**8. Asymptotics of Solutions in the Case** *b* **= 0 and** *d <* **0**

Firstly, consider initial conditions from *S*−. Then, on the first step, *t* ∈ [0, *T*], Equation (2) has the form of (24) and solution has the form

$$
\mu(t) = p\_L e^{-t}.\tag{36}
$$

It follows from (36) that in the segment *t* ∈ [0, *T*], function *u*(*t*) satisfies the inequality *pL* < *u*(*t*) < 0, which is why in the second step, *t* ∈ [*T*, 2*T*], the solution has the form

$$u(t) = p\_L e^{-t} + \lambda \int\_T^t e^{s-t} f(p\_L e^{T-s}) ds. \tag{37}$$

If function *f*(*u*) satisfies the condition

$$\int\_{T}^{t} e^{s-t} f(p\_L e^{T-s}) ds < 0 \text{ for all } t \in (T, 2T], \tag{38}$$

then there exists an asymptotically small by *λ* value *δ* > 0 such that Expression (37) is less than *pL* on the interval *t* ∈ (*T* + *δ*, 2*T*]. It is easy to see that on the segment *t* ∈ [2*T*, 3*T*], the leading term of the asymptotics of the solution to Equation (2) coincides with the leading term of the asymptotics of the solution to Equation (24) with the initial condition

$$
\mu(2T) = p\_L e^{-2T} + \lambda \int\_T^{2T} e^{s - 2T} f(p\_L e^{T - s}) ds. \tag{39}
$$

This is why, in the segment *t* ∈ [2*T*, 3*T*], the solution to Equation (2) has the form

$$u(t) = \lambda \left( \int\_{T}^{2T} e^{s - 2T} f(p\_L e^{T - s}) ds + o(1) \right) e^{-(t - 2T)}.\tag{40}$$

Note that Expression (40) is less than *pL* in the segment *t* ∈ [2*T*, 3*T*], which is why Equation (2) has the form of (24) until the Function (40) becomes greater than *pL*. There exists a value *t*<sup>∗</sup> > 3*T* such that *u*(*t*∗) = *pL* and *u*(*t*) < *pL* for all *t* ∈ (2*T*, *t*∗). This is why at the point *t* = *t*∗, we return to the initial situation: the function *u*(*t*<sup>∗</sup> + *s*) (*s* ∈ [−*T*, 0)) belongs to the set *S*−. Therefore, if we consider the function *u*(*t*<sup>∗</sup> + *s*) as the initial conditions of Equation (2), then we get a negative relaxation cycle. Note that it follows from (40) that *t*<sup>∗</sup> − 2*T* = (1 + *o*(1))ln *λ* at *λ* → +∞.

We obtain the following statement.

**Theorem 5.** *Let b* = 0*, d* < 0*, and let condition* (38) *be true. Then, for all sufficiently large λ* > 0*, Equation* (2) *has a negative relaxation cycle with the asymptotics*

$$\begin{aligned} u(t) &= p\_L e^{-(t - nt\_\*)} , & t \in [nt\_\*, nt\_\* + T] \\ u(t) &= p\_L e^{-(t - nt\_\*)} + \lambda \int\_{T + nt\_\*}^t e^{s - t} f(p\_L e^{nt\_\* + T - s}) ds \, & t \in [nt\_\* + T, nt\_\* + 2T] \\ u(t) &= \lambda \binom{nt\_\* + 2T}{\int\_{nt\_\*}^t e^{s - nt\_\* - 2T} f(p\_L e^{nt\_\* + T - s}) ds + o(1) \, & t \in [nt\_\* - 2T] \\ & t \in [nt\_\* + 2T, (n + 1)t\_\*] \end{aligned} \tag{41}$$

*(where n* = 0, 1, 2, ... *represents the number of periods of a cycle) and period t*<sup>∗</sup> = 2*T* + (1 + *o*(1))ln *λ at λ* → +∞*.*

If the function *f*(*u*) satisfies the condition

$$\int\_{T}^{t} e^{s-t} f(p\_L e^{T-s}) ds > 0 \text{ for all } t \in (T, 2T], \tag{42}$$

then there exists an asymptotically small by *λ* value *δ*<sup>1</sup> > 0 such that *u*(*T* + *δ*1) = *pR*, *u*(*t*) > *pR* in the interval (*T* + *δ*1, 2*T*]. Thus, in the segment *t* ∈ [2*T*, 3*T*], the leading term of the asymptotics of the solution to Equation (2) coincides with the leading term of the asymptotics of the solution to Equation (4) with the initial conditions of (39). This is why this time segment solution has the asymptotics

$$u(t) = \lambda \left( \int\_{T}^{2T} e^{s - 2T} f(p\_L e^{T - s}) ds - d + o(1) \right) e^{-(t - 2T)} + \lambda d. \tag{43}$$

Since *d* < 0, Expression (43) is decreasing, and there exists a time value *t*<sup>1</sup> such that *t*<sup>1</sup> > 2*T* and Expression (43) is equal to zero at the point *t*<sup>1</sup> and greater than zero in the interval *t* ∈ (2*T*, *t*1). Note that until *u*(*t*) < *pR*, Equation (2) has the form of (4), and the solution has the form of (43). Since *λ* is sufficiently large, there exists an asymptotically small by *λ* values *δ*<sup>2</sup> < 0 and *δ*<sup>3</sup> > 0 such that *u*(*t*<sup>1</sup> + *δ*2) = *pR* and *u*(*t*<sup>1</sup> + *δ*3) = *pL*. The length of the interval (*T* + *δ*1, *t*<sup>1</sup> + *δ*2) is greater than *T*, and the solution in this interval is greater than *pR*, which is why Equation (2) has the form of (4) in the segment *t* ∈ [*t*<sup>1</sup> + *δ*2, *t*<sup>1</sup> + *δ*<sup>2</sup> + *T*] and the solution has the form of (43) in this interval.

It is easy to see that

$$
\mu(t\_1 + \delta\_3 + T) = \lambda d(1 - e^{-T} + o(1)).\tag{44}
$$

Since the solution is less than *pL* in the interval of the length of delay (*t* ∈ (*t*<sup>1</sup> + *δ*3, *t*<sup>1</sup> + *δ*<sup>3</sup> + *T*]), and *u*(*t*<sup>1</sup> + *δ*<sup>3</sup> + *T*) is negative and has the order *O*(*λ*), Equation (2) has the form of (24) in the segment of the length (1 + *o*(1))ln *λ* (until the solution becomes greater than *pL*), and the solution has the form

$$u(t) = \lambda d(1 - e^{-T} + o(1))e^{-\left(t - (t\_1 + T)\right)}.\tag{45}$$

Expression (45) is negative and increases. There exists a time moment *t*<sup>∗</sup> > *t*<sup>1</sup> + *T* + *δ*<sup>3</sup> such that Expression (45) is less than *pL* for all *t* ∈ [*t*<sup>1</sup> + *T* + *δ*3, *t*∗) and is equal to *pL* at the point *t* = *t*∗. Thus, function *u*(*t*<sup>∗</sup> + *s*) belongs to the set *S*−: *u*(*t*∗) = *pL* and *u*(*t*<sup>∗</sup> + *s*) < *pL* for all *s* ∈ [−*T*, 0). Therefore, if we take this function as the initial conditions of Equation (2), then we get a sign-changing relaxation cycle of this equation with the period *t*<sup>∗</sup> = *t*<sup>1</sup> + *T* + (1 + *o*(1))ln *λ*.

If we take initial functions from *S*+, then at the first step, *t* ∈ [0, *T*], the equation has the form of (4) and solution has the form of (5). Then, there exists an asymptotically small by *λ* value *δ*<sup>4</sup> > 0 such that *u*(*t*) < *pL* for all *t* ∈ (*δ*4, *T*]. Since, for *t* ∈ [*δ*4, *T* + *δ*4], the solution is less than *pL*, Equation (2) has the form of (24), and the solution has the form

$$
\mu(t) = \lambda d(1 - e^{-T} + o(1))e^{-(t-T)} \tag{46}
$$

for *t* > *T* + *δ*4; until then *u*(*t*) > *pL*.

It follows from (46) that there exists a value *tL* such that *u*(*tL*) = *pL* and *u*(*s* + *tL*) < *pL* in the interval *s* ∈ [−*T*, 0). Therefore, oin the segment *t* ∈ [*tL* − *T*, *tL*], the solution belongs to the set *S*−, which is why we have reduced the problem to the previously studied one.

From the above reasoning, we obtain the following statement.

**Theorem 6.** *Let b* = 0*, d* < 0*, and let condition* (42) *hold. Then, for all sufficiently large λ* > 0*, Equation* (2) *has a sign-changing relaxation cycle with the asymptotics*

$$\begin{aligned} u(t) &= p\_L e^{-(t - nt\_\bullet)}, & t \in [nt\_{\ast \ast}nt\_{\ast} + T], \\ u(t) &= p\_L e^{-(t - nt\_\circ)} + \lambda \int\_{T + nt\_\bullet}^t e^{s - t} f(p\_L e^{nt\_\bullet + T - s}) ds, & t \in [nt\_\ast + T, nt\_\ast + 2T], \\ u(t) &= \lambda \left( \int\_{nt\_\ast + T}^{nt\_\ast + 2T} e^{s - nt\_\ast - 2T} f(p\_L e^{nt\_\ast + T - s}) ds - d + o(1) \right) e^{-(t - nt\_\ast - 2T)} + \lambda d, \\ u(t) &= \lambda d (1 - e^{-T} + o(1)) e^{-(t - (nt\_\ast + t\_\ast + T))}, & t \in [nt\_\ast + T, nt\_\ast + t\_\ast + T], \\ u(t) &= \lambda d (1 - e^{-T} + o(1)) e^{-(t - (nt\_\ast + t\_\ast + T))}, & t \in [nt\_\ast + t\_\ast + T, (n + 1)t\_\ast]. \end{aligned}$$

*(where n* = 0, 1, 2, ... *represents the number of periods of cycle) and period t*<sup>∗</sup> = *t*<sup>1</sup> + *T* + (1 + *o*(1))ln *λ at λ* → +∞*.*

The cycles of Equation (2) in the case that *b* = 0 and *d* < 0 are shown in Figure 5.

**Figure 5.** Relaxation cycles of Equation (2) in the case that *b* = 0 and *d* < 0 and function *f*(*u*) satisfies the conditions of (**a**) (38) and (**b**) (42). Values of parameters: *<sup>λ</sup>* <sup>=</sup> <sup>10</sup>4, *pL* <sup>=</sup> <sup>−</sup>1.5, *pR* <sup>=</sup> 2.5, *<sup>b</sup>* <sup>=</sup> 0, (**a**) *T* = 5, *d* = −2, (**b**) *T* = 3, and *d* = −4.
