WENO3

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
\[ \begin{align} u^{*k} &=\sum^2_{r=0}\omega_rp_r^{(k)}(x)|_{x=\frac{-h}{2}}\\ \beta_0 &=h^2\\ \beta_1 &=\sum_{i=1}^2 \int_{-h}^0 h^{(2i-1)}(p^{(i)}_1(x))^2 dx=(u_1-u_0)^2 \\ \beta_2 &=\sum_{i=1}^2 \int_{-h}^0 h^{(2i-1)}(p^{(i)}_2(x))^2 dx\\ &=\frac{1}{12}(61u_0^2+160u_1^2+74u_0u_2+25u_2^2-196u_0u_1+124u_1u_2)\\ \alpha_r&=\frac{d_r}{(\epsilon+\beta_r)^2}\\ \omega_r&=\frac{\alpha_r}{\sum^2_{s=0} \alpha_s}\\ d_0 &=h^2 \\ d_1 &=h\\ d_2 &=1-h-h^2 \end{align} \]

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