I would like to prove the equivalence of the differential and integral forms of Reynold's transport equation. This problem is stated in terms of fluid mechanics.

Problem Statement

Let $V$ be a closed volume in $\mathbb{R}^3$, $A$ be the surface of $V$, and $\mathbf{n}$ the normal to A ($\mathbf{n}$ is thus a function of position). Let $F(\mathbf{x}, t)$ be a scalar function of three-dimensional position and time. Also let $\mathbf{u} \in \mathbb{R}^3$ be the field velocity in $\mathbb{R}^3$.

By letting $V \to 0$, derive this equality:

$$ \frac{DF}{Dt}= \frac{\partial F}{\partial t} + \frac{\partial F}{\partial x_i} u_i $$

from this one: $$ \frac{D}{Dt} \int_V F \ dV= \int_V \frac{\partial F}{\partial t}\ dV + \int_A (\mathbf{u} \cdot \mathbf{n})\ F \ dA, $$ where $D/Dt$ denotes a full derivative and double indices denote summation.

Attempt at a Solution

We first convert the integral over $A$ to one over $V$ by way of the divergence theorem:

$$ \int_A (\mathbf{u} \cdot \mathbf{n})\ F \ dA = \int_A (F \mathbf{u}) \cdot \mathbf{n} \ dA = \int_V \nabla \cdot (F\mathbf{u}) \ dV. $$ Expand the divergence: $$ \int_V \nabla \cdot (F\mathbf{u}) \ dV = \int_V \nabla F \cdot \mathbf{u} \ dV + \int_V (\nabla \cdot \mathbf{u})F \ dV. $$ Call this result (1).

Although I do not understand why it is true, the text frequently uses equalities of the form $$ \lim_{V \to 0} \ \int_V F(\mathbf{x}, t) \ dV = F(\mathbf{x}, t) $$ and so I intend to do so as well here (aside: I guess it's implicit that $V$ on the LHS shrinks to the point $\mathbf{x}$ on the RHS?).

Substituting result (1) into the integral statement and (admittedly blindly) applying the preceding rule almost gives me what I want:

$$ \frac{DF}{Dt} = \frac{\partial F}{\partial t} + \frac{\partial F}{\partial x_i} u_i + \lim_{V \to 0} \int_V (\nabla \cdot \mathbf{u})F \ dV. $$

The remaining limit is the result of the expansion of the gradient in result (1).

Reiteration of the Question

Is the limit in last equation equal to 0, and if so, why, or have I made a mistake?


The statement

$$ \lim_{V \to 0} \int_V F(\mathbf{x}) \,dV = F(\mathbf{x}) $$

is clearly false; I should have caught this error. The correct statement (statement (1)) is $$ \lim_{V \to 0}\frac{1}{V} \int_V F(\mathbf{x}) \,dV = F(\mathbf{x}). $$ Follow the original proof in the question statement until $$ \frac{d}{dt} \int_{V(t)} F \,dV = \int_{V(t)} \frac{\partial F}{\partial t}dV + \int_{V(t)} \nabla F \cdot \mathbf{u} \,dV + \int_{V(t)} F \nabla \cdot \mathbf{u} \,dV, $$ where we have used the divergence theorem and expanded the terms. We will be taking the limit as $V \to 0$; to keep the algebra under control we assume already that we can discard second-order and higher terms in the Taylor expansion of $F$ (formally we would need to write the expansion and discard the terms once we take the limit). In other words, we take $F$ to be constant on $V$. Then we can take $F$ out of the integral on the LHS, and so the equality

$$ \frac{d}{dt} \int_{V(t)} F \,dV = \frac{d}{dt} F \int_{V(t)} \,dV = V\frac{dF}{dt} + F \frac{dV}{dt} $$ will hold in the limit. Putting this in the statement above, dividing by V, and taking the limit $V \to 0$, we have $$ \frac{dF}{dt} + F \lim_{V \to 0} \frac{1}{V}\frac{dV}{dt} = \frac{\partial F}{\partial t} + u_i \frac{\partial F}{\partial t} + F\frac{\partial u_i}{\partial x_i} $$ where we have applied statement (1) to the integrals. Lastly, it is true that $$ \lim_{V \to 0} \frac{1}{V}\frac{dV}{dt} = \frac{\partial u_i}{\partial x_i}, $$ which we do not prove here (straightforward but a little tedious). Substituting this into the equality above and identifying $\frac{dF}{dt} = \frac{DF}{Dt}$ (a statement that makes me a little uncomfortable) gives the desired equality.


There were two errors in the original proof. The identity regarding the limit $V\to 0$ of the integrals was incorrect. Secondly, the time derivative cannot be taken through the integral since $V$ depends on time (the entire reason we have a transport equation at all). Correcting this error gives a second term on the LHS which ultimately cancels the unwanted term on the RHS, by way of another identity introduced here.


I think what you have really shown is that the integral version is only correct for incompressible flows, that is, the limit in that last equation must be zero to get equality, and by evaluating the limit (blindly as you say) you get $(\nabla\cdot{\bf u})F$. Thus the divergence of ${\bf u}$ must be zero to get the result.

For example, take $F=1$, and ${\bf u}={\bf x}$, and take $V$ to be the unit ball. Then it seems to me that the derivative version is true but the integral one is not.

  • $\begingroup$ Please see my comment to Mark E's response above (the problem statement is in fact correct, but I have made a mistake in the proof). $\endgroup$ – Eric Kightley Sep 30 '13 at 15:27

I believe you're missing an important piece of the problem statement. The flow must be irrotational in order for your last term there to go to zero. The flow could be irrotational if, say, the surface velocity vector is equal to the flow velocity vector. My guess is that you need an irrotational flow as part of the given to prove that they are equal.

  • $\begingroup$ The problem statement is correct as stated. I made a mistake in the application of the limit. $V$ depends on time, and so I cannot pass the limit through the time derivative. The approach apparently is to expand $F$, take it to be constant on the infinitesimal volume $V$, and then pull it out of the integral. This should, in theory, give a term on the left-hand-side that cancels the remaining limit on the right. However, I still can't work it out. I will post as soon as I get it. $\endgroup$ – Eric Kightley Sep 30 '13 at 15:25

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