**6. Asymptotics of Solutions in the Case** *b >* **0 and** *d* **= 0**

Firstly, we consider initial conditions from *S*+. Then, on the first step *t* ∈ [0, *T*], Equation (2) has the form

$$
\ddot{u} + \mu = 0.\tag{24}
$$

Therefore, the solution has the form

$$
\mu(t) = p\_R e^{-t}.\tag{25}
$$

Since the solution to (25) belongs to the interval *u* ∈ (0, *pR*) in the interval *t* ∈ (0, *T*], then in the segment *t* ∈ [*T*, 2*T*], it depends on the values of function *f* . It has the form

$$
\mu(t) = p\_R e^{-t} + \lambda \int\_T^t e^{s-t} f(p\_R e^{T-s}) ds. \tag{26}
$$

In this segment, the asymptotics of the solution to (2) crucially depend on the values of the integral in Formula (26). Below, we assume that this integral preserves its sign on the segment *t* ∈ (*T*, 2*T*] (if the integral changes its sign, then we cannot construct the asymptotics of the solution at the segment *t* ∈ [2*T*, 3*T*] for an arbitrary unknown function *f*). Consider the first case:

$$\int\_{T}^{t} e^{s-t} f(p\_R e^{T-s}) ds > 0 \text{ for all } t \in (T, 2T]. \tag{27}$$

Then, expression (26) is greater than *pR* on the segment *t* ∈ [*T* + *δ*1, 2*T*] (here, *δ*<sup>1</sup> ≥ 0 is some asymptotically small by *<sup>λ</sup>* value; it has order *<sup>O</sup>*(*λ*−1) at *<sup>λ</sup>* → +∞). In the segment *t* ∈ [2*T*, 2*T* + *δ*1], 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 conditions

$$
\mu(2T) = p\_R e^{-2T} + \lambda \int\_T^T e^{s-2T} f(p\_R e^{T-s}) ds\_\prime
$$

and in the segment *t* ∈ [2*T* + *δ*1, 3*T*], Equation (2) has the form of (24). This is why in the whole segment *t* ∈ [2*T*, 3*T*], the solution to (2) has the form of

$$u(t) = \lambda \left( \int\_{T}^{2T} e^{s - 2T} f(p\_R e^{T - s}) ds + o(1) \right) e^{-(t - 2T)}.\tag{28}$$

Note that Expression (28) is greater than *pR* in the segment with length *O*(ln *λ*) at *λ* → +∞ (and this is why Equation (2) has the form of (24) in this segment), and this expression decreases and tends to zero at *t* → +∞. Therefore, there exists a time moment *t*<sup>∗</sup> = 2*T* + (1 + *o*(1))ln *λ* > 3*T* such that *u*(*t*∗) = *pR* and *u*(*t*<sup>∗</sup> + *s*) > *pR* on the interval *s* ∈ [−*T*, 0). Thus, we come to the initial situation (the function *u*(*t*<sup>∗</sup> + *s*) belongs to the set *S*+), and if we take this function as the initial conditions to our equation, then we obtain a positive relaxation cycle of Equation (2) with the amplitude *O*(*λ*) and period *t*<sup>∗</sup> = 2*T* + (1 + *o*(1))ln *λ*.

We obtain the following result.

**Theorem 2.** *Let b* > 0 *and d* = 0*, and* (27) *holds. Then, for all sufficiently large λ* > 0*, Equation* (2) *has a positive relaxation cycle with the asymptotics*

$$\begin{aligned} u(t) &= p\_R e^{-(t - nt\_\*)} , & t \in [nt\_\*, nt\_\* + T] \\ u(t) &= p\_R e^{-(t - nt\_\*)} + \lambda \int\_{T + nt\_\*}^t e^{s - t} f(p\_R e^{nt\_\* + T - s}) ds , & t \in [nt\_\* + T, nt\_\* + 2T] \\ u(t) &= \lambda \left( \int\_{nt\_\* + T}^{nt\_\* + 2T} e^{s - nt\_\* - 2T} f(p\_R e^{nt\_\* + T - s}) ds + o(1) \right) e^{-(t - nt\_\* - 2T)} , & t \in [nt\_\* + 2T, (n + 1)t\_\*] \end{aligned} \tag{29}$$

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

Consider the second case:

$$\int\_{T}^{t} e^{s-t} f(p\_{\mathbb{R}} e^{T-s}) ds < 0 \text{ for all } t \in (T, 2T]. \tag{30}$$

Then, expression (26) is less than *pL* in the segment *t* ∈ [*T* + *δ*2, 2*T*] (here, *δ*<sup>2</sup> > 0 denotes an asymptotically small by *λ* value such that *u*(*T* + *δ*2) = *pL*), and this is why Equation (2) in the segment *t* ∈ [2*T* + *δ*2, 3*T*] has the form of (6). Thus, in the segment *t* ∈ [2*T*, 3*T*], the solution has the form

$$u(t) = \lambda \left( \int\_{T}^{2T} e^{s-2T} f(p\_R e^{T-s}) ds - b + o(1) \right) e^{-(t-2T)} + \lambda b. \tag{31}$$

Note that expression (31) is an increasing function and that there exists a time moment *t*<sup>1</sup> > 2*T* + *δ*<sup>2</sup> such that *u*(*t*1) = 0. It is easy to see that

$$e^{-(t\_1 - 2T)} = \frac{b}{b - \int\_T^{2T} e^{s - 2T} f(p\_R e^{T - s}) ds}. \tag{32}$$

On the segment *t* ∈ [2*T* + *δ*2, *t*<sup>1</sup> + *T* + *δ*3] equation has the form of (6), and therefore, Formula (31) holds for the solution in this segment (here, *δ*<sup>3</sup> < 0 is an asymptotically small by *λ* value that denotes a time moment such that *u*(*t*<sup>1</sup> + *δ*3) = *pL*).

It follows from (32) that

$$
\lambda \mu (t\_1 + T + \delta\_3) = \lambda b (1 - e^{-T} + o(1)). \tag{33}
$$

Since the value (33) is positive and has order *O*(*λ*) at *λ* → +∞, then Equation (2) has the form of (24) in the time interval *t* ∈ [*t*<sup>1</sup> + *δ*<sup>4</sup> + *T*, *t*∗) (here, *δ*<sup>4</sup> > 0 denotes an asymptotically small by *λ* value such that *u*(*t*<sup>1</sup> + *δ*4) = *pR*, and *t*<sup>∗</sup> denotes a first time moment such that *t*<sup>∗</sup> > *t*<sup>1</sup> + *T* + *δ*<sup>4</sup> and *u*(*t*∗) = *pR*). Therefore, the solution has the form

$$u(t) = \lambda b(1 - e^{-T} + o(1))e^{-(t - (t\_1 + T))}.\tag{34}$$

Note that *t*<sup>∗</sup> = *t*<sup>1</sup> + *T* + (1 + *o*(1))ln *λ*. This is why *u*(*t*<sup>∗</sup> + *s*) > *pR* for all *s* ∈ [−*T*, 0). Thus, *u*(*t*<sup>∗</sup> + *s*) belongs to the set *S*+, and therefore, if we take this function as the initial condition, we get a sign-changing relaxation cycle with the amplitude of the order *O*(*λ*) and period *O*(ln *λ*) at *λ* → +∞.

If we consider initial conditions from the set *S*−, then on the first step *t* ∈ [0, *T*], the equation has the form of (6) and the solution has the form of (7). Since *b* > 0, there exists an asymptotically small by *λ* value *δ* > 0 such that *u*(*δ*) = *pR* and *u*(*t*) > *pR* for all *t* ∈ [*δ*, *T*]. Then, for all *t* > *T* + *δ*, until then, the *u*(*t*) = *pR* equation has the form of (24) and the solution has the form

$$
\mu(t) = \lambda b(1 - e^{-T} + o(1))e^{-(t-T)}.\tag{35}
$$

We denote as *tR* a time moment such that *tR* > *T* and *u*(*tR*) = *pR*. This value exists because Expression (35) decreases and tends to zero at *t* → +∞. Note that *tR* = *O*(ln *λ*) at *λ* → +∞. Therefore, function *u*(*tR* + *s*) (*s* ∈ [−*T*, 0]) belongs to the set *S*+, and we return to a problem considered earlier in this section.

From the results of this section we obtain the following statement.

**Theorem 3.** *Let b* > 0*, d* = 0*, and condition* (30) *holds. Then, for all sufficiently large λ* > 0*, Equation* (2) *has a sign-changing relaxation cycle with the asymptotics*

$$\begin{aligned} u(t) &= p\_R e^{-(t - nt\_\*)}, & t \in [nt\_\*, nt\_\* + T], \\ u(t) &= p\_R e^{-(t - nt\_\*)} + \lambda \int\_0^t e^{s - t} f(p\_R e^{nt\_\* + T - s}) ds, & t \in [nt\_\* + T, nt\_\* + 2T], \\ u(t) &= \lambda \left( \int\_{nt\_\* + T}^{nt\_\* + 2T} e^{s - nt\_\* - 2T} f(p\_R e^{nt\_\* + T - s}) ds - b + o(1) \right) e^{-(t - nt\_\* - 2T)} + \lambda b, \\ u(t) &= \lambda b (1 - e^{-T} + o(1)) e^{-(t - (nt\_\* + t\_1 + T))}, & t \in [nt\_\* + 2T, nt\_\* + t\_1 + T], \\ u(t) &= \lambda b (1 - e^{-T} + o(1)) e^{-(t - (nt\_\* + t\_1 + T))}, & t \in [nt\_\* + t\_1 + T, (n + 1)t\_1]. \end{aligned}$$

*(where n* = 0, 1, 2, ... *represents the number of periods of a 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 4.

**Figure 4.** Relaxation cycles of Equation (2) in the case that *b* > 0 and *d* = 0; function *f*(*u*) satisfies the condition (**a**) (27) (**b**) (30). Values of parameters: *<sup>λ</sup>* <sup>=</sup> 104, *<sup>T</sup>* <sup>=</sup> 3, *pL* <sup>=</sup> <sup>−</sup>1, *pR* <sup>=</sup> 2, *<sup>b</sup>* <sup>=</sup> 2, and *<sup>d</sup>* <sup>=</sup> 0.
