**5. Asymptotics of Solutions in the Case** *b >* **0 and** *d <* **0**

In this section, we study behaviour of solutions with initial conditions from sets *S*+ and *S*<sup>−</sup> under the assumption that *b* > 0 and *d* < 0.

Firstly, we take initial conditions from *S*+ and begin to construct asymptotics of solutions. Then, on the first step *t* ∈ [0, *T*], Equation (2) has the form of (4) and solution has the form of (5) and

$$
\mu(T) = \lambda d(1 - e^{-T} + o(1)).\tag{18}
$$

Since *d* < 0, there exists an asymptotically small *λ* value *δ*<sup>1</sup> > 0 such that *pL* < *u*(*t*) < *pR* for *t* ∈ (0, *δ*1) and *u*(*t*) < *pL* for all *t* ∈ (*δ*1, *T*]. Therefore, on the segment *t* ∈ [*T* + *δ*1, 2*T*], Equation (2) has the form of (6). In the segment *t* ∈ [*T*, *T* + *δ*1], the solution to Equation (2) depends on the values of the function *f* , but 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 (6) with initial conditions from (18). This is why in the whole segment *t* ∈ [*T*, 2*T*], the solution has the form of (17). Note that in the case that *b* > 0 and *d* < 0, Expression (17) is an increasing function.

Since *u*(*T*) < 0 and (17) increases to the positive value, there exists an asymptotically small by *λ* value *δ*<sup>2</sup> < 0 and value *t*<sup>1</sup> > *T* + *δ*1, such that *u*(*t*1) = 0 and *u*(*t*<sup>1</sup> + *δ*2) = *pL*. It follows from the definition of *t*<sup>1</sup> and *δ*<sup>2</sup> that Equation (2) has the form of (6) on the segment *t* ∈ [*T* + *δ*1, *t*<sup>1</sup> + *δ*<sup>2</sup> + *T*]. It easily follows from (17) that

$$e^{-(t\_1 - T)} = b / (b - d(1 - e^{-T})),\tag{19}$$

and, consequently,

$$
\mu(t\_1 + T) = \lambda b (1 - e^{-T} + o(1)). \tag{20}
$$

Note that expression (20) is greater than *pR* when *λ* 1. Thus, for *t* > *t*<sup>1</sup> + *T* + *δ*<sup>3</sup> (where *δ*<sup>3</sup> > 0 denotes an asymptotically small by *λ* value such that *u*(*t*<sup>1</sup> + *δ*3) = *pR*), until then, the *u*(*t*) < *pR* solution Equation (2) has the form of (4). Therefore, for *t* > *t*<sup>1</sup> + *T* + *δ*3, until then, the *u*(*t*) < *pR* solution to Equation (2) has the form of

$$u(t) = \left(\lambda b(1 - e^{-T}) - \lambda d\right)e^{-\left(t - (t\_1 + T)\right)} + \lambda d + o(\lambda). \tag{21}$$

Expression (21) is a decreasing function, and there exists *t* = *t*<sup>2</sup> such that (21) is equal to zero. Additionally, for *t*<sup>∗</sup> = *t*<sup>2</sup> + *o*(1), it is true that *u*(*t*∗) = *pR* and *u*(*t*<sup>∗</sup> + *s*) > *pR* for all *s* ∈ [−*T*, 0). Thus, at the point *t* = *t*∗, we return to the initial situation (the function *u*(*t*<sup>∗</sup> + *s*) belongs to the set *S*+). This is why if we take this function as the initial conditions for Equation (2), we obtain a periodic solution to this equation with an amplitude of the order *O*(*λ*) (see Formulas (18) and (20)) and period

$$t\_\* = 2T + \ln\left(\frac{(b(1 - e^{-T}) - d)(b - d(1 - e^{-T}))}{-bd}\right) + o(1). \tag{22}$$

We mention that the logarithm in Formula (22) is positive because its argument is greater than 1 for all *b* > 0, *d* < 0, and *T* > 0.

Note that there exists a point *tL* = *t*<sup>1</sup> + *δ*<sup>2</sup> such that *u*(*tL*) = *pL* and *u*(*s* + *tL*) < *pL* on the segment *s* ∈ [−*T*, 0). Additionally, we stress that if we take an initial function such that it is less than or equal to *pL* in some segment of the length *<sup>T</sup>*: *<sup>s</sup>* ∈ [˜*<sup>t</sup>* − *<sup>T</sup>*, ˜*t*) and is equal to *pL* at the point ˜*t*, then the solution to Equation (2) on the next interval *<sup>t</sup>* ∈ [˜*t*, ˜*<sup>t</sup>* + *<sup>T</sup>*] does not depend on the "history" values of *<sup>u</sup>*(*<sup>s</sup>* + ˜*t*) on *<sup>s</sup>* ∈ [−*T*, 0). This is why if we consider the initial conditions from the set *S*<sup>−</sup> and construct the asymptotics of the solution to Equation (2), we obtain the periodic solution obtained earlier in this section, but this solution will be shifted.

From the results of Sections 2–5, we derive the following theorem.

**Theorem 1.** *Let bd* = 0*. Then, Equation* (2) *with sufficiently large λ* > 0 *has a cycle with initial conditions from S*<sup>+</sup> *or S*<sup>−</sup> *if and only if b* > 0 *and d* < 0*. This sign-changing cycle u*∗(*t*) *has asymptotics*

$$\begin{array}{ll} \mu\_{\*}(t) = p\_{R}e^{-(t-nt\_{\ast})} + \lambda d(1 - e^{-(t-nt\_{\ast})}), & t \in [nt\_{\ast}nt\_{\ast} + T], \\ \mu\_{\*}(t) = \lambda d(1 - e^{-T})e^{-\left(t - (T+nt\_{\ast})\right)} + \lambda b(1 - e^{-\left(t - (T+nt\_{\ast})\right)}) + o(\lambda), \\ & t \in [nt\_{\ast} + T, nt\_{\ast} + t\_{1} + T], \\ \mu\_{\*}(t) = (\lambda b(1 - e^{-T}) - \lambda d)e^{-\left(t - (t\_{1} + T + nt\_{\ast})\right)} + \lambda d + o(\lambda), \\ & t \in [nt\_{\ast} + t\_{1} + T, (n+1)t\_{\ast}]. \end{array} \tag{23}$$

*at λ* → +∞ *(where n* = 0, 1, 2, ... *represents the number of periods of the cycle), and the period of this cycle t*<sup>∗</sup> *is given in* (22)*.*

Note that Formula (23) was obtained from Formulas (5), (17), and (21) using a shift in the time variable *t* by *n* periods *t*<sup>∗</sup> of solution *u*∗(*t*).

It should also be noted that all shifts of cycle *<sup>u</sup>*∗(*<sup>t</sup>* <sup>+</sup> *<sup>C</sup>*) where *<sup>C</sup>* <sup>∈</sup> <sup>R</sup> are solutions to Equation (2), but we consider them as a single object.

A cycle of Equation (2) in the case that *b* > 0 and *d* < 0 is shown in Figure 3.

**Figure 3.** Cycle of Equation (2) in the case that *b* > 0 and *d* < 0. Values of parameters: *λ* = 104, *T* = 5.5, *pL* = −2, *pR* = 3, *b* = 1.5, and *d* = −1.
