# richardson extrapolation

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 March 11, 2013, 16:31 #21 Senior Member     Alex Join Date: Jun 2012 Location: Germany Posts: 1,534 Rep Power: 25 Either you hava a look at Ferzigers book (not freely available online as far as I know) or you can derive the formula yourself. Use the Ansatz F(h_i) = F(h=0) + a*(h_i)^p (where h_i is the grid spacing, f(h_i) is the value obtained on this grid, F(h=0) is the solution without discretization error, a is a constant and p is the error order. With solutions on three different grids with the same refinement ratio r (r= h_2/h_1 = h_3/h_2) you get 3 equations. (I) F(h_1) = F(h=0) + a*(h_1)^p (II) F(h_2) = F(h=0) + a*(h_2)^p (III) F(h_3) = F(h=0) + a*(h_3)^p (I')=(I)-(II) F(h_1)-F(h_2)= a*[(h_1)^p-(h_2)^p] (II')=(I)-(III) F(h_1)-F(h_3)= a*[(h_1)^p-(h_3)^p] (I'')=(I')/(II') [F(h_1)-F(h_2)]/[F(h_1)-F(h_3)]=[(h_1)^p-(h_2)^p]/[(h_1)^p-(h_3)^p] now using the definition of r, the right hand side becomes (1-r^p)/(1-r^2p)=1/(1+r^p) after some more algebra we arrive at p=log[F(h_1)-F(h_3)]/[F(h_1)-F(h_2)-1]/log(r)

 March 12, 2013, 11:30 #22 New Member   Albert Puig Join Date: Jul 2012 Posts: 24 Rep Power: 7 Hi folks again first of all, thanks for the responses! ther were really useful to get a plain understanding of what is the richardson extrapolation and how to apply it to my problem. I have another doubt. If I only have the values of f(h20),f(30) and f(40) because the coarsened grid (f10) does not work, can I manage to get the order of error "p"? for example. in some case I can run all teh simulations with all of the grid, including the coarsened. But there are some conditions that when I apply them to the model, they make it more unstable, to the point that the coarsened grid (f10) falls off. for example case 1 all of them work: I get p and F for richardson extrapolated value x y h -> F_h h/2-> F_h/2 h/4-> F_h/4 case 2 Can I use the value "p" for richardson extrapolated value of CASE 1 if the trends of the two results of F and G are more or less the same? how can I assess the similarity of the trends of F and G to validate the value of the order of the error "p"? x y (h doesn't work) h/2-> G_h/2 h/4-> G_h/4 thank you again for paying attention! It has been a great favour

 March 12, 2013, 11:49 #23 Senior Member     Alex Join Date: Jun 2012 Location: Germany Posts: 1,534 Rep Power: 25 With solutions on only two different grids, there is no way in estimating the error order. Even worse, if the convergence is not monotonic and you just guess a value for p, you end up with a completely wrong extrapolated value. So 3 solutions (with three different element sizes) are always necessary. If you did not refine with the same ratio, you can still try to solve [F(h_1)-F(h_2)]/[F(h_1)-F(h_3)]=[(h_1)^p-(h_2)^p]/[(h_1)^p-(h_3)^p] some way. The easier way is to keep the same ratio in both refinements. Take the coarsest mesh on which you get a reasonable result as a starting point. Now the finer mesh doesnt have to be h/2 if the solution at h/4 becomes too expensive. A refinement factor of e.g. 1.5 is still enough to do a reasonable extrapolation in most cases.

March 12, 2013, 12:07
#24
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Albert Puig
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Quote:
 Originally Posted by flotus1 With solutions on only two different grids, there is no way in estimating the error order. Even worse, if the convergence is not monotonic and you just guess a value for p, you end up with a completely wrong extrapolated value. So 3 solutions (with three different element sizes) are always necessary. If you did not refine with the same ratio, you can still try to solve [F(h_1)-F(h_2)]/[F(h_1)-F(h_3)]=[(h_1)^p-(h_2)^p]/[(h_1)^p-(h_3)^p] some way. The easier way is to keep the same ratio in both refinements. Take the coarsest mesh on which you get a reasonable result as a starting point. Now the finer mesh doesnt have to be h/2 if the solution at h/4 becomes too expensive. A refinement factor of e.g. 1.5 is still enough to do a reasonable extrapolation in most cases.
so I can get a p value from taking for example h/2 h/3 and h/4? even if they are not doubled? Just I have to apply the ansatz formula again with these refinements to get "p"?

 March 12, 2013, 12:32 #25 Senior Member     Alex Join Date: Jun 2012 Location: Germany Posts: 1,534 Rep Power: 25 The thing is that you connot solve for p explicitly if the refinement ratios are not identical. So h/2, h/3 and h/4 is not a good choice. And yes, the meshes dont need to be refined with a factor of 2 (or 0.5 in the definition I used above) Any factor far enough from 1 will do.

March 13, 2013, 10:06
#26
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Albert Puig
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Quote:
 Originally Posted by flotus1 The thing is that you connot solve for p explicitly if the refinement ratios are not identical. So h/2, h/3 and h/4 is not a good choice. And yes, the meshes dont need to be refined with a factor of 2 (or 0.5 in the definition I used above) Any factor far enough from 1 will do.
So then? If it is not explicitly done, how can I solve it?

I did not get how was this step done when getting the three equation solving of "p"
Quote:
 (1-r^p)/(1-r^2p)=1/(1+r^p)

 March 13, 2013, 10:53 #27 Senior Member     Alex Join Date: Jun 2012 Location: Germany Posts: 1,534 Rep Power: 25

March 13, 2013, 11:44
#28
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Albert Puig
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Quote:
 Originally Posted by flotus1
Sorry for being sooo lame..
I just don't get how you are introducing r here.

 March 13, 2013, 13:18 #29 Senior Member     Alex Join Date: Jun 2012 Location: Germany Posts: 1,534 Rep Power: 25 r is the ratio of the element sizes and was introduced a few posts ago.

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