Self-adjoint boundary conditions for f(b)=cf(a), f′(b)=c′f′(a)

[warwick]

Under what condition on the constants c and c' are the boundary conditions

$\displaystyle f(b) = cf(a)$
$\displaystyle f'(b) = c'f'(a)$

self-adjoint for the operator $\displaystyle L(f) = (rf')' + pf$ on [a,b]? Assume that r and p are real.

Boundary conditions are self-adjoint if

$\displaystyle [r(f'\bar g - f\bar g')]^b_a = 0$

Using this evaluation and boundary conditions, I end up with

$\displaystyle f'(a)\bar g [c'r(b)-r(a)] + f(a)\bar g'[r(a) - cr(b)] = 0$


which implies that $\displaystyle c' = c = r(a)/r(b)$


However, the back of the book has $\displaystyle c \bar c'= r(a)/r(b).$ What am I doing wrong?