# Directional derivatives

• Nov 17th 2009, 11:47 AM
letshin
Directional derivatives
Heya! Was just wondering how to do this question:

Got a little stuck and any help would be great :)

Tried looking around for some resources online, but couldn't find any, so here I go. . .

Given that u = (x^2 y, y^2 z , z^2 x), calculate
(a) ∇(∇ · u)
(b) ∇ × (∇ × u)
(c) ∇^2 u

Thanks!
• Nov 17th 2009, 06:58 PM
novice
Quote:

Originally Posted by letshin
Heya! Was just wondering how to do this question:

Got a little stuck and any help would be great :)

Tried looking around for some resources online, but couldn't find any, so here I go. . .

Given that u = (x^2 y, y^2 z , z^2 x), calculate
(a) ∇(∇ · u)
(b) ∇ × (∇ × u)
(c) ∇^2 u

Thanks!

$\bigtriangledown = \frac{\partial}{\partial x} \overrightarrow i+ \frac{\partial}{\partial y} \overrightarrow j + \frac{\partial}{\partial z} \overrightarrow k$

$\bigtriangledown \cdot u = \frac{\partial x^2}{\partial x} + \frac{\partial y^2}{\partial y} + \frac{\partial z^2}{\partial z}$

$\bigtriangledown (\bigtriangledown \cdot u) = (\frac{\partial}{\partial x} \overrightarrow i+ \frac{\partial}{\partial y} \overrightarrow j + \frac{\partial}{\partial z} \overrightarrow k) \cdot (\frac{\partial x^2}{\partial x} + \frac{\partial y^2}{\partial y} + \frac{\partial z^2}{\partial z})$

$\bigtriangledown^2 u= \frac{\partial^{2} x^2}{\partial x^2} \overrightarrow i+ \frac{\partial^{2} y^2}{\partial y^2} \overrightarrow j + \frac{\partial^{2} z^2}{\partial z^2} \overrightarrow k$

$\bigtriangledown \times u = (\frac{\partial z^2}{\partial y} - \frac{\partial y^2}{\partial z}) \overrightarrow i+ (\frac{\partial z^2}{\partial x} - \frac{\partial x^2}{\partial z}) \overrightarrow j + (\frac{\partial y^2}{\partial x}-\frac{\partial x^2}{\partial y}) \overrightarrow k$

$\bigtriangledown \times (\bigtriangledown \times u) = \bigtriangledown (\bigtriangledown \cdot u) - \bigtriangledown^2 u$
• Nov 18th 2009, 03:42 AM
letshin
Hey thanks a bunch! But I still don't really get how
$

\bigtriangledown \times (\bigtriangledown \times u) = \bigtriangledown (\bigtriangledown \cdot u) - \bigtriangledown^2 u
$

I've done up to

∇(∇.U) - ∇^2U = ( $
\frac{\partial^{2} x^2}{\partial x^2} + \frac{\partial^{2} y^2}{\partial y^2} + \frac{\partial^{2} z^2}{\partial z^2}
$
) - ( $
\frac{\partial^{2} x^2}{\partial x^2} \overrightarrow i+ \frac{\partial^{2} y^2}{\partial y^2} \overrightarrow j + \frac{\partial^{2} z^2}{\partial z^2} \overrightarrow k
$
)

and

$

\bigtriangledown \times (\bigtriangledown \times u) =

$
( $

\frac{\partial}{\partial x} \overrightarrow i+ \frac{\partial}{\partial y} \overrightarrow j + \frac{\partial}{\partial z} \overrightarrow k
$
) x ( $

(\frac{\partial z^2}{\partial y} - \frac{\partial y^2}{\partial z}) \overrightarrow i+ (\frac{\partial z^2}{\partial x} - \frac{\partial x^2}{\partial z}) \overrightarrow j + (\frac{\partial y^2}{\partial x}-\frac{\partial x^2}{\partial y}) \overrightarrow k
$
)

From there I get something complicated like
$( \frac{\partial}{\partial y} (\frac{\partial y^2}{\partial x} - \frac{\partial x^2}{\partial y}) - \frac{\partial}{\partial z} (\frac{\partial x^2}{\partial z} - \frac{\partial z^2}{\partial x}) )\overrightarrow i
$

Not sure where to go from here. . . >< Sorry for asking so much~
• Nov 18th 2009, 08:35 AM
novice
$\bigtriangledown = \frac{\partial}{\partial x} \overrightarrow i+ \frac{\partial}{\partial y} \overrightarrow j + \frac{\partial}{\partial z} \overrightarrow k$
$\bigtriangledown \cdot u = \frac{\partial x^2}{\partial x} + \frac{\partial y^2}{\partial y} + \frac{\partial z^2}{\partial z}$
$\bigtriangledown (\bigtriangledown \cdot u) = (\frac{\partial}{\partial x} \overrightarrow i+ \frac{\partial}{\partial y} \overrightarrow j + \frac{\partial}{\partial z} \overrightarrow k) \cdot (\frac{\partial x^2}{\partial x} + \frac{\partial y^2}{\partial y} + \frac{\partial z^2}{\partial z})=6 \overrightarrow i+ 6 \overrightarrow j + 6 \overrightarrow k$------------------------eq(1)
$\bigtriangledown^2 u= \frac{\partial^{2} x^2}{\partial x^2} \overrightarrow i+ \frac{\partial^{2} y^2}{\partial y^2} \overrightarrow j + \frac{\partial^{2} z^2}{\partial z^2} \overrightarrow k= 2 \overrightarrow i+ 2\overrightarrow j + 2 \overrightarrow k$------------------------------eq(2)
$\bigtriangledown \times u = (\frac{\partial z^2}{\partial y} - \frac{\partial y^2}{\partial z}) \overrightarrow i+ (\frac{\partial z^2}{\partial x} - \frac{\partial x^2}{\partial z}) \overrightarrow j + (\frac{\partial y^2}{\partial x}-\frac{\partial x^2}{\partial y}) \overrightarrow k$

$\bigtriangledown \times (\bigtriangledown \times u) = \bigtriangledown (\bigtriangledown \cdot u) - \bigtriangledown^2 u$-------------------------------------eq(3)

Substitue eq(1) into the first term on the RHS of eq(3), and eq(2) into the second term:

$\bigtriangledown \times (\bigtriangledown \times u) = \bigtriangledown (\bigtriangledown \cdot u) - \bigtriangledown^2 u = 4 \overrightarrow i+ 4 \overrightarrow j + 4 \overrightarrow k$
• Nov 18th 2009, 04:18 PM
letshin
I got it now, thanks a lot!