Two-triangle hybrid RT0 local elimination

Each cell has area $1\mathrm{/}2$. The field ${\theta }_1=(x-1,y)$ has zero normal component on $x=1$ and $y=0$, and unit integrated outward flux on the diagonal from $K_1$; ${\theta }_2=(x,y-1)$ has zero normal component on $x=0$ and $y=1$, and unit integrated outward flux on the diagonal from $K_2$. Their divergences are both $2$, so ${\int }_{K_i}\mathrm{div}{\theta }_i=1$ by [F3]. The two zero-boundary conditions and the single free facet degree of freedom make each ${\theta }_i$ a basis of its local space by [F4]. Direct integration gives ${\int }_{K_1}\mathrm{\mid }{\theta }_1{\mathrm{\mid }}^2={\int }_0^1{\int }_0^x((x-1{)}^2+y^2)dydx=\frac{1}{6},$ and reflection symmetry gives ${\int }_{K_2}\mathrm{\mid }{\theta }_2{\mathrm{\mid }}^2=1\mathrm{/}6$.

Write $u_i=t_i{\theta }_i$. The cell balances in [F1] give $t_1={\int }_{K_1}f=1\mathrm{/}2$ and $t_2={\int }_{K_2}f=-1\mathrm{/}2$. For a local test $v_i=s{\theta }_i$, its integrated divergence is $s$ and its interior outward flux is $s$. The first hybrid equation is therefore $t_{i}s\mathrm{/}6-p_{i}s+\lambda s=0$ for every real $s$, equivalently $t_i\mathrm{/}6-p_i+\lambda =0$. The jump equation is $t_1+t_2=0$, which the balance values satisfy.

Solving the local equations gives $p_1=1\mathrm{/}12+\lambda$ and $p_2=-1\mathrm{/}12+\lambda$. Since the cells have equal area, their global mean is $\lambda$; choose the mean-zero gauge $\lambda =0$ to obtain the stated pressures. After local elimination the single skeleton equation is $t_1+t_2=0$, independent of $\lambda$, so its one-dimensional nullspace consists exactly of common constant shifts. This is the constant gauge described in [F2].