If we evaluate the first derivative of a function F(x) on a 1D grid {$x_i$} by central difference at $x=x_i$ as

$$ \frac{dF}{dx} \approx \frac{F_{i+1} - F_{i-1}}{x_{i+1} -x_{i-1}} $$

then it is known that on a uniform grid it is second order accurate but on a non-uniform grid it is only first order accurate.

Now, suppose we have a non-uniform 1D grid $x_i = g(i)$ where $g$ is a smooth function for which the derivative can be calculated analytically. Then we can use the chain rule to evaluate at $x=x_i$

$$ \frac{dF}{dx} = \frac{dF/di}{dx/di} \approx \frac{F_{i+1}-F_{i-1}}{2 g'(i)} $$

The grid in the $i$ index is obviously uniform so the central difference approximation for $dF/di$ must be second order accurate; and $g'=dx/di$ is just exact. So, using the chain rule, the result for $dF/dx$ on this non-uniform grid should be second order accurate, contrary to the common statement that evaluating the first derivative by central difference on a non-uniform grid would be only first order accurate. Of course for any non-uniform grid in 1D one can find a smooth mapping to the uniform grid index, e.g., by the Lagrange interpolating polynomial. Then, is it correct that evaluating the derivatives by central difference on a non-uniform grid, using the chain rule as proposed here, one can always achieve a second-order accurate approximation? Then one would have immediate generalizations, e.g., a second-order accurate approximation for the second derivative,

$$ \frac{d^2F}{dx^2} \approx \frac{F_{i+1}+F_{i-1}-2 F_i}{(g')^2} - \frac{1}{2} \frac{F_{i+1}-F_{i-1}}{(g')^3} g'', $$

and various higher order approximations using larger stencils.

Is this discussed somewhere in computational science literature?


1 Answer 1


Let $x = x(\xi)$ be a smooth, invertible map and we make a uniform grid in $\xi$-space. This induces a grid in $x$-space $$ x_i = x(\xi_i) $$ Method 1: The approximation $$ \frac{F_{i+1} - F_{i-1}}{x_{i+1} - x_{i-1}} = F'(x_i) + O(\Delta x_i) $$ is first order accurate as can be checked from Taylor expansion.

Method 2: The approximation $$ \xi'(x_i) \frac{F_{i+1} - F_{i-1}}{\xi_{i+1} - \xi_{i-1}} = \xi'(x_i) \left[ F'(\xi_i) + O(\Delta\xi)^2 \right] = F'(x_i) + \xi'(x_i) O(\Delta\xi)^2 $$ is second order accurate.

We can also look at Method 1 like this $$ \frac{F_{i+1} - F_{i-1}}{x_{i+1} - x_{i-1}} = \frac{\xi_{i+1} - \xi_{i-1}}{x_{i+1} - x_{i-1}} \frac{F_{i+1} - F_{i-1}}{\xi_{i+1} - \xi_{i-1}} = [\xi'(x_i) + O(\Delta x_i)] [ F'(\xi_i) + O(\Delta\xi)^2] $$ $$ = F'(x_i) + O(\Delta x_i) + O(\Delta\xi)^2 $$ Method 2 becomes Method 1 if we approximate the mapping derivative with finite differences which are first order accurate only, so the method is only first order accurate.

  • $\begingroup$ These are not different things, we are comparing approximations to dF/dx, calculated by two different methods. So, if we use a mapping to a uniform coordinate, then from the same information, $F_{i+1}$ and $F_{i-1}$ we can get a more accurate approximation for $dF/dx$ at $x_i$, (asymptotically, for large number of grid points)? $\endgroup$ May 16, 2020 at 16:45
  • $\begingroup$ I updated the answer. $\endgroup$
    – cfdlab
    May 17, 2020 at 0:58
  • $\begingroup$ In the statement of the question the mapping is given as an analytic function so its derivative is exact. I update the question to make this unambiguous. $\endgroup$ May 17, 2020 at 1:50
  • $\begingroup$ If you use exact mapping derivative, the two methods are different. One is more accurate than other, there is no paradox. $\endgroup$
    – cfdlab
    May 17, 2020 at 2:17
  • $\begingroup$ So you agree that on a non-uniform grid one can indeed obtain second-order accuracy, if the grid can be mapped to a uniform one by a smooth mapping function - correct? $\endgroup$ May 17, 2020 at 2:54

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