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Thread: Canonical Form - Second Order PDE

  1. #1
    Newbie mukmar's Avatar
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    Canonical Form - Second Order PDE

    For the equation:

    $\displaystyle 2x^2u_{xx} + 5xyu_{xy} + 2y^2u_{yy} + 8xu_{x} + 5yu_y = 0$

    I'm just trying to understand how to put it in canonical form.

    I've understood till the point where the characteristics are:

    $\displaystyle \xi = \frac{y}{x^2} $

    AND

    $\displaystyle \eta = \frac{y^2}{x}$

    I'm just not sure how that leads to the canonical form:

    $\displaystyle \eta u_{\xi\eta} + u_{\xi} = 0$
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  2. #2
    Senior Member
    Joined
    Mar 2010
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    To begin with

    $\displaystyle
    u_x=u_\xi \; \xi_x+u_\eta \; \eta_x
    $

    and the second $\displaystyle x$ derivative is

    $\displaystyle
    u_{xx}=(u_\xi \; \xi_x)_x+(u_\eta \; \eta_x)_x=u_{\xi x}\; \xi_x+u_\xi \; \xi_{xx}+u_{\eta x}\; \eta_x+u_\eta \; \eta_{xx}=
    $

    $\displaystyle
    =(u_{\xi \xi} \; \xi_x+u_{\xi \eta} \; \eta_x) \; \xi_x+u_\xi \; \xi_{xx}+(u_{\eta \xi} \xi_x+u_{\eta \eta} \eta_x) \; \eta_x+u_\eta \; \eta_{xx}=
    $

    $\displaystyle
    =u_{\xi \xi} \; \xi_x^2+2u_{\xi \eta} \; \xi_x \eta_x+u_{\eta \eta} \eta_x^2+u_\xi \; \xi_{xx}+u_\eta \; \eta_{xx}.
    $

    Known functions are
    $\displaystyle
    \xi_x=-2 \frac{y}{x^3}
    $
    and
    $\displaystyle
    \eta_x=- \frac{y^2}{x^2},
    $
    also
    $\displaystyle
    \xi_{xx} \; and \; \eta_{xx}.
    $

    In the same way with
    $\displaystyle
    u_{xy} \; \; and \; \; u_{yy}.
    $

    When we have everything then inserting this to initial de.
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