Material Derivative
Definition 1
$$ \mathbf{u} = \mathbf{u} \left( t ; \mathbf{x} \right) = \left( u_{1} \left( t ; \mathbf{x} \right) , u_{2} \left( t ; \mathbf{x} \right) , u_{3} \left( t ; \mathbf{x} \right) \right) $$ In particular, in three-dimensional space let the viewpoint $t$ and the spatial coordinates $\mathbf{x} = \left( x_{1} , x_{2} , x_{3} \right)$’s velocity vector be represented as above.
$$ {\frac{ D }{ D t }} = {\frac{ \partial }{ \partial t }} + u_{1} {\frac{ \partial }{ \partial x_{1} }} + u_{2} {\frac{ \partial }{ \partial x_{2} }} + u_{3} {\frac{ \partial }{ \partial x_{3} }} $$
The differential operator $D$, which is expressed as the sum of the time derivative term and the divergence term as above, is called the material derivative. For the vector form $\mathbf{u}$ it is written as follows. $$ {\frac{ D \mathbf{u} }{ D t }} = {\frac{ \partial \mathbf{u} }{ \partial t }} + \left( \mathbf{u} \cdot \nabla \right) \mathbf{u} $$
Explanation 1
If one expands the material derivative written in vector form into each component, one obtains the following. $$ \begin{align*} {\frac{ D u_{1} }{ D t }} =& {\frac{ \partial u_{1} }{ \partial t }} + \left( \mathbf{u} \cdot \nabla \right) u_{1} \\ {\frac{ D u_{2} }{ D t }} =& {\frac{ \partial u_{2} }{ \partial t }} + \left( \mathbf{u} \cdot \nabla \right) u_{2} \\ {\frac{ D u_{3} }{ D t }} =& {\frac{ \partial u_{3} }{ \partial t }} + \left( \mathbf{u} \cdot \nabla \right) u_{3} \end{align*} $$ Finally, to aid understanding once more, written component-wise it reads as follows. $$ \begin{align*} {\frac{ D u_{1} }{ D t }} =& {\frac{ \partial u_{1} }{ \partial t }} + u_{1} {\frac{ \partial u_{1} }{ \partial x_{1} }} + u_{2} {\frac{ \partial u_{1} }{ \partial x_{2} }} + u_{3} {\frac{ \partial u_{1} }{ \partial x_{3} }} \\ {\frac{ D u_{2} }{ D t }} =& {\frac{ \partial u_{2} }{ \partial t }} + u_{1} {\frac{ \partial u_{2} }{ \partial x_{1} }} + u_{2} {\frac{ \partial u_{2} }{ \partial x_{2} }} + u_{3} {\frac{ \partial u_{2} }{ \partial x_{3} }} \\ {\frac{ D u_{3} }{ D t }} =& {\frac{ \partial u_{3} }{ \partial t }} + u_{1} {\frac{ \partial u_{3} }{ \partial x_{1} }} + u_{2} {\frac{ \partial u_{3} }{ \partial x_{2} }} + u_{3} {\frac{ \partial u_{3} }{ \partial x_{3} }} \end{align*} $$
Combination of Eulerian and Lagrangian descriptions
Returning to the concise form, the right-hand side of the material derivative contains two types of terms as follows. $$ {\frac{ D \mathbf{u} }{ D t }} = {\color{red} {\frac{ \partial \mathbf{u} }{ \partial t }}} + {\color{blue} \left( \mathbf{u} \cdot \nabla \right) \mathbf{u}} $$ Here the first red $\color{red} \partial \mathbf{u} / \partial t$ may be called local acceleration or more simply the inertial term. The second blue $\color{blue} \left( \mathbf{u} \cdot \nabla \right) \mathbf{u}$ is called convective acceleration or more simply the convective term.
Eulerian and Lagrangian descriptions: In fluid dynamics, because a fluid has no fixed shape and its detailed state of motion is hard to know, one imagines a fluid particle. There are two kinds of descriptions for describing the motion of a fluid particle. $$ {\frac{ \partial u }{ \partial t }} $$ The Eulerian description observes the state of the fluid at every point while keeping a fixed location in time. $$ {\frac{ \partial u }{ \partial x }} , {\frac{ \partial u }{ \partial y }} , {\frac{ \partial u }{ \partial z }} $$ The Lagrangian description follows individual fluid particles and observes their state of motion.
The inertial term originates from the Eulerian description, and the convective term originates from the Lagrangian description. As such, both the names of the terms and the material derivative itself inevitably appear throughout fluid mechanics.
$$ u_{t} + u u_{x} = 0 $$ For example, the inviscid Burgers equation can be written more compactly using the material derivative as follows. $$ {\frac{ D u }{ D t }} = 0 $$
Derivation
As those familiar with vector calculus will readily see, the material derivative is merely a special case of the total derivative. Let us look at the derivation directly to understand it.
Chain rule for multivariable vector functions: Let two functions $\mathbf{g} : D \subset \mathbb{R}^{m} \to \mathbb{R}^{k}$ and $\mathbf{f} : \mathbf{g}(\mathbb{R}^{k}) \subset \mathbb{R}^{k} \to \mathbb{R}^{n}$ be differentiable. Then their composition $\mathbf{F} = \mathbf{f} \circ \mathbf{g} : \mathbb{R}^{m} \to \mathbb{R}^{n}$ is also differentiable, and the (total) derivative of $\mathbf{F}$ satisfies the following. $$ \mathbf{F}^{\prime}(\mathbf{x}) = \mathbf{f}^{\prime}\left( \mathbf{g}(\mathbf{x}) \right) \mathbf{g}^{\prime}(\mathbf{x}) $$
Since $\partial t / \partial t = 1$ and $k = 1,2,3$ imply $u_{k} = d x_{k} / dt$, the chain rule for vector functions yields the material derivative.