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authorluccul <luccul@gmail.com>2010-07-06 06:02:58 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-06 06:02:58 +0000
commit802c90d926a3ffc36dec7b4c792a36419eb6f7aa (patch)
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parent6597f0b4cc3e8bc179b78e467eb8310ac578e114 (diff)
downloadafterklein-wiki-802c90d926a3ffc36dec7b4c792a36419eb6f7aa.tar.gz
afterklein-wiki-802c90d926a3ffc36dec7b4c792a36419eb6f7aa.zip
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@@ -58,7 +58,7 @@ $$ \lambda = \frac{\pi n}{L} $$
So, the most general solution we can generate in this manner is:
$$ u(x,t) = \sum_{n = 1}^{\infty} c_n e^{-(\frac{\pi n}{L})^2t} \sin(\frac{\pi n x}{L}) $$
-We would like to assert that any solution takes this form. One way to prove this assertion would be to show that any function $f:[0,L] \to \R$ satisfying $f(0) = f(L) = 0$ has a unique ``[Fourier sine expansion](http://mathworld.wolfram.com/FourierSineSeries.html)'':
+We would like to assert that any solution takes this form. One way to prove this assertion would be to show that any function $f:[0,L] \to \mathbb{R}$ satisfying $f(0) = f(L) = 0$ has a unique ``[Fourier sine expansion](http://mathworld.wolfram.com/FourierSineSeries.html)'':
$$ f(x) = \sum_{n = 1}^{\infty} c_n \sin(\frac{\pi n x}{L}) $$
One could then allow the coefficients $c_n$ to vary with $t$ and apply the same method of solution that we used in the case of periodic boundary conditions.