Differentiation under the integral sign - line integral? In the proof of Kelvin's circulation theorem it is common to do the following manipulation:
$$\newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\f}[2]{\frac{ #1}{ #2}} \newcommand{\l}[0]{\left(} \newcommand{\r}[0]{\right)} 
\f{D \Gamma}{Dt}=\f{D}{Dt} \oint_{C(t)} \vec u \cdot d\vec x$$
$$=\oint_{c(t)} \f{D\vec u}{Dt}\cdot d\vec x+\oint_{c(t)}\vec u\cdot d\l \f{D\vec x}{Dt}\r$$
I am uneasy about the validity of this expression. Please can someone explain the logic and maths behind it. 
 A: Sometimes non-rigorous derivations in physics are both expedient and illuminating, but this is definitely not the case with the linked "proof" of Kelvin's theorem.
First you can dispense with the outer material derivative and replace it with an ordinary derivative with respect to $t$.  The circulation around a closed material curve $C(t)$ -- one that is convected with the flow and contains the same fluid particles -- has no explicit dependence on the spatial coordinates. Hence,
$$\frac{\partial \Gamma}{\partial t}+ \mathbf{u}\cdot \nabla\Gamma = \frac{D\Gamma}{Dt} = \frac{d\Gamma}{dt}.$$
Consider the one-dimensional analog
$$\left(\frac{\partial}{\partial t} + u\frac{\partial}{\partial x}\right)\int_a^{C(t)} u(x',t)\, dx' = \frac{d}{dt}\int_a^{C(t)} u(x',t)\, dx',$$
Assuming all functions are sufficiently smooth we can apply Leibniz's rule to obtain
$$\frac{d}{dt}\int_a^{C(t)} u(x',t)\, dx' = \int_a^{C(t)} \frac{\partial}{\partial t}u(x',t)\, dx' + u(C(t),t) C'(t).$$
This gives you a hint that you cannot merely interchange the derivative and the integral and that some type of "boundary term" is going to appear.
In the Lagrangian description of fluid flow there is a family of smooth maps
$$\mathbf{\xi}(\cdot,t) : \mathbb{R}^3 \to \mathbb{R}^3$$
for which a fluid particle located at $\mathbf{x}_0$ at time $t =0$ is located at $\mathbf{x} = \mathbf{\xi}(\mathbf{x}_0,t)$ at time $t$. We obtain the Eulerian velocity field according to $\mathbf{u}(\mathbf{x},t) = \mathbf{u}(\mathbf{\xi}(\mathbf{x}_0,t),t) = \frac{\partial}{\partial t} \mathbf{\xi}( \mathbf{x}_0,t) = \mathbf{\xi}_t( \mathbf{x}_0,t).$
At time $t$, the closed material curve $C(t)$ can be parametrized as $C(t) = \{\mathbf{x}(s,t) : 0 \leqslant s \leqslant 1\}$ and is the image of a curve $C(0)$ under the map $\mathbb{\xi}(\cdot,t) : \mathbf{x}_0(s) \mapsto \mathbf{x}(s,t).$
The circulation is given by
$$\Gamma(t) = \oint_{C(t)} \mathbf{u}(\mathbf{x},t) \cdot d \mathbf{x}= \int_0^1 \mathbf{u}(\mathbf{x}(s,t),t) \cdot \frac{\partial}{ \partial s} \mathbf{x}(s,t)\,ds.$$
With a change of variable $\mathbb{x} = \mathbb{\xi}(\mathbb{x}_0,t),$ we can express the circulation as
$$\Gamma(t) = \int_0^1 \mathbf{u}(\mathbf{\xi}(\mathbf{x}_0(s),t),t) \cdot \frac{\partial }{ \partial s} \mathbf{\xi}(\mathbf{x}_0(s),t)\,ds.$$
Now assuming the integrand is sufficiently smooth, we can take the derivative with respect to $t$ under the integral sign and split into two integrals to obtain
$$\frac{d\Gamma}{dt} = \int_0^1 \left[\frac{\partial}{\partial t}\mathbf{u}(\mathbf{\xi}(\mathbf{x}_0(s),t),t) + \nabla \mathbf{u}(\mathbf{\xi}(\mathbf{x}_0(s),t),t) \cdot \mathbf{\xi}_t( \mathbf{x}_0(s),t)\right] \cdot \frac{\partial}{ \partial s} \mathbf{\xi}(\mathbf{x}_0(s),t)\,ds \\ + \int_0^1 \mathbf{u}(\mathbf{\xi}(\mathbf{x}_0(s),t),t) \cdot \frac{\partial}{\partial t}\frac{\partial}{ \partial s} \mathbf{\xi}(\mathbf{x}_0(s),t)\,ds.$$
Reversing the change of variables and interchanging the $t-$ and $s-$derivatives in the second integral we get
$$\frac{d\Gamma}{dt} = \oint_{C(t)} \left[\frac{\partial}{\partial t}\mathbf{u}(\mathbf{x},t) + \mathbf{u}(\mathbf{x},t) \cdot \nabla \mathbf{u}(\mathbf{x},t) \right] \cdot d \mathbf{x} + \int_0^1 \mathbf{u}(\mathbf{\xi}(\mathbf{x}_0(s),t),t) \cdot \frac{\partial }{\partial s} \mathbf{\xi}_t(\mathbf{x}_0(s),t)\,ds \\ = \oint_{C(t)} \frac{D\mathbf{u}}{Dt}\cdot d \mathbf{x} + \int_0^1 \mathbf{u}\cdot \frac{\partial}{\partial s} \mathbf{u}(\mathbf{x}(s,t),t)\,ds .$$
I prefer to leave the second integral in this comprehensible parametric form rather than introduce the cryptic notation
$$\oint_{C(t)} \mathbf{u} \cdot d \left(\frac{D\mathbf{x}}{Dt}\right)$$
