Material derivative of a material vector field 
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
$$
    \left. \frac{\partial \mathbf v}{\partial t} \right |_{\mathbf a} =
    \mathbf v\cdot\nabla\mathbf u
$$

My confusion has been resolved. I was assuming that the material derivative of $\mathbf v$ should be equal to 0. 
However, by definition,
$$\frac{\mathbf D \mathbf v}{\mathbf D t} = \left. \frac{\partial \mathbf v}{\partial t} \right|_{\mathbf a}$$
Therefore Eqn. (1.23) holds.
