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.

  1. Is there a general solution to $c(r,t)$ for this problem, i.e. is this a well-posed problem?
  2. 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!
  3. 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 $-D \frac{\partial c}{\partial t}$.


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.


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 $$C(q)=\int_{\mathbb{R}}\Big(rf(r)-\frac{\eta{R}}{2\sqrt{{D}\gamma}}\exp\Big(-\Big{|}\sqrt{\frac{\gamma}{D}}(r-R)\Big{|}\Big)\Big)e^{-iqr}dr=$$ $$\int_{\mathbb{R}}rf(r)e^{-iqr}dr-\frac{\eta{R}}{2\pi{D}}\frac{e^{iq(r-{R})}}{(q^{2}+\gamma/D)}$$

  • $\begingroup$ Thanks! I follow and agree with the steps up till the last calculation (the integral should be over $(R, \infty)$ or at least $(0,\infty)$. The only problem is that the boundary condition $-D \frac{\partial c}{\partial r} = \eta$ is hard to check in this case, because of imaginary terms. I will retry it myself with Laplace transforms instead of Fourier transforms. $\endgroup$ – Yiteng Jul 26 '17 at 16:03

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