On equations with generalized periodic right-hand side
Periodic solutions are studied for second-order differential equations with generalized forcing. Analytical bifurcation results are derived with application to forced harmonic and Duffing oscillators.
Saved in:
| Date: | 2018 |
|---|---|
| Main Authors: | , , , |
| Format: | Article |
| Language: | English |
| Published: |
Institute of Mathematics, NAS of Ukraine
2018
|
| Online Access: | https://umj.imath.kiev.ua/index.php/umj/article/view/1555 |
| Tags: |
Add Tag
No Tags, Be the first to tag this record!
|
| Journal Title: | Ukrains’kyi Matematychnyi Zhurnal |
| Download file: |  |
Institution
Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860507355456733184 |
|---|---|
| author | Fečkan, M. Pospíšil, M. Фечкан, М. Поспісіль, М. |
| author_facet | Fečkan, M. Pospíšil, M. Фечкан, М. Поспісіль, М. |
| author_sort | Fečkan, M. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2019-12-05T09:18:03Z |
| description | Periodic solutions are studied for second-order differential equations with generalized forcing. Analytical bifurcation results
are derived with application to forced harmonic and Duffing oscillators.
|
| first_indexed | 2026-03-24T02:08:00Z |
| format | Article |
| fulltext |
UDC 517.9
M. Fečkan, M. Pospı́šil (Comenius Univ. Bratislava and Math. Inst. Slovak Acad. Sci., Slovak Republic)
ON EQUATIONS WITH GENERALIZED PERIODIC RIGHT-HAND SIDE*
ПРО РIВНЯННЯ З УЗАГАЛЬНЕНОЮ ПЕРIОДИЧНОЮ
ПРАВОЮ ЧАСТИНОЮ
Periodic solutions are studied for second-order differential equations with generalized forcing. Analytical bifurcation results
are derived with application to forced harmonic and Duffing oscillators.
Вивчаються перiодичнi розв’язки для диференцiальних рiвнянь другого порядку з узагальненою примушуючою
силою. Аналiтичнi результати для бiфуркацiй отримано та застосовано до вимушених гармонiчних коливань та
осцилятора Даффiнга.
1. Introduction. In this paper we shall investigate a weakly forced second-order differential equation
\"x(t) + h(x(t)) = \varepsilon F (t), (1.1)
where h \in C(\BbbR n,\BbbR n) is an analytic function, \varepsilon \in \BbbR is a small parameter and F : \BbbR \rightarrow \BbbR n is a
4a-periodic generalized function which can behave like Dirac \delta -function at the points \{ (1 + 2k)a |
k \in \BbbZ \} . We apply a method of nonsmooth transformation of time proposed by Pilipchuk in [6] but we
do not use shooting and numerical computations [7]. Instead, we use the implicit function theorem
and Lyapunov – Schmidt method to obtain analytical results on the existence of periodic solutions
of (1.1). Later, we investigate particular functions h, concretely a linear and cubic case. So, we
consider a weakly forced harmonic oscillator equation
\"x(t) + b2x(t) = \varepsilon F (t) (1.2)
and a weakly forced Duffing equation
\"x(t) + b2x3(t) = \varepsilon F (t). (1.3)
Related results are also given in [8]. Finally, we note that the theory on the existence of periodic
solutions in evolution equations is well developed [2] and our paper is a contribution to this nice
theory.
2. General results. In this section we consider general equation (1.1) and look for a continuous
solution x with possible finite jumps in \.x and a generalized function \"x. Now we recall some results
of [7], for the reader’s convenience. First, we suppose the transformation
x(t) = X
\biggl(
\tau
\biggl(
t
a
\biggr) \biggr)
+ Y
\biggl(
\tau
\biggl(
t
a
\biggr) \biggr)
\tau \prime
\biggl(
t
a
\biggr)
(2.1)
for sufficiently smooth functions X, Y and
\tau (s) =
2
\pi
\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}
\biggl(
\mathrm{s}\mathrm{i}\mathrm{n}
\pi s
2
\biggr)
. (2.2)
Note that \tau is a 4-periodic piecewise-linear saw-tooth function. Moreover,
\tau (4k + 1) = 1, \tau (4k + 3) = - 1, k \in \BbbZ . (2.3)
The following lemma describes some properties of the derivatives of \tau . Throughout the paper, we
* This paper was supported by the Slovak Grant Agency VEGA No. 2/0153/16 and No. 1/0078/17 and the Slovak
Research and Development Agency under the contract No. APVV-14-0378.
c\bigcirc M. FEČKAN, M. POSPÍŠIL, 2018
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2 255
256 M. FEČKAN, M. POSPÍŠIL
shall use a so-called test function \chi : \BbbR \rightarrow \BbbR , which is a sufficiently smooth function with a compact
support.
Lemma 2.1. For \tau given by (2.2) the following holds in the sense of distributions:
\tau \prime (s) = \mathrm{s}\mathrm{g}\mathrm{n} \mathrm{c}\mathrm{o}\mathrm{s}
\biggl(
\pi s
2
\biggr)
,
\tau \prime \prime (s) = 2
\sum
k\in \BbbZ
\delta (s - 4k - 3) - \delta (s - 4k - 1)
with the Dirac \delta -function.
Proof. Taking the test function \chi ,
\infty \int
- \infty
\tau \prime (s)\chi (s) ds = -
\infty \int
- \infty
\tau (s)\chi \prime (s) ds =
= - 2
\pi
\infty \int
- \infty
\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}
\biggl(
\mathrm{s}\mathrm{i}\mathrm{n}
\pi s
2
\biggr)
\chi \prime (s) ds =
\infty \int
- \infty
\mathrm{s}\mathrm{g}\mathrm{n} \mathrm{c}\mathrm{o}\mathrm{s}
\biggl(
\pi s
2
\biggr)
\chi (s) ds.
For the second statement we have
\infty \int
- \infty
\tau \prime \prime (s)\chi (s) ds = -
\infty \int
- \infty
\mathrm{s}\mathrm{g}\mathrm{n} \mathrm{c}\mathrm{o}\mathrm{s}
\biggl(
\pi s
2
\biggr)
\chi \prime (s) ds =
=
\sum
k\in \BbbZ
\left( 3+4k\int
1+4k
\chi \prime (s) ds -
1+4k\int
- 1+4k
\chi \prime (s) ds
\right) = 2
\sum
k\in \BbbZ
\bigl(
\chi (3 + 4k) - \chi (1 + 4k)
\bigr)
=
= 2
\sum
k\in \BbbZ
\left( \infty \int
- \infty
\bigl(
\delta (s - 4k - 3) - \delta (s - 4k - 1)
\bigr)
\chi (s) ds
\right) .
The proof is complete by changing the order of the sum and integral due to a finite number of nonzero
summands.
Lemma 2.1 is proved.
Since \tau is not invertible on the whole \BbbR , the change of coordinates is considered on subintervals
and its periodicity is used. So, (2.1) means
x(t) =
\left\{
X
\biggl(
\tau
\biggl(
t
a
\biggr) \biggr)
+ Y
\biggl(
\tau
\biggl(
t
a
\biggr) \biggr)
, t \in
\bigcup
k\in \BbbZ ((4k - 1)a, (4k + 1)a),
X
\biggl(
\tau
\biggl(
t
a
\biggr) \biggr)
- Y
\biggl(
\tau
\biggl(
t
a
\biggr) \biggr)
, t \in
\bigcup
k\in \BbbZ ((4k + 1)a, (4k + 3)a).
In particular, for t \in ( - a, a), \tau (s) = s and \tau \prime (s) = 1, i.e.,
x(a\tau ) = X(\tau ) + Y (\tau ), \tau \in ( - 1, 1). (2.4)
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
ON EQUATIONS WITH GENERALIZED PERIODIC RIGHT-HAND SIDE 257
On the other side, for t \in (a, 3a), \tau (s) = 2 - s and \tau \prime (s) = - 1, i.e.,
x(a(2 - \tau )) = X(\tau ) - Y (\tau ), \tau \in ( - 1, 1). (2.5)
Summing and subtracting equations (2.4), (2.5) we obtain the inverse transformation
X(\tau ) =
1
2
\bigl(
x(a\tau ) + x(a(2 - \tau ))
\bigr)
,
Y (\tau ) =
1
2
\bigl(
x(a\tau ) - x(a(2 - \tau ))
\bigr) (2.6)
for \tau \in ( - 1, 1). Note that the values of \tau \prime at the points of \Lambda := \{ 1 + 2k | k \in \BbbZ \} are not known.
Hence, the continuity of x on \BbbR gives a necessary condition Y (\pm 1) = 0. The generalized derivative
of (2.1) is given in the next lemma. For simplicity, we omit the argument
t
a
of \tau unless it makes
confusion.
Lemma 2.2. Let x be given by (2.1). Then
\.x(t) = a - 1
\bigl(
X \prime (\tau )\tau \prime + Y \prime (\tau )
\bigr)
in the sense of distributions.
Proof. Let \chi be an arbitrary test function. Then applying Lemma 2.1 we obtain
\infty \int
- \infty
\.x(s)\chi (s) ds = -
\infty \int
- \infty
\biggl(
X
\biggl(
\tau
\biggl(
s
a
\biggr) \biggr)
+ Y
\biggl(
\tau
\biggl(
s
a
\biggr) \biggr)
\tau \prime
\biggl(
s
a
\biggr) \biggr)
\chi \prime (s) ds =
=
1
a
\infty \int
- \infty
X \prime
\biggl(
\tau
\biggl(
s
a
\biggr) \biggr)
\tau \prime
\biggl(
s
a
\biggr)
\chi (s) ds+
1
a
\infty \int
- \infty
Y \prime
\biggl(
\tau
\biggl(
s
a
\biggr) \biggr) \biggl(
\tau \prime
\biggl(
s
a
\biggr) \biggr) 2
\chi (s) ds+
+
2
a
\sum
k\in \BbbZ
\infty \int
- \infty
Y
\biggl(
\tau
\biggl(
s
a
\biggr) \biggr) \biggl(
\delta
\biggl(
s
a
- 4k - 3
\biggr)
- \delta
\biggl(
s
a
- 4k - 1
\biggr) \biggr)
\chi (s) ds =
=
1
a
\infty \int
- \infty
\biggl(
X \prime
\biggl(
\tau
\biggl(
s
a
\biggr) \biggr)
\tau \prime
\biggl(
s
a
\biggr)
+ Y \prime
\biggl(
\tau
\biggl(
s
a
\biggr) \biggr) \biggr)
\chi (s) ds,
where the last identity follows from \tau \prime \in \{ \pm 1\} on \BbbR \setminus \Lambda , (2.3) and Y (\pm 1) = 0.
Lemma 2.2 is proved.
The statement on the second derivative of x follows.
Lemma 2.3. Let x be given by (2.1). Then
\"x(t) = a - 2
\bigl(
X \prime \prime (\tau ) +X \prime (\tau )\tau \prime \prime + Y \prime \prime (\tau )\tau \prime
\bigr)
in the sense of distributions.
Proof. Taking an arbitrary test function \chi we get
\infty \int
- \infty
\"x(s)\chi (s) ds = - 1
a
\infty \int
- \infty
\biggl(
X \prime
\biggl(
\tau
\biggl(
s
a
\biggr) \biggr)
\tau \prime
\biggl(
s
a
\biggr)
+ Y \prime
\biggl(
\tau
\biggl(
s
a
\biggr) \biggr) \biggr)
\chi \prime (s) ds =
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
258 M. FEČKAN, M. POSPÍŠIL
=
1
a2
\infty \int
- \infty
\biggl(
X \prime \prime
\biggl(
\tau
\biggl(
s
a
\biggr) \biggr) \biggl(
\tau \prime
\biggl(
s
a
\biggr) \biggr) 2
+
+X \prime
\biggl(
\tau
\biggl(
s
a
\biggr) \biggr)
\tau \prime \prime
\biggl(
s
a
\biggr)
+ Y \prime \prime
\biggl(
\tau
\biggl(
s
a
\biggr) \biggr)
\tau \prime
\biggl(
s
a
\biggr) \biggr)
\chi (s) ds =
=
1
a2
\infty \int
- \infty
\biggl(
X \prime \prime
\biggl(
\tau
\biggl(
s
a
\biggr) \biggr)
+X \prime
\biggl(
\tau
\biggl(
s
a
\biggr) \biggr)
\tau \prime \prime
\biggl(
s
a
\biggr)
+
+Y \prime \prime
\biggl(
\tau
\biggl(
s
a
\biggr) \biggr)
\tau \prime
\biggl(
s
a
\biggr) \biggr)
\chi (s) ds.
Lemma 2.3 is proved.
We assume that the function F can be written as
F (t) = Q
\biggl(
\tau
\biggl(
t
a
\biggr) \biggr)
+ P
\biggl(
\tau
\biggl(
t
a
\biggr) \biggr)
\tau \prime
\biggl(
t
a
\biggr)
+ f
\biggl(
\tau
\biggl(
t
a
\biggr) \biggr)
\tau \prime \prime
\biggl(
t
a
\biggr)
for sufficiently smooth functions Q, P, f : \BbbR \rightarrow \BbbR n. Therefore after transformation (2.1), if h(x) =
=
\sum \infty
k=0
hkx
k, equation (1.1) becomes
a - 2
\bigl(
X \prime \prime (\tau ) +X \prime (\tau )\tau \prime \prime + Y \prime \prime (\tau )\tau \prime
\bigr)
+
\infty \sum
k=0
hk
\bigl(
X(\tau ) + Y (\tau )\tau \prime
\bigr) k
=
= \varepsilon
\bigl(
Q(\tau ) + P (\tau )\tau \prime + f(\tau )\tau \prime \prime
\bigr)
. (2.7)
Note that
h(x) =
\infty \sum
k=0
hk
k\sum
j=0
\biggl(
k
j
\biggr)
Xk - j(\tau )Y j(\tau )(\tau \prime )j .
Since the functions 1, \tau \prime , \tau \prime \prime are of different smoothness on \BbbR , and using (\tau \prime )2 = 1 on ( - 1, 1),
equation (2.7) is split into the system
a - 2X \prime \prime (\tau ) +
\infty \sum
k=0
hk
\lfloor k
2\rfloor \sum
j=0
\biggl(
k
2j
\biggr)
Xk - 2j(\tau )Y 2j(\tau ) = \varepsilon Q(\tau ),
a - 2Y \prime \prime (\tau ) +
\infty \sum
k=0
hk
\lfloor k - 1
2 \rfloor \sum
j=0
\biggl(
k
2j + 1
\biggr)
Xk - 2j - 1(\tau )Y 2j+1(\tau ) = \varepsilon P (\tau )
(2.8)
for \tau \in ( - 1, 1) with the boundary conditions
a - 2X \prime (\pm 1) = \varepsilon f(\pm 1), Y (\pm 1) = 0.
Denoting K := X + Y, L := X - Y we separate equations (2.8) to get
a - 2K \prime \prime (\tau ) + h(K(\tau )) = \varepsilon
\bigl(
Q(\tau ) + P (\tau )
\bigr)
,
a - 2L\prime \prime (\tau ) + h(L(\tau )) = \varepsilon
\bigl(
Q(\tau ) - P (\tau )
\bigr) (2.9)
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
ON EQUATIONS WITH GENERALIZED PERIODIC RIGHT-HAND SIDE 259
for \tau \in ( - 1, 1) with the mixed boundary conditions
K(\pm 1) - L(\pm 1) = 0, K \prime (\pm 1) + L\prime (\pm 1) = 2\varepsilon a2f(\pm 1). (2.10)
Now we recall the nonlinear variation of constants formula of Alekseev [1].
Lemma 2.4. If \varphi (c, t) is a solution of
\.x(t) = f(x(t)), t \in \BbbR , (2.11)
such that \varphi \prime (c, 0) is regular, then the nonautonomous problem
\.x(t) = f(x(t)) + g(t), t \in \BbbR , (2.12)
has a solution \varphi (c(t), t) with c(t) satisfying
c(t) = c+
t\int
0
\varphi \prime (c(s), s) - 1g(s) ds, t \in \BbbR , (2.13)
for some c \in \BbbR n, where the prime and the dot denote derivatives with respect to c and t, respectively.
Proof. Variational equation corresponding to (2.11) along the solution \varphi (c, t) is
( \.\varphi (c, t))\prime = f \prime (\varphi (c, t))\varphi \prime (c, t), t \in \BbbR .
By Liouville theorem [4] (Theorem 1.2), \varphi \prime (c, t) is nonsingular for any t \in \BbbR . Now suppose that
the solution of (2.12) has the form \varphi (c(t), t) with c(t) to be determined. Then differentiating with
respect to t we obtain
\.x(t) = \varphi \prime (c(t), t) \.c(t) + \.\varphi (c(t), t) =
= \varphi \prime (c(t), t) \.c(t) + f(\varphi (c(t), t)) = f(\varphi (c(t), t)) + g(t).
Therefore
\.c(t) =
\bigl(
\varphi \prime (c(t), t)
\bigr) - 1
g(t),
and we get (2.13). Then the statement follows.
Lemma 2.4 is proved.
Note that nonhomogeneous problem (2.12) does not have to possess a unique solution. Next, let
us consider
\.x(t) = f(x(t)) + \varepsilon g(t), t \in [a1, a2],
Ax(a1) = \varepsilon b1,
Bx(a2) = \varepsilon b2
(2.14)
for a1 < a2, \=b = (b\ast 1, b
\ast
2)
\ast \in \BbbR n and matrices A \in \BbbR k\times n, B \in \BbbR (n - k)\times n, 0 < k < n. Then by
Lemma 2.4, a general solution of the differential equation of (2.14) is given by \varphi (c(\varepsilon , \xi , t), t) with
c(\varepsilon , \xi , t) satisfying c(\varepsilon , \xi , t) = \xi + \varepsilon
\int t
a1
\varphi \prime (c(\varepsilon , \xi , s), s) - 1g(s) ds. So we need to solve
\beta (\varepsilon , \xi ) :=
\left\{ A\varphi (\xi , a1) = \varepsilon b1,
B\varphi (c(\varepsilon , \xi , a2), a2) = \varepsilon b2.
(2.15)
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
260 M. FEČKAN, M. POSPÍŠIL
For \varepsilon = 0, (2.15) is reduced to
\beta (\xi ) := \beta (0, \xi ) = 0. (2.16)
Thus we have to suppose that (2.16) has a solution \xi 0. If it is nondegenerate, i.e., D\beta (\xi 0) is
regular, then we can directly apply the implicit function theorem to (2.15). If D\beta (\xi 0) is singular,
then Lyapunov – Schmidt method should be used. When f(x) in (2.14) is an affine mapping, i.e.,
f(x) = Mx + m for a matrix M and a vector m, then \beta (\varepsilon , \xi , a1, a2) is linear in \xi and matrix
analysis is applied. In general nonlinear case of (2.15), the most simple case is, when we suppose:
(C1) There is a nondegenerate 1-parametric family of solutions \xi \in C3(\BbbR ,\BbbR n) of (2.16), i.e., it
holds \beta (\xi (\alpha )) = 0, \mathrm{k}\mathrm{e}\mathrm{r}D\beta (\xi (\alpha )) = [\xi \prime (\alpha )]. Moreover, \varphi (\xi (\alpha +T ), \cdot ) = \varphi (\xi (\alpha ), \cdot ) for any \alpha \in \BbbR .
Then we introduce the orthogonal projection
P (\alpha ) : \BbbR n \rightarrow \mathrm{i}\mathrm{m}D\beta (\xi (\alpha )) =
\bigl(
\mathrm{k}\mathrm{e}\mathrm{r}D\beta (\xi (\alpha ))\ast
\bigr) \bot
= [\psi (\alpha )]\bot ,
and we split (2.15) as follows:
P (\alpha )
\bigl(
\beta (\varepsilon , \xi (\alpha ) + \varsigma ) - \varepsilon \=b
\bigr)
= 0, \varsigma \in [\xi \prime (\alpha )]\bot ,
\psi \ast (\alpha )
\bigl(
\beta (\varepsilon , \xi (\alpha ) + \varsigma ) - \varepsilon \=b
\bigr)
= 0.
(2.17)
By the implicit function theorem, we can solve the first equation of (2.17) to get \varsigma = \varsigma (\alpha , \varepsilon ) for any
\varepsilon small. Clearly \varsigma (\alpha , 0) = 0. Then inserting \varsigma (\alpha , \varepsilon ) into the second equation of (2.17), we get
\widetilde B(\varepsilon , \alpha ) := \psi \ast (\alpha )
\bigl(
\beta (\varepsilon , \xi (\alpha ) + \varsigma (\alpha , \varepsilon )) - \varepsilon \=b
\bigr)
= 0. (2.18)
We compute
\widetilde B(0, \alpha ) = \psi \ast (\alpha )\beta (\xi (\alpha )) = 0,
M(\alpha ) := \widetilde B\varepsilon (0, \alpha ) = \psi \ast (\alpha )
\bigl(
\beta \varepsilon (0, \xi (\alpha )) - \=b
\bigr)
.
Hence instead of (2.18), we consider
B(\varepsilon , \alpha ) :=
\left\{
M(\alpha ), \varepsilon = 0,\widetilde B(\varepsilon , \alpha )
\varepsilon
, \varepsilon \not = 0.
(2.19)
From (2.15), we derive
\beta \varepsilon (0, \xi ) =
\left\{
0,
B\varphi \prime (\xi , a2)
\int a2
a1
\varphi \prime (\xi , s) - 1g(s) ds
and
\beta \prime (0, \xi ) =
\left\{ A\varphi
\prime (\xi , a1),
B\varphi \prime (\xi , a2).
Consequently, we obtain
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
ON EQUATIONS WITH GENERALIZED PERIODIC RIGHT-HAND SIDE 261
M(\alpha ) = - \psi \ast
1(\alpha )b1 + \psi \ast
2(\alpha )
\left( B\varphi \prime (\xi (\alpha ), a2)
a2\int
a1
\varphi \prime (\xi (\alpha ), s) - 1g(s) ds - b2
\right)
for
\varphi \prime (\xi (\alpha ), a1)
\ast A\ast \psi 1(\alpha ) + \varphi \prime (\xi (\alpha ), a2)
\ast B\ast \psi 2(\alpha ) = 0. (2.20)
Setting
\theta (\alpha , s) =
\bigl(
\varphi \prime (\xi (\alpha ), s) - 1
\bigr) \ast
\varphi \prime (\xi (\alpha ), a2)
\ast B\ast \psi 2(\alpha ), (2.21)
we get
M(\alpha ) =
a2\int
a1
\theta \ast (\alpha , s)g(s) ds - \psi \ast
1(\alpha )b1 - \psi \ast
2(\alpha )b2. (2.22)
Summarizing, we arrive at the following result.
Theorem 2.1. Assume f \in C3(\BbbR n,\BbbR n), g \in C3
\bigl(
[a1, a2],\BbbR n
\bigr)
and (C1) holds. If there is
a simple zero \alpha 0 of (2.22), i.e., M(\alpha 0) = 0 and DM(\alpha 0) is regular, then there is an \alpha \in
\in C1(( - \delta , \delta ),\BbbR ) for some \delta > 0 such that \alpha (0) = \alpha 0 and (2.14) has a unique solution x(\varepsilon , t) =
= \varphi (\xi (\alpha (\varepsilon )), t) +O(\varepsilon ) for any \varepsilon \in ( - \delta , \delta ).
Proof. The result follows from the implicit function theorem applying to B(\varepsilon , \alpha ) = 0 given
by (2.19).
Remark 2.1. Under assumptions of Theorem 2.1, (2.14) has a unique solution x(\varepsilon , t) = \varphi (\xi (\alpha 0),
t) +O(\varepsilon ) for any \varepsilon \in ( - \delta , \delta ).
Remark 2.2. Clearly M(\alpha ) is T -periodic, so if it is changing the sign over its period, then we
can apply the Brouwer degree method to solve B(\varepsilon , \alpha ) = 0, and we get a solution of (2.14) for \varepsilon
small.
Note (2.21) satisfies the adjoint variational linear equation
\.w(t) = - Df(\varphi (\xi (\alpha ), t))\ast w(t), t \geq 0, (2.23)
along with
\theta (\alpha , a1) = (\varphi \prime (\xi (\alpha ), a1)
- 1)\ast \varphi \prime (\xi (\alpha ), a2)
\ast B\ast \psi 2(\alpha ) =
= - (\varphi \prime (\xi (\alpha ), a1)
- 1)\ast \varphi \prime (\xi (\alpha ), a1)
\ast A\ast \psi 1(\alpha ) =
= - A\ast \psi 1(\alpha ),
(2.24)
\theta (\alpha , a2) = B\ast \psi 2(\alpha ),
where we apply (2.20). Assuming
(C2) A and B are surjective,
then A\ast and B\ast are injective, and (2.24) is equivalent to
\theta (\alpha , a1) \in \mathrm{k}\mathrm{e}\mathrm{r}A\bot , \psi 1(\alpha ) = - A\ast - 1\theta (\alpha , a1),
\theta (\alpha , a2) \in \mathrm{k}\mathrm{e}\mathrm{r}B\bot , \psi 2(\alpha ) = B\ast - 1\theta (\alpha , a2),
(2.25)
since \mathrm{i}\mathrm{m}A\ast = \mathrm{k}\mathrm{e}\mathrm{r}A\bot and \mathrm{i}\mathrm{m}B\ast = \mathrm{k}\mathrm{e}\mathrm{r}B\bot .
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
262 M. FEČKAN, M. POSPÍŠIL
Finally, we consider the unperturbed (2.9) and (2.10)
a - 2K \prime \prime (\tau ) + h(K(\tau )) = 0,
a - 2L\prime \prime (\tau ) + h(L(\tau )) = 0
(2.26)
for \tau \in ( - 1, 1) with the mixed boundary conditions
K(\pm 1) - L(\pm 1) = 0, K \prime (\pm 1) + L\prime (\pm 1) = 0. (2.27)
Then we claim that
(i) L(\tau ) = K( - \tau + 2),
(ii) K is 4-periodic.
To prove (i), we note that any solution of (2.26) is defined on \BbbR . So we fix K(\tau ) solving
a - 2K \prime \prime (\tau ) + h(K(\tau )) = 0 and take \widetilde L(\tau ) = K( - \tau + 2). Then \widetilde L(\tau ) satisfies the 2nd equation of
(2.26), and (2.27) implies
\widetilde L(1) = K(1) = L(1), \widetilde L\prime (1) = - K \prime (1) = L\prime (1),
so the uniqueness of solutions gives \widetilde L(\tau ) = L(\tau ). This proves (i). To prove (ii), using also (i) we
compute
K( - 1) = L( - 1) = K(3), K \prime ( - 1) = - L\prime ( - 1) = K \prime (3),
which gives (ii). Reversibly, if K(\tau ) solves the 1st equation of (2.26) and it is 4-periodic, then
taking L(\tau ) by (i), we get a solution of the 2nd equation of (2.26) with (2.27). We note then we
have a family K(\tau + \alpha ) satisfying (i) and (ii).
Moreover, we consider the linearization of (2.26),
a - 2U \prime \prime (\tau ) + h\prime (K(\tau ))U(\tau ) = 0,
a - 2V \prime \prime (\tau ) + h\prime (L(\tau ))V (\tau ) = 0
(2.28)
along K and L satisfying (i), (ii) with the mixed boundary conditions
U(\pm 1) - V (\pm 1) = 0, U \prime (\pm 1) + V \prime (\pm 1) = 0. (2.29)
Then again it holds
(i\prime ) V (\tau ) = U( - \tau + 2),
(ii\prime ) U is 4-periodic.
Indeed, we take \widetilde V (\tau ) = U( - \tau + 2). Then
a - 2 \widetilde V \prime \prime (\tau ) + h\prime (L(\tau ))\widetilde V (\tau ) =
= a - 2U \prime \prime ( - \tau + 2) + h\prime (K( - \tau + 2))U( - \tau + 2) = 0,
so \widetilde V (\tau ) satisfies the 2nd equation of (2.28), and (2.29) implies
\widetilde V (1) = U(1) = V (1), \widetilde V \prime (1) = - U \prime (1) = V \prime (1),
thus we get \widetilde V (\tau ) = V (\tau ). This proves (i\prime ). To prove (ii\prime ), we take \widetilde U(\tau ) = U(\tau + 4) and derive
a - 2 \widetilde U \prime \prime (\tau ) + h\prime (K(\tau ))\widetilde U(\tau ) =
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
ON EQUATIONS WITH GENERALIZED PERIODIC RIGHT-HAND SIDE 263
= a - 2U \prime \prime (\tau + 4) + h\prime (K(\tau + 4))U(\tau + 4) = 0,
so \widetilde U(\tau ) satisfies the 1st equation of (2.28), and using (2.29) with (i\prime ), we obtain
\widetilde U( - 1) = U(3) = V ( - 1) = U( - 1),
\widetilde U \prime ( - 1) = U \prime (3) = - V ( - 1) = U( - 1),
which gives (ii\prime ). Reversibly, if U(\tau ) solves the 1st equation of (2.28) under (i), (ii) and it is
4-periodic, then taking V (\tau ) by (i\prime ), we get a solution of the 2nd equation of (2.28) with (2.29).
Furthermore, assuming (ii), K(\tau ) is surrounded by periodic solutions K(r, t) with periods T (r),
i.e., K(0, t) = K(t), K(r, \tau + T (r)) = K(r, \tau ) and T (0) = 4. Then W (\tau ) = \partial rK(0, \tau ) solves
a - 2W \prime \prime (\tau ) + h\prime (K(\tau ))W (\tau ) = 0,
W (\tau + 4) + T \prime (0)K \prime (\tau ) =W (\tau ).
Hence if
T \prime (0) \not = 0, (2.30)
then W (\tau ) is a non 4-periodic solution of the first equation of (2.28). But K \prime (\tau ) is a 4-periodic
solution of the first equation of (2.28). Summarizing, under (2.30), the dimension of solutions of
BVP (2.28), (2.29) is 1.
3. Harmonic oscillator. In this section we consider equation (1.2) under transformation (2.1).
Following the previous section we derive
a - 2K \prime \prime (\tau ) + b2K(\tau ) = \varepsilon
\bigl(
Q(\tau ) + P (\tau )
\bigr)
,
a - 2L\prime \prime (\tau ) + b2L(\tau ) = \varepsilon
\bigl(
Q(\tau ) - P (\tau )
\bigr) (3.1)
for \tau \in ( - 1, 1) with the boundary conditions
K(\pm 1) - L(\pm 1) = 0, K \prime (\pm 1) + L\prime (\pm 1) = 2\varepsilon a2f(\pm 1). (3.2)
Denoting K1 = K, L1 = L we rewrite (3.1), (3.2) as
K \prime
1(\tau ) = K2(\tau ),
K \prime
2(\tau ) = - a2b2K1(\tau ) + \varepsilon a2
\bigl(
Q(\tau ) + P (\tau )
\bigr)
,
L\prime
1(\tau ) = L2(\tau ),
L\prime
2(\tau ) = - a2b2L1(\tau ) + \varepsilon a2
\bigl(
Q(\tau ) - P (\tau )
\bigr)
(3.3)
along with
K1(\pm 1) - L1(\pm 1) = 0, K2(\pm 1) + L2(\pm 1) = 2\varepsilon a2f(\pm 1). (3.4)
The unperturbed problem has the general solution
\varphi (c, \tau ) :=
\bigl(
c1 \mathrm{s}\mathrm{i}\mathrm{n} ab\tau + c2 \mathrm{c}\mathrm{o}\mathrm{s} ab\tau , c1ab \mathrm{c}\mathrm{o}\mathrm{s} ab\tau - c2ab \mathrm{s}\mathrm{i}\mathrm{n} ab\tau ,
c3 \mathrm{s}\mathrm{i}\mathrm{n} ab\tau + c4 \mathrm{c}\mathrm{o}\mathrm{s} ab\tau , c3ab \mathrm{c}\mathrm{o}\mathrm{s} ab\tau - c4ab \mathrm{s}\mathrm{i}\mathrm{n} ab\tau
\bigr) \ast
(3.5)
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
264 M. FEČKAN, M. POSPÍŠIL
with c = (c1, c2, c3, c4)
\ast . Note that \varphi \prime (c, \tau ) =
\biggl(
\Phi 0
0 \Phi
\biggr)
with
\Phi =
\Biggl(
\mathrm{s}\mathrm{i}\mathrm{n} ab\tau \mathrm{c}\mathrm{o}\mathrm{s} ab\tau
ab \mathrm{c}\mathrm{o}\mathrm{s} ab\tau - ab \mathrm{s}\mathrm{i}\mathrm{n} ab\tau
\Biggr)
does not depend on c. Hence, \mathrm{d}\mathrm{e}\mathrm{t}\varphi \prime (c, \tau ) = a2b2 \not = 0. We set a1 = - 1, a2 = 1 and denote
\beta 1(c) := A\varphi (c, - 1) =
\Biggl(
- c1S + c2C + c3S - c4C
ab(c1C + c2S + c3C + c4S)
\Biggr)
,
\beta 2(c) := B\varphi (c, 1) =
\Biggl(
c1S + c2C - c3S - c4C
ab(c1C - c2S + c3C - c4S)
\Biggr)
with S = \mathrm{s}\mathrm{i}\mathrm{n} ab, C = \mathrm{c}\mathrm{o}\mathrm{s} ab and
A = B =
\Biggl(
1 0 - 1 0
0 1 0 1
\Biggr)
. (3.6)
Hence,
\Biggl(
\beta 1(c)
\beta 2(c)
\Biggr)
=Wc, W :=
\left(
- S C S - C
abC abS abC abS
S C - S - C
abC - abS abC - abS
\right) .
The following lemma describes the null space of W, \mathrm{k}\mathrm{e}\mathrm{r}W, and image of W, \mathrm{i}\mathrm{m}W, with respect to
a and b.
Lemma 3.1. Let a, b > 0 be fixed. Then
\mathrm{k}\mathrm{e}\mathrm{r}W =
\left\{
0,
2ab
\pi
/\in \BbbZ ,\bigl[
(1, 0, - 1, 0)\ast , (0, 1, 0, 1)\ast
\bigr]
,
ab
\pi
\in \BbbZ ,\bigl[
(1, 0, 1, 0)\ast , (0, 1, 0, - 1)\ast
\bigr]
,
ab
\pi
- 1
2
\in \BbbZ ,
\mathrm{i}\mathrm{m}W =
\left\{
\BbbR 4,
2ab
\pi
/\in \BbbZ ,\bigl[
(0, 1, 0, 1)\ast , (1, 0, 1, 0)\ast
\bigr]
,
ab
\pi
\in \BbbZ ,\bigl[
(1, 0, - 1, 0)\ast , (0, 1, 0, - 1)\ast
\bigr]
,
ab
\pi
- 1
2
\in \BbbZ ,
where [v1, v2] denotes the linear span of vectors v1, v2.
Proof. Note that \mathrm{d}\mathrm{e}\mathrm{t}W = - 16a2b2S2C2. Thus for S \not = 0 \not = C, W is nonsingular. Investigating
the cases S \not = 0 = C and S = 0 \not = C separately gives the statement.
Now we consider particular cases of a and b as distinguished in Lemma 3.1.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
ON EQUATIONS WITH GENERALIZED PERIODIC RIGHT-HAND SIDE 265
Theorem 3.1. Let a, b > 0 be fixed, Q, P, f be sufficiently smooth functions, and \varphi be given
by (3.5). Then \varphi (c(\tau ), \tau ) is a solution of (3.3), (3.4), if
c(\tau ) = c( - 1) +
\varepsilon a
b
\tau \int
- 1
v(\sigma ) d\sigma , (3.7)
where
v(\tau ) =
\left(
(Q(\tau ) + P (\tau )) \mathrm{c}\mathrm{o}\mathrm{s} ab\tau
- (Q(\tau ) + P (\tau )) \mathrm{s}\mathrm{i}\mathrm{n} ab\tau
(Q(\tau ) - P (\tau )) \mathrm{c}\mathrm{o}\mathrm{s} ab\tau
- (Q(\tau ) - P (\tau )) \mathrm{s}\mathrm{i}\mathrm{n} ab\tau
\right)
and
if
2ab
\pi
/\in \BbbZ , then
c( - 1) = 2\varepsilon a2W - 1
\left( v0 - 1\int
- 1
v1(\sigma )d\sigma
\right) , (3.8)
if
ab
\pi
\in \BbbZ and
1\int
- 1
P (\sigma ) \mathrm{s}\mathrm{i}\mathrm{n}(ab(1 - \sigma )) d\sigma = 0,
f(1) -
1\int
- 1
Q(\sigma ) \mathrm{c}\mathrm{o}\mathrm{s}(ab(1 - \sigma )) d\sigma = f( - 1),
(3.9)
then
c( - 1) = 2\varepsilon a2W | - 1
(kerW )\bot
\left( v0 - 1\int
- 1
v1(\sigma )d\sigma
\right) + w, (3.10)
for some w \in \mathrm{k}\mathrm{e}\mathrm{r}W,
if
ab
\pi
- 1
2
\in \BbbZ and
1\int
- 1
P (\sigma ) \mathrm{s}\mathrm{i}\mathrm{n}(ab(1 - \sigma )) d\sigma = 0,
f(1) -
1\int
- 1
Q(\sigma ) \mathrm{c}\mathrm{o}\mathrm{s}(ab(1 - \sigma )) d\sigma = - f( - 1),
then c( - 1) is given by (3.10) for some w \in \mathrm{k}\mathrm{e}\mathrm{r}W,
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
266 M. FEČKAN, M. POSPÍŠIL
where
v0 =
\bigl(
0, f( - 1), 0, f(1)
\bigr) \ast
,
v1(\tau ) =
\biggl(
0, 0,
P (\tau )
ab
\mathrm{s}\mathrm{i}\mathrm{n}(ab(1 - \tau )), Q(\tau ) \mathrm{c}\mathrm{o}\mathrm{s}(ab(1 - \tau ))
\biggr) \ast
and | (kerW )\bot denotes the restriction onto the orthogonal complement to \mathrm{k}\mathrm{e}\mathrm{r}W.
Proof. Equation (3.7) follows from Lemma 2.4. It only remains to determine c( - 1).
From (3.7) and (3.4) we get that c( - 1) has to satisfy
\beta 1(c( - 1)) =
\Biggl(
0
2\varepsilon a2f( - 1)
\Biggr)
, \beta 2(c(1)) =
\Biggl(
0
2\varepsilon a2f(1)
\Biggr)
,
where
\beta 2(c(1)) =
\Biggl(
S C - S - C
abC - abS abC - abS
\Biggr) \left( c( - 1) +
\varepsilon a
b
1\int
- 1
v(\sigma )d\sigma
\right) .
In other words,
Wc( - 1) +
\varepsilon a
b
1\int
- 1
W1v(\sigma )d\sigma = 2\varepsilon a2v0, (3.11)
where
W1 =
\left(
0 0 0 0
0 0 0 0
S C - S - C
abC - abS abC - abS
\right) .
If
2ab
\pi
/\in \BbbZ , (3.11) is equivalent to (3.8) by Lemma 3.1. If
ab
\pi
\in \BbbZ , c( - 1) exists if and only if
v0 -
1\int
- 1
v1(\sigma ) d\sigma \in \mathrm{i}\mathrm{m}W.
Note that the first coordinate of the left-hand side of the above inclusion is zero. Hence by Lemma 3.1,
first condition of (3.9) has to be satisfied. Moreover, the second coordinate has to be equal to the
fourth, which gives the second condition. Then taking the inverse of W | (kerW )\bot one obtains the
second statement. The third case follows analogously.
Theorem 3.1 is proved.
For the original problem (1.2) we immediately obtain the next statement.
Corollary 3.1. Let the assumptions of Theorem 3.1 be fulfilled. Equation (1.2) has the solution
x(t) =
\left\{
\varphi 1
\biggl(
c
\biggl(
\tau
\biggl(
t
a
\biggr) \biggr)
, \tau
\biggl(
t
a
\biggr) \biggr)
, t \in
\bigcup
k\in \BbbZ ((4k - 1)a, (4k + 1)a),
\varphi 3
\biggl(
c
\biggl(
\tau
\biggl(
t
a
\biggr) \biggr)
, \tau
\biggl(
t
a
\biggr) \biggr)
, t \in
\bigcup
k\in \BbbZ ((4k + 1)a, (4k + 3)a),
where \varphi has coordinates \varphi i, i = 1, . . . , 4.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
ON EQUATIONS WITH GENERALIZED PERIODIC RIGHT-HAND SIDE 267
Proof. Using (2.1) with X =
1
2
(K +L), Y =
1
2
(K - L) and Theorem 3.1 gives the statement.
Example 3.1. Let us consider the equation
\"x(t) + b2x(t) = \varepsilon
\biggl(
\mathrm{s}\mathrm{i}\mathrm{n}2
\pi t
2
+ ( - 1)\lfloor
t+1
2 \rfloor +
\sum
k\in \BbbZ
\delta (t - 4k - 1)
\biggr)
. (3.12)
In this case, the equality
F (t) = \mathrm{s}\mathrm{i}\mathrm{n}2
\pi \tau (t)
2
+ \tau \prime (t) +
1
4
\biggl(
1 + \mathrm{s}\mathrm{i}\mathrm{n}
\pi \tau (t)
2
\biggr)
\tau \prime \prime (t) (3.13)
holds almost everywhere, since
\tau (t) =
\left\{ t - 4k, t \in (4k - 1, 4k + 1], k \in \BbbZ ,
2 - t+ 4k, t \in (4k + 1, 4k + 3], k \in \BbbZ ,
and
\mathrm{s}\mathrm{i}\mathrm{n}
\pi \tau (t)
2
=
\left\{
\mathrm{s}\mathrm{i}\mathrm{n}
\biggl(
\pi t
2
- 2k\pi
\biggr)
= \mathrm{s}\mathrm{i}\mathrm{n}
\pi t
2
, t \in (4k - 1, 4k + 1], k \in \BbbZ ,
\mathrm{s}\mathrm{i}\mathrm{n}
\biggl(
\pi (2k + 1) - \pi t
2
\biggr)
= \mathrm{s}\mathrm{i}\mathrm{n}
\pi t
2
, t \in (4k + 1, 4k + 3], k \in \BbbZ .
Hence
a = 1, Q(\tau ) = \mathrm{s}\mathrm{i}\mathrm{n}2
\pi \tau
2
, P (\tau ) = 1, f(\tau ) =
1
4
\biggl(
1 + \mathrm{s}\mathrm{i}\mathrm{n}
\pi \tau
2
\biggr)
. (3.14)
Function F is sketched in Fig. 1.
For simplicity, we take b =
\pi
4
. Then applying Theorem 3.1,
c( - 1) =
\surd
2\varepsilon
15\pi 2
\bigl(
15\pi - 176, - (15\pi + 64), 15\pi + 64, - 15\pi + 176
\bigr) \ast
.
Consequently,
c(\tau ) =
\varepsilon
15\pi 2
\left(
15
\surd
2\pi + 360S1 - 20S3 - 12S5
- (15\pi + 240)
\surd
2 + 360C1 + 20C3 - 12C5
15
\surd
2\pi - 120S1 - 20S3 - 12S5
- (15\pi - 240)
\surd
2 - 120C1 + 20C3 - 12C5
\right) ,
where Si = \mathrm{s}\mathrm{i}\mathrm{n}
i\pi \tau
4
, Ci = \mathrm{c}\mathrm{o}\mathrm{s}
i\pi \tau
4
for i = 1, 3, 5. By Corollary 3.1, the solution of (3.12) has the
form
x(t) =
\left\{
\varepsilon
15\pi 2
\biggl(
360 + 8 \mathrm{c}\mathrm{o}\mathrm{s}\pi \tau (t) + 15
\surd
2\pi \mathrm{s}\mathrm{i}\mathrm{n}
\pi \tau (t)
4
-
- (15\pi + 240)
\surd
2 \mathrm{c}\mathrm{o}\mathrm{s}
\pi \tau (t)
4
\biggr)
, t \in
\bigcup
k\in \BbbZ (4k - 1, 4k + 1),
\varepsilon
15\pi 2
\biggl(
- 120 + 8 \mathrm{c}\mathrm{o}\mathrm{s}\pi \tau (t) + 15
\surd
2\pi \mathrm{s}\mathrm{i}\mathrm{n}
\pi \tau (t)
4
-
- (15\pi - 240)
\surd
2 \mathrm{c}\mathrm{o}\mathrm{s}
\pi \tau (t)
4
\biggr)
, t \in
\bigcup
k\in \BbbZ (4k + 1, 4k + 3)
(3.15)
(see Fig. 1) with \tau (t) given by (2.2).
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
268 M. FEČKAN, M. POSPÍŠIL
Fig. 1. Sketch of F given by (3.13), and the solution x of (1.2) given by (3.15).
4. Duffing equation. Here we consider equation (1.3) under transformation (2.1). In this case,
system (2.9), (2.10) has the form
a - 2K \prime \prime (\tau ) + b2K3(\tau ) = \varepsilon (Q(\tau ) + P (\tau )),
a - 2L\prime \prime (\tau ) + b2L3(\tau ) = \varepsilon (Q(\tau ) - P (\tau ))
(4.1)
for \tau \in ( - 1, 1) with the mixed boundary conditions
K(\pm 1) - L(\pm 1) = 0, K \prime (\pm 1) + L\prime (\pm 1) = 2\varepsilon a2f(\pm 1). (4.2)
Denoting K1 = K, L1 = L we rewrite (4.1), (4.2) as
K \prime
1(\tau ) = K2(\tau ),
K \prime
2(\tau ) = - a2b2K3
1 (\tau ) + \varepsilon a2(Q(\tau ) + P (\tau )),
L\prime
1(\tau ) = L2(\tau ),
L\prime
2(\tau ) = - a2b2L3
1(\tau ) + \varepsilon a2(Q(\tau ) - P (\tau ))
(4.3)
along with
K1(\pm 1) - L1(\pm 1) = 0, K2(\pm 1) + L2(\pm 1) = 2\varepsilon a2f(\pm 1). (4.4)
Corresponding unperturbed system has the solution
\varphi (c, \tau ) :=
\bigl(
c1 \mathrm{c}\mathrm{n}1(\tau ), - abc21 \mathrm{s}\mathrm{n}1(\tau ) \mathrm{d}\mathrm{n}1(\tau ), c3 \mathrm{c}\mathrm{n}3(\tau ), - abc23 \mathrm{s}\mathrm{n}3(\tau ) \mathrm{d}\mathrm{n}3(\tau )
\bigr) \ast
, (4.5)
where
c = (c1, c2, c3, c4),
\mathrm{c}\mathrm{n}i(\tau ) = \mathrm{c}\mathrm{n}
\bigl(
(ab\tau + ci+1)ci, 1/
\surd
2
\bigr)
,
\mathrm{s}\mathrm{n}i(\tau ) = \mathrm{s}\mathrm{n}
\bigl(
(ab\tau + ci+1)ci, 1/
\surd
2
\bigr)
,
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
ON EQUATIONS WITH GENERALIZED PERIODIC RIGHT-HAND SIDE 269
\mathrm{d}\mathrm{n}i(\tau ) = \mathrm{d}\mathrm{n}
\bigl(
(ab\tau + ci+1)ci, 1/
\surd
2
\bigr)
for i = 1, 3 and \mathrm{c}\mathrm{n}, \mathrm{s}\mathrm{n}, \mathrm{d}\mathrm{n} are Jacobi elliptic functions [5]. The derivative of \varphi (c, \tau ) satisfies
\varphi \prime (c, \tau ) =
\biggl(
\Phi 1 0
0 \Phi 3
\biggr)
, where
\Phi i = \Phi i(c, \tau ) =
\Biggl(
\mathrm{c}\mathrm{n}i - ci(ab\tau + ci+1) \mathrm{s}\mathrm{n}i \mathrm{d}\mathrm{n}i - c2i \mathrm{s}\mathrm{n}i \mathrm{d}\mathrm{n}i
- abci(2 \mathrm{s}\mathrm{n}i \mathrm{d}\mathrm{n}i+ci(ab\tau + ci+1) \mathrm{c}\mathrm{n}
3
i ) - abc3i \mathrm{c}\mathrm{n}3i
\Biggr)
(4.6)
for i = 1, 3. Hence, \mathrm{d}\mathrm{e}\mathrm{t}\varphi \prime (c, \tau ) = a2b2c31c
3
3 \not = 0 if and only if c1 \not = 0 \not = c3. Let us denote
\varphi \prime (c, \tau ) - 1
\left(
0
a2(Q(\tau ) + P (\tau ))
0
a2(Q(\tau ) - P (\tau ))
\right) =
=
\left(
- a(Q(\tau ) + P (\tau )) \mathrm{s}\mathrm{n}1 \mathrm{d}\mathrm{n}1
bc1
a(Q(\tau ) + P (\tau ))( - \mathrm{c}\mathrm{n}1+c1(ab\tau + c2) \mathrm{s}\mathrm{n}1 \mathrm{d}\mathrm{n}1)
bc31
- a(Q(\tau ) - P (\tau )) \mathrm{s}\mathrm{n}3 \mathrm{d}\mathrm{n}3
bc3
a(Q(\tau ) - P (\tau ))( - \mathrm{c}\mathrm{n}3+c3(ab\tau + c4) \mathrm{s}\mathrm{n}3 \mathrm{d}\mathrm{n}3)
bc33
\right)
=: v(c, \tau ).
We want to apply Theorem 2.1, thus we take a1 = - 1, a2 = 1 and consider matrices A, B given
by (3.6).
First, we look for a solution of the unperturbed problem (4.3), (4.4), i.e., of the corresponding
unperturbed problem satisfying zero boundary conditions. So we only have to find c \in \BbbR 4 such that
equations
c1 \mathrm{c}\mathrm{n}1(\pm 1) - c3 \mathrm{c}\mathrm{n}3(\pm 1) = 0,
c21 \mathrm{s}\mathrm{n}1(\pm 1) \mathrm{d}\mathrm{n}1(\pm 1) + c23 \mathrm{s}\mathrm{n}3(\pm 1) \mathrm{d}\mathrm{n}3(\pm 1) = 0
are satisfied for both signs. Immediately, we obtain the trivial solution:
Lemma 4.1. Let a, b > 0 be arbitrary and fixed, and \varphi be defined by (4.5). Then \varphi (0, c2, 0, c4,
\tau ) \equiv 0 for any c2, c4 \in \BbbR is a solution of (4.3), (4.4) with \varepsilon = 0.
Let c1 = c3. Denoting U := (\pm ab + c2)c1, V := (\pm ab + c4)c3 and omitting the argument
k =
1\surd
2
of the elliptic functions we have
\mathrm{c}\mathrm{n}U - \mathrm{c}\mathrm{n}V = 0,
\mathrm{s}\mathrm{n}U \mathrm{d}\mathrm{n}U + \mathrm{s}\mathrm{n}V \mathrm{d}\mathrm{n}V = 0.
(4.7)
Summing the squares of these equations we get
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
270 M. FEČKAN, M. POSPÍŠIL
1 - k2(1 - \mathrm{c}\mathrm{n}2 U)2 - (\mathrm{c}\mathrm{n}U \mathrm{c}\mathrm{n}V - \mathrm{s}\mathrm{n}U \mathrm{s}\mathrm{n}V \mathrm{d}\mathrm{n}U \mathrm{d}\mathrm{n}V ) = 0,
where we used
\mathrm{d}\mathrm{n}2 U = 1 - k2 \mathrm{s}\mathrm{n}2 U = 1 - k2(1 - \mathrm{c}\mathrm{n}2 U) = \mathrm{d}\mathrm{n}2 V
following from the first equation of (4.7). Next, applying [9] (§ 22.21),
\mathrm{c}\mathrm{n}(U + V ) =
\mathrm{c}\mathrm{n}U \mathrm{c}\mathrm{n}V - \mathrm{s}\mathrm{n}U \mathrm{s}\mathrm{n}V \mathrm{d}\mathrm{n}U \mathrm{d}\mathrm{n}V
1 - k2 \mathrm{s}\mathrm{n}2 U \mathrm{s}\mathrm{n}2 V
, (4.8)
we derive
1 - k2(1 - \mathrm{c}\mathrm{n}2 U)2 - \mathrm{c}\mathrm{n}(U + V )
\bigl(
1 - k2 \mathrm{s}\mathrm{n}2 U \mathrm{s}\mathrm{n}2 V
\bigr)
= 0,
i.e., \bigl(
1 - k2(1 - \mathrm{c}\mathrm{n}2 U)2
\bigr) \bigl(
1 - \mathrm{c}\mathrm{n}(U + V )
\bigr)
= 0.
Since the first bracket is nonzero, \mathrm{c}\mathrm{n}(U + V ) = 1. Now we apply the definition of \mathrm{c}\mathrm{n} saying that
\mathrm{c}\mathrm{n}u = \mathrm{c}\mathrm{o}\mathrm{s}\phi , where
u =
\phi \int
0
dt\sqrt{}
1 - k2 \mathrm{s}\mathrm{i}\mathrm{n}2 t
. (4.9)
Hence, \mathrm{c}\mathrm{n}(U + V ) = 1 if and only if
U + V =
2j\pi \int
0
dt\sqrt{}
1 - k2 \mathrm{s}\mathrm{i}\mathrm{n}2 t
= 4j
\pi /2\int
0
dt\sqrt{}
1 - k2 \mathrm{s}\mathrm{i}\mathrm{n}2 t
= 4jK
for some j \in \BbbZ , where K = K(k) is the complete elliptic integral of the first kind. That means,
(2ab+ c2 + c4)c1 = 4iK,
( - 2ab+ c2 + c4)c1 = 4jK
(4.10)
for some integers i, j. Therefore c1 =
(i - j)K
ab
which has to be nonzero. Thus c1 =
mK
ab
for some
m \in \BbbZ \setminus \{ 0\} . Then, from system (4.10),
c4 = 2ab
\biggl(
2i
m
- 1
\biggr)
- c2
for some i \in \BbbZ . Concluding the above, we get the following lemma.
Lemma 4.2. Let a, b > 0, m \in \BbbZ , m \not = 0 be arbitrary and fixed, and \varphi be defined by (4.5).
Then
\varphi
\left( mK
\biggl(
1\surd
2
\biggr)
ab
, c2,
mK
\biggl(
1\surd
2
\biggr)
ab
, - 2ab - c2, \tau
\right)
for any c2 \in \BbbR is a solution of (4.3), (4.4) with \varepsilon = 0.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
ON EQUATIONS WITH GENERALIZED PERIODIC RIGHT-HAND SIDE 271
Proof. Let i \in \BbbZ be arbitrary and fixed. From the above computations we get the solution
\varphi
\left( mK
\biggl(
1\surd
2
\biggr)
ab
, c2,
mK
\biggl(
1\surd
2
\biggr)
ab
, 2ab
\biggl(
2i
m
- 1
\biggr)
- c2, \tau
\right) .
Then using
\mathrm{c}\mathrm{n}(u+ 2jK) = ( - 1)j \mathrm{c}\mathrm{n}(u),
\mathrm{s}\mathrm{n}(u+ 2jK) = ( - 1)j \mathrm{s}\mathrm{n}(u),
\mathrm{d}\mathrm{n}(u+ 2jK) = \mathrm{d}\mathrm{n}(u)
(4.11)
for each j \in \BbbZ , u \in \BbbR , one can prove the statement.
On the other side, let c1 = - c3. Note that by [9] (§ 22.12) functions \mathrm{c}\mathrm{n}, \mathrm{d}\mathrm{n} are even and \mathrm{s}\mathrm{n} is
odd. Hence, from (4.8) we have
\mathrm{c}\mathrm{n}(U - V ) =
\mathrm{c}\mathrm{n}U \mathrm{c}\mathrm{n}V + \mathrm{s}\mathrm{n}U \mathrm{s}\mathrm{n}V \mathrm{d}\mathrm{n}U \mathrm{d}\mathrm{n}V
1 - k2 \mathrm{s}\mathrm{n}2 U \mathrm{s}\mathrm{n}2 V
.
Following the above arguments we derive \mathrm{c}\mathrm{n}(U - V ) = - 1 which holds if and only if U - V =
= 2(1 + 2j)K. Consequently, c1 =
mK
ab
for some m \in \BbbZ \setminus \{ 0\} , and
c4 = 2ab
\biggl(
1 + 2i
m
- 1
\biggr)
- c2.
We summarize this result to a lemma.
Lemma 4.3. Let a, b > 0, m \in \BbbZ , m \not = 0 be arbitrary and fixed, and \varphi be defined by (4.5).
Then
\varphi
\left( mK
\biggl(
1\surd
2
\biggr)
ab
, c2, -
mK
\biggl(
1\surd
2
\biggr)
ab
, 2ab
\biggl(
1
m
- 1
\biggr)
- c2, \tau
\right)
for any c2 \in \BbbR is a solution of (4.3), (4.4) with \varepsilon = 0.
Proof. The statement can be proved as Lemma 4.2.
Remark 4.1. Using the properties (4.11), for the third coordinate of \varphi from Lemma 4.3 we
obtain
\varphi 3
\biggl(
mK
ab
, c2, -
mK
ab
, 2ab
\biggl(
1
m
- 1
\biggr)
- c2, \tau
\biggr)
=
= - mK
ab
\mathrm{c}\mathrm{n}
\biggl( \biggl(
ab\tau + 2ab
\biggl(
1
m
- 1
\biggr)
- c2
\biggr) \biggl(
- mK
ab
\biggr) \biggr)
=
=
mK
ab
\mathrm{c}\mathrm{n}
\biggl( \bigl(
ab\tau - 2ab - c2
\bigr) mK
ab
\biggr)
=
= \varphi 3
\biggl(
mK
ab
, c2,
mK
ab
, - 2ab - c2, \tau
\biggr)
.
Analogously, it can be shown that the fourth coordinates of \varphi s from Lemmas 4.2 and 4.3 are equal.
That means that Lemmas 4.2 and 4.3 give the same solutions.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
272 M. FEČKAN, M. POSPÍŠIL
Denoting \xi (\alpha ) :=
\biggl(
mK
ab
, \alpha ,
mK
ab
, - 2ab - \alpha
\biggr) \ast
, Lemma 4.2 implies that \beta (\xi (\alpha )) = 0 for all
\alpha \in \BbbR . Moreover, \varphi (\xi (\alpha + T ), \cdot ) = \varphi (\xi (\alpha ), \cdot ) for all \alpha \in \BbbR and T =
4ab
| m|
. To verify (C1) one only
needs to investigate \mathrm{k}\mathrm{e}\mathrm{r}D\beta (\xi (\alpha )). Note that since
\beta (\xi ) =
\biggl(
A
0
\biggr)
\varphi (\xi , - 1) +
\biggl(
0
A
\biggr)
\varphi (\xi , 1),
we get
D\beta (\xi ) =
\biggl(
A
0
\biggr)
D\varphi (\xi , - 1) +
\biggl(
0
A
\biggr)
D\varphi (\xi , 1).
Using that D\varphi (\xi , \tau ) is the fundamental matrix solution of the variational equation of unper-
turbed (4.3),
U \prime
1(\tau ) = U2(\tau ),
U \prime
2(\tau ) = - 3a2b2K2
1 (\tau )U1(\tau ),
V \prime
1(\tau ) = V2(\tau ),
V \prime
2(\tau ) = - 3a2b2L2
1(\tau )V1(\tau ),
(4.12)
which leads to (2.28), we see that \mathrm{k}\mathrm{e}\mathrm{r}D\beta (\xi (\alpha )) is given by (2.28) with (2.29), so we can use results
of Section 2. Taking
c1 =
mK
\biggl(
1\surd
2
\biggr)
ab
+ r
in (4.5), we get that its minimal period is Tmin(r) =
4K
\biggl(
1\surd
2
\biggr)
mK
\biggl(
1\surd
2
\biggr)
+ abr
. Thus we take
T (r) = mTmin(r) =
4mK
\biggl(
1\surd
2
\biggr)
mK
\biggl(
1\surd
2
\biggr)
+ abr
.
Clearly T (0) = 4 and T \prime (0) \not = 0, so (2.30) is satisfied, and consequently, assumption (C1) is verified.
Now, instead of calculating \theta (\alpha , \tau ) from (2.21), we derive it as a solution of the adjoint variational
equation. That is the adjoint system to (4.12),
U \prime
1(\tau ) = 3a2b2K2
1 (\tau )U2(\tau ),
U \prime
2(\tau ) = - U1(\tau ),
V \prime
1(\tau ) = 3a2b2L2
1(\tau )V2(\tau ),
V \prime
2(\tau ) = - V1(\tau ),
(4.13)
which leads to (2.28) of the form
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
ON EQUATIONS WITH GENERALIZED PERIODIC RIGHT-HAND SIDE 273
U \prime \prime
2 (\tau ) + 3a2b2K2
1 (\tau )U2(\tau ) = 0,
V \prime \prime
2 (\tau ) + 3a2b2L2
1(\tau )V2(\tau ) = 0.
(4.14)
Furthermore, we derive (see (3.6))
\mathrm{k}\mathrm{e}\mathrm{r}A = \mathrm{k}\mathrm{e}\mathrm{r}B =
\bigl[
(0, - 1, 0, 1)\ast , (1, 0, 1, 0)\ast
\bigr]
.
Hence (2.25) leads to
- U2(\pm 1) + V2(\pm 1) = 0, U1(\pm 1) + V1(\pm 1) = 0,
i.e.,
U2(\pm 1) - V2(\pm 1) = 0, U \prime
2(\pm 1) + V \prime
2(\pm 1) = 0,
which is just (2.29). So by Section 2, we can take U2(\tau ) = K \prime
1(\tau ), V2(\tau ) = K \prime
1( - \tau +2) = - L\prime
1(\tau ).
Using (4.13), we also have U1(\tau ) = - K \prime \prime
1 (\tau ) and V1(\tau ) = L\prime \prime
1(\tau ). We recall that in the notation of
(4.5), we have
\varphi 1(c, \tau ) = K1(\tau ), \varphi 2(c, \tau ) = K \prime
1(\tau ),
\varphi 3(c, \tau ) = L1(\tau ), \varphi 4(c, \tau ) = L\prime
1(\tau ).
Summarizing, in the notation of Section 2, we obtain a1 = - 1, a2 = 1 and
\theta (\alpha , \tau ) =
\bigl(
- \varphi \prime
2(\xi (\alpha ), \tau ), \varphi 2(\xi (\alpha ), \tau ), \varphi
\prime
4(\xi (\alpha ), \tau ), - \varphi 4(\xi (\alpha ), \tau )
\bigr) \ast
,
b1 = 2a(0, f( - 1))\ast , b2 = 2a(0, f(1))\ast ,
g(\tau ) = a2(0, Q(\tau ) + P (\tau ), 0, Q(\tau ) - P (\tau ))\ast ,
A\ast - 1(x1, x2, x3, x4)
\ast = (x1, x2)
\ast for (x1, x2, x3, x4)
\ast \in \mathrm{i}\mathrm{m}A\ast .
(4.15)
Then by (2.25), we derive
\psi 1(\alpha ) =
\Biggl(
\varphi \prime
2(\xi (\alpha ), - 1)
- \varphi 2(\xi (\alpha ), - 1)
\Biggr)
, \psi 2(\alpha ) =
\Biggl(
- \varphi \prime
2(\xi (\alpha ), 1)
\varphi 2(\xi (\alpha ), 1)
\Biggr)
.
Thus formula (2.22) possesses the form
M(\alpha ) =
1\int
- 1
\bigl(
\varphi 2(\xi (\alpha ), s)(Q(s) + P (s)) - \varphi 4(\xi (\alpha ), s)(Q(s) - P (s))
\bigr)
ds+
+2a\varphi 2(\xi (\alpha ), - 1)f( - 1) - 2a\varphi 2(\xi (\alpha ), 1)f(1). (4.16)
Example 4.1. Let us consider the equation
\"x(t) + b2x3(t) = \varepsilon
\biggl(
\mathrm{s}\mathrm{i}\mathrm{n}2
\pi t
2
+ ( - 1)\lfloor
t+1
2 \rfloor +
\sum
k\in \BbbZ
\delta (t - 4k - 1)
\biggr)
. (4.17)
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
274 M. FEČKAN, M. POSPÍŠIL
Then F is given by (3.13) and sketched in Fig. 1; a, Q, P and f are given by (3.14). Since
\varphi i(c, \tau ) = \.\varphi i - 1(c, \tau ) for i = 2, 4 and any c \in \BbbR 4, t \in \BbbR , we obtain
M(\alpha ) =
1\int
- 1
\.\varphi 1(\xi (\alpha ), s)
\biggl(
\mathrm{s}\mathrm{i}\mathrm{n}2
\pi s
2
+ 1
\biggr)
- \.\varphi 3(\xi (\alpha ), s)
\biggl(
\mathrm{s}\mathrm{i}\mathrm{n}2
\pi s
2
- 1
\biggr)
ds - \varphi 2(\xi (\alpha ), 1) =
= 2
\biggl(
\varphi 1(\xi (\alpha ), 1) - \varphi 1(\xi (\alpha ), - 1)
\biggr)
-
- \pi
2
1\int
- 1
\bigl(
\varphi 1(\xi (\alpha ), s) - \varphi 3(\xi (\alpha ), s)
\bigr)
\mathrm{s}\mathrm{i}\mathrm{n}\pi s ds - \varphi 2(\xi (\alpha ), 1). (4.18)
Using
\varphi 3(\xi (\alpha ), t) =
mK
\biggl(
1\surd
2
\biggr)
b
\mathrm{c}\mathrm{n}
\left( (bt - 2b - \alpha )
mK
\biggl(
1\surd
2
\biggr)
b
,
1\surd
2
\right) =
= ( - 1)m
mK
\biggl(
1\surd
2
\biggr)
b
\mathrm{c}\mathrm{n}
\left( (bt - \alpha )
mK
\biggl(
1\surd
2
\biggr)
b
,
1\surd
2
\right)
we can write M as
M(\alpha ) = 2
\bigl(
\varphi 1(\xi (\alpha ), 1) - \varphi 1(\xi (\alpha ), - 1)
\bigr)
-
-
m\pi K
\biggl(
1\surd
2
\biggr)
2b
\bigl(
I+ - ( - 1)mI -
\bigr)
- \varphi 2(\xi (\alpha ), 1)
for
I\pm :=
1\int
- 1
\mathrm{c}\mathrm{n}
\biggl(
(bs\pm \alpha )
mK
\biggl(
1\surd
2
\biggr)
b
,
1\surd
2
\biggr)
\mathrm{s}\mathrm{i}\mathrm{n}\pi s ds =
=
1\pm \alpha
b\int
- 1\pm \alpha
b
\mathrm{c}\mathrm{n}
\biggl(
zmK
\biggl(
1\surd
2
\biggr)
,
1\surd
2
\biggr)
\mathrm{s}\mathrm{i}\mathrm{n}
\biggl(
\pi
\biggl(
z \mp \alpha
b
\biggr) \biggr)
dz.
Now, if m \in \BbbZ \setminus \{ 0\} is even, then \mathrm{c}\mathrm{n} in I\pm is integrated over an integer multiple of its period, i.e.,
I\pm =
1\int
- 1
\mathrm{c}\mathrm{n}
\biggl(
zmK
\biggl(
1\surd
2
\biggr)
,
1\surd
2
\biggr)
\mathrm{s}\mathrm{i}\mathrm{n}
\biggl(
\pi
\biggl(
z \mp \alpha
b
\biggr) \biggr)
dz =
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
ON EQUATIONS WITH GENERALIZED PERIODIC RIGHT-HAND SIDE 275
= \mp \mathrm{s}\mathrm{i}\mathrm{n}
\pi \alpha
b
1\int
- 1
\mathrm{c}\mathrm{n}
\biggl(
zmK
\biggl(
1\surd
2
\biggr)
,
1\surd
2
\biggr)
\mathrm{c}\mathrm{o}\mathrm{s}(\pi z)dz.
To evaluate this integral, we make use of Fourier series expansion of \mathrm{c}\mathrm{n} function from [3] (8.146),
\mathrm{c}\mathrm{n}(zmK) =
2\pi
kK
\infty \sum
j=1
e - \pi (j - 1/2)
1 + e - \pi (2j - 1)
\mathrm{c}\mathrm{o}\mathrm{s}
\biggl(
(2j - 1)\pi mz
2
\biggr)
.
Multiplying this identity by \mathrm{c}\mathrm{o}\mathrm{s}(\pi z) and integrating over ( - 1, 1), we get
I\pm = \mp \mathrm{s}\mathrm{i}\mathrm{n}
\biggl(
\pi \alpha
b
\biggr)
2
\surd
2\pi
K
\biggl(
1\surd
2
\biggr) \infty \sum
j=1
e - \pi (j - 1/2)
1 + e - \pi (2j - 1)
\delta
1,
(2j - 1)| m|
2
=
= \mp \mathrm{s}\mathrm{i}\mathrm{n}
\biggl(
\pi \alpha
b
\biggr) \surd
2\pi
K
\biggl(
1\surd
2
\biggr)
\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}
\biggl(
\pi
2
\biggr) \delta 2,| m| ,
where \delta i,j is the Kronecker symbol. Furthermore,
\varphi 1(\xi (\alpha ), 1) - \varphi 1(\xi (\alpha ), - 1) =
mK
\biggl(
1\surd
2
\biggr)
b
\biggl(
\mathrm{c}\mathrm{n}
\biggl( \biggl(
1 +
\alpha
b
\biggr)
mK
\biggl(
1\surd
2
\biggr) \biggr)
-
- \mathrm{c}\mathrm{n}
\biggl( \biggl(
- 1 +
\alpha
b
\biggr)
mK
\biggl(
1\surd
2
\biggr) \biggr) \biggr)
=
=
mK
\biggl(
1\surd
2
\biggr)
b
(1 - ( - 1)m) \mathrm{c}\mathrm{n}
\biggl( \biggl(
1 +
\alpha
b
\biggr)
mK
\biggl(
1\surd
2
\biggr) \biggr)
= 0.
Therefore,
M(\alpha ) =
\surd
2m\pi 2 \mathrm{s}\mathrm{i}\mathrm{n}
\biggl(
\pi \alpha
b
\biggr)
b \mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}
\biggl(
\pi
2
\biggr) \delta 2,| m| - \varphi 2(\xi (\alpha ), 1). (4.19)
Next, since
d
d\alpha
\varphi i(\xi (\alpha ), t) =
1
b
\.\varphi i(\xi (\alpha ), t) for i = 1, 2 and any \alpha , t \in \BbbR , we get
M \prime (\alpha ) =
\surd
2m\pi 3 \mathrm{c}\mathrm{o}\mathrm{s}
\biggl(
\pi \alpha
b
\biggr)
b2 \mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}
\biggl(
\pi
2
\biggr) \delta 2,| m| -
1
b
\.\varphi 2(\xi (\alpha ), 1).
Note that the periodicity of Jacobi elliptic functions yields that M of (4.19) is
4b
| m|
-periodic. This
means that \alpha +
4jb
| m|
is a root of M whenever j \in \BbbZ and \alpha is a root of M.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
276 M. FEČKAN, M. POSPÍŠIL
Now, we look for roots of M given by (4.19). If m \not = \pm 2, then
M(\alpha ) =
m2K2
\biggl(
1\surd
2
\biggr)
b
\mathrm{s}\mathrm{n}
\biggl( \biggl(
1 +
\alpha
b
\biggr)
mK
\biggl(
1\surd
2
\biggr)
,
1\surd
2
\biggr)
\times
\times \mathrm{d}\mathrm{n}
\biggl( \biggl(
1 +
\alpha
b
\biggr)
mK
\biggl(
1\surd
2
\biggr)
,
1\surd
2
\biggr)
.
Since
\mathrm{d}\mathrm{n}2
\biggl(
t,
1\surd
2
\biggr)
= 1 - 1
2
\mathrm{s}\mathrm{n}2
\biggl(
t,
1\surd
2
\biggr)
\geq 1
2
for all t \in \BbbR , the roots of M are precisely the roots of \mathrm{s}\mathrm{n}
\biggl( \biggl(
1 +
\alpha
b
\biggr)
mK
\biggl(
1\surd
2
\biggr)
,
1\surd
2
\biggr)
. Applying
the definition of \mathrm{s}\mathrm{n} saying that \mathrm{s}\mathrm{n}u = \mathrm{s}\mathrm{i}\mathrm{n}\phi where u is given by (4.9), one can see that \mathrm{s}\mathrm{n}u = 0 if
and only if u = 2jK for j \in \BbbZ . Therefore, M
\bigl(
\alpha 0
j,m
\bigr)
= 0 for
\alpha 0
j,m =
\biggl(
2j
m
- 1
\biggr)
b, j \in \BbbZ , (4.20)
and m \in \BbbZ \setminus \{ 0,\pm 2\} even. Note that all \alpha 0
j,m with j even are just \alpha 0
0,m shifted by an integer multiple
of period
4b
| m|
. Analogously, \alpha 0
j,m with j odd are shifted \alpha 0
1,m. Using \mathrm{c}\mathrm{n}2 u + \mathrm{s}\mathrm{n}2 u = 1 and that
\mathrm{c}\mathrm{n}(t, k) solves the equation
\"x(t) = (2k2 - 1)x(t) - 2k2x3(t), t \in \BbbR , (4.21)
we get at these points
M \prime (\alpha 0
j,m) =
m3K3
\biggl(
1\surd
2
\biggr)
b2
\mathrm{c}\mathrm{n}
\biggl( \biggl(
1 +
\alpha
b
\biggr)
mK
\biggl(
1\surd
2
\biggr)
,
1\surd
2
\biggr)
\not = 0.
Proposition 4.1. For each m \in \BbbZ \setminus \{ 0,\pm 2\} even and j = 0, 1, there exists \delta > 0 such that
(4.17) has a unique solution xj,m given by
xj,m(\varepsilon , t) =
\left\{
mK
\biggl(
1\surd
2
\biggr)
b
\mathrm{c}\mathrm{n}
\left( (b\tau (t) + \alpha 0
j,m)
mK
\biggl(
1\surd
2
\biggr)
b
,
1\surd
2
\right) +O(\varepsilon ),
t \in
\bigcup
k\in \BbbZ (4k - 1, 4k + 1),
mK
\biggl(
1\surd
2
\biggr)
b
\mathrm{c}\mathrm{n}
\biggl(
(b\tau (t) - 2b - \alpha 0
j,m)
mK
\biggl(
1\surd
2
\biggr)
b
,
1\surd
2
\biggr)
+O(\varepsilon ),
t \in
\bigcup
k\in \BbbZ (4k + 1, 4k + 3),
(4.22)
for any \varepsilon \in ( - \delta , \delta ), where \alpha 0
j,m is given by (4.20).
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
ON EQUATIONS WITH GENERALIZED PERIODIC RIGHT-HAND SIDE 277
Proof. By Theorem 2.1 we obtain a solution
\varphi
\left( mK
\biggl(
1\surd
2
\biggr)
b
, \alpha j,m(\varepsilon ),
mK
\biggl(
1\surd
2
\biggr)
b
, - 2b - \alpha j,m(\varepsilon ), \tau
\right) , \alpha j,m(0) = \alpha 0
j,m,
of the corresponding form of (4.3) for \varphi given by (4.5). Then the statement is proved as Corollary 3.1
along with Remark 2.1.
Finally, if | m| = 2, it is not possible to find analytically the roots of
M(\alpha ) =
\surd
2m\pi 2 \mathrm{s}\mathrm{i}\mathrm{n}
\biggl(
\pi \alpha
b
\biggr)
b \mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}
\biggl(
\pi
2
\biggr) +
m2K2
\biggl(
1\surd
2
\biggr)
b
\times
\times \mathrm{s}\mathrm{n}
\biggl( \biggl(
1 +
\alpha
b
\biggr)
mK
\biggl(
1\surd
2
\biggr)
,
1\surd
2
\biggr)
\mathrm{d}\mathrm{n}
\biggl( \biggl(
1 +
\alpha
b
\biggr)
mK
\biggl(
1\surd
2
\biggr)
,
1\surd
2
\biggr)
. (4.23)
However, we can determine the number of its simple roots. Clearly, \alpha is a simple root of M if and
only if it is a simple root of the function \widetilde M(\alpha ) := bM(\alpha b) which is 2-periodic and independent of
b. The graphs of \widetilde M for m = \pm 2 are given in Fig. 2. One can see that there are six simple roots
\alpha 0
j,m, j = 0, 1, . . . , 5, of M in [0, 2b) .
(a) (b)
Fig. 2. Graphs of \widetilde M for m = - 2 (a) and m = 2 (b), respectively.
Proposition 4.2. For each m = \pm 2 and j = 0, 1, . . . , 5, there exists \delta > 0 such that (4.17) has
a unique solution xj,m given by (4.22) for any \varepsilon \in ( - \delta , \delta ) with \alpha 0
j,m numerically computed roots of
M given by (4.23).
Remark 4.2. 1. The unperturbed solutions of (4.17) for m = 2, 4 are sketched in Fig. 3.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
278 M. FEČKAN, M. POSPÍŠIL
Fig. 3. Sketch of solutions x0,m(0, t) of (4.17) for m = 2, 4.
2. If m \not = \pm 2 is even, then we analytically showed that 0 is a simple root of M. Next, from
(4.11), if m = \pm 2 then \mathrm{s}\mathrm{n}(mK) = ( - 1)m/2 \mathrm{s}\mathrm{n}(0) = 0. Hence by (4.23), 0 is a root of M. Moreover,
M \prime (0) =
m
b2
\left(
\surd
2\pi 3
\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}
\biggl(
\pi
2
\biggr) - 4K3
\biggl(
1\surd
2
\biggr) \right) .
= - 8.01858m
b2
\not = 0,
where we used \mathrm{c}\mathrm{n}(mK) = ( - 1)m/2 \mathrm{c}\mathrm{n}(0) = ( - 1)m/2 and (4.21). So, 0 is a simple root of M
whenever m is even.
On the other hand, if m \in \BbbZ is odd, then directly from (4.18),
M(0) = - \varphi 2(\xi (0), 1) =
=
m2K2
\biggl(
1\surd
2
\biggr)
b
\mathrm{s}\mathrm{n}
\biggl(
mK
\biggl(
1\surd
2
\biggr)
,
1\surd
2
\biggr)
\mathrm{d}\mathrm{n}
\biggl(
mK
\biggl(
1\surd
2
\biggr)
,
1\surd
2
\biggr)
=
= ( - 1)(m - 1)/2
m2K2
\biggl(
1\surd
2
\biggr)
\surd
2b
\not = 0,
where we used
\mathrm{s}\mathrm{n}(mK) = \mathrm{s}\mathrm{n}(K + (m - 1)K) = ( - 1)(m - 1)/2 \mathrm{s}\mathrm{n}(K) = ( - 1)(m - 1)/2,
\mathrm{d}\mathrm{n}(K) =
\sqrt{}
1 - k2 =
1\surd
2
(see [9]). So there is no solution persisting from \varphi (\xi (0), t) if m \in \BbbZ is odd.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
ON EQUATIONS WITH GENERALIZED PERIODIC RIGHT-HAND SIDE 279
3. Integrating (4.18) over its period we get
4b
| m| \int
0
M(\alpha )d\alpha =
1\int
- 1
b
\left( \biggl( \mathrm{s}\mathrm{i}\mathrm{n}2
\pi s
2
+ 1
\biggr) 4b/| m| \int
0
d
d\alpha
\varphi 1(\xi (\alpha ), s) d\alpha -
-
\biggl(
\mathrm{s}\mathrm{i}\mathrm{n}2
\pi s
2
- 1
\biggr) 4b/| m| \int
0
d
d\alpha
\varphi 3(\xi (\alpha ), s)d\alpha
\right) ds -
- b
4b/| m| \int
0
d
d\alpha
\varphi 1(\xi (\alpha ), 1) d\alpha = 0.
Since M is not identically zero, it is changing its sign over the period interval, and Remark 2.2 can
be applied. This justifies the above analytical results on the existence of at least one root of M, and
also proves the existence of a solution of (4.17) for \varepsilon close to 0.
References
1. Alekseev V. M. An estimate for the perturbations of the solutions of ordinary differential equations // Vestnik Moskov.
Univ. – 1961. – № 2 (in Russian).
2. Farkas M. Periodic motions. – New York: Springer-Verlag, 1994.
3. Gradshteyn I. S., Ryzhik I. M. Table of integrals, series, and products. – 7th ed. – Acad. Press, 2007.
4. Hartman P. Ordinary differential equations. – New York: John Wiley & Sons, Inc., 1964.
5. Lawden D. F. Elliptic functions and applications. – New York: Springer, 1989.
6. Pilipchuk V. N. Transformation of oscillating systems by means of a pair of nonsmooth periodic functions // Dokl.
Akad. Nauk Ukr. SSR. Ser. A. – 1988. – P. 37 – 40.
7. Pilipchuk V. N. Nonlinear dynamics: between linear and impact limits. – Berlin: Springer-Verlag, 2010.
8. Samoilenko A. M., Boichuk A. A., Zhuravlev V. F. Weakly nonlinear boundary value problems for operator equations
with impulse action // Ukr. Math. J. – 1997. – 49, № 2. – P. 272 – 288.
9. Whittaker E. T., Watson G. N. A course of modern analysis. – 4th ed. – Cambridge: Cambridge Univ. Press, 1927.
Received 31.07.17
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 2
|
| id | umjimathkievua-article-1555 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:08:00Z |
| publishDate | 2018 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/35/57d8632bfcd14ca9629f2f46d7902435.pdf |
| spelling | umjimathkievua-article-15552019-12-05T09:18:03Z On equations with generalized periodic right-hand side Про рiвняння з узагальненою перiодичною правою частиною Fečkan, M. Pospíšil, M. Фечкан, М. Поспісіль, М. Periodic solutions are studied for second-order differential equations with generalized forcing. Analytical bifurcation results are derived with application to forced harmonic and Duffing oscillators. Вивчаються перiодичнi розв’язки для диференцiальних рiвнянь другого порядку з узагальненою примушуючою силою. Аналiтичнi результати для бiфуркацiй отримано та застосовано до вимушених гармонiчних коливань та осцилятора Даффiнга. Institute of Mathematics, NAS of Ukraine 2018-02-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1555 Ukrains’kyi Matematychnyi Zhurnal; Vol. 70 No. 2 (2018); 255-279 Український математичний журнал; Том 70 № 2 (2018); 255-279 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1555/537 Copyright (c) 2018 Fečkan M.; Pospíšil M. |
| spellingShingle | Fečkan, M. Pospíšil, M. Фечкан, М. Поспісіль, М. On equations with generalized periodic right-hand side |
| title | On equations with generalized periodic right-hand side |
| title_alt | Про рiвняння з узагальненою перiодичною
правою частиною |
| title_full | On equations with generalized periodic right-hand side |
| title_fullStr | On equations with generalized periodic right-hand side |
| title_full_unstemmed | On equations with generalized periodic right-hand side |
| title_short | On equations with generalized periodic right-hand side |
| title_sort | on equations with generalized periodic right-hand side |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1555 |
| work_keys_str_mv | AT feckanm onequationswithgeneralizedperiodicrighthandside AT pospisilm onequationswithgeneralizedperiodicrighthandside AT fečkanm onequationswithgeneralizedperiodicrighthandside AT pospísílʹm onequationswithgeneralizedperiodicrighthandside AT feckanm prorivnânnâzuzagalʹnenoûperiodičnoûpravoûčastinoû AT pospisilm prorivnânnâzuzagalʹnenoûperiodičnoûpravoûčastinoû AT fečkanm prorivnânnâzuzagalʹnenoûperiodičnoûpravoûčastinoû AT pospísílʹm prorivnânnâzuzagalʹnenoûperiodičnoûpravoûčastinoû |