Notation in Crank's "The Mathematics of Diffusion"

Question

I am reading the book The Mathematics of Diffusion by J. Crank (cf. Crank_1975_Diffusion) and, at pages 13-16, I could not understand the relation between $h$ and $l$. I am beginner yet and on my short thought they do not have any relation, I mean, I imagine the column between $0$ and $l$. Where is the $h$ here? So, I could not understand the derivation of $(2.17)$. Moreover, I could not understand why $n=-\infty$ instead of $n=0$ at the bounty of the sum.

Many thanks in advance.

Answer 1

Just before (2.15):

In the same way, we can study the diffusion of a substance initially confined to the region $-h < x < +h$ as in Fig. 2.4.

The idea is that the diffusing substance is initially in a spot (in a 1-dimensional domain) stripe (in a 2-dimensional domain that is translationally invariant along the direction of the strip -- a common situation for laminar planar flow) or of a "slab" (in a 3-dimensional domain) having half-width or half-thickness $h$.

Immediately after (2.9):

This is the initial distribution, for example, when a long column of clear water rests on a long column of solution, or when two long metal bars are placed in contact end to end.

The idea here is that the material is diffusing out into such a large volume, that there is no need to concern ourselves with any of the material reaching the ends of the volume. However, if the system is short enough that the diffusing material can reach the boundary of the volume, we must prevent the diffusing material leaving the volume. Let $\ell$ be the length of the enclosure, then we require, (2.16), $\partial C / \partial x = 0$ at $x = \ell$. That is, we require that the flow of concentration ($C$) through the plane at $x = \ell$ is zero.

The index, $n$ in the sum is tracking the reflections of the diffusing material off the close ends of the enclosing volume. There is the material that has never bounced off either boundary ($n = 0$). There is material that bounces off the right boundary once and does not bounce off either boundary ever again ($n = 1$). There is material that bounces off the left boundary once and does not bounce off either boundary ever again ($n = -1$). Continuing to track right bounce, left bounce ($n = 2$) and left bounce, right bounce ($n = -2$), and so on, we see that we need both positive an negative $n$s to get the contributions from the right-bounce-first and left-bounce-first paths.