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General equation for change of variable in a differential equation

  1. Oct 15, 2011 #1
    I had a second order differential equation where [itex]\psi[/itex] is the unknown function and it is a function of [itex]x[/itex]. We then introduced the following change of variable [itex]x = \sqrt{\frac{\hbar}{m \omega}} \xi[/itex]. When all was said and done I found that,

    [tex] \frac{d^2 \psi}{d \xi^2} = \bigg(\frac{dx}{d \xi}\bigg)^2 \frac{d^2\psi}{dx^2} [/tex]

    My question is, given an arbitrary change of variable for x and given an arbitrary order of the differential equation will the following formula always work?

    [tex] \frac{d^n \psi}{d \xi^n} = \bigg(\frac{dx}{d\xi}\bigg)^n \frac{d^n \psi}{dx^n} [/tex]
     
  2. jcsd
  3. Oct 16, 2011 #2

    lurflurf

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    Homework Helper

    That will very not work. What you might have wanted is

    [tex] \frac{d^n \psi}{d \xi^n} = \bigg(\frac{dx}{d\xi} \frac{d}{dx}\bigg)^n \psi[/tex]

    The trouble is that in general d/dx and dx/dxi do not commute and cannot be reordered. Try some examples like
    x=exp(u)
    x^2=u
    and so on to see this.
     
  4. Oct 16, 2011 #3
    Thanks for the response, but I don't seem to see the difference.

    [tex]
    \frac{d^n \psi}{d \xi^n} = \bigg(\frac{dx}{d \xi} \frac{d}{dx} \bigg)^n \psi
    = \bigg(\bigg(\frac{dx}{d \xi} \bigg)^n \bigg(\frac{d}{dx} \bigg)^n \bigg) \psi
    = \bigg(\frac{dx}{d \xi} \bigg)^n \frac{d^n \psi}{dx^n}
    [/tex]

    Did I make any mistakes in the above? I'm not doing any reordering.
     
  5. Oct 16, 2011 #4

    lurflurf

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    Homework Helper

    You are reordering
    when you write

    (a b)^3=a^3 b^3
    a b a b a b=a a a b b b

    you are making an implicit assumption that the order is not important
    your change of variable equation holds when one variable is a constant multiple of the other

    suppose
    d/du=x d/dx
    (d/du)^n=(x d/dx)^n
    (d/du)^2=(x d/dx)^2=(x d/dx)(x d/dx)=x^2 (d/dx)^2+x d/dx
    this is not equal to x^2 (d/dx)^2 because x d/dx is not 0
     
  6. Oct 20, 2011 #5
    I see now. Thank you for clearing this up for me.
     
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