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?

  • $\begingroup$ 2nd term on the LHS of what equation? $\endgroup$ – David Oct 8 '16 at 21:24

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.