Consider the time dependent vector field F related to some vector field B by F(x, $t)=$B(tx). I am asked to prove $\displaystyle(x \cdot \nabla){\bf F}=t\frac{\partial {\bf F}}{\partial t}$.
I'm not sure how F being related to B helps proving the result which involves only F. Can anyone please point me in the right direction?
Consider
\begin{eqnarray} ({\bf x}\cdot \nabla){\bf F}({\bf x}, t) &=& \sum_{k}x_k \frac{\partial}{\partial x_k}{\bf F}({\bf x}, t) = \sum_{k,l}x_k \frac{\partial}{\partial x_k}\hat{e}_l F_l({\bf x}, t) \\ &=& \sum_{k,l}x_k \hat{e}_l\frac{\partial}{\partial x_k}B_l(t{\bf x}) = \sum_{k,l}x_k \hat{e}_lB_l'(t{\bf x})~t\tag{1} \end{eqnarray}
On the other hand
\begin{eqnarray} \frac{\partial }{\partial t} {\bf F}({\bf x}, t) &=&\sum_{l}\hat{e}_l \frac{\partial }{\partial t}F_l({\bf x}, t)= \sum_{l} \hat{e}_l \frac{\partial }{\partial t}B_l(t{\bf x}) \\ &=& \sum_{k,l} \hat{e}_l B_l'(t{\bf x}) x_k \tag{2} \end{eqnarray}
Which means that
$$ ({\bf x}\cdot \nabla){\bf F}({\bf x}, t) = \frac{\partial }{\partial t} {\bf F}({\bf x}, t) $$