Dynamical Systems and C₀-Semigroups

Abstract

In this section, we are concerned with the behavior of asymptotically almost automorphic semigroups of linear operators T = (T(t))_t≥0 at t →∞. We present some topological and asymptotic properties based on the Nemytskii and Stepanov theory of dynamical systems.

In this section, we are concerned with the behavior of asymptotically almost automorphic semigroups of linear operators T = (T(t))_t≥0 at t →∞. We present some topological and asymptotic properties based on the Nemytskii and Stepanov theory of dynamical systems.

First of all, we present a connection between abstract dynamical systems and C ₀-semigroups of linear operators. $\mathbb X$ is a (real or complex) Banach space.

1 Abstract Dynamical Systems

Definition 7.1

A mapping $u:\mathbb R^+\times \mathbb X\to \mathbb X$ is called an (abstract) dynamical system if

• (i) u(0, x) = x, for every $x\in \mathbb X$;

• (ii) $u(\cdot ,x):\mathbb R^+\to \mathbb X$ is continuous for any t > 0 and right-continuous at t = 0, for each $x\in \mathbb X$;

• (iii) $u(t,\cdot ):\mathbb X\to \mathbb X$ is continuous for each $t\in \mathbb R^+$;

• (iv) u(t + s, x) = u(t, u(s, x)) for all $t,s\in \mathbb R^+$ and $x\in \mathbb X$.

If $u:\mathbb R^+\times \mathbb X \to \mathbb X$ is a dynamical system, the mapping $u(\cdot ,x):\mathbb R^+\to \mathbb X$ will be called a motion originating at $x\in \mathbb X$.

Now, we state and prove the following.

Theorem 7.2

Every C ₀ -semigroup $(T(t))_{t\in \mathbb R^+}$ determines a dynamical system and conversely by defining $u(t,x):=T(t)x,\;t\in \mathbb R^+,\;x\in \mathbb X$.

Proof

Let u(t, x) be a dynamical system in the sense of Definition 7.1 and consider

$$\displaystyle \begin{aligned}T(t)x=u(t,x),\;t\in \mathbb R^+,\;x\in\mathbb X.\end{aligned}$$

Then, obviously T(0) = I, the identity operator on $\mathbb X$, since for every $x\in \mathbb X$, T(0)x = u(0, x) = x.

Let $t,s\in \mathbb R^+$ and $x\in \mathbb X$; then, we have

$$\displaystyle \begin{aligned}T(t+s)x=u(t+s,x)=u(t,u(s,x))\end{aligned}$$

by property (iv) of Definition 7.1. But we have also

$$\displaystyle \begin{aligned}T(t)T(s)x=T(t)u(s,x)=T(t,u(s,x))\end{aligned}$$

using the definition of T(t)x. Therefore,

$$\displaystyle \begin{aligned}T(t+s)x=T(t)T(s)x,\end{aligned}$$

for every $t,s\in \mathbb R^+$ and $x\in \mathbb X$, which proves the semigroup property

$$\displaystyle \begin{aligned}T(t+s)x=T(t)T(s)x,\end{aligned}$$

for all $t,s\in \mathbb R^+$, $x\in \mathbb X$.

The continuity of $T(t)x:\mathbb X\to \mathbb X$ follows readily from property (iii) of Definition 7.1 for every $t\in \mathbb R^+$.

Now, we have

$$\displaystyle \begin{aligned}\displaystyle\lim_{t\to 0^+}T(t)x=\displaystyle\lim_{t\to o^+}u(t,x)=u(0,x)=x\end{aligned}$$

using property (ii) and then property (i) in Definition 7.1. We have proved that $(T(t))_{t\in \mathbb R^+}$ is a C ₀-semigroup.

Conversely, suppose we have a C ₀-semigroup $(T(t))_{t\in \mathbb R^+}$ and define

$$\displaystyle \begin{aligned}u(t,x)=T(t)x,\;t\in\mathbb R^+,\;x\in\mathbb X.\end{aligned}$$

Then, all the properties (i)–(iv) in Definition 7.1 are obviously true. Therefore, the mapping u is a dynamical system. □

Remark 7.3

The above result tells us that the notions of abstract dynamical systems and C ₀-semigroups are equivalent. This fact provides a solid ground to study C ₀-semigroups of linear operators as an independent topic.

2 Complete Trajectories

In this section, we will consider a C ₀-semigroup of linear operators $(T(t))_{t\in \mathbb R+}$ such that the motion $T(t)x_0:\mathbb R^+\to \mathbb X$ is an asymptotically almost automorphic function with principal term f(t).

Let us now introduce some notations and definitions. We recall that x ₀ is some fixed element in $\mathbb X$.

Definition 7.4

A function $\varphi :\mathbb R\to \mathbb X$ is said to be a complete trajectory of T if it satisfies the functional equation φ(t) = T(t − a)φ(a) for all $a\in \mathbb R$ and all t ≥ a.

We have the following properties.

Theorem 7.5

The principal term of T(t)x ₀ is a complete trajectory for T.

Proof

We have $T(t)x_0=f(t)+g(t),\;t\in \mathbb R^+$. Since $f\in AA(\mathbb X)$, there exists a subsequence $(n_k)\subset (n)=\mathbb N$ such that

$$\displaystyle \begin{aligned}\displaystyle\lim_{k\to\infty}f(t+n_k)=g(t)\end{aligned}$$

and

$$\displaystyle \begin{aligned}\displaystyle\lim_{k\to\infty}g(t-n_k)=f(t)\end{aligned}$$

pointwise on $\mathbb R$.

Put φ(t) = T(t)x ₀. Then, φ(0) = x ₀. Let us fix $a\in \mathbb R$ and choose k large enough so that a + n _k ≥ 0. If s ≥ 0, then

$$\displaystyle \begin{aligned} \begin{array}{rlll} \varphi(a+s+n_k)&=&T(a+s+n_k)\varphi(0)\\ &=&T(s)T(a+n_k)\varphi(0)\\ &=&T(s)\varphi(a+n_k). \end{array} \end{aligned}$$

Consequently,

$$\displaystyle \begin{aligned}f(a+s+n_k)+h(a+s+n_k)=T(s)\varphi(a+n_k),\end{aligned}$$

where s ≥ 0 and a + n _k ≥ 0. But we have

$$\displaystyle \begin{aligned}\displaystyle\lim_{k\to\infty}f(a+s+n_k)=g(a+s)\end{aligned}$$

and

$$\displaystyle \begin{aligned}\displaystyle\lim_{k\to\infty}h(a+s+n_k)=0,\end{aligned}$$

so

$$\displaystyle \begin{aligned}\displaystyle\lim_{k\to\infty}\varphi(a+s+n_k) =\displaystyle\lim_{k\to\infty}T(s)\varphi(a+n_k)=g(a+s).\end{aligned}$$

We also have

$$\displaystyle \begin{aligned}\displaystyle\lim_{k\to\infty}\varphi(a+n_k)=g(a).\end{aligned}$$

Using the continuity of T(t), we get

$$\displaystyle \begin{aligned}\displaystyle\lim_{k\to\infty}T(s)\varphi(a+n_k)=T(s)g(a).\end{aligned}$$

We can establish the following equality:

$$\displaystyle \begin{aligned}T(s)g(a)=g(a+s),\;\forall a\in\mathbb R,\;\forall s\geq 0.\end{aligned}$$

But we have

$$\displaystyle \begin{aligned}\displaystyle\lim_{k\to\infty}g(t-n_k)=f(t),\;t\in\mathbb R\end{aligned}$$

and

$$\displaystyle \begin{aligned}g(a-n_k+s)=T(s)g(a-n_k),\;\forall a\in\mathbb R,\;\forall s\geq 0.\end{aligned}$$

Therefore,

$$\displaystyle \begin{aligned}\displaystyle\lim_{k\to\infty}g(a-n_k+s)=T(s)f(a),\;\forall a\in\mathbb R,\;\forall s\geq 0,\end{aligned}$$

so that

$$\displaystyle \begin{aligned}f(a+s)=T(s)f(a),\;\forall a\in\mathbb R,\;\forall s\geq 0.\end{aligned}$$

Finally, let us put s = t − a with t ≥ 0. Then,

$$\displaystyle \begin{aligned}f(t)=T(t-a)f(a),\;\forall a\in\mathbb R,\;\forall t\geq a.\end{aligned}$$

The proof is complete. □

Definition 7.6

The set

$$\displaystyle \begin{aligned}\omega^+(x_0)=\{y\in \mathbb X\;:\;\exists\; 0\leq t_n\to\infty\; s.t. \displaystyle\lim_{n\to\infty}T(t_n)x_0=y\}\end{aligned}$$

is called the ω-limit of f(t), the principal term of T(t)x ₀.

$$\displaystyle \begin{aligned}\gamma^+(x_0)=\{T(t)x_0\;/\;t\in\mathbb R^+\}\end{aligned}$$

is called the trajectory of T(t)x ₀.

Theorem 7.7

ω ⁺(x ₀) ≠ ∅.

Proof

We let t _n = n, n = 1, 2, … Since $f\in AA(\mathbb X)$, there exists a subsequence $(t_{n_k})\subset (t_n)$ such that

$$\displaystyle \begin{aligned}\displaystyle\lim_{k\to\infty}f(t_{n_k})=g(0).\end{aligned}$$

But

$$\displaystyle \begin{aligned}\displaystyle\lim_{k\to\infty}T(t_{n_k})x_0=\displaystyle\lim_{k\to\infty}f(t_{n_k}).\end{aligned}$$

Thus, we get

$$\displaystyle \begin{aligned}\displaystyle\lim_{k\to\infty}T(t_{n_k})x_0=g(0).\end{aligned}$$

Consequently, g(0) ∈ ω ⁺(x ₀), since $t_{n_k} \to \infty$ as k →∞. So, ω ⁺(x ₀) ≠ ∅.

This completes the proof. □

Theorem 7.8

$\omega ^+(x_0)=\omega ^+_f(x_0)$.

Proof

To see that T(t)x ₀ and its principal term have the same ω-limit set, it suffices to observe that

$$\displaystyle \begin{aligned}\displaystyle\lim_{t\to\infty}T(t)x_0=\displaystyle\lim_{t\to\infty}f(t).\end{aligned}$$

□

Definition 7.9

A set $B\subset \mathbb X$ is said to be invariant under the semigroup $T=(T(t))_{t\in \mathbb R^+}$ if T(t)y ∈ B for every y ∈ B and $t\in \mathbb R^+$.

Theorem 7.10

ω ⁺(x ₀) is invariant under T.

Proof

Let y ∈ ω ⁺(x ₀), so there exists 0 ≤ t _n →∞ such that lim_t→∞ T(t _n)x ₀ = y. Consider the sequence (s _n) where s _n = t + t _n, n = 1, 2, …, for a given $t\in \mathbb R^+$. Then, s _n →∞ as n →∞. We have

$$\displaystyle \begin{aligned}T(s_n)x_0=T(t)T(t_n)x_0,\;n=1,2,\ldots\end{aligned}$$

and lim_n→∞ T(s _n)x ₀ = T(t)y, using the continuity of T(t). Therefore, T(t)y ∈ ω ⁺(x ₀).

The proof is complete. □

Theorem 7.11

ω ⁺(x ₀) is closed.

Proof

Let $y\in \overline {\omega ^+(x_0)}$, the closure of ω ⁺(x ₀). Then, there exists a sequence of elements y _m ∈ ω ⁺(x ₀), m = 1, 2, …, with y _m → y. For each y _m, there exists 0 ≤ t _m,n →∞ such that lim_n→∞ T(t _m,n)x ₀ = y _m.

Recursively choose

$$\displaystyle \begin{aligned} t_{1,n_1} >&\ \ 1& \text{such that}\ \ \ & \|y_1-T(t_{1,n_1})x_0\|<\frac{1}{2}\\ t_{2,n_2} >&\ \ max(2,t_{1,n_1})& \text{such that}\ \ \ & \|y_2-T(t_{2,n_2})x_0\|<\frac{1}{2^2}\\ t_{3,n_3} >&\ \ max(3,t_{2,n_2})& \text{such that}\ \ \ & \|y_3-T(t_{3,n_3})x_0\|<\frac{1}{2^3}\\ t_{k,n_k}>&\ \ max(k,t_{k-1,n_{k-1}})& \text{such that}\ \ \ & \|y_k-T(t_{k,n_k})x_0\|<\frac{1}{2^k}.\end{aligned} $$

Let $s_k=t_{k,n_k},\;k=1,2,\ldots$ Clearly, 0 < s _k →∞ as k →∞, and we have

$$\displaystyle \begin{aligned}\|T(s_k)x_0-y\| & \leq \|T(s_k)x_0-y_k\|+\|y_k-y\| \\ & <\frac{1}{2^k}+\|y_k-y\|.\end{aligned} $$

Since lim_k→∞ y _k = y, we have y ∈ ω ⁺(x ₀).

This achieves the proof. □

Theorem 7.12

ω ⁺(x ₀) is compact if γ ⁺(x ₀) is relatively compact.

Proof

It is obvious that $\omega ^+(x_0)\subset \overline {\gamma ^+(x_0)}$. But $\overline {\gamma ^+(x_0)}$ is compact by assumption and ω ⁺(x ₀) is a closed subset (cf. Theorem 7.11). Therefore, ω ⁺(x ₀) is itself compact. □

Theorem 7.13

$\gamma _f(x_0):=\{f(t)\;/\;t\in \mathbb R\}$ is invariant under the semigroup $T=(T(t))_{t\in \mathbb R^+}$.

Proof

We recall that γ _f(x ₀) is relatively compact since $f\in AA(\mathbb X)$. Let y ∈ γ _f(x ₀). So, there exists $\sigma \in \mathbb R$ such that y = f(σ). For arbitrary $a\in \mathbb R$ such that σ ≥ a, we can write

$$\displaystyle \begin{aligned}y=f(\sigma)=T(\sigma-a)f(a),\end{aligned}$$

since f is a complete trajectory (cf. Theorem 7.5). Now, let t ≥ 0. Then,

$$\displaystyle \begin{aligned} \begin{array}{rlll} T(t)y&=&T(t+\sigma-a)f(a)\\ &=&f(t+\sigma), \end{array} \end{aligned}$$

i.e. T(t)y ∈ γ _f(x ₀), for every t ≥ 0, which shows that γ _f(x ₀) is indeed invariant under the semigroup T. □

Theorem 7.14

Let $\nu (t):=\inf \limits _{y\in \omega ^+(x_0)}\|T(t)x_0-y\|$ . Then,

$$\displaystyle \begin{aligned}\displaystyle\lim_{t\to\infty}\nu(t)=0.\end{aligned}$$

Proof

Suppose lim_t→∞ ν(t) ≠ 0. Then, there exists ε > 0 such that for every n = 1, 2, …, there exists $t^{\prime }_n\geq n$ such that $\nu (t^{\prime }_n)\geq \varepsilon$, i.e.

$$\displaystyle \begin{aligned}\exists t^{\prime}_n\geq n,\;\;\|T(t^{\prime}_n)x_0-y\|\geq \varepsilon\;\;\;\forall y\in\omega^+(x_0),\;\;\forall n=1,2,\ldots\end{aligned}$$

Let (t _n) be a subsequence of $(t^{\prime }_n)$ such that (f(t _n)) converges, say, to $\overline {y}$, as guaranteed by the relative compactness of γ _f(x ₀).

Now, since t _n →∞ as n →∞, we get

$$\displaystyle \begin{aligned}\displaystyle\lim_{n\to\infty}T(t_n)x_0=\displaystyle\lim_{n\to\infty}f(t_n)=\overline{y}.\end{aligned}$$

Therefore, $\overline {y}\in \omega ^+(x_0)$, which is a contradiction. □

Remark 7.15

The minimality property above shows that the ω-limit set ω ⁺(x ₀) is the smallest closed set toward which the asymptotically almost automorphic function T(t)x ₀ tends to as t →∞.

Definition 7.16

$e\in \mathbb X$ is called a rest point for the semigroup $T=(T(t))_{t\in \mathbb R^+}$ if $T(t)e=e,\;\forall t\in \mathbb R^+$.

Theorem 7.17

If x ₀ is a rest point for the semigroup $T=(T(t))_{t\in \mathbb R^+}$ , then ω ⁺(x ₀) = {x ₀}.

Proof

Since T(t)x ₀ = x ₀, for every t ≥ 0, then for every sequence of real numbers (t _n) such that 0 ≤ t _n →∞, we get lim_n→∞ T(t _n)x ₀ = x ₀, i.e. x ₀ ∈ ω ⁺(x ₀).

Now, let y ∈ ω ⁺(x ₀). There exists 0 ≤ s _n →∞ such that lim_n→∞ T(s _n)x ₀ = y. But T(s _n)x ₀ = x ₀ for all n = 1, 2, … Therefore, x ₀ = y.

The proof is complete. □

Bibliographical Notes

The results in this chapter are due to N’Guérékata and published for the first time in the first edition of this book.