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3. Investigation of Two-Frequency Systems

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In this section, we consider the case where system (2.1) is a two-frequency

system, i.e., ϕ = (ϕ1, ϕ2) and ω(τ) = (ω1(τ ), ω2(τ )), and study the problem

of the justification of the averaging method on an asymptotically large time interval

[0, T(ε)], where T(ε) → ∞ as ε → 0, and on the infinite time interval

[0,∞) = R+ under assumptions for ω(τ ) weaker than those in Section 2.

Assume that ω(τ ) ∈ C1

[0,∞) and

ω2(τ ) ≥ d1,

___

dτ

_ω1(τ )

ω2(τ )

____ ≥ d1 ∀τ ∈ R+, (3.1)

where d1 is a positive constant. For τ ∈ [0, L], condition (3.1) is the Arnol’d

condition [Arn2], by using which Arnol’d obtained an estimate for the error of the

averaging method on a finite time interval.

We also require that the function ω2(τ ) satisfy at least one of the following

conditions:

(i)

___

−2

2 (τ ) dω2(τ )

dτ

___

≤ d2 = const ∀τ ∈ R+;

(ii) ω2(τ ) is nondecreasing or nonincreasing on R+.

Denote by hd(τ ), d = const > 0, the following even function continuously

differentiable on ∀τ ∈ R :

hd(τ) =

⎧⎪⎪⎨

⎪⎪⎩

1, τ∈ [0, d],

d−4τ 2(2d − τ )2, τ ∈ (d, 2d),

0, τ∈ [2d,∞).

40 Averaging Method in Systems with Variable Frequencies Chapter 1

It is easy to verify that, for all τ ∈ R, the function hd(τ ) satisfies the inequalities

0 ≤ hd(τ ) ≤ 1,

___

dτ

hd(τ )

___

≤ 16

fd(τ ),

where fd(τ) = 1 for d < |τ | < 2d and fd(τ) = 0 for |τ| ≤ d and |τ| ≥ 2d.

The statement below gives an estimate of the time for which the two-frequency

system (2.1) passes through the resonance zone.

Lemma 3.1. Let conditions (3.1) be satisfied and let k = (k1, k2) _= 0 be an

arbitrary vector with integer-valued coordinates. Then, for all τ ∈ R+ except,

possibly, a time interval whose length does not exceed 2μ, μ ≤ d

−1

1 , the function

(k, ω(τ )) = k1ω1(τ) + k2ω2(τ ) satisfies the inequality |k, ω(τ )| ≥ d21

μ.

Proof. If (k, ω(τk)) = 0, then it follows from (3.1) that k1 _= 0 and the

function ω(τ, k) ≡ k2

+ ω1(τ )

ω2(τ )

is monotone. Hence,

|(k, ω(τ ))| = |k1ω2(τ )||ω(τk, k) − ω(τ, k)| ≥ d21μ

for |τ − τk| ≥ μ. Now let (k, ω(τ )) _= 0 ∀τ ∈ R+. If k1 = 0, then

|(k, ω(τ ))| = |k2ω2(τ )| ≥ d1 ≥ d21

μ. If k1 _= 0, then we obtain the estimate

|(k, ω(τ ))| ≥ d21

μ for all τ ∈ [μ,∞). Lemma 3.1 is proved.

Lemma 3.2. Suppose that inequalities (3.1) and at least one of conditions (i)

and (ii) are satisfied. Then, for every γ > 0, one can find a constant d3 =

d3(γ) > 0 such that, for any T ∈ R+, the following estimate is true:

A(T) ≡

−γ(T−τ)

___

dτ

1 − hμ(τ − τk)

(k, ω(τ ))

___

dτ ≤ d3

, k_= 0, (3.2)

where τk ∈ R+ is a point at which (k, ω(τ )) turns into zero; if (k, ω(τ )) _= 0

∀τ ∈ R+, then τk = 0.

Section 3 Investigation of Two-Frequency Systems 41

Proof. Let k1 = 0. Then |k2| ≥ 1, |(k, ω(τ ))| ≥ d1, and

A(T) ≤

−γ(T−τ)

___

dτ

hμ(τ )

___

ω2(τ )) dτ

−γ(T−τ)

___

dτ

_ 1

ω2(τ ))

____(1 − hμ(τ )) dτ

≤ 16

d1μ

fμ(τ )dτ +

−γ(T−τ)

___

dω2(τ )

dτ

___

ω2

2(τ ) dτ.

If condition (i) is satisfied, then

A(T) ≤ 16

d1μ

+ d2

−γ(T−τ)dτ ≤ 16

+ d2

and if condition (ii) is satisfied, then A(T) can be estimated as follows:

A(T) ≤ 16

___

dω2

dτ

ω2

2(τ )

___

dτ =

___

dτ

_ 1

ω2(τ )

dτ

___

≤ 18

Thus, in the case k1 = 0, we get A(T) <

d2.

Now let k1 _= 0. Then

A(T) ≤

−γ(T−τ) 16

fμ(τ − τk)

|(k, ω(τ ))| dτ

|k1|

−γ(T−τ)(1 − hμ(τ − τk))

___

dτ

_ 1

ω2(τ )

____

|ω(τ, k)| dτ

−γ(T−τ)(1 − hμ(τ − τk))

|k1|ω2(τ )

___

dτ

_ 1

ω(τ, k)

____ dτ. (3.3)

According to the definition of the function fμ(τ ) and Lemma 3.1, the first integral

on the right-hand side of (3.3) can be estimated from above by the constant

42 Averaging Method in Systems with Variable Frequencies Chapter 1

32d

−2

1 μ−1. In view of (3.1), the function ω(τ, k) is monotone. Therefore, the

last integral on the right-hand side of (3.3) is estimated as follows:

___

dτ

_ 1

ω(τ, k)

(1 − hμ(τ − τk)) dτ

___

≤ 1

____

τk_−μ

dτ

_ 1

ω_____________(τ, k)

dτ

___

τk+μ

dτ

_ 1

ω(τ, k)

dτ

___

≤ 4

d21

where τk − μ ≥ 0 and τk + μ ≤ T. If τk < μ or τk > T − μ, then the last

inequality remains the same.

Consider the second integral on the right-hand side of (3.3). In case (i), it can

be estimated by the value

d1μ

−γ(T−τ)dτ ≤ d2

d1γμ

;

in case (ii), it can be estimated by the value

d1μ

___

dτ

_ 1

ω2(τ )

____ dτ =

d1μ

___

dτ

_ 1

ω2(τ )

dτ

___

≤ 2

d21

Thus, it follows from (3.3) for k1 _= 0 that

A(T) ≤ 36

d21μ

+ d2

d1γμ

Combining the estimates for A(T) in the cases k1 = 0 and k1 _= 0, we obtain

inequality (3.2) with the constant d2 = 36d

−2

1 + d2(d1γ)−1 for d1μ ≤ 1. Lemma

3.2 is proved.

Consider a two-frequency system of the form

dτ

= a (x, ϕ, τ, ε),

dϕ

dτ

= ω(τ )

+ b (x, ϕ, τ, ε), (3.4)

Section 3 Investigation of Two-Frequency Systems 43

where the functions a, b, and ω are defined for (x, ϕ, τ, ε) ∈ D×R2 × R+ ×

[0, ε0], 2π-periodic in ϕν, ν = 1, 2, continuously differentiable with respect to

x, ϕ, and τ for every fixed ε, and such that

_b(x, ϕ, τ, ε)_ +

___

∂

∂x

b(x, ϕ, τ, ε)

___

∂

∂ϕ

b(x, ϕ, τ, ε)

___

≤ d4

∀(x, ϕ, τ, ε) ∈ G,

(1 + _k_) sup

_ak_ + sup

___

∂ak

∂x

___

+ sup

___

∂ak

∂τ

___

≤ d4. (3.5)

Here, G = D × R+ × [0, ε0], D is a bounded domain from Rn, and ak =

ak(x, τ, ε) are the Fourier coefficients of the function a(x, ϕ, τ, ε).

As in the previous section, by (x(τ, y, ψ, ε); ϕ(τ, y, ψ, ε)), x(0, y, ψ, ε) = y,

ϕ(0, y, ψ, ε) = ψ, we denote a solution of system (3.4) and by x = x(τ ) a

solution of the averaged system of the first approximation (2.27), which is defined

for all τ ∈ R+.

Theorem 3.1. Suppose that the following conditions are satisfied:

(I) inequalities (3.1) and (3.5) and at least one of restrictions (i) and (ii) are

satisfied;

(II) conditions (b) and (c) of Theorem 2.4 and relation (2.26) for δ ≥ 1

are

satisfied.

Then, for all τ ∈ R+, ψ ∈ R2, and ε ∈ (0, ε0] (ε0 is sufficiently small),

the following inequality is true:

_x(τ, .x(0), ψ, ε) − .x(τ )_ ≤ d5

√

ε, (3.6)

where the constant d5 is independent of τ, ψ, and ε.

Prior to the proof of Theorem 3.1, we indicate its difference from Theorem

2.4. In Theorem 2.4, the condition of the uniform continuity of the functions

ω(j)

ν (τ ), ν = 1,m, j = 0, p − 1, on the semiaxis R+ is imposed. This

condition substantially restricts the growth of the functions ων(τ ); in particular,

ων(τ) = τ l for l > 1 is not uniformly continuous on R+. Moreover, in Theorem

2.4, the boundedness of the norm of the matrix (WT

p (τ )Wp(τ ))−1WT

p (τ ) is

44 Averaging Method in Systems with Variable Frequencies Chapter 1

an essential assumption. In Theorem 3.1, conditions (3.1), (i), and (ii) do not require

such strong restrictions on the components ω1(τ ) and ω2(τ ) of the vector

ω(τ ). For example, if ω1(τ) = τ 2 +τ and ω2(τ) = 1, then all conditions (3.1),

(i), and (ii) are satisfied for τ ∈ R+. It is easy to see that ω1(τ) = τ 2 + τ is

not a uniformly continuous function on R+. Moreover, for these frequencies, we

have

_(WT

2 (τ )W2(τ ))−1WT

2 (τ )_ = _W

−1

2 (τ )_ = τ 2 + 3τ + 2

2τ + 1 > 1 +

2τ,

i.e., the indicated norm is not bounded on the semiaxis. It is easy to verify that,

for p > 2, the value _(WT

2 (τ )W2(τ ))−1WT

2 (τ )_ is also not bounded for all

τ ∈ R+. Therefore, for this collection of frequencies ω1(τ ) and ω2(τ ), we

cannot use Theorem 2.4.

Proof of Theorem 3.1. Assume that, for t ∈ [0, T), T = T(ψ, ε), the curve

x = x(τ, .x(0), ψ, ε) does not leave a ρ1-neighborhood of the curve x = x(τ ).

We fix the constant ρ1 ∈ (0, ρ) below. For τ ∈ [0, T), ψ ∈ R2, and ε ∈ (0, ε0],

we consider the function

y(τ,ψ, ε) = x(τ, x(0), ψ, ε) − x (τ )

− εU(x(τ, x(0), ψ, ε), ϕ(τ, x(0), ψ, ε), τ, ε), (3.7)

where

U(x, ϕ, τ, ε) =

k_=0

ak(x, τ, ε)

1 − hμ(τ − τk)

i(k, ω(τ ))

exp{i(k,ϕ)}

and τk ∈ R+ is a point at which (k, ω(τ )) turns into zero; if (k, ω(τ )) _= 0

∀τ ∈ R+, then τk = 0.

By direct differentiation, one can easily verify that y = y(τ,ψ, ε) satisfies

the equation

dτ

= H(τ )y + F(y, τ) + δ_____________(x, ϕ, τ, ε)

− ε

∂

∂τ

U(x, ϕ, τ, ε) + Y (x, y, ϕ, τ, ε), (3.8)

Section 3 Investigation of Two-Frequency Systems 45

where x = x(τ, x(0), ψ, ε), ϕ = ϕ(τ, x(0), ψ, ε),

δ(x, ϕ, τ, ε) =

k_=0

ak(x, τ, ε)hμ(τ − τk) exp{i(k,ϕ)},

F(y, τ) = a(x(τ) + y, τ, 0) − a(x(τ ), τ, 0) − H(τ )y,

H(τ) = ∂

∂x

a(x(τ ), τ, 0),

Y (x, y, ϕ, τ, ε) = a(x, τ, ε) − a(x(τ) + y, τ, 0) − ε

∂

∂ϕ

U(x, ϕ, τ, ε)b(x, ϕ, τ, ε)

− ε

∂

∂x

U(x, ϕ, τ, ε)a(x, ϕ, τ, ε).

Using the definition of the function hμ(τ ), Lemma 3.1, and inequalities

(2.26) and (3.5), we obtain

_F(y, τ)_ ≤ n2σ10_y_2, _Y (x, y, ϕ, τ, ε)_ ≤ σ10εδ + ε

d6,

d6 = n2(1 + 2d4) d24

−2

1 .

Equation (3.8) yields the following representation of the function y = y(τ,ψ, ε)

for all τ ∈ [0, T):

y = Q(τ, 0)y(0, ψ, ε) +

_τ

Q(τ, t)

F(y, t) + δ(x, ϕ, t, ε)

− ε

∂

∂t

U(x, ϕ, t, ε) + Y (x, y, ϕ, t, ε)

dt.

Using this representation, we get

sup

[0,T )

_y(τ,ψ, ε)_

≤ K_y(0, ψ, ε)_ +

σ10εδ + d6

+ n2σ10 sup

[0,T )

_y(ψ, t, ε)_2

_ K

46 Averaging Method in Systems with Variable Frequencies Chapter 1

+ K

k_=0

sup

_ak(x, τ, ε)_ sup

___

∂

∂τ

ak(x, τ, ε)

___

sup

[0,T )

_τ

hμ(t − τk)

+ ε

1 − hμ(t − τk)

|(k, ω(t))| + ε

___

∂

∂t

1 − hμ(t − τk)

(k, ω(t))

___

−γ(τ−t)dt

. (3.9)

Since

_y(τ,ψ, ε)_ ≤ _x(τ, x(0), ψ, ε) − x(τ )_ + ε sup

_U(x, ϕ, τ, ε)_

< ρ1 + ε

d4d

−2

1 ,

_y(0, ψ, ε)_ ≤ ε

d4d

−2

1 ,

for

ρ1 ≤

Kn2σ10

_−1

and

≤ γd21

4Kn2σ10d4

relation (3.9) yields

sup

[0,T )

_y(τ,ψ, ε)_ ≤ 2

d21

d4 +

d21

_ ε

2σ10

Kεδ

+ 2K

k_=0

sup

_ak_ + sup

___

∂ak

∂τ

___

sup

[0,T )

_τ

hμ(t − τk)

+ ε

1 − hμ(t − τk)

|(k, ω(t))| + ε

___

1 − hμ(t − τk)

(k, ω(t))

___

−γ(τ−t)dt

To estimate the integral on the right-hand side of the last inequality, we use

Lemmas 3.1 and 3.2. Then, taking (3.5) into account, we get

sup

[0,T )

_y(τ,ψ, ε)_

≤ 2

−2

1 d4[γ +1+d21

−1

4 d6 + γd21

d3] ε

+ 8Kd4μ +

Kσ10εδ,

which (for δ ≥ 1

2 and μ =

√

ε) yields

_x(τ, x(0), ψ, ε) − x(τ )_ ≤

_ 2

−2

1 d4K(γ +1+γd21

+ d21

−1

4 d6) +

Kσ10 + 8Kd4 + d

−2

1 d4

_√

≡ d5

√

ε (3.10)

Section 3 Investigation of Two-Frequency Systems 47

for any τ ∈ [0, T), ψ ∈ R2, and ε ∈ (0, ε0]. We now choose ε0 > 0 so small

that

ε1/2

≤ min

_ 1

2σ5

ρ1; γd21

4Kn2d4σ10

, ρ1 = min

_ρ

;

Kn2σ10

_−1_

According to estimate (3.10), the curve x = x(τ, x(0), ψ, ε) does not leave the

2ρ1 -neighborhood of the curve x = x(τ ) for all τ ∈ [0, T). This implies that

T = ∞ and inequality (3.10) holds for any τ ∈ R+. Theorem 3.1 is proved.

Now assume that the function a(x, ϕ, τ, ε) averaged with respect to ϕ over

the cube of periods is identically equal to zero, i.e.,

a (x, τ, ε) ≡ 0 ∀(x, τ, ε) ∈ D×R+ × [0, ε0].

In this case, the solutions x(τ ) ≡ x0 = const ∀τ ∈ R+ of the averaged

equations for slow variables are stationary, and, therefore, condition (II) of Theorem

3.1 is not satisfied. This is the case, in particular, for Hamiltonian systems

[Arn4]. Nekhoroshev [Nek1, Nek2] established that, for time exp{ε_____________−α},

α = const > 0, the slow variable x of the solution (x; ϕ) of a Hamiltonian system

deviates from its initial value by a value not greater than cεβ, c = const > 0,

β = const > 0. In what follows, we obtain an analogous result for a twofrequency

system under certain additional assumptions concerning the frequency

vector ω(τ ). The following statement can be proved by analogy with Lemma 3.2:

Lemma 3.3. Suppose that conditions (3.1) are satisfied and the following inequality

is true:

_ 1

ω2(τ )

___

dτ

_ 1

ω2(τ )

____

dτ ≤ d7 ln T + d8 ∀T ≥ 1, (3.11)

where d7 and d8 are nonnegative constants. Then, for k _= 0, 0 < μ <

min

_ 1

;

, and T > 1, the following estimate is true:

B(T) ≡

_1 − hμ(τ − τk)

|(k, ω(τ ))| +

___

dτ

1 − hμ(τ − τk)

(k, ω(τ ))

___

dτ

≤ 36

d21

d1μ

(d7 ln T + d8); (3.12)

48 Averaging Method in Systems with Variable Frequencies Chapter 1

for T ∈ [0, 1], the following estimate is true:

B(T) ≤ d9

, d9 = d

−2

37 + d

−1

1 max

[0,1]

___

dτ

_ 1

ω2(τ )

____

. (3.13)

We denote by Dρ the set of points belonging to D together with their ρ -

neighborhoods and choose ρ > 0 so small that Dρ _= ∅.

Theorem 3.2. Suppose that a(x, τ, ε) ≡ 0 ∀(x, τ, ε) ∈ G and conditions

(3.1), (3.5), and (3.11) are satisfied. Then, for all x0 ∈ Dρ, ψ ∈ R2, and

ε ∈ (0, ε0] (ε0 is sufficiently small), the following estimates are true:

(a) if d7 > 0, then

_x(τ, x0, ψ, ε) − x0_ < d10εβ ∀τ ∈ [0, exp{ε

−(1−2β)}], (3.14)

where β is an arbitrary number from the interval

;

(b) if d7 = 0, then

_x(τ, x0, ψ, ε) − x0_ < d10

√

ε ∀τ ∈ [0,∞), (3.15)

where d10 is a constant independent of ε, x0, and ψ.

Proof. We use the method proposed in the proof of Theorem 3.1. The function

y(τ,ψ, ε) defined by equality (3.7) for x(τ ) ≡ x0 admits the representation

y(τ,ψ, ε) = y(0, ψ, ε) +

_τ

δ(x, ϕ, t, ε) − ε

∂

∂t

U(x, ϕ, t, ε)

− ε

∂

∂x

U(x, ϕ, t, ε)a(x, ϕ, t, ε) − ε

∂

∂ϕ

U(x, ϕ, t, ε)b(x, ϕ, t, ε)

dt,

which yields

_y(τ,ψ, ε)_ ≤ _y(0, ψ, ε)_ +

1 + sup

_a_ + sup

_b_

k_=0

_k_ sup

_ak_ + sup

___

∂ak

∂τ

___

+ sup

___

∂ak

∂x

___

___τ

[hμ(t − τk)]dt

+ ε

_τ

_1 − hμ(t − τk)

|(k, ω(t))| +

___

1 − hμ(t − τk)

(k, ω(t))

___

. (3.16)

Section 3 Investigation of Two-Frequency Systems 49

If τ ∈ [0, 1], then inequalities (3.5), (3.13), and (3.16) for μ =

√

ε yield

_y(τ,ψ, ε)_ ≤ d11μ, d11 = d4d

−2

1 + (1 + 2d4)d4(4 + d9). (3.17)

If τ > 1, then, in view of (3.12), estimate (3.16) takes the form

_y(τ,ψ, ε)_

≤ ε

−2

1 d4 + (1 + 2d4)d4

4μ + (36d

−2

1 + d

−1

1 (d8 + d7 ln τ )) ε

. (3.18)

Let d7 = 0. Then relations (3.17) and (3.18) for μ =

√

ε yield

_y(τ,ψ, ε)_ ≤ d12

√

ε,

d12 = max{d11; d

−2

1 d4 + (1 + 2d4)d4(4 + 36d

−2

1 + d

−1

1 d8)},

_x(τ, x0, ψ, ε) − x0_ ≤ _y(τ,ψ, ε)_ + ε sup

_U_ ≤ (d

−2

1 d4 + d12)

√

for all τ ∈ [0,∞), ψ ∈ R2, ε ∈ (0, ε0], and x0 ∈ Dρ. Thus, estimate (3.15) is

proved.

Consider the case d7 > 0. We fix an arbitrary β ∈

and set εβ = μ.

Analyzing inequality (3.18), we establish that y(τ,ψ, ε) satisfies an estimate of

the form _y(τ,ψ, ε)_ ≤ cεβ on the maximum (in order with respect to ε ) time

interval [1, T(ε)] if εμ−1 ln T(ε) = μ, i.e., T(ε) = exp{ε−(1−2β)}. Hence, for

all τ ∈ [0,+∞), ψ ∈ R2, ε ∈ (0, ε0], and x0 ∈ Dρ, relations (3.17) and

(3.18) yield

_y(τ,ψ, ε)_ ≤ d13εβ, _x(τ, x0, ψ, ε) − x0_ ≤ (d13 + d

−2

1 d4)εβ,

d13 = max{d11; d

−2

1 d4 + (1 + 2d4)d4(4 + 36d

−2

1 + d

−1

1 (d7 + d8))}.

To complete the proof of the theorem, we set d10 = d13 + d

−2

1 d4 and choose ε0

so small that

d10εβ

≤ 1

2 ρ, εβ

≤ min

;

The first of these inequalities guarantees that the curve x = x(τ, x0, ψ, ε) lies

in D together with its

2ρ -neighborhood for any τ ∈ R+ if d7 = 0 and for

any τ ∈ [0, exp{ε−(1−2β)}] if d7 > 0. The second inequality follows from

Lemma 3.3. Theorem 3.2 is proved.

50 Averaging Method in Systems with Variable Frequencies Chapter 1

Remark 4. Restrictions (3.1), (i), (ii), and (3.11) imposed on the frequencies

ω1(τ ) and ω2(τ ) of system (3.4) are sufficient and do not exhaust all possibilities

of establishing the results presented in this section. For example, Theorem 3.1

remains true if, instead of condition (ii), one assumes that

dτ

ω2(τ ) does not

change its sign on finitely many intervals that cover [0,∞), and Theorem 3.2

remains true if, on the left-hand side of (3.11), the integral over the segment [0, T]

is replaced by the integral over [τ0, T], where τ0 is positive and fixed. However,

as follows from the example presented below, the restrictions indicated above are

essential.

Consider the problem

dτ

= x cos ϕ2 + sin ϕ2,

dϕ1

dτ

= τ

dϕ2

dτ

x (0) = ϕ1 (0) = ϕ2 (0) = 0,

for which all conditions of Theorem 3.2 except inequality (3.11) are satisfied.

Below, we show that estimate (3.14) is not true for τ ∼ 1

. Indeed, the x -

component of the solution of this problem is determined by the relation

x = x(τ, ε) = εeε sin(τ/ε)

_τ/ε

−ε sin τ sin τdτ.

The integrand e−ε sin τ sin τ is 2π-periodic. Therefore, we first estimate the integral

over the segment [0, 2π]. We have

_2π

−ε sin τ sin τdτ =

_π

−ε sin τ − eε sin τ ) sin τdτ

≤ −

_π

ε sin2 τdτ = −επ

Section 4 Justification of Averaging Method for Systems with ω = ω (x, τ ) 51

We set τ = 2πεE{ε−2}, where E{α} is the integer part of the number α. Then

x(2πεE{ε

−2}, ε) = ε

2πE_{ε−2}

−ε sin τ sin τdτ

= εE{ε

−2}

_2π

−ε sin τ sin τdτ ≤ −π

(1 − ε2) ≤ − π

for ε2 ≤ 1

2. Hence,

|x(τ, ε) − 0| ≥ π

for τ = 2πεE{ε

−2} ∼ 1

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3. Investigation of Two-Frequency Systems

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