3d heat equation with constant point source

acme37
Messages
23
Reaction score
0
Hi all,

I'd like to solve the following problem in 3 dimensions:

\partial_t u(r,t) = D\Delta u(r,t)
u(r,0) = 0
u(0,t) = C_o

In words, I am looking at a point 'source' that is turned on at t=0 and held at constant temperature. The ultimate goal is to then convolve this solution with constant sources distributed arbitrarily in space.

In 1D, I can find the solution:
u(x,t) = C_o \mathrm{erfc}\left(\frac{x^2}{\sqrt{4Dt}}\right)
In 3D, I proceed by taking the Laplace transform and solving the resulting Helmholtz equation. After applying the r\to\infty boundary condition,
u(r,s)=A\frac{e^{-\sqrt{\frac{s}{D}}r}}{r}
But I can't apply the boundary condition at u(0,s) due to the singularity at the origin. So instead I assume we fix the constant temperature condition on a ball of radius a, for a\ll r. If I do that my solution isn't so bad,
u(r,t)=C_o\frac{a}{r}\mathrm{erfc}\left(\frac{(a-r)^2}{4Dt}\right)
This matches my simulations rather well, when I add appropriate image sources for my particular geometry. Of course, the solution vanishes as I shrink a\to0. So my question is, is there a way to solve the original problem of a point source? Otherwise using this as a sort of Green's Function for a distributed constant temperature boundary seems suspect.

I feel like this issue comes up a bunch in EM but I'm blanking on how to deal with it here. By the way, a second approach I've tried is to integrate the normal Green's Function over time (i.e. convolving with a step source). There again I get a singularity at r=0.

Thanks!
 
Physics news on Phys.org
An ideal point source does not transmit heat in 3 dimensions, so your solution for the ball with a finite size looks reasonable.
 
Last edited:
Thanks, but I'm a bit confused. What else is the Green's Function if not the response to an ideal point source?
 
There is no response (read: zero temperature change) to an ideal point source.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top