momentum balance equation for low reynolds number flow This is a flow in a 2-D microchannel (h/L<<1).
Low Reynolds number flow.
The mass - conservation equation states 
$$\frac{\partial (\rho u)}{\partial x}+\frac{\partial (\rho v)}{\partial y}=0$$
Where, $$\rho=\rho(x)$$
How to get this equation  $$\frac{\partial p}{\partial x}=\mu \frac{\partial ^2u}{\partial y^2}$$  from the momentum balance equation i.e. $$\rho( {\partial{\bf u}\over{\partial t}} + ({\bf u} \cdot \nabla) {\bf u}) = -  \nabla p + \mu\nabla^2{\bf u} + {\mu}{1\over3}\nabla (\nabla \cdot {\bf u})$$ 
Taking scales can eliminate the left- hand side as this a low Reynolds number flow. But I can't vanish the last term on the right- hand side. 
Is it possible $$\rho{\bf u} \cdot \nabla {\bf u}={\bf u} \cdot \nabla (\rho{\bf u})=0 $$ even after considering $$\rho=\rho(x)$$
Or should  I use this form of equation $$ {\partial{\bf u}\over{\partial t}} + ({\bf u} \cdot \nabla) {\bf u} = - {1\over\rho} \nabla p + \gamma\nabla^2{\bf u} + {g_{x}} $$ as $$\rho g_{x}=0$$ 
please help.
 A: If we apply the same coordinate transformation to the momentum equation that we employed in the other forum (Physics Forums), we obtain for the X component of the momentum equation:
$$U\frac{\partial U}{\partial X}+V\frac{\partial U}{\partial Y}=-\frac{1}{\rho}\frac{\partial p}{\partial X}+\frac{\mu}{\rho}\left[\frac{\partial^2U}{\partial X^2}+\frac{\partial^2U}{\partial Y^2}\right]+\frac{\mu}{3\rho}\left[\frac{\partial ^2U}{\partial X^2}+\frac{\partial^2V}{\partial X\partial Y}\right]\tag{1}$$
We next define the following dimensionless parameters:
$$\bar{U}=\frac{U}{c}$$
$$\bar{V}=\frac{\lambda}{h}\frac{V}{c}$$$$\bar{X}=\frac{X}{\lambda}$$
$$\bar{Y}=\frac{Y}{h}$$$$\bar{\rho}=\frac{\rho}{\rho_0}$$
$$\bar{p}=\frac{h^2p}{\mu c \lambda}$$
If we substitute these dimensionless variables into Eqn. 1, we obtain:
$$Re\left(\frac{h}{\lambda}\right)^2\bar{\rho}\left[\bar{U}\frac{\partial \bar{U}}{\partial \bar{X}}+\bar{V}\frac{\partial \bar{U}}{\partial \bar{Y}}\right]=-\frac{\partial \bar{p}}{\partial \bar{X}}+\left(\frac{h}{\lambda}\right)^2\frac{\partial^2\bar{U}}{\partial \bar{X}^2}+\frac{\partial^2\bar{U}}{\partial \bar{Y}^2}+\frac{1}{3}\left(\frac{h}{\lambda}\right)^2\left[\frac{\partial ^2\bar{U}}{\partial \bar{X}^2}+\frac{\partial^2\bar{V}}{\partial \bar{X}\partial \bar{Y}}\right]\tag{2}$$where the Reynolds number Re is given by:
$$Re=\frac{\rho_0 c \lambda}{\mu}$$In the limit of $Re\rightarrow 0$ and $(h/\lambda)^2\rightarrow 0$, Eqn. 2 becomes:
$$0=-\frac{\partial \bar{p}}{\partial \bar{X}}+\frac{\partial^2\bar{U}}{\partial \bar{Y}^2}\tag{3}$$In terms of the original dimensional variables, this now becomes:$$0=-\frac{\partial p}{\partial X}+\mu \frac{\partial^2U}{\partial Y^2}\tag{3}$$
