On page 12 of An Introduction to Theoretical Fluid Dynamics, following the introduction of a material vector field $v_i(\mathbf a,t)=J_{ij}(\mathbf a,t)V_j(\mathbf a)$ the author wrote:
$$
\frac{\mathrm D \mathbf v}{\mathrm D t}
=
\left. \frac{\partial \mathbf v}{\partial t} \right| _ {\mathbf x}
+ \mathbf u \cdot \nabla \mathbf v
– \mathbf v \cdot \nabla \mathbf u
\equiv
v_t+\mathcal L_{\mathbf u} \mathbf v
= 0
$$
Question: Shouldn't the material derivative of $\mathbf v$ be the following? Where is the "extra" term with the negative sign from?
$$
\frac{\mathrm D \mathbf v}{\mathrm D t}
=
\left. \frac{\partial \mathbf v}{\partial t} \right| _ {\mathbf x}
+ \mathbf u \cdot \nabla \mathbf v
$$
Update: I believe it has something to do with Eqn. (1.22) which states that
$$
\left. \frac{\partial \mathbf v}{\partial t} \right |_{\mathbf a} =
\mathbf v\cdot\nabla\mathbf u
$$
Best Answer
For some clarity the author has made the following calculation (I will explicitly give the variables that $\mathbf{v}$ depends on in each equation to avoid confusion) $$\dfrac{\mathbf{Dv}}{\mathbf{D}t} = \dfrac{\text{d}\mathbf{v}(\mathbf{x}(t),t)}{\text{d}t}= \dfrac{\text{d}\mathbf{v}(\mathbf{a},t)}{\text{d}t} = \dfrac{\partial\mathbf{v}(\mathbf{a},t)}{\partial t}\tag{1}$$ But since $\mathbf{v}$ is a material vector field it satisfies the following differential equation $$\dfrac{\partial\mathbf{v}(\mathbf{a},t)}{\partial t} = \mathbf{v}\cdot\nabla \mathbf{u}\tag{2}$$ By definition the material derivative (as it is really just a total derivative) is $$\dfrac{\mathbf{Dv}}{\mathbf{D}t} = \dfrac{\partial\mathbf{v}(\mathbf{x}(t),t)}{\partial t} + \mathbf{u}\cdot\nabla\mathbf{v}(\mathbf{x}(t),t)\tag{3}$$ Then equation $(1)$ implies that $$\dfrac{\mathbf{Dv}}{\mathbf{D}t}-\left.\dfrac{\partial\mathbf{v}}{\partial t}\right\vert_{\mathbf{a}} = 0$$ Rewriting this using equations $(2)$ and $(3)$ we arrive at $$\left.\dfrac{\partial\mathbf{v}}{\partial t}\right\vert_{\mathbf{x}} + \mathbf{u}\cdot\nabla\mathbf{v}-\mathbf{v}\cdot\nabla \mathbf{u}=0$$ The fancy expression you see in this equation has a name, the lie derivative $$\mathcal{L}_\mathbf{u}\mathbf{v} = \mathbf{u}\cdot\nabla\mathbf{v}-\mathbf{v}\cdot\nabla \mathbf{u}$$ So finally we have the result $$\mathbf{v}_t + \mathcal{L}_\mathbf{u}\mathbf{v} = 0$$