Is the following simplification correct in Vector calculus

Question

In vector calculus is $(A \cdot \nabla)A$ equivalent to $A \cdot \nabla A$?

In other words is finding the dot product of $A$ with the gradient of $A$ the same as the term in brackets?

It seems correct but I'm not certain

Answer 1

How do you define $$A\cdot\nabla A,$$ for it doesn't make sense if $A$ is a scalar field. If not, exactly then what do you mean by the gradient of a vector field?

Answer 2

Using Einstein summation notation, $$((A\cdot \nabla )A)_j = A_i\partial_i A_j$$ If like me, you write the Jacobian matrix $\nabla A$ as $$ (\nabla A)_{ij} =\partial_jA_i = \begin{bmatrix}(\nabla A_1)^T \\ (\nabla A_2)^T \\\vdots\\ (\nabla A_N)^T \end{bmatrix}$$ Then if $A\cdot M$ is the contraction $(A\cdot M)_j := A_i M_{ij}$, then $$(A\cdot \nabla A)_j = A_i(\nabla A)_{ij}=A_i\partial_jA_i ≠ ((A\cdot\nabla)A)_j $$ Instead we have $$ (A\cdot \nabla) A=\nabla A\cdot A=(\nabla A)^TA$$ Compare with chain rule. Indeed, this term should pop out when computing the $t$ derivative of $A(X(t))$, where $X'=A(X)$.

A last comment, if you were to compute $A\cdot \nabla f$ where $f$ is a scalar, then $A\cdot \nabla f = (A\cdot \nabla )f$.