Enthalpy vorticity and restricted material circulation

In each chart meeting ${\gamma }_0([0,1])$, the coordinate components of the smooth field $u$ are $C^2$. The local-flow suppliers [F5] therefore give a jointly $C^2$ flow on a neighborhood of each point of the curve. Compactness of the curve image gives a finite subcover; taking the minimum of the finitely many positive flow times gives a common interval about $\lambda =0$. Coordinate changes carry solutions to solutions, and local uniqueness makes the chartwise flows agree on overlaps. Thus ${\gamma }_{\lambda }$ is a $C^2$ closed-curve family on a common local interval.

By [F1] and the chain rule, $n,s,ϵ$ are $C^2$, while $p,T,H$ are $C^1$; hence the stress and $P$ are $C^1$ and $J$ is $C^2$. For these tensors use [F4]'s coordinate derivative formula: differentiate the coefficients and add one Christoffel term for each upper index and subtract one for each lower index. It defines an intrinsic derivative also for $C^1$ coefficients, since differentiating their tensor transformation law uses only the first-order chain rule and the same cancellation of derivatives of the smooth coordinate changes. The product and contraction rules follow from the ordinary product rule, and $\mathrm{\nabla }g=0$ remains valid. Thus all the following derivatives exist classically. From [F1], $de=-pdv+Td{\sigma }_p$, so $dH=Td{\sigma }_p+vdp$ and, using $dv=-n^{-2}dn$, $dϵ=Hdn+nTd{\sigma }_p$. Differentiating $g(u,u)=-1$ gives $u_b{a}^b=0$ and ${\mathrm{\nabla }}_a{\mathrm{\Delta }}^{ab}=\theta u^b+a^b$. The product rule now gives $0={\mathrm{\nabla }}_a{T}^{ab}=(Dϵ+h\theta )u^b+h{a}^b+{\mathrm{\Delta }}^{ab}{\mathrm{\nabla }}_{a}p,0={\mathrm{\nabla }}_a{J}^a=Dn+n\theta \mathrm{.}$ Contracting the stress equation with $u_b$ gives $Dϵ+h\theta =0$; applying ${\mathrm{\Delta }}^c{}_b={\delta }^c{}_b+u^c{u}_b$ gives $h{a}^c+{\mathrm{\Delta }}^{ca}{\mathrm{\nabla }}_{a}p=0$. Since $h=nH$, the differential identity for $ϵ$ gives $0=Dϵ+h\theta =H(Dn+n\theta )+nTD{\sigma }_p=nTD{\sigma }_p$. Positivity of $n,T$ yields $D{\sigma }_p=0$; lowering the projected stress equation and dividing by $n$ yields $H{a}^{\mathrm{♭}}+v\mathrm{\Delta }dp=0$. No division by $H$ is used.

Torsion freedom and the product rule yield, in any chart, $({\iota }_{u}d(H{u}^{\mathrm{♭}}){)}_b=u^a{\mathrm{\nabla }}_a(H{u}_b)-u^a{\mathrm{\nabla }}_b(H{u}_a)=(DH)u_b+H{a}_b+{\mathrm{\nabla }}_{b}H=H{a}_b+{\mathrm{\Delta }}_b{}^a{\mathrm{\nabla }}_{a}H\mathrm{.}$ Here $u^a{\mathrm{\nabla }}_b{u}_a=\frac{1}{2}{\mathrm{\nabla }}_b(u^a{u}_a)=0$, and the last equality is the definition of the spatial covector projection. Substituting the expression for $dH$ from step 1.2 and using $\mathrm{\Delta }d{\sigma }_p=d{\sigma }_p$ from $D{\sigma }_p=0$ gives ${\iota }_{u}d(H{u}^{\mathrm{♭}})=H{a}^{\mathrm{♭}}+Td{\sigma }_p+v\mathrm{\Delta }dp=Td{\sigma }_p$.

Subdivide $[0,1]$ into finitely many closed subintervals whose curve images lie in coordinate charts; compactness gives such a subdivision, and endpoint terms will telescope. On each chart subarc put $F(\lambda ,s)=P_a({\gamma }_{\lambda }(s)){\mathrm{\partial }}_s{\gamma }_{\lambda }^a(s)$. It is $C^1$ on each compact parameter rectangle. The one-variable fundamental theorem of calculus gives $[F(\lambda +h,s)-F(\lambda ,s)]\mathrm{/}h=h^{-1}{\int }_{\lambda }^{\lambda +h}{\mathrm{\partial }}_{r}F(r,s)dr$; uniform continuity of ${\mathrm{\partial }}_{\lambda }F$ on the compact rectangle makes this quotient converge uniformly in $s$ to ${\mathrm{\partial }}_{\lambda }F(\lambda ,s)$. Integrating that uniform limit proves differentiation under the line integral. Since ${\mathrm{\partial }}_{\lambda }\gamma =u$, ${\mathrm{\partial }}_{\lambda }F=(u^b{\mathrm{\partial }}_b{P}_a+P_b{\mathrm{\partial }}_a{u}^b){\mathrm{\partial }}_s{\gamma }_{\lambda }^a\mathrm{.}$ Thus ${\mathcal{C}}^{\mathrm{\prime }}={\int }_{{\gamma }_{\lambda }}{\mathcal{L}}_{u}P$, where in coordinates $({\mathcal{L}}_{u}P{)}_a=u^b{\mathrm{\partial }}_b{P}_a+P_b{\mathrm{\partial }}_a{u}^b$. Direct expansion gives $({\iota }_{u}dP+d(P(u)){)}_a=u^b({\mathrm{\partial }}_b{P}_a-{\mathrm{\partial }}_a{P}_b)+{\mathrm{\partial }}_a(u^b{P}_b)$, so direct expansion proves ${\mathcal{L}}_{u}P={\iota }_{u}dP+d(P(u))$. The exact one-form integrates on each subarc to its endpoint difference, and these differences telescope to zero because ${\gamma }_{\lambda }(0)={\gamma }_{\lambda }(1)$. Since $dP=(1\mathrm{/}c)d(H{u}^{\mathrm{♭}})$, step 2.1 yields ${\mathcal{C}}^{\mathrm{\prime }}=c^{-1}{\int }_{{\gamma }_{\lambda }}Td{\sigma }_p$. If ${\sigma }_p$ is constant throughout the swept region, then for each fixed $\lambda$ the composition $s\mapsto {\sigma }_p({\gamma }_{\lambda }(s))$ is constant. The chain rule therefore gives $d{\sigma }_p({\mathrm{\partial }}_s{\gamma }_{\lambda })={\mathrm{\partial }}_s({\sigma }_p\circ {\gamma }_{\lambda })=0$, so the displayed line integral and hence ${\mathcal{C}}^{\mathrm{\prime }}$ vanish. No vanishing of the ambient covector $d{\sigma }_p$ transverse to the swept image is needed.