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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 $r\in\mathbb{R}_+$?

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  • $\begingroup$ Just remove all that $i$ and conjugation junk by first multiplying (or dividing?) $\eta$ by $e^{i\frac\pi 4}$ and then shifting the time by $\pi/2$ in the appropriate direction. You will get a purely real equation on the Fourier side that amounts to showing that a certain unbounded operator with compact inverse has at least one real eigenvalue and eigenvector. That is relatively easy, but is getting just one solution what you really want? Also, don't hope for small $r$; it tends to some constant as $\varepsilon\to 0$ at best. $\endgroup$ – fedja Sep 11 '17 at 11:55
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    $\begingroup$ The conjugation is harmless in the sesquilinear systems with real coefficients because if you have a complex solution, the real parts of all variables will solve the same system and the imaginary parts will solve a slightly modified system, so the complex equation splits completely into two real ones. BTW, I took one determinant with a wrong sign when computing, so I am no longer sure a solution exists (but still sure that the technique works one way or another). I'll post the details when I have some free time; too busy now... $\endgroup$ – fedja Sep 11 '17 at 20:48
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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.

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  • $\begingroup$ @ChristianRemling Of course. I'm not trying anything non-standard. That's why I thought my first comment would be enough to set the OP on the right track (except, as I said, my algebra was screwed up, so my conclusions in that comment were completely preposterous). $\endgroup$ – fedja Sep 12 '17 at 22:23
  • $\begingroup$ I see Christian removed his comment to which I was replying. I'll still keep mine :-) $\endgroup$ – fedja Sep 12 '17 at 22:24
  • $\begingroup$ I hope to absorb this info and keep it in my arsenal for future related problems. If everything goes as you say, so there is a unique such $r=r_\ast$, I'm very happy. If I understood correctly, since we arbitrarily scaled $\eta$ by $\zeta$ of modulus 1, the solution set of the ODE at $r_\ast$ is 1-dimensional. Quick question: How exactly is it clear that $r_\ast\approx\sqrt\varepsilon$ (and not $\varepsilon^{1/3}$ for example)? $\endgroup$ – Chris Gerig Sep 12 '17 at 23:15
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    $\begingroup$ @ChrisGerig Because we figured out that the large eigenvalues of $S$ are in the disks of radius $0.1\varepsilon^{-1/2}$ around something like $\pm \varepsilon^{-1/2}$ (you can improve this rough bound considerably, of course). $\eta$ is also essentially unique (the scaling was not arbitrary but an appropriate one to bring the equation to the purely real form) $\endgroup$ – fedja Sep 13 '17 at 0:21
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    $\begingroup$ @fedja: You are right, as always. I can do it too now. The miracle happens in fourth order, quite unexpectedly (to me); one should never give up hope. Sorry for the false alarm. $\endgroup$ – Christian Remling Sep 13 '17 at 1:39
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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.

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  • $\begingroup$ I should clarify in my post, but I have an argument to show that there must exist such an $r$. What I'm really asking for are explicit values, and the total number of them. $\endgroup$ – Chris Gerig Sep 12 '17 at 23:26
  • $\begingroup$ Thank you both, I will honestly learn a lot from both responses. My perturbative approach (using the Conley-Zehnder index, a $\mathbb{Z}$-valued homotopy invariant for such operators) does not even give me the magnitude of $r$. $\endgroup$ – Chris Gerig Sep 13 '17 at 9:19
  • $\begingroup$ @ChrisGerig: (1) $T_0$ does solve $LT_0=0$, $T_0(0)=1$, or you could say it's the transfer matrix of $Lu=0$. And yes, everything is at $E=0$; (2) My notation was not well chosen, $D$ means two different things; I'll edit; (3) The term with $d_2-a_2$ comes from combining the linear in $\epsilon$ terms (from $e^{\pm 2\pi\epsilon}$ with the quadratic in $r$ terms of $a,d$; (4) To see that this term vanishes, I need to compute $a_2,d_2$ from the recursion for the $M_n$; however, since $r^2\epsilon=O(\epsilon^2)$ already, it suffices to do this for $\epsilon=0$. $\endgroup$ – Christian Remling Sep 13 '17 at 17:10
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    $\begingroup$ (cont'd) (5) I also expand $a_4,d_4$ wrt $\epsilon$, and then I can actually stop after the constant term (that is, $\epsilon=0$) because the $r^4$ is already $O(\epsilon^2)$ and everything smaller than that goes into the remainder. (It's the same procedure that let me discard $a_3+d_3$.) $\endgroup$ – Christian Remling Sep 13 '17 at 17:17
  • $\begingroup$ The calculations are messy, and I'm aware that what I wrote must be a nightmare to read. If you really want to understand the details, perhaps the best way is to just start from the recursion for the $M_n$'s and do the calculations your own way. You already know you're going to take $r\simeq \epsilon^{1/2}$, you use Taylor's theorem wrt both $r$ and $\epsilon$, and everything $o(\epsilon^2)$ goes into the error term. From that point on, it's just brute force calculation, no ideas involved. $\endgroup$ – Christian Remling Sep 13 '17 at 17:20

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