Understanding PDE Classification: Parabolic Equations and Quadratic Forms

[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 $u_{yy}$?

 

[blue_raver22]

I would first clarify the definitions of the terms being used in this content. A parabolic PDE is one that can be written in the form of a second-order partial differential equation with a quadratic form, as shown in the given equation. The term "quadratic form" refers to an expression that contains terms with degrees of two or less. In this case, the given equation has three terms with degrees of two or less, making it a parabolic PDE.

The statement in the book about "fewer than n squares" refers to the number of squares present in the quadratic form, not the overall equation. In this case, there are two squares present in the quadratic form, making it a parabolic PDE.

Additionally, the equation given in the content is already in its simplest form and does not require any further reduction. The term (y \frac{∂u}{∂x} + x \frac{∂u}{∂y} )^2 = 0 is not a reduction of the original equation, but rather a different representation of the same equation.

In summary, the given equation is a parabolic PDE based on its form and the number of squares present in the quadratic form. It is not necessary to reduce the equation or change its representation to determine its classification.