Mass diffusion from a point source in spherical coordinates Is an analytical solution for the mass diffusion from a point source in spherical coordinates even possible? I posted what I thought was a valid solution here but the plot doesn't align with my expectation.
Any ideas?
 A: You seems to have gone wrong in the section - Casting the Diffusion Equation in Terms of $y(r,t)$, from this step onwards:
$$\Rightarrow \frac{\partial^2 C}{\partial y^2}+\underbrace{\left(\frac{4\sqrt{Dt}}{r}+r\sqrt{\frac{t}{D}}\right)}_{\text{A}} \frac{\partial C}{\partial y}=0$$
$$\Rightarrow \frac{\partial^2 C}{\partial y^2}+A \frac{\partial C}{\partial y}=0$$
From here on you seems to have treated $A$ to be constant, when clearly it is not. $A\equiv A(r,t)=A(y)$ since $y\equiv y(r,t)$.
More specifically, if $$y\equiv\frac{r}{2\sqrt{Dt}}$$ then $$A=\frac{4\sqrt{Dt}}{r}+r\sqrt{\frac{t}{D}}=\frac{2}{y}+ \cdots =A(y)$$

To solve it, I don't think that particular substitution of $y$ will work, the reason being although you have a spherically symmetric system, it should be remembered that you are still in 3 dimensions and not 1 dimension. More specifically, here $r=x\hat{i}+y\hat{j}+z\hat{k}$. So you should be looking for some substitution that has to do with Spherical harmonics.
A: So you have the PDE:
$$\frac{\partial C}{\partial t} = D \nabla^2C = D\frac{1}{r^2}\frac{\partial}{\partial r} \left[ r^2 \frac{\partial C}{\partial r} \right]$$
We're looking for a function $C(r,t)$,
with Boundary Conditions:
$$C(r,0)=C_b$$
$$C(0,t)=C_0$$
Note already that we are missing one value boundary condition because the PDE is second order in $r$.
Developing the first PDE we get (in 'PDE shorthand'):
$$\frac1D C_t=\frac2rC_r+C_{rr}$$
This type of second order, linear PDE is solved by means of Separation of Variables. Assume ('Ansatz') that:
$$C(r,t)=R(r)T(t)$$
Insert to get:
$$\frac1D RT'=\frac2r TR'+TR''$$
Divide by $RT$ to get:
$$\frac1D\frac{T'}{T}=\frac2r \frac{R'}{R}+\frac{R''}{R}=-m^2$$
$$\frac1D\frac{T'}{T}=-m^2\tag{1}$$
$$\frac2r \frac{R'}{R}+\frac{R''}{R}=-m^2\tag{2}$$
where $-m^2$ is a separation constant and it is a Real number.
$(1)$ is of course very easily solved to:
$$T=c_3\exp(-m^2 D t)$$
where $c_3$ is an integration constant.

Slightly reworked from $(2)$, we get:
$$rR''+2R'+m^2 rR=0$$
It can be rewritten as:
$$(rR(r))''+m^2(rR(r))=0$$
Let $S(r)=rR(r)$, then we have:
$$S''+m^2S=0$$
Which has the simple solution:
$$S(r)=c_1\sin mr+c_2\cos mr$$
Back-substitute to get $R(r)$:
$$R(r)=c_1\frac{\sin mr}{r}+c_2\frac{\cos mr}{r}$$
Now apply the boundary condition $C(0,t)=C_0$ as:
$$R(0)=C_0$$
$$\Rightarrow C_0=c_1\frac{\sin m0}{0}+c_2\frac{\cos m0}{0}$$
Because:
$$\lim_{r\to 0}\frac{\cos mr}{r}=\pm\infty$$
it follows that $c_2=0$.
However, you're still stuck with the problem of lacking one BC in $r$. Perhaps $r \to \infty$ then $R(r) \to 0$ could be one?
Another possible BC could be:
$$R(r_o)=0$$
If $r_o$ is very large then it resembles the suggestion above. Inserting into $R(r)=c_1\frac{\sin mr}{r}$ we get:
$$0=c_1\frac{\sin mr_o}{r_o}$$
which boils down to:
$$\sin mr_o=0$$
or:
$$m_ir_o=\frac{i\pi}{2}$$
For $i=0,1,2,3,...$
Thus:
$$m_i=\frac{i\pi}{2r_o}$$
The eigenvalues are the $m_i$.
Your final solution will look like:
$$C(r,t)=\Sigma_0^{\infty}\left(B_i\exp(-m_i^2 D t)R_i(r)\right)$$
Where the $B_i$ are determined from the initial condition:
$${B_i} = \frac{2}{r_o}\int_{{\,0}}^{{\,r_o}}{{ {C_b} \sin \left( {\frac{{i\,\pi r}}{r_o}} \right)\,\mathrm{d}x}}\,\,\,\,\,\,\,\hspace{0.25in}i = 0,1,2,3, \ldots$$
Hope this helps.
