In addition what Wolfgang said: 1. It would be indeed interesting to see whether neglecting z_h really yields the same error, the same effectivity indices, and/or the same mesh.

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2. Going back to your initial questions: Inserting z_h is the key when classical a posteriori bounds in terms of the mesh size h are of interest. The final goal is usually to obtain an error estimate in terms of the mesh size h in order to quantify the order of convergence of your scheme. For this reason you need to insert z_h such that || z - z_h|| to apply interpolation estimates that give you some h^{a} on the right hand side: || z-z_h || = O(h^a) with the order a. 3. In practice you have indeed different choices how to evaluate J(u) - J(u_h). Also some people do not integrate back into the strong form and work with a weak form of the error estimator, which has the advantage that no second-order operators and partial integration needs to be applied. From this point of view, it may be that neglecting z_h could work in practice. But as said above, a careful study for some model problems would be useful. Best Thomas W. -- ++--------------------------------------------++ Prof. Dr. Thomas Wick Institut für Angewandte Mathematik (IfAM) Leibniz Universität Hannover Welfengarten 1 30167 Hannover, Germany Tel.: +49 511 762 3360 Email: thomas.w...@ifam.uni-hannover.de www: http://www.ifam.uni-hannover.de/wick www: http://www.cmap.polytechnique.fr/~wick/ ++--------------------------------------------++ -- On 03/10/2018 11:03 AM, Wolfgang Bangerth wrote:

On 03/06/2018 08:40 AM, 曾元圆 wrote:Now I understand why we need to rewrite the error formula on a cellas residual times dual weight. But I'm still a little confused withthe reason why we must introduce z_h.Just as you mentioned, if we introduce z_h, then z-z_h is a quantitythat is only large where the dual solution is rough. But why do weneed to care about the accuracy of z here? I think the only thing weneed to care about is the value of z on that cell, because z is aquantity that represents how important the residual on that cell is.No. z tells you how important the *locally generated error is for theglobal error functional*. (That is because z is the Green's functionassociated with your error functional.) But you don't have the localerror. All you have is the local residual.My understanding is: now the dual_weight z-z_h does not onlyrepresent how important the residual on a certain cell is, but alsotells us some information about how good the dual solution on thatcell is. But another problem is, does z-z_h still has the sametendency as z?Almost. Think of it as z-phi_h where you can choose phi_h as you want.For example, on each cell you can think of choosing phi_h so that itcancels the constant and linear term of the Taylor expansion of z.Then z-z_h would contain the quadratic and higher order Taylor terms,i.e. something like z''*(x-x0)^2 where x0 can be chosen as a point onthe cell.If not, how z-z_h can represent the importance of a certan cell as zcan?I'm not sure if my understanding is correct. I tried to run the codeusing only z as dual_weights, and I found the result almost the sameas that using z-z_h.Nice idea to try this out. Do you get "almost the same" overall errorestimate, or "almost the same" mesh?I think all of these are good questions to ask. Although I have workedon this for a long time, I can not actually give you a particularlygood answer for all of this. I am sure others who are more versed inthe theory of errors, residuals, etc could tell you the precise reasonfor why it is in fact necessary to subtract z_h. The best I can say isthat that's the way I've always seen it done, and while I have a vagueidea why that is so (see above), I can't say that I can describe itwell enough to explain it.Finally, I am certainly glad to submit patches to deal.II and make myown contribution. But I didn't fork deal.ii on my github account yet,and this is relatively a small issue, so I will be glad if you can doit for the moment.OK, I will take care of this then. Best W.

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