Part 3 - Approximate Solutions of Differential Equations

Faculty Name

Prof. A. A. Saati

Part 3 - Approximate Solutions of Differential Equations

1. Introductory Remarks 2. Taylor Series Expansion

3. Solutions of Differential Equations

1. Introductory Remarks

 The ODE & PDE must be expressed as

approximate expressions, so that a digital computer can be employed to obtain a

solution

 There are two methods for approximating the differentials of the function f

 First method of approximation often used is the Taylor series

 Second method is the use of a polynomial of degree n.

2. Taylor Series Expansion:

 Given a function f(x), which is analytical, can be expanded in a Taylor series about x as

)

(x x

f  

   

 

n n

n

n

x f n

f(x) x

x f x

x f x

x x f

x f x

x f





 



 



 





 



 











^

1

3 3 3

2 2 2

!

! ...

3

! ) 2

( ) ( )

(

 Forward Difference Formulations

 Solving for one obtains

 The truncation error of order x

f





   

 

^{x ,}

) o ( )

(

! ...

3

! 2 )

( )

(

3 3 3

2 2 2



 





 





 





 







 







 





x

x f x

x f x

f

x f x

x x

f x

x x

x f x

x f x

f

) ( x o 

) ( x o 

   

! ...

3

! ) 2

( ) ( )

( ₃

3 3 2

2 2

 



 





 



 







 x

f x

x f x

x x f x

f x

x f

Finite Difference Formulations

 If the subscript index i is used to represent the discrete point in the x-direction

 The truncation error of order

 This equation is known as forward difference of order

 

^{x ,}

1  o 



 



 _

x f f

x

f _{i i}

) ( x o  )

( x o 

1 i i

 Backward Difference Formulations

 Now consider the Taylor series expansion of about x.

 Solving for one obtains

 This is known as backward difference of order x

f



 )

(x x

f  

   

   

ⁿ _n

n

n n

x f n

f(x) x

x f x

x f x

x x f

x f x

x f





 





 



 





 



 











^

1

3 3 3

2 2 2

1 !

! ...

3

! ) 2

( ) ( )

(

 

 

^{x ,}

o

, x ) o

( )

(

1  



 







 





 







x f f

x f

x

x x

f x

f x

f

i i

) ( x o 

 Central Difference Formulations

 Now consider the Taylor series

expansion of and about x.

 Subtracting the above equations, one obtains:

 Solving for

x

f 

 /

)

(x x

f  

   

   

! ...

3

! ) 2

( ) ( )

(

! ...

3

! ) 2

( ) ( )

(

3 3 3

2 2 2

3 3 3

2 2 2

 



 





 



 









 



 





 



 









x f x

x f x

x x f

x f x

x f

x f x

x f x

x x f

x f x

x f

)

(x x

f  

 

....

! 2 3

) (

2 )

( )

( ₃

3 3

 



 



 











 x

f x

x x f

x x

f x

x f

) 2 (

) (

)

( ₂

x x o

x x

f x

x f x

f  











 







 

f f f

) ( x² o 

 Forward & Backward Difference Formulations

 Again consider the Taylor series expansion of and about x.

 Multiply the first equation by 2 and subtract it from the second equation:

 Solving for )

(x x

f  

   

   

! ...

3 2

! 2 ) 2

2 ( ) ( )

2 (

! ...

3

! ) 2

( ) ( )

(

3 3 3

2 2 2

3 3 3

2 2 2

 



 





 



 









 



 





 



 









x f x

x f x

x x f

x f x

x f

x f x

x f x

x x f

x f x

x f

) 2

(x x

f  

 

^....

) (

) ( )

2 (

) (

2 ₃

3 3 2

2

2 



 

 

 

















 x

x f x

x f x

f x

x f x

x f

2 f



) ( x o 

 Forward Difference Formulations

 Solving for

2 2

x f





 

 

^{2 ( )}

) ) (

( )

( 2 )

2 (

2 1 2

2 2

2 2 2

x x o

f f

f x

f or

x x o

x f x

x f x

x f x

f

i i

i  





 







 











 









) ( x o 

 Backward Difference Formulations

 A similar approximation for B.D. using the Taylor series expansions of

and about x. The result is )

(x x

f   )

2

(x x

f  

 

 

^{( )}

2

) ) (

2 (

) (

2 )

(

2

2 1

2 2

2 2 2

x x o

f f

f x

f or

x x o

x x

f x

x f x

f x

f

i i

i  





 







 













 









) ( x o 

 Center Difference Formulations

 Approximation expression for higher order derivatives

 Now consider the Taylor series expansion of and about x.

 Add the above equations, one obtains the center difference:

   

   

! ...

3

! ) 2

( ) ( )

(

! ...

3

! ) 2

( ) ( )

(

3 3 3

2 2 2

3 3 3

2 2 2

 



 





 



 









 



 





 



 









x f x

x f x

x x f

x f x

x f

x f x

x f x

x x f

x f x

x f

2

) ) (

( )

( 2 )

( ₂

2 2

2

f f

f

x x o

x x

f x

f x

x f x

f







 











 





)

(x x

f  f (x x)

) ( x² o 

 Forward Difference Formulations

 Now by considering additional terms in the Taylor series expansion, a more accurate

approximation of the derivatives is produced.

 Consider the Taylor series expansion,

 Solving for

   

! ...

3

! ) 2

( ) ( )

( ₃

3 3 2

2 2

 



 





 



 







 x

f x

x f x

x x f

x f x

x f

! ...

3 ) (

2 )

( )

(

3 3 2 2

2 





 





 







 





x f x

x f x

x

x f x

x f x

f

x f





) ( x² o 

 Substitute a forward difference expression for

 And 0ne obtains a second-order for

! ...

3 ) (

2 )

( )

(

3 3 3 2

2 





 





 







 





x f x

x f x

x

x f x

x f x

f

2 2

x f





) ) (

(

) 2

( )

( 2 )

(

2 2

2

x x o

x x

f x

x f x

f x

f  













 





3 3 2

2

) ( 3 ) (

4 ) 2 (

or 6 ...

) (

) ) (

(

) ( )

( 2 ) 2 (

2 )

( )

(

x f x

x f x

x f f

x f x

x x o

x f x

x f x

x f x x

x f x

x f x

f

















 



 



 



  















 







 





x f





 Forward & Backward Difference Formulations

 The finite difference approximation to the time derivative is expressed for a forward and

backward difference as t

f f

t

f _iⁿ_{j i}ⁿ_j



 



 _, ^1 _,

t f f

t

f _iⁿ_{j i}ⁿ_j



 



 _{, ,} ^1

) ( t o 

1 –FD

2 – BD

3 – CD

4 – FD

5 - BD

Finite Difference Formulations

) ( x o 

1

x f f

x

f _{i i}



 



 _

x f f

x

f _{i i}



 



 _1

) ( x² o  x

f f

x

f _{i i}



 



 _{ }

2

1 1

) ( x o 

x

f f

f x

f _{i i i}





 



 _{ }

2

3

4 ₁

2 o(x²)

) ( x² o 

x

f f

f x

f _{i i i}







 



 _{ }

2

3

4 ₁

2

6 - FD

7 – BD

8 – CD

9 – FD

10 - BD

) ( x o 

) ( t o 

) ( x o 

) ( x² o 

 

²¹

2 2

2 2

x

f f

f x

f _{i i i}





 



 _{ }

 

^{12 2}

2

2 2

x

f f

f x

f _{i i i}





 



 _{ }

2

1 1

2

2 2

x

f f

f x

f _{i i i}





 



 _{ }

t f f

t

f _iⁿ_{j i}ⁿ_j



 



 _, ^1 _{, o(t)}

t f f

t

f _iⁿ_{j i}ⁿ_j



 



 _{, ,} ^1

Finite Difference Formulations

 Read Example:

 2.1 , 2.2, 2.3, & 2.4

 2.6, 2.7, & 2.8

3. Solutions of Differential Equations

 Example:

 Given the function compute the first derivative of f at x = 2 using forward and back ward difference of order

 Compare the results with a central differencing of order and the exact analytical value. Use a step size of

 Solution:

 Form Eq.1

 With

4 / )

(x x²

f 

) (x

) (x²

1 .

 0

x

)

1 o( x

x f f

x

f _{i i}



 

 



 _

1 .

 0

x

 Example (cont.)

 The back ward of order

 ^{1 o( x)} x

f f

x

f _{i i}



 

 



 _

) 1 . 0 ( 975

. 0 )

1 . 0 1 (

. 0

) 9 . 1 ( )

2

( f o o

f x

f    

 



) (x

 Example (cont.)

 The central differencing of order

 The exact value is

22

) (x² )

2 (

1 2

1 o x

x f f

x

f _{i i}



 

 



 _{ }

) 01 . 0 ( 0

. 1 )

1 . 0 ) (

1 . 0 ( 2

) 9 . 1 ( )

1 . 2

( f o o

f x

f    

 



. 1 /

is 2 at x

which ,

4 / ) 2 (

/     

f x x f x

 Example (cont.)

 Read Example:

 2.6, 2.7, & 2.8

 Home work

 Solve problems:

 2.7

 2.12

 Finite Difference Equations

( FDE ) next……….

 Finite Difference Equations

 The finite difference approximations are replace the derivatives that appear in the PDEs.

 Consider the following example, where f is f = f(t,x,y).

 Assume is constant.

 Let represent

 Assume are constant step.

 Now use forward difference in time

 And use center difference in space.



 







 



 





2 2 2

2

y f x

f t

f



 n j

i, , x, y, t

t y x  

 , ,

 Finite Difference Equations

 Now use forward difference in time

 And use center difference in space.



 







 



 





2 2 2

2

y f x

f t

f



)

, (

1

, O t

t f f

t

f _iⁿ_{j i}ⁿ_j  



 



 ^

) 2 (

) 2 (

2 2

1 , ,

1 , 2

2

2 2

, 1 ,

, 1 2

2

y f o

f f f

x x o

f f

f x

f

n j i n

j i n

j i

n j i n

j i n

j i



 



 







 



 













 Finite Difference Equations

 The finite difference formulation of PDE is:

 Note that in this formulation, the spatial approximations are applied at time level n

 This lead to one unknown

 This equation is classified as explicit formulation



^{( ),( ),( )}



2 2

2 2

2

1 , ,

1 , 2

, 1 ,

, 1 ,

1 ,

y x

t o

y

f f

f x

f f

f t

f

f_iⁿ_{j i}ⁿ_{j i}ⁿ _{j i}ⁿ_{j i}ⁿ _{j i}ⁿ_{j i}ⁿ_{j i}ⁿ_j











 









 





 



 _{   }





1 ,

 n

j

fi

 Finite Difference Equations

 The second case evaluated the spatial approximations at ⁿ⁺¹ time level.

 Therefore, the first-order backward difference approximation in time is employed

 The finite difference formulation for PDE takes the form:

 This lead to 5 unknown



^{( ),( ),( )}



2 2

2 2

2

1 1 , 1

, 1

1 , 2

1 , 1 1

, 1

, 1 ,

1 ,

y x

t o

y

f f

f x

f f

f t

f

f_iⁿ_{j i}ⁿ_{j i}ⁿ _{j i}ⁿ_{j i}ⁿ _{j i}ⁿ_{j i}ⁿ_{j i}ⁿ_j











 









 





 



 _^{ } _^{ }_ ^{ }_

 

1 1 , 1

1 , 1 , 1 1

, 1 1

,ⁿ_j^ , _i^n _j, _i^n_j, _iⁿ_j^_ and _iⁿ_j^_

i f f f f

f

 Applications

 Applications

 Example 2.2

 Finite Difference Approximation

of Mixed Partial Derivatives

 Finite Difference Approximation of Mixed Partial Derivatives

 Approximating mixed partial derivatives can be performed by using Taylor series expansion for two variables

 Consider

 The Taylor series expansion for two variables x and y, become as

)

2 /(

y x f  



) ,

(x x y y

f    

 

 ^{2 2} ₂ ²



   ^{3 3}



2 2 2

! , 2 2

! 2

! ) 2

, ( )

, (

y x

y O x

f y

x y

f y

x f x

y y f x

x f y

x f y

y x x

f





 







 





 





 



 

 

 













 Taylor Series Expansion

 Using indices to represent a grid point at x, y.

 Similarly, the expansion j i and

 

^{2 2} ₂

 

^{2 2} ₂

    

^{3 3}



2 ,

1 , 1

! , 2

! 2

O x y

y f y

x f x

y x y f y x

y f x

x f f

f_{i j i j}





 



 





 





 



 

 

 

 



 



) ,

(x x y y

f    

 

^{2 2}

 

^{2 2}

    

^{3 3}



2 ,

1 , 1

,

x f y f O x y

y x y f y x

y f x

x f f

f_{i j i j}





 



 





 





 



 

 

 

 



 



 Taylor Series Expansion

 And the expansion

 And the expansion

 

^{2 2} ₂

 

^{2 2} ₂

    

^{3 3}



2 ,

1 , 1

! , 2

! 2

O x y

y f y

x f x

y x y f y x

y f x

x f f

f_{i j i j}





 



 





 





 



 

 

 

 



 



) ,

(x x y y

f    

 

^{2 2}

 

^{2 2}

    

^{3 3}



2 ,

1 , 1

,

x f y f O x y

y x y f y x

y f x

x f f

f_{i j i}<