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\scriptsize
\frametitle{Laplace Transform of $\Psi_\ell(t,r)$}
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Therefore, for $\color{eqncolor}r=r_B$,
$$\color{eqncolor}
\widehat{\Psi}_{\ell}(s,r) = \sum_{k=0}^\ell \frac{c_{\ell k}}{r^k}\big(s^{\ell-k}e^{-sr}a(s)\big)
$$
$$\color{eqncolor}
= a(s)s^{\ell}e^{-sr}W_\ell (sr), \hspace{2em} W_\ell (sr)=\sum_{k=0}^\ell \frac{c_{\ell k}}{z^k}
$$
$$\color{eqncolor}
s\widehat{\Psi}_\ell(s,r) + \partial_r \widehat{\Psi}_\ell(s,r) = \frac{1}{r}(sr)\frac{W'_\ell(s,r)}{W_\ell(s,r)}\widehat{\Psi}_\ell=\frac{1}{r}\sum_{j=1}^\ell \frac{b_{\ell j}/r}{s-b_{\ell j}/r}\widehat{\Psi}_\ell,
$$
Laplace Convolution Theorem:
$$\color{eqncolor}
\partial_t \Psi_\ell(t,r) + \partial_r \Psi_\ell(t,r) = \frac{1}{r} \int_0^t \Omega_\ell(t-t',r)\Psi_\ell(t',r)dt', \hspace{2em} \Omega_\ell(t,r) = \sum_{k=1}^\ell \frac{b_{\ell k}}{r}e^{\frac{b_{\ell k}}{r}t}
$$
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