Mixed H(div) fluxes, Darcy flow and local conservation — Examples

Let $K_1=\{0<y<x<1\}$ and $K_2=\{0<x<y<1\}$. The affine formulas $p_1(x,y)=y$ and $p_2(x,y)=x$ agree on $x=y$ and attain the four stated vertex values, so they define a continuous piecewise-P1 function. Their negative gradients are $q_1=(0,-1)$ and $q_2=(-1,0)$, each with classical divergence zero. With fixed diagonal normal $n=(1,-1)\mathrm{/}\sqrt{2}$, $q_1\cdot n=1\mathrm{/}\sqrt{2}$ and $q_2\cdot n=-1\mathrm{/}\sqrt{2}$, so the jump magnitude is $\sqrt{2}$.

For $\varphi \in C_c^{\mathrm{\infty }}(D)$, integration by parts on the two triangles shows that the distributional divergence of the piecewise field is the nonzero line measure on the diagonal with density $q_1\cdot n-q_2\cdot n=\sqrt{2}$ in this orientation (the distributional pairing is the negative of the summed outward boundary flux). To see it cannot be an $L^2$ function, choose a nonnegative smooth $\psi$ supported in the open diagonal with positive line integral and a smooth transverse cutoff $\eta$ with $\eta (0)=1$; set ${\varphi }_{\epsilon }(y,t)=\psi (y)\eta (t\mathrm{/}\epsilon )$ in tubular coordinates, extended by zero. The surface pairing is the fixed nonzero number $\sqrt{2}\int \psi$, whereas $\mathrm{\parallel }{\varphi }_{\epsilon }{\mathrm{\parallel }}_{L^2(D)}=O({\epsilon }^{1\mathrm{/}2})$. If the distribution were represented by $g\in L^2(D)$, [F2] would give $\mathrm{\mid }\int g{\varphi }_{\epsilon }\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_2\mathrm{\parallel }{\varphi }_{\epsilon }{\mathrm{\parallel }}_2\to 0$, a contradiction. By [F1] the field is not in H(div).

The gradient is $\mathrm{\nabla }p=(-1,0)$, so $q+\mathrm{\nabla }p=0$ pointwise. The three outward unit normals are $(-1,0)$ on $x=0$, $(0,-1)$ on $y=0$, and $(1,1)\mathrm{/}\sqrt{2}$ on the hypotenuse. Their facet lengths are $1,1,\sqrt{2}$, respectively. Therefore the integrated fluxes are $-1$, $0$, and $(1\mathrm{/}\sqrt{2})\sqrt{2}=1$.

Their sum is zero, equal to ${\int }_K\mathrm{div}q=0$ by [F2]. Since $q$ is in RT0 and the fluxes are its facet degrees of freedom by [F1], the data also uniquely determine this constant flux among RT0 fields.

The two positive scalar values are $1$ and ${10}^6$, so their minimum and maximum are exactly ${\kappa }_{\min }=1$ and ${\kappa }_{\max }={10}^6$. Their reciprocals in $K^{-1}$ are $1$ and ${10}^{-6}$; substituting into the energy definition gives $\mathrm{\parallel }v{\mathrm{\parallel }}_{K^{-1}}^2={\int }_{D_1}\mathrm{\mid }v{\mathrm{\mid }}^2+{10}^{-6}{\int }_{D_{{10}^6}}\mathrm{\mid }v{\mathrm{\mid }}^2$.

The error theorem constants are $M_a=1$, ${\alpha }_{\ast }={10}^{-6}$, so ${\alpha }_{\ast }^{-1}={10}^6$ and $M_a\mathrm{/}{\alpha }_{\ast }={10}^6$. The inf-sup lower bound $\widehat{\beta }=(1+C_R^2{)}^{-1\mathrm{/}2}$ has no permeability parameter because the divergence lift is geometric; this example therefore records an explicit contrast dependence and supplies no contrast-independent error claim.

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].