= 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 $\lambda\operatorname{tr}e+2\mu e_{ii}$, 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 gives
$$
\boxed{\sigma_{ij}=-p\delta_{ij}+2\mu e_{ij}.}
$$
At 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 $\mu_\pm u_\pm''=-G$, 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 $C$:
$$
u_\pm(y)=U_0+\frac C{\mu_\pm}y-\frac G{2\mu_\pm}y^2.
$$
Imposing $u_-(-a)=u_+(a)=0$ gives
$$
\boxed{U_0=\frac{Ga^2}{\mu_++\mu_-},\qquad C=\frac{Ga(\mu_--\mu_+)}{2(\mu_++\mu_-)}.}
$$
The 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>.
Back to article page
