Navier-Stokes
equations
1 Let ω
be a domain in R3
, complement of a compact set B
. Consider the following boundary value problem in ω
:
ν△v=(v?ξ?ω×x)??v+?p+fdiv v=0}in ω,v|?ω=0, lim∣∣x∣∣→∞v(x)=0.(1)
We say that v:ω→R3
is a weak solution of (1)
iff the following conditions are satisfied:
(1)v∈D1,20(ω)
;
(2)v
obeys the following equation:
ν(?v,?φ)+((v?ξ?ω×x)??v,φ)+(f,φ)=0, ? φ∈D(ω),
where, we recall,
D(ω)={φ∈C∞0(ω); div φ=0},
D1,20(ω)= completion of D(ω) in the norm ∥?φ∥2,
and (g,h)=∫ωg?hdx
. Show that for any ξ, ω∈R3
and for every f
such that the map
is a bounded (linear) functional, problem (1)
has, at least, one weak solution.
2 Let
C(ω)={u:ω→R3; u is a weak solution to (1) with ξ=ω=0 andwith
given f, |u(x)|≤M/|x|, for some M>0}
Show that there exists γ>0
such that if v∈C(ω)
and
then v
is the unique weak solution in the class C
.
Hint. Prove and use the inequality
∫ω|u(x)|2(|x|+1)2dx≤4∫ω|ν(x)|2dx, u∈D1,20(ω).
Nonlinear
Elliptic PDEs
1 Let ω
be a bounded open set in Rn
. Let V
be the space of distributions on ω
, X=L2(ω)
, Y=H1(ω)
. Assume that u
is a solution of the equation
in X
for some F∈X∩Y
. Suppose that
(1)T:X→X
is contracting:
∥Tu?Tv∥X<θ∥v?w∥X, for some 0<θ<1, ? v,w∈X;
(2)T:Y→Y
is shrinking:
∥Tv∥Y<θ∥v∥Y, for some 0<θ<1, ? v∈Y;
(3)T?+F:X∩Y→X∩Y
. Prove that u∈Y
.
Note. You can not directly use the regularity
lifting theorem II. However, you can use the idea of proof there.
2 Let α
be a real number satisfying 0<α<n
and τ=n+αn?α
. Assume that u
is a positive solution of the integral equation
u(x)=∫Rn1|x?y|n?αuτ(y)|y|sdy(2)
with s≥0
. Also assume that u
is continous and
∫Rn[uτ?1(y)|y|s]n/αdy<∞, u∈Lq(Rn) for some q>nn?α.
Define
Let xλ=(2λ?x1,x2,?,xn)
be the reflection of x
in the plane
Prove that for λ
sufficiently negative, we have
Hint. You may use the fact that
u(x)?u(xλ)=∫Σλ[1|x?y|n?α?1∣∣xλ?y∣∣n?α][uτ(y)|y|s?uτ(yτ)|yτ|s]dy.
Note. If you have difficulty in proving (3)
for s≥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(?,t):Mn×[0,T)→Rn+1
是平均曲率流方程的解, 其中 Mn
是紧流形且初始超曲面 X0=X(?,0)
具有非负的平均曲率. 试证: 对于每个 t∈(0,T)
, X(?,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(?,t):Mn×[0,T)→Rn+1
满足
??tX(x,t)=H(x,t)n(x,t)+X(x,t), x∈Mn, t>0.
试证:
(1)??tgij(x,t)=2gij(x,t)?2Hhij(x,t)
;
(2)??tdμt=(n?H2)dμt
, 其中 dμt=det(gij(x,t))???????????√dx1?dxn
是超曲面 X(?,t)
的体积元.
Nonlinear
Conservation Laws
1证明:
u(x,t)=????23(t+3x+t2??????√),0,if 4x+t2>0,if 4x+t2<0.
是方程 ut+uux=0
的熵解 (entropy solution).
2考虑如下方程的 Riemann
问题
????????????????t+(?u)x=0,(?u)t+(?u2+?γ)x=0 (γ>1),(?,u)|t=0={(??,u?),(?+,u+),x<0,x>0.
如果 ??=0
, ?+>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 m∈R
, Λm(ξ)=(1+|ξ|2)m/2, ξ∈Rn
. 证明: Λ(ξ)∈Sm(Rn)
; 且对 ? s∈R
, Λm(D): Hs(Rn)→Hs?m(Rn)
是连续的.
2已知引理. 令 K∈C(Rn×Rn)
满足
supy∫Rn|K(x,y)|dx≤C, supx∫Rn|K(x,y)|dy≤C.
若 Ku(x)=∫RnK(x,y)u(y)dy
, 则 ∥Ku∥2≤C∥u∥2
.
假设 a(x,ξ)∈S?n?1(Rn)
. 证明: K(x,x?y)=∫Rne?i(x?y)?ξa(x,ξ)dξ
满足引理的条件.
3若 a(x,ξ)∈S?n?1(Rn)
. 证明: a(x,D):L2→L2
是连续的, 并由此导出对 ? a(x,ξ)∈S?1(Rn)
, a(x,D):L2(Rn)→L2(Rn)
连续.
4 设 a(x,ξ)∈S0(Rn)
. 证明: a(x,D):L2(Rn)→L2(Rn)
连续.
5设 a(x,ξ)∈Sm(Rn)
. 证明: a(x,D): Hs(Rn)→Hs?m(Rn)
连续.
[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,布布扣,bubuko.com
[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
原文:http://www.cnblogs.com/zhangzujin/p/3550479.html