Dense Short Solution Segments from Monotonic Delayed Arguments

Abstract

We construct a delay functional d on an open subset of the space $C^1_r=C^1([-r,0],\mathbb {R})$ and find $h\in (0,r)$ so that the equation

$$\begin{aligned} x'(t)=-x(t-d(x_t)) \end{aligned}$$

defines a continuous semiflow of continuously differentiable solution operators on the solution manifold

$$\begin{aligned} X=\{\phi \in C^1_r:\phi '(0)=-\phi (-d(\phi ))\}, \end{aligned}$$

and along each solution the delayed argument $t-d(x_t)$ is strictly increasing, and there exists a solution whose short segments

$$\begin{aligned} x_{t,short}=x(t+\cdot )\in C^2_h,\quad t\ge 0, \end{aligned}$$

are dense in an infinite-dimensional subset of the space $C^2_h$. The result supplements earlier work on complicated motion caused by state-dependent delay with oscillatory delayed arguments.

1 Introduction

The present paper continues the studies [6, 10,11,12,13,14] of how time lags which are state-dependent affect the behaviour of feedback systems. The basic equation considered is

with $\alpha >0$ and constant time lag $r>0$. This is the simplest delay differential equation modelling negative feedback with respect to the zero solution. Let $C^0_r$ denote the Banach space of continuous functions $[-r,0]\rightarrow \mathbb {R}$ with the maximum norm, $|\phi |_{0,r}=\max _{-r\le t\le 0}|\phi (t)|$. The solutions $x:[-r,\infty )\rightarrow \mathbb {R}$ of Eq. $(\alpha ,r)$, which are continuous and have differentiable restrictions to $[0,\infty )$ which satisfy Eq. $(\alpha ,r)$, define a strongly continuous semigroup on $C^0_r$ by the equations $T(t)x_0=x_t$ with the solution segments

$$\begin{aligned} x_t:[-r,0]\ni s\mapsto x(t+s)\in \mathbb {R}\quad \text {for}\quad t\ge 0, \end{aligned}$$

see [2]. Except for $\alpha =\frac{\pi }{2}+2k\pi$, $k\in \mathbb {N}_0$ the zero solution is hyperbolic [2, 15].

Let $C^1_r$ denote the Banach space of continuously differentiable functions $\phi :[-r,0]\rightarrow \mathbb {R}$, with the norm given by $|\phi |_{1,r}=|\phi |_{0,r}+|\phi '|_{0,r}$. In [6, 10,11,12] delay functionals $d:C^1_r\supset U\rightarrow [0,r]$ were constructed so that for certain $\alpha >0$ the modified equation

has homoclinic solutions, with chaotic motion nearby.

The results in [13, 14] established another kind of complicated solution behaviour, namely, the existence of delay functionals d and parameters $\alpha >0$ so that for a positive number $h<r$ there are solutions whose short solution segments

$$\begin{aligned} x_{t,short}:[-h,0]\ni s\mapsto x(t+s)\in \mathbb {R}, t\ge 0, \end{aligned}$$

are dense in open subsets of the space $C^1_h$.

In [13] density of short segments in the whole space $C^1_h$ was achieved for a continuous delay functional on a set $Y\subset C^1_r$ which is large in some sense but not open, nor a differentiable submanifold. Because of this lack of regularity results from [8, 9] on well-posedness of initial value problems and on differentiability of solutions with respect to initial data do not apply.

In [14] we constructed a continuously differentiable delay functional $d: U\rightarrow [0,r]$, $U\subset C^1_r$ open, so that the results from [8] apply, and found $h\in (0,r)$ so that the previous equation with $\alpha =1$, namely,

$$\begin{aligned} x'(t)=-x(t-d(x_t)) \end{aligned}$$

(1.1)

has a solution $x:[-r,\infty )\rightarrow \mathbb {R}$ whose short segments are dense in an open subset of the space $C^1_h$. The construction involves that the delayed argument function

$$\begin{aligned}{}[0,\infty )\ni t\mapsto t-d(x_t)\in \mathbb {R}\end{aligned}$$

along the solution x is not monotonic, and this oscillatory behaviour seems crucial for density of short segments in an open subset of the space $C^1_h$.

Before stating the result of the present paper let us mention that equations with non-constant, state-dependent delay are not covered by the theory with state space $C^0_r$ which is familiar from monographs on delay differential equations [1,2,3]. We recall what was shown in [8] for delay differential equations in the general form

$$\begin{aligned} x'(t)=f(x_t) \end{aligned}$$

(f)

under hypotheses designed for applications to examples with state-dependent delay. Let $C^0_{r,n}$ and $C^1_{r,n}$ denote the analogues of the spaces $C^0_r$ and $C^1_r$, for maps $[-r,0]\rightarrow \mathbb {R}^n$. Assume $f:U\rightarrow \mathbb {R}^n$, $U\subset C^1_{r,n}$ open, is continuously differentiable so that

(e) each derivative $Df(\phi ):C^1_{r,n}\rightarrow \mathbb {R}^n$, $\phi \in U$, has a linear extension $D_ef(\phi ):C^0_{r,n}\rightarrow \mathbb {R}^n$ and the map

$$\begin{aligned} U\times C^0_{r,n}\ni (\phi ,\chi )\mapsto D_ef(\phi )\chi \in \mathbb {R}^n \end{aligned}$$

is continuous.

The extension property (e) is a variant of the notion of being almost Fréchet differentiable for maps $C^0_{r,n}\supset V\rightarrow \mathbb {R}^n$ which was introduced in [7].

Suppose also there exists $\phi \in U$ with $\phi '(0)=f(\phi )$. Then the nonempty set

$$\begin{aligned} X_f=\{\phi \in U:\phi '(0)=f(\phi )\} \end{aligned}$$

is a continuously differentiable submanifold with codimension n in $C^1_{r,n}$, and each initial value problem

$$\begin{aligned} x'(t)=f(x_t)\quad \text {for}\quad t>0,\quad x_0=\phi \in X_f, \end{aligned}$$

has a unique maximal solution $x:[-r,t_{\phi })\rightarrow \mathbb {R}^n$, $0<t_{\phi }\le \infty$, which is continuously differentiable with $x'(t)=f(x_t)$ for all $t\in [0,t_{\phi })$. The arrow

$$\begin{aligned} (t,\phi )\mapsto x^{\phi }_t, \end{aligned}$$

with the said maximal solution $x=x^{\phi }$, defines a continuous semiflow of continuously differentiable solution operators

$$\begin{aligned} \{\phi \in X_f: t_{\phi }>t\}\ni \phi \mapsto x^{\phi }_t\in X_f,\quad t\ge 0. \end{aligned}$$

In the present paper we prove the following result on complicated motion caused by a delay functional so that the delayed argument functions along solutions of Eq. (1.1) are monotonic.

Theorem 1.1

There exist $r>h>0$ and a continuously differentiable delay functional $d:N\rightarrow (0,r)$, $N\subset C^1_r$ open, and an open subset A of a closed affine subspace of codimension 6 in $C^2_h$ so that Eq. (1.1) has a twice continuously differentiable solution $x^{(d)}:[-r,\infty )\rightarrow \mathbb {R}$ whose short segments $x^{(d)}_{t,short}$, $t\ge 0$, are dense in $A\cup (-A)$.

The functional $f:N\ni \phi \mapsto -\phi (-d(\phi ))\in \mathbb {R}$ is continuously differentiable and has property (e), and for each $\phi \in X_f$ the delayed argument function

$$\begin{aligned}{}[0,t_{\phi })\ni t\mapsto t-d(x^{\phi }_t)\in \mathbb {R}\end{aligned}$$

along the maximal continuously differentiable solution $x^{\phi }:[-r,t_{\phi })\rightarrow \mathbb {R}$ of the initial value problem

$$\begin{aligned} x'(t)=f(x_t)=-x(t-d(x_t))\quad \text {for}\quad t>0,\quad x_0=\phi \in X_f, \end{aligned}$$

is strictly increasing.

Here $C^2_h$ denotes the Banach space of twice continuously differentiable functions $\psi :[-h,0]\rightarrow \mathbb {R}$, with the norm given by $|\psi |_{2,r}=\sum _{k=0}^2\max _{-r\le t\le 0}|\psi ^{(k)}(t)|$.

A different result on complicated motion caused by state-dependent delay with monotonic delayed argument functions has recently been obtained in [5].

The proof of Theorem 1.1 begins in Sect. 2 below with the choice of subsets $A=A_h\subset C^2_h$ as in the theorem, for arbitrary $h>0$. For arbitrary $s>0$ Sect. 3 prepares a sequence of twice continuously differentiable functions $\kappa _{s,n}:[-s,s]\rightarrow \mathbb {R}$ so that certain translates of $\kappa _{s,n}$ and $\kappa _{s,k}$, $n\ne k$, keep a minimal distance from each other, in the sense that there is a constant $a>0$ with

$$\begin{aligned} |(\kappa _{s,n})'(t+u)-(\kappa _{s,k})'(u)|\ge \frac{a}{4} \end{aligned}$$

for small t and some u.

Section 4 is the core of the proof of Theorem 1.1. For suitably chosen $t_b<0<t_5$, $h>0$, $s>0$, a sequence of continuously differentiable delay functions $\Delta _n:[0,t_5]\rightarrow (0,\infty )$ together with a sequence of twice continuously differentiable functions $x_{(n)}:[t_b,t_5]\rightarrow \mathbb {R}$ and a subset $A=A_h\subset C^2_h$ as in Sect. 2 are constructed so that for each $n\in \mathbb {N}$ - the linear nonautonomous equation

$$\begin{aligned} (x_{(n)})'(t)=-x_{(n)}(t-\Delta _n(t)) \end{aligned}$$

holds for $0\le t\le t_5$ ,

• the delayed argument function $[0,t_5]\ni t\mapsto t-\Delta _n(t)\in \mathbb {R}$ along the delay function $\Delta _n$ is strictly increasing,

• on some subinterval of length h in $[0,t_5]$ the function $x_{(n)}$ coincides with a translate of a member $p_n$ of a sequence which is dense in A,

• on some subinterval of length 2s in $[0,t_5]$ the function $x_{(n)}$ coincides with a translate of $\kappa _n=\kappa _{s,n}$.

In Sect. 5 shifted copies of the functions $\Delta _n$ and of the functions $\pm x_{(n)}$ are concatenated, respectively, and yield a twice continuously differentiable function $x:[t_b,\infty )\rightarrow \mathbb {R}$ and a continuously differentiable delay function $\Delta$ on $[0,\infty )$ which is bounded by some $r>\max \{h,-t_b\}$. A twice continuously differentiable extension of the function x to the ray $[-r,\infty )\rightarrow \mathbb {R}$ satisfies the linear equation

$$\begin{aligned} x'(t)=-x(t-\Delta (t)) \end{aligned}$$

(1.2)

for all $t\ge 0$. Proposition 5.1 states that the curve $[r,\infty )\ni t\mapsto x_t\in C^1_r$ is injective, hence the equation

$$\begin{aligned} d(x_t)=\Delta (t) \end{aligned}$$

converts the delay function into a delay functional d on the trace $\{x_t\in C^2_r:t\ge r\}$.

Sections 6, 7, and 8 prepare the extension of this functional to an open neighbourhood N of the trace $\{x_t\in C^2_r:(j_r-1)t_5\le t\}$ in the space $C^1_r$, with an integer $j_r\ge 2$ so that $r<(j_r-1)t_5$. Section 6 contains an ingredient of the construction which will be used in the final Sect. 9, namely, separation of nonadjacent arcs

$$\begin{aligned} \{x_t\in C^2_r:(n-1)t_5\le t\le nt_5\}&\text {and}&\{x_t\in C^2_r:(j-1)t_5\le t\le jt_5\},\\ 2\le n\in \mathbb {N}\quad \text {and}\quad 2\le j\in \mathbb {N}&\text {with}&|n-j|>1, \end{aligned}$$

in the space $C^1_r$. The separation result is based on the properties of the functions $\kappa _{s,n}$ from Sect. 3 whose translates appear as restrictions of x on a sequence of mutually disjoint intervals tending to infinity.

The constructions in Sects. 2, 3, 4, 5, and 6 are to some extent parallel to constructions in [14]. The next steps in Sects. 7 and 8 are rather different from their counterparts in [14]. The new tool, introduced in Sect. 7, is a bundle of transversal hyperplanes $K_t$, $t>0$, along the curve $(0,\infty )\ni t\mapsto x_t\in C^0_r$. Working with the bundle allows for an extension of the delay functional from an arc $\{x_t\in C^2_r:(k-1)t_5\le t\le kt_5\}$, $j_r\le k\in \mathbb {N}$, to a kind of tubular neighbourhood $U_k\subset C^0_r$ (Sect. 8), and for the arrangement of compatibility relations on overlapping domains $U_k\cap U_{k+1}$, in ways which are simpler than corresponding procedures in [14].

Section 9 begins with the definition of the domain $N\subset C^1_r$ and the functional $d:N\rightarrow (0,r)$, and completes the proof of Theorem 1.1. The verification that the functional $f:N\rightarrow \mathbb {R}$ in Theorem 1.1 has property (e) uses that the delay functional $d:N\rightarrow (0,r)$ has property (e). The latter is achieved by means of the following proposition whose statement involves the injective linear continuous inclusion map

$$\begin{aligned} J:C^1_r\ni \phi \mapsto \phi \in C^0_r. \end{aligned}$$

Proposition 1.2

[14, Proposition 1.2] Suppose $d:C^1_r\supset N\rightarrow \mathbb {R}$ is continuously differentiable and for every $\phi \in N$ there exist an open neighbourhood V of $J\phi$ in $C^0_r$ and a continuously differentiable map $d_V:C^0_r\supset V\rightarrow \mathbb {R}$ with $d(\psi )=d_V(J\psi )$ for all $\psi \in N\cap J^{-1}(V)$. Then d has property (e), with

$$\begin{aligned} D_ed(\phi )\chi =Dd_V(J\phi )\chi \quad \text {for all}\quad \phi \in N\cap J^{-1}(V)\quad \text {and}\quad \chi \in C^0_r. \end{aligned}$$

Notation, preliminaries. A sequence in a metric space is called dense if each point of the metric space is an accumulation point of the sequence. A metric space is called separable if it contains a dense sequence.

For $\epsilon >0$ the open $\epsilon$-neighbourhoods of a point x in a normed space X and of a subset $S\subset X$ are given by

$$\begin{aligned} U_{\epsilon }(x)=\{y\in X:|y-x|<\epsilon \}. \end{aligned}$$

and

$$\begin{aligned} U_{\epsilon }(S)=\{y\in X:\text {dist}(y,S)<\epsilon \}, \end{aligned}$$

respectively, with

$$\begin{aligned} \text {dist}(y,S)=\inf _{x\in S}|y-x|. \end{aligned}$$

For $a<b$ in $\mathbb {R}$ and $j\in \mathbb {N}$ let $C^j_{a,b}$ denote the Banach space of j times continuously differentiable functions $\phi :[a,b]\rightarrow \mathbb {R}$, with the norm given by

$$\begin{aligned} |\phi |_{j,a,b}=\sum _0^j\max _{a\le t\le b}|\phi ^{j}(t)|, \end{aligned}$$

and let $C^0_{a,b}$ denote the Banach space of continuous functions $\phi :[a,b]\rightarrow \mathbb {R}$, with the norm given by

$$\begin{aligned} |\phi |_{0,a,b}=\max _{a\le t\le b}|\phi (t)|. \end{aligned}$$

In case $a=-r$ and $b=0$, the abbreviations

$$\begin{aligned} C^j_r=C^j_{-r,0}\quad \text {and}\quad |\cdot |_{j,r}=|\cdot |_{j,-r,0} \end{aligned}$$

are used. If functions $\phi \in C^2_r$ and $\phi \in C^1_r$ are considered as elements of the ambient space $C^0_r$ then we use $\phi \in C^0_r$ or $J\phi \in C^0_r$, depending on which form makes an argument more transparent.

For $r>0$ the evaluation map

$$\begin{aligned} C^0_r\times [-r,0]\ni (\phi ,t)\mapsto \phi (t)\in \mathbb {R}\end{aligned}$$

is continuous but not locally Lipschitz continuous, and the evaluation map

$$\begin{aligned} ev^1_r:C^1_r\times (-r,0)\ni (\phi ,t)\mapsto \phi (t)\in \mathbb {R}\end{aligned}$$

is continuously differentiable with

$$\begin{aligned} D\,ev^1_r(\phi ,t)(\hat{\phi },\hat{t})=D_1ev^1_h(\phi ,t)\hat{\phi }+D_2ev^1_r(\phi ,t)\hat{t}=\hat{\phi }(t)+\hat{t}\phi '(t), \end{aligned}$$

see e. g. [4, 8].

In Sect. 8 below the following is used.

Proposition 1.3

Let B be a Banach space. Let reals $a<b$, a continuous injective map $c:[a,b]\rightarrow B$, some $t\in (a,b)$, and $\epsilon >0$ be given. Then there exists $\rho >0$ with

$$\begin{aligned} U_{\rho }(c([a,t])\cap U_{\rho }(c([t,b])\subset U_{\epsilon }(c(t)). \end{aligned}$$

Proof

By continuity there exists $t_a\in (a,t)$ with $c([t_a,t])\subset U_{\epsilon /2}(c(t))$. The compact sets $c([a,t_a])$ and c([t, b]) are disjoint, which gives

$$\begin{aligned} 0<\min _{a\le u\le t_a}dist(c(u),c([t,b])). \end{aligned}$$

Choose $\rho \in \left( 0,\frac{\epsilon }{2}\right)$ with

$$\begin{aligned} 2\,\rho <\min _{a\le u\le t_a}dist(c(u),c([t,b])). \end{aligned}$$

Consider $z\in U_{\rho }(c([a,t]))\cap U_{\rho }(c([t,b]))$. There exist $u_a\in [a,t]$ and $u_b\in [t,b]$ with

$$\begin{aligned} |z-c(u_a)|<\rho \quad \text {and}\quad |z-c(u_b)|<\rho , \end{aligned}$$

hence $|c(u_a)-c(u_b)|<2\rho$. The assumption $u_a<t_a$ yields a contradiction to the inequality $2\rho <\min _{a\le u\le t_a}dist(c(u),c([t,b]))$. It follows that $u_a\in [t_a,t_b]$. Consequently,

$$\begin{aligned} |z-c(t)|\le |z-c(u_a)|+|c(u_a)-c(t)<\rho +\frac{\epsilon }{2}<\epsilon , \end{aligned}$$

which means $z\in U_{\epsilon }(c(t))$.

2 Separability

Let $h>0$ be given. The restrictions of polynomials $\mathbb {R}\rightarrow \mathbb {R}$ to the interval $[-h,0]$ are dense in $C^2_h$, which is an easy consequence of the Weierstraß approximation theorem. Let $P_5\subset C^2_h$ denote the subspace of restrictions of polynomials of degree not larger than 5 and let $C^2_{h-0}\subset C^2_h$ denote the closed subspace given by the equations

$$\begin{aligned} \phi ^{(j)}(-h)=0=\phi ^{(j)}(0)\quad \text {for}\quad j\in \{0,1,2\}. \end{aligned}$$

Then $\dim \,P_5=6$ and

$$\begin{aligned} C^2_h=C^2_{h-0}\oplus P_5, \end{aligned}$$

which follows from the fact that given $\phi \in C^2_h$ there exists a unique $p\in P_5$ satisfying

$$\begin{aligned} p^{(j)}(-h)=\phi ^{(j)}(-h)\,\,\text {and}\,\,p^{(j)}(0)=\phi ^{(j)}(0)\quad \text {for}\quad j\in \{0,1,2\}, \end{aligned}$$

or, $\phi -p\in C^2_{h-0}$.

Proposition 2.1

Let an open set $U\subset C^2_h$ and $p_{*}\in C^2_h$ with $A=U\cap (p_{*}+C^2_{h-0})\ne \emptyset$ be given. The open subset A of the affine space $p_{*}+C^2_{h-0}$ contains a sequence which is dense in A.

Proof

The restricted polynomials with rational coefficients form a sequence which is dense in $C^2_h$. Projection along $P_5$ onto $C^2_{h-0}$ yields a sequence which is dense in $C^2_{h-0}$, and translation by adding $p_{*}$ results in a sequence which is dense in $p_{*}+C^2_{h-0}$. The members of this sequence which belong to U form a sequence which is dense in A.

Example 2.2

For given reals $w_0<u_0<0$, $u_1<w_1<0$, $u_2>0$, $w_2>0$ let $p_{*}\in P_5$ denote the unique restricted polynomial which satisfies

$$\begin{aligned} p_ {*}^{(j)}(-h)=u_j,\,\,p_ {*}^{(j)}(0)=w_j\quad \text {for}\quad j\in \{0,1,2\}, \end{aligned}$$

and take

$$\begin{aligned} U=\{\phi \in C^2_h:0<\phi ''(t)\,\,\text {on}\,\,[-h,0]\}. \end{aligned}$$

Notice that

$$\begin{aligned} A= & {} U\cap (p_{*}+C^2_{h-0})\\= & {} \{\phi \in C^2_h:0<\phi ''(t)\,\,\text {on}\,\,[-h,0],\\&\phi ^{(j)}(-h)=u_j,\,\,\phi ^{(j)}(0)=w_j\quad \text {for}\quad j\in \{0,1,2\}\}. \end{aligned}$$

We add the obvious fact that the dense sequence provided by Proposition 2.1 is dense in $A\subset C^2_h\subset C^1_h$ also with respect to the norm $|\cdot |_{1,h}$.

3 Differentiable Functions with Separated Shifted Copies

Let $s>0$ be given. We construct a sequence of functions $\kappa _n\in C^2_{-s,s}$, $n\in \mathbb {N}$, so that shifted copies of these functions keep a positive minimal distance from each other with espect to the norm $|\cdot |_{1,-s,s}$.

Let also positive reals $a,\xi ,\eta$ be given and choose $\epsilon \in \left( 0,\frac{a}{4}\right)$. There exists $\chi \in C^1_{-s,0}$ with

$$\begin{aligned} \chi (-s)= & {} -a,\\ \chi ([-s,0])\subset & {} [-a,-a+\epsilon ],\\ \chi '(-s)=\eta ,\\ \chi '(t)> & {} 0\quad \text {on}\quad [-s,0]. \end{aligned}$$

For every $n\in \mathbb {N}$ there exists $\rho _n\in C^1_{-s,s}$ with

$$\begin{aligned} \rho _n(t)=-\rho _n(-t)\quad \text {on}\quad [-s,s] \end{aligned}$$

and

$$\begin{aligned} \rho _n(t)= & {} \chi (t)\quad \text {on}\quad \left[ -s,-\frac{s}{2^n}\right] ,\\ \rho _n\left( -\frac{s}{2^{n+1}}\right)= & {} -\frac{a}{2},\\ \rho _n(0)= & {} 0,\\ (\rho _n)'(t)= & {} (\rho _n)'(0)\,\,\text {constant on}\,\,\left[ -\frac{s}{2^{n+1}},0\right] ,\\ (\rho _n)'(t)> & {} 0\quad \text {on}\quad [-s,0]. \end{aligned}$$

Fig. 1

The function $\rho _1$ for $-s\le t\le 0$

Proposition 3.1

For all integers $n\ne k$ in $\mathbb {N}$ and for each $t\in \left[ -\frac{s}{2},0\right]$ there exists $u\in [-s,s]$ with $t+u\in [-s,s]$ and

$$\begin{aligned} |\rho _n(t+u)-\rho _k(u)|\ge \frac{a}{2}-\epsilon . \end{aligned}$$

Proof

Let positive integers $n\ne k$ and $t\in \left[ -\frac{s}{2},0\right]$ be given. In case $n>k$ consider $u=-\frac{s}{2^{k+1}}$. Then $u\in \left[ -\frac{s}{4},0\right]$ and

$$\begin{aligned} -s\le -\frac{s}{2}-\frac{s}{2^{k+1}}\le t+u\le u=-\frac{s}{2^{k+1}}\le -\frac{s}{2^n}, \end{aligned}$$

hence

$$\begin{aligned} \rho _n(t+u)-\rho _k(u)= & {} \chi (t+u)-\left( -\frac{a}{2}\right) \\\in & {} [-a,-a+\epsilon ]+\frac{a}{2}=\left[ -\frac{a}{2},-\frac{a}{2}+\epsilon \right] . \end{aligned}$$

In case $k>n$ set $u=-t+\frac{s}{2^{n+1}}$. Then

$$\begin{aligned} 0<\frac{s}{2^k}\le \frac{s}{2^{n+1}}=u+t\le u\,\,\left( \,\le \frac{s}{2}+\frac{s}{2^{n+1}}\le s\right) , \end{aligned}$$

hence

$$\begin{aligned} |\rho _n(t+u)-\rho _k(u)|= & {} \left| \rho _n\left( \frac{s}{2^{n+1}}\right) -\rho _k\left( -t+\frac{s}{2^{n+1}}\right) \right| \\= & {} \left| -\rho _n\left( -\frac{s}{2^{n+1}}\right) +\rho _k\left( -\frac{s}{2^{n+1}}+t\right) \right| \\= & {} \left| -\left( -\frac{a}{2}\right) +\chi \left( -\frac{s}{2^{n+1}}+t\right) \right| \\\ge & {} \left| \chi \left( -\frac{s}{2^{n+1}}+t\right) \right| -\frac{a}{2}\ge a-\epsilon -\frac{a}{2}\\= & {} \frac{a}{2}-\epsilon . \end{aligned}$$

For $n\in \mathbb {N}$ define $\kappa _n\in C^2_{-s,s}$ by

$$\begin{aligned} \kappa _n(t)=-\xi +\int _{-s}^t\rho _n(u)du \end{aligned}$$

and observe that

$$\begin{aligned} \kappa _n(-t)&= \kappa _n(t)\quad \text {on}\quad [-s,s],\\ \kappa _n(-s)=&-\xi =\kappa _n(s),\\ (\kappa _n)'(t)&< 0\quad \text {on}\quad [-s,0),\\ (\kappa _n)'(t)&> 0\quad \text {on}\quad (0,s],\\ (\kappa _n)'(-s)&= -a,\\ (\kappa _n)'(s)&= a,\\ (\kappa _n)''(t)&> 0\quad \text {on}\quad [-s,s],\\ (\kappa _n)''(-s)&= \eta . \end{aligned}$$

Using Proposition 3.1 and $\epsilon <\frac{a}{4}$ we get the following result.

Corollary 3.2

For all integers $n\ne k$ in $\mathbb {N}$ and for each $t\in \left[ -\frac{s}{2},0\right]$ there exists $u\in [-s,s]$ with $t+u\in [-s,s]$ and

$$\begin{aligned} |(\kappa _n)'(t+u)-(\kappa _k)'(u)|\ge \frac{a}{4}. \end{aligned}$$

4 The Delay Function on a Compact Interval

In this section we find $h>0$, a set $A\subset C^2_h$, constants $t_b<0$ and $t_5<-t_b$, and functions

$$\begin{aligned} \Delta _n:[0,t_5]\rightarrow (0,\infty )\quad \text{ and }\quad x_{(n)}:[t_b,t_5]\rightarrow \mathbb {R},\quad n\in \mathbb {N}, \end{aligned}$$

which in the next section will be used to form a solution of Eq. (1.2) whose short segments are dense in the set $A\cup (-A)$. Choose reals

$$\begin{aligned} \xi>b>a>0\quad \text {with}\quad \xi -a>b \end{aligned}$$

such that there exists $t_2>1$ with

$$\begin{aligned} bt_2>\xi >at_2, \end{aligned}$$

and choose $t_b\in (-1,0)$ with

$$\begin{aligned} b<(-t_b)\xi . \end{aligned}$$

Choose $v\in C^1_{t_b,0}$ with

$$\begin{aligned} v(t)< & {} 0\quad \text {on}\quad [t_b,0],\\ v'(t)> & {} 0\quad \text {on}\quad [t_b,0],\\ v(t_b)= & {} -\xi ,\\ v(0)= & {} -b,\\ v'(t_b)= & {} \frac{a}{2}, \end{aligned}$$

Because of $v([t_b,0])=[-\xi ,-b]$ and

$$\begin{aligned} b+t_bb>0>b+t_b\xi \end{aligned}$$

we can choose v in such a way that also

$$\begin{aligned} b+\int _{t_b}^0v(t)dt=0. \end{aligned}$$

The equation

$$\begin{aligned} x(t)=b+\int _{t_b}^tv(u)du \end{aligned}$$

defines a strictly decreasing function $x\in C^2_{t_b,0}$ with

$$\begin{aligned} x(t_b)=b\quad \text {and}\quad x(0)=0. \end{aligned}$$

Let $t_a\in (t_b,0)$ be given by $x(t_a)=a$.

Fig. 2

The function $x\in C^2_{t_b,0}$

Fig. 3

The function $v\in C^1_{t_b,t_2}$

Extend $v\in C^1_{t_b,0}$ to a function in $C^1_{t_b,t_2}$ with

$$\begin{aligned} v(t)< & {} 0\quad \text {on}\quad [0,t_2],\\ v(t_2)= & {} -a,\\ v'(t)> & {} 0\quad \text {on}\quad [0,t_2]. \end{aligned}$$

Because of $v([0,t_2])=[-b,-a]$ and

$$\begin{aligned} -bt_2<-\xi <-at_2 \end{aligned}$$

we can choose $v\in C^1_{t_a,t_2}$ in such a way that also

$$\begin{aligned} \int _0^{t_2}v(t)dt=-\xi . \end{aligned}$$

Set

$$\begin{aligned} \eta =v'(t_2)>0. \end{aligned}$$

Extend $x\in C^2_{t_b,0}$ to a strictly decreasing function in $C^2_{t_b,t_2}$ by

$$\begin{aligned} x(t)=\int _0^tv(u)du=b+\int _{t_b}^tv(u)du\quad \text {on}\quad (0,t_2], \end{aligned}$$

so that $x(t_2)=-\xi$, and let $t_1\in (0,t_2)$ be given by $x(t_1)=-a$.

Fig. 4

The function $x\in C^2_{t_b,t_2}$

Fix $t_d\in (t_1,t_2)$ and

$$\begin{aligned} h>0\quad \text{ with }\quad t_1<t_d-h \end{aligned}$$

and define

$$\begin{aligned} u_j=x^{(j)}(t_d-h)\,\,\text {and}\,\,w_j=x^{(j)}(t_d)\,\,\text {for}\,\,j\in \{0,1,2\}. \end{aligned}$$

Then $0>u_0>w_0,u_1<w_1<0,0<u_2,0<w_2$. Consider the set $A\subset C^2_h$ from Example 2.2. The functions in A are negative and strictly decreasing, with the derivative strictly increasing. Proposition 2.1 guarantees a sequence $(p_n)_{n\in \mathbb {N}}$ in A which is dense in A. For $n\in \mathbb {N}$ define $x_{(n)}\in C^2_{t_b,t_2}$ by

$$\begin{aligned} x_{(n)}(t)= & {} x(t)\quad \text{ on }\quad [t_b,t_d-h]\cup [t_d,t_2],\\ x_{(n)}(t)= & {} p_{\frac{n+1}{2}}(t-t_d)\quad \text{ on }\quad [t_d-h,t_d]\\&\text {in case}\quad n\quad \text {odd},\\ x_{(n)}(t)= & {} p_{\frac{n}{2}}(t-t_d)\quad \text{ on }\quad [t_d-h,t_d]\\&\text {in case}\quad n\quad \text {even}. \end{aligned}$$

Notice that $x_{(n)}$ is strictly decreasing on $[t_b,t_2]$ with

$$\begin{aligned} x_{(n)}(t_b)= & {} b,\,\,x_{(n)}(0)=0,\,\,x_{(n)}(t_1)=-a,\,\,x_{(n)}(t_2)=-\xi ,\\ (x_{(n)})'(t)< & {} 0\quad \text {on}\quad [t_b,t_2],\\ (x_{(n)})'(t_2)= & {} x'(t_2)=v(t_2)=-a,\\ (x_{(n)})''(t)> & {} 0\quad \text {on}\quad [t_b,t_2],\\ (x_{(n)})''(t_2)= & {} v'(t_2)=\eta . \end{aligned}$$

The inverse $y_n=(x_{(n)})^{-1}\in C^2_{-\xi ,b}$ maps its domain $[-\xi ,b]$ onto the interval $[t_b,t_2]$, with

$$\begin{aligned} (y_n)'(u)=\frac{1}{(x_{(n)})'(y_n(u))}<0\quad \text {for all}\quad u\in [-\xi ,b]. \end{aligned}$$

Obviously,

$$\begin{aligned} (x_{(n)})'([0,t_2])= & {} [(x_{(n)})'(0),(x_{(n)})'(t_2)]=[-b,-a],\\ -(x_{(n)})'([0,t_2])= & {} [a,b]\subset [-\xi ,b]. \end{aligned}$$

It follows that the equation

$$\begin{aligned} y_n(-(x_{(n)})'(t))=t-\Delta _n(t) \end{aligned}$$

defines a function $\Delta _n\in C^1_{0,t_2}$ with

$$\begin{aligned} 1-(\Delta _n)'(t)= & {} (y_n)'(-(x_{(n)})'(t))[-(x_{(n)})''(t)]>0\quad \text {on}\quad [0,t_2],\\ 0-\Delta _n(0)= & {} y_n(-(x_{(n)})'(0))=y_n(b)=t_b,\\ t_2-\Delta _n(t_2)= & {} y_n(-(x_{(n)})'(t_2))=y_n(a)=t_a,\\ (x_{(n)})'(t)= & {} -x_{(n)}(t-\Delta _n(t))\quad \text {on}\quad [0,t_2]. \end{aligned}$$

In particular,

$$\begin{aligned} (id-\Delta _n)([0,t_2])=[t_b,t_a]. \end{aligned}$$

The estimate $t-\Delta _n(t)\le t_a$ on $[0,t_2]$ yields

$$\begin{aligned} \Delta _n(t)\ge t-t_a\ge t\ge 0\quad \text{ on }\quad [0,t_2]. \end{aligned}$$

Fix some $s>0$ and recall $\kappa _n\in C^2_{-s,s}$ from Sect. 3, with $a,\xi ,\eta$ from the present section. Then

$$\begin{aligned} (\kappa _n)^{(j)}(-s)=(x_{(n)})^{(j)}(t_2)\quad \text {for}\quad j\in \{0,1,2\}. \end{aligned}$$

Set

$$\begin{aligned} t_3=t_2+2s \end{aligned}$$

and define an extension of $x_{(n)}$ to a map in $C^2_{t_b,t_3}$ by

$$\begin{aligned} x_{(n)}(t)=\kappa _n(t-t_3+s)\quad \text {on}\quad [t_2,t_3]. \end{aligned}$$

By the symmetry of $\kappa _n$,

$$\begin{aligned} x_{(n)}(t_3)= & {} x_{(n)}(t_2)=-\xi ,\\ (x_{(n)})'(t_3)= & {} -(x_{(n)})'(t_2)=a,\\ (x_{(n)})''(t_3)= & {} (x_{(n)})''(t_2)=\eta , \end{aligned}$$

Fig. 5

The function $x_{(n)}\in C^2_{t_b,t_3}$

and

$$\begin{aligned} (x_{(n)})'([t_2,t_3])=(\kappa _n)'([-s,s])=[-a,a]=x_{(n)}([t_a,t_1]). \end{aligned}$$

It follows that the equation

$$\begin{aligned} y_n(-(x_{(n)})'(t))=t-\delta _n(t)\quad \text {on}\quad [t_2,t_3] \end{aligned}$$

defines a map $\delta _n\in C^1_{t_2,t_3}$, with

$$\begin{aligned} t_2-\delta _n(t_2)= & {} y_n(-(x_{(n)})'(t_2))=y_n(a)=t_a,\\ t_3-\delta _n(t_3)= & {} y_n(-(x_{(n)})'(t_3))=y_n((x_{(n)})'(t_2))=y_n(-a)=t_1,\\ 1-(\delta _n)'(t)= & {} (y_n)'(-(x_{(n)})'(t))[-(x_{(n)})''(t)]>0\,\,\text {on}\,\,[t_2,t_3],\\ (x_{(n)})'(t)= & {} -x_{(n)}(t-\delta _n(t))\quad \text {on}\quad [t_2,t_3]. \end{aligned}$$

Notice that $\delta _n(t_2)=t_2-t_a=\Delta _n(t_2)$ and

$$\begin{aligned} 1-(\delta _n)'(t_2)=(y_n)'(-(x_{(n)})'(t_2))[-(x_{(n)})''(t_2)]=1-(\Delta _n)'(t_2). \end{aligned}$$

The estimate $t-\delta _n(t)\le t_1$ on $[t_2,t_3]$ yields

$$\begin{aligned} \delta _n(t)\ge t-t_1\ge t_2-t_1>0\quad \text{ on }\quad [t_2,t_3]. \end{aligned}$$

Setting

$$\begin{aligned} \Delta _n(t)=\delta _n(t)\quad \text {on}\quad [t_2,t_3] \end{aligned}$$

we get an extension of $\Delta _n\in C^1_{0,t_2}$ to a nonnegative map in $C^1_{0,t_3}$, with

$$\begin{aligned} 1-(\Delta _n)'(t)>0\,\,\text {on}\,\,[0,t_3]\,\,\text {and}\,\,(id-\Delta _n)([t_2,t_3])=[t_a,t_1]. \end{aligned}$$

Because of $a<\xi -b$ there exists $t_4>t_3$ with

$$\begin{aligned} a(t_4-t_3)<\xi -b<\xi (t_4-t_3), \end{aligned}$$

for example, $t_4=t_3+1$.

Proposition 4.1

There exists $\delta _{n*}\in C^1_{t_3,t_4}$ with

$$\begin{aligned} 1-(\delta _{n*})'(t)> & {} 0\quad \text {in}\quad [t_3,t_4],\\ t_3-\delta _{n*}(t_3)= & {} t_1,\\ t_4-\delta _{n*}(t_4)= & {} t_3,\\ 1-(\delta _{n*})'(t_3)= & {} 1-(\Delta _n)'(t_3),\\ 1-(\delta _{n*})'(t_4)= & {} \frac{1}{2},\quad \text{ and }\\ -\xi +\int _{t_3}^{t_4}x_{(n)}(t-\delta _{n*}(t))dt= & {} -b. \end{aligned}$$

Proof

Consider the discontinuous function $g_0:[t_3,t_4]\rightarrow \mathbb {R}$ given by $g_0(t_3)=t_1$ and $g_0(t)=t_3$ for $t_3<t\le t_4$. There is a sequence of functions $g_j\in C^1_{t_3,t_4}$, $j\in \mathbb {N}$, with

$$\begin{aligned} (g_j)'(t)> & {} 0\quad \text {on}\quad [t_3,t_4],\\ g_j(t_3)= & {} t_1,\\ g_j(t_4)= & {} t_3,\\ (g_j)'(t_3)= & {} 1-(\Delta _n)'(t_3),\\ (g_j)'(t_4)= & {} \frac{1}{2}. \end{aligned}$$

which converge pointwise to $g_0$. For every $j\in \mathbb {N}$, $g_j([t_3,t_4])=[t_1,t_3]$, and the Lebesgue dominated convergence theorem yields

$$\begin{aligned} G_j=\int _{t_3}^{t_4}[-x_{(n)}(g_j(t))]dt\rightarrow -\int _{t_3}^{t_4}x_{(n)}(t_3)dt=\xi (t_4-t_3)\quad \text {as}\quad j\rightarrow \infty . \end{aligned}$$

Similarly there is a sequence of functions $h_j\in C^1_{t_3,t_4}$ with the same properties as $g_j$ which converge pointwise to $h_0:[t_3,t_4]\rightarrow \mathbb {R}$ given by $h_0(t_4)=t_3$ and $h_0(t)=t_1$ for $t_3\le t<t_4$, and

$$\begin{aligned} H_j=\int _{t_3}^{t_4}[-x_{(n)}(h_j(t))]dt\rightarrow -\int _{t_3}^{t_4}x_{(n)}(t_1)dt=a(t_4-t_3)\quad \text {as}\quad j\rightarrow \infty . \end{aligned}$$

The limits satisfy

$$\begin{aligned} a(t_4-t_3)<\xi -b<\xi (t_4-t_3), \end{aligned}$$

due to the choice of $t_4$. So there exists $j\in \mathbb {N}$ with

$$\begin{aligned} H_j<\xi -b<G_j. \end{aligned}$$

The function

$$\begin{aligned} k:[0,1 ]\times [t_3,t_4]\ni (\theta ,t)\mapsto g_j(t)+\theta (h_j(t)-g_j(t))\in \mathbb {R}\end{aligned}$$

is continuous. Using the intermediate value theorem we find some $\theta \in (0,1)$ with

$$\begin{aligned} \int _{t_3}^{t_4}x_{(n)}(k(\theta ,t))dt=(1-\theta )G_j+\theta H_j=\xi -b. \end{aligned}$$

Notice that the convex combination $k(\theta ,\cdot )\in C^1_{t_3,t_4}$ shares the properties of $g_j$ and $h_j$. Define $\delta _{n*}$ by

$$\begin{aligned} t-\delta _{n*}(t)=k(\theta ,t). \end{aligned}$$

The estimate $t-\delta _{n*}(t)\le t_3$ on $[t_3,t_4]$ yields

$$\begin{aligned} \delta _{n*}(t)\ge t-t_3\ge 0\quad \text{ on }\quad [t_3,t_4]. \end{aligned}$$

It follows that the equation

$$\begin{aligned} \Delta _n(t)=\delta _{n*}(t)\quad \text {for}\quad t_3<t\le t_4 \end{aligned}$$

extends $\Delta _n\in C^1_{0,t_3}$ to a nonnegative function in $C^1_{0,t_4}$ which satisfies

$$\begin{aligned} 1-(\Delta _n)'(t)> & {} 0\quad \text {on}\quad [0,t_4],\\ t_4-\Delta _n(t_4)= & {} t_3,\\ t-\Delta _n(t)\in & {} [t_1,t_3]\quad \text {for}\quad t_3\le t\le t_4,\\ 1-(\Delta _n)'(t_4)= & {} \frac{1}{2}. \end{aligned}$$

The function $x_{n*}\in C^2_{t_3,t_4}$ given by

$$\begin{aligned} x_{n*}(t)=-\xi +\int _{t_3}^t[-x_{(n)}(u-\Delta _n(u))]du \end{aligned}$$

satisfies

$$\begin{aligned} x_{n*}(t_3)= & {} -\xi =x_{(n)}(t_3),\\ x_{n*}(t_4)= & {} -b,\\ (x_{n*})'(t)= & {} -x_{(n)}(t-\Delta _n(t))\quad \text{ on }\quad [t_3,t_4],\\ (x_{n*})'(t_3)= & {} -x_{(n)}(t_3-\Delta _n(t_3))=-x_{(n)}(t_1)=a=x_{(n)}'(t_3),\\ (x_{n*})'(t_4)= & {} -x_{(n)}(t_4-\Delta _n(t_4))=-x_{(n)}(t_3)=\xi ,\\ (x_{n*})''(t_3)= & {} -(x_{(n)})'(t_3-\Delta _n(t_3))[1-(\Delta _n)'(t_3)]=(x_{(n)})''(t_3)\\ (x_{n*})''(t_4)= & {} -(x_{(n)})'(t_4-\Delta _n(t_4))[1-(\Delta _n)'(t_4)]\\&-(x_{(n)})'(t_3)\frac{1}{2}=-\frac{a}{2}. \end{aligned}$$

Fig. 6

The function $x_{(n)}\in C^2_{t_b,t_4}$

Therefore the equation

$$\begin{aligned} x_{(n)}(t)=x_{n*}(t)\quad \text {for}\quad t_3<t\le t_4 \end{aligned}$$

defines a continuation of $x_{(n)}\in C^2_{t_b,t_3}$ to a function in $C^2_{t_b,t_4}$ which satisfies Eq. (1.2) on $[0,t_4]$ and maps the interval $[t_3,t_4]$ onto $[-\xi ,-b]$, with positive derivative and

$$\begin{aligned} x_{(n)}(t_4)= & {} -b=-x_{(n)}(t_b),\\ (x_{(n)})'(t_4)= & {} \xi =-v(t_b)=-(x_{(n)})'(t_b),\\ (x_{(n)})''(t_4)= & {} -\frac{a}{2}=-v'(t_b)=-(x_{(n)})''(t_b). \end{aligned}$$

We set $t_5=t_4-t_b$ and extend $x_{(n)}\in C^2_{t_b,t_4}$ to a function in $C^2_{t_b,t_5}$ by

$$\begin{aligned} x_{(n)}(t)=-x_{n}(t-t_5)\quad \text {on}\quad [t_4,t_5]. \end{aligned}$$

Then

$$\begin{aligned} -(x_{(n)})'([t_4,t_5])=(x_{(n)})'([t_b,0])=[-\xi ,-b]=x_{(n)}([t_3,t_4]). \end{aligned}$$

The derivative of the function

$$\begin{aligned} y_{n,5}=(x_{(n)}|_{[t_3,t_4]})^{-1}\in C^2_{-\xi ,-b} \end{aligned}$$

is strictly positive, due to $(x_{(n)})'(t)>0$ on $[t_3,t_4]$. The equation

$$\begin{aligned} y_{n,5}(-(x_{(n)})'(t))=t-\delta _{n,5}(t)\quad \text {for}\quad t_4\le t\le t_5 \end{aligned}$$

defines a function $\delta _{n,5}\in C^1_{t_4,t_5}$ which satisfies

$$\begin{aligned} t_4-\delta _{n,5}(t_4)= & {} y_{n,5}(-(x_{(n)})'(t_4))=y_{n,5}((x_{(n)})'(t_4-t_5))\\= & {} y_{n,5}((x_{(n)})'(t_b))=y_{n,5}(-\xi )=t_3=t_4-\Delta _n(t_4),\\ t_5-\delta _{n,5}(t_5)= & {} y_n(-(x_{(n)})'(t_5))=y_n((x_{(n)})'(0))=y_n(-b)=t_4,\\ 1-(\delta _{n,5})'(t)= & {} (y_{n,5})'(\ldots )[-(x_{(n)})''(t)]\\= & {} (y_{n,5})'(\ldots )[(x_{(n)})''(t-t_5)]>0\quad \text {on}\quad [t_4,t_5],\\ 1-(\delta _{n,5})'(t_4)= & {} (y_{n,5})'(-(x_{(n)})'(t_4))[-(x_{(n)})''(t_4)]\\= & {} (y_{n,5})'((x_{(n)})'(t_b))[(x_{(n)})''(t_b)]=(y_{n,5})'(-\xi )\frac{a}{2}\\= & {} (y_{n,5})'(x_{(n)}(t_3))\frac{a}{2}=\frac{1}{(x_{(n)})'(t_3)}\frac{a}{2}=\frac{1}{a}\frac{a}{2}\\= & {} \frac{1}{2}=1-(\Delta _n)'(t_4). \end{aligned}$$

The estimate $t-\delta _{n,5}(t)\le t_4$ on $[t_4,t_5]$ yields

$$\begin{aligned} \delta _{n,5}(t)\ge t-t_4\ge 0\quad \text{ on }\quad [t_4,t_5]. \end{aligned}$$

It follows that the equation

$$\begin{aligned} \Delta _n(t)=\delta _{n,5}(t)\quad \text {for}\quad t_4<t\le t_5 \end{aligned}$$

defines a continuation of $\Delta _n\in C^1_{0,t_4}$ to a nonnegative function in $C^1_{0,t_5}$ so that we have

$$\begin{aligned} t_4-\Delta _n(t_4)= & {} t_3,\\ t_5-\Delta _n(t_5)= & {} t_4,\quad \text {or equivalently},\\ \Delta _n(t_5)= & {} t_5-t_4=-t_b=\Delta _n(0),\\ 1-(\Delta _n)'(t)> & {} 0\quad \text {on}\quad [0,t_5],\\ (x_{(n)})'(t)= & {} -x_{(n)}(t-\Delta _n(t))\quad \text {on}\quad [0,t_5]. \end{aligned}$$

Also,

$$\begin{aligned} (\Delta _n)'(t_5)=(\Delta _n)'(0) \end{aligned}$$

because of

$$\begin{aligned} 1-(\Delta _n)'(t_5)= & {} 1-(\delta _{n,5})'(t_5)=(y_{n,5})'(-(x_{(n)})'(t_5))[-(x_{(n)})''(t_5)]\\= & {} (y_{n,5})'((x_{(n)})'(0))(x_{(n)})''(0)= (y_{n,5})'(-b)(x_{(n)})''(0)\\= & {} \frac{1}{(x_{(n)})'(t_4)}(x_{(n)})''(0)=\frac{1}{-(x_{(n)})'(t_b)}(x_{(n)})''(0) \end{aligned}$$

and

$$\begin{aligned} (x_{(n)})''(0)=-(x_{(n)})'(0-\Delta _n(0)[1-(\Delta _n)'(0)]=-(x_{(n)})'(t_b)[1-(\Delta _n)'(0)]. \end{aligned}$$

Fig. 7

The function $x_{(n)}\in C^2_{t_b,t_5}$

5 Concatenation

All functions $x_{(n)}\in C^2_{t_b,t_5}$, $n\in \mathbb {N}$, coincide on the set

$$\begin{aligned}{}[t_b,t_d-h]\cup [t_d,t_2]\cup [t_4,t_5], \end{aligned}$$

we have $t_4=t_5+t_b$, and for every $n\in \mathbb {N}$,

$$\begin{aligned} x_{(n)}(t)=-x_{(n)}(t-t_5)\quad \text {for all}\quad t\in [t_4,t_5]. \end{aligned}$$

Moreover, for every $n\in \mathbb {N}$ the nonnegative function $\Delta _n\in C^1_{0,t_5}$ satisfies

$$\begin{aligned} \Delta _n(t_5)= & {} \Delta _n(0)=-t_b,\\ (\Delta _n)'(t_5)= & {} (\Delta _n)'(0),\\ 1-(\Delta _n)'(t)> & {} 0 \quad \text {for all}\quad t\in [0,t_5], \end{aligned}$$

and we have

$$\begin{aligned} (x_{(n)})'(t)=-x_{(n)}(t-\Delta _n(t))\quad \text {for all}\quad t\in [0,t_5]. \end{aligned}$$

Therefore the relations

$$\begin{aligned} x(t)= & {} (-1)^{n-1}x_{(n)}(t-(n-1)t_5)\,\,\text {for}\,\,n\in \mathbb {N},\,\,(n-1)t_5+t_b\le t\le nt_5,\\ \Delta (t)= & {} \Delta _n(t-(n-1)t_5)\,\,\text {for}\,\,n\in \mathbb {N},\,\,(n-1)t_5\le t\le nt_5 \end{aligned}$$

define a twice continuously differentiable function $x:[t_b,\infty )\rightarrow \mathbb {R}$ and a continuously differentiable nonnegative function $\Delta :[0,\infty )\rightarrow \mathbb {R}$ so that Eq. (1.2) holds for all $t\ge 0$, $\Delta (0)=-t_b$, and

$$\begin{aligned} 1-\Delta '(t)>0\quad \text {for all}\quad t\ge 0. \end{aligned}$$

The short segments $x_{(n-1)t_5+t_d,short}=p_{\frac{n+1}{2}}\in C^2_h$, $n\in \mathbb {N}$ odd, which are given by

$$\begin{aligned} x_{(n-1)t_5+t_d,short}(u)=x((n-1)t_5+t_d+u)\quad \text {for}\quad -h\le u\le 0, \end{aligned}$$

are dense in the infinite-dimensional set $A\subset C^2_h\subset C^1_h$ with respect to the norm $|\cdot |_{1,h}$.

Recall

$$\begin{aligned} t_b\le & {} t-\Delta _n(t)\quad \text {in}\quad [0,t_2],\\ t_a\le & {} t-\Delta _n(t)\quad \text {in}\quad [t_2,t_3],\\ t_1\le & {} t-\Delta _n(t)\quad \text {in}\quad [t_3,t_4],\\ t_3\le & {} t-\Delta _n(t)\quad \text {in}\quad [t_4,t_5]=[t_4,t_4-t_b]\\ \end{aligned}$$

for each $n\in \mathbb {N}$ and set

$$\begin{aligned} r=\max \{t_2-t_b,t_3-t_a,t_4-t_1,t_4-t_b-t_3,t_5+3s\}. \end{aligned}$$

Then

$$\begin{aligned} \Delta (t)\le r\quad \text {for all}\quad t\ge 0. \end{aligned}$$

Extend $x:[t_b,\infty )\rightarrow \mathbb {R}$ backward to a twice continuously differentiable function $x:[-r,\infty )\rightarrow \mathbb {R}$, with long segments $x_t\in C^2_r\subset C^1_r$, $t\ge 0$, given by

$$\begin{aligned} x_t(u)=x(t+u)\quad \text {for}\quad -r\le u\le 0. \end{aligned}$$

The curve

$$\begin{aligned} \hat{x}:(0,\infty )\ni t\mapsto x_t\in C^1_r \end{aligned}$$

is continuously differentiable with

$$\begin{aligned} D\hat{x}(t)1=(x_t)'=(x')_t\in C^1_r\quad \text {for all}\quad t>0, \end{aligned}$$

compare [13, Proposition 4.1]. As $\frac{t_2+t_3}{2}$ is the only zero of $(x_{(n)})':[t_b,t_5]\rightarrow \mathbb {R}$, for any $n\in \mathbb {N}$, we have

$$\begin{aligned} D\hat{x}(t)1=(x_t)'\ne 0\quad \text {for all}\quad t>0. \end{aligned}$$

Proposition 5.1

The restriction of the curve $\hat{x}$ to the ray $[r,\infty )$ is injective.

Proof

Assume $r\le t\le u$ and $\hat{x}(t)=\hat{x}(u)$. Then

$$\begin{aligned} x(t+v)=x(u+v)\quad \text {for all}\quad v\in [-r,0]. \end{aligned}$$

There are $n\in \mathbb {N}$ and $k\in \mathbb {N}$ with

$$\begin{aligned} (n-1)t_5\le t<nt_5\quad \text {and}\quad (k-1)t_5\le u<kt_5. \end{aligned}$$

From $t_5<r\le t$ we have $n\ge 2$, and from $t\le u$ we have $n\le k$.

1. Proof of $t-(n-1)t_5=u-(k-1)t_5$. The argument $w=(n-1)t_5-t$ is contained in $(-t_5,0]\subset [-r,0]$, and

$$\begin{aligned} 0=x((n-1)t_5)=x(t+w)=x(u+w). \end{aligned}$$

As the interval $(u-t_5,u]$ contains exactly one zero of x, situated at $(k-1)t_5$, we get $u+w=(k-1)t_5$, hence

$$\begin{aligned} u-(k-1)t_5=-w=t-(n-1)t_5. \end{aligned}$$

2. The case $(n-1)t_5+t_3\le t\,\,(<nt_5)$. Using Part 1 of the proof we get

$$\begin{aligned} (k-1)t_5+t_3\le u. \end{aligned}$$

For every $w\in [-s,s]$ we obtain

$$\begin{aligned} \kappa _n(w)= & {} (-1)^{n-1}x_{(n)}(t_3-s+w)=x((n-1)t_5+t_3-s+w)\\= & {} x(t+[-t+(n-1)t_5+t_3-s+w])\\= & {} x(u+[-t+(n-1)t_5+t_3-s+w])\\&\text {(with}\quad [-t+(n-1)t_5+t_3-s+w]\in [-t_5,0]\subset [-r,0])\\= & {} x(u+[-u+(k-1)t_5+t_3-s+w])\\&\text {(with Part 1)}\\= & {} x((k-1)t_5+t_3-s+w)= (-1)^{k-1}x_{(k)}(t_3-s+w)=\kappa _k(w), \end{aligned}$$

and it follows that $n=k$. By Part 1, $t=u$.

3. The case $((n-1)t_5\le )\,\, t<(n-1)t_5+t_3$. Using Part 1 of the proof we get

$$\begin{aligned} ((k-1)t_5\le )\,\,u<(k-1)t_5+t_3. \end{aligned}$$

For every $w\in [-s,s]$ we have

$$\begin{aligned} -t+(n-2)t_5+t_3-s+w> & {} -[(n-1)t_5+t_3]+(n-2)t_5+t_3-s+w\\= & {} -t_5-s+w\ge -t_5-2s\ge -r\\&\text {and}\\ -t+(n-2)t_5+t_3-s+w\le & {} -(n-1)t_5+(n-2)t_5+t_3-s+w\\\le & {} -t_5+t_3-s+s\le 0, \end{aligned}$$

hence $[-t+(n-2)t_5+t_3-s+w]\in [-r,0]$. It follows that

$$\begin{aligned} \kappa _{n-1}(w)= & {} (-1)^{n-2}x_{(n-1)}(t_3-s+w)=x((n-2)t_5+t_3-s+w)\\= & {} x(t+[-t+(n-2)t_5+t_3-s+w])\\= & {} x(u+[-t+(n-2)t_5+t_3-s+w])\\&\text {(with}\quad [-t+(n-2)t_5+t_3-s+w]\in [-r,0])\\= & {} x(u+[-u+(k-2)t_5+t_3-s+w])\\&\text {(with Part 1)}\\= & {} x((k-2)t_5+t_3-s+w)= (-1)^{k-2}x_{(k-1)}(t_3-s+w)=\kappa _{k-1}(w). \end{aligned}$$

Hence $n-1=k-1$, and by Part 1, $t=u$.

6 Separation of Arcs

Proposition 6.1

There exists $\hat{a}>0$ so that for all integers $n\ge 2,j\ge 2$ with $|n-j|>1$ and for all $t\in [(n-1)t_5,nt_5],u\in [(j-1)t_5,jt_5]$ we have

$$\begin{aligned} |\hat{x}(t)-\hat{x}(u)|_{1,r}\ge \hat{a}. \end{aligned}$$

Proof

1. Recall from Sect. 4 the function $v\in C^1_{t_b,t_2}$. Let $n\in \mathbb {N}$. Notice that

$$\begin{aligned} (x_{(n)})'(t)=v(t)<0\quad \text {on}\quad [t_a,t_1]. \end{aligned}$$

With $v_m=-\max _{t_a\le t\le t_1}v(t)$ and $x_{(n)}(0)=0$ we obtain

$$\begin{aligned} |x_{(n)}(t)|\ge |t|v_m\quad \text {on}\quad [t_a,t_1]. \end{aligned}$$

On $[t_1,t_5+t_a]$ we have $x_{(n)}(t)\le -a$.

2. Let $n\in \mathbb {N},j\in \mathbb {N}$ and $t\in [(n-1)t_5,nt_5],u\in [(j-1)t_5,jt_5]$ be given. Then

$$\begin{aligned} t=(n-1)t_5+t_{*}\quad \text {with}\quad 0\le t_{*}\le t_5\quad \text {and}\quad u=(j-1)t_5+u_{*}\quad \text {with}\quad 0\le u_{*}\le t_5. \end{aligned}$$

We may assume $u_{*}\le t_{*}$. Set $w=u_{*}-t_{*}\in [-t_5,0]$.

3. In case $t_1\le -w\le t_5+t_a$ Part 1 yields the estimate

$$\begin{aligned} |\hat{x}(t)-\hat{x}(u)|_{1,r}\ge & {} |x_t(-u_{*})-x_u(-u_{*})|\\= & {} |x((n-1)t_5+t_{*}-u_{*})-x((j-1)t_5+u_{*}-u_{*})|\\= & {} |x((n-1)t_5-w)|=|x_{(n)}(-w)|\ge a. \end{aligned}$$

4. In case $\min \left\{ t_1,\frac{s}{2}\right\} \le -w\le t_1$ Part 1 yields the estimate

$$\begin{aligned} |\hat{x}(t)-\hat{x}(u)|_{1,r}\ge & {} |x_t(-u_{*})-x_u(-u_{*})|\\= & {} |x((n-1)t_5-w)|=|x_{(n)}(-w)|\ge (-w)v_m\ge v_m\cdot \min \left\{ t_1,\frac{s}{2}\right\} . \end{aligned}$$

5. In case $t_5+t_a\le -w\le t_5-\min \left\{ -t_a,\frac{s}{2}\right\}$ we have $t_a\le -w-t_5\le -\min \left\{ -t_a,\frac{s}{2}\right\}$. Using Part 1 we infer

$$\begin{aligned} |\hat{x}(t)-\hat{x}(u)|_{1,r}\ge & {} |x_t(-u_{*})-x_u(-u_{*})|\\= & {} |x((n-1)t_5-w)|=|x(nt_5-w-t_5)|\\= & {} |x_{(n+1)}(-w-t_5)|\ge |-w-t_5|v_m\ge v_m\cdot \min \{-t_a,s\}. \end{aligned}$$

6. The case $n-j\in 2\mathbb {Z}+1$, $-w\le s$, and $t_3\le u_{*}$.Then $t_2+s-t_{*}\in [-t_5,0]\subset [-r,0]$ since

$$\begin{aligned} -t_5\le -t_{*}\le t_2+s-t_{*}\le t_3-t_{*}\le t_3-u_{*}\le 0. \end{aligned}$$

Using $x_{(m)}(t)\le -\xi$ for all $m\in \mathbb {N}$ and all $t\in [t_2,t_3]=[t_2,t_2+2s]$ we infer

$$\begin{aligned} |\hat{x}(t)-\hat{x}(u)|_{1,r}\ge & {} |x_t(t_2+s-t_{*})-x_u(t_2+s-t_{*})|\\= & {} |x((n-1)t_5+t_{*}+t_2+s-t_{*})-\\&x((j-1)t_5+u_{*}+t_2+s-t_{*})|\\= & {} |x((n-1)t_5+t_2+s)-x((j-1))t_5+t_2+s+w)|\\= & {} |(-1)^{n-1}x_{(n)}(t_2+s)-(-1)^{j-1}x_{(j)}(t_2+s+w)|\\= & {} |(-1)^{n-j}x_{(n)}(t_2+s)-x_{(j)}(t_2+s+w)|\\\ge & {} 2\xi . \end{aligned}$$

7. The case $0\ne n-j\in 2\mathbb {Z}$, $-w\le \frac{s}{2}$, and $t_3\le u_{*}$. Corollary 3.2 yields some $v\in [-s,s]$ so that $w+v\in [-s,s]$ and

$$\begin{aligned} |(\kappa _j)'(w+v)-(\kappa _n)'(v)|\ge \frac{a}{4}. \end{aligned}$$

We have $t_2+s+v-t_{*}\in [-t_5,0]\subset [-r,0]$ since

$$\begin{aligned} -t_5\le -t_{*}\le t_2+s+v-t_{*}\le t_3-t_{*}\le t_3-u_{*}\le 0. \end{aligned}$$

Hence

$$\begin{aligned} |\hat{x}(t)-\hat{x}(u)|_{1,r}\ge & {} |(x_t)'(t_2+s+v-t_{*})-(x_u)'(t_2+s+v-t_{*})|\\= & {} |x'((n-1)t_5+t_{*}+t_2+s+v-t_{*})-\\&x'((j-1)t_5+u_{*}+t_2+s+v-t_{*})|\\= & {} |x'((n-1)t_5+t_2+s+v)-x'((j-1))t_5+t_2+s+w+v)|\\= & {} |(-1)^{n-1}(x_{(n)})'(t_2+s+v)-(-1)^{j-1}(x_{(j)})'(t_2+s+w+v)|\\= & {} |(-1)^{n-j}(x_{(n)})'(t_2+s+v)-(x_{(j)})'(t_2+s+w+v)|\\= & {} |(\kappa _n)'(v)-(\kappa _j)'(w+v)|\ge \frac{a}{4}. \end{aligned}$$

8. The case $n-j\in 2\mathbb {Z}+1$, $2\le n$, $2\le j$, $-w\le s$, and $u_{*}<t_3$. Then $t_5+t_{*}-t_2-s\in [0,t_5+2s]\subset [0,r]$ since

$$\begin{aligned} 0\le & {} t_5-t_3+u_{*}\le t_5-(t_2+s)+t_{*}=t_5+t_{*}-t_2-s\\\le & {} t_5+(u_{*}+s)-t_2-s\le t_5+t_3-t_2=t_5+2s\le r. \end{aligned}$$

Hence

$$\begin{aligned} |\hat{x}(t)-\hat{x}(u)|_{1,r}\ge & {} |x_t(-t_5-t_{*}+t_2+s)-x_u(-t_5-t_{*}+t_2+s)|\\= & {} |x((n-1)t_5+t_{*}-t_5-t_{*}+t_2+s)-\\&x((j-1)t_5+u_{*}-t_5-t_{*}+t_2+s)|\\= & {} |x((n-2)t_5+t_2+s)-x((j-2)t_5+w+t_2+s)|\\= & {} |(-1)^{n-2}x_{(n-1)}(t_2+s)-(-1)^{j-2}x_{(j)}(w+t_2+s)|\\= & {} |(-1)^{n-j}x_{(n-1)}(t_2+s)-x_{(j)}(w+t_2+s)|\ge 2\xi . \end{aligned}$$

9. The case $0\ne n-j\in 2\mathbb {Z}$, $2\le n$, $2\le j$, $-w\le \frac{s}{2}$, and $u_{*}<t_3$. Corollary 3.2 yields some $v\in [-s,s]$ so that $w+v\in [-s,s]$ and

$$\begin{aligned} |(\kappa _{j-1})'(w+v)-(\kappa _{n-1})'(v)|\ge \frac{a}{4}. \end{aligned}$$

We have $t_5+t_{*}-t_2-s-v\in [0,t_5+3s]\subset [0,r]$ since

$$\begin{aligned} 0\le & {} t_5-t_3+u_{*}\le t_5-(t_2+2s)+t_{*}\le t_5+t_{*}-t_2-s-v\\\le & {} t_5+\left( u_{*}+\frac{s}{2}\right) -t_2-s-v<t_5+t_3-t_2-v=t_5+2s-v\le r. \end{aligned}$$

Hence

$$\begin{aligned} |\hat{x}(t)-\hat{x}(u)|_{1,r}\ge & {} |(x_t)'(-t_5-t_{*}+t_2+s+v)-\\&(x_u)'(-t_5-t_{*}+t_2+s+v)|\\= & {} |x'((n-1)t_5+t_{*}-t_5-t_{*}+t_2+s+v)-\\&x'((j-1)t_5+u_{*}-t_5-t_{*}+t_2+s+v)|\\= & {} |x'((n-2)t_5+t_2+s+v)-x'((j-2))t_5+t_2+s+w+v)|\\= & {} |(-1)^{n-2}(x_{(n-1)})'(t_2+s+v)-(-1)^{j-2}(x_{(j-1)})'(t_2+s+w+v)|\\= & {} |(-1)^{n-j}(x_{(n-1)})'(t_2+s+v)-(x_{(j-1)})'(t_2+s+w+v)|\\= & {} |(\kappa _{n-1})'(v)-(\kappa _{j-1})'(w+v)|\ge \frac{a}{4}. \end{aligned}$$

10. The case $0\ne n-j\in 2\mathbb {Z}$, $2\le j$, $t_5-\min \{-t_a,s\}\le -w=t_{*}-u_{*}\le t_5$. Then

$$\begin{aligned} u_{*}\le t_{*}-t_5+s\le s, \end{aligned}$$

and $w_{*}=t_{*}-u_{*}-t_5$ satisfies $w_{*}\in [-s,0]$. We have $t_5+u_{*}-t_2-s\in [0,t_5]\subset [0,r]$ since

$$\begin{aligned} 0 \le t_5-t_3\le t_5-t_2-s\le t_5+u_{*}-t_2-s\le t_5+s-t_2-s\le t_5\le r. \end{aligned}$$

Hence

$$\begin{aligned} |\hat{x}(t)-\hat{x}(u)|_{1,r}\ge & {} |x_t(-t_5-u_{*}+t_2+s)-x_u(-t_5-u_{*}+t_2+s)|\\= & {} |x((n-1)t_5+t_{*}-t_5-u_{*}+t_2+s)-\\&x((j-1)t_5+u_{*}-t_5-u_{*}+t_2+s)|\\= & {} |x((n-1)t_5+w_{*}+t_2+s)-x((j-2)t_5+t_2+s)|\\= & {} |(-1)^{n-1}x_{(n)}(w_{*}+t_2+s)-(-1)^{j-2}x_{(j-1)}(t_2+s)|\\= & {} |(-1)^{n-j+1}x_{(n)}(w_ {*}+t_2+s)-x_{(j-1)}(t_2+s)|\ge 2\xi . \end{aligned}$$

11. The case $n-j\in 2\mathbb {Z}+1$, $2\le j$, $j-1\ne n$, $t_5-\min \left\{ -t_a,\frac{s}{2}\right\} \le -w=t_{*}-u_{*}\le t_5$. Now

$$\begin{aligned} u_{*}\le t_{*}-t_5+\frac{s}{2}\le \frac{s}{2}, \end{aligned}$$

and $w_{*}=t_{*}-u_{*}-t_5$ belongs to $\left[ -\frac{s}{2},0\right]$. Corollary 3.2 yields $v\in [-s,s]$ so that $w_{*}+v\in [-s,s]$ and

$$\begin{aligned} |(\kappa _{j-1})'(v)-(\kappa _n)'(w_{*}+v)|\ge \frac{a}{4}. \end{aligned}$$

We have $t_5+u_{*}-t_2-s-v\in [0,t_5+s]\subset [0,r]$ since

$$\begin{aligned} 0\le & {} t_5-t_3=t_5-t_2-2s\le t_5+u_{*}-t_2-2s\le t_5+\frac{s}{2}-t_2-s-v\\\le & {} t_5-v\le t_5+s\le r. \end{aligned}$$

Hence

$$\begin{aligned} |\hat{x}(t)-\hat{x}(u)|_{1,r}\ge & {} |(x_t)'(-t_5-u_{*}+t_2+s+v)-\\&(x_u)'(-t_5-u_{*}+t_2+s+v)|\\= & {} |x'((n-1)t_5+t_{*}-t_5-u_{*}+t_2+s+v)-\\&x'((j-1)t_5+u_{*}-t_5-u_{*}+t_2+s+v)|\\= & {} |x'((n-1)t_5+w_{*}+t_2+s+v)-x'((j-2))t_5+t_2+s+v)|\\= & {} |(-1)^{n-1}(x_{(n)})'(t_2+s+w_{*}+v)-(-1)^{j-2}(x_{(j-1)})'(t_2+s+v)|\\= & {} |(-1)^{n-j+1}(x_{(n)})'(t_2+s+w_{*}+v)-(x_{(j-1)})'(t_2+s+v)|\\= & {} |(\kappa _n)'(w_{*}+v)-(\kappa _{j-1})'(v)|\ge \frac{a}{4}. \end{aligned}$$

12. Combining the results of Parts 3-11 and the relation $\xi >a$ we arrive at the estimate

$$\begin{aligned} |\hat{x}(t)-\hat{x}(u)|_{1,r}\ge \min \left\{ \frac{a}{4},v_m\cdot \min \{t_1,s\},v_m\cdot \min \{-t_a,s\}\right\} \end{aligned}$$

for all integers $n\ge 2,j\ge 2$ with $|n-j|>1$ and all $t\in [(n-1)t_5,nt_5], u\in [(j-1)t_5,jt_5]$.

7 Delay Functionals on $C^0_r$-Neighbourhoods of Compact Arcs

For $t>0$ define $x'_t\in C^0_r$ by $x'_t(u)=x'(t+u)$, $-r\le u\le 0$. Then

$$\begin{aligned} x'_t=J(x')_t=JD\hat{x}(t)1. \end{aligned}$$

The curve

$$\begin{aligned} \hat{x}':(0,\infty )\ni t\mapsto x'_t\in C^0_r \end{aligned}$$

is continuously differentiable since the derivative $x':[-r,\infty )\rightarrow \mathbb {R}$ is continuously differentiable, compare [13, Proposition 4.1]. Consider the map

$$\begin{aligned} L:(0,\infty )\times C^0_r\rightarrow \mathbb {R}\end{aligned}$$

given by

$$\begin{aligned} L(t,\phi )=\phi (0)x'(t)+\phi (t_b)x'(t+t_b). \end{aligned}$$

We have

$$\begin{aligned} L=m\circ ((ev_0\circ \hat{x}'\circ pr_1)\times (ev_0\circ pr_2))+m\circ ((ev_{t_b}\circ \hat{x}'\circ pr_1) \times (ev_{t_b}\circ pr_2)) \end{aligned}$$

with the projections

$$\begin{aligned} pr_1:(0,\infty )\times C^0_r\rightarrow \mathbb {R},\quad pr_2:(0,\infty )\times C^0_r\rightarrow C^0_r \end{aligned}$$

onto the first and second component, respectively, with the continuous linear evaluation maps

$$\begin{aligned} ev_0:C^0_r\ni \phi \mapsto \phi (0)\in \mathbb {R},\quad ev_{t_b}:C^0_r\ni \phi \mapsto \phi (t_b)\in \mathbb {R}, \end{aligned}$$

and with the multiplication $m:\mathbb {R}\times \mathbb {R}\rightarrow \mathbb {R}$. So L is continuously differentiable.

Each map $L(t,\cdot ):C^0_r\rightarrow \mathbb {R}$, $t>0$, is linear. For the nullspace

$$\begin{aligned} K_t=\{\phi \in C^0_r:L(t,\phi )=0\} \end{aligned}$$

of $L(t,\cdot )$ we have

$$\begin{aligned} x'_t\notin K_t \end{aligned}$$

since

$$\begin{aligned} L(t,\hat{x}'(t))=(x'(t))^2+(x'(t+t_b))^2>0, \end{aligned}$$

which follows from the fact that the zeros of $x'$ in $[t_b,\infty )$ are given by $\frac{1}{2}(t_2+t_3)+jt_5$, $j\in \mathbb {N}_0$. We infer

$$\begin{aligned} C^0_r=\mathbb {R}x'_t\oplus K_t\quad \text {for all}\quad t>0. \end{aligned}$$

In the sequel we show that every compact arc $J\hat{x}([u,v])\subset C^0_r$, $r<u<v$, has a neighbourhood U in $C^0_r$ on which the representation

$$\begin{aligned} \phi =x_t+\kappa \,\,\text {with}\,\,\kappa \in K_t,\,\,t\,\,\text {close to}\,\,[u,v],\,\,\text {and}\,\,\kappa =\phi -x_t\,\,\text {small in}\,\,C^0_r \end{aligned}$$

is unique. Knowing this we shall define a delay functional $d_U:C^0_r\supset U\rightarrow \mathbb {R}$ by

$$\begin{aligned} d(\phi )=\Delta (x_t). \end{aligned}$$

Then d is constant along each fibre $(x_t+K_t)\cap U$, with t close to [u, v].

Obviously,

$$\begin{aligned} \phi -x_t\in K_t\quad \Leftrightarrow \quad L(t,\phi -x_t)=0 \end{aligned}$$

for all $\phi \in C^0_r$ and all $\sigma >0$.

Proposition 7.1

[Local fibre representation] For every $t>0$ there exist $\delta \in (0,t)$, $\epsilon \in (0,\delta ]$, and a continuously differentiable map

$$\begin{aligned} \tau :C^0_r\supset U_{\epsilon }(x_t)\rightarrow (t-\delta ,t+\delta )\subset \mathbb {R}\end{aligned}$$

with $\tau (x_t)=t$ so that for every $(\sigma ,\phi )\in (t-\delta ,t+\delta )\times U_{\epsilon }(x_t)$,

$$\begin{aligned} L(\sigma ,\phi -x_{\sigma })=0 \Leftrightarrow \sigma =\tau (\phi )\}. \end{aligned}$$

For every $\phi \in U_{\epsilon }(x_t)$ and for $\sigma =\tau (\phi )$,

$$\begin{aligned} |\phi -x_{\sigma }|_{0,r}\le \left( 1+\sup _{t-\delta \le u\le t+\delta }|x'_u|_{0,r}\right) \delta . \end{aligned}$$

Proof

Let $t>0$ be given. The map

$$\begin{aligned} f:(0,\infty )\times C^0_r\ni (\sigma ,\phi )\mapsto L(\sigma ,\phi -J\hat{x}(\sigma ))\in \mathbb {R}\end{aligned}$$

is continuously differentiable and satisfies $f(t,x_t)=0$. Using the formula defining the map L we infer

$$\begin{aligned} D_1f(t,\phi )1= & {} \phi (0)x''(t)+\phi (t_b)x''(t+t_b)\\&-((x'(t))^2+x(t)x''(t))-((x'(t+t_b))^2+x(t+t_b)x''(t+t_b)), \end{aligned}$$

hence

$$\begin{aligned} D_1f(t,x_t)1=-(x'(t))^2-(x'(t+t_b))^2<0. \end{aligned}$$

Apply the Implicit Function Theorem and obtain $\delta \in (0,t)$, $\epsilon >0$, and a continuously differentiable map $\tau$ with the properties stated in the first sentence of the proposition. Notice that one can achieve $\epsilon \le \delta$. For $\phi \in U_{\epsilon }(x_t)$ and $\sigma =\tau (\phi )$ we get

$$\begin{aligned} |\phi -x_{\sigma }|_{0,r}\le & {} |\phi -x_t|_{0,r}+|x_t-x_{\sigma }|_{0,r}\\= & {} |\phi -x_t|_{0,r}+|J\hat{x}(t)-J\hat{x}(\sigma )|_{0,r}\\\le & {} \epsilon +\sup _{t-\delta \le u\le t+\delta }|DJ\hat{x}(u)1|_{0,r}|t-\sigma |\\= & {} \epsilon +\sup _{t-\delta \le u\le t+\delta }|JD\hat{x}(u)1|_{0,r}|t-\sigma |\\= & {} \epsilon +\sup _{t-\delta \le u\le t+\delta }|x'_u|_{0,r}|t-\sigma |\\\le & {} (1+\sup _{t-\delta \le u\le t+\delta }|x'_u|_{0,r})\delta . \end{aligned}$$

Proposition 7.2

(Fibre representation along compact arcs) Let reals $u<v$ in $(r,\infty )$ and $n\in \mathbb {N}$ be given. There exist positive $\rho =\rho (u,v,n)\le \frac{1}{n}$ so that for every $\phi \in U_{\rho }(J\hat{x}([u,v]))$ there is one and only one

$$\begin{aligned} \sigma \in \left[ u-\frac{1}{n},v+\frac{1}{n}\right] \cap (0,\infty ) \end{aligned}$$

such that

$$\begin{aligned} L(\sigma ,\phi -x_{\sigma })=0\quad \text {and}\quad |\phi -x_{\sigma }|_{0,r}\le \frac{1}{n}. \end{aligned}$$

In case $\phi =x_t$ with $t\in [u,v]$ we have $\sigma =t$.

Proof

1. Let reals $u<v$ in $(r,\infty )$ be given. As the curve $J\circ \hat{x}$ is continuously differentiable with $DJ\hat{x}(w)1=x'_w\in C^0_r$ for all $w>0$ we obtain

$$\begin{aligned} |x_t-x_{\sigma })|_{0,r}\le \,c|t-\sigma |\quad \text {for all}\quad t,\sigma \,\,\text {in}\,\,\left[ \frac{u}{2},v+1\right] \end{aligned}$$

with

$$\begin{aligned} c=\max _{\frac{u}{2}\le w\le v+1}|x'_w|_{0,r}. \end{aligned}$$

2. Apply Proposition 7.1 to each $w\in [u,v]$, and obtain $\epsilon =\epsilon _w$ and $\delta =\delta _w$ and $\tau =\tau _w$ according to Proposition 7.1. Notice that one my assume

$$\begin{aligned} \frac{u}{2}\le w-\delta _w,\quad (1+c)\delta _w\le \frac{1}{n}. \end{aligned}$$

Using the compactness of $J\hat{x}([u,v])\subset C^0_r$ one finds a strictly increasing finite sequence $(w_j)_1^{\bar{j}}$ in [u, v] so that the associated neighbourhoods $U_{\epsilon _{w_j}}(\hat{x}(w_j))$, $j\in \{1,\ldots ,\bar{j}\}$, form a covering of $J\hat{x}([u,v])$. There exists a positive real number

$$\begin{aligned} \rho =\rho (u,v,n)\le \min _{j=1,\ldots ,\bar{j}}\epsilon _{w_j} \end{aligned}$$

with

$$\begin{aligned} U_{\rho }(J\hat{x}([u,v])\subset \bigcup _{j=1}^{\bar{j}}U_{\epsilon _{w_j}}(\hat{x}(w_j)). \end{aligned}$$

Notice that

$$\begin{aligned} \rho \le \min _{j=1,\ldots ,\bar{j}}\epsilon _{w_j}\le \max _{j=1,\ldots ,\bar{j}}\delta _{w_j}\le \frac{1}{n}. \end{aligned}$$

For every $\phi \in U_{\rho }(J\hat{x}([u,v])$ we obtain (at least one)

$$\begin{aligned} \sigma\in & {} \bigcup _{j=1}^{\bar{j}}(w_j-\delta _{w_j},w_j+\delta _{w_j})\\\subset & {} \left[ \max \left\{ \frac{u}{2},u-\frac{1}{n}\right\} ,v+\frac{1}{n}\right] \end{aligned}$$

with

$$\begin{aligned} L(\sigma ,\phi -x_{\sigma })=0\quad \text {and}\quad |\phi -x_{\sigma }|_{0,r}\le (1+c)\max _{j=1,\ldots ,\bar{j}}\delta _{w_j}\le \frac{1}{n}. \end{aligned}$$

Or, the set $R_n\subset (0,\infty )$ of all $\rho \in \left( 0,\frac{1}{n}\right]$ such that for every $\phi \in U_{\rho }(J\hat{x}([u,v]))$ there exist $\sigma \in \left[ u-\frac{1}{n},v+\frac{1}{n}\right] \cap (0,\infty )$ with

$$\begin{aligned} L(\sigma ,\phi -x_{\sigma })=0\quad \text {and}\quad |\phi -x_{\sigma }|_{0,r}\le \frac{1}{n} \end{aligned}$$

is nonempty. Observe that

$$\begin{aligned} \rho _n=\frac{1}{2}\sup \,R_n \end{aligned}$$

belongs to $R_n$.

3. Assume that the set I of all $n\in \mathbb {N}$ such that $U_{\rho _n}(J\hat{x}([u,v]))$ contains $\phi$ with

$$\begin{aligned} 2\le & {} \#\,\left\{ \sigma \in \left[ u-\frac{1}{n},v+\frac{1}{n}\right] \cap (0,\infty ):L(\sigma ,\phi -x_{\sigma })=0\right. \\&\left. \text {and}\,\,|\phi -x_{\sigma }|_{0,r}\le \frac{1}{n}\right\} \end{aligned}$$

is unbounded. We derive a contradiction. The elements of I form a strictly increasing sequence $(n_k)_1^{\infty }$. For every $k\in \mathbb {N}$ select some $\phi _k$ in $U_{\rho }(J\hat{x}([u,v]))$ with $\rho =\rho _{n_k}$ and $\sigma _k^{(1)}<\sigma _k^{(2)}$ in $\left[ u-\frac{1}{n_k},v+\frac{1}{n_k}\right] \cap (0,\infty )$ with

$$\begin{aligned} L(\sigma _k^{(m)},\phi _k-x_{{\sigma }_k^{(m)}})=0\quad \text {and}\quad |\phi _k-x_{{\sigma }_k^{(m)}}|_{0,r}\le \frac{1}{n_k}\quad \text {for}\quad m\in \{1,2\}. \end{aligned}$$

Using the compactness of, say, $[0,v+1]$, and successively choosing subsequences we find a strictly increasing sequence $(k_{\kappa })_1^{\infty }$ so that the equations

$$\begin{aligned} z^{(m)}_{\kappa }=\sigma ^{(m)}_{k_{\kappa }}\quad \text{ for }\quad \kappa \in \mathbb {N}\quad \text{ and }\quad m\in \{1,2\} \end{aligned}$$

define two sequences which converge to $z^{(1)}\le z^{(2)}$ in $[0,v+1]$, respectively. Necessarily, $u\le z^{(1)}\le z^{(2)}\le v$. The continuity of $J\circ \hat{x}$ yields $x_{z^{(m)}_{\kappa }}\rightarrow x_{z^{(m)}}$ in $C^0_r$ as $\kappa \rightarrow \infty$, for $m\in \{1,2\}$. Using the inequalities

$$\begin{aligned} |\phi _k-x_{{\sigma }_k^{(m)}}|_{0,r}\le \frac{1}{n_k}\quad \text {for}\quad m\in \{1,2\}\quad \text {and}\quad k\in \mathbb {N}\end{aligned}$$

we obtain $\phi _{k_{\kappa }}\rightarrow x_{z^{(1)}}=x_{z^{(2)}}$ as $\kappa \rightarrow \infty$. As $\hat{x}$ is injective on $[r,\infty )\supset [u,v]$, $z^{(1)}=z^{(2)}$. Apply Proposition 7.1 to $t=z^{(1)}=z^{(2)}$ and choose positive $\epsilon \le \delta$ according to this proposition. For $\kappa \in \mathbb {N}$ sufficiently large we have

$$\begin{aligned} \phi _{k_{\kappa }}\in U_{\epsilon }(x_t), \end{aligned}$$

both $z^{(1)}_{\kappa }< z^{(2)}_{\kappa }$ belong to $(t-\delta ,t+\delta )$, and

$$\begin{aligned} L(\sigma ,\phi _{k_{\kappa }}-x_{\sigma })=0\,\,\text {for}\,\,\sigma =z^{(1)}_{\kappa }\,\,\text {and for}\,\,\sigma =z^{(2)}_{\kappa }. \end{aligned}$$

This yields a contradiction to the first part of Proposition 7.1.

4. Combining the results of Parts 1 and 2 we obtain $n(u,v)\in \mathbb {N}$ such that for every integer $n\ge n(u,v)$ and for every $\phi \in U_{\rho _n}(J\hat{x}([u,v]))$ there exists one and only one $\sigma \in \left[ u-\frac{1}{n},v+\frac{1}{n}\right] \cap (0,\infty )$ with $L(\sigma ,\phi -x_{\sigma })=0$ and $|\phi -x_{\sigma }|_{0,r}\le \frac{1}{n}$. Now the assertion of Proposition 7.2 follows easily.

Proposition 7.2 yields that for $u<v$ in $(r, \infty )$ and $n\in \mathbb {N}$ there exists $\rho \le \frac{1}{n}$ so that the relations

$$\begin{aligned} \phi \in U_{\rho }(J\hat{x}([u,v])),&\,&\sigma \in \left[ u-\frac{1}{n},v+\frac{1}{n}\right] \cap (0,\infty ),\\ L(\sigma ,\phi -x_{\sigma })=0,&\,&|\phi -x_{\sigma }|_{0,r}\le \frac{1}{n} \end{aligned}$$

define a map

$$\begin{aligned} s_{u,v,\rho }:C^0_r\supset U_{\rho }(J\hat{x}([u,v]))\rightarrow (0,\infty ) \end{aligned}$$

with

$$\begin{aligned} |\phi -x_{s_{u,v,\rho }(\phi )}|_{0,r}\le \frac{1}{n}\quad \text {for all}\quad \phi \in U_{\rho }(J\hat{x}([u,v])). \end{aligned}$$

Proposition 7.3

Let reals $u<v$ in $(r,\infty )$ and $n\in N$ be given and choose $\rho =\rho (u,v,n)$ according to Proposition 7.2. There exist $\eta =\eta (u,v,n)\in (0,\rho ]$ so that the restriction $s_{u,v,\eta }$ of $s_{u,v,\rho }$ to $U_{\eta }(J\hat{x}([u,v]))$ is continuously differentiable.

For every $\phi \in U_{\eta }(J\hat{x}([u,v]))$ and for every $\sigma \in \left[ u-\frac{1}{n},v+\frac{1}{n}\right] \cap (0,\infty )$,

$$\begin{aligned} \sigma =s_{u,v,\eta }(\phi )\quad \Leftrightarrow \quad \left( L(\sigma ,\phi -x_{\sigma })=0\,\,\text {and}\,\, |\phi -x_{\sigma }|_{0,r}\le \frac{1}{n}\right) . \end{aligned}$$

For every $\sigma \in [u,v]$, $s_{u,v,\eta }(x_{\sigma })=\sigma$.

Proof

For each $t\in [u,v]$ choose $\epsilon =\epsilon _t\le \delta _t=\delta$ and $\tau =\tau _t$ according to Proposition 7.1. Observe that we may assume that $\delta _t$ satisfies

$$\begin{aligned} \max \left\{ 0,u-\frac{1}{n}\right\}<t-\delta _t,\quad t+\delta _t<v+\frac{1}{n} \end{aligned}$$

and

$$\begin{aligned} (1+\sup _{t-\delta _t\le w\le t+\delta _t}|x'_w|_{0,r})\delta _t<\frac{1}{n}. \end{aligned}$$

For every $\phi \in U_{\rho }(J\hat{x}([u,v]))\cap U_{\epsilon _t}(x_t)$ we have that

$$\begin{aligned} \sigma =\tau _t(\phi )\in (t-\delta _t,t+\delta _t)\subset \left[ u-\frac{1}{n},v+\frac{1}{n}\right] \cap (0,\infty ) \end{aligned}$$

satisphies $L(\sigma ,\phi -x_{\sigma })=0$ and

$$\begin{aligned} |\phi -x_{\sigma }|_{0,r}\le (1+\sup _{t-\delta _t\le w\le t+\delta _t}|x'_w|_{0,r})\delta _t<\frac{1}{n}. \end{aligned}$$

By the definition of $s_{u,v,\rho }$,

$$\begin{aligned} s_{u,v,\rho }(\phi )=\sigma =\tau _t(\phi ). \end{aligned}$$

It follows that the restriction of $s_{u,v,\rho }$ to $U_{\rho }(J\hat{x}([u,v]))\cap U_{\epsilon _t}(x_t)$ is continuously differentiable. There exists $\eta \in (0,\rho )$ with

$$\begin{aligned} U_{\eta }(J\hat{x}([u,v]))\subset \bigcup _{u\le t\le v}U_{\rho }(J\hat{x}([u,v]))\cap U_{\epsilon _t}(x_t). \end{aligned}$$

The last statement in Proposition 7.3 is obvious from Proposition 7.2.

Using continuous differentiability of the delay function $\Delta$ we infer that the delay functional

$$\begin{aligned} d_{u,v,\eta }=\Delta \circ s_{u.v,\eta } \end{aligned}$$

defined on the open neighbourhood $U_{\eta }(J\hat{x}([u,v]))$ of the arc $J\hat{x}([u,v])$ is continuously differentiable (with respect to the topology of $C^0_r$). For every $\sigma \in [u,v]$ we have $s_{u,v,\eta }(x_{\sigma })=\sigma$, hence

$$\begin{aligned} d_{u,v,\eta }(x_{\sigma })=\Delta (s_{u,v,\eta }(x_{\sigma }))=\Delta (\sigma ). \end{aligned}$$

8 Compatibility on $C^0_r$-Neighbourhoods of Adjacent Arcs

Let $j=j_r\ge 2$ denote the smallest integer with $r<(j-1)t_5$. For $j\le k\in \mathbb {N}$ set

$$\begin{aligned} X_k=\hat{x}([(k-1)t_5,kt_5])\subset C^1_r. \end{aligned}$$

In the sequel we construct open neighbourhoods $U_k$ of $JX_k$ in $C^0_r$ and continuously differentiable delay functionals $d_k:C^0_r\supset U_k\rightarrow (0,r)$ with $d_k(x_t)=\Delta (t)$ for all $t\in [(k-1)t_5,kt_5]$ so that for every integer $k\ge j$ we have

$$\begin{aligned} d_k(\phi )=d_{k+1}(\phi )\quad \text {for all}\quad \phi \in U_k\cap U_{k+1}. \end{aligned}$$

(8.1)

The construction is iterative. We carry out the initial step and the step thereafter. This second step is the model for the step from statements for general $k\ge j$ to statements for $k+1$.

1. The initial step for $k=j$.

1.1. Apply Proposition 7.1 with $t=jt_5$ at $\hat{x}(t)$, choose $\delta =\delta (j)>0$, $\epsilon =\epsilon (j)\in (0,\delta ]$, and a map $\tau =\tau _j$ from $U_{\epsilon }(\hat{x}(t))\subset C^0_r$ into $(t-\delta , t+\delta )$ accordingly. By continuity there are $n=n(j)\in \mathbb {N}$ with

$$\begin{aligned} \hat{x}\left( \left[ t-\frac{1}{n},t+\frac{1}{n}\right] \right) \subset U_{\epsilon }(\hat{x}(t))\quad \text {and}\quad r<(j-1)t_5-\frac{1}{n}, \end{aligned}$$

and $\epsilon _j\in (0,\epsilon (j)]$ with

$$\begin{aligned} \tau (U_{\epsilon _j}(\hat{x}(t)))\subset \left[ t-\frac{1}{n},t+\frac{1}{n}\right] . \end{aligned}$$

An application of Proposition 1.3 with $a=(j-1)t_5$, $b=(j+1)t_5$, $t=jt_5$ yields $\rho =\rho (j)>0$ with

$$\begin{aligned} U_{\rho }(JX_j)\cap U_{\rho }(JX_{j+1})\subset U_{\epsilon _j}(\hat{x}(t)); \end{aligned}$$

notice that $X_j=\hat{x}([a,t])$ and $X_{j+1}=\hat{x}([t,b])$.

1.2. We apply Proposition 7.3 twice, first with $u=(j-1)t_5$, $v=jt_5$, and $n=n(j)$. This yields $\eta >0$ and a continuously differentiable map

$$\begin{aligned} s_{u,v,\eta }:U_{\eta }(JX_j)\rightarrow \left[ u-\frac{1}{n},v+\frac{1}{n}\right] \subset \mathbb {R}\end{aligned}$$

so that for every $\phi \in U_{\eta }(JX_j)$ we have

$$\begin{aligned} \left( \sigma \in \left[ u-\frac{1}{n},v+\frac{1}{n}\right] \,\text {and}\,L(\sigma ,\phi -x_{\sigma })=0\right) \Leftrightarrow \sigma =s_{u,v,\eta }(\phi ). \end{aligned}$$

Also, $s_{u,v,\eta }(x_w)=w$ for all $w\in [u,v]$. We may assume

$$\begin{aligned} \eta <\rho =\rho (j). \end{aligned}$$

Set

$$\begin{aligned} U_j=U_{\eta }(JX_j)\quad \text {and}\quad s_j=s_{u,v,\eta }. \end{aligned}$$

The map

$$\begin{aligned} d_j:U_j\ni \phi \mapsto \Delta (s_j(\phi ))\in (0,r) \end{aligned}$$

is continuously differentiable with $d_j(x_w)=\Delta (w)$ for all $w\in [(j-1)t_5,jt_5]$.

The second application of Proposition 7.3, with $\hat{u}=(j+1)-1)t_5=jt_5$, $\hat{v}=(j+1)t_5$, and $n=n(j)$ yields $\hat{\eta }>0$ and a continuously differentiable map $s_{\hat{u},\hat{v},\hat{\eta }}:U_{\hat{\eta }}(JX_{j+1})\rightarrow \left[ \hat{u}-\frac{1}{n},\hat{v}+\frac{1}{n}\right] \subset \mathbb {R}$ such that for every $\phi \in U_{\hat{\eta }}(JX_{j+1})$ we have

$$\begin{aligned} \left( \sigma \in \left[ \hat{u}-\frac{1}{n},\hat{v}+\frac{1}{n}\right] \,\text {and}\,L(\sigma ,\phi -x_{\sigma })=0\right) \Leftrightarrow \sigma =s_{\hat{u},\hat{v},\hat{\eta }}(\phi ). \end{aligned}$$

Also, $s_{\hat{u},\hat{v},\hat{\eta }}(x_w)=w$ for all $w\in [\hat{u},\hat{v}]$. We may assume

$$\begin{aligned} \hat{\eta }<\rho =\rho (j). \end{aligned}$$

Set

$$\begin{aligned} \hat{U}_{j+1}=U_{\hat{\eta }}(JX_{j+1})\quad \text {and}\quad \hat{s}_{j+1}=s_{\hat{u},\hat{v},\hat{\eta }}. \end{aligned}$$

1.3. Let $\phi \in U_j\cap \hat{U}_{j+1}$. Proof of $s_j(\phi )=\hat{s}_{j+1}(\phi )$.

We have $\phi \in U_{\epsilon _j}(\hat{x}(t))$, due to Part 1.1 and to $\max \{\eta ,\hat{\eta }\}\le \rho (j)$. Hence

$$\begin{aligned} \tau (\phi )\in \left[ t-\frac{1}{n},t+\frac{1}{n}\right] . \end{aligned}$$

Notice that $t=v=\hat{u}$, and thereby

$$\begin{aligned} \left[ t-\frac{1}{n},t+\frac{1}{n}\right] \subset \left[ u-\frac{1}{n},v+\frac{1}{n}\right] \cap \left[ \hat{u}-\frac{1}{n},\hat{v}+\frac{1}{n}\right] . \end{aligned}$$

For $\sigma =\tau (\phi )$ we have $L(\sigma ,\phi -x_{\sigma })=0$, see Proposition 7.1. Now the properties of $s_j$ and of $s_{\hat{u},\hat{v},\hat{\eta }}$ from Part 1.2 yield

$$\begin{aligned} s_j(\phi )=\sigma =s_{\hat{u},\hat{v},\hat{\eta }}(\phi )=\hat{s}_{j+1}(\phi ). \end{aligned}$$

2. The second step, which includes the definitions of $U_{j+1}\subset \hat{U}_{j+1}$, of $s_{j+1}$, and of $d_{j+1}$, and contains the proof of $d_j(\phi )=d_{j+1}(\phi )$ on $U_j\cap U_{j+1}$.

2.1. Apply Proposition 7.1, now at $\hat{x}(t)$ with $t=(j+1)t_5$, and choose $\delta =\delta (j+1)>0$, $\epsilon =\epsilon (j+1)\in (0,\delta ]$, and a map $\tau =\tau _{j+1}$ from $U_{\epsilon }(\hat{x}(t))$ into $(t-\delta , t+\delta )$ accordingly. By continuity there is an integer $n=n(j+1)\ge n(j)$ with

$$\begin{aligned} J\hat{x}\left( \left[ t-\frac{1}{n},t+\frac{1}{n}\right] \right) \subset U_{\epsilon }(\hat{x}(t))\quad \left( \text {and}\quad r<((j+1)-1)t_5-\frac{1}{n}\right) , \end{aligned}$$

and there exists $\epsilon _{j+1}\in (0,\epsilon (j+1)]$ with

$$\begin{aligned} \tau (U_{\epsilon _{j+1}}(\hat{x}(t)))\subset \left[ t-\frac{1}{n},t+\frac{1}{n}\right] . \end{aligned}$$

An application of Proposition 1.3 with $a=((j+1)-1)t_5=jt_5$, $b=((j+1)+1)t_5=(j+2)t_5$, $t=(j+1)t_5$ yields $\rho =\rho (j+1)>0$ with

$$\begin{aligned} U_{\rho }(JX_{j+1})\cap U_{\rho }(JX_{j+2})\subset U_{\epsilon _{j+1}}(\hat{x}(t)); \end{aligned}$$

notice that $X_{j+1}=\hat{x}([a,t])$ and $X_{j+2}=\hat{x}([t,b])$.

2.2. First we restrict $\hat{s}_{j+1}$ from Part 1.2. As $\hat{s}_{j+1}$ maps $JX_{j+1}$ onto $[(jt_5,(j+1)t_5]$ continuity yields $\tilde{\eta }\in (0,\rho (j+1)]$ such that

$$\begin{aligned} U_{j+1}=U_{\tilde{\eta }}(JX_{j+1}) \end{aligned}$$

is contained in $\hat{U}_{j+1}$ and

$$\begin{aligned} \hat{s}_{j+1}(U_{j+1})\subset \left[ jt_5-\frac{1}{n},(j+1)t_5+\frac{1}{n}\right] , \end{aligned}$$

with $n=n(j+1)$. Set $s_{j+1}=\hat{s}_{j+1}|_{U_{j+1}}$. Part I.3 gives

$$\begin{aligned} s_{j+1}(\phi )=s_j(\phi )\quad \text {for all}\quad \phi \in U_{j+1}\cap U_j, \end{aligned}$$

and it follows that the continuously differentiable map

$$\begin{aligned} d_{j+1}:U_{j+1}\ni \phi \mapsto \Delta (s_{j+1}(\phi ))\in (0,r) \end{aligned}$$

satisfies $d_{j+1}(\phi )=\Delta (s_{j+1}(\phi ))=\Delta (s_j(\phi ))=d_j(\phi )$ for all $\phi \in U_{j+1}\cap U_j$. Also, $d_{j+1}(x_w)=\Delta (s_{j+1}(x_w))=\Delta (w)$ for all $w\in [jt_5,(j+1)t_5]$.

Next we apply Proposition 7.3, with $\check{u}=(j+2)-1)t_5=(j+1)t_5$, $\check{v}=(j+2)t_5$, and $n=n(j+1)$. This yields $\check{\eta }>0$ and a continuously differentiable map $s_{\check{u},\check{v},\check{\eta }}:U_{\check{\eta }}(JX_{j+2})\rightarrow \left[ \check{u}-\frac{1}{n},\check{v}+\frac{1}{n}\right] \subset \mathbb {R}$ such that for every $\phi \in U_{\check{\eta }}(JX_{j+2})$ we have

$$\begin{aligned} \left( \sigma \in \left[ \check{u}-\frac{1}{n},\check{v}+\frac{1}{n}\right] \,\text {and}\,L(\sigma ,\phi -x_{\sigma })=0\right) \Leftrightarrow \sigma =s_{\check{u},\check{v},\check{\eta }}(\phi ). \end{aligned}$$

Also, $s_{\check{u},\check{v},\check{\eta }}(x_w)=w$ for all $w\in [\check{u},\check{v}]$. Again we may assume

$$\begin{aligned} \check{\eta }<\rho =\rho (j+1). \end{aligned}$$

Set

$$\begin{aligned} \hat{U}_{j+2}=U_{\check{\eta }}(JX_{j+2})\quad \text {and}\quad \hat{s}_{j+2}=s_{\check{u},\check{v},\check{\eta }}. \end{aligned}$$

2.3. Proof of $s_{j+1}(\phi )=\hat{s}_{j+2}(\phi )$ for all $\phi \in U_{j+1}\cap \hat{U}_{j+2}$. Such $\phi$ belong to $U_{\epsilon _{j+1}}(\hat{x}((j+1)t_5)$, due to Part 2.1 and to the inequality $\max \{\tilde{\eta },\check{\eta }\}\le \rho (j+1)$. Hence $\sigma =\tau (\phi )$ is contained in $\left[ t-\frac{1}{n},t+\frac{1}{n}\right]$, for $n=n(j+1)$. Notice that $t=(j+1)t_5=\check{u}$, and thereby

$$\begin{aligned} \left[ t-\frac{1}{n},t+\frac{1}{n}\right]\subset & {} \left[ jt_5-\frac{1}{n},(j+1)t_5+\frac{1}{n}\right] \cap \left[ \check{u}-\frac{1}{n},\check{v}+\frac{1}{n}\right] \\\subset & {} \left[ jt_5-\frac{1}{n(j)},(j+1)t_5+\frac{1}{n(j)}\right] \cap \left[ \check{u}-\frac{1}{n(j+1)},\check{v}+\frac{1}{n(j+1)}\right] . \end{aligned}$$

We also have $L(\sigma ,\phi -x_{\sigma })=0$, see Proposition 7.1. Now the properties of $\hat{s}_{j+1}$ from Part 1.2 and of $\hat{s}_{j+2}=s_{\check{u},\check{v},\check{\eta }}$ from Part 2.2 yield

$$\begin{aligned} \hat{s}_{j+1}(\phi )=\sigma =\hat{s}_{j+2}(\phi ), \end{aligned}$$

which is $s_{j+1}(\phi )=\hat{s}_{j+2}(\phi )$.

This ends the second step.

9 A Functional on a $C^1_r$-Neighbourhood of the Trace $\hat{x}([(j_r-1)t_5,\infty ))$

In this section the constructions from Sects. 2–8 are used to prove Theorem 1.1. Let an integer $k\ge j_r$ be given. On the open set of all reals $t>0$ with $J\hat{x}(t)\in U_k$ we have that the map given by $t\mapsto d_k(J\hat{x}(t))$ is continuously differentiable, with the derivatives given by

$$\begin{aligned} Dd_k(Jx_t)JD\hat{x}(t)1=Dd_k(Jx_t)x'_t\in \mathbb {R}. \end{aligned}$$

On $[(k-1)t_5,kt_5]$ we have $\Delta (t)=d_k(J\hat{x}(t))$. It follows that on this interval,

$$\begin{aligned} 1>\Delta '(t)=Dd_k(Jx_t)x'_t. \end{aligned}$$

Recall the constant $\hat{a}$ from Proposition 6.1. The subset

$$\begin{aligned} N_k= & {} \{\phi \in C^1_r\cap J^{-1}(U_k):\quad Dd_k(J\phi )\phi '<1\quad \text {and there exists}\\&\left. t\in [(k-1)t_5,kt_5]\quad \text {with}\quad |\phi -x_t|_{1,r}<\frac{\hat{a}}{2}\right\} \end{aligned}$$

of the space $C^1_r$ is open. Proposition 6.1 yields $N_k\cap N_m=\emptyset$ for all integers $k\ge j_r$ and $m\ge j_r$ with $|k-m|>1$. Also, $N_k\cap N_{k+1}\subset J^{-1}(U_k)\cap J^{-1}(U_{k+1})$ for $j_r\le k\in \mathbb {N}$. Using the relations (8.1) we obtain that on the open set

$$\begin{aligned} N=\bigcup _{k\ge j_r}N_k\quad \supset \hat{x}([(j_r-1)t_5,\infty )) \end{aligned}$$

the equations

$$\begin{aligned} d(\phi )=d_k(J\phi )\quad \text {for}\quad \phi \in N_k\quad \text {and}\quad j_r\le k\in \mathbb {N}\end{aligned}$$

define a map $d:C^1_r\supset N\rightarrow (0,r)$. It follows that

$$\begin{aligned} d(x_t)=\Delta (t)\quad \text {for all}\quad t\ge (j_r-1)t_5 \end{aligned}$$

(9.1)

since for such t there exists $k\ge j_r$ with $t\in [(k-1)t_5,kt_5]$, hence $x_t\in N_k$, and thereby $d(x_t)=d_k(Jx_t)=d_k(x_t)=\Delta (t)$, see Sect. 8.

Proposition 1.2 applies and yields that the functional d is continuously differentiable and has property (e).

Proposition 9.1

The functional

$$\begin{aligned} f:C^1_r\supset N\ni \phi \mapsto -\phi (-d(\phi ))\in \mathbb {R}\end{aligned}$$

is continuously differentiable and has the extension property (e).

This is analogous to [14, Proposition 11.1]. We include the proof for convenience.

Proof

We have

$$\begin{aligned} f(\phi )=-ev^1_r(\phi ,-d(\phi ))=-(ev^1_r\circ (id\times (-d)))(\phi )\quad \text {for all}\quad \phi \in N, \end{aligned}$$

which shows that f is continuously differentiable. Recall $D_1ev^1_r(\phi ,t)\hat{\phi }=\hat{\phi }(t)$ and $D_2ev^1_r(\phi ,t)\hat{t}=\hat{t}\phi '(t)$. The chain rule yields

$$\begin{aligned} Df(\phi )\hat{\phi }=-\hat{\phi }(-d(\phi ))-\phi '(-d(\phi ))[-Dd(\phi )\hat{\phi }]=\phi '(-d(\phi ))Dd(\phi )\hat{\phi }-\hat{\phi }(-d(\phi )). \end{aligned}$$

For $\phi \in N$ the equation

$$\begin{aligned} D_ef(\phi )\chi =\phi '(-d(\phi ))D_ed(\phi )\chi -\chi (-d(\phi )). \end{aligned}$$

defines a linear extension $D_ef(\phi ):C^0_r\rightarrow \mathbb {R}$ of the derivative $Df(\phi ):C^1_r\rightarrow \mathbb {R}$. Using the continuity of the evaluation map $C^0_r\times [-r,0]\ni (\chi ,t)\mapsto \chi (t)\in \mathbb {R}$ and property (e) of d one finds that the map $N\times C^0_r\ni (\phi ,\chi )\mapsto D_ef(\phi )\chi \in \mathbb {R}$ is continuous.

For $t\ge j_rt_5$ we have $x_t\in N$ and, due to Eq. (9.1),

$$\begin{aligned} x'(t)=-x(t-\Delta (t))=-x(t-d(x_t))=f(x_t). \end{aligned}$$

This implies that the twice continuously differentiable function

$$\begin{aligned} x^{(d)}:[-r,\infty )\ni t\mapsto x(t+j_rt_5)\in \mathbb {R}\end{aligned}$$

is a solution of the equation

$$\begin{aligned} y'(t)=f(y_t) \end{aligned}$$

with the flowline $[0,\infty )\ni t\mapsto x^{(d)}_t\in C^1_r$ in the solution manifold

$$\begin{aligned} X_f=\{\phi \in N:\phi '(0)=f(\phi )\}. \end{aligned}$$

Recall the non-empty set $A\subset C^2_h$ chosen in Sect. 4 as a special case of the sets from Example 2.2. The set A is open in the affine space $p_{*}+C^2_{h-0}$ of codimension 6 in $C^2_{h-0}$. Recall the choice of x on $[t_d-h,t_d]\subset [0,t_5]$ in Sect. 4. The short segments $x^{(d)}_{t_d+(n-1)t_5,short}\in C^2_h$, $n\in \mathbb {N}$, are dense in $A\cup (-A)$.

Finally we show that for each $\phi \in X_f$ the delayed argument function

$$\begin{aligned}{}[0,t_{\phi })\ni t\mapsto t-d(x^{\phi }_t)\in \mathbb {R}\end{aligned}$$

is strictly increasing. Let $\phi \in X_f$ and $t\in (0,t_{\phi })$ be given and set $y=x^{\phi }$. As $y:[-r,t_{\phi })\rightarrow \mathbb {R}$ is continuously differentiable the curve $\tilde{y}:[0,t_{\phi })\ni t\mapsto Jy_t\in C^0_r$ is continuously differentiable with $D\tilde{y}(u)1=y'_u$ for all $u>0$, compare [13, Proposition 4.1]. The segment $y_t\in X_f\subset N$ is contained in $N_k$ for some integer $k\ge j_r$. By continuity of the flowline $[0,t_{\phi })\ni u\mapsto y_u\in X_f\subset N\subset C^1_r$, there is $\epsilon >0$ with $y_u\in N_k$ for all $u\in (t-\epsilon ,t+\epsilon )$. Then $d(y_u)=d_k(Jy_u)=d_k(\tilde{y}(u))$ on $(t-\epsilon ,t+\epsilon )$. It follows that the curve

$$\begin{aligned} (t-\epsilon ,t+\epsilon )\ni u\mapsto d(y_u)\in \mathbb {R}\end{aligned}$$

is differentiable with derivatives given by $Dd_k(Jy_u)y'_u<1$. This implies that on $(0,t_{\phi })$ the delayed argument function is differentiable with positive derivative, from which the assertion follows.