Classification of PDE

[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?

[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.

[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?

[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}?