Given $A=\langle yz^2,-3xz^2, 2xyz\rangle$ and $\phi = xyz$, evaluate $A .\nabla $ and $(A \times \nabla)\phi$

Question

I know $\nabla = \langle \frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\rangle$ and $\nabla . A $ and $\nabla \times A$ makes sense to me but how do we evaluate $A.\nabla$ and $A\times \nabla $?

Answer 1

Things become clearer when we focuse on the nature of the object.

Configuring

$$ f_1: (x,y,z)\to yz^2 $$ $$ f_2: (x,y,z)\to -3xz^2 $$ $$ f_3: (x,y,z)\to 2xyz $$ $$ \phi : (x,y,z) \to xyz$$ So A is a function

$$A\in{\mathbb{R}^3}^{\mathbb{R}^3}$$

$$ A=(f_1,f_2,f_3) $$

Thus the operators are well defined

$$ A\cdot \nabla $$ $$ A\times \nabla $$

is a function of function i.e. and generally speaking an operator. Applying to $\phi$

$$ A\cdot \nabla(\phi)=f_1 \partial_1\phi + f_2 \partial_2\phi + f_3 \partial_3\phi $$ $$ (A \times \nabla)(\phi)=<(f_2 \partial_3 - f_3 \partial_2)\phi \ ; \ (f_3 \partial_1 - f_1 \partial_3)\phi \ ; \ (f_2 \partial_1 - f_1 \partial_2)\phi > $$

There you go !

Comment

As you see that , for instance $$A \times \nabla $$ is an operator , so it is an object, a process, an operator ready to be applied to a function.