and now ''all'' we have to do is deal with the right hand side (RHS). We begin by Taylor expansion of h(x) around the intermediate pointy, using subscript notation so that hy(y) is the first derivative of h(x) evaluated at the point x = y. The Taylor expansion of h(x) is h(x) = h(y) + hy(y)(x -y)+1 hyy(y)(x -y)2 + O((x -y)3) (7.68)

The right hand side of Eq. (7.65) is now (with multiplication intended from the top line to the bottom one): 1

RHS = limdi!0 -T dt q(y, t, z, s)q(x, t + dt, y, t)dy — q(x, t, z, s) |h(y)+ hy(y)(x — y)+2 hyy(y)(x — y)2 + O((x — y)3 ) J dx

Now, we do the following: we multiply through all of the terms and re-order them. In addition, we will replace the Taylor expansion by h(x) itself when we multiply by q(x, t, z, s). The result of this process is a long expression, but an intelligible one: 1

q(x, t + dt, y, t) j dxdy (x — y)q(x, t + dt, y, t) | dxdy (x — y)2q(x, t + dt, y, t) j dxdy q(y, t, z, s q(x, t,z, s)h(x)dx

Now let us consider the terms on the right hand side of Eq. (7.70). First, we know that J q(x, t + dt, y, t)}dx = 1 because from y at time t, the process must move to somewhere at time t + dt, so that when we integrate over all values of x, the result is 1. Thus, the first term on the right hand side of Eq. (7.70) and the fifth term are the same, except that in the former the variable of integration is y and in the latter it is x. So these terms cancel. The other three terms are defined in terms of the transition moments for the diffusion process given by Eqs. (7.53) in a slightly different form

(x — y) q(x, t + dt,y, t)}dxdy = a(y, t) + o(dt) O((x — y)3)q(x, t + dt,y, t)}dx = o(dt)

Although they appear to be different, Eqs. (7.53) and (7.71) are really the same, since they both deal with the average of the transition. In Eqs. (7.53), we move from value z to value y + z, so that the transition size is y. In Eqs. (7.71), we move from y to x, so that the transition size is x — y. Now, we can clearly write the LHS, Eq. (7.67), as qt (y, t, z, s)h(y)dy through the simple change of variables of replacing x by y as the integration variable. We have already agreed that the first and fifth terms of Eq. (7.70) cancel, so that if we divide through by dt and apply Eq. (7.71), in Eq. (7.70), we are left with

0 0

Post a comment