First-order least-squares FEM and practical DPG

Writing $R\sigma =(-{\sigma }_2,{\sigma }_1)$ gives $\mathrm{curl}(R\sigma )={\mathrm{\partial }}_1{\sigma }_1+{\mathrm{\partial }}_2{\sigma }_2=\mathrm{div}\sigma$ as distributions; [F1] places it in $L^2$.

Orthogonality gives $R\sigma \cdot R\mathrm{\nabla }\varphi =\sigma \cdot \mathrm{\nabla }\varphi$, so step 1.1 makes the trace integral exactly [F2]. Also $R\sigma \cdot t=R\sigma \cdot Rn=\sigma \cdot n$.

Approximate $\delta u$ by compactly supported smooth functions. The distributional divergence identity and $L^2$ continuity give $(\delta \sigma ,\mathrm{\nabla }\delta u)=-(div\delta \sigma ,\delta u)$. Substitute $\delta \sigma =R_{1}z-\mathrm{\nabla }\delta u$ to get $g^2=(R_{1}z,\mathrm{\nabla }\delta u)+(R_{2}z,\delta u)$.

Cauchy–Schwarz and [F3] imply $g^2\le ag+C_{P}bg$. If $g>0$, divide by $g$; if $g=0$, the bound is immediate. Poincare and $\delta \sigma =R_{1}z-\mathrm{\nabla }\delta u$ give the stated component estimates, controlling the product graph norm.

The first residual square is at most $2\mathrm{\parallel }\delta \sigma {\mathrm{\parallel }}_2^2+2\mathrm{\parallel }\mathrm{\nabla }\delta u{\mathrm{\parallel }}_2^2$ and the second is the divergence component. If $Rz=0$, step 2.1 and Poincare give $\delta u=0$, then $\delta \sigma =0$. This proves reverse boundedness and injectivity.

The space $H^1(D)$ is Hilbert for $(z,w{)}_{H^1}=(z,w{)}_{L^2}+(\mathrm{\nabla }z,\mathrm{\nabla }w{)}_{L^2}$. Indeed, a Cauchy sequence of functions and their weak derivatives converges componentwise in $L^2$ by [F2]; testing against $C_c^{\mathrm{\infty }}(D)$ and passing both integrals to the limits shows the derivative limits are the weak derivatives of the function limit. The quotient map $q:H^1(D)\to T_D$ has norm at most one, so $\mathrm{\Lambda }(w):=L(qw)$ is bounded on $H^1(D)$ with $\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }\le \mathrm{\parallel }L\mathrm{\parallel }$. By [F3] choose its Riesz representative $z\in H^1(D)$, so $\mathrm{\Lambda }(w)=(\mathrm{\nabla }z,\mathrm{\nabla }w)+(z,w)$.

For every $\varphi \in C_c^{\mathrm{\infty }}(D)$, $q\varphi =0$ because $\varphi \in H_0^1(D)$; hence $0=\mathrm{\Lambda }(\varphi )=(\mathrm{\nabla }z,\mathrm{\nabla }\varphi )+(z,\varphi )$. By the distributional definition of divergence, this means $\mathrm{div}(\mathrm{\nabla }z)=z$. Since $z\in L^2(D)$ and $\mathrm{\nabla }z\in L^2(D{)}^d$, [F5] gives $\tau :=\mathrm{\nabla }z\in H(\mathrm{div};D)$, with $\mathrm{\parallel }\tau {\mathrm{\parallel }}_{H(\mathrm{div})}^2=\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_2^2+\mathrm{\parallel }z{\mathrm{\parallel }}_2^2=\mathrm{\parallel }z{\mathrm{\parallel }}_{H^1}^2=\mathrm{\parallel }\mathrm{\Lambda }{\mathrm{\parallel }}^2\le \mathrm{\parallel }L{\mathrm{\parallel }}^2$.

For any $[w]\in T_D$, [F4] and step 2.1 give $⟨{\gamma }_n\tau ,[w]⟩=(\mathrm{\nabla }z,\mathrm{\nabla }w)+(z,w)=\mathrm{\Lambda }(w)=L([w])$. Thus ${\gamma }_n\tau =L$, and step 2.1 gives the claimed norm bound. Since $L$ was arbitrary, ${\gamma }_n$ is onto.

H1 is complete: completeness of real $L^2$ [F6] gives limits in $L^2$ of a Cauchy sequence and each weak derivative satisfy distributional derivative identities by testing against compactly supported smooth functions and passing both integrals to their limits. H1_0 is closed by its closure definition. The scalar trace-quotient map is continuous because $\mathrm{\parallel }[u{\mathrm{\mid }}_K]{\mathrm{\parallel }}_{H^1(K)\mathrm{/}{H}_0^1(K)}\le \mathrm{\parallel }u{\mathrm{\mid }}_K{\mathrm{\parallel }}_{H^1(K)}$ on each cell; the finite product preserves continuity. The normal map is continuous by [F2], again with a finite product. Thus both kernels are closed.

Let $\tau \in H(\mathrm{div};\mathrm{\Omega })$ and let an interior facet $F=K_{+}\cap K_{-}$ be fixed. For $\psi \in C_c^{\mathrm{\infty }}(\mathrm{relint}F)$, choose a smooth function $\varphi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ supported in a small tubular neighborhood of $\mathrm{supp}\psi$ meeting only $K_{+}$ and $K_{-}$ and having trace $\psi$ on $F$. The global distributional-divergence identity [F8] gives ${\int }_{\mathrm{\Omega }}(\tau \cdot \mathrm{\nabla }\varphi +(\mathrm{div}\tau )\varphi )=0$. Applying the cell Green identity [F2] to the two restrictions, with all other cell traces zero, yields $⟨{\gamma }_{n_{+},K_{+}}(\tau {\mathrm{\mid }}_{K_{+}}),[\varphi {\mathrm{\mid }}_{K_{+}}]⟩+⟨{\gamma }_{n_{-},K_{-}}(\tau {\mathrm{\mid }}_{K_{-}}),[\varphi {\mathrm{\mid }}_{K_{-}}]⟩=0.$ The two cell trace classes in this equation restrict to the same $\psi$ on $F$ and vanish on the other facets. Since this holds for every such $\psi$, the two outward weak normal traces cancel as distributions on the relative interior of each interior facet. This is the conservative facet relation for a global H(div) lift; it uses the distributional divergence identity and weak Green formula, not pointwise normal values.

For a closed subspace $M$ of a Hilbert space, a distance-minimizing sequence is Cauchy by the parallelogram identity; its limit is nearest, and variation along $M$ makes the residual orthogonal. Thus $H=M\oplus M^{\perp }$ and $H\mathrm{/}M$ is isometric to $M^{\perp }$. Applying this to both kernels in step 1.1 proves the quotient ranges are Hilbert with their infimum norms.

Global lifts restrict to broken lifts; conversely the admissible broken extensions in the Statement are exactly those gluing to a global lift, so the two infima have the same set. Cell scalar traces are bounded by [F3] and those of a global H1 lift match across interior facets by [F4]; normal traces of a global H(div) lift satisfy the cancellation proved in step 1.2. Thus range membership is precisely the stated compatibility, with the global lift infimum norms.

Perturbations and rounded arithmetic are absent from [F1], so it supplies no error estimate for them.

Under the stated size and rank hypotheses, [F2] gives the displayed identity; this is a conditioning statement, not a solver-specific error bound.

Cauchy–Schwarz bounds the three volume pairings in [F1] by the trial/test product norms.

Lift $\widehat{u}$ to global H1_0 and use [F3] on $\tau \cdot n$. Lift ${\widehat{\sigma }}_n$ to global H(div) and use cell Green identities on broken $v_K$. Cauchy–Schwarz across cells and infimizing by [F2] bound each skeleton term by the same product.

Each of five terms is bounded by the product norm by steps 1.1 and 2.1, so the triangle inequality gives constant $5$, independent of the mesh.

Compactly supported vector tests on each cell give $\sigma +\mathrm{\nabla }u=0$ distributionally; compactly supported scalar tests give $div\sigma =0$. Thus $u{\mathrm{\mid }}_K\in H^1(K)$, $\sigma {\mathrm{\mid }}_K\in H(div;K)$, and both identities hold cellwise.

With $\sigma =-\mathrm{\nabla }u$, the first local equation and [F3] give $⟨\widehat{u}-u{\mathrm{\mid }}_{\mathrm{\partial }K},{\gamma }_n\tau ⟩=0$ for every $\tau \in H(div;K)$. Surjectivity [F4] makes the scalar trace classes equal. The second local equation gives $⟨{\widehat{\sigma }}_n-{\gamma }_n\sigma ,v⟩=0$ for every $v\in H^1(K)$, so those normal-trace functionals are equal as well.

Take a global H1_0 lift $u_0$ of $\widehat{u}$. On each cell, $u-u_0$ has zero scalar trace, hence belongs to $H_0^1(K)$; zero-extending compactly supported approximants and summing over the finite mesh shows $u-u_0\in H_0^1(\mathrm{\Omega })$, so $u\in H_0^1(\mathrm{\Omega })$. Take a global H(div) lift ${\sigma }_0$ of ${\widehat{\sigma }}_n$. The cell field $\sigma -{\sigma }_0$ has zero weak normal trace on every cell. Summing its local Green identities against restrictions of $C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ shows its piecewise $L^2$ divergence is its global distributional divergence; thus it glues to H(div), and $\sigma \in H(div;\mathrm{\Omega })$. Now $\sigma =-\mathrm{\nabla }u$ and $div\sigma =0$; the weak Green identity tested with $u\in H_0^1$ gives $\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2^2=0$. By [F6], $u=0$, then $\sigma$ and both trace components vanish.

Local RT0 fields and their divergences are polynomials; shared normal flux degrees of freedom agree by [F2]. Their global distributional divergence is therefore the piecewise $L^2$ divergence, so the flux factor lies in H(div).

The scalar factor is globally continuous piecewise affine and has H1_0 boundary condition; [F3] gives the interior trace convention. The finite mesh has finitely many polynomial coefficients, so the product space is finite dimensional.

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.

The weak solution supplier in [A1] applies because a bounded polygon is open and bounded in one direction, and $v\mapsto (f,v)$ is bounded on $H_0^1$ by Cauchy–Schwarz and Poincare. It gives $u\in H_0^1$; the distributional equation gives $\sigma =-\mathrm{\nabla }u\in H(div)$ and $div\sigma =f$.

For any $({\sigma }_h,u_h)$ in the trial family, the residual of its difference from $(\sigma ,u)$ equals $({\sigma }_h+\mathrm{\nabla }{u}_h,div{\sigma }_h-f)$ by [F3]. Thus $J_h=\mathrm{\parallel }R(({\sigma }_h,u_h)-(\sigma ,u)){\mathrm{\parallel }}^2$. By [F2], $(z,w{)}_R=(Rz,Rw{)}_{L^2\times L^2}$ is an inner product on the ambient error space.

By Gram–Schmidt choose an orthonormal basis $e_1,\dots ,e_N$ of $R(X_h)$, and define $Pq={\sum }_{j=1}^N(q,e_j)e_j$. Then $q-Pq$ is orthogonal to every $e_j$, so $Pq$ uniquely minimizes distance to $R(X_h)$ by Pythagoras. The finite-dimensional image is closed; injectivity in [F2] pulls $P(R(\sigma ,u))$ back to a unique minimizer and the same identity gives best approximation. The scalar component stays in the stated H1_0 trial factor.

By [F7], $B:U\to V^{\mathrm{\prime }}$ given by $(Bx)(y)=b(x,y)$ is well defined and bounded. By [F2] and [F3], there is a unique $(\rho ,z)\in V$ such that $F(\tau ,v)=(\rho ,\tau )+({\mathrm{div}}_h\rho ,{\mathrm{div}}_h\tau )+(z,v)+({\mathrm{\nabla }}_{h}z,{\mathrm{\nabla }}_{h}v)\mathrm{.}$ Set $g=\rho$, $r={\mathrm{div}}_h\rho$, $f=z$, and $s={\mathrm{\nabla }}_{h}z$, viewed as global $L^2$ classes by the finite mesh partition. If $D=\mathrm{\parallel }F{\mathrm{\parallel }}_{V^{\mathrm{\prime }}}$, then $\mathrm{\parallel }g{\mathrm{\parallel }}_2^2+\mathrm{\parallel }r{\mathrm{\parallel }}_2^2+\mathrm{\parallel }f{\mathrm{\parallel }}_2^2+\mathrm{\parallel }s{\mathrm{\parallel }}_2^2=D^2$.