The transport theorem for a $C^2$ family of diffeomorphisms: $\frac{d}{dt}{\int }_{{\psi }_t(B)}f={\int }_{{\psi }_t(B)}({\mathrm{\partial }}_{t}f+\mathrm{div}(fw))$

Fix $\tau \in J_0$. Choose a bounded open interval $J_1$ with $\tau \in J_1$ and $\overline{J_1}\subseteq J_0$. Its swept set ${\mathrm{\Sigma }}_1:=\{(t,x):t\in \overline{J_1},x\in {\psi }_t(\bar{B})\}$ is compact and lies in $U$, since it is contained in $\mathrm{\Sigma }$. The continuous map $(t,X)\mapsto (t,\psi (t,X))$ therefore maps $\overline{J_1}\times \bar{B}$ into $U$. Compactness gives an open neighbourhood $O_1$ of $\bar{B}$ in $O$ such that $(t,\psi (t,X))\in U$ for $t\in \overline{J_1}$, $X\in O_1$: finitely many product neighbourhoods inside the inverse image of $U$ cover the compact cylinder, or equivalently its positive distance from the closed complement supplies a uniform spatial margin. Define $g:J_1\times O_1⟶\mathbb{R},g(t,X):=f(t,\psi (t,X))J(t,X)\mathrm{.}$ By [F1] and [F5], $g$ is $C^1$, and ${\mathrm{\partial }}_{t}g$ is bounded on the compact cylinder $\overline{J_1}\times \bar{B}$ by [F6]. The argument below is applied on this local interval; compactness of the whole $\overline{J_0}$ is not required.

For $t\in J_1$, [F2] applied to the $C^1$ diffeomorphism ${\psi }_t:B\to {\psi }_t(B)$ and the integrable function $f(t,\cdot )$ restricted to ${\psi }_t(B)$ gives $F(t)={\int }_{{\psi }_t(B)}f(t,x)dx={\int }_{B}f(t,\psi (t,X))J(t,X)dX={\int }_{B}g(t,X)dX\mathrm{.}$ These integrals are finite because $B$ and ${\psi }_t(B)$ are bounded measurable sets of finite measure by [F8], while the integrands are continuous and bounded on their compact closures.

Differentiate $g$ in time at $(t,X)$. By [F5], ${\mathrm{\partial }}_t[f(t,\psi (t,X))]=({\mathrm{\partial }}_{t}f)(t,\psi (t,X))+D_{x}f(t,\psi (t,X))({\mathrm{\partial }}_t\psi (t,X))=({\mathrm{\partial }}_{t}f)(t,\psi (t,X))+⟨\mathrm{\nabla }f,w⟩(t,\psi (t,X))$, where the second form uses $w(t,\psi (t,X))={\mathrm{\partial }}_t\psi (t,X)$. For $A(t)=D_X{\psi }_t(X)$, [F7] gives $A^{\mathrm{\prime }}(t)=D_X{\mathrm{\partial }}_t\psi (t,X)$. Differentiating $w(t,\psi (t,X))={\mathrm{\partial }}_t\psi (t,X)$ with respect to $X$ and applying the chain rule yields $D_{x}w(t,\psi (t,X))A(t)=D_X{\mathrm{\partial }}_t\psi (t,X)=A^{\mathrm{\prime }}(t)$. By [F3], $({\mathrm{\partial }}_{t}J)(t,X)=J(t,X)\mathrm{tr}(D_{x}w(t,\psi (t,X)))=J(t,X)(\mathrm{div}w)(t,\psi (t,X))$. Hence $$\begin{array}{cccc}& {\mathrm{\partial }}_{t}g(t,X)=\left[\right({\mathrm{\partial }}_{t}f+⟨\mathrm{\nabla }f,w⟩+f\mathrm{div}w)\circ (t,\psi (t,X))]J(t,X)\mathrm{.}& & \text{(*)}\end{array}$$

The product rule for a scalar field $f$ and a vector field $w$, read off componentwise from the definitions of gradient and divergence, gives $\mathrm{div}(fw)={\sum }_i{\mathrm{\partial }}_i(f{w}_i)={\sum }_i({\mathrm{\partial }}_{i}f)w_i+f{\sum }_i{\mathrm{\partial }}_i{w}_i=⟨\mathrm{\nabla }f,w⟩+f\mathrm{div}w\mathrm{.}$ Therefore ($\ast$) reads ${\mathrm{\partial }}_{t}g(t,X)=({\mathrm{\partial }}_{t}f+\mathrm{div}(fw))(t,\psi (t,X))J(t,X)$.

Apply [F4] to $g$ on $X=B$ over the interval $J_1$ with the integrable dominating constant ${\sup }_{\overline{J_1}\times \bar{B}}\mathrm{\mid }{\mathrm{\partial }}_{t}g\mathrm{\mid }<+\mathrm{\infty }$ (legitimate because $B$ has finite Lebesgue measure and the supremum is finite by step 1.1 and [F6]): the function $t\mapsto {\int }_{B}g(t,X)dX=F(t)$ is differentiable on $J_1$ with $F^{\mathrm{\prime }}(t)={\int }_B{\mathrm{\partial }}_{t}g(t,X)dX={\int }_B({\mathrm{\partial }}_{t}f+\mathrm{div}(fw))(t,\psi (t,X))J(t,X)dX\mathrm{.}$

Change variables back with [F2], applied to $h:=({\mathrm{\partial }}_{t}f+\mathrm{div}(fw))(t,\cdot )$ restricted to ${\psi }_t(B)$, which is integrable by continuity on the compact closure and the diffeomorphism ${\psi }_t$: ${\int }_B({\mathrm{\partial }}_{t}f+\mathrm{div}(fw))(t,\psi (t,X))J(t,X)dX={\int }_{{\psi }_t(B)}({\mathrm{\partial }}_{t}f+\mathrm{div}(fw))(t,x)dx\mathrm{.}$ Combining with step 4.1 gives the stated formula at $\tau$; since $\tau \in J_0$ was arbitrary, it holds on all of $J_0$.