Research Article Geometric Singularities of the Stokes Problem
Transcript of 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
4 Abstract and Applied Analysis
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)
Abstract and Applied Analysis 5
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)
6 Abstract and Applied Analysis
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)
Abstract and Applied Analysis 7
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)
where Γ is the boundary of Ω
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
times 119867119904minus1
(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)
8 Abstract and Applied Analysis
(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
times 119867119904minus1
(Ω) for 119904 lt 1908529 when 120596 =3120587
2
[119867119904(Ω)]2
times 119867119904minus1
(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)
(3) If f belongs to [119867119904minus2
(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
times 119867119904minus1
(Ω) for 119904 lt 1205782(3120587
2) when 120596 =
3120587
2
[119867119904(Ω)]2
times 119867119904minus1
(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
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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
4 Abstract and Applied Analysis
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)
Abstract and Applied Analysis 5
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)
6 Abstract and Applied Analysis
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)
Abstract and Applied Analysis 7
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)
where Γ is the boundary of Ω
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
times 119867119904minus1
(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)
8 Abstract and Applied Analysis
(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
times 119867119904minus1
(Ω) for 119904 lt 1908529 when 120596 =3120587
2
[119867119904(Ω)]2
times 119867119904minus1
(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)
(3) If f belongs to [119867119904minus2
(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
times 119867119904minus1
(Ω) for 119904 lt 1205782(3120587
2) when 120596 =
3120587
2
[119867119904(Ω)]2
times 119867119904minus1
(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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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
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
4 Abstract and Applied Analysis
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)
Abstract and Applied Analysis 5
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)
6 Abstract and Applied Analysis
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)
Abstract and Applied Analysis 7
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)
where Γ is the boundary of Ω
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
times 119867119904minus1
(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)
8 Abstract and Applied Analysis
(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
times 119867119904minus1
(Ω) for 119904 lt 1908529 when 120596 =3120587
2
[119867119904(Ω)]2
times 119867119904minus1
(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)
(3) If f belongs to [119867119904minus2
(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
times 119867119904minus1
(Ω) for 119904 lt 1205782(3120587
2) when 120596 =
3120587
2
[119867119904(Ω)]2
times 119867119904minus1
(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
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 Abstract and Applied Analysis
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)
Abstract and Applied Analysis 5
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)
6 Abstract and Applied Analysis
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)
Abstract and Applied Analysis 7
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)
where Γ is the boundary of Ω
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
times 119867119904minus1
(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)
8 Abstract and Applied Analysis
(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
times 119867119904minus1
(Ω) for 119904 lt 1908529 when 120596 =3120587
2
[119867119904(Ω)]2
times 119867119904minus1
(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)
(3) If f belongs to [119867119904minus2
(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
times 119867119904minus1
(Ω) for 119904 lt 1205782(3120587
2) when 120596 =
3120587
2
[119867119904(Ω)]2
times 119867119904minus1
(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
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Discrete Dynamics in Nature and Society
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 5
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)
6 Abstract and Applied Analysis
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)
Abstract and Applied Analysis 7
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)
where Γ is the boundary of Ω
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
times 119867119904minus1
(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)
8 Abstract and Applied Analysis
(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
times 119867119904minus1
(Ω) for 119904 lt 1908529 when 120596 =3120587
2
[119867119904(Ω)]2
times 119867119904minus1
(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)
(3) If f belongs to [119867119904minus2
(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
times 119867119904minus1
(Ω) for 119904 lt 1205782(3120587
2) when 120596 =
3120587
2
[119867119904(Ω)]2
times 119867119904minus1
(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
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Abstract and Applied Analysis
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)
Abstract and Applied Analysis 7
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)
where Γ is the boundary of Ω
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
times 119867119904minus1
(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)
8 Abstract and Applied Analysis
(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
times 119867119904minus1
(Ω) for 119904 lt 1908529 when 120596 =3120587
2
[119867119904(Ω)]2
times 119867119904minus1
(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)
(3) If f belongs to [119867119904minus2
(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
times 119867119904minus1
(Ω) for 119904 lt 1205782(3120587
2) when 120596 =
3120587
2
[119867119904(Ω)]2
times 119867119904minus1
(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
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 7
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)
where Γ is the boundary of Ω
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
times 119867119904minus1
(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)
8 Abstract and Applied Analysis
(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
times 119867119904minus1
(Ω) for 119904 lt 1908529 when 120596 =3120587
2
[119867119904(Ω)]2
times 119867119904minus1
(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)
(3) If f belongs to [119867119904minus2
(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
times 119867119904minus1
(Ω) for 119904 lt 1205782(3120587
2) when 120596 =
3120587
2
[119867119904(Ω)]2
times 119867119904minus1
(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
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
8 Abstract and Applied Analysis
(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
times 119867119904minus1
(Ω) for 119904 lt 1908529 when 120596 =3120587
2
[119867119904(Ω)]2
times 119867119904minus1
(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)
(3) If f belongs to [119867119904minus2
(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
times 119867119904minus1
(Ω) for 119904 lt 1205782(3120587
2) when 120596 =
3120587
2
[119867119904(Ω)]2
times 119867119904minus1
(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
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of