# QUICK scheme derivation

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?

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.