Section 3.1 Reynolds’ Transport Theorem

Subsection 3.1.1 Jacobian of the Lagrangian-to-Eulerian map

Consider a volume of fluid that, at time $t = 0\text{,}$ occupies the region $V(0) \subseteq D$ and is dyed a certain colour. As the fluid evolves, this packet of fluid occupies the region $V(t)$ for $t > 0\text{.}$

As an example, we may express the density of the fluid as

\begin{equation*} \rho = \rho(\bx, t), \qquad \textrm{$\bx \in V(t)$} \end{equation*}

i.e. the density is given as a function of position in space and time, for points in the (possibly moving) region $V(t)\text{.}$ This is the natural description for an observer at a fixed point in space, and is the Eulerian viewpoint.

Alternatively, we can write

\begin{equation*} \rho = \rho(\bX, t), \qquad \textrm{$\bX \in V(0)$} \end{equation*}

which is the density of fluid for those particles making up an originally chosen volume, $V(0)\text{.}$ This is a natural quantity to study if we were to move with the fluid; for instance, if we were to colour the volume $V(0)$ with a dye and track its density in time. This is the Lagrangian viewpoint.

Strictly, $\rho(\bX, t)$ and $\rho(\bx, t)$ are different functions of their arguments; we follow the common convention of using the same symbol for both, and the arguments make clear which is meant.

The correspondence between the Eulerian and Lagrangian coordinates is written as

\begin{equation*} \bx = \bx(\bX, t). \end{equation*}

Theorem 3.1.1 Jacobian of the Lagrangian-to-Eulerian map.

Since we assume the medium is continuous, we would generally require that the mapping from Lagrangian to Eulerian coordinates is continuous and one-to-one; then the map assigns every element (label) in the original reference configuration, denoted $\bX\text{,}$ a unique position, $\bx\text{,}$ in the deformed state.

From Multivariable Calculus, the map is locally invertible provided the Jacobian of the transformation,

\begin{equation*} J = \pd{\bx}{\bX} = \pd{(x, y, z)}{(X, Y, Z)} = \begin{vmatrix} \pd{x}{X} & \pd{x}{Y} & \pd{x}{Z} \\ \pd{y}{X} & \pd{y}{Y} & \pd{y}{Z} \\ \pd{z}{X} & \pd{z}{Y} & \pd{z}{Z} \end{vmatrix} \end{equation*}

is finite and non-zero. Since $J = 1$ at $t = 0$ (when $\bx = \bX$) and $J$ varies continuously in time, this means $J > 0\text{,}$ i.e.

\begin{equation*} 0 \lt J \lt \infty. \end{equation*}

The following requires a bit of algebra, and you are not required to prove the result.

Theorem 3.1.2 Euler’s identity.

The material derivative of the Jacobian of the transformation $\bx = \bx(\bX, t)$ is given by

\begin{equation} \DD{J}{t} = J \nabla \cdot \bu,\tag{3.1.1} \end{equation}

where $\bu = \DD{\bx}{t}\text{.}$

Proof.

(Not examinable) The material derivative is the time derivative with $\bX$ held fixed, so $\pd{x}{t} = u\text{,}$ $\pd{y}{t} = v$ and $\pd{z}{t} = w\text{.}$ By the product rule, $\DD{J}{t}$ is a sum of three determinants, each with one row of $J$ differentiated.

Consider the first. A typical entry of its differentiated row is

\begin{equation*} \DD{}{t}\pd{x}{X} = \pd{u}{X} = \pd{u}{x}\pd{x}{X} + \pd{u}{y}\pd{y}{X} + \pd{u}{z}\pd{z}{X}, \end{equation*}

using the chain rule; the entries for the $Y$ and $Z$ columns are the same, with $X$ replaced by $Y$ and $Z$ respectively. So the new first row is

\begin{equation*} \pd{u}{x}\,(\text{row 1}) + \pd{u}{y}\,(\text{row 2}) + \pd{u}{z}\,(\text{row 3}). \end{equation*}

A determinant with two equal rows vanishes, so only the first term survives, and this determinant equals $\pd{u}{x}J\text{.}$

In the same way, the other two determinants give $\pd{v}{y}J$ and $\pd{w}{z}J\text{.}$ Adding them gives $\DD{J}{t} = J\,\nabla\cdot\bu\text{.}$

Subsection 3.1.2 Reynolds’ Transport Theorem

We are now ready to derive a key result that eases our path towards developing the governing equations for a fluid. The result is as follows.

Theorem 3.1.3 Reynolds’ Transport Theorem.

Consider a time-dependent volume, $V(t)\text{,}$ that is transported by the fluid so that it always consists of the same fluid particles. Then, for any function, $f(\bx, t)\text{,}$ continuously differentiable with respect to its arguments, Reynolds’ Transport Theorem is as follows:

\begin{equation} \dd{}{t} \iiint_{V(t)} f \, \de{x} \, \de{y} \, \de{z} = \iiint_{V(t)} \left[ \pd{f}{t} + \nabla \cdot (f\bu)\right] \, \de{x} \, \de{y} \, \de{z}.\tag{3.1.2} \end{equation}

Proof.

We transform the integral in Eulerian coordinates to Lagrangian coordinates, integrating in the label space:

\begin{equation*} I(t) \equiv \iiint_{V(t)} f \, \de{x} \de{y} \de{z} = \iiint_{V(0)} f J \, \de{X} \de{Y} \de{Z}, \end{equation*}

and notice that we now only need to integrate over the fixed volume as defined in Lagrangian space, at the expense of adding the Jacobian factor. We now write

\begin{align*} \dd{I}{t} &= \iiint_{V(0)} \DD{(fJ)}{t} \, \de{X}\de{Y}\de{Z} \end{align*}

and the material derivative passes through since the domain is fixed. The material derivative theorem, Theorem 2.1.5, now allows us to convert the material derivative to regular partial derivatives. By the product rule:

\begin{align*} \DD{(fJ)}{t} &= \DD{f}{t}J + f\DD{J}{t} \\ &= \left[ \pd{f}{t} + \bu \cdot \nabla f\right]J + f \left[J \nabla \cdot \bu\right] \\ &= \left[ \pd{f}{t} + \nabla \cdot (f\bu) \right]J \end{align*}

and we have differentiated the Jacobian via Euler’s identity Theorem 3.1.2 in the second line. The last line follows from the vector identity:

\begin{equation*} \nabla \cdot (f\bu) = f \nabla \cdot \bu + \nabla f \cdot \bu. \end{equation*}

We can now revert from Lagrangian to Eulerian integration, and this completes the proof of Reynolds’ Transport Theorem.

Note 3.1.4.

It is helpful for you to convince yourself that the vector identity used in the proof is the only possible arrangement of operations that makes sense, i.e. in order for $\nabla \cdot (f\bu)$ to return a scalar.

In summary, Reynolds’ Transport Theorem gives an identity for how time differentiation can pass through the integral when the domain of integration is changing in time!

Example 3.1.5 Reynolds’ Transport Theorem for a stretching cube.

Aside: New in 2026.

Consider the steady velocity field

\begin{equation*} \bu = \left(x, \; -\tfrac12 y, \; 0\right). \end{equation*}

At $t = 0$ the fluid occupying the unit cube $V(0) = [0,1]^3$ is dyed, and $V(t)$ denotes the material volume occupied by this dyed fluid at later times. Let

\begin{equation*} f(\bx) = x^2 + y, \qquad I(t) = \iiint_{V(t)} f \, \de{x} \, \de{y} \, \de{z}. \end{equation*}

Find the pathlines $\bx = \bx(\bX, t)$ and hence show that $V(t)$ remains a cuboid. Find its dimensions.
Compute the Jacobian $J = \pd{(x, y, z)}{(X, Y, Z)}$ and verify Euler’s identity (3.1.1). Is the flow incompressible?

Hint.

The natural next step is to compute $I(t)$ and $\de{I}/\de{t}\text{.}$ Since $V(t)$ is a box with known dimensions, $I(t)$ can be found by integrating $f$ directly over $V(t)\text{.}$ Alternatively, map back to the fixed cube $V(0)$ using the Jacobian from part (b), exactly as in the proof of Theorem 3.1.3. To verify Reynolds’ Transport Theorem, compute $\nabla \cdot (f\bu)$ (note that $\pd{f}{t} = 0$) and integrate it over $V(t)\text{;}$ the two answers should agree. You will carry this out in Exercise 3.6.7.

Solution.

(a) The pathlines satisfy $\dot{x} = x\text{,}$ $\dot{y} = -y/2\text{,}$ $\dot{z} = 0$ with $(x, y, z) = (X, Y, Z)$ at $t = 0\text{,}$ so

\begin{equation*} x = X\e^{t}, \qquad y = Y\e^{-t/2}, \qquad z = Z. \end{equation*}

Each coordinate is mapped independently of the others, so the faces of the cube remain planes and

\begin{equation*} V(t) = [0, \, \e^{t}] \times [0, \, \e^{-t/2}] \times [0, \, 1]. \end{equation*}

The cube is stretched in $x$ and squashed in $y\text{,}$ as shown in Figure 3.2.

Three overlapping rectangles showing the dyed volume stretching in x and shrinking in y over time. — Figure 3.2. The material volume $V(t)$ at $t = 0, 0.5, 1\text{;}$ the dotted line is the pathline of the corner initially at $(1, 1, 0)\text{.}$

(b) The Jacobian matrix is diagonal, so

\begin{equation*} J = \begin{vmatrix} \e^{t} & 0 & 0 \\ 0 & \e^{-t/2} & 0 \\ 0 & 0 & 1 \end{vmatrix} = \e^{t/2}. \end{equation*}

Since $J$ does not depend on $\bX\text{,}$ we have $\DD{J}{t} = \dd{J}{t} = \tfrac12 \e^{t/2}\text{.}$ Meanwhile, $\nabla \cdot \bu = 1 - \tfrac12 = \tfrac12\text{,}$ so

\begin{equation*} J \nabla \cdot \bu = \tfrac12 \e^{t/2} = \DD{J}{t}, \end{equation*}

as required by Euler’s identity Theorem 3.1.2. The flow is not incompressible: the volume of the box is $|V(t)| = J|V(0)| = \e^{t/2}\text{,}$ which grows in time.

Remark 3.1.6 A 1D version of Reynolds’ Transport Theorem.

In 1D, Reynolds’ Transport Theorem reduces to an identity known as Leibniz’s rule. This is presented as an exercise in Exercise 3.6.2.

Remark 3.1.7 RTT and conserved quantities.

Reynolds’ Transport Theorem is particularly useful when the integral quantity

\begin{equation*} B = \iiint_{V(t)} f \, \de{x} \, \de{y} \, \de{z} \end{equation*}

is a conserved quantity. A conserved quantity is something for which we can apply a principle of conservation. In practice for this course, this is usually mass or linear momentum:

Mass: $f=\rho$ and the left-hand side of (3.1.2) is zero because mass is conserved.
Linear momentum: $\bf=\rho\bu$ is now a vector quantity and, by Newton’s second law, the left-hand side of (3.1.2) equals the sum of all the forces acting on the fluid in the material volume $V(t)\text{,}$ which are:
- Surface forces: Forces acting on the surfaces of the fluid, which could be reaction forces from the walls of a container, or forces due to pressure or stress from surrounding fluid. (Note that we will cover stress in Chapter 6.)
- Body forces: Forces acting over the interior of the fluid, and in this course you will usually only need to consider the force of gravity, although in principle other body forces could act such as an electromagnetic force. In addition, if working in a non-inertial frame of reference, other apparent forces need to be added (forces due to linear acceleration, angular acceleration and the Coriolis and centrifugal forces).
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