A uniform, heat conductive sphere of radius $\rho$ is perfectly insulated and is heated by a uniform, internal (and constant) heat source $q$.
The PDE is:
$$\frac{\partial T}{\partial t}=\alpha\left(\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial T}{\partial r}\right)\right)+q$$
where $T$ is temperature, $r$ is the radial coordinate and $\alpha$ is the thermal diffusivity.
Developed and using PDE shorthand:
$$T_t=\alpha \frac{T_{r}}{r}+\alpha T_{rr}+q$$
Perfect insulation, using Fourier's first law:
$$T_r(\rho,t)=0$$
Initial ($t=0$) radial temperature distribution:
$$T(r,0)=f(r)$$
We attempt to solve it with:
$$T(r,t)=T_E(r)+T_p(r,t)$$
where $T_E$ is the steady state solution and $T_P$ is a particular solution. Both need to satisfy the original PDE and its BC/IC.
For the steady state solution we have:
$$\alpha \frac{T_{r}}{r}+\alpha T_{rr}+q=0$$
Slightly restated:
$$\alpha rT''(r)+\alpha T'(r)+qr=0$$
This solves to:
$$T(r)=c_1\ln r +c_2-\frac{qr^2}{4\alpha}$$
And here's the rub. For:
$$r \to 0\text{ then } T(r) \to -\infty \Rightarrow c_1=0$$
So $c_1$ must be $0$.
But this causes a problem, because:
$$T_r(\rho,t)=0 \Rightarrow T'(\rho)=0$$
and $c_2$ drops out…
I've a feeling I'll be having lots of egg on my face when this conundrum will be resolved but right now I can't see what's causing the problem?
Best Answer
I'm pretty sure no steady-state solution exists to this problem unless $q=0$. A perfectly insulated sphere with a constant internal heat source will only ever get hotter, no matter how much time has passed.