Steady Flow Between Two Parallel Plates

Consider the steady flow of a compressible Navier-Stokes fluid between two parallel horizontal plates with the lower plate kept stationary and the upper one moved in the positive $x_1$-direction at a uniform speed of $v_0\ m/s$. Assume that the fluid sticks to the plates and that $v_i(\mathbf{x},t) = v(x_2)\delta_{i1}$ where the $x_2$-axis is perpendicular to the plate surfaces, and the fluid extends to infinity in the $x_3$-direction. Thus we have a two-dimensional problem in the $x_1-x_2$ plane with the fluid flowing only in the $x_1$-direction. It is therefore reasonable to assume that the mass density and the pressure field also depend only upon $x_1$ and $x_2$. For the presumed velocity field, $v_{i,i}=0$. Thus the motion is isochoric and the mass density of a material particle does not change. Also, both the local and the convective parts of the acceleration identically vanish.

Nontrivial equations governing the flow of the fluid are

$$\frac{\partial\rho}{\partial x_1}v =\ 0,$$ (6.3.1)

$$0 =\ - \frac{\partial p}{\partial \rho}\frac{\partial\rho}{\partial x_1} + \mu\frac{\partial^2v}{\partial x^2_2},$$ (6.3.2)

$$0 =\ -\frac{\partial p}{\partial\rho}\frac{\partial\rho}{\partial x_2} + \rho g,$$ (6.3.3)

$$v(x_1,0)=\ 0,\ v(x_1,h) = v_0,$$ (6.3.4)

$$T_{ij}n_j\bigg\vert _{x_1=-L}=\ \bar p(x_2)_{\delta_{i1}},\ T_{ij}n_j\bigg\vert _{x_1=L} = -p_0(x_2)\delta_{i1}$$ (6.3.5)

Whereas equation (6.3.1) expresses the balance of mass, equations (6.3.2) and (6.3.3) are the reduced forms of the balance of linear momentum in the $x_1$ and $x_2$ directions. Equation (6.3.4) states the essential boundary conditions on the surfaces of the two plates, and equation (6.3.5) gives natural boundary conditions on the vertical surfaces $x_1 = \pm L$. It follows from (6.3.1) that $\rho = \rho (x_2)$. Equations (6.3.2) and (6.3.4) give

$$v = \frac{x_2}{h} v_0\, .$$ (6.3.6)

Thus the velocity field is known. It follows from (6.3.3) that

$$\frac{d\rho}{dx_2} = \frac{\rho g}{\partial p/\partial\rho}\, .$$ (6.3.7)

The determination of $\rho$ as a function of $x_2$ requires that the compressibility of the fluid be known. Said differently, the constitutive equation $p = p(\rho)$ must be given. As an example, we take

$$p = c\rho$$ (6.3.8)

where $c$ is a constant. Then, equations (6.3.7) and (6.3.8) yield

$$\rho = de^{\frac{g}{c}x_2}$$ (6.3.9)

where $d$ is a constant. Equations (6.3.8) and (6.3.9) imply that $p$ can depend only on $x_2$. The stress tensor in the fluid computed from equations (6.1.3), (6.3.6), (6.3.8) and (6.3.9) is given by

$$T_{ij} = -cde^{gx_2/c}\delta_{ij} + \mu \frac{v_0}{h} \left(\delta_{2j}\delta_{i1} + \delta_{2i}\delta_{j1}\right),$$ (6.3.10)

and is only a function of $x_2$. Thus $\bar p(x_2)$ must equal $p_0(x_2)$ in equation (6.3.5); otherwise there is no solution for the problem. That is, the pressure distribution must be same on every plane $x_1 =$   const. From equations (6.3.5) and (6.3.10) we conclude that

$$\bar p(x_2) = cde^{gx_2/c}\, .$$ (6.3.11)

That is, for the fluid being studied, surface tractions on planes $x_1 = \pm L$ can not be arbitrarily prescribed, but must be of the form (6.3.11). From the given value of $\bar p(x_2)$, we find the constant of integration $d$. The fields of density and velocity are respectively given by (6.3.9) and (6.3.6).

In linear elasticity, the uniquenes theorem guarantees that the solution obtained by a semi-inverse method is the only solution of the problem. However, in fluid mechanics, a uniqueness theorem can not be proved because of the nonlinear governing equations. It implies that there may be other solutions for the problem studied.