The Leibniz integral rule, in its multivariate form, deals with differentiation of the following sort: $$ \frac{\partial}{\partial t} \int_{D(t)} F({\bf x}, t) \, d{\bf x} \, , \qquad D(t)\in \mathbb{R}^d \, .$$

I am looking for a fully rigorous formulation of this theorem, as well as a proper proof. So far, I could only find:

  1. The one-dimensional case (see e.g., Courant calculus book).
  2. Physics-flavored proofs, where the normal speed $v_n ({\bf x})$ is not well defined (see e.g., here). While insightful, this is not what I need.
  3. Differential-geometry and form-based proofs (see the above paper). While these are valid, they are far more general than what I need.

My question: Are you familiar with a multivariate-calculus version and proof of this theorem?

I searched most of the textbooks mentioned here and here, but still no luck. Any ideas? I believe it is not usually covered in graduate curses of advanced multivariate analysis, and so it is pretty difficult to find it in the textbooks.

(This is cross posted from math.se, since a week has passed and it recieved little-to-no-answer. Also, this is not to be confused with the Leibniz rule $(fg)'=f'g +fg'$, as in this post.)

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    $\begingroup$ Why do you consider proofs using differential forms far more general than what you need? Or put differently, what is the level you need? $\endgroup$ – Michael Bächtold May 16 '18 at 13:05
  • $\begingroup$ There is a proof of this in the appendix to the book of Evans in PDEs, or have you already checked that? $\endgroup$ – Giuseppe Negro May 16 '18 at 13:53
  • $\begingroup$ @MichaelBächtold You're on point here, but I wonder if there is a more "elementary" way to do it, given that (a) the differential forms gives much more (b) At least naivly, the theorem can be stated in an elementary way.. $\endgroup$ – Amir Sagiv May 16 '18 at 14:44
  • $\begingroup$ @GiuseppeNegro Thanks! I'm looking at it now, but can't find the theorem. Do you remember where is it? $\endgroup$ – Amir Sagiv May 16 '18 at 14:45
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    $\begingroup$ It's §C.4, pag.713 of the second edition, but now that I see it I think that's not what you are looking for, as there's no proof. Sorry. I wrote a proof on my personal notes, based on the derivative of the determinant formula (search "Reynolds transport theorem", towards the end of the page). However, these notes are not meant to be read by anyone except myself. I hope this helps. $\endgroup$ – Giuseppe Negro May 16 '18 at 17:16

This is probably just a more streamlined version of the proof in the article you linked to, mentioning somewhat less differential forms. I don't know a more elementary proof, but hope this is of use anyway.

I'll write $D_t \subset \mathbb{R}^d$ for the domain of integration and I'll assume that the map $t\mapsto D_t$ is sufficiently regular in the sense that for sufficiently short times, the motion of the domain can be described by the flow of an ambient vector field $X$ on $\mathbb{R}^d$. I.e. for each $t\in \mathbb{R}$ there should exist a vector field $X$ on $\mathbb{R}^d$ with flow $\Phi_h:\mathbb{R}^d \to \mathbb{R}^d$ such that $$ D_{t+h}=\Phi_{h}(D_{t}) \qquad \forall h \in (-\epsilon,\epsilon). $$

(The field $X$ is not unique, one may add any vector field that's tangent to the boundary of $D_t$.)

With such a vector field $X$ given we compute \begin{align} \frac{d}{dt} \int_{D_t}F(\mathbf{x},t)d\mathbf{x} &= \lim_{h\to 0}\frac{1}{h}\left( \int_{D_{t+h}}F(\mathbf{x},t+h)d\mathbf{x} - \int_{D_{t}}F(\mathbf{x},t)d\mathbf{x}\right)\\ &=\lim_{h\to 0}\frac{1}{h}\left( \int_{\Phi_h(D_{t})}F(\mathbf{x},t+h)d\mathbf{x} - \int_{D_{t}}F(\mathbf{x},t)d\mathbf{x}\right)\\ &=\lim_{h\to 0}\frac{1}{h}\left( \int_{D_{t}}\Phi^*_h(F(\mathbf{x},t+h)d\mathbf{x}) - \int_{D_{t}}F(\mathbf{x},t)d\mathbf{x}\right)\\ \end{align}

(The previous equality is just the substitution rule for integrals in several variables. So $\Phi^*_h\left(F(\mathbf{x},t+h)d\mathbf{x}\right)$ denotes the pullback of the differential form $F(\mathbf{x},t+h)d\mathbf{x}$ along $\Phi_h$. It's what you get by replacing every $\mathbf{x}$ with $\Phi_h(\mathbf{x})$ in the expression $F(\mathbf{x},t+h)d\mathbf{x}$. We continue by swapping limit and integral)

\begin{align} &=\int_{D_{t}}\lim_{h\to 0} \frac{\Phi^*_h(F(\mathbf{x},t+h)d\mathbf{x}) - F(\mathbf{x},t)d\mathbf{x}}{h} \end{align}

(To go on use Taylor $F(\mathbf{x},t+h)=F(\mathbf{x},t)+\frac{\partial F(\mathbf{x},t)}{\partial t}\cdot h+O(h^2)$ as well as linearity of $\Phi^*$ and $\Phi^*_0=\text{id}_{\mathbb{R}^n}$.) $$ =\int_{D_{t}} \left(\frac{\partial F(\mathbf{x},t)}{\partial t}d\mathbf{x} +\lim_{h\to 0}\frac{\Phi^*_h(F(\mathbf{x},t)d\mathbf{x}) - F(\mathbf{x},t)d\mathbf{x}}{h}\right). $$ Now that limit in the integral is per definition $L_X(F(\mathbf{x},t)d\mathbf{x})$, the Lie derivative of the form $\omega=F(\mathbf{x},t)d\mathbf{x}$ with respect to $X$. Using Cartan's magic formula $L_x \omega = d(i_X \omega)+i_X(d\omega)$, which in our case simplifies to $L_x \omega = d(i_X \omega)$ since $F(\mathbf{x},t)d\mathbf{x}$ is of top degree, we get $$ =\int_{D_{t}}\frac{\partial F(\mathbf{x},t)}{\partial t}d\mathbf{x} +\int_{D_{t}} d \left(i_X(F(\mathbf{x},t)d\mathbf{x})\right) $$ Applying Stokes theorem to the second integral we finally arrive at $$ \frac{d}{dt} \int_{D_t}F(\mathbf{x},t)d\mathbf{x} = \int_{D_{t}}\frac{\partial F(\mathbf{x},t)}{\partial t}d\mathbf{x} +\int_{\partial D_{t}} i_X(F(\mathbf{x},t)d\mathbf{x}). $$ For those who want to get rid of that insertion in the second integral, a direct computation shows that $\int_\Gamma i_X( f(\mathbf{x})d\mathbf{x})=\int_\Gamma f(\mathbf{x}) X\cdot d\vec{A}$ where $d\vec{A}$ denotes the usual "surface element" with respect to the standard metric. So you can write the Leibniz integral rule as $$ \frac{d}{dt} \int_{D_t}F(\mathbf{x},t)d\mathbf{x} = \int_{D_{t}}\frac{\partial F(\mathbf{x},t)}{\partial t}d\mathbf{x} +\int_{\partial D_{t}} F(\mathbf{x},t)\, X\cdot d\vec{A}. $$

  • $\begingroup$ Thanks! A couple of questions : (a) In the last paragraph, shouldn't the second integral be on $\partial \Gamma$? This is some sort of a Green theorem, right? (b) I get why $F({\bf x},t)d{\bf x}$ is of top rank, but why does that imply that the $i_X(d\omega)$ part vanishes? $\endgroup$ – Amir Sagiv May 17 '18 at 11:14
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    $\begingroup$ (a) no, $\Gamma$ was already supposed to be a codimension 1 surface in $\mathbb{R}^d$. The insertion of a field reduces the order of a differential form by 1, so that integrand is really a $d-1$ form. (b) The differential increases the degree of a form, so if $\omega$ is of top degree then $d\omega=0$ since the only $d+1$ form on $\mathbb{R}^d$ is zero. Btw. one can probably avoid those steps with the insertion of fields and the Cartan rule, and use the divergence of a field and Gauss theorem instead. Probably that is how it's done in the notes of Giuseppe Negro. $\endgroup$ – Michael Bächtold May 17 '18 at 12:19
  • $\begingroup$ Did we really use the fact that $D_t$ are smooth? If they were only piecewise smooth, say squares of side-length $t$, would the result be any different? $\endgroup$ – Amir Sagiv May 22 '18 at 6:03
  • $\begingroup$ No, I think I didn't use the fact that the set $D_t$ has smooth boundary. What I used is that the motion of $D_t$ is smooth, or put differently: that the dependence of $D_t $ on $t$ is smooth (or maybe $C^1$ is enough). I formalized this with the assumption that for any $t\in\mathbb{R}$ there exist an $\epsilon$ and a $C^1$ (?) ambient vector field such that $D_{t+h}=\Phi_h(D_t)$. $\endgroup$ – Michael Bächtold May 22 '18 at 6:53
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    $\begingroup$ Your example of a square with side length $t$ (say centered around the origin) should satisfy this condition, since the homotethy $X=-x\partial_x -y\partial_y$ is a vector field satisfying the condition. Maybe you are worry if Stokes is valid for non smooth boundaries. Here are some references $\endgroup$ – Michael Bächtold May 22 '18 at 8:29

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