How to derive Walen's equation? [closed]

Question

This is perhaps a simple math problem, therefore no books ever given detail calculation. Could someone help to give a proof?

We have continuity eqution as

$\frac{\partial\rho}{\partial t} = -\rho\nabla\cdot v -v\cdot\nabla\rho$

and induction equation as

$\frac{\partial B}{\partial t} = -(\nabla\cdot v)B - (v\cdot\nabla)B+(B\cdot\nabla)v + (\nabla\cdot B)v$

where we can assue $\nabla\cdot B=0$.

Then combine above two equations together into Walen's equation:

$\frac{d}{dt}(\frac{B}{\rho}) = (\frac{B}{\rho}\cdot\nabla)v$

Answer 1

Start by using the convective derivative defined as: $$ \frac{ d }{ dt } = \frac{ \partial }{ \partial t } + \nabla \cdot \mathbf{v} \tag{1} $$ then you can see that the following is true for the density: $$ \frac{ d \rho }{ dt } = - \rho \ \nabla \cdot \mathbf{v} \tag{2} $$ and similarly the magnetic field: $$ \frac{ d \mathbf{B} }{ dt } = - \mathbf{B} \nabla \cdot \mathbf{v} + \left( \mathbf{B} \cdot \nabla \right) \mathbf{v} \tag{3} $$

We then solve Equation 2 for $\nabla \cdot \mathbf{v}$ and insert the result into Equation 3 to get: $$ \frac{ d \mathbf{B} }{ dt } = \frac{ \mathbf{B} }{ \rho } \frac{ d \rho }{ dt } + \left( \mathbf{B} \cdot \nabla \right) \mathbf{v} \tag{4} $$

The last step is found by working backwards from the answer where we find: $$ \frac{ d }{ dt } \left( \frac{ \mathbf{B} }{ \rho } \right) = \frac{ 1 }{ \rho } \frac{ d \mathbf{B} }{ dt } - \frac{ \mathbf{B} }{ \rho^{2} } \frac{ d \rho }{ dt } \tag{5} $$

The rest is just a little algebra.