[家里蹲大学数学杂志]第049期2011年广州偏微分方程暑期班试题---随机PDE-可压NS-几何

 

随机偏微分方程

 

Throughout this section, let $(\Omega, \calF, \calF_t,\ P)$ be a complete filtered probability space satisfying the usual conditions.  

1. Recall the following results:  

a)         The Doob maximal inequality: if $(N_t)$ is a non-negative $\calF_t$-submartingale with $N_0=0$, then for $1<p<\infty$, $$\bex E\sez{\sup_{0\leq t\leq T}\sev{N_t}^p} \leq \sex{\frac{p}{p-1}}^p E\sez{\sev{N_T}^p}. \eex$$

b)        The set $\calS$ of simple processes is dense in the Hilbert space $\sex{\calH,\ \sen{\cdot}_{\calH}}$, where $$\bex \calS:=\left\{\xi_t=\sum_{k=0}^n \xi_k\chi_{[t_k,t_{k+1}]}(t):\ 0=t_0<t_1<\cdots<t_n\leq T,\right.\\ \left.\xi_k\in\calF_{t_k},\ \sup_k\sen{\xi_k}_\infty<\infty\right\}, \eex$$ and $$\bex \calH:=\left\{H:\ [0,T]\times\Omega \to \bbR \mbox{ is continuous and } \calF_t\mbox{-adapted}:\right.\\ \left. \sen{H}_{\calH}^2 := E\sez{\int_0^T\sev{H(s)}^2\rd s}<\infty\right\}. \eex$$  Set $$\bex \calM:=\left\{ M=(M_t)_{t\in [0,T]} \mbox{ is continuous } \calF_t\mbox{-martingales such that } \right.\\ \left. \sen{M}_\calM^2 :=\sup_{0\leq t\leq T} E\sez{\sev{M_t}^2} <+\infty \right\}. \eex$$ Then $(\calM,\sen{\cdot}_\calM)$ is a Hilbert space.  Let $\xi:\ [0,T]\times \Omega\to \bbR$ be the simple process given by $$\bex \xi_t=\sum_{k=0}^n \xi_k\chi_{[t_k,t_{k+1}]}(t), \eex$$ where $0=t_0<t_1<\cdots<t_n=T$, and $\xi_k\in \calF_{t_k}$ such that $\dps{\sup_k \sev{\xi_k}<\infty}$. Define $$\bex M_t=\int_0^t\xi_k\rd W_s :=\sum_{k=0}^n \xi_k\sex{W_{t_{k+1}\wedge t-W_{t_k\wedge t}}}, \eex$$  

a)          Prove that $M_t$ is a continuous $\calF_t$-martingale.

b)         Prove the It\^o's isometry identity: $$\bex E\sez{\sev{M_t}^2} = E\sez{\int_0^t\sev{\xi_s}^2\rd s}. \eex$$

c)        Using the Doob maximal inequality, prove that $$\bex E\sez{\sup_{0\leq t\leq T} \sev{M_t}^2} \leq 4 E\sez{\int_0^T \sev{\xi_s}^2\rd s}. \eex$$

d)        Given $H\in \calH$, let $H_n\in \calS$ be a sequence such that $\sen{H_n-H}_{\calH}\to 0$ as $n\to\infty$. Prove that $\dps{M_t^n =\int_0^t H_n(s)\rd W_s}$ is a Cauchy sequence in $\sex{\calM,\sen{\cdot}_\calM}$. Let $M$ be the limit of $\sed{M_n(t);\ t\in [0,T]}$ in $\sex{\calM,\sen{\cdot}_\calM}$. Prove that this limit does not depend on the choice of the sequence $H_n$ which tends to $H$ in $\sex{\calH,\sen{\cdot}_\calH}$. Denote by $\dps{M_t:=\int_0^t H(s)\rd W_s}$, i.e. $$\bex \int_0^t H(s)\rd W_s =\lim_{n\to\infty} \int_0^t H_n(s)\rd W_s,\mbox{ in } \sex{\calM,\sen{\cdot}_\calM}. \eex$$

e)         Prove that $\dps{M_t=\int_0^t H(s)\rd W_s}$ is a $\calF_t$-martingale and satisfies $$\bex E\sez{\sev{M_t}^2} = E\sez{\int_0^t \sev{H(s)}^2\rd s}, \eex$$ and $$\bex E\sez{\sup_{0\leq t\leq T}\sev{M_t}^2} \leq 4 E\sez{\int_0^T \sev{H(s)}^2\rd s}. \eex$$

f)         Using the Borel-Cantelli lemma, prove that $ P$-a.s., $M=(M_t)\in C([0,T];\bbR)$.

 

2. Consider the following SDE on $\bbR^m$: $$\bex \rd X_t=\rd W_t-\n V(X_t)\rd t,\quad X_0=x, \eex$$ where $V\in C_b^2(\bbR^m)$. Fix $T>0$. Suppose that $u(t,x)\in C_b^{1,2}([0,T]\times\bbR^m,\bbR)$ is a solution of the heat equation $$\bex \left\{\ba{ll} \frac{\p u}{\p t}(t,x) =\frac{1}{2}\lap u(t,x) -\sef{\n V(x),\n u(t,x)},&\mbox{in }[0,T)\times \bbR^m,\\ u(0,x)=f(x),&x\in \bbR^m, \ea\right. \eex$$ where $f\in C_b(\bbR^m)$. Applying It\^o's formula to $u(T-t,X_t)$, prove that $$\bex u(t,x)= E_x\sez{f(X_t)},\quad \forall\ t\geq 0,\ x\in \bbR^m. \eex$$  

 

3. Consider the following SPDE on $[0,T]\times S^1$: $$\bee\label{1} \frac{\p}{\p t}u(t,x) =\lap u+\dot W(t,x), \eee$$ where $t\in [0,\infty)$ and $x\in S^1=[0,2\pi]$, $\dps{\lap=\frac{\p^2}{\p x^2}}$ is the Laplace operator on $S^1$, and $W(t,x)$ is the space-time white noise on $[0,\infty)\times S^1$.  Recall that $\lap$ is a compact operator on $L^2(S^1,\rd x)$ and the spectral of $\lap$ is given by $$\bex \mbox{Sp}(\lap)=\sed{-n^2;\ n\in \bbN}. \eex$$ Indeed, let $$\bex e_{2n}(x)=\frac{1}{\sqrt{\pi}}\cos(nx),\quad e_{2n+1}(x)=\frac{1}{\sqrt{\pi}} \sin (nx),\quad n\in\bbN,\ x\in S^1. \eex$$  Then $$\bex \lap e_{2n}=-n^2 e_{2n},\quad \lap e_{2n+1}=-n^2e_{2n+1},\quad \forall\ n\in\bbN. \eex$$ The set $\sed{e_n}$ consists of a complete orthonormal basis of $L^2(S^1,\rd x)$. Write $$\bex W(t,x)=\sum_{n=1}^\infty W_n(t)e_n(x), \eex$$ where $W_n(t)$ are i.i.d Brownian motion on $\bbR^1$.

(a) Let $$\bex X_t(\cdot) =u(t,\cdot)\in L^2(S^1,\rd x). \eex$$ Prove that $X_t$ satisfies the Ornstein-Uhlenbeck SDE on $L^2(S^1,\rd x)$: $$\bex \rd X_t=\lap X_t+\rd W_t, \eex$$ and $\dps{W_t=\sum_{n=0}^\infty W_n(t)e_n}$ is the cylinder Brownian motion on $L^2(S^1,\rd x)$.  

(b) Let $\dps{u(t,x)=\sum_{n\in \bbN} u_n(t)e_n(x)}$ be the orthogonal decomposition of $u(t,\cdot)$ in $L^2(S^1,\rd x)$. Prove that $u_n(t)$ satisfies the Langevin SDE on $\bbR$: $$\bex \rd u_n(t)=-n^2 u_n(t)\rd t+\rd W_n(t), \eex$$ and solve this Langevin SDE with initial condition $u_n(0)=u_n\in \bbR$.  

© Find the mild solution to the SPDE \eqref{1} with initial condition $\dps{u(0,x)=\sum_{n=0}^\infty u_ne_n(x)}$ for $\dps{\sum_{n=0}^\infty \sev{u_n}^2<+\infty}$.  

(d) Recall that the domain of $\lap$ is given by $$\bex H_0=\left\{u=\sum_{n=1}^\infty u_ne_n\in L^2(S^1,\rd x);\ u_n=\sef{u,e_n},\right.\\ \left.\mbox{ and } \sum_{n=1}^\infty n^2 \sev{u_n}^2<\infty\right\} . \eex$$ Let $$\bex \rd \mu(u)=\prod_{n=1}^\infty \frac{n}{\sqrt{2\pi}} \mbox{exp}\sez{-\frac{n^2\sev{u_n}^2}{2}}\rd u_n. \eex$$ Prove that $\mu$ is a Gaussian measure on $(H,\calB(H))$ with mean zero and with covariance matrix $Q=\sex{q_{ij}}_{\bbN\times\bbN}$ with $$\bex q_{ij}=\frac{1}{i^2}\delta_{ij}, \eex$$ i.e., $\mu=\calN(0,Q)$.  Formally we write $$\bex Q=\sex{-\lap}^{-1},\quad \mu=\calN(0,\sex{-\lap}^{-1}). \eex$$

(e) Prove that $\mu$ is an invariant measure for the Ornstein-Uhlenbeck processs $X_t$ on $L^2(S^1,\rd x)$.  

(f) (Not required) Prove that $\mu$ is the unique invariant measure for the Ornstein-Uhlenbeck process $X_t$ on $L^2(S^1,\rd x)$.  

 

可压 Navier-Stokes 方程

 

1. Consider the compressible fluid flow with damping: $$\bex \left\{\ba{ll} \p_t\rho+\Div(\rho\bbu)=0,\\ \p_t(\rho\bbu)+\Div\sex{\rho\bbu\otimes\bbu} +\n p=-\rho\bbu. \ea\right. \eex$$ Can this system satisfy Kawashima's condition?

2. Follow the similar analysis for Lemma 2.1 to prove (2.18) in Proposition 2.2.

3. Give the details of the proof of Lemma 3.1.

4. Give the details of the proof of Theorem 5.3.

5. Give the complete proof of Lemmas 6.4 and 6.5.

 

几何分析 [参考答案链接]

1.(15') 设 $R(X,Y):\ \calX(M)\to \calX(M)$ 为曲率, 求证:  

(1) $R(X,Y)(fZ_1+gZ_2) =fR(X,Y)Z_1+gR(X,Y)Z_2$,  $\forall\ X,Y,Z_1,Z_2\in \calX(M), f,h\in C^\infty (M)$;

(2) $R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0$,  $\forall\ X,Y,Z\in \calX(M)$.  

 

2.(10') 设 $V(t),\ J(t)$ 是沿最短测地线 $\gamma(t),\ t\in [0,1]$ 的向量场, 它们满足 $$\bex V(t)\perp \dot\gamma(t),\quad J(t)\perp \dot\gamma(t),\quad V(0)=J(0),\quad V(1)=J(1), \eex$$ 且 $J(t)$ 是 Jacobi 场, 求证: $$\bex I(J,J)\leq I(V,V), \eex$$ 其中 $I$ 为 $\gamma$ 上的指标形式.  

 

3.(10') 设 $\gamma(t):\ (-\infty,+\infty)\to M$ 为一条测地直线, 相应地记 $$\bex \gamma_+=\gamma|_{[0,+\infty)},\quad \gamma_-=\gamma|_{(-\infty,0]} \eex$$ 及两 Busemann 函数 $$\bex B_{\gamma_+}(x)=\lim_{t\to+\infty}\sez{d(x,\gamma(t))-t}; \eex$$ $$\bex B_{\gamma_-}(x)=\lim_{t\to-\infty}\sez{d(x,\gamma(t))+t}. \eex$$ 求证: $$\bex B_{\gamma_+}+B_{\gamma_-}=0,\quad\mbox{在 } \gamma \mbox{ 上}; \eex$$ $$\bex B_{\gamma_+}+B_{\gamma_-}\geq 0,\quad\mbox{在 } M \mbox{ 上}. \eex$$  

 

4.(15') 设 $M$ 为紧流形, 再设 $g_{ij}(t)$ 满足 Ricci 流, 且 $f(t),\tau(t)$ 满足 $$\bex \frac{\p }{\p t}f=-\lap f+\sev{\n f}^2-R+\frac{n}{2\tau},\quad \frac{\p }{\p t}\tau =-1. \eex$$ 求证:  

(1) $$\bex \frac{\rd}{\rd t}\int_M \sex{4\pi \tau}^{-\frac{n}{2}} e^{-f}\, \rd vol_{g_{ij}}=0; \eex$$

(2) $$\bex & &\frac{\rd }{\rd x}\int_M  \sez{\tau \sex{R+\sev{\n f}^2} +f-n}(4\pi^\tau)^{-\frac{n}{2}} e^{-f}\,\rd vol_{g_{ij}}\\ & &=\int_M 2\tau \sev{R_{ij}+\n_i\n_j f-\frac{1}{2\tau}g_{ij}}^2 (4\pi \tau)^{-\frac{n}{2}} e^{-f}\,\rd vol_{g_{ij}}. \eex$$ 

张祖锦 赣南师范大学数学教师 微信: zhangzujin361 微信公众账号: 跟锦数学 E-mail:zhangzujin361@163.com
posted @ 2014-06-09 16:27  张祖锦  阅读(667)  评论(0编辑  收藏  举报