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紧嵌入定理
下面证明嵌入
$$\Vert f\Vert_{L^p(\Omega)}\le C\Vert f\Vert_{H^{m,q}(\Omega)},\quad \,\forall\,\frac{n}{p}\ge \frac{n}{q}-m,$$
当不等式严格时, 即$\frac{n}{p}>\frac{n}{q}-m,$ 嵌入是紧的,
将有界集映到列紧集.
首先我们来简化证明的情形. 假设$m=1$时已证好, 那么$\,\forall\,m>1,$
取$q_1>q,$ 使得 $\frac{n}{q_1}-1=\frac{n}{q}-m<\frac{n}{p},$
从而由Sobolev不等式, 我们有
$$H^{m,q}\hookrightarrow H^{1,q_1}\stackrel{c}\hookrightarrow L^p$$
因此嵌入将有界集映成有界集再映至列紧集. 从而只需证$m=1$时的情形.
假设$p>q$的情形已证好, 那么当$p\le q$时, $\,\exists\,p_1>q\ge p,$
$\frac{n}{q}>\frac{n}{p_1}>\frac{n}{q}-1.$ 有嵌入
$$H^{1,q}\stackrel{c}\hookrightarrow L^{p_1}\hookrightarrow L^p.$$
这将有界集映至列紧集再映成列紧集, 因此只需证$p>q$的情形.
此时$\frac{n}{q}>\frac{n}{p}>\frac{n}{q}-1.$
首先我们考虑如下引理:
引理 1. $\,\forall\,f\in C_c^\infty(\mathbb{R}^n),$ $\frac{n}{p}=\theta+\frac{n}{q}-1,$ $\theta\in(0,1),$ 我们有如下不等式:
$$\Vert f(x+h)-f(x)\Vert_{L^p(\mathbb{R}^n)}\le C|h|^{\theta}\Vert f\Vert_{H^{1,q}(\mathbb{R}^n)}.$$
证: 事实上右侧可以取齐次模.
记$\frac{n}{p}>\frac{n}{q}-1=\frac{n}{p_1}.$ $q<p<p_1,$
那么$\frac{1}{p}=\frac{\theta}{q}+\frac{1-\theta}{p_1}.$ 从而
$$\begin{aligned}
\Vert f(x+h)-f(x)\Vert_{L^p}&\le \Vert f(x+h)-f(x)\Vert_{L^q}^\theta \Vert f(x+h)-f(x)\Vert_{L^{p_1} }^{1-\theta}\\
&\le C\left\Vert\int_0^1 \nabla f(x+sh)\cdot hds\right\Vert_{L^{q} }^\theta\Vert f\Vert_{\dot{H}^{1,q} }^{1-\theta}\\
&\le C |h|^\theta \Vert f\Vert^{1-\theta}_{\dot{H}^{1,q} }
\end{aligned}$$
推论 2. $\Vert\alpha_\varepsilon\ast f-f\Vert_{L^p(\mathbb{R}^n)}\le C\varepsilon^{\theta}\Vert f\Vert_{H^{1,q}(\mathbb{R}^n)}$
因此设$\{f_{\nu}\}$为$H^{1,q}(\mathbb{R}^n)$有界函数列,
那么$\,\forall\,\varepsilon>0,$
$\{\alpha_\varepsilon\ast f_\nu\}$为$H^{m,q}$有界函数列,
$\,\forall\,m\ge 1.$
因为对$\alpha_\varepsilon\ast f_{\nu}$的求导可归结到$\alpha_\varepsilon$上.
设对足够大的$m$有$\frac{n}{q}-m<-1,$ 那么由负指数情形的Sobolev不等式,
$\{\alpha_\varepsilon\ast f_\nu\}$是$C_{loc}^1$有界的.
从而由Arzela-Ascoli引理, 其有在任意紧集上$C^0$收敛的子列,
即$C_{loc}^{0}$紧. 自然地, 紧集上收敛列在$L^{p}$模意义下也收敛,
因此是$L^{p}_{loc}$紧的.
由推论,
对$\nu$一致地有$\alpha_\varepsilon\ast f_{\nu} \xrightarrow{L^p} f_{\nu}.$
由对角线法可证$\{f_{\nu}\}$有子列在任意紧集上成为Cauchy序列,
从而是$L^p_{loc}$紧的.
因此$H^{1,q}(\mathbb{R}^n)$到$L^p_{loc}(\mathbb{R}^n)$的嵌入是紧的.
PDE导论
称$\sum_{|\alpha|\le m}a_\alpha(x)\partial^\alpha u=f(x)$为$m$阶线性PDE,
$a_\alpha,f\in C^\infty.$ 若解$u\in C^m,$ 称其为经典解;
若$u\in C^{m-1},$ 在曲面$\{\psi(x)=0\}$上仅沿法向$m$阶第一类间断,
其它$C^m,$ 则称$u$为弱间断解. 该曲面称为弱间断面.
若曲面$\psi(x)=x_1=0,$ 则法向导数为$\frac{\partial {} }{\partial {}x_1}.$
对一弱间断解$u,$ 倘若对于$\beta=(m,0,\cdots,0),$ 在曲面$\psi(x)=0$上,
有$a_\beta(x)\neq 0,$ 那么在该点附近有小邻域,
使得$$\partial_{x_1}^m u(x)=\sum\limits_{\begin{subarray}{c}
|\alpha|\le m\\
\alpha\neq \beta
\end{subarray} } \frac{a_\alpha(x)}{a_\beta(x)}\partial^\alpha u(x)\in C^0$$
与$u$为弱间断解矛盾. 因此$a_\beta(x)=0$沿弱间断面发生.
对任意曲面$\psi(x)=0,$ 通过展平技巧, 令$y_1=\psi,$ $y_i=x_i,$ 则
$$\left\{\begin{aligned}
\frac{\partial {}u}{\partial {}x_1}&=\frac{\partial {}u}{\partial {}y_1}\frac{\partial {}\psi}{\partial {}x_1}\\
\frac{\partial {}u}{\partial {}x_i}&=\frac{\partial {}u}{\partial {}y_1}\frac{\partial {}\psi}{\partial {}x_i}+\frac{\partial {}u}{\partial {}y_i}
\end{aligned}\right.$$
此时弱间断面上满足$\widetilde{a}_\beta(x)=\sum_{|\alpha|\le m}a_\alpha(x)(\nabla \psi)^\alpha=0.$
称$\psi(x)=0$为特征曲面,
若其满足$\sum_{|\alpha|\le m}a_\alpha(x)(\nabla \psi)^\alpha=0.$
称其法向为特征方向. 因此弱间断面必为特征曲面.
PDE理论中, 一个经典的存在性定理是Cauchy-Kovalevskaya定理.
它说明若参数函数都是实解析的, 那么给定任意解析初值,
解析解总是唯一存在的. 然而该定理使用的技术较为久远,
我们采用现代化的方法来说明.
我们考虑解$\frac{\partial {}u}{\partial {}t}=F(t,x,u,\nabla u),$
化作积分方程$u(t,x)=u_0(x)+\int_0^t F(\tau,x,u,\nabla u)d\tau.$
在ODE理论中, 由于$F$中不再涉及导数,
我们可以采用Picard迭代的方法来得到唯一解.
倘若我们可以取到合适的关于$x$的模$\Vert\cdot\Vert$,
使得$\Vert\nabla u\Vert$可以被较好地控制, 那么对于
$$\Vert u(t,x)\Vert\le \Vert u_0\Vert+\int_0^t \Vert F(\tau,x,u,\nabla u)\Vert d\tau,$$
可以从ODE视角来分析处理. 特别地通过这种方式可以给出解的理论估计.
对一般的函数, 显然无法用$\Vert u\Vert$控制$\Vert\nabla u\Vert.$
但是对于解析函数, 导数与函数值是一体的, 因此可以取到这样的模.
对解析函数$f(x)=\sum_\alpha \frac{\partial^\alpha f(0)}{\alpha!}x^\alpha,$
记其收敛半径为$R.$ 那么$\,\forall\,r\le R,$ 定义
$$\Vert f\Vert_r=\sum_\alpha \frac{|\partial^\alpha f(0)|}{\alpha!}r^{|\alpha|}<\infty.$$
显然它是一个模, 且满足$\Vert fg\Vert_r\le \Vert f\Vert_r\Vert g\Vert_r.$
特别地$\Vert f^n\Vert_r\le \Vert f\Vert_r^n.$
函数$f$是解析的当且仅当存在$r$使得$\Vert f\Vert_r<\infty.$
由于导数项只在积分中出现, 我们只需在积分意义下控制$\Vert\nabla u\Vert:$
$$\begin{aligned}
\left\Vert\int_{0}^t \partial_{x_1} u(\tau,x)d\tau\right\Vert_{R-MT}&\le \int_0^t \sum_\alpha \frac{|\partial^\alpha \partial_{x_1}u(\tau,0)|}{\alpha! (\alpha_1+1)}(\alpha_1+1)(R-MT)^{|\alpha|}d\tau\\
&\le \int_0^t \sum_{\alpha} \frac{|\partial^\alpha u(\tau,0)|}{\alpha!}(R-M\tau)^{|\alpha|} \frac{\alpha_1}{R-MT}\left(\frac{R-MT}{R-M\tau}\right)^{|\alpha|}d\tau\\
&\le \sup_{\tau\in[0,t]}\Vert u(\tau,x)\Vert_{R-M\tau} \int_0^t\sum_k \frac{k}{R-MT}\left(\frac{R-MT}{R-M\tau}\right)^kd\tau\\
&=\sup_{\tau\in[0,t]}\Vert u(\tau,x)\Vert_{R-M\tau} \int_0^t \frac{R-M\tau}{(MT-M\tau)^2} d\tau\\
&=C(M,R,T,t)\sup_{\tau\in[0,t]}\Vert u(\tau,x)\Vert_{R-M\tau}
\end{aligned}$$
注意解的解析半径不见得是固定的, 随时间$t$可能有衰减,
因此分析时取的模也不能在固定$r$上考察.
然而, C-K定理只说明了解的存在性与唯一性,
Hadamard指出解的稳定性也是需要考量的. 也就是要关注解对初值的依赖性.
文章最后更新于 2021-11-18 20:12:53
compact embedding theorem
The following proves that embedding
$$\Vert f\Vert_{L^p(\Omega)}\le C\Vert f\Vert_{H^{m,q}(\Omega)},\quad \,\forall\,\frac{n}{p}\ge \frac{n}{q}-m,$$
When the inequality is strict, that is, the $\frac{n}{p}>\frac{n}{q}-m,$ embedding is compact,
Map bounded sets to compact sets.
First, let us simplify the proof situation. Assume that $m=1$ has been proved, then $\,\forall\,m>1,$
Take $q_1>q,$ such that $\frac{n}{q_1}-1=\frac{n}{q}-m<\frac{n}{p},$
Thus from Sobolev's inequality, we have
$$H^{m,q}\hookrightarrow H^{1,q_1}\stackrel{c}\hookrightarrow L^p$$
Therefore, embedding maps bounded sets to bounded sets and then to compact sets. Therefore, we only need to prove the situation when $m=1$.
Assuming that the situation of $p>q$ has been proved, then when $p\le q$, $\,\exists\,p_1>q\ge p,$
$\frac{n}{q}>\frac{n}{p_1}>\frac{n}{q}-1.$ Embedded
$$H^{1,q}\stackrel{c}\hookrightarrow L^{p_1}\hookrightarrow L^p.$$
This maps bounded sets to column compact sets and back to column compact sets, so we only need to prove the case of $p>q$.
At this time $\frac{n}{q}>\frac{n}{p}>\frac{n}{q}-1.$
First we consider the following lemma:
Lemma 1. $\,\forall\,f\in C_c^\infty(\mathbb{R}^n),$ $\frac{n}{p}=\theta+\frac{n}{q}-1,$ $\theta\in(0,1),$ We have the following inequality:
$$\Vert f(x+h)-f(x)\Vert_{L^p(\mathbb{R}^n)}\le C|h|^{\theta}\Vert f\Vert_{H^{1,q}(\mathbb{R}^n)}.$$
Certificate: In fact, the right side can take the homogeneous module.
Note $\frac{n}{p}>\frac{n}{q}-1=\frac{n}{p_1}.$ $q<p<p_1,$
Then $\frac{1}{p}=\frac{\theta}{q}+\frac{1-\theta}{p_1}.$ thus
$$\begin{aligned}
\Vert f(x+h)-f(x)\Vert_{L^p}&\le \Vert f(x+h)-f(x)\Vert_{L^q}^\theta \Vert f(x+h)-f(x)\Vert_{L^{p_1} }^{1-\theta}\\
&\le C\left\Vert\int_0^1 \nabla f(x+sh)\cdot hds\right\Vert_{L^{q} }^\theta\Vert f\Vert_{\dot{H}^{1,q} }^{1-\theta}\\
&\le C |h|^\theta \Vert f\Vert^{1-\theta}_{\dot{H}^{1,q} }
\end{aligned}$$
Corollary 2. $\Vert\alpha_\varepsilon\ast f-f\Vert_{L^p(\mathbb{R}^n)}\le C\varepsilon^{\theta}\Vert f\Vert_{H^{1,q}(\mathbb{R}^n)}$
Therefore, let $\{f_{\nu}\}$ be $H^{1,q}(\mathbb{R}^n)$ a bounded function sequence,
Then $\,\forall\,\varepsilon>0,$
$\{\alpha_\varepsilon\ast f_\nu\}$ is $H^{m,q}$ bounded function sequence,
$\,\forall\,m\ge 1.$
Because the derivation of $\alpha_\varepsilon\ast f_{\nu}$ can be attributed to $\alpha_\varepsilon$.
Suppose there is $\frac{n}{q}-m<-1,$ for $m$ that is large enough, then from the Sobolev inequality in the negative exponential case,
$\{\alpha_\varepsilon\ast f_\nu\}$ is bounded by $C_{loc}^1$.
Therefore, according to the Arzela-Ascoli lemma, which has a subsequence that converges on any compact set $C^0$,
That is, $C_{loc}^{0}$ is compact. Naturally, a convergent sequence on a compact set also converges in the modular sense of $L^{p}$,
Therefore it is $L^{p}_{loc}$ tight.
By inference,
Consistently have $\alpha_\varepsilon\ast f_{\nu} \xrightarrow{L^p} f_{\nu}.$ for $\nu$
It can be proved by the diagonal method that $\{f_{\nu}\}$ has a subsequence that becomes a Cauchy sequence on any compact set,
Therefore it is $L^p_{loc}$ tight.
Therefore the embedding from $H^{1,q}(\mathbb{R}^n)$ to $L^p_{loc}(\mathbb{R}^n)$ is tight.
Introduction to PDE
Call $\sum_{|\alpha|\le m}a_\alpha(x)\partial^\alpha u=f(x)$ a linear PDE of order $m$,
$a_\alpha,f\in C^\infty.$ If $u\in C^m,$ is solved, call it Classic solution;
If $u\in C^{m-1},$ is a discontinuity of the first type only along the normal direction $m$ on the surface $\{\psi(x)=0\}$,
The other $C^m,$ is called $u$. Weak discontinuity solution. This surface is called weak discontinuity.
If the surface is $\psi(x)=x_1=0,$, then the normal derivative is $\frac{\partial {} }{\partial {}x_1}.$
For a weakly discontinuous solution $u,$, if $\beta=(m,0,\cdots,0),$ is on the surface $\psi(x)=0$,
If there is $a_\beta(x)\neq 0,$, then there is a small neighborhood near the point,
Make$$\partial_{x_1}^m u(x)=\sum\limits_{\begin{subarray}{c}
|\alpha|\le m\\
\alpha\neq \beta
\end{subarray} } \frac{a_\alpha(x)}{a_\beta(x)}\partial^\alpha u(x)\in C^0$$
It is inconsistent with $u$ as a weak discontinuity solution. Therefore $a_\beta(x)=0$ occurs along the weak discontinuity surface.
For any surface $\psi(x)=0,$ through the flattening technique, let $y_1=\psi,$ $y_i=x_i,$ be
$$\left\{\begin{aligned}
\frac{\partial {}u}{\partial {}x_1}&=\frac{\partial {}u}{\partial {}y_1}\frac{\partial {}\psi}{\partial {}x_1}\\
\frac{\partial {}u}{\partial {}x_i}&=\frac{\partial {}u}{\partial {}y_1}\frac{\partial {}\psi}{\partial {}x_i}+\frac{\partial {}u}{\partial {}y_i}
\end{aligned}\right.$$
At this time, the weak discontinuity surface satisfies $\widetilde{a}_\beta(x)=\sum_{|\alpha|\le m}a_\alpha(x)(\nabla \psi)^\alpha=0.$
Call $\psi(x)=0$ Feature surface,
If it satisfies $\sum_{|\alpha|\le m}a_\alpha(x)(\nabla \psi)^\alpha=0.$
Call its normal direction Feature direction. Therefore The weak discontinuity surface must be a characteristic surface.
In PDE theory, a classic existence theorem is the Cauchy-Kovalevskaya theorem.
It shows that if the parameter functions are all real analytic, then given any analytic initial value,
Analytical solutions are always the only ones that exist. However, the technology used by this theorem is older,
We use a modern approach to illustrate.
We consider the solution $\frac{\partial {}u}{\partial {}t}=F(t,x,u,\nabla u),$
Transformed into an integral equation $u(t,x)=u_0(x)+\int_0^t F(\tau,x,u,\nabla u)d\tau.$
In ODE theory, since derivatives are no longer involved in $F$,
We can use Picard iteration method to get the unique solution.
If we can get the appropriate module $\Vert\cdot\Vert$ about $x$,
So that $\Vert\nabla u\Vert$ can be better controlled, then for
$$\Vert u(t,x)\Vert\le \Vert u_0\Vert+\int_0^t \Vert F(\tau,x,u,\nabla u)\Vert d\tau,$$
It can be analyzed and processed from the ODE perspective. In particular, a theoretical estimate of the solution can be given in this way.
For general functions, it is obviously impossible to use $\Vert u\Vert$ to control $\Vert\nabla u\Vert.$
But for analytic functions, the derivative and the function value are integrated, so such a module can be obtained.
For analytic function $f(x)=\sum_\alpha \frac{\partial^\alpha f(0)}{\alpha!}x^\alpha,$
Record its convergence radius as $R.$, then $\,\forall\,r\le R,$ is defined
$$\Vert f\Vert_r=\sum_\alpha \frac{|\partial^\alpha f(0)|}{\alpha!}r^{|\alpha|}<\infty.$$
Obviously it is a module and satisfies $\Vert fg\Vert_r\le \Vert f\Vert_r\Vert g\Vert_r.$
Especially $\Vert f^n\Vert_r\le \Vert f\Vert_r^n.$
A function $f$ is analytic if and only if there exists $r$ such that $\Vert f\Vert_r<\infty.$
Since the derivative term only appears in the integral, we only need to control $\Vert\nabla u\Vert:$ in the integral sense
$$\begin{aligned}
\left\Vert\int_{0}^t \partial_{x_1} u(\tau,x)d\tau\right\Vert_{R-MT}&\le \int_0^t \sum_\alpha \frac{|\partial^\alpha \partial_{x_1}u(\tau,0)|}{\alpha! (\alpha_1+1)}(\alpha_1+1)(R-MT)^{|\alpha|}d\tau\\
&\le \int_0^t \sum_{\alpha} \frac{|\partial^\alpha u(\tau,0)|}{\alpha!}(R-M\tau)^{|\alpha|} \frac{\alpha_1}{R-MT}\left(\frac{R-MT}{R-M\tau}\right)^{|\alpha|}d\tau\\
&\le \sup_{\tau\in[0,t]}\Vert u(\tau,x)\Vert_{R-M\tau} \int_0^t\sum_k \frac{k}{R-MT}\left(\frac{R-MT}{R-M\tau}\right)^kd\tau\\
&=\sup_{\tau\in[0,t]}\Vert u(\tau,x)\Vert_{R-M\tau} \int_0^t \frac{R-M\tau}{(MT-M\tau)^2} d\tau\\
&=C(M,R,T,t)\sup_{\tau\in[0,t]}\Vert u(\tau,x)\Vert_{R-M\tau}
\end{aligned}$$
Note that the analytical radius of the solution is not necessarily fixed, and may decay over time $t$.
Therefore, the model taken during analysis cannot be examined on the fixed $r$.
However, the C-K theorem only explains the existence and uniqueness of understanding.
Hadamard pointed out that the solution Stability It also needs to be considered. That is to say, we must pay attention to the dependence of the solution on the initial value.
The article was last updated on 2021-11-18 20:12:53
