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Math Help - Classification of Linear 2nd-Order PDEs

  1. #1
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    Classification of Linear 2nd-Order PDEs

    Hey,
    I'm reading a textbook that tells me that if we have a general linear homogeneous pde of 2nd order:

    a_{11} u_{xx} + 2a_{12}u_{xy} + a_{22}u_{yy} + a_1 u_x + a_2 u_y + a_0 u = 0

    we can create a matrix A, where

    A = \left[ \begin {array}{cc} a_{{11}}&a_{{12}}\\ \noalign{\medskip}a_{{<br />
12}}&a_{{22}}\end {array} \right] <br /> <br />

    and use it to classify the PDE as follows:

    Elliptic: Det A > 0 i.e. a_{11}a_{22} > a_{12}^2

    Hyperbolic: Det A < 0

    Parabolic: Det A = 0

    which seems fairly simple enough. But in the examples they have, I get something completely different to the solutions they provide.

    (a) u_{xx} - 3u_{xy} = 0. I get \left[ \begin {array}{cc} 1&-3\\ \noalign{\medskip}-3&0\end {array}<br />
 \right] so Det A = -9.

    (b) 3u_{xx} - 6u_{xy} + 3u_{yy} + u_x = 0.

    I get \left[ \begin {array}{cc} 3&-6\\ \noalign{\medskip}-6&3\end {array}<br />
 \right] and Det A = -27

    (c) 2u_{xx} + 2u_{xy} + 3u_{yy} = 0

    I get  \left[ \begin {array}{cc} 2&2\\ \noalign{\medskip}2&3\end {array}<br />
 \right] <br />
and det A = 2

    According to the text these should be

    (a) Det A = -9/4

    (b) Det A = 0

    (c) Det A = 5

    but I get something different.

    Also, as another example, we have

    xu_{xx} - u_{xy} + yu_{yy} = 0

    and we should have Det A = a_{11}a_{22} - a_{12}^2 = xy - \frac{1}{4}. Where is the - \frac{1}{4} coming from?

    Thanks
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  2. #2
    MHF Contributor
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    Actually, the way that I was taught (and teach) is to consider the form

    Au_{xx} + B u_{xy} + C u_{yy} + lots = 0 (lots - lower order terms)

    B^2 - 4AC > 0 - hyperbolic
    B^2-4AC = 0 - parabolic
    B^2 - 4AC < 0 - elliptic.

    Just like classifying quadratics

    Ax^2 + Bxy + Cy^2 + lots = 0.
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