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During the derivation of the momentum equation we have

$$\frac{\partial}{\partial t}(\rho u_i)+\frac{\partial}{\partial x_j}(\rho u_iu_j)=f_i+\frac{\partial}{\partial x_j}(\sigma_{ij})$$

Using the product rule we have

$$\rho\frac{\partial u_i}{\partial t} + u_i\frac{\partial \rho}{\partial t} + u_i\frac{\partial}{\partial x_j}(\rho u_j) + u_j\frac{\partial}{\partial x_j}(\rho u_i) = f_i + \frac{\partial}{\partial x_j}(\sigma_{ij})$$

$$\implies \rho \left(\frac{\partial u_i}{\partial t}+u_j\frac{\partial u_i}{\partial x_j}\right)+u_i\left(\frac{\partial \rho}{\partial t}+\frac{\partial}{\partial x_j}(\rho u_j)\right) + u_iu_j\frac{\partial \rho}{\partial x_j} = f_i + \frac{\partial}{\partial x_j}(\sigma_{ij})$$

By mass conservation the second large bracket is zero.

$$\implies \rho \frac{{\rm D}u_i}{{\rm D}t} + u_i(\mathbf{u}\cdot\nabla)\rho=f_i+\frac{\partial}{\partial x_j}(\sigma_{ij})$$

$$\implies \rho\frac{{\rm D}\mathbf{u}}{{\rm D}t} + \mathbf{u}(\mathbf{u}\cdot\nabla)\rho = \mathbf{f}+\nabla\cdot\boldsymbol\sigma$$

Now how do we get rid of the second term on the LHS?

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  • $\begingroup$ 2nd term on the LHS of what equation? $\endgroup$ – David Oct 8 '16 at 21:24
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You made an error in deriving the second equation, by incorrectly applying the product rule for derivatives to obtain

$$\frac{\partial}{\partial x_j}(\rho u_iu_j)= u_i \frac{\partial}{\partial x_j}(\rho u_j) + u_j \frac{\partial}{\partial x_j}(\rho u_i)$$

A correct application of the product rule yields

$$\frac{\partial}{\partial x_j}(\rho u_iu_j)= u_i \frac{\partial}{\partial x_j}(\rho u_j) + \rho u_j \frac{\partial u_i}{\partial x_j}$$

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