When it comes to a second-order accurate finite volume discretization of Navier-Stokes equations, which one of the two following rationales is adopted?
1- Second-order accuracy is a direct consequence of how we eliminate higher order terms in the Taylor expansion of all the terms;
2- From the beginning, we assume that the dependent variable varies linearly between a pair of neighboring cell centers; with such an assumption, high-order terms won't appear at all.
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1$\begingroup$ With #2, why would it be second-order accurate, not third-order or higher? $\endgroup$– Maxim UmanskyJun 24, 2021 at 17:07
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$\begingroup$ Piggybacking on @MaximUmansky, if you feed a linear MMS into a 2nd order algorithm you should find no error, so with assumption 2, any algorithm above 1st order would return 0 error. $\endgroup$– EMPJun 19, 2022 at 21:59
2 Answers
Usually you speak of a $n$'th order (accurate) method if your Taylor truncation error is of order $n+1$. This means your approximation is accurate up to order $n$ terms, and your errors are of order $n+1$. However, in FVM methods you often have no easy way of obtaining the truncation error of your formulation, since you reconstruct the numerical fluxes $F$ based on some procedure, which are in turn based on the your reconstructed trace/edge values $u_L, u_R$.
I guess you are referring here to the linear reconstruction of the face values. This reconstruction is second order accurate since you have a first order approximation with truncation error $\mathcal{O}(\Delta x^2)$ on an interval of size $\mathcal{O}(\Delta x)$, resulting in an overall truncation order of $\mathcal{O}(\Delta x^3)$ and thus a second order accurate scheme.
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$\begingroup$ Your second point is incorrect. Linear reconstruction of face values in FV is typically 2nd order accurate, giving TE of 2nd order accuracy as well. The exception is node-centered schemes using quadratically exact (2nd order) gradient computation algorithms which returns a 3rd order accurate reconstruction and therefore 3rd order TE on the interior nodes of regular meshes. $\endgroup$– EMPJun 19, 2022 at 22:09
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$\begingroup$ Note that I speak of the truncation error, not the order. As usual, subtract one power from the truncation error to get the order of the scheme. $\endgroup$– Dan DoeJun 20, 2022 at 6:27
The answer is that it's a consequence of how we eliminate the high order terms in the taylor series expansion of the discretization error, not the truncation error.