Forced frequency locking for differential equations with distributional forcings
This paper deals with forced frequency locking, i.e., the behavior of periodic solutions to autonomous differential equations under the influence of small periodic forcings. We show that, although the forcings are allowed to be discontinuous (e.g., step-function-like) or even distributional (e.g., D...
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2018
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Репозитарії
Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860507341767573504 |
|---|---|
| author | Recke, L. Реке, Л. |
| author_facet | Recke, L. Реке, Л. |
| author_sort | Recke, L. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2019-12-05T09:17:34Z |
| description | This paper deals with forced frequency locking, i.e., the behavior of periodic solutions to autonomous differential equations
under the influence of small periodic forcings. We show that, although the forcings are allowed to be discontinuous (e.g.,
step-function-like) or even distributional (e.g., Dirac-function-like), the forced frequency locking happens as in the case
of smooth forcings, and we derive formulas for the locking cones and for the asymptotic phases as in the case of smooth
forcings. |
| first_indexed | 2026-03-24T02:07:47Z |
| format | Article |
| fulltext |
UDC 517.9
L. Recke (Humboldt Univ. Berlin, Germany)
FORCED FREQUENCY LOCKING FOR DIFFERENTIAL EQUATIONS
WITH DISTRIBUTIONAL FORCINGS
ВИМУШЕНЕ ЗАМИКАННЯ ЧАСТОТИ ДЛЯ ДИФЕРЕНЦIАЛЬНИХ РIВНЯНЬ
З ДИСТРИБУТИВНИМИ ФОРСУВАННЯМИ
This paper deals with forced frequency locking, i.e., the behavior of periodic solutions to autonomous differential equations
under the influence of small periodic forcings. We show that, although the forcings are allowed to be discontinuous (e.g.,
step-function-like) or even distributional (e.g., Dirac-function-like), the forced frequency locking happens as in the case
of smooth forcings, and we derive formulas for the locking cones and for the asymptotic phases as in the case of smooth
forcings.
Розглянуто вимушене замикання частоти, тобто поведiнку перiодичних розв’язкiв автономних диференцiальних
рiвнянь пiд впливом малих перiодичних форсувань. Показано, що, незважаючи на той факт, що цi форсування
можуть бути розривними (типу схiдчастих функцiй) або навiть дистрибутивними (типу дельта-функцiй), вимуше-
нi замикання частоти вiдбуваються, як i у випадку гладких форсувань, i можна отримати формули для конусiв
замикання та асимптотичних фаз, як i у випадку гладких форсувань.
1. Introduction and main results. This paper is dedicated to A. M. Samoilenko on the occasion
of his 80th birthday. The author gratefully acknowledges many years of friendship and scientific
cooperation with Anatolii Mykhailovych, in particular of scientific cooperation concerning forced
frequency locking [9, 11 – 13].
The following phenomenon is usually called forced frequency locking (or injection locking
or master-slave synchronization or master-slave entrainment): If x0 is a T0-periodic solution to an
autonomous evolution equation and if this equation is forced by a periodic forcing with intensity \varepsilon \approx 0
and period T \approx T0, then generically the following is true: If the pair (\varepsilon , T ) belongs to a certain open
conus-like subset of the plane, then there exist T -periodic solutions x(t) \approx x0(tT0/T + \varphi ) to the
forced equation. Moreover, if x0 is an exponentially orbitally stable periodic solution to the unforced
equation, then at least one of the locked periodic solutions to the forced equation is exponentially
stable.
Forced frequency locking appears in many areas of natural sciences, and it is used in diverse
applications in technology. Moreover, since a long time it is mathematically rigorously described for
ODEs and parabolic PDEs with smooth forcings (see, e.g., [5 – 7, 14, 15, 17]). It turns out that this
phenomenon appears also in dissipative hyperbolic PDEs, functional-differential equations (cf. [8]) as
well as in evolution equations with discontinuous or even Dirac-function-like forcings, but for those
cases there is no rigorous mathematical description available up to now.
As an example, in the present paper we describe forced frequency locking for smooth differential
equations with possibly distributional forcings of the type
\.x = f(x) + \varepsilon g(T ). (1.1)
Here x : \BbbR \rightarrow \BbbR n is a T -periodic function to be determined, f : \BbbR n \rightarrow \BbbR n is a C2-smooth vector
field, \varepsilon > 0 and T > 0 are the intensity and the period of the forcing, respectively, and g(T ) is a
linear functional on the vector space of continuous T -periodic functions \BbbR \rightarrow \BbbR n, which works as
c\bigcirc L. RECKE, 2018
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 1 115
116 L. RECKE
\langle g(T ), y\rangle :=
T\int
0
h(t/T ) \cdot y(t) dt+ d \cdot y(0) for all continuous T -periodic y : \BbbR \rightarrow \BbbR n
with h \in L1((0, 1);\BbbR n) and d \in \BbbR n, where “\cdot ” is the Euclidean scalar product in \BbbR n. Hence, we
suppose that the Dirac-function-like part of the forcing is supported in the times t = 0, \pm 1, \pm 2, . . . ,
and the function h describes the shape of the regular part of the forcing (over one period).
Of course, (1.1) has to be understood in the weak sense: A T -periodic function x \in L\infty (\BbbR ;\BbbR n)
is called a solution to (1.1) if
T\int
0
\bigl(
x(t) \cdot \.y(t) +
\bigl(
f(x(t)) + \varepsilon h(t/T )
\bigr)
\cdot y(t)
\bigr)
dt+ \varepsilon d \cdot y(0) = 0
for all smooth T -periodic y : \BbbR \rightarrow \BbbR n.
(1.2)
In order to describe our results we introduce a new scaled time and new scaled unknown functions
as follows:
tnew :=
1
T
told, xnew(tnew) := xold(told).
Then (1.2) is transformed into
1\int
0
\bigl(
x(t) \cdot \.y(t) + T
\bigl(
f(x(t)) + \varepsilon h(t)
\bigr)
\cdot y(t)
\bigr)
dt+ \varepsilon d \cdot y(0) = 0
for all smooth 1-periodic y : \BbbR \rightarrow \BbbR n.
(1.3)
Problem (1.3) can be formally written in the form
\.x(t) = T
\bigl(
f(x(t)) + \varepsilon h(t)
\bigr)
+ \varepsilon d
\sum
k\in \BbbZ
\delta (t+ k),
where \delta is the Dirac function supported in zero.
Let us formulate our assumptions. We suppose that there exists a non-constant 1-periodic solution
x0 to the unforced problem, i.e., to (1.1) with \varepsilon = 0, or, what is the same, to (1.3) with \varepsilon = 0 and
T = 1:
\.x0(t) = f(x0(t)), x0(t+ 1) = x0(t), \.x0 \not = 0. (1.4)
We are going to show that generically the following is true: For all (\varepsilon , T ) \approx (0, 1), which belong to
a certain open conus-like subset of the plane, there exist 1-periodic solutions
x(t) \approx x0(t+ \varphi ) (1.5)
to (1.3). Moreover, we describe how this conus-like subset of the plane looks like and how the
asymptotic phase \varphi depends on the period T. Finally, we prove certain local uniqueness result
for (1.3). We solve (1.3) for \varepsilon \approx 0 and T \approx 1 and (1.5) by means of a Liapunov – Schmidt reduction,
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 1
FORCED FREQUENCY LOCKING FOR DIFFERENTIAL EQUATIONS WITH DISTRIBUTIONAL FORCINGS 117
by scaling techniques and by means of the implicit function theorem. Remark that we do not use any
results about an initial value problem corresponding to (1.1).
From assumption (1.4) it follows that x = \.x0 is a nontrivial solution to the linear homogeneous
boundary-value problem
\.x(t) = f \prime (x0(t))x(t), x(t+ 1) = x(t). (1.6)
We suppose that
any solution to (1.6) is a scalar multiple of \.x0. (1.7)
In other words, we suppose that the vector space of all solutions to (1.6) is one-dimensional. Therefore
the vector space of all solutions to the adjoint linear homogeneous boundary-value problem
- \.x(t) = f \prime (x0(t))
Tx(t), x(t+ 1) = x(t) (1.8)
is one-dimensional also. Here f \prime (x0(t))T is the transposed to f \prime
\bigl(
x0(t)
\bigr)
matrix. We suppose that
there exists a solution x = x\ast to (1.8) with
1\int
0
x\ast (t) \cdot \.x0(t) dt = 1. (1.9)
In other words, we suppose that the eigenvalue \lambda = 0 to the eigenvalue problem \.x(t) =
\bigl(
\lambda +
+ f \prime (x0(t))
\bigr)
x(t), x(t + 1) = x(t), is not only geometrically simple, but also algebraically simple.
Finally, we introduce a 1-periodic function \Phi : \BbbR \rightarrow \BbbR by
\Phi (\varphi ) := -
1\int
0
x\ast (t+ \varphi ) \cdot h(t) dt - x\ast (\varphi ) \cdot d (1.10)
and, for given \varepsilon 0 > 0 and \tau 0 \in \BbbR , open conus-like sets
K(\varepsilon 0, \tau 0) :=
\bigl\{
(\varepsilon , T ) \in \BbbR 2 : \varepsilon \in (0, \varepsilon 0), T = 1 + \varepsilon \tau , \tau \in (\tau 0 - \varepsilon 0, \tau 0 + \varepsilon 0)
\bigr\}
and the Banach space L\infty
per := \{ x \in L\infty (\BbbR ;\BbbR n) : x(t + 1) = x(t) for almost all t \in \BbbR \} with its
norm \| x\| \infty := esssup
\bigl\{
\| x(t)\| : t \in \BbbR
\bigr\}
. Here \| \cdot \| is the Euclidean norm in \BbbR n.
Now we formulate our results:
Theorem 1.1. Suppose (1.4), (1.7) and (1.9), and let (\varepsilon k, Tk, xk) \in (0,\infty )2 \times L\infty
per, k \in \BbbN , be
a sequence of solutions to (1.3) with
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
\biggl(
\varepsilon k + | Tk - 1| + \mathrm{i}\mathrm{n}\mathrm{f}
\varphi \in \BbbR
\| xk - x0(\cdot - \varphi )\| \infty
\biggr)
= 0.
Then there exist \varphi 0 \in [0, 1] and a subsequence (\varepsilon kl , Tkl , xkl), l \in \BbbN , such that
\mathrm{l}\mathrm{i}\mathrm{m}
l\rightarrow \infty
Tkl - 1
\varepsilon kl
= \Phi (\varphi 0) (1.11)
and
\mathrm{l}\mathrm{i}\mathrm{m}
l\rightarrow \infty
\bigm\| \bigm\| xkl - x0(\cdot + \varphi 0)
\bigm\| \bigm\|
\infty = 0. (1.12)
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 1
118 L. RECKE
Theorem 1.2. Suppose (1.4), (1.7) and (1.9), and let \varphi 0 \in [0, 1] and \tau 0 \in \BbbR be given such that
\Phi (\varphi 0) = \tau 0, \Phi \prime (\varphi 0) \not = 0. (1.13)
Then the following is true:
(i) existence and local uniqueness: there exist \varepsilon 0 > 0 and \delta > 0 such that for all (\varepsilon , T ) \in
\in K(\varepsilon 0, \tau 0) there exists exactly one solution x = \^x(\varepsilon , T ) to (1.3) with\bigm\| \bigm\| x - x0(\cdot + \varphi 0)
\bigm\| \bigm\|
\infty < \delta ;
(ii) smooth dependence: the map (\varepsilon , T ) \in K(\varepsilon 0, \tau 0) \mapsto \rightarrow \^x(\varepsilon , T ) \in L\infty
per is C1-smooth;
(iii) asymptotic behavior: it holds
\mathrm{s}\mathrm{u}\mathrm{p}
(\varepsilon ,T )\in K(\varepsilon 0,\tau 0)
1
\varepsilon
\mathrm{i}\mathrm{n}\mathrm{f}
\varphi \in \BbbR
\bigm\| \bigm\| \^x(\varepsilon , T ) - x0(\cdot + \varphi )
\bigm\| \bigm\|
\infty <\infty ; (1.14)
(iv) asymptotic phases: there exists \^\varphi \in C2
\bigl(
[\tau 0 - \varepsilon 0, \tau 0 + \varepsilon 0];\BbbR
\bigr)
with \^\varphi (\tau 0) = \varphi 0 such that
for all \tau \in (\tau 0 - \varepsilon 0, \tau 0 + \varepsilon ) it holds
\mathrm{l}\mathrm{i}\mathrm{m}
\varepsilon \rightarrow 0
\bigm\| \bigm\| \^x(\varepsilon , 1 + \varepsilon \tau ) - x0(\cdot + \^\varphi (\tau ))
\bigm\| \bigm\|
\infty = 0 and \Phi
\bigl(
\^\varphi (\tau )
\bigr)
= \tau . (1.15)
Remark 1.1. If (\varepsilon , T ) \in K(\varepsilon 0, \tau 0), then T = 1+ \varepsilon \tau with \varepsilon \in (0, \varepsilon 0) and \tau \in (\tau 0 - \varepsilon 0, \tau 0 + \varepsilon ),
i.e.,
\tau =
T - 1
\varepsilon
is a scaled period parameter. Hence, the so-called phase equation \Phi (\varphi ) = \tau describes the relationship
between the scaled period \tau and the corresponding asymptotic phase \varphi = \^\varphi (\tau ) (cf. (1.15)) of the
solution family \^x(\varepsilon , T ).
Remark 1.2. If d = 0 and h \in C1
\bigl(
[0, 1];\BbbR n
\bigr)
with h(0) = h(1) and h\prime (0) = h\prime (1) (i.e., if
the forcing is C1-smooth), and if x0 is an exponentially orbitally stable periodic solution to \.x(t) =
= f(x(t)), then the following is true: if \Phi \prime (\varphi 0) > 0 (or \Phi \prime (\varphi 0) < 0), then \^x(\varepsilon , T ) is an exponentially
stable (or unstable) periodic solution to \.x(t) = T
\bigl(
f(x(t)) + \varepsilon h(t)
\bigr)
(for all (\varepsilon , T ) \in K(\varepsilon 0, \tau 0)
with sufficiently small \varepsilon 0) (cf., e.g., [7], Theorem 3, or [10], Theorem 5.1). In other words: the
phase equation \Phi (\varphi ) = \tau describes not only the relationship between the scaled period and the
corresponding asymptotic phase, but also the stability of the locked periodic solutions \^x(\varepsilon , T ). It is
an open problem if a similar result is true in the case of general distributional forcings.
Remark 1.3. Assertion (ii) of Theorem 1.2 claims that the data-to-solution map \^x of the prob-
lem (1.3) is C1-smooth from K(\varepsilon 0, \tau 0) into L\infty
per. But the corresponding data-to-solution map
(\varepsilon , T ) \in K(\varepsilon 0, \tau 0) \mapsto \rightarrow \~x(\varepsilon , T ) \in L\infty
per with \~x(\varepsilon , T )(t) := \^x(\varepsilon , T )(t/T )
to (1.2) is not smooth, even not continuous, in general!
Remark 1.4. Assertion (iii) of Theorem 1.2 claims that \^x(\varepsilon , 1 + \varepsilon \tau ) tends, for \varepsilon \rightarrow 0, to a
phase shift of x0, and this phase shift depends on \tau . In particular, \^x(\varepsilon , T ) does not converge, for
(\varepsilon , T ) \rightarrow (0, 1) with (\varepsilon , T ) \in K(\varepsilon 0, \tau 0), in L\infty
per.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 1
FORCED FREQUENCY LOCKING FOR DIFFERENTIAL EQUATIONS WITH DISTRIBUTIONAL FORCINGS 119
2. Transformation of (1.3) into an equation in \bfitL \infty
per . In this section we show that the
variational problem (1.3) is equivalent to a smooth equation in the Banach space L\infty
per. For that
reason we introduce a function G : \BbbR 2 \rightarrow \BbbR , which is 1-periodic with respect to both its variables,
by defining it in the square [0, 1]\times [0, 1] as below and then extending it 1-periodically on \BbbR 2 :
G(t, s) :=
\left\{
e
1 - e
et - s for 0 \leq t \leq s \leq 1,
1
1 - e
et - s for 0 \leq s < t \leq 1.
It is easy to verify that G is a Green’s function, i.e., for any 1-periodic functions y, z : \BbbR \rightarrow \BbbR n it
holds
\.y - y = z and only if y(t) =
1\int
0
G(t, s)z(s) ds (2.1)
and
1\int
0
\left( ( \.y(t) - y(t))
1\int
0
G(s, t)z(s) ds
\right) dt = 1\int
0
y(t)z(t) dt. (2.2)
Lemma 2.1. A function x \in L\infty
per is a solution to (1.3) if and only if for almost all t \in \BbbR it holds
x(t) +
1\int
0
G(s, t) (x(s) + T (f(x(s)) + \varepsilon h(s))) ds+ \varepsilon G(0, t) d = 0. (2.3)
Proof. Take x \in L\infty
per and smooth 1-periodic functions y, z : \BbbR \rightarrow \BbbR n with \.y - y = z. Then (2.1)
yields
1\int
0
\Bigl(
x(t) \cdot \.y(t) + T
\bigl(
f(x(t)) + \varepsilon h(t)
\bigr)
\cdot y(t)
\Bigr)
dt+ \varepsilon d \cdot y(0) =
=
1\int
0
\Bigl(
x(t) \cdot
\bigl(
\.y(t) + y(t)
\bigr)
+
\bigl(
x(t) + T (f(x(t)) + \varepsilon h(t))
\bigr)
\cdot y(t)
\Bigr)
dt+ \varepsilon d \cdot y(0) =
=
1\int
0
\left( x(t) \cdot z(t) + \bigl(
x(t) + T
\bigl(
f(x(t)) + \varepsilon h(t)
\bigr) \bigr)
\cdot
1\int
0
G(t, s)z(s) ds
\right) dt+ \varepsilon d \cdot y(0) =
=
1\int
0
z(s) \cdot
\left( x(s) + 1\int
0
G(t, s)
\bigl(
x(t) + T
\bigl(
f(x(t)) + \varepsilon h(t)
\bigr) \bigr)
ds+ \varepsilon G(0, s)d
\right) ds.
Hence, x is a solution to (1.3) iff the left-hand side vanishes for all smooth 1-periodic y : \BbbR \rightarrow \BbbR n,
and this is the case iff the right-hand side vanishes for all smooth 1-periodic z : \BbbR \rightarrow \BbbR n, i.e., iff (2.3)
holds for almost all t \in \BbbR .
Lemma 2.1 is proved.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 1
120 L. RECKE
It follows from Lemma 2.1 that any solution x \in L\infty
per to (1.3), i.e., to (2.3), is continuous up to
the points t = k, k \in \BbbZ , where it jumps from x(k - 0) to
x(k + 0) = x(k - 0) - \varepsilon d.
In order to analyze equation (2.3), let us introduce maps \scrF \in C2(\BbbR \times L\infty
per;L
\infty
per) and \scrG \in
\in C\infty (\BbbR ;L\infty
per) by defining for almost all t \in \BbbR
\scrF (T, x)(t) :=
1\int
0
G(s, t)
\bigl(
x(s) + Tf(x(s))
\bigr)
ds,
\scrG (T )(t) := T
1\int
0
G(s, t)h(s) ds+G(0, t)d.
Then the variational problem (1.3), which is equivalent to the integral equation (2.3), is equivalent to
the abstract equation
x+ \scrF (T, x) + \varepsilon \scrG (T ) = 0. (2.4)
For \varphi \in \BbbR we define S\varphi \in \scrL (L\infty
per) by
S\varphi x(t) := x(t+ \varphi ) for almost all t \in \BbbR .
The map \varphi \mapsto \rightarrow S\varphi is a representation of the rotation group SO(2) on the vector space L\infty
per, but this
representation is not strongly continuous because the map \varphi \in \BbbR \mapsto \rightarrow S\varphi x \in L\infty
per is continuous if and
only if the function x is continuous, i.e., not for all x \in L\infty
per. But the maps \varphi \in \BbbR \mapsto \rightarrow S\varphi x0 \in L\infty
per
and \varphi \in \BbbR \mapsto \rightarrow S\varphi x\ast \in L\infty
per are C3-smooth and C2-smooth, respectively, because the functions x0
and x\ast are C3-smooth and C2-smooth, respectively (because the vector field f is supposed to be
C2-smooth). This will be used repeatedly in what follows.
It is easy to verify that
S\varphi \scrF (T, x) = \scrF (T, S\varphi x) for all T, \varphi \in \BbbR and x \in L\infty
per. (2.5)
Because of (2.5) equation (2.4) is a symmetry breaking problem. If x is a solution to (2.4) with
\varepsilon = 0, then S\varphi x is a solution also for all \varphi \in \BbbR . But for \varepsilon \not = 0 this is not the case, in general.
We are going to use well-known techniques for treating symmetry breaking problems, which are
developed, e.g., in [1 – 3, 10, 16]. The main ingredient for that is the Fredholm property of the
operator I + \partial x\scrF (1, x0).
Lemma 2.2. The operator \partial x\scrF (1, x0) is completely continuous from L\infty
per into L\infty
per, and it
holds
\mathrm{k}\mathrm{e}\mathrm{r}
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
= \mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\{ \.\mathrm{x}0\} , (2.6)
\mathrm{i}\mathrm{m}
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
=
\left\{ x \in L\infty
per :
1\int
0
( \.x\ast (t) - x\ast (t)) \cdot x(t) dt = 0
\right\} . (2.7)
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 1
FORCED FREQUENCY LOCKING FOR DIFFERENTIAL EQUATIONS WITH DISTRIBUTIONAL FORCINGS 121
Proof. We have
\bigl(
\partial x\scrF (1, x0)x
\bigr)
(t) =
\int 1
0
G(s, t)
\bigl(
I + f \prime (x0(s))
\bigr)
x(s) ds. Therefore (2.1) yields
d
dt
\partial x\scrF (1, x0)x = \partial x\scrF (1, x0)x+
\bigl(
I + f \prime (x0)
\bigr)
x. (2.8)
Hence, \partial x\scrF (1, x0) is a linear bounded operator from L\infty
per into the function space
W 1,\infty
per :=
\bigl\{
x \in L\infty
per : \.x \in L\infty
per
\bigr\}
with norm \| x\| \infty + \| \.x\| \infty .
But W 1,\infty
per is compactly embedded into L\infty
per. Hence, \partial x\scrF (1, x0) is completely continuous from L\infty
per
into L\infty
per.
Because of x0 + \scrF (1, x0) = 0 and of (2.5) we have S\varphi x0 + \scrF (1, S\varphi x0) = 0 for all \varphi \in \BbbR .
Differentiating this identity with respect to \varphi in \varphi = 0 we get \.x0 \in ker (I + \partial x\scrF (1, x0)).
Now, take x \in ker
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
, i.e., x + \partial x\scrF (1, x0)x = 0. Then x \in W 1,\infty
per , and (2.8)
yields
d
dt
\partial x\scrF (1, x0)x = \partial x\scrF (1, x0)x+
\bigl(
I + f \prime (x0)
\bigr)
x = f \prime (x0)x.
Hence, assumption (1.7) yields (2.6).
In order to prove (2.7) we consider the Hilbert space L2
per with its scalar product (x, y) :=
:=
\int 1
0
x(t) \cdot y(t) dt. Because of the continuous and dense embedding L\infty
per \lhook \rightarrow L2
per it follows that
L2
per is continuously and densily embedded into the dual space (L\infty
per)
\ast , where a function x \in L2
per
has to be understood as an element of (L\infty
per)
\ast by means of
\langle x, y\rangle := (x, y) for all y \in L\infty
per,
where \langle \cdot , \cdot \rangle : (L\infty
per)
\ast \times L\infty
per \rightarrow \BbbR is the dual pairing. The Fredholmness of I + \partial x\scrF (1, x0) yields
\mathrm{i}\mathrm{m}
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
=
\Bigl\{
x \in L\infty
per : \langle \phi , y\rangle = 0 for all \phi \in \mathrm{k}\mathrm{e}\mathrm{r}
\bigl(
I + \partial x\scrF (1, x0)
\bigr) \ast \Bigr\}
,
\mathrm{d}\mathrm{i}\mathrm{m}\mathrm{k}\mathrm{e}\mathrm{r}
\bigl(
I + \partial xF (1, x0)
\bigr) \ast
= 1.
Hence, in oder to prove (2.7) we have to prove that \.x\ast - x\ast \in \mathrm{k}\mathrm{e}\mathrm{r}
\bigl(
I+\partial x\scrF (1, x0)
\bigr) \ast
. But this is easy
to verify because for any smooth 1-periodic function y : \BbbR \rightarrow \BbbR n we have\bigl\langle
(I + \partial x\scrF (1, x0))
\ast ( \.x\ast - x\ast ), y
\bigr\rangle
=
\bigl\langle
\.x\ast - x\ast , (I + \partial x\scrF (1, x0)y
\bigr\rangle
=
=
\bigl(
\.x\ast - x\ast , (I + \partial x\scrF (1, x0)y
\bigr)
=
=
1\int
0
\bigl(
\.x\ast (t) - x\ast (t)
\bigr)
\cdot
\left( y(t) + 1\int
0
G(s, t)
\bigl(
y(s) + f \prime (x0(s))y(s)
\bigr)
ds
\right) dt =
=
1\int
0
x\ast (t) \cdot
\bigl(
- \.y(t) + f \prime (x0(t))y(t)
\bigr)
dt =
1\int
0
y(t) \cdot
\bigl(
\.x\ast (t) + f \prime (x0(s))
Tx\ast (t)
\bigr)
dt = 0.
Here we used (1.9) and (2.2).
Lemma 2.2 is proved.
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122 L. RECKE
3. A parametrization of a neighbourhood of \bfitx 0 . Let us define a closed codimension one
subspace Y of L\infty
per by
Y :=
\left\{ y \in L\infty
per :
1\int
0
y(t) \cdot x\ast (t) dt = 0
\right\} .
In this section we show that any small neighbourhood of x0 in L\infty
per can be smoothly parametrized
by small (\varphi , y) \in \BbbR \times Y via x = S\varphi x0 + y.
Lemma 3.1. The map (\varphi , y) \in \BbbR \times Y \mapsto \rightarrow f(\varphi , y) := S\varphi x0 + y \in L\infty
per is a diffeomorphism of a
neighbourhood of zero in \BbbR \times Y onto a neighbourhood of x0 in L\infty
per.
Proof. It holds f(0, 0) = x0 and f \prime (0, 0)(\varphi , y) = - \varphi \.x0 + y for all (\varphi , y) \in \BbbR \times Y. On the
other hand, assumption (1.9) yields that \.x0 /\in Y. Hence,
L\infty
per = span\{ \.x0\} \oplus Y. (3.1)
Therefore f \prime (0, 0) is bijective from \BbbR \times Y onto L\infty
per. Hence, the local diffeomorphism theorem
yields the claim.
Lemma 3.1 is proved.
4. An a priori estimate. If x \approx \{ S\psi x0 : \psi \in \BbbR \} is a solution to (2.4), then there exists
\psi \in [0, 1] with x \approx S\psi x0 and, hence, S - \psi x \approx x0. Therefore Lemma 3.1 yields that there exist
\varphi \approx 0 and y \in Y with y \approx 0 such that S - \psi x = S\varphi x0 + y, i.e.,
x = S\psi (S\varphi x0 + y).
Inserting this into (2.4) and using (2.5) we get
S\varphi x0 + y + \scrF (T, S\varphi x0 + y) + \varepsilon S - \psi \scrG (T ) = 0. (4.1)
Lemma 4.1. For all \varphi , T \in \BbbR and y \in L\infty
per it holds
S\varphi x0 + y + \scrF (T, S\varphi x0 + y) =
\left( I + 1\int
0
\partial x\scrF (1, S\varphi x0 + ry)dr
\right) y + (T - 1)\partial T\scrF (1, S\varphi x0 + y).
Proof. We have S\varphi x0 + \scrF (T, S\varphi x0) = 0 for all \varphi \in \BbbR . Therefore
S\varphi x0 + y + \scrF (T, S\varphi x0 + y) =
\left( I + 1\int
0
\partial x\scrF (1, S\varphi x0 + ry)dr
\right) y+
+(T - 1)
1\int
0
\partial T\scrF (1 + r(T - 1), S\varphi x0 + y)dr.
But \scrF (\cdot , S\varphi x0 + y) is affine, hence
1\int
0
\partial T\scrF
\bigl(
1 + r(T - 1), S\varphi x0 + y
\bigr)
dr = \partial T\scrF (1, S\varphi x0 + y).
Lemma 4.1 is proved.
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FORCED FREQUENCY LOCKING FOR DIFFERENTIAL EQUATIONS WITH DISTRIBUTIONAL FORCINGS 123
Lemma 4.2. There exist \delta > 0 and c > 0 such that for all solutions (\varepsilon , T, \varphi , \psi , y) \in (0,\infty )2 \times
\times [0, 1]2 \times Y to (4.1) with \varepsilon + | T - 1| + \| y\| \infty < \delta it holds
| T - 1| + \| y\| \infty < c\varepsilon .
Proof. Suppose the contrary. Then there exist solutions (\varepsilon k, Tk, \varphi k, \psi k, yk) \in (0,\infty )2\times [0, 1]2\times
\times Y, k = 1, 2, . . . , to (4.1) with
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
\biggl(
\varepsilon k + | \varphi k| + | Tk - 1| + \| yk\| \infty +
\varepsilon k
| Tk - 1| + \| yk\| \infty
\biggr)
= 0. (4.2)
Because of Lemma 4.1 it holds\left( I + 1\int
0
\partial x\scrF (1, S\varphi k
x0 + ryk)dr
\right) yk
| Tk - 1| + \| yk\| \infty
+
+
Tk - 1
| Tk - 1| + \| yk\| \infty
\partial T\scrF (1, S\varphi k
x0 + yk) = - \varepsilon k
| Tk - 1| + \| yk\| \infty
S - \psi k
\scrG (Tk). (4.3)
Without loss of generality we may assume that there exists \tau \in \BbbR with
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
Tk - 1
| Tk - 1| + \| yk\| \infty
= \tau . (4.4)
Moreover, because the operator \partial x\scrF (1, x0) is completely continuous, without loss of generality
we may assume that the sequence \partial x\scrF (1, x0)y/
\bigl(
| Tk - 1| + \| yk\| \infty
\bigr)
, k \in \BbbN , converges in L\infty
per.
Hence, (4.2) – (4.4) yield that there exists y \in Y with
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
yk
| Tk - 1| + \| yk\| \infty
= y (4.5)
and \left( I + 1\int
0
\partial x\scrF (1, x0)dr
\right) y + \tau \partial T\scrF (1, x0) = 0. (4.6)
Because of (1.9), (2.2) and (2.7) it follows
0 = \tau
1\int
0
\partial T\scrF (1, x0) \cdot ( \.x\ast - x\ast ) dt =
= \tau
1\int
0
\left( 1\int
0
G(s, t)f(x0(s)) ds
\right) \cdot
\bigl(
\.x\ast (t) - x\ast (t)
\bigr)
dt =
= \tau
1\int
0
f(x0(t)) \cdot x\ast (t) dt = \tau
1\int
0
\.x0(t) \cdot x\ast (t) dt = \tau . (4.7)
Therefore (2.6) and (4.6) imply that y \in Y \cap \mathrm{k}\mathrm{e}\mathrm{r}(I + \partial x\scrF (1, x0)) = Y \cap span \{ \.x0\} , and (3.1) yields
y = 0.
Let us summarize: We got \tau = 0 and y = 0. But from (4.4) and (4.5) it follows | \tau | + \| y\| \infty = 1,
this is the needed contradiction.
Lemma 4.2 is proved.
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124 L. RECKE
5. Proof of Theorem 1.1. Suppose (1.4), (1.7) and (1.9), and take a sequence (\varepsilon k, Tk, xk) \in
\in (0,\infty )2 \times L\infty
per, k \in \BbbN , with
xk + \scrF (Tk, xk) + \varepsilon k\scrG (Tk) = 0 (5.1)
and
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
\biggl(
\varepsilon k + | Tk - 1| + \mathrm{i}\mathrm{n}\mathrm{f}
\psi \in \BbbR
\| xk - S\psi x0\| \infty
\biggr)
= 0. (5.2)
For any k there exist \psi k \in [0, 1] with
\mathrm{i}\mathrm{n}\mathrm{f}
\psi \in \BbbR
\| xk - S\psi x0\| \infty = \| xk - S\psi k
x0\| \infty = \| S - \psi k
xk - x0\| \infty , (5.3)
and without loss of generality we may assume that there exists \varphi 0 \in [0, 1] with
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
\psi k = \varphi 0. (5.4)
Hence, (5.3) yields \| xk - S\varphi 0\| \infty \rightarrow 0 for k \rightarrow \infty , i.e., assertion (1.12) of Theorem 1.1 is proved.
Because of Lemma 3.1 and of (5.3) for large k there exist \varphi k \in [0, 1] and yk \in Y with
S - \psi k
xk = S\varphi k
x0 + yk and \mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
=
\bigl(
| \varphi k| + \| yk\| \infty
\bigr)
= 0.
Inserting xk = S\psi k
(S\varphi k
x0 + yk) into (5.1) and using (2.5) we get
S\varphi k
x0 + yk + \scrF (Tk, S\varphi k
x0 + yk) + \varepsilon kS - \psi k
\scrG (Tk) = 0. (5.5)
Further, from Lemma 4.2 it follows that there exists a bounded sequence (\tau k, zk) \in \BbbR \times Y, k \in \BbbN ,
such that Tk = 1+\varepsilon k\tau k and yk = \varepsilon kzk. Inserting this into (5.5), dividing by \varepsilon k and using Lemma 4.1
we get\left( I + 1\int
0
\partial x\scrF (1, S\varphi k
x0 + r\varepsilon kzk)dr
\right) zk + \tau k\partial T\scrF (1, S\varphi k
x0 + \varepsilon kzk) = - S - \psi k
\scrG (1 + \varepsilon k\tau k). (5.6)
Without loss of generality we may assume that there exists \tau 0 \in \BbbR such that \tau k \rightarrow \tau 0 for k \rightarrow \infty .
Moreover, (2.7) yields
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
1\int
0
( \.x\ast - x\ast ) \cdot
\left( zk + 1\int
0
\partial x\scrF (1, S\varphi k
x0 + r\varepsilon kzk) drzk
\right) dt = 0,
and (4.7) implies
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
1\int
0
( \.x\ast - x\ast ) \cdot \partial T\scrF (1, S\varphi k
x0 + \varepsilon kzk) dt = 1.
Hence, from (1.10), (2.2), (5.4) and (5.6) it follows
\tau 0 = - \mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
1\int
0
S - \psi k
\scrG (1 + \varepsilon k\tau k) \cdot ( \.x\ast - x\ast ) dt =
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FORCED FREQUENCY LOCKING FOR DIFFERENTIAL EQUATIONS WITH DISTRIBUTIONAL FORCINGS 125
= - \mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
1\int
0
\scrG (1 + \varepsilon k\tau k) \cdot S\psi k
( \.x\ast - x\ast ) dt = -
1\int
0
\scrG (1) \cdot S\varphi 0( \.x\ast - x\ast ) dt =
= -
1\int
0
\left( 1\int
0
G(s, t)h(s) ds+G(0, t)d
\right) \cdot
\bigl(
\.x\ast (t+ \varphi 0) - x\ast (t \ast \varphi 0)
\bigr)
dt =
= -
1\int
0
h(t) \cdot x\ast (t) dt -
1\int
0
e
1 - e
e - td \cdot
\bigl(
\.x\ast (t+ \varphi 0) - x\ast (t+ \varphi 0)
\bigr)
dt =
= -
1\int
0
h(t) \cdot x\ast (t) dt - d \cdot x\ast (\varphi 0) = \Phi (\varphi 0). (5.7)
Therefore, assertion (1.11) of Theorem 1.1 is proved.
6. Proof of Theorem 1.2. Suppose (1.4), (1.7) and (1.9), and let \varphi 0 \in [0, 1] and \tau 0 \in \BbbR be
given such that (1.13) is true.
We have to determine all solutions x \approx S\varphi 0x0 to (2.4) with (\varepsilon , T ) \in K(\varepsilon 0, \tau 0) and \varepsilon 0 \approx 0.
Because of Lemma 3.1 we are allowed to make the ansatz
x = S\varphi 0(S\varphi x0 + y) with \varphi \approx 0, y \approx 0, y \in Y.
Inserting this ansatz into (2.4) we get
S\varphi x0 + y + \scrF (T, S\varphi x0 + y) + \varepsilon S - \varphi 0\scrG (T ) = 0. (6.1)
Further, because of Lemma 4.2 we are allowed to make the ansatz
T = 1 + \varepsilon \tau , y = \varepsilon z,
and we get, after deviding by \varepsilon and using Lemma 4.1,\left( I + 1\int
0
\partial x\scrF (1, S\varphi x0 + r\varepsilon z)dr
\right) z + \tau \partial T\scrF (1, S\varphi x0 + \varepsilon z) + S - \varphi 0\scrG (1 + \varepsilon \tau ) = 0. (6.2)
We are going to solve equation (6.2) with respect to (\varphi , z) \approx (\varphi 0, z0) (z0 \in Y is defined below,
see (6.6)) for given (\varepsilon , \tau ) \approx (0, \tau 0) by means of the implicit function theorem. Remark that \varphi 0 and
\tau 0 are given by assumption (1.13).
Let us define x1 \in L\infty
per by x1(t) :=
\int 1
0
G(s, t) \.x0(s) ds. Then (1.9) and (2.2) imply
1\int
0
x1(t) \cdot
\bigl(
\.x\ast (t) - x\ast (t)
\bigr)
dt =
1\int
0
\.x0 \cdot x\ast (t) dt = 1.
Hence, (2.7) yields
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126 L. RECKE
L\infty
per = \mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n} \{ x1\} \oplus \mathrm{i}\mathrm{m}
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
, (6.3)
and P = P 2 \in \scrL (L\infty
per), defined by Px :=
\int 1
0
x(t) \cdot ( \.x\ast (t) - x\ast (t)) dtx1, is the projection cor-
responding to the topological sum (6.3), i.e.,
\mathrm{k}\mathrm{e}\mathrm{r}P = \mathrm{i}\mathrm{m}
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
, \mathrm{i}\mathrm{m}P = span \{ x1\} .
Finally, for \varphi \in \BbbR we define L\varphi \in \scrL
\bigl(
L\infty
per;\BbbR \times \mathrm{i}\mathrm{m}(I + \partial x\scrF (1, x0))
\bigr)
by
L\varphi x :=
\left( 1\int
0
x \cdot S\varphi ( \.x\ast - x\ast ) dt, (I - P )x
\right) .
Lemma 6.1. (i) The operator I + \partial x\scrF (1, x0) is bijective from Y onto \mathrm{i}\mathrm{m}
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
.
(ii) There exists \delta > 0 such that for all \varphi \in \BbbR with | \varphi | < \delta the operator L\varphi is bijective from
L\infty
per onto \BbbR \times \mathrm{i}\mathrm{m}
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
.
Proof. (i) If for y \in Y we have
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
y = 0, i.e., y \in Y \cap \mathrm{k}\mathrm{e}\mathrm{r}
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
=
= Y \cap span \{ \.x0\} , then (3.1) yields y = 0. Hence, I + \partial x\scrF (1, x0) is injective.
If \~x \in \mathrm{i}\mathrm{m}
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
is given, then there exists x \in L\infty
per with \~x =
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
x.
Moreover, because of (3.1) there exist \xi \in \BbbR and y \in Y with x = \xi \.x0 + y. Because of (2.6) it
follows
\~x =
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
(\xi \.x0 + y) =
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
y.
Hence, I + \partial x\scrF (1, x0) is surjective from Y onto \mathrm{i}\mathrm{m}
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
.
(ii) If L0x = 0, then Px = (I - P )x = 0, i.e., x = 0. Hence, L0 is injective. Moreover, for
arbitrary (\xi , \~x) \in \BbbR \times \mathrm{i}\mathrm{m}
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
it holds
L0(\xi x1 + \~x) =
\left( 1\int
0
(\xi x1 + \~x) \cdot ( \.x\ast - x\ast ) dt, (I - P )(\xi x1 + \~x)
\right) = (\xi , \~x).
Hence, L0 is surjective from L\infty
per onto \BbbR \times \mathrm{i}\mathrm{m}
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
.
But the map \varphi \in \BbbR \mapsto \rightarrow L\varphi \in \scrL
\bigl(
(L\infty
per;\BbbR \times \mathrm{i}\mathrm{m}(I+\partial x\scrF (1, x0))
\bigr)
is continuous (even C1-smooth)
with respect to the operator norm, and the set of all bijective maps is open in \scrL
\bigl(
(L\infty
per;\BbbR \times \mathrm{i}\mathrm{m}(I +
+ \partial x\scrF (1, x0))
\bigr)
. Hence, assertion (ii) is proved.
Lemma 6.1 is proved.
Define by \scrH (\varepsilon , \tau , \varphi , z) the left-hand side of (6.2). Because of Lemma 6.1 (ii) the equation (6.2)
with \varphi \approx 0 is equivalent to
L\varphi \scrH (\varepsilon , \tau , \varphi , z) =:
\bigl(
\scrH 1(\varepsilon , \tau , \varphi , z),\scrH 2(\varepsilon , \tau , \varphi , z)
\bigr)
= 0. (6.4)
The maps \scrH 1 \in C1
\bigl(
(0,\infty ) \times \BbbR 2 \times Y ;\BbbR
\bigr)
and \scrH 2 \in C1
\bigl(
(0,\infty ) \times \BbbR 2 \times Y ; \mathrm{i}\mathrm{m}(I + \partial x\scrF (1, x0))
\bigr)
,
which are defined in (6.4), satisfy
\scrH 1(0, \tau , \varphi , z) =
1\int
0
\Bigl( \bigl(
I + \partial x\scrF (1, S\varphi x0)
\bigr)
z + \tau \partial T\scrF (1, S\varphi x0) + S - \varphi 0\scrG (1)
\Bigr)
\cdot S\varphi ( \.x\ast - x\ast ) dt,
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FORCED FREQUENCY LOCKING FOR DIFFERENTIAL EQUATIONS WITH DISTRIBUTIONAL FORCINGS 127
\scrH 2(0, \tau , \varphi , z) = (I - P )
\Bigl( \bigl(
I + \partial x\scrF (1, S\varphi x0)
\bigr)
z + \tau \partial T\scrF (1, S\varphi x0) + S - \varphi 0\scrG (1)
\Bigr)
.
But (2.5) yields S\varphi \partial x\scrF (1, x0) = \partial x\scrF (1, S\varphi x0)S\varphi and \partial T\scrF (1, S\varphi x0) = S\varphi \partial T\scrF (1, x0). Therefore
1\int
0
\bigl(
I + \partial x\scrF (1, S\varphi x0)
\bigr)
z \cdot S\varphi ( \.x\ast - x\ast ) dt =
=
1\int
0
S - \varphi
\bigl(
I + \partial x\scrF (1, S\varphi x0)
\bigr)
z \cdot ( \.x\ast - x\ast ) dt =
=
1\int
0
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
S\varphi z \cdot ( \.x\ast - x\ast ) dt = 0
and (cf. (4.7))
1\int
0
\partial T\scrF (1, S\varphi x0) \cdot S\varphi ( \.x\ast - x\ast ) dt =
=
1\int
0
S - \varphi \partial T\scrF (1, S\varphi x0) \cdot ( \.x\ast - x\ast ) dt =
=
1\int
0
\partial T\scrF (1, x0) \cdot ( \.x\ast - x\ast ) dt = 1
and (cf. (5.7))
1\int
0
S - \varphi 0\scrG (1) \cdot S\varphi ( \.x\ast - x\ast ) dt = - \Phi (\varphi 0 + \varphi ).
Hence,
\scrH 1(0, \tau , \varphi , z) = \tau - \Phi (\varphi 0 + \varphi ). (6.5)
Therefore \varepsilon = 0, \tau = \tau 0, \varphi = 0, z = z0 is a solution to (6.4), i.e., to (6.2) with
z0 := -
\bigl(
I + \partial x\scrF (1, x0)
\bigr) - 1
(I - P )
\bigl(
\tau 0\partial T\scrF (1, x0) + S - \varphi 0\scrG (1)
\bigr)
. (6.6)
Here
\bigl(
I + \partial x\scrF (1, x0)
\bigr) - 1 \in \scrL (\mathrm{i}\mathrm{m}
\bigl(
I + \partial x\scrF (1, x0));Y
\bigr)
is the inverse operator to the operator
I + \partial x\scrF (1, x0) \in \scrL
\bigl(
Y ; \mathrm{i}\mathrm{m}(I + \partial x\scrF (1, x0))
\bigr)
, cf. Lemma 6.1 (i).
Further, we have
\partial \varphi \scrH 1(0, \tau 0, 0, z0) = - \Phi \prime (\varphi 0),
\partial z\scrH 1(0, \tau 0, 0, z0) = 0,
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128 L. RECKE
\partial z\scrH 2(0, \tau 0, 0, z0) = I + \partial x\scrF (1, x0).
Hence, assumption \Phi \prime (\varphi 0) \not = 0 (cf. (1.13)) and Lemma 6.1 (i) imply that the partial derivative with
respect to (\varphi , z) of L\varphi \scrH (\varepsilon , \tau , \varphi , z) in the point \varepsilon = 0, \tau = \tau 0, \varphi = 0, z = z0 is bijective from
\BbbR \times \mathrm{i}\mathrm{m}
\bigl(
I + \partial x\scrF (1, x0)
\bigr)
onto L\infty
per. Therefore the implicit function theorem yields that there exist
\varepsilon 0 > 0 and \delta > 0 such that for all \varepsilon \in [0, \varepsilon 0] and \tau \in [\tau 0 - \varepsilon 0, [\tau 0 + \varepsilon 0] there exists exactly one
solution
\varphi = \~\varphi (\varepsilon , \tau ), z = \~z(\varepsilon , \tau )
to (6.4) with| \varphi | + \| z\| \infty < \delta . Moreover, the data-to-solution maps \~\varphi and \~z are C1-smooth. Hence,
for all (\varepsilon , T ) \in K(\varepsilon 0, \tau 0) we have a solution x = \^x(\varepsilon , T ) to (2.4), where the data-to-solution map \^x
is defined as
\^x(\varepsilon , 1 + \varepsilon \tau ) := S\varphi 0
\bigl(
S \~\varphi (\varepsilon ,\tau )x0 + \varepsilon \~z(\varepsilon , \tau )
\bigr)
.
The map \^x is C1-smooth because the maps maps \~\varphi and \~z are C1-smooth. Moreover,
\mathrm{i}\mathrm{n}\mathrm{f}
\psi \in \BbbR
\| \^x(\varepsilon , 1 + \varepsilon \tau ) - S\psi x0\| \infty \leq \mathrm{i}\mathrm{n}\mathrm{f}
\psi \in \BbbR
\| (S\varphi 0+\~\varphi (\varepsilon ,\tau ) - S\psi )x0\| \infty + \varepsilon \| \~z(\varepsilon , \tau )\| \infty = \varepsilon \| \~z(\varepsilon , \tau )\| \infty ,
i.e., assertion (1.14) of Theorem 1.2 is proved.
Let us fix \tau \in [\tau 0 - \varepsilon 0, \tau 0 + \varepsilon 0] and define \^\varphi (\tau ) := \varphi 0 + \~\varphi (0, \tau ). Then \Phi ( \^\varphi (\tau )) = \tau (cf. (6.5)),
and\bigm\| \bigm\| \^x(\varepsilon , 1 + \varepsilon \tau ) - S \^\varphi (\tau )x0
\bigm\| \bigm\|
\infty =
\bigm\| \bigm\| \bigm\| \bigl( S\varphi 0+\~\varphi (\varepsilon ,\tau ) - S\varphi 0+\~\varphi (0,\tau )
\bigr)
x0 + \varepsilon \~z(\varepsilon , \tau )
\bigm\| \bigm\| \bigm\|
\infty
\rightarrow 0 for \varepsilon \rightarrow 0,
i.e., assertion (1.15) of Theorem 1.2 is proved also.
Finally, let us prove the uniqueness assertion of Theorem 1.2 (i). We have to show that for any
solution (\varepsilon , T, x) \in (0,\infty )2\times L\infty
per to (2.4) with T = 1+ \varepsilon \tau and \varepsilon \approx 0, \tau \approx \tau 0, x \approx S\varphi 0x0 it holds
x = \^x(\varepsilon , T ).
Let (\varepsilon k, Tk, xk) \in (0,\infty )2\times L\infty
per, k \in \BbbN , be a sequence of solutions to (2.4) with Tk = 1+ \varepsilon k\tau k
and
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
\bigl(
\varepsilon k + | \tau k - \tau 0| + \| xk + S\varphi 0x0\| \infty
\bigr)
= 0.
Then for large k we have xk = S\varphi 0 (S\varphi k
x0 + \varepsilon kzk) with zk \in Y and \| zk\| \infty \leq const and \varphi k \rightarrow 0
(cf. Lemmas 3.1 and 4.2). Hence, for large k we get \scrH (\varepsilon k, \tau k, \varphi k, zk) = 0 (cf. (6.2)), i.e.,
L\varphi k
\scrH (\varepsilon k, \tau k, \varphi k, zk) = 0.
In particular, the equation \scrH 2(\varepsilon k, \tau k, \varphi k, zk) = 0 yields\left( (I - P )
\left( I + 1\int
0
\partial x\scrF (1, S\varphi k
x0 + r\varepsilon kzk)dr
\right) \right) zk =
= - (I - P )
\bigl(
\tau k\partial T\scrF (1, S\varphi k
x0 + \varepsilon kzk) + S - \varphi 0\scrG (1 + \varepsilon k\tau k)
\bigr)
,
i.e., \| zk - z0\| \infty \rightarrow 0 (cf. (6.6)). Here we used that
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 1
FORCED FREQUENCY LOCKING FOR DIFFERENTIAL EQUATIONS WITH DISTRIBUTIONAL FORCINGS 129
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\| (I - P )
\left( I + 1\int
0
\partial x\scrF (1, S\varphi k
x0 + r\varepsilon kzk)dr
\right) - I - \partial x\scrF (1, x0)
\bigm\| \bigm\| \bigm\| \bigm\| \bigm\| \bigm\|
\scrL (L\infty
per)
= 0,
and, hence, for large k the operator (I - P )
\biggl(
I +
\int 1
0
\partial x\scrF (1, S\varphi k
x0 + r\varepsilon kzk)dr
\biggr)
is bijective from
Y onto \mathrm{i}\mathrm{m}
\bigl(
I+\partial x\scrF (1, x0)
\bigr)
, and its inverse is bounded with respect to the operator norm (uniformly
with respect to k).
Let us summarize: We got that (\varepsilon k, \tau k, \varphi k, zk) is a solution to (6.4), which is, for large k, close
to the solution (0, \tau 0, 0, z0). Hence, the uniqueness assertion of the implicit function theorem yields
\varphi k = \~\varphi (\varepsilon k, \tau k), zk = \~z(\varepsilon k, \tau k), i.e., xk = \^x(\varepsilon k, Tk).
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Received 30.10.17
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 1
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| id | umjimathkievua-article-1545 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:07:47Z |
| publishDate | 2018 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/84/7fa014272654c5d1fdd434ff0f053f84.pdf |
| spelling | umjimathkievua-article-15452019-12-05T09:17:34Z Forced frequency locking for differential equations with distributional forcings Вимушене замикання частоти для диференцiальних рiвнянь з дистрибутивними форсуваннями Recke, L. Реке, Л. This paper deals with forced frequency locking, i.e., the behavior of periodic solutions to autonomous differential equations under the influence of small periodic forcings. We show that, although the forcings are allowed to be discontinuous (e.g., step-function-like) or even distributional (e.g., Dirac-function-like), the forced frequency locking happens as in the case of smooth forcings, and we derive formulas for the locking cones and for the asymptotic phases as in the case of smooth forcings. Розглянуто вимушене замикання частоти, тобто поведiнку перiодичних розв’язкiв автономних диференцiальних рiвнянь пiд впливом малих перiодичних форсувань. Показано, що, незважаючи на той факт, що цi форсування можуть бути розривними (типу схiдчастих функцiй) або навiть дистрибутивними (типу дельта-функцiй), вимуше- нi замикання частоти вiдбуваються, як i у випадку гладких форсувань, i можна отримати формули для конусiв замикання та асимптотичних фаз, як i у випадку гладких форсувань. Institute of Mathematics, NAS of Ukraine 2018-01-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1545 Ukrains’kyi Matematychnyi Zhurnal; Vol. 70 No. 1 (2018); 115-129 Український математичний журнал; Том 70 № 1 (2018); 115-129 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1545/527 Copyright (c) 2018 Recke L. |
| spellingShingle | Recke, L. Реке, Л. Forced frequency locking for differential equations with distributional forcings |
| title | Forced frequency locking for differential equations with distributional forcings |
| title_alt | Вимушене замикання частоти для диференцiальних рiвнянь
з дистрибутивними форсуваннями |
| title_full | Forced frequency locking for differential equations with distributional forcings |
| title_fullStr | Forced frequency locking for differential equations with distributional forcings |
| title_full_unstemmed | Forced frequency locking for differential equations with distributional forcings |
| title_short | Forced frequency locking for differential equations with distributional forcings |
| title_sort | forced frequency locking for differential equations with distributional forcings |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1545 |
| work_keys_str_mv | AT reckel forcedfrequencylockingfordifferentialequationswithdistributionalforcings AT rekel forcedfrequencylockingfordifferentialequationswithdistributionalforcings AT reckel vimušenezamikannâčastotidlâdiferencialʹnihrivnânʹzdistributivnimiforsuvannâmi AT rekel vimušenezamikannâčastotidlâdiferencialʹnihrivnânʹzdistributivnimiforsuvannâmi |