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Let there is triangular mesh for a simple domain (domain is not important).

If this mesh is not uniform (the mesh is produced by adaptive technique), what is the mesh size?

Is there a mesh size for each triangle? or this is a number relate to the whole mesh on the domain? How do I compute it?

Thank you,

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    $\begingroup$ It depends what you want to do with mesh size. If for example you need it for local refinement - you probably need to have seperate values for each element. If you need it to determinate time step - you need only one value - for example maximum, but it depends on a problem you are trying to solve and equation to determinate timestep. $\endgroup$ – Krzysztof Bzowski Nov 1 '15 at 11:16
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    $\begingroup$ On the off chance you're doing a solution convergence analysis, you might be better off to plot the error as a function of the DoF count (instead of the mesh size). $\endgroup$ – GoHokies Nov 2 '15 at 9:23
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For each element the so-called local mesh size can be defined e.g. as the maximum edge length. This has applications at least in residual a posteriori error analysis where the elementwise error indicator is often multiplied by the local mesh size.

Instead of computing the maximum edge length one can show that the Jacobian determinant of the affine mapping from the reference element to a global element is proportional to the square of the local mesh size. This is a property that I often use when programming adaptive mesh refinement using a posteriori estimators since the code for computing the Jacobian determinant is already there (assuming that the finite element code is implemented using reference elements).

If the mesh is not uniform in the sense that for each pair of triangles their local mesh sizes can be bounded by each other (multiplied by some constant), then the so-called 'global mesh size'—i.e. the maximum of local mesh sizes—may not be a meaningful quantity. For example, in adaptive refinement one usually is concerned about the error versus number of degrees-of-freedom—as indicated in the comments.

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  • $\begingroup$ To add to @knl's comment, another approach is to define $h$ as the ratio of surface area to volume of an element. $\endgroup$ – Jesse Chan Nov 7 '15 at 8:30

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