hi nerds could someone explain to me how the LHS is expanded

sorry i mean LHS to the middle form, i get that the second part of the middle equation is 0 for conserved phi and hence the first part is equivalent to RHS

Use the product rule for each term and then rearrange.

i see product rule on the first term expands into the 2nd equation, but what happens to the div term?

I'm guessing div(row*phi*u) = 0 because conservation?

[math]\frac{\partial\left(\rho \phi\right)}{\partial t}+\mathbf{\nabla}\cdot\left(\rho\phi\mathbf{u}\right)=\left[\rho\frac{\partial \phi}{\partial t}+\phi\frac{\partial\rho}{\partial t}\right]+\left[\rho\mathbf{u}\cdot\mathbf{\nabla}\phi+\phi\mathbf{\nabla}\cdot\left(\rho \phi\right)\right][/math]

I put the expansion of each term in square brackets for you.

>>8577667
[math]\frac{\partial\left(\rho \phi\right)}{\partial t}+\mathbf{\nabla}\cdot\left(\rho\phi\mathbf{u}\right)=\left[\rho\frac{\partial \phi}{\partial t}+\phi\frac{\partial\rho}{\partial t}\right]+\left[\rho\mathbf{u}\cdot\mathbf{\nabla}\phi+\phi\mathbf{\nabla}\cdot\left(\rho \mathbf{u}\right)\right][/math]

*Oops

here's a hint, compute this while treating phi as a function of space and rho as only a function of time

[math]
\nabla \cdot(\rho \phi \textbf{u})

[/math]

>>8577668
is that a latex code?

ah yes managed to process the code. once i can compute the hint ill be able to do the main equation, but ill have to go back over the div rules.anyway thanks for your time!