Proof of vector calculus identities

Question

Here is the all identities : http://en.wikipedia.org/wiki/Vector_calculus_identities

I need help concerning vector functions and indexing notations.

Let $\overrightarrow{a}$ be a (smooth) vector field and $\varphi$ be a (smooth) scalar function. Show $$ \overrightarrow {\nabla }\cdot \left( \varphi\,\overrightarrow {a}\right) = \varphi \left( \overrightarrow {\nabla }\cdot \overrightarrow {a}\right) +\overrightarrow {a} \cdot \overrightarrow {\nabla }\varphi.$$

I have to use this notation to prove this, but how?

I don't really understand.

My second identity is ; $$ \overrightarrow {\nabla }\times \left( \phi \cdot \overrightarrow {a}\right) $$

Answer 1

Here's what's happening in $\mathbb{R}^3$ with rectangular coordinates. You can tweak as needed.

Let $g(x,y,z)$ be a smooth scalar function and $\mathbf{F}(x,y,z)=(F_1(x,y,z),F_2(x,y,z),F_3(x,y,z))$ be a smooth vector field. Then \begin{align} \nabla \cdot (g\,\mathbf{F})&=\nabla\cdot((gF_1,gF_2,gF_3))\\ &=(gF_1)_x+(gF_2)_y+(gF_3)_z\\ &=g_xF_1+g(F_1)_x+g_yF_2+g(F_2)_y+g_zF_3+g(F_3)_z, \end{align} while \begin{align} \nabla g\cdot \mathbf{F}&=g_x F_1+g_y F_2+g_z F_3,\\ g\,(\nabla\cdot \mathbf{F})&=g\,((F_1)_x+(F_2)_y+(F_3)_z)=g(F_1)_x+g(F_2)_y+g(F_3)_z. \end{align} Adding these last two yields the first.

Answer 2

If JohnD has interpreted the problem correctly, then here's how you would work it using index notation. Here, $i$ is an index running from 1 to 3 ($a^1$ might be the x-component of $a$, $a^2$ the y-component, and so on).

$$\nabla \cdot (\varphi a) = \nabla_i (\varphi a^i)$$

Since these are all components (not vectors), you can attack this with the product rule.

$$\nabla_i (\varphi a^i) = (\nabla_i \varphi) a^i + \varphi (\nabla_i a^i)$$

The first term is $a \cdot \nabla \varphi$ and the latter is $\varphi \nabla \cdot a$.

Answer 3

You probably know the product rule $(uv)'=u'v +uv'$.

I have been taught to use such formula (which follows from derivative-like nature of $\nabla$):

$\nabla(\underline{uv}) = \nabla(\underline{u}v) + \nabla(u\underline{v})$

where derivatives work on the underlined part of the brackets. In your case you can factor $\varphi$ and $\vec{a}$ when they are not under effect of nabla, but in other cases (like $\nabla (\vec{a}\cdot\vec{b})$) it can't be done that easily, but you can derive a formula for dot product starting from formula with underlines.

Answer 4

Putting a smooth vector field U in the UNIT VECTORS, and a scalar variable ϕ $$\boldsymbol{U}=u_1\boldsymbol{i}+u_2\boldsymbol{j}+u_3\boldsymbol{k}$$ Now showing that

$$\boldsymbol{\nabla}\times(\boldsymbol{U}\phi)=\phi\boldsymbol{(\nabla\times U)}+\boldsymbol{\nabla}\phi\times\boldsymbol{U}$$

$$\boldsymbol{\nabla}\times(\boldsymbol{U}\phi)=\begin{vmatrix} \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ u_1\phi&u_2\phi&u_3\phi \end{vmatrix}\\ =\left(\boldsymbol{i}\frac{\partial(u_3\phi)}{\partial y}+\boldsymbol{j}\frac{\partial(u_1\phi)}{\partial z}+\boldsymbol{k}\frac{\partial(u_2\phi)}{\partial x}\right)-\left(\boldsymbol{i}\frac{\partial(u_2\phi)}{\partial z}+\boldsymbol{j}\frac{\partial(u_3\phi)}{\partial x}+\boldsymbol{k}\frac{\partial(u_1\phi)}{\partial y}\right)\\ =\phi\left(\boldsymbol{i}\left(\frac{\partial u_3}{\partial y}-\frac{\partial u_2}{\partial z}\right)+\boldsymbol{j}\left(\frac{\partial u_1}{\partial z}-\frac{\partial u_3}{\partial x}\right)+\boldsymbol{k}\left(\frac{\partial u_2}{\partial x}-\frac{\partial u_1}{\partial y}\right)\right)+\left(\boldsymbol{i}\left(u_3\frac{\partial\phi}{\partial y}-u_2\frac{\partial\phi}{\partial z}\right)+\boldsymbol{j}\left(u_1\frac{\partial\phi}{\partial z}-u_3\frac{\partial\phi}{\partial x}\right)+\boldsymbol{k}\left(u_2\frac{\partial\phi}{\partial x}-u_1\frac{\partial\phi}{\partial y}\right)\right)\\ =\phi\begin{vmatrix} \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ u_1&u_2&u_3 \end{vmatrix}+\begin{vmatrix} \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\ \frac{\partial\phi}{\partial x}&\frac{\partial\phi}{\partial y}&\frac{\partial\phi}{\partial z}\\ u_1&u_2&u_3 \end{vmatrix}=\phi\boldsymbol{(\nabla\times U)}+\boldsymbol{\nabla}\phi\times\boldsymbol{U}$$ Having funs with the vector calaulus!! O:DDD