Averaging Principle for Multiscale Stochastic Klein–Gordon-Heat System

Peng Gao

doi:10.1007/s00332-019-09529-4

Journal of Nonlinear Science

pp 1–59 | Cite as

Averaging Principle for Multiscale Stochastic Klein–Gordon-Heat System

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Peng Gao

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First Online: 17 January 2019

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Abstract

This paper investigates multiscale stochastic Klein–Gordon-heat system. We establish the well-posedness and two kinds of stochastic averaging principle for stochastic Klein–Gordon-heat system with two timescales. To be more precise, under suitable conditions, two kinds of averaging principle (the autonomous case and the nonautonomous case) are proved, and as a consequence, the multiscale stochastic Klein–Gordon-heat system can be reduced to a single stochastic Klein–Gordon equation (averaged equation) with a modified coefficient, the slow component of multiscale stochastic system toward the solution of the averaged equation in moment (the autonomous case) and in probability (the nonautonomous case).

Keywords

Multiscale stochastic partial differential equations Stochastic averaging principle Stochastic Klein–Gordon-heat system Stochastic Klein–Gordon equation Effective dynamics

Mathematics Subject Classification

60H15 70K65 70K70 37H10 37L55

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Notes

Acknowledgements

I would like to thank the referees and the editor for their careful comments and useful suggestions. I sincerely thank Professor Yong Li for many useful suggestions and help.

Appendix

Proof of Proposition 3.1

Proof

Let $u=\sum \limits _{n=1}^{\infty }u_{n}e_{n},f=\sum \limits _{n=1}^{\infty }f_{n}e_{n},g=\sum \limits _{n=1}^{\infty }g_{n}e_{n}$, then we have

$$\begin{aligned} \left\{ \begin{array}{llll} \mathrm{d}\dot{u}_{n}+\lambda _{n}u_{n}\mathrm{d}t=f_{n}\mathrm{d}t+g_{n}\mathrm{d}W_{1}, \\ u_{n}(0)=(u_{0},e_{n}), \\ \dot{u}_{n}(0)=(u_{1},e_{n}). \end{array} \right. \end{aligned}$$

By applying the Itô formula, we have

$$\begin{aligned} \mathrm{d}(\dot{u}_{n}^{2}+\lambda _{n}u_{n}^{2}) =2\dot{u}_{n}f_{n}\mathrm{d}t+2\dot{u}_{n}g_{n}\mathrm{d}W_{1}+g_{n}^{2}\mathrm{d}t. \end{aligned}$$

Integrating between 0 and t, we have

$$\begin{aligned} \dot{u}_{n}^{2}(t)+\lambda _{n}u_{n}^{2}(t) =\dot{u}_{n}^{2}(0)+\lambda _{n}u_{n}^{2}(0)+\int _{0}^{t}2\dot{u}_{n}f_{n}\mathrm{d}s+\int _{0}^{t}2\dot{u}_{n}g_{n}\mathrm{d}W_{1}+\int _{0}^{t}g_{n}^{2}\mathrm{d}s.\nonumber \\ \end{aligned}$$

(8.1)

Noting the fact that $ \Vert u(\cdot ,t)\Vert =\sum \nolimits _{n=1}^{\infty }u_{n}^{2}(t),~ \Vert u_{t}(\cdot ,t)\Vert =\sum \nolimits _{n=1}^{\infty }\dot{u}_{n}^{2}(t),~ \Vert u_{x}(\cdot ,t)\Vert =\sum \nolimits _{n=1}^{\infty }\lambda _{n}u_{n}^{2}(t) $ and taking the sum on n in formula (8.1), we can prove (3.1).

For any $\varphi \in C^{\infty }(Q),\varphi (0,t)=\varphi (1,t)=0$, according to $\mathrm{d}(u_{t}\varphi )=\varphi \mathrm{d}u_{t}+ u_{t} \mathrm{d}\varphi + \mathrm{d}u_{t} \mathrm{d}\varphi =\varphi \mathrm{d}u_{t}+ u_{t} \varphi _{t}\mathrm{d}t$, we have

$$\begin{aligned} \begin{aligned} (u_{t},\varphi )(s)&=(u_{t},\varphi )(0)+\int _{0}^{s}\mathrm{d}(u_{t},\varphi ) \\&=(u_{1},\varphi (0))+\int _{0}^{s}(u_{t},\varphi _{t})\mathrm{d}\tau +\int _{0}^{s}(u_{xx}+f,\varphi )\mathrm{d}\tau +\int _{0}^{s}(g,\varphi )\mathrm{d}W_{1}. \end{aligned} \end{aligned}$$

(8.2)

Since $\mathrm{d}(u\varphi _{t})=\varphi _{t} \mathrm{d}u+ u \mathrm{d}\varphi _{t}+ \mathrm{d}u \mathrm{d}\varphi _{t}=\varphi _{t} u_{t}\mathrm{d}t+ u \varphi _{tt}\mathrm{d}t$, we have

$$\begin{aligned} \begin{aligned} \int _{0}^{s}(u_{t},\varphi _{t})\mathrm{d}\tau&=\int _{0}^{s}\mathrm{d}(u,\varphi _{t})-\int _{0}^{s}(u, \varphi _{tt})\mathrm{d}\tau \\&=(u,\varphi _{t})(s)-(u_{0},\varphi _{t}(0))-\int _{0}^{s}(u, \varphi _{tt})\mathrm{d}\tau . \end{aligned} \end{aligned}$$

(8.3)

It follows from (8.2) and (8.3) that

$$\begin{aligned} \begin{aligned}&(u_{t},\varphi )(s) \\&\quad =(u_{1},\varphi (0))+(u,\varphi _{t})(s)-(u_{0},\varphi _{t}(0))-\int _{0}^{s}(u, \varphi _{tt})\mathrm{d}\tau \\&\qquad +\int _{0}^{s}(u_{xx}+f,\varphi )\mathrm{d}\tau +\int _{0}^{s}(g,\varphi )\mathrm{d}W_{1} \\&\quad =(u_{1},\varphi (0))-(u_{0},\varphi _{t}(0))+(u,\varphi _{t})(s)+\int _{0}^{s}(u, \varphi _{xx}-\varphi _{tt})\mathrm{d}\tau \\&\qquad +\int _{0}^{s}(f,\varphi )\mathrm{d}\tau +\int _{0}^{s}(g,\varphi )\mathrm{d}W_{1} , \end{aligned} \end{aligned}$$

thus, we can obtain (3.2).

According to (3.2), we have

$$\begin{aligned} \begin{aligned} (u,\varphi )(t)=&\,(u,\varphi )(0)+\int _{0}^{t}(u,\varphi )^{\prime }\mathrm{d}s \\ =&\,(u,\varphi )(0)+\int _{0}^{t}(u,\varphi _{t})\mathrm{d}s+\int _{0}^{t}(u_{t},\varphi )\mathrm{d}s \\ =&\,(u_{0},\varphi (0))+2\int _{0}^{t}(u,\varphi _{t})\mathrm{d}s+t(u_{1},\varphi (0))-t(u_{0},\varphi _{t}(0))\\&+\int _{0}^{t}\int _{0}^{s}(u, \varphi _{xx}-\varphi _{tt})\mathrm{d}\tau \mathrm{d}s \\&+\int _{0}^{t}\int _{0}^{s}(f,\varphi )\mathrm{d}\tau \mathrm{d}s+\int _{0}^{t}\int _{0}^{s}(g,\varphi )\mathrm{d}W_{1}\mathrm{d}s, \end{aligned} \end{aligned}$$

this implies (3.3). $\square $

Proof of Proposition 3.2

Proof

The proof is inspired from Chow (2014, P133, Lemma 3.1, Lemma 3.2), (Gao 2016, Proposition 2.1).

We know that

$$\begin{aligned} u(t)=G^{\prime }(t)u_{0}+G(t)u_{1}+\int _{0}^{t}G(t-s)f(s) \mathrm{d}s+\int _{0}^{t}G(t-s)g\mathrm{d}W_{1} \end{aligned}$$

is the solution of

$$\begin{aligned} \begin{array}{l} \left\{ \begin{array}{llll} \mathrm{d}u_{t}-u_{xx}\mathrm{d}t=f(t)\mathrm{d}t+g(t)\mathrm{d}W_{1} \\ u(0,t)=0=u(1,t) \\ u(x,0)=u_{0}(x) \\ u_{t}(x,0)=u_{1}(x) \end{array} \right. \end{array} \begin{array}{lll} {\mathrm{in}}~Q,\\ {\mathrm{in}}~(0,T),\\ {\mathrm{in}}~I,\\ {\mathrm{in}}~I. \end{array} \end{aligned}$$

1) If we set $f=0,g=0$, we can know that $G^{\prime }(t)u_{0}+G(t)u_{1}$ is the solution to

$$\begin{aligned} \begin{array}{l} \left\{ \begin{array}{llll} u_{tt}-u_{xx}=0 \\ u(0,t)=0=u(1,t) \\ u(x,0)=u_{0}(x) \\ u_{t}(x,0)=u_{1}(x) \end{array} \right. \end{array} \begin{array}{lll} {\mathrm{in}}~Q,\\ {\mathrm{in}}~(0,T),\\ {\mathrm{in}}~I,\\ {\mathrm{in}}~I, \end{array} \end{aligned}$$

it follows from (8.1) that

$$\begin{aligned} \dot{u}_{n}^{2}(t)+\lambda _{n}u_{n}^{2}(t) =\dot{u}_{n}^{2}(0)+\lambda _{n}u_{n}^{2}(0), \end{aligned}$$

(8.4)

taking the sum form 1 to $+\infty $, we have

$$\begin{aligned} \Vert u_{t}(t)\Vert ^{2}+\Vert u_{x}(t)\Vert ^{2} =\Vert u_{t}(0)\Vert ^{2}+\Vert u_{x}(0)\Vert ^{2}, \end{aligned}$$

thus, we have

$$\begin{aligned} \Vert u_{t}(t)\Vert ^{2}+\Vert u(t)\Vert ^{2}_{H^{1}} \le C(\Vert u_{1}\Vert ^{2}+\Vert u_{0}\Vert ^{2}_{H^{1}}). \end{aligned}$$

Multiplying (8.4) by $\lambda _{n}$, we have

$$\begin{aligned} \lambda _{n}\dot{u}_{n}^{2}(t)+\lambda _{n}^{2}u_{n}^{2}(t) =\lambda _{n}\dot{u}_{n}^{2}(0)+\lambda _{n}^{2}u_{n}^{2}(0), \end{aligned}$$

taking the sum form 1 to $+\infty $, we have

$$\begin{aligned} \Vert u_{xt}(t)\Vert ^{2}+\Vert u_{xx}(t)\Vert ^{2} =\Vert u_{xt}(0)\Vert ^{2}+\Vert u_{xx}(0)\Vert ^{2}, \end{aligned}$$

thus, it holds that

$$\begin{aligned} \Vert u_{t}(t)\Vert _{H^{1}}^{2}+\Vert u(t)\Vert ^{2}_{H^{2}} \le C(\Vert u_{1}\Vert _{H^{1}}^{2}+\Vert u_{0}\Vert ^{2}_{H^{2}}). \end{aligned}$$

2) If we set $g=0,u_{0}=0,u_{1}=0$, we can know that $\int _{0}^{t}G(t-s)f(s)\mathrm{d}s$ is the solution to

$$\begin{aligned} \begin{array}{l} \left\{ \begin{array}{llll} u_{tt}-u_{xx}=f \\ u(0,t)=0=u(1,t) \\ u(x,0)=0 \\ u_{t}(x,0)=0 \end{array} \right. \end{array} \begin{array}{lll} {\mathrm{in}}~Q,\\ {\mathrm{in}}~(0,T),\\ {\mathrm{in}}~I,\\ {\mathrm{in}}~I. \end{array} \end{aligned}$$

It follows from (8.1) that

$$\begin{aligned} \dot{u}_{n}^{2}(t)+\lambda _{n}u_{n}^{2}(t) =\int _{0}^{t}2\dot{u}_{n}f_{n}\mathrm{d}s, \end{aligned}$$

then, it holds that

$$\begin{aligned} \begin{aligned} \sup \limits _{0\le t\le T}(\dot{u}_{n}^{2}(t)+\lambda _{n}u_{n}^{2}(t))&\le \sup \limits _{0\le t\le T}\dot{u}_{n}\cdot \int _{0}^{T}2|f_{n}|\mathrm{d}s \\&\le \frac{1}{2}\sup \limits _{0\le t\le T}\dot{u}_{n}^{2}+C(T)\int _{0}^{T}f_{n}^{2}\mathrm{d}t , \end{aligned} \end{aligned}$$

thus, we have

$$\begin{aligned} \begin{aligned} \sup \limits _{0\le t\le T}(\dot{u}_{n}^{2}(t)+\lambda _{n}u_{n}^{2}(t)) \le C(T)\int _{0}^{T}f_{n}^{2}\mathrm{d}t, \end{aligned} \end{aligned}$$

(8.5)

taking the sum form 1 to $+\infty $, we have (3.4) with $m=0$.

If we multiply (8.5) by $\lambda _{n}$ and take the sum form 1 to $+\infty $, we have (3.4) with $m=1$.

3) If we set $f=0,u_{0}=0,u_{1}=0$, we can know that $\int _{0}^{t}G(t-s)g(s)\mathrm{d}W_{1}$ is the solution to

$$\begin{aligned} \begin{array}{l} \left\{ \begin{array}{llll} \mathrm{d}u_{t}-u_{xx}\mathrm{d}t=g\mathrm{d}W_{1} \\ u(0,t)=0=u(1,t) \\ u(x,0)=0 \\ u_{t}(x,0)=0 \end{array} \right. \end{array} \begin{array}{lll} {\mathrm{in}}~Q,\\ {\mathrm{in}}~(0,T),\\ {\mathrm{in}}~I,\\ {\mathrm{in}}~I, \end{array} \end{aligned}$$

it follows from (8.1) that

$$\begin{aligned} \dot{u}_{n}^{2}(t)+\lambda _{n}u_{n}^{2}(t) =\int _{0}^{t}2\dot{u}_{n}g_{n}\mathrm{d}W_{1}+\int _{0}^{t}g_{n}^{2}\mathrm{d}s, \end{aligned}$$

(8.6)

taking the sum form 1 to $+\infty $, we have

$$\begin{aligned} \Vert u_{t}(t)\Vert ^{2}+\Vert u_{x}(t)\Vert ^{2} =2\int _{0}^{t}(u_{t}(s),g(s))\mathrm{d}W_{1}+\int _{0}^{t}\Vert g(s)\Vert ^{2}\mathrm{d}s. \end{aligned}$$

In view of the Burkholder–Davis–Gundy inequality, it holds that

$$\begin{aligned} \begin{aligned} \mathbb {E}\sup \limits _{0\le t\le T}\left| \int _{0}^{t}(u_{t}(s),g(s))\mathrm{d}W_{1}\right| ^{p}&\le C(p)\mathbb {E}\left( \int _{0}^{T}|(u_{t},g)|^{2}\mathrm{d}t\right) ^{\frac{p}{2}} \\&\le C(p)\mathbb {E}\sup \limits _{0\le t\le T}\Vert u_{t}\Vert ^{p} \left( \int _{0}^{T}\Vert g\Vert ^{2}\mathrm{d}t\right) ^{\frac{p}{2}} \\&\le \eta \mathbb {E}\sup \limits _{0\le t\le T} \Vert u_{t}\Vert ^{2p}+C(\eta ,p)\mathbb {E}\left( \int _{0}^{T}\Vert g\Vert ^{2}\mathrm{d}t\right) ^{p}. \end{aligned} \end{aligned}$$

Thus, we can obtain that

$$\begin{aligned}&\mathbb {E}\sup \limits _{0\le t\le T}(\Vert u_{t}(t)\Vert ^{2}+\Vert u_{x}(t)\Vert ^{2})^{p} \\&\quad \le \eta \mathbb {E}\sup \limits _{0\le t\le T} \Vert u_{t}\Vert ^{2p}+C(\eta ,p)\mathbb {E} \left( \int _{0}^{T}\Vert g(t)\Vert ^{2}\mathrm{d}t\right) ^{p} +C(p)\mathbb {E}\left( \int _{0}^{T}\Vert g(t)\Vert ^{2}\mathrm{d}t\right) ^{p}. \end{aligned}$$

If we take $\eta<<1$, we have (3.5) with $m=0$.

If we multiply (8.6) by $\lambda _{n}$ and take the sum form 1 to $+\infty $, by the above same method as above, we have (3.5) with $m=1$. $\square $

Proof of Proposition 3.4

Proof

It follows from the factorization formula (see Da Prato and Zabczyk 2014, P130, Theorem 5.10) that

$$\begin{aligned} \int _{0}^{t}S(t-s)f(s)\mathrm{d}W_{2}=\frac{\sin \frac{\pi }{4}}{\pi }\int _{0}^{t}S(t-s)(t-s)^{-\frac{3}{4}}U(s)\mathrm{d}s, \end{aligned}$$

where

$$\begin{aligned} U(s)=\int _{0}^{s}S(s-r)(s-r)^{-\frac{1}{4}}f(r)\mathrm{d}W_{2}. \end{aligned}$$

Thus, we have

$$\begin{aligned} \begin{aligned}&\mathbb {E}\sup _{0\le t\le T}\left\| \int _{0}^{t}S(t-s)f(s)\mathrm{d}W_{2}\right\| _{H^{\beta }}^{2p} \\&\quad \le C\mathbb {E}\sup _{0\le t\le T}\left\| \int _{0}^{t}S(t-s)(t-s)^{-\frac{3}{4}}U(s)\mathrm{d}s\right\| _{H^{\beta }}^{2p} \\&\quad \le C\mathbb {E}\sup _{0\le t\le T} \left( \int _{0}^{t}\Vert S(t-s)(t-s)^{-\frac{3}{4}}U(s)\Vert _{H^{\beta }}\mathrm{d}s\right) ^{2p}. \end{aligned} \end{aligned}$$

Due to Proposition 3.3, it holds that

$$\begin{aligned} \Vert S(t-s)(t-s)^{-\frac{3}{4}}U(s)\Vert _{H^{\beta }}\le C(t-s)^{-\frac{3+\beta }{4}}\Vert U(s)\Vert _{H^{\frac{\beta }{2}}}, \end{aligned}$$

then, we can obtain

$$\begin{aligned} \begin{aligned}&\mathbb {E}\sup _{0\le t\le T}\left\| \int _{0}^{t}S(t-s)f(s)\mathrm{d}W_{2}\right\| _{H^{\beta }}^{2p}\\&\quad \le C\mathbb {E}\sup _{0\le t\le T}\left( \int _{0}^{t}(t-s)^{-\frac{3+\beta }{4}}\Vert U(s)\Vert _{H^{\frac{\beta }{2}}}\mathrm{d}s\right) ^{2p} \\&\quad \le C\mathbb {E}\sup _{0\le t\le T} \left[ \left( \int _{0}^{t}(t-s)^{\left( -\frac{3+\beta }{4}\right) \frac{2p}{2p-1}}\mathrm{d}s\right) ^{2p-1} \left( \int _{0}^{t}\Vert U(s)\Vert _{H^{\frac{\beta }{2}}}^{2p}\mathrm{d}s\right) \right] . \end{aligned} \end{aligned}$$

For any $p>\frac{2}{1-\beta }$, it holds that

$$\begin{aligned} \left( \int _{0}^{t}(t-s)^{ \left( -\frac{3+\beta }{4}\right) \frac{2p}{2p-1}}\mathrm{d}s\right) ^{2p-1}\le C(p,T), \end{aligned}$$

then, we can obtain that

$$\begin{aligned} \begin{aligned} \mathbb {E}\sup _{0\le t\le T}\left\| \int _{0}^{t}S(t-s)f(s)\mathrm{d}W_{2}\right\| _{H^{\beta }}^{2p}&\le C\mathbb {E}\sup _{0\le t\le T}\left( \int _{0}^{t}\Vert U(s)\Vert _{H^{\frac{\beta }{2}}}^{2p}\mathrm{d}s\right) \\&\le C\mathbb {E}\int _{0}^{T}\Vert U(s)\Vert _{H^{\frac{\beta }{2}}}^{2p}\mathrm{d}s \\&= C\int _{0}^{T}\mathbb {E}\Vert U(s)\Vert _{H^{\frac{\beta }{2}}}^{2p}\mathrm{d}s . \end{aligned} \end{aligned}$$

Due to the Burkholder–Davis–Gundy inequality, it can be deduced that

$$\begin{aligned} \begin{aligned} \mathbb {E}\Vert U(s)\Vert _{H^{\frac{\beta }{2}}}^{2p}&\le C\mathbb {E}\Vert \int _{0}^{s}S(s-r)(s-r)^{-\frac{1}{4}}f(r)\mathrm{d}W_{2}\Vert _{H^{\frac{\beta }{2}}}^{2p} \\&\le C\mathbb {E} \left( \int _{0}^{s}(s-r)^{-\frac{\beta +1}{2}}\Vert f(r)\Vert ^{2}dr\right) ^{p} \\&\le C\mathbb {E}\left( \int _{0}^{s}(s-r)^{ \left( -\frac{\beta +1}{2}\right) \frac{p}{p-1}}dr\right) ^{p-1} \left( \int _{0}^{s}\Vert f(r)\Vert ^{2p}dr\right) \\&\le C\mathbb {E}\int _{0}^{T}\Vert f(t)\Vert ^{2p}\mathrm{d}t. \end{aligned} \end{aligned}$$

Thus, we can obtain (3.7). $\square $

Proof of Proposition 5.2

Proof

1) It follows from the energy method that

$$\begin{aligned} \begin{aligned} \frac{1}{2}\mathbb {E}\Vert B^{A,X}\Vert ^{2}=&\,\frac{1}{2}\mathbb {E}\Vert X\Vert ^{2}-\int _{0}^{t}\mathbb {E}\Vert B^{A,X}_{x}\Vert ^{2}\mathrm{d}s +\int _{0}^{t}\mathbb {E}(B^{A,X},\mathcal {G}(B^{A,X})\\&+g(A,B^{A,X}))\mathrm{d}s +\frac{1}{2}\int _{0}^{t}\mathbb {E}\Vert \sigma _{2}(A,B^{A,X})\Vert ^{2}\mathrm{d}s, \end{aligned} \end{aligned}$$

namely

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}\mathbb {E}\Vert B^{A,X}\Vert ^{2}= & {} -2\mathbb {E}\Vert B^{A,X}_{x}\Vert ^{2}\\&+2\mathbb {E}(B^{A,X},\mathcal {G}(B^{A,X})+g(A,B^{A,X}))+\mathbb {E}\Vert \sigma _{2}(A,B^{A,X})\Vert ^{2}\\&\le -2\mathbb {E}\Vert B^{A,X}_{x}\Vert ^{2}+2\mathbb {E}(B^{A,X},g(A,B^{A,X}))+\mathbb {E}\Vert \sigma _{2}(A,B^{A,X})\Vert ^{2}. \end{aligned}$$

With the help of the Young inequality and choosing a suitable $\eta $, we can obtain

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}\mathbb {E}\Vert B^{A,X}\Vert ^{2} \le -2\lambda \mathbb {E}\Vert B^{A,X}\Vert ^{2}+3L_{g}\mathbb {E}\Vert B^{A,X}\Vert ^{2}+C\Vert A\Vert ^{2}+C. \end{aligned}$$

Thus, it holds that

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}\mathbb {E}\Vert B^{A,X}\Vert ^{2} \le -\lambda \mathbb {E}\Vert B^{A,X}\Vert ^{2}+C\Vert A\Vert ^{2}+C. \end{aligned}$$

By applying Lemma 3.1 with $\mathbb {E}\Vert B^{A,X}(t)\Vert ^{2}$, we have

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

It follows from the energy method that

$$\begin{aligned}&\mathbb {E}\Vert B^{A,X}-B^{A,Y}\Vert ^{2}\\&\quad =\Vert X-Y\Vert ^{2}+2\mathbb {E}\int _{0}^{t}(B^{A,X}-B^{A,Y},(B^{A,X}-B^{A,Y})_{xx}\\&\qquad +\,g(A,B^{A,X})-g(A,B^{A,Y}))\mathrm{d}s\\&\qquad +\,2\mathbb {E}\int _{0}^{t}(B^{A,X}-B^{A,Y},\mathcal {G}(B^{A,X})-\mathcal {G}(B^{A,Y}))\mathrm{d}s\\&\qquad +\,\mathbb {E}\int _{0}^{t}\Vert \sigma _{2}(A,B^{A,X})-\sigma _{2}(A,B^{A,Y})\Vert ^{2}\mathrm{d}s, \end{aligned}$$

namely

$$\begin{aligned} \begin{aligned}&\frac{\hbox {d}}{\hbox {d}t}\mathbb {E}\Vert B^{A,X}-B^{A,Y}\Vert ^{2}\\&\quad =2\mathbb {E}(B^{A,X}-B^{A,Y},(B^{A,X}-B^{A,Y})_{xx}+g(A,B^{A,X})-g(A,B^{A,Y}))\\&\qquad +2\mathbb {E}(B^{A,X}-B^{A,Y},\mathcal {G}(B^{A,X})-\mathcal {G}(B^{A,Y}))+\mathbb {E}\Vert \sigma _{2}(A,B^{A,X})-\sigma _{2}(A,B^{A,Y})\Vert ^{2}\\&\quad \le -2\mathbb {E}\Vert (B^{A,X}-B^{A,Y})_{x}\Vert ^{2}+2L_{g}\mathbb {E}\Vert B^{A,X}-B^{A,Y}\Vert ^{2}+L_{\sigma _{2}}^{2}\mathbb {E}\Vert B^{A,X}-B^{A,Y}\Vert ^{2}\\&\quad =(-2\lambda +2L_{g}+L_{\sigma _{2}}^{2})\mathbb {E}\Vert B^{A,X}-B^{A,Y}\Vert ^{2}, \end{aligned} \end{aligned}$$

here we have used Corollary 3.1.

It follows from (H) that

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}\mathbb {E}\Vert B^{A,X}-B^{A,Y}\Vert ^{2} \le -\lambda \mathbb {E}\Vert B^{A,X}-B^{A,Y}\Vert ^{2}, \end{aligned}$$

this yields

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

2) (5.2) implies that for any $A\in L^{2}(I)$, there is a unique invariant measure $\mu ^{A}$ for the Markov semigroup $P_{t}^{A}$ associated with system (5.1) in $L^{2}(I)$ such that

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

for any $\varphi \in B_{b}(L^{2}(I))$ the space of bounded functions on $L^{2}(I)$.

Then, by repeating the standard argument as in Cerrai (2011, Proposition 4.2) and Cerrai and Freidlin (2009, Lemma 3.4), the invariant measure satisfies

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

3) According to the invariant property of $\mu ^{A}$, 2) and (5.2), we have

$$\begin{aligned} \Vert \mathbb {E}f(A,B^{A,X})-\bar{f}(A)\Vert ^{2}&=\left\| \mathbb {E}f(A,B^{A,X})-\int _{L^{2}(I)}f(A,Y)\mu ^{A}(\mathrm{d}Y)\right\| ^{2}\\&=\left\| \mathbb {E}f(A,B^{A,X})-\mathbb {E}\int _{L^{2}(I)}f(A,B^{A,Y})\mu ^{A}(\mathrm{d}Y)\right\| ^{2}\\&=\left\| \int _{L^{2}(I)}\mathbb {E}[f(A,B^{A,X})-f(A,B^{A,Y})]\mu ^{A}(\mathrm{d}Y)\right\| ^{2}\\&\le C\int _{L^{2}(I)}\mathbb {E}\Vert B^{A,X}-B^{A,Y}\Vert ^{2}\mu ^{A}(\mathrm{d}Y)\\&\le C\int _{L^{2}(I)}\Vert X-Y\Vert ^{2}e^{-\lambda t}\mu ^{A}(\mathrm{d}Y) \\&\le C(1+\Vert X\Vert ^{2}+\Vert A\Vert ^{2})e^{-\lambda t}.\square \end{aligned}$$

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Peng Gao
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1.School of Mathematics and Statistics, and Center for Mathematics and Interdisciplinary SciencesNortheast Normal UniversityChangchunPeople’s Republic of China

Averaging Principle for Multiscale Stochastic Klein–Gordon-Heat System

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