Understanding PDE Classification: Parabolic Equations and Quadratic Forms

In summary, the author is trying to determine the parabolic form of an equation, but is having trouble because there are two squares in the equation. He solves for the equation to get A^*=0, and then uses the chain rule to determine the contents of F^*.
  • #1
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Ok, so I got this equation:

[itex]y^2 \frac{∂^2 u}{∂x^2} + 2xy \frac{∂^2 u}{∂x∂y} + x^2 \frac{∂^2 u}{∂y^2} = 0 [/itex]

[itex]A = y^2[/itex]
[itex]B = xy[/itex]
[itex]C = x^2[/itex]

Now I want to see what type it is, so I compute [itex]B^2 - A C = 0[/itex] 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 [itex] (y \frac{∂u}{∂x} + x \frac{∂u}{∂y} )^2 = 0 [/itex] and saying it contains less than n squares after dropping the square?
 
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  • #2
No, that's not the reason The quadratic [itex]Ax^2+ 2Bxy+ Cx^2[/itex], with [itex]B^2- AC= 0[/itex] is "parabolic" Completing the square, [itex]A(x^2+ (2B/A)xy+ (B^2/A^2)y^2)- (B^2/A)y^2+ Cy^2[/itex][itex]= A(x- (B/A)y)^2- (B^2- 4AC)y^2/A= A(x- (B/A)x)^2[/itex], a "perfect square". That's the reason.
 
  • #3
Doing what I think you do I get this:

[itex]A (x^2 + 2 \frac{B}{A} xy + \frac{C}{A}y^2)=[/itex]
[itex]A ((x + \frac{B}{A} y)^2 - \frac{B^2}{A^2}y^2 + \frac{C}{A}y^2)[/itex]

Using [itex]C=\frac{B^2}{A}[/itex] (i.e. parabolic)

[itex]A (x + \frac{B}{A} y)^2 - \frac{B^2}{A^2}y^2 + \frac{B^2}{A^2}y^2)=[/itex]
[itex]A (x + \frac{B}{A} y)^2[/itex]

This counts as one square (?), while a [itex]C \neq \frac{B^2}{A}[/itex] would generate an additional [itex]y^2 [/itex] term?
 
  • #4
Now when I know the type, how do I reduce this one to canonical form?


[itex]y^2 \frac{∂^2 u}{∂x^2} + 2xy \frac{∂^2 u}{∂x∂y} + x^2 \frac{∂^2 u}{∂y^2} = 0 [/itex]

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

[itex]\xi = \xi(x,y)[/itex]
[itex]\eta = \eta(x,y)[/itex]

By the chain rule:
[itex]u_x = u_\xi \xi_x + u_\eta \eta_x[/itex]
[itex]u_y = u_\xi \xi_y + u_\eta \eta_y[/itex]

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

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

where

[itex]A^* = A \xi_x \xi_x + 2B \xi_x \xi_y + C \xi_y \xi_y[/itex]
[itex]C^* = A \eta_x \eta_x + 2B \eta_x \eta_y + C \eta_y \eta_y[/itex]
[itex]B^* = A \xi_x \eta_x + B (\xi_x \eta_y + \xi_y \eta_x) + C \xi_y \eta_y[/itex]

I want [itex]A^* = 0[/itex], so I solve that and get

[itex]A( \xi_x + \frac{B}{A} \xi_y)^2 = 0[/itex]

i.e.

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

[itex]B^*[/itex] disappears as well here, which leaves [itex]C^*[/itex]

I'm left with

[itex] \frac{∂^2 u}{∂\eta^2} = - F^* / C^* [/itex]

Is this correct?

However, I still need to determine the contents of [itex]F^*[/itex], will I have to compute lower order terms each time I do the chain rule to get [itex]u_{xx}[/itex] [itex]u_{xy}[/itex] and [itex]u_{yy}[/itex]?
 
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  • #5


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.
 

1. What is a Partial Differential Equation (PDE)?

A Partial Differential Equation (PDE) is a type of mathematical equation that involves multiple independent variables and their partial derivatives. It is used to describe the behavior of a system or phenomenon that changes over multiple dimensions, such as time and space.

2. How are PDEs classified?

PDEs are classified based on their order, linearity, and type. The order of a PDE refers to the highest order of the partial derivatives present in the equation. Linearity refers to whether the equation is linear or nonlinear, while the type refers to the specific form of the equation, such as elliptic, hyperbolic, or parabolic.

3. What is the difference between an ordinary differential equation (ODE) and a PDE?

While both ODEs and PDEs involve derivatives, ODEs only involve one independent variable and its derivatives, while PDEs involve multiple independent variables and their partial derivatives. This means that ODEs only describe phenomena that vary in one dimension, while PDEs can describe phenomena that vary in multiple dimensions.

4. What are some applications of PDEs?

PDEs have many applications in physics, engineering, and other fields. They are commonly used to describe heat transfer, fluid mechanics, electromagnetism, and quantum mechanics, among others. They are also used in financial modeling, image processing, and computer graphics.

5. How are PDEs solved?

The solution of a PDE depends on the specific type of PDE and its boundary conditions. Some PDEs have analytical solutions, while others require numerical methods for approximation. Common methods for solving PDEs include separation of variables, finite difference methods, and finite element methods.

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