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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.

atapaka
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    No you cannot simply "forget" about all lateral directions. There is no coordinate transformation from your original 3D equation that bring it to the equation you have put up so this equation is not equivalent to the one you started with. – Winther Mar 25 '17 at 14:03
  • @Winther For example: Solving in spherical coordinates in a ball (assuming $T$ is only radially dependent) and all boundary conditions are simply a constant as well as the initial condition. Then how does the case differ from simply 1D solution in Cartesian coordinates? (you can imagine that you make up the ball of 1D rods and solve the equation for each such rod but due to the boundary condition, it is sufficient to do this only for a single such rod). And if you do it along $x$-axis, $r$ and $x$ need to have the same physical meaning but you get different solutions. – atapaka Mar 25 '17 at 15:25
  • If we think of $T$ as representing a wave with some energy then in the 1D case (your equation with $x$) this energy is being spread out in only one direction. For the spherical symmetric case the energy is being spread out across the surface of a 3D-sphere which has surface area $4\pi r^2$. The factors of $r^2$ and $\frac{1}{r^2}$ in the correct equation above is accounting for this fact. – Winther Mar 25 '17 at 16:53
  • @Winther so the fact which equation is correct depends on the description of the problem, if the boundary condition is on the surface and at the centre, then I would think the 1D Cartesian case is correct, if the boundary condition is given only at the surface then the spherical version is correct? – atapaka Mar 25 '17 at 19:37
  • Here is what I'm trying to say: if you have spherical symmetry then there is only one correct version of the equation for $T$ in terms of the radial variable $r$ and that is the one with the $r^2$ factors in it. If you want then you can from this equation make a substitution to get an equation that looks identical to the normal 1D-cartesian equation. In this equation $r$ plays the same role of a cartesian coordinate, but it is still the radial coordinate, and the equation is in terms of the redefined field $\Gamma$. Above I tried to argue why your proposed equation fails to be correct. – Winther Mar 26 '17 at 00:00

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