## XyMqCy z Cy tdy d

and the term o(dt)/dt will disappear as dt goes to 0. We now equate the LHS and the RHS:

hy (y)q(y, t, Z, s)b(y, t)+ - hyy(y)q(y, t, z, s)a(y, t)

and this is a good, but not especially useful, equation, because we have h(y) on the left hand side, but its derivatives on the right hand side. We will deal with this difficulty by integrating by parts, recalling the basic formula f1 u dv = uvl ^ — f1 v du. I have included the limits j—i i—i j—i of integration here because they will be important. However, for simplicity, I am going to suppress the indices on h(y), q(y, t, z, s) and b(y, t)

in what follows. Thus, integrating the term involving b(y, t) by parts and using the notation (qb)y = (8 / 8y)(qb) we have hy4b = -

and, for the term involving a(y, t), we integrate by parts not once, but twice:

If we now assume that h(y) and its derivative approach 0 as | y| !i, Eqs. (7.74) and (7.75) simplify (and putting the indices back in):

We return to Eq. (7.71) and conclude q((y, t, z, s)h(y)dy =

so that the entire equation now only involves h(y). In fact, we could bring everything to the same side of the equation and factor h(y) through to obtain h(y)

q(y, t,z, s) + (q(y, t,z, s)b(y, t))y - 2 (q(y, t,z, s)«(y, t))y dy = 0 (7.78)

but this equation is supposed to hold for almost any choice of h(y) -remember that we made minimal assumptions about it. We thus conclude that q(y, t, z, s) satisfies (formally we say ''in a weak sense'') the equation q(y, t, z, s) = (q(y, t,z, s)«(y, t)) -(q(y, t, z, s)b(y, t)L

If we replace y by x, the form of the equation remains unchanged.

As before, we have the initial condition that t, z, s) ! — z) as s ! t, and the boundary condition that q(y, t, z, s) ! 0 as |y| !i. The forward equation is often used in population genetics (see Connections). In the mathematical literature, Eqs. (7.59) and (7.81) are said to be adjoints of each other (Haberman (1998) is a good source for more mathematical detail). We shall employ both backward and forward equations in the next chapter. Before we get to that next chapter, however, I would like to show a couple more backward equations, which have various interesting uses. And before that, I offer an exercise which I hope may clarify some of the differences between backward and forward equations.

Exercise 7.12 (M)

Remember that in Chapter 2 we reached the diffusion equation as the limit of a random walk. Let us reconsider such a random walk. Thus, X(t) represents the position of the process at time t on a lattice with separation e between sites. We assume that steps take place in time interval At and let r(x) and l(x) represent the probabilities of moving to the right or left in the next interval, given that X(t) = x. To get to a backward equation, let u(x, t) = Prfthe process has left [A, B] by time t|X(0) = x} (7.80)

(a) Show that u(x, t) satisfies the equation u(x, t) = (1 — r(x) — /(x))w(x, t — At) + r(x)u(x + e, t — At) + /(x)w(x — e, t — At)

and be certain that you can explain why the right hand side involves t — At.

(b) To get to the forward equation, let v(x, t) = Pr{X(t) = x}. Show that v(x, t) satisfies the equation v(x, t + Dt) =(1 — r(x) — 1(x))v(x, t)+ r(x — e)v(x — e, t) + 1(x + e)v(x + e, t)

(c) Now Taylor expand Eqs. (7.81) and (7.82) to first order in At and to second order in e and compare those results with the backward and forward equations that we have derived. A word of warning: this exercise is heuristic and ignoring all of the higher order terms in the Taylor expansion is not a generally safe thing to do, rather one needs to carefully take limits as we did in Chapter 2.

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