Differentiation under the integral sign for volumes in higher dimensions

Question

Consider a smooth convex/compact domain $D\subset \mathbb{R}^n$ and a smooth, concave function $F:D\to \mathbb{R}$. Then we can define the function that simply takes the volume of the upper contour sets determined by the argument:

$$G(t) = \int_{\{x\in D \; : \; F(x) \ge t\}} d\lambda$$

where $\lambda$ denotes the Lebesgue measure. I'm trying to figure out an expression for $\frac{d}{dt}G(t)$.

This seems like nothing more than a special case of a higher-dimensional Leibniz Integral Rule, but wikipedia gives me a substantially more general formula than I suspect I need for this case (for definitions of terms see the link):

$$\frac{d}{dt} \int_{\Omega(t)} \omega = \int_{\Omega(t)} i_{\vec{v}}(d_x \omega) + \int_{\partial \Omega(t)} i_{\vec{v}}\omega + \int_{\Omega(t)} \dot{\omega}.$$

I have almost no background in differential forms, but immediately I know, for starters, the volume form I'm integrating is time invariant so the last term drops out here. Moreover, given I'm just concerned with a uniform density, I'd imagine the first term should be zero too? (This corresponding to the intuition that all that really matters here is how much 'volume bleeds out of the bag $\Omega(t)$' as I cinch it shut by increasing $t$, and hence I need only be concerned with the incremental flow of volume across the boundary.) But that may be wildly incorrect.

Ideally if someone could help guide me (ideally both intuitively and analytically) to be able to understand and describe this derivative I'd be very grateful! In particular an expression for what the Leibniz rule reduces to in this case would be most welcome.

Answer 1

If you're just considering volumes in $\mathbb R^n$, there's no need to get differential forms involved - just use the "Reynolds transport theorem" from that same Wikipedia page, which in this case just gives the boundary integral

$$\frac d{dt}G(t) = \int_{\{ F = t\}} \mathbf{v}\cdot \mathbf{\nu}\ dA$$

where $\mathbf{v} = \nabla F / |\nabla F|^2$ is the velocity vector field of the boundary, $\mathbf{\nu}$ is the outwards unit normal and $dA$ is the hyperarea element on $\{F =t\} = \partial \{F\le t\}$. So we have two things to justify: why is this formula true, and why is that the correct expression for the velocity?

The intuition behind the integral formula is pretty simple: if the boundary moves in the outwards normal direction with speed $s$, then over a time interval $dt$, part of the boundary with hyperarea $dA$ will sweep out the extra volume $s\ dt\ dA$; so the whole boundary will sweep out the extra volume $dG = dt \int s\ dA$. Since moving the boundary with velocity $\bf v$ is equivalent to moving it normally with velocity $\bf v \cdot \nu$, we get $dG/dt = \int{\bf v \cdot \nu}\ dA$ as desired.

To justify this rigorously is somewhat difficult - in this smooth case I think we need to introduce the idea of flows and Lie derivatives, which I won't try to do here. More generally we could cite something like the co-area formula with $u=F$, $g=|\nabla F|^{-1}$.

For the velocity expression, note that there is some freedom in how we choose $\bf v$, since we could reparametrize the boundary as we move through time. (What remains invariant is the normal velocity $\bf v \cdot \nu$.) Thus we just need to find some path $\mathbf{x}(t)$ that stays in $\{F = t\}$, and the velocity of this path will give us a velocity of the boundary. The guess $\mathbf{v}=\nabla F / |\nabla F|^2$ should be somewhat intuitive - greater $|\nabla F|$ means $F$ has a steeper slope, so you have to move slower in order to achieve the same rate of change of $F$. If $\mathbf{x}'(t)=\nabla F/|\nabla F|^2$ then from the chain rule we can compute $\frac d{dt}F(\mathbf{x}(t)) = \nabla F \cdot \mathbf{x}'(t)=1$, so if $F(\mathbf{x}(t))=t$ then $F(\mathbf{x}(s))= t+\int_t^{s} 1 = s$ for all later times $s$; i.e. $\mathbf{x}(t)$ stays on the boundary as desired.