Suppose for a grid-based calculation a grid is used such that the grid Jacobian is discontinuous. For example, in 1D, for a domain $x \in$ [0,1], one half of the domain is covered uniformly by twice as many grid points as the other side of the domain (and this always remains so as the number of grid points is increased). This grid can be illustrated by this plot, showing the x coordinate of the grid point vs. its index normalized to the total number of grid points.

enter image description here

The Jacobian for this grid, representing the transformation of the x coordinate to the grid index is clearly discontinuous at x=1/2. What would be the implications of using such grids for finite-difference of finite-volume calculations such as solving ODEs or PDEs? Would it reduce the grid convergence to the first order? Or not necessarily? Can it lead to losing grid convergence altogether? Would a finite-element (or spectral element) method have an advantage on discontinuous grids, compared to finite difference or finite volume based methods?

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    $\begingroup$ The physical picture is that you would run into aliasing effects, if a high-frequent wave from the dense region enters the sparse region, where it can't be resolved by the wavelength $2\pi/\Delta x$. With regard to orders -- it depends on what you do with the grid, but the order usually goes with the number of gridpoints, and not their distribution (you'll be able to reproduce an N-1 polynomial in any case). However, different distributions have an impact on the condition. Basic example in this direction: the Runge phenomenon on an equally spaced vs. a Chebyshev grid. $\endgroup$
    – davidhigh
    Commented Oct 28, 2021 at 21:27

1 Answer 1


I am not an expert on this question (I used finite elements mostly as a student) but I did use finite differences with a 2D uneven grid recently. My intuition is that, if you use the right discretization scheme then there is no loss of precision - in fact you might have a very good reason to have more points in one region. If you blindly use the "uniform grid" discretization schemes then for pretty sure you will lose precision.

If we assume for simplicity that you're working with finite differences in 1D, then what you really want is to discretize the operator on your grid. This is done for example here: https://math.stackexchange.com/questions/2470702/finite-difference-method-for-non-uniform-grid

With the proper discretization, you should be able to get the correct order of convergence.


I was looking for a similar problem today and I came across this answer, the article cited in it is really worth a read. https://scicomp.stackexchange.com/a/481/41372

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    $\begingroup$ Thank you for this answer! Yes, thinking further about this I tend to agree that a finite-difference (finite-volume etc.) formulation should be always possible with a desired formal accuracy order on arbitrarily non-smooth grids. Of course the size of the approximation error depends on both the accuracy order and the coefficient in front, and that coefficient in general depends on the grid smoothness, and on the solution itself. I found some fascinating reading on this topic here core.ac.uk/download/pdf/12923402.pdf. $\endgroup$ Commented Oct 28, 2021 at 17:07
  • $\begingroup$ The paper you linked is an interesting example of misleading asymptotic arguments. The link below (pp 18) analyzes the truncation error of the central difference scheme on a uniformly expanding grid. While the leading term of the truncation error is first-order (with respect to grid spacing), in practice it is second-order if you keep your expansion ratio close to 1. ocw.mit.edu/courses/mechanical-engineering/… $\endgroup$
    – Charlie S
    Commented Oct 28, 2021 at 19:55

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