How to find the coefficients of this expansion?

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I have solved a differential equation whose solutions is $$u = B + \sum_{n=1} C_{n }e^{-\lambda_{n}² q² t} J_{0}(\lambda_{n}r)$$

Where ##(\lambda_{n}r)## is such that ##J_{0}'(\lambda_{n}a) = 0##. So i should now try to satisfy the condition that, at t=0, u = ##f(r)##.

The problem is that i don't know what is the orthogonality here. If ##\lambda_{n}## were such that ##J_{0}(\lambda_{n}a)=0##, i would use the normal orthogonaltiy generally used, namely ##\int_{0}^{a} r J_{0}(\lambda_{n}r)J_{0}(\lambda_{q}r)##

But this dosen't work here, since ##\lambda_{q}a## is not a zero of J0, but it is a zero in fact of its derivative.

Not just it, what orthogornality i would use to find the B?
 
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Believe it or not, orthogonality, exactly the way you wrote it, is already guaranteed. My source is Jackson Classical Electrodynamics 3rd Edition, bottom of page 115:

But it will be noted that an alternative expansion is possible in a series of functions ##\sqrt{\rho} J_\nu (y_{\nu n}\rho / a)## where ##y_{\nu n}## is the ##n##th root of the equation ##[dJ_\nu(x)]/dx = 0##. The reason is that, in proving the orthogonality of functions, all that is demanded is that thte quantity ##[\rho J_\nu (k\rho) (d/d\rho) J_\nu (k' \rho) - \rho J_\nu (k' \rho) (d/d\rho) J_\nu (k\rho)]## vanish at the endpoints ##\rho = 0## and ##\rho = a##.
 
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