1 Introduction
Averaging methods are important for describing and investigating the asymptotic behavior of dynamical systems. Therefore, the theory of the averaging principle for stochastic differential equations is a fascinating modern topic and many mathematicians work quite actively within this field. For instance, a weak order in averaging for wave equations with ${L^{2}}$-valued Wiener processes is studied in [14]. Bao et al. [3] considered two-time-scale equations with α-stable noises. Strong and weak orders in averaging for stochastic partial differential equations with Wiener processes are given in [10, 11, 13].
The averaging principle for fractional differential equations driven by Lévy noise is established by Shen et al. [31]. Wang and Xu [34] investigated the stochastic averaging method for neutral stochastic delay equations driven by fractional Brownian motion with Hurst parameter $H\in (1/2,1)$. Other interesting examples of studying of averaging for stochastic differential equations can be found in papers [1, 12, 15, 18].
The averaging principle for equations driven by general stochastic measures is considered in [6, 22, 23, 26, 30].
Our aim here is to establish the averaging principle for the cable equation driven by stochastic measure studied in [25]. For this purpose we also improve the orders of the Hölder condition for the mild solution with respect to space and time variables obtained in [25, Theorem 5.1] (see Theorem 1 below).
Properties of the mild solutions to stochastic partial differential equations are studied in a number of papers. In particular, the existence and uniqueness of a solution for the class of non-autonomous parabolic stochastic partial differential equations defined on a bounded open subset $D\subset {\mathbb{R}^{d}}$ and driven by an ${L^{2}}(D)$-valued fractional Brownian motion with the Hurst index $H>1/2$ are proved in [28]. In [2] the ergodic property of the solution to a fractional stochastic heat equation is established. Wave equations with general stochastic measures and α-stable distributions are investigated in the papers [5, 7, 8, 24] and [20, 29, 19], respectively.
Let ${L_{0}}(\Omega ,\mathcal{F},\mathsf{P})$ be the set of all real-valued random variables defined on a complete probability space $(\Omega ,\mathcal{F},\mathsf{P})$, $\mathsf{X}$ be an arbitrary set and $\mathcal{B}(\mathsf{X})$ be a σ-algebra of Borel subsets of $\mathsf{X}$. Let μ be a stochastic measure on $\mathcal{B}(\mathsf{X})$, i.e. a σ-additive mapping $\mu :\hspace{0.1667em}\mathcal{B}(\mathsf{X})\to {L_{0}}(\Omega ,\mathcal{F},\mathsf{P})$. Such μ is also called a general stochastic measure (see, for example, [17, Section 7]). Examples of stochastic measures can be found in [17, 21, 30].
Consider the mild solution to the following equation:
where $(t,x)\in [0,T]\times [0,L]$, $T>0$, $L>0$, $\varepsilon >0$, and μ is a stochastic measure defined on the Borel σ-algebra $\mathcal{B}([0,L])$.
(1)
\[ \left\{\begin{array}{l}\frac{\partial {u_{\varepsilon }}(t,x)}{\partial t}=\frac{{\partial ^{2}}{u_{\varepsilon }}(t,x)}{\partial {x^{2}}}-{u_{\varepsilon }}(t,x)+\sigma (t/\varepsilon ,x)\hspace{0.1667em}\dot{\mu }(x),\\ {} {u_{\varepsilon }}(0,x)={u_{0}}(x),\hspace{1em}\frac{\partial {u_{\varepsilon }}(t,0)}{\partial x}=\frac{\partial {u_{\varepsilon }}(t,L)}{\partial x}=0,\end{array}\right.\]Let G be the fundamental solution of the homogeneous cable equation, that is
(see, for example, [33, p. 312] or [32, equality (5.69B)]). Then the mild solution of problem (1) is given by the formula
(2)
\[ G(t,x,y)=\frac{{e^{-t}}}{\sqrt{4\pi t}}{\sum \limits_{n=-\infty }^{\infty }}\left({e^{-\frac{{(y-x-2nL)^{2}}}{4t}}}+{e^{-\frac{{(y+x-2nL)^{2}}}{4t}}}\right)\](3)
\[ {u_{\varepsilon }}(t,x)=\hspace{-0.1667em}\hspace{-0.1667em}{\int _{0}^{L}}\hspace{-0.1667em}\hspace{-0.1667em}G(t,x,y){u_{0}}(y)\hspace{0.1667em}dy+{\int _{[0,L]}}\hspace{-0.1667em}d\mu (y)\hspace{-0.1667em}{\int _{0}^{t}}\hspace{-0.1667em}G(t-s,x,y)\sigma (s/\varepsilon ,y)\hspace{0.1667em}ds\hspace{0.1667em}.\]We study the convergence
\[ {u_{\varepsilon }}(t,x)\to \bar{u}(t,x)\hspace{0.1667em},\hspace{1em}\varepsilon \to 0\hspace{0.1667em},\]
where $\bar{u}(t,x)$ is the mild solution of the averaged equation, that is,
and
The rest of the paper is organized as follows. Section 2 contains some basic facts concerning the estimates of stochastic integrals with respect to general stochastic measures. In Section 3, we study the Hölder regularity of the mild solution of the cable equation with respect to the set of all variables. The averaging principle for the cable equation is established in Section 4.
2 Preliminaries
To prove the convergence of solutions, we will apply an estimate of a stochastic integral using the norm of the Besov space ${B_{22}^{\alpha }}([b,c])$, $\alpha \in (1/2,1)$, $b,c\in \mathbb{R}$ (see, for example, [16]). ${B_{22}^{\alpha }}([b,c])$ is the space of functions $g:[b,c]\to \mathbb{R}$ such that the norm
is finite. Here
Denote
Let Z be an arbitrary set and a function $g(y,z):[0,L]\times Z\to \mathbb{R}$ be such that $g(\operatorname{\mathbf{\cdot }},z)$ is continuous on $[0,L]$ for all $z\in Z$. Put
By [27, Lemma 3], the random function
has a version
\[\begin{aligned}{}\widetilde{\eta }(z)& ={\int _{[0,L]}}{g_{0}}(y,z)d\mu (y)\\ {} & +\sum \limits_{n\ge 1}\left({\int _{[0,L]}}{g_{n}}(y,z)d\mu (y)-{\int _{[0,L]}}{g_{n-1}}(y,z)d\mu (y)\right),\end{aligned}\]
such that, for all $\varepsilon >0$, $\omega \in \Omega $, $z\in Z$,
\[\begin{aligned}{}\left|\widetilde{\eta }(z)\right|& \le \left|g(0,z)\mu ([0,L])\right|\\ {} & +\hspace{-0.1667em}{\left\{\sum \limits_{n\ge 1}{2^{n\varepsilon }}\sum \limits_{1\le k\le {2^{n}}}{\left|g(k{2^{-n}}L,z)-g((k-1){2^{-n}}L,z)\right|^{2}}\right\}^{\frac{1}{2}}}\\ {} & \times {\left\{\sum \limits_{n\ge 1}{2^{-n\varepsilon }}\sum \limits_{1\le k\le {2^{n}}}{\left|\mu \left({\Delta _{kn}^{(L)}}\right)\right|^{2}}\right\}^{\frac{1}{2}}}.\end{aligned}\]
This version is the same for all $z\in Z$.According to [16, Theorem 1.2], [21, Lemma 3.2] and [9, inequality (6)], we have
where $\alpha =\varepsilon /2+1/2$, the constant C depends on $\alpha ,t$ and does not depend on $z,\omega $.
(7)
\[ \begin{aligned}{}\left|\widetilde{\eta }(z)\right|& \le |g(0,z)\mu ([0,L])|\\ {} & +C\| g(\operatorname{\mathbf{\cdot }},z){\| _{{B_{22}^{\alpha }}([0,L])}}{\left\{\sum \limits_{n\ge 1}{2^{-n\varepsilon }}\sum \limits_{1\le k\le {2^{n}}}{\left|\mu \left({\Delta _{kn}^{(L)}}\right)\right|^{2}}\right\}^{\frac{1}{2}}},\end{aligned}\]Here and in what follows the same symbol C denotes some positive constants that may be different in different places of the paper. The precise values of these constants are not important for our purposes.
3 Regularity of the mild solution of a cable equation
Regularity of the mild solution
of a cable equation driven by a general stochastic measure is studied in [25]. It was proved there that the paths of the solution are Hölder continuous. The following conditions was considered.
(8)
\[ u(t,x)=\hspace{-0.1667em}\hspace{-0.1667em}{\int _{0}^{L}}\hspace{-0.1667em}\hspace{-0.1667em}G(t,x,y){u_{0}}(y)\hspace{0.1667em}dy+{\int _{[0,L]}}\hspace{-0.1667em}d\mu (y)\hspace{-0.1667em}{\int _{0}^{t}}\hspace{-0.1667em}G(t-s,x,y)\sigma (s,y)\hspace{0.1667em}ds\hspace{0.1667em},\]Condition 1. The function $\sigma (s,y):[0,T]\times [0,L]\to \mathbb{R}$ has the derivative $\frac{{\partial ^{2}}\sigma }{\partial t\partial x}$, which is continuous with respect to the pair of arguments.
Condition 2. The function ${u_{0}}(y)={u_{0}}(y,\omega ):[0,L]\times \Omega \to \mathbb{R}$ is measurable and has the derivative $\frac{\partial {u_{0}}}{\partial y}$, which is continuous with respect to y and bounded for all fixed $\omega \in \Omega $.
By [25, Theorem 5.1], if Conditions 1–2 hold, then for all fixed $\delta >0$, ${\gamma _{2}}<1/18$, and ${\gamma _{1}}<1/6$, function (8) has a version $\tilde{u}(t,x)$ such that
for all ${t_{1}},{t_{2}}\in [\delta ,T]$, ${x_{1}},{x_{2}}\in [0,L]$ and for some ${L_{\tilde{u}}}(\omega )>0$.
(9)
\[ |\tilde{u}({t_{1}},{x_{1}})-\tilde{u}({t_{2}},{x_{2}})|\le {L_{\tilde{u}}}(\omega )\left(|{t_{1}}-{t_{2}}{|^{{\gamma _{2}}}}+|{x_{1}}-{x_{2}}{|^{{\gamma _{1}}}}\right),\]The proof of Theorem 5.1 ([25]) uses the Hölder regularity of the mild solution of a heat equation that is established in [21]. Restrictions ${\gamma _{1}}<1/6$ and ${\gamma _{2}}<1/18$ for relation (9) are due to the restrictions for the corresponding orders of the Hölder condition for the stochastic integrals from the heat equation (see [21, Lemma 5.1 and Lemma 6.1]). But now we can use the results of [4], which improve Hölder continuity orders obtained in [21]. Namely, the Hölder regularity with ${\gamma _{1}}<1/2$ and ${\gamma _{2}}<1/4$ for the mild solution of a parabolic equation with a stochastic measure is proved in [4, Lemma 1 and Lemma 2].
The following assumptions are used in the rest of the paper.
A1. The function ${u_{0}}(y)={u_{0}}(y,\omega ):[0,L]\times \Omega \to \mathbb{R}$ is measurable and Hölder continuous, that is
A2. $\sigma (s,y):{\mathbb{R}_{+}}\times [0,L]\to \mathbb{R}$ is measurable, bounded and Hölder continuous in $s\in {\mathbb{R}_{+}}$, $y\in [0,L]$, that is
\[\begin{aligned}{}|\sigma (s,y)|& \le {C_{\sigma }}\hspace{0.1667em},\\ {} |\sigma ({s_{1}},{y_{1}})-\sigma ({s_{2}},{y_{2}})|& \le {L_{\sigma }}\Big(|{s_{1}}-{s_{2}}{|^{\beta (\sigma )}}+|{y_{1}}-{y_{2}}{|^{\beta (\sigma )}}\Big)\hspace{0.1667em},\hspace{0.2778em}1/2<\beta (\sigma )<1.\end{aligned}\]
A3. Limit (5) exists and $\Sigma (r,y)={\textstyle\int _{0}^{t}}[\sigma (s,y)-\bar{\sigma }(y)]\hspace{0.1667em}ds$ is bounded for all $t\in {\mathbb{R}_{+}}$, $y\in [0,L]$.
Theorem 1.
Let Assumptions A1–A2 hold. Then there exists a version $\tilde{u}(t,x)$ of the random function $u(t,x)$ defined by (8) such that for all fixed $\delta >0$, ${\widetilde{\gamma }_{2}}<1/4\wedge (\beta (\sigma )-1/2)$ and ${\widetilde{\gamma }_{1}}<1-1/(2\beta (\sigma ))$, ${\widetilde{\gamma }_{1}}\le \beta ({u_{0}})$, we have
for all ${t_{1}},{t_{2}}\in [\delta ,T]$ and ${x_{1}},{x_{2}}\in [0,L]$ and for some constant ${C_{\tilde{u}}}(\omega )>0$.
(10)
\[ |\tilde{u}({t_{1}},{x_{1}})-\tilde{u}({t_{2}},{x_{2}})|\le {C_{\tilde{u}}}(\omega )\left(|{t_{1}}-{t_{2}}{|^{{\widetilde{\gamma }_{2}}}}+|{x_{1}}-{x_{2}}{|^{{\widetilde{\gamma }_{1}}}}\right),\]Proof.
The reasoning is the same as that used in [25] with some differences. We have divided the proof into 3 steps: the Hölder continuity of the stochastic integral with respect to the space variable; the Hölder continuity of the stochastic integral with respect to the time variable and the Hölder continuity of the function $u(t,x)$.
Step 1. By obtaining the Hölder condition for the stochastic integral with respect to the space variable we get the analogues of [25, inequalities (3.1) and (3.2)]. We use Assumption A2 instead of Condition 1.
For any fixed ${x_{1}},{x_{2}}\in [0,L]$, and for $y\in [0,L]$, $h\in [0,L-y]$, put
\[\begin{aligned}{}F& =|{f_{n}}(s,{x_{2}},y+h)-{f_{n}}(s,{x_{1}},y+h)-{f_{n}}(s,{x_{2}},y)+{f_{n}}(s,{x_{1}},y)|\\ {} & \le |{f_{n}}(s,{x_{2}},y+h)-{f_{n}}(s,{x_{1}},y+h)|+|{f_{n}}(s,{x_{2}},y)-{f_{n}}(s,{x_{1}},y)|\\ {} & ={F_{1}}+{F_{2}},\end{aligned}\]
where notation
\[ {f_{n}}(s,x,y)=\frac{{e^{-(t-s)}}}{\sqrt{4\pi (t-s)}}{e^{-\frac{{(y-x-2nL)^{2}}}{4(t-s)}}}\sigma (s,y)\]
is used (see [25]).Consider $n\ne 0,1$. By A2, we have
\[\begin{aligned}{}{F_{2}}& \le \frac{{C_{\sigma }}}{\sqrt{\pi }{(4(t-s))^{3/2}}}|{x_{1}}-{x_{2}}|2(|n|+1)L{e^{-\frac{\min \left\{{(y-{x_{1}}-2n)^{2}},{(y-{x_{2}}-2n)^{2}}\right\}{L^{2}}}{4(t-s)}}}\\ {} & \le \frac{C}{{(t-s)^{3/2}}}|{x_{1}}-{x_{2}}|\cdot |n|{e^{-\frac{{(2|n|-1)^{2}}{L^{2}}}{4(t-s)}}}.\end{aligned}\]
For details see the method used for obtaining bound (35) below. The same estimation holds for term ${F_{1}}$. Therefore,
On the other hand,
\[\begin{aligned}{}F& \le |{f_{n}}(s,{x_{1}},y+h)-{f_{n}}(s,{x_{1}},y)|+|{f_{n}}(s,{x_{2}},y+h)-{f_{n}}(s,{x_{2}},y)|\\ {} & ={\widetilde{F}_{1}}+{\widetilde{F}_{2}}.\end{aligned}\]
Since the function σ is bounded and Hölder continuous (by Assumption A2), and ${(t-s)^{-1/2}}\le T{(t-s)^{-3/2}}$, we obtain
\[\begin{aligned}{}{\widetilde{F}_{1}}& \le \frac{1}{\sqrt{4\pi (t-s)}}{e^{-\frac{{(y-{x_{1}}-2nL)^{2}}}{4(t-s)}}}|\sigma (s,y+h)-\sigma (s,y)|\\ {} & +\frac{1}{\sqrt{4\pi (t-s)}}\left|{e^{-\frac{{(y+h-{x_{1}}-2nL)^{2}}}{4(t-s)}}}-{e^{-\frac{{(y-{x_{1}}-2nL)^{2}}}{4(t-s)}}}\right||\sigma (s,y+h)|\\ {} & \le \frac{C}{{(t-s)^{3/2}}}\left(\hspace{-0.1667em}\hspace{-0.1667em}{L_{\sigma }}{h^{\beta (\sigma )}}\hspace{-0.1667em}+{C_{\sigma }}h2(|n|+1)L{e^{-\frac{\min \left\{{(y-{x_{1}}-2n)^{2}}\hspace{-0.1667em},{(y+h-{x_{1}}-2n)^{2}}\right\}{L^{2}}}{4(t-s)}}}\hspace{-0.1667em}\right)\\ {} & \le \frac{C}{{(t-s)^{3/2}}}{h^{\beta (\sigma )}}\cdot |n|{e^{-\frac{{(2|n|-1)^{2}}{L^{2}}}{4(t-s)}}}.\end{aligned}\]
The same estimate holds for term ${\widetilde{F}_{2}}$. Hence,
Raise the inequality (12) to the power ${\theta _{1}}$ and multiply by inequality (11) raised to the power $1-{\theta _{1}}$, for an arbitrary ${\theta _{1}}\in (0,1)$. We have
for all $n\ne 0$.
(13)
\[ F\le \frac{C}{{(t-s)^{3/2}}}|{x_{1}}-{x_{2}}{|^{1-{\theta _{1}}}}{h^{{\theta _{1}}\beta (\sigma )}}\cdot |n|{e^{-\frac{{(2|n|-1)^{2}}{L^{2}}}{4(t-s)}}},\]For the same reason,
for all $n\ne 0,1$.
(14)
\[ \begin{aligned}{}|{f_{n}}(s,-{x_{2}},y+h)& -{f_{n}}(s,-{x_{1}},y+h)-{f_{n}}(s,-{x_{2}},y)+{f_{n}}(s,-{x_{1}},y)|\\ {} & \le \frac{C}{{(t-s)^{3/2}}}|{x_{1}}-{x_{2}}{|^{1-{\theta _{1}}}}{h^{{\theta _{1}}\beta (\sigma )}}\cdot |n|{e^{-\frac{\min \left\{{(2n)^{2}},{(2n-2)^{2}}\right\}{L^{2}}}{4(t-s)}}},\end{aligned}\]Further, we can repeat the proof of Theorem 3.1 ([25]). Instead of using bounds (3.1) and (3.2) of [25], we can use inequalities (13) and (14) above. The integral from (6) is finite if ${\theta _{1}}\beta (\sigma )>1/2$. Thus we get the Hölder condition with respect to the space variable of the order
Besides, as mentioned above, we use [4, Lemma 1 and Lemma 2] instead of [21, Lemma 5.1 and Lemma 6.1] in the case of $n=0,1$.
Step 2. The reasoning is the same in the case of the Hölder property with respect to the time variable. Given x and ${t_{1}}<{t_{2}}$ use the notation of paper [25], that is
and
Moreover, denote
Now we consider the case of $n\ne 0,1$, since the case of $n=0,1$ is the same as in the [25] with reference to [4, Lemma 1 and Lemma 2] instead of [21, Lemma 5.1 and Lemma 6.1] respectively. Namely, we get Hölder continuity of the order 1/4 for these terms.
We use the change of variables $s\to s+{t_{2}}-{t_{1}}$ in the first integral of (15) and obtain
\[\begin{aligned}{}\bar{g}(y)& ={\int _{{t_{1}}-{t_{2}}}^{{t_{1}}}}{\bar{G}_{1}}({t_{1}},s,y)\sigma (s+{t_{2}}-{t_{1}},y)ds-{\int _{0}^{{t_{1}}}}{\bar{G}_{1}}({t_{1}},s,y)\sigma (s,y)ds\\ {} & ={\int _{{t_{1}}-{t_{2}}}^{0}}{\bar{G}^{(1)}}({t_{2}},s+{t_{2}}-{t_{1}},y)\sigma (s+{t_{2}}-{t_{1}},y)ds\\ {} & +{\int _{0}^{{t_{1}}}}{\bar{G}_{1}}({t_{1}},s,y)\left(\sigma (s+{t_{2}}-{t_{1}},y)-\sigma (s,y)\right)ds\\ {} & ={\bar{g}_{1}}(y)+{\bar{g}_{2}}(y).\end{aligned}\]
Similarly to (12), by A2, we have
and
Therefore,
where we used the estimate
for s in the domain of integration.
Then, by assumption A2, we get
On the other hand, the same reasoning as that used in obtaining bound (12) proves
Now we raise the inequality (19) to the power ${\theta _{2}}$ and multiply by inequality (18) raised to the power $1-{\theta _{2}}$, for an arbitrary ${\theta _{2}}\in (0,1)$. Thus we see that
Consequently, taking to consideration (17),
From this point we can repeat the proof of Theorem 4.1 ([25]). Instead of using (4.2) of [25], we use bounds (20) and (21) above. Since the integral from (6) is finite for ${\theta _{2}}\beta (\sigma )>1/2$,
Therefore, we obtain the Hölder condition with respect to the time variable of the order
where we also count the case $n=0,1$.
Step 3. The last part of the proof is analogous to the proof of Theorem 5.1 in [25]. The integral
satisfies condition (10). To prove the Hölder regularity with respect to the space variable we consider ${x_{1}}<{x_{2}}$ and denote
Then we use the change of variables $y\to y+{x_{2}}-{x_{1}}$ and $y\to y-{x_{2}}+{x_{1}}$ in the integrals involving ${x_{2}}$ and ${G_{-}}(t,x,y)$, ${G_{+}}(t,x,y)$ respectively. Thus
\[\begin{aligned}{}\Bigg|& \left.{\int _{0}^{L}}G(t,{x_{1}},y){u_{0}}(y)\hspace{0.1667em}dy-{\int _{0}^{L}}G(t,{x_{2}},y){u_{0}}(y)\hspace{0.1667em}dy\right|\\ {} & =\left|{\int _{0}^{L-{x_{2}}+{x_{1}}}}{G_{-}}(t,{x_{1}},y)\big({u_{0}}(y)-{u_{0}}(y+{x_{2}}-{x_{1}})\big)\hspace{0.1667em}dy\right.\\ {} & +{\int _{{x_{2}}-{x_{1}}}^{L}}{G_{+}}(t,{x_{1}},y)\big({u_{0}}(y)-{u_{0}}(y-{x_{2}}+{x_{1}})\big)\hspace{0.1667em}dy\\ {} & -{\int _{-{x_{2}}+{x_{1}}}^{0}}{G_{-}}(t,{x_{1}},y){u_{0}}(y+{x_{2}}-{x_{1}})\hspace{0.1667em}dy+{\int _{L-{x_{2}}+{x_{1}}}^{L}}{G_{-}}(t,{x_{1}},y){u_{0}}(y)\hspace{0.1667em}dy\\ {} & -\left.{\int _{L}^{L+{x_{2}}-{x_{1}}}}{G_{+}}(t,{x_{1}},y){u_{0}}(y-{x_{2}}+{x_{1}})\hspace{0.1667em}dy+{\int _{0}^{{x_{2}}-{x_{1}}}}{G_{+}}(t,{x_{1}},y){u_{0}}(y)\hspace{0.1667em}dy\right|\\ {} & \le {L_{{u_{0}}}}(\omega )|{x_{1}}-{x_{2}}{|^{\beta ({u_{0}})}}\hspace{-0.1667em}\left({\int _{0}^{L-{x_{2}}+{x_{1}}}}\hspace{-0.1667em}\hspace{-0.1667em}\hspace{-0.1667em}|{G_{-}}(t,{x_{1}},y)|dy+\hspace{-0.1667em}{\int _{{x_{2}}-{x_{1}}}^{L}}\hspace{-0.1667em}\hspace{-0.1667em}\hspace{-0.1667em}|{G_{+}}(t,{x_{1}},y)|dy\right)\\ {} & +C(\omega )|{x_{1}}-{x_{2}}|\underset{t\in [\delta ,L],\hspace{0.1667em}{x_{1}},y\in [0,L]}{\sup }|G(t,x,y)|\le C(\omega )|{x_{1}}-{x_{2}}{|^{\beta ({u_{0}})}},\end{aligned}\]
where we use the boundedness of the function ${u_{0}}$ due to its Hölder continuity on $[0,L]$.To prove the Hölder regularity with respect to the time variable we consider $\delta \le {t_{1}}<{t_{2}}\le T$. We get
\[\begin{array}{l}\displaystyle \begin{aligned}{}\big|{G_{+}}({t_{1}},x,y)& -{G_{+}}({t_{2}},x,y)\big|=\Bigg|\frac{{e^{-{t_{1}}}}}{\sqrt{4\pi {t_{1}}}}{\sum \limits_{n=-\infty }^{\infty }}{e^{-\frac{{(y+x-2nL)^{2}}}{4{t_{1}}}}}\\ {} & +\frac{{e^{-{t_{2}}}}}{\sqrt{4\pi {t_{2}}}}{\sum \limits_{n=-\infty }^{\infty }}{e^{-\frac{{(y+x-2nL)^{2}}}{4{t_{2}}}}}\Bigg|\\ {} & \le \frac{{e^{-{t_{1}}}}-{e^{-{t_{2}}}}}{\sqrt{4\pi {t_{1}}}}{\sum \limits_{n=-\infty }^{\infty }}{e^{-\frac{{(y+x-2nL)^{2}}}{4{t_{1}}}}}\end{aligned}\\ {} \displaystyle \begin{aligned}{}& +\frac{{e^{-{t_{2}}}}}{\sqrt{4\pi {t_{1}}}}{\sum \limits_{n=-\infty }^{\infty }}\Bigg|{e^{-\frac{{(y+x-2nL)^{2}}}{4{t_{1}}}}}-{e^{-\frac{{(y+x-2nL)^{2}}}{4{t_{2}}}}}\Bigg|\\ {} & +{e^{-{t_{2}}}}\left(\frac{1}{\sqrt{4\pi {t_{1}}}}-\frac{1}{\sqrt{4\pi {t_{2}}}}\right){\sum \limits_{n=-\infty }^{\infty }}{e^{-\frac{{(y+x-2nL)^{2}}}{4{t_{2}}}}}\\ {} & ={J_{1}}+{J_{2}}+{J_{3}}.\end{aligned}\end{array}\]
According to estimates (34), (25) and (27) from Section 4,
\[\begin{aligned}{}{J_{1}}& \le \frac{|{t_{1}}-{t_{2}}|}{\sqrt{4\pi {t_{1}}}}{\sum \limits_{n=-\infty }^{\infty }}{e^{-\frac{{(y+x-2nL)^{2}}}{4{t_{1}}}}}\\ {} & \le \frac{|{t_{1}}-{t_{2}}|}{\sqrt{4\pi \delta }}\left(\sum \limits_{n\ne 0}{e^{-\frac{\min \left\{{(2n)^{2}},\hspace{0.1667em}{(2n-2)^{2}}\right\}{L^{2}}}{4{t_{1}}}}}+1\right)\\ {} & \le C|{t_{1}}-{t_{2}}|.\end{aligned}\]
The similar arguments yield
\[\begin{aligned}{}{J_{2}}& \le \frac{1}{\sqrt{4\pi {t_{1}}}}{\sum \limits_{n=-\infty }^{\infty }}\Bigg|\frac{{(y+x-2nL)^{2}}}{4{t_{1}}}-\frac{{(y+x-2nL)^{2}}}{4{t_{2}}}\Bigg|\hspace{0.1667em}{e^{-\frac{{(y+x-2nL)^{2}}}{4{t_{1}}}}}\\ {} & \le C|{t_{1}}-{t_{2}}|{\sum \limits_{n=-\infty }^{\infty }}\frac{{(y+x-2nL)^{2}}}{4{t_{1}}}\hspace{0.1667em}{e^{-\frac{{(y+x-2nL)^{2}}}{4{t_{1}}}}}\\ {} & \le C|{t_{1}}-{t_{2}}|,\end{aligned}\]
where we use bound (30) (see Section 4 below).Finally,
\[\begin{aligned}{}{J_{3}}& \le \frac{C}{\sqrt{4\pi {t_{1}}{t_{2}}}}\left(\sqrt{{t_{2}}}-\sqrt{{t_{1}}}\right)\hspace{-0.1667em}{\sum \limits_{n=-\infty }^{\infty }}{e^{-\frac{{(y+x-2nL)^{2}}}{4{t_{2}}}}}\le \frac{C}{\delta \sqrt{4\pi }}\sqrt{{t_{2}}-{t_{1}}}\le C\sqrt{{t_{2}}-{t_{1}}}.\end{aligned}\]
In consequence,
Likewise, we get the same estimate for $\big|{G_{+}}({t_{1}},x,y)-{G_{+}}({t_{2}},x,y)\big|$. Hence,
4 Averaging principle
In this section we consider the random functions ${u_{\varepsilon }}$ and $\bar{u}$ given by equations (3) and (4).
Theorem 2.
Assume that Assumptions A1–A3 hold. Then there exist versions of ${u_{\varepsilon }}$ and $\bar{u}$ such that for any $\gamma <\frac{1}{2}\Big(1-\frac{1}{2\beta (\sigma )}\Big)$, we have
Proof.
By Theorem 1, if Assumptions A1–A2 hold, functions (3) and (4) have continuous in $(t,x)$ versions. We consider these versions.
Let $\varepsilon >0$, $t\in [0,T]$, $x\in [0,L]$ be fixed. Put
where the fundamental solution $G(t,x,y)$ is given by (2).
(22)
\[ g(t,x,y)={\int _{0}^{t}}G(t-s,x,y)[\sigma (s/\varepsilon ,y)-\bar{\sigma }(y)]\hspace{0.1667em}ds\hspace{0.1667em},\]Then
\[ |{u_{\varepsilon }}(t,x)-\bar{u}(t,x)|=\left|{\int _{[0,L]}}g(t,x,y)d\mu (y)\right|\hspace{0.1667em}.\]
With the aim of using inequality (7), we estimate $|g(t,x,y)|,\hspace{0.1667em}y\in [0,L]$, and $|g(t,x,y+h)-g(t,x,y)|,\hspace{0.1667em}h\in [0,L-y]$, in terms of some positive powers of h and ε.Denote
and
\[ {\Sigma _{\varepsilon }}(r)={\int _{0}^{r}}[\sigma (t/\varepsilon -\tau ,y)-\bar{\sigma }(y)]\hspace{0.1667em}d\tau \hspace{0.1667em},\hspace{1em}0\le r\le t/\varepsilon \hspace{0.1667em}.\]
Then $|{\Sigma _{\varepsilon }}(r)|\le {C_{\Sigma }}$, where ${C_{\Sigma }}$ does not depend on ε, while ${\textstyle\int _{0}^{t}}[\sigma (s,y)-\bar{\sigma }(y)]\hspace{0.1667em}ds$ is bounded by A3.In the integral from (22) we substitute $\tau =(t-s)/\varepsilon $ and obtain
\[\begin{aligned}{}|g(t,x,y)|& =\varepsilon \left|{\int _{0}^{t/\varepsilon }}G(\tau \varepsilon ,x,y)[\sigma (t/\varepsilon -\tau ,y)-\bar{\sigma }(y)]\hspace{0.1667em}d\tau \right|\\ {} & =\sqrt{\varepsilon }\bigg|{\int _{0}^{t/\varepsilon }}\frac{{e^{-\tau \varepsilon }}}{\sqrt{4\pi \tau }}{\sum \limits_{n=-\infty }^{\infty }}\left({e^{-\frac{{(y-x-2nL)^{2}}}{4\tau \varepsilon }}}+{e^{-\frac{{(y+x-2nL)^{2}}}{4\tau \varepsilon }}}\right)\\ {} & \times [\sigma (t/\varepsilon -\tau ,y)-\bar{\sigma }(y)]\hspace{0.1667em}d\tau \bigg|\\ {} & \le \frac{\sqrt{\varepsilon }}{\sqrt{4\pi }}\left|{\sum \limits_{n=-\infty }^{\infty }}{\int _{0}^{t/\varepsilon }}\frac{{e^{-\frac{{D_{n-}}}{\tau \varepsilon }}}}{\sqrt{\tau }}[\sigma (t/\varepsilon -\tau ,y)-\bar{\sigma }(y)]\hspace{0.1667em}d\tau \right.\\ {} & +\left.{\sum \limits_{n=-\infty }^{\infty }}{\int _{0}^{t/\varepsilon }}\frac{{e^{-\frac{{D_{n+}}}{\tau \varepsilon }}}}{\sqrt{\tau }}[\sigma (t/\varepsilon -\tau ,y)-\bar{\sigma }(y)]\hspace{0.1667em}d\tau \right|\hspace{0.1667em}.\end{aligned}\]
We have
and
(23)
\[ \begin{aligned}{}\left|{\sum \limits_{n=-\infty }^{\infty }}{\int _{0}^{1}}\frac{{e^{-\frac{{D_{n\pm }}}{\tau \varepsilon }}}}{\sqrt{\tau }}[\sigma (t/\varepsilon -\tau ,\right.& \left.y)-\bar{\sigma }(y)]\hspace{0.1667em}d\tau \right|\le 2{C_{\sigma }}{\int _{0}^{1}}{\sum \limits_{n=-\infty }^{\infty }}\frac{{e^{-\frac{{D_{n\pm }}}{\tau \varepsilon }}}}{\sqrt{\tau }}\hspace{0.1667em}d\tau \\ {} & \le 4{C_{\sigma }}\underset{\tau \le 1}{\sup }{\sum \limits_{n=-\infty }^{\infty }}{e^{-\frac{{D_{n\pm }}}{\tau \varepsilon }}}=4{C_{\sigma }}{\sum \limits_{n=-\infty }^{\infty }}{e^{-\frac{{D_{n\pm }}}{\varepsilon }}}\hspace{0.1667em}\end{aligned}\](24)
\[ \begin{aligned}{}\Bigg|& \left.{\sum \limits_{n=-\infty }^{\infty }}{\int _{1}^{t/\varepsilon }}\frac{{e^{-\frac{{D_{n\pm }}}{\tau \varepsilon }}}}{\sqrt{\tau }}[\sigma (t/\varepsilon -\tau ,y)-\bar{\sigma }(y)]\hspace{0.1667em}d\tau \right|=\left|{\sum \limits_{n=-\infty }^{\infty }}{\int _{1}^{t/\varepsilon }}\frac{{e^{-\frac{{D_{n\pm }}}{\tau \varepsilon }}}}{\sqrt{\tau }}\hspace{0.1667em}d{\Sigma _{\varepsilon }}(\tau )\right|\\ {} & =\left|{\sum \limits_{n=-\infty }^{\infty }}\frac{{e^{-\frac{{D_{n\pm }}}{\tau \varepsilon }}}}{\sqrt{\tau }}\hspace{0.1667em}{\Sigma _{\varepsilon }}(\tau ){\Big|_{1}^{t/\varepsilon }}-{\sum \limits_{n=-\infty }^{\infty }}{\int _{1}^{t/\varepsilon }}{\Sigma _{\varepsilon }}(\tau ){e^{-\frac{{D_{n\pm }}}{\tau \varepsilon }}}\left(\frac{{D_{n\pm }}}{\varepsilon \sqrt{{\tau ^{5}}}}-\frac{1}{2\sqrt{{\tau ^{3}}}}\right)\hspace{0.1667em}d\tau \right|\\ {} & \le 2{C_{\Sigma }}\underset{1\le \tau \le t/\varepsilon }{\sup }{\sum \limits_{n=-\infty }^{\infty }}{e^{-\frac{{D_{n\pm }}}{\tau \varepsilon }}}+{C_{\Sigma }}{\sum \limits_{n=-\infty }^{\infty }}{\int _{1}^{t/\varepsilon }}\frac{{D_{n\pm }}{e^{-\frac{{D_{n\pm }}}{\tau \varepsilon }}}}{\varepsilon \sqrt{{\tau ^{5}}}}d\tau \\ {} & \le 2{C_{\Sigma }}{\sum \limits_{n=-\infty }^{\infty }}{e^{-\frac{{D_{n\pm }}}{T}}}+{C_{\Sigma }}{\int _{1}^{t/\varepsilon }}\frac{{\textstyle\textstyle\sum _{n=-\infty }^{\infty }}{D_{n\pm }}{e^{-\frac{{D_{n\pm }}}{\tau \varepsilon }}}}{\varepsilon \sqrt{{\tau ^{5}}}}d\tau \hspace{0.1667em}.\end{aligned}\]Now we estimate ${e^{-\frac{{D_{n\pm }}}{s}}}\hspace{0.1667em},\hspace{0.1667em}s\in [0,T]$. For $n=0$,
Consider $n\ne 0$. In this case
Since
and
(26)
\[ \sum \limits_{n\ge 0}{e^{-{n^{2}}a}}\le \sum \limits_{n\ge 0}{e^{-na}}={(1-{e^{-a}})^{-1}}\le {(1-{e^{-b}})^{-1}},\hspace{1em}a\ge b>0,\]
\[ \sum \limits_{n\ge 1}{e^{-{n^{2}}a}}\le {e^{-a}}{(1-{e^{-a}})^{-1}}\le {e^{-b}}{(1-{e^{-b}})^{-1}},\hspace{1em}a\ge b>0,\]
we conclude that, for $s\in [0,T]$,
\[\begin{aligned}{}{\sum \limits_{n\ne 0,n=-\infty }^{\infty }}{e^{-\frac{{D_{n-}}}{s}}}& \le {\sum \limits_{n\ne 0,n=-\infty }^{\infty }}{e^{-\frac{{(2|n|-1)^{2}}{L^{2}}}{4s}}}\le 2{e^{-\frac{{L^{2}}}{4s}}}\sum \limits_{n\ge 0}{e^{-\frac{{n^{2}}{L^{2}}}{s}}}\\ {} & \le 2{e^{-\frac{{L^{2}}}{4s}}}{\left(1-{e^{-\frac{{L^{2}}}{s}}}\right)^{-1}}\le 2{\left(1-{e^{-\frac{{L^{2}}}{T}}}\right)^{-1}}.\end{aligned}\]
Analogously,
(27)
\[ \begin{aligned}{}{\sum \limits_{n\ne 0,n=-\infty }^{\infty }}{e^{-\frac{{D_{n+}}}{s}}}& \le {\sum \limits_{n\ne 0,n=-\infty }^{\infty }}{e^{-\frac{\min \left\{{(2n)^{2}},\hspace{0.1667em}{(2n-2)^{2}}\right\}{L^{2}}}{4s}}}\\ {} & =\sum \limits_{n\ge 1}{e^{-\frac{{(2n)^{2}}{L^{2}}}{4s}}}+\sum \limits_{n\ge 1}{e^{-\frac{{(2n-2)^{2}}{L^{2}}}{4s}}}\le 2{\left(1-{e^{-\frac{{L^{2}}}{T}}}\right)^{-1}}.\end{aligned}\]Substitute obtained bounds to (23) and (24). We thus get
\[ \left|{\sum \limits_{n=-\infty }^{\infty }}{\int _{0}^{1}}\frac{{e^{-\frac{{D_{n\pm }}}{\tau \varepsilon }}}}{\sqrt{\tau }}[\sigma (t/\varepsilon -\tau ,y)-\bar{\sigma }(y)]\hspace{0.1667em}d\tau \right|\le 4{C_{\sigma }}\left(1+2{\left(1-{e^{-\frac{{L^{2}}}{T}}}\right)^{-1}}\right)\]
and
\[\begin{aligned}{}\left|{\sum \limits_{n=-\infty }^{\infty }}\right.& \left.{\int _{1}^{t/\varepsilon }}\frac{{e^{-\frac{{D_{n\pm }}}{\tau \varepsilon }}}}{\sqrt{\tau }}[\sigma (t/\varepsilon -\tau ,y)-\bar{\sigma }(y)]\hspace{0.1667em}d\tau \right|\\ {} & \le 2{C_{\Sigma }}\left(1+2{\left(1-{e^{-\frac{{L^{2}}}{T}}}\right)^{-1}}\right)+{C_{\Sigma }}{\int _{1}^{t/\varepsilon }}\frac{{\textstyle\textstyle\sum _{n=-\infty }^{\infty }}{D_{n\pm }}{e^{-\frac{{D_{n\pm }}}{\tau \varepsilon }}}}{\varepsilon \sqrt{{\tau ^{5}}}}d\tau \hspace{0.1667em}.\end{aligned}\]
Hence,
\[\begin{aligned}{}|& g(t,x,y)|\\ {} & \le C\sqrt{\varepsilon }\Bigg[C\hspace{-0.1667em}+\hspace{-0.1667em}\bigg|\hspace{-0.1667em}{\int _{1}^{t/\varepsilon }}\hspace{-0.1667em}\frac{{\textstyle\textstyle\sum _{n=-\infty }^{\infty }}{D_{n-}}{e^{-\frac{{D_{n-}}}{\tau \varepsilon }}}}{\varepsilon \sqrt{{\tau ^{5}}}}d\tau \bigg|\hspace{-0.1667em}+\hspace{-0.1667em}\bigg|\hspace{-0.1667em}{\int _{1}^{t/\varepsilon }}\hspace{-0.1667em}\frac{{\textstyle\textstyle\sum _{n=-\infty }^{\infty }}{D_{n+}}{e^{-\frac{{D_{n+}}}{\tau \varepsilon }}}}{\varepsilon \sqrt{{\tau ^{5}}}}d\tau \hspace{0.1667em}\bigg|\Bigg].\end{aligned}\]
Next we estimate the integrals from the last relation. For $|y|\le L$ and $|x|\le L$ it is clear that
Therefore,
\[\begin{aligned}{}{\sum \limits_{n=-\infty }^{\infty }}{D_{n-}}{e^{-\frac{{D_{n-}}}{\tau \varepsilon }}}& ={D_{0-}}{e^{-\frac{{D_{0-}}}{\tau \varepsilon }}}+{\sum \limits_{n\ne 0,n=-\infty }^{\infty }}{D_{n-}}{e^{-\frac{{D_{n-}}}{\tau \varepsilon }}}\\ {} & \le {D_{0-}}{e^{-\frac{{D_{0-}}}{\tau \varepsilon }}}+2{L^{2}}{e^{-\frac{{L^{2}}}{4\tau \varepsilon }}}\sum \limits_{n\ge 0}{(n+2)^{2}}{e^{-\frac{n{L^{2}}}{\tau \varepsilon }}}\\ {} & \le {D_{0-}}{e^{-\frac{{D_{0-}}}{\tau \varepsilon }}}+C{L^{2}}{e^{-\frac{{L^{2}}}{4\tau \varepsilon }}}{(1-{e^{-\frac{{L^{2}}}{T}}})^{-1}}.\end{aligned}\]
where we use the relations
\[\begin{aligned}{}\sum \limits_{n\ge 1}{n^{2}}{e^{-na}}& =({e^{-a}}+{e^{-2a}}){(1-{e^{-a}})^{-3}}\le 2{e^{-a}}{(1-{e^{-a}})^{-3}}\le 2{(1-{e^{-a}})^{-1}},\\ {} \sum \limits_{n\ge 1}n{e^{-na}}& \le \sum \limits_{n\ge 1}{n^{2}}{e^{-na}}\le 2{(1-{e^{-a}})^{-1}},\hspace{1em}a>0\hspace{0.1667em},\end{aligned}\]
and estimate (26). Then, due to the fact that $\underset{D>0}{\max }D{e^{-\frac{D}{\tau \varepsilon }}}=\tau \varepsilon {e^{-1}}$, we have
Consequently,
In the same way we obtain
which yields
(30)
\[ \begin{aligned}{}{\sum \limits_{n=-\infty }^{\infty }}{D_{n+}}\hspace{0.1667em}& {e^{-\frac{{D_{n+}}}{\tau \varepsilon }}}={D_{0+}}{e^{-\frac{{D_{0+}}}{\tau \varepsilon }}}+{D_{1+}}{e^{-\frac{{D_{1+}}}{\tau \varepsilon }}}+{\sum \limits_{n\ne 0,1,\hspace{0.1667em}n=-\infty }^{\infty }}{D_{n+}}{e^{-\frac{{D_{n+}}}{\tau \varepsilon }}}\\ {} & \le 2\tau \varepsilon {e^{-1}}+{L^{2}}\sum \limits_{n\ge 1}{(n+1)^{2}}{e^{-\frac{{(2n)^{2}}{L^{2}}}{4\tau \varepsilon }}}+{L^{2}}\sum \limits_{n\ge 2}{(n+1)^{2}}{e^{-\frac{{(2n-2)^{2}}{L^{2}}}{4\tau \varepsilon }}}\\ {} & \le C\tau \varepsilon {(1-{e^{-\frac{{L^{2}}}{T}}})^{-1}}\le C\tau \varepsilon \hspace{0.1667em},\end{aligned}\]Therefore, we get
and so
Now we estimate $|g(t,x,y+h)-g(t,x,y)|$ and then ${w_{2,[0,L]}^{2}}(g,r)$ in terms of some positive power of h. Thus,
\[\begin{aligned}{}|g(t,& x,y+h)-g(t,x,y)|\\ {} & =\left|{\int _{0}^{t}}\left(G(t-s,x,y+h)-G(t-s,x,y)\right)[\sigma (s/\varepsilon ,y+h)-\bar{\sigma }(y+h)]\hspace{0.1667em}ds\right.\\ {} & +\left.{\int _{0}^{t}}G(t-s,x,y)\left([\sigma (s/\varepsilon ,y+h)-\sigma (s/\varepsilon ,y)]-[\bar{\sigma }(y+h)-\bar{\sigma }(y)]\right)\hspace{0.1667em}ds\right|\\ {} & =|{I_{1}}+{I_{2}}|\hspace{0.1667em},\end{aligned}\]
and then
By condition A3 and finiteness of ${\textstyle\sum _{n=-\infty }^{\infty }}{e^{-\frac{{D_{n\pm }}}{s}}}$, we obtain
(33)
\[ \begin{aligned}{}|{I_{2}}|\le & C2{L_{\sigma }}{h^{\beta (\sigma )}}{\int _{0}^{t}}\frac{{e^{-(t-s)}}}{\sqrt{4\pi (t-s)}}\hspace{0.1667em}ds\le C2{L_{\sigma }}{h^{\beta (\sigma )}}\sqrt{t}\le C{h^{\beta (\sigma )}}\hspace{0.1667em},\\ {} {\int _{0}^{L-h}}{I_{2}^{2}}\hspace{0.1667em}dy& \le C{h^{2\beta (\sigma )}}\hspace{0.1667em}.\end{aligned}\]Denote
Recall that here $h\in [0,L]$ and we consider the integrals in (32) such that $y+h\le L$. So
\[\begin{aligned}{}|{I_{1}}|& =\left|{\int _{0}^{t}}\frac{{e^{-(t-s)}}}{\sqrt{4\pi (t-s)}}{\sum \limits_{n=-\infty }^{\infty }}\left({e^{-\frac{{D_{n-}^{h}}}{t-s}}}-{e^{-\frac{{D_{n-}}}{t-s}}}\right)[\sigma (s/\varepsilon ,y+h)-\bar{\sigma }(y+h)]ds\right.\\ {} & +\left.{\int _{0}^{t}}\frac{{e^{-(t-s)}}}{\sqrt{4\pi (t-s)}}{\sum \limits_{n=-\infty }^{\infty }}\left({e^{-\frac{{D_{n+}^{h}}}{t-s}}}-{e^{-\frac{{D_{n+}}}{t-s}}}\right)[\sigma (s/\varepsilon ,y+h)-\bar{\sigma }(y+h)]\hspace{0.1667em}ds\right|\\ {} & =|{I_{1-}}+{I_{1+}}|\hspace{0.1667em}.\end{aligned}\]
Since ${e^{-(t-s)}}\le 1$, for $t-s\ge 0$, we conclude by Assumption A2 that
\[\begin{aligned}{}|{I_{1-}}|& \le 2{C_{\sigma }}\left|{\int _{0}^{t}}\frac{1}{\sqrt{4\pi (t-s)}}{\sum \limits_{n=-\infty }^{\infty }}\left({e^{-\frac{{D_{n-}^{h}}}{t-s}}}-{e^{-\frac{{D_{n-}}}{t-s}}}\right)\hspace{0.1667em}ds\right|\\ {} & =2{C_{\sigma }}\Bigg|\sum \limits_{n=0,\pm 1}{\int _{0}^{t}}\frac{1}{\sqrt{4\pi (t-s)}}\left({e^{-\frac{{D_{n-}^{h}}}{t-s}}}-{e^{-\frac{{D_{n-}}}{t-s}}}\right)ds\\ {} & +{\int _{0}^{t}}\frac{1}{\sqrt{4\pi (t-s)}}\sum \limits_{|n|\ge 2}\left({e^{-\frac{{D_{n-}^{h}}}{t-s}}}-{e^{-\frac{{D_{n-}}}{t-s}}}\right)ds\Bigg|.\end{aligned}\]
For $n=0,\pm 1$, the values of ${D_{n-}^{h}}$ and ${D_{n-}}$ are bounded by $4{L^{2}}$. Thus, the same reasoning as that used in obtaining the bound of ${I_{2}}$ in [22, inequality (13)] proves
Consider the second integral. We have
where we use the following relation
Hence,
(35)
\[ \begin{aligned}{}\left|{\int _{0}^{t}}\frac{1}{\sqrt{4\pi (t-s)}}\right.& \left.\sum \limits_{|n|\ge 2}\left({e^{-\frac{{D_{n-}^{h}}}{t-s}}}-{e^{-\frac{{D_{n-}}}{t-s}}}\right)ds\right|\\ {} & \le Ch{\int _{0}^{t}}\frac{1}{4{(t-s)^{3/2}}}\sum \limits_{n\ge 2}{(1+n)^{2}}{L^{2}}{e^{-\frac{{(2n-1)^{2}}{L^{2}}}{4(t-s)}}}\hspace{0.1667em}ds.\end{aligned}\]Then we use the same reasoning as in getting (29) and have
It follows that
and finally that
The same estimate holds for term $|{I_{1+}}|\le C{h^{\beta (\sigma )}}$. Hence,
Taking into account (33), we deduce that
Raise the latter inequality to the power θ and multiply by inequality (31) raised to the power $1-\theta $, for an arbitrary $\theta \in (0,1)$. We have
Thus, for any $\gamma <\frac{1}{2}\Big(1-\frac{1}{2\beta (\sigma )}\Big)$, there exists $\alpha <\theta \beta (\sigma )$ such that
Consequently, by relation (7),
\[\begin{aligned}{}|{u_{\varepsilon }}(& t,x)-\bar{u}(t,x)|=\left|{\int _{[0,L]}}g(t,x,y)d\mu (y)\right|\\ {} & \le |g(0,z)\mu ([0,L])|+C\| g(\operatorname{\mathbf{\cdot }},z){\| _{{B_{22}^{\alpha }}([0,L])}}{\left\{\sum \limits_{n\ge 1}{2^{-n\varepsilon }}\sum \limits_{1\le k\le {2^{n}}}{\left|\mu \left({\Delta _{kn}^{(L)}}\right)\right|^{2}}\right\}^{\frac{1}{2}}}\\ {} & \le C{\varepsilon ^{\gamma }}\left(|\mu ([0,L])|+{\left\{\sum \limits_{n\ge 1}{2^{-n\varepsilon }}\sum \limits_{1\le k\le {2^{n}}}{\left|\mu \left({\Delta _{kn}^{(L)}}\right)\right|^{2}}\right\}^{\frac{1}{2}}}\right)\le C(\omega ){\varepsilon ^{\gamma }}\hspace{0.1667em},\end{aligned}\]
where the sum with the stochastic measure is finite in view of [21, Lemma 3.1], and the constant $C(\omega )$ does not depend on $t,x$ and ε. □
