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Non-exact differential equation

  1. Dec 29, 2012 #1
    I try to show, that equation
    [itex]\frac{-y}{ x^{2}+y ^{2} } + \frac{x}{ x^{2}+y ^{2}}y'=0[/itex]
    is not exact in [itex]\mathbb{R^{2}} \setminus \{(0,0)\}[/itex].
    It's obvious that I have to use the fact, that the set is not simply connected, but I don't know how to do it.
     
  2. jcsd
  3. Dec 29, 2012 #2

    mfb

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    2016 Award

    Staff: Mentor

    I would try this:
    Assume that it is exact and that there is a potential function F. Integrate its derivative (as given by your problem) in a circle around (0,0). If such a function F exists, the result has to be 0.
     
  4. Dec 30, 2012 #3
    I'm not sure, if we can use the line integral here. We are trying to show, that there doesn't exist function F such that F is exact differential, that is
    [tex]\frac{ \partial F}{ \partial x}=\frac{-y}{ x^{2}+y ^{2} }[/tex] and [tex]\frac{ \partial F}{ \partial y}=\frac{x}{ x^{2}+y ^{2}}[/tex]
    Existence of such function is equivalent to [tex]\int_{L}\frac{-y}{ x^{2}+y ^{2} } \mbox{d}x +\frac{x}{ x^{2}+y ^{2}} \mbox{d}y=0[/tex] for every closed curve L. Although, this statement is true only if the domain is simply connected; and our domain isn't of this kind.
     
  5. Dec 30, 2012 #4

    lurflurf

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    consider
    (y/x)'
    or
    (Arctan(y/x))'
     
  6. Dec 30, 2012 #5
    I know how to solve this equation and how to find F in any "regular" domain, for example in real plane [tex]\mathbb{R^{2}}[/tex]. Problems appear in the neighbourhood of point (0,0) in our domain, because all methods of solving this kind of equation, I know are valid only in simply connected domain.
     
  7. Dec 30, 2012 #6

    lurflurf

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    I do not know why you could not use integrals.

    http://en.wikipedia.org/wiki/Atan2

    Suppose
    F=C+Arctan2(y,x)
    F'=0
    but consider the nonexistence of the limit (by inequality of directional limits)

    [itex]\lim_{(x,y) \rightarrow (0,0)} F(x,y)[/itex]

    F cannot be continuous and the equation is not exact

    Note that if we had excluded a path to infinity along with 0 we would have a simply connected region and an exact equation
     
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