Navier-Stokes equations
1 Let $omega$ be a domain in $bR^3$, complement of a compact set $mathcal{B}$. Consider the following boundary value problem in $omega$: $$eelabel{NS:1} left. a{cc} left.a{ll} u lap v=(v-xi-omega imes x) cdot v+ p+f\ Div v=0 ea ight}mbox{in }omega,\ v|_{p omega}=0, dps{lim_{sev{x} o infty}v(x)=0.} ea ight. eee$$We say that $v:omega o bR^3$ is a weak solution of eqref{NS:1} iff the following conditions are satisfied:
(1)$vin mathcal{D}^{1,2}_0(omega)$;
(2)$v$ obeys the following equation: $$ex usex{ v, varphi} +sex{sex{v-xi-omega imes x}cdot v,varphi} +sex{f,varphi}=0, forall varphiin mathcal{D}(omega), eex$$ where, we recall, $$ex mathcal{D}(omega)=sed{varphiin C^infty_0(omega); Div varphi=0}, eex$$ $$ex mathcal{D}^{1,2}_0(omega)=mbox{ completion of }mathcal{D}(omega)mbox{ in the norm }sen{ varphi}_2, eex$$ and $dps{sex{g,h}=int_omega gcdot h d x}$. Show that for any $xi, omegain bR^3$ and for every $f$ such that the map $$ex mathcal{D}^{1,2}_0(omega) i varphi mapsto sex{f,varphi}in bR eex$$ is a bounded (linear) functional, problem eqref{NS:1} has, at least, one weak solution.
2 Let $$ex mathcal{C}(omega)=sed{u:omega o bR^3; left.a{ll} umbox{ is a weak solution to } eqref{NS:1}mbox{ with }xi=omega=0mbox{ and}\ mbox{with given }f, sev{u(x)}leq M/sev{x},mbox{ for some }M>0 ea ight.} eex$$ Show that there exists $gamma>0$ such that if $vin mathcal{C}(omega)$ and $$ex sup_{xin omega} sez{(sev{x}+1)sev{v(x)}}<gamma, eex$$ then $v$ is the unique weak solution in the class $mathcal{C}$.
Hint. Prove and use the inequality $$ex int_omega frac{sev{u(x)}^2}{sex{sev{x}+1}^2} d x leq 4 int_omega sev{ u(x)}^2 d x, uin mathcal{D}^{1,2}_0(omega). eex$$
Nonlinear Elliptic PDEs
1 Let $omega$ be a bounded open set in $bR^n$. 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 $$ex u=Tu+F eex$$ in $X$ for some $Fin Xcap Y$. Suppose that
(1)$T:X o X$ is contracting: $$ex sen{Tu-Tv}_X< hetasen{v-w}_X,mbox{ for some }0< heta<1, forall v,win X; eex$$
(2)$T:Y o Y$ is shrinking: $$ex sen{Tv}_Y< hetasen{v}_Y, mbox{for some }0< heta<1, forall vin Y; eex$$
(3)$Tcdot +F:Xcap Y o Xcap Y$. Prove that $uin Y$.
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 $dps{ au=frac{n+alpha}{n-alpha}}$. Assume that $u$ is a positive solution of the integral equation $$eelabel{NEPde:1} u(x)=int_{bR^n} frac{1}{sev{x-y}^{n-alpha}}frac{u^ au(y)}{sev{y}^s}dy eee$$with $sgeq 0$. Also assume that $u$ is continous and $$ex int_{bR^n} sez{ frac{u^{ au-1}(y)}{sev{y}^s} }^{n/alpha}dy<infty, uin L^q(bR^n)mbox{ for some }q>frac{n}{n-alpha}. eex$$ Define $$ex varSigma_lambda =sed{x=(x_1,cdots,x_n); x_1<lambda}. eex$$ Let $x^lambda=(2lambda-x_1,x_2,cdots,x_n)$ be the reflection of $x$ in the plane $$ex T_lambda=sed{x; x_1=lambda}. eex$$ Prove that for $lambda$ sufficiently negative, we have $$eelabel{NEPde:2} u(x^lambda)geq u(x), forall xin varSigma_lambda. eee$$
Hint. You may use the fact that $$ex u(x)-u(x^lambda) =int_{varSigma_lambda} sez{frac{1}{sev{x-y}^{n-alpha}}- frac{1}{sev{x^lambda-y}^{n-alpha}}} sez{ frac{u^ au(y)}{sev{y}^s} -frac{u^ au(y^ au)}{sev{y^ au}^s} } dy. eex$$
Note. If you have difficulty in proving eqref{NEPde:2} for $sgeq 0$, 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 $X(cdot,t):M^n imes [0,T) o bR^{n+1}$ 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 $tin (0,T)$, $X(cdot,t)$ 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 $X(cdot,t):M^n imes [0,T) o bR^{n+1}$ 满足 $$ex frac{p }{p t}X(x,t) =H(x,t)n(x,t)+X(x,t), xin M^n, t>0. eex$$ 试证:
(1)$dps{frac{p}{p t}g_{ij}(x,t)=2g_{ij}(x,t)-2Hh_{ij}(x,t)}$;
(2)$dps{frac{p }{p t}dmu_t =(n-H^2)dmu_t}$, 其中 $dps{dmu_t=sqrt{detsex{g_{ij}(x,t)}} d x^1cdots d x^n}$ 是超曲面 $X(cdot,t)$ 的体积元.
Nonlinear Conservation Laws
1证明: $$ex u(x,t)=left{a{ll} -frac{2}{3}sex{t+sqrt{3x+t^2}},&mbox{if }4x+t^2>0,\ 0,&mbox{if }4x+t^2<0. ea ight. eex$$ 是方程 $u_t+uu_x=0$ 的熵解 (entropy solution).
2考虑如下方程的 $Riemann$ 问题 $$ex left{a{lll} varrho_t+sex{varrho u}_x=0,\ sex{varrho u}_t +sex{varrho u^2+varrho^gamma}_x=0 sex{gamma>1},\ sex{varrho, u}|_{t=0} =left{a{ll} sex{varrho_-,u_-},&x<0,\ sex{varrho_+,u_+},&x>0. ea ight. ea ight. eex$$ 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
注. 下列各题任选四题, 记分独立, 可以直接互相引用.
1设 $min bR$, $varLambda^m(xi)=sex{1+sev{xi}^2}^{m/2}, xiin bR^n$. 证明: $varLambda(xi)in S^m(bR^n)$; 且对 $forall sin bR$, $varLambda^m(D): H^s(bR^n) o H^{s-m}(bR^n)$ 是连续的.
2已知引理. 令 $Kin C(bR^n imes bR^n)$ 满足 $$ex sup_yint_{bR^n}sev{K(x,y)} d xleq C, sup_xint_{bR^n}sev{K(x,y)}dyleq C. eex$$ 若 $dps{Ku(x)=int_{bR^n}K(x,y)u(y)dy}$, 则 $dps{sen{Ku}_2leq Csen{u}_2}$.
假设 $a(x,xi)in S^{-n-1}(bR^n)$. 证明: $dps{K(x,x-y)= int_{bR^n}e^{-isex{x-y}cdotxi}a(x,xi)dxi}$ 满足引理的条件.
3若 $a(x,xi)in S^{-n-1}(bR^n)$. 证明: $a(x,D):L^2 o L^2$ 是连续的, 并由此导出对 $forall a(x,xi)in S^{-1}(bR^n)$, $a(x,D):L^2(bR^n) o L^2(bR^n)$ 连续.
4 设 $a(x,xi)in S^0(bR^n)$. 证明: $a(x,D):L^2(bR^n) o L^2(bR^n)$ 连续.
5设 $a(x,xi)in S^m(bR^n)$. 证明: $a(x,D): H^s(bR^n) o H^{s-m}(bR^n)$ 连续.