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

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

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

• Use \langle and \rangle for $\langle$ and $\rangle$. – mattos Jan 8 at 4:33
• Thanks, sorry for that. I am (still) learning LaTeX commands.. – LordVader007 Jan 8 at 4:35
• 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. – mattos Jan 8 at 4:37
• 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 $v(T)=v(0)=0$? – Dreamer123 Jan 8 at 9:24
• 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. – MaoWao Jan 8 at 13:34