# Re: [deal.II] some problems concerning step-14

```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.```
```
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 cell as residual times dual weight. But I'm still a little confused with the reason why we must introduce z_h. Just as you mentioned, if we introduce z_h, then z-z_h is a quantity that is only large where the dual solution is rough. But why do we need to care about the accuracy of z here? I think the only thing we need to care about is the value of z on that cell, because z is a quantity that represents how important the residual on that cell is.
```
```
No. z tells you how important the *locally generated error is for the global error functional*. (That is because z is the Green's function associated with your error functional.) But you don't have the local error. All you have is the local residual.
```

```
My understanding is: now the dual_weight z-z_h does not only represent how important the residual on a certain cell is, but also tells us some information about how good the dual solution on that cell is. But another problem is, does z-z_h still has the same tendency 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 it cancels 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 on the cell.
```

```
If not, how z-z_h can represent the importance of a certan cell as z can? I'm not sure if my understanding is correct. I tried to run the code using only z as dual_weights, and I found the result almost the same as that using z-z_h.
```
```
Nice idea to try this out. Do you get "almost the same" overall error estimate, or "almost the same" mesh?
```
```
I think all of these are good questions to ask. Although I have worked on this for a long time, I can not actually give you a particularly good answer for all of this. I am sure others who are more versed in the theory of errors, residuals, etc could tell you the precise reason for why it is in fact necessary to subtract z_h. The best I can say is that that's the way I've always seen it done, and while I have a vague idea why that is so (see above), I can't say that I can describe it well enough to explain it.
```

```
Finally, I am certainly glad to submit patches to deal.II and make my own 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 do it for the moment.
```
OK, I will take care of this then.

Best
W.

```
```
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