# Heat Equation: Why Sigma?

1. Dec 11, 2005

### crepincdotcom

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:

http://www-solar.mcs.st-and.ac.uk/~alan/MT2003/PDE/node21.html

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 ($$k=-p^2$$), what is $$p$$ and where did it come from?

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

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
http://www.crepinc.com/

2. Dec 11, 2005

### Tom Mattson

Staff Emeritus
That's nothing more than a definition to make the solution look nice. If they hadn't defined $k=-p^2[/tex] then the solution in 2.56 would look like this: $$X(x)=A\cos\left(\sqrt{-k}x\right)+B\sin\left(\sqrt{-k}x\right)$$ 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$. What are the solutions? They are $x=$ (any integer multiple of $\pi$). In other words, $x=n\pi$, $n\in\mathbb{Z}$.

Just replace $x$ in the above equation with $pl$, 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).

3. Dec 11, 2005

### crepincdotcom

Oh well..... guess I'm not solving that one any time soon.

Thanks for the help,

-Jack Carrozzo
http://www.crepinc.com/