Preamble

DIFFERENTIAL EQUATIONS 1967, Vol. III, No. 3
BRIEF COMMUNICATIONS

An Example of a Complex Equilibrium Point

A. D. MYSHKIS, L. E. REIZIN

In the article \cite{1}, the question was raised regarding the possibility of constructing an example of an isolated equilibrium point that belongs to the limit set of some other trajectory, but does not serve as either a positive or negative limit set for any trajectory other than the point itself. In the present article, such an example is constructed. Consider the system of differential equations:

$$\frac{d\zeta}{dt} = P(\zeta, \eta, \zeta),$$

$$\frac{d\eta}{dt} = Q(\zeta, \eta, \zeta),$$ (1)

$$\frac{d\zeta}{dt} = R(\zeta, \eta, \zeta),$$ where

$P(\zeta, \eta, \zeta) = \dots$, $Q(\zeta, \eta, \zeta) = \dots$. The functions are defined as $P(\zeta, \eta, \zeta) = \sum (2 \zeta_i \dots)$ and $Q(\zeta, \eta, \zeta) = \sum (S_i \dots)$, where the sums are taken over all integer values of $k$.

$f_k = f(\rho; 2^{-k} \zeta, 2^{-k} \eta)$,

£ (((6 cos - : ( - l ) 0, sin (*/36)), 3-2-W+i))— + 2 ) , - 0 ,

$$ \exp \left( 2 - 2 (b - a)^2 (t - a)^{-1} (b - t)^{-1} + 1 \right), \text{ if } a < t < b, $$

$$ f(t; a, b) = \dots, \text{ if } t \in (a, b), $$

$$ g(t; a, b) = \int_a^t f(\tau; a, b) \, d\tau \bigg/ \int_a^b f(\tau; a, b) \, d\tau, $$

ft a cp = £(a 2 /p 2 ; sin 2 (2^/9), 1/2),

P = (£ 2 + ^ 2 + C 2 ) 1 / 2 ,

As shown in [FIGURE:N], it is easy to see that at any given point, no more than two terms under the summation sign in expressions (2) are non-zero. Consequently, the functions $P(\xi, \eta, \zeta)$ and $Q(\xi, \eta, \zeta)$ possess continuous derivatives of all orders throughout the entire space. This ensures that the theorems regarding the existence and uniqueness of solutions for system (1) are applicable at every point in the space. Furthermore, it is straightforward to verify that this system possesses a unique equilibrium point at the origin $O(0, 0, 0)$.

In the region defined by $a^2 / p^2 > 1/2$, system (1) transforms into the following system:

$$d\xi/dt = -\eta \exp(-1/p),$$

$$d\eta/dt = \xi \exp(-1/p). \quad (3)$$

d Vdt = 0.

An Example of a Single Complex Equilibrium Point: The trajectories in this region are circles centered on the axis. Within the cone $a (2\pi/9)$ for $\rho = 0$, the system takes the following form:

$$d\xi/dt = \xi \exp(-1/\rho),$$

$$d\eta/dt = \eta \exp(-1/\rho),$$

$$d\zeta/dt = \zeta \exp(-1/\rho),$$

That is, its trajectories intersect the sphere along the normals. The graphs of the functions $f$ and $g$ are provided. For $\rho = (z + 2)$, the system is expressed as:
$$\begin{aligned} d\xi/dt &= ((-1)^k \eta - \xi(1-\zeta)) \exp(-1/\rho), \\ d\eta/dt &= (\xi \cos(\pi/18) - (-1)^k \eta \sin(\pi/18) + \xi(1-\zeta)) \exp(-1/\rho), \end{aligned}$$

$$d\zeta/dt = ((-1)^k \eta \sin(\pi/18) + \zeta \eta) \exp(-1/\rho),$$

From this, since $q = 1$ when $p \to 0$, we obtain $\frac{dq}{dt} = 2 \frac{d \ln t}{dt} \frac{dq}{dt} \exp(1/p)$, where $\frac{dp}{dt}$ is non-zero only in specific regions. If we perform a change of variables,

$l_i = 5 \cos(m\pi/18) - (-1)^k C \sin(m\pi/18)$,

$(-1)^k S \sin(\pi/18) + \cos(\pi/18)$, when $a = (\pi/36)$. The only region where the system trajectories can pass through the sphere is $1/2$. We shall show that between the spheres $\rho = 3 \cdot 2^{k+1}$ and $\rho = 2^k$, the trajectories of the system are contained within circular cones with a common axis. If $c < \sin(\alpha)$, the system transforms into the following system within this region:

$\frac{d \ln d}{dt} = ((-m_{ij} + b_i g_j) f_t + \delta_{ij} h_j) \exp(-1/p)$,

$\frac{d M_i}{dt} = ((\xi_{ij} + \eta_j g_j) f_t + q_{ij}) \exp(-1/p)$,

d k/dt = (C; ft/,- + C; A/) exp ( - 1 / p ) ,

$y = i$, if $3 \cdot 2^{-(i+2)} < 2^{-i}$, and $j = i + 1$, $(i+1)(i+2)$. Consequently, we obtain $a = \text{const}$. From points 2° and 3°, it follows that those trajectories passing through the sphere $r = 3 \cdot 2^{-i}$ reach the sphere and fall into the region defined on it.

a? + ,/p*