Two time derivatives of kinetic energy of fluid Suppose $D$ is a smooth domain, $\rho > 0$ is fluid density (constant) and $u \in C^1([0,1];D)$ is the fluid velocity.  Let $K(t) = \frac{1}{2}\int_D \rho \vert u \vert^2 dV,~0\le t \le 1,$ be the kinetic energy of the incompressible fluid.
On the one hand, by Reynolds Transport Theorem,
\begin{align}
\label{1}
\frac{d}{dt} K(t) = \int_D \rho \dot{u} \cdot u dV,
\end{align}
where $\dot{u} = \frac{\partial u}{\partial t} + (\nabla u)u$ is the material derivative.
On the other hand, by Lebesgue Dominated Convergence Theorem,
\begin{align}
\label{2}
\frac{d}{dt} K(t) = \int_D \rho \frac{\partial u}{\partial t} \cdot u dV. 
\end{align}
The first calculation is used in Equation (6.42) page 254 of Oscar Gonzalez and Andrew M. Stuart - A First Course in Continuum Mechanics, while the second one is used in Equation (1.3.2) page 7 of Charles R. Doering and J. D. Gibbon - Applied Analysis of the Navier-Stokes Equations.
Which one is correct, or am I missing something? Thank you.
 A: This really depends on the nature of the domain $D$. If the domain $D(t)$ is time-dependent where $\mathbf{u}^{(b)}(\mathbf{x})$ is the velocity of an area element at the boundary point $\mathbf{x} \in \partial D(t)$, then according to Reynolds transport theorem (proved here) and an application of the divergence theorem, we have
$$\tag{1}\frac{dK}{dt} = \int_{D(t)} \frac{\partial K}{\partial t}\, dV + \int_{ D(t)}\nabla \cdot \left(K\mathbf{u}^{(b)}\right)  \, dV \\=  \int_{D(t)} \frac{\partial K}{\partial t}\, dV + \int_{\partial D(t)}K\left(\mathbf{u}^{(b)}\cdot \mathbf{n}\right)   \, dA$$
There are three possibilities.
(i) We have an arbitrary moving domain with some prescribed boundary velocity $\mathbf{u}^{(b)}$ and (1) cannot be simplified further without more information.
(ii) We have a solid stationary boundary where $\mathbf{u}^{(b)} \cdot \mathbf{n}$ holds at every boundary point either as a consequence of the no-slip condition for a viscous fluid or the impermeability condition for an inviscid fluid.  In this case (1) reduces to
$$\tag{2}\frac{dK}{dt} = \int_{D(t)} \frac{\partial K}{\partial t}\, dV $$
(iii) The domain is a material element where $\mathbf{u}^{(b)}= \mathbf{u}$, that is
the velocity at the boundary is the fluid velocity. In this case,
$$\tag{3}\frac{dK}{dt} = \int_{D(t)} \frac{\partial K}{\partial t}\, dV + \int_{\partial D(t)}K\left(\mathbf{u}\cdot \mathbf{n}\right)   \, dA$$
Again,  the ambiguity of $\mathbf{u} \cdot \mathbf{n}$ remains.

In the above discussion nothing was said about the incompressiblity of the fluid and the condition $\nabla \cdot \mathbf{u} = 0$, because it was not necessary to derive either form of the Reynolds transport theorem shown in (1).
What you show in your first equation
$$\frac{d}{dt} K(t) = \int_D \rho \dot{u} \cdot u \, dV,$$
is a consequence of the Reynolds transport theorem given both incompressiblity and the specific form $K = \frac{1}{2} \rho |\mathbf{u}|^2$. This follows form
$$ \frac{\partial K}{\partial t} +\nabla \cdot (K\mathbf{u}) = \frac{\partial K}{\partial t}  + \underbrace{K \nabla \cdot \mathbf{u}}_{= 0} + (\mathbf{u} \cdot\nabla)K \\ = \frac{\rho}{2} \frac{\partial }{\partial t} (\mathbf{u} \cdot \mathbf{u}) + \frac{\rho}{2}(\mathbf{u} \cdot \nabla) (\mathbf{u} \cdot \mathbf{u}) = \rho \frac{\partial \mathbf{u}}{\partial t} \cdot \mathbf{u} + \rho \left[(\mathbf{u} \cdot \nabla) \mathbf{u} \right] \cdot \mathbf{u} \\ := \rho \dot{u}\cdot u$$
However, this special form does not make it any clearer why the two forms you present should be equal in all circumstances.
