Solving a PDE with source and degradation I am studying the following PDE for a function $c(r,t)$:
$$
\frac{\partial c}{\partial t} = 
D \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{ \partial c}{\partial r} \right) - \gamma c + \eta \delta(r-R)
$$
The steady-state solution of this equation has the form
$$
c(r) = \frac{c_R R}{r} \exp{\left(\frac{R-r}{\lambda}\right)},
$$
where $\lambda=\sqrt{\frac{D}{\gamma}}$ and $c_R = \frac{\eta \gamma R}{\lambda (\lambda+R)}$. 
I am looking for the time-dependent solution $c(r,t)$ on the domain $(r,t) \in (R,\infty) \times (0, \infty)$ without resorting to numerical computation. 
As boundary conditions I would like to take
$$
-D\frac{\partial c}{\partial r} = \eta, \\
 \lim\limits_{r\rightarrow \infty} c(r) = 0
$$
I still have to think about the initial condition, but let's suppose it is $c(r,0) = f(r)$ for some function $f(r)$ that satisfies the first boundary condition.


*

*Is there a general solution to $c(r,t)$ for this problem, i.e. is this a well-posed problem?

*What are general approaches that will help me in studying this problem? I have only basic background in solving PDEs, so any suggestion for how to approach this problem will be appreciated!

*It is perhaps possible to leave out the last term $\eta \delta(r-R)$ since production is already implied by taking the first boundary condition $-D \frac{\partial c}{\partial t}$.


Note
Without the degradation term $-\gamma c$, this reduces to the heat equation with a constant source. The solution of this equation with the given boundary conditions is known, see e.g. this paper (in the context of signaling molecules too). However, I couldn't find any straightforward generalization to the system with degradation term.
Some physical background
The equation is a so-called reaction-diffusion equation meant to describe the concentration of a molecule secreted by a cell. The cell has radius $R$ and secretes with a rate $\eta$. The diffusion constant is $D$ and the molecules are degraded at a rate $\gamma$.  
$$
\frac{\partial c}{\partial t} = 
\nabla \cdot (D \nabla c) - \gamma c + \eta \delta(r-R)
$$
Since the problem is spherically symmetric, we can limit ourselves to finding a solution for $c(r,t)$. Here is a paper for a (complicated) system of interacting cells that uses the steady-state solution I gave above.
 A: You have
$$\partial_{t}c=D\frac{1}{r^{2}}\partial_{r}(r^{2}\partial_{r}c)-\gamma{c}+\eta\delta(r-R)$$
Let
$$c(t, r)=\frac{g(t, r)}{r}$$
The equation becomes
$$\partial_{t}g=D\partial^{2}_{r}g-\gamma{g}+\eta{r}\delta(r-R)$$
Fourier transform the equation in $r-q$
$$\partial_{t}\hat{g}(t, q)+(Dq^{2}+\gamma)\hat{g}(t, q)=\eta{R}e^{-iq{R}}$$
Thus
$$\hat{g}(t, q)=C(q)\exp\Big(-(Dq^{2}+\gamma)t\Big)+\frac{\eta{R}e^{-iq{R}}}{(Dq^{2}+\gamma)}$$
Hence
$$g(t, r)=\frac{1}{2\pi}\int_{\mathbb{R}}C(q)\exp\Big(iq{r}-(Dq^{2}+\gamma)t\Big)dq+\frac{\eta{R}}{2\pi{D}}\int_{\mathbb{R}}\frac{e^{iq(r-{R})}}{(q^{2}+\gamma/D)}dq=$$
$$=\frac{1}{2\pi}\int_{\mathbb{R}}C(q)\exp\Big(iq{r}-(Dq^{2}+\gamma)t\Big)dq+\frac{\eta{R}}{2\sqrt{{D}\gamma}}\exp\Big(-\Big{|}\sqrt{\frac{\gamma}{D}}(r-R)\Big{|}\Big)$$
Thus
$$c(t, r)=\frac{1}{2\pi{r}}\int_{\mathbb{R}}C(q)\exp\Big(iq{r}-(Dq^{2}+\gamma)t\Big)dq+\frac{\eta{R}}{2\sqrt{{D}\gamma}r}\exp\Big(-\Big{|}\sqrt{\frac{\gamma}{D}}(r-R)\Big{|}\Big)$$
Note, the second term already satisfies the conditions at infinity. At $t=0$
$$c(0, r)=f(r)=\frac{1}{2\pi{r}}\int_{\mathbb{R}}C(q)e^{iq{r}}dq+\frac{\eta{R}}{2\sqrt{{D}\gamma}r}\exp\Big(-\Big{|}\sqrt{\frac{\gamma}{D}}(r-R)\Big{|}\Big)$$
Or
$$rf(r)-\frac{\eta{R}}{2\sqrt{{D}\gamma}}\exp\Big(-\Big{|}\sqrt{\frac{\gamma}{D}}(r-R)\Big{|}\Big)=\frac{1}{2\pi}\int_{\mathbb{R}}C(q)e^{iq{r}}dq$$
Thus
$$C(q)=\int_{\mathbb{R}}\Big(rf(r)-\frac{\eta{R}}{2\sqrt{{D}\gamma}}\exp\Big(-\Big{|}\sqrt{\frac{\gamma}{D}}(r-R)\Big{|}\Big)\Big)e^{-iqr}dr=$$
$$\int_{\mathbb{R}}rf(r)e^{-iqr}dr-\frac{\eta{R}}{2\pi{D}}\frac{e^{iq(r-{R})}}{(q^{2}+\gamma/D)}$$
