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In the event that I'm solving a partial differential equation through separation of variables, if I end up with an eigenvalue of zero, what do I do with the corresponding eigenfunction?

That is to say, if I end up with an eigenvalue of zero, am I going to add it to the sum that I get from solving for the other eigenvalues? If so, am I going to multiply it by my T(t) first? Something like:

$X_0(x)T(t) + \sum a_nT(t)X_n(x)$ or do we just list 0 as one of our eigenvalues, and then we don't make any changes to the way the function is written out?

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It's hard to say in abstract, but if there is nonzero eigenfunction, then the zero eigenvalue is a legitimate one. Note that the temporal part should also depend on the eigenvalue, i.e. $a_0 X_0(x) T_\color{red}{0}(t) + \sum a_n T_\color{red}{n}(t) X_n(x)$. – Pragabhava Nov 14 '12 at 18:48
True. We haven't reached the point where we're dealing with $T_n$ cases yet in my class, but it is a good point. I assume $a_0$ would calculated the same was as any other $a_n$ then also? That is to say that $a_0 = \int T(x)X_0(x)$ where $T(x)$ is an initial condition function? – Nicholas Hunsicker Nov 14 '12 at 18:56
For an initial condition $u(x,0) = f(x)$, then $$u(x,0) = f(x) = a_0 T_0(0) X_0(x) + \sum a_n T_n(0) X_n(x)$$ and you'll determine the $a_n$'s the same way as a regular fourier series. i.e $$ a_n = \frac{1}{T_n(0) \|X_n\|^2} \langle f(x), X_n(x) \rangle$$ where the scalar product is the proper one. – Pragabhava Nov 14 '12 at 19:06

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