Smooth transport, Jacobian and advected reference density

Set $H(t,a)=(t,\chi (t,a))$. It is $C^2$ and injective: equality of two images first gives equal times and then equal labels by injectivity of that time slice. Its derivative is block triangular with determinant $J(t,a)>0$, so [F2] gives local $C^1$ inverses; injectivity makes these local inverses agree on overlaps and yields the joint $C^1$ inverse on $\mathrm{\Omega }$.

For fixed $t$, the image ${\chi }_t(K)$ is compact Jordan by [F11]. The continuous function $q(t,\cdot )$ is Riemann integrable there by [F6]. The map ${\chi }_t$ is injective and $C^1$ with invertible derivative, so applying [F5] and using $J(t,a)>0$ gives ${\int }_{{\chi }_t(K)}q(t,x)dx={\int }_K{G}_1(t,a)da$, where $G_1(t,a):=q(t,\chi (t,a))J(t,a)$.

Differentiating $H\circ H^{-1}=\mathrm{i}\mathrm{d}$ and using [F3] gives $D(H^{-1})=(DH\circ H^{-1}{)}^{-1}$. Since $DH$ and $H^{-1}$ are $C^1$, their composition is $C^1$. Each entry of its inverse is a cofactor divided by its nonzero determinant; cofactors are determinants of minors and are polynomials by [F4]. Thus $D(H^{-1})$ is $C^1$, so $H^{-1}$ is $C^2$ and in particular $A$ is jointly $C^2$.

Differentiate $A(t,\chi (t,a))=a$ in $t$ and in $a$. The chain rule gives ${\mathrm{\partial }}_{t}A+(D_{x}A)u=0$ and $(D_{x}A)F=I_3$, hence $D_{x}A=F^{-1}$. For every $C^1$ scalar field $q$, it also gives ${\mathrm{\partial }}_t(q(t,\chi (t,a)))=(D_{t}q)(t,\chi (t,a))$.

To justify commuting the mixed partials, fix a component $g$ of $\chi$ and a reference coordinate $a_i$. For sufficiently small $h,k$, the rectangular increment $g(t+h,a+k{e}_i)-g(t,a+k{e}_i)-g(t+h,a)+g(t,a)$ equals both ${\int }_t^{t+h}\text{\hspace{0.17em}⁣}{\int }_0^k{\mathrm{\partial }}_{a_i}{\mathrm{\partial }}_{s}g(s,a+r{e}_i)drds$ and ${\int }_0^k\text{\hspace{0.17em}⁣}{\int }_t^{t+h}{\mathrm{\partial }}_s{\mathrm{\partial }}_{a_i}g(s,a+r{e}_i)dsdr$, by the one-variable fundamental theorem of calculus applied in the two orders. Divide by $hk$ and let $h,k\to 0$; continuity of both second partials gives ${\mathrm{\partial }}_t{\mathrm{\partial }}_{a_i}g={\mathrm{\partial }}_{a_i}{\mathrm{\partial }}_{t}g$. Hence ${\mathrm{\partial }}_{t}F=D_a{\mathrm{\partial }}_t\chi$. Now use ${\mathrm{\partial }}_t\chi =u(t,\chi )$ and [F3] to obtain ${\mathrm{\partial }}_{t}F=(D_{x}u)(t,\chi )F=L(t,\chi )F$.

Put $C=F^{-1}{\mathrm{\partial }}_{t}F$. The permutation formula in [F4] is a finite sum with one entry from each column; the product rule therefore differentiates $J$ by replacing one column at a time: ${\mathrm{\partial }}_{t}J={\sum }_{j=1}^3\det (F_1,\dots ,{\mathrm{\partial }}_t{F}_j,\dots ,F_3)$. Since ${\mathrm{\partial }}_{t}F=FC$, multilinearity expands ${\mathrm{\partial }}_t{F}_j={\sum }_i{F}_i{C}_{ij}$ in the $j$th term. For $i\mathrm{\ne }j$, the resulting determinant has two equal columns and its permutation terms cancel in pairs under the transposition of those columns; only $i=j$ remains. Thus ${\mathrm{\partial }}_{t}J=J{\sum }_j{C}_{jj}=J\mathrm{tr}C$. By step 4.1, $C=F^{-1}L(t,\chi )F$, and the finite-sum calculation ${\sum }_i(F^{-1}LF{)}_{ii}={\sum }_{j,k}{L}_{jk}{\sum }_i{F}_{ki}(F^{-1}{)}_{ij}={\sum }_j{L}_{jj}$ gives $\mathrm{tr}(F^{-1}LF)=\mathrm{tr}L$. Hence ${\mathrm{\partial }}_{t}J=J\theta (t,\chi )$.

For fixed $a$, set $h(s)=\theta (s,\chi (s,a))$ and $G(t)={\int }_0^{t}h(s)ds$. Continuity gives Riemann integrability on each compact subinterval by [F6]. For small nonzero $k$, the absolute integral estimate [F8] yields $\mid (G(t+k)-G(t))\mathrm{/}k-h(t)\mid \le {\sup }_{s\text{ between }t\text{ and }t+k}\mathrm{\mid }h(s)-h(t)\mathrm{\mid }$, which tends to zero. Thus $G^{\mathrm{\prime }}=h$. By step 5.1, $(J{e}^{-G}{)}^{\mathrm{\prime }}=0$; [F10] makes this product constant on the interval $I$, and evaluating at $t=0$ gives the stated exponential formula.

The function $G_1$ from step 1.2 is $C^1$. By the chain and product rules and steps 3.1 and 5.1, ${\mathrm{\partial }}_t{G}_1(t,a)=(D_{t}q+q\theta )(t,\chi (t,a))J(t,a)$.

If $K=\mathrm{\varnothing }$, the fixed-domain integral in step 1.2 and its derivative are zero, and step 1.2 identifies the moving-domain integral with it; thus the differentiation assertion holds. Assume $K\mathrm{\ne }\mathrm{\varnothing }$. Fix $t$ and choose a compact interval $T\subset I$ around it with $T\times K\subset I\times U$. For any $\epsilon >0$ and each $a_0\in K$, continuity of ${\mathrm{\partial }}_t{G}_1$ at $(t,a_0)$ supplies a product neighborhood $T_{a_0}\times V_{a_0}$ on which ${\mathrm{\partial }}_t{G}_1$ differs from ${\mathrm{\partial }}_t{G}_1(t,a_0)$ by less than $\epsilon \mathrm{/}3$. A finite subcover of $K$ and the minimum of its finitely many time widths show that $\mathrm{\mid }{\mathrm{\partial }}_t{G}_1(s,a)-{\mathrm{\partial }}_t{G}_1(t,a)\mathrm{\mid }<2\epsilon \mathrm{/}3$ uniformly for $a\in K$ when $s$ is sufficiently close to $t$. Choose an open interval $V$ containing $t$, contained in $\mathrm{int}T$ and in this time neighborhood, and take $k\mathrm{\ne }0$ small enough that the segment from $t$ to $t+k$ lies in $V$. For each fixed $a\in K$, the $C^1$ function ${\varphi }_a(s):=G_1(s,a)-G_1(t,a)-(s-t){\mathrm{\partial }}_t{G}_1(t,a)$ has derivative ${\varphi }_a^{\mathrm{\prime }}(s)={\mathrm{\partial }}_t{G}_1(s,a)-{\mathrm{\partial }}_t{G}_1(t,a)$, whose absolute value is less than $2\epsilon \mathrm{/}3$ throughout $V$. Applying [F9] to ${\varphi }_a{\mathrm{\mid }}_V$ gives the same uniform bound for the difference-quotient error. Thus the difference quotients converge uniformly on $K$ to ${\mathrm{\partial }}_t{G}_1(t,\cdot )$.

The integrands $G_1(t,\cdot )$ and ${\mathrm{\partial }}_t{G}_1(t,\cdot )$ are continuous on $K$ and hence integrable by [F6]. Use [F7] to represent the integrals as integrals of zero extensions on one fixed nondegenerate containing rectangle. Outside $K$ the extended difference quotients are zero, so [F8]'s absolute-value bound moves the difference quotient through the rectangle integral and bounds its error by the rectangle's volume times the uniform error. The uniform convergence from step 7.1 proves $\frac{d}{dt}{\int }_K{G}_1(t,a)da={\int }_K{\mathrm{\partial }}_t{G}_1(t,a)da$.

Substitute step 6.2 into step 8.1 and use step 1.2 to return to current coordinates. The coordinate product rule gives ${\mathrm{\nabla }}_x\text{\hspace{0.17em}⁣}\cdot (qu)=u\cdot {\mathrm{\nabla }}_{x}q+q\theta$, proving both transport identities.

Set $R:=D_t\rho +\rho \theta$. It is continuous on $\mathrm{\Omega }$ because $u,\rho \in C^1$ and $\theta$ is continuous. Step 9.1 gives $M_K^{\mathrm{\prime }}(t)={\int }_{{\chi }_t(K)}R(t,x)dx$. If $R=0$, each $M_K$ has zero derivative and is constant by [F10] on $I$.

Conversely, if every $M_K$ is constant, then its derivative is zero. For each fixed $t$, $R(t,\cdot )$ is continuous, and [F11] makes ${\chi }_t(K)$ compact Jordan; [F6] therefore makes $R(t,\cdot )$ Riemann integrable there, so [F5] and step 9.1 give ${\int }_{K}R(t,\chi (t,a))J(t,a)da=0$ for every compact Jordan $K\subset U$. If the continuous integrand were positive at some $a_0\in U$, continuity and $J>0$ would give a small closed cube $C\subset U$ on which it is at least half its positive value at $a_0$. Monotonicity in [F8] would then give a strictly positive integral over $C$, since $C$ has positive volume, contradicting the zero integral. If it were negative, apply the same argument to its negative and use linearity in [F8]. Thus $R(t,\chi (t,a))J(t,a)=0$ for every $a\in U$; since $J>0$ and ${\chi }_t$ maps $U$ onto ${\chi }_t(U)$, $R=0$ throughout the swept open set.

For fixed $a\in U$, product and chain rules give $\frac{d}{dt}[\rho (t,\chi (t,a))J(t,a)]=J(t,a)R(t,\chi (t,a))$. If $R=0$, [F10] makes this product constant on $I$, with value $\rho (0,\chi (0,a))J(0,a)$. Conversely, that invariance implies its derivative is zero, and $J>0$ gives $R=0$. Together with steps 10.1 and 10.2 this proves all equivalences.