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Reynolds transport theorem

Let Ω(t) be a region in Euclidean space with boundary $\partial \Omega (t)$. Let $\mathbf{x}(t)$ be the positions of points in the region and let $\mathbf{v}(\mathbf{x},t)$ be the velocity field in the region. Let $\mathbf{n}(\mathbf{x},t)$ be the outward unit normal to the boundary. Let $\mathbf{f}(\mathbf{x},t)$ be a vector field in the region (it may also be a scalar field). Show that

$\cfrac{d}{dt}\left(\int_{\Omega(t)} \mathbf{f}~\text{dV}\right) = \int_{\Omega(t)} \frac{\partial \mathbf{f}}{\partial t}~\text{dV} + \int_{\partial \Omega(t)} (\mathbf{v}\cdot\mathbf{n})\mathbf{f}~\text{dA} ~.$

This relation is also known as the Reynold's Transport Theorem and is a generalization of the Leibniz rule. Content of example.

Proof:

Let Ω0 be reference configuration of the region Ω(t). Let the motion and the deformation gradient be given by

$\mathbf{x} = \boldsymbol{\varphi}(\mathbf{X}, t)~; \qquad\implies\qquad \boldsymbol{F}(\mathbf{X},t) = \boldsymbol{\nabla}_{\circ} \boldsymbol{\varphi} ~.$

Let $J(\mathbf{X},t) = \det[\boldsymbol{F}(\mathbf{X},t)]$. Then, integrals in the current and the reference configurations are related by

$\int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV} = \int_{\Omega_0} \mathbf{f}[\boldsymbol{\varphi}(\mathbf{X},t),t]~J(\mathbf{X},t)~\text{dV}_0 = \int_{\Omega_0} \hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)~\text{dV}_0 ~.$

The time derivative of an integral over a volume is defined as

$\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) = \lim_{\Delta t \rightarrow 0} \cfrac{1}{\Delta t} \left(\int_{\Omega(t + \Delta t)} \mathbf{f}(\mathbf{x},t+\Delta t)~\text{dV} - \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) ~.$

Converting into integrals over the reference configuration, we get

$\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) = \lim_{\Delta t \rightarrow 0} \cfrac{1}{\Delta t} \left(\int_{\Omega_0} \hat{\mathbf{f}}(\mathbf{X},t+\Delta t)~J(\mathbf{X},t+\Delta t)~\text{dV}_0 - \int_{\Omega_0} \hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)~\text{dV}_0\right) ~.$

Since Ω0 is independent of time, we have

\begin{align} \cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) & = \int_{\Omega_0} \left[\lim_{\Delta t \rightarrow 0} \cfrac{ \hat{\mathbf{f}}(\mathbf{X},t+\Delta t)~J(\mathbf{X},t+\Delta t) - \hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)}{\Delta t} \right]~\text{dV}_0 \ & = \int_{\Omega_0} \frac{\partial }{\partial t}[\hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)]~\text{dV}_0 \ & = \int_{\Omega_0} \left( \frac{\partial }{\partial t}[\hat{\mathbf{f}}(\mathbf{X},t)]~J(\mathbf{X},t)+ \hat{\mathbf{f}}(\mathbf{X},t)~\frac{\partial }{\partial t}[J(\mathbf{X},t)]\right) ~\text{dV}_0 \end{align}

Now, the time derivative of $\det\boldsymbol{F}$ is given by (see Gurtin: 1981, p. 77)

$\frac{\partial J(\mathbf{X},t)}{\partial t} = \frac{\partial }{\partial t}(\det\boldsymbol{F}) = (\det\boldsymbol{F})(\boldsymbol{\nabla} \cdot \mathbf{v}) = J(\mathbf{X},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\boldsymbol{\varphi}(\mathbf{X},t),t) = J(\mathbf{X},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t) ~.$

Therefore,

\begin{align} \cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) & = \int_{\Omega_0} \left( \frac{\partial }{\partial t}[\hat{\mathbf{f}}(\mathbf{X},t)]~J(\mathbf{X},t)+ \hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)\right) ~\text{dV}_0 \ & = \int_{\Omega_0} \left(\frac{\partial }{\partial t}[\hat{\mathbf{f}}(\mathbf{X},t)]+ \hat{\mathbf{f}}(\mathbf{X},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)\right)~J(\mathbf{X},t) ~\text{dV}_0 \ & = \int_{\Omega(t)} \left(\dot{\mathbf{f}}(\mathbf{x},t)+ \mathbf{f}(\mathbf{x},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)\right)~\text{dV} \end{align}

where $\dot{\mathbf{f}}$ is the material time derivative of $\mathbf{f}$. Now, the material derivative is given by

$\dot{\mathbf{f}}(\mathbf{x},t) = \frac{\partial \mathbf{f}(\mathbf{x},t)}{\partial t} + [\boldsymbol{\nabla} \mathbf{f}(\mathbf{x},t)]\cdot\mathbf{v}(\mathbf{x},t) ~.$

Therefore,

$\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) = \int_{\Omega(t)} \left( \frac{\partial \mathbf{f}(\mathbf{x},t)}{\partial t} + [\boldsymbol{\nabla} \mathbf{f}(\mathbf{x},t)]\cdot\mathbf{v}(\mathbf{x},t) + \mathbf{f}(\mathbf{x},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)\right)~\text{dV}$

or,

$\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}~\text{dV}\right) = \int_{\Omega(t)} \left( \frac{\partial \mathbf{f}}{\partial t} + \boldsymbol{\nabla} \mathbf{f}\cdot\mathbf{v} + \mathbf{f}~\boldsymbol{\nabla} \cdot \mathbf{v}\right)~\text{dV} ~.$

Using the identity

$\boldsymbol{\nabla} \cdot (\mathbf{v}\otimes\mathbf{w}) = \mathbf{v}(\boldsymbol{\nabla} \cdot \mathbf{w}) + \boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{w}$

we then have

$\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}~\text{dV}\right) = \int_{\Omega(t)} \left(\frac{\partial \mathbf{f}}{\partial t} + \boldsymbol{\nabla} \cdot (\mathbf{f}\otimes\mathbf{v})\right)~\text{dV} ~.$

Using the divergence theorem and the identity $(\mathbf{a}\otimes\mathbf{b})\cdot\mathbf{n} = (\mathbf{b}\cdot\mathbf{n})\mathbf{a}$ we have

${ \cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}~\text{dV}\right) = \int_{\Omega(t)}\frac{\partial \mathbf{f}}{\partial t}~\text{dV} + \int_{\partial \Omega(t)}(\mathbf{f}\otimes\mathbf{v})\cdot\mathbf{n}~\text{dV} = \int_{\Omega(t)}\frac{\partial \mathbf{f}}{\partial t}~\text{dV} + \int_{\partial \Omega(t)}(\mathbf{v}\cdot\mathbf{n})\mathbf{f}~\text{dV} ~. }$

References

1. M.E. Gurtin. An Introduction to Continuum Mechanics. Academic Press, New York, 1981.
2. T. Belytschko, W. K. Liu, and B. Moran. Nonlinear Finite Elements for Continua and Structures. John Wiley and Sons, Ltd., New York, 2000.