$\frac{\partial T}{\partial t} = \alpha \nabla ^2r$ for spherically symmetric problems

Question

The standard one dimensional partial differential diffusion equation in Cartesian coordinates has the form;

$$\frac{\partial T}{\partial t} = \alpha \frac{\partial ^2 T}{\partial x^2} \tag{1}$$

For a spherically symmetric diffusion problem, we can use spherical polars to get the following diffusion equation:

$$\frac{\partial T}{\partial t} = \alpha \nabla ^2r$$ $$\implies\frac{\partial T}{\partial t} = \alpha \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial T}{\partial r}\right) \tag{2}$$

But there is a trick we can use to make equation $(2)$ look like equation $(1)$ - if we define

$$T(r,t) = \frac{\Gamma(r,t)}{r} \tag{3}$$

then

$$\frac{\partial \Gamma}{\partial t} = \alpha \frac{\partial ^2 \Gamma}{\partial r^2} \tag{4}$$

Which simplifies the problem to the form shown in $(1)$ and the standard techniques used for $(1)$ can be applied to $(2)$.

This is all well and good if you spot the trick, but how can you derive, without just guessing, that substituting $(3)$ into $(2)$ will make it of the form in $(1)$?

What I am asking for is a method to find $(3)$ from looking at $(1)$ and $(2)$ in a non heuristic way. Hopefully the technique to do this should generalise to other PDEs (if a suitable $(3)$ exists).

Answer 1

I think the reason you're looking for is that there are actually two equivalent ways to write the spherically symmetric part of the Laplacian in three dimensions: $$ (\nabla^2 f)_{\text{radial}} = \frac{\partial^2 f}{\partial r^2} + \frac{2}{r} \frac{\partial f}{\partial r} = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial f}{\partial r} \right) = \frac{1}{r} \frac{\partial^2}{\partial r^2} \left( r f \right), $$ and then you can notice that the group $rf$ appears on both sides of your equation if you rearrange it.

This trick only works in three dimensions because having any higher power in the brackets in the last form would give you three terms, and any other power would stop the numbers working out in the multiplied-out version.

Answer 2

The non-heuristic and generalizable answer is unfortunately quite a bit more complicated than the heuristic answer. The key idea here is that of the Green's function, also called the fundamental solution. This might be expected to exist in linear problems; indeed the method I'm going to describe is essentially a generalization of the principle of superposition.

The Green's function is usually not a function in the classical sense on its whole domain, but is rather a so-called distribution. Formally this means that it is a continuous linear function from the space of smooth, compactly supported functions, denoted by $C^\infty_c$, to $\mathbb{R}$. (Here "continuous" is relative to the topology on $C^\infty_c$, which I won't describe here.) The basic examples of distributions are given by integration against a smooth compactly supported function. That is, if $g \in C^\infty_c$ then we have a distribution $I_g$ defined by

$$I_g(f) = \int_{-\infty}^\infty g(x) f(x) dx.$$

The Green's function $G$ for the operator $L$ satisfies $L(G)=\delta$ where $\delta$ is the Dirac delta. The Dirac delta is rigorously defined as a distribution so that $\delta(f)=f(0)$. Given the distributions from smooth functions I just mentioned, we intuitively think of this as meaning that

$$\int_{-\infty}^\infty \delta(x) f(x) dx = f(0)$$

even though there is no true function with this property. Using this intuition, we define the convolution of a distribution with a smooth compactly supported function, and we find $\delta * f = f$, at least for $f \in C^\infty_c$.

When $L$ is a linear differential operator, this is quite convenient, because if we have

$$L(G)=\delta$$

then we can convolve with $f$ on both sides to get

$$L(G)*f=\delta*f=f.$$

Then we can justify moving the derivatives so that we get

$$L(G*f)=f.$$

This means that to solve $L(u)=f$ we only need to find the fundamental solution and then write down a convolution.

In your situation, suppose we had the fundamental solution for the Laplace operator, a $G$ such that $\Delta(G)=\delta$. Well, we know that the solution to $\Delta u = f$ would be $G*f$, so let's try plugging in $T=G *_x A$ for some $A$. (Here $*_x$ means convolution only in the $x$ variables, not in time.) We get

$$\partial_t (G *_x A) = \Delta(G *_x A)$$

Moving some derivatives around we have

$$\partial_t (G *_x A) = G *_x \partial_t A \\ \Delta(G *_x A) = \Delta(G) *_x A = \delta *_x A = A.$$

So $G *_x \partial_t A = A$. But we said that the solution to $\Delta(u)=f$ was $G * f$, so we have

$$\Delta A = \partial_t A.$$

The catch is that it turns out that the fundamental solution for the Laplace operator in $\mathbb{R}^3$ is, up to a constant factor, $\frac{1}{r}$. So your assumption $T=\frac{\Gamma}{r}$ exploits the Green's function in the background. If you calculate the Green's function for other dimensions $n \geq 3$, you find that it is $\frac{1}{r^{n-2}}$, up to a constant factor which depends on the dimension. So in spherically symmetric diffusion in 4 dimensions, you would have a division by $r^2$ instead of $r$.