笔记

\(F(x)\)
\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:$$\frac{d^2B(x)}{dx2}=-k^2B$$

\[\frac{\partial^2 u}{\partial x^2}=Re[A(t)\frac{d^2B(x)}{dx^2}]=-k^2Re[A(t)B(x)] \]

$(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)\]