RT0/P0 Darcy stability and flux/pressure error bound

Suppose first there are at least two cells. Their independent cell constants, subject to one mean constraint, give $Q_h\mathrm{\ne }\{0\}$. The stable lift [F2] then gives $V_h\mathrm{\ne }\{0\}$ and a positive discrete inf-sup constant. If $z_h$ belongs to the discrete kernel, [F3] gives $\mathrm{div}{z}_h=0$; hence its graph norm is its $L^2$ norm and the coefficient lower bound [F4] gives discrete coercivity with ${\alpha }_{\ast }={\kappa }_{\max }^{}$. Write $m_a=\mathrm{\parallel }a\mathrm{\parallel }$ and $m_b=\mathrm{\parallel }b\mathrm{\parallel }$ for the exact norms on $V\times V$ and $V\times Q$. By [F4], $m_a\le M_a={\kappa }_{\min }^{}$ and $m_b\le M_b=1$.

Apply [F5] to the conforming spaces $V_h\subset V$, $Q_h\subset Q$, continuous solution [F1], and the exact norms $m_a,m_b$ from step 1.1. This gives a unique Galerkin pair and the error bounds with ${\beta }_h,m_a,m_b$. The supplied expressions are nondecreasing in the nonnegative quantities ${\beta }_h^{-1},m_a,m_b$: $R_h$ is nondecreasing first, then $U_h$, then $P_h$. Since ${\beta }_h^{-1}\le {\widehat{\beta }}^{-1}$, $m_a\le M_a$, and $m_b\le M_b$, these replacements give exactly the stated upper bounds $U_h,P_h$. For $e=u-u_h$, the weighted bound follows from [F4] and $\mathrm{\parallel }e{\mathrm{\parallel }}_2\le \mathrm{\parallel }e{\mathrm{\parallel }}_{H(\mathrm{div})}$, since $\mathrm{\parallel }e{\mathrm{\parallel }}_{K^{-1}}^2\le {\kappa }_{\min }^{}\mathrm{\parallel }e{\mathrm{\parallel }}_2^2$.

If the mesh has one cell, every facet is a physical-boundary facet. The boundary-zero facet moments and [F6] force $V_h=\{0\}$, while a constant with zero global mean is zero, so $Q_h=\{0\}$. The zero pair is the sole discrete pair. Finally $M_a\mathrm{/}{\alpha }_{\ast }={\kappa }_{\max }\mathrm{/}{\kappa }_{\min }$, making the permeability-contrast dependence explicit in the formulas rather than uniform in that ratio.