I am also reading the same book as the OP in this question - Material derivative of a material vector field namely - An Introduction To Theoretical Fluid Dynamics
and I have a couple of questions on the methodology of the derivation
Let me start with the first set of equations (1). I understand $$\dfrac{\mathbf{Dv}}{\mathbf{D}t} = \dfrac{\text{d}\mathbf{v}(\mathbf{x}(t),t)}{\text{d}t}= \dfrac{\text{d}\mathbf{v}(\mathbf{a},t)}{\text{d}t}\tag{1}$$
That is just the total derivative or the material derivative.
How is $$\dfrac{\text{d}\mathbf{v}(\mathbf{a},t)}{\text{d}t}=\dfrac{\partial\mathbf{v}(\mathbf{a},t)}{\partial t}$$
This would mean that the advection term of the material derivative is zero or is there something I am missing ?
I do not really understand how if $v$ is a material vector field it has to satisfy (2) because according to me the only way this would be true if the material derivative or the total derivative is zero.
$$\dfrac{\partial\mathbf{v}(\mathbf{a},t)}{\partial t} = \mathbf{v}\cdot\nabla \mathbf{u}\tag{2}$$
Notations and Context
$\mathbf a$ — a vector, that is...
$\mathbf x$ —
(will act as a variable, p. 4)
(will act as a constant, p. 4)
The Confusion
\begin{align} \dfrac{\mathbf{Dv}}{\mathbf Dt} &= \dfrac{\mathbb d\mathbf v(\mathbf x(t),t)}{\mathbb dt} = \dfrac{\color{blue}{\partial}\mathbf v(\color{blue}{\mathbf x(t)},t)}{\color{blue}{\partial x_i}}\dfrac{\mathbb dx_i}{\mathbb dt} + \dfrac{\color{green}{\partial}\mathbf v(\mathbf x(t),\color{green}{t})}{\color{green}{\partial t}} = \mathbf u\cdot\color{blue}{\nabla}\mathbf v + \dfrac{\color{green}{\partial}\mathbf v}{\color{green}{\partial t}}\tag{3}\\ &= \dfrac{\mathbb d\mathbf v(\mathbf a,t)}{\mathbb dt} = \dfrac{\partial\mathbf v(\mathbf a,t)}{\partial t} \color{red}{\ne} \dfrac{\partial\mathbf v(\mathbf x(t),t)}{\partial t} = \dfrac{\partial\mathbf v}{\partial t}. \tag{4} \end{align}
So, what equation $(1)$ means is that the two perspectives below are equivalent in describing a material vector field:
Now, let's see why $\dfrac{\text{d}\mathbf{v}(\mathbf{a},t)}{\text{d}t}=\dfrac{\partial\mathbf{v}(\mathbf{a},t)}{\partial t}$.
In the situation where the spatial dimension is $n$, We view $\mathbf v(\mathbf x,t)$ as a vector function of $n+1$ variables. Therefore, $\mathbf v:\mathbb R^{n+1}\to\mathbb R^n$ which maps $(\mathbf x,t)\mapsto\text{some vector}$. Therefore, $\mathbf v(\mathbf a,t)$ will fix the first $n$ variables. Imagine if $n=3$, then we have $\mathbf v(x,y,z,t)$. Then imagine if the displacement is fixed by a constant $\mathbf a=(a,b,c)$ to get $\mathbf v(a,b,c,t)$.
Then, what is $\dfrac{\mathbb d\mathbf v(a,b,c,t)}{\mathbb dt}$? This total derivative should be $$ \dfrac{\mathbb da}{\mathbb dt}\dfrac{\partial\mathbf v(a,b,c,t)}{\partial x} + \dfrac{\mathbb db}{\mathbb dt}\dfrac{\partial\mathbf v(a,b,c,t)}{\partial y} + \dfrac{\mathbb dc}{\mathbb dt}\dfrac{\partial\mathbf v(a,b,c,t)}{\partial z} + \dfrac{\mathbb dt}{\mathbb dt}\dfrac{\partial\mathbf v(a,b,c,t)}{\partial t}, \tag{5} $$ and you would recognize that $\dfrac{\mathbb da}{\mathbb dt}=\dfrac{\mathbb db}{\mathbb dt}=\dfrac{\mathbb dc}{\mathbb dt}=0$. On the other hand, see why this inequality is true: $$\dfrac{\mathbb d\mathbf v(a,b,c,t)}{\mathbb dt} \ne \dfrac{\mathbb d\mathbf v(x,y,z,t)}{\mathbb dt}.\tag{6}$$
It's because $\mathbf x=(x(t),y(t),z(t))$ and all $x$, $y$ and $z$ are dependent on $t$, unlike $\mathbf a=(a, b, c)$.
This observation generalizes to all spatial dimensions.
About Equation $(2)$
The process of deriving Equation $(2)$ is on page 12 of the book. Related equations are $(1.21)$ and $(1.22)$ which starts off from the partial derivative of the definition of the material vector field $$\mathbf v = (\mathbf J)_{ij}(\mathbf V(\mathbf a))_j,$$ which will eventually lead to $$\dfrac{\partial\mathbf v(\mathbf a,t)}{\partial t} = \mathbf v\cdot\nabla\mathbf u.$$
A few relations being used here are:
$\dfrac{\partial(\mathbf J)_{ij}}{\partial t} = \dfrac{\partial}{\partial t}\dfrac{\partial\mathbf x_i}{\partial\mathbf a_j} = \dfrac{\partial\mathbf u_i}{\partial\mathbf a_j} = \dfrac{\partial\mathbf u_i}{\partial\mathbf x_k} \dfrac{\partial\mathbf x_k}{\partial\mathbf a_j}$,
$\dfrac{\partial\mathbf x_k}{\partial\mathbf a_j} \mathbf V_j = \mathbf v_k$.
See if you can bridge these gaps, comment to let me know if this helps or not.