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Canonical Transformation of Parabolic PDEs

  1. Nov 21, 2007 #1
    Hi again

    I am studying PDEs and came across a solved problem in my textbook, which describes the transformation of a parabolic second order PDE to canonical form. I want to know how to find the second canonical substitution when one has been computed from the characteristic equation.

    (PS--This is not a homework problem.)

    For instance, suppose the given equation is

    y^{2}u_{xx} - 2xyu_{xy} + x^{2}u_{yy} = \frac{y^2}{x}u_{x} + \frac{x^2}{y}u_{y}

    The solution is as follows:

    Compare it with the 'standard' semi-linear second order PDE:

    [tex]a(x,y)u_{xx} + 2b(x,y)u_{xy} + c(x,y)u_{yy} = \phi(x,y,u,u_{x},u_{y})[/tex]

    to get [itex]a(x,y) = y^{2}[/itex], [itex]b(x,y) = -xy[/itex], [itex]c(x,y) = x^{2}[/itex]. Since [itex]b^{2}-ac = 0[/itex], the equation is parabolic. Considering level curves

    [tex]\zeta(x,y) = c_{1}[/tex]
    [tex]\eta(x,y) = c_{2}[/tex]

    corresponding to the new independent variables [itex](\zeta,\eta)[/itex], the characteristic equation is

    [tex]a\left(\frac{dy}{dx}\right)^{2} - 2b\left(\frac{dy}{dx}\right) + c = 0[/tex]

    It has a double root [itex]y^{2}+x^{2} = c_{1}[/itex]. Thus

    [tex]\zeta(x,y) = x^{2} + y^{2}[/tex]

    But this determines only one of the canonical variables. The only condition on [itex]\eta[/itex] is that

    [tex]\frac{\partial(\zeta,\eta)}{\partial(x,y)} \neq 0[/tex]

    which means that [itex]\eta[/itex] should not be explicitly dependent on [itex]\zeta[/itex] or conversely.

    Here, it seems "natural" to take [itex]\eta(x,y) = x^{2}-y^{2}[/itex]. But how does one find a [itex]\eta[/itex] in the general case?

    Last edited: Nov 21, 2007
  2. jcsd
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