# Heat equation and Bessel's function

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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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
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