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Using High Order Finite Differences/Third Order Method

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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.

.








.










.







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  

.

edit end

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