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Change of variable for 2nd partial differentiation and higher.

  1. Aug 19, 2013 #1
    Hello,
    the question I have arises from the 4th Edition of the book "Advanced Engineering Mathematics" written by K.A. Stroud. For those who owns the book, it is the example #2 starting at page 379. More precisely, the example is separated into two parts but the first one is very straight forward and does not require any attention.

    Furthermore, while the problem will be stated in full, it will not be solved completely as my problem is at one of the intermediary step. The technique that the author uses was not, as far as I know, presented or proven in the book and does not appear to be equivalent to the one that is used for that step (the chain rule). I might simply be missing one concept...

    So let us begin:

    1. The problem statement, all variables and given/known data

    [itex] \text{If}\ \ z=f(x,y)\text{, and}\ \ x = \frac{1}{2}(u^2 - v^2)\ \text{and}\ \ y = uv\text{, show that:} [/itex]

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


    2. Relevant equations

    1. [itex] \text{Author's technique is best presented in the context as I do not know how to make it concise.}\ [/itex]
      [itex] [/itex]
    2. [itex] \text{The chain rule for the second partial derivatives:}\ \frac{\partial }{\partial u}(\frac{\partial z}{\partial x}) = \frac{\partial }{\partial x}(\frac{\partial z}{\partial x})\frac{\partial x}{\partial u} = \frac{\partial^2 z}{\partial x^2}\frac{\partial x}{\partial u}\ [/itex]


    3. The attempt at a solution

    [itex] \text{The chain rule for the first partial derivative:}\ \frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u}\ [/itex]


    [itex] \text{∴}\hspace{10mm}\frac{\partial z}{\partial u} = u\frac{\partial z}{\partial x} + v\frac{\partial z}{\partial y}\hspace{5mm}\text{;}\hspace{5mm}\frac{\partial z}{\partial v} = -v\frac{\partial z}{\partial x} + u\frac{\partial z}{\partial y}\ [/itex]


    Author's steps:

    [itex] \text{Because:}\hspace{30mm}\frac{\partial }{\partial u} = u\frac{\partial }{\partial x} + v\frac{\partial }{\partial y}\hspace{5mm}\text{;}\hspace{5mm}\frac{\partial }{\partial v} = -v\frac{\partial }{\partial x} + u\frac{\partial }{\partial y}\ [/itex]


    [itex] \hspace{34.5mm}\frac{\partial^2 z}{\partial u^2} = \frac{\partial }{\partial u}(\frac{\partial z}{\partial u}) = \frac{\partial }{\partial u}(u\frac{\partial z}{\partial x} + v\frac{\partial z}{\partial v})= \frac{\partial z}{\partial x}+u\frac{\partial }{\partial u}(\frac{\partial z}{\partial x})+ v\frac{\partial }{\partial u}(\frac{\partial z}{\partial y})\ [/itex]


    [itex] \text{Problematic step:}\hspace{5mm}= \frac{\partial z}{\partial x}+u(u\frac{\partial }{\partial x} + v\frac{\partial }{\partial y})(\frac{\partial z}{\partial x})+ v(u\frac{\partial }{\partial x} + v\frac{\partial }{\partial y})(\frac{\partial z}{\partial y}) = \frac{\partial z}{\partial x}+ u^2\frac{\partial^2 z}{\partial x^2}+uv\frac{\partial^2 z}{\partial y\partial x})+vu\frac{\partial^2 z}{\partial x\partial y}+v^2\frac{\partial^2 z}{\partial y^2}\ [/itex]


    [itex] \hspace{25.5mm}\text{∴}\hspace{5mm}\frac{\partial^2 z}{\partial u^2} = \frac{\partial z}{\partial x}+ u^2\frac{\partial^2 z}{\partial x^2}+2uv\frac{\partial^2 z}{\partial x\partial y})+v^2\frac{\partial^2 z}{\partial y^2}\ [/itex]



    My steps are all the same until the problematic step. I use the chain rule presented in the relevant equations instead.


    [itex] \text{Problematic step:}\hspace{5mm} = \frac{\partial z}{\partial x}+u\frac{\partial }{\partial u}(\frac{\partial z}{\partial x})+ v\frac{\partial }{\partial u}(\frac{\partial z}{\partial y}) = \frac{\partial z}{\partial x}+u\frac{\partial }{\partial x}(\frac{\partial z}{\partial x})\frac{\partial x}{\partial u}+ v\frac{\partial }{\partial y}(\frac{\partial z}{\partial y})\frac{\partial y}{\partial u}\ = \frac{\partial z}{\partial x}+u\frac{\partial^2 z}{\partial x^2}\frac{\partial x}{\partial u}+ v\frac{\partial^2 z}{\partial y^2}\frac{\partial y}{\partial u}\ [/itex]


    [itex] \text{Now since}\hspace{5mm}\frac{\partial x}{\partial u}=u\hspace{5mm}\text{and}\hspace{5mm}\frac{\partial y}{\partial u}=v\ [/itex]


    [itex] \text{Then}\hspace{5mm}\frac{\partial z}{\partial x}+u\frac{\partial^2 z}{\partial x^2}\frac{\partial x}{\partial u}+ v\frac{\partial^2 z}{\partial y^2}\frac{\partial y}{\partial u}=\frac{\partial z}{\partial x}+u^2\frac{\partial^2 z}{\partial x^2} + v^2\frac{\partial^2 z}{\partial y^2}\ [/itex]


    [itex] \hspace{25.5mm}\text{∴}\hspace{5mm}\frac{\partial^2 z}{\partial u^2}= \frac{\partial z}{\partial x}+u^2\frac{\partial^2 z}{\partial x^2} + v^2\frac{\partial^2 z}{\partial y^2}\ [/itex]


    Obviously, the author and I do not agree what should be the "answer" for that intermediary step and I wonder why.While I understand what the author did, he did not provide a proof or presented the technique prior to using it in the example.

    Therefore, the help that I need is to be explained why the technique used by the author is the appropriate one and, if possible, to be provided with the name of the technique so that I could find further information about it. What additional step was missing in the chain rule that I use so that it would provide the same "answer" as the author? You may provide anything that you may consider appropriate for this post as well.

    Thank you in advance.
     
  2. jcsd
  3. Aug 19, 2013 #2

    lurflurf

    User Avatar
    Homework Helper

    your
    $$\text{The chain rule for the second partial derivatives:}\ \frac{\partial }{\partial u}(\frac{\partial z}{\partial x}) = \frac{\partial }{\partial x}(\frac{\partial z}{\partial x})\frac{\partial x}{\partial u} = \frac{\partial^2 z}{\partial x^2}\frac{\partial x}{\partial u}$$

    should be
    $$ \dfrac{\partial ^2}{\partial u \partial x}=\dfrac{\partial x}{\partial u} \dfrac{\partial ^2}{\partial x^2} + \dfrac{\partial y}{\partial u} \dfrac{\partial ^2}{\partial x \partial y}$$
    I don't know why you are using it though.
     
  4. Aug 19, 2013 #3
    My bad indeed; I misread the other book that I had and was missing the term with the partial derivative with respect to y...

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


    After verification, the author and I have the same answer.

    Thanks lurflurf for pointing out that what was my problem. :)
     
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