1. Sep 30, 2015

### Incand

1. The problem statement, all variables and given/known data
Under what condition on the constant $c$ and $c'$ are the boundary conditions $f(b) = cf(a)$ and $f'(b)=c'f'(a)$ self-adjoint for the operator $L(f) = (rf')'+pf$ on $[a,b]$? (Assume that $r,p$ are real.)

2. Relevant equations
The boundary conditions are self-adjoint (relative L) if
$\left[r(f'\bar g-f\bar g ')\right]_a^b=0$
for all $f,g$ satisfying the boundary conditions.
(Assuming that $L$ is formally self-adjoint as it is in this case.)

3. The attempt at a solution
$\left[r(f'\bar g-f\bar g ')\right]_a^b = r(b)\left( f'(b)\bar g (b) - f(b)\bar g '(b)\right) -r(a)\left( f'(a)\bar g(a)-f(a)\bar g '(a)\right) = f'(a)\bar g (a)\left(c'\bar c r(b)-r(a)\right) + f(a)\bar g(a)\left(-r(b)c\bar c'+r(a)\right)$

So a solution would be $c'\bar c = c\bar c' = \frac{r(a)}{r(b)}$.

The answer to the question says only $c\bar c' = \frac{r(a)}{r(b)}$ but I got that $c'\bar c$ also has to have the same value. Am I correct here or did i make a mistake?

2. Sep 30, 2015

### pasmith

$c'\bar c$ is the complex conjugate of $\bar c' c$. The two are equal as $\frac{r(a)}{r(b)}$ is real and therefore equal to its complex conjugate.

3. Sep 30, 2015

Thanks!