1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ With #2, why would it be second-order accurate, not third-order or higher? $\endgroup$ Jun 24, 2021 at 17:07

1 Answer 1

0
$\begingroup$
  1. 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 $u_L, u_R$.

  2. 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 $\mathcal{O}(\Delta x^3)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.