Is the 1D Heat Equation Parabolic?

  • Thread starter Thread starter Saladsamurai
  • Start date Start date
  • Tags Tags
    Pde
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 3K views
Saladsamurai
Messages
3,009
Reaction score
7

Homework Statement



I am learning about PDE classification from a text on CFD (by Anderson). This section is not complete enough to be able to extend his example problems into more general cases. I read that to classify a system of PDEs as being parabolic, elliptic, or hyperbolic, I need to do some crazy stuff with Cramer's rule. However, the examples that he has shown are 1st order PDEs and like I said, are *systems* of PDEs. Now I have been asked to show that the 1D heat equation is parabolic and I am not sure how to apply what I have learned since a) it is 2nd order and b) it is only 1 eqaution:

[tex]\frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2} \qquad(1)[/tex]

Homework Equations



Cramers rule

The Attempt at a Solution



I thought that I could form a system by forming the total differential of T(x, t)

[tex]dT = \frac{\partial T}{\partial x} \,dx + \frac{\partial T}{\partial t} \,dt \qquad(2)[/tex]However I am not sure if this is helpful. Any hints?
 
Physics news on Phys.org
Saladsamurai said:

Homework Statement



I am learning about PDE classification from a text on CFD (by Anderson). This section is not complete enough to be able to extend his example problems into more general cases. I read that to classify a system of PDEs as being parabolic, elliptic, or hyperbolic, I need to do some crazy stuff with Cramer's rule. However, the examples that he has shown are 1st order PDEs and like I said, are *systems* of PDEs. Now I have been asked to show that the 1D heat equation is parabolic and I am not sure how to apply what I have learned since a) it is 2nd order and b) it is only 1 eqaution:

[tex]\frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2} \qquad(1)[/tex]

Use the definition at:
http://en.wikipedia.org/wiki/Parabolic_partial_differential_equation
 
LCKurtz said:

Hi LCKurtz :smile: I actually found that at wolfram as well. The problem is that he does not use that in the text (explicitly). I think what he has done is kind of derived that equation, but in a less general sense. I am just trying to learn a little more about where that equation at the wiki comes from. Let me show you in broad terms what Anderson did on his example:

The starting point:
Screenshot2012-01-07at114911PM.png


Putting in matrix form:

Screenshot2012-01-07at114924PM.png


He then set's det(A) = 0 in order to get a quadratic algebraic equation in (dy/dx). If the discriminant of this quadratic meets certain criteria, we can classify the PDE as parabolic. I am just having trouble putting everything into matrix form since I don't have a system of PDEs (unless I was right by taking the total differentials in post #1).
 
I don't know if I can help you or not with this. I haven't done that much with PDE's. But I have one thought that occurs to me that you might or might not find useful. I'm thinking about how in ordinary DE's you can take a second order equation and write it as a first order linear system. For example, given the DE ##ay''+ by' + cy = 0## you make the substitution ##y_1=y, y_2= y', y_2' = y'' = \frac 1 a (-cy-by')=\frac 1 a(-cy_1-by_2)##

This gives you a first order linear system$$
\pmatrix{y_1\\y_2}'=\pmatrix{0&1\\-\frac c a &-\frac b a}\pmatrix{y_1\\y_2}$$

Maybe you can try something like that beginning with$$
U = T, V = U_x, V_x = U_{xx} = \frac 1 \alpha T_t=\frac 1 \alpha U_t$$Or maybe not, who knows?