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Homework Help: Heat Equation: Why Sigma?

  1. Dec 11, 2005 #1
    Hey all,

    I've been working on learning to solve some PDE's. To do this I've been reading other people's tutorials. Here's one on the heat equation:


    This is pretty much the same as the others I've read on the heat equation, but it's explained farily well. However there are a few points I don't understand, and I was hoping someone might clarify them for me.

    At 2.54 ([tex]k=-p^2[/tex]), what is [tex]p[/tex] and where did it come from?

    2.57: What happened here? I see that at [tex]X(0)=0[/tex] and [tex]X(l)=0[/tex] to set boundary conditions... I also see that this step is involved in plugging in to find the constants of the equation [tex]X(x)=Acos(px) + Bsin(px)[/tex] but..... I'm lost :frown:

    Finally, we get to 2.58 and everything explodes. Why a sigma? Generally for this type of problem don't we separate the variables into two ODEs, then BAM using some assumed equation forms solve for the constants and you have a solution? More or less?

    Thanks a lot for any insight you can give,

    -Jack Carrozzo
  2. jcsd
  3. Dec 11, 2005 #2

    Tom Mattson

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    That's nothing more than a definition to make the solution look nice. If they hadn't defined [itex]k=-p^2[/tex] then the solution in 2.56 would look like this:


    Yuck! :yuck:

    In 2.57 they are using knowledge that would be picked up in a precalculus course.

    Consider the equation [itex]sin(x)=0[/itex]. What are the solutions? They are [itex]x=[/itex] (any integer multiple of [itex]\pi[/itex]). In other words, [itex]x=n\pi[/itex], [itex]n\in\mathbb{Z}[/itex].

    Just replace [itex]x[/itex] in the above equation with [itex]pl[/itex], and you have 2.57.

    Because it is a basic fact of the theory of linear differential equations that the sum of two solutions is also a solution. This is sometimes called the principle of superposition. So in order to have the complete solution, you have to add up all of the "basis" solutions (to borrow a term from the theory of vector spaces).
  4. Dec 11, 2005 #3
    Oh well..... guess I'm not solving that one any time soon.

    Thanks for the help,

    -Jack Carrozzo
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