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Finite Fractal Dimension of Uniform Attractors for Non-Autonomous Dynamical Systems with Infinite-Dimensional Symbol Space

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Abstract

The aim of this paper is to find an upper bound for the box-counting dimension of uniform attractors for non-autonomous dynamical systems. Contrary to the results in the literature, we do not require the symbol space to have finite box-counting dimension. Instead, we ask a condition on the semi-continuity of pullback attractors of the system as time goes to infinity. This semi-continuity can be achieved if we suppose the existence of finite-dimensional exponential uniform attractors for the limit symbols, that is, the symbols associated with the asymptotic vector fields of the non-autonomous differential equation. After showing these new results, we apply them to study the box-counting dimension of the uniform attractor for a non-autonomous reaction-diffusion equation, and we find a class of forcing terms for this equation such that their associated symbol spaces have infinite box-counting dimension, but the uniform attractors of the dynamical systems generated by these forcing terms have finite box-counting dimension anyway.

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No datasets were generated or analyzed during the current study.

Notes

  1. The Hausdorff semi-distance \(\text{ dist}_H(A,B)\) shows how close an arbitrary point of A is to some point of B. If \(\text{ dist}_H(A,B) < \epsilon \), then the set A is in an \(\epsilon \) neighborhood of the set B. Note that the Hausdorff semi-distance is generally not symmetric; it measures closeness in only one direction.

  2. See Ott and Yorke (2005). A prevalent set is a measure-theoretic analog of “almost every” in infinite-dimensional spaces. In particular, prevalent sets are dense, and the countable intersection of prevalent sets is also prevalent.

  3. If we have a pair of Banach spaces \(Y\subset X\) with Y compactly embedded in X, the Kolmogorov \(\epsilon \)-entropy (see Triebel (1978)) of this embedding is defined as \(\varvec{H}_{\epsilon }(Y;X) = \log _2 N_{\epsilon }\), where \(N_{\epsilon }\) is the minimal number of balls of radius \(\epsilon \) in X necessary to cover the unit ball \(B_Y(0,1)\) in Y. This represents the fractal complexity of the set \(B_Y(0,1)\) in X.

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Funding

A. N. CARVALHO.: Partially supported by FAPESP Grant # 20/14075-6 and by CNPq Grant # 308902/2023-8, Brazil. J. A. LANGA.: Partially supported by Grant # PID2021-122991NB-C21 Ministerio de Ciencia e Innovación, Spain. R. O. MOURA.: Supported by Grants #2022/04886-2 and #2023/11798-5, São Paulo Research foundation (FAPESP)

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Correspondence to R. O. Moura.

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Communicated by Tiago Pereira.

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Appendix A. Proof of Theorem 3.1

Appendix A. Proof of Theorem 3.1

Proof of Theorem 3.1

First part: Let B be a closed, bounded, uniformly absorbing set for the system \(\{U_\sigma (t,s):t\geqslant s\}_{\sigma \in \Sigma }\) and consider the lifted set \(\mathbb {B} = B\times \Sigma \). We will prove that the skew-product semigroup in \(X\times \Xi \) has an exponential global attractor, and the existence of an exponential uniform attractor for \(\{U_\sigma (t,s):t\geqslant s\}\) will follow readily.

Then it is easy to see that if \(\{S(t): t\in {\mathbb R}\}\) is the skew-product semigroup generated by the system \(\{U_\sigma (t,s)\}_{\sigma \in \Sigma }\), then \(S(\tau ) \mathbb {B} \subset \mathbb {B}\).

For all \(n\in \mathbb N\), consider the covering of the finite-dimensional set \(\Sigma \) with a minimum number of open balls of \(\Xi \) of centers \(\{\Sigma _-^n, \dots , \sigma _N^n\}\) and radius \(R_n = R\nu ^n/P e^{\zeta n \tau } L(n\tau )\). We name these balls \(B_i^n = B_\Xi (\sigma _i^n, R_n)\). The number \(N = N_n\) depends on the box counting dimension of \(\Sigma \). More precisely, for every \(\delta > 0\), there exists \(n_0 = n_0(\delta )\) such that

$$\begin{aligned} N_n \leqslant \left( \frac{Pe^{\zeta n \tau } L(n \tau )}{R \nu ^n} \right) ^{d+\delta }, \quad \forall \, n\geqslant n_0 \end{aligned}$$
(35)

Let \(\nu \in (0,1)\) and \(R> 0\), \(x_0 \in \mathcal {B}\) be such that \(\mathcal {B} = B_X(x_0, R) \cap \mathcal {B}\).

The following inductive argument is the famous smoothing technique that was used to prove finite dimensionality of attractors in many settings Cholewa et al. (2008); Carvalho and Sonner (2013); Efendiev et al. (2004). It consists of two steps. First we cover the unitary ball in Y with finitely many balls in X (we can do this because Y is compactly embedded in X). Then we apply the evolution process in these X balls and use the smoothing condition (\(H_2\)) to show that the image is bounded in Y. But since bounded sets in Y are precompact in X, we can start again and cover it with balls in X, keeping track of the number of balls needed as the radius decrease.

To start with, let \(\kappa = \kappa (\tau )\). Since Y is compactly embedded in X, the unit ball \(B_Y(0,1)\) in Y can be covered by \(N_{\frac{\nu }{2\kappa }}\) balls of radius \(\frac{\nu }{2\kappa }\) in X, that is:

$$\begin{aligned} B_Y(0,1) \subset \bigcup _{i=1}^{N_{\frac{\nu }{2\kappa }}} B_X\left( x_i, \frac{\nu }{2\kappa } \right) , \quad x_i \in B_Y(0,1). \end{aligned}$$
(36)

Now, let \(\sigma \in \Sigma \) be arbitrary, it follows from \((H_{2})\) that:

$$\begin{aligned} \begin{aligned} U_\sigma (\tau ,0) \mathcal {B}&= U_\sigma (\tau ,0) [B_X(x_0, R) \cap \mathcal {B}] \\&\subset B_Y(U_\sigma (\tau ,0) x_0, \kappa R) \cap U_\sigma (\tau , 0) \mathcal {B}\end{aligned} \end{aligned}$$
(37)

Let \(y_\sigma = U_\sigma (\tau ,0)x_0\). Then it follows from (16), (36) and (37) that:

$$\begin{aligned} U_\sigma (\tau ,0) \mathcal {B}\subset \bigcup _{i=1}^{N_{\frac{\nu }{2\kappa }}} B_X\left( y_\sigma + \kappa R x_i, \frac{R \nu }{2}\right) \cap \mathcal {B}, \end{aligned}$$
(38)

so there are \(q^\sigma _i \in \mathcal {B}\), \(i = 1, \cdots , N_{\frac{\nu }{2\kappa }}\) such that:

$$\begin{aligned} U_\sigma (\tau ,0) \mathcal {B}\subset \bigcup _{i=1}^{N_{\frac{\nu }{2\kappa }}} B_X\left( q^\sigma _i, R\nu \right) \cap \mathcal {B}, \end{aligned}$$
(39)

Suppose by induction that for some \(k \geqslant 1\), we have \(N_X(U_\sigma (k \tau ,0) \mathcal {B}, R\nu ^k) \leqslant N_{\frac{\nu }{2\kappa }}^k\) (we just proved that this is true for \(k = 1\)). Then, there exist \(q_i^\sigma \subset \mathcal {B}\), \(i = 1, \cdots , N_{\frac{\nu }{2\kappa }}^k\) such that:

$$\begin{aligned} U_\sigma (k \tau ,0) \mathcal {B}\subset \bigcup _{i=1}^{N_{\frac{\nu }{2\kappa }}^k} B_X\left( q^\sigma _i, R\nu ^k \right) \cap \mathcal {B}, \end{aligned}$$
(40)

Now, by the system of processes property:

$$\begin{aligned} \begin{aligned} U_\sigma ((k+1) \tau ,0) \mathcal {B}&= U_\sigma ((k+1) \tau , k \tau ) U_\sigma (k \tau , 0) \mathcal {B}\\&\subset U_{\theta _{k \tau }\sigma }(\tau ,0) U_\sigma (k \tau , 0) \mathcal {B}\end{aligned} \end{aligned}$$

Then, by (40),

$$\begin{aligned} U_\sigma ((k+1) \tau ,0) \mathcal {B}\subset \bigcup _{i=1}^{N_{\frac{\nu }{2\kappa }}^k} U_{\theta _{k \tau }\sigma }(\tau ,0) \left[ B_X\left( q^\sigma _i, R\nu ^k \right) \cap \mathcal {B}\right] \end{aligned}$$
(41)

and for each i, we have by the smoothing property \((H_{2})\) and self absorption of \(\mathcal {B}\) that:

$$\begin{aligned} \begin{aligned} U_{\theta _{k \tau }\sigma }(\tau ,0) \left[ B_X\left( q^\sigma _i, R\nu ^k \right) \cap \mathcal {B}\right]&\subset B_Y\left( U_{\theta _{k \tau }\sigma }(\tau ,0) q^\sigma _i, \kappa R \nu ^k\right) \cap \mathcal {B}\\&\subset \bigcup _{j=1}^{N_{\frac{\nu }{2\kappa }}} B_X\left( p_{i,j}, R\nu ^{k+1} \right) \cap \mathcal {B}, \end{aligned} \end{aligned}$$
(42)

for \(p_{i, j} \in \mathcal {B}\). Then, using (41) and (42), we deduce \(N_X(U_\sigma ((k+1) \tau ,0) \mathcal {B}, R\nu ^{k+1}) \leqslant N_{\frac{\nu }{2\kappa }}^{k+1}\).

Then, we proved by induction that for any \(\sigma \in \Sigma \),

$$\begin{aligned} N_X(U_\sigma (n \tau ,0) \mathcal {B}, R\nu ^n) \leqslant N_{\frac{\nu }{2\kappa }}^n, \quad \forall \, n\in \mathbb N. \end{aligned}$$
(43)

For each \(\sigma \in \Sigma \), chose a set of centers \(W^n(\sigma ) \subset X\) such that \(\# W^n(\sigma ) \leqslant N^n_{\frac{\nu }{2\kappa }}\) and:

$$\begin{aligned} U_\sigma (n\tau , 0) \mathcal {B} \subset \bigcup _{u\in W^n(\sigma )} B_X(u, R\nu ^n) \cap \mathcal {B} \end{aligned}$$
(44)

Now, for each \(n\in \mathbb N\) and \(j\in \{1, \dots N(n)\}\), we define:

$$\begin{aligned} \mathcal {U}^j(n) = W^n(\sigma _j^n), \end{aligned}$$
$$\begin{aligned} \mathcal {U}(n) = \bigcup _{j=1}^{N(n)} \mathcal {U}^j(n). \end{aligned}$$

Second part:

It follows from the definition of skew-product semigroup that if \(\Pi _1: X\times \Xi \rightarrow X\) is the projection onto the first coordinate, then:

$$\begin{aligned} \Pi _1 S(n\tau ) \mathbb {B} = U_\Sigma (n\tau , 0) B \end{aligned}$$

Next we will show that:

$$\begin{aligned} \text{ dist}_H\left( \Pi _1 S(n\tau ) \mathbb {B}, \mathcal {U}(n) \right) \leqslant 2R \nu ^n. \end{aligned}$$
(45)

Indeed, if \(y\in \Pi _1 S(n\tau ) \mathbb {B}\), then \(y = U_\sigma (n\tau ,0)x\), for some \(\sigma \in \Sigma \), \(x\in B\).

By the covering of \(\Sigma \), there exists \(i\in \{1, \dots , N(n)\}\) such that

$$\begin{aligned} d_\Xi (\sigma , \sigma _i^n) \leqslant \frac{R \nu ^n}{L(n \tau )}. \end{aligned}$$

And by the definition of the sets \(\mathcal {U}^i(n)\), we have:

$$\begin{aligned} d\left( U_{\sigma _i^n}(n\tau ,0) x, \mathcal {U}^i(n) \right) \leqslant R\nu ^n. \end{aligned}$$
(46)

Moreover, by \((H_5)\), we have:

$$\begin{aligned} \Vert y - U_{\sigma _i^n}(n\tau ,0) x\Vert \leqslant L(n\tau ) \frac{R \nu ^n}{L(n\tau )} = R\nu ^n \end{aligned}$$
(47)

And the estimates (46) and (47) imply (45). Notice that, more precisely, we showed that:

$$\begin{aligned} \text{ dist}_H\left( U_{B_i^n}(n\tau , 0)B, \mathcal {U}^i(n) \right) \leqslant 2R \nu ^n \end{aligned}$$
(48)

Now, for each \(v\in \mathcal {U}^i(n)\) take, if existing, \((u_v, \xi _v) \in \mathbb {B}\) such that \(\xi _v\in B_i^n\) and

$$\begin{aligned} \Vert v - U_{\xi _v}(n\tau , 0) u_v\Vert \leqslant 2R\nu ^n. \end{aligned}$$
(49)

If no such pair \((u_v, \xi _v)\) exists, we can remove the element v from \(\mathcal {U}^i(n)\) still preserving the estimate (48).

Now we define \(\mathbb {U}^i(n) = \{S(n\tau ) (u_v, \xi _v): v\in \mathcal {U}^i(n)\}\), and

$$\begin{aligned} \mathbb {U}(n) = \bigcup _{i=1}^{N(n)}\mathbb {U}^i(n). \end{aligned}$$

Now, using (45) and the definition of \(\mathbb {U}(n)\), it is not hard to see that:

$$\begin{aligned} \text{ dist}_H \left( \Pi _1 S(n\tau ) \mathbb {B}, \Pi _1 \mathbb {U}(n) \right) \leqslant 4R\nu ^n. \end{aligned}$$

We also notice that \(\mathbb {U}(n) \subset S(n\tau ) \mathbb {B}\) and

$$\begin{aligned} \# \mathbb {U}(n) \leqslant \# \mathcal {U}(n) \leqslant N(n) N^n_{\frac{\nu }{2\kappa }}. \end{aligned}$$

Third part:

We define now the sets:

$$\begin{aligned} \mathbb {E}(0) = \mathbb {U}(0), \quad \mathbb {E}(k+1) = S(\tau ) \mathbb {E}(k) \cup \mathbb {U}(k+1) \end{aligned}$$

The set \(\mathbb {M}^d: = \overline{\bigcup _{k\in \mathbb {N}} \mathbb {E}(k)}\) will be the exponential attractor for the discrete semigroup \(\{S(n\tau ): n\in \mathbb N\}\). We will be able to explicitly calculate the cardinality of the sets \(\mathbb {E}(n)\), which will help us calculate the box-counting dimension of the global attractor \(\mathbb {M}^d\).

First, notice that \(S(\tau ) \mathbb {M}^d \subset \mathbb {M}^d\). Now if \(n\in \mathbb N\), we have:

$$\begin{aligned} \text{ dist}_H(S(n\tau ) \mathbb {B}, \mathbb {M}^d) \leqslant \underset{(x,\sigma ) \in \mathbb {B}}{\sup } \text{ dist }[(U_\sigma (n\tau , 0) x, \theta _{n\tau } \sigma ), \mathbb {U}(n)] \end{aligned}$$

For each \((x,\sigma ) \in \mathbb {B}\), there exists a \(i\in \{1, \dots , N(n)\}\) such that \(\sigma \in B_i^n\), and \(v\in \mathcal {U}^i(n)\) such that

$$\begin{aligned} \Vert U_\sigma (n\tau ,0) x - v\Vert \leqslant 2R\nu ^n, \end{aligned}$$

and there exist \((u_v, \xi _v)\) such that \(\xi _v \in B^n_i\), and

$$\begin{aligned} \Vert v - \Pi _1 S(n\tau ) (u_v, \xi _v)\Vert \leqslant 2R\nu ^n \end{aligned}$$

And we have \(u_n = S(n\tau ) (u_v, \xi _v) \in \mathbb {U}(n)\) and:

$$\begin{aligned} \begin{aligned} \text{ dist }[(U_\sigma (n\tau , 0) x, \theta _{n\tau } \sigma ), u_n]&\leqslant \Vert U_\sigma (n\tau ,0) x - \Pi _1 S(n\tau ) (u_v, \xi _v)\Vert + \text{ dist}_\Xi (\theta _{n\tau }\sigma , \theta _{n\tau } \xi _v) \\&4R\nu ^n + Pe^{\zeta n\tau } 2 \frac{R\nu ^n}{Pe^{\zeta n \tau } L(n\tau )} \leqslant 6R\nu ^n \end{aligned} \end{aligned}$$
(50)

which implies that

$$\begin{aligned} \text{ dist}_H(S(n\tau ) \mathbb {B}, \mathbb {M}^d) \leqslant 6R\nu ^n \end{aligned}$$
(51)

which gives the exponential attraction of \(\mathbb {B}\) by \(\mathbb {M}^d\) under the action of \(\{S(n\tau ): n\in \mathbb N\}\). Since \(\mathbb {B}\) is an absorbing set for the semigroup, the exponential attraction extends to every bounded set \(\mathbb {D} \subset \mathbb {X}\).

Fourth part: Now let us prove that \(\mathbb {M}^d\) has finite box-counting dimension. Since \(\mathbb {U}(n) \subset S(n\tau )\mathbb {B}\), it can be shown by induction that \(\mathbb {E}((n+j)\tau ) \subset S((n+j)\tau ) \mathbb {B} \subset S(n\tau ) \mathbb {B}\), which implies:

$$\begin{aligned} \mathbb {M}^d \subset \bigcup _{j=1}^{n-1} \mathbb {E}(n) \cup \overline{S(n\tau ) \mathbb {B}} \end{aligned}$$
(52)

From the estimates on the cardinality of the sets \(\mathbb {U}(n)\), \(n\in \mathbb N\) and the construction of the sets \(\mathbb {E}(n)\), \(n\in \mathbb N\), we obtain:

$$\begin{aligned} \# \mathbb {E}_0 \cup \cdots \cup \mathbb {E}_{n-1} \leqslant \frac{n(n+1)}{2} N(n) N^n_{\frac{\nu }{2\kappa }} \end{aligned}$$

It has been proved before that

$$\begin{aligned} \text{ dist}_H\left( S(n\tau ) \mathbb {B}, \mathbb {U}(n)\right) \leqslant 6 R\nu ^n \end{aligned}$$

Which implies that:

$$\begin{aligned} N(S(n\tau ) \mathbb {B}, 7R\nu ^n) \leqslant \# \mathbb {U}(n) \leqslant N(n) N^n_{\frac{\nu }{2\kappa }} \end{aligned}$$

Then, we have:

$$\begin{aligned} N(\mathbb {M}^d, 7R\nu ^n) \leqslant \frac{n^2+n+2}{2} N(n) N^n_{\frac{\nu }{2\kappa }} \end{aligned}$$

Which implies that

$$\begin{aligned} \begin{aligned} d_B(\mathbb {M}^d) \leqslant \underset{n\rightarrow \infty }{\limsup } \frac{\log \left( \frac{n^2+n+2}{2} N(n) N^n_{\frac{\nu }{2\kappa }} \right) }{- \log (7R\nu ^n)}. \end{aligned} \end{aligned}$$
(53)

To estimate this, we notice that

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim }\ \frac{\log \frac{n^2+n+2}{2}}{-\log (7R\nu ^n)} = 0 \end{aligned}$$
$$\begin{aligned} \underset{n\rightarrow \infty }{\lim }\ \frac{\log N^n_{\frac{\nu }{2\kappa }}}{-\log (7R\nu ^n)} = \frac{\log N_{\frac{\nu }{2\kappa }}}{-\log \nu } \end{aligned}$$

Finally, by definition of N(n), we have:

$$\begin{aligned} \underset{n\rightarrow \infty }{\limsup }\ \frac{\log N(n)}{-\log (7R\nu ^n)} \leqslant \underset{n\rightarrow \infty }{\limsup }\ \frac{\log N_\Xi \left( \Sigma , \frac{R}{Pc_1} \left( \frac{\nu }{e^{(\zeta + \beta ) \tau }} \right) ^n\right) }{-\log (7R \nu ^n)} \leqslant d_\Sigma \left( \frac{(\beta + \zeta ) \tau }{-\log \nu } +1 \right) \end{aligned}$$

where we used the box-counting dimension \(d_\Sigma \) of \(\Sigma \) in \(\Xi \).

Then, we conclude that:

$$\begin{aligned} d_B(\mathbb {M}^d) \leqslant \frac{\log N_{\frac{\nu }{2\kappa }}}{-\log \nu } + d_\Sigma \left( \frac{(\beta + \zeta ) \tau }{-\log \nu } +1 \right) . \end{aligned}$$

Fifth part: Now, we will extend the exponential attractor \(\mathbb {M}^d\) to an exponential attractor of the continuous semigroup \(\{S(t): t\geqslant 0\}\). We define:

$$\begin{aligned} \mathbb {M} = \bigcup _{t\in [0,\tau ]} S(t) \mathbb {M}^d. \end{aligned}$$

The positive semi-invariance of \(\mathbb {M}\) by \(\{S(t):t\in {\mathbb R}\}\) follows from the positive semi-invariance of \(\mathbb {M}^d\) by \(\{S(n\tau ): n\in \mathbb N\}\).

For the exponential attraction, let \(t\geqslant 0\) and we write \(t = n\tau + s\) with \(s\in [0,\tau )\). Using \((H_3)\), \((H_4)\) and \((H_5)\) we can show that S(t) is \(\gamma \)-Hölder continuous in \(\mathcal {B} \times \Sigma \) with a constant we call K(t), and K(t) can be taken uniform \((=K)\) for \(t\in [0,\tau ]\). Then we have:

$$\begin{aligned} \begin{aligned} \text{ dist}_H(S(t)\mathbb {B}, \mathbb {M})&\leqslant \text{ dist}_H(S(s) S(n\tau ) \mathbb {B}, S(s) \mathbb {M}^d) \leqslant K (\text{ dist}_H(S(n\tau ) \mathbb {B}, \mathbb {M}^d))^\gamma \\&\leqslant K (6R\nu ^n)^\gamma \leqslant C e^{\log (\nu )\gamma \frac{t}{\tau }} \end{aligned} \end{aligned}$$
(54)

This implies that \(\mathbb {M}\) attracts any bounded set in \(\mathbb {X}\) exponentially under the action of \(\{S(t):t\in \mathbb {R}\}\).

Finally, we estimate the box-counting dimension of \(\mathbb {M}\). Notice that:

$$\begin{aligned} \mathbb {M} = \Phi \left( [0,\tau ] \times \mathbb {M}^d\right) , \end{aligned}$$

where \(\Phi : [0,\tau ] \times \mathbb {M}^d \rightarrow \mathbb {M}\) is given by \(\Phi (t,(x,\sigma )) = S(t) (x,\sigma )\). Since \(\Phi \) is \(\theta \)-Hölder continuous in time and \(\gamma \)-Hölder continuous in the \(\mathbb {X}\) variable, it follows that:

$$\begin{aligned} d_B(\mathbb {M}) \leqslant \frac{1}{\theta } + \frac{1}{\gamma } d_B(\mathbb {M}^d) \leqslant \frac{1}{\theta } + \frac{1}{\gamma } \left[ \frac{\log N_{\frac{\nu }{2\kappa }}}{-\log \nu } + d_\Sigma \left( \frac{(\beta + \zeta ) \tau }{-\log \nu } +1 \right) \right] . \end{aligned}$$

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Carvalho, A.N., Langa, J.A. & Moura, R.O. Finite Fractal Dimension of Uniform Attractors for Non-Autonomous Dynamical Systems with Infinite-Dimensional Symbol Space. J Nonlinear Sci 35, 70 (2025). https://doi.org/10.1007/s00332-025-10169-0

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