Identities involving grad, div and curl

7 Identities involving grad, div and curl

There are numerous identities involving the vector derivatives; a selection are given in Table 1.

Table 1

1	$div (ϕ \underset{̲}{A}) = grad ϕ \cdot \underset{̲}{A} + ϕ div \underset{̲}{A}$	or	$\underset{̲}{\nabla} \cdot (ϕ \underset{̲}{A}) = (\underset{̲}{\nabla} ϕ) \cdot \underset{̲}{A} + ϕ (\underset{̲}{\nabla} \cdot \underset{̲}{A})$
2	$curl (ϕ \underset{̲}{A}) = grad ϕ \times \underset{̲}{A} + ϕ curl \underset{̲}{A}$	or	$\underset{̲}{\nabla} \times (ϕ \underset{̲}{A}) = (\underset{̲}{\nabla} ϕ) \times \underset{̲}{A} + ϕ (\underset{̲}{\nabla} \times \underset{̲}{A})$
3	$div (\underset{̲}{A} \times \underset{̲}{B}) = \underset{̲}{B} \cdot curl \underset{̲}{A} - \underset{̲}{A} \cdot curl \underset{̲}{B}$	or	$\underset{̲}{\nabla} \cdot (\underset{̲}{A} \times \underset{̲}{B}) = \underset{̲}{B} \cdot (\underset{̲}{\nabla} \times \underset{̲}{A}) - \underset{̲}{A} \cdot (\underset{̲}{\nabla} \times \underset{̲}{B})$
4	$curl (\underset{̲}{A} \times \underset{̲}{B}) = (\underset{̲}{B} \cdot grad) \underset{̲}{A} - (\underset{̲}{A} \cdot grad) \underset{̲}{B}$	or	$\underset{̲}{\nabla} \times (\underset{̲}{A} \times \underset{̲}{B}) = (\underset{̲}{B} \cdot \underset{̲}{\nabla}) \underset{̲}{A} - (\underset{̲}{A} \cdot \underset{̲}{\nabla}) \underset{̲}{B}$
	$+ \underset{̲}{A} div \underset{̲}{B} - \underset{̲}{B} div \underset{̲}{A}$		$+ \underset{̲}{A} \underset{̲}{\nabla} \cdot \underset{̲}{B} - \underset{̲}{B} \underset{̲}{\nabla} \cdot \underset{̲}{A}$
5	$grad (\underset{̲}{A} \cdot \underset{̲}{B}) = (\underset{̲}{B} \cdot grad) \underset{̲}{A} + (\underset{̲}{A} \cdot grad) \underset{̲}{B}$	or	$\underset{̲}{\nabla} (\underset{̲}{A} \cdot \underset{̲}{B}) = (\underset{̲}{B} \cdot \underset{̲}{\nabla}) \underset{̲}{A} + (\underset{̲}{A} \cdot \underset{̲}{\nabla}) \underset{̲}{B}$
	$+ \underset{̲}{A} \times curl \underset{̲}{B} + \underset{̲}{B} \times curl \underset{̲}{A}$		$+ \underset{̲}{A} \times (\underset{̲}{\nabla} \times \underset{̲}{B}) + \underset{̲}{B} \times (\underset{̲}{\nabla} \times \underset{̲}{A})$
6	$curl (grad ϕ) = \underset{̲}{0}$	or	$\underset{̲}{\nabla} \times (\underset{̲}{\nabla} ϕ) = \underset{̲}{0}$
7	$div (curl \underset{̲}{A}) = \underset{̲}{0}$	or	$\underset{̲}{\nabla} \cdot (\underset{̲}{\nabla} \times \underset{̲}{A}) = \underset{̲}{0}$

Example 18

Show for any vector field $\underset{̲}{A} = A_{1} \underset{̲}{i} + A_{2} \underset{̲}{j} + A_{3} \underset{̲}{k}$ , that div curl $\underset{̲}{A} = \underset{̲}{0}$ .

Solution

\begin{array}{rcl} div curl \underset{̲}{A} & = & div |\begin{matrix} \underset{̲}{i} & \underset{̲}{j} & \underset{̲}{j} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_{1} & A_{2} & A_{3} \end{matrix}| \\ = & div [(\frac{\partial A_{3}}{\partial y} - \frac{\partial A_{2}}{\partial z}) \underset{̲}{i} + (\frac{\partial A_{1}}{\partial z} - \frac{\partial A_{3}}{\partial x}) \underset{̲}{j} + (\frac{\partial A_{2}}{\partial x} - \frac{\partial A_{1}}{\partial y}) \underset{̲}{k}] \\ = & \frac{\partial}{\partial x} (\frac{\partial A_{3}}{\partial y} - \frac{\partial A_{2}}{\partial z}) + \frac{\partial}{\partial y} (\frac{\partial A_{1}}{\partial z} - \frac{\partial A_{3}}{\partial x}) + \frac{\partial}{\partial z} (\frac{\partial A_{2}}{\partial x} - \frac{\partial A_{1}}{\partial y}) \\ = & \frac{\partial^{2} A_{3}}{\partial x \partial y} - \frac{\partial^{2} A_{2}}{\partial z \partial x} + \frac{\partial^{2} A_{1}}{\partial y \partial z} - \frac{\partial^{2} A_{3}}{\partial y \partial x} + \frac{\partial^{2} A_{2}}{\partial z \partial x} - \frac{\partial^{2} A_{1}}{\partial z \partial y} \\ = & 0 \end{array}

N.B. This assumes $\frac{\partial^{2} A_{3}}{\partial x \partial y} = \frac{\partial^{2} A_{3}}{\partial y \partial x}$ etc.

Example 19

Verify identity 1 for the vector $\underset{̲}{A} = 2 x y \underset{̲}{i} - 3 z \underset{̲}{k}$ and the function $ϕ = x y^{2}$ .

$ϕ \underset{̲}{A} = 2 x^{2} y^{3} \underset{̲}{i} - 3 x y^{2} z \underset{̲}{k}$ so
$\underset{̲}{\nabla} \cdot ϕ \underset{̲}{A} = \underset{̲}{\nabla} \cdot (2 x^{2} y^{3} \underset{̲}{i} - 3 x y^{2} z \underset{̲}{k}) = \frac{\partial}{\partial x} (2 x^{2} y^{3}) + \frac{\partial}{\partial z} (- 3 x y^{2} z) = 4 x y^{3} - 3 x y^{2}$
So LHS $= 4 x y^{3} - 3 x y^{2}$ .

$\underset{̲}{\nabla} ϕ = \frac{\partial}{\partial x} (x y^{2}) \underset{̲}{i} + \frac{\partial}{\partial y} (x y^{2}) \underset{̲}{j} + \frac{\partial}{\partial z} (x y^{2}) \underset{̲}{k} = y^{2} \underset{̲}{i} + 2 x y \underset{̲}{j}$ so
$(\underset{̲}{\nabla} ϕ) \cdot \underset{̲}{A} = (y^{2} \underset{̲}{i} + 2 x y \underset{̲}{j}) \cdot (2 x y \underset{̲}{i} - 3 z \underset{̲}{k}) = 2 x y^{3}$
$\underset{̲}{\nabla} \cdot \underset{̲}{A} = \underset{̲}{\nabla} \cdot (2 x y \underset{̲}{i} - 3 z \underset{̲}{k}) = 2 y - 3$ so $ϕ \underset{̲}{\nabla} \cdot \underset{̲}{A} = 2 x y^{3} - 3 x y^{2}$ giving
$(\underset{̲}{\nabla} ϕ) \cdot \underset{̲}{A} + ϕ (\underset{̲}{\nabla} \cdot \underset{̲}{A}) = 2 x y^{3} + (2 x y^{3} - 3 x y^{2}) = 4 x y^{3} - 3 x y^{2}$
So RHS $= 4 x y^{3} - 3 x y^{2} =$ LHS.

So $\underset{̲}{\nabla} \cdot (ϕ \underset{̲}{A}) = (\underset{̲}{\nabla} ϕ) \cdot \underset{̲}{A} + ϕ (\underset{̲}{\nabla} \cdot \underset{̲}{A})$ in this case.

Task!

If $\underset{̲}{F} = x^{2} y \underset{̲}{i} - 2 x z \underset{̲}{j} + 2 y z \underset{̲}{k}$ , find

$\underset{̲}{\nabla} \cdot \underset{̲}{F}$
$\underset{̲}{\nabla} \times \underset{̲}{F}$
$\underset{̲}{\nabla} (\underset{̲}{\nabla} \cdot \underset{̲}{F})$
$\underset{̲}{\nabla} \cdot (\underset{̲}{\nabla} \times \underset{̲}{F})$
$\underset{̲}{\nabla} \times (\underset{̲}{\nabla} \times \underset{̲}{F})$

$2 x y + 2 y,$
$(2 x + 2 z) \underset{̲}{i} - (x^{2} + 2 z) \underset{̲}{k}$ ,
$2 y \underset{̲}{i} + (2 + 2 x) \underset{̲}{j}$ (using answer to (1)),
$0$ (using answer to (2)),
$(2 + 2 x) \underset{̲}{j}$ (using answer to (2))

Task!

If $ϕ = 2 x z - y^{2} z$ , find

$\underset{̲}{\nabla} ϕ$
$\nabla^{2} ϕ = \underset{̲}{\nabla} \cdot (\underset{̲}{\nabla} ϕ)$
$\underset{̲}{\nabla} \times (\underset{̲}{\nabla} ϕ)$

$2 z \underset{̲}{i} - 2 y z \underset{̲}{j} + (2 x - y^{2}) \underset{̲}{k}$ ,
$- 2 z$ ,
$\underset{̲}{0}$ where (b) and (c) use the answer to (a).

Exercise

Which of the following combinations of grad, div and curl can be formed? If a quantity can be formed, state whether it is a scalar or a vector.

div (grad $ϕ$ )
div (div $\underset{̲}{A}$ )
curl (curl $\underset{̲}{F}$ )
div (curl $\underset{̲}{F}$ )
curl (grad $ϕ$ )
curl (div $\underset{̲}{A}$ )
div ( $\underset{̲}{A} \cdot \underset{̲}{B}$ )
grad ( $ϕ_{1} ϕ_{2}$ )
curl (div ( $\underset{̲}{A} \times$ grad $ϕ$ ))

(1), (4) are scalars;

(3), (5), (8) are vectors;

(2), (6), (7) and (9) are not defined.

7 Identities involving grad, div and curl

Example 18

Solution

Example 19

Solution

Task!

Answer

Task!

Answer

Exercise

Answer