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Help with 1st order quasilinear PDE

  1. Sep 24, 2009 #1
    I have to solve:

    [tex]
    x(y^2 - z^2) \frac{\partial z}{\partial x} + y(z^2 - x^2) \frac{\partial z}{\partial y} + z(x^2 - y^2)
    [/tex]

    So, I write out the characteristic system of ODEs:

    [tex]
    \frac{dx}{x(y^2 - z^2)} = \frac{dy}{y(z^2 - x^2)} = \frac{dz}{z(x^2 - y^2)}
    [/tex]

    Now, the variables aren't seperated so I can't integrate two pairs seperately, so what I did was use the componendo et dividendo rule for fractions to write :

    [tex]
    \frac{dx}{x(y^2 - z^2)} = \frac{dy}{y(z^2 - x^2)} = \frac{dz}{z(x^2 - y^2)} =
    \frac{yzdx+xzdy+xydz}{xyz(y^2 - z^2)+xyz(z^2 - x^2)+xyz(x^2 - y^2)} =
    \frac{yzdx+xzdy+xydz}{0}
    [/tex]

    Thus
    [tex]
    yzdx+xzdy+xydz=0
    [/tex]

    which is in integrable Pfaffian Equation, and its integration yields:

    [tex]
    xyz=C_1
    [/tex]

    ie. I have one of the constants of integration of the characteristic system of ODEs.
    But now I'm stuck because I don't know how to obtain the other :S

    I thought to use one pair of the ODE system, eg,

    [tex]
    \frac{dx}{x(y^2 - z^2)} = \frac{dy}{y(z^2 - x^2)}
    [/tex]

    and substitute
    [tex]
    z=C_1/(xy)
    [/tex]

    from the first integration, so that I'd be left with an ode:

    [tex]
    \frac{dx}{x(y^2 - C_1^2/(xy)^2)} = \frac{dy}{y(C_1^2/(xy)^2) - x^2)}
    [/tex]

    from which in theory we could obtain a second integration constant, but this is too hard to solve (I think), so there must be an easier way to get the second integration constant, ie.

    [tex]
    g(x,y,z)=C_2
    [/tex]

    And the general solution of the PDE would be

    [tex]
    F[xyz,g(x,y,z)]=0,
    [/tex]

    with F an arbitrary function.

    Anyone have any ideas please?:P

    Thank you :)
     
  2. jcsd
  3. Sep 25, 2009 #2
    It seems to me that the general solution in implicit form to your PDE is as follows

    \int_{-\infty}^ \infty F(z^2-x^2-y^2,c)(\frac{xy}{z})^c dc=0 ,

    where F(a,c) is an arbitray function.
     
  4. Sep 25, 2009 #3
    Oh dear, how in the world did you get that?
     
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