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Discrete symmetries of equations of dynamics with polynomial integrals of higher degrees V. V. Kozlov Steklov Mathematical Institute of Russian Academy of Sciences, Moscow
Russian version:
Izvestiya Rossiiskoi Akademii Nauk. Seriya Matematicheskaya, 2023, Volume 87, Issue 5, DOI: https://doi.org/10.4213/im9378 
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    § 1. IntroductionThe present paper is mainly concerned with a Hamiltonian system
$$
\begin{equation}
\dot{x}_k=\frac{\partial H}{\partial y_k},\quad \dot{y}_k=-\frac{\partial H}{\partial x_k},\qquad k=1,2,
\end{equation}
\tag{1.1}
$$
where
$$
\begin{equation}
H=\frac{1}{2}\sum_{i,j=1}^2 a_{ij} y_i y_j+V(x_1,x_2)
\end{equation}
\tag{1.2}
$$
is the Hamiltonian function. Here, $\|a_{ij}\|$ is a positive definite constant matrix, which defines the kinetic energy of the system, and $V$ is its potential energy (a smooth function $2\pi$-periodic in $x_1$ and $x_2$). The two-dimensional torus $\mathbb{T}^2=\{x_1,x_2\,\operatorname{mod} 2\pi\}$ is the configuration space. The kinetic energy (defining the intrinsic metric) generates the inner product Below, we will study conditions for the existence of “single-valued” first integrals $\Phi(x,y)$ of the Hamiltonian system (1.1), (1.2) which are functionally independent of the energy integral $H$ and are polynomial in the momenta $y_1$, $y_2$ with smooth coefficients which are $2\pi$-periodic in $x_1$, $x_2$. The following conjecture is well known: if the Hamiltonian equations (1.1) with Hamiltonian (1.2) admit an additional polynomial integral, then there is a momentum-polynomial integral of degree $\leqslant 2$. This conjecture was perhaps first stated in [1], [2]. In [1], the “standard” metric (where $a_{ij}=\delta_{ij}$) was considered, and in [2], systems with an arbitrary metric were studied. This conjecture can be reformulated in a constructive form. Namely, let be the Fourier series of the potential energy. The set will be called the spectrum of the potential energy. Since $V$ is a real-valued function, $\mathfrak{M}$ is carried into itself under the involution $m \mapsto -m$. Apparently an additional polynomial integral exists if and only if the spectrum $\mathfrak{M}$ lies on one or two lines passing through the origin and intersecting each other orthogonally (in the intrinsic metric $\langle\, \cdot , \cdot \rangle$). If the spectrum lies on a single line, then there is a momentum-linear first integral, and if it lies on two (orthogonally intersecting) lines, then there is an additional quadratic integral. In the first case, the integral is a Noether integral (due to the presence of a symmetry group), and in the second case the variables are simply separated.Despite the easy formulation, there is still no complete proof of this conjecture. The advances are made in two ways: an increase in the degree of the anticipated integral and consideration of more involved spectra. These two approaches have the following common point: if system (1.1) admits an additional polynomial integral of degree $m$, then the spectrum of the potential energy lies on at most $m$ different lines passing through the origin (see [2]). The above conjecture was proved for polynomial integrals of degree $m \leqslant 4$ and also for integrals of degree five in the particular case where $\|a_{ij}\|$ is the identity matrix. The cases $m=3$ and $m=4$ were considered in full generality in [3] and [4]. These studies were preceded by the paper of Byalyi [1], in which this conjecture was proved for $m=3$ under the assumption that $a_{ij}=\delta_{ij}$; he also showed that, for $m=4$, the spectrum $\mathfrak{M}$ lies on four lines that determine angles which are multiples of $\pi/4$. In [5], the case $m=5$ was considered under the same assumption. We also mention another result in [3]: if the Hamiltonian system under consideration has a polynomial integral of degree $m$ independent of $H$ and if the spectrum $\mathfrak{M}$ lies precisely on $m$ lines, then these lines are at the angles
$$
\begin{equation*}
\frac{\pi}{m},\ \frac{2\pi}{m},\ \dots,\ \frac{(m-1)\pi}{m}.
\end{equation*}
\notag
$$
In particular, let $\|a_{ij}\|$ be an identity matrix. Since, for $m \geqslant 3$, $\tan(\pi/m)$ is rational if and only if $m=4$, it follows (in view of one result in [4] on integrals of degree four) that there is no additional integral in this case. Now let us mention the available results on relations between the structure of the spectrum $\mathfrak{M}$ and the existence of an additional polynomial integral. First, the above conjecture is true for finite spectrum $\mathfrak{M}$ (see [2]). In addition, if the spectrum lies precisely on two lines, then these lines are orthogonal if there is an additional polynomial integral [2]. The conjecture is also valid if the infinite part of the spectrum $\mathfrak{M}$ lies on a single line (passing through the origin) [6]. The purpose of the present paper is to supplement these results on the structure of the spectrum of integrable Hamiltonian systems by some new propositions. The key point here is related to the presence of discrete symmetries of the spectrum of Hamiltonian systems (1.1), (1.2) which admit polynomial integrals. Our main emphasis will be on the cases of “higher” degrees with $m=5$ and $m=6$. In what follows, it is assumed that the intrinsic metric has the “standard” form ($\|a_{ij}\|$ is the identity matrix). The general case of a Euclidean metric requires special consideration. Under this assumption, the case $m=5$ was already considered in [5]. We will provide a different approach towards the proof of A. E. Mironov’s theorem. § 2. Two lemmas on three resonancesThe problem on additional polynomial integrals is directly related to the perturbation theory of Hamiltonian systems and to the Poincaré theory on the existence of “single-valued” first integrals of equations of dynamics. Indeed, we introduce the small parameter $\varepsilon>0 $ via the substitution
$$
\begin{equation}
t \mapsto \sqrt{\varepsilon}\,t,\qquad x \mapsto x,\qquad y \mapsto y\sqrt{\varepsilon}.
\end{equation}
\tag{2.1}
$$
After applying this change, the Hamiltonian equations (1.1) remain to be Hamiltonian, while the Hamiltonian function (1.2) changes to
$$
\begin{equation}
H_\varepsilon=\frac{1}{2}\sum y_k^2+\varepsilon V(x).
\end{equation}
\tag{2.2}
$$
So, the potential energy $V$ plays the role of the “perturbing” function. For $\varepsilon=0$, we have a completely integrable Hamiltonian system, the variables $y$, $x$ playing the role of the action-angle variables. According to Poincaré [7], the study of (generally non-integrable) Hamiltonian systems with Hamiltonian (2.2) is a part of the “main problem of dynamics”. Let equations (1.1) admit a momentum-polynomial integral of degree $m$, where $\Phi_k$ is a homogeneous form with respect to $y_1$, $y_2$ of degree $k$. It is easily checked that in (2.3) the sums of the terms of even and odd degrees are also first integrals of system (1.1). Using substitution (2.1) and multiplying by $(\sqrt{\varepsilon}\,)^m$, the integral $\Phi_m+\Phi_{m-2}+\Phi_{m-4}+\cdots$ becomesIn general, the Poincaré problem on integrals analytic with respect to the parameter $\varepsilon$ and the problem on momentum-polynomial first integrals are equivalent (see [2]). The relations
$$
\begin{equation*}
\biggl(\frac{\partial H_0}{\partial y},k\biggr)=(y,k)=0,
\end{equation*}
\notag
$$
where $k \in \mathbb{Z}^2 \setminus \{0\}$ and $k \in \mathfrak{M}$, are known as resonance relations. The corresponding line is a resonance line. This line is orthogonal to the line passing through the origin and through point $k$ of the spectrum $\mathfrak{M}$. This orthogonal line, which we call coresonance line, is dual to the corresponding resonance line. According to Poincaré (see [7], Chap. V), the function $\Phi_m$ does not depend on $x_1$, $x_2$, while the functions $\Phi_m$ and $H_0=(y_1^2+y_2^2)/2$ are dependent at all points of the resonance lines. These results underlie the proof of the non-integrability of perturbed Hamiltonian systems. In particular, if there are infinitely many different resonance lines, then the Hamiltonian equation does not admit any additional polynomial integrals independent of the quadratic integral of the energy. So, it what follows, we will assume that the number of resonance lines is finite Let us now assume that the Hamiltonian equations admit an additional integral of degree five. We leave aside the trivial case where the spectrum $\mathfrak{M}$ lies on a single line, and discarding the simple case of two coresonance lines (in this case, these lines are orthogonal and the equations admit an additional momentum-quadratic integral). In general, if there is an integral of degree five, then the spectrum of the potential energy lies on at most five lines passing through the origin. Let $l_0$, $l_1$ and $l_2$ be three of these lines. We will assume that the line $l_0$ is “vertical”; let $\varphi_k\pmod{\pi}$ be the angles counted from the “horizontal line” to the lines $l_k$ as they rotate counterclockwise. In particular, $\varphi_0=\pi/2$. If $z_k=\tan\varphi_k$ ($k=1,2$), then, according to [3]), one of the following conditions holds:
$$
\begin{equation}
{\rm a)}\ z_1=-z_2;\qquad {\rm b)}\ 2z_1z_2+1=z_1^2\ \text{ or }\ 2z_1z_2+1=z_2^2.
\end{equation}
\tag{2.4}
$$
In case a), the lines $l_1$ and $l_2$ are symmetric with respect to $l_0$. It turns out that in case b) equalities (2.4) have a similar geometric meaning. Lemma 1. If the Hamiltonian equations admit an integral of degree five independent of the energy integral, then, among any three resonance (coresonance) lines, there are two ones which are symmetric with respect to the third one. Proof. The first equality in case b) in (2.4) can be written as $\varphi_2 \pm \pi/2=2\varphi_1$ (this was noted in [3]), which is equivalent to The last equality means that $l_1$ is a bisector (one of the two) of the angle between $l_0$ and $l_2$, or that $l_0$ and $l_2$ are symmetric with respect to the line $l_1$. The meaning of the second equality is similar. This proves the lemma. Lemma 2. If the Hamiltonian equations admit an integral of degree six independent of the total energy, then any three resonance (coresonance) lines either contain two orthogonal ones or two lines which are symmetric with respect to the third one. Proof. According to [3], in the case of an integral of degree six, conditions (2.4) should be augmented with the two relations: $z_1z_2=0$ and $z_1z_2=-1$, of which the first one means that one of the lines $l_1$ or $l_2$ is “horizontal”, and the second condition means that the lines $l_1$ and $l_2$ are orthogonal, proving the claim.
So, if there are first integrals of higher degrees, any three resonance (coresonance) lines have the following property: when the momentum plane is reflected with respect to these lines, either one of the remaining lines is mapped onto itself, of the remaining resonance lines are mapped onto each other. This property of the resonance lines (and of the spectrum of the potential energy), which is by no means evident right from the start, is indicative of the hidden discrete symmetries in integrable Hamiltonian differential equations. This fact will be used for classification of the mutual arrangement of the remaining resonance (coresonance) lines in the cases where polynomial integrals degree five and six exist. Assume that we have lines passing through the origin and containing other points of the standard lattice $\mathbb{Z}^2$ (for example, two coresonance lines). The question is whether the bisector of the angle between these lines also contains points of the lattice $\mathbb{Z}^2$ different from the origin. In the general case, the answer is negative. It turns out that the possibility of existence of the bisector qua coresonance line is closely related to Pythagorean triples. Indeed, let points of the lattice with integer coordinates $(a_1,a_2)$ and $(b_1,b_2)$ lie on the lines $l_1$ and $l_2$. If $\varphi$ is the angle between these lines, then is a rational number, which we denote by $\alpha=\alpha_1/\alpha_2$ ($\alpha_j$ are integers). If the bisector of the angle between $l_1$ and $l_2$ also passes through points from $\mathbb{Z}^2\setminus \{0\}$, then $\tan\varphi/2$ should also be rational. By the well-known trigonometric formula, $z=\tan\varphi/2$ satisfies the quadratic equation Therefore, $\sqrt{1+\alpha^2}$ is a rational number. But then $\alpha_1^2+\alpha_2^2=\alpha_3^2$, where $\alpha_3$ is an integer number.So, $\alpha_1$, $\alpha_2$ and $\alpha_3$ form Pythagorean triples of integer numbers. It is known that where $p$ and $q$ are coprime, $p>q$, and $p$, $q$ are of different parity. As a result, we have Therefore, the slope between the coresonance lines with the bisector (equal to the coresonance line) cannot be an arbitrary rational number.§ 3. Integrals of degree fiveTheorem 1. Let a Hamiltonian system admit an additional integral of degree five independent of the energy integral, and let its spectrum lie on four coresonance lines. Then there are just two cases of their mutual arrangement: in the first one the angles between the lines are multiples of $\pi/4$ (see Figure 1), and in the second case they are multiples of $\pi/5$ (see Figure 2). Note that the arrangement of the coresonance lines in Figure 1 is the most difficult in the proof of the absence of irreducible integrals of degree four. Moreover, since the kinetic energy has the “standard” form (the coefficient matrix $\|a_{ij}\|$ is the identity matrix), the second case of mutual arrangement of the coresonance lines (as shown in Figure 2) is indeed impossible (since $\tan(\pi/5)$ is an irrational number). Proof of Theorem 1. Let $l_1$, $l_2$ and $l_3$ be three coresonance lines, and $l_2$ is the bisector of the angle between $l_1$ and $l_3$ (see Lemma 1). We need to find out the possible arrangements of the fourth line $l_4$ in order that each tuple of three lines of the four lines $l_1$, $l_2$, $l_3$, $l_4$ would satisfy the conclusion of Lemma 1. There are several cases to consider.
Assume first that $l_4$ lies in the angle between the lines $l_2$ and $l_3$ (see Figure 3). The case where $l_4$ lies between $l_1$ and $l_2$ is dealt with similarly. Consider the three lines $l_2$, $l_3$ and $l_4$. By Lemma 1, each of these lies is the symmetry axis for the two other lines. Only the line $l_4$ can be this axis. Indeed, let, for example, $l_2$ be the symmetry axis for the lines $l_3$ and $l_4$. In this case, the angle between $l_2$ and $l_4$ (in the counterclockwise direction) would be greater than that between $l_1$ and $l_2$, which is, of course, impossible. So, $l_4$ is the bisector of the angle between $l_2$ and $l_3$ (see Figure 3). Now let us consider the three lines $l_1$, $l_2$ and $l_4$. It is clear that $l_2$ cannot be the symmetry axis (see Figure 3). If $l_1$ is the symmetry axis, then $2\alpha=\pi-3\alpha$, and, therefore, $\alpha=\pi/5$. Thus, we arrive at the configuration of the four lines depicted in Figure 2. If $l_4$ were the symmetry axis, then we would get $\alpha=\pi-3\alpha$ and $\alpha=\pi/4$. But then the lines $l_1$ and $l_3$ would coincide, which is impossible. It remains to consider the case where $l_4$ is not contained in the above angle between the lines $l_2$ and $l_3$ (see Figure 4). It is easily verified that in this case $l_3$ is the bisector of the angle between $l_2$ and $l_4$. Consider now the three lines $l_1$, $l_2$ and $l_4$. It is clear that $l_2$ cannot be the symmetry axis of $l_1$ and $l_4$. If $l_1$ is the symmetry axis of $l_2$ and $l_4$, then $\alpha=\pi-3\alpha$, and, therefore, $\alpha=\pi/4$. This gives us the configuration of the four lines from Theorem 1 shown in Figure 1. If $l_4$ is the symmetry axis, then $\pi-3\alpha=2\alpha$, and hence $\alpha=\pi/5$. This case, which corresponds to the configuration in Figure 2, was already considered. Theorem 1 is proved. Theorem 2. Let a Hamiltonian system admit an additional polynomial integral of degree five, and let its spectrum lie on five coresonance lines. Then these lines are successively rotated by the angle $\pi/5$. This theorem is a particular case of a more general result on polynomial integrals of degree $m$ for Hamiltonian systems whose spectrum lies precisely on $m$ coresonance lines (see [3]). The proof of this result from [3] is purely analytic and is based on properties of binomial coefficients. Our aim here is to prove Theorem 2 using only Lemma 1. Proof of Theorem 2. Let the spectrum of the potential energy lie on five lines $l_1,\dots,l_5$, which are numbered counterclockwise starting from $l_1$. Let $\varphi_{ij}$ be the rotation angle (ranging from $0$ to $\pi$) from the line $l_i$ to the line $l_j$ measured counterclockwise. Consider the smallest of the angles $\varphi_{ij}$; we assume, for definiteness, that this is the angle $\varphi_{1k}$. We necessarily have $\varphi_{1k}=\varphi_{12}$, since otherwise $\varphi_{12}<\cdots<\varphi_{15}$. Given a fixed pair $l_1$ and $l_2$, let us consider all possible variants of arrangement for the remaining lines $l_3$, $l_4$ and $l_5$ so that the triple of lines $l_1$, $l_2$ and $l_i$ would satisfy the conclusion of Lemma 1. The following variants for the line $l_i$ are possible: the line $l'_1$ is symmetric to $l_1$ relative to $l_2$, the line $l'_2$ is symmetric to $l_2$ relative to $l_1$, or $l_i$ is the bisector $l$ of the angle $\varphi_{21}$ (clearly, $l_i$ cannot be the bisector of the angle $\varphi_{12}$). Hence $\{l_2,l_3,l_4\} \subset \{l'_1,l'_2,l\}$. Consequently, each of the lines $l'_1$, $l'_2$ and $l$ is some line $l_i$. Since $\varphi_{23} \geqslant \varphi_{12}$ and since the angle from $l_2$ to $l'_1$ is $\varphi_{12}$, we have $l'_1=l_3$ and $\varphi_{23}=\varphi_{12}$. Next, considering the angle $\varphi_{23}=\varphi_{12}$ as the smallest angle among $\varphi_{ij}$’s, we find, as above, that $\varphi_{34}=\varphi_{23}$. Proceeding in this way, we get, $\varphi_{12}=\varphi_{23}=\varphi_{34}=\varphi_{45}=\varphi_{51}$, the result required.
§ 4. Integrals of degree sixTheorem 3. Let a Hamiltonian system admit an additional polynomial integral of degree six, and let its spectrum lie on $n \leqslant 6$ different coresonance lines. Then 1) for $n=4$, there are four cases of their mutual arrangement shown in Figs. 2 and 5–7; 2) for $n=5$, there are two cases of their mutual arrangement: in the first case, the lines are successively rotated by the angle $\pi/5$, and the configuration in the second case is shown in Figure 8; 3) for $n=6$, the lines are successively rotated by the angle $\pi/6$. In Figure 5, the lines $l_2$ and $l_4$ are orthogonal, and $l_1$ and $l_3$ are symmetric with respect to the line $l_2$ (and, of course, with respect to $l_4$). The configuration in Figure 1 is, of course, a particular case of the configuration in Figure 5. Here, we show a whole family of configurations, which is naturally parameterized by the set of rational numbers: the slope of the symmetric lines $l_1$ and $l_3$ to the “horizontal” line $l_2$ can assume arbitrary rational values. In Figure 6, the lines $l_1$, $l_3$ and $l_2$, $l_4$ are orthogonal. The following result is immediate from Theorem 3 in view of the assumption $\alpha_{ij}=\delta_{ij}$. Corollary. If a Hamiltonian system admits an additional polynomial integral of degree six, then the number of resonance (and coresonance) lines is $\leqslant 4$. Assertions 1 and 2 in Theorem 3 are proved similarly to Theorem 1. One considers three lines satisfying Lemma 2: either one of these lines is the symmetry axis for the other two lines, or there are two orthogonal lines among these three lines. The problem is to add another line so that any three of these four lines would again satisfy the conclusion of Lemma 2. As in Theorem 1, there is a configuration depicted in Figure 2. The configuration in Figure 5 (which is a generalization of that shown in Figure 1) is obtained by adding the line $l_4$ orthogonal to the bisector $l_2$. In Figure 6, the lines $l_1$ and $ l_3$ are orthogonal, and a line orthogonal to $l_2$ is added. The configuration in Figure 7 is obtained by an addition of a coresonance line which is not orthogonal to the bisector. Figure 8 depicts the mutual position of the lines which is obtained from the most “symmetric” configuration of the six lines determine angles which are multiples of $\pi/6$. That there are no other configurations is proved by considering all possible cases for which the conclusion of Lemma 2 is satisfied. We skip these elementary (but rather cumbersome) steps. Let us prove assertion 3 following the general idea of the proof of Theorem 2. Let the spectrum $\mathfrak{M}$ lie on six lines $l_1,\dots,l_6$, and let, as above, $\varphi_{ij}$ be the counterclockwise angle between the line $l_i$ and the line $l_j$. We will assume without loss of generality that $\varphi_{12}$ is the smallest of the angles $\{\varphi_{ij}\}$. We fix one pair of lines $l_1$, $l_2$. Let us see which options are possible for the remaining lines $l_i$, $3 \leqslant i \leqslant 6$, so that the triple $l_1$, $l_2$ and $l_i$ would satisfy the conclusion of Lemma 2. There are five possibilities to consider: the line $l_1^\bot$ is orthogonal to $l_1$, the line $l_2^\bot$ is orthogonal to $l_2$, the line $l'_1$ is symmetric to $l_1$ relative to $l_2$, the line $l'_2$ is symmetric to $l_2$ relative $l_1$, or the line $l_1^\bot$ is the bisector $l$ of the angle $\varphi_{21}$ ($l$ cannot be the bisector of the angle $\varphi_{12}$). Hence $\{l_2,l_3,l_4,l_5\} \subset \{l_1^\bot,l_2^\bot,l'_1,l'_2,l\}$. Note that $l_1^\bot$ and $l$ cannot be both equal to $l_i$, because the angle from $l_1^\bot$ to $l$ is smaller than $\varphi_{12}$. The same argument shows that $l_2^\bot$ and $l$ cannot be both equal to $l_i$, since the angle from $l$ to $l_2^\bot$ is smaller than $\varphi_{12}$. So, the sets of lines $\{l_2,l_3,l_4,l_5\}$ and $\{l_1^\bot,l_2^\bot,l'_1,l'_2\}$ are equal. Since $\varphi_{23} \geqslant \varphi_{12}$ and since the angle from $l_2$ to $l'_1$ is $\varphi_{12}$, we have $l'_1=l_3$ and $\varphi_{23}=\varphi_{12}$. Next, considering $\varphi_{23}=\varphi_{12}$ as the smallest angle among the remaining angles $\varphi_{ij}$, we have, by a similar argument, $\varphi_{34}=\varphi_{23}$. Proceeding in this way, we find that $\varphi_{12}=\varphi_{23}=\varphi_{34}=\varphi_{45}=\varphi_{51}$, which proves assertion 3. § 5. Non-integrability conditionsAssume that the spectrum $\mathfrak{M}$ lies on a finite number of lines passing through the origin. This is a necessary condition for the existence of an additional momentum-polynomial integral. Following [2], we introduce the “second-order” spectral set $\mathfrak{M}'$ (in the second order of perturbation theory for Hamiltonian systems). We set
$$
\begin{equation}
v'_k=\sum_{\tau+\sigma=k}\frac{(\tau_1\sigma_1+\tau_2\sigma_2)v_\tau v_\sigma} {(\tau_2\sigma_1-\tau_1\sigma_2)^2}.
\end{equation}
\tag{5.1}
$$
Here $k$, $\sigma=(\sigma_1,\sigma_2)$, $\tau=(\tau_1,\tau_2)$ are vectors from $\mathbb{Z}^2$, and $k \ne 0$, $\sigma$, $\tau$ are not collinear. It is clear that the vectors $\tau$ and $\sigma$ should be taken from the set $\mathfrak{M}$, since otherwise they do not contribute to the sum (5.1). By the assumption, the number of coresonance lines is finite, and hence the sum (5.1) is also finite. We now set
$$
\begin{equation*}
\mathfrak{M}'=\{k \in\mathbb{Z}^2\colon v'_k \ne 0\}.
\end{equation*}
\notag
$$
Lemma 3. If $\mathfrak{M}' \subset \mathbb{Z}^2\setminus \{0\}$ lies on an infinite number of lines passing through the origin, then the Hamiltonian system does not admit momentum-polynomial first integrals independent of the total energy. This result was proved in [2] by means of the perturbation theory. However, the statement of the corresponding result from [2] differs insignificantly from that of Lemma 3; the reader can easily verify that these statements are equivalent. We also note that formula (5.1) is written under the assumption that $\|a_{ij}\|$ is the identity matrix. Note that if the set $\mathfrak{M}$ is finite (that is, $V$ is a trigonometric polynomial), then $\mathfrak{M}'$ is also finite, and hence Lemma 3 does not apply. The proof of the conjecture from § 1 in this case requires an introduction and analysis of spectral sets for all orders of the perturbation theory (see [2]). As an application of Lemma 3, we indicate one sufficient condition for non-integrability with three resonances. Theorem 4. Let the spectrum $\mathfrak{M}$ lie on three lines passing through the origin and one of them contain only a finite number of points of the spectrum. Then the Hamiltonian system does not admit additional polynomial integrals. Proof. If each of the coresonance lines contains a finite number of points from $\mathfrak{M}$, then the conclusion of the theorem follows from [2]. If the infinite part of the spectrum lies only on a single line, then the non-integrability of the Hamiltonian system follows from the general result in [6]. So, it remains to consider the case where the infinite part of the spectrum lies on two lines ($l_1$ and $l_2$, say), and the third line $l_3$ contains only a finite number of points from $\mathfrak{M}$.
Let us first show that if there exists a non-trivial polynomial integral, then the lines $l_1$ and $l_2$ are orthogonal. Indeed, let points $\sigma$ and $\tau$ of the lattice lie on $l_1$ and $l_2$, respectively. If $|\sigma|$ and $|\tau|$ are sufficiently large, then $k=\sigma+\tau \in \mathfrak{M}'$ (since in (5.1) there are no points of the lattice from $l_3$) if, of course, the lines $l_1$ and $l_2$ are not orthogonal (because in this case $v'_k \ne 0$). Therefore, the number of second-order coresonance lines of the perturbation theory is infinite, and hence, by Lemma 3, the Hamiltonian system is non-integrable. But this contradicts our assumption. Now let $m\in l_3 \cap \mathfrak{M}$ be a farthest point from the origin (there are two such points, since $-m \in \mathfrak{M}$; we take any of them). Consider the points $k=m+\sigma$, where $\sigma \in l_1 \cap \mathfrak{M}$. Since $l_1 \mathbin{\bot} l_2$ and since $m$ is a farthest point of the spectrum among the points lying on the line $l_3$, it follows that, for sufficiently large $|\sigma|$, the sum (5.1) involves only one non-zero term. Hence $m+\sigma \in \mathfrak{M}'$, and now the non-integrability is again secured by Lemma 3. Remark. More generally, if the infinite part of the spectrum $\mathfrak{M}$ lies just on two lines, then these lines are orthogonal. § 6. Integrability conditions with three resonancesIn this section, we will discuss non-integrability conditions of Hamiltonian systems in the presence of three coresonance lines satisfying Lemmas 1 and 2 from § 2. Theorem 5. Let there be three coresonance lines, of which two are orthogonal. Then the Hamiltonian system does not admit additional polynomial integrals. Proof. Let $l_1$, $l_2$, $l_3$ be the coresonance lines, $l_1 \mathbin{\bot} l_3$. Next, let $e_j$ be a primitive vector of the sublattice $l_j \cap \mathbb{Z}^2$; this vector defines a lattice point on $l_j$ which is nearest to the origin (clearly, there are two such points; $e_j$ corresponds to one of these points).
We fix one of the points $\sigma=me_1$ ($m \in \mathbb{N}$) of the spectrum $\mathfrak{M}$ (among the lattice points on $l_1$). Consider the points of the lattice $\mathbb{Z}^2$ defined by the vectors We will assume that the line $l_2$ contains infinitely many points of the spectrum $\mathfrak{M}$, for otherwise the non-integrability of the Hamiltonian system is secured by Theorem 4. We need to show that there are infinitely many non-zero numbers $v'_k$ (which are defined by the sum (5.1)). Once this is proved, we can employ Lemma 3 to show that the Hamiltonian system is non-integrable. We note at once that the sum (5.1) contains at most two non-zero terms (because $l_1 \bot \,l_3$).Under a certain condition, a lattice point (6.1) may also be the sum of lattice points from the lines $l_2$ and $l_3$. This condition can be reduced to the equality which should hold for some integer $l$ and $n_1$. This condition is equivalent to the equality in which the coefficients on the right are fixed rational numbers. Therefore, the integer $l$ and the difference $n-n_1$ are defined uniquely.There are two cases to consider. In the first case, equality (6.2) cannot hold for some integer $n$ (there are infinitely many such $n$’s in this case). It was assumed that $me_1 \in \mathfrak{M}$. If $ne_2$ also lies in the spectrum $\mathfrak{M}$, then $k \in \mathfrak{M}'$ (by (5.1)). Therefore, in this case the line passing through the origin and point (6.1) is a second-order coresonance line. In the second case it assumed that (6.2) holds with fixed $l$ and some $n_1$, which differs from $n$ by a constant quantity $a$ (as mentioned above). We again assume that $me_1$ and $ne_2$ lie in the spectrum $\mathfrak{M}$. If $le_3 \notin \mathfrak{M}$, then we in fact have the case already considered: the sum (5.1) has precisely one non-zero term, and hence $v'_k\neq0$. If $le_3 \in \mathfrak{M}$, but $n_1e_2 \notin \mathfrak{M}$, then the situation is similar. A new possibility appears if $n_1e_2$ also lies in the spectrum $\mathfrak{M}$. Under this assumption, the equality $v'_k=0$ assumes the form where the non-zero complex constant $\lambda$ does not depend on $n$. Here, we have omitted the elementary intervening steps in the derivation of (6.3); the explicit form of $\lambda$ will not be required in what follows.We next proceed with the induction step, replacing $n$ by $n+1$ in (6.1). By the same argument, we either directly get another coresonance second-order line, or arrive an additional relation (6.3) with $n+1$ for $n$. As a result, we either have infinitely many different second-order coresonance lines (and, in this case, the non-integrability of the Hamiltonian system is secured by Lemma 3) or we have an infinite chain of equalities (6.3), which hold for $|n| \geqslant n_0$. If $|\lambda| \geqslant 1$, then, clearly, the ratio does not tend to zero as $n\to+\infty$. But this is impossible by the smoothness assumption on the potential energy. If $|\lambda| < 1$, then $|\lambda^{-1}| > 1$, and hence ratio (6.4) cannot be bounded as $n\to-\infty$, but this is not the case. This completes the proof of the theorem.The case where one of the coresonance lines is the symmetry axis of two other lines is more involved. In this case, there is still no complete proof of the absence of additional polynomial integrals. We will consider a particular case of this problem, which illustrates the ensuing difficulties. Assume that the primitive vectors of the lattice (along which the coresonance lines $l_1$ and $l_3$ are directed) have the following components: ($p$ and $q$ are coprime integers). The point with coordinates $(p,0)$, which is on the symmetry axis $l_2$, lies in the lattice. In particular, the lattice contains all the points $(rp,0)$, $r \in \mathbb{Z}$.Theorem 6. Let, under the above assumptions, among the points of the lattice $(rp,0)$, where $r$ is odd, there be at least one point of the spectrum $\mathfrak{M}$ of the potential energy. Then the Hamiltonian system does not admit additional polynomial first integrals. For a proof, consider the points on the lattice here, $m$ is fixed, and $(rp,0) \in \mathfrak{M}$. On the right, we have the sum of two vectors, which define the points on the lines $l_1$ and $l_2$.Similarly, we have On the right, we have the sum of vectors defining the points of the lattice on the lines $l_3$ and $l_2$.We claim that by summation of vectors parallel to the lines $l_1$ and $l_3$ one cannot obtain points (6.5). Indeed, assume that with integer $k$, $s$. Since $p,q \ne 0$, we have Hence $r=2s$, which contradicts the assumption of the theorem that $r$ is odd.Now we can employ Lemma 3. Assume that the Hamiltonian system has an additional momentum-polynomial first integral. The potential energy $V$ is the sum of three periodic functions $f$, $g$ and $h$, where
$$
\begin{equation*}
f=\sum f_ne^{in(px_1+qx_2)},\quad g=\sum g_ne^{in(px_1-qx_2)},\quad h=\sum h_ne^{inx_1}.
\end{equation*}
\notag
$$
The vanishing of sum (5.1) implies the following relation on the Fourier coefficients: (in the derivation of this formula we also used (6.5) and (6.6)). By Lemma 3 and the assumption that there exists an additional integral, it follows that equality (6.7) holds for all $|n| \geqslant n_0$. Let us use the fact that $l_2$ is the bisector of the angle between the lines $l_1$ and $l_3$. Considering the points of the lattice we get the formula which is similar to (6.7). Multiplying (6.7) and (6.8), we get
$$
\begin{equation}
|f_ng_n|\,\biggl|\frac{h_{rp}}{r}\biggr|=|f_{-n}g_{-n}|\, \biggl|\frac{h_{(2n+r)p}}{2n+r}\biggr|.
\end{equation}
\tag{6.9}
$$
Since the potential energy is a real-valued function, we have $f_{-n}=\overline{f}_n$, $g_{-n}=\overline{g}_n$, and, therefore, Let $n \geqslant n_0$ and $f_n \ne 0$. There are infinitely many of these $n$’s, for otherwise the function $f$ would be a trigonometric polynomial, and in this case the conclusion of the theorem follows from Theorem 4. If, for these $n$’s (except, possibly, finitely many), we have $g_n=0$, then equality (6.7) does not hold (because $h_{rp} \ne 0$ by the assumption of the theorem). Hence the Hamiltonian system is not integrable by Lemma 3. So, it remains to consider the case where $f_ng_n \ne 0$ for infinitely many integer $n$’s. But, in this case, from (6.9) (in view of (6.10)) we have, for infinitely many indexes,
$$
\begin{equation*}
\biggl|\frac{h_{(2n+r)p}}{2n+r}\biggr|=\biggl|\frac{f_{rp}}{r}\biggr|= {\rm const} \ne 0.
\end{equation*}
\notag
$$
In particular, the ratio $h_{(2n+r)p/(2n+r)}$ does not tend to zero as $|n| \to \infty$. However, this contradicts the smoothness of the periodic function $h$. This proves the theorem. § 7. Notes and problems$1^\circ$. An additional integral of degree five may exist only in two cases: a) there are three coresonance lines, and one of these lines is the symmetry axis of the other two; b) there are four coresonance lines (arranged as in Figure 1). Case b) seems simpler: in [8], it was shown that there are no non-trivial polynomial integrals under the additional assumption that all the Fourier coefficients of the potential energy are real. The proof also depends on Lemma 3. It is not improbable that the additional assumptions that the coefficients are real is superfluous and the result can be obtained by the same ideas. It would be desirable to give a proof of Theorem 6 in the general case without additional assumptions. By clarifying these aspects, we will get a different proof of Mironov’s theorem [5] based on new ideas. $2^\circ$. Regarding conditions for the existence of integrals of degree six, we have two more possibilities in addition to case a) from $1^\circ$, where the four coresonance lines are arranged as in Figs. 5 and 6. $3^\circ$. It seems that the problem on the existence of non-trivial polynomial integrals of degree $\geqslant 7$ is also related to the analysis of cases where there exist only three or four coresonance lines. $4^\circ$. The situation becomes much more involved if the matrix of the kinetic energy $\|a_{ij}\|$ is not an identity matrix anymore. The problem of the existence of integrals of third and fourth degrees in the general case, which was considered in [3], [4], gives an idea of the difficulties in this way. $5^\circ$. It is also worth considering the problem on polynomial integrals of the Hamiltonian system (1.1) with Hamiltonian (1.2) in which the kinetic energy is a non-degenerate quadratic form with signature $+-$ (the pseudo-Euclidean metric). The first results in this directions were obtained in [2]. As an example, consider a system in a non-potential magnetic field
$$
\begin{equation}
\ddot{x}_1=\frac{\partial W}{\partial x_2},\qquad \ddot{x}_2=\frac{\partial W}{\partial x_1},
\end{equation}
\tag{7.1}
$$
where $W$ is a smooth function on the two-dimensional torus $\mathbb{T}^2=\{x_1,x_2\mod 2\pi\}$. This system can be found in Appell’s book [9]. Equations (7.1) have a quadratic integral
$$
\begin{equation}
H=y_1y_2-W(x_1,x_2),\quad\text{where}\quad y_1=\dot{x}_2,\quad y_2=\dot{x}_1.
\end{equation}
\tag{7.2}
$$
It is easily checked that equations (7.1) are equivalent to the Hamiltonian system
$$
\begin{equation}
\dot{x}_k=\frac{\partial H}{\partial y_k},\quad \dot{y}_k=-\frac{\partial H}{\partial x_k},\qquad k=1,2.
\end{equation}
\tag{7.3}
$$
In this example, the “kinetic energy” is defined just via the pseudo-Euclidean metric. It is also worth noting that the Hamiltonian system (7.3) with Hamiltonian (7.2) was studied by Drach [10] from the point of view of searching the first integrals of degree three in momenta. A corrected version of the list of integrable cases for the Drach system can be found in [11]. In this list, there are no systems with periodic potential energy $W$. The author is grateful to P. A. Kozhevnikov for a friendly discussion of the combinatorial problems considered in this manuscript. 
Citation in format AMSBIB
\Bibitem{Koz23} |
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