*4.2. Construction of More Complicated Solutions*

We shall consider the construction of rapidly oscillating periodic solutions. We arbitrarily fix *τ*, satisfying condition

$$\tau \in \left( \max\left( h, 1 - h - h(1 - \alpha A)(A - 1)^{-1} \right), 1 \right) \tag{42}$$

and consider the set

$$\begin{aligned} S(\tau) &= \{\varrho(s) \in \mathbb{C}\_{[-1,0]} : 0 \le \varrho(s) \le A; \\ \varrho(-\tau) &= \varrho(0) = 1, \\ \varrho(s) &\le \max\left(\exp\left(-\lambda\delta(s+\tau)\right), \exp(\lambda\delta s)\right), \quad s \in [-\tau, 0]; \\ \varrho(s) &\ge \min\left(A\left[1 + (A-1)\exp\left(-\lambda\delta(s+1)\right)\right]^{-1}, A\left[1 + (A-1)\exp(\lambda\delta(s+\tau))\right]^{-1}\right), \\ &s \in [-1, -\tau] \}. \end{aligned}$$

We shall give the final asymptotic formulas for *u*(*t*, *ϕ*) (*ϕ* ∈ *S*(*τ*)). In them we denote by *g*(*x*) the function

$$\lg(\mathbf{x}) = \begin{cases} \mathbf{x}, & \text{if } \mathbf{x} > 0; \\ 0, & \text{if } \mathbf{x} \le 0. \end{cases}$$

We shall assume that the inequalities

$$1 - \alpha A > 0, \quad 1 - (1 - \alpha)A < 0,\tag{43}$$

and

$$h < 1 - h \tag{44}$$

are fulfilled. Let *t* ∈ (0, *h*]. Then *u*(*t*, *ϕ*) = *A* + *o*(1), and

$$\mu(h,\varphi) = A \left[ 1 + (A - 1) \exp \left( -\lambda A \left[ h - \alpha A \min(h, 1 - \tau) + o(1) \right] \right) \right]^{-1}.$$

It follows from condition (42), that for *t* ∈ [*h*, max(*h*, 1 − *τ*)] we also have *u*(*t*, *ϕ*) = *A* + *o*(1), and

$$\begin{aligned} u(\max(h, 1-\tau), q) &= A \left[ 1 + (A - 1) \exp \left( -\lambda A \left[ \max(h, 1 - \tau) - \kappa A \min(h, 1 - \tau) \right] \right) \right] \\ &- A g (1 - \tau - h) + o(1) \left[ \right]^{-1}. \end{aligned}$$

We set

$$t\_1^0 = \frac{\alpha A (1 - \tau + h) - Ah}{1 - (1 - \alpha)A}.$$

From (42) and (43) we get that *t* 0 <sup>1</sup> > max(*h*, 1 − *τ*). Then *t*1(*ϕ*) = *t* 0 <sup>1</sup> + *o*(1), and for every *t* ∈ (0, *t*1(*ϕ*)) we have the relation *u*(*t*, *ϕ*) = *A* + *o*(1).

Suppose that the following inequality is fulfilled:

$$t\_1^0 < 1 - h.\tag{45}$$

Then on the interval (*t*1(*ϕ*), *t*1(*ϕ*) + *h*] we have the equality *u*(*t*, *ϕ*) = *o*(1), and

$$u\left(t\_1(\varphi) + h, \varphi\right) = A \left[1 + (A - 1) \exp\left(-\lambda A \left[(1 - (1 - a)A)h + o(1)\right]\right)\right]^{-1}.$$

If condition

$$(1 - (1 - \mathfrak{a})A)h + 1 - t\_1^0 - h < 0\tag{46}$$

is true, then on the interval (*t*1(*ϕ*) + *h*, 1] equality *u*(*t*, *ϕ*) = *o*(1) holds, and

$$u(1, \boldsymbol{\varrho}) = A \left[ 1 + (A - 1) \exp \left( -\lambda A \left[ (1 - (1 - \boldsymbol{a})A) h + (1 - t\_1^0 - h) + o(1) \right] \right) \right]^{-1}.$$

We set

$$t\_2^0 = 1 + \left( (1 - \alpha) A h + t\_1^0 - 1 \right) [1 - \alpha A]^{-1}.$$

*t* 0

Suppose

$$t\_2^0 < t\_1^0 + 1,\tag{47}$$

then *t*2(*ϕ*) = *t* 0 <sup>2</sup> + *o*(1), and for every *t* ∈ [1, *t*2(*ϕ*)) the equality *u*(*t*, *ϕ*) = *o*(1) is fulfilled. Finally, we introduce the quantity *τ*¯ = *t*2(*ϕ*) − *t*1(*ϕ*). Then, to order *o*(1) (for *λ* → ∞), we obtain the equality

$$
\mathfrak{x} = f(\mathfrak{x}),
\tag{48}
$$

where *f*(*τ*) = *t* 0 <sup>2</sup> − *t* 0 1.

After *τ*¯ has been determined, the situation is repeated, i.e., while for current value of *τ*¯ inequalities (42), (45)–(47) are fulfilled, using (48) we calculate *τ*¯¯ and so on.

The one-dimensional mapping (48) plays a central role in the study of the solutions from *S*(*τ*). When the conditions (42)–(47) are fulfilled its attractors determine the structure of the periodic solutions from *S*(*τ*). For example, to the equilibrium state *τ*0:

$$
\pi\_0 = f(\pi\_0) \tag{49}
$$

there corresponds, under the condition | *f* (*τ*0)| < 1, a stable one-step periodic solution *u*0(*t*, *λ*) with the initial conditions in *S*(*τ*0).

**Lemma 7.** *Let the inequality*

$$1 - A + A^2 \mathfrak{a} < 0 \tag{50}$$

*hold. Then there exist values of h such that the map* (48) *has a stable fixed point τ* = *τ*∗*, where*

$$\tau\_\* = \frac{A(A-1)(a + (a-1)h)}{1 - A + A^2 a},\tag{51}$$

*such that inequalities* (42)*–*(47) *are fulfilled.*

**Proof.** First, let us show, that if inequality (50) holds, then inequalities (43) are true.

It is easy to see, that inequalities (43) are equivalent to inequality *α* < min(1/*A*, 1 − 1/*A*). Inequality (50) is equivalent to *<sup>α</sup>* < 1/*<sup>A</sup>* − 1/*A*2. This value is less than min(1/*A*, 1 − 1/*A*) (parameter *A* > 1), therefore, if (50) is true, then conditions (43) hold.

Second, let us substitute the values of *t* 0 <sup>1</sup> and *t* 0 <sup>2</sup> into the map (48):

$$\text{f1} = \frac{A^2 a^2}{(-1 + aA)(1 - (1 - a)A)} \text{r} + \frac{A(A - 1)(h(1 - a) - a)}{(-1 + aA)(1 - (1 - a)A)}.\tag{52}$$

This map has a fixed point *τ*<sup>∗</sup> satisfying formula (51). We should check whether this value *τ*<sup>∗</sup> satisfies conditions (42), (45)–(47), because map (52) corresponds to the initial equation only when all these inequalities hold, and we should study stability of the fixed point (51).

Let us begin with the study of stability. The fixed point *τ*<sup>∗</sup> is stable if and only if

$$-1 < \frac{A^2 \alpha^2}{(-1 + \alpha A)(1 - (1 - \alpha)A)} < 1.$$

From the condition (43) we get that this value is greater than zero, and this value is less than 1 if and only if inequality (50) holds. Therefore, this fixed point is stable.

Now let us prove that if *τ* = *τ*<sup>∗</sup> satisfies (51), then conditions (42), (45)–(47) hold. Below, we always assume that inequalities (43) and (50) hold.

Substituting *τ*<sup>∗</sup> in inequality *τ* < 1 from (42) we obtain, that this condition is equivalent to

$$h < \frac{(1 - \alpha)A - 1}{A(A - 1)(1 - \alpha)} =: \gamma\_{1\prime} \tag{53}$$

inequality *h* < *τ* from (42) is equivalent to

$$h > \frac{\alpha A (A - 1)}{(A - 1)^2 + \alpha A} =: \beta\_{11}$$

and condition *<sup>τ</sup>* > <sup>1</sup> − *<sup>h</sup>* − *<sup>h</sup>*(<sup>1</sup> − *<sup>α</sup>A*)(*<sup>A</sup>* − <sup>1</sup>)−<sup>1</sup> from (42) is equivalent to

$$h > \frac{A - 1}{A^2 (1 - \alpha)} =: \beta\_2. \tag{54}$$

Condition (45) at *τ* = *τ*<sup>∗</sup> can be rewritten in the form

$$h < \frac{(A - 1)(1 - \varkappa A)}{(A(1 - \varkappa A) - (1 - (1 - \varkappa)A))} =: \gamma\_2.$$

Condition (46) holds, if *τ* = *τ*<sup>∗</sup> and (54) is true, and inequality (47) holds, if *τ*¯ = *τ*<sup>∗</sup> < 1 (the last inequality is equivalent to (53)).

Thus, at *τ* = *τ*<sup>∗</sup> and under condition (50) the system of inequalities (42), (43), (45)–(47) is equivalent to

$$
\max(\beta\_{1\prime}\beta\_2) < h < \min(\gamma\_{1\prime}\gamma\_2).
$$

Parameter *h* must satisfy condition *h* < 1/2 (we obtain this inequality from condition (44)).

Let us prove that under condition (50)interval(max(*β*1, *β*2), min(1/2, *γ*1, *γ*2))is not empty.

$$
\beta\_1 - \beta\_2 = \frac{(A-1)(1 - (1-\alpha)A)(1 - A + A^2 \alpha)}{A^2(-1+\alpha)((A-1)^2 + \alpha A)} < 0,
$$

therefore max(*β*1, *β*2) = *β*2. Let us estimate the values of *γ*<sup>1</sup> − *β*2, *γ*<sup>2</sup> − *β*2, 1/2 − *β*2.

$$\begin{aligned} \gamma\_1 - \beta\_2 &= \frac{1 - A + A^2 \alpha}{(A - 1)A^2 (-1 + \alpha)} > 0, \\ \gamma\_2 - \beta\_2 &= \frac{(A - 1)(1 - A + A\alpha)(1 - A + A^2 \alpha)}{A^2 (1 - \alpha) \left(A(1 - \alpha A) - (1 - (1 - \alpha)A)\right)} > 0. \end{aligned}$$

Inequality

$$1/2 - \beta\_2 = \frac{2 - 2A + A^2 - A^2 a}{2A^2 (1 - a)} > 0$$

is equivalent to inequality 2 − <sup>2</sup>*<sup>A</sup>* + *<sup>A</sup>*<sup>2</sup> − *<sup>A</sup>*2*<sup>α</sup>* > 0. Function *<sup>q</sup>*(*α*) = <sup>2</sup> − <sup>2</sup>*<sup>A</sup>* + *<sup>A</sup>*<sup>2</sup> − *<sup>A</sup>*2*<sup>α</sup>* is decreasing on the interval *<sup>α</sup>* ∈ (0, 1/*<sup>A</sup>* − 1/*A*2). Therefore, *<sup>q</sup>*(1/*<sup>A</sup>* − 1/*A*2) ≤ *<sup>q</sup>*(*α*) for all *<sup>α</sup>* ∈ (0, 1/*<sup>A</sup>* − 1/*A*2).

$$q(1/A - 1/A^2) = 2 - 2A + A^2 - A^2(1/A - 1/A^2) = A^2 - 3A + 3 > 0,$$

thus inequality 1/2 − *<sup>β</sup>*<sup>2</sup> > 0 holds for all *<sup>α</sup>* ∈ (0, 1/*<sup>A</sup>* − 1/*A*2). The Lemma is proven.

From the constructions and the reasoning given above we obtain the following statement.

**Theorem 5.** *Let the inequality* (50) *hold. Then for all sufficiently large λ there exist h and τ* = *τ*∗*, satisfying* (51)*, such that equation* (37) *has an asymptotically orbitally stable rapidly oscillating periodic (with period T*(*λ*)*) solution u*∗(*t*, *λ*)*, for which u*∗(0, *λ*) = *u*∗(*t*1(*λ*), *λ*) = *u*∗(*t*2(*λ*), *λ*) = 1*, where tj are successive positive roots of the equation u*∗(*t*, *λ*) = 1*, and*

$$\begin{aligned} t\_1(\lambda) &= \frac{A(\mathfrak{a} + (\mathfrak{a} - 1)h)}{1 - A + A^2 \mathfrak{a}} + o(1), \\ T(\lambda) &= t\_2(\lambda) = \frac{A^2(\mathfrak{a} + (\mathfrak{a} - 1)h)}{1 - A + A^2 \mathfrak{a}} + o(1), \\ t\_2(\lambda) - t\_1(\lambda) &= \mathfrak{r}\_\* + o(1) < 1. \end{aligned}$$

*For every t from intervals* (0, *t*1(*λ*)) *and* (*t*1(*λ*), *t*2(*λ*))*, respectively, the following equalities are fulfilled:*

$$u\_\*(t, \lambda) = \begin{cases} A + o(1), & t \in (0, t\_1(\lambda)), \\ o(1), & t \in (t\_1(\lambda), t\_2(\lambda)). \end{cases}$$

In a similar way, one can determine the existence conditions and find the asymptotics of multi-step periodic solutions on the segment [−1, 0] and obtain an *n*-dimensional (in terms of the number of steps on the segment [−1, 0]) mapping for their description.
