Unknown Difference Scheme
 

Hallo!

I try to analyse a numerical program and found an unknown difference scheme to interpolate the value between point x_0 and x_1.

(u_{1/2})= [5 (u_0) + 8 (u_1) - (u_2)] /12h

(u_{1/2})= [-(u_-2) + 13 (u_-1) + 13 (u_0) - (u_+1)] /24h

(u_{1/2})= [-(u_-2) + 8 (u_-1) + 5 (u_0)] /12h

Point x_-2, x_-1, x_0, x_1 and x_2 are equidistant of distant h.

Can anybody tell me, how to derive this scheme?

Thank you!

jc

Re: Unknown Difference Scheme
 

I guess it is dervied from second order polynomial for first and third approximation and third order polynomial for the second approximation.

Re: Unknown Difference Scheme
 

I tried to figure out but I came to result that in case the formulars are derived from polynomials the coefficients shouldn't be independent of the u values, should they? So to me it seems not to be the polynomial approch.

JC

Re: Unknown Difference Scheme
 

You are right, by the way, where did you get those difference schemes?

Re: Unknown Difference Scheme
 

A former colleague wrote a C program for droplet combustion. And I (2 generations later) shall extend this program without anyone who knows exactly what's going on.

JC

Re: Unknown Difference Scheme
 

(u_{1/2})= [5 (u_0) + 8 (u_1) - (u_2)] /12h seems missing something because if u_0, u_1, u_2 are constant, say, u, then u_{1/2} = u/h ?

Re: Unknown Difference Scheme
 

You are right here ;-) In my program h=1.

JC

Re: Unknown Difference Scheme
 

May I suggest that you read the book by Clive Fletcher Computational Techniques for fluid Dynamics, Volume 1, chapter 3 and especially 3.2.2 the general Technique for developing approximation to derivatives. You can also check the book by J.C.Tannehill, Anderson and Pletcher Computational Fluid Mechanics and Heat Transfer. Normally mid point values are needed when developing Quadratic interpolations like the Quick Scheme. The important point to remembere here is to use the correct order of polynomials for developing your scheme.

Re: Unknown Difference Scheme
 

That you very much! The book of Fletcher seems to give me the right answer.

The mentioned sheme seems to be a combination of 4th and 2nd order mean value in the center and 3rd and 2nd order at the edge. Weight for 4th and 3rd order respectively is 2/3 and 1/3 for the 2nd order approximation:

4th order scheme: (-1, 9, 9, -1)/16 2nd order scheme: (0, 1/2, 1/2, 0) leads to (2/3*4th Order + 1/3*2nd Order) (-1, 13, 13, -1)/24

3rd order scheme: (3, 6, -1)/8 2nd order scheme: (1/2, 1/2, 0) leads to (5, 8, -1)/12

Thanks again!

Christian