Looking for analytical solution methods to PDE in radial coordinates (reaction-diffusion equation) I am attempting to find a solution for a PDE (reaction-diffusion equation) in radial coordinates with a specific set of boundary conditions:
\begin{align}
\frac{\partial C(r,t)}{\partial t}&=D\left(\frac{\partial^2 C(r,t)}{\partial r^2}+\frac{1}{r}\frac{\partial C(r,t)}{\partial r}\right)-R C(r,t)\,,\\
\frac{\partial C(r=0,t)}{\partial r}&=0\,,\\
C(r=R,t)&=C_{o}\,,\\
C(r,t=0)&=0\,.
\end{align}
I have been looking for a solution in literature in order to save time and/or find out if the problem even has an analytical solution.
I know that Danckwertz (1951) proposed a solution to this problem in one-dimension, but I am having trouble understanding whether this applies to radial coordinates as well. I would like to know if an analytic solution is even possible.
 A: The original problem probably contains typos in the left hand side ($\partial_r c$ should be $\partial_t c$) and reaction term, where probably the prefactor is not the same as the size $R$ of the domain.
So I'm interested in
\begin{align}
\partial_t c &= D \left(\partial_r^2 c+\frac{1}{r}\partial_r c\right)-  n^2 c\,,\\
\partial_r c(0,t)&=0\,,\\
c(R,t)&=c_0\,,\\
c(r,0)&=0\,.
\end{align}
Let's clean up the notation first: $c'=c/c_0$, $r'=r/R$, $t'=tD/R^2$, and $(n')^2=n^2R^2/R$, and drop primes,
\begin{align}
\partial_t c &= \partial_r^2 c+\frac{1}{r}\partial_r c-n^2 c\,,\\
\partial_r c(0,t)&=0\,,\\
c(1,t)&=1\,,\\
c(r,0)&=0\,.
\end{align}
A Laplace transformation $\hat{c}(r,s)=\mathcal{L}\left\{c(r,t)\right\}=\int_{0}^{\infty}{\rm d}t\, e^{st}c(r,t)$ takes the above equations to
\begin{align}
s\hat{c}(r,s) -c(r,0)&=  \partial_r^2 \hat{c}(r,s)+ \frac{1}{r}\partial_r \hat{c}(r,s)-n^2 \hat{c}(r,s)\nonumber\\
\partial_r^2 \hat{c}(r,s) + \frac{1}{r}\partial_r \hat{c}(r,s)&=  m^2\hat{c}(r,s)\,,
\end{align}
where $m^2=n^2+s$.
This equation is solved by
\begin{align}
\hat{c}(r,s)=a I_0(mr)+b K_0(mr)\,,
\end{align}
with $I_0$ and $K_0$ being modified Bessel functions of the first and second kind.
We fix the constants $a$ and $b$ through the boundary conditions, which, in Laplace space, read
\begin{align}
\partial_r \hat{c}(0,s)&=0\,,\\
\hat{c}(1,s)&=\frac{1}{s}\,.
\end{align}
The first condition yields
\begin{align}
\partial_r \hat{c}(0,s)&=a m I_0(0)-b m K_0(0)=0
\end{align}
from which we conclude that $b=0$ (as $K_0(0)=\infty$).
The other boundary contitions yields
\begin{align}
\hat{c}(1,s)&=a I_0(m)=\frac{1}{s}\Rightarrow a=\frac{1}{s I_0(m)}\,,
\end{align}
hence
\begin{align}
\hat{c}(r,s)&=\frac{I_0(r m)}{s I_0(m)}\,.
\end{align}
Determining $c(r,t)=\mathcal{L}^{-1}\left\{\hat{c}(r,s)\right\}$ now requires performing the inverse Laplace transformation
\begin{align}
c(r,t)=\mathcal{L}^{-1}\left\{\frac{I_0(r\sqrt{n^2+s})}{s I_0(\sqrt{n^2+s})} \right\}\,,
\end{align}
for which I have posted a new question here.
