does anyone answer or give some comment about my questions???
Dear all CFD Web friends,
Would you please answer the following questions? For given differential eqution: Lu=f where L is a nonlinear operator. In twolevel Vcycle Multigrid algorithm(i.e. smoothing on fine mesh,correctin on coarse mesh) 1) when performing coarsegrid correction, why one always choose the same oprator as original differencial equation to solve the defect equation? if choose a different nonlinear operator(e.g. lower order operator G), what difference can be caused(e.g. magnitude of correction from defect equation and efficiency of the whole twolevel Vcycle multigrid approach)? 2) what are their definitons? the trancation error, defect/residual and relative trancation error(between fine and coarse grid). what is the difference between the trancation error and defect/residual ? Thanks! Z. Chang 
Re: does anyone answer or give some comment about my questions???
Multigrid is based on the method of relaxation. The simple relaxation on a given grid first smoothes out the short wavelengths, and as more relaxations are carried out the longer wavelength are smooth out. In order terms the errors are first reduced on the smallest spacing of the given grid and only after many relaxations they are reduced on the size of the whole domain. THe basic idea of multigrid, is to relax the solution (smooth out the errors) on a coarse grid to reduce the errors on the long wavelengths, and then to relax the solution on a finer and finer grid in order to reduce the error on the smallest scale. Since it take much less time to reduce the errors on the long wavelengths on a coarse grid than on the fine grid, the method is much faster, but eventually the accuracy obtained is of the order of the fine grid. This is why you are solving always the same operator equations at the different levels of grids.
Have a look at Press, Flannery, Teukolsky and Vetterling, Numerical Recipes: The Art of Scientific Computing (Cambridge University Press, New York), 1989 I think, or just take the latest version. It should be in Chap. 19., section 6., page 862. If you have Lu=f and the discretization of the equation leads to L_h u_h = f_h , where the subscript h denotes the discretization, then obviously L_h u_h f_h =0. And you are looking for u_h. Initially you have only a guess of the solution (an approximation) say u'_h, and the error is v_h = u_h u'_h The residual is L_h u'_h  f_h and it is not zero because u'_h is not exactly equal to u_h. The residual can be written: L_h u'_h  L_h u_h or L_h v_h etc... see the reference book above. PG 
Re: does anyone answer or give some comment about my questions???
First thank you for your comments.
My question is: If I use operator G instead of L for defect equation,i.e G_H v_H=I_h_to_H d_h (defect equation) where G is operator whose order is lower than L's and relatively be easily solved,H stands for coarse mesh, d_h is defect from fine mesh, I_h_to_H represents restriction, what difference can be caused(e.g. magnitude of correction from defect equation(i.e. v_H ) and efficiency of the whole twolevel Vcycle multigrid approach)? Now I am looking for the book you mentioned. Whether could you give some explaination for my the second question? 
Re: does anyone answer or give some comment about my questions???
For the first question, I mentioned that you need to solve the same equations with the same operator, because you are solving the same problem. On each grid you are solving a different length scale, but it is still the same problem that you want to solve. If you want to use G, then fine, but use it everywhere. I never tried what you suggest, so the best way for you to find out is to try what you are saying. So I can only suggest you to try, but I am not sure the results you will obtain with that.
Concerning the second question, I mentioned (and that's my answer) that the correction (error) is the difference between the approximation and the solution, and the residual is the when the operator L is applied to the correction (error). In any method one tries to have the residual going to zero (or its projection on a given basis in space  in other words one tries to have the discretized operator applied to the error tends to be very small). The book I mentioned is also on the web at <hr>[*] http://www.nr.com <hr> cheers, PG 
Re: does anyone answer or give some comment about my questions???
You can use a lower order approximation on your coarse grid. Actually the coarse grid discretization has a different truncation error anyway, due to the coarser grid. We are using an FAS multigrid together with unstructured (agglomerated) grids. To obtain a higher than first order discretization on an agglomerated grid is very tough. What we do is slightly different from what you said. We compute a difference between a target and driving (2nd and 1st order) discretization on the finest grid. Then we compute the whole Vcycle with the driving discretization (including the finest grid).
Oliver 
Re: does anyone answer or give some comment about my questions???
I do not know if first order or higher order Oliver mentioned in his mail has the same meaning as yours. Could you tell us your equations? What do your operators look like?

Re: does anyone answer or give some comment about my questions???
The equations are the compressible Euler or NavierStokes equations. With order I meant the truncation error of the discretization, meaning that the same equations are discretized everywhere. I don't know if there is any combination which uses different equations on the coarse levels. In principle this should, if it converges at all, converge to the finegrid discretization. I did try to use completely different discretizations, though (e.g. AUSM fluxsplitting as target and a Roesplitting as driver). This worked quite nicely, whereas it was very difficult to use the AUSM on all levels.

Re: does anyone answer or give some comment about my questions???
Oliver,
I am sorry that my mail was for Chang, not for you. What I wanted to ask Chang is perhaps he means something else. 
All times are GMT 4. The time now is 00:01. 