TITLE: Solution to at least one ODE in a family of ODE's
QUESTION [7 upvotes]: In my research I have stumbled across the following 1st order complex differential equation for smooth functions $\eta:\mathbb{R}/2\pi\mathbb{Z}\to\mathbb{C}-\lbrace0\rbrace$ defined on the circle,
$$i\frac{\partial\eta}{\partial t}+(re^{it}+\varepsilon i)\bar\eta=0$$
where $\varepsilon\in\mathbb{R}_+$ is sufficiently small and fixed (so choose $0<\varepsilon<<1$ but maybe not $\varepsilon=\frac{1}{2}$) and $r\in\mathbb{R}_+$ is a nonzero positive real parameter. Motivated by some perturbation theory and functional analysis, I believe there must exist a nontrivial solution $\eta$ to this ODE for at least one choice of $r$. What is such an explicit pair $(r,\eta)$?
This problem arises when studying the asymptotics of punctured $J$-holomorphic curves, and I need the solutions to do what I want to do (when perturbing $J$). Ultimately I desire the set of all such distinct $r$ and the dimension of the kernel of the corresponding differential operators. 
$\underline{\text{Attempt}}$ I originally posted this on StackExchange a month ago with many edits but not much luck. My attempt was to Fourier expand $\eta(t)=\sum_{k\in\mathbb{Z}}a_ke^{ikt}$ and obtain the recurrence relation
$$-ka_k+r\bar a_{1-k}+\varepsilon i\bar a_{-k}=0$$
I need a specific collection $\lbrace a_k\rbrace\subset\mathbb{C}$ which solves this (for some $r>0$). What I get at the least is $r=\varepsilon i\frac{a_0}{a_1}$ (and subsequently $\frac{a_0}{a_1}$ must be purely complex and nonzero). But there is still a good chance that the coefficients $a_k$ will "blow up" as $k\to\infty$ if not chosen carefully. I've done some manipulations and I cannot parse whether they are helpful or harmful.
Also, this complex ODE is equivalent to two coupled real ODEs, but I don't think it helps. Decompose $\eta=x+iy$ and attempt to solve the equivalent system:
$$\dot y(t) -r\cdot\cos(t)\cdot x(t) -[\varepsilon +r\cdot\sin(t)]\cdot y(t) = 0$$
$$\dot x(t) -r\cdot\cos(t)\cdot y(t) +[\varepsilon +r\cdot\sin(t)]\cdot x(t) = 0$$
Perhaps there are numerical methods to find "approximate" periodic solutions, or some software to plot $(x,y)$ for various values of $r\in\mathbb{R}_+$?

REPLY [3 votes]: This is an expanded version of my comments, to make them somewhat less cryptic. I'm not answering the question, and, I don't really expect to ever make further progress.

Update: After I learned from fedja's answer that an $r\simeq \epsilon^{1/2}$ works (much to my surprise), I can now show that too by a perturbation expansion of $D$ below, and this is now included in this new version.

If we write $u=(x,y)^t$, then we can rephrase the question as: Does the Dirac operator $L(\epsilon, r)$,
$$
Lu = Ju'+\epsilon A u + r R(t) u, \quad J=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix},\: A =\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\: R= \begin{pmatrix} \cos t & \sin t \\ \sin t & -\cos t \end{pmatrix} 
$$
on $L^2(0,2\pi)$ have $E=0$ as a periodic eigenvalue (that is, with boundary conditions $u(0)=u(2\pi)$), for suitable $r$?
The transfer matrix $T(t,E)$ is defined as the $2\times 2$ matrix solution of $Lu=Eu$ with initial value $T(0,E)=1$. Since the coefficient matrices $JA$ and $JR$ have trace zero, it follows that $\det T=1$, so the eigenvalues of $T(2\pi,E)$ are determined by its trace $D(E):=\textrm{tr }T(2\pi, E)$. In particular, the periodic eigenvalues of $L$ are the solutions of $D(E)=2$.
Now $D(E=0,\epsilon,r=0)=2\cosh 2\pi\epsilon$, so existence of an $r$ as desired could be established by showing that $D(0,\epsilon,r)<2$ for some $r$. Moreover, since we now know from fedja's answer that this happens at $r\simeq \epsilon^{1/2}$, we can obtain this from a Taylor expansion of (the entire function) $D$ about $r=0$.
Denote the transfer matrix of $L(\epsilon,0)u=0$ by
$$
T_0(t) = \begin{pmatrix} e^{-\epsilon t} & 0 \\ 0 & e^{\epsilon t} \end{pmatrix} .
$$
I'll do variation of constants about $r=0$, so write $T(t,0)=T_0 M$. Then $M$ solves
$M'= r T_0^{-1}JRT_0 M$, and solving this by iteration delivers the Taylor coefficients one after the next. So $M(t,r)=\sum r^n M_n(t)$, with $M_0=1$,
$$
M_{n+1}(t) = \int_0^t \begin{pmatrix} -\sin s & e^{2\epsilon s}\cos s \\ e^{-2\epsilon s}\cos s & \sin s \end{pmatrix} M_n(s)\, ds .
$$
Everything can be worked out explicitly, though at some risk to one's sanity. We are ultimately interested in
$$
D(0,\epsilon,r)=\textrm{tr }T(2\pi,0) = e^{-2\pi\epsilon}a(2\pi, r) + e^{2\pi\epsilon}d(2\pi, r) , \quad \quad M\equiv \begin{pmatrix} a & b\\ c & d \end{pmatrix} .
$$
Expanding everything, this becomes
$$
D(0) = 2 + 4\pi^2\epsilon^2 + 2r^2\pi\epsilon (d_2-a_2) + \sum_{n=1}^4 r^n (a_n+d_n) + O(r^5) + O(\epsilon^3) \quad\quad\quad\quad (1)
$$
(and everything taken at $t=2\pi$, of course).
Brute force calculation will show that $d_2-a_2=0$ and $a_1+d_1=0$ (in fact, $a_1=d_1=0$) and $a_2+d_2=O(\epsilon^2)$. As for $a_3+d_3$, maybe the best way is to check that this vanishes when $\epsilon=0$, so we'll pick up at least one extra $\epsilon$ when we reintroduce this, so the third order term is $O(r^3\epsilon)$.
The big surprise now (for me) is that after everything vanished at $\epsilon=0$ or went the wrong way, out of the blue comes $a_4+d_4=-4\pi^2$ at $\epsilon=0$. So the bottom line is that when $r=\epsilon^{1/2}$, then this term cancels the $4\pi^2\epsilon^2$ exactly, and everything else in (1) is smaller.

Conclusion: There is a solution $r=r(\epsilon)$ of $D(0,\epsilon,r)=2$, and $\epsilon^{-1/2}r\to 1$ as $\epsilon\to0$.

Of course, with this perturbative approach, we cannot get the other half of fedja's result, namely that this is the only solution.