1 Introduction
In this paper, we consider the stochastic heat equation with the Dirichlet boundary condition, which can formally be written as
where $(π‘,π₯)\in [0,π]\times [-πΏ,πΏ]$, and π is a stochastic measure defined on $\mathcal{B}(\mathrm{R})$ (Borel π-algebra in R). For π we assume only π-additivity in probability, assumptions for $π,\hspace{2.5pt}π$ and ${π’_{0}}$ are given in Section 3. We study the solution to the formal equation (1.1) in the mild form (see (3.1) below).
(1.1)
\[\begin{array}{c}\displaystyle \frac{π{π’^{πΏ}}}{ππ‘}\displaystyle =\frac{{π^{2}}{π’^{πΏ}}}{π{π₯^{2}}}\displaystyle +π\displaystyle \Bigl(π‘\displaystyle ,π₯\displaystyle ,{π’^{πΏ}}\displaystyle (π‘\displaystyle ,π₯\displaystyle )\displaystyle \Bigr)\displaystyle +π\displaystyle (π‘\displaystyle ,π₯\displaystyle )\frac{ππ}{ππ₯}\displaystyle ,\\ {} \displaystyle {π’^{πΏ}}\displaystyle (π‘\displaystyle ,\displaystyle -πΏ\displaystyle )\displaystyle ={π’^{πΏ}}\displaystyle (π‘\displaystyle ,πΏ\displaystyle )\displaystyle =0\displaystyle ,\hspace{10.0pt}{π’^{πΏ}}\displaystyle (0\displaystyle ,π₯\displaystyle )\displaystyle ={π’_{0}}\displaystyle (π₯\displaystyle )\displaystyle ,\end{array}\]We study the convergence as $πΏ\to \mathrm{\infty }$, and prove that ${π’^{πΏ}}$ tends to the solution to the equation
where $(π‘,π₯)\in [0,π]\times \mathrm{R}$. The solution to equation (1.2) is also considered in the mild form (see (3.2) below).
(1.2)
\[\frac{ππ’}{ππ‘}=\frac{{π^{2}}π’}{π{π₯^{2}}}+π\Bigl(π‘,π₯,π’(π‘,π₯)\Bigr)+π(π‘,π₯)\frac{ππ}{ππ₯},\hspace{10.0pt}π’(0,π₯)={π’_{0}}(π₯),\]As the main result of the paper, we obtain an exponential rate for the pointwise convergence ${π’^{πΏ}}\to π’$. This is a natural result that follows from the form of the Greenβs functions. Zero Dirichlet conditions represent absorbing boundaries whose effect decays exponentially as the domain expands.
For an equation driven by space-time white noise, this problem was studied in [2, 3], where the Neumann and mixed boundary conditions were also considered.
The difference of the form $|{π’^{πΏ}}-π’|$ is often called a localization error and, in different settings, has been studied in [1, 6]. The impact of localization on some computational problems in financial mathematics is considered in [8].
Theory of equations driven by stochastic measures is developed in [15]. Existence and uniqueness of mild solutions to (1.1) and (1.2), their boundedness and HΓΆlder regularity, are proved in [18] and [17] respectively. Also, generalization of equation (1.2) is considered in [16], where the ${\mathrm{L}_{2}}$-continuity of the solution was proved. For the heat equation on a bounded domain in ${\mathrm{R}^{π}}$, we can refer to [10].
Different classes of equations driven by space-time white noise are considered in detail in [4], those driven by spatially colored Gaussian noise in [5], and those driven by a LΓ©vy process in [12].
The rest of the paper is organized as follows. Section 2 contains the basic information concerning the integrals with respect to stochastic measures. In Section 3, we formulate the problem and our main assumptions. In Section 4 we obtain an estimate for the difference of the stochastic integrals from our equations. Section 5 contains the main result of the paper.
2 Preliminaries
Let ${\mathrm{L}_{0}}={\mathrm{L}_{0}}(\Omega ,\mathcal{F},\mathrm{P})$ be the set of all real-valued random variables defined on the complete probability space $(\Omega ,\mathcal{F},\mathrm{P})$. Convergence in ${\mathrm{L}_{0}}$ means the convergence in probability. Let X be an arbitrary set and $\mathcal{B}$ a π-algebra of subsets of X.
We do not assume the moment existence or martingale properties for SM. We can say that π is an ${\mathrm{L}_{0}}$-valued measure.
Important examples of SMs are random orthogonal measure with finite structural function (see [7, Section 5.3]), πΌ-stable random measures defined on a π-algebra for $πΌ\in (0,1)\cup (1,2]$ (see [19, Chapter 3]). Many examples of the SMs on $\mathcal{B}(\mathrm{R})$ may be given by the Wiener-type integral
where π is a deterministic function, integrable on R. For instance, we can take in (2.1) any square integrable martingale ${π_{π₯}}$, or ${π_{π₯}}={π_{π₯}^{π»}}$ β the fractional Brownian motion with Hurst index $π»\gt 1/2$ (see Corollary 1.9.4 in [11]).
(2.1)
\[π(π΄)={\int _{\mathrm{R}}}{\mathbf{1}_{π΄}}(π₯)π(π₯)\hspace{1.67pt}π{π_{π₯}},\]Other examples of SMs can be found in Section 1.2.1 [15].
For deterministic measurable functions $π:\mathrm{X}\to \mathrm{R}$, an integral of the form ${\int _{\mathrm{X}}}π\hspace{1.67pt}ππ$ is defined and studied in [9, Chapter 7], [15, Chapter 1]. In particular, every bounded measurable π is integrable with respect to any π.
Below, we will use the following statement.
Lemma 1 (Lemma 3.1 [14], Lemma 2.3 [15]).
Let π be an SM on $(\mathrm{X},\mathcal{B})$, ${π_{π}}:\hspace{2.5pt}\mathrm{X}\to \mathrm{R},\hspace{2.5pt}π\ge 1$, be measurable functions such that $\hat{π}(π₯)={\sum _{π=1}^{\infty }}|{π_{π}}(π₯)|$ is integrable with respect to π. Then
For any $π\in \mathrm{Z}$ and all $π\ge 0$, put
The following lemma is a key tool for estimating a version of the stochastic integral. In the sequel, we assume that SM π is defined on Borel subsets of R.
Lemma 2 (Lemma 2.8 [15]).
Let π be an arbitrary set, and the function $π(π§,π₯):π\times [π,π+1]\to \mathrm{R}$ be such that all paths $π(π§,\cdot )$ are continuous on $[π,π+1]$, $\mathrm{A}\subset (π,π+1]$ be a Borel set. Denote
such that for all $π½\gt 0$, $π\in \Omega $, $π§\in π$
\[{π_{π}}(π§,π₯)=\sum \limits_{1\le π\le {2^{π}}}π\Bigl(π§,{π_{(π-1)π}^{(π)}}\Bigr){\mathbf{1}_{{\Delta _{ππ}^{(π)}}}}(π₯).\]
Then the random function
\[π(π§)={\int _{\mathrm{A}}}π(π§,π₯)\hspace{1.67pt}ππ(π₯),\hspace{10.0pt}π§\in π,\]
has a version
(2.2)
\[\tilde{π}(π§)={\int _{A}}{π_{0}}(π§,π₯)\hspace{1.67pt}ππ(π₯)+\sum \limits_{π\ge 1}\Biggl({\int _{A}}{π_{π}}(π§,π₯)\hspace{1.67pt}ππ(π₯)-{\int _{A}}{π_{π-1}}(π§,π₯)\hspace{1.67pt}ππ(π₯)\Biggr)\](2.3)
\[\begin{array}{r@{\hskip0pt}l@{\hskip10pt}r}\displaystyle \Bigl|\tilde{π}(π§)\Bigr|& \displaystyle \le \Bigl|π(π§,π)π(\mathrm{A})\Bigr|& \displaystyle \hspace{-10.0pt}\\ {} & \displaystyle \hspace{10.0pt}+{\Biggl\{\sum \limits_{π\ge 1}{2^{ππ½}}\sum \limits_{1\le π\le {2^{π}}}{\Bigl|π\Bigl(π§,{π_{ππ}^{(π)}}\Bigr)-π\Bigl(π§,{π_{(π-1)π}^{(π)}}\Bigr)\Bigr|^{2}}\Biggr\}^{1/2}}& \\ {} & \displaystyle \hspace{10.0pt}\times {\Biggl\{\sum \limits_{π\ge 1}{2^{-ππ½}}\sum \limits_{1\le π\le {2^{π}}}{\Bigl|π\Bigl({\Delta _{ππ}^{(π)}}\cap \mathrm{A}\Bigr)\Bigr|^{2}}\Biggr\}^{1/2}}.& \end{array}\]In the sequel, by πΆ and $πΆ(π)$ we denote the constants and random constants, respectively, whose exact values are not important. Notations of the kind ${πΆ_{π}}$, ${πΆ_{πΏ}}$ means that these constants may depend of π, πΏ respectively.
3 The problem
We consider (1.1) in the mild form
where equality holds a.s. for each $(π‘,π₯)\in [0,π]\times [-πΏ,πΏ]$. Here ${πΊ^{πΏ}}(π‘,π₯,π¦)$ is the fundamental solution of the heat equation with the Dirichlet boundary condition
(3.1)
\[\begin{array}{r@{\hskip0pt}l@{\hskip10pt}r}\displaystyle {π’^{πΏ}}(π‘,π₯)& \displaystyle ={\int _{(-πΏ,πΏ]}}{πΊ^{πΏ}}(π‘,π₯,π¦){π’_{0}}(π¦)\hspace{1.67pt}ππ¦& \displaystyle \hspace{-10.0pt}\\ {} & \displaystyle \hspace{10.0pt}+{\int _{0}^{π‘}}{\int _{(-πΏ,πΏ]}}{πΊ^{πΏ}}(π‘-π ,π₯,π¦)π\Bigl(π ,π¦,{π’^{πΏ}}(π ,π¦)\Bigr)\hspace{1.67pt}ππ¦\hspace{1.67pt}ππ & \\ {} & \displaystyle \hspace{10.0pt}+{\int _{(-πΏ,πΏ]}}{\int _{0}^{π‘}}{πΊ^{πΏ}}(π‘-π ,π₯,π¦)π(π ,π¦)\hspace{1.67pt}ππ \hspace{1.67pt}ππ(π¦),& \end{array}\]
\[\frac{ππ’}{ππ‘}=\displaystyle \frac{{π^{2}}π’}{π{π₯^{2}}},\hspace{10.0pt}(π‘,π₯)\in [0,π]\times [-πΏ,πΏ],\hspace{10.0pt}π’(π‘,-πΏ)=π’(π‘,πΏ)=0.\]
It is known that
\[{πΊ^{πΏ}}(π‘,π₯,π¦)={\sum \limits_{π=-\mathrm{\infty }}^{\mathrm{\infty }}}\Bigl(π(π‘,π₯-π¦+4ππΏ)-π\Bigl(π‘,π₯+π¦+(4π+2)πΏ\Bigr)\Bigr),\]
where
\[π(π‘,π₯)=\displaystyle \frac{1}{\sqrt{4ππ‘}}\exp \Biggl\{-\displaystyle \frac{{π₯^{2}}}{4π‘}\Biggr\}\]
is a standard heat kernel (see, for example, [13, Section 8.4.2], [4, Lemma 1.4.1]).We will refer to the following assumptions on elements of (3.1).
AssumptionΒ A3.
$π(π ,π¦,π£)$ is globally Lipschitz in $π£,π¦$:
\[\Bigl|π(π ,{π¦_{1}},{π£_{1}})-π(π ,{π¦_{2}},{π£_{2}})\Bigr|\le {πΏ_{π}}\Bigl(|{π£_{1}}-{π£_{2}}|+|{π¦_{1}}-{π¦_{2}}|\Bigr),\hspace{10.0pt}{π¦_{π}},\hspace{2.5pt}{π£_{π}}\in \mathrm{R},\]
for some constant ${πΏ_{π}}$.AssumptionΒ A4.
$π(π ,π¦):[0,π]\times \mathrm{R}\to \mathrm{R}$ is measurable, and
\[\Bigl|π(π ,π¦)\Bigr|\le {πΆ_{π}},\hspace{10.0pt}\Bigl|π(π ,{π¦_{1}})-π(π ,{π¦_{2}})\Bigr|\le {πΏ_{π}}{|{π¦_{1}}-{π¦_{2}}|^{π½(π)}}\]
for some $1/2\lt π½(π)\lt 1$ and constants ${πΆ_{π}},\hspace{2.5pt}{πΏ_{π}}$.Also, we consider (1.2) in the mild form
(3.2)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle π’(π‘,π₯)& \displaystyle ={\int _{\mathrm{R}}}π(π‘,π₯-π¦){π’_{0}}(π¦)\hspace{1.67pt}ππ¦+{\int _{0}^{π‘}}{\int _{\mathrm{R}}}π(π‘-π ,π₯-π¦)π\Bigl(π ,π¦,π’(π ,π¦)\Bigr)\hspace{1.67pt}ππ¦\hspace{1.67pt}ππ \\ {} & \displaystyle \hspace{10.0pt}+{\int _{\mathrm{R}}}{\int _{0}^{π‘}}π(π‘-π ,π₯-π¦)π(π ,π¦)\hspace{1.67pt}ππ \hspace{1.67pt}ππ(π¦).\end{array}\]This paper aims to prove that
for each $(π‘,π₯)\in [0,π]\times \mathrm{R}$. Mainly, we will consider $πΏ=ππΏ$ for some fixed $πΏ\gt 0$, $π\ge 1$, and obtain the estimate of $|{π’^{πΏ}}(π‘,π₯)-π’(π‘,π₯)|$ for some versions of the solutions.
4 Auxiliary lemma
In the sequel, we will use the notations ${π·_{πΏ}}=(-πΏ,πΏ]$, ${π·_{πΏ}^{π}}=\mathrm{R}\backslash (-πΏ,πΏ]$, ${\mathrm{Z}^{\ast }}=\mathrm{Z}\backslash \{0\}$.
Consider the stochastic terms
\[\begin{array}{r@{\hskip0pt}l@{\hskip10pt}r}\displaystyle {π^{πΏ}}(π‘,π₯)& \displaystyle ={\int _{{π·_{πΏ}}}}{\int _{0}^{π‘}}{πΊ^{πΏ}}(π‘-π ,π₯,π¦)π(π ,π¦)\hspace{1.67pt}ππ \hspace{1.67pt}ππ(π¦),& \displaystyle \hspace{-10.0pt}\\ {} \displaystyle π(π‘,π₯)& \displaystyle ={\int _{\mathrm{R}}}{\int _{0}^{π‘}}π(π‘-π ,π₯-π¦)π(π ,π¦)\hspace{1.67pt}ππ \hspace{1.67pt}ππ(π¦).& \end{array}\]
In the following statement, we fix any $πΏ\gt 0$, and consider $πΏ=ππΏ$, $π\in {\mathrm{Z}_{+}}$.
Proof.
For $π₯\in {π·_{πΏ}}$, we will estimate
\[\begin{array}{r@{\hskip0pt}l@{\hskip10pt}r}\displaystyle {π^{πΏ}}(π‘,π₯)-π(π‘,π₯)& \displaystyle ={\int _{{π·_{πΏ}}}}{\int _{0}^{π‘}}{πΊ^{πΏ}}(π‘-π ,π₯,π¦)π(π ,π¦)\hspace{1.67pt}ππ \hspace{1.67pt}ππ(π¦)& \displaystyle \hspace{-10.0pt}\\ {} & \displaystyle \hspace{10.0pt}-{\int _{\mathrm{R}}}{\int _{0}^{π‘}}π(π‘-π ,π₯-π¦)π(π ,π¦)\hspace{1.67pt}ππ \hspace{1.67pt}ππ(π¦)& \\ {} & \displaystyle ={\int _{{π·_{πΏ}}}}{\int _{0}^{π‘}}\Bigl({πΊ^{πΏ}}(π‘-π ,π₯,π¦)-π(π‘-π ,π₯-π¦)\Bigr)π(π ,π¦)\hspace{1.67pt}ππ \hspace{1.67pt}ππ(π¦)& \\ {} & \displaystyle \hspace{10.0pt}-{\int _{{π·_{πΏ}^{π}}}}{\int _{0}^{π‘}}π(π‘-π ,π₯-π¦)π(π ,π¦)\hspace{1.67pt}ππ \hspace{1.67pt}ππ(π¦):={πΌ_{1}}-{πΌ_{2}}.& \end{array}\]
First, consider ${πΌ_{1}}$. For $π₯,π¦\in {π·_{πΏ}}$,
(4.2)
\[\begin{array}{r@{\hskip0pt}l@{\hskip10pt}r}& \displaystyle {πΊ^{πΏ}}(π‘-π ,π₯,π¦)-π(π‘-π ,π₯-π¦)& \displaystyle \hspace{-10.0pt}\\ {} & \displaystyle \hspace{10.0pt}=\sum \limits_{π\in {\mathrm{Z}^{\ast }}}π(π‘-π ,π₯-π¦+4ππΏ)& \\ {} & \displaystyle \hspace{20.0pt}-\sum \limits_{π\in {\mathrm{Z}^{\ast }}}π\Bigl(π‘-π ,π₯+π¦+(4π+2)πΏ\Bigr)-π(π‘-π ,π₯+π¦+2πΏ).& \end{array}\]For $π₯,π¦\in {π·_{πΏ}}$ we have that $|π₯+π¦+2πΏ|\ge πΏ-|π₯|$.
We will take π large enough such that $|π₯-π¦+4ππΏ|,|π₯+π¦+(4π+2)πΏ|\gt π\ge π‘-π ,\hspace{2.5pt}π\in {\mathrm{Z}^{\ast }}$ here. (By Lemma 4.2 [17], π is bounded. By (4.9) of [18], each ${π^{ππΏ}}$ is bounded. Thus, excluding the finite number of π does not influence (4.1).)
For fixed $π‘,π₯$ set
Our considerations will imply that this function is continuous in π¦; thus, the conditions of Lemma 2 hold (for this purpose, we can use estimate (4.6) below).
(4.3)
\[π(π§,π¦)={\int _{0}^{π‘}}π(π‘-π ,π₯-π¦)π(π ,π¦)\hspace{1.67pt}ππ ,\hspace{10.0pt}π§=(π‘,π₯),\hspace{2.5pt}π¦\in {π·_{πΏ}}.\]Consider
\[\begin{array}{r@{\hskip0pt}l}\displaystyle π(π§,π¦+β)-π(π§,π¦)& \displaystyle ={\int _{0}^{π‘}}π(π‘-π ,π₯-π¦)\Bigl(π(π ,π¦+β)-π(π ,π¦)\Bigr)\hspace{1.67pt}ππ \\ {} & \displaystyle \hspace{10.0pt}+{\int _{0}^{π‘}}\Bigl(π(π‘-π ,π₯-π¦-β)-π(π‘-π ,π₯-π¦)\Bigr)π(π ,π¦+β)\hspace{1.67pt}ππ \\ {} & \displaystyle :={πΌ_{11}}+{πΌ_{12}}.\end{array}\]
We will use the following simple estimate for $π/π‘\ge 1$.
We get for $|π₯-π¦|,|π₯-π¦-β|\ge ππΏ-|π₯|\gt π$, $β\lt πΏ$
(4.4)
\[{\int _{0}^{π‘}}\frac{1}{π}{π^{-\frac{π}{π}}}ππ\stackrel{π/π=π§}{=}{\int _{π/π‘}^{\mathrm{\infty }}}\frac{1}{π§}{π^{-π§}}ππ§\le {\int _{π/π‘}^{\mathrm{\infty }}}{π^{-π§}}ππ§={π^{-π/π‘}}.\]
\[\begin{array}{r@{\hskip0pt}l@{\hskip10pt}r}\displaystyle |{πΌ_{12}}|& \displaystyle \stackrel{\text{A4}}{\le }{πΆ_{π}}{\int _{0}^{π‘}}\Bigl|π(π‘-π ,π₯-β-π¦)-π(π‘-π ,π₯-π¦)\Bigr|ππ & \displaystyle \hspace{-10.0pt}\\ {} & \displaystyle \le {πΆ_{π}}{\int _{0}^{π‘}}\Biggl|{\int _{π₯-β}^{π₯}}\displaystyle \frac{ππ(π‘-π ,π£-π¦)}{ππ£}\hspace{1.67pt}ππ£\Biggr|ππ \stackrel{\text{(3.3)}}{\le }{πΆ_{π}}{\int _{0}^{π‘}}\Biggl(\frac{{πΆ_{π}}}{π‘-π }\hspace{-1.67pt}{\int _{π₯-β}^{π₯}}{π^{-\frac{π{|π£-π¦|^{2}}}{π‘-π }}}ππ£\Biggr)ππ & \\ {} & \displaystyle \hspace{-6.68pt}\stackrel{π‘-π =π}{=}{πΆ_{π}}{πΆ_{π}}{\int _{π₯-β}^{π₯}}ππ£{\int _{0}^{π‘}}\frac{1}{π}{π^{-\frac{π{|π£-π¦|^{2}}}{π}}}ππ\stackrel{\text{(4.4)}}{\le }{πΆ_{π}}{πΆ_{π}}{\int _{π₯-β}^{π₯}}{π^{-\frac{π{|π£-π¦|^{2}}}{π‘}}}\hspace{1.67pt}ππ£& \\ {} & \displaystyle \le {πΆ_{π}}{πΆ_{π}}|β|{π^{-\frac{π{(ππΏ-|π₯|)^{2}}}{π‘}}}.& \end{array}\]
Also,
(The constants ${πΆ_{π}}$ and πΆ may also depend on π, though this plays no role in our considerations.) Thus, for any $0\lt π\lt \frac{1}{4}$
Similarly, consider $π(π‘-π ,π₯+π¦+(4π+2)πΏ)$, $π\in {\mathrm{Z}^{\ast }}$, $π(π‘-π ,π₯+π¦+2πΏ)$, and for
As in (4.5), we obtain
Therefore,
(4.5)
\[\begin{array}{r@{\hskip0pt}l@{\hskip10pt}r}\displaystyle |{πΌ_{11}}|& \displaystyle \le {\int _{0}^{π‘}}π(π‘-π ,π₯-π¦)\Bigl|π(π ,π¦+β)-π(π ,π¦)\Bigr|\hspace{1.67pt}ππ & \displaystyle \hspace{-10.0pt}\\ {} & \displaystyle \stackrel{\text{A4}}{\le }πΆ{πΏ_{π}}{β^{π½(π)}}{\int _{0}^{π‘}}\displaystyle \displaystyle \frac{1}{\sqrt{π‘-π }}{π^{-\frac{{|π₯-π¦|^{2}}}{4(π‘-π )}}}ππ \le πΆ{β^{π½(π)}}{\int _{0}^{π‘}}\displaystyle \displaystyle \frac{1}{\sqrt{π‘-π }}{π^{-\frac{{(ππΏ-|π₯|)^{2}}}{4(π‘-π )}}}ππ & \\ {} & \displaystyle \le πΆ{β^{π½(π)}}{π^{-\frac{{(ππΏ-|π₯|)^{2}}}{4π‘}}}{\int _{0}^{π‘}}\displaystyle \displaystyle \frac{1}{\sqrt{π‘-π }}ππ \le πΆ{β^{π½(π)}}{π^{-\frac{{(ππΏ-|π₯|)^{2}}}{4π‘}}}.& \end{array}\](4.6)
\[\Bigl|π(π§,π¦+β)-π(π§,π¦)\Bigr|\le {πΆ_{π}}{β^{π½(π)}}{π^{-\frac{π{(ππΏ-|π₯|)^{2}}}{π‘}}}.\]
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \tilde{π}(π§,π¦)& \displaystyle ={\int _{0}^{π‘}}\Biggl(\sum \limits_{π\in {\mathrm{Z}^{\ast }}}π(π‘-π ,π₯-π¦+4ππΏ)-\sum \limits_{π\in {\mathrm{Z}^{\ast }}}π\Bigl(π‘-π ,π₯+π¦+(4π+2)πΏ\Bigr)\\ {} & \displaystyle \hspace{10.0pt}-π(π‘-π ,π₯+π¦+2πΏ)\Biggr)π(π ,π¦)\hspace{1.67pt}ππ \end{array}\]
we get
(4.7)
\[\Bigl|\tilde{π}(π§,π¦+β)-\tilde{π}(π§,π¦)\Bigr|\le πΆ{πΆ_{π}}{β^{π½(π)}}{π^{-\frac{π{(πΏ-|π₯|)^{2}}}{π‘}}},\hspace{10.0pt}π§=(π‘,π₯),\hspace{2.5pt}π¦,\hspace{2.5pt}π¦+β\in {π·_{πΏ}}.\]We take $π\gt 1/2$ from A5 and some $π½\gt 0$ such that $π½+1-2π½(π)\lt 0$. From (2.3), for the version (2.2) of the stochastic integral (taking intervals $[ππΏ,(π+1)πΏ]$ instead of $[π,π+1]$ and ${π_{ππ}^{(π)}}=(π+π{2^{-π}})πΏ$), we get
(4.9)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle |{πΌ_{1}}|& \displaystyle =\Biggl|{\int _{{π·_{πΏ}}}}{\int _{0}^{π‘}}\Bigl({πΊ^{πΏ}}(π‘-π ,π₯,π¦)-π(π‘-π ,π₯-π¦)\Bigr)π(π ,π¦)\hspace{1.67pt}ππ \hspace{1.67pt}ππ(π¦)\Biggr|\\ {} & \displaystyle =\Biggl|{\int _{{π·_{πΏ}}}}\tilde{π}(π§,π¦)\hspace{1.67pt}ππ(π¦)\Biggr|\le \sum \limits_{π\in \mathrm{Z}}\Biggl|{\int _{(ππΏ,(π+1)πΏ]\cap {π·_{πΏ}}}}\tilde{π}(π§,π¦)\hspace{1.67pt}ππ(π¦)\Biggr|\\ {} & \displaystyle \hspace{-13.36pt}\stackrel{\text{(2.3)},πΏ=ππΏ}{\le }\sum \limits_{π:(ππΏ,(π+1)πΏ]\subset {π·_{πΏ}}}\Bigl|\tilde{π}(π§,ππΏ)π\Bigl(\Bigl(ππΏ,(π+1)πΏ\Bigr]\Bigr)\Bigr|\\ {} & \displaystyle \hspace{10.0pt}+{\Biggl\{\sum \limits_{π\ge 1}{2^{ππ½}}\sum \limits_{1\le π\le {2^{π}}}{\Bigl|\tilde{π}\Bigl(π§,{π_{ππ}^{(π)}}\Bigr)-\tilde{π}\Bigl(π§,{π_{(π-1)π}^{(π)}}\Bigr)\Bigr|^{2}}\Biggr\}^{1/2}}\\ {} & \displaystyle \hspace{10.0pt}\times {\Biggl\{\sum \limits_{π\ge 1}{2^{-ππ½}}\sum \limits_{1\le π\le {2^{π}}}{\Bigl|π\Bigl({\Delta _{ππ}^{(π)}}\Bigr)\Bigr|^{2}}\Biggr\}^{1/2}}\\ {} & \displaystyle \hspace{-11.69pt}\stackrel{\text{(4.8)},\text{(4.7)}}{\le }πΆ{πΆ_{π}}{π^{-\frac{π{(πΏ-|π₯|)^{2}}}{π‘}}}\Biggl[\sum \limits_{π:(ππΏ,(π+1)πΏ]\subset {π·_{πΏ}}}\Bigl|π\Bigl(ππΏ,(π+1)πΏ\Bigr]\Bigr)\Bigr|\\ {} & \displaystyle \hspace{10.0pt}+\sum \limits_{π:(ππΏ,(π+1)πΏ]\subset {π·_{πΏ}}}{\Biggl\{\sum \limits_{π\ge 1}{2^{π(π½+1-2π½(π))}}\sum \limits_{1\le π\le {2^{π}}}{\Bigl|π\Bigl({\Delta _{ππ}^{(π)}}\Bigr)\Bigr|^{2}}\Biggr\}^{1/2}}\Biggr]\\ {} & \displaystyle \le πΆ{πΆ_{π}}{π^{-\frac{π{(πΏ-|π₯|)^{2}}}{π‘}}}\Biggl[\hspace{-1.67pt}{\Biggl(\sum \limits_{π\in \mathrm{Z}}{\Bigl(|π|\hspace{-1.67pt}+\hspace{-1.67pt}1\Bigr)^{2π}}\Bigl(π{\Bigl(\Bigl(ππΏ,(π\hspace{-1.67pt}+\hspace{-1.67pt}1)πΏ\Bigr]\Bigr)\Bigr)^{2}}\Biggr)^{1/2}}{\Biggl(\sum \limits_{π\in \mathrm{Z}}{\Bigl(|π|+1\Bigr)^{-2π}}\Biggr)^{\hspace{-1.67pt}1/2}}\\ {} & \displaystyle \hspace{7.5pt}+{\Biggl(\sum \limits_{π\in \mathrm{Z}}{\Bigl(|π|\hspace{-1.67pt}+\hspace{-1.67pt}1\Bigr)^{2π}}\hspace{-1.67pt}\sum \limits_{π\ge 1}{2^{π(π½+1-2π½(π))}}\hspace{-3.34pt}\sum \limits_{1\le π\le {2^{π}}}{\Bigl|π\Bigl({\Delta _{ππ}^{(π)}}\Bigr)\Bigr|^{2}}\Biggr)^{\hspace{-1.67pt}1/2}}\hspace{-1.67pt}{\Biggl(\sum \limits_{π\in \mathrm{Z}}{\Bigl(|π|+1\Bigr)^{-2π}}\Biggr)^{\hspace{-1.67pt}1/2}}\Biggr].\end{array}\]Here, the sums with π have a form ${\sum _{π=1}^{\mathrm{\infty }}}{({\int _{\mathrm{R}}}{π_{π}}\hspace{1.67pt}ππ)^{2}}$, where
holds for all $πΏ=ππΏ$, $π\in {\mathrm{Z}_{+}}$, and $π\in {\Omega _{1}}$. The set ${\Omega _{1}}$ depends only on π, πΏ and π½.
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \Bigl\{{π_{π}}(π¦),\hspace{2.5pt}π\ge 1\Bigr\}& \displaystyle =\Bigl\{{\Bigl(|π|+1\Bigr)^{π}}\hspace{1.67pt}{\mathbf{1}_{(ππΏ,(π+1)πΏ]}}(π¦),\hspace{2.5pt}π\in \mathrm{Z}\Bigr\},\\ {} \displaystyle \Bigl\{{π_{π}}(π¦),\hspace{2.5pt}π\ge 1\Bigr\}& \displaystyle =\Bigl\{{\Bigl(|π|+1\Bigr)^{π}}{2^{π(π½+1-2π½(π))/2}}{\mathbf{1}_{{\Delta _{ππ}^{(π)}}}}(π¦),\hspace{2.5pt}π\in \mathrm{Z},\hspace{2.5pt}π\ge 1,\hspace{2.5pt}1\le π\le {2^{π}}\Bigr\}.\end{array}\]
We have ${\sum _{π=1}^{\mathrm{\infty }}}|{π_{π}}(π¦)|\le πΆ{(|π¦|+1)^{π}}$ in both cases. From Lemma 1 it follows that
\[{\sum \limits_{π=1}^{\mathrm{\infty }}}{\Biggl({\int _{\mathrm{X}}}{π_{π}}\hspace{1.67pt}ππ\Biggr)^{2}}\lt +\mathrm{\infty }\hspace{2.5pt}\text{a.s.}\]
We take a set ${\Omega _{1}}\subset \Omega $ such that for $π\in {\Omega _{1}}$, this series is finite for both sequences $\{{π_{π}}\}$. Thus $\mathrm{P}({\Omega _{1}})=1$ and
(4.10)
\[|{πΌ_{1}}|\le {πΆ_{πΏ}}(π){πΆ_{π}}{π^{-\frac{π{(πΏ-|π₯|)^{2}}}{π‘}}}\]In ${πΌ_{2}}$ we have $|π₯-π¦|\ge πΏ-|π₯|$. Again we consider $π(π§,π¦)$ from (4.3), now with $|π¦|\ge πΏ$. Repeating the consideration from (4.3) to (4.6), we get here
\[\begin{array}{r@{\hskip0pt}l@{\hskip10pt}r}\displaystyle \Bigl|π(π§,π¦+β)-π(π§,π¦)\Bigr|& \displaystyle \le {πΆ_{π}}{β^{π½(π)}}{π^{-\frac{π{(πΏ-|π₯|)^{2}}}{π‘}}},& \displaystyle \hspace{-10.0pt}\\ {} \displaystyle \Bigl|π(π§,π¦)\Bigr|& \displaystyle \le πΆ{πΆ_{π}}{π^{-\frac{π{(ππΏ-|π₯|)^{2}}}{π‘}}},\hspace{10.0pt}|π¦|\ge πΏ.& \end{array}\]
Similarly to (4.9), we get
(4.11)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle |{πΌ_{2}}|& \displaystyle =\Biggl|{\int _{{π·_{πΏ}^{π}}}}{\int _{0}^{π‘}}π(π‘-π ,π₯-π¦)π(π ,π¦)\hspace{1.67pt}ππ \hspace{1.67pt}ππ(π¦)\Biggr|\\ {} & \displaystyle =\Biggl|{\int _{{π·_{πΏ}^{π}}}}π(π§,π¦)\hspace{1.67pt}ππ(π¦)\Biggr|\le \sum \limits_{π\in \mathrm{Z}}\Biggl|{\int _{(ππΏ,(π+1)πΏ]\cap {π·_{πΏ}^{π}}}}π(π§,π¦)\hspace{1.67pt}ππ(π¦)\Biggr|\\ {} & \displaystyle \hspace{-13.36pt}\stackrel{\text{(2.3)},πΏ=ππΏ}{\le }\sum \limits_{π:(ππΏ,(π+1)πΏ]\subset {π·_{πΏ}^{π}}}\Bigl|π(π§,ππΏ)π\Bigl(\Bigl(ππΏ,(π+1)πΏ\Bigr]\Bigr)\Bigr|\\ {} & \displaystyle \hspace{10.0pt}+{\Biggl\{\sum \limits_{π\ge 1}{2^{ππ½}}\sum \limits_{1\le π\le {2^{π}}}{\Bigl|π\Bigl(π§,{π_{ππ}^{(π)}}\Bigr)-π\Bigl(π§,{π_{(π-1)π}^{(π)}}\Bigr)\Bigr|^{2}}\Biggr\}^{1/2}}\\ {} & \displaystyle \hspace{10.0pt}\times {\Biggl\{\sum \limits_{π\ge 1}{2^{-ππ½}}\sum \limits_{1\le π\le {2^{π}}}{\Bigl|π\Bigl({\Delta _{ππ}^{(π)}}\Bigr)\Bigr|^{2}}\Biggr\}^{1/2}}\\ {} & \displaystyle \hspace{-16.7pt}\stackrel{\text{(4.8)},\text{(4.7)},πΏ=ππΏ}{\le }πΆ{πΆ_{π}}{π^{-\frac{π{(πΏ-|π₯|)^{2}}}{π‘}}}\Biggl[\sum \limits_{π:(ππΏ,(π+1)πΏ]\subset {π·_{πΏ}^{π}}}\Bigl|π\Bigl(ππΏ,(π+1)πΏ\Bigr]\Bigr)\Bigr|\\ {} & \displaystyle \hspace{10.0pt}+\sum \limits_{π:(ππΏ,(π+1)πΏ]\subset {π·_{πΏ}^{π}}}{\Biggl\{\sum \limits_{π\ge 1}{2^{π(π½+1-2π½(π))}}\sum \limits_{1\le π\le {2^{π}}}{\Bigl|π\Bigl({\Delta _{ππ}^{(π)}}\Bigr)\Bigr|^{2}}\Biggr\}^{1/2}}\Biggr]\\ {} & \displaystyle \le πΆ{πΆ_{π}}{π^{-\frac{π{(πΏ-|π₯|)^{2}}}{π‘}}}\Biggl[{\Biggl(\sum \limits_{π\in \mathrm{Z}}{\Bigl(|π|\hspace{-1.67pt}+\hspace{-1.67pt}1\Bigr)^{2π}}\Bigl(π{\Bigl(\Bigl(ππΏ,(π+1)πΏ\Bigr]\Bigr)\Bigr)^{2}}\Biggr)^{\hspace{-1.67pt}1/2}}{\Biggl(\sum \limits_{π\in \mathrm{Z}}{\Bigl(|π|\hspace{-1.67pt}+\hspace{-1.67pt}1\Bigr)^{-2π}}\Biggr)^{\hspace{-1.67pt}1/2}}\\ {} & \displaystyle \hspace{7.5pt}+{\Biggl(\sum \limits_{π\in \mathrm{Z}}{\Bigl(|π|\hspace{-1.67pt}+\hspace{-1.67pt}1\Bigr)^{2π}}\hspace{-1.67pt}\sum \limits_{π\ge 1}{2^{π(π½+1-2π½(π))}}\hspace{-1.67pt}\sum \limits_{1\le π\le {2^{π}}}{\Bigl|π\Bigl({\Delta _{ππ}^{(π)}}\Bigr)\Bigr|^{2}}\Biggr)^{\hspace{-1.67pt}1/2}}\hspace{-1.67pt}{\Biggl(\sum \limits_{π\in \mathrm{Z}}{\Bigl(|π|+1\Bigr)^{-2π}}\Biggr)^{\hspace{-1.67pt}1/2}}\Biggr].\end{array}\]RemarkΒ 1.
Let SM π be such that $π((π,π])\le πΆ(π)$, $π,\hspace{2.5pt}π\in \mathrm{R}$. Then we can obtain (4.1) for arbitrary $πΏ\in {\mathrm{R}_{+}}$, and get
\[\Bigl|{π^{πΏ}}(π‘,π₯)-π(π‘,π₯)\Bigr|\le πΆ(π){πΆ_{π}}{π^{-\frac{π{(πΏ-|π₯|)^{2}}}{π‘}}},\]
where $πΆ(π)$ is independent of πΏ. To prove this, we can fix $πΏ=1$, and note that in this case in (4.9) we get additional summands
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle \Bigl|π(({π_{0}},πΏ]\cap {π·_{πΏ}})\Bigr|+\Bigl|π\Bigl((-πΏ,-{π_{0}}]\Bigr)\Bigr|\\ {} & \displaystyle \hspace{10.0pt}+\sum \limits_{π\ge 1}{2^{-ππ½}}\Bigl|π\Bigl(({π_{{π_{π}}π}^{({π_{0}})}},πΏ]\Bigr)\Bigr|+\hspace{1.67pt}\sum \limits_{π\ge 1}{2^{-ππ½}}\Bigl|π\Bigl(\Bigl(-πΏ,{π_{{π_{π}}π}^{(-{π_{0}})}}\Bigr]\Bigr)\Bigr|\end{array}\]
for some ${π_{0}}$, ${π_{π}}$ that depend on πΏ (and similar summands in (4.11)). If $π((π,π])\le πΆ(π)$, then these sums are uniformly bounded.5 Main result
TheoremΒ 1.
Let Assumptions A1βA5 hold, ${π’^{πΏ}}$ is the solution to (3.1), π’ is the solution to (3.2). Then, for each $πΏ\gt 0$ there exist versions of ${π’^{πΏ}},π’$ such that for all $π\in \Omega $ for each $π₯\in {π·_{πΏ}}$, $πΏ=ππΏ$, $π\in {\mathrm{Z}_{+}}$, $0\lt π‘\le π$, for any $0\lt π\lt \frac{1}{4}$ holds
Proof.
Consider
\[\begin{array}{r@{\hskip0pt}l}\displaystyle {π’^{πΏ}}(π‘,π₯)-π’(π‘,π₯)& \displaystyle =\Biggl({\int _{{π·_{πΏ}}}}{πΊ^{πΏ}}(π‘,π₯,π¦){π’_{0}}(π¦)\hspace{1.67pt}ππ¦-{\int _{\mathrm{R}}}π(π‘,π₯-π¦){π’_{0}}(π¦)\hspace{1.67pt}ππ¦\Biggr)\\ {} & \displaystyle \hspace{10.0pt}+\Biggl({\int _{0}^{π‘}}{\int _{{π·_{πΏ}}}}{πΊ^{πΏ}}(π‘-π ,π₯,π¦)π\Bigl(π ,π¦,{π’^{πΏ}}(π ,π¦)\Bigr)\hspace{1.67pt}ππ¦\hspace{1.67pt}ππ \\ {} & \displaystyle \hspace{10.0pt}-{\int _{0}^{π‘}}{\int _{\mathrm{R}}}π(π‘-π ,π₯-π¦)π\Bigl(π ,π¦,π’(π ,π¦)\Bigr)\hspace{1.67pt}ππ¦\hspace{1.67pt}ππ \Biggr)\\ {} & \displaystyle \hspace{10.0pt}+\Biggl({\int _{{π·_{πΏ}}}}{\int _{0}^{π‘}}{πΊ^{πΏ}}(π‘-π ,π₯,π¦)π(π ,π¦)\hspace{1.67pt}ππ \hspace{1.67pt}ππ(π¦)\\ {} & \displaystyle \hspace{10.0pt}-{\int _{\mathrm{R}}}{\int _{0}^{π‘}}π(π‘-π ,π₯-π¦)π(π ,π¦)\hspace{1.67pt}ππ \hspace{1.67pt}ππ(π¦)\Biggr)\\ {} & \displaystyle :={π½_{1}}+{π½_{2}}+{π½_{3}},\end{array}\]
by Lemma 3, for some versions of the integrals, holds
For ${π½_{1}}$, we have
\[\begin{array}{r@{\hskip0pt}l}\displaystyle |{π½_{1}}|& \displaystyle \le {\int _{{π·_{πΏ}}}}\Bigl|{πΊ^{πΏ}}(π‘,π₯,π¦)-π(π‘,π₯-π¦)\Bigr|\Bigl|{π’_{0}}(π¦)\Bigr|\hspace{1.67pt}ππ¦+{\int _{{π·_{πΏ}^{π}}}}π(π‘,π₯-π¦)\Bigl|{π’_{0}}(π¦)\Bigr|\hspace{1.67pt}ππ¦\\ {} & \displaystyle \stackrel{\text{A1}}{\le }{πΆ_{{π’_{0}}}}(π){\int _{{π·_{πΏ}}}}\Bigl|{πΊ^{πΏ}}(π‘,π₯,π¦)-π(π‘,π₯-π¦)\Bigr|\hspace{1.67pt}ππ¦+{πΆ_{{π’_{0}}}}(π){\int _{{π·_{πΏ}^{π}}}}π(π‘,π₯-π¦)\hspace{1.67pt}ππ¦\\ {} & \displaystyle :={π½_{11}}+{π½_{12}}.\end{array}\]
We recall the Chernoff inequality (see, for instance, Example 2.1 [20]). For random variable $π\sim π(π,{π^{2}})$ and $π\ge 0$ holds
We have that $π(π‘,π¦-π₯)$, as a function of π¦, is the density function of the distribution $π(π₯,2π‘)$.
(5.2)
\[\mathrm{P}\{π\ge π+π\}\le \exp \Biggl\{-\displaystyle \frac{{π^{2}}}{2{π^{2}}}\Biggr\}.\]Thus, for $π₯\in {π·_{πΏ}}$, we obtain
\[\begin{array}{r@{\hskip0pt}l}\displaystyle {\int _{{π·_{πΏ}}}}π(π‘,π₯-π¦+4ππΏ)\hspace{1.67pt}ππ¦& \displaystyle \stackrel{π£=π¦-4ππΏ}{=}{\int _{(-(4π+1)πΏ,-(4π-1)πΏ]}}π(π‘,π₯-π£)\hspace{1.67pt}ππ£\\ {} & \displaystyle \hspace{10.0pt}\stackrel{\text{(5.2)}}{\le }\exp \Biggl\{-\displaystyle \frac{{((4|π|-1)πΏ-|π₯|)^{2}}}{4π‘}\Biggr\}.\end{array}\]
In a similar way, we consider $π(π‘,π₯+π¦+(4π+2)πΏ)$, $π(π‘,π₯+π¦+2πΏ)$ in (4.2), and obtain that
Also, (5.2) implies that
For ${π½_{2}}$, we get
\[\begin{array}{r@{\hskip0pt}l}\displaystyle |{π½_{2}}|& \displaystyle \le {\int _{0}^{π‘}}{\int _{{π·_{πΏ}}}}\Bigl|{πΊ^{πΏ}}(π‘-π ,π₯,π¦)-π(π‘-π ,π₯-π¦)\Bigr|\Bigl|π\Bigl(π ,π¦,{π’^{πΏ}}(π ,π¦)\Bigr)\Bigr|\hspace{1.67pt}ππ¦\hspace{1.67pt}ππ \\ {} & \displaystyle \hspace{10.0pt}+{\int _{0}^{π‘}}{\int _{{π·_{πΏ}}}}π(π‘-π ,π₯-π¦)\Bigl|π\Bigl(π ,π¦,{π’^{πΏ}}(π ,π¦)\Bigr)-π\Bigl(π ,π¦,π’(π ,π¦)\Bigr)\Bigr|\hspace{1.67pt}ππ¦\hspace{1.67pt}ππ \\ {} & \displaystyle \hspace{10.0pt}+{\int _{0}^{π‘}}{\int _{{π·_{πΏ}^{π}}}}π(π‘-π ,π₯-π¦)\Bigl|π\Bigl(π ,π¦,{π’^{πΏ}}(π ,π¦)\Bigr)\Bigr|\hspace{1.67pt}ππ¦\hspace{1.67pt}ππ \\ {} & \displaystyle :={π½_{21}}+{π½_{22}}+{π½_{23}}.\end{array}\]
Similarly to (5.3), we obtain
\[\begin{array}{r@{\hskip0pt}l}\displaystyle {π½_{21}}& \displaystyle \stackrel{\text{A2}}{\le }{\int _{0}^{π‘}}{πΆ_{π}}{\int _{{π·_{πΏ}}}}\Bigl|{πΊ^{πΏ}}(π‘-π ,π₯,π¦)-π(π‘-π ,π₯-π¦)\Bigr|\hspace{1.67pt}ππ¦\hspace{1.67pt}ππ \\ {} & \displaystyle \le πΆ(π){\int _{0}^{π‘}}\exp \Biggl\{-\displaystyle \frac{{(πΏ-|π₯|)^{2}}}{4(π‘-π )}\Biggr\}\hspace{1.67pt}ππ \le πΆ(π)\exp \Biggl\{-\displaystyle \frac{{(πΏ-|π₯|)^{2}}}{4π‘}\Biggr\}.\end{array}\]
As in (5.4), we get
Also
Gathering together all estimates of ${π½_{1}}$, ${π½_{2}}$, and ${π½_{3}}$, we obtain that for any $π\lt \frac{1}{4}$
\[\begin{array}{r@{\hskip0pt}l@{\hskip10pt}r}& \displaystyle \Bigl|{π’^{πΏ}}(π‘,π₯)-π’(π‘,π₯)\Bigr|{\mathbf{1}_{\{π₯\in {π·_{πΏ}}\}}}& \displaystyle \hspace{-10.0pt}\\ {} & \displaystyle \hspace{10.0pt}\le {πΆ_{πΏ}}(π){πΆ_{π}}\exp \Biggl\{-\displaystyle \frac{π{(πΏ-|π₯|)^{2}}}{π‘}\Biggr\}& \\ {} & \displaystyle \hspace{20.0pt}+πΆ(π){\int _{0}^{π‘}}{\int _{\mathrm{R}}}π(π‘-π ,π₯-π¦)\Bigl|{π’^{πΏ}}(π ,π¦)-π’(π ,π¦)\Bigr|{\mathbf{1}_{\{π¦\in {π·_{πΏ}}\}}}\hspace{1.67pt}ππ¦\hspace{1.67pt}ππ .& \end{array}\]
For $|{π’^{πΏ}}(π‘,π₯))-π’(π‘,π₯)|{\mathbf{1}_{|π₯|\le πΏ}}$, we will use the space-time Gronwall-type inequality (see Lemma C.1.7 (d) [4]), and obtain
where $\mathcal{K}$ denotes the series with iterated space-time convolutions
(see more details in Section C.1.3 [4]) and
(5.5)
\[\Bigl|{π’^{πΏ}}(π‘,π₯)-π’(π‘,π₯)\Bigr|\le {π_{πΏ,π,πΏ}}(π‘,π₯)+πΆ(π){\int _{0}^{π‘}}{\int _{\mathrm{R}}}\mathcal{K}(π‘-π ,π₯-π¦){π_{πΏ,π,πΏ}}(π ,π¦)\hspace{1.67pt}ππ¦\hspace{1.67pt}ππ \]
\[{π_{πΏ,π,πΏ}}(π‘,π₯)={πΆ_{πΏ}}(π){πΆ_{π}}\exp \Biggl\{-\displaystyle \frac{π{(πΏ-|π₯|)^{2}}}{π‘}\Biggr\}{\mathbf{1}_{\{π₯\in {π·_{πΏ}}\}}}.\]
It is well-known that
\[{\int _{\mathrm{R}}}π({π‘_{1}},π₯-π¦)π({π‘_{2}},π¦)\hspace{1.67pt}ππ¦=π({π‘_{1}}+{π‘_{2}},π₯),\]
therefore,
and applying induction, we easily obtain that
Further, we get
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle {\int _{0}^{π‘}}{\int _{\mathrm{R}}}\mathcal{K}(π‘-π ,π₯-π¦){π_{πΏ,π,πΏ}}(π ,π¦)\hspace{1.67pt}ππ¦\hspace{1.67pt}ππ \\ {} & \displaystyle \hspace{10.0pt}=πΆ(π){\int _{0}^{π‘}}{\int _{\mathrm{R}}}{π^{π‘-π }}π(π‘-π ,π₯-π¦)\exp \Biggl\{-\displaystyle \frac{π{(πΏ-|π¦|)^{2}}}{π }\Biggr\}{\mathbf{1}_{\{π¦\in {π·_{πΏ}}\}}}\hspace{1.67pt}ππ¦\hspace{1.67pt}ππ \\ {} & \displaystyle \hspace{10.0pt}=πΆ(π)\hspace{-1.67pt}{\int _{0}^{π‘}}\hspace{-3.34pt}{\int _{\mathrm{R}}}\hspace{-1.67pt}{π^{π‘-π }}\displaystyle \frac{1}{2\sqrt{π(π‘\hspace{-1.67pt}-\hspace{-1.67pt}π )}}\exp \Biggl\{-\displaystyle \frac{{(π₯-π¦)^{2}}}{4(π‘-π )}\Biggr\}\exp \Biggl\{-\displaystyle \frac{π{(πΏ\hspace{-1.67pt}-\hspace{-1.67pt}|π¦|)^{2}}}{π }\Biggr\}{\mathbf{1}_{\{π¦\in {π·_{πΏ}}\}}}ππ¦\hspace{1.67pt}ππ \\ {} & \displaystyle \hspace{10.0pt}\le πΆ(π)\displaystyle \frac{2}{\sqrt{π}}{\int _{0}^{π‘}}\displaystyle \displaystyle \frac{\sqrt{ππ }}{\sqrt{π}}{\int _{\mathrm{R}}}\displaystyle \displaystyle \frac{\sqrt{π}}{\sqrt{π(π‘-π )}}\exp \Biggl\{-\displaystyle \frac{π{(π₯-π¦)^{2}}}{π‘-π }\Biggr\}\displaystyle \frac{\sqrt{π}}{\sqrt{ππ }}\Biggl(\exp \Biggl\{-\displaystyle \frac{π{(π¦-πΏ)^{2}}}{π }\Biggr\}\\ {} & \displaystyle \hspace{20.0pt}+\exp \Biggl\{-\displaystyle \frac{π{(π¦+πΏ)^{2}}}{π }\Biggr\}\Biggr)\hspace{1.67pt}ππ¦\hspace{1.67pt}ππ .\end{array}\]
From the properties of the convolution of density functions of normal distributions, we get
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle {\int _{\mathrm{R}}}\displaystyle \displaystyle \frac{\sqrt{π}}{\sqrt{π(π‘-π )}}\exp \Biggl\{-\displaystyle \frac{π{(π₯-π¦)^{2}}}{π‘-π }\Biggr\}\displaystyle \frac{\sqrt{π}}{\sqrt{ππ }}\exp \Biggl\{-\displaystyle \frac{π{(π¦-πΏ)^{2}}}{π }\Biggr\}\hspace{1.67pt}ππ¦\\ {} & \displaystyle \hspace{10.0pt}=\displaystyle \frac{\sqrt{π}}{\sqrt{ππ‘}}\exp \Biggl\{-\displaystyle \frac{π{(π₯-πΏ)^{2}}}{π‘}\Biggr\},\\ {} & \displaystyle {\int _{\mathrm{R}}}\displaystyle \displaystyle \frac{\sqrt{π}}{\sqrt{π(π‘-π )}}\exp \Biggl\{-\displaystyle \frac{π{(π₯-π¦)^{2}}}{π‘-π }\Biggr\}\displaystyle \frac{\sqrt{π}}{\sqrt{ππ }}\exp \Biggl\{-\displaystyle \frac{π{(π¦+πΏ)^{2}}}{π }\Biggr\}\hspace{1.67pt}ππ¦\\ {} & \displaystyle \hspace{10.0pt}=\displaystyle \frac{\sqrt{π}}{\sqrt{ππ‘}}\exp \Biggl\{-\displaystyle \frac{π{(π₯+πΏ)^{2}}}{π‘}\Biggr\},\end{array}\]
therefore
\[{\int _{0}^{π‘}}{\int _{\mathrm{R}}}\mathcal{K}(π‘-π ,π₯-π¦){π_{πΏ,π,πΏ}}(π ,π¦)\hspace{1.67pt}ππ¦\hspace{1.67pt}ππ \le {πΆ_{π,πΏ}}(π)\exp \Biggl\{-\displaystyle \frac{π{(πΏ-|π₯|)^{2}}}{π‘}\Biggr\},\]
and (5.5) implies the result of our Theorem. β‘RemarkΒ 2.
Let SM π be such that $π((π,π])\le πΆ(π)$, $π,\hspace{2.5pt}π\in \mathrm{R}$. Then we can obtain (5.1) for arbitrary $πΏ\in {\mathrm{R}_{+}}$, i.e.
\[\Bigl|{π’^{πΏ}}(π‘,π₯)-π’(π‘,π₯)\Bigr|\le πΆ(π){πΆ_{π}}{π^{-\frac{π{(πΏ-|π₯|)^{2}}}{π‘}}},\]
where $πΆ(π)$ is independent of πΏ. This follows from Remark 1, because we used form $πΏ=ππΏ$ only in Lemma 3.RemarkΒ 3.
For the heat equation with a multiplicative noise driven by the Wiener process, under some assumptions, Theorem 3.1 [2] states that for all $(π‘,π₯)\in [0,π]\times {π·_{πΏ}}$, $π\ge 1$ holds
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle {\Bigl\|{π’^{πΏ}}(π‘,π₯)-π’(π‘,π₯)\Bigr\|_{{\mathrm{L}_{\mathrm{p}}}(\Omega )}}\\ {} & \displaystyle \hspace{10.0pt}\le πΆ\Biggl(1+\sup \limits_{π₯\in \mathrm{R}}\Bigl|{π’_{0}}(π₯)\Bigr|\Biggr)\Biggl(\exp \Biggl\{-\frac{{(πΏ-π₯)^{2}}}{8π‘}\Biggr\}+\exp \Biggl\{-\frac{{(πΏ+π₯)^{2}}}{8π‘}\Biggr\}\Biggr),\end{array}\]
where πΆ is independent of $πΏ,π₯,π‘,{π’_{0}}$. In our paper, we have a similar convergence rate for ${π’^{πΏ}}\to π’$.Also, note that Theorem 3.1 [2] includes the cases of the mixed boundary conditions and the Neumann boundary conditions for the equation.
