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Vol. 13 (2008), Paper no. 45, pages 1283–1306.
Journal URL
http://www.math.washington.edu/~ejpecp/
Semiclassical analysis and a new result for Poisson Lévy excursion measures
Ian M Davies
Department of Mathematics, Swansea University,
Singleton Park, Swansea, SA2 8PP, Wales, UK
Email: I.M.Davies@Swansea.ac.uk
Abstract
The Poisson-Lévy excursion measure for
√ the diffusion process with small noise satisfying
the Itô equation dX ε = b(X ε (t))dt + ε dB(t) is studied and the asymptotic behaviour
in ε is investigated. The leading order term is obtained exactly and it is shown that at an
equilibrium point there are only two possible forms for this term - Lévy or Hawkes – Truman.
We also compute the next to leading order.
Key words: excursion measures; asymptotic expansions.
AMS 2000 Subject Classification: Primary 60H10, 41A60.
Submitted to EJP on March 22, 2007, final version accepted June 5, 2008.
1283
1
Introduction
Consider a one-dimensional diffusion process defined by
dX(t) = b(X(t)) dt + dB(t),
X(0) = a,
where b is a Lipschitz-continuous function and B(t) is a standard Brownian motion. The generator, G, of the above diffusion is
d
1 d2
+ b(x) ,
G=
2 dx2
dx
and the putative invariant density is
¶
µ Z x
b(u) du .
ρ0 (x) = exp 2
If ρ0 ∈ L1 (R, dx) the boundary {−∞, ∞} is inaccessible. We assume this in what follows. The
transition density
pt (x, y) = P(X(t) ∈ dy|X(0) = x)/dy,
satisfies
¶
µ
∂pt (x, y)
∂
1 ∂pt (x, y)
=
− b(y)pt (x, y) ,
∂t
∂y 2
∂y
¢
¡
= G∗y pt (x, y),
lim pt (x, y) = δx (y),
t↓0
G∗y being the L2 adjoint of Gy , and δ being the Dirac delta function. The density of the diffusion
Z
t
ρ (y) = ρ0 (x)pt (x, y) dx
therefore satisfies
Evidently,
so ρ0 is the invariant density.
∂ρt ¡ ∗ t ¢
= Gy ρ (y).
∂t
¡ ∗ ¢
Gy ρ0 (y) = 0,
∂ρ0 (y)
= 0,
∂t
Crucial in what follows is the operator identity for any well-behaved f
³
´
−1/2
1/2
Gf = − ρ0 Hρ0
f,
or
and
³
´
−1/2
1/2
G = − ρ0 Hρ0
³
´
1/2
−1/2
G∗ = − ρ0 Hρ0
,
1284
where H is the one dimensional Schrödinger operator with potential
V = 12 (b2 + b′ ),
1 d2
H=−
+ V (x).
2 dx2
³
´
1/2
1/2
This follows because Hρ0
≡ 0, i.e. ρ0 is the ground state of H. For convenience, we
will assume V ∈ C 2 (R), V bounded below together with V ′′ , V polynomially bounded with
derivatives.
2
Excursion Theory
The map s 7→ X(s) is continuous and so {s > 0 : X(s) ≷ a} is an open subset of R. Therefore,
upward
{s > 0 : X(s) ≷ a} can be decomposed into a countable union of open intervals – downward
excursion intervals. Define
L± (t) = Leb{s ∈ [0, t] : X(s) ≷ a},
and the local time at a
La (t) = lim h−1 Leb{s ∈ [0, t] : X(s) ∈ (a − h/2, a + h/2)}.
h↓0
La (t) has inverse γ a (t), the time required to wait until La equals t. It can be seen that γ a (t) is
a stopping time with X(γ a (t)) = a. Moreover, as is intuitively obvious,
Jumps in γ a (t) = Excursions of X from a up to La equals t.
Example 1 Lévy [1954]
Lévy proved that for b ≡ 0, for each λ > 0,
½ Z
Ea exp(−λγ (t)) = exp −t
a
∞
0
−λs
(1 − e
¾
) dνa (s) ,
with Poisson-Lévy excursion measure
µ ¶1/2
2
s−1/2 , s > 0.
νa [s, ∞) =
π
(2.1)
Equating powers of λ in the above, we conclude that
♯(s, t)= Number of excursions of duration exceeding s up to La equals t
is Poisson with
P(♯(s, t) = N ) = exp (−tνa [s, ∞)) (tνa [s, ∞))N /N !,
for N = 0, 1, 2, . . . , and so the expected number of excursions of duration exceeding s per unit
local time at a is νa [s, ∞), the Poisson-Lévy excursion measure.
1285
Example 2 Hawkes and Truman [1991]
For the Ornstein-Uhlenbeck process b(x) = −kx, where k is a positive constant, the Hamiltonian
is just
¶
µ
1
d2
2 2
H=
− 2 +k x −k
2
dx
¡
¢
and ρ0 (x) = C exp −kx2 . This leads to
¾
½ Z ∞
−λs
a
(1 − e ) dν0 (s) , λ > 0,
E0 exp(−λγ (t)) = exp −t
0
with
ν0 [s, ∞) =
´−1/2
2k 1/2 ³ 2ks
.
e
−
1
π 1/2
(2.2)
upward
We discuss generalisations of the above to upward and downward excursions. Note that downward
excursions can only be affected by values of b(x) for x ≷ a. Therefore it is natural to define the
symmetrised potential
(
V (x),
x > a,
+
Vsymm =
V (2a − x), x < a.
−
with Vsymm
being defined in a similar manner. In an analogous manner we also define
H± = −
1 d2
±
+ Vsymm
(x).
2 dx2
We now have the result due to Truman and Williams [1991]
Proposition 1. Modulo the above assumptions
½ Z
¡
¢
±
a
Ea exp −λL (γ (t)) = exp −t
0
with
νa± [t, ∞)
=
Z
y≷a
Remarks
∞
−λs
(1 − e
¾
) dνa± (s)
¯
1/2
¡
¢
ρ0 (y) ∂ ¯¯
exp −tH ± (x, y).
dy 1/2
¯
ρ0 (a) ∂x x=a
1. L± (γ a (t)) are independent with L+ (γ a (t)) + L− (γ a (t)) = γ a (t).
2. Jumps in L± (γ a (t)) =
upward
downward
excursions from a up to La equals t.
3. νa± [s, ∞) is the expected number of
local time at a.
upward
downward
1286
excursions of duration exceeding s per unit
Proof. (Outline) The proof uses the result of Lévy [1954]
Ea exp(−λγ a (t)) = exp (−t/p̃λ (a, a)) ,
where p̃λ (x, y) =
R∞
0
We can deduce that
e−λs ps (x, y) ds and ps (·, ·) is the transition density.
p̃−1
λ (a, a)
=λ
Z
∞
−∞
ρ0 (x) −λτx (a)
Ee
dx,
ρ0 (a)
where τx (a) = inf{s > 0 : X(s) = a|X(0) = x}. Here the point is that for any point a
intermediate to x and y
Z ∞
P(τx (a) ∈ du)pt−u (a, y).
pt (x, y) =
0
Since the right hand side is a convolutional product, taking Laplace transforms and letting y → a
gives
Ee−λτx (a) = p̃λ (x, a)/p̃λ (a, a).
Now multiply both sides by ρ0 (x) and integrate with respect to x (using the fact that ρ0 is the
invariant density) to get the desired result for p̃−1
λ . Some elementary computation then leads to
the result in Proposition 1.
3
The Poisson-Lévy Excursion Measure for Small Noise
upward
We will now consider the downward
excursions from the equilibrium point 0 for the onedimensional time-homogeneous diffusion process with small noise, X ε (t), where
√
dX ε (t) = b(X ε (t)) dt + ε dB(t).
Introducing the small noise term into the Truman-Williams Law seen in the previous section,
we get:
Proposition 2. The expected number of
per unit local time at 0 is given by
upward
downward
excursions from 0 of duration exceeding s,
1
ν0± [s, ∞)
=±
Z
y≷0
¯
µ
¶
sH ±
∂ ¯¯
exp −
ε
(x, y)dy,
1
¯
ε
ρ02 (0) ∂x x=0
ρ02 (y)
where ρ0 is the invariant density and H ± is the symmetrized Hamiltonian for
V = 12 (b2 + εb′ ).
One should note the form of V , in particular the presence of ε as a multiplier of b′ . This rather
specific dependence originates from the Shrödinger operator mentioned earlier. Consequently,
we are unable to resort to the usual methods for resolving such a dependence.
We now give a result due to Davies and Truman Davies and Truman [1982].
1287
path for the classical action
Proposition
´
³R 3. Let XminR(·) be the minimising
t 2
t 2
−1
A(z) = 2
0 ż (s) ds + 0 b (z(s)) ds with z(0) = x, z(t) = y.
hSet 2A(Xmin )i = A(x, y, t). Then for the self-adjoint quantum mechanical Hamiltonian H(ε) =
− ε2 △ +Vε , where Vε = 12 (b2 + εb′ ) ∈ C ∞ (R) and is convex (where Vε = V0 + εV1 ∈ C 4 ,
1
bounded below with V0′′ ≥ −|β|), then for each finite time t ≥ 0 ( for t ≤ π/|β| 2 ).
µ
¶
tH(ε)
exp −
(x, y)
ε
)
¯1 ³
µ
¶ (¯ 2
´
2
¯
¯
1
∂
A(x,
y,
t)
A(x,
y,
t)
2
¯
¯
.
= (2πε)− 2 exp −
¯ ∂x∂y ¯ 1 + εK + O(ε )
ε
K is a rather complicated expression with many terms involving sums and products of b (and
its derivatives), V (and its derivatives) and the Feynman-Green function G(τ, σ) of the Sturmd2
′′
Liouville differential operator dσ
2 − V (Xmin (τ )) with zero boundary conditions i.e. G(0, τ ) =
G(t, τ ) = 0, and discontinuity of derivative across τ = σ of 1.
For a proof of this result see Davies and Truman [1982].
Henceforth, for simplicity we assume that b2 (x) is an even function of x so that ν0+ = ν0− .
Theorem 1. Using the notation and assumptions of Proposition 3, the leading term of the
Poisson-Lévy excursion measure, for excursions away from the position of stable equilibrium 0,
where b(0) = 0, and b′ (0) ≤ 0 is given by
ν0+ [t, ∞)
∞
½ 2µ 2
¾
¶
y
∂ A(0, 0, t)
A(0, 0, t)
′
∼ (2πε)
dy exp −
− b (0) −
2ε
∂y 2
ε
0
)
(¯
1
¯
µ
¶
¶
µ
Z
¯
¯ ∂2A ¯ 2
1 t ¯¯ ′
∂A
¯
¯
,
exp
b (Xmin (s))¯ ds
×
−
¯ ∂x∂y ¯
∂x x=0
2 0
x=0
Z
− 12
with the action A(x, y, t) =
1
2
Rt
0
ż 2 (s) ds +
Rt
0
¡
¢
V z(s) ds, where V = 21 b2 .
Proof. As usual, the classical path Xmin (t) = X(x, y, t) satisfies, correct to first order in ε
Ẍmin ∼ V0′ (Xmin ) + ε V1′ (Xmin ).
For V0 = 12 b2 (assumed to be convex with V0 (0) = 0, V ′ (0) = 0, and V ′′ (0) > 0, for example
V0 (x) = 12 x2 , b(x) = −x) , and V1 = 12 b′ , then to leading order Ẍmin = V0′ (Xmin ).
¢
Rt ¡
Rt
The contribution to the action A(x, y, t) = 12 0 ż 2 ds + 0 V z(s) ds from V1 is to leading order
ε
Z
0
t
V1 (Xmin (s)) ds =
ε
2
Z
ε
= −
2
1288
t
b′ (Xmin (s)) ds
0
Z
0
t¯
¯
¯b′ (Xmin (s))¯ ds,
introducing | · | for convenience. Therefore, from Proposition 3,
µ
tH(ε)
exp −
ε
¶
µ
A(x, y, t)
(x, y) ∼ (2πε)
exp −
ε
³
´
and so, the contribution to term exp − A(x,y,t)
from V1 is
ε
exp
− 12
µ Z t
¶
¯ ′
¯
1
¯b (Xmin (s))¯ ds ,
2 0
¯1
¶¯ 2
¯ ∂ A(x, y, t) ¯ 2
¯
¯
¯ ∂x∂y ¯ ,
the Zero Point Energy term.
Therefore, we have using Proposition 2 the leading order term in the Poisson-Lévy excursion
measure, for upward excursions from stable equilibrium point 0 given by
ν0+ [t, ∞)
∞
µ Z y
¶
1
dy exp
∼ (2πε)
b(u)du
ε 0
0
"
¯
µ Z t
¶
µ
¶ ¯ 2 ¯ 12 #
¯
¯ ′
¯∂ A¯
A(x,
y,
t)
1
∂ ¯¯
¯
¯b (Xmin (s))¯ ds exp −
¯
exp
×ε
¯
¯ ∂x∂y ¯ .
∂x x=0
2 0
ε
− 12
Z
Hence, for small ε, the leading order term is
¶¶
µ µZ y
Z ∞
1
+
− 12
b(u)du − A(0, y, t)
ν0 [t, ∞) ∼ (2πε)
dy exp
ε
0
0
"¯
¯1 µ
¶¶#
¶µ
µ Z t
¯ ∂2A ¯ 2
¯
¯ ′
∂A
1
¯
¯b (Xmin (s))¯ ds
−
× ¯¯
exp
∂x∂y ¯
∂x
2 0
(3.1)
+ O(ε).
x=0
Comparing this to the Laplace Integral
I(ε) =
Z
b
e−
φ(x)
ε
θ(x) dx,
a
where the main contribution comes from the asymptotic behaviour at points xi ∈ [a, b] with
φ′ (xi ) = 0, we can see that the main contribution to the integral in equation 3.1 comes from
those y(t, 0) satisfying
∂A(0, y, t)
b(y) =
.
∂y
Ry
If we expand φ(y) = 0 b(u)du − A(0, y, t) in a Taylor series about y(t, x) = 0 we get
1
φ(y) = φ(0) + (y − y(t, x))φ′ (0) + (y − y(t, x))2 φ′′ (0) + · · ·
2
1
2 ′′
= φ(0) + (y − y(t, x)) φ (0) + · · ·
2
1289
Hence,
( µ
1
ν0+ [s, ∞) ∼ (2πε)
dy exp
− A(0, 0, s)
ε
0
)
¸¶
·Z y
2
1
2 ∂
+
b(u) du − A(0, y, s)
(y − y(s, x))
2
∂y 2 0
y=y(s,x)=0
(Ã
¶ ¯ 2 ¯ 21 µ
¶ !)
µ Z t
¯∂ A¯
¯ ′
¯
1
∂A
¯
¯b (Xmin (s))¯ ds ¯
,
×
exp
−
¯ ∂x∂y ¯
2 0
∂x x=0
x=0
− 12
giving,
ν0+ [s, ∞)
∞
Z
∞
¾
¸
·
−A(0, 0, s)
1 ∂ 2 A(0, y, s)
2
′
∼ (2πε)
dy exp
y
−
− b (y)
ε
2ε
∂y 2
0
y=0
(Ã
¶ !)
¶ µ 2 ¶ 12 µ
µ Z t
¯
¯ ′
∂A
∂
A
1
¯b (Xmin (s))¯ ds
−
,
×
exp
2 0
∂x∂y x=0
∂x x=0
− 21
Z
½
and so the result follows.
4
Poisson-Lévy Excursion Measure – leading order behaviour
We have seen in the previous section that in order to calculate the Poisson-Lévy excursion
measure ν0+ [t, ∞) for a general process X(t) = X[x, y, t], we require expressions for the following
derivatives of the action A(x, y, t).
¯
∂A
∂A(x, y, t) ¯¯
where
p0 = −
,
¯
∂x
∂x
x=0
¯ 2 ¯ 1 ¯¯
µ 2 ¶−1
¯ ∂ A ¯2 ¯
∂ A
∂X(t)
¯
¯
=
(the Van Vleck identity),
where
¯ ∂x∂y ¯ ¯¯
∂p0
∂x∂y
¯x=0
∂ 2 A ¯¯
∂A
where
b(y) =
(x, y, t).
¯
2
∂y x=0
∂y
These expressions are evaluated in the following propositions :Proposition 4. For p(0) = b(0) = 0, |b′ (0)| =
6 0,
−
∂A
(0, y, t)
∂x
∼
|b′ (0)| y
,
sinh |b′ (0)| t
as
y → 0.
Proof. Observe that px (y, t) = − ∂A(x,y,t)
= initial momentum at x needed to reach y in time
∂x
t, satisfies
Z y
du
t=
³
´1 .
2
x
p20 (y, t) + b2 (u) − b2 (x)
1290
Therefore, p0 (y, t) satisfies
t=
Z
y
0
Changing integration variable u = y v,
Z
t=
du
³
p20 (y, t)
1
0
+
´1 .
b2 (u)
dv
³
p20 (y,t)
y2
+
b2 (yv)
y2
(4.1)
2
´1 ,
2
since to first order V = 12 b2 and V (0) = 0 by assumption,
p0 (y) − p0 (0)
→ p′0 (0),
y
Therefore,
t=
Z
1
0
¯ ′ ¯
¯ p (0) ¯
Letting v = ¯ b0′ (0) ¯ sinh w we get
giving, for b′ (0) 6= 0,
as
y → 0.
dv
³
´1 ,
2
2
p′0 (0, t) + v 2 b′2 (0)
as y → 0.
¯ ′ ¯
¯
¯
1
−1 ¯ b (0) ¯
sinh ¯ ′ ¯ ,
t= ′
|b (0)|
p (0)
|p′0 (0, t)| =
(4.2)
|b′ (0)|
.
sinh |b′ (0)| t
(4.3)
Note that for b′ (0) = 0, we get p′0 (0, t) = ±1/t and so equation 4.2 has the correct limiting
behaviour.
Proposition 5. For |b′ (0)| =
6 0,
∂2A
(0, y, t) ∼
∂x∂y
µ
sinh |b′ (0)|t
|b′ (0)|
¶−1
,
as
y → 0.
Proof. Using the fact that p0 (y) = −∂A/∂x and equation 4.1 we quickly get
Z
¡
¢1 y
du
−1
2
2 2
= p0 (y) p0 (y) + b(y)
3 .
′
p0 (y)
0 (p0 (y)2 + b(u)2 ) 2
Again, changing the variable of integration u = yv, we get
!1 Z
Ã
p0 (y) ³ p0 (y) ´2 ³ b(y) ´2 2 1
−1
+
=
Ã
p′0 (y)
y
y
y
³
0
1291
dv
p0 (y)
y
´2
+
³
b(yv)
y
´2
!3 .
2
Now, following the previous argument, as y → 0,
¡ ′
¢1
−1
′
2
′
2 2
→
p
(0,
t)
p
(0,
t)
+
b
(0)
0
0
p′0 (y)
¯ ′ ¯
¯ p (0) ¯
Letting v = ¯ b0′ (0) ¯ sinh w in the above equation gives
−1
p′0 (y)
p′0 (0)2 + b′ (0)2
|b′ (0)| p′0 (0)2
sinh |b′ (0)|t
¡
→
=
|b′ (0)|
¢1 Z
2
Z
1
0
dv
(p′0 (0)2
˛
˛
˛ |b′ (0)| ˛
sinh−1 ˛˛ p′ (0) ˛˛
0
0
3
+ b′ (0)2 v 2 ) 2
1
dw
cosh2 w
,
using
|p′0 (0)| =
|b′ (0)|
.
sinh |b′ (0)|t
Proposition 6. For b′ (0) ≤ 0, we have
b′ (0) −
∂ 2 A(0, 0, t)
= −|b′ (0)|(1 + coth |b′ (0)| t).
∂y 2
( ∂A
∂y is the momentum at y given that y is reached from x in time t.)
Proof. From
∂2A
∂y 2
∂2A
b (y) −
∂y 2
′
=
=
∂
p(y) =
∂y
′
³
´1
2
∂ p20 (y, t) + b2 (y)
∂y
∂p0 (y)
∂y
b (y) − ³
1+
+
³
,
b(y) ′
p0 (y) b (y)
b(y)
p0 (y)
′
´2 ´ 1
2
b′ (0)
p′0 (0) + pb′ (0)
′
0 (0)
→ b (0) − ³
³ ′ ´2 ´ 1
2
1 + pb′ (0)
(0)
as
y → 0,
0
=
=
′
b (0) −
|b′ (0)|
sinh |b′ (0)|t
³
sinh |b′ (0)|t
|b′ (0)|
1 + sinh2 |b′ (0)|t
b′ (0) − |b′ (0)|
1292
+ b′ (0)2
cosh |b′ (0)|t
sinh |b′ (0)|t
´1
2
.
Therefore
and result follows.
½
¾¯
2 A(0, y, t) ¯
∂
¯
b′ (y) −
¯
2
¯
∂y
y=0
¡
¢
−→ −|b′ (0)| 1 + coth |b′ (0)|t ,
We now come to our main result for excursions from an equilibrium point 0.
Theorem 2. For the diffusion X ε with small noise satisfying
¡
¢
√
dX ε (t) = b X ε (t) dt + ε dB(t),
denote the Poisson-Lévy measure for excursions from 0 by ν0± .
Assuming b is continuous, b having right and left derivatives at 0, with b(0± ) ≶ 0 and b(0) = 0,
then if V0± = 12 b2 satisfies
′′
π
V0± ≥ −|β± |, for t ≤
1 ,
|β± | 2
ν0± [s, ∞)
∼
µ
ε k±
π
¶1 ³
2
e2k± t − 1
´− 1
2
with k± = |b′ (0± )|. When |b′ (0± )| = 0 the limiting behaviour is correct and yields
ν0± [t, ∞)
³ ε ´1
1
2
∼
t− 2 .
2π
Proof. Using Theorem 1, and the expressions obtained in Propositions 4, 5 and 6, for the derivatives of the action as y → 0, we get as the leading order term for excursions from the stable
equilibrium position 0, (dropping ± again for convenience),
ν0+ [s, ∞)
¯
¯ 1µ
¶
¯ sinh |b′ (0)|t ¯− 2
|b′ (0)|
¯
¯
∼ (2πε) e
¯ |b′ (0)| ¯
sinh |b′ (0)|t
!
Ã
Z ∞
y2 ′
×
dy y exp − |b (0)|(1 + coth |b′ (0)|t)
2ε
0
¶
µ
′
1 t|b (0)|
1
1
ε
= (2πε)− 2 e 2 |b′ (0)| 2 (sinh |b′ (0)|t)− 2
cosh |b′ (0)|t + sinh |b′ (0)|t
µ
¶− 1
2
t|b′ (0)|
1
1 t|b′ (0)|
′ (0)|
− 12 12 ′
−
−t|b
2
= (2π) ε |b (0)| 2 e
.
(e
−e
)
2
− 12
s|b′ (0)|
2
These results correspond to the Poisson-Lévy excursion measures for the examples seen earlier.
1293
5
Poisson-Lévy Excursion Measure – higher order behaviour
In calculating higher order terms in the Poisson-Lévy excursion measure ν0+ [s, ∞), we obtain
the surprising result that the next order term is identically zero. We now write the leading term
1
as ε 2 ν +
1 .
2
Once again we must emphasise the particular dependence of Vε on ε and how this requires us
to follow a rather complicated route in determining the higher order dependencies on ε. This
arises due to our study originating from stochastic mechanics where the Schrödinger equation
and operator hold sway.
Theorem 3. For the diffusion process with small noise, assuming b(x) ≤ 0 for all x, the
Poisson-Lévy excursion measure is given by
1
3
2
νε+ ∼
= ε 2 ν+
1 + O(ε )
2
i. e. the second order term is identically zero.
Proof - First part. For the derivation of the next order term of the Poisson- Levy excursion
measure ν0+³[s, ∞) ´
about x = 0, we must include the second order term in the expression for
kernel exp − tH(ε)
(x, y) given in Proposition 3. Hence, from Propositions 2 and 3
ε
1
ν0+ [s, ∞)
¯
·
µ
¶
∂ ¯¯
A(x, y, t)
ε
exp −
∼ (2πε)
1
∂x ¯x=0
ε
y<0 ρ 2 (0)
0
#
µ Z t
¶ ¯ 2 ¯ 21
¯
¯
1
∂
A
¯ [1 + εK] dy,
× exp
|b′ (Xmin (s))| ds ¯¯
2 0
∂x∂y ¯
− 12
Z
ρ02 (y)
where K, recall, is a very complicated expression involving the Feynman-Green function.
1294
Therefore, up to order ε, we have assuming
ν0+ [s, ∞)
∂2A
∂x∂y
6= 0
¶
µ Z y
1
b(u)du
= (2πε)
dy exp
ε 0
0
( "µ
µ
¶ ¯ 2 ¯ 21 #
¶
¶
µ Z t
¯∂ A¯
1
A(0, y, t)
1
∂A
′
¯
×ε
exp −
|b (Xmin (s))| ds ¯¯
exp
−
ε
∂x x=0
ε
2 0
∂x∂y ¯x=0
¯
" ¯
¯ 1
¶
µ
¶
µ Z t
µ 2 ¶
2 A ¯− 2 ∂ ¯
¯
1
1
∂
∂ A
A(0, y, t)
¯
0
′
¯
¯
+ε
|b (Xmin (s))| ds
exp −
exp
¯
2 ¯ ∂x∂y ¯x=0 ∂x ¯
∂x∂y
ε
2 0
x=0
¯
¯ 2 ¯1
¶
¶
µ Z t
µ
Z t
¯ ∂ A ¯2
1
A(0, y, t)
1 ∂ ¯¯
′
′
¯
¯
|b
(X
(s))|
ds
b
(X
(s))
ds
exp
exp
−
−
¯
min
min
¯ ∂x∂y ¯
2 ∂x ¯
ε
2 0
0
x=0
x=0
#)
¯ 2 ¯1
¶
µ Z t
¶
¶µ
µ
¯ ∂ A ¯2
1
∂A
A(0,
y,
t)
¯
+ ¯¯
exp
K
|b′ (Xmin (s))| ds
−
exp −
∂x∂y ¯x=0
ε
2 0
∂x x=0
− 12
Z
∞
+ O(ε2 ).
(5.1)
We now use the result Olver [1974].
Proposition 7.
¶
µ
Z ∞
f (y)
g(y) dy
exp −
ε
0
µ
¶
Z ∞
f (y)
ε2
exp −
[g0 (y) + ε g1 (y) +
g2 (y) + · · · ] dy
=
ε
2!
0
Ã
!
f ′′′
g0′ 1
1 3 g1
1
3 3 g0′′
1
′
6
∼ Γ(1) ε
+ Γ( ) ε 2
− 3 f ′′ g0 ³ ´ 3 + Γ( ) ε 2
³ ′′ ´ 1 + · · · .
2 f ′′
2
4
2
2
′′
2
2
4
f
f
2
2
2
(5.2)
(5.3)
2
Proof. For a proof of this standard result on asymptotic approximations see Olver [1974] .
Comparing equation 5.2 with our expression for ν0+ [s, ∞) to second order in equation 5.1, we
get
− 12
g0 (y) = (2πε)
(¯
¯1
µ Z t
¶µ
¶)
¯ ∂ 2 A(x, y, t) ¯ 2
1
∂A(x,
y,
t)
¯
¯
|b′ (X(s))| ds
−
¯ ∂x∂y ¯ exp 2
∂x
0
and
,
x=0
¯ 2
¯1
¶
µ Z t
¯ ∂ A(0, y, t) ¯ 2
1
′
¯
¯
|b (X(s))|ds
g1 (y) = (2πε) ¯
exp
∂x∂y ¯
2 0
)
(
¯
¯−1
µ
¶
Z t
1 ¯¯ ∂ 2 A(x, y, t) ¯¯
∂A(x, y, t)
∂ ∂ 2 A(x, y, t)
1 ∂
′
|b (X(s))| ds + ¯
−K
×
2 ∂x 0
2
∂x∂y ¯ ∂x
∂x∂y
∂x
− 12
x=0
1295
.
Since we know for x = 0
¯
∂A ¯
∂x ¯x=0
giving
¯
¯
= −p0 (y), so − ∂A
∂x ¯
x=0
= p0 (y) (> 0), and
¯ 2 ¯
¯ ∂ A ¯ ∂p0 (y)
¯
¯
¯ ∂x∂y ¯ = ∂y
(> 0),
¯
∂2A ¯
∂x∂y ¯x=0
0 (y)
,
= − ∂p∂y
¯
¯
∂ ¯¯ ∂ 2 A ¯¯ ∂ 2 p0 (y)
=
,
∂x ¯ ∂x∂y ¯
∂x∂y
we can write the expressions for g0 (y) and g1 (y) as
)
(µ
¶
¶1
µ Z t
2
1
1
∂p
(y)
0
|b′ (X(s))| ds (p0 (y))
exp
g0 (y) = (2πε)− 2
∂y
2 0
,
x=0
and
¶1
µ Z t
¶
2
1
g1 (y) = (2πε)
exp
|b′ (X(s))|ds
2 0
)
(
µ
¶−1 µ 2
¶
Z t
∂p
(y)
∂
p
(y)
1
1 ∂
0
0
− Kp0 (y)
|b′ (X(s))| ds +
×
2 ∂x 0
2
∂y
∂x∂y
− 12
µ
∂p0 (y)
∂y
.
x=0
If we now expand each term in the expression for g0 (y) in a Taylor series we get
¯
à Z
(
¯
t
¡ ′
¢1
1
¯
− 12
′′
′
2
g0 (y) = (2πε)
exp
p0 (0) + yp0 (0) + · · ·
|b (X(s))| ds¯
¯
2 0
y=0
¯
!
)
Z t
¯
1 ∂
¯
+ · · · (p0 (0) + yp′0 (0) + · · · ) .
|b′ (X(s))| ds¯
+ y
¯
2 ∂y 0
y=0
Therefore, we can now see that in order to obtain the first order expressions for g0 (y) and g1 (y),
the following terms need to be evaluated :
µ
Z
∂p0 (y)
∂y
0
t
¶−1
|b′ (X(s))| ds
∂ 2 p0 (y)
∂y 2
Z t
∂
|b′ (X(s))| ds
∂x 0
∂ 2 p0 (y)
∂x∂y
Z t
∂
|b′ (X(s))| ds.
∂y 0
Let us be very thankful that an evaluation of K is not needed in this rather complicated computation.
Each of these terms is evaluated in the following Propositions. Recall that we have already seen
in equation 4.3
|b′ (0)|
∂p0 (y)
=
.
|p′0 (0)| =
∂y
sinh |b′ (0)|t
1296
Proposition 8. For p0 (0) = b(0) = 0 , as y → 0,
∂ 2 p0 (y)
b′′ (0) (cosh |b′ (0)|t − 1)2
.
=
∂y 2
sinh3 |b′ (0)|t
Proof. In order to calculate
∂ 2 p0 (y)
∂y 2
(5.4)
we return to the identity,
t=
Z
0
y
du
¡
¢1 .
p20 (y) + b2 (u) 2
Differentiating the equation above w.r.t. y, and then using the change of variable u = yv, gives
dropping some inessential modulus signs for ease of presentation
Z
¡ 2
¢1 y
1
du
2
2
(5.5)
= p0 (y) p0 (y) + b (y)
¡
¢3
′
p0 (y)
0
p20 (y) + b2 (u) 2
õ
µ
¶
¶ !1 Z
p0 (y)
dv
p0 (y) 2
b(y) 2 2 1
=
+
µ
³
³
´
´2 ¶ 23
y
y
y
2
0
p0 (y)
b(yv)
+ yv
v2
y
Z
¡ ′
¢1 1
dv
′
2
′
2 2
→ p0 (0) p0 (0) + b (0)
as y → 0,
3
0 (p′ (0)2 + b′ (0)2 v 2 ) 2
0
using the same argument as seen in Proposition 4
Expanding the r.h.s. of equation 5.5 in a Taylor Series, using for simplicity the notation p = p0 (0)
and b = b(0), gives
´ µ³
´2 ³
´2 ¶ 21
y ′′
y ′′
y ′′
′
′
p + p + ··· + b + b + ···
p + p + ···
2
2
2
Z 1
dv
×
³¡
¢
¡
¢2 ´ 23
2
0
y ′′
yv ′′
′
′
p + 2 p + · · · + b + 2 b + · · · v2
¶1
´³ 2
´1 µ
³
2
p′ p′′ + b′ b′′
y ′′
′
′2 2
′
1 + y ′2
p +b
+
·
·
·
→ p + p + ···
2
2
p + b′
¶− 32
µ
Z 1
³ 2
´− 3
p′ p′′ + b′ b′′ v 3
2
′
′2 2
+ ···
dv p + b v
1 + y ′2
×
p + b′2 v 2
0




³
´ ³ 2
´1 y
′ p′′ + b′ b′′
y
p
2
2


= p′ + p′′ + · · ·  p′ + b′
+ ¡
¢1 + · · ·
2
2
p′2 + b′2 2




Z 1
3
³ 2
´
′
′′
′
′′
3
−2
3
pp +bb v 
2

− y ¡
dv  p′ + b′ v 2
×
¢5 + · · ·
2
2 2 2
2
′
′
0
p +b v
³
′
= zero order term + f.y + higher order terms(y 2 · · · ).
1297
The zero order term is
2
1
2
p′ (p′ + b′ ) 2
Z
1
0
³ 2
´− 3
2
2
.
dv p′ + b′ v 2
Now the integral term in the equation above can be written as
Z 1
dv
1
´3 .
b′3 0 ³ p′2
2
2
+v
′2
b
¯ ′¯
¯ ¯
Using the change of variable v = ¯ pb′ ¯ sinh w in the equation above gives
1
b′3
˛ ′˛
˛ ˛
sinh−1 ˛ pb ′ ˛
Z
0
1
= ′ ′2
b p
Z
dw ³
˛ ′˛
˛ ˛
sinh−1 ˛ pb ′ ˛
2
¯ ′¯
¯p ¯
¯ b′ ¯ cosh w
2
p′
2
b′
´3
p′
2
b′
+
dw
1
1
1
= ′3·
2
|p | cosh |b′ (0)|t.
cosh w
0
sinh2 w
2
Therefore, the zero order term is (because of equation 4.3)
2
2
1
p′ (p′ + b′ ) 2 ·
1
1
1
=
.
3 ·
′
′|
′
cosh
|b
(0)|
t
|p
p
The coefficient of y is given by
³ 2
³ 2
´− 3 p′ p′ p′′ + b′ b′′ Z 1
´− 3
´1 Z 1
p′′ ³ ′2
2
2
′
′2 2
′
′2 2
′2 2
f=
dv p + b v
+ ¡
dv
p
+
b
v
p +b
1
¢
2
2
2
2 p ′ + b′ 2 0
0
Z
´1 1
p′ p′′ + b′ b′′ v 3
3 ³ 2
2 2
dv ¡
− p ′ p ′ + b′
¢5 .
2
0
p′2 + b′2 v 2 2
Therefore, we have that
p′
1
∂p
=
= ′−1
= p′ (1 − yp′ f − · · · ).
∂y
1 + yp′ f + · · ·
p + yf + · · ·
′
Now, by letting |p′ | = − |ba | giving a = − sinh |b′ (0)|t, we can write
3
2
p′ f
=
−
+
1
p′′ p′ p′ p′′ + b′ b′′ ′−3
+
(1 + a2 )− 2
1 p
2
2 p′ (1 + a2 ) 2
!
(
Ã
′ p′′
1
1
p
3 ′3 ′
2
p p (1 + a2 ) 2
+
2
3(1 + a2 ) 3(1 + a2 ) 32
p′5
!)
Ã
b′ b′′
2 + 3a2
2
,
−
p′5 3a2 3a4 (1 + a2 ) 32
1298
(5.6)
which simplifies to
′2
p f
=
−
¶
µ
p′′ 1 p′′ − ab′′
3 ′′ 2
1
+
− p
+
2
2 (1 + a2 )
2
3 3(1 + a2 )
!
Ã
1
3
2 + 3a2
2(1 + a2 ) 2
′′
(−a) b
− 4
.
2
3a4
3a (1 + a2 )
Hence, from equation 5.6
p′′
3ab′′
p′′ 1 p′′ − ab′′
′′
+
−
p
−
+
−p =
2
2 (1 + a2 )
2(1 + a2 )
2
′′
giving
Ã
1
2 + 3a2
2(1 + a2 ) 2
−
3a4
3a4 (1 + a2 )
1
1 + 23 a2 ′′
p′′
ab′′
b′′ (1 + a2 ) 2
0=
−
+
−
b .
2
2(1 + a2 )
a3
a3 (1 + a2 )
′
|b |
Now since a = − |p
′|
and
|b′ (0)|
sinh |b′ (0)|t ,
|p′ | =
s = sinh |b′ (0)| t
and again using the obvious notation
and c = cosh |b′ (0)| t,
we can write the equation above as
µ
¶
s
c
2 + 3s2
0 = p′′ + b′′ − 2 + 2 3 − 3 2
c
s
s c
µ 4
¶
′′
b
s
2 + 3s2
′′
= p + 3 − 2 + 2c −
s
c
c2
µ
¶
b′′
(c4 − 2c2 + 1) 2c3 2 + 3(c2 − 1)
′′
= p + 3 −
+ 2 −
s
c2
c
c2
µ
¶
b′′ −c4 + 2c3 − c2
= p′′ + 3
.
s
c2
Hence, we get the result
p′′0 (0)
b′′ (0) (cosh |b′ (0)|t − 1)2
.
=
sinh3 |b′ (0)|t
Proposition 9. As y → 0,
∂ 2 p0
b′′ (0) (cosh |b′ (0)|t − 1)2
= p′′0 (0).
→
∂x∂y
sinh3 |b′ (0)|t
Proof. We begin with
t=
Z
x
y
du
1
(p2 (x, y) + b2 (u) − b2 (x)) 2
.
Differentiating both sides w.r.t. x gives,
Z y
∂p
p ∂x
− b(x) ∂b(x)
1
∂x
du
−
0 = −
3 .
|p(x, y)|
2
2
x
(p (x, y) + b (u) − b2 (x)) 2
1299
!
,
Therefore, as x → 0
¯
1
∂p ¯¯
0 =
+ p0 (y) ¯
|p0 (y)|
∂x ¯
Hence,
x=0
y
Z
du
¡
0
¢3 .
p20 (y) + b2 (u) 2
¯
∂p ¯¯
−1
=
¡
¢−3/2 .
R
¯
y
∂x x=0
du
p20 (y) 0 p20 (y) + b2 (u)
(5.7)
If we consider the quotient term on the r.h.s. of equation 5.7, with a change of variable u = yv
and again letting y → 0, we get
r.h.s. =
−1
¢−3/2 .
R y ¡ ′2
dv
p0 (0) 0 p0 (0) + b′2 (0)v 2
′2
Expanding the denominator of equation 5.7 in a Taylor series [p(0)
p′ (0) 6= 0], using the same notation as in the previous Proposition
y
(p + p′′ + · · · )2
2
′
Z
1
dv
0
¡
(p′ +
Z 1
y
2
p′′ + · · · )2 + (b′ +
yv
2
b′′ + · · · )2 v 2
¢3
2
dv
¡ 2
¢3
0
p′ + b′2 v 2 + y(p′ p′′ + v 3 b′ b′′ ) + · · · 2
¾
½
Z 1
1
3 p′ p′′ + v 3 b′ b′′
′2
′ ′′
= (p + y p p + · · · )
¡ 2
¢ 3 1 − 2 y (p′ + b′2 v 2 ) − · · · dv.
0
p′ + b′2 v 2 2
2
= (p′ + y p′ p′′ + · · · )
Therefore,
µ
∂p
∂x
¶−1
x=0
= zero order term + f y + · · · ,
where f is the coefficient of the y term as shown below
µ
∂p
∂x
¶−1
= p
x=0
−
′2
Z
1
′2
′2
(p + b v 2 )
0
3 ′2
p
2
dv
Z
1
0
3
2
Ã
′ ′′
+y p p
p′ p′′ + b′ b′′ v 3
5
(p′2 + b′2 v 2 ) 2
dv
!
Z
0
1
dv
′2
3
(p + b′2 v 2 ) 2
+ ···
Inverting this equation gives
!−1 (
Ã
à Z
¯
1
dv
∂p ¯¯
′2
1−y
= p
3
2
2
∂x ¯x=0
0 (p′ + b′ v 2 ) 2
R1
R
dv
3 ′2 1 p′ p′′ +b′ b′′ v 3
!
)
p′ p′′ 0
3 − 2p
0 (p′2 +b′2 v 2 ) 25 dv
(p′2 +b′2 v2 ) 2
− ··· .
R1
p′2 0 2 dv2 3
(p′ +b′ v 2 ) 2
1300
=
0,
Now since
¯
¯
¯
∂p(x, y) ¯¯
∂p(x, 0) ¯¯
∂ 2 p(x, 0) ¯¯
=
+y
+ ··· .
∂x ¯x=0
∂x ¯x=0
∂x∂y ¯x=0
Comparing y-terms yields
p′ p′′
¯
2
¯
∂ p(x, 0) ¯
=
∂x∂y ¯x=0
¯ ′¯
¯ ¯
Letting a = − ¯ pb′ ¯ ,
¯
∂ 2 p ¯¯
=
∂x∂y ¯x=0
p′′
2
p′
R1
R1
0
dv
0 (1+a2 v 2 ) 23
µ
1
p′
2
dv
− 32 p′
3
2
2
p′ +b′ v 2 2
)
(
µ
R1
p′2 0
−
3 1
2 p′3
R1
R1
p′ p′′ +b′ b′′ v 3
0 (p′2 +b′2 v 2 ) 25
R1
dv
2
3
2
(p′ +b′ v 2 ) 2
p′ p′′ +b′ b′′ v 3
5
0
dv
0 (1+a2 v 2 ) 32
(1+a2 v 2 ) 2
¶2
¶2
dv
.
dv
1
1
1
1 b′ b′′ (1 + a2 ) 2
1 b′ b′′
1 p′′
−
+
.
=−
2
2
′
2
2
2
′
′
2 p (1 + a )
2 p
a
2 p a (1 + a2 )
Now, substituting for p′ and p′′ gives the result.
Remark. A by product of the above is
∂p0 (y)
→ −p′0 cosh |b′ (0)| t.
∂x
Proposition 10. As y → 0,
Z
0
t
|b′ (Xmin (s))| ds → t|b′ (0)|
with Xmin satisfying
Ẍmin (s) = b(Xmin (s))b′ (Xmin (s))
and
Xmin (0) = x, Xmin (t) = y.
Proof. For u = Xmin (s), du = Ẋmin (s) ds, and considering
Z t
¯t Z t
¯
s b′′ (Xmin (s)) Ẋmin (s) ds
|b′ (Xmin (s))| ds = s |b′ (Xmin (s))|¯ +
0
0
0
Z y
′′
′
s(u) b (u) du
= t |b (y)| +
x
Z y Z u
b′′ (u)
′
dv
du
= t |b (y)| +
1
x
x
(p2 (x, y) + b2 (v) − b2 (x)) 2
since,
t(u) =
Z
x
y
du
1
(p2 (x, y) + b2 (u) − b2 (x)) 2
(5.8)
.
Therefore, the integral term on the r.h.s. of equation 5.8 → 0 as y → 0, and so the result
follows (recall y > x > 0).
1301
Proposition 11.
¯
¯
Z t
¯
∂ ¯¯
∂p
¯
|b′ (Xmin (s))| ds → − b′′ (0) p′0 (0)
¯
¯
¯
∂x ¯
∂x
0
x=0
=
x=0
Z
1
du
1
u
0
b′′ (0) cosh |b′ (0)|t − 1
,
|b′ (0)| sinh |b′ (0)|t
Z
as
dv
′2
3
(p0 (0) + b′2 (0)v 2 ) 2
y → 0.
(5.9)
Proof. Using equations 5.7 and 5.8, a calculation along the lines of the proof of Theorem 4, using
integration by parts, yields


¯
¯
Z y
Z t
Z u


¯
¯
∂p ¯
dv
∂ ¯
′
′′
du p0 (y)
|b (Xmin (s))| ds =
¯
¯
¡ 2
¢ 3  b (u)

¯
∂x ¯
∂x
0
0
0
p0 (y) + b2 (v) 2
x=0
x=0
¯
Z 1
Z 1
dv
∂p ¯¯
′′
′
du
→ −b (0) p0 (0)
¯
¢ 3 as y → 0
¡ ′2
∂x ¯
u
0
p0 (0) + b′2 (0) v 2 2
y=x=0
and result follows.
Proposition 12.
Z u
Z 1
Z t
dv
∂
′
′′
′
2
du
|b (X(s))| ds → b (0)(p0 (0))
3
2
2
′
∂y 0
0 (p0 (0) + b′ (0)v 2 ) 2
0
b′′ (0) cosh |b′ (0)|t − 1
=
, as y → 0.
|b′ (0)| sinh |b′ (0)|t
(5.10)
Proof. The proof of this is similar to that of Proposition 11.
Proof - Second part. Therefore, by substituting equations 4.3, 5.4, 5.7, 5.9, 5.10 and Proposition
9 into the expressions obtained for g0 (y) and g1 (y), and using Proposition 7, we can complete
the proof of Theorem 3.
µ
¡ ′
¢1
1 ′
− 12
′′
2
g0 (y) ∼ (2πε)
|b (X(0))|t
exp
p + p y + ···
2
¶³
´
1 b′′ (0) cosh |b′ (0)|t − 1
y 2 ′′
′
+ y ′
+
·
·
·
p
(0)
+
·
·
·
, as y → 0
p(0)
+
yp
(0)
+
2 |b (0)| sinh |b′ (0)|t
2
¶µ
¶1
µ ′
2
p′′
t|b (0)|
− 12 ′ 32
1 + ′ y + ···
= (2πε) (p0 ) exp
2
p
¶
µ
′
2
′′
′′
b (0) cosh |b (0)|t − 1 2
b′′ (0)p′′ cosh |b′ (0)|t − 1 3
p y
−
y −
y + ···
× y+ ′
p 2
2|b′ (0)| sinh |b′ (0)|t
4|b′ (0)|p′ sinh |b′ (0)|t
µ ′′
¶·
¶
µ ′
p
b′′ (0) cosh |b′ (0)|t − 1
1 2
t|b (0)|
− 12 ′ 23
= (2πε) (p ) exp
y+ y 2 ′ − ′
+ ···
2
2
p
|b (0)| sinh |b′ (0)|t
Similarly,
¶
t|b′ (0)|
g1 (y) ∼ (2πε)
exp
2
µ
¶
′′
′
1 b (0) cosh |b (0)|t − 1 1 b′′ (0) (cosh |b′ (0)|t − 1)2
×
−
+
+ y(· · · ) + · · · .
2 |b′ (0)| sinh |b′ (0)|t
2 |b′ (0)| (sinh |b′ (0)|t)2
− 12
1
(p′0 ) 2
µ
1302
Therefore,
g0′ (0)
− 12
= (2πε)
and
g0′′ (0)
− 12
= (2πε)
′
3
2
(p ) exp
µ
′
3
2
(p ) exp
t|b′ (0)|
2
µ
t|b′ (0)|
2
¶
,
¶ µ ′′
¶
b′′ (0) cosh |b′ (0)|t − 1
p
2 ′ − ′
.
p
|b (0)| sinh |b′ (0)|t
Now, in our case, the expression for f (y) in Proposition 7 is
Z y
b(u) du,
f (y) = A(0, y, t) −
0
so
f (0) = A(0, 0, t) −
Also, since p(t) =
f ′ (y) =
p
p20 (y) + b2 (y),
∂
∂A(0, y, t)
−
∂y
∂y
so
Z
Z
0
b(u) du = 0.
0
y
0
b(u)du = p0 (t) − b(y) =
q
p20 (y) + b2 (y) − b(y),
q
f (0) = p20 (0) + b2 (0) − b(0) = 0,
′
since p0 (0) = 0, (the momentum required to go from x = 0 to y = 0).
p
∂
p(0, y, t) − b′ (y), so differentiating p(t) = p20 (y) + b2 (y) twice with respect to
Now, f ′′ (y) = ∂y
y, and using the fact that b(0) = 0 and p(t) → 0 as y → 0, we get
(µ
)¯
¶
∂p(t) 2
∂ 2 p(t) ¯¯
+ p(t)
= p′0 (y)2 + p0 (y) p′′0 (y) + b′ (y)2 + b(y) b′′ (y).
(5.11)
¯
¯
∂y
∂y 2
x=0
As y → 0,
µ
∂p0 (t)
∂y
¶2
→ p′0 (0)2 + b′ (0)2 ,
2
p0 (t)
→ 0, as y → 0. Hence,
since p0 (0) = b(0) = 0, and p0 (t) ∂ ∂y
2
µ
∂p0 (t)
∂y
¶2 ¯¯
¯
¯
¯
= p′0 (0)2 + b′ (0)2
x=0
(b′ (0))2
+ b′ (0)2
(sinh |b′ (0)|t)2
¶
µ
2 ′
′
2 1 + sinh |b (0)|t
= (b (0))
sinh2 |b′ (0)|t
= |b′ (0)|2 coth2 |b′ (0)|t.
=
Therefore,
f ′′ (y) → b′ (0) coth |b′ (0)|t − b′ (0) = |b′ (0)| (coth |b′ (0)|t + 1),
1303
since we are dealing with b(x) < 0.
Similarly,
f ′′′ (y) =
∂2p
(0, y, t) − b′′ (y).
∂y 2
Differentiating equation 5.11 again w.r.t. y gives,
½
∂ 3 p(t)
∂p(t) ∂ 2 p(t)
p(t)
+
3
∂y 3
∂y
∂y 2
¾ ¯¯
¯
¯
¯
=
3 p′0 (y) p′′0 (y) + p0 (y) p′′′ (y)
x=0
+ 3 b′ (y) b′′′ (y) + b(y) b′′′ (y).
And again, since p0 (0) = b(0) = 0, and p0 (t) → 0 as y → 0, we get
½
¾¯
∂p0 (t) ∂ 2 p0 (t) ¯¯
= p′0 (0) p′′0 (0) + b′ (0) b′′ (0),
¯
¯
∂y
∂y 2
y=0
giving, after a little calculation
½
Therefore, as y → 0,
f ′′′ (y) →
=
∂ 2 p0 (t)
∂y 2
¾ ¯¯
¯
¯
¯
=
y=0
p′0 (0) p′′0 (0) + |b′ (0)| b′′ (0)
.
|b′ (0)| coth |b′ (0)| t
p′0 p′′0 + |b′ (0)|b′′ (0)
− b′′ (0)
|b′ (0)| coth |b′ (0)|t
¶
µ
sinh |b′ (0)|t
(cosh |b′ (0)|t − 1)2
′′
−
−1 .
b (0)
cosh |b′ (0)|t(sinh |b′ (0)|t)3 cosh |b′ (0)|t
Finally, we conclude the proof of Theorem 3 by substituting into equation 5.3 to get
µ Z y
¶
Z ∞
1
+
dy exp
ν0 [t, ∞) =
b(u) du − A(0, y, t) (g0 (y) + εg1 (y) + · · · ) dy
ε 0
0
µ ′
¶
t |b (0)| 1
ε
′ 32
− 12
∼
(2πε) (p ) exp
f ′′
2
2
2
(
¶ µ ′′
¶
µ ′
1
r
−
3
2
2p0
b′′ (0) cosh |b′ (0)|t − 1
π 3 (2πε)
t |b (0)|
′ 2
2
− ′
ε
(p )0 exp
+
2f ′′
4
2
p′0
|b (0)| sinh |b′ (0)|t
)
¶
µ ′
f ′′′
3
3 6
t |b (0)|
1
′ 2
− 12
(p )0 exp
−
′′ (2πε)
′′
f
f
4
2
2
2
¶
µ
r
1
1
π 3
t |b′ (0)|
−
′
2
ε 2 (2πε) 2 (p )0 exp −
+
2f ′′
2
¶
µ
′′
′
′′
1 b (0) cosh |b (0)|t − 1 1 b (0) (cosh |b′ (0)|t − 1)2
.
+
×
−
2 |b′ (0)| sinh |b′ (0)|t
2 |b′ (0)| (sinh |b′ (0)|t)2
1304
Substituting for p′ and p′′ eventually gives, using an obvious shorthand notation
µ ′
¶
1
3
t|b (0)|
1
ν0+ [t, ∞) ∼ ε(2πε)− 2 (p′ ) 2 exp
2
f ′′
µ ′
¶µ
¶ 1 (· ′′
2
3
1
3
t|b
(0)|
π
b (0)(c − 1)2 b′′ (0)(c − 1)
−
+ ε 2 (2πε)− 2 (p′ ) 2 exp
2
2f ′′
2p′ s3
4|b′ (0)|s
)
µ
¶¸
b′′ (c − 1)2 − s4 − cs3
b′′ (c − 1) b′′ (c − 1)2
2s
−
−
+
4
b′ cs2 (c + s)
b′ (c + s)
2p′ b′ s
2p′ b′ s2
¶
¶
µ ′
µ ′
3
t|b (0)| 1
t|b (0)|
− 12
′ 32
− 12
′ 12
2
+ ε (2πε) (p ) exp
= ε (2πε) (p ) exp
2
f ′′
2
"
µ
¶
¶
µ
r
1
b′′ (0)
c − 1 (c − 1)2
c − 1 2(c − 1)2
π
−
+
−
×
− ′
2f ′′
2|b (0)|
s
s2
1 + sc
s2
s3
#
¶
µ
1
(c − 1)2 s
+
− −1
¢2
¡
cs3
c
s 1 + sc
µ ′
¶
1
3
t|b (0)| 1
= ε (2πε)− 2 (p′ ) 2 exp
2
f ′′
µ ′
µ
¶
¶·
¸
t|b (0)|
b′′ (0)
ε ′ 1
1
− ′
(p ) 2 exp
+
······ .
1
2
2
2|b (0)|
(f ′′ ) 2
A tedious calculation shows that the terms within the
ν0+ [t, ∞)
·
···
¸
cancel leaving the result,
¶
3
t|b′ (0)| 1
exp
+ O(ε 2 ),
= ε (2πε)
′′
2
f
µ ′
¶1 ³
1
´
−2
3
ε|b (0)| 2 2|b′ (0)|t
=
+ O(ε 2 ).
e
−1
π
− 12
3
(p′0 ) 2
µ
Acknowledgements
The author wishes to thank Aubrey Truman for his crucial input into an earlier joint collaboration concentrating on the leading order behaviour of the Poisson - Lévy excursion measure
Davies and Truman [1998]. I am , as always, indebted to David Williams for his help and
never-ending enthusiasm for probability.
References
P. Lévy, Processus Stochastiques et Mouvement Brownien, (Gauthier Villars, Paris, 1954)
MR0190953
1305
J. Hawkes and A. Truman, Statistics of Local Time and Excursions for the Ornstein-Uhlenbeck
Process, Lecture Notes of the London Mathematical Society 167, 91 (1991) MR1166408
A. Truman and D. Williams, Excursions and Itô Calculus in Nelson’s Stochastic Mechanics,
“Recent Developments in Quantum Mechanics”, (Kluwer Academic Publishers, Netherlands,
1991)
I. M. Davies and A. Truman, Laplace asymptotic expansions of conditional Wiener Integrals
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I. M. Davies and A. Truman, Semiclassical analysis and a new result in in excursion theory,
in “Probability Theory and Mathematical Statistics, Proceedings of the Seventh Vilnius Conference (1998)”, edited by B. Grigelionis et al, (Utrecht: VSP, Vilnius: TEV, 1999), 701 –
706
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