Fixed-epsilon WENO-JS and WENO-Z face reconstructions are fifth order

Each first-difference combination in [F1] is $O(h)$ on C1 data, and each second-difference combination is $O(h^2)$ on C2 data. Squaring shows ${\beta }_k=O(h^2)$ for all three stencils; consequently ${\tau }_5=O(h^2)$. For the constant data case all these differences vanish and both schemes return the ideal weights exactly.

For WENO-JS, ${\alpha }_k^J=d_k{ϵ}_0^{-2}(1+{\beta }_k\mathrm{/}{ϵ}_0{)}^{-2}=d_k{ϵ}_0^{-2}(1+O(h^2))$. Since $\sum d_k=1$, normalization gives ${\omega }_k^J=d_k+O(h^2)$. For WENO-Z, ${ϵ}_0+{\beta }_k\ge {ϵ}_0$, so $({\tau }_5\mathrm{/}({ϵ}_0+{\beta }_k){)}^2=O(h^4)$; hence ${\alpha }_k^Z=d_k(1+O(h^4))$ and normalization gives ${\omega }_k^Z=d_k+O(h^4)$. The positive fixed ${ϵ}_0$ makes all constants uniform for sufficiently small $h$.

For either normalized weight vector $\omega$, ${\sum }_k({\omega }_k-d_k)=0$. Therefore ${\sum }_k{\omega }_k{p}_k-{\sum }_k{d}_k{p}_k={\sum }_k({\omega }_k-d_k)(p_k-u(x_{i+1\mathrm{/}2}))$. By [F2] the right side is $O(h^5)$ for WENO-JS and $O(h^7)$ for WENO-Z. The ideal combination has error $O(h^5)$ by [F2], so both left states have error $O(h^5)$. Reflection changes only the stencil orientation and preserves the same estimates.