# Self-adjointness of $D=\frac{d^2}{dx^2}-1$ with boundary conditions $u'(0) = 0 = u'(a)$ on $[0,a]$.

Im trying to show that $$D=\frac{d^2}{dx^2}-1$$ is self adjoint on $[0,a]$ subject to $u'(0)=u'(a)=0$. I think I need to use integration by parts but I'm not sure how to do that.

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What inner product are you using? – Jp McCarthy Jan 24 '13 at 22:31
What is the source of the problem? – Jonas Meyer Jan 24 '13 at 22:34
Alex's reply to Jp McCarthy was also deleted. I think it said something like $(u,v)=\int_0^a\overline{u(x)}v(x)dx$. – Jonas Meyer Jan 25 '13 at 6:48
@Alex: You are no longer wondering about how to find the eigenvalues and eigenfunctions of $D$? – Jonas Meyer Jan 25 '13 at 21:54
Show that $D$ is self-adjoint is equivalent to showing the integral identity $$\int_0^a \left( \frac{d^2 \bar{u}}{dx^2} - \bar{u} \right) v \, dx = \int_0^a \bar{u} \left( \frac{d^2 v}{dx^2} - v \right) \, dx$$ when both $u,v$ have derivatives vanishing at $x= 0,a$. As you yourself suggested, you should be able to show this using integration by parts, since the whole purpose of integration by parts is to move a derivative from one function to another. Notice that you don't want to do integration by parts on the whole integral as it is written above, since the term $\int_0^a - \bar{u} v \, dx$ already matches on both sides of the equation.