FUNDAMENTAL SOLUTIONS
19
where
d
=
1
4
(2

n
+
p
(
n

2)
2
+ 8
μ
)
,
leads to the Laplace trans
form
R
∞
0
y
d
p
(
x, y, t
)
e

λy
dy
=
x
d
(1+2
λt
)
2
d
+
n/
2
exp
'

λx
1+2
λt
“
.
Inversion gives
the fundamental solution
p
(
x, y, t
) =
1
2
t
x
y
¶
1
2
(1

n
2
)
I
2
d
+
n
2

1
√
xy
t
¶
exp
‰

(
x
+
y
)
2
t
.
(3.9)
Now
E
x
h
e

λX
t

μ
R
t
0
ds
Xs
i
=
Z
∞
0
e

λy
p
(
x, y, t
)
dy
=
e

x
2
t
‡
x
2
t
·
d
Γ(
α
)
1
F
1
(
α, β,
x
2
t
+4
t
2
λ
)
Γ(
β
)(1 + 2
λt
)
α
,
with
α
=
d
+
n
2
, β
= 2
d
+
n
2
.
Here
1
F
1
is Kummer’s confluent hyper
geometric function. This is the Laplace transform of the joint density
of (
X
t
,
R
t
0
ds
X
s
).
See [8] for more on this example and applications of
symmetries to the calculation of joint densities.
Example 3.6.
Let us now consider a two dimensional problem. We
will solve
u
t
=
u
xx
+
u
yy

A
x
2
+
y
2
u,
(
x, y
)
∈
R
2
, A >
0
(3.10)
u
(
x, y,
0) =
f
(
x, y
)
.
We convert the problem to polar coordinates.
So we let
u
(
x, y, t
) =
U
(
p
x
2
+
y
2
,
tan

1
(
y
x
)
, t
), where
U
(
r, θ, t
) satisfies
U
t
=
U
rr
+
1
r
U
r
+
1
r
2
U
θθ

A
r
2
U,
(3.11)
with
U
(
r, θ,
0) =
f
(
r
cos
θ, r
sin
θ
) =
F
(
r, θ
). Taking the Fourier trans
form in
θ
, where
b
f
(
n
) =
R
2
π
0
f
(
θ
)
e

inθ
dθ
, this becomes
b
U
t
=
b
U
rr
+
1
r
b
U
r

n
2
+
A
r
2
b
U.
(3.12)
With initial data
b
U
(
r, n,
0) =
b
F
(
r, n,
0), this has solution
b
U
(
r, n, t
) =
Z
∞
0
b
F
(
ρ, n,
0)
p
(
r, ρ, n, t
)
dρ,
with
p
(
r, ρ, n, t
) a fundamental solution of (3.12). Stationary solutions
are
u
0
(
r
) =
r
√
n
2
+
A
and
u
1
(
r
) =
r

√
n
2
+
A
.
Applying Theorem 2.3 with
u
0
we find that there is a fundamental solution of (3.12) such that
Z
∞
0
e

λρ
2
u
0
(
ρ
)
p
(
r, ρ, n, t
)
dρ
=
r
√
n
2
+
A
(1 + 4
λt
)
1+
√
n
2
+
A
exp

λr
2
1 + 4
λt
¶
.