# Classification of PDE

1. Nov 24, 2011

### S. Moger

Ok, so I got this equation:

$y^2 \frac{∂^2 u}{∂x^2} + 2xy \frac{∂^2 u}{∂x∂y} + x^2 \frac{∂^2 u}{∂y^2} = 0$

$A = y^2$
$B = xy$
$C = x^2$

Now I want to see what type it is, so I compute $B^2 - A C = 0$ 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 $(y \frac{∂u}{∂x} + x \frac{∂u}{∂y} )^2 = 0$ and saying it contains less than n squares after dropping the square?

Last edited: Nov 24, 2011
2. Nov 25, 2011

### HallsofIvy

No, that's not the reason The quadratic $Ax^2+ 2Bxy+ Cx^2$, with $B^2- AC= 0$ is "parabolic" Completing the square, $A(x^2+ (2B/A)xy+ (B^2/A^2)y^2)- (B^2/A)y^2+ Cy^2$$= A(x- (B/A)y)^2- (B^2- 4AC)y^2/A= A(x- (B/A)x)^2$, a "perfect square". That's the reason.

3. Nov 29, 2011

### S. Moger

Doing what I think you do I get this:

$A (x^2 + 2 \frac{B}{A} xy + \frac{C}{A}y^2)=$
$A ((x + \frac{B}{A} y)^2 - \frac{B^2}{A^2}y^2 + \frac{C}{A}y^2)$

Using $C=\frac{B^2}{A}$ (i.e. parabolic)

$A (x + \frac{B}{A} y)^2 - \frac{B^2}{A^2}y^2 + \frac{B^2}{A^2}y^2)=$
$A (x + \frac{B}{A} y)^2$

This counts as one square (?), while a $C \neq \frac{B^2}{A}$ would generate an additional $y^2$ term?

4. Nov 30, 2011

### S. Moger

Now when I know the type, how do I reduce this one to canonical form?

$y^2 \frac{∂^2 u}{∂x^2} + 2xy \frac{∂^2 u}{∂x∂y} + x^2 \frac{∂^2 u}{∂y^2} = 0$

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

$\xi = \xi(x,y)$
$\eta = \eta(x,y)$

By the chain rule:
$u_x = u_\xi \xi_x + u_\eta \eta_x$
$u_y = u_\xi \xi_y + u_\eta \eta_y$

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

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

where

$A^* = A \xi_x \xi_x + 2B \xi_x \xi_y + C \xi_y \xi_y$
$C^* = A \eta_x \eta_x + 2B \eta_x \eta_y + C \eta_y \eta_y$
$B^* = A \xi_x \eta_x + B (\xi_x \eta_y + \xi_y \eta_x) + C \xi_y \eta_y$

I want $A^* = 0$, so I solve that and get

$A( \xi_x + \frac{B}{A} \xi_y)^2 = 0$

i.e.

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

$B^*$ disappears as well here, which leaves $C^*$

I'm left with

$\frac{∂^2 u}{∂\eta^2} = - F^* / C^*$

Is this correct?

However, I still need to determine the contents of $F^*$, will I have to compute lower order terms each time I do the chain rule to get $u_{xx}$ $u_{xy}$ and $u_{yy}$?

Last edited: Nov 30, 2011