# Thread: Classification of Linear 2nd-Order PDEs

1. ## 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:

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

we can create a matrix A, where

\displaystyle 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 $\displaystyle A > 0$ i.e. $\displaystyle a_{11}a_{22} > a_{12}^2$

Hyperbolic: Det $\displaystyle A < 0$

Parabolic: Det $\displaystyle A = 0$

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

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

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

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

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

I get \displaystyle \left[ \begin {array}{cc} 2&2\\ \noalign{\medskip}2&3\end {array} \right] and det $\displaystyle A = 2$

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

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

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

Thanks

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

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

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

Just like classifying quadratics

$\displaystyle Ax^2 + Bxy + Cy^2 + lots = 0$.