I wonder, for the diffusion equation in the form:
$$\frac{\partial}{\partial t} T = \nabla\cdot(\alpha \nabla T)$$
let say, that I want to solve this on a ball. Assuming that $f$ is spherically symmetric, then $f=f(r)$ and the above equation reduces to
$$\frac{\partial}{\partial t} T = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \alpha(r) \frac{\partial T}{\partial r}\right)$$
But I can also think about the problem in a different way, because the function depends only on radial direction, I can "forget" all lateral directions and treat the problem 1 dimensionally, but then the diffusion equation would look like:
$$\frac{\partial}{\partial t} T = \frac{\partial}{\partial x}\left( \alpha(x) \frac{\partial T}{\partial x}\right)$$
In this post, it was shown that if one defines $\Gamma(t,r) \equiv \frac{T}{r}$ I can get the form from spherical coordinates to look like the 1 dimensional version.
But I cannot reconcile the meaning of that. Because, if I solve both equations, in one case I get $T(t,x) = f(t,x)$ in the other I get $T(t,r)=r\Gamma(t,r)$ but the $\Gamma(t,r)$ is also $\Gamma(t,r) = f(t,r)$ and thus $T(t,r)=r f(t,r)$. But $r$ and $x$ are just notations, so I recast the second result in even more suggestive form: $T(t,x) = x f(t,x)$ and I got two different solutions.