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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

$$ \left. \frac{\partial \mathbf v}{\partial t} \right |_{\mathbf a} = \mathbf v\cdot\nabla\mathbf u $$

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up vote 0 down vote accepted

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.

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