If we want what happens at x = 0 and x = L to be independent of each other, the way to do it is to require that each of the terms on the right hand side are 0, so that we obtain the boundary conditions — vpL, t) + (c2/2)p(L, t) = 0 and —vp(0, t) + (c2/2)p(0, t) = 0. In the literature these are often called ''no flux'' (from chemical analogies) or "reflecting" boundary conditions. The latter makes sense: the walker is constrained to stay between 0 and L, so that when it comes up against x = 0 or x = L, it must be reflected back into the region, much as a ball bouncing around in a room will bounce off the walls of the room.

Before moving on, I want to introduce one more concept, which we will use in a slightly different way in the next section, but which you will encounter frequently in the literature. To illustrate these ideas, let us continue with the case of reflecting boundary conditions, and for simplicity set v = 0. Our problem is then to solve pt = (o-2/2)px, given some initial condition px, t) = ^0(x) and subject to the boundary conditions that p{x, 0) = 0) = 0. To do this, let us guess that px, t) is the product of a function of time and a function of space. How do we know to make this guess? Well, first, generations of other scientists and mathematicians have tried it and found that it worked. So, we have history on our side. Second, new things are often discovered by good guessing. In this book, of course, I am not going to take you down too many blind roads (that is, bad guesses). The movie The Co/or of Money begins with an off-screen voice describing 9 ball pool and continues "which is to say that in 9 ball luck plays a part... but for some players, luck itself is an art.'' The same is true of applying mathematical methods to understand scientific questions. We need good guesses and good luck that the guess is correct, but sometimes we create our own luck through experience and thought. The Czech chess instructor Jan Amos Komensky once said, regarding chess, "Through play, knowledge'' (Pandolfini, 1989, p. xix). It works here too.

Accepting this guess, which is called the method of separation of variables, means that px, t) = T(t)S(x), where T(t) is a function depending only upon time and S(x) is a function depending only upon space. The diffusion equation then becomes Tt(t)S(x) = (^2/2)T(t)Sxx(x), where I still use subscripts to denote derivatives, although these are now ordinary (not partial) derivatives. Dividing both sides by T(t)S(x) we obtain

Now, the left hand side of Eq. (2.55) depends only upon time and the right hand side depends only upon space. What does this mean? It means that they both had better be independent of both time and space -each side should be constant. The left hand side also implies that T(t) must be an exponential function. We do not want the solution of the diffusion equation to grow without bound in time, because that makes no sense, so the constant must be negative (or at least not positive). A way of writing a non-positive number is -n2, where n = 0, 1, 2, etc. Then we know that T(t) ~ exp(—n t). The equation that S(x) satisfies will then become Sxx(x) = —(2n2/o-2)S(x). Since the second derivative of S(x) is a negative number times S(x), we know that S(x) must involve sines or cosines. The diffusion equation is a linear equation, so Exercise 2.10 (on the linear combination of solutions) tell us that the most general form of the solution for the equation that S(x) satisfies will be a mixture of sines and cosines. In particular, we can write where the An and Bn are constants, which we must somehow determine. The way this is done is explained in the next section.

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