Surprising property of Lyapunov dimension for invariant sets of dynamical systems^{1}

G. A. Leonov^{2} and S.A. Ljashko

leonov@math.spbu.ru root@bsfgu.balashov.san.ru

1 Introduction

Lyapunov dimension of an attractor is an upper estimate of its Hausdorff dimension and of its fractal dimension [1, 2, 3, 4]. Lyapunov dimension can also be considered as a certain integrated numerical characteristic of inner instability of strange attractors.

Surprising property of Lyapunov dimension for various classes of strange attractors is its exact computability by analytical methods. The base of these computations is created by means of Lyapunov-type functions which “annihilate” the difference between estimates of local Lyapunov dimensions of various trajectories. In this case we can analyse neighbourhoods of stationary points only and that will be enough for the computation of Lyapunov dimension. In this paper such an approach is developed for well known dynamical systems — the Hénon mapping and the Lorenz system.

We also continue here the development of direct Lyapunov method in the dimension theory which was started in articles [5, 6, 7].

Let us consider a continuously differentiable mapping F : U → ℝ^{n}, U ⊂ ℝ^{n} an open set. Here ℝ^{n} is the standard Euclidean space of dimension n.

Let us designate by T_{x}F the Jacobi matrix of the mapping F in the point x. Since F is continuously differentiable the relation

F (x + h) − F (x) = (T_{x}F)h + o(x, h)

is true, where x + h ∈ U.

Further we shall suppose that the set K ⊂ U is invariant with respect to the transformation F, F (K) = K.

Let us introduce singular values of the n × n-matrix A:

α_{1}(A) ≥ … ≥ α_{n}(A).

We remind here those definitions of Lyapunov dimension which use the singular values of the matrix T_{x}F [4].

Definition 1. We shall call the number

dim_{L}(F, x) = j + s,

where j is the maximum integer from [1, n] for which

α_{1}(T_{x}F) … α_{j}(T_{x}F) ≥ 1

1Based on the lectures “Estimates of the Lyapunov dimension of the Lorenz attractor” and “Dimensions of the Hénon attractor” hold at the DFG-Workshop “Dimensions of attractors”, St. Petersburg, Russia, October 11-12, 1999

2Partly supported by the Russian Research Fund 98-01-01005

and s ∈ [0, 1) with

α_{1}(T_{x}F) … α_{j}(T_{x}F)α_{j+1}(T_{x}F)^{s} = 1,

the Lyapunov local dimension of the mapping F at the point x ∈ K. In case α_{1}(T_{x}F) < 1 we put dim_{L}(F, x) = 0 and in case

α_{1}(T_{x}F) … α_{n}(T_{x}F) ≥ 1

we put dim_{L}(F, x) = n. Definition 2. We shall call the number

dim_{L}(F, K) = sup_{x∈K} dim_{L}(F, x)

the Lyapunov dimension of the mapping F on the set K. Definition 3. We shall call the number

dim_{L} x = lim sup_{i→+∞} dim_{L}(F^{i}, x)

the local Lyapunov dimension of the sequence of mappings F^{i} at the point x ∈ K. Definition 4. We shall call the number

dim_{L} K = sup_{x∈K} dim_{L} x

the Lyapunov dimension of the sequence of mappings F^{i} on the set K.

Let us denote by Λ the stationary set of the mapping F in the set K: F (z) = z ∀ z ∈ Λ. Definition 5. We shall say that the sequence of mappings F^{i} on the set K is Lyapunov stationary if there exists a non-singular linear mapping Q: K→ℝ^{n} such that for the Lyapunov dimensions of the sequence of mappings QF^{i}Q^{−1} on the sets QK and QΛ the equality

dim_{L} QK = dim_{L} QΛ  (1)

is true.

So the property of Lyapunov stationarity gives the opportunity to find Lyapunov dimension in a certain basis carrying out the calculations on stationary sets only. Linear operators Q as well as Lyapunov functions mentioned above will be constructed specially for every concrete case. Further we shall prove that Lyapunov stationarity takes place for the Hénon mapping and for the Lorenz system.

2 Estimates of Lyapunov dimension

Let us introduce a continuously differentiable function p : U → ℝ^{1}. We suppose that the following requirement is satisfied:

p_{1} ≤ p(x) ≤ p_{2}, ∀ x ∈ U

with positive numbers p_{1} and p_{2}.

Let j be an integer from [1, n], s ∈ [0, 1), d = j +s, A be an n×n-matrix. We introduce the designations

\begin{matrix}ω_{d}(A) = α_{1}(A) … α_{j}(A)α_{j+1}(A)^{s}, \\ ω(d, K, p, F) = sup_{x∈K} ω_{d} (\frac{p(F (x))}{p(x)} T_{x}F ) . \end{matrix}

Theorem 2.1 Suppose F (K) = K and

ω(d, K, p, F) < 1.  (2)

Then for sufficiently large i the estimate

dim_{L}(F^{i}, K) ≤ d  (3)

is valid.

Proof. Since

T_{x}F^{i} = (T_{F^{i−1}(x)}F)(T_{F^{i−2}(x)}F) … (T_{x}F)

we have

T_{x}F^{i} = \frac{p(x)}{p(F^{i}(x))} (\frac{p(F^{i}(x))}{p(F^{i−1}(x))} T_{F^{i−1}(x)}F ) … (\frac{p(F (x))}{p(x)} T_{x}F )

and, consequently ([1, 7]) it is true that

ω_{d}(T_{x}F^{i}) ≤ sup_{x∈K} (\frac{p(x)}{p(F^{i}(x))})^{d} ω(d, K, p, F)^{i}.

Hence, from (2) and from the definition of Lyapunov dimension we receive the estimate (3). Consider now the system

\frac{dx}{dt} = f(t, x), t ∈ ℝ^{1}, x ∈ ℝ^{n},  (4)

with continuously differentiable vector-function f : ℝ^{1} × ℝ^{n} → ℝ^{n}. Let us denote by J(t, x) the Jacobi matrix of f(t, x):

J(t, x) = \frac{∂f(t, x)}{∂x} .

Let λ_{1}(t, x) ≥ … ≥ λ_{n}(t, x) be the eigenvalues of the matrix

\frac{1}{2} (J(t, x)^{∗} + J(t, x)).

Here ∗ means the operation of conjugation.

Let us denote by G_{τ} the shift operator along the trajectories of (4):

G_{τ}(q) = x(τ, q).

Here τ is a number and x(t, q) is the solution of (4) with initial data x(0, q) = q. We suppose that for a certain set K the relation G_{τ}(K) = K is fulfilled.

Theorem 2.2 Suppose that there exist an integer j ∈ [1, n], a number s ∈ [0, 1), and a continuously differentiable function v(x) such that

sup_{q∈K} ∫_{0}^{τ}(λ_{1}(t, x(t, q)) + … + λ_{j}(t, x(t, q)) + sλ_{j+1}(t, x(t, q)) + v̇(x(t, q))) dt < 0. (5)

Then for sufficiently large natural number i the inequality

dim_{L}(G_{τ}^{i}, K) ≤ d = j + s  (6)

is valid.

When applying to concrete systems the idea of Lyapunov direct method it is convenient to replace (5) by a more rough requirement

sup_{q∈K} [λ_{1}(t, x(t, q)) + … + λ_{j}(t, x(t, q)) + sλ_{j+1}(t, x(t, q)) + v̇(x(t, q))] < 0, ∀ t ∈ ℝ^{1}. (7)

In order to prove Theorem 2.2 we shall need the following Lemma.

Let us consider the system

\frac{dX}{dt} = A(t) · X,  (8)

where A(t) is a real n × n-matrix function, continuous for t ∈ [0, τ]. Let us denote the eigenvalues of the matrix [A(t) + A^{∗}(t)]/2 by λ_{1}(t) ≥ λ_{2}(t) ≥ … ≥ λ_{n}(t). Let X(t) be the solution of (8) on [0, τ] which satisfies the condition X(0) = I_{n}. We denote by σ_{1}(t) ≥ … ≥ σ_{n}(t) the singular values of the matrix X(t), which are ordered with respect to their size and algebraic multiplicity.

Lemma 2.1 [1] If 1 ≤ r ≤ n is a natural number then

σ_{1}(τ)σ_{2}(τ) … σ_{r}(τ) ≤ exp ( ∫_{0}^{τ}(λ_{1}(t) + λ_{2}(t) + … + λ_{r}(t)) dt).  (9)

Proof of Theorem 2.2. Let us introduce the designation

H(t, q) = \frac{∂x(t, q)}{∂q} .

Substituting x(t, q) in (4) and differentiating with respect to q we obtain

\frac{dH(t, q)}{dt} = J[t, x(t, q)] · H(t, q).

Then according to (9) for fixed q the singular values σ_{1}(t) ≥ … ≥ σ_{n}(t) of the matrix H(t, q) satisfy with σ_{i} = σ_{i}(τ) and λ_{i}(t) = λ_{i}(t, x(t, q)) the relation

σ_{1}σ_{2} … σ_{r} ≤ exp ( ∫_{0}^{τ}(λ_{1}(t, x(t, q)) + λ_{2}(t, x(t, q)) + … + λ_{r}(t, x(t, q))) dt)

for 1 ≤ r ≤ n.

From this and from the equality

σ_{1}σ_{2} … σ_{j}σ_{j+1}^{s} = (σ_{1}σ_{2} … σ_{j})^{1−s}(σ_{1}σ_{2} … σ_{j}σ_{j+1})^{s}

we obtain

\begin{matrix}σ_{1}σ_{2} … σ_{j}σ_{j+1}^{s} ≤ \\ exp ( ∫_{0}^{τ}(λ_{1}(t, x(t, q)) + λ_{2}(t, x(t, q)) + … + λ_{j}(t, x(t, q)) + sλ_{j+1}(t, x(t, q))) dt). \end{matrix} (10)

Let us put p(x) = (exp v(x))^{1/(j+s)} and multiply both sides of (10) by the expression

(\frac{p(x(τ, q))}{p(q)})^{j+s} = exp (v(x(τ, q)) − v(q)) = exp ( ∫_{0}^{τ} v̇(x(t, q)) dt).

As a result we have

\begin{matrix}(\frac{p(x(τ, q))}{p(q)})^{j+s} σ_{1}σ_{2} … σ_{j}σ_{j+1}^{s} ≤ \\ exp ( ∫_{0}^{τ}(λ_{1}(t, x(t, q)) + λ_{2}(t, x(t, q)) + … + λ_{j}(t, x(t, q)) + sλ_{j+1}(t, x(t, q)) + v̇(x(t, q)) dt). \end{matrix}

Hence the estimate

\begin{matrix}g_{1}g_{2} … g_{j}g_{j+1}^{s} ≤ \\ exp ( ∫_{0}^{τ}(λ_{1}(t, x(t, q)) + λ_{2}(t, x(t, q)) + … + λ_{j}(t, x(t, q)) + sλ_{j+1}(t, x(t, q)) + v̇(x(t, q))) dt) \end{matrix}

(11)

follows, where g_{1} ≥ g_{2} ≥ … ≥ g_{n} are singular values of the operator

\frac{p(G_{τ}(q))}{p(q)} · \frac{∂G_{τ}(q)}{∂q} = \frac{p(x(τ, q))}{p(q)} · H(τ, q).

From (5) and (11) the inequality

g_{1}g_{2} … g_{j}g_{j+1}^{s} ≤ exp(α) < 1

follows, where

α = sup_{q∈K} ∫_{0}^{τ}(λ_{1}(t, x(t, q)) + … + λ_{j}(t, x(t, q)) + sλ_{j+1}(t, x(t, q)) + v̇(x(t, q))) dt.

Consequently the hypotheses of Theorem 2.1 are true and the shift operator G_{τ}(q) takes the part of F (q). Theorem 2.2 is proved.

3 Hénon map

Consider the Hénon mapping F: ℝ^{2} → ℝ^{2}

\binom{x}{y} ↦ \binom{a + by − x^{2}}{x},  (12)

where a > 0, 0 < b < 1 are parameters.

The change of variables x → x, y → y/\sqrt{b} transforms (12) into the mapping F_{1}

\binom{x}{y} ↦ \binom{a + \sqrt{b}y − x^{2}}{\sqrt{b} x}.  (13)

Theorem 3.1 Let K ⊂ ℝ^{2} be an invariant set with respect to F_{1} and let Λ ⊂ K be the stationary set of (13). Then

dim_{L} K = 1 + \frac{1}{1 − ln b/ ln α_{1}(x_{−})}  (14)

where α_{1}(x_{−}) = \sqrt{x_{−}^{2} + b} − x_{−}, x_{−} = \frac{1}{2}(b − 1 − \sqrt{(1 − b)^{2} + 4a}).

Proof of Theorem 3.1. The singular values α_{1}(x) ≥ α_{2}(x) of the derivative of (13) have the form

\begin{matrix}α_{1}(x) = \sqrt{x^{2} + b} + |x|, \\ α_{2}(x) = \sqrt{x^{2} + b} − |x| = \frac{b}{α_{1}(x)} . \end{matrix}

It follows from Theorem 2.1, that if there exists a number 0 ≤ s < 1 and a continuous positive on K function p(ξ) (ξ = col (x, y)), for which the inequality

α_{1}(x)α_{2}^{s}(x)\frac{p(F_{1}(ξ))}{p(ξ)} < 1, ∀ ξ ∈ K  (15)

is valid, then

dim_{L} K ≤ 1 + s.

We put p(ξ) = e^{(1−s)γ(x+\sqrt{b} y)}, where γ > 0 is a varied parameter. Then we have

p(F_{1}(ξ)) = e^{(1−s)γ(a+\sqrt{b} y+bx−x^{2})}

and consequently

\frac{p(F_{1}(ξ))}{p(ξ)} = e^{(1−s)γ(a+(b−1)x−x^{2})}.

The condition (15) after we have taken the logarithm takes the form

ln α_{1}(x) + s ln α_{2}(x) + (1 − s)γ(a + (b − 1)x − x^{2}) =
= (1 − s) ln α_{1}(x) + s ln b + (1 − s)γ(a + (b − 1)x − x^{2}) < 0.

This inequality will be fulfilled if

s ln b + (1 − s)φ(x) < 0,  (16)

where φ(x) = ln(\sqrt{x^{2} + b} + |x|) + γ(−x^{2} + (b − 1)x + a). Let us evaluate φ(x) from above. Since

for x ≥ 0 φ(x) = φ_{1}(x) = ln(\sqrt{x^{2} + b} + x) + γ(−x^{2} + (b − 1)x + a),
for x ≤ 0 φ(x) = φ_{2}(x) = ln(\sqrt{x^{2} + b} − x) + γ(−x^{2} + (b − 1)x + a),

we can note in virtue of b − 1 < 0, γ > 0 that φ_{2}(−x) ≥ φ_{1}(x) for x ≥ 0. Therefore it is sufficient to evaluate from above the value

ψ(x) = φ_{2}(−x) = ln (\sqrt{x^{2} + b} + x) + γ(−x^{2} − (b − 1)x + a) for x ≥ 0.

We have

ψ′(x) = \frac{1}{\sqrt{x^{2} + b}} − γ(2x + b − 1), ψ′′(x) < 0 for x ≥ 0.

Consequently if the equation ψ′ = 0 has a positive root then the function ψ(x) takes the maximal value for x ≥ 0 in this point (hence the function φ(x) has in this point the maximal value for x ∈ (−∞, +∞)).

If we choose

γ = \frac{1}{(−2x_{−} + b − 1)\sqrt{x_{−}^{2} + b}} > 0

we have ψ′(−x_{−}) = 0.

Hence it follows that

\begin{matrix}φ(x) ≤ ln (\sqrt{x_{−}^{2} + b} − x_{−}) + γ(−x_{−}^{2} + (b − 1)x_{−} + a) = \\ = ln (\sqrt{x_{−}^{2} + b} + |x_{−}|) = ln α_{1}(x_{−}). \end{matrix}

Thus,

dim_{L} K ≤ 1 + s

holds for all s satisfying the condition

s > \frac{1}{1 − ln b/ ln α_{1}(x_{−})} .

From this and from the definition of dim_{L} K it follows that

dim_{L} K ≤ 1 + \frac{1}{1 − ln b/ ln α_{1}(x_{−})} .  (17)

Note now that the point x = x_{−}, y = x_{−} is a fixed point of the mapping F_{1}. And

α_{1}(x_{−})α_{2}(x_{−})^{s} = 1,

where

s = \frac{1}{1 − ln b/ ln α_{1}(x_{−})} .

Hence it follows that local Lyapunov dimension of the stationary point is

1 + \frac{1}{1 − ln b/ ln α_{1}(x_{−})} .  (18)

From (17) and (18) the equality (14) follows.

For a = 1.4, b = 0.3 the formula (14) gives

dim_{L} K = 1.4953...

4 Lorenz system

Let us consider the Lorenz system

\begin{matrix}ẋ = −σx + σy, \\ ẏ = rx − y − xz, \\ ż = −bz + xy, \end{matrix}  (19)

where r, b, σ are positive numbers.

Suppose that the following inequalities are true

\begin{matrix}b ≥ 2, r ≥ 1, σ + 1 − 2b ≥ 0, \\ rσ^{2}(4 − b) + 2σ(b − 1)(2σ − 3b) − b(b − 1)^{2} ≥ 0. \end{matrix}  (20)

The parameters r = 28, b = 8/3, σ = 10 and r = 40, b = 4, σ = 16 are most often considered in numerical experiments. Note that these values satisfy (20).

We put

A = \frac{σ}{\sqrt{rσ + (b − 1)(σ − b)}} .

The linear substitution

x → Ax, y → \frac{A(b − 1)x}{σ} + y, z → z

reduces the system (19) to the system

\begin{matrix}ẋ = (b − σ − 1)x + \frac{σ}{A} y, \\ ẏ = A (r + \frac{(b − 1)(σ − b)}{σ}) x − by − Axz, \\ ż = −bz + \frac{A^{2}(b − 1)}{σ} x^{2} + Axy. \end{matrix}  (21)

Let us introduce the shift operator G_{τ} along the trajectories of the system (21) and the Lyapunov dimension dim_{L} K of the sequence of the mappings G_{τ}^{i} on the set K. Here τ is some positive number.

Theorem 4.1 Suppose that the inequalities (20) are true, 0 ∈ K, and the set K is invariant with respect to the shift operator G_{τ} along the trajectories of the system (21). Then the equality

dim_{L} K = 3 − \frac{2(σ + b + 1)}{σ + 1 + \sqrt{(σ − 1)^{2} + 4rσ}}  (22)

is true.

Proof of Theorem 4.1. Let us calculate the eigenvalues λ_{1}(x, y, z) ≥ λ_{2}(x, y, z) ≥ λ_{3}(x, y, z) of the symmetrized Jacobi matrix for the right-hand side of (21):

\begin{matrix}λ_{2}(x, y, z) = −b, λ_{1,3}(x, y, z) = −\frac{σ + 1}{2}± \\ ±\frac{1}{2}\sqrt{(σ + 1 − 2b)^{2} + (\frac{2A^{2}(b − 1)}{σ} x + Ay)^{2} + (A (r + \frac{(b − 1)(σ − b)}{σ}) − Az + \frac{σ}{A})^{2}} . \end{matrix}

Let us choose a Lyapunov function v(x, y, z) in the form

v(x, y, z) = \frac{1}{2} Aθ^{2}(1 − s) (T_{1}x^{2} + T_{2} (y^{2} + z^{2} + \frac{2A(b − 1)xy}{σ} ) + T_{3}z) ,

where

θ^{2} = (2\sqrt{(σ + 1 − 2b)^{2} + (A (r + \frac{(b − 1)(σ − b)}{σ}) + \frac{σ}{A})^{2}})^{−1} ,

T_{3} = −\frac{2}{b} (\frac{σ}{A} + A (r + \frac{(b − 1)(σ − b)}{σ})) , T_{1} = \frac{A^{2} (T_{3} + 4A(b − 1)/σ) (b − 1)^{2}(σ + 1)}{2σ(2b(σ + 1) − (rσ^{2} + 2σ + b^{2} + 1))},

T_{2} = \frac{σ^{2} (T_{3} + 4A(b − 1)/σ)}{2(2b(σ + 1) − (rσ^{2} + 2σ + b^{2} + 1))}, s ∈ [0, 1).

We get

\begin{matrix}2(λ_{1}(x, y, z) + λ_{2}(x, y, z) + sλ_{3}(x, y, z) + v̇(x, y, z)) = \\ = −(σ + 2b + 1) − s(σ + 1) + (1 − s)φ(x, y, z), \end{matrix}  (23)

where

φ(x, y, z) =

= \sqrt{(σ + 1 − 2b)^{2} + (\frac{2A^{2}(b − 1)}{σ} x + Ay)^{2} + (A (r + \frac{(b − 1)(σ − b)}{σ}) − Az + \frac{σ}{A})^{2}}+

+θ^{2}( (2AT_{1}(b − σ − 1) + 2A^{3}T_{2}(b − 1) \frac{rσ + (b − 1)(σ − b)}{σ^{2}} + T_{3}A^{3} \frac{b − 1}{σ}) x^{2}−

−2AT_{2}y^{2} − 2bAT_{2}z^{2} + (2σT_{1} + 2T_{2}A^{2} \frac{rσ − b^{2} + 1}{σ} + T_{3}A^{2}) xy − bAT_{3}z).

It is clear that the inequality

φ(x, y, z) ≤

≤ \frac{1}{4θ^{2}} + θ^{2}((σ + 1 − 2b)^{2} + (A (r + \frac{(b − 1)(σ − b)}{σ}) + \frac{σ}{A})^{2} +

+(2AT_{1}(b − σ − 1) + 2A^{3}T_{2}(b − 1) \frac{rσ + (b − 1)(σ − b)}{σ^{2}}+

+T_{3}A^{3} \frac{b − 1}{σ} + 4A^{4} \frac{(b − 1)^{2}}{σ^{2}} )x^{2} + (A^{2} − 2AT_{2})y^{2} + (A^{2} − 2bAT_{2})z^{2}+

+ (2σT_{1} + 2T_{2}A^{2} \frac{rσ − b^{2} + 1}{σ} + T_{3}A^{2} + 4A^{3} \frac{b − 1}{σ}) xy−

− (bAT_{3} + 2A (A (r + \frac{(b − 1)(σ − b)}{σ}) + \frac{σ}{A})) z)

holds.

With the help of the expressions for A, T_{1}, T_{2}, T_{3}, θ^{2} and the condition (20) we obtain now that in the latter inequality the coefficients at x^{2}, xy and z are equal to zero and the coefficients at y^{2} and z^{2} are nonpositive. Thus,

φ(x, y, z) ≤ q = \sqrt{(σ − 1)^{2} + 4rσ}

for all real numbers x, y, z. Then it follows from (23) that

λ_{1}(x, y, z) + λ_{2}(x, y, z) + sλ_{3}(x, y, z) + v̇(x, y, z) < 0

for any

s < \frac{q − 2b − σ − 1}{q + σ + 1} .

Let s tend to (q − 2b − σ − 1)/(q + σ + 1) from the right. Then from Theorem 2.2 we have the estimate

dim_{L} K ≤ 3 − \frac{2(σ + b + 1)}{σ + 1 + \sqrt{(σ − 1)^{2} + 4rσ}} .  (24)

It is not difficult to see that local Lyapunov dimension of the stationary point x = y = z = 0 of the system (21) is equal to

3 − \frac{2(σ + b + 1)}{σ + 1 + \sqrt{(σ − 1)^{2} + 4rσ}} .  (25)

The formula (22) follows from the relations (24) and (25).

References

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