FUNDAMENTAL SOLUTIONS
23
Since
U
λ
(
x,
0) =
e

iλx
we seek a fundamental solution
p
(
x, y, t
) such
that
R
∞
∞
e

iλy
p
(
x, y, t
)
dy
=
U
λ
(
x, t
)
.
The Fourier inversion immedi
ately gives the result.
The integrals are standard Gaussians and we
have
p
(
x, y, t
) =
1
2
π
Z
∞
∞
e
iλy
e

λ
2
t

iλ
(
x

2
at
)
1 +
ce
2
ax

4
aiλt
1 +
ce
2
ax
¶
dλ
=
K
(
x

2
at

y, t
) +
ce
2
ax
K
(
x
+ 2
at

y, t
)
1 +
ce
2
ax
,
(4.14)
where
K
(
x

y, t
) =
1
√
4
πt
e

(
x

y
)
2
4
t
. The reader can check that this is a
fundamental solution and since
R
∞
∞
p
(
x, y, t
)
dy
=
U
0
(
x, t
) = 1, it is a
density. In fact it is the transition density for the process.
An interesting feature of this Fourier transform analysis is that it
is not always optimal to use stationary solutions.
In the next set of
examples, we use nonstationary solutions, for the simple reason that
they lead to Fourier transforms which are easier to invert.
Example 4.2.
We will obtain the transition density of a
mean revert
ing OrnsteinUhlenbeck
process
X
=
{
X
t
:
t
≥
0
}
,
dX
t
= (
a

bX
t
)
dt
+
√
2
σdW
t
,
(4.15)
with
a
≥
0
, b >
0.
If
a
= 0, then we have the regular Ornstein
Uhlenbeck process. The Kolmogorov forward equation is
u
t
=
σu
xx
+ (
a

bx
)
u
x
, x
∈
R
.
(4.16)
Observe that (4.16) has a symmetry of the form
˜
u
²
(
x, t
) =
e
a
b

a ²
σ
+
σ
2
b
+
²
(
a

b x
+
b ²
sinh(
b t
))
e
b t
σ
u
(
x

2
²
sinh(
b t
)
, t
)
.
(4.17)
Now ˜
u
²
(
x,
0) =
e
x

b x ²
σ
u
(
x,
0). We will let
²
→
iσλ
b
.
Then we look for a
fundamental solution
p
(
x, y, t
) and a solution
u
such that
Z
∞
∞
e

iλy
u
(
y,
0)
e
y
p
(
x, y, t
)
dy
= ˜
u
iσλ
b
(
x, t
)
.
(4.18)
We could use
u
= 1, but as illustration, instead we seek a solution of
(4.16) of the form
u
(
x, t
) =
e
m
(
t
)
x
+
z
(
t
)
.
Substitution into the PDE gives
u
(
x, t
) =
e

2
e
b t
(
a

b x
)+
σ
2
b e
2
b t
.
Using this solution and the given symmetry,
we seek a fundamental solution such that
Z
∞
∞
e

iλy
e
y
p
(
x, y, t
)
dy
=
e

λ
2
σ
2
sinh(
b t
)
b
+
i λ
(
σ
+
e
b t
(
a

b x

e
b t
(
a
+
σ
)
))
b e
2
b t
+
r
(
x,t
)
with
r
(
x, t
) =

2
e
b t
(
a

b x
)

σ
2
b e
2
b t
.
We can now recover
p
by taking the inverse
Fourier transform. The integrals are standard Gaussians and we leave