# Preventing Finite-Time Blowup in a Constrainted Potential for Reaction-
Diffusion Equations

John Ivanhoe and Michael Salins  Boston University

##  1 Introduction

We are interested in studying how dissipative forcing prevents explosion of
mild solutions to the stochastic reaction-diffusion equation (SRDE) defined on
an open bounded domain $D\subset\mathbb{R}^{d}$ with appropriately-smooth
boundary. Our equation of interest is

$\begin{cases}\frac{\partial u}{\partial
t}(t,x)=\mathcal{A}u(t,x)+f(u(t,x))+\sigma(u(t,x))\dot{W}(t,x)\\\
u(t,x)=0,\;\;x\in\partial D,\hskip 18.06749ptu(0,x)=u_{0}(x)\end{cases}$ (1)

Here, $\mathcal{A}$ is a second-order elliptic differential linear operator
and $\dot{W}$ is a Gaussian noise. The function $f(u(t,x))$ represents a
state-dependent external forcing and the multiplicative noise
$\sigma(u(t,x))\dot{W}$ represents state-dependent stochastic forcing. We are
particularly interested when $f$ models a constrained potential force, where
$\lim_{|x|\to 1}f(x)\text{sign}(x)=-\infty$. We also allow $\sigma$ to be
unbounded near $|x|\to 1$ and identify a sufficient condition on the relative
growth rates of $f$ and $\sigma$ as $|x|\to 1$ that guarantees the existence
of global solutions.

Specifically, we assume there are constants $c_{0}\in(0,1)$, $C,K,\beta>0$,
and $\gamma\geq 0$ such that for $c_{0}<|w|<1$

$\begin{cases}|\sigma(w)|\leq C(1-|w|)^{-\gamma}\\\
f(w)\;\text{sign}(w)\leq-K(1-|w|)^{-\beta}\end{cases}$ (2)

With $\gamma=0$, we allow for additive noise. Analogous to [34], the
stochastic forcing pushes solutions toward the endpoints with growing
intensity, counteracting the dissipative force of $f$.

This set-up is motivated by ecological models. In population growth models, a
common assumption is letting $f$ and $\sigma$ obey logistic growth. One such
example is the FKPP equation on the circle [36]:

$\partial_{t}u=\frac{\alpha}{2}\Delta u+\beta u(1-u)+\sqrt{\gamma
u(1-u)}\dot{W}$

This FKPP model has been studied by various authors [15], [14], [18], [37],
[27], [28], with applications recorded in the survey [31]. The terms $u(1-u)$
assume logistic growth, where a population approaching its carrying capacity
sees the forcing terms vanish. However, extreme downward forcing is also
reasonable for a population nearing its upper limit (for example, if there are
more elk than what a forest can sustain, many elk should die off). The
stochastic forcing should also be large, reflecting the large, yet random,
amount of offspring expected from a large population, as well as modeling
uncertainty as we near physical limits. Together, these ideas form our
assumptions in (2).

With these assumptions, the SRDE (1) is not well-defined whenever $|u(t,x)|$
$\geq 1$, motivating our interest in preventing finite-time blowup, which we
intuitively understand as any point $(t,x)\in(0,\infty)\times D$ such that
$|u(t,x)|=1$, assuming $||u(0,x)||_{L^{\infty}(D)}<1$. We will show that,
among other assumptions, the inequality

$\gamma+1<\frac{(1-\eta)(\beta+1)}{2}$ (3)

prevents finite-time blowup, where $\gamma,\beta$ are our constants from (2).
The constant $\eta\in[0,1)$ was introduced by Cerrai (2003) and describes the
balance between the roughness of the noise and smoothing of the elliptic
operator [4]. A larger $\eta$ corresponds to rougher solutions, with
$\frac{1-\eta}{2}$ being the temporal Hölder continuity of the paths. This is
reflected in inequality (3), where rougher solutions require larger $\beta$
values to prevent blow-up for a fixed $\gamma$. For an explicit definition of
$\eta$, see Assumption 3.

This result is analogous to the results in [34], where solutions to (1) are
allowed to grow arbitrarily large under superlinear forcing. There, the key
inequality preventing finite-time blowup is
$\gamma-1<\frac{(1-\eta)(\beta-1)}{2}$. The two results are similar, differing
largely due to the different exponents one gets from integrating $u^{-\beta}$
compared to $u^{\beta}$. However, our inequality presents a strange question
in the case of additive noise. When $\gamma=0$, our main result shows that
finite-time blow-up is prevented for a strictly positive
$\beta>\frac{2}{1-\eta}-1>0$. It is unknown if this $\beta$ is critical in
general, though criticality is known in the case of space-time white noise in
one spatial dimension [29]. Criticality would imply that there exist exponents
$\beta>0$ such that finite-time blow up is possible for additive noise. In
Mueller’s investigation of $\eqref{eq:SRDE}$ with
$\mathcal{A}=\Delta,f(u)=u^{-\beta}$, additive noise, and spatial dimension
$d=1$, it was shown that solutions couldn’t decrease to $0$ in finite time
(what we refer to as blow-up) when $\beta>3$ [25]. In Mueller’s case,
$\eta=\frac{1}{2}$ and $\gamma=0$, making his result equivalent to (3).
Mueller and Pardoux later showed that $\beta=3$ was critical: for any
$0<\beta<3$, there is a positive probability for the solution to hit $0$ in
finite time.

Another example where solutions can blow-up in finite time is the $1$
dimensional SDE

$dX(t)=X(t)^{-\beta}dt+X(t)^{-\gamma}dB_{t}$

It can be shown (see [5]) that finite-time blowup is prevented so long as

$\gamma+1<\frac{\beta+1}{2},$

For Brownian motion, $\eta=0$, so this condition aligns with our inequality
(3). It is currently unknown if (3) is the optimal condition for superlinear
forcing terms and colored noise, and this problem is left for future work.

In the study of $\eqref{eq:SRDE}$, it is known that stochastic forcing can
result in solutions growing arbitrarily large with positive probability.
Mueller and collaborators [24, 25, 26, 30] investigated the case where
$\mathcal{A}=\Delta,f(u)=0$, and the spatial dimension $d=1$. It was found
that solutions can exhibit finite-time blowup if $|\sigma(u)|>c|u|^{\gamma}$
for some $c>0$ and $\gamma>\frac{3}{2}$. It was also observed that explosion
is prevented if $|\sigma(u)|\leq C(1+|u|^{\gamma})$ for some $C>0$ and
$\gamma<\frac{3}{2}$. See [1, 8, 9, 16] for investigations of explosion by
other authors.

Early work showed that (1) has unique solutions if $f$ and $\sigma$ are
globally Lipschitz continuous with at most linear growth [11, 12, 13, 33, 20,
32, 35, 38, 39]. However, in the case of superlinearly growing $\sigma$, a
linearly growing $f$ is no longer sufficient to prevent explosive growth.
Cerrai demonstrated the existence of global solutions for when $\sigma$ is
locally Lipschitz continuous with linear growth and $f$ is strongly
dissipative with polynomial growth [4]. Assuming polynomial growth for $f$ is
common [2, 6, 19, 21], but Da Prato and Röckner [10] and Marinelli and Röckner
[23] proved that this polynomial growth restriction can be relaxed, such that
existence and uniqueness of solutions is implied by a monotonicity condition
on $f$. See also [7, 17, 22, 23]. Recently, it was shown that superlinearly-
growing dissipative forcing $f$ can prevent blow-up of solutions in the case
of superlinearly-growing $\sigma$ [34]. All of these references consider the
case where $f:\mathbb{R}\to\mathbb{R}$ and $u(t,x)$ can take any value in
$\mathbb{R}$.

Since we assume our constraining force $f$ is of the form
$f(w)=-\text{sign}(w)(1-|w|)^{-\beta}$ for some $\beta>0$, we naturally assume
$||u_{0}||_{L^{\infty}(D)}<1$. In the absence of stochastic forcing, $f$
restricts $u(t,x)$ to remain in $(-1,1)$. Since equation (1) breaks down
whenever $|u(t,x)|=1$, this motivates the following definition of explosion:
for each $n\in\mathbb{N}$, we define the stopping times

$T_{n}=\inf\left\\{t>0:||u(t)||_{L^{\infty}(D)}>1-\frac{1}{3^{n}}\right\\}$
(4)

Under Assumption 2, which can be found in Section 2, below, the realization
$A$ of $\mathcal{A}$ in $L^{2}(D)$ generates a $C_{0}$-semigroup $S(t)$.

######  Definition 1.1.

A $C_{0}(\overline{D})$-valued process $u(t)$ is a local mild solution to
$\eqref{eq:SRDE}$ if

$u(t)=S(t)u_{0}+\int_{0}^{t}S(t-s)f(u(s))ds+\int_{0}^{t}S(t-s)\sigma(u(s))dW(s)$
(5)

for all $t\in[0,T_{n}]$ for any $n$.

######  Definition 1.2.

A mild solution $u$ is global if $u(t)$ solves $\eqref{eq:MildSol}$ for all
$t>0$ with probability $1$.

Local mild solutions $u$ are continuous by definition, so $T_{n}\leq T_{n+1}$
for all $n\in\mathbb{N}$. Therefore, a local mild solution $u$ solves
$\eqref{eq:MildSol}$ on the time interval $[0,\sup_{n}T_{n})$ with probability
$1$. Thus, a local mild solution $u$ is a global solution if and only if
$\sup_{n}T_{n}=\infty$ with probability 1. We define the explosion time as
$\sup_{n}T_{n}$, as mild solutions to (1) do not make sense after
$\sup_{n}T_{n}$, and can now state our key result:

######  Theorem 1.1.

Let $\frac{1-\eta}{2}$ be the time Hölder continuity of our solution to (1),
with $\eta$ defined in Section 2. If Assumptions 1-4 are satisfied and
$\gamma+1<\frac{(1-\eta)(\beta+1)}{2}$, then any local mild solution is a
global solution.

In Section 2, we describe our notation, main assumptions, and main result. In
Section 3, we highlight some crucial estimates and prove Lemma
$\eqref{lem:Limgrowthlemma}$, which is central to limiting the growth of our
solutions. In Section 4, we prove our main theorem.

##  2 Assumptions and Main Result

Let $D\subset\mathbb{R}^{d}$ denote an open, bounded domain. For
$p\in[1,\infty)$, define $L^{p}(D)$ to be the Banach space of functions
$v:D\to\mathbb{R}$ such that the norm

$|v|_{L^{p}(D)}=\left(\int_{D}|v(x)|^{p}dx\right)^{\frac{1}{p}}$

is finite. When $p=\infty$, the $L^{\infty}(D)$ norm is

$|v|_{L^{\infty}(D)}=\sup_{x\in D}|v(x)|$

Define $C_{0}(\overline{D})$ to be the subset of $L^{\infty}(D)$ of continuous
functions $v:\overline{D}\to\mathbb{R}$ such that $v(x)=0$ for $x\in\partial
D$. Define $C_{0}([0,T]\times\overline{D})$ to be the set of continuous
functions $v:[0,T]\times\overline{D}\to\mathbb{R}$ such that $v(t,x)=0$ for
$x\in\partial D$, endowed with the supremum norm

$|v|_{C_{0}([0,T]\times\overline{D})}=\sup_{t\in[0,T]}\sup_{x\in\overline{D}}|v(t,x)|$

We make the following assumptions about the differential operator
$\mathcal{A}$, the noise $\dot{W}$, and the forcing terms $f$ and $\sigma$.

Assumption 1 The initial data $u_{0}\in C_{0}(\overline{D})$ and
$||u_{0}||_{\infty}<1$

Assumption 2 $\mathcal{A}$ is a self-adjoint second-order elliptic
differential operator

$\mathcal{A}\phi(x)=\sum_{i=1}^{d}\sum_{j=1}^{d}\frac{\partial}{\partial
x_{j}}\left(a_{ij}(x)\frac{\partial\phi}{\partial x_{j}}(x)\right)$

where $a_{ij}$ are continuously differentiable on $\overline{D}$, symmetric,
and elliptic. Let $A$ be the realization of $\mathcal{A}$ in $L^{2}(D)$ with
Dirichlet boundary conditions. Then there exists a sequence of eigenvalues
$0\leq\alpha_{1}\leq\alpha_{2}\leq\ldots$ and eigenfunctions $e_{k}\in
L^{2}(D)\cap C_{0}(D)$ such that

$Ae_{k}=-\alpha_{k}e_{k}\hskip 18.06749pt|e_{k}|_{L^{2}(D)}=1$

We believe our results hold for more general second-order elliptic
differential operators. However, for convenience, we assume self-adjoint
operators. Using Assumption 2 and estimates similar to those in Cerrai 2003
[4], we have an elliptic contraction semigroup $S(t)$ generated by this
operator $A$, where $S(t)$ acts on functions $u(t,x)$ through the spatial
integral

$(S(t)u(s))(x)=\int_{D}K(t,x,y)u(s,y)dy$ (6)

where $K(t,x,y)$ is the kernel of this contraction semigroup described by

$K(t,x,y)=\sum_{k=1}^{\infty}e^{-\alpha_{k}t}e_{k}(x)e_{k}(y)$ (7)

Assumption 3 (See [4], [5]) There exists a sequence of numbers
$\lambda_{j}\geq 0$ and a sequence of i.i.d one-dimensional Brownian motions
$\\{B_{j}(t)\\}_{j\in\mathbb{N}}$ such that, formally,

$\dot{W}(t,x)=\sum_{k=1}^{\infty}\lambda_{j}e_{j}(x)dB_{j}(t)$

Furthermore, there exist exponents $\theta>0$ and $\rho\in[2,\infty)$ such
that

$\displaystyle\begin{cases}\left(\sum_{j=1}^{\infty}\lambda_{j}^{\rho}|e_{j}|_{L^{\infty}(D)}^{2}\right)^{\frac{2}{\rho}}<\infty&\text{if
$\rho\in[2,\infty)$}\\\ \sup_{j}\lambda_{j}<\infty&\text{if
$\rho=\infty$}\end{cases}$
$\displaystyle\sum_{k=1}^{\infty}\alpha_{k}^{-\theta}|e_{k}|_{L^{\infty}(D)}^{2}<\infty$

and

$\eta\coloneqq\frac{\theta(\rho-2)}{\rho}<1$

This constant $\eta<1$ represents how the noise $\dot{W}$ and elliptic
operator $\mathcal{A}$ relate to each other, where $\eta<1$ represents the
smoothing of our elliptic operator being strong enough to counteract the
irregularity of the noise so that solutions are function-valued. A trace-class
noise $\eta=0$ corresponds to $\rho=2$. For space-time white noise on a one-
dimensional spatial interval, it is known that $\alpha_{k}\sim k^{2}$, so we
can take $\rho=\infty$ and $\eta=\theta$ can be any number larger than
$\frac{1}{2}$.

Assumption 4 The functions $f:(-1,1)\to\mathbb{R}$ and
$\sigma:(-1,1)\to\mathbb{R}$ are locally Lipschitz continuous functions. There
exist powers $\beta>0$, $\gamma\geq 0$ and constants $c_{0},C,K>0$ such that
for $|u|\in(c_{0},1)$

$\displaystyle|\sigma(u)|$ $\displaystyle\leq C(1-|u|)^{-\gamma}$
$\displaystyle f(u)\text{sign(u)}$ $\displaystyle\leq-K(1-|u|)^{-\beta}$

Our main goal is in showing that local mild solutions can be global solutions,
but this is only meaningful if local mild solutions exist. To show this, for
each $n\in\mathbb{N}$, define functions
$f_{n},\sigma_{n}:\mathbb{R}\to\mathbb{R}$ by

$\displaystyle f_{n}(w)$
$\displaystyle=\begin{cases}f(-1+\frac{1}{3^{n}})&\text{if
$w<-1+\frac{1}{3^{n}}$}\\\ f(w)&\text{if $|w|<1-\frac{1}{3^{n}}$}\\\
f(1-\frac{1}{3^{n}})&\text{if $w>1-\frac{1}{3^{n}}$}\end{cases}$
$\displaystyle\sigma_{n}(w)$
$\displaystyle=\begin{cases}\sigma(-1+\frac{1}{3^{n}})&\text{if
$w<-1+\frac{1}{3^{n}}$}\\\ \sigma(w)&\text{if $|w|<1-\frac{1}{3^{n}}$}\\\
\sigma(1-\frac{1}{3^{n}})&\text{if $w>1-\frac{1}{3^{n}}$}\end{cases}$

These functions $f_{n},\sigma_{n}$ are globally Lipschitz continuous
functions. Thus, the SRDE

$\begin{cases}\frac{\partial u_{n}}{\partial
t}(t,x)=\mathcal{A}u_{n}(t,x)+f_{n}(u_{n}(t,x))+\sigma_{n}(u_{n}(t,x))\dot{W}(t,x)\\\
u_{n}(t,x)=0,\;\;x\in\partial D,\hskip
18.06749ptu_{n}(0,x)=u_{0}(x)\end{cases}$ (8)

has a unique solution by the usual existence and uniqueness theorems [11].
Since $u_{n}$ solves (8), it solves (1) up to $T_{n}$ due to our definitions
of $f_{n},\sigma_{n}$. Since $u_{n}$ uniquely solves (8), it is a unique local
solution. Furthermore, by this uniqueness and our construction of
$f_{n},\sigma_{n}$, the solutions $u_{n}$ are consistent, where $u_{n}=u_{m}$
until $T_{n}$ whenever $n<m$.

Let us recall the definition of a mild solution:

######  Definition 2.1.

A $C_{0}(\overline{D})$-valued process $u(t)$ is a local mild solution to
$\eqref{eq:SRDE}$ if

$u(t)=S(t)u_{0}+\int_{0}^{t}S(t-s)f(u(s))ds+\int_{0}^{t}S(t-s)\sigma(u(s))dW(s)$

for all $t\in[0,T_{n}]$ for any $n$, where

$T_{n}=\inf\left\\{t>0:||u(t)||_{L^{\infty}(D)}>1-\frac{1}{3^{n}}\right\\}$

Under our assumptions, these stopping times are well-defined and we can state
our final result

######  Theorem 2.1.

Under Assumptions 1-4, there exists a unique, global solution to
$\eqref{eq:SRDE}$.

##  3 Estimates

Using the factorization method of Da Prato and Zabczyk [Chapter 5.3.1], see
also Cerrai (2003), the stochastic integral

$Z(t)=\int_{0}^{t}S(t-s)\sigma(u(s))dW(s),$ (9)

where $S(t)$ is the elliptic semigroup defined in (6), can be written as

$Z(t)=\frac{\sin(\pi\alpha)}{\pi}\int_{0}^{t}(t-s)^{\alpha-1}S(t-s)Z_{\alpha}(y)dy$
(10)

for $\alpha\in(0,1)$ and

$Z_{\alpha}(s)=\int_{0}^{t}(t-s)^{-\alpha}S(t-s)\sigma(u(s))dW(s)$ (11)

It is then possible to write $Z(t\wedge\tau)$, where $\tau$ is any stopping
time with respect to the natural filtration of $W(t)$, as

$\displaystyle Z(t\wedge\tau)$
$\displaystyle=\frac{\sin(\pi\alpha)}{\pi}\int_{0}^{t\wedge\tau}(t\wedge\tau-s)^{\alpha-1}S(t\wedge\tau-s)Z_{\alpha}(s)ds$
$\displaystyle=\frac{\sin(\pi\alpha)}{\pi}\int_{0}^{t\wedge\tau}(t\wedge\tau-s)^{\alpha-1}S(t\wedge\tau-s)\tilde{Z}_{\alpha}(s)ds$
(12)

where

$\tilde{Z}_{\alpha}(t)=\int_{0}^{t}(t-s)^{-\alpha}S(t-s)\sigma(u(s))1_{\\{s\leq\tau\\}}dW(s)$
(13)

We will make use of the following propositions

######  Proposition 1.

Let $\alpha\in\left(0,\frac{1-\eta}{2}\right)$ and $p\geq 2$. For any $t>0$,

$\mathbb{E}|\tilde{Z}_{\alpha}(t)|_{L^{p}(D)}^{p}\leq
C_{\alpha,p}\mathbb{E}\left(\int_{0}^{t}(t-s)^{-\eta-2\alpha}|\sigma(u(s))|_{L^{\infty}(D)}^{2}1_{\\{s\leq\tau\\}}ds\right)^{\frac{p}{2}}$
(14)

Proposition 1 and its proof can be found in [34] under Proposition (3.1), with
its proof in that paper’s Appendix.

######  Proposition 2.

For $p,\alpha,\beta$ such that $p(\alpha-\frac{1}{\beta+1})-\frac{d}{2}-1>0$,
we have

$\mathbb{E}\left(\sup_{t\in[0,T]}\sup_{x\in
D}\frac{|Z(t,x)|^{p}}{t^{\frac{p}{\beta+1}}}\right)\leq
C_{\alpha,p,d}T^{p(\alpha-\frac{1}{\beta+1})-\frac{d}{2}-1}\int_{0}^{T}\mathbb{E}|\tilde{Z}_{\alpha}(s)|_{L^{p}(D)}^{p}ds$
(15)

_Proof of Proposition 2_ Consider the factorized form of our stochastic
convolution,

$Z(t,x)=\frac{\sin(\pi\alpha)}{\pi}\int_{0}^{t}\int_{D}(t-s)^{\alpha-1}K(t-s,x,y)\tilde{Z}_{\alpha}(s,y)dyds$
(16)

where $K(t,x,y)$ is the kernel of the semigroup $S(t)$ defined in (7).
Applying Holder’s Inequality with $p$ and $\frac{p}{p-1}$ in both space and
time implies

$\displaystyle|Z(t,x)|$
$\displaystyle\leq\frac{\sin(\pi\alpha)}{\pi}\left(\int_{0}^{t}\int_{D}(t-s)^{\frac{(\alpha-1)p}{p-1}}|K(t-s,x,y)|^{\frac{p}{p-1}}dyds\right)^{\frac{p-1}{p}}$
$\displaystyle\hskip
36.135pt\cdot\left(\int_{0}^{t}\int_{D}|\tilde{Z}_{\alpha}(s,y)|^{p}dyds\right)^{\frac{1}{p}}$
$\displaystyle=\frac{\sin(\pi\alpha)}{\pi}\left(\int_{0}^{t}(t-s)^{\frac{(\alpha-1)p}{p-1}}||K(t-s,x,\cdot)||_{L^{\frac{p}{p-1}}(D)}^{\frac{p}{p-1}}ds\right)^{\frac{p-1}{p}}$
(17) $\displaystyle\hskip
36.135pt\cdot\left(\int_{0}^{t}\int_{D}|\tilde{Z}_{\alpha}(s,y)|^{p}dyds\right)^{\frac{1}{p}}$

Using Holder’s Inequality with the semigroup norm implies

$\displaystyle\leq\frac{\sin(\pi\alpha)}{\pi}\left(\int_{0}^{t}(t-s)^{\frac{(\alpha-1)p}{p-1}}||K(t-s,x,\cdot)||_{L^{1}(D)}||K(t-s,x,\cdot)||_{L^{\infty}(D)}^{\frac{1}{p-1}}ds\right)^{\frac{p-1}{p}}$
(18) $\displaystyle\hskip
36.135pt\cdot\left(\int_{0}^{t}\int_{D}|\tilde{Z}_{\alpha}(s,y)|^{p}dyds\right)^{\frac{1}{p}}$

Properties of the semigroup kernel norm imply for any $t>0$ and $x\in D$ that

$\displaystyle\leq\frac{\sin(\pi\alpha)}{\pi}\left(\int_{0}^{t}(t-s)^{\frac{(\alpha-1)p}{p-1}}\cdot
1\cdot(t-s)^{-\frac{d}{2(p-1)}}ds\right)^{\frac{p-1}{p}}$ $\displaystyle\hskip
36.135pt\cdot\left(\int_{0}^{t}\int_{D}|\tilde{Z}_{\alpha}(s,y)|^{p}dyds\right)^{\frac{1}{p}}$
$\displaystyle=\frac{\sin(\pi\alpha)}{\pi}\left(\int_{0}^{t}s^{\frac{(\alpha-1)p-\frac{d}{2}}{p-1}}ds\right)^{\frac{p-1}{p}}\cdot\left(\int_{0}^{t}\int_{D}|\tilde{Z}_{\alpha}(s,y)|^{p}dyds\right)^{\frac{1}{p}}$
$\displaystyle=C_{\alpha,p,d}t^{\frac{p\alpha-\frac{d}{2}-1}{p}}\left(\int_{0}^{t}\int_{D}|\tilde{Z}_{\alpha}(s,y)|^{p}dyds\right)^{\frac{1}{p}}$
$\displaystyle|Z(t,x)|^{p}$ $\displaystyle\leq
C_{\alpha,p,d}t^{p\alpha-\frac{d}{2}-1}\int_{0}^{t}|\tilde{Z}_{\alpha}(s)|_{L^{p}(D)}^{p}ds$
(19)

Since $\eqref{eq:PolynomialZBound}$ holds for any $t>0$ and $x\in D$, we can
divide both sides by $t^{\frac{p}{\beta+1}}$ to see that, for any $t>0$,

$\frac{|Z(t,x)|^{p}}{t^{\frac{p}{\beta+1}}}\leq
C_{\alpha,p,d}t^{p(\alpha-\frac{1}{\beta+1})-\frac{d}{2}-1}\int_{0}^{t}|\tilde{Z}_{\alpha}(s)|_{L^{p}(D)}^{p}ds$

Since supremums preserve inequalities, we can take expectations to get

$\displaystyle\mathbb{E}\left(\sup_{t\in[0,T]}\sup_{x\in
D}\frac{|Z(t,x)|^{p}}{t^{\frac{p}{\beta+1}}}\right)$ $\displaystyle\leq
C_{\alpha,p,d}\sup_{t\in[0,T]}\left(t^{p(\alpha-\frac{1}{\beta+1})-\frac{d}{2}-1}\int_{0}^{t}\mathbb{E}|\tilde{Z}_{\alpha}(s)|_{L^{p}(D)}^{p}ds\right)$
(20)
$\displaystyle=C_{\alpha,p,d}T^{p(\alpha-\frac{1}{\beta+1})-\frac{d}{2}-1}\int_{0}^{T}\mathbb{E}|\tilde{Z}_{\alpha}(s)|_{L^{p}(D)}^{p}ds$
(21)

so long as we choose $\beta,p$ such that
$p(\alpha-\frac{1}{\beta+1})>1+\frac{d}{2}$, causing the right-hand side to be
increasing in $t$. ∎

###  3.1 Uniform Bounds

To analyze the mild solution to the SPDE, we first demonstrate how the growth
of the mild solution can be controlled by a deterministic function. Consider
our mild solution to the integral equation

$u(t)=S(t)u_{0}+\int_{0}^{t}S(t-s)f(u(s))ds+Z(t)$ (22)

where we have suppressed the spatial variable for convenience. Here, $Z\in
C_{0}([0,T]\times\overline{D})$ is the continuous stochastic convolution
integral from $\eqref{eq:Zdef}$. For notational convenience, we also define

$\displaystyle e(t)\coloneqq 1-|u(t)|_{L^{\infty}(D)},$ (23)

so that $e(0)=1-|u_{0}|_{L^{\infty}(D)}$. Under this notation, explosion
occurs if $e(t)$ ever hits 0.

######  Lemma 3.1.

Let $T>0$ and assume that $Z,u\in C_{0}([0,T]\times D)$ solves
$\eqref{eq:SPDESolution}$ such that, for all $t\in[0,T]$,

$\displaystyle|Z(t)|_{L^{\infty}(D)}\leq\frac{1}{3}e(t)$ (24) $\displaystyle
e(t)\in\left(0,\frac{1}{3^{N}}\right]$ (25)

Then for all $t\in[0,T]$,

$e(t)\geq\frac{3}{4}\left(e(0)^{\beta+1}+K\left(\frac{2}{5}\right)^{\beta}(1+\beta)t\right)^{\frac{1}{\beta+1}}>K_{\beta}t^{\frac{1}{\beta+1}}$
(26)

where $K_{\beta}>0$ is a constant depending solely on $\beta$ and
$N\in\mathbb{N}$ is the minimal integer such that $\frac{2}{3^{N}}<1-c_{0}$,
where $c_{0}$ is from Assumption 4.

_Proof_. If we define $v(t,x)=u(t,x)-Z(t,x)$ for $u,Z\in C_{0}([0,T]\times
D)$, then $v$ is weakly differentiable and weakly solves the PDE

$\frac{\partial v}{\partial t}(t,x)=\mathcal{A}v(t,x)+f(v(t,x)+Z(t,x))$

By a Yosida approximation (via Proposition 6.2.2 of [3] or Theorem 7.7 of
[11]), we can assume that $v$ is a strong solution. By Proposition D.4 in the
appendix of [11], $t\mapsto|v(t)|_{L^{\infty}(D)}$ is left-differentiable such
that

$\frac{d^{-}}{dt}|v(t)|_{L^{\infty}(D)}\leq\mathcal{A}v(t,x_{t})\text{sign}(v(t,x_{t}))+f(v(t,x_{t})+Z(t,x_{t}))\text{sign}(v(t,x_{t}))$
(27)

where $x_{t}\in D$ is a maximizer satisfying

$|v(t)|_{L^{\infty}(D)}=|v(t,x_{t})|=v(t,x_{t})\text{sign}(v(t,x_{t}))$ (28)

The ellipticity of $\mathcal{A}$ and convexity at a maximizer implies

$\mathcal{A}v(t,x_{t})\text{sign}(v(t,x_{t}))\leq 0$

Thus we need only examine

$\displaystyle\frac{d^{-}}{dt}|v(t)|_{L^{\infty}(D)}\leq
f(v(t,x_{t})+Z(t,x_{t}))\text{sign}(v(t,x_{t}))$ (29)

Since $v(t)=u(t)-Z(t)$, using both versions of the triangle inequality
combined with $\eqref{eq:smallZ}$ implies

$\displaystyle|u(t)|_{L^{\infty}(D)}-|Z(t)|_{L^{\infty}(D)}\leq$
$\displaystyle|v(t)|_{L^{\infty}(D)}\leq|u(t)|_{L^{\infty}(D)}+|Z(t)|_{L^{\infty}(D)}$
$\displaystyle 1-|u(t)|_{L^{\infty}(D)}-|Z(t)|_{L^{\infty}(D)}\leq 1$
$\displaystyle-|v(t)|_{L^{\infty}(D)}\leq
1-|u(t)|_{L^{\infty}(D)}+|Z(t)|_{L^{\infty}(D)}$ $\displaystyle
e(t)-|Z(t)|_{L^{\infty}(D)}\leq 1$ $\displaystyle-|v(t)|_{L^{\infty}(D)}\leq
e(t)+|Z(t)|_{L^{\infty}(D)}$

By Assumption $\eqref{eq:smallZ}$, this implies

$\displaystyle 1-|v(t)|_{L^{\infty}(D)}$ $\displaystyle\leq\frac{4}{3}e(t)$
(30) $\displaystyle 1-|v(t)|_{L^{\infty}(D)}$
$\displaystyle\geq\frac{2}{3}e(t)$ (31)

Using the fact that $|Z(t,x_{t})|\leq|Z(t)|_{L^{\infty}(D)}$ by definition of
supremum, and $|v(t,x_{t})|=|v(t)|_{L^{\infty}(D)}$, we similarly find that

$\displaystyle|v(t)|_{L^{\infty}(D)}-|Z(t)|_{L^{\infty}(D)}\leq$
$\displaystyle|v(t,x_{t})+Z(t,x_{t})|\leq|v(t)|_{L^{\infty}(D)}+|Z(t)|_{L^{\infty}(D)}$
$\displaystyle 1-|v(t)|_{L^{\infty}(D)}-|Z(t)|_{L^{\infty}(D)}\leq 1$
$\displaystyle-|v(t,x_{t})+Z(t,x_{t})|\leq
1-|v(t)|_{L^{\infty}(D)}+|Z(t)|_{L^{\infty}(D)}$ $\displaystyle
1-|v(t)|_{L^{\infty}(D)}-\frac{1}{3}e(t)\leq 1$
$\displaystyle-|v(t,x_{t})+Z(t,x_{t})|\leq
1-|v(t)|_{L^{\infty}(D)}+\frac{1}{3}e(t)$

Using inequalities $\eqref{eq:VUpperBound}$ and $\eqref{eq:VLowerBound}$ on
the last series of inequalities implies

$\displaystyle 1-|v(t,x_{t})+Z(t,x_{t})|$ $\displaystyle\leq\frac{5}{3}e(t)$
(32) $\displaystyle 1-|v(t,x_{t})+Z(t,x_{t})|$
$\displaystyle\geq\frac{1}{3}e(t)$ (33)

Since $0<e(t)\leq\frac{1}{3^{N}}$ by $\eqref{eq:largeu}$, inequality
$\eqref{eq:VUpperBound}$ implies
$1-|v(t)|_{L^{\infty}(D)}\leq\frac{4}{3^{N+1}}$ and $\eqref{eq:VLowerBound}$
implies $1-|v(t)|_{L^{\infty}(D)}>0$. Furthermore,

$|v(t)|_{L^{\infty}(D)}>c_{0}\iff 1-|v(t)|_{L^{\infty}(D)}<1-c_{0}$

Since $\frac{2}{3^{N}}<1-c_{0}$, this implies

$1-|v(t)|_{L^{\infty}(D)}\leq\frac{4}{3^{N+1}}=\frac{2}{3}\frac{2}{3^{N}}<\frac{2}{3}(1-c_{0})$

Thus, $|v(t)|_{L^{\infty}(D)}\in(c_{0},1)$ for $t\in[0,T]$. A similar argument
implies $|v(t,x_{t})+z(t,x_{t})|\in(c_{0},1)$ for $t\in[0,T]$. Furthermore,

$\text{sign}(v(t,x_{t}))=\text{sign}(v(t,x_{t})+z(t,x_{t})),$

as $|z(t)|_{L^{\infty}(D)}$ is sufficiently small relative to
$|u(t)|_{L^{\infty}(D)}$ and $|v(t)|_{L^{\infty}(D)}$. By
$\eqref{eq:VLowerBound}$ and $\eqref{eq:vzUpperBound}$, we get

$\displaystyle
1-|v(t,x_{t})+z(t,x_{t})|\leq\frac{5}{2}(1-|v(t)|_{L^{\infty}(D)})$ (34)

Applying $\eqref{eq:vzMainBound}$ to $\eqref{eq:vbound}$ implies

$\displaystyle\frac{d^{-}}{dt}|v(t)|_{L^{\infty}(D)}\leq-K\left(\frac{2}{5}\right)^{\beta}(1-|v(t)|_{L^{\infty}(D)})^{-\beta}$
(35)

Letting $F(x)=\frac{1}{1+\beta}(1-x)^{\beta+1}$ and
$F^{\prime}(x)=-(1-x)^{\beta}$, this implies

$\frac{d^{-}}{dt}F(|v(t)|_{L^{\infty}(D)})\geq
K\left(\frac{2}{5}\right)^{\beta}$ (36)

Therefore,

$F(|v(t)|_{L^{\infty}(D)})\geq
F(|u_{0}|_{L^{\infty}(D)})+K\left(\frac{2}{5}\right)^{\beta}t$ (37)

and

$|v(t)|_{L^{\infty}(D)}\leq
1-\left(e(0)^{\beta+1}+K\left(\frac{2}{5}\right)^{\beta}(1+\beta)t\right)^{\frac{1}{\beta+1}}$
(38)

with $\eqref{eq:BoundOnV}$ implying
$|v(t)|_{L^{\infty}(D)}\leq|u_{0}|_{L^{\infty}(D)}$ for $t\in[0,T]$. Combining
this result with $\eqref{eq:VUpperBound}$ yields our conclusion (26):

$e(t)\geq\frac{3}{4}\left(e(0)^{\beta+1}+K\left(\frac{2}{5}\right)^{\beta}(1+\beta)t\right)^{\frac{1}{\beta+1}}$

∎

Inequality $\eqref{eq:Limgrowtheq}$ implies $|u(t)|_{L^{\infty}(D)}\in(0,1)$
for $t\in[0,T]$. In addition, if we assume
$1-|u_{0}|_{L^{\infty}(D)}=\frac{1}{3^{n}}$ for some $n\in\mathbb{N}$, then

$\displaystyle 1-|u(t)|_{L^{\infty}(D)}\geq\frac{1}{4\cdot
3^{n-1}}>\frac{1}{3^{n+1}}$ (39)

Thus, $e(t)>\frac{1}{3^{n+1}}$ for $t\in[0,T]$, meaning the distance between
$1$ and $|u(t)|_{L^{\infty}(D)}$ can’t decrease below $\frac{1}{3^{n+1}}$ so
long as conditions $\eqref{eq:smallZ}$ and $\eqref{eq:largeu}$ hold.

##  4 Proof of Theorem 1.1/ Main Result

To establish theorem 1.1, we first define a sequence of stopping times.
Recalling $e(t)=1-|u(t)|_{L^{\infty}(D)}$, we define

$\displaystyle\tau_{0}=\text{inf}\\{t\geq 0:e(t)\leq\frac{1}{3^{n}}\;\text{for
some $n\in\\{N,N+1,N+2,\ldots\\}$ }\\}$
$\displaystyle\tau_{k+1}=\begin{cases}\text{inf}\\{t\geq\tau_{k}:e(t)\leq\frac{1}{3^{N+1}}\\}&\text{if
$e(\tau_{k})=\frac{1}{3^{N}}$}\\\
\text{inf}\\{t\geq\tau_{k}:e(t)\leq\frac{1}{3}e(\tau_{k})\\\ \hskip
36.135pt\text{or}\;e(t)\geq 3e(\tau_{k})\\}&\text{if
$e(\tau_{k})\leq\frac{1}{3^{N+1}}$}\end{cases}$

Here, $N$ is the constant defined in Lemma 3.1. While these $\tau_{k}$ aren’t
the same as the stopping times $T_{k}$ defined in (4), it is clear that
$\sup_{k}\tau_{k}<\infty$ if and only if $\sup_{k}T_{k}<\infty$.

######  Lemma 4.1.

There exist constants $C>0$ and $q>1$, independent of $n,k,$ and $\epsilon>0$,
such that for any $\epsilon>0$, any $k\in\mathbb{N}$, and any
$n\in\\{N+1,N+2,\ldots\\}$,

$\mathbb{P}\left(e(\tau_{k+1})=\frac{1}{3}e(\tau_{k})\;\text{and}\;\tau_{k+1}-\tau_{k}<\epsilon\;\middle|\;e(\tau_{k})=\frac{1}{3^{n}}\right)\leq
C\epsilon^{q}$

where $N$ is the constant defined in Lemma 3.1.

_Proof_ As discussed in $\bf{Lemma\;4.1}$ from [34], we may assume $k=0$, so
$\tau_{0}=0$, without loss of generality by the Markov property. Therefore,

$\displaystyle\mathbb{P}\left(e(\tau_{k+1})=\frac{1}{3}e(\tau_{k})\;\text{and}\;\tau_{k+1}-\tau_{k}<\epsilon\;\middle|\;e(\tau_{k})=\frac{1}{3^{n}}\right)$
$\displaystyle=\mathbb{P}\left(e(\tau_{1})=\frac{1}{3}e(0)\;\text{and}\;\tau_{1}<\epsilon\;\middle|\;e(0)=\frac{1}{3^{n}}\right)$

Since $n\geq N+1$, by the definition of $\tau_{1}$, if $e(0)=\frac{1}{3^{n}}$,
then $e(t)\in(\frac{1}{3^{n+1}},\frac{1}{3^{n-1}})$ on $[0,\tau_{1})$,
equalling one of the endpoints at $t=\tau_{1}$. Since
$\frac{1}{3^{n-1}}\leq\frac{1}{3^{N}}$, assumption $\eqref{eq:largeu}$ is
satisfied on the interval $t\in[0,\tau_{1}]$.

Let $Z(t)$ be the stochastic integral in (22). We introduce the additional
stopping time

$\tilde{\tau}=\text{inf}\\{t>0:|Z(t)|_{L^{\infty}(D)}>\frac{1}{3}e(t)\\}$ (41)

By definition, $|Z(t)|_{L^{\infty}(D)}\leq\frac{1}{3}e(t)$ for
$t\in[0,\tilde{\tau}]$. Lemma 3.1 is satisfied up to $\tilde{\tau}$, therefore
if $e(0)=\frac{1}{3^{n}}$, then $e(t)>\frac{1}{3^{n+1}}$ on
$[0,\tilde{\tau}\wedge\tau_{1}]$. Thus, in the event
$e(\tau_{1})=\frac{1}{3}e(0)$, this implies that $\tilde{\tau}\leq\tau_{1}$.
If we were to assume $\tau_{1}<\tilde{\tau}$ and
$e(\tau_{1})=\frac{1}{3}e(0)$, then $e(t)$ decreases to $\frac{1}{3^{n+1}}$
within the time interval $[0,\tilde{\tau}]$, which is impossible by Lemma 3.1.

Consequently,

$\displaystyle\mathbb{P}\left(e(\tau_{1})=\frac{1}{3}e(0)\;\text{and}\;\tau_{1}<\epsilon\;\middle|\;e(0)=\frac{1}{3^{n}}\right)$
$\displaystyle\leq\mathbb{P}\left(\tilde{\tau}\leq\tau_{1}<\epsilon\;|\;e(0)=\frac{1}{3^{n}}\right)$
$\displaystyle\leq\mathbb{P}\left(\sup_{t\in[0,\tilde{\tau}\wedge\tau_{1}\wedge\epsilon]}\frac{|Z(t)|_{L^{\infty}(D)}}{e(t)}\geq\frac{1}{3}\;\middle|\;e(0)=\frac{1}{3^{n}}\right)$
$\displaystyle\leq\mathbb{P}\left(\sup_{t\in[0,\tilde{\tau}\wedge\tau_{1}\wedge\epsilon]}\frac{|Z(t)|_{L^{\infty}(D)}}{K_{\beta}t^{\frac{1}{\beta+1}}}\geq\frac{1}{3}\;\middle|\;e(0)=\frac{1}{3^{n}}\right),$
(42)

where $\eqref{eq:preChebyshev}$ uses the fact that $e(t)\geq
K_{\beta}t^{\frac{1}{\beta+1}}$ on
$[0,\tilde{\tau}\wedge\tau_{1}\wedge\epsilon]$ by Lemma 3.1. Chebyshev’s
Inequality yields

$\leq\frac{\mathbb{E}\left(\sup_{t\in[0,\tilde{\tau}\wedge\tau_{1}\wedge\epsilon]}\frac{|Z(t)|_{L^{\infty}(D)}^{p}}{t^{\frac{p}{\beta+1}}}\right)}{K_{\beta}^{p}\frac{1}{3^{p}}}$
(43)

Using $\eqref{eq:ModifiedZMomentBound}$, by Assumption 4, the above is bounded
by

$\leq
K_{\beta,p}C_{\alpha,p,d}\epsilon^{p(\alpha-\frac{1}{\beta+1})-\frac{d}{2}-1}\int_{0}^{T}\mathbb{E}|\tilde{Z}_{\alpha}(s)|_{L^{p}(D)}^{p}ds$
(44)

From Lemma 3.1, on the time interval
$t\in[0,\epsilon\wedge\tilde{\tau}\wedge\tau_{1}]$

$\displaystyle\mathbb{E}|\tilde{Z}_{\alpha}(t)|_{L^{p}(D)}^{p}$
$\displaystyle\leq
C\mathbb{E}\left(\int_{0}^{t}(t-s)^{-2\alpha-\eta}|\sigma(u(s))|_{L^{\infty}(D)}^{2}1_{\\{s\leq\tau\\}}ds\right)^{\frac{p}{2}}$
$\displaystyle\leq
C\mathbb{E}\left(\int_{0}^{t}(t-s)^{-2\alpha-\eta}(1-|u(s)|_{L^{\infty}(D)})^{-2\gamma}1_{\\{s\leq\tau\\}}ds\right)^{\frac{p}{2}}$
$\displaystyle\leq
C_{\gamma,\beta}\left(\int_{0}^{t}(t-s)^{-2\alpha-\eta}s^{-\frac{2\gamma}{\beta+1}}ds\right)^{\frac{p}{2}}$
(45)

We require $\beta>0$ to satisfy

$\eta+\frac{2\gamma}{\beta+1}<1\iff\gamma<\frac{(1-\eta)(\beta+1)}{2},$ (46)

which is guaranteed by Assumption 4. Choose
$2\alpha=1-\eta-\frac{2\gamma}{\beta+1}$, so the properties of the Beta
Integral ensure that $\eqref{eq:ZTildeModifiedBound}$ equals a constant

$\left(\frac{\pi}{\sin(\pi(2\alpha+\eta))}\right)^{\frac{p}{2}}$ (47)

Therefore, $\mathbb{E}|\tilde{Z}_{\alpha}(t)|_{L^{p}(D)}^{p}\leq
C_{\alpha,\gamma,\eta,p,\beta}$, a constant independent of $\epsilon,n,k$.
This result gives an upper bound for $\eqref{eq: SmallProbability}$:

$\leq
C_{\alpha,p,d,\gamma,\eta}\epsilon^{p(\frac{1-\eta}{2}-\frac{\gamma+1}{\beta+1})-\frac{d}{2}}=\tilde{C}\epsilon^{q}$

by defining $q=p(\frac{1-\eta}{2}-\frac{\gamma+1}{\beta+1})-\frac{d}{2}$.
Assumption $4$ guarantees that

$\frac{1-\eta}{2}-\frac{\gamma+1}{\beta+1}>0$

Choose $p$ sufficiently large to ensure $q>1$. Thus,

$\displaystyle\mathbb{P}\left(e(\tau_{k+1})=\frac{1}{3}e(\tau_{k})\;\text{and}\;\tau_{k+1}-\tau_{k}<\epsilon\;\middle|\;e(\tau_{k})=\frac{1}{3^{n}}\right)\leq
C\epsilon^{q}$ (48)

where $C>0,q>1$ independent of $\epsilon,k,n$ and $\epsilon$ independent of
$k,n$, proving Lemma 4.1. ∎

Now we prove Theorem 1.1.

_Proof of Theorem 1.1_ Using the stopping times $\tau_{k}$ defined above and
Lemma $\eqref{lem:BoundedProbabilityLemma}$, it is known that there exists
$C>0$ and $q>1$ such that for any $k\in\mathbb{N}$ and small $\epsilon>0$

$\mathbb{P}\left(e(\tau_{k+1})=\frac{1}{3}e(\tau_{k})\;\text{and}\;\tau_{k+1}-\tau_{k}<\epsilon\right)\leq
C\epsilon^{q}$

If we pick $\epsilon=\frac{1}{k}$ for sufficiently-large $k$, then

$\mathbb{P}\left(e(\tau_{k+1})=\frac{1}{3}e(\tau_{k})\;\text{and}\;\tau_{k+1}-\tau_{k}<\frac{1}{k}\right)\leq
C\frac{1}{k^{q}}$

Since $q>1$,

$\sum_{k=1}^{\infty}\mathbb{P}\left(e(\tau_{k+1})=\frac{1}{3}e(\tau_{k})\;\text{and}\;\tau_{k+1}-\tau_{k}<\epsilon\right)<\infty$

Thus, by the Borel-Cantelli Lemma, with probability one there exists a random
index $N_{0}(\omega)>0$ such that for all $k\geq N_{0}(\omega)$, either

$\tau_{k+1}-\tau_{k}\geq\frac{1}{k}\hskip 10.84006pt\text{or}\hskip
10.84006pte(\tau_{k+1})=3e(\tau_{k})$ (49)

It is possible that $\tau_{K}=\infty$ for some $K\in\mathbb{N}$, which would
imply $\tau_{k}=\infty$ for $k\geq K$ by our definition (4) of $\tau_{k}$.
This would imply that $e(\tau_{k+1})=\frac{1}{3}e(\tau_{k})$ finitely-many
times, making it impossible for $||u(t)||_{L^{\infty}(D)}$ to hit $1$.
Clearly, this implies mild solutions are global. If
$e(\tau_{k+1})=3e(\tau_{k})$ for all $k\geq N_{0}(\omega)$, then
$||u(t)||_{L^{\infty}(D)}$ is distancing from $1$, again implying mild
solutions are global. Thus, we are left with examining the case when
$\tau_{k+1}-\tau_{k}\geq\frac{1}{k}$ for finite $\tau_{k}$ and
$e(\tau_{k+1})=\frac{1}{3}e(\tau_{k})$ occurring infinitely-many times.

From the definition of our $\tau_{k}$, $\max_{k\geq N_{0}(\omega)}e(\tau_{k})$
is attained. Thus, we may choose $N_{1}(\omega)>N_{0}(\omega)$ such that for
$k\geq N_{1}(\omega)$, we have $e(\tau_{k})\leq e(\tau_{N_{1}(\omega)})$.

Now, for any $m\geq N_{1}(\omega)>N_{0}(\omega)$, we have

$\sum_{k=N_{1}(\omega)}^{m}(\tau_{k+1}-\tau_{k})\geq\sum_{k=N_{1}(\omega)}^{m}\frac{1}{k}1_{\\{e(\tau_{k+1})=\frac{1}{3}e(\tau_{k})\\}}$
(50)

By the definition of our $N_{1}$, there must be more steps where $e(\tau_{k})$
decreases than steps where it increases, such that for $m\geq N_{1}(\omega)$:

$U_{m}(\omega)=\sum_{k=N_{1}(\omega)}^{m-1}1_{\\{e(\tau_{k+1})=\frac{1}{3}e(\tau_{k})\\}}\geq\frac{m-N_{1}(\omega)}{2}$
(51)

At this point, our argument follows the rest of [34], starting from equation
(4.31). Using the summation by parts formula and (51), we find

$\displaystyle\sum_{k=N_{1}(\omega)}^{m}(\tau_{k+1}-\tau_{k})$
$\displaystyle\geq\sum_{k=N_{1}(\omega)}^{m}\frac{1}{k}1_{\\{e(\tau_{k+1})=\frac{1}{3}e(\tau_{k})\\}}$
$\displaystyle=\sum_{k=N_{1}(\omega)}^{m}\frac{1}{k}(U_{k+1}(\omega)-U_{k}(\omega))$
$\displaystyle=\frac{U_{m+1}(\omega)}{m}-\sum_{k=N_{1}(\omega)+1}^{m}U_{k}(\omega)\left(\frac{1}{k}-\frac{1}{k-1}\right)$
$\displaystyle=\frac{U_{m+1}(\omega)}{m}+\sum_{k=N_{1}(\omega)+1}^{m}U_{k}(\omega)\left(\frac{1}{k(k-1)}\right)$
$\displaystyle\geq\frac{m-N_{1}(\omega)}{2m}+\sum_{k=N_{1}(\omega)+1}^{m}\left(\frac{k-N_{1}(\omega)}{2k(k-1)}\right)$

As $m$ tends to infinity, the sum diverges. Therefore, with probability one,

$\sum_{k=N_{1}(\omega)}^{\infty}(\tau_{k+1}-\tau_{k})=\infty$

and solutions cannot explode in finite time. ∎

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