# An equivalent definition of self-adjoint operator?

Suppose the linear operator: $$\begin{array}{rcll} L:&C^2[a,b]&\longrightarrow& C[a,b]\\ &u&\longmapsto&Lu=p_0\ddot u+p_1\dot u+p_2u \end{array}$$ with $p_0,p_1,p_2\in C[a,b]$ and $p_0\ne 0$ in $[a,b]$.

I thought that by analogy to the matrix, the operator $L$ should be called self-adjoint if $\langle Lu,v\rangle =\langle u,Lv\rangle$ for any $u,v\in C^2$, however in a lot of books I have found that $L$ is self-adjoint if: $$Lu=-\frac{d}{dt}\Big(p_0(t)\dot u(t)\Big) +p_2u$$ I can't prove why my definition of self-adjoint is equivalent to de last one.

Any hint? Many thanks!

• It is ODE jargon. In most places, "self-adjoint" means what you mean. However, in the case of differential operators, things tend to be more complicated than that. You need to specify boundary conditions, to begin with. With appropriate boundary conditions, an "ODE-selfadjoint" operator is self-adjoint, as you can see by integration by parts. – Giuseppe Negro Mar 29 '16 at 15:18

Some remarks. The term "adjoint" is a rather general one in mathematics, and is usually related to the following formula: $$\tag{1} \langle Au, v\rangle = \langle u, A^\star v\rangle,$$ where $\langle, \rangle$ is a pairing of some kind. Therefore, "self-adjointness" of $A$ always means $$A=A^\star.$$
In operator theory, the context is Hilbert spaces and the pairing is the scalar product. If $A$ is a bounded linear operator, then (1) is the true definition of "adjoint operator" and there's nothing more to be said.
However, linear differential operators are rarely considered to be bounded operators. (Here's an example). There are lots of technicalities here, but very roughly speaking, one implements differential operators in the abstract framework of operator theory by considering the ambient Hilbert space $L^2(\Omega)$ (here $\Omega$ is either an interval or an open set in $\mathbb{R}^n$) and then considering the differential operator as a mapping defined in a dense subspace of $L^2$, the domain of the operator: $$A\colon \mathrm{Dom}(A)\subset L^2\to L^2.$$ The domain is made of all functions with enough regularity to withstand the application of the differential operator $A$ and that satisfy the required boundary conditions.
It is only after all of this "blah-blah-blah" that one can speak of "self-adjointness" for a differential operator. In the case of ODEs, the blah-blah-blah is much less needed, so many authors take the shortcut to just define "self-adjoint" this particular class of operators: $$Lu(t)=-\frac d{dt}\left( p_0(t)\frac{du}{dt}\right) + p_2 u.$$ A more precise name for them would be "Sturm-Liouville operators". After one has specified appropriate boundary conditions, those operators turn out to be self-adjoint in the "blah-blah-blah" sense. $^{[1]}$
${}^{[1]}$ Essentially because they have complete orthonormal systems of eigenfunctions with real eigenvalues.