Finding Eigenvalues of First Order ODEs with Periodic BC? Suppose I have the eigenproblem:
$$A(x)\frac{df(x)}{dx} + B(x) f(x) = \lambda f(x), x\in[a,b]$$
$$f(a)=f(b)$$
where $A$, $B$ are some smooth real functions for simplicity. If we would have also an extra $\frac{d^2f(x)}{dx^2}$ and some extra BC condition, this would be a Sturm-Liouville problem and there is a lot of theory on that. But how about this first order problem? At first glance it should be easier to handle, but on a close inspection I found it quite troublesome and I also couldn't find any theory about it so far.
 A: First, we can try integrating directly (summarizing comments). This yields
$$
f(x)=f(a)\exp \left(\int\limits_a^x dx' \frac{\lambda-B(x')}{A(x')} \right)
$$
The requirement that $f(a)=f(b)$ sets $\int\limits_a^b dx \frac{\lambda-B(x)}{A(x)}=2\pi in$ with $n \in \mathbb{Z}$. If everything is real, and all the integrals exist, then $\lambda=\frac{\int dx \ B/A}{\int dx \ 1/A}$ so that there is just one eigenfunction with this eigenvalue.
To study the problem more generally, let us ask: "when is there a complete set of eigenfunctions?". This will be true if the operator is (truly) self-adjoint. Let
$$
L=A(x)\partial_x+B(x)
$$
Define an inner product with real weight $w(x)>0$
$$
\left<v|u\right>=\int\limits_a^b dx \ v^*(x)u(x)w(x)
$$
By direct substitution, we find that for real $A$, there is no weight such that $L$ is even formally self-adjoint. This is a bit of a surprise compared to the second order operators. Let $A$, $B$, and $f$ be complex. We will find conditions on $A$ and $B$ as well as $w$ such that $L$ is self-adjoint.
$$
\left<v|Lu\right>=\int\limits_a^b dx v^*Au'w+\int\limits_a^bv^*Buw
$$
Integrating by parts
$$
\left<v|Lu\right>=v^*Auw \bigg|_a^b+\int\limits_a^b dx \ uw \left[(vB^*)^*-(v'A^*)^*- (vA{^*}')^*-(vA^*w'/w)^*\right]
$$
From the terms in $[...]$ we find the formal adjoint
$$
L^\dagger = B^*-A^*\partial_x-A{^*}'-A^*w'/w \stackrel{?}{=} A \partial_x +B
$$
So we require $-A^*=A$, which means $A$ must be pure imaginary. With this, the other terms read
$$
B^*+A'+Aw'/w=B
$$
This may be solved for $w$
$$
w(x)=K\exp\left(\int\limits^x dx' \ \frac{B(x')-B^*(x')-A'(x')}{A(x')} \right)
$$
With the integration constant $K$ which we may choose. In the special case $B \in \mathbb{R}$, we have $w(x)=\pm i/A(x)$, with sign chosen so $w>0$, which is only possible if $iA$ does not change sign over $a<x<b$. One may object to division by a possibly vanishing $A$ here: but now we can see that if the integral does not exist, or if $K$ cannot be chosen such that $w>0$, then there will be no $w$ such that $L$ is self-adjoint. Now the boundary terms. We require
$$
v^*Auw \bigg|_a^b=0
$$
For $u(a)=u(b)$ and those $v$ such that $v^*(a)=v^*(b)$.
$$
v^*(a)A(a)u(a)w(a)=v^*(b)A(b)u(b)w(b)
$$
For these self-adjoint boundary conditions, none of the terms above may vanish, and we require
$$
A(a)w(a)=A(b)w(b)
$$
This is automatically satisfied in the case $B \in \mathbb{R}$, $w=\pm i/A$, so long as $A$ does not vanish at either $a$ or $b$. I think this is as far as we can go in full generality: we have found conditions on $A$ and $B$ (and consequently $w$) for $L$ to possess a complete set of eigenfunctions.
