A Neumann pressure correction can break tangential no-slip

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.