Examples 585equation into a simple algebraic equation. Taking the Laplace transform of both sides
of eqn. (15.3.12), and solving for ̃v(s), we obtain
v ̃(s) =v(0)(s+λ)
s^2 +sλ+λa+
f ̃(s)(s+λ)
s^2 +sλ+λa, (15.3.13)
where the fact that the memory integral is a convolution has been used to give its
Laplace transform as a product of Laplace transforms ofv(t) andζ(t). For Laplace
inversion, the poles of the the function (s+λ)/(s^2 +sλ+λa) are needed. These occur
wheres^2 +sλ+λa= 0, which yields two poless±given by
s±=−λ
2±
√
λ^2 − 4 λa
2. (15.3.14)
The poles will be purely real ifλ≥ 4 aand complex ifλ < 4 a. Performing the Laplace
inversion gives the solution in the form
v(t) =v(0)[
(s++λ)es+t
(s+−s−)+
(s−+λ)es−t
(s−−s+)]
+
∫t0dτ f(t−τ)[
(s++λ)es+τ
(s+−s−)+
(s−+λ)es−τ
(s−−s+)]
. (15.3.15)
Since〈v(0)f(t)〉= 0, the velocity autocorrelation function becomes
〈v(t)v(0)〉=〈
(v(0))^2〉
e−λt/^2[
cos Ωt+λ
2Ωsin Ωt]
, (15.3.16)
where Ω =
√
λa−λ^2 /4 for complex roots and〈v(t)v(0)〉=〈
(v(0))^2〉
e−λt/^2[
coshαt+
λ
2 αsinhαt]
, (15.3.17)
whereα=
√
λ^2 / 4 −λa. For both cases, the diffusion constant obtained from eqn.
(13.3.32) is
D=kT
A. (15.3.18)
Since
∫∞
0 ζ(t)dt=A, eqn. (15.3.18) is consistent with eqn. (15.3.9) for the free Brow-
nian particle. AsA→ ∞, the bath becomes highly dissipative andD→0. Again,
the overall decay is exponential, which means that the long-time algebraic decay of
characteristic of such an autocorrelation function is not properlydescribed.
15.3.3 The harmonic oscillator in a bath with memory
As a final example of a GLE model, consider a harmonic reaction coordinate described
by a bare potentialV(q) =μω^2 q^2 /2. According to eqn. (15.2.13), the potential of mean
forceW(q) is also a harmonic potential but with a different frequency given by