Full Text
UDC 517.917
ON RECURRENT MOTIONS OF PERIODIC SYSTEMS OF TWO DIFFERENTIAL EQUATIONS
V. A. PLISS
In [1] M. L. Cartwright formulated the following problem. Consider a system of differential equations of the form
\[ \frac{dx}{dt}=X(x,y,t),\qquad \frac{dy}{dt}=Y(x,y,t), \tag{1} \]
where the functions \(X\) and \(Y\) are defined for all \(x,y,t\), are sufficiently smooth, and have period \(\omega\) in \(t\).
Let \(x(t,x_0,y_0)\), \(y(t,x_0,y_0)\) be the solution of system (1) with initial data \(t=0,\ x=x_0,\ y=y_0\). Suppose that all such solutions of system (1) are defined for \(0\le t\le \omega\).
Let \(T\) be the transformation of the \(xy\)-plane into itself that associates with the point \((x_0,y_0)\) the point \((x(\omega,x_0,y_0), y(\omega,x_0,y_0))\). Suppose that every bounded set invariant with respect to the transformation \(T\) has measure zero.
The question is posed whether system (1), under the formulated conditions, can have recurrent motions distinct from periodic ones.
In the present paper an example is given of a dissipative system of the form (1) with the following properties: its right-hand sides are infinitely differentiable in all their arguments, every bounded set invariant with respect to \(T\) has measure zero, and system (1) has recurrent motions distinct from periodic ones.
- Introduce the functions \(\alpha(z)\), \(f(z)\), and \(g(z)\). The function \(\alpha(z)\) is defined and has derivatives of all orders for \(-\infty<z<+\infty\), \(\alpha(-z)=-\alpha(z)\), \(\alpha(z)=0.5\) for \(z\ge 0.5\), \(\alpha'(z)>0\) for \(0\le z<0.5\),
\[ f(z)=0.5+\alpha(z-0.5), \tag{1.1} \]
\[ g(z)=f(-z)=0.5+\alpha(-z-0.5). \tag{1.2} \]
Define \(X_1(x,y)\) by the following formulas:
\[ \begin{aligned} X_1(x,y)&=-10xg(x+3) && \text{for } x\le 3,\\ X_1(x,y)&=-10xf(x-3) && \text{for } x\ge 3. \end{aligned} \tag{1.3} \]
The function \(X_1(x,y)\) is infinitely differentiable for all \(x\) and \(y\); \(\dfrac{\partial X_1}{\partial x}\le 0\); for \(|x|\le 3\), \(X_1(x,y)\equiv 0\), and for \(|x|\ge 4\), \(X_1(x,y)=-10x\).
Put
\[ Y_1(x,y)=-2\left(y-10a\left(\frac{x}{2}\right)\right). \tag{1.4} \]
The function \(Y_1(x,y)\) is infinitely differentiable for all \(x,y\);
\[ \frac{\partial Y_1}{\partial y}=-2 \]
for all \(x\) and \(y\); and for \(x\leq -1\), \(Y_1=-2(y+5)\), while for \(x\geq 1\), \(Y_1=-2(y-5)\),
\[ \frac{\partial X_1}{\partial x}+\frac{\partial Y_1}{\partial y}\leq -2 \tag{1.5} \]
for all \(x,y\).
Define the function \(P(x,y)\) by the formulas:
\[ P(x,y)=-10x+f(y+7)(11x+2)\quad \text{for }-\infty<y\leq -4, \]
\[ P(x,y)=x+2-f(y+4)(11x+2)\quad \text{for }-4\leq y\leq 3, \]
\[ P(x,y)=-10x+f(y-3)(11x-2)\quad \text{for }3\leq y\leq 6, \]
\[ P(x,y)=x-2-f(y-6)(11x-2)\quad \text{for }6\leq y<+\infty. \]
The function \(P(x,y)\) is infinitely differentiable and \(P(x,y)=x+2\) for
\(-6\leq y\leq -4\), \(P(x,y)=x-2\) for \(4\leq y\leq 6\).
Put
\[ X_2(x,y)=P(x,y)\quad \text{for }-3\leq x\leq 3, \]
\[ X_2(x,y)=-10xg(x+3)+P(x,y)f(x+4)\quad \text{for }x\leq -3, \]
\[ X_2(x,y)=-10xf(x-3)+P(x,y)g(x-4)\quad \text{for }x\geq 3. \]
The function \(X_2(x,y)\) is infinitely differentiable and \(X_2(x,y)=-10x\) for \(|x|\geq 4\).
Define \(Y_2(y)\) as a function infinitely differentiable for all \(y\), possessing the following properties:
\(Y_2(-y)=-Y_2(y)\); \(Y_2(y)=-2(y-5)\) for \(y\geq 3\);
\(-2\leq Y_2'(y)\leq 0\) for \(2\leq y\leq 3\);
\(Y_2'(2)=0\);
\(0\leq Y_2'(y)\leq 5\) for \(0\leq y\leq 2\).
We shall show that for all \(x\) and \(y\) the inequality
\[ \frac{\partial P}{\partial x}+\frac{\partial Y_2}{\partial y}<0 \tag{1.6} \]
holds.
For \(y\leq -3\), by the definition of \(Y_2\) we have:
\[ \frac{\partial Y_2}{\partial y}=-2. \tag{1.7} \]
For \(y\leq -4\), by the definition of \(P\) we have:
\[ \frac{\partial P}{\partial x}=-10+f(y+7)\cdot 11\leq 1. \tag{1.8} \]
For \(-4\leq y\leq -3\) we have
\[ \frac{\partial P}{\partial x}=1-f(y+4)\cdot 11\leq 1. \tag{1.9} \]
From (1.7), (1.8), and (1.9) it follows that for \(y \leqslant -3\) inequality (1.6) is satisfied.
For \(-3 \leqslant y \leqslant 3\), \(P(x,y)=-10x\); consequently,
\[ \frac{\partial P}{\partial x}=-10 . \tag{1.10} \]
For \(-3 \leqslant y \leqslant 3\),
\[ \frac{\partial Y_2}{\partial y}\leqslant 5 . \tag{1.11} \]
Hence, and from (1.10), it follows that for \(|y|\leqslant 3\) inequality (1.6) is valid.
It is easy to verify that for \(y\geqslant 3\) the inequality
\[ \frac{\partial P}{\partial x}\leqslant 1 \tag{1.12} \]
is satisfied.
On the other hand, for \(y\geqslant 3\), \(Y_2=-2(y-5)\); consequently,
\[ \frac{\partial Y_2}{\partial y}=-2 . \tag{1.13} \]
From inequalities (1.12) and (1.13) it follows that for \(y\geqslant 3\) inequality (1.6) is satisfied. Thus, inequality (1.6) is valid for \(-\infty<y<+\infty\).
Let us compute the derivative \(\dfrac{\partial X_2}{\partial x}\) for \(x\leqslant -3\):
\[ \begin{aligned} \frac{\partial X_2}{\partial x} &=-10g(x+3)-10xg'(x+3)+\frac{\partial P}{\partial x}\,f(x+4)+{}\\ &\quad +P(x,y)f'(x+4) \\ &=-10[g(x+3)+f(x+4)]+{}\\ &\quad +\left[10+\frac{\partial P}{\partial x}\right]f(x+4) -10[g'(x+3)+f'(x+4)]+{}\\ &\quad +[10x+P(x,y)]f'(x+4). \end{aligned} \]
Since, by the definition of the functions \(f\) and \(g\), we have
\[ g(x+3)+f(x+4)=1 \quad\text{and}\quad g'(x+3)+f'(x+4)=0, \]
from the last equality we obtain the following:
\[ \frac{\partial X_2}{\partial x} =-10+\left(10+\frac{\partial P}{\partial x}\right)f(x+4) +[10x+P(x,y)]f'(x+4). \tag{1.14} \]
By the definition of \(P(x,y)\) we have: \(\dfrac{\partial P}{\partial x}\geqslant -10\) and \(P(x,y)\leqslant -10x\) for \(x\leqslant -3\); therefore, from equality (1.14) it follows that for \(x\leqslant -3\) the inequality
\[ \frac{\partial X_2}{\partial x}\leqslant \frac{\partial P}{\partial x}. \tag{1.15} \]
Similarly it is proved that inequality (1.15) is also satisfied for \(x \geqslant 3\). Since for \(|x|\leqslant 3\) \(X_2(x,y)=P(x,y)\), it is clear that for all \(x,y\) the inequality
\[ \frac{\partial X_2}{\partial x}+\frac{\partial Y_2}{\partial y}<0. \tag{1.16} \]
Put \(t_1=\dfrac{1}{2}\ln 11,\quad t_2=\ln 5,\quad t_1+t_2+4=\omega\).
Define the function \(\tau(t)\) by the formula
\[ \tau(t)=f(t)\,g(t-t_1-2)+f(t-t_1-2)\,g(t-t_1-t_2-4) \]
for \(0\leqslant t\leqslant \omega\). For \(t\leqslant 0\) and \(t\geqslant \omega\) we define the function \(\tau(t)\) as an \(\omega\)-periodic function of \(t\). It is clear that \(\tau(t)\) has derivatives of arbitrary order for all \(t\).
Define the functions \(X(x,y,t)\) and \(Y(x,y,t)\) by the formulas:
\[ X(x,y,t)= \begin{cases} X_1(x,y)\tau(t) & \text{for } k\omega\leqslant t\leqslant k\omega+t_1+2,\\ X_2(x,y)\tau(t) & \text{for } k\omega+t_1+2\leqslant t\leqslant (k+1)\omega, \end{cases} \]
\[ Y(x,y,t)= \begin{cases} Y_1(x,y)\tau(t) & \text{for } k\omega\leqslant t\leqslant k\omega+t_1+2,\\ Y_2(x,y)\tau(t) & \text{for } k\omega+t_1+2\leqslant t\leqslant (k+1)\omega, \end{cases} \]
where \(k=\ldots,-2,-1,0,1,2,\ldots\).
It follows from the preceding that the functions \(X\) and \(Y\) are defined, continuous, and have continuous partial derivatives with respect to all their arguments for all \(x,y,t\). These functions have period \(\omega\) in \(t\) and, as follows from (1.5) and (1.16), satisfy the inequality
\[ \frac{\partial X}{\partial x}+\frac{\partial Y}{\partial y}<0 \tag{1.17} \]
for \(t\ne k\omega,\quad t\ne k\omega+t_1+2\quad (k=\ldots,-2,-1,0,1,2,\ldots)\).
2. Consider the system of differential equations
\[ \frac{dx}{dt}=X(x,y,t),\qquad \frac{dy}{dt}=Y(x,y,t). \tag{2.1} \]
For sufficiently large \(|x|\) and \(|y|\), the functions \(X\) and \(Y\) behave as linear functions of \(x\) and \(y\), respectively, with negative coefficients multiplying the argument; hence it follows that system (2.1) is dissipative.
Denote by \(x(t,x_0,y_0,t_0)\), \(y(t,x_0,y_0,t_0)\) the solution of system (2.1) with initial data \(t=t_0,\ x=x_0,\ y=y_0\).
Let \(T\) be the transformation that assigns to the point \((x_0,y_0)\) the point \((x(\omega,x_0,y_0,0),\,y(\omega,x_0,y_0,0))\). From relation (1.17) it follows that every bounded set invariant with respect to \(T\) has measure zero. (See, in this connection, [1] and [2].)
We shall prove that system (2.1) has recurrent motions different from periodic ones.
Introduce the following domains of the \(xy\)-plane:
\[ A=\{|x|\leqslant 3,\ -6\leqslant y\leqslant -4\},\qquad B=\{1\leqslant x\leqslant 3,\ |y|\leqslant 6\}, \]
\[ C=\{|x|\leqslant 3,\ 4\leqslant y\leqslant 6\},\qquad D=\{-3\leqslant x\leqslant -1,\ |y|\leqslant 6\}. \]
\[ a=\{-3\le x\le -1,\ -6\le y\le -4\}, \]
\[ b=\{1\le x\le 3,\ -6\le y\le -4\}, \]
\[ c=\{1\le x\le 3,\ 4\le y\le 6\}, \]
\[ d=\{-3\le x\le -1,\ 4\le y\le 6\}. \]
It is clear that \(a=A\cap D,\quad b=A\cap B,\quad c=B\cap C,\quad d=C\cap D\).
Introduce the following transformations of the \(xy\)-plane into itself: the transformation \(\Theta\) assigns to the point \((x_0,y_0)\) the point
\[
\bigl(x(t_1+2,x_0,y_0,0),\ y(t_1+2,x_0,y_0,0)\bigr),
\]
and the transformation \(\Delta\) assigns to the point \((x_0,y_0)\) the point
\[
\bigl(x(\omega,x_0,y_0,t_1+2),\ y(\omega,x_0,y_0,t_1+2)\bigr).
\]
It is not hard to see that
\[
T=\Delta\Theta.
\]
From formulas (1.3) and (1.4) it follows that for \(0\le t\le t_1+2\) system (2.1) has, on the rectangle \(D\), the form
\[
\frac{dx}{dt}=0,\qquad \frac{dy}{dt}=-2(y+5)\tau(t),
\tag{2.2}
\]
and on the rectangle \(B\) the form
\[
\frac{dx}{dt}=0,\qquad \frac{dy}{dt}=-2(y-5)\tau(t).
\tag{2.3}
\]
Integrating system (2.2), we obtain that if \((x_0,y_0)\in D\) and \((x_1,y_1)=\Theta(x_0,y_0)\), then
\[
x_1=x_0,\qquad y_1=(y_0+5)\beta_1-5,
\tag{2.4}
\]
where \(\beta_1\) is a positive number satisfying the inequality
\[
\beta_1<\frac{1}{11}.
\tag{2.5}
\]
Similarly, if \((x_0,y_0)\in B\), \((x_1,y_1)=\Theta(x_0,y_0)\), then
\[
x_1=x_0,\qquad y_1=(y_0-5)\beta_1+5.
\tag{2.6}
\]
From relations (2.4), (2.5), and (2.6) it follows that \(\Theta D\subset a,\ \Theta B\subset c\).
The segments of the lines \(\{y=-5,\ -3\le x\le -1\}\) and \(\{y=5,\ 1\le x\le 3\}\) remain invariant with respect to the transformation \(\Theta\). The segment \(\{x=-2,\ |y|\le 6\}\) is mapped into the segment \(\{x=-2,\ |y+5|\le 1\}\), and the segment \(\{x=2,\ |y|\le 6\}\) into the segment \(\{x=2,\ |y-5|\le 1\}\). Moreover, every horizontal segment of a straight line located in the rectangle \(D\) or \(B\) is transformed under \(\Theta\) into a horizontal segment, and every vertical segment into a vertical one.
We now describe the transformation \(\Delta\) on the squares \(a\) and \(c\). For \(t_1+2\le t\le \omega\), system (2.1) on the rectangle \(A\) has the form
\[
\frac{dx}{dt}=(x+2)\tau(t),\qquad \frac{dy}{dt}=-2(y+5)\tau(t),
\tag{2.7}
\]
and on the rectangle \(C\) system (2.1) is written in the form
\[ \frac{dx}{dt}=-(x-2)\tau(t),\qquad \frac{dy}{dt}=-2(y-5)\tau(t). \tag{2.8} \]
Hence, by direct integration, we find that if \((x_0,y_0)\in a\), \((x_1,y_1)=\Delta(x_0,y_0)\), and if on the time interval \(t_1+2\leq t\leq \omega\) the point
\[
\bigl(x(t,x_0,y_0,t_1+2),\,y(t,x_0,y_0,t_1+2)\bigr)
\]
is situated in \(A\), then
\[ x_1=(x_0+2)\beta_2-2,\qquad y_1=(y_0+5)\beta_3-5, \tag{2.9} \]
where \(\beta_2\) and \(\beta_3\) are certain numbers satisfying the inequalities
\[ \beta_2>5,\qquad 0<\beta_3<\frac{1}{25}. \tag{2.10} \]
If, however, \((x_0,y_0)\in c\), \((x_1,y_1)=\Delta(x_0,y_0)\), and if on the interval \(t_1+2\leq t\leq \omega\) the point
\[
\bigl(x(t,x_0,y_0,t_1+2),\,y(t,x_0,y_0,t_1+2)\bigr)
\]
does not leave \(C\), then
\[ x_1=(x_0-2)\beta_2+2,\qquad y_1=(y_0-5)\beta_3+5. \tag{2.11} \]
It follows from these formulas that the intersection \(F=\Delta a\cap A\) is a rectangle of the form
\[
F=\{|x|\leq 3,\ |y+5|\leq \beta_3\};
\]
\[
G=\Delta c\cap C=\{|x|\leq 3,\ |y-5|\leq \beta_3\}.
\]
It is clear that any vertical segment situated in \(\Delta^{-1}F\) or \(\Delta^{-1}G\) is transformed into a vertical segment in the region \(F\) or \(G\), respectively, and any horizontal segment into a horizontal one. Moreover, the segments
\[
\{x=-2,\ |y+5|\leq 1\}
\]
and
\[
\{x=2,\ |y-5|\leq 1\}
\]
are transformed under \(\Delta\) into their own parts, while the segments
\[
\{|x|\leq 3,\ y=-5\}
\]
and
\[
\{|x|\leq 3,\ y=5\}
\]
are transformed under \(\Delta^{-1}\) into their own parts.
The regions \(\Delta^{-1}F\) and \(\Delta^{-1}G\) are rectangles of the form:
\[
\Delta^{-1}F=\{-\gamma_1\leq x+2\leq -\gamma_2,\ |y+5|\leq 1\},
\]
\[
\Delta^{-1}G=\{\gamma_2\leq x-2\leq \gamma_1,\ |y-5|\leq 1\},
\]
where \(\gamma_1\) and \(\gamma_2\) are positive numbers less than one.
Consider the rectangles
\[
\widetilde D=\{-\gamma_1\leq x+2\leq -\gamma_2,\ |y|\leq 6\}
\]
and
\[
\widetilde B=\{\gamma_2\leq x-2\leq \gamma_1,\ |y|\leq 6\}.
\]
It is not difficult to see that on \(\widetilde B\) and \(\widetilde D\) the transformation \(T=\Delta\Theta\) is defined and
\[
T\widetilde D=F,\qquad T\widetilde B=G.
\]
Under the transformation \(T\), all vertical and horizontal line segments contained in the rectangles \(\widetilde B\) and \(\widetilde D\) go into such segments. The segments
\[
\{x=-2,\ |y|\leq 6\}
\]
and
\[
\{x=2,\ |y|\leq 6\}
\]
are transformed into their own parts, while the segments
\[
\{-\gamma_1\leq x+2\leq -\gamma_2,\ y=-5\}
\]
and
\[
\{\gamma_2\leq x-2\leq \gamma_1,\ y=5\}
\]
are transformed into the segments
\[
\{|x|\leq 3,\ y=-5\}
\]
and
\[
\{|x|\leq 3,\ y=5\},
\]
respectively. The points \(x=-2,\ y=-5\) and \(x=2,\ y=5\) remain fixed under the transformation \(T\).
In what follows we shall consider rectangles with sides parallel to the coordinate axes, situated in the rectangle \(A\); such rectangles will be denoted by the letter \(s\). Let
\[
s=\{\delta_1\leq x\leq \delta_2,\ \delta_3\leq y\leq \delta_4\};
\]
then by \(\sigma\) we shall denote the segment
\[
\{\delta_1\leq x\leq \delta_2,\ y=-5\}
\]
of the line \(L_1=\{y=-5\}\), and by \(S\) the rectangle
\[
\{\delta_1\leq x\leq \delta_2,\ |y+5|\leq 1\}
\]
corresponding to the rectangle \(s\) and the segment \(\sigma\). Analogously, by the letter \(r\) we shall denote rectangles with sides parallel to the coordinate axes, situated in the rectangle \(C\); by \(\rho\) we denote the corresponding segments of the line
\[
L_2=\{y=5\},
\]
and by \(\mathscr R\) the corresponding rectangles with horizontal sides situated on the lines \(y=5\pm 1\).
Consider the segment \(\sigma'=\{1\le x\le 3,\ y=-5\}\). From formulas (2.4) and (2.9) it follows that the transformation \(\Delta^{-1}\) is defined on this segment, that \(T^{-1}=\Delta^{-1}\) on this segment, and that the segment \(\bar\sigma=T^{-1}\sigma'\) is situated on \(L_1\) inside the square \(a\).
Denote by \(\bar S\) the rectangle of type \(S\) corresponding to the segment \(\bar\sigma\), i.e.,
\[
\bar S=\{x\in\bar\sigma,\ |y+5|\le 1\}.
\]
Similarly,
\[
\rho'=\{-3\le x\le -1,\ y=5\},\qquad T^{-1}\rho'=\bar\rho\subset c
\]
and
\[
\bar R=\{x\in\bar\rho,\ |y-5|\le 1\}.
\]
- Consider a collection of arbitrary natural numbers
\[ n_1,\ n_2,\ \ldots,\ n_{2m}. \tag{3.1} \]
We shall show that there exists a rectangle \(S_{2m}\) with the following properties:
\[
T^{\sum_{i=1}^{2k} n_i+l}S_{2m}\subset a
\tag{3.2}
\]
for \(k=0,1,\ldots,m-1,\ l=1,2,\ldots,n_{2k+1}-1\);
\[
T^{\sum_{i=1}^{2k} n_i}S_{2m}\subset d
\tag{3.3}
\]
for \(k=1,2,\ldots,m\);
\[
T^{\sum_{i=1}^{2k-1} n_i+l}S_{2m}\subset c
\tag{3.4}
\]
for \(k=1,2,\ldots,m,\ l=1,2,\ldots,n_{2k}-1\);
\[
T^{\sum_{i=1}^{2k-1} n_i}S_{2m}\subset b
\tag{3.5}
\]
for \(k=1,2,\ldots,m\); the rectangle
\[
T^{\sum_{i=1}^{2m} n_i}S_{2m}
\]
is a rectangle of type \(r\), and the corresponding segment of the line \(L_2\) is
\[
\rho'=\{-3\le x\le -1,\ y=5\}.
\]
As was already noted above, on the segment \(\sigma'=\{1\le x\le 3,\ y=-5\}\) the transformation \(T^{-1}\) is defined and \(T^{-1}\sigma'=\bar\sigma\subset a\). Since \(\bar\sigma\) is situated on \(L_1\), any negative power of the transformation \(T\) is defined on this segment.
Put
\[
\sigma_1=T^{-n_1+1}\bar\sigma.
\]
Let the rectangle \(S_1\) correspond to the segment \(\sigma_1\). Then it is clear that
\[
T^lS_1\subset a\qquad \text{for } l=1,2,\ldots,n_1-1.
\]
Moreover, if \(s^{(1)}=T^{n_1}S_1\), then, by the very choice of \(S_1\), we obtain that the rectangle \(s^{(1)}\) corresponds to a segment \(\sigma^{(1)}\) of the line \(L_1\) of the form
\[
\sigma^{(1)}=T^{n_1}\sigma_1=T^{n_1}T^{-n_1+1}\bar\sigma=T\bar\sigma=\sigma'.
\]
Now take an arbitrary natural number \(j<m\) and consider the collection of numbers \(n_1,n_2,\ldots,n_j\), which is a part of the collection (3.1). For
For definiteness we shall assume that \(j\) is an odd number (the case of even \(j\), by virtue of the symmetry of the system, is exhausted analogously). Suppose that it has been possible to construct a rectangle \(S_j\) with the following properties:
\[ T^{\sum_{i=1}^{2k} n_i+l} S_j \subset a \tag{3.6} \]
for \(k=0,1,\ldots,\dfrac{j-1}{2}\), \(l=1,2,\ldots,n_{2k+1}-1\);
\[ T^{\sum_{i=1}^{2k} n_i} S_j \subset d \tag{3.7} \]
for \(k=1,2,\ldots,\dfrac{j-1}{2}\);
\[ T^{\sum_{i=1}^{2k-1} n_i+l} S_j \subset c \tag{3.8} \]
for \(k=1,2,\ldots,\dfrac{j-1}{2}\), \(l=1,2,\ldots,n_{2k}-1\);
\[ T^{\sum_{i=1}^{2k-1} n_i} S_j \subset b \tag{3.9} \]
for \(k=1,2,\ldots,\dfrac{j+1}{2}\); the rectangle
\[
T^{\sum_{i=1}^{j} n_i} S_j
\]
is a rectangle of type \(s\), and the corresponding segment of the straight line \(L_1\) is
\[
\sigma'=\{1\le x\le 3,\ y=-5\}.
\]
Put \(\rho^{(j+1)}=T^{-n_{j+1}+1}\bar\rho\) and consider the segment
\[
\sigma^{(j+1)}=\{x\in \rho^{(j+1)},\ y=-5\}
\]
and the rectangle
\[
S^{(j+1)}=\{x\in \sigma^{(j+1)},\ |y+5|\le 1\};
\]
then the rectangle \(\Theta S^{(j+1)}=r^{(j+1)}\) corresponds to the segment \(\rho^{(j+1)}\), and the rectangle
\[
\Delta\Theta S^{(j+1)}=T S^{(j+1)}=\Delta r^{(j+1)}
\]
corresponds to the segment \(T^{-n_{j+1}+2}\bar\rho\) of type \(\rho\). Hence it follows that \(T^l S^{(j+1)}\), for \(l=1,2,\ldots,n_{j+1}-1\), lies in \(c\), and to the rectangle
\[
r=T^{n_{j+1}-1}S^{(j+1)}
\]
there corresponds the segment \(\rho\), while to the rectangle
\[
r'=T^{n_{j+1}}S^{(j+1)}
\]
there corresponds the segment \(\rho'\).
Consider the rectangle
\[
s^{(j)}=T^{\sum_{i=1}^{j} n_i} S_j
\]
and put
\[
s^{(j+1)}=S^{(j+1)}\cap s^{(j)}.
\]
Put
\[
S_{j+1}=T^{-\sum_{i=1}^{j} n_i}\,s^{(j+1)}
\]
(from the very definition of \(s^{(j+1)}\) it follows that this will be a rectangle of type \(S\)). It is clear that \(S_{j+1}\subset S_j\),
\[
T^{\sum_{i=1}^{j} n_i} S_{j+1}=s^{(j+1)}
\]
by the definition of \(S_{j+1}\), and \(s^{(j+1)}\) is a rectangle narrowed only with respect to the coord-
the part \(y\) of the rectangle \(S^{(j+1)}\); consequently, to the rectangle \(T S^{(j+1)}\) of type \(r\) there corresponds the segment \(T^{-n_{j+1}+2}\rho\) of type \(\rho\). Hence it follows that the rectangles \(T^l S^{(j+1)}\), for \(l=1,2,\ldots,n_{j+1}-1\), are rectangles of type \(r\) containing \(c\), and to the rectangle \(T^{n_{j+1}-1} S^{(j+1)}\) there corresponds the segment \(\rho\), while to the rectangle \(T^{n_{j+1}} S^{(j+1)}\) there corresponds the segment \(\rho'\).
Since
\[ s^{(j+1)}=T^{\sum_{i=1}^{j} n_i} S_{j+1}, \]
it follows from what has been said that the rectangle \(S_{j+1}\) has all the properties of the rectangle \(S_{2m}\), if in the formulation of these properties \(2m\) is replaced by \(j+1\). This proves the existence of a rectangle \(S_{2m}\) with the listed properties.
We now take an arbitrary sequence of natural numbers
\[ n_1,\ n_2,\ n_3,\ \ldots \tag{3.10} \]
Let \(m_1\) and \(m_2\) be arbitrary numbers, \(m_1<m_2\). Form the collections \(n_1,n_2,\ldots,n_{2m_1}\) and \(n_1,n_2,\ldots,n_{2m_2}\), which are parts of the sequence (3.10). To these collections there correspond the rectangles \(S_{2m_1}\) and \(S_{2m_2}\). From the preceding arguments it follows that \(S_{2m_2}\subset S_{2m_1}\).
Put
\[ Q=\bigcap_{m=1}^{\infty} S_{2m}. \tag{3.11} \]
Let us note, incidentally, that \(Q\) is a segment of the form \(\{x=x_0,\ |y+5|\leqslant 1\}\), where \(-3\leqslant x_0\leqslant -1\). Take an arbitrary point \(q\in Q\). The point \(q\) has the following properties:
\[ T^{\sum_{i=1}^{2k} n_i} q\in a \tag{3.12} \]
for \(k=0,1,2,\ldots,\quad l=1,2,\ldots,n_{2k+1}-1\);
\[ T^{\sum_{i=1}^{2k} n_i} q\in d \tag{3.13} \]
for \(k=1,2,\ldots\);
\[ T^{\sum_{i=1}^{2k-1} n_i} q\in c \tag{3.14} \]
for \(k=1,2,\ldots,\quad l=1,2,\ldots,n_{2k}-1\);
\[ T^{\sum_{i=1}^{2k-1} n_i} q\in b \tag{3.15} \]
for \(k=1,2,\ldots\).
- Choose a sequence (3.10) having the following properties: (I) the numbers \(n_i\) are bounded, i.e. for all \(i\), \(n_i<N\); (II) whatever collection of numbers
\[ n_{k+1},\ n_{k+2},\ \ldots,\ n_{k+m}, \tag{4.1} \]
which is part of the sequence (3.10), there exists an \(M\) such that the block (4.1) cannot be repeated in the sequence (3.10) more than \(M-1\) times in succession, i.e., whatever the number \(l\), the block of numbers
\[ n_{l+1},\ n_{l+2},\ \ldots,\ n_{l+Mm} \tag{4.2} \]
is not an \(M\)-fold repetition of the block (4.1).
Let \(Q\) be the segment defined by equality (3.11) and corresponding to the sequence (3.10), possessing properties (I) and (II).
Consider the sequence of points \(q,\ Tq,\ T^{2}q,\ \ldots,\ q \in Q\) in the \(x,y\)-plane. Denote by \(\Omega\) the limit set of this sequence. By Birkhoff [3], the set \(\Omega\) contains at least one recurrent point. Let \(p\) be such a point. We shall show that the point \(p\) cannot be periodic.
Suppose, on the contrary, that this is not so, i.e., assume that there exists a number \(\nu\) such that \(T^{\nu}p=p\). From relations (3.12)—(3.15) it follows that, for natural \(j\), the point \(T^{j}q\) lies in one of the squares \(a,\ b,\ c\), or \(d\). Consequently, all points of the form \(T^{j}p\) are also located in one of these squares.
First we shall assume that \(p \in d\). From the properties of the transformation \(T\), defined on the squares \(a,\ b,\ c,\ d\), it follows that there exists such a block of natural numbers
\[ \bar n_{1},\ \bar n_{2},\ \ldots,\ \bar n_{2\mu}, \tag{4.3} \]
that
\[ T^{\sum_{i=1}^{2k}\bar n_i+l}p \in a \tag{4.4} \]
for \(k=0,1,\ldots,\mu-1,\quad l=1,2,\ldots,n_{2k+1}-1;\)
\[ T^{\sum_{i=1}^{2k}\bar n_i}p \in d \tag{4.5} \]
for \(k=0,1,\ldots,\mu;\)
\[ T^{\sum_{i=1}^{2k-1}\bar n_i+l}p \in c \tag{4.6} \]
for \(k=1,2,\ldots,\mu,\quad l=1,2,\ldots,n_{2k}-1;\)
\[ T^{\sum_{i=1}^{2k-1}\bar n_i}p \in b \tag{4.7} \]
for \(k=1,2,\ldots,\mu\), and
\[ \sum_{i=1}^{2\mu}\bar n_i=\nu. \tag{4.8} \]
By property (II) of the chosen sequence (3.10), there exists a number \(M\) such that no segment of this sequence of length \(2\mu M\) is an \(M\)-fold repetition of the set (4.3). Since the point \(p\) is a limit point for the sequence \(T^j q\), there exists a number \(k\) such that the points \(T^{k+j}q\) and \(T^j p\) lie in the same squares \(a, b, c\), or \(d\) for \(j=0,1,2,\ldots,2\mu M\). But then, by virtue of the relations (4.4)—(4.7), it must turn out that the segment \(n_{k+1}, n_{k+2}, \ldots, n_{k+2\mu M}\) of the sequence (3.10) is an \(M\)-fold repetition of the set (4.3). This is impossible by the choice of the number \(M\).
The case in which the point \(p\) is located in the square \(b\) is considered analogously.
Let now \(p \in a\). If for some natural number \(l\) the point \(T^l p \notin a\), then there exists an \(m\) such that \(T^m p \in d\). In this case one should simply take \(T^m p\) as the point \(p\). Suppose now that for all natural \(l\), \(T^l p \in a\).
Since the point \(p\) is a limit point for the sequence \(T^j q\), there exists a \(k\) such that the point \(T^{k+j}q\) is located in \(a\) for \(j=1,2,\ldots,N+1\) (where \(N\) is the number appearing in the formulation of property (I) of our sequence). This, by virtue of the properties (3.12)—(3.15) of the point \(q\), contradicts the boundedness of the terms of the sequence (3.10) by the number \(N\).
The case in which for all natural \(l\), \(T^l p \in c\), is exhausted analogously. We have obtained contradictions which prove that the point \(p\) cannot be periodic.
Thus it has been established that the system (2.1) has a recurrent solution distinct from a periodic one.
References
- Cartwright M. L. Forced oscillations in nonlinear systems. Contributions to the theory of nonlinear oscillations, v. 1, 1950.
- Pliss V. A. Nonlocal Problems in the Theory of Oscillations. “Nauka,” 1964.
- Birkhoff G. D. Dynamical Systems. Gostekhizdat, 1941.
Received by the editors
December 20, 1966
Leningrad State University
named after A. A. Zhdanov