Boundary Layer Solution Obtained with an Initial Value Technique
The problem to be solved for boundary layer flow past a flat plate is
This problem is a boundary value problem, since the conditions on f are at two different positions. The problem is also complicated by the fact that one of the boundary conditions is at infinity. Since infinity is hard to represent in the computer, we must set the outer limit to a large number and iteratively make sure that the location of that boundary does not affect the solution in the region of interest. Thus, we really have two nested iterations: the location of the interface and the iterations needed to solve the two-point boundary value problem.
Let us change the boundary condition at infinity to the following:
The value of 'long' must be determined by trial and error. From other information, we suspect that it is of order one, and we start with the value of 3, and increase it or decrease it as needed. We must change the boundary value problem into an initial value problem. Define the variables
The equations are then
We cannot impose the condition
directly; rather, we must choose a value of y3(0) = c, and adjust c so that the condition is satisfied. The boundary conditions are then
We use the Newton-Raphson method to adjust the value of c, and this requires we define some sensitivity equations. We differentiate each dependent variable by c, since the ultimate solution depends on c. Thus, define
The equations governing these additional variables are obtained by differentiating the initial value equations by c.
The initial conditions are
The condition we would like to satisfy is
The derivative of this is
We use the following Newton-Raphson iteration scheme.
Thus, we solve the six ordinary differential equations from h = 0 to h = c, evaluate y and dy/dc and get a new value of c. The code is straightforward.
It took only 6 iterations to reduce y to less than 10-12 with the Newton-Raphson method. The solutions for various values of 'long' are shown in the table. Using a value of c = 6 gives results accurate to 2.5 x 10-4 per cent. The difference between using 9 and 12 is only one digit in the 9th significant digit. The solution using c = 9 is shown in the figure. This figure shows the expected behavior: the velocity is linear in distance from the wall, near the wall, and then levels out to a flat velocity profile far from the wall. The value of c defines the word 'far'.
Constant from Boundary Layer Solution, Flat Plate
| long | c |
|---|---|
| 3 | 0.490140 |
| 6 | 0.469595 |
| 9 | 0.469583802 |
| 12 | 0.469583801 |
The shear stress constant is
Take Home Message: The initial value method is an easy way to solve boundary layer equations, including on infinite domains.