# Heat equation and Bessel's function [closed]

Could someone please explain why if the time-independent heat equation can, via changing of variables, take the form of Bessel's equation that the $\sqrt\lambda$ should take the values of the zeros of $J_0$ from slides 20-21 of this powerpoint printout? Thank you. (Also, what is that $a$?)

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## closed as off-topic by Emilio Pisanty, Qmechanic♦Sep 5 '13 at 18:21

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The $a$ is the radius of the disc (see slide 18). The need for $\sqrt J a$ to be a zero is because of that boundary condition ($u(a,t) = 0$; apparently the edge is in contact with a heat bath). – genneth Sep 26 '11 at 13:41
@genneth: Thanks! – Sillybilly Sep 27 '11 at 10:08
The link is now broken. – Emilio Pisanty Sep 5 '13 at 17:03
Sep 5th 2013: The question (which is not necessary off-topic) unfortunately had to be closed because of link rot. Let me use this opportunity to remind everyone in the future to give reference (i.e. mention author, title, etc.) of links so that we can reconstruct it in case of link rot. – Qmechanic Sep 5 '13 at 18:21