Tensor Product between Nabla operator and a vector or tensor field I am recently studying solid mechanics and I met a problem regarding Nabla operator. I am trying to prove the following relation:
$$\nabla \otimes\textbf{u}=\frac{\partial\textbf{u}}{\partial x_{i}} \otimes \mathbf{e}_{i} \tag{1}$$
where $\nabla$ is the Nabla operator and $(\bullet)$ represents a smooth vector or tensor field. It is defined as follows,
$$
\nabla(\bullet) = \frac{\partial(\bullet)}{\partial x_{j}}\textbf{e}_j
$$
and $\textbf{u}$ denotes a vector field and we have $\textbf{u}=u_i\textbf{e}_i$ and $\textbf{e}_i(i=1,2,3)$ is a set of orthonomal Cartesian basis vectors.
To prove equation(1), I compute both sides of the equation and get
$$
LHS=\nabla \otimes\textbf{u} = \frac{\partial(\bullet)}{\partial x_{j}}\textbf{e}_j\otimes u_i\textbf{e}_i = \frac{\partial u_i}{\partial x_{j}} \textbf{e}_j\otimes \textbf{e}_i
$$
and
$$
RHS=\frac{\partial \textbf{u}}{\partial x_{i}} \otimes \mathbf{e}_{i} = \frac{\partial u_j}{\partial x_{i}}\textbf{e}_j\otimes\textbf{e}_i
$$
It seems that $LHS \neq RHS$. I am wondering if there is any step where a mistake happens. Can you help me?
 A: You can not prove that relation by applying a tensor product between the nabla operator and a tensor or vector field, because the nabla operator is not a vector or tensor field. You must start the proof by calculating the directional derivative of $\mathbf{u}(\mathbf{x})$ in an arbitrary direction $\mathbf{d}$:
\begin{align*}
(\nabla\otimes\mathbf{u})\cdot\mathbf{d}&=\lim_{a\to 0}\frac{\text{d}}{\text{d}a}\mathbf{u}(\mathbf{x}+a\mathbf{d})\\
&=\lim_{a\to 0}\frac{\partial\mathbf{u}}{\partial{y_{i}}}\frac{\partial y_{i}}{\partial a}\quad\text{where }y_{i}\equiv x_{i}+ad_{i}\\
&=\lim_{a\to 0}\frac{\partial\mathbf{u}}{\partial{y_{i}}}d_{i}\\
&=\frac{\partial\mathbf{u}}{\partial{x_{i}}}d_{i}\\
&=\frac{\partial\mathbf{u}}{\partial{x_{i}}}(\mathbf{e}_{i}\cdot\mathbf{d})\\
&=\left(\frac{\partial\mathbf{u}}{\partial{x_{i}}}\otimes\mathbf{e}_{i}\right)\cdot\mathbf{d}
\end{align*}
Therefore $$\nabla\otimes\mathbf{u}=\frac{\partial\mathbf{u}}{\partial{x_{i}}}\otimes\mathbf{e}_{i}$$
Note that $\mathbf{u}$ may be a smooth vector or tensor field.
