# [家里蹲大学数学杂志]第013期2010年西安偏微分方程暑期班试题---NSE,非线性椭圆,平均曲率流,非线性守恒律,拟微分算子

Navier-Stokes equations

1 Let $\omega$ be a domain in $\bbR^3$, complement of a compact set $\mathcal{B}$. Consider the following boundary value problem in $\omega$: $$\bee\label{NS:1} \left. \ba{cc} \left.\ba{ll} \nu \lap v=(v-\xi-\omega\times x) \cdot\n v+\n p+f\\ \Div v=0 \ea\right\}\mbox{in }\omega,\\ v|_{\p \omega}=0,\ \dps{\lim_{\sev{x}\to \infty}v(x)=0.} \ea \right. \eee$$We say that $v:\omega\to \bbR^3$ is a weak solution of \eqref{NS:1} iff the following conditions are satisfied:

(1)$v\in \mathcal{D}^{1,2}_0(\omega)$;

(2)$v$ obeys the following equation: $$\bex \nu\sex{\n v,\n \varphi} +\sex{\sex{v-\xi-\omega\times x}\cdot\n v,\varphi} +\sex{f,\varphi}=0,\ \forall\ \varphi\in \mathcal{D}(\omega), \eex$$ where, we recall, $$\bex \mathcal{D}(\omega)=\sed{\varphi\in C^\infty_0(\omega);\ \Div \varphi=0}, \eex$$ $$\bex \mathcal{D}^{1,2}_0(\omega)=\mbox{ completion of }\mathcal{D}(\omega)\mbox{ in the norm }\sen{\n \varphi}_2, \eex$$ and $\dps{\sex{g,h}=\int_\omega g\cdot h\rd x}$. Show that for any $\xi,\ \omega\in \bbR^3$ and for every $f$ such that the map $$\bex \mathcal{D}^{1,2}_0(\omega)\ni \varphi \mapsto \sex{f,\varphi}\in \bbR \eex$$ is a bounded (linear) functional, problem \eqref{NS:1} has, at least, one weak solution.

2 Let $$\bex \mathcal{C}(\omega)=\sed{u:\omega\to \bbR^3;\ \left.\ba{ll} u\mbox{ is a weak solution to } \eqref{NS:1}\mbox{ with }\xi=\omega=0\mbox{ and}\\ \mbox{with given }f,\ \sev{u(x)}\leq M/\sev{x},\mbox{ for some }M>0 \ea\right.} \eex$$ Show that there exists $\gamma>0$ such that if $v\in \mathcal{C}(\omega)$ and $$\bex \sup_{x\in \omega} \sez{(\sev{x}+1)\sev{v(x)}}<\gamma, \eex$$ then $v$ is the unique weak solution in the class $\mathcal{C}$.

Hint. Prove and use the inequality $$\bex \int_\omega \frac{\sev{u(x)}^2}{\sex{\sev{x}+1}^2} d x \leq 4 \int_\omega \sev{\nu(x)}^2 d x,\ u\in \mathcal{D}^{1,2}_0(\omega). \eex$$

Nonlinear Elliptic PDEs

1 Let $\omega$ be a bounded open set in $\bbR^n$. Let $V$ be the space of distributions on $\omega$, $X=L^2(\omega)$, $Y=H^1(\omega)$. Assume that $u$ is a solution of the equation $$\bex u=Tu+F \eex$$ in $X$ for some $F\in X\cap Y$. Suppose that

(1)$T:X\to X$ is contracting: $$\bex \sen{Tu-Tv}_X<\theta\sen{v-w}_X,\mbox{ for some }0<\theta<1,\ \forall\ v,w\in X; \eex$$

(2)$T:Y\to Y$ is shrinking: $$\bex \sen{Tv}_Y<\theta\sen{v}_Y,\ \mbox{for some }0<\theta<1,\ \forall\ v\in Y; \eex$$

(3)$T\cdot +F:X\cap Y\to X\cap Y$. Prove that $u\in Y$.

Note. You can not directly use the regularity lifting theorem II. However, you can use the idea of proof there.

2 Let $\alpha$ be a real number satisfying $0<\alpha<n$ and　$\dps{\tau=\frac{n+\alpha}{n-\alpha}}$. Assume that $u$ is a positive solution of the integral equation $$\bee\label{NEPde:1} u(x)=\int_{\bbR^n} \frac{1}{\sev{x-y}^{n-\alpha}}\frac{u^\tau(y)}{\sev{y}^s}dy \eee$$with $s\geq 0$. Also assume that $u$ is continous and $$\bex \int_{\bbR^n} \sez{ \frac{u^{\tau-1}(y)}{\sev{y}^s} }^{n/\alpha}dy<\infty,\ u\in L^q(\bbR^n)\mbox{ for some }q>\frac{n}{n-\alpha}. \eex$$ Define $$\bex \varSigma_\lambda =\sed{x=(x_1,\cdots,x_n);\ x_1<\lambda}. \eex$$ Let $x^\lambda=(2\lambda-x_1,x_2,\cdots,x_n)$ be the reflection of $x$ in the plane $$\bex T_\lambda=\sed{x;\ x_1=\lambda}. \eex$$ Prove that for $\lambda$ sufficiently negative, we have $$\bee\label{NEPde:2} u(x^\lambda)\geq u(x),\ \forall\ x\in \varSigma_\lambda. \eee$$

Hint. You may use the fact that $$\bex u(x)-u(x^\lambda) =\int_{\varSigma_\lambda} \sez{\frac{1}{\sev{x-y}^{n-\alpha}}- \frac{1}{\sev{x^\lambda-y}^{n-\alpha}}} \sez{ \frac{u^\tau(y)}{\sev{y}^s} -\frac{u^\tau(y^\tau)}{\sev{y^\tau}^s} } dy. \eex$$

Note. If you have difficulty in proving \eqref{NEPde:2} for $s\geq 0$, then you can prove it in the special case when $s=0$, and you can still get $80\%$ of the credit.

Mean Curvature Flow

1假设 $X(\cdot,t):M^n\times [0,T)\to \bbR^{n+1}$ 是平均曲率流方程的解, 其中 $M^n$ 是紧流形且初始超曲面 $X_0=X(\cdot,0)$ 具有非负的平均曲率. 试证: 对于每个 $t\in (0,T)$, $X(\cdot,t)$ 或者是极小曲面 (即平均曲率为零), 或者具有正平均曲率.

2如果一族超曲面 $X(\cdot,t):M^n\times [0,T)\to \bbR^{n+1}$ 满足 $$\bex \frac{\p }{\p t}X(x,t) =H(x,t)n(x,t)+X(x,t),\ x\in M^n,\ t>0. \eex$$ 试证:

(1)$\dps{\frac{\p}{\p t}g_{ij}(x,t)=2g_{ij}(x,t)-2Hh_{ij}(x,t)}$;

(2)$\dps{\frac{\p }{\p t}d\mu_t =(n-H^2)d\mu_t}$, 其中 $\dps{d\mu_t=\sqrt{\det\sex{g_{ij}(x,t)}}\rd x^1\cdots \rd x^n}$ 是超曲面 $X(\cdot,t)$ 的体积元.

Nonlinear Conservation Laws

1证明: $$\bex u(x,t)=\left\{\ba{ll} -\frac{2}{3}\sex{t+\sqrt{3x+t^2}},&\mbox{if }4x+t^2>0,\\ 0,&\mbox{if }4x+t^2<0. \ea\right. \eex$$ 是方程 $u_t+uu_x=0$ 的熵解 (entropy solution).

2考虑如下方程的 $Riemann$ 问题 $$\bex \left\{\ba{lll} \varrho_t+\sex{\varrho u}_x=0,\\ \sex{\varrho u}_t +\sex{\varrho u^2+\varrho^\gamma}_x=0\ \sex{\gamma>1},\\ \sex{\varrho, u}|_{t=0} =\left\{\ba{ll} \sex{\varrho_-,u_-},&x<0,\\ \sex{\varrho_+,u_+},&x>0. \ea\right. \ea\right. \eex$$ 如果 $\varrho_-=0$, $\varrho_+>0$, 上述 $Riemann$ 问题能用激波直接连接么? 请说明原因. 如果不能用激波直接连接, 用稀疏波能够连接真空么? 如果能, 请写出解答表达式.

Pseudo-differential Operators

1设 $m\in \bbR$, $\varLambda^m(\xi)=\sex{1+\sev{\xi}^2}^{m/2},\ \xi\in \bbR^n$. 证明: $\varLambda(\xi)\in S^m(\bbR^n)$; 且对 $\forall\ s\in \bbR$, $\varLambda^m(D):\ H^s(\bbR^n)\to H^{s-m}(\bbR^n)$ 是连续的.

2已知引理. 令 $K\in C(\bbR^n\times \bbR^n)$ 满足 $$\bex \sup_y\int_{\bbR^n}\sev{K(x,y)}\rd x\leq C,\ \sup_x\int_{\bbR^n}\sev{K(x,y)}dy\leq C. \eex$$ 若 $\dps{Ku(x)=\int_{\bbR^n}K(x,y)u(y)dy}$, 则 $\dps{\sen{Ku}_2\leq C\sen{u}_2}$.

3若 $a(x,\xi)\in S^{-n-1}(\bbR^n)$. 证明: $a(x,D):L^2\to L^2$ 是连续的, 并由此导出对 $\forall\ a(x,\xi)\in S^{-1}(\bbR^n)$, $a(x,D):L^2(\bbR^n)\to L^2(\bbR^n)$ 连续.

4 设 $a(x,\xi)\in S^0(\bbR^n)$. 证明: $a(x,D):L^2(\bbR^n)\to L^2(\bbR^n)$ 连续.

5设 $a(x,\xi)\in S^m(\bbR^n)$. 证明: $a(x,D):\ H^s(\bbR^n)\to H^{s-m}(\bbR^n)$ 连续.

posted @ 2014-05-18 16:33  张祖锦  阅读(665)  评论(0编辑  收藏  举报