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Math Help - [SOLVED] Calculus with two variables

  1. #1
    RET7
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    [SOLVED] Calculus with two variables

    Hi,

    Could someone show me a proof that the two-dimensional del-squared operator is invariant under a rotation of axes in Cartesian co-ordinates?

    Thanks.
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  2. #2
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    Quote Originally Posted by RET7 View Post
    Hi,

    Could someone show me a proof that the two-dimensional del-squared operator is invariant under a rotation of axes in Cartesian co-ordinates?

    Thanks.
    Under a counter-clockwise rotation by angle \theta:

    x' = x \cos \theta - y \sin \theta

    y' = x \sin \theta + y \cos \theta

    where x' and y' are the new Cartesian coordinates. Then:


    \frac{\partial f}{\partial x} = \frac{\partial f}{\partial x'} \cdot \frac{\partial x'}{\partial x} + \frac{\partial f}{\partial y'} \cdot \frac{\partial y'}{\partial x}


    = \frac{\partial f}{\partial x'} \cdot \cos \theta + \frac{\partial f}{\partial y'} \cdot \sin \theta.


    Therefore:


    \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x'} \cdot \cos \theta + \frac{\partial f}{\partial y'} \cdot \sin \theta \right)


    = \frac{\partial}{\partial x'} \left( \frac{\partial f}{\partial x'} \cdot \cos \theta + \frac{\partial f}{\partial y'} \cdot \sin \theta \right) \cdot \frac{\partial x'}{\partial x} + \frac{\partial}{\partial y'} \left( \frac{\partial f}{\partial x'} \cdot \cos \theta + \frac{\partial f}{\partial y'} \cdot \sin \theta \right) \cdot \frac{\partial y'}{\partial x}


    = \left( \frac{\partial^2 f}{\partial x'^2} \cdot \cos \theta + \frac{\partial^2 f}{\partial y' \partial x'} \cdot \sin \theta \right) \cdot \cos \theta + \left( \frac{\partial^2 f}{\partial x' \partial y'} \cdot \cos \theta + \frac{\partial^2 f}{\partial y'^2} \cdot \sin \theta \right) \cdot \sin \theta


    = \frac{\partial^2 f}{\partial x'^2} \cdot \cos^2 \theta + 2 \frac{\partial^2 f}{\partial y' \partial x'} \cdot \sin \theta \cos \theta + \frac{\partial^2 f}{\partial y'^2} \cdot \sin^2 \theta.



    In a similar way you get:


    \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial y'^2} \cdot \cos^2 \theta - 2 \frac{\partial^2 f}{\partial y' \partial x'} \cdot \sin \theta \cos \theta + \frac{\partial^2 f}{\partial x'^2} \cdot \sin^2 \theta.


    Now add the two expressions together and you have \frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial x'^2} +\frac{\partial^2 f}{\partial y'^2} \, ,

    since \cos^2 \theta + \sin^2 \theta = 1.
    Last edited by mr fantastic; February 13th 2008 at 06:25 PM.
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