Prove that $(Au)(t)=\frac{d^{2}u(t)}{dt^{2}}$ is self-adjoint

Question

Let $D(A)=\{ u \in L_2(0,T)| u, \frac{du}{dt}$ are absolutly continuous with $\frac{du}{dt} \in L_2(0,T)$, $u(0)=u(T)=0\}$

and, $(Au)(t)=\frac{d^{2}u}{dt^{2}}$ prove that $A$ is self-adjoint.

Trial

Consider,

$\langle Au(t), v(t) \rangle$= $\int_{0}^{T}\frac{d^{2}}{dt^{2}} u(t)v(t) dt$= $\int_{0}^{T}\frac{d^{2}}{dt^{2}}(u(t)v(t))dt -2\int_{0}^{T}\frac{du(t)}{dt} \frac{dv(t)}{dt} dt- \int_{0}^{T}(\frac{d^2}{dt^2}v(t))u(t)$ $dt$

=$\frac{dv(t)u(t)}{dt}|_{t=T}-\frac{dv(t)u(t)}{dt}|_{t=0} -2\int_{0}^{T}\frac{du(t)}{dt} \frac{dv(t)}{dt} dt -\langle u(t), Av(t) \rangle$

Then I'm stuck

Answer 1

I think you are missing a term in the second to last line. That being said, yes the operator is self-adjoint.

This is how I would have done it. Using integration by parts, we get that:

\begin{align} \langle Au,v \rangle &= \int_0^T u''(t)v(t)dt \\ &= u'(t)v(t) \bigg|_{0}^{T} - \int_0^T u'(t)v'(t)dt \\ &= u'(t)v(t) \bigg|_{0}^{T} - u(t)v'(t) \bigg|_{0}^{T} +\int_0^T u(t)v''(t)dt \end{align}

Apply the BC's $u(0)=u(T)=0$ and so the first two terms disappear. Note that it also works with another function $v(t)$ living inside $D(A)$. Thus,

$\langle Au,v \rangle = \langle u,Av \rangle$ indeed.

The fun part is now recognizing that any arbitrary differential operator may or may not be self adjoint. For example, try an operator like:

$Lu = iu'''(t)$, with BCs of $u(0)=u'(0) = u''(1)=0$. Is it self-adjoint?