From af9714204bd568e6fd73ed35b349ea1ca472d9b5 Mon Sep 17 00:00:00 2001 From: luccul Date: Tue, 6 Jul 2010 06:04:12 +0000 Subject: more more formatting --- ClassJuly5.page | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/ClassJuly5.page b/ClassJuly5.page index ff7cfb3..664c440 100644 --- a/ClassJuly5.page +++ b/ClassJuly5.page @@ -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. -- cgit v1.2.3