Response of an oscillatory differential delay equation to a periodic stimulus

Abstract

Periodic hematological diseases such as cyclical neutropenia or cyclical thrombocytopenia, with their characteristic oscillations of circulating neutrophils or platelets, may pose grave problems for patients. Likewise, periodically administered chemotherapy has the unintended side effect of establishing periodic fluctuations in circulating white cells, red cell precursors and/or platelets. These fluctuations, either spontaneous or induced, often have serious consequences for the patient (e.g. neutropenia, anemia, or thrombocytopenia respectively) which exogenously administered cytokines can partially correct. The question of when and how to administer these drugs is a difficult one for clinicians and not easily answered. In this paper we use a simple model consisting of a delay differential equation with a piecewise linear nonlinearity, that has a periodic solution, to model the effect of a periodic disease or periodic chemotherapy. We then examine the response of this toy model to both single and periodic perturbations, meant to mimic the drug administration, as a function of the drug dose and the duration and frequency of its administration to best determine how to avoid side effects.

Appendix: Proof of the results

Here we present the proofs of Remarks and Propositions from Sects. 3 and 4.

Proof of Remark 3.1

First it is shown that $a>\beta _{U}$ holds. For $\varDelta \leqslant t \leqslant \varDelta +\sigma$ we have

$$\begin{aligned} x^{(\varDelta )}(t) = -\beta _{U} + a + \left( x^{(\varDelta )}(\varDelta )+\beta _{U}-a\right) \mathrm {e}^{-(t-\varDelta )}, \end{aligned}$$

(41)

with $x^{(\varDelta )}(\varDelta ) = -\beta _{U} + \beta _{U}\mathrm {e}^{\tilde{z}_{2}-\varDelta }$ and

$$\begin{aligned} x^{(\varDelta )}(\varDelta +\sigma )= & {} -\beta _{U} + a + \left( x^{(\varDelta )}(\varDelta )+\beta _{U}-a\right) \mathrm {e}^{-\sigma } \end{aligned}$$

(42)

$$\begin{aligned}= & {} \beta _{U}(\mathrm {e}^{\tilde{z}_{2}-\varDelta -\sigma }-1) + a(1-\mathrm {e}^{-\sigma }). \end{aligned}$$

(43)

From (42) the condition $x^{(\varDelta )}(\varDelta +\sigma )> 0$ can be written as

$$\begin{aligned} x^{(\varDelta )}(\varDelta +\sigma )=(a-\beta _{U})(1-\mathrm {e}^{-\sigma }) + x^{(\varDelta )}(\varDelta )\mathrm {e}^{-\sigma }, \end{aligned}$$

but $x^{(\varDelta )}(\varDelta )=-\beta _{U}(1-\mathrm {e}^{\tilde{z}_{2}-\varDelta })<0$ since $\varDelta >\tilde{z}_{2}$, then we must have $a>\beta _{U}$.

For $\varDelta +\sigma \leqslant t \leqslant \tilde{T}$ we have

$$\begin{aligned} x^{(\varDelta )}(t) = -\beta _{U} + \left( x^{(\varDelta )}(\varDelta +\sigma )+\beta _{U}\right) \mathrm {e}^{-(t-\varDelta -\sigma )}. \end{aligned}$$

(44)

Equation (43) gives $x^{(\varDelta )}(\varDelta +\sigma ) + \beta _{U} = \beta _{U}\mathrm {e}^{\tilde{z}_{2}-\varDelta -\sigma } + a(1-\mathrm {e}^{-\sigma })$ and this combined with $\tilde{T}=\tilde{z}_{2}+\tau$ from (9) in the solution (44) computed at $t=\tilde{T}$ yields

$$\begin{aligned} x^{(\varDelta )}(\tilde{T})= & {} -\beta _{U} + \left( \beta _{U}\mathrm {e}^{\tilde{z}_{2}-\varDelta -\sigma } + a(1-\mathrm {e}^{-\sigma })\right) \mathrm {e}^{-{\tilde{T}}+\varDelta +\sigma } \nonumber \\= & {} -\beta _{U}(1-\mathrm {e}^{-\tau }) + a(\mathrm {e}^{\sigma }-1)\mathrm {e}^{\varDelta -{\tilde{T}}}. \end{aligned}$$

(45)

Hence the inequality $x^{(\varDelta )}(\tilde{T})<0$ holds if and only if

$$\begin{aligned} a(\mathrm {e}^{\sigma }-1)\mathrm {e}^{\varDelta } < \beta _{U}(\mathrm {e}^{\tau }-1)\mathrm {e}^{\tilde{z}_{2}}. \end{aligned}$$

Defining $\delta _{4}$ as the constant such that $x^{(\varDelta )}(\tilde{T})=0$ for $\varDelta =\delta _{4}$ leads to (15).

Then $x^{(\varDelta )}(\tilde{T})<0$ if, and only if, $\varDelta <\delta _{4}$ while $x^{(\varDelta )}(\tilde{T})\geqslant 0$ if, and only if, $\varDelta \geqslant \delta _{4}$. The interval ${ I_{FNFP} }=(\tilde{z}_{2},\tilde{T}-\sigma )\cap (-\infty ,\delta _{2})$ combined with the condition $x^{(\varDelta )}(\tilde{T})\geqslant 0$ (see the examples in Fig. 2) gives the interval ${ I_{FNFP1} }$ given by (13), while ${ I_{FNFP} }$ combined with the condition $x^{(\varDelta )}(\tilde{T})<0$ (see the examples in Fig. 3) yields the interval ${ I_{FNFP2} }$ given by (14). $\square$

Proof of Remark 3.2

FNFP3: for this subcase we have to assume $x^{(\varDelta )}(z_{\varDelta ,3}+\tau )< 0$ (see the examples in Fig. 4).

From $x^{(\varDelta )}(\varDelta )<0$ and $x^{(\varDelta )}(\varDelta +\sigma )>0$ we obtain a zero $z_{\varDelta ,3}$ of $x^{(\varDelta )}$ in $(\varDelta ,\varDelta +\sigma )$ given by

$$\begin{aligned} 0= & {} x^{(\varDelta )}(z_{\varDelta ,3})=-\beta _{U} + a + \left( x^{(\varDelta )}(\varDelta )+\beta _{U}-a\right) \mathrm {e}^{-(z_{\varDelta ,3}-\varDelta )},\nonumber \\ \mathrm {e}^{z_{\varDelta ,3}}= & {} \frac{(a-\beta _{U}-x^{(\varDelta )}(\varDelta ))\mathrm {e}^{\varDelta }}{a-\beta _{U}} = \frac{a\mathrm {e}^{\varDelta }-\beta _{U}\mathrm {e}^{\tilde{z}_{2}}}{a-\beta _{U}}. \end{aligned}$$

(46)

From (44) and $(x^{(\varDelta )}(\varDelta +\sigma )+\beta _{U})>0$ it follows that $x^{(\varDelta )}(t)$ is strictly decreasing for $t\in [\varDelta +\sigma ,\tilde{T}]$. This together with $x^{(\varDelta )}(\varDelta +\sigma )>0$ and $x^{(\varDelta )}(\tilde{T})<0$ yields that there is a zero $z_{\varDelta ,4}$ of $x^{(\varDelta )}$ in $(\varDelta +\sigma ,\tilde{T})$ given by

$$\begin{aligned} 0= & {} x^{(\varDelta )}(z_{\varDelta ,4})=-\beta _{U} + \left( x^{(\varDelta )}(\varDelta +\sigma )+\beta _{U}\right) \mathrm {e}^{-(z_{\varDelta ,4}-\varDelta -\sigma )},\nonumber \\ \mathrm {e}^{z_{\varDelta ,4}}= & {} \frac{\left( x^{(\varDelta )}(\varDelta +\sigma )+\beta _{U}\right) \mathrm {e}^{\varDelta +\sigma }}{\beta _{U}} = \frac{\beta _{U}\mathrm {e}^{\tilde{z}_{2}} + a(\mathrm {e}^{\sigma }-1)\mathrm {e}^{\varDelta }}{\beta _{U}}. \end{aligned}$$

(47)

For $t\in (\tilde{T},z_{\varDelta ,3}+\tau )$ we have $\tilde{z}_{2}<t-\tau <z_{\varDelta ,3}$. Hence $x^{(\varDelta )}(t-\tau )<0$ and

$$\begin{aligned} x^{(\varDelta )}(t) = \beta _{L} + \left( x^{(\varDelta )}(\tilde{T})-\beta _{L}\right) \mathrm {e}^{-(t-\tilde{T})}. \end{aligned}$$

(48)

From (48) and $(x^{(\varDelta )}(\tilde{T})-\beta _{L})<0$ we obtain that $x^{(\varDelta )}(t)$ is strictly decreasing for $t\in (\tilde{T},z_{\varDelta ,3}+\tau )$. At $t=(z_{\varDelta ,3}+\tau )$ Eq. (48) gives

$$\begin{aligned} x^{(\varDelta )}(z_{\varDelta ,3}+\tau ) = \beta _{L} + \left( x^{(\varDelta )}(\tilde{T})-\beta _{L}\right) \mathrm {e}^{-(z_{\varDelta ,3}-\tilde{z}_{2})}, \end{aligned}$$

(49)

which is negative by assumption.

The condition $x^{(\varDelta )}(z_{\varDelta ,3}+\tau )< 0$ together with Eqs. (49), (46), (45) and $a>\beta _{U}$ yield

$$\begin{aligned} \beta _{L}\mathrm {e}^{z_{\varDelta ,3}-\tilde{z}_{2}}< & {} \beta _{L}-x^{(\varDelta )}(\tilde{T}), \nonumber \\ \beta _{L}(a\mathrm {e}^{\varDelta -\tilde{z}_{2}}-\beta _{U})< & {} (a-\beta _{U})[\beta _{L} + \beta _{U}(1-\mathrm {e}^{-\tau }) - a(\mathrm {e}^{\sigma }-1)\mathrm {e}^{\varDelta -\tilde{T}}], \nonumber \\ \varDelta< & {} \ln {\frac{a\beta _{L}+\beta _{U}(a-\beta _{U})(1-\mathrm {e}^{-\tau })}{a[\beta _{L}\mathrm {e}^{-\tilde{z}_{2}}+(a-\beta _{U})(\mathrm {e}^{\sigma }-1)\mathrm {e}^{-\tilde{T}}]}}=:\hat{\delta }_{4}. \end{aligned}$$

(50)

Thus the interval ${ I_{FNFP2} }$, given by (14), together with the extra condition $x^{(\varDelta )}(z_{\varDelta ,3}+\tau )<0$, written as (50), yield the interval ${ I_{FNFP3} }$ given by (16).

FNFP4: in this subcase we also have to assume $x^{(\varDelta )}(z_{\varDelta ,4}+\tau )<0$ (see the examples in Fig. 3).

For $t\in [z_{\varDelta ,3}+\tau ,z_{\varDelta ,4}+\tau ]$ we have $z_{\varDelta ,3}\leqslant t-\tau \leqslant z_{\varDelta ,4}$. Hence $x^{(\varDelta )}(t-\tau )\geqslant 0$ and

$$\begin{aligned} x^{(\varDelta )}(t) = -\beta _{U} + \left( x^{(\varDelta )}(z_{\varDelta ,3}+\tau )+\beta _{U}\right) \mathrm {e}^{-(t-z_{\varDelta ,3}-\tau )}. \end{aligned}$$

From the condition $x^{(\varDelta )}(z_{\varDelta ,3}+\tau )< 0$ we have $x^{(\varDelta )}(z_{\varDelta ,3}+\tau )+\beta _{U}>0$, so $x^{(\varDelta )}$ is strictly decreasing on $[z_{\varDelta ,3}+\tau ,z_{\varDelta ,4}+\tau ]$, and

$$\begin{aligned} x^{(\varDelta )}(z_{\varDelta ,4}+\tau ) = -\beta _{U} + \left( x^{(\varDelta )}(z_{\varDelta ,3}+\tau )+\beta _{U}\right) \mathrm {e}^{-(z_{\varDelta ,4}-z_{\varDelta ,3})}. \end{aligned}$$

(51)

The condition $x^{(\varDelta )}(z_{\varDelta ,4}+\tau )<0$ along with Eqs. (51), (47) and (49) yield

$$\begin{aligned}&\beta _{U}\mathrm {e}^{z_{\varDelta ,4}}> \left( x^{(\varDelta )}(z_{\varDelta ,3}+\tau )+\beta _{U}\right) \mathrm {e}^{z_{\varDelta ,3}}, \nonumber \\&\beta _{U}\mathrm {e}^{z_{\varDelta ,4}}> (\beta _{U} + \beta _{L})\mathrm {e}^{z_{\varDelta ,3}} - (\beta _{L}-x^{(\varDelta )}(\tilde{T}))\mathrm {e}^{\tilde{z}_{2}}, \nonumber \\&\left[ \beta _{U}\frac{\beta _{U}+\beta _{L}}{a-\beta _{U}} + \beta _{L}+\beta _{U}(2-\mathrm {e}^{-\tau })\right] \mathrm {e}^{\tilde{z}_{2}} \nonumber \\&\quad > a\left[ \frac{\beta _{U} + \beta _{L}}{a-\beta _{U}} - (\mathrm {e}^{\sigma }-1)(1-\mathrm {e}^{-\tau })\right] \mathrm {e}^{\varDelta },\nonumber \\&\varDelta < \tilde{z}_{2} + \ln {\frac{\beta _{U}(\beta _{U}+\beta _{L})+\left[ \beta _{L}+\beta _{U}(2-\mathrm {e}^{-\tau })\right] (a-\beta _{U})}{a(\beta _{U}+\beta _{L})-a(\mathrm {e}^{\sigma }-1)(1-\mathrm {e}^{-\tau })(a-\beta _{U})}} =:\delta _{5}. \end{aligned}$$

(52)

From (50) we infer that the condition $x^{(\varDelta )}(z_{\varDelta ,3}+\tau )\geqslant 0$ implies $\varDelta \geqslant \hat{\delta }_{4}$. This condition together with the interval ${ I_{FNFP2} }$, given by (14), plus the extra condition $x^{(\varDelta )}(z_{\varDelta ,4}+\tau )<0$, written as (52), yield the interval ${ I_{FNFP4} }$ given by (17). $\square$

Proof of Remark 3.3

For each case (11) we compute the resetting time $F(\varDelta )$ as follows:

RNRN:$F(\varDelta )=\sigma$, since $x^{(\varDelta )}(t)=\tilde{x}(t+(\tilde{z}_{1}-z_{\varDelta ,1}))$ for all $t\geqslant \varDelta +\sigma$ and $x^{(\varDelta )}(z_{\varDelta ,1}+\tau )=\tilde{x}(\tilde{z}_{1}+\tau )=\bar{x}$, where $z_{\varDelta ,1}=\tilde{z}_{1}+T(\varDelta )-\tilde{T}$ and $T(\varDelta )$ is as in Mackey et al. (2017, Proposition 5.1);

RNRP:$F(\varDelta )=\tilde{z}_{1}+\tau +(z_{\varDelta ,2}-\tilde{z}_{2})-\varDelta$, since $x^{(\varDelta )}(t)=\tilde{x}(t-(z_{\varDelta ,2}-\tilde{z}_{2}))$ for all $t\geqslant \tilde{z}_{1}+\tau +(z_{\varDelta ,2}-\tilde{z}_{2})$ and $x^{(\varDelta )}(z_{\varDelta ,2}+\tau )=\tilde{x}(\tilde{z}_{2}+\tau )={\underline{x}}$, where $z_{\varDelta ,2}=\tilde{z}_{2}+T(\varDelta )-\tilde{T}$ and $T(\varDelta )$ is as in Mackey et al. (2017, Proposition 5.2);

RPRP:$F(\varDelta )=\tilde{z}_{1}+\tau +(z_{\varDelta ,2}-\tilde{z}_{2})-\varDelta$, since $x^{(\varDelta )}(t)=\tilde{x}(t-(z_{\varDelta ,2}-\tilde{z}_{2}))$ for all $t\geqslant \tilde{z}_{1}+\tau +(z_{\varDelta ,2}-\tilde{z}_{2})$ and $x^{(\varDelta )}(z_{\varDelta ,2}+\tau )=\tilde{x}(\tilde{z}_{2}+\tau )={\underline{x}}$, where $z_{\varDelta ,2}=\tilde{z}_{2}+T(\varDelta )-\tilde{T}$ and $T(\varDelta )$ is as in Mackey et al. (2017, Proposition 5.3);

RPFP:$F(\varDelta )=\sigma$, since $x^{(\varDelta )}(t)=\tilde{x}(t-(z_{\varDelta ,2}-\tilde{z}_{2}))$ for all $t\geqslant \varDelta +\sigma$ and $x^{(\varDelta )}(z_{\varDelta ,2}+\tau )=\tilde{x}(\tilde{z}_{2}+\tau )={\underline{x}}$, where $z_{\varDelta ,2}=\tilde{z}_{2}+T(\varDelta )-\tilde{T}$ and $T(\varDelta )$ is as in Mackey et al. (2017, Proposition 5.4);

RPFN:$F(\varDelta )=z_{\varDelta ,2}+\tau -\varDelta$, since $x^{(\varDelta )}(t)=\tilde{x}(t-(z_{\varDelta ,3}-\tilde{z}_{3}))$ for all $t\geqslant z_{\varDelta ,2}+\tau$ and $x^{(\varDelta )}(z_{\varDelta ,3}+\tau )=\tilde{x}(\tilde{z}_{3}+\tau )=\bar{x}$, where $z_{\varDelta ,3}=\tilde{z}_{3}+T(\varDelta )-\tilde{T}$ and $T(\varDelta )$ is as in Mackey et al. (2017, Proposition 5.5);

FPFP:$F(\varDelta )=\sigma$, since $x^{(\varDelta )}(t)=\tilde{x}(t-(z_{\varDelta ,2}-\tilde{z}_{2}))$ for all $t\geqslant \varDelta +\sigma$ and $x^{(\varDelta )}(z_{\varDelta ,2}+\tau )=\tilde{x}(\tilde{z}_{2}+\tau )={\underline{x}}$, where $z_{\varDelta ,2}=\tilde{z}_{2}+T(\varDelta )-\tilde{T}$ and $T(\varDelta )$ is as in Mackey et al. (2017, Proposition 5.6);

FPFN:$F(\varDelta )=z_{\varDelta ,2}+\tau -\varDelta$, since $x^{(\varDelta )}(t)=\tilde{x}(t-(z_{\varDelta ,3}-\tilde{z}_{3}))$ for all $t\geqslant z_{\varDelta ,2}+\tau$ and $x^{(\varDelta )}(z_{\varDelta ,3}+\tau )=\tilde{x}(\tilde{z}_{3}+\tau )=\bar{x}$, where $z_{\varDelta ,2}=\tilde{z}_{2}+T(\varDelta )-\tilde{T}$ and $z_{\varDelta ,3}=\tilde{z}_{3}+T(\varDelta )-\tilde{T}$ with $T(\varDelta )$ and $z_{\varDelta ,2}$ given by Mackey et al. (2017, Proposition 5.7 and its proof);

FNFP1:$F(\varDelta )=z_{\varDelta ,3}+\tau -\varDelta$, since $x^{(\varDelta )}(t)=\tilde{x}(t-(z_{\varDelta ,4}-\tilde{z}_{4}))$ for all $t\geqslant z_{\varDelta ,3}+\tau$ and $x^{(\varDelta )}(z_{\varDelta ,4}+\tau )=\tilde{x}(\tilde{z}_{4}+\tau )={\underline{x}}$, where $z_{\varDelta ,3}$ and $z_{\varDelta ,4}=\tilde{z}_{4}$ are computed as follows. From $x^{(\varDelta )}(\varDelta )<0<x^{(\varDelta )}(\varDelta +\sigma )$ we obtain a zero $z_{\varDelta ,3}$ of $x^{(\varDelta )}$ in $(\varDelta ,\varDelta +\sigma )$ given by

$$\begin{aligned} x^{(\varDelta )}(z_{\varDelta ,3})=-\beta _{U}+a+\left( x^{(\varDelta )}(\varDelta )+\beta _{U}-a\right) \mathrm {e}^{-(z_{\varDelta ,3}-\varDelta )}=0, \end{aligned}$$

where $x^{(\varDelta )}(\varDelta )=-\beta _{U}+\beta _{U}\mathrm {e}^{-(\tilde{z}_{2}-\varDelta )}$. For $\tilde{T}<t<z_{\varDelta ,3}+\tau$ we have $\tilde{z}_{2}<t-\tau <z_{\varDelta ,3}$. Hence $x^{(\varDelta )}(t-\tau )<0$, and

$$\begin{aligned} x^{(\varDelta )}(t)=\beta _{L}+\left( x^{(\varDelta )}(\tilde{T})-\beta _{L}\right) \mathrm {e}^{-(t-\tilde{T})}. \end{aligned}$$

Since $\tilde{z}_{2}<z_{\varDelta ,3}$ and from the proof of Remark 3.1 the inequality $\varDelta \geqslant \delta _{4}$ implies $x^{(\varDelta )}(\tilde{T})\geqslant 0$, we obtain

$$\begin{aligned} x^{(\varDelta )}(z_{\varDelta ,3}+\tau )=\beta _{L}+\left( x^{(\varDelta )}(\tilde{T})-\beta _{L}\right) \mathrm {e}^{-(z_{\varDelta ,3}+\tau -\tilde{T})}>\beta _{L}(1-\mathrm {e}^{{\tilde{z}}_{2}-z_{\varDelta ,3}})>0. \end{aligned}$$

(53)

Notice that $x^{(\varDelta )}(t)>0$ on $(z_{\varDelta ,3},z_{\varDelta ,3}+\tau ]$. Using this and $z_{\varDelta ,3}<\varDelta +\sigma<\tilde{z}_{2}+\tau <z_{\varDelta ,3}+\tau$ we obtain that $x^{(\varDelta )}(t)$ is strictly decreasing on $[z_{\varDelta ,3}+\tau ,\infty )$ as long as $x^{(\varDelta )}(t-\tau )\geqslant 0$. It follows that there is a first zero $z_{\varDelta ,4}$ of $x^{(\varDelta )}(t)$ in $[z_{\varDelta ,3}+\tau ,\infty )$ given by

$$\begin{aligned} x^{(\varDelta )}(z_{\varDelta ,4})=-\beta _{U}+\left( x^{(\varDelta )}(z_{\varDelta ,3}+\tau )-\beta _{U}\right) \mathrm {e}^{-(z_{\varDelta ,4}-(z_{\varDelta ,3}+\tau ))}=0, \end{aligned}$$

with $x^{(\varDelta )}(z_{\varDelta ,3}+\tau )$ given by (53), where $x^{(\varDelta )}(\tilde{T})$ is given by (45);

FNFN:$F(\varDelta )=\tilde{z}_{2}+\tau -\varDelta$, since $x^{(\varDelta )}(t)=\tilde{x}(t+(\tilde{z}_{3}-z_{\varDelta ,3}))$ for all $t\geqslant \tilde{z}_{2}+\tau$ and $x^{(\varDelta )}(z_{\varDelta ,3}+\tau )=\tilde{x}(\tilde{z}_{3}+\tau )=\bar{x}$, where $z_{\varDelta ,3}=\tilde{z}_{3}+T(\varDelta )-\tilde{T}$ and $T(\varDelta )$ is as in Mackey et al. (2017, Proposition 5.8);

FNRN:$F(\varDelta )=\sigma$, since $x^{(\varDelta )}(t)=\tilde{x}(t+(\tilde{z}_{3}-z_{\varDelta ,3}))$ for all $t\geqslant \varDelta +\sigma$ and $x^{(\varDelta )}(z_{\varDelta ,3}+\tau )=\tilde{x}(\tilde{z}_{3}+\tau )=\bar{x}$, where $z_{\varDelta ,3}=\tilde{z}_{3}+T(\varDelta )-\tilde{T}$ and $T(\varDelta )$ is as in Mackey et al. (2017, Proposition 5.10);

FNRP:$F(\varDelta )=\tilde{z}_{3}+\tau -(\tilde{z}_{4}-z_{\varDelta ,4})-\varDelta$, since $x^{(\varDelta )}(t)=\tilde{x}(t+(\tilde{z}_{4}-z_{\varDelta ,4}))$ for all $t\geqslant \tilde{z}_{3}+\tau -(\tilde{z}_{4}-z_{\varDelta ,4})$ and $x^{(\varDelta )}(z_{\varDelta ,4}+\tau )=\tilde{x}(\tilde{z}_{4}+\tau )={\underline{x}}$, where $z_{\varDelta ,4}=\tilde{z}_{4}+T(\varDelta )-\tilde{T}$ and $T(\varDelta )$ is as in Mackey et al. (2017, Proposition 5.10). $\square$

Proof of Remark 3.4

Define the constants $\delta _{1}$ as in Mackey et al. (2017, Eq. (5.6))

$$\begin{aligned} \delta _{1} :=\tilde{z}_{1} - \sigma - \ln \left( \frac{\beta _{L}+a(1-\mathrm {e}^{-\sigma })}{\beta _{L}}\right) <\tilde{z}_{1}. \end{aligned}$$

(54)

For each case (11) we consider the corresponding $\varDelta$ interval as computed in Mackey et al. (2017, Section 5) and listed in (11). Recalling that $\sigma \in (0,\tau ]$, $\tilde{T}=\tilde{z}_{2}+\tau$, $t_{max}=\tilde{z}_{1}+\tau$, $\tilde{z}_{n}=\tilde{T}+\tilde{z}_{n-2}$ for $n\in \{2,3,4\ldots \}$, $\tilde{z}_{j+1}>\tilde{z}_{j}+\tau$ for all $j\in \mathbb {N}$, and from Mackey et al. (2017, Proposition 4.2) we see that $J=j_\varDelta \in \{0,1,2\}$ implies $z_{\varDelta ,J+1}>\varDelta$ and $z_{\varDelta ,J+1}>z_{\varDelta ,J+1}+\tau$. Thus, we show that for each case (11) we have $F(\varDelta )<T(\varDelta )$ as follows:

RNRN:$\varDelta \in { I_{RNRN} }=[0,\delta _{1}]$: since $\tilde{z}_{2}>\tilde{z}_{1}+\tau >\tilde{z}_{1}$,

$$\begin{aligned} T(\varDelta )=\tilde{T}+z_{\varDelta ,1}-\tilde{z}_{1}=\tilde{z}_{2}+\tau +z_{\varDelta ,1}-\tilde{z}_{1}>\tau +z_{\varDelta ,1}>\tau \geqslant \sigma =F(\varDelta ); \end{aligned}$$

RNRP and RPRP:$\varDelta \in { I_{RNRP} }=[\max \{0,\delta _{1}\},\tilde{z}_{1})$ with $\delta _{1}$ given by (54) and $\varDelta \in { I_{RPRP} }=[\tilde{z}_{1},t_{max}-\sigma ]$: using that $\tilde{z}_{2}>\tilde{z}_{1}+\tau >\tilde{z}_{1}$ we infer

$$\begin{aligned} T(\varDelta )=\tilde{T}+z_{\varDelta ,2}-\tilde{z}_{2}=\tau +z_{\varDelta ,2}>\tilde{z}_{1}+\tau +z_{\varDelta ,2}-\tilde{z}_{2}-\varDelta =F(\varDelta ); \end{aligned}$$

RPFP:$\varDelta \in { I_{RPFP} }=(t_{\max }-\sigma ,t_{\max }]\cap (-\infty ,\delta _{2}]$:

$$\begin{aligned} T(\varDelta )=\tilde{T}+z_{\varDelta ,2}-\tilde{z}_{2}=\tau +z_{\varDelta ,2}>\tau \geqslant \sigma =F(\varDelta ); \end{aligned}$$

RPFN:$\varDelta \in { I_{RPFN} }=(t_{\max }-\sigma ,t_{\max }]\,\cap \,(\delta _{2},\infty )$: from Mackey et al. (2017, Proposition 4.2) we have $z_{\varDelta ,3}>z_{\varDelta ,2}+\tau$, and thereby

$$\begin{aligned} T(\varDelta )=\tilde{T}+z_{\varDelta ,3}-\tilde{z}_{3}=z_{\varDelta ,3}-\tilde{z}_{1}>z_{\varDelta ,2}+\tau -\tilde{z}_{1}\geqslant z_{\varDelta ,2}+\sigma -\tilde{z}_{1}, \end{aligned}$$

and using the lower bound of $\varDelta$ in ${ I_{RPFN} }$ we conclude that

$$\begin{aligned} T(\varDelta )>z_{\varDelta ,2}+\sigma -\tilde{z}_{1}= z_{\varDelta ,2}+\tau -(t_{max}-\sigma ) \geqslant z_{\varDelta ,2}+\tau -\varDelta =F(\varDelta ); \end{aligned}$$

FPFP:$\varDelta \in { I_{FPFP} }=[t_{\max },\tilde{z}_{2}]\,\cap \,(-\infty ,\delta _{2}]$:

$$\begin{aligned} T(\varDelta )=\tilde{T}+z_{\varDelta ,2}-\tilde{z}_{2}=\tau +z_{\varDelta ,2}>\tau \geqslant \sigma =F(\varDelta ); \end{aligned}$$

FPFN:$\varDelta \in { I_{FPFN} }=[t_{\max },\tilde{z}_{2}]\,\cap \,(\delta _{2},\infty )$: using $z_{\varDelta ,3}>z_{\varDelta ,2}$ and using the lower bound of $\varDelta$ in ${ I_{FPFN} }$ we have

$$\begin{aligned} T(\varDelta )= & {} \tilde{T}+z_{\varDelta ,3}-\tilde{z}_{3}=z_{\varDelta ,3}-\tilde{z}_{1}>z_{\varDelta ,2}-\tilde{z}_{1}= z_{\varDelta ,2}+\tau -t_{max}\\\geqslant & {} z_{\varDelta ,2}+\tau -\varDelta =F(\varDelta ); \end{aligned}$$

FNFP1:$\varDelta \in { I_{FNFP1} }=\varDelta \in (\tilde{z}_{2},\tilde{T}-\sigma )\,\cap \,(-\infty ,\delta _{2})\,\cap \,[\delta _{4},\infty )$: here we have $x^{(\varDelta )}(z_{\varDelta ,4}+t)=\tilde{x}(\tilde{z}_{2}+t)$ for all $t\geqslant 0$, which gives $T(\varDelta )= z_{\varDelta ,4}-\tilde{z}_{2}$. Noting that $z_{\varDelta ,4}>z_{\varDelta ,3}$ and using the lower bound of $\varDelta$ in ${ I_{FNFP1} }$ we obtain

$$\begin{aligned} T(\varDelta )=z_{\varDelta ,4}-\tilde{z}_{2}>z_{\varDelta ,3}+\tau -\tilde{z}_{2}>z_{\varDelta ,3}+\tau -\varDelta =F(\varDelta ); \end{aligned}$$

FNFP2: Recall that ${ I_{FNFP2} }=\emptyset$ since $\delta _{4}<\tilde{z}_{2}$;

FNFN:$\varDelta \in { I_{FNFN} }=(\tilde{z}_{2},\tilde{z}_{2}+\tau -\sigma )\,\cap \,[\delta _{2},\infty )$: since $z_{\varDelta ,3}>z_{\varDelta ,2}+\tau =\tilde{z}_{2}+\tau$ we have

$$\begin{aligned} T(\varDelta )=\tilde{T}+z_{\varDelta ,3}-\tilde{z}_{3}=z_{\varDelta ,3}-\tilde{z}_{1}>\tilde{z}_{2}+\tau -\tilde{z}_{1}> \tau > \tilde{z}_{2}+\tau -\varDelta =F(\varDelta ); \end{aligned}$$

FNRN:$\varDelta \in { I_{FNRN} }=[\tilde{T}-\sigma ,\tilde{T})\,\cap \,(-\infty ,\tilde{T}+\delta _{1})$: using the fact that $z_{\varDelta ,3}>z_{\varDelta ,2}+\tau =\tilde{z}_{2}+\tau$ we obtain

$$\begin{aligned} T(\varDelta )=\tilde{T}+z_{\varDelta ,3}-\tilde{z}_{3}=z_{\varDelta ,3}-\tilde{z}_{1}>\tilde{z}_{2}+\tau -\tilde{z}_{1}>\tau \geqslant \sigma =F(\varDelta ); \end{aligned}$$

FNRP:$\varDelta \in { I_{FNRP} }=[\tilde{T}-\sigma ,\tilde{T})\,\cap \,[\tilde{T}+\delta _{1},\infty )$: using that $\tilde{z}_{2}+\tau>\tilde{z}_{1}+2\tau >\tilde{z}_{1}+\tau +\sigma$ we have $\tilde{z}_{1}+\tau -(\tilde{T}-\sigma )<0$, and thus

$$\begin{aligned} T(\varDelta )= & {} \tilde{T}+z_{\varDelta ,4}-\tilde{z}_{4}\geqslant \tilde{T}+\tilde{z}_{1}+\tau -(\tilde{T}-\sigma )+z_{\varDelta ,4}-\tilde{z}_{4}\\> & {} \tilde{z}_{3}+\tau -(\tilde{z}_{4}-z_{\varDelta ,4})-\varDelta =F(\varDelta ). \end{aligned}$$

$\square$

Proof of Remark 3.5

Each set of parameters $(\tau ,\beta _{U},\beta _{L},\sigma ,a,\varDelta )$ defines a sequence of cases (11) along $\varDelta \in [0,\tilde{T})$. Thus we show that for each case (11) we have $F(\varDelta )\geqslant \sigma$:

RNRN, RPFP, FPFP, FNRN:$F(\varDelta )=\sigma$;

RNRP: from Mackey et al. (2017, Proposition 5.2) we have $x^{(\varDelta )}(z_{\varDelta ,1}+\tau )>\bar{x}$. Using this and $\bar{x}+\beta _{U}=\beta _{U}\mathrm {e}^{{\tilde{z}}_{2}-t_{max}}$ in $0=x^{(\varDelta )}(z_{\varDelta ,2})=-\beta _{U}+(x^{(\varDelta )}(z_{\varDelta ,1}+\tau )+\beta _{U})\mathrm {e}^{z_{\varDelta ,1}+\tau -z_{\varDelta ,2}}$ we obtain

$$\begin{aligned} \beta _{U}\mathrm {e}^{z_{\varDelta ,2}}= & {} \left( x^{(\varDelta )}(z_{\varDelta ,1}+\tau )+\beta _{U}\right) \mathrm {e}^{z_{\varDelta ,1}+\tau } \\> & {} \beta _{U}\mathrm {e}^{{\tilde{z}}_{2}-t_{max}}\mathrm {e}^{z_{\varDelta ,1}+\tau }. \\= & {} \beta _{U}\mathrm {e}^{{\tilde{z}}_{2}+z_{\varDelta ,1}-\tilde{z}_{1}}. \end{aligned}$$

So $(z_{\varDelta ,2}-\tilde{z}_{2})>(z_{\varDelta ,1}-\tilde{z}_{1})$ and $z_{\varDelta ,1}>\varDelta$ lead to

$$\begin{aligned} F(\varDelta )=\tilde{z}_{1}+\tau +(z_{\varDelta ,2}-\tilde{z}_{2})-\varDelta>\tilde{z}_{1}+\tau +(z_{\varDelta ,1}-\tilde{z}_{1})-\varDelta =z_{\varDelta ,1}-\varDelta +\tau >\tau \geqslant \sigma ; \end{aligned}$$

RPRP: taking the upper bound of $\varDelta$ in ${ I_{RPRP} }=[\tilde{z}_{1},t_{max}-\sigma ]$ and using that $z_{\varDelta ,2}>\tilde{z}_{2}$, see Mackey et al. (2017, Proposition 5.3), we find

$$\begin{aligned} F(\varDelta )= & {} \tilde{z}_{1}+\tau +(z_{\varDelta ,2}-\tilde{z}_{2})-\varDelta>\tilde{z}_{1}+\tau +z_{\varDelta ,2}-\tilde{z}_{2}-t_{max}+\sigma \\= & {} \sigma +z_{\varDelta ,2}-\tilde{z}_{2}>\tau \geqslant \sigma ; \end{aligned}$$

RPFN: taking the upper bound of $\varDelta$ in ${ I_{RPFN} }=(t_{\max }-\sigma ,t_{\max }]\,\cap \,(\delta _{2},\infty )$ and using that $z_{\varDelta ,2}\geqslant \tilde{z}_{2}$ Mackey et al. (2017, Proof of Proposition 5.5, Eq. (9.5)) we obtain

$$\begin{aligned} F(\varDelta )=z_{\varDelta ,2}+\tau -\varDelta \geqslant z_{\varDelta ,2}+\tau -t_{\max }>\tilde{z}_{2}+\tau -t_{\max }\geqslant \tau \geqslant \sigma ; \end{aligned}$$

FPFN: from Mackey et al. (2017, Proof of Proposition 5.7) we have $z_{\varDelta ,2}\geqslant \tilde{z}_{2}$, then

$$\begin{aligned} F(\varDelta )=z_{\varDelta ,2}+\tau -\varDelta \geqslant z_{\varDelta ,2}+\tau -\tilde{z}_{2} \geqslant \tau \geqslant \sigma ; \end{aligned}$$

FNFP1:$F(\varDelta )=z_{\varDelta ,3}+\tau -\varDelta>z_{\varDelta ,3}+\tau -(\tilde{z}_{2}+\tau -\sigma )=\sigma +z_{\varDelta ,3}-\tilde{z}_{2}>\sigma$;

FNFP2: Recall that ${ I_{FNFP2} }=\emptyset$ since $\delta _{4}<\tilde{z}_{2}$;

FNFN:$F(\varDelta )=\tilde{z}_{2}+\tau -\varDelta >\tilde{z}_{2}+\tau -(\tilde{z}_{2}+\tau -\sigma )=\sigma$;

FNRP: from Mackey et al. (2017, Proof of Proposition 5.10) we have $x^{(\varDelta )}(z_{\varDelta ,3}+\tau )>\bar{x}$. Using this and $\bar{x}+\beta _{U}=\beta _{U}\mathrm {e}^{\tilde{z}_{2}-t_{max}}$ in $x^{(\varDelta )}(z_{\varDelta ,4})=0$ we have

$$\begin{aligned} \beta _{U}\mathrm {e}^{z_{\varDelta ,4}}= & {} \left( x^{(\varDelta )}(z_{\varDelta ,3}+\tau )+\beta _{U}\right) \mathrm {e}^{-(z_{\varDelta ,4}-(z_{\varDelta ,3}+\tau ))} \\> & {} \beta _{U}\mathrm {e}^{\tilde{z}_{2}-t_{max}}\mathrm {e}^{z_{\varDelta ,3}+\tau }. \\= & {} \beta _{U}\mathrm {e}^{\tilde{z}_{4}+z_{\varDelta ,3}-\tilde{z}_{3}}. \end{aligned}$$

So $(z_{\varDelta ,4}-\tilde{z}_{4})>(z_{\varDelta ,3}-\tilde{z}_{3})$ and $z_{\varDelta ,3}>\varDelta$ lead to

$$\begin{aligned} F(\varDelta )=\tilde{z}_{3}+\tau +(z_{\varDelta ,4}-\tilde{z}_{4})-\varDelta> \tilde{z}_{3}+\tau +(z_{\varDelta ,3}-\tilde{z}_{3})-\varDelta =z_{\varDelta ,3}+\tau -\varDelta >\tau \geqslant \sigma . \end{aligned}$$

$\square$

Proof of Remark 3.6

Recall that $\delta _{2}$ is given by (12) and is defined for $\beta _{U}>a(1-\mathrm {e}^{-\sigma })$. Thus the definition (12) also holds for $a<\beta _{U}$. For the case RPFN of Mackey et al. (2017, Table 2) we have $\varDelta \in { I_{RPFN} }=(t_{\max }-\sigma ,t_{\max }]\cap (\delta _{2},\infty )$ and $\delta _{2}>t_{max}$, then it follows that ${ I_{RPFN} }=\emptyset$. $\square$

Proof of Remark 3.7

From Mackey et al. (2017, Corollary 4.2) it follows that the cycle length map $T(\varDelta )$ is continuous for $a<\beta _{U}$. The proof is divided into two subcases, $\delta _{2}<t_{max}$ and $\delta _{2}\geqslant t_{max}$ as follows.

The condition $\delta _{2}<t_{max}$ implies that $a<\beta _{U}$ and there exists a sequence of cases from (11) as is shown in Mackey et al. (2017, Table 1). Once the cycle length map is continuous we see from Mackey et al. (2017, Third row of Table 1) that $T(\varDelta )$ is strictly increasing on $[0,\delta _{2}]$ and strictly decreasing on $[\delta _{2},\tilde{T}]$. Thus the maximum of $T(\varDelta )$ occurs for the case RPFP with $\varDelta =\delta _{2}$ and we have $\bar{T}=T(\delta _{2})$ with the cycle length map given by Mackey et al. (2017, Proposition 5.4), i.e.

$$\begin{aligned} \bar{T} = \tilde{T} + \ln \left( 1+\frac{a(\mathrm {e}^{\sigma }+1)}{\beta _{U}}\mathrm {e}^{\delta _{2}-\tilde{z}_{2}}\right) = \tilde{T} + \ln \left( \frac{\beta _{U}}{\beta _{U}-a(1-\mathrm {e}^{-\sigma })}\right) > \tilde{T}, \end{aligned}$$

(55)

where $\tilde{T}$ is defined by (9).

For $\delta _{2}\geqslant t_{max}$ there exists a sequence of cases from (11) as is shown in Mackey et al. (2017, Table 2) and it follows from Remark 3.6 that $a<\beta _{U}$ holds. For the case RPFN of Mackey et al. (2017, Table 2) we have $\delta _{2}\geqslant t_{max}$ and from Remark 3.6 it follows that ${ I_{RPFN} }=\emptyset$. Once the cycle length map is continuous and ${ I_{RPFN} }=\emptyset$, we see from Mackey et al. (2017, Third row of Table 2) that again $T(\varDelta )$ is strictly increasing on $[0,\delta _{2}]$ and strictly decreasing on $[\delta _{2},\tilde{T}]$. Thus the maximum of $T(\varDelta )$ occurs for the case FPFP with $\varDelta =\delta _{2}$ and we have $\bar{T}=T(\delta _{2})$ with the cycle length map given by Mackey et al. (2017, Proposition 5.6), which is equal to (55). $\square$

Proof of Remark 3.8

For $\varDelta \leqslant t \leqslant \varDelta +\sigma$, Eq. (41) together with $x^{(\varDelta )}(\varDelta ) = -\beta _{U} + \beta _{U}\mathrm {e}^{\tilde{z}_{2}-\varDelta }$ yields

$$\begin{aligned} x^{(\varDelta )}(t) = -\beta _{U} + a + (\beta _{U}\mathrm {e}^{\tilde{z}_{2}-\varDelta }-a)\mathrm {e}^{-(t-\varDelta )}. \end{aligned}$$

(56)

The conditions $a>\beta _{U}$ and $\varDelta >\tilde{z}_{2}$ (see Remark 3.1) combined yield $a>\beta _{U}\mathrm {e}^{\tilde{z}_{2}-\varDelta }$. Hence $x^{(\varDelta )}(t)$ is strictly increasing on $[\varDelta ,\varDelta +\sigma ]$. From $x^{(\varDelta )}(\varDelta )<0$ and $x^{(\varDelta )}(\varDelta +\sigma )>0$ we obtain a zero $z_{\varDelta ,3}$ of $x^{(\varDelta )}$ in $(\varDelta ,\varDelta +\sigma )$ given by (46).

For $\varDelta +\sigma \leqslant t \leqslant \tilde{T}$, Eq. (44) together with $(x^{(\varDelta )}(\varDelta +\sigma )+\beta _{U})>0$ shows that $x^{(\varDelta )}(t)$ is strictly decreasing on $[\varDelta +\sigma ,\tilde{T}]$.

For $\tilde{T}<t<z_{\varDelta ,3}+\tau$ we have $\tilde{z}_{2}<t-\tau <z_{\varDelta ,3}$. Hence $x^{(\varDelta )}(t-\tau )<0$, and

$$\begin{aligned} x^{(\varDelta )}(t) = \beta _{L} + \left( x^{(\varDelta )}(\tilde{T})-\beta _{L}\right) \mathrm {e}^{-(t-\tilde{T})}. \end{aligned}$$

Thus $x^{(\varDelta )}$ is strictly increasing on $[\tilde{T},z_{\varDelta ,3}+\tau ]$ since $x^{(\varDelta )}(\tilde{T})<0$. For $t=z_{\varDelta ,3}+\tau$

$$\begin{aligned} x^{(\varDelta )}(z_{\varDelta ,3}+\tau ) = \beta _{L} + \left( x^{(\varDelta )}(\tilde{T})-\beta _{L}\right) \mathrm {e}^{-(z_{\varDelta ,3}-\tilde{z}_{2})}. \end{aligned}$$

(57)

A rapidly oscillating periodic solution occurs if $x^{(\varDelta )}(\varDelta )=x^{(\varDelta )}(\tilde{T})$, $x^{(\varDelta )}(\varDelta +\sigma )=x^{(\varDelta )}(z_{\varDelta ,3}+\tau )$ and if the solution $x^{(\varDelta )}$ for $\varDelta \leqslant t\leqslant \varDelta +\sigma$ is equal to the solution $x^{(\varDelta )}$ for $\tilde{T}\leqslant t\leqslant z_{\varDelta ,3}+\tau$, i.e, $a=\beta _{L}+\beta _{U}$. Then, the necessary conditions for the existence of a rapid oscillation are

$$\begin{aligned} \left\{ \begin{array}{ll} a = \beta _{L}+\beta _{U}, \\ x^{(\varDelta )}(\varDelta )=x^{(\varDelta )}(\tilde{T}), \\ x^{(\varDelta )}(\varDelta +\sigma )=x^{(\varDelta )}(z_{\varDelta ,3}+\tau ). \end{array}\right. \end{aligned}$$

(58)

Combining Eq. (58) with (57) we get

$$\begin{aligned} \left\{ \begin{array}{ll} \beta _{L} = a - \beta _{U}, \\ x^{(\varDelta )}(\tilde{T}) = x^{(\varDelta )}(\varDelta ), \\ x^{(\varDelta )}(\varDelta +\sigma ) = \beta _{L} + \left( x^{(\varDelta )}(\tilde{T})-\beta _{L}\right) \mathrm {e}^{-(z_{\varDelta ,3}-\tilde{z}_{2})}. \end{array}\right. \end{aligned}$$

(59)

Using the first and second relation in the third line of Eq. (59) gives

$$\begin{aligned} x^{(\varDelta )}(\varDelta +\sigma ) + \beta _{U} - a = \left( x^{(\varDelta )}(\varDelta ) + \beta _{U} - a\right) \mathrm {e}^{-(z_{\varDelta ,3}-\tilde{z}_{2})}, \end{aligned}$$

and combining this with (42) yields

$$\begin{aligned} \left( x^{(\varDelta )}(\varDelta ) + \beta _{U} - a\right) \mathrm {e}^{-\sigma } = \left( x^{(\varDelta )}(\varDelta ) + \beta _{U} - a\right) \mathrm {e}^{-(z_{\varDelta ,3}-\tilde{z}_{2})}. \end{aligned}$$

From $a>\beta _{U}$ and $x^{(\varDelta )}(\varDelta ) = -\beta _{U} + \beta _{U}\mathrm {e}^{\tilde{z}_{2}-\varDelta }$ it follows that $(x^{(\varDelta )}(\varDelta ) + \beta _{U} - a)<0$. Hence the conditions (58) are reduced to

$$\begin{aligned} \mathrm {e}^{z_{\varDelta ,3}} = \mathrm {e}^{\tilde{z}_{2}+\sigma }. \end{aligned}$$

(60)

The relation (60) combined with (46) yields

$$\begin{aligned} a\mathrm {e}^{\varDelta } = \beta _{U}\mathrm {e}^{\tilde{z}_{2}} + (a-\beta _{U})\mathrm {e}^{\tilde{z}_{2}+\sigma }, \end{aligned}$$

(61)

where $\tilde{z}_{2}$ is given by (9). Substituting $\varDelta =\delta _{\infty }$ in (61) gives the constant defined $\delta _{\infty }$ in (19) and

$$\begin{aligned} \delta _{\infty }= & {} \tilde{z}_{2} + \sigma + \ln {\frac{a-\beta _{U}(1-\mathrm {e}^{-\sigma })}{a}} < \tilde{z}_{2} + \sigma \\= & {} \tilde{z}_{2} + \ln {\frac{a\mathrm {e}^{\sigma }-\beta _{U}(\mathrm {e}^{\sigma }-1)}{a\mathrm {e}^{\sigma }-a(\mathrm {e}^{\sigma }-1)}} > \tilde{z}_{2}. \end{aligned}$$

So the conditions (58) yield $\varDelta =\delta _{\infty }$ with $\tilde{z}_{2}<\delta _{\infty }<\tilde{z}_{2}+\sigma$.

The period of the unstable periodic solution is given by $\tilde{T}^{(\infty )}=z_{\varDelta ,3}+\tau -(\varDelta +\sigma )$ (see the example from Fig. 5). Computing $z_{\varDelta ,3}$ from (60) and using $\varDelta =\delta _{\infty }$ gives $\tilde{T}^{(\infty )}=\tilde{T}-\delta _{\infty }$, and this together with $\tilde{z}_{2}<\delta _{\infty }<\tilde{z}_{2}+\sigma$ yields $\tau -\sigma<\tilde{T}^{(\infty )}<\tau$.

Recalling that $x^{(\varDelta )}(t)$ is strictly increasing on $[\varDelta ,\varDelta +\sigma ]$, strictly decreasing on $[\varDelta +\sigma ,\tilde{T}]$ and strictly increasing on $[\tilde{T},z_{\varDelta ,3}+\tau ]$, we infer that for $\varDelta =\delta _{\infty }$ the minimum and maximum of the rapid limit cycle are respectively given by ${\underline{x}}^{(\varDelta )}=x^{(\varDelta )}(\delta _{\infty })$ and $\bar{x}^{(\varDelta )}=x^{(\varDelta )}(\delta _{\infty }+\sigma )$. From $x^{(\varDelta )}(\varDelta ) = -\beta _{U} + \beta _{U}\mathrm {e}^{\tilde{z}_{2}-\varDelta }$ with $\varDelta =\delta _{\infty }$ and $0<\delta _{\infty }-\tilde{z}_{2}<\sigma <\tau$ it follows that

$$\begin{aligned} {\underline{x}}^{(\varDelta )}= & {} -\beta _{U}(1-\mathrm {e}^{-(\delta _{\infty }-\tilde{z}_{2})}) \\> & {} -\beta _{U}(1-\mathrm {e}^{-\sigma }) \\> & {} -\beta _{U}(1-\mathrm {e}^{-\tau }) = {\underline{x}}. \end{aligned}$$

From (56) with $t=\delta _{\infty }+\sigma$ and $\varDelta =\delta _{\infty }$ together with $a=\beta _{L}+\beta _{U}$ and $0<\delta _{\infty }-\tilde{z}_{2}<\sigma <\tau$ it follows that

$$\begin{aligned} \bar{x}^{(\varDelta )}= & {} a(1-\mathrm {e}^{-\sigma }) + \beta _{U}(\mathrm {e}^{{\tilde{z}}_{2}-\delta _{\infty }-\sigma } - 1) \\= & {} (\beta _{L}+\beta _{U})(1-\mathrm {e}^{-\sigma }) + \beta _{U}(\mathrm {e}^{{\tilde{z}}_{2}-\delta _{\infty }-\sigma } - 1) \\= & {} \beta _{L}(1-\mathrm {e}^{-\sigma }) - \beta _{U}(1-\mathrm {e}^{-(\delta _{\infty }-\tilde{z}_{2})})\mathrm {e}^{-\sigma } \\< & {} \beta _{L}(1-\mathrm {e}^{-\sigma }) \\< & {} \beta _{L}(1-\mathrm {e}^{-\tau }) = \bar{x}. \end{aligned}$$

$\square$

Proof of Proposition 4.1

Using the fact that $x^{(p)}(\varDelta _{0})=x^{(p)}(z_{2})=0$, for $t\in [\varDelta _{0},\varDelta _{0}+\sigma ]$ we have

$$\begin{aligned} x^{(p)}(t) = -\beta _{U}+a+(\beta _{U}-a)\mathrm {e}^{-(t-\varDelta _{0})}. \end{aligned}$$

The condition $a\geqslant a_{1}$ implies $a>\beta _{U}$, so $x^{(p)}(t)$ is increasing and

$$\begin{aligned} x^{(p)}(\varDelta _{0}+\sigma )= & {} -\beta _{U}+a+(\beta _{U}-a)\mathrm {e}^{-\sigma } \\= & {} \beta _{U}(\mathrm {e}^{-\sigma }-1)+a(1-\mathrm {e}^{-\sigma }), \end{aligned}$$

thus $x^{(p)}(\varDelta _{0}+\sigma ) = (a-\beta _{U})(1-\mathrm {e}^{-\sigma })>0$.

For $t\in [\varDelta _{0}+\sigma ,\varDelta _{1}]$ we have

$$\begin{aligned} x^{(p)}(t) = -\beta _{U}+\left( x^{(p)}(\varDelta _{0}+\sigma ) + \beta _{U}\right) \mathrm {e}^{-(t-\varDelta _{0}-\sigma )}, \end{aligned}$$

so $x^{(p)}(t)$ is decreasing, since $(x^{(p)}(\varDelta _{0}+\sigma ) + \beta _{U}) = \beta _{U}\mathrm {e}^{-\sigma }+a(1-\mathrm {e}^{-\sigma })>0$, and

$$\begin{aligned} x^{(p)}(\varDelta _{1})= & {} -\beta _{U}+\left( x^{(p)}(\varDelta _{0}+\sigma ) + \beta _{U}\right) \mathrm {e}^{-\alpha } \\= & {} \beta _{U}(\mathrm {e}^{-T_{p}}-1)+a(1-\mathrm {e}^{-\sigma })\mathrm {e}^{-\alpha }. \end{aligned}$$

The condition $a\geqslant a_{1}$ implies $x^{(p)}(\varDelta _{1}) \geqslant 0$.

For $t\in [\varDelta _{1},\varDelta _{1}+\sigma ]$ it follows that

$$\begin{aligned} x^{(p)}(t) = -\beta _{U}+a+\left( x^{(p)}(\varDelta _{1})+\beta _{U}-a\right) \mathrm {e}^{-(t-\varDelta _{1})}, \end{aligned}$$

so $x^{(p)}(t)$ is increasing, $(x^{(p)}(\varDelta _{1})+\beta _{U}-a) = (\beta _{U}-a)\mathrm {e}^{-T_{p}}+a(\mathrm {e}^{-\alpha }-1)<0$, and

$$\begin{aligned} x^{(p)}(\varDelta _{1}+\sigma )= & {} -\beta _{U}+a+\left( x^{(p)}(\varDelta _{1})+\beta _{U}-a\right) \mathrm {e}^{-\sigma } \\= & {} \beta _{U}(\mathrm {e}^{-T_{p}-\sigma }-1)+a(1-\mathrm {e}^{-\sigma })(1+\mathrm {e}^{-T_{p}}). \end{aligned}$$

Since $x^{(p)}(t)$ is increasing for $t\in [\varDelta _{1},\varDelta _{1}+\sigma ]$ and $x^{(p)}(\varDelta _{1}) \geqslant 0$, then $x^{(p)}(\varDelta _{1}+\sigma )>x^{(p)}(\varDelta _{1})\geqslant 0$.

For $t\in [\varDelta _{1}+\sigma ,\varDelta _{2}]$ the solution is given by

$$\begin{aligned} x^{(p)}(t) = -\beta _{U}+\left( x^{(p)}(\varDelta _{1}+\sigma )+\beta _{U}\right) \mathrm {e}^{-(t-\varDelta _{1}-\sigma )}, \end{aligned}$$

so $x^{(p)}(t)$ is decreasing, $(x^{(p)}(\varDelta _{1}+\sigma )+\beta _{U}) = \beta _{U}\mathrm {e}^{-T_{p}-\sigma }+a(1-\mathrm {e}^{-\sigma })(1+\mathrm {e}^{-T_{p}})>0$, and

$$\begin{aligned} x^{(p)}(\varDelta _{2})= & {} -\beta _{U}+\left( x^{(p)}(\varDelta _{1}+\sigma )+\beta _{U}\right) \mathrm {e}^{-\alpha }\\= & {} \beta _{U}(\mathrm {e}^{-2T_{p}}-1)+a(1-\mathrm {e}^{-\sigma })(1+\mathrm {e}^{-T_{p}})\mathrm {e}^{-\alpha }. \end{aligned}$$

Thus $x^{(p)}(\varDelta _{2}) > \beta _{U}(\mathrm {e}^{-T_{p}}-1)+a(1-\mathrm {e}^{-\sigma })\mathrm {e}^{-\alpha } = x^{(p)}(\varDelta _{1})\geqslant 0$.

For $t\in [\varDelta _{2},\varDelta _{2}+\sigma ]$ we have

$$\begin{aligned} x^{(p)}(t) = -\beta _{U}+a+\left( x^{(p)}(\varDelta _{2})+\beta _{U}-a\right) \mathrm {e}^{-(t-\varDelta _{2})}, \end{aligned}$$

so $x^{(p)}(t)$ is increasing, $(x^{(p)}(\varDelta _{2})+\beta _{U}-a) = (\beta _{U}-a)\mathrm {e}^{-2T_{p}}+a(\mathrm {e}^{-\alpha }-1)+(\mathrm {e}^{-\alpha }-1)\mathrm {e}^{-2\alpha }<0$, and

$$\begin{aligned} x^{(p)}(\varDelta _{2}+\sigma )= & {} -\beta _{U}+a+\left( x^{(p)}(\varDelta _{2})+\beta _{U}-a\right) \mathrm {e}^{-\sigma } \\= & {} \beta _{U}(\mathrm {e}^{-2T_{p}-\sigma }-1)+a(1-\mathrm {e}^{-\sigma })[1+\mathrm {e}^{-T_{p}}+\mathrm {e}^{-2T_{p}}]. \end{aligned}$$

Since $x^{(p)}(t)$ is increasing for $t\in [\varDelta _{2},\varDelta _{2}+\sigma ]$ and $x^{(p)}(\varDelta _{2})\geqslant 0$, then $x^{(p)}(\varDelta _{2}+\sigma )>x^{(p)}(\varDelta _{2})>0$.

For $t\in [\varDelta _{2}+\sigma ,\varDelta _{3}]$ it follows that

$$\begin{aligned} x^{(p)}(t) = -\beta _{U}+\left( x^{(p)}(\varDelta _{2}+\sigma )+\beta _{U}\right) \mathrm {e}^{-(t-\varDelta _{2}-\sigma )}, \end{aligned}$$

so $x^{(p)}(t)$ is decreasing, $(x^{(p)}(\varDelta _{2}+\sigma )+\beta _{U}) = \beta _{U}\mathrm {e}^{-2T_{p}-\sigma }+a(1-\mathrm {e}^{-\sigma })[1+\mathrm {e}^{-T_{p}}+\mathrm {e}^{-2T_{p}}]>0$, and

$$\begin{aligned} x^{(p)}(\varDelta _{3})= & {} -\beta _{U}+\left( x^{(p)}(\varDelta _{2}+\sigma )+\beta _{U}\right) \mathrm {e}^{-\alpha } \\= & {} \beta _{U}(\mathrm {e}^{-3T_{p}}-1)+a(1-\mathrm {e}^{-\sigma })[1+\mathrm {e}^{-T_{p}}+\mathrm {e}^{-2T_{p}}]\mathrm {e}^{-\alpha }. \end{aligned}$$

Thus $x^{(p)}(\varDelta _{3})> \beta _{U}(\mathrm {e}^{-T_{p}}-1)+a(1-\mathrm {e}^{-\sigma })\mathrm {e}^{-\alpha } = x^{(p)}(\varDelta _{2})>0$.

For $t\in [\varDelta _{3},\varDelta _{3}+\sigma ]$ the solution is given by

$$\begin{aligned} x^{(p)}(t) = -\beta _{U}+a+\left( x^{(p)}(\varDelta _{3})+\beta _{U}-a\right) \mathrm {e}^{-(t-\varDelta _{3})}, \end{aligned}$$

so $x^{(p)}(t)$ is increasing, $(x^{(p)}(\varDelta _{3})+\beta _{U}-a) = (\beta _{U}-a)\mathrm {e}^{-3T_{p}}+a(\mathrm {e}^{-\alpha }-1)+(\mathrm {e}^{-\alpha }-1)\mathrm {e}^{-3\alpha }<0$, and

$$\begin{aligned} x^{(p)}(\varDelta _{3}+\sigma )= & {} -\beta _{U}+a+\left( x^{(p)}(\varDelta _{3})+\beta _{U}-a\right) \mathrm {e}^{-\sigma } \\= & {} \beta _{U}(\mathrm {e}^{-3T_{p}-\sigma }-1)+a(1-\mathrm {e}^{-\sigma })[1+\mathrm {e}^{-T_{p}}+\mathrm {e}^{-3T_{p}}]. \end{aligned}$$

Since $x^{(p)}(t)$ is increasing for $t\in [\varDelta _{3},\varDelta _{3}+\sigma ]$ and $x^{(p)}(\varDelta _{3})\geqslant 0$, then $x^{(p)}(\varDelta _{3}+\sigma )>x^{(p)}(\varDelta _{3})>0$.

Generalizing this procedure, we see that $x^{(p)}(\varDelta _{n})$ and $x^{(p)}(\varDelta _{n}+\sigma )$ can be written according to Eqs. (25) and (26). The proof is completed inductively for (25) and (26) with $n=1,2,3$ and $\varDelta _{n}=nT_{p}+\varDelta _{0}$. Hence, we see that for $t\in [\varDelta _{n},\varDelta _{n}+\sigma ]$ the solution $x^{(p)}(t)$ is given by Eq. (23) and for $t\in [\varDelta _{n}+\sigma ,\varDelta _{n+1}]$ the solution $x^{(p)}(t)$ is given by Eq. (24). $\square$

Proof of Proposition 4.2

(i): Since $a\geqslant a_{1}$ we can take the limit $n\longrightarrow \infty$ in Eqs. (25) and (26). We know that $\sum _{k=0}^{\infty }y^{k}=1/(1-y)$ for $|y|<1$, then

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty } x^{(p)}(\varDelta _{n}) = -\beta _{U}+a\frac{(1-\mathrm {e}^{-\sigma })}{(1-\mathrm {e}^{-T_{p}})}\mathrm {e}^{-\alpha }= {\underline{x}}^{(p)}, \end{aligned}$$

and

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty } x^{(p)}(\varDelta _{n}+\sigma ) = -\beta _{U}+a\frac{(1-\mathrm {e}^{-\sigma })}{(1-\mathrm {e}^{-T_{p}})} = \bar{x}^{(p)}. \end{aligned}$$

(ii): Using that $x(0)=\varphi (0)={\underline{x}}^{(p)}$ and $\varDelta _{0}=0$, for $t\in [0,\sigma ]$ the solution is given by

$$\begin{aligned} x^{(p)}(t) = -\beta _{U}+a+({\underline{x}}^{(p)}+\beta _{U}-a)\mathrm {e}^{-t}, \end{aligned}$$

once that $x(t-\tau )=\varphi (t)\geqslant 0$. So $x^{(p)}(t)$ is increasing, since $({\underline{x}}^{(p)}+\beta _{U}-a)=a\mathrm {e}^{-\alpha }(1-\mathrm {e}^{-\sigma })/(1-\mathrm {e}^{-\sigma -\alpha })-a<0$, and

$$\begin{aligned} x^{(p)}(\sigma )= & {} -\beta _{U}+a+\left( {\underline{x}}^{(p)}+\beta _{U}-a\right) \mathrm {e}^{-\sigma }, \\= & {} -\beta _{U}+a(1-\mathrm {e}^{-\sigma })+a\dfrac{(1-\mathrm {e}^{-\sigma })}{(1-\mathrm {e}^{-T_{p}})}\mathrm {e}^{-T_{p}}= \bar{x}^{(p)}. \end{aligned}$$

Using that $x^{(p)}(\sigma ) = \bar{x}^{(p)}$, for $t\in [\sigma ,\varDelta _{1}]$ we have

$$\begin{aligned} x^{(p)}(t) = -\beta _{U}+\left( \bar{x}^{(p)} + \beta _{U}\right) \mathrm {e}^{-(t-\sigma )}, \end{aligned}$$

so $x^{(p)}(t)$ is decreasing, since $(\bar{x}^{(p)} + \beta _{U}) = a(1-\mathrm {e}^{-\sigma })/(1-\mathrm {e}^{-T_{p}})>0$, and

$$\begin{aligned} x^{(p)}(\varDelta _{1})= & {} -\beta _{U}+\left( \bar{x}^{(p)} + \beta _{U}\right) \mathrm {e}^{-\alpha }, \\= & {} -\beta _{U}+a\dfrac{(1-\mathrm {e}^{-\sigma })}{(1-\mathrm {e}^{-T_{p}})}\mathrm {e}^{-\alpha } = {\underline{x}}^{(p)}. \end{aligned}$$

Furthermore, once $x^{(p)}(\varDelta _{1}) = {\underline{x}}^{(p)}$, for $t\in [\varDelta _{1},\varDelta _{1}+\sigma ]$ it follows that

$$\begin{aligned} x^{(p)}(t) = -\beta _{U}+a+({\underline{x}}^{(p)}+\beta _{U}-a)\mathrm {e}^{-(t-\varDelta _{1})}, \end{aligned}$$

and thus $x^{(p)}(\varDelta _{1}+\sigma ) = \bar{x}^{(p)}$. For $t\in [\varDelta _{1}+\sigma ,\varDelta _{2}]$ we have

$$\begin{aligned} x^{(p)}(t) = -\beta _{U}+\left( \bar{x}^{(p)} + \beta _{U}\right) \mathrm {e}^{-(t-\varDelta _{1}-\sigma )}. \end{aligned}$$

Hence, repeating this process for $t\in [\varDelta _{2},\varDelta _{2}+\sigma ]$, $[\varDelta _{2}+\sigma ,\varDelta _{2}]$, $[\varDelta _{3},\varDelta _{3}+\sigma ]$ and so forth we see that the solution $x^{(p)}(t)$ is given by (27). The proof is completed by checking that the Principle of Mathematical Induction holds for (27) with $n=1,2,3$ and $\varDelta _{n}=nT_{p}+\varDelta _{0}$. $\square$

Proof of Proposition 4.3

Consider the maximum (29) and define an initial perturbation phase $\varDelta _{l}$ such that $x^{(p)}(\varDelta _{l})={\underline{x}}^{(p)}$. The solution for the initial pulse is given by (41) (Mackey et al. 2017, Section 5.3) with $x^{(\varDelta )}(\varDelta )=-\beta _{U}+\beta _{U}\mathrm {e}^{{\tilde{z}}_{2}-\varDelta }$. Thus $x^{(p)}(\varDelta _{l}) = -\beta _{U} + \beta _{U}\mathrm {e}^{{\tilde{z}}_{2}-\varDelta _{l}}={\underline{x}}^{(p)}$. This with ${\underline{x}}^{(p)}\geqslant 0$, since $a\geqslant a_{1}$, gives

$$\begin{aligned} \varDelta _{l} = \tilde{z}_{2} + \ln \left( \frac{\beta _{U}}{\beta _{U}+{\underline{x}}^{(p)}}\right) \leqslant \tilde{z}_{2}. \end{aligned}$$

(62)

The proof is divided between the four cases shown in Fig. 9, where each $\varDelta _{0}$ interval is given by: (a) $(\tilde{z}_{1},t_{max})$, (b) $[t_{max},\varDelta _{l})$, (c) $[\varDelta _{l},\tilde{z}_{2})$, (d) $[\tilde{z}_{2},\tilde{T}+\tilde{z}_{1}]$.

First, we prove case (b) by showing that for $\varDelta _{0}\in [t_{max},\varDelta _{l})$ the points $x^{(p)}(\varDelta _{n})$ converge to (28) and the points $x^{(p)}(\varDelta _{n}+\sigma )$ converge to (29). For this case $x^{(p)}(t_{max})\geqslant x^{(p)}(\varDelta _{0})>{\underline{x}}^{(p)}\geqslant 0$ and $x^{(p)}(\varDelta _{0}-\tau )\geqslant 0$. Thus for $t=\varDelta _{0}$ we have

$$\begin{aligned} x^{(p)}(\varDelta _{0})=-\beta _{U}+\beta _{U}\mathrm {e}^{{\tilde{z}}_{2}-\varDelta _{0}}. \end{aligned}$$

For $t\in [\varDelta _{0},\varDelta _{0}+\sigma ]$ the solution is given by $x^{(p)}(t)=-\beta _{U}+a+(x^{(p)}(\varDelta _{0})+\beta _{U}-a)\mathrm {e}^{-(t-\varDelta _{0})}$, so

$$\begin{aligned} x^{(p)}(\varDelta _{0}+\sigma )=-\beta _{U}+a+\left( x^{(p)}(\varDelta _{0})+\beta _{U}-a\right) \mathrm {e}^{-\sigma }. \end{aligned}$$

For $t\in [\varDelta _{0}+\sigma ,\varDelta _{1}]$ we have $x^{(p)}(t)=-\beta _{U}+(x^{(p)}(\varDelta _{0}+\sigma )+\beta _{U})\mathrm {e}^{-(t-\varDelta _{0}-\sigma )}$, then

$$\begin{aligned} x^{(p)}(\varDelta _{1})=-\beta _{U}+\left( x^{(p)}(\varDelta _{0}+\sigma )+\beta _{U}\right) \mathrm {e}^{-\alpha }. \end{aligned}$$

Repeating the previous steps for $t\in [\varDelta _{1},\varDelta _{1}+\sigma ]$, $t\in [\varDelta _{1}+\sigma ,\varDelta _{2}]$ and so forth, we see that for $t\in [\varDelta _{n},\varDelta _{n}+\sigma ]$ and $n\in \mathbb {N}_{>0}$ the solution is given by $x^{(p)}(t)=-\beta _{U}+a+(x^{(p)}(\varDelta _{n})+\beta _{U}-a)\mathrm {e}^{-(t-\varDelta _{n})}$, so

$$\begin{aligned} x^{(p)}(\varDelta _{n}+\sigma )=-\beta _{U}+a+\left( x^{(p)}(\varDelta _{n})+\beta _{U}-a\right) \mathrm {e}^{-\sigma }, \end{aligned}$$

(63)

and for $t\in [\varDelta _{n}+\sigma ,\varDelta _{n+1}]$ and $n\in \mathbb {N}_{>0}$ we have $x^{(p)}(t)=-\beta _{U}+(x^{(p)}(\varDelta _{n}+\sigma )+\beta _{U})\mathrm {e}^{-(t-\varDelta _{n}-\sigma )}$, then

$$\begin{aligned} x^{(p)}(\varDelta _{n+1})=-\beta _{U}+\left( x^{(p)}(\varDelta _{n}+\sigma )+\beta _{U}\right) \mathrm {e}^{-\alpha }. \end{aligned}$$

(64)

From (63), (64) and $T_{p}=\alpha +\sigma$ it follows that

$$\begin{aligned} x^{(p)}(\varDelta _{n+1})=-\beta _{U}+a\mathrm {e}^{-\alpha }(1-\mathrm {e}^{-\sigma })+\left( x^{(p)}(\varDelta _{n})+\beta _{U}\right) \mathrm {e}^{-T_{p}}. \end{aligned}$$

(65)

Equation (65) is recursive and can be rewritten as

$$\begin{aligned} x^{(p)}(\varDelta _{n})=-\beta _{U}+a\mathrm {e}^{-\alpha }(1-\mathrm {e}^{-\sigma })\sum _{k=0}^{n-1}\mathrm {e}^{-kT_{p}}+\left( x^{(p)}(\varDelta _{0})+\beta _{U}\right) \mathrm {e}^{-nT_{p}}. \end{aligned}$$

(66)

It is known that $\sum _{k=0}^{\infty }y^{k}=1/(1-y)$ for $|y|<1$, thus taking the limit $n\longrightarrow \infty$ of (66) gives

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty } x^{(p)}(\varDelta _{n}) = -\beta _{U}+a\frac{(1-\mathrm {e}^{-\sigma })}{(1-\mathrm {e}^{-T_{p}})}\mathrm {e}^{-\alpha }= {\underline{x}}^{(p)}, \end{aligned}$$

(67)

and taking the limit $n\longrightarrow \infty$ of (63) and using (67) yields

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty } x^{(p)}(\varDelta _{n}+\sigma ) = -\beta _{U}+a\frac{(1-\mathrm {e}^{-\sigma })}{(1-\mathrm {e}^{-T_{p}})} = \bar{x}^{(p)}. \end{aligned}$$

Then in the limit $n\longrightarrow \infty$ we see that the solution of (21) converges to the limit cycle given by (27).

For case (c) the proof is the same as the case (b), but with $\varDelta _{0}\in [\varDelta _{l},\tilde{z}_{2})$ and hence $x^{(p)}(\varDelta _{l})\geqslant x^{(p)}(\varDelta _{0})>x^{(p)}(\tilde{z}_{2})\geqslant 0$.

To prove case (a) we first note that $x^{(p)}(t-\tau )<0$ for $t\in (\tilde{z}_{1},t_{max})$. Given an initial perturbation phase $\varDelta _{0}\in (\tilde{z}_{1},t_{max})$, the solution on $(\tilde{z}_{1},t_{max})$ alternates between $x^{(p)}(t)=\beta _{L}+a+(x^{(p)}(\varDelta _{n})-\beta _{L}-a)\mathrm {e}^{-(t-\varDelta _{n})}$ for $[\varDelta _{n},\varDelta _{n}+\sigma ]$ and $x^{(p)}(t)=\beta _{L}+(x^{(p)}(\varDelta _{0}+\sigma )-\beta _{L})\mathrm {e}^{-(t-\varDelta _{0}-\sigma )}$ for $[\varDelta _{n}+\sigma ,\varDelta _{n}]$ with $n=0,1,2,\ldots ,k$, where the k-th index is such that $\varDelta _{k-1}< t_{max} \leqslant \varDelta _{k}$. Along the interval $(\tilde{z}_{1},t_{max})$ the solution $x^{(p)}(t)$ might have a maximum point if it reaches the value $(\beta _{L}+a)$ in the intervals $[\varDelta _{n},\varDelta _{n}+\sigma ]$ or if it reaches the value $\beta _{L}$ in the intervals $[\varDelta _{n}+\sigma ,\varDelta _{n}]$, otherwise, it will be strictly increasing for $t\in (\tilde{z}_{1},t_{max})$, as is in the examples of Fig. 9a. For $t\geqslant t_{max}$ the proof is the same as the case (b), but using $x^{(p)}(t_{max})$ as initial point $x^{(p)}(\varDelta _{0})$.

For case (d) we need to distinguish it between two subcases, the interval $t\in [\varDelta _{0},\tilde{T}]$, for which $x^{(p)}(t-\tau )\geqslant 0$, and the interval $t\in (\tilde{T},z_{p,1}+\tau )$, where $x^{(p)}(t-\tau )<0$ and $z_{p,1}$ is the first zero of $x^{(p)}(t)$ with $t>\varDelta _{0}$. For the first interval the solution alternates between $x^{(p)}(t)=-\beta _{U}+a+(x^{(p)}(\varDelta _{n})+\beta _{U}-a)\mathrm {e}^{-(t-\varDelta _{n})}$ for $[\varDelta _{n},\varDelta _{n}+\sigma ]$ and $x^{(p)}(t)=-\beta _{U}+(x^{(p)}(\varDelta _{0}+\sigma )+\beta _{U})\mathrm {e}^{-(t-\varDelta _{0}-\sigma )}$ for $[\varDelta _{n}+\sigma ,\varDelta _{n}]$ with $n=0,1,2,\ldots ,k$, where the k-th interval is such that $\varDelta _{k-1}<(z_{p,1}+\tau )\leqslant \varDelta _{k}$. For the second interval, the solution alternates between $x^{(p)}(t)=\beta _{L}+a+(x^{(p)}(\varDelta _{n})-\beta _{L}-a)\mathrm {e}^{-(t-\varDelta _{n})}$ for $[\varDelta _{n},\varDelta _{n}+\sigma ]$ and $x^{(p)}(t)=\beta _{L}+(x^{(p)}(\varDelta _{0}+\sigma )-\beta _{L})\mathrm {e}^{-(t-\varDelta _{0}-\sigma )}$ for $[\varDelta _{n}+\sigma ,\varDelta _{n}]$ with $n=0,1,2,\ldots ,k$, where the k-th index is such that $\varDelta _{k-1}<(\tilde{T}+\tilde{z}_{1})\leqslant \varDelta _{k}$. Along the interval $[\varDelta _{0},\tilde{T}]$ the solution $x^{(p)}(t)$ might oscillate if it reaches the value $(a-\beta _{U})$ in the intervals $[\varDelta _{n},\varDelta _{n}+\sigma ]$ or if it reaches the value $-\beta _{U}$ in the intervals $[\varDelta _{n}+\sigma ,\varDelta _{n}]$, otherwise, it will be strictly increasing for $t\in [\varDelta _{0},\tilde{T}]$, as is in the examples of Fig. 9d. For $t>(z_{p,1}+\tau )$ the proof is the same as the case (a), but using $x^{(p)}(z_{p,1}+\tau )$ as initial point $x^{(p)}(\varDelta _{0})$. $\square$