Periodic pressure Laplacian gauge, solvability, and uniform Poincare bound

Reindexing the finite periodic sums gives, for every cell array $q$ and face array $v$, $(G_{h}q,v{)}_{M_h}=h^d{\sum }_{i,j}\frac{(q_{j+e_i}-q_j)v_{i,j}}{h}=-h^d{\sum }_j{q}_j{\sum }_i\frac{v_{i,j}-v_{i,j-e_i}}{h}=-h^d{\sum }_j{q}_j(D_{h}v{)}_j\mathrm{.}$ In particular, when $q\in Q_h$, this is $(G_{h}q,v{)}_{M_h}=-(q,D_{h}v{)}_{Q_h}$, the negative-adjoint identity in the stated mass pairings.

Let $q\in Q_h$. Since $q$ has mean zero, $q_x=N^{-d}{\sum }_{y\in J_N}(q_x-q_y),$ and [F3]–[F4] give the finite-sum Cauchy--Schwarz inequality $\mathrm{\parallel }q{\mathrm{\parallel }}_{Q_h}^2\le N^{-2d}{\sum }_{x,y\in J_N}\mathrm{\mid }{q}_x-q_y{\mathrm{\mid }}^2\mathrm{.}$ For each ordered pair $(x,y)$ choose the path that changes coordinates in the order $1,\dots ,d$, moving forward modulo $N$ in each coordinate. It has at most $dN$ directed edges. Cauchy--Schwarz along the path bounds $\mathrm{\mid }{q}_x-q_y{\mathrm{\mid }}^2$ by $dN$ times the sum of squared edge differences on it.

For any cell array $q$, applying the explicit weighted-sum identity of step 1.1 with $v=G_{h}q$ gives $h^d{\sum }_j{q}_j({\mathrm{\Delta }}_{h}q{)}_j=-(G_{h}q,G_{h}q{)}_{M_h}\le 0.$ If ${\mathrm{\Delta }}_{h}q=0$, then $G_{h}q=0$, so $q_{j+e_i}=q_j$ for every $i,j$. Coordinate steps connect all cells of the periodic grid, hence $q$ is constant. Every constant has zero gradient, so the kernel on all cell arrays is exactly the constants.

For every $v$, summing $D_{h}v$ over cells cancels each face value once with its opposite occurrence, so $D_{h}v$ has zero mean. The same is true of ${\mathrm{\Delta }}_{h}q=D_h(G_{h}q)$. Therefore ${\mathrm{\Delta }}_h$ restricts to a linear map $Q_h\to Q_h$. Its kernel there is zero by step 2.1, and its domain and codomain are the same finite-dimensional space.

Rank-nullity [F2] now gives $\dim \mathrm{im}{\mathrm{\Delta }}_h=\dim Q_h$; since the image is a subspace of $Q_h$, the restriction is onto as well as one-to-one. Thus each mean-zero right-hand side has a unique mean-zero pressure solution.

Fix one directed edge in coordinate $i$. For each coordinate $r<i$, its value along the path is $y_r$; to use the fixed edge, $y_r$ is fixed and $x_r$ has at most $N$ choices. For each $r>i$, the path still has value $x_r$; then $x_r$ is fixed and $y_r$ has at most $N$ choices. In coordinate $i$ there are at most $N^2$ ordered start/end pairs. Hence this edge occurs in at most $N^{d+1}$ paths. Summing the path bounds and dividing by $N^{2d}$ yields $\mathrm{\parallel }q{\mathrm{\parallel }}_{Q_h}^2\le d{N}^{2-d}{\sum }_{i,j}\mathrm{\mid }{q}_{j+e_i}-q_j{\mathrm{\mid }}^2=d\mathrm{\parallel }{G}_{h}q{\mathrm{\parallel }}_{M_h}^2\mathrm{.}$ Taking nonnegative square roots proves the stated inequality.