Let's solve what we can.
Start with the middle of the fluid. We can first get a general solution to this equation.
![[Graphics:../Images/lubricatedflow_gr_17.gif]](../Images/lubricatedflow_gr_17.gif)
It is convenient to examine this solution to see if an obvious simplification arises. It does. At r=0, we don't anticipate any reason for the velocity to be infinite. Since there is no other term to cancel log(r), we must make
=0. We can't do anything with the other constant at this moment. Recall however that in the single phase example,
was adjusted to make the velocity = 0 at the fixed wall.
This real ugly command sets
=0, and changes the name of
to f1 (to sort of resemble Middleman's notation.). It also picks out just the equation without the braces.
![[Graphics:../Images/lubricatedflow_gr_23.gif]](../Images/lubricatedflow_gr_23.gif)
We can solve the second equation the same way. This time the domain of r does not go to 0 and there is no need (or reason) to exclude the log(r) term -- so it stays.
![[Graphics:../Images/lubricatedflow_gr_25.gif]](../Images/lubricatedflow_gr_25.gif)
Change the name of the constants to e2 and f2 and extract the equation as "eq2a"
![[Graphics:../Images/lubricatedflow_gr_27.gif]](../Images/lubricatedflow_gr_27.gif)
We can take stock at this point. We don't know the three constants, f1, e2, f2 and we expect to need three boundary conditions. We also worry about and will need to resolve if dpdx1 = dpdx2. (They do, but we should discuss this -- see page 80).