Crouzeix–Raviart finite elements

First let $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ and extend it by zero to ${\mathbb{R}}^d$. Write $x=x_{\perp }+te$ with $x_{\perp }\perp e$. For fixed $x_{\perp }$ and $a<t<b$, the fundamental theorem [F1] gives $\phi (x_{\perp }+te)={\int }_a^t{\mathrm{\partial }}_e\phi (x_{\perp }+se)ds$, since the zero extension vanishes at $t=a$. Holder's inequality [F4], applied on $(a,t)$ and then enlarged to $(a,b)$, yields $\mathrm{\mid }\phi (x_{\perp }+te){\mathrm{\mid }}^2\le (b-a){\int }_a^b\mathrm{\mid }{\mathrm{\partial }}_e\phi (x_{\perp }+se){\mathrm{\mid }}^{2}ds$.

Integrate step 1.1 over $t\in (a,b)$ and $x_{\perp }\in e^{\perp }$. Choose an orthonormal basis of $e^{\perp }$ and let $Q$ have those basis vectors followed by $e$ as its columns. Then $T(y)=Qy$ sends $(y_{\perp },t)$ to $x_{\perp }+te$; Tonelli [F2] and $Q^{T}Q=I$, $\mathrm{\mid }\det Q\mathrm{\mid }=1$ in [F3] give the coordinate integral over the slab. The zero extension of $\phi$ and its derivative vanish off $\mathrm{\Omega }$, so those slab integrals equal their integrals over $\mathrm{\Omega }$. Thus $\mathrm{\parallel }\phi {\mathrm{\parallel }}_{L^2(\mathrm{\Omega })}^2\le (b-a{)}^2\mathrm{\parallel }{\mathrm{\partial }}_e\phi {\mathrm{\parallel }}_{L^2(\mathrm{\Omega })}^2\le (b-a{)}^2\mathrm{\parallel }\mathrm{\nabla }\phi {\mathrm{\parallel }}_{L^2(\mathrm{\Omega })}^2$.

Let $\delta >0$. By the closure definition of $H_0^1(\mathrm{\Omega })$, choose a single $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ with $\mathrm{\parallel }v-\phi {\mathrm{\parallel }}_{H^1}<\delta$. The $L^2$ triangle inequality follows by expanding $\mathrm{\parallel }f+g{\mathrm{\parallel }}_2^2$ and applying [F4] to the cross term; hence [F4] and step 2.1 give $\mathrm{\parallel }v{\mathrm{\parallel }}_2\le \mathrm{\parallel }\phi {\mathrm{\parallel }}_2+\delta \le (b-a)\mathrm{\parallel }\mathrm{\nabla }\phi {\mathrm{\parallel }}_2+\delta \le (b-a)\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_2+(b-a+1)\delta$. This holds for every $\delta >0$, so letting $\delta \downarrow 0$ proves the statement without selecting a sequence of approximants.

Let $b_j$ be the barycenter of $F_j$. By [F2], the face average of an affine polynomial $p$ on $F_j$ is $p(b_j)$. At $b_j$, the barycentric coordinates of the $d$ vertices of $F_j$ are all $1\mathrm{/}d$, while the coordinate opposite $F_j$ is zero. Therefore ${\lambda }_i(b_j)=0$ if $i=j$ and ${\lambda }_i(b_j)=1\mathrm{/}d$ if $i\mathrm{\ne }j$. Hence $(1-d{\lambda }_i)(b_j)={\delta }_{ij}$, which proves the displayed duality identities.

The $d+1$ polynomials ${\varphi }_{K,F_i}$ are linearly independent: applying ${\sigma }_{K,F_j}$ to ${\sum }_i{c}_i{\varphi }_{K,F_i}=0$ gives $c_j=0$ by step 1.1. Since ${\mathbb{P}}_1(K)$ has dimension $d+1$, they form a basis. Applying the same functionals to $p-{\sum }_i{\sigma }_{K,F_i}(p){\varphi }_{K,F_i}$ gives zero for every face; expansion in this basis then gives zero coefficients. This proves existence and uniqueness of the stated expansion.

Fractional Poincare inequality. Put $m_K:=\mathrm{\mid }K{\mathrm{\mid }}^{-1}{\int }_{K}q$. Since $\mathrm{\mid }x-y\mathrm{\mid }\le h_K$ on $K\times K$ and $\mathrm{\mid }K\mathrm{\mid }\ge c{h}_K^d$ by [F4], $\mathrm{\parallel }q-m_K{\mathrm{\parallel }}_{L^2(K)}^2=\frac{1}{2\mathrm{\mid }K\mathrm{\mid }}{\int }_K{\int }_K\mathrm{\mid }q(x)-q(y){\mathrm{\mid }}^{2}dxdy\le C{h}_K^{2r}[q{]}_{H^r(K)}^2\mathrm{.}$ The equality follows by expanding the square and evaluating the two cross terms; the inequality follows by multiplying the Gagliardo integrand by $\mathrm{\mid }x-y{\mathrm{\mid }}^{d+2r}\le h_K^{d+2r}$.

Fractional extension from the reference simplex. Cover its closed simplex by finitely many affine coordinate neighborhoods: at a point on a face of codimension $m$, the $m$ vanishing barycentric coordinates are independent and the simplex is locally the orthant where those coordinates are positive. Compactness gives a finite cover with smaller compact subsets still covering the closed simplex. By [F5], choose smooth cutoffs supported in the corresponding neighborhoods and equal to one on those smaller sets; divide them by their positive sum near the closed simplex to obtain a finite smooth partition of unity. Multiplication by a smooth cutoff is bounded on $W^{r,2}$: split $\eta (x)q(x)-\eta (y)q(y)$ into $\eta (x)(q(x)-q(y))$ and $(\eta (x)-\eta (y))q(y)$, use $\mathrm{\mid }\eta (x)-\eta (y)\mathrm{\mid }\le \min (\mathrm{\parallel }D\eta {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }x-y\mathrm{\mid },2\mathrm{\parallel }\eta {\mathrm{\parallel }}_{\mathrm{\infty }})$, and integrate the latter radial kernel, finite exactly for $0<r<1$. Each localized piece is supported away from the artificial edge of its chart, so extending by zero there adds at most $C\mathrm{\parallel }q{\mathrm{\parallel }}_{L^2}^2$ to the seminorm: all cross-pairs have distance at least the support-to-edge distance. In orthant coordinates extend across each active coordinate hyperplane by even reflection. Same-side pairs reproduce the original integral; for opposite-side pairs $(x^{\mathrm{\prime }},t),(y^{\mathrm{\prime }},-s)$ with $t,s>0$, $\mathrm{\mid }(x^{\mathrm{\prime }}-y^{\mathrm{\prime }},t+s)\mathrm{\mid }\ge \mathrm{\mid }(x^{\mathrm{\prime }}-y^{\mathrm{\prime }},t-s)\mathrm{\mid }$, so their Gagliardo integrand is bounded by the corresponding original-side integrand. Iterating over the finitely many active coordinates, pulling back the nonsingular affine charts, and summing the finitely many extensions gives a bounded linear extension $E:W^{r,2}({\widehat{S}}_d)\to W^{r,2}({\mathbb{R}}^d)$. The same construction for $r=1$ uses the ordinary even-reflection weak derivative formula; the zero extension of each compactly supported $W^{1,2}$ localized piece is justified by [F16].

The $H^1$ trace bound. First take smooth $q$ on the closed reference simplex. Fix a facet $F$ and its opposite vertex $z_F$, and set $X(x)=x-z_F$. Then $X\cdot n_G=0$ on every other facet $G$, while $X\cdot n_F=H_F>0$ on $F$. Gauss–Green applied to $\mathrm{\mid }q{\mathrm{\mid }}^{2}X$ gives $H_F\mathrm{\parallel }q{\mathrm{\parallel }}_{L^2(F)}^2={\int }_K(d\mathrm{\mid }q{\mathrm{\mid }}^2+2qX\cdot \mathrm{\nabla }q)dx\mathrm{.}$ On the fixed reference simplex, $H_F>0$ and $\mathrm{\mid }X\mathrm{\mid }$ is bounded, so Holder [F15] gives $\mathrm{\parallel }q{\mathrm{\parallel }}_{L^2(F)}\le C\mathrm{\parallel }q{\mathrm{\parallel }}_{H^1(K)}$. By the $H^1$ density in [F2], this defines the trace for every $q\in H^1(K)$ and proves its boundedness. On a physical cell, apply the affine scaling in [F4] to obtain $\mathrm{\parallel }{\gamma }_{F}q{\mathrm{\parallel }}_{L^2(F)}\le C(h_K^{-1\mathrm{/}2}\mathrm{\parallel }q{\mathrm{\parallel }}_{L^2(K)}+h_K^{1\mathrm{/}2}\mathrm{\parallel }\mathrm{\nabla }q{\mathrm{\parallel }}_{L^2(K)})\mathrm{.}$ The constants are uniform over shape-regular cells.

Trace on a hyperplane for $0<r<1$. For a smooth $v$ on ${\mathbb{R}}^d$, let $a_k(x^{\mathrm{\prime }}):=2^k{\int }_0^{2^{-k}}v(x^{\mathrm{\prime }},t)dt$. For each $x^{\mathrm{\prime }}$, Cauchy–Schwarz and the identity ${\int }_I\mathrm{\mid }v-a_I{\mathrm{\mid }}^{2}dt=\frac{1}{2\mathrm{\mid }I\mathrm{\mid }}{\int }_I{\int }_I\mathrm{\mid }v(t)-v(s){\mathrm{\mid }}^{2}dtds$ give $\mathrm{\mid }{a}_{k+1}(x^{\mathrm{\prime }})-a_k(x^{\mathrm{\prime }})\mathrm{\mid }\le C{2}^{-k(r-1\mathrm{/}2)}[v(x^{\mathrm{\prime }},\cdot ){]}_{H^r(0,1)}$. The geometric series converges because $r>1\mathrm{/}2$, and $a_k(x^{\mathrm{\prime }})\to v(x^{\mathrm{\prime }},0)$. Integrating in $x^{\mathrm{\prime }}$ and using [F1] on the normal direction yields $\mathrm{\parallel }v(\cdot ,0){\mathrm{\parallel }}_{L^2({\mathbb{R}}^{d-1})}\le C(\mathrm{\parallel }v{\mathrm{\parallel }}_{L^2({\mathbb{R}}^d)}+[v{]}_{W^{r,2}({\mathbb{R}}^d)})\mathrm{.}$ By density [F2] this restriction extends uniquely to $W^{r,2}({\mathbb{R}}^d)$. Apply it to the extension in step 1.2, after an orthogonal change of coordinates, to define ${\gamma }_{F}q$ and obtain a bounded trace on each reference facet. If two extensions agree on the simplex, their difference vanishes on the one-sided normal slices; the same dyadic averages show its trace is zero, so the trace is independent of the extension.

Density in the graph norm. Let $q\in W^{r,2}(K;{\mathbb{R}}^d)$ with $\mathrm{div}q\in L^2(K)$, where $1\mathrm{/}2<r\le 1$, and write $g=\mathrm{div}q$. Contract the simplex toward its barycenter by $T_{ϵ}x=c+(1-ϵ)(x-c)$. For each fixed $ϵ>0$, $T_{ϵ}$ maps an open neighborhood $U_{ϵ}$ of $\bar{K}$ into $K$, so $q_{ϵ}=q\circ T_{ϵ}$ is defined there. Apply the affine pullback rule [F13] on $U_{ϵ}$ (with image $T_{ϵ}(U_{ϵ})\subset K$) to obtain $\mathrm{div}{q}_{ϵ}=(1-ϵ)g\circ T_{ϵ}$ distributionally. For $0<r<1$, pull $q$ back to the reference simplex by the affine reference map; change of variables in the double integral and the fixed positive affine Jacobian bounds give equivalent local $W^{r,2}$ norms. Transfer the finite-chart reflection extension of step 1.2 to $K$ by the same affine change of variables. Euclidean smooth density [F2] then supplies functions $q_j$ smooth near $\bar{K}$ with $q_j\to q$ in $W^{r,2}(K)$. Put $c=1-ϵ$. For $c\ge 1\mathrm{/}2$, changing variables $u=T_{ϵ}x$, $v=T_{ϵ}y$ gives $\mathrm{\parallel }{q}_j\circ T_{ϵ}{\mathrm{\parallel }}_{L^2(K)}\le c^{-d\mathrm{/}2}\mathrm{\parallel }{q}_j{\mathrm{\parallel }}_{L^2(K)}$ and $[q_j\circ T_{ϵ}{]}_{H^r(K)}\le c^{-d\mathrm{/}2+r}[q_j{]}_{H^r(K)}$, so these pullbacks are uniformly bounded on $W^{r,2}(K)$. The same change-of-variables estimate holds for every input in $W^{r,2}(K)$, so the pullback operator norms are uniformly bounded for $c\ge 1\mathrm{/}2$. For a fixed smooth $q_j$, $q_j\circ T_{ϵ}-q_j$ and its Lipschitz constant tend uniformly to zero; the estimate $[f{]}_{H^r(K)}^2\le \mathrm{Lip}(f{)}^2{\int }_{K\times K}\mathrm{\mid }x-y{\mathrm{\mid }}^{2-d-2r}dxdy$ and finiteness of that integral for $r<1$ give convergence in $W^{r,2}$. Approximation and the uniform pullback bound therefore give $q_{ϵ}\to q$. For $r=1$, transfer the reference-simplex smooth density [F2] to $K$ using [F13]; its affine derivative rule and change of variables give a uniform $W^{1,2}$ pullback bound when $c\ge 1\mathrm{/}2$, and smooth approximants converge under $T_{ϵ}$ in $W^{1,2}$, hence so does $q$. By [F12], $g\circ T_{ϵ}\to g$ in $L^2(K)$, so $\mathrm{div}{q}_{ϵ}\to g$ there as well. For each fixed $ϵ$, choose by [F5] a smooth cutoff $\eta \in C_c^{\mathrm{\infty }}(U_{ϵ})$ equal to one near $\bar{K}$. The weak product rule [F14] gives $\mathrm{div}(\eta q_{ϵ})=\eta \mathrm{div}{q}_{ϵ}+\mathrm{\nabla }\eta \cdot q_{ϵ}$. The zero extension $Q_{ϵ}$ of $\eta q_{ϵ}$ belongs to $W^{r,2}({\mathbb{R}}^d)$: the fractional multiplier and separated-boundary estimates are those already proved in step 1.2, and for $r=1$ the product rule and compact support give the same conclusion. Its distributional divergence is the zero extension of $\eta \mathrm{div}{q}_{ϵ}+\mathrm{\nabla }\eta \cdot q_{ϵ}$: for a test function on ${\mathbb{R}}^d$, insert a cutoff equal to one near the compact support of $\eta q_{ϵ}$ and use the weak divergence identity in $U_{ϵ}$. This divergence is in $L^2({\mathbb{R}}^d)$. Mollify $Q_{ϵ}$. The fractional density proof [F2] gives convergence in $W^{r,2}({\mathbb{R}}^d)$ when $r<1$; the $W^{1,2}$ convergence follows from [F11] when $r=1$. For a test function, the divergence of each mollification equals the mollification of $\mathrm{div}{Q}_{ϵ}$: expand the convolution integral and use the distributional divergence identity; all integrals are absolutely integrable by [F15] on the compact supports, so Tonelli applied to the absolute values justifies the interchange in [F6]. By [F11], these divergences converge in $L^2$. For small radii the cutoff is identically one throughout every convolution neighborhood of points of $K$, so the smooth fields restrict to approximations of $q_{ϵ}$ and $\mathrm{div}{q}_{ϵ}$ on $K$. Choose for each ${ϵ}_n\downarrow 0$ a radius small enough that both local errors are below $1\mathrm{/}n$. The resulting smooth $q_n$ converge to $q$ in $W^{r,2}(K)$ and their divergences converge to $g$ in $L^2(K)$. This is convergence in the graph norm of $H(\mathrm{div},K)\cap W^{r,2}(K)$.

Face-mean fluctuation. On the reference simplex, for $0<r<1$, apply step 2.1 to $q-m_K$ and use step 1.1; the trace seminorm is unchanged by subtracting a constant. Since the face average is the best $L^2(F)$ constant, $\mathrm{\parallel }{\gamma }_{F}q-{\mathrm{avg}}_F{\gamma }_{F}q{\mathrm{\parallel }}_{L^2(F)}\le \mathrm{\parallel }{\gamma }_F(q-m_K){\mathrm{\parallel }}_{L^2(F)}\le C[q{]}_{H^r(K)}\mathrm{.}$ For $r=1$, take smooth $q$ and apply [F3] to $\mathrm{\mid }q-m_K{\mathrm{\mid }}^{2}X$ with the vector field $X$ from step 1.3. Holder [F15] and [F8] give $\mathrm{\parallel }{\gamma }_F(q-m_K){\mathrm{\parallel }}_{L^2(F)}^2\le C\mathrm{\parallel }\mathrm{\nabla }q{\mathrm{\parallel }}_{L^2(K)}^2$ on the reference simplex; density and step 1.3 extend this to $H^1(K)$. The affine scaling factors in [F4] give the stated physical powers $h_K^{r-1\mathrm{/}2}$ and $h_K^{1\mathrm{/}2}$. For $0<r<1$, the full trace bound follows by applying step 2.1 to $q-m_K$ and adding the trace of the cell mean, $\mathrm{\mid }{\mathrm{avg}}_{K}q\mathrm{\mid }\mathrm{\mid }F{\mathrm{\mid }}^{1\mathrm{/}2}\le C{h}_K^{-1\mathrm{/}2}\mathrm{\parallel }q{\mathrm{\parallel }}_{L^2(K)}$. For $r=1$, the full bound is step 1.3.

Green identity. For each smooth field $q_n$ from step 2.2 and smooth $w$ on a neighborhood of $\bar{K}$, [F3] applied to $w{q}_n$ gives $-{\int }_K(\mathrm{div}{q}_n)w={\int }_K{q}_n\cdot \mathrm{\nabla }w-{\sum }_{F\subset \mathrm{\partial }K}{\int }_F(q_n\cdot n_{K,F})w\mathrm{.}$ For arbitrary $w\in H^1(K)$, transfer $w$ to the reference simplex by an affine map, using [F13] to get norm equivalence; approximate there by functions smooth near its closure using [F2], then compose with the inverse affine map. This gives smooth-near-$\bar{K}$ approximants converging to $w$ in $H^1(K)$. Holder's inequality [F15] makes the volume products integrable, and the bounded $H^1$ traces of step 1.3 and [F15] justify passage in both volume terms and the $L^2(F)$ trace pairings. Now pass $q_n$ to $q$ using step 2.2: the volume terms converge by $W^{r,2}\subset L^2$ and graph convergence, while the boundary terms converge by the $r>1\mathrm{/}2$ trace estimate of step 2.1 applied to $q_n-q$. Taking $q=\mathrm{\nabla }u$, $\mathrm{div}q=\mathrm{\Delta }u=-f$, proves the asserted identity. Interior trace compatibility for a global H^1 function follows as follows. Approximate it in H^1(Ω) by smooth functions on Ω using [F7]. On each compact subfacet G in the relative interior of a shared facet F, G lies inside Ω, so each approximant is a single smooth function near G and its two restrictions have the same trace there. The bounded cell trace map of step 1.3 makes both traces converge in L^2(G); hence the limits agree on G. Exhaust the relative interior of F by countably many compact subfacets; the remaining facet boundary is a finite union of lower-dimensional affine faces and has (d−1)-measure zero by a grid cover. Thus the traces agree almost everywhere on F. For a boundary facet, approximate an H^1_0(Ω) function by smooth compactly supported functions and use the same trace bound to see its trace is zero.

For $q\in H^r(\mathrm{\Omega })$ and a compact subfacet $G$ in the relative interior of an interior mesh facet, choose by [F9] a smooth cutoff $\eta$ supported in $\mathrm{\Omega }$ and equal to one near $G$. The product $\eta q$ extends by zero to $W^{r,2}({\mathbb{R}}^d)$: within $\mathrm{\Omega }$ the smooth multiplier estimate follows by splitting the difference as in step 1.2, and pairs across $\mathrm{\partial }\mathrm{\Omega }$ have one point a positive distance from the compact support. Approximate this extension in $W^{r,2}({\mathbb{R}}^d)$ by smooth functions using [F9]. The trace estimate of step 2.1 on each incident simplex shows that both restrictions converge in $L^2(G)$ to the same limit, because each approximant is one smooth function across $G$. Exhaust the relative interior of the facet by a countable increasing sequence of compact subfacets. Its boundary is a finite union of lower-dimensional affine faces and has $(d-1)$-dimensional measure zero by covering those faces with arbitrarily thin grids. The exceptional sets on the compact subfacets are null, so [F10] gives equality almost everywhere on the full facet.

The two traces of $v\in V_h^{CR,0}$ on an interior facet have equal integrals by definition, so ${\bar{v}}_F$ is single-valued. For an interior facet $F$, the function ${\varphi }_F$ has mean one on $F$, mean zero on every other facet of each incident cell by [F1], and is zero on cells not incident to $F$. Thus ${\varphi }_F\in V_h^{CR,0}$ and ${\bar{\varphi }}_{F,G}={\delta }_{FG}$ for all interior facets $G$.

Let $v\in V_h^{CR,0}$ and set $w:={\sum }_{F\in {\mathcal{F}}^{\circ }}{\bar{v}}_F{\varphi }_F$. On each cell, $v_K$ and $w_K$ have the same mean on every interior facet and both have mean zero on every boundary facet. Local unisolvence [F1] therefore gives $v_K=w_K$ for every cell. Hence the global functions span. Their linear independence follows by applying each single-valued face mean to a linear combination and using step 1.1. The number of basis functions is the number of interior facets, which proves the dimension formula.

Let $F$ be a facet of $K$ and let $b_F$ be its barycenter. For any affine $j_F$ on $F$, $j_F(x)=j_F(b_F)+{\mathrm{\nabla }}_F{j}_F\cdot (x-b_F)$ and ${\int }_F(x-b_F)dS=0$ by the definition $b_F=\mathrm{\mid }F{\mathrm{\mid }}^{-1}{\int }_{F}xdS$; hence its face average is exactly $j_F(b_F)$. On an interior facet, the jump $j_F=[v_h{]}_F$ is affine with zero mean by [F1], hence $j_F(b_F)=0$. On a boundary facet, $j_F=v_K$ has zero mean and likewise $j_F(b_F)=0$. Therefore for every $x\in F$, $\mathrm{\mid }{j}_F(x)\mathrm{\mid }=\mathrm{\mid }{\mathrm{\nabla }}_F{j}_F\cdot (x-b_F)\mathrm{\mid }\le h_F(\mathrm{\mid }\mathrm{\nabla }{v}_{K^{+}}\mathrm{\mid }+\mathrm{\mid }\mathrm{\nabla }{v}_{K^{-}}\mathrm{\mid }),$ with the absent cell term omitted on the boundary.

Squaring step 1.1 and integrating on $F$ gives $h_F^{-1}\mathrm{\parallel }{j}_F{\mathrm{\parallel }}_{L^2(F)}^2\le C{h}_F^d(\mathrm{\mid }\mathrm{\nabla }{v}_{K^{+}}{\mathrm{\mid }}^2+\mathrm{\mid }\mathrm{\nabla }{v}_{K^{-}}{\mathrm{\mid }}^2)$, where the factor $h_F^d$ is $h_F^{-1}\cdot h_F^2\cdot \mathrm{\mid }F\mathrm{\mid }$. By [F2], $h_F^d\mathrm{\mid }\mathrm{\nabla }{v}_K{\mathrm{\mid }}^2\le C\mathrm{\mid }K\mathrm{\mid }\mathrm{\mid }\mathrm{\nabla }{v}_K{\mathrm{\mid }}^2=C\mathrm{\parallel }\mathrm{\nabla }{v}_K{\mathrm{\parallel }}_{L^2(K)}^2$ for each incident cell. Summing over facets counts each cell at most $d+1$ times, proving the estimate.