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Show that PDE is Parabolic

  1. Jan 7, 2012 #1
    1. The problem statement, all variables and given/known data

    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]


    2. Relevant equations

    Cramers rule

    3. 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?
     
  2. jcsd
  3. Jan 7, 2012 #2

    LCKurtz

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    Use the definition at:
    http://en.wikipedia.org/wiki/Parabolic_partial_differential_equation
     
  4. Jan 7, 2012 #3
    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).
     
  5. Jan 8, 2012 #4

    LCKurtz

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    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?
     
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