One of the few situations that the Navier-Stokes equations can be solved for is steady laminar
flow between plates. Were the fluid particles move in the x-direction there is no velocity,
and the y or the z direction it is fully developed flow, and so what we end up with for Navier-Stokes
is that 0 equals minus (dp/dx) plus mu, d^2u/dy^2. So we can rewrite this as d^2u/dy^2 equals
1, over mu, (dp/dx). So in order to solve for u or for our velocity we have to integrate
this twice. So our first integration gives du/dy equals 1 over mu, (dp/dx), times y plus
our constant 1. Now we integrated again, and we get that u equals 1 over 2mu, (dp/dx),
y^2, plus C1y, plus C2, and what we need to do here is solve for C1, and C2 by using boundary
conditions. So if we take a look at what this looks like here. Here are our 2 plates, and
we will have our origin in the middle. So each half of the plate is a length h. So our
2 boundary conditions are at y equals h, the velocity equals 0 because this is the no slip
condition, and the plate is not moving. We can also say at y equals negative h u also
equals 0. Again the purpose of a boundary condition is to find a velocity at a certain
length or boundary. So here we are looking at our y equaling h, and our y equaling negative
h and knowing what the velocities are there. So this allows us to rewrite our original
equations using the two boundary conditions. So the first one. Since u equals 0. Zero equals
(1/(2mu)). dp/dx, times h^2, plus C1,times h, plus C2. When we use the second boundary
conditions. Same thing we get 0 equals 1/2mu, dp/dx, now even though it is negative x when
we square it it becomes h^2, but now we have a minus C1 times h, plus C2. Now if we add
these two together what we will end up with is something that we could us to solve for
C2. So this is going to be 0 equals we add the 2 of these together. So it is 1mu. Instead
of 1/2mu. dp/dx times h^2, but now the C1 cancels out, plus 2C2, and now we can solve
four our C2, which equals minus 1 over 2mu, dp/dx, times h^2. Now to solve for C1. We
just put our expression for C2 into either one of the expression. So 0 equals 1 over
2mu, dp/dx, h^2, minus 1/2mu, dp/dx, h^2, minus C1 times h, and what that shows us is
that C1 equals 0. So this is the way to solve for boundary conditions, and in the end we
can rewrite our original equation, which is our u. By just replacing our C1 with 0, and
our C2 with what we had calculated before. So lets write our final equation. U equals
1 over 2 times mu, dp/dx, times y^2 minus 1 over 2mu, dp/dxh^2. So that is our C2, and
we can simplify this by saying that u equals 1 over 2 times mu, dp/dx, y^2, minus h62.
One thing to point out is that in this we have treated dp/dx, in the integration as
a constant, because it is not a function of y,