Multiply partial differential equation I do not know how to carry out the multiplications in order to get equation (5.103). Can anybody help me out?

 A: Multiplying the LHS of (5.101) and (5.102) by $\Theta$ and adding yields
$$\Theta\left(\frac{\partial\rho\Theta}{\partial t} + \frac{\partial}{\partial x_i}(\rho u_i\Theta)\right) + \Theta\left(\rho\frac{\partial\Theta}{\partial t} + \rho u_i\frac{\partial\Theta}{\partial x_i} \right) \\= \left(\Theta\frac{\partial(\rho\Theta)}{\partial t}+\rho\Theta\frac{\partial\Theta}{\partial t}\right) + \left(\Theta\frac{\partial}{\partial x_i}(\rho u_i\Theta) + \rho u_i\Theta\frac{\partial\Theta}{\partial x_i}\right) \\ = \frac{\partial}{\partial t}(\Theta\cdot\rho\Theta) + \frac{\partial}{\partial x_i}(\Theta\cdot\rho u_i\Theta) \\
= \frac{\partial(\rho\Theta^2)}{\partial t} + \frac{\partial}{\partial x_i}(\rho u_i\Theta^2)$$
which is the LHS of (5.103). Note that the equality from the second to the third line follows from the product rule.
The RHS of (5.101) and (5.102) are the same, so multiplying by $\Theta$ and adding yields
$$ 2\Theta\frac{\partial}{\partial x_i}\left(\rho D\frac{\partial \Theta}{\partial x_i}\right) + 2\Theta\dot{\omega}_\Theta. $$
Now
$$ 2\Theta\frac{\partial}{\partial x_i}\left(\rho D\frac{\partial\Theta}{\partial x_i}\right) = 2\left(\Theta\frac{\partial}{\partial x_i}\left(\rho D\frac{\partial\Theta}{\partial x_i}\right)+\left(\rho D\frac{\partial\Theta}{\partial x_i}\right)\frac{\partial\Theta}{\partial x_i}\right) - 2\rho D\frac{\partial\Theta}{\partial x_i}\frac{\partial\Theta}{\partial x_i} \\
=2\frac{\partial}{\partial x_i}\left(\Theta\cdot\rho D\frac{\partial\Theta}{\partial x_i} \right) - 2\rho D\frac{\partial\Theta}{\partial x_i}\frac{\partial\Theta}{\partial x_i} \\
=\frac{\partial}{\partial x_i}\left(\rho D\left(2\Theta\frac{\partial\Theta}{\partial x_i}\right)\right) - 2\rho D\frac{\partial\Theta}{\partial x_i}\frac{\partial\Theta}{\partial x_i} \\
= \frac{\partial}{\partial x_i}\left(\rho D\frac{\partial(\Theta^2)}{\partial x_i}\right) - 2\rho D\frac{\partial\Theta}{\partial x_i}\frac{\partial\Theta}{\partial x_i}$$
and hence
$$ 2\Theta\frac{\partial}{\partial x_i}\left(\rho D\frac{\partial\Theta}{\partial x_i}\right) + 2\Theta\dot{\omega}_\Theta = \frac{\partial}{\partial x_i}\left(\rho D\frac{\partial(\Theta^2)}{\partial x_i}\right) - 2\rho D\frac{\partial\Theta}{\partial x_i}\frac{\partial\Theta}{\partial x_i} + 2\Theta\dot{\omega}_\Theta $$
which is the RHS of (5.103).
