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](https://math.stackexchange.com/questions/2391598/solution-to-at-least-one-ode-in-a-family-of-odes) 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 $r\in\mathbb{R}\_+$?



Solution to at least one ODE in a family of ODE's
As I said, let's get rid of some junk like $i$ and conjugation first. Note that in the original equation we can replace $\eta$ by $\zeta\eta$ for any complex number $\zeta$ with $|\zeta|=1$ and can replace $t$ by $t+\tau$. These two degrees of freedom allow us to rotate the coefficients at $\varepsilon$ and $r$ independently, so I will rotate the equation to 
$$
i\eta\_t=(re^{it}+\varepsilon)\bar\eta\,.
$$
Now we go to the Fourier side, as Chris suggested, and get the system
$$
-ka\_k=r\bar a\_{1-k}+\varepsilon \bar a\_{-k}
$$
with *real* coefficients. The key point is that if this system has a complex solution $a\_k\in\mathbb C$, then $\Re a\_k$ solve exactly the same system and $\Im a\_k$ solve the similar system with $-r$ and $-\varepsilon$ instead of $r,\varepsilon$. Conversely, if we have a real solution of one of those syestems, we can turn it into a real or purely imaginary solution of the original system respectively. Thus, we can forget about the conjugation altogether and just consider the linear system  

$$
-ka\_k=ra\_{1-k}+\varepsilon a\_{-k}
$$
with small $\epsilon$ of arbitrary sign and $r\in\mathbb R$.
Rewrite it as the eigenvalue problem
$Ta=ra$
where 
$$
(Ta)\_m=(m-1)a\_{1-m}-\varepsilon a\_{m-1}
$$
The operator $T$ is easy to invert: If $b=Ta$ we get 
$$
b\_{m+1}=ma\_{-m}-\varepsilon a\_m,\qquad b\_{1-m}=-ma\_m-\varepsilon a\_{-m}
$$
whence 
$$
a\_m=-\frac{m}{m^2+\varepsilon^2}b\_{1-m}
-\frac{\varepsilon}{m^2+\varepsilon^2}b\_{m+1}
$$
So $a=Sb$ where $S=T^{-1}$ is a *compact* operator. Now we know that eigenvectors of $S$ and $T$ are the same and all we need is to figure out if $S$ has any non-zero real eigenvalues for small $\varepsilon$. 


The next part is to look at the possible size of a real eigenvalue of $T$.
Write $T-rI=P\_r-\varepsilon Q$ where $(Qa)\_m=a\_{m-1}$. Note that $P\_r$ is a block operator with blocks $P\_{r,m}$ of size $2$ corresponding to the variables $A\_m=(a\_m,a\_{1-m})$ ($m=0,2,3,4,\dots$). The matrix of the block $P\_{r,m}$ is $\begin{bmatrix} -r& m-1\\-m&-r \end{bmatrix}$. It is easily seen that this block is not contracting too much if $r$ is not too small. An accurate computation of the inverse $2\times 2$ matrix shows that if $r\ge C\sqrt{|\varepsilon|}$ with some sufficiently large $C>0$, then $\|P\_{r,m}A\_m\|\ge 4\varepsilon\|A\_m\|$ ($P\_{r,0}$ is the only block that can contract that much; every other block has an inverse bounded by some constant independent of $r$). On the other hand, $\|Q\|=1$ and $Q$ acts only either within the block (again, that is possible only if $m=0$), or between adjacent blocks. Thus, if $r>\sqrt{|\varepsilon|}$ and we have any non-trivial $\ell^2$ eigenvector, we can choose $m$ such that $\|A\_m\|$ is the largest, in which case the norm of the $m$-th block in the image under $T-rI$ is at least $4\varepsilon\|A\_m\|-3\varepsilon\|A\_m\|>0$, which is a contradiction. Thus, our only chance is to have a real eigenvalue of $T$ less than $C\sqrt{|\varepsilon|}$ in absolute value. So we only need to look for large real eigenvalues $\rho$ of $S$.


Now $S-\rho I$ also has an $\varepsilon$-perturbation of a block operator, call it $S\_{\rho}$, with the same $2\times 2$ blocks. A typical ($m\ge 2$) block is now $\begin{bmatrix}-\rho &-\frac{m}{m^2+\varepsilon^2}\\ \frac{m-1}{(m-1)^2+\varepsilon^2} & -\rho \end{bmatrix}$, which has a bounded (and even small) inverse if $\rho$ is large and, say, the imaginary part of $\rho$ does not exceed the real part of $\rho$ in the absolute value (to guarantee that $\Re(\rho^2)\ge 0$ so that two terms in the determinant of the block essentially add up in absolute value). However, there is one exceptional block corresponding to $m=0$ with the matrix $\begin{bmatrix}-\rho &-\varepsilon^{-1} \\ -\frac{1}{1+\varepsilon^2} & -\rho \end{bmatrix}$.


The rest of the operator $S-\rho I$ is $-\varepsilon V\_{\varepsilon}$ where $V\_\varepsilon$ is a compact operator of norm at most $1$ (so we write $S-\rho I=S\_\rho-\varepsilon V\_{\varepsilon}$). Thus, we can apply the perturbation theory considering the continuous family $S\_\rho-\delta V\_{\varepsilon}$ with $0\le \delta\le \varepsilon$. Note that $S\_\rho$ is invertible with bounded inverse as long as $\rho$ is large, $|\Im\rho|\le|\Re\rho|$, *and* $|\rho^2-\frac{\varepsilon^{-1}}{1+\varepsilon^2}|$ is approximately the sum of the absolute values of the individual terms under the absolute value sign, which is always the case if $\varepsilon<0$ and is so under the condition $|\rho\pm \varepsilon^{-1/2}|\ge 0.1\varepsilon^{-1/2}$, say, if $\varepsilon>0$. Thus $S\_{\rho}-\delta V\_{\varepsilon}$ is invertible for such $\rho$ for all $\delta\in[0,\varepsilon]$. This means that the only chance to get some real eigenvalues is to assume that $\varepsilon>0$ and they can be only in the disks $|\rho\pm \varepsilon^{-1/2}|< 0.1\varepsilon^{-1/2}$. Moreover, since nothing can travel through the boundaries of those disks, we have the same number of (complex) eigenvalues in each of those disks for the operator $S$ we are interested in and the block operator $S\_{0}$. The latter number is $1$. Since the non-real eigenvalues of $S$ come in complex conjugate pairs, we conclude that the eigenvalues of $S$ in those disks are also real. The rest of the story about the existence and uniqueness of the solution must be clear. It is also clear that $r\approx \sqrt\varepsilon$. If you want more information, you should apply the perturbation theory more carefully.  

Unfortunately, explicit formulae seem to be out of question, so I cannot really answer the question "What is $\eta$?" in a meaningful way (though we can say that its Fourier coefficients decay fast enough, so it is real analytic, etc.).


Please accept my apologies for a) not posting earlier and b) changing my opinion about what the answer should be several times (my mental Algebra skills are totally dismal and that is one of the reasons why I prefer Analysis, where the correctness of a properly laid out proof is independent of the choices of signs and constant coefficients in the formulae :-) ).


The end. 


