A uniformly bounded Fortin map for the practical P1/R3 DPG family

On the reference triangle $B_3^{grad}=\{p\in P_3:p(a_i)=0\}$ has dimension $7$. Define ${\mathrm{\Pi }}_3^{0}w\in B_3^{grad}$ by matching one cell mean and two edge $P_1$ moments on each edge. In the homogeneous system an edge restriction is $t(1-t)(a+bt)$; testing against $a+bt$ forces its weighted square integral to vanish. Each edge restriction is zero, leaving $c{\lambda }_0{\lambda }_1{\lambda }_2$, whose mean forces $c=0$. Hence the seven scalar moments are unisolvent.

On the reference triangle $R{T}_2=P_2^2+x{\tilde{P}}_2$ has dimension $15$. Its normal trace map is onto broken P2: for an edge $F$ opposite vertex $a_i$, extend any $q\in P_2(F)$ polynomially in the other two barycentric coordinates and set $\tau =(x-a_i)q(x)\mathrm{/}{H}_i$, where $H_i$ is the altitude to $F$. This vector is tangent to the other two edges and has normal trace $q$ on $F$; it lies in $R{T}_2$ because its cubic part is $x$ times a homogeneous degree-two polynomial and its remaining terms lie in $P_2^2$. Thus the broken normal trace image has dimension $9$. Let $B_3^{div}$ be its inverse image of continuous P2 boundary traces, a $12$-dimensional subspace. Define ${\mathrm{\Pi }}_3^{div}\tau \in B_3^{div}$ by matching six volume moments against $P_1^2$ and six continuous boundary P2 normal moments. In the homogeneous system the normal trace is continuous P2 and orthogonal to itself, hence zero. Green’s identity and $\mathrm{\nabla }{P}_2\subset P_1^2$ then give $div\tau =0$; its homogeneous cubic part $x{q}_2$ has divergence $4q_2$, so $\tau \in P_2^2$. Write $\tau =R\mathrm{\nabla }\varphi$: explicitly, for polynomial $\tau =({\tau }_1,{\tau }_2)$ use $\varphi (x,y)={\int }_0^x{\tau }_2(s,0)ds-{\int }_0^y{\tau }_1(x,t)dt$. The identity ${\mathrm{\partial }}_x{\tau }_1+{\mathrm{\partial }}_y{\tau }_2=0$ gives ${\mathrm{\partial }}_x\varphi ={\tau }_2$ and ${\mathrm{\partial }}_y\varphi =-{\tau }_1$, and $\mathrm{deg}\varphi \le 3$. Zero normal trace makes $\varphi$ constant on each edge; vertex continuity makes these constants equal. After subtracting the common constant, $\varphi =c{\lambda }_0{\lambda }_1{\lambda }_2$. The volume moment against $Rx$ is ${\int }_K\mathrm{\nabla }\varphi \cdot x=-2c{\int }_K{\lambda }_0{\lambda }_1{\lambda }_2$ by Green’s identity and $\varphi {\mathrm{\mid }}_{\mathrm{\partial }K}=0$, so $c=0$. Thus the 12 moments are unisolvent.

On the reference triangle set ${\widehat{\mathrm{\Pi }}}_3^{grad}\widehat{v}=\bar{\widehat{v}}+{\mathrm{\Pi }}_3^0(\widehat{v}-\bar{\widehat{v}})$; for a physical cell let $F_K(\widehat{x})=A_K\widehat{x}+b_K$ be its orientation-preserving affine map and define ${\mathrm{\Pi }}_{3,K}^{grad}v=({\widehat{\mathrm{\Pi }}}_3^{grad}(v\circ F_K))\circ F_K^{-1}$. Affine change of variables preserves the cell mean and, since each edge Jacobian is constant, the edge $P_1$ moments. The bounded inverse reference moment matrix, [F5], mean-zero Poincare–Wirtinger [F7], and shape scaling [F6] yield $\mathrm{\parallel }{\mathrm{\Pi }}_{3,K}^{grad}v{\mathrm{\parallel }}_{H^1(K)}\le C\mathrm{\parallel }v{\mathrm{\parallel }}_{H^1(K)}$ and $\mathrm{\parallel }v-{\mathrm{\Pi }}_{3,K}^{grad}v{\mathrm{\parallel }}_2\le C{h}_K\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_2$.

Let $F_K(\widehat{x})=A_K\widehat{x}+b_K$, $J_K=\det A_K>0$, and $\widehat{\tau }=J_K{A}_K^{-1}(\tau \circ F_K)$. Applying [F8] to the affine inverse shows $\widehat{\tau }\in H(\mathrm{div};\widehat{K})$ with the transformed divergence. Define the physical projection by ${\mathrm{\Pi }}_{3,K}^{div}\tau :={\mathcal{P}}_K({\widehat{\mathrm{\Pi }}}_3^{div}\widehat{\tau })$. For every $w\in H^1(K)$, $w\circ F_K\in H^1(\widehat{K})$ by [F8], and the weak Green functional, affine weak-derivative/divergence transformations, and change of variables give $⟨{\gamma }_{n,K}\tau ,[w]⟩={\int }_{\widehat{K}}(\widehat{\tau }\cdot \widehat{\mathrm{\nabla }}(w\circ F_K)+(\stackrel{}{\mathrm{div}}\widehat{\tau })(w\circ F_K))=⟨{\gamma }_{\widehat{n},\widehat{K}}\widehat{\tau },[w\circ F_K]⟩\mathrm{.}$ The volume moments also transport: for $p\in P_1(K{)}^2$, affine change of variables gives $({\mathrm{\Pi }}_3^{div}\tau ,p{)}_K=({\widehat{\mathrm{\Pi }}}_3^{div}\widehat{\tau },A_K^{\mathsf{T}}(p\circ F_K){)}_{\widehat{K}},A_K^{\mathsf{T}}(p\circ F_K)\in P_1(\widehat{K}{)}^2,$ and the same identity with ${\widehat{\mathrm{\Pi }}}_3^{div}\widehat{\tau }$ replaced by $\widehat{\tau }$; hence every physical $P_1(K{)}^2$ volume moment is matched. This derives the weak normal-trace transport for arbitrary $H(\mathrm{div})$ inputs; it does not infer it from the smooth-field flux identity. Each continuous boundary $P_2$ function on a triangle is the trace of a polynomial in $P_2(K)$ (the six vertex and edge-midpoint values determine that polynomial), so the identity transports exactly the six boundary moments used in step 1.2. By [F8], the Piola-conjugated reference projection lands in ${\mathrm{RT}}_2(K)\subset P_3(K{)}^2$. Its reference moment vector is bounded by $C\mathrm{\parallel }\widehat{\tau }{\mathrm{\parallel }}_{H(\mathrm{div};\widehat{K})}$: volume moments are controlled by $L^2$, and boundary moments by [F9] and the fixed polynomial trace norms. The unisolvent reference moment matrix therefore gives $\mathrm{\parallel }{\widehat{\mathrm{\Pi }}}_3^{div}\widehat{\tau }{\mathrm{\parallel }}_{L^2(\widehat{K})}\le C\mathrm{\parallel }\widehat{\tau }{\mathrm{\parallel }}_{H(\mathrm{div};\widehat{K})}$. In dimension two, [F6] implies $J_K^{1\mathrm{/}2}\mathrm{\parallel }{A}_K^{-1}\mathrm{\parallel }\le C$ and $\mathrm{\parallel }{A}_K\mathrm{\parallel }{J}_K^{-1\mathrm{/}2}\le C$, while $\mathrm{\parallel }\widehat{\tau }{\mathrm{\parallel }}_{L^2}\le C\mathrm{\parallel }\tau {\mathrm{\parallel }}_{L^2(K)}$ and $\mathrm{\parallel }\stackrel{}{\mathrm{div}}\widehat{\tau }{\mathrm{\parallel }}_{L^2}\le C{h}_K\mathrm{\parallel }\mathrm{div}\tau {\mathrm{\parallel }}_{L^2(K)}$. Since $h_K\le \mathrm{diam}\mathrm{\Omega }$ on the fixed bounded domain, Piola scaling gives $\mathrm{\parallel }{\mathrm{\Pi }}_3^{div}\tau {\mathrm{\parallel }}_{L^2(K)}\le C_{\mathrm{\Omega }}\mathrm{\parallel }\tau {\mathrm{\parallel }}_{H(\mathrm{div};K)}$. Finally, for every $q\in P_2(K)$, Green’s formula, the matched volume moments against $\mathrm{\nabla }q\in P_1(K{)}^2$, and the matched continuous boundary moments imply $(\mathrm{div}{\mathrm{\Pi }}_3^{div}\tau ,q{)}_K=(\mathrm{div}\tau ,q{)}_K$. Thus $\mathrm{div}{\mathrm{\Pi }}_3^{div}\tau =P_2(\mathrm{div}\tau )$ and its $L^2$ norm is at most $\mathrm{\parallel }\mathrm{div}\tau {\mathrm{\parallel }}_{L^2(K)}$. This proves the uniform broken H(div) graph bound.

Vector moments match ${\sigma }_h\in P_1^2$, and the divergence projection from step 2.2 matches $u_h\in P_1$. The continuous boundary $P_2$ moments match ${\widehat{u}}_h$, while the scalar edge $P_1$ moments match ${\widehat{\sigma }}_{n,h}$. For the remaining volume term, put $q=v-{\mathrm{\Pi }}_3^{grad}v$ and integrate $-({\sigma }_h,\mathrm{\nabla }q{)}_K$ by parts: $\mathrm{div}{\sigma }_h\in P_0(K)$ pairs to zero with $q$ by cell-mean preservation, and ${\sigma }_h\cdot n\in P_1(F)$ pairs to zero with $q$ on each edge by the scalar edge moments. Hence every term of [F1] vanishes locally, and summing gives $b(w_h,y-\mathrm{\Pi }y)=0$. The cell bounds add in broken graph norms.