I have solved a differential equation whose solutions is $$u = B + \sum_{n=1} C_{n }e^{-\lambda_{n}^2 q^2 t} J_{0}(\lambda_{n}r),$$
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 orthogonality generally used, namely $\int_{0}^{a} r J_{0}(\lambda_{n}r)J_{0}(\lambda_{q}r)dr$.
But this dosen't work here, since $\lambda_{q}a$ is not a zero of $J_0$, but it is a zero in fact of its derivative.
Not just it, what orthogonality I would use to find the $B$?
