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.



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

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

Then I'm stuck

  • $\begingroup$ What is your question? Is it whether your attempt is correct? $\endgroup$ – LordVader007 Jan 8 at 4:06
  • $\begingroup$ yes, I forgot to mention it $\endgroup$ – Dreamer123 Jan 8 at 4:12

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?

  • 1
    $\begingroup$ Use \langle and \rangle for $\langle$ and $\rangle$. $\endgroup$ – mattos Jan 8 at 4:33
  • $\begingroup$ Thanks, sorry for that. I am (still) learning LaTeX commands.. $\endgroup$ – LordVader007 Jan 8 at 4:35
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    $\begingroup$ No worries, it looks much better now. Also, you can use \begin {align}, \end {align} and &= to format equal signs underneath each other (like I just edited for you). It makes it easier to read. $\endgroup$ – mattos Jan 8 at 4:37
  • $\begingroup$ In the 5th line, we have $u'(T)v(T) - u'(0)v(0) - u(T)v'(T) +u(0)v'(0)$, $u(0)=u(T)=0$ leaves us with two terms $u'(T)v(T)-u'(0)v(0)$ so for A to be self-adjoint we need $v(t)$ to be living in $D(A)$ so that $v(T)=v(0)=0$? $\endgroup$ – Dreamer123 Jan 8 at 9:24
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    $\begingroup$ In addition two Dreamer123's comment, this argument only shows the symmetry of $A$. Proving $D(A)=D(A^\ast)$ is (as usual) the more delicate part. $\endgroup$ – MaoWao Jan 8 at 13:34

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