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Help needed with an elliptic PDE

  1. Nov 2, 2007 #1

    In our math course, we encountered the following elliptic PDE:

    y^{2}u_{xx} + u_{yy} = 0

    In order to solve it, we converted it to the characteristic equation,

    y^{2}\left(\frac{dy}{dx}\right)^{2} + 1 = 0

    Next, we wrote:

    [tex]\frac{dy}{dx} = \frac{i}{y}[/tex]

    My question is: the characteristic equation has no solution in [itex]\mathbb{R}[/itex] but we went ahead and mechanically wrote the expression for [itex]dy/dx[/itex]. Does this mean that we should regard [itex]x[/itex] and [itex]y[/itex] as complex variables? If so, how does one reconcile with the fact that some solution to the PDE as [itex]u(x,y) = c[/itex] is a surface in [itex](x,y,u)[/itex] space? Perhaps this is a trivial question, but I'm just starting to learn PDEs. Does this also mean that we should not ascribe a physical significance to x and y as coordinates in [itex]\mathbb{R}^2[/itex]?

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  3. Nov 2, 2007 #2


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    Yes, the fact that it is an elliptic equation tells you that the characteristic equation does not have real roots.
  4. Nov 2, 2007 #3
    Oops yes, of course...I didn't see that.

    Also, in such a case, do y and x lose their "physical significance" of being real variables in real space?
  5. Nov 3, 2007 #4


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    Mmm..no, it means that the characteristic curve itself doesn't lie in real space.
    That's quite a different thing from saying that x and y cannot be regarded as real variables.
  6. Nov 4, 2007 #5
    Could you please elaborate? And where can I read more about such issues?
  7. Nov 4, 2007 #6


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    Well, my memory on characteristics has gone hazy, so it would be helpful if you posted the precise procedure utilized in the particular example.

    However, as a general trait, the method of characteristics is a trick whereby we get a family of curves along everyone of which the u-signal propagates in a simple manner (say, by being conserved).

    If therefore that family of curves lie in the complex plane, it means that there aren't a set of real curves y(x) along which u propagates. For example, y cannot be solved entirely as a function of x when we constrict ourselves to the real plane.

    Please post a few details about the specific procedure.
  8. Nov 21, 2007 #7
    Ok, so the elliptic equation is

    [tex]au_{xx} + 2bu_{xy} + cu_{yy} = 0[/tex]

    and its characteristic equation is

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

    Here, [itex]b^{2}-ac <0[/itex] so it has complex roots, and the characteristic curves are:

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