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

Question

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.

Answer 1

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.