I am reading, the following paper for my research:
http://www.sciencedirect.com/science/article/pii/S0022247X00973438
I need to know what are the precise definitions of the following terms, since I don't have any book regarding dynamical systems. (Any recommendation would be useful).
First they define $\omega(\partial Y_0)\subset \partial Y_0$ to be the union of sets $\omega(u)$ for $u \in \partial Y_0$. I am assuming that $\omega$ refers to the $\omega-$limit set. (They say that they differ, but I don't see the difference.) Is $\omega(\partial Y_0)$ the set of all equilibrium points?
THEOREM:
Suppose that $Y$ is a complete metric space with $Y = Y_0\cup\partial Y_0$, where $Y_0$ is open. Suppose that a semi-flow on $Y$ leaves both $Y_0$ and $\partial Y_0$ forward invariant, maps bounded sets in $Y$ to pre-compact sets for $t>0$ and is dissipative. If in addition, $$\omega(\partial Y_0) \text{ is isolated and acyclic,}$$ $$W^s(M_k)\cap Y_0 = \emptyset,$$ for all $k$, where $\bigcup_{k=1}^NM_k$ is the isolated covering used in the definition of acyclicity of $\partial Y_0$, then the semi-flow is permanent; i.e., there exists $\epsilon>0$ such that any trajectory with initial data in $Y_0$ will be bounded away from $\partial Y_0$ by a distance greater that $\epsilon$ for $t$ sufficiently large. $W^s$ denotes the stable manifold.
- What is a flow or semi-flow?
I was wondering if It is a solution of a trajectory in $Y$.
- Permenent = there exists $\epsilon>0$ such that any trajectory with initial data in $Y_0$ will be bounded away from $\partial Y_0$ by a distance greater that $\epsilon$ for $t$ sufficiently large.
This means that a the solution don't touch the boundaries? IN particular, never approach zero?
- Dissipative: A system is said to be dissipative if there is a fixed bounded set $X_0\subseteq Y$, such that if $y(t)$ is any trajectory, $y(t) \in X_0$ for all sufficiently large $t$. How large $t$ must depends on $y(0)$.
This means that after sometime, the solution remains in $X_0$?
- Invariant set: $S\subset \mathbb{R}^n$ is an invariant set for a flow $\phi_t$ or a map $G$ on $\mathbb{R}^n$ if $\phi_t(x) = \phi(x,t) \in S$ (or $G(x) \in S$) for $x \in S$ and for all $t \in \mathbb{R}$.
This means that solutions don't escape $S$ right? Forward invariant means solutions in $\mathbb{R}^+$ or solutions that goes to the right?
- Acyclic = The set $\omega(\partial Y_0)$ is said to be acyclic if there exists an isolated covering $\bigcup_{k=1}^N M_k$ such that no subset $\{M_k\}$ is a cycle.
No equilibrium in $\omega(\partial Y_0)$ are cycles, right?
- Isolated = An invariant set $S$ for the flow or semi-flow is said to be isolated if it has a neighborhood $U$ such that $S$ is the maximal invariant subset of $U$.
This means that $S \subseteq U$ and no other $S'$ invariant set have the property $S \subseteq S' \subseteq U$?
This is the application of the theorem to my research. First, under the assumption $\frac{A}{c} > \frac{b}{\delta}$ implies that the equilibrium point $(0,\frac{b}{\delta})$ is a saddle point. I already prove that the system is dissipative.
Theorem: Let us assume that $$\frac{A}{c} > \frac{b}{\delta}$$ then the system is permanent.
Proof: Let $Y_0 = \mathbb{R}^2$. Then, $\omega(\partial Y_0)$ consist of the equilibruim: $$(0,0) \text{ is an unstable node},$$ $$(1,0) \text{ is an saddle point},$$ $$(0,\frac{b}{\delta}) \text{ is a saddle point.}$$ The equilibrium $(1,0)$ has the $x$-axis as the stable manifold and the $y$-axis as the unstable manifold. Similarly, the equilibrium $(0,\frac{b}{\delta})$ has the $y$-axis as the stable manifold and the $x$-axis as the unstable manifold. Now, all the trajectories on the $x-$ axis other than $(0,0)$ approach $(1,0)$ and all the trajectories on the $y$-axis other than $(0,0)$ approach $(0,\frac{b}{\delta})$. It follows from the structural features that $\omega(\partial Y_0)$ is acyclic, Thus, $\omega(\partial Y_0)$ is isolated and acyclic. Since the stable manifold of $(1,0)$ is the $x$-axis and the stable manifold of $(0,\frac{b}{\delta})$ is the $y$-axis, they dont intersect $Y_0$ and since the system is dissipative, then we have that the system is permanent.
I am not understanding the argument, It follows from the structural features that $\omega(\partial Y_0)$ is acyclic, Thus, $\omega(\partial Y_0)$ is isolated and acyclic.