# Volume Discrepancy Factor

The volume discrepancy term is used in the constraint equation to force the system back to the equation of state. We write our velocity constraint equation as

(108)$\nablab \cdotb (\beta_0 \Ub) = \beta_0 \left(S - \frac{1}{\gammabar p_0} \frac{\partial p_0}{\partial t} - \frac{f}{\gammabar p_0} \frac{p_0 - p_\mathrm{EOS}}{\Delta t} \right ) .$

Here, $$f$$ is the volume discrepancy factor and ranges from 0 to 1, and $$p_\mathrm{EOS}$$ is the thermodynamic pressure as returned by the EOS, using the full state as input. In practice we evaluate this as

(109)$p_\mathrm{EOS} = p(\rho,h,X_k)$

The idea behind this forcing term is that if $$p_\mathrm{EOS} > p_0$$ then $$\nablab \cdotb (\beta_0 \Ub) > 0$$, and the cell will evacuate. This has the effect of returning us to $$p_\mathrm{EOS} = p_0$$.

In MAESTROeX, we decomponse the velocity into a base state component and a local component. The base state constraint equation is simply the horizontal average of the full constraint. Starting with $$\Ub = \Ubt + w_0 \er$$ in Eq.108, we have

(110)$\nablab \cdotb (\beta_0 w_0 \er) + \nablab \cdotb (\beta_0 \Ubt) = \beta_0 \left(S - \frac{1}{\gammabar p_0} \frac{\partial p_0}{\partial t} - \frac{f}{\gammabar p_0} \frac{p_0 - p_\mathrm{EOS}}{\Delta t} \right ) .$

Averaging this over a layer, we note that $$\overline{\nablab \cdotb (\beta_0 \Ubt)} = 0$$, yielding

(111)$\nablab \cdotb (\beta_0 w_0 \er) = \beta_0 \left(\overline{S} - \frac{1}{\gammabar p_0} \frac{\partial p_0}{\partial t} - \frac{f}{\gammabar p_0} \frac{p_0 - \overline{p_\mathrm{EOS}}}{\Delta t} \right )$

and

(112)$\nablab \cdotb (\beta_0 \Ubt) = \beta_0 \left(S - \overline{S} + \frac{f}{\gammabar p_0} \frac{p_\mathrm{EOS} - \overline{p_\mathrm{EOS}}}{\Delta t} \right ) .$

In solving the $$w_0$$ evolution Eq.111, we expand the divergence, giving

(113)$\nablab \cdotb (w_0 \er) = \overline{S} - \frac{1}{\gammabar p_0} \frac{\partial p_0}{\partial t} - w_0 \er \cdotb \frac{1}{\beta_0} \nablab \beta_0 - \frac{f}{\gammabar p_0} \frac{p_0 - \overline{p_\mathrm{EOS}}}{\Delta t} .$

Recalling that

(114)$\frac{1}{\gammabar p_0} \frac{\partial p_0}{\partial r} = \frac{1}{\beta_0} \frac{\partial \beta_0}{\partial r}$

(see Paper I), and that $$\psi \equiv D_0 p_0 / Dt \equiv \partial p_0 / \partial t + w_0 \partial p_0 / \partial r$$, we have

(115)$\nablab \cdotb (w_0 \er) = \overline{S} - \frac{1}{\gammabar p_0} \psi - \frac{f}{\gammabar p_0} \frac{p_0 - \overline{p_\mathrm{EOS}}}{\Delta t} .$

This is the form that is solved in make_w0.