 Solution

Let $e_{ij}=({\partial }_i{u}_j+{\partial }_j{u}_i)/2$ be the rate-of-strain tensor. A Newtonian fluid has a linear isotropic constitutive law for its viscous stress tensor. To determine its form, diagonalize the symmetric tensor $e$. Rotations by $\pi$ about its principal axes leave $e$ fixed, forcing its stress response to be diagonal as well. Linearity and symmetry under permutations of the axes force each diagonal entry to be $\lambda \mathrm{tr}e+2\mu e_{ii}$, with the same constants for each axis. Rotating back gives ${\tau }_{ij}=\lambda e_{kk}{\delta }_{ij}+2\mu e_{ij}$. Since incompressibility means $e_{kk}=0$, adding the arbitrary isotropic pressure givesAt a material interface between ordinary viscous fluids without interfacial slip, velocity is continuous, and its normal component equals the interface's normal speed. Tangential traction is continuous when surface tension is constant. Normal traction has the curvature-dependent surface tension jump; in a convention defining signed curvature $\kappa$ by the balance, $({\sigma }_{+}-{\sigma }_{-})n=\gamma \kappa n$. With no surface tension, the full traction is continuous. At rigid stationary walls the no-slip boundary condition imposes zero velocity.

Put $G=-dp/dx>0$ and seek $u=u_{\pm }(y)e_x$. The convective acceleration vanishes, and the Navier-Stokes equations reduce to ${\mu }_{\pm }{u}_{\pm }^{\prime \prime }=-G$, with $p=p_0-Gx$. continuity of velocity and tangential traction at $y=0$ gives a common interface velocity $U_0$ and a common shear integration constant $C$:Imposing $u_{-}(-a)=u_{+}(a)=0$ givesThe velocity has no normal component, so the flat interface remains stationary, and its normal stresses match because pressure is common. These expressions give the two-layer plane Poiseuille flow with unequal viscosities.