I'm working on an unassessed course problem. The setup might be unnecessary for my question, but I give it in case it helps.
An incompressible fluid of constant density $\rho$ has velocity $\boldsymbol{u}(x,t)$.
For fluid occupying a closed volume $V$ bounded by a surface $S$, with outward normal unit vector $\boldsymbol{n}$, deduce that $$\frac{\text{d}}{\text{d}t}\int_V\rho u_i\text{ d}V=\int_S\sigma_{ij}n_j\text{ d}S+\int_V\rho F_i\text{ d}V,$$ where $\sigma_{ij}$ is the stress tensor, and $\boldsymbol{F}$ is the body force per unit mass.
Use Gauss' divergence theorem with $f_j=\sigma_{ij}a_i$, where $\boldsymbol{a}$ is an arbitrary constant vector, to deduce that $$\rho\frac{\text{D}u_i}{\text{D}t}=\frac{\partial\sigma_{ij}}{\partial x_j}+\rho F_i,$$ throughout the fluid.
The solution booklet contains the line,
$\rho\text{ d}V$ is the mass of a fluid element, and does not change as the element moves around, so $$\frac{\text{d}}{\text{d}t}\int_V\rho u_i\text{ d}V=\int_V\left(\frac{\text{D}u_i}{\text{D}t}\right)\rho\text{ d}V.$$
I don't follow. I would interpret the verbal statement to mean $$\frac{\text{D}}{\text{D}t}(\rho\text{ d}V)=0,$$ and I could follow the symbolic step $$\frac{\text{d}}{\text{d}t}\int_V\rho u_i\text{ d}V=\int_V\left(\frac{\text{d}u_i}{\text{d}t}\right)\rho\text{ d}V$$ but I don't get how the verbal statement implies the symbolic step given. Doesn't this require $$(\boldsymbol{u}\cdot\nabla)\boldsymbol{u}=\boldsymbol{0}?$$ and how should we know that's true?
A related question I asked a few months ago yielded a link to a more substantial answer than I can take in without giving significant time to background which is outside the scope of my module.
Best Answer
Pedantic, I know, but fluids are my area of expertise so I really can't let this go.
$\frac{\mathrm d}{\mathrm dt}$ and $\frac{\mathrm D}{\mathrm Dt}$ are not equivalent operators. The operator $\frac{\mathrm d}{\mathrm dt}$ is a single-variable operator defined for functions $\mathbf F:t\mapsto \mathbf F(t)$ only, whereas the material derivative $\frac{\mathrm D}{\mathrm Dt}$ acts on functions of a time and a space variable, $\mathbf F:(t,\boldsymbol x) \mapsto \mathbf F(t,\boldsymbol x)$. The material derivative is defined as
$$\frac{\mathrm D}{\mathrm Dt}=\partial_t+\boldsymbol u\cdot \nabla$$
The two are only "equivalent" in the case that we consider the streamlines of the vector field $\boldsymbol u$, that is, if we consider the trajectory $\boldsymbol x(t)$ of a particle in the velocity field $\boldsymbol u$, i.e, $\boldsymbol x$ satisfies $$\dot{\boldsymbol x}(t)=\frac{\mathrm d}{\mathrm dt}\boldsymbol x(t)=\boldsymbol u(t,\boldsymbol x(t))\tag{1}$$
Then it is true that
$$\left(\frac{\mathrm d}{\mathrm dt}\dot{\boldsymbol x}\right)(t)=\left(\frac{\mathrm D}{\mathrm Dt}\boldsymbol u\right)(t,\boldsymbol x(t))$$
Where the right hand side should be properly interpreted as the function $\frac{\mathrm D}{\mathrm Dt}\boldsymbol u$ evaluated at the point $(t,\boldsymbol x(t))$ in its domain, $\mathbb R_{\geq 0}\times\mathbb R^3$.
More generally, for any possibly tensor valued function $\mathbf F$, given a trajectory $\boldsymbol x(t)$ that satisfies $(1)$, then defining $\mathbf f(t)=\mathbf F(t,\boldsymbol x(t))$, it is true that $$\left(\frac{\mathrm d}{\mathrm dt}\mathbf f\right)(t)=\left(\frac{\mathrm D}{\mathrm Dt}\mathbf F\right)(t,\boldsymbol x(t))$$
Via a simple application of the chain rule. Only in this sense are the two operators "equivalent".