text: Consider a rectangular channel of width much longer than the height. Let the distance between the top and bottom wall be  and choose the coordinates such that  lies in the midway between the two walls, with  points perpendicular to the planes. Let both walls be porous with equal velocity . Then the continuity equation and Navier–Stokes equations for incompressible fluid become
  with boundary conditions
  The boundary conditions at the center is due to symmetry. Since the solution is symmetric above the plane , it is enough to describe only half of the flow, say for . If we look for  a solution, that is independent of , the continuity equation dictates that the horizontal velocity  can at most be a linear function of . Therefore, Berman introduced the following form,
  where  is the average value (averaged cross-sectionally) of  at , that is to say
  This constant will be eliminated out of the problem and will have no influence on the solution. Substituting this into the momentum equation leads to 
  Differentiating the second equation with respect to  gives  this can substituted into the first equation after taking the derivative with respect to  which leads to 
  where  is the Reynolds number. Integrating once, we get
  with boundary conditions
  This third order nonlinear ordinary differential equation requires three boundary condition and the fourth boundary condition is to determine the constant . and this equation is found to possess multiple solutions. The figure shows the numerical solution for low Reynolds number, solving the equation for large Reynolds number is not a trivial computation.