I am trying to solve the radially symmetric 2D heat equation with an absorbing point at the origin in radial coordinates, i.e. (assuming no dependence on the angular variable)
$$u_t = k (u_{rr}+\frac{1}{r}u_r)$$ with boundary condition $u(t,r\!=\!0) = 0$ and some initial condition $u(0,r) = f(r)$.
The standard approach is seperation of temporal and radial part, $u(t,r) = T(t)R(r)$. The temporal part gives the usual exponential: $$ T(t) = C e^{-k \lambda^2 t} $$ However, I cannot get the radial part to fulfil the boundary condition $u(t,0)=0$: The general solution of the radial equation is $$R(r) = A(\lambda) J_0(r \lambda) + B(\lambda) Y_0(r \lambda)$$ where $J_n(x)$ is the Bessel function of the first kind, and $Y_n(x)$ is the Bessel function of the second kind. At $r=0$, $Y_0(r)$ tends to $-\infty$, so $B$ has to be $0$. However, $J_0(0)=1$, so cannot fulfill the boundary condition $u(t,0)=0$ either. So it seems that there is no non-trivial way of satisfying the boundary conditions.
Where is my mistake? Is the point as a boundary condition in 2D ill-defined? Would I have to consider an absorbing disk and shrink that to $0$? But what would be the solutions to the radial equation then, and why would that even be necessary? (Note that in 1D, there is no problem since the general solution is a sum of sines and cosines, and the sines are automatically zero at the origin.)
Indeed there is no solution (other than the trivial $u=0$) with this boundary condition. Essentially, the origin is unable to absorb heat fast enough to maintain its temperature at $0$.
EDIT: Consider the steady-state equation $u_{rr} + \dfrac{1}{r} u_r = 0$. The general solution is $u(r) = a \log(r) + b$. The only steady-state solutions with a finite limit as $r \to 0+$ are constant. With boundary conditions $u(\epsilon) = 0$, $u(1) = 1$, where $\epsilon > 0$, the solution is $$ u_\epsilon(r) = \dfrac{\ln(r)}{\ln(1/\epsilon)} + 1$$ As $\epsilon \to 0+$, $\ln(1/\epsilon) \to \infty$ and the solution approaches $u_0(r) = 1$.