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Chain rule: partial derivatives transformation

  1. Jan 10, 2009 #1

    Let g(x,y) be a function that has second order partial derivatives. Transform the differential equation

    [tex]\frac{\delta ^{2}g}{\delta x^{2}}-\frac{\delta ^{2}g}{\delta y^{2}}=xyg[/tex]

    by chaning to the new variables [tex]u=x^2-y^2[/tex] and [tex]v=xy[/tex]
    The equation doesn't have to be solved.

    Okay, so this is basically an exercise in using the chain rule.
    The first part is pretty easy.

    [tex]\frac{\delta g}{\delta x}=\frac{\delta g}{\delta u}\cdot \frac{\delta u}{\delta x}+\frac{\delta g}{\delta v}\cdot \frac{\delta v}{\delta x}=2x\cdot \frac{\delta g}{\delta u}+y\frac{\delta g}{\delta v}[/tex]
    [tex]\frac{\delta g}{\delta y}=\frac{\delta g}{\delta u}\cdot \frac{\delta u}{\delta y}+\frac{\delta g}{\delta v}\cdot \frac{\delta v}{\delta y}=-2y\cdot \frac{\delta g}{\delta u}+x\frac{\delta g}{\delta v}[/tex]

    Now I try to find [tex]\frac{\delta ^{2}g}{\delta x^{2}}[/tex] using the information from A. I realize I'm supposed to use the chain rule again, but I just can't get the right answer, so I guess I'm missing some fundamental understanding of what's going on here.

    Here's what the first part of the problem eventually should boil down to:
    [tex]\frac{\delta ^{2}g}{\delta x^{2}}=2\frac{\delta g}{\delta u}+4y^{2}\cdot \frac{\delta^{2} g}{\delta u^{2}}+4xy\cdot \frac{\delta^{2} g}{\delta u\delta v}+y^{2}\cdot \frac{\delta^{2} g}{\delta v^{2}}[/tex]

    This part above is what I need help with, if I understand that part, I think I can manage to solve the rest of the problem.

    Help is much appreciated. If I find out how to do it I will post here later. Thanks.
  2. jcsd
  3. Jan 10, 2009 #2


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    There is a rather old textbook, "Advanced Calculus" by Angus Taylor that addresses the kind of problem you are experiencing very carefully in Chapter 6 in a section titled "Second derivatives by the chain rule". Your library may have a copy. As I recall, when I first ran into this problem, the issue was as much notation as anything else.
  4. Jan 10, 2009 #3
    I see where you are stumbling:

    do you know how to do:

    d/dx [ dg/du ] ?

    Are you correctly using the product rule as:

    it's d/dx[ 2x . dg / du]?

    Do you know that dg/du consists of two variables (u,v) so you need to take care of that .. ?

    I got the first part.. apparently, there aren't any tough work involved. You just need to understand:
    d/dx [df (x,y) /dx] = d^2.f / (dx.dy) + d^2 . f/ (dx^2)

    Use trees - they help: write f, and then two children u,v .. both have x,y children
    Last edited: Jan 10, 2009
  5. Jan 13, 2009 #4
    [tex]\frac{\delta ^2 g}{\delta x^2}=\frac{\delta }{\delta x} \left(\frac{\delta g}{\delta x}\right)[/tex]

    [tex]= \frac{\delta }{\delta x} \left( \frac{\delta g}{\delta u}\cdot \frac{\delta u}{\delta x} \right) + \frac{\delta }{\delta x} \left( \frac{\delta g}{\delta v}\cdot \frac{\delta v}{\delta x} \right) [/tex]

    Then use the product rule on each. then the chain rule again when differentiating a partial differential of g. It's long winded but it'll do the trick!
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