1
$\begingroup$

I'm trying to determine when the differential operator $L[f] = - \frac{d^2f}{dx^2} = - f''(x)$ is Hermitian, for different types of boundary conditions on $f$; however, I'm unclear how to interpret the boundary conditions in this context.

I'm not sure if this is because the question is loosely phrased and requires interpretation, or if it is just my own ignorance, so here is the original question:

Verify that $L[f] = -\frac{d^2f}{dx^2}$ is Hermitian on $[a,b]$ if $f$ satisfies (a) Dirichlet, (b) Neumann, [c] Periodic or (d) Robin boundary conditions.

I was able to show that

\begin{eqnarray*} \left<\, L[y_1], \,y_2 \right> = \left<\, y_1, \, L[y_2]\right> - \underbrace{\vphantom{\bigg|}y_1\mkern{-0.16em}' \,y_2\big|_a^b + y_1 \,y_2\mkern{-0.16em}'\big|_a^b}_{\text{boundary-dependent}} \end{eqnarray*}

however, I'm not clear what it would mean for ``$f$ ... satisfies (a) Dirichlet ... boundary conditions'' at this point in my work? Does that mean:

a) $y_1(a) = c_1$, $y_1(b) = c_2$ and $y_2(a) = k_1$, $y_2(b) = k_2$

or

b) $y_1(a) = y_2(a) = c_1$ and $y_1(b) = y_2(b) = c_2$

or something else / additional?

The boundary-dependent terms don't seem to vanish in either of the above interpretations, and the question is asking us to verify the statement, so I feel like I must be missing something. Any insight would be greatly appreciated.

$\endgroup$
  • $\begingroup$ (a) $f = 0$ on the boundaries (b) $f' = 0$ on the boundaries (c) $f' - \alpha f = 0$ for $\alpha \neq 0$ (d) $f(a) = f(b)$ and $f'(a) = f'(b)$, You should be able to easily verify whether or not the boundary term indeed vanishes for each of these cases. $\endgroup$ – Gregory Jun 2 '17 at 0:51
  • $\begingroup$ Except that those are homogeneous versions of the B.C. whereas "Dirichlet B.C." doesn't necessarily mean homogeneous Dirichlet and the question doesn't specify to use the homogeneous versions. Still, after reading several sites, this is the only plausible conclusion I can come up with: the question was just poorly posed. $\endgroup$ – Rax Adaam Jun 3 '17 at 15:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.