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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