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.
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.