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Exact Differential Equation

  1. Dec 5, 2009 #1
    We can use Green's theorem to understand why the Exact Differential Equation satisfy the conditions it should have .....

    How about a DE for more than two variables ?


    IF the equation above is an Exact Differential Equation , what condition it would satisfy ?
  2. jcsd
  3. Dec 5, 2009 #2


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    An equation of the form f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz= 0 is "exact" if and only if there exist some F(x,y,z) such that
    [tex]dF= \frac{\partial F}{\p
    artial x}dx+ \frac{\partial F}{\partial y}dy+ \frac{\partial F}{\partial z}= f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz[/tex]

    That means that we must have
    [tex]\frac{\partial F}{\partial x}= f(x,y,z)[/tex]
    [tex]\frac{\partial F}{\partial y}= g(x,y,z)[/tex]
    [tex]\frac{\partial F}{\partial z}= h(x,y,z)[/tex]

    So, as long as those functions are continuous, we must have the "mixed partials" equal:
    [tex]\frac{\partial^2 F}{\partial x\partial y}= \frac{\partial f}{\partial y}= \frac{\partial g}{\partial x}= \frac{\partial^2 F}{\partial y\partial x}[/tex]

    Since you mention "Green's theorem" (for two variables) you might want to look at [itex]\nabla\times f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}[/itex] and Stoke's theorem.
    Last edited: Dec 5, 2009
  4. Dec 5, 2009 #3
    What u mentioned contains three variables ~and I've seen such a case in the book I read...

    Suppose [tex]\vec{F}[/tex]=P(x,y,z)[tex]\hat{i}[/tex]+Q(x,y,z)[tex]\hat{j}[/tex]+R(x,y,z)[tex]\hat{k}[/tex]

    dT=P(x,y,z)dx+Q(x,y,z)dy+R(x,y,z)dz is exact if only if [tex]\vec{F}[/tex][tex]\bullet[/tex]([tex]\nabla[/tex][tex]\times[/tex][tex]\vec{F}[/tex])=0

    I'm now requesting for the case which contains more than three variables~

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