# Stochastic Kimura Equations

Roland Riachi† and  Linan Chen‡

###### Abstract.

In this work we study the one-dimensional stochastic Kimura equation
$\partial_{t}u\left(z,t\right)=z\partial_{z}^{2}u\left(z,t\right)+u\left(z,t\right)\dot{W}\left(z,t\right)$
for $z,t>0$ equipped with a Dirichlet boundary condition at $0$, with
$\dot{W}$ being a Gaussian space-time noise. This equation can be seen as a
degenerate analog of the parabolic Anderson model. We combine the Wiener chaos
theory from Malliavin calculus, the Duhamel perturbation technique from PDEs,
and the kernel analysis of (deterministic) degenerate diffusion equations to
develop a solution theory for the stochastic Kimura equation. We establish
results on existence, uniqueness, moments, and continuity for the solution
$u\left(z,t\right)$. In particular, we investigate how the stochastic
potential and the degeneracy in the diffusion operator jointly affect the
properties of $u\left(z,t\right)$ near the boundary. We also derive explicit
estimates on the comparison under the $L^{2}-$ norm between
$u\left(z,t\right)$ and its deterministic counterpart for $\left(z,t\right)$
within a proper range.

###### Key words and phrases:

Kimura equation, parabolic Anderson model, stochastic Kimura equation,
stochastic degenerate diffusion equation

###### 2000 Mathematics Subject Classification:

60H15, 60H30, 35K65

Part of this work is supported by the NSERC Discovery Grant of the second
author (No. 241023).

$\dagger$<EMAIL_ADDRESS>
<EMAIL_ADDRESS>
###### Contents

  1. 1 Introduction
    1. 1.1 Kimura Diffusion and Parabolic Anderson Model
    2. 1.2 Stochastic Kimura Equation and Main Results
    3. 1.3 Contents and Notations
  2. 2 Background
    1. 2.1 The Malliavin Calculus
    2. 2.2 The Kimura Equation
  3. 3 Stochastic Kimura Equation
    1. 3.1 Existence and Uniqueness of a Solution
      1. 3.1.1 Space-Time White Noise
      2. 3.1.2 Colored Noise
    2. 3.2 Ratio Comparison in $L^{2}$
      1. 3.2.1 Space-Time White Noise
      2. 3.2.2 Colored Noise
    3. 3.3 $L^{p}-$ Bounds
      1. 3.3.1 Space-Time White Noise
      2. 3.3.2 Colored Noise
    4. 3.4 Hölder Continuity up to Boundary
      1. 3.4.1 Space-Time White Noise
      2. 3.4.2 Colored Noise
  4. 4 Conclusion and Further Questions
  5. 5 Appendix
  6. 6 References

##  1\. Introduction

In this work we study a class of one-dimensional stochastic degenerate
diffusion equations, which we call _stochastic Kimura equations_ , that are
diffusion equations on $\mathbb{R}_{+}$ with the diffusion coefficient
degenerating linearly at the boundary $0$ and subject to a Gaussian noise
potential on $\mathbb{R}_{+}$. This work is a marriage between the study of
degenerate diffusion, which is a classical topic with a long-celebrated
history, and the techniques from stochastic partial differential equations
(SPDEs), which is a fast developing field with many recent breakthroughs. By
combining analytic methods and stochastic analysis, we derive both qualitative
(e.g., existence and uniqueness) and quantitative (e.g., moments and
regularity) results on the solution to the stochastic Kimura equation. We put
a particular emphasis on the short-term behavior of the solution near the
boundary, dissecting the joint effect of the degenerate diffusion operator and
the space-time noise potential. To the best of our knowledge, these aspects of
stochastic degenerate diffusion equations have not been previously studied.

###  1.1. Kimura Diffusion and Parabolic Anderson Model

In this subsection we give a coarse review of the two fields, degenerate
diffusion equation and SPDEs, at the intersection of which our work lies. Both
fields have rich and vast literature. Our reviews will be limited as we will
only mention the results that are most relevant to the problem concerned in
this work. In particular, the two pillars based on which our work is carried
out are the _Kimura diffusion_ , a classical example of degenerate diffusion,
and the _parabolic Anderson model_ , a well studied SPDE model. We now give a
brief introduction to both models, but leave the technical details for the
next section.

#### 1.1.1. Kimura Diffusion

The Kimura diffusion is an important mathematical tool in population genetics
as a model of the propagation of a certain genotype among a population [15].
In the classical setting, it is described by the stochastic differential
equation (SDE)
$\mathrm{d}Z_{t}=\sqrt{2Z_{t}}\mathrm{d}B_{t}+b\left(Z_{t}\right)\mathrm{d}t$
with $Z_{0}=z>0$, $b$ being a drift coefficient and $0$ being an absorbing
boundary. Correspondingly, the _Kimura equation_ is the partial differential
equation (PDE)

(1.1)
$\partial_{t}u\left(z,t\right)=z\partial_{z}^{2}u\left(z,t\right)+b\left(z\right)\partial_{z}u\left(z,t\right),$

for $z,t>0$, equipped with the Dirichlet boundary condition at $0$.
Characterized by the degeneracy in diffusion, this pair of equations has been
of great interest to analysts and probabilists. Properties of the solutions
(e.g., existence, uniqueness, regularity, estimation, etc.) have been
extensively studied for the Kimura diffusion, as well as certain of its
generalizations, in various contexts [23, 24, 12, 21, 1]. Both analytic and
probabilistic tool-kits have been developed to treat degenerate diffusion
operators, replacing classical techniques that are restricted to strict
elliptic/parabolic operators.

The specific Kimura diffusion that will be examined later under a stochastic
potential is the model with $b\equiv 0$, but the methods developed in this
study are extendable to cases with general $b$. Let $q_{0}\left(z,w,t\right)$
denote the fundamental solution to (1.1) with $b\equiv 0$, explicitly
determined in terms of Bessel function [22, 13, 14]:

(1.2)
$q_{0}\left(z,w,t\right):=\sqrt{\mathchoice{\frac{z}{w}}{z/w}{\frac{z}{w}}{\frac{z}{w}}}\mathchoice{\frac{e^{-\mathchoice{\frac{z+w}{t}}{z+w/t}{\frac{z+w}{t}}{\frac{z+w}{t}}}}{t}}{e^{-\mathchoice{\frac{z+w}{t}}{z+w/t}{\frac{z+w}{t}}{\frac{z+w}{t}}}/t}{\frac{e^{-\mathchoice{\frac{z+w}{t}}{z+w/t}{\frac{z+w}{t}}{\frac{z+w}{t}}}}{t}}{\frac{e^{-\mathchoice{\frac{z+w}{t}}{z+w/t}{\frac{z+w}{t}}{\frac{z+w}{t}}}}{t}}I_{1}\left(\mathchoice{\frac{2\sqrt{zw}}{t}}{2\sqrt{zw}/t}{\frac{2\sqrt{zw}}{t}}{\frac{2\sqrt{zw}}{t}}\right)$

for every $z,w,t>0$. That is, if the Kimura equation is equipped with the
initial data $u_{0}\left(z\right)$, then the solution is given by

(1.3)
$u_{0}\left(z,t\right)=\int_{0}^{\infty}u_{0}\left(w\right)q_{0}\left(z,w,t\right)dw,$

for every $z,t>0$. We will refer to 1.3 as the _deterministic solution_ ,
which, as we will see later, serves as the starting point in the construction
of the solution to the stochastic Kimura equation.

In latter sections, a perturbation of the Kimura equation will also be
considered:

$\partial_{t}u\left(z,t\right)=z\partial_{z}^{2}u\left(z,t\right)+u\left(z,t\right)V\left(z,t\right),$

where $V$ is a potential function whose fundamental solution will be denoted
by $q_{0}^{V}\left(z,w,t\right)$. When $V$ is bounded and time-independent, a
sharp regularity theory on $q_{0}^{V}\left(z,w,t\right)$ is developed in [14]
via the Duhamel perturbation method, which produces accurate estimates on the
comparison between $q_{0}^{V}\left(z,w,t\right)$ and $q_{0}\left(z,w,t\right)$
near the boundary for small times. Details of these results will be presented
in $\mathsection\ref{ssec:bgKim}$.

#### 1.1.2. Parabolic Anderson Model

SPDEs have become a novel and popular tool in modeling phenomena observed in
scientific experiments in random environments. Following the seminal papers
[16, 33], many classical deterministic PDEs have been re-visited when the
domain is subject to certain random noise, and a fast-growing set of theories
is being developed for solutions to such SPDEs [16, 33, 5, 29, 31, 35, 8].
Most notably, the parabolic Anderson model (PAM) is the classical heat
equation perturbed by a Gaussian random potential. The one-dimensional model
can be written as

(1.4)
$\partial_{t}u\left(z,t\right)=\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}\partial_{z}^{2}u\left(z,t\right)+u\left(z,t\right)\dot{W}\left(z,t\right),$

for $z\in\mathbb{R}$ and $t>0$, where $\dot{W}\left(z,t\right)$ denotes a
Gaussian space-time noise on some filtered probability space
$\left(\Omega,\mathcal{F},\mathbb{P}\right)$. Formally speaking,
$\\{\dot{W}\left(z,t\right)\mid z\in\mathbb{R},t>0\\}$ can be viewed as a
centered Gaussian family with covariance

(1.5)
$\mathbb{E}\left[\dot{W}\left(z_{1},t_{1}\right)\dot{W}\left(z_{2},t_{2}\right)\right]=f\left(z_{1}-z_{2}\right)\gamma\left(t_{1}-t_{2}\right),$

for $z_{1},z_{2}\in\mathbb{R}$ and $t_{1},t_{2}>0$, where
$f:\mathbb{R}\rightarrow\mathbb{R}^{+}$ and
$\gamma:\mathbb{R}\rightarrow\mathbb{R}^{+}$ are two non-negative definite,
symmetric, and locally integrable kernels. To solve (1.4) with (deterministic)
initial data $u_{0}\left(z\right)$, we seek a process
$\left\\{u\left(z,t\right)\mid z\in\mathbb{R},t\geq 0\right\\}\subseteq
L^{2}\left(\Omega\right)$, to which we refer as a _(mild) solution_ to (1.4),
such that for every $z\in\mathbb{R}$ and $t>0$,

$u\left(z,t\right)=\int_{\mathbb{R}}g_{t}\left(z,w\right)u_{0}\left(w\right)dw+\int_{0}^{t}\int_{\mathbb{R}}g_{t-\tau}\left(z,w\right)u\left(w,\tau\right)W\left(dw,d\tau\right)\text{
a.s.},$

where $g_{t}$$\left(z,w\right)=e^{-\left|z-w\right|^{2}/2}/\sqrt{2\pi t}$ is
the standard heat kernel, and the stochastic integral with “respect to
$W\left(dw,d\tau\right)$” is defined as a generalized Itô integral, following
the theory of Malliavin calculus. In $\mathsection\ref{ssec:bgMalCal}$, we
review all the necessary elements from Malliavin calculus, including the
construction and the properties of such stochastic integrals.

Common approaches toward the study of PAM combine Malliavin calculus with
classical heat kernel analysis, yielding abundant results on the solution [16,
33, 6, 4, 3, 10]. Alternatively, another approach is the theory of rough paths
developed by Hairer [27, 30]. The PAM has also been studied in other settings,
e.g., on $\mathbb{Z}^{d}$ [26], on the hypercube [2], and on Heisenberg groups
[7], as well as subject to Lévy random noise [8]. For degenerate parabolic
SPDEs, the Cauchy problem has been studied in the semilinear and quasilinear
cases [32, 18].

###  1.2. Stochastic Kimura Equation and Main Results

Inspired by the works reviewed in the previous subsection, we decide to bring
the two models, the Kimura diffusion and the PAM, together and investigate the
following stochastic Kimura equation:

(1.6)
$\begin{cases}&\partial_{t}u\left(z,t\right)=z\partial_{z}^{2}u\left(z,t\right)+u\left(z,t\right)\dot{W}\left(z,t\right)\quad\quad
z,t>0,\\\ &\hskip 9.10486ptu(z,0)=u_{0}(z)\hfill z>0,\\\ &\hskip
10.81204ptu(0,t)=0\hfill t>0.\end{cases}$

where $\dot{W}$ is the same as in (1.5). Similarly as in the PAM case, a
(mild) solution to (1.6) is a process $\left\\{u\left(z,t\right)\mid z,t\geq
0\right\\}\subseteq L^{2}\left(\Omega\right)$ such that for every $z,t>0$,

$\displaystyle u\left(z,t\right)=\int_{0}^{\infty}q_{0}\left(z,w,t\right)$
$\displaystyle u_{0}\left(w\right)dw$ (1.7)
$\displaystyle+\int_{0}^{t}\int_{0}^{\infty}q_{0}\left(z,w,t-\tau\right)u\left(w,\tau\right)W\left(dw,d\tau\right)\text{
a.s.}$

The purpose of studying (1.6) has multiple folds. First, we aim to develop a
solution theory for stochastic degenerate diffusion equations, which would
yield applications in modeling diffusion processes in random mediums with
absorbing boundaries, e.g., genotype propagation with random
hibernation/revival. The model of the stochastic Kimura equation is a good
starting point since the deterministic equation is well studied with abundant
results on the fundamental solution $q_{0}\left(z,w,t\right)$. Second, we want
to extend the existing literature on the PAM to the settings of degenerate
diffusions and investigate how the current methods in constructing the
solution to the PAM, as well as the properties of the solution, may change in
the presence of degeneracy in the diffusion. Furthermore, we are interested in
studying the interaction between the degenerate diffusion operator and the
stochastic potential in order to analyze their “joint effect” on the solution;
to this end, we will pay particular attention to the behavior of the solution
near the boundary $0$, and compare the solution to the deterministic solution
in a proper sense.

In $\mathsection\ref{sec:eu}$ we prove that (1.6) admits a solution
$\left\\{u\left(z,t\right)\mid z,t\geq 0\right\\}$ that satisfies (1.7) and
that this solution is unique up to modification (Theorems 3.1 and 3.2). We
adopt a strategy that is similar to the one invoked in [5, 33], which is to
combine the Wiener chaos expansion from the Malliavin calculus and the Duhamel
method from PDEs. However, the techniques we apply are substantially different
from those utilized in the PAM case, because (1) although
$q_{0}\left(z,w,t\right)$ is explicitly determined, unlike
$g_{t}\left(z,w\right)$, $q_{0}\left(z,w,t\right)$ does not have a closed-form
expression, nor certain desirable features such as symmetry; and (2) since our
model is restricted on $\mathbb{R}_{+}$, some convenient technical elements
based on Fourier analysis are not applicable in our case. We establish our
result in two steps: first we solve (1.6) when $\dot{W}$ is the space-time
white noise, i.e., when both $f$ and $\gamma$ in (1.5) are Dirac delta
$\delta_{0}$, and next extend the construction of the solution to the colored
noise setting.

Taking a close look at the behaviors of $u\left(z,t\right)$, we observe that
the impact of the stochastic potential is most significant near the boundary,
which makes the behavior of $u\left(z,t\right)$ for small $z$ fundamentally
different from that of the solution to the deterministic Kimura equation
$u_{0}\left(z,t\right)$. To better understand the role of $\dot{W}$ near the
boundary, in $\mathsection\ref{sec:ratio}$ we also introduce a degenerate
coefficient to $\dot{W}$ and consider the following stochastic Kimura equation
with a degenerate noise. That is, for some $\beta>0$

(1.8)
$\partial_{t}u\left(z,t\right)=z\partial_{z}^{2}u\left(z,t\right)+u\left(z,t\right)\hat{z}^{\beta}\dot{W}\left(z,t\right),$

where $\hat{z}:=z\wedge 1$. Similarly, (1.8) also admits a unique solution
$\left\\{u\left(z,t\right)\mid z,t\geq 0\right\\}$. It turns out that so long
as $\beta>0$, the degeneracy in the stochastic potential will “tame”
$u\left(z,t\right)$ enough so that it remains comparable to
$u_{0}\left(z,t\right)$ under the $L^{2}-$ norm for some finite time $t$ and
for $z$ all the way to the boundary.

To be specific, we prove in Theorem 3.4 that, when $u_{0}\equiv 1$ and $f$
satisfies some additional assumptions, there exists $T\in(0,1)$ such that

(1.9) $\sup_{\begin{subarray}{c}z>0\\\
t\in[0,T]\end{subarray}}\mathbb{E}\left[\left(\mathchoice{\frac{u\left(z,t\right)}{u_{0}\left(z,t\right)}}{u\left(z,t\right)/u_{0}\left(z,t\right)}{\frac{u\left(z,t\right)}{u_{0}\left(z,t\right)}}{\frac{u\left(z,t\right)}{u_{0}\left(z,t\right)}}\right)^{2}\right]<\infty.$

In addition, we also obtain that at every fixed $z>0$,

(1.10) $\lim_{t\searrow
0}\mathbb{E}\left[\left(\mathchoice{\frac{u\left(z,t\right)}{u_{0}\left(z,t\right)}}{u\left(z,t\right)/u_{0}\left(z,t\right)}{\frac{u\left(z,t\right)}{u_{0}\left(z,t\right)}}{\frac{u\left(z,t\right)}{u_{0}\left(z,t\right)}}-1\right)^{2}\right]=0,$

i.e., $u\left(z,t\right)$ stays close to $u_{0}\left(z,t\right)$, again, in
the $L^{2}-$ sense, for sufficiently small $t$, which is consistent with
(3.10) in [14], a result established for the deterministic equation but with a
bounded potential.

In $\mathsection\ref{sec:cont}$ we examine the regularity of
$u\left(z,t\right)$, particularly its continuity in $z$ at the boundary. We
discover that, in order to acquire boundary continuity, we need to impose
further assumptions on the rate of degeneracy of the stochastic potential.
Indeed, when $\beta>1/4$ we are able to prove a Hölder continuity result on
$u\left(z,t\right)$ in $z$ all the way to the boundary in Theorem 3.8. Along
the way, in $\mathsection\ref{sec:moment}$ we also obtain results on $L^{p}-$
integrability for all $p\geq 2$, as well as higher moment estimates, for
$u\left(z,t\right)$ for all $z,t>0$.

The aforementioned results are only part of our study of the stochastic Kimura
equation as the first step in an attempt to advance our general knowledge on
stochastic degenerate diffusion equations. There are other aspects of the work
that are still ongoing and will not be addressed here. For example, our
methods can be extended to study the following two variations of (1.6) without
substantial difficulty: (1) with an order $\alpha$ of degeneracy in the
diffusion operator for $\alpha$ from some proper range, i.e.,

$\partial_{t}u\left(z,t\right)=z^{\alpha}\partial_{z}^{2}u\left(z,t\right)+u\left(z,t\right)\dot{W}\left(z,t\right),$

and (2) with a drift coefficient $b$ satisfying some proper conditions, i.e.,

$\partial_{t}u\left(z,t\right)=z\partial_{z}^{2}u\left(z,t\right)+b\left(z\right)\partial_{z}u\left(z,t\right)+u\left(z,t\right)\dot{W}\left(z,t\right).$

In addition, the framework developed in this paper also allows us to study
long-term asymptotics of $u\left(z,t\right)$ and potentially establish results
on intermittency.

###  1.3. Contents and Notations

In $\mathsection\ref{sec:bg}$ we review the necessary elements from the
Malliavin calculus and technical results from the study of the deterministic
Kimura equation. In $\mathsection\ref{sec:results}$, we set up the model of
the stochastic Kimura equation then proceed to study various aspects of the
solution and prove results on existence, uniqueness, estimates, and continuity
of the solution. $\mathsection\ref{sec:conclusion}$ outlines some related
problems and future directions. Useful formulas on Bessel functions are
deposited in $\mathsection\ref{sec:appendix}$.

We write $\mathbb{R}_{+}$ to mean the interval $(0,\infty)$. For
$a,b\in\mathbb{R}$, we write $a\vee b:=\max\left\\{a,b\right\\}$ and $a\wedge
b:=\min\left\\{a,b\right\\}$. For $z>0$, we denote $\hat{z}:=z\wedge 1$. We
denote by $\mathcal{D}(\mathbb{R}_{+}^{2})$ the set of all smooth, compactly
supported functions defined on $\mathbb{R}_{+}^{2}$. We denote by
$C_{b}(\mathbb{R}_{+})$ the space of all bounded continuous functions defined
on $\mathbb{R}_{+}$, equipped with the uniform norm

$\|f\|_{u}:=\sup_{x>0}|f(x)|.$

For $\alpha\in\mathbb{C}$, we denote by $I_{\alpha}$ and $J_{\alpha}$ the
(modified) Bessel functions of the first kind of order $\alpha$. For $q\geq
p\geq 0$, $a_{1},\ldots,a_{p},b_{1},\ldots,b_{q}\in\mathbb{R}$, and
$x\in\mathbb{C}$, we denote by

${}_{p}F_{q}\left[\begin{array}[]{cc}a_{1},\ldots,a_{p}\\\
b_{1},\ldots,b_{q}\end{array}\mathrel{\Big{|}}x\right]$

the generalized Hypergeometric function. The definitions of these special
functions are reviewed in $\mathsection\ref{sec:appendix}$.

##  2\. Background

###  2.1. The Malliavin Calculus

In this section, we present some prerequisite material regarding the Malliavin
calculus which is essential for discussing the model of the stochastic Kimura
equation. For details, we refer the reader to [34, 17, 28, 36].

Let $f$ and $\gamma$ be as in (1.5), and consider the inner product

(2.1)
$\left\langle\varphi,\psi\right\rangle_{H}:=\int_{\mathbb{R}_{+}^{4}}f(x-y)\gamma(r-s)\varphi(x,r)\psi(y,s)\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s,$

for $\varphi,\psi\in\mathcal{D}(\mathbb{R}_{+}^{2})$. We define the Hilbert
space $(H,\left\langle\cdot,\cdot\right\rangle_{H})$ as the completion of
$\mathcal{D}(\mathbb{R}_{+}^{2})$ with respect to
$\left\langle\cdot,\cdot\right\rangle_{H}$. It can be shown that there exists
an isonormal Gaussian process $W=\left\\{W(h)\mid h\in H\right\\}$, on an
appropriate probability space $(\Omega,\mathcal{F},\mathbb{P})$, such that

$\displaystyle\mathbb{E}[W(h)]\equiv 0$
$\displaystyle\mathbb{E}[W(h_{1})W(h_{2})]=\left\langle
h_{1},h_{2}\right\rangle_{H}.$

When viewed as a map between Hilbert spaces, $W:H\to
W(H)=:\mathcal{H}\subseteq L^{2}(\Omega)$, is an isometry. This framework is
convenient because it allows us to study the noise process $W$ through well-
behaved deterministic functions.

As it turns out, if $\mathcal{F}$ is the $\sigma$-algebra generated by $W$,
then $L^{2}(\Omega)$ admits a decomposition into countably many closed images
of $W$ under a special class of polynomials. Let $n\geq 1$ and $H_{n}$ denote
the $n$th Hermite polynomial. The _$n$ th Wiener chaos space_
$\mathcal{H}_{n}$ is defined as

(2.2) $\mathcal{H}_{n}:=\overline{\text{span}\left\\{H_{n}(W(h))\mid h\in
H,\|h\|_{H}=1\right\\}},$

where the closure is taken in $L^{2}(\Omega)$. Then,

(2.3) $L^{2}(\Omega)=\bigoplus_{n=0}^{\infty}\mathcal{H}_{n},$

i.e., for any $X\in L^{2}(\Omega)$, for each $n\geq 0$ there exists a unique
$X_{n}\in\mathcal{H}_{n}$ such that

(2.4) $X=\sum_{n=0}^{\infty}X_{n},$

where the series converges in $L^{2}(\Omega)$. We remark that since
$H_{0}\equiv 1$, $\mathcal{H}_{0}\cong\mathbb{R}$ is a space of constants and
it can be shown that $X_{0}=\mathbb{E}[X]$. By [33] (see the last line of page
62), for every $n\geq 1$ and $p\geq 2$, $\mathcal{H}_{n}\subseteq
L^{p}(\Omega)$. In fact, if $Y\in\mathcal{H}_{n}$, then

(2.5)
$\mathbb{E}[|Y|^{p}]^{\mathchoice{\frac{1}{p}}{1/p}{\frac{1}{p}}{\frac{1}{p}}}\leq(p-1)^{\mathchoice{\frac{n}{2}}{n/2}{\frac{n}{2}}{\frac{n}{2}}}\mathbb{E}[Y^{2}]^{\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}.$

Below, we will demonstrate representations for these projections $X_{n}$ in
terms of stochastic integrals in the sense of Skorohod, whose moments are
related to the inner product $\left\langle\cdot,\cdot\right\rangle_{H}$.

To this end, we first define a suitable notion of derivative. Let
$C_{p}^{\infty}(\mathbb{R}^{N})$ denote the space of infinitely differentiable
functions $F:\mathbb{R}^{N}\to\mathbb{R}$ which (along with their partial
derivatives) have at most polynomial growth. We say a random variable $X\in
L^{2}(\Omega)$ is _smooth_ if there exists $F\in
C_{p}^{\infty}(\mathbb{R}^{N})$ and $h_{1},\ldots,h_{N}\in H$ such that
$X=F(W(h_{1}),\ldots,W(h_{N}))$. Moreover, we denote by $\mathcal{P}\subseteq
L^{2}(\Omega)$ the set of all such smooth random variables.

Consequently, for some $X=F(W(h_{1}),\ldots,W(h_{N}))\in\mathcal{P}$, the
_Malliavin derivative_ is an unbounded, closed linear operator
$D:\mathcal{P}\to L^{2}(\Omega;H)$ given by

(2.6) $DX:=\sum_{j=1}^{N}\partial_{j}F(W(h_{1}),\ldots,W(h_{N}))h_{j},$

where for $z,t>0$,

$D_{z,t}X=\sum_{j=1}^{N}\partial_{j}F(W(h_{1}),\ldots,W(h_{N}))h_{j}(z,t),$

and without loss of generality, we may assume that $h_{1},\ldots,h_{N}$ are
orthonormal under $\left\langle\cdot,\cdot\right\rangle_{H}$.

As with classical Sobolev spaces, we define the inner product

(2.7) $\left\langle
X,Y\right\rangle_{1,2}:=\mathbb{E}[XY]+\mathbb{E}[\left\langle
DX,DY\right\rangle_{H}]$

for $X,Y\in\mathcal{P}$, and let $\mathbb{D}^{1,2}$ denote the closure of
$\mathcal{P}$ with respect to $\left\langle\cdot,\cdot\right\rangle_{1,2}$.
Note that $\mathcal{P}\subseteq\mathbb{D}^{1,2}\subseteq L^{2}(\Omega)$, and
since $\mathcal{P}$ is dense in $L^{2}(\Omega)$, $\mathcal{P}$ is dense in
$\mathbb{D}^{1,2}$ and hence $D$ can be extended to $\mathbb{D}^{1,2}$.

Viewing $D:\mathbb{D}^{1,2}\to L^{2}(\Omega;H)$, we are ready to introduce the
notion of stochastic integral relevant to (1.6). Let the _divergence operator_
$\delta$ be the adjoint operator of $D$. Namely, the domain
$\text{Dom}(\delta)$ of $\delta$ is the set of random variables $u\in
L^{2}(\Omega;H)$ for which there exists $c_{u}>0$ such that,

$|\mathbb{E}[\left\langle DX,u\right\rangle_{H}]|\leq c_{u}\|X\|_{1,2},$

for all $X\in\mathbb{D}^{1,2}$; in this case, $\delta(u)$ is the unique
element of $L^{2}(\Omega)$ such that

(2.8) $\mathbb{E}[\delta(u)X]=\mathbb{E}[\left\langle DX,u\right\rangle_{H}].$

for all $X\in\mathbb{D}^{1,2}$. As the adjoint of an unbounded and densely
defined operator, $\delta$ is in turn also closed. Furthermore, we use the
notation

(2.9) $\delta(u):=\int_{\mathbb{R}_{+}^{2}}u(z,t)W(\mathrm{d}z,\mathrm{d}t),$

where $W(\mathrm{d}z,\mathrm{d}t)=\dot{W}(z,t)\mathrm{d}z\mathrm{d}t$
(formally, $\dot{W}(z,t)=\partial_{t}\partial_{z}W(z,t)$). This nomenclature
is motivated by the fact that the divergence operator possesses many desirable
properties of a typical integral and coincides with the Itô stochastic
integral in certain cases. For this reason, we also refer to $\delta$ as the
_Skorohod integral_ , and if $u\in\text{Dom}(\delta)$, we say that $u$ is
_Skorohod integrable_.

Continuing along these lines, we define the multiple Skorohod integral to
arrive at representations for the random variables $X_{n}$ in (2.4). For
$n\geq 1$, let $H^{\otimes n}$ denote the $n$th tensor product of $H$, with
the corresponding inner product and norm denoted by
$\left\langle\cdot,\cdot\right\rangle_{n}$ and $\|\cdot\|_{n}$, respectively.
Then for $f_{n}\in H^{\otimes n}$, iteratively applying the Skorohod integral
yields

(2.10)
$\mathcal{I}_{n}(u):=\int_{\mathbb{R}_{+}^{2n}}f_{n}(z_{1},t_{1},\ldots,z_{n},t_{n})W(\mathrm{d}z_{1},\mathrm{d}t_{1})\cdots
W(\mathrm{d}z_{n},\mathrm{d}t_{n}).$

Given $f_{n}\in H^{\otimes n}$ and $g_{m}\in H^{\otimes m}$, we denote by
$\tilde{f}_{n}$ the symmetrization of $f_{n}$, given by

(2.11)
$\tilde{f}_{n}(z_{1},t_{1},\ldots,z_{n},t_{n}):=\mathchoice{\frac{1}{n!}}{1/n!}{\frac{1}{n!}}{\frac{1}{n!}}\sum_{\sigma\in
S_{n}}f(z_{\sigma(1)},t_{\sigma(1)},\ldots,z_{\sigma(n)},t_{\sigma(n)}),$

where $S_{n}$ is the set of all permutations on $n$ elements. Then
$\mathcal{I}_{n}(f_{n})=\mathcal{I}_{n}(\tilde{f}_{n})$ and

(2.12)
$\mathbb{E}[\mathcal{I}_{n}(f_{n})\mathcal{I}_{m}(g_{m})]=\begin{cases}n!\langle\tilde{f}_{n},\tilde{g}_{m}\rangle_{n}&\text{
if }n=m\\\ 0&\text{ otherwise }\end{cases}.$

Conversely, for every $n\geq 1$, $X_{n}\in\mathcal{H}_{n}$ and there exists a
unique $f_{n}\in\mathbb{R}_{+}^{2n}\to\mathbb{R}$ symmetric such that
$X_{n}=\mathcal{I}_{n}(f_{n})$. In light of this, (2.4) becomes

(2.13) $X=\sum_{n=0}^{\infty}\mathcal{I}_{n}(f_{n}),$

where $\mathcal{I}_{0}(f_{0})=\mathbb{E}[X]$. Similarly, if $u\in
L^{2}(\Omega;H)$, then for every $z,t>0$, there exists a unique
$f_{n}(\cdot;z,t)\in H^{\otimes n}$ symmetric in it first $n$ pairs of
variables such that

(2.14) $u(z,t)=\sum_{n=0}^{\infty}\mathcal{I}_{n}(f_{n}(\cdot;z,t)),$

and $\mathcal{I}_{0}(f_{0}(z,t))=\mathbb{E}[u(z,t)]$.

We close this section by presenting the following equivalent criterion for
Skorohod integrability.

######  Proposition 2.1 (Proposition 1.3.7 of [36]).

Let $u\in L^{2}(\Omega;H)$ with the expansion (2.14). As a function
$\mathbb{R}_{+}^{2(n+1)}\to\mathbb{R}$, $f_{n}\in H^{\otimes(n+1)}$. Then
$u\in\text{Dom}(\delta)$ if and only if the series

(2.15) $\delta(u)=\sum_{n=0}^{\infty}\mathcal{I}_{n+1}(f_{n})$

converges in $L^{2}(\Omega)$.

To check convergence of the series (2.15) in $L^{2}(\Omega)$, we often make
use of the property (2.12) of the multiple Skorohod integral.

###  2.2. The Kimura Equation

In this section, we discuss the _deterministic_ Kimura equation, which is the
other prerequisite for studying (1.6). We will rely on the results on the
deterministic Kimura equation to investigate the stochastic analog. Recall
that we are interested in the Kimura equation (1.1) with $b\equiv 0$. In this
section, following [14], we present an introduction to the slightly more
general case of a constant drift $b\equiv\nu<1$. That is, we explore the
equation

(2.16) $\partial_{t}u(z,t)=z\partial_{z}^{2}u(z,t)+\nu\partial_{z}u(z,t),$

for $z,t>0$, which has the fundamental solution

(2.17)
$q_{\nu}(z,w,t):=\mathchoice{\frac{z^{\mathchoice{\frac{1-\nu}{2}}{1-\nu/2}{\frac{1-\nu}{2}}{\frac{1-\nu}{2}}}w^{\mathchoice{\frac{\nu-1}{2}}{\nu-1/2}{\frac{\nu-1}{2}}{\frac{\nu-1}{2}}}}{t}}{z^{\mathchoice{\frac{1-\nu}{2}}{1-\nu/2}{\frac{1-\nu}{2}}{\frac{1-\nu}{2}}}w^{\mathchoice{\frac{\nu-1}{2}}{\nu-1/2}{\frac{\nu-1}{2}}{\frac{\nu-1}{2}}}/t}{\frac{z^{\mathchoice{\frac{1-\nu}{2}}{1-\nu/2}{\frac{1-\nu}{2}}{\frac{1-\nu}{2}}}w^{\mathchoice{\frac{\nu-1}{2}}{\nu-1/2}{\frac{\nu-1}{2}}{\frac{\nu-1}{2}}}}{t}}{\frac{z^{\mathchoice{\frac{1-\nu}{2}}{1-\nu/2}{\frac{1-\nu}{2}}{\frac{1-\nu}{2}}}w^{\mathchoice{\frac{\nu-1}{2}}{\nu-1/2}{\frac{\nu-1}{2}}{\frac{\nu-1}{2}}}}{t}}e^{-\mathchoice{\frac{z+w}{t}}{z+w/t}{\frac{z+w}{t}}{\frac{z+w}{t}}}I_{1-\nu}\left(\mathchoice{\frac{2\sqrt{zw}}{t}}{2\sqrt{zw}/t}{\frac{2\sqrt{zw}}{t}}{\frac{2\sqrt{zw}}{t}}\right).$

In particular, for $\nu=0$, we recover (1.2). Furthermore, when $\nu=0$, we
have the following estimate.

######  Proposition 2.2.

For every $z,w,t>0$ we have

(2.18)
$q_{0}(z,w,t)\leq\mathchoice{\frac{z}{t^{2}}}{z/t^{2}}{\frac{z}{t^{2}}}{\frac{z}{t^{2}}}e^{-\mathchoice{\frac{(\sqrt{z}-\sqrt{w})^{2}}{t}}{(\sqrt{z}-\sqrt{w})^{2}/t}{\frac{(\sqrt{z}-\sqrt{w})^{2}}{t}}{\frac{(\sqrt{z}-\sqrt{w})^{2}}{t}}}$

and further if $zw\geq t^{2}$, then there exists a constant $C>0$ such that

(2.19) $q_{0}(z,x,t)\leq
C\cdot\mathchoice{\frac{z^{\mathchoice{\frac{1}{4}}{1/4}{\frac{1}{4}}{\frac{1}{4}}}w^{-\mathchoice{\frac{3}{4}}{3/4}{\frac{3}{4}}{\frac{3}{4}}}}{\sqrt{t}}}{z^{\mathchoice{\frac{1}{4}}{1/4}{\frac{1}{4}}{\frac{1}{4}}}w^{-\mathchoice{\frac{3}{4}}{3/4}{\frac{3}{4}}{\frac{3}{4}}}/\sqrt{t}}{\frac{z^{\mathchoice{\frac{1}{4}}{1/4}{\frac{1}{4}}{\frac{1}{4}}}w^{-\mathchoice{\frac{3}{4}}{3/4}{\frac{3}{4}}{\frac{3}{4}}}}{\sqrt{t}}}{\frac{z^{\mathchoice{\frac{1}{4}}{1/4}{\frac{1}{4}}{\frac{1}{4}}}w^{-\mathchoice{\frac{3}{4}}{3/4}{\frac{3}{4}}{\frac{3}{4}}}}{\sqrt{t}}}e^{-\mathchoice{\frac{(\sqrt{z}-\sqrt{w})^{2}}{t}}{(\sqrt{z}-\sqrt{w})^{2}/t}{\frac{(\sqrt{z}-\sqrt{w})^{2}}{t}}{\frac{(\sqrt{z}-\sqrt{w})^{2}}{t}}}.$

In addition, the authors of [14] compare the behavior of $q_{\nu}(z,w,t)$ with
that of the solution to (2.16) subject to a bounded, deterministic potential
$V$:

(2.20)
$\partial_{t}u^{V}(z,t)=z\partial_{z}^{2}u^{V}(z,t)+\nu\partial_{z}u^{V}(z,t)+u^{V}(z,t)V(z).$

Namely, if $q_{\nu}^{V}(z,w,t)$ denotes the fundamental solution to (2.20),
then the authors show that

(2.21)
$\sup_{z,w>0}\absolutevalue{\mathchoice{\frac{q_{\nu}^{V}(z,w,t)}{q_{\nu}(z,w,t)}}{q_{\nu}^{V}(z,w,t)/q_{\nu}(z,w,t)}{\frac{q_{\nu}^{V}(z,w,t)}{q_{\nu}(z,w,t)}}{\frac{q_{\nu}^{V}(z,w,t)}{q_{\nu}(z,w,t)}}-1}\leq
e^{t\|V\|_{u}}-1,$

which allows them to establish a sharp regularity theory on
$q_{\nu}^{V}(z,w,t)$ for small $t$.

Following the Duhamel perturbation method, it is possible to express the
fundamental solution $q_{\nu}^{V}(z,w,t)$ of (2.20) in terms of
$q_{\nu}(z,w,t)$. In $\mathsection\ref{sec:eu}$, we apply a similar method to
(1.6); therefore, as a primer we include the Duhamel perturbation method
derivation in [14].

For $z,w,t>0$, $q_{\nu}^{V}(z,w,t)$ satisfies the integral equation

(2.22)
$q_{\nu}^{V}(z,w,t)=q_{\nu}(z,w,t)+\int_{0}^{t}\int_{0}^{\infty}q_{\nu}(z,x,t-\tau)q_{\nu}^{V}(x,w,\tau)\mathrm{d}x\mathrm{d}\tau.$

Notice that $q_{\nu}^{V}$ appears recursively in the integrand. Formally
unfolding (2.22) once, we obtain

$\displaystyle q_{\nu}^{V}(z,w,t)$
$\displaystyle=q_{\nu}(z,w,t)+\int_{0}^{t}\int_{0}^{\infty}q_{\nu}(z,x,t-\tau)q_{\nu}^{V}(x,w,\tau)V(x)\mathrm{d}x\mathrm{d}\tau$
$\displaystyle=q_{\nu}(z,w,t)+\int_{0}^{t}\int_{0}^{\infty}q_{\nu}(z,x,t-\tau)\biggr{(}q_{\nu}(x,w,\tau)$
$\displaystyle\hskip
71.13188pt\left.+\int_{0}^{\tau}\int_{0}^{\infty}q_{\nu}(x,x^{\prime},\tau-\tau^{\prime})q_{\nu}(x^{\prime},w,\tau^{\prime})V(x^{\prime})\mathrm{d}x^{\prime}\mathrm{d}\tau^{\prime}\right)V(x)\mathrm{d}x\mathrm{d}\tau$
$\displaystyle=q_{\nu}(z,w,t)+\int_{0}^{t}\int_{0}^{\infty}q_{\nu}(z,x,t-\tau)q_{\nu}(x,w,\tau)V(x)\mathrm{d}x\mathrm{d}\tau$
$\displaystyle+\int_{0}^{t}\int_{0}^{\tau}\int_{\mathbb{R}_{+}^{2}}q_{\nu}(z,x,t-\tau)q_{\nu}(x,x^{\prime},\tau-\tau^{\prime})q_{\nu}^{V}(x^{\prime},w,\tau^{\prime})V(x^{\prime})V(x)\mathrm{d}x^{\prime}\mathrm{d}x\mathrm{d}\tau^{\prime}\mathrm{d}\tau.$

Continuing this procedure, set $q_{\nu,0}(z,w,t):=q_{\nu}(z,w,t)$ and for
$n\geq 1$, if we inductively define

$q_{\nu,n}(z,w,t):=\int_{0}^{t}\int_{0}^{\infty}q_{\nu}(z,z_{n},t-t_{n})q_{\nu,n-1}(z_{n},w,t_{n})V(z_{n})\mathrm{d}z_{n}\mathrm{d}t_{n},$

then after unfolding $n$ times, we obtain

$\displaystyle
q_{\nu}^{V}(z,w,t)=\sum_{k=0}^{n}q_{\nu,k}(z,w,t)+\int_{\Delta_{n+1}t}\int_{\mathbb{R}_{+}^{n+1}}q_{\nu}^{V}(z_{0},w,t_{0})\prod_{i=1}^{n+1}q_{\nu}(z_{i},z_{i-1},t_{i}-t_{i-1})V(z_{i})\mathrm{d}\mathbf{z}\mathrm{d}\mathbf{t},$

where $\Delta_{n+1}t:=\left\\{(t_{0},\ldots,t_{n})\in(0,t)^{n+1}\mid
0<t_{0}<\cdots<t_{n}<t\right\\}$, $\mathbf{z}=(z_{0},\ldots,z_{n})$,
$\mathbf{t}=(t_{0},\ldots,t_{n})$, and we put $z_{n+1}=z$ and $t_{n+1}=t$.
Iterating this process at infinitum, it can be shown that

(2.23) $q_{\nu}^{V}(z,w,t):=\sum_{n=0}^{\infty}q_{\nu,n}(z,w,t),$

converges absolutely and satisfies (2.22). Using the fact that $q_{\nu}$
satisfies the Kolmogorov-Chapman equations, by a simple inductive argument,

$q_{\nu,n}(z,w,t)\leq\mathchoice{\frac{t^{n}\|V\|_{u}^{n}}{n!}}{t^{n}\|V\|_{u}^{n}/n!}{\frac{t^{n}\|V\|_{u}^{n}}{n!}}{\frac{t^{n}\|V\|_{u}^{n}}{n!}}q_{\nu}(z,w,t),$

from which (2.21) directly follows.

Intuitively, when $t$ is small, the potential $V$ has not yet generated enough
“perturbation”, so $q_{\nu}^{V}(z,w,t)$ is expected to be close to
$q_{\nu}(z,w,t)$. (2.21) not only gives a precise interpretation of this fact,
since it implies that the difference

$\absolutevalue{\mathchoice{\frac{q_{\nu}^{V}(z,w,t)}{q_{\nu}(z,w,t)}}{q_{\nu}^{V}(z,w,t)/q_{\nu}(z,w,t)}{\frac{q_{\nu}^{V}(z,w,t)}{q_{\nu}(z,w,t)}}{\frac{q_{\nu}^{V}(z,w,t)}{q_{\nu}(z,w,t)}}-1}=O(t),$

for small $t$, but the estimate is also uniform in the spatial variables
$(z,w)$ which is sharper and more accurate than the standard heat kernel
estimate.

##  3\. Stochastic Kimura Equation

In this section, we combine the methods from [5, 4, 33] and the results from
[14, 13] to study various aspects of the stochastic Kimura equation

$\begin{cases}&\partial_{t}u\left(z,t\right)=z\partial_{z}^{2}u\left(z,t\right)+u\left(z,t\right)\dot{W}\left(z,t\right)\quad\quad
z,t>0,\\\ &\hskip 9.10486ptu(z,0)=u_{0}(z)\hfill z>0,\\\ &\hskip
10.81204ptu(0,t)=0\hfill t>0,\end{cases}.$

such as existence and uniqueness of a solution, the moment bounds, and the
regularity of its sample paths. We also consider the variation wherein a
degeneracy is introduced to the noise in the equation,

(3.1)
$\begin{cases}&\partial_{t}u\left(z,t\right)=z\partial_{z}^{2}u\left(z,t\right)+\hat{z}^{\beta}u\left(z,t\right)\dot{W}\left(z,t\right)\quad\quad
z,t>0,\\\ &\hskip 9.10486ptu(z,0)=u_{0}(z)\hfill z>0,\\\ &\hskip
10.81204ptu(0,t)=0\hfill t>0,\end{cases}$

and investigate how the choice of $\beta>0$ affects the various properties
studied. Moreover, we compare the behaviors of the solutions of (1.6) and
(3.1) to that of (2.16) with $\nu=0$ near the boundary 0 and for small times.

###  3.1. Existence and Uniqueness of a Solution

In this section, we prove that (1.6) admits a uniqueness mild solution, up to
a modification.

######  Definition 3.1.

A square-integrable random field $u=\left\\{u(z,t)\mid z,t\geq 0\right\\}$ is
a _mild solution_ to the stochastic Kimura equation (1.6) with deterministic
initial condition $u_{0}\in C_{b}(\mathbb{R}_{+})$ if

  1. (a)

$u$ has a jointly measurable modification (again denoted by $u$) and for all
$z,t\geq 0$,

(3.2) $\mathbb{E}[u^{2}(z,t)]<\infty;$

  2. (b)

For all $z,t>0$ the process
$\left\\{\mathbbm{1}_{(0,t)}(\tau)q_{0}(z,w,t-\tau)u(w,\tau)\mid
w,\tau>0\right\\}$ is Skorohod integrable and

(3.3)
$u(z,t)=\int_{0}^{\infty}q_{0}(z,w,t)u_{0}(w)\mathrm{d}w+\int_{0}^{t}\int_{0}^{\infty}q_{0}(z,w,t-\tau)u(w,\tau)W(\mathrm{d}w,\mathrm{d}\tau).$

Suppose $u$ is a mild solution to (1.6), then for every $z,t>0$, $u$ admits
the following Wiener chaos expansion

(3.4) $u(z,t)=\sum_{n=0}^{\infty}\mathcal{I}_{n}(f_{n}(\cdot;z,t)),$

for some $f_{n}(\cdot;z,t)\in H^{\otimes n}$. Consequently, to compute
$\mathbb{E}[u^{2}(z,t)]$ we now obtain

$\mathbb{E}[u^{2}(z,t)]=\sum_{n=0}^{\infty}\mathbb{E}[\mathcal{I}_{n}^{2}(f_{n}(\cdot;z,t))]=\sum_{n=0}^{\infty}n!\|\tilde{f}_{n}(\cdot;z,t)\|_{n}^{2},$

so we turn our attention to finding an expression for $f_{n}(\cdot;z,t)$ and
computing its norm.

Recall that by virtue of being a mild solution to (1.6), $u$ satisfies the
integral equation (3.3). Following the method of Duhamel, unfolding (3.3)
recursively in the Skorohod integral and denoting
$u_{0}(z,t)=\mathbb{E}[u(z,t)]$ yields

$\displaystyle
u(z,t)=u_{0}(z,t)+\int_{0}^{t}\int_{0}^{\infty}q_{0}(z,z_{1},t_{1}-\tau)u(z_{1},t_{1})W(\mathrm{d}z_{1},\mathrm{d}t_{1})$
$\displaystyle\hskip
28.45274pt=u_{0}(z,t)+\int_{0}^{t}\int_{0}^{\infty}q_{0}(z,z_{1},t_{1}-\tau)u_{0}(z_{1},t_{1})W(\mathrm{d}z_{1},\mathrm{d}t_{1})$
$\displaystyle+\int_{0}^{t}\int_{0}^{t_{1}}\int_{0}^{\infty}\int_{0}^{\infty}q_{0}(z,z_{1},t-t_{1})q_{0}(z_{1},z_{2},t_{1}-t_{2})u(z_{2},t_{2})W(\mathrm{d}z_{1},\mathrm{d}t_{1})W(\mathrm{d}z_{2},\mathrm{d}t_{2}),$

which can be repeated indefinitely. In other words, setting
$f_{0}^{\prime}=u_{0}$ then recursively defining

(3.5) $\displaystyle
f_{n}^{\prime}(z_{1},t_{1},\ldots,z_{n},t_{n};z,t)=\mathbbm{1}_{(0,t)}(t_{n})$
$\displaystyle q_{0}(z,z_{n},t-t_{n})$ $\displaystyle\cdot
f_{n-1}^{\prime}(z_{1},t_{1},\ldots,z_{n-1},t_{n-1};z_{n},t_{n})$

for $n\geq 1$, we obtain (formally) that

$u(z,t)=\sum_{n=0}^{\infty}\mathcal{I}_{n}(f_{n}^{\prime}(\cdot;z,t)),$

and $f_{n}(\cdot;z,t)=f_{n}^{\prime}(\cdot;z,t)$. For these functions, we will
slightly abuse the notation and write $n!\|\tilde{f}_{n}(\cdot;z,t)\|_{n}^{2}$
as $\|f_{n}(\cdot;z,t)\|_{n}^{2}$, where by direct computation

$\displaystyle\|f_{n}(\cdot;z,t)\|_{n}^{2}=\int_{\Delta_{n}t}\int_{\mathbb{R}_{+}^{n}}\int_{\Delta_{n}t}\int_{\mathbb{R}_{+}^{n}}\prod_{i=0}^{n-1}$
$\displaystyle
q_{0}(x_{i+1},x_{i},r_{i+1}-r_{i})q_{0}(y_{i+1},y_{i},s_{i+1}-s_{i})$
$\displaystyle\cdot
f(x_{i}-y_{i})\gamma(r_{i}-s_{i})\mathrm{d}\mathbf{x}\mathrm{d}\mathbf{y}\mathrm{d}\mathbf{r}\mathrm{d}\mathbf{s},$

where $x_{n}=y_{n}=z$ and $r_{n}=s_{n}=t$.

Of course, it remains to rigorously show that (3.4) with the derived
expression in (3.5) for $f_{n}$ is in fact the unique mild solution to (1.6).
We build towards the proof by first studying the case of space-time white
noise via a treatment similar to [5, 33]. Henceforth, without loss of
generality we may assume constant initial condition. Because $u_{0}\in
C_{b}(\mathbb{R}_{+})$, all subsequent results may be recovered by first
applying $u_{0}(z)\leq\|u_{0}\|_{u}$.

####  3.1.1. Space-Time White Noise

In the space-time white noise case, we are interested in (1.6) when the
centered Gaussian noise $\\{\dot{W}(z,t)\mid z,t>0\\}$ formally has covariance

$\mathbb{E}[W(z_{1},t_{1})W(z_{2},t_{2})]=\delta_{0}(z_{1}-z_{2})\delta_{0}(t_{1}-t_{2}),$

where $\delta_{0}$ denotes the Dirac delta. In other words,
$\left\langle\cdot,\cdot\right\rangle_{H}$ coincides with the standard $L^{2}$
inner product.

Given the kernels $f_{n}$ defined above, we first aim to prove that for every
$z,t>0$, $f_{n}(\cdot;z,t)\in H^{\otimes n}$. The proof for the base case
$n=0$ is straightforward, but we will need the following technical lemma on
$u_{0}(z,t)$.

######  Lemma 3.1.

Let $z,t>0$, then the following equality holds

(3.6)
$u_{0}(z,t)=\int_{0}^{\infty}q_{0}(z,w,t)\mathrm{d}w=1-e^{-\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}},$

and we have the bounds

(3.7) $e^{-1}\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}\leq
u_{0}(z,t)\leq\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}\wedge
1,$

where the lower bound holds for $z<t$.

###### Proof.

Applying $(\ref{eqn:watsonu0Int})$ with $\nu=0$ and $a=1$ and (5.3),
respectively, we have

$\displaystyle\int_{0}^{\infty}q_{0}(z,w,t)\mathrm{d}w$
$\displaystyle=\int_{0}^{\infty}\sqrt{\mathchoice{\frac{z}{w}}{z/w}{\frac{z}{w}}{\frac{z}{w}}}\mathchoice{\frac{e^{-\mathchoice{\frac{z+w}{t}}{z+w/t}{\frac{z+w}{t}}{\frac{z+w}{t}}}}{t}}{e^{-\mathchoice{\frac{z+w}{t}}{z+w/t}{\frac{z+w}{t}}{\frac{z+w}{t}}}/t}{\frac{e^{-\mathchoice{\frac{z+w}{t}}{z+w/t}{\frac{z+w}{t}}{\frac{z+w}{t}}}}{t}}{\frac{e^{-\mathchoice{\frac{z+w}{t}}{z+w/t}{\frac{z+w}{t}}{\frac{z+w}{t}}}}{t}}I_{1}\left(2\mathchoice{\frac{\sqrt{zw}}{t}}{\sqrt{zw}/t}{\frac{\sqrt{zw}}{t}}{\frac{\sqrt{zw}}{t}}\right)\mathrm{d}w$
$\displaystyle=2\mathchoice{\frac{\sqrt{z}e^{-\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}}{t}}{\sqrt{z}e^{-\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}/t}{\frac{\sqrt{z}e^{-\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}}{t}}{\frac{\sqrt{z}e^{-\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}}{t}}i^{-1}\int_{0}^{\infty}e^{-\mathchoice{\frac{u^{2}}{t}}{u^{2}/t}{\frac{u^{2}}{t}}{\frac{u^{2}}{t}}}J_{1}\left(2\mathchoice{\frac{\sqrt{z}i}{t}}{\sqrt{z}i/t}{\frac{\sqrt{z}i}{t}}{\frac{\sqrt{z}i}{t}}u\right)\mathrm{d}u$
$\displaystyle=2\mathchoice{\frac{\sqrt{z}e^{-\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}}{t}}{\sqrt{z}e^{-\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}/t}{\frac{\sqrt{z}e^{-\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}}{t}}{\frac{\sqrt{z}e^{-\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}}{t}}i^{-1}\cdot\mathchoice{\frac{\sqrt{z}i}{2}}{\sqrt{z}i/2}{\frac{\sqrt{z}i}{2}}{\frac{\sqrt{z}i}{2}}{}_{1}F_{1}\left[\begin{array}[]{cc}1\\\
2\end{array}\mathrel{\Big{|}}\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}\right]$
$\displaystyle=\mathchoice{\frac{ze^{-\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}}{t}}{ze^{-\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}/t}{\frac{ze^{-\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}}{t}}{\frac{ze^{-\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}}{t}}\mathchoice{\frac{(e^{\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}-1)t}{z}}{(e^{\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}-1)t/z}{\frac{(e^{\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}-1)t}{z}}{\frac{(e^{\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}-1)t}{z}}$
$\displaystyle=1-e^{-\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}.$

The bounds follow by taking the first-order Taylor expansion. ∎

For $n\geq 1$, due to the recursive definition of $f_{n}(\cdot;z,t)$, in order
to show that $f_{n}(\cdot;z,t)\in H^{\otimes n}$, it suffices to verify that
$\|f_{n}(\cdot;z,t)\|_{n}^{2}<\infty$ which we can in turn recursively
compute:

(3.8)
$\|f_{n}(\cdot;z,t)\|_{n}^{2}=\int_{0}^{t}\int_{0}^{\infty}q_{0}^{2}(z,w,t-\tau)\|f_{n-1}(\cdot;w,\tau)\|_{n-1}^{2}\mathrm{d}w\mathrm{d}\tau.$

Using (3.7), for $n=1$ we observe

$\|f_{1}(\cdot;z,t)\|_{1}^{2}=\int_{0}^{t}\int_{0}^{\infty}q_{0}^{2}(z,w,t-\tau)u_{0}^{2}(w,\tau)\mathrm{d}w\mathrm{d}\tau\leq\int_{0}^{t}\int_{0}^{\infty}q_{0}^{2}(z,w,\tau)\mathrm{d}w\mathrm{d}\tau,$

which can once again be direct computed and bounded.

######  Lemma 3.2.

Define $U:\mathbb{R}_{+}\to\mathbb{R}_{+}$ by

(3.9)
$U(x):=e^{-x}I_{0}(x)-\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}e^{-2x},$

then for $z,t>0$,

(3.10)
$\int_{0}^{t}\int_{0}^{\infty}q_{0}^{2}(z,w,\tau)\mathrm{d}w\mathrm{d}\tau=U\left(\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}\right),$

and moreover $U(z/t)\leq 1/2$.

###### Proof.

Per the definition of $q_{0}$ and (5.3), we have

$\displaystyle\int_{0}^{\infty}q_{0}^{2}(z,w,\tau)\mathrm{d}w$
$\displaystyle=\int_{0}^{\infty}\mathchoice{\frac{zw^{-1}}{\tau^{2}}}{zw^{-1}/\tau^{2}}{\frac{zw^{-1}}{\tau^{2}}}{\frac{zw^{-1}}{\tau^{2}}}e^{-\mathchoice{\frac{2(z+w)}{\tau}}{2(z+w)/\tau}{\frac{2(z+w)}{\tau}}{\frac{2(z+w)}{\tau}}}I_{1}^{2}\left(2\mathchoice{\frac{\sqrt{zw}}{\tau}}{\sqrt{zw}/\tau}{\frac{\sqrt{zw}}{\tau}}{\frac{\sqrt{zw}}{\tau}}\right)\mathrm{d}w=\mathchoice{\frac{ze^{-\mathchoice{\frac{2z}{\tau}}{2z/\tau}{\frac{2z}{\tau}}{\frac{2z}{\tau}}}}{\tau^{2}i^{2}}}{ze^{-\mathchoice{\frac{2z}{\tau}}{2z/\tau}{\frac{2z}{\tau}}{\frac{2z}{\tau}}}/\tau^{2}i^{2}}{\frac{ze^{-\mathchoice{\frac{2z}{\tau}}{2z/\tau}{\frac{2z}{\tau}}{\frac{2z}{\tau}}}}{\tau^{2}i^{2}}}{\frac{ze^{-\mathchoice{\frac{2z}{\tau}}{2z/\tau}{\frac{2z}{\tau}}{\frac{2z}{\tau}}}}{\tau^{2}i^{2}}}\int_{0}^{\infty}w^{-1}e^{-\mathchoice{\frac{2w}{\tau}}{2w/\tau}{\frac{2w}{\tau}}{\frac{2w}{\tau}}}J_{1}^{2}\left(\mathchoice{\frac{2i\sqrt{zw}}{\tau}}{2i\sqrt{zw}/\tau}{\frac{2i\sqrt{zw}}{\tau}}{\frac{2i\sqrt{zw}}{\tau}}\right)\mathrm{d}w.$

Next, we apply Proposition 5.6 to obtain

$i^{-2}\int_{0}^{\infty}w^{-1}e^{-\mathchoice{\frac{2w}{\tau}}{2w/\tau}{\frac{2w}{\tau}}{\frac{2w}{\tau}}}J_{1}^{2}\left(\mathchoice{\frac{2i\sqrt{zw}}{\tau}}{2i\sqrt{zw}/\tau}{\frac{2i\sqrt{zw}}{\tau}}{\frac{2i\sqrt{zw}}{\tau}}\right)\mathrm{d}w=e^{\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}}\left(I_{0}\left(\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}\right)-I_{1}\left(\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}\right)\right)-1,$

and hence we get

(3.11)
$\int_{0}^{t}\int_{0}^{\infty}q_{0}^{2}(z,w,\tau)\mathrm{d}w=\int_{0}^{t}\mathchoice{\frac{ze^{-\mathchoice{\frac{2z}{\tau}}{2z/\tau}{\frac{2z}{\tau}}{\frac{2z}{\tau}}}}{\tau^{2}}}{ze^{-\mathchoice{\frac{2z}{\tau}}{2z/\tau}{\frac{2z}{\tau}}{\frac{2z}{\tau}}}/\tau^{2}}{\frac{ze^{-\mathchoice{\frac{2z}{\tau}}{2z/\tau}{\frac{2z}{\tau}}{\frac{2z}{\tau}}}}{\tau^{2}}}{\frac{ze^{-\mathchoice{\frac{2z}{\tau}}{2z/\tau}{\frac{2z}{\tau}}{\frac{2z}{\tau}}}}{\tau^{2}}}\left(e^{\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}}\left(I_{0}\left(\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}\right)-I_{1}\left(\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}\right)\right)-1\right)\mathrm{d}\tau.$

Via integration by parts,

$\displaystyle
z\int_{0}^{t}\mathchoice{\frac{e^{-\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}}}{\tau^{2}}}{e^{-\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}}/\tau^{2}}{\frac{e^{-\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}}}{\tau^{2}}}{\frac{e^{-\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}}}{\tau^{2}}}I_{1}\left(\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}\right)\mathrm{d}r$
$\displaystyle=-\int_{0}^{t}e^{-\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}}\derivative{\tau}\left[I_{0}\left(\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}\right)\right]\mathrm{d}\tau$
$\displaystyle=-e^{-\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}}I_{0}\left(\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}\right)\Big{|}_{0}^{t}+z\int_{0}^{t}\mathchoice{\frac{e^{-\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}}}{\tau^{2}}}{e^{-\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}}/\tau^{2}}{\frac{e^{-\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}}}{\tau^{2}}}{\frac{e^{-\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}}}{\tau^{2}}}I_{0}\left(\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}\right)\mathrm{d}\tau$
$\displaystyle=-e^{-\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}I_{0}\left(\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}\right)+z\int_{0}^{t}\mathchoice{\frac{e^{-\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}}}{\tau^{2}}}{e^{-\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}}/\tau^{2}}{\frac{e^{-\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}}}{\tau^{2}}}{\frac{e^{-\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}}}{\tau^{2}}}I_{0}\left(\mathchoice{\frac{z}{\tau}}{z/\tau}{\frac{z}{\tau}}{\frac{z}{\tau}}\right)\mathrm{d}\tau,$

and substituting this last expression into (3.11), (3.10) directly follows.
Finally, $\|U\|_{u}=1/2$ follows from straightforward calculations. ∎

Once again, we have a uniform bound for the norm. This reveals the main pillar
of our strategy - using induction, and Lemmas 3.1 and 3.2 to conclude that

(3.12)
$\|f_{n}(\cdot;z,t)\|_{n}^{2}\leq\mathchoice{\frac{1}{2^{n}}}{1/2^{n}}{\frac{1}{2^{n}}}{\frac{1}{2^{n}}}.$

Indeed, the case $n=0$ is confirmed by Lemma 3.1. Next, assuming (3.12) holds
for $n-1$ for some $n\geq 1$, then by (3.8),

$\displaystyle\|f_{n}(\cdot;z,t)\|_{n}^{2}$
$\displaystyle=\int_{0}^{t}\int_{0}^{\infty}q_{0}^{2}(z,w,t-\tau)\|f_{n-1}(\cdot;w,\tau)\|_{n}^{2}\mathrm{d}w\mathrm{d}\tau,$
(3.13) $\displaystyle\leq
2^{-n+1}\int_{0}^{t}\int_{0}^{\infty}q_{0}^{2}(z,w,t-\tau)\mathrm{d}w\mathrm{d}\tau=\mathchoice{\frac{1}{2^{n}}}{1/2^{n}}{\frac{1}{2^{n}}}{\frac{1}{2^{n}}}.$

Hence it directly follows that $f_{n}(\cdot;z,t)\in H^{\otimes n}$ and we may
set $u_{n}(z,t):=\mathcal{I}_{n}(f_{n}(\cdot;z,t))$ for notational simplicity.
We are now equipped to present the main result of this section.

######  Theorem 3.1.

When $W$ is the space-time white noise, $u=\left\\{u(z,t)\mid z,t\geq
0\right\\}$ as in (3.4) is well-defined as a convergent series in
$L^{2}(\Omega)$. Moreover,

(3.14) $\sup_{z,t\geq 0}\mathbb{E}[u^{2}(z,t)]\leq 2,$

and $u$ is the unique mild solution to (1.6).

###### Proof.

The convergence of the series in (3.4) to $u(z,t)$ in $L^{2}(\Omega)$ and
(3.14) follow by the summability of the bounds (3.12) over $n\geq 0$.

To see that $u$ has a jointly measurable modification, we first show that
$(z,t)\mapsto u(z,t)$ is continuous in $L^{2}(\Omega)$. To this end, we have

$\displaystyle\mathbb{E}[|u_{n}(z_{1},t_{1})-u_{n}(z_{2},t_{2})|^{2}]=$
$\displaystyle\int_{0}^{t_{1}\wedge
t_{2}}\int_{0}^{\infty}\absolutevalue{q_{0}(z_{1},w,t_{1}-\tau)-q_{0}(z_{2},w,t_{2}-\tau)}^{2}\mathbb{E}[u_{n-1}^{2}(w,\tau)]\mathrm{d}w\mathrm{d}\tau$
$\displaystyle\leq 2^{-n+1}\int_{0}^{t_{1}\wedge
t_{2}}\int_{0}^{\infty}\absolutevalue{q_{0}(z_{1},w,t_{1}-\tau)-q_{0}(z_{2},w,t_{2}-\tau)}^{2}\mathrm{d}w\mathrm{d}\tau,$

where the last bound vanishes as $(z_{1},t_{1})\to(z_{2},t_{2})$ by the
dominated convergence theorem and continuity of the fundamental solution
$q_{0}$. By the continuity of $u_{n}$ in $L^{2}(\Omega)$ and (3.14), it
directly follows that $u$ is continuous in $L^{2}(\Omega)$, and in turn by
Theorem 30 in Chapter IV of [19], $u$ possesses a jointly measurable
modification. We denote this modification again by $u$. We have verified
condition $(\textbf{a})$ in Definition 3.1.

Next, we check condition $(\textbf{b})$. For $z,w,t-\tau,\tau>0$, let

(3.15)
$u^{(z,t)}(w,\tau):=\mathbbm{1}_{(0,t)}(\tau)q_{0}(z,w,t-\tau)u(w,\tau).$

By measurability of $u$, it is immediate that $u^{(z,t)}(w,\tau)$ is adapted
to $W$. Moreover by (3.14), $\mathbb{E}[(u^{(z,t)}(w,\tau))^{2}]<\infty$ and
so $u^{(z,t)}(w,\tau)$ admits the Wiener chaos expansion

$u^{(z,t)}(w,\tau)=\sum_{n=0}^{\infty}\mathcal{I}_{n}\left(g_{n}^{(z,t)}(\cdot;w,\tau)\right),$

where
$g_{n}^{(z,t)}(w,\tau)=\mathbbm{1}_{(0,t)}(\tau)q_{0}(z,w,t-\tau)f_{n}(\cdot;w,\tau)=f_{n+1}(\cdot,w,\tau;z,t)$.
Clearly, $g_{n}^{(z,t)}\in H^{\otimes(n+1)}$ by the corresponding properties
for $u$ and $f_{n+1}$, respectively, and

$\delta\left(u^{(z,t)}\right)=\sum_{n=0}^{\infty}\mathcal{I}_{n+1}\left(g_{n}^{(z,t)}\right)=\sum_{n=0}^{\infty}\mathcal{I}_{n+1}\left(f_{n+1}(\cdot;z,t)\right)=\sum_{n=0}^{\infty}u_{n}(z,t)$

converges in $L^{2}(\Omega)$.

Therefore $u^{(z,t)}$ is indeed Skorohod integrable and by Proposition 2.1
(1.3.7 of [36]),

$\displaystyle\int_{0}^{t}\int_{0}^{\infty}q_{0}(z,w,t-\tau)u(w,\tau)W(\mathrm{d}w,\mathrm{d}\tau)$
$\displaystyle=\delta\left(u^{(z,t)}\right)=\sum_{n=0}^{\infty}\mathcal{I}_{n+1}\left(f_{n+1}(\cdot;z,t)\right)=u(z,t)-u_{0}(z,t).$

Thus, $u$ also satisfies condition $(\textbf{b})$ and is in fact a mild
solution of (1.6).

Next, we prove that $u$ is the unique mild solution to (1.6). Let
$v=\left\\{v(z,t):z,t\geq 0\right\\}$ be another solution to (1.6) with Wiener
chaos expansion

$v(z,t)=\sum_{n=0}^{\infty}\mathcal{I}_{n}(k_{n}(\cdot;z,t))$

for some symmetric non-negative functions $k_{n}(\cdot;z,t)\in H^{\otimes n}$
and $z,t>0$ fixed. We will show that
$k_{n}(\cdot;z,t)=\tilde{f}_{n}(\cdot,z,t)$ for every $n\geq 1$.

Let $v^{(z,t)}(w,\tau)$ be defined similarly to (3.15). Then by definition of
a solution, the process
$v^{(z,t)}:=\left\\{v^{(z,t)}(w,\tau)|w,\tau>0\right\\}$ is Skorohod
integrable and admits the Wiener chaos expansion

$v^{(z,t)}(w,\tau)=\sum_{n=0}^{\infty}\mathcal{I}_{n}\left(h_{n}^{(z,t)}(\cdot;w,\tau)\right),$

where
$h_{n}^{(z,t)}(\cdot;w,\tau)=\mathbbm{1}_{(0,t)}(\tau)q_{0}(z,w,t-\tau)k_{n}(\cdot;w,\tau)$.
By Proposition 2.1,

$\sum_{n=0}^{\infty}\mathcal{I}_{n+1}\left(k_{n+1}(\cdot;z,t)\right)=v(z,t)-\mathbb{E}[v(z,t)]=\delta\left(v^{(z,t)}\right)=\sum_{n=0}^{\infty}\mathcal{I}_{n+1}\left(\widetilde{h_{n}^{(z,t)}}\right),$

from which it follows that $k_{n+1}(\cdot,z,t)=\widetilde{h_{n}^{(z,t)}}$ by
the uniqueness of the Wiener chaos expansion with symmetric kernels.

We now express $k_{n}(\cdot,z,t)$ recursively. For $n=0$, we obtain

$k_{1}(z_{1},t_{1},z,t)=\widetilde{h_{0}^{(z,t)}}(z_{1},t_{1})=h_{0}^{(z,t)}(z_{1},t_{1})=\mathbbm{1}_{(0,t)}(t_{1})q_{0}(z,z_{1},t-t_{1})\left(1-e^{-\mathchoice{\frac{z_{1}}{t_{1}}}{z_{1}/t_{1}}{\frac{z_{1}}{t_{1}}}{\frac{z_{1}}{t_{1}}}}\right).$

For $n=1$, we obtain

$\displaystyle
k_{2}(z_{1},t_{1},z_{2},t_{2},z,t)=\widetilde{h_{1}^{(z,t)}}(z_{1},t_{1},z_{2},t_{2})=\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}\left[h_{1}^{(z,t)}(z_{1},t_{1},z_{2},t_{2})+h_{1}^{(z,t)}(z_{2},t_{2},z_{1},t_{1})\right]=$
$\displaystyle\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}\left[\mathbbm{1}_{(0,t)}(t_{2})q_{0}(z,z_{2},t-t_{2})k_{1}(z_{1},t_{1},z_{2},t_{2})+\mathbbm{1}_{(0,t)}(t_{1})q_{0}(z,z_{1},t-t_{1})k_{1}(z_{2},t_{2},z_{1},t_{1})\right].$

Using the expression for $k_{1}$, we conclude

$\displaystyle k_{2}(z_{1},t_{1},z_{2},t_{2},z,t)=$
$\displaystyle\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}\left[\mathbbm{1}_{(0,t)}(t_{2})q_{0}(z,z_{2},t-t_{2})\mathbbm{1}_{(0,t_{2})}(t_{1})q_{0}(z_{2},z_{1},t_{2}-t_{1})\left(1-e^{-\mathchoice{\frac{z_{1}}{t_{1}}}{z_{1}/t_{1}}{\frac{z_{1}}{t_{1}}}{\frac{z_{1}}{t_{1}}}}\right)\right.$
$\displaystyle+\left.\mathbbm{1}_{(0,t)}(t_{1})q_{0}(z,z_{1},t-t_{1})\mathbbm{1}_{(0,t_{1})}(t_{2})q_{0}(z_{1},z_{2},t_{1}-t_{2})\left(1-e^{-\mathchoice{\frac{z_{2}}{t_{2}}}{z_{2}/t_{2}}{\frac{z_{2}}{t_{2}}}{\frac{z_{2}}{t_{2}}}}\right)\right],$

that is, $k_{2}(\cdot,z,t)=\tilde{f}_{2}(\cdot,z,t)$. Iterating this
procedure, we infer that $k_{n}(\cdot,z,t)=\tilde{f}_{n}(\cdot,z,t)$ for every
$n\geq 1$. Hence,

$v(z,t)=\sum_{n=0}^{\infty}\mathcal{I}_{n}(k_{n}(\cdot,z,t))=\sum_{n=0}^{\infty}\mathcal{I}_{n}(\tilde{f}_{n}(\cdot,z,t))=\sum_{n=0}^{\infty}\mathcal{I}_{n}(f_{n}(\cdot,z,t))=u(z,t).$

That is, the solution $u$ given by (3.4) is the unique mild solution to (1.6),
up to a modification. ∎

####  3.1.2. Colored Noise

We continue to the case of colored noise. That is, $W=\\{\dot{W}(z,t)\mid
z,t>0\\}$ is a centered Gaussian process whose covariance is formally given by

$\mathbb{E}\left[\dot{W}(z_{1},t_{1})\dot{W}(z_{2},t_{2})\right]=f(z_{1}-z_{2})\gamma(t_{1}-t_{2}),$

for $z_{1},z_{2},t_{1},t_{2}>0$, where $f,\gamma:\mathbb{R}\to\mathbb{R}_{+}$
are non-negative definite, symmetric, and locally integrable kernels.

As in the previous section, we seek to show that the process $u$ defined in
(3.4) is the unique mild solution to (1.6) when $W$ is a colored noise. To
this end, we show that the series in (3.4) converges in $L^{2}(\Omega)$ for
all $z,t>0$ and we recursively compute $\left\langle
f_{n}(\cdot;z_{1},t_{1}),f_{n}(\cdot;z_{2},t_{2})\right\rangle_{n}$:

(3.16) $\displaystyle\left\langle
f_{n}(\cdot;z_{1},t_{1}),f_{n}(\cdot;z_{2},t_{2})\right\rangle_{n}=\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)\int_{\mathbb{R}_{+}^{2}}f(x-y)$
$\displaystyle\cdot q_{0}(z_{1},x,t_{1}-r)q_{0}(z_{2},y,t_{2}-s)\left\langle
f_{n-1}(\cdot;x,r),f_{n-1}(\cdot;y,s)\right\rangle_{n-1}\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s.$

for $z_{1},z_{2},t_{1},t_{2}>0$ and $n\geq 1$.

However, here the main challenge arises, unlike in (3.13), after we apply
Lemma 3.1 for $n=1$, we arrive at

$\left\langle
f_{1}(\cdot;z_{1},t_{1}),f_{1}(\cdot;z_{2},t_{2})\right\rangle_{1}\leq\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)\int_{\mathbb{R}_{+}^{2}}f(x-y)q_{0}(z_{1},x,t_{1}-r)q_{0}(z_{2},y,t_{2}-s)\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s,$

which cannot be directly computed as in Lemma 3.2. Instead, we seek to extract
$\gamma$ and $f$ from the integrals by imposing the following additional
conditions:

  1. (i)

$f$ is non-increasing on $\mathbb{R}_{+}$,

  2. (ii)

for all $\varepsilon>0$,
$\int_{-\varepsilon}^{\varepsilon}f(x)\mathrm{d}x<\infty$ and

$\lim_{\varepsilon\searrow
0}\int_{-\varepsilon}^{\varepsilon}f(x)\mathrm{d}x=0.$

In addition, for $\varepsilon,t>0$ we introduce the following notation

(3.17) $\displaystyle
A_{\varepsilon}:=\left\\{(x,y)\in\mathbb{R}_{+}^{2}\mid|x-y|<\varepsilon\right\\},$
(3.18) $\displaystyle
F_{\varepsilon}:=\int_{-\varepsilon}^{\varepsilon}f(x)\mathrm{d}x=2\int_{0}^{\varepsilon}f(x)\mathrm{d}x,$
(3.19)
$\displaystyle\Gamma_{t}:=\int_{-t}^{t}\gamma(\tau)\mathrm{d}\tau=2\int_{0}^{t}\gamma(\tau)\mathrm{d}\tau,.$

Note these quantities are well-defined by local-integrability and symmetry of
$f$ and $\gamma$. We proceed to make our strategy concrete.

######  Lemma 3.3.

Let $z_{1},z_{2},t_{1},t_{2}>0$ and let $U$ be as in (3.9). Then for any
$\varepsilon>0$ we have

(3.20) $\left\langle
f_{1}(\cdot;z_{1},t_{1}),f_{1}(\cdot;z_{2},t_{2})\right\rangle_{1}\leq\Gamma_{t_{1}\vee
t_{2}}\left(\mathchoice{\frac{F_{\varepsilon}}{2}}{F_{\varepsilon}/2}{\frac{F_{\varepsilon}}{2}}{\frac{F_{\varepsilon}}{2}}+f(\varepsilon)\sqrt{t_{1}t_{2}}\right).$

###### Proof.

Let $\varepsilon>0$ and define

(3.21) $\displaystyle L_{1,0}$
$\displaystyle:=\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)\int_{A_{\varepsilon}}f(x-y)q_{0}(z_{1},x,t_{1}-r)q_{0}(z_{2},y,t_{2}-s)\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s,$
(3.22) $\displaystyle L_{2,0}$
$\displaystyle:=\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)\int_{A_{\varepsilon}^{c}}f(x-y)q_{0}(z_{1},x,t_{1}-r)q_{0}(z_{2},y,t_{2}-s)\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s.$

Then by (3.7), we’ll treat the two terms (3.21) and (3.22) individually. By
Proposition 5.2 applied twice (once for the spatial variables, once for the
temporal variables), we get

$\displaystyle L_{1,0}$
$\displaystyle\leq\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)F_{\varepsilon}\left(\int_{0}^{\infty}q_{0}^{2}(z_{1},x,t_{1}-r)\mathrm{d}x\int_{0}^{\infty}q_{0}^{2}(z_{2},y,t_{2}-s)\mathrm{d}y\right)^{\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}\mathrm{d}r\mathrm{d}s$
$\displaystyle\leq\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}\left(\int_{0}^{t_{1}}\int_{0}^{\infty}q_{0}^{2}(z_{1},x,t_{1}-r)\mathrm{d}x\mathrm{d}r\int_{0}^{t_{2}}\int_{0}^{\infty}q_{0}^{2}(z_{2},y,t_{2}-s)\mathrm{d}y\mathrm{d}s\right)^{\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}.$

Notice that these last integrals are precisely quantities studied in the white
noise case. Therefore, Lemma 3.2 immediately implies that

(3.23) $L_{1,0}\leq\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}\sqrt{U\left(\mathchoice{\frac{z_{1}}{t_{1}}}{z_{1}/t_{1}}{\frac{z_{1}}{t_{1}}}{\frac{z_{1}}{t_{1}}}\right)U\left(\mathchoice{\frac{z_{2}}{t_{2}}}{z_{2}/t_{2}}{\frac{z_{2}}{t_{2}}}{\frac{z_{2}}{t_{2}}}\right)}.$

On the other hand, by our assumptions on $f$, we have

$\displaystyle L_{2,0}$ $\displaystyle\leq
f(\varepsilon)\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)\int_{\mathbb{R}_{+}^{2}}q_{0}(z_{1},x,t_{1}-r)q_{0}(z_{2},y,t_{2}-s)\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s$
$\displaystyle=f(\varepsilon)\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)\int_{0}^{\infty}q_{0}(z_{1},x,t_{1}-r)\mathrm{d}x\int_{0}^{\infty}q_{0}(z_{2},y,t_{2}-s)\mathrm{d}y\mathrm{d}r\mathrm{d}s$
$\displaystyle=f(\varepsilon)\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)u_{0}(z_{1},t_{1}-r)u_{0}(z_{2},t_{2}-s)\mathrm{d}r\mathrm{d}s.$

Again, the above integrals are also quantities studied in the white noise
case. This time by Lemma 3.1, it follows that

$L_{2,0}\leq
f(\varepsilon)\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)\left(1-e^{-\mathchoice{\frac{z_{1}}{t_{1}-r}}{z_{1}/t_{1}-r}{\frac{z_{1}}{t_{1}-r}}{\frac{z_{1}}{t_{1}-r}}}\right)\left(1-e^{-\mathchoice{\frac{z_{2}}{t_{2}-s}}{z_{2}/t_{2}-s}{\frac{z_{2}}{t_{2}-s}}{\frac{z_{2}}{t_{2}-s}}}\right)\mathrm{d}r\mathrm{d}s\leq
f(\varepsilon)\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)\mathrm{d}r\mathrm{d}s.$

By applying Proposition 5.2 once again,

(3.24) $L_{2,0}\leq\Gamma_{t_{1}\vee t_{2}}f(\varepsilon)\sqrt{t_{1}t_{2}}.$

Substituting the bounds (3.23) and (3.24), we achieve the stated result
(3.20). ∎

As in the space-time white noise case, we obtain a bound uniform in $z_{1}$
and $z_{2}$. Moreover, the bound in (3.20) is increasing with respect to the
time variables. Hence, it is tempting to replicate (3.12) and obtain a
geometric bound

$\displaystyle\|f_{n}(\cdot;z,t)\|_{n}^{2}$
$\displaystyle=\int_{[0,t]^{2}}\gamma(r-s)\int_{\mathbb{R}_{+}^{2}}f(x-y)q_{0}(z,x,t-r)q_{0}(z,y,t-s)\left\langle
f_{n-1}(\cdot;x,r),f_{n-1}(\cdot;y,s)\right\rangle_{n-1}\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s,$
$\displaystyle\leq\int_{[0,t]^{2}}\gamma(r-s)\int_{\mathbb{R}_{+}^{2}}f(x-y)q_{0}(z,x,t-r)q_{0}(z,y,t-s)\left(\Gamma_{r\vee
s}\left(\mathchoice{\frac{F_{\varepsilon}}{2}}{F_{\varepsilon}/2}{\frac{F_{\varepsilon}}{2}}{\frac{F_{\varepsilon}}{2}}+f(\varepsilon)\sqrt{rs}\right)\right)^{n-1}\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s,$
$\displaystyle\leq\Gamma_{t}^{n-1}\left(\mathchoice{\frac{F_{\varepsilon}}{2}}{F_{\varepsilon}/2}{\frac{F_{\varepsilon}}{2}}{\frac{F_{\varepsilon}}{2}}+f(\varepsilon)t\right)^{n-1}\int_{[0,t]^{2}}\gamma(r-s)\int_{\mathbb{R}_{+}^{2}}f(x-y)q_{0}(z,x,t-r)q_{0}(z,y,t-s)\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s,$

Thus, it follows from here that

(3.25)
$\|f_{n}(\cdot;z,t)\|_{n}^{2}\leq\Gamma_{t}^{n}\left(\mathchoice{\frac{F_{\varepsilon}}{2}}{F_{\varepsilon}/2}{\frac{F_{\varepsilon}}{2}}{\frac{F_{\varepsilon}}{2}}+f(\varepsilon)t\right)^{n},$

for the $L^{2}$-norm, but the following remark demonstrates that such an
argument may potentially fail to hold for general $f$.

######  Remark 3.1.

Suppose $\gamma(z)=f(z)=|z|^{-1/2}$, then

$\sum_{n=0}^{\infty}\|f_{n}(\cdot;z,t)\|_{n}^{2}\leq\sum_{n=0}^{\infty}4^{n}t^{\mathchoice{\frac{n}{2}}{n/2}{\frac{n}{2}}{\frac{n}{2}}}\left(2\sqrt{\varepsilon}+\mathchoice{\frac{t}{\sqrt{\varepsilon}}}{t/\sqrt{\varepsilon}}{\frac{t}{\sqrt{\varepsilon}}}{\frac{t}{\sqrt{\varepsilon}}}\right),$

and whose right hand side converges if and only if

$4^{n}t^{\mathchoice{\frac{n}{2}}{n/2}{\frac{n}{2}}{\frac{n}{2}}}\left(2\sqrt{\varepsilon}+\mathchoice{\frac{t}{\sqrt{\varepsilon}}}{t/\sqrt{\varepsilon}}{\frac{t}{\sqrt{\varepsilon}}}{\frac{t}{\sqrt{\varepsilon}}}\right)<1.$

Yet for $t$ sufficiently large, no choice of $\varepsilon$ exists for which
this condition holds.

In light of Remark 3.1, we are forced to derive tighter bounds. As an example,
let us examine the case $n=2$. As before, we arrive at

$\displaystyle\left\langle
f_{2}(\cdot;z_{1},t_{1}),f_{2}(\cdot;z_{2},t_{2})\right\rangle_{2}$
$\displaystyle\leq\Gamma_{t_{1}\vee
t_{2}}\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)\left(\mathchoice{\frac{F_{\varepsilon}}{2}}{F_{\varepsilon}/2}{\frac{F_{\varepsilon}}{2}}{\frac{F_{\varepsilon}}{2}}+f(\varepsilon)\sqrt{rs}\right)\int_{\mathbb{R}_{+}^{2}}f(x-y)q_{0}(z,x,t-r)q_{0}(z,y,t-s)\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s.$

Define

(3.26) $\displaystyle\hskip 28.45274ptL_{1,0}^{\prime}$
$\displaystyle:=\Gamma_{t_{1}\vee
t_{2}}\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)\left(\mathchoice{\frac{F_{\varepsilon}}{2}}{F_{\varepsilon}/2}{\frac{F_{\varepsilon}}{2}}{\frac{F_{\varepsilon}}{2}}+f(\varepsilon)\sqrt{rs}\right)\int_{A_{\varepsilon}}f(x-y)q_{0}(z,x,t-r)q_{0}(z,y,t-s)\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s,$
(3.27) $\displaystyle\hskip 28.45274ptL_{2,0}^{\prime}$
$\displaystyle:=\Gamma_{t_{1}\vee
t_{2}}\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)\left(\mathchoice{\frac{F_{\varepsilon}}{2}}{F_{\varepsilon}/2}{\frac{F_{\varepsilon}}{2}}{\frac{F_{\varepsilon}}{2}}+f(\varepsilon)\sqrt{rs}\right)\int_{A_{\varepsilon}^{c}}f(x-y)q_{0}(z,x,t-r)q_{0}(z,y,t-s)\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s.$

Then by similar arguments to those in the proof of Lemma 3.3, we obtain

$\displaystyle L_{1,0}^{\prime}$ $\displaystyle\leq\Gamma_{t_{1}\vee
t_{2}}\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)\left(\mathchoice{\frac{F_{\varepsilon}}{2}}{F_{\varepsilon}/2}{\frac{F_{\varepsilon}}{2}}{\frac{F_{\varepsilon}}{2}}+f(\varepsilon)\sqrt{t_{1}t_{2}}\right)\int_{A_{\varepsilon}}f(x-y)q_{0}(z,x,t-r)q_{0}(z,y,t-s)\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s$
$\displaystyle\leq\Gamma_{t_{1}\vee
t_{2}}^{2}\left(\mathchoice{\frac{F_{\varepsilon}^{2}}{4}}{F_{\varepsilon}^{2}/4}{\frac{F_{\varepsilon}^{2}}{4}}{\frac{F_{\varepsilon}^{2}}{4}}+\mathchoice{\frac{F_{\varepsilon}}{2}}{F_{\varepsilon}/2}{\frac{F_{\varepsilon}}{2}}{\frac{F_{\varepsilon}}{2}}f(\varepsilon)\sqrt{t_{1}t_{2}}\right),$

and

$L_{2,0}^{\prime}\leq\Gamma_{t_{1}\vee
t_{2}}^{2}\left(\mathchoice{\frac{F_{\varepsilon}}{2}}{F_{\varepsilon}/2}{\frac{F_{\varepsilon}}{2}}{\frac{F_{\varepsilon}}{2}}f(\varepsilon)\sqrt{t_{1}t_{2}}+f(\varepsilon)^{2}\mathchoice{\frac{t_{1}t_{2}}{2}}{t_{1}t_{2}/2}{\frac{t_{1}t_{2}}{2}}{\frac{t_{1}t_{2}}{2}}\right).$

Putting it all together, we have

(3.28) $\displaystyle\hskip 28.45274pt\left\langle
f_{2}(\cdot;z_{1},t_{1}),f_{2}(\cdot;z_{2},t_{2})\right\rangle_{2}\leq
L_{1,0}^{\prime}+L_{2,0}^{\prime}\leq\Gamma_{t_{1}\vee
t_{2}}^{2}\left(\mathchoice{\frac{F_{\varepsilon}^{2}}{4}}{F_{\varepsilon}^{2}/4}{\frac{F_{\varepsilon}^{2}}{4}}{\frac{F_{\varepsilon}^{2}}{4}}+F_{\varepsilon}f(\varepsilon)\sqrt{t_{1}t_{2}}+f(\varepsilon)^{2}\mathchoice{\frac{t_{1}t_{2}}{2}}{t_{1}t_{2}/2}{\frac{t_{1}t_{2}}{2}}{\frac{t_{1}t_{2}}{2}}\right).$

Again, we obtain a bound uniform in $z_{1}$ and $z_{2}$ and increasing in
$t_{1}$ and $t_{2}$, with powers of $\Gamma_{t_{1}\vee t_{2}},F(\varepsilon),$
and $f(\varepsilon)$ appearing throughout. The key insight for generalizing to
higher order Wiener chaos spaces is that this approach, which we have already
applied twice, can be reiterated indefinitely to obtain a bound which
maintains its general pattern. As it turns out, it can be shown that this
pattern possesses a tree-like structure (see Figure 3.1).

$n=0$$n=1$$n=2$1$\displaystyle\mathchoice{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\Gamma_{t}F_{\varepsilon}/2}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}$$\displaystyle\mathchoice{\frac{\Gamma_{t}^{2}F_{\varepsilon}^{2}}{4}}{\Gamma_{t}^{2}F_{\varepsilon}^{2}/4}{\frac{\Gamma_{t}^{2}F_{\varepsilon}^{2}}{4}}{\frac{\Gamma_{t}^{2}F_{\varepsilon}^{2}}{4}}$$\displaystyle\mathchoice{\frac{\Gamma_{t}^{n}F_{\varepsilon}^{n}}{2^{n}}}{\Gamma_{t}^{n}F_{\varepsilon}^{n}/2^{n}}{\frac{\Gamma_{t}^{n}F_{\varepsilon}^{n}}{2^{n}}}{\frac{\Gamma_{t}^{n}F_{\varepsilon}^{n}}{2^{n}}}$$\Gamma_{t}^{2}\displaystyle\mathchoice{\frac{F_{\varepsilon}}{2}}{F_{\varepsilon}/2}{\frac{F_{\varepsilon}}{2}}{\frac{F_{\varepsilon}}{2}}f(\varepsilon)t$$\Gamma_{t}\displaystyle
f(\varepsilon)t$$\Gamma_{t}^{2}\displaystyle\mathchoice{\frac{F_{\varepsilon}}{2}}{F_{\varepsilon}/2}{\frac{F_{\varepsilon}}{2}}{\frac{F_{\varepsilon}}{2}}f(\varepsilon)t$$\Gamma_{t}^{2}\displaystyle\mathchoice{\frac{(f(\varepsilon)t)^{2}}{2}}{(f(\varepsilon)t)^{2}/2}{\frac{(f(\varepsilon)t)^{2}}{2}}{\frac{(f(\varepsilon)t)^{2}}{2}}$$\Gamma_{t}^{n}\displaystyle\mathchoice{\frac{(f(\varepsilon)t)^{n}}{n!}}{(f(\varepsilon)t)^{n}/n!}{\frac{(f(\varepsilon)t)^{n}}{n!}}{\frac{(f(\varepsilon)t)^{n}}{n!}}$$\Gamma_{t}^{n}\displaystyle\binom{m-1}{k}\left(\mathchoice{\frac{F_{\varepsilon}}{2}}{F_{\varepsilon}/2}{\frac{F_{\varepsilon}}{2}}{\frac{F_{\varepsilon}}{2}}\right)^{n-k-1}\mathchoice{\frac{(f(\varepsilon)t)^{k+1}}{(k+1)!}}{(f(\varepsilon)t)^{k+1}/(k+1)!}{\frac{(f(\varepsilon)t)^{k+1}}{(k+1)!}}{\frac{(f(\varepsilon)t)^{k+1}}{(k+1)!}}$++++$+\cdots+$$+\cdots+$
Figure 3.1. The left child of each vertex corresponds to the bound obtained by
handling the integral over $A_{\varepsilon}$ while the right child corresponds
to the integral over $A_{\varepsilon}^{c}$. Each $1\leq m\leq n$ corresponds
to the $m$th subtree from the left whose respective roots are the right
children of the leftmost vertex at each depth. Moreover, for any given path in
a subtree, each $0\leq k\leq m-1$ corresponds to the number of rights in the
path.

Therefore, in order to rigorously prove an analog to (3.12), we need the
following technical lemma.

######  Lemma 3.4.

Let $z_{1},z_{2},t_{1},t_{2}>0$ and let $k\geq 0$. Then for $U$ as in (3.9)
and any $\varepsilon>0$ we have

$\displaystyle\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)$
$\displaystyle\int_{\mathbb{R}_{+}^{2}}f(x-y)q_{0}(z_{1},x,t_{1}-r)q_{0}(z_{2},y,t_{2}-s)(rs)^{\mathchoice{\frac{k}{2}}{k/2}{\frac{k}{2}}{\frac{k}{2}}}\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s$
(3.29) $\displaystyle\leq\Gamma_{t_{1}\vee
t_{2}}\left(F_{\varepsilon}\sqrt{U\left(\mathchoice{\frac{z_{1}}{t_{1}}}{z_{1}/t_{1}}{\frac{z_{1}}{t_{1}}}{\frac{z_{1}}{t_{1}}}\right)U\left(\mathchoice{\frac{z_{2}}{t_{2}}}{z_{2}/t_{2}}{\frac{z_{2}}{t_{2}}}{\frac{z_{2}}{t_{2}}}\right)t_{1}^{k}t_{2}^{k}}+\mathchoice{\frac{f(\varepsilon)(t_{1}t_{2})^{\mathchoice{\frac{k+1}{2}}{k+1/2}{\frac{k+1}{2}}{\frac{k+1}{2}}}}{k+1}}{f(\varepsilon)(t_{1}t_{2})^{\mathchoice{\frac{k+1}{2}}{k+1/2}{\frac{k+1}{2}}{\frac{k+1}{2}}}/k+1}{\frac{f(\varepsilon)(t_{1}t_{2})^{\mathchoice{\frac{k+1}{2}}{k+1/2}{\frac{k+1}{2}}{\frac{k+1}{2}}}}{k+1}}{\frac{f(\varepsilon)(t_{1}t_{2})^{\mathchoice{\frac{k+1}{2}}{k+1/2}{\frac{k+1}{2}}{\frac{k+1}{2}}}}{k+1}}\right).$

###### Proof.

Following a similar strategy to Lemma 3.3, define $A_{\varepsilon}$ as before
and

(3.30) $\displaystyle\hskip 28.45274ptL_{1,k}$
$\displaystyle:=\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)\int_{A_{\varepsilon}}f(x-y)q_{0}(z_{1},x,t_{1}-r)q_{0}(z_{2},y,t_{2}-s)(rs)^{\mathchoice{\frac{k}{2}}{k/2}{\frac{k}{2}}{\frac{k}{2}}}\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s,$
(3.31) $\displaystyle\hskip 28.45274ptL_{2,k}$
$\displaystyle:=\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)\int_{A_{\varepsilon}^{c}}f(x-y)q_{0}(z_{1},x,t_{1}-r)q_{0}(z_{2},y,t_{2}-s)(rs)^{\mathchoice{\frac{k}{2}}{k/2}{\frac{k}{2}}{\frac{k}{2}}}\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s.$

We easily treat (3.30) by reusing the bound for $L_{1,0}$ in (3.21):

$\displaystyle\hskip 28.45274ptL_{1,k}$
$\displaystyle\leq(t_{1}t_{2})^{\mathchoice{\frac{k}{2}}{k/2}{\frac{k}{2}}{\frac{k}{2}}}\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)\int_{A_{\varepsilon}}f(x-y)q_{0}(z_{1},x,t_{1}-r)q_{0}(z_{2},y,t_{2}-s)\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s$
(3.32) $\displaystyle\leq\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}\sqrt{U\left(\mathchoice{\frac{z_{1}}{t_{1}}}{z_{1}/t_{1}}{\frac{z_{1}}{t_{1}}}{\frac{z_{1}}{t_{1}}}\right)U\left(\mathchoice{\frac{z_{2}}{t_{2}}}{z_{2}/t_{2}}{\frac{z_{2}}{t_{2}}}{\frac{z_{2}}{t_{2}}}\right)t_{1}^{k}t_{2}^{k}}.$

To bound (3.31), we again apply a similar strategy used to obtain the bound
for $L_{2,0}$ in (3.22):

$\displaystyle L_{2,k}$ $\displaystyle\leq
f(\varepsilon)\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)(rs)^{\mathchoice{\frac{k}{2}}{k/2}{\frac{k}{2}}{\frac{k}{2}}}u_{0}(z_{1},t_{1}-r)u_{0}(z_{2},t_{2}-s)\mathrm{d}r\mathrm{d}s$
$\displaystyle\leq
f(\varepsilon)\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)(rs)^{\mathchoice{\frac{k}{2}}{k/2}{\frac{k}{2}}{\frac{k}{2}}}\mathrm{d}r\mathrm{d}s$
$\displaystyle\leq\Gamma_{t_{1}\vee
t_{2}}f(\varepsilon)\left(\int_{0}^{t_{1}}r^{k}\mathrm{d}r\int_{0}^{t_{2}}s^{k}\mathrm{d}s\right)^{\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}$
(3.33) $\displaystyle=\mathchoice{\frac{\Gamma_{t_{1}\vee
t_{2}}f(\varepsilon)(t_{1}t_{2})^{\mathchoice{\frac{k+1}{2}}{k+1/2}{\frac{k+1}{2}}{\frac{k+1}{2}}}}{k+1}}{\Gamma_{t_{1}\vee
t_{2}}f(\varepsilon)(t_{1}t_{2})^{\mathchoice{\frac{k+1}{2}}{k+1/2}{\frac{k+1}{2}}{\frac{k+1}{2}}}/k+1}{\frac{\Gamma_{t_{1}\vee
t_{2}}f(\varepsilon)(t_{1}t_{2})^{\mathchoice{\frac{k+1}{2}}{k+1/2}{\frac{k+1}{2}}{\frac{k+1}{2}}}}{k+1}}{\frac{\Gamma_{t_{1}\vee
t_{2}}f(\varepsilon)(t_{1}t_{2})^{\mathchoice{\frac{k+1}{2}}{k+1/2}{\frac{k+1}{2}}{\frac{k+1}{2}}}}{k+1}},$

where the penultimate inequality follows per usual by Proposition 5.2. Hence,
the result of the lemma following by summing (3.32) and (3.33). ∎

We are now equipped to present the solution to (1.6).

######  Theorem 3.2.

Suppose $f$ satisfies conditions (i) and (ii). Then for every $z,t>0$,
$u(z,t)$ as in (3.4) is well-defined as a convergent series in
$L^{2}(\Omega)$. Moreover for $\varepsilon>0$ sufficiently small
($\varepsilon$ depends on $t)$,

$\displaystyle\mathbb{E}[u^{2}(z,t)]\leq u_{0}(z,t)^{2}+$ $\displaystyle
U\left(\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}\right)\mathchoice{\frac{2\Gamma_{t}F_{\varepsilon}}{2-\Gamma_{t}F_{\varepsilon}}}{2\Gamma_{t}F_{\varepsilon}/2-\Gamma_{t}F_{\varepsilon}}{\frac{2\Gamma_{t}F_{\varepsilon}}{2-\Gamma_{t}F_{\varepsilon}}}{\frac{2\Gamma_{t}F_{\varepsilon}}{2-\Gamma_{t}F_{\varepsilon}}}$
(3.34) $\displaystyle\hskip
28.45274pt+\mathchoice{\frac{\sqrt{2}\Gamma_{t}f(\varepsilon)t}{\sqrt{2}-\sqrt{\Gamma_{t}F_{\varepsilon}}}}{\sqrt{2}\Gamma_{t}f(\varepsilon)t/\sqrt{2}-\sqrt{\Gamma_{t}F_{\varepsilon}}}{\frac{\sqrt{2}\Gamma_{t}f(\varepsilon)t}{\sqrt{2}-\sqrt{\Gamma_{t}F_{\varepsilon}}}}{\frac{\sqrt{2}\Gamma_{t}f(\varepsilon)t}{\sqrt{2}-\sqrt{\Gamma_{t}F_{\varepsilon}}}}\exp\left\\{\mathchoice{\frac{2f(\varepsilon)t\sqrt{\Gamma_{t}F_{\varepsilon}}}{F_{\varepsilon}(\sqrt{2}-\sqrt{\Gamma_{t}F_{\varepsilon}})}}{2f(\varepsilon)t\sqrt{\Gamma_{t}F_{\varepsilon}}/F_{\varepsilon}(\sqrt{2}-\sqrt{\Gamma_{t}F_{\varepsilon}})}{\frac{2f(\varepsilon)t\sqrt{\Gamma_{t}F_{\varepsilon}}}{F_{\varepsilon}(\sqrt{2}-\sqrt{\Gamma_{t}F_{\varepsilon}})}}{\frac{2f(\varepsilon)t\sqrt{\Gamma_{t}F_{\varepsilon}}}{F_{\varepsilon}(\sqrt{2}-\sqrt{\Gamma_{t}F_{\varepsilon}})}}\right\\}.$

$u=\left\\{u(z,t)\mid z,t\geq 0\right\\}$ is the unique mild solution to
(1.6).

###### Proof.

We focus on the convergence of the series in (3.4) to $u(z,t)$ in
$L^{2}(\Omega)$. The rest of the theorem follows similarly as in the proof of
Theorem 3.1 for the case of space-time white noise.

Let $z_{1},z_{2},t_{1},t_{2},\varepsilon>0$, we begin by inductively showing

$\displaystyle\left\langle
f_{n}(\cdot;z_{1},t_{1}),f_{n}(\cdot;z_{2},t_{2})\right\rangle_{n}\leq\left(\mathchoice{\frac{\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}}{2}}{\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}/2}{\frac{\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}}{2}}{\frac{\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}}{2}}\right)^{n}\left[2\sqrt{U\left(\mathchoice{\frac{z_{1}}{t_{1}}}{z_{1}/t_{1}}{\frac{z_{1}}{t_{1}}}{\frac{z_{1}}{t_{1}}}\right)U\left(\mathchoice{\frac{z_{2}}{t_{2}}}{z_{2}/t_{2}}{\frac{z_{2}}{t_{2}}}{\frac{z_{2}}{t_{2}}}\right)}\right.$
(3.35) $\displaystyle\hskip
156.49014pt+\left.\sum_{m=1}^{n}\sum_{k=0}^{m-1}\binom{m-1}{k}\mathchoice{\frac{1}{(k+1)!}}{1/(k+1)!}{\frac{1}{(k+1)!}}{\frac{1}{(k+1)!}}\left(\mathchoice{\frac{2f(\varepsilon)\sqrt{t_{1}t_{2}}}{F_{\varepsilon}}}{2f(\varepsilon)\sqrt{t_{1}t_{2}}/F_{\varepsilon}}{\frac{2f(\varepsilon)\sqrt{t_{1}t_{2}}}{F_{\varepsilon}}}{\frac{2f(\varepsilon)\sqrt{t_{1}t_{2}}}{F_{\varepsilon}}}\right)^{k+1}\right]$

for $n\geq 1$. The base case $n=1$ is proved in Lemma 3.3. Next, let $n\geq 2$
and suppose (3.35) holds for $n-1$. Then by (3.16), we obtain

$\displaystyle\left\langle
f_{n}(\cdot;z_{1},t_{1}),f_{n}(\cdot;z_{2},t_{2})\right\rangle_{n}$
$\displaystyle\leq\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)\int_{\mathbb{R}_{+}^{2}}f(x-y)q_{0}(z_{1},x,t_{1}-r)q_{0}(z_{2},y,t_{2}-s)$
$\displaystyle\hskip 14.22636pt\left(\mathchoice{\frac{\Gamma_{r\vee
s}F_{\varepsilon}}{2}}{\Gamma_{r\vee s}F_{\varepsilon}/2}{\frac{\Gamma_{r\vee
s}F_{\varepsilon}}{2}}{\frac{\Gamma_{r\vee
s}F_{\varepsilon}}{2}}\right)^{n-1}\left[1+\sum_{m=1}^{n-1}\sum_{k=0}^{m-1}\binom{m-1}{k}\mathchoice{\frac{1}{(k+1)!}}{1/(k+1)!}{\frac{1}{(k+1)!}}{\frac{1}{(k+1)!}}\left(\mathchoice{\frac{2f(\varepsilon)\sqrt{rs}}{F_{\varepsilon}}}{2f(\varepsilon)\sqrt{rs}/F_{\varepsilon}}{\frac{2f(\varepsilon)\sqrt{rs}}{F_{\varepsilon}}}{\frac{2f(\varepsilon)\sqrt{rs}}{F_{\varepsilon}}}\right)^{k+1}\right]\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s$
$\displaystyle=\left(\mathchoice{\frac{\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}}{2}}{\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}/2}{\frac{\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}}{2}}{\frac{\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}}{2}}\right)^{n-1}\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)\int_{\mathbb{R}_{+}^{2}}f(x-y)q_{0}(z_{1},x,t_{1}-r)q_{0}(z_{2},y,t_{2}-s)\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s$
$\displaystyle\hskip 14.22636pt+\left(\mathchoice{\frac{\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}}{2}}{\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}/2}{\frac{\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}}{2}}{\frac{\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}}{2}}\right)^{n-1}\sum_{m=1}^{n-1}\sum_{k=0}^{m-1}\binom{m-1}{k}\mathchoice{\frac{1}{(k+1)!}}{1/(k+1)!}{\frac{1}{(k+1)!}}{\frac{1}{(k+1)!}}\left(\mathchoice{\frac{2f(\varepsilon)}{F_{\varepsilon}}}{2f(\varepsilon)/F_{\varepsilon}}{\frac{2f(\varepsilon)}{F_{\varepsilon}}}{\frac{2f(\varepsilon)}{F_{\varepsilon}}}\right)^{k+1}$
$\displaystyle\hskip
28.45274pt\cdot\int_{[0,t_{1}]\times[0,t_{2}]}\gamma(r-s)\int_{\mathbb{R}_{+}^{2}}f(x-y)q_{0}(z_{1},x,t_{1}-r)q_{0}(z_{2},y,t_{2}-s)(rs)^{\mathchoice{\frac{k+1}{2}}{k+1/2}{\frac{k+1}{2}}{\frac{k+1}{2}}}\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s.$

By applying Lemma 3.4 to both integrals, it then follows that

$\displaystyle\mathbb{E}\left[u_{n}(z_{1},t_{1})u_{n}(z_{2},t_{2})\right]\leq\left(\mathchoice{\frac{\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}}{2}}{\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}/2}{\frac{\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}}{2}}{\frac{\Gamma_{t_{1}\vee
t_{2}}F_{\varepsilon}}{2}}\right)^{n}\left[2\sqrt{U\left(\mathchoice{\frac{z_{1}}{t_{1}}}{z_{1}/t_{1}}{\frac{z_{1}}{t_{1}}}{\frac{z_{1}}{t_{1}}}\right)U\left(\mathchoice{\frac{z_{2}}{t_{2}}}{z_{2}/t_{2}}{\frac{z_{2}}{t_{2}}}{\frac{z_{2}}{t_{2}}}\right)}+\mathchoice{\frac{2f(\varepsilon)\sqrt{t_{1}t_{2}}}{F_{\varepsilon}}}{2f(\varepsilon)\sqrt{t_{1}t_{2}}/F_{\varepsilon}}{\frac{2f(\varepsilon)\sqrt{t_{1}t_{2}}}{F_{\varepsilon}}}{\frac{2f(\varepsilon)\sqrt{t_{1}t_{2}}}{F_{\varepsilon}}}\right.$
$\displaystyle\left.+\sum_{m=1}^{n-1}\sum_{k=0}^{m-1}\binom{m-1}{k}\left(\left(\mathchoice{\frac{F_{\varepsilon}}{2}}{F_{\varepsilon}/2}{\frac{F_{\varepsilon}}{2}}{\frac{F_{\varepsilon}}{2}}\right)^{-k-1}\mathchoice{\frac{(f(\varepsilon)\sqrt{t_{1}t_{2}})^{k+1}}{(k+1)!}}{(f(\varepsilon)\sqrt{t_{1}t_{2}})^{k+1}/(k+1)!}{\frac{(f(\varepsilon)\sqrt{t_{1}t_{2}})^{k+1}}{(k+1)!}}{\frac{(f(\varepsilon)\sqrt{t_{1}t_{2}})^{k+1}}{(k+1)!}}+\left(\mathchoice{\frac{F_{\varepsilon}}{2}}{F_{\varepsilon}/2}{\frac{F_{\varepsilon}}{2}}{\frac{F_{\varepsilon}}{2}}\right)^{-k-2}\mathchoice{\frac{(f(\varepsilon)\sqrt{t_{1}t_{2}})^{k+2}}{(k+2)!}}{(f(\varepsilon)\sqrt{t_{1}t_{2}})^{k+2}/(k+2)!}{\frac{(f(\varepsilon)\sqrt{t_{1}t_{2}})^{k+2}}{(k+2)!}}{\frac{(f(\varepsilon)\sqrt{t_{1}t_{2}})^{k+2}}{(k+2)!}}\right)\right].$

Regrouping terms, we see that

$\displaystyle\sum_{m=1}^{n-1}\sum_{k=0}^{m-1}\binom{m-1}{k}$
$\displaystyle\left(\left(\mathchoice{\frac{F_{\varepsilon}}{2}}{F_{\varepsilon}/2}{\frac{F_{\varepsilon}}{2}}{\frac{F_{\varepsilon}}{2}}\right)^{-k-1}\mathchoice{\frac{(f(\varepsilon)\sqrt{t_{1}t_{2}})^{k+1}}{(k+1)!}}{(f(\varepsilon)\sqrt{t_{1}t_{2}})^{k+1}/(k+1)!}{\frac{(f(\varepsilon)\sqrt{t_{1}t_{2}})^{k+1}}{(k+1)!}}{\frac{(f(\varepsilon)\sqrt{t_{1}t_{2}})^{k+1}}{(k+1)!}}+\left(\mathchoice{\frac{F_{\varepsilon}}{2}}{F_{\varepsilon}/2}{\frac{F_{\varepsilon}}{2}}{\frac{F_{\varepsilon}}{2}}\right)^{-k-2}\mathchoice{\frac{(f(\varepsilon)\sqrt{t_{1}t_{2}})^{k+2}}{(k+2)!}}{(f(\varepsilon)\sqrt{t_{1}t_{2}})^{k+2}/(k+2)!}{\frac{(f(\varepsilon)\sqrt{t_{1}t_{2}})^{k+2}}{(k+2)!}}{\frac{(f(\varepsilon)\sqrt{t_{1}t_{2}})^{k+2}}{(k+2)!}}\right)$
$\displaystyle=\sum_{m=2}^{n}\sum_{k=0}^{m-1}\binom{m-1}{k}\mathchoice{\frac{1}{(k+1)!}}{1/(k+1)!}{\frac{1}{(k+1)!}}{\frac{1}{(k+1)!}}\left(\mathchoice{\frac{2f(\varepsilon)\sqrt{t_{1}t_{2}}}{F_{\varepsilon}}}{2f(\varepsilon)\sqrt{t_{1}t_{2}}/F_{\varepsilon}}{\frac{2f(\varepsilon)\sqrt{t_{1}t_{2}}}{F_{\varepsilon}}}{\frac{2f(\varepsilon)\sqrt{t_{1}t_{2}}}{F_{\varepsilon}}}\right)^{k+1},$

and

$\displaystyle\mathchoice{\frac{2f(\varepsilon)\sqrt{t_{1}t_{2}}}{F_{\varepsilon}}}{2f(\varepsilon)\sqrt{t_{1}t_{2}}/F_{\varepsilon}}{\frac{2f(\varepsilon)\sqrt{t_{1}t_{2}}}{F_{\varepsilon}}}{\frac{2f(\varepsilon)\sqrt{t_{1}t_{2}}}{F_{\varepsilon}}}$
$\displaystyle+\sum_{m=2}^{n}\sum_{k=0}^{m-1}\binom{m-1}{k}\mathchoice{\frac{1}{(k+1)!}}{1/(k+1)!}{\frac{1}{(k+1)!}}{\frac{1}{(k+1)!}}\left(\mathchoice{\frac{2f(\varepsilon)\sqrt{t_{1}t_{2}}}{F_{\varepsilon}}}{2f(\varepsilon)\sqrt{t_{1}t_{2}}/F_{\varepsilon}}{\frac{2f(\varepsilon)\sqrt{t_{1}t_{2}}}{F_{\varepsilon}}}{\frac{2f(\varepsilon)\sqrt{t_{1}t_{2}}}{F_{\varepsilon}}}\right)^{k+1}$
$\displaystyle=\sum_{m=1}^{n}\sum_{k=0}^{m-1}\binom{m-1}{k}\mathchoice{\frac{1}{(k+1)!}}{1/(k+1)!}{\frac{1}{(k+1)!}}{\frac{1}{(k+1)!}}\left(\mathchoice{\frac{2f(\varepsilon)\sqrt{t_{1}t_{2}}}{F_{\varepsilon}}}{2f(\varepsilon)\sqrt{t_{1}t_{2}}/F_{\varepsilon}}{\frac{2f(\varepsilon)\sqrt{t_{1}t_{2}}}{F_{\varepsilon}}}{\frac{2f(\varepsilon)\sqrt{t_{1}t_{2}}}{F_{\varepsilon}}}\right)^{k+1},$

from which (3.35) follows.

Setting $z_{1}=z_{2}=z,t_{1}=t_{2}=t>0$, (3.35) simplifies to

(3.36) $\displaystyle\hskip
28.45274pt\|f_{n}(\cdot;z,t)\|_{n}^{2}\leq\left(\mathchoice{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\Gamma_{t}F_{\varepsilon}/2}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}\right)^{n}\left[2U\left(\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}\right)+\sum_{m=1}^{n}\sum_{k=0}^{m-1}\binom{m-1}{k}\mathchoice{\frac{1}{(k+1)!}}{1/(k+1)!}{\frac{1}{(k+1)!}}{\frac{1}{(k+1)!}}\left(\mathchoice{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{2f(\varepsilon)t/F_{\varepsilon}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}\right)^{k+1}\right].$

We proceed to prove (3.34). For every $z,t>0$, choose $\varepsilon>0$ such
that $\Gamma_{t}F_{\varepsilon}<1$. By (3.36), we have

$\displaystyle\sum_{n=0}^{\infty}\|f_{n}(\cdot;z,t)\|_{n}^{2}=u_{0}(z,t)^{2}+\sum_{n=1}^{\infty}\|f_{n}(\cdot;z,t)\|_{n}^{2}$
$\displaystyle\leq
u_{0}(z,t)^{2}+U\left(\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}\right)\sum_{n=1}^{\infty}\mathchoice{\frac{\left(\Gamma_{t}F_{\varepsilon}\right)^{n}}{2^{n-1}}}{\left(\Gamma_{t}F_{\varepsilon}\right)^{n}/2^{n-1}}{\frac{\left(\Gamma_{t}F_{\varepsilon}\right)^{n}}{2^{n-1}}}{\frac{\left(\Gamma_{t}F_{\varepsilon}\right)^{n}}{2^{n-1}}}+\sum_{n=1}^{\infty}\left(\mathchoice{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\Gamma_{t}F_{\varepsilon}/2}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}\right)^{n}\sum_{m=1}^{n}\sum_{k=0}^{m-1}\binom{m-1}{k}\mathchoice{\frac{1}{(k+1)!}}{1/(k+1)!}{\frac{1}{(k+1)!}}{\frac{1}{(k+1)!}}\left(\mathchoice{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{2f(\varepsilon)t/F_{\varepsilon}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}\right)^{k+1}.$

By the choice of $\varepsilon$, we have
$\mathchoice{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\Gamma_{t}F_{\varepsilon}/2}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}<1$
and hence

$\sum_{n=1}^{\infty}\mathchoice{\frac{\Gamma_{t}^{n}F_{\varepsilon}^{n}}{2^{n-1}}}{\Gamma_{t}^{n}F_{\varepsilon}^{n}/2^{n-1}}{\frac{\Gamma_{t}^{n}F_{\varepsilon}^{n}}{2^{n-1}}}{\frac{\Gamma_{t}^{n}F_{\varepsilon}^{n}}{2^{n-1}}}=\mathchoice{\frac{\Gamma_{t}F_{\varepsilon}}{1-\left(\displaystyle\mathchoice{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\Gamma_{t}F_{\varepsilon}/2}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}\right)}}{\Gamma_{t}F_{\varepsilon}/1-\left(\displaystyle\mathchoice{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\Gamma_{t}F_{\varepsilon}/2}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}\right)}{\frac{\Gamma_{t}F_{\varepsilon}}{1-\left(\displaystyle\mathchoice{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\Gamma_{t}F_{\varepsilon}/2}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}\right)}}{\frac{\Gamma_{t}F_{\varepsilon}}{1-\left(\displaystyle\mathchoice{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\Gamma_{t}F_{\varepsilon}/2}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}\right)}}=\mathchoice{\frac{2\Gamma_{t}F_{\varepsilon}}{2-\Gamma_{t}F_{\varepsilon}}}{2\Gamma_{t}F_{\varepsilon}/2-\Gamma_{t}F_{\varepsilon}}{\frac{2\Gamma_{t}F_{\varepsilon}}{2-\Gamma_{t}F_{\varepsilon}}}{\frac{2\Gamma_{t}F_{\varepsilon}}{2-\Gamma_{t}F_{\varepsilon}}}<\infty.$

It remains to show the last series converges. First, we have

$\displaystyle\sum_{k=0}^{m-1}\binom{m-1}{k}\mathchoice{\frac{1}{(k+1)!}}{1/(k+1)!}{\frac{1}{(k+1)!}}{\frac{1}{(k+1)!}}\left(\mathchoice{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{2f(\varepsilon)t/F_{\varepsilon}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}\right)^{k+1}$
$\displaystyle=\mathchoice{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{2f(\varepsilon)t/F_{\varepsilon}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}\sum_{k=0}^{m-1}\binom{m-1}{k}\mathchoice{\frac{1}{(k+1)!}}{1/(k+1)!}{\frac{1}{(k+1)!}}{\frac{1}{(k+1)!}}\left(\mathchoice{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{2f(\varepsilon)t/F_{\varepsilon}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}\right)^{k}$
$\displaystyle\leq\mathchoice{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{2f(\varepsilon)t/F_{\varepsilon}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}\sum_{k=0}^{n-1}\binom{n-1}{k}\mathchoice{\frac{1}{(k+1)!}}{1/(k+1)!}{\frac{1}{(k+1)!}}{\frac{1}{(k+1)!}}\left(\mathchoice{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{2f(\varepsilon)t/F_{\varepsilon}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}\right)^{k},$

from which it follows that

$\sum_{m=1}^{n}\sum_{k=0}^{m-1}\binom{m-1}{k}\mathchoice{\frac{1}{(k+1)!}}{1/(k+1)!}{\frac{1}{(k+1)!}}{\frac{1}{(k+1)!}}\left(\mathchoice{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{2f(\varepsilon)t/F_{\varepsilon}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}\right)^{k+1}\leq\mathchoice{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{2f(\varepsilon)t/F_{\varepsilon}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}n\sum_{k=0}^{n-1}\binom{n-1}{k}\mathchoice{\frac{1}{(k+1)!}}{1/(k+1)!}{\frac{1}{(k+1)!}}{\frac{1}{(k+1)!}}\left(\mathchoice{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{2f(\varepsilon)t/F_{\varepsilon}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}\right)^{k},$

and so

$\displaystyle\sum_{n=1}^{\infty}\left(\mathchoice{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\Gamma_{t}F_{\varepsilon}/2}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}\right)^{n}$
$\displaystyle\sum_{m=1}^{n}\sum_{k=0}^{m-1}\binom{m-1}{k}\mathchoice{\frac{1}{(k+1)!}}{1/(k+1)!}{\frac{1}{(k+1)!}}{\frac{1}{(k+1)!}}\left(\mathchoice{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{2f(\varepsilon)t/F_{\varepsilon}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}\right)^{k+1}$
$\displaystyle\leq\Gamma_{t}f(\varepsilon)t\sum_{n=1}^{\infty}\left(\mathchoice{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\Gamma_{t}F_{\varepsilon}/2}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}\right)^{n-1}n\sum_{k=0}^{n-1}\binom{n-1}{k}\mathchoice{\frac{1}{k!}}{1/k!}{\frac{1}{k!}}{\frac{1}{k!}}\left(\mathchoice{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{2f(\varepsilon)t/F_{\varepsilon}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}\right)^{k}$
$\displaystyle=\Gamma_{t}f(\varepsilon)t\sum_{n=0}^{\infty}\left(\mathchoice{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\Gamma_{t}F_{\varepsilon}/2}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}\right)^{n}(n+1)\sum_{k=0}^{n}\binom{n}{k}\mathchoice{\frac{1}{k!}}{1/k!}{\frac{1}{k!}}{\frac{1}{k!}}\left(\mathchoice{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{2f(\varepsilon)t/F_{\varepsilon}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}\right)^{k}$
$\displaystyle\leq\Gamma_{t}f(\varepsilon)t\sum_{n=0}^{\infty}\left(\mathchoice{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\Gamma_{t}F_{\varepsilon}/2}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}{\frac{\Gamma_{t}F_{\varepsilon}}{2}}\right)^{n}(n+1)\sum_{k=0}^{n}2^{n}\mathchoice{\frac{1}{k!}}{1/k!}{\frac{1}{k!}}{\frac{1}{k!}}\left(\mathchoice{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{2f(\varepsilon)t/F_{\varepsilon}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}\right)^{k}$
$\displaystyle\leq\Gamma_{t}f(\varepsilon)te^{\mathchoice{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{2f(\varepsilon)t/F_{\varepsilon}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}}\sum_{n=0}^{\infty}(\Gamma_{t}F_{\varepsilon})^{n}(n+1)$
$\displaystyle=\mathchoice{\frac{\Gamma_{t}f(\varepsilon)t}{(\Gamma_{t}F_{\varepsilon}-1)^{2}}}{\Gamma_{t}f(\varepsilon)t/(\Gamma_{t}F_{\varepsilon}-1)^{2}}{\frac{\Gamma_{t}f(\varepsilon)t}{(\Gamma_{t}F_{\varepsilon}-1)^{2}}}{\frac{\Gamma_{t}f(\varepsilon)t}{(\Gamma_{t}F_{\varepsilon}-1)^{2}}}e^{\mathchoice{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{2f(\varepsilon)t/F_{\varepsilon}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}{\frac{2f(\varepsilon)t}{F_{\varepsilon}}}},$

where convergence in the last equality follows by the fact that
$\Gamma_{t}F_{\varepsilon}<1$. Hence for all $z,t>0$ we have (3.34) and so the
series in (3.4) converges to $u(z,t)$ in $L^{2}(\Omega)$. The rest of the
theorem follows similarly as in the proof of Theorem 3.1 for the case of
space-time white noise. ∎

###  3.2. Ratio Comparison in $L^{2}$

With existence and uniqueness established, next we study the behavior of the
solution near the boundary of the domain. As in (3.10) in Lemma 3.4 of [14],
it is insightful to study how the solution to the Kimura equation with a
potential compares to the unperturbed solution. In our setting the solution
$u$ is a random field, so instead of a point-wise comparison, we examine the
ratio between $u(z,t)$ and the unperturbed, deterministic solution
$u_{0}(z,t)$ in $L^{2}(\Omega)$. Specifically, recalling that
$u_{n}(z,t):=\mathcal{I}_{n}(f_{n}(\cdot;z,t))$ with $f_{n}$ as defined in
(3.5), we study

(3.37)
$\mathbb{E}\left[\left(\mathchoice{\frac{u(z,t)}{u_{0}(z,t)}}{u(z,t)/u_{0}(z,t)}{\frac{u(z,t)}{u_{0}(z,t)}}{\frac{u(z,t)}{u_{0}(z,t)}}\right)^{2}\right]=\sum_{n=0}^{\infty}\mathbb{E}\left[\left(\mathchoice{\frac{u_{n}(z,t)}{u_{0}(z,t)}}{u_{n}(z,t)/u_{0}(z,t)}{\frac{u_{n}(z,t)}{u_{0}(z,t)}}{\frac{u_{n}(z,t)}{u_{0}(z,t)}}\right)^{2}\right]$

for $z$ near 0 and $t$ small.

In the case of (1.6) with space-time white noise, we see that the solution is
quantifiably different at the boundary. Indeed, using the fact that
$I_{1}(w)\geq w/2$ for every $w>0$, by the definition of
$\mathbb{E}[u_{1}^{2}(z,t)]$, we get

$\displaystyle\mathbb{E}\left[u_{1}^{2}(z,t)\right]$
$\displaystyle=\int_{0}^{t}\int_{0}^{\infty}q_{0}^{2}(z,w,t-r)(1-e^{-\mathchoice{\frac{w}{r}}{w/r}{\frac{w}{r}}{\frac{w}{r}}})^{2}\mathrm{d}w\mathrm{d}r,$
$\displaystyle\geq
z^{2}\int_{0}^{t}\mathchoice{\frac{e^{-\mathchoice{\frac{2z}{t-r}}{2z/t-r}{\frac{2z}{t-r}}{\frac{2z}{t-r}}}}{(t-r)^{4}}}{e^{-\mathchoice{\frac{2z}{t-r}}{2z/t-r}{\frac{2z}{t-r}}{\frac{2z}{t-r}}}/(t-r)^{4}}{\frac{e^{-\mathchoice{\frac{2z}{t-r}}{2z/t-r}{\frac{2z}{t-r}}{\frac{2z}{t-r}}}}{(t-r)^{4}}}{\frac{e^{-\mathchoice{\frac{2z}{t-r}}{2z/t-r}{\frac{2z}{t-r}}{\frac{2z}{t-r}}}}{(t-r)^{4}}}\int_{0}^{\infty}e^{-\mathchoice{\frac{2w}{t-r}}{2w/t-r}{\frac{2w}{t-r}}{\frac{2w}{t-r}}}(1-e^{-\mathchoice{\frac{w}{r}}{w/r}{\frac{w}{r}}{\frac{w}{r}}})^{2}\mathrm{d}w\mathrm{d}r,$

and so by direction computations it follows that

$\displaystyle\mathbb{E}\left[u_{1}^{2}(z,t)\right]$
$\displaystyle=z^{2}\int_{0}^{t}\mathchoice{\frac{e^{-\mathchoice{\frac{2z}{t-r}}{2z/t-r}{\frac{2z}{t-r}}{\frac{2z}{t-r}}}}{2t(t-r)(t+r)}}{e^{-\mathchoice{\frac{2z}{t-r}}{2z/t-r}{\frac{2z}{t-r}}{\frac{2z}{t-r}}}/2t(t-r)(t+r)}{\frac{e^{-\mathchoice{\frac{2z}{t-r}}{2z/t-r}{\frac{2z}{t-r}}{\frac{2z}{t-r}}}}{2t(t-r)(t+r)}}{\frac{e^{-\mathchoice{\frac{2z}{t-r}}{2z/t-r}{\frac{2z}{t-r}}{\frac{2z}{t-r}}}}{2t(t-r)(t+r)}}\mathrm{d}r$
$\displaystyle\geq\mathchoice{\frac{z^{2}}{4t^{2}}}{z^{2}/4t^{2}}{\frac{z^{2}}{4t^{2}}}{\frac{z^{2}}{4t^{2}}}\int_{0}^{t}\mathchoice{\frac{e^{-\mathchoice{\frac{2z}{t-r}}{2z/t-r}{\frac{2z}{t-r}}{\frac{2z}{t-r}}}}{(t-r)}}{e^{-\mathchoice{\frac{2z}{t-r}}{2z/t-r}{\frac{2z}{t-r}}{\frac{2z}{t-r}}}/(t-r)}{\frac{e^{-\mathchoice{\frac{2z}{t-r}}{2z/t-r}{\frac{2z}{t-r}}{\frac{2z}{t-r}}}}{(t-r)}}{\frac{e^{-\mathchoice{\frac{2z}{t-r}}{2z/t-r}{\frac{2z}{t-r}}{\frac{2z}{t-r}}}}{(t-r)}}\mathrm{d}r=\mathchoice{\frac{z^{2}}{4t^{2}}}{z^{2}/4t^{2}}{\frac{z^{2}}{4t^{2}}}{\frac{z^{2}}{4t^{2}}}\int_{\mathchoice{\frac{2z}{t}}{2z/t}{\frac{2z}{t}}{\frac{2z}{t}}}^{\infty}\mathchoice{\frac{e^{-\tau}}{\tau}}{e^{-\tau}/\tau}{\frac{e^{-\tau}}{\tau}}{\frac{e^{-\tau}}{\tau}}\mathrm{d}\tau.$

Since $1-e^{-\mathchoice{\frac{z}{t}}{z/t}{\frac{z}{t}}{\frac{z}{t}}}\approx
z/t$ for fixed $t$ and sufficiently small $z$, we in turn have

$\displaystyle\lim_{z\searrow
0}\mathbb{E}\left[\left(\mathchoice{\frac{u_{1}(z,t)}{u_{0}(z,t)}}{u_{1}(z,t)/u_{0}(z,t)}{\frac{u_{1}(z,t)}{u_{0}(z,t)}}{\frac{u_{1}(z,t)}{u_{0}(z,t)}}\right)^{2}\right]\geq\lim_{z\searrow
0}\mathchoice{\frac{1}{4}}{1/4}{\frac{1}{4}}{\frac{1}{4}}\int_{\mathchoice{\frac{2z}{t}}{2z/t}{\frac{2z}{t}}{\frac{2z}{t}}}^{\infty}\mathchoice{\frac{e^{-\tau}}{\tau}}{e^{-\tau}/\tau}{\frac{e^{-\tau}}{\tau}}{\frac{e^{-\tau}}{\tau}}\mathrm{d}\tau=\mathchoice{\frac{1}{4}}{1/4}{\frac{1}{4}}{\frac{1}{4}}\int_{0}^{\infty}\mathchoice{\frac{e^{-\tau}}{\tau}}{e^{-\tau}/\tau}{\frac{e^{-\tau}}{\tau}}{\frac{e^{-\tau}}{\tau}}\mathrm{d}\tau=\infty,$

where the last integral diverges as it is the limit of the gamma function at
zero.

To remedy this, we introduce a degenerate coefficient of order $\beta>0$ to
$\dot{W}$ and consider the following stochastic Kimura equation:

$\begin{cases}&\partial_{t}u\left(z,t\right)=z\partial_{z}^{2}u\left(z,t\right)+\hat{z}^{\beta}u\left(z,t\right)\dot{W}\left(z,t\right)\quad\quad
z,t>0,\\\ &\hskip 9.10486ptu(z,0)=u_{0}(z)\hfill z>0,\\\ &\hskip
10.81204ptu(0,t)=0\hfill t>0,\end{cases}$

where recall that $\hat{z}:=z\wedge 1$. It can be shown similarly to
$\mathsection\ref{sec:eu}$ that (3.1) admits a unique mild solution. In
addition, the second moments of the solution can be recursively computed by

$\displaystyle\mathbb{E}[u_{n}^{2}(z,t)]=\int_{[0,t]^{2}}$
$\displaystyle\gamma(r-s)\int_{\mathbb{R}_{+}^{2}}f(x-y)\hat{x}^{\beta}\hat{y}^{\beta}$
(3.38) $\displaystyle\cdot
q_{0}(z,x,t-r)q_{0}(z,y,t-s)\mathbb{E}[u_{n-1}(x,r)u_{n-1}(y,s)]\mathrm{d}x\mathrm{d}y\mathrm{d}r\mathrm{d}s.$

In the following subsections, we show that so long as $\beta>0$, the
stochastic and deterministic solutions are “comparable” in a proper sense. As
before, we begin our analysis in the case of space-time white noise which we
bootstrap to the case of colored noise.

####  3.2.1. Space-Time White Noise

Our strategy for the case of space-time white noise is similar to the one used
to show convergence of (3.4) in $L^{2}(\Omega)$. Namely, we use (3.8) to
recursively compute a bound with some repetitive pattern. We first make use of
the following technical lemma for $n=1$.

######  Lemma 3.5.

For $\beta,z,t-\tau,\tau>0$, define

(3.39)
$K(z,\tau;2\beta):=\int_{0}^{\infty}\hat{w}^{2\beta}q_{0}^{2}(z,w,t-\tau)\mathchoice{\frac{u_{0}^{2}(w,\tau)}{u_{0}^{2}(z,t)}}{u_{0}^{2}(w,\tau)/u_{0}^{2}(z,t)}{\frac{u_{0}^{2}(w,\tau)}{u_{0}^{2}(z,t)}}{\frac{u_{0}^{2}(w,\tau)}{u_{0}^{2}(z,t)}}\mathrm{d}w.$

Then, there exists $C:=C_{\beta}>0$, such that for all $z,t,t-\tau,\tau>0$,

(3.40)
$K(z,\tau;2\beta)\leq\mathchoice{\frac{4C}{(t-\tau)^{1-\alpha}}}{4C/(t-\tau)^{1-\alpha}}{\frac{4C}{(t-\tau)^{1-\alpha}}}{\frac{4C}{(t-\tau)^{1-\alpha}}}(\hat{z}\vee
t)^{2\beta-\alpha},$

whenever $0<\alpha\leq 2\beta\wedge 1/2$.

###### Proof.

We further split $K(z,\tau;2\beta)$ into two terms:

$\displaystyle K_{1}$
$\displaystyle:=\int_{0}^{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}\hat{w}^{2\beta}q_{0}^{2}(z,w,t-\tau)\mathchoice{\frac{u_{0}^{2}(w,\tau)}{u_{0}^{2}(z,t)}}{u_{0}^{2}(w,\tau)/u_{0}^{2}(z,t)}{\frac{u_{0}^{2}(w,\tau)}{u_{0}^{2}(z,t)}}{\frac{u_{0}^{2}(w,\tau)}{u_{0}^{2}(z,t)}}\mathrm{d}w,$
$\displaystyle K_{2}$
$\displaystyle:=\int_{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}^{\infty}\hat{w}^{2\beta}q_{0}^{2}(z,w,t-\tau)\mathchoice{\frac{u_{0}^{2}(w,\tau)}{u_{0}^{2}(z,t)}}{u_{0}^{2}(w,\tau)/u_{0}^{2}(z,t)}{\frac{u_{0}^{2}(w,\tau)}{u_{0}^{2}(z,t)}}{\frac{u_{0}^{2}(w,\tau)}{u_{0}^{2}(z,t)}}\mathrm{d}w.$

We will treat $K_{1}$ and $K_{2}$ in two separate cases.

_Case $\tau\leq t/2$._ In this case, we apply Proposition 2.2 and (3.7) to
bound $K_{1}$ as

$\displaystyle K_{1}$
$\displaystyle\leq\mathchoice{\frac{\displaystyle\mathchoice{\frac{z^{2}}{(t-\tau)^{4}}}{z^{2}/(t-\tau)^{4}}{\frac{z^{2}}{(t-\tau)^{4}}}{\frac{z^{2}}{(t-\tau)^{4}}}\int_{0}^{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}\hat{w}^{2\beta}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{(t-\tau)}}{2(\sqrt{z}-\sqrt{w})^{2}/(t-\tau)}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{(t-\tau)}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{(t-\tau)}}}\mathrm{d}w}{\displaystyle
e^{-2}\mathchoice{\frac{z^{2}}{t^{2}}}{z^{2}/t^{2}}{\frac{z^{2}}{t^{2}}}{\frac{z^{2}}{t^{2}}}}}{\displaystyle\mathchoice{\frac{z^{2}}{(t-\tau)^{4}}}{z^{2}/(t-\tau)^{4}}{\frac{z^{2}}{(t-\tau)^{4}}}{\frac{z^{2}}{(t-\tau)^{4}}}\int_{0}^{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}\hat{w}^{2\beta}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{(t-\tau)}}{2(\sqrt{z}-\sqrt{w})^{2}/(t-\tau)}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{(t-\tau)}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{(t-\tau)}}}\mathrm{d}w/\displaystyle
e^{-2}\mathchoice{\frac{z^{2}}{t^{2}}}{z^{2}/t^{2}}{\frac{z^{2}}{t^{2}}}{\frac{z^{2}}{t^{2}}}}{\frac{\displaystyle\mathchoice{\frac{z^{2}}{(t-\tau)^{4}}}{z^{2}/(t-\tau)^{4}}{\frac{z^{2}}{(t-\tau)^{4}}}{\frac{z^{2}}{(t-\tau)^{4}}}\int_{0}^{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}\hat{w}^{2\beta}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{(t-\tau)}}{2(\sqrt{z}-\sqrt{w})^{2}/(t-\tau)}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{(t-\tau)}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{(t-\tau)}}}\mathrm{d}w}{\displaystyle
e^{-2}\mathchoice{\frac{z^{2}}{t^{2}}}{z^{2}/t^{2}}{\frac{z^{2}}{t^{2}}}{\frac{z^{2}}{t^{2}}}}}{\frac{\displaystyle\mathchoice{\frac{z^{2}}{(t-\tau)^{4}}}{z^{2}/(t-\tau)^{4}}{\frac{z^{2}}{(t-\tau)^{4}}}{\frac{z^{2}}{(t-\tau)^{4}}}\int_{0}^{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}\hat{w}^{2\beta}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{(t-\tau)}}{2(\sqrt{z}-\sqrt{w})^{2}/(t-\tau)}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{(t-\tau)}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{(t-\tau)}}}\mathrm{d}w}{\displaystyle
e^{-2}\mathchoice{\frac{z^{2}}{t^{2}}}{z^{2}/t^{2}}{\frac{z^{2}}{t^{2}}}{\frac{z^{2}}{t^{2}}}}}$
$\displaystyle\left(t-\tau\geq\mathchoice{\frac{t}{2}}{t/2}{\frac{t}{2}}{\frac{t}{2}}\right)$
$\displaystyle\leq\mathchoice{\frac{C}{(t-\tau)^{2}}}{C/(t-\tau)^{2}}{\frac{C}{(t-\tau)^{2}}}{\frac{C}{(t-\tau)^{2}}}\int_{0}^{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}\hat{w}^{2\beta}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{(t-\tau)}}{2(\sqrt{z}-\sqrt{w})^{2}/(t-\tau)}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{(t-\tau)}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{(t-\tau)}}}\mathrm{d}w$
$\displaystyle=\mathchoice{\frac{C}{(t-\tau)^{\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}}{C/(t-\tau)^{\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}{\frac{C}{(t-\tau)^{\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}}{\frac{C}{(t-\tau)^{\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}}\left(\mathchoice{\frac{1}{\sqrt{\pi(t-\tau)}}}{1/\sqrt{\pi(t-\tau)}}{\frac{1}{\sqrt{\pi(t-\tau)}}}{\frac{1}{\sqrt{\pi(t-\tau)}}}\int_{0}^{\mathchoice{\frac{t-\tau}{\sqrt{z}}}{t-\tau/\sqrt{z}}{\frac{t-\tau}{\sqrt{z}}}{\frac{t-\tau}{\sqrt{z}}}}\hat{u}^{4\beta}ue^{-\mathchoice{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{2(\sqrt{z}-u)^{2}/t-\tau}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}}\mathrm{d}u\right).$

Furthermore, if $z\leq t-\tau$ then by Proposition 5.1,

$\displaystyle K_{1}$
$\displaystyle\leq\mathchoice{\frac{C}{(t-\tau)^{\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}}{C/(t-\tau)^{\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}{\frac{C}{(t-\tau)^{\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}}{\frac{C}{(t-\tau)^{\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}}\left(\mathchoice{\frac{1}{\sqrt{\pi(t-\tau)}}}{1/\sqrt{\pi(t-\tau)}}{\frac{1}{\sqrt{\pi(t-\tau)}}}{\frac{1}{\sqrt{\pi(t-\tau)}}}\int_{0}^{\mathchoice{\frac{t-\tau}{\sqrt{z}}}{t-\tau/\sqrt{z}}{\frac{t-\tau}{\sqrt{z}}}{\frac{t-\tau}{\sqrt{z}}}}u^{4\beta+1}e^{-\mathchoice{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{2(\sqrt{z}-u)^{2}/t-\tau}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}}\mathrm{d}u\right)$
$\displaystyle\leq\mathchoice{\frac{2C}{(t-\tau)^{\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}}{2C/(t-\tau)^{\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}{\frac{2C}{(t-\tau)^{\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}}{\frac{2C}{(t-\tau)^{\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}}(t-\tau)^{2\beta+\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}=2C(t-\tau)^{2\beta-1}.$

Meanwhile, if $z>t-\tau$, again by Proposition 5.1, we get

$\displaystyle K_{1}$ $\displaystyle\leq
C(t-\tau)^{-\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}(t-\tau)^{-1}\left(\mathchoice{\frac{1}{\sqrt{\pi(t-\tau)}}}{1/\sqrt{\pi(t-\tau)}}{\frac{1}{\sqrt{\pi(t-\tau)}}}{\frac{1}{\sqrt{\pi(t-\tau)}}}\int_{0}^{\mathchoice{\frac{t-\tau}{\sqrt{z}}}{t-\tau/\sqrt{z}}{\frac{t-\tau}{\sqrt{z}}}{\frac{t-\tau}{\sqrt{z}}}}\hat{u}^{4\beta}ue^{-\mathchoice{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{2(\sqrt{z}-u)^{2}/t-\tau}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}}\mathrm{d}u\right)$
$\displaystyle\left((t-\tau)^{-1}\leq
u^{-1}z^{-\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}\right)$
$\displaystyle\leq
Cz^{-\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}(t-\tau)^{-\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}\left(\mathchoice{\frac{1}{\sqrt{\pi(t-\tau)}}}{1/\sqrt{\pi(t-\tau)}}{\frac{1}{\sqrt{\pi(t-\tau)}}}{\frac{1}{\sqrt{\pi(t-\tau)}}}\int_{0}^{\mathchoice{\frac{t-\tau}{\sqrt{z}}}{t-\tau/\sqrt{z}}{\frac{t-\tau}{\sqrt{z}}}{\frac{t-\tau}{\sqrt{z}}}}\hat{u}^{4\beta}e^{-\mathchoice{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{2(\sqrt{z}-u)^{2}/t-\tau}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}}\mathrm{d}u\right)$
$\displaystyle\leq\mathchoice{\frac{2Cz^{-\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}}{\sqrt{t-\tau}}}{2Cz^{-\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}/\sqrt{t-\tau}}{\frac{2Cz^{-\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}}{\sqrt{t-\tau}}}{\frac{2Cz^{-\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}}{\sqrt{t-\tau}}}\hat{z}^{2\beta}\leq
2C\hat{z}^{2\beta}\mathchoice{\frac{z^{-\alpha}}{(t-\tau)^{1-\alpha}}}{z^{-\alpha}/(t-\tau)^{1-\alpha}}{\frac{z^{-\alpha}}{(t-\tau)^{1-\alpha}}}{\frac{z^{-\alpha}}{(t-\tau)^{1-\alpha}}}\leq
2C\mathchoice{\frac{\hat{z}^{2\beta-\alpha}}{(t-\tau)^{1-\alpha}}}{\hat{z}^{2\beta-\alpha}/(t-\tau)^{1-\alpha}}{\frac{\hat{z}^{2\beta-\alpha}}{(t-\tau)^{1-\alpha}}}{\frac{\hat{z}^{2\beta-\alpha}}{(t-\tau)^{1-\alpha}}},$

for $0\leq\alpha\leq 2\beta\wedge 1/2$. For $K_{2}$, we apply (2.19) from
Proposition 2.2 and (3.7) to get

$\displaystyle K_{2}$
$\displaystyle\leq\mathchoice{\frac{\displaystyle\mathchoice{\frac{\sqrt{z}}{t-\tau}}{\sqrt{z}/t-\tau}{\frac{\sqrt{z}}{t-\tau}}{\frac{\sqrt{z}}{t-\tau}}\int_{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}^{\infty}\hat{w}^{2\beta}w^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{2(\sqrt{z}-\sqrt{w})^{2}/t-\tau}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}}\mathrm{d}w}{\displaystyle
e^{-2}\mathchoice{\frac{z^{2}}{t^{2}}}{z^{2}/t^{2}}{\frac{z^{2}}{t^{2}}}{\frac{z^{2}}{t^{2}}}}}{\displaystyle\mathchoice{\frac{\sqrt{z}}{t-\tau}}{\sqrt{z}/t-\tau}{\frac{\sqrt{z}}{t-\tau}}{\frac{\sqrt{z}}{t-\tau}}\int_{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}^{\infty}\hat{w}^{2\beta}w^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{2(\sqrt{z}-\sqrt{w})^{2}/t-\tau}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}}\mathrm{d}w/\displaystyle
e^{-2}\mathchoice{\frac{z^{2}}{t^{2}}}{z^{2}/t^{2}}{\frac{z^{2}}{t^{2}}}{\frac{z^{2}}{t^{2}}}}{\frac{\displaystyle\mathchoice{\frac{\sqrt{z}}{t-\tau}}{\sqrt{z}/t-\tau}{\frac{\sqrt{z}}{t-\tau}}{\frac{\sqrt{z}}{t-\tau}}\int_{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}^{\infty}\hat{w}^{2\beta}w^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{2(\sqrt{z}-\sqrt{w})^{2}/t-\tau}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}}\mathrm{d}w}{\displaystyle
e^{-2}\mathchoice{\frac{z^{2}}{t^{2}}}{z^{2}/t^{2}}{\frac{z^{2}}{t^{2}}}{\frac{z^{2}}{t^{2}}}}}{\frac{\displaystyle\mathchoice{\frac{\sqrt{z}}{t-\tau}}{\sqrt{z}/t-\tau}{\frac{\sqrt{z}}{t-\tau}}{\frac{\sqrt{z}}{t-\tau}}\int_{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}^{\infty}\hat{w}^{2\beta}w^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{2(\sqrt{z}-\sqrt{w})^{2}/t-\tau}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}}\mathrm{d}w}{\displaystyle
e^{-2}\mathchoice{\frac{z^{2}}{t^{2}}}{z^{2}/t^{2}}{\frac{z^{2}}{t^{2}}}{\frac{z^{2}}{t^{2}}}}}$
$\displaystyle\left(t-\tau\geq\mathchoice{\frac{t}{2}}{t/2}{\frac{t}{2}}{\frac{t}{2}}\right)$
$\displaystyle\leq
Cz^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}(t-\tau)\int_{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}^{\infty}\hat{w}^{2\beta}w^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{2(\sqrt{z}-\sqrt{w})^{2}/t-\tau}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}}\mathrm{d}w$
$\displaystyle=Cz^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}(t-\tau)^{\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}\left(\mathchoice{\frac{1}{\sqrt{\pi(t-\tau)}}}{1/\sqrt{\pi(t-\tau)}}{\frac{1}{\sqrt{\pi(t-\tau)}}}{\frac{1}{\sqrt{\pi(t-\tau)}}}\int_{\mathchoice{\frac{t-\tau}{\sqrt{z}}}{t-\tau/\sqrt{z}}{\frac{t-\tau}{\sqrt{z}}}{\frac{t-\tau}{\sqrt{z}}}}^{\infty}\hat{u}^{4\beta}u^{-2}e^{-\mathchoice{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{2(\sqrt{z}-u)^{2}/t-\tau}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}}\mathrm{d}u\right)$
$\displaystyle\left(u^{-1}\leq\mathchoice{\frac{\sqrt{z}}{t-\tau}}{\sqrt{z}/t-\tau}{\frac{\sqrt{z}}{t-\tau}}{\frac{\sqrt{z}}{t-\tau}}\right)$
$\displaystyle\leq
Cz^{-1}(t-\tau)^{\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}\left(\mathchoice{\frac{1}{\sqrt{\pi(t-\tau)}}}{1/\sqrt{\pi(t-\tau)}}{\frac{1}{\sqrt{\pi(t-\tau)}}}{\frac{1}{\sqrt{\pi(t-\tau)}}}\int_{\mathchoice{\frac{t-\tau}{\sqrt{z}}}{t-\tau/\sqrt{z}}{\frac{t-\tau}{\sqrt{z}}}{\frac{t-\tau}{\sqrt{z}}}}^{\infty}\hat{u}^{4\beta}e^{-\mathchoice{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{2(\sqrt{z}-u)^{2}/t-\tau}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}}\mathrm{d}u\right).$

Furthermore, if $z>t-\tau$, then again by Proposition 5.1, for
$0\leq\alpha\leq 2\beta\wedge 1/2$, $K_{2}\leq
2Cz^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}(t-\tau)^{\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}\hat{z}^{2\beta}\leq
2C\hat{z}^{2\beta-\alpha}(t-\tau)^{\alpha-1}$. Meanwhile, if $z\leq t-\tau$ we
get

$\displaystyle K_{2}$ $\displaystyle\leq
Cz^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}\mathchoice{\frac{(t-\tau)^{3}}{(t-\tau)^{\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}}{(t-\tau)^{3}/(t-\tau)^{\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}{\frac{(t-\tau)^{3}}{(t-\tau)^{\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}}{\frac{(t-\tau)^{3}}{(t-\tau)^{\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}}\left(\mathchoice{\frac{1}{\sqrt{\pi(t-\tau)}}}{1/\sqrt{\pi(t-\tau)}}{\frac{1}{\sqrt{\pi(t-\tau)}}}{\frac{1}{\sqrt{\pi(t-\tau)}}}\int_{\mathchoice{\frac{t-\tau}{\sqrt{z}}}{t-\tau/\sqrt{z}}{\frac{t-\tau}{\sqrt{z}}}{\frac{t-\tau}{\sqrt{z}}}}^{\infty}u^{4\beta-2}e^{-\mathchoice{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{2(\sqrt{z}-u)^{2}/t-\tau}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}}\mathrm{d}u\right)$
$\displaystyle\left((t-\tau)^{-3}\geq
u^{-3}z^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}\right)$
$\displaystyle\leq
C(t-\tau)^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}\left(\mathchoice{\frac{1}{\sqrt{\pi(t-\tau)}}}{1/\sqrt{\pi(t-\tau)}}{\frac{1}{\sqrt{\pi(t-\tau)}}}{\frac{1}{\sqrt{\pi(t-\tau)}}}\int_{\mathchoice{\frac{t-\tau}{\sqrt{z}}}{t-\tau/\sqrt{z}}{\frac{t-\tau}{\sqrt{z}}}{\frac{t-\tau}{\sqrt{z}}}}^{\infty}u^{4\beta+1}e^{-\mathchoice{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{2(\sqrt{z}-u)^{2}/t-\tau}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}}\mathrm{d}u\right)$
$\displaystyle\leq 2C(t-\tau)^{2\beta-1}.$

_Case $\tau>t/2$_. For $K_{1}$, we apply (2.18) from Proposition 2.2 and (3.7)
again to get

$\displaystyle K_{1}$
$\displaystyle\leq\mathchoice{\frac{\displaystyle\mathchoice{\frac{z^{2}}{(t-\tau)^{4}}}{z^{2}/(t-\tau)^{4}}{\frac{z^{2}}{(t-\tau)^{4}}}{\frac{z^{2}}{(t-\tau)^{4}}}\int_{0}^{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}\hat{w}^{2\beta}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{2(\sqrt{z}-\sqrt{w})^{2}/t-\tau}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}}u_{0}^{2}(w,\tau)\mathrm{d}w}{\displaystyle
e^{-2}\mathchoice{\frac{z^{2}}{t^{2}}}{z^{2}/t^{2}}{\frac{z^{2}}{t^{2}}}{\frac{z^{2}}{t^{2}}}}}{\displaystyle\mathchoice{\frac{z^{2}}{(t-\tau)^{4}}}{z^{2}/(t-\tau)^{4}}{\frac{z^{2}}{(t-\tau)^{4}}}{\frac{z^{2}}{(t-\tau)^{4}}}\int_{0}^{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}\hat{w}^{2\beta}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{2(\sqrt{z}-\sqrt{w})^{2}/t-\tau}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}}u_{0}^{2}(w,\tau)\mathrm{d}w/\displaystyle
e^{-2}\mathchoice{\frac{z^{2}}{t^{2}}}{z^{2}/t^{2}}{\frac{z^{2}}{t^{2}}}{\frac{z^{2}}{t^{2}}}}{\frac{\displaystyle\mathchoice{\frac{z^{2}}{(t-\tau)^{4}}}{z^{2}/(t-\tau)^{4}}{\frac{z^{2}}{(t-\tau)^{4}}}{\frac{z^{2}}{(t-\tau)^{4}}}\int_{0}^{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}\hat{w}^{2\beta}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{2(\sqrt{z}-\sqrt{w})^{2}/t-\tau}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}}u_{0}^{2}(w,\tau)\mathrm{d}w}{\displaystyle
e^{-2}\mathchoice{\frac{z^{2}}{t^{2}}}{z^{2}/t^{2}}{\frac{z^{2}}{t^{2}}}{\frac{z^{2}}{t^{2}}}}}{\frac{\displaystyle\mathchoice{\frac{z^{2}}{(t-\tau)^{4}}}{z^{2}/(t-\tau)^{4}}{\frac{z^{2}}{(t-\tau)^{4}}}{\frac{z^{2}}{(t-\tau)^{4}}}\int_{0}^{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}\hat{w}^{2\beta}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{2(\sqrt{z}-\sqrt{w})^{2}/t-\tau}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}}u_{0}^{2}(w,\tau)\mathrm{d}w}{\displaystyle
e^{-2}\mathchoice{\frac{z^{2}}{t^{2}}}{z^{2}/t^{2}}{\frac{z^{2}}{t^{2}}}{\frac{z^{2}}{t^{2}}}}}$
$\displaystyle\leq\mathchoice{\frac{t^{2}}{(t-\tau)^{4}}}{t^{2}/(t-\tau)^{4}}{\frac{t^{2}}{(t-\tau)^{4}}}{\frac{t^{2}}{(t-\tau)^{4}}}\int_{0}^{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}\hat{w}^{2\beta}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{2(\sqrt{z}-\sqrt{w})^{2}/t-\tau}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}}\mathchoice{\frac{w^{2}}{\tau^{2}}}{w^{2}/\tau^{2}}{\frac{w^{2}}{\tau^{2}}}{\frac{w^{2}}{\tau^{2}}}\mathrm{d}w$
$\displaystyle\left(\tau>\mathchoice{\frac{t}{2}}{t/2}{\frac{t}{2}}{\frac{t}{2}}\right)$
$\displaystyle\leq\mathchoice{\frac{C}{(t-\tau)^{4}}}{C/(t-\tau)^{4}}{\frac{C}{(t-\tau)^{4}}}{\frac{C}{(t-\tau)^{4}}}\int_{0}^{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}\hat{w}^{2\beta}w^{2}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{2(\sqrt{z}-\sqrt{w})^{2}/t-\tau}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}\mathrm{d}w}$
$\displaystyle=\mathchoice{\frac{C}{(t-\tau)^{\mathchoice{\frac{7}{2}}{7/2}{\frac{7}{2}}{\frac{7}{2}}}}}{C/(t-\tau)^{\mathchoice{\frac{7}{2}}{7/2}{\frac{7}{2}}{\frac{7}{2}}}}{\frac{C}{(t-\tau)^{\mathchoice{\frac{7}{2}}{7/2}{\frac{7}{2}}{\frac{7}{2}}}}}{\frac{C}{(t-\tau)^{\mathchoice{\frac{7}{2}}{7/2}{\frac{7}{2}}{\frac{7}{2}}}}}\left(\mathchoice{\frac{1}{\sqrt{\pi(t-\tau)}}}{1/\sqrt{\pi(t-\tau)}}{\frac{1}{\sqrt{\pi(t-\tau)}}}{\frac{1}{\sqrt{\pi(t-\tau)}}}\int_{0}^{\mathchoice{\frac{t-\tau}{\sqrt{z}}}{t-\tau/\sqrt{z}}{\frac{t-\tau}{\sqrt{z}}}{\frac{t-\tau}{\sqrt{z}}}}\hat{u}^{4\beta}u^{5}e^{-\mathchoice{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{2(\sqrt{z}-u)^{2}/t-\tau}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}\mathrm{d}u}\right).$

Furthermore, if $z\leq t-\tau$ then $K_{1}\leq 2C(t-\tau)^{2\beta-1}$.
Meanwhile, if $z>t-\tau$ we get

$\displaystyle K_{1}$
$\displaystyle\leq\mathchoice{\frac{C}{(t-\tau)^{\mathchoice{\frac{7}{2}}{7/2}{\frac{7}{2}}{\frac{7}{2}}}}}{C/(t-\tau)^{\mathchoice{\frac{7}{2}}{7/2}{\frac{7}{2}}{\frac{7}{2}}}}{\frac{C}{(t-\tau)^{\mathchoice{\frac{7}{2}}{7/2}{\frac{7}{2}}{\frac{7}{2}}}}}{\frac{C}{(t-\tau)^{\mathchoice{\frac{7}{2}}{7/2}{\frac{7}{2}}{\frac{7}{2}}}}}\left(\mathchoice{\frac{1}{\sqrt{\pi(t-\tau)}}}{1/\sqrt{\pi(t-\tau)}}{\frac{1}{\sqrt{\pi(t-\tau)}}}{\frac{1}{\sqrt{\pi(t-\tau)}}}\int_{0}^{\mathchoice{\frac{t-\tau}{\sqrt{z}}}{t-\tau/\sqrt{z}}{\frac{t-\tau}{\sqrt{z}}}{\frac{t-\tau}{\sqrt{z}}}}\hat{u}^{4\beta}u^{5}e^{-\mathchoice{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{2(\sqrt{z}-u)^{2}/t-\tau}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}\mathrm{d}u}\right)$
$\displaystyle=C(t-\tau)^{-\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}(t-\tau)^{-3}\left(\mathchoice{\frac{1}{\sqrt{\pi(t-\tau)}}}{1/\sqrt{\pi(t-\tau)}}{\frac{1}{\sqrt{\pi(t-\tau)}}}{\frac{1}{\sqrt{\pi(t-\tau)}}}\int_{0}^{\mathchoice{\frac{t-\tau}{\sqrt{z}}}{t-\tau/\sqrt{z}}{\frac{t-\tau}{\sqrt{z}}}{\frac{t-\tau}{\sqrt{z}}}}\hat{u}^{4\beta}u^{5}e^{-\mathchoice{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{2(\sqrt{z}-u)^{2}/t-\tau}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}\mathrm{d}u}\right)$
$\displaystyle\left((t-\tau)^{-3}\leq
z^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}u^{-3}\right)$
$\displaystyle\leq
Cz^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}(t-\tau)^{-\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}\left(\mathchoice{\frac{1}{\sqrt{\pi(t-\tau)}}}{1/\sqrt{\pi(t-\tau)}}{\frac{1}{\sqrt{\pi(t-\tau)}}}{\frac{1}{\sqrt{\pi(t-\tau)}}}\int_{0}^{\mathchoice{\frac{t-\tau}{\sqrt{z}}}{t-\tau/\sqrt{z}}{\frac{t-\tau}{\sqrt{z}}}{\frac{t-\tau}{\sqrt{z}}}}\hat{u}^{4\beta}u^{2}e^{-\mathchoice{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{2(\sqrt{z}-u)^{2}/t-\tau}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}\mathrm{d}u}\right)$
$\displaystyle\leq\mathchoice{\frac{2C}{\sqrt{t-\tau}}}{2C/\sqrt{t-\tau}}{\frac{2C}{\sqrt{t-\tau}}}{\frac{2C}{\sqrt{t-\tau}}}\hat{z}^{2\beta}z^{-\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}\leq
2C\hat{z}^{2\beta}\mathchoice{\frac{z^{-\alpha}}{(t-\tau)^{1-\alpha}}}{z^{-\alpha}/(t-\tau)^{1-\alpha}}{\frac{z^{-\alpha}}{(t-\tau)^{1-\alpha}}}{\frac{z^{-\alpha}}{(t-\tau)^{1-\alpha}}}\leq
2C\mathchoice{\frac{\hat{z}^{2\beta-\alpha}}{(t-\tau)^{1-\alpha}}}{\hat{z}^{2\beta-\alpha}/(t-\tau)^{1-\alpha}}{\frac{\hat{z}^{2\beta-\alpha}}{(t-\tau)^{1-\alpha}}}{\frac{\hat{z}^{2\beta-\alpha}}{(t-\tau)^{1-\alpha}}}.$

Similarly as above, for $K_{2}$ we apply (2.19) from Proposition 2.2 and (3.7)
to get

$\displaystyle K_{2}$
$\displaystyle\leq\mathchoice{\frac{\displaystyle\mathchoice{\frac{\sqrt{z}}{t-\tau}}{\sqrt{z}/t-\tau}{\frac{\sqrt{z}}{t-\tau}}{\frac{\sqrt{z}}{t-\tau}}\int_{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}^{\infty}\hat{w}^{2\beta}w^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{2(\sqrt{z}-\sqrt{w})^{2}/t-\tau}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}}u_{0}^{2}(w,\tau)\mathrm{d}w}{\displaystyle
e^{-2}\mathchoice{\frac{z^{2}}{t^{2}}}{z^{2}/t^{2}}{\frac{z^{2}}{t^{2}}}{\frac{z^{2}}{t^{2}}}}}{\displaystyle\mathchoice{\frac{\sqrt{z}}{t-\tau}}{\sqrt{z}/t-\tau}{\frac{\sqrt{z}}{t-\tau}}{\frac{\sqrt{z}}{t-\tau}}\int_{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}^{\infty}\hat{w}^{2\beta}w^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{2(\sqrt{z}-\sqrt{w})^{2}/t-\tau}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}}u_{0}^{2}(w,\tau)\mathrm{d}w/\displaystyle
e^{-2}\mathchoice{\frac{z^{2}}{t^{2}}}{z^{2}/t^{2}}{\frac{z^{2}}{t^{2}}}{\frac{z^{2}}{t^{2}}}}{\frac{\displaystyle\mathchoice{\frac{\sqrt{z}}{t-\tau}}{\sqrt{z}/t-\tau}{\frac{\sqrt{z}}{t-\tau}}{\frac{\sqrt{z}}{t-\tau}}\int_{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}^{\infty}\hat{w}^{2\beta}w^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{2(\sqrt{z}-\sqrt{w})^{2}/t-\tau}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}}u_{0}^{2}(w,\tau)\mathrm{d}w}{\displaystyle
e^{-2}\mathchoice{\frac{z^{2}}{t^{2}}}{z^{2}/t^{2}}{\frac{z^{2}}{t^{2}}}{\frac{z^{2}}{t^{2}}}}}{\frac{\displaystyle\mathchoice{\frac{\sqrt{z}}{t-\tau}}{\sqrt{z}/t-\tau}{\frac{\sqrt{z}}{t-\tau}}{\frac{\sqrt{z}}{t-\tau}}\int_{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}^{\infty}\hat{w}^{2\beta}w^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{2(\sqrt{z}-\sqrt{w})^{2}/t-\tau}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}}u_{0}^{2}(w,\tau)\mathrm{d}w}{\displaystyle
e^{-2}\mathchoice{\frac{z^{2}}{t^{2}}}{z^{2}/t^{2}}{\frac{z^{2}}{t^{2}}}{\frac{z^{2}}{t^{2}}}}}$
$\displaystyle\leq\mathchoice{\frac{z^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}t^{2}}{t-\tau}}{z^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}t^{2}/t-\tau}{\frac{z^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}t^{2}}{t-\tau}}{\frac{z^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}t^{2}}{t-\tau}}\int_{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}^{\infty}\mathchoice{\frac{\hat{w}^{2\beta}w^{\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}}{\tau^{2}}}{\hat{w}^{2\beta}w^{\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}/\tau^{2}}{\frac{\hat{w}^{2\beta}w^{\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}}{\tau^{2}}}{\frac{\hat{w}^{2\beta}w^{\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}}{\tau^{2}}}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{2(\sqrt{z}-\sqrt{w})^{2}/t-\tau}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}}\mathrm{d}w$
$\displaystyle\left(\tau>\mathchoice{\frac{t}{2}}{t/2}{\frac{t}{2}}{\frac{t}{2}}\right)$
$\displaystyle\leq\mathchoice{\frac{Cz^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}{t-\tau}}{Cz^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}/t-\tau}{\frac{Cz^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}{t-\tau}}{\frac{Cz^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}{t-\tau}}\int_{\mathchoice{\frac{(t-\tau)^{2}}{z}}{(t-\tau)^{2}/z}{\frac{(t-\tau)^{2}}{z}}{\frac{(t-\tau)^{2}}{z}}}^{\infty}\hat{w}^{2\beta}w^{\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}e^{-\mathchoice{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{2(\sqrt{z}-\sqrt{w})^{2}/t-\tau}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}{\frac{2(\sqrt{z}-\sqrt{w})^{2}}{t-\tau}}}\mathrm{d}w$
$\displaystyle=\mathchoice{\frac{Cz^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}{\sqrt{t-\tau}}}{Cz^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}/\sqrt{t-\tau}}{\frac{Cz^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}{\sqrt{t-\tau}}}{\frac{Cz^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}}{\sqrt{t-\tau}}}\left(\mathchoice{\frac{1}{\sqrt{\pi(t-\tau)}}}{1/\sqrt{\pi(t-\tau)}}{\frac{1}{\sqrt{\pi(t-\tau)}}}{\frac{1}{\sqrt{\pi(t-\tau)}}}\int_{\mathchoice{\frac{t-\tau}{\sqrt{z}}}{t-\tau/\sqrt{z}}{\frac{t-\tau}{\sqrt{z}}}{\frac{t-\tau}{\sqrt{z}}}}^{\infty}\hat{u}^{4\beta}u^{2}e^{-\mathchoice{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{2(\sqrt{z}-u)^{2}/t-\tau}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}}\mathrm{d}u\right).$

Furthermore, if $z>t-\tau$ then $K_{2}\leq
2C(t-\tau)^{-\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}\hat{z}^{2\beta}z^{-\mathchoice{\frac{1}{2}}{1/2}{\frac{1}{2}}{\frac{1}{2}}}\leq
2C\hat{z}^{2\beta-\alpha}(t-\tau)^{-1+\alpha}$. Meanwhile, if $z\leq t-\tau$
we get

$\displaystyle K_{2}$ $\displaystyle\leq
Cz^{-\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}(t-\tau)^{-\mathchoice{\frac{7}{2}}{7/2}{\frac{7}{2}}{\frac{7}{2}}}\cdot(t-\tau)^{3}\left(\mathchoice{\frac{1}{\sqrt{\pi(t-\tau)}}}{1/\sqrt{\pi(t-\tau)}}{\frac{1}{\sqrt{\pi(t-\tau)}}}{\frac{1}{\sqrt{\pi(t-\tau)}}}\int_{\mathchoice{\frac{t-\tau}{\sqrt{z}}}{t-\tau/\sqrt{z}}{\frac{t-\tau}{\sqrt{z}}}{\frac{t-\tau}{\sqrt{z}}}}^{\infty}u^{4\beta+2}e^{-\mathchoice{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{2(\sqrt{z}-u)^{2}/t-\tau}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}}\mathrm{d}u\right)$
$\displaystyle\left((t-\tau)^{3}\leq
z^{\mathchoice{\frac{3}{2}}{3/2}{\frac{3}{2}}{\frac{3}{2}}}u^{3}\right)$
$\displaystyle\leq
C(t-\tau)^{-\mathchoice{\frac{7}{2}}{7/2}{\frac{7}{2}}{\frac{7}{2}}}\left(\mathchoice{\frac{1}{\sqrt{\pi(t-\tau)}}}{1/\sqrt{\pi(t-\tau)}}{\frac{1}{\sqrt{\pi(t-\tau)}}}{\frac{1}{\sqrt{\pi(t-\tau)}}}\int_{\mathchoice{\frac{t-\tau}{\sqrt{z}}}{t-\tau/\sqrt{z}}{\frac{t-\tau}{\sqrt{z}}}{\frac{t-\tau}{\sqrt{z}}}}^{\infty}u^{4\beta+5}e^{-\mathchoice{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{2(\sqrt{z}-u)^{2}/t-\tau}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}{\frac{2(\sqrt{z}-u)^{2}}{t-\tau}}}\mathrm{d}u\right)$
$\displaystyle\leq 2C(t-\tau)^{2\beta-1}.$

Combining all these cases, we have shown that

$K(z,\tau;2\beta)=K_{1}+K_{2}\leq 4C(t-\tau)^{\alpha-1}(\hat{z}\vee
t)^{2\beta-\alpha}.$

∎

Note that if $\beta\leq\beta^{\prime}$, then for every $w>0$,
$\hat{w}^{2\beta^{\prime}}\leq\hat{w}^{2\beta}$ and hence

(3.41) $K(z,\tau;2\beta^{\prime})\leq K(z,\tau;2\beta).$

######  Remark 3.2.

We observe that for $z,t>0$ and $\beta>0$, if we define

(3.42)
$\tilde{K}(z,\tau;\beta):=\int_{0}^{\infty}\hat{w}^{\beta}q_{0}(z,w,t-\tau)\mathchoice{\frac{u_{0}(w,\tau)}{u_{0}(w,\tau)}}{u_{0}(w,\tau)/u_{0}(w,\tau)}{\frac{u_{0}(w,\tau)}{u_{0}(w,\tau)}}{\frac{u_{0}(w,\tau)}{u_{0}(w,\tau)}}\mathrm{d}w,$

then following the proof of Lemma 3.5 almost step-by-step (with minor
changes), we can get that

(3.43) $\tilde{K}(z,\tau;\beta)\leq 4C(\hat{z}\vee(t-\tau))^{\beta}.$

This observation on $\tilde{K}(z,\tau;\beta)$ will be useful in
$\mathsection\ref{ssec:ratioColored}$ in treating the ratio
$u(z,t)/u_{0}(z,t)$ under colored noise.

From this, the ratio follows straightforwardly by computing the integral in
time.

######  Corollary 3.1.

Let $\beta,C,$ and $\alpha$ be the same as above and fixed. Set
$C_{\alpha,\beta}:=4C/\alpha$. Then for all $z,t>0$,

(3.44)
$\mathbb{E}\left[\left(\mathchoice{\frac{u_{1}(z,t)}{u_{0}(z,t)}}{u_{1}(z,t)/u_{0}(z,t)}{\frac{u_{1}(z,t)}{u_{0}(z,t)}}{\frac{u_{1}(z,t)}{u_{0}(z,t)}}\right)^{2}\right]\leq
C_{\alpha,\beta}(\hat{z}\vee t)^{2\beta-\alpha}.$

###### Proof.

Follows directly by integrating the result from Lemma 3.5 in $\tau$ over
$[0,t]$. ∎

In light of Corollary 3.1, for every $w,\tau>0$,

$\mathbb{E}[u_{1}^{2}(w,\tau)]\leq
C_{\alpha,\beta}\tau^{\alpha}(\hat{w}\vee\tau)^{2\beta-\alpha}u_{0}^{2}(w,\tau),$

and hence it becomes possible for us to re-use the bound for the function
$K(z,\tau;\cdot)$ in Lemma 3.5. As it turns out, this is the key ingredient
for the main result for this section.

######  Theorem 3.3.

Let $\beta>0$. When $W$ is a space-time white noise, there exists $T\in(0,1)$
such that for every $t\in[0,T]$, there exists $C_{t,\beta}>0$ such that

(3.45)
$\sup_{z>0}\mathbb{E}\left[\left(\mathchoice{\frac{u(z,t)}{u_{0}(z,t)}}{u(z,t)/u_{0}(z,t)}{\frac{u(z,t)}{u_{0}(z,t)}}{\frac{u(z,t)}{u_{0}(z,t)}}\right)^{2}\right]<C_{t,\beta},$

where $T$ and $C_{t,\beta}$ can be explicitly determined. Furthermore, for
every fixed $z>0$,

(3.46) $\lim_{t\searrow
0}\mathbb{E}\left[\left(\mathchoice{\frac{u(z,t)}{u_{0}(z,t)}}{u(z,t)/u_{0}(z,t)}{\frac{u(z,t)}{u_{0}(z,t)}}{\frac{u(z,t)}{u_{0}(z,t)}}-1\right)^{2}\right]=0.$

###### Proof.

In light of (3.37) we proceed by induction on $n\geq 1$ to show that for
$\beta,z,t>0$ and $0<\alpha\leq 2\beta\wedge 1/2$,

(3.47)
$\mathbb{E}\left[\left(\mathchoice{\frac{u_{n}(z,t)}{u_{0}(z,t)}}{u_{n}(z,t)/u_{0}(z,t)}{\frac{u_{n}(z,t)}{u_{0}(z,t)}}{\frac{u_{n}(z,t)}{u_{0}(z,t)}}\right)^{2}\right]\leq
C_{\alpha,\beta}^{n}t^{n\alpha}(\hat{z}\vee t)^{(2\beta-\alpha)}.$

The base case is given by Corollary 3.1. Next, assume that (3.47) holds for
$n-1$ with $n\geq 2$. Then by using (3.47) and (3.41), we have