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Let $\dot{x} = f(x)$ be a dynamical system on $\mathbb{R}^n$ with corresponding flow $\phi(t,x)$ and let $U \subseteq \mathbb{R}^n$ be Lebesgue measurable. Then $$\frac{d}{dt}\int_{\phi(t,U)}dx = \int_{\phi(t,U)}\text{div}(f(x))dx$$

Proof. By the change of variable formula for multiple integrals we have $$\int_{\phi(t,U)}dx = \int_{U}\det(D\phi_t(x))dx$$

This I do not really understand. I mean, we could define a mapping $$U \to \phi(t,x) \qquad x \mapsto\phi(t,x)$$ for which $U$ has to be open (we want a $C^1$-diffeomorphism). But wouldn't we get then $$\int_{\phi(t,U)}dx = \int_{U}|\det(D\phi_t(x))|dx$$ instead of the above? I do not quite see why the determinant should be positive.

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    Because $Dϕ_t(x)=I$ at $t=0$ and the determinant can not become zero because of uniqueness? – 2017-01-10
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    @LutzL Ah nice point. I overlooked that for fixed $x$ the differential $D\phi_t(x)$ fulfills the the IVP $$\begin{cases} \frac{d}{dt}D\phi_t(x) = Df(\phi_t(x))D\phi_t(x)\\D\phi_t(0) = \mathbb{I}\end{cases}$$ Shall I write the answer or do you want to do it? – 2017-01-10

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We know that $$\frac{d}{dt}\det(A(t)) =tr(A(t)^{-1}A'(t))·\det(A(t)) $$

For $A(t)=Dϕ_t(x)$ we know the differential equation $$ A'(t)=D(f(ϕ_t(x)))=Df(ϕ_t(x))·Dϕ_t(x)=Df(ϕ_t(x))·A(t) $$ so that $$ \frac{d}{dt}\det(A(t))=tr(Df(ϕ_t(x)))·\det(A(t))=\text{div}f(ϕ_t(x))·\det(A(t)) $$ This leads to the exponential solution $$ \det(A(t))=\exp\left(\int_0^t\text{div}f(ϕ_s(x))\,ds\right)·\det(A(0)) $$ which first demonstrates that the sign of the determinant is constant, and as $A(0)=I$ it is positive. And second it already contains the divergence of $f$ that occurs in the claimed formula.