I am trying to calculate the smoothness indicators for the WENO methods using the method given by Jiang and Shu.

$$\beta_k = \sum_{l=1}^k \Delta x^{2l-1} \int_{x_{i-1/2}}^{x_{i+1/2}} \Big(\frac{\partial^l}{\partial x^l} p_j(x)\Big)^2 dx$$

where $p_j(x)$ is the interpolation polynomial in the $k$-th sub-stencil. I believe he uses the Lagrangian interpolation, but I am not getting the exact coefficients that he gets here which is a review, and the smoothness indicators are given explicitly in equation (2.9) of the paper.

Therefore, I would like to ask if anyone knows of a reference that shows the calculation or if anyone has explicitly calculated the smoothness indicators from the polynomials.

  • $\begingroup$ What coefficient values are you getting? It might help find where you're going wrong. $\endgroup$
    – origimbo
    Commented Mar 14, 2017 at 12:59

1 Answer 1


After struggling with the same derivation, I realized what they are doing to get expressions for $\beta$'s. On the uniform grid, the problem may be solved for normalized variable, $\tilde x = (x - x_i-2)/\Delta$, so that the long stencil is (0, 1, 2, 3, 4). I'll Now the first Lagrange parabola $p_0$ is defined on $\tilde x \in (0, 1, 2)$ as

$$p_0(\tilde x) = \frac{1}{2}\left[(\tilde x^2 - 3\tilde x + 2) u_0 - 2(\tilde x^2 - 2\tilde x)u_1 + (\tilde x^2 - \tilde x) u_2\right]$$

$$p_0(\tilde x) = \frac{1}{2}\left[(u_0 - 2u_1 + u_2)\tilde x^2 + (- 3 u_0 + 4u_1 - u_2) x+ 2u_0\right]$$

$$p_0(\tilde x) = \frac{1}{2}\left[2a \tilde x^2 + b \tilde x + c\right]$$

with $a = u_0 - 2u_1 + u_2$, $b = - 3 u_0 + 4u_1 - u_2$ and $c = 2u_0$, and with derivatives

$$p'_0(\tilde x) = \frac{1}{2}\left[2a\tilde x + b \right]$$

$$p''_0(\tilde x) = a$$

$\beta_1$ is then, by definition, equal:

$$\beta_1 = \frac{1}{4} \int_{3/2}^{5/2} (4a^2 \tilde x^2 + 4ab \tilde x + b^2) d\tilde x + \int_{3/2}^{5/2} a^2 d\tilde x $$

After solving the integral and sorting out the terms, we get the form that paper.

$$\beta_1 = \frac{1}{3}(4u_0^2 + 25u_1^2 + 10 u_2^2 - 19 u_0 u_1 + 11 u_0 u_2 - 31 u_1 u_2)$$

This derivation is somewhat simplified as $\Delta$ is set to 1. However, $\beta$'s are relative weights, so in the uniform grid they can be determined for any three consecutive points. Their ratio is what matters in the end.

Equivalent derivation for the other two segments and their parabolas gives the other two $\beta$'s.


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