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Classification of PDE

  1. Nov 24, 2011 #1
    Ok, so I got this equation:

    [itex]y^2 \frac{∂^2 u}{∂x^2} + 2xy \frac{∂^2 u}{∂x∂y} + x^2 \frac{∂^2 u}{∂y^2} = 0 [/itex]

    [itex]A = y^2[/itex]
    [itex]B = xy[/itex]
    [itex]C = x^2[/itex]

    Now I want to see what type it is, so I compute [itex]B^2 - A C = 0[/itex] which by definition is parabolic. However, according to an earlier statement in my book a parabolic PDE is one that has a quadratic form (at some point) that consists of fewer than n squares, not necessarily all of the same sign. n signifies the space we're in (that would be n=2). However, to me it seems like there are two squares in the equation, so it can't be parabolic?

    Are they "reducing" it by writing [itex] (y \frac{∂u}{∂x} + x \frac{∂u}{∂y} )^2 = 0 [/itex] and saying it contains less than n squares after dropping the square?
    Last edited: Nov 24, 2011
  2. jcsd
  3. Nov 25, 2011 #2


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    No, that's not the reason The quadratic [itex]Ax^2+ 2Bxy+ Cx^2[/itex], with [itex]B^2- AC= 0[/itex] is "parabolic" Completing the square, [itex]A(x^2+ (2B/A)xy+ (B^2/A^2)y^2)- (B^2/A)y^2+ Cy^2[/itex][itex]= A(x- (B/A)y)^2- (B^2- 4AC)y^2/A= A(x- (B/A)x)^2[/itex], a "perfect square". That's the reason.
  4. Nov 29, 2011 #3
    Doing what I think you do I get this:

    [itex]A (x^2 + 2 \frac{B}{A} xy + \frac{C}{A}y^2)=[/itex]
    [itex]A ((x + \frac{B}{A} y)^2 - \frac{B^2}{A^2}y^2 + \frac{C}{A}y^2)[/itex]

    Using [itex]C=\frac{B^2}{A}[/itex] (i.e. parabolic)

    [itex]A (x + \frac{B}{A} y)^2 - \frac{B^2}{A^2}y^2 + \frac{B^2}{A^2}y^2)=[/itex]
    [itex]A (x + \frac{B}{A} y)^2[/itex]

    This counts as one square (?), while a [itex]C \neq \frac{B^2}{A}[/itex] would generate an additional [itex]y^2 [/itex] term?
  5. Nov 30, 2011 #4
    Now when I know the type, how do I reduce this one to canonical form?

    [itex]y^2 \frac{∂^2 u}{∂x^2} + 2xy \frac{∂^2 u}{∂x∂y} + x^2 \frac{∂^2 u}{∂y^2} = 0 [/itex]

    The strategy seems to be to make a change of variables to make most high order terms vanish.

    [itex]\xi = \xi(x,y)[/itex]
    [itex]\eta = \eta(x,y)[/itex]

    By the chain rule:
    [itex]u_x = u_\xi \xi_x + u_\eta \eta_x[/itex]
    [itex]u_y = u_\xi \xi_y + u_\eta \eta_y[/itex]

    Then it is applied again, so I get expressions for [itex]u_{xx}, u_{xy}, u_{yy}[/itex]. I reorder them and get

    [itex]A^* \frac{∂^2 u}{∂ \xi ^2} + 2 B^* \frac{∂^2 u}{∂ \xi ∂ \eta} + C^* \frac{∂^2 u}{∂ \eta^2} + F^* = 0 [/itex] with [itex]F^*[/itex] containing lower order elements.


    [itex]A^* = A \xi_x \xi_x + 2B \xi_x \xi_y + C \xi_y \xi_y[/itex]
    [itex]C^* = A \eta_x \eta_x + 2B \eta_x \eta_y + C \eta_y \eta_y[/itex]
    [itex]B^* = A \xi_x \eta_x + B (\xi_x \eta_y + \xi_y \eta_x) + C \xi_y \eta_y[/itex]

    I want [itex]A^* = 0[/itex], so I solve that and get

    [itex]A( \xi_x + \frac{B}{A} \xi_y)^2 = 0[/itex]


    [itex]A \xi_x + B \xi_y = 0[/itex]
    [itex] y^2 \xi_x + xy \xi_y = 0[/itex]
    [itex] ydy = xdx[/itex]
    [itex] y^2 - x^2 = \gamma = \xi[/itex]
    Then I have to choose [itex]\eta[/itex] with respect to the Jacobian (it mustn't vanish), so I pick [itex]\eta=x^2[/itex] .

    [itex]B^*[/itex] disappears as well here, which leaves [itex]C^*[/itex]

    I'm left with

    [itex] \frac{∂^2 u}{∂\eta^2} = - F^* / C^* [/itex]

    Is this correct?

    However, I still need to determine the contents of [itex]F^*[/itex], will I have to compute lower order terms each time I do the chain rule to get [itex]u_{xx}[/itex] [itex]u_{xy}[/itex] and [itex]u_{yy}[/itex]?
    Last edited: Nov 30, 2011
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