The strategy is to pull back the [integral](/page/Integral) over the moving domain $V_t$ to the fixed reference domain $V$ using the change of variables $x = \Phi(t,y)$, differentiate in time on the fixed domain using the product rule, and then push forward again. The main ingredients are the [Change Of Variables](/theorems/22), the [Jacobi Formula For The Flow Map](/theorems/650), and the product rule. The proof proceeds as follows: a preliminary claim establishes that the Jacobian determinant is strictly positive, which ensures $\Phi(t,\cdot)$ is a $C^1$ diffeomorphism so that the change of variables is valid; Step 1 performs the pull-back; Step 2 differentiates under the integral; Step 3 evaluates the resulting terms using the flow equation and the Jacobi formula; and Step 4 pushes forward to recover the Eulerian integral. Define the spatial Jacobian matrix $\hat{J}(t,y) \in \mathbb{R}^{n \times n}$ by $\hat{J}_{ij}(t,y) = \frac{\partial \Phi_i}{\partial y_j}(t,y)$, and the Jacobian determinant $\mathcal{J}(t,y) = \det(\hat{J}(t,y))$. [claim:Positivity Of The Jacobian Determinant] $\mathcal{J}(t,y) > 0$ for all $(t,y) \in [0,T] \times M$. [/claim] [proof] The [Jacobi Formula For The Flow Map](/theorems/650) gives $\partial_t \mathcal{J} = \mathcal{J}\,(\nabla \cdot u)(t, \Phi(t,y))$, which is a linear scalar ODE in $\mathcal{J}$ with [continuous](/page/Continuity) coefficient $t \mapsto (\nabla \cdot u)(t, \Phi(t,y))$ and initial condition $\mathcal{J}(0,y) = \det(I) = 1$. Its unique solution is \begin{align*} \mathcal{J}(t,y) &= \exp\!\Bigl(\int_0^t (\nabla \cdot u)(s, \Phi(s,y))\,ds\Bigr) > 0. \end{align*} [/proof] **Step 1: Change to Lagrangian coordinates.** By the claim, $\mathcal{J}(t,y) > 0$, so $\hat{J}(t,y)$ is invertible for all $t$. The [inverse function theorem](/page/Inverse%20Function%20Theorem) gives that $\Phi(t,\cdot)$ is a local $C^1$ diffeomorphism. Injectivity follows from the uniqueness part of the [Picard Lindelof](/theorems/69) theorem: if $\Phi(t,x_1) = \Phi(t,x_2)$, define $v(s) = -u(t - s, \cdot)$ and solve the ODE $\frac{d}{ds}\Psi(s,z) = v(s, \Psi(s,z))$, $\Psi(0,z) = \Phi(t,x_1) = \Phi(t,x_2)$ on $[0,t]$. By uniqueness, $\Psi(t,\cdot)$ has a single output, but $\Psi(t, \Phi(t,x_i)) = x_i$ for $i = 1,2$, so $x_1 = x_2$. Hence $\Phi(t,\cdot)$ is a $C^1$ diffeomorphism. The [Change Of Variables](/theorems/22) therefore gives \begin{align*} \int_{V_t} f(t,x)\,d\mathcal{L}^n(x) &= \int_V f(t, \Phi(t,y))\,|\mathcal{J}(t,y)|\,d\mathcal{L}^n(y) = \int_V f(t, \Phi(t,y))\,\mathcal{J}(t,y)\,d\mathcal{L}^n(y), \end{align*} where the second equality uses $\mathcal{J}(t,y) > 0$. **Step 2: [Differentiation](/page/Derivative) under the integral.** Since $f$ and $u$ are $C^1$ and the domain $V$ is fixed, we may differentiate under the integral sign: \begin{align*} \frac{d}{dt}\int_{V_t} f(t,x)\,d\mathcal{L}^n(x) &= \int_V \partial_t\bigl[f(t, \Phi(t,y))\bigr]\,\mathcal{J}(t,y)\,d\mathcal{L}^n(y) + \int_V f(t, \Phi(t,y))\,\partial_t \mathcal{J}(t,y)\,d\mathcal{L}^n(y). \end{align*} **Step 3: Evaluating each term.** For the first integral, the chain rule and the flow equation $\frac{d}{dt}\Phi(t,y) = u(t, \Phi(t,y))$ yield \begin{align*} \partial_t\bigl[f(t, \Phi(t,y))\bigr] &= (\partial_t f)(t, \Phi(t,y)) + \nabla f(t, \Phi(t,y)) \cdot \frac{d}{dt}\Phi(t,y) = (\partial_t f + u \cdot \nabla f)(t, \Phi(t,y)). \end{align*} For the second integral, the [Jacobi Formula For The Flow Map](/theorems/650) gives $\partial_t \mathcal{J} = \mathcal{J}\,(\nabla \cdot u)(t, \Phi(t,y))$. Substituting both expressions into Step 2 and factoring out $\mathcal{J}(t,y)$: \begin{align*} \frac{d}{dt}\int_{V_t} f\,d\mathcal{L}^n &= \int_V \bigl(\partial_t f + u \cdot \nabla f + f\,\nabla \cdot u\bigr)(t, \Phi(t,y))\,\mathcal{J}(t,y)\,d\mathcal{L}^n(y). \end{align*} The integrand simplifies via the product rule: $\partial_t f + u \cdot \nabla f + f\,\nabla \cdot u = \partial_t f + \nabla \cdot (fu)$, since $\nabla \cdot (fu) = f\,\nabla \cdot u + u \cdot \nabla f$. **Step 4: Return to Eulerian coordinates.** Applying the [Change Of Variables](/theorems/22) again via $x = \Phi(t,y)$ yields \begin{align*} \frac{d}{dt}\int_{V_t} f(t,x)\,d\mathcal{L}^n(x) &= \int_{V_t} \bigl(\partial_t f + \nabla \cdot (fu)\bigr)(t,x)\,d\mathcal{L}^n(x), \end{align*} which is the desired identity.