Second-order consistency of staggered periodic sampling

For $g\in C^3({\mathbb{T}}^d)$, Taylor's formula with Lagrange remainder at the midpoint and [F3] with $m=2$ give $g(x+h{e}_i\mathrm{/}2)-g(x-h{e}_i\mathrm{/}2)=h{\mathrm{\partial }}_{i}g(x)+r,\mathrm{\mid }r\mathrm{\mid }\le h^3\mathrm{\parallel }{\mathrm{\partial }}_i^{3}g{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}24.$ The same midpoint formula, with the two cell centres surrounding face $x_j+h{e}_i\mathrm{/}2$, gives $(G_h{R}_{h}p{)}_{i,j}={\mathrm{\partial }}_{i}p(x_j+h{e}_i\mathrm{/}2)+O(h^2),$ with the pointwise error bounded by $h^2\mathrm{\parallel }{\mathrm{\partial }}_i^{3}p{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}24$.

For $g\in C^4({\mathbb{T}}^d)$, [F3] with $m=3$ at $x$ gives $g(x+h{e}_i)-2g(x)+g(x-h{e}_i)=h^2{\mathrm{\partial }}_i^{2}g(x)+r,$ where the two fourth-order Lagrange remainders have total absolute value at most $h^4\mathrm{\parallel }{\mathrm{\partial }}_i^{4}g{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}12$. Apply this in every coordinate to $u_i$ at its face centre. It follows that $(A_h{R}_{h}u{)}_{i,j}=-(\mathrm{\Delta }{u}_i)(x_j+h{e}_i\mathrm{/}2)+O(h^2)$, with the error bounded by $h^2{\sum }_{\mathrm{\ell }}\mathrm{\parallel }{\mathrm{\partial }}_{\mathrm{\ell }}^4{u}_i{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}12$.

Applying the midpoint formula to component $u_i$ at cell centre $x_j$ shows, by [F3] with $m=2$, $(D_h{R}_{h}u{)}_j={\sum }_i{\mathrm{\partial }}_i{u}_i(x_j)+r_j,\mathrm{\mid }{r}_j\mathrm{\mid }\le \frac{h^2}{24}{\sum }_i\mathrm{\parallel }{\mathrm{\partial }}_i^3{u}_i{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$ Thus the discrete divergence differs pointwise from the sampled continuous divergence by $O(h^2)$.

Each pointwise error bound in steps 1.1–1.2 is uniform. The weighted mass of the grid contains $h^d{N}^d=1$ total cell volume and there are $d$ velocity components. Therefore the corresponding mass norms are bounded by the same constants (or their finite $d$-component sum) times $h^2$. Combining the divergence, vector Laplacian and pressure gradient bounds proves the first claim.

On each cell centred at $x_j$, fix $x$ and put $z=x-x_j$ and $\psi (t)=g(x_j+tz)$ for $0\le t\le 1$. By [F3] with $m=1$ and [F4], $g(x)=g(x_j)+\mathrm{\nabla }g(x_j)\cdot z+r_j(x)$ with $\mathrm{\mid }{r}_j(x)\mathrm{\mid }\le C_d\mathrm{\parallel }{D}^{2}g{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }z{\mathrm{\mid }}^2$. The linear term integrates to zero on the centred cell, so the difference between $h^{d}g(x_j)$ and ${\int }_{C_j}g$ is at most $C_d{h}^{d+2}\mathrm{\parallel }{D}^{2}g{\mathrm{\parallel }}_{\mathrm{\infty }}$. Summing the $N^d$ cells yields $\mid h^d{\sum }_{j}g(x_j)-{\int }_{{\mathbb{T}}^d}g(x)dx\mid \le C_d{h}^2\mathrm{\parallel }{D}^{2}g{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$ The face centres for any fixed component likewise centre a translated partition into cells of side $h$, so the same estimate holds for their samples. If the integral vanishes, this gives the stated discrete-mean bound.