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An Example of a Vector Field with the Oriented Shadowing Property

Abstract

We consider shadowing properties for vector fields corresponding to different type of reparametrizations. We give an example of a vector field which has the oriented shadowing properties, but does not have the standard shadowing property.

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Acknowledgments

The work is partially supported by Chebyshev Laboratory (Department of Mathematics and Mechanics, St. Petersburg State University) under RF Government grant 11.G34.31.0026, JSC “Gazprom neft,” by the St. Petersburg State University in the framework of project 6.38.223.2014, Russian Foundation of Basic Research 15-01-03797a, and by the German-Russian Interdisciplinary Science Center (G-RISC) funded by the German Federal Foreign Office via the German Academic Exchange Service (DAAD).

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Correspondence to Sergey Tikhomirov.

Appendix

Appendix

Proof of Lemma 1

Note that

$$ \psi(t, (r, \varphi)) = \left(e^{at}r, \varphi+ {\int}_{\!\!\!0}^{t} b(e^{a\tau} r) d \tau\right). $$
(10)

Item (i). Let us show that YC 1( 2). Since b(r)∈C 1(0, + ), it is enough to prove continuity of D Y(x) at x = 0. Assume that \(\sqrt {{x_{1}^{2}} + {x_{2}^{2}}} < 2l\). The following holds:

$$b^{\prime}(r) = \frac{1}{r \ln^{2} r}, \quad r \in (0, 2l); $$
$$\frac{\partial Y_{1}}{\partial x_{1}} = a + b^{\prime}\left(\sqrt{{x_{1}^{2}} + {x_{2}^{2}}}\right)\frac{x_{1} x_{2}}{\sqrt{{x_{1}^{2}} + {x_{2}^{2}}}}; \quad \frac{\partial Y_{1}}{\partial x_{2}} = b^{\prime}\left(\sqrt{{x_{1}^{2}} + {x_{2}^{2}}}\right)\frac{ {x^{2}_{2}}}{\sqrt{{x_{1}^{2}} + {x_{2}^{2}}}}. $$

Since

$$\frac{|x_{1} x_{2}|}{\sqrt{{x_{1}^{2}} + {x_{2}^{2}}}}, \; \frac{{x_{2}^{2}}}{\sqrt{{x_{1}^{2}} + {x_{2}^{2}}}} < \sqrt{{x_{1}^{2}} + {x_{2}^{2}}} $$

and r b (r) → 0 as r → 0, the following holds

$$\lim_{|x| \to 0}\frac{\partial Y_{1}}{\partial x_{1}}(x) = a, \quad \lim_{|x| \to 0} \frac{\partial Y_{1}}{\partial x_{2}}(x) = 0 $$

Arguing similarly for \(\frac {\partial Y_{2}}{\partial x_{1}}, \frac {\partial Y_{2}}{\partial x_{2}}\), we conclude that

$$\lim_{|x| \to 0} D Y (x) = \left( \begin{array}{cc} a & 0 \\ 0 & a \end{array} \right). $$

Note that

$$\left| Y(x) - \left( \begin{array}{cc} a & 0 \\ 0 & a \end{array} \right) x \right| = \left| \left( \begin{array}{cc} 0 & b(|x|) \\ b(|x|) & a \end{array} \right) x \right| \leq \frac{|x|}{|\ln(|x|)|}, $$

which implies that

$$D Y (0) = \left( \begin{array}{cc} a & 0 \\ 0 & a \end{array} \right). $$

and completes the proof of item (i).

Item (ii). By the equality (10), it is enough to show that for r > 0, T 0 < 0 holds the inequality

$${\int}_{-\infty}^{T_{0}} b(e^{a\tau}r)d \tau > 2\pi. $$

Without loss of generality, we can assume that r < 2l. The following holds

$${\int}_{\!\!\!-\infty}^{T_{0}} b(e^{a\tau}r)d \tau = {\int}_{\!\!\!-\infty}^{T_{0}} -\frac{1}{a\tau + \ln r}d \tau = \left.-\frac{1}{a} \ln (|a\tau + \ln r|)\right|_{-\infty}^{T_{0}} = + \infty. $$

Item (ii) is proved.

Item (iii). Note that

$$ -\ln\frac{K}{K + 1} < 1, \quad K \sin \pi/8 > 1. $$
(11)

Fix a > 0. Choose small enough l, satisfying the following inequalities

$$ Kl < 1, \quad l | \ln(Kl)| < 1, \quad -\ln l > 4, $$
(12)
$$ \frac{2}{a}\left(l - \frac{2}{\ln l}\right) < \frac{\pi}{8}. $$
(13)

Let x 0 = (r 0, φ 0), x 1 = (r 1, φ 1) and h(t)∈R e p(l) satisfy assumptions of the lemma. The following holds

$$ r_{0} < l,\quad r_{1} < 2l. $$
(14)

Let us consider T > 0 and Δ∈ such that

$$ e^{aT}r_{0} = Kl, \quad e^{a{\Delta}}r_{0} = r_{1}. $$
(15)

Consider points x 2 = ψ(T, x 0) = (r 2, φ 2) x 3 = ψ(h(T),x 1) = (r 3, φ 3). Note that r 2 = K l. Inequality (2) implies

$$ \text{dist}(x_{2}, x_{3}) < l. $$
(16)

Using inequalities (11), we conclude that

$$ |\varphi_{2} - \varphi_{3}| < \pi/8, \quad r_{3} \in [(K-1)l, (K+1)l]. $$
(17)

Equality (10) implies that

$$r_{3} = e^{ah(T)}r_{1}, \quad \varphi_{2} = \varphi_{0} + {\int}_{\!\!\!0}^{T}b(e^{a\tau}r_{0})d \tau, \quad \varphi_{3} = \varphi_{1} + {\int}_{\!\!\!0}^{h(T)}b(e^{a\tau}r_{1})d\tau. $$

Relations (2) and (15) implies

$$ \frac{K}{K+1}e^{a(h(T)+{\Delta})}r_{0} = \frac{K}{K+1}e^{ah(T)}r_{1} < e^{aT}r_{0}. $$
(18)

The following holds

$$\begin{array}{@{}rcl@{}} \varphi_{2} - \varphi_{3} &=& (\varphi_{0} - \varphi_{1}) + {\int}_{\!\!\!0}^{T} b(e^{a\tau}r_{0})d \tau - {\int}_{\!\!\!0}^{h(T)}b(e^{a\tau}r_{1}) d \tau = \\ &=& (\varphi_{0} - \varphi_{1}) + {\int}_{\!\!\!0}^{T} b(e^{a\tau}r_{0})d \tau - {\int}_{\!\!\!{\Delta}}^{h(T) + {\Delta}} b(e^{a\tau}r_{0})d \tau = \\ &=& (\varphi_{0} - \varphi_{1}) + {\int}_{\!\!\!0}^{\Delta} b(e^{a\tau}r_{0})d \tau - {\int}_{\!\!\!T}^{h(T) + {\Delta}} b(e^{a\tau}r_{0})d \tau. \end{array} $$
(19)

Relations (2) and (15) imply that e a(h(T)+Δ) r 0 = e ah(T) r 1 > (K−1)l and hence

$$ b(e^{a\tau}r_{0}) = 0, \quad \tau \in [T, h(T) + {\Delta}]. $$
(20)

Relations (18) imply inequalities

$$\ln \frac{K}{K+1} + a (h(T) + {\Delta}) < aT, $$

and hence

$${\Delta} < (T - h(T)) - \frac{1}{a}\ln \frac{K}{K+1}. $$

Since h(t)∈Rep(l) and T = (ln(K l)− lnr 0)/a using inequalities (11), (12), we conclude that

$$ {\Delta} < \frac{1}{a}\left(l| \ln (K l) - \ln r_{0}| - \ln \frac{K}{K+1}\right)< \frac{1}{a}(2-l \ln r_{0}). $$
(21)

Inequalities (14) imply that for τ ∈ [0,Δ] holds the inequality e aτ r 0 < 2l, hence b(e aτ r 0) = 1/ ln(e aτ r 0). Inequalities (12), (13), (21) imply that a|Δ|<−(lnr 0)/2, which implies |b(e aτ r 0)| < 2b(r 0) = −2/ lnr 0 and

$$\left|{\int}_{\!\!\!0}^{\Delta} b(e^{a\tau}r_{0})d \tau\right| < - \frac{2 |{\Delta}|}{\ln r_{0}} < - \frac{2}{a}\left(2 - l \ln r_{0}\right)\frac{1}{\ln r_{0}} < \frac{2}{a}\left(l - \frac{2}{\ln l}\right) < \frac{\pi}{8}. $$

Combining this with relations (19), (20), we conclude that

$$\left|(\varphi_{2} - \varphi_{3})-(\varphi_{0} - \varphi_{1})\right| < \pi/8. $$

and hence, (17) implies |φ 0φ 1|<π/4. Item (iii) is proved.

Proof of Lemma 2

Let us fix a linear map Q and a number D > 0. Consider the lines l 1S 1, l 2S 2 defined by x 2 = x 3 = 0, y 2 = y 3 = 0, respectively. Note that Q l 2l 1. Let us consider plane VS 1 containing l 1 and Q l 2. Consider a parralelogram PV, symmetric with respect to 0 with sides parralel to l 1 and Q l 2, satisfying the relation

$$ P \subset \{|x_{1}| < D\} \cap Q(\{|y_{1}|<D\}). $$
(22)

Let us choose R > 0, such that the following inclusions hold

$$ B(R, 0) \cap V \subset P \quad \text{and} \quad Q(B(R, 0) \cap Q^{-1}V) \subset P. $$
(23)

Let z 1 be a point of intersection Sp1 and the line V∩{x 1 = 0}. Condition (23) implies that z 1P. Consider the line k 1, containing z 1 and parallel to l 1. Inclusion (22) implies that k 1P⊂Cyl1.

Similarly, let z 2 be a point of intersection of Sp2 and V∩{y 1 = 0}. Condition (23) implies the inclusion Q z 2P. Let k 2 be the line containing Q z 2 and parallel to Q l 2. Inclusion (22) implies that Q −1(k 2V)⊂Cyl2.

Since \(k_{1} \nparallel k_{2}\), there exists a point zk 1k 2. Inclusions z 1, z 2P imply that zP. Hence, z ∈ Cyl1QCyl2. Lemma 2 is proved.

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Tikhomirov, S. An Example of a Vector Field with the Oriented Shadowing Property. J Dyn Control Syst 21, 643–654 (2015). https://doi.org/10.1007/s10883-015-9272-9

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Keywords

  • Shadowing
  • Vector field
  • Reparametrization
  • Structural stability

Mathematics Subject Classification (2010)

  • 37C50
  • 37C10