Thank you so much Bangerth!

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.
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? 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.

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 

在 2018年3月4日星期日 UTC+8上午1:42:27,Wolfgang Bangerth写道:
> On 03/02/2018 03:36 AM, 曾元圆 wrote: 
> > Hi, nowadays I'm learning about the goal oriented error estimator and I 
> > read the tutorial of step-14 
> > ( But I'm 
> > confused about some problems and hope you can help me with these: 1. 
> When 
> > deriving the error with respect to the functional, why must we change 
> > J(e)=a(e,z) to J(e)=a(e,z-z_h) ? I know the dual solution z must be 
> > approximated in a richer space than the primal solution, otherwise J(e) 
> > will be 0. But why not just solve the dual problem in a richer space 
> > without subtracting its interpolation to the primal space? I didn't see 
> the 
> > necessity to introduce z_h into the formula. 
> You are correct: the values you get from both of these formulas are 
> exactly 
> the same. So it is not *necessary* to introduce z_h if you are interested 
> in 
> computing the *error*. 
> We introduce the interpolant because z-z_h is a quantity that is only 
> large 
> where the dual solution is rough, where z may be large also in other 
> places. 
> Doing so ensures that the error estimator is *localized*: It is large 
> exactly 
> where the primal and dual solutions are rough, i.e., where we expect the 
> error 
> to be caused. As mentioned above, if you sum the contributions of all 
> cells, 
> you will get the same value whether you introduce z_h or not, but the 
> contribution of each cell is going to be different. If you want the 
> contributions of each cell to serve as a good mesh refinement criterion, 
> then 
> it turns out that you need to introduce z_h. 
> > 2. Why must we change J(e)=∑(f+△ uh,z-zh)-(∂uh,z-zh) to J(e)=∑(f+△ 
> > uh,z-zh)-(1/2[∂uh],z-zh)? Is it just an implementation consideration for 
> > saving computational effort? 
> For the same kind of reason. For a smooth (primal) solution, the term 
>    (∂uh,z-zh) 
> may be large because the normal derivative of the primal solution may 
> simply 
> be what it is -- think of, for example, a linear exact solution u that can 
> be 
> perfectly approximated by u_h. So if you leave this term as is, this would 
> suggest that the error on this cell is large. But that's wrong -- the 
> error is 
> actually quite small because you can approximate linear functions well. 
> On the other hand, if you do the rewrite (which again leaves the *sum* 
> over 
> all cells the same, but change the contributions of each cell), then 
>    (1/2[∂uh],z-zh) 
> is going to be small because while the normal derivative of u_h may be 
> large, 
> the *jump* of the normal derivative is small if the solution is linear or 
> nearly so. 
> Another way of seeing this is to think of both of the terms in J(e) as 
>    residual times dual weight. 
> The residuals here are f+△ uh and 1/2[∂uh]. You want to define these 
> residuals 
> in such a way that they are zero for the exact solution. That is true for 
> these two residuals: f+△ u is zero because 'u' satisfies the equation, and 
> 1/2[∂u] is zero because the solutions of the Laplace equation have 
> continuous 
> gradients (if f is smooth enough). 
> On the other hand, the term ∂u is not zero, even for the exact solution. 
> > Can this kind of rewriting be generally adopted in other kind of 
> > problems(e.g in advection problem where the face integrals in J(e) 
> relies 
> > on the upstream information)? 
> Yes. 
> Best 
>   W. 
> -- 
> ------------------------------------------------------------------------ 
> Wolfgang Bangerth          email:        
> <javascript:> 
>                             www: 

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