2)1198632

120579+ 1198634

120579 (35)

The function V becomes a solution of the problem

119871 (119911119863120579) V = ℎ

V (119905 0) = V (119905 2120587) = 120597119905V (119905 0) = 120597

120579V (119905 2120587) = 0

(36)

For all 119911 such that Im(119911) ge 0 the operator 119871(119911 119863120579) has

1199014+ 2 (1 minus 119911

2) 1199012+ (1199114+ 21199112+ 1) = 0 (37)

as a characteristic polynomial This polynomial function hastwo roots 119901 = plusmn119911 and 119901 = plusmn119894

(i) If 119911 = 0 and 119911 = plusmn 119894 then the fundamental solutions ofthe fourth order partial differential equation are

sin 120579119904ℎ120579119911 sin 120579119888ℎ119911120579 cos 120579119888ℎ120579119911 et cos 120579119904ℎ119911120579(38)

(ii) If 119911 = 0 the solutions have the form

sin 120579 cos 120579 120579 sin 120579 et 120579 cos 120579 (39)

(iii) If 119911 = plusmn119894 the solutions have the form

1 120579 sin 2120579 et cos 2120579 (40)

Then let the following homogeneous problem

119871 (119911119863120579) 120593 = 0 in ]0 2120587[

120593 (0) = 120593 (2120587) = 120597119905120593 (0) = 120597

120579120593 (2120587) = 0

(41)

We note 119864119911

= 120593119911

isin 1198672

0(]0 2120587[) 120593

119911solution of the

problem (41)

Proposition 8 The set 119864119911satisfies the following properties

(1) if 119911 = 0 and 119911 = plusmn 119894 where 119904ℎ2120587119911 = 0 then

119864119911= 0 (42)

(2) if 119911 plusmn 119894 then

119864119911= 120593119911isin 1198672

0(]0 2120587[) 120593119911 = 120572

119911sin2120579 120572

119911isin C (43)

(3) if 119911 = 0 119911 = plusmn 119894 and 119904ℎ2120587119911 = 0 which implies that 119911 isequal to minus119894(1198962) where 119896 = 0 and 119896 = 2 then

119864119911= 120593119911isin 1198672

0([0 2120587]) 120593

119911= 120572119896119892119896(120579) + 120573

119896ℎ119896(120579)

(120572119896 120573119896) isin C2

119892119896(120579) = sin(1 +

119896

2) 120579 minus (

119896 + 2

119896 minus 2) sin(minus1 +

119896

2) 120579

ℎ119896= cos(1 +

119896

2) 120579 minus cos(minus1 +

119896

2) 120579

(44)

Proof If 119911 = 0 and 119911 = plusmn 119894

120593 = 120572 sin 120597119904ℎ119911120579 + 120573 sin 120579119888ℎ119911120579 + 120574 cos 120579119904ℎ119911120579 + 120575 cos 120579119888ℎ119911120579(45)


is solution of the problem (41) Since 120593(0) = 120593(2120587) =

120597119905120593(0) = 120597120579120593(2120587) = 0 we obtain the following system

120575 = 0

120574119904ℎ2120587119911 + 120575119888ℎ2120587119911 = 0

120573 = 0

120572119904ℎ2120587119911 + 120573119888ℎ2120587119911 + 2120587120574119904ℎ2120587119911 + 2120587120575119888ℎ2120587119911 = 0

(46)

its determinant is equal to 119904ℎ22120587119911 Then if 119904ℎ2120587119911 = 0 119864

119911=

0 Otherwise the system has the following complexes roots119911119896= minus119894(1198962) for 119896 = 0 and 119896 = 2 119896 isin N The space of solutions

of the homogeneous equation is of dimension 2 Indeed for119911 = 119911119896 we obtain

120593 =120572

2119894ℎ119896(120579) + 119894120574 (

119896

2minus 1)119892

119896(120579) (47)

We check that (ℎ119896 119892119896) are two independent solutions of

the problem (41) then

int

2120587

0

1003816100381610038161003816ℎ1198961003816100381610038161003816 119889120579 = int

2120587

0

10038161003816100381610038161198921198961003816100381610038161003816 119889120579 = 1 (48)

This ends the proof of 1 and 3Now we prove property 2If 119911 = plusmn119894

120593 = 120572 + 120573120579 + 120574 sin 2120579 + 120575 cos 2120579 (49)

is solution of the problem (41) which gives the followingsystem

120572 + 120575 = 0

120572 + 2120587120573 + 120575 = 0

120573 + 2120574 = 0

(50)

The determinant of this system is zero then the dimension of119864119911is equal to 1 Indeed

120593 = 120572 (1 minus cos 2120579) = 2120572 sin2 2120579 (51)

This completed the proof

Remark 9 If

119863119896= 119911119896= minus119894

119896

2 for 119896 = 0 119896 = 2 119896 isin N (52)

119863119896is the set of nonregular values Since ℎ is analytic for

Im(119911) gt minus(119904 + 2) then V analytically extends across Im(119911) gt

minus(119904 + 2) cap 119863119896

Is then

119889 (119911) =minus119904ℎ22120587119911

1199112 (1 + 1199112) for 119911 = 0 119911 = plusmn 119894 (53)

We verify that 119889(119911119896) = 120597

119911119889(119911119896) = 0 and 120597

2

119911119889(119911119896) = 0 Hence

the multiplicity of 119911119896is 2 for 119896 = 0 and 119896 = 2 119889(plusmn119894) = 0 then

(plusmn119894) are simple solutions

Theorem 10 (1) The solution V of the problem (36) can bewritten in the neighborhood of minus119894

V (119911 120579) =1205931(120579)

(119911 + 119894)+ 1199081(119911 120579) (54)

where 1205931is the solution of problem (41) with 119911 = (minus119894) and 119908

1

is an analytic function in the neighborhood of minus119894with values in1198672(]0 2120587[)(2) The solution V of the problem (36) is written in the

neighborhood of (minus119894(1198962)) 119896 = 1 or 119896 = 3 as

V (119911 120579) =120595119896(120579)

(119911 + 119894 (1198962))2

+120593119896(120579)

(119911 + 119894 (1198962))+ 119908119896(119911 120579) (55)

where 120595119896and 119908

119896satisfy the same properties as above and 120593

119896

satisfies

119871 (119911119863120579) 120593119896= 2119894119896(1 minus

1198962

4)120595119896minus 2119894119896120595

10158401015840

119896

120593119896(0) = 120593

119896(2120587) = 120597

119905120593119896(0) = 120597

120579120593119896(2120587) = 0

(56)

Proof We verified that any solution to the problem (41) is alinear combination of

1199061(119911 120579) =

sin 120579119904ℎ119911120579

119911for 119911 = 0

1199061(0 120579) = 120579 sin 120579

1199062(119911 120579) =

1

1199112 + 1(cos 120579119904ℎ119911120579

119911minus sin 120579119888ℎ119911120579)

for 119911 = 0 119911 = plusmn 119894

1199062(0 120579) = 120579 cos 120579 minus sin 120579

1199062(plusmn119894 120579) =

1

2(sin 2120579

2minus 120579)

(57)

In fact if 119911 = minus119894(1198962) for 119896 = 1 or 119896 = 3 we have

1199061(minus119894

119896

2 120579) = minus

1

119896ℎ119896(120579)

1199062(minus119894

119896

2 120579) =

2

119896 (119896 + 2)119892119896(120579)

(58)

Therefore any solution of the problem (41) is linear combi-nation of 119906

1and 119906

2 We also verified that 119906

1and 119906

2are whole

solutions with respect to the variable 119911 and if we set

119889119911= 1199061(119911 2120587)119863

1205791199062(119911 2120587) minus 119906

2(119911 2120587)119863

1205791199061(119911 2120587)

(59)

we note that 119889(119911119896)V(119911119896 120579) = 0This implies that V is analytical

in the neighborhood of 119911119896= minus119894(1198962) for 119896 in 1 2 3 (minus119894) is

a single root of 119889 then V admits a simple pole at (minus119894) On theother side minus1198942 and (minus32)119894 are roots of multiplicity two of 119889and poles of multiplicity two of V

In the neighborhood of (minus119894) V(119911 120579) = (1205951(120579)(119911 minus 119894)) +

1199081(119911 120579) and for all 119911

(119911 minus 119894) ℎ = 119871 (minus119894 119863120579) V (119911 minus 119894)

= 119871 (minus119894 119863120579) 1205951+ (119911 minus 119894) 119871 (minus119894 119863

120579) 1199081

(60)


Then for 119911 = minus119894 we have

119871 (minus119894 119863120579) 1205951= 0

1205951(0) = (119911 minus 119894) 119908

1(119911 0)

(61)

Then

1205951(0) = 0 (62)

and even

1205951(2120587) = 0

1205951015840

1(0) = lim

120579rarr0

120595 (120579)

120579= lim120579rarr0

V (119911 120579) minus (119911 + 1)1199081(119911 120579)

120579= 0

(63)

and even for

1205951015840

1(2120587) = 0 (64)

Thus 1205951is a solution of the problem (41) at 119911 = minus119894

The expression of V in the neighborhood of 119911119896= minus119894(1198962)

for 119896 = 1 and 119896 = 3 is

V (119911 120579) =120595119896(120579)

(119911 + 119894 (1198962))2+

120593119896(120579)

(119911 + 119894 (1198962))+ 119908119896(119911 120579)

for 119896 = 1 119896 = 3

(65)

Then for 119911 in the neighborhood of 119911119896

ℎ(119911 minus 119911119896)2

= 119871 (119911119863120579) 120595119896+ (119911 minus 119911

119896) 119871 (119911 119863

120579) 120593119896

+ (119911 minus 119911119896)2

119871 (119911119863120579) 119908119896

(66)

For 119911 = 119911119896 we obtain that 119871(119911

119896 119863120579)120595119896= 0 and we verify that

120595119896is solution of the problem (41) And we have

119871 (119911119863120579) 120593119896= minus

119871 (119911119863120579) + 119871 (119911119896 119863120579)

(119911 minus 119911119896)

120595119896

+ (119911 minus 119911119896) (ℎ minus 119871 (119911 119863120579)119908

119896)

(67)

And if 119911 tends to 119911119896

119871 (119911119863120579) 120593119896= 2119894119896(1 minus

1198962

4)120595119896minus 2119894119896120595

10158401015840

119896(68)

and 120593119896verifies the boundary conditions which completes the

proof

Remark 11 We remark that the operator 119871(119911119863120579) is self-

adjoint

⟨119871120595119896 120593119896⟩ = ⟨120595

119896 119871120593119896⟩

= int

2120587

0

120595119896(2119896(1 minus

1198962

4)120595119896minus 2119894119896120595

10158401015840

119896)119889120579

(69)

For 0 lt 119896 lt 4

int

2120587

0

1205952

119896119889120579 = int

2120587

0

12059510158402

119896119889120579 = 0 (70)

This implies that 120595119896= 0 thus

V (119911 120579) =120593119896(120579)

(119911 + 119894 (1198962))+ 119908119896(119911 120579)

for 119896 = 1 or 119896 = 3

(71)

where 120593119896is solution of problem (41)

Let 120578 be in R such that 0 lt 120578 lt 12 We note

119908119904(119905 120579) =

1

radic2120587

int

+infin

minusinfin

119890119894119905119911V (119911 minus 119894 (119904 + 2) + 119894120578 120579) 119889119911

120583 (120579) = 119894radic2120587119890119894119905119911V (119911 120579) for fixed 119905

(72)

We prove by Cauchyrsquos theorem that

V (119905 120579) = 119890(119904+2minus120598)119905

119908119904(119905 120579) + 2119894120587 sum

minus(119904+2)ltIm(119911)lt0119877119904 (73)

120583 is an analytic function on 119863119896 119877119904is its residue on the poles

119911119896 Then

(i) for 119904 = minus1

V (119905 120579) = 119890(1minus120578)119905

119908minus1

(119905 120579) + 119894radic21205871205951(120579) (74)

(ii) for 119904 = 0

V (119905 120579) = 119890(2minus120578)119905

1199080(119905 120579)

+ 119894radic2120587 (11989031199052

1205953(120579) + 119890

11990521205951(120579) + 119890

1199051205952(120579))

(75)

where 120595119894are solutions of the problem (36)

Notation Using the variables (119903 120579) we introduce the follow-ing notations

For 119904 = minus1 we define the two singular functions

1205911(119903 120579) = 119903

32(sin 3

2120579 minus 3 sin 120579

2)

1205831(119903 120579) = 119903

32(cos 3

2120579 minus cos 120579

2)

(76)

For 119904 = 0 we define four singular functions

1205911(119903 120579) = 119903

32(sin 3

2120579 minus 3 sin 120579

2)

1205831(119903 120579) = 119903

32(cos 3

2120579 minus cos 120579

2)

1205912(119903 120579) = 119903

52(sin 5

2120579 minus 5 sin 120579

2)

1205832(119903 120579) = 119903

52(cos 5

2120579 minus cos 120579

2)

(77)


Proposition 12 For 120598 gt 0 1205911and 120583

1belong to the space

11986752minus120598

(119881) 1205912and 120583

2belong to the space 119867

72minus120598(119881) and They

are solutions to the problem

Δ2120593 = 0 119894119899 119881

120593 =120597120593

120597119899= 0 119900119899 Γ cap 120597119881

(78)


Proof It is the same prove that in the case 120596 = 31205872

For the Stokes problem we handle the case 119904 = minus1 thus fbelongs to the space [119871

2(Ω)]2

Corollary 13 For f in [1198712(Ω)]2 the velocity is written in the

form

u = u119903+ u119904 (79)

where u119903is in [119867

2(Ω) cap 119867

1

0(Ω)]2 and there exist two real

constants 120582 and such that

u119904= 1205821199041+ 1199041

(80)

with

1199041(119903 120579) = 119903

12(3 sin 120579 sin 120579

2 3 (1 minus cos 120579) sin 120579

2)

1199041(119903 120579) = 119903

12(2 sin 120579

2+ sin 120579 cos 120579

2 (1 minus cos 120579) cos 120579

2)

(81)

1199041and 1199041belong to the [11986732minus120598(119881)]

2 for all 120598 positive

For handling the singularities of the pressure we define

120578 (120596) = inf Re (119911) 119911 is solution of (10) (82)

Indeed for 120596 isin]0 2120587] using the Newton method we canobtain a good approximation of roots of (10) See [20] for theapproximation of 120578(120596) Thus

if 120587 lt 120596 lt 2120587 1 minus120587

120596lt 120578 (120596) lt

120587

120596 120578 (120596) gt

1

2 (83)

To find the pressure we note that f minus ]Δu belongs to[119867minus1(Ω)]2cap 119882perp where

119882 = u isin [1198671

0(Ω)]2

such that div (u) = 0 (84)

Then see [15] there exists a unique function 119901 in 1198712(Ω)

defined such that

nabla119901 = f + ]Δu in 1198631015840(Ω) (85)

From this equality we determine the singularities of the pres-sure from the singularities of the velocity

Then from a regular data f

119901 = 119901119903+ 119901119904 119901119903isin 1198671(Ω) 119901

119904= 1205731198781199011 (86)

where 120573 is first singular coefficientThe linearity of (85) givesus

nabla1198781199011

= ]Δ (curl (1199031+120578(120596)120595 (120579))) (87)

and since we explicitly know the singularities of the velocitywe can deduce by simple integration the singularities of thepressure

Consider

1198781199011

(119903 120579)

=

minus119903120578(120596)minus1

((1 + 120578 (120596))2

(120597120595 (120579) 120597120579) + (1205973120595 (120579) 120597120579))

(1 + 120578 (120596))

(88)

in the case of crack

120595 (120579) = 3 sin(120579

2) minus sin(

3

2120579) 120578 (120596) =

1

2 (89)

in the case of 120596 = 31205872

120595 (120579) =sin ((1 + 120578 (120596)) 120579) cos (120578 (120596) 120596)

(1 + 120578 (120596))

minus cos ((1 + 120578 (120596)) 120579)

+sin ((120578 (120596) minus 1) 120579) cos (120578 (120596) 120596)

(1 minus 120578 (120596))

+ cos ((120578 (120596) minus 1) 120579)

(90)

and 120578(120596) = 0544484

Proposition 14 For 120598 gt 0 1198781199011

belongs to 1198670544484

(119881) when120596 = 31205872 and 119878

1199011belongs to 119867

05(119881) when 120596 = 2120587

Proof It is the same proof that in the case of Proposition 3

5 Conclusion

We summarize below the regularity results for the Stokes pro-blem previously obtained

(1) If f belongs to [119867119904minus2

(Ω)]2 where 119904 gt 0 then (u 119901)

belongs to the space [119867119904(Ω)]2times119867119904minus1

(Ω) if 119904 lt 1+120578(120596) Thismeans

[119867119904(Ω)]2

times 119867119904minus1

(Ω) for 119904 lt 1544484 when 120596 =3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 15 when 120596 = 2120587

(91)

(u 119901) satisfies the following stability condition

u[119867119904(Ω)]2 +

10038171003817100381710038171199011003817100381710038171003817119867119904minus1(Ω)

le 119862f[119867119904minus2(Ω)]2 (92)


(2)We know from Corollaries 4 and 13 and formula (86)that if 120596 = 2120587 (u 119901) is written

u = u119903+ 1205821198781 119901 = 119901

119903+ 1205731198781199011 (93)

and if 120596 = 2120587

u = u119903+ 12058211198781+ 11198781 119901 = 119901

119903+ 12057311198781199011

+ 12057311198781199011 (94)

If f belongs to [119867119904minus2

(Ω)]2 (u119903 119901119903) belongs to [119867

119904(Ω)]2times

119867119904minus1

(Ω) for 119904 lt 1 + 1205781(120596) where 120578

1(120596) is the second real

solution of (10) in the band 0 lt Re(119911) lt 119904 hence

[119867119904(Ω)]2


(Ω) for 119904 lt 1908529 when 120596 =3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 25 when 120596 = 2120587

(95)

We have the following stability condition1003817100381710038171003817u119903

1003817100381710038171003817[119867119904(Ω)]2+1003817100381710038171003817119901119903

1003817100381710038171003817119867119904minus1(Ω)+10038161003816100381610038161205821

1003816100381610038161003816 +10038161003816100381610038161205731

1003816100381610038161003816 le 119862f[119867119904minus2(Ω)]2

(96)


(Ω)]2 we can further decompose

the regular part (u119903 119901119903) of the solution as follows

u119903= u119903+ 12058221198782 119901

119903= 119901119903+ 12057321198781199012

(97)

(u119903 119901) isin [119867

119904(Ω)]2times 119867119904minus1

(Ω) for 119904 lt 1 + 1205782(120596) where 120578

2(120596)

is the third reel solution of (10) in the band 0 lt Re(119911) lt 119904then

[119867119904(Ω)]2


(Ω) for 119904 lt 1205782(3120587

2) when 120596 =

3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 3 5 when 120596 = 2120587

(98)

(1205822 1205732) is the singular coefficient associated to the second

singular function (1198782 1198781199012) then

1003817100381710038171003817u1199031003817100381710038171003817[119867119904(Ω)]2

+1003817100381710038171003817119901119903

1003817100381710038171003817119867119904(Ω)+10038161003816100381610038161205821

1003816100381610038161003816 +10038161003816100381610038161205822

1003816100381610038161003816 +10038161003816100381610038161205731

1003816100381610038161003816 +10038161003816100381610038161205732

1003816100381610038161003816

le 119862f[119867119904minus2(Ω)]2

(99)

In general we can decompose (u 119901) as

u = u119903+ 12058211198781+ 12058221198782+ sdot sdot sdot + 120582

119896119878119896

119901 = 119901119903+ 12057311198781199011

+ 12057321198781199012

+ sdot sdot sdot + 120573119896119878119901119896

(100)

where 119896 is an integer numberIf f belongs to [119867

119904minus2(Ω)]2 (u119903 119901119903) belongs to [119867

119904(Ω)]2times

119867119904minus1

(Ω) for 119904 lt 1 + 120578119896(120596) where 120578

119896(120596) 119896-eme reel solution

of (10) in the band 0 lt Re(119911) lt 119904

Conflict of Interests

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

Acknowledgment

This project was supported by King Saud University Dean-ship of Scientific Research College of Science ResearchCenter

References

[1] V A Kondratiev ldquoBoundary value problems for elliptic equa-tions in domains with conical or angular pointsrdquo Transactionsof the Moscow Mathematical Society vol 16 pp 227ndash313 1967

[2] P Grisvard Elliptic Problems in Nonsmooth Domains vol 24Pitman Boston Mass USA 1985

[3] M Dauge Elliptic Boundary Value Problems on CornerDomains vol 1341 of Lecture Notes in Mathematics SpringerBerlin Germany 1988

[4] M Amara C Bernardi and M A Moussaoui ldquoHandlingcorner singularities by the mortar spectral element methodrdquoApplicable Analysis vol 46 no 1-2 pp 25ndash44 1992

[5] S Agmon A Douglis and L Nirenberg ldquoEstimates near theboundary for solutions of elliptic partial differential equationssatisfying general boundary conditionsrdquo Communications onPure and Applied Mathematics vol 12 pp 623ndash727 1959

[6] M Amara andMMoussaoui ldquoApproximation de coefficient desingularitesrdquo Comptes Rendus de lrsquoAcademie des Sciences I vol313 no 5 pp 335ndash338 1991

[7] I Babuska ldquoFinite element method for domains with cornersrdquoComputing vol 6 no 3-4 pp 264ndash273 1970

[8] J Lelievre ldquoSur les elements finis singuliersrdquoComptes Rendus delrsquoAcademie des Sciences A vol 283 no 15 pp 1029ndash1032 1967

[9] M Suri ldquoThe 119901-version of the finite element method for ellipticequations of order 2119897rdquo ESAIM vol 24 no 2 pp 265ndash304 1990

[10] G Strang andG J FixAnAnalysis of the Finite ElementMethodPrentice-Hall Englewood Cliffs NJ USA 1973

[11] N Chorfi ldquoHandling geometric singularities by the mortarspectral element method case of the Laplace equationrdquo SIAMJournal of Scientific Computing vol 18 no 1 pp 25ndash48 2003

[12] F Ben Belgacem C Bernardi N Chorfi and Y Maday ldquoInf-sup conditions for the mortar spectral element discretization ofthe Stokes problemrdquoNumerische Mathematik vol 85 no 2 pp257ndash281 2000

[13] I Babuska ldquoThe finite element method with Lagrangian multi-pliersrdquoNumerischeMathematik vol 20 no 3 pp 179ndash192 1973

[14] O A Ladyzhenskaya The Mathematical Theory of ViscousIncompressible Flow Gordon and Breach Science PublishersNew York NY USA 1962

[15] R Temam Theory and Numerical Analysis of the Navier-StokesEquations North-Holland Amsterdam The Netherlands 1977

[16] O Pironneau Methode des elements finis pour les fluidesMasson Paris France 1988

[17] V Girault and P A Raviart Finite Element Methods for Navier-Stokes Equations Theory and Algorithms vol 5 Springer NewYork NY USA 1986

[18] L Cattabriga ldquoSu un problema al contorno relativo al sistema diequazioni di Stokesrdquo Rendiconti del Seminario Matematico dellaUniversita di Padova vol 31 pp 1ndash33 1961

[19] P Grisvard Singularities in Boundary Value Problems vol 22Springer Amesterdam The Netherlands 1992

[20] C Bernardi and G Raugel ldquoMethodes drsquoelements finis mixtespour les equations de Stokes et de Navier-Stokes dans unpolygone non convexerdquo Calcolo vol 18 no 3 pp 255ndash291 1981

Submit your manuscripts athttpwwwhindawicom

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Stochastic AnalysisInternational Journal of


The problem (1) is equivalent to the following variationalformulation

For f in [119867minus1(Ω)]2 find u in [119867

1

0(Ω)]2 and 119901 in 119871

2(Ω)

such that for all k in [1198671

0(Ω)]2 and for all 119902 in 119871

2(Ω)

119886 (u k) + 119887 (k 119901) = ⟨119891 k⟩

119887 (u 119902) = 0

(3)

where

119886 (u k) = ]intΩ

nablaunablak 119889119909

119887 (u 119902) = minusint

Ω

(div u) 119902 119889119909

(4)

where the ⟨sdot sdot⟩ denotes the duality pairing between 119867minus1(Ω)

and 1198671

0(Ω) The bilinear form 119886(sdot sdot) is continuous on the

space [11986710(Ω)]2times [1198671

0(Ω)]2 and elliptic on [119867

1

0(Ω)]2 also the

bilinear form 119887(sdot sdot) is continuous and verifies the followinginf-sup condition (see [13 14]) there exists a nonnull positiveconstant 120573 such that

forall119902 isin 1198712

0(Ω) sup

kisin[11986710(Ω)]2

119887 (k 119902)k[1198671(Ω)]2

ge 1205731003817100381710038171003817119902

10038171003817100381710038171198712(Ω) (5)

Then we conclude [15] that for all f in the space [119867minus1(Ω)]2

the problem (3) has a unique solution (u 119901) in [1198671

0(Ω)]2times

1198712

0(Ω)This solution verifies the following stability condition

u[1198671(Ω)]2 + 120573

100381710038171003817100381711990110038171003817100381710038171198712(Ω)

le 119862f[119867minus1(Ω)]2 (6)

where 119862 is a positive constantWe refer to the work of Pironneau [16] for themathemati-

calmodeling of problems resulting fromfluidsmechanics andGirault and Raviart [17] for the mathematical analysis of theNavier-Stokes equations

3 Singular Functions and Regularity Results

We recall that in an open simply connected ofR2 the condi-tion of incompressibility div(u) = 0 induces the existence ofa stream function 120593 in the space119867

2

0(Ω) such that

u = curl (120593) (7)

It thus brings to study the regularity of the function 120593solution of Dirichlet problem for bilaplacian

minus]Δ2120593 = curl (f) in Ω

120593 = 0 on Γ

120597120593

120597119899= 0 on Γ

(8)

The regularity of the solution of the problem is related to thegeometry of the domain and its behavior is local Let 119892 =

curl(f)We know see Cattabriga [18] and Ladyzhenskaya [14]the following theorem

Theorem 1 For 119892 in the space 119867119904(Ω) where 119904 ge minus2 then

120593 isin 119867119904+4

(Ω V) (9)

where V is the union of neighborhoods 119881119895of 119886119895for 119895 isin

1 119869

To study the function 120593 in the neighborhood of a fixedvertex 119886

119895 119895 isin 1 119869 it is convenient to introduce the polar

coordinates (119903 120579) centered at 119886119895 We start by enunciating the

characteristic equation of bilaplacian

sin2 120596119895119911 = 1199112sin2 120596

119895 (10)

Then we stated the following theorem We refer to ([17]Chapter 7Theorem 72112) andKondratiev [1] for the proof

Theorem 2 Let 120596119895in ]0 2120587[ For all 119892 in 119867

119904(Ω) 119904 ge minus2 the

solution 120593 of the problem (8) is written as

120593 = 120593119903+ 120593119904 (11)

where 120593119903is in the space (119867119904+4(Ω) cap 119867

2

0(Ω))

We set

120591119896(119903 120579) = 119903

1+119911119896120595119896(120579)

120583119896(119903 120579) = 119903

1+119896 (120590119896(120579) + log (119903) 120578

119896(120579))

(12)

120593119904is given by

120593119904= sum

0ltRe(119911119896)lt119904+2

120582119896120591119896(119903 120579) + sum

0ltRe(119896)lt119904+2

119896120583119896(119903 120579) (13)

where 120582119896and

119896are real constants120595

119896 120590119896 and 120578

119896belong to the

vectorial space Cinfin([0 120596119895]) cap 119867

2

0([0 120596119895]) of finite dimension

119911119896and 119896are respectively the simple and double roots of (10)

in the band 0 lt Re(119911119896) lt 119904 + 2 excepting 1 if 120596

119895= 119905119892 (120596

119895)

without exception if 120596119895= 119905119892(120596

119895)

We called 120596119890the unique solution of equation 120596 = 119905119892 (120596)

in the ]0 2120587[ (120596119890≃ 1430297120587) We prove see ([19] chapter

3 sect 33) that 119911 is a double root of (10) if and only if 119911 = 0 or119911 = plusmnradic(1 sin120596

2

119895) minus (1120596

2

119895) Hence the sufficient condition

on the angle 120596 for a double root is

sin(radic

1205962

119895

(sin120596119895)2minus 1) = plusmnradic(1 minus

(sin120596119895)2

1205962

119895

)

120596119895isin ]0 2120587[

(14)

Then if 120596119895is not a solution of (14) (10) takes the following

simplified form

120593119904(119903 120579) = sum

0ltRe(119911119896)lt119904+2

120582119896120591119896(119903 120579) (15)

In the following we suppose that Ω has a unique vertex 119886

where 120596 is equal to 31205872 or 2120587 and the other angles are 1205872



119890


1≃ 0544484

and 1199112


2are written as

follows

1205911(119903 120579) = 119903

1+11991111205951(120579)

1205912(119903 120579) = 119903

1+11991121205952(120579)

(16)

with

120595119894(120579) = ((119911


3 (119911119894minus 1) 120587

2) minus (119911

119894+ 1)minus1

times sin(3 (119911119894+ 1) 120587

2))


119894+ 1) 120579))


119894minus1) 120579)minus(119911

119894+1)minus1 sin ((119911

119894+1) 120579))

times(cos(3 (119911119894minus 1) 120587

2) minus cos(

3 (119911119894+ 1) 120587

2))

(17)




Δ2120591119894= 0 119894119899 119881

120591119894=

120597120591119894

120597119899= 0 119900119899 Γ cap 120597119881

(18)



1199031+119911119894120595119894(120579) isin 119867

119904(119881) (19)


1199031+119911119894120595119894(120579) isin 119882

119898

119902(119881) sub 119867

119904(119881) (20)


119894(120579) isin 119882

119898

119902(119881) then 1 + 119911



119894lt 1)



(18)


u = u119903+ u119904 (21)

where u119903is in [119867

2(Ω)cap119867

1


119894

such that

u119904= sum

1le119894le2

120582119894119904119894 (22)

where 119904119894(119903 120579) = curl(120591

119894(119903 120579)) The functions 119904

119894belong to the




119866 = 119903119890119894120579 0 lt 120579 lt 2120587 119903 gt 0 (23)




]Δ2120595 = 119892 in 119867119904(119866)

120595 isin 1198672


(24)


119885119904

2(119866) = 120595 isin 119871

2(119866) 119903

minus119904+|120572|119863120572120595 isin 119871

2(119866) |120572| le 119904

119885minus1

2(119866) = 120595 isin D

1015840(119866) 120595 =

1198920

119903+ 1205971199091198921+ 1205971199101198922

such that 1198920 1198921and 119892

2isin 1198712(Ω)

(25)


2(119866) = 119885

2(119866) and 119867

119904

0(119866) sub

119885119904

2(119866) sub 119867

119904(119866)


2(119866)


]Δ2120595 = 119892 in 119867119904(119866)

119906 isin 1198852

2(119866)

(26)


120595 (119903 120579) = 0 for 119903 ge 119877 (27)







2(119861)


2(119866) and let ℎ =



119867119904(119861)




2(119866) then 119892 = 119892

0119903+1205971199091198921+1205971199101198922with 119892

0 1198921

and 1198922in 1198712(119866)

We denote

119865 (119905 120579) = 119892 (119890119905 cos 120579 119890119905 sin 120579)

119866119895(119905 120579) = 119892

119895(119890119905 cos 120579 119890119905 sin 120579) for 119895 isin 0 1 2

119865 (119905 120579) = 119890minus1199051198660+ (119890minus119905 cos 120579120597119866

1

120597119905minus 119890minus119905 sin 120579

1205971198661

120597120579)

+ (119890minus119905 sin 120579

1205971198662

120597119905minus 119890minus119905 cos 120579120597119866

2

120597120579)

(28)

We have

119890minus119905ℎ = 1198902119905119865 (119905 120579) = 119890

1199051198660+ 120597119905(119890119905 cos 120579119866

1+ sin 120579119890

1199051198662)

+ 120597120579(minus119890119905 sin 120579119866

1+ cos 120579119890119905119866

2)

(29)

and if

1198700= 1198660

1198701= cos 120579119866

1+ sin 120579119866

2

1198702= minus sin 120579119866

1+ cos 120579119866

2

(30)

then we conclude

119890minus119905ℎ = 1198901199051198700+ 1205971199051198901199051198701 + 1205971205791198901199051198702 (31)

since 119892119895

isin 1198712(119866) et 119890119905119866

119895isin 1198712(119861) then 119890119905119870

119895isin 1198712(119861) for

119895 isin 0 1 2 So

119890minus119905ℎ isin 119867

minus1(119861) (32)

We end the proof



problem

(1198634

119905minus 1198632

119905+ 1) V + 2 (119863

2

119905+ 1)119863

2

120579V + 119863

4

120579V = ℎ

V (119905 0) = V (119905 2120587) = 120597119905V (119905 0) = 120597

120579V (119905 2120587) = 0

V isin 1198672(119861)

(33)



V (119911 119905) =1

radic2120587

int

R

119890minus119894119905119911V (119905 119911) 119889119905 (34)


119871 (119911119863120579) = (119911

4+ 21199112+ 1) + 2 (1 minus 119911

2)1198632

120579+ 1198634

120579 (35)


119871 (119911119863120579) V = ℎ

V (119905 0) = V (119905 2120587) = 120597119905V (119905 0) = 120597

120579V (119905 2120587) = 0

(36)


1199014+ 2 (1 minus 119911

2) 1199012+ (1199114+ 21199112+ 1) = 0 (37)





sin 120579 cos 120579 120579 sin 120579 et 120579 cos 120579 (39)


1 120579 sin 2120579 et cos 2120579 (40)


119871 (119911119863120579) 120593 = 0 in ]0 2120587[

120593 (0) = 120593 (2120587) = 120597119905120593 (0) = 120597

120579120593 (2120587) = 0

(41)

We note 119864119911

= 120593119911

isin 1198672

0(]0 2120587[) 120593


problem (41)



119864119911= 0 (42)

(2) if 119911 plusmn 119894 then

119864119911= 120593119911isin 1198672

0(]0 2120587[) 120593119911 = 120572

119911sin2120579 120572

119911isin C (43)


119864119911= 120593119911isin 1198672

0([0 2120587]) 120593

119911= 120572119896119892119896(120579) + 120573

119896ℎ119896(120579)

(120572119896 120573119896) isin C2

119892119896(120579) = sin(1 +

119896

2) 120579 minus (

119896 + 2


119896

2) 120579

ℎ119896= cos(1 +

119896


119896

2) 120579

(44)






120575 = 0

120574119904ℎ2120587119911 + 120575119888ℎ2120587119911 = 0

120573 = 0

120572119904ℎ2120587119911 + 120573119888ℎ2120587119911 + 2120587120574119904ℎ2120587119911 + 2120587120575119888ℎ2120587119911 = 0

(46)


119911=



120593 =120572

2119894ℎ119896(120579) + 119894120574 (

119896

2minus 1)119892

119896(120579) (47)



int

2120587

0

1003816100381610038161003816ℎ1198961003816100381610038161003816 119889120579 = int

2120587

0

10038161003816100381610038161198921198961003816100381610038161003816 119889120579 = 1 (48)


120593 = 120572 + 120573120579 + 120574 sin 2120579 + 120575 cos 2120579 (49)


120572 + 120575 = 0

120572 + 2120587120573 + 120575 = 0

120573 + 2120574 = 0

(50)


120593 = 120572 (1 minus cos 2120579) = 2120572 sin2 2120579 (51)


Remark 9 If

119863119896= 119911119896= minus119894

119896

2 for 119896 = 0 119896 = 2 119896 isin N (52)



minus(119904 + 2) cap 119863119896

Is then

119889 (119911) =minus119904ℎ22120587119911

1199112 (1 + 1199112) for 119911 = 0 119911 = plusmn 119894 (53)

We verify that 119889(119911119896) = 120597

119911119889(119911119896) = 0 and 120597

2

119911119889(119911119896) = 0 Hence




V (119911 120579) =1205931(120579)

(119911 + 119894)+ 1199081(119911 120579) (54)


1



V (119911 120579) =120595119896(120579)

(119911 + 119894 (1198962))2

+120593119896(120579)

(119911 + 119894 (1198962))+ 119908119896(119911 120579) (55)

where 120595119896and 119908


119896

satisfies

119871 (119911119863120579) 120593119896= 2119894119896(1 minus

1198962

4)120595119896minus 2119894119896120595

10158401015840

119896

120593119896(0) = 120593

119896(2120587) = 120597

119905120593119896(0) = 120597

120579120593119896(2120587) = 0

(56)


1199061(119911 120579) =

sin 120579119904ℎ119911120579

119911for 119911 = 0

1199061(0 120579) = 120579 sin 120579

1199062(119911 120579) =

1

1199112 + 1(cos 120579119904ℎ119911120579

119911minus sin 120579119888ℎ119911120579)

for 119911 = 0 119911 = plusmn 119894

1199062(0 120579) = 120579 cos 120579 minus sin 120579

1199062(plusmn119894 120579) =

1

2(sin 2120579

2minus 120579)

(57)


1199061(minus119894

119896

2 120579) = minus

1

119896ℎ119896(120579)

1199062(minus119894

119896

2 120579) =

2

119896 (119896 + 2)119892119896(120579)

(58)


1and 119906


1and 119906

2are whole


119889119911= 1199061(119911 2120587)119863

1205791199062(119911 2120587) minus 119906

2(119911 2120587)119863

1205791199061(119911 2120587)

(59)





1199081(119911 120579) and for all 119911



120579) 1199081

(60)



119871 (minus119894 119863120579) 1205951= 0

1205951(0) = (119911 minus 119894) 119908

1(119911 0)

(61)

Then

1205951(0) = 0 (62)

and even

1205951(2120587) = 0

1205951015840

1(0) = lim

120579rarr0

120595 (120579)

120579= lim120579rarr0

V (119911 120579) minus (119911 + 1)1199081(119911 120579)

120579= 0

(63)

and even for

1205951015840

1(2120587) = 0 (64)



for 119896 = 1 and 119896 = 3 is

V (119911 120579) =120595119896(120579)

(119911 + 119894 (1198962))2+

120593119896(120579)

(119911 + 119894 (1198962))+ 119908119896(119911 120579)

for 119896 = 1 119896 = 3

(65)


ℎ(119911 minus 119911119896)2

= 119871 (119911119863120579) 120595119896+ (119911 minus 119911

119896) 119871 (119911 119863

120579) 120593119896

+ (119911 minus 119911119896)2

119871 (119911119863120579) 119908119896

(66)




119871 (119911119863120579) 120593119896= minus

119871 (119911119863120579) + 119871 (119911119896 119863120579)

(119911 minus 119911119896)

120595119896

+ (119911 minus 119911119896) (ℎ minus 119871 (119911 119863120579)119908

119896)

(67)

And if 119911 tends to 119911119896

119871 (119911119863120579) 120593119896= 2119894119896(1 minus

1198962

4)120595119896minus 2119894119896120595

10158401015840

119896(68)


proof


adjoint

⟨119871120595119896 120593119896⟩ = ⟨120595

119896 119871120593119896⟩

= int

2120587

0

120595119896(2119896(1 minus

1198962

4)120595119896minus 2119894119896120595

10158401015840

119896)119889120579

(69)

For 0 lt 119896 lt 4

int

2120587

0

1205952

119896119889120579 = int

2120587

0

12059510158402

119896119889120579 = 0 (70)


V (119911 120579) =120593119896(120579)

(119911 + 119894 (1198962))+ 119908119896(119911 120579)

for 119896 = 1 or 119896 = 3

(71)



119908119904(119905 120579) =

1

radic2120587

int

+infin

minusinfin

119890119894119905119911V (119911 minus 119894 (119904 + 2) + 119894120578 120579) 119889119911

120583 (120579) = 119894radic2120587119890119894119905119911V (119911 120579) for fixed 119905

(72)


V (119905 120579) = 119890(119904+2minus120598)119905

119908119904(119905 120579) + 2119894120587 sum

minus(119904+2)ltIm(119911)lt0119877119904 (73)


119911119896 Then

(i) for 119904 = minus1

V (119905 120579) = 119890(1minus120578)119905

119908minus1

(119905 120579) + 119894radic21205871205951(120579) (74)

(ii) for 119904 = 0

V (119905 120579) = 119890(2minus120578)119905

1199080(119905 120579)

+ 119894radic2120587 (11989031199052

1205953(120579) + 119890

11990521205951(120579) + 119890

1199051205952(120579))

(75)




1205911(119903 120579) = 119903

32(sin 3

2120579 minus 3 sin 120579

2)

1205831(119903 120579) = 119903

32(cos 3

2120579 minus cos 120579

2)

(76)


1205911(119903 120579) = 119903

32(sin 3

2120579 minus 3 sin 120579

2)

1205831(119903 120579) = 119903

32(cos 3

2120579 minus cos 120579

2)

1205912(119903 120579) = 119903

52(sin 5

2120579 minus 5 sin 120579

2)

1205832(119903 120579) = 119903

52(cos 5

2120579 minus cos 120579

2)

(77)




11986752minus120598

(119881) 1205912and 120583




Δ2120593 = 0 119894119899 119881

120593 =120597120593

120597119899= 0 119900119899 Γ cap 120597119881

(78)




2(Ω)]2


form

u = u119903+ u119904 (79)

where u119903is in [119867

2(Ω) cap 119867

1



u119904= 1205821199041+ 1199041

(80)

with

1199041(119903 120579) = 119903

12(3 sin 120579 sin 120579

2 3 (1 minus cos 120579) sin 120579

2)

1199041(119903 120579) = 119903

12(2 sin 120579

2+ sin 120579 cos 120579

2 (1 minus cos 120579) cos 120579

2)

(81)






if 120587 lt 120596 lt 2120587 1 minus120587

120596lt 120578 (120596) lt

120587

120596 120578 (120596) gt

1

2 (83)


119882 = u isin [1198671

0(Ω)]2



defined such that

nabla119901 = f + ]Δu in 1198631015840(Ω) (85)



119901 = 119901119903+ 119901119904 119901119903isin 1198671(Ω) 119901

119904= 1205731198781199011 (86)


nabla1198781199011

= ]Δ (curl (1199031+120578(120596)120595 (120579))) (87)


Consider

1198781199011

(119903 120579)

=

minus119903120578(120596)minus1

((1 + 120578 (120596))2

(120597120595 (120579) 120597120579) + (1205973120595 (120579) 120597120579))

(1 + 120578 (120596))

(88)


120595 (120579) = 3 sin(120579

2) minus sin(

3

2120579) 120578 (120596) =

1

2 (89)

in the case of 120596 = 31205872

120595 (120579) =sin ((1 + 120578 (120596)) 120579) cos (120578 (120596) 120596)

(1 + 120578 (120596))

minus cos ((1 + 120578 (120596)) 120579)

+sin ((120578 (120596) minus 1) 120579) cos (120578 (120596) 120596)

(1 minus 120578 (120596))

+ cos ((120578 (120596) minus 1) 120579)

(90)

and 120578(120596) = 0544484


belongs to 1198670544484

(119881) when120596 = 31205872 and 119878

1199011belongs to 119867

05(119881) when 120596 = 2120587


5 Conclusion



(Ω)]2 where 119904 gt 0 then (u 119901)


(Ω) if 119904 lt 1+120578(120596) Thismeans

[119867119904(Ω)]2


(Ω) for 119904 lt 1544484 when 120596 =3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 15 when 120596 = 2120587

(91)


u[119867119904(Ω)]2 +

10038171003817100381710038171199011003817100381710038171003817119867119904minus1(Ω)

le 119862f[119867119904minus2(Ω)]2 (92)



u = u119903+ 1205821198781 119901 = 119901

119903+ 1205731198781199011 (93)

and if 120596 = 2120587

u = u119903+ 12058211198781+ 11198781 119901 = 119901

119903+ 12057311198781199011

+ 12057311198781199011 (94)


(Ω)]2 (u119903 119901119903) belongs to [119867

119904(Ω)]2times

119867119904minus1

(Ω) for 119904 lt 1 + 1205781(120596) where 120578



[119867119904(Ω)]2


(Ω) for 119904 lt 1908529 when 120596 =3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 25 when 120596 = 2120587

(95)


1003817100381710038171003817[119867119904(Ω)]2+1003817100381710038171003817119901119903

1003817100381710038171003817119867119904minus1(Ω)+10038161003816100381610038161205821

1003816100381610038161003816 +10038161003816100381610038161205731

1003816100381610038161003816 le 119862f[119867119904minus2(Ω)]2

(96)




u119903= u119903+ 12058221198782 119901

119903= 119901119903+ 12057321198781199012

(97)

(u119903 119901) isin [119867

119904(Ω)]2times 119867119904minus1

(Ω) for 119904 lt 1 + 1205782(120596) where 120578

2(120596)


[119867119904(Ω)]2


(Ω) for 119904 lt 1205782(3120587

2) when 120596 =

3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 3 5 when 120596 = 2120587

(98)



1003817100381710038171003817u1199031003817100381710038171003817[119867119904(Ω)]2

+1003817100381710038171003817119901119903

1003817100381710038171003817119867119904(Ω)+10038161003816100381610038161205821

1003816100381610038161003816 +10038161003816100381610038161205822

1003816100381610038161003816 +10038161003816100381610038161205731

1003816100381610038161003816 +10038161003816100381610038161205732

1003816100381610038161003816

le 119862f[119867119904minus2(Ω)]2

(99)


u = u119903+ 12058211198781+ 12058221198782+ sdot sdot sdot + 120582

119896119878119896

119901 = 119901119903+ 12057311198781199011

+ 12057321198781199012

+ sdot sdot sdot + 120573119896119878119901119896

(100)



119904(Ω)]2times

119867119904minus1

(Ω) for 119904 lt 1 + 120578119896(120596) where 120578





Acknowledgment


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119890


1≃ 0544484

and 1199112


2are written as

follows

1205911(119903 120579) = 119903

1+11991111205951(120579)

1205912(119903 120579) = 119903

1+11991121205952(120579)

(16)

with

120595119894(120579) = ((119911


3 (119911119894minus 1) 120587

2) minus (119911

119894+ 1)minus1

times sin(3 (119911119894+ 1) 120587

2))


119894+ 1) 120579))


119894minus1) 120579)minus(119911

119894+1)minus1 sin ((119911

119894+1) 120579))

times(cos(3 (119911119894minus 1) 120587

2) minus cos(

3 (119911119894+ 1) 120587

2))

(17)




Δ2120591119894= 0 119894119899 119881

120591119894=

120597120591119894

120597119899= 0 119900119899 Γ cap 120597119881

(18)



1199031+119911119894120595119894(120579) isin 119867

119904(119881) (19)


1199031+119911119894120595119894(120579) isin 119882

119898

119902(119881) sub 119867

119904(119881) (20)


119894(120579) isin 119882

119898

119902(119881) then 1 + 119911



119894lt 1)



(18)


u = u119903+ u119904 (21)

where u119903is in [119867

2(Ω)cap119867

1


119894

such that

u119904= sum

1le119894le2

120582119894119904119894 (22)

where 119904119894(119903 120579) = curl(120591

119894(119903 120579)) The functions 119904

119894belong to the




119866 = 119903119890119894120579 0 lt 120579 lt 2120587 119903 gt 0 (23)




]Δ2120595 = 119892 in 119867119904(119866)

120595 isin 1198672


(24)


119885119904

2(119866) = 120595 isin 119871

2(119866) 119903

minus119904+|120572|119863120572120595 isin 119871

2(119866) |120572| le 119904

119885minus1

2(119866) = 120595 isin D

1015840(119866) 120595 =

1198920

119903+ 1205971199091198921+ 1205971199101198922

such that 1198920 1198921and 119892

2isin 1198712(Ω)

(25)


2(119866) = 119885

2(119866) and 119867

119904

0(119866) sub

119885119904

2(119866) sub 119867

119904(119866)


2(119866)


]Δ2120595 = 119892 in 119867119904(119866)

119906 isin 1198852

2(119866)

(26)


120595 (119903 120579) = 0 for 119903 ge 119877 (27)







2(119861)


2(119866) and let ℎ =



119867119904(119861)




2(119866) then 119892 = 119892

0119903+1205971199091198921+1205971199101198922with 119892

0 1198921

and 1198922in 1198712(119866)

We denote

119865 (119905 120579) = 119892 (119890119905 cos 120579 119890119905 sin 120579)

119866119895(119905 120579) = 119892

119895(119890119905 cos 120579 119890119905 sin 120579) for 119895 isin 0 1 2

119865 (119905 120579) = 119890minus1199051198660+ (119890minus119905 cos 120579120597119866

1

120597119905minus 119890minus119905 sin 120579

1205971198661

120597120579)

+ (119890minus119905 sin 120579

1205971198662

120597119905minus 119890minus119905 cos 120579120597119866

2

120597120579)

(28)

We have

119890minus119905ℎ = 1198902119905119865 (119905 120579) = 119890

1199051198660+ 120597119905(119890119905 cos 120579119866

1+ sin 120579119890

1199051198662)

+ 120597120579(minus119890119905 sin 120579119866

1+ cos 120579119890119905119866

2)

(29)

and if

1198700= 1198660

1198701= cos 120579119866

1+ sin 120579119866

2

1198702= minus sin 120579119866

1+ cos 120579119866

2

(30)

then we conclude

119890minus119905ℎ = 1198901199051198700+ 1205971199051198901199051198701 + 1205971205791198901199051198702 (31)

since 119892119895

isin 1198712(119866) et 119890119905119866

119895isin 1198712(119861) then 119890119905119870

119895isin 1198712(119861) for

119895 isin 0 1 2 So

119890minus119905ℎ isin 119867

minus1(119861) (32)

We end the proof



problem

(1198634

119905minus 1198632

119905+ 1) V + 2 (119863

2

119905+ 1)119863

2

120579V + 119863

4

120579V = ℎ

V (119905 0) = V (119905 2120587) = 120597119905V (119905 0) = 120597

120579V (119905 2120587) = 0

V isin 1198672(119861)

(33)



V (119911 119905) =1

radic2120587

int

R

119890minus119894119905119911V (119905 119911) 119889119905 (34)


119871 (119911119863120579) = (119911

4+ 21199112+ 1) + 2 (1 minus 119911

2)1198632

120579+ 1198634

120579 (35)


119871 (119911119863120579) V = ℎ

V (119905 0) = V (119905 2120587) = 120597119905V (119905 0) = 120597

120579V (119905 2120587) = 0

(36)


1199014+ 2 (1 minus 119911

2) 1199012+ (1199114+ 21199112+ 1) = 0 (37)





sin 120579 cos 120579 120579 sin 120579 et 120579 cos 120579 (39)


1 120579 sin 2120579 et cos 2120579 (40)


119871 (119911119863120579) 120593 = 0 in ]0 2120587[

120593 (0) = 120593 (2120587) = 120597119905120593 (0) = 120597

120579120593 (2120587) = 0

(41)

We note 119864119911

= 120593119911

isin 1198672

0(]0 2120587[) 120593


problem (41)



119864119911= 0 (42)

(2) if 119911 plusmn 119894 then

119864119911= 120593119911isin 1198672

0(]0 2120587[) 120593119911 = 120572

119911sin2120579 120572

119911isin C (43)


119864119911= 120593119911isin 1198672

0([0 2120587]) 120593

119911= 120572119896119892119896(120579) + 120573

119896ℎ119896(120579)

(120572119896 120573119896) isin C2

119892119896(120579) = sin(1 +

119896

2) 120579 minus (

119896 + 2


119896

2) 120579

ℎ119896= cos(1 +

119896


119896

2) 120579

(44)






120575 = 0

120574119904ℎ2120587119911 + 120575119888ℎ2120587119911 = 0

120573 = 0

120572119904ℎ2120587119911 + 120573119888ℎ2120587119911 + 2120587120574119904ℎ2120587119911 + 2120587120575119888ℎ2120587119911 = 0

(46)


119911=



120593 =120572

2119894ℎ119896(120579) + 119894120574 (

119896

2minus 1)119892

119896(120579) (47)



int

2120587

0

1003816100381610038161003816ℎ1198961003816100381610038161003816 119889120579 = int

2120587

0

10038161003816100381610038161198921198961003816100381610038161003816 119889120579 = 1 (48)


120593 = 120572 + 120573120579 + 120574 sin 2120579 + 120575 cos 2120579 (49)


120572 + 120575 = 0

120572 + 2120587120573 + 120575 = 0

120573 + 2120574 = 0

(50)


120593 = 120572 (1 minus cos 2120579) = 2120572 sin2 2120579 (51)


Remark 9 If

119863119896= 119911119896= minus119894

119896

2 for 119896 = 0 119896 = 2 119896 isin N (52)



minus(119904 + 2) cap 119863119896

Is then

119889 (119911) =minus119904ℎ22120587119911

1199112 (1 + 1199112) for 119911 = 0 119911 = plusmn 119894 (53)

We verify that 119889(119911119896) = 120597

119911119889(119911119896) = 0 and 120597

2

119911119889(119911119896) = 0 Hence




V (119911 120579) =1205931(120579)

(119911 + 119894)+ 1199081(119911 120579) (54)


1



V (119911 120579) =120595119896(120579)

(119911 + 119894 (1198962))2

+120593119896(120579)

(119911 + 119894 (1198962))+ 119908119896(119911 120579) (55)

where 120595119896and 119908


119896

satisfies

119871 (119911119863120579) 120593119896= 2119894119896(1 minus

1198962

4)120595119896minus 2119894119896120595

10158401015840

119896

120593119896(0) = 120593

119896(2120587) = 120597

119905120593119896(0) = 120597

120579120593119896(2120587) = 0

(56)


1199061(119911 120579) =

sin 120579119904ℎ119911120579

119911for 119911 = 0

1199061(0 120579) = 120579 sin 120579

1199062(119911 120579) =

1

1199112 + 1(cos 120579119904ℎ119911120579

119911minus sin 120579119888ℎ119911120579)

for 119911 = 0 119911 = plusmn 119894

1199062(0 120579) = 120579 cos 120579 minus sin 120579

1199062(plusmn119894 120579) =

1

2(sin 2120579

2minus 120579)

(57)


1199061(minus119894

119896

2 120579) = minus

1

119896ℎ119896(120579)

1199062(minus119894

119896

2 120579) =

2

119896 (119896 + 2)119892119896(120579)

(58)


1and 119906


1and 119906

2are whole


119889119911= 1199061(119911 2120587)119863

1205791199062(119911 2120587) minus 119906

2(119911 2120587)119863

1205791199061(119911 2120587)

(59)





1199081(119911 120579) and for all 119911



120579) 1199081

(60)



119871 (minus119894 119863120579) 1205951= 0

1205951(0) = (119911 minus 119894) 119908

1(119911 0)

(61)

Then

1205951(0) = 0 (62)

and even

1205951(2120587) = 0

1205951015840

1(0) = lim

120579rarr0

120595 (120579)

120579= lim120579rarr0

V (119911 120579) minus (119911 + 1)1199081(119911 120579)

120579= 0

(63)

and even for

1205951015840

1(2120587) = 0 (64)



for 119896 = 1 and 119896 = 3 is

V (119911 120579) =120595119896(120579)

(119911 + 119894 (1198962))2+

120593119896(120579)

(119911 + 119894 (1198962))+ 119908119896(119911 120579)

for 119896 = 1 119896 = 3

(65)


ℎ(119911 minus 119911119896)2

= 119871 (119911119863120579) 120595119896+ (119911 minus 119911

119896) 119871 (119911 119863

120579) 120593119896

+ (119911 minus 119911119896)2

119871 (119911119863120579) 119908119896

(66)




119871 (119911119863120579) 120593119896= minus

119871 (119911119863120579) + 119871 (119911119896 119863120579)

(119911 minus 119911119896)

120595119896

+ (119911 minus 119911119896) (ℎ minus 119871 (119911 119863120579)119908

119896)

(67)

And if 119911 tends to 119911119896

119871 (119911119863120579) 120593119896= 2119894119896(1 minus

1198962

4)120595119896minus 2119894119896120595

10158401015840

119896(68)


proof


adjoint

⟨119871120595119896 120593119896⟩ = ⟨120595

119896 119871120593119896⟩

= int

2120587

0

120595119896(2119896(1 minus

1198962

4)120595119896minus 2119894119896120595

10158401015840

119896)119889120579

(69)

For 0 lt 119896 lt 4

int

2120587

0

1205952

119896119889120579 = int

2120587

0

12059510158402

119896119889120579 = 0 (70)


V (119911 120579) =120593119896(120579)

(119911 + 119894 (1198962))+ 119908119896(119911 120579)

for 119896 = 1 or 119896 = 3

(71)



119908119904(119905 120579) =

1

radic2120587

int

+infin

minusinfin

119890119894119905119911V (119911 minus 119894 (119904 + 2) + 119894120578 120579) 119889119911

120583 (120579) = 119894radic2120587119890119894119905119911V (119911 120579) for fixed 119905

(72)


V (119905 120579) = 119890(119904+2minus120598)119905

119908119904(119905 120579) + 2119894120587 sum

minus(119904+2)ltIm(119911)lt0119877119904 (73)


119911119896 Then

(i) for 119904 = minus1

V (119905 120579) = 119890(1minus120578)119905

119908minus1

(119905 120579) + 119894radic21205871205951(120579) (74)

(ii) for 119904 = 0

V (119905 120579) = 119890(2minus120578)119905

1199080(119905 120579)

+ 119894radic2120587 (11989031199052

1205953(120579) + 119890

11990521205951(120579) + 119890

1199051205952(120579))

(75)




1205911(119903 120579) = 119903

32(sin 3

2120579 minus 3 sin 120579

2)

1205831(119903 120579) = 119903

32(cos 3

2120579 minus cos 120579

2)

(76)


1205911(119903 120579) = 119903

32(sin 3

2120579 minus 3 sin 120579

2)

1205831(119903 120579) = 119903

32(cos 3

2120579 minus cos 120579

2)

1205912(119903 120579) = 119903

52(sin 5

2120579 minus 5 sin 120579

2)

1205832(119903 120579) = 119903

52(cos 5

2120579 minus cos 120579

2)

(77)




11986752minus120598

(119881) 1205912and 120583




Δ2120593 = 0 119894119899 119881

120593 =120597120593

120597119899= 0 119900119899 Γ cap 120597119881

(78)




2(Ω)]2


form

u = u119903+ u119904 (79)

where u119903is in [119867

2(Ω) cap 119867

1



u119904= 1205821199041+ 1199041

(80)

with

1199041(119903 120579) = 119903

12(3 sin 120579 sin 120579

2 3 (1 minus cos 120579) sin 120579

2)

1199041(119903 120579) = 119903

12(2 sin 120579

2+ sin 120579 cos 120579

2 (1 minus cos 120579) cos 120579

2)

(81)






if 120587 lt 120596 lt 2120587 1 minus120587

120596lt 120578 (120596) lt

120587

120596 120578 (120596) gt

1

2 (83)


119882 = u isin [1198671

0(Ω)]2



defined such that

nabla119901 = f + ]Δu in 1198631015840(Ω) (85)



119901 = 119901119903+ 119901119904 119901119903isin 1198671(Ω) 119901

119904= 1205731198781199011 (86)


nabla1198781199011

= ]Δ (curl (1199031+120578(120596)120595 (120579))) (87)


Consider

1198781199011

(119903 120579)

=

minus119903120578(120596)minus1

((1 + 120578 (120596))2

(120597120595 (120579) 120597120579) + (1205973120595 (120579) 120597120579))

(1 + 120578 (120596))

(88)


120595 (120579) = 3 sin(120579

2) minus sin(

3

2120579) 120578 (120596) =

1

2 (89)

in the case of 120596 = 31205872

120595 (120579) =sin ((1 + 120578 (120596)) 120579) cos (120578 (120596) 120596)

(1 + 120578 (120596))

minus cos ((1 + 120578 (120596)) 120579)

+sin ((120578 (120596) minus 1) 120579) cos (120578 (120596) 120596)

(1 minus 120578 (120596))

+ cos ((120578 (120596) minus 1) 120579)

(90)

and 120578(120596) = 0544484


belongs to 1198670544484

(119881) when120596 = 31205872 and 119878

1199011belongs to 119867

05(119881) when 120596 = 2120587


5 Conclusion



(Ω)]2 where 119904 gt 0 then (u 119901)


(Ω) if 119904 lt 1+120578(120596) Thismeans

[119867119904(Ω)]2


(Ω) for 119904 lt 1544484 when 120596 =3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 15 when 120596 = 2120587

(91)


u[119867119904(Ω)]2 +

10038171003817100381710038171199011003817100381710038171003817119867119904minus1(Ω)

le 119862f[119867119904minus2(Ω)]2 (92)



u = u119903+ 1205821198781 119901 = 119901

119903+ 1205731198781199011 (93)

and if 120596 = 2120587

u = u119903+ 12058211198781+ 11198781 119901 = 119901

119903+ 12057311198781199011

+ 12057311198781199011 (94)


(Ω)]2 (u119903 119901119903) belongs to [119867

119904(Ω)]2times

119867119904minus1

(Ω) for 119904 lt 1 + 1205781(120596) where 120578



[119867119904(Ω)]2


(Ω) for 119904 lt 1908529 when 120596 =3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 25 when 120596 = 2120587

(95)


1003817100381710038171003817[119867119904(Ω)]2+1003817100381710038171003817119901119903

1003817100381710038171003817119867119904minus1(Ω)+10038161003816100381610038161205821

1003816100381610038161003816 +10038161003816100381610038161205731

1003816100381610038161003816 le 119862f[119867119904minus2(Ω)]2

(96)




u119903= u119903+ 12058221198782 119901

119903= 119901119903+ 12057321198781199012

(97)

(u119903 119901) isin [119867

119904(Ω)]2times 119867119904minus1

(Ω) for 119904 lt 1 + 1205782(120596) where 120578

2(120596)


[119867119904(Ω)]2


(Ω) for 119904 lt 1205782(3120587

2) when 120596 =

3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 3 5 when 120596 = 2120587

(98)



1003817100381710038171003817u1199031003817100381710038171003817[119867119904(Ω)]2

+1003817100381710038171003817119901119903

1003817100381710038171003817119867119904(Ω)+10038161003816100381610038161205821

1003816100381610038161003816 +10038161003816100381610038161205822

1003816100381610038161003816 +10038161003816100381610038161205731

1003816100381610038161003816 +10038161003816100381610038161205732

1003816100381610038161003816

le 119862f[119867119904minus2(Ω)]2

(99)


u = u119903+ 12058211198781+ 12058221198782+ sdot sdot sdot + 120582

119896119878119896

119901 = 119901119903+ 12057311198781199011

+ 12057321198781199012

+ sdot sdot sdot + 120573119896119878119901119896

(100)



119904(Ω)]2times

119867119904minus1

(Ω) for 119904 lt 1 + 120578119896(120596) where 120578





Acknowledgment


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











2(119866) then 119892 = 119892

0119903+1205971199091198921+1205971199101198922with 119892

0 1198921

and 1198922in 1198712(119866)

We denote

119865 (119905 120579) = 119892 (119890119905 cos 120579 119890119905 sin 120579)

119866119895(119905 120579) = 119892

119895(119890119905 cos 120579 119890119905 sin 120579) for 119895 isin 0 1 2

119865 (119905 120579) = 119890minus1199051198660+ (119890minus119905 cos 120579120597119866

1

120597119905minus 119890minus119905 sin 120579

1205971198661

120597120579)

+ (119890minus119905 sin 120579

1205971198662

120597119905minus 119890minus119905 cos 120579120597119866

2

120597120579)

(28)

We have

119890minus119905ℎ = 1198902119905119865 (119905 120579) = 119890

1199051198660+ 120597119905(119890119905 cos 120579119866

1+ sin 120579119890

1199051198662)

+ 120597120579(minus119890119905 sin 120579119866

1+ cos 120579119890119905119866

2)

(29)

and if

1198700= 1198660

1198701= cos 120579119866

1+ sin 120579119866

2

1198702= minus sin 120579119866

1+ cos 120579119866

2

(30)

then we conclude

119890minus119905ℎ = 1198901199051198700+ 1205971199051198901199051198701 + 1205971205791198901199051198702 (31)

since 119892119895

isin 1198712(119866) et 119890119905119866

119895isin 1198712(119861) then 119890119905119870

119895isin 1198712(119861) for

119895 isin 0 1 2 So

119890minus119905ℎ isin 119867

minus1(119861) (32)

We end the proof



problem

(1198634

119905minus 1198632

119905+ 1) V + 2 (119863

2

119905+ 1)119863

2

120579V + 119863

4

120579V = ℎ

V (119905 0) = V (119905 2120587) = 120597119905V (119905 0) = 120597

120579V (119905 2120587) = 0

V isin 1198672(119861)

(33)



V (119911 119905) =1

radic2120587

int

R

119890minus119894119905119911V (119905 119911) 119889119905 (34)


119871 (119911119863120579) = (119911

4+ 21199112+ 1) + 2 (1 minus 119911

2)1198632

120579+ 1198634

120579 (35)


119871 (119911119863120579) V = ℎ

V (119905 0) = V (119905 2120587) = 120597119905V (119905 0) = 120597

120579V (119905 2120587) = 0

(36)


1199014+ 2 (1 minus 119911

2) 1199012+ (1199114+ 21199112+ 1) = 0 (37)





sin 120579 cos 120579 120579 sin 120579 et 120579 cos 120579 (39)


1 120579 sin 2120579 et cos 2120579 (40)


119871 (119911119863120579) 120593 = 0 in ]0 2120587[

120593 (0) = 120593 (2120587) = 120597119905120593 (0) = 120597

120579120593 (2120587) = 0

(41)

We note 119864119911

= 120593119911

isin 1198672

0(]0 2120587[) 120593


problem (41)



119864119911= 0 (42)

(2) if 119911 plusmn 119894 then

119864119911= 120593119911isin 1198672

0(]0 2120587[) 120593119911 = 120572

119911sin2120579 120572

119911isin C (43)


119864119911= 120593119911isin 1198672

0([0 2120587]) 120593

119911= 120572119896119892119896(120579) + 120573

119896ℎ119896(120579)

(120572119896 120573119896) isin C2

119892119896(120579) = sin(1 +

119896

2) 120579 minus (

119896 + 2


119896

2) 120579

ℎ119896= cos(1 +

119896


119896

2) 120579

(44)






120575 = 0

120574119904ℎ2120587119911 + 120575119888ℎ2120587119911 = 0

120573 = 0

120572119904ℎ2120587119911 + 120573119888ℎ2120587119911 + 2120587120574119904ℎ2120587119911 + 2120587120575119888ℎ2120587119911 = 0

(46)


119911=



120593 =120572

2119894ℎ119896(120579) + 119894120574 (

119896

2minus 1)119892

119896(120579) (47)



int

2120587

0

1003816100381610038161003816ℎ1198961003816100381610038161003816 119889120579 = int

2120587

0

10038161003816100381610038161198921198961003816100381610038161003816 119889120579 = 1 (48)


120593 = 120572 + 120573120579 + 120574 sin 2120579 + 120575 cos 2120579 (49)


120572 + 120575 = 0

120572 + 2120587120573 + 120575 = 0

120573 + 2120574 = 0

(50)


120593 = 120572 (1 minus cos 2120579) = 2120572 sin2 2120579 (51)


Remark 9 If

119863119896= 119911119896= minus119894

119896

2 for 119896 = 0 119896 = 2 119896 isin N (52)



minus(119904 + 2) cap 119863119896

Is then

119889 (119911) =minus119904ℎ22120587119911

1199112 (1 + 1199112) for 119911 = 0 119911 = plusmn 119894 (53)

We verify that 119889(119911119896) = 120597

119911119889(119911119896) = 0 and 120597

2

119911119889(119911119896) = 0 Hence




V (119911 120579) =1205931(120579)

(119911 + 119894)+ 1199081(119911 120579) (54)


1



V (119911 120579) =120595119896(120579)

(119911 + 119894 (1198962))2

+120593119896(120579)

(119911 + 119894 (1198962))+ 119908119896(119911 120579) (55)

where 120595119896and 119908


119896

satisfies

119871 (119911119863120579) 120593119896= 2119894119896(1 minus

1198962

4)120595119896minus 2119894119896120595

10158401015840

119896

120593119896(0) = 120593

119896(2120587) = 120597

119905120593119896(0) = 120597

120579120593119896(2120587) = 0

(56)


1199061(119911 120579) =

sin 120579119904ℎ119911120579

119911for 119911 = 0

1199061(0 120579) = 120579 sin 120579

1199062(119911 120579) =

1

1199112 + 1(cos 120579119904ℎ119911120579

119911minus sin 120579119888ℎ119911120579)

for 119911 = 0 119911 = plusmn 119894

1199062(0 120579) = 120579 cos 120579 minus sin 120579

1199062(plusmn119894 120579) =

1

2(sin 2120579

2minus 120579)

(57)


1199061(minus119894

119896

2 120579) = minus

1

119896ℎ119896(120579)

1199062(minus119894

119896

2 120579) =

2

119896 (119896 + 2)119892119896(120579)

(58)


1and 119906


1and 119906

2are whole


119889119911= 1199061(119911 2120587)119863

1205791199062(119911 2120587) minus 119906

2(119911 2120587)119863

1205791199061(119911 2120587)

(59)





1199081(119911 120579) and for all 119911



120579) 1199081

(60)



119871 (minus119894 119863120579) 1205951= 0

1205951(0) = (119911 minus 119894) 119908

1(119911 0)

(61)

Then

1205951(0) = 0 (62)

and even

1205951(2120587) = 0

1205951015840

1(0) = lim

120579rarr0

120595 (120579)

120579= lim120579rarr0

V (119911 120579) minus (119911 + 1)1199081(119911 120579)

120579= 0

(63)

and even for

1205951015840

1(2120587) = 0 (64)



for 119896 = 1 and 119896 = 3 is

V (119911 120579) =120595119896(120579)

(119911 + 119894 (1198962))2+

120593119896(120579)

(119911 + 119894 (1198962))+ 119908119896(119911 120579)

for 119896 = 1 119896 = 3

(65)


ℎ(119911 minus 119911119896)2

= 119871 (119911119863120579) 120595119896+ (119911 minus 119911

119896) 119871 (119911 119863

120579) 120593119896

+ (119911 minus 119911119896)2

119871 (119911119863120579) 119908119896

(66)




119871 (119911119863120579) 120593119896= minus

119871 (119911119863120579) + 119871 (119911119896 119863120579)

(119911 minus 119911119896)

120595119896

+ (119911 minus 119911119896) (ℎ minus 119871 (119911 119863120579)119908

119896)

(67)

And if 119911 tends to 119911119896

119871 (119911119863120579) 120593119896= 2119894119896(1 minus

1198962

4)120595119896minus 2119894119896120595

10158401015840

119896(68)


proof


adjoint

⟨119871120595119896 120593119896⟩ = ⟨120595

119896 119871120593119896⟩

= int

2120587

0

120595119896(2119896(1 minus

1198962

4)120595119896minus 2119894119896120595

10158401015840

119896)119889120579

(69)

For 0 lt 119896 lt 4

int

2120587

0

1205952

119896119889120579 = int

2120587

0

12059510158402

119896119889120579 = 0 (70)


V (119911 120579) =120593119896(120579)

(119911 + 119894 (1198962))+ 119908119896(119911 120579)

for 119896 = 1 or 119896 = 3

(71)



119908119904(119905 120579) =

1

radic2120587

int

+infin

minusinfin

119890119894119905119911V (119911 minus 119894 (119904 + 2) + 119894120578 120579) 119889119911

120583 (120579) = 119894radic2120587119890119894119905119911V (119911 120579) for fixed 119905

(72)


V (119905 120579) = 119890(119904+2minus120598)119905

119908119904(119905 120579) + 2119894120587 sum

minus(119904+2)ltIm(119911)lt0119877119904 (73)


119911119896 Then

(i) for 119904 = minus1

V (119905 120579) = 119890(1minus120578)119905

119908minus1

(119905 120579) + 119894radic21205871205951(120579) (74)

(ii) for 119904 = 0

V (119905 120579) = 119890(2minus120578)119905

1199080(119905 120579)

+ 119894radic2120587 (11989031199052

1205953(120579) + 119890

11990521205951(120579) + 119890

1199051205952(120579))

(75)




1205911(119903 120579) = 119903

32(sin 3

2120579 minus 3 sin 120579

2)

1205831(119903 120579) = 119903

32(cos 3

2120579 minus cos 120579

2)

(76)


1205911(119903 120579) = 119903

32(sin 3

2120579 minus 3 sin 120579

2)

1205831(119903 120579) = 119903

32(cos 3

2120579 minus cos 120579

2)

1205912(119903 120579) = 119903

52(sin 5

2120579 minus 5 sin 120579

2)

1205832(119903 120579) = 119903

52(cos 5

2120579 minus cos 120579

2)

(77)




11986752minus120598

(119881) 1205912and 120583




Δ2120593 = 0 119894119899 119881

120593 =120597120593

120597119899= 0 119900119899 Γ cap 120597119881

(78)




2(Ω)]2


form

u = u119903+ u119904 (79)

where u119903is in [119867

2(Ω) cap 119867

1



u119904= 1205821199041+ 1199041

(80)

with

1199041(119903 120579) = 119903

12(3 sin 120579 sin 120579

2 3 (1 minus cos 120579) sin 120579

2)

1199041(119903 120579) = 119903

12(2 sin 120579

2+ sin 120579 cos 120579

2 (1 minus cos 120579) cos 120579

2)

(81)






if 120587 lt 120596 lt 2120587 1 minus120587

120596lt 120578 (120596) lt

120587

120596 120578 (120596) gt

1

2 (83)


119882 = u isin [1198671

0(Ω)]2



defined such that

nabla119901 = f + ]Δu in 1198631015840(Ω) (85)



119901 = 119901119903+ 119901119904 119901119903isin 1198671(Ω) 119901

119904= 1205731198781199011 (86)


nabla1198781199011

= ]Δ (curl (1199031+120578(120596)120595 (120579))) (87)


Consider

1198781199011

(119903 120579)

=

minus119903120578(120596)minus1

((1 + 120578 (120596))2

(120597120595 (120579) 120597120579) + (1205973120595 (120579) 120597120579))

(1 + 120578 (120596))

(88)


120595 (120579) = 3 sin(120579

2) minus sin(

3

2120579) 120578 (120596) =

1

2 (89)

in the case of 120596 = 31205872

120595 (120579) =sin ((1 + 120578 (120596)) 120579) cos (120578 (120596) 120596)

(1 + 120578 (120596))

minus cos ((1 + 120578 (120596)) 120579)

+sin ((120578 (120596) minus 1) 120579) cos (120578 (120596) 120596)

(1 minus 120578 (120596))

+ cos ((120578 (120596) minus 1) 120579)

(90)

and 120578(120596) = 0544484


belongs to 1198670544484

(119881) when120596 = 31205872 and 119878

1199011belongs to 119867

05(119881) when 120596 = 2120587


5 Conclusion



(Ω)]2 where 119904 gt 0 then (u 119901)


(Ω) if 119904 lt 1+120578(120596) Thismeans

[119867119904(Ω)]2


(Ω) for 119904 lt 1544484 when 120596 =3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 15 when 120596 = 2120587

(91)


u[119867119904(Ω)]2 +

10038171003817100381710038171199011003817100381710038171003817119867119904minus1(Ω)

le 119862f[119867119904minus2(Ω)]2 (92)



u = u119903+ 1205821198781 119901 = 119901

119903+ 1205731198781199011 (93)

and if 120596 = 2120587

u = u119903+ 12058211198781+ 11198781 119901 = 119901

119903+ 12057311198781199011

+ 12057311198781199011 (94)


(Ω)]2 (u119903 119901119903) belongs to [119867

119904(Ω)]2times

119867119904minus1

(Ω) for 119904 lt 1 + 1205781(120596) where 120578



[119867119904(Ω)]2


(Ω) for 119904 lt 1908529 when 120596 =3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 25 when 120596 = 2120587

(95)


1003817100381710038171003817[119867119904(Ω)]2+1003817100381710038171003817119901119903

1003817100381710038171003817119867119904minus1(Ω)+10038161003816100381610038161205821

1003816100381610038161003816 +10038161003816100381610038161205731

1003816100381610038161003816 le 119862f[119867119904minus2(Ω)]2

(96)




u119903= u119903+ 12058221198782 119901

119903= 119901119903+ 12057321198781199012

(97)

(u119903 119901) isin [119867

119904(Ω)]2times 119867119904minus1

(Ω) for 119904 lt 1 + 1205782(120596) where 120578

2(120596)


[119867119904(Ω)]2


(Ω) for 119904 lt 1205782(3120587

2) when 120596 =

3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 3 5 when 120596 = 2120587

(98)



1003817100381710038171003817u1199031003817100381710038171003817[119867119904(Ω)]2

+1003817100381710038171003817119901119903

1003817100381710038171003817119867119904(Ω)+10038161003816100381610038161205821

1003816100381610038161003816 +10038161003816100381610038161205822

1003816100381610038161003816 +10038161003816100381610038161205731

1003816100381610038161003816 +10038161003816100381610038161205732

1003816100381610038161003816

le 119862f[119867119904minus2(Ω)]2

(99)


u = u119903+ 12058211198781+ 12058221198782+ sdot sdot sdot + 120582

119896119878119896

119901 = 119901119903+ 12057311198781199011

+ 12057321198781199012

+ sdot sdot sdot + 120573119896119878119901119896

(100)



119904(Ω)]2times

119867119904minus1

(Ω) for 119904 lt 1 + 120578119896(120596) where 120578





Acknowledgment


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












120575 = 0

120574119904ℎ2120587119911 + 120575119888ℎ2120587119911 = 0

120573 = 0

120572119904ℎ2120587119911 + 120573119888ℎ2120587119911 + 2120587120574119904ℎ2120587119911 + 2120587120575119888ℎ2120587119911 = 0

(46)


119911=



120593 =120572

2119894ℎ119896(120579) + 119894120574 (

119896

2minus 1)119892

119896(120579) (47)



int

2120587

0

1003816100381610038161003816ℎ1198961003816100381610038161003816 119889120579 = int

2120587

0

10038161003816100381610038161198921198961003816100381610038161003816 119889120579 = 1 (48)


120593 = 120572 + 120573120579 + 120574 sin 2120579 + 120575 cos 2120579 (49)


120572 + 120575 = 0

120572 + 2120587120573 + 120575 = 0

120573 + 2120574 = 0

(50)


120593 = 120572 (1 minus cos 2120579) = 2120572 sin2 2120579 (51)


Remark 9 If

119863119896= 119911119896= minus119894

119896

2 for 119896 = 0 119896 = 2 119896 isin N (52)



minus(119904 + 2) cap 119863119896

Is then

119889 (119911) =minus119904ℎ22120587119911

1199112 (1 + 1199112) for 119911 = 0 119911 = plusmn 119894 (53)

We verify that 119889(119911119896) = 120597

119911119889(119911119896) = 0 and 120597

2

119911119889(119911119896) = 0 Hence




V (119911 120579) =1205931(120579)

(119911 + 119894)+ 1199081(119911 120579) (54)


1



V (119911 120579) =120595119896(120579)

(119911 + 119894 (1198962))2

+120593119896(120579)

(119911 + 119894 (1198962))+ 119908119896(119911 120579) (55)

where 120595119896and 119908


119896

satisfies

119871 (119911119863120579) 120593119896= 2119894119896(1 minus

1198962

4)120595119896minus 2119894119896120595

10158401015840

119896

120593119896(0) = 120593

119896(2120587) = 120597

119905120593119896(0) = 120597

120579120593119896(2120587) = 0

(56)


1199061(119911 120579) =

sin 120579119904ℎ119911120579

119911for 119911 = 0

1199061(0 120579) = 120579 sin 120579

1199062(119911 120579) =

1

1199112 + 1(cos 120579119904ℎ119911120579

119911minus sin 120579119888ℎ119911120579)

for 119911 = 0 119911 = plusmn 119894

1199062(0 120579) = 120579 cos 120579 minus sin 120579

1199062(plusmn119894 120579) =

1

2(sin 2120579

2minus 120579)

(57)


1199061(minus119894

119896

2 120579) = minus

1

119896ℎ119896(120579)

1199062(minus119894

119896

2 120579) =

2

119896 (119896 + 2)119892119896(120579)

(58)


1and 119906


1and 119906

2are whole


119889119911= 1199061(119911 2120587)119863

1205791199062(119911 2120587) minus 119906

2(119911 2120587)119863

1205791199061(119911 2120587)

(59)





1199081(119911 120579) and for all 119911



120579) 1199081

(60)



119871 (minus119894 119863120579) 1205951= 0

1205951(0) = (119911 minus 119894) 119908

1(119911 0)

(61)

Then

1205951(0) = 0 (62)

and even

1205951(2120587) = 0

1205951015840

1(0) = lim

120579rarr0

120595 (120579)

120579= lim120579rarr0

V (119911 120579) minus (119911 + 1)1199081(119911 120579)

120579= 0

(63)

and even for

1205951015840

1(2120587) = 0 (64)



for 119896 = 1 and 119896 = 3 is

V (119911 120579) =120595119896(120579)

(119911 + 119894 (1198962))2+

120593119896(120579)

(119911 + 119894 (1198962))+ 119908119896(119911 120579)

for 119896 = 1 119896 = 3

(65)


ℎ(119911 minus 119911119896)2

= 119871 (119911119863120579) 120595119896+ (119911 minus 119911

119896) 119871 (119911 119863

120579) 120593119896

+ (119911 minus 119911119896)2

119871 (119911119863120579) 119908119896

(66)




119871 (119911119863120579) 120593119896= minus

119871 (119911119863120579) + 119871 (119911119896 119863120579)

(119911 minus 119911119896)

120595119896

+ (119911 minus 119911119896) (ℎ minus 119871 (119911 119863120579)119908

119896)

(67)

And if 119911 tends to 119911119896

119871 (119911119863120579) 120593119896= 2119894119896(1 minus

1198962

4)120595119896minus 2119894119896120595

10158401015840

119896(68)


proof


adjoint

⟨119871120595119896 120593119896⟩ = ⟨120595

119896 119871120593119896⟩

= int

2120587

0

120595119896(2119896(1 minus

1198962

4)120595119896minus 2119894119896120595

10158401015840

119896)119889120579

(69)

For 0 lt 119896 lt 4

int

2120587

0

1205952

119896119889120579 = int

2120587

0

12059510158402

119896119889120579 = 0 (70)


V (119911 120579) =120593119896(120579)

(119911 + 119894 (1198962))+ 119908119896(119911 120579)

for 119896 = 1 or 119896 = 3

(71)



119908119904(119905 120579) =

1

radic2120587

int

+infin

minusinfin

119890119894119905119911V (119911 minus 119894 (119904 + 2) + 119894120578 120579) 119889119911

120583 (120579) = 119894radic2120587119890119894119905119911V (119911 120579) for fixed 119905

(72)


V (119905 120579) = 119890(119904+2minus120598)119905

119908119904(119905 120579) + 2119894120587 sum

minus(119904+2)ltIm(119911)lt0119877119904 (73)


119911119896 Then

(i) for 119904 = minus1

V (119905 120579) = 119890(1minus120578)119905

119908minus1

(119905 120579) + 119894radic21205871205951(120579) (74)

(ii) for 119904 = 0

V (119905 120579) = 119890(2minus120578)119905

1199080(119905 120579)

+ 119894radic2120587 (11989031199052

1205953(120579) + 119890

11990521205951(120579) + 119890

1199051205952(120579))

(75)




1205911(119903 120579) = 119903

32(sin 3

2120579 minus 3 sin 120579

2)

1205831(119903 120579) = 119903

32(cos 3

2120579 minus cos 120579

2)

(76)


1205911(119903 120579) = 119903

32(sin 3

2120579 minus 3 sin 120579

2)

1205831(119903 120579) = 119903

32(cos 3

2120579 minus cos 120579

2)

1205912(119903 120579) = 119903

52(sin 5

2120579 minus 5 sin 120579

2)

1205832(119903 120579) = 119903

52(cos 5

2120579 minus cos 120579

2)

(77)




11986752minus120598

(119881) 1205912and 120583




Δ2120593 = 0 119894119899 119881

120593 =120597120593

120597119899= 0 119900119899 Γ cap 120597119881

(78)




2(Ω)]2


form

u = u119903+ u119904 (79)

where u119903is in [119867

2(Ω) cap 119867

1



u119904= 1205821199041+ 1199041

(80)

with

1199041(119903 120579) = 119903

12(3 sin 120579 sin 120579

2 3 (1 minus cos 120579) sin 120579

2)

1199041(119903 120579) = 119903

12(2 sin 120579

2+ sin 120579 cos 120579

2 (1 minus cos 120579) cos 120579

2)

(81)






if 120587 lt 120596 lt 2120587 1 minus120587

120596lt 120578 (120596) lt

120587

120596 120578 (120596) gt

1

2 (83)


119882 = u isin [1198671

0(Ω)]2



defined such that

nabla119901 = f + ]Δu in 1198631015840(Ω) (85)



119901 = 119901119903+ 119901119904 119901119903isin 1198671(Ω) 119901

119904= 1205731198781199011 (86)


nabla1198781199011

= ]Δ (curl (1199031+120578(120596)120595 (120579))) (87)


Consider

1198781199011

(119903 120579)

=

minus119903120578(120596)minus1

((1 + 120578 (120596))2

(120597120595 (120579) 120597120579) + (1205973120595 (120579) 120597120579))

(1 + 120578 (120596))

(88)


120595 (120579) = 3 sin(120579

2) minus sin(

3

2120579) 120578 (120596) =

1

2 (89)

in the case of 120596 = 31205872

120595 (120579) =sin ((1 + 120578 (120596)) 120579) cos (120578 (120596) 120596)

(1 + 120578 (120596))

minus cos ((1 + 120578 (120596)) 120579)

+sin ((120578 (120596) minus 1) 120579) cos (120578 (120596) 120596)

(1 minus 120578 (120596))

+ cos ((120578 (120596) minus 1) 120579)

(90)

and 120578(120596) = 0544484


belongs to 1198670544484

(119881) when120596 = 31205872 and 119878

1199011belongs to 119867

05(119881) when 120596 = 2120587


5 Conclusion



(Ω)]2 where 119904 gt 0 then (u 119901)


(Ω) if 119904 lt 1+120578(120596) Thismeans

[119867119904(Ω)]2


(Ω) for 119904 lt 1544484 when 120596 =3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 15 when 120596 = 2120587

(91)


u[119867119904(Ω)]2 +

10038171003817100381710038171199011003817100381710038171003817119867119904minus1(Ω)

le 119862f[119867119904minus2(Ω)]2 (92)



u = u119903+ 1205821198781 119901 = 119901

119903+ 1205731198781199011 (93)

and if 120596 = 2120587

u = u119903+ 12058211198781+ 11198781 119901 = 119901

119903+ 12057311198781199011

+ 12057311198781199011 (94)


(Ω)]2 (u119903 119901119903) belongs to [119867

119904(Ω)]2times

119867119904minus1

(Ω) for 119904 lt 1 + 1205781(120596) where 120578



[119867119904(Ω)]2


(Ω) for 119904 lt 1908529 when 120596 =3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 25 when 120596 = 2120587

(95)


1003817100381710038171003817[119867119904(Ω)]2+1003817100381710038171003817119901119903

1003817100381710038171003817119867119904minus1(Ω)+10038161003816100381610038161205821

1003816100381610038161003816 +10038161003816100381610038161205731

1003816100381610038161003816 le 119862f[119867119904minus2(Ω)]2

(96)




u119903= u119903+ 12058221198782 119901

119903= 119901119903+ 12057321198781199012

(97)

(u119903 119901) isin [119867

119904(Ω)]2times 119867119904minus1

(Ω) for 119904 lt 1 + 1205782(120596) where 120578

2(120596)


[119867119904(Ω)]2


(Ω) for 119904 lt 1205782(3120587

2) when 120596 =

3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 3 5 when 120596 = 2120587

(98)



1003817100381710038171003817u1199031003817100381710038171003817[119867119904(Ω)]2

+1003817100381710038171003817119901119903

1003817100381710038171003817119867119904(Ω)+10038161003816100381610038161205821

1003816100381610038161003816 +10038161003816100381610038161205822

1003816100381610038161003816 +10038161003816100381610038161205731

1003816100381610038161003816 +10038161003816100381610038161205732

1003816100381610038161003816

le 119862f[119867119904minus2(Ω)]2

(99)


u = u119903+ 12058211198781+ 12058221198782+ sdot sdot sdot + 120582

119896119878119896

119901 = 119901119903+ 12057311198781199011

+ 12057321198781199012

+ sdot sdot sdot + 120573119896119878119901119896

(100)



119904(Ω)]2times

119867119904minus1

(Ω) for 119904 lt 1 + 120578119896(120596) where 120578





Acknowledgment


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119871 (minus119894 119863120579) 1205951= 0

1205951(0) = (119911 minus 119894) 119908

1(119911 0)

(61)

Then

1205951(0) = 0 (62)

and even

1205951(2120587) = 0

1205951015840

1(0) = lim

120579rarr0

120595 (120579)

120579= lim120579rarr0

V (119911 120579) minus (119911 + 1)1199081(119911 120579)

120579= 0

(63)

and even for

1205951015840

1(2120587) = 0 (64)



for 119896 = 1 and 119896 = 3 is

V (119911 120579) =120595119896(120579)

(119911 + 119894 (1198962))2+

120593119896(120579)

(119911 + 119894 (1198962))+ 119908119896(119911 120579)

for 119896 = 1 119896 = 3

(65)


ℎ(119911 minus 119911119896)2

= 119871 (119911119863120579) 120595119896+ (119911 minus 119911

119896) 119871 (119911 119863

120579) 120593119896

+ (119911 minus 119911119896)2

119871 (119911119863120579) 119908119896

(66)




119871 (119911119863120579) 120593119896= minus

119871 (119911119863120579) + 119871 (119911119896 119863120579)

(119911 minus 119911119896)

120595119896

+ (119911 minus 119911119896) (ℎ minus 119871 (119911 119863120579)119908

119896)

(67)

And if 119911 tends to 119911119896

119871 (119911119863120579) 120593119896= 2119894119896(1 minus

1198962

4)120595119896minus 2119894119896120595

10158401015840

119896(68)


proof


adjoint

⟨119871120595119896 120593119896⟩ = ⟨120595

119896 119871120593119896⟩

= int

2120587

0

120595119896(2119896(1 minus

1198962

4)120595119896minus 2119894119896120595

10158401015840

119896)119889120579

(69)

For 0 lt 119896 lt 4

int

2120587

0

1205952

119896119889120579 = int

2120587

0

12059510158402

119896119889120579 = 0 (70)


V (119911 120579) =120593119896(120579)

(119911 + 119894 (1198962))+ 119908119896(119911 120579)

for 119896 = 1 or 119896 = 3

(71)



119908119904(119905 120579) =

1

radic2120587

int

+infin

minusinfin

119890119894119905119911V (119911 minus 119894 (119904 + 2) + 119894120578 120579) 119889119911

120583 (120579) = 119894radic2120587119890119894119905119911V (119911 120579) for fixed 119905

(72)


V (119905 120579) = 119890(119904+2minus120598)119905

119908119904(119905 120579) + 2119894120587 sum

minus(119904+2)ltIm(119911)lt0119877119904 (73)


119911119896 Then

(i) for 119904 = minus1

V (119905 120579) = 119890(1minus120578)119905

119908minus1

(119905 120579) + 119894radic21205871205951(120579) (74)

(ii) for 119904 = 0

V (119905 120579) = 119890(2minus120578)119905

1199080(119905 120579)

+ 119894radic2120587 (11989031199052

1205953(120579) + 119890

11990521205951(120579) + 119890

1199051205952(120579))

(75)




1205911(119903 120579) = 119903

32(sin 3

2120579 minus 3 sin 120579

2)

1205831(119903 120579) = 119903

32(cos 3

2120579 minus cos 120579

2)

(76)


1205911(119903 120579) = 119903

32(sin 3

2120579 minus 3 sin 120579

2)

1205831(119903 120579) = 119903

32(cos 3

2120579 minus cos 120579

2)

1205912(119903 120579) = 119903

52(sin 5

2120579 minus 5 sin 120579

2)

1205832(119903 120579) = 119903

52(cos 5

2120579 minus cos 120579

2)

(77)




11986752minus120598

(119881) 1205912and 120583




Δ2120593 = 0 119894119899 119881

120593 =120597120593

120597119899= 0 119900119899 Γ cap 120597119881

(78)




2(Ω)]2


form

u = u119903+ u119904 (79)

where u119903is in [119867

2(Ω) cap 119867

1



u119904= 1205821199041+ 1199041

(80)

with

1199041(119903 120579) = 119903

12(3 sin 120579 sin 120579

2 3 (1 minus cos 120579) sin 120579

2)

1199041(119903 120579) = 119903

12(2 sin 120579

2+ sin 120579 cos 120579

2 (1 minus cos 120579) cos 120579

2)

(81)






if 120587 lt 120596 lt 2120587 1 minus120587

120596lt 120578 (120596) lt

120587

120596 120578 (120596) gt

1

2 (83)


119882 = u isin [1198671

0(Ω)]2



defined such that

nabla119901 = f + ]Δu in 1198631015840(Ω) (85)



119901 = 119901119903+ 119901119904 119901119903isin 1198671(Ω) 119901

119904= 1205731198781199011 (86)


nabla1198781199011

= ]Δ (curl (1199031+120578(120596)120595 (120579))) (87)


Consider

1198781199011

(119903 120579)

=

minus119903120578(120596)minus1

((1 + 120578 (120596))2

(120597120595 (120579) 120597120579) + (1205973120595 (120579) 120597120579))

(1 + 120578 (120596))

(88)


120595 (120579) = 3 sin(120579

2) minus sin(

3

2120579) 120578 (120596) =

1

2 (89)

in the case of 120596 = 31205872

120595 (120579) =sin ((1 + 120578 (120596)) 120579) cos (120578 (120596) 120596)

(1 + 120578 (120596))

minus cos ((1 + 120578 (120596)) 120579)

+sin ((120578 (120596) minus 1) 120579) cos (120578 (120596) 120596)

(1 minus 120578 (120596))

+ cos ((120578 (120596) minus 1) 120579)

(90)

and 120578(120596) = 0544484


belongs to 1198670544484

(119881) when120596 = 31205872 and 119878

1199011belongs to 119867

05(119881) when 120596 = 2120587


5 Conclusion



(Ω)]2 where 119904 gt 0 then (u 119901)


(Ω) if 119904 lt 1+120578(120596) Thismeans

[119867119904(Ω)]2


(Ω) for 119904 lt 1544484 when 120596 =3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 15 when 120596 = 2120587

(91)


u[119867119904(Ω)]2 +

10038171003817100381710038171199011003817100381710038171003817119867119904minus1(Ω)

le 119862f[119867119904minus2(Ω)]2 (92)



u = u119903+ 1205821198781 119901 = 119901

119903+ 1205731198781199011 (93)

and if 120596 = 2120587

u = u119903+ 12058211198781+ 11198781 119901 = 119901

119903+ 12057311198781199011

+ 12057311198781199011 (94)


(Ω)]2 (u119903 119901119903) belongs to [119867

119904(Ω)]2times

119867119904minus1

(Ω) for 119904 lt 1 + 1205781(120596) where 120578



[119867119904(Ω)]2


(Ω) for 119904 lt 1908529 when 120596 =3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 25 when 120596 = 2120587

(95)


1003817100381710038171003817[119867119904(Ω)]2+1003817100381710038171003817119901119903

1003817100381710038171003817119867119904minus1(Ω)+10038161003816100381610038161205821

1003816100381610038161003816 +10038161003816100381610038161205731

1003816100381610038161003816 le 119862f[119867119904minus2(Ω)]2

(96)




u119903= u119903+ 12058221198782 119901

119903= 119901119903+ 12057321198781199012

(97)

(u119903 119901) isin [119867

119904(Ω)]2times 119867119904minus1

(Ω) for 119904 lt 1 + 1205782(120596) where 120578

2(120596)


[119867119904(Ω)]2


(Ω) for 119904 lt 1205782(3120587

2) when 120596 =

3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 3 5 when 120596 = 2120587

(98)



1003817100381710038171003817u1199031003817100381710038171003817[119867119904(Ω)]2

+1003817100381710038171003817119901119903

1003817100381710038171003817119867119904(Ω)+10038161003816100381610038161205821

1003816100381610038161003816 +10038161003816100381610038161205822

1003816100381610038161003816 +10038161003816100381610038161205731

1003816100381610038161003816 +10038161003816100381610038161205732

1003816100381610038161003816

le 119862f[119867119904minus2(Ω)]2

(99)


u = u119903+ 12058211198781+ 12058221198782+ sdot sdot sdot + 120582

119896119878119896

119901 = 119901119903+ 12057311198781199011

+ 12057321198781199012

+ sdot sdot sdot + 120573119896119878119901119896

(100)



119904(Ω)]2times

119867119904minus1

(Ω) for 119904 lt 1 + 120578119896(120596) where 120578





Acknowledgment


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












11986752minus120598

(119881) 1205912and 120583




Δ2120593 = 0 119894119899 119881

120593 =120597120593

120597119899= 0 119900119899 Γ cap 120597119881

(78)




2(Ω)]2


form

u = u119903+ u119904 (79)

where u119903is in [119867

2(Ω) cap 119867

1



u119904= 1205821199041+ 1199041

(80)

with

1199041(119903 120579) = 119903

12(3 sin 120579 sin 120579

2 3 (1 minus cos 120579) sin 120579

2)

1199041(119903 120579) = 119903

12(2 sin 120579

2+ sin 120579 cos 120579

2 (1 minus cos 120579) cos 120579

2)

(81)






if 120587 lt 120596 lt 2120587 1 minus120587

120596lt 120578 (120596) lt

120587

120596 120578 (120596) gt

1

2 (83)


119882 = u isin [1198671

0(Ω)]2



defined such that

nabla119901 = f + ]Δu in 1198631015840(Ω) (85)



119901 = 119901119903+ 119901119904 119901119903isin 1198671(Ω) 119901

119904= 1205731198781199011 (86)


nabla1198781199011

= ]Δ (curl (1199031+120578(120596)120595 (120579))) (87)


Consider

1198781199011

(119903 120579)

=

minus119903120578(120596)minus1

((1 + 120578 (120596))2

(120597120595 (120579) 120597120579) + (1205973120595 (120579) 120597120579))

(1 + 120578 (120596))

(88)


120595 (120579) = 3 sin(120579

2) minus sin(

3

2120579) 120578 (120596) =

1

2 (89)

in the case of 120596 = 31205872

120595 (120579) =sin ((1 + 120578 (120596)) 120579) cos (120578 (120596) 120596)

(1 + 120578 (120596))

minus cos ((1 + 120578 (120596)) 120579)

+sin ((120578 (120596) minus 1) 120579) cos (120578 (120596) 120596)

(1 minus 120578 (120596))

+ cos ((120578 (120596) minus 1) 120579)

(90)

and 120578(120596) = 0544484


belongs to 1198670544484

(119881) when120596 = 31205872 and 119878

1199011belongs to 119867

05(119881) when 120596 = 2120587


5 Conclusion



(Ω)]2 where 119904 gt 0 then (u 119901)


(Ω) if 119904 lt 1+120578(120596) Thismeans

[119867119904(Ω)]2


(Ω) for 119904 lt 1544484 when 120596 =3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 15 when 120596 = 2120587

(91)


u[119867119904(Ω)]2 +

10038171003817100381710038171199011003817100381710038171003817119867119904minus1(Ω)

le 119862f[119867119904minus2(Ω)]2 (92)



u = u119903+ 1205821198781 119901 = 119901

119903+ 1205731198781199011 (93)

and if 120596 = 2120587

u = u119903+ 12058211198781+ 11198781 119901 = 119901

119903+ 12057311198781199011

+ 12057311198781199011 (94)


(Ω)]2 (u119903 119901119903) belongs to [119867

119904(Ω)]2times

119867119904minus1

(Ω) for 119904 lt 1 + 1205781(120596) where 120578



[119867119904(Ω)]2


(Ω) for 119904 lt 1908529 when 120596 =3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 25 when 120596 = 2120587

(95)


1003817100381710038171003817[119867119904(Ω)]2+1003817100381710038171003817119901119903

1003817100381710038171003817119867119904minus1(Ω)+10038161003816100381610038161205821

1003816100381610038161003816 +10038161003816100381610038161205731

1003816100381610038161003816 le 119862f[119867119904minus2(Ω)]2

(96)




u119903= u119903+ 12058221198782 119901

119903= 119901119903+ 12057321198781199012

(97)

(u119903 119901) isin [119867

119904(Ω)]2times 119867119904minus1

(Ω) for 119904 lt 1 + 1205782(120596) where 120578

2(120596)


[119867119904(Ω)]2


(Ω) for 119904 lt 1205782(3120587

2) when 120596 =

3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 3 5 when 120596 = 2120587

(98)



1003817100381710038171003817u1199031003817100381710038171003817[119867119904(Ω)]2

+1003817100381710038171003817119901119903

1003817100381710038171003817119867119904(Ω)+10038161003816100381610038161205821

1003816100381610038161003816 +10038161003816100381610038161205822

1003816100381610038161003816 +10038161003816100381610038161205731

1003816100381610038161003816 +10038161003816100381610038161205732

1003816100381610038161003816

le 119862f[119867119904minus2(Ω)]2

(99)


u = u119903+ 12058211198781+ 12058221198782+ sdot sdot sdot + 120582

119896119878119896

119901 = 119901119903+ 12057311198781199011

+ 12057321198781199012

+ sdot sdot sdot + 120573119896119878119901119896

(100)



119904(Ω)]2times

119867119904minus1

(Ω) for 119904 lt 1 + 120578119896(120596) where 120578





Acknowledgment


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











u = u119903+ 1205821198781 119901 = 119901

119903+ 1205731198781199011 (93)

and if 120596 = 2120587

u = u119903+ 12058211198781+ 11198781 119901 = 119901

119903+ 12057311198781199011

+ 12057311198781199011 (94)


(Ω)]2 (u119903 119901119903) belongs to [119867

119904(Ω)]2times

119867119904minus1

(Ω) for 119904 lt 1 + 1205781(120596) where 120578



[119867119904(Ω)]2


(Ω) for 119904 lt 1908529 when 120596 =3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 25 when 120596 = 2120587

(95)


1003817100381710038171003817[119867119904(Ω)]2+1003817100381710038171003817119901119903

1003817100381710038171003817119867119904minus1(Ω)+10038161003816100381610038161205821

1003816100381610038161003816 +10038161003816100381610038161205731

1003816100381610038161003816 le 119862f[119867119904minus2(Ω)]2

(96)




u119903= u119903+ 12058221198782 119901

119903= 119901119903+ 12057321198781199012

(97)

(u119903 119901) isin [119867

119904(Ω)]2times 119867119904minus1

(Ω) for 119904 lt 1 + 1205782(120596) where 120578

2(120596)


[119867119904(Ω)]2


(Ω) for 119904 lt 1205782(3120587

2) when 120596 =

3120587

2

[119867119904(Ω)]2


(Ω) for 119904 lt 3 5 when 120596 = 2120587

(98)



1003817100381710038171003817u1199031003817100381710038171003817[119867119904(Ω)]2

+1003817100381710038171003817119901119903

1003817100381710038171003817119867119904(Ω)+10038161003816100381610038161205821

1003816100381610038161003816 +10038161003816100381610038161205822

1003816100381610038161003816 +10038161003816100381610038161205731

1003816100381610038161003816 +10038161003816100381610038161205732

1003816100381610038161003816

le 119862f[119867119904minus2(Ω)]2

(99)


u = u119903+ 12058211198781+ 12058221198782+ sdot sdot sdot + 120582

119896119878119896

119901 = 119901119903+ 12057311198781199011

+ 12057321198781199012

+ sdot sdot sdot + 120573119896119878119901119896

(100)



119904(Ω)]2times

119867119904minus1

(Ω) for 119904 lt 1 + 120578119896(120596) where 120578





Acknowledgment


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









research article geometric singularities of the stokes problem

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