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\textbf{\boxed{\text{0基本概念}}}\
\boxed{\textbf{0.1Weak Formulation of Boundary Value problem}}\
考虑两点边值问题\
{\begin{align}
-\frac{d2u}{dx2}&=fin(0,1)\
u(0)&=0\
u^{'}(1)&=0
\end{align}}(0.1.1)\设\(f\)是解，\(v(0)=0\)是正规的任意函数，定义：\
\begin{align}
(f,v):=\int_0 ^1f(x)v(x)dx&=\int_0 1-u(x)v(x)dx\
&=\int_0 1u(x)v'(x)dx:=a(u,v)
\end{align*}(0.1.2)\
define:\
\[V=\{v\ \in L^2(0,1):a(v,v)<\infty and v(0)=0 \}
\]
我们称（0.1.1）的解\(u\)是典型的(\emph{\textbf{characterized}})，假如$$a(u,v)=(f,v),\forall v\in V$$
被称为（0.1.1）的weak formulation\
(0.1.3)被称为variational,因为函数\(v(x)\)可以是任意的。一个解释就是，在HIlbert空间\(L^2(0,1)\)里面有内积\((.,.)\)，我们可以看到空间\(V\)可以被视为Hilbert space with inner product \(a(,)\),就是在（0.1.2）里面定义的。\
是哪一中导数被用于定义\(a(,)\) ?\
(0.1.4)\boxed{\textbf{\textbf{Theorem}}\ suppose\ f \in C^0([0,1]) and\ u\in C^2([0,1]) s.t (0.1.3), Then\ u\ be the solves of (0.1.1)}\
proof：\
\((f,v)=a(u,v)\Longleftrightarrow \int_0 ^1 fv=\int_0 ^1 u^{'}v^{'}\)\
欲证明\(u^{''}=-f\),但是问题是\(\int_0 ^1 u^{'}v^{'}=\int_0 ^1(-u^{''})v+u^{'}(1)v(1)\)剩余部分留作\textcolor[rgb]{1.00,0.00,0.00}{\textbf{\emph{练习1}}}.\
\boxed{\emph{\textbf{0.2\ Ritz-Galerkin Approximation}}}\
\textbf{\textcolor[rgb]{0.00,0.50,0.00}{(0.2.2)Theorem. }}Given \(f\in L^2(0,1)\) (0.2.1)has a unique solution\
Remark: (0.2.1)\(u_s \in S\) such that \(a(u_s,v)=(f,v) \forall\ S\)\
\boxed{\emph{\textbf{0.3\ Error\ Estimates}}}\
\textbf{$$a(u_s-u,w)=0\ \forall\ w \in S\eqno(0.3.1)$$}\
Equation(0.3.1) and its subsequent variations are the key to the success of all Ritz-Galerkin methods.Now define\
\[\| v \|_{E}=\sqrt{a(v,v)}
\]
the energy norm.\
\[|a(v,w)| \leq \|v \|_{E}\|w\|_{E} \ Schwarz\ inequality\ \eqno(0.3.2)$$\\
\begin{align*}
\|u-u_s\|^2 &=a(u-u_s,u-u_s)\\
&=a(u-u_s,u-v)+a(u-u_s,v-u_s)\\
&=a(u-u_s,u-v)\ (from\ (0.3.1)\ with\ w=v-u_s)\\
&\leq \| u-u_s\|_{E} \|u-v\|_{E}\\we\ can\ divide\ by\ it\ to\ obtain\ \ \
\|u-u_s\| &\leq \|u-v\| \\ taking\ the\ infimum\ over\ v\in S\ yield\
\|u-u_s\| &\leq inf\{\|u-v\| :v \in S\} \ \ since\ the\ u_s \in \ S \ ,we\ have \\ inf\{\|u-v\| :v \in S\}
&\leq \|u-u_s\|_E \\Therefore\ \\
\|u-u_s\|_E&=inf\{\|u-v\| :v \in S\}
\end{align*}
\textbf{\emph{(0.3.3)theorem}}$$\emph{\textbf{\boxed{\|u-u_s\|_E=min\{\|u-v\|_E :v \in S\}}}}\eqno(0.3.3)$$\\
this is the basic error estimate for the Ritz-Galerkin method,the error is optimal in the energy norm. \\
Define $\|v\|=(v,v)^{1/2}=(\int_0 ^1 v^2(x) dx)^{1/2}$ the $L^2(0,1)$ norma,we wish to consider the size of the error $u-u_s$ in this norm. in fact it is considerably smaller.To estimate $\|u-u_s\|$, we use "duality" argument. Let $w$ be the solution of
$$-w''=u-u_s\ with\ w(0)=w'(1)=0$$\\
integrating by parts, we find \\
$$\| u-u_s\|^2 =a(u-u_s,w-v)\text{\textcolor[rgb]{1.00,0.00,0.00}{留作练习2}}$$\\由施瓦茨不等式，可得：\\
\begin{align}
\| u-u_s\| &\leq \|u-u_s\|_{E} \|w-v \|_{E}/{\| u-u_s\| }\\\
&=\|u-u_s\|_{E} \|w-v \|_{E}/{\|w''\|}\\ we\ taking\ inf\
\| u-u_s\| &\leq \|u-u_s\|_{E}inf \|w-v \|_{E}/{{\|w''\|}}
\end{align}
Thus, we see that the $L^2-norm$ of the error can be much smaller than the energy norm, provide that $w$ can be approximated well by some function in $S$,we take $v \in S$ close to the $w$ ,which we formalize in the following $approximation \ assumption:$\\
$$inf\|w-v \|_E \leq \varepsilon\| w''\| \eqno(0.3.4)$$ Applying the (0.3.4) yields
$$\textbf{\boxed{\| u-u_s\| \leq \varepsilon \| u-u_s\|_E}}$$\\
combining the (0.3.4) again,with $w$ replaced by $u$ ,using (0.3.3) give \\
$$\|u-u_s\|_E \leq \varepsilon \| u''\|$$ combining this estimate and recalling (0.1.1 ) yield:\\
\emph{\textbf{\textbf{ (0.3.5) Theorem}}} Assumption (0.3.4) implies that $$\| u-u_s\| \leq \varepsilon \|u-u_s \|_E \leq \varepsilon^2 \|u''\|=\varepsilon^2\|f\|$$\\ it is to says that
\begin{align}
\|u-u_s\| &\leq \varepsilon^2 \|u''\|\\
\|u-u_s\|_E &\leq \varepsilon \|u''\|_E
\end{align}
\emph{\textbf{\boxed{\textbf{0.4 \ Piecewise\ polynomial\ space-the\ finite\ element\ method}}}}\\
let $0=x_0 <x_1<x_2...<x_n=1$ be a partition of $[0,1]$ ,and let $S$ be the linear space of functions $v$ such that \\
(1) $v\ \in C^0([0,1])$\\
(2) $v|[x_{i-1},x_i]$ is a linear polynomial, $i=1,..n$ and \\
(3) $v(0)=0$\\
we will see that $S\subset V $ ,for each $i=1...,n$ define $\phi_i$ by the requirement that $\phi_i(x_j)=\delta_{ij}$\\
\textbf{(0.4.1)}Lemma $ \{\phi_i \ 1\leq i \leq n\}$ is a basis for $S$\\
\textbf{(0.4.2)} Remark $\{ \phi_i\}$ is called nodal basis for $S$ ,and ${v(x_i)}$ are the nodal values of a function $v$. the point $x_i$ are called the Nodal.\\
\textbf{(0.4.3) Definition.} Given $v \in C^0([0,1])$ ,the interpolant $v_I \in S$ of $v$ is determined by $v_I:=\Sigma_{i=1} ^{n} v(x_{i})\phi_i$ \\
\textbf{(0.4.4)Lemma} $v\in S\Rightarrow v=v_I$. Now we will prov the folling approximation theorem for the interpolant.\\
\textbf{(0.4.5) Theorem} Let $h=max_{1\leq i \leq n} (x_i-x_{i-1})$ Then$$\| u-u_I \|_E \leq Ch \|u''\|$$ for all $u \in V$. where C is independent of $h$ and $u$ .\\
\textbf{(0.4.7) Corollary} $$\| u-u_s \|+Ch\| u-u_s \|_E \leq 2(Ch)^2\| u''\|$$\\
\textbf{(0.4.8) Remark} the interpolant defines a linear operator $I:C^0([0,1])\rightarrow S$ where the $Iv=v_I$, lemma(0.4.4) says that $I$ is a projection(i.e$I^2=I$). the estimate (0.4.6) for $w'$ in the proof of (0.4.5) is an example of Sobolev's inequatiality.\\
\boxed{\textbf{0.5\ Realationship\ to \ Difference \ Methods}}\\
注：stiffness Matrix\ $K$ 取一组$S$ 的基$\{\phi_i :1\leq i\leq n\}$,let $u_s=\Sigma_{j=1}^{n} U_j \phi_j$ ;Let $K_{ij}=a(\phi_i,\phi_j),F_i=(f,\phi_i)$,for $i,j=1,..n$, set $U=(U_j),K=(K_{ij}),F=(F_j)$ then the relationship $a(u_s,v)=(f,v)$ equa to $KU=F$\\
The stiffness matrix$K$ as defined in (0.2.3)，using the basis ${\phi_i}$ can be interpreted as a difference operator.let$h_i=x_i-x_{i-1}$. Then then matrix \\
$K_{ij}=a(\phi_i,\phi_j)$ can be calculated to be $$K_{ij}=h_i^{-1} +h^{-1}_{i+1},K_{i,i+1}=K_{i+1,i}=-h^{-1}_{i+1}\eqno(0.5.1)$$\\ and the $$K_{nn}=h_n ^{-1}\]
\textcolor[rgb]{1.00,0.00,0.00}{这里貌似是有问题的：}
\begin{align}
K_{ij}&=a(\phi_i,\phi_j)=\int_0 ^1 \phi_i {'}\phi_{j} dx\
&=\int_0 1-(h_{i+1})(h_{i+1})^{-1}dx\
&=-h_{i+1}^{-2}\
K_{ii}&=a(\phi_i,\phi_i)=\int_0 ^1 \phi_i {'}\phi_{i} dx\
&=\int_0 ^1(h_i {-2}+h_{i+1}) \
&=h_i {-2}+h_{i+1}
\end{align}
similarly, the entries \(F\) can be approximated if \(f\) is sufficiently smooth:\
\[(f,\phi_i)=\frac{1}{2}(h_i+h_{i+1})(f(x_i)+O(h))\ where \ h=max h_i\eqno(0.5.2)
\]
\boxed{\textbf{0.6 Computer\ implementation \ of\ Finite\ Element\ Methods}}\
\begin{center}
\boxed{\textbf{\emph{Chapter 1 Sobolev\ space}}}
\end{center}
\boxed{\textbf{1.1 Lebesgue\ integration\ Theory}}\
given a real valued functions \(f\), on a given domain \(\Omega\) that are Lebesgue measurable; by\
\[\int_{\Omega} f(x) dx$$\\
we denote the Le integral of $f$ .for $1\leq p \leq \infty$ ,let\\
$$\| f\|_p:=(\int_{\Omega}|f(x)|^{p} dx)^{1/p}$$\\
and for the $p=\infty$ set\\
$$\|f(x)\|_{\infty}:=see sup\{ |f(x)|:x\in \Omega \}$$\\
we define the Lebesgue space\\
$$ L^p (\Omega):=\{ f:\| f\|_p < \infty\}\eqno(1.1.1)$$\\
we identify two functions $f,g$ that satisfy $\| f-g\|_p=0$ .For example ,take $n=1,\Omega=[-1,1]$ and \\
\[f(x)=
\begin{cases}
1& x\geq 0\\
0& x<0
\end{cases}\] and \[g(x)=
\begin{cases}
1& x> 0\\
0& x\leq 0
\end{cases}\] since the $f \ and \ g$ only on a set of measure zero,we view them as the same function.\\
\emph{\textbf{Minkowski inequality }} For $1\leq p \leq \infty$ and $f,g \int L^p(\Omega)$ ,we have\\
$$\| f+g\|_p \leq \| f\|_p +\| g\|_p \eqno(1.1.3)$$\\
\boxed{\text{\textbf{补充内积知识}：}}\\
\textbf{\emph{命题1.6.13 }} 在$B^*$ （赋范线性空间）$(X,\|\cdot\|)$ 中，为了在$X$中引进内积$(\cdot,\cdot)$ 使得$(x,x)^{1/2}=\|x\|\ \ (\forall \ x \in X)$, 当且仅当范数满足以下平心四边形等式：$$\|x-y\|^2 +\|x+y\|^2=2(\|x\|^2 +\|y\|^2) \ \ (\forall x, y \ \in X)\]
\[(x,y)=\frac{1}{4}(\|x+y\|^2 - \|x-y\|^2)$$\\
\emph{\textbf{Example 1.6.6}} $L^2(\Omega, \mu)$ 是内积空间，规定内积是$$ (u,v)= \int_{\Omega} u(x)\overline{{v(x)}} dx$$\\
\emph{\textbf{Example 1.6.7}} 在空间$C^k(\bar{\Omega})$ ，规定内积$$(u,v)=\Sigma_{|\alpha| \leq k} \int_{\Omega} \partial ^{\alpha} u(x)\partial ^{\alpha} v(x) dx$$,那么$(C^k(\bar\Omega),(\cdot,\cdot))$ 是内积空间。\\
反过来问：$L^p$空间里给定范数，求出内积么？
\\
\textbf{\emph{Holder inequality}} For $1\leq p,q \leq \infty$ and $1=1/p +1/q$ ,if $f\in L^p (\Omega), and\ g \in L^q(\Omega)$ then $fg \in L^1(\Omega)$ and \\
$$\|fg\|_{1} \leq \|f\|_p \|g\|_q \eqno(1.1.4)$$\\
\emph{\textbf{Schwarz inequality }}if $f,g \in L^2(\Omega)$ then $fg \in L^1(\Omega)$ and \\
$$\int_{\Omega} |f(x)g(x)|dx \leq \| f\|_2 \| g\|_2 \eqno(1.1.5)$$\\
\emph{\textbf{Definition(1.1.6)}} Given a linear space V, a norm$\| \cdot\|$\\
\emph{\textbf{Definition(1.1.7)}}Banach space\\
\emph{\textbf{Theorem(1.1.8)}} For $1 \leq p \leq \infty, L^p(\Omega)$ is a Banach space.\\
\boxed{\emph{\textbf{1.2\ Weak\ Derivatives}}} \\
the multi index: $\alpha=(\alpha_i),\ and\ the\ lenght\ of\ \alpha \ is\ given\ by\ |\alpha|=\Sigma_{i=1}^{n} \alpha_i$\\
For $\phi \in C^{\infty}$, denote by $D^{\alpha} \phi$ ,the usaul partial derivative $(\frac{\partial}{\partial x_1})^{\alpha_1}..(\frac{\partial}{\partial x_n})^{\alpha_n}\phi$ i.e
\begin{align*}
D^{\alpha} \phi:&=(\frac{\partial}{\partial x_1})^{\alpha_1}..(\frac{\partial}{\partial x_n})^{\alpha_n}\phi\\
x^{\alpha} :&=x_1 ^{\alpha_1} \cdot x_2 ^{\alpha_2} \cdot \cdot x_n ^{\alpha_n}\\
\partial/ \partial x:&=(\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2},...)
\end{align*}
\emph{\textbf{Definition(1.2.1)}} Let $\Omega$ be a domain in $\mathbb{R}$ .Denote by $D(\Omega)\ or \ C_0^\infty (\Omega)$ the set of $C^\infty(\Omega)$ function with compact support in $\Omega$.\\
\textbf{\emph{Example(1.2.2)}} Define\\
\[f(x)=
\begin{cases}
e^{1/(|x^2|-1)}& |x|<1\\
0& |x|\geq 1
\end{cases}\]\\
we claim that, for any $\alpha$ ,$\phi^{({\alpha})}(x)=P_{\alpha}(x)\phi(x)/(1-|x|^2)^{|\alpha|}$ for some polynomial $P_{\alpha}$, as we now show.for $|x|<1$\emph{\textcolor[rgb]{1.00,0.00,0.00}{验证留作练习3}}
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we now use the space to extend the notion of pointwise derivative to a class of function larger than the $C^{\infty}$\\
\textbf{\emph{Definition (1.2.3)}} Given a domain $\Omega$ ,the set of locally integrable functions is denoted by\\
$$L^1 _{loc}(\Omega):=\{ f:f\in L^1(K) \forall\ compact\ K\ \subset interior \Omega \}$$\textcolor[rgb]{1.00,0.00,0.00}{举个例子看看？作为练习4}\\
\emph{\textbf{Definition（1.2.4）}}$f\in L^1_{loc} (\Omega)$ has a weak derivative $D^{alpha}_w f$,there exists a function $g\in L^1_{loc}$ such that \\
$$\int_{\Omega} g(x)\phi(x)dx=(-1)^{|{\alpha}|} \int_{\Omega}f(x)\phi^{(\alpha)}(x)dx\ \ \forall\phi \in D(\Omega)$$ we define $D^{\alpha}_{w}f=g$\\
Example(1.2.5) taking $n=1,\Omega=[-1,1], and\ f(x)=1-|x|$. we claim that $D^1_{w}f$ exists and given by
\[g(x)=
\begin{cases}
1 & x<0\\
-1& x>0
\end{cases}\]\\
To see this ,break [-1,1] into the two parts in which $f$ is smooth ,and we integrate by parts.Let $\phi\in D(\Omega)$ then:\\
\begin{align*}
\int_{-1}^{1} f(x)\phi'(x) dx&=\int_{-1}^{0}f(x)\phi'(x) dx +\int_{0}^{1}f(x)\phi'(x) dx\\
&=f\phi|^{0}_{-1} -\int_{-1}^{0} (+1)\phi(x)dx +f\phi|^{1}_{0}-\int_{0}^{1} (-1)\phi(x) dx\\
&=-\int_{-1}^{1}g(x)\phi(x) dx +(f\phi)(0-)-(f\phi)(0+)\\
&=-\int_{-1}^{1}g(x)\phi(x) dx
\end{align*}\\
\textbf{\textbf{Example(1.2.6)}} Let$\rho$ be a smooth function define for $0<r\leq 1$ satisfying $$\int_{0}^{1} |\rho'(r)|r^{(n-1)}dr <\infty$$ \\define $f$ on $\Omega={x\in \mathbb{R^n}:|x|<1}$ via $f(x)=\rho(|x|)$ where the weak derivative exists for all$|\alpha|=1$ and is given by \\
$$g(x)=\rho'(|x|)x^{\alpha}/|x|$$\textcolor[rgb]{1.00,0.00,0.00}{留着做练习5}\\
\emph{\textbf{Proposition(1.2.7)}} Let $\alpha $ be a arbitrary and $\psi \in C^{|\alpha|}(\Omega)$.Then te weak derivative $D\psi$ exists and is given by $D^{\alpha} \psi$\\
\boxed{\emph{\textbf{\textbf{1.3\ Solbolev Norms and Associated Space}}}}\\
Definition(1.3.1) .设$K$是非负整数，$f$ 局部可积，假设对所有的$|\alpha|\leq k$ 弱导数都存在，定义索伯列夫范数:\\
$$\|f\|_{w} :=(\Sigma_{|\alpha \leq k|}\|D^{\alpha}_w f\|_{L_p}^{p})^{1/p}$$\\
索伯列夫空间an就是局部可积函数空间赋上索伯列夫范数：$$W_p^k(\Omega)=\{f\in L^1_{loc}(\Omega) :\parallel f\parallel_w <\infty \}$$\\
Sobolev space can be related in special cases to other space .For example recall Lipschitz norm\\
$$\parallel f \parallel_{lip}=\parallel f \parallel_{L^{\infty}} +sup{\frac{f(x)-f(y)}{|x-y|}: x,y \in \Omega,x\neq y}$$\\
it corresponding the space of Lipschitz functions \\
$$Lip(\Omega)=\{f\in L^{\infty}(\Omega): \parallel f\parallel _{L} <\infty \}$$\\
\textbf{\textbf{注：这里补充上几个sobolev空间的几个基本性质}}\\
\emph{\textbf{Theorem 1}} 如果区域$\Omega$ 的边界$\partial \Omega$ 是$Lipschitz$ 连续的曲面，$1 \leq <\infty$ ,则$C^{\infty}(\bar{\Omega})$在$W^{m,p}$ 中稠密。($m$代表指标最大长度，$p$代表用的是p范数来定义的sobolov范数)\\
所谓的Lipschitz连续就是说：边界在局部看起来是Lipschitz函数的图\\
由此可知，对于具有$Lipschitz$连续边界的区域$\Omega$, 其上面的$C^{\infty}(\bar{\Omega})$在索伯列夫范数下的完备化就是$W^{m,p}$。\\
空间$C_0^{\infty}(\Omega)$ 在$\|\|_{m,p}$意义下的闭包是$W^{m,p}$的子空间 记作\textcolor[rgb]{1.00,0.00,0.00}{$W_{0}^{m,p}$},其中$C_0^{\infty}(\Omega)\ $ 指的是无穷光滑紧致支集函数空间。$W_{0}^{m,p}$在$\|\|_{m,p}$ 下的完备化是Banach 特别：$p=2,it is a Hilbert space \ \ \textcolor[rgb]{1.00,0.00,0.00}{H_0^m } $ \\
\emph{\textbf{Theorem 2 上半范数控制范数}} 设$\Omega$ 位于两个平行超曲面之间，则存在值 依赖于空间维数$n$ 以及deriver 阶数$m$,平行间距$d$, sobolev 指数$p$ 的常数$K(n,m,d,p)$ 使得：\\
$$|u|_{m,p\Omega} \leq \| u\|_{m,p,\Omega} \leq K(n,m,d,p)|u|_{m,p\Omega}\eqno(1)$$\\
$(1)\ be\ called\ Poincare Friendichs$ 不等式 \\
\textbf{紧嵌入算子} \\
\begin{align*}
I: X &\rightarrow Y and\ I(x)=x \ if \\
\|x\|_Y & \leq C\|x\|_X \text{称其为嵌入算子}\\
if \ I \text{将任意有界闭集映成有界闭集，则称其是紧算子}\\
\text{紧算子如果还是嵌入算子的话，呢就称之为 紧嵌入算子}\\
note \ that\ X\hookrightarrow ^c Y
\end{align*}\\
\emph{\textbf{Theorem 3 嵌入定理}} 如果有界连通区域$\Omega$ 的边界是$Lip$ 连续的，则：\\
\begin{align*}
(1)if\ &m < n/p, & &W^{m+k,p}(\Omega)\hookrightarrow W^{k,q}(\Omega),& &1 \leq \frac{np}{n-mp}, k\geq 0&\\
(2)if\ &m < n/p, & &W^{m+k,p}(\Omega)\hookrightarrow^{\textcolor[rgb]{1.00,0.00,0.00}{c}} W^{k,q}(\Omega),& &1 \leq \frac{np}{n-mp}, k\geq 0&\\
(3)if\ &m = n/p,& &W^{m+k,p}(\Omega)\hookrightarrow^{c} W^{k,q}(\Omega),& &1 \leq \infty, k\geq 0&\\
(4)if\ &m > n/p ,& &W^{m+k,p}(\Omega)\hookrightarrow^{c} C^k(\bar{\Omega}),& &k\geq 0&
\end{align*}\\
\boxed{\textbf{1.4\ Inclusion\ Relations \ and \ Sobolev\ inequality}}\\
\textbf{(1.4.1) Proposition }suppose that $\Omega$ is nany domain, $k\ and\ m$ are non negative integers satisfying $k \leq m$ ,and $p$ s a real number satisfying $1 \leq p\leq \infty$ ,then $W^{m,p} \subset W^{k,p}$.\\
\textbf{(1.4.2) Proposition} suppose that $\Omega$ is a bounded domain, $k \in \mathbb{N}$, and $1 \leq p \leq q \leq \infty$ .Then $W^{k,q} \subset W^{k,p}$\\
\textbf{(1.4.3) Example} let $n >2$ ,let $\Omega=\{x\in \mathbb{R}: |x|<1/2\}$, consider the function $f(x)=log|log|x||$ we see that $f$ has a first order weak derivatives $$D^{\alpha} f(x)=x^{\alpha}/(|x|^2 log|x|),where \ |\alpha|=1$$\\ from the exercise 1.X.5 ,we see that $D^{\alpha} f(x) \in L^{p}\Omega, provoded p\leq n$, For example $$|D^{\alpha} f(x)|^n \leq \rho (|x|) := 1/(|x|^n |log|x||^n)$$\\
satisfy the condition of exercise 1.x.5 because $\rho (r)r^{n-1}= 1/(r|log r|^n)$ is integrable for all $n\geq 2$ on [0,1/2]. in fact that ,yhe change of variable $r=e^{-t}$ given $$\int_0 ^{1/2}\frac{dr}{(r|log r|^n)}= \int_{log 2} ^{\infty} \frac{dt}{t^n} <\infty$$\\
\textbf{(1.4.4) Definition} we may say $\Omega$ has a $LIP\ boundary\ \partial\Omega$ provided ther exists a collection of open set $O_i$ ,$\varepsilon>0, N\in \mathbb{Z}, a\ finite\ number\ M$, such that for all $x \in \partial \Omega$ the ball of radius $\varepsilon$ centerted at the $x$ is contained in some $O_i$ ,no more than $N$ of the set $O_i$ intersect nontrivially ,and each domain $O_i \cap \Omega =O_i \cap \Omega_i$ ,where the domain whose boundary is a graph of Lipschitz function $\phi _i$.\\one consequence of this definition is that we can now relate Sobolev space on a given domain to those on all of $\mathbb{R}^n$， 就是说边界局部可以看成是Lipschit函数的像\\
\begin{center}%此处绘图
\begin{tikzpicture}[yscale=5]
\draw [->,ultra thick] (-1,0) -- (2,0);
\draw [->,ultra thick] (0,-0.3) -- (0,0.3);
\draw [red,ultra thick,domain=0.01:2] plot (\x, {(\x)^(3/2)*sin(1/(\x))});
\end{tikzpicture}$f(x)=x^{3/2}sin(1/x)$就是非Lipschitz的例子，直观上就是导数没有界。
\end{center}
\textbf{(1.4.5) Theorem} suppose that $\Omega$ has a Lipschitz boundary . Then there is an extention mapping $E:W^{k,p}(\Omega)\rightarrow W^{k,p}(\mathbb{R}^n)$ satisfying tha $Ev|_{\Omega}=v$ for all $v\in W^{k,p}$ and$$\|Ev\| \leq C\| v\|$$ where the $C$ is independent of $v$\\
For a proof of the this result ,as well as more detail concering other material in this section~\\
(1.4.6) Theorem (Sobolev Inequality) Let $\Omega$ be an $n-dimmensional$ domain with $Lipschitz\ boundary $ , let $k$ be a positive integer and let $p$ be a real number n the range $1 \leq p < \infty$ such that\\
\begin{align*}
k & \leq n, &when& &p=1&\\
k & \leq n/p,&when& &p>1&.There\ there\ is\ a\ constant\ C\ for\ all\ u\ \in W^{k,p}(\Omega)\\
\|u\|_{L^{\infty}(\Omega)} &\leq C\|u\|_{W^{k,p}(\Omega)}
\end{align*}
This result says that any function wth suitably regular weak derivatves may be vewed as a continuous ,bounded function.Noted that \textcolor[rgb]{1.00,0.00,0.00}{Example 1.4.3 show that the result is sharp, namely the condition $k>n/p$ cannot be relaxed}\\
\textbf{(1.4.8)Remark } if $\partial \Omega$ s not Lipschitz continuous ,then neither theorem 1.4.5 nor theorem 1.4.6 need hold. For example, let \\
$$\Omega=\{ (x,y)\in \mathbb{R}^{2}: 0<x <1,|y|< x^r \}$$\\
where $r>1$ ,and let $u(x,y)=x^{1\varepsilon/p}$, where $0< \varepsilon <r$ ,Then $$\Sigma_{|\alpha=1|} \int_{\Omega} |D^{\alpha}u|^pdx dy=c_{\varepsilon,p}\int_0^1 x^{-\varepsilon-p+r} dx <\infty$$\\
\textcolor[rgb]{1.00,0.00,0.00}{provided that $p<1+r-\varepsilon$} in this case $u \in W^{1,p}$, but $u$ s not essentially bounded on $\Omega\ if\ \varepsilon>0$. choosing $\varepsilon $ so that it is possible to have $p>2$, we found that LIpschitz boundary is necessary for Sololev inequality to held. since Sobolev inequality dose hold on $\mathbb{R}^n$, it is not possilble to extend $u$ to an element of $W^{1,2}(\mathbb{R}^2)$,thus, the extension theorem for sobolev functions can not hold if $\partial \Omega$ is not Lip.\\
\boxed{\ \emph{\textbf{1.5\ Review\ of chapter 0}}}\\
$$V:=\{ v\in W^{1,2}(\Omega) v(0)=0 \}, where\ the \Omega =[0,1]$$ and that makes sense because\textcolor[rgb]{1.00,0.00,0.00}{ Sobolev inequality guarantees that pointwise values are well} defined for functions in $W^{1,2}(\Omega)$, The derivation of the variational formulation(0.1.3) for solution(0.1.1) can now be made rigorous .we have stated in SEct1.3 that function in $W^{1,1}(\Omega)$ and a fortiori those in $W^1_2(\Omega)$ are abosolutely continuous.we saw that piecewise linear function have weak derivatives constant ,thus ,$S \subset W^{1,\infty}(\Omega),and\ therefore\ that\ S\subset V $ \\ since Sobolev inequality implies that $V \subset C_b\Omega$
\\
\boxed{\textbf{1.6\ Trace\ Theorems}}\\
we begin with a simple example to explain the ideas. Let$\Omega$ denote the $$\Omega=\{(x,y): x^2+y^2 <1\}=\{ (r,\theta) :0\leq r <1; 0\leq \theta <2\pi \}$$\\
Let $u \in C^1(\bar{\Omega})$ and consider it restriction to the $\partial \Omega$ as follows:\\
\begin{align*}
u(1,\theta)^2&=\int_0^1 \frac{\partial}{\partial r}(r^2 u(r,\theta)^2)dr\\
&=\int_0^1 2(r^2 u u_r+ru^2)(r,\theta)dr\\
&=\int_0^1 2(r^2u\nabla u\cdot \frac{(x,y)}{r}+ru^2)(r,\theta)dr\\
&\leq \int_0^1 2(|u||\nabla u|+u^2)rdr \\
\int_{\partial\Omega}u^2 d \theta &\leq 2\int_{\Omega}(|u||\nabla u|+u^2)dxdy\\
\int_{\partial\Omega}u^2 d \theta&:=\|u\|^2_{L^2(\partial\Omega)} \\
\|u\|^2_{L^2(\partial\Omega)} &\leq 2\| u\|_{L^2} (\int_{\Omega} |\nabla u|^2 dx dy)^{1/2} + 2\int_{\Omega} u^2 dx dy\\
Remark:\ Schwarz's\ nequality\ is\ \\
\int_{\Omega}|f(x)g(x)| &\leq \|f\|_{L^2} \|g\|_{L^2} \\
(\int_{\Omega} |\nabla u|^2)^{1/2} + (\int_u^2)^{1/2} & \leq (2\int(|\nabla u|^2 + u^2))^{1/2}(\text{几何算数平均不等式})\\
Therefore\ \\
\|u\|_{L^2(\partial \Omega)} &\leq \sqrt[4]{8} \| u\|^{1/2}_{L^2(\Omega)} \|u\|^{1/2} _{W^{1,2}(\Omega)}\ \ \ (\texttt{\textbf{1.6.2}})
\end{align*}\\
\emph{\textbf{(1.6.3) Proposition }}.Let $\Omega$ denote the unite disk in $\mathbb{R}^2$, For all $u\in W^{1,2}(\Omega) $, the restriction $u|_{\partial \Omega}$ may be interpreted as a function in $L^2(\partial\Omega)$ satisfying (1.6.2)\\
\emph{\textbf{Proof:}}
pick up a sequence $u_i \in C^1(\bar\Omega)$ such that$ \|u- u_i\|_{1,2} \leq 1/i$ .By (1.6.2) and triangle inequality $$\|u_k- u_j\|_{L^2(\partial\Omega)}\leq \sqrt[4]{8} (\frac{1}{j} +\frac{1}{k})$$\\
Q: \textcolor[rgb]{1.00,0.00,0.00}{$$\| u\|_{L^2} \leq \| u\|_{w^{1,2}}$$}吗？\\
so that $u_j$ s a Cauchy sequence in $L^2(\partial \partial{\Omega})$. Since \textcolor[rgb]{1.00,0.00,0.00}{this space is compact} ,there must be a limit $v \in L^2(\partial \Omega)$ such that $\|u_j - v\|_{L^2(\partial \omega)}\rightarrow 0$. we define $$u|_{\partial\Omega}:=v $$\\
the first thing we need to check that this definition dose not depend on the sequence .so suppose that the $\|v_i -u\|_{1,2} \rightarrow 0$\\
and $$\|v_i -v\|_{L^2({\partial \Omega})} \rightarrow 0$$\\
thus $u|_{\partial \Omega}$ s well defined in $L^2(\partial \Omega)$ \\
again ,we use the validity of it for smooth functions:\\
$$\|u\|_{L^2(\partial \Omega)}=\|v\|_{L^2(\partial \Omega)}\leq\sqrt[4]{8}\lim_{j\rightarrow \infty } \|u_j\|_{L^2(\Omega)}^{1/2} \|u_j\|_{1,2}^{1/2}\eqno{\square}$$\\
Now let us describe a generalization of p(1.6.3) to more complex domains .if $\partial \Omega$ s given as the graph of a function $\phi$ (c.f. 1.4.4) we can define the integral on $\partial \Omega$ as
\begin{center}
\boxed{$$\int_{\partial \Omega} f dS=\int_{\mathbb{R}^{n-1}} f(x,\phi (x)) \sqrt{1+|\nabla \phi|^2}dx$$ } Q
\end{center}
\textbf{\emph{(1.6.6) Theorem.}} suppose that $\Omega$ has a Lipschitz boundary ,and that $p$ s a real number in the range $1 \leq p \leq \infty$ .Then there is a constant $C$, such that $$\| v\|_{L^p(\partial \Omega)}=C\| v\|_{L^p(\Omega)} ^{1- 1/p} \|v\|^{1/p}_{1,2}\ \ \forall v \in W^{1,p}(\Omega)\]
we will use the notation \(\dot{W}^{1,p}\) to denote the subset of \(W^{1,p}\), consisting of functions whose trace on \(\partial \Omega\) s zero, that is\
\[\dot{W}^{1,p}(\Omega)=\{v\in W^{1,p}(\Omega) , v|_{\partial \Omega}=0\ \ in\ L^2(\partial \Omega) \}
\]
\emph{\boxed{1.7\ Negative\ Normal\ and\ Duality}}\
\begin{center}
\emph{\textbf{Chapter 2 Variational Formulation of Elliptic Boundary Value Problem }}
\end{center}
\textbf{\emph{2.1 Inner-Product Space}}\
(2.1.1)Definition \ \ bilinear form\
(2.1.2)Definition inner-product space\
\begin{tcolorbox}[colback=yellow!20,colframe=red!75!green,title=(2.1.3)Example]
(1)\(V:=\mathbb{R}^n,(x,y):=\Sigma_{i=1}^{n} x_i y_i\)\
(2)\(V:=L^2(\Omega), \Omega \subset \mathbb{R}^n, (u,v):=\int_{\Omega} u(x)v(x) dx\)\
(3)\(V:=W_2 ^k(\Omega), \Omega\subset \mathbb{R}^n, (u,v):=\Sigma_{|\alpha| \leq k}(D^{\alpha}u,D^{\alpha}v)_{L^2(\Omega)}\)\
note: The inner product space (3) is often denoted by \(H^k(\Omega)\), Thus\(H^k=W^k _2\)
\end{tcolorbox}
\begin{tcolorbox}[colback=yellow!20,colframe=red!75!green,title=(2.1.4)Theorem (Schwarz inequality)]
if \((V,(,))\) is an inner product space,then$$|(u,v)|\leq (u,u)^{1/2} (v,v)^{1/2}\eqno(2.15)$$
The equality holds iff \(u\ and\ v\) are linearly dependent.
\end{tcolorbox}
\boxed{(2.18)\textcolor[rgb]{1.00,0.00,0.00}{Remark}}
\ \ (2.15) was proved without the property \boxed{(v,v)=0 \ iff\ v=0}, An example is $on\ H^1(\Omega)\ ,a(u,v)=\int_{\Omega} \nabla u\cdot\nabla v dx $, thus we proved that $$a(u,v)\leq a(u,u)a(v,v)$$, even thought \(a(,)\) is not an inner product.\
\begin{tcolorbox}[colback=yellow!20,colframe=red!75!green,title=(2.1.9)Proposition (内积可以定义范数)]
\[\|v\|:=\sqrt{(v,v)}
\]
\end{tcolorbox}
\textbf{\emph{2.2 Hilbert Space}}\
(2.2.1)Definition Hilbert space\
\begin{tcolorbox}[colback=yellow!20,colframe=red!75!green,title=Example(2.2.2)]
(2.1.3)的例子都是Hilbert空间，特别的，\(W^k_2\) 定义的内积\((u,v):=\Sigma_{|\alpha| \leq k}(D^{\alpha}u,D^{\alpha}v)_{L^2(\Omega)}\)对应的范数就是之前定义的索伯列夫范数\(\| f\|_{w^k _2((\Omega)}=(\Sigma_{|\alpha \leq k|}\| D^{\alpha} f \|_{L^2}^2)^{1/2}\)
\end{tcolorbox}
(2.2.3)Definition if \(S \subset H\) is a close linear subset of \(H\) ,we call the \(S\) is a subspace of \(H\).\
\begin{tcolorbox}[colback=blue!30,colframe=red!75!green,title=Example(2.2.5)\ Example of subspace of Hilbert Space]
(1)\(H\) and {0} \
(2) \(T:H\rightarrow K\) be a continuous linear map,Then the \(kerT\) is a subspace of \(H\)\
(3) Let \(x \in H\) and define \(x^{\perp} :=\{ v\in H: (v,x)=0 \}\). Then the \(x^{\perp}\) s a subspace of \(H\). To see this .note that \(x^{\perp}= ker L_{x}\),where \(L_x : v\rightarrow (v,x)\), \(|L_x(v)| \leq \|x\| \|v\|\) This prove that \(L_x\) is continuous ,then \(x^{\perp}\) is a subspace of \(H\)\
(5) Let \(M\subset H\) de a subset and define that \(M^{\perp} :=\{ v\in H: (v,x)=0, \forall x \in M \}\) note that $M^{\perp}=\bigcap_{x \in M} x^{\perp} \(
\end{tcolorbox}
\begin{tcolorbox}[colback=blue!30,colframe=red!75!green,title=Propostion(2.2.7)\ 直交补的性质]
(1)\) M \subset N \subset H \Rightarrow N^{\perp} \subset M^{\perp}$\
(2) \(M \bigcap M^{\perp}={0}\)\
(3)\(H^{\perp}={0}\)\
(4) ${0}^{\perp}=H \(
\end{tcolorbox}
\begin{tcolorbox}[colback=blue!30,colframe=red!75!green,title=Theorem(2.2.8)\ 平行四边形等式]
设\)| \cdot|$ 是内积空间\(H\) 由内积诱导的的范数，则有平行四边形等式$$|x-y|^2 +|x+y|2=2(|x|2 + |y|^2)$$
\end{tcolorbox}
2.3子空间投影理论\
\begin{tcolorbox}[colback=blue!30,colframe=red!75!green,title=Theorem(2.3.1)\ 最短距离可达]
Let \(M \subset H\) ,\(v \in H /M\), define \(\delta := inf\{ \|v-u\| \} where\ u \in M\). Then there exisit a $w_0 \in M $ such that \
(1)$ | v-w_0|=\delta $ \
(2) \(v-w_0 \in M^{\perp}\)
\end{tcolorbox}
proposition(2.3.1) says that, given a subspace $M \subset H $ and $v \in H $ we can write that \(v=w_0 +w_1\) where $ w_0 \in M and\ w_1=v-w_0 \in M^{\perp}$.Let us show that this decomposition is unique.\
\begin{align}
w_0+w_1 &=v=Z_0 +z_1 \
w_0 -z_0 &=z_1 -w_1 \
Since\ M^{\perp} \bigcap M &=0 \
then w_0&=z_0\
w_1&=z_1 this\ show \ that\ the \ decomposition \ is \ unique\ \
defin\ P_M:H &\rightarrow M \
P_{M^{\perp}}:H& \rightarrow M^{\perp}
\end{align}
\begin{numcases}{P_M v= }
v \ if\ v\in M \label{1} \
w_0 \ if\ v \in H/M \label{2}
\end{numcases} \
\begin{numcases}{P_{m^{\perp}} v=}
0 \ if v \in M\
v-W_0\ \ if\ v \in \ H/M
\end{numcases}
\begin{tcolorbox}[colback=blue!30,colframe=red!75!green,title=Theorem(2.3.5)\ 唯一分解 ]
Given subspace \(M \subset H,\ and\ v\in \ H\), there is a unique decomposition $$v=P_M v +P_{M^{\perp}} v$$\
in other words,$$H=M^{\perp}\bigoplus M$$
\end{tcolorbox}
\begin{itemize}
\item \emph{(2.3.9) Definition} An operator \(P\) on a linear space \(V\) is a projection if $$P^2=I$$ i.e $$Pz=z\ \forall z \in image\ P$$
\item (2.3.10)Remark：
\item \boxed{\emph{\textbf{2.4 Riesz Representation Theorem}}}
\begin{tcolorbox}[colback=blue!30,colframe=red!75!green,title=Theorem(2.3.5)\ Riesz 表示定理 ]
Any continuous linear functional \(L\ \in Hilbert\ space\ H\) can be represented uniquely as$$L(v)=(u,v)$$, where $$|L|{H'} = |u|H$$
\end{tcolorbox}
\item(2.4.6) Remark:\
According to the Riesz Representation Theorem , there is a nature isometry between \(H\ and \ H'\) which is $$L_u \longleftrightarrow u$$
\end{itemize}
\boxed{2.5 Formulation \ of\ Symmetric\ Variational\ Problems}
\begin{itemize}
\item (2.5.1) Example \
Recall that\(H^1(0,1)=W^1_2(0,1)\) is a Hilbert space under the inner product$$(u,v)=\int_0 ^1 uv dx + \int_0 ^1 u'v' dx$$
we defined \(V=\{ v\in H^1(0,1): v(0)=0\}\) .To see that \(V\) is a subspace of \(H\) ,Let \(\delta_0: H^1(0,1)\longrightarrow \mathbb{R}\) by \(\delta_0(v)=v(0)\).and \(sup\{ |\delta_0 (x)|: x \in \Omega\} \leq C(\delta_0,\delta_0)_{H^1}\)\
and that \(\delta_0\) is a bounded linear functional on \(H^1\). Hence the \(ker \delta_0=V\) is closed in \(H^1\)\
we define \(a(u,v)=\int_0 ^1 u'v' dx\), Note that \(a(,)\) is symmetric bilinear ,But \(a(1,1)=0\), ao it is not an inner product on \(H^1\). However ,it dose satisfy the Coercivity property(1.5.1) on \(V\) \boxed{| v| \leq \frac{3}{2} a(v,v)\ where\ a(,) is\ bilinear\ form}
\item (2.5.2) Definition\
A bilinear form \(a(,)\) on a normed linear space \(H\) is called to be bounded, if $\exists C < \infty $ such that
$$|a(v,w)| \leq C | v|H | w|H, \forall v, w \in H$$
and coercive on $V \subset H $ if \(\exists \alpha > 0\) such that $$a(v,v) \geq \alpha | v|^2$$
\item(2.5.3) Proposition \
\begin{tcolorbox}[colback=blue!30,colframe=red!75!green,title=双线性形式构成的Hilbert空间]
Let \(H\) be a Hilbert space , and \(a(,)\) is a symmertirc bilinear form that continuous on \(H\) and coercive on a subspace \(V\) of H, then \((V,a(,))\) is a Hilbert space
\end{tcolorbox}
\item (2.5.4) symmetric variational problem\
\begin{tcolorbox}
\begin{numcases}{ }
(1) (H,(,)) is\ v\ a\ Hilbert\ space \
(2) V\ is\ a\ close\ subspace\ of\ H\
(3) a(,) is\ a\ bounded, symmetric, coercive \ on V
\end{numcases}
\end{tcolorbox}
\item (2.5.5) \begin{tcolorbox}
Given \(F \in V'\) , find $u \in V $ such that \(a(u,v)=F(v), \forall v \in V\)\
i.e $$F\rightarrow u$$
\end{tcolorbox}
\item (2.5.6)\
\begin{tcolorbox}[colback=blue!30,colframe=red!75!green,title=变分问题解存在唯一性]
假设(2.5.4)里的条件(1)-(3) 都满足，则（2.5.5）的解存在唯一。
\end{tcolorbox}
\item \emph{\textbf{椭圆边值问题的变分与弱解}}
假设区域\(\Omega\) 连同有界，且\(\partial \Omega\)是\(Lipschitz\)连续的曲面。考虑\(Possion\) 方程的\(Dirichlet\)班值问题：\
[\begin{cases}
-\Delta u=f,x\in \Omega\
u=\bar{u}0, x\in \partial \Omega \tag{5.2.4}
\end{cases}]
\(\forall v \in C_0 ^{infty} (\Omega)\),将其乘在方程两边，在\(\Omega\)上积分，并利用Green 公式可得
\begin{align*}
-\Delta u v &=fv \
\int -\Delta u v dx &=\int \nabla u \cdot \nabla v dx -\int_{\partial \Omega} v \partial v u dx\
\int \nabla u \cdot \nabla v dx &=\int_{\Omega} -\Delta u v dx\
then: \int_{\Omega} fv &=\int_{\Omega} \nabla u\cdot \nabla v dx\tag{5.2.5}\
Let: a(u,v)&= \int_{\Omega} \nabla u\nabla v dx=(f,v)\
then: a(u,v)&=(f,v)\ \forall v\ in \mathbb{H}0 ^1({\Omega})\tag{5.2.6}
\end{align}
(5.2.6)称为是虚功原理，即使内力与外力在任何允许的虚位移上做的功之和是0.由于边界位移是给定的，所以允许虚位移在边界上的取值只能是0.\
\item \emph{\textbf{定义5.5}}\
若\(u\in V(\bar{u}_0,\Omega):=\{u \in H^1 (\Omega):u|_{\partial \Omega}=\bar{u}_0 \}\) 满足变分方程（5.2.6）,则称\(u\)是椭圆边值问题\((5.2.4)\)的弱解，称相应的变分问题为边值问题的变分问题或者若形式。\
\item \emph{\textbf{定理5.7}} \
\begin{tcolorbox}
设\(f \in C{\bar{\Omega}},\bar{u}_0 \in C(\partial \Omega), u \in C^2 (\bar{\Omega})\), 则\(u\)是古典解 \(\Leftrightarrow\) \(u\)是弱解。
\end{tcolorbox}
\end{itemize}
\begin{center}
Introduce to DG
\end{center}
\begin{itemize}
\item 最简单的DG(1973)\
[\begin{cases}
u_x&=f(x) 0 \ \leq x \leq 1\
u(0)&=u_0
\end{cases}]
传统的差分格式： \begin{align}
\frac{u_j -u{j-1}}{\Delta x} &=f(x_j) \
u_j &= u_{j-1}+\Delta x f(x_j),j=1,2..,N\
u_1 &= u_0+ \Delta x f(x_1)\
u_2 &=u_1+\Delta x f(x_2)\
\end{align}
缺点 ：只有一阶精度（精度是怎么计算出来的？）\
换一种格式：\begin{align}
\frac{au_j +bu_{j-1}+cu_{j-2}}{\Delta x} &=f(x_j) \
\end{align*}
具有二阶精度（？），但是\(u_1=?\)
\item 原始的DG\
[\begin{cases}
x_{j+1/2} -x_{j-1/2}=\Delta x_{j},max \Delta x_j =\Delta x =h\
V_h:= { v: v|{I_i} \in P^k (I_i),i=1,2..N },Note: in the point x many be not continue.\
Find \ u_h \in V_h \ s.t\ \
-\int_{I_j} u_h v_x dx + u_{h}(x_{j+1/2})v(x_{j+1/2}) -u_h (x_{j-1/2})v(x_{j-1/2})=\int_{I_j} fv dx\
\end{cases}]
\item Ideal\
$ \int_{I_j} u_x v = \int_{I_j} fv dx\(,假设\)v\(一阶可微，\)uv(j+1/2) -uv(j-1/2) -\int_{I-j} uv_x dx =\int_{I_j} fvdx\(\\ 所以有\\
\begin{tcolorbox}
\textcolor[rgb]{1.00,0.00,0.00}{\)\(uv(j+1/2) -uv(j-1/2) -\int_{I_j} uv_x dx =\int_{I_j} fvdx \eqno(\ast)\)\(}
\end{tcolorbox}
\item 最简单的情况：\)k=0$\
取 [v=\begin{cases}
1\ x\in I_{j}\
0\ other
\end{cases}]
\[ u_h(-_{j+1/2})v(-_{j-1/2}) - u_h(-_{j-1/2}) v(+_{j+1/2})=u_j-u_{j-1}=f_j \Delta x_j$$这就回到了传统的差分格式\\
\item $u_h$的存在唯一性： \\ 令：$u_h(x_{1/2})=u_0$
$$ u_hv(x_{3/2}) -\int_{I_1} u_hv_x dx =u_hv(x_{1/2})+\int_{I_1} fvdx \eqno(\ast\ast)$$这是一个$(k+1)\times(k+1)$的方程组
\item 唯一性的证明：\\
take $v=u_h \in V_h$
\begin{align*}
-\int_{I_1}&u_h (u_h)_x dx +u^2_h(-_{3/2})&=&0 (where\ Ax=0\ only\ 0 \Longrightarrow Ax=B \ have\ only \ one \ solution) \\
-\frac{1}{2}\int& (u_h)^2_x dx+u^2_h(-_{3/2})&=&0\\
-\frac{1}{2}[u^2_h(-_{3/2})-u^2_h(+_{1/2})]&+u^2_h(-_{3/2})&=&0\\
\frac{1}{2}u^2_h(-_{3/2})+&\frac{1}{2}u^2_h(+_{1/2})&=&0\\
then\\
u_h(x^+_{1/2})&=u(x^-_{3/2})&=&0\\
the (\ast\ast) become\\
-\int_{I_1}u_hv_x dx =0\\
take\\ u_h=(x-x_{1/2})\hat{u}(x),\hat{u}(x) \in P^{k-1}(I_1)\\
\end{align*}
take \[v=\begin{cases}
\int_{x_{1/2}} ^{x}\hat{u}(s) ds \ x\in I_{1}\\
0\ other
\end{cases}\]
then \\
$-\int_{I_1} (x-x_{1/2})\hat{u}(x)\hat{u}(x) dx =0\Longrightarrow\hat{u}=0\Longrightarrow u_{h}=0$
\item \textcolor[rgb]{0.68,0.00,0.00}{练习 1 }\\
\begin{tcolorbox} 用原始DG求解
\[\begin{cases}
u_x&=2\pi cos2\pi x \\
k&=1,2
\end{cases}\] 给出误差表$\| u-u_h\|^2=\int_0 ^1 |u-u_h|^2dx,e=\|u-u_h\|$
\end{tcolorbox}
解：\\
\boxed{k=1}:\ $I_j=[\frac{j-1}{N},\frac{j}{N}],$取基为$\{\phi_1,\phi_2\}=\{ N(x-\frac{j-1}{N}) , N(-x+\frac{j}{N}) \}$,取检验函数$v=\phi_1,\phi_2$，$u_h=a_1\phi_1+a_2\phi_2$
\begin{align*}
u_hv(x_{3/2}) -\int_{I_1} u_hv_x dx &=u_hv(x_{1/2})+\int_{I_1} fvdx \\
-\int_{I_1}a_1\phi_1v_x dx +a_1\phi_1(x^-_{3/2})v(x^-_{3/2})-\int_{I_1}a_2\phi_2v_x dx +a_2\phi_2(x^-_{3/2})v(x^-_{3/2})&=u_hv(x_{1/2})+\int_{I_1} fvdx\\
\end{align*}
set:\ $$ \langle v,\phi_1\rangle =-\int_{I_1}\phi_1v_x dx +\phi_1(x^-_{3/2})v(x^-_{3/2}),$$\\particularly\ Let\ $$v=\phi_i,k_{ji}:= \langle \phi_j,\phi_i \rangle=
-\int_{I_1}\phi_i(\phi_j)_x dx +\phi_i(x^-_{3/2})\phi_j(x^-_{3/2}),F_j=u_0\phi_{j}(x^-_{1/2})+\int_{I_1} f\phi_j(x^-_{1/2}) dx\]
那么方程就变成方程组：\
\[\left(
\begin{array}{cc}
k_{11} & k_{12} \\
k_{21}& k_{22} \\
\end{array}
\right)
\left(
\begin{array}{cc}
a_1 \\
a_2 \\
\end{array}
\right)=
\left(
\begin{array}{cc}
F_1 \\
F_2 \\
\end{array}
\right)
\]
\end{itemize}
\boxed{\textbf{Dispersive and nondispersive waves}}
\[\frac{\partial^2 u}{\partial t^2}-c^2_0\frac{\partial^2 u}{\partial x^2}=0 \eqno(3)
\]
where \(c_0\) is constant,is the speed of wave,or pressure perturbations in a compressible fluid, with \(c_0\) the sound speed, or waves on shallow water.\
Solution(3) by:$$u(x,t)=Re[A(t)B(x)]$$,\(where\ B(x)=e^{ikx},k\ is\ real\ wavenumber\ ,\frac{2\pi}{k}\ is\ wavelenght\)\
then:$$\frac{d2B(x)}{dx2}=-k^2B$$
\[\frac{\partial^2 u}{\partial x^2}=Re[A(t)\frac{d^2B(x)}{dx^2}]=-k^2Re[A(t)B(x)]
\]
$(3)\Longrightarrow $\
\[\frac{d^2A(t)}{dt^2}+k^2c^2_0A=0
\]
solution$$A=u_+e^{-i\omega t};A=u_-e^{+i\omega t}$$ where $u_\pm $ are constantamplitudes.\
\[U(x,t)=Re[u_+e^{i(-\omega t +kx)}+u_-e^{+i(\omega t+kx)}] \eqno(5)$$\\
where $\omega$ is frequency $\omega^2=k^2c^2_0 $,it propagates with phase speed $c=\omega/k $ Note from (5)
that the first term of (5) is constant along characteristics with $x-\frac{\omega}{k} t=constant$\\
The wave are nondispersive: i.e their phase speed is independent of wavenumber\\
$$u(x,t)=F(x\pm c_0t)\eqno(6)$$ is a solution for (3)
}
\end{CJK*}
\end{document}\]
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