Zero differential overlap

Reynolds' transport theorem (also known as the Leibniz-Reynolds' transport theorem), or in short Reynolds theorem, is a three-dimensional generalization of the Leibniz integral rule which is also known as differentiation under the integral sign. The theorem is named after Osborne Reynolds (1842–1912). It is used to recast derivatives of integrated quantities and is useful in formulating the basic equations of continuum mechanics.

Consider integrating $𝐟=𝐟(𝐱,t)$ over the time-dependent region $\Omega (t)$ that has boundary $\partial \Omega (t)$, then taking the derivative with respect to time:

$\frac{\mathrm{d}}{\mathrm{d}t}{\int }_{\Omega (t)}𝐟\text{dV}.$

If we wish to move the derivative under the integral sign, there are two issues: the time dependence of $𝐟$, and the introduction of and removal of space from $\Omega$ due to its dynamic boundary. Reynolds' transport theorem provides the necessary framework.

General form

Reynolds' transport theorem, derived in, is:

$\frac{\mathrm{d}}{\mathrm{d}t}{\int }_{\Omega (t)}𝐟\text{dV}={\int }_{\Omega (t)}\frac{\partial 𝐟}{\partial t}\text{dV}+{\int }_{\partial \Omega (t)}({𝐯}^b\cdot 𝐧)𝐟\text{dA}$

in which $𝐧(𝐱,t)$ is the outward-pointing unit-normal, $𝐱$ is a point in the region and is the variable of integration, $\text{dV}$ and $\text{dA}$ are volume and surface elements at $𝐱$, and ${𝐯}^b(𝐱,t)$ is the velocity of the area element – so not necessarily the flow velocity. The function $𝐟$ may be tensor, vector or scalar valued. Note that the integral on the left hand side is a function solely of time, and so the total derivative has been used.

Form for a material element

In continuum mechanics this theorem is often used for material elements, which are parcels of fluids or solids which no material enters or leaves. If $\Omega (t)$ is a material element then there is a velocity function $𝐯=𝐯(𝐱,t)$ and the boundary elements obey

${𝐯}^b\cdot 𝐧=𝐯\cdot 𝐧.$

This condition may be substituted to obtain

$\frac{\mathrm{d}}{\mathrm{d}t}\left({\int }_{\Omega (t)}𝐟\text{dV}\right)={\int }_{\Omega (t)}\frac{\partial 𝐟}{\partial t}\text{dV}+{\int }_{\partial \Omega (t)}(𝐯\cdot 𝐧)𝐟\text{dA}.$

Proof for a material element

Let ${\Omega }_0$ be reference configuration of the region $\Omega (t)$. Let the motion and the deformation gradient be given by $𝐱=𝝋(𝐗,t);⟹𝑭(𝐗,t)={\mathrm{\nabla }}_{\circ }𝝋.$ Let $J(𝐗,t)=det[𝑭(𝐗,t)]$. Then, integrals in the current and the reference configurations are related by ${\int }_{\Omega (t)}𝐟(𝐱,t)\text{dV}={\int }_{{\Omega }_0}𝐟[𝝋(𝐗,t),t]J(𝐗,t){\text{dV}}_0={\int }_{{\Omega }_0}\widehat{𝐟}(𝐗,t)J(𝐗,t){\text{dV}}_0.$ That this derivation is for a material element is implicit in the time constancy of the reference configuration: it is constant in material coordinates. The time derivative of an integral over a volume is defined as $\frac{d}{dt}({\int }_{\Omega (t)}𝐟(𝐱,t)\text{dV})=\underset{\Delta t\to 0}{lim}\frac{1}{\Delta t}({\int }_{\Omega (t+\Delta t)}𝐟(𝐱,t+\Delta t)\text{dV}-{\int }_{\Omega (t)}𝐟(𝐱,t)\text{dV}).$ Converting into integrals over the reference configuration, we get $\frac{d}{dt}({\int }_{\Omega (t)}𝐟(𝐱,t)\text{dV})=\underset{\Delta t\to 0}{lim}\frac{1}{\Delta t}({\int }_{{\Omega }_0}\widehat{𝐟}(𝐗,t+\Delta t)J(𝐗,t+\Delta t){\text{dV}}_0-{\int }_{{\Omega }_0}\widehat{𝐟}(𝐗,t)J(𝐗,t){\text{dV}}_0).$ Since ${\Omega }_0$ is independent of time, we have $\begin{array}{cc}\frac{d}{dt}({\int }_{\Omega (t)}𝐟(𝐱,t)\text{dV})& ={\int }_{{\Omega }_0}\left[\underset{\Delta t\to 0}{lim}\frac{\widehat{𝐟}(𝐗,t+\Delta t)J(𝐗,t+\Delta t)-\widehat{𝐟}(𝐗,t)J(𝐗,t)}{\Delta t}\right]{\text{dV}}_0\\ & ={\int }_{{\Omega }_0}\frac{\partial }{\partial t}[\widehat{𝐟}(𝐗,t)J(𝐗,t)]{\text{dV}}_0\\ & ={\int }_{{\Omega }_0}(\frac{\partial }{\partial t}[\widehat{𝐟}(𝐗,t)]J(𝐗,t)+\widehat{𝐟}(𝐗,t)\frac{\partial }{\partial t}[J(𝐗,t)]){\text{dV}}_0\end{array}$ Now, the time derivative of $det𝑭$ is given by $\frac{\partial J(𝐗,t)}{\partial t}=\frac{\partial }{\partial t}(det𝑭)=(det𝑭)(\mathrm{\nabla }\cdot 𝐯)=J(𝐗,t)\mathrm{\nabla }\cdot 𝐯(𝝋(𝐗,t),t)=J(𝐗,t)\mathrm{\nabla }\cdot 𝐯(𝐱,t).$ Therefore, $\begin{array}{cc}\frac{d}{dt}({\int }_{\Omega (t)}𝐟(𝐱,t)\text{dV})& ={\int }_{{\Omega }_0}(\frac{\partial }{\partial t}[\widehat{𝐟}(𝐗,t)]J(𝐗,t)+\widehat{𝐟}(𝐗,t)J(𝐗,t)\mathrm{\nabla }\cdot 𝐯(𝐱,t)){\text{dV}}_0\\ & ={\int }_{{\Omega }_0}(\frac{\partial }{\partial t}[\widehat{𝐟}(𝐗,t)]+\widehat{𝐟}(𝐗,t)\mathrm{\nabla }\cdot 𝐯(𝐱,t))J(𝐗,t){\text{dV}}_0\\ & ={\int }_{\Omega (t)}(\dot{𝐟}(𝐱,t)+𝐟(𝐱,t)\mathrm{\nabla }\cdot 𝐯(𝐱,t))\text{dV}\end{array}$ where $\dot{𝐟}$ is the material time derivative of $𝐟$. Now, the material derivative is given by $\dot{𝐟}(𝐱,t)=\frac{\partial 𝐟(𝐱,t)}{\partial t}+[\mathrm{\nabla }𝐟(𝐱,t)]\cdot 𝐯(𝐱,t).$ Therefore, $\frac{d}{dt}({\int }_{\Omega (t)}𝐟(𝐱,t)\text{dV})={\int }_{\Omega (t)}(\frac{\partial 𝐟(𝐱,t)}{\partial t}+[\mathrm{\nabla }𝐟(𝐱,t)]\cdot 𝐯(𝐱,t)+𝐟(𝐱,t)\mathrm{\nabla }\cdot 𝐯(𝐱,t))\text{dV}$ or, $\frac{d}{dt}\left({\int }_{\Omega (t)}𝐟\text{dV}\right)={\int }_{\Omega (t)}(\frac{\partial 𝐟}{\partial t}+\mathrm{\nabla }𝐟\cdot 𝐯+𝐟\mathrm{\nabla }\cdot 𝐯)\text{dV}.$ Using the identity $\mathrm{\nabla }\cdot (𝐯\otimes 𝐰)=𝐯(\mathrm{\nabla }\cdot 𝐰)+\mathrm{\nabla }𝐯\cdot 𝐰$ we then have $\frac{d}{dt}\left({\int }_{\Omega (t)}𝐟\text{dV}\right)={\int }_{\Omega (t)}(\frac{\partial 𝐟}{\partial t}+\mathrm{\nabla }\cdot (𝐟\otimes 𝐯))\text{dV}.$ Using the divergence theorem and the identity $(𝐚\otimes 𝐛)\cdot 𝐧=(𝐛\cdot 𝐧)𝐚$, we have $\frac{d}{dt}\left({\int }_{\Omega (t)}𝐟\text{dV}\right)={\int }_{\Omega (t)}\frac{\partial 𝐟}{\partial t}\text{dV}+{\int }_{\partial \Omega (t)}(𝐟\otimes 𝐯)\cdot 𝐧\text{dA}={\int }_{\Omega (t)}\frac{\partial 𝐟}{\partial t}\text{dV}+{\int }_{\partial \Omega (t)}(𝐯\cdot 𝐧)𝐟\text{dA}◻$

Erroneous sources

This theorem is widely quoted, incorrectly, as being the form specific to material volumes. See the planetmath external link below for an example. Clearly, if the material volume form is applied to regions other than material volumes, errors will ensue.

A special case

If we take $\Omega$ to be constant with respect to time, then ${𝐯}_b=0$ and the identity reduces to

$\frac{\mathrm{d}}{\mathrm{d}t}{\int }_{\Omega }f\text{dV}={\int }_{\Omega }\frac{\partial f}{\partial t}\text{dV},$

as expected. This simplification is not possible if an incorrect form of the Reynolds transport theorem is used.

Interpretation and reduction to one dimension

The theorem is the higher dimensional extension of Differentiation under the integral sign and should reduce to that expression in some cases. Suppose $f$ is independent of $y$ & $z$, and that $\Omega (t)$ is a unit square in the $y-z$ plane and has $x$ limits $a(t)$ and $b(t)$. Then Reynolds transport theorem reduces to

$\frac{\mathrm{d}}{\mathrm{d}t}{\int }_{a(t)}^{b(t)}f\text{dx}={\int }_{a(t)}^{b(t)}\frac{\partial f}{\partial t}\text{dx}+\frac{\partial b(t)}{\partial t}f(b(t),t)-\frac{\partial a(t)}{\partial t}f(a(t),t),$

which is the expression given on Differentiation under the integral sign, except that there the variables x and t have been swapped.

See also

• Differentiation under the integral sign
• Leibniz integral rule

References

L. G. Leal, 2007, Advanced transport phenomena: fluid mechanics and convective transport processes, Cambridge University Press, p. 912.

O. Reynolds, 1903, Papers on Mechanical and Physical Subjects, Vol. 3, The Sub-Mechanics of the Universe, Cambridge University Press, Cambridge.

J. E. Marsden and A. Tromba, 2003, Vector Calculus, 5th ed., W. H. Freeman .

External links

• Osborne Reynolds, Collected Papers on Mechanical and Physical Subjects, in three volumes, published circa 1903, now fully and freely available in digital format:

Volume 1

Volume 2

Volume 3

• http://www.catea.org/grade/mecheng/mod6/mod6.html#slide1
• http://planetmath.org/reynoldstransporttheorem