Response of an oscillatory differential delay equation to a single stimulus

Abstract

Here we analytically examine the response of a limit cycle solution to a simple differential delay equation to a single pulse perturbation of the piecewise linear nonlinearity. We construct the unperturbed limit cycle analytically, and are able to completely characterize the perturbed response to a pulse of positive amplitude and duration with onset at different points in the limit cycle. We determine the perturbed minima and maxima and period of the limit cycle and show how the pulse modifies these from the unperturbed case.

Appendices

Appendix: Proofs of the results from Sects. 3 and 4

Proof of Proposition 3.1

• 1. We begin with continuity of the time- $\tau$-map

  $$\begin{aligned} S(\tau ,\cdot ):Z\ni \phi \mapsto x^{\phi }_{\tau }\in Z. \end{aligned}$$

  Observe that for $0\le t\le \tau$,

  $$\begin{aligned} x^{\phi }(t)=e^{-t}\phi (0)+\int _0^te^{-(t-s)}f(\phi (s-\tau ))ds. \end{aligned}$$

  For $\psi$ and $\phi$ in Z and $0\le t\le \tau$ we have

  $$\begin{aligned} |x^{\psi }(t)-x^{\phi }(t)|\le |\phi (0)-\psi (0)|+\int _{-\tau }^0|f(\psi (s))-f(\phi (s))|ds \end{aligned}$$

  where the integrand is nonzero only on the set

  $$\begin{aligned} N(\psi ,\phi )=\{t\in [-\tau , 0]{:} \ \mathrm { sign}(\psi (t))\ne \mathrm {sign}(\phi (t))\}. \end{aligned}$$

  It follows that

  $$\begin{aligned} |x^{\psi }_{\tau }-x^{\phi }_{\tau }|_C\le |\psi (0)-\phi (0)|+\beta \lambda (N(\psi ,\phi )), \end{aligned}$$

  with the Lebesgue measure $\lambda$ and a positive constant $\beta$. It is easy to see that

  $$\begin{aligned} \lim _{Z\ni \psi \rightarrow \phi \in Z}\lambda (N(\psi ,\phi ))=0. \end{aligned}$$

  (Proof of this in case $\phi \in Z$ has zeros $z_1<z_2<\cdots <z_J$. Let $\epsilon >0$ be given. The complement of the set

  $$\begin{aligned} \bigcup _{j=1}^J\left( z_j-\dfrac{\epsilon }{2J},z_j+\dfrac{\epsilon }{2J}\right) \end{aligned}$$

  in $[-\tau ,0]$ is the finite union of compact intervals on each of which $\phi$ is either strictly positive, or strictly negative. There exists $\delta >0$ so that for every $\psi \in Z$ with $|\psi -\phi |<\delta$ the signs of $\psi (t)$ and $\phi (t)$ coincide on each of the compact intervals. This yields

  $$\begin{aligned} \lambda (N(\psi ,\phi ))\le \sum _{j=1}^J2\dfrac{\epsilon }{2J}=\epsilon .) \end{aligned}$$

  Then it follows easily that

  $$\begin{aligned} \lim _{Z\ni \psi \rightarrow \phi \in Z}(S(\tau ,\psi )-S(\tau ,\phi ))=\lim _{Z\ni \psi \rightarrow \phi \in Z}(x^{\psi }_{\tau }-x^{\phi }_{\tau })=0. \end{aligned}$$

• 2. Iterating we find that for every integer $n>0$ the time-$n\tau$-map $S(n\tau ,\cdot )$ is continuous. Having this we obtain continuous dependence on initial data in the sense that for every $t\ge 0$ and $\phi \in Z$,

  $$\begin{aligned} \lim _{Z\ni \psi \rightarrow \phi \in Z}\max _{-\tau \le s\le t}|x^{\psi }(s)-x^{\phi }(s)|=0. \end{aligned}$$

  Finally, the continuity of S at $(t,\phi )\in [0,\infty )\times Z$ follows by means of the estimate

  $$\begin{aligned} |S(s,\psi )-S(t,\phi )|&\le |S(s,\psi )-S(s,\phi )|+|S(s,\phi )-S(t,\phi )|\\&\le \max _{-\tau \le v\le t+1}|x^{\psi }(v)-x^{\phi }(v)|\\&\quad +\max _{-\tau \le w\le 0}|x^{\phi }(s+w)-x^{\phi }(t+w)| \end{aligned}$$

  for $0\le s\le t+1$ and $\psi \in Z$ from continuous dependence on initial data as before in combination with the uniform continuity of $x^{\phi }$ on $[-\tau ,t+1]$. $\square$

Proof of Proposition 4.2

By definition the value of the cycle length map at $\Delta$ is $T(\Delta )=z-\tilde{z}_J=z-z_{\Delta ,J}$ where z is the smallest zero of $x^{(\Delta )}$ in $(\tilde{z}_J,\infty )$ such that $x^{(\Delta )}(z+t)=\tilde{x}(\tilde{z}_J+t)$ for all $t\ge 0$. We have

$$\begin{aligned} \text {sign}(\tilde{x}(\tilde{z}_J+t))=\text {sign}(x^{(\Delta )}(z_{\Delta ,J}+t))=-\text {sign}(x^{(\Delta )}(z_{\Delta ,J+1}+t)) \end{aligned}$$

for $0<t\le \tau$ since $x^{(\Delta )}$ changes sign at each zero. We infer that $z>z_{\Delta ,J+1}$. Notice that the definition of $J=j_{\Delta }$ implies $\Delta <z_{\Delta ,J+1}$. Hence the next zero $z_{\Delta ,J+2}$ satisfies $z_{\Delta ,J+2}>z_{\Delta ,J+1}+\tau >\Delta +\tau \ge \Delta +\sigma$. Therefore on $(z_{\Delta ,J+2},\infty )$ the function $x^{(\Delta )}$ is given by Eq. (2.5), and satisfies

$$\begin{aligned} \text {sign}(x^{(\Delta )}(z_{\Delta ,J+2}+t))=-\text {sign}(x^{(\Delta )}(z_{\Delta ,J+1}+t))=\text {sign}(\tilde{x}(\tilde{z}_J+t)) \end{aligned}$$

for $0<t\le \tau$. This yields $x^{(\Delta )}(z_{\Delta ,J+2}+t)=\tilde{x}(\tilde{z}_J+t)$ for all $t\ge 0$.

Proposition 9.1

For $\Delta =\tilde{z}_J$, $J=1$ or $J=2$, we have $T(\Delta )=z_{\Delta ,J+1}-z_{\Delta ,J-1}$.

Proof

From $\Delta =\tilde{z}_J$ we obtain $\Delta +\sigma \le \Delta +\tau <z_{\Delta ,J+1}$. This implies that for $t\ge z_{\Delta ,J+1}$ the function $x^{(\Delta )}$ satisfies Eq. (2.5). Using this and the fact that $x^{(\Delta )}$ and $\tilde{x}$ change sign at $\tilde{z}_{J-1}=z_{\Delta ,J-1}$ and at $\tilde{z}_J=z_{\Delta ,J}$ respectively we infer that for all $t\ge z_{\Delta ,J+1}$ we have $x^{(\Delta )}(z_{\Delta ,J+1}+t)=\tilde{x}(\tilde{z}_{J-1}+t)$. It follows that

$$\begin{aligned} z_{\Delta ,J+2}=z_{\Delta ,J+1}+(\tilde{z}_J-\tilde{z}_{J-1}). \end{aligned}$$

Combining this with Proposition 4.2 we find

$$\begin{aligned} T(\Delta )=z_{\Delta ,J+2}-\tilde{z}_J=z_{\Delta ,J+1}-\tilde{z}_{J-1}=z_{\Delta ,J+1}-z_{\Delta ,J-1}. \end{aligned}$$

$\square$

Proof of Corollary 4.2

Let $\Delta _0\in [0,\tilde{T})$ be given and set $J=j(\Delta _0)$. Then $\Delta _0<z_{\Delta _0,J+1}$. Corollary 4.1 yields a neighbourhood N of $\Delta _0$ in $[0,\tilde{T})$ such that for all $\Delta \in N$ we have $\Delta <z_{\Delta ,J+1}$.

• 1. The case $\tilde{z}_J<\Delta _0$. Then by Corollary 4.1, $\tilde{z}_J<\Delta <z_{\Delta ,J+1}$ for all $\Delta$ in a neighbourhood $V\subset N$ of $\Delta _0$ in $[0,\tilde{T})$. For $\Delta \in V$ we get $j(\Delta )=J$, hence $T(\Delta )=z_{\Delta ,J+2}-\tilde{z}_J$, and Corollary 4.1 yields continuity at $\Delta _0$.

• 2. The case $\tilde{z}_J=\Delta _0$. There is a neighbourhood $U\subset N$ of $\Delta _0$ in $[0,\tilde{T})$ with $\tilde{z}_{J-1}<\Delta$ for all $\Delta \in U$. For all $\Delta \in U$ with $\Delta <\tilde{z}_J$ this yields $j(\Delta )=J-1$ and $T(\Delta )=z_{\Delta ,J-1+2}-\tilde{z}_{J-1}$. At $\Delta =\Delta _0$ we have

  $$\begin{aligned} T(\Delta )(\Delta _0)&= z_{\Delta _0,J+2}-\tilde{z}_J\\&= z_{\Delta _0,J+1}-z_{\Delta _0,J-1}\quad \text {(see Proposition 9.1)}\\&= z_{\Delta _0,J+1}-\tilde{z}_{J-1}. \end{aligned}$$

  The continuity of the map $\Delta \mapsto z_{\Delta ,J+1}$ due to Corollary 4.1 now shows that the restriction of the cycle length map to the set $[0,\tilde{z}_J]\cap U$ is continuous. For $\tilde{z}_J\le \Delta \in U\subset N$ we have $\tilde{z}_J\le \Delta <z_{\Delta ,J+1}$, hence $j(\Delta )=J$, and thereby $T(\Delta )=z_{\Delta ,J+2}-\tilde{z}_J$. The continuity of the map $\Delta \mapsto z_{\Delta ,J+2}$ due to Corollary 4.1 shows that the restriction of the cycle length map to the set $U\cap [\tilde{z}_J,\tilde{T})$ is continuous. As both restrictions coincide at $\tilde{z}_J=\Delta _0$ we obtain continuity of the cycle length map at $\Delta _0$. $\square$

Proof of Proposition 4.3

• 1. Let $\Delta _0\in [0,\tilde{T})$ be given. Set $J=j(\Delta )$. Then $\tilde{z}_J\le \Delta _0<z_{\Delta _0,J+1}$. Using Corollary 4.1 we find a neighbourhood N of $\Delta _0$ in $[0,\tilde{T})$ such that for every $\Delta \in N$ we have

  $$\begin{aligned} \tilde{z}_{J-1}<\Delta <z_{\Delta ,J+1}. \end{aligned}$$

  In the following we show continuity of the map $[0,\tilde{T})\ni \Delta \mapsto \overline{x}_{\Delta }\in \mathbb {R}$. The proof for the other map is analogous.

• 2. For $\Delta \in N\cap [\tilde{z}_J,\infty )$ we have $J=j(\Delta )$, hence

  $$\begin{aligned} \overline{x}_{\Delta }=\max _{\tilde{z}_J\le t\le z_{\Delta ,J+2}}x^{(\Delta )}(t). \end{aligned}$$

  Using this, the uniform continuity of the map

  $$\begin{aligned}{}[0,\tilde{T})\times [0,\infty )\ni (\Delta ,t)\mapsto x^{(\Delta )}(t)\in \mathbb {R} \end{aligned}$$

  on compact sets, and the continuity of the map $\Delta \mapsto z_{\Delta ,J+2}$ (see Corollary 4.1) one can easily show that the map

  $$\begin{aligned} N\cap [\tilde{z}_J,\infty )\ni \Delta \mapsto \overline{x}_{\Delta }\in \mathbb {R} \end{aligned}$$

  is continuous. Similarly we have for $\Delta \in N\cap (-\infty ,\tilde{z}_J)$ that $J-1=j(\Delta )$, hence

  $$\begin{aligned} \overline{x}_{\Delta }=\max _{\tilde{z}_{J-1}\le t\le z_{\Delta ,J+1}}x^{(\Delta )}(t). \end{aligned}$$

  As before one can then easily show that the map

  $$\begin{aligned} N\cap (-\infty ,\tilde{z}_J)\ni \Delta \mapsto \overline{x}_{\Delta }\in \mathbb {R} \end{aligned}$$

  is continuous.

• 3. It remains to prove that in case $\Delta _0=\tilde{z}_J$ (where $N\cap (-\infty ,\tilde{z}_J)\ne \emptyset$) we have $\overline{x}_{\Delta }\rightarrow \overline{x}_{\Delta _0}$ as $\Delta \nearrow \Delta _0$.

  • (a) The case $\Delta _0=\tilde{z}_J$ and $\tilde{x}'(\tilde{z}_J)<0$. Using the fact that $x^{(\Delta _0)}$ changes sign at each zero we obtain $x^{(\Delta _0)}(t)\le 0$ on $[\tilde{z}_j,z_{\Delta _0,J+1}]=[\Delta _0,z_{\Delta _0,J+1}]$, and

    $$\begin{aligned} \overline{x}_{\Delta _0}&= \max _{\tilde{z}_J\le t\le z_{\Delta _0,J+2}}x^{(\Delta _0)}(t)\\&= \max _{z_{\Delta _0,J+1}\le t\le z_{\Delta _0,J+2}}x^{(\Delta _0)}(t)\\&= \overline{x} \end{aligned}$$

    where the last equation holds because $\Delta _0=\tilde{z}_J$ implies $\Delta _0+\sigma<\Delta _0+\tau <z_{\Delta _0,J+1}$ and thereby $x^{(\Delta _0)}(z_{\Delta _0,J+1}+t)=\tilde{x}(\tilde{z}_{J-1}+t)$ for all $t\ge 0$. Using continuity as in Part 2 above we find a neighbourhood $U\subset N$ of $\Delta _0$ in $[0,\tilde{T})$ such that for each $\Delta \in U$ we have

    $$\begin{aligned} x^{(\Delta )}(t)<\dfrac{1}{2}\overline{x}\quad \text {on}\quad [\Delta ,z_{\Delta ,J+1}] \end{aligned}$$

    and $\tilde{z}_{J-1}+\tau <\Delta$. For $\Delta \in U$ with $\Delta <\tilde{z}_J$, we have $J-1=j(\Delta )$, and the preceding inequality yields

    $$\begin{aligned} x^{(\Delta )}(\tilde{z}_{J-1}+\tau )=\tilde{x}(\tilde{z}_{J-1}+\tau )=\overline{x}\quad \left( >\dfrac{1}{2}\overline{x}\right) . \end{aligned}$$

    It follows that

    $$\begin{aligned} \overline{x}_{\Delta }&= \max _{\tilde{z}_{J-1}\le t\le z_{\Delta ,J+1}}x^{(\Delta )}(t)\\&= \max _{\tilde{z}_{J-1}\le t\le \Delta }x^{(\Delta )}(t)\\&= \overline{x}, \end{aligned}$$

    so the map $U\cap (-\infty ,\tilde{z}_J)\ni \Delta \rightarrow \overline{x}_{\Delta }\in \mathbb {R}$ is constant with value $\overline{x}=\overline{x}_{\Delta _0}$.

  • (b) The case $\Delta _0=\tilde{z}_J$ and $\tilde{x}'(\tilde{z}_J)>0$. Then $x^{(\Delta _0)}$ is negative on $(\tilde{z}_{J-1},\tilde{z}_J)$, positive on $(\tilde{z}_J,z_{\Delta _0,J+1})$ and negative on $(z_{\Delta _0,J+1},z_{\Delta _0,J+2})$, and

    $$\begin{aligned} \overline{x}_{\Delta _0}=\max _{\tilde{z}_J\le t\le z_{\Delta _0,J+2}}x^{(\Delta )}(t)=\max _{\tilde{z}_J\le t\le z_{\Delta _0,J+1}}x^{(\Delta )}(t)>0. \end{aligned}$$

    Choose $t_0\in (\tilde{z}_J,z_{\Delta _0,J+1})$ with

    $$\begin{aligned} x^{(\Delta )}(t_0)=\overline{x}_{\Delta _0}>0. \end{aligned}$$

    By continuity there exists a neighbourhood $V\subset N$ of $\Delta _0$ in $[0,\tilde{T})$ such that for every $\Delta \in V$ we have

    $$\begin{aligned} x^{(\Delta )}(t)< \dfrac{1}{2}\overline{x}_{\Delta _0}\quad \text {on}\quad [\tilde{z}_{J-1},z_{\Delta ,J}],\\ z_{\Delta ,J}< t_0<z_{\Delta ,J+1},\\ \dfrac{1}{2}\overline{x}_{\Delta _0}&< x^{(\Delta )}(t_0). \end{aligned}$$

    For $\Delta \in V$ with $\Delta <\tilde{z}_J$ we have $j(\Delta )=J-1$, and we conclude that

    $$\begin{aligned} \overline{x}_{\Delta }&= \max _{\tilde{z}_{J-1}\le t\le z_{\Delta ,J+1}}x^{(\Delta )}(t)\\&= \max _{z_{\Delta ,J}\le t\le z_{\Delta ,J+1}}x^{(\Delta )}(t) \end{aligned}$$

    where by continuity the last term converges to

    $$\begin{aligned} \max _{z_{\Delta _0,J}\le t\le z_{\Delta _0,J+1}}x^{(\Delta _0)}(t)=\overline{x}_{\Delta _0} \end{aligned}$$

    as $V\ni \Delta \nearrow \Delta _0$.

$\square$

Proofs of the results from Sect. 5

Proof of Proposition 5.1

First we show that $x^{(\Delta )}$ has a first positive zero $z_{\Delta ,1}<\tilde{z}_1$. We have $x^{(\Delta )}(t)<0$ for $t \in (-\tau ,\Delta +\sigma )$. For $t \in [\Delta +\sigma ,\infty )$, $x^{(\Delta )}(t)$ is given by (5.4) as long as $x^{(\Delta )}(t-\tau )<0$. Compute

$$\begin{aligned} z_{\Delta ,1} = \Delta +\sigma +\ln \dfrac{\beta _L-x^{(\Delta )}(\Delta +\sigma )}{\beta _L} \ge \Delta +\sigma \end{aligned}$$

from the condition $x^{(\Delta )}(z_{\Delta ,1})=0$. Similarly since $\tilde{x}(\tilde{z}_1) = 0$ we obtain

$$\begin{aligned} \tilde{z}_1=\Delta +\sigma +\ln \dfrac{\beta _L-\tilde{x}(\Delta +\sigma )}{\beta _L}. \end{aligned}$$

Since $\tilde{x}(\Delta +\sigma )<x^{(\Delta )}(\Delta +\sigma )$ we have $z_{\Delta ,1}<\tilde{z}_1$.

The largest zero of $\tilde{x}$ on $(-\infty ,\Delta ]$ is $\tilde{z}_0=-\tau$. Hence the minimal value of $x^{(\Delta )}$ on $[-\tau ,z_{\Delta ,1}]$ is equal to $\underline{x}$. On the interval $[z_{\Delta ,1},z_{\Delta ,1}+\tau ]$, $x^{(\Delta )}(t)$ is given by (5.4). This yields $x^{(\Delta )}(t+z_{\Delta ,1})=\tilde{x}(t+\tilde{z}_1)$ for all $t\ge 0$. It follows that $\overline{x}_{\Delta }=\overline{x}$ and $\underline{x}_{\Delta }=\underline{x}$ and

$$\begin{aligned} T(\Delta )&= \bigl (z_{\Delta ,1}+(\tilde{z}_2-\tilde{z}_1)\bigr )-\tilde{z}_{0}=z_{\Delta ,1}-\tilde{z}_1+\tilde{T}\\&= \tilde{T}+\ln \dfrac{\beta _L-x^{(\Delta )}(\Delta +\sigma )}{\beta _L}-\ln \dfrac{\beta _L-\tilde{x}(\Delta +\sigma )}{\beta _L}\\&= \tilde{T}+\ln \dfrac{\beta _L-x^{(\Delta )}(\Delta +\sigma )}{\beta _L-\tilde{x}(\Delta +\sigma )}. \end{aligned}$$

The formula (5.5) yields

$$\begin{aligned} x^{(\Delta )}(\Delta + \sigma )-\beta _L= -\beta _L e^{\tilde{z}_1-\Delta -\sigma }+a(1-e^{-\sigma }) \end{aligned}$$

and with

$$\begin{aligned} \tilde{x}(\Delta +\sigma )-\beta _L=(\underline{x}-\beta _L)e^{-(\Delta +\sigma )}=-\beta _Le^{\tilde{z}_1-(\Delta +\sigma )} \end{aligned}$$

we find

$$\begin{aligned} T(\Delta )&= \tilde{T}+\ln \dfrac{ \beta _L e^{\tilde{z}_1-\Delta -\sigma }-a(1-e^{-\sigma })}{\beta _Le^{\tilde{z}_1-(\Delta +\sigma )}}\\&= \tilde{T}+\ln \left( 1-\dfrac{a(1-e^{-\sigma })}{\beta _L}e^{\Delta +\sigma -\tilde{z}_1}\right) . \end{aligned}$$

Thus, $T(\Delta ) < \tilde{T}$, and the restriction of $T(\Delta )$ to $I_{RNRN}$ is strictly decreasing. $\square$

From (5.4) it follows that $x^{(\Delta )}$ is strictly decreasing right after $t=\Delta +\sigma$ if and only if $x^{(\Delta )}(\Delta +\sigma )-\beta _L>0$. We have, by (5.5),

$$\begin{aligned} x^{(\Delta )}(\Delta +\sigma )-\beta _L=-\beta _Le^{\tilde{z}_1-\Delta -\sigma }+a(1-e^{-\sigma }). \end{aligned}$$

Hence, $x^{(\Delta )}(\Delta +\sigma )-\beta _L>0$ if and only if

$$\begin{aligned} e^{\Delta }>\dfrac{\beta _Le^{\tilde{z}_1-\sigma }}{a(1-e^{-\sigma })}. \end{aligned}$$

Let us define

$$\begin{aligned} \hat{\delta }_1=\ln \dfrac{\beta _Le^{\tilde{z}_1-\sigma }}{a(1-e^{-\sigma })}. \end{aligned}$$

(9.1)

We have

$$\begin{aligned} \hat{\delta }_1=\tilde{z}_1-\sigma +\ln \dfrac{\beta _L}{a(1-e^{-\sigma })}>\delta _1. \end{aligned}$$

Thus we divide the case RNRP when $x^{(\Delta )}(\Delta +\sigma )\ge 0$ into the two subcases $0\le x^{(\Delta )}(\Delta +\sigma )\le \beta _L$ and $x^{(\Delta )}(\Delta +\sigma )>\beta _L$ and we consider these two subcases separately as $\mathbf {RNRP1}$ and $\mathbf {RNRP2}$, see Figs. 3 and 4, respectively. The $\Delta$-intervals are of the form

$$\begin{aligned} I_{RNRP1}=[\max \{0,\delta _1\},\tilde{z}_1)\cap [\delta _1,\hat{\delta }_1]\quad \text {and}\quad I_{RNRP2}=[\max \{0,\delta _1\},\tilde{z}_1)\cap (\hat{\delta }_1,\infty ). \end{aligned}$$

• Case $\mathbf {RNRP1}$ The pulse parameters $(a,\Delta ,\sigma )$ are such that $\Delta <\tilde{z}_1$ and $x^{(\Delta )}(\Delta +\sigma )\in [0,\beta _L]$. Then $x^{(\Delta )}$ is increasing right after the pulse.

• Case $\mathbf {RNRP2}$ The pulse parameters $(a,\Delta ,\sigma )$ are such that $\Delta <\tilde{z}_1$ and $x^{(\Delta )}(\Delta +\sigma )>\beta _L$. Then $x^{(\Delta )}$ is strictly decreasing right after the pulse.

Proof of Proposition 5.2

The equation

$$\begin{aligned} 0=x^{(\Delta )}(z_{\Delta ,1})=\beta _L+a+\bigl (x^{(\Delta )}(\Delta )-(\beta _L+a)\bigr )e^{-(z_{\Delta ,1}-\Delta )} \end{aligned}$$

together with (5.1) yields

$$\begin{aligned} \nonumber (\beta _L+a)e^{z_{\Delta ,1}}&=\bigl (\beta _L+a-x^{(\Delta )}(\Delta )\bigr )e^{\Delta }\\&=\beta _Le^{\tilde{z}_1}+ae^{\Delta }. \end{aligned}$$

(9.2)

Use $\Delta <z_{\Delta ,1}$ and $a>0,\beta _L>0$ to obtain

$$\begin{aligned} e^{z_{\Delta ,1}}-e^{\tilde{z}_1}=\dfrac{a}{\beta _L}(e^{\Delta }-e^{z_{\Delta ,1}})<0 \end{aligned}$$

and to conclude that $z_{\Delta ,1}<\tilde{z}_1$.

We have

$$\begin{aligned} x^{(\Delta )}(z_{\Delta ,1}+\tau )=\beta _L+(x^{(\Delta )}(\Delta +\sigma )-\beta _L)e^{-\left( z_{\Delta ,1}+\tau - (\Delta +\sigma )\right) }, \end{aligned}$$

which by (5.5) can be rewritten as

$$\begin{aligned} x^{(\Delta )}(z_{\Delta ,1}+\tau )&=\beta _L+\bigl (-\beta _Le^{\tilde{z}_1-\Delta -\sigma }+a(1-e^{-\sigma })\bigr ) e^{-\left( z_{\Delta ,1}+\tau - (\Delta +\sigma )\right) }\\&=\beta _L-(\beta _Le^{\tilde{z}_1}+ae^{\Delta })e^{-(z_{\Delta ,1}+\tau )}+ae^{-\left( z_{\Delta ,1}+\tau - (\Delta +\sigma )\right) }. \end{aligned}$$

From (9.2) it follows that

$$\begin{aligned} x^{(\Delta )}(z_{\Delta ,1}+\tau )=\beta _L-(\beta _L+a)e^{-\tau }+ae^{-\tau }e^{\sigma +\Delta -z_{\Delta ,1}} \end{aligned}$$

(9.3)

and that the function $[\max \{0,\delta _1\},\tilde{z}_1)\ni \Delta \mapsto e^{z_{\Delta ,1}-\Delta }\in \mathbb {R}$ is strictly decreasing, which shows that $x^{(\Delta )}(z_{\Delta ,1}+\tau )$ is strictly increasing with respect to $\Delta \in [\max \{0,\delta _1\},\tilde{z}_1)$. Since $\Delta +\sigma \ge z_{\Delta ,1}$ and $ae^{-\tau }>0$, we get

$$\begin{aligned} x^{(\Delta )}(z_{\Delta ,1}+\tau )\ge \beta _L-\beta _Le^{-\tau }=\overline{x}. \end{aligned}$$

Also the function $[\max \{0,\delta _1\},\tilde{z}_1)\ni \Delta \mapsto x^{(\Delta )}(\Delta +\sigma )\in \mathbb {R}$ is increasing.

In subcase $\mathbf {RNRP1}$ the function $x^{(\Delta )}$ is increasing on $[\Delta +\sigma ,z_{\Delta ,1}+\tau ]$ while in subcase $\mathbf {RNRP2}$ it is decreasing on that interval. It follows that in subcase $\mathbf {RNRP1}$,

$$\begin{aligned} \max _{z_{\Delta ,1}\le t\le z_{\Delta ,1}+\tau }x^{(\Delta )}(t)=x^{(\Delta )}( z_{\Delta ,1}+\tau )\ge \overline{x} \end{aligned}$$

while in subcase $\mathbf {RNRP2}$,

$$\begin{aligned} \max _{z_{\Delta ,1}\le t\le z_{\Delta ,1}+\tau }x^{(\Delta )}(t)=x^{(\Delta )}( \Delta +\sigma )\ge x^{(\Delta )}(z_{\Delta ,1}+\tau )\ge \overline{x}. \end{aligned}$$

In both subcases $\max _{z_{\Delta ,1}\le t\le z_{\Delta ,1}+\tau }x^{(\Delta )}(t)$ is increasing with respect to $\Delta$.

Also in both subcases we have $0<x^{(\Delta )}(t)$ for $z_{\Delta ,1}<t\le z_{\Delta ,1}+\tau$. It follows that after $t=z_{\Delta ,1}+\tau \ge \Delta +\sigma$ the function $x^{(\Delta )}$ is given by

$$\begin{aligned} x^{(\Delta )}(t)=-\beta _U+\bigl (x^{(\Delta )}(z_{\Delta ,1}+\tau )+\beta _U\bigr )e^{-\left( t-(z_{\Delta ,1}+\tau )\right) } \end{aligned}$$

as long as $x^{(\Delta )}(t-\tau )>0$. We obtain a first zero $z_{\Delta ,2}$ of $x^{(\Delta )}$ in $(z_{\Delta ,1}+\tau ,\infty )$, and for all $t\ge 0$, $x^{(\Delta )}(z_{\Delta ,2}+t)=\tilde{x}(\tilde{z}_0+t)$ (recall $\tilde{z}_0=-\tau$). Then

$$\begin{aligned} T(\Delta )=z_{\Delta ,2}-\tilde{z}_0=z_{\Delta ,2}+\tau . \end{aligned}$$

Moreover,

$$\begin{aligned} \overline{x}_{\Delta }&= \max _{-\tau \le t\le z_{\Delta ,2}}x^{(\Delta )}(t)=\max _{z_{\Delta ,1}\le t\le z_{\Delta ,1}+\tau }x^{(\Delta )}(t)\\&= \max \{x^{(\Delta )}(\Delta +\sigma ),x^{(\Delta )}(z_{\Delta ,1}+\tau )\}\\&\ge \overline{x} \end{aligned}$$

is increasing with respect to $\Delta$ in both subcases.

Now the equation

$$\begin{aligned} 0= x^{(\Delta )}(z_{\Delta ,2}) = -\beta _U+\bigl (x^{(\Delta )}(z_{\Delta ,1}+\tau )+\beta _U\bigr )e^{z_{\Delta ,1}+\tau -z_{\Delta ,2}} \end{aligned}$$

yields

$$\begin{aligned} \beta _U e^{z_{\Delta ,2}}=\bigl (x^{(\Delta )}(z_{\Delta ,1}+\tau )+\beta _U\bigr )e^{z_{\Delta ,1}+\tau }. \end{aligned}$$

Also, from (9.2) and (9.3) we obtain

$$\begin{aligned} \beta _U e^{z_{\Delta ,2}}=(\beta _L+\beta _U)e^{z_{\Delta ,1}+\tau }-\beta _L e^{\tilde{z}_1}+a(1-e^{-\sigma }) e^{\Delta +\sigma }. \end{aligned}$$

Since

$$\begin{aligned} \beta _U+\beta _L-\beta _{L}e^{-\tau }=\beta _U+\overline{x}=\beta _Ue^{\tilde{z}_2-t_{\max }} \end{aligned}$$

and $\tilde{z}_2-t_{\max }+\tau =\tilde{z}_2-\tilde{z}_1$, we arrive at

$$\begin{aligned} \beta _L e^{\tilde{z}_1}+\beta _U e^{\tilde{z}_2}=(\beta _L+\beta _U)e^{\tau +\tilde{z}_1}. \end{aligned}$$

(9.4)

Thus

$$\begin{aligned} \begin{aligned} \beta _U e^{z_{\Delta ,2}}&=\beta _Ue^{\tilde{z}_2}+(\beta _L+\beta _U)e^{\tau }(e^{z_{\Delta ,1}}-e^{\tilde{z}_1})+a(e^{\sigma }-1) e^{\Delta } \\&=\beta _Ue^{\tilde{z}_2}+\dfrac{a(\beta _L+\beta _U)e^{\tau }}{\beta _L+a}(e^{\Delta }-e^{\tilde{z}_1})+a(e^{\sigma }-1) e^{\Delta }\quad \text {(with (9.2))}, \end{aligned} \end{aligned}$$

which implies the formula for $T(\Delta )=z_{\Delta ,2}+\tau$, since $\tilde{z}_2+\tau =\tilde{T}$. $\square$

In the case $\mathbf {RPRP}$ we always have $x^{(\Delta )}(\Delta +\sigma )>0$, but in the following proof we need to distinguish between two cases

• Case $\mathbf {RPRP1}$ $0< x^{(\Delta )}(\Delta +\sigma )\le \beta _L$, and

• Case $\mathbf {RPRP2}$ $x^{(\Delta )}(\Delta +\sigma )> \beta _L$.

Proof of Proposition 5.3

Using (5.4) for $\Delta +\sigma \le t\le z_{\Delta ,1}+\tau =\tilde{z}_1+\tau =t_{\max }$ and (5.5) we obtain

$$\begin{aligned} x^{(\Delta )}(z_{\Delta ,1}+\tau )&= \beta _L+\bigl (x^{(\Delta )}(\Delta +\sigma )-\beta _L\bigr )e^{-\left( \tilde{z}_1+\tau -(\Delta +\sigma )\right) }\\&= \beta _L-\beta _Le^{-\tau } +a(1-e^{-\sigma })e^{-\left( \tilde{z}_1+\tau -(\Delta +\sigma )\right) }\\&= \overline{x} +a(e^{\sigma }-1)e^{-(\tilde{z}_1+\tau )+\Delta }>\overline{x}, \end{aligned}$$

and $x^{(\Delta )}(z_{\Delta ,1}+\tau )$ is increasing as a function of $\Delta \in I_{RPRP}$. Also $x^{(\Delta )}(\Delta +\sigma )$ is increasing as a function of $\Delta \in I_{RPRP}$. In case $\mathbf {RPRP1}$ the function $x^{(\Delta )}$ is increasing on $[\tilde{z}_1,\tilde{z}_1+\tau ]$ with $x^{(\Delta )}(t)\le \beta _L$ on this interval, hence $\max _{\tilde{z}_1\le t\le \tilde{z}_1+\tau }x^{(\Delta )}(t)=x^{(\Delta )}(\tilde{z}_1+\tau )\le \beta _L$. In case $\mathbf {RPRP2}$ the function $x^{(\Delta )}$ is increasing on $[\tilde{z}_1,\Delta +\sigma ]$ and decreasing on $[\Delta +\sigma ,\tilde{z}_1+\tau ]$, with $x^{(\Delta )}(t)>\beta _L$ on this interval, hence $\max _{\tilde{z}_1\le t\le \tilde{z}_1+\tau }x^{(\Delta )}(t)=x^{(\Delta )}(\Delta +\sigma )>\beta _L$. In both subcases,

$$\begin{aligned} \max _{\tilde{z}_1\le t\le \tilde{z}_1+\tau }x^{(\Delta )}(t)=\max \{x^{(\Delta )}(\Delta +\sigma ),x^{(\Delta )}(\tilde{z}_1+\tau )\}\ge x^{(\Delta )}(\tilde{z}_1+\tau )>\overline{x}, \end{aligned}$$

and $\max _{\tilde{z}_1\le t\le \tilde{z}_1+\tau }x^{(\Delta )}(t)$ is increasing as a function of $\Delta \in I_{RPRP}$.

As $x^{(\Delta )}(t)>0$ on $(\tilde{z}_1,\tilde{z}_1+\tau ]$ we have

$$\begin{aligned} x^{(\Delta )}(t)=-\beta _U+\bigl (x^{(\Delta )}(\tilde{z}_1+\tau )+\beta _U\bigr )e^{-\left( t-(\tilde{z}_1+\tau )\right) } \end{aligned}$$

for $t\ge \tilde{z}_1+\tau$ as long as $x^{(\Delta )}(t-\tau )>0$. It follows that there is a smallest zero $z_{\Delta ,2}$ of $x^{(\Delta )}$ in $(\tilde{z}_1+\tau ,\infty )$, and

$$\begin{aligned} x^{(\Delta )}(z_{\Delta ,2}+t)=\tilde{x}(\tilde{z}_0+t)\quad \text {for all}\quad t\ge 0. \end{aligned}$$

This yields

$$\begin{aligned} T(\Delta )=\bigl (z_{\Delta ,2}+(\tilde{z}_1-\tilde{z}_0)\bigr )-\tilde{z}_1=z_{\Delta ,2}+\tau . \end{aligned}$$

Moreover,

$$\begin{aligned} \overline{x}_{\Delta }&= \max _{\tilde{z}_1\le t\le z_{\Delta ,2}+(\tilde{z}_1-\tilde{z}_0)}x^{(\Delta )}(t)\\&= \max _{\tilde{z}_1\le t\le \tilde{z}_1+\tau }x^{(\Delta )}(t)\\&= \max \{x^{(\Delta )}(\Delta +\sigma ),x^{(\Delta )}(\tilde{z}_1+\tau )\}\ge \overline{x} \end{aligned}$$

is increasing as a function of $\Delta \in I_{RPRP}$, and $\underline{x}_{\Delta }=\underline{x}$.

Recall $t_{\max }=\tilde{z}_1+\tau$. From

$$\begin{aligned} 0=x^{(\Delta )}(z_{\Delta ,2})=-\beta _U+\bigl (x^{(\Delta )}(t_{\max })+\beta _U\bigr )e^{-(z_{\Delta ,2}-t_{\max })} \end{aligned}$$

we get

$$\begin{aligned} \beta _U e^{z_{\Delta ,2}}=\bigl (x^{(\Delta )}(t_{\max })+\beta _U\bigr )e^{t_{\max }}=(\overline{x}+\beta _U)e^{t_{\max }}+a(e^{\sigma }-1)e^{\Delta }. \end{aligned}$$

Since $(\overline{x}+\beta _U)e^{t_{\max }}=\beta _{U}e^{\tilde{z}_2}$, we conclude that

$$\begin{aligned} \beta _U e^{z_{\Delta ,2}}=\beta _U e^{\tilde{z}_2} +a(e^{\sigma }-1)e^{\Delta }. \end{aligned}$$

For the cycle length we obtain $T(\Delta )=z_{\Delta ,2}+\tau >\tilde{z}_2+\tau =\tilde{T}$ and the formula for $T(\Delta )$ follows. $\square$

Proof of Remark 5.1

• 1. Let $\delta _1>0$. We first show that the expressions defining $T(\Delta )$ in Proposition 5.1 and in Proposition 5.2 yield the same value for $\Delta =\delta _1$. Consider the argument of $\ln$ in (5.8). We have

  $$\begin{aligned}&\dfrac{a(e^{\sigma }-1)}{\beta _U} e^{\Delta -\tilde{z}_2}+\dfrac{a(\beta _L+\beta _U)e^{\tau +\tilde{z}_1-\tilde{z}_2}}{\beta _U(\beta _L+a)} (e^{\Delta -\tilde{z}_1}-1)\\&\quad =\dfrac{ae^{-\tilde{z}_2}}{\beta _U(\beta _L+a)} \Bigl (\bigl ((\beta _L+a)(e^{\sigma }-1)+(\beta _L+\beta _U)e^{\tau }\bigr )e^{\Delta } -(\beta _L+\beta _U)e^{\tau +\tilde{z}_1}\Bigr ). \end{aligned}$$

  From (5.6) for $\delta _1$ it follows that

  $$\begin{aligned} \beta _Le^\sigma +a(e^\sigma -1)=\beta _Le^{\tilde{z}_1-\delta _1}, \end{aligned}$$

  which gives

  $$\begin{aligned} (\beta _L+a)(e^{\sigma }-1)=\beta _Le^{\tilde{z}_1-\delta _1}-\beta _L. \end{aligned}$$

  Since

  $$\begin{aligned} -\beta _L+(\beta _L+\beta _U)e^{\tau }=e^{\tau }(\beta _U+\overline{x})=\beta _Ue^{\tilde{z}_2-\tilde{z}_1}, \end{aligned}$$

  we obtain

  $$\begin{aligned}&\dfrac{a(e^{\sigma }-1)}{\beta _U} e^{\Delta -\tilde{z}_2}+\dfrac{a(\beta _L+\beta _U)e^{\tau +\tilde{z}_1-\tilde{z}_2}}{\beta _U(\beta _L+a)} (e^{\Delta -\tilde{z}_1}-1)\\&\quad =\dfrac{ae^{-\tilde{z}_2}}{\beta _U(\beta _L+a)} \left( (\beta _Le^{\tilde{z}_1-\delta _1}+\beta _Ue^{\tilde{z}_2-\tilde{z}_1})e^{\Delta } -(\beta _Le^{\tilde{z}_1}+\beta _Ue^{\tilde{z}_2})\right) , \end{aligned}$$

  which for $\Delta =\delta _1$ becomes

  $$\begin{aligned} \dfrac{ae^{-\tilde{z}_2}}{\beta _U(\beta _L+a)} \left( (\beta _Le^{\tilde{z}_1-\delta _1}+\beta _Ue^{\tilde{z}_2-\tilde{z}_1})e^{\Delta } -(\beta _Le^{\tilde{z}_1}+\beta _Ue^{\tilde{z}_2})\right) =\dfrac{a(e^{\delta _1-\tilde{z}_1}-1)}{\beta _L+a}. \end{aligned}$$

  We have

  $$\begin{aligned} \begin{aligned} e^{\delta _1-\tilde{z}_1}-1&=\dfrac{\beta _L}{\beta _Le^{\sigma }+a(e^{\sigma }-1)}-1 =-\dfrac{(\beta _L+a)(e^{\sigma }-1)}{\beta _Le^{\sigma }+a(e^{\sigma }-1)}, \end{aligned} \end{aligned}$$

  which leads to

  $$\begin{aligned} \begin{aligned} \dfrac{a(e^{\delta _1-\tilde{z}_1}-1)}{\beta _L+a}&=-\dfrac{a(e^{\sigma }-1)}{\beta _Le^{\sigma }+a(e^{\sigma }-1)} =-\dfrac{a(e^{\sigma }-1)}{\beta _L}e^{\delta _1-\tilde{z}_1} \end{aligned} \end{aligned}$$

  and shows that the formulae for $T(\Delta )$ from Propositions 5.1 and 5.2 yield the same value for $\Delta =\delta _1>0$. From (5.7) of Proposition 5.1, this value is strictly less than $\tilde{T}$.

• 2. By continuity, we infer $T(\Delta )<\tilde{T}$ for $\Delta$ close to $\delta _1>0$.

$\square$

Proof of Proposition 5.4

Since $x^{(\Delta )}(t)>0$ for $\tilde{z}_1<t<\Delta +\sigma$ and $\tilde{z}_1+\tau <\Delta +\sigma$ we obtain that on $[\Delta +\sigma ,\infty )$,

$$\begin{aligned} x^{(\Delta )}(t)=-\beta _U+\bigl (x^{(\Delta )}(\Delta +\sigma )+\beta _U\bigr )e^{-\left( t-(\Delta +\sigma )\right) }\quad \text {as long as}\quad 0<x^{(\Delta )}(t-\tau ). \end{aligned}$$

As $-\beta _U<0$ there is a smallest zero $z_{\Delta ,2}$ of $x^{(\Delta )}$ in $[\Delta +\sigma ,\infty )$, and

$$\begin{aligned} x^{(\Delta )}(z_{\Delta ,2}+t)=\tilde{x}(\tilde{z}_2+t)\quad \text {for all}\quad t\ge 0. \end{aligned}$$

It follows that

$$\begin{aligned} T(\Delta )=z_{\Delta ,2}+(\tilde{z}_3-\tilde{z}_2)-\tilde{z}_1=z_{\Delta ,2}+\tilde{T}-\tilde{z}_2=z_{\Delta ,2}-\tilde{z}_0=z_{\Delta ,2}+\tau . \end{aligned}$$

Moreover, $\overline{x}_{\Delta }=x^{(\Delta )}(t_{\max })$ if $\Delta \le \hat{\delta }_2$ (in which case $x^{(\Delta )}$ is decreasing on $[t_{\max },\Delta +\sigma ]$), while for $\Delta >\hat{\delta }_2$ the function $x^{(\Delta )}$ is increasing on $[t_{\max },\Delta +\sigma ]$ and $\overline{x}_{\Delta }=x^{(\Delta )}(\Delta +\sigma )>x^{(\Delta )}(t_{\max })\ge \overline{x}$. Hence $\overline{x}_{\Delta }\ge \overline{x}$. Obviously, $\underline{x}_{\Delta }=\underline{x}$. Also, $\overline{x}_{\Delta }$ is strictly decreasing as a function of $\Delta \in I_{RPFP}$, see (5.10).

From

$$\begin{aligned} 0=x^{(\Delta )}(z_{\Delta ,2})=-\beta _U+\bigl (x^{(\Delta )}(\Delta +\sigma )+\beta _U\bigr ) e^{-(z_{\Delta ,2}-(\Delta +\sigma ))} \end{aligned}$$

it follows that

$$\begin{aligned} \beta _U e^{z_{\Delta ,2}}&= \bigl (x^{(\Delta )}(\Delta +\sigma )+\beta _U\bigr )e^{ \Delta +\sigma }\\&= \bigl (\beta _Ue^{\tilde{z}_2-( \Delta +\sigma )} +a(1-e^{-\sigma })\bigr )e^{ \Delta +\sigma }\quad \text {(see (5.10))}\\&= \beta _Ue^{\tilde{z}_2}+a(e^{\sigma }-1)e^{\Delta }. \end{aligned}$$

Hence $z_{\Delta ,2}>\tilde{z}_2$, and thereby

$$\begin{aligned} T(\Delta )=z_{\Delta ,2}+\tau >\tilde{z}_2+\tau =\tilde{T} \end{aligned}$$

Furthermore,

$$\begin{aligned} T(\Delta )&= z_{\Delta ,2}+\tau =\tilde{T}+z_{\Delta ,2}-\tilde{z}_2\\&= \tilde{T}+\ln \left( 1+\dfrac{a(e^{\sigma }-1)e^{\Delta -\tilde{z}_2}}{\beta _U}\right) , \end{aligned}$$

and the map

$$\begin{aligned} I_{RPFP}\ni \Delta \mapsto T(\Delta )\in \mathbb {R} \end{aligned}$$

is strictly increasing. $\square$

Proof of Proposition 5.5

From $x^{(\Delta )}(\Delta +\sigma )<0<x^{(\Delta )}(t_{\max })$ we know that $x^{(\Delta )}$ is strictly decreasing on $[t_{\max },\Delta +\sigma ]$. It follows that there is a single zero in this interval, which is given by

$$\begin{aligned} 0=x^{(\Delta )}(z_{\Delta ,2})=-\beta _U+a+\bigl (x^{(\Delta )}(t_{\max })+\beta _U-a\bigr )e^{-(z_{\Delta ,2}-t_{\max })}, \end{aligned}$$

or equivalently,

$$\begin{aligned} (\beta _U-a)e^{z_{\Delta ,2}} = \bigl (x^{(\Delta )}(t_{\max })+\beta _U-a\bigr )e^{t_{\max }}. \end{aligned}$$

Also,

$$\begin{aligned} (\beta _U-a)e^{z_{\Delta ,2}}&= \bigl (\overline{x}+a(1-e^{\Delta -t_{\max }})+\beta _U-a\bigr )e^{t_{\max }}\\&= (\overline{x}+\beta _U)e^{t_{\max }}-ae^{\Delta }, \end{aligned}$$

and we arrive at

$$\begin{aligned} (\beta _U-a)e^{z_{\Delta ,2}}= \beta _U e^{\tilde{z}_2}-ae^{\Delta }. \end{aligned}$$

(9.5)

Since $\beta _U>a$ we infer that the map $I_{RPFN}\ni \Delta \mapsto e^{z_{\Delta ,2}-\Delta }\in \mathbb {R}$ is strictly decreasing.

We have $z_{\Delta ,2}<\Delta +\sigma \le z_{\Delta ,2}+\tau$, and on $[\Delta +\sigma ,z_{\Delta ,2}+\tau ]$,

$$\begin{aligned} x^{(\Delta )}(t)=-\beta _U+\bigl (x^{(\Delta )}(\Delta +\sigma )+\beta _U\bigr )e^{-(t-(\Delta +\tau ))} \end{aligned}$$

is strictly decreasing and negative because we have $x^{(\Delta )}(\Delta +\sigma )+\beta _U>0$ from (5.10). It follows that $x^{(\Delta )}$ is strictly increasing on $[z_{\Delta ,2}+\tau ,\infty )$ as long as $x^{(\Delta )}(t-\tau )<0$. There is a smallest zero $z_{\Delta ,3}$ of $x^{(\Delta )}$ in this interval, and

$$\begin{aligned} x^{(\Delta )}(z_{\Delta ,3}+t)&= \tilde{x}(\tilde{z}_1+t)\quad \text {for all}\quad t\ge 0,\\ T(\Delta )&= z_{\Delta ,3}-\tilde{z}_1,\\ \overline{x}_{\Delta }&= x^{(\Delta )}(t_{\max })\ge \overline{x},\\ \underline{x}_{\Delta }&= x^{(\Delta )}(z_{\Delta ,2}+\tau ). \end{aligned}$$

We compute

$$\begin{aligned} \underline{x}_{\Delta }=x^{(\Delta )}(z_{\Delta ,2}+\tau ) = -\beta _U+\bigl (x^{(\Delta )}(\Delta +\sigma )+\beta _U\bigr ) e^{-\left( z_{\Delta ,2}+\tau -(\Delta +\sigma )\right) } \end{aligned}$$

and use

$$\begin{aligned} x^{(\Delta )}(\Delta +\sigma )+\beta _U=\beta _U e^{\tilde{z}_2-(\Delta +\sigma )}+a(1-e^{-\sigma })>0 \end{aligned}$$

from (5.10). This gives

$$\begin{aligned} \underline{x}_{\Delta }=-\beta _U+\bigl (\beta _U e^{\tilde{z}_2} -a e^{\Delta }\bigr )e^{-(z_{\Delta ,2}+\tau )} +ae^{-\left( z_{\Delta ,2}+\tau -(\Delta +\sigma )\right) }. \end{aligned}$$

With (9.5) we obtain

$$\begin{aligned} \underline{x}_{\Delta }&= -\beta _U+(\beta _U-a)e^{-\tau } +ae^{-\left( z_{\Delta ,2}+\tau -(\Delta +\sigma )\right) }\\&= -\beta _U+\beta _U e^{-\tau }+ae^{-\tau }(e^{-z_{\Delta ,2}+\Delta +\sigma }-1). \end{aligned}$$

Since $\Delta +\sigma >z_{\Delta ,2}$ we conclude that

$$\begin{aligned} \underline{x}_{\Delta }>-\beta _U+\beta _U e^{-\tau }=\underline{x}. \end{aligned}$$

We turn to the cycle length $T(\Delta )=z_{\Delta ,3}-\tilde{z}_1$. The equation for $z_{\Delta ,3}$, namely,

$$\begin{aligned} 0&= x^{(\Delta )}(z_{\Delta ,3})=\beta _L+\bigl (x^{(\Delta )}(z_{\Delta ,2}+\tau )-\beta _L\bigr )e^{-\left( z_{\Delta ,3}-(z_{\Delta ,2}+\tau )\right) }\\&= \beta _L+(\underline{x}_{\Delta }-\beta _L)e^{-\left( z_{\Delta ,3}-(z_{\Delta ,2}+\tau )\right) } \end{aligned}$$

yields

$$\begin{aligned} \beta _Le^{z_{\Delta ,3}}&= (\beta _L-\underline{x}_{\Delta })e^{z_{\Delta ,2}+\tau }\\&= (\beta _L+\beta _U)e^{z_{\Delta ,2}+\tau }-(\beta _U e^{\tilde{z}_2} -a e^{\Delta })- ae^{\Delta +\sigma }\\&\qquad \text {(with the formula for}\quad \underline{x}_{\Delta }\quad \text {and (9.5))}\\&= (\beta _L+\beta _U)e^{z_{\Delta ,2}+\tau }-\beta _Ue^{\tilde{z}_2}-a(e^{\sigma }-1)e^{\Delta }. \end{aligned}$$

Since $\tilde{z}_3=\tilde{z}_2+\tau +\tilde{z}_1$, we have

$$\begin{aligned} \begin{aligned} \beta _Le^{\tilde{z}_3}+\beta _Ue^{\tilde{z}_2}&=(\beta _L e^{\tilde{z}_1}+\beta _U e^{-\tau })e^{\tilde{z}_2+\tau }\\&=(\beta _L+\beta _U)e^{\tilde{z}_2+\tau }, \end{aligned} \end{aligned}$$

which gives

$$\begin{aligned} \begin{aligned} \beta _Le^{z_{\Delta ,3}}&= \beta _Le^{\tilde{z}_3}+(\beta _L+\beta _U)e^{\tau }(e^{z_{\Delta ,2}}-e^{\tilde{z}_2})-a(e^{\sigma }-1)e^{\Delta }. \end{aligned} \end{aligned}$$

Now use (9.5) again to obtain

$$\begin{aligned} \begin{aligned} \beta _Le^{z_{\Delta ,3}}&= \beta _Le^{\tilde{z}_3}-\dfrac{a(\beta _L+\beta _U)e^{\tau }}{\beta _U-a} (e^{\Delta }-e^{\tilde{z}_2})-a(e^{\sigma }-1)e^{\Delta }, \end{aligned} \end{aligned}$$

which implies the formula for $T(\Delta )$. We have

$$\begin{aligned} \dfrac{(\beta _L+\beta _U)}{\beta _U-a}e^{\tau }+e^{\sigma }-1 =\dfrac{\beta _Le^{\tilde{z}_1+\tau }+(\beta _U-a)e^{\sigma }+a}{\beta _U-a}. \end{aligned}$$

Thus, the map $I_{RPFN} \ni \Delta \mapsto T(\Delta )\in \mathbb {R}$ is strictly decreasing, since $\beta _U>a$. $\square$

Proof of Remark 5.2

• 1. We first show that the expressions defining $T(\Delta )$ in Proposition 5.3 and in Proposition 5.5 (Eqs. 5.9 and 5.13 respectively) yield the same value for $\Delta =\delta _2$. Consider the argument of $\ln$ in (5.13) of Proposition 5.5. We have

  $$\begin{aligned}&\dfrac{a(e^{\sigma }-1)}{\beta _L}e^{\Delta -\tilde{z}_1-\tilde{T}}+\dfrac{a(\beta _L+\beta _U)e^{-\tilde{z}_1}}{\beta _L(\beta _U-a)}e^{\Delta -\tilde{z}_2}\\&\quad = \dfrac{ae^{\Delta -\tilde{z}_1-\tilde{T}}}{\beta _L(\beta _U-a)}\Bigl ((\beta _U-a)(e^{\sigma }-1)+(\beta _L+\beta _U)e^{\tau }\Bigr ) \end{aligned}$$

  Note that from the definition of $\delta _2$ it follows that

  $$\begin{aligned} \beta _Ue^{\sigma }-a(e^{\sigma }-1)=\beta _Ue^{\tilde{z}_2-\delta _2}, \end{aligned}$$

  which gives

  $$\begin{aligned} \begin{aligned} (\beta _U-a)(e^{\sigma }-1)+(\beta _L+\beta _U)e^{\tau }&=\beta _Ue^{\sigma }-a(e^{\sigma }-1)+(\beta _L-\underline{x})e^{\tau }\\&=\beta _Ue^{\tilde{z}_2-\delta _2}+\beta _L e^{t_{\max }} \end{aligned} \end{aligned}$$

  and leads to

  $$\begin{aligned}&\dfrac{a(e^{\sigma }-1)}{\beta _L}e^{\Delta -\tilde{z}_1-\tilde{T}}+\dfrac{a(\beta _L+\beta _U)e^{-\tilde{z}_1}}{\beta _L(\beta _U-a)}(e^{\Delta -\tilde{z}_2}-1)\\&\quad = \dfrac{ae^{-\tilde{z}_1}}{\beta _L(\beta _U-a)}\Bigl ((\beta _Ue^{-\tau -\delta _2}+\beta _L e^{\tilde{z}_1-\tilde{z}_2})e^{\Delta }-(\beta _L+\beta _U)\Bigr ). \end{aligned}$$

  Observe that for $\Delta =\delta _2$ we have

  $$\begin{aligned} (\beta _Ue^{-\tau -\delta _2}+\beta _L e^{\tilde{z}_1-\tilde{z}_2})e^{\Delta }= \beta _U e^{-\tau }+\beta _Le^{\tilde{z}_1+\delta _2-\tilde{z}_2}. \end{aligned}$$

  Using this we obtain that for $\Delta =\delta _2$,

  $$\begin{aligned} \dfrac{a(e^{\sigma }-1)}{\beta _L}e^{\Delta -\tilde{z}_1-\tilde{T}}+\dfrac{a(\beta _L+\beta _U)e^{-\tilde{z}_1}}{\beta _L(\beta _U-a)}(e^{\Delta -\tilde{z}_2}-1)= \dfrac{a}{(\beta _U-a)}(e^{\delta _2-\tilde{z}_2}-1). \end{aligned}$$

  We have

  $$\begin{aligned} \dfrac{a}{(\beta _U-a)}(e^{\delta _2-\tilde{z}_2}-1)&=\dfrac{a}{(\beta _U-a)}\left( \dfrac{\beta _U}{\beta _Ue^{\sigma }-a(e^{\sigma }-1)}-1\right) \\&=-\dfrac{a(e^{\sigma }-1)}{\beta _Ue^{\sigma }-a(e^{\sigma }-1)}\\&=-\dfrac{a(e^{\sigma }-1)}{\beta _U}e^{\delta _2-\tilde{z}_2}. \end{aligned}$$

  Thus, the formulae (5.9) and (5.13) from Propositions 5.3 and 5.5 are the same for $\Delta =\delta _2$. Since

  $$\begin{aligned} -\dfrac{a(e^{\sigma }-1)}{\beta _U}e^{\delta _2-\tilde{z}_2}<0 \end{aligned}$$

  we also deduce that the value given by both equations for $\Delta =\delta _2$ is strictly larger than $\tilde{T}$.

• 2. By continuity, we infer $T(\Delta )>\tilde{T}$ for $\Delta$ close to $\delta _2$. $\square$

Proof of Proposition 5.6

From $0\le x^{(\Delta )}(t)$ on $[t_{\max },\Delta +\sigma ]$ we infer that $x^{(\Delta )}$ decreases on $[\Delta +\sigma ,\infty )$ as long as $x^{(\Delta )}(t-\tau )>0$. This yields the existence of a smallest zero $z_{\Delta ,2}$ in $[\Delta +\sigma ,\infty )$, and $x^{(\Delta )}(t+z_{\Delta ,2})=\tilde{x}(t+\tilde{z}_2)$ for all $t\ge 0$.

Hence

$$\begin{aligned} T(\Delta )=z_{\Delta ,2}+\tilde{z}_3-\tilde{z}_2-\tilde{z}_1=z_{\Delta ,2}+\tilde{T}-\tilde{z}_2=z_{\Delta ,2}+\tau . \end{aligned}$$

We have

$$\begin{aligned} -\beta _U+\bigl (x^{(\Delta )}(\Delta +\sigma )+\beta _U\bigr )e^{-(z_{\Delta ,2}-(\Delta +\sigma ))}=0 \end{aligned}$$

and conclude that

$$\begin{aligned} \begin{aligned} \beta _U e^{z_{\Delta ,2}}&=\bigl (x^{(\Delta )}(\Delta +\sigma )+\beta _U\bigr )e^{\Delta +\sigma }\\&=\bigl (\beta _Ue^{\tilde{z}_2-\sigma -\Delta }+a(1-e^{- \sigma })\bigr )e^{\Delta +\sigma }\\&=\beta _Ue^{\tilde{z}_2}+a(e^{\sigma }-1)e^{\Delta }, \end{aligned} \end{aligned}$$

which implies (5.9), the desired formula for $T(\Delta )$. $\square$

Proof of Proposition 5.7

Since $[\Delta ,\Delta +\sigma ]\subset (t_{\max },\tilde{T})$ we have

$$\begin{aligned} x^{(\Delta )}(t)=-\beta _U+a+\bigl (x^{(\Delta )}(\Delta )+\beta _U-a\bigr )e^{-(t-\Delta )}\quad \text {on}\quad [\Delta ,\Delta +\sigma ]. \end{aligned}$$

Using $x^{(\Delta )}(\Delta )+\beta _U-a>0$ we see that $x^{(\Delta )}$ is strictly decreasing on $[\Delta ,\Delta +\sigma ]$, and has a unique zero $z_{\Delta ,2}$ in $[\Delta ,\Delta +\sigma )$, which is given implicitly by

$$\begin{aligned} 0=x^{(\Delta )}(z_{\Delta ,2})=-\beta _U+a+\bigl (x^{(\Delta )}(\Delta )+\beta _U-a\bigr ) e^{-(z_{\Delta ,2}-\Delta )}. \end{aligned}$$

Combining this with (5.2) gives

$$\begin{aligned} (\beta _U-a)e^{z_{\Delta ,2}}=\beta _Ue^{\tilde{z}_2}-ae^{\Delta }, \end{aligned}$$

which is the same as (9.5). As $z_{\Delta ,2}\ge \Delta$ we infer

$$\begin{aligned} \beta _U(e^{z_{\Delta ,2}}-e^{\tilde{z}_2})=a(e^{z_{\Delta ,2}}-e^{\Delta })\ge 0. \end{aligned}$$

Hence $z_{\Delta ,2}\ge \tilde{z}_2$. Since $x^{(\Delta )}(\Delta )\ge 0$ and $\beta _U>a$, the map $[\Delta ,\Delta +\sigma ] \ni t\mapsto x^{(\Delta )}(t)$ is strictly decreasing. The rest of the proof is the same as the proof of Proposition 5.5 starting after (9.5). $\square$

Proof of Proposition 5.8

• 1. We have $x^{(\Delta )}(\tilde{z}_2)=0$, and the function $x^{(\Delta )}$ is strictly decreasing on $[\tilde{z}_2,\Delta ]$, monotone on $[\Delta ,\Delta +\sigma ]$ with $x^{(\Delta )}(t)<0$ for $\Delta \le t<\Delta +\sigma$, and strictly decreasing on $[\Delta +\sigma ,\tilde{T}]$. It follows that $x^{(\Delta )}$ is strictly increasing on $[\tilde{T},\infty )$ as long as $x^{(\Delta )}(t-\tau )<0$. This yields a first zero $z_{\Delta ,3}$ of $x^{(\Delta )}$ in $[\tilde{T},\infty )$ and $x^{(\Delta )}(z_{\Delta ,3}+t)=\tilde{x}(\tilde{z}_1+t)$ for all $t\ge 0$, and thus

  $$\begin{aligned} T(\Delta )= & {} z_{\Delta ,3}+(\tilde{z}_2-\tilde{z}_1)-\tilde{z}_2=z_{\Delta ,3}-\tilde{z}_1,\\ \underline{x}_{\Delta }= & {} \min \{x^{(\Delta )}(\Delta ),x^{(\Delta )}(\tilde{T})\}, \end{aligned}$$

  and

  $$\begin{aligned} \overline{x}_{\Delta }=\overline{x}. \end{aligned}$$

  Observe that

  $$\begin{aligned} {\begin{matrix} x^{(\Delta )}(\tilde{T})&{}=-\beta _U+\beta _U e^{-\tau }+a ( e^{ \sigma }-1)e^{\Delta -\tilde{T}}\\ &{}=\underline{x}+a ( e^{ \sigma }-1)e^{\Delta -\tilde{T}}>\underline{x}. \end{matrix}} \end{aligned}$$

  Using this and $x^{(\Delta )}(\Delta )=\tilde{x}(\Delta )>\underline{x}$ we have $\underline{x}_{\Delta }>\underline{x}$. The equation

  $$\begin{aligned} 0=x^{(\Delta )}(z_{\Delta ,3})=\beta _L+\bigl (x^{(\Delta )}(\tilde{T})-\beta _L\bigr )e^{-(z_{\Delta ,3}-\tilde{T})} \end{aligned}$$

  is equivalent to

  $$\begin{aligned} \beta _L e^{z_{\Delta ,3}}=\bigl (\beta _L-x^{(\Delta )}(\tilde{T})\bigr )e^{\tilde{T}}, \end{aligned}$$

  which gives

  $$\begin{aligned} {\begin{matrix} \beta _L e^{z_{\Delta ,3}}&{}=(\beta _L+\beta _U)e^{\tilde{T}}-\bigl (\beta _U e^{-\tau }+a ( e^{ \sigma }-1)e^{\Delta -\tilde{T}}\bigr )e^{\tilde{T}}\\ &{}=(\beta _L+\beta _U-\beta _Ue^{-\tau }) e^{\tilde{T}}-a ( e^{ \sigma }-1)e^{\Delta }. \end{matrix}} \end{aligned}$$

  We have $\beta _L+\beta _U-\beta _Ue^{-\tau }=\beta _L e^{\tilde{z}_1}$ and we conclude that

  $$\begin{aligned} \beta _L e^{z_{\Delta ,3}}=\beta _Le^{\tilde{z}_1+\tau +\tilde{z}_2}-a(e^{\sigma }-1)e^{\Delta }, \end{aligned}$$

  from which we obtain (5.14) for $T(\Delta )=z_{\Delta ,3}-\tilde{z}_1$.

• 2. Observe that for every $\Delta \in I_{FNFN}$ we have

  $$\begin{aligned} x^{(\Delta )}(\tilde{T})-\tilde{x}(\Delta )=g(\Delta ) \end{aligned}$$

  with the strictly increasing function

  $$\begin{aligned} g:\mathbb {R}\ni \Delta \mapsto a(e^{\sigma }-1)e^{\Delta -\tilde{T}}+\beta _Ue^{-\tau }-\beta _Ue^{\tilde{z}_2-\Delta }\in \mathbb {R} \end{aligned}$$

  which has a single zero at $\Delta =\overline{\delta }$ since $g(\Delta )\rightarrow -\infty$ as $\Delta \rightarrow -\infty$, $g(\Delta )\rightarrow \infty$ as $\Delta \rightarrow \infty$, and $g(\Delta )=0$ if and only if

  $$\begin{aligned} \beta _U(e^{\tilde{T}-\Delta })^2-\beta _Ue^{\tilde{T}-\Delta }-a(e^{\sigma }-1)e^{\tau }=0, \end{aligned}$$

  or equivalently,

  $$\begin{aligned} e^{\tilde{T}-\Delta }&= \dfrac{1}{2}+\sqrt{\dfrac{1}{4}+\dfrac{a(e^{\sigma }-1)e^{\tau }}{\beta _U}}\\&= \dfrac{\beta _U+\sqrt{\beta _U ^2 +4a\beta _U(e^{\sigma }-1)e^{\tau }}}{2\beta _U}. \end{aligned}$$

  For $\Delta <\overline{\delta }$ in $I_{FNFN}$ we have $g(\Delta )<0$, hence

  $$\begin{aligned} \underline{x}_{\Delta }=\min \{\tilde{x}(\Delta ),x^{(\Delta )}(\tilde{T})\}=x^{(\Delta )}(\tilde{T}), \end{aligned}$$

  and the formula (5.10) for $x^{(\Delta )}(\tilde{T})$ in Part 1 above shows that in case $I_{FNFN}\cap (-\infty ,\overline{\delta })\ne \emptyset$ the map

  $$\begin{aligned} I_{ FNFN}\cap (-\infty ,\overline{\delta })\ni \Delta \mapsto \underline{x}_{\Delta }\in \mathbb {R} \end{aligned}$$

  is strictly increasing. For $\Delta >\overline{\delta }$ in $I_{ FNFN}$ we get $0<g(\Delta )$, hence

  $$\begin{aligned} \underline{x}_{\Delta }=\min \{\tilde{x}(\Delta ),x^{(\Delta )}(\tilde{T})\}=\tilde{x}(\Delta ), \end{aligned}$$

  and we see that in case $I_{ FNFN}\cap (\overline{\delta },\infty )\ne \emptyset$ the map

  $$\begin{aligned} I_{ FNFN}\cap (\overline{\delta },\infty )\ni \Delta \mapsto \underline{x}_{\Delta }\in \mathbb {R} \end{aligned}$$

  is strictly decreasing. $\square$

Proof of Remark 5.3

The strictly increasing function g from the previous proof satisfies

and

$$\begin{aligned} g(\tilde{T}-\sigma )=a(e^{\sigma }-1)e^{-\sigma }+\beta _Ue^{-\tau }-\beta _Ue^{\tilde{z}_2-\tilde{T}+\sigma }>0 \end{aligned}$$

if and only if $\beta _U e^{\sigma - \tau } < a$. By the intermediate value theorem the only zero $\overline{\delta }$ of g belongs to the interval $(\tilde{z}_2,\tilde{T}-\sigma )$ in case $\beta _U e^{\sigma - \tau } < a$. Otherwise $\overline{\delta }\ge \tilde{T}-\sigma$. $\square$

Proof of Proposition 5.9

From $0>x^{(\Delta )}(\Delta +\sigma )>x^{(\Delta )}(\tilde{T})$ we obtain $x^{(\Delta )}(t)<0$ on $(\tilde{z}_2,\Delta +\sigma ]$. It follows that on $[\Delta +\sigma ,\infty )$ the function $x^{(\Delta )}$ is strictly increasing as long as $x^{(\Delta )}(t-\tau )<0$, and there is a smallest zero $z_{\Delta ,3}$ of $x^{(\Delta )}$ in $[\Delta +\sigma ,\infty )$. Moreover, $x^{(\Delta )}(z_{\Delta ,3}+t)=\tilde{x}(\tilde{z}_1+t)$ for all $t\ge 0$, hence

$$\begin{aligned} T(\Delta )=z_{\Delta ,3}+(\tilde{z}_2-\tilde{z}_1)-\tilde{z}_2=z_{\Delta ,3}-\tilde{z}_1, \end{aligned}$$

and

$$\begin{aligned} \underline{x}_{\Delta }=x^{(\Delta )}(\tilde{T})>\underline{x}, \end{aligned}$$

and $\overline{x}_{\Delta }=\overline{x}$. As the map $[\tilde{T}-\sigma ,\tilde{T})\ni \Delta \mapsto x^{(\Delta )}(\tilde{T})\in \mathbb {R}$ is strictly decreasing we infer that also the map $I_{FNRN}\ni \Delta \mapsto \underline{x}_{\Delta }\in \mathbb {R}$ is strictly decreasing.

From

$$\begin{aligned} 0=x^{(\Delta )}(z_{\Delta ,3})=\beta _L+\bigl (x^{(\Delta )}(\Delta +\sigma )-\beta _L\bigr ) e^{-(z_{\Delta ,3}-(\Delta +\sigma ))} \end{aligned}$$

we obtain

$$\begin{aligned} \beta _L e^{z_{\Delta ,3}}&= \bigl (\beta _L- x^{(\Delta )}(\Delta +\sigma )\bigr )e^{ \Delta +\sigma }\\&= \bigl (\beta _L-\beta _L+\beta _Le^{\tilde{z}_1-( \Delta +\sigma -\tilde{T})}-a(1-e^{-\sigma })\bigr )e^{ \Delta +\sigma }\\&= \beta _Le^{\tilde{z}_1+\tilde{T}}- a(e^{\sigma }-1)e^{ \Delta }, \end{aligned}$$

and the formula for $T(\Delta )=z_{\Delta ,3}-\tilde{z}_1$ follows. $\square$

Proof of Proposition 5.10

We have $x^{(\Delta )}(t)<0$ on $(\tilde{z}_2,\tilde{T}]$, and there is a first zero of $x^{(\Delta )}$ in $(\tilde{T},\Delta +\sigma ]$, which is given by

$$\begin{aligned} 0=x^{(\Delta )}(z_{\Delta ,3})=\beta _L+a+\bigl (x^{(\Delta )}(\tilde{T})-(\beta _L+a)\bigr )e^{-(z_{\Delta ,3}-\tilde{T})}, \end{aligned}$$

or equivalently,

$$\begin{aligned} (\beta _L+a)e^{z_{\Delta ,3}}&= \bigl (\beta _L+a-x^{(\Delta )}(\tilde{T})\bigr )e^{\tilde{T}}\\&= \bigl (\beta _L+a-(\underline{x}+a-ae^{\Delta -\tilde{T} })\bigr )e^{\tilde{T}}\\&= (\beta _L-\underline{x})e^{\tilde{T}}+ae^{\Delta }\\&=\beta _L e^{\tilde{z}_1+\tilde{T}}+ae^{\Delta }. \end{aligned}$$

Incidentally, this shows that the map $I_{FNRP}\ni \Delta \mapsto z_{\Delta ,3}-\Delta \in \mathbb {R}$ is strictly decreasing.

On $[z_{\Delta ,3},\Delta +\sigma ]$ the function $x^{(\Delta )}$ is strictly increasing. Since $x^{(\Delta )}(t-\tau )<0$ on $(\tilde{z}_2,z_{\Delta ,3})$, we infer that $x^{(\Delta )}$ is strictly increasing on $[z_{\Delta ,3},z_{\Delta ,3}+\tau ]$. Hence $0<x^{(\Delta )}(t)$ on $(z_{\Delta ,3},z_{\Delta ,3}+\tau ]$. This implies that $x^{(\Delta )}$ is strictly decreasing on $[z_{\Delta ,3}+\tau ,\infty )$ as long as $x^{(\Delta )}(t-\tau )>0$, and that there is a first zero $z_{\Delta ,4}$ in $(z_{\Delta ,3}+\tau ,\infty )$. We obtain $x^{(\Delta )}(z_{\Delta ,4}+t)=\tilde{x}(\tilde{z}_2+t)$ for all $t\ge 0$, which yields

$$\begin{aligned} T(\Delta )=z_{\Delta ,4}-\tilde{z}_2, \end{aligned}$$

$\underline{x}_{\Delta }=\min \{\tilde{x}(\Delta ),x^{(\Delta )}(\tilde{T})\}>\underline{x}$, and

$$\begin{aligned} \overline{x}_{\Delta }=x^{(\Delta )}(z_{\Delta ,3}+\tau ). \end{aligned}$$

As in the proof of Proposition 5.9 we infer that the map $I_{FNRP}\ni \Delta \mapsto \underline{x}_{\Delta }\in \mathbb {R}$ is strictly decreasing.

From the formula for $x^{(\Delta )}(\Delta +\sigma )$ we obtain

$$\begin{aligned} x^{(\Delta )}(z_{\Delta ,3}+\tau )&= \beta _L+\bigl (x^{(\Delta )}(\Delta +\sigma )-\beta _L\bigr )e^{-(z_{\Delta ,3}+\tau -(\Delta +\sigma ))}\\&= \beta _L-\beta _Le^{\tilde{z}_1+\tilde{T}-(z_{\Delta ,3}+\tau )} + a(e^{\sigma }-1)e^{\Delta }e^{-(z_{\Delta ,3}+\tau )}, \end{aligned}$$

which can be rewritten as

$$\begin{aligned} \overline{x}_{\Delta }=\beta _L-\bigl (\beta _Le^{\tilde{z}_1+\tilde{T}}+ae^{\Delta }\bigr )e^{-(z_{\Delta ,3}+\tau )} + ae^{\sigma +\Delta }e^{-(z_{\Delta ,3}+\tau )}. \end{aligned}$$

Hence,

$$\begin{aligned} \begin{aligned} \overline{x}_{\Delta }&=\beta _L-(\beta _L+a)e^{-\tau }+ae^{\sigma +\Delta }e^{-(z_{\Delta ,3}+\tau )}\\&=\overline{x} +ae^{-\tau }(e^{\sigma +\Delta -z_{\Delta ,3}}-1)\\&>\overline{x}. \end{aligned} \end{aligned}$$

We also know that the map $I_{FNRP}\ni \Delta \mapsto \overline{x}_{\Delta }\in \mathbb {R}$ is strictly increasing.

From

$$\begin{aligned} 0=x^{(\Delta )}(z_{\Delta ,4})=-\beta _U+\bigl (x^{(\Delta )}(z_{\Delta ,3}+\tau )+\beta _U\bigr )e^{-(z_{\Delta ,4}-(z_{\Delta ,3}+\tau ))} \end{aligned}$$

we have

$$\begin{aligned} \beta _Ue^{z_{\Delta ,4}}= (\beta _L+\beta _U)e^{z_{\Delta ,3}+\tau }-\beta _Le^{\tilde{z}_1+\tilde{T}}+a(e^\sigma -1)e^{\Delta }. \end{aligned}$$

Use (9.4) to obtain

$$\begin{aligned} \beta _Le^{\tilde{z}_1+\tilde{T}}+\beta _U e^{\tilde{z}_2+\tilde{T}}=(\beta _L+\beta _U)e^{\tau +\tilde{z}_1+\tilde{T}} \end{aligned}$$

which, combined with the previous equation, gives

$$\begin{aligned} \beta _Ue^{z_{\Delta ,4}}=\beta _U e^{\tilde{z}_2+\tilde{T}}+ (\beta _L+\beta _U)e^{\tau }\bigl (e^{z_{\Delta ,3}}-e^{\tilde{z}_1+\tilde{T}}\bigr )+a(e^\sigma -1)e^{\Delta }. \end{aligned}$$

Now the formula for $z_{\Delta ,3}$ leads to

$$\begin{aligned} \beta _Ue^{z_{\Delta ,4}}=\beta _U e^{\tilde{z}_2+\tilde{T}}+\dfrac{a(\beta _L+\beta _U)e^{\tau }}{\beta _L+a} \bigl (e^{\Delta }-e^{\tilde{z}_1+\tilde{T}}\bigr )+a(e^\sigma -1)e^{\Delta } \end{aligned}$$

which yields the formula for $T(\Delta )=z_{\Delta ,4}-\tilde{z}_2$. $\square$