Int. J.IndustrialMathematisVol. 3,No. 1(2011)1-7
Exat and Numerial Solutions for Nonlinear
Dierential Equation of Jerey-Hamel Flow
R.Ellahi
Departmentof Mathematis,Faulty ofBasiandAppliedSienesIIUI,H-10 Setor,Islamabad,
Pakistan.
Reeived16February 2011;aepted5May2011.
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ThispaperlooksattheanalysisofJeeryHamelow. Theinvestigationismainlyaimedto
determineanexatanalytisolutionforanonlinearproblem. Tothebestofmyknowledge,
no suhanalysis isavailablein theliteraturewhihan desribe theexat solutionofthe
Jerey-Hamel ow. Besides this a omparative study between the numerial and exat
solutionsis presented. The eetsof thevariousparameters intrinsito theproblemsare
analyzed and depitedvia graphs.
Keywords: Nonlinearproblem; Exatanalytisolution;NumerialsolutionJerey-Hamelow.
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1 Introdution
The ow between two planes whih meet at an angle was rst analyzed by Jeery [7 ℄
and Hamel [6℄. Under suitable assumptions, the problem an be reduedto an ordinary
dierentialequation. The inompressiblevisousuidowthroughonvergent{divergent
hannelsis one of the mostappliable ases inuid mehanis, ivil, environmental,
me-hanialandbio-mehanialengineering. AlotofinformationandreferenesaboutJeery
Hamel ow an be found in the refs. [1, 4, 5 , 13 , 14 ℄. Most sienti problems suh as
Jeery Hamel ows are inherently of nonlinearity. Exept a limited number of these
problem, most of them do not have exat analytial solution. Therefore, these
nonlin-ear equations have been solved either numerially [2 , 15℄ or by perturbation methods
[10 , 11 , 12 ℄. Very little [3 , 8 , 9 ℄ has been yet said in the regime of exat solutions for
nonlinearproblems. Theonvergeneofthesolutionandthelargeparameterare
deien-iesof numerialandtheperturbationmethodsrespetively. We onneourselveshereto
present a general exat analytial solutionand theomparison of numerial solutionas
well. It is also worth mentioning that our exat analytial solution is not only validfor
smallbutalso forlargevaluesof emergingparameters.
2 Problem formulation
Considerthesteadytwo-dimensionalowof aninompressiblevisousuidfromasoure
or sinkat the intersetion between two rigid plane walls that theangle betweenthem is
2asshowninFig. 1 givenbelow:
Fig. 1. Geometryoftheproblem.
We assume that the veloity is only along radial diretion and depend upon r and
h, i.e., V(u(r;h);0) [10 , 11 ℄. Using ontinuity and the Navier{Stokes equations in polar
oordinateswe have
rr
(ru(r;))=0; (2.1)
u(r;) u(r;) r = 1 p r + 2 u(r;) r 2 + 1 r u(r;) r + 1 r 2 2 u(r;) 2 u(r;) r 2 (2.2) 1 r p + 2 r 2 u(r;) =0:
Equation (2.1) yields
f()ru(r;): (2.3)
Introduing
F( ) f() f max ;= (2.4)
and eliminating p in Eqs. (2.2) and (2.3), we obtainthe following equation for the
nor-malizedfuntionF() as[11℄
F 000
( )+2R e F ()F 0
()+4 2
F 0
()=0: (2.5)
The subjeted orrespondingboundaryonditionsare
F(0)=1;F 0
( 0)=0;F (1)=0; (2.6)
3 Exat solution
A rstintegralof equation (2.5) is
F 00
+R e F 2 +4 2 F = 1 ; (3.7) where 1
isanarbitraryonstant. Equation(3.7) hasthetranslationalsymmetryinand
its order an be reduedas
1
2 dF
02
dF
+R e F 2 +4 2 F = 1 : (3.8)
Hene we have
F 0 = r 2 1 F 2 3 R eF 3 4 2 F 2 +2 2 ; (3.9) where 2
isa furtheronstant. The boundaryonditions((2.6),b,) then requirethat
1 + 2 = 1 3
R e+2 2
: (3.10)
Sine we want F 0
>0;we omitthenegative signin (3.9). Thus
Z F 1 dE q 2 1 E 2 3 R e E
3 4 2 E 2 + 2 3
R e+4 2
2
1
= (3.11)
subjetto
F (1)=0: (3.12)
isan exat solutionof JeeryHamelow.
4 Numerial Solution
In this setion we present the numerial solution of Jeery Hamel ow by a so alled
method"Shooting method". To applyshootingmethodon Eqs. (2.5) and(2.6), wewrite
ourthirdorder equation inthree rst orderequations
F 0
( )=v( ) (4.13)
F 00
()=u()=v 0
( ) (4.14)
u 0
()= 2R e F()v()+4 2
v()
(4.15)
F (0)=1;v( 0)=0;F (1)=0 (4.16)
and missingonditionis
F 00
( 0)=s or u( 0)=s (4.17)
Equations(4.13)-(4.17) an be dierentiatedwithrespetto sto obtain
F 0
()=V () (4.18)
U 0
()= 2R e(F
()v()+F()V () )+4 2
V ()
(4.20)
F
(0)=0;V (0)=0;U(0)=1 (4.21)
where
F
= F
s ;V =
v
s ;U =
u
s
(4.22)
and sisan initialguess and hangeiterativelyafter eahstep byNewton'sformula
s n+1
=s n
F (L;s n
) A
F(L;s n
)
s
(4.23)
orhere A=0;L=1 andF=s=F
thentheequation is
s n+1
=s n
F (1;s n
)
F
( 1;s n
)
(4.24)
wheres 1
istaken to be 1:
5 Graphs and omparison of results
In order to illustratethe inuenes ofR
e
and on F; we have plotted theFigures 2and
3 respetively. The obtained analytial solution is ompared with numerial solution in
Figures4 and 5respetively.
Fig. 3. ProlesofF forvarious valuesofR
e
whenisxed.
6 Conluding remarks
The present studyontributesexat solutionfortheJerey-Hamel ow. In addition,the
numerialanalysis isalso presentedfor theJerey-Hamelow. Asa result,thefollowing
observationsare made.
InreaseinR
e
resultsintheinreaseof boundarylayer.
An inreasein leadsto anderease intheboundarylayerthikness:
Theonstant >0and<0giveadivergentandonvergent hannelrespetively.
It is also worth mentioning that our exat solutions are more general and suh
solutionshave beenpresentedrst timeintheliterature.
Aknowledgment
RE also thanks Higher Eduation Commission of Pakistan (HEC) for NRPU and
UnitedState EduationFoundationPakistanto honoredhimbyFulbrightSholarAward
fortheyear 2011-2012.
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