In differential calculus, the Reynolds transport theorem (also known as the Leibniz–Reynolds transport theorem), or simply the Reynolds theorem, named after Osborne Reynolds (1842–1912), is a three-dimensional generalization of the Leibniz integral rule. It is used to recast time 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 Ω(t) that has boundary ∂Ω(t), then taking the derivative with respect to time:

${\displaystyle {\frac {d}{dt))\int _{\Omega (t)}\mathbf {f} \,dV.}$
If we wish to move the derivative into the integral, there are two issues: the time dependence of f, and the introduction of and removal of space from Ω due to its dynamic boundary. Reynolds transport theorem provides the necessary framework.

## General form

Reynolds transport theorem can be expressed as follows:[1][2][3]

${\displaystyle {\frac {d}{dt))\int _{\Omega (t)}\mathbf {f} \,dV=\int _{\Omega (t)}{\frac {\partial \mathbf {f} }{\partial t))\,dV+\int _{\partial \Omega (t)}\left(\mathbf {v} _{b}\cdot \mathbf {n} \right)\mathbf {f} \,dA}$
in which n(x,t) is the outward-pointing unit normal vector, x is a point in the region and t is the variable of integration, dV and dA are volume and surface elements at x, and vb(x,t) is the velocity of the area element (not the flow velocity). The function f may be tensor-, vector- or scalar-valued.[4] 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. These are parcels of fluids or solids which no material enters or leaves. If Ω(t) is a material element then there is a velocity function v = v(x,t), and the boundary elements obey

${\displaystyle \mathbf {v} _{b}\cdot \mathbf {n} =\mathbf {v} \cdot \mathbf {n} .}$
This condition may be substituted to obtain:[5]
${\displaystyle {\frac {d}{dt))\left(\int _{\Omega (t)}\mathbf {f} \,dV\right)=\int _{\Omega (t)}{\frac {\partial \mathbf {f} }{\partial t))\,dV+\int _{\partial \Omega (t)}(\mathbf {v} \cdot \mathbf {n} )\mathbf {f} \,dA.}$

Proof for a material element

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

${\displaystyle \mathbf {x} ={\boldsymbol {\varphi ))(\mathbf {X} ,t),}$
${\displaystyle {\boldsymbol {F))(\mathbf {X} ,t)={\boldsymbol {\nabla )){\boldsymbol {\varphi )).}$

Let J(X,t) = det F(X,t). Define

${\displaystyle {\hat {\mathbf {f} ))(\mathbf {X} ,t)=\mathbf {f} ({\boldsymbol {\varphi ))(\mathbf {X} ,t),t).}$
Then the integrals in the current and the reference configurations are related by
${\displaystyle \int _{\Omega (t)}\mathbf {f} (\mathbf {x} ,t)\,dV=\int _{\Omega _{0))\mathbf {f} ({\boldsymbol {\varphi ))(\mathbf {X} ,t),t)\,J(\mathbf {X} ,t)\,dV_{0}=\int _{\Omega _{0)){\hat {\mathbf {f} ))(\mathbf {X} ,t)\,J(\mathbf {X} ,t)\,dV_{0}.}$

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

${\displaystyle {\frac {d}{dt))\left(\int _{\Omega (t)}\mathbf {f} (\mathbf {x} ,t)\,dV\right)=\lim _{\Delta t\to 0}{\frac {1}{\Delta t))\left(\int _{\Omega (t+\Delta t)}\mathbf {f} (\mathbf {x} ,t+\Delta t)\,dV-\int _{\Omega (t)}\mathbf {f} (\mathbf {x} ,t)\,dV\right).}$

Converting into integrals over the reference configuration, we get

${\displaystyle {\frac {d}{dt))\left(\int _{\Omega (t)}\mathbf {f} (\mathbf {x} ,t)\,dV\right)=\lim _{\Delta t\to 0}{\frac {1}{\Delta t))\left(\int _{\Omega _{0)){\hat {\mathbf {f} ))(\mathbf {X} ,t+\Delta t)\,J(\mathbf {X} ,t+\Delta t)\,dV_{0}-\int _{\Omega _{0)){\hat {\mathbf {f} ))(\mathbf {X} ,t)\,J(\mathbf {X} ,t)\,dV_{0}\right).}$

Since Ω0 is independent of time, we have

{\displaystyle {\begin{aligned}{\frac {d}{dt))\left(\int _{\Omega (t)}\mathbf {f} (\mathbf {x} ,t)\,dV\right)&=\int _{\Omega _{0))\left(\lim _{\Delta t\to 0}{\frac ((\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)\,dV_{0}\\&=\int _{\Omega _{0)){\frac {\partial }{\partial t))\left({\hat {\mathbf {f} ))(\mathbf {X} ,t)\,J(\mathbf {X} ,t)\right)\,dV_{0}\\&=\int _{\Omega _{0))\left({\frac {\partial }{\partial t)){\big (}{\hat {\mathbf {f} ))(\mathbf {X} ,t){\big )}\,J(\mathbf {X} ,t)+{\hat {\mathbf {f} ))(\mathbf {X} ,t)\,{\frac {\partial }{\partial t)){\big (}J(\mathbf {X} ,t){\big )}\right)\,dV_{0}.\end{aligned))}

The time derivative of J is given by:[6]

{\displaystyle {\begin{aligned}{\frac {\partial J(\mathbf {X} ,t)}{\partial t))&={\frac {\partial }{\partial t))(\det {\boldsymbol {F)))\\&=(\det {\boldsymbol {F)))\operatorname {tr} \left({\boldsymbol {F))^{-1}{\frac {\partial {\boldsymbol {F))}{\partial t))\right)\\&=(\det {\boldsymbol {F)))\operatorname {tr} \left({\frac {\partial {\boldsymbol {X))}{\partial {\boldsymbol {\varphi )))){\frac {\partial }{\partial t))\left({\frac {\partial {\boldsymbol {\varphi ))}{\partial {\boldsymbol {X))))\right)\right)\\&=(\det {\boldsymbol {F)))\operatorname {tr} \left({\frac {\partial {\boldsymbol {X))}{\partial {\boldsymbol {\varphi )))){\frac {\partial }{\partial {\boldsymbol {X))))\left({\frac {\partial {\boldsymbol {\varphi ))}{\partial t))\right)\right)\\&=(\det {\boldsymbol {F)))\operatorname {tr} \left({\frac {\partial }{\partial {\boldsymbol {x))))\left({\frac {\partial {\boldsymbol {\varphi ))}{\partial t))\right)\right)\\&=(\det {\boldsymbol {F)))({\boldsymbol {\nabla ))\cdot \mathbf {v} )\\&=J(\mathbf {X} ,t)\,{\boldsymbol {\nabla ))\cdot \mathbf {v} {\big (}{\boldsymbol {\varphi ))(\mathbf {X} ,t),t{\big )}\\&=J(\mathbf {X} ,t)\,{\boldsymbol {\nabla ))\cdot \mathbf {v} (\mathbf {x} ,t).\end{aligned))}

Therefore,

{\displaystyle {\begin{aligned}{\frac {d}{dt))\left(\int _{\Omega (t)}\mathbf {f} (\mathbf {x} ,t)\,dV\right)&=\int _{\Omega _{0))\left({\frac {\partial }{\partial t))\left({\hat {\mathbf {f} ))(\mathbf {X} ,t)\right)\,J(\mathbf {X} ,t)+{\hat {\mathbf {f} ))(\mathbf {X} ,t)\,J(\mathbf {X} ,t)\,{\boldsymbol {\nabla ))\cdot \mathbf {v} (\mathbf {x} ,t)\right)\,dV_{0}\\&=\int _{\Omega _{0))\left({\frac {\partial }{\partial t))\left({\hat {\mathbf {f} ))(\mathbf {X} ,t)\right)+{\hat {\mathbf {f} ))(\mathbf {X} ,t)\,{\boldsymbol {\nabla ))\cdot \mathbf {v} (\mathbf {x} ,t)\right)\,J(\mathbf {X} ,t)\,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)\,dV.\end{aligned))}
where ${\displaystyle {\dot {\mathbf {f} ))}$ is the material time derivative of f. The material derivative is given by
${\displaystyle {\dot {\mathbf {f} ))(\mathbf {x} ,t)={\frac {\partial \mathbf {f} (\mathbf {x} ,t)}{\partial t))+{\big (}{\boldsymbol {\nabla ))\mathbf {f} (\mathbf {x} ,t){\big )}\cdot \mathbf {v} (\mathbf {x} ,t).}$

Therefore,

${\displaystyle {\frac {d}{dt))\left(\int _{\Omega (t)}\mathbf {f} (\mathbf {x} ,t)\,dV\right)=\int _{\Omega (t)}\left({\frac {\partial \mathbf {f} (\mathbf {x} ,t)}{\partial t))+{\big (}{\boldsymbol {\nabla ))\mathbf {f} (\mathbf {x} ,t){\big )}\cdot \mathbf {v} (\mathbf {x} ,t)+\mathbf {f} (\mathbf {x} ,t)\,{\boldsymbol {\nabla ))\cdot \mathbf {v} (\mathbf {x} ,t)\right)\,dV,}$
or,
${\displaystyle {\frac {d}{dt))\left(\int _{\Omega (t)}\mathbf {f} \,dV\right)=\int _{\Omega (t)}\left({\frac {\partial \mathbf {f} }{\partial t))+{\boldsymbol {\nabla ))\mathbf {f} \cdot \mathbf {v} +\mathbf {f} \,{\boldsymbol {\nabla ))\cdot \mathbf {v} \right)\,dV.}$

Using the identity

${\displaystyle {\boldsymbol {\nabla ))\cdot (\mathbf {v} \otimes \mathbf {w} )=\mathbf {v} ({\boldsymbol {\nabla ))\cdot \mathbf {w} )+{\boldsymbol {\nabla ))\mathbf {v} \cdot \mathbf {w} ,}$
we then have
${\displaystyle {\frac {d}{dt))\left(\int _{\Omega (t)}\mathbf {f} \,dV\right)=\int _{\Omega (t)}\left({\frac {\partial \mathbf {f} }{\partial t))+{\boldsymbol {\nabla ))\cdot (\mathbf {f} \otimes \mathbf {v} )\right)\,dV.}$

Using the divergence theorem and the identity (ab) · n = (b · n)a, we have

{\displaystyle {\begin{aligned}{\frac {d}{dt))\left(\int _{\Omega (t)}\mathbf {f} \,dV\right)&=\int _{\Omega (t)}{\frac {\partial \mathbf {f} }{\partial t))\,dV+\int _{\partial \Omega (t)}(\mathbf {f} \otimes \mathbf {v} )\cdot \mathbf {n} \,dA\\&=\int _{\Omega (t)}{\frac {\partial \mathbf {f} }{\partial t))\,dV+\int _{\partial \Omega (t)}(\mathbf {v} \cdot \mathbf {n} )\mathbf {f} \,dA.\end{aligned))}
Q.E.D.

## A special case

If we take Ω to be constant with respect to time, then vb = 0 and the identity reduces to

${\displaystyle {\frac {d}{dt))\int _{\Omega }f\,dV=\int _{\Omega }{\frac {\partial f}{\partial t))\,dV.}$
as expected. (This simplification is not possible if the flow velocity is incorrectly used in place of the velocity of an area element.)

### Interpretation and reduction to one dimension

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

${\displaystyle {\frac {d}{dt))\int _{a(t)}^{b(t)}f(x,t)\,dx=\int _{a(t)}^{b(t)}{\frac {\partial f}{\partial t))\,dx+{\frac {\partial b(t)}{\partial t))f{\big (}b(t),t{\big )}-{\frac {\partial a(t)}{\partial t))f{\big (}a(t),t{\big )}\,,}$
which, up to swapping x and t, is the standard expression for differentiation under the integral sign.