Affine Piola maps commute with differential operators and preserve moments

For $\widehat{v}\in H(\mathrm{div},{\widehat{S}}_d)$ and $\varphi \in C_c^{\mathrm{\infty }}(K)$, put $\widehat{\varphi }=\varphi \circ F$. Then [F1] and the determinant substitution give ${\int }_K({\mathcal{P}}_F^{\mathrm{d}\mathrm{i}\mathrm{v}}\widehat{v})\cdot \mathrm{\nabla }\varphi dx={\int }_{{\widehat{S}}_d}\widehat{v}\cdot \mathrm{\nabla }\widehat{\varphi }d\widehat{x}=-{\int }_{{\widehat{S}}_d}(\stackrel{}{\mathrm{div}}\widehat{v})\widehat{\varphi }d\widehat{x}\mathrm{.}$ The final expression equals $-{\int }_K[(\det A{)}^{-1}\stackrel{}{\mathrm{div}}\widehat{v}\circ F^{-1}]\varphi dx$. This is the distributional divergence identity.

For smooth $\widehat{v}$, differentiate $v=A^{-T}\widehat{v}\circ F^{-1}$ componentwise. In three dimensions, write $C=A^{-1}$. For any $z\in {\mathbb{R}}^3$ and indices $m,\mathrm{\ell }$, the scalar triple product gives $z\cdot ((C^T{e}_m)\times (C^T{e}_{\mathrm{\ell }}))=\det [z,C^T{e}_m,C^T{e}_{\mathrm{\ell }}]=\det (C)\det [C^{-T}z,e_m,e_{\mathrm{\ell }}]=\det (C)z\cdot (C^{-1}(e_m\times e_{\mathrm{\ell }}))\mathrm{.}$ Since this holds for every $z$, comparing components yields ${ϵ}_{ijk}{C}_{mj}{C}_{\mathrm{\ell }k}=(\det A{)}^{-1}{A}_{ir}{ϵ}_{rm\mathrm{\ell }}$. The chain rule then gives ${\mathrm{curl}}_{x}v=(\det A{)}^{-1}A({\mathrm{curl}}_{\widehat{x}}\widehat{v})\circ F^{-1}$. In two dimensions write $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ and $\mathrm{\Delta }=ad-bc>0$. Then $v_1=(d{\widehat{v}}_1-c{\widehat{v}}_2)\mathrm{/}\mathrm{\Delta }$, $v_2=(-b{\widehat{v}}_1+a{\widehat{v}}_2)\mathrm{/}\mathrm{\Delta }$, ${\mathrm{\partial }}_{x_1}=(d{\mathrm{\partial }}_{{\widehat{x}}_1}-c{\mathrm{\partial }}_{{\widehat{x}}_2})\mathrm{/}\mathrm{\Delta }$, and ${\mathrm{\partial }}_{x_2}=(-b{\mathrm{\partial }}_{{\widehat{x}}_1}+a{\mathrm{\partial }}_{{\widehat{x}}_2})\mathrm{/}\mathrm{\Delta }$. Substituting these four expressions and cancelling the two mixed terms gives ${\mathrm{\partial }}_{x_1}{v}_2-{\mathrm{\partial }}_{x_2}{v}_1=(\det A{)}^{-1}({\mathrm{\partial }}_{{\widehat{x}}_1}{\widehat{v}}_2-{\mathrm{\partial }}_{{\widehat{x}}_2}{\widehat{v}}_1)\circ F^{-1}$.

The physical field and candidate divergence in step 1.1 lie in $L^2(K)$: affine change of variables makes their squared norms finite constant multiples of the reference squared norms, and multiplication by the fixed matrix $A\mathrm{/}(\det A)$ preserves finiteness. The test identity in step 1.1 therefore gives membership in $H(\mathrm{div},K)$ and the asserted divergence formula.

For $d=3$, take $\psi \in C_c^{\mathrm{\infty }}(K;{\mathbb{R}}^3)$ and set $\widehat{\psi }=A^T(\psi \circ F)$. The component calculation in step 1.2 gives ${\mathrm{curl}}_{\widehat{x}}\widehat{\psi }=\det (A)A^{-1}({\mathrm{curl}}_x\psi )\circ F$. Change variables and use the weak curl identity on the reference cell: ${\int }_{K}v\cdot \mathrm{curl}\psi dx={\int }_{{\widehat{S}}_d}\widehat{v}\cdot \mathrm{curl}\widehat{\psi }d\widehat{x}={\int }_{{\widehat{S}}_d}(\stackrel{}{\mathrm{curl}}\widehat{v})\cdot \widehat{\psi }d\widehat{x}={\int }_K[(\det A{)}^{-1}A(\stackrel{}{\mathrm{curl}}\widehat{v})\circ F^{-1}]\cdot \psi dx\mathrm{.}$ For $d=2$, take $\varphi \in C_c^{\mathrm{\infty }}(K)$ and set $\widehat{\varphi }=\varphi \circ F$. The determinant identity $\det (A)\det (A^{-T}\widehat{v},A^{-T}\mathrm{\nabla }\widehat{\varphi })=\det (\widehat{v},\mathrm{\nabla }\widehat{\varphi })$ transforms the weak scalar-curl pairing into ${\int }_{{\widehat{S}}_d}({\widehat{v}}_1{\mathrm{\partial }}_2\widehat{\varphi }-{\widehat{v}}_2{\mathrm{\partial }}_1\widehat{\varphi })$, which by the weak reference curl identity equals ${\int }_{{\widehat{S}}_d}(\stackrel{}{\mathrm{curl}}\widehat{v})\widehat{\varphi }$ and hence the pairing with the asserted physical scalar curl. In both dimensions the mapped field and candidate curl lie in $L^2$ by the fixed-matrix and determinant bounds. Thus the weak definition gives the claimed $H(\mathrm{curl})$ mapping.

The normal flux identity follows by dotting $v=(\det A{)}^{-1}A\widehat{v}$ with the area vector [F2]: $v\cdot ndS=(\det A{)}^{-1}A\widehat{v}\cdot \det (A)A^{-T}\widehat{n}d\widehat{S}=\widehat{v}\cdot \widehat{n}d\widehat{S}\mathrm{.}$ For a reference tangent $\widehat{\tau }$, $v(F)\cdot A\widehat{\tau }=(A^{-T}\widehat{v})\cdot A\widehat{\tau }=\widehat{v}\cdot \widehat{\tau }$. Along an edge this equality is the line-form identity $v\cdot dx=\widehat{v}\cdot d\widehat{x}$. On a facet, integration against a pulled-back tangent vector and scalar weight gives ${\int }_{F}v\cdot (A\widehat{\tau })(\widehat{q}\circ F^{-1})dS={\int }_{\widehat{F}}(\widehat{v}\cdot \widehat{\tau })\widehat{q}{J}_{F}d\widehat{S},$ where $J_F$ is the affine facet surface Jacobian. The normal-flux moment with scalar weight $\widehat{q}$ equals ${\int }_{\widehat{F}}(\widehat{v}\cdot \widehat{n})\widehat{q}d\widehat{S}$. The signs are those of the selected outward facet orientation; reversing it reverses the signed normal moment.

For $\widehat{u}\in W^{1,2}({\widehat{S}}_d;\mathbb{R})$, apply [F5] to $F^{-1}$. For each $i$, $D_{x_i}(\widehat{u}\circ F^{-1})={\sum }_j(A^{-1}{)}_{ji}(D_{{\widehat{x}}_j}\widehat{u})\circ F^{-1}\mathrm{.}$ In vector notation this is ${\mathrm{\nabla }}_x(\widehat{u}\circ F^{-1})=A^{-T}(\widehat{\mathrm{\nabla }}\widehat{u})\circ F^{-1}={\mathcal{P}}_F^{\mathrm{c}\mathrm{u}\mathrm{r}\mathrm{l}}(\widehat{\mathrm{\nabla }}\widehat{u}),$ which proves the scalar-gradient identity.