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I Question About Exact Differential Form

  1. Jun 13, 2017 #1

    Drakkith

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    My book is going through a proof on exact differential forms and the test to see if they're exact, and I'm lost on one part of it.

    It says:

    If $$M(x,y)dx + N(x,y)dy = \frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy$$ then the calculus theorem concerning the equality of continuous mixed partial derivatives $$\frac{\partial }{\partial y}\frac{\partial F}{\partial x}=\frac{\partial }{\partial x}\frac{\partial F}{\partial y}$$ would dictate a "compatibility condition" on the functions ##M## and ##N##: $$\frac{\partial}{\partial y}M(x,y)=\frac{\partial}{\partial x}N(x,y)$$

    What does this mean? What is the "calculus theorem concerning the equality of continuous mixed partial derivatives" it talks about?
     
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  3. Jun 13, 2017 #2

    jedishrfu

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  4. Jun 13, 2017 #3

    mathwonk

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    the calculus theorem says that if F has 2 continuous partial derivatives then ∂^2F/∂x∂y = ∂^2F/∂y∂x. Thus a necessary condition for M to equal ∂F/∂x and for N to equal ∂F/∂y, is that we must have ∂M/∂y = ∂^2F/∂x∂y = d^2F/∂y∂x = ∂N/∂x.
     
  5. Jun 13, 2017 #4

    Drakkith

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    So, this is basically saying that given a function ##F(x,y)##, the 2nd order mixed partial derivatives of that function are equal even if you flip the order in which you take the derivatives?

    If so, then the compatibility condition that ##\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}## makes sense.
    It helps when you understand what a 2nd order mixed partial derivative is. :rolleyes:
     
  6. Jun 14, 2017 #5

    mathwonk

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    the theorem does not say that the second order partials must always be equal in both orders, but it does hold if they are continuous. i.e. there do exist functions whose second order partials exist but are not continuous, and then the mixed order partials do not need to be equal.
     
  7. Jun 14, 2017 #6

    Drakkith

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    If the 2nd order mixed partials are not continuous, I assume the theorem is still true over the range in which they are continuous?
     
  8. Jun 14, 2017 #7

    fresh_42

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    Yes, because differentiability as well as continuity are local properties. If you narrow down the domain to an open neighborhood where the conditions are met, the theorem is applicable.
     
  9. Jun 14, 2017 #8

    Drakkith

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    Got it. Thanks all!
     
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