Using High Order Finite Differences/Third Order Method

A Third Order Accurate Method

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Statement of the Problem

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Let D be the rectangle

 

and let C be the boundary of D.

The operator

 

is the usual Laplacian. The problem, determine a function u(x, y) such that

 

is called a Poisson problem.

discrete approximation

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To approximate u(x, y) numerically, use the grid

 .
with
 
and
 

The second partial derivative
 
can be approximated on the grid by difference quotients
 .

These difference quotients are given by

 .

 

 .

 

 

 .

 


The second partial derivative
 
can be approximated on the grid by difference quotients
 .

These difference quotients are given by

 .

 

 .

 

 

 .

 

truncation errors

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The difference quotients       are third order accurate with truncation errors:

 

with

  ,

for some      ,

  ,

for some     

      

When      is continuous, these estimates also hold.

    with

 

for some         

 

The case for     is

  ,

for some      .


The difference quotients       are third order accurate with truncation errors:

 

with

  ,

for some      ,

  ,

for some     

      

When      is continuous, these estimates also hold.

    with

 

for some         

 

The case for     is

  ,

for some      .

The Laplacian     then can be approximated on the interior of the grid by

 

The truncation error

 

is given by

 .

finite difference operations defined

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For the grid vector

 

define the finite difference operations

 

by the following.

 .

 

 .

 

 

 .

 

 .

 

 .

 

 

 .

 

simulation of the problem

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To simulate the problem  (1.0)   let

 


Then solve the non-singular linear system

   ;

for the remaining    

error estimation

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The error

  ,

satisfies

  .

for

   ,

and

  .

proof of truncation error estimates

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The truncation error estimates for     are done under the assumption that     is sufficiently smooth so that     is continuous. For notational convenience let

 

Expand     in it's Taylor expansion about   ,

 .

where     is some number between     and   . Then

 

 

 

where

 .

Since

 

 

from the intermediate value property

 

 .

This gives

 

which is

 



For  

 

 

 

where

 .

Reasoning as before, combining terms with like signs and using the intermediate value property,

 

 .

This gives

 

which is

 

Under the assumption that     is continuous, in the preceding argument, the expression

 

can be replaced by

 

This gives

 

with         which is

 

The remaining truncation error estimates are done in the same way.

end working

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Let the error

 

be defined by

  .

  is the solution of the finite difference scheme (xx) and   is the solution to (1.0).

Since

  
we get that

  .

Next it will be shown that the operator     is positive definite for   ,  in particular that

  

        ,

with

  .


Begin with   
 
 

The sum     will be estimated first.

work

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 .

 



 

 




 

 .

 


 

 

 





 

 

 .

 

 



 

 



end work

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The  summation by parts formula  is now stated so it can be used.


 

 

 

 

 

 .

 

 

   

Taking into account that     it follows

 

 

 

working here

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Collect like terms in the expression immediately above as follows.

 

 

 

 

Now, rewrite the expression after making cancellations.

 

The following simple inequality will be used to bound terms.

 
and also

 .

 

 

 

 

 

 

Now, substitute all the inequalities into the expression.

 


 

 

The choice

 

bounds all of the coefficients in the     and     by      and yields the long sought inequality

 .

and

 .

Reasoning in the exact same manner for the dimension in   

 .

and

 .

Applying    

 

leads to the inequality    

 .

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