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Math Help - Transforming a 2nd order PDE into canonical form

  1. #1
    Senior Member bkarpuz's Avatar
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    Exclamation Transforming 2nd order PDE into canonical form

    Dear MHF members,

    first of all glad to see the form back again.
    I have the following problem $yz_{xx}+z_{yy}=0$.
    When I check the discriminant, I see that $\Delta=0^2-4y=-4y$.
    Hence, the following cases occur.
    (a) For $y<0$, the equation is of hyperbolic type.
    (b) For $y=0$, the equation is of parabolic type.
    (c) For $y>0$, the equation is of elliptic type.

    I have no problem with the cases (b) and (c).
    The equation is already in its canonical form in (b),
    and in (c) I can transform the equation in this case
    into its canonical form $z_{\xi\xi}+z_{\eta\eta}+\frac{1}{3\eta}z_{\eta}=0 $
    by the change of variables $\xi=\frac{2}{3}x$ and $\eta=y^{3/2}$.
    But I really stuck at the part (a). I don't know why
    I cannot transform the equation into its canonical form.
    There must to be a trick in this case, which I miss.

    Thank you.
    bkarpuz
    Last edited by bkarpuz; March 4th 2012 at 12:44 AM.
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  2. #2
    Senior Member bkarpuz's Avatar
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    Re: Transforming 2nd order PDE into canonical form

    Quote Originally Posted by bkarpuz View Post
    I have the following problem $yz_{xx}+z_{yy}=0$.
    When I check the discriminant, I see that $\Delta=0^2-4y=-4y$.
    Hence, the following cases occur.
    (a) For $y<0$, the equation is of hyperbolic type.
    ...
    But I really stuck at the part (a). I don't know why
    I cannot transform the equation into its canonical form.
    There must to be a trick in this case, which I miss.
    Okay, I got the solution. While obtaining the characteristics I, in general, used
    $\frac{\mathrm{d}y}{\mathrm{d}x}=\pm i\frac{1}{\sqry{y}}$ but here it makes problem.
    Thus I took the reciprocal equation $\frac{\mathrm{d}y}{\mathrm{d}x}=\pm i\sqrt{y}$,
    which gives $\xi=\frac{3}{2}x+(-y)^{3/2}$ and $\eta=\frac{3}{2}x-(-y)^{3/2}$.
    Thus, the equation transforms into $z_{\xi\eta}+\frac{1}{6(\xi-\eta)}(z_{\xi}-z_{\eta})=0$.

    Have a good day.
    bkarpuz
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