I have a little question that might be basic for some experts, but right now, its not clear for me. I want to implement temperature depending viscosity in a finite difference scheme (incompressible fluid). Therefore, I get for the viscous term
$\frac{\partial}{\partial x_j} \left( \nu \left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right) \right) $
Now, I equate this using the product rule (noting that $\nu$ is spatial variable):
$\frac{\partial \nu}{\partial x_j} \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_i}{\partial x_j} \right) + \nu \left( \frac{\partial^2 u_i}{\partial x_j \partial x_j} + \frac{\partial^2 u_j}{\partial x_j \partial x_i} \right)$
Making use of Schwarz's theorem, I can switch the cross derivatives in the last term. Because of the incompressibility condition ($\frac{\partial u_i}{\partial x_i} = 0$), I expect the last term to vanish, i.e. ending up with
$\frac{\partial \nu}{\partial x_j} \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_i}{\partial x_j} \right) + \nu \frac{\partial^2 u_i}{\partial x_j \partial x_j} $
My question now is rather simple: Is this correct?? I had some doubts since I remember some discussions that in case of temperature dependent viscosity one needs to implement cross-derivatives of the velocity field, and that these cross-derivatives are somehow tricky regarding the order of the numerical scheme. However, I don't need them here and this confuses me.
I would be happy if someone could help me.
Thanks Marius