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I am running a mesh independence study. I start with Mesh 1 and proceed up to Mesh 4, each time doubling the number cells in the mesh. In parallel, I am comparing my computational results to experimental data. M. 1 shows poor results. M. 2 shows a significant improvement and a good match to experimetnal results. M. 3 and M. 4 produce identical results, that are only slightly different from M. 2. It would then seem sensible to pick M. 3 as my final mesh. But it seems that the results are too smooth, loosing some of the fine details, produced by M. 2 (and observed in the experiments).

Can there actually be some sort of overcorrection? Could it be that the mesh independent solution is not necesarily the best one?

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2 Answers 2

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No. As the mesh size $h\rightarrow 0$, the solution on a given mesh converges to the solution of the differential equation (assuming a well-posed PDE and a suitable discretization). Consequently, if your discrete solution on M2 is closer to experimental results than that on M3 or M4, then there are only two possibilities:

  • The differential equation does not appropriately describe the real world because its solution is not close to what you measured.
  • Your measured data are not correct.

Essentially, what you are seeing is that the exact solution of the differential equation (apparently well approximated on meshes M3 and M4) does not match your measurements. Which one is wrong is for you to find out now.

(All this was written under the assumption that your numerical scheme is indeed correct and, thus, that your numerical solution on fine meshes is a good approximation to the exact solution.)

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  • $\begingroup$ Thank you Wolfgang. Probably there is a bit of truth to both explanations: (i) I am using a multiphase model that is not necessarily fully capturing complex fluid-particle interactions, (ii) in the experiments 3D data was obtained using 2D technology. Irrespective, from what I gather the independent mesh is the best one. $\endgroup$ Jun 3, 2013 at 7:00
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I claim that the independent mesh is the best one. Say the actual solution is $U$ and your solver delivers an $u_h$ depending on a mesh parameter $h$. Then you can do the estimate for the distance of $u_h$ to the actual solution $$\|U-u_h \| \leq \|U-u_m \| + \|u_m - u_h\|,$$ where $u_m$ is the solution of the model used to describe the problem mathematically.

Assume the modelling error $\| U - u_m\|$, that is independent of the mesh, is of constant size $C_m$, while the numerical error satisfies an estimate of type $\|u_m - u_h\| < C_h h^p$, with $p\in \mathbb N$. Thus, the error measured can be expressed as $$\|U-u_h\| \approx C_m + C_hh^p.$$ Reducing the mesh size, i.e. $h\to 0$, there will be a point, when the numerical error goes below the modelling error, and the differences to the actual solution does not change anymore.

At this point your solution is called mesh independent, since the modelling error dominates over the numerical error.

Thus, your observation is probably due to a lucky cancellation of errors. But I would not call on this, since I don't see a mathematical justification for using coarser discretizations, unless one faces instabilities.

I would rather consider a remodelling...

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  • $\begingroup$ Oh, as @WolfgangBangerth has mentioned, the measurements are also subjected to error sources. $\endgroup$
    – Jan
    Jun 2, 2013 at 18:50

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