Consider the 1D nonstationary convection-diffusion PDE $$ \begin{alignat}{2} \partial_t u &= -a \partial_x u + D \partial_{xx}u, &\qquad x \in (0,1), t \in (0,T), \\ f(t) &= \left.\left( a u - D\partial_x u \right) \right|_{x=0} & \qquad t \in (0,T), \\ 0 &= \left. -D\partial_x u \right|_{x=1} & \qquad t \in (0,T), \\ u_0 &= \left. u \right|_{t=0} & \qquad x \in (0,1), \end{alignat} $$ where $a, D > 0$ are constant and $f,u_0$ are given functions. The problem is convection-dominated (i.e., $a \gg D$).

A cell-centered finite volume method with uniform grid is used to discretize the PDE in space. In the interior, central differences for the diffusive flux and a second-order accurate convective flux (e.g., QUICK) can be used. Crank-Nicolson is used in time to obtain an overall second-order convergent scheme.

The boundary conditions eliminate the diffusive fluxes in the discretization at the boundaries. For the right faces of the first and last cell, however, there are not enough neighbouring cells to apply the second-order convective flux (i.e., the QUICK scheme requires the left and right cells of a face and one additional upstream cell).

Now, for these two faces, an upwind flux is used, which is only first-order accurate. Theoretically, the whole method is, hence, reduced to first-order consistency.

In computational experiments, second-order convergence is observed, though. This suggests the reasoning that an $O(h)$ error committed in a region of size $O(h)$ does itself result in an $O(h^2)$ error, which does not impede overall $O(h^2)$ convergence.

Is this reasoning correct? How can it be proved / are there any references?

According to (Randall LeVeque, Finite Difference Methods for Ordinary and Partial Differential Equations , Sec. 2.12), this does not hold for Poisson's equation with central and one-sided finite differences. So it may not hold in general (i.e., general equation type, general discretization method).

  • $\begingroup$ What is the role of $f(t)$? $\endgroup$ – Maxim Umansky Jun 20 '20 at 4:20
  • $\begingroup$ $f(t)$ is the inlet profile. Often, it's a step function $f(t) = u_{\mathrm{in}}\chi_{(0, T_{\mathrm{in}})}(t)$ with characteristic function $\chi$ and constants $T_{\mathrm{in}} \in (0, T)$ and $u_{\mathrm{in}} > 0$. But for this question, it might as well be considered smooth. $\endgroup$ – cos_theta Jun 20 '20 at 6:52

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