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Partial Differential problem

  1. Nov 23, 2011 #1
    If [itex]z = f(x,y)[/itex] and [itex]x = e^{u}, y =e^{v} [/itex] Prove:

    [itex]x^{2}\frac{\partial^{2}z}{\partial x^{2}} + y^{2}\frac{\partial^{2}z}{\partial y^{2}} + x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = \frac{\partial^{2}z}{\partial u^{2}} + \frac{\partial^{2}z}{\partial v^{2}} [/itex]

    I used [itex] u = ln(x), v = ln(y) [/itex] and the following partial differential set ups:


    [itex]
    \frac{\partial z}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x}[/itex]
    [itex]
    \frac{\partial z}{\partial y} = \frac{\partial f}{\partial v}\frac{\partial v}{\partial y} [/itex]
    [itex]
    \frac{\partial z}{\partial u} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial u}
    [/itex]
    [itex]
    \frac{\partial z}{\partial u} = \frac{\partial f}{\partial y} \frac{\partial y}{\partial v}[/itex]

    Then:

    [itex]
    \frac{\partial z}{\partial x} = \frac{\partial f}{\partial u} \frac{1}{x}[/itex]
    [itex]
    \frac{\partial z}{\partial y} = \frac{\partial f}{\partial v} \frac{1}{y}
    [/itex]
    [itex]
    \frac{\partial f}{\partial u} = \frac{\partial z}{\partial u} e^{u}
    [/itex]
    [itex]
    \frac{\partial f}{\partial v} = \frac{\partial z}{\partial v} e^{v}
    [/itex]

    For the second derivatives:

    [itex]
    \frac{\partial^{2} z}{\partial x^{2}} = \frac{\partial^{2} f}{\partial u^{2}} \frac{-1}{x^{2}}
    [/itex]
    [itex]

    \frac{\partial^{2} z}{\partial x^{2}} = \frac{\partial^{2} f}{\partial v^{2}} \frac{-1}{y^{2}}
    [/itex]
    [itex]
    \frac{\partial^{2} z}{\partial u^{2}} = \frac{\partial^{2} f}{\partial x^{2}} e^{u}
    [/itex]
    [itex]
    \frac{\partial^{2} z}{\partial v^{2}} = \frac{\partial^{2} f}{\partial y^{2}} e^{v}
    [/itex]

    I cannot get these terms to equal the statement that is to be proved. Am I doing something wrong?

    Thanks!
     
  2. jcsd
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