Classification of Linear 2nd-Order PDEs

**JavaJunkie** · August 29th 2010, 03:50 AM

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_{{ 12}}&a_{{22}}\end {array} \right] $

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 $\left[ \begin {array}{cc} 1&-3\\ \noalign{\medskip}-3&0\end {array} \right]$ so Det $A = -9.$

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

I get $\left[ \begin {array}{cc} 3&-6\\ \noalign{\medskip}-6&3\end {array} \right]$ 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} \right] $ 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

**Danny** · August 29th 2010, 05:44 AM

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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