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I Mixed partial differentials

  1. Sep 18, 2016 #1
    Hi! Can someone give me an example of a function ##f(x,y)## for which the mixed partial differentials are not equal, i.e. $$\frac{\partial^2 f}{\partial x \partial y} \neq \frac{\partial^2f}{\partial y \partial x}$$
    It says in Boas that these mixed differentials are equal only if the first and second order partial differentials of ##x## and ##y## are continuous. I couldn't come up with any. Thanks :smile:
     
  2. jcsd
  3. Sep 18, 2016 #2

    PeroK

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    Note that it should be easy to find a function for which only one of the mixed partials exists. Can you do that?

    But, finding a function where both partials exist but are not equal is harder. To do this you need a function with a discontinuous derivative. Do you know any?
     
  4. Sep 18, 2016 #3
    Yes, I'm aware of this. But in such a case, one of the mixed deferential wouldn't exist ,right? I wanted to find a function where both the differentials would exist and be unequal to each other.
    That's exactly what I thought, but couldn't come up with such a function .
    .
    Edit: Maybe we could use a piecewise function? ( not too sure about it though)
     
  5. Sep 18, 2016 #4

    PeroK

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    Try ##f(x) = x^2 \sin(\frac{1}{x})##
     
  6. Sep 18, 2016 #5
    This function did come to mind!! But I didn't "know" that its derivative was discontinuous (only that it has infinite extrema as ##x## goes to zero). So,
    $$\frac{d}{dx}\left[x^2 sin\frac{1}{x}\right] = 2xsin\frac{1}{x}-cos\frac{1}{x}$$
    It's discontinuous at ##x=0## right?
    But even then, I'm getting the mixed partial fractions to be equal, with ##f(x,y)= yx^2sin\frac{1}{x}## or ##f(x,y)= y+ x^2sin\frac{1}{x}## ..
    .
    (Sorry if the questions are too simple. I didn't learn these concepts back in school properly. It still haunts me)
     
  7. Sep 18, 2016 #6

    PeroK

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    This is not simple at all.

    To show that the derivative is discontinuous at ##0## you need to define ##f(0) = 0## and check:

    ##f'(0) = lim_{h \rightarrow 0} \frac{f(h) - f(0)}{h} = lim_{h \rightarrow 0} hsin(\frac{1}{h}) = 0##

    But:

    ##lim_{x \rightarrow 0} f'(x) = lim_{x \rightarrow 0} (2xsin\frac{1}{x}-cos\frac{1}{x})##

    is undefined. So, ##f'## is discontinuous at 0.

    How to use this for the multiple variable case? Let me have a think!
     
  8. Sep 18, 2016 #7
    Phew! I got that right. :redface:
     
  9. Sep 18, 2016 #8

    PeroK

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  10. Sep 18, 2016 #9
    Thanks for the link. Still have a small confusion though. When you use the definition $$f_{xy}(0,0)= \lim_{h\rightarrow 0} \frac{f_x (0,h)-f_x (0,0)}{h}$$
    ,we find them to be unequal.
    But if you find it simply as $$f_{xy}= \frac{\partial}{\partial y} \left(\frac{\partial f}{\partial x}\right)$$ they come out to be the same. Obviously we shouldn't have this inconsistency, so where am I wrong?
    I'll remember that :smile:
     
  11. Sep 18, 2016 #10

    PeroK

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    When you have a function defined by a formula that is not defined at a certain point then you need to be careful when differentiating. You can only use the formulaic differentiation for points where the formula holds. For points where the formula does not hold, you must use the function value at that point. This is what I did in post #6. We have:

    ##f(x) = \begin{cases} x^2 \sin(\frac{1}{x}) \ \ (x \ne 0) \\ 0 \ \ (x = 0) \end{cases}##

    That function is differentiable and its derivative is continuous for ##x \ne 0##, as you can simply differentiate it using the formula for all ##x \ne 0##.

    But, to get the derivative at ##x = 0## you must use the definition of the derivative and the function value at ##0##, as I did in post #6. In this case you find that ##f'(0) = 0##

    This leads to the conclusion that although ##f## is differentiable at all points the derivative is not continuous at ##x = 0## as:

    ##\lim_{x \rightarrow 0} f'(x) \ne f'(0)##
     
  12. Sep 18, 2016 #11
    Right. Got it. Thanks for your time!
     
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