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Proof of function existence

  1. Aug 18, 2014 #1
    Suppose all second partial derivatives of [itex]F = F (x, y)[/itex] are continuous and [itex]F_{xx} + F_{yy} = 0[/itex] on an open rectangle [itex]R[/itex].
    Show that [tex]F_ydx - F_xdy = 0[/tex] is exact on [itex]R[/itex], and therefore there’s a function [itex]G[/itex] such that
    [tex]G_x = −F_y[/tex] and [tex]Gy = F_x[/tex] in [itex]R[/itex].
    ≈≈≈≈≈≈≈≈
    To prove that [itex]F_ydx + F_xdy = 0[/itex] is exact on [itex]R[/itex],
    I have [tex]F_{xx} + F_{yy} = 0[/tex]
    which is [tex]F_{xx}=-F_{yy}[/tex]
    Integrating both sides and cancel out the constants I obtain
    [tex]F_x=-F_y[/tex]
    This proves the function [itex]F_ydx - F_xdy = 0[/itex] is exact on [itex]R[/itex]​
    Could you help me prove the existence of [itex]G[/itex] ? Thank you.....
     
  2. jcsd
  3. Aug 19, 2014 #2

    WWGD

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    Science Advisor
    Gold Member

    If your rectangle lives in the plane, then just show the form [tex]F_ydx - F_xdy = 0[/tex] is closed, since in a rectangle, every closed form is exact. If you haven't seen this, what results can you use?
     
  4. Aug 19, 2014 #3

    Borek

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    Staff: Mentor

    How do you know constants are identical?
     
  5. Aug 19, 2014 #4
    I answered so, I meant to choose 2 exact constants to give both a go.
    I'm thinking because I already answered the first which is also the main part of the problem, I may continue to put "the given function has become exact on R, so there must be a function G that satisfies both of the given conditions in R"

    Is this a correct solution ?
     
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