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Characterisitics of a Parabolic PDE

by Tohiko
Tags: characterisitics, parabolic
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Tohiko
#1
Jul29-11, 02:22 AM
P: 8
Greetings,

I want to find the characteristics of the following parabolic PDE
[itex]u_t + v u_x + w u_y + a(t, x,y,v,w, u) u_v + b(t, x,y,v,w, u) u_w - u_{vv} - u_{ww} = c(t,x,y,v,w,u)[/itex]
Where [itex]u=u(t,x,y,v,w)[/itex]
I know how to find the characteristics of a 2nd-order one-dimensional PDE. I also know how to find the Riemann invariants of a hyperbolic multidimensional PDE.
But how do I find the characteristics of a 2nd-order, nonlinear, multidimensional, parabolic PDE?

Any pointers or references are much appreciated.
Thanks
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hunt_mat
#2
Jul29-11, 04:19 AM
HW Helper
P: 1,583
I am unsure what exactly you mean here, you can if you wish find the standard form for your equation by finding the characteristic directions in your second order system.

From what I can gather about your system you have one variable [itex]u[/itex] which is a function of [itex]t,x,y,v,w[/itex] Is this the case?
Tohiko
#3
Jul29-11, 06:44 AM
P: 8
That's exactly the case
And as you said, what I want to find are the characteristic directions for this PDE.
It's just that I don't know how to generalize what I already know in 1D case to this case.

hunt_mat
#4
Jul29-11, 07:02 AM
HW Helper
P: 1,583
Characterisitics of a Parabolic PDE

Characteristics are generally defined as where the second order derivatives are not uniquely defined, so you could start with this idea, you will get a 5x5 determinant that when expanded would give you a polynomial that you would have to solve.
Tohiko
#5
Jul29-11, 07:14 AM
P: 8
Thank you hunt_mat,
I think I understand your idea. I will try it out and see what I'd get.

Thank you again
hunt_mat
#6
Jul29-11, 07:21 AM
HW Helper
P: 1,583
I got this from the book applied partial differential equation be Ockendon et al. They even have some discussion about high dimensional systems at the end of chapter 3. It comes down to looking at quadratic forms.
HallsofIvy
#7
Jul29-11, 09:22 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,566
A hyperbolic pde has two independent characteristics.

A parabolic pde has only one.

An ellipitic pde has none.
hunt_mat
#8
Jul29-11, 09:51 AM
HW Helper
P: 1,583
for a system with two variables, however this is not the case in question.
Tohiko
#9
Jul29-11, 02:11 PM
P: 8
I'm trying out this idea. And I'm a little perplexed
I don't have access to the book that you mentioned but I'm following these notes
http://www2.imperial.ac.uk/~jdg/AE2MAPDE.PDF
Section 2.1, pages 8 and 9

Following similar ideas to what these notes have I wrote the differentials: dt, dx, dy, dv, dw
And set up a 6x15 matrix multiplying a vector of 2nd order derivatives (mixed and otherwise). But then I don't know what other rows to add to the matrix.

You said that I should obtain a 5x5 determinant, but frankly I don't how I would obtain it.
hunt_mat
#10
Jul29-11, 06:09 PM
HW Helper
P: 1,583
My mistake, it should be a 15x15 matrix. Let me give an example with 2 variables. Differentiate the first derivative, say [itex]\partial_{x}u[/itex] with respect to the characteristic variable to obtain:
[tex]
\left( \frac{\partial u}{\partial x}\right) '(s)=\frac{\partial^{2}u}{\partial x^{2}}\dot{x} +\frac{\partial^{2}u}{\partial x\partial y}\dot{y}
[/tex]
Likewise:
[tex]
\left( \frac{\partial u}{\partial y}\right) '(s)=\frac{\partial^{2}u}{\partial x\partial y}\dot{x} +\frac{\partial^{2}u}{\partial y^{2}}\dot{y}
[/tex]
Along with the differential equation itself:
[tex]
a\frac{\partial^{2}u}{\partial x^{2}}+b\frac{\partial^{2}u}{\partial x\partial y}+c\frac{\partial^{2}u}{\partial y^{2}}=d
[/tex]
Now as we sais that the definition of characteristics were when the second order derivative were not unique, that you mean that the determinant:
[tex]
\left| \begin{array}{ccc}
a & b & c \\
\dot{x} & \dot{y} & 0 \\
0 & \dot{x} & \dot{y}
\end{array}\right| =0
[/tex]
From here one obtains the quadratic necessary to find the characteristics. Do you this for your system.
Tohiko
#11
Jul30-11, 01:12 AM
P: 8
That's what I did, but as I said I don't have enough rows.
I differentiated dt,dx,dy,dv and dw with respect to the characteristic variable s. These gave me 5 rows. Plus one row from the differential equation itself. So I obtain 6 rows of the 15x15 matrix.
But what about the other 9 rows?
Tohiko
#12
Jul30-11, 09:22 AM
P: 8
I think I understand what I'm missing. I read a paper about finding the characteristics of a 3D PDE here (http://aerade.cranfield.ac.uk/ara/arc/rm/2615.pdf).

In there since the PDE is a function of 3 variables the characteristics would be a function of 2 variables \alpha and \beta. Then they differentiate the differentials dx1, dx2 and dx3 wrt \alpha and \beta obtaining 6 equations of which 5 are independent (since the PDE defines a relation between 2nd order differentials). These 5 equations along with the original PDE give a 6x6 matrix whose determinant should equal 0.

In my case since I have 5 independent variables I'm guessing the characteristics would be a function of 4 variables p,q,r and s
Then I differentiate the differentials dt,dx,dy,dv,dw wrt to p,q,r and s obtaining 20 equations. Of which I'm guessing only 14 are independent. These 14 equations along with the original PDE give another PDE of the characteristic surfaces.

Does this sound right? If it is, do you know why only 14 of the equations are independent?


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