Averaging Principle for Multiscale Stochastic Fractional Schrödinger–Korteweg-de Vries System

Abstract

This work concerns the averaging principle for multiscale stochastic fractional Schrödinger–Korteweg-de Vries system. Stochastic averaging principle is a powerful tool for studying qualitative analysis of stochastic dynamical systems with different time-scales. To be more precise, under suitable conditions, the multiscale system can be reduced to a single stochastic fractional Korteweg-de Vries equation (averaged equation) with a modified coefficient, the slow component of multiscale system towards to the solution of the averaged equation in moment. In order to prove the averaging principle, there are two key points: uniform bounds for the solution and Hölder continuity of time variable for the slow component. For this, we need a crucial property—smoothing effect of the fractional Korteweg-de Vries semigroup, but the traditional method can’t help us obtain this property. Such analysis does not follow in a straightforward manner from results already available in the mathematical literature. On the contrary, it requires the introduction of some new ideas and techniques. We try to overcome this difficulty by the duality method, the interpolation arguments, the energy estimate method and refined inequality technique.

Appendix

1.1 Proof of Proposition 2.1

Proof

It is easy to see $v^{u,X}$ satisfies

$$\begin{aligned} \begin{array}{l} \left\{ \begin{array}{lll} dv=[-i(-\partial _{xx})^{\theta }v-\lambda v-ig(u,v)]dt-i\sigma _{2}(u,v)dB_{2}, \\ v(x,0)=X(x) \end{array} \right. \end{array} \begin{array}{lll} \text {in}~{\mathbb {T}}\times [0,T], \\ \text {in}~{\mathbb {T}}. \end{array} \end{aligned}$$

(1) $\bullet$ For the sake of simplicity, and $v^{u,X}$ is denoted by v. By applying the generalized Itô formula with $\frac{1}{2}\Vert v^{u,X}\Vert ^{2},$ we can obtain that

$$\begin{aligned} \begin{aligned} \Vert v\Vert ^{2} =&\Vert X\Vert ^{2}+2\int _{0}^{t}(v,-i(-\partial _{xx})^{\beta }v-\lambda v-ig(u,v))ds \\&+2\int _{0}^{t}(v,i\sigma _{2}(u,v))dB_{2}+\int _{0}^{t} \Vert \sigma _{2}(u,v)\Vert ^{2}ds \\ =&\Vert X\Vert ^{2}-2\lambda \int _{0}^{t}\Vert v\Vert ^{2}ds-2\int _{0}^{t}(v,ig(u,v))ds\\&+2\int _{0}^{t}(v,i\sigma _{2}(u,v))dB_{2}+\int _{0}^{t}\Vert \sigma _{2}(u,v)\Vert ^{2}ds. \end{aligned} \end{aligned}$$

Taking mathematical expectation from both sides of above equation, we have

$$\begin{aligned} {\mathbb {E}}\Vert v\Vert ^{2} =\Vert X\Vert ^{2}-2\lambda {\mathbb {E}}\int _{0}^{t}\Vert v\Vert ^{2}ds -2{\mathbb {E}}\int _{0}^{t}(v,ig(u,v))ds+{\mathbb {E}} \int _{0}^{t}\Vert \sigma _{2}(u,v)\Vert ^{2}ds, \end{aligned}$$

namely, $\frac{d}{dt}{\mathbb {E}}\Vert v\Vert ^{2}=-2\lambda {\mathbb {E}}\Vert v\Vert ^{2}-2{\mathbb {E}}(v,ig(u,v))+{\mathbb {E}}\Vert \sigma _{2}(u,v)\Vert ^{2}.$ With the help of the Young’s inequality (Lemma 2.3) and choosing the suitable $\eta ,$ we can obtain

$$\begin{aligned} \frac{d}{dt}{\mathbb {E}}\Vert v^{u,X}\Vert ^{2} \le -2\lambda {\mathbb {E}}\Vert v^{u,X}\Vert ^{2}+3L_{g}{\mathbb {E}}\Vert v^{u,X}\Vert ^{2}+C\Vert u\Vert ^{2}+C . \end{aligned}$$

Thus $\frac{d}{dt}{\mathbb {E}}\Vert v^{u,X}\Vert ^{2} \le -\lambda {\mathbb {E}}\Vert v^{u,X}\Vert ^{2}+C\Vert u\Vert ^{2}+C.$ Hence, by applying Lemma 2.4 with ${\mathbb {E}}\Vert v^{u,X}(t)\Vert ^{2}$, we have

$$\begin{aligned} {\mathbb {E}}\Vert v^{u,X}(t)\Vert ^{2}\le e^{-2\lambda t}\Vert X\Vert ^{2}+C(\Vert u\Vert ^{2}+1). \end{aligned}$$

$\bullet$ It follows from the energy method that

$$\begin{aligned} \begin{aligned}&{\mathbb {E}}\Vert (v^{u,X}-v^{u,Y})(t)\Vert ^{2} \\&\quad =\Vert X-Y\Vert ^{2}+2{\mathbb {E}}\int _{k\delta }^{t}(v^{u,X}-v^{u,Y}, -i(-\partial _{xx})^{\theta }(v^{u,X}-v^{u,Y}) \\&\qquad -\lambda v^{u,X}+\lambda v^{u,Y}-ig(u,v^{u,X})+ig(u,v^{u,Y}))ds \\&\qquad +{\mathbb {E}}\int _{0}^{t} \Vert \sigma _{2}(u,v^{u,X})-\sigma _{2}(u,v^{u,Y})\Vert ^{2}ds , \end{aligned} \end{aligned}$$

namely,

$$\begin{aligned} \begin{aligned}&\frac{d}{dt}{\mathbb {E}}\Vert v^{u,X}-v^{u,Y}\Vert ^{2}\\&\quad =2{\mathbb {E}}(v^{u,X}-v^{u,Y},-i(-\partial _{xx})^{\theta }(v^{u,X}-v^{u,Y}) -\lambda v^{u,X}+\lambda v^{u,Y}) \\&\qquad +2{\mathbb {E}}(v^{u,X}-v^{u,Y},-ig(u,v^{u,X})+ig(u,v^{u,Y})) +{\mathbb {E}}\Vert \sigma _{2}(u,v^{u,X})-\sigma _{2}(u,v^{u,Y})\Vert ^{2}\\&\quad \le -2\lambda {\mathbb {E}}\Vert v^{u,X}-v^{u,Y}\Vert ^{2}+2L_{g} {\mathbb {E}}\Vert v^{u,X}-v^{u,Y}\Vert ^{2}+L_{\sigma _{2}}^{2}{\mathbb {E}}\Vert v^{u,X}-v^{u,Y}\Vert ^{2}\\&\quad = (-2\lambda +2L_{g}+L_{\sigma _{2}}^{2}){\mathbb {E}}\Vert v^{u,X}-v^{u,Y}\Vert ^{2} . \end{aligned} \end{aligned}$$

It follows from assumptions (H) that

$$\begin{aligned} \frac{d}{dt}{\mathbb {E}}\Vert v^{u,X}-v^{u,Y}\Vert ^{2} \le -\lambda {\mathbb {E}}\Vert v^{u,X}-v^{u,Y}\Vert ^{2}, \end{aligned}$$

this yields

$$\begin{aligned} {\mathbb {E}}\Vert v^{u,X}-v^{u,Y}\Vert ^{2}\le \Vert X-Y\Vert ^{2}e^{-\lambda t}. \end{aligned}$$

(2) We shall adopt the remote start method, which is also called a dissipative method, to show the existence of an invariant measure of (4.3). Let v(t, s, x) be the solution to

$$\begin{aligned} \begin{array}{l} \left\{ \begin{array}{lll} dv=[-i(-\partial _{xx})^{\beta }v-\lambda v-ig(u,v)]dt-i\sigma _{2}(u,v)dB_{2}, \\ v(s)=x, \end{array} \right. \end{array} \begin{array}{lll} \text {in}~{\mathbb {T}}\times [s,T],\\ \text {in}~{\mathbb {T}}, \end{array} \end{aligned}$$

by the same method as in (2.2), we can obtain

$$\begin{aligned} {\mathbb {E}}\Vert v(t,s_{1},x)-v(t,s_{2},x)\Vert ^{2}\le e^{-\lambda (t-s_{2})}{\mathbb {E}}\Vert v(s_{2},s_{1},x)-x\Vert ^{2}, \end{aligned}$$

where $s_{1}\le s_{2}\le t,$ note the fact ${\mathbb {E}}\Vert v(t,s_{1},x)\Vert ^{2}\le C(1+\Vert x\Vert ^{2}),$ it holds that

$$\begin{aligned} {\mathbb {E}}\Vert v(t,s_{1},x)-v(t,s_{2},x)\Vert ^{2}\le C(1+\Vert x\Vert ^{2})e^{-\lambda (t-s_{2})}. \end{aligned}$$

Letting $s_{2}$(hence $s_{1}$) tend to $-\infty ,$ it follows that there exists $\xi ^{u}(t)\in L^{2}(\Omega ;L^{2}({\mathbb {T}}))$ such that

$$\begin{aligned} \lim \limits _{s\rightarrow -\infty }v(t,s,x)=\xi ^{u}(t). \end{aligned}$$

It follows from (2.2) that $\xi ^{u}(t)$ is independent of $x\in L^{2}({\mathbb {T}}).$ For any $u\in L^{2}({\mathbb {T}}),$ let $\mu ^{u}={\mathcal {L}}(\xi ^{u}(0)),$ then it is unique invariant measure $\mu ^{u}$ for the Markov semigroup $P_{t}^{u}$ associated with the system (4.3) in $L^{2}({\mathbb {T}})$ such that

$$\begin{aligned} \int _{L^{2}({\mathbb {T}})}P_{t}^{u}\varphi d\mu ^{u}=\int _{L^{2}({\mathbb {T}})}\varphi d\mu ^{u},~~t\ge 0 \end{aligned}$$

for any $\varphi \in B_{b}(L^{2}({\mathbb {T}}))$ the space of bounded functions on $L^{2}({\mathbb {T}}).$ It is easy to see that the invariant measure satisfies

$$\begin{aligned} \int _{L^{2}({\mathbb {T}})}\Vert z\Vert ^{2}\mu ^{u}(dz)\le C(1+\Vert u\Vert ^{2}). \end{aligned}$$

(3) According to the invariant property of $\mu ^{u},$ (2) and (2.2), we have

$$\begin{aligned} \begin{aligned} \Vert {\mathbb {E}}f(u,v^{u,X})-{\bar{f}}(u)\Vert ^{2}=&\Vert {\mathbb {E}}f(u,v^{u,X})-\int _{L^{2}({\mathbb {T}})}f(u,Y)\mu ^{u}(dY)\Vert ^{2} \\ =&\Vert {\mathbb {E}}f(u,v^{u,X})-{\mathbb {E}}\int _{L^{2}({\mathbb {T}})}f(u,v^{u,Y}) \mu ^{u}(dY)\Vert ^{2} \\ =&\Vert \int _{L^{2}({\mathbb {T}})}{\mathbb {E}}[f(u,v^{u,X})-f(u,v^{u,Y})]\mu ^{u}(dY)\Vert ^{2} \\ \le&C\int _{L^{2}({\mathbb {T}})}{\mathbb {E}}\Vert v^{u,X}-v^{u,Y}\Vert ^{2}\mu ^{u}(dY) \\ \le&C\int _{L^{2}({\mathbb {T}})}\Vert X-Y\Vert ^{2}e^{-2\eta t}\mu ^{u}(dY) \\ \le&C(1+\Vert X\Vert ^{2}+\Vert u\Vert ^{2})e^{-2\eta t} . \end{aligned} \end{aligned}$$

$\square$

1.2 Proof of Proposition 2.2

Lemma 5.1

For any $(u_{0},v_{0})\in H^{l}({\mathbb {T}})\times L^{2}({\mathbb {T}}),$ (1.1) admits a unique mild solution $(u^{\varepsilon },v^{\varepsilon })\in Y_{\tau _{\infty }},$ where $\tau _{\infty }$ is stopping time. Moreover, if $\tau _{\infty }<+\infty ,$ then ${\mathbb {P}}-$a.s.

$$\begin{aligned} \lim \limits _{t\rightarrow \tau _{\infty }}\Vert (u^{\varepsilon },v^{\varepsilon })\Vert _{X_{t}}=+\infty . \end{aligned}$$

Proof

In this part, we will take

$$\begin{aligned} \varepsilon =1 \end{aligned}$$

for the sake of simplicity, and $(u^{\varepsilon },v^{\varepsilon })$ is denoted by (u, v). All the results can be extended without difficulty to the general case. We set

$$\begin{aligned} \begin{aligned}&X_{\tau }=C([0,\tau ];H^{l}({\mathbb {T}}))\times C([0,\tau ];L^{2}({\mathbb {T}})),\\&Y_{\tau }=L^{2}(\Omega ;C([0,\tau ];H^{l}({\mathbb {T}})))\times L^{2}(\Omega ; C([0,\tau ];L^{2}({\mathbb {T}}))). \end{aligned} \end{aligned}$$

Let $\rho \in C^{\infty }_{0}({\mathbb {R}})$ be a cut-off function such that $\rho (r)=1$ for $r\in [0,1]$ and $\rho (r)=0$ for $r\ge 2.$ For any $R>0,y\in H^{l}({\mathbb {T}}),$ we set $\rho _{R}(y)=\rho (\frac{\Vert y\Vert _{H^{l}({\mathbb {T}})}}{R}).$ The truncated equation corresponding to (1.1) is the following stochastic partial differential equations:

$$\begin{aligned} \begin{array}{l} \left\{ \begin{array}{llll} du+(u_{x}+u_{xxx}+\rho _{R}(u)uu_{x})dt=f(u,v)dt+\sigma _{1}(u)dB_{1}\\ dv+i(-\partial _{xx})^{\theta }v dt=[-\lambda v-ig(u,v)]dt-i\sigma _{2}(u,v)dB_{2}\\ u(x,0)=u_{0}(x), ~v(x,0)=v_{0}(x) \end{array} \right. \end{array} \begin{array}{lll} \mathrm{{in}}~{\mathbb {T}}\times (0,T)\\ \mathrm{{in}}~{\mathbb {T}}\times (0,T)\\ \mathrm{{in}}~{\mathbb {T}}. \end{array} \end{aligned}$$

(5.1)

We define

$$\begin{aligned}&\Phi _{R}(u(t),v(t)) \\&\quad =\left( \begin{array}{c} \Phi _{R}^{1}(u(t),v(t))\\ \Phi _{R}^{2}(u(t),v(t)) \end{array}\right) \\&\quad =\left( \begin{array}{c} S_{\alpha }(t)u_{0}+\int _{0}^{t}S_{\alpha }(t-s)[-\rho _{R}(u)(s)u(s) u_{x}(s)+f(u(s),v(s))]ds +\int _{0}^{t}S_{\alpha }(t-s)\sigma _{1} (u(s))dB_{1} \\ G_{\theta }(t)v_{0}+\int _{0}^{t}G_{\theta }(t-s)[-\lambda v-ig(u(s),v(s))] ds-\int _{0}^{t}G_{\theta }(t-s)i\sigma _{2}(u,v)dB_{2} \end{array}\right) . \end{aligned}$$

It is easy to see that for any $T_{0}>0,$ operator $\Phi _{R}(u,v)$ maps $X_{T_{0}}$ into itself.

$\bullet$ The estimate of ${\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert \Phi _{R}^{1}(u_{1},v_{1})(t)-\Phi _{R}^{1}(u_{2},v_{2})(t)\Vert _{H^{l}}^{2}.$

Indeed, we set $F(u)=uu_{x}.$ Noting the fact for

$$\begin{aligned} \begin{aligned} \Vert u_{1}u_{1x}-u_{2}u_{2x}\Vert =&\Vert u_{1}(u_{1}-u_{2})_{x}+(u_{1}-u_{2})u_{2x}\Vert \\ \le&C(\Vert u_{1}\Vert _{H^{1}}+\Vert u_{2}\Vert _{H^{1}})\Vert u_{1}-u_{2}\Vert _{H^{1}} \\ \le&C(\Vert u_{1}\Vert _{H^{l}}+\Vert u_{2}\Vert _{H^{l}})\Vert u_{1}-u_{2}\Vert _{H^{l}}. \end{aligned} \end{aligned}$$

Set $F_{R}(u)=F(u)\rho _{R}(u).$ Assume that $\Vert u_{1}\Vert _{H^{l}}\ge \Vert u_{2}\Vert _{H^{l}},$ we have without any loss of generality,

$$\begin{aligned}&\Vert F_{R}(u_{1})-F_{R}(u_{2})\Vert \\&\quad =\Vert (F(u_{1})-F(u_{2}))\rho _{R}(u_{1})+(\rho _{R}(u_{1})-\rho _{R}(u_{2}))F(u_{2})\Vert \\&\quad \le \Vert (F(u_{1})-F(u_{2}))\rho _{R}(u_{1})\Vert +\Vert (\rho _{R}(u_{1})-\rho _{R}(u_{2}))F(u_{2})\Vert \\&\quad =\Vert (F(u_{1})-F(u_{2}))\rho _{R}(u_{1})\Vert +\Vert (\rho _{R}(u_{1})-\rho _{R}(u_{2}))F(u_{2})\Vert \chi _{\{\Vert u_{2}\Vert _{H^{l}}\le 2R\}}\\&\quad \le \Vert (F(u_{1})-F(u_{2}))\rho _{R}(u_{1})\Vert +|\rho _{R}(u_{1})-\rho _{R}(u_{2})|\Vert F(u_{2})\Vert \chi _{\{\Vert u_{2}\Vert _{H^{l}}\le 2R\}}\\&\quad \le C\Vert u_{1}-u_{2}\Vert _{H^{l}}(\Vert u_{1}\Vert _{H^{l}}+\Vert u_{2}\Vert _{H^{1}})\chi _{\{\Vert u_{1}\Vert _{H^{l}}\le 2R\}}\\&\quad \quad +\frac{C}{R}\Vert u_{1}-u_{2}\Vert _{L^{\infty }}\Vert F(u_{2})\Vert \chi _{\{\Vert u_{2}\Vert _{H^{l}}\le 2R\}}\\&\quad \le CR\Vert u_{1}-u_{2}\Vert _{H^{l}}+\frac{C}{R}\Vert u_{1}-u_{2}\Vert _{H^{l}}\Vert u_{2}u_{2x}\Vert \chi _{\{\Vert u_{2}\Vert _{H^{l}}\le 2R\}}\\&\quad \le CR\Vert u_{1}-u_{2}\Vert _{H^{l}} +\frac{C}{R}\Vert u_{1}-u_{2}\Vert _{H^{l}}\Vert u_{2}\Vert _{H^{1}}^{2}\chi _{\{\Vert u_{2}\Vert _{H^{l}}\le 2R\}}\\&\quad \le CR\Vert u_{1}-u_{2}\Vert _{H^{l}} +\frac{C}{R}\Vert u_{1}-u_{2}\Vert _{H^{l}}\Vert u_{2}\Vert _{H^{l}}^{2}\chi _{\{\Vert u_{2}\Vert _{H^{l}}\le 2R\}}\\&\quad \le CR\Vert u_{1}-u_{2}\Vert _{H^{l}}. \end{aligned}$$

It follows from (3.1) with $\gamma =\beta =l$ that

$$\begin{aligned} \begin{aligned}&{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert \int _{0}^{t}S_{\alpha }(t-s)(F_{R}(u_{1})(s) -F_{R}(u_{2})(s))ds\Vert _{H^{l}}^{2} \\&\quad \le C{\mathbb {E}}\sup _{0\le t\le T_{0}}\Big (\int _{0}^{t} (t-s)^{-\frac{l}{2\alpha }}\Vert (F_{R}(u_{1})(s)-F_{R}(u_{2})(s))\Vert ds\Big )^{2} \\&\quad \le C{\mathbb {E}}\sup _{0\le t\le T_{0}}\Big (\int _{0}^{t} (t-s)^{-\frac{l}{2\alpha }}R\Vert u_{1}(s)-u_{2}(s)\Vert _{H^{l}}ds\Big )^{2} \\&\quad \le C R^{2}\sup _{0\le t\le T_{0}}\Big (\int _{0}^{t} (t-s)^{-\frac{l}{2\alpha }}ds\Big )^{2}{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert u_{1}(t)-u_{2}(t)\Vert _{H^{l}}^{2} \\&\quad \le C R^{2}T_{0}^{2-\frac{l}{\alpha }}{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert u_{1}(t)-u_{2}(t)\Vert _{H^{l}}^{2} \end{aligned} \end{aligned}$$

(5.2)

and

$$\begin{aligned}&{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert \int _{0}^{t}S_{\alpha }(t-s) (f(u_{1}(s),v_{1}(s))-f(u_{2}(s),v_{2}(s)))ds\Vert _{H^{l}({\mathbb {T}}) }^{2}\nonumber \\&\quad \le {\mathbb {E}}\sup _{0\le t\le T_{0}}\Big (\int _{0}^{t}\Vert S_{\alpha } (t-s)(f(u_{1}(s),v_{1}(s))-f(u_{2}(s),v_{2}(s)))\Vert _{H^{l}}ds\Big )^{2}\nonumber \\&\quad \le C{\mathbb {E}}\sup _{0\le t\le T_{0}}\Big (\int _{0}^{t} (t-s)^{-\frac{l}{2\alpha }}\Vert (f(u_{1}(s),v_{1}(s))-f(u_{2}(s),v_{2}(s)))\Vert ds\Big )^{2}\nonumber \\&\quad \le C{\mathbb {E}}\sup _{0\le t\le T_{0}}\Big (\int _{0}^{t} (t-s)^{-\frac{l}{2\alpha }}(\Vert u_{1}(s)-u_{2}(s)\Vert +\Vert v_{1}(s)-v_{2}(s)\Vert )ds\Big )^{2} \nonumber \\&\quad \le C \sup _{0\le t\le T_{0}}\Big (\int _{0}^{t} (t-s)^{-\frac{l}{2\alpha }}ds\Big )^{2}\Big ({\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert u_{1}(t)-u_{2}(t)\Vert ^{2}+{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert v_{1}(t)-v_{2}(t)\Vert ^{2}\Big )\nonumber \\&\quad \le C T_{0}^{2-\frac{l}{\alpha }} \Big ({\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert u_{1}(t)-u_{2}(t)\Vert ^{2}+{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert v_{1}(t)-v_{2}(t)\Vert ^{2}\Big ). \end{aligned}$$

(5.3)

It follows from (3.15) with $\rho =l,p=1$ that

$$\begin{aligned}&{\mathbb {E}}\sup \limits _{t\in [0,T_{0}]}\Vert \int _{0}^{t}S_{\alpha }(t-s) [\sigma _{1}(u_{1}(s))-\sigma _{1}(u_{2}(s))]dB_{1}\Vert _{H^{l}}^{2}\nonumber \\&\quad \le C(p,T){\mathbb {E}}\int _{0}^{T_{0}}\Vert \sigma _{1}(u_{1}(s))- \sigma _{1}(u_{2}(s))\Vert _{H^{l}}^{2}dt\nonumber \\&\quad \le C(p,T){\mathbb {E}}\int _{0}^{T_{0}}\Vert u_{1}(s)-u_{2}(s)\Vert _{H^{l}}^{2}dt\nonumber \\&\quad \le C(p,T)T_{0}{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert u_{1}(t)-u_{2}(t)\Vert _{H^{l}}^{2}. \end{aligned}$$

(5.4)

Finally, collecting the above estimates (5.2)–(5.4), we get

$$\begin{aligned} \begin{aligned}&{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert \Phi _{R}^{1}(u_{1},v_{1})(t) -\Phi _{R}^{1}(u_{2},v_{2})(t)\Vert _{H^{l}}^{2}\\&\quad \le C(R^{2}T_{0}^{2-\frac{l}{\alpha }} +T_{0}^{2-\frac{l}{\alpha }}+T_{0}) \\&\qquad \Big ({\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert u_{1}(t)-u_{2}(t)\Vert _{H^{l}}^{2}+{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert v_{1}(t)-v_{2}(t)\Vert ^{2}\Big ). \end{aligned} \end{aligned}$$

(5.5)

$\bullet$ The estimate of ${\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert \Phi _{R}^{2}(u_{1},v_{1})(t)-\Phi _{R}^{2}(u_{2},v_{2})(t)\Vert ^{2}.$

Indeed, we have

$$\begin{aligned}&{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert \int _{0}^{t}G_{\theta }(t-s) (-\lambda v_{1}-ig(u_{1}(s),v_{1}(s))+\lambda v_{2}+ig(u_{2}(s),v_{2}(s)))ds\Vert ^{2}\\&\quad \le {\mathbb {E}}\sup _{0\le t\le T_{0}}\Big (\int _{0}^{t}\Vert G_{\theta } (t-s)(-\lambda v_{1}-ig(u_{1}(s),v_{1}(s))+\lambda v_{2}+ig(u_{2}(s),v_{2}(s)))\Vert ds\Big )^{2} \\&\quad \le C{\mathbb {E}}\sup _{0\le t\le T_{0}}\Big (\int _{0}^{t}\Vert -\lambda v_{1}-ig(u_{1}(s),v_{1}(s))+\lambda v_{2}+ig(u_{2}(s),v_{2}(s))\Vert ds\Big )^{2}\\&\quad \le C{\mathbb {E}}\sup _{0\le t\le T_{0}}\Big (\int _{0}^{t} (\Vert u_{1}(s)-u_{2}(s)\Vert +\Vert v_{1}(s)-v_{2}(s)\Vert )ds\Big )^{2}\\&\quad \le C \int _{0}^{T_{0}}\Big ({\mathbb {E}}\sup _{0\le t\le T_{0}} \Vert u_{1}(t)-u_{2}(t)\Vert ^{2}+{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert v_{1}(t)-v_{2}(t)\Vert ^{2}\Big )ds\\&\quad \le C T_{0}\Big ({\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert u_{1}(t) -u_{2}(t)\Vert ^{2}+{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert v_{1}(t)-v_{2}(t)\Vert ^{2}\Big ). \end{aligned}$$

It follows from Proposition 3.3 with $\rho =0$ that

$$\begin{aligned} \begin{aligned}&{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert \int _{0}^{t}G_{\theta }(t-s) (-i\sigma _{2}(u_{1}(s),v_{1}(s))+i\sigma _{2}(u_{2}(s),v_{2}(s)))dB_{2}\Vert ^{2}\\&\quad \le C{\mathbb {E}}\int _{0}^{T_{0}}\Vert G_{\theta }(t-s)(\sigma _{2} (u_{1}(s),v_{1}(s))-\sigma _{2}(u_{2}(s),v_{2}(s)))\Vert ^{2}ds\\&\quad \le C{\mathbb {E}}\int _{0}^{T_{0}}\Vert \sigma _{2}(u_{1}(s),v_{1}(s)) -\sigma _{2}(u_{2}(s),v_{2}(s))\Vert ^{2}ds\\&\quad \le C{\mathbb {E}}\int _{0}^{T_{0}}(\Vert u_{1}(s)-u_{2}(s)\Vert ^{2}+ \Vert v_{1}(s)-v_{2}(s)\Vert ^{2})ds\\&\quad \le C T_{0}\Big ({\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert u_{1}(t) -u_{2}(t)\Vert ^{2}+{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert v_{1}(t)-v_{2}(t)\Vert ^{2}\Big ). \end{aligned} \end{aligned}$$

This implies that

$$\begin{aligned} \begin{aligned}&{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert \Phi _{R}^{2}(u_{1},v_{1})(t)- \Phi _{R}^{2}(u_{2},v_{2})(t)\Vert ^{2}\\&\quad \quad \le CT_{0}\Big ({\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert u_{1}(t)-u_{2}(t)\Vert _{H^{l}}^{2}+{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert v_{1}(t)-v_{2}(t)\Vert ^{2}\Big ). \end{aligned} \end{aligned}$$

(5.6)

$\bullet$ The estimate of $\Vert \Phi _{R}(u_{1},v_{1})-\Phi _{R}(u_{2},v_{2})\Vert _{Y_{T_{0}}}.$

Indeed, it follows from (5.5) and (5.6) that

$$\begin{aligned} \begin{aligned}&{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert \Phi _{R}^{1}(u_{1},v_{1}) (t)-\Phi _{R}^{1}(u_{2},v_{2})(t)\Vert _{H^{l}}^{2}+{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert \Phi _{R}^{2}(u_{1},v_{1})(t)-\Phi _{R}^{2}(u_{2},v_{2})(t)\Vert ^{2}\\&\quad \le C(R^{2}T_{0}^{2-\frac{l}{\alpha }} +T_{0}^{2-\frac{l}{\alpha }}+ T_{0})\Big ({\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert u_{1}(t)-u_{2}(t)\Vert _{H^{l}}^{2}+{\mathbb {E}}\sup _{0\le t\le T_{0}}\Vert v_{1}(t)-v_{2}(t)\Vert ^{2}\Big ), \end{aligned} \end{aligned}$$

namely, we have

$$\begin{aligned} \begin{aligned} \Vert \Phi _{R}(u_{1},v_{1})-\Phi _{R}(u_{2},v_{2})\Vert _{Y_{T_{0}}} \le C(R^{2}T_{0}^{2-\frac{l}{\alpha }} +T_{0}^{2-\frac{l}{\alpha }} +T_{0})^{\frac{1}{2}}\Vert (u_{1},v_{1})-(u_{2},v_{2})\Vert _{Y_{T_{0}}}. \end{aligned}\nonumber \\ \end{aligned}$$

(5.7)

For a sufficiently small $T_{0},$ $\Phi _{R}(u,v)$ is a contraction mapping on $Y_{T_{0}}.$

Hence, by applying the Banach contraction principle, $\Phi _{R}(u,v)$ has a unique fixed point in $Y_{T_{0}},$ which is the unique local solution to (5.1) on the interval $[0,T_{0}].$ Since $T_{0}$ does not depend on the initial value $(u_{0},v_{0}),$ this solution may be extended to the whole interval [0, T].

We denote by $(u_{R},v_{R})$ this unique mild solution and let

$$\begin{aligned} \begin{aligned} \tau _{R}=\inf \{t\ge 0:\Vert (u_{R},v_{R})\Vert _{X_{t}}\ge R\}, \end{aligned} \end{aligned}$$

with the usual convention that $\inf \emptyset =\infty .$

Since $R_{1}\le R_{2},$ $\tau _{{R}_{1}}\le \tau _{{R}_{2}},$ we can put $\tau _{\infty }=\lim \limits _{R\rightarrow +\infty }\tau _{R}.$ Set $\tau =\tau _{R_{1}}\wedge \tau _{R_{2}}.$ We define a local solution to (1.1) as follows

$$ u(t)=u_{R}(t),~\forall ~ t\in [0,\tau _{R}],~~v(t)=v_{R}(t),~\forall ~t\in [0,\tau _{R}]. $$

Proceeding as in the proof of (5.7), we can obtain that $\Vert (u_{R_{1}},v_{R_{1}})-(u_{R_{2}},v_{R_{2}})\Vert _{Y_{t}}\le C(t)\Vert (u_{R_{1}},v_{R_{1}})-(u_{R_{2}},v_{R_{2}})\Vert _{Y_{t}}$, where C(t) is a monotonically increasing function and $C(0)=0.$ If we take t sufficiently small, we can obtain $u_{R_{1}}(t)=u_{R_{2}}(t),v_{R_{1}}(t)=v_{R_{2}}(t).$ Repeating the same argument for the interval [t, 2t] and so on yields $u_{R_{1}}(t)=u_{R_{2}}(t),v_{R_{1}}(t)=v_{R_{2}}(t),$ for the whole interval $[0,\tau ].$ According to this, we can know the above definition of local solution to (1.1) is well defined.

If $\tau _{\infty }<+\infty ,$ the definition of (u, v) yields ${\mathbb {P}}-$a.s.

$$\begin{aligned} \lim \limits _{t\rightarrow \tau _{\infty }}\Vert (u,v)\Vert _{X_{t}}=+\infty , \end{aligned}$$

which shows that (u, v) is the unique local solution to (1.1) on the interval $[0,\tau _{\infty }).$

This completes the proof of Lemma 5.1. $\square$

First, we shall prove uniform bounds with respect to $\varepsilon \in (0,1)$ for $p-$moment of the solution for (1.1).

Lemma 5.2

For any $u_{0},v_{0}\in L^{2}({\mathbb {T}})$, $p>0$, there exists a constant C(p, T) such that

$$\begin{aligned} \sup \limits _{\varepsilon \in (0,1)}\sup \limits _{t\in [0,T]}{\mathbb {E}} \Vert u^{\varepsilon }(t)\Vert ^{2p}\le C(p,T),~~\sup \limits _{\varepsilon \in (0,1)} \sup \limits _{t\in [0,T]}{\mathbb {E}}\Vert v^{\varepsilon }(t)\Vert ^{2p}\le C(p,T). \end{aligned}$$

Proof

It is sufficient to prove this proposition when p is large enough. It is easy to see

$$\begin{aligned} \begin{aligned} dv^{\varepsilon }&=-\frac{i}{\varepsilon }[(-\partial _{xx})^{\theta } v^{\varepsilon }-i\lambda v^{\varepsilon }+g(u^{\varepsilon }, v^{\varepsilon })]dt-\frac{i}{\sqrt{\varepsilon }}\sigma _{2} (u^{\varepsilon },v^{\varepsilon })dB_{2}\\&=-\frac{i}{\varepsilon }(-\partial _{xx})^{\theta }v^{\varepsilon } dt-\frac{1}{\varepsilon }\lambda v^{\varepsilon }dt -\frac{i}{\varepsilon }g(u^{\varepsilon },v^{\varepsilon })dt -\frac{i}{\sqrt{\varepsilon }}\sigma _{2}(u^{\varepsilon },v^{\varepsilon })dB_{2}. \end{aligned} \end{aligned}$$

For $p\ge 2$, applying Itô formula with $\Vert v^{\varepsilon }(t)\Vert ^{2p}$, we have

$$\begin{aligned}&{\mathbb {E}}\Vert v^{\varepsilon }(t)\Vert ^{2p} \nonumber \\&\quad =\Vert v_{0}\Vert ^{2p}+\frac{2p}{\varepsilon }{\mathbb {E}}\int _{0}^{t}\Vert v^{\varepsilon }(s)\Vert ^{2p-2}(-i(-\partial _{xx})^{\theta }v^{\varepsilon } (s),v^{\varepsilon }(s)) ds \nonumber \\&\qquad +\frac{2p}{\varepsilon }{\mathbb {E}} \int _{0}^{t}\Vert v^{\varepsilon }(s)\Vert ^{2p-2}(-\lambda v^{\varepsilon } (s),v^{\varepsilon }(s)) ds\nonumber \\&\qquad +\frac{2p}{\varepsilon }{\mathbb {E}}\int _{0}^{t}\Vert v^{\varepsilon }(s) \Vert ^{2p-2}(-ig(u^{\varepsilon }(s),v^{\varepsilon }(s)),v^{\varepsilon }(s))ds\nonumber \\&\qquad +\frac{p}{\varepsilon }{\mathbb {E}}\int _{0}^{t}\Vert v^{\varepsilon }(s) \Vert ^{2p-2}\Vert \sigma _{2}(u^{\varepsilon },v^{\varepsilon })\Vert ^{2}ds\nonumber \\&\qquad +\frac{2p(p-1)}{\varepsilon }{\mathbb {E}}\int _{0}^{t}\Vert v^{\varepsilon } (s)\Vert ^{2p-4} |(\sigma _{2}(u^{\varepsilon },v^{\varepsilon }),v^{\varepsilon }(s))|^{2}ds, \end{aligned}$$

(5.8)

where Itô formula can be understood in the way that we first use the Galerkin approximation to get (5.8) in the finite dimensional setting, then we take the limit of the dimension to obtain (5.8) in the infinite dimensional setting.

It follows from Lemma 2.1 that $(-i(-\partial _{xx})^{\theta }v^{\varepsilon }(s),v^{\varepsilon }(s))=0,$ thus, we have

$$\begin{aligned} \begin{aligned} \frac{d}{dt}{\mathbb {E}}\Vert v^{\varepsilon }\Vert ^{2p} =&-\frac{2p\lambda }{\varepsilon }{\mathbb {E}}\Vert v^{\varepsilon }\Vert ^{2p} +\frac{2p}{\varepsilon }{\mathbb {E}}\Vert v^{\varepsilon }\Vert ^{2p-2} (-ig(u^{\varepsilon },v^{\varepsilon }),v^{\varepsilon }) \\&+\frac{p}{\varepsilon }{\mathbb {E}}\Vert v^{\varepsilon }\Vert ^{2p-2}\Vert \sigma _{2}(u^{\varepsilon },v^{\varepsilon })\Vert ^{2} +\frac{2p(p-1)}{\varepsilon }{\mathbb {E}}\Vert v^{\varepsilon }\Vert ^{2p-4} |(\sigma _{2}(u^{\varepsilon },v^{\varepsilon }),v^{\varepsilon })|^{2}. \end{aligned} \end{aligned}$$

It follows from the property of g that

$$\begin{aligned} \begin{aligned} 2\Vert v^{\varepsilon }(s)\Vert ^{2p-2}(-ig(u^{\varepsilon }(s),v^{\varepsilon }(s)), v^{\varepsilon }(s))&\le 2\Vert v^{\varepsilon }\Vert ^{2p-1}\Vert g(u^{\varepsilon },v^{\varepsilon })\Vert \\&\le 2L_{g}\Vert v^{\varepsilon }(t)\Vert ^{2p-1}(1+\Vert u^{\varepsilon }(t)\Vert +\Vert v^{\varepsilon }(t)\Vert )\\&=2L_{g}\Vert v^{\varepsilon }(t)\Vert ^{2p-1}+2L_{g}\Vert v^{\varepsilon }(t)\Vert ^{2p-1} \Vert u^{\varepsilon }(t)\Vert \\&\quad +2L_{g}\Vert v^{\varepsilon }(t)\Vert ^{2p}. \end{aligned} \end{aligned}$$

According to the Young inequality (see Lemma 2.3), it holds that

$$\begin{aligned} 2L_{g}\Vert v^{\varepsilon }(t)\Vert ^{2p-1}\le 2L_{g}\eta \Vert v^{\varepsilon }(t)\Vert ^{2p}+C(\eta ,L_{g},p), \end{aligned}$$

if we choose $\eta =\frac{1}{6},$ we have

$$\begin{aligned} 2L_{g}\Vert v^{\varepsilon }(t)\Vert ^{2p-1}\le \frac{ L_{g}}{3}\Vert v^{\varepsilon }(t)\Vert ^{2p}+C(L_{g},p). \end{aligned}$$

By the same method, we have

$$\begin{aligned} 2L_{g}\Vert v^{\varepsilon }(t)\Vert ^{2p-1}\Vert u^{\varepsilon }\Vert \le \frac{ L_{g}}{3}\Vert v^{\varepsilon }(t)\Vert ^{2p}+C(L_{g},p)\Vert u^{\varepsilon }(t)\Vert ^{2p}. \end{aligned}$$

It follows from the property of $\sigma _{2}$ and the same method as above that

$$\begin{aligned}&\Vert v^{\varepsilon }(t)\Vert ^{2p-2}\Vert \sigma _{2}(u^{\varepsilon }, v^{\varepsilon })\Vert ^{2}+2(p-1)\Vert v^{\varepsilon }(t)\Vert ^{2p-4} (v^{\varepsilon },\sigma _{2}(u^{\varepsilon },v^{\varepsilon }))^{2}\\&\quad \le \frac{ L_{g}}{3}\Vert v^{\varepsilon }(t)\Vert ^{2p}+C(M,p). \end{aligned}$$

Thus, it holds that

$$\begin{aligned} \begin{aligned} \frac{d}{dt}{\mathbb {E}}\Vert v^{\varepsilon }(t)\Vert ^{2p}&\le -\frac{2p\lambda }{\varepsilon }{\mathbb {E}}\Vert v^{\varepsilon }(t)\Vert ^{2p} +\frac{3L_{g}p}{\varepsilon }{\mathbb {E}}\Vert v^{\varepsilon }(t)\Vert ^{2p} +\frac{C}{\varepsilon }{\mathbb {E}}\Vert u^{\varepsilon }(t)\Vert ^{2p} +\frac{C}{\varepsilon } \\&\le -\frac{ p \lambda }{\varepsilon }{\mathbb {E}}\Vert v^{\varepsilon }(t)\Vert ^{2p} +\frac{C}{\varepsilon }{\mathbb {E}}\Vert u^{\varepsilon }(t)\Vert ^{2p}+\frac{C}{\varepsilon } , \end{aligned} \end{aligned}$$

hence, by comparison theorem (see Lemma 2.4), we have

$$\begin{aligned} {\mathbb {E}}\Vert v^{\varepsilon }(t)\Vert ^{2p} \le \Vert v_{0}\Vert ^{2p}e^{-\frac{\lambda p}{\varepsilon } t}+\frac{C}{\varepsilon }\int _{0}^{t}e^{-\frac{\lambda p}{\varepsilon } (t-s)}({\mathbb {E}}\Vert u^{\varepsilon }(s)\Vert ^{2p}+1)ds. \end{aligned}$$

(5.9)

For $\Vert u^{\varepsilon }(t)\Vert ^{2p}$, by Itô formula again and using Lemma 2.1, the following holds,

$$\begin{aligned} \begin{aligned}&{\mathbb {E}}\Vert u^{\varepsilon }(t)\Vert ^{2p}\\&\quad =\Vert u_{0}\Vert ^{2p}+{\mathbb {E}} \int _{0}^{t}[2p\Vert u^{\varepsilon }\Vert ^{2p-2}(-u^{\varepsilon }_{xxx} -(-\partial _{xx})^{\alpha }u^{\varepsilon }-u^{\varepsilon }u_{x}^{\varepsilon } +f(u^{\varepsilon },v^{\varepsilon }),u^{\varepsilon })\\&\quad \quad +p\Vert u^{\varepsilon }\Vert ^{2p-2}\Vert \sigma _{1}(u^{\varepsilon })\Vert ^{2} +2p(p-1)\Vert u^{\varepsilon }\Vert ^{2p-4}(u^{\varepsilon },\sigma _{1}(u^{\varepsilon }))^{2}]ds\\&\quad =\Vert u_{0}\Vert ^{2p}+{\mathbb {E}}\int _{0}^{t}[2p\Vert u^{\varepsilon }\Vert ^{2p-2} (-\Vert (-\partial _{xx})^{\frac{\alpha }{2}}u^{\varepsilon }\Vert ^{2})+2p\Vert u^{\varepsilon }\Vert ^{2p-2}(f(u^{\varepsilon },v^{\varepsilon }),u^{\varepsilon })\\&\quad \quad +p\Vert u^{\varepsilon }\Vert ^{2p-2}\Vert \sigma _{1}(u^{\varepsilon })\Vert ^{2}+2p(p-1) \Vert u^{\varepsilon }\Vert ^{2p-4}(u^{\varepsilon },\sigma _{1}(u^{\varepsilon }))^{2}]ds, \end{aligned} \end{aligned}$$

then

$$\begin{aligned} \begin{aligned} \frac{d}{dt}{\mathbb {E}}\Vert u^{\varepsilon }(t)\Vert ^{2p}&=2p{\mathbb {E}}\Vert u^{\varepsilon }\Vert ^{2p-2}(-\Vert (-\partial _{xx})^{\frac{\alpha }{2}}u^{\varepsilon }\Vert ^{2}) +2p{\mathbb {E}}\Vert u^{\varepsilon }\Vert ^{2p-2}(f(u^{\varepsilon }, v^{\varepsilon }),u^{\varepsilon }) \\&\quad +p{\mathbb {E}}\Vert u^{\varepsilon }\Vert ^{2p-2}\Vert \sigma _{1}(u^{\varepsilon })\Vert ^{2} \\&\quad +2p(p-1){\mathbb {E}}\Vert u^{\varepsilon }\Vert ^{2p-4}(u^{\varepsilon },\sigma _{1} (u^{\varepsilon }))^{2} \\&\le 2p{\mathbb {E}}\Vert u^{\varepsilon }\Vert ^{2p-2}(C\Vert u^{\varepsilon } \Vert ^{2}+C\Vert v^{\varepsilon }\Vert ^{2}+C)+C(p){\mathbb {E}}\Vert u^{\varepsilon } \Vert ^{2p-2}\Vert \sigma _{1}(u^{\varepsilon })\Vert ^{2}, \end{aligned} \end{aligned}$$

thus, it follows from Young inequality that

$$\begin{aligned} \frac{d}{dt}{\mathbb {E}}\Vert u^{\varepsilon }(t)\Vert ^{2p} \le C({\mathbb {E}}\Vert u^{\varepsilon }(t)\Vert ^{2p}+ {\mathbb {E}}\Vert v^{\varepsilon }(t)\Vert ^{2p}+1), \end{aligned}$$

hence, by comparison theorem (see Lemma 2.4), we have

$$\begin{aligned} {\mathbb {E}}\Vert u^{\varepsilon }(t)\Vert ^{2p} \le e^{Ct}\Vert u_{0}\Vert ^{2p}+C\int _{0}^{t}e^{C (t-s)}({\mathbb {E}}\Vert v^{\varepsilon }(s)\Vert ^{2p}+1)ds. \end{aligned}$$

Plug this inequality into (5.9), we have

$$\begin{aligned} {\mathbb {E}}\Vert v^{\varepsilon }(t)\Vert ^{2p}&\le \Vert v_{0}\Vert ^{2p}e^{-\frac{\lambda p}{\varepsilon } t}+\frac{C}{\varepsilon }\int _{0}^{t}e^{-\frac{\lambda p}{\varepsilon } (t-s)}({\mathbb {E}}\Vert u^{\varepsilon }(s)\Vert ^{2p}+1)ds \\&\le C(1+\Vert v_{0}\Vert ^{2p})+\frac{C}{\varepsilon }\int _{0}^{t}e^{-\frac{\lambda p}{\varepsilon } (t-s)}{\mathbb {E}}\Vert u^{\varepsilon }(s)\Vert ^{2p}ds \\&\le C(1+\Vert v_{0}\Vert ^{2p})+\frac{C}{\varepsilon }\int _{0}^{t}e^{-\frac{\lambda p}{\varepsilon } (t-s)}[e^{Cs}\Vert u_{0}\Vert ^{2p} \\&\quad +C\int _{0}^{s}e^{C (s-\tau )}({\mathbb {E}}\Vert v^{\varepsilon }(\tau )\Vert ^{2p}+1)d\tau ]ds \\&\le C(1+\Vert u_{0}\Vert ^{2p}+\Vert v_{0}\Vert ^{2p})+ \frac{C}{\varepsilon }\int _{0}^{t}e^{-\frac{\lambda p}{\varepsilon } (t-s)} \int _{0}^{s}{\mathbb {E}}\Vert v^{\varepsilon }(\tau )\Vert ^{2p}d\tau ds \\&\le C(p,T,u_{0},v_{0})+\frac{C}{\lambda p}\int _{0}^{t} (1-e^{-\frac{\lambda p}{\varepsilon } (t-\tau )}){\mathbb {E}} \Vert v^{\varepsilon }(\tau )\Vert ^{2p}d\tau \\&\le C(p,T,u_{0},v_{0})+C\int _{0}^{t}{\mathbb {E}}\Vert v^{\varepsilon }(\tau )\Vert ^{2p}d\tau , \end{aligned}$$

it follows from Gronwall inequality that $\sup \limits _{0\le t\le T}{\mathbb {E}} \Vert v^{\varepsilon }(t)\Vert ^{2p}\le C(p,T,u_{0},v_{0})$, thus, we have

$$ \sup \limits _{0\le t\le T}{\mathbb {E}}\Vert u^{\varepsilon }(t)\Vert ^{2p}\le C(p,T,u_{0},v_{0}). $$

$\square$

Next, we will present the estimate on the slow motion $u^{\varepsilon }$ as a process valued in $H^{l}({\mathbb {T}})$.

Lemma 5.3

Let $\tau =\tau _{\infty }\wedge T.$ For any $(u_{0},v_{0})\in H^{l}({\mathbb {T}})\times L^{2}({\mathbb {T}})$ and $p>0$, we have

$$\begin{aligned}&\sup \limits _{\varepsilon \in (0,1)}{\mathbb {E}}\sup \limits _{t\in [0,\tau ]} \Vert u^{\varepsilon }(t)\Vert _{H^{l}}^{2p}\le C(p,T) \end{aligned}$$

(5.10)

$$\begin{aligned}&{\mathbb {E}}\sup \limits _{t\in [0,\tau ]}\Vert v^{\varepsilon }(t)\Vert ^{2p}\le C(\varepsilon ,p,T). \end{aligned}$$

(5.11)

Proof

It is sufficient to prove (5.10) and (5.11) when p is large enough.

$\bullet$ We can write the first equation in system (1.1) in its integral form

$$\begin{aligned} \begin{aligned} u^{\varepsilon }(t)=&S_{\alpha }(t)u_{0}-\int _{0}^{t}S_{\alpha }(t-s) (u^{\varepsilon }u^{\varepsilon }_{x})(s)ds+\int _{0}^{t}S_{\alpha } (t-s)f(u^{\varepsilon }(s),v^{\varepsilon }(s))ds \\&+\int _{0}^{t}S_{\alpha }(t-s)\sigma _{1}(u^{\varepsilon }(s))dB_{1} \\ =&I_{1}+I_{2}+I_{3}+I_{4}. \end{aligned} \end{aligned}$$

Indeed, it follows from (3.3) with $s=l$ that

$$\begin{aligned} \Vert I_{1}\Vert _{H^{l}}^{2p}\le C(p)\Vert u_{0}\Vert _{H^{l}}^{2p}. \end{aligned}$$

It follows from (3.1) with $\gamma =l,\beta =\frac{l}{4}+\frac{3}{2}$ that

$$\begin{aligned} \Vert I_{2}\Vert _{H^{l}}\le & {} \int _{0}^{t}\Vert S_{\alpha }(t-s)(u^{\varepsilon } u^{\varepsilon }_{x})(s)\Vert _{H^{l}}ds\\\le & {} C(T)\int _{0}^{t}(t-s)^{-\frac{l+6}{8\alpha }}\Vert (u^{\varepsilon } u^{\varepsilon }_{x})(s)\Vert _{H^{\frac{3(l-2)}{4}}}ds\\\le & {} C(T)\int _{0}^{t}(t-s)^{-\frac{l+6}{8\alpha }}\Vert u^{\varepsilon } (s)\Vert _{H^{\frac{l-1}{4}}}\Vert u^{\varepsilon }(s)\Vert _{H^{\frac{l+3}{4}}}ds\\\le & {} C(T)\int _{0}^{t}(t-s)^{-\frac{l+6}{8\alpha }}\Vert u^{\varepsilon } (s)\Vert ^{\frac{3l+1}{4l}}\Vert u^{\varepsilon }(s)\Vert _{H^{l}}^{\frac{l-1}{4l}} \Vert u^{\varepsilon }(s)\Vert ^{\frac{3l-3}{4l}}\Vert u^{\varepsilon } (s)\Vert _{H^{l}}^{\frac{l+3}{4l}}ds\\= & {} C(T)\int _{0}^{t}(t-s)^{-\frac{l+6}{8\alpha }}\Vert u^{\varepsilon } (s)\Vert ^{\frac{3l-1}{2l}}\Vert u^{\varepsilon }(s)\Vert ^{\frac{l+1}{2l}}_{H^{l}}ds, \end{aligned}$$

where we have used Lemma 2.5 for $\alpha _{1}=\alpha _{2}=\frac{l-1}{4},~\alpha _{3}=\frac{3(2-l)}{4}$ and interpolation inequality

$$\begin{aligned} \Vert u^{\varepsilon }(s)\Vert _{H^{a}}\le C\Vert u^{\varepsilon }(s)\Vert ^{\frac{b-a}{b}} \Vert u^{\varepsilon }(s)\Vert ^{\frac{a}{b}}_{H^{b}}~~\forall ~0\le a\le b. \end{aligned}$$

Then, we can deduce that

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\sup \limits _{t\in [0,\tau ]}\Vert I_{2}\Vert ^{2p}_{H^{l}} \le&C(p,T){\mathbb {E}}\Big (\sup \limits _{t\in [0,\tau ]}\int _{0}^{t} (t-s)^{-\frac{l+6}{8\alpha }} \Vert u^{\varepsilon }(s)\Vert ^{\frac{3l-1}{2l}}\Vert u^{\varepsilon }(s) \Vert ^{\frac{l+1}{2l}}_{H^{l}}ds\Big )^{2p}\\ \le&C(p,T){\mathbb {E}}\Big [\Big (\sup \limits _{t\in [0,\tau ]}\int _{0}^{t} (t-s)^{-\frac{(l+6)p}{4\alpha (2p-1)}}ds\Big )^{2p-1} \\&\qquad \qquad \Big (\sup \limits _{t\in [0,\tau ]}\int _{0}^{t} \Vert u^{\varepsilon }(s)\Vert ^{\frac{(3l-1)p}{l}}\Vert u^{\varepsilon }(s) \Vert ^{\frac{(l+1)p}{l}}_{H^{l}}ds\Big )\Big ]\\ \le&C(p,T)\Big (\int _{0}^{T}s^{-\frac{(l+6)p}{4\alpha (2p-1)}}ds \Big )^{2p-1} \\&\Big (\int _{0}^{T} {\mathbb {E}}\Vert u^{\varepsilon }(s)\Vert ^{\frac{2(3l-1)p}{l-1}}ds +\int _{0}^{T}{\mathbb {E}}\Vert u^{\varepsilon }(s)\Vert ^{2p}_{H^{l}}ds\Big ). \end{aligned} \end{aligned}$$

Taking $p>\frac{4\alpha }{8\alpha -l-6}$ and using Proposition 2.2, we have

$$\begin{aligned} {\mathbb {E}}\sup \limits _{t\in [0,\tau ]}\Vert I_{2}\Vert ^{2p}_{H^{l}}\le C(p,T)\Big (1+\int _{0}^{T}{\mathbb {E}}\Vert u^{\varepsilon }(s)\Vert ^{2p}_{H^{l}}ds\Big ). \end{aligned}$$

According to (3.1) for $\beta =\gamma =l$, if $p>\frac{\alpha }{2\alpha -l},$ we can obtain that

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\sup \limits _{t\in [0,\tau ]}\Vert I_{3}\Vert ^{2p}_{H^{l}} \le&C(p,T){\mathbb {E}}\Big (\sup \limits _{t\in [0,\tau ]}\int _{0}^{t} (t-s)^{-\frac{l}{2\alpha }}\Vert f(u^{\varepsilon }(s),v^{\varepsilon }(s))\Vert ds\Big )^{2p}\\ \le&C(p,T){\mathbb {E}}\Big (\sup \limits _{t\in [0,\tau ]}\int _{0}^{t} (t-s)^{-\frac{l}{2\alpha }}(1+\Vert u^{\varepsilon }(s)\Vert +\Vert v^{\varepsilon }(s)\Vert )ds\Big )^{2p}\\ \le&C(p,T)\Big (\int _{0}^{T}s^{-\frac{ lp}{\alpha (2p-1)}}ds\Big )^{2p-1} \int _{0}^{T}(1+{\mathbb {E}}\Vert u^{\varepsilon }(s)\Vert ^{2p}+{\mathbb {E}} \Vert v^{\varepsilon }(s)\Vert ^{2p})ds\\ \le&C(p,T). \end{aligned} \end{aligned}$$

It follows from Proposition 3.2 with $\rho =l$ that

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\sup \limits _{t\in [0,T]}\Vert \int _{0}^{t}S_{\alpha }(t-s) \sigma _{1}(u^{\varepsilon }(s))dB_{1}\Vert _{H^{l}}^{2p}&\le C(p){\mathbb {E}}\int _{0}^{T}\Vert \sigma _{1}(u^{\varepsilon }(s)) \Vert _{H^{l}}^{2p}dt\\&\le C(p){\mathbb {E}}\int _{0}^{T}(1+\Vert u^{\varepsilon }(s)\Vert _{H^{l}})^{2p}dt\\&\le C(p,T)\Big (1+{\mathbb {E}}\int _{0}^{T}\Vert u^{\varepsilon } (s)\Vert ^{2p}_{H^{l}}ds\Big ). \end{aligned} \end{aligned}$$

Consequently, we have

$$\begin{aligned} {\mathbb {E}}\sup \limits _{t\in [0,\tau ]}\Vert u^{\varepsilon }(t)\Vert _{H^{l}}^{2p}\le C(p,T)\Big (1+{\mathbb {E}}\int _{0}^{T}\Vert u^{\varepsilon }(s)\Vert ^{2p}_{H^{l}}ds\Big ). \end{aligned}$$

By applying Gronwall’s inequality, we can obtain (5.10).

$\bullet$ To prove (5.11), noting that

$$\begin{aligned} \begin{aligned} v^{\varepsilon }(t)=&G_{\theta }(\frac{t}{\varepsilon })v_{0} +\frac{i}{\varepsilon }\int _{0}^{t}G_{\theta }\left( \frac{t-s}{\varepsilon }\right) [-\lambda v-ig(u^{\varepsilon }(s),v^{\varepsilon }(s))]ds \\&-\frac{1}{\sqrt{\varepsilon }}\int _{0}^{t}G_{\theta } \left( \frac{t-s}{\varepsilon }\right) i\sigma _{2} (u^{\varepsilon }(s),v^{\varepsilon }(s))dB_{2}\\ =&I_{1}^{\prime }+I_{2}^{\prime }+I_{3}^{\prime }. \end{aligned} \end{aligned}$$

It is clear that

$$\begin{aligned} \begin{aligned}&{\mathbb {E}}\sup _{0\le t\le \tau }\Vert I_{1}^{\prime }\Vert ^{2p}\le C(p,T)\Vert v_{0}\Vert ^{2p}\le C(p,T),\\&{\mathbb {E}}\sup _{0\le t\le \tau }\Vert I_{3}^{\prime }\Vert ^{2p}\le C(\varepsilon ,p,T). \end{aligned} \end{aligned}$$

Now, it is sufficient to estimate $I_{2}^{\prime }$. Applying Proposition 2.2, we can deduce that

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\sup _{0\le t\le \tau }\Vert I_{2}^{\prime }\Vert ^{2p}\le&C(\varepsilon ,p,T){\mathbb {E}}\sup _{0\le t\le T}\left( \int _{0}^{t}(1+\Vert u^{\varepsilon }(s)\Vert +\Vert v^{\varepsilon }(s)\Vert )ds\right) ^{2p}\\ \le&C(\varepsilon ,p,T)\int _{0}^{T}(1+{\mathbb {E}} \Vert u^{\varepsilon }(s)\Vert ^{2p}+E\Vert v^{\varepsilon }(s)\Vert ^{2p})ds\\ \le&C(\varepsilon ,p,T). \end{aligned} \end{aligned}$$

With the help of the above estimates, we arrive at ${\mathbb {E}}\sup _{0\le t\le \tau }\Vert v^{\varepsilon }(t)\Vert ^{2p}\le C(\varepsilon ,p,T).$

This completes the proof of (5.11). $\square$

Now, we prove Proposition 2.2.

Proof of Proposition 2.2

By Chebyshev inequality, Lemma 5.3 and the definition of $(u^{\varepsilon },v^{\varepsilon })$, we have

$$\begin{aligned} \begin{aligned} {\mathbb {P}}(\{\omega \in \Omega ~|~ \tau _{\infty }(\omega )<+\infty \}) =&\lim \limits _{T\rightarrow +\infty }{\mathbb {P}}(\{\omega \in \Omega ~|~ \tau _{\infty }(\omega )\le T\}) \\ =&\lim \limits _{T\rightarrow +\infty }{\mathbb {P}}(\{\omega \in \Omega ~|~ \tau (\omega )=\tau _{\infty }(\omega )\}) \\ =&\lim \limits _{T\rightarrow +\infty }\lim \limits _{R\rightarrow +\infty }{\mathbb {P}}(\{\omega \in \Omega ~|~ \tau _{R}(\omega )\le \tau (\omega )\}) \\ =&\lim \limits _{T\rightarrow +\infty }\lim \limits _{R\rightarrow +\infty }{\mathbb {P}}(\{\omega \in \Omega ~|~ \Vert (u^{\varepsilon },v^{\varepsilon })\Vert _{X_{\tau }}\ge \Vert (u^{\varepsilon },v^{\varepsilon })\Vert _{X_{\tau _{R}}}\}) \\ =&\lim \limits _{T\rightarrow +\infty }\lim \limits _{R\rightarrow +\infty }{\mathbb {P}}(\{\omega \in \Omega ~|~ \Vert (u^{\varepsilon },v^{\varepsilon })\Vert _{X_{\tau }}\ge R\}) \\ \le&\lim \limits _{T\rightarrow +\infty }\lim \limits _{R\rightarrow +\infty }\frac{\mathbb {{\mathbb {E}}}\Vert (u^{\varepsilon },v^{\varepsilon }) \Vert _{X_{\tau }}^{2}}{R^{2}}=0, \end{aligned} \end{aligned}$$

this shows that ${\mathbb {P}}(\{\omega \in \Omega ~|~ \tau _{\infty }(\omega )=+\infty \})=1,$ namely, $\tau _{\infty }=+\infty$ ${\mathbb {P}}$-a.s. $\square$