Electronic Journal of Differential Equations, Vol. 2025 (2025), No. 134, pp. 1–62.

ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu

DOI: 10.58997/ejde.2025.14

PROPERTIES OF THE DIRICHLET GREEN’S FUNCTION FOR

LINEAR DIFFUSIONS ON A HALF LINE

JOSEPH G. CONLON, MICHAEL DABKOWSKI

Communicated by Nestor D. Guillen

Abstract. This article concerns the study of Green’s functions for one di-
mensional diffusions with constant diffusion coefficient and linear time inho-

mogeneous drift. It is well know that the whole line Green’s function is given

by a Gaussian. Formulas for the Dirichlet Green’s function on the half line
are only known in special cases. The main object of study in the paper is

the ratio of the Dirichlet to whole line Green’s functions. Bounds, asymptotic

behavior in the limit as the diffusion coefficient vanishes, and a log concavity
result are obtained for this ratio. These results have been used in the proof of

asymptotic behavior for a simple model of Ostwald ripening.

Contents

1. Introduction 1
2. Representation and convergence of the function qε 5
3. Regularity and bounds on the function qε 15
4. Estimating solutions of the Hamilton-Jacobi PDE 22
5. Uniform bounds on qε and its derivatives 36
6. Convergence of the function ∂qε(x, y, T )/∂x as ε → 0 49
References 60

1. Introduction

In this article we prove some results for diffusions on the line with affine time
dependent drift, which are used in [5] to study the large time behavior of solutions to
a nonlinear nonlocal diffusion problem occurring in the theory of Ostwald ripening
[14]. This theory describes the time evolution of crystals in a solute, whereby smaller
crystals dissolve and then depos it onto larger crystals. An important quantity is
the coarsening rate, which is the rate of increase of the average crystal volume with
time. Mean field models of Ostwald ripening were developed by Becker-Döring
[3], Lifshitz-Slyozov [13] and Wagner [19]. The LSW model introduced in [13],

2020 Mathematics Subject Classification. 35F21, 35K20, 49N10.

Key words and phrases. Nonlinear PDE; coarsening.
©2025. This work is licensed under a CC BY 4.0 license.
Submitted December 13, 2023. Published February 19, 2025.

1



2 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

and independently in [19], consists of a first order linear transport PDE on the
half line with a linear constraint corresponding to conservation of volume, making
the model nonlinear and nonlocal. The Becker-Döring (BD) model is a linear
transport equation on the positive integers with a linear conservation of volume
constraint. The transport equation on the positive integers can be interpreted as
the discretization of a linear diffusion equation [18]. Furthermore, solutions of the
BD model at large time are expected to be approximate solutions to the LSW model
[15].

The LSWmodel has a family of self-similar solutions, which may be parametrized
by a real number β with 0 < β ≤ 1. Already in [13, 19] it was conjectured that the
only physically relevant self-similar solution is the β = 1 solution. Therefore the
large time asymptotic coarsening rate in Ostwald ripening may be obtained from
the β = 1 solution. The Carr-Penrose (CP) model introduced in [4] is a simplified
version of the LSW model in which the transportation vector field is affine. It also
has a family of self-similar solutions parametrized by β with 0 < β ≤ 1. It is
shown in [4] that for each β with 0 < β ≤ 1 there exists a large class of initial data
with compact support for the CP model which asymptotically converges to the self-
similar solution with parameter β. The main result of [5] is that all solutions to a
diffusive CP model with initial data of compact support asymptotically converge to
the β = 1 self-similar solution of the CP model. Thus diffusion acts as a selection
principle on the one parameter family of self-similar solutions of the CP model.
The selection principle was initially established for a semi-classical approximation
to the diffusive CP model [6].

The key difficulty in going from proving the selection principle for the semi-
classical diffusive CP model to proving it for the diffusive model is controlling the
ratio of the Dirichlet Green’s function for the CP diffusion equation on the half line
to the Green’s function on the whole line, which is Gaussian. This is the subject of
the present paper. The results most relevant for the diffusive CP problem are the
log concavity property (1.17) of the ratio of Green’s functions and the convergence
(1.19) of the logarithmic derivative of the ratio as the diffusion constant goes to
zero.

To specify the diffusion equation we are interested in, let b : R × R → R be a
continuous function (y, t) → b(y, t), which is linear in the space variable y. For
ε > 0, the terminal value problem

∂uε(y, t)

∂t
+ b(y, t)

∂uε(y, t)

∂y
+

ε

2

∂2uε(y, t)

∂y2
= 0, y ∈ R, t < T, (1.1)

uε(y, T ) = uT (y), y ∈ R , (1.2)

has a unique solution uε which has the representation

uε(y, t) =

∫ ∞

−∞
Gε(x, y, t, T )uT (x) dx, y ∈ R, t < T, (1.3)

where Gε is the Green’s function for the problem. The adjoint problem to (1.1),
(1.2) is the initial value problem

∂vε(x, t)

∂t
+

∂

∂x
[b(x, t)vε(x, t)] =

ε

2

∂2vε(x, t)

∂x2
, x ∈ R, t > 0, (1.4)

vε(x, 0) = v0(x), y ∈ R. (1.5)



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 3

The solution to (1.4), (1.5) is given by the formula

vε(x, T ) =

∫ ∞

−∞
Gε(x, y, 0, T )v0(y) dy, x ∈ R, T > 0. (1.6)

Since the drift b(·, ·) is linear and Gε is Gaussian, (x, y) → logGε(x, y, t, T ) is a
quadratic function in (x, y). Here we shall obtain properties of the corresponding
Dirichlet Green’s function (x, y) → Gε,D(x, y, t, T ) on the half line x, y > 0. Thus

uε,D(y, t) =

∫ ∞

0

Gε,D(x, y, t, T )uT (x) dx, y > 0, t < T, (1.7)

is the solution to (1.1), (1.2) in the domain {(y, t) : y > 0, t < T} with Dirichlet
boundary condition uε,D(0, t) = 0, t < T .

The drifts b(·, ·) we consider are of the form

b(y, t) = A(t)y − 1, where A : R → R is a continuous function, (1.8)

but the methods of the paper may be extended to more general linear drifts. In the
case A(·) ≡ 0 there are simple explicit formulas for Gε and Gε,D. These are given
by

Gε(x, y, t, T ) =
1√

2πε(T − t)
exp

[
− (x+ T − t− y)2

2ε(T − t)

]
, (1.9)

Gε,D(x, y, t, T ) =
{
1− exp

[
− 2xy

ε(T − t)

]}
Gε(x, y, t, T ) . (1.10)

For non-trivial A(·), we write

Gε,D(x, y, t, T ) =
{
1− exp

[
− qε(x, y, t, T )

ε

]}
Gε(x, y, t, T ) , (1.11)

and study the properties of the function qε. When A(·) ≡ 0 we have from (1.10)
that the function (x, y) → qε(x, y, t, T ) is independent of ε and bilinear. One can
also obtain explicit formulas for qε in some other cases of linear drift, in particular
for the Ornstein-Uhlenbeck process where b(y, t) = −γy with constant γ (see [17,
Prop. 20] and Remark 3.4 of the present paper). However there appears not to be
an explicit formula for qε in the case of the general function A(·).

We are able to obtain linear bounds on the function x → qε(x, y, t, T ), x > 0,
and its first two x derivatives, which are uniform in ε > 0, when the function A(·)
is assumed to be non-negative:

Theorem 1.1. Assume the function A(·) of (1.8) is continuous and non-negative,
and qε is defined by (1.11). Then there exists a continuous positive function α1(·)
and continuous non-negative functions β1(·), β2(·), with domain {(t, T ) : t, T ∈
R, t < T}, which bound qε as follows:

α1(t, T )xy ≤ qε(x, y, t, T ) ≤ [α1(t, T )x+ β1(t, T )]y , x, y > 0, t < T, (1.12)

lim
x→∞

{[α1(t, T )x+ β1(t, T )]y − qε(x, y, t, T )} = 0 , y > 0, t < T, (1.13)

qε(x, y, t, T ) ≤ [α1(t, T )y + β2(t, T )]x , x, y > 0, t < T, (1.14)

α1(t, T )y ≤ ∂qε(x, y, t, T )

∂x
≤ α1(t, T )y + β2(t, T ) , x, y > 0, t < T, (1.15)

lim
x→∞

{∂qε(x, y, t, T )

∂x
− α1(t, T )y

}
= 0 , y > 0, t < T, (1.16)

The functionx → qε(x, y, t, T ), x, y > 0 t < T , is concave. (1.17)



4 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

Since qε(0, y, t, T ) = 0 the lower bound in (1.12) is implied by the lower bound
in (1.15). Similarly the upper bound in (1.15) implies the upper bound in (1.14).
However our proof of (1.15) in Proposition 5.1 uses the inequalities (1.12), (1.14),
which have previously been established in Proposition 3.3. The proof of (1.13) is
given in Proposition 5.2, the proof of (1.16) in Proposition 5.4, and the proof of
(1.17) in Theorem 5.7.

Theorem 1.1 tells us that the graph of the function x → qε(x, y, t, T ), x > 0, lies
between two parallel lines and is asymptotic to the upper line at large x. The graph
also lies in a wedge formed by two lines through the origin and is concave. This
geometric picture gives us a rather precise understanding of the global behavior
of the function x → qε(x, y, 0, T ), x > 0. The significance of the upper bounds
(1.14), (1.15) can be understood by considering the situation when ε → 0. The
function [x, T ] → qε(x, y, 0, T ), x, T > 0, is a solution to the Hamilton-Jacobi-
Bellman equation (2.22). One expects then that limε→0 qε(x, y, t, T ) = q0(x, y, t, T )
exists, and in the case t = 0 is a solution to the Hamilton-Jacobi equation (2.29).
In §2 we show that the limit does exist and q0(x, y, 0, T ) is given by the variational
formula (2.28) corresponding to the Hamilton-Jacobi equation. In §4 we study
this variational problem in great detail, establishing in particular that if A(·) is
continuous and non-negative then the function x → q0(x, y, t, T ) is differentiable
in a neighborhood of x = 0 and ∂q0(0, y, t, T )/∂x = α1(t, T )y + β2(t, T ). More
precisely we have the following theorem.

Theorem 1.2. Assume the function A(·) of (1.8) is continuous and and qε is
defined by (1.11). Then there is a continuous function [x, y, t, T ] → q0(x, y, t, T )
with domain {[x, y, t, T ] : x, y ≥ 0, t, T ∈ R, t < T} such that limε→0 qε(x, y, t, T ) =
q0(x, y, t, T ) for all x, y ≥ 0, t < T . If A(·) is non-negative then the function
x → q0(x, y, t, T ) is differentiable at x = 0 and

∂q0(x, y, t, T )

∂x

∣∣∣
x=0

= α1(t, T )y + β2(t, T ) . (1.18)

Furthermore, for any t ∈ R, T0 > 0, there are constants C1, C2 > 0, depending only
on T0 and supt≤s≤t+T0

A(s), such that∣∣∣∂qε(0, y, t, T )
∂x

− [α1(t, T )y + β2(t, T )]
∣∣∣ ≤ εC2(T − t)2

y2
(1.19)

for y ≥ C1(T − t)2, T − t ≤ T0, provided ε ≤ (T − t)3.

In Theorem 2.6 we prove that limε→0 qε(x, y, t, T ) = q0(x, y, t, T ) exists and is
the solution to the variational problem (2.28). In Proposition 4.2 we show if A(·)
is non-negative that q0(x, y, 0, T ) may be obtained by the method of characteristics
in a subdomain of {[x, T ] : x > 0, T > 0}, which includes a neighborhood of the
boundary {[0, T ] : T > 0}. The function [x, T ] → q0(x, y, 0, T ) is then a classical
solution of the Hamilton-Jacobi equation (2.29) in this region, and the derivative
∂q0(x, y, 0, T )/∂x is given by the formula (4.26).

We use the methods of stochastic control theory to prove that limε→0 qε = q0
and (1.19). The Bellman equation (2.22) corresponds to the stochastic variational
problem (2.26). The proofs of limε→0 qε = q0 and (1.19) are then obtained by
comparing the solution of this stochastic variational problem to the solution of the
classical variational problem (2.28). The proof of (1.19) is given in Proposition



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 5

6.2, and uses in a crucial way the regularity of the function x → q0(x, y, 0, T ) in a
neighborhood of x = 0.

The ratio of Green’s functions Gε,D(x, y, t, T )/Gε(x, y, t, T ) given in (1.11) is the
probability that a generalized Brownian bridge, beginning at y at time t and ending
at x at time T , lies entirely in the positive half line. We may therefore try to estimate
qε(x, y, t, T ) by comparing this generalized bridge to the standard Brownian bridge.
This method of bridge comparison is used in the proof of Proposition 3.2. However
for the most part we use the fact that the bridge process is a Gaussian Markov
process with the linear drift (2.15) in order to estimate qε(x, y, t, T ). There is a
considerable literature on the study of bridges. In [8, 9] Conforti et al study bridges
associated to diffusions with a gradient drift, using the fact that it is the reciprocal
characteristics which determine the bridge uniquely. In particular, diffusions with
differing drifts may have the same bridge processes. A simple example of this is the
case of the drift (1.8) with A(·) ≡ 0. The bridge process associated with the constant
drift is the same as the Brownian bridge. In Proposition [12, Prop. 3] formulas for
first passage time for diffusions with time-inhomogeneous drift are given. However
in these cases there needs to be a relation between the graph of the boundary
and the drift and diffusion coefficients of the process. The first passage time for
the half line is given in terms of the Dirichlet Green’s function by the function
t →

∫∞
0

Gε,D(x, y, t, T ) dx, t < T . An alternative approach to understanding the
limit limε→0 qε may be taken using the techniques of large deviation theory [11].
This is the approach in Baldi et al [1, 2], which considers the asymptotic behavior
of the ratio of Green’s functions in the limit T − t → 0 for time homogeneous
diffusions.

2. Representation and convergence of the function qε

For any t ∈ R let Yε(s), s > t, be the solution to the initial value problem for
the stochastic differential equation (SDE)

dYε(s) = b(Yε(s), s)ds+
√
ε dB(s), Yε(t) = y, (2.1)

where B(·) is Brownian motion. Then the Green’s function Gε(·, y, t, T ) defined by
(1.3) is the probability density for the random variable Yε(T ). In the case when the
function (y, t) → b(y, t) is linear in y it is easy to see that (2.1) can be explicitly
solved. The solution to (2.1) with b(y, t) = A(t)y − 1 as in (1.8) is given by

Yε(s) = exp
[ ∫ s

t

A(s′)ds′
]
y −

∫ s

t

exp
[ ∫ s

s′
A(s′′)ds′′

]
ds′

+
√
ε

∫ s

t

exp
[ ∫ s

s′
A(s′′)ds′′

]
dB(s′) .

(2.2)

Hence the random variable Yε(T ) conditioned on Yε(0) = y is Gaussian with mean
m1,A(T )y −m2,A(T ) and variance εσ2

A(T ), where

m1,A(T ) = exp
[ ∫ T

0

A(s′)ds′
]
, m2,A(T ) =

∫ T

0

exp
[ ∫ T

s

A(s′)ds′
]
ds , (2.3)

σ2
A(T ) =

∫ T

0

exp
[
2

∫ T

s

A(s′)ds′
]
ds . (2.4)



6 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

The Green’s function Gε(x, y, 0, T ) is therefore explicitly given by the formula

Gε(x, y, 0, T ) =
1√

2πεσ2
A(T )

exp

[
−{x+m2,A(T )−m1,A(T )y}2

2εσ2
A(T )

]
. (2.5)

It is useful to recall that the ratio Gε,D/Gε is a probability for a generalized
Brownian bridge process. Thus

Gε,D(x, y, 0, T )

Gε(x, y, 0, T )
= P ( inf

0<s<T
Yε(s) > 0 | Yε(0) = y, Yε(T ) = x) , (2.6)

where Yε(·) is the solution to the SDE (2.1). The process Yε(·) of (2.1), conditioned
on Yε(0) = y, Yε(T ) = x is Gaussian and has variance independent of x, y. We may
obtain a formula by extending the functions m1,A,m2,A, σ

2
A of (2.3), (2.4), defined

with respect to the interval [0, T ], to any interval [t, T ] with t < T . Thus we define
m1,A(t, T ),m2,A(t, T ) by

m1,A(t, T ) = exp
[ ∫ T

t

A(s′)ds′
]
, m2,A(t, T ) =

∫ T

t

exp
[ ∫ T

s

A(s′)ds′
]
ds , (2.7)

and σ2
A(t, T ) by

σ2
A(t, T ) =

∫ T

t

exp
[
2

∫ T

s

A(s′)ds′
]
ds . (2.8)

The variance of Yε(s) is then given by the formula

Var[Yε(s) | Yε(0) = y, Yε(T ) = x] = εσ2
A(s)σ

2
A(s, T )/σ

2
A(T ) . (2.9)

More generally, the covariance of Yε(·) is also independent of x, y and is given by
the formula

Covar[Yε(s1), Yε(s2) | Yε(0) = y, Yε(T ) = x] = εΓA(s1, s2) , 0 ≤ s1, s2 ≤ T,
(2.10)

where the symmetric function Γ : [0, T ]× [0, T ] → R is defined by

ΓA(s1, s2) =
m1,A(s1, s2)σ

2
A(s1)σ

2
A(s2, T )

σ2
A(T )

, 0 ≤ s1 ≤ s2 ≤ T . (2.11)

Let yclass(s), 0 ≤ s ≤ T , be the path going from y at s = 0 to x at s = T defined
by

σ2
A(T )yclass(s) = xm1,A(s, T )σ

2
A(s) + ym1,A(s)σ

2
A(s, T )

+m1,A(s, T )m2,A(s, T )σ
2
A(s)−m2,A(s)σ

2
A(s, T ) .

(2.12)

Then the mean of Yε(·) conditioned on Yε(0) = y, Yε(T ) = x is given by the formula

E[Yε(s) | Yε(0) = y, Yε(T ) = x] = yclass(s), 0 ≤ s ≤ T . (2.13)

In the case A(·) ≡ 0 the process Yε(·), conditioned on Yε(0) = y, Yε(T ) = x is the
standard Brownian Bridge (BB) from y at time 0 to x at time T .

It is well known that the conditioned process Yε(·) is also Markovian. In [6] we
showed that it is the solution to an SDE with a linear drift depending on x and
with initial condition Yε(0) = y (see [6, (4.43), (4.46)]). The SDE in this case is run
forwards in time. Here we observe that the conditioned process is also the solution
of an SDE with a linear drift depending on y, which is run backwards in time.
Denoting by Xε(s), 0 < s < T , the solution to this SDE with terminal condition
Xε(T ) = x, we have that

dXε(s) = λ(Xε(s), y, s) ds+
√
ε dB(s), 0 < s < T, Xε(T ) = x . (2.14)



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 7

The function λ(x, y, s) is given by the formula

λ(x, y, s) =
[
A(s) +

1

σ2
A(s)

]
x− 1 +

m2,A(s)

σ2
A(s)

− m1,A(s)y

σ2
A(s)

. (2.15)

Integrating (2.14), (2.15), we have that

Xε(s) = yclass(s)−
√
ε

σ2
A(s)

m1,A(s)
Z(s), with Z(s) =

∫ T

s

m1,A(s
′) dB(s′)

σ2
A(s

′)
, (2.16)

where yclass(·) is given in (2.12). Note that the drift λ(·) is independent of ε and
λ(x, y, s) becomes singular as s → 0. The singularity is necessary in order to ensure
that Xε(0) = y with probability 1,

We define now

vε(x, y, T ) = P
(

inf
0≤s≤T

Xε(s) < 0
∣∣ Xε(T ) = x

)
. (2.17)

Comparing (2.6), (2.17) we see that

vε(x, y, T ) = 1− Gε,D(x, y, 0, T )

Gε(x, y, 0, T )
. (2.18)

The function vε is a solution to the PDE

∂vε(x, y, T )

∂T
= −λ(x, y, T )

∂vε(x, y, T )

∂x
+

ε

2

∂2vε(x, y, T )

∂x2
, T > 0, x > 0, (2.19)

with initial and boundary conditions

lim
x→0

vε(x, y, T ) = 1, T > 0, lim
T→0

vε(x, y, T ) = 0, x > 0 . (2.20)

Next we set qε to be

qε(x, y, T ) = −ε log vε(x, y, T ) . (2.21)

Then qε is a solution to the PDE

∂qε(x, y, T )

∂T
= −λ(x, y, T )

∂qε(x, y, T )

∂x
− 1

2

[∂qε(x, y, T )
∂x

]2
+
ε

2

∂2qε(x, y, T )

∂x2
, (2.22)

for T > 0 and x > 0, with initial and boundary conditions

qε(0, y, T ) = 0, T > 0, qε(x, y, 0) = +∞, x > 0 . (2.23)

In the case A(·) ≡ 0 it is easy to see that the drift λ and solution qε to (2.22), (2.23)
are given by the formulae

λ(x, y, s) =
(x− y)

s
, qε(x, y, T ) =

2xy

T
. (2.24)

Evidently (2.24) is consistent with (1.10).
The PDE (2.22) is the Hamilton-Jacobi -Bellman (HJB) equation for a stochastic

control problem. Thus consider solutions Xε(·) to the SDE

dXε(s) = µε(Xε(s), y, s) ds+
√
εdB(s) , (2.25)

run backwards in time with controller µε(·) and given terminal data. For x, y, T > 0
define qε(x, y, T ) by

qε(x, y, T ) = min
µε

E
[1
2

∫ T

τ

[µε(Xε(s), y, s)− λ(Xε(s), y, s)]
2 ds :

Xε(T ) = x, 0 < τ < T, Xε(·) > 0, Xε(τ) = 0
]
,

(2.26)



8 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

where the function λ(·) is given by (2.15). The class of controllers µε(·) in (2.25)
are those which have the property that paths Xε(s), s < T , with Xε(T ) = x > 0,
exit the half line (0,∞) before time 0 with probability 1. The HJB equation for qε
is then given by (2.22), with initial and boundary conditions (2.23). The optimal
controller µ∗

ε(·) in (2.25), (2.26) is given by the formula

µ∗
ε(x, y, T ) = λ(x, y, T ) +

∂qε(x, y, T )

∂x
. (2.27)

The zero noise limit ε → 0 of (2.25), (2.26) yields the classical variational formula

q0(x, y, T ) = min
{1

2

∫ T

τ

[dx(s)
ds

− λ(x(s), y, s)
]2

ds :

0 < τ < T, x(T ) = x, x(·) > 0, x(τ) = 0
}
.

(2.28)

At least formally, the function q0 is the solution to the Hamilton-Jacobi (HJ) equa-
tion

∂q0(x, y, T )

∂T
= −λ(x, y, T )

∂q0(x, y, T )

∂x
− 1

2

[∂q0(x, y, T )
∂x

]2
, T > 0, x > 0, (2.29)

with initial and boundary conditions

q0(0, y, T ) = 0, T > 0, q0(x, y, 0) = +∞, x > 0 . (2.30)

When A(·) ≡ 0 the function q0(x, y, T ) = 2xy/T is a classical C1 solution to (2.29),
(2.30). However in general we can only expect q0 to be a viscosity solution of the
HJ equation (see [10, Chapter 10]).

To obtain an upper bound on qε by q0 plus a constant which vanishes as ε → 0,
we observe that the variational problem (2.28) for fixed τ with 0 < τ < T , without
the positivity constraint x(·) > 0, is quadratic with a linear constraint, which may
be easily solved. The Euler-Lagrange equation for the minimization problem with
fixed τ is { d

ds
+

∂λ(x(s), y, s)

∂x

}[dx(s)
ds

− λ(x(s), y, s)
]
= 0 . (2.31)

The minimizing trajectory is then the solution to (2.31) with initial and terminal
conditions x(τ) = 0, x(T ) = x. To obtain a formula for this trajectory we observe
that the solution to the equation

dϕ(s)

ds
+

∂λ(x(s), y, s)

∂x
ϕ(s) = 0 , (2.32)

is given by the formula

ϕ(s) =
C1m1,A(s)

σ2
A(s)

where C1 is a constant. (2.33)

We then need to obtain the solution to

dx(s)

ds
− λ(x(s), y, s) = ϕ(s) , (2.34)

with initial and terminal conditions x(τ) = 0, x(T ) = x, and this determines the
constant C1 in (2.33). Observe that the function x(s) = yclass(s), 0 < s < T ,
where yclass(·) is defined by (2.12), is the solution to (2.34) with terminal condition



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 9

x(T ) = x in the case ϕ(·) ≡ 0. Let yp(·) be the solution to the terminal value
problem

dyp(s)

ds
−
[
A(s) +

1

σ2
A(s)

]
yp(s) +

m1,A(s)

σ2
A(s)

= 0 , yp(T ) = 0 . (2.35)

The solution to (2.34) with initial and terminal conditions x(τ) = 0, x(T ) = x, is
then x(s) = yclass(s)− C1yp(s), where C1 is chosen so that x(τ) = 0. The solution
to (2.35) is given by

σ2
A(T )yp(s) = m1,A(s)σ

2
A(s, T ) . (2.36)

We have then from (2.35), (2.36) that

σ2
A(T )x(s) = xm1,A(s, T )σ

2
A(s) + [y − γ(τ)]m1,A(s)σ

2
A(s, T )

+m1,A(s, T )m2,A(s, T )σ
2
A(s)−m2,A(s)σ

2
A(s, T ) .

(2.37)

where γ(τ) is chosen so that x(τ) = 0. The optimal controller µ∗
0,τ for the variational

problem with fixed τ is obtained by evaluating dx(s)/ds at s = T . Thus

µ∗
0,τ (x, y, T ) = λ(x, y, T ) +

γ(τ)m1,A(T )

σ2
A(T )

= λ(x, y, T ) +
m1,A(T )

σ2
A(T )

[y + g1,A(τ, T )x+ g2,A(τ, T )] ,

(2.38)

where the functions g1,A, g2,A are given by the formulae

g1,A(s, T ) =
m1,A(s, T )σ

2
A(s)

m1,A(s)σ2
A(s, T )

, s < T , (2.39)

g2,A(s, T ) =
m1,A(s, T )m2,A(s, T )σ

2
A(s)−m2,A(s)σ

2
A(s, T )

m1,A(s)σ2
A(s, T )

, s < T . (2.40)

Lemma 2.1. Let τ > 0 and Xε(s), s > τ , be the solution to the SDE (2.25) with
µε given by µε(x, y, s) = µ∗

0,τ (x, y, s), s > τ . For x > 0, T > τ let τε,x,T be the
first exit time from the interval (0,∞) of Xε(·) with terminal condition Xε(T ) = x.
Then τε,x,T > τ with probability 1 and

qε(x, y, T ) ≤ E
[1
2

∫ T

τε,x,T

[µε(Xε(s), y, s)−λ(Xε(s), y, s)]
2 ds

∣∣∣ Xε(T ) = x
]
. (2.41)

Proof. Since the function A(·) is continuous, we have from (2.38)-(2.40) that

µε(x, y, s) =
[ 1

s− τ
+Aτ (s)

]
x+ (s− τ)Bτ (s) , s > τ, (2.42)

where Aτ , Bτ are continuous functions on the closed interval [τ,∞). Let m1,Aτ

be defined as in (2.7). The solution to (2.25) with µε as in (2.42) and terminal
condition Xε(T ) = x is given by

Xε(s) =
s− τ

(T − τ)m1,Aτ (s, T )

[
Xclass(s)−

√
εZ(s)

]
, τ < s < T , (2.43)

where Xclass(·), Z(·) are given by the formulae

Xclass(s) = x− (T − τ)

∫ T

s

m1,Aτ (s
′, T )Bτ (s

′) ds′ , (2.44)

Z(s) = (T − τ)

∫ T

s

m1,Aτ
(s′, T )

s′ − τ
dB(s′) . (2.45)



10 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

Since Z(·) is by a change of variable equivalent to Brownian motion, the reflection
principle applies to it. Hence for any a > 0, τ < s < T ,

P
(

sup
s<s′<T

Z(s′) > a
)
= 2P (Z(s) > a) . (2.46)

From (2.45) we see that Z(s) is Gaussian with mean zero. The variance Var[Z(s)]
satisfies the inequality

c1(T − τ)(T − s)

s− τ
≤ Var[Z(s)] ≤ C1(T − τ)(T − s)

s− τ
, τ < s < T , (2.47)

for some positive constants c1, C1. From (2.46), (2.47) we conclude that

P ( sup
s<s′<T

Z(s′) > a) ≥ 1− C2a

√
s− τ√

(T − τ)(T − s)
, τ < s < T , (2.48)

where C2 > 0 is a constant. We also have from (2.44) there is a constant C3 such
that supτ<s<T Xclass(s) ≤ C3. Choosing a = ε−1/2C3 in (2.48), we conclude from
(2.43), (2.48) that lims→τ P (τε,x,T > s) = 1. Hence τε,x,T > τ with probability 1.

To prove (2.41) we first observe from Ito’s lemma that the mapping s → M(s)
on the interval τ < s ≤ T , where

M(s) = qε(x, y, T )− qε(Xε(s), y, s)−
∫ T

s

∂qε(Xε(s
′), y, s′)

∂s′

+
∂qε(Xε(s

′), y, s′)

∂x
µε(Xε(s

′), y, s′)− ε

2

∂2qε(Xε(s
′), y, s′)

∂x2
ds′

(2.49)

is a (backwards in time) stochastic integral. From (2.22) we see that M(s) can be
written as

M(s) = qε(x, y, T )− qε(Xε(s), y, s)

−
∫ T

s

∂qε(Xε(s
′), y, s′)

∂x

[
µε(Xε(s

′), y, s′)− λ(Xε(s
′), y, s′)

]
− 1

2

[∂qε(Xε(s
′), y, s′)

∂x

]2
ds′ .

(2.50)

For K > 0 and 0 < x < K let τε,x,T,K be the first exit time of Xε(s), s < T , with
Xε(T ) = x from the interval (0,K). By the optional sampling theorem we have
that E[M(s ∨ τε,x,T,K)] = 0 for all s in the interval τ < s ≤ T . Hence on using the
Schwarz inequality in (2.50) we have that

qε(x, y, T )

≤ E[qε(Xε(s ∨ τε,x,T,K), y, s ∨ τε,x,T,K)]

+ E
[1
2

∫ T

s∨τε,x,T,K

[µε(Xε(s), y, s)− λ(Xε(s), y, s)]
2 ds | Xε(T ) = x

]
,

(2.51)

for τ < s ≤ T .
Observe that for τ < s ≤ T we have s ∨ τε,x,T,K ≥ τε,x,T,K ≥ τε,x,T > τ with

probability 1. Hence the second expectation on the RHS of (2.51) is bounded above
by the expectation on the RHS of (2.41). Thus it is sufficient to show that the first
term on the RHS of (2.51) converges to 0 as s → τ and K → ∞. We first consider
the limit s → τ . Since τε,x,T,K > τ with probability 1, we have by path continuity



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 11

of Xε(·) that lims→τ Xε(s ∨ τε,x,T,K) = Xε(τε,x,T,K) with probability 1. Using the
continuity of the function qε, it follows by dominated convergence that

lim
s→τ

E[qε(Xε(s ∨ τε,x,T,K), y, s ∨ τε,x,T,K)]

= E[qε(Xε(τε,x,T,K), y, τε,x,T,K)]

≤ sup
τ<s≤T

qε(K, y, s)P (Xε(τε,x,T,K) = K) .

(2.52)

From (2.46) and (2.47) we have for some constants C2, c2 > 0 that

P ( inf
s<s′<T

Z(s′) < −a) ≤ C2

a
√
s− τ

exp
[
−c2a

2(s− τ)
]
, a > 0 . (2.53)

We have from (2.43) there is a constant c3 > 0 such that for all large K,

P (Xε(τε,x,T,K) = K) ≤ P ( inf
τ<s≤T

(s− τ)Z(s) < −c3K/
√
ε) . (2.54)

From (2.53) we see that

P
(

inf
τ<s≤T

(s− τ)Z(s) < −c3K/
√
ε
)

≤
∞∑

n=0

P
(

inf
2n<(T−τ)/(s−τ)≤2n+1

Z(s) < − 2nc3K√
ε(T − τ)

)
≤ C2

∞∑
n=0

exp
[
− c2c

2
3K

22n−1

√
ε(T − τ)

]
,

(2.55)

if K is large. From our bound (3.9) on qε we conclude from (2.54), (2.55) that

lim
K→∞

sup
τ<s≤T

qε(K, y, s)P (Xε(τε,x,T,K) = K) = 0,

whence (2.41) follows from (2.51). □

Lemma 2.2. For x, y, T positive one has lim supε→0[qε(x, y, T )− q0(x, y, T )] = 0.

Proof. We first observe that a minimizing τ = τ0,x,T (which may not be unique)
in (2.28) satisfies 0 < τ0,x,T < T . To see this we use that fact that the minimizing
trajectory for fixed τ is given by (2.43), (2.44) with ε = 0. From (2.15) we have
that λ(x, y, s) ≃ (x− y)/s as s → 0, whence the minimum action integral diverges
as τ → 0. When τ is close to T we have from (2.42), (2.43) that µ∗

0,τ (X0(s), y, s) ≃
x/(T − τ), whence the minimum action integral again diverges as τ → T .

From (2.43)-(2.45) we have that

Xε(s)−X0(s) = − s− τ

(T − τ)m1,Aτ
(s, T )

√
εZ(s) , τ < s ≤ T . (2.56)

From (2.15), (2.41), (2.42) we have that

qε(x, y, T )

≤
√
εE

[ ∫ T

τε,x,T

fτ (s)Z(s) ds
]
+ C(τ)εE

[1
2

∫ T

τε,x,T

Z(s)2 ds
]

+ E
[1
2

∫ T

τε,x,T

[µ∗
0,τ (X0(s), y, s)− λ(X0(s), y, s)]

2 ds
∣∣∣ X0(T ) = x

]
,

(2.57)

where the function fτ : [τ, T ] → R is a continuous function depending on τ , and
C(τ) is a constant also depending on τ . We choose now τ to be a minimizer of



12 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

the action (2.28). Since τε,x,T > τ with probability 1, it follows that the last
expectation on the RHS of (2.57) is bounded above by q0(x, y, T ).

To bound the first term on the RHS of (2.57) we note from the optional sampling
theorem that E[Z(s∨τε,x,T )] = 0. Hence E[Z(s); τε,x,T < s] = −E[Z(τε,x,T ); τε,x,T >
s]. Letting M = supτ<s≤T Xclass(s), we see from (2.43) that 0 < Z(τε,x,T ) <

M/
√
ε. Hence we have that

√
εE

[ ∫ T

τε,x,T

fτ (s)Z(s) ds
]
≤ C1E[τε,x,T − τ ] , (2.58)

for some constant C1 independent of ε. For 0 < λ < T − τ we let γ(λ) =
infτ+λ<s<T Xclass(s), whence γ : (0, T − τ ] → R is a positive decreasing function.
We have from (2.53) and reflection symmetry that

P (τε,x,T − τ > λ) ≤ P
(

sup
τ+λ<s<T

Z(s) > γ(λ)/
√
ε
)

≤ C2
√
ε√

λγ(λ)
exp

[
− c2λγ(λ)

2

ε

]
.

(2.59)

We see from (2.59) that limε→0 P (τε,x,T − τ > λ) = 0 for all λ > 0, whence
limε→0 E[τε,x,T − τ ] = 0 by dominated convergence.

We see from (2.47) that the second term on the RHS of (2.57) diverges if we
replace τε,x,T by τ . Therefore it is again necessary to estimate the distribution of
the variable τε,x,T − τ > 0 as ε → 0. With M as in the previous paragraph, and
using (2.46), (2.47) we have from (2.43) that

P
(
τε,x,T − τ < λ

)
≤ P

(
sup

τ+λ<s<T
Z(s) < M/

√
ε
)

= P
(
|Z(τ + λ)| < M/

√
ε
)

≤ C2(λ/ε)
1/2, 0 < λ < T − τ ,

(2.60)

where c2, C2 > 0 are constants. We write now

E
[ ∫ T

τε,x,T

Z(s)2 ds
]
≤ E

[ ∫ T

τ+ε

Z(s)2 ds
]
+

∞∑
n=0

an , (2.61)

where

an = E
[ ∫ τ+2−nε

τ+2−(n+1)ε

Z(s)2 ds; τε,x,T − τ < 2−nε
]
, n = 0, 1, . . . . (2.62)

It follows from (2.47) that the first term on the RHS of (2.61) is bounded by
C3| log ε| for some constant C3. From the Schwarz inequality we have that

an ≤ 2−(n+1)/2
√
εP

(
τε,x,T − τ < 2−nε

)1/2
E
[ ∫ τ+2−nε

τ+2−(n+1)ε

Z(s)4 ds
]1/2

. (2.63)

It follows from (2.47), (2.60), (2.63) that an ≤ C4P (τε,x,T − τ < 2−nε)
1/2 ≤ C52

−n/2

for some constants C4, C5. We have shown that the second term on the RHS of
(2.57) converges to 0 as ε → 0. □

Remark 2.3. One can obtain a rate of convergence ε log ε in Lemma 2.2 as ε → 0
by making a further assumption that the classical trajectory X0(·) has the property
X ′

0(τ) > 0. In that case limλ→0 γ(λ) ≥ c1 for some constant c1 > 0, whence (2.59)
implies that E[τε,x,T − τ ] ≤ C2ε for some constant C2.



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 13

To obtain a lower bound for qε by q0 plus a constant which vanishes as ε → 0 we
need to show that the variational formula (2.26) yields a lower bound when µε is
chosen to be the optimal controller µ∗

ε given by (2.27). We have from propositions
3.1 and 3.2 that the function (x, T ) → µ∗

ε(x, y, T ) is C
1 on the domain x, T > 0 and

µ∗
ε(x, y, T ) ≥ λ(x, y, T ) for x, y, T > 0. Hence the SDE (2.25) with µε = µ∗

ε may
be solved backwards in time. Letting X∗

ε (s), s ≤ T , be the solution with terminal
condition X∗

ε (T ) = x, then X∗
ε (s) ≤ Xε(s), 0 < s ≤ T , where Xε(·) is given by

(2.16).

Lemma 2.4. For x, T > 0 and paths X∗
ε (s), s < T , with X∗

ε (T ) = x we define
τ∗ε,x,T = inf{s > 0 : X∗

ε (s
′) > 0, s ≤ s′ ≤ T}. Then τ∗ε,x,T > 0 with probability 1

and

E
[( 1

τ∗ε,x,T

)1/2−ν]
< ∞ for all ν with 0 < ν ≤ 1/2 , (2.64)

qε(x, y, T ) ≥ E
[1
2

∫ T

τ∗
ε,x,T

[µ∗
ε(X

∗
ε (s), y, s)− λ(X∗

ε (s), y, s)]
2 ds | X∗

ε (T ) = x
]
.

(2.65)

Proof. We consider the stochastic integral s → M(s), 0 < s ≤ T , defined by (2.49),
(2.50) with µε = µ∗

ε. For K > 0 and 0 < x < K let τ∗ε,x,T,K be the first exit time of

X∗
ε (s), s < T , with X∗

ε (T ) = x from the interval (0,K). Since qε is non-negative,
we have from (2.27) and the optional sampling theorem that

qε(x, y, T )

≥ E
[1
2

∫ T

s∨τ∗
ε,x,T,K

[µ∗
ε(X

∗
ε (s), y, s)− λ(X∗

ε (s), y, s)]
2 ds

∣∣∣ X∗
ε (T ) = x

] (2.66)

for 0 < s ≤ T . Since X∗
ε (·) ≤ Xε(·), it follows that s ∨ τ∗ε,x,T,K → s ∨ τ∗ε,x,T with

probability 1 as K → ∞, whence the inequality (2.66) holds with τ∗ε,x,T in place of
τ∗ε,x,T,K .

Letting Xε(·) be the solution to (2.25) with terminal condition Xε(T ) = x > 0,
we have from (2.15), (2.25) that

m1,A(s)

σ2
A(s)

[µε(Xε(s), y, s)− λ(Xε(s), y, s)] ds

= d
[m1,A(s)Xε(s)

σ2
A(s)

]
−
√
ε
m1,A(s)

σ2
A(s)

dB(s)− d
[m1,A(s)

2σ2
A(s, T )

σ2
A(T )σ

2
A(s)

]
y

− d
[m1,A(T )m2,A(s, T )

σ2
A(T )

− m1,A(s)m2,A(s)σ
2
A(s, T )

σ2
A(T )σ

2
A(s)

]
, 0 < s ≤ T .

(2.67)



14 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

Let δ satisfy 0 < δ ≤ T and τδ be the stopping time τδ = δ ∨ τ∗ε,x,T . On integrating

(2.67) with µε = µ∗
ε over the interval τδ < s < T , we obtain the identity∫ T

τδ

m1,A(s)

σ2
A(s)

[µ∗
ε(X

∗
ε (s), y, s)− λ(X∗

ε (s), y, s)] ds

=
m1,A(T )x

σ2
A(T )

− m1,A(τδ)X
∗
ε (τδ)

σ2
A(τδ)

−
√
εZ(τδ)

+
m1,A(τδ)

2σ2
A(τδ, T )

σ2
A(T )σ

2
A(τδ)

y +
m1,A(T )m2,A(τδ, T )

σ2
A(T )

− m1,A(τδ)m2,A(τδ)σ
2
A(τδ, T )

σ2
A(T )σ

2
A(τδ)

,

(2.68)

where the martingale Z(·) is defined in (2.16). From the Schwarz inequality we see
that the LHS of (2.68) is bounded above by

αy

2

∫ T

τδ

m1,A(s)
2

σ4
A(s)

ds+
1

2αy

∫ T

τδ

[µ∗
ε(X

∗
ε (s), y, s)− λ(X∗

ε (s), y, s)]
2
ds , (2.69)

for any α > 0. Choosing α sufficiently small, we conclude from (2.68) that

y

τ∗ε,x,T
≤ C1

[
1 +

√
ε |Z(τδ)|

]
+

C1

2αy

∫ T

τ∗
ε,x,T

[
µ∗
ε(X

∗
ε (s), y, s)− λ(X∗

ε (s), y, s)
]2
ds

(2.70)

if δ ≤ τ∗ε,x,T ≤ T/2, for some constants C1, α depending only on T . Multiplying

(2.70) by (τ∗ε,x,T )
1/2+ν and taking the expected value, we conclude from the limit

of (2.66) as K → ∞ that

yE
[( 1

τ∗ε,x,T

)1/2−ν

; δ < τ∗ε,x,T < T/2
]

≤ C1

[
T 1/2+ν +

√
εE

[
τ
1/2+ν
δ |Z(τδ)|

]]
+

C1T
1/2+νqε(x, y, T )

αy
.

(2.71)

Let τ be a stopping time for the martingale s → Z(s), 0 < s < T , of (2.16).
Then for ν > 0 there is a constant Cν , depending on ν, such that

E
[
τ1/2+ν ; |Z(τ)| > a

]
≤ Cν

a1+2ν
for a > 0 . (2.72)

To show (2.72), let τa = sup{s : 0 < s < T, |Z(s)| ≥ a}. Then any stopping time τ
for Z(·) has the property

{τ > s, |Z(τ)| > a} ⊂ {τa > s} , (2.73)

since {τa > s} is the largest Borel set in Fs,T , the σ-field generated by B(s′), s <
s′ < T , on which sups<s′<T |Z(s′)| > a. Hence

E[τ1/2+ν ; |Z(τ)| > a] =
(1
2
+ ν

) ∫ T

0

sν−1/2P (τ > s, |Z(τ)| > a) ds

≤
(1
2
+ ν

) ∫ T

0

sν−1/2P (τa > s) ds

= E[τ1/2+ν
a ] .

(2.74)



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 15

The reflection principle applies to Z(·) and also Var[Z(s)] satisfies an inequality
(2.47) with τ = 0. We have therefore from (2.53) that

P (τa > n/a2) = P
(

sup
n/a2<s<T

|Z(s)| ≥ a
)
≤ 2C2√

n
e−c2n , n = 1, 2, . . . (2.75)

The inequality (2.72) follows now from (2.74) and (2.75).

We easily see from (2.72) that E[τ
1/2+ν
δ |Z(τδ)|] is bounded by a constant, uni-

formly in δ as δ → 0. Thus for any a > 0,

E
[
τ
1/2+ν
δ |Z(τδ)|

]
≤ aE

[
τ
1/2+ν
δ

]
+

∞∑
n=0

2n+1aE[τ
1/2+ν
δ ; |Z(τδ)| > 2na]

≤ aT 1/2+ν +
C1,ν

a2ν
,

(2.76)

where C1,ν depends on ν > 0, but not on δ. Choosing a to minimize the RHS of

(2.76) we conclude that E
[
τ
1/2+ν
δ |Z(τδ)|

]
≤ C2,νT

ν for some constant depending

only on ν. Hence the RHS of (2.71) is bounded by a constant independent of δ.
Letting δ → 0 we conclude that τ∗ε,x,T > 0 with probability 1 and (2.64) holds.

The inequality (2.65) follows from (2.66) and the monotone convergence theorem
by letting K → ∞ first and then s → 0. □

Lemma 2.5. For x, y, T positive one has lim infε→0[qε(x, y, T )− q0(x, y, T )] = 0.

Proof. Let X∗
ε (s), s ≤ T , be a solution to (2.25) with µε = µ∗

ε and terminal
condition X∗

ε (T ) = x. We associate with X∗
ε (·) the differentiable path X∗

ε,c(·)
defined by

dX∗
ε,c(s)

ds
= λ(X∗

ε,c(s), y, s) + [µ∗
ε(X

∗
ε (s), y, s)− λ(X∗

ε (s), y, s)] , s < T , (2.77)

with terminal condition X∗
ε,c(T ) = x. From (2.15), (2.25), and (2.77) we see that

d
{
X∗

ε (s)−X∗
ε,c(s)

}
=

[
A(s) +

1

σ2
A(s)

] {
X∗

ε (s)−X∗
ε,c(s)

}
+

√
ε dB(s) , s < T ,

(2.78)
with zero terminal condition at s = T . Comparing (2.78) to (2.14), we conclude
from (2.16) that

X∗
ε,c(s) = X∗

ε (s) +
√
ε

σ2
A(s)

m1,A(s)
Z(s) , s < T , (2.79)

with Z(·) as given in (2.16). We define a classical action which generalizes (2.28).
Thus for x, y, T > 0 and z ∈ R we define q0(x, y, z, T ) just as in (2.28) but with ter-
minal condition x(τ) = z instead of x(τ) = 0, and without the positivity constraint
x(·) > 0. It follows from (2.77), (2.79) and (2.65) of Lemma 2.4 that

qε(x, y, T ) ≥ E
[
q0(x, y,

√
εZε, T )

]
, Zε =

σ2
A(τ

∗
ε,x,T )

m1,A(τ∗ε,x,T )
Z(τ∗ε,x,T ) , (2.80)

where Z(·) is as in (2.16).
We have already observed that a minimizing τ in (2.28) satisfies 0 < τ < T . It

follows from this there exist constants C, δ > 0, depending on x, y, T , such that if
|z| < δ then |q0(x, y, z, T )− q0(x, y, T )| ≤ C|z|. Hence from (2.80) we have that

qε(x, y, T ) ≥ q0(x, y, T )
[
1− P (|Zε| > δ/

√
ε)
]
− C

√
εE [|Zε|] . (2.81)



16 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

From (2.3), (2.4) we see that |Zε| ≤ C1τ
∗
ε,x,T |Z(τ∗ε,x,T )| for some constant C1. Hence

from the proof of Lemma 2.4 we have that E[|Zε|] ≤ C2 for some constant C2

independent of ε as ε → 0. We conclude from (2.81) and the Chebyshev inequality
that qε(x, y, T ) ≥ q0(x, y, T )− C3

√
ε for some constant C3. □

We summarize the main result of this section.

Theorem 2.6. Assume A(·) is continuous and the function qε is defined by (2.18),
(2.21). Then For x, y, T positive one has limε→0 qε(x, y, T ) = q0(x, y, T ), where the
function q0 is defined by (2.28).

3. Regularity and bounds on the function qε

We first prove a regularity result for the function (x, y, t, T ) → Gε,D(x, y, t, T ),
which will imply the regularity results for the function (x, y, T ) → qε(x, y, T ) we
shall need.

Proposition 3.1. Let A : [0,∞) → R be a continuous function and Gε,D(x, y, t, T ),
x, y,> 0, 0 ≤ t < T < ∞, be the Dirichlet Green’s function for the PDE (1.1) with
drift (1.8). Then the derivatives

∂n

∂xn

∂m

∂ym
Gε,D(x, y, t, T ) with 0 ≤ n,m ≤ 2, n+m ≤ 3 , (3.1)

∂k

∂tk
∂l

∂T l

∂n

∂xn

∂m

∂ym
Gε,D(x, y, t, T ) with 0 ≤ k + l, k +m, l + n ≤ 1 , (3.2)

exist and are continuous in the region D = {(x, y, t, T ) : x, y ≥ 0, 0 ≤ t < T < ∞}.
Let G(x, t) be the Gaussian distribution with mean 0 and variance t,

G(x, t) =
1√
2πt

exp
[
− x2

2t

]
, x ∈ R, t > 0 . (3.3)

For any L0, T0 > 0 define DL0,T0
to be the region DL0,T0

= {(x, y, t, T ) : 0 ≤ x, y,≤
L0, 0 ≤ t < T ≤ T0, T − t ≤ L2

0}. Then there is a constant C(L0, T0, ε) such that
if m,n satisfy the conditions of (3.1), then∣∣ ∂n

∂xn

∂m

∂ym
Gε,D(x, y, t, T )

∣∣ ≤ C(L0, T0, ε)

(T − t)(n+m)/2
G(x− y, 2ε(T − t)) , (3.4)

for (x, y, t, T ) ∈ DL0,T0 . Similarly if k, l,m, n satisfies the conditions of (3.2), then∣∣ ∂k

∂tk
∂l

∂T l

∂n

∂xn

∂m

∂ym
Gε,D(x, y, t, T )

∣∣ ≤ C(L0, T0, ε)

(T − t)k+l+(n+m)/2
G(x−y, 2ε(T − t)) , (3.5)

for (x, y, t, T ) ∈ DL0,T0
.

For each y > 0 the function (x, T ) → ∂2Gε,D(x, y, 0, T )/∂x2, with domain
{(x, T ) : x, T > 0} is continuous up to the boundary x = 0, and is also contin-
uously differentiable in T , twice continuously differentiable in x.

Proof. Since the drift b(·, ·) is continuous and satisfies for each T0 > 0 the bound
sup{|∂b(y, t)/∂y| : y ≥ 0, 0 ≤ t ≤ T0} < ∞ we may apply the perturbation
argument of [7, Lemma3.4]. From this we see that the derivatives (3.1) with
n ≤ 1,m ≤ 2 are continuous and satisfy the inequality (3.4). In making this
conclusion we are using the backwards in time PDE (1.1). Since the adjoint PDE
(1.4) is similar to (1.1) except run forwards in time, we conclude that (3.4) also
holds with n ≤ 2,m ≤ 1. The continuity of the derivatives (3.2) and bounds (3.5)



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 17

follow from the fact that Gε,D is a solution to the PDEs (1.1), (1.4). Hence the
derivatives in (3.2) can be expressed as a sum of the derivatives in (3.1).

Letting vε(x, T ) = exp[2
∫ T

0
A(s) ds]∂2Gε,D(x, y, 0, T )/∂x2, we see by differenti-

ating twice the PDE (1.4) that vε is also a solution to (1.4). It is also continuous up
to the boundary x = 0. For any L0 > 0 we consider vε to be a solution to (1.4) in
the region {(x, T ) : 0 < x < L0, T > 0}. Let Gε,D,L0

(x, y, t, T ) be the correspond-
ing Dirichlet Green’s function. The function Gε,D,L0 has the same differentiability
properties as Gε,D given in (3.1)-(3.5). We also have the integral representation

vε(x, T )

=

∫ L0

0

Gε,D,L0(x, y, t, T )vε(y, t) dy + ε

∫ T

t

∂Gε,D,L0
(x, 0, s, T )

∂y
vε(0, s) ds

− ε

∫ T

t

∂Gε,D,L0
(x, L0, s, T )

∂y
vε(L0, s) ds , 0 < x < L0, 0 ≤ t < T .

(3.6)

The differentiability properties of the function vε follow from (3.6) and the differ-
entiability properties of Gε,D,L0

by differentiating under the integral and using the
estimates (3.4), (3.5). Note that while the function (x, T ) → vε(x, T ) itself is con-
tinuous up to the boundary x = 0, our proof does not establish that the derivatives
are continuous up to the boundary. □

In the case A(·) ≡ 0 the process Xε(·) of (2.16) is the standard Brownian Bridge
(BB) process from x at time T to y at time 0. In that case

Xε(s) =
sx+ (T − s)y

T
−

√
ε s

∫ T

s

dB(s′)

s′
, 0 < s < T , (3.7)

and from (1.10) we have that

P
(

inf
0≤s≤T

Xε(s) < 0
∣∣ Xε(T ) = x

)
= exp

[
− 2xy

εT

]
. (3.8)

We can obtain a linear upper bound on the function x → qε(x, y, T ) in the case of
non-trivial A(·) by comparing Xε(·) to the BB process.

Proposition 3.2. Let A : [0,∞) → R be continuous and qε be defined by (2.18),
(2.21). Then for any y, T > 0 the function x → qε(x, y, T ), x ≥ 0, is continuous
increasing with qε(0, y, T ) = 0. For any T0 > 0 there exists a constant CA(T0),
depending only on T0 and sup0≤t≤T0

|A(t)|, such that qε satisfies the inequality∣∣qε(x, y, T )− 2m1,A(T )xy

σ2
A(T )

∣∣ ≤ CA(T0)Ty , x, y > 0, 0 < T ≤ T0 . (3.9)

Proof. The monotonicity of the function x → qε(x, y, T ) follows from (2.17), (2.21).
Since Xε(s), s ≤ T , is the solution to (2.14), it follows from the non-intersection
of paths property that the function x → vε(x, y, T ) is decreasing.

To prove (3.9) we make the change of variable s ↔ t in (2.16) defined by

t2
ds

dt
= g(s)2 , g(s) =

σ2
A(s)

m1,A(s)
, 0 < s < T, s(T ) = T . (3.10)

Since σA(·) is strictly positive and lims→0 g(s)/s = 1, the function s(·) is continuous,
strictly monotonic and limt→0 s(t)/t = 1. We see that the stochastic integral in



18 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

(2.16) becomes ∫ T

s

m1,A(s
′) dB(s′)

σ2
A(s

′)
=

∫ T

t

dB̃(t′)

t′
, (3.11)

where B̃(·) is a Brownian motion. We define the stochastic process X̃ε(t), 0 < t < T ,
by

X̃ε(t) =
m1,A(s(t))

σ2
A(s(t))

t yclass(s(t))−
√
εt

∫ T

t

dB̃(t′)

t′
, 0 < t ≤ T , (3.12)

so that the events {inf0<s<T Xε(s) < 0} and {inf0<t<T X̃ε(t) < 0} are the same.
Observe that

g3,A(s, T ) =

∫ T

s

m1,A(s
′)2 ds′

σ4
A(s

′)

=
m1,A(s)

2

σ2
A(s)

− m1,A(T )
2

σ2
A(T )

=
m1,A(s)

2σ2
A(s, T )

σ2
A(T )σ

2
A(s)

.

(3.13)

From (3.10) we see that g3,A(s(t), T ) = (T − t)/tT , 0 < t < T . We conclude then
from (2.12), (3.10), (3.13) that

m1,A(s(t))

σ2
A(s(t))

tyclass(s(t)) = t
m1,A(T )

σ2
A(T )

{
x+

g2,A(s(t), T )

g1,A(s(t), T )

}
+ (T − t)

y

T
, (3.14)

for 0 < t < T , where g1,A, g2,A are defined by (2.39), (2.40).
The expression (2.40) for g2,A can be simplified by observing that

g2,A(s, T ) =
m2,A(s, T )

m1,A(T )

[σ2
A(T )− σ2

A(s, T )

σ2
A(s, T )

]
− m2,A(s)

m1,A(s)

=
σ2
A(T )

m1,A(T )

m2,A(s, T )

σ2
A(s, T )

− m2,A(T )

m1,A(T )
.

(3.15)

We have then from (2.39) and (3.15) that

g2,A(s, T )

g1,A(s, T )
= m2,A(T )−

σ2
A(T )m2,A(s)

m1,A(s, T )σ2
A(s)

. (3.16)

We have from (3.16) that

lim
s→T

g2,A(s, T )

g1,A(s, T )
= 0,

d

ds

[g2,A(s, T )
g1,A(s, T )

]
= − σ2

A(T )

m1,A(s, T )

[
σ2
A(s)−m2,A(s)

]
σ4
A(s)

. (3.17)

It is easy to see from (3.17) that for any T0 > 0,

∣∣g2,A(s, T )
g1,A(s, T )

∣∣ ≤ CA,1(T0)T (T − s) , 0 < s < T ≤ T0 , (3.18)

where the constant CA,1(T0) depends only on T0 and sup0≤t≤T0
|A(t)|.



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 19

It follows from (3.12), (3.14), and (3.18) that

P
(

inf
0≤t≤T

X̃ε(t) < 0
∣∣ X̃ε(T ) = x

)
≥ P

(
inf

0≤t≤T

[ tx1 + (T − t)y

T
−

√
εt

∫ T

t

dB̃(t′)

t′

]
< 0

)
,

x1 =
Tm1,A(T )

σ2
A(T )

[
x+ CA,1(T0)T

2
]
.

(3.19)

We conclude from (3.7), (3.8), and (3.19) that

qε(x, y, T ) ≤
2x1y

T
≤ 2m1,A(T )xy

σ2
A(T )

+ CA,2(T0)Ty , 0 < T ≤ T0 , (3.20)

for some constant CA,2(T0) depending only on T0 and sup0≤t≤T0
|A(t)|. A similar

argument yields a lower bound corresponding to (3.20), whence (3.9) follows. □

For general A(·) one can construct a linear solution to the HJ equation (2.29),
which is therefore also a solution to the HJB equation (2.22). To find it we set

q0(x, y, T ) = a(y, T ) + b(y, T )x . (3.21)

Equating the coefficients of x in (2.29), we obtain the ODE

db(y, T )

dT
= −

[
A(T ) +

1

σ2
A(T )

]
b(y, T ) . (3.22)

Integrating (3.22), we conclude that

b(y, T ) = C(y)
m1,A(T )

σ2
A(T )

(3.23)

for a constant C(y) depending on y. We choose the constant C(y) in (3.23) so that
our linear solution gives the function x → 2xy/T , corresponding to (1.10), when
A(·) ≡ 0. Thus we choose C(y) = 2y. Equating the terms independent of x in
(2.29), we obtain the ODE

da(y, T )

dT
=

[
1− m2,A(T )

σ2
A(T )

+
m1,A(T )y

σ2
A(T )

]
b(y, T )− 1

2
b(y, T )2 . (3.24)

With the choice of C(y) = 2y in (3.23), this reduces (3.24) to the equation

da(y, T )

dT
=

[
1− m2,A(T )

σ2
A(T )

]
b(y, T ) . (3.25)

Integrating (3.25) with initial condition a(y, 0) = 0 yields the solution

a(y, T ) =
2y

σ2
A(T )

[
m1,A(T )m2,A(T )− σ2

A(T )
]
. (3.26)

We conclude from (3.23)-(3.26) that

qlinear(x, y, T ) =
2y

σ2
A(T )

[
m1,A(T )m2,A(T )− σ2

A(T ) +m1,A(T )x
]

(3.27)

is a linear solution to (2.29). We shall show that the linear solution (3.27) tightly
bounds the function qε defined by (2.18), (2.21) when A(·) is non-negative.



20 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

Proposition 3.3. Assume the function A(·) is continuous non-negative, and let
qε(x, y, T ) be defined by (2.18), (2.21). Then{∂qlinear(x

′, y, T )

∂x′

∣∣∣
x′=0

}
x ≤ qε(x, y, T ) ≤ qlinear(x, y, T ) for x, y, T > 0 , (3.28)

and

qε(x, y, T ) ≤ −2λ(0, y, T )x for x, y, T > 0 . (3.29)

Proof. We see using the formula (2.5) for Gε(x, y, 0, T ) and the fact that the func-
tion (x, t) → Gε,D(x, y, 0, t) is a solution to the PDE (1.4) with drift b(x, t) =
A(t)x − 1, that the function (x, T ) → vε(x, y, T ) defined by (2.18) is a solu-
tion to the PDE (2.19). Furthermore, vε satisfies the initial and boundary con-
ditions (2.20). Since the function qlinear is a solution of (2.22) it follows that
vε,1(x, y, T ) = exp[−qlinear(x, y, T )/ε] is a solution to the PDE (2.19). From the
non-negativity of A(·) we also have that qlinear(0, y, T ) ≥ 0 for T > 0. In addition,
one has for T small that qlinear(x, y, T ) ≃ 2xy/T . We conclude that vε,1 satisfies
the initial and boundary conditions

vε,1(0, y, T ) ≤ 1, T > 0, vε,1(x, y, 0) = 0, x > 0 . (3.30)

Comparing (2.20) and (3.30), we expect that an application of the maximum prin-
ciple for linear parabolic PDE [16] implies that vε,1(x, y, T ) ≤ vε(x, y, T ) for all
x, T > 0, whence the upper bound in (3.28).

In the application of the maximum principle we need to take account of the
fact that the domain {x ∈ R : x > 0} is unbounded, and that the drift λ(x, y, T )
of (2.15) becomes unbounded as T → 0. To deal with this we apply for any
M,T0 > 0, 0 < δ < T0, the maximum principle to a bounded domain DM,T0,δ =
{(x, T ) : 0 < x < M, δ < T < T0} on which the drift is continuous and bounded.
Then we let M → ∞, δ → 0.

We first consider the case M → ∞. It is evident from (3.27) that

lim
M→∞

sup
0<T<T0

vε,1(M,y, T ) = 0.

It follows from (3.9) of Proposition 3.2 that also limM→∞ sup0<T<T0
vε(M,y, T ) =

0. Next we consider the case δ → 0. We see from (3.27) that for any m > 0 then
limδ→0 supx≥m vε,1(x, y, δ) = 0. Observe from (2.3), (2.4) that since the function
s → A(s) is continuous at s = 0 then

σ2
A(s)

m1,A(s)
= s[1 + so(s)] ,

m2,A(s)

m1,A(s)1/2
= s[1 + so(s)] . (3.31)

It follows from (3.27) and (3.31) that

qlinear(x, y, T ) =
2xy

T
+ o(T ) as T → 0 . (3.32)

We conclude from (3.32) that

lim
δ→0

sup
0<x≤m

{
vε,1(x, y, δ)− exp

[
− 2xy

εδ

]}
= 0 . (3.33)

From Proposition 3.2 we see that limδ→0 supx≥m vε(x, y, δ) = 0 and a similar result
to (3.33) holds for vε. We conclude that

lim
δ→0

sup
0<x<∞

{
vε,1(x, y, δ)− vε(x, y, δ)

}
= 0 . (3.34)



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 21

From Proposition 3.1 it follows that the function (x, T ) → u(x, T ) = vε,1(x, y, T )−
vε(x, y, T ) is continuously differentiable in T and twice continuously differentiable
in x on the domain DM,T0,δ. It is also continuous up to the boundaries x = 0,
x = M , T = δ. It follows then from the maximum principle [16, Theorem 2,
Chapter 3] applied to the solution u(x, T ) of (2.19) that the maximum of u(·, ·)
on DM,T0,δ occurs on one of those boundaries. The upper bound in (3.28) fol-
lows by letting δ → 0,M → ∞ and using (2.20), (3.30), (3.34) and the fact that
limM→∞ sup0<T<T0

u(M,T ) = 0..
The lower bound in (3.28) can be established similarly. Thus we define the

function vε,2 by

vε,2(x, y, T ) = exp
[
− 2m1,A(T )xy

εσ2
A(T )

]
. (3.35)

Let L be the linear differential operator L = −∂/∂T + · · · such that the PDE (2.19)
is Lvε = 0. Then we have from (3.35) that

Lvε,2(x, y, T ) = −2m1,A(T )y

εσ2
A(T )

[
1− m2,A(T

σ2
A(T )

]
vε,2(x, y, T ) . (3.36)

Since A(·) is non-negative the RHS of (3.36) is less than or equal to 0. Arguing as
in the previous paragraphs we also see that

lim
M→∞

sup
0<T<T0

[vε(M,y, T )− vε,2(M,y, T )] = 0 ,

lim
δ→0

sup
0<x<∞

{vε(x, y, δ)− vε,2(x, y, δ)} = 0 .
(3.37)

Setting u(x, T ) = vε(M,y, T ) − vε,2(M,y, T ), we see from (2.19) and (3.36) that
Lu(x, T ) ≥ 0 for (x, T ) ∈ DM,T0,δ. Applying [16, Theorem 2, Chapter 3] again, we
conclude that u(·, ·) takes its maximum on the boundaries x = 0, x = M , T = δ of
DM,T0,δ. The lower bound in (3.28) then follows from (3.37) upon letting δ → 0,
M → ∞.

The upper bound (3.29) also follows in a similar way. We define the function

vε,3(x, y, T ) = exp
[2λ(0, y, T )x

ε

]
= exp

[
− 2x

ε

{
1− m2,A(T )

σ2
A(T )

+
m1,A(T )y

σ2
A(T )

}]
.

(3.38)

Taking derivatives in (3.38) we see that

Lvε,3(x, y, T ) = 2xε−1A(T )vε,3(x, y, T ) . (3.39)

Setting u(x, T ) = vε,3(x, y, T ) − vε(x, y, T ), we have from (2.19) and (3.39) that
Lu(x, T ) ≥ 0 for (x, T ) ∈ DM,T0,δ. As with (3.37), we have that

lim
M→∞

sup
0<T<T0

u(M,T ) = 0, lim
δ→0

sup
0<x<∞

u(x, δ) = 0 . (3.40)

Applying [16, Theorem 2 Chapter 3] once more, and taking the limits δ → 0,
M → ∞ using (3.40), then yields the upper bound (3.29). □

Remark 3.4. In the case A(·) ≡ 0 the upper and lower bounds in (3.28) are
identical, yielding the function qε(x, y, T ) of (2.24). A similar situation also occurs



22 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

when the drift in (2.1) has the form b(y, s) = −γy, where γ is constant. In that
case the solution to (2.1) is

Yε(s) = e−γ(s−t)y +
√
ε

∫ s

t

e−γ(s−s′) dB(s′) . (3.41)

Hence the random variable Yε(T ) conditioned on Yε(t) = y is Gaussian with mean
m(T − t)y and variance εσ2(T − t), where

m(T ) = e−γT , σ2(T ) =
1

2γ

[
1− e−2γT

]
. (3.42)

The whole line Green’s function Gε(x, y, t, T ) is explicitly given by the formula

Gε(x, y, t, T ) =
1√

2πεσ2(T − t)
exp

[
− {x−m(T − t)y}2

2εσ2(T − t)

]
. (3.43)

We again define qε(x, y, T ) in terms of the Dirichlet Green’s function by (2.18),
(2.21). Then the function [x, T ] → qε(x, y, T ), x, T > 0, is a solution to (2.22) and
(2.23) with

λ(x, y, T ) = −γx− ε
∂

∂x
logGε(x, y, 0, T ) = −γx+

x−m(T )y

σ2(T )
. (3.44)

We may solve (2.22) and (2.23) in the case of (3.42), (3.44) by looking for a solution
of the form qε(x, y, T ) = a(T )xy. Then a(·) is given by the formula

a(T ) =
2m(T )

σ2(T )
=

2γ

sinh γT
. (3.45)

The relation with the function p̂ of [17, Proposition 20] is

p̂(T, x, y) = exp
[γx2

ε

]
Gε,D(x, y, 0, T )

= exp
[γx2

ε

]
Gε(x, y, 0, T ) [1− vε(x, y, T )] with ε = 1 ,

(3.46)

where Gε is given by (3.43), and vε, qε are related by (2.21).

Remark 3.5. The upper bound (3.29) suggests that the function x → q0(x, y, T )
is concave. To see this consider solutions to the HJ equation (2.29) with the initial
and boundary conditions given by (2.23). In view of the boundary condition at
x = 0 we have that ∂q0(0, y, T )/∂T = 0 for T > 0. It follows then from the PDE
(2.29) that

∂q0(x, y, T )

∂x

∣∣∣
x=0

= −2λ(0, y, T ) . (3.47)

Letting ε → 0 in the inequality (3.29) we see that the graph of the function x →
q0(x, y, T ) lies below the line through the origin with slope (3.47).

Corollary 3.6. Assume A(·) is continuous non-negative, and let qε(x, y, T ) be
defined by (2.18), (2.21). Then the function x → qε(x, y, T ) is twice continuously
differentiable in x for x ≥ 0 and ∂2qε(x, y, T )/∂x

2 ≤ 0 at x = 0 and y, T > 0.

Proof. The regularity of qε follows from Proposition 3.1. Since qε(0, y, T ) = 0 and
the function x → qε(x, y, T ) is non-negative, we have using the inequality (3.29)



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 23

that 0 ≤ ∂qε(x, y, T )/∂x ≤ −2λ(0, y, T ) at x = 0. Observing that qε satisfies the
PDE (2.22) and ∂qε(x, y, T )/∂T = 0 at x = 0, we also have that

ε
∂2qε(x, y, T )

∂x2

∣∣∣
x=0

=
∂qε(x, y, T )

∂x

∣∣∣
x=0

[
2λ(0, y, T ) +

∂qε(x, y, T )

∂x

∣∣∣
x=0

]
. (3.48)

We conclude from (3.48) and our bounds on ∂qε(x, y, T )/∂x at x = 0 that

∂2qε(x, y, T )/∂x
2 ≤ 0

at x = 0. □

4. Estimating solutions of the Hamilton-Jacobi PDE

In §2 we already observed that the infinite dimensional variational problem (2.28)
may be reduced to a single variable variational problem in the first hitting time
parameter τ , 0 < τ < T . From (2.33), (2.34) we have that

q0(x, y, T ) = min
0<τ<T

γ(τ)2

2

∫ T

τ

m1,A(s)
2 ds

σ4
A(s)

, (4.1)

where the function γ(·) is defined by (2.37). The RHS of (4.1) can be expressed in
terms of the functions g1,A, g2,A, g3,A of (2.39), (2.40), (3.13). Thus we have that

q0(x, y, T ) = min
0<τ<T

g3,A(τ, T )

2
[y + g1,A(τ, T )x+ g2,A(τ, T )]

2
. (4.2)

One sees from (2.39) that limτ→0 g1,A(τ, T ) = 0 and limτ→T g1,A(τ, T ) = ∞. From
(3.15), the function τ → g2,A(τ, T ), 0 < τ < T , satisfies limτ→0 g2,A(τ, T ) = 0
and limτ→T g2,A(τ, T ) = [σ2

A(T )−m2,A(T )]/m1,A(T ). From (3.13) we see that the
function τ → g3,A(τ, T ), 0 < τ < T , has the properties limτ→0 g3,A(τ, T ) = ∞ and
limτ→T g3,A(τ, T ) = 0. It follows then from (4.2) that q0(x, 0, T ) = q0(0, y, T ) = 0.
In the case of x → 0 with fixed y > 0, the minimizer τ(x, y, T ) in (4.2) satisfies
τ(x, y, T ) → T , with the minimum in (4.2) converging to 0. In the case of y → 0
with fixed x > 0, the minimizer τ(x, y, T ) satisfies τ(x, y, T ) → 0, with the minimum
in (4.2) also converging to 0. For general x, y > 0, there may not be a unique
minimizer τ(x, y, T ), so one does not expect the function q0(x, y, T ) of (4.2) to be
a C1 solution to the HJ equation (2.29). Note however from (4.2) that the function

x →
√

q0(x, y, T ), x > 0, is concave for all y, T > 0. This is a simple consequence
of the fact that the function is the minimum of a set of linear functions. Concavity
of the function x → q0(x, y, T ) implies concavity of the function x →

√
q0(x, y, T ).

We shall prove concavity of x → q0(x, y, T ) in the case when A(·) is non-negative.
When A(·) ≡ 0 the formula (4.2) becomes

q0(x, y, T ) = min
0<τ<T

T − τ

2τT

[
y +

τx

(T − τ)

]2
=

1

2T

[
2xy +min

α>0
{αx2 + y2/α}

]
=

2xy

T

(4.3)

with τ(x, y, T ) = yT
x+y . In this case the minimization problem (4.3) is convex in

α, but one does not expect for general A(·) that (4.2) is a convex minimization
problem.

The solution of the variational problem (2.28) with fixed τ and without the
positivity constraint on x(·) is given by the expression on the RHS of (4.2). In the
case when the function A(·) is non-negative this is also the solution to the fixed τ



24 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

variational problem with the positivity constraint on x(·). We see this by observing
that the optimizing trajectory (2.37) for the unconstrained problem is positive.
This follows from the fact that the functions s → g1,A(s, T ) and s → g2,A(s, T ),
0 < s < T , are increasing if A(·) is non-negative. The monotonicity of g1,A follows
by noting that it may be written as

g1,A(s, T ) =
1

m1,A(T )

[ σ2
A(T )

σ2
A(s, T )

− 1
]
. (4.4)

To show monotonicity of g2,A we differentiate (3.15) to obtain the formula

∂g2,A(s, T )

∂s
=

σ2
A(T )

m1,A(s)σ4
A(s, T )

[
m1,A(s, T )m2,A(s, T )− σ2

A(s, T )
]
. (4.5)

It is easy to see that m1,A(s, T )m2,A(s, T ) − σ2
A(s, T ) ≥ 0 for 0 < s < T if the

function A(·) is non-negative.
For fixed y > 0 and x large, minimizers τ(x, y, T ) for (4.2) are close to 0. We can

use this observation to show that when x is large, q0(x, y, T ) is well approximated by
qlinear(x, y, T ) of (3.27). To see this first observe from (3.13) that limτ→0 τg3(τ, T ) =
1. We also have on differentiating (4.4) that

∂g1,A(s, T )

∂s
=

σ2
A(T )m1,A(s, T )

2

m1,A(T )σ4
A(s, T )

. (4.6)

Upon setting s = 0 in (4.5), (4.6) we see that

g1,A(0, T ) = 0,
∂g1,A(0, T )

∂s
=

m1,A(T )

σ2
A(T )

,

g2,A(0, T ) = 0,
∂g2,A(0, T )

∂s
=

m1,A(T )m2,A(T )

σ2
A(T )

− 1 .

(4.7)

Hence if the minimizer τ(x, y, T ) of (4.2) is close to 0, the minimization problem is
given to leading order by

q0(x, y, T ) ≃
1

2

{
min
τ>0

[ y√
τ
+

√
τ
{∂g1,A(0, T )

∂s
x+

∂g2,A(0, T )

∂s

}]}2

= qlinear(x, y, T ) .

(4.8)

The minimizer in (4.8) gives the leading order term in an expansion of τ(x, y, T ),
whence

τ(x, y, T ) ≃ 2y2

qlinear(x, y, T )
≃ Ty

x
for large x. (4.9)

Note from (4.9) that τ(x, y, T ) = O(1/x) as x → ∞. We make this argument
precise in the following.

Proposition 4.1. Assume the function A : [0,∞) → R is continuous and non-
negative. Then for any T0 > 0 there exists a constant CA(T0), depending only
on T0 and sup0≤t≤T0

A(t), such that the function (x, T ) → q0(x, y, T ) satisfies the
inequalities

−CA(T0)Ty
2

x
≤ q0(x, y, T )− qlinear(x, y, T ) ≤ 0 (4.10)

for x ≥ max{2y, T 2} and 0 < T < T0, and

0 ≤ q0(x, y, T )−
2m1,A(T )xy

σ2
A(T )

≤ CA(T0)Tx for x, y > 0, 0 < T < T0 . (4.11)



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 25

Proof. All the constants C1, C2, . . . , in the following can be chosen to depend only
on T0 and sup0≤t≤T0

A(t). We first observe from (2.39) and (3.13) that

g3,A(s, T )g1,A(s, T ) =
m1,A(T )

σ2
A(T )

, 0 < s < T . (4.12)

Next we note from (2.39) and (4.5) that for any T0 > 0, there are constants
C1, C2, C3 > 0 such that

C1s

T − s
≤ g1,A(s, T ) ≤

C2s

T − s
, 0 ≤ ∂g2,A(s, T )

∂s
≤ C3T ,

0 ≤ g2,A(s, T ) ≤ C3sT , for 0 < s < T, 0 < T ≤ T0 .
(4.13)

Evaluating the functional on the RHS of (4.2) at τ = Ty/x we conclude from (4.12),
(4.13) that

q0(x, y, T ) ≤
m1,A(T0)

2C1
[1+2C2+C3]

2xy

T
, 0 < T ≤ T0, x ≥ max{2y, T 2} . (4.14)

We also have from (4.12) and (4.13) that

g3,A(τ, T )

2
[y + g1(τ, T )x+ g2,A(τ, T )]

2 ≥ m1,A(T )

2σ2
A(T )

g1,A(τ, T )x
2

≥ C1τx
2

2m1,A(T0)T 2
.

(4.15)

It follows from (4.14) and (4.15) that there is a constant C4 such that any mini-
mizing τ = τ(x, y, T ) in (4.2) satisfies the inequality

0 < τ(x, y, T ) ≤ C4Ty

x
, 0 < T ≤ T0, x ≥ max{2y, T 2} . (4.16)

We have from (3.16) that

lim
τ→0

g2,A(τ, T )

g1,A(τ, T )
= m2,A(T )−

σ2
A(T )

m1,A(T )
. (4.17)

We also see from (3.17) that if A(·) is non-negative, the derivative of the function
s → g2,A(s, T )/g1,A(s, T ), 0 < s < T , is less than or equal to zero and bounded by
a constant times T . We conclude therefore from (4.17) there is a constant C5 such
that

−C5τT ≤ g2,A(τ, T )

g1,A(τ, T )
−
{
m2,A(T )−

σ2
A(T )

m1,A(T )

}
≤ 0 , 0 < τ < T ≤ T0 . (4.18)

The inequality (4.10) follows from (4.12), (4.16), and (4.18). Thus from the
upper bound in (4.18) we have that

q0(x, y, T ) ≤ inf
0<τ<T

g3,A(τ, T )

2

[
y +

1

g3,A(τ, T )

qlinear(x, y, T )

2y

]2
= qlinear(x, y, T ) .

(4.19)



26 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

Similarly from the lower bound in (4.18) we have upon using (4.16) that

q0(x, y, T ) ≥ inf
0<τ<T

g3,A(τ, T )

2

[
y

+
1

g3,A(τ, T )

{qlinear(x, y, T )

2y
− m1,A(T )

σ2
A(T )

C5C4T
2y

x

}]2
= qlinear(x, y, T )−

m1,A(T )

σ2
A(T )

2C5C4T
2y2

x
,

(4.20)

provided the term inside the curly braces is positive. Observe that the lower bound
in (4.10) is trivial if qlinear(x, y, T ) − CTy2/x ≤ 0, whence we need only consider
situations where qlinear(x, y, T )−CTy2/x is positive. Upon choosing C sufficiently
large, we see that the term of (4.20) inside the curly braces is then positive. Hence
we obtain the lower bound (4.10) for all x ≥ max{2y, T 2}.

We can argue similarly to obtain the bound (4.11). Thus since g2,A(·, ·) is non-
negative, we have that

q0(x, y, T ) ≥ inf
0<τ<T

g3,A(τ, T )

2

[
y +

1

g3,A(τ, T )

m1,A(T )x

σ2
A(T )

]2
=

2m1,A(T )xy

σ2
A(T )

. (4.21)

From (4.13) we have that g2,A(s, T ) ≤ C3T
2, 0 < s < T ≤ T0, for some constant

C3. Hence from (4.2) and (4.12) we have that

q0(x, y, T ) ≤ inf
0<τ<T

g3,A(τ, T )

2

[
y + C3T

2 +
1

g3,A(τ, T )

m1,A(T )x

σ2
A(T )

]2
=

2m1,A(T )x(y + C3T
2)

σ2
A(T )

.

(4.22)

□

We wish also to understand the behavior of ∂q0(x, y, T )/∂x and ∂2q0(x, y, T )/∂x
2

as x → ∞ and as T → 0. To do this we will show that the function [x, T ] →
q0(x, y, T ), defined by (2.28) or equivalently (4.2), can be obtained for [x, T ] in a
certain domain by the method of characteristics applied to solving the HJ equation
(2.29) with boundary condition (2.30). To implement the method of characteristics,
we first observe from (2.30) that ∂q0(x, y, T )/∂T = 0 at x = 0, whence (2.29) yields
the formula

u0(0, y, T ) =
∂q0(x, y, T )

∂x

∣∣∣
x=0

= −2λ(0, y, T ) =
2m1,A(T )

σ2
A(T )

[y + g2,A(T, T )] , (4.23)

where from (3.15) we define

g2,A(T, T ) = lim
s→T

g2,A(s, T ) = [σ2
A(T )−m2,A(T )]/m1,A(T ).

Next we differentiate the PDE (2.29) with respect to x to obtain the Burgers’
equation

∂u0(x, y, T )

∂T
+ [λ(x, y, T ) + u0(x, y, T )]

∂u0(x, y, T )

∂x

+
[
A(T ) +

1

σ2
A(T )

]
u0(x, y, T ) = 0 ,

(4.24)

for the function u0(x, y, T ) = ∂q0(x, y, T )/∂x. We seek to solve (4.24), with the
boundary condition u0(0, y, T ) = ∂q0(0, y, T )/∂x given by (4.23), by using the



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 27

method of characteristics. If s → [x(s), s], s > τ , is a characteristic with initial
condition x(τ) = 0, then (4.23) and (4.24) yield the ODE initial value problem

d

ds
u0(x(s), y, s) +

[
A(s) +

1

σ2
A(s)

]
u0(x(s), y, s) = 0 , s > τ,

u0(x(τ), y, τ) =
2m1,A(τ)

σ2
A(τ)

[y + g2,A(τ, τ)] .

(4.25)

The solution to (4.25) is

u0(x(s), y, s) =
2m1,A(s)[y + g2,A(τ, τ)]

σ2
A(s)

, s > τ . (4.26)

It follows from (4.24) and (4.26) that the characteristics are solutions to the ODE
initial value problem

dx(s)

ds
= λ(x(s), y, s) +

2m1,A(s)[y + g2,A(τ, τ)]

σ2
A(s)

, s > τ, x(τ) = 0 . (4.27)

From (2.15) we see that (4.27) is the same as

dx(s)

ds
=

[
A(s) +

1

σ2
A(s)

]
x(s) +

m1,A(s)

σ2
A(s)

[y + 2g2,A(τ, τ)− g2,A(s, s)] (4.28)

for s > τ , x(τ) = 0. The general solution to the ODE (4.28) is

x(s) = C
σ2
A(s)

m1,A(s)
− [y + 2g2,A(τ, τ)]m1,A(s)−m2,A(s) , (4.29)

where C is an arbitrary constant. The constant C is determined for the character-
istic by the initial condition x(τ) = 0. In the case A(·) ≡ 0 this yields the formula
x(s) = [s/τ − 1]y, s > τ , for the characteristic.

We have that
d

ds
g2,A(s, s) =

A(s)σ2
A(s)

m1,A(s)
. (4.30)

If we assume A(·) non-negative, it follows from (4.30) that the function s →
g2,A(s, s) is increasing. This implies the characteristics that are solutions to (4.28)
may meet, whence one cannot expect the HJ equation (2.29) to have a classical
(continuously differentiable) solution. We can make this more precise by consider-
ing characteristics s → x(τ, s), s > τ > 0, which are solutions to (4.28) with initial
condition x(τ, τ) = 0. The first variation Dτx(τ, s) = ∂x(τ, s)/∂τ , s > τ > 0, is
from (4.28), (4.30) the solution to the initial value problem

d

ds
Dτx(τ, s) =

[
A(s) +

1

σ2
A(s)

]
Dτx(τ, s) +

2m1,A(s)

σ2
A(s)

A(τ)σ2
A(τ)

m1,A(τ)
, s > τ ,

Dτx(τ, s)
∣∣∣
s=τ

= −m1,A(τ)

σ2
A(τ)

[y + g2(τ, τ)] .

(4.31)

We note that (4.31) is equivalent to

d

ds

[m1,A(s)

σ2
A(s)

Dτx(τ, s)
]
=

2m1,A(s)
2

σ4
A(s)

A(τ)σ2
A(τ)

m1,A(τ)
, s > τ ,

Dτx(τ, s)
∣∣∣
s=τ

= −m1,A(τ)

σ2
A(τ)

[y + g2(τ, τ)] .

(4.32)



28 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

Since Dτx(τ, s) < 0 at s = τ and the derivative on the LHS of (4.32) is non-
negative, we can have Dτx(τ, s) = 0 for some s > τ , from which point the solution
to (4.24) cannot be continued by using the method of characteristics.

When the method of characteristics does apply to obtain the solution of (4.24)
with boundary data (4.23), we may obtain a formula for ∂u0(x, y, T )/∂x along
characteristics similarly to how we obtained (4.26) for u0(x, y, T ). To see this first
note from (4.23) and (4.30) that

∂u0(x, y, T )

∂T

∣∣∣
x=0

=
d

dT
u0(0, y, T )

= −
[
A(T ) +

1

σ2
A(T )

]
u0(0, y, T ) + 2A(T ) .

(4.33)

Setting x = 0 in (4.24) and using (4.33) we conclude that

∂u0(x, y, T )

∂x

∣∣∣
x=0

= − 2A(T )σ2
A(T )

m1,A(T )[y + g2,A(T, T )]
. (4.34)

Differentiating (4.24) with respect to x, we obtain a partial differential equation for
v0(x, y, T ) = ∂u0(x, y, T )/∂x,

∂v0(x, y, T )

∂T
+ [λ(x, y, T ) + u0(x, y, T )]

∂v0(x, y, T )

∂x

+ v0(x, y, T )
2 + 2

[
A(T ) +

1

σ2
A(T )

]
v0(x, y, T ) = 0 .

(4.35)

From the method of characteristics applied to (4.35), we obtain using (4.34) the
ODE initial value problem

d

ds
v0(x(s), y, s) + v(x(s), y, s)2 + 2

[
A(s) +

1

σ2
A(s)

]
v0(x(s), y, s) = 0 , s > τ,

v0(x(τ), y, τ) = − 2A(τ)σ2
A(τ)

m1,A(τ)[y + g2(τ, τ)]
,

(4.36)
where s → x(s), s > τ , is the characteristic defined by (4.28). It follows from (4.36)
that the function s → 1/v0(x(s), y, s) is a solution to a linear differential equation,
whence we conclude that 1/v0(x(s), y, s), s > τ , is of the form

1

v0(x(s), y, s)
= C

σ4
A(s)

m1,A(s)2
− σ2

A(s) , s > τ , (4.37)

for some constant C. Choosing C in (4.37) to satisfy the initial condition (4.36),
we have then that

∂u0(x(s), y, s)

∂x
= −K(τ)

m1,A(s)
2

σ4
A(s)

/{
1 +K(τ)

[m1,A(s)
2

σ2
A(s)

− m1,A(τ)
2

σ2
A(τ)

]}
for s > τ , where K(τ) =

2A(τ)σ6
A(τ)

m1,A(τ)3[y + g2,A(τ, τ)]
.

(4.38)

Since the function s → m1,A(s)/σ
2
A(s) is decreasing, we see that the formula (4.38)

for ∂u0(x, y, T )/∂x can blow up to −∞. This is again a consequence of the fact
that we cannot in general expect a classical solution to (4.23), (4.24) when A(·) is
non-negative.



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 29

We assume that A(·) is non-negative. Observe that the condition x(τ) = 0 in
(4.28) implies that the constant C in (4.29) is given by the formula

C =
[y + 2g2(τ, τ)]m1,A(τ)

2

σ2
A(τ)

+
m1,A(τ)m2,A(τ)

σ2
A(τ)

=
[y + g2(τ, τ)]m1,A(τ)

2

σ2
A(τ)

+m1,A(τ) .

(4.39)

Substituting (4.39) into (4.29) gives the formula for the characteristic,

x(s) = [y + g2(τ, τ)]
{m1,A(τ)

2

σ2
A(τ)

σ2
A(s)

m1,A(s)
−m1,A(s)

}
+m1,A(s)m1,A(τ)

{ σ2
A(s)

m1,A(s)2
− σ2

A(τ)

m1,A(τ)2
}

−m1,A(s)
{m2,A(s)

m1,A(s)
− m2,A(τ)

m1,A(τ)

}
.

(4.40)

The first term on the RHS of (4.40) is bounded below by c1 [s/τ − 1] [y+g2,A(τ, τ)]
for 0 < τ < s ≤ T0, where constant c1 > 0 depends only on T0 and sup0≤t≤T0

A(t).
The remaining terms can be expressed as an integral over the interval [τ, s],

m1,A(s)

∫ s

τ

[m1,A(τ)

m1,A(s′)
− 1

] ds′

m1,A(s′)
= m1,A(s)f(s) . (4.41)

The function s → f(s), s ≥ τ , is decreasing and f(τ) = f ′(τ) = 0. We conclude
that the characteristic s → x(τ, s), s > τ , is an increasing function of s for s > τ
such that s− τ is sufficiently small. However it could decrease for s large, even to
0. We see from (4.40), (4.41) that

c1(s− τ)
[ [y + g2,A(τ, τ)]

τ
− C1(s− τ)

]
≤ x(τ, s) ≤ C2(s− τ)

[ [y + g2,A(τ, τ)]

τ
− c2(s− τ)

]
, 0 < τ < s ≤ T0 ,

(4.42)

where c1, c2, C1, C2 > 0 depend only on T0 and sup0≤t≤T0
A(t). It follows from

(4.42) that x(τ, s) > 0 for 0 < τ < s < min{τ + [y + g2,A(τ, τ)]/C1τ, T0}.
Next we obtain from the variation equation (4.31) conditions that imply charac-

teristics do not intersect. Thus setting y(τ, s) = m1,A(s)Dτx(τ, s)/σ
2
A(s), we have

from (4.31), (4.32)

∂

∂s
y(τ, s) ≤ C3τ

s2
, τ < s ≤ T0 , y(τ, τ) ≤ −C4

y + g2,A(τ, τ)

τ2
, (4.43)

for some positive constants C3 and C4 depending only on T0 and sup0≤t≤T0
A(t).

Integrating (4.43) we conclude that

y(τ, s) ≤ C3

[
1− τ

s

]
− C4

y + g2,A(τ, τ)

τ2
for τ < s ≤ T0 . (4.44)

It follows from (4.44) that Dτx(τ, s) < 0 for 0 < τ < s < min{τ + C4[y +
g2,A(τ, τ)]/C3τ, T0}.

Let Λ0 > 0 be a constant such that Λ0 < 1/C1, Λ0 < C4/C3 and consider the

function Ty(τ) = τ + Λ0[y + g2,A(τ, τ)]/τ , 0 < τ ≤ T0. Evidently Ty(·) ≥ T̃y(·),
where T̃y is the convex function T̃y(τ) = τ +Λ0y/τ . The infimum of T̃y is attained

at τ =
√
Λ0y and inf0<τ<∞ T̃y(τ) = 2

√
Λ0y. Since (4.30) implies that g2,A(τ, τ) ≤



30 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

Cτ2, we see that if 2
√
Λ0y ≤ T0 then inf0<τ<T0

Ty(τ) ≤ 2[1 + C ′]
√
Λ0y, where

C ′ ≥ 0 depends only on T0 and sup0≤t≤T0
A(t).

We wish to identify the largest domain Dy,T0 contained in {[x, T ] : x > 0, 0 <
T < T0} such that characteristics do not intersect within the domain. We have
already seen that if 2

√
Λ0y ≥ T0 then we may take Dy,T0

= {[x, T ] : x > 0, 0 <
T < T0}, so let us assume that 2

√
Λ0y < T0. Then Dy,T0

contains {[x, T ] : x >
0, 0 < T < 2

√
Λ0y}, so we just need to consider the situation 2

√
Λ0y < T < T0.

Then the equation

τ +
Λ[y + g2,A(T, T )]

τ
= T (4.45)

has two solutions provided that 2
√

Λ[y + g2,A(T, T )] < T . Since 2
√
Λ0y < T < T0

we may choose Λ1 < Λ0, depending only on T0 and sup0≤t≤T0
A(t), such that

4
√
Λ1[y + g2,A(T, T )] < T . The larger solution to (4.45) is

τ1,Λ,y(T ) =
T

2
+

T

2

{
1− 4Λ[y + g2,A(T, T )]

T 2

}1/2

. (4.46)

If Λ ≤ Λ1 then τ1,Λ,y(T ) satisfy the inequality

Λ[y + g2,A(T, T )]

T
≤ T − τ1,Λ,y(T ) ≤

2Λ[y + g2,A(T, T )]√
3T

. (4.47)

It follows from (4.30), (4.47) that

0 ≤ g2,A(T, T )− g2,A(τ1,Λ,y(T ), τ1,Λ,y(T )) ≤ CΛ[y + g2,A(T, T )] , (4.48)

for some constant C depending only on T0 and sup0≤t≤T0
A(t). We conclude from

(4.42), (4.48) there is a constant Λ2, depending only on T0 and sup0≤t≤T0
A(t), such

that if

2
√
Λ0y < T < T0 and 0 < x ≤ Λ2[y + g2,A(T, T )]

2

T 2
then [x, T ] ∈ Dy,T0

. (4.49)

We can make a similar argument to show that if 2
√
Λ0y < T < T0, then [x, T ] ∈

Dy,T0
for x sufficiently large. In that case we define τ2,Λ,y as the smaller solution

to the equation τ + Λy/τ = T . If Λ < Λ0, then

τ2,Λ,y(T ) =
T

2
− T

2

{
1− 4Λy

T 2

}1/2
. (4.50)

If Λ1 satisfies 4
√
Λ1y < T , then for Λ < Λ1 we have that

Λy

T
≤ τ2,Λ,y(T ) ≤

2Λy√
3T

. (4.51)

From (4.42) there is a constant Λ3, depending only on T0 and sup0≤t≤T0
A(t), such

that

if 2
√
Λ0y < T < T0 and x ≥ Λ3T

2, then [x, T ] ∈ Dy,T0 . (4.52)

We define the domain

Uy,T0 = {[τ, s] : 0 < τ < T0, τ < s < min[Ty(τ), T0]} . (4.53)

The mapping [τ, s] → [x(τ, s), s] is a diffeomorphism from Uy,T0
onto a domain

Dy,T0 , which has the properties (4.49), (4.52). The fact that the mapping is
onto follows from the intermediate value theorem since we see from (4.42) that
limτ→0 x(τ, s) = ∞ for all 0 < s < T0. It is one-one since Dτx(τ, s) < 0 for
[τ, s] ∈ Uy,T0

.



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 31

Proposition 4.2. Assume the function A : [0,∞) → R is continuous and non-
negative. For [x, T ] ∈ Dy,T0 let [τ, T ] ∈ Uy,T0 be such that x(τ, T ) = x, and define
q0(x, y, T ) by

q0(x, y, T ) =
g3,A(τ, T )

2
[y + g1,A(τ, T )x+ g2,A(τ, T )]

2
. (4.54)

Then the function [x, T ] → q0(x, y, T ) is a C1 solution of the HJ equation (2.29)
on Dy,T0

and satisfies the boundary condition limx→0 q0(x, y, T ) = 0, 0 < T <
T0. Furthermore, the function x → q0(x, y, T ) is C2 on Dy,T0 and the derivatives
∂q0(x, y, T )/∂x, ∂

2q0(x, y, T )/∂x
2 are given respectively by the formulas (4.26) and

(4.38).
Also τ = τ(x, y, T ) in (4.54) is the unique minimizer in the variational problems

(2.28), (4.2) for [x, T ] with x > 0 and 0 < T < T0, in the following regions: (a) all
x > 0 if 2

√
Λ0y ≥ T , otherwise (b) 0 < x ≤ Λy[y + g2,A(T, T )]/T

2, (c) x ≥ T 2/Λ,
where Λ > 0 is chosen sufficiently small depending only on T0 and sup0≤t≤T0

A(t).
Therefore if [x, T ] is in one of the regions (a), (b), (c) the functions (4.2) and
(4.54) are identical.

Proof. All constants in the following can be chosen to depend only on T0 and
sup0≤t≤T0

A(t). To show regularity of the function [x, T ] → q0(x, y, T ) we first
differentiate (4.54) with respect to x. The resulting formula for ∂q0(x, y, T )/∂x
involves gj,A(τ, T ), j = 1, 2, 3, and their first derivatives with respect to τ . It
also involves ∂τ(x, y, T )/∂x = [Dτx(τ, T )]

−1, which we see from (4.31), (4.32) is
a continuous function of [τ, T ] and hence of [x, T ]. We conclude that the function
x → q0(x, y, T ) is differentiable and the function [x, T ] → ∂q0(x, y, T )/∂x continu-
ous. We can make a similar argument to see that the function T → q0(x, y, T ) is
differentiable and the function [x, T ] → ∂q0(x, y, T )/∂T continuous. In that case
we need to show the continuity of the function [x, T ] → ∂τ(x, y, T )/∂T , which is
given by the formula

∂τ(x, y, T )

∂T
= −DTx(τ, T )

Dτx(τ, T )
. (4.55)

Evidently DTx(τ, T ) is given by the RHS of (4.28) with s = T, x(s) = x and hence
is a continuous function of [x, T ]. We conclude that the function [x, T ] → q0(x, y, T )
is C1 on Dy,T0

.
To show that the function (4.54) is a solution to the HJ equation (2.29), we

proceed by the standard method [10], writing (2.29) as

∂q0(x, y, T )

∂T
+H

(
x, y,

∂q0
∂x

, T
)
= 0 , (4.56)

where the Hamiltonian is

H(x, y, p, T ) = λ(x, y, T )p+
1

2
p2 . (4.57)

The corresponding Hamiltonian equations of motion are

dx

ds
=

∂H(x, y, p, s)

∂p
,

dp

ds
= −∂H(x, y, p, s)

∂x
. (4.58)

We solve (4.58) with initial conditions

x(τ) = 0 , H(0, y, p(τ), τ) = 0 . (4.59)



32 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

Note that the initial condition (4.59) for p(·) is the same as in (4.25). If we solve
the second equation in (4.58) with initial condition (4.59) we obtain

p(τ, s) =
2m1,A(s)

σ2
A(s)

[y + g2(τ, τ)] , s > τ , (4.60)

corresponding to (4.26). Taking p(τ, s) to be given by (4.60), the first equation
in (4.58) becomes identical to the characteristic equation (4.27). We define the
function w : Uy,T0

→ R by

∂

∂s
w(τ, s) = −H(x(τ, s), p(τ, s), s) + p(τ, s)

∂H(x(τ, s), p(τ, s), s)

∂p
(4.61)

for s > τ and initial condition w(τ, τ) = 0. Then by standard theory the function
q0(·, y, ·) defined on Dy,T0

by q0(x(τ, s), y, s) = w(τ, s) is a solution to (4.56) and
p(τ, s) = ∂q0(x(τ, s), y, s)/∂x.

We see that

w(τ, s) =
g3,A(τ, s)

2
[y + g1,A(τ, s)x(τ, s) + g2,A(τ, s)]

2
, (4.62)

by verifying that the RHS of (4.62) is a solution to the differential equation (4.61),
where x(τ, s) and p(τ, s) are given by (4.29), (4.39), (4.60) and with initial condition
w(τ, τ) = 0. To obtain the formula (4.38) for ∂2q0/∂x

2 we observe that

∂

∂τ

∂q0(x(τ, s), y, s)

∂x
=

∂2q0(x(τ, s), y, s)

∂x2

∂x(τ, s)

∂τ
=

∂p(τ, s)

∂τ
, (4.63)

and use the formulas (4.40), (4.60). Note that we may choose Λ0 sufficiently small,
depending only on T0, such that the denominator in the formula (4.38) is positive
if [τ, s] ∈ Uy,T0

.
Finally we consider for which [x, T ] ∈ Dy,T0 that τ = τ(x, y, T ) is the minimizer

for (4.2). We apply the standard verification theorem to paths [x(s), s] and τ <
s ≤ T , with x(τ) = 0 and x(T ) = x, which lie in Dy,T0

. Using the fact that q0 is a
C1 solution of (4.56), (4.57) on Dy,T0

and q(0, y, τ) = 0, we have that

q0(x, y, T )

=

∫ T

τ

d

ds
q0(x(s), y, s) ds

=

∫ T

τ

∂q0(x(s), y, s)

∂x

[dx(s)
ds

− λ(x(s), y, s)
]
− 1

2

∫ T

τ

[∂q0(x(s), y, s)
∂x

]2
ds

≤ 1

2

∫ T

τ

[dx(s)
ds

− λ(x(s), y, s)
]2

ds .

(4.64)

If τ = τ(x, y, T ) and x(s) = x(τ, s), τ < s < T , then from (4.26) and (4.27)
we obtain equality in (4.64). Let F0(x, y, τ, T ) be the function on the RHS of
(4.2). We wish to find [x, T ] such that q0, defined by (4.54), satisfies q0(x, y, T ) =
inf0<τ<T F0(x, y, τ, T ). We shall identify a subset Sx,y,T of (0, T ) such that
F0(x, y, τ, T ) > F0(x, y, τ(x, y, T ), T ) = q0(x, y, T ) for τ ∈ Sx,y,T . Hence it is
necessary only to consider τ ∈ (0, T )− Sx,y,T . We have already observed that the
variational problem (2.28) with fixed τ is quadratic and has the unique solution



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 33

(2.37) given by Γ(τ, s, T, x) = a(τ, s, T )x+ b(τ, s, T ), τ < s < T , where

σ2
A(T )a(τ, s, T ) = m1,A(s)σ

2
A(s, T ) [g1,A(s, T )− g1,A(τ, T )] ,

σ2
A(T )b(τ, s, T ) = m1,A(s)σ

2
A(s, T ) [g2,A(s, T )− g2,A(τ, T )] .

(4.65)

In view of the verification result (4.64), if we show that the path s → Γ(τ, s, T, x),
τ < s < T , lies in Dy,T0

when τ ∈ (0, T )−Sx,y,T , then it follows that τ = τ(x, y, T )
is the unique minimizer for (4.2). If T < 2

√
Λ0y then [x, T ] ∈ Dy,T0

for all x > 0.
Since the functions τ → g1,A(τ, T ), g2,A(τ, T ), 0 < τ < T , are increasing when A(·)
is non-negative, we see from (4.65) that the path s → Γ(τ, s, T, x), τ < s < T , lies
in Dy,T0 when 0 < τ < T . Hence τ = τ(x, y, T ) is the unique minimizer for (4.2)
when x > 0 and 0 < T < 2

√
Λ0y.

For x, y ≥ 0 let

q̃0(x, y, T ) = min
0<τ<T

g3,A(τ, T )

2
[y + g1,A(τ, T )x]

2
. (4.66)

We have from (4.2) that

q̃0(x, y, T ) ≤ inf
0<τ<T

F0(x, y, τ, T ) ≤ q̃0(x, y + g2,A(T, T ), T ) . (4.67)

Using identity (4.12) we see that the minimizing τ for the RHS of (4.66) is

g3,A(τ, T ) =
m1,A(T )x

σ2
A(T )y

. (4.68)

Substituting (4.68) into the RHS of (4.66) yields the formula

q̃0(x, y, T ) =
2m1,A(T )xy

σ2
A(T )

. (4.69)

It is intuitively clear from (4.67) that the function q0 defined by (4.54) on Dy,T0

should satisfy the inequality

q0(x, y, T ) ≤ q̃0(x, y + g2,A(T, T ), T ) , [x, T ] ∈ Dy,T0 . (4.70)

To see this observe that the function v(·, ·) defined by v(τ, s) = q̃0(x(τ, s), y +
g2,A(s, s), s) − w(τ, s), 0 < τ < s < T0, with w(·, ·) as in (4.62) satisfies the differ-
ential equation

∂v(τ, s)

∂s
= 2A(s)x(τ, s)− 2m1,A(s)

2

σ4
A(s)

[g2,A(s, s)− g2,A(τ, τ)]
2 . (4.71)

In deriving (4.71) we use the fact that w(τ, s) = q0(x(τ, s), y, s) and that the func-
tion [x, T ] → q0(x, y, T ) is a solution to the HJ equation (2.29). Integrating (4.71)
over an interval [τ, T ] with 0 < τ < T ≤ T0, we obtain using (4.30), (4.42) the
inequality,

v(τ, T ) ≥ c1
y + g2,A(τ, τ)

τ

∫ T

τ

(s− τ)A(s) ds− 2

∫ T

τ

[ ∫ s

τ

A(s′) ds′
]2

ds , (4.72)

provided 0 < s− τ ≤ [y + g2,A(τ, τ)]/2C1τ . Using the inequality[ ∫ s

τ

A(s′) ds′
]2

≤ 2
[

sup
τ<s′<s

A(s′)
] ∫ s

τ

(s′ − τ)A(s′) ds′ , (4.73)

we see that the RHS of (4.72) is non-negative for 0 < T − τ ≤ [y+ g2,A(τ, τ)]/C2τ ,
provided C2 is chosen sufficiently large.



34 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

Consider now the function f(·) defined by

f(λ) =
λ

2
[y1 + x1/λ]

2 , where x1, y1 > 0. (4.74)

Evidently the minimizing λ = λmin and minimizer are

λmin = x1/y1, f(λmin) = 2x1y1 . (4.75)

We have furthermore that

f(λmin/8) = f(8λmin) ≥ 5x1y1 . (4.76)

We consider [x, T ] ∈ Dy,T0 which satisfies (4.49). Let us assume now that
g2,A(T, T ) ≤ y. Then we have from (4.70) that

q0(x, y, T ) ≤ 2q̃0(x, y, T ) . (4.77)

Suppose τ ∈ (0, T ) lies outside the region

m1,A(T )x

8σ2
A(T )y

≤ g3,A(τ, T ) ≤
8m1,A(T )x

σ2
A(T )y

. (4.78)

Then from (4.68), (4.75), upon setting y1 = y and x1 = m1,A(T )x/σ
2
A(T ) in (4.76),

we obtain the inequality

F0(x, y, τ, T ) ≥ 5q̃0(x, y, T )/2 . (4.79)

It follows from (4.77), (4.79) that τ ∈ Sx,y,T if τ does not satisfy (4.78). Observe
from (4.49) that x ≤ 4Λ2y

2/T 2, which implies that Tx/y ≤ 4Λ2y/T ≤ Λ2T/Λ0.
Using (4.12), (4.13), (4.78) it follows on choosing Λ2 sufficiently small, that τ ∈
(0, T )− Sx,y,T satisfies the inequalities

T

2
< τ < T and C1

Tx

y
≤ T − τ ≤ C2

Tx

y
, (4.80)

for some constants C1, C2 > 0. We have from (4.5), (4.6), (4.65) there are constants
c3, C3 > 0, depending only on T0, such that

c3(s− τ)
x

T − τ
≤ Γ(τ, s, T, x) ≤ C3(s− τ)

{ x

T − τ
+ T − s

}
, (4.81)

for 0 < τ < s < T . Hence we have from (4.80), (4.81) that if τ ∈ (0, T ) − Sx,y,T

then

Γ(τ, s, T, x) ≤ C3

{
x+

1

4
(T − τ)2

}
≤ C3x

{
1 +

C2
2T

2x

4y2

}
≤ C3x{1 + C2

2Λ2} .

(4.82)

We conclude there exists Λ > 0, depending only on T0, such that if x ≤ Λy2/T 2 and
τ ∈ (0, T )− Sx,y,T then the path s → Γ(τ, s, T, x), τ < s < T , lies in Dy,T0

. It fol-
lows that τ = τ(x, y, T ) is the unique minimizer for the function τ → F0(x, y, τ, T ),
0 < τ < T , when x ≤ Λy2/T 2.

Next we consider the case g2,A(T, T ) > y and consider [x, T ] ∈ Dy,T0 which
satisfies (4.49). We may estimate τ(x, y, T ) from (4.42). Thus we have from (4.42),



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 35

(4.48) that

Tx

2C2[y + g2,A(T, T )]
≤ T − τ(x, y, T ) ≤ 2Tx

c1[y + g2,A(T, T )]

if x ≤ Λ[y + g2,A(T, T )]
2

T 2
,

(4.83)

provided Λ ≤ Λ2 is chosen sufficiently small. From (4.12), (4.48), (4.70) we see that
(4.83) implies

q̃0(x, {y + g2,A(T, T )}/3, T ) ≤ F0(x, y, τ(x, y, T ), T )

≤ q̃0(x, y + g2,A(T, T ), T ) ,
(4.84)

if Λ is sufficiently small. We have already observed that if q0 is defined by (4.2)
then limy→0 q0(x, y, T ) = 0. It follows from the lower bound (4.84) that

inf
0<τ<T

F0(x, y, τ, T ) ̸= F0(x, y, τ(x, y, T ), T )

if y is sufficiently small.
We show for sufficiently small Λ > 0 that

if x ≤ Λy[y + g2,A(T, T )]

T 2
, then

inf
0<τ<T

F0(x, y, τ, T ) = F0(x, y, τ(x, y, T ), T ) .
(4.85)

To see this observe from (4.5) that

c3T

(T − s)2

∫ T

s

(T − s′)A(s′) ds′ ≤ ∂g2,A(s, T )

∂s

≤ C3T

(T − s)2

∫ T

s

(T − s′)A(s′) ds′,

(4.86)

where c3, C3 are constants. Similarly we have from (4.30) that

c4

∫ T

0

sA(s) ds ≤ g2,A(T, T ) ≤ C4

∫ T

0

sA(s) ds . (4.87)

We define δ, 0 < δ < 1, by∫ (1−δ)T

0

sA(s) ds =
1

2

∫ T

0

sA(s) ds . (4.88)

It follows from (4.87), (4.88) that δ satisfies the inequality

δ ≥ g2,A(T, T )

2∥A∥∞C4T 2
. (4.89)

Integrating (4.86) over the interval 0 < s < τ we have from the lower bound the
inequality,

g2,A(τ, T ) ≥ c3

[ ∫ τ

0

sA(s) ds+
τ

T − τ

∫ T

τ

(T − s)A(s) ds
]
. (4.90)

We assume first that δ ≥ 1/2, whence (4.87), (4.90) imply that g2,A(τ, T ) ≥
c3g2,A(T, T )/2C4 if τ ≥ T/2. We have then from (4.12), (4.13) that

F0(x, y, τ, T ) ≥
g3,A(τ, T )g2,A(τ, T )y

2
≥ c5(T − τ)g2,A(T, T )y

T 2
(4.91)



36 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

if τ ≥ T/2, for some constant c5 > 0. If 0 < τ < T/2 then we again see from
(4.87), (4.90) that g2,A(τ, T ) ≥ c3τg2,A(T, T )/2C4(T − τ). In this case we obtain
the inequality

F0(x, y, τ, T ) ≥
c6g2,A(T, T )y

T
if τ < T/2 , (4.92)

where c6 > 0 is constant. It is easy to see from (4.69), the upper bound (4.84)
and (4.92) that if 0 < τ < T/2 then F0(x, y, τ, T ) > F0(x, y, τ(x, y, T ), T ) provided
x satisfies (4.85) with Λ > 0 sufficiently small. Similarly from (4.91) we conclude
that F0(x, y, τ, T ) > F0(x, y, τ(x, y, T ), T ) if T − τ ≥ C7Tx/y, for some constant
C7. Hence if we show that the paths s → Γ(τ, s, T, x), τ < s < T , lie in Dy,T0

if
T−τ < C7Tx/y then it follows that inf0<τ<T F0(x, y, τ, T ) = F0(x, y, τ(x, y, T ), T ).
We see from (4.49) (4.81), (4.82) that this is the case provided x satisfies (4.85)
with Λ > 0 sufficiently small.

Next we assume that δ < 1/2, whence we have from (4.88) and (4.90) that

g2,A(τ, T ) ≥ c3
τδ

2(T − τ)

∫ T

0

sA(s) ds if T − τ ≥ δT . (4.93)

Then using (4.12), (4.13) again together with (4.87), (4.89) we conclude from (4.93)
the inequality

F0(x, y, τ, T ) ≥ c8
g2,A(T, T )

2y

T 3
if T − τ ≥ δT , (4.94)

where c8 > 0 is constant. It follows from (4.69), the upper bound (4.84) and
(4.94) that if T − τ ≥ δT then F0(x, y, τ, T ) > F0(x, y, τ(x, y, T ), T ) provided x
satisfies (4.85) with Λ > 0 sufficiently small. If T − τ < δT then g2,A(τ, T ) ≥
c3g2,A(T, T )/2C4, whence the inequality (4.91) holds provided T−τ < δT . We may
argue now as in the previous paragraph to conclude that inf0<τ<T F0(x, y, τ, T ) =
F0(x, y, τ(x, y, T ), T ) if x satisfies (4.85) with Λ > 0 sufficiently small.

Finally we show that for 2
√
Λ0y < T and Λ > 0 sufficiently small, if x ≥ T 2/Λ

then inf0<τ<T F0(x, y, τ, T ) = F0(x, y, τ(x, y, T ), T ). Observe that in this case x ≥
max{2y, T 2}, whence (4.16) gives a bound on any minimizing τ for the function
τ → F0(x, y, τ, T ), 0 < τ < T . Since x ≥ T 2/Λ, this implies 0 < τ ≤ C4Λy/T .
Hence if we show that all paths s → Γ(τ, s, T, x), τ < s < T , lie in Dy,T0

if
0 < τ ≤ C4Λy/T it follows that inf0<τ<T F0(x, y, τ, T ) = F0(x, y, τ(x, y, T ), T ).
We see from (4.42) there exists Λ3 > 0 such that that if 0 < τ ≤ Λ3y/T = τ∗ then
the characteristic s → x(τ, s), τ < s ≤ T , lies in Dy,T0

. Furthermore x(τ∗, s) ≤
C5T (s− τ), where C5 is constant. Since x ≥ T 2/Λ the lower bound (4.81) implies
that Γ(τ, s, T, x) ≥ c3T (s − τ)/Λ, τ < s < T . We conclude that if Λ < c3/C5 then
the path s → Γ(τ, s, T, x), τ < s < T , lies in Dy,T0 . □

Remark 4.3. The interval (b) in the statement of Proposition 4.2 is more or less
an optimal interval for which the method of characteristics yields the minimizer in
(4.2). One can see this by choosing A(·) to have support in a small neighborhood
of T . Thus for any δ, 0 < δ < 1/2, we set the function A(·) = Aδ(·), where
Aδ(s) = A(T )[1 − (T − s)/δT ] for 0 ≤ T − s ≤ δT and Aδ(s) = 0 for T − s > δT .
Then the ratio g2,A(τ, T )/g2,A(T, T ) ≃ 1/N if T − τ ≃ NδT .

Corollary 4.4. Assume the function A : [0,∞) → R is continuous and non-
negative. Then for any T0 > 0, there exist constants C1, C2 > 0, depending only on



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 37

T0 and sup0≤t≤T0
A(t), such that the function (x, T ) → q0(x, y, T ) defined by (4.54)

satisfies the inequalities

0 ≤ ∂q0(x, y, T )

∂x
− 2m1,A(T )y

σ2
A(T )

≤ C1Ty
2

x2
, −C1Ty

2

x3
≤ ∂2q0(x, y, T )

∂x2
≤ 0 , (4.95)

for [x, T ] ∈ Dy,T0 , x ≥ C2 max{2y, T 2}, 0 < T < T0. In addition for each T0 > 0
there is a constant C > 0 such that

0 ≤ ∂q0(x, y, T )

∂x
− 2m1,A(T )y

σ2
A(T )

≤ CT, −CT

y
≤ ∂2q0(x, y, T )

∂x2
≤ 0, (4.96)

for [x, T ] ∈ Dy,T0
, x, y > 0, 0 < T < T0.

Proof. Since [x, T ] ∈ Dy,T0
we may use the formulas (4.26), (4.38) to show (4.95),

(4.96). For (4.95) we use the inequality (4.16) for τ(x, y, T ). The first inequality
follows from (4.26) and the fact that 0 ≤ g2,A(τ, τ) ≤ Cτ2, where C is constant. To
obtain the second inequality we observe that the function K(τ) of (4.38) satisfies
an inequality K(τ) ≤ C3τ

3/y, where C3 is constant. Hence for τ = τ(x, y, T ) we
have using (4.16) that

K(τ)m1,A(τ)
2

σ2
A(τ)

≤ C3m1,A(T0)
2τ2

y

≤ C3m1,A(T0)
2C2

4T
2y

x2

≤ C3m1,A(T0)
2C2

4

2C2
2

(4.97)

for x ≥ C2 max{2y, T 2}, 0 < T ≤ T0. We choose C2 large enough so that the final
expression on the RHS of (4.97) is less than 1/2. We have then from (4.38) the
lower bound

∂2q(x, y, T )

∂x2
≥ −2C3m1,A(T0)

2τ3

T 2y
, (4.98)

whence the second inequality of (4.95) follows on using the bound (4.16) for τ =
τ(x, y, T ) in (4.98). To prove (4.96) we again use the formulas (4.26), (4.38), ob-
serving that τ(x, y, T ) < T . □

5. Uniform bounds on qε and its derivatives

In this section our goal is to show that the bounds on q0 and its first two space
derivatives obtained in Proposition 4.1 and Corollary 4.4 can be extended to qε
with ε > 0. First we prove results for ∂qε(x, y, T )/∂x analogous to the bounds on
qε(x, y, T ) obtained in Proposition 3.3. In fact the lower bound in (5.1) implies the
lower bound in (3.28), and the upper bound in (5.1) implies (3.29).

Proposition 5.1. Assume the function A : [0,∞) → R is continuous non-negative,
and let qε(x, y, T ) be defined by (2.18) and (2.21). Then

2m1,A(T )y

σ2
A(T )

≤ ∂qε(x, y, T )

∂x
≤ 2

[
1− m2,A(T )

σ2
A(T )

]
+

2m1,A(T )y

σ2
A(T )

, x, y, T > 0 . (5.1)

Proof. Letting uε(x, y, T ) = ∂qε(x, y, T )/∂x, we note from Proposition 3.1 that the
function (x, T ) → uε(x, y, T ) is continuous in the domain {(x, T ) : x, T > 0}, with
continuous derivatives ∂uε(x, y, T )/∂x, ∂uε(x, y, T )/∂T , ∂2uε(x, y, T )/∂x

2. Fur-
thermore (x, T ) → uε(x, y, T ) is continuous up to the boundary {(x, T ) : x = 0, T >



38 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

0}. Differentiating (2.22) with respect to x, we see using (2.15) that uε(x, y, T ) is
a solution to the diffusive Burger’s PDE

∂uε(x, y, T )

∂T
+ [λ(x, y, T ) + uε(x, y, T )]

∂uε(x, y, T )

∂x

+
[
A(T ) +

1

σ2
A(T )

]
uε(x, y, T )

=
ε

2

∂2uε(x, y, T )

∂x2
, x, y, T > 0 .

(5.2)

It follows from (5.2) that

∂vε(x, y, T )

∂T
+ [λ(x, y, T ) + uε(x, y, T )]

∂vε(x, y, T )

∂x
=

ε

2

∂2vε(x, y, T )

∂x2
,

where vε(x, y, T ) =
σ2
A(T )uε(x, y, T )

m1,A(T )
.

(5.3)

From Proposition 3.2 it follows that the function (x, T ) → vε(x, y, T ) is non-
negative. Since qε(0, y, T ) = 0 we also have from the lower bound (3.28) of Propo-
sition 3.3 that vε(0, y, T ) ≥ 2y, y, T > 0.

Using Ito’s lemma and the martingale optional sampling theorem, we have that

vε(x, y, T ) = E
[
vε(X

∗
ε (s ∨ τ∗ε,x,T,K), y, s ∨ τ∗ε,x,T,K) | X∗

ε (T ) = x
]

(5.4)

for 0 < s < T , where the stopping time τ∗ε,x,T,K is defined in the proof of Lemma

2.4. In view of the non-negativity of vε and lower bound on vε(0, y, ·), we conclude
from (5.4) that

vε(x, y, T ) ≥ E
[
vε(0, y, τ

∗
ε,x,T ); τ

∗
ε,x,T = τ∗ε,x,T,K > s

]
≥ 2yP

(
τ∗ε,x,T = τ∗ε,x,T,K > s

)
, 0 < s < T .

(5.5)

Observe that

{τ∗ε,x,T > s} = {τ∗ε,x,T = τ∗ε,x,T,K > s} ∪ {τ∗ε,x,T,K > s, X∗
ε (τ

∗
ε,x,T,K) = K} . (5.6)

As in the proof of Lemma 2.4 we use the inequality X∗
ε (·) ≤ Xε(·), where Xε(·)

is given by (2.16), to show that lim supK→∞ P (τ∗ε,x,T,K > s, X∗
ε (τ

∗
ε,x,T,K) = K) ≤

lim supK→∞ P (sups<s′<T Xε(s
′) > K) = 0. Letting K → ∞ in (5.5) we then have

from (5.6) that vε(x, y, T ) ≥ 2yP (τ∗ε,x,T > s) for 0 < s < T . Since Lemma 2.4

implies that lims→0 P (τ∗ε,x,T > s) = 1 we conclude that vε(x, y, T ) ≥ 2y, whence

the lower bound in (5.1).
To obtain the upper bound in (5.1) we define for h > 0 a function qε,h by

qε,h(x, y, T ) = qε

(
x+

σ2
A(T )

m1,A(T )
h, y, T

)
x, T > 0 . (5.7)

The function (x, T ) → qε,h(x, y, T ) is also a solution to (2.22), whence the function
vε,h(x, y, T ) = [qε,h(x, y, T )− qε(x, y, T )] /h is a solution to the PDE

∂vε,h(x, y, T )

∂T
+ λ(x, y, T )

∂vε,h(x, y, T )

∂x

+
1

2

[∂qε,h(x, y, T )
∂x

+
∂qε(x, y, T )

∂x

]∂vε,h(x, y, T )
∂x

=
ε

2

∂2vε,h(x, y, T )

∂x2
, x, T > 0 .

(5.8)



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 39

Let X∗
ε,h(·) denote solutions to the backwards in time SDE (2.25) with µε given by

µε(x, y, T ) = λ(x, y, T ) +
1

2

[∂qε,h(x, y, T )
∂x

+
∂qε(x, y, T )

∂x

]
. (5.9)

Letting τ∗ε,h,x,T,K be the first exit time for X∗
ε,h(s), s < T , with terminal condition

X∗
ε,h(T ) = x from the interval (0,K) we have again from Ito’s lemma and the

martingale optional sampling theorem the identity

vε,h(x, y, T ) = E
[
vε,h(X

∗
ε,h(s ∨ τ∗ε,h,x,T,K), y, s ∨ τ∗ε,h,x,T,K) | X∗

ε,h(T ) = x
]

(5.10)

for 0 < s < T . We can simplify the expression (5.10) by taking K → ∞. Observe
that∣∣E [

vε,h(K, y, τ∗ε,h,x,T,K); τ∗ε,h,x,T,K > s,X∗
ε (τ

∗
ε,h,x,T,K) = K | X∗

ε,h(T ) = x
]∣∣

≤ sup
s<s′<T

|vε,h(K, y, s′)|P
(

sup
s<s′<T

X∗
ε,h(s

′) > K | X∗
ε,h(T ) = x

)
.

(5.11)
Since the drift µε of (5.9) satisfies µε(x, y, T ) ≥ λ(x, y, T ), we see that the proba-
bility on the RHS of (5.11) is bounded by P (sups<s′<T Xε(s

′) > K | Xε(T ) = x),
where Xε(·) is given by (2.16). We may bound this latter probability by us-
ing the reflection principle (2.46), whence the probability converges to zero as
K → ∞ like exp[−cK2] for some constant c > 0. From Proposition 3.3 we see
that sups<s′<T |vε,h(K, y, s′)| is bounded linearly in K as K → ∞. We conclude
that the RHS of (5.11) converges to 0 as K → ∞. Letting K → ∞ in (5.10) we
have then that

vε,h(x, y, T ) = E
[
vε,h(0, y, τ

∗
ε,h,x,T ); τ

∗
ε,h,x,T > s | X∗

ε,h(T ) = x
]

+ E
[
vε,h(X

∗
ε,h(s), y, s); τ

∗
ε,h,x,T < s | X∗

ε,h(T ) = x
]
,

(5.12)

for 0 < s < T , where τ∗ε,h,x,T is the first hitting time at 0 for the diffusion X∗
ε,h(s),

s < T , with terminal condition X∗
ε,h(T ) = x.

It is easy to bound from above the first expectation on the RHS of (5.12) by
using Proposition 3.3. Thus we have from the upper bound (3.29) the inequality
vε,h(0, y, τ) ≤ 2g2,A(τ, τ)+2y, τ > 0, where g2,A(τ, τ) = [σ2

A(τ)−m2,A(τ)]/m1,A(τ)
has derivative (4.30). Since A(·) is non-negative, whence the function τ → g2,A(τ, τ)
is increasing, we see that the first expectation on the RHS of (5.12) is bounded above
by 2g2,A(T, T ) + 2y for all s satisfying 0 < s < T . We shall show that the limit of
the second expectation on the RHS of (5.12) converges to 0 as s → 0. The upper
bound in (5.1) follow then by first letting s → 0 in (5.12) and then h → 0.

Using the bound (3.9) of Proposition 3.2, we see that the second expectation on
the RHS of (5.12) is bounded in absolute value by

C(T )y

sh
E
[
X∗

ε,h(s) + sh+ s2; τ∗ε,h,x,T < s | X∗
ε,h(T ) = x

]
, 0 < s < T, (5.13)

where the constant C(T ) depends only on T . To estimate the expression (5.13) we
use the lower bound (5.1) which has been already proven. Let Xε,linear(·) denote
solutions to the SDE (2.25) with µε given by

µε(x, y, T ) = λ(x, y, T ) +
∂qlinear(x, y, T )

∂x
= λ(x,−y, T ) . (5.14)

If X∗
ε,h(T ) = Xε,linear(T ) = x then X∗

ε,h(s) ≤ Xε,linear(s) for all τ∗ε,h,x,T < s <

T . Letting τε,linear,x,T be the first hitting time at 0 for Xε,linear(s), s < T , with



40 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

Xε,linear(T ) = x, we see that τ∗ε,h,x,T ≥ τε,linear,x,T . Hence we have that

E
[
X∗

ε,h(s); τ
∗
ε,h,x,T < s | X∗

ε,h(T ) = x
]

≤ E [Xε,linear(s); τε,linear,x,T < s | Xε,linear(T ) = x] .
(5.15)

Since the drift (5.14) is linear the SDE (2.25) can be explicitly solved in this case
and the solution is obtained by replacing y by −y in the formula (2.16). Thus we
have that

Xε(s) = xclass(s)−
√
ε

σ2
A(s)

m1,A(s)
Z(s), s < T , (5.16)

where Z(·) is defined in (2.16), and from (2.12) we have that

σ2
A(T )xclass(s) = xm1,A(s, T )σ

2
A(s)− ym1,A(s)σ

2
A(s, T )

+m1,A(s, T )m2,A(s, T )σ
2
A(s)−m2,A(s)σ

2
A(s, T ) .

(5.17)

We see from (5.17) that lims→0 xclass(s) = −y, whence there exists s0 with 0 <
s0 < T such that xclass(s) ≤ −y/2 for 0 < s ≤ s0. It follows then from (5.16) that

P (τε,linear,x,T < s) ≤ P
(

inf
s<s′<s0

s′Z(s′) > cy/
√
ε
)
, 0 < s ≤ s0 , (5.18)

for some constant c > 0 depending only on s0. The variables sZ(s), s < T , are
Gaussian with zero mean and variance bounded above by C(T )s for some con-
stant C(T ) depending only on T . Hence the probability on the RHS of (5.18)
is bounded above by exp[−c/s] for some constant c > 0. Observing also that
sup0<s<T E

[
Xε,linear(s)

2 | Xε,linear(T ) = x
]
< ∞, we conclude from (5.18) and the

Schwarz inequality applied to the RHS of (5.15) that the expression (5.13) converges
to 0 as s → 0. □

Proposition 5.2. Assume the function A : [0,∞) → R is continuous and non-
negative. Then for any T0 > 0, there exists a constant C > 0, depending only
on T0 and sup0≤t≤T0

A(t), such that the function (x, T ) → qε(x, y, T ) satisfies the
inequality

−CTy2

x
≤ qε(x, y, T )− qlinear(x, y, T ) (5.19)

for xy ≥ εT , x ≥ max{2y, T 2}, 0 < T < T0, and the inequality

qε(x, y, T )−
2m1,A(T )xy

σ2
A(T )

≤ CTx for x > 0, 0 < T < T0 . (5.20)

Proof. All constants in the following can be chosen to depend only on T0 and
sup0≤t≤T0

A(t). We consider the stochastic integral s → M(s) defined similarly to
(2.49) but with qlinear in place of qε and µε in (2.25) given by µε = µ∗

ε of (2.27).
Then similarly to (2.51) we obtain the inequality

qlinear(x, y, T )

≤ E[qlinear(X
∗
ε (s ∨ τ∗ε,x,T,K), y, s ∨ τ∗ε,x,T,K)]

+ E
[1
2

∫ T

s∨τ∗
ε,x,T,K

[µ∗
ε(X

∗
ε (s), y, s)− λ(X∗

ε (s), y, s)]
2 ds

∣∣∣X∗
ε (T ) = x

] (5.21)

for 0 < s ≤ T , where X∗
ε (·) is the solution to the SDE (2.25) with µε = µ∗

ε.
The stopping time τ∗ε,x,T,K in (5.21) is the first exit time of X∗

ε (s), s < T, with

X∗
ε (T ) = x from the interval (0,K). As in the proof of Lemma 2.4 we use the



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 41

inequality X∗
ε (·) ≤ Xε(·), where Xε(·) is given by (2.16). Letting K → ∞ in (5.21)

we have then using (2.65) of Lemma 2.4 the inequality

qlinear(x, y, T ) ≤ E[qlinear(X
∗
ε (s ∨ τ∗ε,x,T ), y, s ∨ τ∗ε,x,T )] + qε(x, y, T ) (5.22)

for 0 < s < T , with τ∗ε,x,T the stopping time defined in Lemma 2.4. Using the

inequality qlinear(x, y, τ) ≤ Cy[x/τ + τ ], 0 < τ ≤ T0, where C is constant, we have
from (5.22) the inequality

qlinear(x, y, T )− qε(x, y, T ) ≤ CyE[τ∗ε,x,T ] +
Cy

s
E
[
X∗

ε (s) + s2; τ∗ε,x,T < s
]
, (5.23)

for 0 < s < T . Just as in the proof of Proposition 5.1 we use the lower bound (5.1)
to show that the second expectation on the RHS of (5.23) converges to 0 as s → 0.

To bound E[τ∗ε,x,T ] we use the identity (2.68), observing since A(·) is non-negative
that the sum of the last two terms on the RHS are non-negative. We have then
upon applying the Schwarz inequality to the RHS of (2.68) that for any T0 > 0 one
has

C1√
τ∗ε,x,T

{1

2

∫ T

τ∗
ε,x,T

[µ∗
ε(X

∗
ε (s), y, s)− λ(X∗

ε (s), y, s)]
2 ds

}1/2

≥ c1x

T
−
√
εZ(τ∗ε,x,T ) for 0 < T ≤ T0 ,

(5.24)

where C1, c1 > 0 are constants. It follows from (5.24) that if |Z(τ∗ε,x,T )| ≤ c1x/2
√
εT

then

τ∗ε,x,T ≤
(2TC1

c1x

)2 1

2

∫ T

τ∗
ε,x,T

[µ∗
ε(X

∗
ε (s), y, s)− λ(X∗

ε (s), y, s)]
2 ds . (5.25)

We conclude from (2.65) of Lemma 2.4 and (5.25) that

E
[
τ∗ε,x,T ; |Z(τ∗ε,x,T )| ≤ c1x/2

√
εT

]
≤

(2TC1

c1x

)2

qε(x, y, T ) . (5.26)

From (2.74), (2.75) we have that

E
[
τ∗ε,x,T ; |Z(τ∗ε,x,T )| > a

]
≤ E[τa] ≤

C2

a2
, a > 0, 0 < T ≤ T0, (5.27)

where C2 is constant. It follows from (5.27) that

E[τ∗ε,x,T ; |Z(τ∗ε,x,T )| > c1x/2
√
εT ] ≤ 4C2εT

2

c21x
2

, 0 < T ≤ T0 . (5.28)

We conclude from (5.26), (5.28) and Proposition 3.2 that

E
[
τ∗ε,x,T

]
≤ C3Ty

x
if x ≥ 2y, x ≥ T 2, xy > εT, 0 < T ≤ T0, (5.29)

where C3 is constant. The lower bound (5.19) follows from (5.23), (5.29) on letting
s → 0 in (5.23).

To prove (5.20), we first observe that if x ≥ y the inequality follows from the
inequality qε(x, y, T ) ≤ qlinear(x, y, T ). Hence we may assume 0 < x < y. We
consider the stochastic integral s → M(s) defined similarly to (2.49) but with



42 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

µε in (2.25) given by (5.14). Arguing as in the proof of Proposition 5.1 we have
analogously to (2.51) the inequality

qε(x, y, T )

≤ E
[1
2

∫ T

τε,linear,x,T

[µε(Xε(s), y, s)− λ(Xε(s), y, s)]
2 ds | Xε(T ) = x

]
.

(5.30)

Next we consider the stochastic integral s → M(s) defined similarly to (2.49) but
with qlinear in place of qε and µε in (2.25) again given by (5.14). Then using Ito’s
formula and the martingale optional sampling theorem we have the identity

qlinear(x, y, T )

= E [qlinear (0, y, τε,linear,x,T )]

+ E
[1
2

∫ T

τε,linear,x,T

[µε(Xε(s), y, s)− λ(Xε(s), y, s)]
2 ds | Xε(T ) = x

]
.

(5.31)

Observe next that

qlinear(x, y, T )− E [qlinear (0, y, τε,linear,x,T )]

≤ 2m1,A(T )xy

σ2
A(T )

+ C4yE [T − τε,linear,x,T ] , 0 < T ≤ T0 ,
(5.32)

for some constant C4. The inequality (5.20) follows from (5.30)-(5.32) if we can
show that

E [T − τε,linear,x,T ] ≤
C5Tx

y
, 0 < x < y, 0 < T ≤ T0 , (5.33)

for a constant C5.
We show that (5.33) holds if y ≥ C6T

2 for some constant C6. We choose C6

such that the drift µε defined by (5.14) satisfies the inequality µε(x, y, s) ≥ y/C5s,
0 < s ≤ T0 for some constant C5 > 0. This follows from (2.15) since σ2

A(s) −
m2,A(s) ≤ C7s

2, 0 < s ≤ T0, where C7 is constant. Then the LHS of (5.33) is

bounded above by uε(x) = E
[
T − τ∗ε,linear,x,T

]
, where τ∗ε,linear,x,T is the first exit

time from (0,∞) for the diffusion Xε(s), s < T , which is the solution to (2.25)
with terminal condition Xε(T ) = x and drift µε(x, y, T ) = y/C5T . Now uε(·) is the
solution to the boundary value problem,

−ε

2

d2uε(x)

dx2
+

y

C5T

duε(x)

dx
= 1 , x > 0, uε(0) = 0 . (5.34)

The solution to (5.34) is the linear function uε(x) = C5Tx/y, whence we obtain
the upper bound (5.33).

To finish the proof of (5.20) we need to deal with the case 0 < x < y < C6T
2. In

this case (5.20) reduces to the inequality qε(x, y, T ) ≤ C8Tx, 0 < T ≤ T0, where C8

is constant. We consider again the stochastic integral s → M(s) defined similarly to
(2.49) but with µε in (2.25) given by µε(x, y, s) = λ(x, y, s)+C9T, 0 < s < T ≤ T0,
where the constant C9 > 0 is chosen sufficiently large. Let τε,x,T be the first exit
time for the diffusion Xε(s), s < T , of (2.25) with Xε(T ) = x. Then a similar
inequality to (5.30) holds, whence we have

qε(x, y, T ) ≤
C2

9T
2

2
E[T − τε,x,T ] , 0 < x ≤ y ≤ C6T

2, 0 < T ≤ T0 . (5.35)



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 43

We also see as before that E[T − τε,x,T ] ≤ C10x/T , 0 < T ≤ T0,, where C10 is
constant. The result follows. □

Remark 5.3. The inequality (5.20) also follows from the upper bound in (5.1) by
integration over the interval [0, x]. In our proof in Proposition 5.2 we use the fact
that the optimizing τ in (4.2) is close to T as x → 0.

We have already shown in Proposition 5.1 that all of the bounds on the derivative
∂qε(x, y, T )/∂x with ε = 0 in Corollary 4.4, with the exception of the upper bound
(4.95), extend to ε > 0. Next we extend the upper bound (4.95) to ε > 0.

Proposition 5.4. Assume the function A : [0,∞) → R is continuous and non-
negative. Then for any T0 > 0 there exists a constant C > 0, depending only
on T0 and sup0≤t≤T0

A(t), such that the function (x, T ) → qε(x, y, T ) satisfies the
inequality

∂qε(x, y, T )

∂x
− 2m1,A(T )y

σ2
A(T )

≤ CTy2

x2
, (5.36)

for xy ≥ εT , x ≥ max{2y, CT 2}, and 0 < T ≤ T0.

Proof. We use the identity (5.4). The upper bound (5.1) implies that for any T0 > 0
there is a constant C1 such that vε(x, y, τ) ≤ 2y + C1τ

2, 0 < τ ≤ T0. Hence we
may let K → ∞ and s → 0 in (5.4) to obtain the identity

vε(x, y, T ) = E
[
vε(0, y, τ

∗
ε,x,T ) | X∗

ε (T ) = x
]
, 0 < s < T . (5.37)

Inequality (5.36) follows then from (5.37) if we can show that

E
[
(τ∗ε,x,T )

2
]
≤ C2T

2y2

x2
, x ≥ max{2y, C2T

2}, 0 < T ≤ T0 , (5.38)

for a constant C2. To prove (5.38) we use the upper bound (5.1). Analogously to
(5.14) we consider solutions X∗

ε,linear(·) to the SDE (2.25) with µε given by

µε(x, y, T ) = λ(x,−y, T ) + 2
[
1− m2,A(T )

σ2
A(T )

]
. (5.39)

Letting τ∗ε,linear,x,T be the first hitting time at 0 for X∗
ε,linear(s), s < T , with

X∗
ε,linear(T ) = x, we see that τ∗ε,x,T ≤ τ∗ε,linear,x,T . To prove (5.38) it will be sufficient

therefore to estimate E[(τ∗ε,linear,x,T )
2].

Since the drift (5.39) is linear the SDE (2.25) can be again explicitly solved, and
the solution is

Xε(s) = x∗
class(s)−

√
ε

σ2
A(s)

m1,A(s)
Z(s), s < T , (5.40)

where Z(·) is defined in (2.16), and x∗
class(·) is obtained from (5.17) by switching

the signs of the terms which do not involved x or y. Thus we have that

σ2
A(T )x

∗
class(s) = xm1,A(s, T )σ

2
A(s)− ym1,A(s)σ

2
A(s, T )

− m1,A(s, T )m2,A(s, T )σ
2
A(s) +m2,A(s)σ

2
A(s, T ) .

(5.41)

We have from (2.40), (4.5) and (5.41) that

x∗
class(s) ≥ c3

sx

T
− [1− s

T
] {C3y + C4sT} , 0 < s < T, 0 < T ≤ T0, (5.42)



44 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

for some constants C3, c3, C4 > 0. We conclude from (5.42) that

τ∗0,linear,x,T ≤ T
2C3y

c3x+ 2C3y
if x ≥ 2C4T

2

c3
. (5.43)

We may extend the inequality (5.43) to ε > 0 by considering for n = 1, 2, . . . ,
events An where An = {τ∗ε,linear,x,T > nTy/x}. Assuming x ≥ max{2y, 2C4T

2/c3},
we have from (5.42) the inequality x∗

class(τ
∗
ε,linear,x,T ) ≥ c3τ

∗
ε,linear,x,T (x/4T ) on the

event An provided n ≥ 4C3/c3. Hence from (5.40) we have on An with n ≥ 4C3/c3
the inequality Z

(
τ∗ε,linear,x,T

)
< −c5x/

√
εT , where c5 > 0 depends only on T0.

From (2.75) we have that

P ( sup
s<s′<T

|Z(s′)| > a) ≤ C6

(a2s)4
, a > 0, 0 < s < T ≤ T0 , (5.44)

where C6 > 0 depends only on T0. Choosing integers n0, n1 such that n0 ≥ 4C3/c3
and n1 ≥ x/y, we have from (5.44) that

E
[ (

τ∗ε,linear,x,T
)2 ] ≤ T 2y2

x2

[
n2
0 +

∑
n0≤n≤n1

(n+ 1)2P (An)
]

≤ T 2y2

x2

[
n2
0 +

C6

c85

(εT
xy

)4 ∑
n0≤n≤n1

(n+ 1)2

n4

]
.

(5.45)

The inequality (5.38) follows from (5.45) provided xy ≥ εT . □

Finally we show that the function x → qε(x, y, T ) is concave, thereby extending
the upper bound on ∂2qε(x, y, T )/∂x

2 with ε = 0 in Corollary 4.4 to ε > 0. Because
of the singularity in the drift [x, T ] → λ(x, y, T ), x, y, T > 0, of (2.15) as T → 0,
we use an approximation method.

Lemma 5.5. Assume the function A : [0,∞) → R is continuous and non-negative.
Then for any δ > 0 there is a unique classical solution [x, T ] → qε,δ(x, y, T ), x > 0,
T > δ, to the PDE (2.22) with boundary and initial conditions

qε,δ(0, y, T ) = 0, T > δ, qε,δ(x, y, δ) =
2m1,A(δ)xy

σ2
A(δ)

, x > 0 . (5.46)

Furthermore, the function [x, T ] → qε,δ(x, y, T ) satisfies the inequalities

2m1,A(T )xy

σ2
A(T )

≤ qε,δ(x, y, T ) ≤ −2λ(0, y, T )x , x > 0, T > δ , (5.47)

2m1,A(T )y

σ2
A(T )

≤ ∂qε,δ(x, y, T )

∂x
≤ 2

[
1− m2,A(T )

σ2
A(T )

]
+

2m1,A(T )y

σ2
A(T )

(5.48)

for x > 0 and T > δ. Let qε(x, y, T ) be defined by (2.18) and (2.21). Then

lim
δ→0

[qε,δ(x, y, T )− qε(x, y, T )] = 0,

and the limit is uniform in all sets {[x, T ] : x > 0, T0 < T < T1} with 0 < T0 <
T1 < ∞.

Proof. We proceed as in the proof of Proposition 3.3 by setting vε,δ(x, y, T ) =
exp [−qε,δ(x, y, T )/ε]. If [x, T ] → vε,δ(x, y, T ) is a solution to the PDE (2.19) in the
region x > 0, T > δ with boundary and initial conditions

vε,δ(0, y, T ) = 1, T > δ, vε,δ(x, y, δ) = exp
[
− 2m1,A(δ)xy

σ2
A(δ)ε

]
, x > 0 , (5.49)



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 45

then qε,δ(x, y, T ) = −ε log vε,δ(x, y, T ) is a solution to (2.22) with boundary and
initial conditions (5.46). Since the drift [x, T ] → λ(x, y, T ) is linear in x and contin-
uous in T for T ≥ δ, standard regularity theory implies that [x, T ] → vε,δ(x, y, T )
is a classical solution to (2.19), (5.49), from whence we conclude that [x, T ] →
qε,δ(x, y, T ) is a classical solution to (2.22), (5.46). The proof of (5.47) proceeds
as in the proof of Proposition 3.3 by using the maximum principle. With (5.47)
established, the proof of (5.48) then follows along the same lines as the proof of
Proposition 5.1.

To prove the convergence of qε,δ as δ → 0 we define the function uε,δ(x, y, T ) =
qε(x, y, T ) − qε,δ(x, y, T ) . Since both functions [x, T ] → qε(x, y, T ) and [x, T ] →
qε,δ(x, y, T ) are solutions to (2.22) it follows that the function [x, T ] → uε,δ(x, y, T ) =
qε(x, y, T )− qε,δ(x, y, T ) is a solution to the PDE

∂uε,δ(x, y, T )

∂T
=

ε

2

∂2uε,δ(x, y, T )

∂x2
−

{
λ(x, y, T )

+
1

2

[∂qε(x, y, T )
∂x

+
∂qε,δ(x, y, T )

∂x

]}∂uε,δ(x, y, T )

∂x
,

(5.50)

in the region{[x, T ] : x > 0, T > δ}. It follows from (5.46) and the upper bound
(3.28) of Proposition 3.3 that the boundary and initial conditions satisfy

uε,δ(0, y, T ) = 0, T > δ, 0 ≤ uε(x, y, δ) ≤ Cδy , x > 0 , (5.51)

where the constant C may be chosen uniformly in any interval 0 < δ < δ0 < ∞.
From (5.51) and the maximum principle applied to (5.50) we conclude that 0 ≤
uε,δ(x, y, T ) ≤ Cδy for x > 0, T > δ. □

Lemma 5.6. Assume the function A : [0,∞) → R is continuous non-negative, and
qε,δ(x, y, T ), x, y > 0, T > δ, the function defined in Lemma 5.5. Then for any
T0 > δ there exist constants C,M > 0, depending on ε, δ, y, T0 and sup0≤t≤T0

A(t),
such that ∣∣∂2qε,δ(x, y, T )

∂x2

∣∣ ≤ C for x ≥ M, δ < T ≤ T0 . (5.52)

Proof. Similarly to the derivation of (5.3) we see that the function

vε,δ(x, y, T ) =
σ2(T )

m1,A(T )

∂qε,δ(x, y, T )

∂x
(5.53)

is a solution to the PDE

∂vε,δ(x, y, T )

∂T
+
[
λ(x, y, T ) +

m1,A(T )

σ2
A(T )

vε,δ(x, y, T )
]∂vε,δ(x, y, T )

∂x

=
ε

2

∂2vε,δ(x, y, T )

∂x2

(5.54)

in the region {[x, T ] : x > 0, T > δ} with constant initial condition 2y on the half
line {[x, δ] : x > 0}. We make a change of variable to eliminate the linear drift
λ(·, ·, ·) from (5.54). To see this consider the PDE

∂w(x, T )

∂T
+ [α(T )x+ β(T )]

∂w(x, T )

∂x
=

ε

2

∂2w(x, T )

∂x2
, x ∈ R, T > δ . (5.55)



46 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

If we make the transformation w(x, T ) = u(z, t) where

z = exp
[
−
∫ T

δ

α(s) ds
]
x−

∫ T

δ

β(s) exp
[
−
∫ s

δ

α(s′) ds′
]
ds ,

t =

∫ T

δ

exp
[
− 2

∫ s

δ

α(s′) ds′
]
ds .

(5.56)

then u is a solution to the heat equation

∂u(z, t)

∂t
=

ε

2

∂2u(z, t)

∂2z
, z ∈ R, t > 0 . (5.57)

Now writing λ(x, y, T ) = α(T )x+β(T ) and setting vε,δ(x, y, T ) = u(z, t) according
to the change of variables (5.56), we see from (5.54) that u is a solution to the
Burgers’ equation

∂u(z, t)

∂t
+ γ(t)u(z, t)

∂u(z, t)

∂z
=

ε

2

∂2u(z, t)

∂2z
, (5.58)

where

γ(t) = exp
[ ∫ T

δ

α(s) ds
]m1,A(T )

σ2
A(T )

, T ≥ δ . (5.59)

For each z0 ∈ R, t0 > 0 we define the domain D(z0, ε) = {[z, t] : |z− z0| <
√
εt0,

0 < t < t0}. The Dirichlet Green’s function for the heat equation (5.57) on the
domain D(z0, ε) is simply a space translation and dilation of the Green’s function
on the domain D(0, 1). This latter Green’s function can be obtained by the method
of images. Thus for t > 0 let z → G(z, t) be the pdf of the Gaussian variable with
mean 0 and variance t, so

G(z, t) =
1√
2πt

exp[−z2

2t
] , z ∈ R . (5.60)

Then the Dirichlet Green’s function GD(z, z′, t) for D(0, 1) is given by the series

GD(z, z′, t) =

∞∑
m=0

p(m)G(z − zm, t) , (5.61)

where z0 = z′ and zm,m = 1, 2, . . . , are reflections of z′ in the boundaries z′ = ±
√
t0

with parities p(m) = ±1. The function

u(z, t) =

∫ √
t0

−
√
t0

GD(z, z′, t)u0(z
′) dz′ , [z, t] ∈ D(0, 1) (5.62)

is then a solution to (5.57) with ε = 1. It satisfies the initial condition u(z, 0) =
u0(z), |z| <

√
t0, and boundary condition u(z, t) = 0, z = ±

√
t0, 0 < t < t0.

Letting t = t0 correspond to T = T0 in (5.56), we see from (5.48) of Lemma 5.5
there exist a constant C0 > 0, depending only on T0, and a constant M0, depending
only on δ, T0, such that the solution u to the Burgers’ equation (5.58) satisfies

|u(z, t)| ≤ C0 + 2y for z ≥ M0, 0 < t ≤ t0 . (5.63)

We can integrate (5.58) on the domain D(z0, ε) with z0 > M0 +
√
εt0 by using the

Green’s function (5.61). We obtain the integral equation

u(z, t)

=

∫ √
t0

−
√
t0

GD((z − z0)/
√
ε, z′, t)u(

√
εz′ + z0, 0) dz

′



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 47

+ 2
∑

z′=±
√
t0

p(z′)

∫ t

0

∂GD((z − z0)/
√
ε, z′, (t− s))

∂z′
u(
√
εz′ + z0, s) ds (5.64)

+
1

2
√
ε

∫ t

0

∫ √
t0

−
√
t0

∂GD((z − z0)/
√
ε, z′, (t− s))

∂z′
γ(s)u(

√
εz′ + z0, s)

2dz′ ds

for [z, t] ∈ D(z0, ε), where p(z′) = −1 if z′ =
√
t0 and p(z′) = 1 if z′ = −

√
t0.

We can use the representation (5.64) and the bound (5.63) to obtain a bound on
∂u(z, t)/∂z at z = z0, 0 < t ≤ t0, which is independent of z0 as z0 → ∞. Observe
from (5.46) that u(·, 0) ≡ 2y is constant. Hence if u1(z, t) denotes the first term on
the RHS of (5.64) we have from (5.60), (5.61) the inequality∣∣∂u1(z, t)

∂z

∣∣ ≤ C1y√
ε
, |z − z0| <

√
εt0
2

, 0 < t ≤ t0 , (5.65)

where C1 depends only on t0. Letting u2(z, t) be the second (boundary) term on
the RHS of (5.64), we may use (5.63) to bound ∂u2(z, t)/∂z at z = z0. Thus we
have that ∣∣∂u2(z, t)

∂z

∣∣ ≤ C2(C0 + 2y)√
ε

, |z − z0| <
√
εt0
2

, 0 < t ≤ t0 , (5.66)

where C2 depends only on t0.
To bound the derivative of the third term on the RHS of (5.64) we define an

operator L on functions w : D(0, 1) → R by

Lw(z, t) =
∫ t

0

∫ √
t0

−
√
t0

∂GD(z, z′, t− s)

∂z′
γ(s)w(z′, s) dz′ ds , [z, t] ∈ D(0, 1) (5.67)

We see from (5.60), (5.61) there is a constant C3, depending only on t0 such that∣∣∣∣∂2GD(z, z′, t)

∂z∂z′

∣∣∣∣ ≤ C3

t
G(z − z′, 2t) , [z, t], [z′, t] ∈ D(0, 1) . (5.68)

It follows from (5.68) that∣∣∂GD(z1, z
′, t)

∂z′
− ∂GD(z2, z

′, t)

∂z′
∣∣

≤ C3|z1 − z2|
t

∫ 1

0

G(λ(z1 − z′) + (1− λ)(z2 − z′), 2t) dλ .

(5.69)

Let Uz1,z2 = {z′ ∈ R : |z′| <
√
t0, |z1 − z′| ≥ 2|z1 − z2|}. It is evident that

|z1 − z2| ≤ |λ(z1 − z′) + (1− λ)(z2 − z′)| for 0 < λ < 1, z′ ∈ Uz1,z2 . (5.70)

Next we write

Lw(z1, t)− Lw(z2, t) = F1(z1, z2, t) + F2(z1, z2, t) , (5.71)

where

F1(z1, z2, t)

=

∫ t

0

∫
Uz1,z2

[∂GD(z1, z
′, t− s)

∂z′
− ∂GD(z2, z

′, t− s)

∂z′

]
γ(s)w(z′, s) dz′ ds .

(5.72)



48 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

It follows from (5.69), (5.70) that for all α satisfying 0 < α < 1 there is a constant
C4 such that

|F1(z1, z2, t)| ≤ C4|z1 − z2|α∥w∥∞
∫ t

0

ds

s(1+α)/2
=

2C4

1− α
|w∥∞

√
t
( |z1 − z2|√

t

)α

.

(5.73)
To bound F2(z1, z2, t) we use the inequalities∣∣∂GD(z, z′, t)

∂z′
∣∣ ≤ C5√

t
G(z − z′, 2t) , [z, t], [z′, t] ∈ D(0, 1) , (5.74)∫

|z′|<a

G(z′, 2t) dz′ ≤ C6 min

{
a√
t
, 1

}
, (5.75)

where C5, C6 are constants depending only on t0. We have from (5.74), (5.75) that∫ t

0

∫
[−

√
t0,

√
t0]−Uz1,z2

∣∣∂GD(z1, z
′, t− s)

∂z′
− ∂GD(z2, z

′, t− s)

∂z′
∣∣ dz′ ds

≤ 2C5C6

∫ t

0

1√
t− s)

min
{3|z1 − z2|√

t− s
, 1
}
ds

= 6C5C6|z1 − z2|
∫ t/9|z1−z2|2

0

min
{ 1√

s′
, 1
} ds′√

s′

≤ C7

1− α

√
t
( |z1 − z2|√

t

)α

if
|z1 − z2|√

t
≤ 1 .

(5.76)

We conclude from (5.71), (5.73), (5.76) that

|Lw(z1, t)− Lw(z2, t)| ≤ Cα∥w∥∞
√
t
( |z1 − z2|√

t

)α

if
|z1 − z2|√

t
≤ 1 , (5.77)

where the constant Cα depends only on α, t0. It follows from (5.67), (5.74) that
∥Lw∥∞ ≤ C

√
t∥w∥∞ for some universal constant C. Hence (5.77) holds for all

[z1, t], [z2, t] ∈ D(0, 1).
Next for a continuous function f : [−

√
t0,

√
t0] → R and α satisfying 0 < α < 1

define the α Hölder norm of f by

∥f∥0,α = sup{|f(z)| : z ∈ [−
√
t0,

√
t0]}

+ sup
{ |f(z1)− f(z2)|

|z1 − z2|α
: z ∈ [−

√
t0,

√
t0]

}
.

(5.78)

We may bound the derivative of (∂/∂z)Lw(z, t) in terms of the norms (5.78) for
w(·, s), 0 < s < t. To see this observe from (5.67) that

Lw(z + h, t)− Lw(z, t)
h

=

∫ t

0

∫ √
t0

−
√
t0

∫ 1

0

∂2GD(z + λh, z′, t− s)

∂z∂z′
dλγ(s)w(z′, s) dz′ ds (5.79)

=

∫ t

0

∫ √
t0

−
√
t0

∫ 1

0

∂2GD(z + λh, z′, t− s)

∂z∂z′
γ(s)[w(z′, s)− w(z + λh, s)] dλ dz′ ds .

We have from (5.68), (5.79) that∣∣Lw(z + h, t)− Lw(z, t)
h

∣∣ ≤ C

∫ t

0

ds

(t− s)1−α/2
∥w(·, s∥0,α ds , (5.80)



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 49

for 0 < α < 1 0 < t ≤ t0, where C8 is a constatn depending on t0.
Let u3(z, t) with [z, t] ∈ D(z0, ε) be the third term on the RHS of (5.64). Then

u3(z, t) = Lw1((z − z0)/
√
ε, t) with [z, t] ∈ D(z0, ε), where w1 is given by the

formula

w1(z, t) =
1

2
√
ε
u2(

√
εz + z0, t) , [z, t] ∈ D(0, 1) . (5.81)

We have then from (5.63), (5.77) that

∥Lw1(·, t)∥0,α ≤ Cα

2
√
ε
[C0 + 2y]2t(1−α)/2 , 0 < t < t0, 0 < α < 1 . (5.82)

Now defining w2;D(0, 1) → R by w2((z−z0)/
√
ε, t) = u(z0+(z−z0)/2, t), we have

from (5.65), (5.66), (5.82) the bound

∥w2(·, t)∥0,α ≤ C9[C0 + 2y] +
Cα

2
√
ε
[C0 + 2y]2t(1−α)/2 , (5.83)

for 0 < t < t0, 0 < α < 1, and a constant C9 depending only on t0.
The inequality (5.83) shows that the solution u(·, ·) of (5.58) is Hölder contin-

uous in the domain D(z0, ε/4). We represent u(z, t) again as in (5.64) but with
t0 replaced by t0/4, whence (5.83) implies that u(·, ·) is Hölder continuous in the
domain D(z0, ε). Letting w3(z, t) = w2(z, t)

2/2
√
ε, then we see that

u3(z, t) = Lw3((z − z0)/
√
ε, t) , ∥w3(·, t)∥0,α ≤ 1√

ε
∥w2(·, t)∥20,α . (5.84)

It follows from (5.80), (5.83), (5.84) that∣∣∂u3(z, t)

∂z

∣∣ ≤ C for |z − z0| <
√
εt0/2, 0 < t < t0/2 , (5.85)

for some constant depending on ε, y, t0, but not on z0. The inequality (5.52) follows
from (5.65), (5.66), and (5.85). □

Theorem 5.7. Assume the function A : [0,∞) → R is continuous non-negative,
and let qε(x, y, T ) be defined by (2.18), (2.21). Then for all y, T > 0 the function
x → qε(x, y, T ) is concave.

Proof. We show that the function x → qε,δ(x, y, T ) is concave, and then the result
follows from Lemma 5.5 by letting δ → 0. We define the function wε,δ by

wε,δ(x, y, T ) =
σ4
A(T )

m1,A(T )2
∂2qε,δ(x, y, T )

∂x2
. (5.86)

By differentiating the PDE (5.3) we see from Proposition 3.1 that the function
[x, T ] → wε,δ(x, y, T ) is a classical solution of the PDE

∂wε,δ(x, y, T )

∂T
+
[
λ(x, y, T ) +

∂qε,δ(x, y, T )

∂x

]∂wε,δ(x, y, T )

∂x

+
m1,A(T )

2

σ4
A(T )

wε,δ(x, y, T )
2

=
ε

2

∂2wε,δ(x, y, T )

∂x2
.

(5.87)



50 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

We proceed as in the proof of Proposition 5.1 using Ito’s lemma and the martingale
optional sampling theorem. Thus similarly to (5.4) we have the representation

wε,δ(x, y, T ) = E
[
wε,δ(X

∗
ε,δ(δ ∨ τ∗ε,δ,x,T,K), y, δ ∨ τ∗ε,δ,x,T,K) | X∗

ε,δ(T ) = x
]

− E
[ ∫ T

δ∨τ∗
ε,δ,x,T,K

m1,A(s)
2

σ4
A(s)

wε,δ(X
∗
ε,δ(s), y, s)

2 ds
]
,

(5.88)

where X∗
ε,δ(·) is the solution to the SDE (2.25) with drift µε(x, y, T ) given by the

coefficient of ∂wε,δ(x, y, T )/∂x in (5.87). The stopping time τ∗ε,δ,x,T,K is the first

exit time of X∗
ε,δ(s), s ≤ T , with X∗

ε,δ(T ) = x from the interval (0,K). From (5.46)

we have that wε,δ(·, y, δ) ≡ 0, whence the first term on the RHS of (5.88) can be
written as

E
[
wε,δ(0, y, τ

∗
ε,δ,x,T,K); τ∗ε,δ,x,T,K > δ,Xε,δ(τ

∗
ε,δ,x,T,K) = 0

]
+ E

[
wε,δ(K, y, τ∗ε,δ,x,T,K); τ∗ε,δ,x,T,K > δ,Xε,δ(τ

∗
ε,δ,x,T,K) = K

]
.

(5.89)

Using Lemma 5.5 and arguing as in Corollary 3.6, we see that wε,δ(0, y, s) ≤ 0
for s > δ, y > 0. Hence the first term in (5.89) is non-positive. The second term
converges to 0 as K → ∞. To prove this we use Lemma 5.6, which yields a uniform
upper bound on |wε,δ(K, y, s)|, δ < s < T , as K → ∞. Then we follow the
corresponding argument around (5.11) in the proof of Proposition 5.1. By letting
K → ∞ in (5.88) we conclude that wε,δ(x, y, T ) ≤ 0 for x ≥ 0, y > 0, T > δ. □

6. Convergence of the function ∂qε(x, y, T )/∂x as ε → 0

In this section we assume the function A(·) is non-negative, whence the results of
Proposition 4.2 and Corollary 4.4 imply that the function x → q0(x, y, T ) of (2.28),
(4.2) is C1 for certain ranges of [x, T ]. We will show that limε→0 ∂qε(0, y, T )/∂x =
∂q0(0, y, T )/∂x. In view of the upper bound (3.29), we only need to prove for
small x a lower bound for qε(x, y, T ) in terms of q0(x, y, T ) and a correction term
which goes to 0 as ε → 0. We already obtained such a lower bound in Lemma 2.5.
Our starting point was the inequality (2.80), which leads to the inequality (2.81).
However the second term on the RHS of (2.81) is not sufficient for our purposes
since we need the correction to be bounded by a constant times x as x → 0. Instead
of (2.80) we observe from (2.77)-(2.80) that

1

2

∫ T

τ∗
ε,x,T

[µ∗
ε(X

∗
ε (s), y, s)− λ(X∗

ε (s), y, s)]
2
ds

≥ q0(x, y, T )− q0(
√
εZε, y, τ

∗
ε,x,T ) .

(6.1)

The function [x, T ] → q0(x, y, T ) is defined by (2.28) for x, T > 0, and for x <
0, T > 0 by

q0(x, y, T ) = −min
{1

2

∫ τ

T

[dx(s)
ds

− λ(x(s), y, s)
]2

ds : τ > T,

x(T ) = x, x(·) < 0, x(τ) = 0
}
.

(6.2)



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 51

Assuming the function [x, T ] → q0(x, y, T ) defined by (2.28), (6.2) is sufficiently
differentiable at [0, T ], we can do a Taylor expansion,

q0(
√
εZε, y, τ

∗
ε,x,T )

=
∂q0(0, y, T )

∂x

√
εZε + εZ2

ε

∫ 1

0

∫ 1

0

λ dλ dµ
∂2q0(λµ

√
εZε, y, T )

∂x2

− [T − τ∗ε,x,T ]

∫ 1

0

dλ
∂q0(

√
εZε, y, λτ

∗
ε,x,T + (1− λ)T )

∂t
.

(6.3)

Then we can estimate the expectation of the correction term in (6.1) by estimating
the expectation of each term on the RHS of (6.3).

We may obtain a formula similar to (4.2) for q0(x, y, T ), x < 0, defined by (6.2).
If the minimization in (6.2) is for fixed τ > T then there is a unique minimizing
trajectory x(s), T < s < τ , given by (2.37), where γ(τ) is chosen so that x(τ) = 0.
The functions s → g1,A(s, T ), g2,A(s, T ), which were defined in (2.39), (2.40) for
0 < s < T may be extended by the same formulas to s > T . Similarly we may
extend the function s → g3,A(s, T ) by using (4.12). Note that the functions s →
g1,A(s, T ), g3,A(s, T ), s > T , are negative. We have then from (6.2) that

q0(x, y, T ) = −min
τ>T

|g3,A(τ, T )|
2

[y + g1,A(τ, T )x+ g2,A(τ, T )]
2 . (6.4)

When A(·) ≡ 0 the formula (6.4) becomes

q0(x, y, T ) = −min
τ>T

τ − T

2τT

[
y+

τx

(T − τ)

]2
= − 1

2T

[
−2xy+min

α>1
{αx2+y2/α}

]
. (6.5)

Hence we have that

q0(x, y, T ) =
2xy

T
if − y < x < 0, (6.6)

q0(x, y, T ) = − (x− y)2

2T
if x < −y . (6.7)

For −y < x < 0 the minimizing τ > T in (6.4) is given by τ(x, y, T ) = yT/(x+ y).
Otherwise the minimum is obtained by letting τ → ∞. The function [x, T ] →
q0(x, y, T ) defined by (6.6), (6.7) is a C1 solution to the HJ equation (2.29) in the
region {[x, T ] : x < 0, T > 0}. However the second derivative ∂2q0(x, y, T )/∂x

2 is
discontinuous across the boundary {[x, T ] : x = −y, T > 0}. The characteristics
which yield the function (6.6) are the same as in the situation x > 0 studied in §4,
and are given by x(τ, s) = (s− τ)y/τ, s > 0. Then we have

q0(x, y, T ) =
1

2

∫ T

τ

p(τ, s)2 ds = 2

∫ T

τ

y2

s2
ds =

2xy

T
, x = x(τ, T ) . (6.8)

This set of characteristics covers the region {[x, T ] : x > −y, T > 0} without
intersecting, but all characteristics converge to the point [−y, 0]. The characteristics
which yield the function (6.7) are given by x(λ, s) = λ, λ < −y, s > 0. In that case

q0(x, y, T ) =
1

2

∫ T

∞
p(λ, s)2 ds =

1

2

∫ T

∞

(λ− y)2

s2
ds = − (x− y)2

2T
, (6.9)

where x = x(λ, T ). This set of characteristics covers the region {[x, T ] : x <
−y, T > 0}, also without intersecting.

We shall show that the function [x, T ] → q0(x, y, T ) defined by (2.28), (6.2)
is differentiable for [x, T ] in a neighborhood of the initial line {[0, T ] : T > 0},



52 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

by proving it may be obtained via the method of characteristics. To do this we
extend the domain Dy,T0 defined just prior to (4.45). It follows from (4.32) that the
characteristics s → x(τ, s) do not meet if 0 < s < τ ≤ T0. Hence we may extend
the domain Dy,T0

of §4 to include the set {[x, T ] : 0 < T < T0, x(T0, T ) < x ≤ 0},
and similarly extend the region Uy,T0

. The inequality (4.42) for the characteristic
s → x(τ, s), τ ≤ s ≤ T0, continues to hold for 0 < s < τ . More precisely, we have
from (4.40), (4.41) that

C1(s− τ)
[ [y + g2,A(τ, τ)]

τ
+ (τ − s)

]
≤ x(τ, s)

≤ c1(s− τ)
[ [y + g2,A(τ, τ)]

τ
+ c2(τ − s)

]
, 0 < s < τ ≤ T0 ,

(6.10)

where C1, c1, c2 depend only on T0 and sup0≤t≤T0
A(t). Hence there is a constant

Λ3 > 0, depending only on T0 and sup0≤t≤T0
A(t), such that

if 0 < T ≤ 3T0/4 and − Λ3[y + g2,A(T0, T0)] < x ≤ 0,

then [x, T ] ∈ Dy,T0 .
(6.11)

We extend the results of Proposition 4.2 to include [x, T ] in the region (6.11).

Lemma 6.1. The results of Proposition 4.2, with q0(x, y, T ) defined by (4.54),
continue to hold in the extended region Dy,T0

. Also τ = τ(x, y, T ) in (4.54) is the
unique minimizer in the variational problems (6.2), (6.4) for [x, T ] with x < 0,
0 < T < T0/2, in the following regions:

(a) −Λ[y + g2,A(T, T )] < x < 0 if 4
√
Λ0y/3 ≥ T , otherwise

(b) −Λ[y + g2,A(T, T )]
2/T 2 < x < 0, where Λ > 0 is chosen sufficiently small

depending only on T0 and sup0≤t≤T0
A(t).

Therefore if [x, T ] is in one of the regions (a), (b), the functions (6.4) and (4.54)
are identical.

Proof. All constants in the following can be chosen to depend only on T0 and
sup0≤t≤T0

A(t). It is clear from (6.10) that the characteristics s → x(τ, s), 0 < s <
τ , satisfy x(τ, s) < 0, and from (4.32) that Dτx(τ, s) < 0. The differentiability
properties of the function [x, T ] → q0(x, y, T ) and the fact that it is a solution to
the HJ equation (2.29) follow as in the proof of Proposition 4.2. We also have from
(4.64) that

−q0(x, y, T ) ≤
1

2

∫ τ

T

[dx(s)
ds

− λ(x(s), y, s)
]2

ds , (6.12)

for any path s → x(s), T < s < τ < T0, in Dy,T0
with x(T ) = x < 0, x(τ) = 0.

Equality holds in (6.12) if x(·) is the characteristic.
Let F0(x, y, τ, T ) be the function on the RHS of (6.4). We wish to find [x, T ] ∈

Dy,T0 with x < 0 such that q0, defined by (4.54), satisfying

−q0(x, y, T ) = inf
τ>T

F0(x, y, τ, T ).

To do this we first observe from (4.5) that since A(·) is non-negative, the func-
tion s → g2,A(s, T ), s > 0, is increasing and hence non-negative. We also have
from (2.8) that the function s → σ2

A(s, T ), s > T , is negative and decreasing
with lims→T σ2

A(s, T ) = 0. Letting lims→∞ σ2
A(s, T ) = σ2

A(∞, T ), one sees in



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 53

the case A(·) ≡ 0 that σ2
A(∞, T ) = −∞. It is however possible for some non-

negative A(·) that σ2
A(∞, T ) > −∞. We have then from (4.4) that the func-

tion s → g1,A(s, T ), s > T , is increasing with lims→T g1,A(s, T ) = −∞, and
g1,A(s, T ) < −1/m1,A(T ), s > T . It follows that lims→∞ g1,A(s, T ) = g1,A(∞, T ) ≤
−1/m1,A(T ).

Similarly to (4.66) we consider [x, T ] ∈ Dy,T0
which satisfies (6.11), and define

q̃0(x, y, T ) by

−q̃0(x, y, T ) = min
τ>T

|g3,A(τ, T )|
2

[y + g1,A(τ, T )x]
2
. (6.13)

Using the identity (4.12) we see that the minimizing τ for the RHS of (6.13) is
given by

g3,A(τ, T ) =
m1,A(T )x

σ2
A(T )y

, if
y

g1,A(∞, T )
< x < 0 . (6.14)

Substituting (6.14) into the RHS of (6.13) then yields the formula

q̃0(x, y, T ) =
2m1,A(T )xy

σ2
A(T )

, (6.15)

which is the same as (4.69). We should however note that the RHS of (6.15) is
negative in this case. Following the argument in the proof of Proposition 4.2, we
see that if τ > T lies outside the region

8m1,A(T )x

σ2
A(T )[y + g2,A(T, T )]

≤ g3,A(τ, T ) ≤
m1,A(T )x

8σ2
A(T )[y + g2,A(T, T )]

, (6.16)

F0 satisfies the inequality

F0(x, y, τ, T ) ≥ −5q̃0(x, y + g2,A(T, T ), T )/2 . (6.17)

In concluding (6.17) we have used the fact that the function s → g2,A(s, T ), s > T ,
is increasing.

We require x < 0 to be sufficiently close to 0 so that if τ > T with 0 < T ≤ T0/2
lies in the region (6.16) then τ ≤ 3T/2 ≤ 3T0/4. This is the case if x > −Λ4[y +
g2,A(T, T )], where Λ4 > 0 is constant. Next we determine how small |x| needs
to be so that the minimizing paths s → Γ(τ, s, T, x) = a(τ, s, T )x + b(τ, s, T ),
T < s < τ , for fixed τ defined by (4.65) lie in Dy,T0

. To see this first note that the
functions s → a(τ, s, T ), b(τ, s, T ), ; T < s < τ , are non-negative. Hence if x < 0
is sufficiently small one may have Γ(τ, s, T, x) > 0 for some s ∈ (T, τ). We see from
(4.5), (4.6), (4.65) there are constants C1, c1, C2 > 0 such that

C1(τ − s)
x

τ − T
≤ Γ(τ, s, T, x) ≤ (τ − s)

[ c1x

τ − T
+ C2(s− T )

]
, (6.18)

for T < s < τ < T0. Let Λ0 be the constant defined just after (4.44), whence if
T < 4

√
Λ0y/3 then Dy,T0

contains the domain {[x, s] : x > 0, 0 < s < 3T/2}. We
assume that τ > T lies in the region (6.16) and that −Λ4[y + g2,A(T, T )] < x < 0,
whence τ ≤ 3T0/4. Choosing Λ4 to also satisfy the inequality C1Λ4 < Λ3, we see
from (6.11), (6.18) that if T < 4

√
Λ0y/3 then the path s → Γ(τ, s, T, x), T < s < τ ,

lies in Dy,T0
. In the case T ≥ 4

√
Λ0y/3 we observe that if τ > T satisfies (6.16)

and −Λ4[y + g2,A(T, T )] < x < 0 then τ − T satisfies an inequality

τ − T ≤ C3Tx

[y + g2,A(T, T )]
, where C3 is constant. (6.19)



54 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

Hence the RHS of (6.18) is negative provided x < 0 satisfies the inequality

|x| ≤ c1[y + g2,A(T, T )]
2

C2C2
3T

2
. (6.20)

We conclude in this case that the path s → Γ(τ, s, T, x), T < s < τ , lies in Dy,T0

provided [x, T ] satisfies (6.20).
Finally we need to show that if τ = τ(x, y, T ), then F0(x, y, τ, T ) < −5q̃0(x, y +

g2,A(T, T ), T )/2, which is the same as q0(x, y, T ) ≥ 5q̃0(x, y+ g2,A(T, T ), T )/2. We
first observe from (4.71) and (6.10) that since x(τ, s) < 0, 0 < s < τ , we have
∂v(τ, s)/∂s ≤ 0 for T < s < τ and v(τ, τ) = 0. We conclude that

q0(x, y, T ) ≤ q̃0(x, y + g2,A(T, T ), T ) , [x, T ] ∈ Dy,T0 , x < 0 . (6.21)

We assume now that [x, T ] lies in the domain {[x, T ] : 0 < T < T0/2,−Λ[y +
g2,A(T, T )] < x < 0}, where Λ satisfies 0 < Λ ≤ Λ3. It follows from (6.11) that
[x, T ] ∈ Dy,T0 . Letting x = x(τ, T ) we have from (6.10) that

τ − T ≤ τ |x|
c1[y + g2,A(τ, τ)]

≤ Λτ

c1
. (6.22)

Choosing Λ ≤ c1/3 we see from (6.22) that τ − T ≤ ΛT/[c1 − Λ] ≤ T/2. Observe
from (4.30) that

1 ≤ y + g2,A(τ, τ)

y + g2,A(T, T )
≤ 1 + C4Λ (6.23)

if T < 4
√
Λ0y/3, where C4 is constant. It follows then from (4.71), (6.10), (6.23),

upon using a lower bound for the integral of the RHS of (4.71) on the interval
T < s < τ similar to the one in (4.72), that

q0(x, y, T ) ≥ [1 + C5Λ]q̃0(x, y + g2,A(T, T ), T ) , (6.24)

where C5 is constant. We assume that T ≥ 4
√
Λ0y/3 and that [x, T ] satisfies the

inequality

0 < T < T0/2, −Λ[y + g2,A(T, T )]
2

T 2
< x < 0 . (6.25)

Then inequality (6.22) continues to hold, whence we see from (4.30) and (6.25) that
(6.23) also holds. Similarly to before we see that (6.24) holds in this case also. Now
we choose Λ so that C5Λ ≤ 1. □

Observe from (3.48) that we expect

∂qε(0, y, T )

∂x
− ∂q0(0, y, T )

∂x
= ε

∂2q0(0, y, T )

∂x2

/∂q0(0, y, T )

∂x
+o(ε) as ε → 0. (6.26)

This evidently suggests that the LHS of (6.26) isO(ε) as ε → 0. Furthermore, we see
from (4.23) that ∂q0(0, y, T )/∂x ≃ y/T , and from (4.34) that ∂2q0(0, y, T )/∂x

2 ≃
T/y if y ≥ T 2. Hence we expect the LHS of (6.26) to be bounded by a constant
times εT 2/y2 + o(ε).

Proposition 6.2. Assume the function A : [0,∞) → R is continuous non-negative,
and let qε(x, y, T ) be defined by (2.18), (2.21), and q0(x, y, T ) by (2.28), (4.2).
Then for each T0 > 0 there are constants C1, C2, depending only on T0 and
sup0≤t≤T0

A(t), such that∣∣∂qε(0, y, T )
∂x

− ∂q0(0, y, T )

∂x

∣∣ ≤ C2ε

y
[
T 2

y
+

εT

y2
] , 0 < T ≤ T0 , (6.27)



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 55

provided 0 < T ≤ T0, y ≥ C1T
2, εT < y2.

Proof. All constants in the following can be chosen to depend only on T0 and
sup0≤t≤T0

A(t). We define stopping times for the martingale s → Z(s), s < T , of
(2.16). For δ > 0 let τδ be given by

τδ = inf
{
s : 0 < s < T,

∣∣ σ2
A(T )

m1,A(T )
Z(s′)

∣∣ < δ for s < s′ < T
}
. (6.28)

It is easy to see that τδ > 0 with probability 1. Since τ∗ε,x,T , defined in the statement

of Lemma 2.4, is also a stopping time for the martingale (2.16), it follows that
τ̃ = τ̃ε,δ,x,T = τδ ∨τ∗ε,x,T ∨ (T/2) is a stopping time. On taking expectations in (6.1)

we have from (2.65) of Lemma 2.4 that for any δ > 0,

qε(x, y, T ) ≥
[
1− P

(
τ∗ε,x,T < T/2

)
− P

(
τ∗ε,x,T < τδ

)]
q0(x, y, T )

− E
[
q0(

√
εZε, y, τ

∗
ε,x,T ); τ

∗
ε,x,T > max{τδ, T/2}

]
.

(6.29)

Since the function s → σ2
A(s)/m1,A(s), s > 0, is increasing, we see from (2.80)

that |Zε| ≤ σ2
A(T )|Z(τ∗ε,x,T )|/m1,A(T ). Hence if δ is small enough we may use the

Taylor expansion (6.3) to estimate the second expectation on the RHS of (6.29). It
follows from Proposition 4.2 and Lemma 6.1 that it is sufficient for δ to lie in the
interval

√
εδ ≤ Λmin

{ y

T 2
, 1
}
[y + g2,A(T/2, T/2)] , (6.30)

where Λ depends only on T0.
We show that

lim
x→0

P
(
τ∗ε,x,T < T/2

)
= 0, lim

x→0
P
(
τ∗ε,x,T < τδ

)
= 0 . (6.31)

To prove the first limit in (6.31) we use the inequality ∂qε(x, y, T )/∂x ≥ 0, whence it
follows that X∗

ε (s) ≤ Xε(s), 0 < s < T , where Xε(·) is defined by (2.16). This was
already observed just prior to Lemma 2.4. We have from (2.12) that one can choose
a constant ν with 0 < ν < 1/2, such that yclass(s) < 2x if 0 < T−s < Tνx/(y+T 2),
provided x < y + T 2 0 < T ≤ T0. The first limit in (6.31) follows if we can show
that

lim
x→0

P
(√

ε inf
T−Tνx/(y+T 2)<s<T

Z(s) > −2m1,A(T/2)x

σ2
A(T/2)

)
= 0 , (6.32)

since P
(
τ∗ε,x,T < T − Tνx/(y + T 2)

)
is smaller than the probability in (6.32). The

limit in (6.32) follows from the reflection principle. Similarly to (2.75) we have
that the probability in (6.32) is bounded in terms of a probability for the standard
normal variable Y by

P
(
|Y | < 2x/

√
εσ(x)

)
,where σ(x)2 ≥ cνTx

(y + T 2)
, (6.33)

with c > 0 a constant. Since the probability (6.33) is bounded by a constant times√
x the limit (6.32) follows. To prove the second limit in (6.31) we argue similarly,

using the inequality P (τ∗ε,x,T < τδ) ≤ P (τ∗ε,x,T < T − Tνx/(y + T 2)) + P (τδ >

T − Tνx/(y + T 2)). Using the reflection principle again we see that

lim
x→0

P
(
τδ > T − Tνx/(y + T 2)

)
= 0.



56 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

We estimate the contribution of the first term in the Taylor expansion (6.3) to the
expectation in (6.29). To do this we use the inequality∣∣Zε −

σ2
A(T )

m1,A(T )
Z(τ∗ε,x,T )

∣∣ ≤ C1

[
T − τ∗ε,x,T

] ∣∣Z(τ∗ε,x,T )
∣∣ , (6.34)

where C1 is a constant. Using that

s → Z(s)2 −
∫ T

s

m1,A(s
′)2

σ4
A(s

′)
ds′ , 0 < s < T , (6.35)

is a martingale, we have from (6.35) and the optional stopping theorem that

E
[
Z(τ∗ε,x,T )

2; τ∗ε,x,T > T/2
]
≤ C2

T 2

{
E
[
T − τ∗ε,x,T

]
+ TP

(
τ∗ε,x,T < T/2

)}
≤ 3C2

T 2
E
[
T − τ∗ε,x,T

]
,

(6.36)

where C2 is constant. To bound the RHS of (6.36) we use the lower bound (5.1)
of Proposition 5.1. Recalling the definition of τε,linear,x,T after (5.14), we have that
E[T − τ∗ε,x,T ] ≤ E[T − τε,linear,x,T ]. We obtain then from (5.33) an upper bound

for E
[
T − τ∗ε,x,T

]
, provided y ≥ C3T

2 where C3 is constant. We may also obtain

an inequality E
[
(T − τ∗ε,x,T )

2
]
≤ vε(x), where vε(·) is the solution to a boundary

value problem. Thus

−ε

2

d2vε(x)

dx2
+

y

C5T

dvε(x)

dx
= 2uε(x) , x > 0, vε(0) = 0 , (6.37)

where uε(·) is the solution to (5.34). Evidently we have that

vε(x) =
(C5T

y

)3
x[ε+

xy

C5T
] . (6.38)

It follows from (6.34), upon using the Schwarz inequality and (6.36)-(6.38), that

√
ε lim sup

x→0

1

x
E
[∣∣Zε −

σ2
A(T )

m1,A(T )
Z(τ∗ε,x,T )

∣∣; τ∗ε,x,T > T/2
]
≤ C4ε

T

y2
, (6.39)

where C4 is a constant.
Applying the optional sampling theorem to the martingale s → Z(s), 0 < s < T ,

we have that

σ2
A(T )

m1,A(T )

∣∣E [
Z(τ∗ε,x,T ); τ

∗
ε,x,T > max{T/2, τδ}

]∣∣
≤ σ2

A(T )

m1,A(T )
E[|Z(T/2)|; τ∗ε,x,T < T/2] + δP

(
τ∗ε,x,T < τδ

)
.

(6.40)

To bound the first term on the RHS of (6.40) we observe that

√
ε

σ2
A(T )

m1,A(T )
E[|Z(T/2)|; τ∗ε,x,T < T/2]

≤ C5E [|Xε(T/2)− xclass(T/2)|; τε,linear,x,T < T/2] ,

(6.41)

where Xε(·) is given by (5.16) and C5 is a constant. From (5.17) and the Chebyshev
inequality we have, using the inequality y ≥ C3T

2, that

|xclass(T/2)| |P (τε,linear,x,T < T/2) ≤ C6yvε(x)

T 2
, (6.42)



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 57

where C6 is a constant. LettingX
∗
ε (s), s < T , be the diffusion with drift µε(x, y, T ) =

y/C5T defined just after (5.33), we have that

E [Xε(T/2); τε,linear,x,T < T/2] ≤ E
[
X∗

ε (T/2); τ
∗
ε,linear,x,T < T/2

]
. (6.43)

Let [x, t] → uε(x, t), x > 0, t > 0, be the solution to the PDE

∂uε(x, t)

∂t
= −µ

∂uε(x, t)

∂x
+

ε

2

∂2uε(x, t)

∂x2
, x > 0, t > 0, (6.44)

with boundary and initial conditions

uε(0, t) = 0, t > 0, uε(x, 0) = x, x > 0 . (6.45)

Then one has

uε(x, T − t) = E
[
X∗

ε (t); τ
∗
ε,linear,x,T < t

]
, t < T, when µ =

y

C5T
. (6.46)

The solution to (6.44), (6.45) is

uε(x, t) =

∫ ∞

0

Gε,D(x, x′, t)x′ dx′ , (6.47)

where the Green’s function Gε,D has the formula

Gε,D(x, x′, t) =
1√
2πεt

exp
[
− (x− x′ − µt)2

2εt

]{
1− exp

[
− 2xx′

εt

]}
. (6.48)

Using the inequality 1− e−z ≤ z, z ≥ 0, we have from (6.47), (6.48) that

lim sup
x→0

1

x
uε(x, t) ≤

2√
2π(εt)3/2

exp[−µ2t

2ε
]

∫ ∞

0

e−µx′/εx′2 dx′

=
4√
2π

( ε

µ2t

)3/2
exp[−µ2t

2ε
] .

(6.49)

We conclude from (6.38) and (6.42)-(6.49), that the first term on the RHS of (6.40)
is bounded as

√
ε lim sup

x→0

1

x

σ2
A(T )

m1,A(T )
E[|Z(T/2)|; τ∗ε,x,T < T/2] ≤ C7εT

y2
if εT ≤ y2, (6.50)

where C7 is a constant.
To find a bound for the second term on the RHS of (6.40) we use the inequality

P
(
τ∗ε,x,T < τδ

)
≤ P

(
τ∗ε,x,T < νT

)
+ P

(
νT < τ∗ε,x,T < τδ

)
, where 1/2 < ν < 1 and

ν is a suitably chosen constant. We may then bound
√
εδP

(
τ∗ε,x,T < νT

)
from the

inequalities obtained in the previous paragraph. Assuming y ≥ T 2, it follows from
(6.30) that we may take

√
εδ = Λy, whence lim supx→0 x

−1
√
εδP

(
τ∗ε,x,T < νT

)
≤

C8εT/y
2 for some constant C8. We estimate the second probability as

P
(
νT < τ∗ε,x,T < τδ

)
≤ P (τδ > νT, τε,linear,x,T < τδ)

≤ P
(

sup
τε,linear,x,T∨(νT )<s<T

|Xε(s)− xclass(s)| > c8
√
εδ
)
,

(6.51)

where Xε(·), xclass(·) are given in (5.16), (5.17), and c8 > 0 is a constant. We
choose ν so that −c8Λy/2 ≤ xclass(s) ≤ x for νT < s < T . Hence if νT < s <
T, 0 < x < c8Λy and Xε(s)− xclass(s) < −c8Λy then Xε(s) < 0. Since Xε(s) > 0
for τε,linear,x,T < s < T it follows that if s satisfies τε,linear,x,T ∨ (νT ) < s < T
and |Xε(s) − xclass(s)| > c8Λy, then Xε(s) > c8Λy/2. We see therefore, using the



58 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

inequality X∗
ε (·) ≥ Xε(·), that the probability on the RHS of (6.51) is bounded

above by

wε(x) = P
(

sup
τ∗
ε,linear,x,T<s<T

X∗
ε (s) > c8Λy/2

)
. (6.52)

The function wε : [0, c8Λy/2] → [0, 1] is the solution to the boundary value problem,

−ε

2

d2wε(x)

dx2
+

y

C5T

dwε(x)

dx
= 0 ,

0 < x < c8Λy/2, wε(0) = 0, wε(c8Λy/2) = 1 ,

(6.53)

which has solution

wε(x) =
{
exp[

2xy

C5εT
]− 1

}/{
exp[

c8Λy
2

C5εT
]− 1

}
. (6.54)

Taking limits in (6.54) we see that

Λy lim
x→0

1

x
wε(x) =

2Λy2

C5εT

/{
exp[

c8Λy
2

C5εT
]− 1

}
≤ C9εT

y2
, (6.55)

for some constant C9, provided εT < y2.
We conclude now a bound on the contribution of the first term in the Taylor

expansion (6.3) to the RHS of (6.29). Using the formula (4.23) and the inequality
y ≥ C3T

2, we have that

lim sup
x→0

1

x

∂q0(0, y, T )

∂x

√
ε|E[Zε : τδ ∨ (T/2) < τ∗ε,x,T ]| ≤

C10ε

y
, (6.56)

where δ = Λy and C10 is a constant. To bound the contribution of the sec-
ond term in the Taylor expansion (6.3) we use (4.38) to obtain the inequality∣∣∂2q0(x, y, T )/∂x

2
∣∣ ≤ C11T/y, provided |x| < Λy and y ≥ C3T

2. We have then
from (5.33), (6.36) that

lim sup
x→0

1

x
ε
∣∣∣E[

Z2
ε

∫ 1

0

∫ 1

0

λ dλ dµ
∂2q0(λµ

√
εZε, y, T )

∂x2
; τδ ∨ (T/2) < τ∗ε,x,T

]∣∣∣
≤ C11εT

y
lim sup

x→0

1

x
E
[
Z2
ε ; τδ ∨ (T/2) < τ∗ε,x,T

]
≤ C12εT

2

y2
,

(6.57)

where C12 is a constant. To bound the contribution of the final term in the Taylor
expansion (6.3) we use the fact that ∂q0(0, y, t)/∂t = 0, whence

∂q0(x, y, t)

∂t
= x

∫ 1

0

∂2q0(µx, y, t)

∂t∂x′ dµ , (6.58)

where we have from (2.29) that

∂2q0(x
′, y, t)

∂t∂x′ = −
[
A(t) +

1

σ2
A(t)

]∂q0(x′, y, t)

∂x′

−
[
λ(x′, y, t) +

∂q0(x
′, y, t)

∂x′

]∂2q0(x
′, y, t)

∂x′2 .

(6.59)

It follows from (4.26) and (4.38) there are constants C11, C12 such that∣∣[A(t) +
1

σ2
A(t)

]∂q0(x′, y, t)

∂x′

∣∣ ≤ C11y

T 2
, (6.60)



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 59

∣∣[λ(x′, y, t) +
∂q0(x

′, y, t)

∂x′

]∂2q0(x
′, y, t)

∂x′2

∣∣ ≤ C12 , (6.61)

provided 0 < x′ < Λy, T/2 < t < T , and y ≥ C3T
2. Hence we may estimate the

expectation of the final term in the Taylor expansion by combining our estimate on
the expectation of the RHS of (6.34) with (6.58)-(6.61). We obtain an inequality

lim sup
x→0

1

x

∣∣∣E[
(T − τ∗ε,x,T )

∫ 1

0

dλ
∂q0(

√
εZε, y, λτ

∗
ε,x,T + (1− λ)T )

∂t
;

τδ ∨ (T/2) < τ∗ε,x,T

]∣∣∣
≤ C13ε

y
for δ = Λy, y ≥ C3T

2 .

(6.62)

It follows then from (6.56), (6.57), and (6.62) that the LHS of (6.27) is bounded by
C14ε/y provided y ≥ C3T

2 and εT ≤ y2.
This bound on the LHS of (6.27) may be improved by noting a cancellation in

the Taylor expansion (6.3). To see this we use the identity

σ2
A(T )

m1,A(T )
− σ2

A(τ)

m1,A(τ)
=

∫ T

τ

[A(t) +
1

σ2
A(t)

]
σ2
A(t)

m1,A(t)
dt . (6.63)

Hence using (6.58), (6.59), and (6.63) we have that the sum of the first and third
terms in the Taylor expansion (6.3) may be written as

∂q0(0, y, T )

∂x

√
εZε −

∫ T

τ

dt
∂q0(

√
εZε, y, t)

∂t

=
∂q0(0, y, T )

∂x

√
ε

σ2
A(T )

m1,A(T )
Z(τ)−

√
ε

∫ T

τ

dt
[
A(t) +

1

σ2
A(t)

]
×
{ σ2

A(t)

m1,A(t)

∂q0(0, y, T )

∂x
− σ2

A(τ)

m1,A(τ)

∫ 1

0

dµ
∂q0(µ

√
εZε, y, t)

∂x

}
Z(τ)

+
√
εZε

∫ T

τ

dt

∫ 1

0

dµ
[
λ(µ

√
εZε, y, t)

+
∂q0(µ

√
εZε, y, t)

∂x′

]∂2q0(µ
√
εZε, y, t)

∂x′2 ,

(6.64)

where τ = τ∗ε,x,T .

To bound the expectation of the first term on the RHS of (6.64) we observe that

lim sup
x→0

1

x

√
ε

σ2
A(T )

m1,A(T )

∣∣E[Z(τ∗ε,x,T ) : τδ ∨ (T/2) < τ∗ε,x,T ]
∣∣

≤ lim sup
x→0

1

x

[
uε(x, T/2) + C15y

{
wε(x) + P

(
τ∗ε,linear,x,T < νT

)}]
,

(6.65)

where we assume y ≥ C3T
2, εT ≤ y2,

√
εδ = Λy. Then from (6.49) and (6.55)

we see that lim supx→0 x
−1[uε(x, T/2) + ywε(x)] ≤ C16(εT/y

2)2. Instead of using

the bound P
(
τ∗ε,linear,x,T < νT

)
≤ vε(x)/(1− ν)2T 2 as in (6.42), we use the iden-

tity P
(
τ∗ε,linear,x,T < νT

)
= uε(x, (1 − ν)T ), where [x, t] → uε(x, t), x, t > 0, is

the solution to the PDE (6.44) with boundary and initial conditions uε(0, t) = 0,



60 J. G. CONLON, M. DABKOWSKI EJDE-2025/14

uε(x.0) = 1. Similarly to (6.49) we have now that

lim sup
x→0

1

x
uε(x, t) ≤

2√
2π(εt)3/2

exp[−µ2t

2ε
]

∫ ∞

0

e−µx′/εx′ dx′

=
2

µt
√
2π

( ε

µ2t

)1/2
exp[−µ2t

2ε
] .

(6.66)

It follows from (6.66) that y lim supx→0 x
−1P (τ∗ε,linear,x,T < νT ) ≤ C17(εT/y

2)2.

We conclude from (6.65) that the lim sup as x → 0 of x−1 times the expectation of
the first term on the RHS of (6.64) is bounded by C18ε

2T/y3.
Using (6.61) and arguing as in the previous paragraph, we see from (6.36), (6.38)

and the Schwarz inequality that the lim sup as x → 0 of x−1 times the expectation
of the third term on the RHS of (6.64) is bounded by C19εT

2/y2. The expectation
of the second term on the RHS of (6.64) is bounded by

C20
√
ε

T

∂q0(0, y, T )

∂x
E
[
(T − τ∗ε,x,T )

2|Z(τ∗ε,x,T )|; τδ ∨ (T/2) < τ∗ε,x,T
]

+ C21εT sup
0<x′<

√
εδ,T/2<t<T

∣∣∂2q0(x
′, y, t)

∂x′2

∣∣
× E[(T − τ∗ε,x,T )Z(τ∗ε,x,T )

2; τδ ∨ (T/2) < τ∗ε,x,T ] .

(6.67)

We bound the first expectation in (6.67) by using the Schwarz inequality and the

inequality E
[
(T − τ∗ε,x,T )

4
]
≤ E

[
(T − τ∗ε,linear,x,T )

4
]
. This latter expectation can

be estimated by considering the function

uε,α(x) = E
[
exp

{
−α(T − τ∗ε,linear,x,T )

}]
, x, α > 0 . (6.68)

Then uε,α(·) is the solution to a boundary value problem

ε

2

d2uε,α(x)

dx2
− µ

duε,α(x)

dx
= αuε,α(x) , x > 0, uε,α(0) = 1 , (6.69)

with µ = y/C5T . Evidently we have that

uε,α(x) = exp
[
−x

ε

{√
µ2 + 2εα− µ

}]
. (6.70)

From (6.70) we conclude that

E
[
(T − τ∗ε,linear,x,T )

4
]
=

( ∂

∂α

)4

uε,α(x)
∣∣∣
α=0

=
15ε3x

µ7
+

15ε2x2

µ6
+

6εx3

µ5
+

x4

µ4
.

(6.71)

It follows from (6.36), (6.71) that the lim sup of x−1 times the first expectation in
(6.67) as x → 0 is bounded by C22ε

2T/y3.
We also use the Schwarz inequality to bound the second expectation in (6.67)

by using the fact that for α ∈ R the function

s → exp
[
αZ(s)− α2

2

∫ T

s

m1,A(s
′)2

σ4
A(s

′)
ds′

]
, 0 < s < T , (6.72)



EJDE-2025/14 GREEN’S FUNCTION ON HALF LINE 61

is also a martingale. On differentiating (6.72) twice with respect to α and setting
α = 0 we see that (6.35) is a martingale. On differentiating four times we have that

s → Z(s)4 − 6Z(s)2
∫ T

s

m1,A(s
′)2

σ4
A(s

′)
ds′ + 3

(∫ T

s

m1,A(s
′)2

σ4
A(s

′)
ds′

)2

, (6.73)

for 0 < s < T , is a martingale. Hence

E
[
Z(τ∗ε,x,T )

4 ; τ∗ε,x,T > T/2
]

≤ C23

{ 1

T 4
E[(T − τ∗ε,x,T )

2] +
1

T
E[Z(T/2)2; τ∗ε,x,T < T/2]

}
.

(6.74)

We have already seen that the first expectation on the RHS of (6.74) is bounded
by a constant times vε(x) of (6.38). To bound the second expectation we proceed
in a similar way to how we bounded the expectation in (6.41). We have that

ε
σ4
A(T )

m1,A(T )2
E[Z(T/2)2; τ∗ε,x,T < T/2]

≤ C24E
[
[Xε(T/2)− xclass(T/2)]

2; τε,linear,x,T < T/2
]
,

(6.75)

and the expectation on the RHS of (6.75) can be bounded using solutions to the
PDE (6.44). Thus we have that the lim sup of x−1 times the RHS of (6.75) as x → 0
is bounded by C24ε

2T 2/y3, whence the lim sup of x−1 times the RHS of (6.74) as
x → 0 is bounded by C25ε/Ty

3. We conclude that the lim sup of x−1 times the
second expectation in (6.67) as x → 0 is bounded by C26ε

2T 3/y4 ≤ C27ε
2T/y3. □

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Joseph G. Conlon

University of Michigan, Department of Mathematics, Ann Arbor, MI 48109-1109, USA
Email address: conlon@umich.edu

Michael Dabkowski

University of Michigan-Dearborn, Department of Mathematics and Statistics, Dear-
born, MI 48128, USA

Email address: mgdabkow@umich.edu


	1. Introduction
	2. Representation and convergence of the function q
	3. Regularity and bounds on the function q
	4. Estimating solutions of the Hamilton-Jacobi PDE
	5. Uniform bounds on q and its derivatives
	6. Convergence of the function q(x,y,T)/x as 0
	References