Page 259 - Computational Fluid Dynamics for Engineers

8.2 Solution of the Orr-Sommerfeld Equation 249

We note from the difference equations defined by Eqs. (8.2.3) and (8.2.5)
that the solution of Eq. (4.4.29) depends upon the three parameters a, u and
R and we can denote this dependence by writing

6 = ^(a,u;,R) (8.2.10)

Recalling that a is complex, and that u is real in spatial-amplification theory,
the above relation implies that the solution of Eq. (4.4.29) depends upon four
scalars. With any two of these scalars fixed, the remaining scalars can be com-
puted in such a way that the missing boundary condition 0'(O) = 0 is satisfied.
In our finite-difference notation, this corresponds to the condition

f 0(a,uj,R) = 0 (8.2.11)

8.2.2 Eigenvalue Procedure
To discuss the solution procedure for Eq. (8.2.11), let us consider the computa-
tion of a r , u and R for a specified value of c^. In this case since UJ is real, the
solution of Eq. (8.2.11) can be obtained by fixing one parameter and solving
for the remaining two. The choice of the fixed parameter depends on the slope
of da r/dR so that, for example, it is more convenient to fix R and solve for a r
and uj away from the critical Reynolds number where da r/dR is small. In the
region of critical Reynolds number, da r/dR is large and increases as R —* R cr
and it is necessary to specify a r and solve for u and R. To explain these points
further, consider the eigenvalue problem corresponding to small da r/dR. Since
/o is complex, OLI is given and R is fixed, Eq. (8.2.11) represents two equa-
tions with two unknowns (a r,u) and the equations can be solved by Newton's
v
method. Specifically, if (a u ,uo ) are the i/-th iterates, the (z/-f- l)-th iterates are
determined by using
v
u
< + 1 =a r+6a r (8.2.12a)
u+1 / v
u> =(J + 8u (8.2.12b)
v
V
in Eq. (8.2.11), expanding /o about a T and UJ and retaining only linear terms
in the expansion. This gives the linear system by taking ^ + 1 = 0 and f" + = 0
/

+ f a + & / , <8 2 i3b)
^ (^)" f (^)" =° - -
Here, for convenience, we have dropped the subscript 0 on / and used r and i
to denote the real and imaginary parts of / at the wall. The solution of Eqs.
(8.2.13) is, (8214a)
^ = iHi)"^(i)l -