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Partial derivatives (question I am grading).

  1. Mar 30, 2012 #1

    MathematicalPhysicist

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    We have a function f:R^2->R and it has partial derivative of 2nd order.
    Show that [itex]f_{xy}=0 \forall (x,y)\in \mathbb{R}^2 \Leftrightarrow f(x,y)=g(x)+h(y)[/itex]

    The <= is self explanatory, the => I am not sure I got the right reasoning.

    I mean we know that from the above we have: [itex]f_x=F(x)[/itex] (it's a question before this one), but now besides taking an integral I don't see how to show the consequent.

    Any thoughts how to show this without invoking integration?
     
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  3. Mar 30, 2012 #2

    hunt_mat

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    My first thought is the mean value theorem which states that if [itex]f'(x)=0[/itex] then [itex]f\equiv[/itex]constant., apply this to [itex]x[/itex] when [itex]y[/itex] yields [itex]\partial_{y}f\equiv[/itex]constant, but this constant will be dependent on [itex]y[/itex] and therefor is a function of [itex]y[/itex].

    Get the general idea now?
     
  4. Mar 30, 2012 #3

    MathematicalPhysicist

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    That's basically what I have written, so I know that f_x =F(x) and f_y=G(y). But this is where I am not sure how to procceed.

    I mean I know that: if F(x)=h'(x), then G(x,y)= f(x,y)-h(x) then G_x=0 and then G=g(y).

    The problem is how do I know that F(x)=h'(x) I am assuming that F is of this form, aren't I?
     
  5. Mar 30, 2012 #4

    hunt_mat

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    I think that as you effectively have the equation [itex]f'(x)=g(x)[/itex] then you want to show that there is a function [itex]F(x)[/itex] with the property [itex]F'(x)=f(x)[/itex], and I think that this follows from a version of the fundamental theorem of calculus. By extension [itex]G(x)=F(x)+C[/itex] will also satisfy this equation. I think you can say this because [itex]f\in C^{1}[/itex] as you have [itex]f'(x)[/itex].
     
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