Page 360 - Numerical Methods for Chemical Engineering

Brownian dynamics and SDEs 349

Taking the expectation of (7.184) over a very small time step, the stochastic contribution
cancels out as dW t = 0. Thus,

∂F 1 ∂ F 2
2
L t F(x) = a(t, x) + [b(t, x)] (7.186)
∂x 2 ∂x 2

Let p(t, x|t , x ) be the transition probability that if the particle is at x at time t , then at

time t,itisat x. Then, we could write (7.185) as
1 ' +∞ 1

L t F(x) = lim F(x )p(t + δt, x |t, x)dx − F(x) (7.187)
δt→0 δt
−∞

Let us multiply (7.187) by p(t, x|t , x ) with t < t and integrate over x:

+∞
'

[L t F(x)]p(t, x|t , x )dx
−∞
1 ' +∞ ' +∞

= lim F(x )p(t + δt, x |t, x)dx p(t, x|t , x )dx
δt→0 δt
−∞ −∞
+∞
' 1

− F(x)p(t, x|t , x )dx (7.188)
−∞
We now use the Chapman–Kolmogorov equation,
+∞
'

p(t + δt, x |t, x)p(t, x|t , x )dx = p(t + δt, x |t , x ) (7.189)
−∞
to write (7.188) as
+∞
'

[L t F(x)]p(t, x|t , x )dx
−∞
1 ' +∞ ' +∞ 1

= lim F(x )p(t + δt, x |t , x )dx − F(x)p(t, x|t , x )dx

δt→0 δt
−∞ −∞
(7.190)
In the first integral on the right-hand side x is only a dummy variable of integration, and

we are free to replace it with x,
'
+∞

[L t F(x)]p(t, x|t , x )dx

−∞
1 ' +∞ ' +∞ 1

= lim F(x)p(t + δt, x|t , x )dx − F(x)p(t, x|t , x )dx
δt→0 δt
−∞ −∞
(7.191)
Collecting the integrals on the right-hand side, taking the limit δt → 0, and using the finite
difference approximation
∂ 1

p(t, x|t , x ) ≈ [p(t + δt, x|t , x ) − p(t, x|t , x )] (7.192)
∂t δt
we have
+∞ +∞ ∂
' '

[L t F(x)]p(t, x|t , x )dx = F(x) p(t, x|t , x ) dx (7.193)
∂t
−∞ −∞