## Info

We now know the equation that n(x, t) must satisfy. But how do we find the solution? One method would be to make the substitution n(x, t) = w(x, t)ert, in which case we find that w(x, t) satisfies the diffusion equation wt = (o-2/2)wxx (you can consider this an optional exercise). One could say that we've reduced this problem to the previous one, which could be solved by separation of variables. But let's go on, to see what new insights can be gained.

Another route is the following. Given the result in Eq. (2.56), let us guess that n(x, t) can be represented by a mixture of sines and cosines, in which case we might guess a form such as n(x, t) = A(t)sin(wx) + B(t)cos(wx) where the coefficients A(t) and B(t) and the frequency w need to be determined. Let's think about the boundary conditions, which apply to nx(x, t) = A(t)wcos(wx) — B(t)wsin(wx). Since cos(0) = 1 and sin(0) = 0, we can satisfy the boundary condition at x = 0 by picking A(t) = 0 for all time. Then to satisfy the boundary condition at x = L, we must have sin(wx) = 0. This will be true if wL = 0, wL = p, wL = 2p, etc. Now, if wL = 0, w must be 0, in which case cos(wx) = 1, independent of space. But we have already taken account of the spatially independent aspects of the solution, so we can ignore w = 0. We thus conclude that wL = kp, for k = 1, 2, 3, and so forth. Because of our result on the linear superposition of solutions, this means that the solution must be n(x , t) = Bk (t) cos x j

But we still do not know the values of the different Bk(t), which we call the amplitude of the kth mode. We will find them in two steps. First, we will derive an equation for each Bk(t). Second, we will find the initial value Bk(0); together these will tell us the entire solution. For the first step, we take the partial derivatives of n(x, t) given by Eq. (2.58)

= Bk (t) (y) cos(^"x where I have now used d/dt to denote the time derivative of each of the Bk(t). If we now substitute these back into the equation for n(x, t), we obtain

This equation will be satisfied if we choose the coefficients Bk(t) to satisfy

Equation (2.59) provides us with the main intuition about the interaction of diffusion and population growth. We see from this equation that Bk(t) is an exponential function of time. It is exponentially growing constant if equality holds, and exponentially We have thus derived a very precise relationship between the rate of population growth and the rate of diffusion. This relationship tells us how the amplitude of the kth mode if r > (a2/2) (k"/L)2 declining if r < (a2/2) (k"/L)2

n xx grows or declines. The result has nice intuitive appeal: if diffusion is very strong (so that r < (ct2/2) (kp/L)2 even when k = 1) then all of the modes will decline and initial fluctuations in n(x, 0) will smear out over time. On the other hand, if the diffusion coefficient is not too big, then for some values of k we will have r > (c2/2) (kp/L)2 and the amplitude of those modes will grow in time; for other values of k the amplitudes will decline (and there may be one value of k where exact equality holds, in which case the amplitude will remain constant). In this case, the more slowly varying amplitudes will be accentuated and small deviations in n(x, 0) will be enhanced in time (Figure 2.19c).

To understand what is happening, and to find that pesky value of Bk(0), it is helpful to think geometrically about the cosine function (we will get mathematical details in a minute). In Figure 2.20a, I have plotted y = cos(px/L) and y = cos(6px/L) for L = 10 (the choices k = 1, k = 6, and L = 10 are arbitrary, and you might want to make your own similar plots with different values of k). Notice the shape of the plot for k = 1: the curve starts at 1 when x = 0, smoothly decreases, passing through 0 when x = 5 and reaches — 1 when x = 10. The curve is symmetric around the line y = 0 when x = 5: each value of x < 5 has a certain value of y = cos(px/L) and there is a value of x > 5 with exactly the opposite value. Thus, for example, the integral ofy = cos(px/L) from x = 0 to x = L will be 0 (you could do this, of course, by simply integrating, but understanding the geometry is important). The same is true of the curve y = cos(6px/L), which fluctuates much more between x = 0 and x = L but is also symmetrical. In Figure 2.20b, I have plotted the product y = cos(px/L)cos(6px/L), which is also symmetric around the line y = 0 at x = 5. So, the integral of this product will also be 0. The situation is different, however, if we consider the squares y = cos2(px/L) or y = cos2(6px/L). In this case, of course, there are no negative values. The more slowly fluctuating cos2(px/L) has much less frenetic changes, but a remarkable fact from calculus is that their integrals are the same.

To be completely general, I report results for both sine and cosine. Suppose that j and k are any two integers greater than or equal to 1.

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