Write $a=c=2\pi$. Direct differentiation gives $\mathrm{\nabla }\cdot \tilde{u}=-c\sin (ax)y(1-y)$. Define $g(y)=-\frac{c}{a^2}{y}^2+\frac{c}{a^2}y-\frac{2c}{a^4}+\frac{c}{a^3\sinh (a\mathrm{/}2)}\cosh (a(y-1\mathrm{/}2)),\varphi (x,y)=\sin (ax)g(y)\mathrm{.}$ Differentiating the polynomial and hyperbolic terms gives $g^{\mathrm{\prime }\mathrm{\prime }}-a^{2}g=-cy(1-y)$ and $g^{\mathrm{\prime }}(0)=g^{\mathrm{\prime }}(1)=0$. By [F2], this $\varphi$ solves the stated Poisson problem and satisfies both Neumann conditions. Since ${\tilde{u}}_y=0$ and ${\mathrm{\partial }}_y\varphi =0$ at the walls, the corrected normal velocity remains zero.

For uniqueness of its gradient, if two classical periodic Neumann solutions are given, their difference $w$ is harmonic with ${\mathrm{\partial }}_{y}w=0$ on both walls. For each fixed $y$, one-dimensional integration by parts in $x$ gives ${\int }_{\mathbb{T}}w{w}_{xx}=-{\int }_{\mathbb{T}}{w}_x^2$ because the periodic endpoint terms cancel; integration by parts in $y$ gives ${\int }_0^{1}w{w}_{yy}=[w{w}_y{]}_0^1-{\int }_0^1{w}_y^2=-{\int }_0^1{w}_y^2$ by the Neumann data. Integrating the latter equality in $x$ and using $\mathrm{\Delta }w=0$ yields ${\int }_{\mathrm{\Omega }}\mathrm{\mid }\mathrm{\nabla }w{\mathrm{\mid }}^2=0$. Hence $\mathrm{\nabla }w=0$, so the correction is independent of the pressure constant.

Since ${\tilde{u}}_x(x,0)=0$, the corrected tangential trace at the lower wall is $u_x(x,0)=-{\mathrm{\partial }}_x\varphi (x,0)=-a\cos (ax)g(0)\mathrm{.}$ At $x=0$, $g(0)=\frac{c}{a^4}(a\coth (a\mathrm{/}2)-2)>0,$ because $\coth (a\mathrm{/}2)>1$ and $a=2\pi >2$. Therefore $u_x(0,0)=-ag(0)\mathrm{\ne }0$, which proves the tangential no-slip defect.

Since $N$ is even, summing $(-1{)}^{j_1}$ over its $N$ values gives zero; the other coordinates simply repeat this sum. Hence $q$ has zero discrete mean. It is nonconstant because its values at $j_1=0$ and $j_1=1$ differ.

For $i=1$, [F2] gives $q_{j+e_1}=q_{j-e_1}$, so the first component of $G_{c}q$ is zero. For every $i>1$, the array is independent of coordinate $i$, hence again $q_{j+e_i}=q_{j-e_i}$. The defining centred difference therefore gives $G_{c}q=0$ in every component, proving the nonconstant zero-mean pressure mode is invisible.

The scalar matrix $A=[1]$ is SPD, $B^{T}q=0$ implies $q=0$, and the Schur matrix is $S=B{A}^{-1}{B}^T=[1]$. Thus the normalized Schur eigenvalue is $\mu =1$ and $P_1=\mathrm{diag}(1,1)=I$ by [F1].

Applying [F2] at $\mu =1$ gives the characteristic polynomial ${\lambda }^2-\lambda -1$ and roots $(1\pm \sqrt{5})\mathrm{/}2$. Direct substitution in $u+p=1$ and $u=0$ yields $(u,p)=(0,1)$; the coefficient matrix is invertible because its determinant is $-1$, so this solution is unique.

Since $N$ is even, $q$ has zero mean. The staggered gradient is not zero: $(G_{h}q{)}_{1,j}=-2q_j\mathrm{/}h$. Also ${\mathrm{\Delta }}_{h}q=-4q\mathrm{/}{h}^2$, so [F1] gives $D_h{G}_{h}q={\mathrm{\Delta }}_{h}q$. Periodic summation shows both $G_{h}q$ and $A_h{G}_{h}q$ have zero component means, hence the force is admissible.

Substituting ${\tilde{u}}^{n+1}=G_{h}q$ and $u^n=p^n=0$ into the predictor equation [F2] gives $k^{-1}{G}_{h}q+\nu A_h{G}_{h}q=f^{n+1}$, so uniqueness [F3] identifies it as the predictor. For the pressure equation, ${\mathrm{\Delta }}_h(k^{-1}q)=k^{-1}{D}_h{G}_{h}q=k^{-1}{D}_h{\tilde{u}}^{n+1};$ both sides have mean zero and [F3] gives ${\varphi }^{n+1}=k^{-1}q$.

The velocity update [F2] now gives $u^{n+1}=G_{h}q-k{G}_h(k^{-1}q)=0$, while the pressure update gives $p^{n+1}=k^{-1}q$. This is exact one-step removal of the selected mode.