Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I would like some help understanding the proof of lemma 6.1 given here, for the case of an autonomous Retarded Functional Differential Equation (RFDE).

The problem for the autonomous case (In the proof of the book, drop the dependence on $t$ in the first component and suppose $\sigma=0$) is as follows.

Background:

Suppose $r$ is fixed. Let $C=C([-r,0],\mathbb{R})$.

An autonomous RFDE is a differential equation of the form $$x'(t)=f(x_t)$$ where $x_t \in C$ defined by $x_t(\theta)=x(t+\theta)$ for $\theta \in [-r,0]$.

$f$ is a continuous function from $C$ to $\mathbb R$.

Given an initial function $\phi \in C$ suppose there exists a function $x \in C[-r,A)$ such that $x_0=\phi$ and $x$ satisfies the RFDE given above. Then $x$ is called a solution of the RFDE with initial function $\phi$.

In such a case when there is a unique solution of the RFDE, define the solution map $T_t:C\to C$ by $T_t(\phi)=x_t$, where $x$ is the solution of the RFDE with initial function $\phi$.

Question

Show that $T_t$ is locally completely continuous. That is for every $\phi \in C$ there is a neighbourhood $V_t$ of $\phi$ such that $T_t(V_t)$ is contained in a compact set of $C$.

What I don't understand in the proof is how they conclude that there is a nbd $V_t$ of $\phi$ and a constant $M$ such that $|T_\tau V_t|_C \leq M$ and $|f(T_\tau V_t)|<M$ for all $0 \leq \tau \leq t$?

How does one get such $V_t$ and $M$ that is uniform for all $0 \leq \tau \leq t$.

If someone with some experience in this area reads the question, I would appreciate it if you can suggest me some references to learn about Autonomous RFDE's.

share|improve this question
3  
Is "retarded" politically correct? You dont want to upset a differential equation. Believe me. –  CogitoErgoCogitoSum Feb 24 '13 at 19:38
2  
:) I didn't name these equations. –  Cantor Feb 24 '13 at 19:40
1  
Differently-abled function. –  Eric Gregor Feb 27 '13 at 2:58

1 Answer 1

up vote 1 down vote accepted

I retyped the argument making the simplifications indicated in the OP.

Lemma 6.1 For any $t\ge r$, the map $T_t$ is locally completely continuous; that is, for any $t\ge r$, $\phi\in C$, there is a neighborhood $V_t$ of $\phi$ such that $T_t(V_t)$ is contained in a compact subset of $C$.

Proof. Since $f$ is continuous and $T_\tau\psi$ is jointly continuous in $\tau,\psi$, for any $t\ge 0 $ there is a neighborhood $V_t$ of $\phi$ and a constant $M$ such that $|T_\tau (V_t)|\le M$, $|f(T_t(V_t))|\le M$ for $0\le \tau\le t$. Thus $|\dot x(V_t(\tau))|\le M$ for $0\le \tau\le t$. This implies the family of functions $\{x_t(\psi):\psi\in V_t\}$ is precompact for $t\ge r$. $\Box$


This is how I understand the above. Since $T$ is jointly continuous in $\tau$ and $\psi$, for every $\tau_0\in [0,t]$ there exists $\delta>0$ such that $$\|T_\tau \psi-T_{\tau_0}\phi\|\le 1\quad \text{whenever }\ |\tau-\tau_0|<\delta \text{ and } \|\psi-\phi\|<\delta \tag1$$ Using the compactness of the interval $[0,t]$, we can cover it by finitely many intervals $(\tau_i-\delta_i ,\tau_i+\delta_i)$ of the above nature. Now let $$M=1+\max_i \|T_{\tau_i}\phi\|,\quad r=\min_i \delta_i \tag2$$ It follows that $\|T_\tau \psi\|\le M$ whenever $\|\psi-\phi\|<r$ and $\tau\in [0,t]$.

The same argument yields a uniform bound for $f(T_\tau \psi)$ in some neighborhood of $\phi$: you only need to replace (1) with $$|f(T_\tau \psi)-f(T_{\tau_0}\phi)|\le 1\quad \text{whenever }\ |\tau-\tau_0|<\delta \text{ and } \|\psi-\phi\|<\delta \tag{3}$$ Then take the intersection of the two neighborhoods of $\phi$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.