a hybrid numerical scheme: an application in fluid mechanics · an application in fluid mechanics...
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A Hybrid Numerical Scheme:An application in fluid
Mechanics
Dr. Tariq JavedDr. Tariq Javed
Assistant Professor
Department of Mathematics
International Islamic University, Islamabad
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Different ProceduresDifferent Procedures……
Solution of Nonlinear Solution of Nonlinear BVPsBVPs
1.1. Analytic MethodsAnalytic Methods
• Perturbation MethodsPerturbation Methods
•• AdomianAdomian Decomposition MethodDecomposition Method
•• HomotopyHomotopy Perturbation MethodPerturbation Method
•• HomotopyHomotopy Analysis MethodAnalysis Method
•• Optimal Optimal HomotopyHomotopy Analysis MethodAnalysis Method
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2.2. Numerical MethodsNumerical Methods
Conversion to IVPConversion to IVP1. Method of Superposition
2. Method of Chasing
3. Method of Adjoint Operator
4. Shooting Methods
Retained as a BVPRetained as a BVPExplicit Finite Difference Schemes
Implicit Finite Difference Schemes
Solution of Nonlinear Solution of Nonlinear BVPsBVPs
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Working RulesWorking Rules
No of B.C = Order of ODENo of B.C = Order of ODE
No Singularity at allNo Singularity at all
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Example with ProblemsExample with Problems
1)(',0)0(')0(0)''''''2('1''''' 22
=∞===+−+−++
ffffffffkffff iv
Two dimensional flow near a stagnation PointTwo dimensional flow near a stagnation Point
Order is 4k=0
No of Boundary conditions =3
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MHD pipe flow of a fourth grade fluidMHD pipe flow of a fourth grade fluid(Due to constant pressure Gradient)(Due to constant pressure Gradient)
( )0)1(,0)0('
0'''3'4''' 223
===−++−+
fffMfffff ξξλξξ
Singularity at ξ=0
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What happens whenWhat happens when
oo Number of boundary conditions are Number of boundary conditions are less then the Order of less then the Order of EqsEqs??
oo Derivative of highest power vanishes?Derivative of highest power vanishes?
oo Singularity occurs?Singularity occurs?
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Hybrid MethodHybrid Method
'',', 321 fyfyfy ===
1,0,0 202
01 === Jyyy
To convert the Example 1 into system of ode in which two are 1st orders and one is 2nd order ode, we set
Solution of Example 1Solution of Example 1
Boundary conditions become
Here the missing boundary condition is
sy =03
10
( )
( ) ,02
22
12
23
13
13
22
133
13
1
2231
13
13
=⎟⎟⎠
⎞⎜⎜⎝
⎛+
−−
+−+
−++−
−+−+
−+
jjj
jjjj
j
jjjjj
yh
yyyh
yyyyk
yyyh
yy
2
1332
12
++ +=
− jjjj yyh
yy
2
1221
11
++ +=
− jjjj yyh
yy
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( )( )( ) ⎥
⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎟⎟⎠
⎞⎜⎜⎝
⎛++
+−−
−+−⎟⎠⎞
⎜⎝⎛ −+=
−−
−−
+2
31
32
133
1
2231
131
311
32222
12221
jjjjj
j
jjjj
jjj
yhyyh
yyyk
yyyhykyy
hky
( ),2
1332
12
++ ++= jjjj yyhyy
( ).2
1221
11
++ ++= jjjj yyhyy
The whole problem reduces to
syyy === 03
02
01 ,0,0
Subject to the boundary conditions
12
( ) ( ) ''2
' 03
203
03
13 yhyhyy ++=
( ),2
13
03
02
12 yyhyy ++=
( ).2
12
02
01
11 yyhyy ++=
syyy === 03
02
01 ,0,0
Subject to the boundary conditions
( )22
13 1...)0(
2)0(''')0('' kshsfhhffy iv +−=+++=
Which can be understood in a simple form as
13
0 1 2 3 4h
0.2
0.4
0.6
0.8
1
f'
k=0.3,0.2,0.1,0.0
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Stagnation point flow of a third grade Stagnation point flow of a third grade fluid with chemically reactive speciesfluid with chemically reactive species
f ff f 2 1 ☺1 2f f ff iv 3☺1 2☺2 f 2 6 Re xf 2f 0,
Scf Sc n,
f 0 f w, f 0 0, f 1,
0 1, 0,
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Backward IntegrationBackward Integration
e c ,
F f and ,
Semi-InfiniteDomian
BoundedDomian
[1,0][[0,∞
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y3i 1 1
2 ci ☺1 c2i c 1 ln i y 1i i 1/2 ci 1 y2
i 3ci Rex y3i 2 1
12 ciy3
i 1 c 1 ln i y1i y3
i 2y2i y2
i 2
☺1c2i c 1 ln i y1
i 2iy3i i 1/2 y3
i 1
ci 1 y2i y3
i 1
3☺1 2☺2 y3i 2 3ci Re x y3
i 2y3i 1
,
y20 0, y3
0 0, y1N f w, y2
N 1,
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zi 1 c2i i 1/2 12 Sc c 1 ln y1 ci
1
c2i 2izi i 1/2 zi 1
12 Sc c 1 ln y1 cizi 1 Sc zi n
.
z0 0, zN 1.
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0 1 2 3 4 50
0.2
0.4
0.6
0.8
1
η
f ’ (η)
fw
= 10, β = 0
α = 0.0, 0.01, 0.1, 1.0, 10.0
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0 20 40 600
0.2
0.4
0.6
0.8
1
η
f ’ (η)
fw
= 10, β =0
α = 10, 20, 50, 75, 100
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α f’’(0) α f’’(0)0 10.193554
9.330114
8.693450
7.442309
6.276552
5.072516
3.629950
2.734932
1.3312220
0.001
0.5
1.0
2.0
5.0
10.0
20.0
50.0
0.960444
0.002 0.688459
0.005 0.440521
0.01 0.313218
0.02 0.222283
0.05 0.140989
0.1 100.0 0.099819
Illustrating the variation of f’’(0) with α for fw=10
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ConclusionsConclusions
SingularitySingularity Can be tackledCan be tackled
No need of No need of augmenting the boundary conditionaugmenting the boundary condition
Still works if the Still works if the higher order derivative get higher order derivative get vanishesvanishes
Not too much initial guess dependentNot too much initial guess dependent
This is This is explicit finite differenceexplicit finite difference--shootingshootingtechniquetechnique
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Future Motivations
100
101
102
10−15
10−10
10−5
100
N
erro
r
Convergence of spectral differentiation
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Thank you Thank you forfor
your attentionyour attention