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authorluccul <luccul@gmail.com>2010-07-06 06:04:12 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-06 06:04:12 +0000
commitaf9714204bd568e6fd73ed35b349ea1ca472d9b5 (patch)
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parent802c90d926a3ffc36dec7b4c792a36419eb6f7aa (diff)
downloadafterklein-wiki-af9714204bd568e6fd73ed35b349ea1ca472d9b5.tar.gz
afterklein-wiki-af9714204bd568e6fd73ed35b349ea1ca472d9b5.zip
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@@ -56,10 +56,10 @@ and to satisfy $b(L) = 0$, we must impose a constraint on $\lambda$:
$$ \lambda = \frac{\pi n}{L} $$
So, the most general solution we can generate in this manner is:
-$$ u(x,t) = \sum_{n = 1}^{\infty} c_n e^{-(\frac{\pi n}{L})^2t} \sin(\frac{\pi n x}{L}) $$
+$$ u(x,t) = \sum_{n = 1}^{\infty} c_n e^{-\frac{\pi^2 n^2 t}{L^2}} \sin \left(\frac{\pi n x}{L} \right) $$
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)'':
-$$ f(x) = \sum_{n = 1}^{\infty} c_n \sin(\frac{\pi n x}{L}) $$
+$$ f(x) = \sum_{n = 1}^{\infty} c_n \sin\left(\frac{\pi n x}{L}\right) $$
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.