I am reading about QUICK scheme for calculating the value of unknown variable $\phi$ in finite volume method. Given a locally one dimensional flow, we assume the value of $\phi$ is computed as a 2nd order polynomial: $$ \phi = k_0 + k_1 x + k_2 x^2 $$

Subject to:

  • $\phi = \phi_U$ at $x = x_U$ (Upwind)
  • $\phi = \phi_C$ at $x = x_C$
  • $\phi = \phi_D$ at $x = x_D$ (Downwind)

For the case of a uniform grid, the value of $\phi$ at cell $C$ face $f$ reduces to: $$\begin{aligned} \phi_f &= \frac{\phi_C + \phi_D}{2} - \frac{\phi_D - 2\phi_C + \phi_U}{8} \\ &= \frac{3}{4}\phi_C - \frac{1}{8}\phi_W + \frac{3}{8}\phi_E \end{aligned}$$

I am trying to understand how the above reduced formula was derived from the 2nd order polynomial, any pointers?


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QUICK utilizes the two upwind nodes, $x_U$ and $x_C$, and the downwind node, $x_D$ for a quadratic interpolation at the control volume face, $x_f$. Utilizing the Lagrange polynomial form of a quadratic:

\begin{multline*} \phi ({x_f}) = \left[ { - \frac{{\left( {{x_f} - {x_C}} \right)\left( {{x_D} - {x_f}} \right)}}{{\left( {{x_C} - {x_U}} \right)\left( {{x_D} - {x_U}} \right)}}} \right]{\phi _U} + \left[ {\frac{{\left( {{x_f} - {x_U}} \right)\left( {{x_D} - {x_f}} \right)}}{{\left( {{x_C} - {x_U}} \right)\left( {{x_D} - {x_C}} \right)}}} \right]{\phi _C} \\ + \left[ {\frac{{\left( {{x_f} - {x_U}} \right)\left( {{x_f} - {x_C}} \right)}}{{\left( {{x_D} - {x_U}} \right)\left( {{x_D} - {x_C}} \right)}}} \right]{\phi _D} \end{multline*}

Noting that $x_C - x_U = \Delta x$, $x_f - x_C = \Delta x/2$ and so-on, your uniform mesh stencil is easily recovered.


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