- WENO3
\[
\begin{align}
\phi^-_{x,i}&=\frac{1}{2h}(\Delta^+\phi_{i-1}+\Delta^+ \phi_i)-\frac{1}{2h}\omega_{-}(\Delta^+ \phi_{i-2}-2\Delta^+\phi_{i-1}+\Delta^+ \phi_i)\\omega^- &=\frac{1}{1+2r^2_{-}}\r_{-} &=\frac{\epsilon+(\Delta^-\Delta^+ \phi_{i-1})^2}{\epsilon+(\Delta^-\Delta^+ \phi_i)^2}\\phi^+_{x,i}&=\frac{1}{2h}(\Delta^+ \phi_{i-1}+\Delta^+\phi_i)-\frac{1}{2h}\omega^+(\Delta^+\phi_{i+1} -2\Delta^+\phi_i+\Delta^+\phi_{i-1})\\omega_{+} &=\frac{1}{1+2r^2_{+}}\r_{+} &=\frac{\epsilon+(\Delta^-\Delta^+ \phi_{i+1})^2}{\epsilon+(\Delta^-\Delta^+ \phi_{i})^2}
\end{align}
\]
- 1D WENO Type Extrapolation
Assume that we have a stencil of three points \(x_0=0,x_1=h,x_2=2h\) with point values \(u_j,j=0,1,2\), We aim to obtain a \(3-k\)th order approximation of
\(\displaystyle \frac{d^k u}{dx^k}|_{x=-h/2},k=0,1,2\) denoted by $ u^{*k}$. We have three candidate substencils given by
\[S_r=\{x_0,..x_r\},r=0,1,2\]
On each stencil \(S_r\) we have a \(r\) degree \(p_r(x)\)
\[
\begin{align}
p_0(x) &=u_0\ p_1(x) &=u_0+\frac{1}{h} x(u_1-u_0)\ p_2(x) &=u_0+\frac{1}{2h}x(-3u_0+4u_1-u_2)+\frac{1}{2h^2}x^2(u_0-2u_1+u_2)
\end{align}
\]
Suppose \(u(x)\) is smooth on \(S_r\), the \(u^{*k}\) can be extrapolated by
\[
\begin{align}
u^{*k} &=\sum^2_{r=0}d_rp^{(k)}(x)|_{x=\frac{-h}{2}}\ d_0 &=h^2 \ d_1 &=h\ d_2 &=1-h-h^2
\end{align}
\]
We now look for WENO type extrapolation in the form
\[
\begin{align}
u^{*k} &=\sum^2_{r=0}\omega_rp_r^{(k)}(x)|_{x=\frac{-h}{2}}\ \beta_0 &=h^2\ \beta_1 &=\sum_{i=1}^2 \int_{-h}^0 h^{(2i-1)}(p^{(i)}_1(x))^2 dx=(u_1-u_0)^2 \ \beta_2 &=\sum_{i=1}^2 \int_{-h}^0 h^{(2i-1)}(p^{(i)}_2(x))^2 dx\ &=\frac{1}{12}(61u_0^2+160u_1^2+74u_0u_2+25u_2^2-196u_0u_1+124u_1u_2)\ \alpha_r&=\frac{d_r}{(\epsilon+\beta_r)^2}\ \omega_r&=\frac{\alpha_r}{\sum^2_{s=0} \alpha_s}\ d_0 &=h^2 \ d_1 &=h\ d_2 &=1-h-h^2
\end{align}
\]
WENO3
原文：https://www.cnblogs.com/yuewen-chen/p/11628976.html
