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Thread: Directional derivatives

  1. #1
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    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!
    Last edited by letshin; Nov 17th 2009 at 11:13 AM.
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  2. #2
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    Quote Originally Posted by letshin View Post
    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!
    $\displaystyle \bigtriangledown = \frac{\partial}{\partial x} \overrightarrow i+ \frac{\partial}{\partial y} \overrightarrow j + \frac{\partial}{\partial z} \overrightarrow k$

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

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

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

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


    $\displaystyle \bigtriangledown \times (\bigtriangledown \times u) = \bigtriangledown (\bigtriangledown \cdot u) - \bigtriangledown^2 u$
    Last edited by novice; Nov 17th 2009 at 06:17 PM.
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  3. #3
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    Hey thanks a bunch! But I still don't really get how
    $\displaystyle

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

    I've done up to

    ∇(∇.U) - ∇^2U = ( $\displaystyle
    \frac{\partial^{2} x^2}{\partial x^2} + \frac{\partial^{2} y^2}{\partial y^2} + \frac{\partial^{2} z^2}{\partial z^2}
    $ ) - ( $\displaystyle
    \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

    $\displaystyle

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

    $( $\displaystyle

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

    (\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
    $\displaystyle ( \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~
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  4. #4
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    $\displaystyle \bigtriangledown = \frac{\partial}{\partial x} \overrightarrow i+ \frac{\partial}{\partial y} \overrightarrow j + \frac{\partial}{\partial z} \overrightarrow k$
    $\displaystyle \bigtriangledown \cdot u = \frac{\partial x^2}{\partial x} + \frac{\partial y^2}{\partial y} + \frac{\partial z^2}{\partial z}$
    $\displaystyle \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)
    $\displaystyle \bigtriangledown^2 u= \frac{\partial^{2} x^2}{\partial x^2} \overrightarrow i+ \frac{\partial^{2} y^2}{\partial y^2} \overrightarrow j + \frac{\partial^{2} z^2}{\partial z^2} \overrightarrow k= 2 \overrightarrow i+ 2\overrightarrow j + 2 \overrightarrow k$------------------------------eq(2)
    $\displaystyle \bigtriangledown \times u = (\frac{\partial z^2}{\partial y} - \frac{\partial y^2}{\partial z}) \overrightarrow i+ (\frac{\partial z^2}{\partial x} - \frac{\partial x^2}{\partial z}) \overrightarrow j + (\frac{\partial y^2}{\partial x}-\frac{\partial x^2}{\partial y}) \overrightarrow k$

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

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

    $\displaystyle \bigtriangledown \times (\bigtriangledown \times u) = \bigtriangledown (\bigtriangledown \cdot u) - \bigtriangledown^2 u = 4 \overrightarrow i+ 4 \overrightarrow j + 4 \overrightarrow k$
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  5. #5
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    I got it now, thanks a lot!
    Last edited by letshin; Nov 18th 2009 at 04:49 PM.
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