On finding periodic solutions of second order difference equations in a Banach spase
With the use of the numerical-analytic method of A.M. Samoilenko and a modification of Newton’s method, we construct an approximation to the periodical solution of a difference equation in pertially ordered Banach spaces with an arbitrary given precision.
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| Cite this: | On finding periodic solutions of second order difference equations in a Banach spase / Y.V. Teplinsky, I.V. Semenyshyna // Нелінійні коливання. — 2001. — Т. 4, № 3. — С. 405-421. — Бібліогр.: 14 назв. — англ. |
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Teplinsky, Y.V. Semenyshyna, I.V. 2021-01-27T11:30:35Z 2021-01-27T11:30:35Z 2001 On finding periodic solutions of second order difference equations in a Banach spase / Y.V. Teplinsky, I.V. Semenyshyna // Нелінійні коливання. — 2001. — Т. 4, № 3. — С. 405-421. — Бібліогр.: 14 назв. — англ. 1562-3076 https://nasplib.isofts.kiev.ua/handle/123456789/174698 With the use of the numerical-analytic method of A.M. Samoilenko and a modification of Newton’s method, we construct an approximation to the periodical solution of a difference equation in pertially ordered Banach spaces with an arbitrary given precision. en Інститут математики НАН України Нелінійні коливання On finding periodic solutions of second order difference equations in a Banach spase Про відшукання періодичних розв'язків різницевих рівнянь другого порядку в банаховому просторі О нахождении периодических решений разностных уравнений второго порядка в банаховом пространстве Article published earlier |
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On finding periodic solutions of second order difference equations in a Banach spase |
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On finding periodic solutions of second order difference equations in a Banach spase Teplinsky, Y.V. Semenyshyna, I.V. |
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On finding periodic solutions of second order difference equations in a Banach spase |
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On finding periodic solutions of second order difference equations in a Banach spase |
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On finding periodic solutions of second order difference equations in a Banach spase |
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On finding periodic solutions of second order difference equations in a Banach spase |
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on finding periodic solutions of second order difference equations in a banach spase |
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Teplinsky, Y.V. Semenyshyna, I.V. |
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Teplinsky, Y.V. Semenyshyna, I.V. |
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Про відшукання періодичних розв'язків різницевих рівнянь другого порядку в банаховому просторі О нахождении периодических решений разностных уравнений второго порядка в банаховом пространстве |
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With the use of the numerical-analytic method of A.M. Samoilenko and a modification of
Newton’s method, we construct an approximation to the periodical solution of a difference
equation in pertially ordered Banach spaces with an arbitrary given precision.
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1562-3076 |
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https://nasplib.isofts.kiev.ua/handle/123456789/174698 |
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On finding periodic solutions of second order difference equations in a Banach spase / Y.V. Teplinsky, I.V. Semenyshyna // Нелінійні коливання. — 2001. — Т. 4, № 3. — С. 405-421. — Бібліогр.: 14 назв. — англ. |
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AT teplinskyyv onfindingperiodicsolutionsofsecondorderdifferenceequationsinabanachspase AT semenyshynaiv onfindingperiodicsolutionsofsecondorderdifferenceequationsinabanachspase AT teplinskyyv provídšukannâperíodičnihrozvâzkívríznicevihrívnânʹdrugogoporâdkuvbanahovomuprostorí AT semenyshynaiv provídšukannâperíodičnihrozvâzkívríznicevihrívnânʹdrugogoporâdkuvbanahovomuprostorí AT teplinskyyv onahoždeniiperiodičeskihrešeniiraznostnyhuravneniivtorogoporâdkavbanahovomprostranstve AT semenyshynaiv onahoždeniiperiodičeskihrešeniiraznostnyhuravneniivtorogoporâdkavbanahovomprostranstve |
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2025-11-25T10:42:49Z |
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2025-11-25T10:42:49Z |
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Nonlinear Oscillations, Vol. 4, No. 3, 2001
ON FINDING PERIODIC SOLUTIONS OF SECOND ORDER DIFFERENCE
EQUATIONS IN A BANACH SPACE
Yu. V. Teplins’ky, I. V. Semenishina
Kam’yanets’-Podil’s’ky State Pedagogic University, Ukraine
With the use of the numerical-analytic method of A.M. Samoilenko and a modification of
Newton’s method, we construct an approximation to the periodical solution of a difference
equation in pertially ordered Banach spaces with an arbitrary given precision.
AMS Subject Classification: 39A11
There where a large number of works published for the last several years that deal with the
research of problems of the reduction, existence of oscillating solutions, and construction of
invariant manifolds of equations of different kinds in infinite dimension Banach spaces. Let’s
mention here some of them, namely [1 – 11]. In this article we use the numerical analytic method
of A.M.Samoilenko and the modified method of Newton to construct an approximation of the
periodical solution of the equation in partially ordered Banach spaces, which enables us to
solve the problem with a given precision. It should be noticed that a similar problem for scalar
equation of the second order is partly solved in [12] and, for the linear equation in Banach
spaces, in [11].
Consider the equation
∆2xn = fn(xn, xn+1), n ∈ Z, (1)
where ∆xn = xn+1−xn, the mapping fn(x, y) : W×W → W,xn ∈ W,W is a real Banach space
with the norm ‖ · ‖, Z the set of the integers. Let’s set Z+
0 = {0, 1, 2, . . .}, Z− = Z\Z+, Z+ =
Z+
0 \{0}.
It is easy to see that for any {x∗0, x∗1} ⊂ W, equation (1) has, on the set Z+
0 , the only solution
xn = xn(x∗0, x
∗
1) such that x0(x∗0, x
∗
1) = x∗0, x1(x∗0, x
∗
1) = x∗1. For n ∈ {2, 3, 4, . . .}, it is given by
the relation
xn = −(n− 1)x∗0 + nx∗1 +
n−1∑
i=1
(n− i)fi−1(xi−1, xi). (2)
Further we shall consider the function fn(x, y), N -periodic in n on the set Z.
Using the last equality, it’s easy to see that the solution xn(x∗0, x
∗
1) of equations (1) is N -
periodic on Z+
0 if and only if its initial conditions, x∗0 and x∗1, satisfy the system of equations
N−1∑
k=0
fk(xk(x
∗
0, x
∗
1), xk+1(x∗0, x
∗
1)) = 0,
(3)
x∗1 = x∗0 −
1
N
N−1∑
i=1
(N − i)fi−1(xi−1, xi).
c© Yu. V. Teplins’ky, I. V. Semenishina, 2001 405
406 Yu.V. TEPLINS’KY, I.V. SEMENISHINA
If such a solution exists, it is easy to continue it to the set Z with the help of periodicity.
Define a linear operator L that acts on the sequence of points {a0, a1, a2, . . . , an, . . .} ⊂ W
by
La0 = 0 ∈ W, Lan =
n−1∑
i=0
(ai − aν), aν =
1
N
N−1∑
ν=0
aν , n ∈ Z+.
Then
L2a0 = 0 ∈ W, L2a1 = − 1
N
N−1∑
i=1
i−1∑
s=0
(as − aν),
L2an =
n−1∑
i=1
i−1∑
k=0
(ak − aν)− n
N
N−1∑
ν=1
ν−1∑
k=0
(ak − aν), n ∈ Z+\{1}.
Set L2fn(x, y) = qn(x, y) and write the equations
∆2xn = fn(xn, xn+1), n ∈ Z+
0 , (4)
xn = x0 + gn(xn, xn+1), n ∈ Z+
0 , (5)
∆2xn = fn(xn, xn+1)− µ, n ∈ Z+
0 , (6)
where µ ∈ W.
Using the relation (2), (3), it is easy to prove the following statements:
1) any N -periodic on Z+
0 solution xn(x0, x1) of equation (4) is an N -periodic solution of
equation (5) on Z+
0 ;
2) if xn(x0, x1) is an N -periodic on Z+
0 solution of equation (5), then it is an N -periodic
solution on Z+
0 of equation (6), where µ = fν(xν , xν+1);
3) any N -periodic solution xn(x0, x1) of equation (6) on Z+
0 is an N -periodic solution of
equation (5) on this set.
The next conditions are referred to as conditions (A):
i)fn(x, y) is a function, N -periodic in n ∈ Z, continuous in x, y and defined in the region
D0 = Z ×D ×D, D = {x ∈ W
∣∣∣‖x‖ ≤ R}, where
‖fn(x, y)‖ ≤ M,∥∥fn(x, y)− fn(x′, y′)
∥∥ ≤ K1
∥∥x− x′∥∥+K2
∥∥y − y′∥∥ ,
and M,R,K1,K2 are positive constants;
ON FINDING PERIODIC SOLUTIONS OF SECOND ORDER DIFFERENCE EQUATIONS. . . 407
ii) the setDf =
{
x ∈ D
∣∣∣‖x‖ ≤ R−N
2M
4
}
⊂ D is nonempty, and γ =
N2
4
(K1+K2) < 1.
Define a sequence
{
x
(m)
n (x0)
}∞
m=0
, by the recurence relations,
x(0)
n ≡ x0, x
(m)
0 (x0) ≡ x0,
(7)
x(m)
n (x0) = x0 + gn(x(m−1)
n (x0), x
(m−1)
n+1 (x0)),m ∈ Z+.
Under the conditions (A) the following statements take place:
4) equation (5) can’t have more than one N -periodic solution xn with the inital value x0 ∈
Df ;
5) the function x̃n(x0) = lim
m→∞
x
(m)
n (x0) is an N -periodic solution of equation (5) in Z+
0 and
also of equation (6), where µ = µ∗ = fν(x̃ν , x̃ν+1), and
∥∥∥x̃n(x0)− x(m)
n (x0)
∥∥∥ ≤ σ∗(m) =
N2
4
m+1
M(K1 +K2)m
1− N2
4
(K1 +K2)
→ 0
as m → ∞;
6) for a function x̃n(x0) to be the only N -periodic on Z+
0 solution of equation (4) with the
initial value x0 ∈ Df , it is necessary and sufficient that µ∗ = 0;
7) if xn = xn(x0, x1) is an N -periodic solution of equation (6) and
yn = yn(x0, y1) is an N -periodic solution of the equation ∆2yn = fn(yn, yn+1) − µ1 on Z+
0 ,
then those solutions coincide on this set and µ1 = µ2.
The proof of the statements 1) – 7) doesn’t essentially depend on the dimension of the space
W and is partly stated in [10].
Thus to find an N -periodic solution of equation (1), it is necessary to find x0 ∈ Df that
satisfies the equation µ∗ = 0. Having continued the function x̃(x0) by periodicity to Z−, we get
an N -periodic solution of equation (1).
Let’s denote by
dφZ(x)
dx
the Frecher of the function Z(x) : W → W, by Z(x)
∣∣x2
x1
the di-
fference Z(x2)− Z(x1), set Φ = {0, 1, 2, . . . , N − 1} and prove first some auxiliary statements.
Lemma 1. Let conditions (A) be satisfied and, moreover, the function fn(x, y) be defined for
all n ∈ Φ in the area Z×D1×D1, whereD1 =
{
x ∈ W
∣∣∣‖x‖ < R+ ρ
}
, ρ is a constant positive,
Frechet differentiable in the area Ω ∈ D1 × D1 with
∥∥∥∥dΦfn(x, y)
d(x, y)
∥∥∥∥ ≤ P, P a positive constant
independent on n ∈ Z, (x, y) ∈ Ω.
If η =
N2P
4
< 1, then the functions x(m)
n (x) defined by relations (7) are Frechet differenti-
able in the region Df
0 =
{
x ∈ Df
∣∣∣‖x‖ < R− N2M
4
}
and
408 Yu.V. TEPLINS’KY, I.V. SEMENISHINA
∥∥∥∥∥dΦx
(m)
n (x)
dx
∥∥∥∥∥ < 1
1− η
(8)
for all n ∈ Φ, m ∈ Z+
0 .
Proof. Write relation (7) as
x(0)
n (x) = x, x
(m)
0 (x) = x, n ∈ Φ, m ∈ Z+
0 , (9)
x
(m)
1 (x) = x− 1
N
N−1∑
i=1
i−1∑
s=0
(fs(x
(m−1)
s (x), x
(m−1)
s+1 (x))
− fn(x
(m−1)
n (x), x
(m−1)
n+1 (x))), m ∈ Z+, (10)
x(m)
n (x) = x+
n−1∑
i=1
i−1∑
s=0
(fs(x
(m−1)
s (x), x
(m−1)
s+1 (x))− fn(x
(m−1)
n (x), x
(m−1)
n+1 (x)))
− n
N
N−1∑
i=1
i−1∑
s=0
(fs(x
m−1
s (x), x
(m−1)
s+1 (x))− fn(x
(m−1)
n (x), x
(m−1)
n+1 (x))), (11)
n ∈ Φ\{0, 1},m ∈ Z+.
By the norm ‖(x, y)‖, (x, y) ∈ Ω, we understand max{‖x‖, ‖y‖}, where ‖x‖, ‖y‖ is the norm
in the region D1. It is obvious that the set Ω is open in W ×W.
For m = 0 it follows from (9) that for all n ∈ Φ,
dΦx
(0)
n (x)
dx
= E,
∥∥∥∥∥dΦx
(0)
n (x)
dx
∥∥∥∥∥ = 1 <
1
1− η
,
where E is the identy operator.
For m ∈ Z+, we prove inequality (8) by induction.
Let m = 1.
Using (10) we write the equalities
dΦx
(1)
1 (x)
dx
= E − 1
N
N−1∑
i=1
i−1∑
s=0
dΦfs(x, x)
dx
− dΦfn(x, x)
dx
= E − 1
N
N−1∑
i=1
i−1∑
s=0
dΦfs(x, x)
d(x, x)
dΦg(x)
dx
− dΦfs(x, x)
d(x, x)
dΦg(x)
dx
,
ON FINDING PERIODIC SOLUTIONS OF SECOND ORDER DIFFERENCE EQUATIONS. . . 409
where g(x) is a mapping of the open set Df
0 to the set Ω with the components, g1(x) and g2(x),
being identity operators, because they map Df
0 in D1.
Then ∥∥∥∥dΦg(x)
dx
∥∥∥∥ = sup
‖h‖=1
∥∥∥∥dΦg(x)
dx
h
∥∥∥∥ = sup
‖h‖=1
max
{∥∥∥∥dΦg
1(x)
dx
h
∥∥∥∥ ,
∥∥∥∥dΦg
2(x)
dx
h
∥∥∥∥} ≤ sup
‖h‖=1
max
{∥∥∥∥dΦg
1(x)
dx
∥∥∥∥ ‖h‖, ∥∥∥∥dΦg
2(x)
dx
∥∥∥∥ ‖h‖}
= max
{∥∥∥∥dΦg
1(x)
dx
∥∥∥∥ ,∥∥∥∥dΦg
2(x)
dx
∥∥∥∥} = 1, h ∈ W.
Using proposition 1 from [11], where the space W is replaced by the space of linear conti-
nuous operators, £(W,W ), with an appropriate norm, we come to the inequalities∥∥∥∥∥dΦx
(1)
1 (x)
dx
∥∥∥∥∥ ≤ 1 +
N − 1
N
N
2
max
s∈Φ
{∥∥∥∥dΦfs(x, x)
d(x, x)
∥∥∥∥ ∥∥∥∥dΦg(x)
dx
∥∥∥∥}
≤ 1 +
P (N − 1)
2
≤ 1 +
PN2
4
<
1
1− η
,
since N − 1 <
N2
2
for all N ∈ Z.
Now let’s estimate the norm of the derivative
dΦx
(1)
n (x)
dx
for n ∈ Φ\{0, 1}. Using (11) we
have ∥∥∥∥∥dΦx
(1)
n (x)
dx
∥∥∥∥∥ ≤ 1 +
N2
4
max
s∈Φ
{∥∥∥∥dΦfs(x, x)
dx
∥∥∥∥} ≤ 1 +
N2P
4
<
1
1− η
.
So, estimation (8), for m = 1, takes place for all n ∈ Φ.
Assume that it holds for all 1 < m ≤ k uniformly in n ∈ Φ and show that it holds for
m = k + 1. If n = 0, estimate (8) is obvious. For n = 1 we have
∥∥∥∥∥dΦx
(k+1)
1 (x)
dx
∥∥∥∥∥ ≤ 1 +
N − 1
2
max
s∈Φ
{∥∥∥∥∥dΦfs(x
(k)
s (x), x
(k)
s+1(x))
dx
∥∥∥∥∥
}
≤ 1 +
N − 1
2
Pmax
s∈Φ
{∥∥∥∥∥dΦx
(k)
s (x)
dx
∥∥∥∥∥ ,
∥∥∥∥∥dΦx
(k)
s+1(x)
dx
∥∥∥∥∥
}
≤ 1 +
η
1− η
=
1
1− η
.
410 Yu.V. TEPLINS’KY, I.V. SEMENISHINA
At n ∈ Φ\{0, 1}∥∥∥∥∥dΦx
(k+1)
n (x)
dx
∥∥∥∥∥ ≤ 1 +
PN2
4
max
s∈Φ
{∥∥∥∥∥dΦx
(k)
s (x)
dx
∥∥∥∥∥ ,
∥∥∥∥∥dΦx
(k)
s+1(x)
dx
∥∥∥∥∥
}
≤ 1
1− η
,
and this finishes the proof of Lemma 1.
Lemma 2. Let, with conditions of Lemma 1, the following inequality hold for all {x1, x2, y1, y2} ⊂
D1 uniformly in n ∈ Φ :∥∥∥∥∥ dΦfn(x, y)
d(x, y)
∣∣∣∣(x1,y1)
(x2,y2)
∥∥∥∥∥ ≤ L0max{‖x1 − x2‖, ‖y1 − y2‖},
where L0 is a positive constant. Then, uniformly in n ∈ Φ,m ∈ Z+
0 , we have
∥∥∥∥∥∥ dΦx
(m)
n (x)
dx
∣∣∣∣∣
(x1)
(x2)
∥∥∥∥∥∥ ≤ L0η
P (1− η)2(1− γ)
‖x1 − x2‖. (12)
Proof. Form = 0, the statement of Lemma 2 is obvious. Form ∈ Z+,we’ll prove inequali-
ty (12) by induction. Let m = 1. If m = 1, n = 0, inequality (12) is obvious.
Let’s estimate
dΦx
(1)
1 (x)
dx
∣∣∣∣∣
x1
x2
in the norm. We have
∥∥∥∥∥ dΦx
(1)
1 (x)
dx
∣∣∣∣∣
x1
x2
∥∥∥∥∥ ≤ 1
N
N−1∑
i=1
∥∥∥∥∥∥
i−1∑
s=0
dΦfs(x, x)
dx
∣∣∣∣∣
x1
x2
−dΦfs(x, x)
dx
∣∣∣∣∣
x1
x2
∥∥∥∥∥∥
≤ N − 1
2
max
s∈Φ
{∥∥∥∥∥ dΦfs(x, x)
dx
∣∣∣∣x1
x2
∥∥∥∥∥
}
≤ N − 1
2
max
s∈Φ
{∥∥∥∥dΦfs(x, x)
d(x, x)
∥∥∥∥
∥∥∥∥∥ dΦg(x)
dx
∣∣∣∣x1
x2
∥∥∥∥∥+
∥∥∥∥∥dΦfs(x, x)
d(x, x)
∣∣∣∣∣
x1
x2
∥∥∥∥∥
∥∥∥∥dΦg(x2)
dx
∥∥∥∥
}
≤ N − 1
2
max
s∈Φ
{P · 0 + L0‖x1 − x2‖} =
=
N − 1
2
L0‖x1 − x2‖ ≤
ηL0
P
‖x1 − x2‖,
that is, inequality (12) takes place.
Now we shall prove this inequality for m = 1, n ∈ Φ\{0, 1},∥∥∥∥∥dΦx
(1)
n (x)
dx
∣∣∣∣∣
x1
x2
∥∥∥∥∥ ≤ N2
4
max
s∈Φ
{∥∥∥∥∥ dΦfs(x, x)
dx
∣∣∣∣x1
x2
∥∥∥∥∥
}
≤ ηL0
P
‖x1 − x2‖.
Thus, estimate (12) holds for m = 1, n ∈ Φ.
ON FINDING PERIODIC SOLUTIONS OF SECOND ORDER DIFFERENCE EQUATIONS. . . 411
Suppose that the inequality hold for all 1 < m ≤ k and n ∈ Φ,
∥∥∥∥∥dΦx
(m)
n (x)
dx
∣∣∣∣∣
x1
x2
∥∥∥∥∥ ≤ L0η
P (1− η)(1− γ)
m−1∑
i=0
ηi‖x1 − x2‖, (13)
from which, certainly, estimate (12) follows. Note that inequality (13) as it was shown above,
for m = 1, takes place for n ∈ Φ. Let’s prove it for m = k + 1.
For x(k+1)
0 (x), inequality (13) is obvious. Taking into account that the set Df
0 is convex and
denoting by g(k)
s (x) the mapping Df
0 → Ω with components g(k)1
s (x) = x
(k)
s (x), g
(k)2
s (x) =
x
(k)
s+1(x), we write the following inequations:
∥∥∥∥∥ dΦx
(k+1)
1 (x)
dx
∣∣∣∣∣
x1
x2
∥∥∥∥∥ ≤ N − 1
2
max
s∈Φ
{∥∥∥∥∥dΦfs(x
(k)
s (x1), x
(k)
s+1(x1))
d(x
(k)
s , x
(k)
s+1
∥∥∥∥∥
∥∥∥∥∥ dΦg
(k)
s (x)
dx
∣∣∣∣∣
x1
x2
∥∥∥∥∥
+
∥∥∥∥∥∥ dΦfs(x
(k)
s (x), x
(k)
s+1(x))
d(x
(k)
s , x
(k)
s+1)
∣∣∣∣∣
x1
x2
∥∥∥∥∥∥
∥∥∥∥∥ dΦg
(k)
s (x2)
dx
∥∥∥∥∥
}
≤ N − 1
2
max
s∈Φ
{
P
∥∥∥∥∥dΦg
(k)
s (x)
dx
∣∣∣∣∣
x1
x2
∥∥∥∥∥
+ L0max
{∥∥∥∥x(k)
s (x)
∣∣∣x1
x2
∥∥∥∥ ,∥∥∥∥x(k)
s+1(x)
∣∣∣x1
x2
∥∥∥∥}
×max
{∥∥∥∥∥dΦx
(k)
s (x2)
dx
∥∥∥∥∥ ,
∥∥∥∥∥ dΦx
(k)
s+1(x2)
dx
∥∥∥∥∥
}}
≤ N − 1
2
max
s∈Φ
{
Pmax
{∥∥∥∥∥dΦx
(k)
s (x)
dx
∣∣∣∣∣
x1
x2
∥∥∥∥∥ ,
∥∥∥∥∥ dΦx
(k)
s+1(x)
dx
∣∣∣∣∣
x1
x2
∥∥∥∥∥
}
+
L0
(1− η)(1− γ)
‖x1 − x2‖
}
≤ η
P
{
P
L0η
P (1− η)(1− γ)
k−1∑
i=0
ηi +
L0
(1− η)(1− γ)
}
‖x1 − x2‖
=
L0η
P (1− η)(1− γ)
k∑
i=0
ηi‖x1 − x2‖.
It remains to estimate
dΦx
(k+1)
n (x)
dx
∣∣∣∣∣
x1
x2
in the norm for n ∈ Φ\{0, 1}.
412 Yu.V. TEPLINS’KY, I.V. SEMENISHINA
Using (11) we get similarly to the above that
∥∥∥∥∥dΦx
(k+1)
n (x)
dx
∣∣∣∣∣
x1
x2
∥∥∥∥∥ ≤ N2
4
max
s∈Φ
{∥∥∥∥∥dΦfs(x
(k)
s (x), x
(k)
s+1(x))
dx
∥∥∥∥∥
}
≤ L0η
P (1− η)(1− γ)
k∑
i=0
ηi‖x1 − x2‖.
Lemma 2 is proved.
Lemma 3. With the conditions of Lemma 2, the sequence
{
dΦx
(m)
n (x)
dx
}∞
m=0
converges, as
m → ∞, uniformly in x ∈ Df
0 for n ∈ Φ.
Proof. This sequence belongs to the space £(W,W ) which is complete. Therefore, it is
enough to prove that it is fundamental. For n = 0, the claim of the lemma is obvious. We’ll
prove that for all m ∈ Z+\{1} and n ∈ Φ\{0}, the following inequality holds:
∥∥∥∥∥dΦx
(m)
n (x)
dx
− dΦx
(m−1)
n (x)
dx
∥∥∥∥∥ ≤ η
P
m−2∑
i=0
β
1− η
γm−2−iηi + Pηm−1
, (14)
where β =
L0N
2M
4
.
Let’s prove estimate (14) for m = 2. Using equalities (4), (5) we have
∥∥∥∥∥dΦx
(1)
1 (x)
dx
− dΦx
(0)
1 (x)
dx
∥∥∥∥∥ ≤ N − 1
2
max
s∈Φ
∥∥∥∥dΦfs(x, x)
dx
∥∥∥∥ ≤ N − 1
2
P ≤ η,
∥∥∥∥∥dΦx
(1)
n (x)
dx
− dΦx
(0)
n (x)
dx
∥∥∥∥∥ ≤ N2
4
max
s∈Φ
∥∥∥∥dΦfs(x, x)
dx
∥∥∥∥ ≤ N2
4
P = η,
n ∈ Φ\{0, 1}.
ON FINDING PERIODIC SOLUTIONS OF SECOND ORDER DIFFERENCE EQUATIONS. . . 413
Now we have the chain of inequalities
∥∥∥∥∥dΦx
(2)
1 (x)
dx
− dΦx
(1)
1 (x)
dx
∥∥∥∥∥ ≤ N − 1
2
max
s∈Φ
{∥∥∥∥∥dΦfs(x
(1)
s (x), x
(1)
s+1(x))
dx
− dΦfs(x, x)
dx
∥∥∥∥∥
}
≤ N − 1
2
max
s∈Φ
{∥∥∥∥∥dΦfs(x
(1)
s (x), x
(1)
s+1(x))
d(x
(1)
s , x
(1)
s+1)
dΦg
(1)
s (x)
dx
− dΦfs(x, x)
d(x, x)
dΦg(x)
dx
∥∥∥∥∥
}
≤ N − 1
2
max
s∈Φ
{∥∥∥∥∥dΦfs(x
(1)
s (x), x
(1)
s+1(x))
d(x
(1)
s , x
(1)
s+1)
− dΦfs(x, x)
d(x, x)
∥∥∥∥∥
∥∥∥∥∥dΦg
(1)
s (x)
dx
∥∥∥∥∥+
∥∥∥∥∥dΦfs(x, x)
d(x, x)
∥∥∥∥∥
∥∥∥∥∥dΦg
(1)
s (x)
dx
− dΦg(x)
dx
∥∥∥∥∥
}
≤ N − 1
2
max
s∈Φ
{
L0max
{∥∥∥x(1)
s (x)− x(0)
s (x)
∥∥∥ , ∥∥∥x(1)
s+1(x)− x
(0)
s+1(x)
∥∥∥}
×max
{∥∥∥∥∥dΦx
(1)
s (x)
dx
∥∥∥∥∥ ,
∥∥∥∥∥dΦx
(1)
s+1(x)
dx
∥∥∥∥∥
}
+ Pmax
{∥∥∥∥∥dΦx
(1)
s (x)
dx
− dΦx
(0)
s (x)
dx
∥∥∥∥∥ ,
∥∥∥∥∥dΦx
(1)
s+1(x)
dx
−
dΦx
(0)
s+1(x)
dx
∥∥∥∥∥
}}
≤ N − 1
2
β
1− η
+ Pη
≤ η
P
β
1− η
+ Pη
.
Taking into account the last inequalities for n ∈ Φ\{0, 1} we have
∥∥∥∥∥dΦx
(2)
n (x)
dx
− dΦx
(1)
n (x)
dx
∥∥∥∥∥ ≤ N2
4
max
s∈Φ
{∥∥∥∥∥dΦfs(x
(1)
s (x), x
(1)
s+1(x))
dx
−
− dΦfs(x, x)
dx
∥∥∥∥∥
}
≤ η
P
β
1− η
+ Pη
,
that is, estimate (14) holds for m = 2, n ∈ Φ\{0}.
414 Yu.V. TEPLINS’KY, I.V. SEMENISHINA
Let’s assume that it holds for all 2 < m ≤ k and prove its validity for m = k + 1.
For all n ∈ Φ\{0}, we have∥∥∥∥∥dΦx
(k+1)
n (x)
dx
− dΦx
(k)
n (x)
dx
∥∥∥∥∥ ≤ η
P
max
s∈Φ
{∥∥∥∥∥dΦfs(x
(k)
s (x), x
(k)
s+1(x))
dx
−
dΦfs(x
(k−1)
s (x), x
(k−1)
s+1 (x))
dx
∥∥∥∥∥
}
≤ η
P
{
β
1− η
γk−1
+ Pmax
{∥∥∥∥∥dΦx
(k)
s (x)
dx
− dΦx
(k−1)
s (x)
dx
∥∥∥∥∥ ,
∥∥∥∥∥dΦx
(k)
s+1(x)
dx
−
dΦx
(k−1)
s+1 (x)
dx
∥∥∥∥∥
}}
≤ η
P
{
β
1− η
γk−1
+ η
k−2∑
i=0
β
1− η
γk−2−iηi + Pηk−1
}
,
which finishes the proof of estimate (14).
If η 6= γ right-hand side of estimate (14) is equal to the expression
ηβ
P (1− η)(η − γ)
(ηm−1 − γm−1) + ηm.
Since η < 1 and γ < 1, the statement of Lemma 3 is proved.
If η = γ, then the right-hand side of estimate (14) is equal to the expression
β
P (1− γ)
γm−1(m− 1) + γm ≤
β
P (1− γ)
+ 1
γm−1(m− 1), m ≥ 2,
and it also shows that the sequence
{dΦx
(m)
n (x)
dx
}∞
m=0
, is fundamental since the series
∞∑
m=2
γm−1(m− 1) converges. Lemma 3 is proved.
Corollary 1. With the conditions of Lemma 2, the function x̃n(x) determined in claim 5), is
Frechet differentiable on Df
0 for all n ∈ Φ.
The proof of this statement follows at once from Theorem 111 [13, p.780].
Lemma 4. With the conditions of Lemma 2, the mapping ∆(x) = fn(x̃n(x), x̃n+1(x)) is
Frechet differentiable on the setDf
0 and, moveover, for all {x1, x2} ⊂ Df
0 the following inequality
holds: ∥∥∥∥∥ dΦ∆(x)
dx
∣∣∣∣x1
x2
∥∥∥∥∥ ≤ L‖x1 − x2‖,
where L = const > 0.
ON FINDING PERIODIC SOLUTIONS OF SECOND ORDER DIFFERENCE EQUATIONS. . . 415
Proof. Taking into account Lemmas 1 – 3, it is not difficult to see that for all n ∈ Φ and
{x1, x2} ⊂ Df
0 , the following relations take place:∥∥∥∥dΦx̃n(x)
dx
∥∥∥∥ ≤ 1
1− η
,
∥∥∥∥∥ dΦx̃n(x)
dx
∣∣∣∣x1
x2
∥∥∥∥∥ ≤ L0η
P (1− η)2(1− γ)
‖x1 − x2‖,
dΦ∆(x)
dx
=
1
N
N−1∑
s=0
dΦfs(x̃s(x), x̃s+1(x))
d(x̃s, x̃s+1)
dg̃s(x)
dx
,
where g̃s(x) is a mapping of the set Df
0 to the set Ω with components g̃1
s(x) = x̃s(x), g̃2
s(x) =
x̃s+1(x).
Taking into account the estimate for ‖ωn‖0 from Theorem 2 of [10], we have∥∥∥∥∥ dΦ∆(x)
dx
∣∣∣∣x1
x2
∥∥∥∥∥ ≤ 1
N
N−1∑
s=0
{∥∥∥∥∥ dΦfs(x̃s(x), x̃s+1(x))
d(x̃s, x̃s+1)
∣∣∣∣x1
x2
∥∥∥∥∥
∥∥∥∥dΦg̃s(x1)
dx
∥∥∥∥
+
∥∥∥∥dΦfs(x̃s(x2), x̃s+1(x2))
d(x̃s, x̃s+1)
∥∥∥∥
∥∥∥∥∥ dΦg̃s(x)
dx
∣∣∣∣x1
x2
∥∥∥∥∥
}
≤ 1
N
N−1∑
s=0
{
L0max
{∥∥∥ x̃s(x)|x1x2
∥∥∥,∥∥∥ x̃s+1(x)|x1x2
∥∥∥}
×max
{∥∥∥∥dΦx̃s(x1)
dx
∥∥∥∥ , ∥∥∥∥ dΦx̃s+1(x1)
dx
∥∥∥∥}
+ Pmax
{∥∥∥∥∥ dΦx̃s(x)
dx
∣∣∣∣x1
x2
∥∥∥∥∥ ,
∥∥∥∥∥ dΦx̃s+1(x)
dx
∣∣∣∣x1
x2
∥∥∥∥∥
}}
≤ L0
(1− η)(1− γ)
1 +
η
1− η
‖x1 − x2‖.
Denoting the constant term
L0
(1− η)(1− γ)
1 +
η
1− η
by L, we finish the proof of
Lemma 4.
Lemma 5. Let the conditions of Lemma 2 be satisfied, and there exist a point x0 ∈ Df
0 and a
sequence of indexes p1 < p2 < p3 < . . . < pk < . . . such that, for s ∈ Z+,∥∥∥∥dΦ∆ps(x
0)
dx0
− E
∥∥∥∥ ≤ l0 < 1.
416 Yu.V. TEPLINS’KY, I.V. SEMENISHINA
Then the mapping
dΦ∆(x0)
dx0
is circulating invertible and, moreover,
∥∥∥∥∥
[
dΦ∆(x0)
dx0
]−1
∥∥∥∥∥ ≤ N∗0 =
1
1− l0
.
Proof. It is obvious that the statement of Lemma 5 follows at once if we pass to the limit,
∥∥∥∥dΦ∆(x)
dx
− dΦ∆m(x)
dx
∥∥∥∥ → 0 as m → ∞. (15)
First, let’s estimate the difference Imn (x) =
dΦx̃n(x)
dx
− dΦx
(m)
n (x)
dx
,
‖Imn (x)‖ = lim
p→∞
∥∥∥∥∥dΦx
m+p
n (x)
dx
− dΦx
(m)
n (x)
dx
∥∥∥∥∥
≤ lim
p→+∞
p∑
i=1
∥∥∥∥∥dΦx
(m+i)
n (x)
dx
− dΦx
(m+i−1)
n (x)
dx
∥∥∥∥∥ .
Using Lemma 3 with m ≥ 2, n ∈ Φ\{0}, and η 6= γ (for example, η > γ), we come to the
inequalities
∥∥∥I(m)
n (x)
∥∥∥ ≤ ∞∑
i=1
ηβ
P (1− η)(η − γ)
+ 1
ηm+i−1
≤ ηm
1− η
β
P (1− η)(η − γ)
+ 1
,
from which it follows that ‖Imn (x)‖ → 0 as m → ∞, since η < 1.
If γ = η, then
∥∥∥I(m)
n (x)
∥∥∥ ≤ ∞∑
i=1
β
P (1− γ)
+ 1
γm+i−1(m+ i− 1)
= γm−1
β
1− γ
+ 1
∞∑
i=1
iγi +
(m− 1)γ
1− γ
→ 0
as m → ∞, since the series
∑∞
i=1 iγ
i converges to a number l∗.
ON FINDING PERIODIC SOLUTIONS OF SECOND ORDER DIFFERENCE EQUATIONS. . . 417
Now we have∥∥∥∥dΦ∆(x)
dx
− dΦ∆m(x)
dx
∥∥∥∥ ≤ 1
N
N−1∑
s=0
{∥∥∥∥∥dΦfs(x̃s(x), x̃s+1(x))
d(x̃s, x̃s+1)
−
dΦfs(x
(m)
s (x), x
(m)
s+1(x))
d(x
(m)
s , x
(m)
s+1)
∥∥∥∥∥
∥∥∥∥dg̃s(x)
dx
∥∥∥∥
+
∥∥∥∥∥dΦfs(x
(m)
s (x), x
(m)
s+1(x))
d(x
(m)
s , x
(m)
s+1)
∥∥∥∥∥
∥∥∥∥∥dΦg̃s(x)
dx
− dg
(m)
s (x)
dx
∥∥∥∥∥
}
≤ 1
N
N−1∑
s=0
{
L0max
{∥∥∥x̃s(x)− x(m)
s (x)
∥∥∥,∥∥∥x̃s+1(x)− x(m)
s+1(x)
∥∥∥}
×max
{∥∥∥∥dΦx̃s(x)
dx
∥∥∥∥ ,∥∥∥∥dΦx̃s+1(x)
dx
∥∥∥∥}
+ Pmax
{∥∥∥Ims (x)
∥∥∥, ∥∥∥Ims+1(x)
∥∥∥}} ≤ L0σ
∗(m)
1
1− η
+ Pσ∗(m) → 0
as m → ∞, because σ∗(m) =
MN2γm
4(1− γ)
defined in the statement 5) and σ∗(m) given for m ≥ 2
by the relation
σ∗(m) =
ηm
1− η
β
P (1− η)(η − γ)
+ 1
if η > γ;
γm−1
β
P (1− γ)
+ 1
l∗ +
(m− 1)γ
1− γ
if η = γ;
γm
1− γ
β
P (1− η)(γ − η)
+ 1
if η < γ,
approach zero as m → ∞. This ends the proof of estimate (15) and Lemma 5.
Let’s introduce the notations
N0 =
∥∥∥∥∥
[
dΦ∆(x0)
dx0
]−1
∥∥∥∥∥ , k =
∥∥∥∥∥
[
dΦ∆(x0)
dx0
]−1
·∆(x0)
∥∥∥∥∥ , h = N0kL,
t0 is the smallest root of the equation ht2 − t+ 1 = 0.
The lemmas proved above allow us to formulate the following statement which gives suffi-
cient conditions for existence of an N -periodic solution of equation (1).
Theorem 1. Let the conditions of Lemmas 4 and 5 hold. Suppose that the constants defined
there, L and N∗0 , are such that h∗ = LN∗
2
0 M < 0, 25 and the closed ball B∗(x0, k∗t∗) ⊂ Df
0 ,
where k∗ = N∗0M, t∗ is the smallest root of the equation h∗t2 − t+ 1 = 0.
418 Yu.V. TEPLINS’KY, I.V. SEMENISHINA
Then there is only one point x∗ in the closed ball B(x0, kt0) ⊂ B∗(x0, k∗t∗) generating an
N -periodic solution x̃n(x∗), x̃0(x∗) = x∗ of equation (1).In such a case we have
∥∥∥x̃n(x∗)− x(p)
n (xp)
∥∥∥ ≤ M
1− γ
N∗0 q∗p
1− q∗
+
N2γp
4
, (16)
where q∗ =
1−
√
1− 4h∗
2
< 0, 5, there function x(p)
n (xp) are defined by relations (7), and {xp}
is a sequence defined by the recurrence relation
x0 = x0, xn+1 = xn −
[
dΦ∆(x0)
dx0
]−1
·∆(xn), n ∈ Z+.
Proof. Since k ≤ k∗, h ≤ h∗, we have t0 ≤ t∗ and the ball B(x0, kt0) is embedded into the
ball B∗(x0, k
∗t∗) and, hence, in the set Df
0 . This allows to use Theorem 1 from [14, p. 430] and
to write the estimate
‖x∗ − xp‖ ≤
q∗p
1− q∗
N∗0M,
which leads to the inequalities∥∥∥x̃n(x∗)− x(p)
n (xp)
∥∥∥ ≤ ∥∥∥x̃n(x∗)− x̃n(xp)
∥∥∥+
∥∥∥x̃n(xp)− x(p)
n (xp)
∥∥∥
≤ ‖x
∗ − xp‖
1− γ
+ σ∗(p) ≤ q∗
p
1− q∗
N∗0M
1
1− γ
+
N2
4
p+1
M(K1 +K2)p
1− γ
. (17)
The last estimate gives inequality (16), and this completes the demonstration of Theorem 1.
Now let’s consider the equation
∆ps(x) = fn
(
x
(ps)
n (x), x
(ps)
n+1(x)
)
= 0. (18)
In the same way as in the proof of Lemma 4, it is not difficult to see that for all {x1, x2} ⊂
Df
0 , the following inequality holds:∥∥∥∥∥ dΦ∆ps(x)
dx
∣∣∣∣x1
x2
∥∥∥∥∥ ≤ L‖x1 − x2‖, s ∈ Z+.
Besides, it is obvious that
∥∥∥∥∥
[
dΦ∆ps(x
0)
dx0
]−1
∥∥∥∥∥ ≤ N∗0 . Therefore, if the conditions of Theorem 1
holds, equation (18), for aech s ∈ Z+ in the closed ball B∗(x0, k
∗t∗), has a solution xps which
ON FINDING PERIODIC SOLUTIONS OF SECOND ORDER DIFFERENCE EQUATIONS. . . 419
is a limit, as k → ∞, of the sequence {x(ps)
k }, x(ps)
0 = x0, constructed using the recurrence
formula
x
(ps)
k+1 = x
(ps)
k −
[
dΦ∆ps(x
0)
dx0
]−1
·∆ps(x
(ps)
k ).
For all s ∈ Z+, k ∈ Z+
0 , the points x(ps)
k , together with the point x∗, belong to the closed ball
B∗(x0, t∗k∗).
Let’s introduce the notations
l∗ =
∞∑
i=1
ili−1
0 , G = 1 +
N∗0 (K1 +K2)
1− γ
,
ε(ps) = l∗M
{
L0σ
∗(ps)
1
1− η
+ Pσ∗(ps)
}
+N∗0 (K1 +K2)σ∗(ps),
δ(ps) = N∗0 (K1 +K2)σ∗(ps) +M
{
L0σ
∗(ps)
1
1− η
+ Pσ∗(ps)
}
.
Since σ∗(ps) and σ∗(ps) approach zero as s → ∞, both ε(ps) and δ(ps) have the same property.
Let’s formulate a statement which allows to approximate the function x̃n(x∗) by the functi-
on x(m)
n (x
(ps)
k ) with an arbitrary precision.
Theorem 2. In the conditions of Theorem 1, the following relations hold:
lim
k→∞
lim
s→∞
x
(ps)
k = x∗, (19)
moreover, if ps and m are greater than two, we have
∥∥∥x̃n(x∗)− x(m)
n (x
(ps)
k )
∥∥∥ ≤
∥∥∥x∗ − x(ps)
k
∥∥∥
1− γ
+ σ∗(m), (20)
∥∥∥x∗ − x(ps)
k
∥∥∥ ≤ q∗k
1− q∗
N∗0M + ε(ps)G
k−1 + δ(ps)
k−2∑
i=0
Gi. (21)
Proof. Estimate (20) is obtained in the same way as estimate (17). We’ll show that, for
k ∈ Z+
0 ,
lim
s→∞
x
(ps)
k = xk, (22)
where {xk} is a sequence of point obtained by using the modified method of Newton which
consists in applying the recurrence formula from Theorem 1 with x0 = x0.
420 Yu.V. TEPLINS’KY, I.V. SEMENISHINA
The inequality∥∥∥∥∥
[
dΦ∆(x0)
dx0
]−1
−
[
dΦ∆ps(x
0)
dx0
]−1
∥∥∥∥∥ ≤
∥∥∥∥dΦ∆(x0)
dx0
− dΦ∆ps(x
0)
dx0
∥∥∥∥ l∗
leads to the estimate∥∥∥x1 − x(ps)
1
∥∥∥ ≤ ∥∥∥∥dΦ∆(x0)
dx0
− dΦ∆ps(x
0)
dx0
∥∥∥∥ l∗M +N∗0 (K1 +K2)σ∗(ps) ≤ ε(ps).
Now write the following recurrence relation:∥∥∥xk − x(ps)
k
∥∥∥ ≤ ∥∥∥xk−1 − x
(ps)
k−1
∥∥∥+
{
L0σ
∗(ps)
1
1− η
+ Pσ∗(ps)
}
M
+N∗0
∥∥∥∆(xk−1)−∆ps(x
(ps)
k−1)
∥∥∥
≤
∥∥∥xk−1 − x
(ps)
k−1
∥∥∥{1 +
N∗0 (K1 +K2)
1− γ
}
+N∗0 (K1 +K2)σ∗(ps)
+M
{
L0σ
∗(ps)
1
1− η
+ Pσ∗(ps)
}
≤
∥∥∥xk−1 − x
(ps)
k−1
∥∥∥G+ δ(ps),
from which we get:
∥∥∥xk − x(ps)
k
∥∥∥ ≤ ε(ps)G
k−1 + δ(ps)
k−2∑
i=0
Gi
obtained by induction and from which relation (22) follows, where passing to the limit is not
uniform in k ∈ Z+
0 .
From the last inequality it is easy to obtain inequality (21) which guarantees the validity of
relation (19).
We must remark that nothing can be said about the validity of interchanging the limits (19).
Theorem 2 is proved.
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Received 19.03.2001
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