WENO3

Posted 2022-12-31 yuewen-chen

• WENO3
  
  \[ \beginalign \phi^-_x,i&=\frac12h(\Delta^+\phi_i-1+\Delta^+ \phi_i)-\frac12h\omega_-(\Delta^+ \phi_i-2-2\Delta^+\phi_i-1+\Delta^+ \phi_i)\\omega^- &=\frac11+2r^2_-\r_- &=\frac\epsilon+(\Delta^-\Delta^+ \phi_i-1)^2\epsilon+(\Delta^-\Delta^+ \phi_i)^2\\phi^+_x,i&=\frac12h(\Delta^+ \phi_i-1+\Delta^+\phi_i)-\frac12h\omega^+(\Delta^+\phi_i+1 -2\Delta^+\phi_i+\Delta^+\phi_i-1)\\omega_+ &=\frac11+2r^2_+\r_+ &=\frac\epsilon+(\Delta^-\Delta^+ \phi_i+1)^2\epsilon+(\Delta^-\Delta^+ \phi_i)^2 \endalign \]
• 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
  \(\displaystyle \fracd^k udx^k|_x=-h/2,k=0,1,2\) 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 \(p_r(x)\)
  \[ \beginalign p_0(x) &=u_0\ p_1(x) &=u_0+\frac1h x(u_1-u_0)\ p_2(x) &=u_0+\frac12hx(-3u_0+4u_1-u_2)+\frac12h^2x^2(u_0-2u_1+u_2) \endalign \]
  Suppose \(u(x)\) is smooth on \(S_r\), the \(u^*k\) can be extrapolated by
  \[ \beginalign u^*k &=\sum^2_r=0d_rp^(k)(x)|_x=\frac-h2\ d_0 &=h^2 \ d_1 &=h\ d_2 &=1-h-h^2 \endalign \]
  We now look for WENO type extrapolation in the form
  \[ \beginalign u^*k &=\sum^2_r=0\omega_rp_r^(k)(x)|_x=\frac-h2\ \beta_0 &=h^2\ \beta_1 &=\sum_i=1^2 \int_-h^0 h^(2i-1)(p^(i)_1(x))^2 dx=(u_1-u_0)^2 \ \beta_2 &=\sum_i=1^2 \int_-h^0 h^(2i-1)(p^(i)_2(x))^2 dx\ &=\frac112(61u_0^2+160u_1^2+74u_0u_2+25u_2^2-196u_0u_1+124u_1u_2)\ \alpha_r&=\fracd_r(\epsilon+\beta_r)^2\ \omega_r&=\frac\alpha_r\sum^2_s=0 \alpha_s\ d_0 &=h^2 \ d_1 &=h\ d_2 &=1-h-h^2 \endalign \]