I am trying to derivative the equations from [1] for a compressible Volume-of-fluids formulation but I am stuck in one of the last steps and would like to request some help to solve it.
The governing equations are:
Mass conservation: $$ \frac{\partial (\alpha_k \rho_k)}{\partial t} + \nabla \cdot (\alpha_k \rho_k \vec{v}) = 0 \: \:, k={1,2} \: \:\:\: (1) $$
Phase transport: $$ \frac{\partial (\alpha_1)}{\partial t} + \nabla \cdot (\alpha_1 \vec{v}) = 0 \:\:\:\: (2) $$
With
$$ \alpha_1 + \alpha_2 = 1 \:\:\:\: (3) $$
I can take equation (1) and expand it into:
$$ \alpha_k \left [\frac{\partial (\rho_k)}{\partial t} + \vec{v} \cdot \nabla(\rho_k) \right ] + \rho_k \left[ \frac{\partial (\alpha_k)}{\partial t} + \nabla \cdot (\alpha_k \vec{v}) \right ] = 0 $$
$$ \frac{\partial (\alpha_k)}{\partial t} + \nabla \cdot (\alpha_k \vec{v}) = -\frac{\alpha_k}{\rho_k} \left [\frac{\partial (\rho_k)}{\partial t} + \vec{v} \cdot \nabla(\rho_k) \right ] \:\:\:\: (4) $$
Given that $\rho$ depends on $p$, and that $\frac{\partial \rho_k}{\partial p_k} = \psi_k$, through the chain derivation rule I get:
$$ \frac{\partial (\alpha_k)}{\partial t} + \nabla \cdot (\alpha_k \vec{v}) = -\frac{\alpha_k \psi_k}{\rho_k} \left [\frac{\partial (p)}{\partial t} + \vec{v} \cdot \nabla(p) \right ] \:\:\:\: (5) $$
Summing equation (5) for both phases, and attending to equation(3), I end up with:
$$ \nabla \cdot \vec{v} = - \left( \frac{\alpha_1 \psi_1}{\rho_1} + \frac{\alpha_2 \psi_2}{\rho_2} \right )\left [\frac{\partial (p)}{\partial t} + \vec{v} \cdot \nabla(p) \right ] \:\:\:\: (6) $$
Here is the part I cannot yet grasp:
The end result should be:
$$ \frac{\partial \alpha_1}{\partial t} + \nabla \cdot (\alpha_1 \vec{v}) - \alpha_1 (\nabla \cdot \vec{v}) = \alpha_1(1-\alpha_1) \left( \frac{\psi_2}{\rho_2} - \frac{\psi_1}{\rho_1} \right ) \left [\frac{\partial (p)}{\partial t} + \vec{v} \cdot \nabla(p) \right ] $$
If I expand equation (2), I can get:
$$ \frac{\partial (\alpha_1)}{\partial t} + \alpha_1 (\nabla \cdot \vec{v}) + \vec{v} \cdot \nabla \alpha_1= 0 $$
Moving $\alpha_1 (\nabla \cdot \vec{v})$ to the RHS and replacing the velocity divergence with Equation (6), I get:
$$ \frac{\partial (\alpha_1)}{\partial t} + \vec{v} \cdot \nabla \alpha_1 = -\alpha_1 \left (- \left( \frac{\alpha_1 \psi_1}{\rho_1} + \frac{\alpha_2 \psi_2}{\rho_2} \right )\left [\frac{\partial (p)}{\partial t} + \vec{v} \cdot \nabla(p) \right ] \right) $$
Knowing that: $$ \frac{\partial (\phi)}{\partial t} + \nabla \cdot (\vec{v} \phi) = \frac{\partial (\phi)}{\partial t} + \phi (\nabla \cdot \vec{v}) + \vec{v} \cdot \nabla \phi $$
$$ \frac{\partial (\phi)}{\partial t} + \nabla \cdot (\vec{v} \phi) - \phi (\nabla \cdot \vec{v}) = \frac{\partial (\phi)}{\partial t} + \vec{v} \cdot \nabla \phi $$
I can replace the LHS with:
$$ \frac{\partial (\alpha_1)}{\partial t} + \nabla \cdot( \alpha_1 \vec{v} ) - \alpha_1(\nabla \cdot \vec{v}) = -\alpha_1 \left (- \left( \frac{\alpha_1 \psi_1}{\rho_1} + \frac{\alpha_2 \psi_2}{\rho_2} \right )\left [\frac{\partial (p)}{\partial t} + \vec{v} \cdot \nabla(p) \right ] \right) \: \: \: \: (7) $$
The question is: How can I get the RHS correctly?
Best Regards!